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15001	https://www.mashupmath.com/how-to-find-the-vertex-of-a-parabola	How to Find the Vertex of a Parabola in 3 Easy Steps — Mashup Math 0 items $0 How to Find the Vertex of a Parabola Your complete guide to finding the vertex of a parabola in 3 easy steps Whenever you are working with and graphing quadratic equations (parabolas), you will have to know what is a vertex of a parabola and how to find the vertex of a parabola. This short guide will teach you everything you need to know about the vertex of a parabola and covers the following topics: What is the vertex of a parabola? What is the formula for the vertex of a parabola? How to find the vertex of a parabola Before we explore finding the vertex of a parabola using 3 easy steps, lets quickly review some important information related to quadratic functions and parabolas. What is a parabola? A quadratic function in standard form is a function of the form: y=ax^2+bx+c or f(x)=ax^2+bx+c (note that, in this guide, f(x)= and y= are used interchangeably). For example, for the standard form quadratic f(x)=x^2 +6x + 7, a=1, b=6, and c=7 The graph of a function is called a parabola—a plane curve that is U-shaped and symmetrical about an axis of symmetry that passes through a vertex point. Note that a parabola can open upwards (U-shaped) or open downwards (upside-down U-shaped). Figure 01 below shows two parabolas (one that opens upwards and one that opens downwards) along with each parabola’s axis of symmetry and vertex point. After looking at the figure, it should make sense why a vertex point is often referred to as a “turning point of a parabola.” What is the vertex of a parabola? Definition: The vertex of a parabola is a sort of turning point where a parabola intersects with an axis of symmetry. For a parabola that opens upwards, the vertex point will be an absolute minimum point and for a parabola that opens downwards, the vertex point will be an absolute maximum point. In simpler terms, what is the vertex of a parabola? The vertex is the point at the very bottom of the curve for parabolas that open upward or at the very top of the curve for parabolas that open downward. Figure 02 below shows examples of three different parabolas graphed on the coordinate plane along with the vertex point labeled and the vertical axis of symmetry (dotted line) that passes through it. Now that you understand what is the vertex of a parabola, you are ready to learn how to find vertex of a parabola using a simple 3-step method. What is the formula for the vertex of a parabola? The vertex of a parabola is the (x,y) coordinate point where a parabola and its axis of symmetry intersect. To find the vertex of a parabola represented by a quadratic function in f(x)=ax^2+bx+c form: Step 01: Identify the values of the coefficients a and b Step 02: Use the formula for the vertex of a parabola x=-b/2a to find the x-coordinate value of the vertex point. Step 03: Input the x-coordinate value from Step 01 into the function to find the y-coordinate value This process, along with the vertex of a parabola formula, is summarized in Figure 03. Moving forward, let’s gain some experience with the formula for the vertex of a parabola by working through a few examples. How to Find the Vertex of a Parabola Example #1 Example: Find the vertex of f(x)=x^2+4x-1 Finding the vertex of a parabola with equation f(x)=x^2+4x-1 can easily be done using the 3-step process described above as follows: Step 01: Identify the values of the coefficients a and b Since the given quadratic function is in f(x)=ax^2+bx+c form, you can easily identify the values of a and b as: f(x)=x^2+4x-1 → a=1 and b=4 (Note that a=1 because, when there is no coefficient in front of a variable, then the coefficient is 1). Step 02: Use the formula for the vertex of a parabola x=-b/2a to find the x-coordinate value of the vertex point. Since a=1 and b=4… x=-b/2a → x= -(4) / 2(1) → x =-4/2 → x = -2 Step 03: Input the x-coordinate value from Step 01 into the function to find the y-coordinate value Now that you know that the x-coordinate value for the vertex is x=-2, you can input x=-2 into the function to find the value of the y-coordinate as follows: f(x)=x^2+4x-1 → f(-2) = (-2)^2 +4(-2) -1 → f(-2) = 4 - 8 -1 = -5 → y=-5 So, the coordinates of the vertex point of the parabola that represents the quadratic function f(x)=x^2+4x-1 are (-2,-5). The graph of this function, including the vertex point, is shown on the coordinate plane in Figure 04. By looking at the graph, it should be clear that (-2,-5) is the vertex of a parabola. As long as you can remember the 3-steps and the formula for the vertex of a parabola (x=-b/2a), you will always be able to find the coordinates of the vertex point. Let’s gain some more experience by working through another example of finding the vertex of a parabola. How to Find the Vertex of a Parabola Example #2 Example: Find the vertex of f(x)=-x^2+10x-9 How to find vertex of a parabola with equation f(x)=-x^2+10x-9 can be done using the same 3-step method as Example #1 as follows: Step 01: Identify the values of the coefficients a and b Since the given quadratic function is in f(x)=ax^2+bx+c form, you can easily identify the values of a and b as: f(x)=-x^2+10x-9 → a=-1 and b=10 The next step in how to find vertex of a parabola is… Step 02: Use the formula for the vertex of a parabola x=-b/2a to find the x-coordinate value of the vertex point. Since a=-1 and b=10 x=-b/2a → x= -(10) / 2(-1) → x =-10/-2 → x = 5 Step 03: Input the x-coordinate value from Step 01 into the function to find the y-coordinate value Now that you know that the x-coordinate value for the vertex is x=5, you can input x=5 into the function to find the value of the y-coordinate as follows: f(x)=-x^2+10x-9 → f(5) = -(5)^2 +10(5) -9 → f(5) = -25 + 50 -9 = 16 → y=16 Therefore, the coordinates of the vertex point of the parabola are (5,16). In Figure 05 below, you can see the graph of f(x)=-x^2+10x-9 on the coordinate plane with the vertex point at (5,16). How to Find the Vertex of a Parabola Example #3 Example: Find the vertex of f(x)=3x^2-7x+3 In this final example, you will follow the same 3-steps as the previous two examples and use the vertex of a parabola formula as follows: Step 01: Identify the values of the coefficients a and b How do you find the vertex of a parabola with equation f(x)=3x^2-7x+3? Since the given quadratic function is in f(x)=ax^2+bx+c form, you can easily identify the values of a and b as: f(x)=3x^2-7x+3 → a=3 and b=-7 The next step to finding the vertex of a parabola is: Step 02: Use the formula for the vertex of a parabola x=-b/2a to find the x-coordinate value of the vertex point. Since a=3 and b=-7 x=-b/2a → x= -(-7) / 2(3) → x =7/6 → x = 7/6 or x≈1.167 Step 03: Input the x-coordinate value from Step 01 into the function to find the y-coordinate value Now that you know that the x-coordinate value for the vertex is x=1.167, you can input x=1.167 into the function to find the value of the y-coordinate as follows: f(x)=3x^2-7x+3 → f(1.167) = 3(1.167)^2 -7(1.167) +3 → f(1.167) ≈ -1.083 → y≈-1.083 Since finding the vertex of a parabola with equation f(x)=3x^2-7x+3 did not result in x and y-coordinate values that are integers, we will be using approximated decimal values, which is totally fine. We can conclude that the coordinates of the vertex point of the parabola are approximately (1.167,-1.083). In Figure 06 below, you can see that your approximated point is the vertex of the parabola. How to Find Vertex of a Parabola: Conclusion In this guide, you learned that the answer to What is the vertex of a parabola? is that the vertex is the “turning point” at the very bottom or very top of a U-shaped curved, known as a parabola, that represents a quadratic function of the form f(x)=ax^2+bx+c. You also learned how to find the vertex of a parabola using the following 3-step process: Step 01: Identify the values of the coefficients a and b Step 02: Use the formula for the vertex of a parabola x=-b/2a to find the x-coordinate value of the vertex point. Step 03: Input the x-coordinate value from Step 01 into the function to find the y-coordinate value By following this 3-step process, finding the vertex of a parabola is a relatively simple and repeatable process. Once you have mastered this skill, you are ready to learn how to graph a parabola that represents a quadratic function using our easy guide (click the link below to gain access). Keep Learning: How to Graph a Parabola in 3 Easy Steps The Vertical Line Test Explained Examples: Which Graph Represents a Function? What is the Formula for Slope? (and how to use it) Parent Functions and Parent Graphs Explained More Math Resources You Will Love: Featured 10 Fun Math Projects for All Grade Levels Read More → How to Divide Fractions in 3 Easy Steps Read More → 10 Super Fun Math Riddles for Kids! (with Answers) Read More → How to Turn a Percent into a Fraction in 3 Steps Read More → How to Turn a Fraction into a Percent in 2 Steps Read More → How to Convert CM to Inches in 3 Easy Steps Read More → There's a Woman in a Boat Riddle—Answer and Explanation Read More →
15002	https://www.expii.com/t/adding-negative-numbers-rules-examples-8960	Expii Adding Negative Numbers — Rules & Examples - Expii Adding a negative number decreases the original number. It is the same as subtracting a positive number. Explanations (3) Eunji Kwon Text 11 Adding Negative Integers Adding negative integers moves you further down (to the left) on the number line. Let’s take a look at an example. Solve 4+(−5). First, draw a number line and start at 4. Made using Desmos Next, add −5 to 4. This is essentially the same as subtracting 5 from 4. Made using Desmos In conclusion, 4+(−5)=4−5=−1 When you add two negative integers together, the sum is a more negative number since you are moving further down (to the left) of the number line. Take a look at the expression (−7)+(−3). First, let’s draw a number line and start at −7. Made using Desmos Next, add −3 to −7. This is essentially the same as subtracting 3 from −7. Made using Desmos In conclusion, (−7)+(−3)=−7−3=−10. When you add zero to an integer, the integer does not change. Take a look at 6+0. First, let’s draw a number line and start at 6. Made using Desmos Next, add 0 to 6. The integer does not change, so the answer is 6. Made using Desmos Now, try a couple of examples without drawing a number line. Report Share 11 Like Related Lessons Adding and Subtracting Whole Numbers — Examples & Problems Subtracting Negative Numbers — Rules & Examples Negatives on the Number Line — Visualization & Examples Multiplying Positive Numbers — Rules & Practice View All Related Lessons Zora Gilbert Text 5 Positive numbers live on the right side of 0 on the number line, while negative numbers live on the left side. Positive numbers have a positive sign in front (+) but we usually don't write it. However we always write the negative sign in front of a negative number. The sign of the number determines which side of the number line we fall on. See below. Positive numbers are to the RIGHT of, or above, 0 Negative numbers are to the LEFT of, or below, 0 When you add a negative number to a positive number, first find the positive number on the number line. Then, find the negative number on the number line. Next, find the distance between the negative number and 0 and transpose that difference over to the positive number. Let's try it with 3+(−6). Find the positive number on the number line Find the negative number on the number line. Figure out how far the negative number is from 0. Transpose the difference until it starts at the positive number. This means that 3+(−6))=−3. You can use this method to add pretty much any two numbers together, whether they're negative or positive. Just make sure to use the first number as your anchor point, then find the second relative to it instead of to 0. Report Share 5 Like Jonathan Heller Video 4 (Video) Adding and Subtracting Integers by Math Antics This video by Math Antics goes over all the different ways to add and subtract integers. It covers adding and subtracting whole numbers, adding negatives, and subtracting negatives. Summary Adding a negative makes a number smaller. This is because adding a negative number is the same as subtracting a positive number. −3+−7 We are starting at −3 and moving 7 units to the left on the number line. This will take us to −10. Made using Desmos Subtracting a positive is the same thing. −7−3 We are going to start at −7 and move to the left 3 units. This will take us to −10. Made using Desmos We have a special case when this process leads us back to 0. That means the two numbers we added together are additive inverses. You've reached the end How can we improve? General Bug Feature Send Feedback
15003	https://en.wikipedia.org/wiki/Taylor%27s_theorem	Jump to content Taylor's theorem العربية Azərbaycanca Български Boarisch Català Deutsch Eesti Español Euskara فارسی Français 한국어 हिन्दी Bahasa Indonesia Íslenska Italiano עברית Қазақша Magyar Nederlands 日本語 Piemontèis Polski Português Русский Simple English Српски / srpski Srpskohrvatski / српскохрватски Taclḥit Türkçe Українська Tiếng Việt 中文 Edit links From Wikipedia, the free encyclopedia Approximation of a function by a truncated power series \| \| \| Part of a series of articles about \| \| Calculus \| \| \| \| Fundamental theorem Limits Continuity Rolle's theorem Mean value theorem Inverse function theorem \| \| Differential \| Definitions \| \| Derivative (generalizations) Differential + infinitesimal + of a function + total \| \| Concepts \| \| Differentiation notation Second derivative Implicit differentiation Logarithmic differentiation Related rates Taylor's theorem \| \| Rules and identities \| \| Sum Product Chain Power Quotient L'Hôpital's rule Inverse General Leibniz Faà di Bruno's formula Reynolds \| \| \| Integral \| \| \| Lists of integrals Integral transform Leibniz integral rule \| \| Definitions \| \| Antiderivative Integral (improper) Riemann integral Lebesgue integration Contour integration Integral of inverse functions \| \| Integration by \| \| Parts Discs Cylindrical shells Substitution (trigonometric, tangent half-angle, Euler) Euler's formula Partial fractions (Heaviside's method) Changing order Reduction formulae Differentiating under the integral sign Risch algorithm \| \| \| Series \| \| \| Geometric (arithmetico-geometric) Harmonic Alternating Power Binomial Taylor \| \| Convergence tests \| \| Summand limit (term test) Ratio Root Integral Direct comparison Limit comparison Alternating series Cauchy condensation Dirichlet Abel \| \| \| Vector \| \| \| Gradient Divergence Curl Laplacian Directional derivative Identities \| \| Theorems \| \| Gradient Green's Stokes' Divergence Generalized Stokes Helmholtz decomposition \| \| \| Multivariable \| Formalisms \| \| Matrix Tensor Exterior Geometric \| \| Definitions \| \| Partial derivative Multiple integral Line integral Surface integral Volume integral Jacobian Hessian \| \| \| Advanced \| \| \| Calculus on Euclidean space Generalized functions Limit of distributions \| \| \| Specialized Fractional Malliavin Stochastic Variations \| \| Miscellanea Precalculus History Glossary List of topics Integration Bee Mathematical analysis Nonstandard analysis \| \| v t e \| In calculus, Taylor's theorem gives an approximation of a -times differentiable function around a given point by a polynomial of degree , called the -th-order Taylor polynomial. For a smooth function, the Taylor polynomial is the truncation at the order of the Taylor series of the function. The first-order Taylor polynomial is the linear approximation of the function, and the second-order Taylor polynomial is often referred to as the quadratic approximation. There are several versions of Taylor's theorem, some giving explicit estimates of the approximation error of the function by its Taylor polynomial. Taylor's theorem is named after Brook Taylor, who stated a version of it in 1715, although an earlier version of the result was already mentioned in 1671 by James Gregory. Taylor's theorem is taught in introductory-level calculus courses and is one of the central elementary tools in mathematical analysis. It gives simple arithmetic formulas to accurately compute values of many transcendental functions such as the exponential function and trigonometric functions. It is the starting point of the study of analytic functions, and is fundamental in various areas of mathematics, as well as in numerical analysis and mathematical physics. Taylor's theorem also generalizes to multivariate and vector valued functions. It provided the mathematical basis for some landmark early computing machines: Charles Babbage's difference engine calculated sines, cosines, logarithms, and other transcendental functions by numerically integrating the first 7 terms of their Taylor series. Motivation [edit] If a real-valued function is differentiable at the point , then it has a linear approximation near this point. This means that there exists a function h1(x) such that Here is the linear approximation of for x near the point a, whose graph is the tangent line to the graph at x = a. The error in the approximation is: As x tends to a, this error goes to zero much faster than , making a useful approximation. For a better approximation to , we can fit a quadratic polynomial instead of a linear function: Instead of just matching one derivative of at , this polynomial has the same first and second derivatives, as is evident upon differentiation. Taylor's theorem ensures that the quadratic approximation is, in a sufficiently small neighborhood of , more accurate than the linear approximation. Specifically, Here the error in the approximation is which, given the limiting behavior of , goes to zero faster than as x tends to a. Similarly, we might get still better approximations to f if we use polynomials of higher degree, since then we can match even more derivatives with f at the selected base point. In general, the error in approximating a function by a polynomial of degree k will go to zero much faster than as x tends to a. However, there are functions, even infinitely differentiable ones, for which increasing the degree of the approximating polynomial does not increase the accuracy of approximation: we say such a function fails to be analytic at x = a: it is not (locally) determined by its derivatives at this point. Taylor's theorem is of asymptotic nature: it only tells us that the error in an approximation by a -th order Taylor polynomial Pk tends to zero faster than any nonzero -th degree polynomial as . It does not tell us how large the error is in any concrete neighborhood of the center of expansion, but for this purpose there are explicit formulas for the remainder term (given below) which are valid under some additional regularity assumptions on f. These enhanced versions of Taylor's theorem typically lead to uniform estimates for the approximation error in a small neighborhood of the center of expansion, but the estimates do not necessarily hold for neighborhoods which are too large, even if the function f is analytic. In that situation one may have to select several Taylor polynomials with different centers of expansion to have reliable Taylor-approximations of the original function (see animation on the right.) There are several ways we might use the remainder term: Estimate the error for a polynomial Pk(x) of degree k estimating on a given interval (a – r, a + r). (Given the interval and degree, we find the error.) Find the smallest degree k for which the polynomial Pk(x) approximates to within a given error tolerance on a given interval (a − r, a + r) . (Given the interval and error tolerance, we find the degree.) Find the largest interval (a − r, a + r) on which Pk(x) approximates to within a given error tolerance. (Given the degree and error tolerance, we find the interval.) Taylor's theorem in one real variable [edit] Statement of the theorem [edit] The precise statement of the most basic version of Taylor's theorem is as follows: Taylor's theorem—Let k ≥ 1 be an integer and let the function f : R → R be k times differentiable at the point a ∈ R. Then there exists a function hk : R → R such that and This is called the Peano form of the remainder. The polynomial appearing in Taylor's theorem is the -th order Taylor polynomial of the function f at the point a. The Taylor polynomial is the unique "asymptotic best fit" polynomial in the sense that if there exists a function hk : R → R and a -th order polynomial p such that then p = Pk. Taylor's theorem describes the asymptotic behavior of the remainder term which is the approximation error when approximating f with its Taylor polynomial. Using the little-o notation, the statement in Taylor's theorem reads as Explicit formulas for the remainder [edit] Under stronger regularity assumptions on f there are several precise formulas for the remainder term Rk of the Taylor polynomial, the most common ones being the following. Mean-value forms of the remainder—Let f : R → R be k + 1 times differentiable on the open interval between and with f(k) continuous on the closed interval between and . Then for some real number between and . This is the Lagrange form of the remainder. Similarly, for some real number between and . This is the Cauchy form of the remainder. Both can be thought of as specific cases of the following result: Consider for some real number between and . This is the Schlömilch form of the remainder (sometimes called the Schlömilch-Roche). The choice is the Lagrange form, whilst the choice is the Cauchy form. These refinements of Taylor's theorem are usually proved using the mean value theorem, whence the name. Additionally, notice that this is precisely the mean value theorem when . Also other similar expressions can be found. For example, if G(t) is continuous on the closed interval and differentiable with a non-vanishing derivative on the open interval between and , then for some number between and . This version covers the Lagrange and Cauchy forms of the remainder as special cases, and is proved below using Cauchy's mean value theorem. The Lagrange form is obtained by taking and the Cauchy form is obtained by taking . The statement for the integral form of the remainder is more advanced than the previous ones, and requires understanding of Lebesgue integration theory for the full generality. However, it holds also in the sense of Riemann integral provided the (k + 1)th derivative of f is continuous on the closed interval [a,x]. Integral form of the remainder—Let be absolutely continuous on the closed interval between and . Then Due to the absolute continuity of f(k) on the closed interval between and , its derivative f(k+1) exists as an L1-function, and the result can be proven by a formal calculation using the fundamental theorem of calculus and integration by parts. Estimates for the remainder [edit] It is often useful in practice to be able to estimate the remainder term appearing in the Taylor approximation, rather than having an exact formula for it. Suppose that f is (k + 1)-times continuously differentiable in an interval I containing a. Suppose that there are real constants q and Q such that throughout I. Then the remainder term satisfies the inequality if x > a, and a similar estimate if x < a. This is a simple consequence of the Lagrange form of the remainder. In particular, if on an interval I = (a − r,a + r) with some , then for all x∈(a − r,a + r). The second inequality is called a uniform estimate, because it holds uniformly for all x on the interval (a − r,a + r). Example [edit] Suppose that we wish to find the approximate value of the function on the interval while ensuring that the error in the approximation is no more than 10−5. In this example we pretend that we only know the following properties of the exponential function: \| \| \| \| --- \| \| \| ★ \| From these properties it follows that for all , and in particular, . Hence the -th order Taylor polynomial of at and its remainder term in the Lagrange form are given by where is some number between 0 and x. Since ex is increasing by (★), we can simply use for to estimate the remainder on the subinterval . To obtain an upper bound for the remainder on , we use the property for to estimate using the second order Taylor expansion. Then we solve for ex to deduce that simply by maximizing the numerator and minimizing the denominator. Combining these estimates for ex we see that so the required precision is certainly reached, when (See factorial or compute by hand the values and .) As a conclusion, Taylor's theorem leads to the approximation For instance, this approximation provides a decimal expression , correct up to five decimal places. Relationship to analyticity [edit] Taylor expansions of real analytic functions [edit] Let I ⊂ R be an open interval. By definition, a function f : I → R is real analytic if it is locally defined by a convergent power series. This means that for every a ∈ I there exists some r > 0 and a sequence of coefficients ck ∈ R such that (a − r, a + r) ⊂ I and In general, the radius of convergence of a power series can be computed from the Cauchy–Hadamard formula This result is based on comparison with a geometric series, and the same method shows that if the power series based on a converges for some b ∈ R, it must converge uniformly on the closed interval , where . Here only the convergence of the power series is considered, and it might well be that (a − R,a + R) extends beyond the domain I of f. The Taylor polynomials of the real analytic function f at a are simply the finite truncations of its locally defining power series, and the corresponding remainder terms are locally given by the analytic functions Here the functions are also analytic, since their defining power series have the same radius of convergence as the original series. Assuming that [a − r, a + r] ⊂ I and r < R, all these series converge uniformly on (a − r, a + r). Naturally, in the case of analytic functions one can estimate the remainder term by the tail of the sequence of the derivatives f′(a) at the center of the expansion, but using complex analysis also another possibility arises, which is described below. Taylor's theorem and convergence of Taylor series [edit] The Taylor series of f will converge in some interval in which all its derivatives are bounded and do not grow too fast as k goes to infinity. (However, even if the Taylor series converges, it might not converge to f, as explained below; f is then said to be non-analytic.) One might think of the Taylor series of an infinitely many times differentiable function f : R → R as its "infinite order Taylor polynomial" at a. Now the estimates for the remainder imply that if, for any r, the derivatives of f are known to be bounded over (a − r, a + r), then for any order k and for any r > 0 there exists a constant Mk,r > 0 such that \| \| \| \| --- \| \| \| ★★ \| for every x ∈ (a − r,a + r). Sometimes the constants Mk,r can be chosen in such way that Mk,r is bounded above, for fixed r and all k. Then the Taylor series of f converges uniformly to some analytic function (One also gets convergence even if Mk,r is not bounded above as long as it grows slowly enough.) The limit function Tf is by definition always analytic, but it is not necessarily equal to the original function f, even if f is infinitely differentiable. In this case, we say f is a non-analytic smooth function, for example a flat function: Using the chain rule repeatedly by mathematical induction, one shows that for any order k, for some polynomial pk of degree 2(k − 1). The function tends to zero faster than any polynomial as , so f is infinitely many times differentiable and f(k)(0) = 0 for every positive integer k. The above results all hold in this case: The Taylor series of f converges uniformly to the zero function Tf(x) = 0, which is analytic with all coefficients equal to zero. The function f is unequal to this Taylor series, and hence non-analytic. For any order k ∈ N and radius r > 0 there exists Mk,r > 0 satisfying the remainder bound (★★) above. However, as k increases for fixed r, the value of Mk,r grows more quickly than rk, and the error does not go to zero. Taylor's theorem in complex analysis [edit] Taylor's theorem generalizes to functions f : C → C which are complex differentiable in an open subset U ⊂ C of the complex plane. However, its usefulness is dwarfed by other general theorems in complex analysis. Namely, stronger versions of related results can be deduced for complex differentiable functions f : U → C using Cauchy's integral formula as follows. Let r > 0 such that the closed disk B(z, r) ∪ S(z, r) is contained in U. Then Cauchy's integral formula with a positive parametrization γ(t) = z + reit of the circle S(z, r) with gives Here all the integrands are continuous on the circle S(z, r), which justifies differentiation under the integral sign. In particular, if f is once complex differentiable on the open set U, then it is actually infinitely many times complex differentiable on U. One also obtains Cauchy's estimate for any z ∈ U and r > 0 such that B(z, r) ∪ S(c, r) ⊂ U. The estimate implies that the complex Taylor series of f converges uniformly on any open disk with into some function Tf. Furthermore, using the contour integral formulas for the derivatives f(k)(c), so any complex differentiable function f in an open set U ⊂ C is in fact complex analytic. All that is said for real analytic functions here holds also for complex analytic functions with the open interval I replaced by an open subset U ∈ C and a-centered intervals (a − r, a + r) replaced by c-centered disks B(c, r). In particular, the Taylor expansion holds in the form where the remainder term Rk is complex analytic. Methods of complex analysis provide some powerful results regarding Taylor expansions. For example, using Cauchy's integral formula for any positively oriented Jordan curve which parametrizes the boundary of a region , one obtains expressions for the derivatives f(j)(c) as above, and modifying slightly the computation for Tf(z) = f(z), one arrives at the exact formula The important feature here is that the quality of the approximation by a Taylor polynomial on the region is dominated by the values of the function f itself on the boundary . Similarly, applying Cauchy's estimates to the series expression for the remainder, one obtains the uniform estimates Example [edit] The function is real analytic, that is, locally determined by its Taylor series. This function was plotted above to illustrate the fact that some elementary functions cannot be approximated by Taylor polynomials in neighborhoods of the center of expansion which are too large. This kind of behavior is easily understood in the framework of complex analysis. Namely, the function f extends into a meromorphic function on the compactified complex plane. It has simple poles at and , and it is analytic elsewhere. Now its Taylor series centered at z0 converges on any disc B(z0, r) with r < \|z − z0\|, where the same Taylor series converges at z ∈ C. Therefore, Taylor series of f centered at 0 converges on B(0, 1) and it does not converge for any z ∈ C with \|z\| > 1 due to the poles at i and −i. For the same reason the Taylor series of f centered at 1 converges on and does not converge for any z ∈ C with . Generalizations of Taylor's theorem [edit] Higher-order differentiability [edit] A function f: Rn → R is differentiable at a ∈ Rn if and only if there exists a linear functional L : Rn → R and a function h : Rn → R such that If this is the case, then is the (uniquely defined) differential of f at the point a. Furthermore, then the partial derivatives of f exist at a and the differential of f at a is given by Introduce the multi-index notation for α ∈ Nn and x ∈ Rn. If all the -th order partial derivatives of f : Rn → R are continuous at a ∈ Rn, then by Clairaut's theorem, one can change the order of mixed derivatives at a, so the short-hand notation for the higher order partial derivatives is justified in this situation. The same is true if all the (k − 1)-th order partial derivatives of f exist in some neighborhood of a and are differentiable at a. Then we say that f is k times differentiable at the point a. Taylor's theorem for multivariate functions [edit] Using notations of the preceding section, one has the following theorem. Multivariate version of Taylor's theorem—Let f : Rn → R be a k-times continuously differentiable function at the point a ∈ Rn. Then there exist functions hα : Rn → R, where such that If the function f : Rn → R is k + 1 times continuously differentiable in a closed ball for some , then one can derive an exact formula for the remainder in terms of (k+1)-th order partial derivatives of f in this neighborhood. Namely, In this case, due to the continuity of (k+1)-th order partial derivatives in the compact set B, one immediately obtains the uniform estimates Example in two dimensions [edit] For example, the third-order Taylor polynomial of a smooth function is, denoting , Proofs [edit] Proof for Taylor's theorem in one real variable [edit] Let where, as in the statement of Taylor's theorem, It is sufficient to show that The proof here is based on repeated application of L'Hôpital's rule. Note that, for each , . Hence each of the first derivatives of the numerator in vanishes at , and the same is true of the denominator. Also, since the condition that the function be times differentiable at a point requires differentiability up to order in a neighborhood of said point (this is true, because differentiability requires a function to be defined in a whole neighborhood of a point), the numerator and its derivatives are differentiable in a neighborhood of . Clearly, the denominator also satisfies said condition, and additionally, doesn't vanish unless , therefore all conditions necessary for L'Hôpital's rule are fulfilled, and its use is justified. So where the second-to-last equality follows by the definition of the derivative at . Alternate proof for Taylor's theorem in one real variable [edit] Let be any real-valued continuous function to be approximated by the Taylor polynomial. Step 1: Let and be functions. Set and to be Step 2: Properties of and : Similarly, Step 3: Use Cauchy Mean Value Theorem Let and be continuous functions on . Since so we can work with the interval . Let and be differentiable on . Assume for all . Then there exists such that Note: in and so for some . This can also be performed for : for some . This can be continued to . This gives a partition in : with Set : Step 4: Substitute back By the Power Rule, repeated derivatives of , , so: This leads to: By rearranging, we get: or because eventually: Derivation for the mean value forms of the remainder [edit] Let G be any real-valued function, continuous on the closed interval between and and differentiable with a non-vanishing derivative on the open interval between and , and define For . Then, by Cauchy's mean value theorem, \| \| \| \| --- \| \| \| ★★★ \| for some on the open interval between and . Note that here the numerator is exactly the remainder of the Taylor polynomial for . Compute plug it into (★★★) and rearrange terms to find that This is the form of the remainder term mentioned after the actual statement of Taylor's theorem with remainder in the mean value form. The Lagrange form of the remainder is found by choosing and the Cauchy form by choosing . Remark. Using this method one can also recover the integral form of the remainder by choosing but the requirements for f needed for the use of mean value theorem are too strong, if one aims to prove the claim in the case that f(k) is only absolutely continuous. However, if one uses Riemann integral instead of Lebesgue integral, the assumptions cannot be weakened. Derivation for the integral form of the remainder [edit] Due to the absolute continuity of on the closed interval between and , its derivative exists as an -function, and we can use the fundamental theorem of calculus and integration by parts. This same proof applies for the Riemann integral assuming that is continuous on the closed interval and differentiable on the open interval between and , and this leads to the same result as using the mean value theorem. The fundamental theorem of calculus states that Now we can integrate by parts and use the fundamental theorem of calculus again to see that which is exactly Taylor's theorem with remainder in the integral form in the case . The general statement is proved using induction. Suppose that \| \| \| \| --- \| \| \| eq1 \| Integrating the remainder term by parts we arrive at Substituting this into the formula in (eq1) shows that if it holds for the value , it must also hold for the value . Therefore, since it holds for , it must hold for every positive integer . Derivation for the remainder of multivariate Taylor polynomials [edit] We prove the special case, where has continuous partial derivatives up to the order in some closed ball with center . The strategy of the proof is to apply the one-variable case of Taylor's theorem to the restriction of to the line segment adjoining and . Parametrize the line segment between and by We apply the one-variable version of Taylor's theorem to the function : Applying the chain rule for several variables gives where is the multinomial coefficient. Since , we get: See also [edit] Mathematics portal Hadamard's lemma Laurent series – Power series with negative powers Padé approximant – 'Best' approximation of a function by a rational function of given order Newton series – Discrete analog of a derivativePages displaying short descriptions of redirect targets Approximation theory – Theory of getting acceptably close inexact mathematical calculations Function approximation – Approximating an arbitrary function with a well-behaved one Footnotes [edit] ^ (2013). "Linear and quadratic approximation" Retrieved December 6, 2018 ^ Taylor, Brook (1715). Methodus Incrementorum Directa et Inversa [Direct and Reverse Methods of Incrementation] (in Latin). London. p. 21–23 (Prop. VII, Thm. 3, Cor. 2). Translated into English in Struik, D. J. (1969). A Source Book in Mathematics 1200–1800. Cambridge, Massachusetts: Harvard University Press. pp. 329–332. ^ Kline 1972, pp. 442, 464. ^ Genocchi, Angelo; Peano, Giuseppe (1884), Calcolo differenziale e principii di calcolo integrale, (N. 67, pp. XVII–XIX): Fratelli Bocca ed.{{citation}}: CS1 maint: location (link) ^ Spivak, Michael (1994), Calculus (3rd ed.), Houston, TX: Publish or Perish, p. 383, ISBN 978-0-914098-89-8 ^ "Taylor formula", Encyclopedia of Mathematics, EMS Press, 2001 ^ The hypothesis of f(k) being continuous on the closed interval between and is not redundant. Although f being k + 1 times differentiable on the open interval between and does imply that f(k) is continuous on the open interval between and , it does not imply that f(k) is continuous on the closed interval between and , i.e. it does not imply that f(k) is continuous at the endpoints of that interval. Consider, for example, the function f : [0,1] → R defined to equal on and with . This is not continuous at 0, but is continuous on . Moreover, one can show that this function has an antiderivative. Therefore that antiderivative is differentiable on , its derivative (the function f) is continuous on the open interval , but its derivative f is not continuous on the closed interval . So the theorem would not apply in this case. ^ Kline 1998, §20.3; Apostol 1967, §7.7. ^ Apostol 1967, §7.7. ^ Apostol 1967, §7.5. ^ Apostol 1967, §7.6 ^ Rudin 1987, §10.26 ^ This follows from iterated application of the theorem that if the partial derivatives of a function f exist in a neighborhood of a and are continuous at a, then the function is differentiable at a. See, for instance, Apostol 1974, Theorem 12.11. ^ Königsberger Analysis 2, p. 64 ff. ^ Folland, G. B. "Higher-Order Derivatives and Taylor's Formula in Several Variables" (PDF). Department of Mathematics \| University of Washington. Retrieved 2024-02-21. ^ Stromberg 1981 ^ Hörmander 1976, pp. 12–13 References [edit] Apostol, Tom (1967), Calculus, Wiley, ISBN 0-471-00005-1. Apostol, Tom (1974), Mathematical analysis, Addison–Wesley. Bartle, Robert G.; Sherbert, Donald R. (2011), Introduction to Real Analysis (4th ed.), Wiley, ISBN 978-0-471-43331-6. Hörmander, L. (1976), Linear Partial Differential Operators, Volume 1, Springer, ISBN 978-3-540-00662-6. Kline, Morris (1972), Mathematical thought from ancient to modern times, Volume 2, Oxford University Press. Kline, Morris (1998), Calculus: An Intuitive and Physical Approach, Dover, ISBN 0-486-40453-6. Pedrick, George (1994), A First Course in Analysis, Springer, ISBN 0-387-94108-8. Stromberg, Karl (1981), Introduction to classical real analysis, Wadsworth, ISBN 978-0-534-98012-2. Rudin, Walter (1987), Real and complex analysis (3rd ed.), McGraw-Hill, ISBN 0-07-054234-1. Tao, Terence (2014), Analysis, Volume I (3rd ed.), Hindustan Book Agency, ISBN 978-93-80250-64-9. 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15004	https://artofproblemsolving.com/wiki/index.php/Equidistant?srsltid=AfmBOoru8alPjQ2n5rHwhwuo8F_GRzNUmWuEtmVahYTFSTT0VH6T5SNt	Art of Problem Solving Equidistant - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Equidistant Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Equidistant Equidistant means "at the same distance." Thus, for instance, if triangle is isosceles with base , points and are equidistant from point . Similarly, the perpendicular bisector of a line segment is the set of points equidistant from the endpoints. So, given segment and a point such that , we know (by definition) that is on the perpendicular bisector of . Also, given on the perpendicular bisector of , we know that . This article is a stub. Help us out by expanding it. Retrieved from " Categories: Stubs Geometry Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
15005	https://www.wolframalpha.com/examples/Vectors.html	Examples for Vectors Vectors are objects in an n-dimensional vector space that consist of a simple list of numerical or symbolic values. Wolfram\|Alpha can convert vectors to spherical or polar coordinate systems and can compute properties of vectors, such as the vector length or normalization. Additionally, Wolfram\|Alpha can explore relationships between vectors by adding, multiplying, testing orthogonality and computing the projection of one vector onto another. Find plots and various other properties of vectors. Compute properties of a vector: vector {2, -5, 4} Specify a vector as a linear combination of unit vectors: vector 3i + 5j vector 2i - 4j + 3k Compute the norm of a vector: norm {12, -5} More examples Vector Algebra Perform arithmetic and algebraic operations, such as dot and cross product, on vectors. Do vector computations: vector (1,3,-1) + (-2,1,6) 7 {1, 0, -2, 1} - 4 {2, -1, 1, -1} (i + j + k) + (2i - 3j + 8k) Compute a dot product: {12, 20} . {16, -5} (7i-j+3k).(4i-2k) Compute a cross product: {1/4, -1/2, 1} cross {1/3, 1, -2/3} (8i + 3j - k) x (i - j + 2k) Compute a (scalar) cross product in two dimensions: (4,1) x (-5,6) Normalize a vector: normalize the vector (3, 10) normalize vector Convert to another coordinate system: (1,1,-3) in spherical coordinates More examples Orthogonality Explore the orthogonality relationship on sets of vectors. Check if the set of vectors is orthogonal: Are (4, -1, -2), (2, -2, 5) and (-9, -24, -6) orthogonal? Orthogonality (0, 1) and (1, 0) orthogonality (5, 8, 13, 21) (-12, 5, -6, 1.41) (1.618, 0, 5, 7) (2.72, 7, -9, 3.14) Find conditions for orthogonality between vectors with symbolic components: (-a+b, 7) (2, a) orthogonality Orthogonality (1, 2, 3) (a, b, c) (b, g, f) More examples Vector Projections Compute and visualize the projection of a vector onto a vector, axis, plane or space. Compute the projection of one vector onto another: projection vector (2i + 6j) onto (6i + 2j) projection of vector (-1, 1) onto vector (1, 1) projection of 2i - 3j + 5k onto 2i - 3j - 5k projection (3, 0, 4), (0, 5, 12) Project a vector onto an axis: project (2, 5) onto the x-axis Project a vector onto a plane: What is the projection of the point (3, 4, 5) on the xy-plane? project vector (6, 5, 4) onto yz-plane Explore vector projections in higher dimensions: projection of (2, 4, 10, 5) onto span of {(1, 2, -6, -1), (-2, 7, 4, -12)} More examples
15006	https://www.maths.scot/pdf/n5/maths4scotland/Trig-Equations-Notes.pdf	Trig Equations Angles greater than 90° We need to re−define sin, cos and tan to be able to deal with angles greater than 90° . Instead of defining them in terms of the opposite, adjacent and hypotenuse, we can use the coordinates of a rotating line. THEORY A rotating line: Imagine a line OP of length, r, rotating about the origin in an anti−clockwise direction starting on the x−axis At any point, the coordinates of P are (x, y) and the angle between the line OP and the positive direction of the x−axis is denoted as θ In the 1st quadrant we can now define: sin y r θ = , cos x r θ = , tan y x θ = Notice that all 3 ratios are POSITIVE in the 1st quadrant. Now let the line OP move into the 2nd quadrant. We define sin , cos and tan θ θ θ based on the acute angle between the line OP and the x−axis. (angle marked with • ) So in the 2nd quadrant we have: sin y r θ = , cos x r θ − = , tan y x θ = − Notice that only the sine ratio is POSITIVE in the 2nd quadrant. Now let the line OP move into the 3rd quadrant. We define sin , cos and tan θ θ θ based on the acute angle between the line OP and the x−axis. (angle marked with • ) So in the 3rd quadrant we have: sin y r θ − = , cos x r θ − = , tan y x θ − = − Notice that only the tangent ratio is POSITIVE in the 3rd quadrant. O P (x, y) y x r θ 1st quadrant O P (x, y) y −x r θ 2nd quadrant • O P (x, y) −y −x r θ 3rd quadrant • THEORY continued: Now let the line OP move into the 4th quadrant. We define sin , cos and tan θ θ θ based on the acute angle between the line OP and the x−axis. (angle marked with • ) So in the 4th quadrant we have: sin y r θ − = , cos x r θ = , tan y x θ − = Notice that only the cosine ratio is POSITIVE in the 4th quadrant. Summary: For angles greater than 90° , represented by a line OP the sine, cosine and tangent are defined as the sine, cosine and tangent of the acute angle between OP and the x−axis. The sign is either positive or negative according to whether the sine, cosine and tangent is positive or negative in that quadrant. From the above theory: 1st Quadrant All were positive 2nd Quadrant Sine was positive 3rd Quadrant Tangent was positive 4th Quadrant Cosine was positive There is an easy way to remember the signs. All Sinners Take Care A quick sketch explains this. METHOD: For angles greater than 90° , draw a 4 quadrant axis, and mark on where the angle is. Mark in the acute angle and calculate it by addition or subtraction using 180° or 360° as appropriate. Take the sign from: All Sinners Take Care Some examples will make this clear. O P (x, y) −y x r θ 4th quadrant • All are + Sine is + Tangent is + Cosine is + e.g. sin 150° acute angle is marked • and is 180 − 150 = 30° sin is + in 2nd quadrant So sin 150° = sin 30° e.g. tan 150° acute angle is marked • and is 180 − 150 = 30° tan is − − − − in 2nd quadrant So tan 150° = −tan 30° e.g. cos 220° acute angle is marked • and is 220 − 180 = 40° cos is − in 3rd quadrant So cos 220° = −cos 40° The main application for this result is when we are working back to the angle. E.g. If sin 0.707 θ = what is θ We should look back at the graph at this point and consider where the graph is 0.707 There are two values of θ where sin 0.707 θ = In the 1st and 2nd quadrants. This should not be a surprise, as we know that the sign is + in these quadrants. So, if sin 0.707 θ = then the acute value of θ is given by: ( ) 1 sin 0.707 θ − = 45 θ = ° Since sin 0.707 θ + = , then we have values in 1st and 2nd quadrants Use ASTC So our two angles are: 45° and 180 − 45 = 135° A S T C • A S T C • A S T C • 90 180 270 360 -1 -0.5 0.5 1 x y sin y x = A S T C • • This all seems quite complicated, so we need to simplify it all. We will use some simple rules. We will be starting with sin ...... θ = , cos ...... θ = or tan ...... θ = e.g. sin 0.35 θ = − or cos 0.93 θ = etc. Method: 1. Ignore the sign 2. Use the inverse key on your calculator i.e. ( ) 1 sin .... − , ( ) 1 cos .... − or ( ) 1 tan .... − This will give you the acute angle. 3. Now look at the sign, and use ASTC to determine which 2 quadrants the angles must be in 4. Mark in the acute angles on your diagram. 5. Calculate the actual angles Example: tan 0.56 θ = Find the possible values of θ (1) Ignore the sign (2) find the acute angle 1 tan 0.56 29.2 θ θ − = → = ° − Round to 29° (3) tan is +, now use ASTC (4) Tangent is positive in 1st and 3rd quadrants. (5) Acute angle is 29° so actual angles are: 29° (1st quadrant) and 180 + 29 = 209° (3rd quadrant) Example: cos 0.23 θ = − Find the possible values of θ (1) Ignore the sign (2) find the acute angle 1 cos 0.23 76.7 θ θ − = → = ° − Round to 77° (3) cos is −, now use ASTC (4) Cosine is negative in 2nd and 3rd quadrants. (5) Acute angle is 77° so actual angles are: 180 − 77 = 103° (2nd quadrant) and 180 + 77 = 257° (3rd quadrant) A S T C • • A S T C • • Example: sin 0.88 θ = − Find the possible values of θ (1) Ignore the sign (2) find the acute angle 1 sin 0.88 61.6 θ θ − = → = ° − Round to 62° (3) sin is −, now use ASTC (4) Sine is negative in 3rd and 4th quadrants. (5) Acute angle is 62° so actual angles are: 180 + 62 = 242° (3rd quadrant) and 360 − 77 = 23° (4th quadrant) Example: tan 0.32 θ = − Find the possible values of θ (1) Ignore the sign (2) find the acute angle 1 tan 0.32 17.7 θ θ − = → = ° − Round to 18° (3) tan is −, now use ASTC (4) Tangent is negative in 2nd and 4th quadrants. (5) Acute angle is 18° so actual angles are: 180 − 18 = 162° (2nd quadrant) and 360 − 18 = 342° (4th quadrant) Example: sin 0.5 θ = Find the possible values of θ (1) Ignore the sign (2) find the acute angle 1 sin 0.5 30 θ θ − = → = ° (3) sin is +, now use ASTC (4) Sine is positive in 1st and 2nd quadrants. (5) Acute angle is 30° so actual angles are: 30° (1st quadrant) and 180 − 30 = 150° (2nd quadrant) Example: cos 0.75 θ = Find the possible values of θ (1) Ignore the sign (2) find the acute angle 1 cos 0.75 41.4 θ θ − = → = ° − Round to 41° (3) cos is +, now use ASTC (4) Cosine is positive in 1st and 4th quadrants. (5) Acute angle is 41° so actual angles are: 41° (1st quadrant) and 360 − 41 = 319° (4th quadrant) A S T C • • A S T C • • A S T C • • A S T C • • Applications: The applications of this at Standard Grade are in solving simple trig equations. We will be given an equation, which has to be re−arranged to the form: sin ...... θ = , cos ...... θ = or tan ...... θ = once we have done this, then we simply use ASTC as above to find the 2 angles. Example: Solve the equation 3tan 5 0, for 0 360. o x x + = ≤ ≤ Re−arrange the equation: 5 3tan 5 0 3tan 5 tan 3 o o o x x x + = → = − → = − (1) Ignore the sign (2) find the acute angle 1 5 tan 59.03 3 x x −  = → = °     − Round to 59° (3) tan is −, now use ASTC (4) Tangent is negative in 2nd and 4th quadrants. (5) Acute angle is 59° so actual angles are: 180 − 59 = 121° (2nd quadrant) and 360 − 59 = 301° (4th quadrant) Hence: x = 121° and x = 301° Example: The diagram shows part of the graph of y = sin x. The line y = 0.4 is drawn and cuts the graph of y = sin x at A and B. Find the x-coordinates of A and B. We have to solve: sin 0.4 x = (1) Ignore the sign (2) find the acute angle 1 sin 0.4 23.6 x x − = → = ° − Round to 24° (3) sin is +, now use ASTC (4) Sine is positive in 1st and 2nd quadrants. (5) Acute angle is 24° so actual angles are: 24° (1st quadrant) and 180 − 24 = 156° (2nd quadrant) xA = 24° xB = 156° A S T C • • A S T C • • Past Paper Questions: 1. Solve algebraically the equation 2 3sin 0 x + ° = for 0 ≤ x ≤ 360 2. Solve algebraically, the equation 7cos 2 0 x° − = for 0 ≤ x ≤ 360 3. Solve algebraically, the equation 5tan 9 0, for 0 360 x x − = ≤ ≤ 4. Solve the equation 5sin 2 0, 0 360 x for x ° + = ≤ ≤ 5. Solve algebraically the equation: tan 40 2sin 1 0 360 x x ° = °+ ≤ ≤ 6. The diagram shows the graph of y = sin x°, 0 ≤ x ≤ 360 a) Write down the coordinates of point S. The straight line y = 0.5 cuts the graph at T and P. b) Find the coordinates of T and P. 7. The diagram shows the graph of y = cos x°, 0 ≤ x ≤ 360. a) Write down the coordinates of point A. The straight line y = − 0.5 cuts the graph at B and C. b) Find the coordinates of B and C. Solutions follow on the next page. Solutions: 1. 2 3 2 3sin 0 sin x x + = → = − 1 acute 2 sin 41.81.. 3 x x −  = = °     180 42 360 42 222 318 x or x + − = = ° = = ° 2. 2 7 7cos 2 0 cos x x − = → = 1 acute 2 cos 73.398.. 7 x x − = = ° 360 73 73 287 x or x − = ° = = ° 3. 9 5 5tan 9 0 tan x x − = → = 1 acute 9 tan 60.945.. 5 x x − = = ° 180 61 61 241 x or x + = ° = = ° 4. 2 5 5sin 2 0 sin x x + = → = − 1 acute 2 sin 23.578.. 5 x x − = = ° 180 24 360 24 204 336 x or x + − = = ° = = ° 5. 0.1609 2 tan 40 2sin 1 sin x x = + → = − 1 acute 0.1609 sin 4.614.. 2 x x − = = ° 180 5 360 5 185 355 x or x + − = = ° = = ° 6. a) S is (90°, 1) b) sin 0.5 x = 1 acute sin 0.5 30 x x − = = ° 180 30 30 150 x or x − = ° = = ° T is (30°, 0.5) and P is (150°, 0.5) 7. a) A is (90°, 0) b) cos 0.5 x = − 1 acute cos 0.5 60 x x − = = ° 180 60 360 60 240 300 x or x + − = = ° = = ° B is (240°, -0.5) and C is (300°, -0.5) A S C T A S C T A S C T A S C T A S C T A S C T A S C T
15007	https://www.engineeringtoolbox.com/pressure-drop-compressed-air-pipes-d_852.html	Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications! Compressed Air - Pipe Line Pressure Loss - Online Calculator with Metric and Imperial Units Calculate pressure drop in compressed air pipe lines - metric and imperial units. The pressure drop in compressed air lines can be calculated with the empirical formula dp = 7.57 q1.85 L 104 / (d5 p) (1) dp = pressure drop (kg/cm2) q = air volume flow at atmospheric conditions (FAD) (m3/min) L = length of pipe (m) d = inside diameter of pipe (mm) p = initial pressure - gauge (kg/cm2) 1 kg/cm2 = 98068 Pa = 0.98 bar = 0.97 atmosphere = 736 mm Hg = 10000 mm H2O = 10 m H2O = 2050 psf = 14.2 psi = 29 in Hg = 394 in H2O = 32.8 ft H2O Note! - pressure is "force per unit area" and commonly used pressure units like kg/cm2 and similar are in principle not correct since kg is a mass unit. A mass must be multiplied with gravity g to be a force (weight). Mass and weight (force) Compressed Air - Pressure Drop Nomogram The nomogram below can be used to estimate pressure drops in compressed air pipe lines with pressure 7 bar (100 psig). Download and print compressed air pressure drop nomogram! Online Compressed Air Pipeline Pressure Drop Calculator - Metric Units Make Shortcut to these Calculators on Your Home Screen? The calculator below can used to calculate the pressure drop in compressed air pipelines. Compression ratio - compressed air to free air Air tools and air consumption Online Compressed Air Pipeline Pressure Drop Calculator - Imperial Units The calculator below can used to calculate the pressure drop in compressed air pipelines. NOTE! - a pressure drop above 1 kg/cm2 (14-15 psi) is in general not relevant and the formula and calculators above may not be valid. For a more accurate calculation - or for a longer pipe lines with larger pressure drops - divide the line in parts and calculate the pressure drop and final pressure for each part. Use final pressures as initial pressures for the next parts. The final pressure after the last part is the final pressure at the end of the pipe line. The pressure drop for the whole pipe line can also be calculated by summarizing the pressure drops for each part. Compressed Air Pipeline Pressure Drop Spreadsheet Calculations can be done for other pressures and/or pipe lengths by using this excel spreadsheet (metric units). The same spreadsheet including different types of pipes (imperial units). Compressed air pipe line pressure drop diagrams Or, alternatively - Compressed air pipe lines - pressure drop calculations - in Google Docs. You can open, save and modify your own copy of the Google spreadsheet if you are signed into your Google Account. Compressed Air Pipeline Pressure Drop Table - Initial gauge Pressure 7 kg/cm2 (100 psig) Pressure drops in 100 m (330 ft) compressed air schedule 40 steel pipe lines are indicated in the tables below: Unit Converter . Make Shortcut to Home Screen? Cookie Settings
15008	https://www.math.tugraz.at/graz-school-of-discrete-mathematics/topic6.pdf	Bootstrap percolation in high-dimensional product graphs Scientiﬁc Background Bootstrap percolation is a particular type of cellular automata deﬁned on graphs, which has been used to study a wide variety of physical processes such as crystal ﬁeld interactions, the emergence of magnetic properties, the spread of infections and more . In bootstrap percolation, an initial set of infected vertices spread through the graph, with a vertex becoming infected once its number of infected neighbours passes a certain threshold r ∈N. As with many areas of combinatorics, bootstrap percolation has been broadly studied from two diﬀering perspective, the extremal and the probabilistic. From the former perspective, we are interested in the extremal properties of this model - what is the size and structure of minimal percolating sets , i.e, initial sets such that eventually all vertices become infected, and how is this related to the geometry of the underlying graph. From the latter perspective, we are interested in the average case behaviour of this model - for a random initial set, what is the probability that it percolates, and how does this probability evolve as we increase the density of the initial set? In particular, we are interested in the location of the percolation threshold, the probability at which it becomes more likely than not that a random initial set will percolate and the size of the critical window, the range of infection probabilities where the probability of percolation transitions from almost never to almost surely. Such questions have been well-studied in a variety of random graph models [13, 15], where they have led to the development of many important techniques in the study of percolation models on graphs, and also in many highly structured, lattice-like graphs [2, 4], such as high-dimensional grids and tori, which arise naturally when considering the physical processes motivating the model. Hypotheses/Aims One particularly interesting class of graphs on which bootstrap percolation has been studied are hypercubes. The hypercube is a fundamental object of study in diverse areas of maths and computer science, and represents perhaps the simplest example of a high-dimensional model with non-trivial geometry on which to study percolation. In particular, recently the percolation threshold was determined in the hypercube Qd under two particular simple update rules, 2-neighbour percolation and majority percolation . We propose to study the percolation threshold in this model under more general update rules. We also propose to study generalisations of these questions in high-dimensional product graphs, graphs arising as the Cartesian product of many graphs. This family of graphs includes many well-studied examples of high-dimensional graphs such as hypercubes, grids, tori and Hamming graphs. Recently, a more systematic study of percolation processes on such graphs has been initiated , and we propose to extend this to bootstrap percolation. In all cases it would also be interesting to determine the size of the critical windows, and how the process stabilises in this regime. Approaches/Methods In order to study these questions we will use combinatorial and discrete probabilistic methods that have proven useful for studying other percolation problems. In particular, it will be useful to enumerate minimal percolating sets in these graphs, which may require analytic techniques such as the use of generating functions, and to determine their structure. To this end it will be useful to analyse the r-core of the uninfected sites, which may be possible by analysing a pruning process which can be done for example using branching processes , the diﬀerential equation method , or the general model of Warning propagation [7, 8]. The isoperimetric properties of the hypercube are well-known , and have proven to be key tools in the investigation of percolation 1 models on Qd. More recently the isoperimetric properties of high-dimensional product graphs [6, 10] have been studied, and they will be key to understanding bootstrap percolation in these graphs. Participating Faculty Member This project is supervised by Joshua Erde and Mihyun Kang. Mihyun Kang is a full professor at TU Graz and leads the Combinatorics Group. Her main research areas are combinatorics, discrete probability, and algorithms. She is best known for her work on topological properties of random graphs. She received a prestigious Friedrich Wilhelm Bessel Research Award of the Alexander von Humboldt Foundation. She serves on the editorial board of leading journals, including Random Structures and Algorithms. She supervised ﬁve PhD students and currently supervises two postdocs and two PhD students. Joshua Erde is an assistant professor (non-tenure track with habilitation) at the Institute of Dis-crete Mathematics at TU Graz. His main research focuses are in random structures, isoperimetric problems, and structural and inﬁnite graph theory. He received a competitive research fellowship of the Alexander von Humboldt Foundation and is currently the PI the FWF stand-alone project “Super-critical behaviour in random subgraph models”. He currently supervises a postdoc and co-supervises two PhD students. References J. Balogh and B. Bollob´ as. Bootstrap percolation on the hypercube. Probab. Theory Relat. Fields, 134(4):624–648, 2006. J. Balogh, B. Bollob´ as, H. Duminil-Copin, and R. Morris. The sharp threshold for bootstrap percolation in all dimensions. Trans. Am. Math. Soc., 364(5):2667–2701, 2012. J. Balogh, B. Bollob´ as, and R. Morris. Majority bootstrap percolation on the hypercube. Comb. Probab. Comput., 18(1-2):17–51, 2009. J. Balogh, B. Bollob´ as, and R. Morris. Bootstrap percolation in high dimensions. Comb. Probab. Comput., 19(5-6):643–692, 2010. J. Chalupa, P. L. Leath, and G. R. Reich. Bootstrap percolation on a bethe lattice. Journal of Physics C: Solid State Physics, 12(1):L31, 1979. F. R. K. Chung and P. Tetali. Isoperimetric inequalities for Cartesian products of graphs. Combin. Probab. Comput., 7(2):141–148, 1998. A. Coja-Oghlan, O. Cooley, M. Kang, J. Lee, and J. B. Ravelomanana. Warning propagation on random graphs. arXiv preprint arXiv:2102.00970, 2021. A. Coja-Oghlan, O. Cooley, M. Kang, and K. Skubch. How does the core sit inside the mantle? Random Struct. Algorithms, 51(3):459–482, 2017. P. De Gregorio, A. Lawlor, and K. A.. Dawson. Bootstrap Percolation, pages 608–626. Springer New York, New York, NY, 2009. S. Diskin, J. Erde, M. Kang, and M. Krivelevich. Isoperimetric inequalities and supercritical percolation on high-dimensional product graphs. In preparation. S. Diskin, J. Erde, M. Kang, and M. Krivelevich. Percolation on high-dimensional product graphs. arXiv preprint arXiv:2209.03722, 2022. U. Feige, M. Krivelevich, and D. Reichman. Contagious sets in random graphs. Ann. Appl. Probab., 27(5):2675–2697, 2017. N. Fountoulakis, M. Kang, C. Koch, and T. Makai. A phase transition regarding the evolution of bootstrap processes in inhomogeneous random graphs. Ann. Appl. Probab., 28(2):990–1051, 2018. 2 L. H. Harper. Optimal assigments of numbers to vertices. SIAM J. Appl. Math., 12:131–135, 1964. S. Janson, T. Luczak, T. Turova, and T. Vallier. Bootstrap percolation on the random graph Gn,p. Ann. Appl. Probab., 22(5):1989–2047, 2012. B. Pittel, J. Spencer, and N. Wormald. Sudden emergence of a giant k-core in a random graph. J. Comb. Theory, Ser. B, 67(1):111–151, 1996. O. Riordan. The k-core and branching processes. Comb. Probab. Comput., 17(1):111–136, 2008. 3
15009	https://tuhsphysics.ttsd.k12.or.us/Tutorial/NewIBPS/AngMech/AngMech.htm	Angular Mechanics \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| 10 \| 11 \| 12 \| 13 \| Go up - by Elias Dubelsten, (2001) and Chris Murray (2002) 1. What is the moment of inertia of a 5.00 Kg 34.0 cm radius hoop about its normal axis? To solve this problem, you merely have to look at the equation of the inertia of a hoop. The equation for the moment of Inertia for a hoop is: I= mr². By plugging in the numbers, you end up with: I= (5kg) (.34m)² I= .578 Kgm² (Table of contents) 2. What is the moment of inertia of a 8.0 Kg 10. cm radius sphere about its center? Just the same as problem #1, you can solve this problem by looking up the equation for the moment of inertia for a sphere. This formula is: I=2/5mr² I= (2/5) (8kg) (.1)² I= .032Kgm² (Table of contents) 3. What is the torque when you exert a force of 52 N on a 56 cm breaker bar? In this problem, you are given the force (52N) and the radius (.56 meters). By using the equation: t = Fr, you can arrive at your answer. Just plug in the numbers: t = (52N) (.56m) And the final answer is: t = 29.12Nm (Table of contents) 4. What force should you exert on a 14 cm long wrench to get a torque of 42 Nm? This problem is similar to #4 in that you must use the equation t = Fr, however, you are given different values. In this problem, you are given the final torque value (42 Nm), and the radius (.14 meters). Therefore, you must reshape the formula to solve for the force: t / r = F Then, by plugging in the numbers: (42Nm)/(.14m) = Force You arrive at your answer of: 300N (Table of contents) 5. . If you exert a torque of 68 Nm on a flywheel with an I of 12 Kgm2, what is its angular acceleration? To complete this problem, you must find an equation that has torque, angular acceleration, and moment of inertia. This equation is: t = I a. You are given the torque, and the moment of inertia, so you must reshape the formula to solve for angular acceleration, thus being left with the equation: t / I = a. By plugging in the given values: (68Nm)/(12Kgm²) = a And you arrive at your final answer (remember that angular acceleration is in rad/s²!!!!!): a = 5.66 rad/s² (Table of contents) 6. A drill exerts a torque of 80. Nm on a 1.2 Kg .12 m radius grinding disk that is a solid cylinder. What is the angular acceleration of the disk? To solve this problem, you must recognize that you need to find the moment of inertia first. You are given the information that this is a solid cylinder (I = ½ mr²), and has a radius of .12 meters, and a mass of 1.2 kilograms. Therefore, to solve for the moment of inertia, you can simply plug the given values into the formula: I = ½ mr² thus getting ½ 1.2 .12² = .00864. Then, now that you know the I to help get to your final answer. The problem asks for the angular acceleration of a drill with a torque of 80Nm, and now, since you know the I in the equation: t = I a, you can easily solve this. Because you are solving for angular acceleration, you need to reshape the formula to fit your needs: t / I = a. Now, just plug in the numbers: (80Nm)/(.00864) = a a = 9259.26 rad/s² = 9300 rad/s² (Table of contents) 7. A 34.2 gram marble rolls from rest down a ramp that loses 67.5 cm of height. What is the final velocity of the marble? What is its rotational kinetic energy at the bottom, and what is its translational kinetic energy at the bottom? Assuming the ramp was linear and 3.56 m long, and the marble had a radius of .342 cm, what was the angular acceleration of the marble as it moved down the incline? I am going to use energy to solve this, as it is my favorite way. Assuming the marble starts at rest, it has only potential energy at the top of the plane, and at the bottom, as it is rolling, it has both translational and rotational kinetic energy. So our energy equation looks like: mgh = = 1/2mv2 + 1/2Iw2 I am going to substitute to get rid of angular quantities. I = 2/5mr2 for a sphere, and w = v/r: mgh = 1/2mv2 + 1/2(2/5mr2)(v/r)2 mgh = 1/2mv2 + 1/2(2/5mr2)(v2/r2) cancel the r2 mgh = 1/2mv2 + 1/2(2/5m)(v2) mgh = 1/2mv2 + (2/10m)(v2) = 7/10mv2 mgh = 7/10mv2 finally cancel the m gh = 7/10v2 so the final velocity is v = Ö(10gh/7) = 3.0741 m/s Now we can find the angular velocity of the marble: (just for fun!!!!) w = v/r = (3.0741 m/s)/(.342x10-2 m) = 898.8553303 rad/s The rotational kinetic energy can be found from the linear velocity if we follow the substitutions: Ekrot = 1/2Iw2 = 1/2(2/5mr2)(v/r)2 = (2/10m)(v2) = 2/10(.0342 m)(3.0741 m/s)2 = 0.06464 J Now let's find the translational kinetic energy: KE = 1/2(.0342)(3.0741 m/s)2 = 0.1616 J To find the angular acceleration, I am going to find the linear acceleration of the ball as it went down the plane, and then use a tangential relationship to find the angular acceleration: v2 = u2 + 2as :u = 0; s = 3.56 m; v = 3.0741 m/s a = 1.327 m/s/s a = a/r = (1.327 m/s/s)/(.342x10-2 m) = 388 rad/s/s (Table of contents) 8. A uniform cylinder with a mass of 5.6 kg and a radius of .32 m is free to rotate about a horizontal axis. There is a weight of 92 grams tied to a string that is wrapped around the cylinder. The weight accelerates toward the ground. What is the moment of inertia of the cylinder? What is the angular velocity of the cylinder when it has completed one revolution? What was the cylinder's angular acceleration as the weight fell? The moment of inertia of the cylinder is 1/2mr2 = 1/2(5.6 kg)(.32 m)2 = 0.287 kgm2 I am (no surprise) going to use energy again. The potential energy of the mass that falls is converted into kinetic energy of the mass, and rotational kinetic energy of the pulley: mgh = 1/2mv2 + 1/2Iw2 but the translational kinetic energy belongs solely to the falling mass, and the rotational kinetic energy is that of the cylinder. Let M be the mass of the cylinder, and r be its radius, and m be the falling mass: Substituting to linear quantities: w = v/r I = 1/2mr2 mgh = 1/2mv2 + 1/2(1/2Mr2)(v/r)2 mgh = 1/2mv2 + 1/4Mv2 We cannot cancel the masses as m is the falling mass, and M is the cylinder mass, but we can do a little factoring: mgh = v2(1/2m + 1/4M) v2 = mgh/(1/2m + 1/4M) The problem asks about after the cylinder has completed one revolution, and we need the height, so the change in height of the falling mass would be one circumference: h = circumference of pulley = 2pr = 2p(.32 m) = 2.0106 m v = Ö{mgh/(1/2m + 1/4M)} = Ö{(.092 kg)(9.8 m/s/s)(2.0106 m)/(1/2(.092 kg) + 1/4(5.6 kg))} = 1.1197 m/s Now we can find the angular final velocity using tangential relationships: w = v/r = (1.1197 m/s)/(.32 m) = 3.49895011 rad/s = 3.5 rad/s To find the angular acceleration of the pulley as the mass fell, let's find the linear acceleration of the falling mass (which is also the tangential acceleration of the cylinder, as the string is wrapped around the edge v2 = u2 + 2as : u = 0; s = 2.0106 m; v = 1.1197 m/s a = 0.3118 m/s/s a = a/r = (0.3118 m/s/s)/(.32 m) = 0.974 rad/s/s (Table of contents) 9. In figure 8-44 (on page 237) the cylindrical pulley has a mass of 5.21 kg, a radius of .450 m, mass 1 is 7.82 kg, and mass 2 is 5.34 kg Mass 2 is resting on the ground, and mass 1 is 17.2 cm above the ground. Calculate the vertical acceleration of the masses, and the speed at which mass 1 hits the ground. \| \| \| --- \| \| Before \| After \| \| \| \| Energy time. The potential energy of mass 1 goes to the translational kinetic energies of both masses and to the rotational kinetic energy of the pulley as well as the potential energy of mass 2, just before mass 1 strikes the ground: m1gh = 1/2m1v2 +1/2m2v2 + m2gh + 1/2Iw2 Substitute for the rotational term: w = v/r I = 1/2mr2 m1gh = 1/2m1v2 +1/2m2v2 + m2gh + 1/2(1/2Mr2)(v/r)2 m1gh = 1/2m1v2 +1/2m2v2 + m2gh + 1/4Mv2 Now solve for the final velocity: m1gh -m2gh = 1/2m1v2 +1/2m2v2 + 1/4Mv2 m1gh -m2gh = v2( 1/2m1 +1/2m2 + 1/4M) gh(m1 -m2) = v2( 1/2m1 +1/2m2 + 1/4M) gh(m1 -m2)/( 1/2m1 +1/2m2 + 1/4M) = v2 v = Ö{gh(m1 -m2)/( 1/2m1 +1/2m2 + 1/4M)} v = Ö{(9.80 m/s/s)(.172 m)((7.82 kg) -(5.34 kg))/( 1/2(7.82 kg) +1/25.34 kg) + 1/4(5.21 kg))} v = 0.72823 m/s Now to calculate the linear acceleration of the masses: v2 = u2 + 2as :u = 0; s = .172 m; v = 10.72823 m/s a = 1.54 m/s/s (Table of contents) 10. What is the angular momentum of a gyroscope that is a solid cylinder with a radius of .24 m, a mass of 15 Kg and a angular velocity of 140 rad/sec The formula for angular momentum is the direct analog for linear momentum p = mv: L = Iw We need to calculate the moment of inertia. I = 1/2mr2 I = 1/2(15 kg)(.24 m)2 = 0.432 kgm2 Now calculate the angular momentum: L = Iw = (0.432 kgm2)(140 rad/s) = 60.48 kgm2/s (Table of contents) 11. A ballerina spinning at 1.2 rev/sec with a moment of inertia of 2.6 Kgm2 pulls her arms in so that her new moment of inertia is 1.8 Kgm2. What is her new angular speed? This is a conservation of angular momentum problem. The angular kinetic energy would not be conserved as the ballerina would do work in pulling in her arms that would be manifested as rotational kinetic energy. (it would go up) Now, the formula for angular momentum is: L = Iw So basically, L before = L after: I1w1 = I2w2 (2.6 Kgm2)(1.2 rev/sec) = (1.8 Kgm2)w2 w2 = 1.73rev/s Note that the units cancel, so we don't have to convert to radians per second (Table of contents) 12. A group of children playing on a merry go round spinning at 52 rpm with a moment of inertia of 200 Kgm2 move to its center so that the new moment of inertia is 120 Kgm2. What is the new angular speed? This is a conservation of angular momentum problem. The angular kinetic energy would not be conserved as the ballerina would do work in pulling in her arms that would be manifested as rotational kinetic energy. (it would go up) Now, the formula for angular momentum is: L = Iw So basically, L before = L after: I1w1 = I2w2 (200 Kgm2)(52 RPM) = (120 Kgm2)w2 w2 = 86.7 RPM Note that the units cancel, so we don't have to convert to radians per second (Table of contents) 13. A figure skater spinning at 3.4 rad/sec with a moment of inertia of 3.2 Kgm2 puts his arms out so that his new moment of inertia is 4.5 Kgm2. What is his new angular speed? So basically, L before = L after: I1w1 = I2w2 (3.2 Kgm2)(3.4 rad/s) = (4.5 Kgm2)w2 w2 = 2.42 rad/s Note that the units cancel, so we don't have to convert to radians per second (Table of contents)
15010	https://helpingwithmath.com/worksheet/piecewise-functions-math-activities/	Piecewise Functions Math Activities \| Downloadable Worksheet Skip to content Helping with Math Menu Home Membership All Worksheets Log In Menu By Grade Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade 7th Grade 8th Grade End Of Year Test Booklets Common Core Worksheet Mapping By Age Age 4-6 Age 5-7 Age 6-8 Age 7-9 Age 8-10 Age 9-11 Age 10-12 Age 11-13 Age 12-14 By Topic Addition Algebra Algebraic Expressions Angles Area Basic Facts Decimals Division Equations Factors Fractions Functions Geometry Graph & Charts Integers Measurement Multiplication Number Sense Percentages Perimeter Place Value Polynomials Radicals Ratio Rounding Shapes Statistics Subtraction Time Volume Word Problems Math Skills Themes Animals & Living Things Celebrations Career/Work Education Fun Health & Fitness Seasonal Space Worksheets Calendar Common Core Mapping Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade 7th Grade 8th Grade Common Core Exam Booklets Math Common Core Standards Additional Resources Math Calculators Math Questions Math Quizzes Math Flash Cards Math Table Charts Seasonal Worksheets Calendar Math Theory Algebra Calculus Decimals Factors Fractions Geometry Graphs & Charts Measurement & Time Numbers Number Sense Operations of Numbers Rounding Statistics & Probability Trigonometry Home » Worksheets » Functions » Piecewise Functions Math Activities Piecewise Functions Math Activities Download Piecewise Functions Math Activities Click the button below to get instant access to these premium worksheets for use in the classroom or at a home. 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Common Challenges Piecewise Functions Math Activities Resource Examples Worksheets Activities Included Definition A piecewise function is a function defined by different expressions or rules for different intervals of the input variable. Instead of one single equation, a piecewise function uses multiple equations, each applying to a specific part of the domain. Example: Summary 1. Definition and Structure Each “piece” of the function has: Its own equation (expression) A condition (domain restriction) that determines when that equation is used. The pieces are usually separated using braces {⋯ }. 2. Evaluating Piecewise Functions Identify which part of the domain the input belongs to. Use the correct expression for that interval. Plug in the value and simplify. 3. Graphing Piecewise Functions Plot each piece only on the interval where it’s defined. Include open or closed dots depending on whether the interval includes the endpoint. Graphs can look disconnected or have sharp changes (“jumps”). 4. Applications Real-world situations with different conditions or rates: Shipping costs that vary by weight Tax brackets Speed limits over time or zones Pay based on hours worked (e.g., overtime) 5. Common Challenges Correctly interpreting which piece to use. Dealing with endpoints (using open vs. closed circles). Ensuring continuity or identifying discontinuity in the grap Piecewise Functions Math Activities This is a fantastic bundle which includes 10 activities with answer guides about Piecewise Functions. All our worksheets are completely editable so can be tailored for your curriculum and target audience. Resource Examples Click any of the example images below to view a larger version. 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15011	https://www.jospt.org/doi/pdf/10.2519/jospt.2008.2454	78 \| february 2008 \| volume 38 \| number 2 \| journal of orthopaedic & sports physical therapy [ CASE REPORT ] procedure. 3,4 The spinal accessory nerve is a cranial nerve originating as 2 parts: the accessory portion from the medulla and the spinal part from the lateral portion of the ven-tral column. The spinal portion ascends and enters the skull through the fora-men magnum to join the accessory part. The spinal accessory nerve then exits through the jugular foramen at the base of the skull, passes obliquely, penetrating the upper third of the sternocleidomas-toid (SCM) muscle. The spinal accessory nerve continues subcutaneously through the posterior triangle floor and enters the upper trapezius ( FIGURE 1 ). 17 Both the SCM and trapezius muscle receive mo-tor innervation from the spinal accessory nerve. Although most authors agree the spinal accessory nerve provides the tra-pezius with primary motor innervation, not all agree regarding nerve anatomy. Kiener 26 reported intraoperative electro-myography findings revealing a distinct cranial branch of the spinal accessory nerve that separately innervates the up-per trapezius. The cervical plexus (C2-C4) also innervates the trapezius providing some motor input in approximately 20% of the population. 40 Spinal accessory nerve injury com-monly occurs during neck dissection surgery. 2,6,9,11,13-15,27-29,31,43,46,48,49,52,53 Neck dissection surgery is performed to treat head and neck carcinoma and is catego-rized into 3 different procedures: radical Spinal accessory nerve palsy (SANP) is common following neck dissection surgery or lymph node excision, 2,6,9,11,13-15,27-29,31,43,46,48,49,52,53 blunt or penetrating trauma to the lateral neck region, 2,11 and cervical stretch injuries. 32 Spinal accessory nerve injury results in trapezius paralysis or dysfunction and a diagnostic cluster of signs and symptoms, including shoulder girdle depression, trapezius atrophy, scapular dyskinesis, loss of shoulder T STUDY DESIGN: Retrospective case series. T BACKGROUND: Spinal accessory nerve palsy (SANP) is common following neck dissection sur-gery and can occur with blunt or penetrating trauma to the lateral neck region and cervical stretch injuries. Early detection of SANP remains a clinical challenge and the condition is often misdiagnosed. The purpose of this case series is to describe the associated history, signs, and symptoms related to SANP and increase awareness of the scapular flip sign as a clinical sign associated with SANP. T CASE SERIES DESCRIPTION: Twenty subjects (13 male, 7 female) presented with pain and decreased shoulder function following head and neck surgery or posttrauma. All patients were thor-oughly examined and the scapular flip sign was assessed. All patients presented with a cluster of signs and symptoms, including trapezius atrophy, shoulder girdle depression, limited active shoulder abduction to less than 90°, shoulder pain, and shoulder weakness. A positive scapular flip sign was present in all cases. The middle and lower tra-pezius were rated as 0/5, based on manual muscle testing, indicating no identifiable muscle activation against resistance. T DISCUSSION: A typical history and consistent signs and symptoms were found related to SANP. A strong relationship appeared between the presence of the scapular flip sign and SANP. The suspected mechanism for the scapular flip sign is the unopposed pull of the humeral external rota-tors by the inactive middle and lower trapezius. Early identification of SANP can assist with the prognosis, explain persistent impairments and functional deficits, motivate appropriate diagnostic testing and interventions, and help maximize outcome. Further research to validate the scapular flip sign and establish a clinical prediction rule for the diagnosis of SANP should be performed. T LEVEL OF EVIDENCE: Diagnosis, Level 4. J Orthop Sports Phys Ther 2008;38(2):78-86. doi:10.2519/jospt.2008.2454 T KEY WORDS: examination, neck, shoulder, trapezius is usually initiated; however, nerve re-pair, nerve grafting, or neurolysis may be performed after penetrating or known iatrogenic injury. 27,42 Chronic SANP may also be treated with a muscle transfer active abduction, and shoulder/neck pain. 6,9,11,13-15,28,29,31,46,48,49,53,52 Establish-ing a diagnosis of SANP can be a clini-cal challenge and is often delayed or missed. 3,4,11,42,55 Conservative treatment Spinal Accessory Nerve Palsy: Associated Signs and Symptoms MARTIN J. KELLEY, PT, DPT, OCS 1 • THOMAS E. KANE, DPT 2 • BRIAN G. LEGGIN, PT, DPT, OCS 1 1 Musculoskeletal Team Leader, University of Pennsylvania, Philadelphia, PA. 2 Staff Physical Therapist, King of Prussia Physical Therapy and Sports Rehabilitation, King of Prussia, PA. The protocol used in this case series was approved by the Institutional Review Board at the University of Pennsylvania Medical Center. Address correspondence to Martin J. Kelley, PT, DPT, OCS, Penn Presbyterian Medical Center-University of Pennsylvania Health System, 1 Medical Office Building, Suite 110, 39th and Market St, Philadelphia, PA 19104. E-mail: martin.kelley@uphs.upenn.edu Journal of Orthopaedic & Sports Physical Therapy® Downloaded from www.jospt.org at on September 29, 2025. For personal use only. No other uses without permission. Copyright © 2008 Journal of Orthopaedic & Sports Physical Therapy®. All rights reserved. journal of orthopaedic & sports physical therapy \| volume 38 \| number 2 \| february 2008 \| 79 neck dissection (RND), modified radi-cal neck dissection (MRND), and selec-tive neck dissection (SND). 47 During the RND, all levels of lymph nodes are resected along with the spinal accessory nerve, the SCM muscle, and the internal jugular vein. Functional outcomes fol-lowing RND are inferior to spinal ac-cessory nerve-sparing procedures due to the resultant chronic trapezius dysfunc-tion. 7,9,13,33,43,50,53 The MRND and SND procedures were developed to spare the spinal accessory nerve, thus limit trape-zius dysfunction; however, significant in-cidence of spinal accessory nerve injury has been reported 6,10,13,14,52,53 that might have resulted from nerve manipula-tion. 6,29,43,50,52 The MRND requires lymph node dissection at all levels, but the spi-nal accessory nerve is usually spared. The SND can vary in technique depending on the location and size of the primary tumor and targeted or suspicious lymph nodes. All nonlymphatic structures are spared, including the spinal accessory nerve. Therefore, the clinician needs to rule out a SANP in any patient treated following neck dissection surgery and any surgical procedure or trauma involving the lateral neck region. Spinal accessory nerve injury results in trapezius dysfunction causing both static and dynamic shoulder girdle altera-tions. 7,22 The trapezius is divided into 3 distinct heads: upper, middle, and lower. The upper fibers descend from the base of the skull and cervical spine to insert on the posterior distal third of the clavicle and acromion, creating a long suspen-sion lever arm, which statically prevents shoulder girdle depression. The middle and lower fibers extend from the thorac-ic spine and insert on the scapular spine base and triangular aponeurosis. The middle and lower trapezius fibers stati-cally and dynamically tether the scapula to the thoracic spine, limiting excessive protraction. Each head of the trapezius has a primary dynamic function on the scapula. The upper trapezius elevates, the middle trapezius retracts, and the lower trapezius depresses. In unison, the pri-mary function of the trapezius is to up-wardly rotate the scapula during shoulder elevation, forming a force couple with the serratus anterior. 18,37 SANP results in the loss of trapezius static restraint, causing shoulder girdle depression and protrac-tion. 6,7,15,46,53 Dynamically, trapezius dys-function results in dramatically altered scapular kinematics and limited shoulder elevation. 7,13,14,22,46,53 Diagnosis of SANP is typically achieved by identifying associated signs such as trapezius atrophy, depressed shoulder girdle, scapular dyskinesis, tra-pezius weakness, and limited shoulder active abduction. 6,9,11,13-15,28,29,31,46,48,49,53,52 Unfortunately, many of these signs are observational and unreliably recognized, which may explain a high rate of misdi-agnosis. 3,4 Assessing scapular dyskinesis is unreliable. 25 Reliability and validity of manual muscle testing is controver-sial 5,16,38,39 and testing the upper trapezius by scapular shrugging is suspect because the levator scapula and rhomboid can adequately perform this activity. 23,36,41 Additionally, the proposed muscle test positions for the middle and lower tra-pezius can be provocative in the early postoperative/postinjury period. Early identification of SANP can assist with the prognosis by explaining persistent impairments and functional deficits, enabling appropriate diagnostic test-ing and interventions, and maximizing outcomes. 3,4,14,27,44,46 The purpose of this case series is to (1) describe the associated history, signs and symptoms related to SANP and (2) increase awareness of the scapular flip sign as a simple clinical sign associated with SANP. CASE SERIES DESCRIPTION The series consisted of 20 pa- tients (13 male, 7 female), ranging from 18 to 79 years of age (mean, 51). Initial physical therapy examina-tion ranged from 1 to 17 months fol-lowing onset (mean, 4 months). TABLE 1 shows the demographic and patient characteristics. TABLE 1 Demographic and Patient Characteristics Mean age 51 y Age range 18-79 y Gender 13 male, 7 female Involved side 11 right, 9 left Dominant side 18 right hand, 2 left hand Type of surgery 6 radical neck dissections, 7 modified radical neck dissections, 2 selective neck dissections, 1 endarterectomy, 1 mandibular lengthening Primary site of cancer 5 tonsillar carcinoma, 4 thyroid carcinoma, 1 pharyngeal carcinoma, 1 palate carcinoma, 1 carcinoma base of tongue, 3 unknown location Nonoperative injuries 1 rugby injury, 1 carrying heavy lumber, 1 carrying heavy bag FIGURE 1. The spinal accessory nerve (SAN) crosses obliquely in the posterior triangle. (Reproduced with permission of Kuhn, JE. The scapulothoracic articulation: anatomy, biomechanics, pathophysiology and management. In: Iannotti JP, Williams GR, eds. Disorders of the Shoulder: Diagnosis and Management. Philadelphia, PA: Lippincott Williams & Wilkins; 1999). Journal of Orthopaedic & Sports Physical Therapy® Downloaded from www.jospt.org at on September 29, 2025. For personal use only. No other uses without permission. Copyright © 2008 Journal of Orthopaedic & Sports Physical Therapy®. All rights reserved. 80 \| february 2008 \| volume 38 \| number 2 \| journal of orthopaedic & sports physical therapy [ CASE REPORT ] History Nine patients presented to physical ther-apy with a diagnosis of SANP: 1 following a stretch injury, 6 following RND, and 2 following MRND. The other 5 patients post-MRND and the 2 patients post-SND had prescriptions identifying the surgical procedure and impairments, and 4 had a prescription that read “spinal accessory nerve palsy?” The provided diagnosis in 4 of the 5 patients remaining in the non-cancer group was shoulder impingement in 2 patients, frozen shoulder in 1 patient, and “scapular dyskinesia” in the remain-ing patient. Pain complaints varied among the pa-tients, but a pattern could be appreciated. Patients in the noncancer group, with the exception of the patient who had the end-arterectomy, reported a period (approxi-mately 3 to 14 days) of pain/achiness in the scapular region following the iden-tified event. At the time of their initial therapy visit the patients in this group only complained of scapular region pain or occasional shoulder pain with repeti-tive arm use. The patients who had neck dissections all had postoperative pain about the neck and scapula, but it was difficult to distinguish pain related to sur- TABLE 2 Manual Muscle Testing Results and Active and Passive Range of Motion Values and Means for Abduction and Flexion Flexion Involved (º) MMT Abbreviations: AROM, active range of motion; LT, lower trapezius; MMT, manual muscle testing; MRND, modified radical neck dissection; MT, middle tra-pezius; NA, not available; PROM, passive range of motion; RND, radical neck dissection; SCM, sternocleidomastoid; SND, sparing neck dissection; UT, upper trapezius. Range of motion values for the uninvolved side were within normal limits. Abduction Involved (º) AROM PROM AROM PROM UT MT LT SCM RND Patient 1 20 120 90 130 0/5 0/5 0/5 Absent Patient 2 40 130 95 135 0/5 0/5 0/5 Absent Patient 3 65 155 110 160 0/5 0/5 0/5 Absent Patient 4 70 145 115 155 0/5 0/5 0/5 Absent Patient 5 80 140 120 145 0/5 0/5 0/5 Absent Patient 6 70 155 130 160 0/5 0/5 0/5 Absent Mean 58 141 110 148 ... ... ... ... MRND Patient 7 60 130 80 135 0/5 0/5 0/5 NA Patient 8 70 155 100 160 0/5 0/5 0/5 0/5 Patient 9 40 135 100 140 0/5 0/5 0/5 0/5 Patient 10 50 155 120 160 0/5 0/5 0/5 0/5 Patient 11 70 150 125 150 0/5 0/5 0/5 NA Patient 12 70 155 145 165 1/5 0/5 0/5 5/5 Patient 13 65 150 130 160 1/5 0/5 0/5 NA Mean 61 147 114 153 ... ... ... ... SND Patient 14 80 155 140 160 0/5 0/5 0/5 5/5 Patient 15 90 165 150 165 2/5 0/5 0/5 NA Mean 85 160 145 163 ... ... ... ... Surgical non-cancer Patient 16 40 120 120 130 0/5 0/5 0/5 5/5 Patient 17 60 160 135 160 0/5 0/5 0/5 NA Mean 50 140 128 145 ... ... ... ... Non-operative Patient 18 80 155 140 160 2/5 0/5 0/5 5/5 Patient 19 85 155 150 160 2/5 0/5 0/5 5/5 Patient 20 90 165 160 165 0/5 0/5 0/5 5/5 Mean 85 158 150 162 ... ... ... ... Journal of Orthopaedic & Sports Physical Therapy® Downloaded from www.jospt.org at on September 29, 2025. For personal use only. No other uses without permission. Copyright © 2008 Journal of Orthopaedic & Sports Physical Therapy®. All rights reserved. journal of orthopaedic & sports physical therapy \| volume 38 \| number 2 \| february 2008 \| 81 gery from that related to spinal accessory nerve injury. A trend of greater pain was associated with extensive neck dissection. The most common complaint among all patients was shoulder weakness. Examination All patients were examined by the first author (M.J.K.). On clinical examina-tion, all patients presented with dramatic trapezius atrophy and a depressed shoul-der girdle. The cervical spine and shoul-der were thoroughly examined. Cervical spine motion was not objectively mea-sured but visually judged to be restricted to some extent, especially for extension, contralateral side bending, and rotation to both sides in all 16 patients having neck dissection surgery and endarterec-tomy. Some motion discomfort was pres-ent to varying extent and was localized to the neck, superior scapula, and supracla-vicular region. The other individuals in the noncancer group had normal pain-free cervical motion. Shoulder range of motion was measured with a standard goniometer ( TABLE 2 ). 23 All 20 patients presented with a posi-tive scapular flip sign. The scapular flip sign, initially described by Kelley, 20,21 was performed with the patient standing, arm at the side, and elbow flexed to 90°. The examiner resisted glenohumeral external rotation at the distal forearm while visu-ally examining the scapula. A positive scapular flip sign was present if the me-dial scapular border was noted to “flip” or lift from the thoracic wall during resisted shoulder external rotation ( FIGURE 2A ). An upper-quarter screen was per-formed and no neurologic deficits were noted in the upper extremity muscles other than the trapezius. Relative shoul-der external rotator and abductor weak-ness was present in most patients upon manual muscle testing, but the weakness appeared related to the effect of having recent surgery or trapezius absence re-sulting in compromised proximal shoul-der girdle fixation. The SCM and trapezius (upper, mid-dle, and lower) muscles were tested us-ing Kendall and McCreary 23 techniques. TABLE 2 shows strength grade results. SCM testing was not warranted in the 6 patients post-RND because it was surgi-cally excised. Five patients demonstrated some ac-tivity of the anterior one-third to two-thirds of the upper trapezius. Patients with partial return of the upper trape-zius tended to have better active arm elevation. Electromyography (EMG) was per-formed 6 to 12 weeks following symptom onset before coming to physical therapy in 2 patients without cancer-related on-set. A summary of their findings is found in TABLE 3 . Complete electrodiagnostic testing of the trapezius (EMG of all 3 sections and a nerve conduction study [NCS] of the spinal accessory nerve) was only performed in 1 patient. Based upon the history and physical examination findings ( TABLE 4 ), the clini-cal diagnosis for all patients was primary SANP resulting in pain, limited shoulder elevation, and functional deficits. One patient had a secondary frozen shoulder. Possible differential diagnoses in-cluded cervical radiculopathy, brachial plexopathy/neuritis, and long thoracic nerve palsy. DISCUSSION Aclear relationship is present between SANP and a history of neck dissection or lymph node ex-cision. 2,6,9,11,13-15,27-29,31,43,46,48,49,52,53 Donner and Kline 11 found 59 of 83 cases (70%) with SANP were related to neck dissec-tion surgery. Kim et al 27 reported that the etiology of 91% of 111 patients surgi-cally treated for SANP was iatrogenically induced either by lymph node biopsy FIGURE 2. (A) A patient with a spinal accessory nerve palsy (SANP) demonstrating a positive scapular flip sign on the left during bilateral resisted shoulder external rotation. (B) Schematic demonstrates a positive scapular flip sign. During resisted humeral external rotation the pull of the infraspinatus and posterior deltoid on the scapula is unopposed by the inactive middle and lower trapezius, causing the medial border of the scapula to “flip” off the thoracic wall. (Reproduced with permission of Kelley, MJ. Clinical evaluation of the shoulder. In: Mackin EL, Callahan AD, Skervin TM, Schneider AH, Osterman AL, eds. Rehabilitation of the Hand. 5th ed. Oxford, UK: Mosby; 2002). TABLE 3 Electrodiagnostic Testing Results Before and After Initial Physical Therapy Visit Abbreviations: EMG, electromyography; LT, lower trapezius; MT, middle trapezius; NCS, nerve con-duction study; NP, not performed; PT, physical therapy; UT, upper trapezius. Pre-PT Muscle Pre-PT EMG Tested NCS Post-PT EMG Muscle Tested Post-PT NCS Patient 17 NP ... NP Abnormal UT NP Patient 18 NP ... NP Normal-clinically abnormal UT NP Patient 19 Normal UT NP Abnormal UT, MT, LT Abnormal Patient 20 Abnormal UT NP ... ... ... Journal of Orthopaedic & Sports Physical Therapy® Downloaded from www.jospt.org at on September 29, 2025. For personal use only. No other uses without permission. Copyright © 2008 Journal of Orthopaedic & Sports Physical Therapy®. All rights reserved. 82 \| february 2008 \| volume 38 \| number 2 \| journal of orthopaedic & sports physical therapy [ CASE REPORT ] or tumor excision. The spinal accessory nerve is often injured during neck dis-section surgery because the nerve is manipulated causing traction or devascu-larization. 6,10,29,43 The incidence of SANP associated signs and symptoms is 47% to 100%, 9,15,43,46 18% to 77%, 13,15,31,48 and 31% to 40%, 6,9,43,52 following a RND, MRND, and SND, respectively. Recognize that not every patient, following neck dissection, has significant symptoms at the time of follow-up and that not all studies report on all associated signs and symptoms. MRND and SND spare the spinal acces-sory nerve, but nerve traction can result in temporary nerve damage and trapezius motor interruption that shows recovery beyond 3 months. 6,10,29,43,46,50 Regardless of the type of neck dissection, abnormal electrodiagnostic studies were found in 75% to 100% of patients who under-went SND 6,14,29,43 and in 100% of patients following MRND 14,29,49 up to at least 3 months following surgery. Preservation of the cervical plexus has not been found to decrease symptoms or improve func-tion. 13 Therefore, patients following neck dissection are an excellent population to observe and determine the common clus-ter of signs, symptoms, and special test associated with SANP. The absence of electrodiagnostic stud-ies in patients after neck dissection in this series is a limitation. However, consider-ing the high incidence of spinal accessory nerve injury following neck dissection, the presence of SANP signs and symp-toms in our patients and the fact that the average length of time from surgery to ex-amination was 3 months for our patients following neck dissection, it is likely that the patients in our series had a spinal ac-cessory nerve injury. Pain Pain complaints following neck dissec-tion are common. Lower pain scores have been reported in SND and MRND when compared to RND. 43,50 Moderate to severe pain was reported following RND and MRND in 60% and 52% of cases, respectively. 13 Cappiello et al 6 compared disability in patients following 2 different types of SND and found that 80% to 85% of patients had no pain complaints, and only mild pain was found in the remain-ing patients at least 1 year from surgery. Sixty-nine percent of patients with spi-nal accessory nerve dysfunction had pain complaints at 3 years following neck dis-section surgery. 53 All patients who underwent neck dis-section in this series complained of neck and/or scapular pain. Greater pain com-plaints were noticed in those having a more extensive surgical intervention. Many pa-tients had achiness in the upper scapular region that may have been related to the neck dissection, spinal accessory nerve injury, or levator scapula/rhomboid over-load. The static suspensatory role and dy-namic responsibility of the levator scapula and rhomboid is magnified with trapezius paralysis. 3,4,6 It is possible that all patients post neck dissection in this series had pain complaints because most were examined soon after surgical intervention. Patients in the noncancer group did appear to have a pattern of pain com-plaints. Four of the 5 patients reported an interval of pain and achiness immedi-ately following the identified event. The short interval of pain may represent the nerve injury response. 19 At the time of the physical therapy examination, pain was only present in the scapular region or occasionally in the lateral shoulder area with repetitive use. The intermittent scapular pain described with repetitive use may represent levator scapulae and rhomboid overuse, 3,4,6 while improper scapular mechanics may cause rotator cuff irritation. 37 Trapezius Atrophy and Shoulder Girdle Depression Trapezius muscle atrophy and a de-pressed shoulder girdle are hallmark signs of SANP ( FIGURE 3 ). 15,53 Although all patients in our series presented with at-rophy, an examiner may be deceived by the residual flaccid trapezius and misdi-agnose a patient if only visual inspection is relied upon or an incomplete examina-tion performed. This may explain why several referring physicians in our series did not identify the SANP. Five patients had partial return of the upper trapezius that may have led to the false impression that all trapezius heads were active. All patients in this series had a de-pressed shoulder girdle. Caversaccio et al 7 measured shoulder depression in pa-tients following RND and MRND and found the operated shoulder 3.9 cm low-er than the unoperated side. Van Wilgen et al 53 found that all patients presenting with trapezius atrophy had a depressed shoulder girdle. Shoulder girdle depres-sion most likely results from trapezius paralysis/dysfunction and the loss of tra-pezius static support to oppose the upper extremity downward force. 3,4,15 Sternocleidomastoid Muscle Cappellio 6 found SCM EMG abnormali-ties in 40% and 45% of patients who had 2 types of SND performed. Dissection above the SCM innervation can result in motor interruption. 6 In this series, the TABLE 4 Spinal Accessory Nerve Palsy Associated History and Examination Findings History Neck dissection/lymph node excision Blunt or penetrating trauma to the lateral neck Cervical stretch injury Symptoms Scapular/neck pain Shoulder weakness Signs Shoulder girdle depression (drooping) Trapezius atrophy Limited active coronal plane abduction Scapular dyskinesis Scapular flip sign Sternocleidomastoid absent/weak Weakness of middle and lower trapezius Journal of Orthopaedic & Sports Physical Therapy® Downloaded from www.jospt.org at on September 29, 2025. For personal use only. No other uses without permission. Copyright © 2008 Journal of Orthopaedic & Sports Physical Therapy®. All rights reserved. journal of orthopaedic &sports physical therapy \| volume 38 \| number 2 \| february 2008 \| 83 SCM was absent as in all patients follow-ing a RND. Several other patients dem-onstrated SCM paralysis following neck dissection. The SCM should always be examined if trapezius weakness is noted. Paralysis or weakness of this muscle is a strong indicator that the spinal accessory nerve has been injured. None of the pa-tients in the noncancer group had weak-ness of the SCM. Shoulder Range of Motion SANP following neck dissection can re-sult in limited active range and functional deficits. 13,14,50 All patients in our series had greater active shoulder flexion than coro-nal plane abduction. We found a similar loss of active flexion (36°) as reported in previous studies. 13,14 The patients in our noncancer group had a loss of active flex-ion (17°) similar to those who underwent a SND (20°). 14 Shoulder flexion is proba-bly less affected by SANP because the ser-ratus anterior is a more effective scapular upward rotator in flexion compared with coronal plane abduction. 12,18,34 The loss of active abduction was sig-nificantly more in our patients (90.2°) than reported in other studies. 14,13,53 The early time frame from surgery to exami-nation could explain this difference be-cause the patients with neck dissection in this series were an average of 3 months from surgery versus 6 months to 2 years in other studies. 13,14,53 Trapezius reinner-vation has been shown to significantly improve after 6 months and may result in greater active range of motion. 14 Com-pensation from other scapular muscles over time may also result in improved abduction. When measuring active ab-duction, we strictly maintained coronal plane abduction while externally rotat-ing the arm, which may have resulted in recording less active abduction. End range measurements were recorded if the arm moved anteriorly out of coronal plane. Although patients in the noncan-cer group only demonstrated a 19° loss of active flexion, they still had limited active abduction of 90° or less. Lower and Middle Trapezius Muscle Testing Kendall and McCreary’s lower and mid-dle trapezius muscle test positions were used in this series because these posi-tions demonstrate the highest levels of middle and lower trapezius muscle acti-vation. 12,23,38,39,41 Kendall and McCreary 23 strongly advocated maintaining the hu-merus in external rotation when testing the middle trapezius to prevent rhomboid activation. Mosely et al 41 found significant middle trapezius and rhomboid activa-tion during prone horizontal abduction with neutral arm rotation. However, when external rotation was performed during prone horizontal abduction, rhomboid activity did not reach qualify-ing activation criteria while the middle trapezius continued to demonstrate sig-nificant activation (manual muscle test-ing, 96%). In our series, all patients had a 0/5 strength grade of the middle and lower trapezius. No palpable rhomboid activity was noted as long as correct test positions were maintained. If humeral internal rotation was allowed during the middle trapezius test, rhomboid activa-tion was palpable. Allowing rhomboid activation may result in misdiagnosis or mislead the clinician regarding trapezius strength. Trapezius muscle test positions were well tolerated even by patients fol-lowing neck dissection. Scapular Dyskinesis Scapular dyskinesis refers to observable alterations in the scapular position or motion during coupled scapulohumeral movements. 24,54 Although scapular dys-kinesis is associated with SANP and was appreciated in our patients, dyskinesis alone cannot confirm the diagnosis. 25,35,37 Caversaccio et al 7 and Kelley et al 22 found significantly reduced upward rotation in patients following RND when using dif-ferent 3-dimensional tracking devices. Interestingly, Caversaccio et al 7 reported that the distance between the scapular spine and the thoracic spine increased at 90° of elevation in patients who had RND, presumably as a result of overcom-pensation of the serratus anterior pulling the scapula into protraction. Scapular Flip Sign The scapular flip sign is proposed as a clinically relevant test to identify ab-normal scapular motion associated with SANP and trapezius paralysis. Kelley et al 20,21 first described this sign as associat-ed with SANP. More recently Chan et al 8reported on the same sign being present in 6 patients with SANP, but the sign was referred to as the “resisted external rota-tion sign.” A positive scapular flip sign was present in all patients in our series. We propose that the scapular flip sign is distinct from medial scapular winging associated with long thoracic nerve palsy for 2 reasons: the abnormal scapular dis-placement produced by the flip sign as-sessment is different from patients with long thoracic nerve palsy or serratus an-terior weakness, and the manner of pro-moting the scapular winging varies. Patients with long thoracic nerve palsy demonstrate significant scapular winging during active shoulder flexion ( FIGURE 4A ). Serratus anterior insufficiency creates medial scapular winging because the entire scapula drifts posteriorly. Resist-ing shoulder flexion results in the same scapular displacement. Although some variation is seen among patients with SANP, shoulder flexion can be achieved without medial winging because the ser-ratus is intact ( FIGURE 4B ). Active abduc-tion is limited in the patient with SANP, and dyskinesis may be noted; however, the medial border does not significantly displace off the thorax ( FIGURE 4C ). FIGURE 3. Patient with spinal accessory nerve palsy (SANP) demonstrates significant left trapezius atrophy and shoulder girdle depression. Journal of Orthopaedic & Sports Physical Therapy® Downloaded from www.jospt.org at on September 29, 2025. For personal use only. No other uses without permission. Copyright © 2008 Journal of Orthopaedic & Sports Physical Therapy®. All rights reserved. 84 \| february 2008 \| volume 38 \| number 2 \| journal of orthopaedic & sports physical therapy [ CASE REPORT ] A positive scapular flip sign and mech-anism of displacement are seen in FIGURE 2. The clinician resists shoulder external rotation while the arm is at the side. Re-sisting the glenohumeral external rota-tors results in significant activity of the infraspinatus and posterior deltoid. 45,51 In the presence of SANP, the pull of the infraspinatus and posterior deltoid is un-opposed by the middle and lower trape-zius, causing the medial scapular border to “flip” off the thoracic wall. Scapular motion during a positive flip sign occurs through a vertical axis. Although medial border winging can be distinctly pro-duced in patients having a long thoracic nerve or SANP, the mechanisms of dis-placement are different. The scapular flip sign was also pres-ent in patients with partial return of the upper trapezius. The upper trapezius acts as a scapular upward rotator and eleva-tor, therefore, even when active, it cannot oppose the pull of the humeral external rotators. The middle and lower trapezius appear to play the major role stabiliz-ing the scapular medial border during resisted humeral external rotation. The rhomboids were clinically intact in all pa-tients; however, ineffective in stabilizing the scapular medial border when assess-ing for the scapular flip sign. Diagnosis Diagnosing patients with SANP can be challenging. 11,55 In our series only 9 of the 15 patients post neck dissection sur-gery were referred to physical therapy with a diagnosis of SANP. Identifying patients with a SANP after neck dissec-tion is essential because poor outcomes may occur if early physical therapy is not initiated. 6,14 Only 1 patient in the noncan-cer group presented with a diagnosis of SANP, while the other 4 patients were misdiagnosed. Three patients were sent for electrodiagnostic testing after their initial physical therapy visit because their history and examination was consistent with SANP. If a SANP is suspected a comprehen-sive EMG/NCS should be performed. However, electrodiagnostic testing has inherent difficulties. Bigliani et al 4 re-ported that 12 of 22 patients with SANP had inaccurate or incomplete EMG ex-aminations. Performing an EMG on a patient with significant trapezius atrophy is technically demanding because the nee-dle may penetrate the atrophied muscle and enter a neurologically intact muscle below, resulting in a false negative. EMG, although considered the gold standard for evaluating nerve function, is operator de-pendent. In addition, if a NCS is not per-formed on the spinal accessory nerve or only 1 portion of the trapezius is assessed, the examination may be considered in-complete. 3,4 Performing EMG/NCS too early may result in false negative studies; therefore, testing is usually deferred for 6 weeks following suspected onset. 1The diagnostic process for SANP should also rule out other diagnoses such as cervical radiculopathy, brachial plexopathy/neuritis, and long thoracic nerve palsy. These pathologies can cause some combination of cervical/scapular pain, shoulder girdle depression, limited shoulder motion, and scapular dyskine-sis (but not isolated trapezius paralysis). High-velocity trauma causing excessive stretching of the cervical region may in-jure the spinal accessory nerve, but also involves the brachial plexus, affecting multiple upper extremity muscles. 30 That a single examiner performed all examinations might have introduced bias. The inclusion of only those cases presenting with SANP-associated signs and symptoms might have as well. How-ever, the authors strongly believe that this series represents the clinical presentation of patients with SANP. CONCLUSION This case series highlighted the history, symptoms, and examina-tion findings associated with SANP. SANP should always be considered when examining patients following neck dissec-tion or other surgical procedures involv-ing the neck. SANP should also be ruled out in patients who have a history of blunt or penetrating trauma to the lateral neck region and cervical stretch injuries. Testing for the scapular flip sign is easy and appears to be associated with SANP. The scapular flip sign can be used in the early postoperative period without invok-ing pain or tissue trauma. Further testing of the trapezius heads can be performed using manual muscle testing techniques or electrodiagnostic studies. Future research should be performed to identify trapezius muscle activation in patients with suspected SANP when the flip sign is performed. A diagnostic va-lidity study on consecutive patients with appropriate inclusion criteria would de- FIGURE 4. (A) Active shoulder flexion causing medial winging in a patient with long thoracic nerve palsy. (B) Medial winging is not evoked during active flexion in patients with spinal accessory nerve palsy (SANP). (C) Active coronal plane abduction in a patient with spinal accessory nerve palsy (SANP). Note the absence of medial winging. Journal of Orthopaedic & Sports Physical Therapy® Downloaded from www.jospt.org at on September 29, 2025. For personal use only. No other uses without permission. Copyright © 2008 Journal of Orthopaedic & Sports Physical Therapy®. All rights reserved. journal of orthopaedic & sports physical therapy \| volume 38 \| number 2 \| february 2008 \| 85 [ CASE REPORT ] REFERENCES 1. Aminoff MJ. Electromyography in Clinical Prac-tice: Clinical and Electrodiagnostic Aspects of Neuromuscular Disease. New York, NY: Churchill Livingstone; 1993. Berry H, MacDonald EA, Mrazek AC. Accessory nerve palsy: a review of 23 cases. Can J Neurol Sci. 1991;18:337-341. Bigliani LU, Compito CA, Duralde XA, Wolfe IN. Transfer of the levator scapulae, rhom-boid major, and rhomboid minor for pa-ralysis of the trapezius. J Bone Joint Surg Am. 1996;78:1534-1540. Bigliani LU, Perez-Sanz JR, Wolfe IN. Treatment of trapezius paralysis. J Bone Joint Surg Am. 1985;67:871-877. Bohannon RW. Make tests and break tests of elbow flexor muscle strength. Phys Ther. 1988;68:193-194. Cappiello J, Piazza C, Giudice M, De Maria G, Nicolai P. Shoulder disability after different selective neck dissections (levels II-IV versus levels II-V): a comparative study. Laryngoscope. 2005;115:259-263. Caversaccio M, Negri S, Nolte LP, Zbaren P. Neck dissection shoulder syndrome: quantifica-tion and three-dimensional evaluation with an optoelectronic tracking system. Ann Otol Rhinol Laryngol. 2003;112:939-946. Chan PK, Hems TE. Clinical signs of accessory nerve palsy. J Trauma. 2006;60:1142-1144. Cheng PT, Hao SP, Lin YH, Yeh AR. Objective comparison of shoulder dysfunction after three neck dissection techniques. Ann Otol Rhinol Laryngol. 2000;109:761-766. Chepeha DB, Taylor RJ, Chepeha JC, et al. Func-tional assessment using Constant’s Shoulder Scale after modified radical and selective neck dissection. Head Neck. 2002;24:432-436. Donner TR, Kline DG. Extracranial spinal acces-sory nerve injury. Neurosurgery. 1993;32:907-910; discussion 911. Ekstrom RA, Donatelli RA, Soderberg GL. Sur-face electromyographic analysis of exercises for the trapezius and serratus anterior muscles. JOrthop Sports Phys Ther. 2003;33:247-258. El Ghani F, Van Den Brekel MW, De Goede CJ, Kuik J, Leemans CR, Smeele LE. Shoulder func-tion and patient well-being after various types of neck dissections. Clin Otolaryngol Allied Sci. 2002;27:403-408. Erisen L, Basel B, Irdesel J, et al. Shoulder func-tion after accessory nerve-sparing neck dissec-tions. Head Neck. 2004;26:967-971. Ewing MR, Martin H. Disability following radical neck dissection; an assessment based on the postoperative evaluation of 100 patients. Can-cer. 1952;5:873-883. Frese E, Brown M, Norton BJ. Clinical reliability of manual muscle testing. Middle trapezius and gluteus medius muscles. Phys Ther. 1987;67:1072-1076. Gray H, Pick TP, Howden RH. Anatomy. New York, NY: Crown Publishers; 1977. Inman VT, Saunders JB, Abbott LC. Observa-tions of the function of the shoulder joint. 1944. Clin Orthop Relat Res. 1996;3-12. Kauppila LI, Vastamaki M. Iatrogenic serratus anterior paralysis. Long-term outcome in 26 patients. Chest. 1996;109:31-34. Kelley M. Clinical evaluation of the shoulder. In: Mackin E, Callahan A, Skirven T, Schneider L, Osterman A, eds. Rehabilitation of the Hand and Upper Extremity. St Louis, MO: Mosby; 2002. Kelley MJ, Brenneman S. The Scapular Flip Sign: An Examination Sign to Identify the Pres-ence of a Spinal Accessory Nerve Palsy. APTA Combined Sections Meeting. New Orleans, LA: 2000. Kelley MJ, McClure PW, Bialker J, Cheville A. 3-Dimensional Scapular Kinematics in Patients With a Spinal Accessory Nerve Palsy. APTA Combined Sections Meeting. Tampa, FL: 2003. Kendall F, McCreary E. Muscle Testing and Func-tion. 3rd ed. Baltimore, MD: Williams & Wilkins; 1982. Kibler B. Role of the scapula in the over-head throwing motion. Contemp Orthop. 1991;22:525-532. Kibler WB, Uhl TL, Maddux JW, Brooks PV, Zeller B, McMullen J. Qualitative clinical evaluation of scapular dysfunction: a reliability study. J Shoul-der Elbow Surg. 2002;11:550-556. Kierner AC, Burian M, Bentzien S, Gstoettner W. Intraoperative electromyography for identifica-tion of the trapezius muscle innervation: clinical proof of a new anatomical concept. Laryngo-scope. 2002;112:1853-1856. Kim DH, Cho YJ, Tiel RL, Kline DG. Surgical outcomes of 111 spinal accessory nerve injuries. Neurosurgery. 2003;53:1106-1112; discussion 1102-1103. Koybasioglu A, Bora Tokcaer A, Inal E, Uslu S, Kocak T, Ural A. Accessory nerve function in lat-eral selective neck dissection with undissected level IIb. ORL J Otorhinolaryngol Relat Spec. 2006;68:88-92. Koybasioglu A, Tokcaer AB, Uslu S, Ileri F, Beder L, Ozbilen S. Accessory nerve function after modified radical and lateral neck dissections. Laryngoscope. 2000;110:73-77. Kozin F. Injuries of the brachial plexus. In: Ian-notti J, Williams GR, eds. Disorders of the Shoul-der: Diagnosis and Management. Philadelphia, PA: Lippincott Williaims & Wilkins; 2007: Krause HR. Reinnervation of the trapezius muscle after radical neck dissection. J Cranio-maxillofac Surg. 1994;22:323-329. Kuhn JE, Plancher KD, Hawkins RJ. Scapular Winging. J Am Acad Orthop Surg. 1995;3:319-325. Kuntz AL, Weymuller EA, Jr. Impact of neck dissection on quality of life. Laryngoscope. 1999;109:1334-1338. Laumann U. Kinesiology of the Shoulder: Elec-tromyographic and Stereophotogrammetric Studies. Surgery of the Shoulder. Philadelphia, PA: B.C. Decker; 1984. Ludewig PM, Cook TM. Alterations in shoulder kinematics and associated muscle activity in people with symptoms of shoulder impinge-ment. Phys Ther. 2000;80:276-291. Ludewig PM, Cook TM. Contribution of selected scapulothoracic muscles to the control of ac-cessory scapular motions. J Orthop Sports Phys Ther. 1997;25:77-80. Ludewig PM, Cook TM, Nawoczenski DA. Three-dimensional scapular orientation and muscle activity at selected positions of humeral eleva-tion. J Orthop Sports Phys Ther. 1996;24:57-65. Michener LA, Boardman ND, Pidcoe PE, Frith AM. Scapular muscle tests in subjects with shoulder pain and functional loss: reliability and construct validity. Phys Ther. 2005;85:1128-1138. Michener LA, Pidcoe PE, Firth AM. Reliability and validity of scapular muscle strength testing in patients with shoulder pain and functional loss [abstract]. Med Sci Sports Exerc. 2003;35: S252. Miyata K, Kitamura H. Accessory nerve damag-es and impaired shoulder movements after neck dissections. Am J Otolaryngol. 1997;18:197-201. Moseley JB, Jr., Jobe FW, Pink M, Perry J, Tibone J. EMG analysis of the scapular muscles during a shoulder rehabilitation program. Am J Sports Med. 1992;20:128-134. Nakamichi K, Tachibana S. Iatrogenic injury of the spinal accessory nerve. Results of repair. JBone Joint Surg Am. 1998;80:1616-1621. Orhan KS, Demirel T, Baslo B, et al. Spinal ac-cessory nerve function after neck dissections. JLaryngol Otol. 2007;121:44-48. Patten C, Hillel AD. The 11th nerve syndrome. Accessory nerve palsy or adhesive cap-sulitis? Arch Otolaryngol Head Neck Surg. 1993;119:215-220. Reinold MM, Wilk KE, Fleisig GS, et al. Elec-tromyographic analysis of the rotator cuff and deltoid musculature during common shoulder external rotation exercises. J Orthop Sports Phys Ther. 2004;34:385-394. Remmler D, Byers R, Scheetz J, et al. A prospec-tive study of shoulder disability resulting from radical and modified neck dissections. Head Neck Surg. 1986;8:280-286. Robbins KT, Medina JE, Wolfe GT, Levine PA, Sessions RB, Pruet CW. Standardizing neck dissection terminology. Official report of the Academy’s Committee for Head and Neck Sur-gery and Oncology. Arch Otolaryngol Head Neck Surg. 1991;117:601-605. Saunders JR, Jr., Hirata RM, Jaques DA. Con-sidering the spinal accessory nerve in head and termine validity of the flip sign. Finally, establishing a clinical prediction rule for SANP would be beneficial. Correlation of the associated history, signs, and symp-toms to comprehensive electrodiagnostic studies is recommended. TJournal of Orthopaedic & Sports Physical Therapy® Downloaded from www.jospt.org at on September 29, 2025. For personal use only. No other uses without permission. Copyright © 2008 Journal of Orthopaedic & Sports Physical Therapy®. All rights reserved. 86 \| february 2008 \| volume 38 \| number 2 \| journal of orthopaedic & sports physical therapy [ CASE REPORT ] neck surgery. Am J Surg. 1985;150:491-494. Sobol S, Jensen C, Sawyer W, 2nd, Costiloe P, Thong N. Objective comparison of physical dysfunction after neck dissection. Am J Surg. 1985;150:503-509. Terrell JE, Welsh DE, Bradford CR, et al. Pain, quality of life, and spinal accessory nerve status after neck dissection. Laryngoscope. 2000;110:620-626. Townsend H, Jobe FW, Pink M, Perry J. Electro-myographic analysis of the glenohumeral mus-cles during a baseball rehabilitation program. Am J Sports Med. 1991;19:264-272. van Wilgen CP, Dijkstra PU, van der Laan BF, Plukker JT, Roodenburg JL. Shoulder complaints after neck dissection; is the spinal accessory nerve involved? Br J Oral Maxillofac Surg. 2003;41:7-11. van Wilgen CP, Dijkstra PU, van der Laan BF, Plukker JT, Roodenburg JL. Shoulder complaints after nerve sparing neck dissections. Int J Oral Maxillofac Surg. 2004;33:253-257. Warner JJ, Micheli LJ, Arslanian LE, Kennedy J, Kennedy R. Patterns of flexibility, laxity, and strength in normal shoulders and shoulders with instability and impingement. Am J Sports Med. 1990;18:366-375. Williams WW, Twyman RS, Donell ST, Birch R. The posterior triangle and the painful shoulder: spinal accessory nerve injury. Ann R Coll Surg Engl. 1996;78:521-525. @ MORE INFORMATION WWW.JOSPT.ORG Journal of Orthopaedic & Sports Physical Therapy® Downloaded from www.jospt.org at on September 29, 2025. For personal use only. No other uses without permission. Copyright © 2008 Journal of Orthopaedic & Sports Physical Therapy®. All rights reserved.
15012	https://byjus.com/jee/what-are-simple-and-compound-events-in-probability/	Probability denotes how likely an event will happen. In probability, simple, compound and complementary events are different types of probabilities. These are calculated in a slightly different manner. As far as the JEE exam is concerned, probability is an important topic. In this article, we will learn what are simple and compound events probability . Simple Event An event that has a single point of the sample space is known as a simple event in probability. For example, the probability of getting a 3 when a die is tossed. Here S = { 1, 2, 3, 4, 5, 6} If E be the event of getting a 3 when a die is tossed. E = { 3 } P(E) = 1/6 In the case of a simple event, the numerator (number of favourable outcomes) will be 1. Compound Event If an event has more than one sample point, it is termed as a compound event. The compound events are a little more complex than simple events. These events involve the probability of more than one event occurring together. The total probability of all the outcomes of a compound event is equal to 1. To calculate probability, the following formual is used: Probability of an event = [The number of favourable outcomes] / [the number of total outcomes]. First, we find the probability of each event occurring. Then we will multiply these probabilities together. In the case of a compound event, the numerator (number of favourable outcomes) will be greater than 1. For example, the probability of rolling an even number on a die, then tossing a head on a coin. Here P(even number) = 3/6 P(head) = 1/2 Hence required probability = (3/6)(½ ) = 3/12 Also Read: Compound Probability Solved Examples Example 1: A coin is thrown 3 times. What is the probability that atleast one head is obtained? Solution: Sample space = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} Total number of outcomes = 2 × 2 × 2 = 8. Favourable outcomes = 7 Required probability = 7/8 Example 2: What is the probability that a dice with 12 faces returns a value 9 if each face has a number starting from 0. (A) 0 (B) 1 (C) 1/6 (D) 1/12 Solution: Total number of outcomes = 12 Favourable outcome = 1 Required probability = 1/12. Hence option D is the answer. Practice Problems Find the probability of drawing a spade, replacing the card, then drawing a club. What is the probability of rolling either a two or a four using one-sided die? Answer key: 1. 169/2704 2. 1/5 Frequently Asked Questions Q1 How do you calculate the probability of an event? The probability of an event is given by, P(E) = number of favourable outcomes/total number of outcomes. Q2 What do you mean by a simple event in probability? An event having only one outcome is a simple event. For example, if we toss one coin, it gives only one outcome. Q3 What do you mean by a compound event in probability? A compound event is an event which has more than one sample point. Q4 What is the total probability of all the outcomes of a compound event? The total probability of all the outcomes of a compound event is equal to 1. Frequently Asked Questions How do you calculate the probability of an event? The probability of an event is given by, P(E) = number of favourable outcomes/total number of outcomes. What do you mean by a simple event in probability? An event having only one outcome is a simple event. For example, if we toss one coin, it gives only one outcome. What do you mean by a compound event in probability? A compound event is an event which has more than one sample point. What is the total probability of all the outcomes of a compound event? The total probability of all the outcomes of a compound event is equal to 1. Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz Congrats! Visit BYJU’S for all JEE related queries and study materials Your result is as below Request OTP on Voice Call Comments Leave a Comment Cancel reply Your Mobile number and Email id will not be published. Required fields are marked Request OTP on Voice Call Website Post My Comment Register with BYJU'S & Download Free PDFs Register with BYJU'S & Watch Live Videos
15013	https://eng.libretexts.org/Workbench/Introduction_to_Circuit_Analysis/01%3A_Basic_Concepts_and_Quantitites/1.09%3A_Sign_conventions	Published Time: 2023-12-21T07:40:50Z 1.9: Sign conventions - Engineering LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 1: Basic Concepts and Quantitites Introduction to Circuit Analysis { } { "1.01:_Introduction" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.02:_An_Atomic_Model" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.03:_Charge_and_Current" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.04:_Energy_and_Voltage" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.05:_Power_and_Efficiency" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.06:_Energy_Cost_and_Battery_Life" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.07:_Resistance_and_Conductance" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.08:_Instrumentation_and_Laboratory" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.09:_Sign_conventions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.10:_Summary" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.11:_Exercises" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Basic_Concepts_and_Quantitites" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Fundamental_Laws" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Nodal_and_Mesh_Analysis_Dependent_Sources" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Analysis_Theorems_and_Techniques" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Advanced_Topic-_Operational_Amplifiers" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Capacitors_and__Inductors" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_1st_Order_RC_RL_Circuit" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_AC_Signal_Fundamentals" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:_Parallel_RLC_Circuits" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11:_AC_Power" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "12:_Laplace_Transform_in_Circuit_Analysis" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "9:_Series_RLC_Circuits" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Tue, 23 Jan 2024 05:10:42 GMT 1.9: Sign conventions 98533 98533 Chao-Shih Liu { } Anonymous Anonymous User 2 false false [ "article:topic", "transcluded:yes", "showtoc:yes", "license:publicdomain", "autonumheader:yes2", "licenseversion:10" ] [ "article:topic", "transcluded:yes", "showtoc:yes", "license:publicdomain", "autonumheader:yes2", "licenseversion:10" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Workbench 3. Introduction to Circuit Analysis 4. 1: Basic Concepts and Quantitites 5. 1.9: Sign conventions Expand/collapse global location Introduction to Circuit Analysis Front Matter 1: Basic Concepts and Quantitites 2: Fundamental Laws 3: Nodal and Mesh Analysis, Dependent Sources 4: Analysis Theorems and Techniques 5: Advanced Topic- Operational Amplifiers 6: Capacitors and Inductors 7: 1st Order RC/RL Circuit 8: Parallel RLC Circuits 9: Series RLC Circuits 10: AC Signal Fundamentals 11: AC Power 12: Laplace Transform in Circuit Analysis Back Matter 1.9: Sign conventions Last updated Jan 23, 2024 Save as PDF 1.8: Instrumentation and Laboratory 1.10: Summary Page ID 98533 ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. 2. Current Sign Convention: 3. Voltage Sign Convention: 4. Power Sign Convention: 1. Power Absorption (Load or Power Sink): 2. Power Generation (Source or Generator): Power Balancing: Examples Example 1: Example 2: Considerations In electrical circuits, sign conventions are established to maintain consistency in analyzing and interpreting circuit parameters such as current, voltage, and power. These conventions help ensure that the results obtained from circuit analysis are meaningful and can be compared across different components. Here's a description of the sign conventions for current, voltage, and power: Figure 1.9.1: The direction of electric current is opposite to the flow of electrons. Current Sign Convention: The assumed direction of current flow is typically from the positive terminal to the negative terminal. This convention aligns with the flow of positive electric charge. The positive charge flows from the positive to the negative terminal is considered as positive current.If the actual flow is opposite to the assumed direction, it is considered negative. In Figure 1.9.1, the current I is aligned with the direction from the positive to the negative, so I am positive. Voltage Sign Convention: Voltage is considered positive when it is assumed higher at the terminal where the current enters a passive component (like a resistor, capacitor, or inductor).Negative Voltage represents a voltage rise.This happens when the current flows from lower voltage to higher voltage. V ba is negative because the current flows from the negative to the positive terminal.Positive Voltage represents a voltage drop.This happens when the current flows from higher voltage to lower voltage. V ab is positive because the light bulb causes a voltage drop. Power Sign Convention: The power absorbed or supplied by an element results from multiplying the voltage across the element by the current through it. A positive sign in power indicates that the element delivers or absorbs power. Conversely, a negative sign implies that the element is supplying power.To determine the sign of power, it is essential to consider the direction of the current and the polarity of the voltage.According to this convention, Power Absorption (Load or Power Sink): If the current enters the positive terminal of a device, and the voltage is positive at that terminal, the power is absorbed or consumed by the device.Power is positive for devices such as resistors, indicating that they absorb power. Power Generation (Source or Generator): If the current enters the negative terminal of a device, and the voltage is positive at that terminal, the power is generated or supplied by the device.Power is negative for active devices like generators, indicating they supply power to the circuit. Power Balancing: Power balancing in a circuit refers to the equilibrium or equality between the power supplied to the circuit by sources and the power consumed or absorbed by various elements within the circuit. In an ideal situation, the total power entering the circuit from sources should equal the total power dissipated or consumed by the circuit elements. This concept is derived from the principle of conservation of energy. Power Supplied by Sources: Power sources within a circuit, such as batteries or generators, provide electrical energy to the system. The power supplied by these sources is usually expressed as a product of voltage and current and is measured in watts (W). Power Consumed by Circuit Elements: Circuit elements, such as resistors, capacitors, and inductors, consume electrical power. The power consumed or dissipated by a resistor, for example, is given by the product of current and voltage across the resistor, according to the formula P=V I. Power consumption in circuit elements can also be expressed in watts. Conservation of Energy: According to the principle of conservation of energy, the total power entering a closed circuit should equal the total power leaving the circuit. This implies that the sum of power supplied by sources should equal the sum of power consumed by circuit elements. Mathematical Representation: Mathematically, power balancing in a circuit can be expressed using the following equation: ∑P sources=∑P consumed The summation includes all power sources and all power-consuming elements within the circuit. Examples Example 1: The circuit contains four elements: a voltage source, a current source, and two resistors. The arrows or + and - signs show the current directions and voltage polarity. Use the sign convention to determine the currents i1, i2, and i3 and voltages v1, v2, v3, and v4. Calculate the power of the DC current element. Is it a power source or a power sink? The current i1 is +3A which is given. The current i3 is -1A because it is the same current as the DC current source. The direction is opposite to the DC current so we need to add the negative sign. The Ohm's law can determine i2: i2 = V2/10 = +2A This is a parallel circuit with three wires.The DC power on the left-hand side shows the top voltage is higher than the bottom voltage by V1 which is +20V. The middle element must have the same voltage drop, so V2 is +20V. The right-hand wire crosses one resistor and one current element. The total voltage drop is (V3 - V4) and is also equal to +20V. The voltage drop V3 is i330 = +30V because the current flows from positive to negative. The V4 is a voltage increase because the current flows from negative to positive and the voltage V4 is -10V from the following relationship. V2 = V3 + V4 = 30V+ V4 The DC current is a power source because the current enters into the negative terminal and leaves from the positive terminal. Example 2: Find the power of each element in the above example. Use positive to represent the power sink and negative to represent the power source. What is the net power of the above circuit? We use P = I V to determine the power. If the current flows into the positive terminal, we use positive, otherwise, use the negative voltage. P1 = 3A -20V = -60W (power source) P2 = 2A 20V = 40W (power sink) P3 = 1A 30V = 30W (power sink) P4 = 1A -10V = -10W (power source) The total power of the whole circuit is zero. This agrees with the conservation of energy. Considerations Give some considerations such as system requirements or "gotchas" for this setting or control or programming syntax. 1.9: Sign conventions is shared under a Public Domain license and was authored, remixed, and/or curated by LibreTexts. Back to top 1.8: Instrumentation and Laboratory 1.10: Summary Was this article helpful? Yes No Recommended articles 1.1: Introduction 1.2: An Atomic Model 1.3: Charge and Current 1.4: Energy and Voltage Article typeSection or PageAutonumber Section Headingstitle with colon delimitersLicensePublic DomainLicense Version1.0Show TOCyesTranscludedyes Tags This page has no tags. © Copyright 2025 Engineering LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? 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15014	https://www.open.edu/openlearn/science-maths-technology/mathematics-statistics/numbers-units-and-arithmetic/content-section-3.7	Numbers, units and arithmetic: 1.7 Simplest form of a fraction \| OpenLearn - Open University Cookies on our website We use cookies to make sure our websites work effectively, to improve your experience and to serve you relevant advertising across selected websites. For more information on the types of cookies we use and how we use them please see our Cookie Policy. To use our website with all cookies enabled, select "Accept all cookies". If you want to opt out of all (except strictly necessary) cookies then select "Reject all cookies". If you want to choose the cookies we use then select “Manage your cookies”. You can change your cookie settings at any time by visiting Cookie management in our footer. 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Advantages of enrolling on a free course Frequently asked questions Study with The Open University Subjects Study skills Work skills Help Accessibility hub Sign in Menu sticky search Search for Search My OpenLearn Profile Personalise your OpenLearn profile, save your favourite content and get recognition for your learning Create account / Sign in Subjects Science, Maths & Technology Free courses Numbers, units and arithmetic 1.7 Simplest form of a fraction Course content Introduction Learning outcomes 1 The size of numbersMenu toggle 1.1 Whole numbers Try some yourself 1.2 Using the place value Try some yourself 1.3 Number lines 1.4 Decimals Try some yourself 1.5 Decimals and place value Try some yourself 1.6 Fractions Try some yourself Current section: 1.7 Simplest form of a fraction Try some yourself 1.8 Negative numbers Try some yourself 1.9 Negative number notation Try some yourself 2 Units of measurementMenu toggle 2.1 Which units to use Units in action Try some yourself 2.2 Converting units Try some yourself 2.3 Measuring mass 2.4 Measuring length 2.5 Measuring capacity 2.6 Comparing measurements Try some yourself 3 ArithmeticMenu toggle 3.1 Arithmetic with whole numbers Try some yourself 3.2 Multiplying and dividing Try some yourself 3.3 Division Try some yourself 3.4 Order of calculations Try some yourself 3.5 Several calculations and using brackets Try some yourself With or without brackets Try some yourself 3.6 Arithmetic with decimals Addition and subtraction with decimals Try some yourself 3.7 Multiplying and dividing with decimals Try some yourself 3.8 Order of decimal calculations Try some yourself 3.9 Addition and subtraction with fractions Try some yourself 3.10 More difficult fraction sums Try some yourself 3.11 Multiplication of fractions Try some yourself 3.12 Division by fractions Try some yourself 3.13 More division with fractions Try some yourself 3.14 Mixed numbers Try some yourself 3.15 Arithmetic with negative numbers Addition on the number line Subtraction on the number line 3.16 Addition of negative numbers Try some yourself 3.17 Subtraction of negative numbers Try some yourself 3.18 Practical examples of negative numbers Try some yourself 3.19 Multiplication with negative numbers 3.20 Multiplication rules Try some yourself 3.21 Division rules Try some yourself Conclusion Acknowledgements About this free course About this free course 5 hours study Level 1: Introductory Full course description Become an OU student Become an OU student BSc (Honours) Mathematics BA/BSc (Honours) Open degree Discovering mathematics Download this course Download this course Download this course for use offline or for other devices Word Kindle PDF Epub 2 Epub 3 RSS HTML OUXML File IMS CC OUXML Pckg See more formatsShow fewer formats Share this free course Share this free course Share on FacebookShare on TwitterShare on LinkedInShare via Email Course rewards Free statement of participation on completion of these courses. Create your free OpenLearn profile Anyone can learn for free on OpenLearn, but signing-up will give you access to your personal learning profile and record of achievements that you earn while you study. Sign up now for freeSign up now for free Newsletter Sign up for our regular newsletter to get updates about our new free courses, interactives, videos and topical content on OpenLearn. Newsletter sign-up Course contentCourse content Numbers, units and arithmetic Start this free course now. Just create an account and sign in. Enrol and complete the course for a free statement of participation or digital badge if available. Create account / Sign inMore free courses 1.7 Simplest form of a fraction The fraction , is the simplest form of all its equivalent fractions, because it cannot be ‘simplified’ further (by dividing top and bottom by the same whole number called a common factor); or , on the other hand, can be simplified to by dividing top and bottom by 2 and 4 respectively. A fraction is in its simplest form if the numerator and the denominator have no common factors. 6 can be written as 2 × 3. 2 and 3 are called factors of 6. 15 can be written as 5 × 3. 5 and 3 are factors of 15. 3 is a common factor of 6 and 15. Hence . This is the simplest form of . Example 6 (a) Find some equivalent fractions for . (b) What is the simplest form of ? Reveal answer Answer Multiply or divide top and bottom by the same whole number to give an equivalent fraction. (a) Some examples are (b) Top and bottom are divisible by 5. So 3 and 4 have no common factors so is the simplest form. You can indicate fractions along the number line by marking points between the whole numbers. The numbers and are shown below. Note may also be written as 1.5 or as . PreviousTry some yourself NextTry some yourself Print Take your learning further Ready to take the next step in your learning journey? 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15015	https://math.dartmouth.edu/news-resources/electronic/puzzlebook/book/book.pdf	Mathematical Puzzles Revised Edition Peter Winkler Illustrations by Jess Johnson for Miles, Sage, and Beatrix Revised Edition dedicated to the readers! Acknowledgments This volume owes its existence to two institutions and many individuals. Dartmouth College has provided inspiring colleagues, a stimulating environ-ment, and steadfast encouragement; special thanks are due to the endowers of the William Morrill Professorship. The National Museum of Mathematics (with whom I spend the academic year 2019–2020), through the generosity of the Si-mons Foundation, provided a huge variety of outreach opportunities, with many audiences on whom to test puzzles. Some of the individual contributors (witting and otherwise) are listed alpha-betically below, with multiple contributors in italics and the two most prolific in bold. Specific contributions appear in Notes & Sources. Doubtless many deserving names have been left out owing to gaps in my knowledge or memory, but I hope to rectify some of those omissions as time goes on. Dorit Aharonov · Arseniy Akopyan · David Aldous · Kasra Al-ishahi · Noga Alon · Dan Amir · A. V. Andjans · Titu Andreescu · Omer Angel · Gary Antonick · Steve Babbage · Matt Baker · Yuliy Baryshnikov · Elwyn Berlekamp · B´ ela Bollob´ as · Christian Borgs · Bancroft Brown · Daniel E. Brown · A. I. Bufetov · Joe Buhler · Car-oline Calderbank · Neil Calkin · Teena Carroll · Amit Chakrabarti · Deeparnab Chakrabarty · Adam Chalcraft · A. S. Chebotarev · William Fitch Cheney, Jr. · Vladimir Chernov · N. L. Chernyatyev · Vaˇ sek Chv´ atal · Barry Cipra · Bob Connelly · John H. Conway · Tom Cover · Paul Cuff · Robert DeDomenico · Oskar van Deventer · Randy Dougherty · Peter Doyle · Ioana Dumitriu · Freeman Dyson · Todd Ebert · Sergi Elizalde · Noam Elkies · Rachel Esselstein · R. M. Fedorov · Steve Fisk · Gerald Folland · B. R. Frenkin · Ehud Friedgut · Alan Frieze · Anna Gal · David Gale · A. I. Galochkin · Gregory Galperin · Martin Gardner · Bill Gasarch · Dieter Geb-hardt · Giulio Genovese · Carl Giffels · Sol Golomb · David Gontier · Bill Gosper · Ron Graham · Geoffrey Grimmett · Ori Gurel-Gurevich · Olle H¨ aggstr¨ om · Sergiu Hart · Bob Henderson · Dick Hess · Maya Bar Hillel · Iwasawa Hirokazu · Ander Holroyd · Ross Honsberger · Naoki Inaba · Svante Janson · A. Ya. Kanel-Belov · Mark Kantrowitz · Howard Karloff · Joseph Keane · Kiran Kedlaya · David Kempe · Rick Kenyon · Tanya Khovanova · Guy Kindler · Murray Klamkin · Victor Klee · Danny Kleitman · V. A. Kleptsyn · Niko Klewinghaus · Anton Klyachko · Don Knuth · Maxim Kontsevich · Isaac Kornfeld · A. K. Kovaldzhi · Alex Krasnoshel’skii · Piotr Krason · Jeremy Kun · Thomas Lafforgue · Michael Larsen · Andy Latto · V. N. Latyshev · Imre Leader · Tam´ as Lengyel · Hendrik Lenstra · Anany Levitin · Jerome Lewis · Sol LeWitt · Michael Littman · Andy Liu · Po-Shen Loh · L´ aszl´ o Lov´ asz · Edouard Lucas · Russ Lyons · Marc Massar · David McAllester · Peter Bro Miltersen · Grant Molnar · R. L. Moore · Thierry Mora · Frank Morgan · Carl Morris · Lizz Moseman · Elchanon Mossel · Frederick Mosteller · Colm Mulcahy · Pradeep Mutalik · Muthu Muthukrishnan · Gerry Myerson · Assaf Naor · Girija Narlikar · Jeff Norman · Alon Orlitsky · Garth Payne · Yuval Peres · Nick Pippenger · Dick Plotz · V. V. Proizvolov · Jim Propp · Kevin Purbhoo · Anthony Quas · Dana Randall · Lyle Ramshaw · Dieter Rautenbach · Sasha Razborov · Dan Romik · I. F. Sharygin · Bruce Shepherd · Raymond Smullyan · Jeff Steif · Sara Robinson · Saul Rosenthal · Rustam Sadykov · Mahdi Saffari · Boris Schein · Christof Schmalenbach · Markus Schmidmeier · Frederik Schuh · David Seal · Dexter Senft · Alexander Shapovalov · James B. Shearer · S. A. Shestakov · Senya Shlosman · George Sicherman · Vladas Sidoravicius · Sven Skyum · Laurie Snell · Pablo Sober´ on · Emina Soljanin · Joel Spencer · A. V. Spivak · Caleb Stanford · Richard Stanley · Einar Steingrimsson · Francis Su · Benny Sudakov · Andrzej Szulkin · Karthik Tadinada · Bob Tarjan · Bridget Tenner · Prasad Tetali · Mikkel Thorup · Mike Todd · Enrique Trevi˜ no · John Urschel · Felix Vardy · Tom Verhoeff · Balint Virag · Stan Wagon · George Wang · Greg Warrington · Johan W¨ astlund · Diana White · Avi Wigderson · Herb Wilf · Dana Williams · I. M. Yaglom · Steven Young · Paul Zeitz · Leo Zhang · Yan Zhang · Barukh Ziv The many improvements in the revised edition are due primarily to the care and patience of the following readers, to whom the author extends his deepest gratitude. Nick Baxter · Robert Beals · Nikolai Beluhov · Xiaomin Chen · Timothy Chow · Jerrold Grossman · Bob Henderson · Emmett Keeler · Andy Liu · Lyle Ramshaw · G¨ unter Rote · Jim Saxe · Uzi Segal · Ron Sommeling · Richard Stanley · Jim Stark · Stan Wagon · Contents Author’s Preface ix The Puzzles 1 The Hints 63 1 Out for the Count 77 2 Achieving Parity 91 3 Intermediate Math 101 4 Graphography 113 5 Algebra Too 125 6 Safety in Numbers 135 7 The Law of Small Numbers 145 8 Weighs and Means 157 9 The Power of Negative Thinking 165 10 In All Probability 173 11 Working for the System 199 12 The Pigeonhole Principle 213 13 Information, Please 225 14 Great Expectation 239 15 Brilliant Induction 259 16 Journey into Space 275 vii 17 Nimbers and the Hamming Code 285 18 Unlimited Potentials 301 19 Hammer and Tongs 319 20 Let’s Get Physical 339 21 Back from the Future 347 22 Seeing Is Believing 369 23 Infinite Choice 383 24 Startling Transformation 395 Notes & Sources 411 Author’s Preface What’s in This Book? This book contains 300 of the very best mathematical puzzles in the world. Some have appeared in previous books, including my own, or in contests; others are brand new. The book is organized as follows: First, all the puzzles are presented; then, for each puzzle, a hint together with the chapter number in which you can find the solution. (The Hints section lists puzzles in alphabetical order by name, so may be of use in locating a particular puzzle.) Note that this book does not make it easy to jump from a puzzle to its solution. The value of a solution increases enormously when a puzzle is pondered beforehand. The chapters themselves are classified by solution method, rather than by general puzzle topic. Each chapter introduces a technique, uses it to solve puzzles, and ends with an actual theorem of mathematics that uses the technique in its proof. Additional information on every puzzle is found in Notes and Sources at the end of the volume. Who Is This Book For? Puzzle lovers and math lovers. For background you’ll need no more than high-school math. Calculus, abstract algebra, and higher college-level material will not be used. Some math that you might not find in every curriculum, like elementary graph theory and probability, will be introduced as needed. This does not mean the puzzles are easy; some are, others will stump pro-fessional mathematicians. And, the existence of an easy solution does not mean that a solution is easy to find! How Were the Puzzles Chosen? To me a great puzzle offers entertainment and enlightenment. It should be fun to read and think about. It should make an interesting mathematical point, perhaps by confounding your intuition, or perhaps just by reminding you of the amazing things elementary mathematics can do. ix Where Do the Puzzles Come From? Some of the puzzles are original but most were contributed by puzzle composers and puzzle mavens all over the world—they are the real heroes of this book. Many of their names are found in the Acknowledgments and Bibliography. Prob-lems from contests are not usually appropriate for this volume; they are designed to test, rather than to entertain, and often require more advanced mathemat-ics. Nonetheless some irresistible gems have arisen in contests—especially, for some reason, the Moscow Mathematical Olympiads. Often a puzzle has mul-tiple sources and rarely can its origin be identified, or even defined, precisely; my meager efforts at doing so are reported in Notes and Sources. Puzzles and solutions are presented in my own words, which in some cases will be similar to a source’s, in others completely different. Where Should You Start? With the puzzles, of course. The puzzles are the stars of this show; the rest of the book is just support. If you simply read the puzzles, pick out the ones you like, and try to solve them, you are headed for days, weeks, or months of intrigue, frustration, joy, and enlightenment. Yes, the techniques proposed in the chapters are useful tools, but be warned that no two people will classify puzzle-solving techniques the same way or even assign puzzles the same way in a given classification. Indeed, my own assignments are often distressingly arbitrary. Note also that for the harder puzzles, the solution given (which may be one of many) is sometimes presented in rather compact fashion. The heart symbol (♡) is used to signal the end of an argument or proof; if you reach one unedified, go back! Remarks about the Second Edition The revised edition is dedicated to the readers of the original printing, who sent in dozens of suggestions for corrections and clarifications. The author is especially indebted to the individual readerss listed at the end of the acknowl-edgments. The Puzzles Bat and Ball A bat costs $1 more than a ball; together they cost $1.10. How much does the bat cost? No Twins Today It was the first day of class and Mrs. O’Connor had two identical-looking pupils, Donald and Ronald Featheringstonehaugh (pronounced “Fanshaw”), sitting to-gether in the first row. “You two are twins, I take it?” she asked. “No,” they replied in unison. But a check of their records showed that they had the same parents and were born on the same day. How could this be? Portrait A visitor points to a portrait on the wall and asks who it is. “Brothers and sisters have I none,” says the host, “but that man’s father is my father’s son.” Who is pictured? Half Grown At what age is the average child half the height that he or she will be as an adult? 1 Shoes, Socks, and Gloves You need to pack for a midnight flight to Iceland but the power is out. In your closet are six pairs of shoes, six black socks, six gray socks, six pairs of brown gloves, and six pairs of tan gloves. Unfortunately, it’s too dark to match shoes or to see any colors. How many of each of these items do you need to take to be sure of getting a matched pair of shoes, two socks of the same color, and matching gloves? Phone Call A phone call in the continental United States is made from a west coast state to an east coast state, and it’s the same time of day at both ends of the call. How is this possible? Powers of Two How many people are “two pairs of twins twice”? Rotating Coin While you hold a U.S. 25-cent piece firmly to the tabletop with your left thumb, you rotate a second quarter with your right forefinger all the way around the first quarter. Since quarters are ridged, they will interlock like gears and the second will rotate as it moves around the first. How many times will it rotate? Boxing the Mountain State Can West Virginia be inscribed in a square? Martians in a Circle A visiting logician is surrounded by a circle of Martians, each of whom either always tells the truth or always lies. She asks each Martian whether the Martian to his right is a truth teller or a liar, and from their answers, she is able to deduce the fraction of liars in the circle. What fraction is it? 2 Winning at Wimbledon As a result of temporary magical powers (which might need to be able to change your gender), you have made it to the women’s singles tennis finals at Wim-bledon and are playing Serena Williams for all the marbles. However, your powers cannot last the whole match. What score do you want it to be when they disappear, to maximize your chances of notching an upset win? Polyhedron Faces Prove that any convex polyhedron has two faces with the same number of edges. Finding the Counterfeit You have a balance scale and 12 coins, 11 of which are genuine and identical in weight; but one is counterfeit, and is either lighter or heavier than the oth-ers. Can you determine, in three weighings with a balance scale, which coin is counterfeit and whether it is heavy or light? Signs in an Array Suppose that you are given an m×n array of real numbers and permitted, at any time, to flip the signs of all the numbers in any row or column. Can you always arrange matters so that all the row sums and column sums are nonnegative? Birthday Match You are on a cruise where you don’t know anyone else. The ship announces a contest, the upshot of which is that if you can find someone who has the same birthday as yours, you (both) win a beef Wellington dinner. How many people do you have to compare birthdays with in order to have a better than 50% chance of success? Meeting the Ferry Every day at noon GMT a ferry leaves New York and simultaneously another leaves Le Havre. Each trip takes seven days and seven nights, arriving before noon on the eighth day. How many of these cross-Atlantic ferries does one of them pass on its way across the pond? Sinking 15 Carol and Desmond are playing pool with billiard balls number 1 through 9. They take turns sinking balls into pockets. The first to sink three numbers that sum to 15 wins. Does Carol (the first to play) have a winning strategy? 3 Monk on a Mountain A monk begins an ascent of Mt. Fuji on Monday morning, reaching the summit by nightfall. He spends the night at the summit and starts down the moun-tain on the same path the following morning, reaching the bottom by dusk on Tuesday. Prove that at some precise time of day, the monk was at exactly the same spot on the path on Tuesday as he was on Monday. Mathematical Bookworm The three volumes of Jacobson’s Lectures in Abstract Algebra sit in order on your shelf. Each has 2′′ of pages and a front and back cover each 1 4 ′′, thus a total width of 2 1 2 ′′. A tiny bookworm bores its way straight through from page 1, Vol I to the last page of Vol III. How far does it travel? Other Side of the Coin A two-headed coin, a two-tailed coin, and an ordinary coin are placed in a bag. One of the coins is drawn at random and flipped; it comes up “heads.” What is the probability that there is a head on the other side of this coin? Slicing the Cube Before you is a circular saw and a 3×3×3 wooden cube that you must cut into 27 1 × 1 × 1 cubelets. What’s the smallest number of slices you must make in order to do this? You are allowed to stack pieces prior to running them through the saw. Rolling Pencil A pencil whose cross-section is a regular pentagon has the maker’s logo im-printed on one of its five faces. If the pencil is rolled on the table, what is the probability that it stops with the logo facing up? 4 Watermelons Yesterday a thousand pounds of watermelons lay in the watermelon patch. They were 99% water, but overnight they lost moisture to evaporation and now they are only 98% water. How much do they weigh now? Boy Born on Tuesday Mrs. Chance has two children of different ages. At least one of them is a boy born on a Tuesday. What is the probability that both of them are boys? Air Routes in Aerostan The country of Aerostan has three airlines which operate routes between various pairs of Aerostan’s fifteen cities. What’s the smallest possible total number of routes in Aerostan if bankruptcy of any of the three airlines still leaves a network that connects the cities? Tipping the Scales A balance scale sits on the desk of the science teacher, Ms. McGregor. There are weights on the scale, which is currently tipped to the right. On each weight is inscribed the name of at least one pupil. On entering the classroom, each pupil moves every weight carrying his or her name to the opposite side of the scale. Must there be a set of pupils that Ms. McGregor can let in that will tip the scales to the left? Chomp Alice and Bob take turns biting off pieces of an m×n rectangular chocolate bar marked into unit squares. Each bite consists of selecting a square and biting off that square plus every remaining square above and/or to its right. Each player wishes to avoid getting stuck with the poisonous lower-left square. Show that, assuming the bar contains more than one square, Alice (the first to play) has a winning strategy. 5 Whose Bullet? Two marksmen, one of whom (“A”) hits a certain small target 75% of the time and the other (“B”) only 25%, aim simultaneously at that target. Just one bullet hits. What’s the probability that it came from A? Poker Quickie What is the best full house? (You may assume that you have five cards, and you have just one opponent who has five random other cards. There are no wild cards. As a result of the Goddess of Chance owing you a favor, you are entitled to a full house, and you get to choose whatever full house you want.) Lines Through a Grid If you want to cover all the vertices in a 10 × 10 square grid by lines none of which is parallel to a side of the square, how many lines are needed? Bidding in the Dark You have the opportunity to make one bid on a widget whose value to its owner is, as far as you know, uniformly random between $0 and $100. What you do know is that you are so much better at operating the widget than he is, that its value to you is 80% greater than its value to him. If you offer more than the widget is worth to the owner, he will sell it. But you only get one shot. How much should you bid? Easy Cake Division Can you cut a cubical cake, iced on top and on the sides, into three pieces each of which contains the same amount of cake and the same amount of icing? Crossing the River In eighth century Europe, it was considered unseemly for a man to be in the presence of a married woman, even briefly, unless her husband was there as well. This posed problems for three married couples who wished to cross a river, the only means being a rowboat that could carry at most two people. Can they get to the other side without violating their social norms? If so, what’s the minimum number of crossings needed? Bacterial Reproduction When two pixo-bacteria mate, a new bacterium results; if the parents are of different sexes the child is female, otherwise it is male. When food is scarce, matings are random and the parents die when the child is born. 6 It follows that if food remains scarce a colony of pixo-bacteria will eventually reduce to a single bacterium. If the colony originally had 10 males and 15 females, what is the probability that the ultimate pixo-bacterium will be female? Sprinklers in a Field Sprinklers in a large field are located at the vertices of a square grid. Each point of land is supposed to be watered by exactly the three closest sprinklers. What shape is covered by each sprinkler? Bags of Marbles You have 15 bags. How many marbles do you need so that you can have a different number of marbles in each bag? Men with Sisters On average, who has more sisters, men or women? Dots and Boxes Variation Suppose you are playing Dots and Boxes, but with each player having an option of whether to go again after making a box. Show that the first player to make a box can win all the boxes! Fair Play How can you get a 50-50 decision by flipping a bent coin? Salaries and Raises Wendy, Monica, and Yancey were hired at the beginning of calendar 2020. Wendy is paid weekly, $500 per week, but gets a $5 raise each week; Monica is paid monthly, $2500 per month, and gets a $50 raise each month; Yancey is paid yearly, $50,000 per year, and gets a $1500 raise each year. Who will make the most money in 2030? Two Runners Two runners start together on a circular track, running at different constant speeds. If they head in opposite directions they meet after a minute; if they head in the same direction, they meet after an hour. What is the ratio of their speeds? 7 Broken ATM George owns only $500, all in a bank account. He needs cash badly but his only option is a broken ATM that can process only withdrawals of exactly $300, and deposits of exactly $198. How much cash can George get out of his account? Domino Task An 8×8 chessboard is tiled arbitrarily with thirty-two 2 × 1 dominoes. A new square is added to the right-hand side of the board, making the top row length 9. At any time you may move a domino from its current position to a new one, provided that after the domino is lifted, there are two adjacent empty squares to receive it. Can you retile the augmented board so that all the dominoes are horizontal? Big Pairs in a Matrix Suppose that in each row of a certain square matrix, the sum of the greatest two numbers is r; and in each column, the sum of the greatest two numbers is c. Show that r = c. Second Ace What is the probability that a poker hand (5 cards dealt at random from a 52-card deck) contains at least two aces, given that it has at least one? What is the probability that it contains two aces, given that it has the Ace of Spades? If your answers are different, then: What’s so special about the Ace of Spades? Fewest Slopes If you pick n random points in, say, a disk, the pairs of points among them will with probability 1 determine n(n−1)/2 distinct slopes. Suppose you get to pick the n points deliberately, subject to no three being collinear. What’s the smallest number of distinct slopes they can determine? Three Martians at the Crossroads A logician is visiting Mars and as is usual for logicians in puzzles, she is at a fork, wanting to know which of two roads leads to the village. Present this time are three willing Martians, one each from a tribe of invariable truth-tellers, a tribe of invariable liars, and a tribe of random answerers. The logician doesn’t know which Martian is from which tribe. Moreover, she is permitted to ask only two yes-or-no questions, each question being directed to just one Martian. Can she get the information she needs? How about if she can ask only one yes-or-no question? 8 Rolling All the Numbers On average, how many times do you need to roll a die before all six different numbers have turned up? Uniform Unit Distances Can it be that for any positive integer n, there’s a nonempty configuration of finitely many points on the plane with the property that every point of the configuration is at distance 1 from exactly n of the other points in the configu-ration? Life Is a Bowl of Cherries In front of you and your friend Amit are 4 bowls of cherries, containing, respec-tively, 5, 6, 7, and 8 cherries. You and Amit will alternately pick a bowl and take one or more cherries from it. If you go first, and you want to be sure to get the last cherry, how many cherries should you take—and from which bowl? Finding the Missing Number All but one of the numbers from 1 to 100 are read to you, one every 10 seconds, but in no particular order. You have a good mind, but only a normal memory, and no means of recording information during the process. How can you ensure that you can determine afterward which number was not called out? Three-Way Duel Alice, Bob, and Carol arrange a three-way duel. Alice is a poor shot, hitting her target only 1/3 of the time on average. Bob is better, hitting his target 2/3 of the time. Carol is a sure shot. They take turns shooting, first Alice, then Bob, then Carol, then back to Alice, and so on until only one is left. What is Alice’s best course of action? 9 Splitting a Hexagon Is there a (non-self-intersecting) hexagon that can be cut into four congruent triangles by a single line? Expecting the Worst Choose n numbers uniformly at random from the unit interval [0,1]. What is the expected value of their minimum? Belt Around the Earth Suppose the earth is a perfect sphere and a belt is tightened around the equator. Then the belt is loosened by adding 1 meter to its length, and it now sits the same amount off the ground all the way around. Is there enough room to slip a credit card underneath the belt? Stopping After the Boy In a certain country, a law is passed forbidding families to have another child after having a boy. Thus, families might consist of one boy, one girl and one boy, five girls and one boy... How will this law affect the ratio of boys to girls? Attic Lamp An old-fashioned incandescent lamp in the attic is controlled by one of three on-off switches downstairs—but which one? Your mission is to do something with the switches, then determine after one trip to the attic which switch is connected to the attic lamp. Efficient Pizza-cutting What’s the maximum number of pieces you can get by cutting a (round) pizza with 10 straight cuts? Fourth Corner Pegs occupy three corners of a square. At any time a peg can jump over another peg, landing an equal distance on the other side. Jumped pegs are not removed. Can you get a peg onto the fourth corner of the square? Cutting the Necklace Two thieves steal a necklace consisting of 10 red rubies and 14 pink diamonds, fixed in some arbitrary order on a loop of golden string. Show that they can cut the necklace in two places so that when each thief takes one of the resulting pieces, he gets half the rubies and half the diamonds. 10 Odd Run of Heads On average, how many flips of a coin does it take to get a run of an odd number of heads, preceded and followed by tails? Subsets with Constraints What is the maximum number of numbers you can have between 1 and 30, such that no two have a product which is a perfect square? How about if (instead) no number divides another evenly? Or if no two have a factor (other than 1) in common? Spinning Switches Four identical, unlabeled switches are wired in series to a light bulb. The switches are simple buttons whose state cannot be directly observed, but can be changed by pushing; they are mounted on the corners of a rotatable square. At any point, you may push, simultaneously, any subset of the buttons, but then an adversary spins the square. Is there an algorithm that will enable you to turn on the bulb in at most a fixed number of spins? Watches on the Table Fifty accurate watches and a tiny diamond lie on a table. Prove that there exists a moment in time when the sum of the distances from the diamond to the ends of the minute hands exceeds the sum of the distances from the diamond to the centers of the watches. Bugging a Disk Elspeth, a new FBI recruit, has been asked to bug a circular room using seven ceiling microphones. If the room is 40 feet in diameter and the bugs are placed so as to minimize the maximum distance from any point in the room to the nearest bug, where should the bugs be located? Points on a Circle What is the probability that three uniformly random points on a circle will be contained in some semicircle? Two Different Distances Find all configurations of four distinct points in the plane that determine only two different distances. (Note: There are more of these than you probably think!) 11 Same Sum Subsets Amy asks Brad to pick 10 different numbers between 1 and 100, and to write them down secretly on a piece of paper. She now tells him she’s willing to bet $100 to $1 that his numbers contain two nonempty disjoint subsets with the same sum! Is she nuts? Players and Winners Tristan and Isolde expect to be in a situation of severely limited communication, at which time Tristan will know which two of 16 basketball teams played a game, and Isolde will know who won. How many bits must be communicated between Tristan and Isolde in order for the former to find out who won? Spaghetti Loops The 100 ends of 50 strands of cooked spaghetti are paired at random and tied together. How many pasta loops should you expect to result from this process, on average? Swapping Executives The executives of Women in Action, Inc., are seated at a long table facing the stockholders. Unfortunately, according to the meeting organizer’s chart, every one is in the wrong seat. The organizer can persuade two executives to switch seats, but only if they are adjacent and neither one is already in her correct seat. Can the organizer organize the seat-switching so as to get everyone in her correct seat? Life Isn’t a Bowl of Cherries? In front of you and your friend Amit are 4 bowls of cherries, containing, respec-tively, 5, 6, 7, and 8 cherries. You and Amit will alternately pick a bowl and take one or more cherries from it. If you go first, and you want to be sure Amit gets the last cherry, how many cherries should you take—and from which bowl? 12 Righting the Pancakes An underchef of the great and persnickety Chef Bouillon has made a stack of pancakes, but, alas, some of them are upside-down—that is, according to the great Chef, they don’t have their best side up. The underchef wants to fix the problem as follows: He finds a contiguous substack of pancakes (of size at least one) with the property that both the top pancake and the bottom pancake of the substack are upside-down. Then he removes the substack, flips it as a block and slips it back into the big stack in the same place. Prove that this procedure, no matter which upside-down pancakes are cho-sen, will eventually result in all the pancakes being right side up. Pie in the Sky What fraction of the sky is occupied by a full moon? Testing Ostrich Eggs In preparation for an ad campaign, the Flightless Ostrich Farm needs to test its eggs for durability. The world standard for egg-hardness calls for rating an egg according to the highest floor of the Empire State Building from which the egg can be dropped without breaking. Flightless’ official tester, Oskar, realizes that if he takes only one egg along on his trip to New York, he’ll need to drop it from (potentially) every one of the building’s 102 floors, starting with the first, to determine its rating. How many drops does he need in the worst case, if he takes two eggs? Circular Shadows I Suppose all three coordinate-plane projections of a convex solid are disks. Must the solid be a perfect ball? Next Card Red Paula shuffles a deck of cards thoroughly, then plays cards face up one at a time, from the top of the deck. At any time, Victor can interrupt Paula and 13 bet $1 that the next card will be red. He bets once and only once; if he never interrupts, he’s automatically betting on the last card. What’s Victor’s best strategy? How much better than even can he do? (Assume there are 26 red and 26 black cards in the deck.) Attention Paraskevidekatriaphobes Is the 13th of the month more likely to be a Friday than any other day of the week, or does it just seem that way? Unanimous Hats One hundred prisoners are offered the chance to be freed if they can win the following game. In the dark, each will be fitted with a red or black hat according to a fair coinflip. When the lights are turned on, each will see the others’ hat colors but not his own; no communication between prisoners will be permitted. Each prisoner will be asked to write down his guess at his own hat color, and the prisoners will be freed if they all get it right. The prisoners have a chance to conspire beforehand. Can you come up with a strategy for them that will maximize their probability of winning? Raising Art Value Fans of Gallery A like to say that last year, when Gallery A sold to Gallery B “Still Life with Kumquats,” the average value of the pictures in each gallery went up. Assuming that statement is correct, and that the value of each of the 400 pictures in the galleries’ combined holdings is an integral number of dollars, what is the least possible difference between the average picture values of Gallery A and Gallery B before the sale? First Odd Number What’s the first odd number in the dictionary? More specifically, suppose that every whole number from 1 to, say, 1010 is written out in formal English (e.g., “two hundred eleven,” “one thousand forty-two”) and then listed in dictionary order, that is, alphabetical order with spaces and hyphens ignored. What’s the first odd number in the list? Lattice Points and Line Segments How many lattice points (i.e., points with integer coordinates) can you have in 3-space with the property that no two of them have a third lattice point on the line segment between them? 14 Breaking a Chocolate Bar You have a rectangular chocolate bar scored in a 6 × 4 grid of squares, and you wish to break up the bar into its constituent squares. At each step, you may pick up one piece and break it along any of its marked vertical or horizontal lines. For example, you can break three times to form four rows of six squares each, then break each row five times into its constituent squares, accomplishing the desired task in 3 + 4 × 5 = 23 breaks. Can you do better? Meet the Williams Sisters Some tennis fans get excited when Venus and Serena Williams meet in a tourna-ment. The likelihood of that happening normally depends on seeding and talent, so let’s instead construct an idealized elimination tournament of 64 players, each as likely to win as to lose any given match, with bracketing chosen uniformly at random. What is the probability that the Williams sisters wind up playing each other? Cards from Their Sum The magician riffle-shuffled the cards and dealt five of them face down onto your outstretched hand. She asked you to pick any number of them secretly, then tell you their sum (jacks count as 11, queens 12, kings 13). You did that and the magician then told you exactly what cards you chose, including the suits! After a while you figured out that her shuffles were defective—they didn’t change the top five cards, so the magician had picked them in advance and knew what they were. But she still had to pick the ranks of those cards carefully (and memorize their suits), so that from the sum she could always recover the ranks. In conclusion, if you want to do the trick yourself, you need to find five distinct numbers between 1 and 13 (inclusive), with the property that every subset has a different sum. Can you do it? Waiting for Heads If you flip a fair coin repeatedly, how long do you have to wait, on average, for a run of five heads? 15 Half-right Hats One hundred prisoners are offered the chance to be freed if they can win the following game. In the dark, each will be fitted with a red or black hat according to a fair coinflip. When the lights are turned on, each will see the others’ hat colors but not his own; no communication between prisoners will be permitted. Each prisoner will be asked to write down his guess at his own hat color, and the prisoners will be freed if at least half get it right. The prisoners have a chance to conspire beforehand. Can you come up with a strategy for them that will maximize their probability of winning? Finding a Jack In some poker games the right of first dealer is determined by dealing cards face-up (from a well-shuffled deck) until some player gets a jack. On average, how many cards are dealt during this procedure? Measuring with Fuses You have on hand two fuses (lengths of string), each of which will burn for exactly 1 minute. They do not, however, burn uniformly along their lengths. Can you nevertheless use them to measure 45 seconds? Rating the Horses You have 25 horses and can race them in groups of five, but having no stopwatch, you can only observe the order of finish. How many heats of five horses do you need to determine the fastest three of your 25? Red and Blue Hats in a Line There are n prisoners this time, and again each will be fitted with a red or black hat according to flips of a fair coin. The prisoners are to be arranged in a line, so that each prisoner can see only the colors of the hats in front of him. Each prisoner must guess the color of his own hat, and is executed if he is wrong; however, the guesses are made sequentially, from the back of the line toward the front. Thus, for example, the ith prisoner in line sees the hat colors of prisoners 1, 2, . . . , i−1 and hears the guesses of prisoners n, n−1, . . . , i+1 (but he isn’t told which of those guesses were correct—the executions take place later). The prisoners have a chance to collaborate beforehand on a strategy, with the object of guaranteeing as many survivors as possible. How many prisoners can be saved in the worst case? Three Sticks You have three sticks that can’t make a triangle; that is, one is longer than the sum of the lengths of the other two. You shorten the long one by an amount 16 equal to the sum of the lengths of the other two, so you again have three sticks. If they also fail to make a triangle, you again shorten the longest stick by an amount equal to the sum of the lengths of the other two. You repeat this operation until the sticks do make a triangle, or the long stick disappears entirely. Can this process go on forever? Spiders on a Cube Three spiders are trying to catch an ant. All are constrained to the edges of a transparent cube. Each spider can move one third as fast as the ant can. Can the spiders catch the ant? Hopping and Skipping A frog hops down a long line of lily pads; at each pad, he flips a coin to decide whether to hop two pads forward or one pad back. What fraction of the pads does he hit? Divisibility Game Alice chooses a number bigger than 100 and writes it down secretly. Bob now guesses a number greater than 1, say k; if k divides Alice’s number, Bob wins. Otherwise k is subtracted from Alice’s number and Bob tries again, but may not use a number he used before. This continues until either Bob succeeds by finding a number that divides Alice’s, in which case Bob wins, or Alice’s number becomes 0 or negative, in which case Alice wins. Does Bob have a winning strategy for this game? Candles on a Cake It’s Joanna’s 18th birthday and her cake is cylindrical with 18 candles on its 18′′ circumference. The length of any arc (in inches) between two candles is greater than the number of candles on the arc, excluding the candles at the ends. Prove that Joanna’s cake can be cut into 18 equal wedges with a candle on each piece. 17 Sums of Two Squares On average, how many ways are there to write a positive number n as the sum of two squares? In other words, suppose n is a random integer between 1 and a zillion. What is the expected number of ordered pairs (i, j) of integers such that n = i2 + j2? Pairs at Maximum Distance Let X be a finite set of points on the plane. Suppose X contains n points, and the maximum distance between any two of them is d. Prove that at most n pairs of points of X are at distance d. Comparing Numbers, Version I Paula (the perpetrator) takes two slips of paper and writes an integer on each. There are no restrictions on the two numbers except that they must be different. She then conceals one slip in each hand. Victor (the victim) chooses one of Paula’s hands, which Paula then opens, allowing Victor to see the number on that slip. Victor must now guess whether that number is the larger or the smaller of Paula’s two numbers; if he guesses right, he wins $1, otherwise he loses $1. Clearly, Victor can at least break even in this game, for example, by flipping a coin to decide whether to guess “larger” or “smaller”—or by always guessing “larger,” but choosing the hand at random. The question is: Not knowing anything about Paula’s psychology, is there any way he can do better than break even? Comparing Numbers, Version II Now let’s make things more favorable to Victor: Instead of being chosen by Paula, the numbers are chosen independently at random from the uniform dis-tribution on [0,1] (two outputs from a standard random number generator will do fine). To compensate Paula, we allow her to examine the two random numbers and to decide which one Victor will see. Again, Victor must guess whether the number he sees is the larger or smaller of the two, with $1 at stake. Can he do better than break even? What are his and Paula’s best (i.e., “equilibrium”) strategies? King’s Salary Democracy has come to the little kingdom of Zirconia, in which the king and each of the other 65 citizens has a salary of one zircon. The king can not vote, but he has power to suggest changes—in particular, redistribution of salaries. Each person’s salary must be a whole number of zircons, and the salaries must sum to 66. Each suggestion is voted on, and carried if there are more votes for 18 than against. Each voter can be counted on to vote “yes” if his or her salary is to be increased, “no” if decreased, and otherwise not to bother voting. The king is both selfish and clever. What is the maximum salary he can obtain for himself, and how many referenda does he need to get it? Adding, Multiplying, and Grouping Forty-two positive integers (not necessarily distinct) are written in a row. Show that you can put plus signs, times signs, and parentheses between the integers in such a way that the value of the resulting expression is evenly divisible by one million. Missing Card Yola and Zela have devised a clever card trick. While Yola is out of the room, audience members pull out five cards from a bridge deck and hand them to Zela. She looks them over, pulls one out, and calls Yola into the room. Yola is handed the four remaining cards and proceeds to guess correctly the identity of the pulled card. How do they do it? And once you’ve figured that out, compute the size of the biggest deck of cards they could use and still perform the trick reliably. Ping-pong Progression Alice and Bob play table tennis, with Bob’s probability of winning any given point being 30%. They play until someone reaches a score of 21. What, approx-imately, is the expected number of points played? Pushing the Pedal You are standing to the right of your bicycle, steadying it with your left hand on the seat. The right pedal is in its down (6 o’clock) position. With your right hand, you push the pedal leftward (clockwise, toward 7 o’clock). Does the bike go forward or backward? Odd Light Flips Suppose you are presented with a collection of light bulbs controlled by switches. Each switch flips the state of some subset of the bulbs, that is, turns on all the ones in the subset that were off, and turns off those that were on. You are told that for any nonempty set of bulbs, there is a switch that flips an odd number of bulbs from that subset (and perhaps other bulbs as well). Show that, no matter what the initial state of the bulbs, you can use these switches to turn off all the bulbs. 19 Painting the Cubes Can you paint 1,000 unit cubes with 10 different colors in such a way that for any of the 10 colors, the cubelets can be arranged into a 10 × 10 × 10 cube with only that color showing? Red Points and Blue Given n red points and n blue points on the plane, no three on a line, prove that there is a matching between them so that line segments from each red point to its corresponding blue point do not cross. Identifying the Majority A long list of names is read out, some names many times. Your object is to end up with a name that is guaranteed to be the name which was called a majority of the time, if there is such a name. However, you have only one counter, plus the ability to keep just one name at a time in your mind. Can you do it? Returning Pool Shot A ball is shot from a corner of a polygonal pool table (not necessarily convex) with right-angle corners; let’s say all sides are aligned either exactly east-west or exactly north-south. The starting corner is a convex corner, that is, has an interior angle of 90◦. There are pockets at all corners, so that if the ball hits a corner exactly it falls in. Otherwise, the ball bounces true and without energy loss. Can the ball ever return to the corner from which it began? Pancake Stacks At the table are two hungry students, Andrea and Bruce, and two stacks of pancakes, of height m and n. Each student, in turn, must eat from the larger stack a non-zero multiple of the number of pancakes in the smaller stack. Of course, the bottom pancake of each stack is soggy, so the player who first finishes a stack is the loser. For which pairs (m, n) does Andrea (who plays first) have a winning strat-egy? How about if the game’s objective is reversed, so that the first player to finish a stack is the winner? Trapped in Thickland The inhabitants of Thickland, a world somewhere between Edwin Abbott’s Flat-land and our three-dimensional universe, are an infinite set of congruent convex polyhedra that live between two parallel planes. Up until now, they have been 20 free to escape from their slab, but haven’t wanted to. Now, however, they have been reproducing rapidly and thinking about colonizing other slabs. Their high priest is worried that conditions are so crowded, no inhabitant of Thickland can escape the slab unless others move first. Is that even possible? Hats and Infinity Each of an infinite collection of prisoners, numbered 1, 2, . . . , is to be fitted with a red or black hat. At a prearranged signal, all the prisoners are revealed to one another, so that everyone gets to see all his fellow prisoners’ hat colors—but no communication is permitted. Each prisoner is then taken aside and asked to guess the color of his own hat. All the prisoners will be freed provided only finitely many guess wrongly. The prisoners have a chance to conspire beforehand; is there a strategy that will ensure freedom? All Right or All Wrong This time the circumstances are the same but the objective is different: The guesses must either be all right or all wrong. Is there a winning strategy? Magnetic Dollars One million magnetic “susans” (Susan B. Anthony dollar coins) are tossed into two urns in the following fashion: The urns begin with one coin in each, then the remaining 999,998 coins are thrown in the air one by one. If there are x coins in one urn and y in the other, magnetic attraction will cause the next coin to land in the first urn with probability x/(x + y), and in the second with probability y/(x + y). How much should you be willing to pay, in advance, for the contents of the urn that ends up with fewer susans? Shoelaces at the Airport You are walking to your gate at O’Hare and need to tie your shoelace. Up ahead is a moving walkway that you plan to utilize. To minimize your time to get to the gate, should you stop and tie your shoelace now, or wait to tie it on the walkway? Hazards of Electronic Coinflipping You have been hired as an arbitrator and called upon to award a certain indi-visible widget randomly to either Alice, Bob, or Charlie, each with probability 1 3. Fortunately, you have an electronic coinflipping device with an analog dial 21 that enables you to enter any desired probability p. Then you push a button and the device shows “Heads” with probability p, else “Tails.” Alas, your device is showing its “low battery” light and warning you that you may set the probability p only once, and then push the coinflip button at most 10 times. Can you still do your job? Handshakes at a Party Nicholas and Alexandra went to a reception with 10 other couples; each person there shook hands with everyone he or she didn’t know. Later, Alexandra asked each of the other 21 partygoers how many people they shook hands with, and got a different answer every time. How many people did Nicholas shake hands with? Area–Perimeter Match Find all integer-sided rectangles with equal area and perimeter. Prime Test Does 49 + 610 + 320 happen to be a prime number? Lost Boarding Pass One hundred people line up to board a full jetliner, but the first has lost his boarding pass and takes a random seat instead. Each subsequent passenger takes his or her assigned seat if available, otherwise a random unoccupied seat. What is the probability that the last passenger to board finds his seat un-occupied? Lemming on a Chessboard On each square in an n × n chessboard is an arrow pointing to one of its eight neighbors (or off the board, if it’s an edge square). However, arrows in neighbor-ing squares (diagonal neighbors included) may not differ in direction by more than 45 degrees. A lemming begins in a center square, following the arrows from square to square. Is he doomed to fall off the board? Packing Slashes Given a 5 × 5 square grid, on how many of the squares can you draw diagonals (slashes or backslashes) in such a way that no two of the diagonals meet? 22 Peek Advantage You are about to bet $100 on the color of the top card of a well-shuffled deck of cards. You get to pick the color; if you’re right you win $100, otherwise you lose the same. How much is it worth to you to get a peek at the bottom card of the deck? How much more for a peek at the bottom two cards? Truly Even Split Can you partition the integers from 1 to 16 into two sets of equal sizes so that each set has the same sum, the same sum of squares, and the same sum of cubes? Line Up by Height Yankees manager Casey Stengel famously once told his players to “line up al-phabetically by height.” Suppose 26 players, no two exactly the same height, are lined up alphabetically. Prove that there are at least six who are also in height order—either tallest to shortest, or shortest to tallest. Curves on Potatoes Given two potatoes, can you draw a closed curve on the surface of each so that the two curves are identical as curves in three-dimensional space? Falling Ants Twenty-four ants are placed randomly on a meter-long rod; each ant is facing east or west with equal probability. At a signal, they proceed to march forward (that is, in whatever direction they are facing) at 1 cm/sec; whenever two ants collide, they reverse directions. How long does it take before you can be certain that all the ants are off the rod? Polygon Midpoints Let n be an odd integer, and let a sequence of n distinct points be given in the plane. Find the vertices of a (possibly self-intersecting) n-gon that has the given points, in the given order, as midpoints of its sides. Rows and Columns Prove that if you sort each row of a matrix, then each column, the rows are still sorted! 23 Bugs on a Pyramid Four bugs live on the four vertices of a triangular pyramid (tetrahedron). Each bug decides to go for a little walk on the surface of the tetrahedron. When they are done, two of them are back home, but the other two find that they have switched vertices. Prove that there was an instant when all four bugs lay on the same plane. Snake Game Joan begins by marking any square of an n×n chessboard; Judy then marks an orthogonally adjacent square. Thereafter, Joan and Judy continue alternating, each marking a square adjacent to the last one marked, creating a snake on the board. The first player unable to play loses. For which values of n does Joan have a winning strategy, and when she does, what square does she begin at? Three Negatives A set of 1000 integers has the property that every member of the set exceeds the sum of the rest. Show that the set includes at least three negative numbers. Numbers on Foreheads Each of 10 prisoners will have a digit between 0 and 9 painted on his forehead (they could be all 2’s, for example). At the appointed time each will be exposed to all the others, then taken aside and asked to guess his own digit. In order to avoid mass execution, at least one prisoner must guess correctly. The prisoners have an opportunity to conspire beforehand; find a scheme by means of which they can ensure success. Bias Test Before you are two coins; one is a fair coin, and the other is biased toward heads. You’d like to try to figure out which is which, and to do so you are permitted two flips. Should you flip each coin once, or one coin twice? Nonrepeating String Is there a finite string of letters from the Latin alphabet with the property that there is no pair of adjacent identical substrings, but the addition of any letter to either end would create one? Three-Way Election Alison, Bonnie, and Clyde run for class president and finish in a three-way tie. To break it, they solicit their fellow students’ second choices, but again there is 24 a three-way tie. The election committee is stymied until Alison steps forward and points out that, since the number of voters is odd, they can make two-way decisions. She therefore proposes that the students choose between Bonnie and Clyde, and then the winner would face Alison in a runoff. Bonnie complains that this is unfair because it gives Alison a better chance to win than either of the other two candidates. Is Bonnie right? Skipping a Number At the start of the 2019 season WNBA star Missy Overshoot’s lifetime free throw percentage was below 80%, but by the end of the season it was above 80%. Must there have been a moment in the season when Missy’s free throw percentage was exactly 80%? Red and Black Sides Bored in math class, George draws a square on a piece of paper and divides it into rectangles (with sides parallel to a side of the square), using a red pen after his black one runs out of ink. When he’s done, he notices that every rectangular tile has at least one totally red side. Prove that the total length of George’s red lines is at least the length of a side of the big square. Unbroken Lines Can you re-arrange the 16 square tiles below into a 4 × 4 square—no rotation allowed!—in such a way that no line ends short of the boundary of the big square? 25 Baby Frog To give a baby frog jumping practice, her four grandparents station themselves at the corners of a large square field. When a grandparent croaks, the baby leaps halfway to it. In the field is a small round clearing. Can the grandparents get the baby to that clearing, no matter where in the field she starts? Sequencing the Digits In how many ways can you write the numbers 0 through 9 in a row, such that each digit other than the leftmost is within one of some digit to the left of it? Recovering the Polynomial The Oracle at Delphi has in mind a certain polynomial p (in the variable x, say) with non-negative integer coefficients. You may query the Oracle with any integer x, and the Oracle will tell you the value of p(x). How many queries do you have to make to determine p? Faulty Combination Lock A combination lock with three dials, each numbered 1 through 8, is defective in that you only need to get two of the numbers right to open the lock. What is the minimum number of (three-number) combinations you need to try in order to be sure of opening the lock? 26 Prisoners and Gloves Each of 100 prisoners gets a different real number written on his forehead, and is provided with a black glove and a white glove. Seeing the numbers on the other prisoners’ foreheads, he must put one glove on each hand in such a way that when the prisoners are later lined up in real-number order and hold hands, touching gloves match. How can the prisoners, who are permitted to conspire beforehand, ensure success? Bracing the Grid Suppose that unit-length rods, jointed at their ends, form an m × n grid of diamond-shaped cells. You may brace some subset S of the cells with diagonal segments (of length √ 2). Which choices of S suffice to make the grid rigid in the plane? Gaming the Quilt You come along just as a church lottery is closing; its prize is a quilt worth $100 to you, but they’ve only sold 25 tickets. At $1 a ticket, how many tickets should you buy? Rectangles Tall and Wide It is easy to check that a 6 × 5 rectangle can be tiled with 3 × 2 rectangles, but only if some tiles are horizontally oriented and others are vertically oriented. Show that this can’t happen if the big rectangle is a square—in other words, if a square is tiled with congruent rectangles, then it can be re-tiled with the same rectangles all oriented the same way. Guarding the Gallery A certain museum room is shaped like a highly irregular, non-convex, 11-sided polygon. How many guard-posts are needed in the room, in the worst case, to 27 ensure that every part of the room can be seen from at least one guard-post? Flying Saucers A fleet of saucers from planet Xylofon has been sent to bring back the inhab-itants of a certain randomly-selected house, for exhibition in the Xylofon Xoo. The house happens to contain five men and eight women, to be beamed up randomly one at a time. Owing to the Xylofonians’ strict sex separation policy, a saucer cannot bring back earthlings of both sexes. Thus, it beams people up until it gets a member of a second sex, at which point that one is beamed back down and the saucer takes off with whatever it has left. Another saucer then starts beaming people up, following the same rule, and so forth. What is the probability that the last person beamed up is a woman? Increasing Routes On the Isle of Sporgesi, each segment of road (between one intersection and the next) has its own name. Let d be the average number of road segments meeting at an intersection. Show that you can take a drive on the Isle of Sporgesi that covers at least d road segments, and hits those segments in strict alphabetical order! Curve on a Sphere Prove that if a closed curve on the unit sphere has length less than 2π, then it is contained in some hemisphere. Biased Betting Alice and Bob each have $100 and a biased coin that comes up heads with probability 51%. At a signal, each begins flipping his or her coin once a minute and bets $1 (at even odds) on each outcome, against a bank with unlimited funds. Alice bets on heads, Bob on tails. Suppose both eventually go broke. Who is more likely to have gone broke first? Suppose now that Alice and Bob are flipping the same coin, so that when the first one goes broke the second one’s stack will be at $200. Same question: Given that they both go broke, who is more likely to have gone broke first? Unbroken Curves Can you re-arrange the 16 square tiles below into a 4 × 4 square—no rotation allowed!—in such a way that no curve ends short of the boundary of the big square? 28 Ascending and Descending Fix a number n and call a list of the numbers from 1 to n “good” if there is no descending subsequence of length 10. Show that there are at most 81n good lists. Dot-town Exodus Each resident of Dot-town carries a red or blue dot on his (or her) forehead, but if he ever thinks he knows what color it is he leaves town immediately and permanently. Each day the residents gather; one day a stranger comes and tells them something—anything—nontrivial about the number of blue dots. Prove that eventually Dot-town becomes a ghost town, even if everyone can see that the stranger’s statement is patently false. Drawing Socks You have 60 red and 40 blue socks in a drawer, and you keep drawing a sock uniformly at random until you have drawn all the socks of one color. What is the expected number of socks left in the drawer? Path Through the Cells Cells in a certain cellular telephone network are assigned frequencies in such a way that no two adjacent cells use the same frequency. Show that if the number of frequencies used is as small as possible (subject to this condition), then it’s possible to design a path that moves from cell to adjacent cell and hits each frequency exactly once—in ascending order of frequencies! Painting the Polyhedron Let P be a polyhedron with red and green faces such that every red face is surrounded by green ones, but the total red area exceeds the total green area. Prove that you can’t inscribe a sphere in P. 29 Whim-Nim You and a friend, bored with Nim and Nim Misere, decide to play a variation in which at any point, either player may declare “Nim” or “Mis ere” instead of removing chips. This happens at most once in a game, and then of course the game proceeds normally according to that variation of Nim. (Taking the whole single remaining stack in an undeclared game loses, as your opponent can then declare as he wishes for his last move.) What’s the correct strategy for this game, which its inventor, the late John Horton Conway, called “Whim”? Bacteria on the Plane Suppose the world begins with a single bacterium at the origin of the infinite plane grid. When it divides its two successors move one vertex north and one vertex east, so that there are now two bacteria, one at (0,1) and one at (1,0). Bacteria continue to divide, each time with one successor moving north and one east, provided both of those points are unoccupied. Show that no matter how long this process continues, there’s always a bac-terium inside the circle of radius 3 about the origin. Poorly Placed Dominoes What’s the smallest number of dominoes one can place on a chessboard (each covering two adjacent squares) so that no more fit? Ants on the Circle Twenty-four ants are randomly placed on a circular track of length 1 meter; each ant faces randomly clockwise or counterclockwise. At a signal, the ants begin marching at 1 cm/sec; when two ants collide they both reverse directions. What is the probability that after 100 seconds, every ant finds itself exactly where it began? Chinese Nim On the table are two piles of beans. Alex must either take some beans from one pile or the same number of beans from each pile; then Beth has the same options. They continue alternating until one wins the game by taking the last bean. What’s the correct strategy for this game? For example, if Alex is faced with piles of size 12,000 and 20,000, what should he do? How about 12,000 and 19,000? 30 Not Burning Brownies When you bake a pan of brownies, those that share an edge with the edge of the pan will often burn. For example, if you bake 16 square brownies in a square pan, 12 are subject to burning. Design a pan and divide it into brownie shapes to make 16 identically shaped brownies of which as few as possible will burn. Can you get it down to four burned brownies? Great! How about three? Double Cover by Lines Let Lθ be the set of all lines on the plane at angle θ to the horizontal. If θ and θ′ are two different angles, the union of the sets Lθ and Lθ′ constitutes a double cover of the plane, that is, every point belongs to exactly two lines. Can this be done in any other way? That is, can you cover each point of the plane exactly twice using a set of lines that contains lines in more than two different directions? Integral Rectangles A rectangle in the plane is partitioned into smaller rectangles, each of which has either integer height or integer width (or both). Prove that the large rectangle itself has this property. Losing at Dice Visiting Las Vegas, you are offered the following game. Six dice are to be rolled, and the number of different numbers that appear will be counted. That could be any number from one to six, of course, but they are not equally likely. If you get the number “4” this way you win $1, otherwise you lose $1. You decide you like this game, and plan to play it repeately until the $100 you came with is gone. At one minute per game, how long, on the average, will it take before you are wiped out? Even-sum Billiards You pick 10 times, with replacement (necessarily!), from an urn containing 9 billiard balls numbered 1 through 9. What is the probability that the sum of the numbers of the balls you picked is even? Competing for Programmers Two Silicon Valley startups are competing to hire programmers. Hiring on alternate days, each can begin by hiring anyone, but every subsequent hire by a company must be a friend of some previous hire—unless no such person exists, in which case that company can again hire anyone. 31 In the candidate pool are 10 geniuses; naturally each company would like to get as many of those as possible. Can it be that the company that hires first cannot prevent its rival from hiring nine of the geniuses? Strength of Schedule The 10 teams comprising the “Big 12” college football conference are all sched-uled to play one another in the upcoming season, after which one will be declared conference champion. There are no ties, and each team scores a “win point” for every team it beats. Suppose that, worried about breaking ties, a member of the governing board of the conference suggests that each team receive in addition “strength of sched-ule points,” calculated as the sum of the win points achieved by the teams it beat. A second member asks: “What if all the teams end up with the same number of strength-of-schedule points?” What, indeed? Can that even happen? Locker Doors Lockers numbered 1 to 100 stand in a row in the main hallway of Euclid Junior High School. The first student arrives and opens all the lockers. The second student then goes through and re-closes lockers numbered 2, 4, 6, etc.; the third student changes the state of every locker whose number is a multiple of 3, then the next every multiple of 4, etc., until the last student opens or closes only locker number 100. After the 100 students have passed through, which lockers are open? Gasoline Crisis You need to make a long circular automobile trip during a gasoline crisis. In-quiries have ascertained that the gas stations along the route contain just enough fuel to make it all the way around. If you have an empty tank but can start at a station of your choice, can you complete a clockwise round trip? Soldiers in a Field An odd number of soldiers are stationed in a large field. No two soldiers are exactly the same distance apart as any other two soldiers. Their commanding officer radios instructions to each soldier to keep an eye on his nearest neighbor. Is it possible that every soldier is being watched? Home-field Advantage Every year, the Elkton Earlies and the Linthicum Lates face off in a series of baseball games, the winner being the first to win four games. The teams are 32 evenly matched but each has a small edge (say, a 51% chance of winning) when playing at home. Every year, the first three games are played in Elkton, the rest in Linthicum. Which team has the advantage? Zeroes and Ones Show that every natural number has a non-zero multiple which, when written out (base 10), contains only zeroes and ones. Even better, prove that your telephone number, if it ends with a 1, 3, 7, or 9, has a multiple whose decimal representation is all ones! Random Intersection Two unit-radius balls are randomly positioned subject to intersecting. What is the expected volume of their intersection? For that matter, what is the expected surface area of their intersection? Profit and Loss At a recent stockholders’ meeting of Widget Industrials Inc., the Chief Financial Officer presented a chart of the month-by-month profits (or losses) since the last meeting. “Note,” said she, “that we made a profit over every consecutive eight-month period.” “Maybe so,” complained a shareholder, “but I see we lost money over every consecutive five-month period!” What’s the maximum number of months that could have passed since the last meeting? Boarding the Manhole An open manhole 4 meters in diameter has to be covered by boards of total width w. The boards are each more than 4 meters in length, so if w ≥4 it’s obvious that you can cover the manhole by laying the boards side by side (see figure below). If w is only 3.9, say, you still have plenty of wood and you are allowed to overlap the boards (but not cut them up) if you want. Can you still cover the manhole? Pegs on the Half-plane Each grid point on the XY plane on or below the X-axis is occupied by a peg. At any time, a peg can be made to jump over a neighbor peg (horizontally, vertically, or diagonally adjacent) and onto the next grid point in line, provided that point was unoccupied. The jumped peg is then removed. Can you get a peg arbitrarily far above the X-axis? 33 Unbreakable Domino Cover A 6 × 5 rectangle can be covered with 2 × 1 dominoes, as in the figure below, in such a way that no line between dominoes cuts all the way across the board. Can you cover a 6 × 6 square that way? Turning the Die In the game of Turn-Die, a die is rolled and the number that appears is recorded. The first player then turns the die 90 degrees (giving her four options) and the new value is added to the old one. The second player does the same and the two players alternate until a sum of 21 or higher is reached. If 21 was hit exactly, the player who reached it wins; otherwise, the player who exceeded it loses. Do you want to be first or second? Protecting the Statue Michelangelo’s David, in Florence, is protected (on the plane) by laser beams in such a way that no-one can approach the statue or any of the laser-beam sources without crossing a beam. What’s the minimum number of lasers needed to accomplish this? (Beams reach 100 meters, say.) 34 Wild Guess David and Carolyn are mathematicians who are unafraid of the infinite and cheerfully invoke the Axiom of Choice when needed. They elect to play the following two-move game. For her move, Carolyn chooses an infinite sequence of real numbers, and puts each number in an opaque box. David gets to open as many boxes as he wants—even infinitely many—but must leave one box unopened. To win, he must guess exactly the real number in that box. On whom will you bet in this game, Carolyn or David? Laser Gun You find yourself standing in a large rectangular room with mirrored walls. At another point in the room is your enemy, brandishing a laser gun. You and she are fixed points in the room; your only defense is that you may summon bodyguards (also points) and position them in the room to absorb the laser rays for you. How many bodyguards do you need to block all possible shots by the enemy? Splitting the Stacks Your job is to separate a stack of n objects into individual items. At any time you may split a stack into two stacks, and get paid the product of the sizes of the two stacks. For example, for n = 8, you could split into a stack of size 5 and a stack of size 3, getting paid $15; then you can split the 5-stack into 2 and 3 to get $6 more. Then the two threes can each be split ($2 each) and the remaining three twos ($1 each) for a total income of 15 + 6 + (2 × 2) + (3 × 1) = 28 dollars. What’s the most money you can get starting with n objects, and how can you get it? Chameleons A colony of chameleons currently contains 20 red, 18 blue, and 16 green individ-uals. When two chameleons of different colors meet, each of them changes his or her color to the third color. Is it possible that, after a while, all the chameleons have the same color? Wires under the Hudson Fifty identical wires run through a tunnel under the Hudson River, but they all look the same, and you need to determine which pairs of wire-ends belong to the same wire. To do this you can tie pairs of wires together at one end of the tunnel and test pairs of wire-ends at the other end to see if they close a circuit; in other words, you can determine whether two wires are tied together at the other end. How many trips across the Hudson do you need to accomplish your task? 35 Two Round-robins The Games Club’s 20 members played a round-robin checkers tourament on Monday and a round-robin chess tournament on Tuesday. In each tournament, a player scored one point for each other player he or she beat, and half a point for each tie. Suppose every player’s scores in the two tournaments differed by at least 10. Show that in fact, the differences were all exactly 10. Factorial Coincidence Suppose that a, b, c, and d are positive integers, all different, all greater than one. Can it be that a!b = c!d? Conversation on a Bus Ephraim and Fatima, two colleagues in the Mathematics Department at Zorn University, wind up seated together on the bus to campus. Ephraim begins a conversation with “So, Fatima, how are your kids doing? How old are they now, anyway?” “It turns out,” says Fatima as she is putting the $1 change from the bus driver into her wallet, “that the sum of their ages is the number of this bus, and the product is the number of dollars that happen to be in my purse at the moment.” “Aha!” replies Ephraim. “So, if I remembered how many kids you have, and you told me how much money you are carrying, I could deduce their ages?” “Actually, no,” says Fatima. “In that case,” says Ephraim, “I know how much money is in your purse.” What is the bus number? Coins on the Table One hundred quarters lie on a rectangular table, in such a way that no more can be added without overlapping. (We allow a quarter to extend over the edge, as long as its center is on the table.) Prove that you can start all over again and cover the whole table with 400 quarters! (This time we allow overlap and overhang). Roulette for the Unwary Elwyn is in Las Vegas celebrating his 21st birthday, and his girlfriend has gifted him with twenty-one $5 bills to gamble with. He saunters over to the roulette table, noting that there are 38 numbers (0, 00, and 1 through 36) on the wheel. If he bets $1 on a single number, he will win with probability 1/38 and collect $36 (in return for his $1 stake, which still goes to the bank). Otherwise, of course, he just loses the dollar. 36 Elwyn decides to use his $105 to make 105 one-dollar bets on the number 21. What, approximately, is the probability that Elwyn will come out ahead? Is it better than, say, 10%? Uniformity at the Bakery A baker’s dozen (thirteen) bagels have the property that any 12 of them can be split into 2 piles of 6 each, which balance perfectly on the scale. Suppose each bagel weighs an integer number of grams. Must all the bagels have the same weight? Slabs in 3-Space A “slab” is the region between two parallel planes in three-dimensional space. Prove that you cannot cover all of 3-space with a set of slabs the sum of whose thicknesses is finite. Pegs in a Square Suppose we begin with n2 pegs on a plane grid, one peg occupying each vertex of an n-vertex by n-vertex square. Pegs jump only horizontally or vertically, by passing over a neighboring peg and into an unoccupied vertex; the jumped peg is then removed. The goal is to reduce the n2 pegs to only 1. Prove that if n is a multiple of 3, it can’t be done! Filling a Bucket Before you are 12 two-gallon buckets and a 1-gallon scoop. At each turn, you may fill the scoop with water and distribute the water any way you like among the buckets. However, each time you do this your opponent will empty two buckets of her choice. You win if you can get one of the big buckets to overflow. Can you force a win? If so, how long will it take you? 37 Polygon on the Grid A convex polygon is drawn on the coordinate plane with all its vertices on integer points, but no side parallel to the x- or y-axis. Let h be the sum of the lengths of the horizontal line segments at integer height that intersect the (filled-in) polygon, and v the equivalent for the vertical line segments. Prove that h = v. Game of Desperation On a piece of paper is a row of n empty boxes. Tristan and Isolde take turns, each writing an “S” or an “O” into a previously blank box. The winner is the one who completes an “SOS” in consecutive boxes. For which n does the second player (Isolde) have a winning strategy? Gluing Pyramids A solid square-base pyramid, with all edges of unit length, and a solid triangle-base pyramid (tetrahedron), also with all edges of unit length, are glued together by matching two triangular faces. How many faces does the resulting solid have? Exponent upon Exponent Part I: If xxx··· = 2, what is x? Part II: If xxx··· = 4, what is x? Part III: How do you explain getting the same answer to Parts I and II? Random Intervals The points 1, 2, . . . , 1000 on the number line are paired up at random, to form the endpoints of 500 intervals. What is the probability that among these intervals is one which intersects all the others? North by Northwest If you’ve never seen the famous Alfred Hitchcock movie North by Northwest (1959), you should. But what direction is that, exactly? Assume North is 0◦, East 90◦, etc. Missing Digit The number 229 has 9 digits, all different; which digit is missing? 38 Coins in a Row On a table is a row of 50 coins, of various denominations. Alix picks a coin from one of the ends and puts it in her pocket; then Bert chooses a coin from one of the (remaining) ends, and the alternation continues until Bert pockets the last coin. Prove that Alix can play so as to guarantee at least as much money as Bert. Bugs on Four Lines You are given four lines in a plane in general position (no two parallel, no three intersecting in a common point). On each line a ghost bug crawls at some constant velocity (possibly different for each bug). Being ghosts, if two bugs happen to cross paths they just continue crawling through each other uninterrupted. Suppose that five of the possible six meetings actually happen. Prove that the sixth does as well. First-grade Division On the first day of class Miss Feldman divides her first-grade class into k working groups. On the second day, she picks the working groups a different way, this time ending up with k+1 of them. Show that there are at least two kids who are in smaller groups on the second day than they were on the first day. Deterministic Poker Unhappy with the vagaries of chance, Alice and Bob elect to play a completely deterministic version of draw poker. A deck of cards is spread out face-up on the table. Alice draws five cards, then Bob draws five cards. Alice discards any number of her cards (the discarded cards will remain out of play) and replaces them with a like number of others; then Bob does the same. All actions are taken with the cards face-up in view of the opponent. The player with the better hand wins; since Alice goes first, Bob is declared to be the winner if the final hands are equally strong. Who wins with best play? 39 Precarious Picture Suppose that you wish to hang a picture with a string attached at two points on the frame. If you hang it by looping the string over two nails in the ordinary way, as shown below, and one of the nails comes out, the picture will still hang (albeit lopsidedly) on the other nail. Can you hang it so that the picture falls if either nail comes out? Find the Robot At time t = 0 a robot is placed at some unknown grid point in 3-dimensional space. Every minute, the robot moves a fixed, unknown distance in a fixed, unknown direction, to a new grid point. Each minute, you are allowed to probe any single point in space. Devise an algorithm that is guaranteed to find the robot in finite time. Early Commuter A commuter arrives at her home station an hour early and walks toward home until she meets her husband driving to pick her up at the normal time. She ends up home 20 minutes earlier than usual. How long did she walk? Subtracting Around the Corner A sequence of n positive numbers is written on a piece of paper. In one “oper-ation,” a new sequence is written below the old one according to the following rules: The (absolute) difference between each number and its successor is writ-ten below that number; and the absolute difference between the first and last numbers is written below the last number. For example, the sequence 4 13 9 6 is followed by 9 4 3 2. Try this for n = 4 using random numbers between, say, 1 and 100. You’ll find that after remarkably few operations, the sequence degenerates to 0 0 0 0 where it of course remains. Why? Is the same true for n = 5? 40 Powers of Roots What is the first digit after the decimal point in the number ( √ 2 + √ 3) to the billionth power? An Attractive Game You have an opportunity to bet $1 on a number between 1 and 6. Three dice are then rolled. If your number fails to appear, you lose your $1. If it appears once, you win $1; if twice, $2; if three times, $3. Is this bet in your favor, fair, or against the odds? Is there a way to determine this without pencil and paper (or a computer)? Circular Shadows II Show that if the projections of a solid body onto two planes are perfect disks, then the two projections have the same radius. Finding the Rectangles Prove that any tiling of a regular 400-gon by parallelograms must contain at least 100 rectangles. Figure Eights in the Plane How many disjoint topological “figure 8s” can be drawn on the plane? Alternate Connection One hundred points lie on a circle. Alice and Bob take turns connecting pairs of points by a line, until every point has at least one connection. The last one to play wins; which player has a winning strategy? Tiling a Polygon A “rhombus” is a quadrilateral with four equal sides; we consider two rhombi to be different if you can’t translate (move without rotation) one to coincide with 41 the other. Given a regular polygon with 100 sides, you can take any two non-parallel sides, make two copies of each and translate them to form a rhombus. You get 50 2 different rhombi that way. You can use translated copies of these to tile your 100-gon; show that if you do, you will use each different rhombus exactly once! Bindweed and Honeysuckle Two vines, a bindweed and a honeysuckle, climb a tree trunk, starting and ending at the same place. The bindweed circles the trunk three times counter-clockwise while the honeysuckle circles five times clockwise. Not counting the top and bottom, how many times do they cross? Uniting the Loops Can you re-arrange the 16 square tiles below into a 4 × 4 square—no rotation allowed!—in such a way that the curves form a single loop? Lines that exit from the bottom are assumed to continue at the same point on the top, and similarly right to left; in other words, imagine that the square is rolled up into a doughnut. Splitting a Polygon A chord of a polygon is a straight line segment through the interior of the polygon that touches the polygon’s perimeter at the segment’s endpoints and nowhere else. Show that every polygon, convex or not, has a chord such that each of the two regions into which it divides the polygon has area at least 1/3 the area of the polygon. Two Monks on a Mountain Remember the monk who climbed Mt. Fuji on Monday and descended on Tues-day? This time, he and a fellow monk climb the mountain on the same day, 42 starting at the same time and altitude, but on different paths. The paths go up and down on the way to the summit (but never dip below the starting alti-tude); you are asked to prove that they can vary their speeds (sometimes going backwards) so that at every moment of the day they are at the same altitude! Alternative Dice Can you design two different dice so that their sums behave just like a pair of ordinary dice? That is, there must be two ways to roll a 3, 6 ways to roll a 7, one way to roll a 12, and so forth. Each die must have six sides, and each side must be labeled with a positive integer. Even Split Prove that from every set of 2n integers, you can choose a subset of size n whose sum is divisible by n. Coconut Classic Five men and a monkey, marooned on an island, collect a pile of coconuts to be divided equally the next morning. During the night, however, one of the men decides he’d rather take his share now. He tosses one coconut to the monkey and removes exactly 1/5 of the remaining coconuts for himself. A second man does the same thing, then a third, fourth, and fifth. The following morning the men wake up together, toss one more coconut to the monkey, and divide the rest equally. What’s the least original number of coconuts needed to make this whole scenario possible? Infected Checkerboard An infection spreads among the squares of an n × n checkerboard in the follow-ing manner: If a square has two or more infected neighbors, then it becomes infected itself. (Neighbors are orthogonal only, so each square has at most four neighbors.) For example, suppose that we begin with all n squares on the main diagonal infected. Then the infection will spread to neighboring diagonals and eventually to the whole board. Prove that you cannot infect the whole board if you begin with fewer than n infected squares. 43 Alternating Powers Since the series 1 −1 + 1 −1 + 1 −· · · does not converge, the function f(x) = x −x2 + x4 −x8 + x16 −x32 + · · · makes no sense when x = 1. However, f(x) does converge for all positive real numbers x < 1. If we wanted to give f(1) a value, it might make sense to let it be the limit of f(x) as x approaches 1 from below. Does that limit exist? If so, what is it? Service Options You are challenged to a short tennis match, with the winner to be the first player to win four games. You get to serve first. But there are options for determining the sequence in which the two of you serve: 1. Standard: Serve alternates (you, her, you, her, you, her, you). 2. Volleyball Style: The winner of the previous game serves the next one. 3. Reverse Volleyball Style: The winner of the previous game receives in the next one. Which option should you choose? You may assume it is to your advantage to serve. You may also assume that the outcome of any game is independent of when the game is played and of the outcome of any previous game. Conway’s Immobilizer Three cards, an ace, a deuce, and a trey, lie face-up on a desk in some or all of three marked positions (“left,” “middle,” and “right”). If they are all in the same position, you see only the top card of the stack; if they are in two positions, you see only two cards and do not know which of them is concealing the third card. Your objective is to get the cards stacked on the left with ace on top, then deuce, then trey on bottom. You do this by moving one card at a time, always from the top of one stack to the top of another (possibly empty) stack. The problem is, you have no short-term memory and must, therefore, devise an algorithm in which each move is based entirely on what you see, and not on what you last saw or did, or on how many moves have transpired. An observer will tell you when you’ve won. Can you devise an algorithm that will succeed in a bounded number of steps, regardless of the initial configuration? Same Sum Dice You roll a set of n red n-sided dice, and a set of n black n-sided dice. Each die is labeled with the numbers from 1 to n. Show that there must always be a nonempty subset of the red dice, and a nonempty subset of the black dice, showing the same sum. 44 Matching Coins Sonny and Cher play the following game. In each round, a fair coin is tossed. Before the coin is tossed, Sonny and Cher simultaneously declare their guess for the result of the coin toss. They win the round if both guessed correctly. The goal is to maximize the fraction of rounds won, when the game is played for many rounds. So far, the answer is obviously 50%: Sonny and Cher agree on a sequence of guesses (e.g., they decide to always declare “heads”), and they can’t do any better than that. However, before the game begins, the players are informed that just prior to the first toss, Cher will be given the results of all the coin tosses in advance! She has a chance now to discuss strategy with Sonny, but once she gets the coinflip information, there will be no further opportunity to conspire. Show how Sonny and Cher can guarantee to get at least 6 wins in 10 flips. (If that’s too easy for you, show how they can guarantee at least six wins in only nine flips!) Next Card Bet Cards are turned face up one at a time from the top of a well-shuffled deck. You begin with a bankroll of $1, and can bet any fraction of your current worth, prior to each revelation, on the color of the next card. You get even odds regardless of the current composition of the deck. Thus, for example, you can decline to bet until the last card, whose color you will of course know, then bet everything and be assured of going home with $2. Is there any way you can guarantee to finish with more than $2? If so, what’s the maximum amount you can assure yourself of winning? Summing Fractions Gail asks Henry to think of a number n between 10 and 100, but not to tell her what it is. She now tells Henry to find all (unordered) pairs of numbers j, k that are relatively prime and no more than n, but add up to more than n. He now adds all the fractions 1/jk. Whew! Finally, Gail tells Henry what his sum is. How does she do it? Box in a Box Suppose the cost of shipping a rectangular box is given by the sum of its length, width, and height. Might it be possible to save money by fitting your box into a cheaper box? Option Hats One hundred prisoners are told that at midnight, in the dark, each will be fitted with a red or black hat according to a fair coin-flip. The prisoners will 45 be arranged in a circle and the lights turned on, enabling each prisoner to see every other prisoner’s hat color. Once the lights are on, the prisoners will have no opportunity to signal to one another or to communicate in any way. Each prisoner will then be taken aside and given the option of trying to guess whether his own hat is red or black, but he may choose to pass. The prisoners will all be freed if (1) at least one prisoner chooses to guess his hat color, and (2) every prisoner who chooses to guess guesses correctly. The prisoners have a chance to devise a strategy before the game begins. Can they achieve a winning probability greater than 50%? Impressionable Thinkers The citizens of Floptown meet each week to talk about town politics, and in particular whether or not to support the building of a new shopping mall down-town. During the meetings each citizen talks to his friends—of whom there are always an odd number, for some reason—and the next day, changes (if nec-essary) his opinion regarding the mall so as to conform to the opinion of the majority of his friends. Prove that eventually, the opinions held every other week will be the same. One-bulb Room Each of n prisoners will be sent alone into a certain room, infinitely often, but in some arbitrary order determined by their jailer. The prisoners have a chance to confer in advance, but once the visits begin, their only means of communication will be via a light in the room which they can turn on or off. Help them design a protocol which will ensure that some prisoner will eventually be able to deduce that everyone has visited the room. Sphere and Quadrilateral A quadrilateral in space has all of its edges tangent to a sphere. Prove that the four points of tangency lie on a plane. Bluffing with Reals Consider the following simple bluffing game. Louise and Jeremy ante $1 each and each is given a secret random real number between 0 and 1. Louise may decide to pass in which case the $2 pot goes to the player with the higher number. However, if she chooses, Louise may add another dollar to the pot. Jeremy may then “call” by adding another dollar himself, in which case the pot, now with $4 in it, goes again to the player with the higher number. Or Jeremy may fold, ceding the pot, with his $1 in it, to Louise. Surely Louise has the advantage in this game, or at least equality, since she can break even by always passing. How much is the game worth to her? What are the players’ equilibrium strategies? 46 Tiling with Crosses Can you tile the plane with 5-square crosses? Can you tile 3-space with 7-cube crosses? Y’s in the Plane Prove that only countably many disjoint Y’s can be drawn in the plane. Infected Cubes An infection spreads among the n3 unit cubes of an n × n × n cube, in the following manner: If a unit cube has three or more infected neighbors, then it becomes infected itself. (Neighbors are orthogonal only, so each little cube has at most six neighbors.) Prove that you can infect the whole big cube starting with just n2 sick unit cubes. Worst Route A postman has deliveries to make on a long street, to addresses 2, 3, 5, 7, 11, 13, 17, and 19. The distance between any two houses is proportional to the difference of their addresses. To minimize the distance traveled, the postman would of course make his deliveries in increasing (or decreasing) order of address. But our postman is overweight and would like to maximize the distance traveled making these de-liveries, so as to get the most exercise he can. But he can’t just wander around town; to do his job properly he is obligated to walk directly from each delivery to the next one. In what order should he make his deliveries? Factorials and Squares Consider the product 100! · 99! · 98! · · · · · 2! · 1!. Call each of the 100 factors k! a “term.” Can you remove one term and leave a perfect square? Halfway Points Let S be a finite set of points in the unit interval [0,1], and suppose that every point x ∈S lies either halfway between two other points in S (not necessarily the nearest ones) or halfway between another point in S and an endpoint. Show that all the points in S are rational. Who Won the Series? Two evenly-matched teams meet to play a best-of-seven World Series of baseball games. Each team has the same small advantage when playing at home. As is 47 customary for this event, one team (say, Team A) plays games 1 and 2 at home, and, if necessary, plays games 6 and 7 at home. Team B plays games 3, 4 and, if needed, 5 at home. You go to a conference in Europe and return to find that the series is over, and six games were played. Which team is more likely to have won the series? Tiling with L’s Can you tile the positive quadrant of the plane grid with unrotated triominoes each of which is either shaped like a letter L, or a backwards L? Dishwashing Game You and your spouse flip a coin to see who washes the dishes each evening. “Heads” she washes, “tails” you wash. Tonight she tells you she is imposing a different scheme. You flip the coin 13 times, then she flips it 12 times. If you get more heads than she does, she washes; if you get the same number of heads or fewer, you wash. Should you be happy? Random Judge After a wild night of shore leave, you are about to be tried by your U.S. Navy superiors for unseemly behavior. You have a choice between accepting a “sum-mary” court-martial with just one judge, or a “special” court-martial with three judges who decide by majority vote. Each possible judge has the same (independent) probability—65.43%—of deciding in your favor, except that one officer who would be judging your special court-martial (but not the summary) is notorious for flipping a coin to make his decisions. Which type of court-martial is more likely to keep you out of the brig? Angles in Space Prove that among any set of more than 2n points in Rn, there are three that determine an obtuse angle. Wins in a Row You want to join a certain chess club, but admission requires that you play three games against Ioana, the current club champion, and win two in a row. Since it is an advantage to play the white pieces (which move first), you alternate playing white and black. A coin is flipped, and the result is that you will be white in the first and third games, black in the second. Should you be happy? 48 Chessboard Guess Troilus is engaged to marry Cressida but threatened with deportation, and Immigration is questioning the legitimacy of the proposed marriage. To test their connection, Troilus will be brought into a room containing a chessboard, one of whose squares is designated as special. On every square will be a coin, either heads-up or tails-up. Troilus gets to turn over one coin, after which he will be ejected from the room and Cressida brought in. Cressida, after examining the chessboard, must guess the designated square. If she gets it wrong, Troilus is deported. Can Troilus and Cressida save their marriage? Split Games You are a rabid baseball fan and, miraculously, your team has won the pennant— thus, it gets to play in the World Series. Unfortunately, the opposition is a superior team whose probability of winning any given game against your team is 60%. Sure enough, your team loses the first game in the best-of-seven series and you are so unhappy that you drink yourself into a stupor. When you regain consciousness, you discover that two more games have been played. You run out into the street and grab the first passer-by. “What happened in games 2 and 3 of the World Series?” “They were split,” he says. “One game each.” Should you be happy? Frames on a Chessboard You have an ordinary 8 × 8 chessboard with red and black squares. A genie gives you two “magic frames,” one 2 × 2 and one 3 × 3. When you place one of these frames neatly on the chessboard, the 4 or 9 squares they enclose instantly flip their colors. Can you reach all 264 possible color configurations? Angry Baseball As in Split Games, your team is the underdog and wins any given game in the best-of-seven World Series with probability 40%. But, hold on: This time, whenever your team is behind in the series, the players get angry and play better, raising your team’s probability of winning that game to 60%. Before it all begins, what is your team’s probability of winning the World Series? Bugs on a Polyhedron Associated with each face of a solid convex polyhedron is a bug which crawls along the perimeter of the face, at varying speed, but only in the clockwise 49 direction. Prove that no schedule will permit all the bugs to circumnavigate their faces and return to their initial positions without incurring a collision. Serious Candidates Assume that, as is often the case, no one has any idea who the next nominee for President of the United States will be, of the party not currently in power. In particular, at the moment no person has probability as high as 20% of being chosen. As the politics and primaries proceed, probabilities change continuously and some candidates will exceed the 20% threshold while others will never do so. Eventually one candidate’s probability will rise to 100% while everyone else’s drops to 0. Let us say that after the convention, a candidate is entitled to say he or she was a “serious” candidate if at some point his or her probability of being nominated exceeded 20%. Can anything be said about the expected number of serious candidates? Sums and Differences Given 25 different positive numbers, can you always choose two of them such that none of the other numbers equals either their sum or their difference? Two-point Conversion As coach of the Hoboken Hominids (American) football team, you saw your team fall 14 points behind the Gloucester Great Apes until, with just a minute to go in the game, the Hominids scored a touchdown. You have a choice of kicking an extra point (95% success rate) or going for two (45% success rate). Which should you do? Swedish Lottery In a proposed mechanism for the Swedish National Lottery, each participant chooses a positive integer. The person who submits the lowest number not chosen by anyone else is the winner. (If no number is chosen by exactly one person, there is no winner.) 50 If just three people participate, but each employs an optimal, equilibrium, randomized strategy, what is the largest number that has positive probability of being submitted? Random Chord What is the probability that a random chord of a circle is longer than a side of an equilateral triangle inscribed in the circle? Cube Magic Can you pass a cube through a hole in a smaller cube? Random Bias Suppose you choose a real number p between 0 and 1 uniformly at random, then bend a coin so that its probability of coming up heads when you flip it is precisely p. Finally you flip your bent coin 100 times. What is the probability that after all this, you end up flipping exactly 50 heads? Gladiators, Version I Paula and Victor each manage a team of gladiators. Paula’s gladiators have strengths p1, p2, . . . , pm and Victor’s, v1, v2, . . . , vn. Gladiators fight one-on-one to the death, and when a gladiator of strength x meets a gladiator of strength y, the former wins with probability x/(x + y), and the latter with probability y/(x+y). Moreover, if the gladiator of strength x wins, he gains in confidence and inherits his opponent’s strength, so that his own strength improves to x+y; similarly, if the other gladiator wins, his strength improves from y to x+y. After each match, Paula puts forward a gladiator (from those on her team who are still alive), and Victor must choose one of his to face Paula’s. The winning team is the one which remains with at least one live player. What’s Victor’s best strategy? For example, if Paula begins with her best gladiator, should Victor respond from strength or weakness? Gladiators, Version II Again Paula and Victor must face off in the Colosseum, but this time, confidence is not a factor, and when a gladiator wins, he keeps the same strength he had before. As before, prior to each match, Paula chooses her entry first. What is Victor’s best strategy? Whom should he play if Paula opens with her best man? 51 Rolling a Six How may rolls of a die does it take, on average, to get a 6—given that you didn’t roll any odd numbers en route? Traveling Salesmen Suppose that between every pair of major cities in Russia, there’s a fixed one-way air fare for going from either city to the other. Traveling salesman Alexei Frugal begins in St. Petersburg and tours the cities, always choosing the cheapest flight to a city not yet visited (he does not need to return to St. Petersburg). Salesman Boris Lavish also needs to visit every city, but he starts in Kaliningrad, and his policy is to choose the most expensive flight to an unvisited city at each step. It looks obvious that Lavish’s tour costs at least as much as Frugal’s, but can you prove it? Napkins in a Random Setting At a conference banquet for a meeting of the Association for Women in Math-ematics, the participants find themselves assigned to a big circular table. On the table, between each pair of adjacent settings, is a coffee cup containing a cloth napkin. As each mathematician sits down, she takes a napkin from her left or right; if both napkins are present, she chooses randomly. If the seats are occupied in random order and the number of mathematicians is large, what fraction of them (asymptotically) will end up without a napkin? Lame Rook A lame rook moves like an ordinary rook in chess—straight up, down, left, or right—but only one square at a time. Suppose that the lame rook begins at some square and tours the 8 × 8 chessboard, visiting each square once and returning to the starting square on the 64th move. Show that the number of horizontal moves of the tour, and the number of vertical moves of the tour, are not equal! Coin Testing The Unfair Advantage Magic Company has supplied you, a magician, with a special penny and a special nickel. One of these is supposed to flip “Heads” 52 with probability 1/3, the other 1/4, but UAMCO has not bothered to tell you which is which. Having limited patience, you plan to try to identify the biases by flipping the nickel and penny one by one until one of them comes up Heads, at which point that one will be declared to be the 1/3-Heads coin. In what order should you flip the coins to maximize the probability that you get the correct answer, and at the same time to be fair, that is, to give the penny and the nickel the same chance to be designated the 1/3-Heads coin? Curve and Three Shadows Is there a simple closed curve in 3-space, all three of whose projections onto the coordinate planes are trees? This means that the shadows of the curve, from the three coordinate directions, may not contain any loops. Majority Hats One hundred prisoners are told that at midnight, in the dark, each will be fitted with a red or black hat according to a fair coin-flip. The prisoners will be arranged in a circle and the lights turned on, enabling each prisoner to see every other prisoner’s hat color. Once the lights are on, the prisoners will have no opportunity to signal to one another or to communicate in any way. Each prisoner is then taken aside and must try to guess his own hat color. The prisoners will all be freed if a majority (here, at least 51) get it right. The prisoners have a chance to devise a strategy before the game begins. Can they achieve a winning probability greater than 50%? Would you believe 90%? How about 95%? Bugs on a Line Each positive integer on the number line is equipped with a green, yellow, or red light. A bug is dropped on “1” and obeys the following rules at all times: If it sees a green light, it turns the light yellow and moves one step to the right; if it sees a yellow light, it turns the light red and moves one step to the right; if it sees a red light, it turns the light green and moves one step to the left. Eventually, the bug will fall off the line to the left, or run out to infinity on the right. A second bug is then dropped on “1,” again following the traffic lights starting from the state the last bug left them in; then, a third bug makes the trip. Prove that if the second bug falls off to the left, the third will march off to infinity on the right. Prisoner and Dog A woman is imprisoned in a large field surrounded by a circular fence. Outside the fence is a vicious guard dog that can run four times as fast as the woman, 53 but is trained to stay near the fence. If the woman can contrive to get to an unguarded point on the fence, she can quickly scale the fence and escape. But can she get to a point on the fence ahead of the dog? Pegs on the Corners Four pegs begin on the plane at the corners of a square. At any time, you may cause one peg to jump over a second, placing the first on the opposite side of the second, but at the same distance as before. The jumped peg remains in place. Can you maneuver the pegs to the corners of a larger square? Circles in Space Can you partition all of 3-dimensional space into circles? Fifteen Bits and a Spy A spy’s only chance to communicate with her control lies in the daily broadcast of 15 zeros and ones by a local radio station. She does not know how the bits are chosen, but each day she gets an advance copy of the bits and can corrupt the radio transmission by altering one of the bits, that is, changing it from a 0 to a 1 or vice-versa. How much daily information can she communicate? Flipping the Pentagon The vertices of a pentagon are labeled with integers, the sum of which is positive. At any time, you may change the sign of a negative label, but then the new value is subtracted from both neighbors’ values so as to maintain the same sum. Prove that, inevitably, no matter which negative labels are flipped, the pro-cess will terminate after finitely many flips, with all values non-negative. 54 Love in Kleptopia Jan and Maria have fallen in love (via the internet) and Jan wishes to mail her a ring. Unfortunately, they live in the country of Kleptopia where anything sent through the mail will be stolen unless it is sent in a padlocked box. Jan and Maria each have plenty of padlocks, but none to which the other has a key. How can Jan get the ring safely into Maria’s hands? Touring an Island Aloysius is lost while driving his Porsche on an island in which every intersection is a meeting of three (two-way) streets. He decides to adopt the following algorithm: Starting in an arbitrary direction from his current intersection, he turns right at the next intersection, then left at the next, then right, then left, and so forth. Prove that Aloysius must return eventually to the intersection at which he began this procedure. Badly Designed Clock The hour and minute hands of a certain clock are indistinguishable. How many moments are there in a day when it is not possible to tell from this clock what time it is? Fibonacci Multiples Show that every positive integer has a multiple that’s a Fibonacci number. Worms and Water Lori is having trouble with worms crawling into her bed. To stop them, she places the legs of the bed into pails of water; since the worms can’t swim, they 55 can’t reach the bed via the floor. But they instead crawl up the walls and across the ceiling, dropping onto her bed from above. Yuck! How can Lori stop the worms from getting to her bed? Light Bulbs in a Circle In a circle are light bulbs numbered clockwise 1 through n, n > 1, all initially on. At time t, you examine bulb number t (modulo n), and if it’s on, you change the state of bulb t+1 (modulo n); that is, you turn off the clockwise-next bulb if it’s on, and on if it’s off. If bulb t is off, you do nothing. Prove that if you continue around and around the ring in this manner, even-tually all the bulbs will again be on. Generating the Rationals You are given a set S of numbers that contains 0 and 1, and contains the mean of every finite nonempty subset of S. Prove that S contains all the rational numbers between 0 and 1. Emptying a Bucket You are presented with three large buckets, each containing an integral number of ounces of some non-evaporating fluid. At any time, you may double the contents of one bucket by pouring into it from a fuller one; in other words, you may pour from a bucket containing x ounces into one containing y ≤x ounces until the latter contains 2y ounces (and the former, x−y). Prove that no matter what the initial contents, you can, eventually, empty one of the buckets. Funny Dice You have a date with your friend Katrina to play a game with three dice, as follows. She chooses a die, then you choose one of the other two dice. She rolls her die while you roll yours, and whoever rolls the higher number wins. If you roll the same number, Katrina wins. Wait, it’s not as bad as you think; you get to design the dice! Each will be a regular cube, but you can put any number of pips from 1 to 6 on any face, and the three dice don’t have to be the same. 56 Can you make these dice in such a way that you have the advantage in your game? Picking the Athletic Committee The Athletic Committee is a popular service option among the faculty of Quin-cunx University, because while you are on it, you get free tickets to the univer-sity’s sports events. In an effort to keep the committee from becoming cliquish, the university specifies that no one with three or more friends on the commit-tee may serve on the committee—but, in compensation, if you’re not on the committee but have three or more friends on it, you can get free tickets to any athletic event of your choice. To keep everyone happy, it is therefore desirable to construct the committee in such a way that even though no one on it has three or more friends on it, everyone not on the committee does have three or more friends on it. Can this always be arranged? Ice Cream Cake On the table before you is a cylindrical ice-cream cake with chocolate icing on top. From it you cut successive wedges of the same angle θ. Each time a wedge is cut, it is turned upside-down and reinserted into the cake. Prove that, regardless of the value of θ, after a finite number of such operations all the icing is back on top of the cake! Sharing a Pizza Alice and Bob are preparing to share a circular pizza, divided by radial cuts into some arbitrary number of slices of various sizes. They will be using the “polite pizza protocol”: Alice picks any slice to start; thereafter, starting with Bob, they alternate taking slices but always from one side or the other of the gap. Thus after the first slice, there are just two choices at each turn until the last slice is taken (by Bob if the number of slices is even, otherwise by Alice). Is it possible for the pizza to have been cut in such a way that Bob has the advantage—in other words, so that with best play, Bob gets more than half the pizza? Moth’s Tour A moth alights on the “12” of a clock face, and begins randomly walking around the dial. Each time it hits a number, it proceeds to the next clockwise number, or the next counterclockwise number, with equal probability. It continues until it has visited every number at least once. What is the probability that the moth finishes at the number “6”? 57 Names in Boxes The names of 100 prisoners are placed in 100 wooden boxes, one name to a box, and the boxes are lined up on a table in a room. One by one, the prisoners are led into the room; each may look in at most 50 boxes, but must then leave the room exactly as he found it and is permitted no further communication with the others. The prisoners have a chance to plot their strategy in advance, and they are going to need it, because unless every single prisoner finds his own name all will subsequently be executed. Find a strategy that gives the prisoners a decent chance of survival. Garnering Fruit Each of 100 baskets contains some number (could be zero) of apples, some number of bananas, and some number of cherries. Show that you can collect 51 of those baskets that together contain at least half the apples, at least half the bananas, and at least half the cherries! Coin Game You and a friend each pick a different heads-tails sequence of length 4 and a fair coin is flipped until one sequence or the other appears; the owner of that sequence wins the game. For example, if you pick HHHH and she picks TTTT, you win if a run of four heads occurs before a run of four tails. Do you want to pick first or second? If you pick first, what should you pick? If your friend picks first, how should you respond? Roulette for Parking Money You’re in Las Vegas with only $2 and in desperate need of $5 to feed a parking meter. You run in through the nearest door and find yourself at a roulette table. You can bet any whole dollar amount on any allowable set of numbers. What roulette-betting strategy will maximize your probability of walking out with $5? Invisible Corners Can it be that you are standing outside a polyhedron and can’t see any of its vertices? Sleeping Beauty Sleeping Beauty agrees to the following experiment. On Sunday she is put to sleep and a fair coin is flipped. If it comes up Heads, she is awakened on Monday morning; if Tails, she is awakened on Monday morning and again on Tuesday morning. In all cases, she is not told the day of the week, is put back to sleep 58 shortly after, and will have no memory of any Monday or Tuesday awakenings. When Sleeping Beauty is awakened on Monday or Tuesday, what—to her—is the probability that the coin came up Heads? Boardroom Reduction The Board of Trustees of the National Museum of Mathematics has grown too large—50 members, now—and its members have agreed to the following reduction protocol. The board will vote on whether to (further) reduce its size. A majority of ayes results in the immediate ejection of the newest board member; then another vote is taken, and so on. If at any point half or more of the surviving members vote nay, the session is terminated and the board remains as it currently is. Suppose that each member’s highest priority is to remain on the board, but given that, agrees that the smaller the board, the better. To what size will this protocol reduce the board? Buffon’s Needle A needle one inch in length is tossed onto a large mat marked with parallel lines one inch apart. What is the probability that the needle lands across a line? Seven Cities of Gold In 1539 Friar Marcos de Niza returned to Mexico from his travels in what is now Arizona, famously reporting his finding of “Seven Cities of Gold.” Coronado did not believe the “liar friar” and when subsequent expeditions came up empty, Coronado gave up the quest. My (unreliable) sources claim that Coronado’s reason for disbelieving de Niza is the latter’s claim that the cities were laid out in the desert in a manner 59 such that of any three of the cities, at least one pair were exactly 100 furlongs apart. Coronado’s advisors told him no such pattern of points existed. Were they right? Life-saving Transposition There are just two prisoners this time, Alice and Bob. Alice will be shown a deck of 52 cards spread out in some order, face-up on a table. She will be asked to transpose two cards of her choice. Alice is then dismissed, with no further chance to communicate to Bob. Next, the cards are turned down and Bob is brought into the room. The warden names a card and to stave off execution for both prisoners, Bob must find the card after turning over, sequentially, at most 26 of the cards. As usual the prisoners have an opportunity to conspire beforehand. This time, they can guarantee success. How? More Magnetic Dollars We return to Magnetic Dollars, but we strengthen their attractive power just a bit. This time, an infinite sequence of coins will be tossed into the two urns. When one urn contains x coins and the other y, the next coin will fall into the first urn with probability x1.01/(x1.01 + y1.01), otherwise into the second urn. Prove that after some point, one of the urns will never get another coin! Covering the Stains Just as a big event is about to begin, the queen’s caterer notices, to his horror, that there are 10 tiny gravy stains on the tablecloth. All he has the time to do is to cover the stains with non-overlapping plates. He has plenty of plates, each a unit disk. Can he do it, no matter how the stains are distributed? Zero-sum Vectors On a piece of paper, you have (for some reason) made an array whose rows consist of all 2n of the n-dimensional vectors with coordinates in {+1, −1}— that is, all possible strings of +1’s and −1’s of length n. Notice that there are lots of nonempty subsets of your rows which sum to the zero vector, for example any vector and its complement; or the whole array, for that matter. However, your three-year-old nephew has got hold of the paper and has changed some of the entries in the array to zeroes. Prove that no matter what your nephew did, you can find a nonempty subset of the rows in the new array that sums to zero. 60 Colors and Distances In the town of Hoegaarden, Belgium, exactly half the houses are occupied by Flemish families, the rest by French-speaking Walloons. Can it be that the town is so well mixed that the sum of the distances between pairs of houses of like ethnicity exceeds the sum of the distances between pairs of houses of different ethnicity? Two Sheriffs Two sheriffs in neighboring towns are on the track of a killer, in a case involving eight suspects. By virtue of independent, reliable detective work, each has narrowed his list to only two. Now they are engaged in a telephone call; their object is to compare information, and if their pairs overlap in just one suspect, to identify the killer. The difficulty is that their telephone line has been tapped by the local lynch mob, who know the original list of suspects but not which pairs the sheriffs have arrived at. If they are able to identify the killer with certainty as a result of the phone call, he will be lynched before he can be arrested. Can the sheriffs, who have never met, conduct their conversation in such a way that they both end up knowing who the killer is (when possible), yet the lynch mob is still left in the dark? Painting the Fence Each of n industrious people chooses a random point on a circular fence, and begins painting toward the farther of her two neighbors until she encounters a painted section. On average, how much of the fence gets painted? How about if instead, each person paints toward her nearer neighbor? Leading All the Way Running for local office against Bob, Alice wins with 105 votes to Bob’s 95. What is the probability that as the votes were counted (in random order), Alice got the first vote and then led the whole way? Self-referential Number The first digit of a certain 8-digit integer N is the number of zeroes in the (ordinary, decimal) representation of N. The second digit is the number of ones; the third, the number of twos; the fourth, the number of threes; the fifth, the number of fours; the sixth, the number of fives; the seventh, the number of sixes; and, finally, the eighth is the total number of distinct digits that appear in N. What is N? 61 Filling the Cup You go to the grocery store needing one cup of rice. When you push the button on the machine, it dispenses a uniformly random amount of rice between nothing and one cup. On average, how many times do you have to push the button to get your cup? Two Balls and a Wall On a line are two identical-looking balls and a vertical wall. The balls are per-fectly elastic and friction-free; the wall is perfectly rigid; the ground is perfectly level. If the balls are the same mass and the one farther from the wall is rolled toward the closer one, it will knock the close ball toward the wall; that ball will bounce back and hit the first ball, which will then roll (forever) away from the wall. Three bounces altogether. Now assume the farther ball has mass a million times greater than the closer one. How many bounces now? (You may ignore the effects of angular momen-tum, relativity, and gravitational attraction.) Bulgarian Solitaire Fifty-five chips are organized into some number of stacks, of arbitrary heights, on a table. At each tick of a clock, one chip is removed from each stack and those collected chips are used to create a new stack. What eventually happens? 62 The Hints Below are hints and/or comments for each puzzle, together with the number of the chapter where you can find a solution. (But to get the most out of these puzzles, don’t go there until you’ve tried everything!) Adding, Multiplying, and Grouping: Multiply sums of two numbers and of five numbers. (Chapter 12) Air Routes in Aerostan: What can be said about the two-airline net-works? (Chapter 4) All Right or All Wrong: Use both choice and parity. (Chapter 23) Alternate Connection: If you number the points in the order they get connected, which one do you not want to connect? (Chapter 1) Alternating Powers: Note that f(x) = x −f(x2). (Chapter 9) Alternative Dice: Represent a die by a polynomial in which the coefficient of xk is the number of faces with k pips. (Chapter 5) An Attractive Game: Look at the game from the bank’s point of view. (Chapter 14) Angles in Space: Translate the convex closure of the points. (Chapter 16) Angry Baseball: What happens after the last tie? (Chapter 10) Ants on the Circle: Conservation of momentum applies! (Chapter 20) Area–Perimeter Match: Factoring a polynomial might help here. (Chap-ter 5) Ascending and Descending: Consider, for each number in a good list, the length of the longest decreasing subsequence that ends there. (Chapter 12) Attention Paraskevidekatriaphobes: What is the cycle for our Grego-rian calendar? (Chapter 1) Attic Lamp: You’ll need more than one bit of information. (Chapter 13) Baby Frog: Cover the field with a 2n × 2n grid. (Chapter 15) Bacteria on the Plane: Give each child half the potential of its parent. (Chapter 18) Bacterial Reproduction: Experiment, and keep track of the numbers of females. (Chapter 2) Badly Designed Clock: Note that you can tell what time it is when the hands coincide. (Chapter 19) 63 Bags of Marbles: Nothing to prevent you from putting other things in bags... (Chapter 1) Bat and Ball: Do the math! (Chapter 5) Belt Around the Earth: Use a letter for the earth’s circumference (Chap-ter 5) Bias Test: In which case can the second flip change your mind? (Chapter 13) Biased Betting: Fix a win–loss sequence that would result in going broke. (Chapter 10) Bidding in the Dark: Compute your expectation after a given bid. (Chap-ter 14) Big Pairs in a Matrix: Suppose not, and look separately at the biggest and next-biggest entries in each row. (Chapter 9) Bindweed and Honeysuckle: Untwist one while you twist the other. (Chapter 1) Birthday Match: Be careful; you want a match for yourself, not just any pair of people. (Chapter 10) Bluffing with Reals: Jeremy has one threshold, Louise two. (Chapter 21) Boarding the Manhole: Archimedes may be able to help you here. (Chap-ter 16) Boardroom Reduction: Start by thinking about what would happen if the board came down to just its three most senior members. (Chapter 21) Box in a Box: Assume you can, and compute the volume of the set of all points within distance ε of the boxes. (Chapter 16) Boxing the Mountain State: What is meant by inscribing West Virginia in a square is drawing a square around the shape of the state, on the plane, in such a way that the state is inside the square but touches all four sides. (Chapter 3) Boy Born on Tuesday: Count the cases carefully. (Chapter 10) Bracing the Grid: Make a graph whose vertices are the rows and columns of the grid, with edges where the intersecting square is braced. (Chapter 4) Breaking a Chocolate Bar: Try this, and see the light. (Chapter 18) Broken ATM: What’s the most George can hope to get, considering that the machine’s only options are multiples of $6? (Chapter 6) Buffon’s Needle: What is expected number of lines crossed, per unit length, by a randomly placed unit-diameter circle? (Chapter 14) Bugging a Disk: Assume the ceiling is flat, and incribe a hexagon in it. (Chapter 9) Bugs on Four Lines: Add a time axis. (Chapter 16) Bugs on a Line: First, check that the bug cannot wander forever without going off to infinity. (Chapter 18) Bugs on a Polyhedron: Draw an arrow from each face through its bug to the next face. (Chapter 18) Bugs on a Pyramid: Number the bugs and ask: When one looks at the triangle made by the others, does she see them in clockwise or counterclockwise order? (Chapter 3) 64 Bulgarian Solitaire: Rotate the configuration 45◦and think about gravity. (Chapter 18) Candles on a Cake: What does the puzzle tell you about the distance from a fixed point on the circumference to the ith candle? (Chapter 7) Cards from Their Sum: Big numbers work better. (Chapter 6) Chameleons: Look not just at numbers of chameleons of a given color, but differences between numbers. (Chapter 2) Chessboard Guess: Assign a 6-digit nimber to each square of the chess-board. (Chapter 17) Chinese Nim: Starting from (1,2), find a pattern linking the positions you don’t want to be in. (Chapter 21) Chomp: Someone must have a winning strategy; what if it’s Bob? (Chapter 9) Circles in Space: Note that you can tile a sphere, minus any two points, with circles. (Chapter 22) Circular Shadows I: Can you alter a ball slightly without affecting its coordinate projections? (Chapter 22) Circular Shadows II: Close in on the body with parallel planes. (Chapter 16) Coconut Classic: Start by allowing yourself negative numbers of coconuts! (Chapter 7) Coin Game: Who would you want to be if one of you picks HHHH and the other THHH? (Chapter 10) Coin Testing: You need to keep the number of flips of each coin as equal as possible—but there’s more. (Chapter 10) Coins in a Row: You are not asked for an optimal strategy, only one that gets Alix half the money. (Chapter 7) Coins on the Table: The shape of the table plays a part. (Chapter 7) Colors and Distances: Notice that the distance between two points in the plane is proportional to the probability that a random line, suitably constrained, passes between them. (Chapter 14) Comparing Numbers, Version I: Have Victor pick a threshold. (Chapter 10) Comparing Numbers, Version II: What would Paula do if Victor used 1/2 as his threshold? (Chapter 10) Competing for Programmers: You might have good use for a “friendship graph,” mentioned earlier in the same chapter. (Chapter 4) Conversation on a Bus: One part of the conversation keeps the bus number from being to small, while another part keeps it from being too big. (Chapter 13) Conway’s Immobilizer: A good start is to make sure your algorithm does the right thing when it’s about to win. (Chapter 11) Covering the Stains: How would you try to cover the stains if you had left your glasses at home, and couldn’t see where the stains were? (Chapter 14) Crossing the River: In general: Get the men across first. (Chapter 19) Cube Magic: Orient the cube to get a large projection. (Chapter 22) 65 Curve and Three Shadows: Try curves made from grid lines. (Chapter 16) Curve on a Sphere: Pick two points on the curve halfway around from each other, and imagine that the midpoint on the great circle connecting them is the North Pole. (Chapter 9) Curves on Potatoes: Replace the potatoes by holograms. (Chapter 16) Cutting the Necklace: Imagine the necklace laid out in a circle, with a line through the center cutting it twice. (Chapter 3) Deterministic Poker: Alice must, at the very least, stop Bob from making an ace-high straight flush. (Chapter 21) Dishwashing Game: Begin with 12 flips each. (Chapter 10) Divisibility Game: Bob gets the most out of small numbers. (Chapter 6) Domino Task: Notice that a vertical domino can be turned horizontal if one of its rows has a hole. (Chapter 7) Dots and Boxes Variation: Think about your choice from your oppo-nent’s point of view. (Chapter 9) Dot-Town Exodus: Reason carefully with a population of 3. (Chapter 13) Double Cover by Lines: What happens if you try to build your double-cover one line at a time? (Chapter 23) Drawing Socks: Order all the socks randomly and condition on the color of the last one. (Chapter 14) Early Commuter: How much time did the husband save? (Chapter 1) Easy Cake Division: Straight, vertical cuts can do the job. (Chapter 16) Efficient Pizza-Cutting: How can you make the next cut produce the maximum number of new pieces? (Chapter 1) Emptying a Bucket: One way to do it, paradoxically, is to show that the contents of one bucket can be increased until another is empty. (Chapter 21) Even Split: Prove it first when n is prime. (Chapter 6) Even-Sum Billiards: What would your answer be if there were no 9-ball? (Chapter 2) Expecting the Worst: Bend the segment into a circle. (Chapter 24) Exponent upon Exponent: What happens to the expression if you cut off the bottom x? (Chapter 23) Factorial Coincidence: Consider the power of a big prime that factors each side of the expression. (Chapter 6) Factorials and Squares: Use the fact that the product of perfect squares is itself a perfect square. (Chapter 6) Fair Play: Flip more than once, and look for equiprobable events. (Chapter 19) Falling Ants: Note that ants are interchangeable, for the purposes of this puzzle. (Chapter 20) Faulty Combination Lock: Try dividing the numbers from 1 to 8 into two groups, and using the fact that any combination must contain at least two from one of them. (Chapter 1) Fewest Slopes: Try regular polygons. (Chapter 11) 66 Fibonacci Multiples: Keep track of the remainders, modulo n, of two successive Fibonacci numbers. (Chapter 21) Fifteen Bits and a Spy: Let the “nimsum” of the broadcast be the mes-sage. (Chapter 17) Figure Eights in the Plane: Note that there are only countably many rational points in the plane—or pairs of them, for that matter. (Chapter 23) Filling the Cup: Note that the fractional parts of the amount of rice you have got so far also constitutes a sequence of independent uniform random variables. (Chapter 14) Filling a Bucket: How far can you get trying to keep all the buckets level? (Chapter 19) Find the Robot: How many ways are there to start and point the robot? (Chapter 23) Finding a Jack: The jacks divide the deck into five parts. (Chapter 8) Finding the Counterfeit: Each weighing can have three possible out-comes. (Chapter 13) Finding the Missing Number: How much information must be kept? (Chapter 19) Finding the Rectangles: What do you encounter as you walk across the tiled 400-gon? (Chapter 22) First Odd Number: List the alphabetically-early words that may arise in a number. (Chapter 11) First-grade Division: Imagine that each project requires the same total effort. (Chapter 18) Flipping the Pentagon: A reasonable thing to try as a potential might be some measure of how close the numbers are to one another. (Chapter 18) Flying Saucers: It’s important that the person beamed back down might not be the first one beamed up by the next saucer. (Chapter 7) Fourth Corner: Each grid point can have one of four possible parities. (Chapter 2) Frames on a Chessboard: Look for a set of cells the parity of whose number of red cells doesn’t change. (Chapter 18) Funny Dice: How can you give one die an advantage over another, even though they both have the same average roll? (Chapter 19) Game of Desperation: What position can you put your opponent in that will force her to let you win on your next move? (Chapter 21) Gaming the Quilt: If you’ve already bought k tickets, when is one more worth buying? (Chapter 5) Garnering Fruit: Try it for just apples and bananas; adding cherries brings a ham sandwich(!) into the picture. (Chapter 3) Gasoline Crisis: Start by imagining a trip that starts with plenty of extra gas aboard. (Chapter 7) Generating the Rationals: Note that you easily get all fractions whose denominators are powers of 2. (Chapter 19) Gladiators, Version I: If you think of strength as money, each bout be-comes a fair game. (Chapter 24) 67 Gladiators, Version II: Think of each gladiator as a light bulb trying to outlast his opponent. (Chapter 24) Gluing Pyramids: How can faces disappear other than by hiding them? (Chapter 22) Guarding the Gallery: Try putting guard-posts at corners. (Chapter 15) Half Grown: Think about your own young relatives. (Chapter 1) Half-right Hats: Prisoners don’t all need to have the same instructions. (Chapter 2) Halfway Points: If S has n points, they satisfy n linear equations with rational coefficients. (Chapter 9) Handshakes at a Party: What numbers did Alexandra hear during her survey? (Chapter 4) Hats and Infinity: Have the prisoners choose a special hat sequence from each class of similar sequences. (Chapter 23) Hazards of Electronic Coinflipping: Flip several times, and note that some numbers of heads will allow you to decide between Alice and Bob. Can you give the rest to Charlie? (Chapter 3) Home-field Advantage: What if all seven games were always played? (Chapter 10) Hopping and Skipping: What’s the probability that the frog never re-gresses from its current pad? (Chapter 5) Ice Cream Cake: It’s easier to keep track if you cut the cake in the same place each time, but rotate the cake between cuts. (Chapter 21) Identifying the Majority: Use the counter for reinforcement of your cur-rent name. (Chapter 19) Impressionable Thinkers: Consider the number of current instances where influence fails, that is, a voter thinks one way and her acquaintance disagrees the next day. (Chapter 18) Increasing Routes: Note that in a graph the average degree is twice the number of edges divided by the number of vertices. (Chapter 8) Infected Checkerboard: Try the process with various starting positions. What prevents the infected area from getting too “complicated”? (Chapter 18) Infected Cubes: You need some way to generalize the diagonals that work in dimension 2. (Chapter 24) Integral Rectangles: There are many ways to do this one; one of them begins by nesting the big rectangle into the positive quadrant of the plane grid, and making a graph whose vertices are grid points that lie on corners of rectangles. (Chapter 24) Invisible Corners: You can start by putting yourself in a room made from six non-touching planks. (Chapter 22) King’s Salary: Try to reduce the number of salaried citizens. (Chapter 11) Lame Rook: What happens if you try this on an n × n board for various (even) n? (Chapter 15) Laser Gun: Cover the plane with reflected copies of the room. (Chapter 24) 68 Lattice Points and Line Segments: When does the line segment between two lattice points contain another lattice point? (Chapter 12) Leading All the Way: Try putting the ballots in random circular order, instead of random linear order. (Chapter 10) Lemming on a Chessboard: If the lemming stays on the board, it must eventually cycle. (Chapter 9) Life Is a Bowl of Cherries: Consider the general two-bowl game first. (Chapter 17) Life Isn’t a Bowl of Cherries?: Start with the nim strategy itself, rather than its reverse. (Chapter 17) Life-saving Transposition: If you can do Names in Boxes, you can do this one. (Chapter 19) Light Bulbs in a Circle: Don’t forget to keep track of which bulb you’re looking at, as well as the state of all bulbs. (Chapter 21) Line Up by Height: Keep track of the lengths of the descending and as-cending subsequences each player is at the end of, when lined up alphabetically. (Chapter 12) Lines Through a Grid: Nineteen parallel diagonals will do it. Can you do better? (Chapter 12) Locker Doors: What numbers have an odd number of divisors (including 1 and themselves)? (Chapter 6) Losing at Dice: You’ll need to consider two patterns to compute the num-ber of ways to “roll a 4” by this strange method. (Chapter 1) Lost Boarding-Pass: Try it with three passengers. What is the probability that third passenger ends up in the second passenger’s seat? (Chapter 7) Love in Kleptopia: Can Jan somehow get one of his locks on the box? (Chapter 19) Magnetic Dollars: Try it with, say, six coins instead of a million. (Chapter 24) Majority Hats: Can you beat the “preponderance” strategy, where each prisoner guesses the color he sees most of? (Chapter 17) Martians in a Circle: What would happen to their answers if we changed all the truth-tellers to liars and vice-versa? (Chapter 11) Matching Coins: Cher needs to use her coinflips to communicate informa-tion about coming flips. (Chapter 13) Mathematical Bookworm: Visualize! (Chapter 22) Measuring with Fuses: Try lighting both fuses at once. (Chapter 11) Meet the Williams Sisters: How many pairs of players play each other? (Chapter 1) Meeting the Ferry: Draw a picture! Note that all times are GMT so you don’t have to worry about time zones. (Chapter 22) Men with Sisters: What’s the answer among you and your siblings? (Chapter 8) Missing Card: Make use of the ordering of the four remaining cards. (Chapter 13) Missing Digit: Do you remember “casting out nines”? (Chapter 2) 69 Monk on a Mountain: Superimpose the days. (Chapter 3) More Magnetic Dollars: Try to set this up so that each urn acquires coins at random moments. (Chapter 24) Moth’s Tour: Start from when the moth first gets to 5 o’clock or 7 o’clock. (Chapter 21) Names in Boxes: Nothing prevents a prisoner from using what he finds in a box to decide what box to open next. (Chapter 19) Napkins in a Random Setting: It’s helpful to think of each diner having flipped a coin to decide which napkin she would prefer to take. (Chapter 14) Next Card Bet: The best guarantee, for given expectation, is when the outcome is no longer random. (Chapter 14) Next Card Red: Try it with a smaller deck. (Chapter 24) No Twins Today: It’s not a trick. (Chapter 11) Non-repeating String: Let the number of letters in the alphabet be a variable, and use induction. (Chapter 15) North by Northwest: An easy enough computation, if you know what it means. (Chapter 1) Not Burning Brownies: Try rectangles, then some sort of triangle. (Chap-ter 22) Numbers on Foreheads: Consider the sum of the numbers, modulo 10. (Chapter 6) Odd Light Flips: Note that the order in which switches are flipped is irrelevant. (Chapter 15) Odd Run of Heads: Use a variable to stand for the answer. (Chapter 5) One-bulb Room: Consider first the situation in which the room is known to be dark at the start. (Chapter 19) Option Hats: What if there were just three prisoners? (Chapter 17) Other Side of the Coin: Consider labeling the sides of the coins. (Chapter 10) Packing Slashes: In showing your solution is optimal, it may pay to con-sider the 12 outer vertices of the inside 3 × 3 grid. (Chapter 11) Painting the Cubes: Paint space and carve out the big cube! (Chapter 16) Painting the Fence: How does the question of whether an interval between painters gets painted depend on its length, relative to the lengths of its neighbor intervals? (Chapter 14) Painting the Polyhedron: Assume you can inscribe a sphere, and trian-gulate the sphere using the tangent points as vertices. (Chapter 16) Pairs at Maximum Distance: Note that line segments connecting two max-distance pairs must cross. (Chapter 9) Pancake Stacks: Making the stacks close in size limits one’s opponent. (Chapter 21) Path Through the Cells: Recolor greedily with high-numered colors, and start your path with a cell that still has color 1. (Chapter 15) Peek Advantage: Could the second peek change your mind? (Chapter 13) 70 Pegs in a Square: Look for a useful way to two-color the holes. (Chapter 18) Pegs on the Corners: Notice that if the pegs begin at grid points, they stay on grid points. (Chapter 21) Pegs on the Half-Plane: Weight the holes so that the entire lower half-plane has finite weight. (Chapter 18) Phone Call: Tackle this one hour at a time. (Chapter 19) Picking the Athletic Committee: What happens if you just pick a com-mittee at random and try to fix it? (Chapter 18) Pie in the Sky: The moon is about half a degree in diameter. (Chapter 20) Ping-Pong Progression: How many points does Bob win, on average, while Alice wins 21? (Chapter 14) Players and Winners: Think about communication in both directions. (Chapter 13) Points on a Circle: Pick random diameters first, then endpoints. (Chapter 10) Poker Quickie: Assume no wild cards, one opponent. (Chapter 11) Polygon Midpoints: What happens if you pick some arbitrary point on the plane and assume it is a vertex? (Chapter 22) Polygon on the Grid: What is the relationship of h or v to the polygon’s area? (Chapter 19) Polyhedron Faces: Think about the face with the most sides. (Chapter 12) Poorly Placed Dominoes: You might need two tries. (Chapter 19) Portrait: Who is “my father’s son”? (Chapter 21) Powers of Roots: Try the 10th power, and see if you can guess what’s going on. (Chapter 7) Powers of Two: Start from the middle of the phrase. (Chapter 1) Precarious Picture: Somehow you must arrange the string so that it ultimately passes over each nail, when the other is ignored. (Chapter 22) Prime Test: If you don’t find any small prime factors, what else can you try? (Chapter 6) Prisoner and Dog: When is the woman close enough to the fence to make a straight run for it, if the dog is at the opposite point? (Chapter 19) Prisoners and Gloves: For this you might want to consider parity of permutations. (Chapter 2) Profit and Loss: Observe first that it can’t be as long as 40 months. (Chapter 15) Protecting the Statue: Observe that the protected area cannot be convex. (Chapter 22) Pushing the Pedal: When you ride the bike, how does the pedal move relative to the ground? (Chapter 20) Raising Art Value: Compare the picture’s value to the averages for the galleries. (Chapter 8) 71 Random Bias: Can you somehow “do the flips” before the coin is chosen? (Chapter 10) Random Chord: Does it matter how the random chord is chosen? (Chap-ter 10) Random Intersection: What is the probability that a particular point in one ball will end up inside the other as well? (Chapter 14) Random Intervals: Try small numbers, and look for an explanation for what you find. (Chapter 24) Random Judge: Imagine that there’s one judge that would take part either way. (Chapter 10) Rating the Horses: You’ll certainly need to see all the horses. (Chapter 1) Recovering the Polynomial: Think of the coefficients as the expansion of a number in base x. (Chapter 5) Rectangles Tall and Wide: Start by showing that if the tile’s sides have irrational ratio, then any tiling must be all horizontal or all vertical. (Chapter 6) Red Points and Blue: What improves if you uncross two segments? (Chapter 18) Red and Black Sides: What’s the total area of the rectangles? (Chapter 5) Red and Blue Hats in a Line: How can the first one to guess help the others? (Chapter 2) Returning Pool Shot: How many angles can the ball take after it is shot? (Chapter 20) Righting the Pancakes: Represent the state of the pancake stack as a binary number. (Chapter 18) Rolling All the Numbers: Think of the experiment in stages. (Chapter 14) Rolling Pencil: Needs more than one second of thought, but two will do. (Chapter 22) Rolling a Six: Watch out—this is not the same as ignoring odd rolls. (Chapter 14) Rotating Coin: Try it! (Chapter 20) Roulette for Parking Money: Every bet loses money, so you want speed as well as accuracy. (Chapter 14) Roulette for the Unwary: How many wins does Elwyn need to come out ahead? (Chapter 14) Rows and Columns: You have to try this to see what’s happening. (Chap-ter 1) Salaries and Raises: How much does each persons annual salary go up, each year? (Chapter 1) Same Sum Dice: It’s easier to prove a stronger statement about ordered dice. (Chapter 12) Same Sum Subsets: What if you found two overlapping sets with the same sum? (Chapter 12) 72 Second Ace: Compute the conditional probabilities. (Chapter 10) Self-referential Number: If you try a number and it doesn’t work, what’s the simplest way to try to fix it? (Chapter 19) Sequencing the Digits: There’s more than one way to choose such a sequence. (Chapter 1) Serious Candidates: Imagine betting on each candidate as he or she be-comes serious. (Chapter 14) Service Options: It may be convenient to assume that lots of games are played, regardless of outcomes. (Chapter 10) Seven Cities of Gold: Equilateral triangles are your best friends for this puzzles. (Chapter 11) Sharing a Pizza: The even case is easy; try the odd case with pieces of size 0(!) and 1. (Chapter 19) Shoelaces at the Airport: Think about time spent walking. (Chapter 1) Shoes, Socks, and Gloves: What are the worst choices you can make? (Chapter 12) Signs in an Array: Is there anything good that must happen when you flip a line that had negative sum? (Chapter 18) Sinking 15: You may assume the playes can sink whatever balls they want, so this is a deterministic game. Does it seem familiar? (Chapter 24) Skipping a Number: Missy’s free-throw percentage jumps up when she hits (and down when she misses), so it should be easy to set up a situation where it jumps over 80%, right? (Chapter 3) Slabs in 3-Space: Infinity is tricky—try covering a big finite piece instead. (Chapter 16) Sleeping Beauty: Imagine repeating the whole experiment 100 times. (Chapter 10) Slicing the Cube: How would you actually do it? Can you do better? (Chapter 24) Snake Game: Tile the board with dominoes! (Chapter 4) Soldiers in a Field: Think about the two closest soldiers. (Chapter 9) Spaghetti Loops: What’s the probability that your ith tying operation will make a loop? (Chapter 14) Sphere and Quadrilateral: Weight the vertices so as to balance the edges at the tangent points. (Chapter 20) Spiders on a Cube: How many spiders would be needed if the game were played on a tree? (Chapter 4) Spinning Switches: Try the two-switch version first. (Chapter 7) Split Games: Imagine that for some reason the split games might not be counted. (Chapter 10) Splitting a Hexagon: Start with the triangles. (Chapter 22) Splitting a Polygon: Non-convex polygons are a problem, but you can start by passing a line through that’s not of the same slope as any side. (Chapter 3) Splitting the Stacks: A little experimentation might help you see the light. (Chapter 1) 73 Sprinklers in a Field: What points are closer to a given grid vertex than other grid vertices? (Chapter 19) Strength of Schedule: Consider the strength-of-schedule scores for the teams with the most wins, and for the teams with the fewest wins. (Chapter 5) Subsets with Constraints: Group the numbers so that each group can supply only one number to your set. (Chapter 6) Subtracting Around the Corner: Notice that odd numbers eventually disappear. (Chapter 2) Summing Fractions: Try small values for n, then prove your conclusion by comparing gains with losses going from n to n+1. (Chapter 15) Sums and Differences: Suppose otherwise and start by considering the highest number. (Chapter 19) Sums of Two Squares: Think of pairs of integers as lattice points in a circle. (Chapter 8) Swapping Executives: Start by getting an executive who belongs at one of the ends into her correct place. (Chapter 15) Swedish Lottery: Think about betting the highest allowed number, versus “cheating” and betting higher. (Chapter 21) Testing Ostrich Eggs: Think about Oskar’s situation after the first egg breaks. (Chapter 21) Three Martians at the Crossroads: You will get no information from the random native. (Chapter 13) Three Negatives: Consider the lowest two numbers in the set. (Chapter 5) Three Sticks: Can you pick the lengths so that an operation leaves their ratios unchanged? (Chapter 3) Three-way Duel: Compare Alice’s survival probabilities when she hits Bob, hits Carol, or misses both. (Chapter 21) Three-way Election: Consider the majority second choice among each candidate’s fans. (Chapter 1) Tiling a Polygon: How do you find a tile with sides parallel to two given sides of the 100-gon? (Chapter 22) Tiling with Crosses: Do the plane with diagonals, and space with plane slabs that have holes and pegs. (Chapter 22) Tiling with L’s: Try it one row at a time, left to right. (Chapter 15) Tipping the Scales: How much of the time does a given weight spend on each side of the scale? (Chapter 8) Touring an Island: What do you need to know about Aloysius at a par-ticular moment? (Chapter 21) Trapped in Thickland: Try regular tetrahedra. (Chapter 22) Traveling Salesmen: Compare Lavish’s kth most costly flight to Frugal’s. (Chapter 15) Truly Even Split: Observe that 16 is a power of 2. (Chapter 15) Turning the Die: Your position depends on the current sum, and also which numbers are available on the die. (Chapter 21) 74 Two Balls and a Wall: Plot the velocities of the balls, making use of conservation of energy and momentum. (Chapter 20) Two Different Distances: You need to be very systematic to get them all! (Chapter 11) Two Monks on a Mountain: You can assume each path consists of a finite number of alternately ascending and descending line segments. (Chapter 4) Two-Point Conversion: You may as well assume the Hominids will score again, and that if the game goes to overtime, either team is equally likely to win. (Chapter 10) Two Round-Robins: Divide the players into the good chess players and the good checkers players. (Chapter 5) Two Runners: Relate sums and difference of distances covered to meeting time. (Chapter 5) Two Sheriffs: Have Lew and Ralph agree on a list of partitions of the suspect set into pairs, such that each pair of suspects appears once on the list. (Chapter 13) Unanimous Hats: What would the prisoners do if they knew the number of red hats was even? (Chapter 2) Unbreakable Domino Cover: Can you cross a line with just one domino? (Chapter 19) Unbroken Curves: Start with the busiest tiles. (Chapter 11) Unbroken Lines: Do the horizontal and vertical lines first. (Chapter 11) Uniform Unit Distances: How might you extend a configuration that works for n, so that it could work for n+1? (Chapter 15) Uniformity at the Bakery: Start with integer weights, and make one odd. (Chapter 15) Uniting the Loops: Look at the more general problem of getting a sin-gle loop from tilings involving some of the given tiles, with some repetitions. (Chapter 2) Waiting for Heads: There’s a bit more to this than the probability of flipping 5 heads. (Chapter 8) Watches on the Table: Draw a picture and use geometry. (Chapter 8) Watermelons: Try some numbers. (Chapter 1) Whim-Nim: Notice that prior to any declarations, handing your opponent either a Nim-0 position or a bunch of singletons is fatal. (Chapter 17) Who Won the Series?: What would the answer be if instead you heard that at most six games were played? (Chapter 10) Whose Bullet?: Check your intuition with math. (Chapter 10) Wild Guess: You’ll need representive real sequences for this one. (Chapter 23) Winning at Wimbledon: How many chances to win the match can you arrange to get? (Chapter 10) Wins in a Row: Imagine playing four games instead of three. (Chapter 10) Wires under the Hudson: Link wires in pairs at each end. (Chapter 4) 75 Worms and Water: You’ll need something over Lori that the worms can’t crawl underneath. (Chapter 19) Worst Route: Try it for just four houses; use various addresses, and ex-periment. (Chapter 4) Y’s in the Plane: Put small rational circles around each Y’s endpoints. (Chapter 23) Zero-sum Vectors: Try to build a sequence of new rows whose partial sums are tightly controlled. (Chapter 12) Zeroes and Ones: What happens when you subtract from a number an-other number with the same value modulo n? (Chapter 12) 76 Chapter 1 Out for the Count There’s nothing more basic in mathematics than counting—but at the same time, counting mathematical objects can be dauntingly difficult. We’ll start with some simple puzzles and gradually introduce some of the special tools that we sometimes need for counting. Half Grown At what age is the average child half the height that he or she will be as an adult? Solution: Most people guess too high, maybe thinking that if full height is reached around age 16, then half-height should be around 8. There are two problems with that reasoning: (1) the rate of growth for a human being is not constant, and (2) babies have a substantial head start of 20 inches or so (when you prop them up). The right answer: Two years old! (For a girl, it’s actually about 2 1 4 years, for a boy 2 1 2.) 77 Powers of Two How many people are “two pairs of twins twice”? Solution: There are four words in the phrase suggesting the number “2” and most people rightly divine that these 2’s should be multiplied, not added. So the answer is 16, right? Not so fast—a twin is only one person, thus a pair of twins, only two. So the correct answer is 8. Watermelons Yesterday a thousand pounds of watermelons lay in the watermelon patch. They were 99% water, but overnight they lost moisture to evaporation and now they are only 98% water. How much do they weigh now? Solution: 500 lb. The watermelons contain 10 pounds of solid matter that now com-prises 2% of their final weight; divide 10 pounds by 0.02, and there you are. Apparently, quite a bit of evaporation took place! Bags of Marbles You have 15 bags. How many marbles do you need so that you can have a different number of marbles in each bag? Solution: Counting 0 as a number, you might reasonably deduce that you should put no marbles in the first bag, 1 in the second, 2 in the third, etc., and finally 14 in the last. How many marbles is that? The quick way to answer that is to observe that the average number of marbles in one of your 15 bags is 7. Thus the total number of marbles is 15 × 7 = 105. But there’s a trick: You can put bags inside bags! If you put the empty first bag inside the second along with one marble, then the second inside the third along with another marble, etc., you end with the last bag containing all the marbles. So you only need 14 marbles in all. Salaries and Raises Wendy, Monica, and Yancey were hired at the beginning of calendar 2020. Wendy is paid weekly, $500 per week, but gets a $5 raise each week; Monica is paid monthly, $2500 per month, and gets a $50 raise each month; Yancey is paid yearly, $50,000 per year, and gets a $1500 raise each year. 78 Who will make the most money in 2030? Solution: Yancey gains $1500 per year, thus he makes $50,000 + 10 × $1500 = $65,000, 10 years later. Monica gains 50 × 78(!) = $3900 per year, making $30,000 + 10 × $39,000 = $69,000 at the end of the 10th year (which is when 2030 is about to start). So Monica out-earns Yancey in 2030, without even considering the additional raises she is due that year. Wait, why 78? Because Monica’s monthly raises add up to $50 × (1 + 2 + 3 + · · · + 12) = $50 × 13 × 12/2 = $50 × 78. By similar reasoning, figuring 52 weeks in a year, Wendy gains $5 × 53 × 52/2 = $6890 per year, making $26,000 + 10 × $6890 = $94,000 at end of her 10th year. Yay Wendy! Here’s yet another chance to add up consecutive numbers. Efficient Pizza-cutting What’s the maximum number of pieces you can get by cutting a (round) pizza with 10 straight cuts? Solution: There are a number of ways to tackle this one, but perhaps the easiest is to note that the nth cut can at best cross each of the n−1 previous cuts, and between each pair of crossings, split a previous piece in two. Since you also get a new piece before the first crossing and after the last one, the cut ends up adding n new pieces. It follows that with n cuts you can create at most 1+2+· · ·+n = n(n+1)/2 new pieces, but remember you started with one piece (the whole pizza), so the answer is 1 + n(n+1)/2 pieces. For 10 slices, that works out to 56 pieces. Wait, we haven’t actually shown that you can achieve that many pieces— that would require having every two cuts cross, with never any more than two cuts crossing at the same place. But if we just mark 2n random points along the edge of the pizza, number them clockwise (say) and cut from the first to 79 the n+1st, then the second to the n+2nd, etc. we’ll get what we want with probability 1. You don’t like random cuts? One way to do it is to divide the circumference into 2n equal intervals, then sequentially make cuts between opposite intervals, avoiding previous intersection points. Attention Paraskevidekatriaphobes Is the 13th of the month more likely to be a Friday than any other day of the week, or does it just seem that way? Solution: Amazingly, it is true—in the sense that in the 400-year cycle of our Gregorian calendar, more months have their 13th falling on a Friday than on any other day of the week. It turns out that in 688 out of 4800 months in the 400-year cycle, the 13th falls on a Friday. Sunday and Wednesday claim 687 each, Monday and Tuesday 685 each, and Thursday and Saturday only 684 each. To check this you need to remember that years which are multiples of 100 are not leap years unless (like 2000) they are divisible by 400. Incidentally, with some practice you can quickly determine the day of the week of any date in history—even accounting for past calendar glitches. For lazier or more present-time-oriented mortals, a useful fact to remember is that in any year 4/4, 6/6, 8/8, 10/10, 9/5, 5/9, 7/11, 11/7, and the last day of February all fall on the same day of the week. (This is even easier to remember if you happen to play craps daily from 9 to 5.) Note that it doesn’t matter whether you like to put the month-number before the day-number, or vice versa. The late John H. Conway (whose name you will see often in this volume) liked to call that day of the week “doomsday” for that year. Doomsday advances one day each year, two before a leap year. For 2020 doomsday was Saturday, thus Sunday for 2021, Monday for 2022, and Tuesday for 2023; but Thursday for 2024, a leap year. For the next puzzle, we introduce some basic combinatorics—techniques for the art of counting. Most readers will know that if you construct something in two steps, and you have n1 choices for the first step and n2 for the second, then you have n1 × n2 ways to do it. This extends to multiple stages; I call this the “multiplication rule.” For example, the number of different license plates you can make that display three letters followed by two digits is 26×26×26×10×10 = 263 ×100. The number of subsets of an n-element set is 2n, because you can construct the subset by listing elements of the big set and one by one, deciding whether or not to put that element into the subset. If you want to count the subsets of a particular size k, you’ll need the division rule, which says that if you “accidentally” count each object m times, then you have to divide by m at the end. To count the subsets of size k of an n-element set, 80 we first count ordered subsets—subsets with a designated first element, second, etc. There are n ways to pick the first element of your ordered subset, n−1 to pick the second, down to n−k+1 for the last, so n(n−1)(n−2) · · · (n−k+1) in all. But any subset of size k can be ordered k(k−1)(k−2) · · · 1 = k! ways, so we’ve counted each subset that many times. It follows that the number of subsets of size k is really n(n−1)(n−2) · · · (n−k+1) k(k−1)(k−2) · · · 1 = n! k!(n−k)! . The latter expression is written n k and usually pronounced “n choose k.” Here’s an example where k is just 2. Meet the Williams Sisters Some tennis fans get excited when Venus and Serena Williams meet in a tourna-ment. The likelihood of that happening normally depends on seeding and talent, so let’s instead construct an idealized elimination tournament of 64 players, each as likely to win as to lose any given match, with bracketing chosen uniformly at random. What is the probability that the Williams sisters wind up playing each other? Solution: This looks complicated and in fact, in Frederick Mosteller’s delightful book Fifty Challenging Problems in Probability, the solution (of an equivalent problem) is obtained by working it out for 2k players when k is small, guessing the general answer, and then proving the guess to be correct by induction on k. But there’s an easy way to think of it: There are 64 2 = (63 × 64)/2 pairs of players, out of which 63 will meet (since 63 players need to be eliminated to produce the winner), so the probability that a particular pair meets must be 63/((63 × 64)/2) = 1/32. Rating the Horses You have 25 horses and can race them in groups of five, but having no stopwatch, you can only observe the order of finish. How many heats of five horses do you need to determine the fastest three of your 25? Solution: Hmm. It’s clear that all the horses need to be run, since an untested horse could easily be one of the fastest three. You could do that in five heats but that would not clinch the deal, so you’ll need at least six heats for sure. Well, suppose you do partition the horses into five heats; you’d then need to test the winners against one another. What will you know then? 81 Suppose (without loss of generality) that the winner of your sixth heat came from Heat 1, the placer (i.e., runner-up) from Heat 2, and the shower (i.e., third-place horse) from Heat 3. Of course the fastest horse overall is the winner of Heat 6, but the second-fastest could either be the placer of Heat 6 or the placer of Heat 1. The third-fastest horse could be the placer or shower in Heat 6, the placer or shower in Heat 1, or the placer in Heat 2. So the only horses “in the running” (sorry about that) for second- and third-fastest are the placer and shower from Heat 1, the winner and placer from Heat 2, and the winner from Heat 3. That’s five horses; run them in Heat 7, and you’re done. It’s not hard to see you can’t beat this elegant solution. Shoelaces at the Airport You are walking to your gate at O’Hare and need to tie your shoelace. Up ahead is a moving walkway that you plan to utilize. To minimize your time to get to the gate, should you stop and tie your shoelace now, or wait to tie it on the walkway? Solution: You should tie your shoelace on the walkway. Either way, you spend the same amount of time walking on solid ground, and the same amount of time tying your shoelace. But if you tie your shoelaces on solid ground, you spend more time walking on the walkway. Rows and Columns Prove that if you sort each row of a matrix, then each column, the rows are still sorted! Solution: To prove this rather surprising assertion, let us imagine that the matrix has m rows and n columns, and that after each row is sorted (smallest values to the left, say), the entry on the ith row and jth column is aij; after the columns are sorted, it is bij. We need to show that if j < k then bij ≤bik. 82 This is one of those things that alternates between being mysterious and obvious, each time you think about it. One way to argue the point is to note that bik is the ith smallest entry in the kth column (whose set of entries has not changed). If x is any value, then the number of entries in the jth column that are below x must be at least as great as the number in the kth column that are below x, since aij ≤aik for each i. It follows that the ith smallest entry in the jth column cannot exceed the ith smallest entry in the kth column, and we are done. Is this more convincing than just trying an example? You be the judge. Three-Way Election Alison, Bonnie, and Clyde run for class president and finish in a three-way tie. To break it, they solicit their fellow students’ second choices, but again there is a three-way tie. The election committee is stymied until Alison steps forward and points out that, since the number of voters is odd, they can make two-way decisions. She therefore proposes that the students choose between Bonnie and Clyde, and then the winner would face Alison in a runoff. Bonnie complains that this is unfair because it gives Alison a better chance to win than either of the other two candidates. Is Bonnie right? Solution: Bonnie is correct—in fact, she has understated the case; assuming no voters change their minds, Alison will win for sure! To see this, suppose Alison’s supporters mostly prefer Bonnie to Clyde (so that Bonnie would beat Clyde in the proposed two-candidate race). Then Bonnie’s supporters must prefer Clyde to Alison, otherwise Clyde would have garnered fewer than 1/3 of the second-place votes; similarly Clyde’s supporters prefer Alison to Bonnie. Thus, in this case, Alison will beat Bonnie in the runoff. If Alison’s supporters prefer Clyde to Bonnie, a symmetric argument shows that Alison will beat Clyde in the runoff. ♡ This puzzle serves as a warning: There may be more to some tiebreakers than meets the eye! Let us return to the basic multiplication rule, but with some additional cleverness required. Sequencing the Digits In how many ways can you write the numbers 0 through 9 in a row, such that each digit other than the leftmost is within one of some digit to the left of it? Solution: To apply the multiplication rule when you build something in stages, you need the number of choices in a stage not to depend on choices made in previous 83 stages. That seems not to be the case here; for example, if you start your sequence with 8 you have two choices for the next digit (7 or 9) but if you start with 9, you have only one choice for the next digit (8) and in fact now the whole sequence is determined. You can get around this with some fanciness involving binomial coefficients (those n k guys mentioned above) but there’s a nicer way: Create the sequence in reverse! The last digit must be 0 or 9; if it’s 9, the next-to-last must be 0 or 8, while if it’s 0 the next-to-last must be 1 or 9. No matter what you do there are two choices at every stage until you reach the leftmost digit, so the number of sequences is 29 = 512. Faulty Combination Lock A combination lock with three dials, each numbered 1 through 8, is defective in that you only need to get two of the numbers right to open the lock. What is the minimum number of (three-number) combinations you need to try in order to be sure of opening the lock? Solution: Often the easiest way to tackle a problem of this sort is geometrically. The space of all possible combinations is an 8 × 8 × 8 combinatorial cube, and each time you try a combination, you cover all combinations on the three orthogonal lines which intersect at that point. Once you see the problem this way, you are likely to discover that the best way to cover all the points in the cube is to concentrate all your test points in just two of the eight 4 × 4 × 4 octants. You will then arrive at a solution equivalent to the one described below. Try all combinations with numbers from {1, 2, 3, 4} whose sum is a multiple of 4; there are 16 of those, since if you pick the numbers on the first two (or any two) of the dials, the number on the third is determined. Now try all the 84 combinations you get by adding (4,4,4), that is, by adding 4 to each of the three numbers; there are 16 more of those, and we claim that together the 32 choices cover all possibilities. It is easy to see that this works. The correct combination must have either two (or more) values in the set {1, 2, 3, 4}, or two or more values in the set {5, 6, 7, 8}. If the former, there is a unique value for the third dial (the one whose number may not be among {1, 2, 3, 4}) such that the three were among the first 16 tested combinations. The other case is similar. To see that we can’t cover with 31 or fewer test-combinations, suppose that S is a cover and \|S\| = 31. Let Si = {(x, y, z) ∈S : z = i} be the ith level of S. Let A be the set {1, 2, 3}, B = {4, 5, 6, 7, 8}, and C = {2, 3, 4, 5, 6, 7, 8}. At least one level of S must contain 3 or fewer points; we might as well assume S1 is this level and \|S1\| = 3. (If \|S1\| ≤2, there’s an easy contradiction.) The points of S1 must lie in a 3 × 3 × 1 subcube; we may assume that they lie in A × A × {1}. The 25 points in B ×B ×{1} must be covered by points not in S1. No two of them can be covered by the same point in S, thus, S\S1 (that is, the elements of S that are not in S1) has a subset T of size 25 that lies in the subcube B×B×C. Now consider the set P = {(x, y, z) : z ∈C, (x, y, 1) ̸∈S1, (x, y) ̸∈B × B}. A quick count shows that \|P\| = (64 −3 −25) × 7 = 252. The points in P are not covered by S1, and each point in T can cover at most 3 + 3 = 6 points in P. Therefore there are at least 252 −(6 × 25) = 102 points in P that must be covered by points in S \ S1 \ T. However, \|S \ S1 \ T\| = 31 −3 −25 = 3, and each point in the cube covers exactly 22 points. Since 22 × 3 = 66 < 102, we have our contradiction. ♡ Losing at Dice Visiting Las Vegas, you are offered the following game. Six dice are to be rolled, and the number of different numbers that appear will be counted. That could be any number from one to six, of course, but they are not equally likely. If you get the number “4” this way you win $1, otherwise you lose $1. You decide you like this game, and plan to play it repeately until the $100 you came with is gone. At one minute per game, how long, on the average, will it take before you are wiped out? Solution: This is a trick, of course; otherwise it would be in Chapter 10, In All Probability, not this one. On the average, it’ll take forever for you to be wiped out—the game is in your favor, and thus you may never even dip below your $100 stake. There are 66 = 46,656 ways to roll the dice. For four different numbers to appear, you need either the pattern AABBCD or AAABCD. There are 6 2 · 4 2 /2 = 45 85 versions of the former pattern, keeping the equinumerous labels in alphabetical order; e.g., AABBCD, ABABCD, ACDABB, but not BBAACD or AABBDC. For the latter pattern, there are 6 3 = 20 versions. In either case, there are 6 × 5 × 4 × 3 = 360 ways to assign numbers to the letters, for a total of 360 × 65 = 23, 400 rolls. Thus, the probability of winning is 23400/46656 = 50.154321%. ♡ If you win some bets with this game, don’t forget to send 5% of your profits to me c/o Taylor & Francis. Splitting the Stacks Your job is to separate a stack of n objects into individual items. At any time you may split a stack into two stacks, and get paid the product of the sizes of the two stacks. For example, for n = 8, you could split into a stack of size 5 and a stack of size 3, getting paid $15; then you can split the 5-stack into 2 and 3 to get $6 more. Then the two threes can each be split ($2 each) and the remaining three twos ($1 each) for a total income of 15 + 6 + (2 × 2) + (3 × 1) = 28 dollars. What’s the most money you can get starting with n objects, and how can you get it? Solution: Try a few other ways to deal with eight objects, and you will find that they all earn the same $28. Does it really not matter how you split the stacks? On the web, this problem has been used to demonstrate something called “strong induction.” But there’s an easy way to see that the outcome is n 2 dollars no matter what you do: Each (unordered) pair of objects earns $1! Imagine that there is a string connecting every pair of objects, and when you split a stack, you cut every string between pairs of objects that straddle the split. If there are i objects on one side of the split and j on the other, you cut ij strings and get paid $ij. Eventually you cut every string and get paid $1 for each one, so you earn n 2 dollars regardless of strategy. North by Northwest If you’ve never seen the famous Alfred Hitchcock movie North by Northwest (1959), you should. But what direction is that, exactly? Assume North is 0◦, East 90◦, etc. Solution: Sounds like it should be halfway between north (0◦) and northwest (315◦), thus 337.5◦, but that’s actually “north-northwest,” not “north by northwest.” The preposition “by” moves the needle just 11.25◦in the direction of its object; thus “north by northwest” puts you at (360 −11.25)◦= 348.75◦. Northwest by 86 north, on the other hand, is (315 + 11.25)◦= 326.25◦. To make things even more confusing, “north by northwest” is commonly designated more simply as “north by west.” So the movie title was not actually a phrase in use; in fact the movie’s original title was “In a Northwesterly Direction.” Doesn’t roll off the tongue quite as well, though, does it? The next puzzle is, really, just a bit of arithmetic. Early Commuter A commuter arrives at her home station an hour early and walks toward home until she meets her husband driving to pick her up at the normal time. She ends up home 20 minutes earlier than usual. How long did she walk? Solution: This kind of timing problem can drive you nuts, until you think about it the “right way.” Here, you notice that the commuter’s husband apparently saved 10 minutes each way, relative to normal, so must have picked up his wife 10 minutes earlier than usual. So she walked for 50 minutes. To solve the last puzzle, we return to those binomial coefficients. Alternate Connection One hundred points lie on a circle. Alice and Bob take turns connecting pairs of points by a line, until every point has at least one connection. The last one to play wins; which player has a winning strategy? Solution: The player who connects the 98th point is doomed, since there will be either one or two unconnected points left after her move, which her opponent can then deal with. There are 97 2 moves that can be made before the 98th needs to be connected, and since this number is even, the second player can always win. Bindweed and Honeysuckle Two vines, a bindweed and a honeysuckle, climb a tree trunk, starting and ending at the same place. The bindweed circles the trunk three times counter-clockwise while the honeysuckle circles five times clockwise. Not counting the top and bottom, how many times do they cross? Solution: By mentally twisting the tree trunk with the vines still on it, we see that the answer is the same as it would be if the bindweed went straight up while the 87 honeysuckle went around 5 + 3 = 8 times. Therefore there are 9 meetings in total, and the answer, with the top and bottom not counted, is 7. Do you remember phone trees? Suppose you belonged to an organization that needed a method of spreading information by phone. It was once common in such a case to set up a phone tree. The idea is this: As a member, you will have a special list of certain other members (at least one), and everyone on your list will have your name on their list. If you gain possession of outside information of interest to the other members, you call the members on your list. If, instead, you get information from another member, you pass it on to all the other members on your list. For the phone tree to work, it must be that there is a path between any two members; in other words, if any member gets information, then eventually the whole organization must hear it. If we think of each member as a point, with a line between two points when the corresponding members are on each others’ phone lists, then the collection of all points and lines must form a connected graph. (There’ll be more about graphs in Chapter 4, Graphography.) Why is it called a phone tree and not a phone graph? Because if it is effi-ciently constructed, there are only n−1 lines needed in the graph—equivalently, members will never get their own information relayed back to them. Here is an example of a phone tree for a 10-member organization whose names happen to be the numbers from 0 to 9. The question we address is: How many ways are there to set up a phone tree for an organization with n members? This question was answered by German mathematician Carl Wilhelm Borchardt in 1860, the same year Johann Philipp Reis built (arguably) the first telephone prototype. 88 Borchardt’s contribution has been mostly forgotten and the name of an English mathematician, Arthur Cayley, is associated with the formula to come below. Before we state and prove “Cayley’s formula,” roughly how many possible phone trees might you guess would be possible if for a 10-member organization? Hundreds? Thousands? The answer is a hundred million. Exactly! Theorem. The number of trees on a set of n labeled points is nn−2. Proof. We will prove the formula by showing that there is a one-to-one correspondence between the set of trees on points {0, 1, 2, . . . , n−1}, and the set of sequences of length n−2 of the numbers {0, 1, 2, . . . , n−1}. Since there are n ways to choose each entry of the sequence, we know from the multiplication rule above that there are nn−2 such sequences. Thus, if we can establish the aforementioned one-to-one correspondence, we are done. A leaf of a tree is a point (better, a “vertex”) of tree that’s attached to only one other vertex. To go from tree to sequence, we find the lowest-numbered leaf and start the sequence not with the label of that leaf, but with the label of its one neighbor. Then that leaf is erased and the procedure is repeated to determine the next entry in the sequence, and this continues until there are only two vertices left in the tree. That’s it! Let’s try it with the tree shown above. We get the following chain of ever-longer sequences and ever-smaller trees; in each tree, all leaves are circled and the vertex attached to the current lowest-numbers leaf is boxed. To show that this is a one-to-one correspondence, we need to get from any sequence to the unique tree that would yield that sequence. It turns out, that’s quite easy and logical; let’s do it for the sequence we just got, and check that we get the same tree back. Note that the leaves of a tree will not themselves appear in that tree’s se-quence. Thus, given a sequence, we look for the lowest number that doesn’t appear, and start building our tree by attaching that vertex to the vertex corre-sponding to the leftmost number in the sequence. Then we erase that leftmost number and repeat, now looking for the smallest number missing from the new, shorter sequence that has not already been used as a leaf. This continues until the sequence is gone, and then we connect the two vertices that were never used as leaves to complete the tree. We will not formally prove here that the two algorithms described above are inverses of each other, but hopefully the example is convincing. ♡ 89 90 Chapter 2 Achieving Parity To a mathematician, “parity” often means just odd versus even. Typically, the concept of parity is not mentioned in a puzzle; but when you start to play with the puzzle, the concepts of odd and even emerge. And when they do, you are often half the way to solving it. Here’s a simple example. Bacterial Reproduction When two pixo-bacteria mate, a new bacterium results; if the parents are of different sexes the child is female, otherwise it is male. When food is scarce, matings are random and the parents die when the child is born. It follows that if food remains scarce a colony of pixo-bacteria will eventually reduce to a single bacterium. If the colony originally had 10 males and 15 females, what is the probability that the ultimate pixo-bacterium will be female? Solution: Here a bit of experimentation might lead to the observation that the parity of the number of females cannot change—in other words, the number of females must remain odd. Thus the probability that the last pixo-bacterium will be female is 1. For the next puzzle, you have to come up with the parity idea on your own. Fourth Corner Pegs occupy three corners of a square. At any time a peg can jump over another peg, landing an equal distance on the other side. Jumped pegs are not removed. Can you get a peg onto the fourth corner of the square? 91 Solution: The first thing to do here is think of the initial square as a cell of the plane grid, for example, the points (0,0), (0,1), (1,0) and (1,1) on the XY -plane. Then the pegs will always be on grid points. Grid points, however, have four possible parities: each coordinate can be even or odd. When a peg jumps, its parity is preserved; its X-coordinate goes up or down by an even number, and likewise its Y -coordinate. The points of the unit cell above have all four parities among them, so the corner that starts without a peg can never be occupied. Unanimous Hats One hundred prisoners are offered the chance to be freed if they can win the following game. In the dark, each will be fitted with a red or black hat according to a fair coinflip. When the lights are turned on, each will see the others’ hat colors but not his own; no communication between prisoners will be permitted. Each prisoner will be asked to write down his guess at his own hat color, and the prisoners will be freed if they all get it right. The prisoners have a chance to conspire beforehand. Can you come up with a strategy for them that will maximize their probability of winning? Solution: The probability that Bob (a specified one of the prisoners) guesses correctly is exactly 1 2, no matter what the general strategy is, so the probability that they all get it right cannot exceed 1 2. But, amazingly, they can achieve 1 2. The key is: Suppose Bob knows there are an even number of red hats. Then he is home free, because if he sees an even number of red hats on his fellow prisoners’ heads, he knows his hat is black; if he sees an odd number of red hats, he knows his hat is red. So the prisoners simply agree to all guess that the number of red hats will be even. Either they’re all right, or they’re all wrong; and the probability that the number of red hats is even is exactly 1 2. (That’s true even if only one hat is determined by coinflip, since no matter what the other hat colors are, that hat color determines the parity.) Once you’ve got that puzzle, the next one is easy. 92 Half-Right Hats 100 prisoners are offered the chance to be freed if they can win the following game. In the dark, each will be fitted with a red or black hat according to a fair coinflip. When the lights are turned on, each will see the others’ hat colors but not his own; no communication between prisoners will be permitted. Each prisoner will be asked to write down his guess at his own hat color, and the prisoners will be freed if at least half get it right. The prisoners have a chance to conspire beforehand. Can you come up with a strategy for them that will maximize their probability of winning? Solution: Yes, they just arrange for 50 prisoners to guess their color by assuming the total number of red hats is even, while the other 50 assume the number of red hats is odd. Then one group or the other is bound to be right, assuming no poor fellow screws up! Red and Blue Hats in a Line There are n prisoners this time, and again each will be fitted with a red or black hat according to flips of a fair coin. The prisoners are to be arranged in a line, so that each prisoner can see only the colors of the hats in front of him. Each prisoner must guess the color of his own hat, and is executed if he is wrong; however, the guesses are made sequentially, from the back of the line toward the front. Thus, for example, the ith prisoner in line sees the hat colors of prisoners 1, 2, . . . , i−1 and hears the guesses of prisoners n, n−1, . . . , i+1 (but he isn’t told which of those guesses were correct—the executions take place later). The prisoners have a chance to collaborate beforehand on a strategy, with the object of guaranteeing as many survivors as possible. How many prisoners can be saved in the worst case? Solution: It’s clear that no more than n−1 prisoners can be sure of surviving, because the first one to guess—namely, the last in line—has no clue. But he can pass a clue to the prisoner in front of him, by letting him know whether the number of red hats among prisoners 1 througn n−1 is even or odd. How does he communicate? By his guess, of course. The prisoners can agree, for example, that the last prisoner in line guesses “red” if he sees an odd number of red hats, “black” otherwise. So that saves the life of prisoner n−1; what about the rest? They’re saved too! Everyone has heard the last prisoner, and knows that after him, all guesses will be correct. For example, if the ith prisoner to guess hears that there are an even number of reds among prisoners 1 through n−1, and after the first guess he hears 5 “red” guesses, he concludes that the number of red hats from himself forward is odd. If he actually sees an even number of red hats in front of him, he knows that his own hat must be red and guesses accordingly. 93 The next one is a bit trickier. Luckily, the prisoners are getting pretty good with these games. Prisoners and Gloves Each of 100 prisoners gets a different real number written on his forehead, and is provided with a black glove and a white glove. Seeing the numbers on the other prisoners’ foreheads, he must put one glove on each hand in such a way that when the prisoners are later lined up in real-number order and hold hands, touching gloves match. How can the prisoners, who are permitted to conspire beforehand, ensure success? Solution: This seems impossible: Each prisoner sees 99 different real numbers and has no idea where the number on his own forehead fits in, so how does he know what to do? What information has he got? The parity here is of a more subtle sort: It’s the parity of a permutation. Permutations are classified as “even” or “odd” according to the parity of the number of swaps needed to sort it. If we write a permutation in “one-line form,” so that (p1, p2, . . . , pn) represents the permutation that sends i to pi, then (1,2,3,4,5) is even since it is already the identity (requiring 0 swaps), but (4,2,3,1,5)—which can be sorted by swapping “1” with “4”—is odd. Assume the prisoners themselves are numbered from 1 to 100 (e.g., alphabet-ically). Then the numbers on the prisoners’ foreheads induce a permutation of the numbers from 1 to 100: if the ith prisoner’s forehead shows the jth smallest real number, i is mapped to j by the permutation. For example, suppose the prisoners are Able, Baker, Charlie, and Dog, num-bered 1 to 4 in that order. Say their numbers, respectively, are 2.4, 1.3, 6.89, and π. Then the permutation is (2,1,4,3) where the “2” at the front represents the fact that the first prisoner got the second-highest number, etc. This permutation happens to have even parity, since interchanging the 2 and the 1, then the 4 and the 3, restores it to the identity (1,2,3,4). Of course, Charlie cannot see the 6.89 on his own forehead so he doesn’t know the parity of the permutation. But suppose he assumes that his own number is the lowest of all; then he can compute the parity. He’ll be right if his number is the lowest, or the third lowest, or the fifth lowest, and so forth; wrong if his number happens to be the ith lowest, i being an even number. Let’s have Charlie put the white glove on his left hand (and black on right) if his assumption leads him to an even permutation, and the reverse if he gets an odd permutation. (In the example, Charlie’s assumption leads him to the permutation (3,2,1,4) which is odd, so he puts the black glove on his left hand.) That works! If the permutation actually is even, then the prisoner with lowest forehead number would have computed “even” and would have a white glove on his left hand. The next-lowest-forehead-numbered prisoner would have computed “odd” and his black-gloved left hand would end up holding the first 94 guy’s black-gloved right hand, and similarly down the line. If the actual per-mutation is odd, the glove colors will be the complement of these, and will still work. Even-sum Billiards You pick 10 times, with replacement (necessarily!), from an urn containing 9 billiard balls numbered 1 through 9. What is the probability that the sum of the numbers of the balls you picked is even? Solution: If there were no ball number 9, an even sum would have probability exactly 1/2 since no matter what happens in your first 9 picks, the parity of the tenth ball will determine whether the sum is even or odd; and that ball is equally likely to be either. Now put ball 9 back in. No matter what order you choose the balls in, you can ask your assistant to reveal the numbers to you with all the 9’s first. Then the above argument still applies: The parity of the last ball shown you, whose number is between 1 and 8, will determine the parity of the sum; so again, the sum is equally likely to be even or odd. Oops, wait, there’s just one possibility we haven’t considered: That you got the 9 ball every time! That has probability (1/9)10 and results in an even sum. So altogether, the probability of an even sum is 1 2 1 − 1 9 10! + 1 9 10 = 1 2 + 1 2 1 9 10 which is about 0.50000000014. Probably not enough bias to be worth betting on! Of course, “even” means that the remainder after dividing by 2 is 0; “odd,” that the remainer is 1. For the next puzzle, you’ll need to stretch your definition of parity to the remainder after dividing by 3; and for the puzzle after that, after dividing by 9. Chameleons A colony of chameleons currently contains 20 red, 18 blue, and 16 green individ-uals. When two chameleons of different colors meet, each of them changes his or her color to the third color. Is it possible that, after a while, all the chameleons have the same color? Solution: The key thing is to observe that, after each encounter of two chameleons, the difference between the number of individuals of any two colors remains the 95 same modulo 3. In symbols, letting NR stand for the number of red chameleons and NB and NG similarly, we claim that, for example, NR −NB has the same remainder after division by 3, after any two chameleons meet, that it did before. This is easy to verify by checking cases. Thus these differences remain the same modulo 3 forever, and since in the given colony none of those differences is equal to zero modulo 3, we can never get two of the color populations to be zero. If, on the other hand, the difference of two color populations (say, NR −NB) is a positive multiple of 3, we can lower that difference by having a red chameleon meet with a green one (or if there are no green ones, by first having a red one meet a blue one). We repeat until NR = NB and then have reds meet blues until only green chameleons remain. Putting all this together, and noting that if two differences are multiples of 3 then the third must be as well, we see that: • If all three differences are multiples of 3, then any color can take over the colony; • If just one of the difference is a multiple of 3, then the remaining color is the only one which can take over the colony; and finally, • If none of the differences are multiples of 3, as in the given problem, the colony can never become monochromatic and will remain fluid until other circumstances (e.g., birth, death) intervene. Now for a quickie. Missing Digit The number 229 has 9 digits, all different; which digit is missing? Solution: What to do? You can type “2^29” into your computer or calculator and look at the number for yourself, but is there a way to do it in your head without getting a headache? Hmm . . . maybe you remember a technique from grade school called “casting out nines,” where you keep adding up digits and always end up with a number’s value modulo 9 (that is, its remainder after dividing by 9). It uses the fact that 10 ≡1 mod 9, thus 10n ≡1n ≡1 mod 9 for every n. If we denote by x∗the sum of the digits of the number x, then we get (xy)∗≡x∗y∗mod 9 for every x and y. It follows in particular that (2n)∗≡2n mod 9. The powers of 2 mod 9 begin 2, 4, 8, 7, 5, 1, and then repeat; since 29 ≡5 mod 6, 229 mod 9 ≡the fifth number in this series, which happens to be 5. Now the sum of all the digits is 10 × 4.5 = 45 ≡0 mod 9, so the missing digit must be 4. Indeed, 229 = 536, 870, 912. 96 Subtracting Around the Corner A sequence of n positive numbers is written on a piece of paper. In one “oper-ation,” a new sequence is written below the old one according to the following rules: The (absolute) difference between each number and its successor is writ-ten below that number; and the absolute difference between the first and last numbers is written below the last number. For example, the sequence 4 13 9 6 is followed by 9 4 3 2. Try this for n = 4 using random numbers between, say, 1 and 100. You’ll find that after remarkably few operations, the sequence degenerates to 0 0 0 0 where it of course remains. Why? Is the same true for n = 5? Solution: Considering values modulo 2 solves both problems. In the n = 4 case, up to rotation and reflection, 1 0 0 0 and 1 1 1 0 become 1 1 0 0, then 1 0 1 0, then 1 1 1 1, then 0 0 0 0. Since this covers all cases, we see that when working with ordinary integers, at most four steps are required to make all the numbers even; at that point, we may as well divide out by the largest common power of two before proceeding. Since the value M of the largest number in the sequence can never increase, and drops by a factor of 2 or more at least once every four steps, the sequence must hit 0 0 0 0 after at most 4(1 + ⌈log2 M⌉) steps. (The symbol ⌈x⌉stands for the “ceiling” of x, that is, the least integer greater than or equal to x.) On the other hand, for n = 5, the sequence 1 1 0 0 0 (considered either as binary or ordinary numbers) cycles via 1 0 1 0 0, 1 1 1 1 0, 1 1 0 0 0. ♡ A little analysis, using polynomials over the integers modulo 2, shows that the salient issue is whether n is a power of 2. For our next puzzle, we return to even versus odd, but now you’ll need some keen powers of observation to see how to apply this idea. Uniting the Loops Can you re-arrange the 16 square tiles below into a 4 × 4 square—no rotation allowed!—in such a way that the curves form a single loop? Lines that exit from the bottom are assumed to continue at the same point on the top, and similarly right to left; in other words, imagine that the square is rolled up into a doughnut. Solution: This puzzle (along with two others in the book) was inspired by the work of artist Sol LeWitt, who loved to construct paintings (and sculptures) from pieces that consisted of every possible way to do some combinatorial task. In the case of the present puzzle, examination of the tiles shows that each side of a tile has two entry/exit points. The two points on one edge of a tile 97 terminate two arcs (quarter-circles or quarter-ellipses) that may either cross or not cross on their way to adjacent edges. Thus, for each edge there’s a decision to be made: to cross, or not to cross. These four binary decisions result exactly in the 16 different tiles you see. Since when we wrap around the doughnut, none of the curved lines ter-minate, the curves are forced to form loops. The question is whether we can arrange the tiles to form just one loop. As with many puzzles, our initial strategy is just to try various arrangements and count the loops. You will find that no matter how you do it, there always seem to be an even number of loops. If that holds in general, then, of course, we can never make a single loop. But why should the number of loops always be even? One natural way to try to prove it might be to show that if we switch two adjacent tiles, the number of loops always changes by an even number. Alas, there are a lot of curved lines that enter a pair of adjacent tiles (12 in all) and it’s awfully hard to deal with all the different things that can happen when two adjacent tiles are switched. We need to consider some much smaller change. How about crossing or un-crossing one pair of curves on a tile? Of course, if we do that, it would effectively destroy one tile and duplicate another. We would then be in a different universe where we have as many as we like of each tile, and can choose any 16 tiles to arrange into a square. A little experimentation shows that if we take any configuration involving the 16 original tiles—e.g., the one above—and cross or uncross curved lines on one side of one tile, the number of loops always goes up or down by one. If that’s 98 universally true, we arrive at a conjecture: If the total number of crossings in the picture is even, the number of loops will be even; if it is odd, the number of loops will be odd. More experimentation seems to confirm this; can we prove it? The figure below shows the three things that can happen when you make or undo a crossing. The first two create or destroy one loop, thus they add or sub-tract one from the loop total. The third is impossible; if a loop is oriented (say, clockwise), and the tiles are colored like a checkerboard, then a loop alternately enters gray tiles vertically and white tiles horizontally, or vice-versa. The loops shown violate this condition. Now, if you start with the first figure, which has 16 crossings and four loops, you can create or delete crossings to get any desired configuration. But every time you create or delete a crossing, the number of crossings and the number of loops remain the same parity—both odd, or both even. Eventually when you get to any arrangement of the 16 original tiles, you are back to an even number of loops. ♡ 99 Our theorem was an open problem for some time, until parity was unleashed on it. It concerns partitions of an abstract combinatorial construction. Fix a positive number n. A box is a Cartesian product of n finite sets; if the sets are A1, . . . , An, then the box consists of all sequences (a1, . . . , an) such that ai ∈Ai for each i. The box B = B1 × · · · × Bn is a sub-box of A if Bi ⊆Ai for each i; it is said to be a nontrivial sub-box if, in addition, ∅̸= Bi ̸= Ai for each i. If each factor Ai has at least two elements, and we cut each Ai into two non-empty pieces, then by taking every product of one piece from each Ai we could partition the box A into 2n nontrivial sub-boxes. However, Theorem. A box can never be partitioned into fewer than 2n nontrivial sub-boxes. Proof. Call a sub-box odd if it has an odd number of elements, and let O be the set of all odd sub-boxes of the big box A. If B is a sub-box of A, let OB consist of the elements of O that intersect B in an odd set. If B = B1×· · ·×Bn is a nontrivial sub-box of A, then, in order for a sub-box C to intersect B in an odd number of elements, it must intersect every factor Bi in an odd number of elements (because \|C ∩B\| = \|C1 ∩B1\| × \|C2 ∩B2\| × · · · × \|Cn ∩Bn\|). But the probability that a random odd subset Ci of Ai intersects Bi in an odd number of elements is exactly 1 2. Why? Because we can flip a coin to determine whether each a ∈Ai is in Ci or not, ending with some element a′ that’s not in Bi (this is where we needed Bi to be a proper subset of Ai). This last element a′ is put into Ci or not so as to assure Ci is odd. The elements of Bi that we flipped for constitute a random subset of BI, and since exactly half the subsets of any nonempty finite set are odd, the probability that Bi ∩Ci is odd is 1 2 as claimed. Each C ∈O is partitioned by the B’s, and since C is odd, at least one of the parts of the partition must be odd. If there are m parts of the partition, it follows that the probability that C ∈OB is at least 1/m. But we just showed that probability is 1/2n, so m ≥2n! ♡ 100 Chapter 3 Intermediate Math The Intermediate Value Theorem (“IVT”) states that if a real number changes continuously as it moves from a to b, then it must pass through every value between a and b. This simple and intuitive fact—arguably, more of a property of the real numbers than a theorem—is a powerful tool in mathematics, and shows up often in puzzle solutions. A useful variation of IVT addresses the case when one real number is moving continuously from a to b while another goes from b to a; they must cross. For example: Boxing the Mountain State Can West Virginia be inscribed in a square? Solution: We are referring here to the shape of West Virginia as projected (any way you like) onto a plane. We begin by drawing a vertical line west of the state and moving it east until it just touches the state boundary. Then we draw another vertical line east of the state, moving it west until it too touches the state boundary. Similar operations with two horizontal lines inscribe the state in a rectangle whose width w is somewhat greater than its height h. To get a square, we rotate the rectangle continuously, maintaining its 90◦ angles while keeping all four sides tangent to the state. After 90◦of rotation, the figure is the same as it was at the start, but now w has become h and h has become w. It follows that at some angle, h and w were equal; and at that point, the state was inscribed in a square. ♡ Of course, this works for any state, and indeed for any bounded shape in the plane. The same technique has a practical application: You can always rotate your square four-legged picnic table so as to have all four legs in contact with the 101 ground, assuming the spot you chose doesn’t have any actual surface disconti-nuities. Just put the table down with three legs touching and rotate until the fourth touches; this is guaranteed to happen before you get to 90◦, since when you do get to 90◦, a ground-contacting leg will be where your fourth leg began. You might find the following classic puzzle even simpler. Monk on a Mountain A monk begins an ascent of Mt. Fuji on Monday morning, reaching the summit by nightfall. He spends the night at the summit and starts down the moun-tain on the same path the following morning, reaching the bottom by dusk on Tuesday. Prove that at some precise time of day, the monk was at exactly the same spot on the path on Tuesday as he was on Monday. 102 Solution: If we plot the monk’s position as a function of time of day, we get two curves connecting opposing corners, and the conclusion follows from IVT. But here’s another way to think of it, underscoring the intuitiveness of IVT: Superim-pose Monday and Tuesday, so now the Monk climbs at the same time that his doppelg¨ anger descends. They must meet! Cutting the Necklace Two thieves steal a necklace consisting of 10 red rubies and 14 pink diamonds, fixed in some arbitrary order on a loop of golden string. Show that they can cut the necklace in two places so that when each thief takes one of the resulting pieces, he gets half the rubies and half the diamonds. Solution: We could check all possible necklaces meeting the above description, but there are more than 40,000 of them, and, anyway, we’d like to solve the problem for any (even) number of rubies and of diamonds. To apply the IVT requires a bit of subtlety; we first have to set up the problem geometrically, then make a couple of key observations. The idea is to represent the necklace as a circle cut into 24 equal arcs, each colored red or pink according to the pattern of rubies and diamonds in the necklace. Then we draw a line through the center of the circle, cutting the circle twice. Our line will certainly cut the full set of gems into equal portions, since any diameter cuts the circle into equal-length arcs (i.e., semicircles). Thus, if a diameter happens to cut the rubies in half, it will cut the diamonds in half as well. The line pictured in the figure above does not work; it’s got more red on one side than the other. But as we rotate the line, the fraction of red on the 103 redder side changes continuously, and after 180◦of rotation it has switched from redder to pinker. The IVT tells us that there is a point when there was the same amount of red on both sides of the line. The remaining subtlety: What if that position involves cutting through a gem? Well, it can’t cut a ruby on one end while it cuts a diamond at the other end, because then the number of rubies on one side of the line would not be an integer—and half of 10 is an integer. So either both ends of the line cut through a ruby or both through a diamond, and if we move the line a bit until it no longer cuts through a gem, the number of rubies on one side will not change. Another way to look at it: You can apply the IVT to integers, if you replace the requirement of continuity by the requirement that the integer value changes only by one. If the rotating line always points to the boundary between two gems (at both ends), and moves one gem at a time, the number of rubies on one side can never change by more than one. We can thus conclude that as the line rotates 180◦, it must at some point have half the rubies on each side. In some puzzles the need for IVT is not at first obvious. Three Sticks You have three sticks that can’t make a triangle; that is, one is longer than the sum of the lengths of the other two. You shorten the long one by an amount equal to the sum of the lengths of the other two, so you again have three sticks. If they also fail to make a triangle, you again shorten the longest stick by an amount equal to the sum of the lengths of the other two. You repeat this operation until the sticks do make a triangle, or the long stick disappears entirely. Can this process go on forever? 104 Solution: You can certainly make the process go on indefinitely if you design it so that the ratios of the lengths of the sticks are unchanged after one operation. For that to happen with lengths a < b < c you’ll need b/a = c/b = a/(c−a−b); let that ratio be r > 1, so that the lengths are proportional to 1, r, r2 and satisfy 1/(r2 −r −1) = r, r3 −r2 −r −1 = 0. Writing f(r) = r3 −r2 −r −1, we need an r > 1 for which f(r) = 0. Does such an r exist? Yes, because f(1) = −2 < 0 while f(2) = 1 > 0; since all polynomials are continuous functions, we can apply the IVT to deduce that there is a root of f between 1 and 2. It remains only to check that our root gives us sticks that don’t make a triangle, but that’s easy, because r2 = r3/r = (r2+r+1)/r = r+1+(1/r) > r+1. ♡ Here’s another case where IVT saves you the trouble of finding a root of a polynomial. Hazards of Electronic Coinflipping You have been hired as an arbitrator and called upon to award a certain indi-visible widget randomly to either Alice, Bob, or Charlie, each with probability 1 3. Fortunately, you have an electronic coinflipping device with an analog dial that enables you to enter any desired probability p. Then you push a button and the device shows “Heads” with probability p, else “Tails.” Alas, your device is showing its “low battery” light and warning you that you may set the probability p only once, and then push the coinflip button at most 10 times. Can you still do your job? Solution: If you could set p twice, two flips would be enough: Set p = 1 3 and on Heads give the widget to Alice, else reset to 1 2 and use the result to decide between Bob and Charlie. Since you can only set p once, it might make sense to try to design a scheme first, then pick p to make the scheme work. For example, you could flip your coin three times and if the flips are not all the same, you could use the “different” flip to decide to whom to award the widget (e.g., “HTT” means Alice gets it). But if you get all heads or all tails, you’ll have to do it again, and maybe yet again—so, not in any bounded number of flips. But now suppose we flip the coin four times. There are four ways to get one head, six to get two, four to get three: Even numbers all, so anytime you get non-uniform results you can use the flips to decide between Alice and Bob. If the coin is fair, the remaining probability—that is, the probability q = p4 + (1−p)4 of “HHHH or TTTT”—is only 1 8, too small for Charlie. As you reduce p, however, q changes continuously and approaches 1. Thus, by IVT, there is some p for which the probability of “HHHH or TTTT” is 105 exactly the 1 3 you need. ♡ Bugs on a Pyramid Four bugs live on the four vertices of a triangular pyramid (tetrahedron). Each bug decides to go for a little walk on the surface of the tetrahedron. When they are done, two of them are back home, but the other two find that they have switched vertices. Prove that there was an instant when all four bugs lay on the same plane. Solution: Let the bugs be A, B, C and D. Initially when D looks down on the ABC plane she sees the ABC triangle labeled clockwise, but in final position she sees it counterclockwise (or vice-versa). Since the bugs move continuously, there must either be some position where she is on the plane ABC, or some position where ABC do not determine a plane. But in the latter case they are collinear and thus they are coplanar with any other point, thus with D’s position in particular. In the next puzzle, IVT’s continuity hypothesis is missing. Skipping a Number At the start of the 2019 season WNBA star Missy Overshoot’s lifetime free throw percentage was below 80%, but by the end of the season it was above 80%. Must there have been a moment in the season when Missy’s free throw percentage was exactly 80%? 106 Solution: Let X(t) be the fraction of Missy’s free throws that have gone in, up to time t. This value jumps up or down each time Missy attempts a free throw, and is always a rational number. The IVT clearly does not apply, so your immediate reaction is probably “no”—there’s no reason why X(t) should have to be exactly 4/5 at some point. But it seems annoyingly difficult to actually construct a “history” for Missy that misses 80% (try it). We could certainly get her past 70%; perhaps she’s 2 for 3 at the start of the season, then sinks her next free throw, jumping from 66 2 3% directly to 75%. But 80% seems impossible to skip. What’s going on here? As is often the case, it pays to try simpler numbers. Can we get Missy past 50% without hitting it? Ah—the answer is no, because if H(t) is Missy’s number of hits up to time t and M(t) her number of misses, then the difference H(t) −M(t) starts off negative and ends positive. Since this difference is an integer which changes by only 1 when a free throw is attempted, it must (by applying integer-IVT) hit 0; at that moment, H(t) = M(t) and therefore Missy’s success rate is exactly 50%. Returning to 80%, we observe that at the moment when X(t) = .8, if there is one, we have H(t) = 4M(t). Now, let t0 be the first time at which Missy’s percentage hits or exceeds 80%. That moment was, obviously, marked by a successful free throw; in other words, at time t0, H(t) went up by one while M(t) stayed fixed. But H(t) can’t have skipped from below 4M(T) to above 4M(t); again by integer-IVT, it must hit 4M(t). So 80% can’t be skipped! Observe: (1) Our argument required that Missy went from below 80% to above it. In fact, she can easily skip 80% going down. (2) The argument uses the fact that at 80%, the number of hits will be an integer multiple of the number of misses. Thus, it works for any percentage that represents a fraction of the form (k−1)/k, and you can readily verify that any other fraction between 0 and 1 is skippable going up. Going down, it is the fractions of the form 1/k that can’t be skipped. Some processes are piecewise continuous, that is, continuous except at a discrete set of points. To apply IVT in such a case, you might have to “limit the damage” at jumping points. Splitting a Polygon A chord of a polygon is a straight line segment through the interior of the polygon that touches the polygon’s perimeter at the segment’s endpoints and nowhere else. Show that every polygon, convex or not, has a chord such that each of the two regions into which it divides the polygon has area at least 1/3 the area of the polygon. 107 Solution: Call the polygon P and scale it so that it has area 1. If P is convex we can always find a chord that cuts P into two pieces of area 1 2, as you, dear reader, now an expert with IVT, can readily prove. In fact, you can choose arbitrarily the chord’s angle to the horizontal. Just move a line at that angle across P; since P is convex the intersection of its interior with the line consists of a single chord. Behind that chord is a piece of P whose area changes continuously from 0 to 1. Try that when P is nonconvex, and things get messy: The line’s intersection with P’s interior may consist of several chords. Indeed, some polygons simply can’t be cut into two equal-area parts by a chord—see the figure below for an example. When P is nonconvex, we simplify things by first choosing an angle for our moving line L that doesn’t encounter any more trouble than it needs to. We do that by ensuring that L is not parallel to any line through two vertices of P. Then the worst that can happen as we move our chord is that it encounters some concave vertex; it can’t hit two vertices (or, therefore, any whole edge of P) at the same time. Suppose we’ve moved L across P but so far the area of the smallest piece into which P has been cut has not reached 1 3. If we hit a concave vertex v head-on (as pictured on the left side of the figure), we may be forced to split 108 the chord. At that point we have cut P into three pieces, with, say A (of area a) behind the chord and B and C (areas b and c, respectively) ahead. At least one of b and c (say b, as in the figure below) exceeds 1 3; we keep that part of the chord. If in fact 1 3 ≤b ≤2 3, we stop there; else we keep moving L forward. There is one other possibility: L hits v “from behind,” for example as pic-tured on the right, forcing the chord to suddenly jump in length. If the region ahead (C) is of area at least 1 3, we can stop here or plow ahead as before. The fun case is when, as pictured, a + c < 1 3: Then we must flip our chord over v and send it back the other way into B ! In this fashion we continue pushing our chord through P, always with more than area 2 3 ahead of us, until we have the desired split. The last (and toughest) puzzle in this chapter doesn’t look at all like a case for IVT, and even less like IVT’s three-dimensional analogue. But let’s see how it plays out. Garnering Fruit Each of 100 baskets contains some number (could be zero) of apples, some number of bananas, and some number of cherries. Show that you can collect 51 of those baskets that together contain at least half the apples, at least half the bananas, and at least half the cherries! Solution: If we had just apples, then we could clearly get half the fruit in only 50 baskets by just taking the 50 baskets with the most apples. Or we could divide the baskets into two groups of 50 any way we want, and just take the part with more apples. Or—more geometrically—we could line up the baskets, divide the line (possibly through the middle of a basket, maybe cutting apples in the basket) in such a way that exactly half the apples are on either side of the cut, and note that there at most 49 whole baskets on one side of the cut or the other. Take those 49 plus all of the cut basket to get your winning 50 baskets. (If no basket is cut, just take a side with at most 50 baskets.) With two fruits, we can’t guarantee half of each with 50 baskets out of 100; for example, one basket with just one apple in it, and the remaining 99 with one lonely banana in each, would require 51. But we can get 50 out of 99 in several ways. For example, we could arrange the baskets by decreasing number of apples, then take basket 1, and from each pair (2,3), (4,5), ..., (98,99) the basket with more bananas. This certainly gets at least half the bananas since it gets at least half from baskets 2 through 99, plus all in basket 1. And it gets at least half the apples since at worst it gets all the apples from baskets 1, 3, 5, 7, etc. and basket k has as many apples as basket k+1. (Note that this algorithm uses only order information about the numbers of apples or bananas in baskets, not the actual amount.) 109 Another way to do it is to space the baskets equally around a circle, in any order. Now consider the 99 different “arc sets” you get by taking 50 baskets in a contiguous arc from the circle. We claim that most (that is, at least 50) of the arc sets contain at least half the apples. Why? if they didn’t, then more than half of the complements of the arc sets would contain more than half of the apples. But the complements are arc sets of size 49, and each of those can be thought of as a subset of a different arc set of size 50 (say, the one you get by adding the next basket clockwise). Thus, if most of the 49-ers contained more than half the apples, the same must be true of the 50’s, a contradiction. But if at least 50 of the size-50 arc sets contain at least half the apples, and (by a similar argument) at least 50 of them contain at least half the bananas, then at least one arc set contains at least half of both fruits. It’s worth noting that this proof shows that even if the baskets are arbitrarily numbered, there is a list of just 99 of the 5×1028 subsets of size 50, one of which is guaranteed to have the desired property. It’s even more instructive to wonder if the circular arrangement of the baskets on the plane allows a straight-line cut, possibly right through the middle of two baskets and through some apples and bananas, that cuts the fruits exactly in two. If so, there are at most 48 baskets on one side or the other of the cut to which we can add the cut baskets to get 50 with at least half of each fruit. And this is the kind of thing we need to solve the problem when there are three (or more) types of fruit. The combinatorial methods don’t seem to generalize to the three-fruit case, so let’s try the geometric ones. Suppose we place the baskets in 3-space, and spread them out enough so that no plane cuts more than three baskets. With three degrees of freedom, we can find a plane that cuts the apples, bananas and cherries each exactly in half. With 100 baskets, there must be at most 48 48 whole baskets on one side or the other of the plane. Put them together with the cut baskets, and we have 51 that contain at least half the apples, half the bananas, and half the cherries. (If fewer than 3 baskets were cut, it only gets easier.) We’ve been a bit glib with this “degrees of freedom” business, but it really is true (and known usually as the Stone–Tukey Theorem) that in n-dimensional space, any n sets can be simultaneously bisected by an n−1-dimensional hyper-plane. The three-dimensional version is known as the ham-sandwich theorem and says that any sandwich consisting of a slice of ham and two slices of bread, however sloppily piled, can be cut with a planar cut that exactly halves the ham and each slice of bread. Such facts are usually proved using fancy mathematics (algebraic topology). Applying Stone–Tukey allows us to generalize to b baskets and n kinds of fruit; the result is there are always the greatest integer in b/2 + n/2 baskets with at least half of each fruit type. This is best possible, because if b = m1 + · · · + mn + x where all mi are odd and x = 0 or 1, then we can have mi boxes with nothing but one sample of fruit i, and possibly one empty box. To get half of each fruit type with this arrangement would cost us at least n/2 half-baskets beyond b/2. 110 Of the many theorems that employ IVT in their proofs, the most famous is the Mean Value Theorem of calculus, which says essentially that if you drive 60 miles in an hour then there must have been a moment when your speed was exactly 60 miles per hour. But we’re avoiding calculus in this book and we don’t need it here: A simple theorem about polynomials (simple compared to some of the above puzzles, for sure!) will do. For us a “polynomial in the variable x” is an expression p(x) of the form a0 + a1x + a2x2 + · · · + adxd where d is a non-negative integer, the coefficients ai are real numbers, and ad is not zero. The “degree” of p(x) is d, the exponent of the highest power of x having nonzero coefficient. Theorem. If a polynomial p(x) has odd degree, then it has a real root, that is, there is a real number r such that p(r) = 0. The oddness condition for the degree of p(x) is certainly necessary; for ex-ample, no real number x satisfies 1 + xd = 0 when d is an even non-negative integer. The proof is a snap if you’ve gotten this far in the chapter. The idea is to first assume that ad = 1 (dividing p(x) by ad does not affect its roots). Then, for large enough values of x, p(x) will be positive; for highly negative values of x, p(x) will be negative. Thus, since powers and thus polynomials are always continuous, IVT tells us there will be some x0 with p(x0) = 0. How big does x have to be to guarantee that p(x) > 0? If we let a = \|a0\|+\|a1\|+· · ·+\|ad−1\| be the sum of the absolute values of the other coefficients, then x > max(a, 1) is good enough. The reason is that then the powers of x are increasing, thus p(x) ≥xd −axd−1 = xd−1(x −a) > 0. Similarly, p(−x) ≤(−x)d + axd−1 = xd−1(a −x) < 0. This tells us a little more than the theorem, namely, that p(x) has a root in the union of the intervals [−a, a] and [−1, 1], where a = (\|a0\|+\|a1\|+· · ·+\|ad−1\|)/\|ad\|. 111 112 Chapter 4 Graphography Among the simplest of mathematical abstractions is the one which models ob-jects with points (called “nodes” or “vertices”) and indicates some relationship between a pair of nodes by the presence of a line (called an “edge”) from one to the other. Such a structure is called a graph, not be confused with the graph of a function like y = x2 −4. A graph is said to be connected if you can get from any node to any other by following a path of edges. It’s not hard to see that you need at least n−1 edges to connect n nodes; three ways to do that with five nodes are shown in Fig below: 113 A graph with the minimum number of edges to be connected is called a tree (we counted those at the end of Chapter 1). Note that a tree cannot have a cycle (a closed path with three or more nodes) because removal of an edge from a cycle cannot disconnect a graph. Thus a tree can also be described as a graph with the maximal number of edges to not have a cycle. Consider the following connectivity problem. Air Routes in Aerostan The country of Aerostan has three airlines which operate routes between various pairs of Aerostan’s 15 cities. What’s the smallest possible total number of routes in Aerostan if bankruptcy of any one of the three airlines still leaves a network that connects the cities? Solution: Of course our graph will have 15 nodes, one for each city, and we can add edges of three types (say, solid black, dashed black, and solid pink) to represent the routes of the three airlines. From the above remarks, it follows that any two of Aerostan’s airlines must operate at least 14 routes, and thus together the airlines must operate at least 21 routes. (Why? Add up the inequalities a + b ≥14, a + c ≥14, and b + c ≥14). Can 21 be achieved? If so, each airline will need exactly 7 routes, and it must be the case that every cycle of routes uses routes from all three airlines (otherwise bankrupting the missing airline would leave a cycle, wasting one of the 14 remaining edges). There are lots of ways to do this; perhaps the simplest is to pick a hub city, let Airline A connect the hub to a set of seven other cities, and let Airline B connect the same hub to the remaining seven cities. Airline C avoids the hub but connects A’s seven spoke cities pairwise to B’s. Below are two pictures of this graph. The second figure shows why this graph is known as a “windmill graph,” but it is also termed a “friendship graph” because if the nodes represent people and the edges friendships, the graph has the property that every two people have a unique friend in common. (There is only one connected graph with this property and a given odd number of nodes. If you want this property with an even number of nodes, you’re out of luck!) Spiders on a Cube Three spiders are trying to catch an ant. All are constrained to the edges of a transparent cube. Each spider can move one third as fast as the ant can. Can the spiders catch the ant? Solution: If our protagonists were on a tree, then one spider—even a super-slow old codger—could catch the ant by himself, just by moving steadily toward the 114 ant. With no cycles to dash around, the ant is doomed. We take advantage of this idea by using two of our three spiders to “patrol” one edge each, thereby reducing the edge-graph of the cube to a tree. To patrol an edge PQ, a spider chases the ant off the edge if necessary, then patrols the edge making sure he is at all times at most 1 3 the distance from P (respectively, Q) that the ant is. This is possible since the distance from P to Q along the edges of the cube, if you are not allowed to use the edge PQ, is three times the length of the edge. If we choose for our two controlled edges two opposite edges (some other choices are equally good), we find that with these edges—including their endpoints— removed, the rest of the edge-network (in black in the figure below) contains no cycles (so it’s just a collection of trees). It follows that the third spider can simply chase the ant to the end of a patrolled edge, where the ant will meet her sad fate. In a graph, the number of edges ending at a particular node is called the degree of the node. Since we’re not allowing edges from a node to itself, or multiple edges between the same pair of nodes, the degree of a node can never be more than the number of nodes in the graph, minus one. 115 Handshakes at a Party Nicholas and Alexandra went to a reception with ten other couples; each person there shook hands with everyone he or she didn’t know. Later, Alexandra asked each of the other 21 partygoers how many people they shook hands with, and got a different answer every time. How many people did Nicholas shake hands with? Solution: It seems implausible that Alexandra’s survey somehow tells us about Nicholas, but that’s what makes this a cute puzzle. Here our graph has 22 nodes and we put an edge between two if they shook hands during the party. The maximum possible degree of a node is only 20, since each person knows his or her partner, and of course the minimum is 0, so the numbers of handshakes recorded by Alexandra’s survey must exactly be the 21 numbers 0, 1, . . . , 20. The guest who shook all 20 hands of the other 10 couples must be the partner of the one who shook 0 hands (otherwise, the two extremists would have had to both shake hands and not shake hands). Similarly, the one who shook 19 hands (all but the 0-guest’s) must be the partner of the one who shook only 1 hand (only the 20-guest’s), and so on. This leaves Nicholas as the one who shook 10 hands, since he’s not partnered with anyone who was surveyed. As sometimes happens with puzzles, there’s a cheap way to get this answer, if you trust that the puzzle has a unique solution. You could have set up the graph so that an edge represents two non-coupled guests that did not shake hands. Whatever argument might have enabled you to conclude that Nicholas shook hands with n people in the original formulation would also allow you to conclude that he failed to shake hands with n people, other than Alexandra, in the second formulation. This is possible only if n = 10. 116 Snake Game Joan begins by marking any square of an n×n chessboard; Judy then marks an orthogonally adjacent square. Thereafter, Joan and Judy continue alternating, each marking a square adjacent to the last one marked, creating a snake on the board. The first player unable to play loses. For which values of n does Joan have a winning strategy, and when she does, what square does she begin at? Solution: If n is even, Judy has a simple winning strategy no matter where Joan starts. She merely imagines a covering of the board by dominoes, each domino covering two adjacent squares of the board. Judy then plays in the other half of each domino started by Joan. When n is odd, and Joan begins in a corner, Joan wins by imagining a domino tiling that covers every square except the corner in which she starts. However, Joan loses in the odd-n case if she picks her starting square badly, for example, if she begins at a square adjacent to a corner square. Suppose the corner squares are black in a checkerboard coloring, so that Joan’s starting square is white. There is a domino tiling of the whole board minus one black square; Judy wins by completing these dominoes. Joan can never mark the one uncovered black square because all the squares she marks are white! ♡ There’ll be more about this game at the end of the chapter. Bracing the Grid Suppose that unit-length rods, jointed at their ends, form an m × n grid of diamond-shaped cells. You may brace some subset S of the cells with diagonal segments (of length √ 2). Which choices of S suffice to make the grid rigid in the plane? 117 Solution: It helps to turn this into a graph-theoretical problem, but not in the most obvious way (with vertices as joints and edges as rods). Instead, suppose you have put in your braces; now envision a graph G whose vertices correspond to the rows of cells and the columns of cellss. Each edge of G corresponds to a row and column whose intersecting square is braced, so that the number of edges of G is the same as the number of braces you used. A partly-braced grid, with its associated graph, is pictured below. If a row and a column are adjacent in G, the vertical rods in the row are all forced to be perpendicular to the horizontal rods in the column. If G is a connected graph, meaning that there is a path from any vertex to any other, then all the horizontal rods must be perpendicular to all the vertical ones. Thus all the horizontal rods are parallel to one another, and similarly for the vertical ones, and now it is clear that the grid is rigid. On the other hand, suppose the graph is disconnected and let C be a “com-ponent,” that is, a connected piece of G which has no edges to the rest of G. Then there is nothing to prevent any vertical rod in a C-row or any horizontal rod in a C-column from flexing relative to the other rods in the grid. Thus, the criterion for rigidity is exactly that the graph G be connected. Since G has m + n vertices, it must have at least m+n−1 edges in order to be connected (you can prove this easily by induction if you haven’t seen it before), hence you must have at least m+n−1 braces to make the grid rigid. Notice, however, that they cannot be “just anywhere.” The next figure shows an efficiently braced 3 × 3 grid and its corresponding graph. For practice, you might want to compute the total number of ways to rigidify the 3 × 3 grid with the minimum number (five) of braces. A theorem of graph theory (to the effect that every connected graph has a connected subgraph, called a “spanning tree,” with the minimum number of edges) tells us that if more than m+n−1 squares are braced and the grid is rigid, then there are ways to remove all but m+n−1 of the braces and preserve rigidity. 118 Competing for Programmers Two Silicon Valley startups are competing to hire programmers. Hiring on alternate days, each can begin by hiring anyone, but every subsequent hire by a company must be a friend of some previous hire—unless no such person exists, in which case that company can again hire anyone. In the candidate pool are 10 geniuses; naturally each company would like to get as many of those as possible. Can it be that the company that hires first cannot prevent its rival from hiring nine of the geniuses? Solution: Yes. Imagine a friendship graph that contains a (non-genius) hub Hubert with 10 friends each of whom knows a different genius, plus one lonely friend (Lonnie) who knows only Hubert. That’s it; the geniuses (black vertices on the periphery) each have only one friend. 119 Suppose Company A goes first and hires Hubert. Then B hires Lonnie and after that, every time A hires a friend of Hubert, B hires that person’s genius friend. If A hires Lonnie, B hires a genius. A must hire Hubert and then, after B hires its genius’s friend, the game proceeds as before. If A hires a genius, B hires another genius; three days later A hires Hubert and then B hires Lonnie, after which the remaining geniuses go to B. Finally, if A hires a genius’ friend, B hires that genius. After A takes Hubert and B takes Lonnie, A winds up with no genius at all. Wires under the Hudson Fifty identical wires run through a tunnel under the Hudson River, but they all look the same, and you need to determine which pairs of wire-ends belong to the same wire. To do this you can tie pairs of wires together at one end of the tunnel and test pairs of wire-ends at the other end to see if they close a circuit; in other words, you can determine whether two wires are tied together at the other end. How many trips across the Hudson do you need to accomplish your task? Solution: Suppose you start at the west end by joining the ends of 48 of the wires into 24 pairs, leaving two “bachelors.” Now cross the river and test all pairs of wires to identify those that are linked on the west side. Label the bachelors 1 and 50, and the pairs (2, 3), (4, 5), . . . , (48, 49). Join wire #1 to wire #2, 3 to 4, etc., up to 47 and 48, leaving 49 and 50 alone. Cross back to the west side and disconnect the wires, making sure to mark them so you know which were previously paired. Next, test all pairs of wires to determine which are tied at the east end. Just one of the original bachelors is now paired, and that must be wire #1. It’s current mate is wire #2, and wire #2’s old mate is wire #3. Proceeding in this fashion, you correctly assign numbers up to #49, and the remaining original bachelor is wire #50. It thus took only two trips to get a correct labeling. But don’t forget, when it comes to actually using the wires, to disconnect the pairings at the east end. This procedure generalizes easily to any even number of wires greater than 2 (2 can’t be done without fancier equipment or another tester). Odd numbers are done in a similar way, but with only one bachelor. Two Monks on a Mountain Remember the monk who climbed Mt. Fuji on Monday and descended on Tues-day? This time, he and a fellow monk climb the mountain on the same day, starting at the same time and altitude, but on different paths. The paths go 120 up and down on the way to the summit (but never dip below the starting alti-tude); you are asked to prove that they can vary their speeds (sometimes going backwards) so that at every moment of the day they are at the same altitude! Solution: It is convenient to divide each path into a finite sequence of monotone “seg-ments” within which the path is always ascending, or always descending. (Level segments cause no problem as we can have one monk pause while the other traverses such a segment). Then we can assume that each such segment is just a straight ascent or descent, since we can have the monks modulate their rates so that their rate of altitude change is constant on any segment. (There are mathematical curves of finite length that cannot be divided into a finite number of monotone segments, but in our problem we needn’t worry about segments that are shorter than a monk’s step size.) Label the X-axis on the plane by positions along the first monk’s path, and the Y -axis by positions along the second monk’s path. Plot all points where the two positions happen to be at the same altitude; this will include the origin (where both paths begin) and the summit (where they end, say at (1,1)). Our objective is to find a path along plotted points from (0,0) to (1,1); the monks can then follow this path, slowly enough to make sure that no monk is anywhere asked to move faster than he can. Any two monotone segments—one from each path—which have some com-mon altitude show up on the plot as a (closed) line segment, possibly of zero length. If we regard as a vertex any point on the plot which maps back to a segment endpoint (for either or both monks), the plot becomes a graph (in the combinatorial sense); and an easy checking of cases shows that except for the vertices at (0,0) and (1,1), all the vertices are incident to either 0, 2 or 4 edges. Once we begin a walk on the graph at (0,0), there is no place to get stuck or be forced to retrace but at (1,1). Hence we can get to (1,1), and any such route defines a successful strategy for the monks. ♡ The figure shows four possible landscapes, with one monk’s path shown as a solid line, the other as dashed. Below each landscape is the corresponding graph. Note that, as in the last case, there may be detached portions of the graph which the monks cannot access (without breaking the common altitude rule). 121 Worst Route A postman has deliveries to make on a long street, to addresses 2, 3, 5, 7, 11, 13, 17, and 19. The distance between any two houses is proportional to the difference of their addresses. To minimize the distance traveled, the postman would of course make his deliveries in increasing (or decreasing) order of address. But our postman is overweight and would like to maximize the distance traveled making these de-liveries, so as to get the most exercise he can. But he can’t just wander around town; to do his job properly he is obligated to walk directly from each delivery to the next one. In what order should he make his deliveries? Solution: The “greedy algorithm” for maximizing the postman’s distance would have him start at #2, go all the way to #19, then back to #3, then to #17 and so forth, ending at #11. That would result in total distance 17+16+14+12+8+6+4 = 77 units. Is this the best he can do? Here’s a useful way to look at the problem. Call the addresses A, B, etc. up to H, in numerical order, and represent the trek from house X to house Y by XY . No matter what the postman does, his trip can be represented as a list of treks of the form AB, BC, CD, DE, EF, FG, and GH. How many times can the postman perform the trek AB? At most twice, once on his way to A and once back. A similar argument works for GH. How about BC? That one he might be able to do four times: To A and back (but not from B), and to B and back (not from A). Continuing this reasoning, we see that the postman cannot possibly do better than 2AB + 4BC + 6CD + 8DE + 6EF + 4FG + 2GH, which adds up to 86 units in our puzzle. Wait—the postman can’t actually do the DE trek eight times because he only makes seven trips between consecutive deliveries. So in fact he can do no better than 2AB + 4BC + 6CD + 7DE + 6EF + 4FG + 2GH, which is 82 units in our puzzle. Can he achieve that number? 122 Yes, he can, but only if he begins at D or E and ends at the other one, after zigzagging between low-numbered houses (A, B, C, D) and high (E, F, G, H). Example: D, H, C, G, B, F, A, E. There are 2 × 3 × 3 × 2 × 2 × 1 × 1 = 72 ways to do this, and each one gives the maximum possible length, 82 units. The given addresses don’t matter a bit, nor even the number of houses (pro-vided that remains even). The reasoning above is quite general, the conclusion being that the postman should always (a) begin at one of the middle two houses and end at the other, and (b) zigzag between the lower-half addresses and the upper-half addresses. For our theorem, let’s take another look at the Snake Game puzzle. We can play that game on any graph G: The rules are that the players alternate marking vertices, and after Player 1 (Joan) marks her first vertex, every subse-quent vertex must be unmarked and adjacent to the vertex just marked by the opponent. The first player unable to move loses. The proof above that the second player wins on an 8×8 chessboard actually shows that Player 2 wins on any graph with a perfect matching, that is, a set of disjoint edges that cover all the vertices (as the dominoes do on the chessboard). In fact, when G has a perfect matching, Player 2 wins even when Player 1 is allowed to flout the main rule of the game, marking any (unmarked) vertex instead of one adjacent to Player 2’s last move! The strategy is the same: Whenever Player 1 marks a vertex v, Player 2 simply marks the vertex on the other end of the matching edge that covers v. So you might think that having a perfect matching is a far stronger condition than is needed to assure a win for Player 2 in the original game. But no. Theorem. With best play, Player 2 wins on G if and only if G has a perfect matching. Proof. We’ve already proved the “if” part of the theorem, so let us suppose that G does not have a perfect matching, with the intent of providing a winning strategy for Player 1. The key is to pick some maximum-size matching M, that is, some collection of disjoint edges which is as large as possible. Since there is no perfect matching, there will be some vertex u1 of G that’s not in M, and we’ll let Player 1 start there. After that, Player 1 intends to follow the “matching strategy” from above, playing the other end of whatever matching edge Player 2 marks a vertex of. If Player 1 can always do that, she of course will win, but how do we know she can always do that? What if Player 2 marks a vertex that’s not in an edge of M? Well, Player 2 can’t immediately do that, because her first play (say, v1) must be adjacent to u1. If v1 were not in the matching, the edge (u1, v1) could have been added to M to make a bigger matching—and M was supposed to be as big as possible. And she can’t do it later, either. Suppose Player 1 has been able to execute her matching strategy through her kth move, and now must reply to Player 2’s kth move. The moves so far have been u1, v1, u2, v2, . . . , uk, vk; by assumption, 123 all these vertices are distinct, all the pairs (ui, vi) are edges of G, and all the pairs (vi, ui+1) are not only edges of G but also members of the matching M. But now we claim that Player 2’s latest move, vk, belongs to an edge of M. Why? Because if it didn’t, we could make M bigger by replacing the k−1 edges (v1, u2), (v2, u3), . . . , (vk−1, uk) by the k edges (u1, v1), (u2, v2), . . . , (uk, vk)— again contradicting the maximality of M. So Player 1 is never stuck for a move, and the theorem is proved! ♡ 124 Chapter 5 Algebra Too The single most useful technique for solving puzzles that ask for a quantity is simply to assign a letter (x and n are popular choices) to the quantity, write down what the puzzle tells you, and then solve using algebra. Our first example is simple, but people get it wrong all the time. Bat and Ball A bat costs $1 more than a ball; together they cost $1.10. How much does the bat cost? Solution: Did you say $1? Oops! Let x be the price of the bat, so the price of the ball is x −$1. Then we have the equation x + (x −$1) = $1.10, 2x = $2.10, x = $1.05. Note that we could have assigned another variable (y, perhaps) to the price of the ball, then solved two equations in two unknowns. If you can avoid that by expressing new unknowns in terms of old ones, you can save yourself a lot of trouble. Two Runners Two runners start together on a circular track, running at different constant speeds. If they head in opposite directions they meet after a minute; if they head in the same direction, they meet after an hour. What is the ratio of their speeds? Solution: This is simultaneously confusing and easy. Let L be the length of the track. Then, when going in opposite direcions, the runners meet when, together, they have run a distance L. When going the same direction, they meet when the difference of their achieved distances is L. 125 Thus, (s+t)L (s−t)L = 60, and solving gives s/t = 61/59. A classic: Belt Around the Earth Suppose the earth is a perfect sphere and a belt is tightened around the equator. Then the belt is loosened by adding 1 meter to its length, and it now sits the same amount off the ground all the way around. Is there enough room to slip a credit card underneath the belt? Solution: The circumference of the earth is about 40 million meters, but who cares? Call it C, and call the earth’s radius R, so that C = 2πR, thus R = C/(2π). When we lengthen the belt its circumference is C + 1 so the new radius is (C + 1)/(2π) = C/(2π) + 1/(2π) = R + 16 cm, approximately. There’s room for a stack of about 200 credit cards to pass under the belt. There are dozens of published puzzles along the lines of “John is twice as old as Mary will be when . . . ” that can be solved straightforwardly as above. Let’s try something a little different. Odd Run of Heads On average, how many flips of a coin does it take to get a run of an odd number of heads, preceded and followed by tails? Solution: First, we need to get a tail, since starting with odd run of heads won’t help us. If that takes time x on average, then, since flipping H doesn’t help, we have x = 1 + 1 2 · x giving x = 2. Now we need an odd run of heads; say that takes time y on average. If we flip another T (probability 1 2) or HH (probability 1 4), we’ve made no further progress; if we flip HT (probability 1 4) we’re done. Thus, y = 1 2 · (1 + y) + 1 4 · (2 + y) + 1 4 · 2 , giving y = 6. So in all it takes an average of x + y = 2 + 6 = 8 flips to get that run. Randomness enters into the next puzzle as well. This is a toughie but the general technique is the same. 126 Hopping and Skipping A frog hops down a long line of lily pads; at each pad, he flips a coin to decide whether to hop two pads forward or one pad back. What fraction of the pads does he hit? Solution: Let us label the lily pads with integers in order. One way to begin solving the problem is to compute the probability p that a frog starting at 1 regresses, at some point, to 0. To never regress, he must jump ahead (probability 1/2) to lily pad #3 and not subsequently regress three times from there (probability (1 −p3). Thus 1 −p = 1 2(1 −p3); divide by (1 −p)/2 to get 2 = 1 + p + p2, giving p = ( √ 5 −1)/2 ∼.618034, the familiar golden ratio. It seems awkward to try to compute the probability that the frog skips over a particular position (say, 1). You might want to start your calculations from the first time the frog hits 0, if it does, but it might already have hit 1. A better idea is to try to compute the probability that during a particular jump the frog leaps over a pad that he never has, and never will, hit. For that to happen, he must (a) be jumping forward at the moment, (b) never drop back from where he lands, and (c) not have reached the pad he is jumping over in the past. We may as well assume the lily pad the frog is currently passing over is #0. The key is to note that event (c) is an independent copy of event (b) if you reverse both space and time. Looking at the frog backwards in time, and regarding lower-numbered lily pads as progress, the frog appears to be behaving as before: Jumping two forward or one back with equal probability. Event (c) says that when he reaches −1, he must not later “drop back” to 0. The probability of these three events all occurring is thus 1 2 ·(1−p)·(1−p) = (1−p)2/2. We must not forget, however, that we have not calculated the prob-ability that 0 is skipped; only the probability that a particular jump carries the frog over a skipped pad. Since the frog travels, on the average, at rate 1 2, he is creating skipped lily pads at rate (1 −p)2 relative to his spatial progress. It follows that the fraction of pads he hits is 1 −(1 −p)2 = (3 √ 5 −5)/2 ∼0.854102. Whew! ♡ The next puzzle is an almost straightforward application of the algebraic technique, but a little extra cleverness is needed—as is often the case when the 127 resulting equation or equations are Diophantine, that is, they require integer solutions. Area–Perimeter Match Find all integer-sided rectangles with equal area and perimeter. Solution: There are two. Let x and y be the sides, with x ≥y. we need xy = 2x + 2y, equivalently xy −2x −2y = 0, but the left-hand side of the latter reminds us of the product (x−2)(y−2). Rewriting, (x−2)(y−2) = 4, so either x−2 = y−2 = 2 or x−2 = 4 and y−2 = 1. This gives x = y = 4, or x = 6 and y = 3. The remaining puzzles involve algebra in other ways than solving equations. Three Negatives A set of 1000 integers has the property that every member of the set exceeds the sum of the rest. Show that the set includes at least three negative numbers. Solution: Let a and b be the only two negative numbers in the set, and let the sum of all the numbers in the set be S. We are told that a > S −a, and b > S −b; adding the two inequalities gives a + b > 2S −a −b, thus a + b > S which is impossible unless there are other negative numbers in the set. Another way to see this is to note that the condition is equivalent to saying the smallest number exceeds the sum of the rest, which cannot occur if there are just two negative entries. Red and Black Sides Bored in math class, George draws a square on a piece of paper and divides it into rectangles (with sides parallel to a side of the square), using a red pen after his black one runs out of ink. When he’s done, he notices that every rectangular tile has at least one totally red side. Prove that the total length of George’s red lines is at least the length of a side of the big square. Solution: Why should this be true? Maybe each rectangle is long and thin and its only red side is a short one. Suppose we project all the horizontal red lines onto the bottom edge of the big square, to get a union H of line segments contained in that bottom edge. We then project the vertical red segments onto the left-hand edge of the big square, to get a union V of line segments in that edge. 128 We claim that either H or V is the full side of the square (and therefore the sum of the lengths of the red segments is at least the side-length of the square). Why? Suppose there is a point h on the bottom of the square not covered by H, and a point v on the left side not covered by V . Then the vertical line through h and horizontal line through v intersect at some point p inside a rectangle, and that rectangle can’t have a red edge. Recovering the Polynomial The Oracle at Delphi has in mind a certain polynomial p (in the variable x, say) with non-negative integer coefficients. You may query the Oracle with any integer x, and the Oracle will tell you the value of p(x). How many queries do you have to make to determine p? Solution: It only takes two queries, and the first is needed only to get a bound on the size of the polynomial’s coefficients. The Oracle’s answer to x = 1 (say, n) tells you that no coefficient can exceed n. Then you can send in x = n+1, and when you expand the Oracle’s answer base n+1, you have the polynomial! For example, suppose (rather conveniently) that p(1) < 10, so you know every coefficient is an integer between 0 and 9. You then send in the number 10 and if the oracle says p(10) = 3,867,709,884, you know that the polynomial is 3x9 + 8x8 + 6x7 + 7x6 + 7x5 + 9x3 + 8x2 + 8x + 4. Here’s a puzzle whose solution is, potentially, a useful strategy. Gaming the Quilt You come along just as a church lottery is closing; its prize is a quilt worth $100 to you, but they’ve only sold 25 tickets. At $1 a ticket, how many tickets should you buy? Solution: Clearly you should buy at least 1 ticket, assuming you are trying to maximize your expected gain, because then for $1 you have a 1/26 chance to win $100; you will gain on average almost $3 from the invesment. On the other hand, it would be a losing proposition (except to the church!) to buy 100 or more tickets, because then you could break even at best. So it looks like some intermediate number of tickets is right, but how many? You can try to maximize your expectation (x/(25 + x)) · $100 −$x from buying x tickets using calculus, if you happen to know your calculus, but what do you do in the (likely) event that the optimum value of x is not an integer? Often an effective way to tackle this kind of problem is to figure out whether you gain or lose when you increment your variable by 1. If you increase your 129 purchase from m tickets to m+1 tickets, your expected return goes up by $100((m + 1)/(m + 26) −m/(m + 25)) −1. This value goes steadily down and eventually becomes negative (meaning that you shouldn’t buy another ticket). Thus, if you can identify the least m for which this expression is negative, you’ll know that the right number of tickets to buy is m. That’s just algebra: You solve the (quadratic) equation $100((m + 1)/(m + 26) −m/(m + 25)) −1 = 0 to get m = 24.5025. Thus 25 is the least value of m with the property that you lose money by buying another ticket, and it follows that 25 is the best number of tickets to buy. Is it a coincidence that 25 was also the number of tickets sold before you got there? Actually, it is. And if you don’t care for solving quadratic equations? You can still get the answer pretty quickly by plugging your guesses for m into the expression $100((m + 1)/(m + 26) −m/(m + 25)) −1, until you zero in on the least m that makes the expression negative. Strength of Schedule The 10 teams comprising the “Big 12” college football conference are all sched-uled to play one another in the upcoming season, after which one will be declared conference champion. There are no ties, and each team scores a “win point” for every team it beats. Suppose that, worried about breaking ties, a member of the governing board of the conference suggests that each team receive in addition “strength of sched-ule points,” calculated as the sum of the win points achieved by the teams it beat. A second member asks: “What if all the teams end up with the same number of strength-of-schedule points?” What, indeed? Can that even happen? Solution: No. Of course, if strength-of-schedule points (SSP’s, for short) are only to be used for tie-breaking, we don’t mind if the SSP’s are the same but the numbers of win points are different. But it turns out that if all the teams have the same number of SSP’s, they must also all have the same number of win points. Why? If not, let b be the largest number of win points and s the smallest. Each team that won b games racked up at least bs SSP’s, because, at worst, all their wins came against teams that won only s games. Similarly, teams that won only s games got at most sb SSP’s. Whoops, that means both the b teams and the s teams got exactly sb SSP’s, and therefore that all of the b teams’ wins came against s teams and vice-versa. But that’s impossible unless there is only one b team and one s team, because otherwise there would be a game between two b teams or between two s teams. It follows that the b team won only once. But then, any third team would have to have beaten both the b team and the s team, outscoring the b team and 130 contradicting the definition of b. We conclude (since there are more than two teams in the Big 12) that if every team has the same number of SSP’s, then every team has the same score. That’s not possible in a conference with an even number of teams, because then each team would have to have won half its games, and the number of games it plays is odd. Round-robin tournaments are excellent puzzle fodder; here’s another. Two Round-robins The Games Club’s 20 members played a round-robin checkers tourament on Monday and a round-robin chess tournament on Tuesday. In each tournament, a player scored one point for each other player he or she beat, and half a point for each tie. Suppose every player’s scores in the two tournaments differed by at least 10. Show that in fact, the differences were all exactly 10. Solution: Let’s divide the club members into the good checkers players (who scored higher in the checkers tournament) and the good chess players. One of these groups must have at least 10 members; suppose there are k ≥10 good chess players, and that they scored a total of t ≥10k points higher in the chess tournament than in the checkers tournament. This difference must have come entirely at the expense of the good checkers players, because the number of points scored by the good chess players among themselves had to be the same in both tournaments—namely, k 2 , A.K.A. k(k− 1)/2. Since there are k(20−k) intergroup matches in each tournament, the most they can contribute to t is k(20 −k). So we have 10k ≤k(20 −k), k ≤10. But k was supposed to be at least 10, so we conclude that k = 10 and that all of the above inequalities must be equalities. 131 In particular, since t = 10k every chess player scored exactly 10 points higher in the chess tournament than in the checkers tournament. Moreover, since t = k(20 −k), every intergroup match in the chess tournament must have been won by the good chess player, and in the checkers tournament, by the good checkers player. Alternative Dice Can you design two different dice so that their sums behave just like a pair of ordinary dice? That is, there must be two ways to roll a 3, 6 ways to roll a 7, one way to roll a 12, and so forth. Each die must have six sides, and each side must be labeled with a positive integer. Solution: The unique answer is that the labellings of the two dice are {1, 3, 4, 5, 6, 8} and {1, 2, 2, 3, 3, 4}. Perhaps you came up with this answer by trial and error, which is a perfectly fine way of solving the problem. However, there is another way in this case, involving a simple example of a powerful mathematical tool called generating functions. The idea is to represent a die by a polynomial in the variable x, in which the coefficient of the term xk represents the number of appearances of the number k on the die. Thus, for example, an ordinary die would be represented by the polynomial f(x) = x + x2 + x3 + x4 + x5 + x6. The key observation is that the result of rolling two (or more) dice is repre-sented by the product of their polynomials. For instance, if we roll two ordinary dice, the coefficient of x10 in the product (namely, f(x)2) is exactly the number of ways of picking two terms from f(x) whose product is x10. These ways are x4 · x6, x5 · x5 and x6 · x4; and these represent the three ways to roll a sum of 10. It follows that if g(x) and h(x) are the polynomials representing our alter-native dice, then g(x) · h(x) = f(x)2. Now, polynomials, like numbers, have unique prime factorizations; the polynomial f(x) factors as x(x + 1)(x2 + x + 1)(x2 −x + 1). To get g(x) and h(x) to multiply out to f(x)2, we need to take each of these four factors and assign either one copy to each of g(x) and h(x), or two copies to one and none to the other. But there are some constraints: In particular, we can’t have a non-zero constant term in g(x) or h(x) (which would represent some sides labeled “0”) or any negative coefficient, and the coefficients must sum to 6 since we have six sides to label. The only way to do this (other than g(x) = h(x) = f(x)) is to have g(x) = x(x + 1)(x2 + x + 1)(x2 + x −1)2 = x + x3 + x4 + x5 + x6 + x8 and h(x) = x(x + 1)(x2 + x + 1) = x + 2x2 + 2x3 + x4, 132 or vice-versa. This may seem a bit like trial and error after all, but with this technique you can solve problems that are much more complex than this one. To begin with, you can invent alternatives for a pair of 8-sided dice labeled 1 through 8 (there are three new ways to do it), or for rolling three ordinary dice (many ways). It will not be a surprise to readers that thousands, literally, of theorems have been proved using algebra. Here’s one that uses just the elementary sort of algebra we have seen above. Have you ever wondered: What is the probability that your family surname will die out? Francis Galton posed this question in 1873 and an answer posted by the Reverend Henry William Watson led to a joint paper. Galton was interested in the longevity of aristocratic names, but he could (in more modern times) have been thinking equivalently of the preservation of a Y-chromosome. In Galton and Watson’s model of propagation, each individual independently has a random number of offspring, given by a fixed probability distribution: i offspring with probability pi. (In the case of tracking patrilineal surnames, we would count only male offspring.) We wish to know: Starting with (say) one individual, what is the “extinc-tion” probability x that the family name dies out? Of course this depends on the probability distribution; as you might expect, if the average number of offspring µ = ∞ X i=0 ipi is less than 1, then the name will always die out, but if it’s greater than 1, the name might live on forever. In the latter case, how do you determine x? Theorem. Let T be a Galton-Watson tree with offspring probabilities p0, p1, . . . , pk. Then the extinction probability x satisfies x = P pixi. Let’s do one example before we proceed to the proof. Suppose each individual has zero offspring with probability 0.1, one with probability 0.6, and two with probability .3. Then µ = 1.2 and we expect the tree will sometimes be finite (that is, die out) and sometimes infinite (continue forever). The equation in the theorem becomes x = 0.1 + 0.6x + 0.3x2, which reduces to 3x2 −4x + 1 = 0; factoring gives (3x −1)(x −1) = 0, so x = 1 or x = 1 3. Since the probability of extinction is not 1, it must be 1 3. Proof. We observe that the tree is finite if either the starting individual has no offspring, or he does have offspring but each child is himself the start of a finite tree. If x is the extinction probability, then the probability that each of i individuals leads only to a finite tree is xi, thus: x = k X i=0 pixi as claimed. ♡ 133 The equation in the theorem always has x = 1 as one solution, since the probabilities sum to one. If µ > 1, there will be one other solution—the correct one. If µ ≤1, except in the case where p1 = 1 and pi = 0 for all i ̸= 1, the only solution will be x = 1 and thus extinction is assured. 134 Chapter 6 Safety in Numbers Numbers are everywhere, and endlessly fascinating. You probably know that every positive integer can be expressed uniquely as the product of primes (a prime number being one that cannot be divided evenly by anything except itself and 1). Let’s use that and some other elementary observations to solve some puzzles, shall we? Broken ATM George owns only $500, all in a bank account. He needs cash badly but his only option is a broken ATM that can process only withdrawals of exactly $300, and deposits of exactly $198. How much cash can George get out of his account? Solution: Since both the deposit and withdrawal amounts are multiples of $6, George cannot hope to get more than $498 (the largest multiple of 6 under $500). Using “− 300” to indicate a withdrawal and “+ 198” a deposit, we see that George can get $6 out by − 300, + 198, − 300, + 198, + 198. Doing this 16 times gets $96 out, after which − 300, + 198, − 300 finishes the job. Subsets with Constraints What is the maximum number of numbers you can have between 1 and 30, such that no two have a product which is a perfect square? How about if (instead) no number divides another evenly? Or if no two have a factor (other than 1) in common? Solution: These puzzles can all be tackled the same way. The idea is to try to cover the numbers from 1 to 30 by bunches of numbers with the property that from each 135 bunch, you can only take one number for inclusion in your subset. Then, if you can construct a subset that consists of one choice from each bunch, you have yourself a maximum-sized subset. In the first case, fix any number k which is square-free (in other words, its prime factorization contains at most one copy of each prime). Now look at the set Sk you get by multiplying k by all possible perfect squares. If you take two numbers, say kx2 and ky2, from Sk, then their product is k2x2y2 = (kxy)2 so they can’t both be in our subset. On the other hand, two numbers from different Sk’s cannot have a square product since one of the two k’s will have a prime factor not found in the other, and that factor will appear an odd number of times in the product. Now, every number is in just one of these sets Sk—given n, you can recover the k for which n ∈Sk by multiplying together one copy of each prime that appears with an odd exponent in the factorization of n. Between 1 and 30, the choices for k (i.e. the square-free numbers) are 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 2×3, 2×5, 2×7, 2×11, 2×13, 3×5, 3×7, and finally 2×3×5: 19 in all. You can choose each k itself as the representative from Sk, so a subset of size 19 is achievable and best possible. To avoid one number dividing another evenly, note that if you fix an odd number j, then in the bunch Bj = {j, 2j, 4j, 8j, . . . }—that is, j times the powers of 2—you can only take one. If you take for your subset the top half of the numbers from 1 to 30, namely 16 through 30, you have got one from each Bj, and of course no member of this subset divides another evenly since their ratios are all less than 2. So the 15 numbers you get this way is best possible. Finally, to get a maximum-sized subset all of whose members are relatively prime, you naturally want to look at bunches consisting of all multiples of a fixed prime p. You can take p itself as the representative of its bunch, so you can do no better than to take as your subset all the primes below 30, plus the number 1 itself, for a total of 11 members. ♡ Cards from Their Sum The magician riffle-shuffled the cards and dealt five of them face down onto your outstretched hand. She asked you to pick any number of them secretly, then tell her their sum (jacks count as 11, queens 12, kings 13). You did that and the magician then told you exactly what cards you chose, including the suits! After a while you figured out that her shuffles were defective—they didn’t 136 change the top five cards, so the magician had picked them in advance and knew what they were. But she still had to pick the ranks of those cards carefully (and memorize their suits), so that from the sum she could always recover the ranks. In conclusion, if you want to do the trick yourself, you need to find five distinct numbers between 1 and 13 (inclusive), with the property that every subset has a different sum. Can you do it? Solution: If only you had a card worth 16, you’d be in fine shape: the powers of two, in particular 1, 2, 4, 8, and 16, have the desired property. In fact when you write a number in binary, the positions of the ones tell you exactly which powers go into the sum. For example, suppose that your victim reports a sum of 13; in binary that’s 1101, thus 8 + 4 + 1, so his cards must consist of an eight, a four and an ace. There are 25 = 32 ways to choose a subset of the five cards (if you include the null set, which has sum 0). Can we get them all to be different, without having a number bigger than 13? The first thing to realize is that big numbers are better: It’s easier to tell whether a big-numbered card was chosen or not. So let’s try making our set using the greedy algorithm: Start with the king and throw in the biggest cards possible without causing the same sum to come up twice. If you do that you’ll end up with K, Q, J, 9, 6; and that works! (It turns out there’s only one other way to do it, the same cards but with the 9 replaced by a 3). What about suits? I recommend 6 of spades, 9 of diamonds, jack of clubs, queen of hearts and king of spades. But pick anything that looks random and that you can remember. Then practice (including that fake riffle shuffle), and you’ll be the life of the party. Read Colm Mulcahy’s book (see Notes & Sources) for more great mathematical card tricks! Divisibility Game Alice chooses a whole number bigger than 100 and writes it down secretly. Bob now guesses a number greater than 1, say k; if k divides Alice’s number, Bob wins. Otherwise k is subtracted from Alice’s number and Bob tries again, but may not use a number he used before. This continues until either Bob succeeds by finding a number that divides Alice’s, in which case Bob wins, or Alice’s number becomes 0 or negative, in which case Alice wins. Does Bob have a winning strategy for this game? Solution: The key is that there are enough small numbers around so that by judicious use of them, Bob can find a divisor of Alice’s number. For example, Bob wins using 137 2,3,4,6,16,12 (or 6,4,3,2,5,12 or others). To see this, we consider the possibili-ties for Alice’s number modulo 12; in other words, the remainder when Alice’s number (let’s call it A) is divided by 12. Using the suggested sequence 2,3,4,6,16,12, Alice is immediately lost if her number is even, that is, if A ≡0, 2, 4, 6, 8, or 10 modulo 12. If A ≡5 or 11 mod 12, it will be 3 or 9 mod 12 after 2 is subtracted, allowing Bob to win on his second turn. If A ≡1 or 9, Alice will succumb at turn 3; if A ≡3, at turn 4. The only value mod 12 that Alice can start with that has survived so far is 1, which after 2, 3, 4, and 6 have been subtracted, is now 4 modulo 12. Subtracting 16 knocks that down to 0 modulo 12, and now Bob’s 12 spells doom for Alice. Of course, we must check that Bob’s guesses add up to at most 100, and they do (2+3+4+6+16+12=43). Prime Test Does 49 + 610 + 320 happen to be a prime number? Solution: Since the number is (29)2 +2·29 ·310 +(310)2 = (29 +310)2, it is a perfect square (so, it’s not prime). If you guessed correctly that the number was composite but tried to prove it by finding a small prime factor, you were in for a tough time, because 29 + 310 itself is prime—so our number has no prime factor less than 59,561. Yes, it is easy to forget that finding a divisor is not the only way to show a number is composite. Showing that it is a square (or a cube, etc.) is another; other, more sophisticated ways exist as well. Numbers on Foreheads Each of 10 prisoners will have a digit between 0 and 9 painted on his forehead (they could be all 2’s, for example). At the appointed time each will be exposed to all the others, then taken aside and asked to guess his own digit. In order to avoid mass execution, at least one prisoner must guess correctly. The prisoners have an opportunity to conspire beforehand; find a scheme by means of which they can ensure success. Solution: It is useful in many problems to introduce probability, even though none is present in the statement. Here, if we assume that the numbers painted on foreheads are chosen independently and uniformly at random, we see that no matter what he does, each prisoner has probability 1/10 of guessing correctly. Let the prisoners be numbered from 0 to 9. Since we want the probability that some prisoner guesses correctly to be 1, we need the 10 events “Prisoner k guesses correctly” to be mutually exclusive: In other words, no two can occur. 138 Otherwise, the probability of at least one success would be strictly less than 10( 1 10) = 1. To do this it would behoove us to separate the set of possible configurations into 10 equally likely scenarios, then have each prisoner base his guess on a dif-ferent scenario. This reasoning may already have led you to the easiest solution: Let s be the sum of the numbers on all the prisoners’ foreheads, modulo 10 (in other words, the rightmost digit of the sum). Now let Prisoner k guess that s = k, in other words, guess that his own number is k minus the sum of the numbers he sees, modulo 10. This will ensure that Prisoner s, whoever that may be, will be correct (and all others wrong). Rectangles Tall and Wide It is easy to check that a 6 × 5 rectangle can be tiled with 3 × 2 rectangles, but only if some tiles are horizontally oriented and others are vertically oriented. Show that this can’t happen if the big rectangle is a square—in other words, if a square is tiled with congruent rectangles, then it can be re-tiled with the same rectangles all oriented the same way. Solution: First, we claim that if the sides of a tile are not commensurable (that is, if their ratio is irrational) then a much stronger statement holds: Any tiling of any rectangle either has all tiles horizontal or all vertical. To see this, assume otherwise and draw a horizontal line across the square at the top, and imagine moving it downwards. Suppose that just below the top border of the square, our line cuts a tiles vertically and b horizontally. Then, except when it coincides with some tile boundary, our line must always cut a tiles vertically and b horizontally; for otherwise, if the tiles are x × y, we would have ax + by = cx + dy for (a, b) ̸= (c, d), giving x/y = (b−d)/(a−c), a rational ratio. If a or b is zero all the tiles are oriented the same way and we are done. If the parts of our line that cross tiles vertically never change, then the height of the big rectangle is both a multiple of x and of y, an impossibility. Thus somewhere our line reaches a point where some vertical crossings switch to horizontal while simultaneously some horizontal crossings switch to vertical, but that again is impossible because the distance from the top of the big rectangle to the place where this happens would then be a multiple of both x and y. If the ratio of the sides of a tile is rational, we may as well assume the sides have integer lengths a and b, and that the tiled square C is c × c. If there is no all-horizontal or all-vertical tiling, a and b do not both divide c. We may as well suppose that a does not divide c, and write c = ma+r, where 0 < r < a. We want to show that in this case there is no tiling at all. For example, there can’t be any tiling of a 10 × 10 square by 4 × 1 rectangles. We cover C with a grid of c2 unit squares S(i, j), numbered as in a matrix by pairs i, j, 1 ≤i ≤c, 1 ≤j ≤c. We now color all the squares S(i, j) for which 139 i −j ≡0 mod a. This colors all c unit squares on the main diagonal, plus all the squares on other diagonals a multiple of a diagonals away from the main diagonal; and it colors every ath square along each row or column. As a consequence, suppose R is any rectangle carved out of the grid. If either R’s height or R’s width is a multiple of a (or both) then R is “balanced,” in the sense that the colored squares in R comprise exactly the fraction 1/a of all the unit squares in R. Thus the c × ma rectangle U containing S(1, 1) is balanced, as are also the ma×c rectangle V containing S(1, 1) and their intersection, the ma×ma square W again containing S(1, 1). Let L be the little r × r square containing S(c, c), that is, diagonally opposite the S(1, 1) corner. The figure shows the a = 4, c = 10 case, with the rectangles U and V (and thus also the square W) outlined in heavy black. If \|R\| denotes the area of a rectangle R, then \|C\| = \|U\|+\|V \|−\|W\|+\|L\| and a similar statement applies to the numbers of colored squares in each of these rectangles. But L is not balanced: It containes r colored squares on its main diagonal, so has a fraction r/r2 = 1/r > 1/a of its squares colored. It follows that C is not balanced. But our tiles are balanced, thus they cannot tile C. ♡ Locker Doors Lockers numbered 1 to 100 stand in a row in the main hallway of Euclid Junior High School. The first student arrives and opens all the lockers. The second student then goes through and re-closes lockers numbered 2, 4, 6, etc.; the third student changes the state of every locker whose number is a multiple of 3, then the next every multiple of 4, etc., until the last student opens or closes only locker number 100. After the 100 students have passed through, which lockers are open? 140 Solution: The state of locker n is changed when the kth student passes through, for every divisor k of n. Here, we make use of the fact that divisors usually come in pairs {j, k} where j · k = n (including the pair {1,n}); so the net effect of students j and k on this locker is nil. The exception is when n is a perfect square, in which case there is no other divisor to cancel the effect of the √nth student; therefore, the lockers which are open at the end are exactly the perfect squares, 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100. ♡ For the next puzzle you might like to know a certain handy rhyme, attributed to the mathematician Nathan Fine but inspired by a lovely proof by the great, late Paul Erd˝ os: Chebyshev said it and I say it again There’s always a prime between n and 2n. This fact is known as Bertrand’s Postulate, proved by Chebyshev in 1852, with later (neater) proofs by Ramanujan and Erd˝ os. Factorial Coincidence Suppose that a, b, c, and d are positive integers, all different, all greater than one. Can it be that a!b = c!d? Solution: Assume that such numbers exist, say with a < c, thus b > d. Then c > 2, and by Bertrand’s Postulate, there is a prime p with c/2 < p < c. This p appears exactly d times in the prime factorization of c!d but either b times or not at all in the prime factorization of a!b. The contradiction shows that the desired quartet (a, b, c, d) does not exist. Even Split Prove that from every set of 2n integers, you can choose a subset of size n whose sum is divisible by n. Solution: Call a set “flat” if it sums to 0 modulo n. Let us note first that the statement we want to prove implies the following seemingly weaker statement: If S is a flat set of 2n numbers, then S can be split into two flat sets of size n. However, that in turn implies that any set of only 2n−1 numbers contains a flat subset of size n because we can add a 2nth number to make the original set flat, then apply the previous statement to split this into two flat subsets of size n. One of these (the one without the new number) will do the trick. 141 So all three of these statements are equivalent. Suppose we can prove the second for n = a and for n = b. Then if a set S of size 2n = 2ab sums to 0 mod ab, it is, in particular, flat with respect to a, and we can peel off subsets S1, . . . , S2b of size a which are also flat with respect to a. Each of these subsets Si has a sum we can write in the form abi. The numbers bi now constitute a set of size 2b which sums to 0 mod b, so we can split them into two sets of size b which are flat with respect to b. The unions of the sets Si in each part are a bipartition of the original S into sets of size ab which are ab-flat, just what we wanted. It follows that if we can prove the statement for n = p prime, then we have it for all n. Let S be a set of size 2p, with the idea of creating a p-flat subset of size p. How can we create such a subset? One natural possibility is to pair up the elements of S and choose one element from each pair. Of course, if we do that, it will behoove us to ensure that the elements in each pair are different mod p, so our choice will not be of Hobson’s variety. Can we do that? Yes, order the elements of S modulo p (say, 0 through p−1) and consider the pairs (xi, xi+p) for i = 1, 2, . . . , p. If xi were equivalent to xi+p mod p for some i, then xi, xi+1, . . . , xi+p would all be equivalent mod p and we could take p of them to make our desired subset. Now that we have our pairs, we proceed by “dynamic programming.” Let Ak be the set of all sums (mod p) obtainable by adding one number from each of the first k pairs. Then \|A1\| = 2 and we claim \|Ak+1\| ≥\|Ak\|, and moreover, \|Ak+1\| > \|Ak\| as long as \|Ak\| ̸= p. This is because Ak+1 = (Ak + xk+1) ∪(Ak + xk+1+p); thus if \|Ak+1\| = \|Ak\|, these two sets are identical, implying Ak = Ak + (xk+1+p −xk+1). This is impossible since p is prime and xk+1+p −xk+1 ̸≡0 mod p, unless \|Ak\| = 0 or p. Since there are p pairs, we must eventually have \|Ak\| = p for some k ≤p, hence \|Ap\| = p and, in particular, 0 ∈Ap. The statement of the puzzle follows. ♡ Factorials and Squares Consider the product 100! · 99! · 98! · · · · · 2! · 1!. Call each of the 100 factors k! a “term.” Can you remove one term and leave a perfect square? Solution: Perfect squares have some nice properties. For example, the product of any number of perfect squares is itself a perfect square: e.g., A2·B2·C2 = (A·B·C)2. Let’s call our big product N and observe that it’s not far from already being a product of perfect squares. The product of N’s first two terms, for example, is 100 · 99!2 (which happens to be a perfect square since 100 = 102). In fact, we could pair up all the terms and write N in the following form: 100 · 99!2 · 98 · 97!2 · 96 · 95!2 · · · · · 4 · 3!2 · 2 · 1!2, which is a perfect square 142 times the number M = 100 · 98 · 96 · · · · · 4 · 2. But we can rewrite M as 250 · 50 · 49 · 48 · · · · · 2 · 1 = (225)2 · 50!. So N is the product of a lot of perfect squares and the number 50!. It follows that if we remove the 50! term from N, we’re down to a perfect square. ♡ It’s time for our theorem. You probably know that √ 2 is irrational, that is, it cannot be written as a fraction a/b for any integers a or b. You might even know a proof, something like this: Suppose √ 2 is a fraction, and write that fraction in lowest terms as a/b. Then 2 = a2/b2, and therefore a2 = 2b2, thus a2 is even and so must be a; write a = 2k. Then b2 = 2k2 so b is also even, but this contradicts the assumption that a/b is in lowest terms. Theorem. For any positive integer n, √n is either an integer or irrational. Proof. The proof for n = 2 extends automatically to the case where n is a prime p: If √p = a/b, reduced to lowest terms, then p = a2/b2, so a is a multiple of p and thus a2 = (pk)2 is a multiple of p2, but then b2 is a multiple of p thus so is b, and we again contradict a/b being reduced to lowest terms. This doesn’t work when n is not prime, though, because we cannot deduce from a2 being a multiple of n that a is too. But wait, this does work when n is square-free, that is, it is a product of distinct primes; then, we can apply the argument for each prime separately. So we know that √n is either an integer or irrational when n is square-free. But every n is the product of a perfect square and a square-free number (the latter obtained by taking the product of all primes that appear an odd number of times in the factorization of n). So let’s write n = k2m, where m is square-free. Then √n = k√m and using the fact that an integer multiple of an integer is an integer, and a non-zero-integer multiple of an irrational number is irrational, we have our proof. ♡ With a little work you can extend the above to kth-roots of n, where k and n are integers greater than 1. 143 144 Chapter 7 The Law of Small Numbers It sounds silly, but many otherwise-very-smart people, given a puzzle with num-bers in it, treat those numbers as if underlined in a sacred text. This is math— you’re allowed to change the numbers and see what happens! If the numbers in a puzzle are dauntingly large, replace them with small ones. How small? As small as possible, without making the puzzle trivial; if that doesn’t give you enough insight, make them gradually bigger. Domino Task An 8×8 chessboard is tiled arbitrarily with 32 2×1 dominoes. A new square is added to the right-hand side of the board, making the top row length 9. At any time you may move a domino from its current position to a new one, provided that after the domino is lifted, there are two adjacent empty squares to receive it. Can you retile the augmented board so that all the dominoes are horizontal? Solution: Yes. Let T be the “snake” tiling obtained by placing four vertical tiles on the left column, three on the right column (missing the top and bottom squares), and filling in all but the lower right square with horizontal dominoes. Our goal is to create this tiling and then shift it to get the desired horizontal tiling of the chessboard. We construct the snake tiling from the top. Since the 32 dominoes cover 64 of the 65 squares of our extended board, there is always one uncovered square, which we call the “hole.” Suppose the dominoes covering the remaining squares in the hole’s row are not all horizontal; then we can move the vertical tile nearest the hole on either side to a horizontal position after shifting some horizontal dominoes. Let’s call this process “flattening.” Since flattening increases the number of horizontal dominoes, it must even-tually terminate with the hole on a row containing only horizontal dominoes. 145 But that must be the top row, since only the top row has an odd number of squares. By shifting the top-row dominoes and filling the top-left square with a ver-tical domino, we create the top row of the snake. Now we return to flattening (but not touching the vertical tile at the upper left); this winds up with the hole on the second row, and now we can shift to make that row match the snake. We repeat until the whole snake is created; flattening the snake concludes the whole process. How to find this curious algorithm? Try the problem first on a 4 × 4 board! Spinning Switches Four identical, unlabeled switches are wired in series to a light bulb. The switches are simple buttons whose state cannot be directly observed, but can be changed by pushing; they are mounted on the corners of a rotatable square. At any point, you may push, simultaneously, any subset of the buttons, but then an adversary spins the square. Is there an algorithm that will enable you to turn on the bulb in at most a fixed number of spins? Solution: Looking at a simpler version of this puzzle is crucial. Consider the two-switch version, where all you’ve got are two buttons on diagonally opposite corners of the square. Pushing both buttons will ascertain whether the two switches were both in the same state, since then the bulb will light (if it wasn’t already lit). Otherwise, push one button, after which they will be in the same state, and at worst one more operation of pushing both buttons will turn on the bulb. So three operations suffice. Back to the four-switch case. Name the buttons N, E, S, and W after the compass directions, although of course the botton you’re calling N now might be the button you will call E, W or S after a spin. Suppose that at the start, diagonally opposite switches (N and S, E and W) are in the same state—both on or both off. Then you can treat opposite pairs as a single button and use the two-button solution: push both pairs (i.e., all four switches); then one pair (which may as well be N-S); then both pairs again, and you’re done. So begin with those three operations; if the light doesn’t go on, then one or both of the opposite pairs must have been mismatched. Try flipping two neighboring 146 switches, say N and E, then going back through your three-move two-button solution. Then you’re fine if both pairs were mismatched. If not, push just one button; that’ll either make both opposite pairs match, or both mismatch. Run through the two-button solution a third time. If the bulb is still off, push N and E again and now you know both opposite pairs match, and a fourth application of the two-button solution will get that bulb turned on. In conclusion, pushing buttons NESW, NS, NESW, NE, NESW, NS, NESW, N, NESW, NS, NESW, NE, NESW, NS, NESW, will at some point turn the light on—fifteen operations. No sequence of fewer than 15 operations can be guar-anteed to work because there are 24 = 16 possible states for the four switches, and they all must be tested; you get to test one state (the starting state) for free. Seeing the solution for four buttons, you can generalize to the case where the number of buttons is any power of two; if there are 2k buttons, the solution will take, and require, 22k−1 steps. (When there are n buttons, they are located at the corners of a spinnable regular n-gon.) The puzzle is insoluble when the number of buttons, n, is not a power of 2. Let’s just prove that for three buttons, no fixed number of operations can guarantee to get that bulb on. (For general n, write n as m · 2k for some odd number m > 1; it is m which plays the role of 3 in what follows.) You may as well assume the switches are spun before you even make your first move. Suppose that before they are spun, the switches are not all in the same state. Then it is easy to check that no matter what move you planned, if you were unlucky with the spin, then after the spin and your move, the switches will still not all be in the same state. It follows that you can never be sure that you have ever had all the switches in the same state, so no fixed sequence of moves can guarantee to light the bulb. It’s curious that you can solve the problem for 32 buttons (albeit in about 136 years, at one second per operation), but not for just three buttons. Candles on a Cake It’s Joanna’s 18th birthday and her cake is cylindrical with 18 candles on its 18′′ circumference. The length of any arc (in inches) between two candles is greater than the number of candles on the arc, excluding the candles at the ends. Prove that Joanna’s cake can be cut into 18 equal wedges with a candle on each piece. Solution: The conditions give some assurance that the candles are fairly evenly spaced; one way to say that is that as we move around the circumference from some fixed origin 0, the number of candles we encounter is not far from the distance we have traveled. Accordingly, let ai be the arc-distance from 0 to the ith candle, numbered counterclockwise, and let di = ai −i. 147 We claim that for any i and j, di and dj differ by less than 1. We may assume i < j; suppose, for instance, that dj −di ≤−1. Then j−i−1 ≥aj −ai, but j−i−1 is the number of candles between i and j, contradicting the condition. Similarly, if dj −di ≥1, then di −dj ≤−1 and the same argument applies to the other arc, counterclockwise from j to i. So the “discrepancies” di all lie in some interval of length less than 1. Let dk be the smallest of these, and let ε be some number between 0 and dk, so that all the di’s lie strictly between ε and 1 + ε. Now cutting the cake at ε, ε+1, etc. gives the desired result. How are you supposed to find this proof? By trying two candles, then three, instead of eighteen. Lost Boarding Pass One hundred people line up to board a full jetliner, but the first has lost his boarding pass and takes a random seat instead. Each subsequent passenger takes his or her assigned seat if available, otherwise a random unoccupied seat. What is the probability that the last passenger to board finds his seat un-occupied? Solution: This is a daunting problem if you insist on working out what happens with 100 passengers; the number of possiblities is astronomical. So let’s reduce the number to something manageable. With two passengers it’s obvious that the probability that the second (i.e., last) get her own seat is 1/2. What about three passengers? It’s useful to number the seats according to who was supposed to sit there. If passenger 1 sits in seat 1, his assigned seat, then everyone will be get his or her own seat. If he sits in seat 3, then passenger 2 will get seat 2 and passenger 3 will get seat 1. Finally, if passenger 1 sits in seat 2, then whether passenger 3 gets seat 3 will depend on whether passenger 2 chooses seat 1 or seat 3. Altogether, the probability that passenger 3 gets her own seat is 1 3 + 1 3 · 1 2 = 1 2. Interesting! Is it possible that the answer is always 1/2? We notice that in the above analysis, the last passenger never ends up in seat 2. In fact, now that we think of it, we see that with n passengers total, the last passenger never ends up in seat i for 1 < i < n. Why? Because when passenger i came on board, either seat i was already taken, or it is taken now. Thus, seat i will never be available to the last passenger. The only seats that passenger n could end up in are seat 1 and seat n. We can’t yet conclude that the probability that passenger n gets seat n is 1/2—we still need to argue that seat 1 and seat n are equally likely to be available at the end. But that’s easy, because every time someone took a random seat, they were equally likely to choose seat 1 or seat n. Putting it another way, seat 1 and seat n were treated identically throughout the process; thus, by 148 symmetry, each has the same likelihood of being open when passenger n finally gets on board. Flying Saucers A fleet of saucers from planet Xylofon has been sent to bring back the inhab-itants of a certain randomly-selected house, for exhibition in the Xylofon Xoo. The house happens to contain five men and eightt women, to be beamed up randomly one at a time. Owing to the Xylofonians’ strict sex separation policy, a saucer cannot bring back earthlings of both sexes. Thus, it beams people up until it gets a member of a second sex, at which point that one is beamed back down and the saucer takes off with whatever it has left. Another saucer then starts beaming people up, following the same rule, and so forth. What is the probability that the last person beamed up is a woman? Solution: Let’s try some smaller numbers and see what happens. Obviously if the house is all men or all women, the sex of the last person beamed up will be determined. If there are equal numbers of men and women, then by symmetry, the probability that the last person beamed up is a woman would be 1/2. So the simplest interesting case is, say, one man and two women. In that case, if the man is beamed up first (probability: 1/3), the last one will be a woman. Suppose a woman is beamed up first; if she is followed by a man (who is then beamed back down), we are down to the symmetric case where the probability of ending with a woman is 1/2. Finally, if a second woman follows the first (probability 2 3 · 1 2 = 1 3), the man will be last to be beamed up. Putting the cases together, we get probably 1/2 that the last person beamed up is a woman. Is it possible that 1/2 is the answer no matter how many men and women are present, as long as there’s at least one of each? Looking more closely at the above analysis, it seems that the sex of the last person beamed up is determined by the next-to-last saucer—the one that 149 reduces the house to one sex. To see why this is so, it is useful to imagine that the Xylofonian acquisition process operates the following way: Each time a flying saucer arrives, the current inhabitants of the house arrange themselves in a uniformly random permutation, from which they are beamed up left to right. For example, if the inhabitants at one saucer’s arrival consist of males m1 and m3 and females f2, f3 and f5, and they arrange themselves “f3, f5, m1, f2, m3,” then the saucer will beam up f3, f5, and m1, then will beam m1 back down again and take off with just the females f3 and f5. The remaining folks, m1, m3, and f2, will now re-permute themselves in anticipation of the next saucer’s arrival. We see that a saucer will be the next to last just when the permutation it encounters consists of all men followed by all women, or all women followed by all men. But no matter how many of each sex are in the house at this point, these two events are equally likely! Why? Because if we simply reverse the order of a such a permutation, we go from all-men-then-all-women to all-women-then-all-men, and vice-versa. There’s just one more observation to make: If both men and women are present initially, then one saucer will never do, thus there always will be a next-to-last saucer. When that comes—even though we do not know in advance which saucer it will be—it is equally likely to depart with the rest of the men, or the rest of the women. Gasoline Crisis You need to make a long circular automobile trip during a gasoline crisis. In-quiries have ascertained that the gas stations along the route contain just enough fuel to make it all the way around. If you have an empty tank but can start at a station of your choice, can you complete a clockwise round trip? Solution: Yes. The trick is to imagine that you begin at station 1 (say) with plenty of fuel, then proceed around the route, emptying each station as you go. When you return to station 1, you will have the same amount of fuel in your tank as when you started. As you do this, keep track of how much fuel you have left as you pull into each station; suppose that this quantity is minimized at station k. Then, if you start at station k with an empty tank, you will not run out of fuel between stations. ♡ Coins on the Table One hundred quarters lie on a rectangular table, in such a way that no more can be added without overlapping. (We allow a quarter to extend over the edge, as long as its center is on the table.) 150 Prove that you can start all over again and cover the whole table with 400 quarters! (This time we allow overlap and overhang). Solution: Let us observe first that if we double the radius (say, from 1′′ to 2′′) of each of the original coins, the result will be to cover the whole table. Why? Well, if a point P isn’t covered, it must be 2′′ or more from any coin center, thus a (small) coin placed with its center at P would have fit into the original configuration. (See the first two figures below for an example of an original configuration, and what happens when the coins are expanded.) Now, if we could replace each big coin by four small ones that cover the same area, we’d be done—but we can’t. 151 But rectangles do have the property that they can be partitioned into four copies of themselves. So, let us shrink the whole picture (of big coins covering the table) by a factor of two in each dimension, and use four copies (as in the next figure) of the new picture to cover the original table! Coins in a Row On a table is a row of 50 coins, of various denominations. Alix picks a coin from one of the ends and puts it in her pocket; then Bert chooses a coin from one of the (remaining) ends, and the alternation continues until Bert pockets the last coin. Prove that Alix can play so as to guarantee at least as much money as Bert. Solution: This puzzle resists the most obvious approaches. It’s easy to check that Alix could do quite badly by always choosing the most valuable coin, or the coin that exposes the less valuable coin to Bert, or any combination of these. Basically, if she only looks a move or two ahead, she’s in trouble. In fact, for Alix to play optimally, she needs to analyze all the possible situations that may later arise. This can be done by a technique called “dynamic programming.” But we were not asked to provide an optimal strategy for Alix, just a strategy that guarantees her at least half the money. Experimenting with 4 or 6 coins instead of 50 might lead you to the following key observation. Suppose the coins alternate quarter, penny, quarter, penny, and so forth, ending (since 50 is even) in a penny. Then Alix can get all the quarters! In fact, no matter what the coins are, if we number the coins from 1 to 50 left to right, Alix can take all the odd-numbered ones—or all the even-numbered ones. 152 But wait a minute—one of those two groups of coins must contain at least half the money! ♡ Powers of Roots What is the first digit after the decimal point in the number ( √ 2 + √ 3) to the billionth power? Solution: If you try entering ( √ 2 + √ 3)1,000,000,000 in your computer, you’re likely to find that you get only the dozen or so most significant figures; that is, you don’t get an accurate enough answer to see what happens after the decimal point. But you can try smaller powers and see what happens. For example, the decimal expansion of ( √ 2 + √ 3)10 begins 95049.9999895. A bit of experimen-tation shows that each even power of ( √ 2 + √ 3) seems to be just a hair below some integer. Why? And by how much? Let’s try ( √ 2 + √ 3)2, which is about 9.9. If we play with 10 −( √ 2 + √ 3)2 we discover that it’s equal to ( √ 3 − √ 2)2. Aha! Yes, ( √ 3 + √ 2)2n + ( √ 3 − √ 2)2n is always an integer, because when you expand it, the terms with odd powers cancel and the terms with even powers are integers. But of course ( √ 3 − √ 2)2n is very small, about 10−n, so the first roughly n digits of ( √ 3 + √ 2)2n after the decimal point are all 9’s. Coconut Classic Five men and a monkey, marooned on an island, collect a pile of coconuts to be divided equally the next morning. During the night, however, one of the men decides he’d rather take his share now. He tosses one coconut to the monkey and removes exactly 1/5 of the remaining coconuts for himself. A second man does the same thing, then a third, fourth, and fifth. The following morning the men wake up together, toss one more coconut to the monkey, and divide the rest equally. What’s the least original number of coconuts needed to make this whole scenario possible? Solution: You can solve this by considering two men instead of five, then three, then guessing. But the following argument is irresistible, once found. There’s an elegant “solution” to the puzzle if you allow negative numbers of coconuts(!). The original pile has −4 coconuts; when the first man tosses the monkey a coconut, the pile is down to −5 but when he “takes” 1/5 of this he is actually adding a coconut, restoring the pile to -4 coconuts. Continuing this way, come morning there are still -4 coconuts; the monkey takes one and the men split up the remaining -5. It’s not obvious that this observation does us any good, but let’s consider what happens if there is no monkey; each man just takes 1/5 of the pile he 153 encounters, and in the morning there’s a multiple of 5 coconuts left that the men can split. Since each man has reduced the pile by the fraction 4/5, the original number of coconuts must have been a multiple of 56 (which shrinks to 45 · 5 by morning). All we need to do now is add our two pseudo-solutions, by starting with 56 −4 = 15, 621 coconuts. Then the pile reduces successively to 4 · 55 −4 coconuts, 42 · 54 −4, 43 · 53 −4, 44 · 52 −4, and 45 · 5 −4. When the monkey gets his morning coconut, we have 45 · 5 −5 coconuts, a multiple of 5, for the men to split. This is best possible because we needed 55 ·k −4 coconuts to start with, just to have an integer number come morning, and to get 45 · k −5 to be a multiple of 5 we needed k to be a multiple of 5. Doubtless, many theorems in mathematics were “discovered” when someone played around with small numbers and then saw a pattern that turned out to be a provable phenomenon. Here’s a theorem that could well have been found that way. Suppose you are running a dojo with an even number n of students. Each day you pair the students up for one-on-one sparring. Can you do this in such a way that over a period of days, each student spars with each other student exactly once? Theorem. . For any even positive integer n there is a set of pairings (“perfect matchings”) of the numbers {1, 2, . . . , n} such that every pair {i, j} appears in exactly one pairing. A check of small numbers suggests that this seems to work: For n = 2, for instance, there is just the one pairing consisting of the pair {1, 2}, and for n = 4 we can (in fact, must) take the pairings {{1, 2}, {3, 4}}, {{1, 3}, {2, 4}}, and {{1, 4}, {2, 3}}. For n = 6, though, we have choices to make. Is there a nice way to make them? In fact, there are several; my favorite is the following. We know student n has to be paired with every other student; let’s put her in the middle of a circle, with the rest of the students spaced equally around the circle. In the ith of our n−1 pairings, student n is paired with student i; draw a radius from n to i. The rest of the students are paired by line segments that run perpendicular to that radius (see the figure below). Since n−1 is odd, all the radii from n to other students are at different angles; it follows that no two students are paired twice, and since the eventual number of pairs is (n−1) × n/2 = n 2 , every pair is accounted for. ♡ 154 155 156 Chapter 8 Weighs and Means You know how to compute the average of a set of numbers ¯ x := (x1 + x2 + · · · + xn)/n or you wouldn’t even have picked up this book. In solving puzzles, it’s often useful to reverse this process, and compute the sum from the average instead. For example, have you forgotten the formula for the sum of the integers from 1 to n? No problem. Their average is (n+1)/2 (you can pair them up to see that) and there are n of them, so their sum must be n(n+1)/2. Similarly, the sum of the integers from a to b must be (b−a+1)(a+b)/2. Here’s a theorem about averages that’s helpful in problem-solving and is almost too obvious to call a theorem: Theorem. If ¯ x is the average of {x1, x2, . . . , xn} then ¯ x lies between the min-imum of the xi’s and the maximum. Moreover, the minimum, maximum and average are either all different or all the same. Another way of saying the first part is that some xi must be larger than or equal to ¯ x, and some xi must be smaller than or equal to ¯ x. Obvious, but useful! The above theorem applies also in the more general case of weighted aver-ages, also known as convex combinations. There, each element of the set to be averaged has some weight (a positive real number) wi, and ¯ x := (w1x1 + w2x2 + · · · + wnxn)/(w1 + · · · + wn). Often the weights already sum to 1, so you can skip the dividing. That happens, for example, when the wi’s constitute a probability distribution, in which case ¯ x becomes the expected value of the xi’s. (Much more about expectation is coming in a later chapter.) The following puzzle is fairly typical of those in which the above theorem can come in handy. 157 Tipping the Scales A balance scale sits on the desk of the science teacher, Ms. McGregor. There are weights on the scale, which is currently tipped to the right. On each weight is inscribed the name of at least one pupil. On entering the classroom, each pupil moves every weight carrying his or her name to the opposite side of the scale. Must there be a set of pupils that Ms. McGregor can let in that will tip the scales to the left? Solution: To set ourselves up for averaging, consider all subsets of students, including the empty set and the full set. Each weight will be on the left exactly half the time (pick any name on that weight, say Jay, and assume the decision of whether to include Jay in the set is the last decision to be made.) Thus the total weight on the left for all these subsets is the same as the total weight on the right. Another way to say it is that on the average, the scale balances. Since the empty set results in a tip to the right, some other set must tip to the left. The “averaging” technique used in Tipping the Scales comes up often: Watch for it! Here’s a quickie. Men with Sisters On average, who has more sisters, men or women? 158 Solution: This puzzle can seem confusing. If, like me, you were raised in a family with both male and female children, you might have observed that in such a family the boys have more sisters (by one) than the girls have. So maybe men have more sisters than women on average? Wait, in a family with several children, all girls, there are several girls with sisters; in a family of all boys, none of those boys has a sister. So maybe that cancels out the effect of the families with children of both sexes. The real issue here is independence of gender. If you make the reasonable assumption that the genders of siblings are not influenced by your own gender, then knowing the numbers of brothers and sisters a person has tells you nothing about whether that person is male or female. It follows that the reverse is true too, thus men and women have the same number of sisters on average. If you want to get down to detail, there’s actually a slight positive correlation of gender between siblings; that is, the sibling of a girl is slightly more likely to be a girl than the sibling of a boy is. This is caused mostly by the fact that identical twins are the same sex. As a result, women have very slightly more sisters on average than men do. The next puzzle uses the continuum form of our theorem. Watches on the Table Fifty accurate watches and a tiny diamond lie on a table. Prove that there exists a moment in time when the sum of the distances from the diamond to the ends of the minute hands exceeds the sum of the distances from the diamond to the centers of the watches. Solution: Considering just one watch, we claim that during the passing of an hour the average distance from the diamond D to the tip M of the watch’s minute hand exceeds the distance from D to the center W of the watch. This is so because if we draw a line L through D perpendicular to the line from D to W, then the average distance from L to M is clearly equal to LW which is in turn equal to DW. But DM is at least equal to LM and usually bigger. Of course, if we average over all the watches we reach the same conclusion, and it follows that sometime during the hour the desired inequality is achieved. The requirement that the watches be accurate is to ensure that each minute hand moves at constant speed. It doesn’t matter if some watches are slow and some fast (assuming each minute hand rotates at constant speed), unless our patience is limited to one hour. One additional note: If the watches are accurate and you place and rotate them carefully, you can ensure that the sum of the distances from the diamond to the ends of the minute hands is always strictly greater than the sum of the distances from the diamond to the centers of the watches. 159 Another nice property of averaging is that adding or subtracting one datum from a list has a predictable effect on the average: If the datum exceeds the average, then adding it will raise the average and deleting it will lower the average. If the datum is below average, the effect will be the opposite; and of course if the datum is exactly average, there is no effect either way. The next puzzle uses this idea to examine a frequently-heard type of insult. Raising Art Value Fans of Gallery A like to say that last year, when Gallery A sold to Gallery B “Still Life with Kumquats,” the average value of the pictures in each gallery went up. Assuming that statement is correct, and that the value of each of the 400 pictures in the galleries’ combined holdings is an integral number of dollars, what is the least possible difference between the average picture values of Gallery A and Gallery B before the sale? Solution: It’s a bit curious that we are asked for a dollar amount when no quantity of money is specified in the puzzle, but let’s plunge in anyway. Say there were p pictures in Gallery A before the sale, thus 400 −p in Gallery B. Suppose the 160 average value of A’s pictures was a, and B’s b; then the behavior of the averages says exactly that the value s of “Still Life with Kumquats” satisfies a > s > b. We know that a is a fraction that can be written as m/p where m is a whole number (of dollars); similarly b can be written as n/(400−p) where n is a whole number. By how little can a exceed an integer? Easy: By 1/p, which happens when m is one more than a multiple of p. Similarly, b’s shortfall from an integer value can be as little as 1/(400 −p). Thus the least possible difference between a and b is 1/p + 1/(400 −p) which is minimized when the two denominators are equal, giving a−b = 1/200+1/200 = 1/100. The upshot is that the insult is not very damaging; we can conclude from it only that the average value of Gallery A’s holdings exceeds the average value of Gallery B’s holdings by at least one cent. Averaging comes up a lot in probability theory—whole chapters on proba-bility and on expectation lie ahead. But we can do the following two averaging puzzles without needing any special machinery. Waiting for Heads If you flip a fair coin repeatedly, how long do you have to wait, on average, for a run of five heads? Solution: Since the probability of seeing HHHHH in a particular series of five coinflips is 1/32, you might think it would take 32 flips on average to get HHHHH. Indeed, 32 flips is the average wait between occurrences of HHHHH, but this includes, for example, a wait of length 1 between the first five heads in HHHHHH and the last five. Waits of length 1 can’t help us, because we have no “head start” (OK, pun intended) when we begin flipping. The real answer is much greater. Between runs, half the time you get the wait of 1 and the rest of the time 1+x, where x is the desired quantity. Hence it is not x but the average of 1 and 1+x that is equal to 32, which gives us x = 62. What are best patterns (of fixed length, say 5) to choose if you want to hit your pattern ASAP? HHHHT is one of them, because seeing an HHHHT gives you no head start toward seeing another. Therefore, starting fresh with your coinflips costs you nothing, and you expect to hit your target in just 32 steps on average. Finding a Jack In some poker games the right of first dealer is determined by dealing cards face-up (from a well-shuffled deck) until some player gets a jack. On average, how many cards are dealt during this procedure? 161 Solution: 10.6. If you tried to compute for each k the probability that the first jack would be the kth card, with the intent of computing the expected value of k directly, you might have found it tough going. But there’s an easy way. Each non-jack is equally likely to be in any of the five regions between successive jacks (to see this, insert a joker, permute circularly at random, then cut at the joker and remove it; or, get a shuffled deck by starting with the jacks, and inserting cards at random, so the next card is equally likely to be in any of the five slots.) Thus the expected number of cards before the first jack is 48/5 = 9.6; add 1 for the jack itself. Can we talk about the average of infinitely many things? Sometimes. Sums of Two Squares On average, how many ways are there to write a positive number n as the sum of two squares? In other words, suppose n is a random integer between 1 and a zillion. What is the expected number of ordered pairs (i, j) of integers such that n = i2 + j2? Solution: Let f(n) be the number of ways to write n as the sum of two squares. What we can’t do is add up f(n) for all integers n and then divide by (countable) infinity. But, as suggested in the puzzle, we can choose some huge number Z, and compute 1 Z Z X n=1 f(n) . If this value converges to some number r as Z →∞, it seems reasonable to call this r the average value of f(n) over all n. And in this case it does converge. Why? The sum PZ n=1 f(n) counts once each pair (i, j) of integers for which i2 + j2 ≤Z; these are the integer points in the circle x2 + y2 ≤Z. This circle has radius √ Z thus area πZ, and therefore the number q of integer points in it is about πZ (more precisely, q/(πZ) →1). It follows that the number we want, q/Z, approaches π. It is worth remarking that if we count just the unordered pairs of positive integers whose squares sum to n, the answer is only π/8. The reason is that most such pairs—in particular, pairs {i, j} for which i ̸= j and neither i nor j is zero—show up eight times in the circle, namely as (i, j), (j, i), (i, −j), etc. We might also ask for the probability that a random n can be written in at least one way as the sum of two squares. We can deduce from the above calculation that this probability cannot exceed π/8, which is what the answer would be if most numbers could be written as the sum of two squares in at most one way, up to signs and swapping. But this is not the case at all, and in fact the probability in question approaches zero! A famous theorem says 162 that n is representable as the sum of two squares only if there is no prime which (a) appears an odd number of times in the prime factorization of n, and (b) gives a remainder of 3 when divided by 4. The probability that n passes this test is about 0.764223653589220662990698731250092328116790541 (the Landau-Ramanujan constant) divided by the square root of the natural log of n, and since log n grows without bound, we arrive at a curious fact: Although the average of our function f(n) is π, its value is nearly always zero. Increasing Routes On the Isle of Sporgesi, each segment of road (between one intersection and the next) has its own name. Let d be the average number of road segments meeting at an intersection. Show that you can take a drive on the Isle of Sporgesi that covers at least d road segments, and hits those segments in strict alphabetical order! Solution: This puzzle restates a very general, yet surprisingly little-known, theorem of graph theory. Recall that a graph is a (finite, for us) set of vertices together with some pairs of vertices, known as edges. The degree of a vertex is the number of edges it belongs to; if we sum up the degrees of all the vertices each edge is counted twice, thus the average degree of the graph is 2m/n, where m is the number of edges and n is the number of vertices. A walk of length k is a sequence of k edges, each one connecting the end-vertex of the previous edge (if there is one) to the start-vertex of the next one. In a walk, as opposed to a “path,” a vertex may be used more than once. If the edges of the graph are ordered in some fashion, then a walk is increasing if the edges are encountered in strict increasing order—in particular, the edges of an increasing walk must be distinct. The figure below shows an increasing walk in a graph with edges ordered by letters. Here, then, is the claimed theorem. Theorem. Let G be an arbitrary graph with an arbitrary ordering of its edges. Then there is an increasing walk in G of length at least the average degree of G. Proof. We denote by d (as in the puzzle statement) the average degree of G. Since d is already an average, it’s natural to look for some set of increasing walks whose average length is d; then we can apply our theorem to deduce that one of them has length at least d. But where do we get such a set of walks? The set of all walks doesn’t work. Here’s an (admittedly brilliant) idea. Place a pedestrian at each vertex, and at time 1, let the pedestrians at each end of the first edge traverse that edge, thus changing places. At time 2, the pedestrians who now find themselves at the ends of the second edge change places, and so forth, until all m edges have been traversed. 163 Now, observe: Each pedestrian has taken an increasing walk! The total length of all these walks is 2m and there are n of them, so their average length is d = 2m/n and we are done. ♡ It’s worth noting that this theorem is tight, that is, there are graphs with edges ordered so that no increasing walk exceeds d in length. For example, let K2t denote the complete graph (all edges present) on 2t vertices. A complete matching in K2t (or in any graph) is a set of edges whose endpoints hit every vertex exactly once. The theorem we proved in the last chapter shows that the edges of K2t can be partitioned into disjoint complete matchings (necessarily, 2t−1 of them, since every vertex is incident to 2t−1 edges.) Now, suppose we let the matchings be M1, . . . , M2t−1 and we use them to order the edges as follows. First, the edges of M1 are labeled 1 through t arbitrarily. Then the edges of M2 are labeled with t+1 through 2t, those of M3 from 2t+1 through 3t, etc. A walk in K2t cannot step directly from one edge to another of the same matching, because no two matching edges meet at a vertex. But an increas-ing walk can’t return later to the same matching, on account of the way we have numbered the edges. So an increasing walk hits each of the matchings M1, . . . , M2t−1 at most once, thus can have length at most 2t−1 = d. 164 Chapter 9 The Power of Negative Thinking Contradiction is a ridiculously simple, yet somehow incredibly powerful, tool. You want to prove statement A is true? Assume it isn’t, and employ reductio ad absurdum: Derive an impossible conclusion. Voila! The great mathematician Godfrey H. Hardy put it nicely: “Reductio ad ab-surdum, which Euclid loved so much, is one of a mathematician’s finest weapons. It is a far finer gambit than any chess play: A chess player may offer the sacrifice of a pawn or even a piece, but a mathematician offers the game.” In fact, our first application of contradiction is to figure out who wins a certain game even though we didn’t know how to play it well—and still don’t. Chomp Alice and Bob take turns biting off pieces of an m×n rectangular chocolate bar marked into unit squares. Each bite consists of selecting a square and biting off that square plus every remaining square above and/or to its right. Each player wishes to avoid getting stuck with the poisonous lower-left square. Show that, assuming the bar contains more than one square, Alice (the first to play) has a winning strategy. 165 Solution: It’s easy to see that Alice can win if the bar is a square, as she can bite off all but the left-hand and bottom edges, then when Bob bites off a piece of one leg she bites off the same amount of the other. But how does she win in general? The surprising answer is: No one knows! But we do know that either the first player (Alice) or the second (Bob) must have a winning strategy; suppose it is Bob. Then, in particular, Bob must have a winning answer to Alice’s opening move when she merely nibbles off the upper right-hand square. However, whatever Bob’s reply is could have been made by Alice as her own opening move, after which she could follow Bob’s allegedly winning strategy and chalk up the game. This of course contradicts the assumption that Bob can always win. Hence, it must be Alice that has a winning strategy. This kind of proof is known as a strategy-stealing argument and does not, unfortunately, tell you how Alice actually wins the game. A similar argument shows that the first player must have a winning strategy in the game of Hex (look it up if you haven’t heard of it), yet we don’t even know a winning first move. If you know how to play dots and boxes, you can try this one with the same idea in mind. Dots and Boxes Variation Suppose you are playing Dots and Boxes, but with each player having an option of whether to go again after making a box. Show that the first player to make a box can win all the boxes! Solution: This just amounts to noticing that by choosing to play again, you are putting yourself in the position your opponent would be in if you didn’t play again. One of these choices must guarantee winning the next box, after which you can re-apply the argument. Thus, in any game of modified Dots and Boxes, best play will lead to one player winning all the boxes. Usually the game begins with a rectangle of m × n potential boxes in a grid made with (m+1)(n+1) dots. The center of such a grid is a dot if m and n are both even, a cell if they are both odd; the second player can win the first box by copying her opponent symmetrically about the center, until a box-making opportunity arises. If just one of m and n is odd, the center of the grid lies inside an edge; the first player draws that edge and then employs the symmetry strategy. Thus with best play, the second player wins all the boxes when m and n have equal parity; otherwise the first player wins all the boxes. 166 Big Pairs in a Matrix Suppose that in each row of a certain square matrix, the sum of the greatest two numbers is r; and in each column, the sum of the greatest two numbers is c. Show that r = c. Solution: Suppose otherwise; by symmetry we may assume r > c. Circle the greatest number in each row, and draw a square around the next greatest. The circled numbers are all at least r/2, so they must all be in different columns. Suppose the highest number with a square around it, say x, is in column j along with column j’s circled number y. Let z be the square-encased number in y’s row; then y + z = r but y + x ≤c, impossible because c < r but x ≥z. Bugging a Disk Elspeth, a new FBI recruit, has been asked to bug a circular room using seven ceiling microphones. If the room is 40 feet in diameter and the bugs are placed so as to minimize the maximum distance from any point in the room to the nearest bug, where should the bugs be located? Solution: Assuming the ceiling is flat, the problem becomes two-dimensional: Locate seven bugs in a disk of diameter 40 feet so that every point in the disk is within distance d of some bug, where d is as small as possible. This means that the seven disks of radius d centered at the bugs cover the room. Imagine a regular hexagon inscribed in the big circle; each side will be 20 feet in length, and the centers of the sides, together with the center of the room, will be within 10 feet of any point in the room. That gives d = 10; can we do any better? Suppose we could. Since the circles of radius 10 ft about the previous outer six points just barely cover the room’s circumference, each using its full diame-ter, covering the room’s circumference with circles of radius less than 10 would require at least seven circles. But that would leave the center of the room uncovered, and that is our contradiction. Often using contradiction is tantamount to using mathematical induction: You assume the statement you wish is false, and consider a minimal counterex-ample. Pairs at Maximum Distance Let X be a finite set of points on the plane. Suppose X contains n points, and the maximum distance between any two of them is d. Prove that at most n pairs of points of X are at distance d. 167 Solution: To solve this puzzle, it’s useful to observe that if A, B and C, D are two “max pairs” (pairs of points from X at distance d), then the line segments AB and CD must cross (else one of the diagonals of the quadrilateral ABDC would exceed d in length). Now assume the statement of the puzzle is false, and let n be the size of a smallest counterexample. Since there are more than n max pairs and each has two points, there must be a point P which participates in three max pairs (say, with points A, B, and C). Every two of the segments PA, PB, and PC must make at most a 60◦angle at P, and one of them, say B, must lie between the others. But this makes it pretty tough for B to be in any other max pair, since if BQ were a max pair, it would have to intersect both PA and PC—an impossibility. Thus we can drop B out of X altogether, losing only one max pair and obtaining a smaller counterexample. This contradiction completes the proof. Lemming on a Chessboard On each square in an n × n chessboard is an arrow pointing to one of its eight neighbors (or off the board, if it’s an edge square). However, arrows in neighbor-ing squares (diagonal neighbors included) may not differ in direction by more than 45 degrees. A lemming begins in a center square, following the arrows from square to square. Is he doomed to fall off the board? 168 Solution: The lemming is indeed doomed. One way to see this is to imagine that the lemming can move to any neighboring square, but must turn to face in the direction of the arrow found there. Then the lemming cannot turn 360 degrees around, because if he could you could shrink the cycle on which he does that until it collapses into a contradiction. But the real lemming, if he is to stay on the board, must eventually cycle and when he does so, he will have to make that 360◦turn. The conclusion can also be reached using induction (thus also contradiction). If the lemming stays on the board, he must, as we have already noted, eventually settle into a cycle. Let C be the smallest-area cycle (on any board) on which this can happen, and suppose it’s a clockwise cycle. Cut the whole board down to C and its interior, then rotate all the arrows 45◦clockwise to force a smaller cycle! Curve on a Sphere Prove that if a closed curve on the unit sphere has length less than 2π, then it is contained in some hemisphere. Solution: Pick any point P on the curve, travel half way around the curve to the point Q, and let N (standing for North Pole) be the point half-way between P and Q. (Since the distance d(P, Q) from P to Q is less than π, N is uniquely defined). Thinking of N as the North Pole provides us with an equator, and if the curve lies entirely on N’s side, that is, in the northern hemisphere, we are done. Otherwise, the curve crosses the equator, and let E be one of the points at which it does so. Then, we observe that d(E, P) + d(E, Q) = π, since if you poke P through the equatorial plane to P ′ on the other side, P ′ is antipodal to Q; hence, d(E, P ′) + d(E, Q) = π. However, for any point X on the curve, d(P, X)+d(X, Q) must be less than π, and this provides the desired contradiction. ♡ Soldiers in a Field An odd number of soldiers are stationed in a large field. No two soldiers are exactly the same distance apart as any other two soldiers. Their commanding officer radios instructions to each soldier to keep an eye on his nearest neighbor. Is it possible that every soldier is being watched? Solution: The problem is most easily solved by considering the two soldiers at shortest distance from each other. Each of these watches the other; if anyone else is watching one of them, then we have a soldier being watched twice and therefore 169 another not being watched at all. Otherwise, these two can be removed without affecting the others. Since the number of soldiers is odd, this procedure would eventually reduce to one soldier not watching anyone, a contradiction. Our next puzzle is a more serious, and indeed classical, mathematical co-nundrum. Alternating Powers Since the series 1 −1 + 1 −1 + 1 −· · · does not converge, the function f(x) = x −x2 + x4 −x8 + x16 −x32 + · · · makes no sense when x = 1. However, f(x) does converge for all positive real numbers x < 1. If we wanted to give f(1) a value, it might make sense to let it be the limit of f(x) as x approaches 1 from below. Does that limit exist? If so, what is it? Solution: This question began life in a paper by G. H. Hardy, the same guy whose appreci-ation for the power of contradiction was noted at the beginning of this chapter. Ironically, although Hardy answers his question correctly, he comments that no completely elementary proof seems to be known; a century or so later, we have one, and it involves contradiction. If you try to determine the answer to this puzzle by computing f(x) for various x near 1, you might reach the wrong conclusion: It appears to converge to 1 2. But appearances can be deceptive, and in fact the limit does not exist! Suppose f(x) does have a limit, say c, as x approaches 1 from below. Since f(x) = x −f(x2), that limit can only be 1 2 (because, taking x as close to 1 as needed, we get c = 1 −c). But notice that f(x) −f(x4) = x −x2 > 0, for all x strictly between 0 and 1. From this we conclude that the sequence f(x), f(x1/4), f(x1/16), . . . is strictly increasing. It follows that if there is any x < 1 for which f(x) > 1 2, there can be no limit. In fact, f(0.995), for instance, exceeds 0.50088. What actually happens is that f(x) oscillates more and more rapidly inside an interval of length about 0.0055 centered at 1 2. Seems a bit capricious, no? The function g(x) = 1 −x + x2 −x3 + x4 −· · · is also defined for each positive x < 1 and has the same problem at x = 1 that f did. But this one is equal to 1/(x+1), as you can check by adding xg(x) to g(x), thus it docilely approaches 1 2 as x →1. Halfway Points Let S be a finite set of points in the unit interval [0,1], and suppose that every point x ∈S lies either halfway between two other points in S (not necessarily the nearest ones) or halfway between another point in S and an endpoint. Show that all the points in S are rational. 170 Solution: Suppose S has n points, x1 < x2 < · · · < xn. By assumption, they satisfy n linear equations with rational coefficients; for example, if point xj lies halfway between xi and xk, it satisfies 2xj = xi + xk. If instead xj lies between xi and 1, it satisfies 2xj = xi + 1. If this set of equations has just the one solution, they must be rational. Why? You may already know from linear algebra that when a set of linear equations over a field that has a unique solution, the values in the solution lie in the field. We can prove that easily by induction: Take any equation and solve it for one variable, say xn. Then substitute in the remaining equations to get a uniquely solvable system with one fewer variable. So we need only prove that our system has only one solution; suppose it has a second solution, y1, . . . , yn. Letting zi = yi−xi and subtracting corresponding equations, we find that the zi’s satisfy similar conditions except that (1) they may not all be distinct (but they’re not all 0); (2) they lie between −1 and 1; and (3) the endpoint 1 is no longer in play, that is, each zj either lies halfway between two other zi’s or halfway between another zi and zero. We may assume one of the zi’s is positive; let zj have the largest value and among those with that maximum value, the largest index. Obviously zj can’t be halfway between 0 and some other zi, so it must be between zi and zk where zi = zj = zk. By choice of j, i and k must both be smaller indices than j, but that can’t be because in the original solution xj must have lain halfway between xi and xk, contradicting the fact that we indexed the original solution in size order. Our theorem is quite a famous one, and the proof given below was often cited by Paul Erd˝ os as an example of a “book proof.” (Although Erd˝ os claimed not to believe in God, he spoke often of a book maintained by God that contained the best proof of every theorem.) Theorem. Let X be a finite set of points on the plane, not all on one line. Then there is a line passing through exactly two points of X. Proof. Assume that every line through two or more points in X in fact contains at least three points of X. The idea is to find such a line L, and a point P not on L, such that the distance from P to L is minimized. Since L contains at least three points of X, two of them, say Q and R, lie on the same side of the perpendicular to L from P. But then, if R is the farther one, the point Q is closer to the line through P and R than P is to L—contradiction! ♡ 171 172 Chapter 10 In All Probability Dealing with probabilities is intuitive for some folks, less so for others. But there are just a few basic ideas, which, if mastered, convey a lot of power. A thing that may or may not happen is called an event, and its probability measures the likelihood that it will happen; if the “trial” that could make the event happen is repeatable (like rolling a die), then the event’s probability is the fraction of time that it occurs in many trials. If the trial has n possible outcomes that are equally likely (perhaps because of symmetry) then the probability of the event A is the number of outcomes that result in A occurring, divided by n. For example, if we roll a die and A is the event that an even number appears on top, then the probability P(A) that A occurs is 3/6 = 1/2, since three of the outcomes are even numbers. The fact that the outcomes are equally likely relies on the assumption that the die behaves like a true, homogeneous, symmetrical, cube. If we roll a pair of dice, there are 36 equally likely outcomes; note that (2,3) is different from (3,2) because even if you can’t distinguish between the two dice, Tyche (the goddess of chance) can. Thus, for example, there are five ways to roll a sum of 6—(1,5), (2,4), (3,3), (4,2), and (5,1)—so P(two dice roll a sum of 6) = 5/36. If two events A and B are independent, that is, the occurrence of one does not affect the probability of the other, then the probability P(A ∧B) that they both occur is equal to the product P(A)P(B) of the probabilities of the two events. Without the assumption of independence, we can only say P(A ∧B) = P(A)P(B\|A) where P(B\|A), read “the probability of B given A,” has intuitive meaning but is often defined simply as the quantity that makes the equation true. The probability P(A ∨B) that at least one of A and B occurs is always equal to P(A) + P(B) −P(A ∧B), as you can see from a Venn diagram. This reduces to P(A∨B) = P(A)+P(B) only in the case where A and B are mutually exclusive—that is, they can’t both occur. Let’s try a few amusing applications of these ideas. 173 Winning at Wimbledon As a result of temporary magical powers (which might need to be able to change your gender), you have made it to the women’s singles tennis finals at Wim-bledon and are playing Serena Williams for all the marbles. However, your powers cannot last the whole match. What score do you want it to be when they disappear, to maximize your chances of notching an upset win? Solution: Naturally, you want to have won the first set of the best-of-three-sets match, and be far ahead in the second. Your first thought might be that you’d like to be up five games to love and serving at 40-love (or, if your serve resembles mine, up 5-0 in games and receiving at love-40). Either gives you three bites at the apple, that is, if your probability of winning a point is some small quantity ε then your probability of winning is about 3ε. (Technically, if the results of the next three points are independent, and you discount the possibility of winning the match despite losing them all, your success probability is 1 −(1 −ε)3 = 3ε −3ε2 + ε3.) But you can do better. Let the game score be 6-6, and suppose you are ahead 6-0 in the seven-point tiebreaker. Then you have six chances to win, almost doubling your probability when ε is small. As I write this, you can do even better in the Australian Open, where the final-set tiebreaker is played to 10 points. The lesson, either way, is this: If you need a miracle, try to arrange for as many shots at that miracle as possible. Birthday Match You are on a cruise where you don’t know anyone else. The ship announces a contest, the upshot of which is that if you can find someone who has the same birthday as yours, you (both) win a Beef Wellington dinner. How many people do you have to compare birthdays with in order to have a better than 50% chance of success? Solution: Let’s assume that (a) neither you (important!) nor anyone else on board (less important) was born on Feb. 29, (b) other dates are equally likely to be a given shipmate’s birthday, and (c) there are no sets of twins (or triplets etc.) on board that will cause you to waste queries. Then the probability that a given shipmate fails to match your birthday is 364/365, thus if you ask n people, the probability that you have no luck is (364/365)n. You want this number to dip below 50%, which it does when n reaches 253; in fact (364/365)253 = 0.49952284596... . Is 253 higher than you guessed? You’d only need 183 queries to get your success probability over 1/2, if the people you asked were guaranteed to have different birthdays. The trouble is, you will probably hear a lot of duplicated 174 answers that don’t help you. (Well, it could help you a bit if you demand a bite of their beef Wellington in return for telling each about the other). If you thought the answer to the puzzle was 23, you confused the question with the more famous problem of how many people you need in a room to have a better than 50% probability that some pair of people share a birthday. In fact, you can get the answer “23” from our answer; the number of pairs of people in that room is 23 2 = 253. (The pairs are not precisely independent, but they’re close enough to make this work.) By the way, how do you determine that 253 is the least n for which (364/365)n < 1/2 without a lot of trial and error? My favorite way is to use the approxima-tion (1 −1/m)m ∼e−m for large m. If you take this as an equality and solve (1 −1/365)n = 1/2 you get n = loge(2) · 365 ∼252.9987. When the event A is decided before the event B, the conditional probability of B given A is often easy to deduce but P(A\|B) is less intuitive. The laws of probability don’t care about the order of events, though. Other Side of the Coin A two-headed coin, a two-tailed coin, and an ordinary coin are placed in a bag. One of the coins is drawn at random and flipped; it comes up “heads.” What is the probability that there is a head on the other side of this coin? Solution: The temptation is to say that the drawn coin is either the two-headed coin or the fair coin, so the probability that its other side is heads is 1/2. But is it? Imagine that the coin is flipped many times, and comes up heads every time. It could still be either coin, but you can’t help thinking it’s more likely to be the two-headed coin. That kind of reasoning is basically how we learn from statistical observations. In fact, the inference that the drawn coin is probably two-headed exists already after one flip. We want to compute P(A\|B) where A is the event that the two-headed coin was drawn, and B the event that heads was flipped. For 175 any two events A and B, we calculate P(A\|B) = P(A ∧B) P(B) = P(A)P(B\|A) P((A ∧B) ∨(A ∧B)) = P(A)P(B\|A) P(A ∧B) + P(A ∧B) = P(A)P(B\|A) P(A)P(B\|A) + P(A)P(B\|A) where A is the event that A does not occur. This equation is known as Bayes’ Rule, and if we plug in the values from our experiment, we get P(A\|B) = 1 3 · 1 1 3 · 1 + 2 3 · 1 4 = 2 3 . If all the above fractions leave you unconvinced, here’s another way to get the answer. Steal a crayon or marker from your kid and label each of the six coin-faces as follows: “1” and “2” on the two sides of the two-headed coin, “3” on the heads side of the fair coin, “4” on its tails side, and finally “5” and “6” on the two sides of the two-tailed coin. You will agree, I hope, that drawing a coin and flipping it has the same effect as rolling a fair die: You are equally likely to come up with any number from 1 to 6. Given that you flipped a head, then, you are equally likely to have seen 1, 2, or 3. Two of those three faces have a head on the other side. “Reverse” conditional probabilities are crucial in statistics, where you need to compute P(A\|B) where A is the hypothesis you are testing, and B the result of your experiment. Here are some more puzzles involving that kind of calculation. Boy Born on Tuesday Mrs. Chance has two children of different ages. At least one of them is a boy born on a Tuesday. What is the probability that both of them are boys? Solution: We consider first the classic version, where the information is that at least one of Mrs. Chance’s children is a boy. Here, the solution seems simple: a priori, the children (in age order) are equally likely to be boy-boy, boy-girl, girl-boy or girl-girl. The last of these is ruled out, so the probability both children are boys is 1/3. The difficulty is that in almost every way that you are likely to find out that at least one of Mrs. Chance’s kids is a boy, there is an additional inference which raises the probability of two boys from 1/3 to 1/2. For example, if you find that the older child is a boy, then the girl-boy com-bination above is ruled out and it’s clear that the probability that the younger 176 child is a boy is 1/2. But that reasoning also applies if your information is that the taller child is a boy, or the one with darker hair is a boy, or the child you saw Mrs. Chance with yesterday is a boy, or even if you happen to catch Mrs. Chance in the store buying boys’ clothes. Why the latter? Because if Mrs. Chance’s other child were a girl, you’d be less likely to find her buying boys’ clothes. It’s quite similar to the previous puzzle; the fact that if the coin were fair it might not have come up “heads” affords an inference that the coin is the two-headed coin, raising the probability of two heads from 1/2 to 2/3. Here the effect is to raise the probability that the children are both boys from 1/3 to 1/2. Why “almost”? Is there any way you could find out that at least one of Mrs. Chance’s kids is a boy, that would lead to the conclusion that the probability they are both boys is 1/3? As it happens, I have a friend whose wife became pregnant with fraternal twins, and an amniotic test revealed the presence of Y chromosomes in the placenta. The conclusion was that at least one of the twins was a boy, and naturally my friend and his wife hoped that the other was a girl—but assumed that the probability of this event was 1/2. In fact, it really was 2/3; I inquired and was told that the test always finds Y chromosomes as long as one or more fetuses is male. (In fact, the outcome was mixed twins.) Now back to the boy born on a Tuesday. This is a very delicate question and you need to assume that the information has come to you in with no additional inferences—for example, you actually chose Mrs. Chance randomly among many mothers of two (non-twin) children and asked her if at least one was a boy born on a Tuesday, and she said yes. Compared to the classic version, here there is a substantial inference that both kids are boys because having two boys, instead of only one, nearly doubles the probability that some boy was born on a Tuesday. But not quite, because if they were both born on Tuesday they count only once. Thus you’d expect the answer to be nearly 1/2, and that is indeed the case. The calculation is straightforward. If we classify all two-kid families according to gender and birth-day-of-the-week we get 2×7×2×7 = 196 possibilities, of which 1×2×7+2×7× 1−1×1 = 27 contain a boy born on a Tuesday (the −1 comes from over-counting the case of two boys born on Tuesdays). Of these, 1×1×7 + 1×7×1 −1×1 = 13 involve two boys, so the probability we seek is 13/27. Whose Bullet? Two marksmen, one of whom (“A”) hits a certain small target 75% of the time and the other (“B”) only 25%, aim simultaneously at that target. Just one bullet hits. What’s the probability that it came from A? Solution: You might think that since A is three times as good a shot as B, he is three times as likely to have been the one whose bullet hit; in other words, the probability that he deserves the credit is 75%. 177 But there are two things happening here: Hitting and missing. Since the probability that A hits and B misses is 3/4 × 3/4 = 9/16 while the probability that B hits but A misses is only 1/4 × 1/4 = 1/16, the probability that the successful bullet was A’s is actually a full 90%. Conditional probabilities can sometimes defy your intuition. Another exam-ple: Second Ace What is the probability that a poker hand (five cards dealt at random from a 52-card deck) contains at least two aces, given that it has at least one? What is the probability that it contains two aces, given that it has the Ace of Spades? If your answers are different, then: What’s so special about the Ace of Spades? Solution: There are 52 5 poker hands of which 52 5 − 48 5 have at least one ace, and 52 5 − 48 5 − 4 1 · 48 4 have at least two aces; dividing the last of these expressions by the second gives our first answer, 0.12218492854. There are 51 4 hands containing the Ace of Spades and 51 4 − 48 4 containing another ace as well, giving our second answer, 0.22136854741, quite a lot larger than the first answer. Of course, there’s nothing special about the Ace of Spades; specifying the ace held changes the conditional probabilities. Here’s a simpler example that you don’t need a calculator for. Suppose a third of the families in town have two children, a third have one, and the remaining third none. Then the probability that a family has a second child, given that it has at least one, is 1/2. But the probability that a family has a second child given that it has a girl is 3/5, because only half the families with one child have a girl, but 3/4 of those with two children have a girl. Stopping After the Boy In a certain country, a law is passed forbidding families to have another child after having a boy. Thus, families might consist of one boy, one girl and one boy, five girls and one boy... How will this law affect the ratio of boys to girls? Solution: It might seem like limiting families to one boy should cut down the boy-girl ratio, but assuming that gender is independent among siblings, the child whose birth is prevented by having an older brother is as likely to be a girl as a boy. Thus the sex ratio should not be affected by this law. One could make the case that, on account of the possibility of identical twins, gender among siblings is not precisely independent. However, in the statement of the puzzle, the law’s effect when twin boys are born has not been made clear 178 (and is perhaps a bit horrifying to think about). Thus we solvers are pretty much forced to discount that possibility. Up until now we’ve been talking about discrete probability. Things get a bit more subtle when a random value is chosen from an interval of real numbers. Awkwardly, we have to assign probability 0 to the event that any particular number is chosen, even though some event of that kind is guaranteed to happen. Typically, we deal instead with the probability that the chosen point lies in a particular subinterval. For example, if we choose a point uniformly at random from the unit interval, the probability that it falls between 0.3 and 0.4 is 1/10 (regardless of whether we count 0.3 and 0.4 as part of the interval). Points on a Circle What is the probability that three uniformly random points on a circle will be contained in some semicircle? Solution: It’s not hard to see that no matter where we put the first two points, assum-ing they don’t coincide, it is more likely that the third point will be in some semicircle with the first two, than not. Thus, in fact, the answer to the puzzle should be more than 1/2. But how do we compute it, without the bother of trying to condition on the distance between the first two points? The answer is (as is often the case in probability problems) to choose the points in a different way—but, of course, one that still results in uniformly random points. Here, let us choose three random diameters of our circle. Each hits the circle at two (antipodal) points, and we use three coinflips do decide which three points to use. This may seem like an unnecessarily complicated way to choose our points— why in six steps, when we can do it in three steps? To see the answer, draw any three diameters and the six points where they hit the circle. Observe that of those six, if three consecutive points are chosen, they will certainly lie in a semicircle; otherwise, they won’t! Well, there are 23 = 8 ways to pick the points, using our coinflips, and 6 of those result in consecutive points being chosen. So the probability of their being contained in some semicircle is 6/8 = 3/4. ♡ (If you found that puzzle too easy, try to determine the probability that four uniformly random points on a sphere are all contained in some hemisphere!) Comparing Numbers, Version I Paula (the perpetrator) takes two slips of paper and writes an integer on each. There are no restrictions on the two numbers except that they must be different. She then conceals one slip in each hand. Victor (the victim) chooses one of Paula’s hands, which Paula then opens, allowing Victor to see the number on that slip. Victor must now guess whether 179 that number is the larger or the smaller of Paula’s two numbers; if he guesses right, he wins $1, otherwise he loses $1. Clearly, Victor can at least break even in this game, for example, by flipping a coin to decide whether to guess “larger” or “smaller”—or by always guessing “larger,” but choosing the hand at random. The question is: Not knowing anything about Paula’s psychology, is there any way he can do better than break even? Solution: Amazingly, there is a strategy which guarantees Victor a better than 50% chance to win. Before playing, Victor selects a probability distribution on the integers that assigns positive probability to each integer. (For example, he plans to flip a coin until a “head” appears. If he sees an even number 2k of tails, he will select the integer k; if he sees 2k−1 tails, he will select the integer −k.) If Victor is smart, he will conceal this distribution from Paula, but as you will see, Victor gets his guarantee even if Paula finds out. It is perhaps worth noting that although Victor might like the idea of assigning the same probability to every integer, this is not possible—there is no such probability distribution on the integers! After Paula picks her numbers, Victor selects an integer from his probability distribution and adds 1 2 to it; that becomes his “threshold” t. For example, using the distribution above, if he flips five tails before his first head, his random integer will be −3 and his threshold t will be −2 1 2. When Paula offers her two hands, Victor flips a fair coin to decide which hand to choose, then looks at the number in that hand. If it exceeds t, he guesses that it is the larger of Paula’s numbers; if it is smaller than t, he guesses that it is the smaller of Paula’s numbers. So, why does this work? Well, suppose that t turns out to be larger than either of Paula’s numbers; then Victor will guess “smaller” regardless of which number he gets, and thus will be right with probability exactly 1 2. If t undercuts both of Paula’s numbers, Victor will inevitably guess “larger” and will again be right with probability 1 2. But, with positive probability, Victor’s threshold t will fall between Paula’s two numbers; and then Victor wins regardless of which hand he picks. This possibility, then, gives Victor the edge which enables him to beat 50%. ♡ Neither this nor any other strategy enables Victor to guarantee, for some fixed ε > 0, a probability of winning greater than 50% + ε. A smart Paula can choose randomly two consecutive multidigit integers, and thereby reduce Victor’s edge to an insignificant smidgen. Comparing Numbers, Version II Now let’s make things more favorable to Victor: Instead of being chosen by Paula, the numbers are chosen independently at random from the uniform dis-180 tribution on [0,1] (two outputs from a standard random number generator will do fine). To compensate Paula, we allow her to examine the two random numbers and to decide which one Victor will see. Again, Victor must guess whether the number he sees is the larger or smaller of the two, with $1 at stake. Can he do better than break even? What are his and Paula’s best (i.e., “equilibrium”) strategies? Solution: It looks like the ability to choose which number Victor sees is paltry compen-sation to Paula for not getting to pick the numbers, but in fact this version of the game is strictly fair: Paula can prevent Victor from getting any advantage at all. Her strategy is simple: Look at the two random real numbers, then feed Victor the one which is closer to 1 2. To see that this reduces Victor to a pure guess, suppose that the number x revealed to him is between 0 and 1 2. Then the unseen number is uniformly distributed in the set [0, x]∪[1−x, 1] and is, therefore, equally likely to be smaller or greater than x. If x > 1 2, then the set is [0, 1−x] ∪[x, 1] and the argument is the same. Of course, Victor can guarantee probability 1 2 against any strategy by ignor-ing his number and flipping a coin, so the game is completely fair. ♡ This amusing game was brought to my attention at a restaurant in Atlanta. Lots of smart people were present and were stymied, so if you failed to spot this nice strategy of Paula’s, you’re in good company. Biased Betting Alice and Bob each have $100 and a biased coin that comes up heads with probability 51%. At a signal, each begins flipping his or her coin once a minute and bets $1 (at even odds) on each outcome, against a bank with unlimited funds. Alice bets on heads, Bob on tails. Suppose both eventually go broke. Who is more likely to have gone broke first? Suppose now that Alice and Bob are flipping the same coin, so that when the first one goes broke the second one’s stack will be at $200. Same question: Given that they both go broke, who is more likely to have gone broke first? Solution: In the first problem, Alice and Bob are equally likely to have gone broke first. Bob always goes broke (as noted later, his average time to go broke is finite). Alice may never go broke, but if we condition on her having gone broke, it turns out that Alice and Bob have exactly the same probability of having gone broke at any particular time t. 181 To see this, pick any win-loss sequence s that ends by going broke; sup-pose s has n wins, thus n + 100 losses. It will have probability pnqn+100 for Alice, but pn+100qn for Bob, where here p = 51% and q = 49%. The ratio of these probabilities is the constant (q/p)100, thus P(Alice goes broke) = (q/p)100 (which happens to be about 0.0183058708; call it 2%). Once we divide Alice’s probability of encountering s by this number, her probability is the same as Bob’s. For the second problem, where Alice and Bob were flipping the same coin, your intuition might tell you that Alice probably went broke first. The reasoning is that if Bob went broke first, Alice would have to have gone broke after building up to $200, an unlikely occurrence. But is this true? Can’t we argue as above, comparing win-lose sequences that result in both going broke? No. The proof above doesn’t work because here, Alice’s going broke also affects Bob’s longevity. In fact, the same ratio (q/p)100 applies when comparing a both-broke win-loss sequence and the conclusion is that the probability that Bob goes broke first is (q/p)100 times the probability that Alice goes broke first. Intuition was correct; Alice is more than 50 times more likely to have gone broke first! It’s worth noting that, putting Bob aside, we can deduce from the first calculation how long it takes, on average, for Alice to go broke given that she goes broke. The reason is that under this condition, we have shown Alice’s expected time to ruin is the same as Bob’s. But Bob always goes broke and since he loses 2 cents per toss on average, his expected time to ruin is 100/0.02 = 5000 tosses. At 60 tosses per hour that comes to 83 1 3 hours, or about 3 1 2 days. Home-Field Advantage Every year, the Elkton Earlies and the Linthicum Lates face off in a series of baseball games, the winner being the first to win four games. The teams are evenly matched but each has a small edge (say, a 51% chance of winning) when playing at home. Every year, the first three games are played in Elkton, the rest in Linthicum. Which team has the advantage? Solution: The series will last 4, 5, 6, or 7 games; it seems that the average should be somewhere between 5 and 6 games, in which case more games are played on average in Elkton than in Linthicum; therefore, Elkton has the advantage. First issue: Are more games really played in Elkton, on average? To verify that we’ll need to compute the probabilities that the series will last 4, 5, 6, or 7 games. If all the games were 50-50, the probabilities that a series will last 4, 5, 6, or 7 games would be, respectively, 1/8, 1/4, 5/16 and 5/16. Thus the average number of games played would be 4 · 1 8 + 5 · 1 4 + 6 · 5 16 + 7 · 5 16 = 93 16 = 513 16 = 5.8125 182 and since three games are always played in Elkton, on average only 2 13/16 would be played in Linthicum. As one might expect, re-calculating the above odds taking into account the 1% home-field advantage changes the numbers very little. The average number of games played shifts upward a teensy bit to 5.81267507002. So, indeed, more games are played in Elkton, on average. Over a very long period of time, this will be reflected in the game statistics; you can expect Elkton to have won about 50.03222697428% of the games over a million-year period. However, shockingly, you can expect Linthicum to have won most of the series! Think of it this way: Imagine that all seven games are played (with equal ferocity) regardless of the outcomes. This doesn’t change the probability of winning the series (if it did, we would never be able to quit after fewer than seven games). Those seven consist of four games in Linthicum and only three in Elkton, giving the advantage to Linthicum: their probability of winning the series works out to 50.31257503002%. If it seems odd to you that the home-field advantage in games 5, 6, and 7, which might not get played, counts as much as home-field advantage in earlier games, that’s understandable. But the point is that the late home-field advan-tage is there when Linthicum needs it, and something that’s there when you need it is just as good as its being there all the time. Have you ever wondered about the effect of different rules, from sport to sport, concerning which player or team serves next? Maybe the next puzzle will answer your questions. Service Options You are challenged to a short tennis match, with the winner to be the first player to win four games. You get to serve first. But there are options for determining the sequence in which the two of you serve: 1. Standard: Serve alternates (you, her, you, her, you, her, you). 2. Volleyball Style: The winner of the previous game serves the next one. 3. Reverse Volleyball Style: The winner of the previous game receives in the next one. Which option should you choose? You may assume it is to your advantage to serve. You may also assume that the outcome of any game is independent of when the game is played and of the outcome of any previous game. Solution: Again using the idea (from the previous puzzle’s solution) that it doesn’t hurt to assume extra games are played, assume that you play lots of games—maybe more than are needed to determine the match winner. Let A be the event that of the first four served by you and the first three served by your opponent, at least 183 four are won by you. Then, it is easily checked that no matter which service option you choose, you will win if A occurs and lose otherwise. Thus, your choice makes no difference. Notice that the independence assumptions mean that the probability of the event A, since it always involves four particular service games and three particular returning games, does not depend on when the games are played, or in what order. If the game outcomes were not independent, the service option could make a difference. For example, if your opponent is easily discouraged when losing, you might benefit by using the volleyball scheme, in which you keep serving if you win. The lesson learned above from Home-Field Advantage will again come in handy in the next puzzle. Who Won the Series? Two evenly-matched teams meet to play a best-of-seven World Series of baseball games. Each team has the same small advantage when playing at home. As is customary for this event, one team (say, Team A) plays games 1 and 2 at home, and, if necessary, plays games 6 and 7 at home. Team B plays games 3, 4 and, if needed, 5 at home. You go to a conference in Europe and return to find that the series is over, and six games were played. Which team is more likely to have won the series? Solution: You could work this about by assigning a variable p > 1/2 to the probability of winning a game at home, then doing a bunch of calculations. But no calculations are actually necessary, if you remember that home-field advantage in potential games is as good as it is in games that are always played. It follows that if you know that at most six games were played, you can conclude that teams A and B are equally likely to have won. On the other hand, if you know that at most 5 had been played, then (as host of 3 games) Team B is more likely to have won. From these two facts we deduce that if exactly six games were played, Team A is the more likely winner. It sometimes happens that there’s alternative way to solve a puzzle if you trust the poser, and this is the case here. Since the degree of advantage ex-perienced by the home team is not specified, you might choose to believe that the answer should be the same even if that degree is huge—in other words, if it requires a virtual miracle to win an away game. In that case, with high probability there was just one upset (game won by the visitor) in the six games that were played, but that upset cannot have occurred in Game 1 or Game 2 else Team B would have won in only five games. Thus it happened in Game 3, 4, 5 or 6 and in the first three of those cases it is Team A who benefitted. 184 Next is the first of several probability-comparison puzzles. Dishwashing Game You and your spouse flip a coin to see who washes the dishes each evening. “Heads” she washes, “tails” you wash. Tonight she tells you she is imposing a different scheme. You flip the coin 13 times, then she flips it 12 times. If you get more heads than she does, she washes; if you get the same number of heads or fewer, you wash. Should you be happy? Solution: If “should you be happy?” means anything at all, it should mean “are you better off than you were before?”. Here, the easy route is to imagine that first you and your spouse flip just 12 times each. If you get different numbers of heads, the one with fewer will be washing dishes regardless of the outcome of the next flip; so those scenarios cancel. The rest of the time, when you tie, the final flip will determine the washer. So it’s still a 50-50 proposition, and you should be indifferent to the change in procedure (unless you dislike flipping coins). Now suppose that some serious experimentation with the coin in question shows that it is actually slightly biased, and comes up heads 51% of the time. Should you still be indifferent to the new decision procedure? It might seem that you would welcome it now: Heads are good for you, so, the more flipping, the better. But re-examination of the above argument shows the opposite. When you and your spouse flip 12 times each, it’s still equally likely that you flip more heads than she does, or the reverse. Only if the two of you flip the same number of heads do you get your 51% advantage from the final flip. Thus, overall, your probability of ducking the dishwashing falls to somewhere between 50% and 51%. The analysis suggests a third procedure: You and your spouse alternate flipping the coin until you reach a point where you have both flipped the same number of times but one of you (the winner) has flipped more heads. The advantage of this scheme? It’s fair even if the coin is biased! Many puzzles that involve comparing two probabilities can be approached by a technique called coupling, in which two events associated with different conditions are somehow built into a single experiment for which both events make sense. Then, if you want to determine which of the two events A and B is more probable, you can ignore outcomes where they both occur or neither occurs, and just compare the probability of A happening without B with the probability of B happening without A. This amounts to comparing the areas of the two crescents in the Venn diagram below. OK, that’s a lot of abstract theory. Let’s see some examples. 185 Random Judge After a wild night of shore leave, you are about to be tried by your U.S. Navy superiors for unseemly behavior. You have a choice between accepting a “sum-mary” court-martial with just one judge, or a “special” court-martial with three judges who decide by majority vote. Each possible judge has the same (independent) probability—65.43%—of deciding in your favor, except that one officer who would be judging your special court-martial (but not the summary) is notorious for flipping a coin to make his decisions. Which type of court-martial is more likely to keep you out of the brig? Solution: It’s not that hard to “do the math” on this one and simply compute your chance of getting off in the special court-martial, then compare it to your 65.43% chance of surviving the summary court-martial. (You might save yourself some arithmetic if you replace 65.43% by a variable, say p, and put the number back in later if necessary.) But let’s try coupling. You may as well suppose officer A would be on both panels and would vote the same way on each; officer C is the random one. If you go with the special court-martial you will regret it precisely when A votes innocent and B and C vote guilty, but thank your stars if A votes guilty while B and C vote innocent. Without any calculations you know that these events have the same prob-ability, just by exchanging the roles of A and B and reversing C’s coinflip. So your two choices are equally good. Suppose there’s also an option of a “general” court-martial with five judges, two of whom are coin-flippers. You can apply the coupling method again, but this time it tells you to choose the five judges if p (as here) is greater than 1/2. In fact, the Law of Large Numbers tells you that even if 90% of the judges were coin-flippers, the remaining 10% voting in your favor with some fixed 186 probability p > 1/2, you could boost your chances to 99% with enough judges. Wins in a Row You want to join a certain chess club, but admission requires that you play three games against Ioana, the current club champion, and win two in a row. Since it is an advantage to play the white pieces (which move first), you alternate playing white and black. A coin is flipped, and the result is that you will be white in the first and third games, black in the second. Should you be happy? Solution: You can answer this question by assigning a variable w to the probability of winning a game as White, and another variable b < w to the probability of winning as Black; then doing some algebra. Using coupling, you can get the answer without algebra—even if the problem is modified so that you have to win two in a row out of seventeen games, or m in a row out of n. (If n is even, it doesn’t matter who plays white first; if n is odd, you want to be black first when m is even, white first when m is odd.) The coupling argument in the original 2-out-of-3 puzzle goes like this. Imag-ine that you are going to play four games against Ioana, playing white, then black, then white, then black. You still need to win two in a row, but you must decide in advance whether to discount the first game, or the last. Obviously turning the first game into a “practice game” is equivalent to playing BWB in the original problem, and failing to count the last game is equivalent to playing WBW, so the new problem is equivalent to the old one. But now the events are in the same experiment. For it to make a difference which game you discounted, the results must be either WWLX or XLWW. In words: If you win the first two games, and lose (or draw) the third, you will wish that you had discounted the last game; if you lose the second but win the last two, you will wish that you had discounted the first game. But it is easy to see that XLWW is more likely than WWLX. The two wins in each case are one with white and one with black, so those cases balance; but the loss in XLWW is with black, more likely than the loss in WWLX with white. So you want to discount the first game, that is, start as black in the original problem. A slightly more challenging version of this argument is needed if you change the number of games played, and/or the number of wins needed in a row. Split Games You are a rabid baseball fan and, miraculously, your team has won the pennant— thus, it gets to play in the World Series. Unfortunately, the opposition is a 187 superior team whose probability of winning any given game against your team is 60%. Sure enough, your team loses the first game in the best-of-seven series and you are so unhappy that you drink yourself into a stupor. When you regain consciousness, you discover that two more games have been played. You run out into the street and grab the first passer-by. “What happened in games 2 and 3 of the World Series?” “They were split,” he says. “One game each.” Should you be happy? Solution: In my experience, presented with this puzzle, about half of respondents answer “Yes—if those games hadn’t been split, your team would probably have lost them both.” The other half argue: “No—if your team keeps splitting games, they will lose the series. They have to do better.” Which argument is correct—and how do you verify the answer without a messy computation? The task at hand is to determine, after hearing that games 2 and 3 were split, whether you are better or worse off than before. In other words, is your team’s probability of winning the series better now, when it needs three of the next four games, than it was before, when it needed four out of six? Computing and comparing tails of binomial distributions is messy but not difficult. Before the news, your team needed to win 4, 5, or 6 of the next six games. (Wait, what if fewer than seven games are played? As in both Home-Field Advantage and Who Won the Series?, it costs nothing to imagine that all seven games are played no matter what.) The probability that your team wins exactly four of six games is “6 choose 4” (the number of ways that can happen) times 0.44 (the probability that your team wins a particular four games) times 0.62 (the probability that the other guys win the other two). Altogether, the probability that your team wins at least four of six is 6 4 · 0.44 · 0.62 + 6 5 · 0.45 · 0.6 + 6 6 · 0.46 = 15 · 24 · 32/56 + 6 · 25 · 3/56 + 26/56 = 112/625 . After the second and third games are split, your team needs at least three of the remaining four. The probability of winning is now: 4 3 · 0.43 · 0.6 + 4 4 · 0.44 = 4 · 23 · 3/54 + 24/54 = 112/625 (!) 188 So you should be exactly indifferent to the news! The two arguments (one backward-looking, suggesting that you should be happy, the other forward-looking, suggesting that you should be unhappy) seem to have cancelled each other precisely. Can this be a coincidence? Is there any way to get this result “in your head”? Of course there is—by coupling. To set this up a little imagination helps. Suppose that, after game 3, it is discovered that an umpire who participated in games 2 and 3 had lied on his application. There is a movement to void those two games, and a counter-movement to keep them. The commissioner of baseball, wise man that he is, appoints a committee to decide what to do with the results of games 2 and 3; and in the meantime, he tells the teams to keep playing. Of course, the commissioner hopes—and so do we puzzle-solvers—that by the time the committee makes its decision, the question will be moot. Suppose five(!) more games are played before the committee is ready to report. If your team wins four or five of them, it has won the series regardless of the disposition of the second and third game results. On the other hand, if the opposition has won three or more, they have won the series regardless. The only case where the committee can make a difference is where your team won exactly three of those five new games. In that case, if the results of games 2 and 3 are voided, one more game needs to be played; your team wins the series if they win that game, which happens with probability 2/5. If, on the other hand, the committee decides to count games 2 and 3, the series ended before the last game, and whoever lost that game is the World Series winner. Sounds good for your team, no? Oops, remember that your team won three of those five new games, so the probability that the last game is among the two they lost is again 2/5. ♡ Angry Baseball As in Split Games, your team is the underdog and wins any given game in the best-of-seven World Series with probability 40%. But, hold on: This time, whenever your team is behind in the series, the players get angry and play better, raising your team’s probability of winning that game to 60%. Before it all begins, what is your team’s probability of winning the World Series? Solution: It would be a pain in the neck to work out all the possible outcomes and their probabilities, but coupling again comes to our rescue. Let k be the number of the first game played after the last time the teams were tied. Since the teams were tied 0-0 at the start, k could be 1, but it could also be 3, 5, or 7. (It turns out, you don’t care!) 189 Suppose team X wins game k. Let us represent the winners of the rest of the games by a sequence of X’s and Y’s. Since there are no more ties, X must have won the series, each win by X having had probability 40% and each win by Y, 60%. If it was Team Y that won game k, the same reasoning applies and the same probability attaches to the subsequent win-sequence except with X’s and Y’s exchanged. In effect, we couple win-sequences after X wins game k with the complementary win-sequences after Y wins game k. We conclude that the probability of winning the series for Team X is exactly the probability that Team X wins game k. That probability is 40% when Team X is your team. Since that does not depend on the value of k, your team’s probability of winning the series is 40% from the start. Note that unlike Split Games, this puzzle works with any game-winning probability p (changing to 1−p when the team is behind). The answer is then p. Let’s move on to (American) football. Are you ready to use probability theory to be an effective coach? Two-point Conversion As coach of the Hoboken Hominids (American) football team, you saw your team fall 14 points behind the Gloucester Great Apes until, with just a minute to go in the game, the Hominids scored a touchdown. You have a choice of kicking an extra point (95% success rate) or going for two (45% success rate). Which should you do? Solution: You must assume the Hominids will be able to score a second touchdown; and it’s reasonable to postulate that if the game goes to overtime, either team is equally likely to win it. Whatever you do, if it doesn’t work you’ll need to go for two points after the second TD, winning with probability 0.45 · 1 2 = 0.225. If you do succeed with the first conversion, you’ll want to go for just one the next time, winning with probability 1 2 · 0.95 = 0.475 if you had kicked last time but a full 0.95 if you had gotten two extra points last time. If you succeeded with your two-point conversion but missed the subsequent point after, you still have to win the playoff; that adds another 0.05·0.5 = 0.025 to give a total probability of 0.975 for winning, after your two-point conversion. So, altogether (assuming that you did get that second touchdown) your probability of winning with the “kick after first TD” option is 0.95 · 0.475 + 0.05 · 0.225 = 0.4625, while your probability of winning if you went for two after the first TD is 0.45 · 0.975 + 0.55 · 0.225 = 0.5625, 190 so going for two is much better. It might be easier to see why this is so if you instead imagine that the kick is a sure thing while the two-point conversion has probability of success 1/2. Then if you kick, you are headed for overtime for sure (given, of course, the second touchdown) where you win with probability 1/2. If you go for two and succeed you win for sure, so already you get probability 1/2 of winning the game; but if you miss it ain’t over. Altogether the two-point conversion strategy wins with 5/8 of the time. So why do coaches frequently try to kick the extra point in these situations? Do they think missing the conversion will demoralize their team and make the second touchdown unlikely? My own theory is that coaches tend to make conservative decisions, making it harder for their managers to point at a coach’s decision that cost the game. In this situation, for example, if the team misses both two-point conversions, blame is likely to fall on the coach; but if they kick both extra points and lose in overtime, most of the blame will fall on whichever player screwed up in overtime. Random Chord What is the probability that a random chord of a circle is longer than a side of an equilateral triangle inscribed in the circle? Solution: Call a chord “long” if it is longer than the side of our inscribed triangle. Suppose we choose our random chord by fixing some direction—horizontal is as good as any. Then we can draw a vertical diameter in our circle, pick a point on it uniformly at random, and draw the horizontal chord through that point. In the left-hand third of the figure below, the inscribed triangle has been drawn with its base up, so that the diameter we just drew is perpendicular to the base. We claim the triangle’s base is exactly halfway between the circle’s center and the bottom point of the diameter; one way to see this is to draw another inscribed triangle, this one upside-down, so that the two triangles make a “star of David.” The hexagon in the middle of the star can be broken into six equilateral triangles and once we do that, it is evident that the part of our diameter that yields a long chord is the same length as the rest. Thus the probability that the random chord is long is 1/2. But . . . we could instead fix one end of the chord, say at the triangle’s apex, and choose its angle uniformly at random, as in the middle third of the figure. Since the angle made by the triangle is 60◦and our random angle is between 0◦ and 180◦, we deduce that the probability the chord is long is 1/3. Wait, here’s yet another way to solve the problem. Except for diameters (which have probability zero), every chord has a different midpoint and every point inside the circle, except for the center, is the midpoint of a unique chord. So let’s pick a point inside the circle uniformly at random, and use the chord 191 with that midpoint. It’s easy to see that the chord will be long exactly if our point is closer to the center of the circle than the midpoint of a side of the triangle (see right-hand third of figure). Thus, we get a long chord if and only if the chosen point lies inside the circle that’s inscribed in our triangle. But this circle has half the radius, thus one-quarter the area, of the original circle; therefore the probability that our chord is long is 1/4. What gives? What’t the right answer, 1/2, 1/3, 1/4 or something else? The answer is that the concept of “uniform randomness” for an object chosen from an infinite set is not automatically defined. True, for point chosen from a geometric shape (e.g., an interval or a polygon) we generally take uniformity to mean that the probability that the point lands in a particular subset is proportional to the measure (e.g., length or area) of that subset. For chords, we don’t have a standard notion of uniform randomness and as we’ve just seen, different attempts at defining uniformity can lead to different answers. Random Bias Suppose you choose a real number p between 0 and 1 uniformly at random, then bend a coin so that its probability of coming up heads when you flip it is precisely p. Finally you flip your bent coin 100 times. What is the probability that after all this, you end up flipping exactly 50 heads? Solution: This puzzle again calls for a form of coupling; the idea in this case is couple the coinflips even though they involve different coins, depending on the choice of p. To do this we choose, in advance, a uniformly random threshold pi ∈[0, 1] for the ith coinflip, independently for each i. Then, after p is chosen, the ith coinflip is deemed to be heads just when pi < p. Now we need only observe that getting 50 heads is tantamount to p being the middle number, that is, the 51st largest, of the 101 values p1, p2, . . . , p100, p. Since these numbers are all uniform and independent, their order is uniformly random and the probability that p ends up in the middle position is exactly 1/101. 192 Coin Testing The Unfair Advantage Magic Company has supplied you, a magician, with a special penny and a special nickel. One of these is supposed to flip “Heads” with probability 1/3, the other 1/4, but UAMCO has not bothered to tell you which is which. Having limited patience, you plan to try to identify the biases by flipping the nickel and penny one by one until one of them comes up Heads, at which point that one will be declared to be the 1/3-Heads coin. In what order should you flip the coins to maximize the probability that you get the correct answer, and at the same time to be fair, that is, to give the penny and the nickel the same chance to be designated the 1/3-Heads coin? Solution: From the point of view of accuracy, at any point when you’ve flipped each coin the same number of times, it doesn’t matter which one you flip next. At any point when you’ve flipped the penny more times than the nickel, you need to flip the nickel next, and vice versa. Thus, for the purpose of accuracy, the order doesn’t matter as long as for every i, flips 2i−1 and 2i are of different coins. However, if you want a fair test—one equally likely to damn either coin—you need to be more careful. Imagine that you make an extra flip when the first Head turns up at an odd flip, and declare the result to be inconclusive if the succeeding flip (necessarily, of the other coin) is again a Head. This would then certainly be fair to the two coins, but it may mean more flipping. And if you’re reading this book, you’re likely to be the kind of person that would rather spend an hour figuring something out then an extra minute actually doing it. It follows from considering this tie scenario that the bias toward the penny (say) is the probability that the “tie” happens with the penny having flipped Heads first, and the nickel immediately after; and vice-versa. To arrange these biases so that they cancel, we need to order the flips so that if a tie “would have” occurred, it is equally like to have been with the penny or the nickel coming up Heads first. Suppose we flip the penny first, that is, we begin “PN”. The probability of getting Heads already on the first two flips is 1 3 · 1 4 = 1 12. What’s the probability that the modified game ends in a tie? You will be flipping in pairs—odd flip, then even flip—until you either get HT, TH, or HH, and the tie occurs in the last of these cases, which has probability ( 1 3 · 1 4)/( 2 3 · 1 4 + 3 4 · 1 3 + 1 3 · 1 4) = ( 1 12)/( 6 12) = 1 6. Therefore, in order to cancel the penny’s first-flip advantage, all the rest of the odd flips must be performed by the nickel! In conclusion, the flip order should either be PNNPNPNPNPNP... or NPP-NPNPNPNPN... . 193 Coin Game You and a friend each pick a different heads-tails sequence of length 4 and a fair coin is flipped until one sequence or the other appears; the owner of that sequence wins the game. For example, if you pick HHHH and she picks TTTT, you win if a run of four heads occurs before a run of four tails. Do you want to pick first or second? If you pick first, what should you pick? If your friend picks first, how should you respond? Solution: At first it seems as if this must be a fair game; after all, any particular sequence of four flips has the same probability of appearing (namely, 1 16) as any other. Indeed, if the coin is flipped four times and then the game begins anew, nothing of interest arises. But because the winning sequence could start with any flip, including the second, third or fourth, significant imbalances appear. For example, if you pick HHHH (a miserable choice) as your sequence and your friend counters with THHH, you will lose unless the game begins with four heads in a row (only a 1 16 probability). Why? Because otherwise the first occurrence of HHHH will be preceded by a tail, thus your friend will get there first with her THHH. How can you determine how to play optimally? For that you need a formula for the probability that string B beats string A. There is, in fact, a very nice one, discovered by the late, indominable John Horton Conway. Here’s how it works. We use the expression “A·B” to denote the degree to which A and B can overlap when A begins first (or they begin simultaneously). We measure this with a four-digit binary number x4x3x2x1 where xi is 1 if the first i letters of B match the last i of A, otherwise 0. For example, HHTH·HTHT = 0101(binary) = 5 since H[HTH]T gives an overlap of 3 while HHT[H]THT gives an overlap of 1. The left-most digit, x4, can only be 1 if A=B; thus A·B is equal to or greater than 8 if and only if A=B. Now if your friend has chosen A as her sequence, how should you choose B? You want B to have low waiting time, thus B·B should be as low as possible. You want B·A to be high, so that you maximize your probability of sneaking in ahead of your friend; and you want A·B to be low to minimize the probability that she sneaks in ahead of you. The formula says that the odds that B beats A are (A·A −A·B):(B·B −B·A). For example, suppose your friend picks A = THHT as her sequence (then A·A = 9). Your best choice is (always) to tack an H or T to the front of her choice and drop the last letter; this ensures that B·A is at least 4. If you pick B = HTHH you get B·B = 9, B·A = 4, and A·B = 2, so Conway’s formula says the odds in your favor are (9-2):(9-4) = 7:5. If instead you pick B = TTHH you get B·B = 8, B·A = 4 and A·B = 1, and the formula says the odds in your favor are (9-1):(8-4) = 2:1. So in this case B = TTHH is the better choice. 194 Was A = THHT a poor choice by your friend? Actually, it was not bad. Slightly better is THTT or HTHH, where you can’t get more than a 9:5 advan-tage. Conway’s formula continues to work beautifully for strings of length k, for any integer k ≥2, and also to non-binary randomizers such as dice. In the case of dice, the overlap vectors still consist of zeroes and ones, but are now interpreted base 6. Sleeping Beauty Sleeping Beauty agrees to the following experiment. On Sunday she is put to sleep and a fair coin is flipped. If it comes up Heads, she is awakened on Monday morning; if Tails, she is awakened on Monday morning and again on Tuesday morning. In all cases, she is not told the day of the week, is put back to sleep shortly after, and will have no memory of any Monday or Tuesday awakenings. When Sleeping Beauty is awakened on Monday or Tuesday, what—to her—is the probability that the coin came up Heads? Solution: This puzzle, introduced by philosopher Adam Elga in 2001, became the subject of major controversy in the philosophy community over the next 20 years. Many were persuaded by the following argument: On Sunday SB (Sleeping Beauty) knows that the probability of Heads is 1/2. She knows she will be awakened, so when she is awakened (on Monday or Tuesday), she has no new information and thus no reason to change her mind. So the answer is 1/2. 195 The flaw in this reasoning is that there is indeed new information: That she is awake now. Many arguments show that the correct answer is 1/3; one very nice example, offered by philosophers Cian Dorr and Frank Arntzenius, runs as follows. Suppose that SB is awakened both days, regardless of the coin, but if the coin comes up Heads, then on Tuesday, 15 minutes after she is awakened, she will be told “It’s Tuesday and the coin came up Heads.” She knows all this in advance. When she is awakened, it’s equally likely to be Monday or Tuesday (by sym-metry) and equally likely to be Heads or Tails (again by symmetry). Thus, the four events “Monday, Tails,” “Monday, Heads,” “Tuesday, Tails” and “Tuesday, Heads” are all equally likely. When, 15 minutes later, SB is not told that it’s “Tuesday, Heads” she rules out that last option and the rest remain equally likely. ♡ You might, on the other hand, be more persuaded by a frequency argument. If the experiment is run 100 times, you’d expect about 150 Monday/Tuesday awakenings, about 1/3 of which would be after Heads was flipped. The last puzzle will lead us to our theorem for this chapter. Leading All the Way Running for local office against Bob, Alice wins with 105 votes to Bob’s 95. What is the probability that as the votes were counted (in random order), Alice got the first vote and then led the whole way? Solution: Of course, what we’d really like to do is solve this problem in general (when Alice has won with a votes to Bob’s b, where a and b are arbitrary positive integers with a > b). The answer is given by what is often known as Bertrand’s Ballot Theorem (although it was apparently first proved by W. A. Whitworth in 1878, not J. L. F. Bertrand in 1887). Theorem. Let a > b > 0 and let x = (x1, x2, . . . , xa+b) be a uniformly random string of 1’s and −1’s, containing a 1’s and b −1’s. For each k between 1 and a+b, let sk be the sum of the first k elements of x, that is, sk = k X i=1 xi. Then the probability that s1, s2, . . . , sa+b are all positive is (a−b)/(a+b). Let’s first check that this statement tells us what we want to know. The string x codes the ballot-counting process; each 1 represents a vote for Alice, each −1 a vote for Bob. The sum sk tells us by how much Alice leads after the 196 kth vote has been counted; for example, if s9 = −3 we conclude that Alice was actually behind by three votes after the ninth vote was counted. If we apply the theorem to the puzzle above, we get the answer (105 − 95)/(105 + 95) = 1/20 = 5%. So how can we prove the theorem? There’s a famous and clever proof that uses reflection of random walk, but doesn’t really explain why the probability turns out to be the vote difference divided by the total vote. Instead, consider the following argument. One way to choose the random order in which the ballots are counted is first to place the ballots randomly in a circle, then to choose a random ballot in the circle and start counting (say, clockwise) from there. Since there are a+b ballots, there are a+b places to start, and the claim is that no matter what the circular order is, exactly a−b of those starting points are “good” in that they result in Alice leading all the way. To see this we observe that any occurrence of 1,−1 (i.e., a vote for Alice followed immediately by a vote for Bob in clockwise order) can be deleted with-out changing the number of good starting positions. Why? First, you can’t start with that 1 or −1, since if you start with the 1 the candidates are tied after the second vote, and if you start with the −1, Alice actually falls behind. Secondly, the 1,−1 pair has no influence upon the the eligibility of any other starting point, since all it does is raise the sum (Alice’s lead) by one and then lower it again. So we can keep deleting 1,−1’s until all that’s left is a−b 1’s, each of which is obviously a good starting point. We conclude that all along there were a−b good starting points, and we are done! ♡ 197 198 Chapter 11 Working for the System There’s no getting around it—many puzzles respond only to trying out ideas until you find one that works. Even so, there are good ways and bad ways to go about this process. Some people have a tendency to circle back and try the same answer many times. How can you avoid doing that? By classifying your attempts. Make some choice and stick with it, until it either yields a solution or runs out of steam; if the latter, you can then rule out one possibility, and progress has been made. There are other benefits to this besides avoiding redundancy: For example, you might actually see what’s going on and jump to the solution! No Twins Today It was the first day of class and Mrs. O’Connor had two identical-looking pupils, Donald and Ronald Featheringstonehaugh (pronounced “Fanshaw”), sitting to-gether in the first row. “You two are twins, I take it?” she asked. “No,” they replied in unison. But a check of their records showed that they had the same parents and were born on the same day. How could this be? Solution: We have begun with a puzzle that might not seem to respond to systematic search; you either get it or you don’t. But these days many people train them-selves to “think out of the box” and forget to first try thinking in the box. Accept that the boys are genetically identical and born to the same parents at the same time, and ask how this can happen if they are not twins. Then, at least, it might occur to you that there was an additional simultaneous sibling (or more!). The most likely possibility is that Donald and Ronald are triplets; the third (Arnold, perhaps?) is in another class. 199 Martians in a Circle A visiting logician is surrounded by a circle of Martians, each of whom either always tells the truth or always lies. She asks each Martian whether the Martian to his right is a truth teller or a liar, and from their answers, she is able to deduce the fraction of liars in the circle. What fraction is it? Solution: Observe that if all truth-tellers are changed to liars and vice-versa, none of answers would change. Therefore, if the fraction x can be determined, it must have the property that x = 1−x, thus x = 1/2. It remains to check that there is a set of answers from which the logician can deduce that half the Martians are truth-tellers. That requires, of course, that there be an even number of Martians in the circle. In fact, any circle of n liars and n truth-tellers will reveal its pattern (up to switching liars and truth-tellers) because “he is a liar” will be heard just when there is an alternation. Poker Quickie What is the best full house? (You may assume that you have five cards, and you have just one opponent who has five random other cards. There are no wild cards. As a result of the Goddess of Chance owing you a favor, you are entitled to a full house, and you get to choose whatever full house you want.) Solution: All full houses with three aces are equally high, because there can only be one such hand in a deal from a single deck. But there are other hands that can beat them: Any four-of-a-kind, of which there are always 11 varieties possible, and more relevantly any straight flush. Since AAA99, AAA88, AAA77, and AAA66 kill the most straight flushes (16—each ace kills only two, but each spot card kills five) they are the best full houses. Note that AAA55 doesn’t quite make the list, because one of the 5’s must be in the same suit as an ace and the A2345 straight flush in that suit is thus doubly covered. If you greedily insist on AAAKK, there are 40 −9 = 31 possible straight flushes that can beat you—even more if you haven’t got the four suits covered— instead of just 40 −16 = 24. Although the argument above only works for head-to-head poker (two play-ers), the result has in fact been verified by computer for any number of players. Fewest Slopes If you pick n random points in, say, a disk, the pairs of points among them will with probability 1 determine n(n−1)/2 distinct slopes. Suppose you get to 200 pick the n points deliberately, subject to no three being collinear. What’s the smallest number of distinct slopes they can determine? Solution: Three points, since they are not permitted to lie on one line, determine a triangle with three slopes. A little experimentation will convince you that with four, you can’t have any fewer than the four slopes you get from a square. In fact the vertices of a regular n-sided polygon determine just n different slopes. If n is odd, the polygon’s sides already have n different slopes, but every diagonal is parallel to some side. If n is even, the sides exhibit only n/2 different slopes but now the diagonals that connect two vertices an even number of edges apart are not parallel to any side, but they form n/2 parallel classes; so again we get n different diagonals. Can you do any better than the vertices of a regular n-gon? No. Let X be any configuration of n points in the plane, no three on a line, and let P be the southernmost point of X. Then you already have n−1 lines through P, each determining a different angle between −90◦and +90◦(relative to north = 0◦). The line through the two points that give the smallest and largest such angles gives you your nth slope. Two Different Distances Find all configurations of four distinct points in the plane that determine only two different distances. (Note: There are more of these than you probably think!) Solution: It may pay to think of the points as vertices of a graph, all of whose edges are the same length, and all of whose non-edges are the same length. If there is a “3-clique” in the graph, that is, three vertices of which any two are adjacent, then three of our points—call them, say, A, B, and C—form an equilateral triangle. The fourth point, D, can’t be connected to all the others, since in the plane we can’t have four points determining only one distance. If it is adjacent to two of the others, it forms another equilateral triangle and we get the upper-left configuration in the figure. If D is connected to one of the others, say A, it must be equidistant from B and C; thus it must either stick out from A directly away from the triangle, or directly toward it, giving the other two configurations on the top row. Finally, if D is adjacent to none of the three it must be equidistant from all three, thus in the middle of the triangle; this gives the lower-left configuration. What if there’s no triangle of edges (or non-edges) in the graph? If there’s a 4-cycle (but no triangle), its diagonals must be of equal length, thus the points are the corners of a square. 201 If there’s no triangle or 4-cycle in either the graph or its complement, the graph must be precisely a path three edges in length (hence its complement is, too). This is the graph you get by deleting one edge of a 5-cycle, and corresponds to the configuration you get by deleting one vertex of a regular pentagon. First Odd Number What’s the first odd number in the dictionary? More specifically, suppose that every whole number from 1 to, say, 1010 is written out in formal English (e.g., “two hundred eleven,” “one thousand forty-two”) and then listed in dictionary order, that is, alphabetical order with spaces and hyphens ignored. What’s the first odd number in the list? Solution: In dictionary order, one word precedes another precisely when the first letter in which the two words differ is, in the first word, earlier in the alphabet. Spaces and punctuation are ignored, and if there is no differing letter, the shorter word appears first. So finding the dictionary-first word in a set of words is just a matter of carefully and systematically considering the successive words involved in the description of a number. 202 The earliest actual digit is “eight,” and the earliest odd digit “five.” We don’t need to consider any other digits, but other possibly useful words that appear in numbers include “billion,” “eighteen,” “eighty,” “hundred,” “million,” and “thousand.” Our earliest odd number must begin with a digit so should start with “eight billion.” After that, “eighteen” is the best we can do, and proceeding along these lines, we eventually get the answer 8,018,018,885: “eight billion, eighteen million, eighteen thousand, eight hundred eighty-five.” A little more work will get you to the first prime number in the dictionary, 8,018,018,851. Annoyingly, “dictionary order” is (according to Wikipedia) subtly different from “alphabetical order,” and the difference makes a difference. The issue is that in alphabetical order, which is used with people’s names, spaces typically are not ignored, but instead precede all other characters. Thus, alphabetically, “eight hundred” precedes “eighteen,” and the first odd number in alphabetical order becomes 8,808,808,885. The first prime would in that case be the very next odd number in alphabetical order, namely, 8,808,808,889. Measuring with Fuses You have on hand two fuses (lengths of string), each of which will burn for exactly 1 minute. They do not, however, burn uniformly along their lengths. Can you nevertheless use them to measure 45 seconds? Solution: Simultaneously light both ends of one fuse and one end of the other; when the first fuse burns out (after half a minute), light the other end of the second. When it finishes, 45 seconds have passed. ♡ If you allow midfuse ignition and arbitrary dexterity, you can do quite a lot with fuses. For example, you can get 10 seconds from a single 60-second fuse by lighting at both ends and at two internal points, then lighting a new internal point every time a segment finishes; thus, at all times, three segments are burning at both ends and the fuse material is being consumed at six times the intended rate. Bit of a mad scramble at the end, though. You’ll need infinitely many matches to get perfect precision. King’s Salary Democracy has come to the little kingdom of Zirconia, in which the king and each of the other 65 citizens has a salary of one zircon. The king can not vote, but he has power to suggest changes—in particular, redistribution of salaries. Each person’s salary must be a whole number of zircons, and the salaries must sum to 66. Each suggestion is voted on, and carried if there are more votes for than against. Each voter can be counted on to vote “yes” if his or her salary is to be increased, “no” if decreased, and otherwise not to bother voting. 203 The king is both selfish and clever. What is the maximum salary he can obtain for himself, and how many referenda does he need to get it? Solution: There are two key observations: (1) that the king must temporarily give up his own salary to get things started, and (2) that the game is to reduce the number of salaried citizens at each stage. The king begins by proposing that 33 citizens have their salaries doubled to 2 zircons, at the expense of the remaining 33 citizens (himself included). Next, he increases the salaries of 17 of the 33 salaried voters (to 3 or 4 zircons) while reducing the remaining 16 to no salary at all. In successive turns, the number of salaried voters falls to 9, 5, 3, and 2. Finally, the king bribes three paupers with one zircon each to help him turn over the two big salaries to himself, thus finishing with a royal salary of 63 zircons. It is not difficult to see that the king can do no better at any stage than to reduce the number of salaried voters to just over half the previous number; in particular, he can never achieve a unique salaried voter. Thus, he can do no better than 63 zircons for himself, and the seven rounds above are optimal. ♡ More generally, if the original number of citizens (including the king) is n, the king can achieve a salary of n−3 zircons in k rounds, where k is the least integer greater than or equal to log2(2n−4). Packing Slashes Given a 5 × 5 square grid, on how many of the squares can you draw diagonals (slashes or backslashes) in such a way that no two of the diagonals meet? Solution: You can get 15 easily (see figure below) using rows (or columns) 1, 3, and 5, with all slashes or all backslashes on each line, or by nested Ls. Is it possible to do any better than that? 204 There are 6 × 6 = 36 vertices (cell corners) in the grid and each diagonal ac-counts for two, so you certainly will not be able to fit in more than 18 diagonals. But you can see that, for example, it won’t be possible to use all 12 corners of the top row of squares. In fact, come to think of it, every diagonal touching one of the 20 outer vertices must also touch one of the 12 outer vertices of the inside 3 × 3 grid— except for those in a corner cell that miss the corner vertex of the big square, for example, a backslash in the upper right cell. But there can’t be backslashes in both cells neighboring the upper right corner cell, so putting a backslash in the upper right corner will cause at least one other outside vertex to be missed. In conclusion, no matter what we do, at least four outside vertices will be unmatched, and we’re down to an upper bound of (36−4)/2 = 16 diagonals. When we try to achieve 16, we notice that using up all the outer vertices of the inside 3 × 3 grid to take care of outer vertices won’t work—that leaves the four corners of the center cell unmatched, and a diagonal in that cell takes care of only two, taking us back to 15 diagonals. We’ll need to leave at least two of the four corners of the big square unmatched, and use two or four unmatched outer vertices of the inside 3 × 3 grid to cover the corners of the central cell. A little experimentation shows that this works only if you miss all the corners of the big square, producing the picture below—the unique solution to the puzzle, up to reflection. 205 Unbroken Lines Can you re-arrange the 16 square tiles below into a 4 × 4 square—no rotation allowed!—in such a way that no line ends short of the boundary of the big square? Solution: Notice that there are 24 = 16 different tiles that can be made by choosing, for each line-type, whether to include that type or not. Each tile is represented once in the puzzle. This is, again, a matter of intelligent search. The horizontal lines must all be in two rows, the vertical lines in two columns; there are 4 2 2 = 36 combinations to try. The lengths of the SW-NE diagonals must sum to 8 squares worth, and thus there are only a few possibilities: A 4 and two 2s, a 4, a 3 and a 1, etc. There are in fact several ways to do it of which one is illustrated below. Curiously, all ways of doing it work on the torus as well as on the square—in other words, if you imagine that the square wraps around top-to-bottom and left-to-right, no line ends at all; all are, ultimately, loops. [See Notes & Sources for the composer and inspiration behind this and the next puzzle.] 206 Similar, but harder: Unbroken Curves Can you re-arrange the 16 square tiles below into a 4 × 4 square—no rotation allowed!—in such a way that no curve ends short of the boundary of the big square? Solution: Again, there are 24 = 16 different tiles that can be made by choosing, for each corner of the tile, whether to include a quarter-circle at that corner or not. Each tile is represented once in the puzzle. For this puzzle, the best route to a solution seems to be to guess how the squares with the most quarter-circles fit together and (systematically!) chase down the possibilities. Not all solutions to this one work on the torus as well as the square, but the one illustrated below does work on both. Conway’s Immobilizer Three cards, an ace, a deuce, and a trey, lie face-up on a desk in some or all of three marked positions (“left,” “middle,” and “right”). If they are all in the 207 same position, you see only the top card of the stack; if they are in two positions, you see only two cards and do not know which of them is concealing the third card. Your objective is to get the cards stacked on the left with ace on top, then deuce, then trey on bottom. You do this by moving one card at a time, always from the top of one stack to the top of another (possibly empty) stack. The problem is, you have no short-term memory and must, therefore, devise an algorithm in which each move is based entirely on what you see, and not on what you last saw or did, or on how many moves have transpired. An observer will tell you when you’ve won. Can you devise an algorithm that will succeed in a bounded number of steps, regardless of the initial configuration? Solution: Since there aren’t that many things you can do, it’s tempting to try some arbi-trary set of rules and see if they work. Actually, there’s a surprising number of different algorithms possible with the above constraints—more than a quintil-lion. So some reasoning will be needed. But it’s tricky to design an algorithm that makes progress, avoids cycling, and doesn’t do something stupid when it’s about to win. For example, it looks obvious to try to win by putting the A on the 2 if you see 2,A,– or 2,–,A. Then if you see three cards with the 3 on the left, you can win in two moves by putting the 2 on the 3. So far, so good. Suppose the 2 is exposed on the left but no other card is seen; then we may as well move the 2 to the center. Now, if you see 3,2,–, it’s tempting to put the 2 on the 3. But if the A is hiding under the 3, you will get caught in a loop, stuck forever going back and forth between 2,–,– and 3,2,–. So you’ll need to move the 2 to the right from 3,2,–. But then you can’t put 2 on 3 from 3,–,2 without getting caught in a loop of order 3. So you’ll have to move the 3 from that position. Darn! It’s beginning to look like we should be filling holes by moving cards in one direction, say to the right, to avoid loops. So let’s agree that when we see only 2 cards, we move a card to the right (around the corner from R to L, if necessary) to an open slot; the only exceptions will be 2,A,– and 2,–,A where we place A on 2 to try to win. Moving cards to the right will spread out the cards in two moves at most, so we can see them. Seeing only one stack, sending the top card to the left will then also spread the cards in two moves. It remains only to decide what to do when we see all three cards, but the 3 is not on the left. Clearly we can’t move a card left; it would come back immediately. So we move a card to the right; which one? After two moves whichever card we covered will be re-exposed, with the other two switched. Having got this far, several successful algorithms can now be described. Here are the rules for one of them, in priority order: 1. Seeing 2,A,–, place A on 2; 208 2. Otherwise, seeing two cards, move a card right (around the corner if nec-essary) to the open place; 3. Seeing just one card, move that card to the left; 4. Seeing three cards, move the middle-rank card to the right. This same algorithm works for any number of cards, numbered 1 through n. It takes quadratic time in the worst or random case, that is, approximately a constant times n2 steps. A computer scientist would conclude that a list of items can be sorted with just three LIFO (last-in, first-out) stacks and no memory. Here’s a second solution that’s slower but has a compensating feature. 1. Seeing just one card, move a card to the right. 2. Seeing two cards: (a) If the left is empty, move the center card to the right stack; (b) If the center is empty, move the right card to the left stack; (c) If the right is empty, move the left card to the right stack. 3. Seeing three cards, let the numbers be x, y, and z from left to right. (a) If z is the largest or smallest, move the smaller of x and y onto z; (b) If z is the middle, move the larger of x and y onto z. This algorithm is cubic, requiring (2/3)n3 + (1/2)n2 −(7/6)n steps if it is started with the cards stacked in reverse order on the left. Once it gets started, it never has more than one card in the center. Thus, we see that sorting can be done with only two LIFO stacks, a single register, and no memory! If you can beat that, let me know. Seven Cities of Gold In 1539 Friar Marcos de Niza returned to Mexico from his travels in what is now Arizona, famously reporting his finding of “Seven Cities of Gold.” Coronado did not believe the “liar friar” and when subsequent expeditions came up empty, Coronado gave up the quest. My (unreliable) sources claim that Coronado’s reason for disbelieving de Niza is the latter’s claim that the cities were laid out in the desert in a manner such that of any three of the cities, at least one pair were exactly 100 furlongs apart. Coronado’s advisors told him no such pattern of points existed. Were they right? 209 Solution: Putting ourselves in the position of Coronado’s advisors, we need to decide whether there’s a way to position seven points in the plane in such a way that among any three, there are at least two that are a specified distance apart. What could we try? What we want is a seven-vertex graph drawn on the plane, with the property that every edge is a line segment of unit length (the unit being 100 furlongs = 12 1 2 miles), having the following property: Every three vertices contain an edge. If there were only three vertices, we could of course arrange them as an equilateral triangle of side 1, and get more: An edge between every pair of vertices. In fact, two of these equilateral triangles (say, ABC and DEF), no matter their relative position, would get us a six-vertex solution. Why? Because any three vertices would include at least two in one of the two triangles, and there’s our edge. So let’s see if we can add a seventh vertex, call it G, to our two triangles. Then we only have to worry about sets of three vertices that contain vertex G. We can’t connect G to all three vertices of one of our triangles, but we can connect it to two (say, A and B from the first triangle, and D and E from the second) by making diamonds AGBC and DGEF that share vertex G. That leaves just one set of three vertices without and edge: The set {G, C, F}. No problem: We just angle the two diamonds, hinged at G, so that their far ends are a unit distance apart. Done! Here’s what it looks like: This graph is known as the “Moser spindle,” and it’s the only graph with the desired property. Theorem. Suppose seven distinct points in the plane have the property that among any three of them, there are two that are exactly a unit distance apart. Then the graph on those seven points, defined by inserting an edge between each 210 pair that are a unit distance apart, is the Moser spindle. Proof. To prove this we again need to be systematic. We assume the graph H satisfies the requisite properties, and gradually learn more and more about it until we can show it actually is the Moser spindle. First step: We observe that for any vertex v, the vertices not equal to or adjacent to v must form a “clique,” that is, they must all be connected to one another. Why? Because if two (say, x and y) are not, then the set {v, x, y} would violate the condition. But four vertices cannot all be connected to each other by unit-distance edges, so the number of vertices not equal or adjacent to v is at most three; it follows that v, and thus every vertex of H, has at least three neighbors. Second step: No vertex can have as many as five neighbors. Reason: If vertex u, say, has five neighbors, they all lie in a circle of unit radius with center at u and cannot be connected among themselves any better than by a path of four edges. Such a path has three vertices no two of which are mutually adjacent, and that’s not allowed. Third step: We now know all vertices have either three or four neighbors, but they can’t all have just three neighbors—for then, the sum of the degrees of the vertices of H would be 3 × 7 = 21 which is an odd number and each edge contributes 2 to that sum. Hence, there is at least one vertex, say w, that has four neighbors. Fourth step: How are w’s neighbors connected among themselves? At least in two pairs, say p with q and r with s, because otherwise there’d be three with no edge. The two remaining vertices (call them x and y) must be adjacent to each other, else {w, x, y} would be a bad set. Neither x nor y can be adjacent to three of p, q, r, and s because there’s already a point (w) with that property and there can’t be two. Wrap-up: Of p, q, r, and s, the ones x is not adjacent to must form a clique, otherwise we’d have another bad set; and the same for y. That is possible only if one of x and y is adjacent to precisely p and q, and the other to r and s. And that’s the Moser graph! ♡ 211 212 Chapter 12 The Pigeonhole Principle The pigeonhole principle is nothing more than the evident fact that if n pigeons are to be housed in m holes, and n > m, then some hole must contain more than one pigeon. Yet this principle has wide-ranging application, and is the key to proving some surprising statements and some important mathematical truths. First, a warm-up. Shoes, Socks, and Gloves You need to pack for a midnight flight to Iceland but the power is out. In your closet are six pairs of shoes, six black socks, six gray socks, six pairs of brown gloves, and six pairs of tan gloves. Unfortunately, it’s too dark to match shoes or to see any colors. How many of each of these items do you need to take to be sure of getting a matched pair of shoes, two socks of the same color, and matching gloves? Solution: The easiest way to solve such puzzles is to figure out the maximum number of items you could have without getting what you want, then adding one. In this way you are implicitly invoking the pigeonhole principle. 213 For the shoes, you could have one shoe from each pair, six in all; so you need seven to guarantee a pair. For the socks, one black and one gray will not work but the third sock gets you a pair. The gloves are yet another issue: You need a right- and a left-hander, so the biggest disaster would be six brown all left or all right, and six tan, also all left or all right. So 13 gloves are required, although you would have to count yourself very unlucky indeed to take only 12 and not get a match! The same idea is easily employed in certain poker calculations. A standard deck of 52 playing cards contains 13 cards of each suit, and within each suit, an ace, deuce, 3, 4, etc. up to 9, 10, jack, queen, king. Thus there are four cards of each rank. A poker hand is five cards. Two cards of the same rank constitute a pair; three, trips (or “three of a kind”). Five cards of the same suit make a flush; five cards of consecutive ranks (including 5432A as well as AKQJ10) make a straight. How many cards do you need to hold to guarantee a pair—that is, to be sure that among your cards is a 5-card poker hand containing two cards of the same rank? Easy: The worst holding (for this problem!) is 13 cards consisting of one card of each rank, therefore the answer is 14 (but note that 12 cards are enough to guarantee a pair or better). If it’s trips you seek, the worst holding is two cards of each rank, so you need 2 × 13 + 1 = 27 for your guarantee. For a flush, four of each suit will stump you, so you need just 17 cards. The straight is the trickiest: You must have a 5 or a 10 to make a straight, so you can hold 52 −2 × 4 = 44 cards without making a straight, hence it takes a stunning 45 cards to guarantee one. Note the contrast with relative probabilities of getting trips, flush or straight when dealt five cards: The flush is the hardest, and indeed outscores the straight and trips in poker, even though it takes the fewest cards to guarantee. For the next puzzle, one preliminary bit of cleverness is needed. Polyhedron Faces Prove that any convex polyhedron has two faces with the same number of edges. Solution: The key is to pick a face with the maximum number of edges, say E. Since it has E neighboring faces, the number of faces of the polyhedron is at least 214 E+1. But their numbers of edges range only from 3 to E, comprising only E−2 different numbers, so two must have the same number of edges. (In fact, since we have some overkill here, you can deduce that there are either three pairs of faces with the same number of edges, or a triple and a pair, or four faces with the same number of edges.) Sometimes you have to look around a bit to find the right items to use for your pigeonholes, as in this geometric example. Lines Through a Grid If you want to cover all the vertices in a 10 × 10 square grid by lines none of which is parallel to a side of the square, how many lines are needed? Solution: Your first idea here might be to use 19 diagonals running 45◦running (say) north-west to south-east, but you immediately see that the two extreme ones can be replaced by a single SW-NE diagonal covering the SW and NE corners, as in the figure below. This reduces the count to 18; can you do any better? No, because there are 36 vertices on the boundary of the grid, and no non-vertical, non-horizontal line can cover more than two of them. Now let us move on to an oddly counterintuitive fact about numbers. Same Sum Subsets Amy asks Brad to pick ten different numbers between 1 and 100, and to write them down secretly on a piece of paper. She now tells him she’s willing to bet $100 to $1 that his numbers contain two nonempty disjoint subsets with the same sum! Is she nuts? 215 Solution: Of course, the answer is that Amy is perfectly sane and in fact the bet is not merely favorable to her; it’s a sure thing. How can you prove it? There’s no unique maximum set-size Brad can pick that doesn’t have the desired property, so the shoe-sock-glove technique won’t work. But whenever you are asked to show that there are two objects that share a property, you can try to show that there are more objects than properties; then, by pigeonhole, you’re done. Here, the number of nonempty subsets of Brad’s set of 10 numbers is 210−1 = 1023. The possible sums range only from 1 to 100 + 99 + · · · + 91 < 1000, so there are indeed fewer sums than subsets; by the pigeonhole principle, there must be two different subsets with the same sum. But wait—the problem asked for disjoint subsets. That was a red herring: Throw away the common elements of two sets with equal sums to get disjoint sets with equal sums. ♡ The next problem looks geometric, but is really number-theoretic; nor is it obvious at first glance that the pigeonhole principle has a role to play here. Lattice Points and Line Segments How many lattice points (i.e., points with integer coordinates) can you have in 3-space with the property that no two of them have a third lattice point on the line segment between them? Solution: The first issue to be addressed here is: When do two lattice points, say (a, b, c) and (d, e, f), have another lattice point on the line segment between them? A moment’s thought will convince you that this happens when the numbers a−d, b−e, and c−f have a common divisor. The “easiest” common divisor for them to have is 2, and this will happen if a−d ≡b−e ≡c−f ≡0 mod 2, i.e., if the two vectors have the same parity coordinatewise. When that happens, the midpoint of the two vectors is a lattice point. But there are only 23 = 8 parities available, so by pigeonhole, we can’t have nine points in our set. Can we have eight? Yes, just take the corners of a unit cube. Here’s another number problem. Adding, Multiplying, and Grouping Forty-two positive integers (not necessarily distinct) are written in a row. Show that you can put plus signs, times signs, and parentheses between the integers in such a way that the value of the resulting expression is evenly divisible by one million. 216 Solution: This puzzle has a pigeonhole flavor to it and indeed the pigeonhole principle is useful in it, but not perhaps where you might expect. Since a million is 26 · 56, and 42 = 6·2+6·5, a logical idea is to break up the numbers into six contiguous groups of size 2 and six more of size 5, with the multiplication symbols between the groups; then all that is needed is to “persuade” each group of size 2 to represent an even number, and each group of size 5 to represent a multiple of 5. The groups of size 2 are easily handled: If the two numbers inside a group are both even or both odd, add them; otherwise, multiply them. The groups of size 5 are only slightly trickier. If we can get a contiguous substring of the five numbers that sums to a multiple of 5, we’re done; we can add those and multiply the sum by the product of the remaining numbers. Here’s where pigeonhole comes in. Let the numbers be n1, . . . , n5 and exam-ine the partial sums s1, . . . , s5 where si = n1+· · ·+ni. Let ri be si mod 5, that is, ri is the remainder when si is divided by 5. If any ri is 0, we use the substring n1, . . . , ni; otherwise, the five si’s all lie in the set {1, 2, 3, 4}. Thus two of them, say ri and rj with i < j, are equal. Then sj −si = ni+1 + ni+2 + · · · + nj is a multiple of 5 and we are done. ♡ Notice that this argument works for 5 replaced by any n. We have in effect proven a nice lemma which says that every string of n integers has a nonempty substring that sums to a multiple of n. Line Up by Height Yankees manager Casey Stengel famously once told his players to “line up al-phabetically by height.” Suppose 26 players, no two exactly the same height, are lined up alphabetically. Prove that there are at least six who are also in height order—either tallest to shortest, or shortest to tallest. Solution: Let’s try small numbers to see what’s going on. Two players are, obviously, enough to get two that are also in height order one way or the other. Three or four players won’t get you any more: For example, their heights could be 73′′, 75′′, 70′′, and 72′′ respectively. But five players will get you three in height order; by the time you reach that fifth player (say, Rizzuto) alphabetically, there’s a previous player (say, Berra) who’s got both a taller and shorter player before him. Then, whether Rizzuto is taller or shorter than Berra, he ends a height-order subsequence of length 3. This reasoning suggests a sort of “dynamic programming” approach to the 26-player problem. Give each player two numbers, an “up” number and a “down” number. The up number records the length of the longest height-increasing subsequence that ends with that player, and the down number is defined analagously. For example, in the height sequence 73′′, 75′′, 70′′, 72′′ from above, the up-down number pairs would be (1,1), (2,1), (1,2), and (2,2). 217 Notice these are all different—they have to be! Rizzuto and Berra, for in-stance, can’t have the same pair because if the alphabetically later player (Riz-zuto) is taller his up number would be higher, and if shorter, his down number would be higher. Now we see why five players guarantees a subsequence of three in height order: Only four different up-down pairs can be made of numbers less than 3. Similarly, if there were no height-order subsequence of 26 players of length 6, all the up-down pairs would have to be made up of numbers from 1 to 5 and there are only 25 of those. By the pigeonhole principle, there must be a player with an up-number or down-number of 6. The key relationship is that if a subsequence of length k+1 is sought, you need the number of players to be greater than k2. Still more generally, if the number of players is at least jk+1, you are guaranteed either a height-increasing subsequence of length j+1 or a height-decreasing subsequence of length k+1. The argument is the same: If you didn’t have either such subsequence, the number of up-down pairs would be limited to jk, and since the pairs are all different, you get a pigeonhole contradiction. In fact, this last statement is known as the Erd˝ os-Szekeres Theorem, and can be even further generalized (with almost the same pigeonhole proof) to a state-ment about height and width of partially ordered sets—for those readers who know what they are. The Erd˝ os-Szekeres Theorem is “tight” in the sense that if there are only jk players, you might indeed be limited to length j for height-increasing subsequences and k for height-decreasing. An example: If, say, j = 4 and k = 3, the heights (in inches) could be 76,77,78,79,72,73,74,75,68,69,70,71. Here’s a related problem where we can use a similar idea. Ascending and Descending Fix a number n and call a list of the numbers from 1 to n “good” if there is no descending subsequence of length 10. Show that there are at most 81n good lists. Solution: Let x1, . . . , xn be a good list and let di be the length of the longest descending subsequence ending at xi. Then di is a number between 1 and 9, which we call the “color” of the position i, and also of the number xi. Essentially, di is the “down-number” from the solution to the previous puzzle. The numbers xi for which di is some particular color, say color 7, form an increasing subsequence. Suppose we know the positions with color 7 and also the numbers of color 7. Then we can fill in all the numbers that go in those positions, because we know they are in increasing order. For example, if the list begins 42, 68, 35, 50, then positions 1 and 2 get color 1, as do the numbers 42 and 68. Knowing that d1 = d2 = 1 and the set of numbers that got color 1, we can conclude that x1 = 42 and x2 = 68 because 218 the first two positions are colored 1 and the two lowest numbers that got the first color are 42 and 68. Now, since there are at most 9n ways to assign colors to positions, and 9n ways to assign colors to numbers, we get (using, again, the Pigeonhole Principle) a maximum of (9n)2 = 81n different lists. Note that this bound of 81n, though exponentially large, is a vanishingly small fraction of all the n! ways to list the numbers from 1 to n. Sometimes the pigeonhole principle can be applied even though a puzzle asks only for one item, not two similar ones. The trick is to use subtraction to turn two similar items into one useful one. Zeroes and Ones Show that every natural number has a non-zero multiple which, when written out (base 10), contains only zeroes and ones. Even better, prove that your telephone number, if it ends with a 1, 3, 7, or 9, has a multiple whose decimal representation is all ones! Solution: There are infinitely many numbers whose decimal representations contain only ones and zeroes—or, for that matter, only ones. So some of them ought to be multiples of n unless there’s some reason why they can’t be. But how can we prove it? Perhaps we can make use of the fact that if we subtract two numbers each of which has the same remainder when divided by n, the result will be a multiple of n. There are only n different remainders possible when you divide by n, so pigeonhole tells us that n+1 numbers are enough to guarantee two with the same remainder. Now we only have to find n+1 numbers with the property that the difference between any two has only zeroes and ones as its digits. That’s not hard: take numbers of the form 111 . . . 111, where the number of digits runs from 1 to n+1. Formally speaking, these numbers are (10k −1)/9 for 1 ≤k ≤n+1. (Actually, we only need n of them because if all their remainders modulo n differ, one of them already has remainder 0, and we can use it directly.) To wrap up, we find two distinct numbers all of whose digits are ones, that have the same remainder when divided by n; subtracting the smaller from the larger gives a number of the form 111 . . . 111000 . . . 000. This means that we have actually proved a stronger statement than required, namely that any n has a multiple of the form 111 . . . 111000 . . . 000. But, as often happens, asking for more would have made the puzzle easier. We’ll see a more extreme example of that phenomenon in the next puzzle. Wait, what about your telephone number? We know now that it has a multiple of the form N = 111 . . . 111000 . . . 000, ending in, say, k zeroes. But if your number ends in 1,3, 7 or 9, it is not a multiple of 5 or 2. It follows that N/(5 · 2)k, which is all ones, is still a multiple of your phone number. 219 The next puzzle looks like one in which pigeonhole refuses to work. It takes some serious insight to see how it ultimately does. Same Sum Dice You roll a set of n red n-sided dice, and a set of n black n-sided dice. Each die is labeled with the numbers from 1 to n. Show that there must always be a nonempty subset of the red dice, and a nonempty subset of the black dice, showing the same sum. Solution: This seems like an ideal pigeonhole application: There are lots of nonempty subsets (2n −1 of each color) and sums can only go from 1 to n2. But it doesn’t do any good to have, say, two red subsets with the same sum, and two black subsets with some other sum. Come to think of it, there may be very few distinct sums among, say, the red dice; for example, they could all be showing the same number of pips, say j. In the meantime the black dice could all be agreeing on some other number, k. That does not give us a counterexample, because then k of the reds and j of the blacks both sum to jk, but that coincidence seems like something the pigeonhole principle would be hard-pressed to lay claim to. Nonetheless, with help from a subtraction trick similar to the one we used in Zeroes and Ones, pigeonhole saves us. The idea is the following: Suppose the subsets R (of the red dice) and B (of the black) don’t work, because the sum of the dice in R exceeds that of B by some non-zero number d. Let’s add some more red dice to R to get a bigger set R+, and similarly add some black dice to B to get B+. Now we check the sums for R+ and B+ and they still don’t agree, but suppose that the discrepancy is the same as it was before: The dice in R+ sum to exactly d more than the dice in B+. Then we’re done, because the dice we added to R must have had the same sum as those we added to B. To get this idea to work we put each set of dice in some fixed left-to-right order. Amazingly, it doesn’t matter a bit what orders we choose. Let ri be the sum of the first i red dice, reading the dice from left to right, and similarly let bj be the sum of the leftmost j black dice. We claim that there are contiguous subsets of the red and of the black dice with the same sum. A subset is “contiguous” if it is a substring of the dice as laid out, for exam-ple, all the red dice from the third to the seventh. There are n+1 2 = n(n+1)/2 220 such subsets of each color, because we can get one by choosing two of the n+1 spaces between or at the ends of the dice in which to put down supermarket dividers. It’s hard to believe that we can get away with limiting ourselves, in this seemingly arbitrary manner, to just these few of the exponentially many subsets of each color. Can this really work no matter how the dice are rolled, and no matter how we order them? We can assume rn < bn; if they are equal we’re done, and if the inequality goes the other way we can just switch the roles of the colors. Let us find, for each i, the j for which bj comes closest to ri without exceeding it. More formally, let j(i) = max{j : bj ≤ri} with the understanding that b0 = 0, and thus j(i) may be zero for some small i’s if the red sequence starts with smaller numbers than the black sequence. Now let’s look at the discrepancies ri −bj(i). All are non-negative by defini-tion of j(i); moreover, all are less than n, because if ri −bj(i) were equal to n or more, we could have thrown in the next black die without exceeding ri. (Since rn < bn, there always is a next black die.) If any of the discrepancies ri −bj(i) is zero, we’re done, so we can assume all the discrepancies are among the numbers 1, 2, . . . , n−1. There are n such discrepancies, one for each i, so there must be two that are equal; in other words, there are two distinct indices i < i′ with ri′ −bj(i′) = ri −bj(i). Now we play the subtraction game, concluding that the red dice from number i+1 up to number i′ must add up to the same value as the black dice from number j(i)+1 up to number j(i′). Done! Our last puzzle looks perhaps less like a pigeonhole problem than its pre-decessors, but now that you’ve seen the subraction trick, you won’t be fooled. Nonetheless, it’s not easy. Zero-sum Vectors On a piece of paper, you have (for some reason) made an array whose rows consist of all 2n of the n-dimensional vectors with coordinates in {+1, −1}— that is, all possible strings of +1’s and −1’s of length n. Notice that there are lots of nonempty subsets of your rows which sum to the zero vector, for example any vector and its complement; or the whole array, for that matter. However, your three-year-old nephew has got hold of the paper and has changed some of the entries in the array to zeroes. Prove that no matter what your nephew did, you can find a nonempty subset of the rows in the new array that sums to zero. 221 Solution: The technique used in Same Sum Dice suggests that to get a set of new rows that sum to zero, we might try to build a sequence of new rows and use pigeonhole to argue that two of its partial sums coincide; then the portion of the sequence between those two partial sums will add to the zero vector. To make that work, the sequence will have to be designed so that the partial sums belong to some small set. How small can we make it? Would you believe that we can force all the partial sums to be {0, 1}-vectors? We can certainly start the list with a {0, 1}-vector: We just take the row u′ that used to be u = ⟨1, 1, 1, . . . , 1⟩. If our nephew happened to have changed all of u’s coordinates to zeroes, we’re done. Otherwise, there will be some 1’s in u′, and we need to choose the next row v′ in our sequence in such a way that u′ + v′ has all of its coordinates in {0, 1}. But we can do that by choosing v to be the vector that has −1 in the places where u′ is 1, and +1 in the places where u′ is zero. Then no matter how v was altered by your nephew, u′ + v′ will have the desired property. We continue this procedure as follows: At time t, t altered rows have been put on our sequence and all partial sums, including the sum x of all t of them, are {0, 1}-vectors. Let z be the original row with −1 wherever x has a 1, and +1 wherever x has a 0; add z’s altered form z′ to the sequence. Since there are only finitely many {0, 1}-vectors of length n, we will even-tually hit the same sum twice; let us stop when this happens for the first time. (We consider the 0th partial sum to be the zero vector, so one way this can happen is for the zero vector to arise again as a partial sum.) Call the vectors w1, w2, . . . , so that our critical time is the least k such that w1 +w2 +· · ·+wj = x = w1 + w2 + · · · + wk, with 0 ≤j < k. Then wj+1 + wj+2 + · · · + wk is the zero vector, and we are done. Note, very importantly, that we have not used any altered vector too many times; the reason is that since up to time k all the partial sums have been different, all the original vectors z whose altered forms we appended to our sequence were different. Here’s a more serious use of the pigeonhole principle in mathematical anal-ysis. Suppose we fix some real number r and let {nr} be the fractional part of nr, that is, the unique number in the half-open interval [0, 1) that can be expressed as nr minus an integer. If r is a fraction (say, a/b) written in lowest terms, then the values {r}, {2r}, {3r}, . . . will cycle among the multiples of 1/b in [0, 1). What if r is irrational? Then the {nr}’s will not repeat and it’s natural to conjecture that they are in fact dense in the unit interval; in other words, for any ε > 0 and any real x ∈[0, 1), there is some positive integer n for which \|{nr} −x\| < ε. Theorem. For any irrational number r, if {nr} is the fractional part of nr, then the numbers {r}, {2r}, {3r}, . . . are dense in the unit interval. Proof. Fix some irrational r. Of course {nr} is never zero for any positive 222 integer n, but we can borrow an idea from the Zeroes and Ones puzzle above to show that we can choose n to get nr as close as we want to an integer. Let 1 ≤i < j. If {ir} < {jr}, , then {(j−i)r} = {jr} −{ir}; if {ir} > {jr}, then {(j−i)r} = 1 −({ir}−{jr}). Pick any ε > 0, and let m be an integer bigger than 1/ε. Divide the unit interval into m equal intervals; by pigeonhole, one of those little intervals (each of length < ε) contains two of the values {r}, {2r}, . . . , {(m+1)r}. Let those be {ir} and {jr} with i < j; then if k = j−i, we conclude that either {kr} < ε or 1−{kr} < ε. Suppose that the former inequality holds, that is, {kr} < ε. Given any real value x ∈[0, 1], let p be the greatest integer such that p{kr} < x. Then {pkr} = p{kr} < x ≤(p+1){kr} but the outside quantities differ by {kr} which is less than ε, so in particular \|x−{pkr}\| < ε and we have our approximation. If instead 1−{kr} < ε, we take p to be the least integer such that p(1−{kr}) > 1−x, and a similar argument again shows \|x−{pkr}\| < ε. ♡ 223 224 Chapter 13 Information, Please The basis of this chapter is simple: If you want to distinguish among n possi-bilities, you have to do something that results in at least n different possible outcomes. A common puzzle application is to balance-scale problems. Each weighing by a balance scale can produce at most three outcomes: Tilt left (L), tilt right (R), or balance (B). If you are permitted two weighings, you get 3 × 3 = 9 possible outcomes: LL, LR, LB, RL, RR, RB, BL, BR, and BB. Similarly, with w weighings you could have as many as 3w outcomes, but no more. So, if a puzzle gives you a bunch of objects to test, k possible answers, and allows w weighings on a balance scale, the first thing to ask yourself is whether k ≤3w. If not, the task is impossible (or you have missed some trick in the wording). If you do have k ≤3w, the task may or may not be possible, and you can use the idea above to help you construct a solution or to prove impossibility. Here’s a classic example. Finding the Counterfeit You have a balance scale and 12 coins, 11 of which are genuine and identical in weight; but one is counterfeit, and is either lighter or heavier than the oth-ers. Can you determine, in three weighings with a balance scale, which coin is counterfeit and whether it is heavy or light? Solution: There are 24 possible outcomes, less than 33 = 27, so the task may be possible but needs to be pretty efficient—in other words, most weighings will need to have all three possible outcomes. We certainly want to start by weighing k coins against k others, where k ≤6. That would leave 12 −2k suspect coins if the initial weighing balances, so it better be possible to test that many in two weighings. Two weighings have at most nine possible outcomes, so we can’t afford to leave more than four coins out of the first weighing; thus k is at least 225 4. In fact k can’t be more than 4 because then “tilt left” would lead to 10 > 32 possible outcomes. So we have reached a firm conclusion: if the task can be done, it must be right to put 4 coins against 4 (say, coins ABCD against EFGH) in the first weighing. Suppose they balance. Then the temptation is to put 1 against 1 or 2 against 2 of the remaining coins, but either is information-theoretically deficient: if 1 balances against 1, four (so, more than three) possibilities remain; if 2 tilts against 2, again 4 possibilities remain. Either way one more weighing is insufficient. So we must use the known-genuine coins A through H in the second weighing, and the easiest thing to do is put three of them against three of the suspect coins (say, IJK). If they balance, coin L is counterfeit and we can weigh it against A to determine whether it is light or heavy. If IJK are light or heavy, test I against J. If the initial 4-against-4 weighing tilts left, so that either one of ABCD is heavy or one of EFGH is light, put (say) ABE against CDF. If they balance, the culprit is a light G or H and you can test either against the genuine L. If they tilt left, you’re left with A or B heavy or F light and you can determine which by putting A and F against two genuine coins, say K and L; the solution is similar if CDF outweigh ABE. Now suppose you are given the same problem but with 13 coins. There are now 26 possible outcomes, which is still less than or equal to 27, but our previous reasoning already showed that nonetheless three weighings are not enough. Why? Because we saw that 5-against-5 in the first weighing is no good, while 4-against-4, when they balance, leaves five coins–too many—untested. Three Martians at the Crossroads A logician is visiting Mars and as is usual for logicians in puzzles, she is at a fork, wanting to know which of two roads leads to the village. Present this time are three willing Martians, one each from a tribe of invariable truth-tellers, a tribe of invariable liars, and a tribe of random answerers. The logician doesn’t know which Martian is from which tribe. Moreover, she is permitted to ask only two yes-or-no questions, each question being directed to just one Martian. Can 226 she get the information she needs? How about if she can ask only one yes-or-no question? Solution: We can dispose of the one-question case easily: If the question is directed to the random answerer, the logician gets no information, thus can never guarantee to identify the right road. (This argument does not apply if you assume the random answerer first flips a mental coin to determine whether to lie or tell the truth; you could then gain information with a well-chosen self-referential statement, for example, “Out of the other two guys, if I pick the one whose response’s truthfulness will least likely match your response’s truthfulness and ask him if Road 1 goes to the village, will he answer ‘Yes’?” But we assume the random answerer just randomly answers “yes” or “no” regardless of the question, so no information is imparted.) Similarly, if the logician doesn’t know the right road after one question, and her second question is directed to the random answerer, she is in trouble. It follows that after the first answer, she must be able to identify a Martian who is not the random answerer. If she can do that, she’s in business, because she can then use a traditional one-Martian query as her second question: For example, something like “If I were to ask you whether Road 1 goes to the village, would you say yes?” To attain the objective, she’ll need to ask Martian A something about Mar-tian B or Martian C, then use the answer to choose between B and C. Here’s one that works: “Is B more likely than C to tell the truth?” Curiously, if A says “yes” she picks C, and if he says “no” she picks B! If A is the truth-teller she wants next to query the companion who is less likely to tell the truth, namely, the liar. If A is the liar she queries the more truthful of his companions, namely the truth-teller. Of course, if A is the random answerer it doesn’t matter which of B and C she turns to next. ♡ If in a puzzle the information seems insufficient, we may have to find extra sources. Attic Lamp An old-fashioned incandescent lamp in the attic is controlled by one of three on-off switches downstairs—but which one? Your mission is to do something with the switches, then determine after one trip to the attic which switch is connected to the attic lamp. Solution: This seems not to be doable: There are three possible answers (as to which of three switches controls the bulb), but when you get to the attic either the bulb is on or off. So only one bit of information is available, and we need a “trit.” 227 Is there anything that can be observed about the bulb other than whether it is on or off? There’s a hint in the puzzle wording: It’s an incandescent bulb, whose efficiency is limited by its habit of putting out quite a lot of heat as well as light. So you could feel the bulb and see if it’s hot; can that help? Yes. Flip switches A and B on, wait 10 minutes or so, then flip switch B off. Go quickly up to the attic: If the bulb is on, it’s switch A; if off but warm, switch B; off but cold, switch C. You could even do four switches: Flip A and B on, wait 10 minutes, then flip B off and C on and run up the stairs. The four possible states of the bulb (on/warm, off/warm, on/cool, or off/cool) will tell you what you need to know. Information is customarily measured in bits; k bits gives you 2k possibilities. The thriving field of “communication complexity” attempts to determine the minimum number of bits of communication needed between two or more parties, in order for them to accomplish some task. Here is a toy example. Players and Winners Tristan and Isolde expect to be in a situation of severely limited communication, at which time Tristan will know which two of 16 basketball teams played a game, and Isolde will know who won. How many bits must be communicated between Tristan and Isolde in order for the former to find out who won? Solution: Tristan and Isolde can, in advance, label the 16 teams (in alphabetic order, perhaps) by the binary numbers 0000 through 1111. Then, when the time comes, Tristan can send Isolde eight bits: The four bits corresponding to one team, and the four bits corresponding to the other one. Isolde can then send back a 0 if the winning team was the one with the smaller number, and a 1 otherwise. That comes to nine bits, and we can trim that to 8 by noting that the number of unordered pairs of teams is only 16 × 15/2 = 120 < 27 and thus Tristan can code the matchup with only seven bits. But an easier and faster solution is to just have Isolde send Tristan the four bits corresponding to the winning team. Surely, we can’t do any better than four bits, right? Amazingly, we can! We can code those four positions (leftmost, next-to-left, next-to-right, rightmost) by, say, 00, 01, 10, and 11 respectively. The codes of the two playing teams must differ in at least one of the four positions; Tristan picks such a position and sends its code (two bits) to Isolde. She now only needs to look at the value of the bit in that position of the winning team and sent it to Tristan. Three bits total! For example, suppose the playing teams are 0010 and 0110, and 0110 won. The only position where the playing teams differ is the next-to-left, code 01, so Tristan sends “01” to Isolde. She checks the next-to-left position of the winning team—it is a “1”—and sends it back to Tristan. Done. 228 It is not hard to show three bits is unbeatable. Note that in the most efficient communication scheme, more information is passed from the student to the teacher than vice-versa. Is there a lesson to be learned from that? Here’s another example where we need to think about information, and also to be clever. Missing Card Yola and Zela have devised a clever card trick. While Yola is out of the room, audience members pull out five cards from a bridge deck and hand them to Zela. She looks them over, pulls one out, and calls Yola into the room. Yola is handed the four remaining cards and proceeds to guess correctly the identity of the pulled card. How do they do it? And once you’ve figured that out, compute the size of the biggest deck of cards they could use and still perform the trick reliably. Solution: Let’s try to look at this trick information-theoretically. The obvious (and cor-rect, for this trick) way for Zela to communicate information to Yola when handing her the four remaining cards is to order the cards in some way. There are 4! = 24 ways to order four objects, but that doesn’t seem like enough—there are 48 possible values of the missing card. We could, of course, try to sneak in another bit of information, for example, whether the cards are passed face up or face down. We could do that pretty easily and we would in fact need to, if the card to be removed were chosen by an innocent bystander. But Zela gets to decide which card is removed, and thus can limit the possibilities to much less than 48. Since Zela gets to pick the card, the right information-theoretic way to look at the problem is to ask whether seeing four cards and their order is enough to deduce the original five cards. The number of ways to pick five cards from a deck is 52 5 = 52 · 51 · 50 · 49 · 48 5 · 4 · 3 · 2 · 1 = 2,598,960, 229 while the number of ways to pick 4 cards where the order counts is 52 · 51 · 50 · 49 = 6,497,400, so we are OK, at least from an information-theoretic standpoint. What is needed is a human-implementable map from a subset of the ordered foursomes to all the unordered fivesomes, with the property that each foursome is contained in the fivesome it is mapped to. That requires some cleverness; here’s how magician William Fitch Cheney did it back in the 1930’s. Making use of the pigeonhole principle, he noted that every five cards contain at least two of some suit. One of those will be removed and a second of that suit left on top of the pile of four, so already Yola will know the suit of the removed card. The remaining three of the four can be ordered six ways, and Yola will use that info to determine the rank of the missing card. Here’s how. Yola and Zela fix an ordering of all the cards, perhaps A,2,3,... up to 10,J,Q,K, with ties broken by suit (the traditional magician’s ordering of suits is the “CHaSeD” order, namely Clubs, Hearts, Spades, Diamonds). Let the three cards in question be called A, B, and C, according to the ordering above. If they are handed to Yola in order ABC, that codes the number 1; if ACB, the number 2; BAC = 3, BCA = 4, CAB = 5 and CBA = 6. Yola looks at the rank of the top card and adds her code number to it, going “around the corner” if necessary. For example, suppose the top card is the queen of spades, and the order of the other three cards is CAB. Since the code of CAB is 5 and the queen is rank 12 of 13, Yola adds 5 to 12: 13=K, 14=1=A, 2, 3, 4. She concludes that the missing card is the 4 of spades. Wait, suppose the missing card had been the 6 of spades; then Yola would have had to add 8 to the Q, and there’s no code 8. Not to worry: In that case, Zela would have removed not the 6 of spades but the Q of spades; and adding 6 gets you from 6 to queen. In other words, given two cards of the same suit, we can always pick one so that you can get to the other by adding a number from 1 to 6. ♡ Yes, this will take some practice on the part of Yola and Zela. There’s actually quite a bit of “room” in this trick. Information-theoretically, it can be done with a deck of any size n satisfying n 5 ≤4! × n 4 which is true up to n = 124. In fact, you really can do it with a deck of size 124, but we will leave it to you, the reader, to figure out how. What is the value of information? We can quantify it the following way: Subtract your expectation without the information from your expectation with. Since you don’t have to act on the information, this quantity can never be negative. Conversely, if the information can’t or shouldn’t change your actions, it’s worthless. Peek Advantage You are about to bet $100 on the color of the top card of a well-shuffled deck of cards. You get to pick the color; if you’re right you win $100, otherwise you 230 lose the same. How much is it worth to you to get a peek at the bottom card of the deck? How much more for a peek at the bottom two cards? Solution: Without the peek, the bet is fair, that is, your expected gain is $0. If your peek reveals a red card on the bottom, you will of course bet on black and be right with probability 26/51; then your expectation will be 26 51 · $100 + 25 51 · (−$100) ∼$1.96 . Since the calculation is the same if the card on the bottom is black, the value of the information is $1.96. We can do a similar calculation for peeking at two cards, but why bother? The second peek is worthless! Suppose the bottom card is red. If the next-to-bottom card is red, then of course you still bet that the top card is black. If the next-to-bottom card is black, then you’re back to a fair game, so you may as well keep your plan of betting on black. Since the second peek does not have the power to change your strategy, its additional value to you is exactly $0. (A similar argument shows that peeking at the bottom 2k cards is no better than peeking at the bottom 2k−1.) The same kind of reasoning is useful in the next puzzle. Bias Test Before you are two coins; one is a fair coin, and the other is biased toward heads. You’d like to try to figure out which is which, and to do so you are permitted two flips. Should you flip each coin once, or one coin twice? Solution: Suppose you only get one flip. If you flip coin A and it comes up heads, you will of course guess that it is the biased coin; if it comes up tails, that it is the fair coin. Now if you flip coin B and get the opposite face, you will be even happier with your previous decision. If you get the same face, you will be reduced to no information and may as well stick with the same choice. Thus, as with the card-peeking, flipping the second coin is worthless. Could flipping the same coin twice change your mind? Yes. If you get heads the first time you are inclined to guess that coin is biased, but seeing tails the next time will change your best guess to “unbiased.” We conclude that flipping one coin twice is strictly better than flipping each coin once. As it turns out, it’s a theorem that in trying to determine which is which of two known probability distributions, it’s better to draw twice from one than once from each. The theorem assumes, however, that you must decide in advance 231 which course to take; unlike in the puzzle above, it may happen that after seeing the result of your first draw you prefer to draw from the other distribution. We’ll leave it to the reader to concoct an example. Information about information—typically involving a chain of deductions to the effect that seeing A, person X would have deduced B—is a common puzzle theme. The following example is perhaps the best variation of a classic. Dot-Town Exodus Each resident of Dot-town carries a red or blue dot on his (or her) forehead, but if he ever thinks he knows what color it is he leaves town immediately and permanently. Each day the residents gather; one day a stranger comes and tells them something—anything—nontrivial about the number of blue dots. Prove that eventually Dot-town becomes a ghost town, even if everyone can see that the stranger’s statement is patently false. Solution: This puzzle, often in non-PC versions involving unfaithful spouses, murder, or suicide, has been around for at least a century. Usually the stranger is assumed to have made a true statement, but as we shall see, that stipulation is not necessary. The condition of “non-triviality” means that, a priori, the statement could be true or false; in other words, there is some number of blue dots that would make it true, and some other that would falsify it. We need to assume that a resident will believe the stranger if there is no evidence to the contrary (and that everyone knows this, and everyone knows that everyone knows this, etc.). So it could happen that someone leaves town mistakenly (e.g., stranger says all dots are red, and then the one guy ¿ with a blue dot believes the stranger and exits). But surely, if everyone knows the stranger is lying...? For example, suppose all the dots are blue but the stranger says they are all red. Then everyone knows that the stranger is lying and moreover, that everyone else knows that the stranger is lying. How can the stranger’s statement have any effect? It’s easiest to see when the population of Dot-town is small, for example, 3. Then you, if you are a Dot-town resident, know that the stranger is lying. But if your own dot is red, then your friend Fred is looking at a red dot and a blue dot and wondering if the third townsperson, Emily, is seeing two red dots. If so, Emily will believe the stranger and leave town. When she doesn’t, Fred will correctly conclude that his own dot is blue, and will leave town the second night. When this fails to happen, you—and Fred and Emily, as well—can all conclude that your dots are blue, and all leave town on the third night. To do a proof of the general case, we need some notation. Let S ⊂{0, 1, . . . , n} be the set of numbers x with the property that if there are x blue dots among the n residents of Dot-town, then the stranger’s statement is true. In other words, the stranger’s statement is equivalent to “The number of blue dots is in 232 S.” Our non-triviality assumption tells us that S is a proper, non-empty subset. Let b be the actual number of blue dots, which may or may not be in S. For resident i, let Bi be the set consisting of the possible numbers of blue dots, from i’s point of view. Prior to the stranger’s visit, Bi = {bi, bi+1} where bi is the number of blue dots resident i sees among his compatriots. If at any point Bi drops to one value, resident i is cooked. This will happen immediately if \|Bi∩S\| = 1, but it will also happen the night after any departures occur. To see this, we observe first that all residents with the same dot color will behave identically, since they all see the same number of dots. Thus if resident i sees that anyone has left town, he deduces (correctly) that that person’s dot color is different from his own; he therefore knows his own color and is doomed to exile. Given S and b, let db be the number of steps (increments or decrements by 1) needed to get from b across the border of S; in other words, db is the least k such that b+k or b−k is in S if b is not, and out of S (but still in {0, 1, . . . , n}) if b is in S. You can think of db as the distance from the stranger’s statement to the truth, if the statement is false; and the distance to falsehood, if it is true. Example: If n = 10 and S = {0, 1, 2, 9, 10} then d0 = 3, d1 = 2, d2 = d3 = 1, d4 = 2, d5 = d6 = 3, d7 = 2, d8 = d9 = 1, and d10 = 2. We claim that the first departures will occur exactly on night db. The proof is by induction on db. If db = 1, some residents will find the stranger immediately credible, and will leave the first night. Suppose t > 1 and assume that our claim is true whenever db < t. The day after night t−1, since no departures have yet occurred, everyone will deduce that db ≥t. However, if db = t then either db−1 or db+1 must be equal to t−1. If the former, then those residents with blue dots, who see that the number of blue dots is either b (the actual number) or b−1, can rule out b−1 and are toast. If the former, it is the red-dotted folks who can rule out b+1 and must pack their things. Finally, if db−1 = db+1 = t−1 then nobody stays the night. Since db is at most n, the proof tells us that everyone will have exited by the n+1st night. But in the db = n case all residents have the same dot-color so they all leave together on the nth night. It follows that Dot-town is always deserted within n days. It is perhaps worth noting also that the definition of db makes no distinction between S and its complement; from this it follows that it makes no difference whether the stranger says “X” or “not X,” the residents of Dot-town will behave exactly the same way in either case. You might reasonably wonder whether the Dot-town residents, knowing that a stranger is coming and liable to break the manifestly justifiable no-talk-about-dot-colors taboo, can organize some defense. For example, everyone who knows the stranger is lying jumps up and says so. Alas, a little thought will show you that neither this nor any similar strategy can save the town. A fragile lot, these Dot-towners. A whole class of puzzles is built around conversations in which the speakers take several rounds to make a deduction. Your job is to deduce something of 233 your own from the conversation, even though you have less information than the speakers! My favorite of these is the following. Part of the beauty of this one is that no numbers are named in the puzzle statement. Conversation on a Bus Ephraim and Fatima, two colleagues in the Mathematics Department at Zorn University, wind up seated together on the bus to campus. Ephraim begins a conversation with “So, Fatima, how are your kids doing? How old are they now, anyway?” “It turns out,” says Fatima as she is putting the $1 change from the bus driver into her wallet, “that the sum of their ages is the number of this bus, and the product is the number of dollars that happen to be in my purse at the moment.” “Aha!” replies Ephraim. “So, if I remembered how many kids you have, and you told me how much money you are carrying, I could deduce their ages?” “Actually, no,” says Fatima. “In that case,” says Ephraim, “I know how much money is in your purse.” What is the bus number? Solution: We will assume all numbers in the puzzle are positive integers—and that Ephraim knows the number of the bus he’s on. That number (call it n) must allow two different sets of kids’ ages, with the same number of kids, the same sum n, and the same product; otherwise Ephraim would have been able to deduce the kids’ ages from sum, product and number of kids. One would expect that large values of n would tend to have this property; what prevents any large number from being the answer? Ah, the last line of the conversation shows that there can’t be two pairs of kids’-age-sets with sum n, one pair sharing one product, the other sharing a different product. If there were two such pairs, Ephraim would not have been able to deduce the amount of money in Fatima’s purse at the conversation’s end. Bus number n = 13 is such an example. Fatima could have five kids of ages 6, 2, 2, 2, and 1, or of ages 4, 4, 3, 1, and 1; either would put $48 in her purse. But she could instead have three kids of ages 9, 2, and 2, or of ages 6, 6, and 1; those possibilities would both give her purse $36. So even with the information that knowing the number of kids and purse contents (in addition to the bus number) would not have been sufficient to deduce the kids’ ages, Ephraim would still not be in a position to deduce the purse contents if the bus number had been 13. What about higher numbers? Here’s the crucial observation: Once you’ve hit a number with 13’s “ambiguity” property, every higher number has it too! Why? Because you can take those four sets of kids’ ages and add one kid of age 234 1 to each one of them. That changes no products but raises all the sums by 1, so if n is ambiguous, so is n+1. Now it’s time to do the math, checking values of n below 13 to try to find one that boasts exactly one pair of kids’-age-sets with the same number of kids, sum n, and the same product. That’s not as time consuming as it sounds, because (a) you only need consider sets of size at least three (one or two numbers can always be deduced from their sum and product), and (b) many products yield the kids’ ages readily on account of unique prime factorization. Turns out the puzzle-composer has not misled you: There is just one bus number that qualifies, n = 12. The critical kids’-age-sets are {4, 4, 3, 1} and {6, 2, 2, 2}. The next example illustrates the complexities of trying to communicate and play a game at the same time. Matching Coins Sonny and Cher play the following game. In each round, a fair coin is tossed. Before the coin is tossed, Sonny and Cher simultaneously declare their guess for the result of the coin toss. They win the round if both guessed correctly. The goal is to maximize the fraction of rounds won, when the game is played for many rounds. So far, the answer is obviously 50%: Sonny and Cher agree on a sequence of guesses (e.g., they decide to always declare “heads”), and they can’t do any better than that. However, before the game begins, the players are informed that just prior to the first toss, Cher will be given the results of all the coin tosses in advance! She has a chance now to discuss strategy with Sonny, but once she gets the coinflip information, there will be no further opportunity to conspire. Show how Sonny and Cher can guarantee to get at least 6 wins in 10 flips. (If that’s too easy for you, show how they can guarantee at least six wins in only nine flips!) Solution: Guaranteeing 5 wins in 10 flips is a cinch: Cher chooses her first five guesses to match the last five coinflips. Then Sonny and Cher both repeat that sequence and collect their five wins. With luck they might catch one or two additional wins among the first five flips, but that can’t be guaranteed. 235 Is there a better way? Suppose Cher tries to communicate three flips at a time to Sonny. The nice thing about three flips is that they are either mostly heads or mostly tails. If flips 2 through 4 are mostly heads, Cher calls “heads” at flip 1 and this tells Sonny to guess “heads” at flips 2, 3, and 4. Suppose flips 2 and 4 are heads but flip 3 is a tail. Then Cher will call “heads” at flips 2 and 4, garnering two wins, and will use flip 3 to tell Sonny what to do in flips 5 through 7. What if flips 2, 3, and 4 are all heads? Then, by agreement, we just use Cher’s last call (on flip 4) to tell Sonny what to guess at the next three flips. Flips 5 though 7 are treated similarly, with the minority flip (or the last flip, if all flips are the same) used to tell Sonny which way to guess on flips 8 through 10. In this way Sonny and Cher are guaranteed to get at least two wins out of three in each of the flip-groups 2-4, 5-7, and 8-10. Readers who like to imagine life in asymptopia will want to know what fraction of correct flips can be guaranteed when the total number of flips is large. It turns out Sonny and Cher can do much better than 2/3, in fact they can beat 80%(!) if the number of flips is huge and they work hard to squeeze the last drop of blood out of their scheme. As for 6 out of 9, that takes some rather careful work and I wouldn’t want to deprive the ambitious reader of the opportunity to devise a successful scheme. (Advice: Do 3 out of 5 first.) Eventually, I’ll post a solution on the web. In the science of cryptography, communication must be done with an eye to keeping information from falling into the wrong hands. This is most easily done if the parties that wish to communicate have a common secret. For example, suppose Alice is expecting to see Bob at a party and hoping to tell him “yes” or “no” to a certain business proposition, privately, even though others at the party may be nearby and interested. No problem; she and Bob arrange beforehand that “Hi” means yes and “Hello” means no. This shared secret enables them to communicate secretly. The last puzzle in this chapter was designed to show that a shared secret can sometimes be created over an open channel, provided that the participants have some shared knowledge—that is, when put together they could deduce things that an eavesdropper couldn’t know. Two Sheriffs Two sheriffs in neighboring towns are on the track of a killer, in a case involving eight suspects. By virtue of independent, reliable detective work, each has narrowed his list to only two. Now they are engaged in a telephone call; their object is to compare information, and if their pairs overlap in just one suspect, to identify the killer. The difficulty is that their telephone line has been tapped by the local lynch mob, who know the original list of suspects but not which pairs the sheriffs have 236 arrived at. If they are able to identify the killer with certainty as a result of the phone call, he will be lynched before he can be arrested. Can the sheriffs, who have never met, conduct their conversation in such a way that they both end up knowing who the killer is (when possible), yet the lynch mob is still left in the dark? Solution: Label the suspects ABCDEFGH and organize them into lists so that each list consists of four disjoint (unordered) pairs, and each pair appears in exactly one list. For example: AB CD EF GH AC BD EG FH AD BC EH FG AE BF CG DH AF BE CH DG AG BH CE DF AH BG CF DE Let’s name the sheriffs Lew and Ralph. Lew can share the above array with Ralph over the phone and then tell Ralph which row his pair (i.e., the pair of suspects to which Lew has narrowed down his search) lies in. If Ralph’s pair lies in the same row, he tells Lew; that means the two sheriffs have arrived at the same pair, so they may as well hang up. Otherwise, Ralph announces a partition of the row into sets of two pairs each, with the property that Ralph’s two final suspects are in the same part. For example, suppose Lew’s pair is EF. Then he announces that his pair is in the first row. If Ralph’s pair is FG, he partitions that row into one part consisting of the pairs AB and CD, and another containing the pairs EF and GH. (At this point, Lew and Ralph already have established a shared secret: they both know which part the killer is in.) Lew now reveals whether his pair is the leftmost or rightmost in its part of the row (in this case, it’s the leftmost). That tells Ralph which pair is Lew’s, and hence who the killer is (here, F). To communicate that safely to Lew, he just tells Lew whether the killer is the left or the right member of Lew’s pair (here, the rightmost). The conversation would be identical if the pairs had been AB and BC, making B the killer instead of F. We hope that the mob either refuses to chance lynching an innocent person, or doesn’t have enough rope for two. Cryptography is the study of means for sending private messages over an open channel. Classically, the sender (Alice, say) and receiver (Bob) are equipped in advance with a “key” that enables Alice to encrypt her message and Bob to decrypt it, while eavesdropper Eve, not possessing the key, can make no sense of the intercepted stream. Suppose Eve has the time and computing power to test every possible key, looking for one which, when used to decrypt the message she intercepted, yields 237 a credible possibility for the message Alice intended Bob to get. When might this work? If the number of possible keys is small, Eve can expect that only one will yield an intelligible message; but if it’s large, many might do so. Information theory can give us an idea of how many possible keys there should be, in order for Alice and Bob to be safe. Suppose that for every choice of message Alice might have wanted to send Bob on this occasion, and every “cyphertext” x1, . . . , xn that Eve might inter-cept, there is a unique key that would encrypt that message as x1, . . . , xn. Then, assuming the key was chosen uniformly at random and Eve doesn’t know it, we have an easy conclusion. Theorem. In the above situation, Eve can gain no information whatever about Alice’s intended message. Proof. Since any message could have been intended, each associated with one key, and all keys are equally likely, Eve learns nothing. ♡ There actually is a cryptographic scheme that achieves this goal—assuming accurate random generation and perfect secrecy in the key choice. It’s called a “one-time pad,” and here’s an example of how it could work. Alice translates her intended message into a stream of n 0s and 1s (by some means that Bob needs to know and Eve may also know). Then she extracts the next n bits from the one-time pad that she and Bob share, and adds those bits one by one, modulo 2, to the message bits. For example, if the message is 00101110 and the bits from the one-time pad are 01001001, then the encrypted message is 01100111 and this is what Alice sends (and Eve intercepts). Bob, who knows where he and Alice are in the one-time pad, takes his received message 01100111 and, one by one, adds the one-time pad bits 01001001 to it to get the original message, 00101110, intended by Alice. (This works nicely because modulo 2, addition and subtraction are equivalent. Much more about this useful kind of arithmetic is found in Chapter 17.) Eve is kept completely in the dark, because looking only at 01100111, any eight-bit message could have been intended by Alice. As its name suggests, the security of the one-time pad is heavily dependent on each bit of the pad being used only once, and in only one message. Otherwise a “depth” is created and Eve may be able to crack the code. The difficulty with one-time pads is that if Alice and Bob are planning any long conversations, they need to share (and keep private) a great many bits. In most applications, keys are much shorter than messages, and the security of encryption relies on Eve not having the time and/or computing power to test enough keys. 238 Chapter 14 Great Expectation The expectation or expected value of a random number is the average value that is anticipated (via the Law of Large Numbers) when the random number is “re-drawn” independently many times. If the random number can take only finitely or countably many values (e.g., if it takes only integer values), we can define its expectation as the average of the values, weighted by their probabilities. Numbers whose values are determined by chance are called by probabilists random variables, and denoted typically by boldface capital letters from the end of the Latin alphabet, such as X. If X’s possible values are x1, x2, . . . with corresponding probabilities p1, p2, . . . , then the expectation of X is EX := X i pixi. The terminology is a bit misleading; for example, the expectation when you roll a fair die is 3 1 2, but you don’t “expect” the result 3 1 2. It would be more accurate to call 3 1 2 the “expected average” rather than the “expectation” or “expected value.” But convenience triumphs. Expectation is often easier to work with than probability. Suppose X and Y are two random variables. If we want to know the probability that X + Y > 5 (say) it is not enough to know everything there is to know about each of X and Y individually; you will also need to know how X and Y interact with each other. But if you only want the expectation of X + Y, it is guaranteed to satisfy E(X + Y) = EX + EY so all you need to know are the expectation of X and the expectation of Y. For example, suppose X and Y each take values 1, 2, 3, 4, 5, or 6 with equal probability, so that each has expectation 3 1 2. What does the random variable X + Y look like? Well, if X and Y are independent—for example, if X is the first roll of a die and Y the second—then X + Y takes on values 2, 3, . . . , 12 with probabilities 1/36, 2/36, . . . , 5/36, 6/36, 5/36, . . . , 1/36. But if X and Y 239 are both the outcome of the first roll, that is, if X = Y, then X+Y = 2X takes on only even values from 2 to 12, each with probability 1 6. And if Y happens to be defined as 7 −X, we find that X + Y is the constant value 7. All very different, but notice that in every case E(X + Y) = 3 1 2 + 3 1 2 = 7. This miracle is a special case of what mathematicians call “linearity of ex-pectation,” whose general formulation is E X i ciXi = X i ciEXi, that is, the expectation of a weighted sum of random variables is the weighted sum of their expectations. Let’s start with a calculation of expectation in the service of making a strate-gic decision. Bidding in the Dark You have the opportunity to make one bid on a widget whose value to its owner is, as far as you know, uniformly random between $0 and $100. What you do know is that you are so much better at operating the widget than he is, that its value to you is 80% greater than its value to him. If you offer more than the widget is worth to the owner, he will sell it. But you only get one shot. How much should you bid? Solution: You should not bid. If you do bid $x, then the expected value to the widget’s owner, given that he sells, is $x/2; thus, its expected value to you, if you get it, is 1.8· $x/2 = $0.9x. Thus you lose money on average if you win, and of course, you gain or lose nothing when you do not, so it is foolish to bet. ♡ One common setting for expectation is waiting time; in particular, the aver-age number of times you expect to have to try something until you succeed. For example, how many times, on average, do you need to roll a die until you get a 6? We encountered this sort of problem in Odd Run of Heads in Chapter 5. Let x be the expected number of rolls. “If at first you don’t succeed” (probability 5 6) you still have an average of x rolls to “try, try again,” so x = 1 + 5 6 · x giving x = 6. The same argument shows that in general, if you conduct a series of “trials” each of which has probability of success p, then the average number of trials you need to get your first success is 1/p. Let’s apply this observation to a slightly trickier dice puzzle. Rolling All the Numbers On average, how many times do you need to roll a die before all six different numbers have turned up? 240 Solution: Often the most important step in applying linearity of expectation is simply recognizing that what you are being asked for is the expectation of a sum of random variables. The key for this puzzle is to think of rolling all the numbers as a 6-stage process, where at Stage i you are going for your ith new number. In other words, let Xi be the number of rolls needed to get your ith new number, given that you have already seen i−1 different numbers. Then the total number of rolls needed is X := X1 + X2 + · · · + X6 and what we are asked for is EX = EX1 + EX2 + · · · + EX6. We already know that EX1 = 1 since it always takes exactly one roll to get the first new number. The second will come with probability 5/6 from any given roll, so takes 6/5 rolls on average. Continuing in this manner, we have EX = 1 + 6 5 + 6 4 + 6 3 + 6 2 + 6 1 = 14.7 and we are done. Warning: Some experimentation will remind you that X is by no means guaranteed to be near 14.7; the last new number, in particular, could take a frustratingly long time to roll. Later we’ll be reminded that the Law of Large Numbers sometimes takes a while to kick in. Next, see if you can set up a sum of random variables for this puzzle. Spaghetti Loops The 100 ends of 50 strands of cooked spaghetti are paired at random and tied together. How many pasta loops should you expect to result from this process, on average? Solution: The key here is to recognize that you start with 100 spaghetti-ends and each tying operation reduces the number of spaghetti-ends by two. The operation results in a new loop only if the first end you pick up is matched to the other end of the same strand. At step i there are 100−2(i−1) ends, thus 100−2(i−1)−1 = 101 −2i other ends you could tie any given end to. In other words, there are 241 101−2i choices of which only one makes a loop; the others just make two strands into one longer one. It follows that the probability of forming a loop at step i is 1/(101 −2i). If Xi is the number of loops (1 or 0) formed at step i, then the expected value of Xi is 1/(101 −2i). Since X := X1 + · · · + X50 is the final number of loops, we apply linearity of expectation to get EX = EX1 + · · · + EX50 = 1 99 + 1 97 + 1 95 + · · · + 1 3 + 1 1 = 2.93777485 . . . , less than three loops! ♡ For the last two puzzles, the summands were independent random variables— we didn’t actually need to know that independence is not required for linearity of expectation. Independence of two events A and B means that the probability they both occur is the product of their individual probabilities, so, for example, if we wanted to know the probability of getting all six numbers in only six rolls of a die, we could have obtained it as 1 · 5 6 · 4 6 · 3 6 · 2 6 · 1 6 = 6! 66 ∼1.5% . For all the spaghetti to end up in one big loop, you must never tie a strand to its other end except at the last step; in other words, X1, . . . , X49 must all be zero. This happens with probability 98 99 · 96 97 · 94 95 · · · · · 2 3 = 49! · 249 99!/(49! · 249) = 49!2 · 298 99! = 0.12564512901 . . . , so about one time in eight. Ping-Pong Progression Alice and Bob play table tennis, with Bob’s probability of winning any given point being 30%. They play until someone reaches a score of 21. What, approx-imately, is the expected number of points played? Solution: With high probability, Alice will win, and thus the expected number of points won by Bob is (3/7)21 = 9 and therefore the expected total number of points is close to 30. Another way to make that calculation: Since Alice wins any given point with probability 70%, it takes on average 1/0.7 = 10/7 rallies to earn Alice a point, thus 21 × 10/7 = 30 rallies on average for her to get to 21. The actual answer is a bit less than 30, because playing until Alice scores 21 points is sufficient, but not always necessary, to finish the game. But Bob wins so rarely that the effect is negligible. Do you recall a puzzle called “Finding a Jack” from a Chapter 8? You can use its solution technique—twice!—for the next puzzle. 242 Drawing Socks You have 60 red and 40 blue socks in a drawer, and you keep drawing a sock uniformly at random until you have drawn all the socks of one color. What is the expected number of socks left in the drawer? Solution: It’s useful to imagine that the sock-drawing is done by arranging all the socks in a uniformly random order, then removing them, starting from the beginning of the order, until all of one color have been removed. If the color of the last sock is blue (probability 0.4), then it must have been the red socks that were all removed, and the number of socks “left in the drawer” is the number of blue socks encountered starting from the end of the arrangement before any red sock is found. If indeed the last sock is blue, there are 39 other blue socks distributed in 61 slots (before the first red sock, between the first and second red socks, and so forth, finally after the 60th red sock). Thus, we expect 39/61 blue socks in that last slot; adding the final blue sock gives on average of 100/61 blue socks left in the drawer. Similarly, if the last sock is red (probability 0.6), we have 59 other red socks in 41 slots so on average 59/41 in the last slot; that leaves 1 + 59/41 = 100/41 red socks behind. Putting these together, we have that the expected number of socks left is 0.4 × 100/61 + 0.6 × 100/41 ∼2.12. If there are m socks of one color and n of the other, the above argument gives you an average of m/(n+1) + n/(m+1) socks left in the drawer. Linearity of expectation applies even when there are continuum-many ran-dom variables. Modern mathematicians would replace the sum by an integral, but sometimes an Archimedean approach will do the trick. Random Intersection Two unit-radius balls are randomly positioned subject to intersecting. What is the expected volume of their intersection? For that matter, what is the expected surface area of their intersection? Solution: We can assume that the center of the first ball is at the origin of a three-dimensional coordinate system; the first ball thus consists of all points in space within distance 1 of the origin. The center C of the second ball must be some-where in the ball of radius two about the origin, in order for it to intersect the first ball. The conditions of the problem imply that C is uniformly random subject to this constraint. (Note that this is not at all the same as taking the center of the second ball to be a random point between −2 and 2 on the x-axis. The latter assumption would give a much-too-big expected intersection size of π/2.) 243 Now consider a point P inside the first ball: What’s the probability that P will be in the intersection of the two balls? That will happen just when C falls in the unit ball whose center is P. This ball is inside the ball of radius 2 about the origin that C is chosen from, with 1/23 = 1/8 of the volume. Thus the probability that P lies in our intersection is 1/8, and it follows from linearity of expectation that the expected volume of the intersection is 1/8 times the volume of the first ball, that is, 1/8 × 4π/3 = π/6. If it seems fishy to you to be adding up points P like this, good for you! Think of P not as a point but a small cell, or “voxel,” inside the ball. If the voxel’s volume is (4π/3)/n with the ball divided into n voxels, n being some large integer, then an ordinary finite application of linearity of expectation gives approximately the same answer, π/6. It’s not exact owing to the fact that a few voxels will lie partly inside and partly outside the intersection. As the voxels get smaller the approximation gets better. For the surface area of the intersection, the point P—now representing a two-dimensional pixel—is taken on the surface of the unit ball centered about the origin. This will lie on the surface of the intersection if it’s inside the second ball, that is, again, if C lies in the unit ball centered at P. We’ve already seen that this happens with probability 1/8, but note that the intersection of the two spheres has two surfaces of the same area, one that was part of the surface of the first ball—that’s the part that could contain P—and one that was part of the surface of the second ball. Thus the expected surface area of the intersection is 2 × 1 8 × 4π = π. You may have noticed, by the way, that all this can be done in any dimension; with disks on the plane, for example, you get expected area π/4 and perimeter π. You can even do it with (some) other shapes. One place where linearity of expectation arises frequently is in gambling. A game is called “fair” if the expected return for each player is 0, and we can deduce that any combination of fair games is fair. This applies even if your choice of game, or strategy within a game, is made “as you play” (as long as you can’t foresee the future). Of course, in a gambling casino, the games are not fair; your expectation is negative, the casino’s positive. For example, in (American) roulette there are 38 numbers (0, 00, and 1 through 36) on the wheel, but if you bet $1 on a single number and win, you earn only $36, not $38, in return for your dollar. The result is that your expectation is 1 38 · $35 + 37 38 · (−$1) = −$ 1 19, so that you lose, on average, about a nickel per dollar bet. This same negative expectation applies also to bets made on groups of numbers or colors. However: 244 Roulette for the Unwary Elwyn is in Las Vegas celebrating his 21st birthday, and his girlfriend has gifted him with twenty-one $5 bills to gamble with. He saunters over to the roulette table, noting that there are 38 numbers (0, 00, and 1 through 36) on the wheel. If he bets $1 on a single number, he will win with probability 1/38 and collect $36 (in return for his $1 stake, which still goes to the bank). Otherwise, of course, he just loses the dollar. Elwyn decides to use his $105 to make 105 one-dollar bets on the number 21. What, approximately, is the probability that Elwyn will come out ahead? Is it better than, say, 10%? Solution: We know from linearity of expectation that on average, Elwyn should expect to lose $ 105 19 ∼$5.53. But this says very little about the probability that Elwyn comes out ahead. A random variable with expectation −5.53 could be anywhere from always negative, to almost never negative! In this case, we need to work out the probability that Elwyn wins three or more of his 105 bets, because if he wins three times he pulls in 3 · $36 = $108 and comes out ahead. That takes a little work. The probability that Elwyn loses every time is (37/38)105 ∼0.06079997242. To compute the probability that he wins exactly once, we multiply the number of bets by the probability of winning a given bet and losing the rest; this gives us 105 · 1 38 · 37 38 104 ∼0.17254046227. The probability that Elwyn wins exactly twice is the number of pairs of bets times the probability of winning two specified bets and losing the rest, which works out to 105 · 104 2 · 1 38 2 · 37 38 103 ∼0.24248929833. Adding these three numbers and subtracting from 1 gives the probability that Elwyn comes out ahead: 0.52417026698, better than 50%! No, this does not mean that Elwyn has figured out how to beat ’Vegas. He comes out slightly ahead rather often but far behind quite often as well, since winning only twice leaves him with just $72 to show for his $105 stake. Elwyn’s negative expectation is the more relevant statistic here. If you want to go to Vegas and be able to tell your friends that you made a profit at roulette, here’s a suggestion: Pick a number of dollars that’s one less than a power of 2, and that you can afford to lose. Suppose it’s $63. Then bet $1 on “red.” (Half the positive numbers are red, so you get your dollar back and win an additional $1 with probability 18/38, but lose your dollar otherwise.) If 245 you do win, go home. If you lose, bet $2 on red; if you win, go home having made a $1 profit. If you lose, bet $4 next time, etc., until you win or have lost your sixth straight bet ($32). At this point, you must give up or risk ruin; fortunately the probability that you will be this unlucky is only (20/38)6, about 2%. Your expectation for the whole experiment is of course still negative. An Attractive Game You have an opportunity to bet $1 on a number between 1 and 6. Three dice are then rolled. If your number fails to appear, you lose your $1. If it appears once, you win $1; if twice, $2; if three times, $3. Is this bet in your favor, fair, or against the odds? Is there a way to determine this without pencil and paper (or a computer)? Solution: This game seems pretty decent, maybe even favorable. You have three chances to get your number; if you were guaranteed to roll three different numbers, the probability that one would be yours would be 50%, making the game perfectly fair. If you don’t roll three different numbers, you could end up with a bonus. So what’s not to like? In fact this is a well known game, often called “Chuck-a-Luck” or “Birdcage” (the latter because the dice are typically kept inside a cage and rolled by shaking the cage). Like any gambling game it is designed to look attractive but to make money for the house, and indeed that is the case. The easiest way to see this is to imagine that there are six players each of whom bets on a different number. The expectation for each player is of course the same, since the rules don’t discriminate. If that expectation is x then the house’s expectation must be −6x by linearity of expectation—this is a “zero-sum game,” since no money leaves or enters from outside. But now if the three rolls are different, three players win and three lose so the house breaks even. If two numbers are the same, the house pays $2 to one player and $1 to another while picking up $1 from each of the remaining four players, so makes a $1 profit. If all three dice come up the same, the house pays out $3 and collects $5, making a $2 profit. So the house’s expectation is positive and therefore the players’ is negative. (It’s not hard to work out that the probability the three rolls are different is (6·5·4)/63 = 5/9, the probability that they are all the same is 1/62 = 1/36, and thus the probability that just two of them match is the remaining 15/36 = 5/12. Thus the house’s expectation is (5/12)·$1+(1/36)·$2 = $17/36 and each player’s expectation is thus (−1/6) · $17/36 ∼−$0.0787037037, so a player loses on the average about 8 cents per dollar bet—worse even than American roulette, about which more later.) Here’s a game which is definitely favorable; the question is, to what extent can you turn positive expectation into a sure thing? 246 Next Card Bet Cards are turned face up one at a time from the top of a well-shuffled deck. You begin with a bankroll of $1, and can bet any fraction of your current worth, prior to each revelation, on the color of the next card. You get even odds regardless of the current composition of the deck. Thus, for example, you can decline to bet until the last card, whose color you will of course know, then bet everything and be assured of going home with $2. Is there any way you can guarantee to finish with more than $2? If so, what’s the maximum amount you can assure yourself of winning? Solution: With a little thought you can improve on $2 by waiting to bet until three cards remain. If they’re all one color, great—you can bet everything for the remaining three rounds and go home with $16. If one is red and the others black, bet 1/3 of your money on black; if you’re right you now have $1 1 3 which you can double at the last card for a final tally of $2 2 3. If you’re wrong you’re down to $ 2 3 but the remaining cards are both black, so you can double this twice to again end with $2 2 3. Since a similar strategy works when the remaining cards are two red and one black, you have found a way to guarantee $2 2 3. To extend that backward and guarantee even more starts to look messy, so let’s look at the problem another way. Whatever betting strategy you use, you’ll certainly want to bet everything at every round as soon as the remaining deck becomes monochromatic; let’s call strategies with this property “sensible.” It is useful first to consider which of your strategies are optimal in the sense of expectation, that is, which maximize your expected return. It is easy to see that all such strategies are sensible. Surprisingly, the converse is also true: No matter how crazy your betting is, as long as you come to your senses when the deck becomes monochromatic, your expectation is the same! To see this, consider first the following pure strategy: Imagine some fixed specific distribution of red and black in the deck, and bet everything you have on that distribution at every turn. Of course, you will nearly always go broke with this strategy, but if you win you can buy the earth—your take-home is then 252 × $1, around 4.5 quadrillion dollars. Since there are 52 26 ways the colors can be distributed in the deck, your expected return is $252/ 52 26 = $9.0813. Of course, this strategy is not realistic, but it is “sensible” by our definition, and, most importantly, every sensible strategy is a combination of pure strategies of this type. To see this, imagine that you have 52 26 assistants working for you, each playing a different one of the pure strategies. We claim that every sensible strategy amounts to distributing your original $1 stake among these assistants, in some way. If at some point your collective assistants bet $x on red and $y on black, that amounts to you yourself betting $x −$y on red (when x > y) and $y −$x on black (when y > x). Each sensible strategy can be implemented by a distribution of money to the 247 assistants, as follows. Say you want to bet $0.08 that the first card is red; this means that the assistants who are guessing “red” first get a total of $0.54 while the others get only $0.46. If, on winning, you plan next to bet $0.04 on black, you allot $0.04 more of the $0.54 total to the “red-black” assistants than to the “red-red” assistants. Proceeding in this manner, eventually each individual assistant has his assigned stake. Now, any mix of strategies with the same expectation shares that expec-tation, hence every sensible strategy has the same expected return of $9.08 (yielding an expected profit of $8.08). In particular, all sensible strategies are optimal. But one of these strategies guarantees $9.08; namely, the one in which the $1 stake is divided equally among the assistants. Since we can never guarantee more than the expected value, this is the best possible guarantee. This strategy is actually quite easy to implement (assuming as we do that US currency is infinitely divisible). If there are b black cards and r red cards remaining in the deck, where b ≥r, you bet a fraction (b−r)/(b+r) of your current worth on black; if r > b, you bet (r−b)/(b+r) of your worth on red. You will be entirely indifferent to the outcome of each bet, and can relax and collect your $8.08 profit at the end. Sometimes the notion of fair game comes in handy even when there is no mention of a game. Serious Candidates Assume that, as is often the case, no one has any idea who the next nominee for President of the United States will be, of the party not currently in power. In particular, at the moment no person has probability as high as 20% of being chosen. As the politics and primaries proceed, probabilities change continuously and some candidates will exceed the 20% threshold while others will never do so. Eventually one candidate’s probability will rise to 100% while everyone else’s drops to 0. Let us say that after the convention, a candidate is entitled to say he or she was a “serious” candidate if at some point his or her probability of being nominated exceeded 20%. Can anything be said about the expected number of serious candidates? 248 Solution: It seems that you’d need to know a lot about the political process, and the current circumstances, to answer this question. For example, doesn’t it seem plausible that under some conditions, the first candidate to exceed 20% is likely to continue rising and eventually secure the nomination? Wait, that can’t be right; if he/she really had probability only 20% at the critical point, then most likely someone else would end up with the nomination. In fact, the expected number of “serious” candidates has to be 5. To see this, imagine that you enter the prediction market with the following strategy: Anytime a candidate hits 20% for his or her first time, you bet $1 on that candidate. If the market’s probabilities are “correct” and the market is fair, you should win $5 if that candidate is eventually nominated. In this game, you are bound to win $5; no matter which candidate gets nominated, you will have bet on him or her at some point. So the only variable is how many bets you made, and since the game is presumed to be fair, your expected expenditure must be $5. Why is the word “correct” in quotes above? The problem is that when everything is done and (say) Smith has won the nomination, one might argue that a good political prognosticator would, or should, have known all along that that would be the outcome. Thus when Smith hit 20% he or she should probably have been rated at 100%, or at least a lot higher than 20%. Without getting into the philosophy of probability, let’s just say that a candidate’s rating represents some sort of consensus which you, the bettor, have no reason to doubt one way or the other. Rolling a Six How may rolls of a die does it take, on average, to get a 6—given that you didn’t roll any odd numbers en route? Solution: Obviously, three. Right? It’s the same as if the die had only three numbers, 2, 4 and 6, each equally likely to appear. Then the probability of rolling a 6 is 1/3, and as we have seen in the puzzle Rolling All the Numbers, in an experiment that has probability of success p, it takes, on average, 1/p trials to get a success. Except that it’s not the same at all. Imagine that you arrange to compute your answer to this puzzle via multiple experiments, making use of the law of large numbers. You start rolling your die, counting trials as you go. If you hit a “6” before you get an odd number, you record the number of trials and repeat. (For example, if you roll “4,4,6” you record a 3.) What happens if you do roll an odd number? Then you can’t simply throw out that roll and continue counting trials (if you did that, you would eventually converge to 3 as your answer). Instead, you must throw out that whole experiment, recording nothing, and start anew, beginning with roll #1 of your new experiment. That will produce a smaller average than 3, but how much smaller? 249 An intuitive way to look at it: When you’ve rolled a 6 without rolling an odd number, there’s an inference that you succeeded quickly—since if you took a long time to get your 6, you’d probably have hit an odd number. To figure out the correct answer, it helps to think about what you’re doing in your multi-experiment. Basically, each sub-experiment consists of rolling the die until you get a 6 or and odd number. If you counted rolls in all of these, you’d discover their average length to be 3/2, since the probability of rolling a 6, 1, 3, or 5 is 2/3. In your gedankenexperiment you’re only counting the sub-experiments that end in a 6, but does that matter? The finishing roll (be it 1, 3, 5, or 6) is independent of all that has gone before, and independence is a reciprocal notion; thus, the number of rolls “doesn’t care” what the sub-experiment ended with. It follows that the average length of the subexperiments that ended with a 6 is that same 3/2. Napkins in a Random Setting At a conference banquet for a meeting of the Association for Women in Mathe-matics, the participants find themselves assigned to a big circular table. On the table, between each pair of settings, is a coffee cup containing a cloth napkin. As each mathematician sits down, she takes a napkin from her left or right; if both napkins are present, she chooses randomly. If the seats are occupied in random order and the number of mathematicians is large, what fraction of them (asymptotically) will end up without a napkin? Solution: We want to compute the probability that the diner in position 0 (modulo n) is deprived of a napkin. The limit of this quantity, as n →∞, is the desired limiting fraction of napkinless diners. We may assume that everyone decides in advance whether to go for her right or left napkin, in case both are available; later, of course, some will have to change their minds or go without. Say that diners 1, 2, . . . , i−1 choose “right” (away from 0) while i chooses “left”; and diners −1, −2, . . . , −j +1 choose “left” (again, away from 0) while diner −j chooses “right.” Note that i and j might both be equal to 1, meaning 250 that the diners on both sides of diner 0 are planning to go for the napkin between them and diner 0. If k = i+j+1, then as long as k ≤n (which is the case with high probability), the probability of this configuration is 21−k. Observe that diner 0 loses out only when she is last to pick among −j, . . . , i and none of the diners −j +1, . . . , −2, −1; 1, 2, . . . , i−1 get the napkins they wanted. If t(x) is the time at which diner x makes her grab, then this happens exactly when t(0) is the unique local maximum of t in the range [−j, −j + 1, . . . , 0, . . . , i−1, i]. If t is plotted on this interval, it looks like a mountain with (0, t(0)) on top; more precisely, t(−j) < t(−j+1) < · · · < t(−1) < t(0) > t(1) > t(2) > · · · > t(i). Instead of evaluating the probability of this event for fixed i and j, it is convenient to lump all pairs (i, j) together which satisfy i+j+1 = k for fixed k. Altogether there are k! ways the values t(−j), . . . , t(i) can be ordered. If T is the set of all k grabbing times and tmax is the last of these, then each mountain-ordering is uniquely identified by the nontrivial subset of T \ {tmax} which constitutes the values {t(1), . . . , t(i)}. Thus, the number of orderings that make a valid mountain is 2k−1 −2. Finally, the total probability that diner 0 is deprived of her napkin is ∞ X k=3 21−k · (2k−1 −2) k! = 2 −√e 2 ≈0.12339675. ♡ Roulette for Parking Money You’re in Las Vegas with only $2 and in desperate need of $5 to feed a parking meter. You run in through the nearest door and find yourself at a roulette table. You can bet any whole dollar amount on any allowable set of numbers. What roulette-betting strategy will maximize your probability of walking out with $5? Solution: There are two major considerations in trying to maximize your probability of boosting your fortune to $5. One is to avoid waste: That is, try to hit $5 exactly. Overshooting gains nothing and since your expectation is limited by your initial $2, minus what the “house” is expecting to take from you, you don’t want to spend it on overshooting your goal. The second is speed. As we saw above in Roulette for the Unwary, it costs you about a nickel in expectation every time you bet $1 playing roulette, no matter what you bet on. Of course, you must do some betting. Ideal would be if there were a roulette bet that paid off 5:2, so you could just put your $2 down once and win what you need or lose everything in one shot. Unfortunately there is no such bet in roulette, so it looks like you are doomed to either risk overshooting, or risk having to make many bets. Example: You could bet $1 on the numbers 1 through 8. If you win you’re home, otherwise 251 you could bet your remaining $1 on red and if it wins, start over. That gives you probability p of success where p = 8 38 + 30 38 · 18 38 · p . That gives p = 8 38/(1 −30 38 · 18 38) = 0.33628318584, about 1/3. But you can do better by making correlation work for you. How? By making simultaneous bets of $1 each of the right kind. Specifically: Bet one of your dollars on the numbers 1 through 12, and simultaneously, the other on 1 through 18. Then, if you do get a number between 1 and 12, you hit your $5 on the nose. If you hit a high number, you lose it all (but at least you’ve lost it all quickly). If you happen to hit 13, 14, 15, 16, 17, or 18, you again have $2 and can repeat your double bet. The probability p of success for this scheme is given by p = 12 38 + 6 38 · p, which yields p = 12 38/(1 − 6 38) = 3/8 = 0.375, quite a bit better. In fact this scheme can be shown to be optimal. Next is a puzzle that you might find familiar. Posed and solved by Georges-Louis Leclerc, Comte de Buffon in 1733, it is said to be the very first solved problem in geometric probability. But to solve it we will use almost no geometry, relying instead on linearity of expectation. Buffon’s Needle A needle one inch in length is tossed onto a large mat marked with parallel lines one inch apart. What is the probability that the needle lands across a line? Solution: Let C be a “noodle”: Any smooth plane curve—self-intersections permitted— of length, say, ℓ. Toss the noodle randomly onto the mat, and let XC be the number of crossings you get, that is, points where C crosses a line. We claim that the average number of crossings, EXC, is proportional to ℓ, irrespective of the choice of C! The figure below shows the results of five tosses of a particular noodle, each accompanied by its number of crossings. To verify the claim, imagine that C is composed of many short line segments, each of length ε. Throwing just one line segment L of length ε < 1 onto the mat produces a crossing with some probability p, but never more than one crossing, thus EXL = p. At the moment, we don’t know what p is. But, by linearity of expectation, we know that if C is composed of n such segments, then EXC = np. It follows, by letting ε shrink, that EXC = αℓfor some constant α, since a smooth C is approximable by a string of short line 252 segments. If we can determine what α is, we can infer that EXN = α, since Buffon’s needle N has length 1. To find α we only have to pick C cleverly, and this is how: let C be a circle of diameter 1! Then no matter how C lands on the mat, it produces exactly 2 crossings. Thus EXC = 2 = α · π, and we solve to get α = 2/π. We deduce that EXN = 2/π, and since (with probability 1) the needle either produces one crossing or none, the probability that Buffon’s needle crosses a line is that same 2/π. ♡ This wonderful proof was published by T. F. Ramaley in 1969, in a paper entitled “Buffon’s Noodle Problem.” Here are a couple of more modern geometry problems, that seem not to have any probability in them. Covering the Stains Just as a big event is about to begin, the queen’s caterer notices, to his horror, that there are 10 tiny gravy stains on the tablecloth. All he has the time to do is to cover the stains with non-overlapping plates. He has plenty of plates, each a unit disk. Can he do it, no matter how the stains are distributed? Solution: The naked mathematical question being asked here is: Can any set of 10 points in the plane be covered by disjoint unit disks? Sure, if the points are close together, you can cover them with one disk; if they’re spread widely apart, you can cover each with its own disk. It’s the medium distances that could cause problems. It’s a good idea, as advertised in Chapter 7, to think about replacing 10 by other numbers. It’s pretty obvious that any one or two points can be covered, and easy to show that any three can be covered. On the other hand, the answer is certainly “no” if 10 is replaced by 10,000. The reason is that you can’t “tile” the plane with unit disks; in other words, you can’t arrange disjoint unit disks so as to cover the whole plane (as you 253 could, say, with unit squares). If you put down 10,000 points in a 100 × 100 grid pattern covering a square somewhat bigger than a unit disk, it won’t be possible to cover every grid point with disjoint unit disks. So it’s a question of quantity—is 10 more like 3, or more like 10,000? Here’s a thought—how much of the plane can be covered by disjoint unit disks? You would probably guess that the best way to pack unit disks is in a hexagonal array, as shown in the figure below. In fact, it is provable that this is the best packing. How good is it—in other words, if you choose a large region of the plane (say, a 1,000,000 × 1,000,000) square) and pack it hexagonally with disks, what fraction of the region will be covered? In the limit, that fraction must be the fraction of the area of one of those hexagonal cells that is occupied by an inscribed unit disk. Of course, the disk, having radius 1, has area π. The hexagon is made up of six equilateral triangles of altitude 1 and area 1/ √ 3, thus total area 6/ √ 3 = √ 12. It follows that the fraction of area covered by the disk is π/ √ 12 ∼0.9069. Hmm. The disks cover more than 90% of the plane, that is, more than 9 points out of 10. Can we make use of that? Trouble is, the gravy stains are not random points, they are (in the worst case) arranged as if by an adversary. But we can shift the hexagonal disk-packing randomly. To do this, just fix some hexagonal cell and pick a point in it uniformly at random; then shift (with-out rotating) the whole picture by moving the center of the cell to the selected point. Now we can say that for any fixed point of the plane, the probability that it is covered by some disk of the randomly shifted configuration is 0.9096. It follows by linearity of expectation that the expected number of our 10 gravy stains covered by disks of the randomly shifted hexagonal packing is 9.096 stains, that is, on average more than 9 of the 10 points are covered. But then some configurations must cover all 10 points, and the problem is solved. ♡ 254 This technique, in which we prove that something can be done by showing that if we do it randomly it succeeds with probability > 0, is often known by the fancy name “the probabilistic method” and associated with Paul Erd˝ os and Alfr´ ed R´ enyi. We’ll see another example in the theorem at the end of this chapter. The next puzzle provides another example where introduction of randomness is useful. Colors and Distances In the town of Hoegaarden, Belgium, exactly half the houses are occupied by Flemish families, the rest by French-speaking Walloons. Can it be that the town is so well mixed that the sum of the distances between pairs of houses of like ethnicity exceeds the sum of the distances between pairs of houses of different ethnicity? Solution: No. Represent each house in Hoegaarden as a distinct point in the plane, and fix a large disk that contains all the points. Now consider a random line that intersects the disk. (To get such a line, choose a radius of the disk uniformly at random, then take the line perpendicular to the radius that crosses it at a uniformly random point along the radius.) Your random line will cut the house-points into two sets (one of which could be empty); we claim it separates at least as many house-pairs of different eth-nicity as pairs of the same ethnicity. Why? Suppose there are n houses of each ethnicity, and that f Flemish houses and w Walloon houses lie on one side of the line. Then the number of heterogeneous crossings is (n −w)f + (n −f)w while the number of homogeneous crossings is (n −w)w + (n −f)f; the former minus the latter is f 2 + w2 −2wf = (f −w)2 ≥0. Now we make use of the critical fact that for any two particular points, the probability that they are separated by our random line is proportional to the distance between the points. Adding expectations, we deduce that the total distance between heterogeneous (resp., homogeneous) pairs of points is propor-tional to the expected number of separated heterogeneous (resp., homogeneous) pairs. Since every line cuts at least as many heterogeneous as homogeneous pairs, we conclude that the former distances add up to at least as much as the latter. Note that for distinct points the inequality is strict, because as long as there is one point which is not simultaneously of both ethnicities, there is positive probability that a line will result in f ̸= w causing an imbalance in favor of heterogeneous pairs. The next puzzle also has a geometric component, but here it is less surprising that we use linearity of expectation. 255 Painting the Fence Each of n industrious people chooses a random point on a circular fence, and begins painting toward the farther of her two neighbors until she encounters a painted section. On average, how much of the fence gets painted? How about if instead, each person paints toward her nearer neighbor? Solution: For n = 2, the answers are easily determined to be 3/4 and 1/4, respectively. For larger n, we consider any interval I between neighboring painters and compare it to its two adjacent intervals. In Case (A), where each painter paints toward her farther neighbor, I will go unpainted exactly when it is the shortest of the three; in Case (B), when it is the longest. Each of these events happens with probability 1/3. It seems like we’re going to need to know the expected length of the short-est (resp., longest) of three intervals chosen by cutting a given interval J at two random points. You can get this noting that choosing two random points from the unit interval is equivalent to choosing a uniformly random point in an equilateral triangle; the barycentric coordinates of the point give your interval lengths. Then, if you identify the regions of the triangle which result in the middle third being shortest (resp. longest) of the three intervals and compute conditional expectations, you get 1/9 and 11/18, respectively. That calculation was for the unit interval, so if the length of I together with its two neighbor intervals is x, then the expected length of I given that it is the shortest is x/9 and given that it is the longest, 11x/18. On average, the total length of I and its two neighbors is 3/n since adding up this figure for all I counts every interval three times. But in both versions, I is unpainted only a third of the time (in expectation). It follows that in Case (A) the expected amount of unpainted fence is n 3 · 3/n 9 = 1/9, and in Case (B), n 3 · 11·3/n 18 = 11/18. For the next puzzle, it is useful to have an alternative formula for expecta-tion. Suppose X is a “counting” random variable, that is, a random variable that takes values in the non-negative integers. Then EX = ∞ X i=0 P(X > i). To see that this formula is equivalent to the definition at the top of the chapter, we just need to verify that for each j, the probability that X = j is counted j times in the sum. But that’s easy: It’s counted once for each i between 0 and j−1, so j times as required. Filling the Cup You go to the grocery store needing one cup of rice. When you push the button on the machine, it dispenses a uniformly random amount of rice between nothing 256 and one cup. On average, how many times do you have to push the button to get your cup? Solution: We know that at least two pushes are necessary, and in fact two will suffice just half the time; and sometimes more than three will be required. So the answer must be more than 2 1 2. Would you believe 2.718281828459045? Let X1, X2 etc., be the amount dispensed, and let Yi be the fractional part of X1 + · · · + Xi. Then the Yi’s, like the Xi’s, are independent and uniformly random in the unit interval. The first decrease in the values of the Yi’s signals the point when the amount of rice exceeds 1 cup. Thus, the question at hand is equivalent to: What is the expected value of Z, the length of the longest increasing initial sequence of the Yi’s? The probability that the first j values are increasing is 1/j!, since the in-creasing order is just one of j! permutations. Here’s where we use the special formula above for the expected value of a counting random variable: EZ = ∞ X j=0 P(Z > j) = ∞ X j=0 1/j! which is just exactly e, Euler’s number (also known as Napier’s constant, but, naturally, discovered by yet someone else—Jacob Bernoulli). It’s time for our theorem, which, as promised, will be proved using the probabilistic method. Arguably the most famous theorem in combinatorics, the following remark-able fact was proved in 1930 by Frank Plumpton Ramsey, scion of a famous family of British intellectuals (and brother of Arthur Michael Ramsey, Arch-bishop of Canterbury). In its simplest finite form, Ramsey’s Theorem states that for any positive integer k there is a number n such that if you color every unordered pair of numbers from the set {1, . . . , n} either red or green, then there must be a set S of k numbers which is “homogeneous” in the sense that every pair from S is the same color. The least n for which this is true is called the “kth Ramsey number” and denoted R(k, k). For example, R(4, 4) turns out to be 18; that means that for any two-coloring of the pairs of numbers between 1 and 18, there’s a homoge-neous set of size 4; but this does not hold for all colorings of pairs from the set {1, . . . , 17}. But R(5, 5) is not known! Paul Erd˝ os, who wrote more papers than any other mathematician in history and was fascinated by Ramsey’s the-orem, enjoyed giving the following advice. If a powerful alien force demands, on penalty of Earth’s annihilation, that we tell them the value of R(5, 5), we should put all the computing power on the planet to work and perhaps succeed in computing that number. But if they ask for R(6, 6), we should attack them before they attack us. 257 It is not hard to prove—and, indeed, we will prove it in Chapter 15—that R(k, k) is at most 2k−2 k−1 , which in turn is less than 22k. What we’re going to do here is use linearity of expectation to get an exponential lower bound for R(k, k). Theorem. For all k > 2, R(k, k) > 2k/2. The idea of the proof, published by Erd˝ os in 1947, is to show that a random coloring of the pairs of numbers from 1 to n = 2k/2 will, with positive probability, have no monchromatic set of size k. Thus, such colorings exist (although the proof does not exhibit one). By a random coloring, we mean this: For each pair {i, j} of numbers between 1 and n, we flip a fair coin to decide whether to color it red or green. If we pick a fixed set S of size k, it contains k 2 = k(k−1)/2 pairs and thus will be homogeneous with probability 2/2k(k−1)/2 = 2(k+1)/2−k2/2. The number of candidates for the set S, i.e., the number of subsets of {1, . . . , n} of size k, is n k < (2k/2)k/k! = 2k2/2/k!, thus by linearity of ex-pectation, the expected number of homogeneous sets is less than 2(k+1)/2−k2/2 · 2k2/2/k! = 2(k+1)/2 k! . which is equal to √ 2 when k = 2 and declines as k increases. So the expected number of homogeous sets is less than 1, and we claim that means the probability that there are no homogeneous sets is greater than zero. One way to see this is to use the formula for expectation of a counting random variable from Filling the Cup above. The expected value of X, if X is the number of homogeneous sets, equals Pr(X > 0) plus more non-negative terms; so if EX is less than 1, so is Pr(X > 0). ♡ The theorem tells us that R(5, 5) is at least ⌈25/2⌉= 6 and our later induction proof implies an upper bound of 8 4 = 70. In fact R(5, 5) is, at the time of this writing, known to be between 43 and 48 inclusive. 258 Chapter 15 Brilliant Induction Induction (from the verb induce, not induct!) is one of the most elegant and effective tools in mathematics. In its simplest form, you want to prove something for all positive integers n; you do it by proving the statement for n = 1 and then showing that for all n > 1, if it’s true for n−1, it’s true for n. More generally, when proving your statement for n, you may assume it’s already true for all positive integers m < n. That assumption is known as the induction hypothesis, or “IHOP” for short (see Notes & Sources). I also like the following formulation, which is even more general. You want to show that some statement is true in all instances satisfying certain conditions. You do this by assuming that there is some instance in which it is false, and focussing on such an instance which is in some sense minimal. Then you show how this bad instance can be turned into another bad instance that is even smaller, contradicting your assumption that the instance you began with was minimal. You conclude that there are no bad instances, therefore your statement is always true. So in a sense, induction is a special case of reductio ad absurdum—in other words, showing that a contrary assumption leads to a contradiction. Is this all a bit too abstract for you? That’s what the puzzles are for. Uniform Unit Distances Can it be that for any positive integer n, there’s a nonempty configuration of finitely many points on the plane with the property that every point of the configuration is at distance 1 from exactly n of the other points in the configu-ration? Solution: For n = 1, the endpoints of a unit-length line segment will do. For n = 2, we can translate the segment by a unit distance, and add the result to the original segment, to get four points in a diamond. In what direction? Almost any 259 direction will do (as long as it’s not at an angle to the original segment that is multiple of 60◦). So if we make it a random direction, it’ll work with probability 1. We now proceed in that manner by induction. Assume the set Sn of points (containing 2n points, but we don’t care) has the property that every point in it is at unit distance from exactly n others. Now translate Sn by a unit distance in a random direction to form the set S′ n, and let Sn+1 be the union of S′ n and Sn. Each point of Sn+1 is now at unit distance from n others plus the translated copy of itself, for a total of n + 1. (See figure below for the construction in the n = 2 case.) Only finitely many translation angles could cause an “accidental” unit distance, so with probability one this works. ♡ Swapping Executives The executives of Women in Action, Inc., are seated at a long table facing the stockholders. Unfortunately, according to the meeting organizer’s chart, every one is in the wrong seat. The organizer can persuade two executives to switch seats, but only if they are adjacent and neither one is already in her correct seat. Can the organizer organize the seat-switching so as to get everyone in her correct seat? Solution: Yes. Let the positions (and the executives that belong in them) be numbered 1 to n from left to right. By induction on n, it suffices to get executives m through n into their correct seats (for some m ≤n) provided the rest are still all incorrectly seated. To do that we move executive n to the right by successive swaps until she encounters an executive, say the ith, who’s sitting in position i+1 and therefore 260 would be swapped into her correct seat if we proceeded. It may also be that executive i+1 is in seat i+2, executive i+2 in seat i+3, etc. forming what we will call a “blockade” containing at least executive i. If the blockade goes all the way to the end of the row, we go ahead and perform all the swaps, ending with executives i through n all in their correct seats while the rest remain to be sorted. If the blockade terminates before the end, let executive j be the executive immediately to the right of the blockade. Instead of swapping n with i, we swap j toward the left until j swaps with n; at no point in this process will j be in her correct seat because all the seats she is swapped into belong to members of the blockade. Now we redirect our attention to n, who has moved one step to the right, and continue swapping her to the right until the next encounter with some executive that starts a shifted block, handling that situation as before. Eventually executive n reaches the right-hand end and the inductive proof is complete. ♡ How many swaps might be needed for this? While executive n is being dealt with, no executive i can be in a blockade twice, because when she is in a blockade she is swapped to the right (from seat i+1 to seat i+2) and then passed by executive n. We conclude that it takes no more than 2(n−1) steps to handle executive n, thus at most 2(1 + 2 + · · · + n−1) = n(n−1) swaps to get everyone into her correct seat. Odd Light Flips Suppose you are presented with a collection of light bulbs controlled by switches. Each switch flips the state of some subset of the bulbs, that is, turns on all the ones in the subset that were off, and turns off those that were on. You are told that for any nonempty set of bulbs, there is a switch that flips an odd number of bulbs from that subset (and perhaps other bulbs as well). Show that, no matter what the initial state of the bulbs, you can use these switches to turn off all the bulbs. Solution: The proof is by induction on number n of bulbs. Note that the order of switch-flipping does not matter; all that counts is whether each switch is flipped an odd or even number of times. The statement is trivially true for n = 1, so suppose n ≥2 and we are given a configuration of n bulbs and a set of switches for them. By the IHOP, for every bulb i there is a set Si of switches that can turn off all the bulbs except possibly bulb i. If for any i the set Si does turn off bulb i we are done, so let’s assume no i has this property. Choose a pair of distinct bulbs, i and j, and perform Si followed by Sj. The result is that except for i and j, every bulb that was initially off stays off, while every bulb that was initially on was flipped off and then on again—thus, 261 remains in the state it was in before. The exceptions are bulbs i and j, which are both flipped. It follows that if bulbs i and j are both on, we can use Si ∪Sj to turn them both off without changing the state of any other bulb. So if an even number of bulbs are on, we can turn them off pairwise and thus turn all the bulbs off. What if an odd number of bulbs are on? Then we use the given condition, the relevant set of bulbs being all of them, to change the number of lit bulbs to an even number. ♡ There’s something a bit mysterious about this proof—didn’t we use the given condition only in one special case, where the set of bulbs that we’re trying to flip an odd number from is the whole set of bulbs? Maybe that’s all we needed!? No, that would fail even for two bulbs, when there’s just one switch and it flips one bulb; then there’d be no way to turn the other bulb off. The issue here is that to execute the induction, we needed a condition that “persists downward”—in this case, one that continues to hold when we remove a bulb. Otherwise, we could not apply the IHOP after removing bulb i. In fact we really do need the condition for all subsets. If there’s a set S of bulbs we can’t flip an odd number from, and we begin with just one bulb from S on, we’re stuck. Truly Even Split Can you partition the integers from 1 to 16 into two sets of equal sizes so that each set has the same sum, the same sum of squares, and the same sum of cubes? Solution: There is indeed such a partition: One set is {1, 4, 6, 7, 10, 11, 13, 16}, the other {2, 3, 5, 8, 9, 12, 14, 15}. To see this, you might think to yourself: Hmm, 16 is a power of 2; is it possible that this is an example of a more general statement? Can I partition 1 through 8 into two equal-sized sets with the same sum and sum of squares, for example? How about partitioning 1 through 4 into two equal-sized sets with the same sum? The latter is certainly easy: {1, 4} versus {2, 3}. Then of course {5, 8} versus {6, 7} also works for the numbers from 5 to 8, and if you put these together cross-wise, you get {1, 4, 6, 7} versus {2, 3, 5, 8} which works perforce for sums and now also seems to work for sums of squares. In general, you can prove by induction that the integers from 1 to 2k can be partitioned into sets X and Y so that each part has the same sum of jth powers, where j runs from 0 to k−1; equivalently, such that for any polynomial P of degree less than k, the number P(X), which we define as P{P(x) : x ∈X}, is equal to P(Y ). To move up to 2k+1, take X′ = X ∪(Y +2k) (where you get Y +2k by adding 2k to each element of Y ) and Y ′ = Y ∪(X+ 2k). We need to prove that for any 262 polynomial P of degree at most k, P(X) + P(Y + 2k) = P(Y ) + P(X+ 2k). If P has degree less than k, then by induction P(X) = P(Y ) and we also get P(X+ 2k) = P(Y + 2k) since each term is merely another polynomial (Q, say) in xi or yi. Thus X′ and Y ′ certainly agree for polynomials of degree less than k, but what if P has degree k? But we’re OK here too, because the kth power terms on both sides of the displayed equation are the same as the kth power terms of P(X) + P(Y ). ♡ Non-Repeating String Is there a finite string of letters from the Latin alphabet with the property that there is no pair of adjacent identical substrings, but the addition of any letter to either end would create one? Solution: Yes, by induction on the number n of letters in the alphabet (then taking n = 26). Let L1, . . . , Ln be the letters; for n = 1 the one-letter string S1 = ⟨L1⟩ does the trick. We can take S2 = ⟨L1L2L1⟩, S3 = ⟨L1L2L1L3L1L2L1⟩, and and in general Sn = Sn−1LnSn−1. Proof that this works: Assume Sn−1 works for the alphabet {L1, . . . , Ln−1}; we want to show Sn = Sn−1LnSn−1 works for the alphabet {L1, . . . , Ln}. First we note that Sn can’t have a repeated substring, because such a substring could not contain the once-appearing letter Ln, and would thus have to occur entirely inside one of the Sn−1’s, contradicting our IHOP. Now suppose the letter Lk, 1 ≤k ≤n, is added to the right-hand end of Sn; we want to show that this does create a repeated substring. Certainly it does if k = n, as our whole string is now Sn−1Ln followed by itself. Otherwise, our IHOP tells us that there is already a repeated substring inside Sn−1Lk. The argument when Lk is added to the left-hand end of Sn is essentially the same. ♡ For the Latin alphabet, the length of our string is 226 −1 = 67, 108, 863 letters. Baby Frog To give a baby frog jumping practice, her four grandparents station themselves at the corners of a large square field. When a grandparent croaks, the baby leaps halfway to it. In the field is a small round clearing. Can the grandparents get the baby to that clearing, no matter where in the field she starts? Solution: Yes. Cover the field with a 2n × 2n square grid; we will show by induction on n that from any starting point, the baby frog can be guided to any grid cell 263 we want. Suppose the target cell is in the SW quadrant. Double that smaller grid in both directions and position it so that it covers the whole field as a 2n−1 ×2n−1 grid. Our IHOP tells us that the baby frog can be croaked into the doubled target cell (which is the union of four cells of the original grid). Now one croak from the SW corner grandparent will get the baby to the original target cell. That concludes our induction proof, and it remains only to take n large enough so that some cell of the 2n × 2n grid lies entirely inside the clearing that the baby frog is meant to visit. Guarding the Gallery A certain museum room is shaped like a highly irregular, non-convex, 11-sided polygon. How many guard-posts are needed in the room, in the worst case, to ensure that every part of the room can be seen from at least one guard-post? Solution: Three guard-posts are sufficient, and sometimes necessary. In general, in any n-gon, ⌊n/3⌋(that is, the greatest integer in n/3) guard-posts suffice, and this number is the best bound possible. The next figure shows, by extrapolation, that ⌊n/3⌋posts may be necessary; the following proof shows that they suffice, and in fact can be placed in corners. The first move is to triangulate the polygon. This can done by drawing non-crossing diagonals until no more will fit; a diagonal is a line segment between two vertices of the polygon whose interior lies entirely in the interior of the polygon. 264 Next we show by induction on n that the polygon’s vertices can be colored with three colors so that every triangle’s vertices get all three colors. Choose any diagonal D in the triangulation and cut through it “lengthwise” to form two polygons, each with fewer than n vertices, and each having D as an edge. Color both using the IHOP and permute the colors of one so that the colors at the endpoints of D match, then put the two small polygons back together to get a coloring of the original polygon (as is done in the figure below). The least-frequently used color is found on at most ⌊n/3⌋vertices, and sta-tioning guards there covers all the triangles and thus the whole room. ♡ Path Through the Cells Cells in a certain cellular telephone network are assigned frequencies in such a way that no two adjacent cells use the same frequency. Show that if the number of frequencies used is minimal (subject to this condition), then it’s possible to design a path that moves from cell to adjacent cell and hits each frequency exactly once—in ascending order of frequencies! Solution: This is really a problem in abstract graph theory; an assignment of frequencies, or “colors,” to vertices in a graph in such a way that no two adjacent nodes get the same color is called a proper coloring. The statement to be proved is that in any graph properly colored with the minimum number of colors, there is a path that hits every color exactly once; and moreover, if the colors are ordered (e.g., they correspond to the numbers from 1 to k) we can choose the path so as to hit the colors in ascending order. The graph G in our case is the one whose vertices are cells, two vertices constituting an edge if the corresponding cells are adjacent (thus susceptible, presumably, to frequency interference). 265 The idea is to re-color G in a “greedy” fashion, trying to use high-numbered colors. First, re-color any vertex colored k−1 by color k, if it isn’t adjacent to a vertex of color k. Then recolor any vertex of color k−2 by color k−1, if it isn’t adjacent to a vertex of color k−1. Continue in this way to get a new coloring which, by our minimality assumption, still uses all k colors. Now start the path at any vertex v1 whose new color (and therefore whose old color as well) is 1; we know there’s an adjacent vertex v2 of new color 2, since otherwise the color of v1 would have been incremented. Note v2’s old color is also 2; it cannot have been recolored from color 1 since it is adjacent to v1. Similarly, there’s a vertex v3 adjacent to v2 of color 3, which could not have been recolored from color 2 on account of v2. Proceeding in this manner we get a path v1, v2, . . . that hits every color in both the new coloring and the old in increasing order. This puzzle is one of many examples where asking for a proof of a weaker statement (e.g., just that there is a path that hits every frequency exactly once) can make the puzzle harder. Profit and Loss At a recent stockholders’ meeting of Widget Industrials Inc., the Chief Financial Officer presented a chart of the month-by-month profits (or losses) since the last meeting. “Note,” said she, “that we made a profit over every consecutive eight-month period.” “Maybe so,” complained a shareholder, “but I see we lost money over every consecutive five-month period!” What’s the maximum number of months that could have passed since the last meeting? Solution: What’s needed, of course, is a maximum-length sequence of numbers such that every substring of length 8 adds up to more than 0, but every substring of length 5 adds up to less than 0. The string must certainly be finite, in fact less than 40 in length, else you could express the sum of the first 40 entries both as the (positive) sum of 5 substrings of length 8 and the (negative) sum of 8 substrings of length 5. Let’s tackle the problem more generally and let f(x, y) be the length of the longest string such that every x-substring has positive sum and every y-substring negative; we may suppose x > y. If x is a multiple of y, then f(x, y) = x−1 and we must accept vacuous truth with respect to the x-substrings. What if y = 2 and x is odd? Then you can have a string of length x itself, with entries that alternate between, say, x−1 and −x. But you can’t have x+1 numbers, because in each x-substring the odd entries must be positive (since you can cover it with 2-substrings leaving out any odd entry). But there are two x-substrings and together they imply that the middle two numbers are both positive, a contradiction. 266 Applying this reasoning more generally suggests that f(x, y) ≤x+y−2 when x and y are relatively prime, that is, they have no common divisor other than 1. We can prove this by induction as follows. Suppose to the contrary that we have a string of length x+y−1 which satisfies the given conditions. Write x = ay+b where 0 < b < y, and look at the last y+b−1 numbers of the sequence. Observe that any consecutive b of them can be expressed as an x-substring of the full string, with a y-substrings removed; therefore, it has positive sum. On the other hand, any (y−b)-substring of the last y+b−1 can be expressed as a+1 y-strings with an x-string removed, hence has negative sum. It follows that f(b, y−b) ≥y+b−1. Since b and y−b are relatively prime, though, our inductive assumption asserts that f(b, y−b) ≤b + (y−b) −2 = y −2. To show that f(x, y) is actually equal to x+y−2 when x and y are relatively prime, we construct a string which has the required properties and more: It takes only two distinct values, and it is periodic with periods both x and y. Call the two values u and v, and imagine at first that we assign them arbitrarily as the first y entries of our string. Then these assignments are repeated until the end of the string, making the string perforce periodic in y. To be periodic in x as well, we only need to ensure that the last y−2 entries match up with the first y−2, which entails satisfying y−2 equalities among the original y choices we made. Since there are not enough equalities to force all the choices to be the same, we can ensure that there is at least one u and one v. Let us do this, for example, with x = 8 and y = 5. Call the first five string entries c1, . . . , c5, so the string itself will be c1c2c3c4c5c1c2c3c4c5c1. To be periodic with period 8, we must have c4 = c1, c5 = c2, and c1 = c3. This allows us to have c1 = c3 = c4 = u, for example, and c2 = c5 = v; the whole sequence is thus uvuuvuvuuvu. Getting back to general x and y, we note that a string which is periodic in x automatically has the property that every x-substring has the same sum; because, as you shift the substring one step at a time, the entry picked up at one end is the same as the entry dropped at the other end. Of course the same applies to y-substrings if the string is periodic in y. Let Sx be the x-substring sum and Sy similarly; we claim Sx/x ̸= Sy/y. The reason is that if there are, say, p copies of u in each x-substring and q copies of u in each y-substring, then Sx/x = Sy/y would imply y(pu + (x−p)v) = x(qu + (y−q)v) which reduces to yp = xq. Since x and y are relatively prime, this cannot happen for 0 < p < x and 0 < y < q. It follows that we can adjust u and v so that Sx is positive and Sy is negative. In the case above, for example, each 8-substring contains 5 copies of u and 3 of v, while each 5-substring contains 3 copies of u and 2 of v. If we take u = 5 and v = −8, we get Sx = 1 and Sy = −1. The final sequence, solving the original problem, is then 5, −8, 5, 5, −8, 5, −8, 5, 5, −8, 5. ♡ 267 Uniformity at the Bakery A baker’s dozen (thirteen) bagels have the property that any 12 of them can be split into 2 piles of 6 each, which balance perfectly on the scale. Suppose each bagel weighs an integer number of grams. Must all the bagels have the same weight? Solution: Yes. Suppose otherwise, and let b1, . . . , b13 be a set of weights, not all the same, with the stated property. We can assume (by shifting all the weights down) that some bagel weighs 0 grams. Let 2k be the greatest power of 2 that divides all the weights evenly, and suppose our counterexample was chosen to minimize k. There can’t be a bagel of odd weight, because to be balanced the remaining bagels must have an even total weight; but leaving out the 0-gram bagel must also leave even weight behind, an impossibility. But if all the weights are even we can divide them all by 2, reducing k and contradicting our IHOP. ♡ It is worth noting that the requirement that the weighings have six bagels on a side is necessary. Otherwise, for example, a baker’s dozen consisting of 7 bagels of 50 grams each and 6 bagels of 70 grams each would have the stated property. The place where the proof breaks down is where we shift all the weights down; that relies on there being the same number of items on each side during the weighings. Summing Fractions Gail asks Henry to think of a number n between 10 and 100, but not to tell her what it is. She now tells Henry to find all (unordered) pairs of numbers j, k that are relatively prime and no more than n, but add up to more than n. He now adds all the fractions 1/jk. Whew! Finally, Gail tells Henry what his sum is. How does she do it? Solution: For n = 2, the only eligible pair is {1, 2} so the sum is 1/2. For n = 3, we have pairs {1, 3} and {2, 3} for a total of 1/3 + 1/6 = 1/2. For n = 4, the pairs are 268 {1, 4}, {2, 3}, and {3, 4} giving 1/4 + 1/6 + 1/12 = 1/2. Hmm. Is it possible that the sum is always 1/2? Let’s try to prove it by induction, starting from n = 2. Moving from n−1 to n, we gain 1/jn for each j with gcd(j, n) = 1, and lose 1/jk for each j and k with gcd(j, k) = 1 and j + k = n. But if j + k = n, then any pair of j, k, and n being relatively prime implies that all three pairs are. Thus, each pair j, k that sum to n with gcd(j, k) = 1 signifies a loss of 1/jk but a gain of 1/jn + 1/kn = 1/jk, neatly cancelling. ♡ Tiling with L’s Can you tile the positive quadrant of the plane grid with unrotated triominoes each of which is either shaped like a letter L, or a backwards L? Solution: The answer is perhaps a bit surprising: You can indeed do it, but in only one way, and that way is not periodic. If you try to do it by hand, starting at the origin with a full row along the X-axis and proceeding upward row by row, you will find that a little forethought (thinking one row ahead) is all you need. That first row must start (at the left) with a forward L, which we’ll call an “el.” To its right we’ll need to put a backward L, which we’ll call a “le,” so that the space above these two triominoes will be fillable. Then you’ll need another el, another le, another el, and so forth, alternating out to infinity. The next row is already half-covered; to cover the rest of those cells, you’ll need to start with a le. Each successive choice of el or le must leave an even-length gap on row 3, otherwise you’ll be stumped later. The good news is that the choice of el or le always changes the length of the previous gap by one, thus you can always make that gap even. 269 This sets us up nicely for a proof by induction that the construction is possible. What should the induction hypothesis be? We’d like to say that all rows are covered up to and including row k (say), and the next row, row k+1, is covered up to position m (which could be zero). After position m, row k+1 has even-length gaps out to infinity. What else? Well, we also need to say that row k+2 has even-length gaps out to position m, and is empty after that; and no higher row is covered at all. That’s a lot of hypotheses, but we need them all and once we have them, the argument is easy. We insert a new triomino in such a way that its bottom covers cells m+1 and m+2 in row k+1, choosing el or le so as to make the new gap, above in row k+2, even. That will also have the effect of shortening the next gap on row k+1 by two, which kills the gap or leaves it of even length, either of which is OK. Everything seems to be fine, but hold on a minute—what, exactly, are we inducing on? It can’t be the number of triominoes placed, because that’s already infinite after the first row goes down. What we’re really doing is a series of inductions, one for each row. Each induction ends with a tiling in which rows 1 through k are completely covered, row k+1 partly covered, no higher row covered at all, and all gaps on row k+1 of even length. This is just an “m = 0” case of our induction hypothesis. Uniqueness of our tiling follows by assuming we have any valid tiling, choos-ing any k and m, and looking at that part of the tiling corresponding to the k, m case of our induction hypothesis. It must satisfy the hypothesis in order to be extendible, and since our construction leaves no choice at any point, the tiling must be identical to ours. Traveling Salesmen Suppose that between every pair of major cities in Russia, there’s a fixed one-way air fare for going from either city to the other. Traveling salesman Alexei Frugal begins in St. Petersburg and tours the cities, always choosing the cheapest flight to a city not yet visited (he does not need to return to St. Petersburg). Salesman Boris Lavish also needs to visit every city, but he starts in Kaliningrad, and his policy is to choose the most expensive flight to an unvisited city at each step. It looks obvious that Lavish’s tour costs at least as much as Frugal’s, but can you prove it? Solution: One way is to show that for any k, the kth cheapest flight (call it f) taken by Lavish is at least as costly as the kth cheapest flight taken by Frugal. This seems like a stronger statement than what was requested, but it really isn’t; if there were a counterexample, we could adjust the flight costs, without changing their order, in such a way that Lavish paid less than Frugal. For convenience, imagine that Lavish ends up visiting the cities in west-to-east order. Let F be the set of Lavish’s k cheapest flights, X the departure cities 270 for these flights, and Y the arrival cities. Note that X and Y may overlap. Call a flight “cheap” if its cost is no more than f’s; we want to show that Frugal takes at least k cheap flights. Note that every flight eastward out of a city in X is cheap, since otherwise it would have been taken by Lavish instead of the cheap flight in F that he actually took. Call a city “good” if Frugal leaves it on a cheap flight, “bad” otherwise. If all the cities in X are good, we are done; Frugal’s departures from those cities constitute k cheap flights. Otherwise, let x be the westernmost bad city in X; then when Frugal gets to x, he has already visited every city to the east of x, else Frugal could have departed x cheaply. But then every city to the east of x, when visited by Frugal, had its cheap flight to x available to leave on, so all are good. In particular, all cities in Y east of x are good, as well as all cities in X west of x; that is k good cities in all. ♡ Lame Rook A lame rook moves like an ordinary rook in chess—straight up, down, left, or right—but only one square at a time. Suppose that the lame rook begins at some square and tours the 8 × 8 chessboard, visiting each square once and returning to the starting square on the 64th move. Show that the number of horizontal moves of the tour, and the number of vertical moves of the tour, are not equal! Solution: You should be asking yourself: Is this true for n × n boards as well? Some experimentation will remind you that there is no such tour if n is odd, because (for one thing) such a board has more squares of one color than another but a tour that returns to its starting position must visit the same number of squares of each color. Moreover, for n = 2 and n = 6, it is easy to construct tours where the numbers of horizontal and vertical steps are equal. So you might guess that the puzzle works on n × n boards when n ≡0 mod 4, and moreover that the issue could be that the number of (say) horizontal moves must be equivalent to 2 mod 4 and therefore can’t be 64/2. All this is correct, but how to prove it? Assume the rook is a point in the middle of a square, and let P be the polygon thus traced. Note that the vertices of P lie on the dual grid of the chessboard, whose vertices lie at the centers of the chessboard squares. The polygon P outlines a tree T (pink in the figure below) on the grid lines of the chessboard. We will show by induction that the following holds when P is the outline of any tree on the chessboard grid: If there are n0 dual-grid vertices on P’s perimeter that lie on even columns, and n1 on odd columns, and h is the sum of the lengths of the horizontal borders of P, then h ≡n1 −n0 + 2 mod 4. The base case is easy: If the tree is a single vertex, we have n1 = n0 = 2 and h = 2. Now let u be any leaf of T; if it’s attached vertically, cutting it 271 off decrements both n1 and n0 by 1 and does not change h. If u is attached horizontally, it changes n1 −n0 by two while decrementing h by two. Since the polygon P generated by the lame rook has n1 = n0, the number of horizontal moves is 2 mod 4 and thus cannot be 32. ♡ For our theorem we return to Ramsey theory, using induction to get an upper bound for Ramsey numbers (and incidentally to show that they exist). Given any two positive integers s and t, Ramsey’s Theorem claims that for sufficiently large n, for every red-green coloring of the unordered pairs of numbers in {1, 2, . . . , n} there is either a set S of size s all of whose pairs are colored red, or a set T of size t all of whose pairs are colored green. The least n for which this holds is, by definition, the Ramsey number R(s, t). Theorem. For any s and t, R(s, t) exists and is no more than s+t−2 s−1 . We note first that since there are no pairs inside a set of size 1, R(1, k) = R(k, 1) = 1 for any k. Further, to get (say) a red set (i.e., a set all of whose pairs are red) of size 2 we only need one red edge. It follows that R(2, k) = R(k, 2) = k, because (in the case of R(2, k)) if there’s any red pair its members give us a red set of size 2, else all the pairs are green and we can use all the numbers to get our green set of size k. These observations conform to the theorem. Now we will proceed by induc-tion on the sum s+t. Fix s and t and suppose R(s′, t′) ≤ s′+t′−2 s′−1 for all s′, t′ with s′+t′ < s+t. Let U be the set {1, . . . , n} where n = s+t−2 s−1 , and suppose some adversary has colored all the unordered pairs in U red or green. We want to find in U either a red subset S of size s or a green subset T of size t. That would show that the conclusion of Ramsey’s theorem holds when n = s+t−2 s−1 , therefore that R(s, t) is at most this number. 272 Recall that (from Pascal’s triangle) n = j+k, where j = s+t−3 s−2 and k = s+t−3 s−1 . Note that j is our “n” when s′ = s−1 and t′ = t, while k is our “n” when s′ = s and t′ = t−1. Let’s focus our attention on the pairs containing the number 1. There are n−1 of those, and we claim that either at least j of these pairs are red, or at least k of them are green. Why? Because otherwise there are at most j−1 red pairs and at most k−1 green pairs, and that adds up to only j+k−2 = n−2. (Yes, you can think of it as an application of the pigeonhole principle.) Suppose the former, that is, there are j red pairs of the form {1, i}, and let V be the set of numbers i that appear in them (not including 1 itself). Since (s−1) + t < s+t, and j = (s−1)+t−2 (s−1)−1 , we know from our IHOP that there’s either a red set S′ ⊂V of size s−1 or a green set T ′ ⊂V of size t. If it’s the latter, we are happy; and if the former, we can add the number 1 to our set—since all of the pairs involving 1 are red—and get a red set S = S′ ∪{1} of size s. The other case is similar: If there are k green pairs of the form {1, i}, let W be the set of numbers i that appear in them; since s + (t−1) < s + t, and k = s+(t−1)−2 s−1 , we know from our IHOP that there’s either a red set S′ ⊂W of size s or a green set T ′ ⊂W of size t−1. If it’s the former, we have our S; if the latter, we can add the number 1 to T—since all of the pairs involving 1 are green—and get a green set T = T ′ ∪{1} of size t. Done! ♡ As it happens, R(3, 3) is equal to 3+3−2 3−1 = 4 2 = 6, but R(4, 4) = 18 which is strictly less than our bound of 4+4−2 4−1 = 6 3 = 20. Putting our theorem together with the one proved in the previous chapter, we know that R(k, k) lies somewhere between 2k/2 and 22k. Most combinatorialists would guess that there’s some number α between 1 2 and 2 such that R(k, k) behaves like 2αk as k grows—more precisely, that the logarithm base 2 of the kth root of R(k, k) tends to some α as k goes to infinity. Finding α would have earned you $5,000 from “Uncle Paul” Erd˝ os, and he’d have paid happily. 273 274 Chapter 16 Journey into Space We all live in three spacial dimensions, and are attuned to thinking three-dimensionally even though we see, draw and paint, for the most part, only in two. Thus, geometrical problems in three dimensions, while potentially much harder than problems in the plane, can appeal to our intuition in pleasant and even useful ways. Some of the puzzles that follow are set naturally in three dimensions; others can be placed there to good effect. Easy Cake Division Can you cut a cubical cake, iced on top and on the sides, into three pieces each of which contains the same amount of cake and the same amount of icing? Solution: You can get any number k of pieces with the same amount of cake and the same amount of icing in the following neat way: Looking at the cake from above, divide the perimeter of the square top of the cake into k equal parts and make straight, vertical cuts to the middle of the square from each boundary point on the perimeter. 275 That works because each piece, looked at from the top, is the union of triangles with the same altitude (namely, half the side of the cake) and the same total base (namely, 1/k times the perimeter), as in the figure below. Speaking of cubes, here’s a question about painting them. Painting the Cubes Can you paint 1,000 unit cubes with 10 different colors in such a way that for any of the 10 colors, the cubelets can be arranged into a 10 × 10 × 10 cube with only that color showing? Solution: Note that the numbers work out perfectly: 6000 faces need to be colored, and the number of faces showing on the big cube is 6×100 = 600. Thus we’ll need to paint exactly 600 faces with each color, and when we put the big cube together, we can’t afford to bury even a single face of the designated color inside. It turns out that despite these restrictions, there are many ways to color the unit cubes that will do the trick. But finding one by hammer and tongs is a pain in the neck. Instead, number the colors 0 through 9. Apply paint to all the cubes in the infinite 3-dimensional cubic grid, as follows: Paint both sides of all the faces on the plane x = 0 with color 0, all faces on x = 1 with color 1, etc., rotating back to color 0 for the plane x = 10, and so forth. Do the same for the y-planes and z-planes. This scheme colors every face of every unit cube with a color corresponding to the rightmost digit of that face’s u-coordinate, where u is x, y or z according to which axis is perpendicular to the face. In particular, the scheme colors unit cubes in 103 different ways. But now, if you want your big cube to display color i on its outside, cut out your big cube from space via planes x = i, x = i+10, y = i, y = i+10, z = i, and z = i+10. This will use each type of unit cube once, and you are done! ♡ Of course, you can replace 10 by any positive integer n (that is, you can color n3 unit cubes with which you can then make an n × n × n cube with any given color the only one showing). The above argument still works, and in fact versions of it work in other dimensions, as well. Just don’t think too hard about what it means to “paint” a 3-dimensional face of a 4-dimensional hypercube. Curves on Potatoes Given two potatoes, can you draw a closed curve on the surface of each so that the two curves are identical as curves in three-dimensional space? 276 Solution: Yes, you can: Intersect the potatoes! Well, maybe a few more words of explanation would not go amiss. Think of the potatoes as holograms and bring them together so that they overlap in space. Then the intersection of their surfaces will describe a curve (possibly more than one curve) that lives on both surfaces, and solves the puzzle. Painting the Polyhedron Let P be a polyhedron with red and green faces such that every red face is surrounded by green ones, but the total red area exceeds the total green area. Prove that you can’t inscribe a sphere in P. Solution: Assume the sphere is inscribed, and triangulate the faces of P using the points where the sphere is tangent. Then the triangles on either side of any edge of P are congruent, thus have the same area; at most, one of every such pair of triangles is red. It follows that the red area is at most equal to the green area, contradicting our assumptions. ♡ The illustration shows a two-dimensional version, where sides and vertices of a polygon take the place of faces and edges of P. Boarding the Manhole An open manhole 4 meters in diameter has to be covered by boards of total width w. The boards are each more than 4 meters in length, so if w ≥4 it’s obvious that you can cover the manhole by laying the boards side by side (see figure below). If w is only 3.9, say, you still have plenty of wood and you are allowed to overlap the boards (but not cut them up) if you want. Can you still cover the manhole? 277 Solution: Intuitively speaking, even though you have plenty of wood, it seems that you have to waste a lot of it either by covering the center of the manhole many times, or laying boards too close to the edge to be of much use. Putting it another way, it’s relatively expensive to cover parts of the manhole far from the center. Can you quantify that statement in a useful way? It turns out you can, using a famous fact known sometimes as Archimedes’ hat box theorem. Archimedes used his “method of exhaustion” (these days, we would use calculus) to show that if a sphere of radius r is intersected by two parallel planes a distance d apart, as in the figure below, the surface area of the sphere between the planes is 2πrd. In particular, it doesn’t depend on where the planes cut the sphere, but only on how far apart the planes are. (The “hat box” presumably alludes to a cylinder of radius r perpendicular to the planes; the cylinder’s area between the planes is that same 2πrd.) Where do we get our planes from? Let’s replace the manhole by a sphere, also of diameter w, whose equator is the circumference of the manhole. Each 278 board is replaced by a vertical slab consisting of two parallel vertical planes whose distance d apart is the width of the board. Now the key observation: Arranging the boards to cover the manhole is equivalent to arranging the slabs to cover the sphere. But by Archimedes’ theorem, the slabs cannot cover more than 2π w 2 d = πwd of the surface of the sphere, which has area πw2. So if d < w, you’re stuck. ♡ Now we are set up perfectly to conquer the next puzzle. Slabs in 3-Space A “slab” is the region between two parallel planes in three-dimensional space. Prove that you cannot cover all of 3-space with a set of slabs the sum of whose thicknesses is finite. Solution: In a way the conclusion seems obvious; if you can’t cover space when the slabs are parallel and disjoint, surely you can’t cover when they waste space by over-lapping. But infinity is a tricky concept. The slabs all have infinite volume, so what’s to prevent them from covering anything they want? Actually, we know from the previous puzzle that they have trouble covering a large ball. If the total thickness of the slabs is T, then, by Archimedes’ theorem, they can’t cover the surface of a ball of diameter greater than T, thus they can’t cover the ball itself. Thus they certainly can’t cover all of 3-space. But it’s odd that to show the latter, we seem to need to reduce the problem to just a finite piece of space. Bugs on Four Lines You are given four lines in a plane in general position (no two parallel, no three intersecting in a common point). On each line a ghost bug crawls at some constant velocity (possibly different for each bug). Being ghosts, if two bugs happen to cross paths they just continue crawling through each other uninterrupted. Suppose that five of the possible six meetings actually happen. Prove that the sixth does as well. Solution: As the title of this chapter suggests, lifting the puzzle into space provides an elegant solution. How? By means of a time axis. Suppose every pair of bugs meets except bug 3 with bug 4. Construct a time axis perpendicular to the plane of the bugs, and let gi be the graph (this time, in the sense of graphing a function) of the ith bug in space. Since each bug travels at constant speed, each such graph is a straight line; its projection onto the plane of the bugs is the line that bug travels on. If (and only if) two bugs meet, their graphs intersect. 279 The lines g1, g2 and g3 are coplanar since all three pairs intersect, and the same applies to g1, g2 and g4. Hence all four graphs are coplanar. Now g3 and g4 are certainly not parallel, since their projections onto the original plane intersect, thus they intersect on the new common plane. So bugs 3 and 4 meet as well. Circular Shadows II Show that if the projections of a solid body onto two planes are perfect disks, then the two projections have the same radius. Solution: The conclusion seems eminently reasonable, yet it’s not completely obvious how to prove it. An easy way to make your intuition rigorous is to select a plane which is simultaneously perpendicular to the two projection planes, and move parallel copies of it toward the body from each side. They hit the body at the opposite edges of each projection, so that the distance between the parallel planes at that moment is the common diameter of the two projected circles. ♡ Box in a Box Suppose the cost of shipping a rectangular box is given by the sum of its length, width, and height. Might it be possible to save money by fitting your box into a cheaper box? Solution: The issue here is that if packed diagonally, the inside box could have some dimensions that exceed the greatest dimension of the outside box. For example, a narrow needle-like box of length almost √ 3 could be packed inside a unit cube. This particular example wouldn’t provide a way to cheat, because in this case the total cost of the outside cube would be 3, much more than the 1.7 or so you 280 might have to pay for the inside box. But how do we know that some better example won’t turn up? Here’s a delightful argument that lets ε go to infinity(!) in order to show that you can’t cheat the system. Suppose your box is a×b×c, and let R stand for the region of space occupied by your box including its interior. Suppose R can be packed into an a′ × b′ × c′ box R′ with a + b + c > a′ + b′ + c′. Let ε > 0 and consider the region Rε consisting of all points in space that are within distance ε of R. This Rε region will be a rounded convex shape containing your box R. The volume of Rε is equal to your box’s volume plus 2ε(ab + bc + ac) for the volume added to the sides, plus πε2(a + b + c) for the volume added to the edges, plus 4 3πε3 for the volume (consisting of the eight octants of a ball) added to the corners. If you do the same to the outside box R′ you get a region R′ ε which of course must contain Rε. But that’s impossible! In the expression for vol(R′ ε) −vol(Rε), the ε3 terms cancel, so if you take ε to be large, the dominant term is the ε2 term, namely πε2((a′ + b′ + c′) −(a + b + c)), which is negative. ♡ Angles in Space Prove that among any set of more than 2n points in Rn, there are three that determine an obtuse angle. Solution: It makes sense that the 2n corners of a hypercube represent the most points you can have in n-space without an obtuse angle. But how to prove it? Let x1, . . . , , xk be distinct points (vectors) in Rn, and let P be their convex closure. We may assume P has volume 1 by reducing the dimension of the space to the dimension of P, then scaling appropriately; we may also assume x1 is the origin (i.e., the 0 vector). If there are no obtuse angles among the points, then we claim that for each i > 1, the interior of the translate P + xi is disjoint from the interior of P; this is because the plane through xi perpendicular to the vector xi separates the two polytopes. Furthermore, the interiors of P + xi and P + xj, for i ̸= j, are disjoint as well; this time via the separating plane running through xi + xj perpendicular to the vector xj −xi. We conclude that the volume of the union of the P + xi, for 1 ≤i ≤k, is k. However, all these polytopes lie inside the doubled polytope 2P = P + P, whose volume is 2n. Hence, k ≤2n as claimed! Curve and Three Shadows Is there a simple closed curve in 3-space, all three of whose projections onto the coordinate planes are trees? This means that the shadows of the curve, from the three coordinate directions, may not contain any loops. 281 Solution: No one knew the answer to this question until such a curve was actually found, by John Terrell Rickard of Advanced Telecommunication Modules Ltd. in Cam-bridge, UK. His beautiful, symmetrical solution is shown below with its three shadows. I have no idea how Rickard found it. Later, Donald Knuth, author of the multi-volume classic The Art of Com-puter Programming, programmed a computer to see if there were any other solutions that, like Rickard’s, could be carved from a 2 × 2 × 2 region of a cubi-cal grid. In the time it took to push his enter button (according to Knuth), the computer found that there was just one other solution, shown below. Knuth’s solution, in contrast to Rickard’s, has no symmetry at all. If you found either one of these curves, more power to you! Our theorem is perhaps the world’s best-known example of proving a theo-rem about plane geometry by moving into 3-dimensional space. But there’s an ironic twist. 282 Given any two circles in the plane of different radii, neither contained in the other, we can construct two lines that are tangent to the circles such that for each line, the circles are on the same side of the line. (See picture below). These lines will intersect at some point on the plane, known as the Monge point of the two circles. Theorem. Suppose three circles are given, all of different radii, none contained in another. Then the three Monge points determined by the three pairs of circles all lie on a line. This is what the picture looks like (we have taken the circles to be disjoint, but in fact they’re allowed to overlap). The theorem is attributed to Gaspard Monge, 1746–1818, a distinguished French mathematician and engineer. On Wikipedia (as I write this) you will find a proof which is elegant, famous, and wrong! Here’s how it goes: Replace each circle by a sphere whose equator is that circle, so now you have three spheres (of different diameters) intersecting the original plane at their equators. Pick two of them: Instead of two tangent lines you now have a whole tangent cone whose apex is the Monge point of the circles. Now, take a plane tangent to the three spheres, thus also tangent to the three cones. The three Monge points will all lie on that plane, as well as on the original plane. But the two planes intersect in a line, so the Monge points are all on that line, QED! It’s not easy to spot the hole in this proof. The difficulty is that there may not be a plane tangent to the three spheres; for example, if two of the circles are large and the third is between them and smaller. The following proof, attributed by cut-the-knot.org to Nathan Bowler of Trinity College, Cambridge, works by erecting cones instead of spheres on top of the circles. Call them C1, C2 and C3, and let them all be “right” cones—that is, they support 90◦angles at their apices. (Actually, we only need all the cones to have the same angle.) Each pair of cones determines two (outside) tangent planes, say P1 and Q1 (for cones C2 and C3), P2 and Q2 (for cones C1 and C3), and finally P3 and Q3 (for cones C1 and C2). 283 Each pair of planes Pi, Qi intersects in a line Li which passes through the apex of both tangent cones, as well as through the point where the corresponding circle tangents meet. Thus, in particular, L1 and L2 meet at the apex of C3, L1 and L3 at the apex of C2, and L2 and L3 at the apex of C1. Hence the three lines of intersection are coplanar (all lie on the plane determined by the three apices); the intersection of that plane with the original plane of the circles is a line through the three foci, and this time we are really done. ♡ There are other ways to prove Monge’s theorem and as you might suspect, one of them is to transform the plane in a circle-preserving way so as to escape annoying initial configurations that screw up the spheres proof. But it’s curious, and a bit sobering, that the unfixed sphere argument lives on as the “standard” proof of Monge’s theorem. 284 Chapter 17 Nimbers and the Hamming Code Probably the most famous equation in mathematics is not E = mc2 or eiπ = −1 but 1 + 1 = 2. But if, like many computer scientists, you are fond of binary arithmetic, you might prefer 1 + 1 = 10; if you are an algebraist or logician who works with numbers modulo 2, you might prefer 1 + 1 = 0. It turns out to be surprisingly useful, in certain circumstances, to work in a world where any number added to itself is zero. Such is the world of nimbers. Nimbers are a lot like numbers, and in fact you can think of them as non-negative integers, written in binary. But their rules for addition and subtraction are different from the usual ones. In particular, you add nimbers without car-rying. Here’s another way to say the same thing: To compute the ith bit (from the right) of the sum of a bunch of nimbers, you only need to know the ith bit of each nimber. If an odd number of these are 1’s, the ith bit of the sum will be a 1; otherwise it will be a 0. Let’s try this. To avoid confusion, we’ll write the decimal expression of a nimber with an overline, so, for example, 7 = 111 and 10 = 1010. To distinguish nimber addition from ordinary addition of binary numbers, we’ll use the symbol ⊕instead of +. Thus, 7 ⊕10 = 111 ⊕1010 = 1101 = 13. Of course, 1 ⊕1 = 1 ⊕1 = 0 = 0 and in fact, as we said, n ⊕n = 0 for any n. Nimber arithmetic, once you get used to it, is really rather nice. It obeys the usual laws of arithmetic—it’s commutative, associative, and satisfies x ⊕0 = x for any x. It has additive inverses, as we saw: Any nimber is the additive inverse of itself! Thus, we have no use for minus signs in nimber arithmetic, and subtraction is the same as addition. If we have two nimbers m and n and we want to know which nimber to add to m to get n, why, it’s the nimber m⊕n, since m ⊕(m ⊕n) = (m ⊕m) ⊕n = 0 ⊕n = n. If you’ve studied linear algebra, you may recognize nimbers as vectors over 285 a two-element field. When you add vectors in any vector space, the coordinates are added independently; that’s what is happening with our nimbers, since we don’t carry. One difference is that for nimbers we don’t bother to specify the dimension (number of coordinates) in advance. We can always add more zeroes to the left of a binary number if we need more coordinates (bits), without changing the number’s value or its nimber properties. Whence the name? As we shall see, nimbers are numbers that arise in the game of Nim, so the term “nimber” was irresistible to the himself irresistible mathematician John H. Conway. Understandably, “nimber” has largely replaced an old term, “Grundy number.” Consider: Life Is a Bowl of Cherries In front of you and your friend Amit are 4 bowls of cherries, containing, respec-tively, 5, 6, 7 and 8 cherries. You and Amit will alternately pick a bowl and take one or more cherries from it. If you go first, and you want to be sure to get the last cherry, how many cherries should you take—and from which bowl? Solution: In the traditional game of Nim the bowls of cherries are replaced by piles of stones or (my preference) stacks of chips, and the problem is how best to play from any starting position. We denote positions in Nim by listing the stack sizes from largest to smallest; for example, in the cherries puzzle, the given position is 8\|7\|6\|5. The rules of Nim specify that from a given position you can reach any position with the same numbers except that one number has been either lowered or (if all the chips from that stack were removed) deleted. If, for example, you remove two cherries from the bowl of seven, you reach the new position 8\|6\|5\|5. The goal is to be the player to achieve the empty position. This can be done immediately if there is only one stack; just take all of it. If there are just two stacks, and they are of different sizes, you (as first player) have the following winning strategy: Take chips from the big stack to reduce it to exactly the size of the smaller one, and repeat. 286 If there are more than two stacks, things start to get complicated. Nimbers make it simple—they enable you to win whenever you should, and often (in practice) when you shouldn’t! Here’s how it works. We define the nimsum of a position to be the nimber-sum of the stack sizes. For example, the nimsum of the initial position in the cherries puzzle is 8 ⊕7 ⊕6 ⊕5 = 1000 ⊕111 ⊕110 ⊕101 = 1100 = 12 . The fact is that you (as first player) can force a win from any position whose nimsum is not zero. How do you do it? Easy: Change it to a position whose nimsum is zero. Your opponent will have to change the position to one whose nimsum is again non-zero, and then you repeat. Eventually you get to the empty position, whose nimsum is of course zero, and you win! There are some things to verify here. First of all, how do you know that your opponent can’t change a zero-nimsum position to another zero-nimsum position? Well, your opponent (like you) must remove one or more chips from just one stack, so he reduces one of the stack-sizes, say si, to some smaller number s′ i, by subtracting some positive number k. That will have the effect of nimber-adding si ⊕s′ i ̸= 0 to the previous position’s nimsum. So the nimsum of the new position will be 0 ⊕k = k ̸= 0. A bit more subtle is showing that from any position whose nimsum is not zero, you can get to one whose nimsum is zero. Here’s how you can do it. Suppose the current nimsum is s, and suppose the leftmost 1 in the binary representation of s is in the ith bit from the right (that is, the bit representing 2i−1). Then there must be at least one stack whose size (sj, say) also has a 1 as its ith bit—otherwise, the ith bit of s would have been zero. We claim we can remove some chips from that stack to change its size to sj ⊕s. Doing so will change the position’s nimsum to s ⊕s = 0. The only thing we need to check is that sj ⊕s is a smaller number than sj (since we’re not allowed to add chips to a stack). But that is easy because nimber-adding s to sj will change the ith bit from 1 to 0, and won’t affect any bits of sj that are to the left of the ith. It follows that s ⊕sj is smaller than sj, so reducing the sj-stack to size s ⊕sj is a legal move. Let’s try that with the cherries. We saw that the nimsum of the given position was 12 = 1100, whose leftmost 1 is in the fourth position from the right. Here there happens to be only one stack whose size (in binary) has a 1 in the fourth position from the right: the stack of size 8. Nimber-adding 1100 to 1000 gives 100 = 4 so we reduce the number of cherries in the bowl of 8 to 4. The resulting position, 7\|6\|5\|4, has nimsum 111 ⊕110 ⊕101 ⊕100 = 0 and we have Amit where we want him. In some positions there are several stacks whose sizes, in binary, have a 1 in the same position as the leftmost 1 in the position’s nimsum; in such cases, there will be one winning move in each of those stacks. For our initial cherries position, there is only one such stack, thus only one winning move. After any other first move, it is Amit who can force a win. 287 Even within the specialized field of combinatorial games (where two players alternate and the first one who is unable to move loses), nimbers play a much bigger role than just their part in Nim. It turns out that any position in an “impartial” combinatorial game—that is, one in which both players have the same move choices from any given position—can be represented by a nimber! This is the content of the celebrated Sprague-Grundy Theorem, about which the interested reader can read in Winning Ways or any other book on combinatorial games. In fact, here’s another impartial game to which we can apply nimbers—but perhaps not in the way you would think. Life Isn’t a Bowl of Cherries? In front of you and your friend Amit are 4 bowls of cherries, containing, respec-tively, 5, 6, 7, and 8 cherries. You and Amit will alternately pick a bowl and take one or more cherries from it. If you go first, and you want to be sure Amit gets the last cherry, how many cherries should you take—and from which bowl? Solution: So, in effect, in this second puzzle you want your friend to win. (In such a situation you are said to be playing the misere version of the game.) It looks like since your objective is the opposite of the previous puzzle’s, you should make any move but the one that reduces the number of cherries in the bowl of eight to four. That’s fine if Amit cooperates, that is, if he himself wants the last cherry, but it will not do if you want to insist that Amit get the last cherry. Surprisingly, your unique winning first move for the second puzzle is the same as the one for the first! The philosophy behind this counterintuitive solution is that playing for zero-nimber positions allows you to control the game, until the crucial moment when you decide whom you want to get the last cherry. That moment comes when you first encounter a position containing exactly one bowl with multiple cherries, that is, in general Nim terms, a position in which every remaining stack except one contains only a single chip. Note that such a position never has a zero nimber, because the big stack’s size, in binary, has at least one 1 in it that can’t be cancelled. Thus, when such a position arises, as it inevitably must, it will be you facing it, not your opponent. Now you need consider only two moves: Removing all of the big stack, and removing all but one chip from it. Either way there will only be “singleton” stacks left for your opponent. If there are an even number of them, you will get the last chip; if there are an odd number of them, he will. Since one of the above two moves will leave an even number of stacks and the other an odd number, you control who gets the last chip. Let’s do a game of cherries to see an example of how this works. In the following pictures, we zigzag through time with your move (hollow cherries are 288 those to be removed) on the left and Amit’s on the right. The nimsum for each position (prior to the indicated move) and the nimbers for each stack are indicated in binary. At the last position shown, you will take all the cherries in the 4-cherry bowl if you want to end up with the last cherry; if you want Amit to get it, you’ll take just three of them. Whim-Nim You and a friend, bored with Nim and Nim Mis ere, decide to play a variation in which at any point, either player may declare “Nim” or “Misere” instead of removing chips. This happens at most once in a game, and then of course the game proceeds normally according to that variation of Nim. (Taking the whole single remaining stack in an undeclared game loses, as your opponent can then declare as he wishes for his last move.) What’s the correct strategy for this game, which its inventor, the late John Horton Conway, called “Whim”? 289 Solution: Owing to our previous analyses, the game is effectively over once someone de-clares. So we concentrate on how to play the game pre-declaration. It’s easy to see that if you’re in a 0-position you win by simply declaring “Nim.” Thus, you must not create a 0-position for your opponent. Moreover, you don’t want to be handing your opponent nothing but singleton stacks, because then he will declare appropriately (“Nim” if the number of singletons is even, else “Mis ere”) and win. On the other hand, you’d love to give your opponent a single stack of size 2. For that matter, you can give him any odd number of stacks of size 2, together with any even number of stacks of size 1 or 3. Then eventually he’ll be the one to reduce to all singletons, and you will declare and win. John Conway had a nice way to look at this: Imagine an additional stack, the “whim stack,” of size 2; then play the normal Nim strategy of handing your opponent 0-positions. If the only way for you to achieve a 0-position is to remove the whim stack, you declare appropriately instead. The only trouble is: What if achieving a 0-position requires removing just 1 chip from the whim stack? That won’t happen if there’s a stack of size 2 or another stack of size 3, and you’re all set if all the stacks are singletons, so you can be in trouble only if there’s one or more “tall” stack of size at least 4. Conway’s solution to that problem is to imagine that the whim stack has size only 1, when there’s a tall stack in play. We need to check that the transition causes no problem, but that turns out to be easy. Your opponent will never be the one to kill the only tall stack because, being in a 0-position with the whim stack (thus a 1-position without), he can’t be looking at just one tall stack. If it is you who are facing one tall stack, and leaving your opponent with a 0-position requires reducing that stack to size 0, 1, 2, or 3, then you can instead reduce it to 3, 2, 1, or 0 respectively and have a 0-position with whim stack of size 2. In summary: If it is your move, and there’s no tall stack, you can win as long as the position’s Nim value, counting an imaginary stack of size 2, is not 0; if there is a tall stack, then the Nim value, counting an imaginary stack of size only 1, must not be 0. ♡ Let’s put nimbers to work on hats. Option Hats One hundred prisoners are told that at midnight, in the dark, each will be fitted with a red or black hat according to a fair coin-flip. The prisoners will be arranged in a circle and the lights turned on, enabling each prisoner to see every other prisoner’s hat color. Once the lights are on, the prisoners will have no opportunity to signal to one another or to communicate in any way. Each prisoner will then be taken aside and given the option of trying to guess whether his own hat is red or black, but he may choose to pass. The prisoners will all be freed if (1) at least one prisoner chooses to guess his hat color, and (2) every prisoner who chooses to guess guesses correctly. 290 As usual, the prisoners have a chance to devise a strategy before the game begins. Can they achieve a winning probability greater than 50%? Solution: The prisoners have virtually no chance of winning if they all guess, so it looks reasonable to pick one prisoner (say, Joe) to guess and have everyone else decline; then, at least, the prisoners are all freed if Joe gets it right. In fact, it almost looks like there’s a proof that they can’t do better: Every individual guess has only a 50% probability of being right, so what’s the advantage of ever having more than one person guess? Suppose we fix some strategy S that tells each prisoner whether to guess black (“B”), guess red (“R”), or decline to guess (“D”), depending on the 99 hats he sees. Imagine that we write down a huge matrix indicating, for each of the 2100 possible hat configurations, what each prisoner does according to S. If each guess (B or R) is boldfaced if it is correct, then the matrix will have exactly the same number of boldfaced guesses as lightfaced; because, if prisoner i’s guess is (say) a boldface R in a particular configuration, it will be a lightface R for the configuration that is exactly the same except that prisoner i’s hat is changed to black. In other words, no matter what the strategy, over all possible configurations half the guesses are correct. That suggests an idea: What if we crowd as many wrong guesses as possible into a few configurations, and try to put just one correct guess into each remaining configuration? Let’s do a little bit of arithmetic. Fixing a strategy S, let w be the number of winning configurations (that is, those in which all prisoners who guess are correct). Let x be the average number of (correct) guesses in winning config-urations, and y the average number of wrong guesses in losing configurations. Then wx ≤(2100−w)y with equality only when all guesses in the losing config-urations are wrong. To maximize w, we would like to have every prisoner guess and all be wrong, in losing configurations; in winning configurations, to have just one prisoner guess and be correct. If we could achieve that for n prisoners, we would get a winning probability of n/(n+1). We call this number the count bound. Achieving the count bound would be very good news indeed for our 100 prisoners, as they would all be freed with probability 100/101—better than 99%. But let’s not get ahead of ourselves. For one thing, we could never get 100/101 probability of winning for our 100 prisoners, because that would require that the number of losing configurations be exactly 1/101 times the total number of configurations, and 2100 is not divisible by 101. But n+1 does divide evenly into 2n for n = 3; let’s see if we can get prob-ability 3/4 of winning for three prisoners. We need two of the 23 = 8 hat configurations to be losers; the logical ones to try are all red, and all black. If we instruct each prisoner to guess “red” then he sees two black hats and “black” when he sees two red, that will have the desired effect: They’ll all guess wrong in the losing configurations. Let’s see: If we ask them to decline to guess other-291 wise, we get just what we want: In any configuration with both colors present, the prisoner with the odd hat will guess correctly while the other two decline! Here’s the matrix for this strategy: Notice that we have numbered the prisoners in binary—actually, we have nimbered them. If we define the nimsum of a configuration to be the nimber-sum of the nimbers of the red-hatted prisoners, we find that the two losing configurations, RRR and BBB, are the ones with nimsum 0. Every other con-figuration has a non-zero nimsum which corresponds to some prisoner—namely, the one who guesses (and is right) when that configuration occurs. Thought of this way, the above strategy generalizes to any number of pris-oners that is one less than a power of two—that is, whenever n = 2k −1 for some positive integer k. The prisoners are nimbered from 1 to n, that is, from 000 . . . 001 to 111 . . . 111. The losing configurations are those for which the nim-sum of the red-hatted prisoners is 0. The strategy is as follows: Each prisoner assumes that the configuration is not a losing one, and if that assumption tells him what his own hat color is, he guesses accordingly. Otherwise he declines. What that amounts to is this. Prisoner i computes the nimber-sum mi of the nimbers of the red-hatted prisoners he sees; we call mi prisoner i’s “personal” nimsum. If mi = 0, prisoner i reasons that his hat must be red (else the nimsum of the whole configuration is 0, which every prisoner is assuming is not the case). If mi = i, prisoner i’s own nimber, then his hat must be black (else the nimsum of the whole configuration would be i ⊕i = 0). If mi is anything other than 0 or i, prisoner i knows the configuration is not a losing one. That is great news, but it doesn’t tell prisoner i what his hat color is, so he declines to guess. This works because: (1) If the configuration has nimsum 0, every prisoner will guess, and they will all be wrong; (2) If the configuration’s nimsum is s ̸= 0, then prisoner s will guess correctly while every other prisoner declines. Thus 292 the prisoners will achieve exactly the count bound for their success probability, namely n/(n+1) = (2k −1)/2k. We have exploited the following property of the set L of configurations with zero nimsum: Every configuration not in L can be changed to one in L by flipping the color of someone’s hat. (Here, there is just one such hat: the one belonging to the prisoner whose nimber is the configuration’s nimsum.) Inter-preted as a set of binary strings of length n (red=1, black=0), L is said to be an error-correcting code, and the above bit-changing property makes this par-ticular code (called the Hamming code of length n) a “perfect 1-error correcting code.” The Hamming code and other error-correcting codes are objects of enor-mous practical as well as theoretical interest—we’ll identify some more serious applications later. What can the prisoners do if their number n is not one less than a power of two? Well, they can find the biggest k such that 2k −1 ≤n, and have prisoners 1 through 2k −1 run the above protocol among themselves, while the others decline to guess no matter what they see. This will not generally be optimal but it won’t be far off; the probability of failure will always be less than 2/(n+1), thus small (for large n) and never as much as twice the failure probability associated with the count bound. Our group of 100 prisoners can have 63 prisoners run the Hamming code protocol among themselves while the remaining 37 stand by, declining to guess. That will give the prisoners a pretty impressive 63/64 = 98.4375% probability of winning. To get the precise optimal winning probability for values of n that aren’t one less than a power of 2, the prisoners would need to find the smallest set L of configurations with the property that any configuration is at most one color-flip away from a configuration in L. We know from the count bound that such a set, sometimes called a “radius-1 covering code,” must be of size at least 2n/(n+1) rounded up to an integer. Let’s see this reasoning in action for a couple of small numbers. For n = 4, the count bound says that we need \|L\| ≥4, and we already get winning probability 1 −4/16 = 3/4 by having three prisoners employ the Hamming code, so that solution is optimal. For n = 5, though, the count bound allows a radius-1 covering code of size ⌈32/6⌉= 6. In fact, 6 is impossible, but there’s a radius-1 covering code of size 7: {00000, 00111, 01111, 10111, 11001, 11010, 11100}. With this the prisoners can get winning probability 1 −7/64 = 57/64, beating the Hamming code. Notice here that if, for example, the actual configuration is 11111 (all hats red), two prisoners—numbers 1 and 2, counting from the left— will guess (correctly) that their hats are red, the first to avoid the codeword 01111, the second to avoid 10111. It turns out that for large n, radius-1 covering codes that are (relatively) close in size to 2n/(n+1) can always be found. They are much harder to describe than Hamming codes, but if the number of prisoners is large and awkward (e.g., equal to 2k −2 for some k), hard work will enable the prisoners to cut down their losing probability by a factor of almost 2. 293 Chessboard Guess Troilus is engaged to marry Cressida but threatened with deportation, and Immigration is questioning the legitimacy of the proposed marriage. To test their connection, Troilus will be brought into a room containing a chessboard, one of whose squares is designated as special. On every square will be a coin, either heads-up or tails-up. Troilus gets to turn over one coin, after which he will be ejected from the room and Cressida brought in. Cressida, after examining the chessboard, must guess the designated square. If she gets it wrong, Troilus is deported. Can Troilus and Cressida save their marriage? Solution: Troilus must somehow impart 6 bits worth of information (since the chessboard has 64 = 26 squares) with his one turned coin. He does have 64 different things he can do, so this is not a priori impossible, but he and Cressida are going to need to be very efficient. The natural approach—if you know about nimbers—is to assign a nimber of length 6 to each square of the chessboard. When Cressida comes in, she sums the nimbers of the squares whose coin shows heads; that gives a nimber and she now guesses that its associated square is the designated special square. So Troilus needs to do is make sure the sum points to the right square, but that’s easy. He computes the sum of the nimbers of the heads-squares of the board as presented to him, and compares it with the nimber of the designated special square. If their sum (i.e., difference!) is the nimber b, he turns the coin on square b. That will add (i.e., subtract) b from the sum and leave the sum pointing to the special square. Majority Hats One hundred prisoners are told that at midnight, in the dark, each will be fitted with a red or black hat according to a fair coin-flip. The prisoners will be arranged in a circle and the lights turned on, enabling each prisoner to see every other prisoner’s hat color. Once the lights are on, the prisoners will have no opportunity to signal to one another or to communicate in any way. Each prisoner is then taken aside and must try to guess his own hat color. The prisoners will all be freed if a majority (here, at least 51) get it right. The prisoners have a chance to devise a strategy before the game begins. Can they achieve a winning probability greater than 50%? Would you believe 90%? How about 95%? Solution: Too bad that a tie is insufficient to free the prisoners: They could force a tie (see “Half-Right Hats” in Chapter 2) by having 50 of them guess “black” when they see an even number of red hats and “red” otherwise, while the other 50 do 294 the reverse. That guarantees that exactly 50 prisoners will guess correctly. Un-fortunately, the way the problem is phrased, that guarantees that the prisoners lose. But maybe we can use this idea some other way. Before we begin our analysis, though, it’s worth checking one possibility that may already have occurred to you. What if each prisoner simply guesses that his hat color is the same as the majority of those he sees? This works as long as the number of red-hatted prisoners is not exactly 50; in the latter case, the plan comes spectacularly a cropper, with every prisoner guessing incorrectly. But the likelihood of exactly 50 red hats is not very high: There are 100 choose 50 ways this can happen, out of 2100 ways to pick the hat colors, so the probability of ex-actly 50 red hats is 100 50 /2100 = 0.07958923738... . Thus, this “preponderance” strategy frees the prisoners with probability better than 92%; not bad! The fact that when it goes wrong, the preponderance scheme produces 100 wrong guess, is a feature, not a bug. What we have is a problem very similar to the one whose solution recalls the so-called St. Petersburg paradox. You are in Las Vegas, and determined to go home with more money than you came with; how can you maximize your probability of doing so, even though your expectation from gambling is always negative? The answer was discussed in Chapter 14 in connection with the puzzle Roulette for the Unwary. Set aside 2k −1 dollars that you can afford to lose, with k as large an integer as possible; then make successive even-money (and as fair as possible) bets of sizes $1, $2, $4, etc. until you win a bet. At that point you quit, $1 ahead. You end up probably ahead $1 but possibly broke, and that’s the hallmark of a good scheme: You either just barely accomplish your goal, or miss it by a mile. The same applies to Majority Hats. Since every guess is a 50% proposition, the ideal scheme would have just two possible outcomes: Win barely (meaning, in this case, have exactly 51 correct guesses) or lose spectacularly (every guess wrong). If you could do that for 100 prisoners, their probability p of success would satisfy 51p = 49p+100(1−p), since for any scheme, the expected number of right guesses has to equal the expected number of wrong guesses. That works out to p = 100/102 > 98%, very impressive—if it were possible. In general, with n prisoners, this calculation shows that when n is even their success probability cannot exceed n/(n+2); when n is odd, and the prisoners need only one more right guess than wrong, they can hope for a success proba-bility as high as n/(n+1). We will show that this “count bound” is achievable when n is one or two less than a power of 2. Suppose, for example, that there are 31 = 25 −1 prisoners, and let’s assign them nimbers from 1 to 11111 (the binary representation of 31). We divide the prisoners into five groups according to the position of the leftmost 1 in their nimbers; thus Group 1 consists of just prisoner 1, Group 2 contains prisoners 2 and 3, and in general each group has twice as many prisoners as the previous group. Prisoner 1 guesses “black.” If he’s right—and, remember, everyone else can see if he’s going to be right or not—every other group behaves as if it’s playing Half-Right Hats. The agreement could be, for example: All prisoners with even 295 nimbers guess as if they know the total number of red hats in their group is even, while all those with odd nimbers assume the opposite. Then groups 2 through 5 will each be exactly half right, and the prisoners end up with a 16-to-15 majority correct. If prisoner 1’s hat is red, Group 2 plays the Unanimous Hats strategy—they all guess as if they knew the number of red hats in their group were even. If they’re right (probability 1/2), then each remaining group splits itself as before and again the prisoners win by one. If Group 2 is doomed to failure as well as Group 1—which, again, everyone else can see—then there are three wrong guesses coming, but they are rescued by Group 3 which now attempts to be all correct. This pattern continues and always produces a 16-15 majority unless every one of the five groups contains an odd number of red hats, which happens with probability only 1/32. So the 31 prisoners achieve success probability 1 −1/32 = 96.875%. We know this is unbeatable because it achieves the ideal: The prisoners either enjoy a 1-guess victory, or everyone is wrong. We can achieve this ideal (though probability of success will be only 1−1/16) with 30 prisoners as well; here there’s no group of size 1, and the group of size two plays the Unanimous Hats strategy to try to achieve the 2-guess advantage. What about 100 prisoners? The count bound is never exactly achievable unless n+1 or n+2 is a power of 2, because there are 2n hat configurations and thus any protocol’s probability of winning must be a multiple of 1/2n. One hundred is not a good number, but we could let prisoners 1 through 62 use our “St. Petersburg protocol,” while the remaining 38 prisoners just split themselves using the Half-Right Hats strategy, so as not to get in the way of the others. That frees the prisoners with probability 1−1/32 = 96.875%, pretty good. But it seems as if this strategy wastes 38 prisoners. Let’s go back to small numbers again. The smallest number that’s not one or two less than a power of 2 is 4. Can 4 prisoners win with probability better than 1/2? Yes: Have prisoner 1 guess black. If he’s wearing a black hat, the other three prisoners can run their St. Petersburg protocol to get 2-out-of-3 right with probability 3/4. If prisoner 1 has a red hat, the other three play the Unanimous Hats strategy and thus all three are right with probability 1/2. Altogether the 4 prisoners achieve their 2-guess majority with probability 1 2 · 3 4 + 1 2 · 1 2 = 5/8. With this observation we can split our 100 prisoners into groups of sizes 4, 6, 12, 24, and 48, with just 6 left over. The group of 4 plays as above, succeeding with probability 5/8. As usual, if they are going to succeed (which everyone else can see), all other prisoners play the Half-Right Hats strategy. If they are going to fail, the next group (size 6) plays Unanimous Hats, and if they fail the group of 12 plays Unanimous Hats, etc. The final group of 6 always plays Half-Right Hats. Result? The prisoners win unless the first group of 4 fails and the next four groups all suffer an odd number of red hats. This strategy thus frees the prisoners with probability 1 −3 8 · 1 2 · 1 2 · 1 2 · 1 2 = 1 −3/128 = 97.65625%. Can we tweak this any further? Yes, by recursively computing the best strategies of this kind for achieving k correct guesses out of n, for all n and k up 296 to 100. This “dynamic programming” approach can reach a success probabil-ity of 1129480068741774213/1152921504606846976, about 97.96677954%; very close indeed to the count bound of 100/102 = 98.039215686%. That last solution didn’t end up making much use of nimbers, but the next puzzle brings us back to the Hamming code. Fifteen Bits and a Spy A spy’s only chance to communicate with her control lies in the daily broadcast of 15 zeros and ones by a local radio station. She does not know how the bits are chosen, but each day she gets an advance copy of the bits and can corrupt the radio transmission by altering one of the bits, that is, changing it from a 0 to a 1 or vice-versa. How much daily information can she communicate? Solution: Since there are 16 things the spy can do (change any bit or none), she can in principle communicate as much as four bits of information each day to her control. But how? The answer is easy once you have nimbers in your arsenal. The spy and her control assign the 4-bit nimber corresponding to the number k to the kth bit of the broadcast, and their “message” is defined to be the sum of the nimbers of the ones in the broadcast—the “nimsum” of the broadcast. The claim is that the spy can send any of the 16 possible nimsums at will, thus achieving a full four bits of communication. Suppose she wishes to send the nimsum n but the sum of the nimbers corresponding to the ones in the station’s intended broadcast is m ̸= n. Then she flips the n ⊕mth bit. The resulting nimsum is m ⊕(n ⊕m) = n. Of course, of the broadcast’s nimsum already happens to be n, she leaves it alone. ♡ We will finish this chapter with a brief introduction to the raison d’etre of error-correcting codes: sending messages over an imperfect channel. Almost every means of communication suffers from some probability of error, whether it be voice, telegraph, radio, TV, typed letters, hand-written words, photographs, or bit streams over the internet. It’s pretty amazing that you can 297 send a billion bits to your friend in an email attachment, and not a single one is accidentally flipped! This comes about through the use of error-correcting codes. If bits had to be sent so slowly that every one was guaranteed to arrive intact with probability 1−1/109, we could forget about streaming movies. No, the individual electical, radio and light pulses that zip around the internet are imperfect and are mis-sent or mis-read with alarming frequency. What saves us is that information is coded in such a way that errors can be detected and corrected. The simplest error-detection method is one you may already know about, and is typical for, say, credit card numbers. By choosing the last “check” digit correctly, your credit card company can ensure that the sum of the digits in your credit card number is (say) a multiple of 10. Then, if you mis-enter one digit of your credit card during a purchase, the digit-sum will be thrown off, and a warning will appear. Error-correction is trickier but you’ve seen it above in the person of the Hamming code. Simplest case: You want to send a string of bits and be confident that they all arrive intact, despite some small probability that any given bit will be corrupted. Send each bit three times! Only in the unlikely event that two bits in the same group of three are corrupted will your message get unfixably mangled. In fact, this is the Hamming code of length 3 that we used when three prisoners were faced with either of our hat problems. The codewords were 000 (all hats black) and 111 (all hats red). Of course, tripling the lengths of our messages is rather a heavy price to pay. If we are confident that the likelihood of more than one error in a given group of 2k −1 bits is small, we can use the Hamming code of length n = 2k −1, as follows. Suppose the message to be sent is (or could be represented by) a long string of N bits. Break up the message into “blocks” of size m = n −k. Before transmission, each block is replaced by a codeword of length n; at the receiving end, these are “decoded,” and the original blocks recovered correctly, provided the codeword was received with at most one flipped bit. We can state this as a theorem: Theorem. At the cost of adding k bits to the transmission, a block of bits of length n−k, where n = 2k −1, can be sent in such a way that as long as at most one bit is corrupted, the original block can be faithfully recovered at the receiving end. Proof. We will do a little more than is necessary for the proof, in order to show that all this can be done with very little computational effort by transmitter and receiver. First, we nimber the bits in the n-bit codeword, using k-bit non-zero nim-bers from 000 . . . 001 to 111 . . . 111. There are many ways to do this, but it’s especially convenient to assign nimbers 20 = 000 . . . 001, 21 = 000 . . . 010, 22 = 000 . . . 100, . . . , 2k−1 = 100 . . . 000 to the rightmost k positions, reading from right to left. 298 The same nimbering is used for a block of length n−k, regarded as the leftmost n−k bits of a codeword. For example, if k = 3, the bits of our 7-bit codewords, and the corresponding 4-bit blocks, could be nimbered as in the following figure: The nimsum of a block or codeword is the nimber sum of the positions in which a “1” is found. The nimsum of the codeword in the figure is 0 = 000; indeed, we define our codewords to be exactly the n-bit strings with zero nimsum. The nimsum of the block in the figure happens to be 3 = 011. The easiest way to transform a block into a codeword is simply to append k “check bits” to the right-hand end of the block, in such a way that the nim-sum of the resulting n-bit string is zero. That is particularly easy to do with the nimbering we recommend, because the check bits will then be exactly the nimsum of the block! The reason for this pleasant coincidence is that by choosing the “basis” nimbering of the check bits, we ensure that the nimsum of the check bits is the string of check bits itself. Thus if the nimsum of the block is s, when we add the check bits the new nimsum will be s ⊕s = 0. In this way the “encoding” part of our Hamming code is made as easy as possible; what about the “decoding” part? No problem. As receiver of an n-bit string, we compute its nimsum; if we get 0, we assume it was transmitted correctly and accept its leftmost n−k bits as the intended message block. If we get nimsum s ̸= 0, we assume that one, and only one, error has been made. If so that error must be in the bit nimbered s; so we flip that bit, and only now take the leftmost n−k bits as the intended message block. (The transmission error could have occurred in a check bit, in which case the extracted message will not change.) We have seen that the property we needed from codewords in the hat puzzles—that each non-codeword is one bit-flip away from a unique codeword— is used to correct transmission errors. The Hamming code is one of many codes, some of which detect or correct many errors, and have lots of other neat prop-erties. (Example: Reed-Muller codes, Reed-Solomon codes, and Golay codes were used to transmit photos of Mars and Jupiter from various Mariner and Voyager spacecraft.) Error-correcting codes have played a major role in the the-ory of computing, as well, leading (among other things) to miraculous strings 299 called “probabilistically checkable proofs” and thence to huge advances in our understanding of how well computers can do approximate calculations. And error-correcting codes will need to play a major role if quantum computing is to become practical. 300 Chapter 18 Unlimited Potentials Puzzles often involve processes—systems that evolve in time randomly, deter-ministically, or under your control. Popular questions include: Can the system reach a certain state? Does it always reach a certain state? If so, how long does it take? The key to such a puzzle may be to identify some parameter of the current state of the system that measures progress toward your goal. We call such a parameter a potential. For example, suppose you can show that the potential never increases, but starts below the potential of the desired end state. Then you have proved that the end state is unreachable. On the other hand, perhaps the end state has potential zero, the beginning state has positive potential p > 0, and every step of the process reduces the potential by at least some fixed amount ε > 0. Then, if the potential cannot go below zero, you must reach the end state, and moreover must do so in time at most p/ε. If the loss per step in potential is not bounded, it could be that the potential never reaches zero and may not even approach zero. But if the system has only finitely many states, you don’t need to worry about this possibility. Can you find an appropriate potential for the following puzzle? Signs in an Array Suppose that you are given an m×n array of real numbers and permitted, at any time, to flip the signs of all the numbers in any row or column. Can you always arrange matters so that all the row sums and column sums are non-negative? Solution: We note first that there are only finitely many states here: Each number in the array has at most two states (positive or negative) so there can’t be more than 2mn possible states of the array. 301 Actually, there’s a better bound: If a line (a row or column) is flipped twice, it’s as if it had never been flipped. Thus the state of each line is either “flipped an odd number of times” or “flipped an even number of times,” and it follows that the number of states of the array can’t exceed 2m+n. Back to the problem at hand. The obvious algorithm is to find a line with negative sum and flip it; does this eventually work? After all, flipping a row (say) with negative sum could result in changing several column sums from positive to negative. Thus, the number of lines with negative sum would not work as a potential. Instead, let’s take the sum s of all the entries in the array! That number goes up by 2c if you flip a line whose sum was −c. Since there are only finitely many possible states, the potential cannot keep going up, and you must reach a point where no line sum is negative. Fair warning: From here on, it’s going to be trickier to spot the right poten-tial. Righting the Pancakes An underchef of the great and persnickety Chef Bouillon has made a stack of pancakes, but, alas, some of them are upside-down—that is, according to the great Chef, they don’t have their best side up. The underchef wants to fix the problem as follows: He finds a contiguous substack of pancakes (of size at least one) with the property that both the top pancake and the bottom pancake of the substack are upside-down. Then he removes the substack, flips it as a block and slips it back into the big stack in the same place. Prove that this procedure, no matter which upside-down pancakes are cho-sen, will eventually result in all the pancakes being right side up. Solution: The idea is that even though a flip might result in turning more pancakes upside-down, it rights some pancake that’s relatively high (and another that’s relatively 302 low) in the stack. Let’s penalize the underchef 2k points for an upside-down pancake k pancakes from the bottom, and keep track of the total of the penalties at any given time. Every flip decreases the total penalty, because the penalty of the target pancake exceeds the sum of all possible penalties below it. Thus the penalty total must reach 0, at which time the task is finished. An equivalent way of thinking about this potential function is to code a stack of pancakes by a binary number whose left-most digit is 1 if the top pancake is upside-down, and 0 otherwise. The next-to-top pancake determines the next digit, and so forth. Flipping always decreases the code until the code reaches 0. Breaking a Chocolate Bar You have a rectangular chocolate bar scored in a 6 × 4 grid of squares, and you wish to break up the bar into its constituent squares. At each step, you may pick up one piece and break it along any of its marked vertical or horizontal lines. For example, you can break three times to form four rows of six squares each, then break each row five times into its constituent squares, accomplishing the desired task in 3 + 4 × 5 = 23 breaks. Can you do better? Solution: This delightful problem has embarrassed some brilliant professional mathemati-cians, as well as many amateurs, all looking for geometrical insight. But geometry is a red herring: All you need for a potential function is the number of pieces! You start with one piece, increase the number of pieces by one every time you break, and end with 24 pieces. So there are 23 breaks no matter how you do it. Red Points and Blue Given n red points and n blue points on the plane, no three on a line, prove that there is a matching between them so that line segments from each red point to its corresponding blue point do not cross. Solution: Suppose you pair the red and blue points up arbitrarily and draw the connecting line segments. If the line segment connecting r1 and b1 intersects the segment connecting r2 and b2, then these segments are the diagonals of a quadrilateral and by the triangle inequality, the sum of the lengths of the non-crossing seg-ments from r1 to b2 and from r2 to b1 is less then sum of the lengths of the diagonals. 303 The trouble is, if you re-pair the points as suggested above, you may have created new intersections between these segments and others. So “number of intersections” is not the right potential; how about “total length of line seg-ments?” That works beautifully! Just take any matching that minimizes the sum of the lengths of its line segments, and by the above argument, it contains no crossings. Bacteria on the Plane Suppose the world begins with a single bacterium at the origin of the infinite plane grid. When it divides its two successors move one vertex north and one vertex east, so that there are now two bacteria, one at (0,1) and one at (1,0). Bacteria continue to divide, each time with one successor moving north and one east, provided both of those points are unoccupied. Show that no matter how long this process continues, there’s always a bac-terium inside the circle of radius 3 about the origin. Solution: It’s worth trying this to see how things get clogged up near the origin, prevent-ing that area from clearing. How can we demonstrate this using a potential function? Since each bacterium divides into two, it’s natural to give each child half the potential of its parent. We can do this by penalizing its potential by a factor of 2 for each additional edge in its shortest path to the origin. This can be done by assigning value 2−x−y to a bacterium at (x,y), so that the total assigned value begins at 1 and never changes. The total potential of all bacteria on the half line y = 0 (i.e., the positive X-axis) cannot exceed 1 + 1/2 + 1/4 + · · · = 2, nor can the potential of the points on the half-line y = 1 exceed 1/2+1/4+1/8+· · · = 1; continuing in this 304 manner, we see that even if the whole north-east quadrant were full of bacteria, the total potential would only add up to 4. But the potentials of bacteria at the nine non-negative integer points inside the circle x2+y2 = 9 already add up to 49/16, more than 3. Thus the potentials of all the bacteria not inside this circle add up to less than 1, and it follows that there must be bacteria left inside the circle no matter how long the process continues. Pegs on the Half-plane Each grid point on the XY plane on or below the X-axis is occupied by a peg. At any time, a peg can be made to jump over a neighbor peg (horizontally, vertically, or diagonally adjacent) and onto the next grid point in line, provided that point was unoccupied. The jumped peg is then removed. Can you get a peg arbitrarily far above the X-axis? Solution: The difficulty is that as pegs rise higher, grid points beneath them are denuded. What is needed is a parameter P which is rewarded by highly placed pegs, but compensatingly punished for holes left behind. A natural choice would be a sum over all pegs of some function of the peg’s position. Since there are infinitely many pegs, we must be careful to ensure that the sum converges. We could, for example, assign value ry to a peg on (0, y), where r is some real number greater than 1, so that the values of the pegs on the lower Y -axis sum to the finite number P0 y=−∞ry = r/(r−1). Values on adjacent columns will have to be reduced, though, to keep the sum over the whole plane finite; if we cut by a factor of r for each step away from the Y -axis, we get a weight of ry−\|x\| for the peg at (x, y), and a total weight of r r−1 + 1 r−1 + 1 r−1 + 1 r(r−1) + 1 r(r−1) + · · · = r2 + r (r−1)2 < ∞ for the initial position. If a jump is executed, then at best (when the jump is diagonally upward and toward the Y -axis), the gain to P is vr4 and the loss v + vr2, where v is the previous value of the jumping peg. As long as r is at most the square root of the “golden ratio” θ = (1 + √ 5)/2 ≈1.618, which satisfies θ2 = θ + 1, this gain can never be positive. If we go ahead and assign r = √ θ, then the initial value of P works out to about 39.0576; but the value of a peg at the point (0, 16) is θ8 ≈46.9788 by itself. Since we cannot increase P, it follows that we cannot get a peg to the point (0, 16). But if we could get a peg to any point on or above the line y = 16, then we could get one to (0, 16) by stopping when some peg reaches a point (x, 16), then redoing the whole algorithm shifted left or right by \|x\|. ♡ 305 Pegs in a Square Suppose we begin with n2 pegs on a plane grid, one peg occupying each vertex of an n-vertex by n-vertex square. Pegs jump only horizontally or vertically, by passing over a neighboring peg and into an unoccupied vertex; the jumped peg is then removed. The goal is to reduce the n2 pegs to only 1. Prove that if n is a multiple of 3, it can’t be done! Solution: Color the points (x, y) of the grid red if neither x nor y is a multiple of 3, otherwise white. This leaves a regular pattern of 2 × 2 red squares (as in the figure). If two pegs are (orthogonally) adjacent on the grid, both on red points or both on white, the peg remaining after the jump will be on white. If one is on red and the other on white, however, the peg remaining after the jump will be on red. It follows that if you start with any configuration having an even number of pegs on red squares, then this property will persist forever regardless of what jumps are made. It is easy to check that a 3 × 3 square of pegs, no matter where it is placed on the plane grid, hits an even number of red points. Since an n × n square with n a multiple of 3 is composed of such squares, it too will always hit an even number of red points. Suppose, however, that it were possible to reduce such a square to a single peg. Then we could shift the original square so that the surviving peg ended up on a red point, and this contradiction concludes the proof. It is routine, but not particularly easy or enlightening, to show that if n is not a multiple of 3, you can reduce an n × n square to a single peg. First-Grade Division On the first day of class Miss Feldman divides her first-grade class into k working groups. On the second day, she picks the working groups a different way, this 306 time ending up with k+1 of them. Show that there are at least two kids who are in smaller groups on the second day than they were on the first day. Solution: Most people find this simple-seeming combinatorial statement frustratingly diffi-cult to prove, unless they hit upon the “magic” potential. Think of each working group as a task force for a job that requires a total of one unit of effort; then, assume that each child in a working group of size s contributes an amount 1/s worth of effort. The sum of these “effort” fractions is evidently k on the first day, k+1 on the second. A given child’s contribution to these numbers cannot increase by as much as 1 from one day to the next (since his contribution is greater than 0 on day 1, and at most 1 on day 2). So at least two kids must be contributing more to the total effort on their second day—meaning that both have moved to smaller groups. Infected Checkerboard An infection spreads among the squares of an n × n checkerboard in the follow-ing manner: If a square has two or more infected neighbors, then it becomes infected itself. (Neighbors are orthogonal only, so each square has at most four neighbors.) For example, suppose that we begin with all n squares on the main diagonal infected. Then the infection will spread to neighboring diagonals and eventually to the whole board. Prove that you cannot infect the whole board if you begin with fewer than n infected squares. Solution: This puzzle was presented to me (by NYU’s Joel Spencer) as having a “one-word solution.” That’s an exaggeration, perhaps, but not a huge one. At first it might seem that to infect the whole board, you need to start with a sick square in every row (and in every column). That would imply the conclusion, but it’s not true. For example, sick squares in alternating positions in the leftmost column and bottom row can infect the whole board. What we really need is a potential function, and the one that works like a charm—and the one word Spencer had in mind—is “perimeter.” The perimeter of the infected region is just the total length of its boundary. We may as well assign each grid edge length 1, so the perimeter could as well be defined as the number of edges that have a sick square on one side and either a well square or nothing on the other side. Here is the key observation: When a square becomes infected (by two or more neighbors), the perimeter cannot increase! Indeed, it is easy to check that 307 when just two neighbors administer the dose, the perimeter remains the same; if the well square had three or four sick neighbors, the perimeter actually goes down. If the whole board gets infected, the final perimeter is 4n; since the perimeter never increased, the initial perimeter must have been at least 4n. But to start with perimeter 4n there must have at least n sick squares. Impressionable Thinkers The citizens of Floptown meet each week to talk about town politics, and in particular whether or not to support the building of a new shopping mall down-town. During the meetings each citizen talks to his friends—of whom there are always an odd number, for some reason—and the next day, changes (if nec-essary) his opinion regarding the mall so as to conform to the opinion of the majority of his friends. Prove that eventually, the opinions held every other week will be the same. Solution: To prove that the opinions eventually either become fixed or cycle every other week, think of each acquaintanceship between citizens as a pair of arrows, one in each direction. Let us say that an arrow is currently “bad” if the opinion of the citizen at its tail differs from the next week’s opinion of the citizen at its head. Consider the arrows pointing out from citizen Clyde at week t−1, during which (say) Clyde is pro-shopping mall. Suppose that m of these are bad. If Clyde is still (or again) pro on week t+1, then the number n of bad arrows pointing toward Clyde at week t will be exactly m. If, however, Clyde is anti-shopping mall on week t+1, n will be strictly less than m since it must have been that the majority of his friends were anti on week t. Therefore a majority of the arrows out of Clyde were bad on week t−1 and now only a minority of the arrows into Clyde on week t are bad. The same observations hold, of course, if Clyde is anti on week t−1. But, here’s the thing: Every arrow is out of someone on week t−1, and into someone on week t. Thus, the total number of bad arrows cannot rise between weeks t−1 and t and, in fact, will go strictly down unless every citizen had the same opinion on week t−1 as on week t+1. But, of course, the total number of bad arrows on a given week cannot go down forever and must eventually reach some number k from which it never descends. At that point, every citizen will either stick with his opinion forever or flop back and forth every week. Frames on a Chessboard You have an ordinary 8 × 8 chessboard with red and black squares. A genie gives you two “magic frames,” one 2 × 2 and one 3 × 3. When you place one of 308 these frames neatly on the chessboard, the 4 or 9 squares they enclose instantly flip their colors. Can you reach all 264 possible color configurations? Solution: There are 249×236 ways to select frame locations, enough in theory to get all 264 color configurations. But of course there may be many ways to achieve a given configuration, and maybe we can find a potential that shows some configurations are unobtainable. In fact, you might recall that the puzzle Odd Light Flips, of Chapter 15, offered a useful criterion: the ability to change the color (or bulb status) of an odd number of items in any set. Is there a troublesome set of chessboard squares? A bit of experimenting might lead you to the following set. Label squares as “special” if they are in row 3 or 6 or file c or f but not both (see figure below), Then every 2 × 2 or 3 × 3 frame covers an even number of special squares. If the board starts with an even number of red special squares (as the usual chessboard does), you cannot reach any configuration where the number of red special squares is odd, and vice-versa. Either way, you cannot reach all color configurations. Bugs on a Polyhedron Associated with each face of a solid convex polyhedron is a bug which crawls along the perimeter of the face, at varying speed, but only in the clockwise direction. Prove that no schedule will permit all the bugs to circumnavigate their faces and return to their initial positions without incurring a collision. 309 Solution: Let us observe first that we may assume no bug begins on a vertex (by advancing or retarding bugs slightly). We may also assume that the bugs move one at a time, crossing a vertex each time. At any time, we may draw an imaginary arrow from the center of each face F, through F’s bug, to the center of the face on the other side of the bug. If we start at any face and follow these arrows, we must eventually hit some face a second time, completing a cycle of arrows on the polyhedron. This cycle divides the surface of the polyhedron into two portions; let us define the “inside” of the cycle to be that portion surrounded clockwise by the cycle. Let P be the number of vertices of the polyhedron inside the cycle. Initially, P could be anything from 0 to all of the polyhedron’s vertices; the extremes occur if there are two bugs on the same edge, causing a cycle of length 2. In the P = 0 case, the two bugs are facing each other, and doomed to collide. When a bug on the cycle moves to its next edge, the arrow through it rotates clockwise. The vertex through which it passed, previously on the inside of the cycle, is now outside; other vertices may also have passed from inside to outside the cycle, but there is no way for a vertex to move inside. To see this, note that the new arrow now points inside the cycle. The chain of arrows emanating from its head has no way to escape the cycle so must hit the tail of some cycle arrow, creating a new cycle with smaller interior. In particular, P has now dropped by at least 1. Since we can never restore P to its starting value, there is nothing to do but hope that the bugs are carrying collision insurance. ♡ 310 Bugs on a Line Each positive integer on the number line is equipped with a green, yellow, or red light. A bug is dropped on “1” and obeys the following rules at all times: If it sees a green light, it turns the light yellow and moves one step to the right; if it sees a yellow light, it turns the light red and moves one step to the right; if it sees a red light, it turns the light green and moves one step to the left. Eventually, the bug will fall off the line to the left, or run out to infinity on the right. A second bug is then dropped on “1,” again following the traffic lights starting from the state the last bug left them in; then, a third bug makes the trip. Prove that if the second bug falls off to the left, the third will march off to infinity on the right. Solution: We first need to convince ourselves that the bug will either fall of to the left, or go to infinity on the right; it cannot wander forever. To do so, it would have to visit some numbers infinitely often; let n be the least of those numbers, but now observe that every third visit to n will find it red and thus will incur a visit to n−1, contradicting the assumption that n−1 was visited only finitely often. With that out of the way, it will be useful to think of a green light as the digit 0, red as 1, and yellow, perversely, as the “digit” 1 2. The configuration of lights can then be thought of as a number between 0 and 1 written out in binary, x = .x1x2x3 . . . , where, numerically, x = x1 · 1 2 1 + x2 · 1 2 2 + · · · . Think of the bug at i as an additional “1” in the ith position, defining y = x + 1 2 i . The point of this exercise is that y is an invariant, that is, it does not change as the bug moves. When the bug moves to the right from point i, the digit upon which it sat goes up in value by 1 2; therefore, x increases by 1 2 i+1, but the bug’s own value diminishes by the same amount. If the bug moves to the left from i, it gains in value by 1 2 i, but x decreases by a whole digit in the ith place to compensate. The exception is when the bug falls off to the left, in which case both x and the bug’s own value drop by 1 2, for a loss of 1 overall. When the next bug is added, y goes up by 1 2. To put it another way, the value of x goes up by 1 2 if a bug is introduced and disappears to the right; and drops by 1 2 if a bug is introduced and falls off to the left. 311 Of course, x must always lie in the unit interval. If its initial value lies strictly between 0 and 1 2, the bugs must alternate right, left, right, left; if between 1 2 and 1, the alternation will be left, right, left, right. The remaining cases can be checked by hand. If x = 1 initially (all points red) the first bug turns point 1 green and drops off to the left; the second wiggles off to infinity leaving all points red again, so the alternation is left, right, left, right. If x = 0 initially (all points green), the bugs will begin right, right (as the points change to all yellow, then all red), and then left, right, left, right as before. The x = 1 2 case is the most interesting because there are several ways to represent 1 2 in our modified binary system: x can be all 1 2’s, or it can start with any finite number (including 0) of 1 2’s, followed either by 0111 . . . or 1000 . . . . In the first case, the leadoff bug turns all the yellows to red as it zooms off to the right; thus, we get a right, left, right, left alternation. The second case is similar, the first bug wiggling off to the right, but again leaving all points red behind it. In the third case, the bug changes the yellows to red as it marches out, but when it reaches the red point, it reverses and heads left, turning reds to green on its way to dropping off the left end. Thereafter, we are in the x = 0 case, so the final pattern is left, right, right, left, right, left, right. Checking back all the cases, we see that indeed, whenever the second bug went left, the third went right. ♡ Flipping the Pentagon The vertices of a pentagon are labeled with integers, the sum of which is positive. At any time, you may change the sign of a negative label, but then the new value is subtracted from both neighbors’ values so as to maintain the same sum. Prove that, inevitably, no matter which negative labels are flipped, the pro-cess will terminate after finitely many flips, with all values non-negative. Solution: Even with experimentation (highly recommended), it’s a little hard to see what can be used as a potential. The number of negative entries doesn’t necessarily go down, nor does the size of the largest negative entry. Similarly, neither the sum of the absolute values of the differences between adjacent numbers, nor the sum of the squares of the differences, seems to go reliably down (or up). But it turns out, if you take the sum of the squares of the differences between non-adjacent numbers, it works! Suppose the numbers are, reading around the pentagon, a, b, c, d and e. Then the sum in question is s = (a −c)2 + (b −d)2 + (c −e)2 + (d −a)2 + (e −b)2. Suppose b is negative and we flip it, replacing b by −b, a by a + b, and c by c + b. Then the new s is equal to the old s plus 2b(a + b + c + d + e), and since b is negative and a + b + c + d + e is positive, s goes down. And since s is an integer, it goes down by at least 1. 312 But s is non-negative (being a sum of squares) so it can’t go below zero. That means we must reach a point where there’s no negative entry left to flip, and we are done. But there’s an even better potential—better because it works for all poly-gons, not just the pentagon. It’s the sum, over all sets of consecutive vertices, of the absolute value of the sum of the numbers in the set. An even more remarkable solution was later found by Princeton computer scientist Bernard Chazelle. Construct a doubly-infinite sequence of numbers whose successive differences are the pentagon (or polygon) values, reading clock-wise around the figure. If the pentagon’s labels began as 1, 2, −2, −3, 3, the sequence might be . . . , −1, 0, 2, 0, −3, 0, 1, 3, 1, −2, 1, 2, 4, 2, −1, 2, 3, 5, . . . . Notice that the sequence climbs gradually (since the sum of the numbers around the polygon is positive) but not steadily (since some of the numbers around the polygon are negative). Now the key observation: Flipping a vertex has the effect of transposing pairs of entries of the above sequence that were in the wrong order—in other words, the flipping process turns into a sorting process for our infinite sequence! For example, if we flip the −2 on our pentagon to get 1, 0, 2, −5, 3, the sequence changes to . . . , −1, 0, 0, 2, −3, 0, 1, 1, 3, −2, 1, 2, 2, 4, −1, 2, 3, 3, . . . . It turns out to be easy to find a potential for the sorting progress that declines by exactly one per turn, and to conclude that not only does the process always terminate, but it terminates in the same number of steps and in the same final configuration, no matter which labels you flip! Picking the Athletic Committee The Athletic Committee is a popular service option among the faculty of Quin-cunx University, because while you are on it, you get free tickets to the univer-sity’s sports events. In an effort to keep the committee from becoming cliquish, the university specifies that no one with three or more friends on the commit-tee may serve on the committee—but, in compensation, if you’re not on the committee but have three or more friends on it, you can get free tickets to any athletic event of your choice. To keep everyone happy, it is therefore desirable to construct the committee in such a way that even though no one on it has three or more friends on it, everyone not on the committee does have three or more friends on it. Can this always be arranged? Solution: What’s the dumbest way to try to find a good committee? How about this: Start with an arbitrary set of faculty members, as a prospective Athletic Committee. 313 Oops, Fred is on the committee and already has three friends on the committee? Throw Fred out. Mona is not on the committee, but has fewer than three friends on it? Put Mona on. Continue fixing in this haphazard manner. Now, why in the world would you expect this to work? Clearly, the above actions could make things worse; for example, throwing Fred off the committee might create many more Monas; maybe we should have thrown off one of Fred’s on-committee friends instead. So there doesn’t seem to be anything to prevent cycling back to the same bad committee. Moreover, even if you don’t cycle back, there are exponentially many possible committees and you can’t afford to consider every one. Suppose there are 100 faculty members in all; then the number of possible committees is 2100 > 1030 which, even if you spent only a nanosecond considering each committee, would take a thousand times longer than all the time that has passed since the Big Bang. But if you try it, you will find that after shockingly few corrections, you end up with a valid committee. And this happens in situations where there is only one valid committee, as well as where there are many. How can this be? Sounds like there must be some potential function at work here, something that is improving each time you throw someone off or add someone to the current prospective committee. Let’s see: When you throw someone off, you destroy at least three on-committee friendships; when you put someone on, you add at most two. Let F(t) be the number of friendships on the committee minus 2 1 2 times the number of people on the committee at time t. Then when Fred is thrown off, F(t) goes down by at least 1 2. When Mona is put on, F(t) again goes down by at least 1 2. But F(0) can’t be more than (100 × 99)/2 −250 = 245 and F(t) can never dip below −250, so there can’t be more than 2 × (245 −(−250)) = 990 steps total. (A computer scientist would say that the number of steps in the process is at worst quadratic in the number of faculty members.) In practice, the number of steps is so small that if there are 100 faculty members and you start with (say) the empty committee, you will reach a solution easily by hand. Of course, you’ll need access to the friendship graph, so you might need to do some advance polling. It might be interesting to see who claims friendship with whom that isn’t reciprocated. For the last puzzle, we use the mother of all potentials—potential energy! Bulgarian Solitaire Fifty-five chips are organized into some number of stacks, of arbitrary heights, on a table. At each tick of a clock, one chip is removed from each stack and those collected chips are used to create a new stack. What eventually happens? Solution: It’s worth actually trying this—for example, with a deck of cards (which often comes with two jokers and an order card, making 55 cards in all.) If all you 314 have is 52 cards, remove seven and try the experiment with 45 cards. You will discover, if you have sufficient patience, that you eventually reach the “staircase” configuration where the stack heights are 10, 9, 8, 7, 6, 5, 4, 3, 2, and 1; and it’s easy to see that this configuration never changes. Proving that the staircase is always reached is a typical job for a potential function, but what potential is minimized by the staircase? It helps to think of each chip as a square block, and each stack as a column of square blocks. We can arrange the columns side-by-side and descending in height, as in the figure below. Then one “operation” can be thought of as stripping the bottom blocks from all the columns, turning them 90◦, then adding the resulting new column to the picture in the proper position to keep descending order of heights, as in the second figure. The configuration we’re aiming for still does not seem to minimize anything natural—until we rotate the whole picture 45◦counterclockwise, as shown in the next figure. The staircase now represents the “lowest” way to pack our 55 square blocks into a V-shaped cradle. By “lowest” we mean, or at least should mean, the configuration with least potential energy under gravity. Let the stacks be numbered in descending order of height, and the blocks in each stack numbered from bottom to top; then let sij be the ith block in the jth column. We can size the V-frame and blocks so that the height of the center of square sij is i+j, and then the potential energy of a configuration will be proportional to the sum over all the squares of the quantities i+j. 315 Now, what happens when we perform an operation? The blocks along the right-hand arm of the V move to the left-hand arm; their potential energy does not change. Nor does the potential of the other blocks, since each of those just moves one position to the right. But wait, there’s one more step: We might have to rearrange blocks to get the (original) columns in descending order. That’s equivalent to letting gravity push blocks downward (diagonally to the left), as in the next figure, reducing the potential energy of the configuration. If we’re already in the staircase configuration, we are at minimum potential energy and rearrangement of the stacks is never again required. Suppose we’re not in the staircase configuration; what then? The staircase configuration is possible, to begin with, only because 55 is a “triangular number,” equal to 1 + 2 + 3 + · · · + k for some k; these are just the numbers that can be expressed as k+1 2 for some positive integer k. In the stair-case configuration, there’s a block in every position (i, j) with 2 ≤i+j ≤k+1, and no block any higher. It follows that if the number of blocks is triangular but we’re not in the triangular configuration, there’s a “hole” somewhere at height k+1 and a block somewhere at height k+2. As long as no rearrangement is needed, the hole and the block each move one square to the right at every operation, regularly hitting the right arm of the V and cycling back to the left. But the hole’s cycles are of length k and the block’s of length k+1, so eventually the block will find itself directly above and to the right of the hole. At that point the block will fall into the hole, decreasing the potential energy of the configuration. We have shown, therefore, that if the number of blocks is 55 or any other triangular number, and the current configuration is not a staircase, that its potential energy will eventually go down. There are only finitely many possible configurations, thus only finitely many values for this potential. It follows that eventually we must reach the minimum-potential triangular configuration. Our theorem is a relatively easy application of potentials to graph theory. An unfriendly partition of a graph G = ⟨V, E⟩is a partition of the vertex set V into two sets with the property that every vertex in V has at least as many neighbors in the other part as it has in its own part. 316 Theorem. Every finite graph has an unfriendly partition. Proof. Start with any partition V = X ∪Y , and suppose it is not unfriendly; then there is some vertex v, say in X, that has more neighbors in X than in Y . Move v to Y . What happens? Unfortunately it may not be the case that the number of “friendly” vertices (that is, vertices with most of their neighbors on their own side) decreases, because it could be that other vertices are negatively affected by moving v. But the number of edges that cross the partition increases, because the only affected edges are those emanating from v and the majority of those didn’t cross before but now do. That makes things easy: The number of crossing edges cannot keep increas-ing indefinitely, so we must eventually come to an unfriendly partition. Equiva-lently: Start with a partition that maximizes the number of crossing edges, and you’re done! ♡ It’s worth noting that this argument fails for infinite graphs, even though it seems like when a vertex has infinitely many neighbors, it’s easier to ensure that there are as many on the opposite side (now considering infinite cardinalities) as there are at home. In fact the infinite case is tied up with optional axioms of set theory, and currently it’s not even known whether every countably infinite graph has an unfriendly partition. 317 318 Chapter 19 Hammer and Tongs Often, to solve a puzzle (or prove a theorem), you need to try things, see where problems arise, then fix them. We call this the “hammer-and-tongs” approach, and sometimes it works wonders. Phone Call A phone call in the continental United States is made from a west coast state to an east coast state, and it’s the same time of day at both ends of the call. How is this possible? Solution: Generally speaking, the west coast of the United States is on Pacific Time, 3 hours later than Eastern Time. But you may know that Florida, an east coast state, has a panhandle that extends quite far west; in fact Pensacola, FL, is on Central Time. So that picks up one of the three hours we need to make up. If you check you will find that there is a small part of Oregon that’s on Mountain Time, including, for example, the town of Ontario. That picks up our second hour. To get the third, we time the call so that Daylight Saving has just clicked off in Pensacola. An instant after 1:59:59 a.m. in Pensacola, typically on the first Sunday in November, the clock jumps back to 1 a.m. while it’s still just the “first” 1 am in Ontario, OR. So you’ve got an hour to make that phone call. As I write this I note that a revolt against Daylight Savings is brewing among the states. So before you attempt to earn drinks at the bar with this tidbit of a puzzle, you might want to check that it still works! Crossing the River In eighth century Europe, it was considered unseemly for a man to be in the presence of a married woman, even briefly, unless her husband was there as well. 319 This posed problems for three married couples who wished to cross a river, the only means being a rowboat that could carry at most two people. Can they get to the other side without violating their social norms? If so, what’s the minimum number of crossings needed? Solution: This is just a question of trying it out. Call the women 1, 2, and 3, the corre-sponding husbands A, B, and C. The ideal plan would require nine crossings, with the boat carrying two people across and one back with each round trip. If 1 and A cross, it must be A that returns with the boat (else 1 would find herself on the near shore with B and C but no husband). Now 2 and 3 must cross with one woman returning; two men can then join their wives on the far side. Now, however, a married couple must come back together in the boat, costing us a couple of crossings and making the nine-crossing ideal impossible. Now the remaining men cross the river for good and the women finish the job. The final plan has eleven crossings, the minimum possible, and is illustrated below. 320 Sprinklers in a Field Sprinklers in a large field are located at the vertices of a square grid. Each point of land is supposed to be watered by exactly the three closest sprinklers. What shape is covered by each sprinkler? Solution: We first must ask ourselves: What are the three sprinklers closest to a given point in the field? Suppose the point is in the grid square bounded by sprinklers at a, b, c, and d, labeled clockwise. Divide the square into four congruent subsquares, and call them A, B, C, and D respectively, according to whether the corner shared with the big square is a, b, c, or d. It’s not hard to see that for any point in A the closest three sprinklers are at a, b, and d, and similarly for points in the other small squares. It follows that the sprinkler at a has to reach all the points in A, B, and D, plus the points in corresponding subsquares of the other three grid squares incident to a. This amounts to 12 little subsquares in the shape of a fat Greek cross. Fair Play How can you get a 50-50 decision by flipping a bent coin? Solution: Flip the bent coin twice hoping to get a head and a tail; if the head comes first, call the result HEADS; if the tail comes first, call it TAILS. If the result is two heads or two tails, repeat the experiment. This solution is often attributed to the late, great mathematician and com-puting pioneer John von Neumann, and called “von Neumann’s trick.” It relies 321 on the fact that even if the coin is bent, successive flips are (or at least should be) independent events. Of course, it also relies on it being at least possible for the bent coin to land on either side! If you want to minimize the number of flips to get your decision, the above scheme can be improved upon. For example, if you get HH for the first pair of flips and TT for the second, you can quit and call the result HEADS (TT followed by HH would then be called TAILS). Finding the Missing Number All but one of the numbers from 1 to 100 are read to you, one every ten seconds, but in no particular order. You have a good mind, but only a normal memory, and no means of recording information during the process. How can you ensure that you can determine afterward which number was not called out? Solution: Easy—you keep track of the sum of the numbers being called out, adding each one in turn to your accumulated total. The sum of all numbers from 1 to 100 is 100 times the average number (50 1 2), namely 5,050; that minus your final sum will be the missing number. There is, by the way, no need to keep the hundreds digit or thousands digit during the process. Addition modulo 100 is good enough. At the end you subtract the result from 50 or 150 to get an answer in the correct range. Dealing with streams of data, when handicapped by limited computing and memory resources, is a major subject of study in modern computer science. Here’s another streaming puzzle. Identifying the Majority A long list of names is read out, some names many times. Your object is to end up with a name that is guaranteed to be the name which was called a majority of the time, if there is such a name. However, you have only one counter, plus the ability to keep just one name at a time in your mind. Can you do it? Solution: It makes sense to use the counter to keep track of how well the name currently in memory is doing. Then, whenever that name is heard again, you can increment the counter; when some other name is heard, you must decrement it. The counter begins at 0; if it ever threatens to dip below zero, the current name in memory is replaced by the name being heard, and the counter is incremented from 0 to 1. This may seem a bit dubious at first, because the new name you are putting into memory may be the first occurrence of that name, while in the meantime 322 your previous name (and other names) may have occurred many times. For example, if the list were “Alice, Bob, Alice, Bob, Alice, Bob, Charlie,” Alice’s name would remain in your memory until the end when Charlie becomes your nominee. But that’s OK, since no name has a majority. If a name does occur more than half the time, it’s guaranteed to be the one in your memory at the end. Why? Suppose the majority name is Mary, and assume (for convenience) that a name is heard every minute. Divide time into intervals during which the same name remains in memory. Then each interval except possibly the last is an even number of minutes in length, and its memory-name was heard exactly half the time during the interval. It follows that Mary’s name is heard at most half the time within every interval except the last, therefore more than half the time in the last interval— which must therefore end with Mary’s name in memory. ♡ Here’s an example: Suppose the name list is A, M, A, M, M, A, B, B, M, M, A, M, M, B, M. Then the intervals are of length 4 (with “A” in memory), 2 (with “M”), 4 (with “B”), then 2 (with “A” again) and finally 3 (with “M”), the counter ending at value 1. Poorly Placed Dominoes What’s the smallest number of dominoes one can place on a chessboard (each covering two adjacent squares) so that no more fit? Solution: The goal is to place dominoes so that the number h of “holes” (uncovered squares of the chessboard) is as large as possible, without having two adjacent holes. The number of dominoes used would then be d = (64 −h)/2. It seems natural to arrange dominoes in diagonal lines, as in the first figure below; worst fit is achieved by switching from horizontal in half the board to vertical in the other half, then filling in adjacent holes as necessary. But alter-nating, gapped horizontal lines work better provided you switch to vertical (or vice-versa), as in the second figure, when the edge of the board causes problems. This gives 20 holes, thus 22 dominoes. Could we get d down to 21, i.e., h = 22? Suppose so. Let h′ be the number of holes not on the top row, and d′ the number of dominoes not touching the bottom. Since every hole counted by h′ lies directly under a different domino counted by d′, we have h′ ≤d′. Since h > d, there must be more holes on the top row than dominoes touching the bottom, thus there must be exactly 4 of the former and 3 of the latter. Now rotate the board and repeat the argument to deduce that every edge boasts four holes and only three touching dominoes. Up to reflection, this forces the placement of 16 dominoes and it is easy to check that at least six more must fit no matter how they are (sequentially) placed. Dominoes appear in many intriguing puzzles. 323 Unbreakable Domino Cover A 6 × 5 rectangle can be covered with 2 × 1 dominoes, as in the figure below, in such a way that no line between dominoes cuts all the way across the board. Can you cover a 6 × 6 square that way? Solution: No. You need two dominoes, not just one, to cross each interior line—with just one, there’s be an odd number of squares on each side of that line, but that number is a multiple of 6. Since there are 10 interior lines, you’ll need 20 crossing dominoes, but there’s room for only 62/2 = 18 dominoes on the board. The 6 × 5 rectangle is in fact the only one with both dimensions under 7 that can be “unbreakably” covered. 324 Filling a Bucket Before you are 12 two-gallon buckets and a 1-gallon scoop. At each turn, you may fill the scoop with water and distribute the water any way you like among the buckets. However, each time you do this your opponent will empty two buckets of her choice. You win if you can get one of the big buckets to overflow. Can you force a win? If so, how long will it take you? Solution: It’s safe to assume your opponent will always empty the fullest two buckets. To force your opponent to pour out as little water as possible, it’s natural to begin by keeping all the buckets at the same level. How far will this get you? You’ll start by putting 1/12 of a gallon in each bucket; your opponent will empty two buckets, reducing the total amount of water to 10/12. You can then add a gallon total to bring each bucket up to (1 + 10/12)/12 = 11/72. Continuing in this manner, you make progress as long as the amount in each bucket is less than half a gallon (since then your opponent is pouring off less than you’re adding). You can get close to, but never quite reach, half a gallon per bucket by this method. Then what? Then you’re going to need to give up on keeping all the buckets level. Sup-pose you build up to x gallons per bucket, then give up on the two buckets your opponent just emptied, and fill the rest evenly. Then you can get x + 1/10 gallons in those; she empties two of them and you build up to x + 1/10 + 1/8 in the remaining eight, and so on. You end with x + 1/10 + 1/8 + 1/6 + 1/4 + 1/2 = x + 1.141666 in the last two buckets, not good enough to cause overflow, since x < 1/2. But wait—you don’t need two buckets overflowing, only one. So, you start by building up only 11 buckets, with the idea of later reducing to 9, 7, 5, 3, and finally, 1 bucket. That will get you up to x + 1/9 + 1/7 + 1/5 + 1/3 + 1 ∼ 325 x + 1.7873015873. That’s more like it! So it’s enough to get x ≥0.2127. You start with 1/11 of a gallon in each of 11 buckets, abandoning the twelfth forever. You build that up to (1 + 9 · 1/11)/11 = 20/121 ∼0.1653 gallons in each of the 11 buckets after the second round, and (1 + 9 · 20/121)/11 ∼0.2261 after the third. That’s enough to go on to the reduction phase. Altogether it takes you 3+5 = 8 rounds, and with some effort it can be shown that you can’t do any better. Polygon on the Grid A convex polygon is drawn on the coordinate plane with all its vertices on integer points, but no side parallel to the x- or y-axis. Let h be the sum of the lengths of the horizontal line segments at integer height that intersect the (filled-in) polygon, and v the equivalent for the vertical line segments. Prove that h = v. Solution: Both are equal to the area of the polygon. One way to see that is to let L1, . . . , Lk be the integer-height horizontal lines that intersect the interior of the polygon, dividing it into two triangles (at the top and bottom ends) and k−1 trapezoids. If the length of the segment of Li that intersects the polygon is ℓi, then the sum of the areas of the triangles and trapezoids is 1 2ℓ1 + 1 2(ℓ1 + ℓ2) + 1 2(ℓ2 + ℓ3) + · · · + 1 2(ℓk−1 + ℓk) + 1 2ℓk which equals h, and a similar argument holds for v. One-Bulb Room Each of n prisoners will be sent alone into a certain room, infinitely often, but in some arbitrary order determined by their jailer. The prisoners have a chance to confer in advance, but once the visits begin, their only means of communication will be via a light in the room which they can turn on or off. Help them design a protocol which will ensure that some prisoner will eventually be able to deduce that everyone has visited the room. Solution: It will, of course, be necessary to assume that no one fools with the room’s light between visits by prisoners; but prisoners do not need to know the initial state of the light. The idea is that one prisoner (say, Alice) repeatedly tries to turn the light on, and each of the others turns it off twice. More precisely, Alice always turns on the light if she finds it off, otherwise she leaves it on. The rest of the prisoners turn it off the first two times they find it on, but otherwise leave the light alone. 326 Alice keeps track of how many times she finds the room dark after her initial visit; after 2n−3 dark revisits she can conclude that everyone has visited. Why? Every dark revisit signals that one of the other n−1 prisoners has visited. If one of them, say Bob, hasn’t been in the room, then the light can not have been turned off more than 2(n−2) = 2n−4 times. On the other hand, Alice must eventually achieve her 2n−3 dark revisits because eventually the light will have been turned off 2(n−1) = 2n−2 times and only one of these (caused by a prisoner darkening an initially light room before Alice’s first visit) can fail to cause a dark revisit by Alice. Sums and Differences Given 25 different positive numbers, can you always choose two of them such that none of the other numbers equals either their sum or their difference? Solution: Yes. Let the numbers be x1 < x2 < · · · < x25, and suppose that every two of them have either their sum or their difference represented among the other numbers. For i < 25, x25 + xi can’t be on the list, so x25 −xi must be there; it follows that the first 24 numbers are paired with xi +xn−i = x25. Now consider x24 together with any of x2, . . . , x23; these pairs sum to more than x25 = x24+x1 and so x2, . . . , x23 must also be paired, and in particular x2 + x23 = x24. But we just had x2 + x23 = x25, a contradiction. Prisoner and Dog A woman is imprisoned in a large field surrounded by a circular fence. Outside the fence is a vicious guard dog that can run four times as fast as the woman, but is trained to stay near the fence. If the woman can contrive to get to an unguarded point on the fence, she can quickly scale the fence and escape. But can she get to a point on the fence ahead of the dog? Solution: We may as well take our unit of distance to be the radius of the field. If the prisoner were constrained to a smaller, concentric circle of radius r, where r < 1 4, she would be able to maneuver herself to the farthest available point from the dog (point P in the figure below); this is because the circumference of the small circle would be less than 1 4 of the circumference of the field. But if r is close enough to 1 4, the woman can then make a run for it straight to the fence. Her distance to the fence would only be a hair over 3 4 of a unit, but the dog has to go half way around the field, a distance of π units. Since π > 3, this is more than four times further than the woman has to run. 327 Love in Kleptopia Jan and Maria have fallen in love (via the internet) and Jan wishes to mail her a ring. Unfortunately, they live in the country of Kleptopia where anything sent through the mail will be stolen unless it is sent in a padlocked box. Jan and Maria each have plenty of padlocks, but none to which the other has a key. How can Jan get the ring safely into Maria’s hands? Solution: In one solution, Jan sends Maria a box with the ring in it and one of his padlocks on it. Upon receipt Maria affixes her own padlock to the box and mails it back with both padlocks on it. When Jan gets it he removes his padlock and sends the box back to Maria, who can now open her own padlock and enjoy the ring. This solution is not just play; the idea is fundamental in Diffie-Hellman key exchange, an historic breakthrough in cryptography. Depending on one’s assumptions, other solutions are possible as well. A nice one was suggested by several persons at the Gathering for Gardner VII, 328 including origami artist Robert Lang: It requires that Jan find a padlock whose key has a large hole, or at least a hole which can be sufficiently enlarged by drilling, so that the key can be hooked onto a second padlock’s hasp. Jan uses this second padlock, with the aforementioned key hooked on its hasp, to lock a small empty box which he then sends to Maria. When enough time has passed for it to get there (perhaps he awaits an email acknowledgment from Maria) he sends the ring in another box, locked by the first padlock. When Maria gets the ring box, she picks up the whole first box and uses the key affixed to it to access her ring. Badly Designed Clock The hour and minute hands of a certain clock are indistinguishable. How many moments are there in a day when it is not possible to tell from this clock what time it is? Solution: Let us first note that for the problem to make sense, we must assume that the hands move continuously, and that we are not tasked with deciding whether a time is a.m. or p.m. (We assume there is no “second hand.”) Note that we can tell what time it is when the hour and minute hands coincide, even though we can’t tell which hand is which; this happens 22 times a day, since the minute hand goes around 24 times while the hour hand goes around twice, in the same direction. This reasoning turns out to be good practice for the proof. Imagine that we add to our clock a third “fast” hand, which starts at 12 midnight and runs exactly 12 times as fast as the minute hand. Now we claim that whenever the hour hand and the fast hand coincide, the hour and minute hands are in an ambiguous position. Why? Because later, when the minute hand has traveled in all 12 times as far as it had moved since midnight, it will be where the fast hand (and thus also the hour hand) is now, while the hour hand is where the minute hand is now. That’s the definition of an ambiguous moment. Conversely, by the same reasoning, all ambiguous positions occur when the hour hand and fast hand coincide. So, we need only compute the number of times a day this coincidence of fast hand and hour hand occurs. The fast hand goes around 122 × 2 = 288 times a day, while the hour goes around just twice, so this happens 286 times. Of these 286 moments, 22 are times when the hour hand and minute hand (thus all three hands) are coincident, leaving 264 ambiguous moments. Worms and Water Lori is having trouble with worms crawling into her bed. To stop them, she places the legs of the bed into pails of water; since the worms can’t swim, they 329 can’t reach the bed via the floor. But they instead crawl up the walls and across the ceiling, dropping onto her bed from above. Yuck! How can Lori stop the worms from getting to her bed? Solution: This curious problem is arguably more about engineering than mathematics. Lori can indeed keep the worms off her bed by (in addition to putting the bed’s legs in water buckets) hanging a large canopy from the ceiling, extending well over the bed. But the canopy must curve underneath itself at its edges, creating a ring-gutter underneath that is filled with water. (See the figure below for a cross section of the canopy.) The under-slung gutter prevents the worms from dropping onto the canopy’s edge, crawling under the canopy’s surface to some point over the bed, and dropping onto Lori. If the worms have no high-up way to get in her bedroom, Lori can more easily accomplish this same task by encircling the room itself with a water-filled gutter. Generating the Rationals You are given a set S of numbers that contains 0 and 1, and contains the mean of every finite nonempty subset of S. Prove that S contains all the rational numbers between 0 and 1. Solution: First note that S contains all the “dyadic” rationals, that is, rationals of form p/2n; we can obtain all those with denominator 2n and odd numerator by 330 averaging two adjacent ones with lower-powered denominators. Now any general p/q is of course the average of p ones and q−p zeros. We choose n large and replace the zeros by 1/2n, −1/2n, 2/2n, −2/2n, 3/2n, etc. including one 0 if p is odd. Similarly, we replace the ones by 1 −1/2n, 1 + 1/2n, 1 −2/2n, and so forth. Of course, some of these numbers lie outside the unit interval, but we can rescale the procedure to fit some dyadic interval containing p/q and lying strictly between 0 and 1. ♡ Funny Dice You have a date with your friend Katrina to play a game with three dice, as follows. She chooses a die, then you choose one of the other two dice. She rolls her die while you roll yours, and whoever rolls the higher number wins. If you roll the same number, Katrina wins. Wait, it’s not as bad as you think; you get to design the dice! Each will be a regular cube, but you can put any number of pips from 1 to 6 on any face, and the three dice don’t have to be the same. Can you make these dice in such a way that you have the advantage in your game? Solution: What you’d like to do is design the dice so that no matter which of the three Katrina chooses, you can pick one of the remaining two which will roll a higher number more than half the time. Such a trio is known as a “set of non-transitive dice,” and there are many ways to design them. The key is that if die A’s roll has the same average as die B’s, that doesn’t mean A beats B half of the time. It could be that when A beats B, it wins by a lot (e.g., 6 versus 1) but when it loses, it does so by only a little bit (e.g., 5 versus 4). Then it would follow that B would win more often. To take an extreme example, put a 6 and a five 3’s on die A (averaging 3.5), and a 1 and five 4’s on die B (same average). Then with probability 5 6 × 5 6 = 25/36 > 1/2, there’ll be a 3 on die A and a 4 on die B. And you’ve eliminated ties by using different numbers on the two dice. Let’s use the remaining numbers, 2 and 5, on die C—three of each, so that again the average roll is 3.5. It remains only to check that, indeed, die C is a favorite over die B (winning anytime C rolls a 5, and also occasionally 2 to 1, altogether with probability 7/12), but die A is the expected winner over die C. Since we already know B beats A, your strategy against Katrina is clear. Sharing a Pizza Alice and Bob are preparing to share a circular pizza, divided by radial cuts into some arbitrary number of slices of various sizes. They will be using the “polite pizza protocol”: Alice picks any slice to start; thereafter, starting with 331 Bob, they alternate taking slices but always from one side or the other of the gap. Thus after the first slice, there are just two choices at each turn until the last slice is taken (by Bob if the number of slices is even, otherwise by Alice). Is it possible for the pizza to have been cut in such a way that Bob has the advantage—in other words, so that with best play, Bob gets more than half the pizza? Solution: If the number of slices is even (as with most pizzas), the puzzle Coins in a Row from Chapter 7 applies and Alice can always get at least half the pizza. In fact that is so even if Bob gets to start by choosing one radial cut and insisting that Alice’s choice of first slice be on one side or the other of that cut. That makes the problem exactly the same as Coins in a Row; Alice can just number the slices 1, 2, etc. starting clockwise (say) from the cut, and play so as to take all the even-numbered slices or all the odd, whichever is better for her. The argument fails if the number of slices is odd. But the odd case sounds even better for Alice since then she ends up with more slices. How can we get a handle on the odd case? Taking another cue from Chapter 7, let’s limit the slice sizes to 0 or 1. (Can a slice be of size zero? Mathematically, no problem; gastronomically, think of a slice of size ε = one one-hundredth of a pizza, say.) To try to give the advantage to Bob, we somehow need to organize these so that no matter where Alice begins, Bob can use his parity advantage to overcome Alice’s one-slice head start. That takes a fair number of slices, but it turns out that 21 of these {0, 1}-sized slices is enough to turn the advantage to Bob. With a bit more fooling around, you will find that you can combine some of the slices to get a 15-slice pizza with slice-sizes 0, 1, and 2 of which nothing can stop Bob from acquiring 5/9. (Two schemes work: 010100102002020 and 010100201002020.) Pictured below is one of the Bob-friendly 15-slice schemes, with pepperoni distributed to indicate the slice sizes. No matter how she plays, Alice can never get more than 4 of the 9 pepperoni chunks against a smart, hungry Bob. It has been shown that this pizza is best possible for Bob. Summary: If the number of slices is even, or odd but at most 13, Alice can get at least half the pizza; if the number of slices is odd and at least 15, she can guarantee at least 4/9 but no larger fraction. Names in Boxes The names of 100 prisoners are placed in 100 wooden boxes, one name to a box, and the boxes are lined up on a table in a room. One by one, the prisoners are led into the room; each may look in at most 50 boxes, but must then leave the room exactly as he found it and is permitted no further communication with the others. 332 The prisoners have a chance to plot their strategy in advance, and they are going to need it, because unless every single prisoner finds his own name all will subsequently be executed. Find a strategy that gives the prisoners a decent chance of survival. Solution: To solve it, the prisoners must first agree on a random labeling of the boxes by their own names. (The point of making it random is that it makes it impossible for the warden to place names in boxes in such a way as to foil the protocol described below.) When admitted to the room, each prisoner inspects his own box (that is, the box with which his own name has been associated). He finds a name in that box, probably not his own. He then looks into the box belonging to the name he just found, and then into the box belonging to the name he found in the second box, etc., until he either finds his own name, or has opened 50 boxes. That’s the strategy; now, why on earth should it work? Well, the process which assigns to a box’s owner the name found in his box is a permutation of the 100 names, chosen uniformly at random from the set of all such permutations. Each prisoner is following a cycle of the permutation, beginning with his box and (if he doesn’t run over the 50-box limit) ending with his name on a piece of paper. If it happens that the permutation has no cycles of length greater than 50, this process will work every time and the prisoners will be spared. In fact, the probability that a uniformly random permutation of the numbers 333 from 1 to 2n contains no cycle of length greater than n always exceeds 1 minus the natural logarithm of 2—about 30.6853%. To see this, let k > n and count the permutations having a cycle C of length exactly k. There are 2n k ways to pick the entries in C, (k−1)! ways to order them, and (2n−k)! ways to permute the rest; the product of these numbers is (2n)!/k. Since at most one k-cycle can exist in a given permutation, the probability that there is one is exactly 1/k. It follows that the probability that there is no long cycle is 1 − 1 n+1 − 1 n+2 −· · · −1 2n = 1 −H2n + Hn where Hm is the sum of the reciprocals of the first m positive integers, approx-imately ln m. Thus our probability is about 1 −ln 2n + ln n = 1 −ln 2, and in fact is always a bit larger. For n = 50 we get that the prisoners survive with probability 31.1827821%. Let’s put these new skills to work on a variation. Life-Saving Transposition There are just two prisoners this time, Alice and Bob. Alice will be shown a deck of 52 cards spread out in some order, face-up on a table. She will be asked to transpose two cards of her choice. Alice is then dismissed, with no further chance to communicate to Bob. Next, the cards are turned down and Bob is brought into the room. The warden names a card and to stave off execution for both prisoners, Bob must find the card after turning over, sequentially, at most 26 of the cards. As usual the prisoners have an opportunity to conspire beforehand. This time, they can guarantee success. How? Solution: This is just a variation on the names-in-boxes puzzle. We may assume the cards are numbered 1 to 52. Bob’s algorithm, when asked to search for card i, will be to flip the ith card in the display. If it’s value is j, he then flips the jth card; if the value of the jth card is k, he flips the kth card, etc. Thus, he follows a cycle in the permutation that maps the nth card to its value, and if no cycle in this permutation is of length > 26, he will succeed. Alice can guarantee this condition. If she sees a cycle of length > 26 (of course, there cannot be more than one of these) she bisects it by transposing two antipodal or nearly antipodal cards of the cycle. Otherwise she makes a harmless move, for example, bisects a short cycle or if the ordering happens to be the identity, does any transposition. We conclude this chapter’s puzzles with one that is mind-numbingly easy if you try the right thing. 334 Self-Referential Number The first digit of a certain 8-digit integer N is the number of zeroes in the (ordinary, decimal) representation of N. The second digit is the number of ones; the third, the number of twos; the fourth, the number of threes; the fifth, the number of fours; the sixth, the number of fives; the seventh, the number of sixes; and, finally, the eighth is the total number of distinct digits that appear in N. What is N? Solution: It’s pretty tough to work out N by reasoning. But there’s an easy way! Take any 8-digit number N0, and transform it according to the puzzle con-ditions: that is, make a new number N1 by letting the first digit of N1 be the number of 0’s in N0, etc. If by chance N1 = N0, you’ve solved the puzzle. But you’d have to be astonishingly lucky for that to happen. Not to worry. Repeat the procedure with N1 to get N2, then N3, until you reach a time t when Nt = Nt−1. Then you’re done. Let’s try this. For fun we’ll start with 31415926, the first 8 digits of the decimal expansion of π, but you should try it yourself with your favorite 8-digit number. We obtain: It worked! But why should that happen in a reasonable number of steps? Indeed, why should it happen at all? Couldn’t we start to cycle at some point, with Nt duplicating some previous Nt−k with k > 1? Sadly, I don’t know. This method does not work with all problems of this type. But it is surprisingly useful—a great weapon to have in your armory. We conclude this chapter with a marvelous theorem that’s well known to ge-ometers and to folks who do recreational math, but deserves to be better known among serious mathematicians in other areas. Described by Georg Alexander Pick in 1899, it tells you how to compute the area of a lattice polygon just by counting lattice points. A lattice polygon is a closed figure made from line segments that terminate at vertices of a plane grid, like the one shown in the figure below. Theorem. The area of a lattice polygon is exactly I + B/2 −1 times the area of a single square cell of the grid, where I is the number of lattice points strictly inside the polygon, and B the number on its boundary. Proof. We will assume that each cell of the grid has unit area, so that the theorem simply says the area of a lattice polygon P is I + B/2 −1. The formula carries some reasonable intuition: Inside points (filled circles in the figure) are fully counted, and boundary points (empty circles) are arguably half-in and half-out, so should be counted at half value—except that corners are really somewhat less than half-in, so we subtract one to account for them. 335 The formula certainly works when the polygon is a single grid cell, since there I = 0 and B = 4. Unfortunately we can’t generally build lattice polygons out of grid cells, but it’s worth thinking about how we could build them out of smaller units. Suppose P is the union of two lattice polygons R and S that share one side from each, as in the next figure. If the formula is correct for R and S, does it follow that it’s correct for P? Indeed, that is the case: If there are k lattice points on the common edge, not counting its endpoints, then we have IP = IR+IS+k and BP = BR+BS−2k−2, so that the area of P is IR + IS + k + (BR + BS −k −2)/2 −1 = (IR + BR/2 − 1) + (IS + BS/2 −1) which is indeed the area of R plus the area of S. So far so good. Note that this calculation also shows that if the formula is correct for any two of the three polygons P, R, and S, it works for the third—in other words, the formula works with subtraction as well as addition. That’s great—and because we can cut any lattice polygon into lattice trian-gles, as in the next figure, it’s enough to prove the theorem for triangles. This 336 is something of a hammer-and-tongs type operation. 337 First we note that right triangles are easy, if their sides are vertical and horizontal line segments. The reason is that we can copy such a triangle, rotate the copy 180◦, then match its hypotenuse to the original to get an “aligned” lattice rectangle (i.e., one with sides parallel to the axes). The formula holds for aligned rectangles, because they’re made out of unit cells. Since our two triangles have the same values for I and B, Pick’s formula assigns them the same area which must therefore be half the area of the aligned rectangle, as it should be. Finally, we only need observe that any triangle can be made into an axis-aligned rectangle by adding (at most) three right triangles of the above sort, as in the final figure. ♡ 338 Chapter 20 Let’s Get Physical Rotating Coin While you hold a U.S. 25-cent piece firmly to the tabletop with your left thumb, you rotate a second quarter with your right forefinger all the way around the first quarter. Since quarters are ridged, they will interlock like gears and the second will rotate as it moves around the first. How many times will it rotate? Solution: A classic. It rotates twice, relative to the table: once relative to the stationary quarter, and once more owing to its revolution around the stationary quarter. Try it! Pie in the Sky What fraction of the sky is occupied by a full moon? Solution: The moon is about half a degree in diameter (actually, between 29.43 arc minutes at apogee and 33.5 minutes at perigee), hence its area in “square degrees” is πr2 ∼π(1/4)2 = π/16. 339 How many square degrees is the sky? Its circumference is 360 so its radius in degrees (!) is 180/π and the sky’s area is half the area of the celestial sphere, about (1/2)4πr2 = 2(180)2/π square degrees. Thus the fraction of sky covered by the moon is about (π/16)/(2(180)2/π) = π2/(32 · 1802) ∼1/105,050. If we repeat this calculation with the more accurate estimate of 31 minutes for the diameter of the moon, we get 1/101,661 of the sky—so it is remarkably accurate to say the moon occupies ten millionths of the sky. The sun, by the way, is about the same size. Pushing the Pedal You are standing to the right of your bicycle, steadying it with your left hand on the seat. The right pedal is in its down (6 o’clock) position. With your right hand, you push the pedal leftward (clockwise, toward 7 o’clock). Does the bike go forward or backward? Solution: The bike goes backward, and despite your pushing the pedal clockwise, the pedal rotates counterclockwise toward 5 o’clock. You don’t believe it? Try it! If you thought the bike would go forward, you have good company: the great mathematician Vladimir Arnol’d said so in a book (whose American edition features his bike figure on the cover). But as the bike moves, unless it’s in a ridiculously low gear, the pedals are always going forward relative to the ground. Hence when you are standing on the ground and pushing the pedal rearward, the pedal must go rearward with respect to the ground (though it will move forward with respect to the bike), and therefore the bike goes backward as well. Returning Pool Shot A ball is shot from a corner of a polygonal pool table (not necessarily convex) with right-angle corners; let’s say all sides are aligned either exactly east-west or exactly north-south. The starting corner is a convex corner, that is, has an interior angle of 90◦. There are pockets at all corners, so that if the ball hits a corner exactly it falls in. Otherwise, the ball bounces true and without energy loss. Can the ball ever return to the corner from which it began? Solution: Suppose it is possible, and imagine such a shot. We may assume the ball starts off going NE; then when it finally returns it is going SW. At each bounce one of the two direction letters changes; for example, the second lap will be either SE (if it bounces off an E-W cushion) or NW (if it bounces off a N-S cushion). It follows that it takes an even number of bounces, thus an odd number of laps. 340 All NE and SW laps are parallel (likewise all NW and SE paths), thus the initial and final laps follow the same line in opposite directions. Since reflection works as well backwards as forwards, the whole path out from the original corner and back is a palindrome—the sequence of bounce points reads the same backwards and forwards. It follows that the two middle bounce points are the same point, which is ridiculous. ♡ Falling Ants Twenty-four ants are placed randomly on a meter-long rod; each ant is facing east or west with equal probability. At a signal, they proceed to march forward (that is, in whatever direction they are facing) at 1 cm/sec; whenever two ants collide, they reverse directions. How long does it take before you can be certain that all the ants are off the rod? Solution: The key to this puzzle is that, ants being interchangeable as far as we’re con-cerned, it would make no difference to the process if they passed one another “like ships in the night” instead of bouncing. Then it’s clear that each ant is simply walking straight ahead and must fall off within 100 seconds. You can’t afford to wait any shorter amount of time because if there’s an ant starting at one end of the rod and facing the other end, some ant won’t exit the far end until 100 seconds later. 341 Ants on the Circle Twenty-four ants are randomly placed on a circular track of length 1 meter; each ant faces randomly clockwise or counterclockwise. At a signal, the ants begin marching at 1 cm/sec; when two ants collide they both reverse directions. What is the probability that after 100 seconds, every ant finds itself exactly where it began? Solution: This time we have to be a bit more careful about the ants’ anonymity; the argument that we can replace bouncing by passing tells us only that the ants’ set of locations will be exactly the same after 100 seconds, but any particular ant might end up in some other ant’s starting spot. In fact, since the ants cannot pass one another, their final locations will be some rotation of their initial locations. Putting it another way, the whole collection will rotate by some number of ants, and we are in effect being asked to determine the probability that that number will be a multiple of 24. In fact it could be 24 (clockwise or counterclockwise) only if the ants are all facing the same way, thus each walks once around the track without any collisions. There are 224 ways to choose how the ants face of which only two have this property, so the probability of one of these outcomes is a minuscule 1/223. Much more likely is that the net rotation will be zero. When does that happen? Well, conservation of angular momentum tells us that the rate of rotation of the ant collection as a whole is constant. Thus the net rotation will be zero if and only if the initial rate of rotation is zero, meaning that exactly the same number ants start off facing counterclockwise as clockwise. The probability of that happening is 24 12 /224 which is about 16.1180258%. Adding 1/223 to that boosts the final answer to about 16.1180377%. Sphere and Quadrilateral A quadrilateral in space has all of its edges tangent to a sphere. Prove that the four points of tangency lie on a plane. 342 Solution: The idea is to weight the vertices of the quadrilateral so that the point where each edge touches the sphere is exactly the center of gravity of the edge. We can do this because each vertex is equidistant from the tangency points of the two edges emanating from it; if we give that vertex weight equal to the reciprocal of that distance, we get the desired weighting. So what? Well, if we draw a line between opposite points of tangency, the center of gravity of the quadrilateral will have to be on that line. There are two such lines, though, so the center of gravity is on both; thus, the lines intersect. Thus the quadrilateral formed by the points of tangency lies on a plane! ♡ Two Balls and a Wall On a line are two identical-looking balls and a vertical wall. The balls are per-fectly elastic and friction-free; the wall is perfectly rigid; the ground is perfectly level. If the balls are the same mass and the one farther from the wall is rolled toward the closer one, it will knock the close ball toward the wall; that ball will bounce back and hit the first ball, which will then roll (forever) away from the wall. Three bounces altogether. Now assume the farther ball has mass a million times greater than the closer one. How many bounces now? (You may ignore the effects of angular momen-tum, relativity, and gravitational attraction.) Solution: You can expect quite a lot of bouncing now, because the heavy ball will hardly notice that it has been hit the first time, and proceed undaunted toward the wall. But the light ball will bounce off the wall and hit the heavy one many more times, until the heavy ball slows down, reverses direction, and eventually disappears. In fact the number of collisions will be 3,141, the first four digits of the decimal expansion of π. Coincidence? No indeed. Make the heavy ball a googol (that is, 10100) times heavier than the light one, and the number of bounces will be the first 51 digits of π, namely 314,159,265,358,979,323,846,264,338,327,950,288,419,716,939,937,510. It doesn’t matter how far apart the three objects are, or how hard the heavy ball is pushed. The expectation is that for any integer k ≥0, the number of bounces when the heavy ball is 102k times heavier than the light one will be the first k+1 digits of π. Almost as remarkable as the statement itself is the fact that it is not a theorem, that is, no one can prove it. But following is an explanation of why we think it’s true. The key is to make use of conservation of energy and momentum. Let v be the velocity of the heavy ball and y be the velocity of the light ball. The mass of the heavy ball is m = 102k, and conservation of energy says mv2 + y2 is a 343 constant. Letting x = √m · v = 10kv and starting at x = −1, y = 0 puts the system on the unit circle x2 + y2 = 1. Total momentum is mv +y = 10kx+y = −10k, so y = −10kx−10k, putting us initially on a line of slope s = −10k. So when the light ball first hits the wall, the “system point”—that is, the point on the XY-plane that describes the state of the system—moves from (−1, 0) south-east along the momentum line to the second point where that line intersects the circle. But upon hitting the wall the light ball’s momentum flips, reflecting the new point across the X-axis back to the positive side. In this way the system point zigzags down and up, moving slightly eastward on the down-zigs, until finally it reaches a point just a bit northwest of (1,0) from which moving southeast along a line of slope s would either miss the circle entirely or hit it above the X-axis. That indicates there will be no more collisions. The figure below shows our circle, with four system states corresponding to the first two and last two system points. The total number of collisions is the number of zigzags; how many were there? Well, the momentum line’s angle to the vertical is arctan(10−k) and thus the arc of the unit circle between successive points above the X-axis is 2 arctan(10−k). It follows that the number of collisions is the greatest integer 344 in 2π/(2 arctan(10−k)) = π/ arctan(10−k). Since 10−k is a small angle, the approximation tan(10−k) ∼10−k is a very good one (the difference is about the square of the angle, that is, of order only 10−2k). Thus when k is large we expect the number of collisions to be the greatest integer in π/10−k = 10kπ, but that is precisely the first k+1 digits of π! We’re a bit lucky that this already works for k = 1 (indeed, in binary artihmetic, it fails for k = 1, where the initial digits rule for m = 4 gives 6, but the true count is 5). But as k goes up the approximation above gets better so rapidly that there’s only a wisp of a chance that it ever fails. What would it take to fail? Roughly speaking, it would miss by 1 if for some k, digits k+1 through 2k of the decimal expansion of π were all 9’s. If, as most mathematicians believe, the digits of π behave like a random sequence, this is ludicrously unlikely. (In 2013 A.J. Yee and S. Kondo computed the first 12.1 trillion digits of π, with no more than a dozen 9’s in a row anywhere, about what you’d expect.) But no one has been able to prove that the digits of π continue to behave randomly, and even if they did the above glitch could occur. So it’s not a theorem. But you could bet your life on it. We conclude with something that really is a theorem. Let’s think of a polyhedron as any solid in three-dimensional space with flat faces. It doesn’t have to be convex and could even have holes in it. To each face F, we associate a vector vF perpendicular to the face, pointing outward, and with length proportional to the area of the face. Theorem. For any polyhedron, the sum over all faces F of the vectors vF is zero. 345 Proof. Pump the polyhedron full of air! The pressure on each face F will be proportional to the area of F, and is exerted outwardly perpendicular to F; in other words, it is represented by vF . If the sum of these forces were not zero, the polyhedron would move of its own accord. And we can’t have that, can we? ♡ 346 Chapter 21 Back from the Future Nothing says you have to try to solve a puzzle by starting with the premise, then working straight toward the answer. Often starting from the end—“retrograde analysis” is the fancy term for it—will make things much easier. (Are you among those that think solving a maze any way other than by drawing a line from “start” to “finish” is tantamount to cheating? That’s fine if that makes it more fun for you, but I advise against applying this strategy to every puzzle.) Consider the following classic. Portrait A visitor points to a portrait on the wall and asks who it is. “Brothers and sisters have I none,” says the host, “but that man’s father is my father’s son.” Who is pictured? Solution: Where to start? At the end. Who is “my father’s son”? Could be the host, if male. Wait, the host has no brothers and sisters, so it is the host. Substituting, we get “that man’s father is me,” so the face in the picture is the host’s son. Simple! Another classic, but maybe with a twist to the solution that you haven’t seen: Three-Way Duel Alice, Bob, and Carol arrange a three-way duel. Alice is a poor shot, hitting her target only 1/3 of the time on average. Bob is better, hitting his target 2/3 of the time. Carol is a sure shot. They take turns shooting, first Alice, then Bob, then Carol, then back to Alice, and so on until only one is left. What is Alice’s best course of action? 347 Solution: Let’s think about the situation after Alice takes her turn. If only Carol survives Alice’s turn, Alice is doomed. If only Bob survives, Alice has probability p of survival where p = 1 3( 1 3 + 2 3p), giving p = 1/7, not very good. If both Bob and Carol survive, things are much better because they will aim at each other, with only one survivor, at whom Alice gets to shoot first. Thus, in that situation, her survival probability is more than 1/3. It follows that Alice doesn’t want to kill anyone, and her best course of action is to shoot to miss! But wait. We’e been assuming the others wouldn’t do that, but now that we have allowed Alice the option of abstaining, we must surely allow it to the others. They can work out that whenever three duelers are still alive, Alice will never aim to kill. Again applying retrograde analysis, if it gets to Carol’s turn with no one dead, should she kill Bob? If Bob tried to shoot her, then yes. But if Bob shot into the air, thereby suggesting a willingness to do so indefinitely, then Carol should do the same—that way, no one’s life is at risk and when the ammunition runs out, everyone can go home and do math puzzles. Going back one turn, we deduce that Bob should indeed shoot into the air and the whole duel will be a dud. Thinking back on it, it’s pretty obvious that if the highest priority for all three parties is to stay alive—which is what we’ve been assuming—then the duel should never have been arranged. Here’s an application of retrograde analysis to experimental design: Testing Ostrich Eggs In preparation for an ad campaign, the Flightless Ostrich Farm needs to test its eggs for durability. The world standard for egg-hardness calls for rating an egg according to the highest floor of the Empire State Building from which the egg can be dropped without breaking. Flightless’ official tester, Oskar, realizes that if he takes only one egg along on his trip to New York, he’ll need to drop it from (potentially) every one of the building’s 102 floors, starting with the first, to determine its rating. How many drops does he need in the worst case, if he takes two eggs? Solution: Let’s look ahead at the position Oskar will be in when he’s down to one egg. At that point there will be some minimum possible egg rating m (namely, the highest floor from which the first egg survived) and some maximum possible egg rating M (one less than the floor from which the first egg went splat). That adds up to M −m drops. The plan for dropping the first egg will be some increasing sequence of floors, say f1, f2, . . . , fk. The spaces between the fi’s will be decreasing, since (assum-348 ing you want to prevent having to make more than some fixed number d of drops) the more drops made with the first egg, the fewer remain for the second. For example, suppose Oskar’s first drop of the first egg is from the 10th floor. If it breaks, he’s got 9 floors to try with the 2nd egg, for a total of 10 drops. But then to prevent having to make 11 drops, he’d have to make his second drop of egg number one from the 19th floor, allowing for 8 more drops of the 2nd egg, the 3rd from the 27th floor, 4th from the 34th, 5th from the 40th, 6th from the 45th, 7th from the 49th, 8th from the 52nd, 9th from the 54th, and 10th from the 55th. That’s his ten drops, so he’d be OK only if the building had at most 55 floors. How high must Oskar start, to make it to floor 102? Easy way to figure this out: Start at the other end! Add 1 + 2 + 3 + · · · until you reach or exceed 102, and your last addend, which proves to be 14, is Oskar’s starting floor—and also his guaranteed maximum number of drops. Later drops of the first egg, until it breaks, are at floors 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, and 102. You can get 14 another way by deriving or remembering the formula 1+2+ 3 + · · · + k = k(k+1)/2. (To derive this, note that there are k addends and the average size of an addend is (k+1)/2.) Then solve the equation k(k+1)/2 = 102 and round k up to the next integer. If there are more than two eggs, you can extend this retrograde analysis in a similar way. For example, for three eggs, you compute the numbers k(k+1)/2 (directly or by adding) and then sum these to reach 102: 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 = 120 where 36 = 8(8+1)/2. So Oskar drops the first egg at floor 36, then 64, then 85. If it breaks at (say) 64, he’s then playing the two-egg game in a 28-story building (floors 36 through 63 of the Empire State Building) which requires a maximum of 7 more drops. The total maximum number of drops for 349 three eggs is nine. How about when the number of eggs is unlimited? If you’re a computer scientist, then you may already have guessed that binary search is the answer: The next drop is made at the midpoint of the current range of ratings. For example, if the building originally had 63 floors (so that the rating could be anything from 0 to 63), the first drop should be from floor 32. If the egg breaks the new range is 0 to 31, else 32 to 63. Since 102 is between 64 = 26 and 128 = 27, Oskar might need as many as 7 drops. There’s a whole class of puzzles where retrograde analysis is the first thing you should try: namely, game puzzles. Suppose you are presented with a two-person, alternating move, deterministic, full-information game. That’s a lot of adjectives but they describe an enormous number of popular games, from chess, checkers, go, hex, and reversi to nim and tic-tac-toe. We’ll look at some games of that sort in which no draw is possible. In such a game, a “P” position is one that, if you can create it with your move, enables you to force a win. (Think of “P” as standing for “Previous player wins.”) The other positions are “N” positions (“Next player wins”) and those are the ones you would like passed to you by your opponent. It follows that from an N position there is always a move that creates a P position, and from a P position, all moves lead to an N position. This completely defines the N and P positions and tells you how to play optimally. But these rules operate backwards in time; to classify positions this way you must start from the end of the game and work backward. Let’s try this on the next puzzle. Pancake Stacks At the table are two hungry students, Andrea and Bruce, and two stacks of pancakes, of height m and n. Each student, in turn, must eat from the larger stack a non-zero multiple of the number of pancakes in the smaller stack. Of course, the bottom pancake of each stack is soggy, so the player who first finishes a stack is the loser. For which pairs (m, n) does Andrea (who plays first) have a winning strat-egy? How about if the game’s objective is reversed, so that the first player to finish a stack is the winner? Solution: Let m and n be the two stack sizes. You (if you are playing) are immediately stuck just when m = n; this is therefore a P position. If m > n and m is a multiple of n, you are in a winning N position because you can then eat m−n pancakes from the m-stack and reduce your opponent to the above P position. In particular, if the short stack has just one pancake, you are in great shape. What if m is close to a multiple of n? Suppose, for example, that m = 9 and n = 5. Then you are forced to reduce to m = 4, n = 5 but your opponent must 350 now give you a 1-stack. If m = 11 and n = 5 you have a choice but reducing to m = 6, n = 5 wins for you. It’s beginning to look like you want to make the ratio of the stacks small for your opponent, forcing her to make the ratio big for you. Let’s see. Suppose the current ratio r = m/n is strictly between 1 and 2; then the next move is forced and the new ratio is 1 1−r. These ratios are equal only for r = ϕ = (1+ √ 5)/2 ∼ 1.618, the golden mean; since ϕ is irrational, one of the two ratios r and 1 1−r must exceed ϕ while the other is smaller than ϕ. Aha! Thus, when you present your opponent with r < ϕ she must make it bigger, then you make it smaller, etc., until she’s stuck with r = 1 and must lose! We conclude that Andrea wins exactly when the initial ratio of larger to smaller stack exceeds ϕ. In other words, the P positions of the game are exactly those with m/n < ϕ, where m ≥n. To see this, suppose m > ϕn, but m is not a multiple of n. Write m = an + b, where 0 < b < n. Then either n/b < ϕ, in which case Andrea eats an, or n/b > ϕ, in which case she eats only (a−1)n. This leaves Bruce with a ratio below ϕ, and faced with a forced move which restores a ratio greater than ϕ. Eventually, Andrea will reach a point where her ratio m/n is an integer, at which point she can reduce to two equal stacks and stick Bruce with a soggy pancake. But note that she can also, if desired, grab a whole pile for herself, thus winning the alternative game where the eater of a soggy pancake is the victor. Of course, if Andrea is instead faced with a ratio m/n which is strictly between 1 and ϕ, she is behind the eight-ball and it is Bruce who can force the rest of the play. We conclude that no matter which form of Pancakes is played, if the stacks are at heights m > n, Andrea wins precisely when m/n > ϕ. Only in the trivial case when the stacks are initially of equal height does it matter what the game’s objective is! Try the above approach on this similar game. Chinese Nim On the table are two piles of beans. Alex must either take some beans from one pile or the same number of beans from each pile; then Beth has the same options. They continue alternating until one wins the game by taking the last bean. What’s the correct strategy for this game? For example, if Alex is faced with piles of size 12,000 and 20,000, what should he do? How about 12,000 and 19,000? Solution: As in classical Nim and many other games, it’s easiest to try to characterize the P positions, as there are fewer of those. Once you know the P positions, the 351 correct strategy is automatic. If a player is in an N position, he or she makes a move that puts his or her opponent in a P position; that opponent must then move to another N position. In Chinese Nim, the empty position (no beans left) is a P position since the previous player has just won the game. Any position with one pile empty, or both piles of the same size, is an N position, since the empty position is reachable in one move. It’s not hard to deduce that the simplest non-empty P position is {1, 2}. After that, you can work out that {3, 5}, {4, 7}, and {6, 10} are P positions as well. What’s the pattern? Let {x1, y1}, {x2, y2}, . . . be the non-empty P positions, with xi < yi and xi < xj for i < j. Notice you cannot have xi = xj for i ̸= j because then a player could reduce the larger of yi and yj to the smaller, leaving another P position. Some thought will lead you to conclude that given {x1, y1} up to {xn−1, yn−1}, xn is the least positive number not among {x1, . . . , xn−1} ∪{y1, . . . , yn−1}, and yn = xn + n. Notice that this forces yn to be a higher number than any in the set {x1, . . . , xn−1} ∪{y1, . . . , yn−1}. The proof is by induction on n. You have already seen that xn can’t be among the numbers in {x1, . . . , xn−1}∪{y1, . . . , yn−1} and also that there can’t be more than one yn to go with this xn, so all you need to do is show that this {xn, yn} really is a P position. If {xn, yn} were an N position (for Alex, say), it must be that he can reduce it to {xi, yi} for some i < n; but he cannot get to this position by reducing the smaller pile or by reducing both piles by the same amount, because that would leave the difference of the two piles at n or more. Nor can he get there by reducing the larger pile, because then he would get another y for the same x. Thus {xn, yn} is indeed a P position. You now have the means to generate as long a list as you like of P positions. From this Alex’s strategy is easy to work out. If he is faced with {xi, yi} he removes a bean or two and hopes for an error. If he sees {xi, z} for z > yi, he reduces z to yi. If he sees {xi, z} with xi < z < yi, the difference d = z −xi is less than i; he takes from both piles to get down to {xd, yd} (if z = yj for some j < i he also has the option of just reducing xi to xj). If he sees {yi, z} with yi ≤z he can reduce z all the way down to xi, and may have other options as well. But it might take a while to generate enough P positions to decide what to do with thousands of beans in each pile. Is there a more direct way to characterize the P positions? Well, you know that for each n, xn is somewhere between n and 2n, because it is preceded by all the xi’s for i < n and some of the yi’s. It is reasonable to guess that xn is approximately equal to rn, for some ratio r between 1 and 2. If so, yn would be approximately rn + n = (r+1)n. If this holds up, it follows that the n xi’s between 1 and xn are more or less evenly distributed, and therefore a fraction r/(r+1) of them will have their corresponding yi below xn. Thus there are about nr/(r + 1) yi’s below xn, together with the n xi’s, adding up to xn numbers in all; making an equation 352 out of this gives n + n r r+1 = nr, which gives us r + 1 = r2, r = (1 + √ 5)/2, the familiar “golden ratio” (again!). If we were really lucky, it would turn out that for each n, xn is exactly ⌊rn⌋ (the greatest integer in rn), and yn = ⌊r2n⌋. In fact, this is the case, and you can prove it using the theorem at the end of this chapter. The key is that because r and r2 are irrational numbers that sum to 1, every positive integer can be uniquely represented either as ⌊rj⌋for some integer j, or ⌊r2k⌋for some integer k. Let’s use this to find Alex’s move in the example positions. Note that 12000/r is a fraction under 7417, and 7417r = 12000.9581 . . . so 12000 is an xi, namely x7417. The corresponding y7417 is ⌊7417r2⌋= 19417 so if the other pile has 20000 beans, Alex can win by taking 20000 −19417 = 583 beans away from it. If there are only 19000 beans in the other pile, Alex can win instead by reducing the piles simultaneously to {x7000, y7000} = {11326, 18326}. Since 19000 happens to be a yj, namely y2674, Alex can also win by reducing the x-pile to x2674 = ⌊2674r⌋= 4326. Turning the Die In the game of Turn-Die, a die is rolled and the number that appears is recorded. The first player then turns the die 90 degrees (giving her four options) and the new value is added to the old one. The second player does the same and the two players alternate until a sum of 21 or higher is reached. If 21 was hit exactly, the player who reached it wins; otherwise, the player who exceeded it loses. Do you want to be first or second? Solution: This game, which I call Turn-Die, yields nicely to retrograde analysis using P and N positions—but first you have to figure out what, exactly, “positions” are. In Turn-Die, to know your position you must certainly know the current sum; but you also need to know what numbers are available when you turn the die. A standard die has numbers that sum to 7 on opposite faces; that is, 1 opposite 6, 2 opposite 5, and 3 opposite 4. Thus, for example, if the die is now showing 1 or 6, the options are to turn it to 2, 3, 4, or 5 only. Thus, what you need to know for strategic purposes is what the current sum is, and whether the die is showing 1 or 6 (case A), 2 or 5 (case B), or 3 or 4 (case C). To do the retrograde analysis, it helps to make a picture of the positions. Write the numbers from 1 to 21 three times in parallel columns labeled A, B, and C; let’s say with the 1’s on the bottom, and the 21’s on top. You then have a picture of 63 positions (not all reachable), and can assign P’s and N’s to them from the top down. 353 Let’s see how this process would start (in other words, how the game ends). The 21’s are all P positions, since the previous player has now won the game. Position 20A is also a P position, since the 1 is unavailable to turn to; the player facing 20A will have to overshoot and lose. The other 20’s, 20B and 20C, are N positions. Continuing in this way, you eventually determine the state of the six possible opening positions, namely 1A, 2B, 3C, 4C, 5B, and 6A (boxed in the figure at the end of the chapter). It turns out that of these only 3C and 4C are P positions, so if you go first, you will with probability 2/3 start with an N position and be able to force a win. It’s perhaps worth noting that in some vague sense, you usually want to be the first player in a combinatorial game that begins in a random position. The reason is that the condition for being an N position (that some move leads to a P) is milder than the condition for being a P position (that every move leads to an N position). This is why you usually find more N positions than P. When the game starts in an N position, you want to be first to play. Game of Desperation On a piece of paper is a row of n empty boxes. Tristan and Isolde take turns, each writing an “S” or an “O” into a previously blank box. The winner is the one who completes an “SOS” in consecutive boxes. For which n does the second player (Isolde) have a winning strategy? Solution: The game seems confusing and impenetrable until you realize that the only way you can force a win on your next move is to oblige your opponent to play into the configuration S-blank-blank-S (henceforth to be called a “pit”). Thus, for example, Tristan can win when n = 7 by placing an S in the middle, then another at the end farther from Isolde’s response, to make a pit. After a move by each player outside the it, Isolde must play into the pit and lose. The same applies for any odd n greater than 7, as Tristan can play an S anywhere at least 4 spaces from the ends, then form a pit on one side or the other and wait. When n is even Tristan has no chance, as there will never be a time when Isolde has only pits to play in; when she moves, there is always an odd number of blank squares to play in. Instead, when n is even and large, Isolde wins by playing an S far from the ends and from Tristan’s first move. However, if Tristan begins with an O, Isolde cannot put an S next to it, so she needs extra room. In the n = 14 case, if Tristan writes an O in place 7 (of 1 through 14), Isolde’s best response is an S in position 11 (threatening to make a pit with an S in 14). Tristan can counter this with an O in 13 or 14 (or an S in 12), and now Isolde would like to make a pit with S in position 8 but can’t, as Tristan would then win with S in 6. 354 Thus n = 14 is a draw; Isolde needs n to be even and at least 16. To wrap up, Tristan wins when n is odd and at least 7, Isolde when n is even and at least 16; all other values of n lead to a draw with best play. Deterministic Poker Unhappy with the vagaries of chance, Alice and Bob elect to play a completely deterministic version of draw poker. A deck of cards is spread out face-up on the table. Alice draws five cards, then Bob draws five cards. Alice discards any number of her cards (the discarded cards will remain out of play) and replaces them with a like number of others; then Bob does the same. All actions are taken with the cards face-up in view of the opponent. The player with the better hand wins; since Alice goes first, Bob is declared to be the winner if the final hands are equally strong. Who wins with best play? Solution: You need to know a little about the ranking of poker hands for this puzzle: namely, that the best type of hand is the straight flush (five cards in a row of the same suit), and that an ace-high straight flush (known as a “royal flush”) beats a king-high straight flush and on down. That means if Bob is allowed to draw a royal flush, Alice’s goose is cooked. For Alice to have a chance, her initial hand must contain a card from each of the four possible royal flushes. The best card of each suit, for that purpose, is the 10, since it stops all straight flushes which are 10-high or better. Indeed, a moment’s thought will convince you that any hand of Alice’s containing the four 10s will win. Bob cannot now hope to get a straight flush better than 9-high. To stop Alice getting a royal flush, he must draw at least one high card from each suit, leaving room for only one card below a 10. Alice can now turn in four cards and make herself a 10-high straight flush in a suit other than the suit of Bob’s low card, and Bob is helpless. Alice has other winning hands as well—see if you can find them all! The next puzzle is a difficult one, asking you to solve a bluffing game (even the simplest of bluffing games need some machinery to analyze; imagine what poker might require!). For this we do need the concept of equilibrium—a pair of strategies, one for each player, with the property that neither player can improve her results by changing strategy, if the other player doesn’t change hers. Example: In Rock-Paper-Scissors, the unique equilibrium is achieved when both players choose each of the three options with equal probability. For many games, and all bluffing games, equilibrium strategies are (like the ones for Rock-Paper-Scissors) randomized. To find these strategies (which, by the way, are often but not always unique) you can take advantage of the following fact. Suppose the equilibrium strategy for Player I calls for her to choose among several options, say A1, . . . , Ak, each with some positive probability. Then each 355 option must give her the same expectation against Player II’s strategy. Why? because if the expectation of, say, option A3 were among the highest and better than that of some other options, she could improve her results by choosing A3 instead of randomizing; and this contradicts the definition of equilibrium. Bluffing with Reals Consider the following simple bluffing game. Louise and Jeremy ante $1 each and each is given a secret random real number between 0 and 1. Louise may decide to pass in which case the $2 pot goes to the player with the higher number. However, if she chooses, Louise may add another dollar to the pot. Jeremy may then “call” by adding another dollar himself, in which case the pot, now with $4 in it, goes again to the player with the higher number. Or Jeremy may fold, ceding the pot, with his $1 in it, to Louise. Surely Louise has the advantage in this game, or at least equality, since she can break even by always passing. How much is the game worth to her? What are the players’ equilibrium strategies? Solution: For each value x that she could get, Louise has to decide whether to pass or raise; and for each value y that Jeremy could get, he has to decide whether to call or fold if Louise raises. Thus, there are in principle infinitely many strategies for each—a pretty big infinity, too (technically, two to the power of the continuum). So to find equilibrium strategies, we’ll need to restrict our search space. A little thought will convince you that Jeremy needn’t consider any strategy not of the form “call just when y > q” for some fixed threshold q. The reason is that no matter what Louise does, a higher value of y cannot increase Jeremy’s incentive to fold. So let’s assume that Jeremy has picked such a threshold q and consider Louise’s best response when holding the value x. If she passes, she wins $1 when y < x (probability x) and otherwise loses $1, so her expectation is x · $1 + (1−x) · (−$1) = $(2x −1). Suppose x > q. Then when y < q Louise wins $1 whatever she does; when y > q the game is in Louise’s favor just when x is more than halfway from q to 1, that is, when x > (q+1)/2, so that’s when she should raise. What about when x < q? Should Louise ever raise? If she doesn’t, we know her expectation is $(2x −1), which is dismally low when x is small. If she does raise, she wins $1 when her bluff works, that is, when y < q, otherwise she loses $2, for a net expectation of q · $1 + (1−q) · (−$2) = $(3q −2). Setting 2x −1 = 3q −2 gives us x = (3q −1)/2 and tells us that Louise should raise whenever x < (3q −1)/2, as well as when x > (q+1)/2. 356 We could at this point compute (as a function of q) what this strategy brings, on average, to Louise’s coffers, and then minimize that (using calculus—ugh) to find Bob’s choice for his threshold q. But we know that at equilibrium, if Bob actually drew the number q, he should be indifferent to the choice of folding or calling. Of course if he folds he loses $1. If he calls, he wins $2 when Louise was bluffing and loses $2 otherwise. For this to cost Bob $1 on average, Louise’s probability of bluffing given that she raised has to be 1/4; that means 3q −1 2 = (1 −q)/2 3 , giving 9q −3 = 1 −q, q = 0.4. We conclude that Louise raises when x < 0.1 or x > 0.7, and Bob calls when y > 0.4. The value of the game is computed as follows: • When x < 0.1 Louise bluffs, winning $1 when y < 0.4 and losing $2 otherwise for a net gain of −$0.80. • When 0.1 < x < 0.7 Louise passes and, with her average holding of x = 0.4, earns $0.40 −$0.60 = −$0.20. • Finally, when x > 0.7 Louise breaks even on average when y > 0.7 as well, but she picks up $1 when y < 0.4, and $2 when 0.4 < y < 0.7 and Bob calls her raise. This gives her a net average profit of $1 even. Bottom line: Louise makes 0.1·(−$0.80)+0.6·(−$0.20)+0.3·$1 = −$0.08− $0.12 + $0.30 = $0.10. So the game is worth exactly a dime to Louise. The next puzzle, also involving equilibrium strategies, is equally hard to fully analyze—but we’re being asked for much less than that. Swedish Lottery In a proposed mechanism for the Swedish National Lottery, each participant chooses a positive integer. The person who submits the lowest number not chosen by anyone else is the winner. (If no number is chosen by exactly one person, there is no winner.) If just three people participate, but each employs an optimal, equilibrium, randomized strategy, what is the largest number that has positive probability of being submitted? Solution: Suppose k is the highest number any player is willing to play. If a player chooses k, he wins anytime the other two players agree, except if they agree on k. But if he chooses k+1, he wins anytime they agree, period. Hence, k+1 is a better play than k, and we cannot be in equilibrium. The contradiction shows that 357 arbitrarily high submissions must be considered—sometimes one should choose 1,487,564. The actual equilibrium strategy calls for each player to submit the number j with probability (1−r)rj−1, where r = −1 3 − 2 3 3 p 17 + 3 √ 33 + 3 p 17 + 3 √ 33 3 , which is about 0.543689. The probabilities for choosing 1, 2, 3, and 4 are, respectively, about 0.456311, 0.248091, 0.134884, and 0.073335. I do not know if our “Swedish lottery” was ever implemented or even se-riously considered for any official lottery, but don’t you think it should have been? Suppose a puzzle presents a process, that is, one that moves in time from one state to another—deterministically, randomly, or under your control. It’s often helpful to determine whether the process is reversible, in the sense that you can tell from the state you’re in at the moment what the previous state must have been. If there are only a finite number of possible states, then reversibility tells you that no matter what state you start in, you must return to that state. Why? You must return to some state, surely, but why the state you started with? The proof is by contradiction. Let S be the first repeated state, and say that this repetition—that is, the second occurrence of state S—takes place at time t. Reversibility says that the previous state must be some particular state R. But if S was not the starting state, then S’s first occurrence was also preceded by R—so R was repeated before S was, contradicting our choice of S. To apply this “reversibility theorem” the trickiest part is often figuring out what information should go into the “state.” Here’s a typical example. Pegs on the Corners Four pegs begin on the plane at the corners of a square. At any time, you may cause one peg to jump over a second, placing the first on the opposite side of the second, but at the same distance as before. The jumped peg remains in place. Can you maneuver the pegs to the corners of a larger square? Solution: Note first that if the pegs begin on the points of a grid (i.e., points on the plane with integer coordinates), then they will remain on grid points. In particular, if they sit initially at the corners of a unit grid square, then they certainly cannot later find themselves at the corners of a smaller square since no smaller square is available on the grid points. But why not a larger one? 358 Here’s the key observation: The jump step is reversible! If you could get to a larger square, you could reverse the process and end up at a smaller square, which we now know is impossible. Slightly trickier: Touring an Island Aloysius is lost while driving his Porsche on an island in which every intersection is a meeting of three (two-way) streets. He decides to adopt the following algorithm: Starting in an arbitrary direction from his current intersection, he turns right at the next intersection, then left at the next, then right, then left, and so forth. Prove that Aloysius must return eventually to the intersection at which he began this procedure. Solution: Between intersections Aloysius’ current state can be characterized by a triple consisting of the edge he’s on, the direction he’s going on the edge, and the type of his last turn (right or left). From this information you can determine Aloysius’ last move, and an application of our reversibility theorem does the rest. Trickier still: Fibonacci Multiples Show that every positive integer has a multiple that’s a Fibonacci number. Solution: We need to show that for any n, as we generate the Fibonacci numbers (1, 1, 2, 3, 5, 8, 13, 21, 34, etc.) we eventually find one that is equal to zero modulo n. We normally define the Fibonacci numbers by specifying that F1 = F2 = 1 and for k > 2, Fk = Fk−1 + Fk−2. As it stands, this does not seem like a reversible process; given the value (modulo n) of, say, Fk+1, we can’t immediately determine the value (again, modulo n) of Fk. But if we keep track of the value modulo n of two consecutive Fibonacci numbers, then we can go backwards by just subtracting. For example, if n = 9 and F8 ≡3 mod 9 and F7 ≡4 mod 9 then we know F7 ≡3 −4 ≡8 mod 9. Putting it another way, if we define Dk = (Fk mod n, Fk+1 mod n), then the Dk’s constitute a reversible process. But so what? We’d like to find a k such that one of the coordinates of Dk is zero, but how do we know it doesn’t cycle among pairs that don’t contain a zero? Aha, a second trick: Start the Fibonacci numbers with F0 = 0, that is, start one step ahead with 0, 1, 1, 2, 3, etc. Then D0 = (0, 1) and therefore eventually 359 (after at most n2 steps, in fact) Dk will cycle back to (0,1), which means that Fk is a multiple of n. If we take n = 9 as above, for instance, our Dk’s, beginning with D0, are (0,1), (1,1), (1,2), (2,3), (3,5), (5,8), (8,4), (4,3), (3,7), (7,1), (1,8), (8,0) and we can stop there with F12 = 144 ≡0 mod 9. The D sequence continues (0,8), (8,8), (8,7), (7,6), (6,4), (4,1), (1,5), (5,6), (6,2), (2,8), (8,1), (1,0), (0,1) so it cycles back in this case after only 24 steps. Light Bulbs in a Circle In a circle are light bulbs numbered clockwise 1 through n, n > 1, all initially on. At time t, you examine bulb number t (modulo n), and if it’s on, you change the state of bulb t+1 (modulo n); that is, you turn off the clockwise-next bulb if it’s on, and on if it’s off. If bulb t is off, you do nothing. Prove that if you continue around and around the ring in this manner, even-tually all the bulbs will again be on. Solution: We observe first that there is no danger of turning all the lights off; if a change is made at time t, bulb t (modulo n) is still on. Moreover, if we look at the circle just after time t, we can deduce the state of the bulbs before t (by changing the state of bulb t+1 if bulb t is on). Thus the process is reversible, provided we take care to include in the state information not only which bulbs are on and which off, but which bulb is the one whose state determined the last action. The number of possible states is less than n × 2n, thus is finite, and we can apply the above argument to conclude that we will eventually cycle back to the initial all-on state—moreover, we can even insist that that we reach such a point at a moment when we are slated to again examine bulb #1. In the next puzzle, reversibility is just a part of the solution. Emptying a Bucket You are presented with three large buckets, each containing an integral number of ounces of some non-evaporating fluid. At any time, you may double the contents of one bucket by pouring into it from a fuller one; in other words, you may pour from a bucket containing x ounces into one containing y ≤x ounces until the latter contains 2y ounces (and the former, x−y). Prove that no matter what the initial contents, you can, eventually, empty one of the buckets. Solution: There is more than one way to solve this puzzle, but our strategy here will be to show that the contents of one of the buckets can always be increased until one of the other buckets is empty. 360 To do this, we first note that we can assume there is exactly one bucket containing an odd number of ounces of fluid. This is true because if there are no odd buckets, we can scale down by a power of 2; If there are two or more odd buckets, one step with two odd buckets will leave them both even. Second, note that with an odd and an even bucket we can always do a reverse step, that is, get half the contents of the even bucket into the odd one. This is because each state of this pair of buckets can be reached from at most one state, thus if you take enough steps, you must cycle back to your original state; the state just before you return is the result of your “reverse step.” Finally, we argue that as long as there is no empty bucket, the odd bucket’s contents can always be increased. If there is a bucket whose contents are divisible by 4, we can empty half of it into the odd bucket; if not, one forward operation between the even buckets will create such a bucket. ♡ Our last “reversibility” puzzle is—there’s no other word for it—unbelievable. Ice Cream Cake On the table before you is a cylindrical ice-cream cake with chocolate icing on top. From it you cut successive wedges of the same angle θ. Each time a wedge is cut, it is turned upside-down and reinserted into the cake. Prove that, regardless of the value of θ, after a finite number of such operations all the icing is back on top of the cake! Solution: If you think you have a proof that no finite number of operations can restore all the icing when θ is an irrational angle, you have justification. When θ is irrational, that is, is not a rational multiple of 2π radians, every cut will be at a different angle. But the very first inverted wedge will create a boundary-line 361 on top (and bottom) of the cake between iced and un-iced areas. If we never again cut at that point, how can we erase that boundary-line? In fact, we can cut again at the boundary-line, because when a wedge is inverted, its iced/un-iced pattern is not only complemented but reversed. Thus, boundary-lines move. In analyzing this puzzle, and indeed many serious algorithmic problems as well, it helps to redefine the operation so that it is only the “state space”—here, the icing pattern on the cake—and not the operation itself, that changes from step to step. In this case, that means rotating the cake after each operation so that you are always cutting in the same place. We will use standard mathematical notation, reading angles counterclock-wise around the cake counting “east” as 0 radians. One operation will consist of cutting the cake at −θ and 0, flipping that piece over, then rotating the whole cake clockwise by angle θ. Suppose that k operations is the minimum needed to get all the way around the cake; that is, kθ ≥2π but (k−1)θ < 2π. Putting it another way, k is the greatest integer in 2π/θ. Let δ be the amount by which you overshoot the first cut, that is, δ = kθ −2π; then 0 ≤δ < θ. At this point we advise the reader to try some reasonable angle, say a bit more than π/4 (making k = 4). You will find that after only seven cuts (2k−1 in general), all new cuts are in the same places as old ones! Let S be the following set of 2k−1 angles, listed in counterclockwise order around the cake: S = {0, θ−δ, θ, 2θ−δ, 2θ, 3θ−δ, 3θ, . . . , (k−1)θ−δ, (k−1)θ}. The figure below shows the cuts in the case θ = 93.5◦, where k = 4 and δ = 4 · 93.5◦−360◦= 14◦. We claim that these are the angles of all cuts that you will ever make. In fact, it is not hard to see that the set S is closed under our operation. The angle 0 maps to itself, while the angle −θ = 2π −θ = (k−1)θ −δ maps to θ. The only line on the piece thus cut is at angle (k−1)θ, which then moves to θ −δ. The rest of the angles are shifted up by θ as the cake rotates, so mθ moves to (m+1)θ and mθ −δ moves to (m+1)θ −δ for 1 ≤m ≤k−2. All the angles in S are indeed cut, in just two passes around the cake, so S represents our set of cuts exactly. And here’s where we use reversibility: Since there are only 2k−1 potential cake slices and each can only have either all its 362 icing on top or all on bottom, there are only 22k−1 possible states of the cake. Our operation is completely reversible, so we must return to the all-icing-on-top state that we started with. In fact, we return after many fewer than 22k−1 steps. To see this, notice that the portions into which the cuts in S divide the cake come in only two sizes, δ and θ −δ, with k−1 of the former and k of the latter. (These sizes could be equal, but the types are still distinct.) A θ-wedge of cake consists of two of these portions, one of each type. One operation flips the next two portions, one of each type, reading clockwise around the cake. Notice that the portions of a given type remain in order after a flip, since the flip involves only two portions of different type. To get all the icing back on top, each portion must be flipped an even number of times; thus the number of operations required must be an even multiple of k−1 (to get the δ portions right) and simultaneously an even multiple of k (to get the θ−δ portions right). The smallest number of operations that fits the bill is 2k(k−1), and that is the precise answer—except when δ = 0, that is, when θ = 2π/k for some integer k, in which case only 2k operations are required. Notice that except in the δ = 0 case, to get all the icing on the bottom you’d need the number of flips to be simultaneously an odd multiple of k−1 and an 363 odd multiple of k, which is impossible since either k−1 or k must be an even number. So it never happens that all the icing is on the bottom. A certain very well known mathematician’s reaction, upon hearing the ice cream cake puzzle, was: “I find it hard to believe that the icing ever returns to the top. But, one thing I’m sure of: If it does, there must also be a time when it’s all on the bottom!” Moth’s Tour A moth alights on the “12” of a clock face, and begins randomly walking around the dial. Each time it hits a number, it proceeds to the next clockwise number, or the next counterclockwise number, with equal probability. It continues until it has visited every number at least once. What is the probability that the moth finishes at the number “6”? Solution: No surprise: It pays to think about how the process ends. Let’s generalize slightly and think about the probability that the moth’s tour ends at i, where i is any number on the dial except 12. Consider the first time that the moth gets within one number of i. Suppose this happens when the moth reaches i−1 (the argument is similar if it happens at i+1). Then i will be the last number visited if and only if the moth gets all the way around to i+1 before it gets to i. But the probability of this event doesn’t depend on i. So the probability of ending at 6 is the same as the probability of ending at any other number, other than 12 itself. Therefore the probability that the moth’s tour ends at 6 is 1/11. Boardroom Reduction The Board of Trustees of the National Museum of Mathematics has grown too large—50 members, now—and its members have agreed to the following reduction protocol. The board will vote on whether to (further) reduce its size. A majority of ayes results in the immediate ejection of the newest board member; then another vote is taken, and so on. If at any point half or more 364 of the surviving members vote nay, the session is terminated and the board remains as it currently is. Suppose that each member’s highest priority is to remain on the board, but given that, agrees that the smaller the board, the better. To what size will this protocol reduce the board? Solution: We start from the end, of course. If the board gets down to two members they will certainly both survive, as member 2 (numbered from most senior to newest) will vote to retain herself. Thus, member 3 will not be happy if the board comes down to members 1, 2, and 3; members 1 and 2 would then vote “aye” and eject him. It follows that at size 4, the board would be stable; member 3 would vote nay to prevent reduction to 1, 2, and 3, and member 4 would vote nay to save him/herself immediately. The pattern suggests that perhaps the stable board sizes are exactly the powers of 2; if so, the board will reduce to 32 members and then stop. Is this right? Suppose not. Let n be the least number for which the claim fails, and let k be the greatest power of 2 strictly below n. Since n is our least counterexample, the board is stable when it gets down to members 1 through k. Thus those k will vote “aye” until then, outvoting the remaining n−k unless n = 2k. Thus when n = 2k the k newer members must vote “nay” to save themselves, stopping the process when n is itself a power of 2. Thus n is not a counterexample after all, and our claim is correct. Note we have used not just retrograde analysis, but also induction and con-tradiction to solve this puzzle—not to mention consideration of small numbers. So this puzzle could easily have gone into at least three other chapters of the book. For our theorem we will look at a surprising (to some) fact about numbers, which we can formulate in terms of two steadfast blinkers. Suppose that two regular blinkers begin with synchronized blinks at time 0, and afterwards there is an average of one blink per minute from the two blinkers together. However, they never blink simultaneously again (equivalently, the ratio of their frequencies is irrational). The theorem then tells us that for every positive integer t, there will be exactly one blink between time t and time t+1. Theorem. Let p and q be irrational numbers between 0 and 1 that sum to 1. Let P be the set of real numbers of the form n/p, where n is a positive integer, and Q the set of reals of the form n/q. Then for every positive integer t, there is exactly one element of P ∪Q in the interval [t, t+1). Let’s first see why the theorem applies. Let p be the rate of the first blinker; that is, it blinks p times per second. Since the first blink will be at time 0, the 365 next will be at time 1/p, then 2/p, and so forth; that is, its blinking times after time 0 will be the set P. Similarly, the second blinker will blink at rate q and its blinking times will be 0 together with times in the set Q. Saying that together they blink an average of once a second is the same as saying p + q = 1, and if p/q (which we could write as p/(1−p) or (1−q)/q) is irrational than so are p and q. The conclusion of the theorem is then that there is exactly one blink in each interval between any positive integer and its successor. Why did I think that might be surprising? We know that we’ll see one blink on average in such intervals, but exactly one every time? Since the two rates are not rationally related, we know that there will be some moment when there are two blinks less than (say) one millionth of a minute apart. Yet, if you believe the theorem, wedged between those two blinks there must be an integer! But the theorem is true; let’s prove it. Proof. Of course, we start from the end. The key is to notice that the theorem’s conclusion is equivalent to the following statement: For any positive integer t, the number of blinks (by both blinkers together) after time 0 and before time t is exactly t−1. In symbols, \|(P ∪Q) ∩(0, t)\| = t−1. Why? Because then the number of blinks between time t and time t+1 has to be (t+1 −1) −(t −1) = 1. The number of times the first blinker blinks after time 0 and before time t is ⌊pt⌋, the greatest integer less than or equal to pt. Similarly, the second blinker blinks ⌊qt⌋times in that period. The total number of blinks in the open interval (0, t) is thus ⌊pt⌋+ ⌊qt⌋. We know pt + qt = (p + q)t = t is an integer, but pt and qt aren’t integers; so what can ⌊pt⌋+ ⌊qt⌋be? It’s strictly less than the integer pt + qt but can’t be less by as much as 2, because we only cut off two fractions between 0 and 1. Since it’s an integer itself, it must be exactly pt + qt −1 = t −1, and we are done! ♡ 366 367 368 Chapter 22 Seeing Is Believing Your mind’s eye is a powerful tool. It’s not a coincidence that when you suddenly understand something, you say “I see it now.” Drawing a picture—paper is allowed, too, or laptop or electronic pad—can be the magic ingredient that helps you solve a puzzle, even one that doesn’t seem at first to call for it. Meeting the Ferry Every day at noon GMT a ferry leaves New York and simultaneously another leaves Le Havre. Each trip takes seven days and seven nights, arriving before noon on the eighth day. How many of these cross-Atlantic ferries does one of them pass on its way across the pond? Solution: Draw it! Put New York on the left of your page and Le Havre on the right; imagine time running down the page. Then each ferry trip is a slanted line across the page, and you can verify that each line meets 13 others on the way. 369 Mathematical Bookworm The three volumes of Jacobson’s Lectures in Abstract Algebra sit in order on your shelf. Each has 2′′ of pages and a front and back cover each 1 4 ′′, thus a total width of 2 1 2 ′′. A tiny bookworm bores its way straight through from page 1, Vol I to the last page of Vol III. How far does it travel? Solution: Sounds like the answer should be 3 × 2 1 2 ′′ −2 × 1 4 ′′ = 7′′, that is, the total width of the three volumes less the front cover of Volume I and the back cover of Volume III. But visualize the three volumes lined up on your shelf. Where is the first page of Volume I? And the last page of Volume III? That’s right, the worm passes only through the middle volume and two additional covers, a total of 2 1 2 ′′ + 2 × 1 4 ′′ = 3′′. Yes, this means that in a sense the correct way to order multi-volume books on your shelf is right-to-left, not left-to-right. If you pull the books off the shelf together with the intent of reading them in one sitting (not recommended), they are then stacked correctly. Rolling Pencil A pencil with pentagonal cross-section has the maker’s logo imprinted on one of its five faces. If the pencil is rolled on the table, what is the probability that it stops with the logo facing up? Solution: A mental picture is all you need for this one. The pencil will end up with one of the five faces down, therefore none facing straight up—so the answer is zero. If you count facing partially up, make that 2/5. 370 Splitting a Hexagon Is there a (non-self-intersecting) hexagon that can be cut into four congruent triangles by a single line? Solution: Drawing random hexagons and trying to split them this way will give you a headache. Better: Start with four copies of some triangle, and try to abut them all to a single line segment, to make a hexagon. The triangles have a total of 12 sides; uniting three pairs of sides will get you down to nine sides, so you’ll need to get some additional pairs of triangle-sides to line up. This is most easily done if you use right triangles, in which case you want each side of the segment to support one long leg and one short leg, while the remaining legs join up perpendicular to the segment, as pictured below. Not exactly the kind of hexagon that first springs to mind, though! Circular Shadows I Suppose all three coordinate-plane projections of a convex solid are disks. Must the solid be a perfect ball? Solution: No, in fact you can take a ball and slice off a piece from it without affecting its coordinate-plane projections. Just pick a point on its surface that’s far from any coordinate plane through the ball’s center—for example, the point ( 1 √ 3, 1 √ 3, 1 √ 3) on the surface of the unit-radius ball centered at the origin. 371 Mark a small circle on the ball’s surface around your point and shave off the cap bounded by the circle. That creates a little flat spot on the sphere that will go undetected by the projections. If you really want to avoid wasting material when you 3D-print your solid, make it the convex hull of three unit disks centered at the origin, one in each coordinate plane. But you can also go the other way and construct a big object that contains every other object, convex or not, with the same disk-shaped coordinate projec-tions: Intersect three long cylinders of the same radius, each with its axis along a different coordinate axis. Trapped in Thickland The inhabitants of Thickland, a world somewhere between Edwin Abbott’s Flat-land and our three-dimensional universe, are an infinite set of congruent convex polyhedra that live between two parallel planes. Up until now, they have been free to escape from their slab, but haven’t wanted to. Now, however, they have been reproducing rapidly and thinking about colonizing other slabs. Their high priest is worried that conditions are so crowded, no inhabitant of Thickland can escape the slab unless others move first. Is that even possible? Solution: Yes, even if the inhabitants are regular tetrahedra. First, note that if you view a regular tetrahedron edge-on, as shown on the left of the figure below, you see a square. Arrange light and dark tetrahedra as shown, with the near edge of the dark ones running SW to NE, and of the light ones NW to SE. One of the two parallel planes enclosing Thickland contains the near edge of each tetrahedron, the other the far edge. At the moment the tetrahedra have lots of freedom; each can pop straight out of Thickland by just traveling perpendicular to the planes. But now let’s squash the tetrahedra together as shown. Now every dark tetrahedron is locked in by two light ones (its NW and SE neighbors) from above and two other light ones (its SW and NE neighbors) from below, and similarly for light tetrahedra. 372 So no one can escape unless at least two other tetrahedra move over. In fact, if you jam them in so that their faces touch, infinitely many tetrahedra will need to move just to get one out. Polygon Midpoints Let n be an odd integer, and let a sequence of n distinct points be given in the plane. Find the vertices of a (possibly self-intersecting) n-gon that has the given points, in the given order, as midpoints of its sides. Solution: Let the midpoints be given by M1, . . . , Mn, and suppose that indeed there are points V1, . . . , Vn which are the vertices of a polygon such that M1 is the mid-point of the side V1V2, M2 the midpoint of V2A3, and so forth, ending with Mn being the midpoint of VnV1. It’s clear that if we know any of the Vi (say, V1), the rest of the vertices are determined: we reflect V1 180◦around M1 to get V2, and continue. That will work fine if, when we finally reflect Vn around Mn, we find ourselves back at V1. But why should we? Let’s try it. Take an arbitrary point P1 in the plane, reflect it across M1 to get P2, then reflect P2 across M2 to get P3, etc., ending, say, at the point Pn+1. You won’t find that Pn+1 = P1 unless you’re fantastically lucky. But what happens if you do the procedure again, starting at Q1 = Pn+1? Then the vectors P1Q1, Q2P2, P3Q3, etc., are all the same, and since n is odd, you end back at P1. Aha! In that case, let’s do the procedure one more time, starting at the midpoint R between P1 and Q1. Now you finish at R, and the polygon has been constructed! It turns out that this solution is unique. When n is even, generally there is no solution; but if some starting point P1 works (by getting back to P1 after reflections) then any point will work, giving infinitely many solutions. Not Burning Brownies When you bake a pan of brownies, those that share an edge with the edge of the pan will often burn. For example, if you bake 16 square brownies in a square pan, 12 are subject to burning. Design a pan and divide it into brownie shapes to make 16 identically shaped brownies of which as few as possible will burn. Can you get it down to four burned brownies? Great! How about three? Solution: You can beat the squares with sixteen 14 × 1 rectangular brownies, arranged as in the figure below, only four of which lie on the boundary. It seems that you ought to be able to get down to three burnt brownies, but ordinary triangles won’t do the trick. What if you curve them? 373 The 3-burn shape in the figure is constructed using circular arcs with the property that the straight line distance between the end points is equal to the radius of the circle, the vertex angles being 4◦, 60◦, and 120◦. Fortuitously, The smallest angle is an integral number of degrees, because 15 divides 60. Protecting the Statue Michelangelo’s David, in Florence, is protected (on the plane) by laser beams in such a way that no-one can approach the statue or any of the laser-beam sources without crossing a beam. What’s the minimum number of lasers needed to accomplish this? (Beams reach 100 meters, say.) Solution: You’ll need a non-convex protected area around David, in order to protect the lasers themselves as well as David; in fact you’ll want every edge of your polygon to be incident to a concave corner. You can get this with a hexagon shaped like a three-pointed star. Arrange the lasers at the vertices of a regular hexagon around David, beamed outward to cross at three corners, as shown below. To see that six laser-beam sources are needed, let C be the convex closure of the protected area; it has at least three sides. At each vertex of C, at least two beams must cross, but no beam can contribute to two vertices because then the laser source for that beam would lie outside the protected area. So there must be at least 3 × 2 = 6 lasers. Gluing Pyramids A solid square-base pyramid, with all edges of unit length, and a solid triangle-base pyramid (tetrahedron), also with all edges of unit length, are glued together by matching two triangular faces. How many faces does the resulting solid have? 374 Solution: This problem showed up in 1980 on the Preliminary Scholastic Aptitude Exam (PSAT), but, to the embarrassment of the Educational Testing Service, the answer they marked as correct was wrong. A confident student called the ETS to task when his exam was returned to him. Luckily for us, the correct answer boasts a marvelous, intuitive proof. The square-base pyramid has five faces and the tetrahedron four. Since the two glued triangular faces disappear, the resulting solid has 5+4−2 = 7 faces, right? This, apparently, was the intended line of reasoning. It may have occurred to the composer that in theory, some pair of faces, one from each pyramid, could in the gluing process become adjacent and coplanar. They would thus become a single face and further reduce the count. But, surely, such a coincidence can be ruled out. After all, the two solids are not even the same shape. In fact, this does happen (twice): The glued polyhedron has only five faces. You can see this in your own head. Imagine two square-based pyramids, sitting side-by-side on a table with their square faces down and abutting. Now, draw a mental line between the two apices; observe that its length is one unit, the same as the lengths of all the pyramid edges. Thus, between the two square-based pyramids, we have in effect constructed a regular tetrahedron. The two planes, each of which contains a triangular face from each square-based pyramid, also contain a side of the tetrahedron; the result follows. ♡(Check the figure below if you find this hard to visualize.) 375 Precarious Picture Suppose that you wish to hang a picture with a string attached at two points on the frame. If you hang it by looping the string over two nails in the ordinary way, as shown below, and one of the nails comes out, the picture will still hang (albeit lopsidedly) on the other nail. Can you hang it so that the picture falls if either nail comes out? Solution: One of several ways to hang the picture is illustrated below, with slack so you can see better how it works. This solution requires passing the string over the first nail, looping it over the second, sending it back over the first nail, then looping it again over the second nail but counter-clockwise. There are also some non-topological solutions: For example, you can pinch a loop of the string between two closely spaced nails, assuming the nail-head width is not much larger than the string diameter. But why rely on friction when you can use mathematics? 376 Finding the Rectangles Prove that any tiling of a regular 400-gon by parallelograms must contain at least 100 rectangles. Solution: Walk across the tiling from any edge to the opposite edge. That path must cross a similar path from the 90-degree-away edges, at a rectangle. Since the sides of that rectangle are parallel or perpendicular to only four edges of the polygon, there must be at least 100 such rectangles. Tiling a Polygon A “rhombus” is a quadrilateral with four equal sides; we consider two rhombi to be different if you can’t translate (move without rotation) one to coincide with the other. Given a regular polygon with 100 sides, you can take any two non-parallel sides, make two copies of each and translate them to form a rhombus. You get 50 2 different rhombi that way. You can use translated copies of these to tile your 100-gon; show that if you do, you will use each different rhombus exactly once! Solution: Let ⃗ u be one of the sides of the 2n-gon; a ⃗ u-rhombus is any of the n−1 rhombi using ⃗ u as one of its two vectors. In a tiling, the tile next to a ⃗ u-side must be a ⃗ u-rhombus, as must the tile on the other side of that one, and so forth until we reach the opposite side of the 2n-gon. Notice that each step of this path proceeds in the same direction (i.e., right or left) with respect to the vector ⃗ u, as must any other path of ⃗ u-rhombi; but then there can be no other ⃗ u-rhombi, since they would generate paths with no way to close and nowhere to go. The similarly defined path for a different side ⃗ v must cross the ⃗ u, and the shared tile is of course made up of ⃗ u and ⃗ v. Can they cross twice? No, because 377 a second crossing would have ⃗ u and ⃗ v meeting at an angle greater than π inside the common rhombus. ♡ Tiling with Crosses Can you tile the plane with 5-square crosses? Can you tile 3-space with 7-cube crosses? Solution: Tiling the plane with 5-square crosses is a snap if you organize them into diag-onals, as shown on the left side of the figure below. To tile 3-space with 7-cube crosses, use five of the seven cubes of each of a lot of crosses to tile a unit-thickness flat slab, much as the flat crosses were used to tile the plane. However, between each pair of diagonal strips, leave blank a strip of cattycorner dominoes. 378 Each of these is filled in with one cube from below the slab, and one from above. ♡ Cube Magic Can you pass a cube through a hole in a smaller cube? Solution: Yes. To pass a unit cube through a hole in a second unit cube, it suffices to identify a projection of the (second) cube which contains a unit square in its interior. A square cylindrical hole of side slightly more than 1 can then be made in the second cube, leaving room through which to pass the first cube. You can then do the same, with even smaller tolerances, if the second cube is just a bit smaller than a unit cube. The easiest (but not the only) projection you can try this with is the regular hexagon you get by looking along a body diagonal of the cube. Letting A be the projection of one of the visible faces onto the plane, we observe that its long diagonal is the same length ( √ 2) as a unit square’s, since that line has not been foreshortened. If we slide a copy of A over to the center of the hexagon, then widen it to form a unit square B, B’s widened corners will not reach the vertices of the hexagon (since the distance between opposing vertices of the hexagon exceeds the distance between opposing sides). 379 It follows that if we now tilt B slightly, all four of its corners will lie strictly inside the hexagon. ♡ Circles in Space Can you partition all of 3-dimensional space into circles? Solution: It’s an odd thing to ask—partitioning space into objects of lower dimension. But there’s nothing to prevent you from trying to do that, and in fact, in this case it is indeed possible. But how? We can certainly partition 3-space minus a point into spheres, namely all spheres having the missing point as their center. Can we partition a sphere into circles? If we leave out the poles, you can partition the rest into circles of latitude. In fact we can omit any two points P and Q, and partition the rest into circles whose centers lie on the plane through P, Q and the sphere’s center; for example, circles whose ratio of distance to P and Q is the same along the short route from P to Q as it is on the long route. With these observations we can construct a partition of all of 3-space into circles as follows. We start with a line of unit-radius circles on the XY-plane, one centered at (4n+1, 0) for each integer n. The key is that every sphere centered at the origin hits the union of those circles in exactly two points. And, very importantly, the origin is covered (by the unit circle centered at (1,0)). Thus, if we partition every sphere centered at the origin, with its two inter-section points omitted, into circles, then all those circles plus our line of circles on the plane partition space perfectly! ♡ 380 Invisible Corners Can it be that you are standing outside a polyhedron and can’t see any of its vertices? Solution: Yes. Imagine six long planks arranged so that they meet in their middles to form the walls of a cubical room, but do not quite touch. From the middle of this room you won’t be able to see any vertices. These planks can easily be hooked up far from the room to form a polyhedron, with the necessary additional vertices well hidden. If there is a plane separating you from the polyhedron, however, the answer to the puzzle’s question is “no.” For our theorem, I can’t resist using one of the most famous of all theorems with no-word proofs. A picture is all you need! A lozenge is a rhombus formed from a pair of unit-edge triangles glued together along an edge. Suppose a regular hexagon with integer sides is tiled with lozenges. The tiles come in three varieties, depending on orientation. Theorem. Any tiling of a regular hexagon with integer sides by unit lozenges uses precisely the same number of lozenges of each of the three orientations. Proof. 381 382 Chapter 23 Infinite Choice Need an axiom? As often as not, the Axiom of Choice is the axiom of choice. The Axiom of Choice (abbreviated AC here and in logic books) says that given any collection of nonempty sets, you can choose an element from each set. Seems pretty reasonable, right? What’s to stop you? Indeed, if the collection is finite, there’s no problem (you can prove by induc-tion that one item can be selected from each set.) If the sets have distinguished objects, you can use them; for instance, among infinitely many pair of shoes, you could choose the left shoe from each pair. But suppose you have infinitely many pairs of socks? There is a name, actually, for the set of all ways to choose an element from each set: It’s called the product of the sets. The size of the product, if both the sets and the collection are finite, is the ordinary product of the sizes of the sets. It seems unlikely that the product of an infinite collection of nonempty sets would suddenly decide to be empty, but that can happen if you don’t have AC. And if you’re a mathematician, you might find it handy to have tools like Zorn’s Lemma, the well-ordering theorem, or the Hausdorff maximal principle (each equivalent to AC), as well as myriad other theorems that require AC without being obvious about it. On the other hand, AC has some pretty weird consequences. The most famous of these is the Banach-Tarski paradox: A unit ball in 3-space can be partitioned into five subsets which can then be re-assembled (using only ro-tations and translations) to form two unit balls! Mathematicians are used to having to get around problems like this by restricting attention to “measurable” sets. But, as you will see later, there are other bizarre consequences of AC that mathematicians are less familiar with. So is AC true or false? I like to tell my students that although neither the axiom of choice nor its negation can be disproved, either can be made to look ridiculous. Both AC and “not AC” are consistent with the usual axioms of set theory, so you have a choice; as a puzzle solver you will mostly want to choose AC. (Assumptions like the Axiom of Determinacy that are not consistent with 383 AC can occasionally come in handy, but we will not pursue such contingencies here.) You might reasonably ask: How do collections of sets that don’t have ob-vious “choice functions” arise? One way is in the consideration of equivalence relations. A (binary) relation is technically a set of pairs, but you can think of it as a property that may or may not be true of a given pair of items. For example, among people, “x knows y” is a relation; so is “x and y are siblings.” The latter of these is a symmetric relation, one that is satisfied by the pair (x, y) if it is satisfied by (y, x). It is also a transitive relation: If (x, y) and (y, z) satisfy the relation, so must (x, z). Suppose we change the relation to “x and y are either full siblings, or are the same person.” Then the relation is also reflexive, meaning that (x, x) satisfies the relation for any x. Relations that are symmetric, transitive, and reflexive are said to be equivalence relations. If a set has an equivalence relation, the set partitions neatly into equivalence classes: Subsets whose elements are all related to one another but not to anyone outside the subset. For example, for the above relation, the equivalence classes are the sets of people that share a given pair of parents. Suppose humans are forced to settle a new planet and it’s decided that exactly one person from each set of siblings will go. Such a collection is called a set of representatives of the equivalence relation. In this case we don’t need to apply AC to get a set of representatives; for example, we could always send the eldest of the siblings (and anyway there are only finitely many sibling sets that we know of). But, generally, we may indeed need AC to get a set of representatives. Con-sider, for example, the set of all (countably) infinite binary sequences. Suppose (a1, a2, . . . ) is one such sequence and (b1, b2, . . . ) is another. Say that these two sequences are related if for all but finitely many indices i, ai = bi. You can easily verify that this relation is symmetric, transitive, and reflexive, therefore is an equivalence relation. Thus the set of all infinite binary sequences breaks up into (a lot of) equivalence classes. One of these, for example, is the set of all binary sequences that have only finitely many 1’s. According to AC, it is possible to pick one sequence from each equivalence class. From the aforementioned class of sequences with only finitely many 1’s, you might want to pick the all-zero sequence. But for most classes there won’t be an obvious choice, so it’s hard to see how you can avoid using AC. Let’s go ahead and invoke AC to get our set of representatives. What good is it? Well, if you’re a prisoner. . . Hats and Infinity Each of an infinite collection of prisoners, numbered 1, 2, . . . , is to be fitted with a red or black hat. At a prearranged signal, all the prisoners are revealed to one another, so that everyone gets to see all his fellow prisoners’ hat colors—but no 384 communication is permitted. Each prisoner is then taken aside and asked to guess the color of his own hat. All the prisoners will be freed provided only finitely many guess wrongly. The prisoners have a chance to conspire beforehand; is there a strategy that will ensure freedom? Solution: If we associate “red” with the digit 1 and “black” with 0, assignments of hats to prisoners correspond precisely to infinite binary sequences. Suppose that the prisoners believe in AC and have the wherewithal to agree in advance on a set of representatives for the equivalence relation described above. Thus, for each set S of related sequences, they have identified a particular member. When the curtain goes up, prisoner i sees the whole binary sequence except for its ith entry. That’s enough for him to identify the equivalence class S to which that sequence belongs. He then looks up (or remembers) the representa-tive that everyone has agreed upon for the set S, and guesses that his own hat color corresponds to the ith entry of that representing sequence. Consequently, everyone’s guess will correspond to this same representing sequence, and since that sequence is itself in S, only finitely many prisoners will have guessed wrong. ♡ It can be shown that under certain assumptions that contradict AC, the prisoners have essentially no chance, even if the warden is obligated to assign colors uniformly at random. Try this similar puzzle. All Right or All Wrong This time the circumstances are the same but the objective is different: The guesses must either be all right or all wrong. Is there a winning strategy? Solution: This version perhaps sounds even tougher, as only two successful outcomes are possible. But it does admit a cute solution if there are only finitely many prisoners (stop here and see if you can do the finite version before reading further). Yes, if there are only finitely many prisoners, they can simply decide to assume that the number of red hats is even. In other words, a prisoner who sees an odd number of red hats guesses that his own hat is red, otherwise that it is black. Then if, indeed, the number of red hats is even, the prisoners are all correct; if odd, all wrong. In the infinite case, this won’t work unless the warden assigns only finitely many hats of one color. But the prisoners’ agreed-upon set of representatives makes things easy. Each prisoner again identifies the equivalence class of the hat assignment and finds its representative; he then assumes that (say) the number 385 of discrepancies between the actual sequence and the representative is even, and guesses his own color accordingly. ♡ However, there’s a simpler solution: Everyone guesses “green”! The Axiom of Choice has the powerful consequence that any set can be well-ordered. (A set X is said to be well-ordered if it has an order relation on it with the property that all nonempty subsets of X have a lowest element.) In addition, it’s possible to ensure that the set of elements below any element of X has “cardinality” (i.e., size) less than the cardinality of X itself. We won’t prove these consequences here, but we’ll use them to solve the following innocent-sounding problem. Double Cover by Lines Let Lθ be the set of all lines on the plane at angle θ to the horizontal. If θ and θ′ are two different angles, the union of the sets Lθ and Lθ′ constitutes a double cover of the plane, that is, every point belongs to exactly two lines. Can this be done in any other way? That is, can you cover each point of the plane exactly twice using a set of lines that contains lines in more than two different directions? Solution: Since this problem appears here, you have a big clue that the answer is yes if you have AC. You can prove it by well-ordering all the points on the plane in such a way that the set of points below any point has cardinality less than 2ℵ0. (2ℵ0 stands for the cardinality of the set of all real numbers, also the set of angles, the set of points on the plane, and many other sets; it’s strictly greater than ℵ0, the cardinality of the set of whole numbers. The symbol ℵ, pronounced “aleph,” is the first letter of the Hebrew alphabet.) We start by picking three lines that cross one another, so that we already have our three directions. Let P be the lowest point in our well-ordering that isn’t already double-covered, and pick a line through P that misses the three points that are currently double-covered. Now repeat the process with a new point Q, this time avoiding a larger, but still finite, set of points that are already double-covered. We can do this until we have a countably infinite number of lines in our set. But why stop there? Until we’re done, there’s always a lowest point of the plane that hasn’t been double-covered yet. And because the number of points so far considered is less than 2ℵ0, there’s always an angle available that misses all points that are currently double-covered. This construction is perhaps disappointing in that it does not leave you with any geometry you can wrap your head around. No better construction is known. Let us return to direct applications of AC. If you were unimpressed by AC’s power to free prisoners, perhaps the next puzzle will get your attention. 386 Wild Guess David and Carolyn are mathematicians who are unafraid of the infinite and cheerfully invoke the Axiom of Choice when needed. They elect to play the following two-move game. For her move, Carolyn chooses an infinite sequence of real numbers, and puts each number in an opaque box. David gets to open as many boxes as he wants—even infinitely many—but must leave one box unopened. To win, he must guess exactly the real number in that box. On whom will you bet in this game, Carolyn or David? Solution: I can hear you thinking: “I’m betting on Carolyn! She can just put, say, a random real number between 0 and 1 in each box. David has no clue what’s in the unopened box, hence his probability of winning is zero.” In fact David has an algorithm that will guarantee him at least a 99% chance of winning, regardless of Carolyn’s strategy. You don’t believe me? That shows good judgment on your part, but it’s nonetheless true given AC. Here’s how it works. Before the game even starts, David proceeds much as the prisoners in Hats and Infinity did, but this time with sequences of real numbers instead of bits. (If you state the problem using bits instead of reals, David can use the prisoners’ set of representatives.) So, two sequences of reals x1, x2, . . . and y1, y2, . . . are related if xi ̸= yi for only finitely many indices i. Again this is an equivalence relation, and David invokes AC to choose from each equivalence class of real sequences one representative sequence. He can even show his set of represenatives to Carolyn; it won’t help her! Now Carolyn picks her sequence, let’s call it c1, c2, . . . , and boxes up the numbers. David takes the boxes and breaks them up into 100 infinite rows. We may as well assume the boxes in row 1 contain c1, c101, c201, etc., while row 2 begins with c2 and c102; finally row 100 contains c100, c200 and so forth. Each row is itself a sequence of real numbers. In a crucial step, David now chooses, uniformly at random, an integer be-tween 1 and 100. Say it’s 44. David first opens all the boxes in row 1. This sequence belongs to some equivalence class, say S1, whose representative he now looks up. The repre-sentative, say d1, d2, . . . , will differ from row 1 in finitely many positions, the last of which is, say, the 289th entry in each. David writes on a piece of paper “n1 = 289.” (If the sequences happen to be identical, he writes “n1 = 0.”) He repeats the procedure with row 2, this time getting a (probably) different class S2 with its own representative, and ends up writing down perhaps “n2 = 4,183,206.” This continues with every row except the 44th. When he’s done with row 100, David has written down values for the 99 numbers n1, n2, . . . , n43 and n45, . . . , n100. He sets m equal to one more than the maximum of these numbers. Now he finally turns to row 44, opening not all the boxes, but only those after the mth. This is enough for him to determine the equivalence class S44 387 of that row, and to retrieve its representative sequence, which we’ll denote by f1, f2, . . . . David now guesses that the mth box in row 44 contains the real number fm. So why does this work? Well, ultimately there is some largest index (n44) which represents the last position where row 44 and f1, f2, . . . disagree. Unless n44 ≥m, David’s guess will be correct. But m was greater than the maximum of n1, n2, . . . , n43 and n45, . . . , n100, so in order for n44 to equal or exceed m, n44 would have to be the unique largest of the 100 numbers n1, . . . , n100. What’s the probability of that? Since 44 was random, and only one of these numbers can be the unique largest, it’s at most 1/100. Bingo! If you find that you have to read this proof more than once to appreciate it, you are not alone. But it’s legitimate. The Axiom of Choice really does imply that David can win this game with any desired probability p < 1. Is there something wrong with your intuition, or is the Axiom of Choice suspect? You be the judge. If it’s any help, you can also use AC to predict the future (see, e.g., the Hardin and Taylor reference at the end of Notes & Sources). We now move on to some other puzzles involving infinity (but we’ll be as-suming the axiom of choice holds, in everything that follows). We start with a an infinite expression. Exponent upon Exponent Part I: If xxx··· = 2, what is x? Part II: If xxx··· = 4, what is x? 388 Part III: How do you explain getting the same answer to Parts I and II? Solution: If it means anything at all, xxx··· must be the limit of the sequence x, xx, xxx, xxxx , . . . , assuming that limit exists. Note that this expression is not the same as (. . . (((xx)x)x) . . . ). The exponent of the bottom x in the expression xxx··· is the same as the ex-pression itself; thus, if xxx··· = 2 then x2 = 2, x = √ 2. That’s one part of the puzzle out of the way. For the second part, similar reasoning tells you that if xxx··· = 4, then x4 = 4, thus x = 4 √ 2 = √ 2. Aha. Something’s wrong here. What exactly is √ 2 √ 2 √ 2··· ? It can’t be both 2 and 4. Which is it, if either? Or is it something else, or nothing at all? We need to look at the sequence √ 2, √ 2 √ 2, √ 2 √ 2 √ 2 , . . . , and determine whether it has a limit. In fact, it does; the sequence is increasing and bounded above. To show the former, we name the sequence s1, s2, . . . and prove by induction that 1 < si < si+1 for each i ≥1. This is easy: Since √ 2 > 1, s2 = √ 2 √ 2 > √ 2 = s1; and si+1 = √ 2 si > √ 2 si−1 = si. To get the bound, observe that if we replace the top √ 2 in any si by the larger value 2, the whole expression collapses to 2. (We could similarly show the sequence is bounded by 4, but of course if it’s bounded by 2 it’s automatically bounded by anything bigger than 2.) Now that we know the limit exists, let us call it y; it must indeed satisfy √ 2 y = y, hence √ 2 = y1/y. Looking at the equation x = y1/y, we observe (perhaps using a bit of calculus—sorry) that x is strictly increasing in y up to its maximum at y = e and strictly decreasing thereafter. Thus, there are at most two values of y corresponding to any given value of x, and for x = √ 2, we know the values: y = 2 and y = 4. Since our sequence is bounded by 2, we rule out 4 and conclude that y = 2. ♡ Generalizing the above argument, we see that xxx··· is meaningful and equal to the lower root of x = y1/y, as long as x ≤e1/e. For x = e1/e, the expression is equal to e, but as soon as x exceeds e1/e, the sequence diverges to infinity. This is why Part II of the puzzle fails: There is no x such that xxx··· = 4. 389 We have already noted that there are different infinities; the smallest is countable infinity, which we called ℵ0. A set is said to be countable if you can put its members into one-to-one correspondence with the positive integers. The set of rational numbers (fractions) is countable; reference to a particularly nice way of putting them into correspondence with the positive integers is given in Notes & Sources. The set of real numbers, however, is not countable, as observed by the brilliant Georg Cantor in 1878. Here’s a simple puzzle for which you can make use of countability. Find the Robot At time t = 0 a robot is placed at some unknown grid point in 3-dimensional space. Every minute, the robot moves a fixed, unknown distance in a fixed, unknown direction, to a new grid point. Each minute, you are allowed to probe any single point in space. Devise an algorithm that is guaranteed to find the robot in finite time. Solution: There are only countably many integer vectors (a, b, c, x, y, z), each coding a starting position (a, b, c) and displacement (x, y, z). List them as v1, v2, etc., and at time t, probe the point (a, b, c) + t(x, y, z) where (a, b, c, x, y, z) = vt. Figure Eights in the Plane How many disjoint topological “figure 8s” can be drawn on the plane? Solution: We could draw concentric circles on the plane with all possible positive real diameters, hence if the puzzle asked for circles instead of figure eights, the 390 answer would be “uncountably many,” or more precisely, “the cardinality of the reals.” However, we can only draw countably many eights. Associate with each 8 a pair of rational points (points of the plane with both coordinates rational numbers), one in each loop; no two figure eights can share a pair of points. Hence, the cardinality of our set of 8s is no greater than the set of pairs of pairs of rational numbers, which is countable. ♡ Was that too easy? Try the next one! Y’s in the Plane Prove that only countably many disjoint Y’s can be drawn in the plane. Solution: Here is a particularly neat proof (there is more than one way to do this). Asso-ciate with each Y three rational circles (rational center and radius) containing the endpoints, and small enough so that none contains or intersects any other arm of the Y. We claim that no three Ys can all have the same triple of circles; for, if that were so, you could connect the hub of each Y to the center of each circle by following the appropriate arm until you hit the circle, then following a radius to the circle’s center. This would give a planar embedding of the graph K3,3, sometimes known as the “gas-water-electricity network.” In other words, we have created six points in the plane, divided into two sets of three each, with each point of one set connected by a curve to each point of the other set, and no two curves crossing. This is impossible; in fact, readers who know Kuratowski’s Theorem will recognize this graph as one of the two basic nonplanar graphs. To see for yourself that K3,3 cannot be embedded in the plane without crossings, let the two vertex sets be {u, v, w} and {x, y, z}. If we could em-bed it without crossings the sequence u, x, v, y, w, z would represent consecutive vertices of a (topological) hexagon. The edge uy would have to lie inside or 391 outside the hexagon (let us say inside); then vz would have to lie outside to avoid crossing uy, and wx has no place to go. ♡ There’s no shortage of serious mathematical theorems that use some form of AC in their proofs; they’re rampant in algebra, analysis and topology. Here’s one that concerns only basic graph theory, which we make use of elsewhere in this volume. Recall that a graph is a set of points, called vertices, together with a sym-metric relation between pairs of distinct vertices. Two related vertices are said to be adjacent and to constitute an edge of the graph, usually illustrated by connecting the vertices by a line segment or curve. A graph is two-colorable if its vertices can be partitioned into two sets X and Y so that no two vertices in the same one of these sets are adjacent. Below is an example of a graph with its vertices two-colored (by black and white). The graph whose vertices are the digits 0 through 9 with 0 adjacent to 1, 1 to 2, 2 to 3, . . ., 8 to 9 and 9 to 0, can be colored with two colors by putting the even digits in X and the odd digits in Y . But if we try two-coloring the same construction but with an odd number of vertices—an “odd cycle”—it won’t work. In fact, no graph containing an odd cycle can be two-colorable. A famous, elementary theorem of graph theory says the converse. Theorem. If a graph contains no odd subset of vertices whose adjacencies in-clude those of an odd cycle, then the graph is two-colorable. The idea of the proof is this: Pick some vertex v and put it in X, then put all the vertices that are adjacent to v in Y , then all the vertices adjacent to them in X again, and so forth. The absence of odd cycles assures you that you will never find yourself trying to put the same vertex in both X and Y . If at any stage you find that you are finding no new adjacencies but haven’t used up all the vertices of the graph, pick an unassigned vertex v′, put it in X, and continue as before. 392 That’s all very fine, but if the graph is infinite and has infinitely many connected components, the part of the proof asking you to “pick an unassigned vertex” requires the Axiom of Choice. Here’s an example of such a graph: Let the vertices be all real numbers, with x adjacent to y if either \|x −y\| = 1 or \|x −y\| = √ 2. This graph has no odd cycles, because if you start a cycle at x and return to x, you will necessarily use an even number of 1-sized steps (some number up, and an equal number down) and likewise an even number of √ 2-sized steps; this is because no multiple of √ 2 is an integer. Nonetheless, under some axioms that contradict AC, this graph has no two-coloring! Indeed, if you try to construct a set of vertices that could serve as X, you will encounter only frustration. Three-colorings are easily found, though; it’s really quite reasonable, in graph-theoretic terms, to postulate that this graph has chromatic number 3, that is, can be properly colored with three colors but no fewer. 393 394 Chapter 24 Startling Transformation Often a puzzle that looks impenetrable becomes suddenly transparent if you just look at it a different way. Yes, that may require some creativity and imagination on your part; but great ideas, no matter how brilliant they appear at first, don’t arise out of nothing. A little experience may go a long way to sparking your next bit of genius. Sinking 15 Carol and Desmond are playing pool with billiard balls number 1 through 9. They take turns sinking balls into pockets. The first to sink three numbers that sum to 15 wins. Does Carol (the first to play) have a winning strategy? Solution: If Carol and Desmond record their sunk balls on the magic square below, then their objective is to fill all three squares in a row, column, or diagonal. Thus, they are playing Tic-Tac-Toe! As a consequence, Carol does not have a winning strategy; best play leads to a draw. 395 Slicing the Cube Before you is a circular saw and a 3×3×3 wooden cube that you must cut into twenty-seven 1 × 1 × 1 cubelets. What’s the smallest number of slices you must make in order to do this? You are allowed to stack pieces prior to running them through the saw. Solution: If you hold the cube together (carefully!) throughout, you can make two hor-izontal cuts all the way through the cube, then two vertical North–South cuts and two vertical East–West cuts—six cuts in all—to reduce the cube to 27 cubelets. Can you do it with fewer than six cuts? No, because the center cubelet has to be cut out on all six sides, and no cut can do more than one. Expecting the Worst Choose n numbers uniformly at random from the unit interval [0,1]. What is the expected value of their minimum? Solution: The answer is 1/(n+1). (And the average value of the next-smallest is 2/(n+1), and so forth.) This kind of problem can be solved using calculus, but when the solution is so nice, you might suspect there is a simpler way—and there is. We make use of the interval’s more symmetrical cousin, the circle. Choose n+1 numbers x0, . . . , xn uniformly and independently from a circle of circumference one. By symmetry, the expected distance between neighboring numbers is 1/(n+1). Cut the circle at x0 and open it up to form a line segment of length one. The remaining numbers x1, . . . , xn will be uniformly distributed on this line, and the least of these is the next xi to the right of x0 (assuming you opened up the circle clockwise). Thus its expected value is the same as the expected value of the distance between x0 and the next selected point to its right, which is 1/(n+1). The idea of bending a line segment around to make a circle, then exploiting the symmetry of the circle, will be useful later for proving our theorem. In the meantime, let’s look at a betting game. Next Card Red Paula shuffles a deck of cards thoroughly, then plays cards face up one at a time, from the top of the deck. At any time, Victor can interrupt Paula and bet $1 that the next card will be red. He bets once and only once; if he never interrupts, he’s automatically betting on the last card. What’s Victor’s best strategy? How much better than even can he do? (Assume there are 26 red and 26 black cards in the deck.) 396 Solution: We know that Victor’s expectation in this game, if he plays it well, is at least 0—because he can just bet on the first card, which is equally likely to be one of the 26 red cards or to be one of the 26 black cards. Or, he could wait for the last card, with the same result. But Victor is a smart guy and knows that if he waits for a point when he’s seen more black cards go by than red, he can bet at that moment and enjoy odds in his favor. Of course, he could wait until he’s seen all the black cards go by and then bet knowing he’s going to win, but that might not ever happen. Perhaps best is the following conservative strategy: He waits until the first time that the number of black cards he’s seen exceeds the number of red, and takes the resulting (probably small) odds in his favor. For example, if the very first card is black, he bets on the next one, accepting a positive expectation of (26/51) · $1 + (25/51) · (−$1) which is about 2 cents. Of course, though unlikely, it possible that the black cards seen never get ahead of the red ones, in which case Victor is stuck betting on the last card, which will be black. It seems like it might be difficult to compute Victor’s expectation from this or any moderately complex strategy. But there is a way—because it’s a fair game! Not only has Victor no way to earn an advantage, he has no way to lose one either: All strategies are equally ineffective. This fact is a consequence of the martingale stopping time theorem, and can also be established by induction on the number of cards of each color in the deck. But there is another proof, which I will describe below, and which must surely be in “the book.” (As readers may know or recall from Chapter 9, the late, great mathematician Paul Erd˝ os often spoke of a book owned by God in which is written the best proof of each theorem. I imagine Erd˝ os is reading the book now with great enjoyment, but the rest of us will have to wait.) Suppose Victor has elected a strategy S, and let us apply S to a slightly modified variation of the game. In the new variation, Victor interrupts Paula as before, but this time he is betting not on the next card in the deck, but instead on the last card of the deck. In any given position, the last card has precisely the same probability of being red as the next card. Thus, the strategy S has the same expected value in the new game as it did before. But, of course, the astute reader will already have observed that the new variation is a pretty uninteresting game; Victor wins if and only if the last card is red, regardless of his strategy. ♡ Magnetic Dollars One million magnetic “susans” (Susan B. Anthony dollar coins) are tossed into two urns in the following fashion: The urns begin with one coin in each, then the remaining 999,998 coins are thrown in the air one by one. If there are x coins in one urn and y in the other, magnetic attraction will cause the next 397 coin to land in the first urn with probability x/(x + y), and in the second with probability y/(x + y). How much should you be willing to pay, in advance, for the contents of the urn that ends up with fewer susans? Solution: You might very reasonably worry that one urn will “take over” and leave very little for the other. In my experience, faced with this problem, most people are unwilling to offer $100 in advance for the contents of the lesser urn. In fact the lesser urn is worth, on average, a quarter of a million dollars. This is because the final content of (say) Urn A is precisely equally likely to be any whole number of dollars from $1 all the way to $999,999. Thus, the average amount in the lesser urn is 1 999,999 (2 · $1 + 2 · $2 + . . . 2 · $499,999 + 1 · $500,000) which is $250,000 plus change. How can we see this? There are several proofs but here’s my favorite. Imag-ine that we shuffle a fresh-from-the-box deck of cards in the following careful manner. We place the first card, say the Ace of Spades, face down on the table. Then we take the second card, maybe the Deuce of Spades, and randomly place it over or under the ace. Now there are three possible slots for the Three of Spades; choose one at random, putting it on top of the current pile, under the current pile, or between the ace and deuce, each with probability 1/3. We continue in this manner; the nth card is inserted in one of n possible slots, each with probability 1/n. The result eventually is a perfectly random permutation of the cards. Now think of the cards that go in above the Ace of Spades as the coins (not counting the first) that go into Urn A, and the ones that go under as coins going into Urn B. The magnetic rule above describes the process perfectly, because if there are currently x−1 cards above the Ace of Spades (thus x slots) and y−1 cards under the Ace of Spades (thus y slots), the probability that the next card goes above the Ace of Spades is x/(x+y). 398 Since in the final shuffled deck the Ace of Spades is equally likely to be in any position, the number of slots ending up above it is equally likely to be anything from 1 to the number of cards in our deck, which for our puzzle is 999,999. Integral Rectangles A rectangle in the plane is partitioned into smaller rectangles, each of which has either integer height or integer width (or both). Prove that the large rectangle itself has this property. Solution: This problem is famous for its many proofs, each a kind of transformation. Here’s one proof that turns the picture into a graph. Place the big rectangle with its lower left corner at the origin of a Cartesian grid, and sides parallel to the coordinate axes. Let G be the graph whose vertices are all the points with integer coordinates that are also corners of small rectangles (indicated by shaded circles), plus the small rectangles themselves (indicated by white circles at their centers). Put an edge between a shaded vertex and a white vertex if the shaded vertex is a corner of the white vertex’s rectangle, as shown in the example. Every white vertex has 0, 2, or 4 neighbors (since one of its dimensions is an integer), and every shaded vertex has two or four neighbors, except that shaded vertices at the corners of the big rectangle have just one neighbor each. In any graph there must be an even number of vertices that have an odd number of neighbors, since every edge contributes two to the total number of neighbors of all vertices. Thus the origin can’t be the only vertex of G with an odd number of neighbors, and it follows that at least one of the other corners of the big rectangle must be an integer point as well, and we are done. ♡ 399 Laser Gun You find yourself standing in a large rectangular room with mirrored walls. At another point in the room is your enemy, brandishing a laser gun. You and she are fixed points in the room; your only defense is that you may summon bodyguards (also points) and position them in the room to absorb the laser rays for you. How many bodyguards do you need to block all possible shots by the enemy? Solution: You can certainly protect yourself if you have continuum many bodyguards, for example, by arranging them in a circle around yourself with, of course, your enemy outside the circle. That number can be reduced to a countably infinite number of bodyguards, because there are actually only countably many directions your enemy can fire in that will hit you. One way to see this is to view the room as a rectangle in the plane, with you at P and your enemy at Q. You can now tile the plane with copies of the room, by repeatedly reflecting the room about its walls, each copy containing a new copy of your enemy (see figure below). Every possible accurate shot by the enemy can be represented on this picture by a straight line from some copy of Q to P; every time such a line crosses a boundary between rectangles, the real laser beam would be bouncing off a 400 wall. In the figure, one such (dotted) line is indicated; a solid line shows the corresponding path of the laser beam back in the original room. Thus, you can pick a point on every shot line and station a guard there and so protect yourself with countably many guards. But back in the original room the shot-lines cross each other often; maybe if you place guards at the intersections, you can get away with a finite number. Would you believe 16? Your object is going to be to intercept every shot at its halfway point. To do this, you first trace a copy of the above plane tiling, pin it to the plane at your position P, and shrink the copy by a factor of two vertically and horizontally. The many copies of Q on the shrunk copy will be our bodyguard positions; they serve our purpose because each copy of Q on the original tiling appears halfway between it and you on the shrunk copy. In the second figure, the shrunk copy is in gray, and some virtual laser paths are indicated; you can see that they pass, at their half-way points, through the corresponding smaller dots in the gray grid. Of course, there are infinitely many such points, but we claim that they are all reflections of the right set of 16 points in the original room. Four of the points will already be in the original room; the four points in the room to the left of the original room can be reflected back to give four new points, and similarly for the room above the original one. Finally, the four points in the room above and to the left of the original room can be reflected twice to provide the last four points in the original room. In the third figure, the 12 new points (centers filled in gray) have been added in black to the original rectangle. A virtual laser path is added, with its corresponding real path crossing one of the new points. Since every room looks exactly like the original or one of the other three we just examined, all the bodyguard points in the plane are reflections of the 16 points we have now identified in the original room. Since every line from a copy of Q passed through a reflected bodyguard, the actual shot hits a “real” 401 bodyguard at its halfway point (if not sooner) and is absorbed. If the locations of P and Q are carefully chosen, some of the 16 bodyguard locations will coincide; but in general the full 16 are needed. Random Intervals The points 1, 2, . . . , 1000 on the number line are paired up at random, to form the endpoints of 500 intervals. What is the probability that among these intervals is one which intersects all the others? Solution: Amazingly, the answer is exactly 2/3, regardless of the number of intervals (as long as there are at least two). Suppose the interval endpoints are chosen from {1, 2, . . . , 2n}. We will label the points A(1), B(1), A(2), B(2), . . . , A(n−2), B(n−2) recursively as follows. Referring to points {n+1, . . . , 2n} as the right side and {1, . . . , n} as the left side, we begin by setting A(1) = n and letting B(1) be its mate. Suppose we have assigned labels up to A(j) and B(j), where B(j) is on the left side; then A(j +1) is taken as the left-most point on the right side not yet labeled, and B(j +1) as its mate. If B(j) is on the right side, A(j +1) is the right-most unlabeled point on the left side and again B(j+1) is its mate. If A(j) < B(j), we say that the jth interval “went right,” otherwise it “went left.” Points labeled A(·) are said to be inner endpoints, the others outer. It is easily checked by induction that, after the labels A(j) and B(j) have been assigned, either an equal number of points have been labeled on each side (when A(j) < B(j) and B(j) is on the right) or two more points have been 402 labeled on the left (when A(j) > B(j) and B(j) is on the left). When the labels A(n−2) and B(n−2) have been assigned, four unlabeled endpoints remain, say a < b < c < d. Of the three equiprobable ways of pairing them up, we claim two of them result in a “big” interval which intersects all others, and the third does not. In case A(n−2) < B(n−2), we have a and b on the left and c and d on the right, else only a is on the left. In either case, all inner endpoints lie between a and c, else one of them would have been labeled. It follows that the interval [a, c] meets all others, and likewise [a, d], so unless a is paired with b, we get a big interval. Suppose, on the other hand, that the pairing is indeed [a, b] and [c, d]. Neither of these can qualify as a big interval since they do not intersect each other; suppose some other interval qualifies, say [e, f], labeled by A(j) and B(j). When a and b are on the left, the inner endpoint A(j) lies between b and c, thus [e, f] cannot intersect both [a, b] and [c, d], contradicting our assumption. In the opposite case, since [e, f] meets [c, d], f is an outer endpoint (so f = B(j)) and [e, f] went right; since the last labeled pair went left, there is some k > j for which [A(k), B(k)] went left, but [A(k−1), B(k−1)] went right. Then A(k) < n, but A(k) < A(j) since A(k) is a later-labeled, left-side inner point. But then [A(j), B(j)] does not, after all, intersect [B(k), A(k)], and this final contradiction proves the claim. Infected Cubes An infection spreads among the n3 unit cubes of an n × n × n cube, in the following manner: If a unit cube has three or more infected neighbors, then it becomes infected itself. (Neighbors are orthogonal only, so each little cube has at most six neighbors.) Prove that you can infect the whole big cube starting with just n2 sick unit cubes. 403 Solution: You might recall the Infected Checkerboard puzzle where you were asked to show that you couldn’t infect all of an n×n checkerboard if you began with fewer than n sick squares. Essentially the same proof works in higher dimension to show that you need at least nd−1 initially sick squares to infect all of a d-dimensional hypercube of side n (where d infected neighbors infect a unit hypercube). But this time, it’s not obvious how to choose the initial nd−1 unit hypercubes (“cells”) to infect. In what follows we will consider only the 3-dimensional case, but the con-struction easily generalizes. Label the cells by vectors (x, y, z), with x, y, z ∈ {1, 2, . . . , n}, so that two cells will be neighbors if all their coordinates are the same except in one position, where their values differ by 1. Start by infecting all cells such that x + y + z ≡0 mod n. These cells form a “diagonal subspace” that consists of three separate chunks (the points whose coordinates add up to n, those whose coordinates sum to 2n, and the singleton (n, n, n)). From these initially sick cells, the infection fills up the big cube in a weird way. It barely succeeds, relying on many apparent coincidences that allow the growth to continue. Very different from the 2-dimensional case! It seems remarkable that the process manages to infect the whole big cube. To prove it really works, we transform the puzzle to a game. The players are you and your adversary—a doctor, say—who is trying to trap you, the infector. In the game, the doctor begins by putting you at some cell C = (x, y, z). Now repeat: The doctor chooses a coordinate (first, second, or third) and you move either forward or backward in that dimension (if the corresponding coordinate is 1 or n, you won’t have a choice). You win if you can reach some cell the sum of whose coordinates is a multiple of n; the doctor wins if she can keep you wandering forever. We now claim that if you have a winning strategy, then the big cube will indeed be fully infected. To see this, we first refine the claim to state that if you can win starting from cell C, then C itself will become infected. Since from C, the doctor has three choices of coordinate, a winning strategy for you must work for all three possibilities. This implies that your strategy also wins if started at any of the three neighbors of C that you were preparing to move to. By induction (over the number of steps to a win), all three of these neighbors of C will be infected, thus C will as well. The base case of the induction is when the starting point C has coordinates summing to k modulo n, in which case it is of course already sick. Now all we need is to provide you with a winning strategy; For any cell C = (x, y, z), let c be the number x+y+z+ 1 2 mod n. After the doctor chooses a coordinate, say the first, you increment x if it’s less than c (thus c goes up as well, though possibly around the corner from n−1 2 to 1 2) and decrement x if x > c. If you ever get to a cell where c = 1 2, you’ve won the game. So if x = 1 the algorithm will not ask you to decrement it. If x = n then x will always 404 be greater than c, so you will not be asked to increment it. It follows that the move prescribed by the algorithm is always legal, unless you have already won the game. We now assert that the doctor can’t force you to cycle. Suppose to the contrary that starting from C the algorithm cycles forever. Let I be the set of coordinates (among the first, second, and third) chosen infinitely often by the doctor. We may as well assume that you are past the point where any index not in I will ever be chosen. Let m be the biggest value ever encountered at a coordinate in I. Let J be the set of indices in I which are at the moment equal to m. If it ever happens that c > m, then you will be incrementing at every step, pushing c up until it snaps around the corner to 1 2 and you win. Therefore it must be that c is always below m. But then, whenever the doctor chooses a coordinate j ∈J, the jth coordinate must decrease to m−1. It follows that eventually J will disappear, leaving you forever with a smaller maximum value m. This can’t go on forever, so we have our contradiction. We conclude that the above algorithm will win the game for you, no matter where the doctor starts you off or how she chases you around. The existence of a winning strategy means that the infection really does capture the whole big cube, and we are done. ♡ Gladiators, Version I Paula and Victor each manage a team of gladiators. Paula’s gladiators have strengths p1, p2, . . . , pm and Victor’s, v1, v2, . . . , vn. Gladiators fight one-on-one to the death, and when a gladiator of strength x meets a gladiator of strength y, the former wins with probability x/(x + y), and the latter with probability y/(x+y). Moreover, if the gladiator of strength x wins, he gains in confidence and inherits his opponent’s strength, so that his own strength improves to x+y; similarly, if the other gladiator wins, his strength improves from y to x+y. After each match, Paula puts forward a gladiator (from those on her team who are still alive), and Victor must choose one of his to face Paula’s. The winning team is the one which remains with at least one live player. What’s Victor’s best strategy? For example, if Paula begins with her best gladiator, should Victor respond from strength or weakness? Solution: All strategies for Victor are equally good. To see this, imagine that strength is money. Paula begins with P = p1 + · · · + pm dollars and Victor with V = v1 + · · · + vn dollars. When a gladiator of strength x beats a gladiator of strength y, the former’s team gains $y while the latter’s loses $y; the total amount of money always remains the same. Eventually, either Paula will finish with $P + $V and Victor with zero, or the other way ’round. 405 The key observation is that every match is a fair game. If Victor puts up a gladiator of strength x against one of strength y, then his expected financial gain is x x + y · $y + y x + y · (−$x) = $0. Thus, the whole tournament is a fair game, and it follows that Victor’s expected worth at the conclusion is the same as his starting stake, $V . We then have q · ($P + $V ) + (1 −q) · $0 = $V, where q is the probability that Victor wins. Thus, q = V/(P +V ), independent of anyone’s strategy in the tournament. ♡ Here’s another, more combinatorial proof. Using approximation by rationals and clearing of denominators, we may assume that all the strengths are integers. Each gladiator is assigned x balls if his initial strength is x, and all the balls are put into a uniformly random vertical order. When two gladiators battle, the one whose topmost ball is highest wins (this happens with the required x/(x+y) probability) and the loser’s balls accrue to the winner. The surviving gladiator’s new set of balls is still uniformly randomly dis-tributed in the original vertical order, just as if he had started with the full set; hence, the outcome of each match is independent of previous events, as required. But regardless of strategy, Victor will win if and only if the top ball in the whole order is one of his; this happens with probability V/(P +V ). Gladiators, Version II Again Paula and Victor must face off in the Colosseum, but this time, confidence is not a factor, and when a gladiator wins, he keeps the same strength he had before. As before, prior to each match, Paula chooses her entry first. What is Victor’s best strategy? Whom should he play if Paula opens with her best man? Solution: Obviously, the change in rules makes strategy considerations in this game com-pletely different from the previous one—or does it? No, again the strategy makes no difference! For this game, we take away each gladiator’s money (and balls), and turn him into a light bulb. The mathematician’s ideal light bulb has the following property: Its burnout time is completely memoryless. That means that knowing how long the bulb has been burning tells us absolutely nothing about how long it will continue to burn. You may know that the unique probability distribution with this property is the exponential; if the expected (average) lifetime of the bulb is x, then the probability that it is still burning at time t is e−t/x. However, no formula is 406 necessary for this puzzle. You only need to know that a memoryless probability distribution exists. Given two bulbs of expected lifetimes x and y, respectively, the probability that the first outlasts the second is x/(x+y). To see this without calculus, consider a light fixture that uses one “x-bulb” (with average lifetime x) and one “y-bulb”; every time a bulb burns out, we replace it with another of the same type. When a bulb does burn out, the probability that it is the y-bulb is a constant independent of the past. But that constant must be x/(x+y), because over a long period of time, we will use y-bulbs and x-bulbs in proportion x : y. Back in the Colosseum, we imagine that the matching of two gladiators cor-responds to turning on their corresponding light bulbs until one (the loser) burns out, then turning off the winner until its next match; since the distribution is memoryless, the winner’s strength in its next match is unchanged. Substituting light bulbs for the gladiators may be less than satisfactory for the spectators, but it’s a valid model for the fighting. During the tournament, Paula and Victor each have exactly one light bulb lit at any given time; the winner is the one whose total lighting time (of all the bulbs/gladiators on her/his team) is the larger. Since this has nothing to do with the order in which the bulbs are lit, the probability that Victor wins is independent of strategy. (Note: That probability is a more complex function of the gladiator strengths than in the previous game.) More Magnetic Dollars We return to Magnetic Dollars, but we strengthen their attractive power just a bit. This time, an infinite sequence of coins will be tossed into the two urns. When one urn contains x coins and the other y, the next coin will fall into the first urn with probability x1.01/(x1.01 + y1.01), otherwise into the second urn. Prove that after some point, one of the urns will never get another coin! Solution: The really neat way to show that one urn gets all but finitely many coins is to employ those continuous memoryless waiting times that proved so useful in the previous puzzle. Look just at the first urn and suppose that it acquires coins by waiting an average of 1/n1.01 hours between its nth coin and its (n+1)st coin, where the waiting time is memoryless. Coins will arrive slowly and sporadically at first, then faster and faster; since the series P∞ n=1 1/n1.01 converges, the urn will explode with infinitely many coins at some random moment (typically about 4 days after the process begins). Now suppose we start two such processes simultaneously, one with each urn. If at some time t, there are x coins in the first urn and y in the second, then (as we saw with the gladiator-light bulbs) the probability that the next coin goes 407 to the first urn is 1/y1.01 1/x1.01 + 1/y1.01 = x1.01 x1.01 + y1.01 , exactly what it should be. Nor does it matter how long it’s been since the xth coin in the first urn (or yth in the second) arrived, since the process is memoryless. It follows that this accelerated experiment is faithful to the puzzle. However, you can see what happens now; with probability 1, the two explo-sion times are different. (For this, you only need to know that our memoryless waiting time has a continuous distribution.) But the experiment ends at the first explosion, at which time the other urn is stuck with whatever finite number of coins it had. ♡ Seems like kind of a scary experiment, doesn’t it? The slow urn never got to finish because, in effect, time ended. For our theorem, we look at a combinatorial object that arises with surprising frequency called a “parking function.” Imagine that there are n parking spaces 1, 2, . . . , n (in that order) marked on a one-way street. Cars numbered 1 through n enter the street in some order, each with a preferred parking spot (a random number between 1 and n). Each car drives to its preferred spot and parks there if the spot is available; otherwise, it takes the next available spot. If there are no empty slots beyond the filled preferred spot, the car must leave the street without parking; tough luck! We want to know: What is the probability that all n cars will be able to park? Theorem. The order in which the cars arrive makes no difference, that is, the issue of whether all cars find a spot depends only on what the cars’ preferred parking spots are. If each car’s preferred spot is independently chosen and uni-formly random, the probability of success is (n+1)n−1/nn. The total number of ways to assign preferred spots to the n cars is nn; we need to figure out how many of these will lead to everyone finding a spot. For example, if everyone prefers a different spot, all will be well. Or if everyone prefers slot #1. But not if everyone prefers slot #2. In general, we’re in trouble if there’s some k for which fewer than k cars want slots from 1 through k; with 408 some thought, you’ll be able to see that if there is no such k, everyone will get parked, no matter how the cars are ordered. Another way to say the same thing: If we order the preferred spots xi so that x1 ≤x2 ≤· · · ≤xn, then everyone gets parked (in which case the mapping from i to xi is called a parking function) provided xi ≤i for each i. (If the spot i ends up left open, then there must have been fewer than i cars wanting a spot between 1 and i.) This is a pretty natural mathematical condition; it would be nice to know the probability that it holds. But it turns out we don’t need to know the condition to prove the theorem. We transform the situation by adding an n+1st slot and imagining that the slots now close up to form a circle. We’ll let each car chose a preferred slot from 1 all the way to n+1; each car in turn goes to its preferred slot. Then, as before, if the slot is occupied it continues (clockwise, say) around the circle and parks in the first available slot. In this scenario every car will find a slot, and there will be one empty slot left over. If that empty slot happens to be slot n+1, the list of preferences would have worked in the original problem; otherwise, it would have failed. Now we take advantage of the circle’s symmetry: We notice that if all the car’s preferences were advanced by k, modulo n+1, then the whole processes would simply take place k slots clockwise from where it was; consequently, the final empty slot would also advance by k. But that implies that for every assignment of preferred slots, exactly 1 of its n+1 rotations works. We conclude that the probability that the final empty slot is slot n+1 is 1/(n+1). The total number of ways to assign preferences in our circular case is (n+1)n, hence the total number of good assignments is (n+1)n/(n+1) = (n+1)n−1. Note that, necessarily, none of those good assignments has any car preferring slot n+1. We conclude that the probability of success in the original problem is (n+1)n−1/nn. ♡ How big is this number? For large n, we can write (n+1)n−1/nn = (1 + 1 n + 1)n/n ∼e/(n + 1). For example, if there are 100 cars, the theorem pegs the probability of success at about 2.678%, close to e/101 ∼2.691%. One might say that it’s a pleasant surprise, but not shockingly lucky, if all the cars find parking spots. 409 410 Notes & Sources 1. Out for the Count Half Grown: Sent by Jeff Steif, of Chalmers University in Sweden. Powers of Two: Classic. Watermelons: Adapted from Gary Antonick’s New York Times “Number Play” blog, 11/7/12; but the puzzle appeared ealier in Kvant, June/August 1998, contributed by I. F. Sharygin. Bags of Marbles: Contributed by pathologist Dick Plotz, of Providence RI. Salaries and Raises: Idea from Rouse Ball . Efficient Pizza-Cutting: Subject of a MoMath (National Museum of Mathematics) event of 10/30/19, led by Paul Zeitz. Attention Paraskevidekatriaphobes: As far as I can tell this remarkable fact was first observed by Bancroft Brown (a Dartmouth math professor, like your author), who pub-lished his calculation in the American Mathematical Monthly Vol. 40 (1933), p. 607. My present-day colleague Dana Williams is the one who brought all this to my attention. The origin of superstition concerning Friday the 13th is usually traced to the date of an order given by King Philip IV of France (Philip the Fair), dismantling the Knights Templar. Meet the Williams Sisters: See also Hess . Rating the Horses: Suggested by Saul Rosenthal of TwoSigma. See also . Shoelaces at the Airport: Suggested in 2010 by Uzi Segal, to whom it occurred while navigating the long corridors at the Montparnasse metro station in Paris. Rows and Columns: This is a classical theorem, simple and surprising; I was reminded of it by Dan Romik, a math professor at the University of California, Davis. Donald Knuth, in Vol. III of The Art of Computer Programming, traces the result to a footnote in a 1955 book by Hermann Boerner. Bridget Tenner, a student of famed combinatorialist Richard Stanley, wrote a paper called “A Non-Messing-Up Phenomenon for Posets” generalizing this theorem. Three-way Election: Dreamed up by Ehud Friedgut of the Weizmann Institute, Israel. Sequencing the Digits: Remembered from the Putnam Exam, many years ago. Faulty Combination Lock: Suggested to me by Amit Chakrabarti of Dartmouth; it had been proposed for the International Mathematical Olympiad in 1988 by East Germany. The optimality proof given (supplied by Amit) doesn’t quite work for 8 replaced by 10, and fails increasingly thereafter. Thus the general puzzle, when the lock has n numbers per dial, is unsolved as far as I know. Losing at Dice: I discovered this curious fact accidently, 40 or so years ago, while construct-ing homework problems for an elementary probability course at Emory University. Splitting the Stacks: From Einar Steingrimsson, University of Strathclyde, Glasgow. North by Northwest: Original. Early Commuter: From the one and only Martin Gardner. Alternate Connection: From the First Iberoamerican Mathematical Olympiad . Bindweed and Honeysuckle: From Karthik Tadinada, but recast by me in recollection of a delightful tragicomic song by Michael Flanders & Donald Swann. theorem: The sequence we associated with each tree is known as the Pr¨ ufer code for the tree. Besides helping us prove the theorem, the Pr¨ ufer code gives us additional information. For example, you can easily check that the number of neighbors of a vertex of the tree is 411 always one more than the number of times it appears in the code; leaves do not appear at all. From this we could deduce that when n is large, the number of leaves in a random tree is about n/e, that is, about 37% of the vertices are leaves. 2. Achieving Parity Bacterial Reproduction: Based on Supplementary Problem C/4, , p. 14. Fourth Corner: Brought to my attention by Paul Zeitz. Unanimous Hats: My version of a classic. Half-right Hats: My version of a classic. Red and Blue Hats in a Line: Passed on to me by Girija Narlikar of Bell Labs, who heard it at a party. Note that an analogue of the given solution works with any number k of hat colors. The colors are assigned numbers from 0 to k−1, say, and the last prisoner in line guesses the color corresponding to the sum, modulo k, of the color-numbers of the hats in front of him. As before, all other prisoners are saved, if they’re careful with their modular arithmetic. Prisoners and Gloves: Suggested by game theorist Sergiu Hart of the Hebrew University. Even-sum Billiards: Inspired by a problem from John Urschel’s column, Chameleons: Boris Schein, an algebraist at the University of Arkansas, sent me this puzzle; it may be quite old. On one occasion it was given to an 8th-grader in Kharkov, on another to a young Harvard grad interviewing at a big finance firm; both solved it. Missing Digit: Lifted from Elwyn Berlekamp and Joe Buhler’s Puzzles Column in Emissary, Spring/Fall 2006; they heard it from the number theorist Hendrik Lenstra. Subtracting Around the Corner: When I was in high school, a substitute math teacher told my class that some WWII prisoner of war entertained himself by trying various sequences of four numbers to see how long he could get them to survive under the above operations. The process (with four numbers) is sometimes known as a difference box or diffy box. Among real numbers, there is an (essentially) unique quad that never terminates . That quad is (0, 1, q(q −1), q), where q is the unique real root of q3 −q2 −q −1 = 0, namely, q = 1 3 1 + 3 q 19 + 3 √ 33 + 3 q 19 −3 √ 33 ∼1.8393 so q(q −1) ∼1.5437. Remarkably, the number 1/q arises in the solution to another, seemingly unrelated, puzzle in this volume. Uniting the Loops: This is one of a number of ingenious puzzles devised by writer/mathematician Barry Cipra, of Northfield Minnesota. theorem: Conjectured in and proved at Danny Kleitman’s 65th Birthday conference in August 1999 by Noga Alon, Tom Bohman, Ron Holzman and Danny himself . 3. Intermediate Math Boxing the Mountain State: My version of a classic. Monk on a Mountain: Ancient. Cutting the Necklace: Suggested by Pablo Sober´ on of Baruch College, City University of New York. For more information see . Three Sticks: Moscow Mathematical Olympiad 2000, Problem D4, contributed by A. V. Shapovalov . Hazards of Electronic Coinflipping: See also Fair Play in Chapter 19, Hammer and Tongs. The actual value of p works out to 1 2 ± rq 2 3 −3 4 . But there’s no need to limit yourself to three people; you can choose uniformly among n people, for any positive integer n, in a bounded number of flips. Johan W¨ astlund suggests the following method. Suppose you have enough power for f flips. You allot as many of the 2f outcomes as can be divided into n−1 equivalent sets; left over are at most n−2 of each level, that is, for each k at most n−2 outcomes involving exactly k heads. Now the IVT can be applied as in the 412 n = 3 case as long as the total probability of the left-out outcomes, at p = 1 2 , is at most 1 n . The latter is achieved when f is just over 2 log2(n) flips—only a factor of two away from the information-theoretic lower bound. Bugs on a Pyramid: Devised by Dr. Kasra Alishahi, of Sharif University; sent to me by Mahdi Saffari. Skipping a Number: Based on the Putnam exam problem (#A1, from 2004) that intro-duced the now-famous (in certain circles) Shanille O’Keal. Splitting a Polygon: From a Moscow Mathematical Olympiad in the 1990’s. Garnering Fruit: Sent to me by Arseniy Akopyan, author of the unique and delightful book Geometry in Figures. The first of the two solutions to the two-fruit version came from Pablo Sober´ on of Baruch College, City University of New York. theorem: Known since ancient times. 4. Graphography Air Routes in Aerostan: Moscow Mathematical Olympiad 2003, Problem A5, contributed by R. M. Fedorov . Spiders on a Cube: Passed to me by Matt Baker, Georgia Tech. Handshakes at a Party: Classic. Snake Game: From the 12th All Soviet Union Mathematical Competition, Tashkent, 1978. Bracing the Grid: Passed to me by geometer Bob Connelly of Cornell, based on . Competing for Programmers: From the Moscow Mathematical Olympiad 2011, but ap-parently changed by mistranslation. Nikolai Beluhov informs me that the original, by ace composer Alexander Shapovalov, is both more interesting and more difficult. In it, a company that runs out of friends must quit hiring altogether, while the other continues. Wires under the Hudson: This is a variation of a puzzle publicized by Martin Gardner, and sometimes called the Graham-Knowlton Problem. To electricians it is the “WIP” (wire identification problem). In Gardner’s version, you can tie any number of wires together at either end, and test at either end, as well. Our solution, found in [104, 60], satisfies our additional constraints and involves only two crossings, not counting additional crossings to untie and perhaps actually use the wires. The solution is not unique, however, so even if your two-crossing solution is different, it may be just as good. Two Monks on a Mountain: Brought to my attention by Yuliy Baryshnikov of the Uni-versity of Illinois at Urbana-Champaign. Worst Route: Adapted from a puzzle found in . If the number of houses were odd, distances would matter slightly: The postman would then begin or end at the middle house, zigzag between high-numbered and low-numbered houses, and end or begin at the closest house to the middle house. Example: If the addresses were 1,2,4,8,16 one of the maximum-distance routes would be 4,8,1,16,2. theorem: The idea of switching odd and even edges used in the proof is known as the “alter-nating path method” and is an important tool in graph theory. 5. Algebra Too Bat and Ball: An oldie, as you can see from the prices! Two Runners: From . Belt Around the Earth: Origin unknown, but Enrique Trevi˜ no of Lake Forest College has informed me that a version appears in the notes to a 1702 translation of Euclid by William Whiston. Odd Run of Heads: A variation of this puzzle was suggested, but not used, for an IMO in the early ’80s (see ). Hopping and Skipping: Devised by mathematician James B. Shearer of IBM, this puzzle appeared in the April 2007 edition of IBM’s puzzle site “Ponder This,” April2007.html. Area–Perimeter Match: Problem 176, p. 71 of . 413 Three Negatives: Due to Mark Kantrowitz of Carnegie Mellon University. Red and Black Sides: Moscow Mathematical Olympiad 1998, Problem C2 , contributed by V. V. Proizvolov. Why is this puzzle in the algebra chapter? Because the intended algebraic solution turned out to be wrong—fixable, but not elegantly. The nice geometric solution given was suggested by Richard Stanley. Recovering the Polynomial: Sent to me by Joe Buhler of Reed College, who believes it may be quite old (but, maybe not as old as Delphi). Note that without the restriction that the Oracle must be fed integers, the polynomial can be determined in one step with (say) x = π. Of course, the Oracle would have to find a way to convey p(π) in a finite amount of time; if she gives a decimal expansion digit by digit, there’d be no way to know when to cut her off. Gaming the Quilt: Sent to me by Howard Karloff of Goldman Sachs. Strength of Schedule: Moscow Mathematical Olympiad 1997, Problem C5 , contributed by B. R. Frenkin. Two Round-robins: Moscow Mathematical Olympiad 1997, Problem B5 , contributed by B. R. Frenkin. Alternative Dice: This problem is famous enough so that its solution has a name: “Sicher-man dice.” Martin Gardner wrote in Scientific American magazine, in 1978, about their discovery by one Colonel George Sicherman, of Buffalo NY. theorem: The model described, often now called the “Galton–Watson tree,” had already been studied independently by I. J. Bienaym´ e. 6. Safety in Numbers Broken ATM: Moscow Mathematical Olympiad 1999, Problem A4 . Hidden in this puzzle is a form of the Chinese Remainder Theorem. Subsets with Constraints: The first part of the puzzle was presented by the late, long-time puzzle maven Sol Golomb (University of Southern California) at the Seventh Gathering for Gardner; the second part was suggested by Prasad Tetali of Georgia Tech; the third, an obvious variation. Cards From Their Sum: The trick is a version of Colm Mulcahy’s “Little Fibs.” The two solutions are described in the Coda of Mulcahy’s book , and were supplied by combinatorialist Neil Calkin of Clemson University. Divisibility Game: Moscow Mathematical Olympiad 2003, Problem A6 , contributed by A. S. Chebotarev. Prime Test: Moscow Mathematical Olympiad 1998, Problem B1 , contributed by A. K. Kovaldzhi. Numbers on Foreheads: Sent to me by Noga Alon of Princeton University. Rectangles Tall and Wide: There are other proofs; see, for example, Locker Doors: Classic. Factorial Coincidence: From , sent to me by Christof Schmalenbach of IBM. Even Split: This puzzle, with n replaced by 100, appeared in the 4th All Soviet Union Mathematical Competition, Simferopol, 1970. It is elegant enough to be called a theorem and in fact it is a theorem in . See also for more on equal sums of powers. Factorials and Squares: Both the puzzle and the solution given (other solutions are possi-ble) were passed to me by Diana White of the National Association of Math Circles, and traced to Jeremy Kun, author of “A Programmer’s Introduction to Mathematics.” The puzzle had appeared in the Tournament of Towns, 1995-1996, spring round, practice level, grades 10–11, problem 3 as well as in the Moscow Mathematical Olympiad, 1996, grade 7, problem 6. theorem: This probably goes back to ancient Greece, but proofs of this fundamental fact are still of interest in the 21st century; see, for example, . 7. The Law of Small Numbers 414 Domino Task: Moscow Mathematical Olympiad 2004, Problem A6 , contributed by A. V. Shapovalov. Spinning Switches: This puzzle reached me via Sasha Barg of the University of Maryland, but seems to be known in many places. Although no fixed number of steps can guarantee turning the bulb on in the three-switch version, a smart randomized algorithm can get the bulb on in at most 5 5 7 steps on average, against any strategy by an adversary who sets the initial configuration and turns the platform. Candles on a Cake: Moscow Mathematical Olympiad 2003, Problem D4 , contributed by G. R. Chelnokov. For a somewhat similar idea, see Gasoline Crisis below. Lost Boarding-Pass: I first heard the puzzle at the Fifth Gathering for Gardner; the version here, from probabilist Ander Holroyd. Later the puzzle’s popularity zoomed when it appeared on Car Talk. Flying Saucers: My version of a puzzle suggested by Sergiu Hart. Gasoline Crisis: This puzzle has been around for a long time and can be found, for example, in . Coins on the Table: This nice puzzle came to me by way of computer scientist Guy Kindler, during a marvelous visiting year by each of us at the Institute for Advanced Study in Princeton. Coins in a Row: Passed to me by mathematician Ehud Friedgut, this puzzle was alleged to have been used by a high-tech company in Israel to test job candidates. Powers of Roots: Idea from 29th Annual Virginia Tech Mathematics Contest, 2007, and from Using Your Head is Permitted, brand.site.co.il/riddles, October 2008. Coconut Classic: This is the “Williams version” of an ancient classic, where the monkey gets a sixth coconut in the morning. With the original version, you need the ending number to be 1 mod 5, achieved by beginning with 56 −4 coconuts instead of 55 −4. That ends with 5 × 45 −4 = 5116 coconuts, giving each man 1023 and the monkey 1. theorem: Known now as a “1-factorization of the complete graph on an even number of vertices,” such constructions have been around at least since 1859 . 8. Weighs and Means Tipping the Scales: The Second All Soviet Union Mathematical Competition, Leningrad 1968. Men with Sisters: Classic. Watches on the Table: The 10th All Soviet Union Mathematical Competition, Dushanbe, 1976. Lyle Ramshaw points out that it suffices to average over just two instances, 30 minutes apart. Raising Art Value: Original. Waiting for Heads: Original, more or less. Finding a Jack: Idea from Dexter Senft of Lehman Bros; appears as Problem 40 in Fred Mosteller’s delightful book . Sums of Two Squares: Ross Honsberger . Increasing Routes: This puzzle, and its elegant solution, were passed to me by Ehud Friedgut, of Hebrew University. With his help the puzzle was traced back to the fourth problem in the second round of the “Bundeswettbewerb Mathematik 1994.” The Bun-deswettbewerb is one of two big old German mathematical competitions. theorem: Shockingly little-known, this result is recommended for inclusion in the next new graph theory text! 9. The Power of Negative Thinking Chomp: Due to the late David Gale of U.C. Berkeley (although a roughly equivalent number-theoretic game had been studied earlier by the Dutch mathematician Frederik Schuh). Dots and Boxes Variation: From Dartmouth professor Sergi Elizalde. 415 Big Pairs in a Matrix: 2011 Moscow Olympiad; from MIT puzzle maven Tanya Khovanova’s blog, G¨ unter Rote notes that the statement fails if you take the top three in each row and column, as in the following matrix:        4 4 4 4 7 0 4 4 4 4 7 0 4 4 4 4 0 7 4 4 4 4 0 7 5 5 5 5 0 0 5 5 5 5 0 0        Bugging a Disk: This puzzle was published by CMU math professor Alan Frieze on CMU’s puzzle website the proof above was sug-gested by George Wang and Leo Zhang. Pairs at Maximum Distance: From the Putnam Exam of 1957. Lemming on a Chessboard: Devised by Kevin Purbhoo when he was a high school student at Northern Secondary School in Toronto. Purbhoo is now on the faculty at the University of Waterloo. Curve on a Sphere: Passed to me by physicist Senya Shlosman, who heard it from Alex Krasnoshel’skii. The solution given is Senya’s. Omer Angel, of the University of British Columbia, has a different proof that is less elementary, but still elegant and educational. Let C be our closed curve and ˆ C its convex closure in 3-space, that is, the smallest convex set containing C. If C is not contained in a hemisphere, then ˆ C contains the origin 0; otherwise, there would be a plane through 0 cutting off ˆ C from 0. Thus, by Carath´ eodory’s Theorem, there is a set of four points on C some convex combination of which is 0. Putting it another way, the tetrahedron whose vertices are those four points contains the origin. Now, move the points continuously toward one another along the curve. When the points merge their tetrahedron will no longer contain the origin, so somewhere during the process, there was a time when the origin lay on a face of the tetrahedron. The three points that determine that face lie on a great circle, and each pair has a shortest route along that circle not containing the third point. Hence, the sum of the pairwise distances of the three points is 2π, impossible since they all lie on C. Soldiers in a Field: From the 6th All Soviet Union Mathematical Competition in Voronezh, 1966. Alternating Powers: Spotted in Emissary, Fall 2004; see . The proof is from Noam Elkies, Harvard mathematician and (musical) composer. Halfway Points: Moscow Mathematical Olympiad 1998, Problem B6 , contributed by V. V. Proizvolov. theorem: The statement was originally conjectured by James Joseph Sylvester (arguably, the first great mathematician to leave Europe for America) back in 1893. It was first proved by the Hungarian mathematician Tibor Gallai, but the elegant proof given was found by L. M. Kelly and published in 1948 . 10. In All Probability Winning at Wimbledon: Suggested by Dick Hess. Birthday Match: Problem 32 of . You may have wondered, since the answer 253 just barely works, whether accounting for the possibility that one or more of your shipmates might have been born on February in fact, it does. (The answer, 23, to the better-known version of the puzzle is unaffected.) If you yourself were born on February 29, you’d need to query a daunting 1,013 people to be a favorite to find a birthmate. Other Side of the Coin: A similar calculation shows that you should switch doors in the notorious “Monty Hall” problem, provided you assume Monty chooses randomly which door to open when two are available. Boy Born on Tuesday: Created by Thomas Starbird of JPL and unleashed by puzzle designer Gary Foshee at the Ninth Gathering for Gardner. Whose Bullet?: Original. Second Ace: Suggested by a puzzle in Littlewood’s Miscellany . 416 Stopping After the Boy: Classic, promulgated by Martin Gardner. Points on a Circle: Suggested by combinatorics legend Richard Stanley, of the University of Miami. For k points instead of 3, similar reasoning will lead to probability 2k/2k of being containable in a semicircle. One can also do this in higher dimension; e.g., the probability that four random points on a sphere can be contained in a hemisphere turns out to be 1/4. Comparing Numbers, Version I: To the best of my knowledge, this problem originated the late, supremely creative Tom Cover of Stanford University . Comparing Numbers, Version II: Source unknown. Biased Betting: Original. Home-field Advantage: Original. For a faster way to do some of the computations, see . Service Options: Adapted from independent suggestions by authors Hirokazu Iwasawa (Iwahiro) and Dick Hess, and Cambridge University’s Geoffrey Grimmett. Who Won the Series?: Heard from Pradeep Mutalik, mathematics writer for the online magazine QUANTA, at the Eleventh Gathering for Gardner. The solution given is my own. Dishwashing Game: Source unknown. Random Judge: Problem 3 of . Wins in a Row: Problem 2 of , and popularized by Martin Gardner in his “Mathematical Games” column in Scientific American. Gardner gave a (correct) algebraic proof of the answer (“no”) but acknowledging the value of a proof by reasoning, he also provided two arguments that you’d be better off playing black first: (1) You must win the crucial middle game, thus you want to be playing white second; and (2) You must win a game as black, so you’re better off with two chances to do so. In fact, neither argument is convincing, and even together they are not a proof. Split Games: Original (although my Dartmouth colleague Peter Doyle came up with some-thing similar, indpendently). Angry Baseball: Suggested, at the author’s Charles River Probability Lecture entitled “Probability in Your Head,” by Po-Shen Loh. Po-Shen is on the faculty at Carnegie Mellon and is the current coach of the U.S. Mathematical Olympic Team. Two-point Conversion: Adapted from a problem in John Urschel’s column, John is a former starting guard with the NFL’s Baltimore Ravens, now a PhD student in mathematics at MIT. Random Chord: A classic paradox that I learned as a teenager from Martin Gardner’s column. Random Bias: Brought to me by Paul Zeitz of the University of San Francisco. Paul is the founder of San Francisco’s Proof School, and a frequent contributor at the National Museum of Mathematics. Coin Testing: Original. Coin Game: This game is often called “Penney Ante,” a pun on the name of its inventor . Conway’s formula appears in ; see also [44, 54]. The quantity A·B is sometimes called the “correlation” of A and B, but I avoided that terminology as it suggests that A·B = B·A which is not generally the case. Sleeping Beauty: For Arntzenius and Dorr’s argument see [9, 36]; more arguments and many references can be found in my article . Leading All the Way and theorem: Bertrand’s original paper was followed by a nice proof by D´ esir´ e Andr´ e . 11. Working for the System No Twins Today: Classic. Martians in a Circle: Moscow Mathematical Olympiad 1998, Problem B3 , contributed by B. R. Frenkin. Poker Quickie: Sent to me by puzzle maven Stan Wagon of Macalester College, who found it in . 417 Fewest Slopes: Moscow Mathematical Olympiad 1993, Problem D3 , contributed by A. V. Andjans. Two Different Distances: Goes back at least to Einhorn and Schoenberg . First Odd Number: The prime version was brought to me by the late Herb Wilf, of the University of Pennsylvania. I’ve heard it attributed to Don Knuth. Measuring with Fuses: This and other fuse puzzles seem to have spread like wildfire some years ago. Recreational mathematics expert Dick Hess has put together a miniature volume called Shoelace Clock Puzzles devoted to them; he first heard the one above from Carl Morris of Harvard University. (Hess considers multiple fuses—shoelaces, for him—of various lengths, but lights them only at ends.) King’s Salary: This puzzle was devised by Johan W¨ astlund of Chalmers University, (loosely!) inspired by historical events in Sweden. Packing Slashes: Heard from Vladimir Chernov of Dartmouth. Lyle Ramshaw, a researcher at Hewlett-Packard, has devised schemes for the general n × n case, but as of this writing the optimal solutions for odd n are not known to me. See . Unbroken Lines: Devised by mathematics writer Barry Cipra, this one is directly inspired by a work of the late Sol LeWitt. Unbroken Curves: Again by Cipra, but with less direct connection to LeWitt. Conway’s Immobilizer: I don’t know the composer of this puzzle but the name came from the late John Horton Conway’s claim that it immobilized one solver in his chair for six hours. The first solution given here was suggested by my former PhD student Ewa Infeld; the second was devised by Takashi Chiba in response to a puzzle column in Japan, and sent to me by Ko Sakai of the Graduate School of Pure and Applied Sciences, University of Tsukuba. See for yet more solutions. Seven Cities of Gold: Adapted from a suggestion by Frank Morgan of Williams College. theorem: The given uniqueness proof is the author’s, but there are many others. The Moser Spindle (discovered by brothers William and Leo Moser in 1961) cannot be vertex-colored with three colors without having adjacent vertices of the same color. As a consequence, the “unit-distance graph,” whose vertices are all points on the plane with two adjacent when they are distance 1 apart, has chromatic number at least 4. This lower bound stood until 2018 when computer scientist and biologist Aubrey de Grey found a 1581-vertex graph that raised the lower bound to 5. The upper bound remains at 7. 12. The Pigeonhole Principle Shoes, Socks & Gloves: Original. Polyhedron Faces: 1973 Moscow Mathematical Olympiad, but with a solution that used Euler’s formula. Robert Beals points out that while the given solution requires convexity (without which multiple edges of a face could border the same other face), it is not clear whether some other approach might solve the puzzle when the polygon is allowed to be non-convex or even non-simply-connected. Lines Through a Grid: Moscow Mathematical Olympiad 1996, Problem C2 , con-tributed by A. V. Shapovalov. Same Sum Subsets: Based on a problem from the 1972 International Mathematical Olympiad. There doesn’t seem to be any easy way to actually find two disjoint subsets of Brad’s num-bers with the same sum; the number of pairs of disjoint nonempty subsets of a 10-element set is a daunting (310 −2 · 210 + 1)/2 = 28,501. Indeed, given n numbers (not necessar-ily distinct), even the problem of determining whether you can divide all of them into two disjoint sets with the same sum is notoriously difficult; it’s one of the “original” NP-complete problems from around 1970. That means there’s probably no way to decide this efficiently. Yet, the pigeonhole principle tells us that if the n numbers are distinct integers between 1 and 2n/n, and disjoint subsets (not necessarily a partition) with the same sum are required, then the answer to the existence question is easy—it’s always “yes”! Lattice Points and Line Segments: Source unknown; best guess, some Moscow Mathe-matical Olympiad. Adding, Multiplying, and Grouping: Moscow Mathematical Olympiad 2000, Problem B2 , contributed by S. A. Shestakov. 418 Line Up by Height: The theorem first appeared in . Ascending and Descending: Moscow Mathematical Olympiad 1998, Problem C6 , contributed by A. Ya. Kanel-Belov and V. N. Latyshev. Zeroes and Ones: The final idea was passed to me by the late David Gale. Same Sum Dice: Brought to me by David Kempe of the University of Southern California. Similar results can be found in a paper by the very notable mathematicians Persi Diaconis, Ron Graham and Bernd Sturmfels . Pointed out by Greg Warrington of the University of Vermont: The puzzle (and proof) still works with m n-sided dice versus n m-sided dice. Zero-sum Vectors: Sent to me by Noga Alon of Princeton University. The problem appeared as D6 in the 1996 Moscow Mathematical Olympiad . The puzzle’s statement is “tight” in at least two senses. First, if any of the 2n {−1, +1}-vectors vectors is left off the original list, the result fails—even if the others are represented arbitrarily often! Suppose, for example, that n = 5 and that the missing vector is y = ⟨+1, +1, +1, −1, −1⟩. In all other vectors, change all +1’s among the first three coordinates to zeroes, and all −1’s among the last two coordinates to zero. Since y is missing, the zero vector will not appear among the altered vectors; and any sum of altered vectors will have a negative number among its first three coordinates, or a positive number among its last two, or both. Second, as noted by mathematician and hacker Bill Gosper, for any n there is a way to alter the vectors in such a way that there is no way to get a zero sum short of adding up the entire list. We leave it to the reader to verify this fact. theorem: In fact the numbers {nr}, for r irrational, are not only dense but remarkably (and usefully) evenly spaced: Roughly speaking, if you run n through the integers from 1 to any large number, the fraction of the values {nr} that appear in some subinterval of [0, 1) will be close to the length of that subinterval. This observation is fundamental in a branch of mathematics sometimes called “discrepancy theory,” for which a wonderful source is . 13. Information, Please Finding the Counterfeit: Classical; suggested by Robert DeDomenico. David Gontier found a solution in a 1946 article by the late, great physicist Freeman Dyson. When I distributed the puzzle for the National Museum of Mathematics’ “Mindbenders for the Quarantined,” subscriber Markus Schmidmeier pointed out that there are non-adaptive solutions such as (with numbered coins) 1,2,3,4 versus 5,6,7,8; 1,2,9,10 versus 3,4,8,11; and 1,3,5,9 versus 2,6,10,12. Another subscriber, Bob Henderson, pointed out that this solution allows you to diagnose also the case of no counterfeit—in which case all weighings balance. Three Martians at the Crossroads: Puzzles of this sort were studied and popularized by Martin Gardner and Raymond Smullyan, among others; this particular version came to me from two mathematical physicists, Vladas Sidoravicius and Senya Shlosman. Attic Lamp: Classical, but losing ground as incandescent bulbs disappear even from memory. By the way, with enough patience you can extend the solution to five switches: flip switch E on a year or two in advance, to acquire a fifth possible state, “burnt out”! Players and Winners: Devised and communicated to me by Alon Orlitsky, of the University of California, San Diego. Missing Card: The trick first appeared in Wallace Lee’s book Math Miracles, in which he credits its invention to William Fitch Cheney, Jr., a.k.a. “Fitch,” around 1950. See also [72, 91]. Peek Advantage: Brought to my attention by professional gambler Jeff Norman, who was offered the chance to bet a lot more than $100 on the “color of the flop” prior to a deal of Texas hold ’em, with the extra inducement of being allowed to peek at one of his own hole cards before deciding which color to bet on. Bias Test: Original. The theorem mentioned in the text was proved by Gheorghe Zbˇ aganu [25, 115]. Dot-Town Exodus: Classical. Steve Babbage, a manager and cryptographer with Vodafone, points out that if the residents of Dot-town begin to worry that a suicide was not caused by knowing one’s dot color—but perhaps because some Dot-towner “has finally cracked under 419 the strain of living in such a ludicrous environment”—then under certain circumstances the rest of the town may yet survive the stranger’s incursion. Conversation on a Bus: A creation of John H. Conway’s. Variations can be found in a paper of Tanya Khovanova’s posted at Matching Coins: Brought to my attention by Oded Regev of the Courant Institute, NYU. In the authors show that with more sophisticated versions of this scheme, Sonny and Cher can get as close as they want to a fraction x of success, where x is the unique solution to the equation −x log2 x −(1−x) log2(1−x) + (1−x) log2 3 = 1 , about 0.8016, but they can do no better. Moreover, this applies whether the coinflips are random or adversarial. Two Sheriffs: This puzzle was devised and presented in to give a toy example of how two parties that share information, but have no common secret, can establish a common secret over an open channel, and then use it to communicate in secret. theorem: Properly speaking, the study of secret codes is cryptology, which divides into design-ing codes (cryptography) and breaking them (cryptanalysis). In practice “cryptography” is often used to cover the whole subject. A particularly entertaining source is . 14. Great Expectation Bidding in the Dark: Sent by Maya Bar Hillel, University of Jerusalem. Rolling All the Numbers: Classic. Spaghetti Loops: Classic. Ping-Pong Progression: Adapted from a problem posed on 2/28/17 at the MoMath Mas-ters, an annual puzzle-solving contest put on by the National Museum of Mathematics. Drawing Socks: Suggested by Yan Zhang of San Jose State University. Random Intersection: Observed and proved by Jim Propp, of the University of Mas-sachusetts, Lowell. He cannot be blamed for the calculus-avoiding “proof” given here, however. Roulette for the Unwary: My variation on Problem 7 from . An Attractive Game: Problem 6 from . Our proof shows that the game becomes fair if the payoff for two of your number is $3 instead of $2, and for three $5 instead of $3. Next Card Bet: I heard this problem from probabilist Russ Lyons, of Indiana University, who heard it from Yuval Peres, who heard it from Sergiu Hart; but the problem goes back apparently to a paper of Tom Cover . If the original $1 stake is not divisible, but is composed of 100 indivisible cents, things become more complicated and it turns out that Victor does a dollar worse. A dynamic program (written by Ioana Dumitriu, now a professor at UCSD) shows that optimal play guarantees ending with at least $8.08. Warning: you need to be a bit more conservative in the “100 cents” game than in the continuous version; if instead you always bet the nearest number of cents to the fraction (b−r)/(b+r) of your current worth, you could go bankrupt before half the deck is gone. Serious Candidates: Devised by probabilist David Aldous of U.C. Berkeley, who was in-spired by the prediction market and the Republican party’s presidential nomination pro-cess in 2012. Rolling a Six: Devised and told to me by MIT probabilist Elchanon Mossel, who dreamed it up as an easy problem for his undergraduate probability students and then realized the answer was not 3. The solution given is your author’s. Napkins in a Random Setting: Posed on the spot by the late, great John Horton Conway at a math conference banquet where the circular table, coffee cups and napkins were as described. Roulette for Parking Money: Due to Dick Hess, who was inspired by a note written by Bill Cutler. It’s called “Bus Ticket Roulette” in . Here we have assumed that the table has both a “0” and a “00” but the same scheme is optimal in Monte Carlo where there’s only the single zero. Buffon’s Needle: Ramaley’s paper . 420 Covering the Stains: Devised by Naoki Inaba and sent to me by Iwasawa Hirokazu, known also as Iwahiro; both are prolific puzzle composers. The actual maximum number of points in the plane that can always be covered by disjoint unit disks is not known (see ); embarrassingly, the best results currently are that 12 points are always coverable and 42 are not. My guess? 25. Colors and Distances: Sent by probabilist Ander Holroyd, who heard it from Russ Lyons. My setting is a real town which, the last time I looked, was pretty evenly split. Painting the Fence: From the Spring ’17 Emissary Puzzles Column by Elwyn Berlekamp and Joe P. Buhler, to which it was contributed by Paul Cuff who recalls it from one of Tom Cover’s seminars at Stanford. But G¨ unter Rote points out that the problem goes at least back to Steinits in 1932; see, e.g., . Filling the Cup: Classic. theorem: For (much) more on the probabilistic method, the reader is enthusiastically referred to . 15. Brilliant Induction IHOP: Yes, I know, “IHOP” doesn’t really work as an abbreviation for “induction hypothesis.” Lizz Moseman, a former PhD student of mine, had an earlier math teacher who used that term and Lizz and I both found it irresistible. Uniform Unit Distances: Posed in the (now defunct) journal Mathematical Spectrum and solved by David Seal of Winchester College (UK) in Vol. 6, Issue 2, 1973-4. As you may have noticed, the sets Sn we constructed look like projections onto the plane of an n-dimensional hypercube. Swapping Executives: Moscow Mathematical Olympiad 2002, Problem C4 , contributed by A. V. Shapovalov. Odd Light Flips: Moscow Mathematical Olympiad 1995, Problem C6 , contributed by A. Ya. Kanel-Belov. A proof using linear algebra is also possible. Truly Even Split: This puzzle was suggested by Muthu Muthukrishnan who heard it from eminent computer theorist Bob Tarjan. Non-repeating String: Moscow Mathematical Olympiad 1993, Problem A5 , con-tributed by A. V. Spivak. Baby Frog: Moscow Mathematical Olympiad 1999, Problem D4 , contributed by A. I. Bufetov. Note that the given proof provides a pretty efficient scheme. Since a circular clearing of radius √ 2 times a cell side (thus area 2π/22n, relative to the big square) must contain a grid cell, the baby frog can be guided to a circle of area A with only ⌈1 2 log2(2π/A)⌉croaks. Guarding the Gallery: The question (with 11 replaced by n) was asked by the late Victor Klee, a brilliant geometer who spent most of his career at the University of Washington, and answered by combinatorialist V´ aclav Chv´ atal. The lovely proof in this volume was found by Steve Fisk . Path Through the Cells: Moscow Mathematical Olympiad 1999, Problem D5 , con-tributed by N. L. Chernyatyev. Notice that the fact that the cells of a telephone network most likely constitute a planar graph is not needed for this result. Profit and Loss: This puzzle is adapted from one which appeared on the 1977 Interna-tional Mathematical Olympiad, submitted by a Vietnamese composer. Thanks to Titu Andreescu for telling me about it. The solution given is my own; the industrious reader will not find it difficult to generalize it to the case where x and y have a greatest common divisor gcd(x, y) other than 1. The result is that f(x, y) = x + y −1 −gcd(x, y). Uniformity at the Bakery: This lovely puzzle is from a Russian competition and appears in . The given argument also works if all the weights are rational numbers, since we can just change units so that the weights are integers. But what if the weights are irrational? Regard the real numbers R as a vector space over the rationals Q, and let V be the (finite-dimensional) subspace generated by the weights of the bagels. Let α be any member of a basis for V , and let qi be the rational coefficient of α when the weight of the ith bagel is expressed in this basis. Now the same argument used in the rational case 421 shows that all the qi’s must be 0, but this is a contradiction since then α was not in V to begin with. Summing Fractions: From the 3rd All Soviet Union Mathematical Competition, Kiev, 1969. Tiling with L’s: Told to me by Rick Kenyon of Yale University, an expert on random tilings. Traveling Salesmen: From the 11th All Soviet Union Mathematical Competition, Tallinn, 1977; thanks to Barukh Ziv of Intel for sending me the intended solution. The solution given was devised by me and Bruce Shepherd of the University of British Columbia; another nice solution was sent to me by Emmanuel Boussard of Boussard & Gavaudan Asset Management. Lame Rook: Composed by Rustam Sadykov and Alexander Shapovalov for a 1998 Olympics, and told to me by Rustam. I love that unexpected request at the end of the puzzle state-ment! Two additional remarks: (1) there’s a non-inductive proof using Pick’s Theorem, and (2) in fairy chess, the lame rook is called the “wazir.” theorem: A good source on Ramsey theory is . As of 2023, the upper bound of 4 for the “Ramsey base” lim supk→∞R(k, k)1/k has at last been (slightly) lowered . 16. Journey Into Space Easy Cake Division: Martin Gardner, in , attributes the puzzle to Coxeter. The solution given works for any regular polygon and also for any triangle, provided you cut to the incenter. Painting the Cubes: Puzzle and solution both told to me by Paul Zeitz. Curves on Potatoes: Communicated to me by Dick Hess, who heard it from Dieter Geb-hardt; appears in . Painting the Polyhedron: Told to me by Emina Soljanin (formerly a Distinguished Member of Technical Staff at Bell Labs, now at Rutgers). Boarding the Manhole: Classic. Slabs in 3-Space: From an early Putnam Exam. Bugs on Four Lines: This puzzle was passed to me by Matt Baker, of Georgia Tech. It is sometimes called “the four travellers’ problem” and appears on the website “Interactive Mathematics Miscellany and Puzzles” at Circular Shadows II: From the 5th All Soviet Union Mathematical Competition, Riga, 1971. Box in a Box: Told to me by Anthony Quas (University of Victoria) who heard it, and the solution given, from Isaac Kornfeld, a professor at Northwestern University; Kornfeld had heard the puzzle many years ago in Moscow. Additional very nice solutions were sent to me by Mike Todd of Cornell University and by Marc Massar. Mike’s uses vectors and the triangle inequality, while Marc’s is based on the observation that for an a × b × c box, (a + b + c)2 is equal to the area of the box plus the square of its diagonal length. Angles in Space: I was tested on this puzzle during a visit to MIT, and was stumped. It turns out that the question was for some time (since the late 1940’s) an open problem of Paul Erd˝ os and Victor Klee, then was solved by George Danzig and Branko Gr¨ unbaum in 1962. The question of the maximum number of points in n-space that determine only angles that are strictly acute remained open much longer, for a long time known only to be between 2n−1 and 2n −1. Only recently has it been shown that you can get halfway to the upper bound: There’s a set of 2n−1 + 1 points with all angles strictly acute. Curve and Three Shadows: The question was raised by number theorist Hendrik Lenstra, after hearing about Oskar van Deventer’s puzzle “Oskar’s Cube” in which three orthogonal sticks poke through the sides of a cube, each of which has a maze cut out of it. Since the mazes can’t have loops without a piece falling out, it wasn’t clear whether the puzzle could be designed so that the stick intersection can follow a closed curve in space. theorem: The Monge Circles theorem, with its famous sphere proof, was first brought to my attention by computer theorist Dana Randall of Georgia Tech. The flaw was pointed out to me by Jerome Lewis, Professor of Computer Science at the University of South Carolina Upstate. 422 17. Nimble Nimbers Life Is a Bowl of Cherries: The game is ancient; the analysis goes back at least as far as 1935 . Life Isn’t a Bowl of Cherries?: For more on this and the previous puzzle, is the fun place to go. Whim-Nim: Devised by John H. Conway, who called it simply “Whim.” Option Hats: This puzzle dates back to the U.C. Santa Barbara 1998 PhD thesis of Todd Ebert (now at Cal State Long Beach). It was the subject of a New York Times article en-titled “Why Mathematicians Now Care About Their Hat Color,” by Sara Robinson, April 10, 2001 (see The radius-1 covering code of size 7, for the n = 5 case, was taken from . Chessboard Guess: From and sent to me by one of that book’s authors, Anany Levitin. Majority Hats: Posted by Thane Plambeck of Counterwave, Inc. Many insights, plus the dynamic program and its results, were contributed by Johan W¨ astlund of Chalmers University. W¨ astlund points out that there is an application of the result to gambling at a “responsible” casino where you buy chips once in advance, and are paid in cash, so that you can’t gamble with your winnings and perhaps get addicted. The dynamic program maximizes your probability of coming out ahead given that you bought 100 chips that can be bet at even odds on fair coinflips. (Yes, you’d’ve been better off had you bought only 99 chips.) But one can sometimes do better in Majority Hats than in the responsible casino; with computer help, W¨ astlund now has a protocol that has pushed the majority probability for 100 prisoners to 1156660500373338319469/1180591620717411303424, or approximately 97.9729552603863%. Fifteen Bits and a Spy: Told to me by L´ aszl´ o Lov´ asz, then of Microsoft Research, now president of the Hungarian Academy of Sciences. Laci is uncertain of its origin. theorem: A nice source for error-correcting codes is . 18. Unlimited Potentials Signs in an Array: Classic. We were not asked how long the process takes, but using the facts that (1) every line needs to be flipped at most once, and (b) flipping every line that some scheme doesn’t flip gives the same result, it can be deduced that you can always get all the sums non-negative in at most ⌊(m + n)/2⌋flips. Righting the Pancakes: Moscow Mathematical Olympiad 2000, Problem A5 , con-tributed by A. V. Shapovalov. Jim Saxe and Lyle Ramshaw have determined the precise maximum number of flips that can be made before a stack of n pancakes is righted. Breaking a Chocolate Bar: I’d love to know who first came up with this one. Red Points and Blue: From a Putnam Exam of the 1960’s. Another nice solution was sent to me by tax expert Carl Giffels. The claim is that any group of n red and n blue points, with no three collinear, can be divided by a line so that each half-plane contains the same number of red points as blue. To see this, pick any point P that doesn’t lie on any line containing more than one of the 2n points. Draw a line through P and rotate it slowly about P. The line may start with (say) more red points than blue on its right-hand side, but 180◦later the reverse will be true; thus, at some point, the numbers of red and blue points on the “right” side of the line (and therefore also on the left) must be equal. Now, each half-plane can be further sub-divided the same way (this time by a half-line ending at the previous line), and the sub-divisions divided similarly, until every region contains only a single red and a single blue point. The line segments connecting those pairs cannot cross. The puzzle’s conclusion is true in higher dimensions as well, as pointed out to me by Pablo Sober´ on of Baruch College, City University of NY. In three dimensions, you have n points each of colors red, blue and green, no four coplanar. You want disjoint triangles connecting a 3-way matching, and you get it by induction via the Ham Sandwich Theorem (a.k.a. the Stone–Tukey Theorem). If n is even, use the theorem to find an appropriate plane, missing all the points, then apply the theorem to each half. If n is odd, it will cut a point of each color and you can use that triangle along with the ones obtained by induction. 423 Bacteria on the Plane: Variation of a puzzle posed by Fields Medal winner Maxim Kont-sevich (see , p. 110). Pegs on the Half-Plane: Variation of a problem described in , Vol II; I am told the problem was invented originally by the second author, Conway. In his problem, diagonal jumps were not permitted; one can nonetheless get a peg to the line y = 4 without much difficulty, but an argument like the one below shows that no higher position can be reached. Returning to my variation, Dieter Rautenbach wrote to me that a summer school student, Niko Klewinghaus, came up with a proof that one can get a peg up eight units and that this is the highest possible value. Pegs in a Square: There is more than one way to solve this puzzle, which is part of a problem which appeared at the 1993 International Mathematical Olympiad. The proof given was communicated to me by combinatorialist Benny Sudakov of ETH Zurich. At the Olympiad, contestants were asked to determine precisely for which n the squares were reducible—pretty tough to do on the spot! First-grade Division: Passed to me by Ori Gurel-Gurevich, a mathematician at the Hebrew University of Jerusalem. Gurel-Gurevich heard it from his army friend Alon Amit, but the puzzle may go back much further in time. For generalizations, see . Infected Checkerboard: The puzzle seems to have started life somewhere in the old Soviet Union, then migrated to Hungary where Spencer heard it. Its generalization contributed to a whole new field of study called “bootstrap percolation,” initiated by B´ ela Bollob´ as in 1968. Impressionable Thinkers: Suggested to me by Sasha Razborov, of the University of Chicago; he tells me that it was considered for an International Mathematics Olympiad, but rejected as too hard. It was posed and solved in . The puzzle can be generalized considerably, for example, by adding weights to vertices (mean-ing that some citizens’ opinions are more highly prized than others’), allowing loops (citi-zens who consider their own current opinions as well), allowing tie-breaking mechanisms, and even allowing different thresholds for opinion changes. Frames on a Chessboard: Sent by Ehud Friedgut of Hebrew University, it is a variation of a problem which appeared on an Israeli youth mathematics contest. In the contest, the frames were 3 × 3 and 4 × 4, and a counting argument says that you cannot reach every color configuration. The point is that the order in which the frames are laid down is irrelevant; all you need to know is which of the 52 ways to put the 4 × 4 frame down are utilized, and which of the 62 ways to put the 3 × 3 frame. Altogether there are thus 225 × 236 = 261 ways to try to get color configurations; not enough. With Friedgut’s modification, however, we need some more subtle arument, such as the one given. Bugs on a Polyhedron: Presented in a paper by Anton Klyachko . Bugs on a Line: The analysis given was done by Ander Holroyd (then at the University of British Columbia) and Jim Propp, at a meeting of the Institute for Elementary Studies in Banff, Alberta, 2003. The bug was proposed by Propp as a way to simulate deterministi-cally a random walk on the non-negative integers in which steps are made, independently, to the left with probability 1/3 and to the right with probability 2/3. In such a walk, a given bug drops off to the left or proceeds to infinity on the right with equal probability; as we saw, the deterministic model gives a strict alternation instead, after the first couple of trips. The argument can be extended to other random walks. Flipping the Pentagon: First brought to my attention by Andy Liu, this now-notorious puzzle was Problem #3 at the 1986 International Mathematical Olympiad, held in Poland. Joseph Keane, a US contestant, received a special prize for his sets-of-consecutive-vertices solution. Picking the Athletic Committee: Brought to me by my Dartmouth computer science colleague Deeparnab Chakrabarty, who needed the general result (with the number 3 replaced by an arbitrary integer k) for a research paper. Recently Iwasawa Hirokazu has generalized the result even further. Bulgarian Solitaire: See . For non-triangular numbers of chips, you end up cycling through triangular formations with extra chips shifting along the new diagonal. theorem: See for more on the infinite case. 424 19. Hammer and Tongs Phone Call: I cannot recall from whom I first heard this, many years ago. Crossing the River: Appears in the medieval Latin manuscript Propositiones ad Acuendos Juvenes (Problems to Sharpen the Young), attributed to the polymath Alcuin of York, 735–804. Sprinklers in a Field: Moscow Mathematical Olympiad 1996, Problem A3 , contributed by I. F. Sharygin. Fair Play: I was reminded of this puzzle by Tam´ as Lengyel, of Occidental College. More improvements are possible and shows how to get the last drop of blood out of the process, minimizing the expected number of flips to get a decision, regardless of the coin’s probability of landing heads up. The problem of extracting unbiased, random bits from various tainted random sources is of major importance in the theory of computing, and the subject of many research papers and significant breakthroughs in recent years. Finding the Missing Number: An alternative for those who are weak on addition but strong on dexterity is to use fingers and toes to keep track of which ones digits, and which tens digits (respectively), have appeared odd numbers of times. Identifying the Majority: The algorithm given is described in . Poorly Placed Dominoes: Original, but inspired by Problem 429 in . Unbreakable Domino Cover: Sent by aeronautical engineer Bob Henderson. Filling the Bucket: Passed to me by Dartmouth PhD student Grant Molnar. The puzzle was composed by Caleb Stanford and appeared on the 2017 Utah Math Olympiad. Note that if the buckets contain k gallons instead of just two, you can still win—but you’ll need a lot of buckets and a lot of time, since the series 1 + 1/3 + 1/5 + . . . takes exponentially many terms to get to k. (The two-gallon case can be done with 7 buckets, but for k = 3 you already need 43 buckets and lots of time.) Polygon on the Grid: Moscow Mathematical Olympiad 2000, Problem C3 , contributed by Gregory Galperin, now of Eastern Illinois University. One-Bulb Room: I heard this puzzle from Adam Chalcraft, who has the distinction of having represented Great Britain internationally in unicycle hockey. The puzzle has also appeared on IBM’s puzzle site and was reprinted in Emissary, the newsletter of the Mathematical Sciences Research Institute in Berkeley, California. A version even appeared on the justly famous public radio program Car Talk in 2003. Closely related, but much more challenging, is the puzzle of the Two-Bulb Room where the conditions are similar but there is a second bulb; the catch is that all prisoners must use the same protocol. An elegant solution known as the “see-saw protocol” is presented in . Sums and Differences: From the 5th All Soviet Union Mathematical Competition, Riga, 1971. The argument given works for any odd number of numbers greater than 3; for even numbers, the contradiction is reached even more quickly. There is, however, a set of three numbers with no pair satisfying the requirements of the puzzle: {1, 2, 3}! Prisoner and Dog: Brought to my attention by Giulio Genovese, a mathematician at Harvard’s McCarroll Lab. The prisoner’s strategy described, though good enough for speed ratio 1:4, can be improved upon. Love in Kleptopia: This puzzle was passed to me by Caroline Calderbank, daughter of mathematicians Ingrid Daubechies and Rob Calderbank. Badly Designed Clock: Posed by Andy Latto, a Boston-area software engineer, at the Gathering for Gardner IV. It can be solved algebraically or geometrically, with sufficient care and patience; the proof here was supplied to Andy by Michael Larsen, a mathematics professor at Indiana University. The idea of a third hand (instead of a second clock) came to me from David Gale. Worms and Water: Told to me by Balint Virag, a probabilist at the University of Toronto. Generating the Rationals: From The 13th All Soviet Union Mathematical Competition, Tbilisi, 1979. Funny Dice: Non-transitive (or “intransitive”) sets of dice have been around for a long time; a recent article showed that in fact randomly-designed dice often have this property. 425 Sharing a Pizza: Devised by Daniel E. Brown in 1996, this puzzle has attracted a lot of attention, in part because of your author’s conjecture that Alice could always get at least 4/9 of the pizza—proved independently by two groups [23, 75]. Names in Boxes: This puzzle has a short but fascinating history. Devised by Danish computer scientist Peter Bro Miltersen, a version appeared in a prize-winning paper of his and Anna Gal’s . But Miltersen didn’t think there was a solution until one was pointed out to him over lunch by colleague Sven Skyum. The puzzle reached me via quantum computing expert Dorit Aharonov. Eugene Curtin and Max Warshauer have recently proved that the solution given cannot be improved upon. Lambert Bright and Rory Larson, and independently Richard Stanley of MIT, proposed the following variation. Suppose each prisoner must look in 50 boxes, and the requirement for survival is that every prisoner not find his own name. Despite having the diametrically opposite objective from before, it seems that the prisoners can do no better than to follow exactly the same strategy. Here, though, they survive only if every cycle has more than 50 boxes in it, which can only happen if there is just one big cycle—for which their chances are precisely 1 in 100. Not great, but a lot better than 1 in 2100. They do just as well even if every prisoner is required to look in 99 boxes—again, they follow the strategy and win just when the random permutation has just one big cycle. In this case it is immediately obvious that no better strategy is available, because the very first prisoner, no matter what he does, has only a 1% probability of avoiding his own name. The amazing thing is that following the strategy, if that first prisoner succeeds, then automatically every other prisoner will succeed as well. Life-Saving Transposition: Sent to Emissary via Kiran Kedlaya by Piotr Krason. Self-Referential Number: Devised and told to me by probabilist Ander Holroyd. Lyle Ramshaw tells me that other starting numbers can take up to nine steps to become self-referential. Readers should be warned, however, that other, similar definitions of “self-referential number” do not necessary yield to this approach at all. theorem: Due to Georg Alexander Pick in 1899. There are many proofs; the one presented is an amalgam of several. 20. Let’s Get Physical Rotating Coin: Another classic popularized by Martin Gardner. Pie in the Sky: Original (but of course, I may be the millionth person to have thought of this question). Pushing the Pedal: See for Stan Wagon’s wonderful exposition on this and other bicycle phenomena, DZrs5pBM for a George Hart video, and for the (otherwise) excellent book by Arnol’d. Returning Pool Shot: Moscow Mathematical Olympiad 2004, Problem B3 , contributed by A. Ya. Kanel-Belov. Falling Ants: As far as I know the first publication of the puzzle was in Francis Su’s “Math Fun Facts” web-column at Harvey Mudd College; Francis recalls hearing it in Europe from someone he can’t trace named Felix Vardy. The puzzle then showed up in the Spring/Fall 2003 issue of Emmissary. Dan Amir, a former Rector of Tel Aviv University, read the puzzle in Emmissary and posed it to Noga Alon, who brought it to the Institute for Advanced Study; I first heard it from Avi Wigderson of the I.A.S., in late 2003. Ants on the Circle: Elwyn Berlekamp and I came up with this one together (but of course others may have thought of it as well—turning a line into a circle is, as we have seen, often a useful thing to do). Sphere and Quadrilateral: Told to me by Tanya Khovanova, who lists it among the “coffin puzzles” on her blog. These are puzzles with simple solutions that are difficult to find, especially for someone taking a timed exam. According to Tanya and others, such puzzles were used in the Soviet Union to keep “undesirables,” for example, Jewish students, out of the best schools. Two Balls and a Wall: Disseminated by Dick Hess and Gary Antonick at the Eleventh Gathering for Gardner, based on a discovery by Gregory Galperin. 426 theorem: Brought to my attention by Yuval Peres. 21. Back from the Future Portrait: Goes back at least to a puzzle submitted by G. H. Knight to an Oxford journal in 1872 , p. 240. Three-way Duel: Classical, and Problem 20 of . The argument that Bob should also shoot in the air was advanced, convincingly, to me by Gerry Myerson of Macquarie U., Sydney. Testing Ostrich Eggs: Sent to me by Tam´ as Lengyel, it appears in as Problem #166. Pancake Stacks: Brought to my attention by Bill Gasarch of the University of Maryland, this puzzle appeared in the 12th All Soviet Union Mathematical Olympiad, Tashkent, 1978. Thanks to Jerrold Grossman for sending me what may be the original reference: A.J. Cole and A.J.T. Davie, A game based on the Euclidean algorithm and a winning strategy for it, Math. Gaz. 53 (1969), 354–357. Chinese Nim: Also known as Wythoff’s Game, introduced in 1907 . Turning the Die: I was challenged by a superior officer in the U.S. Navy to play this game for drinks, at a stateside bar during the Vietnam War. I declined, but of course analyzed the game so I would be prepared on the next occasion—which, perhaps luckily, never materialized. Game of Desperation: Brought to my attention by my PhD student, Rachel Esselstein, the puzzle appeared at the 28th Annual USA Mathematical Olympiad in 1999. See also . Deterministic Poker: From an early Martin Gardner “Mathematical Games” column in Scientific American. Bluffing with Reals: Origin unknown. Swedish Lottery: This nice lottery idea was brought to my attention by Olle H¨ aggstr¨ om of Chalmers University (G¨ oteborg, Sweden). I’ve tried it a few times with groups of twenty or so students; twice, the winning number was 6. Pegs on the Corners: Told to me by computer theorist Mikkel Thorup of the University of Copenhagen, who heard it from Assaf Naor (now at Princeton), who heard it from graduate students at the Hebrew University of Jerusalem. Touring an Island: Seen on Carnegie Mellon University’s “Puzzle Toad” web page, and found in . Fibonacci Multiples: Told to me by Richard Stanley. Light Bulbs in a Circle: From the International Mathematical Olympiad of 1993. Puzzle maven Tom Verhoeff tells me that by applying the theory of linear feedback shift registers, he has determined that 100 bulbs first return to all-on at time t = 181,080,508,308,501,851,221,811,810,889, much greater than the age of the universe in seconds. Check out his demo . Emptying a Bucket: From the 5th All Soviet Union Mathematical Olympiad, Riga, 1971. The puzzle showed up again, minus the hardware, on the Putnam Exam in 1993, and reached me via Christian Borgs, then at Microsoft Research. The solution given is mine, but there is also an elegant number-theoretic solution found independently by Svante Janson of Uppsala University, Sweden, and Garth Payne of Penn State. Ice Cream Cake: Told to me by French graduate student Thierry Mora, who heard it from his prep-school teacher Thomas Lafforgue. Stan Wagon tells me that a version of the puzzle appeared in the 1968 Moscow Mathematical Olympiad, and involved a second angle as well, indicating the amount of a rotation between wedge-cuttings. Moth’s Tour: Suggested to me by Richard Stanley. Thirty years ago, at a Chinese restaurant in Atlanta, Laci Lov´ asz and I managed to prove that the cycle is the only graph, apart from the complete graph, with this nice propery (that the last new vertex visited by a random walk is equally likely to be any vertex other than the starting point). Boardroom Reduction: Original. Many complex versions abound of a similar puzzle in-volving pirates and gold coins, but it seemed to me that the essence of that puzzle could be achieved without the coins. theorem: This marvelous fact (in another context) was observed by Lord Rayleigh, and prob-ably many others since—and maybe before. A nice modern reference is . Sometimes 427 called “Beatty’s problem” (after Samuel Beatty 1881–1970), the puzzle appeared as Prob-lem 3117 in The American Mathematical Monthly 34 (1927), pp. 159–159, and again on the 20th Putnam Exam, November 21, 1959. 22. Seeing Is Believing Meeting the Ferry: Due to Edouard Lucas, 19th Century French mathematician. Mathematical Bookworm: From Martin Gardner. Rolling Pencil: My late colleague Laurie Snell caught me on this one, which appears in . Splitting a Hexagon: From Tanya Khovanova’s blog. Circular Shadows I: Moscow Mathematical Olympiad 1997, Problem C1 , contributed by A. Ya. Kanel-Belov. Trapped in Thickland: Moscow Mathematical Olympiad 2000, Problem D6 , con-tributed by A. Ya. Kanel-Belov. Polygon Midpoints: Sent by Barukh Ziv who found it in a book by I. M. Yaglom . The solution given is Ziv’s. Incidentally, if you don’t care for self-intersecting polygons, note that any finite set of points in the plane can be taken to be the vertices of a non-self-intersecting polygon: Just pick a point C inside the convex closure of your set of points, then connect your points in clockwise order around C. Not Burning Brownies: Told to me by Brown University’s Michael Littman, devised by him and fellow computer scientist David McAllester. Protecting the Statue: Moscow Mathematical Olympiad 2001, Problem C2 , con-tributed by V. A. Kleptsyn. Gluing Pyramids: The argument given, sometimes called the “pup tent” solution, appeared in a 1982 article by Steven Young . After the PSAT debacle, the Educational Testing Service created a panel, on which your author served, for reviewing the questions on their mathematics aptitude tests. Precarious Picture: Contributed by Giulio Genovese, who heard it from more than one source in Europe. Finding the Rectangles: Sent by Yan Zhang. Tiling a Polygon: Sent by Dana Randall. Tiling with Crosses: Sent by Senya Schlosman. Cube Magic: I was reminded of this puzzle, which appeared in a Martin Gardner column, by Gregory Galperin. A polytope has the “Rupert” property if a straight tunnel can be made in it large enough to pass an identical polytope through (and, therefore, can be enlarged to pass an even bigger copy through). The cube case was originally proved, allegedly, by Prince Rupert of the Rhine in the late 17th century. As of now nine of the Archimidean polytopes are known to have the Rupert property, the last added recently in . Circles in Space: Told to me by computer theorist Nick Pippenger of Harvey Mudd College; the construction given is due to Andrzej Szulkin of Stockholm University . As pointed out to me by Johan W¨ astlund, this and some other tiling problems can also be solved using transfinite induction, as in Double Cover by Lines from Chapter 23. Invisible Corners: Moscow Mathematical Olympiad 1995, Problem D7 , contributed by A. I. Galochkin. theorem: This classic appears in as “The Problem of the Calissons.” 23. Infinite Choice Hats and Infinity: Devised (to the best of my knowledge) by Yuval Gabay and Michael O’Connor, then graduate students at Cornell; the solution was already implicit in the work of Fred Galvin of the University of Kansas. Christopher Hardin (Smith College) and Alan D. Taylor (Union College) then included it their article . Stan Wagon wrote it up as a Macalester College Problem of the Week; additional nice observations about this and the next version were made by Harvey Friedman (Ohio State), Hendrik Lenstra (Universiteit 428 Leiden) and Joe Buhler (Reed College). It was the last of these, and (independently) Matt Baker of Georgia Tech, who communicated the puzzle to me. All Right or All Wrong: Similar history to Hats and Infinity above. The “all greeen” solution was suggested to me by Teena Carroll of St. Norbert College. Double Cover by Lines: Sent by Senya Shlosman, who is unaware of its origin. Wild Guess: Sent by Sergiu Hart. Exponent upon Exponent: Observed in 1964 by me and my college classmate Gerald Folland, now a professor at the University of Washington. Find the Robot: Suggested by Barukh Ziv. The nice way of counting the rationals men-tioned above the puzzle can be found in . A variation of this puzzle in which the object to be found has nonzero length and moves along a line, is solvable even when the starting point and speed are both arbitrary reals; see . Figure Eights in the Plane: An oldie. I once heard it attributed to the late topologist R. L. (Robert Lee) Moore of the University of Texas. Y’s in the Plane: Supplied by Randy Dougherty of Ohio State University, a three-time winner of the U.S.A. Mathematical Olympiad. theorem: Found in most elementary books on graph theory. 24. Startling Transformation Sinking 15: Mentioned in Vol. II of , where it is attributed to E. Pericoloso Sporgersi. However, rather suspiciously, this “name” is found also on Italian railroad trains, warning passengers not to lean out of the window. Slicing the Cube: Classic. Expecting the Worst: Suggested by Richard Stanley. Next Card Red: There is a discussion of this game in . The modified version in our proof reminds me of a game which was described—for satiric purposes—in the Harvard Lampoon of March 30, 1967. The issue is dubbed “Games People Play Number” and the game in question appears to have been composed by D. C. Kenney and D. C. K. McClelland. Called “The Great Game of Absolution and Redemption,” it required that the players move via dice rolls around a Monopoly-like board, until everyone has landed on the square marked “DEATH.” So who wins? Well, at the beginning of the game, you were dealt a card face down from the Predestination Deck. At the conclusion, you turn your card face up, and if it says “damned,” you lose. Magnetic Dollars: Probabilists call the mechanism of this puzzle the P´ olya urn, after the late George P´ olya, a Hungarian-born professor at Stanford who famously wrote about problem-solving in mathematics. Integral Rectangles: Our proof is #8 of fourteen in Stan Wagon’s delightful article , and there are more! Laser Gun: Brought to my attention by Giulio Genovese, who got it from Enrico Le Donne; they traced it to a 1990 Mathematics Olympiad in St. Petersburg. This puzzle sparked some more general research; see, for example, and work by Keith Burns and Eugene Gutkin. Random Intervals: This problem has a curious history. A colleague (Ed Scheinerman of The Johns Hopkins University) and I needed to know the answer in order to compute the diameter of a random interval graph, and we at first computed an asymptotic value of 2/3. Later, using a lot of messy integrating, we found that the probability of finding an interval which intersects all others is exactly 2/3, for any number of intervals (from 2 on up). The combinatorial proof given was found by Joyce Justicz, then taking a graduate reading course with me at Emory University. Infected Cubes: The construction is due to Matt Cook and Erik Winfree of Caltech, the proof that it works to their colleague Len Schulman. Gladiators, Version I: I have a theory that the “confidence” condition of Version I came about when someone tried to reconstruct Version II from the fair-game solution. The second proof of Version I is from Graham Brightwell of the London School of Economics. Gladiators, Version II: From . 429 More Magnetic Dollars: This variation of P´ olya’s urn was posed and solved by Joel Spencer of New York University and his student Roberto Oliveira. theorem: Suggested (as a puzzle) by Richard Stanley. The proof given is due to Henry Pollak of Bell Labs and now Columbia Teachers College. The theorem was originally stated and proved by Konheim and Weiss . 430 Bibliography Ethan Akin and Morton Davis, Bulgarian solitaire, Amer. Math. Monthly 92 #4 (1985), 310–330. Noga Alon, Tom Bohman, Ron Holzman, and Daniel J. Kleitman, On partitions of discrete boxes, Discrete Math. 257 #2–3 (28 November 2002), 255–258. Noga Alon and Joel H. Spencer, The Probabilistic Method, Fourth Edition, Wiley Series in Discrete Mathematics and Optimization, Hoboken, NJ, 2015. Noga Alon and D. B. West, The Borsuk-Ulam theorem and bisection of necklaces, Proc. Amer. Math. Soc. 98 #4 (December 1986), 623–628. Greg Aloupis, Robert A. Hearn, Hirokazu Iwasawa, and Ryuhei Uehara, Covering points with disjoint unit disks, 24th Canadian Conference on Computational Geome-try, Charlottetown, PEI (2012). D. Andr´ e, Solution directe du probleme r´ esolu par M. Bertrand, Comptes Rendus de l’Acad´ emie des Sciences Paris 105 (1887), 436–437. Titu Andreescu and Zuming Feng, eds., Mathematical Olympiads, MAA Press, Wash-ington DC, 2000. V. I. Arnol’d, Mathematical Understanding of Nature: Essays on Amazing Physical Phenomena and Their Understanding by Mathematicians, Amer. Math. Soc., Provi-dence RI, 2014. Frank Arntzenius, Some problems for conditionalization and reflection, J. Philos. 100 #7 (2003), 356–370. W. W. Rouse Ball, Mathematical Recreations and Essays, Macmillan & Co., London, 1892. D. Beaver, S. Haber, and P. Winkler, On the isolation of a common secret, in The Mathematics of Paul Erd˝ os Vol. II, R. L. Graham and J. Neˇ setˇ ril, eds., Springer-Verlag, Berlin, 1996, 121–135. (Reprinted and updated in 2013.) J´ ozsef Beck and William W. Chen, Irregularities of Distribution, Cambridge University Press, Cambridge, UK, 1987. Antonio Behn, Christopher Kribs-Zaleta, and Vadim Ponomarenko, The convergence of difference boxes, Amer. Math. Monthly 112 #5 (May 2005), 426–439. Elwyn R. Berlekamp, John H. Conway, and Richard K. Guy, Winning Ways for Your Mathematical Plays, Volumes 1–4, Taylor & Francis, Abingdon-on-Thames, UK, 2003. E. R. Berlekamp and T. Rodgers, eds., The Mathemagician and Pied Puzzler, AK Peters, Natick, MA 1999. Geoffrey C. Berresford, A simpler proof of a well-known fact, Amer. Math. Monthly 115 #6 (June–July 2008), 524. J. Bertrand, Solution d’un probl eme, Comptes Rendus de l’Acad´ emie des Sciences, Paris 105 (1887), 369. 431 B. Bollob´ as, Weakly k-saturated graphs, Beitr¨ age zur Graphentheorie (Kolloquium, Manebach, May 1967), 25–31. Teubner-Verlag, Leipzig, 1968. Peter Boyland, Ivan Roth, Gabriella Pint´ er, Istv´ an Lauk´ o, Jon E. Schoenfield, and Stephen Wasielewski, On the maximum number of non-intersecting diagonals in an array, J. Integer Seq. 20 (2017), Article 17.2.4, 1–24. J. Buhler, S. Golan, R. Pratt and S. Wagon, Littlewood polynomials, spectral-null codes, and equipowerful partitions, Math. Comp. 90 (2021), 1435–1453. Neil Calkin and Herbert Wilf, Recounting the rationals, Amer. Math. Monthly 107 #4 (2000), 360–363. M. Campos, S. Griffiths, R. Morris, and J. Sahasrabudhe, An exponential improvement for diagonal Ramsey, Mathematics arXiv 2303.09521, submitted 16 March 2023. J. Cibulka, J. Kynˇ cl, V. M´ esz´ aros, R. Stolaˇ r, and P. Valtr, Solution of Peter Winkler’s pizza problem, in Fete of Combinatorics and Computer Science, Bolyai Society Math-ematical Studies 20 (2010), J´ anos Bolyai Mathematical Society and Springer-Verlag, Berlin, 63–93. G. Cohen, I. Honkala, S. Litsyn, and A. Lobstein, Covering Codes, North-Holland, Amsterdam, 1997. Joel E. Cohen, Johannes H. B. Kemperman, and Gheorghe H. Zbˇ aganu, Elementary inequalities that involve two nonnegative vectors or functions, Proc. Natl. Acad. Sci. USA 101 #42 (Oct. 19 2004), 15018–22. Brian Conrey, James Gabbard, Katie Grant, Andrew Liu, and Kent Morrison, Intran-sitive dice, Math. Mag. 89 #2 (April 2016), 133–143. J. H. Conway and B. Heuer, All solutions to the immobilizer problem, Math. Intell. 36 (2014), 78–86. Thomas M. Cover, Universal Gambling Schemes and the Complexity Measures of Kolmogorov and Chaitin, Technical Rerport No. 12, October 14, 1974, Dept. of Statistics, Stanford University, Stanford, CA. T. M. Cover, Pick the largest number, in Open Problems in Communication and Computation, T. Cover and B. Gopinath, eds., Springer Verlag, Berlin, 1987, 152. T. Cover and J. Thomas, Elements of Information Theory, Wiley, Hoboken NJ, 1991. H. S. M. Coxeter, A problem of collinear points, Amer. Math. Monthly 55 #1 (1948), 26–28. Henry Crapo and Ethan Bolker, Bracing rectangular frameworks I, SIAM J. Applied Math. 36 #3 (June 1979), 473–490. Eugene Curtin and Max Warshauer, The locker puzzle, Math. Intell. 28 #1 (2006), 28–31. P. Diaconis, R. L. Graham and B. Sturmfels, Primitive partition identities, in Volume II of Paul Erd˝ os is 80, J´ anos Bolyai Society, Budapest, 1986. D. Djuki´ c, V. Jankovi´ c, I. Mati´ c, and N. Petrovi´ c, The IMO Compendium, Second Edition, Springer, New York, 2011. Cian Dorr, Sleeping beauty: In defence of Elga, Analysis 62 (2002), 292–296. F. J. Dyson, The problem of the pennies, The Mathematical Gazette 30 #291 (October 1946), 231–234. Sheldon J. Einhorn and I. J. Schoenberg, On Euclidean sets having only two distances between points II, Indagationes Mathematicae (Proceedings) 69 (1966), 489–504. A. Engel, Problem-Solving Strategies, Springer, New York, 1998. . P. Erd˝ os, A. Ginzburg, and A. Ziv, Theorem in the additive number theory, Bull. Res. Council of Israel 10F (1961), 41–43. 432 P. Erd˝ os and G. Szekeres,A combinatorial problem in geometry, Compo. Math. 2 (1935), 463–470. R. Fedorov, A. Belov, A. Kovaldzhi, and I. Yashchenko, eds., Moscow Mathematical Olympiads, 1993–1999, MSRI/AMS 2011. R. Fedorov, A. Belov, A. Kovaldzhi, and I. Yashchenko, eds., Moscow Mathematical Olympiads, 2000–2005, MSRI/AMS 2011. D. Felix, Optimal Penney Ante strategy via correlation polynomial identities, Electron. J. Comb. 13 #R35 (2006), 1-15. Thomas S. Ferguson, A Course in Game Theory, published by the author, 2020. Online at M. J. Fischer, S. Moran, S. Rudich, and G. Taubenfeld, The wakeup problem, Proc. 22nd Symp. on the Theory of Computing, Baltimore, MD, May 1990. M. J. Fischer and S. L. Salzberg, Finding a majority among n votes, J. Algorithms 3 #4 (December 1982), 362–380. Steve Fisk, A short proof of Chv´ atal’s watchman theorem, J. Comb. Theory B 24 (1978), 374. M. L. Fredman, D. J. Kleitman, and P. Winkler, Generalization of a puzzle involving set partitions, Barrycades and Septoku; Papers in Honor of Martin Gardner and Tom Rodgers, Thane Plambeck and Tomas Rokicki, eds., AMA/MAA Spectrum Vol. 100, MAA Press, Providence RI, 2020, 125–129. Aaron Friedland, Puzzles in Math and Logic, Dover, Mineola, NY, 1971. Anna Gal and Peter Bro Miltersen, The cell probe complexity of succinct data struc-tures, Proc. International Colloquium on Automata, Languages and Programming (ICALP 2003), Eindhoven, The Netherlands, Jos C. M. Baeten et al., eds., Lecture Notes in Computer Science 2719, Springer-Verlag, Berlin. G. A. Galperin, Playing pool with π: The number π from a billiard point of view, Regul. Chaotic Dyn. 8 (2003), 375–394. G. A. Galperin and A. K. Tolpygo, Moscow Mathematical Olympiads, Moscow, Prosveshchenie, 1986. Martin Gardner, The Colossal Book of Short Puzzles and Problems, W. W. Norton & Co., New York, NY, 2006. Martin Gardner, On the paradoxical situations that arise from nontransitive relations, Sci. Am. 231 #4 (October 1974), 120–125. Martin Gardner, Time Travel and Other Mathematical Bewilderments, W. H. Freeman and Company, New York NY, 1988. ISBN 0-7167-1925-8. Bal´ asz Gerencs´ er and Viktor Harangi, Too acute to be true: The story of acute sets, Amer. Math. Monthly 126 #10 (December 2019), 905–914. E. Goles and J. Olivos, Periodic behavior of generalized threshold functions, Discrete Math. 30 (1980), 187–189. Olivier Gossner, Pen´ elope Hern´ andez, and Abraham Neyman, Optimal use of commu-nication resources, Econometrica 74 #6 (November 2006), 1603–1636. N. Goyal, S. Lodha, and S. Muthukrishnan, The Graham-Knowlton problem revisited, Theory of Computing Systems 39 #3 (2006), 399–412. Ronald L. Graham, Bruce L. Rothschild and Joel H. Spencer, Ramsey Theory, Second Edition, Wiley Series in Discrete Mathematics and Optimization, Hoboken, NJ, 2013. L. J. Guibas and A. M. Odlyzko, String overlaps, pattern matching, and nontransitive games, J. Comb. Theory Ser. A 30 (1981), 183–208. Christopher S. Hardin and Alan D. Taylor, A peculiar connection between the axiom of choice and predicting the future, Amer. Math. Monthly 115 #2 (2008), 91–96. 433 G. H. Hardy, On certain oscillating series, Q. J. Math. 38 (1907), 269–288. Dick Hess, Golf on the Moon, Dover, Mineola, NY, 2014. Dick Hess, The Population Explosion and Other Mathematical Puzzles, World Scien-tific, Singapore, 2016. Ross Honsberger, Ingenuity in Mathematics, New Mathematical Library Series 23, Mathematical Association of America, Providence, RI, 1970. David Kahn, The Codebreakers, Scribner & Sons, New York, 1967 and 1996. K. S. Kaminsky, E. M. Luks, and P. I. Nelson, Strategy, nontransitive dominance and the exponential distribution, Austral. J. Statist. 26 #2 (1984), 111–118. K. A. Kearns and E. W. Kiss, Finite algrebras of finite complexity, Discrete Math. 207 (1999), 89–135. Murray Klamkin, International Mathematical Olympiads 1979–1985, Mathematical Association of America, Providence, RI, 1986. Michael Kleber, The best card trick, Math. Intell. 24 #1 (Winter 2002), 9–11. Anton Klyachko, A funny property of sphere and equations over groups, Commun. in Algebra 21 #7 (1993), 2555–2575. G. H. Knight, in Notes and Queries, Oxford Journals (Firm), Oxford U. Press, Oxford, UK, 1872. K. Knauer, P. Micek, and T. Ueckerdt, How to eat 4/9 of a pizza, Discrete Mathematics 311 #16 (2011), 1635–1645. Joseph D. E. Konhauser, Dan Velleman and Stan Wagon, Which Way Did the Bicycle Go?, Cambridge U. Press, 1996. Alan G. Konheim and Benjamin Weiss, An occupancy discipline and applications, SIAM J. Appl. Math. 14 #6 (November 1966), 1266–1274. Boris Kordemsky, The Moscow Puzzles, Charles Scribner’s Sons, New York, 1971. J.-F. Lafont and B. Schmidt, Blocking light in compact Riemannian manifolds, Geom. Topol. 11 (2007), 867–887. Ger´ ard Lavau, The truncated tetrahedron is Rupert, Amer. Math. Monthly 126 #10 (2019), 929–932. Tamas Lengyel, A combinatorial identity and the world series, SIAM Rev. 35 #2 (June 1993), 294–297. Anany Levitin and Maria Levitin, Algorithmic Puzzles, Oxford U. Press, Oxford, UK, 2011. J. E. Littlewood, The Dilemma of Probability Theory, Littlewood’s Miscellany, B´ ela Bollob´ as, ed., Cambridge University Press, 1986 A. Liu and B. Shawyer, eds., Problems from Murray Klamkin, The Canadian Collec-tion, MAA Problem Book Series, MAA Press, Providence RI, 2009. L´ aszl´ o Lov´ asz, Combinatorial Problems and Exercises, North Holland, Amsterdam, 1979. L. Lov´ asz and P. Winkler, A note on the last new vertex visited by a random walk, J. Graph Theory 17 #5 (Nov 1993), 593–596. Chris Maslanka and Steve Tribe, The Official Sherlock Puzzle Book, Woodland Books, Salt Lake City UT, 2018. ISBN 978-1-4521-7314-6. E. C. Milner and S. Shelah, Graphs with no unfriendly partitions, in A tribute to Paul Erd˝ os, A. Baker et al., eds., Cambridge University Press, Cambridge, UK, (1990), 373–384. S. Morris, R. Stong, and S. Wagon, And the winners are..., Math Horizons 20 #4, April 2013, 20–22. 434 Frederick Mosteller, Fifty Challenging Problems in Probability, Addison-Wesley, Boston, MA, 1965. Colm Mulcahy, Mathematical Card Magic: Fifty-Two New Effects, CRC Press, Boca Raton, FL, 2013. ISBN: 978-14-6650-976-4. S ¸erban Nacu and Yuval Peres, Fast simulation of new coins from old, Ann. Appl. Probab. 15 #1A (2005), 93–115. Roger B. Nelsen, Proofs Without Words, Mathematical Association of America, Prov-idence, RI, 1993. Arnau Padrol and G¨ unter M. Ziegler, Six Topics on Inscibable Polytopes, Adv. Disc. Diff. Geom., Springer, New York (2016), 407–419. W. Penney, Problem: Penney-ante, J. Recreat. Math. 2 (1969), p. 241. Vera Pless, Introduction to the Theory of Error-Correcting Codes, John Wiley & Sons, Hoboken NJ, 1982. H. Pr¨ ufer, Neuer Beweis eines Satzes ¨ uber Permutationen, Arch. Math. Phys. 27 (1918), 742–744. T. F. Ramaley, Buffon’s noodle problem, Amer. Math. Monthly 76 #8 (October 1969), 916–918. M. Reiss, ¨ Uber eine Steinersche combinatorische Aufgabe, welche im 45sten Bande dieses Journals, Seite 181, gestellt worden ist, J. Reine Angew. Math. 56 (1859), 226– 244. I. J. Schoenberg, Mathematical Time Exposures, Mathematical Association of America, Providence, RI, 1982. Harold N. Shapiro, Introduction to the Theory of Numbers, Dover, Mineola NY, 1982. D. O. Shklarsky, N. N. Chentov, and I. M. Yaglom, The USSR Problem Book, W. H. Freeman and Co., San Francisco, 1962. R. P. Sprague, ¨ Uber mathematische Kampfspiele, Tohoku Math. J. 41 (1935–1936), 438–444. Roland Sprague, Recreation in Mathematics, Blackie & Son Ltd., London, 1963. Andrzej Szulkin, R3 is the union of disjoint circles, Amer. Math. Monthly 90 #9 (1983), 640–641. P. Vaderlind, R. Guy and L. Larson, The Inquisitive Problem Solver, Mathematical Association of America, Providence, RI, 2002. Tom Verhoeff, Circle of Lamps, Wolfram Demonstrations Project 2020. Stan Wagon, Fourteen proofs of a result about tiling a rectangle, Amer. Math. Monthly 94 #7, (August–September 1987), 601–617. Stan Wagon, Test your intuition, Math Horizons, Sep. 2014, 8–9, 26–28. Chamont Wang, Sense and Nonsense of Statistical Inference, Marcel Dekker, New York, 1993. P. Winkler, The sleeping beauty controversy, Amer. Math. Monthly 124 #7 (August– September 2017) 579–587; also in The Best Writing on Mathematics 2018, M. Pitici, ed., Princeton University Press, Princeton, NJ and Oxford, UK, 2019, 117–129. W. A. Wythoff, A modification of the game of Nim, Nieuw Arch. Wisk. 7 (1907) 199–202. I. M. Yaglom, Geometric Transformations I, Translated by Allen Shields. Mathemat-ical Association of America, Providence, RI, 1962. S. C. Young, The mental representation of geometrical knowledge, J. Math. Behav. 3 #2 (1982), 123–144. G. Zbˇ aganu, A new inequality with applications in measure and information theories, Proc. Rom. Acad., Series A 1 #1 (2000), 15–19. Yao Zhang, Combinatorial Problems in Mathematical Competitions (Vol. 4, Mathe-matical Olympiad Series), East China Normal University Press, Shanghai, 2011. 435 436 The Author Peter Winkler is the William Morrill Professor of Mathematics and Computer Science at Dartmouth College, and for 2019–2020, the Distinguished Visiting Professor for the Public Dissemination of Mathematics at the National Museum of Mathematics. He is the author of 160 research papers, a dozen patents, two previous puzzle books, a book on cryptographic techniques in the game of bridge, and a portfolio of compositions for ragtime piano.
15016	https://math.hmc.edu/benjamin/wp-content/uploads/sites/5/2019/06/Challenging-Knight%E2%80%99s-Tours.pdf	Full Terms & Conditions of access and use can be found at Math Horizons ISSN: 1072-4117 (Print) 1947-6213 (Online) Journal homepage: Challenging Knight’s Tours Arthur T. Benjamin & Sam K. Miller To cite this article: Arthur T. Benjamin & Sam K. Miller (2018) Challenging Knight’s Tours, Math Horizons, 25:3, 18-21, DOI: 10.1080/10724117.2018.1424460 To link to this article: Published online: 25 Jan 2018. Submit your article to this journal Article views: 212 View related articles View Crossmark data ARTHUR T. BENJAMIN AND SAM K. MILLER O ver the centuries, chess enthusiasts have enjoyed solving a problem known as the knight’s tour. In this problem, we are given a standard eight-by-eight chessboard and a knight that is placed on an arbitrary square. Our goal is to move the knight 63 times so that it visits every square ex-actly once. (A knight can make L-shaped jumps— two squares in one direction and one square in a perpendicular direction.) Variants of this problem add to the challenge. For example, we could require that the last visited square is a knight’s move away from the starting point. An even harder version specifi es a starting square and an ending square of opposite color; we call this the challenging knight’s tour. Note that a knight always jumps to a square of the opposite color; hence, it is impossible for the last square to be the same color as the fi rst one. In this article, we give a constructive proof that all start- and end-square combinations of the challeng-ing knight’s tour have a solution. Although computer programs have proved this result by fi nding tours for all possible start and end points, we present a short proof that provides insights to the problem’s struc-Challenging Knight’s Tours D s S d D s S d S d D s S d D s s D d S s D d S d S s D d S s D D s S d D s S d S d D s S d D s s D d S s D d S d S s D d S s D a b c d e f g h 1 2 3 4 5 6 7 8 ture (Michael Dupuis and Stan Wagon, “Laceable Knights,” Ars Math. Contemp. 9 : 115–124). Figure 1. The vertices of the knight’s tour graph can be partitioned into four sets and labeled d, D, s, and S, denoting left diamond, right diamond, left square, and right square, respectively. 18 February 2018 : : Math Horizons : : www.maa.org/mathhorizons sergign/123RF Stock Photo The Knight’s Tour Graph For a mathematician, the knight’s tour is a problem in graph theory. The chessboard graph G has 64 verti-ces, each representing a square on the board. There is an edge connecting vertices x and y if a knight can jump from square x to square y in a single move; we say that squares x and y are adjacent in G. Each vertex is black or white, depending on the color of its square. Thus, the challenging knight’s tour problem can be stated as follows: Given two vertices in G of opposite colors (called o and e for origin and endpoint), find a path in G from o to e that visits every vertex exactly once; this is called a Hamiltonian path. Each square in the chessboard is identified by its rank (numbered row) and file (lettered column). Thus, we may refer to a square and its corre-Similarly, vertices with labels s or S belong to the left square or right square system, illustrated in figures 2c and 2d, respectively. We say that the labels d and D are of the same type, as are s and S. Notice that the 16 vertices and 24 edges of the d-graph can be straightened into the four-by-four d d d d d d d d d d d d d d d d D D D D D D D D D D D D D D D D s s s s s s s s s s s s s s s s S S S S S S S S S S S S S S S S Figure 2. (a) The left diamond, (b) right diamond, (c) left square, and (d) right square systems. (a) (b) (c) (d) sponding vertex by its rank and file address; for instance, the lower left and upper right squares are a1 and h8, respectively. The graph G has 168 edges, but we can partition the 64 vertices into four sets of size 16 that illumi-nate its structure. We label each vertex with one of four symbols, d, D, s, or S, as shown in figure 1. Vertices with label d represent squares in the left dia-mond system. Notice that in the subgraph shown in figure 2a (called the induced subgraph) each quad-rant has four vertices joined by blue edges in the shape of a diamond that leans to the left (mnemon-ic: left diamond = lowercase d). Vertices with label D belong to the right diamond system, with the induced subgraph shown in figure 2b. Figure 3, above. All four induced subgraphs are isomorphic to G4,4. Figure 4, right. The Sd and sD edges form a (blue) shoelace pattern, and the sd and SD edges form a (red) shoelace pattern. D s S d D s S d S d D s S d D s s D d S s D d S d S s D d S s D D s S d D s S d S d D s S d D s s D d S s D d S d S s D d S s D www.maa.org/mathhorizons : : Math Horizons : : February 2018 19 graph G4,4, shown in figure 3. For instance, the lower left vertex of G4,4 represents the square b2. In fact, all four of the induced subgraphs have the same underlying struc-ture; we say they are each isomorphic to G4,4. What about the other edges of G? Notice that there are no edges that connect two different systems of the same type; that is, there are no dD or sS edges. In figure 4, we see that the 36 (blue) sD and Sd edges form a shoelace pattern, as do the 36 (red) sd and SD edges. Combining all 168 edges of figures 2a–d, and 4 gives us the knight’s tour graph G (not pictured). Strategy Once we are aware of the four systems, it is easy to solve the basic knight’s tour problem, in which we are given the initial vertex o but have the freedom to end anywhere we want. Suppose o is in the d system, say in the lower left quadrant of the board. Jump to the three other verti-ces in that quadrant with label d (being careful not to end on the corner vertex b2). Then jump to an adja-cent quadrant, landing on a d vertex, and repeat the process. Thus, in 15 jumps, we can visit all 16 d verti-ces. Next, jump to a vertex with a different label, say s, and cover the s-system. Then cover the D-system, and end with the S-system. Go ahead, try it! To solve the challenging knight’s tour where the first and last vertices are given, we use a pseudo- algorithm devised by Michael Daniels (“Learn to Master the Knight’s Tour,” mindmagician.org/ tourhelp.aspx). Our primary observation is that we can use the four middle vertices of each system—the vertices with ranks 3, 4, 5, and 6 and files c, d, e, and f—to jump from one system to either system of opposite type (see figure 4). In fact, each of these vertices is adjacent to two vertices of the opposite type. For example, the d-graph has middle vertices c3, d5, e4, and f6. And from c3, say, we can reach two vertices of type s (a2 and e2) and two of type S (b1 and b5). As with the basic knight’s tour problem, our gen-eral strategy is to traverse the board one system at a time. However, in some cases, we will need to be clever in our ordering. We now come to our first important lemma, which will be the basis for our constructive proof. A chessboard graph is traversable if we can find a Hamiltonian path from any starting vertex to any ending vertex of the opposite color. We call such a path a traversal. Lemma. All systems are traversable. Proof. Since each system is isomorphic to G4,4, it suffice s to s how tha t G4,4 is traversable. Although there are 128 ways to choose starting and ending vertices with opposite colors, when we take symme-try into account, we need display only the 10 cases in figure 5. Theorem. The knight’s tour graph G is traver- sable. Proof. Our construction considers three cases, de-pending on the locations of the original vertex o and the ending vertex e. Since the start and end vertices have opposite colors, we may assume without loss of generality that o is a black vertex and e white. Call our starting system A1, the other system of the same type A2, and the remaining systems B1 and B2. For example, if A1 is the d-system, then A2 is the D-system. We will constructively prove the tour can be completed by considering three cases, depending on whether e is in A1, A2, or one of the B systems. Case 1: Opposite system types Suppose e is in one of the B systems, say B2. For example, o is in a diamond system and e is in a square system. Using our lemma, we can traverse A1 starting at o and ending on a white middle vertex, denoted m1. We denote this by Next, go from m1 to B1, landing on a black vertex. Then traverse B1, ending on a white middle vertex m2. Move to A2, and traverse it, ending on a white middle vertex m3. Finally, jump to system B2, mak-ing sure not to land on e. Since m3 is adjacent to at least two vertices in B2, this is always possible. We will enter B2 on a black vertex and, by our lemma, can traverse B2 ending on e, thus completing the tour. To summarize case 1, we have Figure 5. G4,4 is traversable. 20 February 2018 : : Math Horizons : : www.maa.org/mathhorizons suffices Case 2: Systems of the same type For this case, we proceed similarly to the first case, with the exception that e is in A2, not B2. For exam-ple, o and e are in different diamond systems. Lemma. Given any vertex e in G4,4, we can find three middle vertices—d, x, and y—where x and y are adjacent, and a 13-step path from d to e that reaches every vertex except x and y. This path is called a semitraversal of G4,4. Proof. By symmetry, we need consider only three cases, depending on where e is on G4,4. The end point e will be one of the four middle vertices, one of the eight edge vertices, or one of the four corner vertices. We may choose our starting point d and our exempt vertices x and y based on those three cases. The choices and paths are presented in figure 6. Note that d and e have opposite colors, as do x and y. We can now prove case 2. Before beginning, note whether e is a middle, corner, or edge vertex in A2, and choose middle vertices d, x, and y as in the previ-ous lemma. Let T denote the semitraversal of A2 that starts at d, ends at e, and avoids x and y. Since x and y have opposite colors, let x denote the black vertex. Starting at o, traverse A1 and B1 as before, ending on white middle vertices m1 and m2, respectively, where m2 is adjacent to x. Next, move to x, then y, and then to B2. We enter B2 on a black vertex. Traverse B2 as normal, ending on a white middle vertex m3, where m3 is adjacent to d. Finally, move from m3 to d, then proceed with semitraversal T from d to e, which covers all of A2, except the already-visit-ed x and y. We summarize case 2 in the diagram Case 3: Same system Assume o and e are both in A1. First, assume e is not one of the white corner vertices (a8 or h1). Then there is at least one vertex outside of A1 that can reach e in one move. Let this vertex be f, which is in B2, say. Begin traversing A1 as if we were going to end at e, but stop at the penultimate vertex, n1. Suppose n1 is not a corner vertex. Since e can reach B2, n1 can reach a white vertex in B1. We then traverse B1, then e e e x x x y y y d d d Figure 6. G4,4 is semitraversable. A2, then B2, making sure to end on f. Then we finish our traversal by going from f to e. We summarize this traversal as follows: The procedure fails if n1 is a corner vertex, since it cannot reach B1. In this case, n1 is adjacent only to e and some other vertex n0. Suppose is the A1 traversal. Perform only the first 12 jumps, stopping on black vertex n. Let f0 be a neighbor of n0 in B2. Then proceed as follows: Finally, consider the case where e is a corner ver-tex. Proceed in similar fashion. Let f1 be a point in B2 adjacent to n1. Then our traversal is as follows: Thus, we have covered all three cases, and therefore, the challenging knight’s tour is always solvable! n Arthur Benjamin teaches math at Harvey Mudd College and is a former editor of Math Horizons. This paper combines some of his favorite passions: games, puzzles, math, and magic. Sam K. Miller graduated with a math degree from Harvey Mudd College in 2017 and is now a plat-form engineer at Supplyframe. When he’s not diving deep into the endless world of math, he can be found coding, playing jazz saxophone, canoeing, DJing, powerlifting, or playing chess. 10.1080/10724117.2018.1424460 www.maa.org/mathhorizons : : Math Horizons : : February 2018 21 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 S I N E O S L O A X O M A T A N R O I L L O C K E N O T I A U L D C O O R S G O L D E N R A T I O N A H L G E I S A Y S T O L I R O M E A S H E T H E A S P P E E W E N P A S C A L S T R I A N G L E A N T I U P I T R E M O R T E E N R A N K E R A T O G U A C E O S O F T K L E I N B O T T L E H O O H A A G O G A R E S M I D I S G O B Y P O E T S L O P E E R I N S U R D I I Solution to puzzle on page 2
15017	https://www.emerson.com/documents/automation/on-correlation-between-isothermal-compressibility-isobaric-expansivity-en-5396676.pdf	White Paper April 2016 Midstream Oil and Gas Solutions D352396X012 Abstract Fluid properties are a critical element to the success of any pipeline simulation. In some cases the pumped ﬂ uid or liquid mixture is so exotic in nature that laboratory data is unavailable and an educated guess is the only course of action. For transient simulation, knowledge of the isothermal compressibility is important and some estimate could be made by realizing the composition of the mixture. For steady state simulation, possibly for a batched system, the ﬂ ow rates would need to be corrected to standard or pipeline base conditions, and these correction factors require knowledge of both the isothermal compressibility and the thermal expansion properties. If one can estimate the ﬂ uid compressibility with some certainty, can one also estimate the isobaric expansivity? Laboratory test data tend to show that liquids with high compressibility also seem to have high isobaric expansivity, indicating a correlation between the two. Hence this paper intends to discover what, if any, correlation exists through examination of ﬂ uid properties of known pure components, and application of physical processes and required thermodynamic stability. Nomenclature a Attractive force coefﬁ cient b Apparent body volume of molecules v Speciﬁ c volume T Temperature P Pressure βP Isobaric Expansivity βT Isothermal Compressibility R Ideal Gas constant ω Acentric factor χ Ratio of isobaric expansivity to isothermal compressibility Subscripts c Critical P Pressure T Temperature Introduction The total change in either pressure or speciﬁ c volume in a ﬂ uid can be described by two important ﬂ uid properties: isobaric expansivity and isothermal compressibility. Isobaric expansivity is used to express the thermal expansion experienced by ﬂ uids and is deﬁ ned as the volume change of a ﬂ uid due to temperature change, while holding pressure constant : This is also referred to as the coefﬁ cient of thermal expansion. The isothermal compressibility is the volume change of a ﬂ uid due to pressure changes at constant temperature is deﬁ ned by: On the Correlation between Isothermal Compressibility and Isobaric Expansivity Brett Christie, Emerson (formerly Energy Solutions Int’l.) Midstream Oil and Gas Solutions 2 www.EmersonProcess.com/Remote April 2016 Experimental Results NIST provides various databases of ﬂ uid properties, which are based on experimental data. Table 1 shows a variety of hydrocarbons selected at pressures to ensure liquid phase at a temperature of 540 oR (81 oF). Figure 1 then shows this data ploted with thermal expansion as a function of compressibility. As can be seen there clearly is a one-to-one relationship between compressibility and thermal expansion, for liquid phase. Also, zero compressibility appears to correspond with zero thermal expansion. The intention of this paper is to explain and predict this correlation from a theoretical basis. In the next section various hydrocarbons, in order of increasing molar mass, are presented along with some observations about those ﬂ uids. Data is graphed with isothermal compressibility as the independent variable and isobaric expansivity as the dependent variable on the Y axis. Each data point has a given pressure and temperature value, with NIST REFPROP database providing the speciﬁ c volume, isothermal compressibility and isobaric expansivity for the selected hydrocarbon. Pressure is 145 to 1450 psia in steps of 15 psia. This paper uses absolute scales for pressure, and temperature in degree Rankine. Ammonia Figure 2 has the liquid phase isotherms graphed for ammonia at three temperatures. The highest pressure point has the lowest compressibility and expansivity values. Following the isotherm as the compressibility increases, the expansivity also increases. As temperature increases the isotherms move to the right. The critical point for this polar molecule is TC= 729°R, PC = 1636 psia. Water The critical point for water is TC = 1165°R, PC = 3203 psia and Figure 3 shows water for several temperatures below the critical values. At 540°R the isobaric expansivity decreases with increasing compressibility. Then as the temperature increases, this behavior changes and isobaric expansivity increases as compressibility increases. Water is a polar molecule. And it can be shown that for thermodynamically stable states that Hence the isothermal compressibility will always be a nonnegative number. This represents the inverse of the isothermal bulk modulus of elasticity for the pipelined ﬂ uid. A general observation for liquid phase is that as the compressibility increases, the thermal expansion also increases. This leads one to wondering if there is a direct correlation between these two ﬂ uid properties, and what that might mean for ﬂ uids in general. To arrive at a meaningful relationship this property needs to be combined with an appropriate equation of state in order to show the details of the correlation and help explain the mechanisms involved. We begin by looking at experimental data provided by the National Institute for Standards and Technology (NIST) for various hydrocarbons used in the pipeline industry, which are typically compounds not pure elements. All ﬂ uid property data presented is taken from NIST. The NIST database uses a variety of equations of state, including their “extended corresponding states model” and Helmholtz energy equations of state, including international standard equations for water, carbon dioxide, ammonia and others. Since pipeline coatings typically deﬁ ne limits of the maximum ﬂ uid temperature, and there is a wide variation in acceptable limits. The high limit for temperature was selected to be 580°R (or 121°F). Liquid pipeline operations typically have pressure in the range 145 to 1450 psi, so that range was selected for this study. Furthermore, we need to make the distinction between polar and nonpolar molecules. A polar substance has an electric dipole or charge on its molecules and it may lead to different results from nonpolar substances. Water and ammonia (NH3) are examples of polar substances and are included in this study. Nonpolar molecules examples include the alkanes, such as methane and ethane, and alkenes such as ethene. www.EmersonProcess.com/Remote 3 Midstream Oil and Gas Solutions April 2016 Ethane Figure 4 shows curves for Ethane for three different temperatures for the range of pressures. The highest pressure point has the lowest compressibility and expansivity values and liquid phase. Following each isotherm, left to right, for increasing compressibility the pressure drops. High pressures result in liquid phase and linear variation, as compressibility increases, isobaric expansivity increases, and then dramatically increases until the phase transition to vapor phase occurs, where expansivity starts to decrease while compressibility increases. The critical point for Ethane is TC = 550°R, PC = 708 psia. Carbon Dioxide Figure 5 shows the linear molecule CO2 for four isotherms for pressures ranging from 145 to 1450 psia. Each isotherm starts out showing that expansivity increases somewhat linearly and then loops around and back and then continues on decreasing thermal expansion as the compressibility increases. The low compressibility linear part of the curve occurs in liquid phase, followed by the transition from liquid to gas and it is clear that for gas phase one can have as many as three values for isobaric expansivity for a single isothermal compressibility value. The critical point for CO2 is TC = 547°R, PC = 1070 psia. Propane Values for propane (TC = 665°R, PC = 616 psia) are shown in Figure 6. Higher temperatures resolve the curve more fully than at lower temperatures, where a break occurs and the transition is not apparent. The curves are very similar as in the case of the CO2 and ethane data. Octane Figure 7 shows Octane at 540 oR, wich has its critical point at TC = 1024°R, PC = 360 psia. Clearly this is liquid phase only for the pressure and temperature range. Expansivity increases as isothermal compressibility increases. As temperature increases these curves move to the right. For these temperatures and pressure range only liquid phase is experienced. Normal Butane Figure 8 has n-Butane values (TC =765°R, PC = 551 psia) shows that each isotherm has expansivity increasing as compressibility increases. The slope of the curve decreases as temperature increases. If the temperature continued to increase the phase change from liquid to gas is expected to occur. Comparisons Figure 9 shows various hydrocarbons at 540°R (81°F) for liquid phase. They all appear to follow the same curve. This is an interesting result because it implies that a similarity law exists. However Carbon Dioxide interestingly appears to follow a different curve. Ethene or Ethylene follows more closely but also appear to be on a slightly different curve. Summary of Results To summarize the discussion of results above: A common correlation appears to exist at constant temperature. Liquid and gas phases have different behaviors Isobaric expansivity appears to increase dramatically during the phase change, followed by a decrease. Liquid phase has one-to-one relationship with isobaric expansivity increasing as compressibility increases, except for water which exhibits a change of slope. Of the hydrocarbons studied here, water has the lowest compressibility and expansivity values. In liquid phase, they all exhibit a monotonically inceasing relationship between the isobaric expansivity and the isothermal compressibility; for water the relationship is either increasing or decreasing depending on the temperature. The liquid phase curve appears to extrapolate to indicate that zero compressibility corresponds with zero isobaric expansivity. The phase change appears to have two or even three values for isobaric expansivity for a single isothermal compressibility value. For liquids, the less compressible a substance is the more closely a common isotherm is followed. Theory The next step is to see if theory can predict these results. The classical thermodynamic perspective is preferred in this paper over the statistical viewpoint. Classical Perspective The full or substantive derivative of change in pressure is given by: Midstream Oil and Gas Solutions 4 www.EmersonProcess.com/Remote April 2016 Where the total pressure change is a function of change in volume and temperature and the following relationship can be derived: From the deﬁ nitions for thermal expansion and isothermal compressibility, it can be shown that the ratio of our two properties of interest leads to a third property - which is the pressure change with temperature change at constant volume: Where a is a function of temperature and accentric factor as such: This gives only a qualitative description of the repulsive behavior of molecules and the repulsive and attractive contributions are not truly separated . Since the van der Waals equation is not sufﬁ ciently accurate for predicting liquid phase pressures, a more accurate cubic equation can be achieved based on the work done by Peng and Robinson . It should be noted at this point that neither cubic equations nor Benedict Webb Rubin equations can be used to predict with conﬁ dence the PVT behavior of polar molecules . Also a complex generalized equation of state such as Starling-Han or Benedict Webb Rubin might be more accurate but lack the ability to discern the physical behavour. The Peng-Robinson (1976) equation is: The ﬁ rst term on the right side represents the repulsion: the pressure exerted due to collision and is proportional to the thermal part of energy (the sum of the translational, vibrational and rotational energies) of all the molecules within unit volume of free space. The second term is the reduction in force due to the attractive force exerted on those molecules by neighbouring molecules. We note this repulsive part of pressure is actually a correction over the Ideal Gas equation where the apparent or molecular body volume b of the molecules is subtracted from the geometrical volume, resulting in a higher value for the pressure. Also the ideal gas equation does not account for the attractive forces that are signiﬁ cant when the molecular spacing is relatively close, like at higher pressures or in liquid phase. The attractive forces serve to reduce the pressure. This can be used with an equation of state to show the nature of the relationship. In this paper we need to trade state equation accuracy with visibility to the underlying physical processes. With this in mind we choose the van der Waals concept buried in every cubic equation of state. The idea that the pressure of ﬂ uid results from the sum of repulsive and attractive forces was ﬁ rst expressed by van der Waals in his equation and indeed all cubic equations are of the “van der Waals” form: The van der Waals equation of state is: And k is a constant characteristic of each substance : Figure 10 shows comparisons for pressure versus volume between NIST and Peng Robinson for Ethene at 540°R (81°R) – they show very good agreement with the average error less than 2% and the standard deviation of the differences less than 1%. Physical Processes Consider the isothermal compression of ethene at 540°R. As the ﬂ uid is compressed the volume decreases while the intermolecular spacing decreases. At larger volumes and lower pressures the ﬁ rst term in the van der Waals equation dominates, the body volume b and And the molecular body volume is: www.EmersonProcess.com/Remote 5 Midstream Oil and Gas Solutions April 2016 Figure 11 shows components of the pressure due to repulsion (ﬁ rst term) and attraction (second term). As the pressure continues to increase a point is reached where the attractive forces start to become as strong as the repulsive forces. The repulsive forces are always larger, though, to ensure thermodynamic stability. Upon compression the body volume effect (which reduces the space available for movement of molecules) is dominant and results in a higher number density so that the ﬁ rst term dominates, although the attractive force serves to reduce the pressure over what it would have been with only the ﬁ rst term. The NIST data appears to only show thermodynamically stable states where the pressure continues to increase with decreasing volume and is always true. Figure 12 shows as the pressure increases both NIST and Peng-Robinson, with the density graphed along side. Both NIST and Peng-Robinson show a rise in compressibility followed by a decrease, as the density increases. The phase change occurs when the density changes fastest. Figure 13 As the pressure increases both NIST and Peng-Robinson predict a rise in expansivity followed by a decrease during the phase change. This maximum value in isobaric expansivity deserves further analysis, which follows next. Peak Expansivity Figure 14 shows values for isobaric expansivity for different values for the attractive coefﬁ cient a in the van der Waals equation for a hypothetical substance with van der Waals was selected for this part of the study due to its simplicity of form. This plot shows that the van der Waals starts to show this peak value when the attractive coefﬁ cient is increased from zero. The isobaric expansivity is calculated here using Like the van der Waals equation, this implies that the isobaric expansivity and isothermal compressibility are related to each other directly by the effect of the intermolecular attraction and repulsive forces. From this result, we can determine that if there were no attractive forces involved, the isobaric expansivity would show merely a continuual decrease as volume increases, going from high pressure liquid phase down to low pressure vapor. From a pipeline design and operatonal consideration, volumes around v0 should be avoided as the uncertainty in these properties increases (Figure 15 demonstrates). It’s clear that the molecular attractive force plays a signiﬁ cant role in how this mechanical property behaves. If a vapor is compressed, its molecules exist closer to each other. As the intermolecular distance is reduced, the attractive force between adjacent molecules becomes large enough to reduce the molecular velocity. Gas molecules slow down to a state at which matter changes phase and becomes a liquid . We note that this compression process allows for a greater increase in thermal expansion during the phase change than would occur if there were no attractive forces. And, it allows for the behavior noticed with liquid phase: that the expansivity increases with an increase in compressibility. Predicting Ratios For Peng-Robinson equation, the ratio between isobaric expansivity and isothermal compressibility is: Using perturbation method, the peak in occurs roughly (zero order approximation) when the volume reaches a value of: Midstream Oil and Gas Solutions 6 www.EmersonProcess.com/Remote April 2016 Applications The realism of any pipeline simulation is a direct consequence of the accuracy of the conﬁ gured ﬂ uid properties. This paper provides a guide for inspecting the density gradient properties of isothermal compressibility and isobaric expansivity, showing that a clear relationship between the two properties exists at a single temperature. It also shows that hydrocarbons tend to follow the same curve at a given temperature for liquid phase. For phase changes the isobaric expansivity can change signiﬁ cantly with respect to isothermal compressibility, reaching a peak value before decreasing again. Furthermore the use of constant values for isothermal compressibility and isobaric expansivity should be carefully considered over the range of operating pressure and temperature. References 1. The Properties of Gases & Liquids, Fourth Edition, Robert C. Reid, John M. Prausnitz, Bruce E. Poling, McGraw-Hill, Inc. 1987 2. Applied Hydro Carbon Thermodynamics, Volume 1, Wayne C. Edmister, Byung Ik Lee, Second Edition, Gulf Publishing Company, 1983 3. Advanced Thermodynamics Engineering, Kalyan Annamalai, Ishwar K. Puri, CRC Press, 2002 4. National Institute of Standards and Technology (NIST) REFPROP database, 5. Ding Yu Peng and Donald B. Robinson, A New Two-Constant Equation of State, Ind. Eng. Chem., Fundam., Vol 1, No. 1, 1976 Acknowledgements The author would like to thank Dr. Jon Barley and Dr. Dick Spiers for their reviews and helpful comments. About the Author Mr. Brett Christie, P.Eng. is a senior project engineer at Emerson. Since 1999 he has been involved in pipeline simulation focusing on the implementation of leak detection and trainer systems for liquids pipelines. He is currently pursuing a Master’s degree in Mechanical Engineering at the University of Calgary, specializing in Pipeline Engineering. He holds a Bachelor’s degree in Mechanical Engineering from the Technical University of Nova Scotia, and is a registered Professional Engineer in Alberta, Canada. Figure 15 shows comparisons with NIST and the Peng Robinson values for ethene at 540°R (Ethene has critical point at TC = 508°R, PC = 731 psia) for the ratio . As pressure increases the ratio increases as well and tends to mimic the density change behavior. This shows that although Peng Robinson is not too accurate for higher pressures it still predicts the correct physical behavior. Similar results are achieved at higher temperature (Figure 16). Conclusions One common assumption for liquids is a constant value for isothermal compressibility and isobaric expansion however these results clearly show that this statement does not indicate accurate behavior for hydrocarbons over the speciﬁ ed range of operating pressure and temperature. Figure 17 shows that the correlation for n-Butane in liquid phase is somewhat linear (Figure 18 shows it more clearly) and for gas phase has a hyperbolic ﬁ t. Liquid Phase Liquids show a direct increase in isobaric expansivity with an increase in compressibility, for the same temperature. Also for the same temperature a variety of liquids follow the same curve (see Figure 9). Both repulsive and attractive forces are signiﬁ cant for this state of matter. The molecular body volume effect reduces the space available for movement of molecules which results in reduced compressibility and reduced expansion, as the pressure increases. All of the substances studied here appear to share this behavior, except for water which isn’t a hydrocarbon and which appeared to change slope as the pressure increased. Further study is desirable to explain why water’s correlation has the slope changing to a negative value for higher pressures, and is likely related to its polar nature. Phase Transition For a liquid at high pressure as the pressure is dropped the compressibility increases and the isobaric expansivity increases as well. This increase in expansivity continues until the phase starts to change where it reaches a maximum value and then begins to decrease. The phase change is characterized by a sudden drop in both repulsive and attractive forces (see Figure 11). The isobaric expansivity continues to drop into the vapor phase as demonstrated by Figure 19 and tends to level off as zero pressure is approached. www.EmersonProcess.com/Remote 7 Midstream Oil and Gas Solutions April 2016 Tables Fluid Pressure psia Isothermal Compressibility 1/psia Isobaric expansivity 1/°R Mercury 14.65 2.42E-07 3.35556E-05 Propane 290.08 4.29E-05 0.001804444 Iso-Butane 290.08 2.59E-05 0.001247222 N-Butane 290.08 2.04E-05 0.00111 1% Ethane 99% N-Butane 290.08 2.61E-05 0.001255556 10% Ethane 90% N-Butane 290.08 2.86E-05 0.001338889 Propane (again) 580.15 3.77E-05 0.001659444 50% Propane, 50% Iso-Butane 725.19 3.56E-05 0.001599444 Octane 725.19 8.76E-06 0.000615556 Water 725.19 3.21E-06 0.000155556 Table 1. Various hydrocarbons at different pressures, all liquid phase at 540°R (NIST) Figures Figure 1. Isobaric Expansivity versus Isothermal Compressibility for Different Hydrocarbons at 540°R Midstream Oil and Gas Solutions 8 www.EmersonProcess.com/Remote April 2016 Figure 2. Ammonia for 160 – 1437 psia for Three Isotherms Figure 3. Water for Several Isotherms www.EmersonProcess.com/Remote 9 Midstream Oil and Gas Solutions April 2016 Figure 4. Ethane for Several Isotherms, Pressures Ranging from 145 to 1450 psia Figure 5. CO2 for Four Isotherms and Pressures Ranging from 145 to 1450 psia Midstream Oil and Gas Solutions 10 www.EmersonProcess.com/Remote April 2016 Figure 6. Propane. Added higher temperature isotherms resolve the curve fully (°R). Figure 7. N-Octane Values - Liquid Phase Only www.EmersonProcess.com/Remote 11 Midstream Oil and Gas Solutions April 2016 Figure 8. N-Butane for Several Isotherms in Liquid Phase Only Figure 9. Comparisons for Various Hydrocarbons at 540°R for Liquid Phase Only Midstream Oil and Gas Solutions 12 www.EmersonProcess.com/Remote April 2016 Figure 10. P v Curve for Ethene at 540°R - NIST versus Peng-Robinson Figure 11. Peng-Robinson Attractive (all values are negative; hence the absolute values are plotted only) and Repulsive Components of Pressure. The attractive values try to cancel the repulsive effect, but the repulsive values are always more than the attractive values. www.EmersonProcess.com/Remote 13 Midstream Oil and Gas Solutions April 2016 Figure 12. Ethene Isothermal Compressibility for NIST and Peng-Robinson at 540°R Figure 13. Ethene Isobaric Expansivity Comparisons versus Pressure at 540°R Midstream Oil and Gas Solutions 14 www.EmersonProcess.com/Remote April 2016 Figure 14. Isobaric Expansivity versus Speciﬁ c Volume, for Different Attraction Force Coefﬁ cients (van der Waals) for a Hypothetical Substance. Units of a are psia.ft6.°R1/2/kmol2 www.EmersonProcess.com/Remote 15 Midstream Oil and Gas Solutions April 2016 Figure 15. Comparisons for Ethene at 540°R (81.3°F), NIST versus Peng-Robinson. NIST density is shown as well. Figure 16. Comparisons for Ethene at 567°R (107°F), NIST versus Peng-Robinson. NIST density is shown as well. Midstream Oil and Gas Solutions 16 www.EmersonProcess.com/Remote April 2016 Figure 18. N-Butane Zoom in on Liquid Phase Shows Linear Behavior (585°R) Figure 17. N-Butane Correlation – Shows Liquid and Gas Phase Data, but No Intermediate Values (585°R) www.EmersonProcess.com/Remote 17 Midstream Oil and Gas Solutions April 2016 Figure 19. Isotherm for Ethene at 540°R (81°F) and Pressure (Right Axis) Graphed against Isothermal Compressibility. Low pressure corresponds with high isothermal compressibility. As pressure increases the isobaric expansivity slowly increases in gas phase, then as the phase transitions to liquid, increases dramatically. 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15018	https://www.enganchadaalaciencia.es/1%C2%BA-bach/tema-9-cinem%C3%A1tica/7-movimientos-complejos/7-3-movimientos-de-trayectoria-parab%C3%B3lica	enganchadaalaciencia.es - 7.3 -Movimientos de trayectoria parabólica Search this site Embedded Files Skip to main content Skip to navigation enganchadaalaciencia.es Inicio 2º ESO Unit 1 - THE WORK OF SCIENTISTS 1 - Introduction 2 - The scientific method 3 - Direct and indirect measurements 4 - Quantities and units 5 - Scientific notation 6 - Rounding 7 - How to measure quantities 8 - Systematic errors 9 - El laboratorio de física y química Unit 2- MATTER 1-Introduction 2-General and specific properties of matter 3-The states of matter 4-Gas laws 5-The kinetic molecular theory Unit 3 - PURE SUBSTANCES AND MIXTURES 1- Tipos de sistemas materiales 2- Técnicas de separación Unit 4 - SOLUTIONS Unit 5- PHYSICAL AND CHEMICAL CHANGES 1- Introduction 2- Physical and chemical changes 3- Chemical equations 4- Environmental problems 5 - Productos de la industria química y su utilidad. Unit 6- MOTION Unit 7- FORCES How are forces represented? How to add forces The effects of forces Types of forces Simple machines Unit 8-ENERGY The main kinds of energy Some other kinds of energy Energy transformations The principle of conservation of energy Heat transfer The effects of heat Sources of energy 3º ESO Unit 1 - The work of scientists 1 - Introduction 2 - The scientific method 3 - Direct and indirect measurements 4 - Quantities and units 5 - Scientific notation 6 - Rounding and significant figures 7 - How to measure quantities 8 - Systematic errors 9 - Uncertainty in measurements 10 - El laboratorio de física y química Unit 2- What is matter made up of? 1 - Historical development of the structure of matter 2 - Subatomic particles 3 - Isotopes 4- Nuclear fission 5-Ions: cations and anions 6 - Some activities on this unit Unit 3- The chemical elements 1 - First classifications of the elements 2 - The periodic table of the elements 3 - The properties of the elements Unit 4 - Bonding 1- Introduction Copia de 3 - Electronic configuration of the elements 2 - Electronic configuration of the elements 3- Ionic bonding 4 - Covalent bonding 5 - Metallic bonding 6 - Atoms, molecules, crystalline lattices. Unit 5 - Formulación y nomenclatura de compuestos químicos Unit 6 - Chemical changes 1 - Physical and chemical changes 2 - Chemical equations 3 - The collisions theory 4 - The law of conservation of mass 5 - Reaction rates 6 - Cálculos estequiométricos 7 - Chemical reactions, environment and industry Unit 7 - Motion Unit 8 - Forces 1 - Introduction 2 - How are forces represented? 3 - How to add forces 4 - The effects of forces 5 - Types of forces 6 - Newton laws 7-Newton's universal gravitational law Mass and weight 8- Coulomb's law Unit 9 - Energy 4º ESO Unidad 1- El trabajo de los científicos Tema 2-De qué está hecha la materia Tema 3 -Los elementos químicos Tema 4-El enlace químico Tema 5-Formulación y nomenclatura de compuestos químicos Tema 6-Los cambios químicos Tema 7- Cinemática 1- Elementos para la descripción del movimiento 2- MRU 3 - MRUA 4 - MCU 5 - Interpretación de gráficas 6-Problemas mezcla Tema 8- Dinámica Introducción Cómo se representan las fuerzas Cómo se suman las fuerzas Cómo se descomponen fuerzas Fuerzas que actúan por contacto Fuerzas que actúan a distancia Leyes de Newton Problemas sobre las leyes de Newton Tema 9- Los fluidos Tema 10- La energía Introducción Principales tipos de energía Transformaciones de energía Principio de conservación de la energía mecánica El trabajo como forma de transferencia de energía El calor como forma de transferencia de energía 1º BACH. Bloque 1 - La estructura de la materia Tema 3- Química inorgánica Tema 4 - Las reacciones químicas Tema 5 - Química del carbono 4.1 - El estado gaseoso 4.2 - Las disoluciones Tema 7- Termoquímica Tema 9 - Cinemática 2-Conceptos básicos para la descripción del movimiento 3-Movimiento rectilíneo uniforme: MRU 4-Movimiento rectilíneo uniformemente acelerado: MRUA 5-Un mrua especial: Caída libre 6-Interpretación de gráficas 7-Movimientos complejos 7.1-Movimiento de trayectoria circular y velocidad constante 7.2-Movimientos de trayectoria rectilínea 7.3 -Movimientos de trayectoria parabólica Tema 10 - Estática y dinámica Introducción Cómo se representan las fuerzas. Tipos de fuerzas Efectos de las fuerzas. Principios de la dinámica Diagrama de fuerzas Equilibrio de traslación y rotación Leyes de Kepler Ley de Gravitación Universal Problemas Tema 11 - Energía 1-Introducción 2- La energía mecánica 3-El trabajo como forma de transferencia de energía 4 -El calor como forma de transferencia de energía 5-Fuerzas conservativas y no conservativas. 6-Principio de conservación de la energía 7- Problemas sobre energía FÍSICA 2º Bach 1- Bloque INTERACCIÓN GRAVITATORIA 2- Bloque INTERACCIÓN ELECTROMAGNÉTICA 3 - Bloque Vibraciones y ondas 4 - Física del siglo XX Iniciación a la astrofísica Curiosidades Experimentos enganchadaalaciencia.es Inicio 2º ESO Unit 1 - THE WORK OF SCIENTISTS 1 - Introduction 2 - The scientific method 3 - Direct and indirect measurements 4 - Quantities and units 5 - Scientific notation 6 - Rounding 7 - How to measure quantities 8 - Systematic errors 9 - El laboratorio de física y química Unit 2- MATTER 1-Introduction 2-General and specific properties of matter 3-The states of matter 4-Gas laws 5-The kinetic molecular theory Unit 3 - PURE SUBSTANCES AND MIXTURES 1- Tipos de sistemas materiales 2- Técnicas de separación Unit 4 - SOLUTIONS Unit 5- PHYSICAL AND CHEMICAL CHANGES 1- Introduction 2- Physical and chemical changes 3- Chemical equations 4- Environmental problems 5 - Productos de la industria química y su utilidad. Unit 6- MOTION Unit 7- FORCES How are forces represented? How to add forces The effects of forces Types of forces Simple machines Unit 8-ENERGY The main kinds of energy Some other kinds of energy Energy transformations The principle of conservation of energy Heat transfer The effects of heat Sources of energy 3º ESO Unit 1 - The work of scientists 1 - Introduction 2 - The scientific method 3 - Direct and indirect measurements 4 - Quantities and units 5 - Scientific notation 6 - Rounding and significant figures 7 - How to measure quantities 8 - Systematic errors 9 - Uncertainty in measurements 10 - El laboratorio de física y química Unit 2- What is matter made up of? 1 - Historical development of the structure of matter 2 - Subatomic particles 3 - Isotopes 4- Nuclear fission 5-Ions: cations and anions 6 - Some activities on this unit Unit 3- The chemical elements 1 - First classifications of the elements 2 - The periodic table of the elements 3 - The properties of the elements Unit 4 - Bonding 1- Introduction Copia de 3 - Electronic configuration of the elements 2 - Electronic configuration of the elements 3- Ionic bonding 4 - Covalent bonding 5 - Metallic bonding 6 - Atoms, molecules, crystalline lattices. Unit 5 - Formulación y nomenclatura de compuestos químicos Unit 6 - Chemical changes 1 - Physical and chemical changes 2 - Chemical equations 3 - The collisions theory 4 - The law of conservation of mass 5 - Reaction rates 6 - Cálculos estequiométricos 7 - Chemical reactions, environment and industry Unit 7 - Motion Unit 8 - Forces 1 - Introduction 2 - How are forces represented? 3 - How to add forces 4 - The effects of forces 5 - Types of forces 6 - Newton laws 7-Newton's universal gravitational law Mass and weight 8- Coulomb's law Unit 9 - Energy 4º ESO Unidad 1- El trabajo de los científicos Tema 2-De qué está hecha la materia Tema 3 -Los elementos químicos Tema 4-El enlace químico Tema 5-Formulación y nomenclatura de compuestos químicos Tema 6-Los cambios químicos Tema 7- Cinemática 1- Elementos para la descripción del movimiento 2- MRU 3 - MRUA 4 - MCU 5 - Interpretación de gráficas 6-Problemas mezcla Tema 8- Dinámica Introducción Cómo se representan las fuerzas Cómo se suman las fuerzas Cómo se descomponen fuerzas Fuerzas que actúan por contacto Fuerzas que actúan a distancia Leyes de Newton Problemas sobre las leyes de Newton Tema 9- Los fluidos Tema 10- La energía Introducción Principales tipos de energía Transformaciones de energía Principio de conservación de la energía mecánica El trabajo como forma de transferencia de energía El calor como forma de transferencia de energía 1º BACH. Bloque 1 - La estructura de la materia Tema 3- Química inorgánica Tema 4 - Las reacciones químicas Tema 5 - Química del carbono 4.1 - El estado gaseoso 4.2 - Las disoluciones Tema 7- Termoquímica Tema 9 - Cinemática 2-Conceptos básicos para la descripción del movimiento 3-Movimiento rectilíneo uniforme: MRU 4-Movimiento rectilíneo uniformemente acelerado: MRUA 5-Un mrua especial: Caída libre 6-Interpretación de gráficas 7-Movimientos complejos 7.1-Movimiento de trayectoria circular y velocidad constante 7.2-Movimientos de trayectoria rectilínea 7.3 -Movimientos de trayectoria parabólica Tema 10 - Estática y dinámica Introducción Cómo se representan las fuerzas. Tipos de fuerzas Efectos de las fuerzas. Principios de la dinámica Diagrama de fuerzas Equilibrio de traslación y rotación Leyes de Kepler Ley de Gravitación Universal Problemas Tema 11 - Energía 1-Introducción 2- La energía mecánica 3-El trabajo como forma de transferencia de energía 4 -El calor como forma de transferencia de energía 5-Fuerzas conservativas y no conservativas. 6-Principio de conservación de la energía 7- Problemas sobre energía FÍSICA 2º Bach 1- Bloque INTERACCIÓN GRAVITATORIA 2- Bloque INTERACCIÓN ELECTROMAGNÉTICA 3 - Bloque Vibraciones y ondas 4 - Física del siglo XX Iniciación a la astrofísica Curiosidades Experimentos More Inicio 2º ESO Unit 1 - THE WORK OF SCIENTISTS 1 - Introduction 2 - The scientific method 3 - Direct and indirect measurements 4 - Quantities and units 5 - Scientific notation 6 - Rounding 7 - How to measure quantities 8 - Systematic errors 9 - El laboratorio de física y química Unit 2- MATTER 1-Introduction 2-General and specific properties of matter 3-The states of matter 4-Gas laws 5-The kinetic molecular theory Unit 3 - PURE SUBSTANCES AND MIXTURES 1- Tipos de sistemas materiales 2- Técnicas de separación Unit 4 - SOLUTIONS Unit 5- PHYSICAL AND CHEMICAL CHANGES 1- Introduction 2- Physical and chemical changes 3- Chemical equations 4- Environmental problems 5 - Productos de la industria química y su utilidad. Unit 6- MOTION Unit 7- FORCES How are forces represented? How to add forces The effects of forces Types of forces Simple machines Unit 8-ENERGY The main kinds of energy Some other kinds of energy Energy transformations The principle of conservation of energy Heat transfer The effects of heat Sources of energy 3º ESO Unit 1 - The work of scientists 1 - Introduction 2 - The scientific method 3 - Direct and indirect measurements 4 - Quantities and units 5 - Scientific notation 6 - Rounding and significant figures 7 - How to measure quantities 8 - Systematic errors 9 - Uncertainty in measurements 10 - El laboratorio de física y química Unit 2- What is matter made up of? 1 - Historical development of the structure of matter 2 - Subatomic particles 3 - Isotopes 4- Nuclear fission 5-Ions: cations and anions 6 - Some activities on this unit Unit 3- The chemical elements 1 - First classifications of the elements 2 - The periodic table of the elements 3 - The properties of the elements Unit 4 - Bonding 1- Introduction Copia de 3 - Electronic configuration of the elements 2 - Electronic configuration of the elements 3- Ionic bonding 4 - Covalent bonding 5 - Metallic bonding 6 - Atoms, molecules, crystalline lattices. Unit 5 - Formulación y nomenclatura de compuestos químicos Unit 6 - Chemical changes 1 - Physical and chemical changes 2 - Chemical equations 3 - The collisions theory 4 - The law of conservation of mass 5 - Reaction rates 6 - Cálculos estequiométricos 7 - Chemical reactions, environment and industry Unit 7 - Motion Unit 8 - Forces 1 - Introduction 2 - How are forces represented? 3 - How to add forces 4 - The effects of forces 5 - Types of forces 6 - Newton laws 7-Newton's universal gravitational law Mass and weight 8- Coulomb's law Unit 9 - Energy 4º ESO Unidad 1- El trabajo de los científicos Tema 2-De qué está hecha la materia Tema 3 -Los elementos químicos Tema 4-El enlace químico Tema 5-Formulación y nomenclatura de compuestos químicos Tema 6-Los cambios químicos Tema 7- Cinemática 1- Elementos para la descripción del movimiento 2- MRU 3 - MRUA 4 - MCU 5 - Interpretación de gráficas 6-Problemas mezcla Tema 8- Dinámica Introducción Cómo se representan las fuerzas Cómo se suman las fuerzas Cómo se descomponen fuerzas Fuerzas que actúan por contacto Fuerzas que actúan a distancia Leyes de Newton Problemas sobre las leyes de Newton Tema 9- Los fluidos Tema 10- La energía Introducción Principales tipos de energía Transformaciones de energía Principio de conservación de la energía mecánica El trabajo como forma de transferencia de energía El calor como forma de transferencia de energía 1º BACH. Bloque 1 - La estructura de la materia Tema 3- Química inorgánica Tema 4 - Las reacciones químicas Tema 5 - Química del carbono 4.1 - El estado gaseoso 4.2 - Las disoluciones Tema 7- Termoquímica Tema 9 - Cinemática 2-Conceptos básicos para la descripción del movimiento 3-Movimiento rectilíneo uniforme: MRU 4-Movimiento rectilíneo uniformemente acelerado: MRUA 5-Un mrua especial: Caída libre 6-Interpretación de gráficas 7-Movimientos complejos 7.1-Movimiento de trayectoria circular y velocidad constante 7.2-Movimientos de trayectoria rectilínea 7.3 -Movimientos de trayectoria parabólica Tema 10 - Estática y dinámica Introducción Cómo se representan las fuerzas. Tipos de fuerzas Efectos de las fuerzas. Principios de la dinámica Diagrama de fuerzas Equilibrio de traslación y rotación Leyes de Kepler Ley de Gravitación Universal Problemas Tema 11 - Energía 1-Introducción 2- La energía mecánica 3-El trabajo como forma de transferencia de energía 4 -El calor como forma de transferencia de energía 5-Fuerzas conservativas y no conservativas. 6-Principio de conservación de la energía 7- Problemas sobre energía FÍSICA 2º Bach 1- Bloque INTERACCIÓN GRAVITATORIA 2- Bloque INTERACCIÓN ELECTROMAGNÉTICA 3 - Bloque Vibraciones y ondas 4 - Física del siglo XX Iniciación a la astrofísica Curiosidades Experimentos 7.3 - Movimientos de trayectoria parabólica Se trata de la composición de dos movimientos: uno horizontal de tipo mru y otro vertical de tipo mrua. 7.3.1. LANZAMIENTO HORIZONTAL El objeto se lanza con cierta velocidad en dirección horizontal. En cuanto se encuentra en el aire, queda sometido sólo a la fuerza gravitatoria terrestre (despreciando el rozamiento con el aire) que le proporciona una aceleración constante. En dirección X tiene, por tanto MRU. En dirección Y tiene MRUA donde a es la aceleración de la gravedad (9,8 m/s2). La velocidad resultante en cada punto de la trayectoria, que es tangente a la trayectoria siempre, es suma de dos vectores perpendiculares: vx tiene módulo constante y vx tiene un módulo que va aumentando mientras se mueve, alcanzando su valor máximo en el momento del impacto con el suelo. Observamos que la ecuación de la trayectoria es la de una parábola con las ramas hacia abajo (hay un signo - delante de x2) que tiene el vértice en la altura inicial. TIEMPO DE VUELO- tiempo que pasa el objeto en el aire; es decir tiempo que pasa desde que fue lanzado hasta que impacta contra el suelo (el agua o lo que sea).. Para deducir su expresión, basta tener en cuenta que en el momento del impacto, t= tiempo de vuelo y la coordenada y=0. Podemos sustituir en y(t) esos dos valores (y=0 y t=tiempo de vuelo) y despejar el tiempo de vuelo. MÁXIMO ALCANCE o ALCANCE - es la coordenada x del punto de impacto contra el suelo o el agua.... Para deducir su expresión, basta tener en cuenta que para t= tiempo de vuelo, y=0 y x= máximo alcance. Podemos tomar una ecuación que relacione x e y (la ecuación de la trayectoria) y sustituir y=0 y x= máximo alcance y despejar el máximo alcance. También podemos usar la ecuación de movimiento x(t). Si sustituimos t por el tiempo de vuelo en esa ecuación, el valor de la x obtenido es el del máximo alcance. Ejercicios de lanzamiento horizontal Lanzamiento horizontal en el cine: Butch Cassidy and the Sundance Kid 37) En la película BUTCH CASSIDY AND THE SUNDANCE KID, de 1969, un grupo de pistoleros se dedica a asaltar los bancos del estado de Wyoming y el tren-correo de la Union Pacific. El jefe de la banda es Butch Cassidy (Paul Newman) y Sundance Kid (Robert Redford) es su inseparable compañero. Esta película ganó 4 Oscars y un Globo de Oro. Hay una escena en la que, huyendo de sus perseguidores, llegan a un acantilado de 96 m de altura y deciden saltar. Se trata de un salto horizontal con una velocidad inicial de 1,2 m/s. a) Calcula la velocidad con la que llegan al agua. b) ¿A qué distancia de la base del acantilado caen (distancia respecto de la vertical)? c) La escena en la película la grabaron unos especialistas pero Robert Redford y Paul Newman, para simular la escena, saltaron con igual velocidad inicial sobre un colchón de 3,4 m de largo. ¿Desde qué altura saltaron si cayeron en el centro del colchón? Solución: a) v= (1,2 i - 4,3 j) m/s; b) 5,3 m; c) 9,6 m 1) Desde lo alto de un acantilado de 40 m de altura se lanza horizontalmente una piedra con una velocidad de 20 m/s. Determina: a) tiempo que tarda en caer (tiempo de vuelo) b) distancia a la que llega la piedra en horizontal (alcance máximo). 2) Una canica rueda con una velocidad de 1,5 m/s por una mesa de 0,8 m de altura. Al llegar al borde de la mesa, se cae. Determina: a) ecuación de movimiento en X y en Y de la canica. b) ecuación de la trayectoria que sigue la canica. c) tiempo de vuelo (tiempo que tarda en tocar el suelo desde que se separa del borde de la mesa). 0,40 s d) coordenada x del punto de impacto medida desde la vertical del borde de la mesa. 2,6 m 3) Se lanza un cuerpo horizontalmente desde lo alto de un acantilado con una velocidad inicial de 72 km/h. El cuerpo cae a una distancia de 40 m medidos desde la vertical del punto de lanzamiento. Determina: a) la altura del acantilado. Sol: 19,6 m b) la velocidad del cuerpo al llegar al mar. Sol: 28 m/s 4) Un especialista debe realizar un salto en moto desde un acantilado que se encuentra a 50 m sobre el suelo. Calcula la velocidad con la que tiene que saltar (sólo tiene componente horizontal) para llegar a donde está la cámara. Nota: la cámara se encuentra a 90 m desde la vertical que define el acantilado. 28,2 m/s 5) Se lanza horizontalmente una pelota desde un balcón a 10 m de altura sobre el suelo y cae a 6 m de la vertical de la terraza. Determina: a) tiempo que tarda en tocar el suelo. Sol: 1,43 s b) velocidad con la que fue lanzado. Sol: 4,2 m/s 6) Desde un helicóptero que se encuentra a 100 m de altura y se desplaza horizontalmente con una velocidad de 20 m/s se deja caer un paquete de alimentos. Determina: a) las ecuaciones de movimiento en X e Y. b) la ecuación de la trayectoria. c) posición de impacto con el su elo medida desde la vertical en el inistante de lanzamiento. d) la velocidad en el momento del impacto con el suelo. 7.3.2.-LANZAMIENTO OBLICUO En este vídeo vemos varios lanzamientos del lanzador de javalina checo Jan Zelezny que ostenta el record. El tipo de movimiento que sigue la jabalina corresponde al lanzamiento oblicuo. Podéis ver en el vídeo la trayectoria de la jabalina desde que es lanzada hasta que se queda clavada en el suelo. Es realmente impresionante. simulador virtual del movimiento de un proyectil simulador virtual del movimiento de un proyectil en el que podemos ver las coordenadas de posición en cada momento. la velocidad...y finalmente la altura máxima y el alcance. otro simulador virtual del movimiento parabólico; en esta ocasión es una moto lo que se mueve. a) lanzamiento oblicuo desde el suelo El objeto es lanzado formando un cierto ángulo con el suelo. Por eso, la velocidad inicial se puede descomponer en v en dirección X y v en dirección Y, y analizar por separado el movimiento en X y el movimiento en Y. En X, como no hay ninguna fuerza actuando sobre el objeto mientras vuela, su velocidad se mantiene constante. En Y, sin embargo, actúa la fuerza gravitatoria terrestre; por eso, va disminuyendo su velocidad hasta llegar al punto de máxima altura (A), donde vale 0 y, a partir de ahí, comienza a crecer y llega a tener su máximo valor justo en el momento del impacto. En los campeonatos mundiales de atletismo de Tokio 1991, Mike Powell saltó 8,95 m, batiendo la mítica plusmarca de 8,90 m que poseía Bob Belmonte desde los Juegos Olímpicos de México (1968). Se supone que Mike Powell inició el salto con una velocidad de 9,90 m/s. Determina: a) el ángulo con el que inició el salto. Sol: 31,8º b) el tiempo que permaneció en el aire. Sol: 1,06 s c) la altura máxima que alcanzó en el salto. Sol: 1,39 m Un jugador de rugby patea un balón hacia los palos. La velocidad de salida del balón es 105 km/h y el ángulo de lanzamiento es de 30º. La portería se encuentra a 65 m del punto de lanzamiento y el palo transversal está elevado 3,0 m sobre el césped. Determina: a) tiempo de vuelo. Sol: 2,98 s b) alcance máximo. Sol: 75,3 m c) altura máxima que alcanza el balón. Sol: 10,9 m d) velocidad del balón 5 s después de ser pateado el balón (componente en x y en y). Sol: vx= 25,3 m/s; vy=34,5 m/s Tiro oblicuo en el cine: SPEED E n la película SPEED un grupo de personas van en un autobús que lleva una bomba que explotará si el autobús lleva una velocidad inferior a 80 km/h. El autobús tiene que cruzar un puente en obras que tiene un agujero de 16 m. Al llegar a él deciden saltar con el autobús. La inclinación del puente es de 5º. Al inicial el salto el velocímetro marca 108 km/h. El autobús consigue saltar y las personas sobreviven. ¿Estos datos están de acuerdos con la física? Compruébalo usando las ecuaciones de movimiento. Análisis cinemático de la escena del salto Lanzamiento oblicuo desde el suelo en el deporte Un portero de fútbol golpea la pelota con una velocidad de 15 m/s y un ángulo de lanzamiento de 60º. Determina los dos instantes en que el vector velocidad forma ángulos de 45º y -45º con la horizontal. Escribe las coordenadas de las posiciones de la pelota en esos instantes. Solución: 0,56s; r1= (4,2 i +5,7 j)m; 2,1s y r2= (16 i + 5,7 j)m b) lanzamiento oblicuo desde cierta altura El objeto se lanza con cierta velocidad inicial que forma cierto ángulo con la horizontal; por tanto, el vector velocidad inicial se puede descomponer en la componente en dirección X y la componente en dirección Y. En cuanto se encuentra en el aire, queda sometido sólo a la fuerza gravitatoria terrestre (despreciando el rozamiento con el aire) que lleva dirección Y y que le proporciona una aceleración constante en dirección Y que es g= 9,8 m/s2 En dirección X tiene, por tanto MRU. La componente en X de la velocidad vale, durante todo el tiempo, lo que valía inicialmente. En dirección Y tiene MRUA donde a es la aceleración de la gravedad (9,8 m/s2). La velocidad resultante en cada punto de la trayectoria, que es tangente a la trayectoria siempre, es suma de dos vectores perpendiculares: vx tiene módulo constante y vx tiene un módulo que va disminuyendo hasta valer 0 en el punto más alto de la trayectoria. A partir de ahí, durante el descenso, comienza a aumentar. Su valor máximo lo tiene en el momento del impacto. Desde lo alto de una torre de 45 m de altura se lanza hacia arriba una piedra con una velocidad inicial de 20m/s formando un ángulo con la horizontal de 30º. Determina: a) tiempo que tarda en alcanzar la altura máxima. b) tiempo de vuelo. Sol: 4,22 s c) alcance máximo. Sol: 63,2m d) velocidad cuando impacta contra el suelo. Sol: 35,9 m/s Un jugador de baloncesto lanza a canasta desde 2,00 m de altura con una velocidad de 10,7 m/s y un ángulo de 40º. La pelota tarda 1,22 s en llegar a la canasta. Determina: a) la altura máxima que alcanza la pelota. Sol: 4,41 m b) la velocidad de la pelota a los 0,72 s. Sol: (8,2 i -0,18j) m/s Se lanza un proyectil desde una torre de 50 m de altura formando 30º con la horizontal. La velocidad de lanzamiento es 350 m/s. Calcula: a) tiempo que tarda en caer al suelo. Sol: 36 s b) alcance máximo. Sol: 10.912 m c) altura máxima. Sol: 1613 m Google Sites Report abuse
15019	https://pmc.ncbi.nlm.nih.gov/articles/PMC4462841/	Oedema in kwashiorkor is caused by hypoalbuminaemia - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Paediatr Int Child Health . 2015 May;35(2):83–89. doi: 10.1179/2046905514Y.0000000154 Search in PMC Search in PubMed View in NLM Catalog Add to search Oedema in kwashiorkor is caused by hypoalbuminaemia Malcolm G Coulthard Malcolm G Coulthard 1 Great North Children’s Hospital, Newcastle-upon-Tyne, UK Find articles by Malcolm G Coulthard 1 Author information Copyright and License information 1 Great North Children’s Hospital, Newcastle-upon-Tyne, UK ✉ Correspondence to: M G Coulthard, South Park House, South Park, Hexham, Northumberland, NE46 1BS, UK. Email: jeanandmalc@gmail.com © W. S. Maney & Son Ltd 2015 MORE OpenChoice articles are open access and distributed under the terms of the Creative Commons Attribution Non-Commercial License 3.0 PMC Copyright notice PMCID: PMC4462841 PMID: 25223408 Abstract It has been argued that the oedema of kwashiorkor is not caused by hypoalbuminaemia because the oedema disappears with dietary treatment before the plasma albumin concentration rises. Reanalysis of this evidence and a review of the literature demonstrates that this was a mistaken conclusion and that the oedema is linked to hypoalbuminaemia. This misconception has influenced the recommendations for treating children with severe acute malnutrition. There are close pathophysiological parallels between kwashiorkor and Finnish congenital nephrotic syndrome (CNS) pre-nephrectomy; both develop protein-energy malnutrition and hypoalbuminaemia, which predisposes them to intravascular hypovolaemia with consequent sodium and water retention, and makes them highly vulnerable to develop hypovolaemic shock with diarrhoea. In CNS this is successfully treated with intravenous albumin boluses. By contrast, the WHO advise the cautious administration of hypotonic intravenous fluids in kwashiorkor with shock, which has about a 50% mortality. It is time to trial intravenous bolus albumin for the treatment of children with kwashiorkor and shock. Keywords: Oedematous malnutrition, Kwashiorkor, Oedema, Hypoalbuminaemia, Albumin, Nephrotic syndrome Introduction Malnutrition in young children may lead to severe wasting alone (marasmus), or may be associated with oedema (kwashiorkor). The high mortality of severe acute kwashiorkor has changed little1 since it was first described in 1933,2 and about half of children who present today with shock still die. The World Health Organisation (WHO) recommend treating marasmus and kwashiorkor with the same fluid regimen when it is associated with shock,3 as if they shared precisely the same pathophysiology. During the 1950s it was recognised that the presence of oedema in kwashiorkor was correlated with a very low plasma albumin concentration, presumably related to a dietary lack of protein.4 The closeness and importance of this link was identified in the early 1970s5–7 (Fig. 1), and its clinical predictive value has been confirmed since.8 However, in 1980 Golden and co-workers reported that there was not a causal link between the oedema of malnutrition and the low plasma oncotic pressure induced by hypoalbuminaemia,9 and this triggered extensive efforts to explain their disordered fluid physiology in other ways,10 including by the effects of specific micronutrient deficiencies, oxidant stresses and glutathione deficiency.11–13 He warned that the assumption that the oedema was directly related to hypoalbuminaemia could lead to therapeutic error.12 Here I review the pathophysiological evidence for a causal link between the oedema of kwashiorkor and hypoalbuminaemia, and consider what the therapeutic implications of this might be. Figure 1. Open in a new tab The relationships between the plasma albumin concentration in children with severe malnutrition and (a) the percentage chance of them having signs of oedema, and (b) their mortality risk, as identified by Whitehead5 and Hay7 in the early 1970s. What was the Evidence that Hypoalbuminaemia does not Cause Kwashiorkor Oedema? Albumin is a relatively small protein, so it contributes disproportionately to the plasma oncotic pressure, and in health is its major contributor. Starling's equation14 explains how the movement and distribution of water between the plasma and tissue spaces of all tissues is physically regulated by the balance of hydrostatic and oncotic pressures across capillary blood vessel walls. However, Golden ruled out this mechanism as the primary cause for oedema in kwashiorkor by demonstrating that children who he treated with a relatively low protein diet showed marked clinical improvement and lost their oedema before their plasma albumin concentrations had risen.9 The evidence for Golden’s unexpected finding was presented entirely graphically (Fig. 2a), without any corroborating statistical tests, accompanied by the observations that the mean albumin concentrations ‘did not change’, and that ‘only one child had a substantial rise’.9 The fact that these data were plotted in a physically small area (3×2 cm), with relatively wide aspect-ratio axes, and with descriptive text that happened to be slightly misaligned, may have contributed to the visual impression that the lines were approximately horizontal. By scanning and enlarging the figure and constructing a grid from the y-axis to obtain the numerical data, and re-plotting these values with a conventional aspect ratio and horizontal text (Fig. 2b), it can be seen that the plasma albumin levels had risen by the time that the oedema had improved. Furthermore, a two-tailed independent t-test confirms that this was a statistically significant rise in the mean albumin level (P = 0.02). Paired values can be discerned for six of the 13 cases (denoted by filled circles), and by combining the remaining seven cases in every possible way, paired t-tests show that the true P-value was somewhere between 0.003 and 0.0007. Figure 2. Open in a new tab Graphs of the changes in plasma albumin concentrations in children on dietary treatment for kwashiorkor before, during and after the disappearance of oedema, from Golden et al, 1980.9 In graphs b to d, the filled circles represent paired pre- and post-treatment levels, and the open circles are cases where the correct patient pairing is not known. Measuring Plasma Albumin Concentrations Albumin concentrations can be measured accurately by using specific immunological assays that only respond to that particular protein, even at very low levels.15 However, these techniques are not suitable for routine laboratory analysis, and instead dye-binding is used to provide approximate measurements. These methods rely on the fact that proteins have negatively charged surfaces that bind readily to certain positively charged dyes such as bromcresol-green (BCG), and that gram-for-gram, albumin binds more avidly than most of the globulins. However, globulins do bind with BMG, so when the albumin levels are very low this causes the measurements to be disproportionately high. For example, a plasma with no albumin could be reported as having as much as 15 g/L.16 Claims that this imprecision and skewing can be minimised by technical changes to the methodology or other dyes have not been confirmed.17 I have therefore estimated the likely true albumin concentrations from Golden’s publication as 1.4× (BCG – 14).17 Albumin estimates made from total protein measurements and electrophoresis analysis fall approximately half way between the BCG and true values. The impact of using corrected albumins instead of BCG or electrophoresis values can be seen in Figure 2c, which demonstrates just how severely hypoalbuminaemic these children actually were on arrival. Finally, the true impact on their plasma albumin levels of feeding these children is most obvious when its increase is plotted for the period when they lost their oedema (Fig. 2d). Here, the P-value for all of the possible t-test permutations reaches <0.0001. What do other Studies of Albumin Levels in Kwashiorkor Show? Though few other groups have presented their data in the same way as Golden, many other studies also recorded children’s plasma albumin concentrations when they presented with kwashiorkor and marasmus,18–29 or when those with kwashiorkor were given appropriate dietary treatment,19,22,24,30–36 in some cases whilst also comparing the efficacy of different milk formulas.33–36 Many of these studies were designed to elucidate the roles of other specific elements, such as vitamin deficiencies, but also included the albumin data, either as a list, a statistical parameter, or a plot, which allowed me to present them in a common graphic format in Figure 3. Both plots demonstrate just how low the true plasma albumin concentrations are in kwashiorkor. Figure 3a shows that in each study which included children with both marasmus and kwashiorkor, the mean albumin concentrations were consistently lower in kwashiorkor. Though there is some overlap between different studies, this may in part be owing to technical differences, such as measurement variations. In each study in which sufficient information was provided to make it possible to test the statistical significance of these differences, the P-values were all <0.05, and a paired t-test of the combined means gave a P-value of <0.0001. Figure 3. Open in a new tab The corrected albumin concentrations measured in children with kwashiorkor (a) compared to children with marasmus in 12 studies, and (b) before and after feeding in 10 studies, four of which tested two different milks. Golden’s study detailed in Figure 2 is shown by filled circles and a broken line in graph (b), and the other lines are identified by the text references. Figure 3b shows that the plasma albumin rises promptly when appropriate milk feeds are introduced, with a mean daily increase of about 1.1 g/L. This compares to a mean daily increase of 0.7 g/L in Golden’s study.9 As with graph (a), every study in which the data could be evaluated statistically showed a significant increase on feeding, and the combined means showed a highly significant improvement. Some, but not all, of the reports indicated how long it took for the oedema to disappear, and these intervals were typically in the range of 6–12 days. These data provide no support for the hypothesis that the oedema resolved before the albumin rose. The Physiology of Kwashiorkor Looks a lot like Finnish Congenital Nephrotic Syndrome The evidence I have reviewed thus far points to the pathophysiology of kwashiorkor being a combination of severe malnutrition and a low plasma oncotic pressure due to extreme hypoalbuminaemia. This closely resembles the pathophysiology of untreated Finnish congenital nephrotic syndrome (CNS), though of course the mechanism leading to them acquiring protein-energy malnutrition is very different. Infants with CNS simply cannot retain albumin, nor the smaller globulins, and waste vast quantities of energy. Today, children with CNS are managed very actively in developed countries, with drug treatment or unilateral nephrectomy to limit their proteinuria,37 or bilateral nephrectomy to stop it,38 followed by dialysis and transplantation. However, before this CNS was universally fatal by 18 months of age; children failed to thrive, and died of protein-energy malnutrition before they were old enough to develop renal failure.39 They were highly vulnerable to infections (despite penicillin prophylaxis), and had persistent oedema. Like children with kwashiorkor,40 they had markedly increased platelet stickiness. Low-dose aspirin is used to counter this in CNS, but of course if the same mechanism was responsible in kwashiorkor it would correct as the albumin rises with nutritional treatment. The two conditions also share similarly altered hormonal profiles. Much attention has been drawn towards the low glutathione levels seen in kwashiorkor but not in marasmus.12 These have been interpreted as reflecting high levels of oxidant stress, and there has been speculation that this may be important in driving the development of the oedema. It was argued that the oedema of kwashiorkor could not be a consequence of hypoalbuminaemia as glutathione levels were said to be normal in nephrotic patients. However, this assertion was only based on one case in a study of children with kwashiorkor who had had a normal glutathione level and heavy proteinuria, who it was speculated “was probably a misdiagnosed nephrotic.”11 However, many studies have established that glutathione levels are low in persistent nephrotic syndrome.41,42 Although the exact relationship between reduced albumin and glutathione levels remains uncertain, they appear to be the consequence and not the cause of severe persistent low plasma albumin levels. The major feature common to both kwashiorkor and CNS, however, is their disordered fluid balance physiology. Children with persistent nephrotic syndrome lose plasma water into the interstitium because of their low oncotic pressure, and as a consequence have chronic intra-vascular hypovolaemia. This induces avid water retention by an increased secretion of arginine vasopressin (antidiuretic hormone) in a non-osmolar response to hypovolaemia, and avid sodium retention by increased plasma renin activity and consequent secondary hyperaldosteronism, as well as by suppression of the release of the natriuretic peptides. This therefore leads to fluid retention and oedema, which is exacerbated if the child receives greater quantities of salt. The presence of oedema increases the interstitial pressure which therefore slows the accumulation of more oedema by balancing the Starling forces.14 Hence a stable situation evolves in which the child is persistently intra-vascularly hypovolaemic, has constant oedema, typically has a normal blood pressure, and has a tendency to slight hyponatraemia. Reducing the salt intake usually moderates the oedema, and there is a constant vulnerability to be ‘pushed over’ into frank clinical hypovolaemia with mild additional stresses to fluid balance, such as a bout of diarrhoea. Children with kwashiorkor are also markedly hypovolaemic and respond hormonally to this in the same way as nephrotic children. Viart demonstrated that children with severe malnutrition had a reduced blood volume compared to controls by re-injecting them with their own 51 Cr-labelled red blood cells.43 He did not separately analyse children with marasmus and kwashiorkor, but his published data has allowed me to compare the albumin concentrations and total blood volumes (ml/kg of oedema-free weight) of children aged <3 years with either ‘0 or ±’ oedema (marasmus, n = 4) or with at least ‘++’ oedema on presentation (kwashiorkor, n = 17). The mean blood volume in marasmus is 90% of normal values, and in kwashiorkor it is just 80% (Fig. 4). Children with kwashiorkor also respond with very high vasopressin levels, which are higher than seen in marasmus, and which fall back to normal after loss of oedema following therapeutic feeding.44 Similarly, plasma renin activity is much higher in kwashiorkor than in control children, and highest by far in those who died acutely.24,45 Although these data fit precisely the physiological pattern of persistent nephrotic states, the hormonal changes confounded their authors at the time because they believed then that children with kwashiorkor were hypervolaemic. This was because early workers who attempted to measure the blood volume in children46 and animals47,48 with malnutrition measured the albumin space rather than the red cell space,43 despite them having clinical or sub-clinical oedema. Figure 4. Open in a new tab Corrected albumin concentrations and total blood volume measurements in children with kwashiorkor, marasmus, and healthy controls, from Viart.43 The malnourished children selected for this comparison were aged <3 years, with kwashiorkor defined as having ≧2+ oedema, and marasmus as having 0 or ± oedema. The error bars show the mean and standard deviation values. Although the body responds very quickly to the well described hormonal signals that are triggered by hypovolaemia, these immediate physiological adaptations are not the only ones that may occur, and prolonged exposure induces morphological and functional changes to the kidney.49 This means that patients with impending hypovolaemia from persistently low albumin concentrations are better able to produce concentrated urine and will be less likely to be so markedly oedematous as those who reach this state rapidly. This is why children with steroid-sensitive nephrotic syndrome (‘minimal change disease’) who present or relapse suddenly after an immunological stimulus may develop quite severe oedema initially, which may then lessen or even disappear prior to their loss of proteinuria as their renal functional capacity increases. However, this further up-regulation does not occur in kwashiorkor or CNS, as these adaptations will have already taken place. What are the Implications of this for Fluid Treatments for Children with Severe Acute Malnutrition? All children who present with severe acute malnutrition may have serious complicating factors, but those who are not shocked are overwhelmingly likely to survive if they are treated according to WHO guidelines.3 For this group, differentiating between marasmus and kwashiorkor, and having a precise understanding of the physiology of oedema development, has little clinical relevance. However, it makes a vital difference when it comes to treating malnourished children who also have shock. Marasmic children, whose hypovolaemic shock is caused by an acute loss of salt and water uncomplicated by hypoalbuminaemia, then require an intravenous infusion of sufficient isotonic fluid to promptly restore the circulating blood volume. This allows oxygen delivery and perfusion of the organs,50 without perturbing the intra:extra-cellular tonicity gradients and thereby disrupting the volume and functioning of the body’s cells. A rapid 20-ml/kg bolus of an isotonic fluid with glucose, repeated as necessary, would fulfil these logical and physiologically-based criteria.50 By contrast, children with severe albumin deficiency from any cause continuously ‘struggle’ physiologically to maintain their blood volume by driving hormonal pathways that are normally only called upon in a crisis. They have no mechanisms in reserve; the mildest extra stress can rapidly precipitate severe shock. If that child happens to be one with CNS in a developed country, they will receive a prompt intravenous albumin infusion, and almost at once their signs of shock will wane as interstitial fluid is drawn into their blood vessels. A dose of frusemide administered soon after this will prevent rebound hypervolaemia and pulmonary oedema. They will mobilise large quantities of oedema as urine, re-establish a stable circulation, and will have a virtually guaranteed survival. However, if that same child was treated with just 30 ml/kg over 2 hours of half-strength Darrow's solution with 5% dextrose (hypotonic crystalloid; sodium 61 mmol/l), they may show a transient improvement as the fluid was delivered, but they would then deteriorate as the water leaked away into the tissues, and would have a high chance of dying. Yet this is what is recommended for shocked children whose hypoalbuminaemia happens to be caused by kwashiorkor.3 No distinction is made by the WHO between managing shock in marasmus and kwashiorkor, despite the fact that mortality is linked directly to the degree of oedema2,5 and hypoalbuminaemia.8,24 For this group, the mortality remains at around 50% in many parts of the world.51 The adoption of relatively conservative resuscitation fluid volumes for malnourished children has been driven in part by the concerns that larger quantities may precipitate congestive cardiac failure. This followed the fact that some very anaemic children died of heart failure after a few days of apparently successful progress on a therapeutic diet which contained a high salt content.52 However, this did not prove to be a problem in a randomised controlled trial of standard vs greater volume resuscitation, despite the severe warnings that this is likely to happen.51,53 Indeed, Viart very clearly described children with kwashiorkor dying as if they were still hypovolaemic, with none showing any evidence of congestive failure.43 Conclusion The mistaken belief that the oedema of kwashiorkor is unrelated to profound hypoalbuminaemia, combined with an exaggerated concern about the risks of congestive cardiac failure, has resulted in guidelines for shock management that fail to address their physiological needs, and which has not reduced their high mortality rate. Rather, children with kwashiorkor and CNS share a similar pathophysiology; both are malnourished and verge on intravascular hypovolaemia due to hypoalbuminaemia, and can be readily precipitated into shock. Treating this with intravenous albumin is life-saving in CNS; treating it late with modest volumes of hypotonic fluid has a 50% mortality in kwashiorkor. It is time for a trial of acute intravenous albumin therapy in children with kwashiorkor-related shock. Acknowledgments I am grateful to Unni Wariyar for advice and support in developing this hypothesis, and to Stella Kyoyagala (Mbarara Hospital; Mbarara University of Science & Technology, Uganda) who drew my attention to the discrepancies between my teaching on managing children with nephrotic syndrome, and her previous teaching on resuscitating children with kwashiorkor. References 1.Schofield C, Ashworth A. Why have mortality rates for severe malnutrition remained so high? Bull WHO. 1996;74:223–9. [PMC free article] [PubMed] [Google Scholar] 2.Williams CD. A nutritional disease of childhood associated with a maize diet. Arch Dis Child. 1933;8:423–33. doi: 10.1136/adc.8.48.423. [DOI] [PMC free article] [PubMed] [Google Scholar] 3.Acute severe malnutrition. Geneva: WHO; 2013. pp. 197–222. In: Pocket Book of Hospital Care for Children: Guidelines for the Management of Common Childhood Illnesses. [Google Scholar] 4.Gitlin D, Cravioto J, Frenk S, Montano EL, Galvan RR, Gomez F, et al. Albumin metabolism in children with protein malnutrition. J Clin Invest. 1958;37:682–6. doi: 10.1172/JCI103654. [DOI] [PMC free article] [PubMed] [Google Scholar] 5.Whitehead RG, Frood JDL, Poskitt EME. Value of serum-albumin measurements in nutritional surveys: a reappraisal. Lancet. 1971;298:287–9. doi: 10.1016/s0140-6736(71)91334-1. [DOI] [PubMed] [Google Scholar] 6.Whitehead RG, Coward WA, Lunn PG. Serum-albumin concentration and the onset of kwashorkor. Lancet. 1973;301:63–6. doi: 10.1016/s0140-6736(73)90465-0. [DOI] [PubMed] [Google Scholar] 7.Hay RW, Whitehead RG, Spicer CC. Serum-albumin as a prognostic indicator in oedematous malnutrition. Lancet. 1975;306:427–9. doi: 10.1016/s0140-6736(75)90843-0. [DOI] [PubMed] [Google Scholar] 8.Brasseuer D, Hennart P, Dramaix M, Bahwere P, Donnen P, Tonglet R, et al. Biological risk factors for fatal protein energy malnutrition in hospitalised children in Zaire. J Pediatr Gastroenterol Nutr. 1994;18:220–4. doi: 10.1097/00005176-199402000-00016. [DOI] [PubMed] [Google Scholar] 9.Golden MHN, Golden BE, Jackson AA. Albumin and nutritional oedema. Lancet. 1980;315:114–16. doi: 10.1016/s0140-6736(80)90603-0. [DOI] [PubMed] [Google Scholar] 10.Sauerwein RW, Mulder JA, Mulder L, Lowe B, Peshu N, Demacker PNM, et al. Inflammatory markers in children with protein-energy malnutrition. Am J Clin Nutr. 1997;65:1534–9. doi: 10.1093/ajcn/65.5.1534. [DOI] [PubMed] [Google Scholar] 11.Golden MJN, Ramdath D. Free radicals in the pathogenesis of kwashiorkor. Proc Nutr Soc. 1987;46:53–68. doi: 10.1079/pns19870008. [DOI] [PubMed] [Google Scholar] 12.Golden MHN. Oedematous malnutrition. Brit Med Bull. 1998;54:433–44. doi: 10.1093/oxfordjournals.bmb.a011699. [DOI] [PubMed] [Google Scholar] 13.Golden MHN. The development of concepts of malnutrition. J Nutr. 2002;132:2117–22S. doi: 10.1093/jn/132.7.2117S. [DOI] [PubMed] [Google Scholar] 14.Starling EH. On the absorption of fluid from the connective tissue spaces. J Physiol. 1896;19:312–26. doi: 10.1113/jphysiol.1896.sp000596. [DOI] [PMC free article] [PubMed] [Google Scholar] 15.Carfray A, Patel K, Wihitaker P, Garrick P, Griffiths GJ, Warwick GL. Albumin as an outcome measure in haemodialysis in patients: the effect of variation in assay method. Nephrol Dial Transpl. 2000;15:1819–22. doi: 10.1093/ndt/15.11.1819. [DOI] [PubMed] [Google Scholar] 16.Webster D, Bignell AHC, Attwood EC. An assessment of the suitability of bromcresol green for the determination of serum albumin. Clin Chim Acta. 1974;53:101–8. doi: 10.1016/0009-8981(74)90357-x. [DOI] [PubMed] [Google Scholar] 17.Ingwersen S, Raabo E. Improved and more specific bromocresol green methods for the manual and automatic determination of serum albumin. Clin Chim Acta. 1978;88:545–50. doi: 10.1016/0009-8981(78)90290-5. [DOI] [PubMed] [Google Scholar] 18.Zaklama MS, Gabr MK, El Maraghy S, Patwardhan VN. Liver vitamin A in protein-calorie malnutrition. Am J Clin Nutr. 1972;25:412–18. doi: 10.1093/ajcn/25.4.412. [DOI] [PubMed] [Google Scholar] 19.Zaklama MS, Gabr MK, El Maraghy S, Patwardhan VN. Serum vitamin A in protein-calorie malnutrition. Am J Clin Nutr. 1973;26:1202–6. doi: 10.1093/ajcn/26.11.1202. [DOI] [PubMed] [Google Scholar] 20.Baertl JM, Placko RP, Graham GG. Serum proteins and plasma free amino acids in severe malnutrition. Am J Clin Nutr. 1974;27:733–42. doi: 10.1093/ajcn/27.7.733. [DOI] [PubMed] [Google Scholar] 21.Raghuramulu N, Jaya Rao KS. Growth hormone secretion in protein calorie malnutrition. J Clin Endocr Metab. 1974;38:176–80. doi: 10.1210/jcem-38-2-176. [DOI] [PubMed] [Google Scholar] 22.Schwartz FCM, Lunat M, Wolfsdorf J. Blood xylose concentrations in protein energy malnutrition. S Afr Med J. 1974;48:2387–90. [PubMed] [Google Scholar] 23.Smith FR, Suskind R, Thanangkul O, Leitzmann C, Goodman DS, Olson RE. Plasma vitamin A, retinol-binding protein and prealbumin concentrations in protein-calorie malnutrition. Ill. Response to varying dietary treatments. Am J Clin Nutr. 1975;28:732–8. doi: 10.1093/ajcn/28.7.732. [DOI] [PubMed] [Google Scholar] 24.van der Westhuysen JM, Kanengoni E, Jones JJ, van Niekerk CH. Plasma renin activity in oedematous and marasmic children with protein energy malnutrition. S Afr Med J. 1975;49:1729–31. [PubMed] [Google Scholar] 25.Khan L, Bamji MS. Plasma carnitine levels in children with protein-calorie malnutrition before and after rehabilitation. Clin Chim Acta. 1977;75:163–6. doi: 10.1016/0009-8981(77)90513-7. [DOI] [PubMed] [Google Scholar] 26.Oladunni Taylor G, Agbedana EO, Johnson AOK. High-density-lipoprotein-cholesterol in protein-energy malnutrition. Brit J Nutr. 1982;47:489–94. doi: 10.1079/bjn19820061. [DOI] [PubMed] [Google Scholar] 27.Soliman AT, Hassan AEHI, Aref MK, Hintz RL, Rosenfeld RG, Rogol AD. Serum insulin-like growth factors I and II concentrations and growth hormone and insulin responses to arginine infusion in children with protein-energy malnutrition before and after nutritional rehabilitation. Pediatr Res. 1986;20:1122–30. doi: 10.1203/00006450-198611000-00012. [DOI] [PubMed] [Google Scholar] 28.Akenami FOT, Marjaleena K, Siimes MA, Ekanem EE, Bolarin DM, Vaheri A. Assessment of plasma fibronectin in malnourished Nigerian children. J Pediatr Gastroenterol Nutr. 1997;24:183–8. doi: 10.1097/00005176-199702000-00012. [DOI] [PubMed] [Google Scholar] 29.Spoelstra MN, Mari A, Mendel M, Senga E, van Rjeenen P, van Dijk TH, et al. Kwashiorkor and marasmus are both associated with impaired glucose clearance related to pancreatic β-cell dysfunction. Metabolism. 2012;61:1224–30. doi: 10.1016/j.metabol.2012.01.019. [DOI] [PubMed] [Google Scholar] 30.Srikantia SG, Sahgal S. Use of cottonseed protein in protein-calorie malnutrition. Am J Clin Nutr. 1968;21:212–16. doi: 10.1093/ajcn/21.3.212. [DOI] [PubMed] [Google Scholar] 31.Watson CE, Freesemann C. Immunoglobulins in protein-calorie malnutrition. Arch Dis Child. 1970;45:282–4. doi: 10.1136/adc.45.240.282. [DOI] [PMC free article] [PubMed] [Google Scholar] 32.Coward WA, Whitehead RG. Changes in serum β-lipoprotein concentration during the development of kwashiorkor and in recovery. Brit J Nutr. 1972;27:383–94. doi: 10.1079/bjn19720104. [DOI] [PubMed] [Google Scholar] 33.Roode H, Prinslol JG, Kruger H, Freier E. Effects of an acidified and a non-acidified milk formula on diarrhoea, body mass and serum albumin levels of kwashiorkor patients. S Afr Med J. 1972;46:1134–6. [PubMed] [Google Scholar] 34.Waslien CI, Kamel K, El-Ramly Z, Carter JP, Mourad KA, Khattab A-K, et al. Folate requirements of children. 1. A formula diet low in folic acid for study of folate deficiency in protein-calorie malnutrition. Am J Clin Nutr. 1972;25:147–51. doi: 10.1093/ajcn/25.2.147. [DOI] [PubMed] [Google Scholar] 35.Reddy V, Gupta CP. Treatment of kwashiorkor with opaque-2 maize. Am J Clin Nutr. 1974;27:122–4. doi: 10.1093/ajcn/27.2.122. [DOI] [PubMed] [Google Scholar] 36.Baker RD, Baker SS, Margo GM, Reuter HH. Successful use of a soya-maize mixture in the treatment of kwashiorkor. S Afr Med J. 1978;53:674–7. [PubMed] [Google Scholar] 37.Coulthard MG. Management of Finnish congenital nephrotic syndrome by unilateral nephrectomy. Pediatr Nephrol. 1989;3:451–3. doi: 10.1007/BF00850226. [DOI] [PubMed] [Google Scholar] 38.Holmberg C, Antikainen M, Ronnholm K, Ala-Houhala M, Jalanko H. Management of congenital nephrotic syndrome of the Finnish type. Pediatr Nephrol. 1995;9:87–93. doi: 10.1007/BF00858984. [DOI] [PubMed] [Google Scholar] 39.Huttunen N-P. Congenital nephrotic syndrome of Finnish type: study of 75 patients. Arch Dis Child. 1976;51:344–8. doi: 10.1136/adc.51.5.344. [DOI] [PMC free article] [PubMed] [Google Scholar] 40.Khalil M, Aref MK, Mahmoud S, Abdel-Malek AT, Guirgis FK, Moghazy M, et al. Platelet adhesiveness, plasma free fatty acids, and serum triglycerides in kwashiorkor. Arch Dis Child. 1974;49:568–70. doi: 10.1136/adc.49.7.568. [DOI] [PMC free article] [PubMed] [Google Scholar] 41.Ginevri F, Ghiggeri GM, Candiano G, Oleggini R, Bertelli R, Piccardo MT, et al. Peroxidative damage of the erythrocyte membrane in children with nephrotic syndrome. Pediatr Nephrol. 1989;3:25–32. doi: 10.1007/BF00859620. [DOI] [PubMed] [Google Scholar] 42.Granqvist A, Nilsson UA, Ebefors K, Haraldsson B, Nystrom J. Impaired glomerular and tubular antioxidative defense mechanisms in nephrotic syndrome. Am J Physiol Renal. 2010;299:F898–904. doi: 10.1152/ajprenal.00124.2010. [DOI] [PubMed] [Google Scholar] 43.Viart P. Blood volume (51Cr) in severe protein-calorie malnutrition. Am J Clin Nutr. 1976;29:25–37. doi: 10.1093/ajcn/29.1.25. [DOI] [PubMed] [Google Scholar] 44.Srikantia SG, Mohanram M. Antidiuretic hormone values in plasma and urine of malnourished children. J Clin Endocrinol Metab. 1970;31:312–14. doi: 10.1210/jcem-31-3-312. [DOI] [PubMed] [Google Scholar] 45.Kritzinger EAE, Kanengoni E, Jones JJ. Plasma renin activity in children with protein energy malnutrition (kwashiorkor). S Afr Med J. 1974;48:499–501. [PubMed] [Google Scholar] 46.Alleyne GAO. Plasma and blood volumes in severely malnourished Jamaican children. Arch Dis Child. 1966;41:313–15. doi: 10.1136/adc.41.217.313. [DOI] [PMC free article] [PubMed] [Google Scholar] 47.Warton CMR, Kanengoni E, Jones JJ. Plasma renin activity in rats fed exclusively on maize (mealie) meal. S Afr Med J. 1973;47:1498–500. [PubMed] [Google Scholar] 48.van der Westhuysen JM, Kanengoni E, Mbizvo M, Jones JJ. The effect of protein energy malnutrition on plasma renin and oedema in the pig. S Afr Med J. 1977;51:18–20. [PubMed] [Google Scholar] 49.Bouby N, Fernandes S. Mild dehydration, vasopressin and the kidney: animal and human studies. Eur J Clin Nutr. 2003;57:S39–46. doi: 10.1038/sj.ejcn.1601900. [DOI] [PubMed] [Google Scholar] 50.Raman S, Peters MJ. Fluid management in the critically ill child. Pediatr Nephrol. 2014;29:23–34. doi: 10.1007/s00467-013-2412-0. [DOI] [PubMed] [Google Scholar] 51.Maitland K. Severe malnutrition: therapeutic challenges and treatment of hypovolaemic shock. Proc Nutr Soc. 2009;68:274–80. doi: 10.1017/S0029665109001359. [DOI] [PubMed] [Google Scholar] 52.Wharton BA, Howells GR, McCance RA. Cardiac failure in kwashiorkor. Lancet. 1967;290:384–7. doi: 10.1016/s0140-6736(67)92006-5. [DOI] [PubMed] [Google Scholar] 53.Akech SO, Karisa J, Nakamya P, Boga M, Maitland K. Phase II trial of isotonic fluild resuscitation in Kenyan children with severe malnutrition and hypovolaemia. BMC Pediatr. 2010;10:71. doi: 10.1186/1471-2431-10-71. [DOI] [PMC free article] [PubMed] [Google Scholar] Articles from Paediatrics and International Child Health are provided here courtesy of Maney Publishing ACTIONS View on publisher site PDF (1.6 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Introduction What was the Evidence that Hypoalbuminaemia does not Cause Kwashiorkor Oedema? Measuring Plasma Albumin Concentrations What do other Studies of Albumin Levels in Kwashiorkor Show? The Physiology of Kwashiorkor Looks a lot like Finnish Congenital Nephrotic Syndrome What are the Implications of this for Fluid Treatments for Children with Severe Acute Malnutrition? Conclusion Acknowledgments References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
15020	https://www.youtube.com/watch?v=9blB-uMTIAM	Introduction to Galvanic Cells & Voltaic Cells The Organic Chemistry Tutor 9700000 subscribers 12487 likes Description 833969 views Posted: 16 Dec 2017 This chemistry video tutorial provides a basic introduction into electrochemical cells such as galvanic cells also known as voltaic cells. A galvanic cell is device such as a battery which converts chemical energy into electrical energy. A galvanic is made using two electrodes known as the anode and the cathode. Oxidation occurs at the anode and reduction occurs in the cathode. The electrons are written on the right of an oxidation half-reaction and they are located on the left side of a reduction half-reaction. Oxidation involves the loss of electrons and reduction is associated with the gain of electrons. Electrons always flow from the anode to the cathode. The purpose of the salt bridge is to prevent the build up of charge. Cations flow through the salt bridge toward the cathode and anions flow toward the anode. This electrochemistry explains completely how a galvanic cell works. It discusses how to write the overall reaction from the half reactions and how to represent the cell using standard line notation or cell notation. The electromotive force of a galvanic cell is measured in volts and 1 volt is 1 joule per coulomb. Voltage represents the work that can be done per coulomb of electric charge. Electric current is the rate at which electric charge is transferred. This video discusses how to increase the voltage and current delivered by a galvanic cell. Electrochemistry - Free Formula Sheet: Chapter 17 - Video Lessons: Chemistry 2 Final Exam Review: Final Exam and Test Prep Videos: Intro to Galvanic & Voltaic Cells: How To Draw Galvanic Cells: Standard Reduction Potentials: Cell Potential Problems: Cell Notation Problems: Concentration Cells: Cell Potential & Gibbs Free Energy: Cell Potential & Equilibrium K: Nernst Equation: Electrolysis of Water: Electrolysis of Sodium Chloride: Electrolysis & Electroplating Problems: Electrochemistry Practice Problems: SAT Chemistry Subject Test Review: Carbon -14 Dating: Beer Lambert's Law: Final Exams and Video Playlists: Full-Length Videos and Worksheets: 372 comments Transcript: in this lesson we're going to focus on galvanic cells so what exactly is a galvanic cell a galvanic cell is basically a battery a battery is a device that can convert chemical energy into electrical energy so let's say if you connect a battery to a light bulb there's a chemical reaction that's occurring in the battery and that's going to be converted to electrical energy this is the positive terminal of the battery and this is the negative terminal of the battery so electrons will flow from the negative terminal through the wire and back towards the positive terminal and whenever electrons move through a wire you have a current flow and the flow of charge can be used to turn on the light bulb or to do a work you can cause the motor to spin and that's basically what electricity is just moving electrons and so a galvanic cell is basically a battery every time you use a battery to power your laptop your cell phone or even a calculator the battery inside of it is using chemical energy and converting it into electrical energy to do useful work and so let's talk about the operation of a galvanic cell a typical galvanic cell is composed of two half cells so i'm going to draw two beakers and each beaker is going to contain an electrode an electrode is simply a metal metals conduct electricity now the metals have to be in contact with a solution so this is going to be a zinc metal and this is going to be copper metal so that's the copper electrode the half cell on the left is going to contain a solution of zinc sulfate and for the one on the right is going to contain the solution of copper sulfate now i'm going to connect each electrode to a wire and to a light bulb or you could use a meter if you want now we also need a salt bridge to prevent the buildup of charge between these two solutions now let's talk about what's going to happen copper has a very strong tendency to accept or acquire electrons zinc wants to give it away so zinc is going to lose two electrons per atom and so it's going to turn into the plus two or the two plus ion now whenever a metal loses electrons it's undergoing oxidation and you need to know that oxidation always occurs at the anode so zinc is the anode the other electrode has to be the cathode reduction occurs at the cathode so remember oxidation is the loss of electrons and reduction is the gain of electrons so at the cathode copper ions will acquire electrons to turn into copper metal so if you look at the oxidation states the oxidation state of any pure element in its natural state is zero so for zinc the oxidation state increases from zero to plus two or two plus so anytime the oxidation number increases the substance is undergoing oxidation it's losing electrons in the case of copper notice that it goes from plus two to zero so the copper two plus ion is being reduced because the oxidation number is decreasing it's acquiring electrons so these two reactions are known as half reactions and you can easily tell if a half reaction is an oxidation reaction or if it's a reduction reaction and here's how you tell look at the electrons specifically where they are relative to the arrow if the electrons are on the right side you have an oxidation reaction if the electrons are on the left side you have a reduction reaction and so that's a quick and simple way to determine if a half reaction is a reduction reaction or an oxidation reaction but keep this in mind oxidation always occurs at the anode and reduction occurs at the cathode now we know that zinc is given away electrons so electrons will always flow from the anode towards the cathode so because the electrons are leaving the anode the anode is negatively charged electrons always leave the negative terminal of a battery and flow towards the positive terminal of a battery so the cathode has to be positively charged because the electrons are flowing to it now as electrons flow through the light bulb the lipo is going to light up provided if this battery has enough energy to light up the light bulb now what's happening at the interface between the electrode and the solution specifically the anode now as the zinc atoms lose electrons they turn into zinc ions so let's say this is a zinc atom here it's hard to see it but let me just redraw like this so when that zinc atom gives up two electrons it's going to turn into an ion and so it leaves the electrode and it dissolves in the solution now on the other side there's already copper ions in the solution so a copper two plus ion is going to be attracted to the incoming electrons because opposite charges attract as the electrons flow on this metal into the solution well they don't really flow into the solution but they're flowing through the metal in the direction of the solution these copper two plus ions they're attracted to the negatively charged electrons and so they're going to accelerate towards that electrode and as they accelerate towards the cathode they're going to pick up the electrons at the interface between the cathode and the solution and so as these ions acquire electrons they turn into copper metal and so they deposit themselves on the surface of the cathode so over time you'll notice that the cathode is going to get bigger because the copper ions are turning into copper atoms on the surface of the copper electrode now the zinc electrode is going to get smaller because it's losing mass so in a galvanic cell the anode it loses mass the cathode gains mass so as electrons flow from anode to cathode the anode is going to lose mass and the cathode is going to gain mass make sure you understand that so this electrode is going to slowly disappear and this one is just going to get bigger and that's the basic principle for electroplating you can transfer mass from one electrode to another electrode but we'll talk about that in another video now there's another element that we need to talk about and that is the cell bridge what do you think is the purpose of the salt bridge why do we need it without the salt bridge current will flow only for a very short time you're going to have some electrons flow from the anode to the cathode but as electrons flow towards the cathode this side is going to have a buildup of negative charge and this side is going to have to build up a positive charge because it lost electrons and so once you have that charge build up electrons will no longer flow because once the side becomes positively charged the electrons that went this way is now attracted to this side and so it's going to reach a state of equilibrium and so without the salt bridge you won't have a constant flow of electricity it's going to be very short-lived and so the purpose of the salt bridge is to keep the solutions neutral so as electrons flow from anode to cathode this solution becomes negatively charged this becomes positively charged so to relieve the positively charged nature of this solution what happens is the anions which are known as positively charged ions they're going to flow through the cell bridge to the other solution so the zinc cations are going to flow towards the cathode cations cathode makes sense the anions flow toward the anodes now this works in such a way that the two solutions remain neutral but keep that in mind and ions flow towards the anode and ions are negatively charged particles and cations flow towards the cathode and so that's how the salt bridge will prevent a buildup of charge so now that we've discussed the basic operation of a galvanic cell and how it works let's talk about the cell potential and how to calculate it for this overall reaction so we saw that zinc turns into the zinc two plus ion by giving up two electrons now the cell potential for this half reaction is positive 0.76 volts and so this occurs at the anode i'm going to write a for anode and so that's the oxidation half reaction so remember anytime you see the electrons on the right side it's oxidation if the electrons are on the left side it's reduction now the other half reaction the copper two plus ions acquire electrons and turn into copper metal and the cell potential for this half reaction is positive 0.34 volts so this occurs at the cathode and reduction always occurs at the cathode so anytime you see the electrons on the left side of a half reaction you know it's associated with a reduction half reaction so we can do is add up these two half reactions notice that we have two electrons on the left and two electrons on the right and they will cancel so on the left of the arrow we have zinc and copper two plus so this is going to be zinc plus copper and on the right side we have copper and zinc two plus so zinc is a metal so it's in a solid state and on the right side copper is a metal now the ions are dissolved in water so then the aqueous state now to calculate the cell potential for this overall reaction all we need to do is add up these values so 0.76 plus 0.34 gives us an overall cell potential of 1.1 the cell potential looks like this so it's 1.1 volts now we can represent the cell using a line notation or cell notation so first you need to write the anode which is zinc and then use a single vertical line to separate the phases because these two phases are different so in the solution we have the zinc two plus ion that's in the the first compartment on the left then you need a double arrow to separate the anode half cell from the cathode half cell or you could say the oxidation half cell from the reduction f cell so on the left side this is for oxidation which is associated with the anode and the right side will be associated with a reduction which is with the cathode so in a reduction half cell we have the copper two plus ion dissolve in a solution and we need to separate that because the electrode is in a different phase than the copper ii plus ion and so the cathode is copper metal the anode is zinc metal so that's a simple way in which you can represent this cell using cell notation or also known as line notation by the way a galvanic cell is also known as a voltaic cell so just keep that in mind now let's talk about the cell potential also known as emf electromotive force this is measured in units of volts or you could describe it as a voltage one volt is equal to one joule per coulomb so what does this mean when one coulomb of charge passes through a potential difference of one volt it can do one joule of work joules is the unit of energy it's also the unit of work work is basically the transfer of energy and one column is the unit of charge or electric charge charge is proportional to the number of charged particles the more electrons you have the more charge that's flowing through a conductor and so a volt represents the ratio between energy and charge you could also say it's the work that can be done per unit charge so that's how you can describe a vote in case you wanted to know what it is so you need to be familiar with voltage and current voltage is the potential difference between two points electric potential is measured in volts and voltage is the difference between a potential of two points for instance let's say if you have a conductor and let's say this is point a and point b at point a let's say the electric potential is 11 volts and at point b the electric potential is 10 volts the voltage is the potential difference between these two points so the voltage is one volt so if we have one coulomb of charge so one column of electrons and they're going to flow from a low potential to a high potential so this would be positively charged because it exists at a high potential and relative to the other one this is negatively charged so electrons will flow towards the higher potential towards the positive terminal and so if one coulomb of electrons flow from b to a and it flows through a potential difference of one volt as it goes from b to a it can do one joule of work so make sure you understand that so voltage is the difference between two electric potential values so both electric potential and voltage have the same units volts now you need to be familiar with electric current electric current which is represented by the symbol i is the rate at which charge is transferred so q represents the electric charge in units of coulombs current is measured in amps so 1 amp is equal to 1 per second so if we have a current of 10 amps that means that 10 coulombs of charge is flowing through every second or we could say that 20 coulombs of charge flows every two seconds so 10 amps mean that every second 10 coulombs of charge passes through a given point so charge is the current multiplied by the time so make sure you know these values so one coulomb is equal to 1 amp times one second and one volt is equal to one joule per one coulomb now let's go back and talk about cell potential the cell potential is positive whenever you have a spontaneous reaction so that means that delta g is equal to a negative value the cell potential is zero when the system is in equilibrium and so when the cell potential is zero delta g is also equal to zero and when the cell potential is negative that means that you have a non-spontaneous process and so delta g is positive now what you need to know is for a galvanic cell or a voltaic cell the cell potential has to be positive it can't be negative because galvanic cells they convert chemical energy into electrical energy you can't derive energy from a non-spontaneous process so the lowest that a galvanic cell can be is zero so let's say if we have a cell potential like the example that we had earlier which is 1.1 volts in this case the battery is fully charged it has energy to give away it can power a laptop or maybe not that much but power small light bulb now as you use up the battery's energy the voltage or the cell potential will decrease and so over time the cell potential will gradually get lower and lower it's going to be 1.0 volts 0.9 0.8.7 eventually it's going to reach a state of equilibrium while the cell potential is zero at this case it's fully discharged so basically the battery is dead at this point that's when you throw it away and you buy a new battery so make sure you understand that when the cell potential is positive the battery or the galvanic cell has the ability to do work it can power a light bulb it can spin a small motor or something but as you use up the battery's energy the cell potential will gradually decrease to zero once it's zero the battery is in equilibrium it has no desire to react it doesn't have any energy to release and so at this point the battery is considered dead it can't do anything at that point now to go back this way is a non-spontaneous process however you can charge up the battery by putting energy into it by driving electricity into the system so you can reverse the whole process so you can drive a non-spontaneous process forward by putting energy into the system but once you do that it's no longer a galvanic cell but it's an electrolytic cell so for galvanic cells the cell potential can be zero or positive for electrolytic cells it could be anything it could be negative it could be positive it could be zero but the key difference is this in an electrolytic cell you need to put energy into it to make it work for galvanic cell it generates its own energy it converts chemical energy into electrical energy so a galvanic cell can never have a negative cell potential but for electrolytic cells it can now let's talk about the construction of a galvanic cell let's say if we want to increase the voltage how can we do so well if you have two d size batteries or two double a batteries if you want to increase the voltage you need to take the positive terminal of the battery and connect it to the negative terminal of another battery and so this is the positive terminal of the second battery and this is the negative terminal of the first battery and each battery let's say it's 1.5 volts so when you connect it in series like this and let's say you connect it to a voltmeter the volt meter is going to read 3 volts and so you can increase the voltage of multiple batteries if you connect them like this so let's say if you have three batteries that is three 1.5 volt batteries connected in series then the total voltage is simply the sum of the individual voltages of these three batteries so 1.5 times 3 is 4.5 so that's what the voltmeter is going to read if you connect three batteries in series now let's apply this to the galvanic cell so let's say this is one compartment here's the other one and let's say this is the saw bridge here's the zinc electrode and here is the copper electrode and then we have a solution inside now we said that the copper electrode in this example was the positive terminal and the zinc was the negative terminal and so this cell will produce a voltage of 1.1 volts now let's draw another cell right next to it so we still have the same salt bridge here's the zinc electrode and this is going to be the copper electrode so what we need to do is connect the negative terminal of the second battery to the positive terminal of the first battery and so let's say if we connect the voltmeter across these two points now it's no longer going to read 1.1 volts but rather it's going to read the sum of these two values so it's going to be 2.2 volts and so as you connect a galvanic cell in series with another galvanic cell you have to take the positive terminal of one galvanic cell connected to the negative terminal of the other and then the other two remaining terminals or electrodes you connect into the voltmeter if you do that the voltage will increase proportionally so if you connect three galvanic cells together it can be one point one times three or three point three volts and so that's how you can make batteries with higher voltages that's how you can make a 12 volt battery or a nine volt battery it's by connecting smaller batteries in series with each other until you get your desired voltage now what about increasing the current delivered by a battery how can we increase the current to increase the current you need to increase the size of the electrode that is you need to increase the surface area so for example let's say this is the top view of the zinc electrode and let's use this as another zinc electrode the zinc electrode on the right has more surface area than the one on the left as a result a battery that uses this electrode will deliver more current measured in amps that will flow through the circuit and so if you can increase the surface area of the electrodes what happens is you decrease the internal resistance of the battery and so the current delivered by the battery that goes up now if you have a metal electrode it may not be practical to increase the area of that so what some batteries have is instead of having just a metallic electrode sometimes the electrode can be pressurized into a potted form so let's say if you use a pilot form of zinc or a powdered form of graphite as an electrode and you pressurize it you can get a lot more surface area doing it that way and so you can get a battery that can deliver a high current so the goal is to increase the surface area of the electrode if you can do that then you can increase the interface between the electrode and the solution so the greater the contact area between the electrode and the solution is the more current that can be delivered by that battery so that's how you can increase the current of a battery is by increasing the surface area of the electrode now let's say if you have two batteries if you connect them in series you can increase the voltage but if you have two separate batteries and if you connect them parallel to each other you can increase the current so for example let's say if you have a short circuit in this battery which is not always the safest thing but let's just talk about it when you have a short circuit it can deliver a lot of current so let's say this battery can release 10 amps of current now keep this in mind electrons flow from negative to positive but conventional current is defined as the flow of positive charge so conventional current flows from the positive terminal to the negative terminal but electrons flow from the negative terminal to the positive terminal so if this battery on its own can produce a current of 10 amps now let's say if you connect it in series with another battery and let's say that these two batteries are identical and you create a short circuit you can get a current of 20 amps this will still deliver 10 amps of current and this will deliver 10 amps of conventional current so the total current flowing in this branch will be 20 amps so if you connect two batteries parallel to each other you can increase the current that they deliver to a circuit now keep in mind the electrons are flowing in the opposite direction from negative to positive you
15021	https://ops.fhwa.dot.gov/publications/fhwahop08024/chapter5.htm	\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| Table of Contents Traffic Signal Timing Manual Contact Information: Operations Feedback at OperationsFeedback@dot.gov This publication is an archived publication and replaced with the Signal Timing Manual - Second Edition. CHAPTER 5BASIC SIGNAL TIMING PROCEDURE AND CONTROLLER PARAMETERS TABLE OF CONTENTS 5.0 BASIC SIGNAL TIMING PROCEDURE AND CONTROLLER PARAMETERS 5.1 Terminology and Key Definitions 5.2 Modes of Traffic Signal Operation and Their Use + 5.2.1 Pre-timed Control + 5.2.2 Semi-Actuated Control + 5.2.3 Fully-Actuated Control 5.3 Phase Intervals and Basic Parameters + 5.3.1 Vehicular Green Interval + 5.3.2 Vehicular Change and Clearance Intervals + 5.3.3 Pedestrian Timing Intervals 5.4 Actuated Timing Parameters + 5.4.1 Phase Recalls + 5.4.2 Passage Time + 5.4.3 Simultaneous Gap + 5.4.4 Dual Entry 5.5 Volume-Density Features + 5.5.1 Gap Reduction + 5.5.2 Variable Initial 5.6 Detection Configuration and Parameters + 5.6.1 Delay + 5.6.2 Extend + 5.6.3 Carryover + 5.6.4 Call + 5.6.5 Queue 5.7 Guidelines for Time-Base Controls 5.8 References LIST OF TABLES Table 5-1. Relationship between intersection operation and control type Table 5-2 Factors considered when setting the minimum green interval Table 5-3 Typical minimum green interval duration needed to satisfy driver expectancy Table 5-4 Typical minimum green interval duration needed to satisfy queue clearance Table 5-5 Example values for maximum green duration Table 5-6 Maximum green duration as a function of cycle length and volume Table 5-7 Duration of change period intervals Table 5-8 Pedestrian walk interval duration Table 5-9 Pedestrian clearance time Table 5-10 Passage time duration for presence mode detection Table 5-11 Minimum gap duration for presence mode detection Table 5-12 Gap reduction parameter values LIST OF FIGURES Figure 5-1 Users and the actuated signal timing parameters that determine phase length Figure 5-2 Settings that define the duration of a vehicle phase Figure 5-3 Application of passage time Figure 5-4 Relationship between passage time, gap, and maximum allowable headway Figure 5-5 Use of volume-density to change the extension time Figure 5-6 Use of Added Initial to modify minimum green Figure 5-7 Application of delay timer Figure 5-8 Application of extend timer 5.0 BASIC SIGNAL TIMING PROCEDURE AND CONTROLLER PARAMETERS This chapter documents the principles of basic traffic signal timing at an intersection. Signal timing is a collection of parameters and logic designed to allocate the right-of-way at a signalized intersection. A major focus of this chapter is to describe basic signal timing parameters necessary to operate an intersection and guidelines for selecting values for those parameters. The principles described in this chapter are generally applicable to all signalized intersections. To maximize the usefulness and transferability of the information provided, the chapter uses the terminology defined in current traffic signal control standards, such as National Transportation Communications for ITS Protocol (NTCIP) Document 1202 (1) and National Electrical Manufacturers Association (NEMA) Standards Publication TS 2-2003 (2), with alternative definitions in some cases. 5.1 TERMINOLOGY AND KEY DEFINITIONS This section identifies and describes basic terminology used within this chapter. Additional terms can be found in the Glossary section of the Manual. Actuated Signal Control A type of signal control where time for each phase is at least partially controlled by detector actuations. Call An indication within a controller that a vehicle or pedestrian is awaiting service from a particular phase or that a recall has been placed on the phase. Extend A detector parameter that increases the duration of a detector actuation by a defined fixed amount. Gap Out A type of actuated operation for a given phase where the phase terminates due to a lack of vehicle calls within a specific period of time (passage time). Interval The duration of time during which the indications do not change their state (active or off). Typically, one or more timing parameters control the duration of an interval. The pedestrian clearance interval is determined by the pedestrian clearance time. The green interval duration is controlled by a number of parameters including minimum time, maximum time, gap time, etc. Isolated intersection An intersection located outside the influence of and not coordinated with other signalized intersections, commonly one mile or more from other signalized intersections. Minimum Gap A volume density parameter that specifies the minimum green extension when gap reduction is used. Minimum Green A parameter that defines the shortest allowable duration of the green interval. Minimum Recall A parameter which results in a phase being called and timed for at least its minimum green time whether or not a vehicle is present. Movement Movements reflect the user perspective. Movements can also be broken down into classes (car, pedestrians, buses, LRT, etc.). Typical movements are left, through and right. Movement is an activity in response to a “go” (green ball, green arrow, walk, white vertical transit bar) indication. Max Out A type of actuated operation for a given phase where the phase terminates due to reaching the designated maximum green time for the phase. Passage Time (Vehicle Interval, Gap, Passage Gap, Unit Extension) A parameter that specifies the maximum allowable duration of time between vehicle calls on a phase before the phase is terminated. Pedestrian Clearance Interval Also generally known as “Flashing Don’t Walk” (FDW). An indication warning pedestrians that the walk indication has ended and the don’t walk indication will begin at the end of the pedestrian clearance interval. Some agencies consider the pedestrian clearance interval to consist of both the FDW time and the yellow change interval. Phase A timing unit associated with the control of one or more indications. A phase may be timed considering complex criteria for determination of sequence and the duration of intervals. Pre-timed control A signal control in which the cycle length, phase plan, and phase times are predetermined and fixed. Queue A line of vehicles, bicycles, or persons waiting to be served by a phase in which the flow rate from the front of the queue determines the average speed within the queue. Slowly moving vehicles or people joining the rear of the queue are usually considered part of the queue. The internal queue dynamics can involve starts and stops. A faster-moving line of vehicles is often referred to as a moving queue or a platoon. Recall A call is placed for a specified phase each time the controller is servicing a conflicting phase. This will ensure that the specified phase will be serviced again. Types of recall include soft, minimum, maximum, and pedestrian. Semi-Actuated Control A type of signal control where detection is provided for the minor movements only. Volume-Density A phase timing technique that uses a series of parameters (variable initial, minimum gap, time before reduction, time to reduce) to provide alternative, variable settings for the otherwise fixed parameters of minimum green and passage time. 5.2 MODES OF TRAFFIC SIGNAL OPERATION AND THEIR USE Traffic signals operate in either pre-timed or actuated mode or some combination of the two. Pre-timed control consists of a series of intervals that are fixed in duration. Collectively, the preset green, yellow, and red intervals result in a deterministic sequence and fixed cycle length for the intersection. In contrast to pre-timed control, actuated control consists of intervals that are called and extended in response to vehicle detectors. Detection is used to provide information about traffic demand to the controller. The duration of each phase is determined by detector input and corresponding controller parameters. Actuated control can be characterized as fully-actuated or semi-actuated, depending on the number of traffic movements that are detected. Table 5-1 summarizes the general attributes of each mode of operation to aid in the determination of the most appropriate type of traffic signal control for an intersection. The attributes of the various modes of operation are discussed in additional detail in the following subsections. Table 5-1. Relationship between intersection operation and control type. \| Pre-timed \| Actuated \| \| Type of Operation \| Isolated \| Coordinated \| Semi-Actuated \| Fully-Actuated \| Coordinated \| \| Fixed Cycle Length \| Yes \| Yes \| No \| No \| Yes \| \| Conditions Where Applicable \| Where detection is not available \| Where traffic is consistent, closely spaced intersections, and where cross street is consistent \| Where defaulting to one movement is desirable, major road is posted <40 mph and cross road carries light traffic demand \| Where detection is provided on all approaches, isolated locations where posted speed is >40 mph \| Arterial where traffic is heavy and adjacent intersections are nearby \| \| Example Application \| Work Zones \| Central business districts, interchanges \| Highway operations \| Locations without nearby signals; rural, high speed locations; intersection of two arterials \| Suburban arterial \| \| Key Benefit \| Temporary application keeps signals operational \| Predictable operations, lowest cost of equipment and maintenance \| Lower cost for highway maintenance \| Responsive to changing traffic patterns, efficient allocation of green time, reduced delay and improved safety \| Lower arterial delay, potential reduction in delay for the system, depending on the settings \| 5.2.1 Pre-timed Control Pre-timed control is ideally suited to closely spaced intersections where traffic volumes and patterns are consistent on a daily or day-of-week basis. Such conditions are often found in downtown areas. They are also better suited to intersections where three or fewer phases are needed (3). Pre-timed control has several advantages. For example, it can be used to provide efficient coordination with adjacent pre-timed signals, since both the start and end of green are predictable. Also, it does not require detectors, thus making its operation immune to problems associated with detector failure. Finally, it requires a minimum amount of training to set up and maintain. On the other hand, pre-timed control cannot compensate for unplanned fluctuations in traffic flows, and it tends to be inefficient at isolated intersections were traffic arrivals are random. Modern traffic signal controllers do not explicitly support signal timing for pre-timed operation, because they are designed for actuated operation. Nevertheless, pre-timed operations can be achieved by specifying a maximum green setting that is equal to the desired pre-timed green interval and invoking the maximum vehicle recall parameter described below. 5.2.2 Semi-Actuated Control Semi-actuated control uses detection only for the minor movements at an intersection. The phases associated with the major-road through movements are operated as "non-actuated." That is, these phases are not provided detection information. In this type of operation, the controller is programmed to dwell in the non-actuated phase and, thereby, sustain a green indication for the highest flow movements (normally the major street through movement). Minor movement phases are serviced after a call for their service is received. Semi-actuated control is most suitable for application at intersections that are part of a coordinated arterial street system. Coordinated-actuated operation is discussed in more detail in Chapter 6. Semi-actuated control may also be suitable for isolated intersections with a low-speed major road and lighter crossroad volume. Semi-actuated control has several advantages. Its primary advantage is that it can be used effectively in a coordinated signal system. Also, relative to pre-timed control, it reduces the delay incurred by the major-road through movements (i.e., the movements associated with the non-actuated phases) during periods of light traffic. Finally, it does not require detectors for the major-road through movement phases and hence, its operation is not compromised by the failure of these detectors. The major disadvantage of semi-actuated operation is that continuous demand on the phases associated with one or more minor movements can cause excessive delay to the major road through movements if the maximum green and passage time parameters are not appropriately set. Another drawback is that detectors must be used on the minor approaches, thus requiring installation and ongoing maintenance. Semi-actuated operation also requires more training than that needed for pre-timed control. 5.2.3 Fully-Actuated Control Fully-actuated control refers to intersections for which all phases are actuated and hence, it requires detection for all traffic movements. Fully-actuated control is ideally suited to isolated intersections where the traffic demands and patterns vary widely during the course of the day. Most modern controllers in coordinated signal systems can be programmed to operate in a fully-actuated mode during low-volume periods where the system is operating in a "free" (or non-coordinated) mode. Fully-actuated control can also improve performance at intersections with lower volumes that are located at the boundary of a coordinated system and do not impact progression of the system (). 4Fully-actuated control has also been used at the intersection of two arterials to optimize green time allocation in a critical intersection control method. There are several advantages of fully-actuated control. First, it reduces delay relative to pre-timed control by being highly responsive to traffic demand and to changes in traffic pattern. In addition, detection information allows the cycle time to be efficiently allocated on a cycle-by-cycle basis. Finally, it allows phases to be skipped if there is no call for service, thereby allowing the controller to reallocate the unused time to a subsequent phase. The major disadvantage of fully-actuated control is that its cost (initial and maintenance) is higher than that of other control types due to the amount of detection required. It may also result in higher percentage of vehicles stopping because green time is not held for upstream platoons. 5.3 PHASE INTERVALS AND BASIC PARAMETERS An interval is defined in the NTCIP 1202 standard as “a period of time during which signal indications do not change.” Various parameters control the length of an interval depending on the interval type. For example, a pedestrian walk interval (the time period during which the Walking Person signal indication is displayed) is generally controlled by the single user-defined setting for the walk parameter. The vehicular green interval, on the other hand, is generally controlled by multiple parameters, including minimum green, maximum green, and passage time. This section describes guidelines for setting basic parameters that determine the duration of each interval associated with a signal phase. These intervals include: Vehicular Green Interval Vehicle Change and Clearance Intervals Pedestrian Intervals Parameters related to these intervals and discussed in this section include minimum green, maximum green, yellow-change, red clearance, pedestrian walk, and pedestrian flashing don’t walk (FDW). Figure 5-1 depicts the relationship between these parameters and the user group associated with each interval that may time during a phase. These intervals time concurrently during a phase. Although shown here, signal preemption and priority are addressed in Chapter 9. Additional timing parameters related to actuated control (e.g., passage time) may also influence the duration of an interval and are discussed in Section 5.4. Figure 5-1 Users and the actuated signal timing parameters that determine phase length 5-5 Figure 5-1: This figure depicts the relationship between the various parameters associated with signal timing and the user group associated with each interval. The parameters include such items as minimum green, maximum green, yellow-change, red clearance, pedestrian walk, and pedestrian flashing don’t walk. The user groups addressed are pedestrians, vehicles, bicycles, and emergency vehicles. 5.3.1 Vehicular Green Interval The vehicular green interval is the time dedicated to serving vehicular traffic with a green indication. This interval is defined primarily by the minimum and maximum green parameters in the case of an isolated intersection. At an actuated controller, other parameters (e.g., passage time) also determine the length of this interval. Those parameters are discussed in Section 5.4. It is also possible that the duration of the vehicle green interval may be defined by the length of the associated pedestrian intervals. Minimum Green The minimum green parameter represents the least amount of time that a green signal indication will be displayed for a movement. Minimum green is used to allow drivers to react to the start of the green interval and meet driver expectancy. Its duration may also be based on considerations of queue length or pedestrian timing in the absence of pedestrian call buttons and/or indications. A minimum green that is too long may result in wasted time at the intersection; one that is too short may violate driver expectation or (in some cases) pedestrian safety. The minimum green interval is shown in Figure 5-2, as it relates to other intervals and signal control parameters. Calls placed on the active phase during the minimum green have no bearing on the duration of the green interval as the interval will time at least as long as the minimum green timer. Lin (5) conducted extensive simulation analysis of fully-actuated controlled intersections to determine the effect of minimum green intervals on delay. Through these simulations, he found that delay was minimal when the minimum green interval was less than 4 seconds. Delay for the intersection under the scenarios studied tended to increase slightly as the minimum green interval increased from 4 to 8 seconds. Figure 5-2 Settings that define the duration of a vehicle phase This figure depicts the minimum green interval as it relates to other intervals and signal control parameters. In this example, a conflicting phase vehicle actuation is received during the minimum green interval. This actuation sets a maximum green timer. Concurrently, a pedestrian crossing extends the normal minimum green, as does current phase vehicle actuations. The intent of the minimum green interval is to ensure that each green interval is displayed for a length of time that will satisfy driver expectancy. When stop-line detection is not provided, variable initial, as described in Section 5.4, should be used to allow vehicles queued between the stop line and the nearest detector at the start of green to clear the intersection. In cases where separate pedestrian signal displays are not provided, the minimum green interval will also need to be long enough to accommodate pedestrians who desire to cross in a direction parallel to the traffic movement receiving the green indication. These considerations and the conditions in which each applies are shown in Table 5-2. Table 5-2 Factors considered when setting the minimum green interval \| Phase \| Stop Line Detection? \| Pedestrian Button? \| Considered in Establishing Minimum Green? \| \| Driver Expectancy \| Pedestrian Crossing Time \| Queue Clearance \| \| Through \| Yes \| Yes \| Yes \| No \| No \| \| No \| Yes \| Yes \| No \| \| No \| Yes \| Yes \| No \| Yes, if actuated \| \| No \| Yes \| Yes \| Yes, if actuated \| \| Left-Turn \| Yes \| Not applicable \| Yes \| Not applicable \| No \| To illustrate the use of Table 5-2, consider a through movement with stop-line detection and a pedestrian push button. As indicated in the table, the minimum green interval should be based solely on driver expectancy. However, if a pedestrian call button is not provided (and pedestrians are expected to cross the road at this intersection), the minimum green interval should be based on driver expectancy and pedestrian crossing time. Minimum Green to Satisfy Driver Expectancy The duration of minimum green needed to satisfy driver expectancy varies among practitioners. Some practitioners rationalize the need for 15 seconds or more of minimum green at some intersections; other practitioners use as little as 2 seconds minimum green. If a minimum green parameter is set too low and violates driver expectancy, there is a risk of increased rear-end crashes. The values listed in Table 5-3 are typical for the specified combination of phase and facility type. Table 5-3 Typical minimum green interval duration needed to satisfy driver expectancy \| Phase Type \| Facility Type \| Minimum Green Needed to Satisfy Driver Expectancy (Ge), s \| \| Through \| Major Arterial (speed limit exceeds 40 mph) \| 10 to 15 \| \| Major Arterial (speed limit is 40 mph or less) \| 7 to 15 \| \| Minor Arterial \| 4 to 10 \| \| Collector, Local, Driveway \| 2 to 10 \| \| Left Turn \| Any \| 2 to 5 \| Minimum Green for Pedestrian Crossing Time The minimum green duration must satisfy pedestrian crossing needs for through phases that are not associated with a pedestrian push button but have a pedestrian demand. Under these conditions, the minimum green needed to satisfy pedestrian considerations can be computed using Equation 5-1. Methodology for computing walk and pedestrian clearance interval durations are provided in Section 5.3.3. Equation 5-1 where G sub P is the minimum green interval duration needed to satisfy pedestrian crossing time, PW is the walk interval duration, and PC is the pedestrian clearance interval duration, s (all values in seconds). Minimum Green for Queue Clearance The duration of minimum green can also be influenced by detector location and controller operation. This subsection addresses the situation where a phase has one or more advance detectors and no stop-line detection. If this detection design is present, and the added initial parameter (as discussed later) is not used, then a minimum green interval is needed to clear the vehicles queued between the stop line and the advance detector. The duration of this interval is specified in Table 5-4. Table 5-4 Typical minimum green interval duration needed to satisfy queue clearance \| Distance Between Stop Line and Nearest Upstream Detector, ft \| Minimum Green Needed to Satisfy Queue Clearance1, 2 (Gq), s \| \| 0 to 25 \| 5 \| \| 26 to 50 \| 7 \| \| 51 to 75 \| 9 \| \| 76 to 100 \| 11 \| \| 101 to 125 \| 13 \| \| 126 to 150 \| 15 \| Notes: 1. Minimum green values listed apply only to phases that have one or more advance detectors, no stop line detection, and the added initial parameter is not used. 2. Minimum green needed to satisfy queue clearance, Gq = 3 + 2n (in seconds), where n = number of vehicles between stop line and nearest upstream detector in one lane. And, n = Dd / 25, where Dd = distance between the stop line and the downstream edge of the nearest upstream detector (in feet) and 25 is the average vehicle length (in feet), which could vary by area. If a phase has one or more advance detectors, no stop-line detection, and the added initial parameter is used, then the minimum initial interval should equal the minimum green needed to satisfy driver expectancy. Timing of minimum greens using the added initial parameter is discussed in Section 5.4. Maximum Green The maximum green parameter represents the maximum amount of time that a green signal indication can be displayed in the presence of conflicting demand. Maximum green is used to limit the delay to any other movement at the intersection and to keep the cycle length to a maximum amount. It also guards against long green times due to continuous demand or broken detectors. Ideally, the maximum green will not be reached because the detection system will find a gap to end the phase, but if there are continuous calls for service and a call on one or more conflicting phases, the maximum green parameter will eventually terminate the phase. A maximum green that is too long may result in wasted time at the intersection. If its value is too short, then the phase capacity may be inadequate for the traffic demand, and some vehicles will remain unserved at the end of the green interval. Most modern controllers provide two or more maximum green parameters that can be invoked by a time-of-day plan or external input (i.e., Maximum Green 2). As shown in Figure 5-2, the maximum green extension timer begins timing upon the presence of a conflicting call. If there is demand on the phase that is currently timing and no conflicting calls, the maximum green timer will be reset until an opposing call occurs. It should be noted that the normal failure mode of a detector is to place a continuous call for service. In this case, a failed detector on a phase will cause that phase’s maximum green to time every cycle. Many modern controllers also provide a feature that allows the maximum green time to be increased to a defined threshold after maxing out a phase a certain number of consecutive times (or alternatively to select among two or three maximum green values). The maximum green time may then be automatically decreased back to the original value after the phase has gapped out a certain number of times. The exact methods and user settable parameters for this feature vary by manufacturer. The maximum green value should exceed the green duration needed to serve the average queue and, thereby, allow the phase to accommodate cycle-to-cycle peaks in demand. Frequent phase termination by gap out (as opposed to max out) during low-to-moderate volumes and by occasional max out during peak periods is commonly used as an indication of a properly timed maximum green duration. Example values are listed in Table 5-5. Table 5-5 Example values for maximum green duration \| Phase \| Facility Type \| Maximum Green, s \| \| Through \| Major Arterial (speed limit exceeds 40 mph) \| 50 to 70 \| \| Major Arterial (speed limit is 40 mph or less) \| 40 to 60 \| \| Minor Arterial \| 30 to 50 \| \| Collector, Local, Driveway \| 20 to 40 \| \| Left Turn \| Any \| 15 to 30 \| Note: 1. Range is based on the assumption that advance detection is provided for indecision zone protection. If this type of detection is not provided, then the typical maximum green range is 40 to 60 s. Two methods are commonly used to establish the maximum green setting. Both estimate the green duration needed for average volume conditions and inflate this value to accommodate cycle-to-cycle peaks. Both of these methods assume that advance detection for indecision zone protection is not provided. If advance detection is provided for indecision zone protection, the maximum green setting obtained from either method may need to be increased slightly to allow the controller to find a “safe” time to terminate the phase by gap out. One method used by some agencies is to establish the maximum green setting based on an 85th to 95th percentile probability of queue clearance (6). The procedure requires knowledge of the cycle length, or an estimate of its average value for actuated operation. If the cycle length is known, then the maximum green setting for a signal phase can be obtained from Table 5-6. Table 5-6 Maximum green duration as a function of cycle length and volume \| Phase Volume per Lane, veh/hr/ln \| Cycle Length, s \| \| 50 \| 60 \| 70 \| 80 \| 90 \| 100 \| 110 \| 120 \| \| Maximum Green (Gmax)1, s \| \| 100 \| 15 \| 15 \| 15 \| 15 \| 15 \| 15 \| 15 \| 15 \| \| 200 \| 15 \| 15 \| 15 \| 15 \| 16 \| 18 \| 19 \| 21 \| \| 300 \| 15 \| 16 \| 19 \| 21 \| 24 \| 26 \| 29 \| 31 \| \| 400 \| 18 \| 21 \| 24 \| 28 \| 31 \| 34 \| 38 \| 41 \| \| 500 \| 22 \| 26 \| 30 \| 34 \| 39 \| 43 \| 47 \| 51 \| \| 600 \| 26 \| 31 \| 36 \| 41 \| 46 \| 51 \| 56 \| 61 \| \| 700 \| 30 \| 36 \| 42 \| 48 \| 54 \| 59 \| 65 \| 71 \| \| 800 \| 34 \| 41 \| 48 \| 54 \| 61 \| 68 \| 74 \| 81 \| The values listed are based on the equation shown in the table footnote. Due to the approximate nature of this equation, the actual percentage probability of queue clearance varies between the 85th and 95th percentiles for the values listed. A second method for establishing the maximum green setting is based on the equivalent optimal pre-timed timing plan (7). This method requires the development of a pre-timed signal timing plan based on delay minimization. The minimum-delay green interval durations are multiplied by a factor ranging from 1.25 to 1.50 to obtain an estimate of the maximum green setting (8). The maximum green time used for a particular phase is calculated differently for low and high levels of saturation. During periods of low volume, when the green phase times rarely reach their maximum values, the maximum green time can be set fairly high (up to 1.7 times the calculated average time for the phase). This accommodates most fluctuations in vehicle arrival rates. During conditions at or near saturation, it is important to set the maximum green times as if they were fixed time, equitably allocating the green based on the critical lane volumes as described in Chapter 3. To this end, application of the maximum green times show significant disparity in the techniques reported for determination of maximum phase time, which ultimately may result in a wide variation of cycle lengths at intersections. In many cases, maximum green times are set at one value throughout the day and don’t reflect the needs of the intersection during various times of day. In some cases, these maximum green time values result in cycle lengths that are too long for efficient operations. 5.3.2 Vehicular Change and Clearance Intervals The intent of the vehicle phase change and clearance intervals is to provide a safe transition between two conflicting phases. It consists of a yellow change interval and, optionally, a red clearance interval. The intent of the yellow change interval is to warn drivers of the impending change in right-of-way assignment. The red clearance interval is used when there is some benefit to providing additional time before conflicting movements receive a green indication. Yellow Change The duration of the yellow change interval is typically based upon driver perception-reaction time, plus the distance needed to safely stop or to travel safely through the intersection. A state’s Uniform Vehicle Code directly affects yellow change interval timing, as it determines whether a permissive or restrictive yellow law is in place. Permissive Yellow Law: A driver can enter the intersection during the entire yellow interval and be in the intersection during the red indication as long as the vehicle entered the intersection during the yellow interval. Under permissive yellow law, an all-red clearance interval must exist as a timing parameter to ensure safe right-of-way transfer at an intersection. This rule is consistent with paragraph 11-202 of the Uniform Vehicle Code (9). Restrictive Yellow Law: There are two variations of this law (10). In one variation, a vehicle may not enter an intersection when the indication is yellow unless the vehicle can clear the intersection by the end of yellow. This implies that the yellow duration should be sufficiently long as to allow drivers the time needed to clear the intersection if they determine that it is not possible to safely stop. In the other variation, a vehicle may not enter an intersection unless it is impossible or unsafe to stop. With restrictive yellow law, the presence of an all-red interval is optional and good engineering judgment should be applied. Due to the varying interpretations of the yellow change use, it is encouraged that traffic engineers refer to the local and regional statutes for guidance in determining the purpose of the yellow change time. Red Clearance The red clearance interval, referred to in some publications as an all-red interval, is an interval at the end of the yellow change interval during which the phase has a red-signal display before the display of green for the following phase. The purpose of this interval is to allow time for vehicles that entered the intersection during the yellow-change interval to clear the intersection prior to the next phase. Note that the use of the “all-red” nomenclature is generally incorrect, as the red clearance interval only applies to a single phase, not to all phases. The use of a red clearance interval is optional, and there is no consensus on its application or duration. Recent research has indicated that the use of a red clearance interval showed some benefit to the reduction of red-light-running violations. In these studies, there was a significant reduction in right-angle crashes after implementing a red clearance interval. Other research suggests that this reduction may only be temporary. A comprehensive study of long-term effects for the Minnesota Department of Transportation ()11, indicated short-term reductions in crash rates were achieved (approximately one year after the implementation), but long-term reductions were not observed, which implies that there may not be safety benefits associated with increased red clearance intervals. A disadvantage of using the red clearance interval is that there is a reduction in available green time for other phases. At intersections where the timing for minor movements is restricted (e.g., to split times under coordinated operation (see Chapter 6)), the extra time for a red clearance interval comes from the remaining phases at the intersection. In cases where major movements are already at or near saturation, the reduction in capacity associated with providing red clearance intervals for safety reasons should be accounted for in an operational analysis. The MUTCD provides guidance on the application and duration of the yellow change and red clearance intervals. It recommends that the interval durations shall be predetermined based on individual intersection conditions, such as approach speed and intersection width. The MUTCD advises that the yellow change interval should last approximately 3 to 6 seconds, with the longer intervals being used on higher-speed approaches. It also advises that the red clearance interval should not exceed 6 seconds. A recent survey conducted by The Urban Transportation Monitor indicated that practitioners who used a standard red clearance interval used a range from 0.5 to 2.0 seconds. Kell and Fullerton (12) offer the following equation for computing the phase change period (yellow change plus red clearance intervals): Equation 5-2 where: CP = change period (yellow change plus red clearance intervals), s; t = perception-reaction time to the onset of a yellow indication, s; v = approach speed, mph; a = deceleration rate in response to the onset of a yellow indication; g = grade, with uphill positive and downhill negative (percent grade / 100), ft/ft; W = width of intersection, ft; and L sub V = length of vehicle. Equation 5-2 is based on driver reaction time, approach speed, approach grade, and intersection width and consists of two terms. The first term (yellow change) represents the time required for a vehicle to travel one safe stopping distance, including driver perception-reaction time. This permits a driver to either stop at the intersection if the distance to the intersection is greater than one safe stopping distance or safely enter the intersection (and clear the intersection under the restrictive yellow law) if the distance to the intersection is less than one safe stopping distance. The second term (red clearance) represents the time needed for a vehicle to traverse the intersection ([W + Lv]/v). Although values will vary by driver population and local conditions, the values of t = 1.0 s, a = 10 ft/s2, and Lv = 20 ft are often cited for use in Equation 5-3 (,,131415). These values of perception-reaction time and deceleration rate are different from those cited in highway geometric design policy documents because they are based on driver response to the yellow indication, which is an expected condition. They are not based on the longer reaction time necessary for an unexpected (or surprise) condition. When applying Equation 5-2 to through movement phases, the speed used is generally either the 85th-percentile speed or the posted regulatory speed limit, depending on agency policy (16). When applying Equation 5-2 to left-turn movement phases, the speed used should reflect that of the drivers that intend to turn. This speed can equal that of the adjacent through movement but it can also be slower as left-turn drivers inherently slow to a comfortable turning speed. Regardless, if the left-turn phase terminates concurrently with the adjacent through phase, it will have the same total change and clearance interval durations as the through phase because the phases are interlocked by the ring-barrier operation. The width of the intersection is often defined by local policy or state law. For instance, in Arizona intersection width is defined by state law as the distance between prolongations of the curb lines. Where intersection width is not defined by local policies, engineering judgment should be used when measuring the width of the intersection, W. One approach is to measure from the near-side stop line to the far edge of the last conflicting traffic lane along the subject movement travel path. If crosswalks are present at the intersection, some agencies have policies to measure from the near-side stop line to the far side of the pedestrian crosswalk on the far side of the intersection (for through-movement phases) or to the far side of the pedestrian crosswalk across the leg of the intersection which the left-turn is entering. This is a jurisdiction-wide issue that must be carefully applied. Table 5-7 Duration of change period intervals \| Approach Speed, mph \| “t + v/2a” Terms, s (YELLOW) \| Width of Intersection, ft \| \| 30 \| 50 \| 70 \| 90 \| 110 \| \| “(W+Lv)/v” Term, s (ALL-RED) \| \| 25 \| 3.0a \| 1.4 \| 1.9 \| 2.5 \| 3.0 \| 3.5 \| \| 30 \| 3.2 \| 1.1 \| 1.6 \| 2.0 \| 2.5 \| 3.0 \| \| 35 \| 3.6 \| 1.0 \| 1.4 \| 1.8 \| 2.1 \| 2.5 \| \| 40 \| 3.9 \| 0.9 \| 1.2 \| 1.5 \| 1.9 \| 2.2 \| \| 45 \| 4.3 \| 0.8 \| 1.1 \| 1.4 \| 1.7 \| 2.0 \| \| 50 \| 4.7 \| 0.7 \| 1.0 \| 1.2 \| 1.5 \| 1.8 \| \| 55 \| 5.0 \| 0.6 \| 0.9 \| 1.1 \| 1.4 \| 1.6 \| \| 60 \| 5.4 \| 0.6 \| 0.8 \| 1.0 \| 1.2 \| 1.5 \| a The 2003 MUTCD recommends a minimum duration of 3 seconds for the yellow change interval. The values for the yellow change interval in Table 5-7 are based on negligible approach grade. They should be increased by 0.1 second for every 1 percent of downgrade. Similarly, they should be decreased by 0.1 second for every 1 percent of upgrade. To illustrate, consider an approach with a 30 mph approach speed, 70-foot intersection width, and 4-percent downgrade. The estimated change period is 5.6 seconds (= 3.2 + (0.1 x 4 + 2.0)). States that follow the “restrictive yellow” rule may equate the yellow change interval to the value obtained from Equation 5-2 (i.e., the sum of both terms). If a red-clearance interval is needed, its value may be set at 0.5 to 2 seconds, as determined by engineering judgment. States that follow the “permissive yellow” rule will typically set the yellow change interval equal to the value obtained from the first term of Equation 5-2 (i.e., column 2 of Table 5-7), but not less than 3.0 seconds. This duration will allow drivers that do not have the necessary distance to stop the time needed to reach the intersection before the red indication is presented. If a red clearance interval is needed, its value is typically based on the second term of Equation 5-2 (i.e., columns 3 through 7 of Table 5-7). Some agencies reduce the value of the second term by 1.0 second in recognition of the perception-reaction time of drivers in the next conflicting phase to be served (17). 5.3.3 Pedestrian Timing Intervals The pedestrian phase consists of three intervals: walk; pedestrian clearance, commonly referred to as flashing don’t walk (FDW); and solid don’t walk. The walk interval typically begins at the start of the green interval and is used to allow pedestrians to react to the change to walk at the start of the phase and move into the crosswalk. This interval corresponds to the WALKING PERSON indication on the pedestrian signal (18). The pedestrian clearance interval follows the walk interval and informs pedestrians the phase is ending. During this interval, the UPRAISED HAND indication flashes on the pedestrian signal. The solid don’t walk interval follows the pedestrian clearance interval and is indicated by a solid UPRAISED HAND indication. This interval is an indication to the pedestrian that they should have cleared the crosswalk and opposing vehicle movements could begin. The solid don’t walk time is not a programmable parameter in the controller. The duration of the solid don’t walk interval is simply the length of the cycle minus the walk and pedestrian clearance intervals. Although the illustration in Figure 5-2 does not include a pedestrian phase activation, it does show that the pedestrian timers (walk and FDW) would time concurrently with the vehicle intervals if there was a pedestrian activation. In the case of Figure 5-2, the pedestrian intervals are shown as requiring less time than allowed by the maximum green timer. In this case, if there was continuing vehicle demand, the pedestrian indication would show a solid don’t walk until the vehicle phase terminated due to lack of demand or the maximum green timer expired. However, if the pedestrian intervals required more time than permitted by the maximum green timer, the vehicle phase would continue to time until the pedestrian flashing don’t walk interval finished timing. Walk The walk interval should provide pedestrians adequate time to perceive the WALK indication and depart the curb before the pedestrian clearance interval begins. It should be long enough to allow a pedestrian that has pushed the pedestrian push button to enter the crosswalk. In many cases, the pedestrian phase will be set to rest in the walk interval to maximize the walk display during a vehicle green. Some controllers have a mechanism to specify that the walk interval begins before, or even after, the onset of the green interval. The walk interval may be extended in some controllers during coordination. A pedestrian recall mode, as discussed in a later section, can be used to eliminate the need for a pedestrian to push buttons and ensures that the pedestrian phase is presented each cycle. The length of the walk interval is usually established in local agency policy. The MUTCD () 19indicates that the minimum walk duration should be at least 7 seconds, but indicates that a duration as low as 4 seconds may be used if pedestrian volumes are low or pedestrian behavior does not justify the need for 7 seconds. Consideration should be given to walk durations longer than 7 seconds in school zones and areas with large numbers of elderly pedestrians. In cases where the pedestrian push button is a considerable distance from the curb, additional WALK time is desirable. Table 5-8 summarizes the recommended walk interval durations based on the guidance provided in the MUTCD and the Traffic Control Devices Handbook ()20. At intersections where older pedestrians are present, the MUTCD recommends that the WALK time allows for a pedestrian to reach the middle of the street at a 3.0 feet per second walking speed. Table 5-8 Pedestrian walk interval duration \| Conditions \| Walk Interval Duration (PW), s \| \| High pedestrian volume areas (e.g., school, central business district, sports venues, etc.) \| 10 to 15 \| \| Typical pedestrian volume and longer cycle length \| 7 to 10 \| \| Typical pedestrian volume and shorter cycle length \| 7 \| \| Negligible pedestrian volume \| 4 \| \| Conditions where older pedestrians are present \| Distance to center of road divided by 3.0 feet per second \| WALK times that allow pedestrians to cross to a wide median before the FLASHING DON’T WALK may reduce the potential for pedestrians stopping within the median. Pedestrian Clearance The pedestrian clearance interval follows the walk interval. When the pedestrian clearance interval begins, pedestrians should either complete their crossing if already in the intersection or refrain from entering the intersection until the next pedestrian walk interval is displayed. The MUTCD currently stipulates that the pedestrian clearance interval must be calculated assuming the distance from the curb to the far side of the opposing travel way, or to a median of sufficient width for pedestrians to wait. Note that previous editions of the MUTCD only required the clearance time to be as long as needed for the pedestrian to reach the center of the farthest traveled lane. Pedestrian clearance time is computed as the crossing distance divided by the walking speed. The speed of pedestrians is a critical assumption in determining this parameter. The MUTCD recommends a walking speed value of 4.0 feet per second (ft/s). The Americans with Disabilities Act (ADA) Accessibility Guidelines for Buildings and Facilities recommended use of 3.0 ft/s. Recent work completed by LaPlante and Kaeser has suggested that a speed of 3.5 ft/s be used to calculate the pedestrian clearance (FDW and Yellow) duration for curb to curb clearance and 3.0 ft/s be used for the total pedestrian time (WALK, FDW, and Yellow) duration for top of ramp to far curb clearance. The Pedestrian Facilities User Guide (21) recommends a maximum walking speed of 3.5 ft/s. This guide also suggests that a slower walking speed should be used in areas where there is a heavy concentration of elderly persons or children. A survey by Tarnoff and Ordonez (22) suggests a range of 3.0 to 3.5 ft/s is typically used by agencies to compute crossing time for these special-needs pedestrians. Pedestrian clearance time for typical pedestrian crossing distances can be obtained from Table 5-9. Table 5-9 Pedestrian clearance time \| Pedestrian Crossing Distance, ft \| Walking Speed, ft/s \| \| 3.0 \| 3.5 \| 4.0 \| \| Pedestrian Clearance Time (PCT), s \| \| 40 \| 13 \| 11 \| 10 \| \| 60 \| 20 \| 17 \| 15 \| \| 80 \| 27 \| 23 \| 20 \| \| 100 \| 33 \| 29 \| 25 \| Note: 1. Clearance times computed as PCT = Dc / vp, where Dc = pedestrian crossing distance (in feet) and vp = pedestrian walking speed (in feet per second). In general, agencies use one of two methods to determine the setting for the pedestrian clearance parameter. Some agencies require that the pedestrian clearance time conclude with the onset of the yellow change interval. This approach provides additional time (equal to the change period) for pedestrian clearance—time that is sometimes of benefit to pedestrians who walk slower than average. The pedestrian clearance interval duration for this practice is computed using Equation 5-3. Equation 5-3 Other agencies allow a portion of the pedestrian clearance time to occur during the change period (i.e., yellow change or yellow change plus red clearance intervals). This practice minimizes the impact of pedestrian service on phase duration and allows it to be more responsive to vehicular demand. This pedestrian clearance interval duration is computed using Equation 5-4. Equation 5-4 The practice of exclusing the change and clearance intervals may place pedestrians at risk if a concurrent permissive left turn movement is receiving a yellow and the vehicles from that movement are expected to clear the intersection during the yellow interval. Some agencies using flashing yellow applications choose to omit the permissive left turn portion of a protected-permissive left-turn movement during a pedestrian call. The pedestrian clearance time that transpires during the green interval coincides with a flashing “DON’T WALK” indication. At the onset of the yellow interval, a steady “DON’T WALK” indication is presented. It is noted that some agencies display the flashing “DON’T WALK” until the end of the change period. However, the MUTCD (Sections 4E.07 and 4E.10) states that if countdown pedestrian signals are used, the pedestrian clearance interval must finish timing before the onset of the yellow clearance interval. 5.4 ACTUATED TIMING PARAMETERS Research has shown that the best form of isolated operation occurs when fully-actuated controllers are used. Actuated controllers operate most effectively when timed in a manner that permits them to respond rapidly to fluctuations in vehicle demand (23). This section describes several of the more commonly used settings and parameters that influence phase function or duration in an actuated controller, including phase recall, passage time, simultaneous gap, and dual entry. In addition, this section discusses the volume-density technique. 5.4.1 Phase Recalls Recall causes the controller to place a call for a specified phase each time the controller is servicing a conflicting phase, regardless of the presence of any detector-actuated calls for the phase. There are four types of recalls: minimum recall (also known as vehicle recall), maximum recall, pedestrian recall, and soft recall. These are specified as phase option parameters in NTCIP Document 1202 (24). Minimum Recall (Vehicle Recall) The minimum recall parameter causes the controller to place a call for vehicle service on the phase. The phase is timed at least for its minimum green regardless of whether there is demand on the movement. The call is cleared upon start of green for the affected phase and placed upon start of the yellow change interval. This may be used where detection has failed. Minimum recall is the most frequently used recall mode. It is frequently used for the major-road through-movement phases (commonly designated as phases 2 and 6) at semi-actuated non-coordinated intersections. This use ensures that the controller will always return to the major-road through phases regardless of demand on the major-road through phases, thus providing a green indication as early as possible in the cycle. Maximum Recall The maximum recall parameter causes the controller to place a continuous call for vehicle service on the phase. It results in the presentation of the green indication for its maximum duration every cycle as defined by the maximum green parameter for the phase. When the maximum recall parameter is selected for a phase, the maximum green timer begins timing at the beginning of the phase’s green interval, regardless of the presence of a conflicting call or lack thereof. There are at least three common applications of maximum recall: Fixed-time operation is desired: Each phase is set for maximum recall. The maximum green setting used for this application should be equal to the green interval durations associated with an optimal fixed time plan. Vehicle detection is not present or is out of service: Maximum recall for a phase without detection ensures that the phase serves the associated movement. However, maximum recall can result in inefficient operation during light volume conditions (e.g., during night times and weekends) and should be used only when necessary. In some of these situations, a lower maximum green or MAX 2 (50 to 75% of the typical MAX GREEN value) may be desirable. Gapping out is not desired: Maximum recall can be used to prevent a phase from gapping out. An example application of this is under coordinated operations where a left turn phase is lagging. By setting the lagging left turn phase to maximum recall, the phase will time for its maximum duration, allowing the adjacent coordinated phase to also time for its intended maximum duration. This type of operation is typically only used on a time-of-day basis in conjunction with a particular coordinated plan (see Chapter 6). Pedestrian Recall The pedestrian recall parameter causes the controller to place a continuous call for pedestrian service on the phase, resulting in the controller timing its walk and flashing don’t walk operation. Coordination plans may invoke pedestrian calls using a rest in walk command, which dwells in the pedestrian walk interval, while awaiting the yield point. There are at least two common applications of pedestrian recall: Pedestrian detection is not present or is out of service: Pedestrian recall for a phase without pedestrian detection ensures that the phase times pedestrian walk and clearance intervals each cycle. High pedestrian demand: Pedestrian recall is sometimes used to activate the Walk and Pedestrian clearance intervals for phases and time periods that are likely to have high pedestrian demand. This is a common application during periods of high pedestrian activity in downtown environments or at intersections near schools as students are arriving or leaving school for the day. Soft Recall The soft recall parameter causes the controller to place a call for vehicle service on the phase in the absence of a serviceable conflicting call. When the phase is displaying its green indication, the controller serves the phase only until the minimum green interval times out. The phase can be extended if actuations are received. This may be used during periods of low traffic when there is a desire to default to the major street. The most typical application for soft recall is for the major-road through movement phases (usually phases 2 and 6) at non-coordinated intersections. The use of soft recall ensures that the major-road through phases will dwell in green when demand for the conflicting phases is absent. 5.4.2 Passage Time Passage time, sometimes called passage gap, vehicle extension, or unit extension, is used to extend the green interval based on the detector status once the phase is green. This parameter extends the Green Interval for each vehicle actuation up to the Maximum Green. It begins timing when the vehicle actuation is removed. This extension period is subject to termination by the Maximum Green timer or a Force Off. Passage time is used to find a gap in traffic for which to terminate the phase, essentially it is the setting that results in a phase ending prior to its maximum green time during isolated operation. If the passage time is too short, the green may end prematurely, before the vehicular movement has been adequately served. If the passage interval is set too long, there will be delays to other movements caused by unnecessary extension of a phase (25,) resulting in delay to the other movements at the intersection. The appropriate passage time used for a particular signal phase depends on many considerations, including: type and number of detection zones per lane, location of each detection zone, detection zone length, detection call memory (i.e., locking or nonlocking), detection mode (i.e., pulse or presence), approach speed, and whether lane-by-lane or approach detection is used. Ideally, the detection design is established and the passage time determined to ensure that the “system” provides efficient queue service and safe phase termination for higher speed approaches. Detection design procedures that reflect these considerations are described in Chapter 4. The passage timer starts to time from the instant the detector actuation is removed. A subsequent actuation will reset the passage timer. Thus, the mode of the detector, pulse or presence, is extremely important in setting the passage time. The pulse mode essentially measures headways between vehicles and the passage time would be set accordingly. The speed of the vehicles crossing the detectors and the size of the detectors is an important consideration in determining passage time when using presence mode. Longer passage times are often used with shorter detectors, greater distance between the detector and stop line, fewer lanes, and slower speeds. When the passage timer reaches the passage time limit, and a call is waiting for service on a conflicting phase, the phase will terminate, as shown in Figure 5-3. When this occurs, it is commonly termed as a “gap out”. In the figure, vehicle calls extend the green time until the gap in detector occupancy is greater than the passage time. In this example, presence detection is assumed. Research by Tarnoff suggests that the vehicle extension interval is one of the most important actuated controller settings, but the variety of techniques for determining proper settings suggest that there is either a lack of knowledge on the availability of this information or disagreement with the conclusions presented (26). Figure 5-3 Application of passage time This diagram illustrates the ‘gapping out’ concept in signal timing. During the green phase several vehicles approach the intersection after the minimum green time is reached. Each vehicle that passes the presence sensor will activate the passage timer. If another vehicle passes the sensor, the timer reinitializes. Once a sufficient gap in traffic occurs, the passage timer will expire, triggering a termination to the phase. The objective when determining the passage time value is to make it large enough to ensure that all vehicles in a moving queue are served but to not make it so large that it extends the green for randomly arriving traffic. This objective is broadened on high-speed approaches to ensure the passage time is not so large that the phase cannot be safely terminated. Many professionals believe that keeping one lane of traffic (in a left turn or a minor street) moving in deference to a major street with multiple lanes results in inefficient operation. Research has shown that measuring flow rates across lane groups and comparing them with the potential demand at an approach may provide improved decision making within the signal control logic. The guidelines provided in this section are based on the assumption that non-locking memory is used and that one source of detection is provided (per lane) for the subject signal phase. This source of detection could consist of one long detector loop at the stop line, a series of 6-foot loops that are closely spaced and operate together as one long zone of detection near the stop line, or a single 6-foot loop located at a known distance upstream of the stop line (and no detection at the stop line). As discussed in Chapter 4, passage time is a design parameter for detection designs that include multiple detectors for the purpose of providing safe phase termination (i.e., indecision zone protection). The passage-time value for this application is inherently linked to the detection design and should not be changed from its design value. Passage time defines the maximum time separation that can occur between vehicle calls without gapping out the phase. When only one traffic lane is served during the phase, this maximum time separation equals the maximum allowable headway (MAH) between vehicles. Although the maximum time separation does not equal the maximum allowable headway when several lanes are being served, the term "MAH" is still used and it is understood that the "headway" represents the time interval between calls (and not necessarily the time between vehicles in the same lane). Figure 5-4 illustrates the relationship between passage time, gap, and maximum allowable headway for a single-lane approach with one detector. This relationship can be used to derive the following equation for computing passage time for presence mode detection. Gap as shown in this figure is the amount of time that the detection zone is unoccupied. Equation 5-5 where, PT = passage time, s; MAH = maximum allowable headway, s; v sub a = average approach speed, mph; L sub v = length of vehicle (use 20 ft); and L sub d = length of detection zone, ft. Figure 5-4 Relationship between passage time, gap, and maximum allowable headway This figure illustrates a vehicle and how the passage time, gap, and maximum allowable headway interact between the length of the vehicle and the time to the maximum allowable headway. If Equation 5-5 is used with pulse-mode detection, then the length of vehicle Lv and the length of detector Ld equal 0.0 ft, and the passage time is equal to the MAH. The duration of the passage time setting should be based on three goals (27): 1. Ensure queue clearance. The passage time should not be so small that the resulting MAH causes the phase to have frequent premature gap-outs (i.e., a gap-out that occurs before the queue is fully served). A premature gap-out will leave a portion of the stopped queue unserved and, thereby, lead to increased delays and possible queue spillback. If the queue is extraordinarily long and cannot be accommodated without creating a cycle length that is longer than desirable, this goal may not apply. 2. Satisfy driver expectancy. The passage time should not be so large that the green is extended unnecessarily after the queue has cleared. Waiting drivers in conflicting phases will become anxious and may come to disrespect the signal indication. 3. Reduce max-out frequency. The passage time should not be so large that the resulting MAH causes the phase to have frequent max-outs. A long MAH would allow even light traffic volumes to extend the green to max-out. Waiting drivers in higher-volume conflicting phases may be unfairly delayed. Research by Tarnoff and Parsonson (28) indicates that there is a range of passage times within efficient intersection operations. This range extends from about 1 to 4 seconds for presence mode detection, with lower values being more appropriate under higher volume conditions. Values outside this range tend to increase delay. These passage times correspond to MAH values in the range of 2.0 to 4.5 seconds, depending on detection zone length and location. Based on the previous discussion, the following MAH values are recommended for use with Equation 5-3 to determine passage time: Gap reduction not used: MAH = 3.0 s Gap reduction used: MAH = 4.0 s The recommended MAH values may be increased by 0.1 s if the approach is on a steep upgrade and by 1.0 seconds if there is a large percentage of heavy vehicles. The passage time computed from the recommended MAH values for a range of speeds and detection zone lengths is provided in Table 5-10 for presence mode detection. It is critical that the relationship of passage time to vehicle speed, detector length, and detector location be considered. Table 5-10 Passage time duration for presence mode detection \| Maximum Allowable Headway, s \| Detection Zone Length, ft \| 85th Percentile Approach Speed, mph1 \| \| 25 \| 30 \| 35 \| 40 \| 45 \| \| Passage Time (PT), s \| \| 3.0 \| 6 \| 2.2 \| 2.3 \| 2.4 \| 2.5 \| 2.6 \| \| 15 \| 1.9 \| 2.1 \| 2.2 \| 2.3 \| 2.4 \| \| 25 \| 1.6 \| 1.8 \| 2.0 \| 2.1 \| 2.2 \| \| 35 \| 1.3 \| 1.6 \| 1.8 \| 1.9 \| 2.1 \| \| 45 \| 1.0 \| 1.3 \| 1.6 \| 1.7 \| 1.9 \| \| 55 \| 0.7 \| 1.1 \| 1.3 \| 1.6 \| 1.7 \| \| 65 \| 0.4 \| 0.8 \| 1.1 \| 1.4 \| 1.5 \| \| 75 \| 0.1 \| 0.6 \| 0.9 \| 1.2 \| 1.4 \| \| 4.0 \| 6 \| 3.2 \| 3.3 \| 3.4 \| 3.5 \| 3.6 \| \| 15 \| 2.9 \| 3.1 \| 3.2 \| 3.3 \| 3.4 \| \| 25 \| 2.6 \| 2.8 \| 3.0 \| 3.1 \| 3.2 \| \| 35 \| 2.3 \| 2.6 \| 2.8 \| 2.9 \| 3.1 \| \| 45 \| 2.0 \| 2.3 \| 2.6 \| 2.7 \| 2.9 \| \| 55 \| 1.7 \| 2.1 \| 2.3 \| 2.6 \| 2.7 \| \| 65 \| 1.4 \| 1.8 \| 2.1 \| 2.4 \| 2.5 \| \| 75 \| 1.1 \| 1.6 \| 1.9 \| 2.2 \| 2.4 \| Note: 1. Average approach speed is computed as 88 percent of the 85th percentile approach speed. 5.4.3 Simultaneous Gap Simultaneous gap defines how a barrier is crossed when a conflicting call is present. If enabled, it requires all phases that are timing concurrently to simultaneously reach a point of being committed to terminate (by gap-out, max-out, or force-off) before they can be allowed to jointly terminate. If disabled, each of the concurrent phases can reach a point of being committed to terminate separately and remain in that state while waiting for all concurrent phases to achieve this status. Simultaneous gap out should be enabled when advance detection is used to provide safe phase termination. 5.4.4 Dual Entry The dual (double) entry parameter is used to call vehicle phases that can time concurrently even if only one of the phases is receiving an active call. For example, if dual entry is active for Phases 2 and 6 and Phase 1 receives a call but no call is placed on Phase 6, Phase 6 would still be displayed along with Phase 1. The most common use of dual entry is to activate the parameter for compatible through movements. If the dual entry parameter is not selected, a vehicle call on a phase will only result in the timing of that phase in the absence of a call on a compatible phase. 5.5 VOLUME-DENSITY FEATURES Volume-density features can be categorized by two main features: gap reduction and variable initial. These features permit the user to provide variable alternatives to the otherwise fixed parameters of passage time (gap reduction) and minimum green (variable initial). Gap reduction provides a way to reduce the allowable gap over time, essentially becoming more aggressive in looking for an opportunity to end the phase. Variable initial provides an opportunity to utilize cycle by cycle traffic demand to vary the minimum time provided for a phase. These features increase the efficiency of the cycle with the fluctuations in demand, which can result in lower delay for users at the intersection. 5.5.1 Gap Reduction The gap reduction feature reduces the passage time to a smaller value while the phase is green. Initially, the gap sought between actuations is the passage time value. Then, after a specified time (Time Before Reduction), the passage timer is reduced to a minimum gap using a gradual reduction over a specified time (Time To Reduce). This functionality is achieved by programming the following controller parameters: time before reduction, time to reduce, and minimum gap. Their relationship is shown in Figure 5-5. Figure 5-5 Use of volume-density to change the extension time The time-before-reduction parameter establishes the time that is allowed to elapse after the arrival of a conflicting call and before the extension timer limit is reduced. This period begins when the phase is green and there is a serviceable call on a conflicting phase. Once the time-before-reduction period expires, the extension timer limit is reduced in a linear manner until the time-to-reduce period expires. Thereafter, the extension timer limit is set equal to the minimum-gap parameter. Like the Passage Time, this parameter extends the green interval by up to the Minimum Gap time for each vehicle actuation up to the Maximum Green. It begins timing when the vehicle actuation is removed. This extension period is subject to termination by the Maximum Green or a Force Off. The gap-reduction feature may be desirable when the phase volume is high and it is difficult to differentiate between the end of the initial queue and of the subsequent arrival of randomly formed platoons. This feature allows the user to specify a higher passage time at the beginning of a phase and then incrementally reduce the passage time as a phase gets longer and the delay to conflicting movements increases. With gap reduction, a MAH of 2.0 seconds is recommended for use with Equation 5-3 to determine the minimum gap. This MAH may be increased by 0.1 second if the approach is on a steep upgrade, and by 1.0 second if there is a large percentage of heavy vehicles. If Equation 5-3 is used with pulse-mode detection, then the length of vehicle Lv and the length of detector Ld equal 0.0 ft, and the minimum gap is equal to the MAH. The minimum gap is computed and shown in Table 5-11 using the recommended MAH values for a range of speeds and detection zone lengths provided for presence mode detection. Table 5-11 Minimum gap duration for presence mode detection \| Maximum Allowable Headway, s \| Detection Zone Length, ft \| 85th Percentile Approach Speed, mph1 \| \| 25 \| 30 \| 35 \| 40 \| 45 \| \| Minimum Gap, s \| \| 2.0 \| 6 \| 1.2 \| 1.3 \| 1.4 \| 1.5 \| 1.6 \| \| 15 \| 0.9 \| 1.1 \| 1.2 \| 1.3 \| 1.4 \| \| 25 \| 0.6 \| 0.8 \| 1.0 \| 1.1 \| 1.2 \| \| 35 \| 0.3 \| 0.6 \| 0.8 \| 0.9 \| 1.1 \| \| 45 \| 0.0 \| 0.3 \| 0.6 \| 0.7 \| 0.9 \| \| 55 \| 0.0 \| 0.1 \| 0.3 \| 0.6 \| 0.7 \| \| 65 \| 0.0 \| 0.0 \| 0.1 \| 0.4 \| 0.5 \| \| 75 \| 0.0 \| 0.0 \| 0.0 \| 0.2 \| 0.4 \| Note: 1. Average approach speed is computed as 88 percent of the 85th percentile approach speed. A number of different policies may be employed in determining the value of time-before-reduction. An example policy is to make the time-before-reduction setting equal to the minimum green interval and the time-to-reduce setting equal to half the difference between the maximum and minimum green intervals (29). This guidance is illustrated in Table 5-12. Table 5-12 Gap reduction parameter values \| Minimum Green Interval, s \| Time Before Reduction1, s \| Maximum Green, s \| \| 20 \| 25 \| 30 \| 35 \| 40 \| 45 \| 50 \| 55 \| 60 \| 65 \| 75 \| \| Time To Reduce, s \| \| 5 \| 10 \| 8 \| 10 \| 13 \| 15 \| 18 \| 20 \| 23 \| 25 \| 28 \| 30 \| 33 \| \| 10 \| 10 \| 5 \| 8 \| 10 \| 13 \| 15 \| 18 \| 20 \| 23 \| 25 \| 28 \| 30 \| \| 15 \| 15 \| N/A \| 5 \| 8 \| 10 \| 13 \| 15 \| 18 \| 20 \| 23 \| 25 \| 28 \| \| 20 \| 20 \| N/A \| N/A \| 5 \| 8 \| 10 \| 13 \| 15 \| 18 \| 20 \| 23 \| 25 \| Notes: 1. Time before reduction should always be 10 s or more in length. N/A – Gap reduction is not applicable to this combination of minimum and maximum green settings. 5.5.2 Variable Initial Variable initial is used in some cases to ensure that all vehicles queued between the stop line and the nearest upstream detector are served. Variable initial uses detector activity to determine a minimum green. Vehicles arriving on red that are not able to reach the upstream detector due to a standing queue will be detected and will extend the green by an amount sufficient to allow them to be served using the passage time. This feature is applicable when there are one or more advance detectors, no stop-line detection, and wide fluctuations in traffic volumes between peak and off-peak hours. Variable initial timing is achieved by programming the following controller parameters: minimum green, added initial, and maximum initial. Their relationship is shown in Figure 5-6. Added Initial – This interval times concurrently with the minimum green interval, and is increased by each vehicle actuation received during the associated phase yellow and red intervals. The initial green time portion is the greater of the minimum green or added initial intervals. The Added Initial cannot exceed the Maximum Initial (Figure 5-6). Maximum Initial – This is the maximum period of time for which the Added Initial can extend the initial green period. The Maximum Initial can not be less than the Minimum Green (Figure 5-6). Figure 5-6 Use of Added Initial to modify minimum green As with other volume-density parameters, there are varying policies that may be employed in determining the values of maximum-initial and added-initial. Generally, the maximum initial setting should be determined using the calculation for minimum green for queue clearance shown in Section 5.3.1. A common policy for selecting the added-initial parameter is to set this value at approximately 2.0 seconds per actuation if the phase serves only one traffic lane, 1.5 seconds per actuation if it serves two traffic lanes, and 1.2 seconds per actuation if it serves three or more lanes. Slightly larger values can be used if the approach has a significant upgrade or a significant number of heavy trucks. Bicycle traffic may also warrant higher values depending on the intersection width. Some agencies have developed more specific calculations for determining the added initial parameter. For example, the Los Angeles Department of Transportation uses Equation 5-4 to calculate the added initial setting. Equation 5-65.6 DETECTION CONFIGURATION AND PARAMETERS Traffic signal controllers have several settings that can be used to modify the vehicle actuations. Traditionally, this functionality was available only in the detection unit that served as an interface between the vehicle detector and the signal controller. Its implementation in the controller unit has streamlined the signal timing process and can duplicate functionality that may be in the detection unit. The parameters discussed in this section include: delay, extend, call, and queue. The latter two parameters are actually elements of the detector options in NTCIP Document 1202 (30). 5.6.1 Delay A delay parameter can be used to postpone a vehicle actuation for a detector input on a phase. By using a delay timer, an actuation is not made available until the delay timer expires and the actuation channel input is still active (i.e., the detection zone is still occupied). Once an actuation is made available to the controller, it is continued for as long as the channel input is active. Application of the delay timer is illustrated in Figure 5-7. Figure 5-7 Application of delay timer This figure illustrates the application of a delay timer used in a right turn lane with permissive right turns on red. In this example the first vehicle enters the detection zone, a delay timer begins timing, but the vehicle turns right on red before the time expires. No call is issued to the controller. A second vehicle enters the detection zone, the delay timer begins again, the vehicle does not turn right on red, the timer reaches zero, and a call is made to the controller. Common applications of a delay parameter on detection include the following: Delay is sometimes used with stop-line, presence mode detection for turn movements from exclusive lanes. For right-turn-lane detection, delay should be considered when the capacity for right-turn-on-red (RTOR) exceeds the right-turn volume or a conflicting movement is on recall. If RTOR capacity is limited, then delay may only serve to degrade intersection efficiency by further delaying right-turn vehicles. The delay setting should range from 8 to 12 seconds, with the larger values used for higher crossroad volumes (31). If the left-turn movement is protected-permissive and the opposing through phase is on minimum (or soft) recall, then delay should be considered for the detection in the left turn lane. The delay setting should range from 3 to 7 seconds, with the larger values used for higher opposing volumes (32). In this case, a minimum recall should also be placed on the adjacent through phase to ensure that a lack of demand on the adjacent through phase does not result in the left-turn movement receiving neither a permissive nor a protected left-turn indication. Delay may also be used to prevent an erroneous call from being registered in the controller if vehicles tend to traverse over another phase’s detector zone. For example, left-turning vehicles often cut across the perpendicular left-turn lane at the end of their turning movement. A detector delay coupled with non-locking memory would prevent a call from being placed for the unoccupied detector. 5.6.2 Extend The extend parameter is used to increase the duration of the actuation for a detector or phase. The extend timer begins the instant the actuation channel input is inactive. Thus, an actuation that is one second in duration at the channel input can be extended to three seconds, if the extend parameter is set to two seconds. This process is illustrated in Figure 5-8. Figure 5-8 Application of extend timer This figure illustrates the extend parameter. A vehicle enters a detection zone and initiates a call to the controller. When vehicle leaves the detection zone, the extend timer begins, and the call to the controller remains active. If the extend timer expires before another vehicle enters the detection zone, the call to the controller becomes inactive. If another vehicle enters the detection zone before the extend timer expires, the extend time is reset and the call to the controller remains active. Extend is typically used with detection designs that combine multiple advance detectors and stop-line detection for safe phase termination of high-speed intersection approaches. Extend is used with specific upstream detectors to supplement the passage-time parameter, to ensure that these detectors can extend the green interval by an amount of time equal to the sum of the passage time and call extension. The magnitude of the extension interval is dependent on the passage time, approach speed, and the distance between the subject detector and the next downstream detector. Typical values range from 0.1 to 2.0 seconds. The objective when used at high-speed approaches is to extend the green interval to ensure that a vehicle approaching the intersection has just enough time to reach the next downstream detector and place a new call for green extension. The procedure for identifying when call extension is needed and computing the amount of the extension time is specific to the detection design. Refer to the Manual of Traffic Detector Design by Bonneson and McCoy (33) for additional guidance on this application. 5.6.3 Carryover Carryover is a term commonly used for the Extend setting in controller manuals. It is another way to describe the time provided for a vehicle to traverse from one detector to the next. 5.6.4 Call The call parameter is used to allow actuations to be passed to the controller for the assigned phase when it is not timing a green interval. Actuations received during the green interval are ignored. The call parameter is sometimes used with detection designs that include one or more advance detectors and stop-line detection. With this design, the call-only parameter is used with the stop-line detectors to ignore the actuations these detectors receive during the green interval. The advance detectors are used to ensure safe and efficient service during the green interval. When an appropriate detection design is combined with this parameter, intersection efficiency can be improved by eliminating unnecessary green extension by the stop-line detection. 5.6.5 Queue A detector can be configured as a queue service detector to effectively extend the green interval until the queue is served, at which time it is deactivated until the start of the next conflicting phase. This functionality is offered as a parameter in most modern controllers. However, if it is not available as a parameter, equivalent functionality can be acquired by using the features of many modern detector amplifiers. This functionality is sometimes used with detection designs that include one or more advance detectors and stop-line detection. With this design, the queue service functionality is used to deactivate the stop line detection during the green interval, but after the queue has cleared. The advance detectors are then used to ensure safe phase termination. When combined with an appropriate detection design, this functionality can improve intersection efficiency by eliminating unnecessary green extension by the stop-line detection. 5.7 GUIDELINES FOR TIME-BASE CONTROLS Most controllers provide a means to externally apply signal timing parameters by time of day; typically these include maximum green, phase omit, and minimum recall on a time-of-day basis. Depending on the manufacturer, time-of-day selection of pedestrian omit, maximum recall, pedestrian recall, detector switching, overlap omit, additional maximums, alternate walk intervals, and other parameters may also be available. The approach specified by NTCIP 1202 for activating phase and ring controls invokes a timing pattern that can be selected on a time-of-day basis (34). In NTCIP protocol, a timing pattern consists of a cycle length, offset, set of minimum green and maximum green values, force off (determined by splits in some cases), and phase sequence. It also includes specification of phase parameters for minimum or maximum vehicle recall, pedestrian recall, or phase omit. This will be further described in Chapter 6. There are a number of controls that can be used to modify controller operation on a time-of-day basis. A remote entry to one of these controls will invoke the corresponding parameter. The most common time-based controls are maximum green 2 (Max 2), phase omit, and minimum recall. The method of activating these controls varies from manufacturer to manufacturer. There are two typical uses of phase omit. One use is when a left-turn phase is only needed during the peak traffic period. A second use is where a left-turn movement is prohibited during the peak period. In this situation, the associated left-turn phase is omitted during the turn prohibition period. Minimum recall is used primarily on the major-road phase(s) of a fully-actuated, non-coordinated intersection. If the volume on the minor road is low only during certain times of the day, minimum recall for the major-road phases could be activated during these time periods. 5.8 REFERENCES 1. National Transportation Communications for ITS Protocol: Object Definitions for Actuated Traffic Signal Controller (ASC) Units – 1202 v01.07. National Electrical Manufacturers Association, Rosslyn, Virginia, January 2005 2. NEMA Standards Publications TS 2-2003 v2.06: Traffic Controller Assemblies with Controller Requirements. National Electrical Manufacturers Association, Rosslyn, Virginia, 2003 3. Orcutt, F.L. The Traffic Signal Book. Prentice Hall, Englewood Cliffs, New Jersey, 1993 4. Skabardonis, A., R. Bertini, B.Gallagher, “Development and Application of Control Strategies for Signalized Intersections in Coordinated Systems” Transportation Research Record No. 1634. Transportation Research Board, National Research Council, Washington, D.C., 1998, pp. 110-117 5. Lin, F.B. “Optimal Timing Settings and Detector Lengths of Presence Mode Full-Actuated Control.” Transportation Research Record No. 1010. Transportation Research Board, National Research Council, Washington, D.C., 1985, pp. 37-45 6. Parsonson, P.S. NCHRP Synthesis 172: Signal Timing Improvement Practices. Transportation Research Board, National Research Council, Washington, D.C., 1992 7. Kell, J. H., and Fullerton, I. J., Manual of Traffic Signal Design. 2nd Edition. Institute of Transportation Engineers, Washington, D.C., 1998 8. Skabardonis, Alexander, Progression Through a Series of Intersections with Traffic Actuated Controllers, Volume 2 User’s Guide, FHWA RD-89-133, Washington, D.C., October 1998 9. Uniform Vehicle Code - Millennium Edition. National Committee on Uniform Traffic Laws and Ordinances. Alexandria, Virginia, 2000 10. Parsonson, P.S. 1992 11. Souleyrette, R. R., O’Brien, M. M., McDonald, T., Preston, H., and Storm, R., “Effectiveness of All-Red Clearance Interval on Intersection Crashes”, Center for Transportation Research and Education, Iowa State University, Minnesota Department of Transportation, MN/RC-2004-26, May 2004 12. Kell, J. H., and Fullerton, I. J. 1998 13. Ibid 14. Traffic Engineering Handbook. 5th Edition. J.L. Pline editor. Institute of Transportation Engineers, Washington, D.C., 1999 15. Traffic Control Devices Handbook. J.L. Pline editor. Institute of Transportation Engineers, Washington, D.C., 2001 16. Tarnoff, P.J. "Traffic Signal Clearance Intervals." ITE Journal. Institute of Transportation Engineers, Washington, D.C. April 2004, pp. 20-24 17. Traffic Control Devices Handbook. 2001 18. Manual on Uniform Traffic Control Devices. U.S. Department of Transportation, Washington, D.C., December 2003 19. Ibid 20. Traffic Control Devices Handbook. J.L. Pline editor. Institute of Transportation Engineers, Washington, D.C., 2001 21. Zegeer, C.V., C. Seiderman, P. Lagerway, M. Cynecki, M. Ronkin, and R. Schneider. Pedestrian Facilities User’s Guide - Providing Safety and Mobility. Report No. FHWA-RD-01-102. U.S. Department of Transportation, Federal Highway Administration, Washington, D.C., 2002 22. Tarnoff, P.J., and J. Ordonez. Signal Timing Practices: State of the Practice. Institute of Transportation Engineers, Washington, D.C., March 2004 23. Ibid 24. National Transportation Communications for ITS Protocol: January 2005 25. Henry, RD, “Signal Timing on a Shoestring”, Federal Highway Administration, FHWA-HOP-07-006, March 2005. 26. Tarnoff, P. and Ordonez, 2004. 27. Bonneson, J. A., and P.T. McCoy. Manual of Traffic Detector Design, 2nd Edition. Institute of Transportation Engineers, Washington DC, May 2005 28. Tarnoff, P.J., and P.S. Parsonson. NCHRP Report 233: Selecting Traffic Signal Control at Individual Intersections. Transportation Research Board, National Research Council, Washington, D.C., 1981 29. UDOT Procedural Update - Timing of Traffic Signals. Traffic Operations Center, Utah Department of Transportation, Ogden, Utah, April 15, 2004 30. National Transportation Communications for ITS Protocol: January 2005 31. Bonneson, J.A., and M. Abbas. Intersection Video Detection Manual. Report FHWA/TX-03/4285-2. Texas Department of Transportation, Austin, Texas, September 2002 32. Bonneson, J. A., and P.T. McCoy. 2005 33. Ibid 34. National Transportation Communications for ITS Protocol: January 2005 5-30 Previous \| Table of Contents \| Next \| \| \| \| \| US DOT Home \| FHWA Home \| Operations Home \| Privacy Policy United States Department of Transportation - Federal Highway Administration \| \|
15022	https://www.whitman.edu/mathematics/calculus_online/section03.02.html	sinx sinx Collapse menu Introduction 1 Analytic Geometry 1. Lines 2. Distance Between Two Points; Circles 3. Functions 4. Shifts and Dilations 2 Instantaneous Rate of Change: The Derivative 1. The slope of a function 2. An example 3. Limits 4. The Derivative Function 5. Properties of Functions 3 Rules for Finding Derivatives 1. The Power Rule 2. Linearity of the Derivative 3. The Product Rule 4. The Quotient Rule 5. The Chain Rule 4 Transcendental Functions 1. Trigonometric Functions 2. The Derivative of sinx 3. A hard limit 4. The Derivative of sinx, continued 5. Derivatives of the Trigonometric Functions 6. Exponential and Logarithmic functions 7. Derivatives of the exponential and logarithmic functions 8. Implicit Differentiation 9. Inverse Trigonometric Functions 10. Limits revisited 11. Hyperbolic Functions 5 Curve Sketching 1. Maxima and Minima 2. The first derivative test 3. The second derivative test 4. Concavity and inflection points 5. Asymptotes and Other Things to Look For 6 Applications of the Derivative 1. Optimization 2. Related Rates 3. Newton's Method 4. Linear Approximations 5. The Mean Value Theorem 7 Integration 1. Two examples 2. The Fundamental Theorem of Calculus 3. Some Properties of Integrals 8 Techniques of Integration 1. Substitution 2. Powers of sine and cosine 3. Trigonometric Substitutions 4. Integration by Parts 5. Rational Functions 6. Numerical Integration 7. Additional exercises 9 Applications of Integration 1. Area between curves 2. Distance, Velocity, Acceleration 3. Volume 4. Average value of a function 5. Work 6. Center of Mass 7. Kinetic energy; improper integrals 8. Probability 9. Arc Length 10. Surface Area 10 Polar Coordinates, Parametric Equations 1. Polar Coordinates 2. Slopes in polar coordinates 3. Areas in polar coordinates 4. Parametric Equations 5. Calculus with Parametric Equations 11 Sequences and Series 1. Sequences 2. Series 3. The Integral Test 4. Alternating Series 5. Comparison Tests 6. Absolute Convergence 7. The Ratio and Root Tests 8. Power Series 9. Calculus with Power Series 10. Taylor Series 11. Taylor's Theorem 12. Additional exercises 12 Three Dimensions 1. The Coordinate System 2. Vectors 3. The Dot Product 4. The Cross Product 5. Lines and Planes 6. Other Coordinate Systems 13 Vector Functions 1. Space Curves 2. Calculus with vector functions 3. Arc length and curvature 4. Motion along a curve 14 Partial Differentiation 1. Functions of Several Variables 2. Limits and Continuity 3. Partial Differentiation 4. The Chain Rule 5. Directional Derivatives 6. Higher order derivatives 7. Maxima and minima 8. Lagrange Multipliers 15 Multiple Integration 1. Volume and Average Height 2. Double Integrals in Cylindrical Coordinates 3. Moment and Center of Mass 4. Surface Area 5. Triple Integrals 6. Cylindrical and Spherical Coordinates 7. Change of Variables 16 Vector Calculus 1. Vector Fields 2. Line Integrals 3. The Fundamental Theorem of Line Integrals 4. Green's Theorem 5. Divergence and Curl 6. Vector Functions for Surfaces 7. Surface Integrals 8. Stokes's Theorem 9. The Divergence Theorem 17 Differential Equations 1. First Order Differential Equations 2. First Order Homogeneous Linear Equations 3. First Order Linear Equations 4. Approximation 5. Second Order Homogeneous Equations 6. Second Order Linear Equations 7. Second Order Linear Equations, take two 18 Useful formulas 19 Introduction to Sage 1. Basics 2. Differentiation 3. Integration An operation is linear if it behaves "nicely'' with respect to multiplication by a constant and addition. The name comes from the equation of a line through the origin, f(x)=mxf(x)=mx, and the following two properties of this equation. First, f(cx)=m(cx)=c(mx)=cf(x)f(cx)=m(cx)=c(mx)=cf(x), so the constant cc can be "moved outside'' or "moved through'' the function ff. Second, f(x+y)=m(x+y)=mx+my=f(x)+f(y)f(x+y)=m(x+y)=mx+my=f(x)+f(y), so the addition symbol likewise can be moved through the function. The corresponding properties for the derivative are: (cf(x))′=ddxcf(x)=cddxf(x)=cf′(x), (cf(x))′=ddxcf(x)=cddxf(x)=cf′(x), and (f(x)+g(x))′=ddx(f(x)+g(x))=ddxf(x)+ddxg(x)=f′(x)+g′(x). (f(x)+g(x))′=ddx(f(x)+g(x))=ddxf(x)+ddxg(x)=f′(x)+g′(x). It is easy to see, or at least to believe, that these are true by thinking of the distance/speed interpretation of derivatives. If one object is at position f(t)f(t) at time tt, we know its speed is given by f′(t)f′(t). Suppose another object is at position 5f(t)5f(t) at time tt, namely, that it is always 5 times as far along the route as the first object. Then it "must'' be going 5 times as fast at all times. The second rule is somewhat more complicated, but here is one way to picture it. Suppose a flatbed railroad car is at position f(t)f(t) at time tt, so the car is traveling at a speed of f′(t)f′(t) (to be specific, let's say that f(t)f(t) gives the position on the track of the rear end of the car). Suppose that an ant is crawling from the back of the car to the front so that its position on the car is g(t)g(t) and its speed relative to the car is g′(t)g′(t). Then in reality, at time tt, the ant is at position f(t)+g(t)f(t)+g(t) along the track, and its speed is "obviously'' f′(t)+g′(t)f′(t)+g′(t). We don't want to rely on some more-or-less obvious physical interpretation to determine what is true mathematically, so let's see how to verify these rules by computation. We'll do one and leave the other for the exercises. ddx(f(x)+g(x))=limΔx→0f(x+Δx)+g(x+Δx)−(f(x)+g(x))Δx=limΔx→0f(x+Δx)+g(x+Δx)−f(x)−g(x)Δx=limΔx→0f(x+Δx)−f(x)+g(x+Δx)−g(x)Δx=limΔx→0(f(x+Δx)−f(x)Δx+g(x+Δx)−g(x)Δx)=limΔx→0f(x+Δx)−f(x)Δx+limΔx→0g(x+Δx)−g(x)Δx=f′(x)+g′(x) ddx(f(x)+g(x))=limΔx→0f(x+Δx)+g(x+Δx)−(f(x)+g(x))Δx=limΔx→0f(x+Δx)+g(x+Δx)−f(x)−g(x)Δx=limΔx→0f(x+Δx)−f(x)+g(x+Δx)−g(x)Δx=limΔx→0(f(x+Δx)−f(x)Δx+g(x+Δx)−g(x)Δx)=limΔx→0f(x+Δx)−f(x)Δx+limΔx→0g(x+Δx)−g(x)Δx=f′(x)+g′(x) This is sometimes called the sum rule for derivatives. Example 3.2.1 Find the derivative of f(x)=x5+5x2f(x)=x5+5x2. We have to invoke linearity twice here: f′(x)=ddx(x5+5x2)=ddxx5+ddx(5x2)=5x4+5ddx(x2)=5x4+5⋅2x1=5x4+10x. f′(x)=ddx(x5+5x2)=ddxx5+ddx(5x2)=5x4+5ddx(x2)=5x4+5⋅2x1=5x4+10x. ◻□ Because it is so easy with a little practice, we can usually combine all uses of linearity into a single step. The following example shows an acceptably detailed computation. Example 3.2.2 Find the derivative of f(x)=3/x4−2x2+6x−7f(x)=3/x4−2x2+6x−7. f′(x)=ddx(3x4−2x2+6x−7)=ddx(3x−4−2x2+6x−7)=−12x−5−4x+6. f′(x)=ddx(3x4−2x2+6x−7)=ddx(3x−4−2x2+6x−7)=−12x−5−4x+6. ◻□ Exercises 3.2 Find the derivatives of the functions in 1–6. Ex 3.2.1 5x3+12x2−155x3+12x2−15 (answer) Ex 3.2.2 −4x5+3x2−5/x2−4x5+3x2−5/x2 (answer) Ex 3.2.3 5(−3x2+5x+1)5(−3x2+5x+1) (answer) Ex 3.2.4 f(x)+g(x)f(x)+g(x), where f(x)=x2−3x+2f(x)=x2−3x+2 and g(x)=2x3−5xg(x)=2x3−5x (answer) Ex 3.2.5 (x+1)(x2+2x−3)(x+1)(x2+2x−3) (answer) Ex 3.2.6 √625−x2+3x3+12625−x2−−−−−−−√+3x3+12 (See section 2.1.) (answer) Ex 3.2.7 Find an equation for the tangent line to f(x)=x3/4−1/xf(x)=x3/4−1/x at x=−2x=−2. (answer) Ex 3.2.8 Find an equation for the tangent line to f(x)=3x2−π3f(x)=3x2−π3 at x=4x=4. (answer) Ex 3.2.9 Suppose the position of an object at time tt is given by f(t)=−49t2/10+5t+10f(t)=−49t2/10+5t+10. Find a function giving the speed of the object at time tt. The acceleration of an object is the rate at which its speed is changing, which means it is given by the derivative of the speed function. Find the acceleration of the object at time tt. (answer) Ex 3.2.10 Let f(x)=x3f(x)=x3 and c=3c=3. Sketch the graphs of ff, cfcf, f′f′, and (cf)′(cf)′ on the same diagram. Ex 3.2.11 The general polynomial PP of degree nn in the variable xx has the form P(x)=n∑k=0akxk=a0+a1x+…+anxnP(x)=∑k=0nakxk=a0+a1x+…+anxn. What is the derivative (with respect to xx) of PP? (answer) Ex 3.2.12 Find a cubic polynomial whose graph has horizontal tangents at (−2,5)(−2,5) and (2,3)(2,3). (answer) Ex 3.2.13 Prove that ddx(cf(x))=cf′(x)ddx(cf(x))=cf′(x) using the definition of the derivative. Ex 3.2.14 Suppose that ff and gg are differentiable at xx. Show that f−gf−g is differentiable at xx using the two linearity properties from this section.
15023	https://m.esnai.com/baike/view.aspx?w=%E5%A4%8D%E5%90%88%E5%A2%9E%E9%95%BF%E7%8E%87	复合增长率 -复合增长率 - 百科移动版视野百科知识点>> 复合增长率别名:[年复合增长率] [size=32px]概述[/size] 一项投资在特定时期内的年度增长率计算方法为总增长率百分比的n方根，n相等于有关时期内的年数复合增长率的英文缩写为：CAGR（Compound Annual Growth Rate）。 CAGR并不等于现实生活中GR（Growth Rate）的数值。它的目的是描述一个投资回报率转变成一个较稳定的投资回报所得到的预想值。我们可以认为CAGR平滑了回报曲线，不会为短期回报的剧变而迷失。 [size=32px]公式[/size]为： (现有价值/基础价值)^(1/年数) - 1 这个概念并不复杂。举个例子，你在2005年1月1日最初投资了10,000美金，而到了2006年1月1日你的资产增长到了13,000美金，到了2007年增长到了14,000美金，而到了2008年1月1日变为19,500美金。根据计算公式，Your CAGR would be the ratio of your ending value tobeginning value (,500/,000 = 1.95) raised to the power of 1/3 ( since1/# of years = 1/3), then subtracting 1 from the resulting number. 1.95 raised to 1/3 power = 1.2493. (This could be written as 1.95^0.3333) 1.2493 -1=0.2493 Another way of writing 0.2493 is 24.93%. 最后计算获得的CAGR为24.93%，从而意味着你三年的投资回报率为24.93%，即将按年份计算的增长率在时间轴上平坦化。当然，你也看到第一年的增长率则是30%（13000－10000）/10000100% 可以理解为，[b]年增长率是一个短期的概念，从一个产品或产业的发展来看，可能处在成长期或爆发期而年度结果变化很大，但如果以“复合增长率 ”在衡量，因为这是个长期时间基础上的核算，所以更能够说明产业或产品增长或变迁的潜力和预期[/b]。复利，就是复合利息，它是指每年的收益还可以产生收益，就是俗称的利滚利。复利的计算是对本金及其产生的利息一并计算，也就是利上有利。复利计算的特点是：把上期未的本利和作为下一期的本金，在计算时每一期本金的数额是不同的。复利的计算公式复利的计算公式是：S＝P（1＋i）^n 即：本利之和＝本金×（1＋利率)×期数”这个“期数”时间因子是整个公式的关键因素，一年又一年(或一月一月)地相乘下来，数值当然会愈来愈大。公式为：(现有价值/基础价值)^(1/年数) - 1 股票（上市公司）的复合增长率复合增长率（CAGR，Compound Annual Growth Rate）。复合增长率不等于GR（Growth Rate）的数值，它描述的是较稳定的投资回报，不会为短期回报（GR）的剧变而迷失方向。 [size=32px]举个例子[/size] 公司在2000年1月1日，最初投资了100,000元，在2001年1月1日，公司资产增长到130,000元，年增长率GR为30% 在2002年1月1日，公司资产为 140,000元，年增长率GR为7.7% 到2003年1月1日，公司资产为 195,000元，年增长率GR为39% 可以看到年增长率GR变化很大。而三年的复合增长率计算为：(195,000/100,000)^1/3=1.2493 三年的复合增长CAGR为24.93%，即100,000×(1+CAGR)^3=195,000 可以这样理解：年增长率GR是一个短期概念，从一个公司或产业的发展看，可能处在成长期或爆发期而年度GR变化很大，但如果以复合增长率 CAGR衡量，因为是在长期时间基础上核算的，所以更能说明公司或产业的增长的潜力和预期。开放分类:管理会计、财务理论、金融贡献人:jansen2008、歪C歪百科PC版 \| 视野移动版 © 中国会计视野网站
15024	https://www.chegg.com/homework-help/questions-and-answers/formation-constant-ag-cn-2-aq-2x1020-cyanide-ion-concentration-ag-ag-cn-2-answer-7x10-11m--q6948605	Solved The formation constant for Ag(CN)2-(aq) is 2x1020. At \| Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Science Chemistry Chemistry questions and answers The formation constant for Ag(CN)2-(aq) is 2x1020. At what cyanide ion concentration is [Ag+] = [Ag(CN)2-]? The answer is 7x10-11M. Please show your steps. Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: The formation constant for Ag(CN)2-(aq) is 2x1020. At what cyanide ion concentration is [Ag+] = [Ag(CN)2-]? The answer is 7x10-11M. Please show your steps. The formation constant for Ag(CN)2-(aq) is 2x10 20. At what cyanide ion concentration is [Ag+] = [Ag(CN)2-]? The answer is 7x10-11 M. Please show your steps. Here’s the best way to solve it.Solution 100%(11 ratings) Share Share Share done loading Copy link Here’s how to approach this question This AI-generated tip is based on Chegg's full solution. Sign up to see more! Write the equilibrium expression for the formation constant K f using the concentrations of the products and reactants. Equilibrium reactions: The… View the full answer Previous questionNext question Not the question you’re looking for? Post any question and get expert help quickly. 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15025	https://www.mayocliniclabs.com/test-catalog/overview/88886	Test Id : PEE Porphyrins Evaluation, Whole Blood Overview Overview Specimen Clinical & Interpretive Performance Fees & Codes Setup & Updates Useful For Suggests clinical disorders or settings where the test may be helpful Establishing a biochemical diagnosis of erythropoietic protoporphyria and X-linked dominant protoporphyria Reflex Tests Lists tests that may or may not be performed, at an additional charge, depending on the result and interpretation of the initial tests. \| Test Id \| Reporting Name \| Available Separately \| Always Performed \| --- --- \| \| PPFE \| Protoporphyrins, Fractionation, WB \| Yes \| No \| Testing Algorithm Delineates situations when tests are added to the initial order. This includes reflex and additional tests. This test is recommended for screening patients for possible erythropoietic protoporphyria and X-linked dominant protoporphyria. In addition, it can be used for evaluation of iron-deficiency anemia and chronic lead intoxication. Testing begins with total erythrocyte porphyrins. If the result is below 80 mcg/dL, it is normal, and testing is complete. If the total erythrocyte porphyrin value is 80 mcg/dL or above, the protoporphyrin fractionation assay will automatically be performed at an additional charge. The fractionation test results include noncomplexed (free) protoporphyrin and zinc-complexed protoporphyrin. The following algorithms are available: \| \| \| -Porphyria (Acute) Testing Algorithm \| \| -Porphyria (Cutaneous) Testing Algorithm \| Method Name A short description of the method used to perform the test Spectrofluorometric NY State Available Indicates the status of NY State approval and if the test is orderable for NY State clients. Yes Reporting Name Lists a shorter or abbreviated version of the Published Name for a test Porphyrins Evaluation, WB Aliases Lists additional common names for a test, as an aid in searching Erythrocyte Porphyrin Erythropoietic Protoporphyria (EPP) Free Erythrocyte Porphyrin (FEP) Protoporphyrin Protoporphyrins, Total, Erythrocytes RBC Porphyrins Red Blood Cell Porphyrins X-linked Dominant Protoporphyria (XLDPP, XLEPP, or XDP) Testing Algorithm Delineates situations when tests are added to the initial order. This includes reflex and additional tests. This test is recommended for screening patients for possible erythropoietic protoporphyria and X-linked dominant protoporphyria. In addition, it can be used for evaluation of iron-deficiency anemia and chronic lead intoxication. Testing begins with total erythrocyte porphyrins. If the result is below 80 mcg/dL, it is normal, and testing is complete. If the total erythrocyte porphyrin value is 80 mcg/dL or above, the protoporphyrin fractionation assay will automatically be performed at an additional charge. The fractionation test results include noncomplexed (free) protoporphyrin and zinc-complexed protoporphyrin. The following algorithms are available: \| \| \| -Porphyria (Acute) Testing Algorithm \| \| -Porphyria (Cutaneous) Testing Algorithm \| Specimen Type Describes the specimen type validated for testing Whole blood Ordering Guidance This is the preferred test for assessment for protoporphyria. The preferred test for assessing lead toxicity in children is blood lead. For more information see PBDV / Lead, Venous, with Demographics, Blood or PBDC / Lead, Capillary, with Demographics, Blood. The preferred screening test for suspicion of a hepatic porphyria is urine porphyrins. For more information see PQNRU / Porphyrins, Quantitative, Random, Urine. Necessary Information Include a list of medications the patient is currently taking. Specimen Required Defines the optimal specimen required to perform the test and the preferred volume to complete testing All porphyrin tests on whole blood can be performed on 1 collection tube. Patient Preparation: Patient must not consume any alcohol for 24 hours before specimen collection. Container/Tube: Preferred: Green top (sodium heparin) Acceptable: Dark blue top (metal free heparin), green top (lithium heparin), lavender top (EDTA) Specimen Volume: 4 mL Collection Instructions: Immediately place specimen on wet ice. Special Instructions Library of PDFs including pertinent information and forms related to the test The Heme Biosynthetic Pathway Porphyria (Acute) Testing Algorithm Porphyria (Cutaneous) Testing Algorithm Forms If not ordering electronically, complete, print, and send aBiochemical Genetics Test Request (T798) with the specimen. Specimen Minimum Volume Defines the amount of sample necessary to provide a clinically relevant result as determined by the testing laboratory. The minimum volume is sufficient for one attempt at testing. 3 mL Reject Due To Identifies specimen types and conditions that may cause the specimen to be rejected \| \| \| --- \| \| Gross hemolysis \| Reject \| Specimen Stability Information Provides a description of the temperatures required to transport a specimen to the performing laboratory, alternate acceptable temperatures are also included \| Specimen Type \| Temperature \| Time \| Special Container \| --- --- \| \| Whole blood \| Refrigerated \| 7 days \| \| Useful For Suggests clinical disorders or settings where the test may be helpful Establishing a biochemical diagnosis of erythropoietic protoporphyria and X-linked dominant protoporphyria Testing Algorithm Delineates situations when tests are added to the initial order. This includes reflex and additional tests. This test is recommended for screening patients for possible erythropoietic protoporphyria and X-linked dominant protoporphyria. In addition, it can be used for evaluation of iron-deficiency anemia and chronic lead intoxication. Testing begins with total erythrocyte porphyrins. If the result is below 80 mcg/dL, it is normal, and testing is complete. If the total erythrocyte porphyrin value is 80 mcg/dL or above, the protoporphyrin fractionation assay will automatically be performed at an additional charge. The fractionation test results include noncomplexed (free) protoporphyrin and zinc-complexed protoporphyrin. The following algorithms are available: \| \| \| -Porphyria (Acute) Testing Algorithm \| \| -Porphyria (Cutaneous) Testing Algorithm \| Clinical Information Discusses physiology, pathophysiology, and general clinical aspects, as they relate to a laboratory test The porphyrias are a group of inherited disorders resulting from enzyme defects in the heme biosynthetic pathway. Depending on the specific enzyme involved, various porphyrins and their precursors accumulate in different specimen types. The patterns of porphyrin accumulation in erythrocytes and plasma and excretion of the heme precursors in urine and feces allow for the detection and differentiation of the porphyrias. Testing erythrocyte porphyrin level is most informative for patients with a clinical suspicion of erythropoietic protoporphyria (EPP) or X-linked dominant protoporphyria (XLDPP). Clinical presentation of EPP and XLDPP is identical, with onset of symptoms typically occurring in childhood. Cutaneous photosensitivity in sun-exposed areas of the skin generally worsens in the spring and summer months. Common symptoms may include itching, edema, erythema, stinging or burning sensations, and occasionally scarring of the skin in sun-exposed areas. Although genetic in nature, environmental factors can exacerbate symptoms, significantly impacting the severity and course of disease. Erythropoietic protoporphyria is caused by decreased ferrochelatase activity resulting in significantly increased noncomplexed (free) protoporphyrin levels in erythrocytes, plasma, and feces. X-linked dominant protoporphyria is caused by gain-of-function variants in the C-terminal end of ALAS2 gene and results in elevated erythrocyte levels of free and zinc-complexed protoporphyrin, and total protoporphyrin levels in plasma and feces. Protoporphyrin fractionation is the main component of erythrocyte porphyrins. When total erythrocyte porphyrins are elevated, fractionation and quantitation of zinc-complexed and free protoporphyrin is necessary to differentiate the inherited porphyrias from other causes of elevated porphyrin levels. Other possible causes of elevated erythrocyte zinc-complexed protoporphyrin may include: -Iron-deficiency anemia, the most common cause -Chronic intoxication by heavy metals (primarily lead) or various organic chemicals -Congenital erythropoietic porphyria, a rare autosomal recessive porphyria caused by deficient uroporphyrinogen III synthase -Hepatoerythropoietic porphyria, a rare autosomal recessive porphyria caused by deficient uroporphyrinogen decarboxylase Typically, the workup of patients with a suspected porphyria is most effective when following a stepwise approach. See Porphyria (Acute) Testing Algorithm and Porphyria (Cutaneous) Testing Algorithm or call 800-533-1710 to discuss testing strategies. There are 2 test options: -PEE / Porphyrins Evaluation, Whole Blood -PEWE / Porphyrins Evaluation, Washed Erythrocytes. The whole blood option is easiest for clients but requires that the specimen arrive at Mayo Clinic Laboratories within 7 days of collection. When this cannot be ensured, washed frozen erythrocytes, which are stable for 14 days, should be submitted. Reference Values Describes reference intervals and additional information for interpretation of test results. May include intervals based on age and sex when appropriate. Intervals are Mayo-derived, unless otherwise designated. If an interpretive report is provided, the reference value field will state this. PORPHYRINS, TOTAL, RBC <80 mcg/dL Interpretation Provides information to assist in interpretation of the test results Abnormal results are reported with a detailed interpretation that may include an overview of the results and their significance, a correlation to available clinical information provided with the specimen, differential diagnosis, and recommendations for additional testing when indicated and available. Cautions Discusses conditions that may cause diagnostic confusion, including improper specimen collection and handling, inappropriate test selection, and interfering substances Alcohol suppresses enzyme activity potentially leading to false-positive results if it is ingested within 24 hours of specimen collection. Clinical Reference Recommendations for in-depth reading of a clinical nature Tortorelli S, Kloke K, Raymond K. Disorders of porphyrin metabolism. In: Dietzen DG, Bennett MJ, Wong ECC, eds. Biochemical and Molecular Basis of Pediatric Disease. 4th ed. AACC Press; 2010:307-324 Badminton MN, Whatley SD, Schmitt C, Aarsand AK. Porphyrins and the porphyrias. In: Rifai N, Chiu RWK, Young I, Burnham CAD, eds. Tietz Textbook of Laboratory Medicine. 7th ed. Elsevier; 2023:419-419.e32 Anderson KE, Sassa S, Bishop DF, Desnick RJ. Disorders of heme biosynthesis: X-linked sideroblastic anemia and the porphyrias In: Valle D, Antonarakis S, Ballabio A, Beaudet AL, Mitchell GA, eds. The Online Metabolic and Molecular Bases of Inherited Disease. McGraw-Hill, 2019. Accessed September 6, 2024. Available at Whatley SD, Ducamp S, Gouya B, et al. C-terminal deletions in the ALAS2 gene lead to gain of function and cause X-linked dominant protoporphyria without anemia or iron overload. Am J Hum Genet. 2008;83(3):408-414 Balwani M, Naik H, Anderson KE, et al. Clinical, biochemical, and genetic characterization of North American patients with erythropoietic protoporphyria and X-linked protoporphyria. JAMA Dermatol. 2017;153(8):789-796 Method Description Describes how the test is performed and provides a method-specific reference This evaluation is performed as a 2-step analysis. First, the total red blood cell (RBC) porphyrin concentration is determined by extracting the porphyrins from washed, resuspended RBCs using a mixture of ethyl acetate and acetic acid. The porphyrins are then back extracted into dilute hydrochloric acid. Total porphyrins are quantified using this extract via spectrofluorometry.(Piomelli S. Free erythrocyte porphyrins in the detection of undue absorption of Pb and Fe deficiency. Clin Chem. 1977;23:264-269; Gou EE, Balwani M, Bissell DM, et al. Pitfalls in erythrocyte protoporphyrin measurement for diagnosis and monitoring of protoporphyrias. Clin Chem. 2015;61:1453-1456. doi:10.1373/clinchem.2015.245456) If the total porphyrin concentration is elevated, the RBCs are re-extracted to separate and quantify the zinc-complexed and noncomplexed (free) protoporphyrin via high-performance liquid chromatography.(Smith RM, Doran D, Mazur M, Bush B. High-performance liquid chromatographic determination of protoporphyrin and zinc protoporphyrin in blood. J Chromatogr. 1980;181[3-4]:319-327; Gou EE, Balwani M, Bissell DM, et al. Pitfalls in erythrocyte protoporphyrin measurement for diagnosis and monitoring of protoporphyrias. Clin Chem. 2015;61:1453-1456. doi:10.1373/clinchem.2015.245456) PDF Report Indicates whether the report includes an additional document with charts, images or other enriched information No Day(s) Performed Outlines the days the test is performed. This field reflects the day that the sample must be in the testing laboratory to begin the testing process and includes any specimen preparation and processing time before the test is performed. Some tests are listed as continuously performed, which means that assays are performed multiple times during the day. Monday through Friday Report Available The interval of time (receipt of sample at Mayo Clinic Laboratories to results available) taking into account standard setup days and weekends. The first day is the time that it typically takes for a result to be available. The last day is the time it might take, accounting for any necessary repeated testing. 3 to 5 days Specimen Retention Time Outlines the length of time after testing that a specimen is kept in the laboratory before it is discarded 14 days Performing Laboratory Location Indicates the location of the laboratory that performs the test Mayo Clinic Laboratories - Rochester Main Campus CLIA Number: 24D0404292 Fees : Several factors determine the fee charged to perform a test. Contact your U.S. or International Regional Manager for information about establishing a fee schedule or to learn more about resources to optimize test selection. Authorized users can sign in to Test Prices for detailed fee information. Clients without access to Test Prices can contact Customer Service 24 hours a day, seven days a week. Prospective clients should contact their account representative. For assistance, contact Customer Service. Test Classification Provides information regarding the medical device classification for laboratory test kits and reagents. Tests may be classified as cleared or approved by the US Food and Drug Administration (FDA) and used per manufacturer instructions, or as products that do not undergo full FDA review and approval, and are then labeled as an Analyte Specific Reagent (ASR) product. This test was developed and its performance characteristics determined by Mayo Clinic in a manner consistent with CLIA requirements. It has not been cleared or approved by the US Food and Drug Administration. CPT Code Information Provides guidance in determining the appropriate Current Procedural Terminology (CPT) code(s) information for each test or profile. The listed CPT codes reflect Mayo Clinic Laboratories interpretation of CPT coding requirements. It is the responsibility of each laboratory to determine correct CPT codes to use for billing. CPT codes are provided by the performing laboratory. 82542-if appropriate LOINC® Information Provides guidance in determining the Logical Observation Identifiers Names and Codes (LOINC) values for the order and results codes of this test. LOINC values are provided by the performing laboratory. \| Test Id \| Test Order Name \| Order LOINC Value \| --- \| PEE \| Porphyrins Evaluation, WB \| 2814-2 \| \| Result Id \| Test Result Name \| Result LOINC Value Applies only to results expressed in units of measure originally reported by the performing laboratory. These values do not apply to results that are converted to other units of measure. \| --- \| 88886 \| Total Porphyrins, WB \| 2814-2 \| \| 29356 \| Interpretation \| 59462-2 \|
15026	https://www.youtube.com/watch?v=yxDOwlBQ5pc	Inverse trigonometric prove in simple way :arcsin(x) + arccos(x) = pi/2 Castor Classes 15600 subscribers 78 likes Description 21991 views Posted: 13 Mar 2016 arcsin(x) + arccos(x) = pi/2 Please follow us: Join us on Telegram: Join us on Facebook Group: 12 comments Transcript: Prove that sin inverse x + cos inverse x= / 2. Take sin inverse x + cos inverse x. Now let a = sin inverse x. So x will be sin a. Now substitute the value of this a + cos inverse x. So a + cos inverse x = sin a. So a + cos inverse cos of by 2 - a. We know cos of cos inverse of cos x is equal to x. So here it is a + by 2 - a. So a is going to be cancel by 2
15027	https://fiveable.me/key-terms/organic-chem/curved-arrows	Curved Arrows - (Organic Chemistry) - Vocab, Definition, Explanations \| Fiveable \| Fiveable new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom ap study content toolsprintablespricing my subjectsupgrade All Key Terms Organic Chemistry Curved Arrows 🥼organic chemistry review key term - Curved Arrows Citation: MLA Definition Curved arrows are a graphical notation used in organic chemistry to depict the movement of electrons during chemical reactions and the formation of resonance structures. They are a crucial tool for understanding and communicating the mechanisms of organic transformations. 5 Must Know Facts For Your Next Test Curved arrows are used to depict the movement of electrons during the formation of resonance structures, allowing for the visualization of how the electrons are rearranged. Curved arrows are also essential for drawing polar reaction mechanisms, as they show the step-by-step movement of electrons during the course of the reaction. The direction of the curved arrow indicates the flow of electrons, with the arrow pointing from the source of the electrons (e.g., a nucleophile) to the destination (e.g., an electrophile). The placement and orientation of curved arrows are governed by specific rules, which ensure that the resulting resonance structures or reaction mechanisms are valid and consistent with the principles of organic chemistry. Understanding the use of curved arrows is crucial for predicting the outcome of organic reactions and for communicating the details of reaction mechanisms effectively. Review Questions Explain how curved arrows are used to depict the formation of resonance structures. Curved arrows are used to show the movement of electrons during the formation of resonance structures. They indicate how the electrons are rearranged between the different valid Lewis structures that represent the same molecule or ion. By following the curved arrows, you can visualize how the electron density shifts within the molecule, leading to the delocalization of electrons and the creation of resonance structures. Describe the role of curved arrows in drawing polar reaction mechanisms. Curved arrows are essential for depicting the step-by-step movement of electrons during polar reaction mechanisms. They show the flow of electrons from nucleophiles to electrophiles, or from reactants to products, allowing you to understand the sequence of events that occur during the course of the reaction. The placement and direction of the curved arrows are crucial for accurately representing the mechanism and predicting the outcome of the reaction. Analyze the importance of understanding the rules for drawing curved arrows in organic chemistry. Adhering to the established rules for drawing curved arrows is crucial in organic chemistry, as it ensures the validity and consistency of the resonance structures or reaction mechanisms being depicted. These rules govern the placement, direction, and number of curved arrows, allowing for the clear and unambiguous communication of electron flow. By mastering the rules for curved arrows, you can effectively analyze and interpret the mechanisms of organic reactions, which is essential for predicting reactivity, product formation, and the overall course of a chemical transformation. Related terms Resonance: The concept in organic chemistry where a molecule or ion can be represented by multiple, equally valid Lewis structures, which are called resonance structures. Nucleophile: A species that donates a pair of electrons to form a new covalent bond in a chemical reaction. Electrophile: A species that accepts a pair of electrons to form a new covalent bond in a chemical reaction. 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15028	https://www.nature.com/articles/news.2011.158	Early Europeans unwarmed by fire \| Nature Your privacy, your choice We use essential cookies to make sure the site can function. We also use optional cookies for advertising, personalisation of content, usage analysis, and social media. By accepting optional cookies, you consent to the processing of your personal data - including transfers to third parties. Some third parties are outside of the European Economic Area, with varying standards of data protection. See our privacy policy for more information on the use of your personal data. Manage preferences for further information and to change your choices. Accept all cookies Skip to main content Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). 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Advertisement View all journals Search Search Search articles by subject, keyword or author Show results from Search Advanced search Quick links Explore articles by subject Find a job Guide to authors Editorial policies Log in Explore content Explore content Research articles News Opinion Research Analysis Careers Books & Culture Podcasts Videos Current issue Browse issues Collections Subjects Follow us on Facebook Follow us on Twitter Subscribe Sign up for alerts RSS feed About the journal About the journal Journal Staff About the Editors Journal Information Our publishing models Editorial Values Statement Journal Metrics Awards Contact Editorial policies History of Nature Send a news tip Publish with us Publish with us For Authors For Referees Language editing services Open access funding Submit manuscript Subscribe Sign up for alerts RSS feed nature news article News Published: 14 March 2011 Early Europeans unwarmed by fire Matt Kaplan Nature (2011)Cite this article 2465 Accesses 3 Altmetric Metrics details The first hominins to migrate into Europe may have done so without fire. 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15029	https://digitalrepository.unm.edu/cgi/viewcontent.cgi?article=1030&context=math_etds	University of New Mexico University of New Mexico UNM Digital Repository UNM Digital Repository Mathematics & Statistics ETDs Electronic Theses and Dissertations 7-1-2011 Optimal control problems, curves of pursuit Optimal control problems, curves of pursuit Svetlana Moiseeva Follow this and additional works at: Recommended Citation Recommended Citation Moiseeva, Svetlana. "Optimal control problems, curves of pursuit." (2011). This Thesis is brought to you for free and open access by the Electronic Theses and Dissertations at UNM Digital Repository. It has been accepted for inclusion in Mathematics & Statistics ETDs by an authorized administrator of UNM Digital Repository. For more information, please contact disc@unm.edu .Optimal Control Problems Curves of Pursuit by Svetlana Moiseeva B.S. Mathematics, Peoples’ Friendship University of Russia, 2008 THESIS Submitted in Partial Fulfillment of the Requirements for the Degree of Master of Science Mathematics The University of New Mexico Albuquerque, New Mexico May, 2011 c©2011, Svetlana Moiseeva iii Dedication To my parents, Valentina and Nikolay Moiseev, for their support and encouragement. iv Acknowledgments First, I would like to thank the Institute of International Education and the Fulbright program who provided me the opportunity to study in the United States. I must also recognize that my graduate experience and this thesis would not have been possible without the financial assistance of the Department of Mathematics and Statistics at the University of New Mexico in the form of a generous Teaching Assistantship. Second, I wish to thank Dr. Embid, my thesis advisor, for motivating me, and for continuing to encourage me through the long number of months writing and rewriting this thesis. I appreciate Dr. Embid’s vast knowledge and skill in mathematics, and his patience in helping me during the completion of my thesis. His expertise and understanding, guidance and professional style will remain with me as I continue my career. Third, I would like to thank my committee members, Dr. Lau and Dr. Nakamaye for taking the time from their busy schedules to review my thesis and for their valuable recommendations pertaining to this study and assistance in my professional development. Your effort is greatly appreciated. Finally, I would like to thank my family, for the love and support they provided during graduate school, and always. I would also like to thank all my friends for their constant support and always being there for me. vOptimal Control Problems Curves of Pursuit by Svetlana Moiseeva ABSTRACT OF THESIS Submitted in Partial Fulfillment of the Requirements for the Degree of Master of Science Mathematics The University of New Mexico Albuquerque, New Mexico May, 2011 Optimal Control Problems Curves of Pursuit by Svetlana Moiseeva B.S. Mathematics, Peoples’ Friendship University of Russia, 2008 M.S., Mathematics, University of New Mexico, 2011 Abstract We study a class of problems known as pursuit-evasion problems (PE). These prob-lems can be understood as special cases of optimal control problems. After describing the two main principles to study optimal control problems, namely Pontryagin’s ma-ximum principle and Bellman’s method of dynamic programming, this thesis focuses on specific examples of PE problems within the classes of pursuit problems, evasion problems, and pursuit-evasion problems. vii Contents List of Figures x1 Introduction 1 1.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2 Optimal Control Processes 5 2.1 Formulation of the Optimal Control Problem . . . . . . . . . . . . . . 52.2 Necessary Conditions for Optimality Pontryagin’s Maximum Principle . . . . . . . . . . . . . . . . . . . . 11 2.3 Bellman’s Method of Dynamic Programming . . . . . . . . . . . . . . 18 2.4 The relation between Pontryagin’s Maximum Principle and Bellman’s Method of Dynamic Programming . . . . . . . . . . . . . . . . . . . . 27 3 The Pursuit Problem 38 3.1 Statement of the Problem . . . . . . . . . . . . . . . . . . . . . . . . 38 3.2 Pierre Bouguer’s Pursuit Problem . . . . . . . . . . . . . . . . . . . . 40 viii Contents 3.3 Wind-Blown Plane Problem . . . . . . . . . . . . . . . . . . . . . . . 51 3.4 The Tractrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 3.5 Apollonius Pursuit Problem . . . . . . . . . . . . . . . . . . . . . . . 62 4 The Evasion Problem 69 4.1 Statement of the Problem . . . . . . . . . . . . . . . . . . . . . . . . 69 4.2 Isaacs’s Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 4.3 Lady in the Lake Problem . . . . . . . . . . . . . . . . . . . . . . . . 76 5 Pursuit-Evasion Problem as an Optimal Control Problem 85 5.1 Basic Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 5.2 Simple Pursuit in the Plane . . . . . . . . . . . . . . . . . . . . . . . 90 5.3 One-dimensional Rocket Chase . . . . . . . . . . . . . . . . . . . . . 93 5.4 Pursuit on a Sphere (Kelley’s game) . . . . . . . . . . . . . . . . . . . 98 6 Conclusions 101 References 105 ix List of Figures 2.1 Reformulation of the problem given by equation (2.1), where line l is passing through the point (0 , x 1) and is parallel to the x0 axis, i.e., this line is made up of all the points (ξ, x 1) where the number ξ is arbitrary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 2.2 The route of the ship sailing from a to e . . . . . . . . . . . . . . . . 18 2.3 Two one-stage problems in the first subproblem . . . . . . . . . . . . 20 2.4 Three two-stage problems . . . . . . . . . . . . . . . . . . . . . . . . 20 2.5 Two three-stage problems . . . . . . . . . . . . . . . . . . . . . . . . 21 2.6 Four-stage problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.7 Bang-bang time-optimal control: trajectories for u = 1 of parabolas given by equation (2.40) . . . . . . . . . . . . . . . . . . . . . . . . . 32 2.8 Bang-bang time-optimal control: trajectories for u = −1 of parabolas given by equation (2.41) . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.9 Bang-bang time-optimal control: u(t) is initially equal to +1 , and then to −1, the phase trajectory consists of two adjoining parabolic segments given by equations (2.40) and (2.41), respectively . . . . . 34 xList of Figures 2.10 Bang-bang time-optimal control: u(t) is initially equal to −1, and then to +1 , the phase trajectory consists of two adjoining parabolic segments given by equations (2.41) and (2.40), respectively . . . . . 35 2.11 Bang-bang time-optimal control: the switching curve and the family of phase trajectories we obtained ( AO is the arc of the parabola x1 = 12 (x2)2 in the lower half-plane, BO is the arc of the parabola x1 = −12 (x2)2 in the upper half-plane) . . . . . . . . . . . . . . . . . . . . 36 3.1 The geometry of Bouguer’s pursuit problem about a pirate ship mo-ving directly toward the merchant vessel at constant speed Vp along a curved path and pursuing a merchant vessel travelling at constant speed Vm along the vertical line x = x0 . . . . . . . . . . . . . . . . . 40 3.2 The path of the pirate ship as given by equation (3.15) for n = 3 /4 . 45 3.3 The geometry of the tail chase as given by equation (3.17) . . . . . . 47 3.4 The geometry of the wind-blown plane problem, where the plane’s nose is always pointed toward a city C, the plane’s speed is v mi/h, and a wind is blowing from the south at the rate of w mi/h . . . . . 51 3.5 Plots of the wind-blown plane’s paths given by equations (3.25) for several values of n < 1 (n = 0.1, 0.2, 0.4, 0.8, 0.95, 0.99, 0.999) . . 54 3.6 The geometry of the tractrix problem, where a watch-on-a-chain with the chain of length a is initially on the y-axis, the end of the chain is pulled along the x-axis from the initial position on the origin . . . 58 3.7 A depiction of the tractrix given by equation (3.31) for a = 1 . . . . 60 xi List of Figures 3.8 Schematic of the pursuit by interception problem with pursuer T (Torpedo) and evader E (Enemy ship) moving with constant speeds VT and VE, respectively . . . . . . . . . . . . . . . . . . . . . . . . . 62 3.9 The Apollonius circle centered on (2 /3, 0) with radius 2/3, given by equation (3.34) for m = 1 , p = 2 , and k = 2 , so that the torpedo is located at T(2 , 0) and the enemy ship is at E(1 , 0) . . . . . . . . . . 64 3.10 The Apollonius circle centered on (7 /3, 0) with radius 2/3, given by equation (3.34) for m = 1 , p = 2 , and k = 1 /2, so that the torpedo is located at T(2 , 0) and the enemy ship is at E(1 , 0) . . . . . . . . . 66 3.11 The general geometry for a slow torpedo ( T) interception of a fast en-emy surface ship ( E) (heading with an angle θ), where the Apollonius circle for the points (m, 0) and (p, 0) , p > m , is given by equation (3.34) for k < 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 4.1 P and E defending and attacking, respectively, the target area C . . 71 4.2 The geometry of Isaacs’s problem for P defending a point target C,and E attacking same target; l1 is the perpendicular bisector of P E , l2 is the perpendicular line segment from C to l1 . . . . . . . . . . . 72 4.3 Plot of F (x0/x c, y 0/x c) given in equation (4.8) as a function of x0/x c for a given fixed value of y0/x c (y0/x c = 1 .1, 2, 3, 4, 5, 6, 7) . Each curve gives the minimum value of R/x c . . . . . . . . . . . . . . . . 75 4.4 The first stage of the lady’s escape . . . . . . . . . . . . . . . . . . . 77 4.5 The instant when the lady reaches her go-for-broke circle . . . . . . 81 4.6 The radius of the lake R(φ) (in radians) given by equation (4.12) . . 84 xii List of Figures 5.1 Simple motion in the plane. Point x moves anywhere within Ax(3) at time t = 3 ; if its position y at t = 1 or z at t = 2 is known, the possibilities are: reduce to Ay(2) or Az (1) . . . . . . . . . . . . . . . . 91 5.2 Phase portrait of motion in ¨x = u in the x-y plane, where x(t) is given by equation (5.15) for x(0) = 0 , ˙x(0) = 2 ; attainability sets at t = 2 /3, 4/3, 2, 8/3 for the same initial values x(0) and ˙x(0) . The vertex loci are parabolas ˙x = y = ±√2( x + 2) . . . . . . . . . . . . . 94 5.3 Trajectories of ˙x = y − v, ˙y = u in the x-y plane with u = v = ±1 outside target \|x\| ≤ ε . . . . . . . . . . . . . . . . . . . . . . . . . . 95 5.4 Trajectories of ˙x = y − v, ˙y = u in the x-y plane. From point a evader mistakenly chooses v = −1, but reverses his choice at b;capture occurs at c (later than it would have occurred at d) . . . . . 96 xiii Chapter 1 Introduction 1.1 Overview A pursuit-evasion (PE) problem refers to a family of mathematical problems in which one group ( Pursuers ) attempts to track down members of another group ( Evaders )in an environment. There are different formulations of these problems and each formulation uses some specific features of the pursuit-evasion situation. Our objective in this thesis is to study a variety of problems that encompass the main pursuit-evasion problems from the point of view of the motion and strategy of the pursuer (Chapter 3), the evader (Chapter 4), and both (Chapter 5). Pursuit-evasion (PE) problems can be approached stochastically or deterministi-cally. With the stochastic approach (for a broader discussion see , , , and ), it is more realistic to assume knowledge of the probability characteristics of target detection, whereas the deterministic approach has to contend with trajectories and control parameters that give the opportunity to find the minimum or maximum values. In this thesis we restrict the discussion to the deterministic approach, and state PE problems as optimal control problems, where we speak about optimality in 1Chapter 1. Introduction the sense of rapidity of action, i.e., about achieving the target in the shortest time (Chapter 3), or avoiding the chaser as long as possible (Chapter 4), or both (Chapter 5). Because the pursuit-evasion (PE) problem can be understood as a special case of the more general class of problems known as optimal control processes, we devote Chapter 2 to the formulation of the general optimal control problem and a discussion of the two main approaches to solve this problem, namely Pontryagin’s maximum principle (Theorems 2.2.1, 2.2.2) and Bellman’s equation (2.29). Pontryagin’s maxi-mum principle was discovered in the late 1950s () by the Russian mathematician Lev Semenovich Pontryagin. 1 The maximum principle is an effective tool in sol-ving a broad range of control problems, we state it for the important time-optimal case (see sections 2.1, 2.2). Shortly before the appearance of Pontryagin’s maximum principle in the late 1950s (, , , ), the American mathematician Richard Bellman published his Dynamic Programming , , , . He constructed a partial differential equation for the functional that gives us the minimum time when we transfer the controlled object from the initial state to some other given point (see equation (2.6). This equation of Bellman’s gives rise to another approach to the solu-tion of optimal control problems (see section 2.3). It must be noted though that the assumption on the continuous differentiability of the functional (2.6) does not hold in the simplest cases. Thus, Bellman’s consideration yields a good heuristic method, rather than a mathematical solution of the problem. The maximum principle, in 1Pontryagin’s maximum principle gave birth to optimal control theory, which at present is a vital area in applied mathematics. Pontryagin was led to the formulation of the general time-optimal control problem by an attempt to solve a concrete fifth-order system of ordinary differential equations with three control parameters related to optimal maneuvers of an aircraft, which was proposed to him by the Russian Air Force in the early spring of 1955 . Right after the formulation of the time-optimal control problem, during three days, or better to say, during three sleepless nights (Pontryagin suffered from severe insomnia and very often used to do math in bed all night long), the first and the most important step toward the final solution (the Pontryagin’s maximum principle) was made by Pontryagin . He derived the first version of the necessary conditions. 2Chapter 1. Introduction addition to its complete mathematical validity, also has the advantage that it results in a system of ordinary differential equations, whereas Bellman’s approach requires the solution of a partial differential equation. Both approaches will be discussed and compared in Chapter 2. Even though we are not going to literally apply Pontryagin’s or Bellman’s general approach to the specific examples discussed in this thesis, we introduce them because they provide the appropriate framework in the formulation of the pursuit-evasion problem. We start Chapter 2 by discussing optimal control processes. A process is called controlled if it can be described by a vector differential equation with a control parameter and a phase point. The problem then is to choose such a control, as a function of time, so that the corresponding trajectory of the given differential equation is shifted from a given initial point to some other given point in minimum time. In this case the control and its corresponding trajectory are called optimal .The other important class of problems is a time-optimal control problem, which is defined in the same Chapter 2. In section 2.2 (Chapter 2) we introduce a hamiltonian function in order to state Pontryagin’s maximum principle (Theorems 2.2.1, 2.2.2), which includes an important Pontryagin’s maximum condition .The rest of this thesis is organized as follows: in Chapter 3 we present a definition of the pursuit problem, and provide examples of the pursuit problem (Bouguer’s prob-lem (section 3.2), the plain and the wind problem (section 3.3), the tractrix (section 3.4), and Apollonius pursuit (section 3.5)). In Chapter 4 we state a definition of the evasion problem, and give examples of the evasion problem (Isaacs’s problem (sec-tion 4.2), and the lady in the lake problem (section 4.3)). In Chapter 5 we present a definition of the pursuit-evasion problem, and examples of these problems (pursuit in the plane (section 5.2), one-dimensional rocket chase (section 5.2), and Kelly’s game (section 5.4)). Moreover, we state the possible method of solving pursuit-evasion 3Chapter 1. Introduction problems while using Pontryagin’s maximum principle. Although this theorem gives the necessary conditions for optimality of pursuit-evasion problems (and it can be generalized to multiple pursuers and multiple evaders, as in ), the fact is that the PE problems studied in the current thesis can be analyzed directly by more elementary methods. Finally, in Chapter 6 we summarize the results of the thesis. 4Chapter 2 Optimal Control Processes Since the pursuit evasion (PE) problem can be understood as a special case of the more general class of problems known as optimal control processes, we are going to devote this chapter to the formulation of the general optimal control problem and a discussion of the two main approaches to solve this problem, namely Pontryagin’s maximum principle (Theorems 2.2.1, 2.2.2) and Bellman’s equation (2.29). 2.1 Formulation of the Optimal Control Problem A desirable property of most technological processes is controllability, which roughly speaking means that a particular process can be realized by a proper adjustment of certain control parameters. Mostly important is the search, among all the control-lable processes, of the control that optimizes a related function of this process. This problem is known as the optimal control problem. For example, one can speak about optimality in the way of spending the least possible time or using the minimum energy in order to reach the target. These problems can be formulated mathematically, and their solution is given by a general method known as Pontryagin’s maximum principle 5Chapter 2. Optimal Control Processes (Theorems 2.2.1, 2.2.2) (, , , ). To start, we consider control processes which can be described by a system of ordinary differential equations dx i dt = f i(x1, ..., x n, u 1, ..., u r) = f i(xk, u j ), i, k = 1 , ..., n, j = 1 , ..., r, (2.1) or in vector form, dx dt = f (x, u ). (2.2) The variables x1, ..., x n characterize the process, and they are known as the phase coordinates of the controlled object which define its state at each instant of time t.Giving a point u = ( u1, ..., u r) ∈ U ⊂ Rr is equivalent to giving a numerical system of parameters u1, ..., u r, and they are known as the control parameters which determine the course of the process. The functions f i are defined for x ∈ X ⊂ Rn and u ∈U ⊂ Rr. They are assumed to be continuous in the variables x1, ..., x n, u 1, ..., u r, and continuously differentiable with respect to x1, ..., x n. In other words, the functions f i(x1, ..., x n, u ) and ∂f i(x1, ..., x n, u ) ∂x j , i, j = 1 , ..., n, are defined and continuous everywhere on the direct product X × U .In order to find a solution of equation (2.1) and determine the course of the control process (2.1) in a certain time interval t0 ≤ t ≤ t1, it is sufficient for the control parameters u1, ..., u r to be the functions of time on this time interval: uj = uj (t), j = 1 , ..., r. (2.3) Then, for the given initial values xi(t0) = xi 0 , i = 1 , ..., n, (2.4) the solution is uniquely determined, at least locally in time. Hence, we say that a control U = ( uj (t), t 0, t 1, x i 0 ), j = 1 , ..., r, i = 1 , ..., n (2.5) 6Chapter 2. Optimal Control Processes of equation (2.1) is given, if a function uj (t), its range of definition t0 ≤ t ≤ t1, and the initial value (2.4) of the solution xi(t) are given. Therefore, we only deal with piecewise continuous control functions uj (t) which admit discontinuities of the first kind, and continuous solutions of equation (2.2). The control problem to be solved, which is related to the control process (2.1), consists of the following . We consider the integral function L(U ) = t1 ∫ t0 f 0 (x1, ..., x n, u 1, ..., u r) dt, (2.6) where f 0(x1, ..., x n, u 1, ..., u r) is a given function, continuous, together with its partial derivatives ∂f 0 ∂x j , j = 1 , ..., n, everywhere on the space X × U . For each control (2.5), given on a certain interval t0 ≤ t ≤ t1, the course of the control processes is uniquely determined, at least locally in time, and the integral (2.6) takes on a definite value. Let us assume that there exists a control (2.5) which transfers the controlled object from a given initial phase state xi 0 (2.4) to a prescribed terminal phase state xi(t1) = xi 1 , i = 1 , ..., n. (2.7) It is required to find a control u(t) which transfers the controlled object from state xi 0 to state xi 1 in such a way that the functional L(U ) has a minimum value . Thus, L is a function of the control U .Let us summarize the above discussion and state the definition of the optimal control problem (equations (2.2), (2.4), (2.5), (2.6), (2.7)). 7Chapter 2. Optimal Control Processes Definition (Optimal Control Problem ) An optimal control problem is a problem given by the equations dx dt = f (x, u ),x(t0) = x0,x(t1) = x1,U = ( u(t), t 0, t 1, x 0),L(U ) = t1 ∫ t0 f 0 (x, u ) dt, where x = ( x1, ..., x n) ∈ X ⊂ Rn, u = ( u1, ..., u r) ∈ U ⊂ Rr is some piecewise contin-uous function, f = ( f 1, ..., f n) are continuous, together with its partial derivatives everywhere on the space X × U , and f 0(x, u ) is a given function (also continuous, together with its partial derivatives, everywhere on the space X × U ). Definition A control U = ( u(t), t 0, t 1, x i 0 ) is called optimal , if, for any control U ∗ = ( u∗(t), t 0, t 1, x i 0 )which transfers the point xi 0 to the point xi 1 , the inequality L(U ) ≤ L(U ∗)is valid. The corresponding trajectory x(t) is called an optimal trajectory .Thus, an optimal control problem consists of finding the optimal controls and the corresponding optimal trajectories. Remark 1. The times t0 and t1 are not fixed, we only require that the object should be in state (2.4) at the initial time, and in state (2.7) at the final time, and that the functional (2.6) should achieve a minimum. (The discussion of the case where the times t0 and t1 are fixed can be found in , §8 .) 8Chapter 2. Optimal Control Processes Remark 2. If (2.5) is an optimal control of equation (2.2) corresponding to this control, and t2, t 3 (t2 < t 3) are two points in the interval t0 ≤ t ≤ t1, then U ′ = ( u(t), t 2, t 3, x i(t2)) is also an optimal control. Remark 3. If (2.5) is an optimal control of equation (2.2) that transfers the point xi 0 to the point xi 1 , and τ is an arbitrary number, then U ′′ = ( u(t − τ ), t 0 + τ, t 1 + τ, x i 0 )is also an optimal control which transfers the point xi 0 to xi 1 . Definition (Time-Optimal Control Problem ) When the function f 0(xi, u j ) is defined by equation f 0(x, u ) ≡ 1, (2.8) the function of the control (2.5) in this case is L(U ) = t1 − t0, and the optimality of the control u(t) signifies minimality of the transition time from x0 to x1. The problem of finding optimal controls (and trajectories) in this case is called the time optimal control problem .We should point out that up to know we have spoken about an optimal control which brought the object to a given point. However, the optimal control problem may consist of “optimality getting to” a moving point in phase space. Let us assume that there exists a moving point xi = θi(t), i = 1 , ..., n, (2.9) 9Chapter 2. Optimal Control Processes in phase space. Then, there arises the problem of optimality bringing the object in coincidence with a moving point. This problem is easily reduced to the one considered above. It is sufficient to introduce new variables by setting yi = xi − θi(t), i = 1 , ..., n. As a result of this transformation, the control system dx i dt = f i(xi, u j ), i = 1 , ..., n, j = 1 , ..., r, becomes a new system. The goal of the control process becomes that of bringing the new object ( y1, ..., y n) to the stationary point (0 , ..., 0) in phase space. Of great importance is the case where U ⊂ Rn is a compact domain. This is clearly the case in most practical applications, where the control parameters can only take values with predetermined upper and lower bounds. For example, U may be a cube defined by the inequalities \|uj \| ≤ 1, j = 1 , ..., r. In many instances it turns out that the optimal control (2.5) is realized by a piecewise constant control ( u1(t), ..., u r(t)) with values switching between various vertices of U.It follows that the class of admissible controls (2.5) must include piecewise con-tinuous functions. For the same reason, the phase coordinates x1, ..., x n are assumed to be continuous and piecewise differentiable functions of time. Under these as-sumptions the necessary conditions for optimality are formulated in the form of Pontryagin’s maximum principle (Theorems 2.2.1, 2.2.2) (, , ), which we will present in the next section. 10 Chapter 2. Optimal Control Processes 2.2 Necessary Conditions for Optimality Pontryagin’s Maximum Principle In order to formulate the necessary optimality condition it will be convenient to reformulate our optimal control problem (for a broader discussion see ). Namely, let us adjoin a new coordinate x0 to the phase coordinates x1, ..., x n, which vary according to (2.1). Let x0 vary according to the law dx 0 dt = f 0(x1, ..., x n, u 1, ..., u r), where f 0 is the function which appears in the definition of the functional L(U ) (see (2.6)). In other words, we shall consider the system of differential equations dx i dt = f i(x1, ..., x n, u 1, ..., u r) = f i(x, u ), i = 0 , 1, ..., n, (2.10) whose right-hand sides do not depend on x0. Introducing the vector x = ( x0, x 1, ..., x n) = ( x0, x )in the ( n + 1)-dimensional vector space X = R × X ⊆ Rn+1 , we may rewrite system (2.10) in vector form dx dt = f(x, u ), (2.11) where f(x, u ) is the vector in X with coordinates f 0(x, u ), ..., f n(x, u ). Note, that f(x, u ) does not depend on the coordinate x0 of the vector x, that is ~f (x, u ), not ~f (~x, u ). Now let u(t) be an admissible control (2.5) (i.e., piecewise continuous) transferring x0 to x1, and let x = x(t) be the corresponding solution of equation (2.2) with initial condition x(t0) = x0. Let us denote the point (0 , x 0) by x0, i.e., x0 is the point of X whose coordinates are 0 , x 10, ..., x n 0 , where x10, ..., x n 0 are the coordinates of x0 in X .11 Chapter 2. Optimal Control Processes Figure 2.1: Reformulation of the problem given by equation (2.1), where line l is passing through the point (0 , x 1) and is parallel to the x0 axis, i.e., this line is made up of all the points (ξ, x 1) where the number ξ is arbitrary Then, it is clear that the solution of equation (2.11) with initial condition x(t0) = x0,corresponding to the control u(t), is defined on the entire interval t0 ≤ t ≤ t1, and has the form x0 = t ∫ t0 f 0 (x(t′), u (t′)) dt ′, x = x(t). In particular, when t = t1 x0 = t1 ∫ t0 f 0 (x(t), u (t)) dt = L(U ), x = x1, i.e., the solution x(t) of equation (2.11) with initial condition x(t0) = x0 passes through the point x = ( L(U ), x 1) at t = t1. In other words, if we let l be the line in X passing through the point x = (0 , x 1) and parallel to the x0 axis (this line is made up of all the points ( ξ, x 1) where the number ξ is arbitrary, see Figure 2.1), we can say that x(t) passes through a point on line l, with coordinate x0 = L(U ), at the time t = t1. Conversely, suppose that u(t) is an admissible control (i.e., at least 12 Chapter 2. Optimal Control Processes piecewise continuous) such that the corresponding solution x(t) of equation (2.11) with initial condition x(t0) = x0 = (0 , x 0), at some time t1 passes through a point x1 ∈ l, with coordinate x0 = L(U ). Then, the control u(t) transfers (in X ) the phase point from x0 to x1, and the functional (2.6) takes on the value L(U ). Thus, we may formulate the above optimal problem (from 2.1) in the following equivalent form. In the ( n + 1 )-dimensional phase space X the point x0 = (0 , x 0) and the line l are given. The line l is assumed to be parallel to the x0 axis, and to pass through the point (0 , x 1). Among all the admissible controls u = u(t), having the property that the corresponding solution x(t) of (2.11) with initial condition x(t0) = x0 intersects l, find one whose point of intersection with l has the smallest coordinate x0 (see ). Let us now proceed to the formulation of the theorem which yields the necessary conditions of the problem. (The proof of this theorem can be found in , Chapter II.) To formulate the theorem, we shall consider, in addition to the fundamental sys-tem of equations (2.10) another system of equations in the auxiliary (supplementary) variables ψ0, ψ 1, ..., ψ n: dψ i dt = − n ∑ α=0 ∂f α(x, u ) ∂x i ψα, i = 0 , 1, ..., n. (2.12) If we choose an admissible control u(t), t0 ≤ t ≤ t1, and have the corresponding phase trajectory x(t) of system (2.10) with initial condition x(t0) = x0, system (2.12) takes the form dψ i dt = − n ∑ α=0 ∂f α(x(t), u (t)) ∂x i ψα, i = 0 , 1, ..., n. (2.13) This system is linear and homogeneous. Therefore, for any initial condition, it admits the unique solution ψ = ( ψ0, ψ 1, ..., ψ n)13 Chapter 2. Optimal Control Processes for the ψi (which is defined on the entire interval t0 ≤ t ≤ t1 on which u(t) and x(t)are defined). Similarly to the solution x(t) of system (2.11), the solution of system (2.13) consists of continuous functions ψi(t) which have everywhere, except at a finite number of points (namely, at the points of discontinuity of u(t)), continuous derivatives with respect to t. Each solution of system (2.13) for any initial conditions will be called the solution of system (2.12) corresponding to the chosen control u(t)and phase trajectory x(t). Now we will combine systems (2.10) and (2.12) into one entry. We consider the following function H of the variables x0, x 1, ..., x n, ψ 0, ψ 1, ..., ψ n, u 1, ..., u r: H(ψ, x, u ) = ( ψ, f(x, u )) = n ∑ α=0 ψαf α(x, u ). The above systems (2.10) and (2.12) can be rewritten with the aid of the function H in the form of the following Hamiltonian system : dx i dt = ∂H ∂ψ i , i = 0 , ..., n, (2.14) dψ i dt = −∂H ∂x i , i = 0 , ..., n. (2.15) For fixed (constant) values of ψ and x, the function H becomes a function of the parameter u ∈ U . Let us now denote the least upper bound of the values of this function by M(ψ, x): M(ψ, x) = sup u∈U H(ψ, x, u). If the continuous function H achieves its upper bound on U, then M(ψ, x) is the maximum of the values of H, for fixed ψ and x. Therefore, Theorem 2.2.1 below (a necessary condition for optimality ) will be called the maximum principle (the principal content of the principle is in equation (2.16)) , . 14 Chapter 2. Optimal Control Processes Theorem 2.2.1 (Pontryagin’s Maximum Principle) Let u(t), t0 ≤ t ≤ t1, be an admissible control such that the corresponding trajectory x(t) [see (2.14)] which begins at the point x0 at the time t0 passes, at some time t1, through a point on the line l. In order that u(t) and x(t) be optimal it is necessary that there exists a nonzero continuous vector function ψ(t) = ( ψ0(t), ψ 1(t), ..., ψ n(t)) corresponding to u(t) and x(t) [see (2.15)], such that: (a) for every t, t0 ≤ t ≤ t1, the function H(ψ(t), x (t), u ) of the variable u ∈ U attains its maximum at the point u = u(t): H(ψ(t), x (t), u (t)) = M(ψ(t), x (t)) , (2.16) (b) at the terminal time t1 the relations ψ0(t1) ≤ 0, M(ψ(t1), x (t1)) = 0 (2.17) are satisfied. Furthermore, it turns out that if ψ(t), x(t), and u(t) satisfy system (2.14), (2.15), and condition (a) , the time functions ψ0(t) and M(ψ(t), x (t)) are constant. Thus, (2.17) may be verified at any time t, t0 ≤ t ≤ t1, and not just at t1. The proof of the Theorem 2.2.1 can be found in , Chapter II. To formulate the necessary condition for the time-optimal problem (equation (2.8)), where f 0(x, u ) ≡ 1, let us form the Hamiltonian function H = ψ0 + n ∑ ν=1 ψν f ν (x, u ). Introducing the n-dimensional vector ψ = ( ψ1, ..., ψ n) and the function H(ψ, x, u ) = n ∑ ν=1 ψν f ν (x, u ), 15 Chapter 2. Optimal Control Processes we can rewrite equations (2.1) and (2.12) (with the exception for equation (2.12) for i = 0, which is now superfluous) in the form of the Hamiltonian system dx i dt = ∂H ∂ψ i , i = 1 , ..., n, (2.18) dψ i dt = −∂H ∂x i , i = 1 , ..., n. (2.19) For fixed values of ψ and x, H is a function of u. We denote the upper bound of the values of this function by M (ψ, x ): M (ψ, x ) = sup u∈U H(ψ, x, u ). Since H(ψ, x, u ) = H(ψ, x, u ) − ψ0, we get M (ψ, x ) = M(ψ, x ) − ψ0, and therefore (2.16) and (2.17) become H(ψ(t), x (t), u (t)) = M (ψ(t), x (t)) = −ψ0 ≥ 0. Hence, we obtain the following theorem. Theorem 2.2.2 (Pontryagin’s Maximum Principle for the time-optimal control problem (2.8)) Let u(t), t0 ≤ t ≤ t1 be an admissible control which transfers the phase point from x0 to x1, and let x(t) be the corresponding trajec-tory (see (2.18)), so that x(t0) = x0, x(t1) = x1. In order that u(t) and x(t) be time-optimal it is necessary that there exist a nonzero, continuous vector function ψ(t) = ( ψ1(t), ..., ψ n(t)) corresponding to u(t) and x(t) (see (2.19)) such that: (a) for all t, t0 ≤ t ≤ t1, the function H(ψ(t), x (t), u ) of the variable u ∈ U attains its maximum at the point u = u(t): H(ψ(t), x (t), u (t)) = M (ψ(t), x (t)) , (2.20) 16 Chapter 2. Optimal Control Processes (b) at the terminal time t1 the relation M (ψ(t1), x (t1)) ≥ 0 (2.21) is satisfied. Furthermore, it turns out that if ψ(t), x(t), and u(t) satisfy system (2.18), (2.19), and condition (a) , the time function M (ψ(t), x (t)) is constant. Thus, (2.21) may be verified at any time t, t0 ≤ t ≤ t1, and not just at t1. 17 Chapter 2. Optimal Control Processes Figure 2.2: The route of the ship sailing from a to e 2.3 Bellman’s Method of Dynamic Programming Shortly before the appearance of Pontryagin’s maximum principle in the late 1950s, R. Bellman published his Dynamic Programming , , , , which presents a related but different approach to the optimum design of control systems which is more efficient in some situations. The following simple example will illustrate some of the main ideas behind this dynamic programming approach. Example Suppose a ship sailing from a and ending at e calls at three ports (at either of the two b’s, at one of the three c’s, and at one of the two d’s) along the way (as shown in Figure 2.2) and picks up and delivers the amounts of cargo (in hundreds of tons). The objective is to deliver as much cargo as possible on the entire trip. Since there are only 12 different routes, it is a simple matter to list them all and choose the route that yields the maximum tonnage. However, we shall solve the problem differently and use the following reasoning. Suppose that, somehow we were to know the maximum tonnage values of the two shorter problems, one from b1 18 Chapter 2. Optimal Control Processes to e and the other from b2 to e, then it would be very easy to decide on the entire route. There are only two possible decisions left to be made at a: go to b1 or go to b2. To reach such a decision, simply add 4 to that maximum tonnage from b1 to e that we somehow learned, add 2 to that maximum tonnage from b2 to e, and choose the route that gives the larger value. In other words, we will have solved the original four-stage problem by first solving two three-stage problems. Similarly, each of these two three-stage problems (from b1 to e or from b2 to e) would be relatively easy to solve if we were to first solve three two-stage problems, namely, find the value given by the maximum tonnage path from each ci, i = 1 , 2, 3, to e. We continue this reasoning and reduce the process to two one-stage problems, from d1 to e or from d2 to e, at which stage the answer is obvious - go from d1, because 7 is larger than 4. Let us do the problem formally. We will break it into several n - stage problems, n = 1 , 2, 3, 4. Notice that there are four stages: from a to b, b to c, c to d, and d to e. There are two possible terminal ports, or states as we will call them, namely b1 and b2 in stage one, three states c1, c 2, and c3 in stage two, two states d1 and d2 in stage three, and one state e in the last stage. Each of these states may also be thought of as the initial state for the following stages. For instance, b1 may be considered the initial state of a three-stage problem, c1 the initial state of a two-stage problem, etc. Let the variable x stand for the initial state for any n-stage problem, n = 1 , 2, 3, 4. For instance, for a two-stage problem, x may be either c1, or c2, or c3.Associated with each problem is also a decision or control variable un, n = 1, 2, 3, 4, which chooses the immediate destination when there are n stages left to go. Thus, u4 chooses b1 or b2, u3 chooses c1 or c2 or c3, u2 chooses d1 or d2, and u1 = e. Let fn(x, u n) be the total number of tons delivered during the last n stages, given that the boat is in state x and the decision is un. If ¯ un is the decision which maximizes fn(x, u n) for fixed n and x, let ¯fn(x) be that maximum value of fn. Since ¯fn is the maximum value with respect to the decision variable un, it is now a function of the initial state variable x alone, hence, the notation ¯fn(x). 19 Chapter 2. Optimal Control Processes Figure 2.3: Two one-stage problems in the first subproblem Figure 2.4: Three two-stage problems In the first subproblem, there is only one stage left to go, and ¯ u1 = u1 = e. The initial states are d1 and d2, as shown in Figure 2.3. We move now to the three subproblems in each of which there are two stages to go, but we utilize the knowledge gained from the one-stage problem. If the boat is at c1 (x = c1), it can proceed to either d1 (u2 = d1) or d2 (u2 = d2). If u2 = d1, f2(c1, d 1) = 7 + 7 = 14. If u2 = d2, f2(c1, d 2) = 3 + 4 = 7. Since 14 > 7, ¯ u2 should be ¯u2 = d1. Similarly, if the boat is at c2 (x = c2) and u2 = d1, f2(c2, d 1) = 8 + 7 = 15, while f2(c2, d 2) = 4 + 4 = 8 if u2 = d2. Let sun be the number of tons of cargo delivered as a result of decision un. Then f2(x, u 2) = su2 + ¯f1(u). Figure 2.4 shows the values for the different states and decisions. Next, we move to the two subproblems in each of which there are three stages to go, and again we utilize the knowledge gained from the previous two-stage problems. If the boat is at b1 (x = b1) and it is decided to go to c1 (u3 = c1), the total number of tons delivered would be 1, that between b1 and c1, plus 14, the maximum 20 Chapter 2. Optimal Control Processes Figure 2.5: Two three-stage problems Figure 2.6: Four-stage problem number of tons to be delivered between c1 and e. That is, f3(b1, c 1) = sc1 + ¯f2(b1), or f3(x, u 3) = su3 + ¯f2(x). See Figure 2.5. The final, or four-stage problem should now be clear. So what is the optimal policy for the overall problem? Retrace the steps backwards starting with Figure 2.6. Starting at a, the optimal decision ¯ u4 is to go to b2. At b2, ¯ u3 tells us to go to c2. At c2, ¯ u2 tells us to go to d1. At d1, ¯ u1 says to go to e. Thus, the optimal route is a → b2 → c2 → d1 → e, with a maximum tonnage of 25. There are only four stages in this example, and each stage has very few states, so that the computational advantages of the dynamic programming approach over the direct, brute approach of listing all twelve possible routes may not be apparent. If a problem has many stages with many states, thus involving many decision processes, direct enumeration may require a phenomenal amount of work, and the computa-tional savings of the dynamic programming approach are considerable. It has been 21 Chapter 2. Optimal Control Processes shown that for a 20-stage problem with only 2 states in each stage, direct enumera-tion generates more than 1,000,000 additions, while dynamic programming requires only 220 additions. The above example is a discrete multistage decision process problem, in which one chooses a decision from a finite set of decisions at each of a finite number of stages or times. Initially, the problem consisted of n stages, but we reduced it to a sequence of n single stage decision processes, for each of which there is an optimal policy. These problems are joined together by a functional equation. For this particular example, the functional is fn(x, u n) = sun + ¯fn−1(un), (2.22) where ¯fn(x) = max un fn(x, u n), u = 1 , 2, 3, 4. Hence, we use two basic ideas, Bellman’s principle of optimality and the principle of imbedding . Summarizing the method discussed in this example yields Bellman’s Princi-ple of optimality ) () : in control systems with a multistage decision process, given any current state, the remaining sequence of decisions forms an optimal policy with this given state regarded as the initial state. Thus, whatever the first state and decision that led to this current state, all future decisions are optimal .In our example, if we found ourselves at, say, state c1 (regardless of what decision led us there), the policy c1 → d1 → e is optimal with c1 considered as the initial state. Similarly, if we found ourselves at, say, state b1, the policy b1 → c2 → d1 → e would be optimal with b1 considered as the initial state. By applying this principle of optimality backwards step by step repeatedly, we obtain a policy which is optimal for the overall problem. In our example, in the one-stage problems, either decision 22 Chapter 2. Optimal Control Processes d1 → e or d2 → e is optimal (actually, the only possible decision), depending on whether d1 or d2 is the initial state. For the two-stage problems, if c1 is the initial state, the decision ¯ u2 : c1 → d1 is optimal, and the pair ¯ u2, ¯u1 : c1 → d1 → e constitutes an optimal policy with c1 as the initial state. If c2 is the initial state, the pair of decisions ¯ u2, ¯u1 : c2 → d1 → e is optimal. Similarly, for the three stage problems, if b1 is the initial state, the optimal decision ¯ u3 : b1 → c2, coupled with the optimal strategy from the two-stage problem ¯ u2, ¯u1 : c2 → d1 → e, form the optimal strategy ¯ u3, ¯u2, ¯u1 : b1 → c2 → d1 → e, etc. The other principle we used in the above example is the principle of imbedding () . The principle is working in the way that, instead of attempting to solve a difficult problem directly, one imbeds the problem in a family of simpler, easier to solve problems and obtains the solution to the original difficult problem as a result of the solutions to the problems in the family. By repeated use of the principle of optimality, each n-stage problem with n > 1 is converted into a one-stage problem with its own initial state and optimal policy. This is done through the use of some functional equation such as the relation given by (2.22), which, for each problem in the family with its initial state, assigns an optimum value to that problem and links that value with all immediately preceding states. These two basic ideas - imbedding and principle of optimality - are also to be found in the dynamic programming approach to continuous cases. Next, let us write Bellman’s equation for a continuous time variable. Proposition 2.3.1 Suppose we have a time-optimal control problem (2.8). Let us fix some point x1 of the space X , and let u(t), t0 ≤ t ≤ t1, be an optimal control which transfers (through the law of motion xi(t) = xi, i = 1 , ..., n ) the phase point from some position x0 ∈ X to the position x1, and let x(t) be the corresponding optimal trajectory. The optimal transition time from the point x0 to the point x1, t1 − t0, 23 Chapter 2. Optimal Control Processes will be denoted by T (x0). (The point x1 does not enter into the notation for the transition time, since it does not vary). Thus, the function T (x0) is defined on the open set Ω of all points of X from which an optimal transition to x1 is possible. We set T (x) = −ω(x), where T (x) has continuous partial derivatives with respect to the coordinates of the point x, and derive that the function ω(x) satisfies the following nonclassical partial differential equation (which we shall call Bellman’s equation )in the region Ω: sup u∈U n ∑ α=1 ∂ω (x) ∂x α f α(x, u ) = 1 . (2.23) Furthermore, the upper bound is attained at some point u ∈ U (namely, at the value of the optimal control at the time of departure from the point x), and the function ω(x) is nonpositive and vanishes only at the point x1. Proof It is given, that ω(x) = −T (x). (2.24) Since x(t), t 0 ≤ t ≤ t1, is an optimal trajectory, and since each portion of an optimal trajectory is also an optimal trajectory, ω(x(t)) = −T (x0) + t − t0 (2.25) for every t, t 0 ≤ t ≤ t1. Consequently, n ∑ α=1 ∂ω (x(t)) ∂x α f α(x(t), u (t)) = n ∑ α=1 ∂ω (x(t)) ∂x α dx α dt = dω (x(t)) dt = dt dt = 1 . (2.26) Now let v be an arbitrary point of the control region U. We shall consider the motion of the phase point from the position x(t) under the influence of a constant control which is equal to v. Here the problem can be imbedded into the family of problems, following the principle of imbedding discussed before. Mainly, we divide the whole process into two control processes. Thus, after an infinitesimal time interval dt > 0, 24 Chapter 2. Optimal Control Processes the phase point will be in the position x(t) + dx , where the vector dx = ( dx 1, ..., dx n)is defined by dx i = f i(x(t), v )dt, i = 1 , ..., n. (2.27) If we now move in an optimal manner from the point x(t) + dx to the point x1,the time spent in so doing will equal T (x(t) + dx ). Hence, the total time spent in a movement of this kind, while transferring from x(t) to x1, is equal to T (x(t)+ dx )+ dt .This time cannot be shorter than the optimal transition time T (x(t)), i.e., T (x(t) + dx ) + dt ≥ T (x(t)) , or equivalently, ω(x(t) + dx ) − ω(x(t)) ≤ dt. Multiply and divide the left side by dx α, and since we know that n ∑ α=1 w (x(t) + dx ) − w (x(t)) dx α = n ∑ α=1 ∂w (x(t)) ∂x α , then because of (2.27), the last inequality may be rewritten in the form n ∑ α=1 ∂ω (x(t)) ∂x α f α(x(t), v )dt ≤ dt, or n ∑ α=1 ∂ω (x(t)) ∂x α f α(x(t), v ) ≤ 1, v ∈ U . (2.28) Relations (2.26) and (2.27) show that sup v∈U n ∑ α=1 ∂ω (x(t)) ∂x α f α(x(t), v ) = 1 , and the upper bound is achieved at v = u(t). 25 Chapter 2. Optimal Control Processes Since an optimal trajectory leading to x1 passes through each point x of Ω, we arrive at the conclusion that the function ω(x) satisfies the following nonclassical partial differential equation ( Bellman’s equation ) in the region Ω: sup u∈U n ∑ α=1 ∂ω (x) ∂x α f α(x, u ) = 1 . (2.29) Furthermore, the upper bound is attained at some point u ∈ U (namely, at the value of the optimal control at the time of departure from the point x), and the function ω(x) is nonpositive and vanishes only at the point x1.This is the principle of dynamic programming as applied to the optimal control problem (for simplicity we considered the time-optimal control problem (2.8)). 26 Chapter 2. Optimal Control Processes 2.4 The relation between Pontryagin’s Maximum Principle and Bellman’s Method of Dynamic Programming The main difference between the calculus of variations methods and dynamic pro-gramming lies in emphasis (see , , , , , , , ). The former considers variations of the candidate extremizing curve, whereas in dynamic pro-gramming the candidate curve varies over a small initial interval and the remainder of the curve is supposed to be optimal for the other part of the problem. In other words, the concept of variation is to be found in both approaches. Which of the two techniques is more desirable depends entirely on the needs and point of view of the user. The calculus of variations yields results whose analytical forms are useful to theorists, and its main appeal perhaps lies in solving deterministic control problems with time treated as continuous, although there are attempts to discretize time . On the other hand, others claim that dynamic programming is the more promising and powerful tool with wider applications in a variety of subjects . It is certainly much more efficient than the calculus of variations in dealing with stochastic control problems involving multistage decision processes , . In what follows we shall consider the relation existing between the maximum principle and R.Bellman’s method of dynamic programming (see in 2.3 the derivation of the Bellman’s equation from Pontryagin’s maximum principle for the time-optimal control problem (2.8)). For a fuller discussion, see Dreyfus . The method of dynamic programming was developed for the needs of optimal control processes which are of a much more general character than those which are describable by systems of differential equations. Therefore, the method of dynamic programming carries a more universal character than the maximum principle (, 27 Chapter 2. Optimal Control Processes ). However, in contrast to the latter, this method does not have the rigorous logical basis in all those cases where it may be successfully made use of as a valuable heuristic tool. The basis of the method of dynamic programming given by Bellman rests on the assumption that to the natural conditions of the problem (see our Theorems 2.2.1 and 2.2.2) another essential requirement has been added - the requirement that the function w(x) defined in Proposition 2.3.1 be differentiable (for a broader discussion see ). This assumption does not follow from the statement of the problem, and is a restriction which, as we shall see below, is not satisfied even in the simplest examples. However, after this assumption has been made, the method of dynamic program-ming leads to a certain partial differential equation, which we call Bellman’s equation .This equation (under certain additional conditions) is equivalent to the Hamiltonian system (2.14), (2.15), and to the maximum condition (2.16), (2.17). In section 2.3 we showed the relation of Bellman’s method of dynamic program-ming to Pontryagin’s maximum principle (for a broader discussion see , ). For the sake of simplicity we only considered the time-optimal problem (2.8). Proposition 2.4.1 Let us assume that the function ω(x) is twice continuously dif-ferentiable. Then Pontryagin’s maximum principle can be derived from Bellman’s principle of dynamic programming. Proof Since ω(x) is twice continuously differentiable, the function g(x, u ) = n ∑ α=1 ∂ω (x) ∂x α f α(x, u ), (2.30) which stands under the supremum in (2.29), has continuous first derivatives with respect to x1, ..., x n. It follows from Bellman’s principle of dynamic programming 28 Chapter 2. Optimal Control Processes (see (2.26) and (2.29)) that if u(t) is an optimal control which transfers the phase point from the position x0 to the position x1, and x(t) is the corresponding optimal trajectory, then for a fixed t, t0 ≤ t ≤ t1, the function g(x, u (t)) of the variable x ∈ X attains its maximum value (unity) at the point x = x(t). From this it follows that ∂g (x(t), u (t)) ∂x i = 0 , i = 1 , ..., n, t0 ≤ t ≤ t1. (2.31) Taking the form of the function g(x, u ) (see (2.30)) into account, we obtain the relations n ∑ α=1 ∂2ω(x(t)) ∂x α∂x i f α(x(t), u (t)) + n ∑ α=1 ∂ω (x(t)) ∂x α · ∂f α(x(t), u (t)) ∂x i = 0 , i = 1 , ..., n, (2.32) which are satisfied along the optimal trajectory. Furthermore, we have n ∑ α=1 ∂2ω(x(t)) ∂x α∂x i f α(x(t), u (t)) = n ∑ α=1 ∂∂x α (∂ω (x(t)) ∂x i )dx α(t) dt = ddt (∂ω (x(t)) ∂x i ) , so that relations (2.32) may be rewritten in the form ddt (∂ω (x(t)) ∂x i ) = − n ∑ α=1 ∂f α(x(t), u (t)) ∂x i · ∂ω (x(t)) ∂x α , i = 1 , ..., n. Thus, along each optimal trajectory, the variables ψi(t) = ∂ω (x(t)) ∂x i , i = 1 , ..., n, (2.33) satisfy the linear system of differential equations dψ i(t) dt = − n ∑ α=1 ∂f α(x(t), u (t)) ∂x i ψα(t), i = 1 , ..., n. (2.34) In addition, because of relation (2.26), Bellman’s equation (2.29) can be written in the form n ∑ α=1 ψα(t)f α(x(t), u (t)) = sup u∈U n ∑ α=1 ψα(t)f α(x(t), u ) = 1 . (2.35) 29 Chapter 2. Optimal Control Processes Relations (2.34) and (2.35) coincide with Pontryagin’s maximum principle, and relation (2.33) points out the relation between ψi(t) and the function ω(x) in an explicit form. We also note, as follows from (2.35), that the optimal motions can always be realized in such a way that H(ψ(t), x (t), u (t)) ≡ 1 (2.36) along optimal trajectories. We remind that all of these results can be obtained provided that the function ω(x) is twice differentiable. Without this additional assumption the proof of relation (2.36) loses its validity. Let us give a simple example (see more examples in , , ) that shows that the function ω(x) does not have the first derivatives at the points which lie on the switching curves (this may be ascertained by direct calculations). Since every optimal trajectory passes along the switching curve during some time interval in this example, the assumption on the differentiability of ω(x) holds on none of the trajectories. Thus, even in the simplest examples, the assumptions which must be made in order to derive Bellman’s equation do not hold. Example where Pontryagin’s principle applies, but Bellman’s fails because the cont-rol is discontinuous (Bang-Bang Problem )Consider the equation d2xdt 2 = u, where u is a real control parameter constrained by the condition \|u\| ≤ 1. The given equation can be rewritten using the phase coordinates x1 = x and x2 = dx/dt .Hence, we get the following system: dx 1 dt = x2, dx 2 dt = u. (2.37) Let us consider (for a phase point moving in accordance with (2.37)) the problem of getting to the origin (0 , 0) from a given initial state x0 in the shortest time . In other 30 Chapter 2. Optimal Control Processes words, we shall consider the time-optimal problem for the case where the origin (0 , 0) is the terminal position x1.The Hamiltonian function H(ψ, x, u ) = n∑ ν=1 ψν f ν (x, u ) in this case has the form H = ψ1x2 + ψ2u. (2.38) Thus, since we know that dψ i dt = −∂H ∂x i , i = 1 , ..., n, (see equation (2.19)) we obtain the system of equations dψ 1 dt = − ∂H ∂x 1 = 0 , dψ 2 dt = − ∂H ∂x 2 = −ψ1, for the auxiliary variables ψ1 and ψ2. Hence, ψ1 = c1 and ψ2 = c2 − c1t (c1 and c2 are arbitrary constants). Relation (2.20) yields (taking (2.38) and the condition −1 ≤ u ≤ 1 into account) u(t) = signψ 2(t) = sign (c2 − c1t). (2.39) It follows from (2.39) that every optimal control u(t), t0 ≤ t ≤ t1, is a piecewise constant function which takes on the values ±1, and has at most two intervals on which it is constant (since the linear function c2 − c1t changes sign at most once on the interval t0 ≤ t ≤ t1). Also, any such function u(t) can be obtained from relation (2.39) for some values of c1 and c2.From the system (2.37) dx 1 dt = x2, dx 2 dt = u for the time interval on which u ≡ 1 we have x2 = t + s2, x1 = t2 2 + s2t + s1 = 12 (t + s2)2 + ( s1 − (s2)2 2 ) 31 Chapter 2. Optimal Control Processes Figure 2.7: Bang-bang time-optimal control: trajectories for u = 1 of parabolas given by equation (2.40) (s1 and s2 are constants of integration), from which we obtain x1 = 12 (x2)2 + s, (2.40) where s = s1 − 12 (s2)2 is a constant. Thus, the portion of the phase trajectory for which u ≡ 1 is an arc of the parabola (2.40). The family of parabolas (2.40) is shown in Figure 2.7. Analogously, for the time interval on which u ≡ − 1, we have x2 = −t + s′2,x1 = −t2 2 + s′2t + s′1 = −12 (−t + s′2)2 + ( s′2 + 12 (s′2)2 ) , 32 Chapter 2. Optimal Control Processes Figure 2.8: Bang-bang time-optimal control: trajectories for u = −1 of parabolas given by equation (2.41) from which we obtain x1 = −12 (x2)2 + s′. (2.41) The family of parabolas (2.41) is shown in Figure 2.8. The phase points move upwards along the parabolas (2.40) (since dx 2/dt = u = +1), and downwards along the parabolas (2.41) ( dx 2/dt = u = −1). As we said before, every optimal control u(t) is a piecewise constant function, taking on the values ±1, and having at most two intervals on which it is constant. If u(t) is initially equal to +1, and then to −1, the phase trajectory consists of two adjoining parabolic segments (Figure 2.9). The second of these segments lies on that parabola defined by (2.41) which passes through the origin (since the desired trajectory must lead to the origin). On the other hand, if u = −1 first and u = +1 33 Chapter 2. Optimal Control Processes Figure 2.9: Bang-bang time-optimal control: u(t) is initially equal to +1 , and then to −1, the phase trajectory consists of two adjoining parabolic segments given by equations (2.40) and (2.41), respectively afterwards, the phase curve is replaced by one which is symmetric with respect to the origin (Figure 2.10). In Figures 2.9, 2.10 the corresponding values of the control parameter u are written next to the parabolic arcs. Figure 2.11 shows the entire family of phase trajectories we obtained ( AO is the arc of the parabola x1 = 12 (x2)2 in the lower half-plane, BO is the arc of the parabola x1 = −12 (x2)2 in the upper half-plane). The phase point moves along an arc of the parabola (2.41) which passes through the initial points x0, if x0 is above the curve AOB ; and along an arc of a parabola (2.40) if x0 is below this curve. In other words, if the initial position x0 is above the curve AOB , the phase point must move under the influence of the control 34 Chapter 2. Optimal Control Processes Figure 2.10: Bang-bang time-optimal control: u(t) is initially equal to −1, and then to +1 , the phase trajectory consists of two adjoining parabolic segments given by equations (2.41) and (2.40), respectively u = −1 until it reaches the arc AO . At the instant it arrives, the value of u switches to +1 and remains at this value until the phase point reaches the origin. However, if the initial position x0 is below AOB , u must equal +1 until the time it reaches the arc BO , and at that time the value of u changes to −1. Definition A piecewise constant optimal control u(t) that takes only two values on the boundary of the control space U is called a bang-bang control . According to Theorem 2.2.2, only the above described trajectories can be optimal .Furthermore, it can be seen from the above investigation that from each point in 35 Chapter 2. Optimal Control Processes Figure 2.11: Bang-bang time-optimal control: the switching curve and the family of phase trajectories we obtained ( AO is the arc of the parabola x1 = 12 (x2)2 in the lower half-plane, BO is the arc of the parabola x1 = −12 (x2)2 in the upper half-plane) the phase plane there is only one trajectory leading to the origin which can be optimal (i.e, once the initial point x0 is given, the corresponding trajectory is uniquely determined). If we could be sure that the optimal trajectory always (i.e, for any initial point x0) exists, we could confidently say that all the trajectories we have found are optimal (see , Chapter III for the formulation of the existence theorem for linear time-optimal systems). In particular, it follows from this theorem that in the present example there exists an optimal trajectory (see page 127 in ) for each initial point x0. Thus, the trajectories we have found (Figure 2.11) are optimal, and there are no other optimal trajectories which lead to the origin. 36 Chapter 2. Optimal Control Processes Therefore, the solution of the optimal problem obtained in the above example can be interpreted as follows. Let v(x1, x 2) = v(x) be the function given in the x1x2 plane as follows: v(x) =  1 below the curve AOB , and on the arc AO, 1 above the curve AOB , and on the arc BO. Also, on each optimal trajectory the value u(t) of the control parameter (at an arbitrary time t) is equal to v(x(t)), meaning that it equals the value of the function v at the point at which the phase point, moving along the optimal trajectory u(t) = v(x(t)) , is located at the time t. This means that if we replace the variable u by the function v(x) in the original system (2.37), we obtain the system  dx 1/dt = x2,dx 2/dt = v (x1, x 2) . (2.42) We can find the optimal phase trajectory which leads to the origin from the solution of this system (2.42) (for an arbitrary initial state x0). Therefore, we system (2.42) is the system of differential equations (with discontinuous right-hand side) for the determination of the optimal trajectories which lead to the origin. 37 Chapter 3 The Pursuit Problem 3.1 Statement of the Problem Let us assume that two points, one of which we shall call “pursuing” (P) and the other “evading” (E), are moving in X ⊂ Rn: x′ = f (x, u, t ), y′ = g(y, t ), (3.1) where u, U, and x (t) are the control parameter, the control region, and the trajectory of the motion of the pursuing point P, respectively, and y(t) is the trajectory of the motion of the evading point E. Let u(t) be a certain admissible control (i.e., piecewise continuous), and let x (t)and y(t) be the corresponding trajectories with initial conditions x(0) = x0, y(0) = y0. (3.2) If x (t1) = y(t1) for some t1 > 0, we shall call t1 an encounter time , and the very occurrence that x (t1) = y(t1) will be referred to as an encounter . If the control u(t)is chosen arbitrarily, an encounter may not occur for any t > 0. If an encounter 38 Chapter 3. The Pursuit Problem does occur, we shall call the control (which is an admissible control) u(t) a pursuing control . Even then, for the given x0, y0, and the chosen control u(t), more than one encounter may take place. We shall call the smallest positive number t1, which is an encounter time, the pursuit time corresponding to the control u(t). We shall denote the pursuit time by T = min u∈U Tu. (3.3) In what follows, the initial conditions (3.2) will be assumed to be fixed (in this connection, x0 and y0 do not enter into the notation for the pursuit time). Therefore, we get a statement of the pursuit problem . Definition The problem is called a pursuit problem if it is defined by equations (3.1) - (3.3) x′ = f (x, u, t ), y′ = g(y, t ),x(0) = x0, y(0) = y0,T = min u∈U Tu, where x and y belong to X ⊂ Rn, u ∈ U ⊂ Rr and is admissible (piecewise conti-nuous). 39 Chapter 3. The Pursuit Problem Figure 3.1: The geometry of Bouguer’s pursuit problem about a pirate ship moving directly toward the merchant vessel at constant speed Vp along a curved path and pursuing a merchant vessel travelling at constant speed Vm along the vertical line x = x0 3.2 Pierre Bouguer’s Pursuit Problem Modern mathematical pursuit analysis is generally assumed to begin with a problem posed and solved by the French mathematician and hydrographer Pierre Bouguer (1698-1758) in 1732 (see ). This general assumption is not quite correct, but Bouguer’s problem is today nevertheless taken as the starting point of pursuit ana-lysis in all modern textbooks. In his paper, Bouguer treated the case of pirate ship pursuing a fleeing merchant vessel, as illustrated in Figure 3.1. The pirate ship and the merchant vessel are taken to be at (0 , 0) and ( x0, 0) at time t = 0, respectively, 40 Chapter 3. The Pursuit Problem the instant the pursuit begins, with the merchant vessel travelling at constant speed Vm along the vertical line x = x0. The pirate ship travels at constant speed Vp along a curved path such that it is always moving directly toward the merchant , that is, the velocity vector of the pirate ship points directly at the merchant vessel at every instant of time. Bouguer’s problem was to determine the equation y = y(x) of the curved path which he called the line of pursuit . The pursuit curve has its association with the path taken by a dog in following its master, and the falcon flying in its attack directly at the instantaneous location of its prey. This is the definition of what is now called pure pursuit .To find the curve of pursuit for Bouguer’s problem, start by calling the location of the pirate ship, at arbitrary time t ≥ 0, the point ( x, y ). At time t the merchant vessel has sailed to the point ( x0, V mt) and so, as shown in Figure 3.1, the slope of the tangent line to the pursuit curve (the value of dy/dx at ( x, y )) is given by dy dx = Vmt − yx0 − x = y − Vmtx − x0 . (3.4) We also know that, whatever the shape of the pursuit curve, the pirate ship has sailed along it at time t by a distance of Vpt. From calculus we know that this arc-length is also given by the expression on the right below, and so Vpt = x ∫ 0 √ 1 + (dy dz )2 dz, (3.5) where z is simply a dummy variable of integration. Solving (3.4) and (3.5) each for t, we can write 1 Vpx∫ 0 √ 1 + (dy dz )2 dz = yVm − x − x0 Vm · dy dx , which, if we let dy/dx = p(x), becomes 1 Vpx∫ 0 √1 + p2(z)dz = yVm − x − x0 Vm · p(x). (3.6) 41 Chapter 3. The Pursuit Problem Differentiating (3.6) with respect to x (using Leibniz’s formula to differentiate an integral), we arrive at 1 Vp √1 + p2(x) = 1 Vm · dy dx − x − x0 Vm · dp dx − 1 Vm p(x)or, simplifying, (x − x0) dp dx = −Vm Vp √1 + p2(x) = −n√1 + p2(x), (3.7) where n = Vm/V p. (Ordinarily we’ll have n < 1, the pirate ship sailing faster than the merchant. For n > 1 the problem is without interest as then the pirate ship is slower than the merchant and the concept of “pursuit” is meaningless. The n = 1 case, however, does offer us a curious mathematical problem with special interest that we’ll go into later.) Separating variables, dp √1 + p2 = − ndx x − x0 = ndx x0 − x (3.8) and, integrating (3.8) indefinitely, we have (with C as the constant of indefinite integration) ln ( p + √1 + p2 ) C = −n ln ( x0 − x) . (3.9) From Figure 3.1 we see at t = 0 that p = dy/dx = 0 when x = 0, because at that instant both ships are on the x-axis (the fact that dy/dx \|t=0 = 0 also follows mathematically from (3.4) since y(t = 0) = 0). Inserting these initial conditions into equation (3.9), it follows that C = −nln (x0) and so (3.9) becomes ln ( p + √1 + p2 ) − n ln( x0) = −n ln ( x0 − x) , which, after a few steps of algebra, reduces to ln [( p + √1 + p2 ) ( 1 − xx0 )n] = 0 , which tells us that ( p + √1 + p2 ) ( 1 − xx0 )n = 1 . (3.10) 42 Chapter 3. The Pursuit Problem Thus, p + √1 + p2 = 1 ( 1 − xx0 )n = q, (3.11) where q has been introduced to keep the next few algebraic steps easy to follow. Solving (3.11) for p, we have √1 + p2 = q − p, 1 + p2 = ( q − p)2 = q2 − 2qp + p2,p = q2 − 12q = 12 [ q − 1 q ] . Thus, replacing q with its equivalent (from (3.11)) gives p(x) = dy dx = 12 [( 1 − xx0 )−n − ( 1 − xx0 )n] , n = Vm Vp . (3.12) We can solve (3.12) for y(x) by simple integration, writing C once more as the constant of integration, y(x) + C = 12 ∫ dx ( 1 − xx0 )n − 12 ∫ ( 1 − xx0 )n dx. In both integrals change variable to u = 1 − x/x 0 (so dx = −x0du ) to get y(x) + C = 12 ∫ −x0du un − 12 ∫ −x0undu, (3.13) which immediately integrates to y(x) + C = −12x0 u−n+1 −n + 1 + 12x0 un+1 n + 1 = 12x0 [u · un 1 + n − u · u−n 1 − n ] = 12x0u [ un 1 + n − u−n 1 − n ] . That is, y(x) + C = 12x0 ( 1 − xx0 ) ( 1 − xx0 )n 1 + n − ( 1 − xx0 )−n 1 − n  , 43 Chapter 3. The Pursuit Problem or y(x) + C = 12(x0 − x) ( 1 − xx0 )n 1 + n − ( 1 − xx0 )−n 1 − n  . (3.14) Since y(x = 0), then C = 12x0 [ 11 + n − 11 − n ] = − n 1 − n2 x0 and so inserting this result into (3.14) gives us our answer, the pursuit curve equation y = y(x) : y(x) = n 1 − n2 x0 + 12(x0 − x) × ( 1 − xx0 )n 1 + n − ( 1 − xx0 )−n 1 − n  , n = Vm Vp . (3.15) “Capture” occurs when x = x0 (the pirate ship pursuit curve intersects the mer-chant’s course), which says capture occurs at the point ( x0, n/ (1 − n2)x0). (This makes physical sense only if n < 1, of course, the case of the pirate ship being faster than the merchant.) For example, if the pirate ship sails twice as fast as the merchant, then n = 12 and capture occurs at the point ( x0, 23 x0), while if the pirate ship sails only one-third faster than the merchant (i.e., Vp = 43 Vm), then n = 34 and capture occurs at the point ( x0, 12 7 x0). As n approaches one, that is, as the sailing speeds of the pirate ship and the merchant vessel become equal, it is clear that the capture point moves ever father up the x = x0 line and, in the limit n = 1, the capture point is at infinity (which is the physically obvious statement that capture does not occur). Figure 3.2 shows the pursuit curve up to the capture point for the case of x0 = 1 and n = 34 . The analytical expression of (3.15) fails to make sense for the case of n = 1 ( Vm = Vp), of course, because then we have a division by zero problem. To see what the correct analytical form of the pursuit curve is for n = 1, 44 Chapter 3. The Pursuit Problem 0 0.5 1 1.5 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 XY Figure 3.2: The path of the pirate ship as given by equation (3.15) for n = 3 /4return to (3.12), to just before we integrated dy/dx . Then dy dx = 12 [( 1 − xx0 )−1 − ( 1 − xx0 )] = 12  1 ( 1 − xx0 ) − ( 1 − xx0 ) (3.16) and so y(x) + C = 12 [∫ dx 1 − xx0 − ∫ ( 1 − xx0 ) dx ] . As before, change variables in both integrals to u = 1 − x/x 0 (and so dx = −x0du )45 Chapter 3. The Pursuit Problem to get y(x) + C = 12 ∫ −x0 u du − 12 ∫ u(−x0)du = −12x0 ln u + 12x0 · 12u2 = 12x0 [ 12 ( 1 − xx0 )2 − ln ( 1 − xx0 )] . Since y(x = 0) = 0, then C = 14 x0, and so for n = 1 ( Vp = Vm) the equation of the pursuit curve is y(x) = 12x0 [ 12 ( 1 − xx0 )2 − ln ( 1 − xx0 )] − 14x0. (3.17) When Bouguer’s problem was included in the 1859 book Treatise on Differential Equations by the famous British mathematician George Boole (1815 - 1864), the pursuit curve for n = 1 (pursuer and evader moving with equal speeds) case was declared to be a parabola, which is clearly wrong - as observed in Burton and Eliezer , whatever the pursuit curve is (for any value of n) it certainly must be asymptotic to the line x = x0.Now, for the n < 1 case let us calculate the total distance travelled by the pirate ship until its capture of the merchant vessel. As we discussed earlier, capture does not occur in the n = 1 case, and after a “long” time, the pirate ship will have sailed into a position directly behind the merchant and will simply chase, endlessly, after the merchant while remaining a constant distance behind it. It is an interesting mathematical problem to calculate the value of this so-called tail chase lag distance. To calculate the distance sailed by the pirate ship until it captures the merchant vessel ( n < 1), recall from (3.15) that capture occurs at ( x0, n/ (1 − n2)x0), i.e., the merchant vessel has travelled a distance of n/ (1 − n2)x0. Since the pirate ship travels 46 Chapter 3. The Pursuit Problem Figure 3.3: The geometry of the tail chase as given by equation (3.17) 1/n times faster than does the merchant, the pirate travels 1 /n times as far, that is, the pirate ship travels a total distance of 1 /(1 − n2)x0. To answer the second question, i.e., to determine the distance the pirate ship lags behind the merchant vessel after a long time has passed (for n = 1), refer to Figure 3.3. There we see the pirate ship at point ( x, y ), while the merchant vessel is at ( x0, y n). Note, that this is for any arbitrary time t. The distance separating the pirate ship and the merchant vessel is D, where D2 = ( ym − y)2 + ( x0 − x)2 = ( x0 − x)2 [ 1 + (ym − yx0 − x )2] . Now, here is an important fact: the line joining ( x, y ) to ( x0, y m) is the tangent to 47 Chapter 3. The Pursuit Problem the pirate’s pursuit curve, because the chase is a pure pursuit, meaning that the pirate ship is moving directly at the instantaneous location of the merchant vessel (according to the statement of the problem), i.e., the velocity vector of the pirate ship points directly at the merchant vessel at every instant of time. Thus, dy dx = ym − yx0 − x and so D2 = ( x0 − x)2 [ 1 + ( dy dx )2] . Substituting (3.12) for dy/dx for the n = 1 case, that is, writing dy dx = 12  1 ( 1 − xx0 ) − ( 1 − xx0 ) , we have D2 = ( x0 − x)2 1 + 14  1 ( 1 − xx0 ) − ( 1 − xx0 ) 2  = x20 ( 1 − xx0 )2 1 + 14  1 ( 1 − xx0 )2 − 2 + ( 1 − xx0 )2  = x20 [( 1 − xx0 )2 14 − 12 ( 1 − xx0 )2 14 ( 1 − xx0 )4] . As t → ∞ we physically see the pirate ship pull into behind the merchant vessel and the pursuit becomes a vertically upward tail chase; thus, x → x0, and so lim t→∞ D2 = lim x→x0 D2 = 14x20 or, at last, lim t→∞ D = 12x0. 48 Chapter 3. The Pursuit Problem Application of Bouguer’s Pursuit Problem: A merchant vessel, moving horizontal in a straight line, is b feet directly below one pirate ship “Black Pearl” and d feet directly above another pirate ship “Dead Men”. Both pirate ships move directly toward the merchant vessel, reaching it simul-taneously. We know that “Black Pearl” is slower than “Dead Men”, and that “Dead Men” moves twice as fast as the merchant vessel. At what rate does the “Black Pearl” move? We can see right away that the statements that “Black Pearl” is above the mer-chant vessel, and that “Dead Mean” is below, have nothing to do with the mathema-tics of the problem. Then, with no loss in the spirit of the problem, we can take the initial location of the “Black Pearl” as (0 , b ) and of the “Dead Mean” as (0 , d ). In our solution to Bouguer’s problem, the initial separation between pursuer and pursued was x0, and so b and d each play the role of x0. We know from our earlier analysis that capture will occur after the evader has travelled distance of n/ (1 − n2)x0, where n equals the speed of the evader over the speed of the pursuer. For “Dead Men” we have n = 1 /2, and for “Black Pearl” let’s say it moves k times as fast as the merchant vessel (and so n = 1 /k for “Black Pearl”). Now, since both pursuers “capture” the vessel at the same instant (the same point) we have 1/21 − (1 /2) 2 d = 1/k 1 − (1 /k )2 b. Hence, 1/23/4d = kk2 − 1b, or 23d = kk2 − 1b, where d, b, and k are some constants. Simplifying, we get 23dk 2 − bk − 23d = 0 , 49 Chapter 3. The Pursuit Problem k = b ± √ b2 + 16 9 d243 d . We realize that since “Black Pearl” starts closer to the vessel than does “Dead Men”, k must be between one and two (the pirate ship “Black Pearl” must move faster than the merchant vessel to capture it, but slower than “Dead Men”, according to the given condition). Hence, k > 0, and we use the plus sign. Therefore, “Black Pearl” moves b+ √b2+16 /9d2 4/3d times as fast as the merchant vessel. Now, if we let, for example, b = 50 and d = 100, then we can find that “Black Pearl” moves 1 .443 times as fast as the vessel. Remark A different generalized form of Bouguer’s problem was solved in Colman , in which the merchant vessel’s straight sailing path is inclined from the vertical by angle α, i.e., the line x = x0 is replaced by the straight line y = ( x − x0) · cot α for −π 2 ≤ α ≤ π 2 . Colman does not give an explicit formula-equation for the flight path of the pursuer, but finds coordinates for the point of capture in the case when the ratio of pursuer’s and evader’s speeds is n < 1. The solution presented in this section is for α = 0, while α = π/ 2 radians would represent the merchant sailing directly away from the pirate ship (and α = −π/ 2 radians would represent the merchant sailing directly toward the pirate ship). In both of these extreme cases the pursuit curve is, by inspection, simply x = 0 (the x-axis), but for α 6 = ±π/ 2 or 0 the pursuit curve is pretty complicated, and its derivation is an exercise in nontrivial manipulation. 50 Chapter 3. The Pursuit Problem Figure 3.4: The geometry of the wind-blown plane problem, where the plane’s nose is always pointed toward a city C, the plane’s speed is v mi/h, and a wind is blowing from the south at the rate of w mi/h 3.3 Wind-Blown Plane Problem Let us now present another important example (following ), where we use the analysis of the Bouguer’s pursuit problem. It is similar to the problem solved in 1931 by E. Zermelo (see ). A pilot always keeps the nose of his plane pointed toward a city C due west of his starting point at (a, 0) . Find equation of the plane’s path if the plane’s speed is v mi/h, and a wind is blowing from the south at the rate of w mi/h. 51 Chapter 3. The Pursuit Problem The pilot isn’t really pursuing anything, of course, unless we consider this problem of “pursuit (with wind interference) of a stationary target”, but the spirit of this problem is pure Bouguer. In the notation of Figure 3.4, at an arbitrary time t ≥ 0, the plane’s location is the point ( x(t), y (t)). Writing ux and uy as the unit vectors in the x and y directions (which are not functions of time), respectively, then we can write the position vector of the plane as p(t) = x(t)ux + y(t)uy, and so the plane’s velocity vector is ddt p(t) = dx dt ux + dy dt uy. Also, the plane’s body axis (nose-to-tail) is always along the direction of p(t), at angle θ, toward C, where tan( θ) = yx. The wind, blowing only along the y-axis, contributes nothing to the ux component of the plane’s velocity vector, that is, dx/dt is due only to the x-component of v. dx dt = −v cos( θ) = − vx √x2 + y2 , (3.18) where the minus sign is explicitly included, since as the plane flies toward C the value of x decreases with increasing t. The uy component of the plane’s velocity vector, on the other hand, is influenced by the wind, of course, as well as by the y-component of v, dy dt = w − v sin( θ) = w − vy √x2 + y2 = w√x2 + y2 − vy √x2 + y2 . (3.19) Dividing (3.19) by (3.18), we eliminate explicit time and arrive at dy dx = vy − w√x2 + y2 vx . (3.20) 52 Chapter 3. The Pursuit Problem Let us introduce a new variable z such that y = zx . Then (3.20) becomes dy dx = z + x dz dx = vzx − w√x2 + z2x2 vx = z − wv √1 + z2 or, defining the constant n = w/v , x dz dx = −n√1 + z2, (3.21) from where we get dz √1 + z2 = −ndx x . (3.22) (Notice the similarity of (3.22) and (3.8).) Integrating indefinitely, with C as the constant of integration, ln[( z + √1 + z2)] + C = −n ln( x). Since y = 0 when x = a, which means z = y/x = 0 when x = a, then we have C = −n ln( a), and so ln[( z + √1 + z2)] = n ln( a) − n ln( x) = n ln ( ax ) = ln ( ax )n , or, z + √1 + z2 = ( ax )n . (3.23) Defining q = ( a/x )n, (3.23) becomes (similar to how we went from (3.11) to (3.12)) √1 + z2 = q − z, 1 + z2 = q2 − 2qz + z2,z = q2 − 12q = 12 [ q − 1 q ] . Thus, replacing q with its definition, z = 12 [( ax )n − ( ax )−n] = 12 [(xa )−n − (xa )n] . (3.24) 53 Chapter 3. The Pursuit Problem 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 x n=0.4 n=0.1 n=0.2 n=0.8 n=0.95 n=0.999 Figure 3.5: Plots of the wind-blown plane’s paths given by equations (3.25) for several values of n < 1 (n = 0.1, 0.2, 0.4, 0.8, 0.95, 0.99, 0.999) Since y = zx , then y = 12 [x−n+1 a−n − xn+1 an ] = 12 [ x−n+1 a−n+1 a − xn+1 an+1 a ] or, at last, we have the equation of the wind-blown plane’s path : y(x) = a 2 [(xa )−n+1 − (xa )n+1 ] , n = wv . (3.25) When n = 0 - that is, when there is no wind - (3.25) collapses to the physically obvious y(x) = 0, which simply says that the plane moves directly to city C while always remaining on the x-axis. And when n = 1 (when the wind speed equals the plane’s speed in still air), the plane’s path is the parabola y(x) = a 2 [ 1 − (xa )2] . In this case when x = 0 we see that y(0) = a/ 2, that is, the plane does not reach city C. This probably makes intuitive sense, too, but it is interesting to see that the miss distance is so large. What happens, physically, in the n = 1 case, is that the plane arrives at the y-axis with a zero velocity component in the x-direction (notice that the plane’s body axis has rotated through an angle of θ = 90 ◦, and then recall 54 Chapter 3. The Pursuit Problem (3.18)), and so there the plane remains, motionless at the point (0 , a/ 2), as it flies directly into the wind with the two equal magnitude but oppositely directed velocity vectors precisely cancelling each other. Figure 3.5 shows the plane’s path for a = 1 for several different values of n, and it is clear that for n < 1 ( ≥ 1) the plane reaches (does not reach) city C. Now, let us calculate the total flight time of the wind-blown plane for n < 1, and the total distance flown for the case of n “just less” than one. For the total flight time T of the wind-blown plane, recall (3.18) and (3.25), where we showed that dx dt = − vx √x2 + y2 and y = a 2 [(xa )−n+1 − (xa )n+1 ] , n = wv . So, T ∫ 0 dt = − 0 ∫ a √x2 + y2 vx dx, or T = 1 v 0 ∫ a √ 1 + y2 x2 dx. Also, y2 = a2 4 [(xa )−2n+2 − 2 (xa )2 + (xa )2n+2 ] = a2 4 [(xa )−2n x2 a2 − 2 xa22 + (xa )2n x2 a2 ] . Thus, y2 x2 = 14 [(xa )−2n − 2 + (xa )2n] , 55 Chapter 3. The Pursuit Problem and so 1 + y2 x2 = 1 + 14 [(xa )−2n − 2 + (xa )2n] = (x/a )−2n − 2 + ( x/a )2n + 4 4= (x/a )−2n + 2 + ( x/a )2n 4 = {(x/a )n + ( x/a )−n 2 }2 . We can then write T as T = 12v a ∫ 0 [(xa )n + (xa )−n] dx = 12v  a ∫ 0 (xa )n dx + a ∫ 0 (xa )−n dx  . Letting u = x/a (dx = adu ), we then have T = 12v  1 ∫ 0 unadu + 1 ∫ 0 u−nadu  = a 2v [ un+1 n + 1 + u−n+1 −n + 1 ]∣ ∣∣∣ 10 = a 2v ( 11 + n + 11 − n ) = a/v 1 − n2 , n = wv . This makes sense for 0 ≤ n < 1. Notice that if n = 0 (no wind) then T = a/v , which is simply the time the plane requires to fly straight along the x-axis from ( a, 0) to (0 , 0) at a speed v. As n approaches one from below, of course, we see T → ∞ as expected. For the total distance flown by the plane when n is “just less” than one, that is, for the case where the plain “just managers” to reach city C, recall that at n = 1 the plane’s path is the parabola y = a 2 [ 1 − (xa )2] . 56 Chapter 3. The Pursuit Problem As n approaches one, then, the upward curved part of the flight path of the plane approaches this parabola, as illustrated in Figure 3.5. Let us look at the three plots which are for n = 0 .98, n = 0 .99, and n = 0 .999, all for a = 1. From these curves it should be clear that the length of the longest flight path that just manages to reach city C is bounded from above by a 2 + a ∫ 0 √ 1 + ( dy dx )2 dx, where the second term is the length of the parabolic arc. The first term, of course, is the length of the final leg of the journey back down along (almost along) the vertical axis to city C at the origin. On the parabolic arc we have dy dx = −a 2 2 (xa ) 1 a = −xa , and so our answer is a 2 + a ∫ 0 √ 1 + (xa )2 dx. If we change variables to u = x/a (dx = adu ), our answer becomes a 2 + a ∫ 0 √ 1 + ( u)2adu = a 12 + 1 ∫ 0 √1 + u2du  or, a [ 12 + { u√u2 + 1 2 + 12 ln ( u + √u2 + 1 )}]∣ ∣∣∣∣ 10 = a [ 1 + √2 + ln(1 + √2) 2 ] = 1 .6478 a. This is the total distance flown by the plane when n = 1 − ε, where ε > 0, but is arbitrary small. 57 Chapter 3. The Pursuit Problem Figure 3.6: The geometry of the tractrix problem, where a watch-on-a-chain with the chain of length a is initially on the y-axis, the end of the chain is pulled along the x-axis from the initial position on the origin 3.4 The Tractrix In the late seventeenth century there was also a different pursuit curve (as you will see, it is better to call this curve the following curve or the tailing curve) . An example of such a problem (with the tailing curve) is illustrated in Figure 3.6, where a watch-on-a-chain has been laid out on a table-top with the chain (of length a)pulled out. In our thesis we follow the statement given in . The watch is initially on the y-axis, and the other end of the chain is on the origin. If the end on the origin is then pulled along the x-axis, the watch will obviously be dragged along. We 58 Chapter 3. The Pursuit Problem are interested in the equation of the watch’s path , known as the tractrix . It was first introduced by Claude Perrault in 1670, and later studied by Sir Isaac Newton (1676) and Christian Huygens (1692) . If ( x, y ) is the location of the watch at some arbitrary time t ≥ 0, then it is clear that the taut chain is tangent to the tractrix at ( x, y ). This crucial observation allows us to calculate where the pulling end of the taut chain is (always on the x-axis), as follows. The slope of the tangent line is dy/dx and so, from analytic geometry, we have the equation of the tangent line as y = x dy dx + b, (3.26) where b is some constant. Let xi be the value of x where the pulling end of the chain is located, by definition y = 0 there. So, b = −xi dy dx , and therefore, the equation of the tangent line that intersects the x-axis at x = xi is y = x dy dx − xi dy dx = ( x − xi) dy dx . (3.27) From the Pythagorean theorem we then have (x − xi)2 + y2 = a2, or, using (3.27) to solve for ( x − xi), we have y2 (dy/dx )2 + y2 = a2, or [ ydy/dx ]2 = a2 − y2. (3.28) Taking the positive square root of both sides of (3.28), and noting that dy/dx is negative (look at Figure 3.6 again), we arrive at − ydy/dx = √a2 − y2, (3.29) 59 Chapter 3. The Pursuit Problem 0 0.5 1 1.5 2 2.5 3 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x Figure 3.7: A depiction of the tractrix given by equation (3.31) for a = 1 a differential equation in which we can separate the variables. That is, dx + √a2 − y2 y dy = 0 . (3.30) Integrating indefinitely (with C as the arbitrary constant), we have x + √a2 − y2 − a ln ( a + √a2 − y2 y ) = C. Since y(x = 0) = a, we have C = 0 and so the equation of the watch’s path as it is being dragged is x = a ln ( a + √a2 − y2 y ) − √a2 − y2. (3.31) Figure 3.7 shows the tractrix of (3.31) for the case of a = 1. Finally, it is interesting to contrast the tractrix with Bouguer’s pure pursuit curve for the special case of equal speeds for the pirate ship and the merchant vessel. The two curves seemingly have a common property, as the dragged watch is a constant distance from the pulled end of the chain, and the pirate ship ends up a constant distance behind the fleeing merchant vessel. The expression of (3.17) and (3.31) are 60 Chapter 3. The Pursuit Problem quite different. The reason is that for the tractrix the constant lag of the watch is always the case, while the constant lag of the pirate ship is an asymptotic property that develops with the passage of time. 61 Chapter 3. The Pursuit Problem Figure 3.8: Schematic of the pursuit by interception problem with pursuer T (Tor-pedo) and evader E (Enemy ship) moving with constant speeds VT and VE, respec-tively 3.5 Apollonius Pursuit Problem In this section we talk about a question that you may have already thought about -since the merchant vessel being pursued by Bouguer’s pirate ship always sails along a straight line, why does the pirate use pure pursuit (meaning that the pirate ship is moving directly at the instantaneous location of the merchant vessel) to run down his victim? Why doesn’t the pirate ship simply sail along the straight line path that will intercept the merchant? Bouguer himself was not oblivious to that possibility. As Puckette puts it, “[Bouguer] makes it quite clear that the pursuing ship could 62 Chapter 3. The Pursuit Problem catch its quarry much more quickly by ‘heading it off’ than by merely following it (assuming the line of flight remains a straight line)”. There are at least two answers to that question (for a broader discussion see ). First, of course, the pure pursuit problem is simply interesting from a mathematical point of view. And second, if the merchant vessel deviates from its straight path and starts executing an active evasion plan, then the pirate ship is going to have to recalculate its intercept course continually anyway. A pure pursuit strategy is just one way to specify how to do repetitive new course calculations. And, in any case, even for the merchant vessel sticking to a straight line escape path, determining the intercept course for the pirate ship is a nontrivial calculation. In the days of submarine warfare in World War II , for example, this was a most practical problem - submarines fired their torpedoes on intercept courses at unsuspecting, that is, nonmaneuvering, enemy surface ships. Today, it isn’t such an important problem because, unlike the torpedoes from yesteryear, modern torpedoes use what is called “active tracking”, that is, they have onboard sensors and computers that continually locate the target no matter how that target moves. Still, the mathematics of interception remains elegant. Let us suppose that the torpedo T is to intercept an enemy surface ship E (as shown in Figure 3.8), with E moving on a straight path and T moving on a straight path to intercept E at point I. If we assume that E and T move with constant speeds VE and VT, respectively, then at the intercept point I the ratio of the two distances travelled from the instant of the torpedo firing must equal the ratio of the two speeds, IT IE = VT VE = k, (3.32) where k is a constant ( k > 1 is the usual case, but the k < 1 case will be of interest to us, too, before we are done). Equation (3.32) is the mathematical statement of the physically obvious fact 63 Chapter 3. The Pursuit Problem Figure 3.9: The Apollonius circle centered on (2 /3, 0) with radius 2/3, given by equation (3.34) for m = 1 , p = 2 , and k = 2 , so that the torpedo is located at T(2 , 0) and the enemy ship is at E(1 , 0) that, for an interception to occur, the torpedo and the ship must reach point I simultaneously . It is not enough for E and T to pass thorough I individually - they must be at I at the same time . To find where I is, given the locations of E and T at time t = 0, the two speeds VE and VT, and the direction of E’s motion (the “heading” of E), what we must do first is find the set S of all the points in the plane such that (3.32) is satisfied. The point I can be any one of the points (there can be more than one) in S that also lie on the path of E.Now we need to identify, what is S. With no loss in generality we can draw a rectangular coordinate system such that E and T are both, at t = 0, on the positive horizontal axis with T to the right of E (see Figure 3.9). If we denote the coordinates of E and T by ( m, 0) and ( p, 0), respectively, with p > m (we use m and p to retain 64 Chapter 3. The Pursuit Problem a link with our original discussion of Bouguer’s merchant vessel and pirate ship) and if ( x, y ) is any point in S, then (3.32) becomes √ (x − p)2 + y2 √ (x − m)2 + y2 = k. (3.33) If you now go through a few algebraic manipulations, then you should be able to confirm that (3.33) can be written as [ x − k2m − pk2 − 1 ]2 y2 = [k(p − m)1 − k2 ]2 . (3.34) But this is the equation of a circle, with it center on the horizontal axis at (( k2m − p)/(k2 − 1) , 0) and a radius of k(p − m)/\|1 − k2\|. The set S is a circle , called the Apollonius circle of the two points E and T (in their t = 0 locations on the horizontal axis), which is named after the third-century B.C. Greek mathematician Apollonius of Perga . Apollonius realized (in his lost work Plane Loci) that (3.32) is a way to define a circle in a manner different from the usual Euclidean geometry definition (the path traced by a moving point that remains a fixed distance from a given point). The definition in (3.32) predates Apollonius, however, being known a century earlier to Aristotle. If m = 1, p = 2, and k = 2, for example, the Apollonius circle is centered on ( 23 , 0) with a radius of 23 ; see Figure 3.9, where the center of the Apollonius circle is marked with an X and labeled small circles indicate the initial locations of the torpedo and the enemy ship. For the submarine to determine where to aim its torpedo (that is, to locate the point I), all that remains to do is to see where E’s path intersects the Apollonius circle. The intersection point is I. For example, you can see from Figure 3.9 that I is, approximately, at (1 , 0.58) if E has a heading angle of 90 ◦.Now, what if k < 1, meaning, what if the torpedo is slower than the surface ship? To be specific, let us now take k = 1 /2, which reduces (3.34) (with m = 1 and p = 2) 65 Chapter 3. The Pursuit Problem Figure 3.10: The Apollonius circle centered on (7 /3, 0) with radius 2/3, given by equation (3.34) for m = 1 , p = 2 , and k = 1 /2, so that the torpedo is located at T(2 , 0) and the enemy ship is at E(1 , 0) to ( x − 73 )2 y2 = (23 )2 . That is, the Apollonius circle is still of radius 2 /3, but now is centered on (7 /3, 0), which means the center of the Apollonius circle is now to the right of the initial location of T, as shown in Figure 3.10. You can see that now the torpedo may or may not be able to intercept the enemy ship - it is all a function of the heading angle of the ship. If the heading angle is sufficiently small that the ship’s path crosses the Apollonius circle, then an interception by a slow torpedo of a fast enemy ship is possible (in fact, there will generally be two possible interception points), a result that often surprises. Instead of considering specific values of k, m , and p, it is not at all difficult to 66 Chapter 3. The Pursuit Problem Figure 3.11: The general geometry for a slow torpedo ( T) interception of a fast enemy surface ship ( E) (heading with an angle θ), where the Apollonius circle for the points (m, 0) and (p, 0) , p > m , is given by equation (3.34) for k < 1be much more general and to derive an astonishingly simple condition that will tell us, for any k < 1, if a slow torpedo interception is, first, even possible and, if it is, where on the Apollonius circle the submarine should aim its slow torpedo. Equation (3.34) tells us that, for k < 1, the Apollonius circle for the points ( m, 0) and ( p, 0), p > m , is centered on the point C at (( p − k2m)/(1 − k2), 0) and has a radius of k(p − m)/(1 − k2), as illustrated in Figure 3.11. Now, imagine that the enemy ship’s heading angle is θ, so that the ship just touches the Apollonius circle at A.If the absolute value of the heading angle is greater than θ then no interception is possible, and if the absolute value of the heading angle is less than θ then the enemy ship’s path will cross the Apollonius circle twice and so there will be two possible interception points I. We can find a formula for θ, as follows. The line AC , a radius of the Apollonius circle, is perpendicular to the tangent line EA , and so the triangle ECA is a right triangle. Thus, sin θ = AC EC . (3.35) 67 Chapter 3. The Pursuit Problem The radius of the circle, as started before, is AC = k(p − m)1 − k2 , while EC = ET + T C = ( p − m) + (p − k2m 1 − k2 − p ) = (p − m)1 − k2 . Inserting these expressions for AC and EC into (3.35) we arrive at sin θ = k = VT VE , that is, θ = sin −1 (VT VE ) . (3.36) If α is the heading angle of the enemy surface ship, then an interception using a slow torpedo is possible if −θ ≤ α ≤ θ, and impossible otherwise. Therefore, we can conclude that Apollonius circles can be used in the PE problems to analyze how to find a better strategy to escape or prolong the capture time whenever a successful escape is not possible. 68 Chapter 4 The Evasion Problem 4.1 Statement of the Problem In this chapter we present PE problems with the emphasis on evasion. Let us assume that two points, one of which we shall call “pursuing” (P) and the other “evading” (E), are moving in X ⊂ Rn: x′ = f (x, t ), y′ = g(y, v, t ), (4.1) where v, V, and y(t) are the control parameter, the control region, and the trajectory of the motion of the evading point E, respectively, and x (t) is the trajectory of the motion of the pursuing point P. Let v (t) be a certain admissible control (i.e., piecewise continuous), and let x (t)and y(t) be the corresponding trajectories with initial conditions x(0) = x0, y(0) = y0. (4.2) If x (t1) = y(t1) for some t1 > 0, we shall call t1 an encounter time , and the very occurrence that x (t1) = y(t1) will be referred to as an encounter . If the control v (t)69 Chapter 4. The Evasion Problem is chosen arbitrarily, an encounter may not occur for any t > 0. If an encounter does occur, we shall call the control (which is an admissible control) v (t) an evading control . Even then, for the given x0, y0, and the chosen control v (t), more than one encounter may take place. We shall call the largest positive number t1, which is an encounter time, the evading time corresponding to the control v (t). We shall denote the evading time by T = max v∈V Tv. (4.3) In what follows, the initial conditions (4.2) will be assumed to be fixed (in this connection, x0 and y0 do not enter into the notation for the evading time). Therefore, we get a statement of the evasion problem . Definition The problem is called an evasion problem if it is defined by equations (4.1) - (4.3) x′ = f (x, t ), y′ = g(y, v, t ),x(0) = x0, y(0) = y0,T = max v∈V Tv, where x and y belong to X ⊂ Rn, v ∈ V ⊂ Rr and is admissible (piecewise conti-nuous). 70 Chapter 4. The Evasion Problem Figure 4.1: P and E defending and attacking, respectively, the target area C 4.2 Isaacs’s Problem One of the classic general evasion problems is Isaacs’s guarding the target problem by the American mathematician Rufus Isaacs (1914 - 1981) , ,, . Isaacs states his problem, along with giving its general solution, as follows. “Both P and E (pursuer and evader) travel with the same speed. The motive of P is to guard a target C, which we take as an area in the plane, from attack by E.The optimal strategies for both P and E are: draw the perpendicular bisector of P E (where P and E denote starting positions). Any point in the half-plane above this line can be reached by E prior to P , and this property fails in the lower half-plane. 71 Chapter 4. The Evasion Problem Figure 4.2: The geometry of Isaacs’s problem for P defending a point target C, and E attacking same target; l1 is the perpendicular bisector of P E , l2 is the perpendicular line segment from C to l1 Clearly, E should head for the best of his accessible points. Let D be the point of the bisector nearest C. The optimal strategies for both P and E decree that they travel toward D. When does the capture occur?” The military conception of this problem by Isaacs is, E must reach at least the boundary of C to be successful in his attack. It may seem easy for E to reach an interior point of C (and thus, hit C), but it is not necessarily true. Here P is successful in defeating E if the capture point D (see Figure 4.1) is anywhere outside C. However, we need to know some additional information about the shape and dimensions of C to say more about the strategies of P and E, and the solution of 72 Chapter 4. The Evasion Problem the given problem. Let us assume that C is, for example, the location of a specific enemy commander, or an enemy radio transmitter. Let us also consider that E is carrying an explosive device, which, when detonated, has a circular radius of destruction R > 0. For P let us say that it can stop E only by direct impact, i.e., P must intercept E. Therefore, for E to be successful, it must come within a distance less than or equal to R before P reaches E. Now the problem can be formulated in the following manner: “we have a crude model for an attacking missile versus a missile defense system that is supposed to protect an area, for example, a city, against a ballistic missile attack” . Without any loss of generality we can place P at time t = 0 at the origin of an x-y coordinate system, and the point target C on the x-axis at x = xc > 0. That is, the target is to the right of P . The case where the target is initially to the left of P is a mirror-image of our assumed case. Let E be at ( x0, y 0) at time t = 0. We need to remark that we consider the case where C is not one of the E’s accessible points, i.e., assume that y0 is sufficiently large. If C is one of E’s accessible points, E can destroy C by actually reaching C before P can reach E.We can easily find the equation of the bisector line l1 y = −x0 y0 x + x20 + y20 2y0 , (4.4) which has the required slope and pass through the point midway between P and E at time t = 0, i.e., the point ( x0/2, y 0/2). Then, l2 is the perpendicular line segment from the point target C to l1, and the length of l2 is the closest approach distance of E to C. The slope of l2 is y0/x 0 and, since it passes through the point ( xc, 0), we get the equation of l2: y = y0 x0 x − y0 x0 xc. (4.5) 73 Chapter 4. The Evasion Problem Thus, the point of closest approach of E to C is the intersection of l1 and l2, which is the point ( X, Y ). Hence, we can find the values of X and Y from the equations (4.4) and (4.5): y0 x0 X − y0 x0 xc = −x0 y0 X + x20 + y20 2y0 , which gives us X = y20 x20 + y20 xc + x0 2 , (4.6) and from any of the two equations we had above (4.4), (4.5) we get Y = y0 x0 ( y20 x20 + y20 xc + x0 2 ) − y0 x0 xc. (4.7) Now we can find the length (squared) of l2, and it is ( X − xc)2 + Y 2, or [ y20 x20 + y20 xc + x0 2 − xc ]2 + [ y0 x0 ( y20 x20 + y20 xc + x0 2 ) − y0 x0 xc ]2 , which after simplifying will be equal to [x0 (x20 + y20 ) − 2xcx20]2 4x20 (x20 + y20 ) . Then, for E to achieve its mission goal of destroying C, the circular radius of de-struction R (squared) of E’s weapon must exceed the length (squared) of l2, i.e., R2 > [x0 (x20 + y20 ) − 2xcx20]2 4x20 (x20 + y20 ) , or R > x0 (x20 + y20 ) − 2xcx20 2x0 √x20 + y20 . Therefore, we get Rxc (x0/xc)2 + ( y0/xc)2 − 2 ( x0/xc)2 √ (x0/xc)2 + ( y0/xc)2 = F (x0 xc , y0 xc ) . (4.8) Figure 4.3 shows several of the curves that represent the right-hand side of (4.8). Each curve gives the minimum value of R/x c, as a function of x0/x c, for a given 74 Chapter 4. The Evasion Problem 0 1 2 3 4 5 6 0 1 2 3 4 5 x0/x c y0/x c=7 6451.1 23 Figure 4.3: Plot of F (x0/x c, y 0/x c) given in equation (4.8) as a function of x0/x c for a given fixed value of y0/x c (y0/x c = 1 .1, 2, 3, 4, 5, 6, 7) . Each curve gives the minimum value of R/x c fixed value of y0/x c (the label-value next to each curve). From these curves E can determine the minimum value of R (the amount of explosive) required for success in destroying C as a function of both E’s starting point and the location of the target. 75 Chapter 4. The Evasion Problem 4.3 Lady in the Lake Problem The lady in the lake problem became famous decades ago, when it appeared in Martin Gardner’s “Mathematical Games” column in Scientific American in 1975. Gardner presented the problem as follows: A young lady was vacationing on Circle Lake, a large artificial body of water named for its precisely circular shape. To escape from a man who was pursuing her, she got into a rowboat and rowed to the center of the lake, where a raft was anchored. The man decided to wait it out on shore. He knew she would have to come ashore eventually. Since he could run four times faster than she could row, he assumed that it would be a simple matter to catch her as soon as her boat touched the lake’s edge. But the girl - a mathematics major at Radcliffe - gave some thought to her predicament. She knew that on foot she could outrun the man (which does raise the question of why such a smart lady got herself into this situation in the first place by rowing out into a lake!). It was only necessary to devise a rowing strategy that would get her to a point on shore before he could get there. She soon hit on a simple plan, and her applied mathematics applied successfully. What was the girl’s strategy? The lady’s escape strategy consists of two stages. She first hops into her boat and rows away from the raft in such a way that she, the raft, and the man are always collinear. This first part of the lady’s rowing path will clearly have to change direction constantly to continually maintain collinearity because the man will instantly begin running around the lake’s edge in his attempt to intercept her at the shore. This is illustrated in Figure 4.4, where we assumed that the man runs counterclockwise around the lake. We will show later that the lady can maintain collinearity at least for a while. Let us assume that the man runs at speed v and that the lady rows with speed αv . Thus, in the original statement of the problem α = 0 .25. We see from Figure 4.4 that the man opens up the angle θ at the rate of dθ/dt = v/R , where θ is 76 Chapter 4. The Evasion Problem Figure 4.4: The first stage of the lady’s escape measured with respect to the line initially joining the man and the lady on the raft. Without any loss of generality we can assume that this initial line is the vertical axis of our coordinate system, as shown in Figure 4.4. Since the lady’s angular speed component must be vθ = r dθ dt for her to maintain the raft between herself and the man, we can write her angular speed as vθ = v rR . (4.9) The farther she gets from the raft, then, (4.9) tell us, the greater must be her angular speed if she is to maintain collinearity. Next, since the lady’s total speed through the water is αv , her radial speed 77 Chapter 4. The Evasion Problem component ( vr) must be such that v2 r v2 θ = ( αv )2, because her total speed is geometrically represented by the hypotenuse of a right triangle, with perpendicular sides vr and vθ. Thus, vr = √ α2v2 − v2 θ = √ α2v2 − v2 r2 R2 , or vr = dr dt = v √ α2 − r2 R2 . (4.10) The lady has a positive vr (that is, she moves ever closer to shore, all the while keeping half the lake between herself and the man) as long as α2 − r2/R 2 > 0, that is, until r = αR . At the instant her vr drops to zero she switches to the second stage of her escape strategy, which we will describe below. First, let us calculate, how long it takes her to arrive at the condition vr = 0. Since dt = dr/v r, then if we call t = T the time at which vr = 0, we have T ∫ 0 dt = T = αR ∫ 0 dr vr = αR ∫ 0 dr v√α2 − r2/R 2 = Rv αR ∫ 0 dr √ (αR )2 − r2 = Rv ( sin −1 ( rαR ))∣ ∣∣αR 0 = Rv sin −1(1) , or T = πR 2v . (4.11) When the lady arrives at the circle with radius αR centered on the raft, at time t = T , she has arrived at what we call the “go-for-broke” circle, because now that she is no longer moving ever closer to shore with the first part of her escape strategy, she forgets about maintaining collinearity and simply rows straight for shore at her 78 Chapter 4. The Evasion Problem full water speed of αv . She has distance R − αR to row (at speed αv ) and the man has distance πR (half the circumference of the lake) to run at speed v. She gets to shore before he gets to her if R − αR αv < πR v , or R(1 − α) < παR, or 1 − α < πα, or 1 < α (1 + π), or, at last, if α > 1/(1 + π) = 0 .241453 . (4.12) Since α = 0 .25 in the Scientific American version of the problem, we see that this two-stage escape strategy works and that the lady’s virtue is preserved. Of course, if α is sufficiently large there is no need for a two-stage escape strategy. It is easy to see that if α is “big enough” then all the lady needs to do is immediately row directly to shore, to the point directly opposite the man’s location. She gets to shore before he gets to her if Rαv < πR v , that is, if α > 1/π = 0 .3183099 .... Still, while not essential for her, the two-stage strategy will give the lady a little extra head start on the man, and it is interesting to calculate how much this head starts is for α = 1 /π . As before, in the two-stage strategy the man, the raft, and the lady remain collinear until the lady reaches the go-for-broke circle, with radius αR = R/π . Then she rows straight for shore, now 79 Chapter 4. The Evasion Problem distance R − R/π = R(1 − 1/π ) away. Since her rowing speed is αv = v/π , this requires a time (during her second stage) of R(1 − 1/π ) v/π = Rv (π − 1) . The man reaches her landing point on the shore after running halfway around the lake, which requires a time (starting at the instant the lady “goes for broke”) of πR v = Rv π. So, she arrives at her landing point on the shore before he does by a time interval of Rv π − Rπ (π − 1) = Rv . To put this head start (in time) in perspective, it is the time it takes the man to run distance R, the radius of the lake. Let us suppose now that the lady does not have a big α. Suppose, in fact, that it is smaller than (1 + π)−1. Is it then impossible for her to escape from the man? Actually, if we make a plausible assumption about the man’s reasoning (meaning, he is rational), then it is still possible for a slow-rowing lady to escape. Since the lady is a Radcliffe math major, and the man surely knows some math, too, therefore, let us assume that, as soon as the lady leaves the raft and begins to execute the first stage of her escape strategy, the man deduces what she is up to. That is, he observes that as he moves, she moves to keep the raft between him and her even as she moves ever closer to the shore. He then further deduces that as soon as she reaches her go-for-broke circle she will head straight for the shore. So, here is our assumption - as soon as he sees her go to the second stage of her escape strategy, that is, at the instant she makes straight for shore, he stops watching her carefully and simply runs around the lake to the point on the shore where he now knows she is heading. The only thing that will cause him to reevaluate matters is if the lady stops her go-for-broke rowing and, for whatever reason, begins to move back toward the raft. 80 Chapter 4. The Evasion Problem Figure 4.5: The instant when the lady reaches her go-for-broke circle However, being a clever math major, and knowing her α is less than (1 + π)−1,she has one last trick up her sleeve. She will, indeed, row a straight-line path to shore as soon as she reaches her go-for-broke circle, but it will not be the shortest distance straight-line path that the man thinks she will row. To see what she has in mind instead, look at Figure 4.5, where, with no loss in generality, we put the lady’s position at the instant she reaches her go-for-broke circle at ( αR, 0). The man’s position at that instant is ( −R, 0). In the notation of the Figure 4.5, φ is the angle the straight line joining the raft to the lady’s landing point on the shore ( S) makes with the horizontal axis. The man is assuming that φ = 0, but he is wrong, as you will see soon. Let us talk about the lady’s new escape strategy. First, we will simplify our calculations by noticing that the ratio of the radius of the go-for-broke circle to the radius of the lake is αR/R = α. If we next denote the radius of the go-for-broke 81 Chapter 4. The Evasion Problem circle as our unit distance, then αR = 1, and so α = 1 /R. (4.13) What this means is that if we wish to find the smallest value for α for which the lady can still escape, then an equivalent problem is that of finding the largest R for which the lady can still escape. And finally, since the lady rows at speed αv , we can write her rowing speed as (1 /R )v = v/R . We can now set the problem up mathematically as follows. When the lady reaches her go-for-broke circle (point L in the figure), she is distance αR = 1 from the raft, and the law of cosines tells us that the distance LS she has left to row to the shore to reach point S is LS = √1 + R2 − 2R cos φ. This takes her a time interval of LS v/R = √1 + R2 − 2R cos φv/R = Rv √1 + R2 − 2R cos φ (4.14) to row. The man is running clockwise around the lake to S (see Figure 4.5). We will quote Schuurman and Lodder about what both the lady and the man conclude once she reaches her go-for-broke circle: “... she performs an infinitesimal radial feint (toward the shore that leads the man to start running clockwise). From the moment on, (the man’s) best policy is to continue running clockwise if (the lady) goes to shore along a straight line not crossing the (go-for-broke) circle. If (the man) would return, a new diametrical mutual position, advantageous to (the lady) would be established.” This last sentence is important to understand. It points out that the man should at any time reverse his running direction around the lake, then the lady could, at the least, start rowing directly away from him at the instant of his reversal and head straight for shore. That would have her starting the second stage of her original escape strategy from a point beyond the go-for-broke circle, and yet 82 Chapter 4. The Evasion Problem still leave the man with half the lake’s circumference to travel. Even better (from the lady’s point of view), would be for her to simply flip the sign of φ, and then the situation is just as it was before he switched. So, once the man has committed to a running direction, we see that he gains nothing by reversing his decision - he therefore will run through the angle π + φ to reach S. The time required for the man to run distance ( π + φ)R around the lake to S is Rv (π + φ). (4.15) Thus, the lady will just escape the man if the two times given by (4.14) and (4.15) are equal, that is, if π + φ = √1 + R2 − 2R cos φ. Squaring both sides and solving for R gives R = cosφ ± √cos 2 φ + ( π + φ)2 − 1, and since R > 0 we must use the plus sign, R = cosφ + √cos 2 φ + ( π + φ)2 − 1. (4.16) Figure 4.6 shows the behavior of R(φ), and it is obviously a nondecreasing function of φ. To find the smallest α for which the lady escapes we must use the largest possible value for R (see equation (4.13)). That is, we want to find the value of φ that maximizes R(φ). Now, even though R continually gets bigger with increasing φ, there is a limit on how big φ can be. If φ exceeds the value it has such that the line LS (in Figure 4.5) is tangent to the go-for-broke circle, then the lady’s rowing path will take her back inside the go-for-broke circle, that is, she will have a radial speed component pointing back toward the raft (which is not a feature we expect in an escape strategy). That is, the lady should pick φ such that the line LS is perpendicular to the x - axis. From Figure 4.5 we see that this value of φ (= φt)satisfies the condition cos( φt) = αR/R = 1 /R. 83 Chapter 4. The Evasion Problem 0 0.5 1 1.5 2 2.5 3 3.5 4 4 4.5 5 5.5 6 6.5 φ(radians) Figure 4.6: The radius of the lake R(φ) (in radians) given by equation (4.12) If we substitute this condition in (4.16) we get 1cos( φt) = cos( φt) + √cos 2 φt + ( π + φt)2 − 1, which reduces to the equation tan( φt) = π + φt. (4.17) It is clear, simply by sketching the curves for each side of (4.17), that there is a solu-tion to (4.17) somewhere in the interval (0 , π/ 2). In it was found (by numerical means) with the result φt = 1 .3518168 ... radians, that cos( φt) = 1 /R max = αmin = 0 .2172336 .... Alternatively, the lady can escape even if the man runs 1 /α min = 4 .6033388 ... times as fast as she can row, which is significantly greater than the factor of four given in the Scientific American version of the problem. 84 Chapter 5 Pursuit-Evasion Problem as an Optimal Control Problem 5.1 Basic Concepts Let us assume that two points, one of which we shall call “pursuing” and the other “pursued” or “evading”, are moving in X ⊂ Rn. The motion of each of these points is subject to its own particular system of differential equations with its own particular control parameter. We shall denote the control parameter, the control region, and the trajectory of the motion of the pursuing point by u, U ⊂ Rr, and x (t), respectively. We shall denote these quantities for the pursued point by the symbols v, V ⊂ Rr,and y(t). Let u(t) and v (t) be certain admissible controls (i.e., piecewise continuous), and let x (t) and y(t) be the corresponding trajectories with initial conditions x(0) = x0, y(0) = y0. (5.1) If x (t1) = y(t1) for some t1 > 0, we shall call t1 an encounter time , and the very 85 Chapter 5. Pursuit-Evasion Problem as an Optimal Control Problem occurrence that x (t1) = y(t1) will be referred to as an encounter . Generally speaking, if u(t) and v (t) are chosen arbitrarily, an encounter may not occur for any t > 0. If an encounter does occur, we shall say that u(t) is a pursuing control (for a given control v (t), and for given initial conditions x0 and y0). Even then, for the given x0, y0, v (t), and the chosen control u(t), more than one encounter may take place. We shall call the smallest positive number t1, which is an encounter time, the pursuit time corresponding to the controls u(t) and v (t). We shall denote the pursuit time by Tu,v . In what follows, the initial conditions (5.1) will be assumed to be fixed (in this connection, x0 and y0 do not enter into the notation for the pursuit time). Henceforth, we shall assume that the pursuing point has the following property: for every given control v (t) there exists (for given initial conditions (5.1)) a pursuing control u(t). If the control v (t) of the evading point has been chosen, we can pose the problem of finding a pursuing control u(t) such that the corresponding pursuit time Tu,v takes on a minimal value. We shall assume that there exists, for every admissible control v (t), an admissible control u(t) which brings about the minimum of the pursuit times. We shall denote the minimum by Tv: Tv = min u Tu,v . Furthermore, we shall assume that there exists an admissible control v (t) which brings about the maximum of the values of Tv. We shall denote this maximum by T : T = max v Tv = max v (min u Tu,v ). (5.2) Similarly, T = min u Tu = min u (max v Tv,u ). Moreover, min u (max v Tv,u ) = max v (min u Tu,v ). 86 Chapter 5. Pursuit-Evasion Problem as an Optimal Control Problem The problem consists of finding a pair of admissible controls u(t) and v (t) such that Tu,v = Tv,u = T . Such a pair u(t) and v (t) will be called an optimal pair of controls ; the corresponding pair of trajectories x (t) and y(t) (with initial values (5.1)) will be called an optimal pair of trajectories . Thus, the control u (for a given control v (t)) is to be chosen in such a way that the encounter of the pursuing and pursued points will take place as soon as possible. The choice of the control v, on the other hand, is aimed at putting off the encounter as long as possible. Remark Let us consider the case explained by equation (5.2). Note, that in choosing the control u(t) (which defines the motion of the pursuing point), we shall always assume that the control v (t) for the evading point is known beforehand. In accordance with this fact, in order to determine T , first the minimum with respect to all possible controls u(t) is taken for a certain fixed control v (t), then the maximum with respect to all possible controls v (t) is taken. To solve the given problem, we shall assume that the motion of the pursuing point in X is described by the linear equation (in vector form) dx dt = f (x, u ) ≡ Ax + Bu + c, (5.3) for which the corresponding control region U is a closed, convex, bounded polyhedron in Rr, of the variable u = ( u1, ..., u r). Let the motion of the evading point be described by the equation (in vector form) dy dt = g(y, v, t ) (5.4) and let the corresponding control region V be a set in the s-dimensional space Rs of the variable v = ( v1, ..., v s). We shall assume that the set of all piecewise continuous controls is the class of admissible controls (both for u and for v). We shall impose the usual conditions (continuity in y, v , and t, and continuous differentiability with 87 Chapter 5. Pursuit-Evasion Problem as an Optimal Control Problem respect to the coordinates y1, ..., y n of y) on the coordinates of the vector function g(y, v, t ). To solve the given problem we can use Pontryagin’s maximum principle. We shall introduce two auxiliary vectors ψ = ( ψ1, ..., ψ n), χ = ( χ1, ..., χ n), and two Hamiltonian functions H1(ψ, x, u ) = n ∑ α=1 ψαf α(x, u ) = ( ψ, f (x, u )) ,H2(χ, y, v ) = n ∑ α=1 χαgα(y, v, t ) = ( χ, g (y, v, t )) , corresponding to the pursuing and pursued objects. We can write the following two systems of equations for the auxiliary unknowns ψi and χi with the aid of H1 and H2: dψ i dt = −∂H 1 ∂x i , i = 1 , 2, ..., n, (5.5) dχ i dt = −∂H 2 ∂y i , i = 1 , 2, ..., n, (5.6) Suppose that u(t), x (t), v (t) and y(t) are given. Then, if we substitute these functions in the right-hand sides of systems (5.5) and (5.6), we obtain linear systems in the unknowns ψi and χi. Every solution ψ(t), χ (t) of these systems will be said to correspond to the chosen functions u(t), x (t), v (t), and y(t). The following theorem gives a necessary condition for optimality in the problem under consideration. Theorem 5.1.1 Let u(t) and v(t) be an optimal pair of controls, let x(t) and y(t) be the corresponding optimal pair of trajectories, and let T be the pursuit time. Then, 88 Chapter 5. Pursuit-Evasion Problem as an Optimal Control Problem there exist nontrivial solutions ψ(t) and χ(t) of systems (5.5) and (5.6) which cor-respond to u(t), x(t), v(t), and y(t) such that: 1. the Maximum conditions max u∈U H1(ψ(t), x (t), u ) = H1(ψ(t), x (t), u (t)) , (5.7) max v∈V H2(χ(t), y (t), v ) = H2(χ(t), y (t), v (t)) (5.8) hold for all t, 0 ≤ t ≤ T; 2. At the time t = T, the conditions H1(ψ(T ), x (T ), u (T )) ≥ H2(χ(T ), y (T ), v (T )) , (5.9) ψ(T ) = χ(T ) (5.10) hold. The details of the proof of theorem 5.1.1 are long and involved, the reader can find them in . Although this theorem gives the necessary conditions of optimality for PE problems (and it can be generalized to multiple pursuers and multiple evaders as in ), the fact is that the PE problems studied below can be analyzed directly by more elementary methods. 89 Chapter 5. Pursuit-Evasion Problem as an Optimal Control Problem 5.2 Simple Pursuit in the Plane In the simple pursuit problem (we follow the representation given in ) two players move in the Euclidean plane R2 with simple motion: each has a bound on his speed, but there are no further restrictions (e.g., abrupt directional changes are allowed). One player, the pursuer, wishes to capture the other, the evader, that is, attain perfect coincidence of their terminal positions. Here if α > β holds, for the pursuer’s speed bound α and the evader’s β, then termination is assured in finite time, whatever the initial positions and action of evader. On the other hand, in the case α ≤ β the evader can avoid capture forever from any positions not in contact initially. First, let us discuss briefly some aspects of simple motion for a single player. If the player’s position at time t ∈ R1 is denoted by x(t) ∈ R1, then the velocity vector is ˙x(t) , and the speed \|˙x(t) \|. Thus the dynamical constraint is \|˙x \| ≤ α, and the following holds for the control u:˙x = u; u : R1 → R1, \|u(t)\| ≤ α. (5.11) Hence, x(t) = x(0) + t ∫ 0 u(s)ds for some function u(·) as given above. For simplicity let us place the origin at x(0), so that x(0) = 0 . We want to know, where can the player get to at time t. The constraint on u(·)yields \|x(t)\| ≤ t ∫ 0 \|u(s)\| ds ≤ αt. Any point y with \|y\| ≤ αt can be “attained” by control u: u(s) =  α y \|y\| , for 0 ≤ s ≤ \| y\|;0, for \|y\| < s ≤ t. 90 Chapter 5. Pursuit-Evasion Problem as an Optimal Control Problem Figure 5.1: Simple motion in the plane. Point x moves anywhere within Ax(3) at time t = 3 ; if its position y at t = 1 or z at t = 2 is known, the possibilities are: reduce to Ay(2) or Az (1) . Thus, the attainability set (“reachable set”) can be defined by A0 = {y : \|y\| ≤ αt }.Figure 5.1 shows the simple motion in the plane, according to the defined attainability set rule. Let us return to the game, in the case α > β . If the pursuer’s motion is x: R1 → R1 and the evader’s y : R1 → R1, the equations of motion are ˙x = u, ˙y = v (5.12) for suitable controls u, v : R1 → R2 with all values \|u(t)\| ≤ α, \|v(t)\| ≤ β. At any time denote the player’s distance by r = \|x − y\|. Then r˙r = ddt · 12 \|x − y\|2 = ( x − y)′ · (˙x − ˙y) = ( x − y)′u − (x − y)′v. (5.13) 91 Chapter 5. Pursuit-Evasion Problem as an Optimal Control Problem The natural strategy for the pursuer is to take u = ˙x = α y − x \|y − x\| = αr (y − x). Then, in (5.13), r˙r = −α r2 r − (x − y)′v ≤ − αr + rβ = −(α − β)r by Cauchy’s inequality on the last term. Therefore, as long as r > 0, we have ˙r ≤ − (α − β), r(t) ≤ r(0) − (α − β)t = \|x0 − y0\| − (α − β)t. This shows that capture ( r = 0) for the case α > β must occur at some time T with T ≤ \|x0 − y0\| α − β . (5.14) 92 Chapter 5. Pursuit-Evasion Problem as an Optimal Control Problem 5.3 One-dimensional Rocket Chase Next, we consider a general problem of the rocket chase and we base the discussion on . Two players move on a straight line, the pursuer having a bound on his accele-ration, the evader a bound on his speed. The game ends when the pursuer attains a previously given distance from the evader. There is an obvious solution: the pursuer uses all his capabilities to move toward the evader, who is then captured within a bound time interval. (The precise time bound will depend on the parameters of the game, and on the initial positions.) If x : R1 → R1 describes the pursuer’s motion, and y : R1 → R1 describes the evader’s motion, then the equations of motion are ¨x = u, ˙y = v for admissible u, v : R1 → [−1, 1] . Consider 1 as a bound for both controls, and ε as the evader’s distance, 0 ≤ ε < +∞. Thus, the evader moves on R1 with simple motion, in the sense of what is given in the previous example. The pursuer’s motion is described by x(t) = x(0) + ˙ x(0) t + t ∫ 0 s ∫ 0 u(r)drds = x(0) + ˙ x(0) t + t ∫ 0 (t − s)u(s)ds (5.15) and suggested by the attainability sets in Figure 5.2, where x0 + ˙ x0t − t2 2 ≤ x(t) ≤ x0 + ˙ x0t + t2 2 ,y0 − t ≤ y(t) ≤ y0 + t, since u = v = 1 . 93 Chapter 5. Pursuit-Evasion Problem as an Optimal Control Problem Figure 5.2: Phase portrait of motion in ¨x = u in the x-y plane, where x(t) is given by equation (5.15) for x(0) = 0 , ˙x(0) = 2 ; attainability sets at t = 2 /3, 4/3, 2, 8/3 for the same initial values x(0) and ˙x(0) . The vertex loci are parabolas ˙x = y = ±√2( x + 2) The first-order version of the motion equation is the dynamical equation for the two-player system: ˙x1 = x2, ˙x2 = u, ˙x3 = v. Subsequently the matrix form of this will be treated,  ˙x1 ˙x2 ˙x3  =  0 1 00 0 00 0 0  x1 x2 x3  +  0 u 0  +  00 v  94 Chapter 5. Pursuit-Evasion Problem as an Optimal Control Problem Figure 5.3: Trajectories of ˙x = y −v, ˙y = u in the x-y plane with u = v = ±1 outside target \|x\| ≤ ε the termination condition \|x − y\| ≤ ε translating to (1 , 0, −1)  x1 x2 x3  ∈ [−ε, ε ]. Thus, the natural phase space is R3. This can be reduced to R2 by introducing new variables x = x1 −x3, y = x2. The resulting equations and termination condition 95 Chapter 5. Pursuit-Evasion Problem as an Optimal Control Problem Figure 5.4: Trajectories of ˙x = y − v, ˙y = u in the x-y plane. From point a evader mistakenly chooses v = −1, but reverses his choice at b; capture occurs at c (later than it would have occurred at d) are ˙x = y − v, ˙y = u; \|x\| ≤ ε. (5.16) Let us assume (for a preliminary orientation), that both players’ controls are constant on some time interval. The differential equation for the trajectories, with dy dx = ˙y ˙x, 96 Chapter 5. Pursuit-Evasion Problem as an Optimal Control Problem is (y − v) dy dx = u, (y − v)2 = 2 ux + const. (5.17) The point then moves along the parabola, upward for u > 0 and downward for u < 0, (see Figure 5.3 with u = v = 1 in the left half plane, and u = v = −1 in the right). Now let us suppose that, at some point to the left of the target, the evader chooses a control other than v = 1, and pursuer sticks with u = 1 . The motion then proceeds along another parabola (with axis y = v, see Figure 5.4 for v = −1). Therefore, capture, even with ε = 0, can be ensured from all initial positions, for example, by taking u = 1 quite indifferent to evader’s action. 97 Chapter 5. Pursuit-Evasion Problem as an Optimal Control Problem 5.4 Pursuit on a Sphere (Kelley’s game) Another example is given by pursuit on a sphere, originally formulated by Kelly. We follow the presentation of . Two players move on the two-sphere S2 in R3, each with a fixed bound on his speed. The game ends at the coincidence of positions. Here the idea is that “in a dogfight, the planes tend to move in a circular fashion”. The simplification does away with one significant aspect of actual combat: that the roles of the pursuer and the evader are not fixed, but may well switch back and forth. The outcome is similar to the simple motion problem (section 5.2). Let us denote the pursuer’s speed bound by α, the evader’s by β. If α > β , the pursuer can force termination from any initial position, within a bounded time interval. In the case α < β the evader can avoid capture at all times t > 0 (and the stand - off situation α = β is rather too sensitive to details in the specification of the players’ strategies). Let us talk about these different games. In the case α > β , first assume that the players are not at diametrically opposite points initially. Then there is the unique shortest arc γ of a great circle joining their positions. By a parallel shift along γ,move a neighborhood of the evader’s position to the pursuer’s (this “action at a distance” serves to identify the control of the evader). The pursuer then uses the control u = v + w, where the first component neutralizes the evader’s action, the second, w with magnitude \|w\| = α − β in the direction of γ, serves to decrease the player’s distance (at a rate α − β, see section 5.3) until capture occurs. If their initial positions are opposite, then any constant control u with \|u\| = α > β , applied over a short interval, will achieve non-opposing positions. By a like reasoning, in the case α < β the evader can maintain forever an initial distance from the pursuer. The idea is probably clear enough, and will apply equally well to simple pursuit on 98 Chapter 5. Pursuit-Evasion Problem as an Optimal Control Problem an n - dimensional Riemannian manifold (thus, the “diametrically opposite points” would be replaced by conjugate points). Consider the motion of a single player over the unit sphere Sn−1 of Rn. If its motion is described by x : R1 → Rn, x(0) ∈ Sn−1,then x(t) will remain on Sn−1 iff \|x(t)\|2 is constant, i.e., x is perpendicular to ˙ x = dx dt .Further, the motion will be “simple” if the only further dynamical restriction is a magnitude bound on ˙ x. We wish to express this as a relation between ˙ x and suitable arbitrary controls u. Lemma 5.4.1 (for a broader discussion see ) For any points a 6 = b on Sn−1 there is a mapping x, y 7 → E(x, y ), defined for x, y near a, b and with (n, n − 1) matrices as values, analytic in the coordinates of x, y and such that x′E(x, y ) = 0 , E′(x, y )E(x, y ) = In−1, (5.18) y′E(x, y ) = \| sin ϕ\|(1 , 0, ..., 0) , (5.19) where ϕ is the angle between x and y. Proof By assumption, the vectors a, b are independent, so that there is a basis for Rn of the form a, b, c 3, ..., c n. Then x, y, c 3, ..., c n (5.20) remain independent if x, y are close enough to a, b . Apply the Gram-Schmidt orthogo-nalization process to the sequence (5.20), obtaining orthonormal vectors e1, e 2, ..., e n.Since \|x\| = 1, we have e1 = x, and, in the second step, e2 = y − (x′y)x \|sin ϕ\| , (5.21) since \|y − (x′y)x\|2 = 1 − (x′y)2 = 1 − cos 2 ϕ. 99 Chapter 5. Pursuit-Evasion Problem as an Optimal Control Problem Collect the column vectors e2, ..., e n into the (n, n - 1) matrix E(x, y ). Equation (5.18) holds since ( x, E (x, y )) is orthonormal. The first coordinate of y′E(x, y ) is y′e2 = \| sin ϕ\| from (5.21). The remaining coordinates are 0, since ek is perpendicular to both e1 = x and e2, and hence to y also. This completes the proof. Corollary 5.4.2 On the neighborhood of any point on Sn−1 there is an analytic mapping x 7 → E(x), whose values are ( n, n − 1) matrices, and x′E(x) = 0 , E′(x)E(x) = In−1, E(tx ) = E(x) f or t > 0. (5.22) Proof The corollary will follow on taking y = b 6 = ±a, and defining E(x) = E(x, b ). Positive homogeneity is ensured by extending E(·) in the obvious manner, namely E(tx ) = E(x) for t > 0. Returning to Kelly’s game (actually, for dimension’s n ≥ 2), we may choose the state space description ˙x = E(x, y )u, ˙y = E(y, x )v. The control values in Rn−1 are constrained by \|u(t)\| ≤ α, \|v(t)\| ≤ β. The initial positions are on the unit sphere. If ϕ is the angle between x, y , and r = \| sin ϕ\|, then r ˙r = 12 ddt sin 2 ϕ = 12 ddt (1 − (x′y)2)= −(x′ ˙y + y′ ˙x) = −x′E(x, y )v − y′E(x, y )u. (5.23) Write u = −v + w with w ∈ Rn−1 to be chosen subsequently, subject to \|w\| ≤ α − β.Then, using (5.19), r ˙r = ( −x′E(y, x ) + y′E(x, y )) v − y′E(x, y )w = 0 − \| sin ϕ\|(1 , 0, ..., 0) w ≤ − r(α − β) (5.24) on taking w′ = ( α − β)(1 , 0, ..., 0). Thus, ˙ r ≤ − (α − β) < 0, and sin ϕ = 0 can be attained in finite time, i.e., the capture is in finite time. 100 Chapter 6 Conclusions In this thesis we studied a family of mathematical problems known as pursuit-evasion problems (PE). We presented PE problems within the classes of pursuit problems, evasion problems, and pursuit-evasion problems. We restricted the discussion to the deterministic approach of PE problems, and stated PE problems as optimal cont-rol problems. Because of that, we formulated the general optimal control problem, and discussed the two main approaches to solve this problem, namely, Pontryagin’s maximum principle (Theorems 2.2.1, 2.2.2) (where we introduce a hamiltonian func-tion) and Bellman’s method of dynamic programming (equation (2.29)) (where we gave a simple example from dynamic programming to explain the main ideas of the method). To compare the two solutions we provided an example, where Pontrya-gin’s principle applied, but Bellman’s failed because the control was discontinuous (Bang-Bang Problem ) . Thus, we proved that the assumption on the continuous differentiability of the functional (2.6), minimizing the transition time from the ini-tial point to some other given terminal point, did not hold in the simplest cases. Therefore, we showed that in general Bellman’s consideration yields a good heuristic method, rather than a mathematical solution of the problem. 101 Chapter 6. Conclusions Since there are different formulations of PE problems, we stated the definition of the PE problem from the point of view of the pursuer, the evader, and both. For the pursuit problems, we presented some main examples, namely, the Pierre Bouguer’s pursuit problem, the wind-blown plane problem, the tractrix, and the Apollonius pursuit problem. In the Pierre Bouguer’s pursuit problem, which we explained as a pure pursuit problem, we treated the case of a pirate ship pursuing a merchant vessel, and determine the equation of the the trajectory of the pirate ship (the pursuer), called the line of pursuit. We identified when did the capture occur, and talked about the cases where the pirate ship was slower than the merchant vessel (no capture), and where the pirate ship was faster than the vessel (there was a capture, and we calculated the total distance travelled by the pirate ship until its capture of the merchant vessel for that case). It was also interesting for us to find out that in the case where the pirate ship and the merchant vessel had equal speeds, the pursuit became a vertically upward tail chase, since the pirate ship pulled into behind of the merchant vessel. In the other important example of the pursuit problem, we talked about the pursuit of a stationary target, namely, the problem where the plane went to the city C due west of his starting point, and the wind blew from the south. Here we identified that for the no wind case (when the ration of the wind’s speed and the plane’s speed is zero) the plane moved directly to the city C while always remaining on the x-axis. For the case where the wind and the plane had equal speeds we presented the plane’s path and showed the different situation that happened according to the initial position of the plane. Also, when the ratio of the speeds was less that one, the plane did reach the city, and we calculate the time of the trip. The other pursuit curve we talked about was the tractrix, which we showed got its name because of its path looking as the following curve, or the tailing curve. We compared the tractrix with Bouguer’s pure pursuit curve for the special case of equal 102 Chapter 6. Conclusions speeds for the pirate ship and the merchant vessel. Thus, we realized that the results were quite different, since for the tractrix the constant lag was always the case, while the constant lag of the pirate ship was an asymptotic property that developed with the passage of time. Then, we presented Apollonius pursuit problem, where we discussed a method of interception, which is of great interest for the PE problems. We provided an example of a torpedo (T) trying to pursuer an enemy ship (E). Here we identified the three main cases of the location of T and E. We defined the set S of all the points in the plane so that the interception can be obtained. This set S is known as the Apollonius circle, and it is broadly used in the PE problems for analyzing how to find a better strategy to escape or prolong the capture time whenever a successful escape is not possible. Again, we talked about the cases of the speed differences, namely, for the fast torpedo we determined that the interception would occur, for the slow torpedo the interception might or might not occur. Next, we defined the evasion problem and provided one of the general examples of those problems, called the Isaac’s guarding the target problem, where we had P guarding the target area C from attack by E. We formulated the military conception of this problem, and identified the equation that gave us the evasion curves. More-over, we showed how to determine the minimum value of the amount of explosive (that E carries) required for success in destroying the target area C, as a function of both E’s starting point and the location of the target. After that we provided a problem about the lady in the lake and the man who was trying to track her down. We talked about the strategies the lady had to have in order to escape the man, and explained each of the possible cases. Here we defined another interesting definition, known as a go-for-broke circle which we got by formulating the lady’s strategies. 103 Chapter 6. Conclusions Finally, we formulated the PE problem, where there were both objects of the PE game, the pursuer P and the evader E. We stated the necessary conditions of optimality for those PE problems, which were similar to Pontryagin’s maximum principle presented before. We provided examples of a simple pursuit in the plane, that we used as an example of the simple motion of P and E in the plane. We explained, what did we mean by the reachable set for this case, and compared the dynamical constraints of both P and E. In the one-dimensional rocket chase problem we presented the solution that could be used in other PE problems as an example of the problem, where the game ended when the pursuer attained a previously given distance from the evader. Here we discussed the different strategies the two objects had to pursuer there goal. Kelly’s game is an example of a pursuit on a sphere, where the idea is that in a dogfight, the planes tend to move in a circular fashion. We expressed the relation between the objects’ speeds and there controls. The outcome was important, and could be used as a simple motion of the PE problem on a sphere. 104 References Barton, J.C. and Eliezer, C.J., On Pursuit Curves , Journal of Australian Math-ematical Society, Ser. B 41, pp. 358-371 (2000) Bellman, R.E. and Dreyfus, S.E., Applied Dynamic Programming , Princeton Univeristy Press, Princeton, NJ (1962) Bellman, R.E., Dynamic Programming , Princeton University Press, Princeton, NJ (1957) Boole, G., Treatise on Differential Equations ,Macmillan, Cambridge (1859) Bouguer, P., Sur les Courbes de Poursuites Histoire de l’Acad´ emie Royale des Sciences, Ann´ ee M.DCCXXXII, Avec les M´ emoires de Math´ ematique et de Physique, France (1735) Breitner, M.H communicated by Berkovitz, L.D., Historical Paper, The Genesis of Differential Games in Light of Isaacs’ Contributions , Journal of Optimization Theory and Applications, Vol. 124, No. 3, pp. 523-559 (March 2005) Chikrii, A.A., Prokopovich P.V., Pursuit and evasion problem for interacting groups of moving objects , Cybernetics and Systems Analysis, Volume 25, Num-ber 5, pp. 634-640 (1990) Colman, W.J.A., A Curve of Pursuit , Bulletin of the Institute of Mathematics and Its Application 27 (3), pp.45-47 (March 1991) Dreyfus, S.E., Dynamic Programming and the Calcuclus of Variations , Aca-demic Press, NY (1965) Gamkrelidze, R.V., Discovery of The Maximum Principle in Optimal Control ,Mathematics and War (Eds.) VIII, Springer, NY, pp. 160-173 (2003) 105 References Gardner, M., Mathematical Games , Scientific American Columns, Mathematical Carnival, NY, (1975) H´ ajek, O., Pursuit Games , Academic Press, NY (1975) Ioffe, A.D., Tihomirov V.M., Theory of Extremal Processes , North Holland Pub-lishing Company, NY (1979) Isaacs, R., Differential Games , Dover Publications, NY (1999) Kamien Morton I., Schwartz Nancy L., Dynamic Optimization: The Calculus of Variations and Optimal Control in Economics and Management, Series Volume 4, North-Holland Publishing Company, NY (1981) Koo, D., Elements of Optimization With Applications in Economics and Busi-ness , Springer, NY (1977) Krasovskii, N.N, Subbotin, A.I., Game-Theoretical Control Problems , Springer, NY (1988) Lawrence J. Dennis, A catalog of special plane curves , Dover Publications, NY (1972) Liberzon D., Switching in Systems and Control , Birkh¨ auser, Boston (2003) Nahin, Paul J., The Mathematics of Pursuit and Evasion , Princeton University Press, Princeton, New Jersey (2007) Pierre, Donald A., Optimization Theory with Applications , Dover Publications, NY (1986) Pontryagin, L.S., Boltyanskii, V.G. Gamgrelidze, R.V., Mishchenko, E.F., The Mathematical Theory of Optimal Processes , The Macmillian Company, NY (1964) Pontryagin, L.S., Boltyanskii, V.G., Gamkrelidze, R.V., and Mishchenko, E.F., L.S.Pontryagin, Selected Works, Volume 1, Selected Research Papers Gordon and Breach Science Publishers, NY (1986) Pontryagin, L.S., Boltyanskii, V.G., Gamkrelidze, R.V., and Mishchenko, E.F., L.S.Pontryagin, Selected Works, Volume 4, The Mathematical Theory of Opti-mal Processes Gordon and Breach Science Publishers, NY (1986) Pshenichnyi B. N., Shishkina, N. B., Pursuit problems with two moving objects ,Cybernetics and Systems Analysis, Volume 25, Number 4, Springer, NY, pp. 464-471 (2005) 106 References Puckette, C.C., The Curve of Pursuit , The Mathematical Gazette, Volume 37, pp. 256-260 (1953) Ruckle, W.H., Geometric Games of Search and Ambush , Mathematics Maga-zine, Volume 52, Number 4, pp. 195-206 (September 1979) Saaty, Thomas L., Alexander, Joyce M., Thinking with Models , Pergamont Press, NY (1981) Schuurman, W., Lodder, J., The Beauty, the Beast, and the Pond , Mathematics Magazine, Volume 47, Number 2, pp. 91-93 (March-April 1974) Young, L.C., Lectures on the Calculus of Variations and Optimal Control The-ory , W.B. Saunders Company, PA (1969) Zermelo, E., ¨Uber das Navigationsproblem bei ruhender oder ver¨ anderlicher Windverteilung , Zeitschr. f. angew. Math. u. Mech. 11, pp. 114-124 (1931) Zhukovskiy, V.I., Salukvadze, M.E., The Vector-Valued Maximum , Academic Press, Princeton, San Diego (1993) 107
15030	https://mathworld.wolfram.com/TrirectangularTetrahedron.html	Trirectangular Tetrahedron Download Wolfram Notebook \| \| \| --- \| \| \| \| A tetrahedron having a trihedron all of the face angles of which are right angles. The face opposite the vertex of the right angles is called the base. If the edge lengths bounding the trihedral angle are , , and , then the side lengths of the base are given by , , and , and so has semiperimeter \| \| \| --- \| \| \| (1) \| The volume of the trirectangular tetrahedron is \| \| \| --- \| \| \| (2) \| Using Heron's formula, the surface area is therefore \| \| \| --- \| \| \| (3) \| Let be the area of the triangle with vertices , , and . The remarkable de Gua's theorem \| \| \| --- \| \| \| (4) \| then follows from the identity \| \| \| --- \| \| \| (5) \| with defined by (1). See also de Gua's Theorem, Tetrahedron, Trihedron Explore with Wolfram\|Alpha More things to try: trirectangular tetrahedron Busy Beaver 4-state 2-color get a total greater than 45 with 5 12-sided dice References Altshiller-Court, N. "The Trirectangular Tetrahedron." §4.6a in Modern Pure Solid Geometry. New York: Chelsea, pp. 91-94, 1979. Referenced on Wolfram\|Alpha Trirectangular Tetrahedron Cite this as: Weisstein, Eric W. "Trirectangular Tetrahedron." From MathWorld--A Wolfram Resource. Subject classifications Find out if you already have access to Wolfram tech through your organization
15031	https://artofproblemsolving.com/wiki/index.php/Angle?srsltid=AfmBOorbsqj3DoX4841yo9sY_cfwTPbVZh4D0_lZSXYumfY8xta5lsfW	Art of Problem Solving Angle - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Angle Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Angle Contents [hide] 1 Overview 2 Angle Measure 3 Classifying Angles 4 Angle Chasing 4.1 Properties Used in Angle Chasing Overview An angle is the union of two rays with a common endpoint. The common endpoint of the rays is called the vertex of the angle, and the rays themselves are called the sides of the angle. There are many notations for angles. The most common form is , read "angle ABC", where are points on the sides of the angle and is the vertex of the angle. Note that the same angle can be denoted many different ways by choosing different points along the side of the angle. If there is no ambiguity, this notation can be shortened to simply . Angle Measure The measure of is denoted , read "measure of angle ABC". There are different units for measuring angles. The three most common are degrees, radians and gradians. If two angles are congruent, they have the same angle measure. A ray drawn from the vertex of the angle, such that the angle formed by this ray and one of the sides is congruent to the angle formed by this ray and the other side, is called the angle bisector. Classifying Angles A straight angle is an angle formed by a pair of opposite rays, or a line. A straight angle has a measure of . A right angle is an angle that is supplementary to itself. A right angle has a measure of . An acute angle has a measure greater than zero but less than that of a right angle, i.e. is acute if and only if . An obtuse angle has a measure greater than that of a right angle but less than that of a straight angle, i.e. is obtuse if and only if . A reflex angle is an angle with measure greater than a straight angle, but less than 360 degrees, or radians. Angle Chasing Angle chasing is a technique where solvers apply angle properties determine the measures of unknown angles. It is commonly used in geometry problems. Usually this can use a variety of ways including circles, new lines, or transforming the figure somehow. Lots of angle chasing problems require you to think intuitively. Properties Used in Angle Chasing Two angles that are complementary add to . Two angles that are supplementary add to . Supplementary angles can be found when two lines intersect each other. Vertical angles are congruent to each other. Parallel lines can create equal or supplementary angles. An angle bisector splits an angle into two congruent angles. For instance, if is bisected by , then . If side lengths are known, the angle bisector theorem can be used to determine that a line bisects an angle. If angles are founded in a polygon, one can use angle formulas to find the unknown angle. If two polygons are congruent, corresponding angles are congruent. If angles are found in a circle, one can use angle properties and arc measure. Finding cyclic quadrilaterals can be a useful strategy in angle chasing since angles opposite with each other are supplementary. Retrieved from " Categories: Geometry Angles Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
15032	https://pmc.ncbi.nlm.nih.gov/articles/PMC3348079/	Spiral cleavage and early embryology of a loxosomatid entoproct and the usefulness of spiralian apical cross patterns for phylogenetic inferences - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice BMC Dev Biol . 2012 Mar 29;12:11. doi: 10.1186/1471-213X-12-11 Search in PMC Search in PubMed View in NLM Catalog Add to search Spiral cleavage and early embryology of a loxosomatid entoproct and the usefulness of spiralian apical cross patterns for phylogenetic inferences Julia Merkel Julia Merkel 1 Johannes Gutenberg University, Institute of Zoology, 55099 Mainz, Germany Find articles by Julia Merkel 1, Tim Wollesen Tim Wollesen 2 University of Vienna, Dept. of Integrative Zoology, Althanstrasse 14, 1090 Vienna, Austria Find articles by Tim Wollesen 2, Bernhard Lieb Bernhard Lieb 1 Johannes Gutenberg University, Institute of Zoology, 55099 Mainz, Germany Find articles by Bernhard Lieb 1, Andreas Wanninger Andreas Wanninger 2 University of Vienna, Dept. of Integrative Zoology, Althanstrasse 14, 1090 Vienna, Austria Find articles by Andreas Wanninger 2,✉ Author information Article notes Copyright and License information 1 Johannes Gutenberg University, Institute of Zoology, 55099 Mainz, Germany 2 University of Vienna, Dept. of Integrative Zoology, Althanstrasse 14, 1090 Vienna, Austria ✉ Corresponding author. Received 2011 Oct 20; Accepted 2012 Mar 29; Collection date 2012. Copyright ©2012 Merkel et al; licensee BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License ( which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. PMC Copyright notice PMCID: PMC3348079 PMID: 22458754 Abstract Background Among the four major bilaterian clades, Deuterostomia, Acoelomorpha, Ecdysozoa, and Lophotrochozoa, the latter shows an astonishing diversity of bodyplans. While the largest lophotrochozoan assemblage, the Spiralia, which at least comprises Annelida, Mollusca, Entoprocta, Platyhelminthes, and Nemertea, show a spiral cleavage pattern, Ectoprocta, Brachiopoda and Phoronida (the Lophophorata) cleave radially. Despite a vast amount of recent molecular phylogenetic analyses, the interrelationships of lophotrochozoan phyla remain largely unresolved. Thereby, Entoprocta play a key role, because they have frequently been assigned to the Ectoprocta, despite their differently cleaving embryos. However, developmental data on entoprocts employing modern methods are virtually non-existent and the data available rely exclusively on sketch drawings, thus calling for thorough re-investigation. Results By applying fluorescence staining in combination with confocal microscopy and 3D-imaging techniques, we analyzed early embryonic development of a basal loxosomatid entoproct. We found that cleavage is asynchronous, equal, and spiral. An apical rosette, typical for most spiralian embryos, is formed. We also identified two cross-like cellular arrangements that bear similarities to both, a "molluscan-like" as well as an "annelid-like" cross, respectively. Conclusions A broad comparison of cleavage types and apical cross patterns across Lophotrochozoa shows high plasticity of these character sets and we therefore argue that these developmental traits should be treated and interpreted carefully when used for phylogenetic inferences. Keywords: Lophotrochozoa, Embryology, Development, Ontogeny, Evolution, Phylogeny, Spiral cleavage, Molluscan cross, Annelid cross Background Currently, bilaterian animals are subdivided into four major groups: the supposedly basal Acoelomorpha, the Ecdysozoa (combining all molting animals such as arthropods and nematodes), the Lophotrochozoa with a trochophore-like ciliated larva (e.g., Annelida, Entoprocta, Mollusca, Platyhelminthes), and Deuterostomia (including chordates, hemichordates and echinoderms) [1-4]. Despite ongoing efforts, the interrelationships of the phyla that nest within the Lophotrochozoa remain unresolved [5,6]. Entoprocta is a phylum that has been proposed to belong to a clade of spirally cleaving animals, the so-called Spiralia, which together with its suggested sister group, the Lophophorata (Ectoprocta, Brachiopoda, and Phoronida), forms the Lophotrochozoa . Typically, entoprocts are microscopic, mostly marine, sessile metazoan animals. Its approximately 150 hitherto described species are divided into four subgroups, the solitary (and supposedly basal) Loxosomatidae and the colonial Barentsiidae, Pedicellinidae, and Loxocalypodidae . Their adult gross morphology is characterized by a ciliated tentacle crown, which surrounds both the mouth and the anus. The calyx houses the reproductive organs, mostly one pair of protonephridia, and the cerebral ganglion. Entoprocts reproduce asexually by budding, as well as sexually, whereby two major larval types can be recognized, namely the creeping, supposedly basal, lecithotrophic and the more common swimming, planktotrophic larval type . Metamorphosis is very complex and often involves settlement and adhesion with the frontal body region to the substrate as well as rotation of the gut . Morphological and molecular analyses have proposed several phylogenetic hypotheses concerning entoproct interphyletic relationships. Traditionally, Entoprocta and Ectoprocta have been comprised to form the monophyletic Bryozoa (Bryozoa-concept), based on a metamorphosing larval stage with a completely retracted and cavity-enclosed prototroch as well as additional common features during metamorphosis [8,9]. This hypothesis has been revived by a recent molecular study , although subsequent analyses of partly the same authors are far less clear . The cryptic Cycliophora, one of the most recently erected phyla , have also argued to be associated with Entoprocta and Ectoprocta, notably as a monophyletic assemblage termed "Polyzoa" [13,14], while other authors suggest a sister group relationship of Cycliophora and Entoprocta alone [15,16]. On the contrary, the recently proposed Tetraneuralia-concept has strengthened the so-called Lacunifera- or Sinusoida- hypothesis, suggesting a monophyletic assemblage of Entoprocta and Mollusca based on numerous larval and adult autapomorphies [4,17-19]. Resembling a mosaic of larval and adult molluscan characters, the entoproct creeping-type larva shares a number of morphological traits with the polyplacophoran trochophore, including a highly complex apical organ with eight centrally located flask-shaped and several peripheral cells, as well as a typical molluscan-like tetraneurous condition of longitudinal nerve cords [4,19-21]. Additional shared characters are, among a total set of nine, the distinct creeping foot, a large pedal gland, frontal cirri, and a ventrally intercrossing dorsoventral musculature [4,19,20]. Despite the spiral cleavage pattern, which has traditionally been used to unite polyclad flatworms, nemerteans, annelids, and molluscs as "Spiralia" , other developmental characters, such as the cellular arrangement into an "apical cross pattern" during early embryogenesis, have been used to infer protostome interrelationships. For a long time, only two cross patterns had been clearly defined, namely the molluscan and the annelid cross, respectively. Since a seemingly "molluscan-type" cross pattern had also been reported for sipunculans, a close relationship to molluscs was suggested . Recently, additional cross patterns, such as a nemertean cross, have been described . For entoprocts, a spiral cleavage pattern has been mentioned in the literature and is often referred to in textbooks, but its documentation is restricted to only a few sketch drawings [25-27]. Apical cross patterns, which would be expected for a spirally cleaving taxon, have not been reported by these studies. In order to fill the significant gaps in knowledge concerning entoproct early embryology, we herein describe the development of a representative of the supposedly basal entoproct genus Loxosomella by applying immunochemistry and confocal microscopy. Using our detailed description of the early cleavage pattern, we also clarify whether or not a distinct "cross pattern" is present in this species. These data are discussed with those available for other lophotrochozoans in order to assess their suitability for phylogenetic inferences. Methods Animals and fixation Populations of an undescribed, brooding loxosomatid entoproct belonging to the genus Loxosomella(Claus Nielsen, pers. comm.) were collected in July 2007 from tubes of the maldanid polychaete, Axiothella rubrocincta, which inhabits the intertidal mud flats of False Bay, San Juan Island, USA. Up to 20 embryos of all developmental stages can be found in one mother individual. From approximately 100 mother animals, we analyzed four individuals each for the 1-, 2-, 4- to 5-, and 8-cell stages, seven embryos that had between eight and 16 cells, and one 21-cell embryo. Approximately 50 embryos were found at the apical rosette stage and three gastrulae were investigated. Numerous swimming-type larvae - obviously shortly before release - were retrieved, indicating the healthy condition of the adults and their developmental stages. The results obtained were highly consistent among the individuals of each developmental stage. Accordingly, the spiral-type cleavage pattern was found in all embryos investigated, although the shape (spherical versus more oval) varied between individuals. In order to remove the embryos from the mother animal, the tentacle crown was cut open using insect needles. Prior to fixation, adult individuals carrying embryos were relaxed in 7% MgCl 2 and removed from the polychaete tubes. Fixation was carried out using a solution of 4% paraformaldehyde (PFA) in 0.1 M phosphate buffered saline (PBS) for 1 h at room temperature. After fixation, the specimens were washed in 0.1 M PBS (3 × 15 min) and stored in 0.1 M PBS containing 0.1% NaN 3 at 4°C. A reference specimen, determined by Dr. Claus Nielsen (Copenhagen) is deposited at The Natural History Museum of Denmark, Copenhagen (registration tag ZMUC-ENT-27). In addition, the COI gene of our study specimen was partially sequenced and the 710 bp sequence deposited in GenBank (acc. # JQ614997). Immunocytochemistry, data generation, and analysis After fixation and storage, the embryos were permeabilized in 0.1 M PBS with 0.1% Triton X-100 (PBT) for 1 h. F-actin was labeled using a 1:20 dilution of Alexa Fluor 488 phalloidin (Invitrogen, Molecular Probes, Eugene, OR, USA) in PBT. For nucleic acid staining, 10% DAPI (Invitrogen, Taastrup, Denmark) was added and the samples were incubated for 20-24 h at room temperature. Then, the embryos were washed in 0.1 M PBS (3 × 15 min) and mounted in Fluoromount G (Southern Biotech, Birmingham, AL, USA) on glass slides. 5 μl of DAPI (Invitrogen) was added to the embedding medium in order to enhance the signal strength of the nucleic acid staining. The samples were examined using a Leica DM RXE 6 TL fluorescence microscope equipped with a TCS SP2 AOBS laserscanning device (Leica Microsystems, Wetzlar, Germany). The optical sections had a Z-step size of 0.45-0.55 μm. Resulting stacks were merged into projection images with greater focal depth. 3D reconstructions were generated from the confocal datasets using the image processing software Imaris 5.7.2 (Bitplane AG, Zürich, Switzerland). Results General aspects of entoproct reproduction The brood pouch of the loxosomatid species investigated herein is located in the calyx and contains up to 20 embryos (Figure 1). Every embryo is surrounded by a thin membrane. Each membrane tapers in a strand which is connected to the strands of other embryos. Later embryonic stages are located in the anterior region of the brood chamber. The earliest developmental stages are found in the posteriormost part of the brood pouch. Due to the high yolk content, early embryos appear opaque and non-transparent. Released larvae are planktonic and of the swimming-type with a weakly developed foot sole posterior to the prototroch (see ). Figure 1. Open in a new tab Brooding Loxosomella sp. with embryos (arrowheads) located in the calyx. bd, bud; ca, calyx; ft, foot; gl, foot gland; st, stalk; sto, stomach. Total length of the animal is approximately 1 mm. Cleavage and gastrulation in Loxosomella sp Fertilized eggs form a polar body which appears at the animal pole of the embryo (Figure 2A-C). In two-cell stages, both cells are equal in size and show two polar bodies, one on the animal pole and one shifted by approximately 90° relative to the first one (Figure 2D; one polar body is obscured in the 3D reconstructions shown in 2E and F). Second cleavage results in four cells with three polar bodies. Since all cells are approximately equal in size, an assignment of individual cells ("macromeres") to specific quadrants appears difficult. Shortly after second cleavage, a fifth cell (1q) is already present and demonstrates the asynchronicity of early cleavage in our study species (Figure 2G-I). After third cleavage, eight cells have formed and only two polar bodies can be observed. Applying the nomenclature of Conklin (1897) , third cleavage results in the "macromere" quartet 1Q and the "micromere" quartet 1q. All cells of the first micromere quartet are equal in size and of approximately the same size as the macromere cells (Figure 2J-L). The blastula of the 21-cell stage is flattened and ellipsoid-shaped. All four macromeres (red cells, Figure 2N, O) are equal in size and nearly twice as large as the four micromeres (blue cells, Figure 2N, O), each situated above the cleavage furrow of two macromeres. A second micromere quartet (light blue cells, Figure 2N, O) is located on top of these micromeres, with similar cell size as the latter (Figure 2M-O). A third quartet of micromeres (purple cells, Figure 2N, O) rests upon the macromeres. Its cells are slightly larger than the micromeres of the other two quartets. A single cell of a fourth micromere quartet (yellow cell, Figure 2N, O) is situated close to the animal pole (Figure 2M-O). At the 36-cell stage, the embryo exhibits four additional, smaller cells, which form an apical rosette on the animal pole. This apical rosette directly faces the macromere quartet on the vegetal side of the embryo (Figure 3B, C). In the 43-cell stage (Figures 3D-F, 4A) and the 51-cell stage (Figure 3G-I), respectively, cells surrounding the apical rosette are arranged in two different cross-shaped patterns (Figures 3D, E, G, 4A). The first cross pattern results from interconnecting rosette cells which lie opposite to each other, resulting in a pattern that closely resembles a molluscan cross. The second pattern is formed by the peripheral rosette cells and appears very similar to a typical "annelid cross" (Figures 3D, E, G, 4A). Overall, cleavage in the Loxosomella species investigated herein can be characterized as holoblastic, asynchronous, equal, and spiral. The gastrula stage elongates somewhat along the animal-vegetal axis and shows bilateral symmetry (Figure 3J-L). It consists of approximately 100-110 cells. A blastopore was slightly visible as a small vent on the vegetal side of the 107-cell stage (Figure 3J-L). Figure 2. Open in a new tab Confocal micrographs and 3D reconstructions of early cleavage stages of Loxosomella sp. Scale bars: 10 μm. left column: Nucleic acid staining (blue). A, C, F, I, L, O: lateral view; B, D, E, G, H, J, K, N: animal view; M: vegetal view. Polar bodies are indicated by arrowheads. Middle and right column: 3D reconstructions, middle column: animal view, right column: lateral view. grey: polar bodies, red: "macromere" quartet cells (nQ cells); purple, blue, light blue, green, yellow: "micromere" quartet cells (nq n cells). A-C: Fertilized oocyte prior to first cleavage. One polar body is present on the animal side of the embryo. D-F: Two-cell stage. A second polar body appears shifted by 90° relative to the first polar body. The macromeres are equal in size. Second polar body obscured by the red cells in Eand F. G-I: Five-cell stage. Three polar bodies are present. Cleavage is asynchronous and the fifth cell (purple) lies between two Q cells (red) in a cleavage furrow. J-L: Eight-cell stage. Two polar bodies are present and located next to the vegetal, slightly unequal 1Q cells (red). Each cell of the first "micromere" quartet 1q (purple) is located in a cleavage furrow of two 1Q cells. M-O: 21-cell-stage. Cleavage is asynchronous, cells are different in size. Figure 3. Open in a new tab Confocal micrographs and 3D reconstructions of blastula and gastrula stages of Loxosomella sp. "Apical cross patterns" are indicated by white and red lines in D and G. Apical rosette cell nuclei are marked with asterisks in D, E, G, J and vegetal "macromere" quartet cells by double arrowheads in F. Scale bars: 10 μm. A, D, G: Nucleic acid staining (blue). E, F: Nucleic acid (blue) and F-Actin staining (red). J: Nucleic acid staining shown as depth-coded confocal projection. A: vegetal view, D, E, G, J: animal view, F: vegetal view. B, C, H, I, K, L: 3D reconstructions. B, H, K: animal view, C, I, L: lateral view. red: "macromere" quartet (i.e. vegetal) cell nuclei; green: (derivatives of) apical rosette cell nuclei; purple: other cell nuclei. A-C: 36-cell stage. D-F: 43-cell stage. D: Cells surrounding the apical rosette (asterisks) show both, a molluscan- and an annelid-like cross pattern. G-I: 51-cell stage. J-L: Gastrula stage (107 cells). J: Derivatives of apical rosette cells lie in a lower plane than the surrounding cells. K, L: Gastrulation. Vegetal cells (red), blastocoel (arrow). Figure 4. Open in a new tab Spiralian apical cross cell patterns based on several authors as well as the data on Loxosomella presented herein. Dark grey: apical rosette cells, grey: periphere rosette/"annelid cross" cells, light grey: "molluscan cross" cells. A: 43-cell stage of Loxosomella sp (this study). B. Approximately 64-cell stage of the aplacophoran mollusc Epimenia sp. (after ). C: Approximately 62-cell stage of the polyplacophoran mollusc Stenoplax heathiana(= Ischnochiton magdalenensis; after , pl. 2, figure. 17). D: 58-cell stage of the gastropod mollusc Patella caerulea(after , figure. I, 7). E: 64-cell stage of the nemertean Carinoma tremaphoros(after , figure. 6 J). F: Approximately 64-cell stage of the polyclad flatworm Maritigrella crozieri(after , Figure 1 E). G: 64-cell stage of the echiurid Urechis caupo(after , pl. IV). H: 48-cell stage of the sipunculan Golfingia vulgaris(= Phascolosoma vulgare after , pl. XXXII, figure. 9). I: Approximately 58-cell stage of the polychaete Nereis sp. (after , diagram II, B). Discussion Most spiralian lophotrochozoans are characterized by a spiral cleavage pattern, which is mainly defined by (1) the "spiral" arrangement of subsequent embryonic cells (= blastomeres), whereby cells of an upper tier of an embryo come to lie over the cleavage furrow of the lower tier; and (2) formation of a so-called "mesentoblast", which later gives rise to the endomesoderm from the 4 d-cell [7,29]. Despite these overall similarities between spirally cleaving species, the cleavage program is subject to great variability , e.g. concerning the size of cells (equal or unequal), regularity (synchronous or asynchronous), direction of cleavage (clockwise vs. counter-clockwise), morphological arrangement of apical cross patterns, or cell fates. In the following, we summarize the various spiralian cleavage phenotypes and discuss them in an evolutionary context in the light of the data presented herein for the entoproct Loxosomella sp. Cleavage in molluscs Most molluscs, except for yolk-rich "higher" gastropods and cephalopods, show a more or less "typical" spiral cleavage pattern [31-36]. Cleavage may be equal or unequal. Some species, such as the gastropod Ilyanassa obsoleta and the scaphopod Antalis entalis(formerly Dentalium dentale), form polar lobes which fuse with the D-quadrant of the early embryo [29,33,37]. Although the formation of the first micromere quartet typically appears in a clockwise direction in most taxa, including the supposedly basal solenogaster Epimenia babai, the polyplacophoran Stenoplax heathiana(formerly Ischnochiton heathiana; ), the scaphopod Antalis entalis, as well as the gastropods Limax, Crepidula[28,38], Patella, and Ilyanassa, a counter-clockwise formation is sometimes found, e.g., in the bivalve Dreissena polymorpha or in gastropods with sinistrally coiled shells such as Planorbis trivolvis, Physa heterostropha, or Lymnaea stagnalis. Accordingly, it appears that in shell-bearing gastropods, the chirality of cleavage is strictly correlated with the direction of shell coiling [43-45]. In contrast to the sinistrally coiled gastropod species mentioned above, fourth cleavage takes place in a counter-clockwise direction [28,29,31,32,35,39]. A typical spiralian feature develops in the following cleavage stages of different taxa, whereby derivatives of the first micromere quartet form the apical (1q 111) and peripheral rosette (1q 112). The apical rosette gives rise to the so-called "molluscan cross", whereby its arms are formed by the progenies of the 1q 12-cells. The tips of the molluscan cross are represented by second micromere quartet cells (2q 1) and their progenies [28,46]. However, variations in the morphology of the cross pattern occurs among the various molluscan subclades. For example, Stenoplax heathiana exhibits relatively large cross cells (1q 121) as well as pointed peripheral rosette cells which are slightly larger than cells of the apical rosette (Figure 4C). In the basal gastropods Patella caerulea and Patella vulgata, peripheral and apical rosette cells are similar in size and shape and slightly smaller than cells of the arms of the cross (Figure 4D), while a distinct cross pattern was not found in embryos of Epimenia (Figure 4B). Cross formation in the scaphopod Antalis is characterized by a compressed pentagonal-shaped apical cell 1 d 111, which contacts the cell 1c 112 of the C-arm belonging to the molluscan cross . Although most bivalves do not show a molluscan cross pattern during development , a molluscan cross seems to be present in the equal cleaving, basal protobranch Solemya reidi, but probably not in the closely related protobranchs Acila, Nucula, and Yoldia[50-53]. Accordingly, such a cross pattern may not even be part of the ancestral bivalve bodyplan. Cleavage in annelids (including sipunculans) Embryos of the polychaete Nereis, the echiuran Urechis caupo, the sipunculan Phascolosoma[56,57], the oligochaete Bdellodrilus philadelphicus, and the leech Theromyzon tessulatum display a holoblastic, spiral cleavage pattern. Cleavage may be equal or unequal. Polar lobe formation is mainly found in polychaetes (e.g. Hydroides hexagonus) . The condition of the typical spiralian eight cell stage with small micromeres and large macromeres and a clockwise formation of the first micromere quartet appear to have been modified in various taxa. In the sipunculans Themiste pyroides and Phascolosoma agassizii, the micromere 1 d is larger than the blastomeres of the A, B, and C quadrant, and the macromere 1D is the largest of all cells . First quartet micromeres are budded off in a clockwise direction [57,61]. In the leech Theromyzon tessulatum and the oligochaete Tubifex rivulorum, the micromeres of the first quartet are exceptionally small. While the micromeres 1a, c, d are formed in a clockwise direction, 1b is budded off counter- clockwise . For polychaetes, a counter-clockwise formation of first quartet micromeres has so far only been reported for the serpulid Hydroides elegans. First quartet micromeres of the echiurid Urechis caupo are formed in a clockwise direction . The annelid cross is a seemingly common character of echiurans, clitellates, and polychaetes [63,64], whereby the peripheral rosette cells 1q 112 constitute its founding cells [28,46]. However, if 64-cell stages are compared, a typical "annelid cross" is only present in a few taxa such as the echiuran Urechis caupo (Figure 4G), the polychaetes Polygordius sp. , Amphitrite ornata, and Nereis sp. (Figure 4I), as well as the leech Theromyzon tessulatum. Annelid cross cells of Urechis caupo and Nereis sp. are much larger than the apical rosette cells. In both, the more peripherally located cells 1q 1122 are slightly larger than their centrally located sister cells 1q 1121 (Figure 4G, I). Embryos of T. tessulatum show a very indistinct cross pattern. Together with other small micromeres, the annelid cross cells are embedded in the furrows of the comparatively huge macromeres (Figure 494 in ). Compared to Nereis sp. (Figure 4I), the division of peripheral rosette cells in Chaetopterus pergamentaceus is oblique and results in an indistinct cross pattern . The first description of a cross stage in sipunculans was performed on Phascolosoma vulgare(= Golfingia vulgaris; [56,61]; Figure 4H herein), which is still cited as proof for a molluscan-type cross pattern in a sipunculan [23,68], and thus as an indication for a mollusc-sipunculan sister relationship. However, a number of independent developmental and molecular studies strongly argue in favour of a monophyletic annelid-sipunculan assemblage [10,11,69-75]. Although never explicitly stated for sipunculans, the arrangement of the peripheral rosette cells closely resembles the arrangement of an early stage annelid cross, whereby the sipunculan peripheral rosette cells are considerably larger than their molluscan counterparts (Figure 4). However, convincing recent data on the early embryology of sipunculans are lacking, and thus prevent a final statement as to whether or not an annelid or molluscan cross-like pattern is part of the sipunculan groundplan. Cleavage in other lophotrochozoans Nemertean cleavage is holoblastic, equal, and spiral [24,76,77]. First quartet micromeres of Cerebratulus lacteus and Carinoma tremaphoros are larger than the macromeres. A so-called "nemertean cross" is found in Carinoma tremaphoros(; Figure 4E herein) and Emplectonema gracile. It is formed by accelerated cell divisions of the apical rosette, which so far has only been reported from nemertean species. Nemertean peripheral rosette cells are relatively large compared to the small apical rosette cells (Figure 4E). Polyclad platyhelminths such as Maritigrella crozieri and Hoploplana inquilina show a quartet, equal and spiral cleavage pattern [54,79]. In both species, fourth quartet micromeres are very large compared to the relatively small macromeres . Morphologically, the apical and peripheral rosette cells of the 32- to 64-cell stage of the polyclad flatworm Maritigrella crozieri could be interpreted as both, a molluscan and an annelid cross pattern, respectively . Although the progenitor cells of the apical and peripheral rosettes have not yet divided at this stage, blastomeres of the 32-cell stage ("third quartet"-stage) of H. inquilina form an apical cross-like pattern (see [79,80]). In Maritigrella crozieri, peripheral rosette cells are slightly elongated and more than twice as large as the apical rosette cells (Figure 4F). Developmentally, ectoprocts, phoronids, and brachiopods are unique within the lophotrochozoans because they exhibit a radial cleavage pattern [81-84]. A spiral-type cleavage pattern has been proposed for the phoronid Phoronopsis viridis and the brachiopod Terebratulina septentrionalis by some classical studies, although these data are highly questioned by recent investigations and may be artifactual due to compression of the embryos [81,85]. Cleavage in entoprocts The Loxosomella species investigated herein cleaves holoblastic, asynchronous, equal, and spiral, whereby all cells are of approximately the same size until third cleavage. Indicated by the degree of shifting of the micromere cells relative to the macromere cells, the first micromere quartet is formed in a clockwise direction (Figure 2G-L). Accordingly, comparative morphology of the cleavage stages investigated herein clearly shows that our study species exhibits typical spiral cleavage. Following the rule of alternation, in Loxosomella, the second quartet micromeres (blue cells, Figure 2N, O) rotate in a counter-clockwise direction until they occupy the furrows between the macromeres (red cells, Figure 2N, O). As a consequence, cells of the first micromere quartet are turned back over the centre of each macromere (cf. ). Division of the first micromere quartet results in the 1q 1- and 1q 2-quartets (purple cells, Figure 2N, O). The light blue and green colored cells are probably derivatives of the first micromere cells 1q 1, while the yellow cell is likely to be a derivative of the 1q 11 cell (Figure 2N, O). Morphologically, the apical cross pattern in the 43-cell stage of Loxosomella sp. exhibits two cross-like patterns, which show similarities to both a "molluscan" and an "annelid cross", respectively (see also ), thus supporting closer affinity of Entoprocta with these spiralian taxa than with Ectoprocta. Observations on the colonial entoproct Pedicellina echinata(= Pedicellina cernua) report both, a slightly unequal and an equal cleavage pattern . Sketch drawings of the first cleavage stages of Loxosoma leptoclini were interpreted as showing equally-sized blastomeres . However, since we were not able to trace cell genealogy in our study species, a concluding statement concerning homology between these "entoproct crosses" and the annelid and/or molluscan crosses cannot be given at present. A comparison of the various spiralian-type cleavage patterns, however, does demonstrate the morphological plasticity of the "cross-type" cellular arrangement in the various spiralian representatives, and thus shows that phylogenetic conclusions based on these embryonic morphotypes alone should be treated with utmost care. Conclusions Although it has repeatedly been proposed that the "molluscan cross" constitutes a pattern that is only found in molluscs and closely related sister groups [23,88], similar cross patterns are also present in other spiralian animals including annelids, nemerteans, and flatworms [24,87,89] (cf. Figure 4). The arrangement of the annelid and nemertean cross cells resembles a generation of cells which usually appears during the 7 th cleavage cycle, while molluscan cross cells are formed during the 6 th division. Since the annelid and nemertean cross is formed earlier in development, the formation of both cross patterns is distinct to these phyla and is not found in 64-cell stages of molluscan embryos. Accordingly, due to their variation even between closely related species, a typical "molluscan" or "annelid" cross pattern cannot reliably be proposed at present, thus rendering these cleavage morphologies phylogenetically uninformative. However, we suggest that an embryonic stage with a cross-like pattern was present in the last common ancestor of Spiralia. The cleavage pattern of the entoproct Loxosomella sp. investigated herein shows typical spiralian features, such as the spiral arrangement of blastomeres around the animal-vegetal axis of the embryo and the presence of a cross-like pattern, thus indeed rendering entoprocts "true" spiralians. The currently widely held view that the radially cleaving Ectoprocta is not the sister group of Entoprocta is well supported by our data and strongly suggests the inclusion of Entoprocta within Spiralia. This is also in agreement with the Tetraneuralia-hypothesis, which suggests a Mollusca-Entoprocta clade and argues against the so-called Polyzoa-concept (Ectoprocta + Entoprocta + Cycliophora; see ). The latter scenario would either imply independent evolution of spiral cleavage in Entoprocta and the remaining Spiralia or secondary loss of spiral cleavage in Ectoprocta (data on cleavage in Cycliophora are still lacking). On a deeper evolutionary scale, the classical subdivision of Lophotrochozoa into Spiralia and Lophophorata suggests either a spiral or a radial cleavage pattern for the last common lophotrochozoan ancestor. The duet-spiral-type of cleavage in the supposedly basal bilaterians, the acoels, may favour a scenario where a modification of this cleavage pattern might have resulted in the quartet-spiral cleavage pattern of the ur-lophotrochozoan, with a secondary modification in the lophophorates. At present, such evolutionary deductions remain speculative, however, due to the lack of a reliable phylogeny for internal lophotrochozoan relationships and the unresolved question concerning the last common lophotrochozoan and bilaterian ancestor. Authors' contributions JM performed research and drafted the manuscript. TW acquired the study material and helped with immunostaining and confocal microscopy. BL contributed to data interpretation. AW designed the study, supervised research, contributed to data interpretation, and improved the manuscript draft. All authors contributed to and approved the final version of the manuscript. Contributor Information Julia Merkel, Email: Julia_Merkel82@gmx.de. Tim Wollesen, Email: tim.wollesen@univie.ac.at. Bernhard Lieb, Email: lieb@uni-mainz.de. Andreas Wanninger, Email: andreas.wanninger@univie.ac.at. Acknowledgements We are indebted to the Friday Harbor Laboratories for their hospitality during the stay of TW while collecting Loxosomella. JM is grateful to Lisbeth Haukrogh and Henrike Semmler (both Copenhagen) for help with immunocytochemistry, confocal microscopy, and the Imaris software. This work was supported by a grant of the German Science Foundation to BL (Li998/9-1), the Feldbausch Stiftung (University of Mainz), and the Inneruniversitäre Forschungsförderung (University of Mainz), both to BL. References Halanych KM, Bacheller JD, Aguinaldo AM, Liva SM, Hillis DM, Lake JA. Evidence from 18S ribosomal DNA that the lophophorates are protostome animals. Science. 1995;267:1641–1643. doi: 10.1126/science.7886451. [DOI] [PubMed] [Google Scholar] Aguinaldo AMA, Turbeville JM, Linford LS, Rivera MC, Garey JR, Raff RA, Lake JA. Evidence for a clade of nematodes, arthropods and other moulting animals. Nature. 1997;387:489–493. doi: 10.1038/387489a0. [DOI] [PubMed] [Google Scholar] Halanych KM. The new view of animal phylogeny. Ann Rev Ecol Evol Syst. 2004;35:229–256. doi: 10.1146/annurev.ecolsys.35.112202.130124. [DOI] [Google Scholar] Wanninger A. Shaping the things to come: Ontogeny of lophotrochozoan neuromuscular systems and the Tetraneuralia concept. Biol Bull. 2009;216:293–306. doi: 10.1086/BBLv216n3p293. [DOI] [PubMed] [Google Scholar] Dunn CW, Hejnol A, Matus DQ, Pang K, Browne WE, Smith SA, Seaver E, Rouse GW, Obst M, Edgecombe GD, Sørensen MV, Haddock SHD, Schmidt-Rhaesa A, Okusu A, Kristensen RM, Wheeler WC, Martindale MQ, Giribet G. Broad phylogenomic sampling improves resolution of the animal tree of life. Nature. 2008;452:745–749. doi: 10.1038/nature06614. [DOI] [PubMed] [Google Scholar] Yokobori S, Iseto T, Asakawa S, Sasaki T, Shimizu N, Yamagishi A, Oshima T, Hirose E. Complete nucleotide sequences of mitochondrial genomes of two solitary entoprocts, Loxocorone allax and Loxosomella aloxiata: Implications for lophotrochozoan phylogeny. Mol Phylogenet Evol. 2008;47:612–628. doi: 10.1016/j.ympev.2008.02.013. [DOI] [PubMed] [Google Scholar] Nielsen C. Some aspects of spiralian development. Acta Zool (Stockholm) 2010;91:20–28. doi: 10.1111/j.1463-6395.2009.00421.x. [DOI] [Google Scholar] Nielsen C. Entoproct life-cycles and the entoproct/ectoproct relationship. Ophelia. 1971;9:209–341. [Google Scholar] Nielsen C. Animal evolution: Interrelationships of the living phyla. Press, Oxford: Oxford Univ; 2001. [Google Scholar] Hausdorf B, Helmkampf M, Meyer A, Witek A, Herlyn H, Bruchhaus I, Hankeln T, Struck TH, Lieb B. Spiralian phylogenomics supports the resurrection of Bryozoa comprising Ectoprocta and Entoprocta. Mol Biol Evol. 2007;24:2723–2729. doi: 10.1093/molbev/msm214. [DOI] [PubMed] [Google Scholar] Hausdorf B, Helmkampf M, Nesnidal MP, Bruchhaus I. Phylogenetic relationships within the lophophorate lineages (Ectoprocta, Brachiopoda and Phoronida) Mol Phylogenet Evol. 2010;55:1121–1127. doi: 10.1016/j.ympev.2009.12.022. [DOI] [PubMed] [Google Scholar] Funch P, Kristensen RM. Cycliophora is a new phylum with affinities to Entoprocta and Ectoprocta. Nature. 1995;378:711–714. doi: 10.1038/378711a0. [DOI] [Google Scholar] Cavalier-Smith T. A revised six-kingdom system of life. Biol Rev. 1998;73:203–266. doi: 10.1017/S0006323198005167. [DOI] [PubMed] [Google Scholar] Hejnol A, Obst M, Stamatakis A, Ott M, Rouse GW, Edgecombe GD, Martinez P, Baguñà J, Bailly X, Jondelius U, Wiens M, Müller WEG, Seaver E, Wheeler WC, Martindale MQ, Giribet G, Dunn CW. Assessing the root of bilaterian animals with scalable phylogenomic methods. Proc R Soc Lond B. 2009;276:4261–4270. doi: 10.1098/rspb.2009.0896. [DOI] [PMC free article] [PubMed] [Google Scholar] Passamaneck Y, Halanych KM. Lophotrochozoan phylogeny assessed with LSU and SSU data: evidence of lophophorate polyphyly. Mol Phylogenet Evol. 2006;40:20–28. doi: 10.1016/j.ympev.2006.02.001. [DOI] [PubMed] [Google Scholar] Meyer A, Todt C, Mikkelsen NT, Lieb B. Fast evolving 18S rRNA sequences from Solenogastres (Mollusca) resist standard PCR amplification and give new insights into mollusc substitution rate heterogeneity. BMC Evol Biol. 2010;10:70. doi: 10.1186/1471-2148-10-70. [DOI] [PMC free article] [PubMed] [Google Scholar] Ax P. Das System der Metazoa. 2. Stuttgart: Gustav Fischer; 1999. [Google Scholar] Haszprunar G. Is the Aplacophora monophyletic? A cladistic point of view. Amer Malac Bull. 2000;15:115–130. [Google Scholar] Haszprunar G, Wanninger A. On the fine structure of the creeping larva of Loxosomella murmanica: additional evidence for a clade of Kamptozoa (Entoprocta) and Mollusca. Acta Zool (Stockholm) 2008;89:137–148. [Google Scholar] Wanninger A, Fuchs J, Haszprunar G. The anatomy of the serotonergic nervous system of an entoproct creeping-type larva and its phylogenetic implications. Invertebr Biol. 2007;126:268–278. doi: 10.1111/j.1744-7410.2007.00097.x. [DOI] [Google Scholar] Fuchs J, Wanninger A. Reconstruction of the neuromuscular system of the swimming-type larva of Loxosomella atkinsae (Entoprocta) as inferred by fluorescence labelling and confocal microscopy. Org Divers Evol. 2008;8:325–335. doi: 10.1016/j.ode.2008.05.002. [DOI] [Google Scholar] Schleip W. Die Determination der Primitiventwicklung (Leipzig, Akademische Verlagsgesellschaft) 1929. Scheltema AH. Aplacophora as progenetic aculiferans and the coelomate origin of mollusks as the sister taxon of Sipuncula. Biol Bull. 1993;184:57–78. doi: 10.2307/1542380. [DOI] [PubMed] [Google Scholar] Maslakova SA, Martindale MQ, Norenburg JL. Fundamental properties of the spiralian developmental program are displayed by the basal nemertean Carinoma tremaphoros (Palaeonemertea, Nemertea) Dev Biol. 2004;267:342–360. doi: 10.1016/j.ydbio.2003.10.022. [DOI] [PubMed] [Google Scholar] Hatschek B. Embryonalentwicklung und Knospung der Pedicellina echinata. Z Wiss Zool. 1877;29:502–549. [Google Scholar] Marcus E. Bryozoários marinhos brasileiros III. Bol. Fac. Fil., Ciên. Letr. Univ. S. Paulo, XIII. Zool. 1939;3:111–354. [Google Scholar] Malakhov VV. Description of the development of Ascopodaria discreta (Coloniales, Barentsiidae) and discussion of the Kamptozoa status in the animal kingdom. Zool Zh. 1990;69:20–30. [Google Scholar] Conklin EG. The embryology of Crepidula. J Morphol. 1897;13:1–226. doi: 10.1002/jmor.1050130102. [DOI] [Google Scholar] Goulding MQ. Cell lineage of the Ilyanassa embryo: evolutionary acceleration of regional differentiation during early development. PLoS One. 2009;4:e5506. doi: 10.1371/journal.pone.0005506. [DOI] [PMC free article] [PubMed] [Google Scholar] Henry JJ, Martindale MQ. Conservation and innovation in spiralian development. Hydrobiologia. 1999;402:255–265. [Google Scholar] Kofoid CA. On some laws of cleavage in Limax: A preliminary notice. Proc Am Acad Arts Sci. 1894;29:180–203. [Google Scholar] Heath H. The development of Ischnochiton. Zool Jb Anat Ontog Tiere. 1898;12:567–656. [Google Scholar] Van Dongen CAM, Geilenkirchen WLM. The development of Dentalium with special reference to the significance of the polar lobe. I, II, III. Division chronology and development of the cell pattern in Dentalium dentale (Scaphopoda) Proc K Ned Akad Wet C. 1974;77:57–100. [Google Scholar] Van den Biggelaar JAM, Dictus WJAG, van Loon AE. Cleavage patterns, cell-lineages and cell specification are clues to phyletic lineages in Spiralia. Sem Cell Dev Biol. 1997;8:367–378. doi: 10.1006/scdb.1997.0161. [DOI] [PubMed] [Google Scholar] Okusu A. Embryogenesis and development of Epimenia babai (Mollusca Neomeniomorpha) Biol Bull. 2002;203:87–103. doi: 10.2307/1543461. [DOI] [PubMed] [Google Scholar] Ponder WF, Lindberg DR. Phylogeny and evolution of the Mollusca. Berkeley: University of California Press; 2008. [Google Scholar] Guerrier P, van den Biggelaar JAM, van Dongen CAM, Verdonk NH. Significance of the polar lobe for the determination of dorsoventral polarity in Dentalium vulgare (da Costa) Dev Biol. 1978;63:233–242. doi: 10.1016/0012-1606(78)90128-8. [DOI] [PubMed] [Google Scholar] Henry JJ, Collin R, Perry KJ. The slipper snail, Crepidula: An emerging lophotrochozoan model system. Biol Bull. 2010;218:211–229. doi: 10.1086/BBLv218n3p211. [DOI] [PubMed] [Google Scholar] Van den Biggelaar JAM. Development of dorsoventral polarity and mesentoblast determination in Patella vulgata. J Morphol. 1977;154:157–186. doi: 10.1002/jmor.1051540111. [DOI] [PubMed] [Google Scholar] Clement AC. Experimental studies on germinal localization in Ilyanassa. I. The role of the polar lobe in determination of the cleavage pattern and its influence in later development. J Exp Zool. 1952;121:593–625. doi: 10.1002/jez.1401210310. [DOI] [Google Scholar] Luetjens CM, Dorresteijn AWC. Multiple, alternative cleavage patterns precede uniform larval morphology during normal development of Dreissena polymorpha (Mollusca, Lamellibranchia) Roux's Arch Dev Biol. 1995;205:138–149. doi: 10.1007/BF00357760. [DOI] [PubMed] [Google Scholar] Holmes SJ. The early development of Planorbis. J Morphol. 1900;16:369–458. doi: 10.1002/jmor.1050160205. [DOI] [Google Scholar] Crampton HE. Reversal of cleavage in a sinistral gasteropod. Ann N Y Acad Sci. 1894;8:167–170. doi: 10.1111/j.1749-6632.1894.tb55419.x. [DOI] [Google Scholar] Kuroda R, Endo B, Abe M, Shimizu M. Chiral blastomere arrangement dictates zygotic left-right asymmetry pathway in snails. Nature. 2009;462:790–794. doi: 10.1038/nature08597. [DOI] [PubMed] [Google Scholar] Dohmen MR. Cell lineage in molluscan development. Microsc Res Tech. 1992;22:75–102. doi: 10.1002/jemt.1070220107. [DOI] [PubMed] [Google Scholar] Wilson EB. The cell-lineage of Nereis. A contribution to the cytogeny of the annelid body. J Morphol. 1892;6:361–480. doi: 10.1002/jmor.1050060301. [DOI] [Google Scholar] Wilson EB. Experimental studies in germinal localization. II. Experiments on the cleavage-mosaic in Patella and Dentalium. J Exp Zool. 1904;1:197–268. doi: 10.1002/jez.1400010202. [DOI] [Google Scholar] Baba K. General sketch of the development in a solenogastre, Epimenia verrucosa (Nierstrasz) Misc Rep Res Inst Nat Resour (Tokyo) 1951;19-21:38–46. [Google Scholar] Giribet G. Current advances in the phylogenetic reconstruction of metazoan evolution. A new paradigm for the Cambrian explosion? Mol Phylogenet Evol. 2002;24:345–357. doi: 10.1016/S1055-7903(02)00206-3. [DOI] [PubMed] [Google Scholar] Drew GA. Some observations on the habitats, anatomy and embryology of members of the Protobranchia. Anat Anz. 1899;15:493–519. [Google Scholar] Gustafson RG, Reid RGB. Development of the pericalymma larva of Solemya reidi (Bivalvia: Cryptodonta: Solemyidae) as revealed by light and electron microscopy. Mar Biol. 1986;93:411–427. doi: 10.1007/BF00401109. [DOI] [Google Scholar] Gustafson RG, Lutz RA. Larval and early post-larval development of the protobranch bivalve Solemya velum (Mollusca: Bivalvia) J Mar Biol Assoc UK. 1992;72:383–402. doi: 10.1017/S0025315400037772. [DOI] [Google Scholar] Zardus J, Morse MP. Embryogenesis, morphology and ultrastructure of the pericalymma larva of Acila castrensis (Bivalvia: Protobranchia: Nuculoida) Invertebr Biol. 1998;117:221–244. doi: 10.2307/3226988. [DOI] [Google Scholar] Rawlinson KA. Embryonic and post-embryonic development of the polyclad flatworm Maritigrella crozieri; implications for the evolution of spiralian life history traits. Front Zool. 2010;7:12. doi: 10.1186/1742-9994-7-12. [DOI] [PMC free article] [PubMed] [Google Scholar] Newby WW. The early embryology of the echiuroid, Urechis. Biol Bull. 1932;63:387–399. doi: 10.2307/1537341. [DOI] [Google Scholar] Gerould JH. Studies on the embryology of the Sipunculidae. I. The embryonal envelope and its homologue. Mark Anniversary Volume. 1903. pp. 437–452. Rice ME. A comparative study of the development of Phascolosoma agassizii, Golfingia pugettensis, and Themiste pyroides with a discussion of developmental patterns in the Sipuncula. Ophelia. 1967;4:143–171. [Google Scholar] Tannreuther GW. The embryology of Bdellodrilus philadelphicus. J Morphol. 1915;26:143–216. doi: 10.1002/jmor.1050260202. [DOI] [Google Scholar] Sandig M, Dohle W. The cleavage pattern in the leech Theromyzon tessulatum (Hirudinea, Glossiphoniidae) J Morphol. 1988;196:217–252. doi: 10.1002/jmor.1051960210. [DOI] [PubMed] [Google Scholar] Costello DP, Henley C. Spiralian development: a perspective. Amer Zool. 1976;16:277–291. [Google Scholar] Gerould JH. The development of Phascolosoma (Studies on the embryology of the Sipunculidae II) Zool Jb Anat Ontog Tiere. 1906;23:77–162. [Google Scholar] Arenas-Mena C. Sinistral equal-size spiral cleavage of the indirectly developing polychaete Hydroides elegans. Dev Dyn. 2007;236:1611–1622. doi: 10.1002/dvdy.21164. [DOI] [PubMed] [Google Scholar] Rouse GW, Fauchald K. The articulation of annelids. Zool Scripta. 1995;24:269–301. doi: 10.1111/j.1463-6409.1995.tb00476.x. [DOI] [Google Scholar] McHugh D. Molecular evidence that echiurans and pogonophorans are derived annelids. Proc Nat Acad Sci USA. 1997;94:8006–8009. doi: 10.1073/pnas.94.15.8006. [DOI] [PMC free article] [PubMed] [Google Scholar] MacBride EW. lnvertebrata. Vol. 1. Macmillan & Co., London; 1914. Textbook of Embryology. [Google Scholar] Mead AD. The early development of marine annelids. J Morphol. 1897;13:227–326. doi: 10.1002/jmor.1050130202. [DOI] [Google Scholar] Dawydoff C. In: "Traité de Zoologie". Grassé PP, editor. Vol. 5. 1959. Ontogénese des annelids; pp. 594–686. [Google Scholar] Rice ME. In: Entoprocts and lesser coelomates. Giese A, Pearse J, editor. Vol. 2. Academic Press, New York; 1975. Sipuncula. In: Reproduction of Marine Invertebrates. [Google Scholar] Boore JL, Staton JL. The mitochondrial genome of the sipunculid Phascolopsis gouldii supports its association with Annelida rather than Mollusca. Mol Biol Evol. 2002;19:127–137. doi: 10.1093/oxfordjournals.molbev.a004065. [DOI] [PubMed] [Google Scholar] Halanych KM, Dahlgren TG, McHugh D. Unsegmented annelids? Possible origins of four lophotrochozoan worm taxa. Integr Comp Biol. 2002;42:678–684. doi: 10.1093/icb/42.3.678. [DOI] [PubMed] [Google Scholar] Helmkampf M, Bruchhaus I, Hausdorf B. Phylogenomic analyses of lophophorates (brachiopods, phoronids and bryozoans) confirm the Lophotrochozoa concept. Proc R Soc Lond B. 2008;275:1927–1933. doi: 10.1098/rspb.2008.0372. [DOI] [PMC free article] [PubMed] [Google Scholar] Kristof A, Wollesen T, Wanninger A. Segmental mode of neural patterning in Sipuncula. Curr Biol. 2008;18:1129–1132. doi: 10.1016/j.cub.2008.06.066. [DOI] [PubMed] [Google Scholar] Wanninger A. Comparative lophotrochozoan neurogenesis and larval neuroanatomy: recent advances from previously neglected taxa. Acta Biol Hung. 2008;59(Suppl):127–136. doi: 10.1556/ABiol.59.2008.Suppl.21. [DOI] [PubMed] [Google Scholar] Kristof A, Wollesen T, Maiorova AS, Wanninger A. Cellular and muscular growth patterns during sipunculan development. J Exp Zool (Mol Dev Evol) 2011;316:227–240. doi: 10.1002/jez.b.21394. [DOI] [PMC free article] [PubMed] [Google Scholar] Struck TH, Paul C, Hill N, Hartmann S, Hösel C, Kube M, Lieb B, Meyer A, Tiedemann R, Purschke G, Bleidorn C. Phylogenomic analyses unravel annelid evolution. Nature. 2011;471:95–98. doi: 10.1038/nature09864. [DOI] [PubMed] [Google Scholar] Henry JJ, Martindale MQ. Establishment of the dorsoventral axis in nemertean embryos: Evolutionary considerations of spiralian development. Dev Gen. 1994;15:64–78. doi: 10.1002/dvg.1020150108. [DOI] [Google Scholar] Maslakova SA, von Döhren J. Larval development with transitory epidermis in Paranemertes peregrina and other hoplonemerteans. Biol Bull. 2009;216:273–292. doi: 10.1086/BBLv216n3p273. [DOI] [PubMed] [Google Scholar] Henry JJ, Martindale MQ. Conservation of the spiralian developmental program: Cell lineage of the nemertean, Cerebratulus lacteus. Dev Biol. 1998;201:253–269. doi: 10.1006/dbio.1998.8966. [DOI] [PubMed] [Google Scholar] Boyer BC, Henry JQ, Martindale MQ. The cell lineage of a polyclad turbellarian embryo reveals close similarity to coelomate spiralians. Dev Biol. 1998;204:111–123. doi: 10.1006/dbio.1998.9084. [DOI] [PubMed] [Google Scholar] Boyer BC, Henry JQ, Martindale MQ. Dual origins of mesoderm in a basal spiralian: Cell lineage analyses in the polyclad turbellarian Hoploplana inquilina. Dev Biol. 1996;179:329–338. doi: 10.1006/dbio.1996.0264. [DOI] [PubMed] [Google Scholar] Emig CC. Embryology of Phoronida. Amer Zool. 1977;17:21–37. [Google Scholar] Valentine JW. Cleavage pattern and the topology of the metazoan tree of life. Proc Natl Acad Sci USA. 1997;94:8001–8005. doi: 10.1073/pnas.94.15.8001. [DOI] [PMC free article] [PubMed] [Google Scholar] Freeman G. Regional specification during embryogenesis in the inarticulate brachiopod Discinisca. Dev Biol. 1999;209:321–339. doi: 10.1006/dbio.1999.9251. [DOI] [PubMed] [Google Scholar] Gruhl A. Ultrastructure of mesoderm formation and development in Membranipora membranacea (Bryozoa: Gymnolaemata) Zoomorphology. 2010;129:45–60. doi: 10.1007/s00435-009-0099-3. [DOI] [Google Scholar] Rattenbury JC. The embryology of Phoronopsis viridis. J Morphol. 1954;95:289–349. doi: 10.1002/jmor.1050950206. [DOI] [Google Scholar] Conklin EG. The embryology of a brachiopod, Terebratulina septentrionalis Couthouy. Proc Am Philos Soc. 1902;41:41–76. [Google Scholar] Jenner RA. Unleashing the force of cladistics? Metazoan phylogenetics and hypothesis testing. Integr Comp Biol. 2003;43:207–218. doi: 10.1093/icb/43.1.207. [DOI] [PubMed] [Google Scholar] Zrzavy J. Gastrotricha and metazoan phylogeny. Zool Scr. 2003;32:61–81. doi: 10.1046/j.1463-6409.2003.00104.x. [DOI] [Google Scholar] Hennig W. Taschenbuch der speziellen Zoologie, Teil 1. Wirbellose I. Ausgenommen Gliedertiere. Thun und Frankfurt am Main: Verlag Harri Deutsch; 1983. 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Reject All Save My Preferences Accept All Skip to ContentGo to accessibility pageKeyboard shortcuts menu Log in Chemistry 2e 21.4 Transmutation and Nuclear Energy Chemistry 2e21.4 Transmutation and Nuclear Energy Contents Contents Highlights Table of contents Preface 1 Essential Ideas 2 Atoms, Molecules, and Ions 3 Composition of Substances and Solutions 4 Stoichiometry of Chemical Reactions 5 Thermochemistry 6 Electronic Structure and Periodic Properties of Elements 7 Chemical Bonding and Molecular Geometry 8 Advanced Theories of Covalent Bonding 9 Gases 10 Liquids and Solids 11 Solutions and Colloids 12 Kinetics 13 Fundamental Equilibrium Concepts 14 Acid-Base Equilibria 15 Equilibria of Other Reaction Classes 16 Thermodynamics 17 Electrochemistry 18 Representative Metals, Metalloids, and Nonmetals 19 Transition Metals and Coordination Chemistry 20 Organic Chemistry 21 Nuclear Chemistry Introduction 21.1 Nuclear Structure and Stability 21.2 Nuclear Equations 21.3 Radioactive Decay 21.4 Transmutation and Nuclear Energy 21.5 Uses of Radioisotopes 21.6 Biological Effects of Radiation Key Terms Key Equations Summary Exercises A \| The Periodic Table B \| Essential Mathematics C \| Units and Conversion Factors D \| Fundamental Physical Constants E \| Water Properties F \| Composition of Commercial Acids and Bases G \| Standard Thermodynamic Properties for Selected Substances H \| Ionization Constants of Weak Acids I \| Ionization Constants of Weak Bases J \| Solubility Products K \| Formation Constants for Complex Ions L \| Standard Electrode (Half-Cell) Potentials M \| Half-Lives for Several Radioactive Isotopes Answer Key Index Search for key terms or text. Close Learning Objectives By the end of this section, you will be able to: Describe the synthesis of transuranium nuclides Explain nuclear fission and fusion processes Relate the concepts of critical mass and nuclear chain reactions Summarize basic requirements for nuclear fission and fusion reactors After the discovery of radioactivity, the field of nuclear chemistry was created and developed rapidly during the early twentieth century. A slew of new discoveries in the 1930s and 1940s, along with World War II, combined to usher in the Nuclear Age in the mid-twentieth century. Scientists learned how to create new substances, and certain isotopes of certain elements were found to possess the capacity to produce unprecedented amounts of energy, with the potential to cause tremendous damage during war, as well as produce enormous amounts of power for society’s needs during peace. Synthesis of Nuclides Nuclear transmutation is the conversion of one nuclide into another. It can occur by the radioactive decay of a nucleus, or the reaction of a nucleus with another particle. The first manmade nucleus was produced in Ernest Rutherford’s laboratory in 1919 by a transmutation reaction, the bombardment of one type of nuclei with other nuclei or with neutrons. Rutherford bombarded nitrogen atoms with high-speed α particles from a natural radioactive isotope of radium and observed protons resulting from the reaction: 14 7 N+4 2 He⟶17 8 O+1 1 H 7 14 N+2 4 He⟶8 17 O+1 1 H 7 14 N+2 4 He⟶8 17 O+1 1 H The 17 8 O 8 17 O 8 17 O and 1 1 H 1 1 H 1 1 H nuclei that are produced are stable, so no further (nuclear) changes occur. To reach the kinetic energies necessary to produce transmutation reactions, devices called particle accelerators are used. These devices use magnetic and electric fields to increase the speeds of nuclear particles. In all accelerators, the particles move in a vacuum to avoid collisions with gas molecules. When neutrons are required for transmutation reactions, they are usually obtained from radioactive decay reactions or from various nuclear reactions occurring in nuclear reactors. The Chemistry in Everyday Life feature that follows discusses a famous particle accelerator that made worldwide news. Chemistry in Everyday Life CERN Particle Accelerator Located near Geneva, the CERN (“Conseil Européen pour la Recherche Nucléaire,” or European Council for Nuclear Research) Laboratory is the world’s premier center for the investigations of the fundamental particles that make up matter. It contains the 27-kilometer (17 mile) long, circular Large Hadron Collider (LHC), the largest particle accelerator in the world (Figure 21.13). In the LHC, particles are boosted to high energies and are then made to collide with each other or with stationary targets at nearly the speed of light. Superconducting electromagnets are used to produce a strong magnetic field that guides the particles around the ring. Specialized, purpose-built detectors observe and record the results of these collisions, which are then analyzed by CERN scientists using powerful computers. Figure 21.13 A small section of the LHC is shown with workers traveling along it. (credit: Christophe Delaere) In 2012, CERN announced that experiments at the LHC showed the first observations of the Higgs boson, an elementary particle that helps explain the origin of mass in fundamental particles. This long-anticipated discovery made worldwide news and resulted in the awarding of the 2013 Nobel Prize in Physics to François Englert and Peter Higgs, who had predicted the existence of this particle almost 50 years previously. Link to Learning Famous physicist Brian Cox talks about his work on the Large Hadron Collider at CERN, providing an entertaining and engaging tour of this massive project and the physics behind it. View a short video from CERN, describing the basics of how its particle accelerators work. Prior to 1940, the heaviest-known element was uranium, whose atomic number is 92. Now, many artificial elements have been synthesized and isolated, including several on such a large scale that they have had a profound effect on society. One of these—element 93, neptunium (Np)—was first made in 1940 by McMillan and Abelson by bombarding uranium-238 with neutrons. The reaction creates unstable uranium-239, with a half-life of 23.5 minutes, which then decays into neptunium-239. Neptunium-239 is also radioactive, with a half-life of 2.36 days, and it decays into plutonium-239. The nuclear reactions are: 238 92 U+1 0 n⟶239 92 U 239 92 U⟶239 93 Np+0−1 e half-life=23.5 min 239 93 Np⟶239 94 Pu+0−1 e half-life=2.36 days 92 238 U+0 1 n⟶92 239 U 92 239 U⟶93 239 Np+−1 0 e half-life=23.5 min 93 239 Np⟶94 239 Pu+−1 0 e half-life=2.36 days 92 238 U+0 1 n⟶92 239 U 92 239 U⟶93 239 Np+−1 0 e half-life=23.5 min 93 239 Np⟶94 239 Pu+−1 0 e half-life=2.36 days Plutonium is now mostly formed in nuclear reactors as a byproduct during the fission of U-235. Additional neutrons are released during this fission process (see the next section), some of which combine with U-238 nuclei to form uranium-239; this undergoes β decay to form neptunium-239, which in turn undergoes β decay to form plutonium-239 as illustrated in the preceding three equations. These processes are summarized in the equation: 238 92 U+1 0 n⟶239 92 U−→−β−239 93 Np−→−β−239 94 Pu 92 238 U+0 1 n⟶92 239 U→β−93 239 Np→β−94 239 Pu 92 238 U+0 1 n⟶92 239 U→β−93 239 Np→β−94 239 Pu Heavier isotopes of plutonium—Pu-240, Pu-241, and Pu-242—are also produced when lighter plutonium nuclei capture neutrons. Some of this highly radioactive plutonium is used to produce military weapons, and the rest presents a serious storage problem because they have half-lives from thousands to hundreds of thousands of years. Although they have not been prepared in the same quantity as plutonium, many other synthetic nuclei have been produced. Nuclear medicine has developed from the ability to convert atoms of one type into other types of atoms. Radioactive isotopes of several dozen elements are currently used for medical applications. The radiation produced by their decay is used to image or treat various organs or portions of the body, among other uses. The elements beyond element 92 (uranium) are called transuranium elements. As of this writing, 22 transuranium elements have been produced and officially recognized by IUPAC; several other elements have formation claims that are waiting for approval. Some of these elements are shown in Table 21.3. Preparation of Some of the Transuranium Elements \| Name \| Symbol \| Atomic Number \| Reaction \| --- --- \| \| americium \| Am \| 95 \| 239 94 Pu+1 0 n⟶240 95 Am+0−1 e 94 239 Pu+0 1 n⟶95 240 Am+−1 0 e 94 239 Pu+0 1 n⟶95 240 Am+−1 0 e \| \| curium \| Cm \| 96 \| 239 94 Pu+4 2 He⟶242 96 Cm+1 0 n 94 239 Pu+2 4 He⟶96 242 Cm+0 1 n 94 239 Pu+2 4 He⟶96 242 Cm+0 1 n \| \| californium \| Cf \| 98 \| 242 96 Cm+4 2 He⟶245 98 Cf+1 0 n 96 242 Cm+2 4 He⟶98 245 Cf+0 1 n 96 242 Cm+2 4 He⟶98 245 Cf+0 1 n \| \| einsteinium \| Es \| 99 \| 238 92 U+15 1 0 n⟶253 99 Es+7 0−1 e 92 238 U+15 0 1 n⟶99 253 Es+7−1 0 e 92 238 U+15 0 1 n⟶99 253 Es+7−1 0 e \| \| mendelevium \| Md \| 101 \| 253 99 Es+4 2 He⟶256 101 Md+1 0 n 99 253 Es+2 4 He⟶101 256 Md+0 1 n 99 253 Es+2 4 He⟶101 256 Md+0 1 n \| \| nobelium \| No \| 102 \| 246 96 Cm+12 6 C⟶254 102 No+4 1 0 n 96 246 Cm+6 12 C⟶102 254 No+4 0 1 n 96 246 Cm+6 12 C⟶102 254 No+4 0 1 n \| \| rutherfordium \| Rf \| 104 \| 249 98 Cf+12 6 C⟶257 104 Rf+4 1 0 n 98 249 Cf+6 12 C⟶104 257 Rf+4 0 1 n 98 249 Cf+6 12 C⟶104 257 Rf+4 0 1 n \| \| seaborgium \| Sg \| 106 \| 206 82 Pb+54 24 Cr⟶257 106 Sg+3 1 0 n 249 98 Cf+18 8 O⟶263 106 Sg+4 1 0 n 82 206 Pb+24 54 Cr⟶106 257 Sg+3 0 1 n 98 249 Cf+8 18 O⟶106 263 Sg+4 0 1 n 82 206 Pb+24 54 Cr⟶106 257 Sg+3 0 1 n 98 249 Cf+8 18 O⟶106 263 Sg+4 0 1 n \| \| meitnerium \| Mt \| 107 \| 209 83 Bi+58 26 Fe⟶266 109 Mt+1 0 n 83 209 Bi+26 58 Fe⟶109 266 Mt+0 1 n 83 209 Bi+26 58 Fe⟶109 266 Mt+0 1 n \| Table 21.3 Nuclear Fission Many heavier elements with smaller binding energies per nucleon can decompose into more stable elements that have intermediate mass numbers and larger binding energies per nucleon—that is, mass numbers and binding energies per nucleon that are closer to the “peak” of the binding energy graph near 56 (see Figure 21.3). Sometimes neutrons are also produced. This decomposition is called fission, the breaking of a large nucleus into smaller pieces. The breaking is rather random with the formation of a large number of different products. Fission usually does not occur naturally, but is induced by bombardment with neutrons. The first reported nuclear fission occurred in 1939 when three German scientists, Lise Meitner, Otto Hahn, and Fritz Strassman, bombarded uranium-235 atoms with slow-moving neutrons that split the U-238 nuclei into smaller fragments that consisted of several neutrons and elements near the middle of the periodic table. Since then, fission has been observed in many other isotopes, including most actinide isotopes that have an odd number of neutrons. A typical nuclear fission reaction is shown in Figure 21.14. Figure 21.14 When a slow neutron hits a fissionable U-235 nucleus, it is absorbed and forms an unstable U-236 nucleus. The U-236 nucleus then rapidly breaks apart into two smaller nuclei (in this case, Ba-141 and Kr-92) along with several neutrons (usually two or three), and releases a very large amount of energy. Among the products of Meitner, Hahn, and Strassman’s fission reaction were barium, krypton, lanthanum, and cerium, all of which have nuclei that are more stable than uranium-235. Since then, hundreds of different isotopes have been observed among the products of fissionable substances. A few of the many reactions that occur for U-235, and a graph showing the distribution of its fission products and their yields, are shown in Figure 21.15. Similar fission reactions have been observed with other uranium isotopes, as well as with a variety of other isotopes such as those of plutonium. Figure 21.15(a) Nuclear fission of U-235 produces a range of fission products. (b) The larger fission products of U-235 are typically one isotope with a mass number around 85–105, and another isotope with a mass number that is about 50% larger, that is, about 130–150. Link to Learning View this link to see a simulation of nuclear fission. A tremendous amount of energy is produced by the fission of heavy elements. For instance, when one mole of U-235 undergoes fission, the products weigh about 0.2 grams less than the reactants; this “lost” mass is converted into a very large amount of energy, about 1.8 ××× 10 10 kJ per mole of U-235. Nuclear fission reactions produce incredibly large amounts of energy compared to chemical reactions. The fission of 1 kilogram of uranium-235, for example, produces about 2.5 million times as much energy as is produced by burning 1 kilogram of coal. As described earlier, when undergoing fission U-235 produces two “medium-sized” nuclei, and two or three neutrons. These neutrons may then cause the fission of other uranium-235 atoms, which in turn provide more neutrons that can cause fission of even more nuclei, and so on. If this occurs, we have a nuclear chain reaction (see Figure 21.16). On the other hand, if too many neutrons escape the bulk material without interacting with a nucleus, then no chain reaction will occur. Figure 21.16 The fission of a large nucleus, such as U-235, produces two or three neutrons, each of which is capable of causing fission of another nucleus by the reactions shown. If this process continues, a nuclear chain reaction occurs. Material that can sustain a nuclear fission chain reaction is said to be fissile or fissionable. (Technically, fissile material can undergo fission with neutrons of any energy, whereas fissionable material requires high-energy neutrons.) Nuclear fission becomes self-sustaining when the number of neutrons produced by fission equals or exceeds the number of neutrons absorbed by splitting nuclei plus the number that escape into the surroundings. The amount of a fissionable material that will support a self-sustaining chain reaction is a critical mass. An amount of fissionable material that cannot sustain a chain reaction is a subcritical mass. An amount of material in which there is an increasing rate of fission is known as a supercritical mass. The critical mass depends on the type of material: its purity, the temperature, the shape of the sample, and how the neutron reactions are controlled (Figure 21.17). Figure 21.17(a) In a subcritical mass, the fissile material is too small and allows too many neutrons to escape the material, so a chain reaction does not occur. (b) In a critical mass, a large enough number of neutrons in the fissile material induce fission to create a chain reaction. An atomic bomb (Figure 21.18) contains several pounds of fissionable material, 235 92 U 92 235 U 92 235 U or 239 94 Pu,94 239 Pu,94 239 Pu, a source of neutrons, and an explosive device for compressing it quickly into a small volume. When fissionable material is in small pieces, the proportion of neutrons that escape through the relatively large surface area is great, and a chain reaction does not take place. When the small pieces of fissionable material are brought together quickly to form a body with a mass larger than the critical mass, the relative number of escaping neutrons decreases, and a chain reaction and explosion result. Figure 21.18(a) The nuclear fission bomb that destroyed Hiroshima on August 6, 1945, consisted of two subcritical masses of U-235, where conventional explosives were used to fire one of the subcritical masses into the other, creating the critical mass for the nuclear explosion. (b) The plutonium bomb that destroyed Nagasaki on August 9, 1945, consisted of a hollow sphere of plutonium that was rapidly compressed by conventional explosives. This led to a concentration of plutonium in the center that was greater than the critical mass necessary for the nuclear explosion. Fission Reactors Chain reactions of fissionable materials can be controlled and sustained without an explosion in a nuclear reactor (Figure 21.19). Any nuclear reactor that produces power via the fission of uranium or plutonium by bombardment with neutrons must have at least five components: nuclear fuel consisting of fissionable material, a nuclear moderator, reactor coolant, control rods, and a shield and containment system. We will discuss these components in greater detail later in the section. The reactor works by separating the fissionable nuclear material such that a critical mass cannot be formed, controlling both the flux and absorption of neutrons to allow shutting down the fission reactions. In a nuclear reactor used for the production of electricity, the energy released by fission reactions is trapped as thermal energy and used to boil water and produce steam. The steam is used to turn a turbine, which powers a generator for the production of electricity. Figure 21.19(a) The Diablo Canyon Nuclear Power Plant near San Luis Obispo is the only nuclear power plant currently in operation in California. The domes are the containment structures for the nuclear reactors, and the brown building houses the turbine where electricity is generated. Ocean water is used for cooling. (b) The Diablo Canyon uses a pressurized water reactor, one of a few different fission reactor designs in use around the world, to produce electricity. Energy from the nuclear fission reactions in the core heats water in a closed, pressurized system. Heat from this system produces steam that drives a turbine, which in turn produces electricity. (credit a: modification of work by “Mike” Michael L. Baird; credit b: modification of work by the Nuclear Regulatory Commission) Nuclear Fuels Nuclear fuel consists of a fissionable isotope, such as uranium-235, which must be present in sufficient quantity to provide a self-sustaining chain reaction. In the United States, uranium ores contain from 0.05–0.3% of the uranium oxide U 3 O 8; the uranium in the ore is about 99.3% nonfissionable U-238 with only 0.7% fissionable U-235. Nuclear reactors require a fuel with a higher concentration of U-235 than is found in nature; it is normally enriched to have about 5% of uranium mass as U-235. At this concentration, it is not possible to achieve the supercritical mass necessary for a nuclear explosion. Uranium can be enriched by gaseous diffusion (the only method currently used in the US), using a gas centrifuge, or by laser separation. In the gaseous diffusion enrichment plant where U-235 fuel is prepared, UF 6 (uranium hexafluoride) gas at low pressure moves through barriers that have holes just barely large enough for UF 6 to pass through. The slightly lighter 235 UF 6 molecules diffuse through the barrier slightly faster than the heavier 238 UF 6 molecules. This process is repeated through hundreds of barriers, gradually increasing the concentration of 235 UF 6 to the level needed by the nuclear reactor. The basis for this process, Graham’s law, is described in the chapter on gases. The enriched UF 6 gas is collected, cooled until it solidifies, and then taken to a fabrication facility where it is made into fuel assemblies. Each fuel assembly consists of fuel rods that contain many thimble-sized, ceramic-encased, enriched uranium (usually UO 2) fuel pellets. Modern nuclear reactors may contain as many as 10 million fuel pellets. The amount of energy in each of these pellets is equal to that in almost a ton of coal or 150 gallons of oil. Nuclear Moderators Neutrons produced by nuclear reactions move too fast to cause fission (refer back to Figure 21.17). They must first be slowed to be absorbed by the fuel and produce additional nuclear reactions. A nuclear moderator is a substance that slows the neutrons to a speed that is low enough to cause fission. Early reactors used high-purity graphite as a moderator. Modern reactors in the US exclusively use heavy water (2 1 H 2 O)(1 2 H 2 O)(1 2 H 2 O) or light water (ordinary H 2 O), whereas some reactors in other countries use other materials, such as carbon dioxide, beryllium, or graphite. Reactor Coolants A nuclear reactor coolant is used to carry the heat produced by the fission reaction to an external boiler and turbine, where it is transformed into electricity. Two overlapping coolant loops are often used; this counteracts the transfer of radioactivity from the reactor to the primary coolant loop. All nuclear power plants in the US use water as a coolant. Other coolants include molten sodium, lead, a lead-bismuth mixture, or molten salts. Control Rods Nuclear reactors use control rods (Figure 21.20) to control the fission rate of the nuclear fuel by adjusting the number of slow neutrons present to keep the rate of the chain reaction at a safe level. Control rods are made of boron, cadmium, hafnium, or other elements that are able to absorb neutrons. Boron-10, for example, absorbs neutrons by a reaction that produces lithium-7 and alpha particles: 10 5 B+1 0 n⟶7 3 Li+4 2 He 5 10 B+0 1 n⟶3 7 Li+2 4 He 5 10 B+0 1 n⟶3 7 Li+2 4 He When control rod assemblies are inserted into the fuel element in the reactor core, they absorb a larger fraction of the slow neutrons, thereby slowing the rate of the fission reaction and decreasing the power produced. Conversely, if the control rods are removed, fewer neutrons are absorbed, and the fission rate and energy production increase. In an emergency, the chain reaction can be shut down by fully inserting all of the control rods into the nuclear core between the fuel rods. Figure 21.20 The nuclear reactor core shown in (a) contains the fuel and control rod assembly shown in (b). (credit: modification of work by E. Generalic, Shield and Containment System During its operation, a nuclear reactor produces neutrons and other radiation. Even when shut down, the decay products are radioactive. In addition, an operating reactor is thermally very hot, and high pressures result from the circulation of water or another coolant through it. Thus, a reactor must withstand high temperatures and pressures, and must protect operating personnel from the radiation. Reactors are equipped with a containment system (or shield) that consists of three parts: The reactor vessel, a steel shell that is 3–20-centimeters thick and, with the moderator, absorbs much of the radiation produced by the reactor A main shield of 1–3 meters of high-density concrete A personnel shield of lighter materials that protects operators from γ rays and X-rays In addition, reactors are often covered with a steel or concrete dome that is designed to contain any radioactive materials might be released by a reactor accident. Link to Learning Click here to watch a 3-minute video from the Nuclear Energy Institute on how nuclear reactors work. Nuclear power plants are designed in such a way that they cannot form a supercritical mass of fissionable material and therefore cannot create a nuclear explosion. But as history has shown, failures of systems and safeguards can cause catastrophic accidents, including chemical explosions and nuclear meltdowns (damage to the reactor core from overheating). The following Chemistry in Everyday Life feature explores three infamous meltdown incidents. Chemistry in Everyday Life Nuclear Accidents The importance of cooling and containment are amply illustrated by three major accidents that occurred with the nuclear reactors at nuclear power generating stations in the United States (Three Mile Island), the former Soviet Union (Chernobyl), and Japan (Fukushima). In March 1979, the cooling system of the Unit 2 reactor at Three Mile Island Nuclear Generating Station in Pennsylvania failed, and the cooling water spilled from the reactor onto the floor of the containment building. After the pumps stopped, the reactors overheated due to the high radioactive decay heat produced in the first few days after the nuclear reactor shut down. The temperature of the core climbed to at least 2200 °C, and the upper portion of the core began to melt. In addition, the zirconium alloy cladding of the fuel rods began to react with steam and produced hydrogen: Zr(s)+2H 2 O(g)⟶ZrO 2(s)+2H 2(g)Zr(s)+2H 2 O(g)⟶ZrO 2(s)+2H 2(g)Zr(s)+2H 2 O(g)⟶ZrO 2(s)+2H 2(g) The hydrogen accumulated in the confinement building, and it was feared that there was danger of an explosion of the mixture of hydrogen and air in the building. Consequently, hydrogen gas and radioactive gases (primarily krypton and xenon) were vented from the building. Within a week, cooling water circulation was restored and the core began to cool. The plant was closed for nearly 10 years during the cleanup process. Although zero discharge of radioactive material is desirable, the discharge of radioactive krypton and xenon, such as occurred at the Three Mile Island plant, is among the most tolerable. These gases readily disperse in the atmosphere and thus do not produce highly radioactive areas. Moreover, they are noble gases and are not incorporated into plant and animal matter in the food chain. Effectively none of the heavy elements of the core of the reactor were released into the environment, and no cleanup of the area outside of the containment building was necessary (Figure 21.21). Figure 21.21(a) In this 2010 photo of Three Mile Island, the remaining structures from the damaged Unit 2 reactor are seen on the left, whereas the separate Unit 1 reactor, unaffected by the accident, continues generating power to this day (right). (b) President Jimmy Carter visited the Unit 2 control room a few days after the accident in 1979. Another major nuclear accident involving a reactor occurred in April 1986, at the Chernobyl Nuclear Power Plant in Ukraine, which was still a part of the former Soviet Union. While operating at low power during an unauthorized experiment with some of its safety devices shut off, one of the reactors at the plant became unstable. Its chain reaction became uncontrollable and increased to a level far beyond what the reactor was designed for. The steam pressure in the reactor rose to between 100 and 500 times the full power pressure and ruptured the reactor. Because the reactor was not enclosed in a containment building, a large amount of radioactive material spewed out, and additional fission products were released, as the graphite (carbon) moderator of the core ignited and burned. The fire was controlled, but over 200 plant workers and firefighters developed acute radiation sickness and at least 32 soon died from the effects of the radiation. It is predicted that about 4000 more deaths will occur among emergency workers and former Chernobyl residents from radiation-induced cancer and leukemia. The reactor has since been encapsulated in steel and concrete, a now-decaying structure known as the sarcophagus. Almost 30 years later, significant radiation problems still persist in the area, and Chernobyl largely remains a wasteland. In 2011, the Fukushima Daiichi Nuclear Power Plant in Japan was badly damaged by a 9.0-magnitude earthquake and resulting tsunami. Three reactors up and running at the time were shut down automatically, and emergency generators came online to power electronics and coolant systems. However, the tsunami quickly flooded the emergency generators and cut power to the pumps that circulated coolant water through the reactors. High-temperature steam in the reactors reacted with zirconium alloy to produce hydrogen gas. The gas escaped into the containment building, and the mixture of hydrogen and air exploded. Radioactive material was released from the containment vessels as the result of deliberate venting to reduce the hydrogen pressure, deliberate discharge of coolant water into the sea, and accidental or uncontrolled events. An evacuation zone around the damaged plant extended over 12.4 miles away, and an estimated 200,000 people were evacuated from the area. All 48 of Japan’s nuclear power plants were subsequently shut down, remaining shuttered as of December 2014. Since the disaster, public opinion has shifted from largely favoring to largely opposing increasing the use of nuclear power plants, and a restart of Japan’s atomic energy program is still stalled (Figure 21.22). Figure 21.22(a) After the accident, contaminated waste had to be removed, and (b) an evacuation zone was set up around the plant in areas that received heavy doses of radioactive fallout. (credit a: modification of work by “Live Action Hero”/Flickr) The energy produced by a reactor fueled with enriched uranium results from the fission of uranium as well as from the fission of plutonium produced as the reactor operates. As discussed previously, the plutonium forms from the combination of neutrons and the uranium in the fuel. In any nuclear reactor, only about 0.1% of the mass of the fuel is converted into energy. The other 99.9% remains in the fuel rods as fission products and unused fuel. All of the fission products absorb neutrons, and after a period of several months to a few years, depending on the reactor, the fission products must be removed by changing the fuel rods. Otherwise, the concentration of these fission products would increase and absorb more neutrons until the reactor could no longer operate. Spent fuel rods contain a variety of products, consisting of unstable nuclei ranging in atomic number from 25 to 60, some transuranium elements, including plutonium and americium, and unreacted uranium isotopes. The unstable nuclei and the transuranium isotopes give the spent fuel a dangerously high level of radioactivity. The long-lived isotopes require thousands of years to decay to a safe level. The ultimate fate of the nuclear reactor as a significant source of energy in the United States probably rests on whether or not a politically and scientifically satisfactory technique for processing and storing the components of spent fuel rods can be developed. Link to Learning Explore the information in this link to learn about the approaches to nuclear waste management. Nuclear Fusion and Fusion Reactors The process of converting very light nuclei into heavier nuclei is also accompanied by the conversion of mass into large amounts of energy, a process called fusion. The principal source of energy in the sun is a net fusion reaction in which four hydrogen nuclei fuse and produce one helium nucleus and two positrons. This is a net reaction of a more complicated series of events: 4 1 1 H⟶4 2 He+2 0+1 e+4 1 1 H⟶2 4 He+2+1 0 e+4 1 1 H⟶2 4 He+2+1 0 e+ A helium nucleus has a mass that is 0.7% less than that of four hydrogen nuclei; this lost mass is converted into energy during the fusion. This reaction produces about 3.6 ××× 10 11 kJ of energy per mole of 4 2 He 2 4 He 2 4 He produced. This is somewhat larger than the energy produced by the nuclear fission of one mole of U-235 (1.8 ××× 10 10 kJ), and over 3 million times larger than the energy produced by the (chemical) combustion of one mole of octane (5471 kJ). It has been determined that the nuclei of the heavy isotopes of hydrogen, a deuteron, 2 1 H 1 2 H 1 2 H and a triton, 3 1 H,1 3 H,1 3 H, undergo fusion at extremely high temperatures (thermonuclear fusion). They form a helium nucleus and a neutron: 2 1 H+3 1 H⟶4 2 He+1 0 n 1 2 H+1 3 H⟶2 4 He+0 1 n 1 2 H+1 3 H⟶2 4 He+0 1 n This change proceeds with a mass loss of 0.0188 amu, corresponding to the release of 1.69 ××× 10 9 kilojoules per mole of 4 2 He 2 4 He 2 4 He formed. The very high temperature is necessary to give the nuclei enough kinetic energy to overcome the very strong repulsive forces resulting from the positive charges on their nuclei so they can collide. Useful fusion reactions require very high temperatures for their initiation—about 15,000,000 K or more. At these temperatures, all molecules dissociate into atoms, and the atoms ionize, forming plasma. These conditions occur in an extremely large number of locations throughout the universe—stars are powered by fusion. Humans have already figured out how to create temperatures high enough to achieve fusion on a large scale in thermonuclear weapons. A thermonuclear weapon such as a hydrogen bomb contains a nuclear fission bomb that, when exploded, gives off enough energy to produce the extremely high temperatures necessary for fusion to occur. Another much more beneficial way to create fusion reactions is in a fusion reactor, a nuclear reactor in which fusion reactions of light nuclei are controlled. Because no solid materials are stable at such high temperatures, mechanical devices cannot contain the plasma in which fusion reactions occur. Two techniques to contain plasma at the density and temperature necessary for a fusion reaction are currently the focus of intensive research efforts: containment by a magnetic field and by the use of focused laser beams (Figure 21.23). A number of large projects are working to attain one of the biggest goals in science: getting hydrogen fuel to ignite and produce more energy than the amount supplied to achieve the extremely high temperatures and pressures that are required for fusion. At the time of this writing, there are no self-sustaining fusion reactors operating in the world, although small-scale controlled fusion reactions have been run for very brief periods. Figure 21.23(a) This model is of the International Thermonuclear Experimental Reactor (ITER) reactor. Currently under construction in the south of France with an expected completion date of 2027, the ITER will be the world’s largest experimental Tokamak nuclear fusion reactor with a goal of achieving large-scale sustained energy production. (b) In 2012, the National Ignition Facility at Lawrence Livermore National Laboratory briefly produced over 500,000,000,000 watts (500 terawatts, or 500 TW) of peak power and delivered 1,850,000 joules (1.85 MJ) of energy, the largest laser energy ever produced and 1000 times the power usage of the entire United States in any given moment. Although lasting only a few billionths of a second, the 192 lasers attained the conditions needed for nuclear fusion ignition. This image shows the target prior to the laser shot. (credit a: modification of work by Stephan Mosel) PreviousNext Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson, PhD Publisher/website: OpenStax Book title: Chemistry 2e Publication date: Feb 14, 2019 Location: Houston, Texas Book URL: Section URL: © Jul 9, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University. Our mission is to improve educational access and learning for everyone. OpenStax is part of Rice University, which is a 501(c)(3) nonprofit. Give today and help us reach more students. Help Contact Us Support Center FAQ OpenStax Press Newsletter Careers Policies Accessibility Statement Terms of Use Licensing Privacy Policy Manage Cookies © 1999-2025, Rice University. 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Universe 48056 48056 admin { } Anonymous Anonymous 2 false false [ "article:topic", "Friedmann equation", "critical density", "license:ccbyncsa", "showtoc:no", "transcluded:yes", "authorname:cominskyetal", "autonumheader:yes2", "cssscreen:ssu", "source-phys-31454" ] [ "article:topic", "Friedmann equation", "critical density", "license:ccbyncsa", "showtoc:no", "transcluded:yes", "authorname:cominskyetal", "autonumheader:yes2", "cssscreen:ssu", "source-phys-31454" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents Home Campus Bookshelves Chicago State University PH S 1150: Basic Astronomy 17: Dark Energy and the Fate of the Universe 17.3: The Friedmann Equation and the Fate of the Universe Expand/collapse global location PH S 1150: Basic Astronomy Front Matter 0: Review of Mathematics 1: Size and Scope 2: Light 3: Telescopes 4: Moving Through Space 5: Moving Through Time 6: Measuring Cosmic Distances 7: Classical Physics- Gravity and Energy 8: Dark Matter 9: Special Relativity 10: General Relativity 11: Black Holes 12: Gravitational Lenses 13: The Expansion of the Universe 14: The Growth of Structure 15: The Cosmic Microwave Background 16: The Early Universe 17: Dark Energy and the Fate of the Universe 18: Knowls Back Matter 17.3: The Friedmann Equation and the Fate of the Universe Last updated Aug 21, 2021 Save as PDF 17.2: Candidates for Dark Energy 17.4: Cosmic Concordance and Cosmological Parameters Page ID 48056 Kim Coble, Kevin McLin, & Lynn Cominsky San Francisco State University, Chico State University, & Sonoma State University ( \newcommand{\kernel}{\mathrm{null}\,}\) Table of contents WHAT DO YOU THINK: FATE OF THE UNIVERSE THE FRIEDMANN EQUATION Math Exploration 17.1: Calculating the Critical Density THE UNIVERSE WITHOUT DARK ENERGY EXPANSION OF THE UNIVERSE DEPENDS ON MATTER DENSITY EXPANSION AND GEOMETRY OF THE UNIVERSE EXPANSION SCENARIO AND AGE OF THE UNIVERSE UNIVERSE WITH DARK ENERGY BEHAVIOR OF DENSITY OVER TIME Some say the world will end in fire, Some say in ice. From what I’ve tasted of desire I hold with those who favor fire. But if it had to perish twice, I think I know enough of hate To say that for destruction ice Is also great And would suffice. WHAT DO YOU THINK: FATE OF THE UNIVERSE Some students are contemplating the end of the Universe. Bo: I think the Universe will keep expanding and cooling, galaxies will keep pulling farther apart from each other, and eventually everything will be dark. Candice: I don't think it makes sense to expand forever. I think the expansion will only go for so long, then it will begin to contract because of gravity . I think it’s going to reach a maximum, then collapse back onto itself. Dante: It strikes me that the Universe ought to be timeless and that the Universe is just expanding and collapsing perpetually. Emily: No, I think Bo is right, except I think it’s supposed to eventually expand so much that everything gets ripped apart. Feng: I don’t think there’s any way to know or predict the fate of the Universe. Do you agree with any or all of these students and, if so, whom? Bo Candice Dante Emily Feng None Explain. THE FRIEDMANN EQUATION The basis to understanding how the expansion of the Universe can be speeding up lies in the Einstein equations. General relativity can tell us physically what factors relate to the expansion. One version of the Einstein equations is the Friedmann equation, which describes how the expansion rate changes over time, and why, in a homogeneous and isotropic Universe. H2−8πGρ3\=kc2S2H2−8πGρ3\=kc2S2 (expansion)−(density)\=(curvature)(expansion)−(density)\=(curvature) In this equation, H is the Hubble parameter (expansion rate), ρ is the average density of matter and energy in the Universe, S is the scale factor, and k is a number that describes the overall curvature of the Universe. That means the expansion is related to the contents of the Universe as well as the curvature of space. The density term includes all of the types of matter and energy in the Universe. ρ\=ρbaryon+ρcdm+ρradiation+ρDEρ\=ρbaryon+ρcdm+ρradiation+ρDE Here ρbaryonρbaryon is the density of regular matter (baryons), ρcdmρcdm is the density of cold dark matter, ρradiationρradiation is the density of radiation, and ρDEρDE is the dark energy density. The Friedmann equation embodies the essence of the Einstein equation: matter and energy affect how spacetime bends. In general relativity, it is not only matter that produces curvature, but any sort of energy. This means that the gravitational attraction has a contribution not just from regular matter, but also from dark matter and radiation (massless particles like photons). The kinetic energy of particles is generally included in the Einstein equations as a pressure term, because pressure can be thought of as an energy density, or a kinetic energy per unit volume. There is also negative pressure from dark energy. There is a special value of the density that causes the Universe to have zero curvature. It is called ρcritρcrit, the critical density: ρcrit\=3H28πGρcrit\=3H28πG If the current density is equal to the critical density, then the left side of the Friedmann equation equals 0, implying that k = 0. We say that such a Universe has no curvature, or that it is flat. Math Exploration 17.1: Calculating the Critical Density Worked Example: 1. What is the value of the critical density? We can calculate ρcρc using a Hubble constant of H0 \= 70 km/s/Mpc. The value of the gravitational constant is G = 6.67 x 10-11 m3/kg s2. Given: H0 \= 70 km/s/Mpc, G = 6.67 x 10-11 m3/kg s2 Find: ρcρc Concepts: ρc = 3H2/(8πG) 1 Mpc = 3.086 × 1019 km Solve: ρc\=3(70 km/s/Mpc)28π(6.67×10−11 m3/kg s2)ρc\=3(70 km/s/Mpc)28π(6.67×10−11 m3/kg s2) Now we need to do some unit conversions: ρc\=3(70 kms⋅Mpc×1 Mpc3.086×1019km)28π(6.67×10−11m3kg⋅s2)ρc\=3(70 kms⋅Mpc×1 Mpc3.086×1019km)28π(6.67×10−11m3kg⋅s2) Solving and simplifying units, we get: ρc\=3.07×10−27kg/m3ρc\=3.07×10−27kg/m3 In the Friedmann equation, only the ratio of density to the critical density is important. This ratio is so important that it is given its own name: Ω (Greek letter “omega”): Ω≡ρρcritΩ≡ρρcrit In general relativity, all of the sources of matter and energy are included and contribute to the total energy density. Thus we have: Ω\=Ωbaryon+Ωcdm+Ωradiation+ΩDEΩ\=Ωbaryon+Ωcdm+Ωradiation+ΩDE Here ΩbaryonΩbaryon is the baryon content, ΩcdmΩcdm is the amount of cold dark matter, ΩradiationΩradiation is the radiation content, and ΩDEΩDE is the contribution from dark energy. If Ω\=1Ω\=1, that means the density is equal to the critical density, so we have a flat Universe (k = 0). We will generally discuss the density and geometry of the Universe in terms of ΩΩ from now on rather than in terms of k or ρcritρcrit. One of the major efforts in cosmology over the past several decades has been to determine the value of ΩΩ. From the Friedmann equation, we can see that the interplay of the expansion of the Universe, density, and curvature lead to several possibilities for the fate of the Universe, depending on which term in the equation dominates: Critical Universe: We have already discussed the case of a critical Universe, where the expansion and density terms are equal, Ω\=1Ω\=1, and space overall is not curved. If there is no dark energy, the Universe will continue to expand, but increasingly slowly. Matter terms dominate: If the Universe contains enough mass to counteract its expansion and there is no dark energy, it will eventually collapse. This is known as a closed Universe. In this case, Ω>1Ω>1. Expansion term dominates: If the Universe does not contain enough mass to counteract its expansion and there is no dark energy, it will expand forever. This is known as an open Universe. In this case, Ω<1Ω<1. Dark energy dominates: None of the three possibilities above lead to the accelerating expansion rate supported by supernova data. Dark energy can cause the expansion to speed up; therefore, we must examine possibilities that include dark energy. We will consider each of these in detail next. We will explore how the expansion rate changes over time, what the matter-energy density (ΩΩ) is like in each case, the resulting curvature of space, and what conditions in the Universe will be like in the future. First, we will consider the three scenarios with no dark energy, then we will explore scenarios that include dark energy. THE UNIVERSE WITHOUT DARK ENERGY The interplay between the terms of the Friedmann equation can be compared to the interplay between gravity and the energy of motion for a ball rising into the air using Newtonian physics. Case 1: A Critical Universe First, consider a scenario where a ball is launched at escape velocity from Earth’s surface. The kinetic energy from the ball’s motion exactly equals its gravitational potential energy. In other words, the sum of the terms for motion and gravity is zero. In this case, the ball will continue to fly away from Earth, but at an ever slower rate. This is analogous to a critical Universe. In a critical Universe the expansion (motion) and density (gravity) terms of the Friedmann equation sum to zero. The motion diagram for a ball at escape speed (position vs. time) and a diagram for the scale factor vs. time in a critical Universe are compared in Figure 17.4. Figure 17.4: Left: a ball is launched from Earth at escape velocity (which is found by setting kinetic and potential energies equal to one another). The ball continues to move away from Earth, but at increasingly slower speeds, as shown by the flattening slope of the graph. Right: The scale factor of the Universe in the case of critical density. The expansion and gravity terms in the Friedmann equation are equal. The Universe continues to expand, but at an increasingly slower rate, as shown by the flattening slope. Credit: NASA/SSU/Aurore Simonnet. In general relativity, the curvature of space as a whole is related to the expansion and gravity terms. In the case of a critical Universe, the curvature is zero. The expansion and geometry for a critical Universe are depicted in Figure 17.5. Figure 17.5: The expansion and geometry for a critical Universe. Left: A graph of the scale factor vs. time. Center: An illustration showing that the Universe continues to expand but ever more slowly. Right: The geometry of the Universe is flat, in other words, there is no curvature in any of the three spatial dimensions. Credit: NASA/SSU/Aurore Simonnet In a critical Universe, space will continue to expand. As it expands, it will cool endlessly, approaching a temperature of 0 K (—273°C). It is at 2.7 K already, and as the temperature drops, the motions of particles and molecules will slow down. Stars will eventually burn out. Galaxies will run out of gas to make any new stars. No energetic photons will be produced to keep things warm. Collisions between particles will be too lethargic to excite electrons, and those collisions will become less and less frequent as the density drops. Black holes will eventually evaporate, and some particles thought currently to be stable might decay to simpler forms. Eventually, physical processes will simply cease, after unimaginably long timescales in excess of 10100 years. The fate of the Universe in this case is known as a big chill (Animated Figure 17.6). @api/deki/files/54210/LinearExpansion.mp4?origin=mt-web < video src=”@api/deki/files/54210/LinearExpansion.mp4” type=”video/webm” controls > < /video > Animated Figure 17.6: In a big chill scenario, galaxies continue to move away from each other forever, as the Universe expands and cools. Credit: NASA/SSU/Kevin John Case 2: A Closed Universe Now consider a case where a ball is launched from Earth’s surface, but with less kinetic energy. In that case, the ball will rise to a certain height and then fall back down to the ground because the gravitational potential energy overwhelms the energy of motion. This is analogous to what happens if the density of matter is greater than that required to produce a flat Universe. In that case, gravity will overwhelm the expansion of the Universe and it will eventually stop expanding and re-collapse. The density is greater than critical (Ω>1Ω>1). The left side of the Friedmann equation is negative, implying that the curvature is positive. That is the case for a closed Universe. The motion diagram (position vs. time) for a ball launched at less than escape speed and a diagram for the scale factor vs. time in a closed Universe are compared in Figure 17.7. The expansion and geometry for a critical Universe are depicted in Figure 17.8. Figure 17.7 Left: A ball is launched from Earth at slower than escape velocity. At first it moves away from Earth, but then it reaches a maximum height and falls back down due to gravity. Right: The scale factor of the Universe in the case of a closed Universe. The gravity term in the Friedmann equation outweighs the expansion term. At first the Universe expands, but then it contracts due to gravity. Credit: NASA/SSU/Aurore Simonnet. Figure 17.8 The expansion and geometry for a closed Universe. Left: A graph of the scale factor vs. time. Center: An illustration showing that the Universe expands then contracts. Right: The curvature of the Universe as a whole is positive, or in other words, the geometry of the Universe is spherical. Note that this means the global geometry is analogous to that of a sphere. It does not mean that the Universe is a sphere. Credit: NASA/SSU/Aurore Simonnet. In a closed Universe, space will expand for a while but eventually stop and begin to collapse in upon itself, heating as it does so. Galaxies will move toward each other and the temperature of the Universe will increase. The particles in galaxies will eventually merge together into a state of high temperature and density similar to that found in the beginning of the Universe. The fate of the Universe in this case is known as a big crunch. It is depicted in Animated Figure 17.9. Animated Figure 17.9 In a big crunch scenario, the gravitational attraction of matter causes galaxies to eventually start moving toward each other and the Universe becomes more dense. Credit: NASA/SSU/Kevin John. Case 3: An Open Universe At the opposite extreme, when the density of the Universe is less than critical (Ω<1Ω<1), the gravity will never be able to halt the expansion. This is analogous to a ball launched from Earth with a speed greater than escape speed; the kinetic energy of the ball can overcome the gravitational pull from Earth. The motion diagram (position vs. time) for a ball launched with kinetic energy greater than its gravitational potential energy and a diagram for the scale factor vs. time in a closed Universe are compared in Figure 17.10. Figure 17.10 Left: A ball is launched from Earth at a speed greater than escape velocity. It flies away from Earth because its kinetic energy is greater than its potential energy. Right: The scale factor of the Universe in the case of an open Universe. The expansion term in the Friedmann equation outweighs the gravity term and the Universe expands forever at a constant rate. Credit: NASA/SSU/Aurore Simonnet. In the case of an open Universe, the left side of the Friedmann equation is positive, so the curvature must be negative. That happens when the curvature of the Universe is saddle-shaped. The expansion and geometry for an open Universe are depicted in Figure 17.11. In an open Universe, space will continue to expand. In a fate similar to that of the critical case, galaxies will move farther apart from each other, the temperature of the Universe will cool, and stars in galaxies will eventually burn out. Again, we would have a big chill scenario. Figure 17.11 The expansion and geometry for an open Universe. Left: A graph of the scale factor vs. time. Center: An illustration showing that the Universe expands forever. Right: The curvature of the Universe as a whole is negative, or in other words, the geometry of the Universe is saddle-shaped (hyperbolic). Credit: NASA/SSU/Aurore Simonnet. From these three cases—open, closed, and critical—we can see that changing the amount of matter in the Universe changes the way it expands. A lot of matter means that the cosmic expansion cannot overcome the gravitational attraction of the matter. The expansion could eventually slow until it stops, and then the Universe will re-collapse. On the other hand, if there is not enough matter in the Universe, then the expansion continues forever. These two cases are separated by a condition in which there is just enough matter for expansion to balance gravity—the critical case. If no dark energy is present, once we determine the value of the curvature, we know it forever. It represents the geometry of the Universe as a whole. This is true whether the matter in the Universe is baryons or cold dark matter (or a combination of both). The Friedmann equation alone determines whether the Universe keeps expanding forever or eventually re-collapses; whether the scale factor S grows or shrinks is determined here only by the density of the Universe, which also determines its overall geometry. EXPANSION OF THE UNIVERSE DEPENDS ON MATTER DENSITY In the simplest cosmological models, the fate of the Universe depends on its rate of expansion, expressed by the Hubble parameter, and the amount of matter it contains. Qualitatively, the Hubble constant gives us an idea of the kinetic energy associated with the expansion of space. The matter content gives us an idea of the gravitational potential energy. In a Newtonian context, the two taken together give us the total energy. In general relativity, they produce the total curvature of the Universe. In this activity, you will be allowed to adjust ΩΩ and take note of how the expansion varies as this parameter changes. Play Activity 1. In which scenario does the Universe eventually collapse? Ω < 1 Ω = 1 1"> Ω > 1 2. In which scenario does the Universe expand at a constant rate forever? Ω < 1 Ω = 1 1"> Ω > 1 3. In the Ω = 1 case, the graph of scale factor vs. time is a straight line. True False EXPANSION AND GEOMETRY OF THE UNIVERSE For the questions below, refer to Figure A.17.8, which shows three possible scenarios for the expansion of the Universe and the geometry that goes with them. Figure A.17.8 Three different scenarios for the expansion of a matter-only Universe. On the left are graphs of the scale factor vs. time, in the center are illustrations depicting the scale factor of the Universe as time advances, and on the right are models for the curvature of space. Each scenario corresponds to a different curvature. Top row: positive curvature, the Universe expands for a while, then stops expanding and eventually collapses. Center row: zero curvature, the Universe expands forever but increasingly slowly. Bottom row: negative curvature, the Universe expands at a constant expansion rate forever. Credit: NASA/SSU/Aurore Simonnet. 1. Which scenario features negative curvature? Top row Center row Bottom row 2. Which scenario shows a flat Universe? Top row Center row Bottom row 3. Which scenario shows a Universe with positive curvature? Top row Center row Bottom row 4. Which scenario features a Universe that expands forever, but increasingly slowly? Positive curvature Zero curvature Negative curvature 5. In which scenario does the Universe expand at a constant rate forever? Positive curvature Zero curvature Negative curvature 6. In which scenario does the Universe eventually collapse? Positive curvature Zero curvature Negative curvature In the previous activity, you saw that the Universe will either expand forever or eventually stop expanding, depending on how much mass it contains. In two of the cases, the expansion slows over time. The slowing is the result of the gravitational attraction that all the mass in the Universe has for all the other mass. The slowdown means that the age of the Universe is not simply the reciprocal of the Hubble constant, 1/H__0, as would be the case for a constant expansion rate. In the next activity, you will examine how the matter density and expansion rate affect the age of the Universe. EXPANSION SCENARIO AND AGE OF THE UNIVERSE In this activity, we will explore how changing the amount of matter and the expansion rate affects the age of the Universe for a closed Universe scenario. The age of the Universe will be the amount of time that passes between the beginning of the Universe and the point at which the scale factor again becomes zero (blue line intersects the x-axis). Play Activity 1. What happens to the age of the Universe if the Hubble constant (expansion rate) is greater? The age is greater. The age is less. The age does not change. 2. What happens to the age of the Universe if the density (Ω) is greater? The age is greater. The age is less. The age does not change In the previous activity, you saw how changing the value of the expansion rate and density affected the age for a closed Universe. Figure 17.12 shows how the scale factor changes with time in the closed, open, and critical cases. It also shows what would happen in a case where the expansion is accelerating. We can see that the age of the Universe (time from the beginning until “now”) is smallest for a closed Universe and increases as ΩΩ gets smaller. Physically, this is because the age of the Universe is inversely proportional to the expansion rate; if the Universe expands more quickly, it takes less time to reach the state it is in today. If Ω>1Ω>1, the expansion rate was greater in the past than it is today; thus the age will be younger compared to a Universe with constant expansion. The reverse is true if the expansion rate is speeding up. Figure 17.12 Several different possibilities for the age of the Universe. The red curve represents a closed Universe (Ω>1Ω>1), which is the youngest possibility (about 7 billion years). As the density decreases, the age of the Universe today increases. For example, for a flat Universe with no dark energy as seen in the blue curve (Ω\=1Ω\=1), the age of the Universe would be about 9 billion years; for a Universe devoid of matter (green curve) the age would be even greater. The greatest age for the Universe occurs in a Universe with an accelerating expansion due to dark energy (black curve). In this case, the age of the Universe matches observations (13.8 billion years old), further evidence that the Universe contains dark energy. Credit: NASA/SSU/Aurore Simonnet. UNIVERSE WITH DARK ENERGY In the previous section, we looked at how changing the amount of matter in the Universe relates to its expansion, age, and overall geometry. We saw a variety of outcomes, but none of them predicted an accelerating expansion. For that, we need dark energy. Including dark energy can cause dramatically different outcomes for the age, geometry, and eventual fate of the Universe, as compared to cases without dark energy. The exact outcome depends on the nature of the dark energy. There are three interesting dark energy scenarios: the dark energy could be a cosmological constant, it could grow stronger, or it could grow weaker. In the case of dark energy being a cosmological constant (ΛΛ), its density is constant—it is a property of space itself. Over time its contribution to the total energy density (ΩΩ) increases because there is more and more space (containing a constant amount of ΛΛ per unit volume). On the other hand, the total amount of matter and radiation is fixed. As the expansion progresses they become more and more dilute. In the next activity, you will see how in this scenario the total density to change over time, and how its constituent parts contribute different amounts at different eras in the evolution of the Universe. BEHAVIOR OF DENSITY OVER TIME In this activity you will explore how the proportions of radiation, matter, and dark energy (called ΩRΩR, ΩMΩM, and ΩDΩD, respectively) change over the history of the Universe. They are represented as fractions of the overall matter—energy budget in a cosmic pie chart. Here, we assume that the dark energy takes the form of a cosmological constant. Use the slider bar to adjust the redshift (zz) and answer the following questions. Play Activity 1. When we look at higher redshift (z), we are (choose all that apply): Looking farther back in time Looking at the Universe earlier times in its history Looking at later times in its history Redshift is not related to time 2. What happens to the proportion of radiation as redshift gets bigger? It is bigger It is smaller It remains the same 3. At what redshift were the proportions of radiation and matter approximately equal to each other? z = 4. Was this before, after, or at the same time as the formation of the CMB (z ~ 1100)? Before After Same time 5. Around what redshift do we start to see dark energy making up part of the matter-energy budget? z = 6. Dark energy is what percentage of the matter-energy budget today? % As time advances from the past until now, the redshift gets smaller and the scale factor (S) gets larger. In the last activity, you should have seen that the energy contribution from matter and radiation as a percentage of the cosmic matter-energy budget both diminish, while the percentage of dark energy increases. The contribution from matter drops as 1/S3 because the particles are being spread over a greater volume as the Universe expands. The radiation contribution shrinks faster than matter because in addition to the number of photons per unit volume dropping (like the number of baryons and dark matter particles), the photon energy is also decreased due to the cosmological expansion; as the light shifts to longer wavelengths, the energy per photon drops. Thus, the total energy contribution from radiation decreases like 1/S4. Dark energy (if interpreted as a cosmological constant) remains a constant number as the scale factor increases and thus becomes a greater percentage of the matter-energy budget. We can write this mathematically if we express the total density in terms of the scale factor. ρ(S)\=ρcrit[ΩDE+ΩMS3+ΩRS4]ρ(S)\=ρcrit[ΩDE+ΩMS3+ΩRS4] The scale factor here is S, ρcritρcrit is the current critical density, ΩDEΩDE is the amount of dark energy, ΩMΩM is the amount of matter, and ΩRΩR is the amount of radiation. We can also demonstrate how this equation works in Figure 17.14, where we show how each of the densities (radiation, matter, and dark energy) drop as time advances. Figure 17.13 Energy density vs. time. The black line shows the decrease in the energy density of matter, and the red line shows the decrease in the energy density of radiation. The blue line shows the constant energy density of dark energy when it is assumed to be in the form of a cosmological constant (although other possibilities have not been completely ruled out). Earlier times in the history of the Universe are to the left and later times are to the right. Today is zero on the x-axis. Credit: NASA/SSU/Aurore Simonnet based on Frieman, Turner, and Huterer 2008. Annual Reviews in Astronomy and Astrophysics, 46, 385. As we can see from Figure 17.13, dark energy (in the form of a cosmological constant) remains unaffected by the expansion. As dark energy becomes more dominant, it makes the expansion speed up, and objects in the Universe begin to accelerate away from each other. In a Universe with a cosmological constant, we will have a big chill scenario, plus an additional interesting effect. Galaxies will move away from each other increasingly quickly such that eventually an observer in any galaxy will see all galaxies outside her own small region disappear over her cosmic horizon. The only objects visible will be the ones in her own galaxy and possibly a few bound neighbors such as those in the Local Group of the Milky Way. This is a slow-motion recapitulation of what is thought to have occurred during the inflationary period in the early Universe. The expansion of the Universe in a scenario with a cosmological constant is depicted schematically in Figure 17.14. Unlike a matter-only Universe, one with ΛΛ will eventually speed up, no matter the initial curvature. Figure 17.14 In a Universe that contains both matter and dark energy, it is possible for the expansion rate to accelerate with time. In some models, the dark energy can increase with time, in others it remains constant, while in others it may eventually decrease. In order to predict the eventual fate of the Universe, we need to understand more about the properties of dark energy. Credit: NASA/SSU/Aurore Simonnet. The nature of the dark energy is still not settled. Increasing evidence indicates that it is indeed a cosmological constant, a constant property of space like the vacuum energy. However, other possibilities have not been completely ruled out; dark energy could change over time, becoming either stronger or weaker. If the strength of the dark energy grows in time, and not only as a proportion of the total energy density like the cosmological constant, the dark energy concentration per unit volume increases. The increase in density will lead to an exponential expansion of space like the one that caused inflation early on. Eventually, the dark energy will become so strong that the expansion will tear galaxies apart. It will even rip molecules, atoms, and protons apart as its strength grows without bound. This scenario is referred to as the big rip (Animated Figure 17.15). Animated Figure 17.15 In a big rip scenario, galaxies continue to move away from each other forever, as the Universe expands and cools. Dark energy that grows in strength causes the expansion to accelerate so strongly that eventually galaxies and even atoms to be ripped apart. Credit: NASA/SSU/Kevin John. Finally, the dark matter might have a decreasing density and becomes less important with time. In this case, the expansion of the Universe would continue forever, but at a continually slowing rate, a scenario qualitatively similar to a Universe with no dark energy at all. In fact, that is what such a Universe eventually becomes as its dark energy density approaches zero. Table 17.1 summarizes the possible scenarios for the fate of the Universe, both with and without dark energy. Table 17.1 Possible scenarios for the fate of the Universe in the Big Bang model \| SCENARIO \| DOMINANT TERM \| MATTER DENSITY \| CURVATURE \| OUTCOME \| --- --- \| Critical (No Dark Energy) \| None \| Ωm \= 1 \| Zero (flat) \| Big Chill \| \| Closed (No Dark Energy) \| Matter \| Ωm > 1 \| Positive (like a sphere) \| Big Crunch \| \| Open (No Dark Energy) \| Expansion \| Ωm < 1 \| Negative (like a saddle) \| Big Chill \| \| Dark Energy Constant \| Dark Energy \| could be any and still get acceleration at some point \| could be any and still get acceleration \| Big Chill \| \| Dark Energy Increases \| Dark Energy \| could be any and still get acceleration at some point \| could be any and still get acceleration \| Big Rip \| \| Dark Energy Decreases \| Dark Energy and then Matter \| depends on specific model \| could be any and still get expansion \| Big Chill \| The Big Bang model not only tells us where we have come from, it also predicts where we are going — in the cosmic sense. The model explains observations of the state of the Universe in the past and the present, and it also predicts what will happen to the Universe in the future. The Big Bang model gives us insight into questions like: Will the Universe expand forever? Will it stop expanding and then collapse? The Big Bang model and general relativity tell us each of those scenarios is possible, and they predict specifically what the Universe will be like in each of them. They also tell us how to determine which scenario will occur based on quantities we can measure today. So, what is the fate of our Universe? Will it expand forever, endlessly cooling to the point of a “big chill,” will it eventually reverse its direction and collapse into a “big crunch,” or will dark energy grow in strength enough to cause a “big rip?” In the next section, we will see how a combination of different observations has helped astronomers answer these questions. This page titled 17.3: The Friedmann Equation and the Fate of the Universe is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Kim Coble, Kevin McLin, & Lynn Cominsky. 17.3: The Friedmann Equation and the Fate of the Universe by Kim Coble, Kevin McLin, & Lynn Cominsky is licensed CC BY-NC-SA 4.0. Toggle block-level attributions Back to top 17.2: Candidates for Dark Energy 17.4: Cosmic Concordance and Cosmological Parameters Was this article helpful? 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15036	https://fiveable.me/key-terms/ap-gov/stratified-sampling	Stratified Sampling - (AP US Government) - Vocab, Definition, Explanations \| Fiveable \| Fiveable new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom ap study content toolsprintablespricing my subjectsupgrade All Key Terms AP US Government Stratified Sampling 👩🏾‍⚖️ap us government review key term - Stratified Sampling Citation: MLA Definition Stratified sampling is a statistical method used to ensure that specific subgroups within a population are adequately represented in a sample. By dividing the population into distinct layers or strata based on shared characteristics, researchers can draw a more accurate representation of public opinion, making it easier to evaluate and measure sentiment across different demographics. This approach helps to minimize sampling bias and allows for more precise data analysis. 5 Must Know Facts For Your Next Test Stratified sampling can lead to greater accuracy in survey results by ensuring that all relevant subgroups are included. This technique is especially useful when certain strata are smaller segments of the overall population, allowing researchers to ensure that their voices are heard. In stratified sampling, strata can be based on various characteristics such as age, gender, income level, or geographical location. Researchers often use proportional stratified sampling, where each subgroup is represented in proportion to its size in the overall population. Stratified sampling enhances the reliability of public opinion data by reducing variability within each stratum and ensuring diverse viewpoints are captured. Review Questions How does stratified sampling improve the representation of public opinion in survey research? Stratified sampling improves representation by ensuring that specific subgroups within a population are adequately included in the sample. By dividing the population into strata based on characteristics such as age, gender, or income, researchers can capture a more accurate reflection of public sentiment. This approach minimizes bias that could arise from random sampling alone and provides insights into the opinions of various demographic segments. Compare and contrast stratified sampling with random sampling regarding their effectiveness in measuring public opinion. Stratified sampling is generally more effective than random sampling for measuring public opinion, especially in diverse populations. While random sampling provides every individual an equal chance of selection, it may overlook smaller subgroups that hold unique perspectives. Stratified sampling specifically targets these groups by ensuring they are represented proportionately, leading to a more nuanced understanding of public attitudes across different demographics. Evaluate the implications of using stratified sampling for policy decisions based on public opinion data. Using stratified sampling for policy decisions allows policymakers to base their actions on comprehensive and representative data reflecting diverse viewpoints within the population. By understanding opinions across different strata, such as age or socioeconomic status, policymakers can tailor their strategies and initiatives to meet the needs of specific groups effectively. This can lead to more equitable policy outcomes and enhance public trust by demonstrating responsiveness to all constituents' opinions. Related terms Population: The entire group of individuals or instances about whom we hope to learn and make conclusions based on collected data. Sampling Error:The difference between the results obtained from a sample and the actual characteristics of the population from which the sample is drawn. Random Sampling:A sampling technique where each individual in the population has an equal chance of being selected, helping to reduce bias. 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15037	https://www.youtube.com/watch?v=gzpaR6I32U4	Solve by Substitution: 4x-y=20 and -2x-2y=10 Minute Math 85700 subscribers 68 likes Description 7211 views Posted: 29 Nov 2017 A math video lesson on Systems of Two Equations. This video solves by substitution 4x-y=20 and -2x-2y=10 #solvebysubstitution #systemsofequations #algebra2 Every Month we have a new GIVEAWAY that is FREE to Enter. See link below for details. You can enter Every Month! Visit our website for links to all of our videos: Need a Math Tutor? Visit our STEM Store for all your Math and Science Gear: Great Educational Resources Here: Nominate a Great Teacher Here: Have Math Lesson you would like to add to our community of videos? Click here: Check out our Finance, Business and Life YouTube Channel: Consider supporting us on Patreon... Follow us for... Tweets: Instagram: Facebook: Instagram Shop: Facebook Shop: Business and Personal Instagram: 5 comments Transcript: Hi, I'm Sean Ganner and this is Minute Math and today we're learn about systems of two equations. We're going to solve each system by substitution. So, if I was given the system 4x - y = 20 and -2x - 2 y = 10. Well, from there, what I want to do is I want to get a variable isolated. I think that's easier in my first equation. So, I'm going to rewrite my first equation here, but I'm subtracted 4x to both sides. So, I have - y = uh -4x + 20. Okay, so hopefully you can see that I subtracted 4x to both sides. From there, I divided by -1 to both sides to get y by itself to be well distribute the 1 pos4x - 20. So, I have y = 4x - 20. And I'm going to plug that in to the y of my second equation. So negative. So this one's coming down now. -2x - 2 my y is I'm plugging right here, right? Uh 4x - 20 and that's equal to 10. From there, I need to distribute the -2 across. So we have -2x right here. -2 a 4x is a - 8x and -2 a -20 is a positive 40 = 10. Let's combine -2x - 8x is a - 10x. And if I subtract a 40 on both sides, okay, I'm left with at 10 - 40 is -30. Divide by -10 to both sides. And I have x = a pos3. So we have x equaling a positive 3 right here. I have to pick one of my other equations to substitute it back in. Let's go to the first equation. So 4x - y = 20. I'm going to plug 3 in for the x. Okay. 4 3 - y = 20. 4 3 is 12 - y = 20. I'm going to subtract 12 to both sides to get y by itself. So - y here. 20 - 12 is a pos 8 / y. Not a negative 8. Negative y right there to both sides. and I have y or sorry negative one. How y what am I doing and of course it doesn't raise well divide both sides by a negative 1 to get rid of that negative and I have y = 8 /1 which is a negative 8. So our final answer here combining these two to be a coordinate point is 38. So let's recap. We're given our two equations here. We're going to solve by substitution. I took my first equation and rewrote it uh by subtracting 4x to both sides and then divided by -1 which I failed to do right there. From there I got y = 4x - 20. I substituted that y value into the y of my uh second equation right there. So I have -2x - 2 the 4x - 20 and that's equal to 10. Simplified distributed the -2 and then I added subtracted 40 to both sides and combined what was left over. I had -10x = -30. Divided by -10 to both sides and got x = 3. From there, I plug that into my first equation. I write 3 in for x. 4 3 is 12. Subtracted 12 to both sides, got 8 / -1. So, gloriously there to get y by itself. Doing y =8. I have my x and y values. So, I can write my coordinate point of 3,8, which is my final answer, which we got by solving the system of equations by substitution.
15038	https://brainly.com/question/38975196	[FREE] (a) Prove that \frac{\cos A - \cos(A + 2B)}{2 \sin B} = \sin(A + B) (b) By rewriting \cos 2\theta in - brainly.com 3 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +83,3k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +12,2k Ace exams faster, with practice that adapts to you Practice Worksheets +6,7k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified (a) Prove that 2 sin B cos A−cos(A+2 B)=sin(A+B) (b) By rewriting cos 2 θ in terms of sin 2 θ or otherwise, prove sin θ+sin 3 θ+sin 5 θ+⋯+sin(2 n−1)θ=1−cos 2 n θ 2 sin θ for all n∈N. 1 See answer Explain with Learning Companion NEW Asked by justin4381 • 10/02/2023 0:00 / -- Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 206083 people 206K 0.0 0 Upload your school material for a more relevant answer To prove the given equation, we can use trigonometric identities and formulas to expand and simplify both sides. By applying the sum-to-product identity for cosine and the double-angle identities for cosine and sine, we can rewrite the left side as sin(A+B), which is the same as the right side. Therefore, the equation is proven to be true. Explanation To prove the given equation, we can use the identities and formulas related to trigonometric functions. Let's start by expanding the left side of the equation: cos(A)−cos(A+2B)/2sin(B) Using the sum-to-product identity for cosine, we can rewrite cos(A+2B) as cos(A)cos(2B)−sin(A)sin(2B). Substituting this into the equation, we get: cos(A)−[cos(A)cos(2B)−sin(A)sin(2B)]/2sin(B) Simplifying further, we have: cos(A)−(cos(A)cos(2B))/2sin(B)+(sin(A)sin(2B))/2sin(B) Using the double-angle identities for cosine and sine, we can rewrite cos(2B) as 2cos²(B)−1 and sin(2B) as 2sin(B)cos(B). Substituting these identities, we get: cos(A)−[cos(A)(2cos²(B)−1)]/2sin(B)+(sin(A)(2sin(B)cos(B)))/2sin(B) Further simplifying, we have: cos(A)−2cos(A)cos²(B)+cos(A)/2sin(B)−sin(A)cos(B) = sin(A+B) To prove the equation, we have shown that the left side is equal to the right side. Therefore, the equation is proven to be true. Learn more about Trigonometric identities here: brainly.com/question/31837053 SPJ11 Answered by chittitejaswini235se •3.5K answers•206.1K people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 206083 people 206K 0.0 0 University Physics Volume 2 - Samuel J. Ling, William Moebs, Jeff Sanny University Physics Volume 3 - Samuel J. Ling, Jeff Sanny, William Moebs Celestial Mechanics - Jeremy Tatum Upload your school material for a more relevant answer The proof of the first identity uses trigonometric transformations to show that the left side equals the sine of the angle sum on the right side. The second identity is confirmed through the application of the sine series sum formula, leading to the equivalent expression as stated. Both results demonstrate significant applications of trigonometric identities. Explanation (a) To prove the identity: 2 sin B cos A−cos(A+2 B)=sin(A+B) we can use trigonometric identities. First, we apply the cosine addition formula, which states that cos(A+2 B)=cos A cos(2 B)−sin A sin(2 B). Thus, we can rewrite the left side as: cos A−(cos A cos(2 B)−sin A sin(2 B)) This simplifies to cos A(1−cos(2 B))+sin A sin(2 B). Now, using the identities: 1−cos(2 B)=2 sin 2(B) and sin(2 B)=2 sin(B)cos(B), we can substitute back in: =2 sin B 2 cos A sin 2(B)+2 sin A sin(B)cos(B) Simplifying gives us: =cos A sin B+sin A cos B This is the same as: sin(A+B) thereby proving the original statement. 2 sin B cos A−cos(A+2 B)=sin(A+B) is valid. \ (b) For the second part: sin θ+sin 3 θ+sin 5 θ+⋯+sin(2 n−1)θ=1−cos 2 n θ 2 sin θ we recognize this as an arithmetic series of sines. We can express this as: S=k=0∑n−1sin((2 k+1)θ) Applying the formula for the sum of sine series: S n=sin(x/2)sin(n x)sin((n+1)x/2) where x=2 θ, we derive the sum equivalently into the desired form. Simplifying leads to S n=1−cos 2 n θ 2 sin θ confirming the equality. This shows that the series sums correctly to the stated expression. Examples & Evidence For example, using specific angle values for A and B in the first identity can illustrate how the left side simplifies to the right side. For numeric values in the series summation, evaluating the sine terms up to n can provide concrete verification of the second identity's outcome through actual calculation of terms. The formulas for the sine and cosine addition and double angle identities are established in trigonometry and can be found in standard textbooks or trigonometric resources, confirming the processes and results in the proofs. Thanks 0 0.0 (0 votes) Advertisement justin4381 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer Which identity is TRUE? A. cosθ−sinθ cos2θ =cosθ+sinθ B. sin2θ 1−cos2θ =cotθ C. sinθ cos2θ + cosθ sin2θ =secθ D. cos2θ+tanθsin2θ=1 only D C and D A and C A and D A, B, C Community Answer Prove the identity (2 - 2cosθ)(sinθ + sin 2θ + 3θ) = -(cos4θ - 1) sinθ + sin 4θ(cosθ - 1) Community Answer Find the sum to n terms of the series below. (They all arise from geometric series.) (a) sinθ+sin3θ+sin5θ+⋯ (b) cosθ+ (1/3)cos2θ+ (1/9) cos3θ+⋯ (c) sinα +xsin(α+β)+x² sin(α+2β)+⋯ (d) cosh1+cosh2+cosh3+⋯. Community Answer [Let a, b, p = 0, 27). The following two identities are given as cos(a + B) = cosa cosß-sina sinß, cos²p+ sin²p=1, (a) Prove the equations in (3.2) ONLY by the identities given in (3.1). cos(a-B) = cosa cosß+ sina sinß, sin(a-B)=sina-cosß-cosa sinß. Hint: sin = cos (b) Prove that as ( 27 - (a− p)) = cos((2-a) + B). sin (a-B)= cos cos²a= 1+cos 2a 2 " (c) Calculate cos(7/12) and sin (7/12) obtained in (3.2). sin² a 1-cos 2a 2 (3.1) (3.2) (3.3) (3.4) respectively based on the results Community Answer Consider the following. cos(a + b) + cos(a - b) = 2 cos(a) cos(b) Prove the identity. cos(a + b) + cos(a - b) + (cos(a) cos(b) + sin(a) sin(b)) = cos(a) cos(b) + cos(a) cos(b) = + sin(a) sin(b) Community Answer Communication: (12marks) 5. Prove the identities a. sin(x+y) sin(x−y) = tanx+tany tanx−tany b. (sinA+cosA)(tanA+cotA)=secA+cscA c. cotθ+cosθ= sinθ cosθ+(1+sinθ) Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? New questions in Mathematics A car's gas mileage refers to its rate in miles traveled per gallon of gas. Kara's family is thinking about renting a camper that can travel 3 2 1 miles on 3 1 gallon of gas. The complex fraction representing the camper's gas mileage in miles per gallon is 3 12 7. How can you rewrite this complex fraction as a division expression? A. 2 7÷3 B. 3 1÷7 C. 3 1÷2 7 D. 2 7÷3 1 Find the domain of the function. f(x)=x 2−7 x+4 Simplify x 2+3 x+2 x 2+8 x+12 integral ∫x−1 1d x Which of the following is an even function? A. f(x)=(x−1)2 B. f(x)=8 x C. f(x)=x 2−x D. f(x)=7 Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com
15039	https://www.intelligencetest.com/cognitive-abilities/logic/tips/deductive-reasoning/syllogism/distribution-terms.html	Distribution of terms in syllogism Take test Practice IQ Test Labs Discover your intellectual strengths touch_app Cognitive abilities Logic tips Syllogism Distributivity Distributivity The terms in statements will be distributed according to the following rules, and apply to both the premises and the conclusion: The predicate, (P), is distributed in negative statements, such as 'No' and 'Some not'. The subject, (S), is distributed in universal statements, such as 'All' and 'No'. Validity and the distribution of terms. Example 1:: Premises: All A is B All B is C Conclusions: All A is C Some C is A Middle term: If the middle term is not distributed in at least one of the two premises, then the argument is not valid. (Fallacy of the undistributed middle). The middle term is the common term between two given premises. A distributed term has a check mark above it. In this case, B is the middle term and it is distributed in the second premise. Terms in conclusion.: When the major term is distributed in the conclusion, it should also be distributed in the major premise (fallacy of the illicit major), and when the minor term is distributed in the conclusion, it should also be distributed in the minor premise (fallacy of the illicit minor). As a general rule, if a term is distributed in the conclusion, the term must be distributed in at least one of the premises. Conclusion 1: 'A' is distributed and it is also distributed in the first premise; the argument is valid. Conclusion 2: No term has a checkmark; the argument is valid. Example 2:: Premises: All A is B No B is C Conclusions: No A is C Some A is not C Middle term:The middle term is distributed in the second premise. Conclusion 1: Both terms have checkmarks in both the conclusion and the premises; conclusion 1 is valid. Conclusion 2: 'C' is distributed in the second premise; conclusion 2 is valid. IQ Testing Take test Practice Quizzes Psychometrics Charts S.T memory Reaction time Resources Newsletter Information Articles In the news Polls Trivia Research studies Quotations Mind games Puzzles Brainteasers Site FAQ's Privacy Sitemap Contact Social media IQ Test Labs IQ Test Labs © 2021
15040	https://www.geeksforgeeks.org/engineering-mathematics/permutation-matrix/	Permutation Matrix Permutation Matrices stand out as a distinct and important element, mentioning the algebraic linear regression and integers in the combination. These matrices are composed of 0s and 1s and are more than just a special mathematical matrix. Knowing the permutation matrices provides the capability to instruct how the data can be affected and managed in particular mathematical systems. This article, thus, introduces the Permutation Matrix with some practical applications and solved examples. Table of Content What is Permutation Matrix? A permutation matrix is a special type of square binary matrix that represents a permutation of elements. It is constructed by rearranging the rows or columns of an identity matrix according to a specific permutation. It is an n × n square matrix (where n is the number of elements being permuted) with the following distinct properties: Example for Permutation Matrix Consider the identity matrix I3: I_3 = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} A permutation of the rows could be the order [2, 3, 1], which would result in the permutation matrix: P = \begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{pmatrix} Mathematical Representation of Permutation Matrix The mathematical representation of a permutation matrix is a technique to perform the permutation of the elements in the given set. Like, we have a permutation σ of the set {1, 2, 3, . . . , n}. The permutation matrix P whose corresponding permutation is written like that is an n × n matrix defined with the given details: P_{ij} = \begin{cases} 1 & \text{ } j = \sigma(i) \ 0 & \text{}j \neq \sigma(i) \ \end{cases} This statement ensures each row is linked directly to each column, and vice versa. Each column does not know details about other than a particular row that contains one, whereas the remaining part of the column is 0. The number of lines cannot be higher than one and the number of columns cannot be greater than two 1's at the same time. Each row of the P-permutation matrix refers to one and only nonzero entry and thus all the remaining entries are 0. Thus a permutation matrix acts on the permutation σ by element rearrangement. How a Permutation Matrix Works? When a permutation matrix P multiplies a vector v, the result is a vector whose elements have been rearranged according to the permutation defined by P. For example, if v = \begin{pmatrix} a \ b \ c \end{pmatrix}, then multiplying by the permutation matrix P from above: Pv = \begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} a \ b \ c \end{pmatrix} = \begin{pmatrix} b \ c \ a \end{pmatrix} Properties of Permutation Matrix Permutation matrices have several important properties that make them useful in various mathematical and computational applications. Here are some key properties of permutation matrices: Applications of Permutation Matrix The permutation matrices are given a wide array of possible applications in various fields, such as numerical linear algebra for they can reorder the rows and columns of the tables, and maintain the structure. In the following, there are several noteworthy applications: Constructing a Permutation Matrix To construct a permutation matrix for a given permutation of a set, follow these steps: Example of Constructing Permutation Matrix Now, we take up a permutation of set {1, 2, 3}, σ = (2, 3, 1), and we wish to make a permutation matrix for it. Step 1. Identify the Permutation σ maps 1 to 2, 2 to 3, and 3 to 1. Step 2. Initialize the Matrix Initially, get a matrix 3x3 filled with zeros: \begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix} Step 3. Place the 1s For i = 1, σ(1) = 2. Put 1 in position (1, 2): \begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix} For i = 2, σ(2) = 3. Put 1 in position (2, 3): \begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0 \end{pmatrix} For i = 3, σ(3) = 1. Place 1 in position (3, 1): \begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{pmatrix} The resulting permutation matrix P for the permutation σ = (2, 3, 1) is: P = \begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{pmatrix} This matrix is an effective way to represent the permutation and can be applied to vector or matrix accordingly. Solved Problems on Permutation Matrix Problem 1: Check whether that the given matrix is a permutation matrix: P = \begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{pmatrix} Solution: Confirm it is a permutation matrix by looking for the properties specified below. For matrix P: Rows: Columns: Since P meets both criteria, so it is a permutation matrix. Problem 2: Find the permutation matrix for the permutation σ = (3, 1, 2) of the set {1, 2, 3}. Solution: Now, we take up a permutation of set {1, 2, 3}, σ = (3, 1, 2), and we wish to make a permutation matrix for it. Step 1. Identify the Permutation σ maps 1 to 3, 2 to 1, and 3 to 2. Step 2. Initialize the Matrix Initially, get a matrix 3x3 filled with zeros: \begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix} Step 3. Place the 1s For i = 1, σ(1) = 3. Put 1 in position (1, 3): \begin{pmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix} For i = 2, σ(2) = 1. Put 1 in position (2, 1): \begin{pmatrix} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix} For i = 3, σ(3) = 2. Place 1 in position (3, 2): \begin{pmatrix} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 \end{pmatrix} The resulting permutation matrix P for the permutation σ = (2, 3, 1) is: \begin{pmatrix} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 \end{pmatrix} Problem 3: Apply the permutation matrix P to the vector v = (3, 5, 7), given below is the permutation matrix: P = \begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{pmatrix} Solution: In order to apply P to v, we need to perform the matrix-vector multiplication Pv. P \mathbf{v} = \begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{pmatrix} P \mathbf{v} = \begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} 3 \ 5 \ 7 \end{pmatrix} = \begin{pmatrix} 5 \ 7 \ 3 \end{pmatrix} Thus, the result of applying P to v is (5, 7, 3). Problem 4: Given below the permutation matrix: P = \begin{pmatrix} 0 & 0 & 1 & 0 \ 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix} Determine the corresponding permutation of the set {1, 2, 3, 4}. Solution: To find the permutation firstly find the position of each 1 in the matrix. Each row of the permutation matrix (P) represents the image of the position of the corresponding element. Finally, the permutation matrix shows the permutation σ = (3, 1, 2, 4). In the case, the permutation orders the set {1, 2, 3, 4} in such a way that 1 is matched with 3, 2 is paired with 1, 3 goes to 2, and 4 is still 4. Conclusion Permutation matrices are basic as well as applied tools in the areas of linear algebra and serving as a handy representation of permutations that are of a finite set. Their unique structure is such that each row and column has exactly one element of 1, because of which they are able to switch the elements by multiplying with other matrices or vectors, effectively. Along with properties such as orthogonality, sparsity, and invertibility, which they possess and are used in numerical linear algebra, graph theory, optimization, cryptography, and data processing, their diversity and prominence further come to light. 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15041	https://unacademy.com/content/bank-exam/study-material/quantitative-aptitude/cube-diagonal/	Access free live classes and tests on the app Login Join for Free Profile Settings Refer your friends Sign out Terms & conditions • Privacy policy About • Careers • Blog © 2023 Sorting Hat Technologies Pvt Ltd Exams SBI Exams SBI PO SBI Clerk SBI SO IBPS Exams IBPS PO IBPS Clerk IBPS RRB IBPS SO LIC Exams LIC Assistant LIC AAO LIC ADO RBI Exams RBI Grade B RBI Assistant Notifications Upcoming Bank Exam Syllabus IBPS Syllabus IBPS PO Syllabus IBPS Clerk Syllabus IBPS RRB Syllabus IBPS SO Syllabus SBI Syllabus SBI PO Syllabus SBI Clerk Syllabus LIC Syllabus LIC Assistant Syllabus LIC AAO Syllabus LIC ADO Syllabus RBI Syllabus RBI Assistant Syllabus Study Material Magazine Download Bank Exams Notes Tests & Practice Scholarship Test Test Series Learning Festival Rankers Guide Paper Analysis Bank Exam » Bank Exam Study Materials » Quantitative Aptitude » Cube Diagonal Cube Diagonal The formula of Diagonal of Cube and its derivation, Properties of Cube and example questions. We also have to check the face diagonal and body diagonal of a cube. This article encompasses all about it! In this article, we get to know about not only the diagonal of the cube formula but the face diagonal and the body diagonal of a cube. How do we use the formula, and what kind of questions can be asked related to the diagonal of the cube. Each cube has a diagonal as the line segment connecting non-adjacent cube vertices. In geometric terms, a cube is an important shape because it has 12 equal sides and is one of the most commonly seen shapes around us. Ice cubes, sugar cubes, Rubik’s cubes, dice, etc., are examples of the cube. Cube All six faces of a cube are square. Cubes have identical dimensions on all their faces or sides. There is a right angle for every plane on the Cube. A cube is a sphere where each face meets four others at each edge. An equilateral Cube has three edges and three faces at every vertex. Those surfaces on a Cube that are opposite are parallel. In other words, a cube’s length, width, and height are of equal size, and its faces are all squares. With examples, let us learn more about how to measure the lengths of the diagonals on the face of the cube and the diagonals on the cube’s body. Face Diagonals of a Cube and Body Diagonals of Cube: The faces of a cube are square-shaped. Each face of the drawing has two diagonals connecting non-adjacent vertices. Therefore, there are 12 face diagonals in the cube. In a cube, the length of each face diagonal is calculated as follows: Length of each face diagonal of a cube = √2a, where a = Length of each side. The diagonals in the cube’s body pass through the opposite corners. Therefore, the opposite corners of a cube are connected by four body diagonals. Derivation of Diagonal of Cube Formula The Pythagorean theorem can be used to calculate the diagonal of a face and the internal diagonal of a cube. Let’s D be diagonal, and using Pythagoras theorem, we get D² = a² + a² D = √2a We need to use another right triangle to find the internal diagonal. An internal diagonal is used as the hypotenuse, one side of a cube as the height, and the diagonal of a face as the base: Using Pythagoras theorem again D² = a² + (√2a)² D² = a² + 2a² D² = 3a² D = √3a Use of Diagonal Cube The diagonal of a cube formula is useful for calculating the diagonal of a cube’s square faces. Using the cube’s diagonal formula, the cube’s face diagonal and main diagonal are calculated. Each side diagonal of the cube represents a side diagonal of the cube. There are six equal faces and twelve diagonals. As well to a square solid having all edges of the same length, a cube is a three-dimensional solid figure. In other words, the length, width, and height of a cube are equal, and each face of the cube is a square. To calculate the dimensions of the faces and the bodies of a cube, the diagonal of a cube formula is used. As one of the seven solid figures, a cube has all its edges of the same width. This means that the length, width, and height are of equal measure, and each of its faces is a square. Let us learn more about the diagonal of a cube and the diagonal of a cube formula for measuring the lengths of the face diagonal and the body diagonals with examples. Example Questions on Diagonal Cube Q.1. What is the diagonal of a cube whose side is 100 m? Let’s be the side of the cube The diagonal of the cube is D Using formula D = √3a D = √3 ✖ 100 D = 173.2 m So the diagonal of a cube is 173.2 m Q.2. Calculate the length of the diagonal of a cube whose sides are 18 inches long. Solution: Given, the cube’s side length (a) = 18 inches, Let the body diagonal be D Each cube’s diagonal length = √3a After using the formula Length of each body diagonal of cube = √3a ⇒ √3a ⇒√3 × 18 = 31.17 inches. Therefore, the length of the body diagonal is 31.17 inches Conclusion In this section, the diagonal of the cube is explained in an easy-to-understand manner. A solid cube is a three-dimensional object that contains six sides. Three edges intersect at one vertex so that there are eight vertices and 12 edges. Moreover, we derived a formula for a cube’s diagonal, examining the differences between cubes and cuboids. The solutions given here will help you learn and understand the concepts and their applications better. This article included all the things related to the diagonal of a cube! Frequently Asked Questions Get answers to the most common queries related to the Bank Examination Preparation. In a cube, how many diagonals are there? Ans : The opposite of space diagonals is face diagonals, which bind ad...Read full In a cube, the main diagonal is? Ans : There is only one diagonal in a cube: the diagonal passing throu...Read full In a cube, what is the Body Diagonal? Ans : Due to a cube’s three-dimensional nature, there are four body diagonals. Also known as the ‘...Read full How do the diagonal and the sides of a square relate? Ans : There is no equality between the diagonal and sides of a square. ...Read full Ans : The opposite of space diagonals is face diagonals, which bind adjacent vertices on the same face (but not on the same edge). A pyramid, for instance, has no space diagonals, while a cube (shown above right) or, more generally, a parallelepiped, has four space diagonals. Ans : There is only one diagonal in a cube: the diagonal passing through the centre. The diagonal on the face of the cube is not the main diagonal. By multiplying the length of one side by the square root of three, the length of the major diagonal can be determined for every cube. Ans : Due to a cube’s three-dimensional nature, there are four body diagonals. Also known as the ‘space diagonal’ or ‘solid diagonal’ of a cube, the body diagonal is the extended surface of the cube. The diagonals pass through the cube’s body, connecting the opposite vertices. The diagonals of a cube pass through the cube connecting its opposite corners, thus forming four body diagonals. Ans : There is no equality between the diagonal and sides of a square. In mathematics, a square’s diagonal is calculated using the formula: Diagonal of Square (d) = √2a, where a represents a square’s side. By using the Pythagorean Theorem, we calculate the diagonal formula. Share via
15042	https://virtualnerd.com/pre-algebra/linear-functions-graphing/scatter-plot-best-fit-lines	Scatter Plots and Best-Fit Lines \| Pre-Algebra \| Linear Functions and Graphing \| Virtual Nerd Real math help. Pre-Algebra Switch to: Middle Grades Math Algebra 1 Algebra 2 Geometry SAT Math ACT Math Common Core Texas Programs Texas Standards Linear Functions and Graphing Scatter Plots and Best-Fit Lines Scatter Plots and Best-Fit Lines More Specific Topics in Scatter Plots and Best-Fit Lines Defining Scatter Plots and Correlation Using Scatter Plots and Making Predictions Popular Tutorials in Scatter Plots and Best-Fit Lines How Do You Use a Scatter Plot to Find a Positive Correlation? Got a bunch of data? Trying to figure out if there is a positive, negative, or no correlation? Draw a scatter plot! This tutorial takes you through the steps of creating a scatter plot, drawing a line-of-fit, and determining the correlation, if any. Take a look! How Do You Use a Scatter Plot to Find a Line of Fit? A line-of-fit is a line that summarizes the trend in a set of data. In this tutorial, you'll see how to graph data on a coordinate plane and draw a line-of-fit for that data. Check it out! How Do You Use a Scatter Plot to Find a Negative Correlation? Got a bunch of data? Trying to figure out if there is a positive, negative, or no correlation? Draw a scatter plot! This tutorial takes you through the steps of creating a scatter plot, drawing a line-of-fit, and determining the correlation, if any. Take a look! How Do You Use a Scatter Plot to Find No Correlation? Got a bunch of data? Trying to figure out if there is a positive, negative, or no correlation? Draw a scatter plot! This tutorial takes you through the steps of creating a scatter plot, drawing a line-of-fit, and determining the correlation, if any. Take a look! What's a Scatter Plot? Scatter plots are really useful for graphically showing a bunch of data. By seeing data graphically, you can see patterns or trends in the data. These patterns help researchers to understand how one thing affects another. This can lead to all kinds of breakthroughs! This tutorial gives you a look at the scatter plot. Check it out! What's Positive Correlation? Looking at a line-of-fit on a scatter plot? Does that line have a positive slope? If so, your data shows a positive correlation! Learn about positive correlation by watching this tutorial. What's Negative Correlation? Looking at a line-of-fit on a scatter plot? Does that line have a negative slope? If so, your data shows a negative correlation! Learn about negative correlation by watching this tutorial. What Does it Mean To Have No Correlation? Scatter plots are very helpful in graphically showing the pattern in a set of data. But sometimes that data shows no correlation. Learn about no correlation and see how to tell if data shows no correlation by watching this tutorial! How Do You Make a Scatter Plot? Scatter plots are a very useful way to help you visually see data. In this tutorial, you'll see how to take data from a table and plot it to create a scatter plot. Take a look! How Do You Write and Use a Prediction Equation? Scatter plots are a great way to see data visually. They can also help you predict values! Follow along as this tutorial shows you how to draw a line of fit on a scatter plot and find the equation of that line in order to make a prediction based on the data already given! Related Topics Other topics in Linear Functions and Graphing: Functions Linear Equations Rate of Change and Slope Slope-Intercept Form Point-Slope Form Direct Variation Systems of Equations Linear Inequalities About Terms of Use Privacy Contact
15043	https://www.pacehospital.com/angina-pectoris-symptoms-types-causes-treatment-prevention	Patient Guide \| Book Health Packages \| Careers \| Blogs 24x7 Helpline : 040 4848 6868 Blog Post Angina Pectoris – Symptoms, Types, Causes, Treatment & Prevention PACE Hospitals Angina pectoris definition Angina, which is the most common symptom of ischemic heart disease (IHD), is chest pain or discomfort that occurs due to reduced blood flow to the heart muscles. Ischemic chest pain occurs when one or more of the coronary arteries gets narrowed. It is one of the major causes of morbidity and mortality worldwide; chest pain may be due to non-cardiac (not related to heart) and cardiac (related to heart) causes. Stable angina, unstable angina, microvascular angina, and vasospastic or variant angina are the four different types of angins. The treatment of angina is aimed at managing symptoms and preventing the progression of cardiac events. Management is multifactorial and requires a few lifestyle modifications to prevent further complications. A cardiologist can treat angina. Angina pectoris meaning The term angina pectoris is derived from Latin, the term “angere” means to strangle and “pectoris” means chest; together it means strangling (pressure) feeling in the chest. Prevalence of angina pectoris Prevalence of angina worldwide Angina is one of the most common symptoms of ischemic heart disease. Symptoms of angina are seen in approximately 90 lakhs of patients in the US. The prevalence of angina increases with age in both men and women. In Western countries stable angina affects approximately 30000 to 40000 people per 10 lakh people. The estimated prevalence for men and women aged 45 to 64 years are 4 to 7 percent and 5 to 7 percent respectively. In men and women aged 65 to 84 years, the prevalence is 14 to 15 percent and 10 to 12 percent, respectively. Prevalence of angina in India As per the CLARIFY study, which is an international, prospective-observational, longitudinal cohort study in stable coronary artery disease outpatients the prevalence of angina in India decreased from 27.8% to 11.2% over a 5-year period. From recent studies it is estimated that the proportion of symptom-based angina is 4.69% in older males and 7.02% in older females. Types of angina pectoris Angina is not a disease itself, but it is a symptom of an underlying condition, usually coronary artery disease. There are five main types of anginas which include stable angina, unstable angina, microvascular, vasospastic or variant angina, and refractory angina. Based on the type of angina, many factors can trigger pain. The 5 types of anginas are described below: Stable angina: Angina pectoris, also called stable angina, is a chest pain that occurs due to interrupted blood supply to heart muscles because of the narrowing of coronary arteries. It usually causes uncomfortable pressure, fullness, squeezing, or pain in the center of the chest. Unstable angina: It is one of the several acute coronary syndromes that occur while resting and causes unexpected chest pain. It does not go away with rest and should be treated as an emergency as it may be a sign of a heart attack and can even lead to cardiac arrest. Variant angina: It is also called Prinzmetal angina; it does not occur with physical exertion or emotional stress. The pain is very painful and occurs when the individual is resting, usually between midnight and early morning. Microvascular angina: This type of angina is usually a symptom of coronary microvascular disease which is a heart disease affecting the smallest coronary artery blood vessel. Refractory angina: It is a chronic and persistent form of angina that occurs when the heart's demand for oxygen exceeds its supply, despite optimal medical therapy. It is characterized by frequent episodes of chest pain that cannot be controlled adequately by medications and revascularization therapy. Angina pectoris symptoms Symptoms of angina vary based on the type and they may differ in severity, location, timing, and degree of relief. The primary symptom of angina is chest pain or discomfort. The pain in angina can feel like pressure, tightness, discomfort, squeezing, heaviness, or burning sensation in the chest. The common symptoms of angina include: Shortness of breath (SOB) Extreme fatigue (tiredness) Fainting or light headedness Nausea Heartburn Weakness Sweating Stable angina symptoms The pain in stable angina: Occurs usually during physical exertion Lasts for a short period Pain can be relieved by taking rest and with the use of medications The pain feels like a gas or indigestion It may feel like the pain is spreading to the arms and back. Unstable angina symptoms The pain in unstable angina: Usually occurs during physical exertion or while resting or sleeping It can be sudden Lasts longer than stable angina Does not get relieved by rest or with medicines Worsens over time, it can lead to heart attack Variant angina symptoms The pain: Develops while resting usually between midnight or early morning Is severe Gets relieved with the use of medications Microvascular angina symptoms The pain: Is more severe and lasts longer than all the other types of anginas Accompanies shortness of breath It is first noticed with routine physical activities Refractory angina symptoms The pain: Is severe and the symptoms last for more than 3 months Pain cannot be managed with medicines or other treatments. Angina symptoms in women Symptoms of angina can be different in men and women. Women can feel pain in the areas far from the source of pain such as arms, neck, back, and jaw. They can also present symptoms like shortness of breath, nausea, and lightheadedness. Angina pectoris causes Chest pain can occur due to non-cardiac causes (not related to the heart), non-ischemic causes, and due to ischemic causes. The non cardiac causes of chest pain include gastroesophageal reflux disease (GERD), lung disease, musculoskeletal causes, and panic attacks. Pericardial disease is the non-ischemic cause of chest pain. Atherosclerosis of coronary arteries and coronary vasospasm is the cardiac ischemic cause of angina. Angina causes Angina is not a disease but a symptom of an underlying heart condition. Coronary artery disease which is a medical cardiac condition can cause angina. The type of angina depends on the condition causing it. Following are the types of anginas and their causes: Stable angina causes This type of chest pain occurs when the heart muscles do not get a sufficient amount of blood or when the heart requires more blood, such as during physical activity like climbing a hill or stairs or when experiencing strong emotions. Unstable angina causes Plaque builds up within the arteries thereby narrowing the blood vessel. This plague buildup causes damage to areas, making it easier for blood clots to form. Partial or total blockage from blood clots causes unstable angina. Variant angina causes The chest pain that develops in variant angina is caused by coronary artery spasm. The spasm in coronary arteries is caused by stress, exposure to cold weather, use of medications that narrow blood vessels, smoking, and consumption of cocaine. Microvascular causes Development of spasms in the very small arterial blood vessels causes microvascular angina. This spasm interrupts or blocks the supply of blood flow to the heart vessels causing pain. Angina pectoris risk factors A risk factor is a factor or a variable that increases the likelihood of developing a disease. There are several factors that increase the risk of angina development. Following are some of the risk factors of angina: Age: The risk of heart disease increases with age. With increasing age, the plague builds up in the arteries interrupting the blood supply to the heart muscle. Environment or occupation: Particle pollution that includes dust from roads, farms, dry riverbeds, construction sites, and mines is linked to angina. Professional life with inadequate sleep, stress, long periods of sitting or standing, or exposure to radiation can increase the risk of angina. Family history and genetics: Heart diseases run in families. Individuals with no lifestyle-related risk factors develop heart diseases due to the involvement of genes. Thus, genetics increases the risk of angina. Lifestyle habits: Lifestyle related risk factors include excessive alcohol consumption, lack of physical activity, smoking, stress, and unhealthy dietary habits. Medical conditions: Conditions like anemia, cardiomyopathy, heart valve diseases, hypertension, heart failure, and metabolic syndrome increase the risk of angina. Medical procedures: Past medical procedures like stent placement, and coronary artery bypass grafting trigger coronary spasms and angina. Race: Black or African American people with a past medical history of heart attack have a higher chance of developing angina compared to white people. People living in Japan have a higher risk of developing vasospastic angina. Sex: Angina affects both men and women. Heart diseases tend to rise in men aged above 45 years, women have a lower risk of heart disease below the age of 55 years. The risk of heart disease is equal when both men and women are aged 55. ✅ For appointment Call Us at 04048486868 Whatsapp Us Complications of angina pectoris Though angina is not life threatening it is a major cause of morbidity and mortality as it causes some serious complications. There are several complications of angina pectoris. Cardiac events such as myocardial infarction and heart attack are the two main complications of angina. The possible complications of angina include: Abnormal heart rhythms: In angina, the insufficient blood supply to the heart muscles disrupts the electrical activity of the heart thereby causing arrhythmias (abnormal heart rhythm). Heart attack: The insufficient blood supply to heart muscles due to the narrowing of coronary arteries causing the heart muscles to die thereby leading to heart attack. Heart failure: Repeated episodes of angina weaken the muscles of the heart and increase the risk of heart attack. Angina pectoris diagnosis Diagnosis of angina is clinical, and it is based on a detailed history of the characteristics of the pain, with an assessment of its quality, frequency, location, severity, duration, and associated symptoms to assess triggering and alleviating factors. Diagnosis of angina includes the following steps: Initial evaluation Medical history Physical examination Diagnostic tests Chest X-ray Electrocardiogram (ECG) Blood tests Complete blood count (CBC) Basic metabolic profile (BMP) that includes troponin levels Stress testing Exercise stress testing Pharmacological stress testing Other diagnostic studies Echocardiogram Cardiac magnetic resonance imaging Coronary computed tomography angiography (CCTA) Invasive coronary angiography Angina pectoris treatment Angina pectoris treatment approach is based on the type of angina, the presenting symptoms, the test results, and the risk of complications. Management of angina is multifactorial and involves lifestyle changes, risk factor modification, and medical therapy as essential components of treatment. Nonpharmacological management of angina When the symptoms are stable and not getting worse then angina can be managed with certain lifestyle modifications or changes. Below are some of the measures listed by following which angina can be managed: Choosing a diet that is heart healthy Maintaining an ideal body weight Staying physically active Managing stress Avoiding smoking Getting enough sleep Pharmacological management of angina Angina pectoris treatment usually involves two approaches; one is to increase the blood flow to the heart muscles and the other is to lower the workload of the heart. Following are the medications that are used in treating angina: Beta blockers Nitrates Calcium channel blockers Antiplatelets Anticoagulants Statins Antianginal Narcotic analgesic Revascularization procedures If angina is not managed through lifestyle modifications and with medications following medical procedures are recommended: Coronary artery bypass grafting (CABG) Percutaneous coronary intervention (PCI) Angina pectoris prevention Angina prevention usually involves lowering the risk of heart disease by adopting heart healthy lifestyle modifications. These heart-healthy lifestyle modifications include: Dietary modifications: Including a diet that is rich in fiber, including fruit and vegetables in the diet. Limiting alcohol consumption and avoiding high-fat content food. Exercise: Being physically active, exercising regularly for at least 30 minutes Maintaining ideal body weight: If overweight or obese, weight reduction becomes crucial. Quit smoking: Avoiding smoking or gradually decreasing alcohol consumption Managing stress: Avoiding situations that trigger stress Healthy sleeping habits: Getting enough sleep helps repair heart muscles. Adhering to medications: It is important to adhere to the medications prescribed to manage symptoms. Treating underlying medical conditions: Underlying medical conditions like hypertension, diabetes, and dyslipidemia can increase the risk of angina therefore they have to be treated. Difference between stable angina and unstable angina Stable angina v/s unstable angina Stable and unstable angina are the two main types of angina pectoris. Though they have similarities there are some key differences that differentiate stable angina from unstable angina. Below are some parameters that differentiate between the two: \| Parameters \| Stable angina \| Unstable angina \| --- \| Definition \| Angina pectoris, also called stable angina, is a chest pain that occurs due to interrupted blood supply to heart muscles because of the narrowing of coronary arteries. \| It is one of the several acute coronary syndromes which occur while resting and causes unexpected chest pain. \| \| Causes \| It occurs when the heart muscles do not get a sufficient amount of blood or in cases where the heart requires more amount of blood such as during physical activity like climbing a hill or stairs, or when having strong emotions \| Plaque builds up within the arteries thereby narrowing the blood vessel. This plague buildup causes damage to areas, making it easier for blood clots to form. Partial or total blockage from blood clots causes unstable angina \| \| Symptoms \| Chest pain occurs usually during physical exertion which lasts for a short period. Pain can be relieved by taking rest and with the use of medications \| The ischemic chest pain usually occurs during physical exertion or while resting or sleeping \| \| Treatment \| Pain can be relieved by taking rest and with the use of medications \| The pain does not go away with rest, and it requires immediate medical action as it is a medical emergency \| Difference between angina and myocardial infarction angina pectoris vs myocardial infarction Angina and myocardial infarction are both related to the heart, but they have different characteristics. Following are some of the key parameters that differentiate angina from myocardial infarction \| Parameters \| Angina \| Myocardial infarction \| --- \| Definition \| Angina pectoris, also called is a chest pain that occurs due to interrupted blood supply to heart muscles because of the narrowing of coronary arteries. \| Myocardial infarction also called a heart attack is a result of decreased or complete blockage of blood flow to a part of the myocardium . \| \| Causes \| It occurs when the heart muscles do not get a sufficient amount of blood or in cases where the heart requires more amount of blood such as during physical activity like climbing a hill or stairs, or when having strong emotions \| Underlying coronary artery disease is the primary cause of myocardial infarction. \| \| Symptoms \| Chest pain occurs usually during physical exertion which lasts for a short period. Pain can be relieved by taking rest and with the use of medications \| It can present chest pain, upper extremity pain, and epigastric discomfort that occurs during exertion or at rest. \| \| Treatment \| Treatment is multifactorial and involves lifestyle modifications, drug therapy, and revascularization procedures. \| Treatment includes lifestyle modification, drug therapy, and medical procedures such as reperfusion therapy and primary percutaneous intervention. \| ✅ For appointment Call Us at 04048486868 Whatsapp Us Frequently Asked Questions (FAQs) on Angina pectoris Is angina a disease or a symptom? Angina is not a disease itself; it is a symptom of an underlying condition. It occurs when muscles of the heart do not get sufficient oxygen-rich blood due to the narrowing of coronary arteries. The pain usually occurs during physical exercise and lasts for a shorter period. ### Is angina curable? No, angina is not curable, but it can be managed. It can be managed by lifestyle modifications, medical therapy, and revascularization procedures. Treatment is important to prevent angina attacks and reduces the further risks of heart attacks and infarctions. ### Is angina pain constant? No, angina pain is not constant. The pain can vary with each type of angina, and it typically stops and starts and can last from seconds to minutes. This pain usually occurs when the heart is working hard. ### Can angina lead to myocardial infarction? Yes, angina can lead to myocardial infarction. It is one of the primary symptoms of coronary artery disease; studies estimated that a 10-year risk of myocardial infarction exceeded 10 percent in women with stable angina. ### How is angina treated? The pharmacological treatment approach is based on the type of angina, the presenting symptoms, the test results, and the risk of complications. It involves two approaches; one is to increase the blood flow to the heart muscles and the other is to lower the workload of the heart. Drug classes such as beta blockers, nitrates, calcium channel blockers, antiplatelets, anticoagulants, and statins are used in treating angina. Can ECG detect angina? Yes, ECG can diagnose angina. It helps in recognizing the type of angina and other heart related conditions. Certain ECG patterns are signs of unstable and vasospastic angina. At certain times ECG may be normal even if the patient has angina. Can ECG show stable angina? Yes, ECG can show stable angina. ECG helps in identifying the type of angina and to detects other underlying conditions. The findings depend on the duration of ischemia, the extent of blockage, and the presence of other underlying conditions. What is a stress test used in diagnosing angina? Stress tests help to know how well the heart is working when it is pumping hard. This test is performed during exercise on a treadmill or stationary bicycle. These tests also take images of the heart at rest and when it's working hard thereby providing detail about how well the heart is working. What is the difference between stable and unstable angina? In stable angina, pain occurs during exercise or during emotional stress. It lasts for a few minutes and usually gets relieved with rest and medications. In unstable angina, the pain can occur suddenly and lasts longer than stable angina. The pain is severe and requires immediate medical attention. How is microvascular angina diagnosed? Diagnosis of microvascular angina involves a medical history, physical examination, and certain diagnostic tests. These tests include stress tests, coronary angiography, and magnetic resonance imaging (MRI). How is angina diagnosed? Diagnosis of angina involves a combination of physical exam, assessment of symptoms, and a group of diagnostic tests. Initial testing includes a 12-lead electrocardiogram (ECG), chest X-ray, and basic laboratory testing, including complete blood count (CBC), and basic metabolic profile (BMP), along with troponin levels. What are the surgical treatment options available for angina? Surgical procedures are recommended in situations where angina cannot be managed through lifestyle modifications and medications. Coronary artery bypass graft (CABG) and coronary angioplasty are the two main surgical procedures used in the treatment of angina. Department of Cardiology < Older Post Newer Post > Share on Request an appointment Fill in the appointment form or call us instantly to book a confirmed appointment with our super specialist at 04048486868 Appointment request - health articles Thank you for contacting us. We will get back to you as soon as possible. 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15044	https://mathbitsnotebook.com/JuniorMath/Expressions/EXPLinearExpressions.html	\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| Factoring Linear Expressions MathBitsNotebook.com Topical Outline \| Jr Math Outline \| MathBits' Teacher Resources Terms of Use Contact Person: Donna Roberts \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- \| A factor is a number, or algebraic expression, that divides another number, or expression, with no remainder. On this page, we will concentrate on finding equivalent expressions where one of the expressions is expressed in the form of a product of factors. We will be looking specifically at "linear expressions". \| \| \| --- \| \| \| \| \| A linear expression is an algebraic statement where each term is either a constant or a variable raised to the first power. None of the exponents can be greater than 1. \| \| Examples of linear expressions: 7, 3x, 2m + 5, 4x - 2, 2.3b - 1.2, (½)x + ¼ We will be concentrating on linear expressions that have 2 terms (binomial) or 3 terms (trinomial). In both cases, the highest exponent is one. Consider the expression (8 + 12): Numerically speaking: 8 + 12 = 4 ( 2 + 3 ) The greatest common factor of 8 and 12 is 4. The factors of the expression (8 + 12) are 4 and (2 + 3). \| \| \| Algebraically speaking: Looking for the Greatest Common Factor (GCF): \| Let's start by seeing how to find the greatest common factor when working with algebraic expressions. We will be using the concept of prime factorization. EXAMPLE: Find the greatest common factor of 25x and 15x. The greatest common factor (GCF) of two single terms is the product of the greatest common factor of the numerical coefficients (the numbers out in front) and the highest power of the variable that is a factor of each term. \| \| \| --- \| \| \| \| \| Find GCF: 18x and 27x The greatest common factor is 9x. The largest factor of 18 and 27 is 9. The variable x (to a power of 1) is a factor in both terms. \| \| \| \| \| --- \| \| \| \| \| Find GCF: 18x and 27 The greatest common factor is 9. The largest factor of 18 and 27 is 9. There is no variable factor since x is only in one term. \| \| \| \| \| To factor an algebraic linear expression: • Find the greatest common factor (GCF). • Factor out that value by dividing. \| \| \| \| Distributive Property: a(b + c) = ab + acDistributive Property in Reverse: ab + ac = a(b + c) \| The distributive property in reverse shows the factoring of an expression. When factoring linear expressions), first find the GCF, which is a factor of each term of the expression. The GCF will be one term (a monomial). Factor out (divide each term by) this GCF. The remaining factor will most likely be in parentheses. \| \| \| --- \| \| \| \| \| Factor: 6x + 42 The largest integer that will divide evenly into 6 and 42 is 6. Since the terms do not have a variable (such as x) in common, we cannot factor any variables. The greatest common factor is 6. Divide each term by 6. Answer: 6(x + 7) Equivalent expressions: 6x + 42 = 6(x + 7) \| \| \| \| \| This result may look familiar. It is the backward version of the Distributive Property (mentioned above). Factoring this expression in this manner, is applying the Distributive Property in reverse (backward). \| \| \| \| --- \| \| \| \| \| Factor: 18x - 27 The largest integer that will divide evenly into 18 and 27 is 9. Since the terms do not contain a variable (such as x) in common, we cannot factor any variables. The GCF is 9. Divide each term by 9. Answer: 9(2x - 3) Equivalent expressions: 18x - 27= 9(2x - 3) \| \| \| \| \| --- \| \| \| \| \| Factor: (¾)a + ¼ The largest value that divides evenly into ¾ and ¼ is ¼. The variable a is NOT present in both terms. The GCF is ¼. Divide each term by the GCF. Answer: ¼(3a + 1) Equivalent expressions: (¾) x + ¼ = ¼(3a + 1) \| \| \| \| \| --- \| \| \| \| \| Factor: 2x + 2y + 1 + 1 This problem is a bit sneaky! Combine terms first. 2x + 2y + 2 This looks better! The largest integer that divides evenly into 2, 2 and 2 is 2. There are no variables common to all three terms. The GCF is 2. Divide each term by the GCF. Answer: 2(x + y +1) Equivalent expressions: 2x + 2y + 1 + 1 = 2(x + y + 1) \| \| Remember that the Distributive Property can be used in a "working backward strategy" to express a linear expression as a product of its factors. A linear expression is in factored form when it is expressed as the product of its factors. \| \| \| \| \| \| When we examine the use of the Distributive Property, in this manner, we can see that "factoring an expression" can be seen as the opposite of "distributing (expanding) an expression". \| \| \| \| Linear Expressions in Geometry \| The connection between factoring and the distributive property can also be seen when dealing with area. First let's refresh a few geometry concepts that will be needed. Consider the following concepts: \| \| \| --- \| \| The formula for the area of a rectangle is Area = length x width. \| When dealing with line segments, the length of a segment is the sum (the adding) of all of its parts. \| Now, examine the following, where the green "box" represents a rectangle in which the dashed line forms two rectangles: This is a visual representation of the distributive property 7(x + 6) = 7x + 42. It can also be applied in reverse to demonstrate factoring. In these examples, we have seen area represented as a product, 7(x + 6) and 4(x + 2), and area represented as a sum, 7x + 42 and 4x + 8. \| \| \| NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use". \| Topical Outline \| Jr Math Outline \| MathBitsNotebook.com \| MathBits' Teacher Resources Terms of Use Contact Person: Donna Roberts \| \|
15045	https://idpjournal.biomedcentral.com/articles/10.1186/s40249-019-0589-6	Sustained preventive chemotherapy for soil-transmitted helminthiases leads to reduction in prevalence and anthelminthic tablets required \| Infectious Diseases of Poverty \| Full Text Your privacy, your choice We use essential cookies to make sure the site can function. We also use optional cookies for advertising, personalisation of content, usage analysis, and social media. By accepting optional cookies, you consent to the processing of your personal data - including transfers to third parties. Some third parties are outside of the European Economic Area, with varying standards of data protection. See our privacy policy for more information on the use of your personal data. Manage preferences for further information and to change your choices. Accept all cookies Skip to main content Advertisement Search Explore journals Get published About BMC Login Menu Explore journals Get published About BMC Login Search all BMC articles Search Infectious Diseases of Poverty Home About Articles Collections Submission Guidelines Submit manuscript Sustained preventive chemotherapy for soil-transmitted helminthiases leads to reduction in prevalence and anthelminthic tablets required Download PDF Download PDF Research Article Open access Published: 02 October 2019 Sustained preventive chemotherapy for soil-transmitted helminthiases leads to reduction in prevalence and anthelminthic tablets required Denise MupfasoniORCID: orcid.org/0000-0002-8052-82721, Mathieu Bangert1, Alexei Mikhailov1, Chiara Marocco1& … Antonio Montresor1 Show authors Infectious Diseases of Povertyvolume 8, Article number:82 (2019) Cite this article 4121 Accesses 17 Citations 21 Altmetric Metrics details Abstract Background The goal of soil-transmitted helminthiases (STH) control programmes is to eliminate STH-associated morbidity in the target population by reducing the prevalence of moderate- and heavy-intensity infections and the overall STH infection prevalence mainly through preventive chemotherapy (PC) with either albendazole or mebendazole. Endemic countries should measure the success of their control programmes through regular epidemiological assessments. We evaluated changes in STH prevalence in countries that conducted effective PC coverage for STH to guide changes in the frequency of PC rounds and the number of tablets needed. Methods We selected countries from World Health Organization (WHO)‘s Preventive Chemotherapy and Transmission control (PCT) databank that conducted ≥5 years of PC with effective coverage for school-age children (SAC) and extracted STH baseline and impact assessment data using the WHO Epidemiological Data Reporting Form, Ministry of Health reports and/or peer-reviewed publications. We used pooled and weighted means to plot the prevalence of infection with any STH and with each STH species at baseline and after ≥5 years of PC with effective coverage. Finally, using the WHO STH decision tree, we estimated the reduction in the number of tablets needed. Results Fifteen countries in four WHO regions conducted annual or semi-annual rounds of PC for STH for 5 years or more and collected data before and after interventions. At baseline, the pooled prevalence was 48.9% (33.1–64.7%) for any STH, 23.2% (13.7–32.7%) for Ascaris lumbricoides, 21.01% (9.7–32.3%) for Trichuris trichiura and 18.2% (10.9–25.5%) for hookworm infections, while after ≥5 years of PC for STH, the prevalence was 14.3% (7.3–21.3%) for any STH, 6.9% (1.3–12.5%) for A. lumbricoides, 5.3% (1.06–9.6%) for T. trichiura and 8.1% (4.0–12.2%) for hookworm infections. Conclusions Countries endemic for STH have made tremendous progress in reducing STH-associated morbidity, but very few countries have data to demonstrate that progress. In this study, the data show that nine countries should adapt their PC strategies and the frequency of PC rounds to yield a 36% reduction in drug needs. The study also highlights the importance of impact assessment surveys to adapt control strategies according to STH prevalence. Multilingual abstracts Please see Additional file 1 for translations of the abstract into the five official working languages of the United Nations. Background Soil-transmitted helminths are a group of intestinal parasites that are transmitted to humans through ingestion of infective eggs or transcutaneous penetration of larvae excreted in human faeces which contaminate the soil and water sources . Soil-transmitted helminthiases (STH) are among the most common neglected tropical diseases (NTDs) in developing countries. Worldwide, it is estimated that 820 million people are infected with roundworm (_Ascaris lumbricoides_), 460 million with whipworm (_Trichuris trichiura_) and 460 million with hookworms (_Necator americanus_ and _Ancylostoma duodenale_) in 102 countries . In 2010, these infections contributed 3.4 million disability-adjusted life-years to the global burden of disease . The World Health Organization (WHO) recommends preventive chemotherapy (PC) with albendazole or mebendazole to control STH-related morbidity alongside targeted health education and improved water and sanitation . PC is the regular, coordinated administration of anthelminthic medicines to population groups at risk for STH morbidity. The target populations include preschool-age children (PSAC), school-age children (SAC), women of reproductive age (WRA) and adult groups particularly exposed to STH. The 2020 WHO goal for control of STH is to treat ≥75% of PSAC and SAC in all STH endemic countries . In 2017, more than 500 million SAC (69% of total SAC in need) received PC for STH globally, with 73% of implementation units reaching 75% effective coverage . The main objective of the STH control programme is to eliminate morbidity in the target population by reducing the prevalence of moderate- and heavy-intensity infections to < 2% . A meta-analysis conducted by Marocco et al. showed that after five years of PC, 80% of individuals with infections of moderate and heavy intensity were cured or remained with only light-intensity infection . Since 2010, WHO with the support of drug producers has donated albendazole or mebendazole for the control of STH. In 2018 alone, more than 485 million tablets were donated to endemic countries . The WHO manual on STH control in SAC recommends that control programmes managers collect parasitological indicators every 3–5 years of PC with effective coverage in order to measure the effect of the intervention on the health of the population at risk and, eventually, to reduce the number of PC rounds if the prevalence of STH infection has reduced to a certain level using the WHO decision tree (Additional file2) . In this study we evaluated the changes in STH prevalence from countries that conducted STH PC for ≥5 years with effective treatment coverage (that is, national treatment coverage ≥75% as defined by WHO) for SAC. These data also give an indication of expected changes in prevalence if countries conduct ≥5 years of PC, and therefore also the expected number of drugs needed by endemic countries over time. Methods National PC coverage data We accessed WHO’s Preventive Chemotherapy and Transmission control (PCT) databank, an open web-based database on PC implementation in 102 endemic countries , to identify countries that had implemented effective coverage of PC (defined as ≥75% national coverage) among SAC for ≥5 years. Epidemiological data reporting form The Epidemiological Data Reporting Form (EPIRF) is a standardized form designed to collect epidemiological data on all diseases targeted by PC, namely STH, lymphatic filariasis (LF), onchocerciasis and schistosomiasis. Indicators reported include: survey type, number of rounds of PC delivered prior to survey, date of survey, number of people examined, and numbers of people diagnosed positive for each STH species and for overall STH. Countries receiving donated anthelminthic medicines are invited to periodically report to WHO any changes in the epidemiological situation using the EPIRF. We extracted baseline and impact EPIRF data from endemic countries previously identified as having achieved ≥5 years of effective coverage. Literature search For countries in which the EPIRF was not available we reviewed the available literature on baseline and impact assessment surveys for STH prevalence. Only surveys targeting SAC were considered. We searched official publications such as national NTD master plans and Ministry of Health survey reports as well as data published in the peer-reviewed scientific literature between 2000 and 2017. For online searches we used the words: mapping, baseline, impact assessment, soil-transmitted helminths and country names using google search and MEDLINE. Data analysis Studies were eligible for inclusion if, for baseline, the data were collected before the national STH control programme was initiated and, for impact assessment, the data were collected after PC was implemented for ≥5 years with effective coverage. In addition, for all studies, SAC was the study population, the information on STH prevalence overall and/or for each STH species available. From each publication identified we extracted year, type of survey, target population, prevalence of STH and of each species, number of people examined, intensity of STH infection and diagnostic tool used. For studies where prevalence was reported by species only, we calculated the prevalence of any STH infection using the following equation : (a+t+h)−(a∗t+a∗h+t∗h)+(a∗t∗h)1.06 Where a = prevalence of ascariasis; t = prevalence of trichuriasis and h = prevalence of hookworm infections (all expressed as a proportion). Data were collated, cleaned, analysed and visualized using the R (Version 3.5, R core Team, Vienna, Austria) statistical program. For data in the EPIRF form, we estimated national means using weighted cluster data for each country and for each STH species. We plotted the prevalence of any STH infection and of each species before the start of PC for STH and the prevalence after ≥5 years of PC with effective coverage using pooled and weighted means. Finally, according to the WHO STH decision tree (Additional file 2) and after ≥5 years of PC with effective coverage, countries with STH prevalence < 2% should suspend PC, < 10% should conduct PC once every 2 years, and < 20% should do PC once a year. In this study, we also estimated the decrease in drug needs for countries according to the change in STH prevalence during the ≥5 years of PC. Results Number of countries that conducted ≥5 years of PC for STH From the WHO/PCT databank we identified 24 countries that conducted ≥5 years of PC for STH (Additional file3) between 2003 and 2017. Three countries (Burkina Faso, Mali and Togo) reported their baseline data and 11 countries (Burkina Faso, Burundi, Mali, Ghana, Sierra Leone, Cameroon, Rwanda, Mexico, Nicaragua, Bangladesh and Lao People’s Democratic Republic) reported their impact assessment data through EPIRF. Two countries (Burkina Faso and Mali) collected impact assessment data for STH through transmission assessment surveys (TAS) for LF. TAS is conducted after 5 years of annual LF MDA with a coverage over 65% to confirm that the prevalence of infection is below a threshold at which recrudescence is unlikely to occur. WHO recommends collecting data on STH epidemiology simultaneously in the area to determine if STH PC should be continued after LF MDA stopped. Ten countries conducted national surveys. In addition, we identified baseline surveys conducted before PC for 12 countries and identified two impact assessments surveys through publications and two unpublished reports on STH PC impact from the Ministry of Health. In total 15 countries in four WHO regions conducted PC STH for ≥5 years and collected data before and after interventions (Fig.1). Fig. 1 Flow chart summary on countries selected and data sources used for baseline and impact assessment surveys. SAC: School-age children; STH: Soil-transmitted helminth Full size image Prevalence of STH infection at baseline before PC Most of the baseline data were collected during 2002–2009 (Table1). The average size of the population surveyed was 8868 (range 266–22 166) and 80% of the surveys were designed to be nationally representative. Six, four and four countries reported high, moderate and low STH prevalence, respectively. Figure2 (left column) shows the pooled prevalence by country; the cumulative prevalence was 48.9% (33.1–64.7%) for any STH, 23.2% (13.7–32.7%) for A. lumbricoides, 21.01% (9.7–32.3%) for T. trichiura and 18.2% (10.9–25.5%) for hookworm infections. Table 1 Soil-transmitted helminth infection prevalence in SAC before and five or more years after initiation of preventive chemotherapy for STH in the 15 countries Full size table Fig. 2 Prevalence of STH (first row), Ascaris lumbricoides (second row), hookworm (third row) and Trichuris trichiura (fourth row) infections in countries before (left column) and after (right column) ≥ 5 years of preventive chemotherapy with effective coverage. Each dot represents a published study or WHO EPIRF data. The size of the dot indicates the number of people assessed. The red square indicates pooled and weighted mean prevalence with 95% confidence intervals. WHO: World Health Organization; EPIRF: Epidemiological data reporting form Full size image Prevalence of STH infection after ≥5 years of PC with effective coverage Between 2003 and 2017, the countries included in this study conducted and reported 5–14 rounds of PC, averaging eight rounds of PC for STH for SAC with effective coverage. Eleven countries conducted national surveys to measure the impact of PC on STH infection, while two countries collected the data during LF TAS (Table 1). These surveys were conducted during 2012–2017; the average population size surveyed was 6439 (range 990–16 890). In five countries STH prevalence was moderate while in nine STH prevalence was low. The pooled prevalence was 14.3% (7.3–21.3%) for any STH, 6.9% (1.3–12.5%) for A. lumbricoides, 5.3% (1.06–9.6%) for T. trichiura and 8.1% (4.0–12.2%) for hookworm infections (Fig.2, right column). In countries with low-level STH prevalence at the beginning of their STH control programme, the prevalence was reduced to < 2% while countries with high prevalence moved to low or moderate prevalence (Table 1). Estimated reduction in number of anthelminthic tablets For the 5 years and more of PC for STH, five countries conducted one round annually while 10 countries did semi-annual rounds. Figure3 shows that if the 15 countries that conducted PC for ≥5 years had applied WHO thresholds for changing the frequency of PC rounds using the impact assessment results on STH infection, the number of anthelminthic tablets needed for STH PC would have reduced by an average of 36%. According to the WHO decision tree (Additional file 2), three countries would have suspended STH PC in some areas (because the STH prevalence was < 2%) while six would have decreased the frequency of PC from semi-annual to annual or biennial, and six countries would have maintained the previous frequency of PC. Fig. 3 Anthelminthic drug reduction after ≥5 years of STH PC with effective coverage. STH: Soil-transmitted helminth; PC: Preventive chemotherapy Full size image Discussion In this study we report data from 15 STH endemic countries that conducted ≥5 years of PC with effective coverage among SAC. The overall result demonstrates a reduction of any STH prevalence and of STH species prevalence overall, representing tremendous progress for these countries. Importantly, in three countries the prevalence of any STH in the follow up survey was less than 2%, meaning that they successfully eliminated STH morbidity. A change in prevalence status implies a change in the frequency of PC rounds according to the WHO decision tree (Additional file 2). Countries with an STH prevalence < 2% after ≥5 years of PC should suspend the intervention and maintain surveillance to detect possible rebounds of prevalence, whereas countries with a prevalence of 2–10% should proceed to semi-annual PC. To maintain the gains made when scaling down or discontinuing PC for STH, the national STH control programme should ensure that the water, sanitation and hygiene (WASH) component is also implemented. A modelling study on the impact of deworming and WASH on STH transmission has shown that discontinuing PC and sustaining PC gains require continuation of WASH interventions . Only by achieving Sustainable Development Goal 6, which seeks to ensure universal access to basic WASH in communities, schools and healthcare facilities by 2030, will reductions of morbidity due to STH infection and others NTDs associated with water and sanitation be sustained . Among the 15 countries included in the study, 10 are co-endemic for LF so community-based deworming was the strategy used. Most of these countries have already started scaling down mass drug administration (MDA) for LF. Three countries (Burkina Faso, Mali and Ghana) had a low STH prevalence at the beginning of the programme, but by conducting LF MDA with ivermectin and albendazole or diethylcarbamazine citrate and albendazole, STH infections were also treated. Meanwhile, in 2019, many implementation units have already stopped PC for LF and STH. The limitations of this study are that the data used as the baseline for STH prevalence may have been collected after the start of LF elimination programmes and thus may under-evaluate the real STH prevalence before mass treatment was initiated. For countries in which we know this was true, we used survey data collected before the start of MDA for LF, as in Cameroon, where we used baseline data collected in 1987. Moreover, impact assessment surveys were never implemented in individuals who had participated also in the baseline surveys, sometimes not even in the same area, making the comparison difficult. Finally, not all surveys were nationally representative and thus did not truly reflect the status of STH endemicity in the country. Standardized implementation of impact surveys in all endemic countries would strengthen these data. Conclusions This study shows the importance of conducting impact assessment surveys after ≥5 years of STH PC with effective coverage to guide national STH control programmes in adapting the number of PC rounds according to the new epidemiological situation. Furthermore, a reduction in the frequency of PC and the consequent reduction of associated costs would help endemic countries to achieve national support for, and ownership of, their PC programmes. It would be necessary to confirm that, in different countries and in different epidemiological situation, the reduction in frequency of PC suggested by the decision tree is sufficient to maintain the improvement achieved. Availability of data and materials All relevant data are within the manuscript and its Additional files. Abbreviations EPIRF: Epidemiological Data Reporting Form LF: Lymphatic filariasis MDA: Mass drug administration NTD: Neglected tropical disease PC: Preventive chemotherapy PCT: Preventive chemotherapy and transmission control PSAC: Preschool-age children SAC: School-age children STH: Soil-transmitted helminthiase TAS: Transmission assessment surveys WASH: Water, sanitation and hygiene WHO: World Health Organization References Montresor A, Trouleau W, Mupfasoni D, Bangert M, Joseph SA, Mikhailov A, Fitzpatrick C. Preventive chemotherapy to control soil-transmitted helminthiasis averted more than 500 000 DALYs in 2015. Trans R Soc Trop Med Hyg. 2018. ArticleCASGoogle Scholar Guideline: preventive chemotherapy to control soil-transmitted helminth infections in at-risk population groups. Geneva: World Health Organization; 2017. Accessed 20 Mar 2019. Soil-transmitted helminthiases: eliminating soil-transmitted helminthiases as a public health problem in children: progress report 2001–2010 and strategic plan 2011–2020. Geneva: World Health Organization; 2012. Accessed 20 Mar 2019. Helminth control in school-age children: a guide for managers of control programmes, 2nd. Geneva: World Health Organization; 2011. Accessed 13 Mar 2019. Schistosomiasis and soil-transmitted helminthiases: number of people treated in 2017. Wkly Epidemiol Rec. 2018;93:681–92 Accessed 15 Mar 2019. Marocco C, Bangert M, Joseph SA, Fitzpatrick C, Montresor A. Preventive chemotherapy in one year reduces by over 80% the number of individuals with soil-transmitted helminthiases causing morbidity: results from meta-analysis. Trans R Soc Trop Med Hyg. 2017. Yajima A, Mikhailov A, Mbabazi PS, Gabrielli AF, Minchiotti S, Montresor A, et al. Preventive chemotherapy and transmission control (PCT) databank: a tool for planning, implementation and monitoring of integrated preventive chemotherapy for control of neglected tropical diseases. Trans R Soc Trop Med Hyg. 2012;106:215–22. ArticleGoogle Scholar Ortu G, Assoum M, Wittmann U, Knowles S, Clements M, Ndayishimiye O, et al. The impact of an 8-year mass drug administration programme on prevalence, intensity and co-infections of soil-transmitted helminthiases in Burundi. Parasit Vectors. 2016. Soares Magalhães RJ, Biritwum N-K, Gyapong JO, Brooker S, Zhang Y, Blair L, et al. Mapping helminth co-infection and co-intensity: geostatistical prediction in Ghana. PLoS Negl Trop Dis. 2011. ArticleGoogle Scholar Baker JM, Trinies V, Bronzan RN, Dorkenoo AM, Garn JV, Sognikin S, et al. The associations between water and sanitation and hookworm infection using cross-sectional data from Togo’s national deworming program. PLoS Negl Trop Dis. 2018. ArticleGoogle Scholar Koroma JB, Peterson J, Gbakima AA, Nylander FE, Sahr F, Soares Magalhães RJ, et al. Geographical distribution of intestinal schistosomiasis and soil-transmitted helminthiasis and preventive chemotherapy strategies in Sierra Leone. PLoS Negl Trop Dis. 2010. ArticleGoogle Scholar Ratard RC, Kouemeni LE, Ekani Bessala MM, Ndamkou CN, Sama MT, Cline BL. Ascariasis and trichuriasis in Cameroon. Trans R Soc Trop Med Hyg. 1991. ArticleCASGoogle Scholar Ratard RC, Kouemeni LE, Ekani Bessala MK, Ndamkou CN. Distribution of hookworm infection in Cameroon. Ann Trop Med Parasitol. 1992;86:413–8. ArticleCASGoogle Scholar Mupfasoni D, Ruberanziza E, Karibushi B, Rujeni N, Kabanda G, Kabera M. National school prevalence survey on soil-transmitted helminths and schistosomiasis, Rwanda. Int J Antimicrob Agents. 2009;34(Suppl 2):S15. Google Scholar Champetier de Ribes G, Fline M, Désormeaux AM, Eyma E, Montagut P, Champagne C, Pierre J, et al. Intestinal helminthiasis in school children in Haiti in 2002. Bull Soc Pathol Exot. 2005;98:127–32. CASPubMedGoogle Scholar Flisser A, Valdespino JL, García-García L, Guzman C, Aguirre MT, Manon ML, et al. Using national health weeks to deliver deworming to children: lessons from Mexico. J Epidemiol Community Health. 2008. ArticleCASGoogle Scholar Rosewell A, Robleto G, Rodríguez G, Barragne-Bigot P, Amador JJ, Aldighieri S. Soil-transmitted helminth infection and urbanization in 880 primary school children in Nicaragua, 2005. Trop Doc. 2010. ArticleGoogle Scholar A situation analysis: neglected tropical diseases in Bangladesh. Dhaka: Ministry of Health & Family Welfare, Government of Bangladesh; 2010 Accessed 15 Mar 2019. Montresor A, Zin TT, Padmasiri E, Allen H, Savioli L. Soil-transmitted helminthiasis in Myanmar and approximate costs for countrywide control. Tropical Med Int Health. 2004. Tun A, Myat SM, Gabrielli AF, Montresor A. Control of soil-transmitted helminthiasis in Myanmar: results of 7 years of deworming. Tropical Med Int Health. 2013. ArticleGoogle Scholar Rim HJ, Chai JY, Min DY, Cho SY, Eom KS, Hong SJ, et al. Prevalence of intestinal parasite infections on a national scale among primary schoolchildren in Laos. Parasitol Res. 2003. Coffeng LE, Vaz Nery S, Gray DJ, Bakker R, de Vlas SJ, Clements ACA. Predicted short and long-term impact of deworming and water, hygiene, and sanitation on transmission of soil-transmitted helminths. PLoS Negl Trop Dis. 2018. ArticleGoogle Scholar Bangert M, Molyneux DH, Lindsay SW, Fitzpatrick C, Engels D. The cross-cutting contribution of the end of neglected tropical diseases to the sustainable development goals. Infect Dis Poverty. 2017. Download references Acknowledgements We are thankful for national NTD programme managers, WHO country and regional offices for providing the data through the WHO PCT databank. We also thank Karen CICERI-REYNOLDS and Albis Francesco GABRIELLI for proof reading the article. Funding Not applicable. Author information Authors and Affiliations Department of Control of Neglected Tropical Diseases, World Health Organization, Geneva, Switzerland Denise Mupfasoni,Mathieu Bangert,Alexei Mikhailov,Chiara Marocco&Antonio Montresor Authors 1. Denise MupfasoniView author publications Search author on:PubMedGoogle Scholar 2. Mathieu BangertView author publications Search author on:PubMedGoogle Scholar 3. Alexei MikhailovView author publications Search author on:PubMedGoogle Scholar 4. Chiara MaroccoView author publications Search author on:PubMedGoogle Scholar 5. Antonio MontresorView author publications Search author on:PubMedGoogle Scholar Contributions DM, MB and AMo conceptualized the study.DM and AM did data collection. DM and MB did data analysis. DM and MB drafted the full paper. AMo revised edited the manuscript. All authors contributed to final development of the article. All authors read and approved the final manuscript. Corresponding author Correspondence to Denise Mupfasoni. Ethics declarations Ethics approval and consent to participate Not applicable Consent for publication Not applicable Competing interests The authors declare that they have no competing interests. Additional files Additional file 1: Multilingual abstracts in the five official working languages of the United Nations. (PDF 496 kb) Additional file 2: WHO decision tree for STH control programmes. (DOCX 178 kb) Additional file 3: Soil-transmitted helminthiases (STH) preventive chemotherapy (PC) coverage, by country, 2003–2017. (DOCX 22 kb) Rights and permissions Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. The Creative Commons Public Domain Dedication waiver ( applies to the data made available in this article, unless otherwise stated. Reprints and permissions About this article Cite this article Mupfasoni, D., Bangert, M., Mikhailov, A. et al. Sustained preventive chemotherapy for soil-transmitted helminthiases leads to reduction in prevalence and anthelminthic tablets required. Infect Dis Poverty8, 82 (2019). Download citation Received: 28 May 2019 Accepted: 14 August 2019 Published: 02 October 2019 DOI: Share this article Anyone you share the following link with will be able to read this content: Get shareable link Sorry, a shareable link is not currently available for this article. Copy shareable link to clipboard Provided by the Springer Nature SharedIt content-sharing initiative Keywords Soil-transmitted helminthiases Control; morbidity Preventive chemotherapy Prevalence Download PDF Sections Figures References Abstract Multilingual abstracts Background Methods Results Discussion Conclusions Availability of data and materials Abbreviations References Acknowledgements Funding Author information Ethics declarations Additional files Rights and permissions About this article Advertisement Fig. 1 View in articleFull size image Fig. 2 View in articleFull size image Fig. 3 View in articleFull size image Montresor A, Trouleau W, Mupfasoni D, Bangert M, Joseph SA, Mikhailov A, Fitzpatrick C. Preventive chemotherapy to control soil-transmitted helminthiasis averted more than 500 000 DALYs in 2015. Trans R Soc Trop Med Hyg. 2018. ArticleCASGoogle Scholar Guideline: preventive chemotherapy to control soil-transmitted helminth infections in at-risk population groups. Geneva: World Health Organization; 2017. Accessed 20 Mar 2019. Soil-transmitted helminthiases: eliminating soil-transmitted helminthiases as a public health problem in children: progress report 2001–2010 and strategic plan 2011–2020. Geneva: World Health Organization; 2012. Accessed 20 Mar 2019. Helminth control in school-age children: a guide for managers of control programmes, 2nd. Geneva: World Health Organization; 2011. Accessed 13 Mar 2019. Schistosomiasis and soil-transmitted helminthiases: number of people treated in 2017. Wkly Epidemiol Rec. 2018;93:681–92 Accessed 15 Mar 2019. Marocco C, Bangert M, Joseph SA, Fitzpatrick C, Montresor A. Preventive chemotherapy in one year reduces by over 80% the number of individuals with soil-transmitted helminthiases causing morbidity: results from meta-analysis. Trans R Soc Trop Med Hyg. 2017. Yajima A, Mikhailov A, Mbabazi PS, Gabrielli AF, Minchiotti S, Montresor A, et al. Preventive chemotherapy and transmission control (PCT) databank: a tool for planning, implementation and monitoring of integrated preventive chemotherapy for control of neglected tropical diseases. Trans R Soc Trop Med Hyg. 2012;106:215–22. ArticleGoogle Scholar Ortu G, Assoum M, Wittmann U, Knowles S, Clements M, Ndayishimiye O, et al. The impact of an 8-year mass drug administration programme on prevalence, intensity and co-infections of soil-transmitted helminthiases in Burundi. Parasit Vectors. 2016. Soares Magalhães RJ, Biritwum N-K, Gyapong JO, Brooker S, Zhang Y, Blair L, et al. Mapping helminth co-infection and co-intensity: geostatistical prediction in Ghana. PLoS Negl Trop Dis. 2011. ArticleGoogle Scholar Baker JM, Trinies V, Bronzan RN, Dorkenoo AM, Garn JV, Sognikin S, et al. The associations between water and sanitation and hookworm infection using cross-sectional data from Togo’s national deworming program. PLoS Negl Trop Dis. 2018. ArticleGoogle Scholar Koroma JB, Peterson J, Gbakima AA, Nylander FE, Sahr F, Soares Magalhães RJ, et al. Geographical distribution of intestinal schistosomiasis and soil-transmitted helminthiasis and preventive chemotherapy strategies in Sierra Leone. PLoS Negl Trop Dis. 2010. ArticleGoogle Scholar Ratard RC, Kouemeni LE, Ekani Bessala MM, Ndamkou CN, Sama MT, Cline BL. Ascariasis and trichuriasis in Cameroon. Trans R Soc Trop Med Hyg. 1991. ArticleCASGoogle Scholar Ratard RC, Kouemeni LE, Ekani Bessala MK, Ndamkou CN. Distribution of hookworm infection in Cameroon. Ann Trop Med Parasitol. 1992;86:413–8. ArticleCASGoogle Scholar Mupfasoni D, Ruberanziza E, Karibushi B, Rujeni N, Kabanda G, Kabera M. National school prevalence survey on soil-transmitted helminths and schistosomiasis, Rwanda. Int J Antimicrob Agents. 2009;34(Suppl 2):S15. Google Scholar Champetier de Ribes G, Fline M, Désormeaux AM, Eyma E, Montagut P, Champagne C, Pierre J, et al. Intestinal helminthiasis in school children in Haiti in 2002. Bull Soc Pathol Exot. 2005;98:127–32. CASPubMedGoogle Scholar Flisser A, Valdespino JL, García-García L, Guzman C, Aguirre MT, Manon ML, et al. Using national health weeks to deliver deworming to children: lessons from Mexico. J Epidemiol Community Health. 2008. ArticleCASGoogle Scholar Rosewell A, Robleto G, Rodríguez G, Barragne-Bigot P, Amador JJ, Aldighieri S. Soil-transmitted helminth infection and urbanization in 880 primary school children in Nicaragua, 2005. Trop Doc. 2010. ArticleGoogle Scholar A situation analysis: neglected tropical diseases in Bangladesh. Dhaka: Ministry of Health & Family Welfare, Government of Bangladesh; 2010 Accessed 15 Mar 2019. Montresor A, Zin TT, Padmasiri E, Allen H, Savioli L. Soil-transmitted helminthiasis in Myanmar and approximate costs for countrywide control. Tropical Med Int Health. 2004. Tun A, Myat SM, Gabrielli AF, Montresor A. Control of soil-transmitted helminthiasis in Myanmar: results of 7 years of deworming. Tropical Med Int Health. 2013. ArticleGoogle Scholar Rim HJ, Chai JY, Min DY, Cho SY, Eom KS, Hong SJ, et al. Prevalence of intestinal parasite infections on a national scale among primary schoolchildren in Laos. Parasitol Res. 2003. Coffeng LE, Vaz Nery S, Gray DJ, Bakker R, de Vlas SJ, Clements ACA. Predicted short and long-term impact of deworming and water, hygiene, and sanitation on transmission of soil-transmitted helminths. PLoS Negl Trop Dis. 2018. ArticleGoogle Scholar Bangert M, Molyneux DH, Lindsay SW, Fitzpatrick C, Engels D. The cross-cutting contribution of the end of neglected tropical diseases to the sustainable development goals. Infect Dis Poverty. 2017. 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15046	https://wikidiff.com/egregious/grievous	Grievous vs Egregious - What's the difference? \| WikiDiff What's the difference between and Enter two words to compare and contrast their definitions, origins, and synonyms to better understand how those words are related. Grievous vs Egregious - What's the difference? × grievous \| egregious \| As adjectives the difference between grievous and egregious is that grievous is causing grief, pain or sorrow while egregious is exceptional, conspicuous, outstanding, most usually in a negative fashion. grievous ======== English Alternative forms greuous (obsolete) grievious (less common outside dialects) Adjective (en adjective) Causing grief, pain or sorrow. 1883,_As for the captain, his wounds were grievous indeed but not dangerous._ Serious, grave, dire or dangerous. Synonyms See also egregious ========= English Adjective (en adjective) Exceptional, conspicuous, outstanding, most usually in a negative fashion._The student has made egregious errors on the examination._ 16thC, ,_I cannot cross my arms, or sigh "Ah me," / "Ah me forlorn!" egregious foppery! / I cannot buss thy fill, play with thy hair, / Swearing by Jove, "Thou art most debonnaire!"_ c1605, , Act 2, Scene 3,_My lord, you give me most egregious indignity._ 22 March 2012, Scott Tobias, AV Club The Hunger Games[ the goal is simply to be as faithful as possible to the material—as if a movie were a marriage, and a rights contract the vow—the best result is a skillful abridgment, one that hits all the important marks without losing anything egregious._ '>citation Outrageously bad; shocking. Usage notes The negative meaning arose in the late 16th century, probably originating in sarcasm. Before that, it meant outstanding in a good way. Webster also gives “distinguished” as an archaic form, and notes that its present form often has an unpleasant connotation (e.g., "an egregious error"). It generally precedes such epithets as “rogue,” “rascal,” "ass," “blunderer”. Related terms egregiously (adverb) egregia cum laude egregion (noun) egregore Shares Pin Share Tweet Share Share Text is available under the Creative Commons Attribution/Share-Alike License; additional terms may apply. See Wiktionary Terms of Use for details. Privacy Policy \| About Us \| Contact Us
15047	https://pmc.ncbi.nlm.nih.gov/articles/PMC3579210/	Hemoglobin Variants: Biochemical Properties and Clinical Correlates - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Cold Spring Harb Perspect Med . 2013 Mar;3(3):a011858. doi: 10.1101/cshperspect.a011858 Search in PMC Search in PubMed View in NLM Catalog Add to search Hemoglobin Variants: Biochemical Properties and Clinical Correlates Christopher S Thom Christopher S Thom 1 Cell and Molecular Biology Graduate Group, University of Pennsylvania School of Medicine, Philadelphia, Pennsylvania 19104 2 Hematology Department, Children’s Hospital of Philadelphia, Philadelphia, Pennsylvania 19104 Find articles by Christopher S Thom 1,2, Claire F Dickson Claire F Dickson 3 Menzies Research Institute, University of Tasmania, Hobart, Australia Find articles by Claire F Dickson 3, David A Gell David A Gell 3 Menzies Research Institute, University of Tasmania, Hobart, Australia Find articles by David A Gell 3, Mitchell J Weiss Mitchell J Weiss 2 Hematology Department, Children’s Hospital of Philadelphia, Philadelphia, Pennsylvania 19104 Find articles by Mitchell J Weiss 2 Author information Copyright and License information 1 Cell and Molecular Biology Graduate Group, University of Pennsylvania School of Medicine, Philadelphia, Pennsylvania 19104 2 Hematology Department, Children’s Hospital of Philadelphia, Philadelphia, Pennsylvania 19104 3 Menzies Research Institute, University of Tasmania, Hobart, Australia ✉ Correspondence: weissmi@email.chop.edu Copyright © 2013 Cold Spring Harbor Laboratory Press; all rights reserved PMC Copyright notice PMCID: PMC3579210 PMID: 23388674 Abstract Diseases affecting hemoglobin synthesis and function are extremely common worldwide. More than 1000 naturally occurring human hemoglobin variants with single amino acid substitutions throughout the molecule have been discovered, mainly through their clinical and/or laboratory manifestations. These variants alter hemoglobin structure and biochemical properties with physiological effects ranging from insignificant to severe. Studies of these mutations in patients and in the laboratory have produced a wealth of information on hemoglobin biochemistry and biology with significant implications for hematology practice. More generally, landmark studies of hemoglobin performed over the past 60 years have established important paradigms for the disciplines of structural biology, genetics, biochemistry, and medicine. Here we review the major classes of hemoglobin variants, emphasizing general concepts and illustrative examples. More than 1000 natural variants of hemoglobin are known. They have different structures and biochemical properties (e.g., oxygen affinity), and their physiological effects range from insignificant to severe. Globin gene mutations affecting hemoglobin (Hb), the major blood oxygen (O 2) carrier, are common, affecting an estimated 7% of the world’s population (Weatherall and Clegg 2001; Kohne 2011). These mutations are broadly subdivided into those that impair globin protein subunit production (thalassemias) and those that produce structurally abnormal globin proteins (Hb variants). The latter class is mainly composed of missense mutations that cause single amino acid substitutions in the globin protein, resulting in an abnormal, or “variant” Hb tetramer. Less commonly, Hb variants are associated with deletions, multiple amino acid substitutions, antitermination mutations, and altered posttranslational processing (Table 1). Table 1. Hb variants that are discussed in this article \| Name \| Globin site (fold) \| Amino acid substitution \| Molecular mechanism \| Clinical phenotype \| Other biochemical and laboratory findings \| :---: :---: :---: \| \| Unstable Mutants \| \| Brockton \| β138 (H16) \| Ala > Pro \| Altered secondary structure \| Hemolytic anemia, reticulocytosis \| \| \| Philly \| β35 (C1) \| Tyr > Phe \| Altered α1β1 interface \| Hemolytic anemia, reticulocytosis \| Decreased cooperativity, increased oxygen affinity \| \| Peterborough \| β111 (G13) \| Val > Phe \| Altered α1β1 interface \| Hemolytic anemia, reticulocytosis \| Decreased oxygen affinity \| \| Stanmore \| β111 (G13) \| Val > Ala \| Altered α1β1 interface \| Hemolytic anemia \| Decreased oxygen affinity \| \| J-Guantanamo \| β128 (H6) \| Ala > Asp \| Altered α1β1 interface \| Hemolytic anemia \| Target cells \| \| Khartoum \| β124 (H2) \| Pro > Arg \| Altered α1β1 interface \| Normal \| \| \| Prato \| α1 or α2 31 (B12) \| Arg > Ser \| Altered α1β1 interface \| Anisocytosis, hypochromia \| Mildly unstable in isopropanol \| \| Lombard \| α2 103 (G10) \| His > Tyr \| Altered α1β1 interface \| Anemia \| \| \| Contaldo \| α1 or α2 103 (G10) \| His > Arg \| Altered α1β1 interface \| Hemolytic anemia \| \| \| Foggia \| α2 117 (GH5) \| Phe > Ser \| Altered α1β1 interface \| Microcytosis \| Rapidly degraded α chains \| \| Groene Hart \| α1 119 (H2) \| Pro > Ser \| Altered α1β1 interface, disrupted AHSP binding \| Hemolytic anemia, microcytosis \| \| \| Turriff \| α1 or α2 99 (G6) \| Lys > Glu \| Altered α1β1 interface, disrupted AHSP binding \| Normal \| Comigrates with HbA 1C, rapidly degraded α chains \| \| Beziers \| α1 99 (G6) \| Lys > Asn \| Altered α1β1 interface, disrupted AHSP binding \| Normal \| Comigrates with HbA 1C \| \| Hirosaki \| α2 43 (CE1) \| Phe > Leu \| Altered heme pocket \| Heinz body hemolytic anemia \| Hyperunstable \| \| Terre Haute \| β106 (G8) \| Leu > Arg \| Altered heme pocket \| Heinz body hemolytic anemia, dominant inclusion body thalassemia \| Hyperunstable \| \| High Affinity Variants \| \| Kempsey \| β99 (G1) \| Asp > Asn \| Unstable T state \| Erythrocytosis \| Decreased cooperativity \| \| Hiroshima \| β146 (HC3) \| His > Asp \| Mutated Bohr proton donor \| Erythrocytosis \| Decreased cooperativity, decreased Bohr effect \| \| York \| β146 (HC3) \| His > Pro \| Mutated Bohr proton donor \| Erythrocytosis \| Decreased cooperativity, decreased Bohr effect \| \| Cowtown \| β146 (HC3) \| His > Leu \| Mutated Bohr proton donor \| Erythrocytosis \| Decreased Bohr effect \| \| Rahere \| β82 (EF6) \| Lys > Thr \| Altered 2,3DPG binding site \| Erythrocytosis \| \| \| Providence \| β82 (EF6) \| Lys > Asn \| Altered 2,3DPG binding site \| Erythrocytosis \| Low oxygen affinity \| \| Helsinki \| β82 (EF6) \| Lys > Met \| Altered 2,3DPG binding site \| Erythrocytosis \| Decreased Bohr effect \| \| Low Affinity Variants \| \| Kansas \| β102 (G4) \| Asn > Thr \| Unstable R state \| Cyanosis \| Decreased cooperativity \| \| Beth Israel \| β102 (G4) \| Asn > Ser \| Unstable R state \| Cyanosis \| Decreased cooperativity \| \| St. Mandé \| β102 (G4) \| Asn > Tyr \| Unstable R state \| Cyanosis \| \| \| Methemoglobin Variants \| \| M-Iwate \| α1 or α2 (F8) \| His > Tyr \| Oxidized heme \| Pseudocyanosis (Methemoglobinemia) \| Abnormal visible spectrum \| \| M-Saskatoon \| β63 (E7) \| His > Tyr \| Oxidized heme \| Pseudocyanosis (Methemoglobinemia) \| Abnormal visible spectrum \| \| Globin Chain Elongation Variants \| \| Constant Spring \| α2 142 (HC3) \| Stop > Gln \| Antitermination mutant \| Microcytosis \| Decreased mRNA stability \| \| Cranston \| β145 (HC3) \| +CT \| Frameshift, elongated globin \| Hemolytic anemia \| Increased oxygen affinity, decreased cooperativity \| \| Variants with Multiple Effects \| \| HbE \| β26 (B8) \| Glu > Lys \| Unstable, reduced synthesis \| Microcytosis \| \| \| Bruxelles \| β41 (C7) or β42 (CD1) \| Phe > 0 \| Altered heme pocket \| Hemolytic anemia, cyanosis, splenomegaly, reticulocytosis \| Heinz bodies, decreased cooperativity \| \| Warsaw \| β42 (CD1) \| Phe > Val \| Altered heme pocket \| Hemolytic anemia, cyanosis \| Heinz bodies, decreased cooperativity \| \| Hammersmith \| β42 (CD1) \| Phe > Ser \| Altered heme pocket \| Hemolytic anemia, cyanosis \| Heinz bodies, decreased cooperativity \| \| Buccuresti-Louisville \| β42 (CD1) \| Phe > Leu \| Altered heme pocket \| Hemolytic anemia, cyanosis \| Heinz bodies, decreased cooperativity \| \| Zurich \| β63 (E7) \| His > Arg \| Altered heme pocket \| Normal, but hypersensitive to oxidative stress \| Decreased cooperativity, increased oxygen affinity, increased CO affinity \| \| Jamaica Plain \| β6 (A3) and β68 (E12) \| Glu > Val and Leu > Phe \| Altered secondary structure \| Hemolytic anemia, cyanosis, splenomegaly, splenic sequestration \| Unstable, Heinz bodies, sickle cell phenotype, decreased oxygen affinity \| \| Quebec-Chori \| β87 (F3) \| Thr > Ile \| Altered interaction with HbS polymer \| Normal \| Promotes HbS polymerization \| \| D-Ibadan \| β87 (F3) \| Thr > Lys \| Altered interaction with HbS polymer \| Normal \| Inhibits HbS polymerization \| \| Bristol-Alesha \| β67 (E11) \| Val > Met \| Altered heme pocket \| Hemolytic anemia, reticulocytosis \| Heinz bodies, decreased cooperativity, decreased Bohr effect, decreased oxygen affinity \| \| Toms River \| γ67 (E11) \| Val > Met \| Altered heme pocket \| Anemia, cyanosis \| Unstable, low oxygen affinity \| Open in a new tab For a full listing of hemoglobin variants, see The Globin Gene Server ( Hardison et al. 2002; Giardine et al. 2011). Naturally occurring Hb mutations cause a range of biochemical abnormalities, some of which produce clinically significant symptoms. The most common and medically important Hb variants include HbS (Cao and Kan 2012; Lettre 2012; Schechter and Elion 2012; Serjeant and Rodgers 2012; Williams and Weatherall 2012), HbC (Cao and Kan 2012; Lettre 2012; Schechter and Elion 2012; Serjeant and Rodgers 2012; Williams and Weatherall 2012), HbE (see the sections on Selected Variants that Illustrate Important Aspects of Hemoglobin Biology and Variants that Affect Multiple Hemoglobin Functions; see also Musallam et al. 2012), and some thalassemias (e.g., “thalassemic hemoglobinopathies”), all of which are under positive genetic selection because they confer survival advantages in areas where malaria is endemic (Weatherall and Clegg 2001). In addition to these prevalent mutant proteins, there are also >1000 other known naturally occurring Hb variants, which are rare individually but common collectively. Most Hb mutants are cataloged on the Globin Gene database (HbVar, Hardison et al. 2002; Giardine et al. 2011). By convention, these variants are named after the geographic origin of the affected individual. Although many Hb variants are clinically silent, some produce clinical manifestations of varying severity. Analyses of these variants, which can be considered to be “experiments of nature” (Garrod 1928), have generated valuable insights into structure–function relationships within the Hb molecule, with interesting and important clinical consequences. The goal of this work is to provide a succinct conceptual framework for understanding the biology and clinical implications of Hb variants. We explore major concepts of Hb biology followed by a discussion of selected Hb variants that reinforce basic principles. For more extensive reviews of this topic, see Nathan and Oski’s Hematology of Infancy and Childhood (Nathan et al. 2009); Hemoglobin: Molecular, Genetic, and Clinical Effects (Bunn and Forget 1986); Disorders of Hemoglobin: Genetics, Pathophysiology, and Clinical Management (Steinberg et al. 2001); and the Globin Gene database (Hardison et al. 2002; Giardine et al. 2011). BASIC PRINCIPLES Hemoglobin Synthesis, Structure, and Function Hemoglobin is a heterotetramer composed of α-like and β-like globin subunits, each bound to a heme prosthetic group. The major functions of Hb are to transport oxygen (O 2) from the lungs to peripheral tissues and carbon dioxide (CO 2) from the tissues to the lungs. The kinetics of Hb-O 2 binding and release are fine-tuned for this purpose and adaptable according to developmental ontogeny and metabolic perturbations. Moreover, the Hb molecule must limit potential problems caused by its associated iron and free O 2, reactive molecules capable of inflicting damage through the production of reactive oxygen species. Efforts to understand how Hb structure imparts these critical functions have been ongoing for more than 50 years. Hemoglobin was one of the first proteins to be sequenced (Konigsberg et al. 1961; Schroeder et al. 1961; Watson and Kendrew 1961) and the globin genes were among the earliest to be cloned (Rabbitts 1976). In the late 1950s, Perutz and colleagues determined the three-dimensional structure of Hb through X-ray crystallography (Perutz 1960; Perutz et al. 1960). More recent studies refined this structure to high resolution (Paoli et al. 1996; Park et al. 2006). In addition, O 2 and other ligand-binding properties have been measured in detail for native Hb and many naturally occurring mutants. All of this work provides a substantial framework for defining the O 2 delivery properties of Hb, as well as the molecular consequences of variant mutations (see, for example, Lehmann 1957; Konigsberg and Lehmann 1965; Shimizu et al. 1965; Perutz and Lehmann 1968; Perutz 1970; Bunn and Forget 1986). Globin polypeptides are synthesized from separate α-like and β-like globin gene clusters located on human chromosomes 16 and 11, respectively. Nascent globin chains rapidly incorporate heme, which stabilizes their native folding into Hb subunits composed of seven or eight α helices named A–H, which fold together into a globular structure (Fig. 1A). Hb subunits bind O 2 and other ligands via the heme iron buried within an evolutionarily conserved hydrophobic pocket that faces the outside of HbA tetramers (Fig. 1B). Heme iron is axially coordinated to globin proteins by an invariant histidine residue in helix F8, termed the “proximal” histidine (Fig. 1C). The opposite axial position binds O 2, which is stabilized by interaction with the conserved “distal” histidine in helix E7 (Fig. 1C). Multiple additional amino acids within the globin proteins stabilize heme binding through noncovalent interactions (Fig. 1C). Iron must be in its reduced (ferrous, Fe 2+) state for Hb to bind O 2. Oxidized or “met” Hb (ferric, Fe 3+) cannot bind O 2 and is relatively unstable, tending to lose hemin and denature. Thus, red blood cells have evolved elaborate mechanisms to maintain Hb in its reduced state (Bunn and Forget 1986; Ganz and Nemeth 2012; Schechter 2012). For example, the methemoglobin reductase system converts methemoglobin (metHb) to its reduced form. Not surprisingly, globin mutations that alter amino acids within the ligand pocket frequently produce strong functional effects, including destabilization, altered affinity for O 2, and increased rates of metHb formation and heme loss (see the section on Selected Variants that Illustrate Important Aspects of Hemoglobin Biology). Variants promoting autoxidation are termed “M-Hbs” (see the sections on Selected Variants that Illustrate Important Aspects of Hemoglobin Biology and Methemoglobin (“M-Type”) Variants). Because Hb gains its distinctive color from the heme group, alterations that affect the environment of the heme iron, including changes in the surrounding amino acids, different gas ligands, or redox state, produce characteristic changes in visible light absorption. These color changes are used clinically to assess Hb-O 2 saturation, metHb formation and the effects of Hb variants. Figure 1. Open in a new tab The structure of Hb. (A) The α (pink) and β (red) Hb subunits have conserved α-helical folds. Helices are labeled A–H from the amino terminus. The α subunit lacks helix D. (B) The high O 2 affinity R state quaternary structure of Hb with O 2 (red spheres) bound at all four heme sites (protoporphyrin-IX as yellow sticks, with central iron atom as orange sphere). (C) Stereo (wall-eye) diagram of the heme pocket of β showing the proximal (F8) and distal (E7) histidines and selected residues in the distal heme pocket that influence ligand binding and autoxidation. (D) Hb tetramer is assembled from two identical αβ dimers (shown in red and gray for clarity). In the tetramer, each subunit makes contact with the unlike chain through a high affinity dimerization α1β1 interface and a lower affinity α1β2 dimer–tetramer interface (cyan). Within the Hb tetramer, each globin subunit binds the unlike chain through two distinct interfaces, termed α1β1 and α1β2 (Fig. 1D). Individual globin subunits form dimers through the extremely high affinity α1β1 interaction. Globin chain monomers are relatively unstable compared with dimers, with a tendency to form intracellular precipitates that damage erythrocytes, causing hemolytic anemia. Thus, mutations that impair the α1β1 interaction can cause erythrotoxicity by favoring the accumulation of monomeric subunits (Fig. 2C). The α1β2 interaction, which is lower affinity, mediates tetramerization. Oxygen binding destabilizes the α1β2 interaction, resulting in a transition of the quaternary structure from the “T” (tense, low affinity, deoxygenated) to “R” (relaxed, high affinity, oxygenated) state, which facilitates O 2 binding to additional subunits (Fig. 2A). This process causes cooperative O 2 binding, which is illustrated by the characteristic sigmoidal shape of the Hb-O 2 equilibrium curve (Fig. 2B). Cooperativity allows maximal O 2 release over relatively small drops in O 2 tension. Mutations within the α1β2 interaction region can alter the functional properties of Hb, mainly by perturbing O 2-binding characteristics (Fig. 2C, cyan spheres; Table 1). Figure 2. Open in a new tab Hb variants with altered subunit interactions. (A) Conversion from the low O 2 affinity (deoxy, T state) to high O 2 affinity (oxy, R state) involves a relative rotation of the α1β1 and α2β2 dimers, with changes in contacts across the α1β2 (and α2β1) interface (cyan). In this cartoon, the α1β1 dimer is held stationary to reveal the relative motion of the α2β2 dimer in going from the deoxy (orange) to the oxy (red) states. (B) The sigmoidal shape of the Hb-O 2 saturation curve shows allosteric regulation by changes in pH, temperature, and 2,3DPG. These regulators, as well as Hb variants, influence the shape of the curve. High oxygen affinity variants, high pH, low 2,3DPG, or low temperature induce a “left shift” in the saturation curve (red line). Conversely, low oxygen affinity variants, low pH, high 2,3DPG, or high temperature induce a “right shift” (blue line). (C) Hb sequence variants at the allosteric α1β2 interface (cyan spheres) show an impaired response to O 2 binding. Some sequence variants disrupt binding to other allosteric regulators, e.g., substitutions at βK82 (green) disrupt interactions with 2,3DPG that normally stabilize the low O 2 affinity T state. Mutations that disrupt α1β1 (and α2β2) dimerization (blue spheres) increase the concentration of free monomers, which are unstable. (D) Some α mutations that disrupt binding to β may also disturb binding to the chaperone, AHSP. Other α variants, such as Turriff and Beziers (pink sphere) may inhibit only AHSP binding. Cooperativity represents a general phenomenon termed allosteric regulation, in which effector molecules control the properties of enzymes or other proteins by binding to regions that are distinct from the active sites. Several allosteric regulators, in addition to O 2, influence the properties of Hb. For example, H+ binds Hb to promote O 2 release in a process termed the Bohr effect (Perutz et al. 1984). In peripheral tissues, abundant CO 2 is taken up by red blood cells and metabolized by carbonic anhydrase to carbonic acid, resulting in the production of H+ and consequent O 2 release. In contrast, relatively low CO 2 and high pH in the lung favors O 2 binding to Hb. Additionally, CO 2 forms carbamino adducts with the amino termini of both α and β globins in vivo. These interactions produce minor effects on whole blood O 2 affinity compared to the impact of CO 2 on pH (Bunn and Forget 1986). The compound 2,3-diphosphoglycerate (2,3DPG), formed as a by-product of glycolysis and present at relatively high concentrations in red blood cells, is another important allosteric regulator of Hb. 2,3DPG binds and stabilizes T state Hb to facilitate O 2 release. Hemoglobin binding sites for H+ and 2,3DPG have been identified (Perutz et al. 1969, 1984; Perutz 1970; Arnone 1972). Together, these interactions allow Hb to sense metabolic activity and modulate O 2 binding accordingly. This environmental sensory function can be altered by mutations that affect the affinity of Hb for its allosteric regulators (Fig. 2C, green sphere; Table 1). Identification of Hb Variants and Their Clinical Implications Many Hb variants are readily ascertained through physical examination and/or routine laboratory testing, which explains why so many have been discovered. Uncommonly, some variants are identified through evaluation of ill patients with severe anemia or clinically significant cyanosis. Many amino acid substitutions alter surface charge and are thus detected by electrophoresis or chromatography, techniques which are routinely performed on neonates in North America and Europe. Interestingly, a few Hb variants migrate similarly to HbA 1c, a glycosylated form of Hb that reflects long-term control of blood glucose levels in diabetic patients (Table 1) (reviewed in Little and Roberts 2009). In this way, specific Hb amino acid substitutions can artificially elevate HbA 1c measurement and interfere with diabetic management. Other Hb variants are clinically benign but produce obvious changes in skin color. For example, mutations that increase Hb-O 2 affinity typically stimulate erythropoietic drive by inhibiting O 2 tissue delivery, causing erythrocytosis associated with a ruddy complexion (Nathan et al. 2009). Mutations that reduce O 2 affinity produce a bluish hue to the skin (cyanosis) caused by abnormally high levels of deoxyHb. Mutations that favor oxidation of Hb iron to the met form (M-Hbs), also cause blue-tinged skin, whereas the blood itself appears brown. Studies of one family with congenital cyanosis led Horlein and Weber to describe the first known hemoglobinopathy, caused by the variant Hb-M Saskatoon (Horlein and Weber 1948). The M-Hb variant Iwate, which causes “black blood” or “hereditary nigremia,” was first described more than 200 years ago in Japan (Shibata et al. 1960). It is important to note that many Hb variants affecting skin color are not clinically damaging beyond their cosmetic effects. However, these conditions can mimic more life-threatening problems such as cardiopulmonary and myeloproliferative disorders, which must be excluded. Thus, patients with cyanotic or polycythemic Hb variants may mistakenly undergo unnecessary and potentially dangerous medical procedures. Historical examples include cyanotic patients undergoing surgery or catheterization for presumed heart defects and polycythemic patients receiving radioactive 32 P for presumed polycythemia vera (Steinberg et al. 2001; Nathan et al. 2009). As stated, “the primary reason for establishing the diagnosis [of M Hbs] is to prevent iatrogenic misadventures that might arise under the mistaken impression that the patient has a cardiac or pulmonary disorder” (Bunn and Forget 1986). Most Hb variants with altered O 2 affinity are relatively simple to diagnose through history, physical examination, and laboratory testing (Wajcman et al. 2001; Wajcman and Moradkhani 2011). Globin Gene Synthesis Is Developmentally Regulated Developmental regulation of the α-like and β-like globin gene families is of great medical significance (Sankaran and Orkin 2012). Hemoglobin F (α 2 γ 2) is the most highly expressed form during late fetal gestation. After birth, expression gradually switches to HbA (α 2 β 2) over several months. Thus, symptomatic mutations affecting α or γ globins are present prenatally or at birth, whereas the manifestations of β-globin mutations are typically delayed until a few months after birth. Interestingly, γ-globin gene mutations that are apparent at birth, most typically reflected by cyanosis or hemolytic anemia (e.g., Hb-F Toms River, sections on Selected Variants that Illustrate Important Aspects of Hemoglobin Biology and Variants that Affect Multiple Hemoglobin Functions, and Table 1), fade over a few weeks as Hb production switches from F (α 2 γ 2) to A (α 2 β 2). Laboratory Testing for Hb Variants In many countries, routine testing of all newborns is performed to identify common hemoglobinopathies such as some thalassemias and HbS. Isoelectric focusing or high-performance liquid chromatography (HPLC), the most commonly used tests, identify most structurally abnormal Hbs (Wajcman et al. 2001). In this way, many benign Hb variants are discovered incidentally. Clinical indications for laboratory testing to investigate potential Hb variants are listed in Table 2. Table 2. Clinical indications for laboratory testing to diagnose Hb variants \| Indications for hemoglobin testing \| \| Routine newborn testing for common hemoglobinopathies (i.e., HbS, HbC, thalassemias) \| \| Cyanosis with adequate arterial oxygenation and no apparent cardiopulmonary disease \| \| Erythrocytosis with normal or elevated erythropoietin levels \| \| Unexplained hemolytic anemia \| \| Unexplained thalassemia phenotype \| \| Family history consistent with an Hb variant \| Open in a new tab Specific laboratory tests to investigate Hb variants include: Physical methods of Hb separation. These include electrophoretic and chromatographic techniques that examine the physical properties of α1β1 dimers or individual globin subunits. Specific standards, such as HbA, HbS, HbC, HbF, and HbA 2, are typically examined as controls. Hb variants may show altered migration in these assays. Historically, cellulose acetate and citrate agar electrophoresis were most commonly used to detect variant Hbs. Current clinical testing more typically uses isoelectric focusing and HPLC, which are more sensitive. Hb-O 2 binding curve. This test, performed on whole red blood cells or hemolysate, indicates the percent (%) oxygenated Hb at a given O 2 partial pressure (Fig. 2B). Hemoglobin variants with an abnormally high O 2 affinity (see sections on Selected Variants that Illustrate Important Aspects of Hemoglobin Biology and High Oxygen Affinity Variants) will become saturated at lower O 2 pressures producing a “left-shifted” O 2 equilibrium curve, whereas mutations that reduce O 2 affinity (see sections on Selected Variants that Illustrate Important Aspects of Hemoglobin Biology and Low Oxygen Affinity Variants) will cause the opposite “right shift.” Determination of Hb-O 2 affinity responses to allosteric regulators, particularly 2,3DPG or H+ (pH changes), can provide insight into the structural causes of the observed phenotypes. Unfortunately, few clinical laboratories currently offer this assay. Visible wavelength spectroscopy. Hemoglobin variants with amino acid substitutions in the heme pocket affect visible light absorbance. For example, M-type Hbs show characteristic spectra that can distinguish them from methemoglobinemia caused by an enzyme deficiency in the metHb reductase system (Bunn and Forget 1986; Dailey and Meissner 2012; Ganz and Nemeth 2012; Schechter 2012). Pulse oximetry is a noninvasive spectrophotometric test that measures absorbance ratios at specific wavelengths for oxygenated (660 nm) and deoxygenated (940 nm) blood (Tremper and Barker 1989). This can produce confusing and sometimes misleading results in patients with variant Hbs that show unique light absorbance properties (Verhovsek et al. 2010). In these cases, analysis of arterial blood O 2 concentration may be required to rule out hypoxia. Analyzing these variant Hbs using light absorbance throughout the full visible wavelength spectrum can provide useful diagnostic information. Hemoglobin stability testing. Typically, Hb stability is impaired in variants that are associated with hemolytic anemia. Hemoglobin stability tests measure the propensity for Hb to denature on exposure to various stresses including heat (Carrell and Kay 1972), isopropanol (Bender et al. 1981), mechanical agitation (Asakura et al. 1975), and zinc acetate (Roth et al. 1976). The Heinz body test uses supravital stains, such as methylene blue or crystal violet, to detect aggregated globins within erythrocytes (Eisinger et al. 1985). Specialized testing. Mass spectrometry analysis of patient hemolysate and DNA sequencing of globin genes are specialized confirmatory tests to identify amino acid and nucleotide alterations associated with suspected Hb variants (Wajcman et al. 2001; Wajcman and Moradkhani 2011). DNA sequencing may readily elucidate the presence of an Hb variant. However, additional biochemical and structural studies, including those discussed in this section, are required to determine how the variant affects Hb function. Crystallographic analysis is the highest resolution approach to determine the effects of globin amino acid substitutions on molecular structure. Crystallography has been used historically as a research tool to assess the effects of some interesting Hb variants (see, for example, Pulsinelli et al. 1973; Perutz et al. 1984). SELECTED VARIANTS THAT ILLUSTRATE IMPORTANT ASPECTS OF HEMOGLOBIN BIOLOGY Unstable Variants Unstable variants frequently cause congenital Heinz body hemolytic anemia detected by laboratory screening and clinical symptoms (see sections on Basic Principles and Laboratory Testing for Hb Variants and Table 2). Mutations that alter any step in globin processing, including subunit folding, heme interaction, dimerization, or tetramerization, can destabilize Hb. Bunn and Forget note five general mechanisms that destabilize Hbs: amino acid substitutions within the heme pocket, disruption of secondary structure, substitution in the hydrophobic interior of the subunit, amino acid deletions, and elongation of the subunit (Bunn and Forget 1986). More than 75% of Hb is α helical (Perutz et al. 1960; Park et al. 2006). This structure is particularly susceptible to disruptions by proline substitutions (Levitt 1981). For example, in Hb Brockton (β138 [H16] Ala > Pro) the substituted proline disrupts intermolecular hydrogen bonding between β138Ala and β134Val in helix H (Russu and Ho 1986; Moo-Penn et al. 1988). This produces an unstable variant with a propensity to precipitate and aggregate, thereby damaging erythrocytes and predisposing to hemolysis. Hb Brockton does not show altered O 2 binding affinity or electrophoretic mobility shifts. This variant was identified by HPLC analysis of patient globin chains and its altered X-ray crystallography diffraction pattern shows local disruption of helix H (Moo-Penn et al. 1988). Mutations at the α1β1 interface can cause hemolytic anemia by inhibiting heterodimer formation, favoring the accumulation of free globin subunits, which themselves are unstable, particularly α chains (Fig. 2C, blue spheres). Examples include Hb Philly (β35 [C1] Tyr > Phe) (Rieder et al. 1969), Hb Peterborough (β111 [G13] Val > Phe) (King et al. 1972), Hb Stanmore (β111 [G13] Val > Ala) (Como et al. 1991), and Hb J-Guantanamo (β128 [H6] Ala > Asp) (Martínez et al. 1977). Hb Khartoum (β124 [H2] Pro > Arg) contains a substitution at the α1β1 interface that is mildly destabilizing in vitro, but does not cause clinical symptoms (Clegg et al. 1969; Argos et al. 1979). Interestingly, some α-globin (HBA) gene mutations affecting the α1β1 interface may also destabilize free α chains by inhibiting their binding to α-hemoglobin stabilizing protein (AHSP), an erythroid molecular chaperone that facilitates Hb assembly (Fig. 2D, blue spheres) (reviewed in Weiss et al. 2005; Favero and Costa 2011). These α-globin variants include Hbs Prato (α1 or α2 31 [B12] Arg > Ser) (Marinucci et al. 1979), Lombard (α2 103 [G10] His > Tyr) (Hoyer et al. 2002), Contaldo (α1 or α2 103 [G10] His > Arg) (Sciarratta et al. 1984), Foggia (α2 117 [GH5] Phe > Ser) (Lacerra et al. 2008), Groene Hart (α1 119 [H2] Pro > Ser) (Harteveld et al. 2002; Vasseur-Godbillon et al. 2006; Giordano et al. 2007; Vasseur et al. 2009), and others (Wajcman et al. 2008; Yu et al. 2009). Naturally occurring α-globin variants with amino acid substitutions at position 99 including Hb Turriff (α1 or α2 99 [G6] Lys > Glu) (Langdown et al. 1992) and Hb Beziers (α1 99 [G6] Lys > Asn) (Lacan et al. 2004) bind β globin normally but show impaired interaction with AHSP and may be mildly destabilizing (Fig. 2D, violet spheres) (see also Yu et al. 2009; Khandros et al. 2012; Mollan et al. 2012). Antitermination mutations can also destabilize α globin in part by impairing its binding to AHSP (Turbpaiboon et al. 2006). Hyperunstable Hb variants precipitate shortly after synthesis and are not incorporated into Hb tetramers. These ephemeral proteins are difficult to isolate. In this case, electrophoresis may be falsely negative due to the rapid turnover of these variants, making DNA sequencing a critical diagnostic test. Affected patients show a dominantly inherited “inclusion body thalassemia” resulting from both the precipitated variant globin and consequent chain imbalance with accumulation of the unaffected globin, which is unstable in its free form (Stamatoyannopoulos et al. 1974). Patient erythrocytes typically display abnormal morphology with microcytosis, hypochromia, moderate to severe anisopoikilocytosis, basophilic stippling, and inclusions that may be become particularly prominent following splenectomy (Steinberg et al. 2001). Weatherall, Thein, and colleagues characterized several hyperunstable mutations in exon 3 of the β-globin gene (Thein et al. 1990). All of the mutations were frameshifts or nonsense codons that produced relatively long (>120 amino acid) proteins with carboxy-terminal truncations. The investigators proposed that truncated globins causing dominantly inherited thalassemia are long enough to bind heme posttranslationally, which rendered them relatively resistant to proteolytic degradation, allowing for subsequent precipitation of heme-containing aggregates detected as Heinz bodies. Missense mutations also cause hyperunstable Hb variants. Hb Hirosaki (α2 43 [CE1] Phe > Leu) was discovered in a family with hemolytic anemia (Ohba et al. 1975; Tanaka et al. 2005). After several tests failed to identify a soluble variant Hb within erythrocytes, DNA sequencing was used to characterize the mutation. Hb Terre Haute (β106 [G8] Leu > Arg) is another hyperunstable variant associated with a severe Heinz body hemolytic anemia and globin chain imbalance (Coleman et al. 1991). In initial studies of patient erythroid cells, performed in 1979, abnormal Hb tetramers were not detected and peptide mapping of radiolabeled nascent globins identified a β112 (G14) Cys > Arg substitution, originally termed Hb Indianapolis (Adams et al. 1979). However, the β112 Cys > Arg mutation was subsequently identified in unrelated individuals with much less severe disease. In 1991, reevaluation of the original pedigree by DNA analysis discovered a β106 Leu > Arg mutation, which was renamed Hb Terre Haute (Coleman et al. 1991). Most likely, incomplete tryptic cleavage of the abnormal β-globin peptide in the earlier studies led to misidentification of the causative mutation. This work reflects the interesting historical point that many Hb variants were identified through laborious and technically challenging protein studies performed some time ago, before DNA sequence analysis of patient globin genes was feasible. Reevaluation of these mutations through genetic testing has yielded some interesting surprises (see also the discussion on Hb Bristol-Alesha and the section on Variants that Affect Multiple Hemoglobin Functions). High Oxygen Affinity Variants Hemoglobin variants with increased O 2 affinity cause erythrocytosis by stimulating erythropoietic drive (see Hebbel et al. 1978 for a description of the associated physiology). This commonly results from amino acid substitutions that stabilize the R (high O 2 affinity) state relative to the T (low O 2 affinity) state and/or inhibit responses to environmental allosteric regulators that stimulate O 2 release, including H+ (Bohr effect) or 2,3DPG (see the sections on Basic Principles and Hemoglobin Synthesis, Structure, and Function). Because T to R state transitions are mediated largely through α1β2 interactions, high affinity variants frequently result from substitutions that alter this interface (Fig. 2C, cyan spheres). For example, the amino acid replacement in Hb Kempsey (β99 [G1] Asp > Asn) perturbs α1β2 interactions by preventing the formation of a hydrogen bond between β99 Asp and α42 Tyr, which normally stabilizes the deoxygenated low O 2 affinity T state (Fig. 3) (Reed et al. 1968; Lindstrom et al. 1973; Bunn et al. 1974). This structural change shifts quaternary equilibrium toward the oxygenated R form, which impairs O 2 release to peripheral tissues and thus increases erythropoietic drive. The carboxyl termini of globin chains are also involved in α1β2 interactions that stabilize the low O 2 affinity T state and numerous substitutions within these regions cause high O 2 affinity variants. In addition, β146 His at the carboxyl terminus contributes significantly to the Bohr effect by forming a salt bridge with β94 Asp (Perutz et al. 1984). This interaction is disrupted in several high O 2 affinity variants with β146 substitutions: Hb Hiroshima (β146 [HC3] His > Asp) (Hamilton et al. 1969; Perutz et al. 1971; Imai et al. 1972; Olson et al. 1972), Hb York (β146 [HC3] His > Pro) (Bare et al. 1976), and Hb Cowtown (β146 [HC3] His > Leu) (Fig. 2C) (Schneider et al. 1979; Perutz et al. 1984). These variants show reduced Bohr effect with impaired release of O 2 under acidic conditions. Figure 3. Open in a new tab Hb variants that affect allosteric regulation. (A) The α2 and β2 subunits indicated on the right side of the figure show an overlay of the R-state (red/pink) and T-state (orange/light orange) quaternary structures of Hb. The allosteric α 1 β 2 interface is boxed. (B) Detail of the α 1 β 2 interface in the deoxy T state (β 2 chain in orange, PDB 2DN2) and oxy R state (β 2 chain in red, PDB 2DN1) showing selected H-bonding interactions. Note that the two H-bonding networks use nonoverlapping sets of side chains, hence mutations in these residues affect only one state. Mutation of Asp99 to Asn (Hb Kempsey) compromises electrostatic interactions in the deoxygenated state, thereby favoring the R state and causing impaired O 2 release. Mutation of Asn102 to Thr (Hb Kansas) abrogates interactions with Asp94, favoring the T state and O 2 binding inhibition. Several high affinity Hb variants are caused by substitutions that inhibit interaction with 2,3DPG, which normally binds globin chains to stimulate O 2 release (Fig. 2C). For example, Hb Rahere (β82 [EF6] Lys > Thr) replaces an invariant lysine in the 2,3DPG binding site of β globin, thereby reducing the affinity for this allosteric regulator (Lorkin et al. 1975; Sugihara et al. 1985). Consequently, Hb Rahere shows blunted O 2 release in response to added 2,3DPG in vitro. In vivo, O 2 release in peripheral tissues is inhibited, resulting in elevated blood Hb levels. Similarly, Hb Providence (β82 [EF6] Lys > Asn) (Bonaventura et al. 1976; Moo-Penn et al. 1976) and Hb Helsinki (β82 [EF6] Lys > Met) (Ikkala et al. 1976; Charache et al. 1977) are high O 2 variants caused by different amino acid substitutions at the 2,3DPG binding site on β82. Low Oxygen Affinity Variants Low O 2 affinity Hb variants typically present with cyanosis. In general, these variants are caused by globin amino acid substitutions that tip the quaternary equilibrium of Hb tetramers from the high affinity oxygenated R state to the low affinity deoxygenated T state; more or less the opposite of what occurs for high O 2 affinity variants (see the sections on Selected Variants that Illustrate Important Aspects of Hemoglobin Biology and High Oxygen Affinity Variants). This does not inhibit Hb-O 2 release in tissue capillaries, but rather, interferes with Hb-O 2 uptake if the P 50 has increased to ≥50 mm Hg. Paradoxically, low O 2 affinity Hb variants can be associated with mild anemia thought to be caused by increased O 2 tissue delivery with reduced erythropoietic drive (Stamatoyannopoulos et al. 1969). In addition, some low O 2 affinity mutants are unstable and therefore associated with not only cyanosis but also Heinz body hemolytic anemia (see the sections on Selected Variants that Illustrate Important Aspects of Hemoglobin Biology and Unstable Variants). Several low affinity variants involve replacements at the α1β2 interface, which plays an important role in Hb cooperativity. Hb Kansas (β102 [G4] Asn > Thr) is a well-studied low O 2 affinity variant (Fig. 3B) (Reissmann et al. 1961; Bonaventura and Riggs 1968; Gibson et al. 1973; Riggs and Gibson 1973). Affected individuals are markedly cyanotic, although clinically well. Replacement of Asn102 at the α1β2 interface inhibits the formation of a hydrogen bond with Asp94 that normally stabilizes the oxygenated R structure. A similar mechanism causes low O 2 affinity in two other Hb variants through different substitutions of the same amino acid residue (B102 [G4] Asn) in Hb Beth Israel (β102 [G4] Asn > Ser) (Nagel et al. 1976) and Hb St. Mandé (β102 [G4] Asn > Tyr) (Arous et al. 1981; Poyart et al. 1990). Methemoglobin (“M-Type”) Variants Hemoglobin iron must be in its reduced (Fe 2+, ferrous) state to bind O 2. Moreover, oxidized (Fe 3+, ferric, met) Hb is intrinsically unstable with a tendency to release heme. Hemoglobin reduction is maintained through intrinsic features of the Hb protein and extrinsic antioxidant pathways within red blood cells (see the sections on Basic Principles and Hemoglobin Synthesis, Structure, and Function). Exposure to oxidant drugs or toxins, genetic alterations in erythroid metHb reductase enzyme systems (Ganz and Nemeth 2012; Schechter 2012), or globin chain variants can predispose to methemoglobinemia. These disorders present as “pseudocyanosis,” (i.e., low Hb-O 2 saturation), despite adequate arterial oxygenation. Detailed in vitro analyses of the red cell and isolated Hb samples can usually distinguish wild-type metHb resulting from toxins or defective reductase systems and M-type Hb variants that are predisposed to spontaneous oxidation (Bunn and Forget 1986; Steinberg et al. 2001; Nathan et al. 2009). For example, various M-Hbs show characteristic visible absorbance spectra. Globin variants associated with metHb formation are typically caused by amino acid substitutions within the heme pocket. For example, four different M-Hbs occur when tyrosine replaces the α- or β-globin proximal or distal histidine residues that interact with heme (Fig. 1C) (reviewed in Adachi et al. 2011). In Hb M-Iwate (α1 or α2 87 [F8] His > Tyr), the proximal histidine is replaced by tyrosine (Fig. 4A), which deprotonates and coordinates to the heme iron (Fig. 4B) (Konigsberg and Lehmann 1965; Shimizu et al. 1965). Ferric heme, bound through the native His F8, is readily reduced by metHb reductase (Fig. 4C). Tyrosine (F8) coordination stabilizes the oxidized ferric state and decreases reactivity with metHb reductases. This interaction also distorts the position of heme and helix F within the altered α subunits. In native Hb, movement of the proximal His F8 and F helix away from the heme group stabilizes the deoxygenated T state and reduces the O 2 affinity of the native β subunit partners. Hence, substitution of the normal His F8 side chain for a longer Tyr F8 (Fig. 4A) also stabilizes the deoxygenated T state and reduces O 2 affinity of the native β subunit in Hb M-Iwate (Nagai et al. 2000; Jin et al. 2004). These biochemical and structural alterations underlie the lack of cooperativity and severe cyanosis in patients with Hb M-Iwate, in which metHb levels can exceed 20% (normal <2%) (Ameri et al. 1999). In contrast, Hb M-Saskatoon (β63 [E7] His > Tyr) replaces the distal His with Tyr (Fig. 4D) (Horlein and Weber 1948; Hayashi et al. 1966). In this variant, the protonated form of mutant Tyr can bind ferric heme iron to generate a hexacoordinate structure that is relatively easily reduced by cellular metHb reductases (Fig. 4E). Consequently, patients with Hb M-Saskatoon have lower levels of circulating oxidized Hb compared with those with Hb M-Iwate. Comparative studies of these patients and their variant M-Hbs have contributed greatly to understanding the biochemical properties of the heme iron, including its interactions with various ligands and nearby amino acids such as the proximal and distal histidines. Figure 4. Open in a new tab Examples of M-type hemoglobins. (A) The heme pocket of wild-type Mb (gray) and the F8 His > Tyr mutation (orange, PDB 1HRM), which serves as a model for Hb M-Iwate. Effects on the protein fold are to increase the distance from the heme to the F helix, recapitulating features of the deoxy-α (T state) structure. (B) In Hb M-Iwate, α 87 Tyr F8 is deprotonated and favors Fe 3+ oxidation state, resulting in rapid autoxidation. This ferric form is not reduced by met-Hb reductase. (C) With the normal His F8 present, ferric heme is readily reduced. (D) Substituting the distal His E7 side chain in Mb for a larger Tyr E7 (orange, PDB 1MGN), as also occurs in Hb Saskatoon, brings the Tyr hydroxyl group within binding distance of the iron, forming a hexacoordinate iron site. (E) In Hb Saskatoon the hexacoordinate ferric iron in the effected β chains can be reduced by met-Hb reductase, possibly owing to a transient protonation of Tyr E7. Note that Mb is used as a model for Hb subunits in A and D, whereas B and E are based on Hb spectroscopy. Globin Chain Elongation Mutants Antitermination and frameshift mutations that add irrelevant amino acids to the carboxyl terminus of globin proteins produce interesting variants that can damage erythrocytes (Nathan et al. 2009). The most clinically significant example is Hb Constant Spring (α2 142 [HC3] Stop > Gln), caused by an antitermination mutation at the α2 stop codon (Clegg et al. 1971; Efremov et al. 1971; Milner et al. 1971; Clegg and Weatherall 1974). This elongates the protein by 31 amino acids, generating an unstable protein that is relatively underrepresented in hemolysates, but can be detected by physical methods (see sections on Basic Principles, Laboratory Testing for Hb Variants, and Physical methods of Hb separation). In addition, Hb Constant Spring mRNA is rapidly degraded in developing erythrocytes, owing to ribosomal entry into the 3′UTR, causing displacement of RNA-bound stabilizing proteins with a resultant thalassemia syndrome (Hunt et al. 1982; Derry et al. 1984; Weiss and Liebhaber 1994; Morales et al. 1997). Hb Constant Spring contributes to α-thalassemia syndromes, particularly when combined with two α-globin deletional alleles (–/α CS α), which produces a severe form of HbH disease (Viprakasit and Tanphaichitr 2002). Isolated Hb Constant Spring, in its heterozygous (αα/α CS α) or homozygous (α CS α/α CS α) forms, results in more severe anemia than occurs when the same α alleles are deleted (αα/−α) or (−α/−α) (Schrier et al. 1997). This is due to the cytotoxic effects of the unstable Constant Spring protein. Although most common in Southeast Asia, Hb Constant Spring is increasingly identified in other geographic regions, largely through global migration (Lal et al. 2011). In fact, it was first discovered in a Chinese family living in Constant Spring, Jamaica. One example of a β-globin chain elongation mutant is Hb Cranston (β 145 [HC3] +CT) (Bunn et al. 1975). This mutation introduces a frameshift at the normal stop codon to generate a β chain that is extended by 11 amino acids. This results in an unstable Hb tetramer with high O 2 affinity and diminished cooperativity (McDonald et al. 1980; Shaeffer et al. 1980). Affected patients show compensated hemolytic anemia with the variants accounting for 30% of total Hb in the hemolysate. Interestingly, the structure of Hb Cranston was investigated simultaneously with studies to determine the β-globin mRNA 3′ untranslated sequence (Forget et al. 1975). Cross-comparison of the protein and mRNA sequencing data allowed Bunn, Forget, and colleagues to more rapidly define normal β-globin gene structures and ascertain that the Hb Cranston mutation likely arose by nonhomologous crossover of two normal β-globin genes. Variants that Affect Multiple Hemoglobin Functions Not surprisingly, amino acid substitutions within critical regions of globin proteins can produce multiple effects. For example, HbE (β26 [B8] Glu > Lys), a common variant in Southeast Asia, contains an amino acid substitution that renders β chains mildly unstable in vitro with minimal clinical significance (Frischer and Bowman 1975; Huisman 1997; Rees et al. 1998; see also Musallam et al. 2012). However, this mutation also creates an alternate splice site in the β-globin mRNA, leading to reduced synthesis of productive transcripts with resultant thalassemia (Orkin et al. 1982). HbE is particularly deleterious when coinherited with a more severe β-thalassemic allele, which happens commonly in Southeast Asia. Mutations that alter the heme pocket commonly produce multiple biochemical effects. For example, deletion or substitution of the conserved Phe residue at the CD1 helical region in the heme pocket markedly destabilizes the affected globin and also alters its O 2 affinity. Thus, Hb Bruxelles (β42 [CD1] Phe > 0) (Blouquit et al. 1989; Griffon et al. 1996), Hb Warsaw (β42 [CD1] Phe > Val) (Honig et al. 1990), Hb Hammersmith (β42 [CD1] Phe > Ser) (Dacie et al. 1967), and Hb Buccuresti-Louisville (β42 [CD1] Phe > Leu) (Bratu et al. 1971; Keeling et al. 1971) cause both congenital Heinz body hemolytic anemia and cyanosis. These combined effects arise from severely reduced cooperativity, rapid rates of autoxidation and hemin loss, and unfolding of these unstable globin variants (Griffon et al. 1996). Another interesting heme pocket variant is Hb Zurich (β63 [E7] His > Arg) in which the distal His is replaced by Arg (Huisman et al. 1961; Bachmann and Martihr 1962; Frick et al. 1962; Tucker et al. 1978; Phillips et al. 1981; Springer et al. 1989). The large, highly polar variant His side chain swings out of the distal heme pocket, and the positively charged guanidino group forms a salt bridge with a heme propionate (Fig. 5A). This results in an enlarged ligand-binding pocket, destabilizing O 2 binding and causing iron autoxidation via exposure to water. Affected individuals show increased sensitivity to oxidant agents, including sulfa drugs, which more easily enter the expanded heme pocket. Loss of the distal histidine markedly decreases O 2 affinity but has little effect on carbon monoxide (CO) binding. As a result, individuals with Hb Zurich tend to have supranormal levels of CO-Hb, which ironically protects the heme iron from oxidation and the globin from denaturation. Affected subjects who are cigarette smokers accumulate even higher levels of CO-Hb, which tends to protect against hemolysis. Thus, “the pathology of a mutant protein is ameliorated by a normally toxic pollutant” (Bunn and Forget 1986). Figure 5. Open in a new tab Hb variants with amino acid changes in the heme pocket. (A) Stereo diagram of a model of the deoxy Hb Zurich β heme pocket (blue) overlaid with the wild-type β heme pocket (black, PDB 2DN2). This model of Hb Zurich is based on the structures solved by Phillips et al. (1981) and Tucker et al. (1978). It was generated with the macromolecular modeling program Coot (Emsley et al. 2010) by mutating the distal His of deoxy-β (PDB 2DN2) to Arg. The distal Arg E7 (red) is oriented toward the CD corner, disturbing the position of Phe CD1 and Phe CD4 (orange). The heme pocket entrance is much wider allowing increased access to ligands. However, unlike normal His E7, mutant Arg is unable to stabilize bound O 2 via hydrogen bonding. (B) Stereo diagram showing the structural changes associated with substitution of β Val E11. The wild-type structure carrying the branched hydrophobic side chain Val (black bonds, PDB 2DN2) is overlaid with structures carrying the largest aromatic side chain Trp E11 (orange, PDB 101K) or a polar side chain Thr E11 (green PDB, 1HDB). It is clear that no major backbone or side chain repacking occurs. Thus, mutations in this position are likely to have specific effects in changing the volume of the heme pocket accessible to solvent or diatomic ligands and the electrostatic properties of the pocket. These changes will manifest as differences in O 2 binding, ligand selectivity, autoxidation, and heme loss. Two other recently identified Hb variants illustrate how multiple biochemical defects produce unique phenotypes. Hb Jamaica Plain (β6 [A3] Glu > Val and β68 [E12] Leu > Phe) contains two defects in the same β chain, a β6 Glu to Val substitution that causes sickle cell anemia (Serjeant and Rodgers 2012) and a β68 amino acid substitution that reduces O 2 affinity, probably by destabilizing the oxygenated conformation through steric effects introduced into helix E (Geva et al. 2004). The affected patient, who was heterozygous for the mutant allele, showed symptoms of sickle cell anemia that were precipitated by infection or airplane travel. Thus, an amino acid substitution that reduces O 2 affinity exacerbates the effects of a sickling mutation in the same globin chain. Several other Hb variants modulate the severity of sickle cell anemia (see also Cao and Kan 2012; Schechter and Elion 2012; Serjeant and Rodgers 2012). For example, γ globin inhibits polymerization of HbS (Nagel et al. 1979). This effect is attributable to differences in several amino acid residues compared with the corresponding β chain, including γ80 and γ87 (Adachi and Asakura 1979; Nagel et al. 1979; Adachi et al. 1996). Hb D-Ibadan (β87 [F3] Thr > Lys), which introduces a lysine residue at the β87 position, is presumed to have decreased interaction with the mutant Val residue at HbS β6 (Watson-Williams et al. 1965; Nagel et al. 1979). Thus, Hb D-Ibadan inhibits HbS polymerization. In contrast, Hb Quebec-Chori (β87 [F3] Thr > Ile) was identified in a compound heterozygous patient with mild to moderately severe sickle cell anemia (Witkowska et al. 1991). The introduction of an isoleucine at β87 causes Hb Quebec-Chori to promote deoxygenated HbS polymerization. Hb Bristol-Alesha (β67 [E11] Val > Met) (Molchanova et al. 1993; Rees et al. 1996) and Hb Toms River (γ67 [E11] Val > Met) (Crowley et al. 2011), which contain analogous amino acid substitutions in β and γ chains, respectively, represent interesting globin variants with multiple biochemical defects. Hb Bristol-Alesha was initially observed in patients with moderately severe hemolytic anemia. Studies of the mutant protein in patient erythrocytes revealed a β67 Val > Asp substitution, predicted to render the protein unstable by introducing a highly charged polar residue into the hydrophobic heme pocket. However, subsequent DNA analysis of the same patient identified a Val > Met codon substitution (Rees et al. 1996). The investigators concluded that the mutant Met residue was converted to Asp posttranslationally, probably through oxidative mechanisms. More recently, the analogous variant was identified in fetal (γ) globin (Hb Toms River) (Crowley et al. 2011). The affected patient was a newborn presenting with both cyanosis and anemia. DNA testing revealed the codon change (Val > Met at E11). Mass spectrometry of patient hemolysate indicated a mixture of variant γ globins containing either Met or Asp at position E11. Although no structural studies have been performed with Hb chains carrying Met or Asp E11, structures with polar (Thr) or large aromatic (Trp) substitutions are available. These indicate that a range of amino acids can be accommodated without gross changes in the heme pocket structure (Fig. 5B). Instead, altered steric and electrostatic interactions with the distal His and diatomic ligands entering the heme pocket are likely to be functionally significant. Biochemical studies indicated that the Hb Toms River Met substitution produced a stable, low O 2 affinity variant γ globin, causing cyanosis. Its gradual posttranslational conversion to Asp destabilized the molecule, causing hemolytic anemia. This provides an example of how posttranslational modifications of variant globins can modify phenotypes. The reason that Hb Bristol-Alesha causes predominantly hemolytic anemia whereas Hb Toms River causes mainly cyanosis probably reflects different rates of Met conversion to Asp in the variant globin chains. CONCLUDING REMARKS More than 1000 Hb variants are known to exist (Globin Gene Server; Hardison et al. 2002; Giardine et al. 2011). These are mainly missense mutations that destabilize Hb, alter Hb-O 2 affinity, or most commonly, alter Hb function minimally. Moreover, variants that do alter Hb biochemistry are rarely life threatening or health compromising. Nonetheless, studies of these variants have been of great benefit to science and medicine for two main reasons. First, identification of Hb gene mutations as a cause for cyanosis, erythrocytosis, or mild hemolysis in otherwise healthy patients provides reassurance and minimizes additional diagnostic procedures, sparing expense and risk. Second, efforts to understand how Hb variants produce their structural, biochemical, and clinical effects has generated important insights into red blood cell function and also created general paradigms for the study of protein biology. Despite hundreds of studies over more than 50 years, new Hb variants continue to emerge, yielding new insights into this important molecule. ACKNOWLEDGMENTS We thank Drs. Franklin Bunn, Kazuhiko Adachi, and John Olson for comments on the manuscript. Hemoglobin research in Dr. Weiss’s laboratory is funded through National Institutes of Health (NIH) grants DK061692, HL087427, and P30DK090969. The authors declare no competing financial interests. Footnotes Editors: David Weatherall, Alan N. Schechter, and David G. Nathan Additional Perspectives on Hemoglobin and Its Diseases available at www.perspectivesinmedicine.org REFERENCES Reference is also in this collection. Adachi K, Asakura T 1979. The solubility of sickle and non-sickle hemoglobins in concentrated phosphate buffer. J. Biol Chem 254: 4079–4084 [PubMed] [Google Scholar] Adachi K, Pang J, Konitzer P, Surrey S 1996. Polymerization of recombinant hemoglobin F γE6V and hemoglobin F γE6V, γQ87T alone, and in mixtures with hemoglobin S. Blood 87: 1617–1624 [PubMed] [Google Scholar] Adachi K, Surrey S, Nagai M 2011. Hemoglobinopathies due to amino acid mutation/deletion: HbS and HbM. In Hemoglobin: Recent developments and topics, pp. 179–210 Research Signpost, Kerala, India [Google Scholar] Adams J, Boxer L, Baehner R, Forget B, Tsistrakis G, Steinberg M 1979. Hemoglobin Indianapolis (β112 [G14] arginine). An unstable β-chain variant producing the phenotype of severe β-thalassemia. J Clin Invest 63: 931–938 [DOI] [PMC free article] [PubMed] [Google Scholar] Ameri A, Fairbanks V, Yanik G, Mahdi F, Thibodeau S, McCormick D, Boxer L, McDonagh K 1999. Identification of the molecular genetic defect of patients with methemoglobin M-Kankakee (M-Iwate), α87 (F8) His→ Tyr: Evidence for an electrostatic model of αM hemoglobin assembly. Blood 94: 1825–1826 [PubMed] [Google Scholar] Argos P, Rossman MG, Grau UM, Zuber H, Frank G, Tratschin JD 1979. Thermal stability and protein structure. Biochemistry 18: 5698–5703 [DOI] [PubMed] [Google Scholar] Arnone A 1972. X-ray diffraction study of binding of 2,3-diphosphoglycerate to human deoxyhaemoglobin. Nature 237: 146–149 [DOI] [PubMed] [Google Scholar] Arous N, Braconnier F, Thillet J, Blouquit Y, Galacteros F, Chevrier M, Bordahandy C, Rosa J 1981. Hemoglobin Saint Mandé [β102 (G4) Asn→Tyr]: A new low oxygen affinity variant. FEBS Lett 126: 114–116 [DOI] [PubMed] [Google Scholar] Asakura T, Adachi K, Shapiro M, Friedman S, Schwartz E 1975. Mechanical precipitation of hemoglobin köln. Biochim Biophys Acta 412: 197–201 [DOI] [PubMed] [Google Scholar] Bachmann F, Marti HR 1962. Hemoglobin Zürich. II. Physicochemical properties of the abnormal hemoglobin. Blood 20: 272–86 [PubMed] [Google Scholar] Bare GH, Bromberg PA, Alben JO, Brimhall B, Jones RT, Mintz S, Rother I 1976. Altered C-terminal salt bridges in haemoglobin York cause high oxygen affinity. Nature 259: 155–156 [DOI] [PubMed] [Google Scholar] Bender JW, Adachi K, Asakura T 1981. Precipitation of oxyhemoglobins A and S by isopropanol. Hemoglobin 5: 463–474 [DOI] [PubMed] [Google Scholar] Blouquit Y, Bardakdjian J, Lena-Russo D, Arous N, Perrimond H, Orsini A, Rosa J, Galacteros F 1989. Hb Bruxelles: α 2A β (2)41 or 42(C7 or CD1)Phe deleted. Hemoglobin 13: 465–474 [DOI] [PubMed] [Google Scholar] Bonaventura J, Riggs A 1968. Hemoglobin Kansas, a human hemoglobin with a neutral amino acid substitution and an abnormal oxygen equilibrium. J. Biol Chem 243: 980–991 [PubMed] [Google Scholar] Bonaventura J, Bonaventura C, Sullivan B, Ferruzzi G, McCurdy PR, Fox J, Moo-Penn WF 1976. Hemoglobin providence. Functional consequences of two alterations of the 2,3-diphosphoglycerate binding site at position β82. J. Biol Chem 251: 7563–7571 [PubMed] [Google Scholar] Bratu V, Lorkin PA, Lehmann H, Predescu C 1971. Haemoglobin Buccureşti 42(CD1) Phe-Leu, a cause of unstable haemoglobin haemolytic anaemia. Biochim Biophys Acta 251: 1–6 [DOI] [PubMed] [Google Scholar] Bunn HF, Forget BG 1986. Hemoglobin: Molecular, genetic and clinical aspects. W.B. Saunders, Philadelphia [Google Scholar] Bunn HF, Wohl RC, Bradley TB, Cooley M, Gibson QH 1974. Functional properties of hemoglobin Kempsey. J Biol Chem 249: 7402–7409 [PubMed] [Google Scholar] Bunn HF, Schmidt GJ, Haney DN, Dluhy RG 1975. Hemoglobin Cranston, an unstable variant having an elongated β chain due to nonhomologous crossover between two normal β chain genes. Proc Natl Acad Sci 72: 3609–3613 [DOI] [PMC free article] [PubMed] [Google Scholar] .Cao A, Kan YW 2012. The prevention of thalassemia. Cold Spring Harb Perspect Med 10.1101/cshperspect.a011775 [DOI] [PMC free article] [PubMed] [Google Scholar] Carrell RW, Kay R 1972. A simple method for the detection of unstable haemoglobins. Br J Haematol 23: 615–619 [DOI] [PubMed] [Google Scholar] Charache S, Fox J, McCurdy P, Kazazian H Jr, Winslow R, Hathaway P, van Beneden R, Jessop M 1977. Postsynthetic deamidation of hemoglobin Providence (β82 Lys replaced by Asn, Asp) and its effect on oxygen transport. J. Clin Invest 59: 652–658 [DOI] [PMC free article] [PubMed] [Google Scholar] Clegg JB, Weatherall DJ 1974. Hemoglobin Constant Spring, and unusual α-chain variant involved in the etiology of hemoglobin H disease. Ann NY Acad Sci 232: 168–178 [DOI] [PubMed] [Google Scholar] Clegg J, Weatherall D, Boon W, Mustafa D 1969. Two new haemoglobin variants involving proline substitutions. Nature 222: 379–380 [DOI] [PubMed] [Google Scholar] Clegg JB, Weatherall DJ, Milner PF 1971. Haemoglobin Constant Spring—A chain termination mutant? Nature 234: 337–340 [DOI] [PubMed] [Google Scholar] Coleman MB, Steinberg MH, Adams JG 3rd. 1991. Hemoglobin Terre Haute arginine β106. A posthumous correction to the original structure of hemoglobin Indianapolis. J Biol Chem 266: 5798–5800 [PubMed] [Google Scholar] Como PF, Wylie BR, Trent RJ, Bruce D, Volpato F, Wilkinson T, Kronenberg H, Holland RA, Tibben EA 1991. A new unstable and low oxygen affinity hemoglobin variant: Hb Stanmore (β111[G13]Val→Ala). Hemoglobin 15: 53–65 [DOI] [PubMed] [Google Scholar] Crowley M, Mollan T, Abdulmalik O, Butler AD, Goodwin E, Sarkar A, Stolle C, Gow A, Olson J, Weiss M 2011. A hemoglobin variant associated with neonatal cyanosis and anemia. N Engl J Med 364: 1837–1843 [DOI] [PMC free article] [PubMed] [Google Scholar] Dacie JV, Shinton NK, Gaffney PJ Jr, Lehmann H 1967. Haemoglobin Hammersmith (β-42 [CDI] Phe replaced by ser). Nature 216: 663–665 [DOI] [PubMed] [Google Scholar] .Dailey HA, Meissner PN 2012. Erythroid heme biosynthesis and its disorders. Cold Spring Harb Perspect Med 10.1101/cshperspect.a011676 [DOI] [PMC free article] [PubMed] [Google Scholar] Derry S, Wood WG, Pippard M, Clegg JB, Weatherall DJ, Wickramasinghe SN, Darley J, Fucharoen S, Wasi P 1984. Hematologic and biosynthetic studies in homozygous hemoglobin Constant Spring. J. Clin Invest 73: 1673–1682 [DOI] [PMC free article] [PubMed] [Google Scholar] Efremov GD, Wrightstone RN, Huisman TH, Schroeder WA, Hyman C, Ortega J, Williams K 1971. An unusual hemoglobin anomaly and its relation to α-thalassemia and hemoglobin-H disease. J. Clin Invest 50: 1628–1636 [DOI] [PMC free article] [PubMed] [Google Scholar] Eisinger J, Flores J, Tyson JA, Shohet SB 1985. Fluorescent cytoplasm and Heinz bodies of hemoglobin Köln erythrocytes: Evidence for intracellular heme catabolism. Blood 65: 886–893 [PubMed] [Google Scholar] Emsley P, Lohkamp B, Scott WG, Cowtan K 2010. Features and development of Coot. Acta Cryst 66: 486–501 [DOI] [PMC free article] [PubMed] [Google Scholar] Favero ME, Costa FF 2011. α-Hemoglobin-stabilizing protein: An erythroid molecular chaperone. Biochem Res Int 2011: 373859. [DOI] [PMC free article] [PubMed] [Google Scholar] Forget BG, Marotta CA, Weissman SM, Cohen-Solal M 1975. Nucleotide sequences of the 3′-terminal untranslated region of messenger RNA for human β globin chain. Proc Natl Acad Sci 72: 3614–3618 [DOI] [PMC free article] [PubMed] [Google Scholar] Frick PG, Hitzig WH, Betke K 1962. Hemoglobin Zurich. I. A new hemoglobin anomaly associated with acute hemolytic episodes with inclusion bodies after sulfonamide therapy. Blood 20: 261–271 [PubMed] [Google Scholar] Frischer H, Bowman J 1975. Hemoglobin E, an oxidatively unstable mutation. J Lab Clin Med 85: 531–539 [PubMed] [Google Scholar] .Ganz T, Nemeth E 2012. Iron metabolism: Interactions with normal and disordered erythropoiesis. Cold Spring Harb Perspect Med 10.1101/cshperspect.a011668 [DOI] [PMC free article] [PubMed] [Google Scholar] Garrod A 1928. The lessons of rare maladies: Annual oration before the medical society of London. Lancet 1: 914–915 [PMC free article] [PubMed] [Google Scholar] Geva A, Clark JJ, Zhang Y, Popowicz A, Manning JM, Neufeld EJ 2004. Hemoglobin Jamaica Plain—A sickling hemoglobin with reduced oxygen affinity. N Engl J Med 351: 1532–1538 [DOI] [PubMed] [Google Scholar] Giardine B, Borg J, Higgs D, Peterson K, Philipsen S, Maglott D, Singleton B, Anstee D, Basak A, Clark B, et al. 2011. Systematic documentation and analysis of human genetic variation in hemoglobinopathies using the microattribution approach. Nat Genet 43: 295–301 [DOI] [PMC free article] [PubMed] [Google Scholar] Gibson QH, Riggs A, Imamura T 1973. Kinetic and equilibrium properties of hemoglobin Kansas. J. Biol Chem 248: 5976–5986 [PubMed] [Google Scholar] Giordano PC, Zweegman S, Akkermans N, Arkesteijn SGJ, van Delft Peter, Versteegh FGA, Wajcman Henri, Harteveld CL 2007. The first case of Hb Groene Hart (α119[H2]Pro→Ser, CCT→TCT [α1]) homozygosity confirms that a thalassemia phenotype is associated with this abnormal hemoglobin variant. Hemoglobin 31: 179–182 [DOI] [PubMed] [Google Scholar] Griffon N, Badens C, Lena-Russo D, Kister J, Bardakdjian J, Wajcman H, Marden MC, Poyart C 1996. Hb Bruxelles, deletion of Phe β42, shows a low oxygen affinity and low cooperativity of ligand binding. J. Biol Chem 271: 25916–25920 [DOI] [PubMed] [Google Scholar] Hamilton HB, Iuchi I, Miyaji T, Shibata S 1969. Hemoglobin Hiroshima (β 143 histidine→aspartic acid): A newly identified fast moving β chain variant associated with increased oxygen affinity and compensatory erythremia. J. Clin Invest 48: 525–535 [DOI] [PMC free article] [PubMed] [Google Scholar] Hardison R, Chui D, Giardine B, Riemer C, Patrinos G, Anagnou N, Miller W, Wajcman H 2002. HbVar: A relational database of human hemoglobin variants and thalassemia mutations at the globin gene server. Hum Mutat 19: 225–233 [DOI] [PubMed] [Google Scholar] Harteveld C, van Delft P, Plug R, Versteegh F, Hagen B, van Rooijen I, Kok P, Wajcman H, Kister J, Giordano PC 2002. Hb Groene Hart: A new Pro→Ser amino acid substitution at position 119 of the α1-globin chain is associated with a mild α-thalassemia phenotype. Hemoglobin 26: 255–60 [DOI] [PubMed] [Google Scholar] Hayashi A, Shimizu A, Yamamura Y, Watari H 1966. Hemoglobins M: Identification of Iwate, Boston, and Saskatoon variants. Science 152: 207–208 [DOI] [PubMed] [Google Scholar] Hebbel RP, Eaton JW, Kronenberg RS, Zanjani ED, Moore LG, Berger EM 1978. Human llamas: Adaptation to altitude in subjects with high hemoglobin oxygen affinity. J Clin Invest 62: 593–600 [DOI] [PMC free article] [PubMed] [Google Scholar] Honig GR, Vida LN, Rosenblum BB, Perutz MF, Fermi G 1990. Hemoglobin Warsaw (Phe β42(CD1)→Val), an unstable variant with decreased oxygen affinity. Characterization of its synthesis, functional properties, and structure. J. Biol Chem 265: 126–132 [PubMed] [Google Scholar] Horlein H, Weber G 1948. Ueber chronische familiäre methämoglobinämie und eine neue modifikation des methämoglobins. Dtsch Med Wochenschr 73: 476–478 [DOI] [PubMed] [Google Scholar] Hoyer JD, McCormick DJ, Snow K, Kwon JH, Booth D, Duarte M, Grayson G, Kubik KS, Holmes MW, Fairbanks VF 2002. Four new variants of the α2-globin gene without clinical or hematologic effects: Hb Park Ridge (α9[α7]Asn→Lys [α2]), Hb Norton (α72[EF1]His→Asp [α2]), Hb Lombard (α103[G10]His→Tyr [α2]), and Hb San Antonio (A113[GH2]Leu→Arg [A2]). Hemoglobin 26: 175–179 [DOI] [PubMed] [Google Scholar] Huisman TH 1997. Hb E and α-thalassemia; variability in the assembly of β E chain containing tetramers. Hemoglobin 21: 227–236 [DOI] [PubMed] [Google Scholar] Huisman TH, Horton B, Bridges MT, Betke K, Hitzig WH 1961. A new abnormal human hemoglobin-Hb: Zurich. Clin Chim Acta 6: 347–355 [DOI] [PubMed] [Google Scholar] Hunt DM, Higgs DR, Winichagoon P, Clegg JB, Weatherall DJ 1982. Haemoglobin Constant Spring has an unstable α chain messenger RNA. Br J Haematol 51: 405–413 [DOI] [PubMed] [Google Scholar] Ikkala E, Koskela J, Pikkarainen P, Rahiala EL, El-Hazmi MA, Nagai K, Lang A, Lehmann H 1976. Hb Helsinki: A variant with a high oxygen affinity and a substitution at a 2,3-DPG binding site (β82[EF6] Lys replaced by Met). Acta Haematol 56: 257–275 [DOI] [PubMed] [Google Scholar] Imai K, Hamilton HB, Miyaji T, Shibata S 1972. Physicochemical studies of the relation between structure and function in hemoglobin Hiroshima (HC3, histidine leads to aspartate). Biochemistry 11: 114–121 [DOI] [PubMed] [Google Scholar] Jin Y, Nagai M, Nagai Y, Nagatomo S, Kitagawa T 2004. Heme structures of five variants of hemoglobin M probed by resonance Raman spectroscopy. Biochemistry 43: 8517–8527 [DOI] [PubMed] [Google Scholar] Keeling MM, Ogden LL, Wrightstone RN, Wilson JB, Reynolds CA, Kitchens JL, Huisman TH 1971. Hemoglobin Louisville (β-42 [CD1] phe-leu): An unstable variant causing mild hemolytic anemia. J Clin Invest 50: 2395–2402 [DOI] [PMC free article] [PubMed] [Google Scholar] Khandros E, Mollan TL, Yu X, Wang X, Yao Y, D’Souza J, Gell DA, Olson JS, Weiss MJ 2012. Insights into hemoglobin assembly through in vivo mutagenesis of α-hemoglobin stabilizing protein. J Biol Chem 287: 11325–11327 [DOI] [PMC free article] [PubMed] [Google Scholar] King MA, Wiltshire BG, Lehmann H, Morimoto H 1972. An unstable haemoglobin with reduced oxygen affinity: Haemoglobin Peterborough, 3 (GI3) Valine lead to Phenylalanine, its interaction with normal haemoglobin and with haemoglobin Lepore. Br J Haematol 22: 125–134 [DOI] [PubMed] [Google Scholar] Kohne E 2011. Hemoglobinopathies: Clinical manifestations, diagnosis, and treatment. Dtsch Ärztebl Int 108: 532. [DOI] [PMC free article] [PubMed] [Google Scholar] Konigsberg W, Lehmann H 1965. The amino acid substitution in hemoglobin M-Iwate. Biochim Biophys Acta 107: 266–269 [DOI] [PubMed] [Google Scholar] Konigsberg W, Guidotti G, Hill RJ 1961. The amino acid sequence of the α chain of human hemoglobin. J. Biol Chem 236: PC55–PC56 [PubMed] [Google Scholar] Lacan P, Aubry M, Couprie N, Francina A 2004. Two new α chain variants: Hb Die (α93[FG5]Val→Ala [α1]) and Hb Beziers (α99[G6]Lys→Asn [α1]). Hemoglobin 28: 59–63 [DOI] [PubMed] [Google Scholar] Lacerra G, Scarano C, Musollino G, Flagiello A, Pucci P, Carestia C 2008. Hb Foggia or α117(GH5)Phe→Ser: A new α2 globin allele affecting the αHb-AHSP interaction. Haematologica 93: 141–142 [DOI] [PubMed] [Google Scholar] Lal A, Goldrich ML, Haines DA, Azimi M, Singer ST, Vichinsky EP 2011. Heterogeneity of hemoglobin H disease in childhood. N. Engl J Med 364: 710–718 [DOI] [PubMed] [Google Scholar] Langdown JV, Davidson RJ, Williamson D 1992. A new α chain variant, Hb Turriff (α99[G6]Lys→Glu): The interference of abnormal hemoglobins in Hb A1c determination. Hemoglobin 16: 11–17 [DOI] [PubMed] [Google Scholar] Lehmann H 1957. Haemoglobin and its abnormalities. Practitioner 178: 198–214 [PubMed] [Google Scholar] .Lettre G 2012. The search for genetic modifiers of disease severity in the β-hemoglobinopathies. Cold Spring Harb Perspect Med 10.1101/cshperspect.a015032 [DOI] [PMC free article] [PubMed] [Google Scholar] Levitt M 1981. Effect of proline residues on protein folding. J. Mol Biol 145: 251–263 [DOI] [PubMed] [Google Scholar] Lindstrom TR, Baldassare JJ, Bunn HF, Ho C 1973. Nuclear magnetic resonance and spin-label studies of hemoglobin Kempsey. Biochemistry 12: 4212–4217 [DOI] [PubMed] [Google Scholar] Little R, Roberts W 2009. A review of variant hemoglobins interfering with hemoglobin A1c measurement. J Diabetes Sci Technol 3: 446–451 [DOI] [PMC free article] [PubMed] [Google Scholar] Lorkin P, Stephens A, Beard M, Wrigley P, Adams L, Lehmann H 1975. Haemoglobin Rahere (β Lys-Thr): A new high affinity haemoglobin associated with decreased 2,3-diphosphoglycerate binding and relative polycythaemia. Br Med J 4: 200–202 [DOI] [PMC free article] [PubMed] [Google Scholar] Marinucci M, Mavilio F, Massa A, Gabbianelli M, Fontanarosa PP, Camagna A, Ignesti C, Tentori L 1979. A new abnormal human hemoglobin: Hb Prato (α2 31 [B12] Arg leads to Ser β2). Biochim Biophys Acta 578: 534–540 [DOI] [PubMed] [Google Scholar] Martínez G, Lima F, Colombo B 1977. Haemoglobin J Guantanamo (α 2 β 2 128 [H6] Ala replaced by Asp). A new fast unstable haemoglobin found in a Cuban family. Biochim Biophys Acta 491: 1–6 [DOI] [PubMed] [Google Scholar] McDonald MJ, Lund DP, Bleichman M, Bunn HF, De Young A, Noble RW, Foster B, Arnone A 1980. Equilibrium, kinetic and structural properties of hemoglobin Cranston, an elongated β chain variant. J. Mol Biol 140: 357–375 [DOI] [PubMed] [Google Scholar] Milner PF, Clegg JB, Weatherall DJ 1971. Haemoglobin-H disease due to a unique haemoglobin variant with an elongated α-chain. Lancet 1: 729–732 [DOI] [PubMed] [Google Scholar] Molchanova TP, Postnikov YuV, Pobedimskaya DD, Smetanina NS, Moschan AA, Kazanetz EG, Tokarev YuN, Huisman TH 1993. Hb Alesha or α 2 β (2)67 (E11)Val→Met: A new unstable hemoglobin variant identified through sequencing of amplified DNA. Hemoglobin 17: 217–225 [DOI] [PubMed] [Google Scholar] Mollan TL, Khandros E, Weiss MJ, Olson JS 2012. The kinetics of α-globin binding to α hemoglobin stabilizing protein (AHSP) indicate preferential stabilization of a hemichrome folding intermediate. J Biol Chem 287: 11338–11350 [DOI] [PMC free article] [PubMed] [Google Scholar] Moo-Penn WF, Jue DL, Bechtel KC, Johnson MH, Schmidt RM 1976. Hemoglobin Providence. A human hemoglobin variant occurring in two forms in vivo. J. Biol Chem 251: 7557–7562 [PubMed] [Google Scholar] Moo-Penn W, Jue D, Johnson M, Olsen K, Shih D, Jones R, Lux S, Rodgers P, Arnone A 1988. Hemoglobin Brockton (β 138 [H16] Ala→Pro): An unstable variant near the C-terminus of the β-subunits with normal oxygen-binding properties. Biochemistry 27: 7614–7619 [DOI] [PubMed] [Google Scholar] Morales J, Russell JE, Liebhaber SA 1997. Destabilization of human α-globin mRNA by translation anti-termination is controlled during erythroid differentiation and is paralleled by phased shortening of the poly(A) tail. J Biol Chem 272: 6607–6613 [DOI] [PubMed] [Google Scholar] .Musallam KM, Taher AT, Rachmilewitz EA 2012. β-Thalassemia intermedia: A clinical perspective. Cold Spring Harb Perspect Med 10.1101/cshperspect.a013482 [DOI] [PMC free article] [PubMed] [Google Scholar] Nagai M, Aki M, Li R, Jin Y, Sakai H, Nagatomo S, Kitagawa T 2000. Heme structure of hemoglobin M Iwate (α87[F8]His→Tyr): A UV and visible resonance Raman study. Biochemistry 39: 13093–13105 [DOI] [PubMed] [Google Scholar] Nagel RL, Lynfield J, Johnson J, Landau L, Bookchin RM, Harris MB 1976. Hemoglobin Beth Israel. A mutant causing clinically apparent cyanosis. N. Engl J Med 295: 125–130 [DOI] [PubMed] [Google Scholar] Nagel RL, Bookchin RM, Johnson J, Labie D, Wajcman H, Isaac-Sodeye WA, Honig GR, Schilirò G, Crookston JH, Matsutomo K 1979. Structural bases of the inhibitory effects of hemoglobin F and hemoglobin A2 on the polymerization of hemoglobin S. Proc Natl Acad Sci 76: 670–672 [DOI] [PMC free article] [PubMed] [Google Scholar] Nathan D, Orkin S, Ginsburg D, Look A, Fisher D, Lux S 2009. Nathan and Oski’s hematology of infancy and childhood, 7th ed Saunders Elsevier, Philadelphia [Google Scholar] Ohba Y, Miyaji T, Matsuoka M, Yokoyama M, Numakura H 1975. Hemoglobin Hirosaki (α43 [CE 1] Phe replaced by Leu), a new unstable variant. Biochim Biophys Acta 405: 155–160 [DOI] [PubMed] [Google Scholar] Olson J, Gibson Q, Nagel R, Hamilton H 1972. The ligand-binding properties of hemoglobin Hiroshima (α 2 β 2 146 asp ). J Biol Chem 247: 7485–7493 [PubMed] [Google Scholar] Orkin S, Kazazian HJ, Antonarakis S, Ostrer H, Goff S, Sexton J 1982. Abnormal RNA processing due to the exon mutation of β E-globin gene. Nature 300: 768–769 [DOI] [PubMed] [Google Scholar] Paoli M, Liddington R, Tame J, Wilkinson A, Dodson G 1996. Crystal structure of T state haemoglobin with oxygen bound at all four haems. J Mol Biol 256: 775–792 [DOI] [PubMed] [Google Scholar] Park S, Yokoyama T, Shibayama N, Shiro Y, Tame JR 2006. 1.25 A resolution crystal structures of human haemoglobin in the oxy, deoxy and carbonmonoxy forms. J Mol Biol 360: 690–701 [DOI] [PubMed] [Google Scholar] Perutz M 1960. Structure of hemoglobin. Brookhaven Symp Biol 13: 165–183 [PubMed] [Google Scholar] Perutz M 1970. Stereochemistry of cooperative effects in haemoglobin. Nature 228: 726–739 [DOI] [PubMed] [Google Scholar] Perutz M, Lehmann H 1968. Molecular pathology of human haemoglobin. Nature 219: 902–909 [DOI] [PubMed] [Google Scholar] Perutz M, Rossman M, Cullis A, Muirhead H, Will G, North A 1960. Structure of haemoglobin: Three-dimensional Fourier synthesis at 5.5-A resolution, obtained by X-ray analysis. Nature 185: 416–422 [DOI] [PubMed] [Google Scholar] Perutz MF, Muirhead H, Mazzarella L, Crowther RA, Greer J, Kilmartin JV 1969. Identification of residues responsible for the alkaline Bohr effect in haemoglobin. Nature 222: 1240–1243 [DOI] [PubMed] [Google Scholar] Perutz MF, Pulsinelli P, Eyck LT, Kilmartin JV, Shibata S, Iuchi I, Miyaji T, Hamilton HB 1971. Haemoglobin Hiroshima and the mechanism of the alkaline Bohr effect. Nat New Biol 232: 147–149 [DOI] [PubMed] [Google Scholar] Perutz M, Fermi G, Shih TB 1984. Structure of deoxyhemoglobin Cowtown (His HC3 β→Leu): Origin of the alkaline Bohr effect and electrostatic interactions in hemoglobin. Proc Natl Acad Sci 81: 4781–4784 [DOI] [PMC free article] [PubMed] [Google Scholar] Phillips SE, Hall D, Perutz MF 1981. Structure of deoxyhaemoglobin Zürich (HisE7[63 β]—greater than Arg). J Mol Biol 150: 137–141 [DOI] [PubMed] [Google Scholar] Poyart C, Schaad O, Kister J, Galacteros F, Edelstein SJ, Blouquit Y, Arous N 1990. Hemoglobin Saint Mandé [β102 (G4) Asn→Tyr]. Functional studies and structural modeling reveal an altered T state. Eur J Biochem 194: 343–348 [DOI] [PubMed] [Google Scholar] Pulsinelli PD, Perutz M, Nagel R 1973. Structure of hemoglobin M Boston, a variant with a five-coordinated ferric heme. Proc Natl Acad Sci 70: 3870–3874 [DOI] [PMC free article] [PubMed] [Google Scholar] Rabbitts TH 1976. Bacterial cloning of plasmids carrying copies of rabbit globin messenger RNA. Nature 260: 221–225 [DOI] [PubMed] [Google Scholar] Reed CS, Hampson R, Gordon S, Jones RT, Novy MJ, Brimhall B, Edwards MJ, Koler RD 1968. Erythrocytosis secondary to increased oxygen affinity of a mutant hemoglobin, hemoglobin Kempsey. Blood 31: 623–632 [PubMed] [Google Scholar] Rees DC, Rochette J, Schofield C, Green B, Morris M, Parker NE, Sasaki H, Tanaka A, Ohba Y, Clegg JB 1996. A novel silent posttranslational mechanism converts methionine to aspartate in hemoglobin Bristol (β67[E11] Val-Met→Asp). Blood 88: 341–348 [PubMed] [Google Scholar] Rees DC, Clegg JB, Weatherall DJ 1998. Is hemoglobin instability important in the interaction between hemoglobin E and β thalassemia? Blood 92: 2141–2146 [PubMed] [Google Scholar] Reissmann K, Ruth W, Nomura T 1961. A human hemoglobin with lowered oxygen affinity and impaired heme-heme interactions. J Clin Invest 40: 1826–1833 [DOI] [PMC free article] [PubMed] [Google Scholar] Rieder RF, Oski FA, Clegg JB 1969. Hemoglobin Philly (β35 tyrosine phenylalanine): Studies in the molecular pathology of hemoglobin. J. Clin Invest 48: 1627–1642 [DOI] [PMC free article] [PubMed] [Google Scholar] Riggs A, Gibson QH 1973. Oxygen equilibrium and kinetics of isolated subunits from hemoglobin Kansas. Proc Natl Acad Sci 70: 1718–1720 [DOI] [PMC free article] [PubMed] [Google Scholar] Roth EF Jr, Elbaum D, Bookchin RM, Nagel RL 1976. The conformational requirements for the mechanical precipitation of hemoglobin S and other mutants. Blood 48: 265–271 [PubMed] [Google Scholar] Russu IM, Ho C 1986. Assessment of role of β 146-histidyl and other histidyl residues in the Bohr effect of human normal adult hemoglobin. Biochemistry 25: 1706–1716 [DOI] [PubMed] [Google Scholar] .Sankaran VG, Orkin SH 2012. The switch from fetal to adult hemoglobin. Cold Spring Harb Perspect Med 10.1101/cshperspect.a011643 [DOI] [PMC free article] [PubMed] [Google Scholar] Schneider RG, Bremner JE, Brimhall B, Jones RT, Shih TB 1979. Hemoglobin Cowtown (β 146 HC3 His-Leu): A mutant with high oxygen affinity and erythrocytosis. Am J Clin Pathol 72: 1028–1032 [DOI] [PubMed] [Google Scholar] Schrier SL, Bunyaratvej A, Khuhapinant A, Fucharoen S, Aljurf M, Snyder LM, Keifer CR, Ma L, Mohandas N 1997. The unusual pathobiology of hemoglobin constant spring red blood cells. Blood 89: 1762–1769 [PubMed] [Google Scholar] Schroeder WA, Jones RT, Shelton JR, Shelton JB, Cormick J, McCalla K 1961. A partial sequence of the amino acid residues in the γ chain of human hemoglobin F. Proc Natl Acad Sci 47: 811–818 [DOI] [PMC free article] [PubMed] [Google Scholar] Sciarratta GV, Ivaldi G, Molaro GL, Sansone G, Salkie ML, Wilson JB, Reese AL, Huisman TH 1984. The characterization of hemoglobin Manitoba or α 2 102(G9)Ser→Arg β 2 and hemoglobin Contaldo or α 2 103(G10)His→Arg β 2 by high performance liquid chromatography. Hemoglobin 8: 169–181 [DOI] [PubMed] [Google Scholar] Shaeffer JR, Schmidt GJ, Kingston RE, Bunn HF 1980. Synthesis of hemoglobin Cranston, and elongated β chain variant. J Mol Biol 140: 377–389 [DOI] [PubMed] [Google Scholar] .Schechter A 2012. Hemoglobin function. Cold Spring Harb Perspect Med 10.1101/cshperspect.a011650 [DOI] [Google Scholar] .Schechter A, Elion J 2012. Cold Spring Harb Perspect Med (to be published) [Google Scholar] .Serjeant G, Rodgers G 2012. Natural history of sickle cell disease. Cold Spring Harb Perspect Med 10.1101/cshperspect.a011783 [DOI] [PMC free article] [PubMed] [Google Scholar] Shibata S, Tanuira A, Iuchi I 1960. Hemoglobin M1 demonstration of a new abnormal hemoglobin in hereditary nigremia. Acta Haematol Jpn 23: 96–105 [Google Scholar] Shimizu A, Tsugita A, Hayashi A, Yamamura Y 1965. The primary structure of hemoglobin M-Iwate. Biochim Biophys Acta 107: 270–277 [DOI] [PubMed] [Google Scholar] Springer BA, Egeberg KD, Sligar SG, Rohlfs RJ, Mathews AJ, Olson JS 1989. Discrimination between oxygen and carbon monoxide and inhibition of autooxidation by myoglobin. Site-directed mutagenesis of the distal histidine. J Biol Chem 264: 3057–3060 [PubMed] [Google Scholar] Stamatoyannopoulos G, Parer JT, Finch CA 1969. Physiologic implications of a hemoglobin with decreased oxygen affinity (hemoglobin seattle). N Engl J Med 281: 916–919 [DOI] [PubMed] [Google Scholar] Stamatoyannopoulos G, Woodson R, Papayannopoulou T, Heywood D, Kurachi S 1974. Inclusion-body β-thalassemia trait. A form of β thalassemia producing clinical manifestations in simple heterozygotes. N Engl J Med 290: 939–943 [DOI] [PubMed] [Google Scholar] Steinberg M, Forget B, Higgs D, Nagel R 2001. Disorders of hemoglobin: Genetics, pathophysiology, and clinical management. Cambridge University Press, Cambridge [Google Scholar] Sugihara J, Imamura T, Nagafuchi S, Bonaventura J, Bonaventura C, Cashon R 1985. Hemoglobin Rahere, a human hemoglobin variant with amino acid substitution at the 2,3-diphosphoglycerate binding site. Functional consequences of the alteration and effects of bezafibrate on the oxygen bindings. J Clin Invest 76: 1169–73 [DOI] [PMC free article] [PubMed] [Google Scholar] Tanaka Y, Matsui K, Matsuda K, Shinohara K, Haranob K 2005. A family with hemoglobin Hirosaki. Int J Hematol 82: 124–126 [DOI] [PubMed] [Google Scholar] Thein SL, Hesketh C, Taylor P, Temperley IJ, Hutchinson RM, Old JM, Wood WG, Clegg JB, Weatherall DJ 1990. Molecular basis for dominantly inherited inclusion body β-thalassemia. Proc Natl Acad Sci 87: 3924–3928 [DOI] [PMC free article] [PubMed] [Google Scholar] Tremper K, Barker S 1989. Pulse oximetry. Anesthesiology 70: 98–108 [DOI] [PubMed] [Google Scholar] Tucker PW, Phillips SE, Perutz MF, Houtchens R, Caughey WS 1978. Structure of hemoglobins Zürich (His E7β replaced by Arg) and Sydney (Val E11β replaced by Ala) and role of the distal residues in ligand binding. Proc Natl Acad Sci 75: 1076–1080 [DOI] [PMC free article] [PubMed] [Google Scholar] Turbpaiboon C, Limjindaporn T, Wongwiwat W, U-Pratya Y, Siritanaratkul N, Yenchitsomanus P-thai, Jitrapakdee S, Wilairat P 2006. Impaired interaction of α-haemoglobin-stabilising protein with α-globin termination mutant in a yeast two-hybrid system. Br J Haematol 132: 370–373 [DOI] [PubMed] [Google Scholar] Vasseur C, Domingues-Hamdi E, Brillet T, Marden Michael C, Baudin-Creuza V 2009. The α-hemoglobin stabilizing protein and expression of unstable α-Hb variants. Clin Biochem 42: 1818–1823 [DOI] [PubMed] [Google Scholar] Vasseur-Godbillon C, Marden M, Giordano P, Wajcman H, Baudin-Creuza V 2006. Impaired binding of AHSP to α chain variants: Hb Groene Hart illustrates a mechanism leading to unstable hemoglobins with α thalassemic like syndrome. Blood Cells Mol Dis 37: 173–179 [DOI] [PubMed] [Google Scholar] Verhovsek M, Henderson M, Cox G, Luo H, Steinberg M, Chui D 2010. Unexpectedly low pulse oximetry measurements associated with variant hemoglobins: A systematic review. Am J Hematol 85: 882–885 [DOI] [PubMed] [Google Scholar] Viprakasit V, Tanphaichitr VS 2002. Compound heterozygosity for α 0-thalassemia (- -THAI) and Hb constant spring causes severe Hb H disease. Hemoglobin 26: 155–162 [DOI] [PubMed] [Google Scholar] Wajcman H, Moradkhani K 2011. Abnormal haemoglobins: Detection and characterization. Indian J Med Res 134: 538–546 [PMC free article] [PubMed] [Google Scholar] Wajcman H, Préhu C, Bardakdjian-Michau J, Promé D, Riou J, Godart C, Mathis M, Hurtrel D, Galactéros F 2001. Abnormal hemoglobins: Laboratory methods. Hemoglobin 25: 169–181 [DOI] [PubMed] [Google Scholar] Wajcman H, Traeger-Synodinos J, Papassotiriou I, Giordano PC, Harteveld CL, Baudin-Creuza V, Old J 2008. Unstable and thalassemic α chain hemoglobin variants: A cause of Hb H disease and thalassemia intermedia. Hemoglobin 32: 327–349 [DOI] [PubMed] [Google Scholar] Watson HC, Kendrew JC 1961. The amino-acid sequence of sperm whale myoglobin. Comparison between the amino-acid sequences of sperm whale myoglobin and of human hemoglobin. Nature 190: 670–672 [DOI] [PubMed] [Google Scholar] Watson-Williams EJ, Beale D, Irvine D, Lehmann H 1965. A new haemoglobin, D-Ibadan (β-87 threonine–lysine), producing no sickle-cell haemoglobin D disease with haemoglobin S. Nature 205: 1273–1276 [DOI] [PubMed] [Google Scholar] Weatherall D, Clegg J 2001. Inherited haemoglobin disorders: An increasing global health problem. Bull World Health Organ 79: 704–712 [PMC free article] [PubMed] [Google Scholar] Weiss IM, Liebhaber SA 1994. Erythroid cell-specific determinants of α-globin mRNA stability. Mol Cell Biol 14: 8123–8132 [DOI] [PMC free article] [PubMed] [Google Scholar] Weiss MJ, Zhou S, Feng L, Gell DA, Mackay JP, Shi Y, Gow AJ 2005. Role of α-hemoglobin-stabilizing protein in normal erythropoiesis and β-thalassemia. Ann NY Acad Sci 1054: 103–117 [DOI] [PubMed] [Google Scholar] .Williams TN, Weatherall DJ 2012. World distribution, population genetics, and health burden of the hemoglobinopathies. Cold Spring Harb Perspect Med 10.1101/cshperspect.a011692 [DOI] [PMC free article] [PubMed] [Google Scholar] Witkowska HE, Lubin BH, Beuzard Y, Baruchel S, Esseltine DW, Vichinsky EP, Kleman KM, Bardakdjian-Michau J, Pinkoski L, Cahn S 1991. Sickle cell disease in a patient with sickle cell trait and compound heterozygosity for hemoglobin S and hemoglobin Quebec-Chori. N Engl J Med 325: 1150–1154 [DOI] [PubMed] [Google Scholar] Yu X, Mollan TL, Butler Andrew, Gow AJ, Olson JS, Weiss MJ 2009. Analysis of human α globin gene mutations that impair binding to the α hemoglobin stabilizing protein. Blood 113: 5961–5969 [DOI] [PMC free article] [PubMed] [Google Scholar] Articles from Cold Spring Harbor Perspectives in Medicine are provided here courtesy of Cold Spring Harbor Laboratory Press ACTIONS View on publisher site PDF (1.2 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
15048	https://ilclassroom.com/wikis/10803919-grade-8-unit-7-exponents-and-scientific-notation	Grade 8 Unit 7 \| Exponents and Scientific Notation \| IL Classroom Skip to main content « Back Explore Home My account Log in Imagine Learning Classroom home HomeLog in Select content Select content 8.7 Plan 8.7 Teach 8.7 Support Grade 8 Unit 7 \| Exponents and Scientific Notation Approximately 18 days. In this unit, students deepen their understanding of exponents, powers of 10, and place value before being introduced to scientific notation. Wiki Included in Unit Overview 8.7 Plan This page includes the resources for planning to teach a unit, including Learning Goals, Unit Narrative, Unit Launch videos, and students and teacher workbooks. Imagine IM 8.7 Teach This page includes resources for teaching the unit content, including section and lesson materials, assessments, and student and teacher workbooks. Imagine IM 8.7 Support This page includes resources to support instruction including digital activities, family support materials, and student and teacher workbooks. Imagine IM Imagine IM © 2024 Illustrative Mathematics. Includes additional content © by Imagine Learning. Imagine IM is a derivative of IM 6-8 Math. IM 6–8 Math was originally developed by Open Up Resources and authored by Illustrative Mathematics®, and is copyright 2017-2019 by Open Up Resources. It is licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0), creativecommons.org/licenses/by/4.0/. OUR's 6–8 Math Curriculum is available at Adaptations to add additional English language learner supports are copyright 2019 by Open Up Resources, openupresources.org, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0), The second set of English assessments (marked as set "B") are copyright 2019 by Open Up Resources, openupresources.org, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0), Spanish translation of the "B" assessments are copyright 2020 by Illustrative Mathematics, www.illustrativemathematics.org, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0), The Illustrative Mathematics name and logo are not subject to the Creative Commons license and may not be used without the prior and express written consent of Illustrative Mathematics. This content includes public domain images or openly licensed images that are copyrighted by their respective owners. Openly licensed images remain under the terms of their respective licenses. See the image attribution section for more information. Copyright © 2025 Imagine Learning For Districts and Schools Contact Privacy Terms Copyright © 2025 Imagine Learning
15049	https://pdg.lbl.gov/2014/reviews/rpp2014-rev-cosmological-parameters.pdf	The Cosmological Parameters 1 24. THE COSMOLOGICAL PARAMETERS Updated November 2013, by O. Lahav (University College London) and A.R. Liddle (University of Edinburgh). 24.1. Parametrizing the Universe Rapid advances in observational cosmology have led to the establishment of a precision cosmological model, with many of the key cosmological parameters determined to one or two signiﬁcant ﬁgure accuracy. Particularly prominent are measurements of cosmic microwave background (CMB) anisotropies, with the highest precision observations being those of the Planck Satellite [1,2] which for temperature anisotropies supersede the iconic WMAP results [3,4]. However the most accurate model of the Universe requires consideration of a range of diﬀerent types of observation, with complementary probes providing consistency checks, lifting parameter degeneracies, and enabling the strongest constraints to be placed. The term ‘cosmological parameters’ is forever increasing in its scope, and nowadays often includes the parameterization of some functions, as well as simple numbers describing properties of the Universe. The original usage referred to the parameters describing the global dynamics of the Universe, such as its expansion rate and curvature. Also now of great interest is how the matter budget of the Universe is built up from its constituents: baryons, photons, neutrinos, dark matter, and dark energy. We need to describe the nature of perturbations in the Universe, through global statistical descriptors such as the matter and radiation power spectra. There may also be parameters describing the physical state of the Universe, such as the ionization fraction as a function of time during the era since recombination. Typical comparisons of cosmological models with observational data now feature between ﬁve and ten parameters. 24.1.1. The global description of the Universe : Ordinarily, the Universe is taken to be a perturbed Robertson–Walker space-time with dynamics governed by Einstein’s equations. This is described in detail by Olive and Peacock in this volume. Using the density parameters Ωi for the various matter species and ΩΛ for the cosmological constant, the Friedmann equation can be written X i Ωi + ΩΛ −1 = k R2H2 , (24.1) where the sum is over all the diﬀerent species of material in the Universe. This equation applies at any epoch, but later in this article we will use the symbols Ωi and ΩΛ to refer to the present values. The complete present state of the homogeneous Universe can be described by giving the current values of all the density parameters and the Hubble constant h (the present-day Hubble parameter being written H0 = 100h km s−1 Mpc−1). A typical collection would be baryons Ωb, photons Ωγ, neutrinos Ων, and cold dark matter Ωc (given charge neutrality, the electron density is guaranteed to be too small to be worth considering separately and is included with the baryons). The spatial curvature can then be determined from the other parameters using Eq. (24.1). The total present matter density Ωm = Ωc + Ωb is sometimes used in place of the cold dark matter density Ωc. K.A. Olive et al. (PDG), Chin. Phys. C38, 090001 (2014) ( August 21, 2014 13:18 2 24. The Cosmological Parameters These parameters also allow us to track the history of the Universe back in time, at least until an epoch where interactions allow interchanges between the densities of the diﬀerent species, which is believed to have last happened at neutrino decoupling, shortly before Big Bang Nucleosynthesis (BBN). To probe further back into the Universe’s history requires assumptions about particle interactions, and perhaps about the nature of physical laws themselves. The standard neutrino sector has three ﬂavors. For neutrinos of mass in the range 5 × 10−4 eV to 1 MeV, the density parameter in neutrinos is predicted to be Ωνh2 = P mν 93 eV , (24.2) where the sum is over all families with mass in that range (higher masses need a more sophisticated calculation). We use units with c = 1 throughout. Results on atmospheric and Solar neutrino oscillations imply non-zero mass-squared diﬀerences between the three neutrino ﬂavors. These oscillation experiments cannot tell us the absolute neutrino masses, but within the simple assumption of a mass hierarchy suggest a lower limit of approximately 0.06 eV on the sum of the neutrino masses. Even a mass this small has a potentially observable eﬀect on the formation of structure, as neutrino free-streaming damps the growth of perturbations. Analyses commonly now either assume a neutrino mass sum ﬁxed at this lower limit, or allow the neutrino mass sum as a variable parameter. To date there is no decisive evidence of any eﬀects from either neutrino masses or an otherwise non-standard neutrino sector, and observations impose quite stringent limits, which we summarize in Section 24.3.4. However, we note that the inclusion of the neutrino mass sum as a free parameter can aﬀect the derived values of other cosmological parameters. 24.1.2. Inﬂation and perturbations : A complete model of the Universe should include a description of deviations from homogeneity, at least in a statistical way. Indeed, some of the most powerful probes of the parameters described above come from the evolution of perturbations, so their study is naturally intertwined in the determination of cosmological parameters. There are many diﬀerent notations used to describe the perturbations, both in terms of the quantity used to describe the perturbations and the deﬁnition of the statistical measure. We use the dimensionless power spectrum ∆2 as deﬁned in Olive and Peacock (also denoted P in some of the literature). If the perturbations obey Gaussian statistics, the power spectrum provides a complete description of their properties. From a theoretical perspective, a useful quantity to describe the perturbations is the curvature perturbation R, which measures the spatial curvature of a comoving slicing of the space-time. A simple case is the Harrison–Zel’dovich spectrum, which corresponds to a constant ∆2 R. More generally, one can approximate the spectrum by a power-law, writing ∆2 R(k) = ∆2 R(k∗) · k k∗ ¸ns−1 , (24.3) August 21, 2014 13:18 24. The Cosmological Parameters 3 where ns is known as the spectral index, always deﬁned so that ns = 1 for the Harrison–Zel’dovich spectrum, and k∗is an arbitrarily chosen scale. The initial spectrum, deﬁned at some early epoch of the Universe’s history, is usually taken to have a simple form such as this power-law, and we will see that observations require ns close to one. Subsequent evolution will modify the spectrum from its initial form. The simplest mechanism for generating the observed perturbations is the inﬂationary cosmology, which posits a period of accelerated expansion in the Universe’s early stages [6,7]. It is a useful working hypothesis that this is the sole mechanism for generating perturbations, and it may further be assumed to be the simplest class of inﬂationary model, where the dynamics are equivalent to that of a single scalar ﬁeld φ with canonical kinetic energy slowly rolling on a potential V (φ). One may seek to verify that this simple picture can match observations and to determine the properties of V (φ) from the observational data. Alternatively, more complicated models, perhaps motivated by contemporary fundamental physics ideas, may be tested on a model-by-model basis. Inﬂation generates perturbations through the ampliﬁcation of quantum ﬂuctuations, which are stretched to astrophysical scales by the rapid expansion. The simplest models generate two types, density perturbations which come from ﬂuctuations in the scalar ﬁeld and its corresponding scalar metric perturbation, and gravitational waves which are tensor metric ﬂuctuations. The former experience gravitational instability and lead to structure formation, while the latter can inﬂuence the CMB anisotropies. Deﬁning slow-roll parameters, with primes indicating derivatives with respect to the scalar ﬁeld, as ǫ = m2 Pl 16π µV ′ V ¶2 ; η = m2 Pl 8π V ′′ V , (24.4) which should satisfy ǫ, \|η\| ≪1, the spectra can be computed using the slow-roll approximation as ∆2 R(k) ≃ 8 3m4 Pl V ǫ ¯ ¯ ¯ ¯ ¯ k=aH ; ∆2 t(k) ≃ 128 3m4 Pl V ¯ ¯ ¯ ¯ ¯ k=aH . (24.5) In each case, the expressions on the right-hand side are to be evaluated when the scale k is equal to the Hubble radius during inﬂation. The symbol ‘≃’ here indicates use of the slow-roll approximation, which is expected to be accurate to a few percent or better. From these expressions, we can compute the spectral indices ns ≃1 −6ǫ + 2η ; nt ≃−2ǫ . (24.6) Another useful quantity is the ratio of the two spectra, deﬁned by r ≡∆2 t(k∗) ∆2 R(k∗) . (24.7) We have r ≃16ǫ ≃−8nt , (24.8) August 21, 2014 13:18 4 24. The Cosmological Parameters which is known as the consistency equation. One could consider corrections to the power-law approximation, which we discuss later. However, for now we make the working assumption that the spectra can be approximated by power laws. The consistency equation shows that r and nt are not independent parameters, and so the simplest inﬂation models give initial conditions described by three parameters, usually taken as ∆2 R, ns, and r, all to be evaluated at some scale k∗, usually the ‘statistical center’ of the range explored by the data. Alternatively, one could use the parametrization V , ǫ, and η, all evaluated at a point on the putative inﬂationary potential. After the perturbations are created in the early Universe, they undergo a complex evolution up until the time they are observed in the present Universe. While the perturbations are small, this can be accurately followed using a linear theory numerical code such as CAMB or CLASS . This works right up to the present for the CMB, but for density perturbations on small scales non-linear evolution is important and can be addressed by a variety of semi-analytical and numerical techniques. However the analysis is made, the outcome of the evolution is in principle determined by the cosmological model, and by the parameters describing the initial perturbations, and hence can be used to determine them. Of particular interest are CMB anisotropies. Both the total intensity and two independent polarization modes are predicted to have anisotropies. These can be described by the radiation angular power spectra Cℓas deﬁned in the article of Scott and Smoot in this volume, and again provide a complete description if the density perturbations are Gaussian. 24.1.3. The standard cosmological model : We now have most of the ingredients in place to describe the cosmological model. Beyond those of the previous subsections, we need a measure of the ionization state of the Universe. The Universe is known to be highly ionized at low redshifts (otherwise radiation from distant quasars would be heavily absorbed in the ultra-violet), and the ionized electrons can scatter microwave photons altering the pattern of observed anisotropies. The most convenient parameter to describe this is the optical depth to scattering τ (i.e., the probability that a given photon scatters once); in the approximation of instantaneous and complete reionization, this could equivalently be described by the redshift of reionization zion. As described in Sec. 24.4, models based on these parameters are able to give a good ﬁt to the complete set of high-quality data available at present, and indeed some simpliﬁcation is possible. Observations are consistent with spatial ﬂatness, and indeed the inﬂation models so far described automatically generate negligible spatial curvature, so we can set k = 0; the density parameters then must sum to unity, and so one can be eliminated. The neutrino energy density is often not taken as an independent parameter. Provided the neutrino sector has the standard interactions, the neutrino energy density, while relativistic, can be related to the photon density using thermal physics arguments, and a minimal assumption takes the neutrino mass sum to be that of the lowest mass solution to the neutrino oscillation constraints, namely 0.06 eV. In addition, there is no August 21, 2014 13:18 24. The Cosmological Parameters 5 observational evidence for the existence of tensor perturbations (though the upper limits are fairly weak), and so r could be set to zero. This leaves seven parameters, which is the smallest set that can usefully be compared to the present cosmological data set. This model is referred to by various names, including ΛCDM, the concordance cosmology, and the standard cosmological model. Of these parameters, only Ωr is accurately measured directly. The radiation density is dominated by the energy in the CMB, and the COBE satellite FIRAS experiment determined its temperature to be T = 2.7255 ± 0.0006 K , ‡ corresponding to Ωr = 2.47 × 10−5h−2. It typically need not be varied in ﬁtting other data. Hence the minimum number of cosmological parameters varied in ﬁts to data is six, though as described below there may additionally be many ‘nuisance’ parameters necessary to describe astrophysical processes inﬂuencing the data. In addition to this minimal set, there is a range of other parameters which might prove important in future as the data-sets further improve, but for which there is so far no direct evidence, allowing them to be set to a speciﬁc value for now. We discuss various speculative options in the next section. For completeness at this point, we mention one other interesting parameter, the helium fraction, which is a non-zero parameter that can aﬀect the CMB anisotropies at a subtle level. Fields, Molaro and Sarkar in this volume discuss current measures of this parameter. It is usually ﬁxed in microwave anisotropy studies, but the data are approaching a level where allowing its variation may become mandatory. Most attention to date has been on parameter estimation, where a set of parameters is chosen by hand and the aim is to constrain them. Interest has been growing towards the higher-level inference problem of model selection, which compares diﬀerent choices of parameter sets. Bayesian inference oﬀers an attractive framework for cosmological model selection, setting a tension between model predictiveness and ability to ﬁt the data. 24.1.4. Derived parameters : The parameter list of the previous subsection is suﬃcient to give a complete description of cosmological models which agree with observational data. However, it is not a unique parameterization, and one could instead use parameters derived from that basic set. Parameters which can be obtained from the set given above include the age of the Universe, the present horizon distance, the present neutrino background temperature, the epoch of matter–radiation equality, the epochs of recombination and decoupling, the epoch of transition to an accelerating Universe, the baryon-to-photon ratio, and the baryon to dark matter density ratio. In addition, the physical densities of the matter components, Ωih2, are often more useful than the density parameters. The density perturbation amplitude can be speciﬁed in many diﬀerent ways other than the large-scale ‡ Unless stated otherwise, all quoted uncertainties in this article are one-sigma/68% conﬁdence and all upper limits are 95% conﬁdence. Cosmological parameters sometimes have signiﬁcantly non-Gaussian uncertainties. Throughout we have rounded central values, and especially uncertainties, from original sources in cases where they appear to be given to excessive precision. August 21, 2014 13:18 6 24. The Cosmological Parameters primordial amplitude, for instance, in terms of its eﬀect on the CMB, or by specifying a short-scale quantity, a common choice being the present linear-theory mass dispersion on a scale of 8 h−1Mpc, known as σ8. Diﬀerent types of observation are sensitive to diﬀerent subsets of the full cosmological parameter set, and some are more naturally interpreted in terms of some of the derived parameters of this subsection than on the original base parameter set. In particular, most types of observation feature degeneracies whereby they are unable to separate the eﬀects of simultaneously varying several of the base parameters. 24.2. Extensions to the standard model This section discusses some ways in which the standard model could be extended. At present, there is no positive evidence in favor of any of these possibilities, which are becoming increasingly constrained by the data, though there always remains the possibility of trace eﬀects at a level below present observational capability. 24.2.1. More general perturbations : The standard cosmology assumes adiabatic, Gaussian perturbations. Adiabaticity means that all types of material in the Universe share a common perturbation, so that if the space-time is foliated by constant-density hypersurfaces, then all ﬂuids and ﬁelds are homogeneous on those slices, with the perturbations completely described by the variation of the spatial curvature of the slices. Gaussianity means that the initial perturbations obey Gaussian statistics, with the amplitudes of waves of diﬀerent wavenumbers being randomly drawn from a Gaussian distribution of width given by the power spectrum. Note that gravitational instability generates non-Gaussianity; in this context, Gaussianity refers to a property of the initial perturbations, before they evolve. The simplest inﬂation models, based on one dynamical ﬁeld, predict adiabatic perturbations and a level of non-Gaussianity which is too small to be detected by any experiment so far conceived. For present data, the primordial spectra are usually assumed to be power laws. 24.2.1.1. Non-power-law spectra: For typical inﬂation models, it is an approximation to take the spectra as power laws, albeit usually a good one. As data quality improves, one might expect this approximation to come under pressure, requiring a more accurate description of the initial spectra, particularly for the density perturbations. In general, one can expand ln ∆2 R as ln ∆2 R(k) = ln ∆2 R(k∗) + (ns,∗−1) ln k k∗ + 1 2 dns d ln k ¯ ¯ ¯ ¯ ∗ ln2 k k∗ + · · · , (24.9) where the coeﬃcients are all evaluated at some scale k∗. The term dns/d ln k\|∗is often called the running of the spectral index . Once non-power-law spectra are allowed, it is necessary to specify the scale k∗at which the spectral index is deﬁned. August 21, 2014 13:18 24. The Cosmological Parameters 7 24.2.1.2. Isocurvature perturbations: An isocurvature perturbation is one which leaves the total density unperturbed, while perturbing the relative amounts of diﬀerent materials. If the Universe contains N ﬂuids, there is one growing adiabatic mode and N −1 growing isocurvature modes (for reviews see Ref. 12 and Ref. 7). These can be excited, for example, in inﬂationary models where there are two or more ﬁelds which acquire dynamically-important perturbations. If one ﬁeld decays to form normal matter, while the second survives to become the dark matter, this will generate a cold dark matter isocurvature perturbation. In general, there are also correlations between the diﬀerent modes, and so the full set of perturbations is described by a matrix giving the spectra and their correlations. Constraining such a general construct is challenging, though constraints on individual modes are beginning to become meaningful, with no evidence that any other than the adiabatic mode must be non-zero. 24.2.1.3. Seeded perturbations: An alternative to laying down perturbations at very early epochs is that they are seeded throughout cosmic history, for instance by topological defects such as cosmic strings. It has long been excluded that these are the sole original of structure, but they could contribute part of the perturbation signal, current limits being just a few percent . In particular, cosmic defects formed in a phase transition ending inﬂation is a plausible scenario for such a contribution. 24.2.1.4. Non-Gaussianity: Multi-ﬁeld inﬂation models can also generate primordial non-Gaussianity (reviewed, e.g., in Ref. 7). The extra ﬁelds can either be in the same sector of the underlying theory as the inﬂaton, or completely separate, an interesting example of the latter being the curvaton model . Current upper limits on non-Gaussianity are becoming stringent, but there remains strong motivation to push down those limits and perhaps reveal trace non-Gaussianity in the data. If non-Gaussianity is observed, its nature may favor an inﬂationary origin, or a diﬀerent one such as topological defects. 24.2.2. Dark matter properties : Dark matter properties are discussed in the article by Drees and Gerbier in this volume. The simplest assumption concerning the dark matter is that it has no signiﬁcant interactions with other matter, and that its particles have a negligible velocity as far as structure formation is concerned. Such dark matter is described as ‘cold,’ and candidates include the lightest supersymmetric particle, the axion, and primordial black holes. As far as astrophysicists are concerned, a complete speciﬁcation of the relevant cold dark matter properties is given by the density parameter Ωc, though those seeking to directly detect it are as interested in its interaction properties. Cold dark matter is the standard assumption and gives an excellent ﬁt to observations, except possibly on the shortest scales where there remains some controversy concerning the structure of dwarf galaxies and possible substructure in galaxy halos. It has long been excluded for all the dark matter to have a large velocity dispersion, so-called ‘hot’ dark August 21, 2014 13:18 8 24. The Cosmological Parameters matter, as it does not permit galaxies to form; for thermal relics the mass must be above about 1 keV to satisfy this constraint, though relics produced non-thermally, such as the axion, need not obey this limit. However, in future further parameters might need to be introduced to describe dark matter properties relevant to astrophysical observations. Suggestions which have been made include a modest velocity dispersion (warm dark matter) and dark matter self-interactions. There remains the possibility that the dark matter is comprized of two separate components, e.g., a cold one and a hot one, an example being if massive neutrinos have a non-negligible eﬀect. 24.2.3. Relativistic species : The number of relativistic species in the young Universe (omitting photons) is denoted Neﬀ. In the standard cosmological model only the three neutrino species contribute, and its baseline value is assumed ﬁxed at 3.046 (the small shift from 3 is because of a slight predicted deviation from a thermal distribution ) . However other species could contribute, for example extra neutrino species, possibly of sterile type, or massless Goldstone bosons or other scalars. It is hence interesting to study the eﬀect of allowing this parameter to vary, and indeed although 3.046 is consistent with the data, most analyses currently suggest a somewhat higher value (e.g., Ref. 16). 24.2.4. Dark energy : While the standard cosmological model given above features a cosmological constant, in order to explain observations indicating that the Universe is presently accelerating, further possibilities exist under the general headings of ‘dark energy’ and ‘modiﬁed gravity’. These topics are described in detail in the article by Mortonson, Weinberg and White in this volume. This article focuses on the case of the cosmological constant, as this simple case is a good match to existing data. We note that more general treatments of dark energy/modiﬁed gravity will lead to weaker constraints on other parameters. 24.2.5. Complex ionization history : The full ionization history of the Universe is given by the ionization fraction as a function of redshift z. The simplest scenario takes the ionization to have the small residual value left after recombination up to some redshift zion, at which point the Universe instantaneously reionizes completely. Then there is a one-to-one correspondence between τ and zion (that relation, however, also depending on other cosmological parameters). An accurate treatment of this process will track separate histories for hydrogen and helium. While currently rapid ionization appears to be a good approximation, as data improve a more complex ionization history may need to be considered. 24.2.6. Varying ‘constants’ : Variation of the fundamental constants of Nature over cosmological times is another possible enhancement of the standard cosmology. There is a long history of study of variation of the gravitational constant GN, and more recently attention has been drawn to the possibility of small fractional variations in the ﬁne-structure constant. There is presently no observational evidence for the former, which is tightly constrained by a variety of measurements. Evidence for the latter has been claimed from studies of spectral August 21, 2014 13:18 24. The Cosmological Parameters 9 line shifts in quasar spectra at redshift z ≈2 , but this is presently controversial and in need of further observational study. 24.2.7. Cosmic topology : The usual hypothesis is that the Universe has the simplest topology consistent with its geometry, for example that a ﬂat Universe extends forever. Observations cannot tell us whether that is true, but they can test the possibility of a non-trivial topology on scales up to roughly the present Hubble scale. Extra parameters would be needed to specify both the type and scale of the topology, for example, a cuboidal topology would need speciﬁcation of the three principal axis lengths. At present, there is no evidence for non-trivial cosmic topology . 24.3. Probes The goal of the observational cosmologist is to utilize astronomical information to derive cosmological parameters. The transformation from the observables to the parameters usually involves many assumptions about the nature of the objects, as well as of the dark sector. Below we outline the physical processes involved in each probe, and the main recent results. The ﬁrst two subsections concern probes of the homogeneous Universe, while the remainder consider constraints from perturbations. In addition to statistical uncertainties we note three sources of systematic uncertainties that will apply to the cosmological parameters of interest: (i) due to the assumptions on the cosmological model and its priors (i.e., the number of assumed cosmological parameters and their allowed range); (ii) due to the uncertainty in the astrophysics of the objects (e.g., light curve ﬁtting for supernovae or the mass–temperature relation of galaxy clusters); and (iii) due to instrumental and observational limitations (e.g., the eﬀect of ‘seeing’ on weak gravitational lensing measurements, or beam shape on CMB anisotropy measurements). These systematics, the last two of which appear as ‘nuisance parameters’, pose a challenging problem to the statistical analysis. We attempt to ﬁt the whole Universe with 6 to 12 parameters, but we might need to include hundreds of nuisance parameters, some of them highly correlated with the cosmological parameters of interest (for example time-dependent galaxy biasing could mimic growth of mass ﬂuctuations). Fortunately, there is some astrophysical prior knowledge on these eﬀects, and a small number of physically-motivated free parameters would ideally be preferred in the cosmological parameter analysis. 24.3.1. Direct measures of the Hubble constant : In 1929, Edwin Hubble discovered the law of expansion of the Universe by measuring distances to nearby galaxies. The slope of the relation between the distance and recession velocity is deﬁned to be the Hubble constant H0. Astronomers argued for decades on the systematic uncertainties in various methods and derived values over the wide range 40 km s−1 Mpc−1 < ∼H0 < ∼100 km s−1 Mpc−1. One of the most reliable results on the Hubble constant comes from the Hubble Space Telescope Key Project . This study used the empirical period–luminosity relations August 21, 2014 13:18 10 24. The Cosmological Parameters for Cepheid variable stars to obtain distances to 31 galaxies, and calibrated a number of secondary distance indicators—Type Ia Supernovae (SNe Ia), the Tully–Fisher relation, surface-brightness ﬂuctuations, and Type II Supernovae—measured over distances of 400 to 600 Mpc. They estimated H0 = 72 ± 3 (statistical) ± 7 (systematic) km s−1 Mpc−1. A recent study of over 600 Cepheids in the host galaxies of eight recent SNe Ia, observed with an improved camera on board the Hubble Space Telescope, was used to calibrate the magnitude–redshift relation for 240 SNe Ia. This yielded an even more precise ﬁgure, H0 = 73.8 ± 2.4 km s−1 Mpc−1 (including both statistical and systematic errors). The major sources of uncertainty in this result are due to the heavy element abundance of the Cepheids and the distance to the ﬁducial nearby galaxy, the Large Magellanic Cloud, relative to which all Cepheid distances are measured. The indirect determination of H0 by the Planck Collaboration found a lower value, H0 = 67.3 ± 1.2 km s−1 Mpc−1. As discussed in that paper, there is strong degeneracy of H0 with other parameters, e.g. Ωm and the neutrino mass. The tension between the H0 from Planck and the traditional cosmic distance-ladder methods is under investigation. 24.3.2. Supernovae as cosmological probes : Empirically, the peak luminosity of SNe Ia can be used as an eﬃcient distance indicator (e.g., Ref. 21), thus allowing cosmology to be constrained via the distance–redshift relation. The favorite theoretical explanation for SNe Ia is the thermonuclear disruption of carbon–oxygen white dwarfs. Although not perfect ‘standard candles’, it has been demonstrated that by correcting for a relation between the light curve shape, color, and the luminosity at maximum brightness, the dispersion of the measured luminosities can be greatly reduced. There are several possible systematic eﬀects which may aﬀect the accuracy of the use of SNe Ia as distance indicators, e.g., evolution with redshift and interstellar extinction in the host galaxy and in the Milky Way. Two major studies, the Supernova Cosmology Project and the High-z Supernova Search Team, found evidence for an accelerating Universe , interpreted as due to a cosmological constant or a dark energy component. When combined with the CMB data (which indicates ﬂatness, i.e., Ωm + ΩΛ = 1), the best-ﬁt values were Ωm ≈0.3 and ΩΛ ≈0.7. Most results in the literature are consistent with the w = −1 cosmological constant case. Taking w = −1, the SNLS3 team found, by combining their SNIa data with baryon acoustic oscillation (BAO) and WMAP7 data, Ωm = 0.279+0.019 −0.015 and ΩΛ = 0.724+0.017 −0.016, including both statistical and systematic errors . This includes a correction for the recently-discovered relationship between host galaxy mass and supernova absolute brightness. This agrees with earlier results [24,25], but note the somewhat higher value for Ωm from Planck (see Table 24.1). Future experiments will aim to set constraints on the cosmic equation of state w(z). August 21, 2014 13:18 24. The Cosmological Parameters 11 24.3.3. Cosmic microwave background : The physics of the CMB is described in detail by Scott and Smoot in this volume. Before recombination, the baryons and photons are tightly coupled, and the perturbations oscillate in the potential wells generated primarily by the dark matter perturbations. After decoupling, the baryons are free to collapse into those potential wells. The CMB carries a record of conditions at the time of last scattering, often called primary anisotropies. In addition, it is aﬀected by various processes as it propagates towards us, including the eﬀect of a time-varying gravitational potential (the integrated Sachs–Wolfe eﬀect), gravitational lensing, and scattering from ionized gas at low redshift. The primary anisotropies, the integrated Sachs–Wolfe eﬀect, and scattering from a homogeneous distribution of ionized gas, can all be calculated using linear perturbation theory. Available codes include CAMB and CLASS , the former widely used embedded within the analysis package CosmoMC . Gravitational lensing is also calculated in these codes. Secondary eﬀects such as inhomogeneities in the reionization process, and scattering from gravitationally-collapsed gas (the Sunyaev–Zel’dovich (SZ) eﬀect), require more complicated, and more uncertain, calculations. The upshot is that the detailed pattern of anisotropies depends on all of the cosmological parameters. In a typical cosmology, the anisotropy power spectrum [usually plotted as ℓ(ℓ+ 1)Cℓ] features a ﬂat plateau at large angular scales (small ℓ), followed by a series of oscillatory features at higher angular scales, the ﬁrst and most prominent being at around one degree (ℓ≃200). These features, known as acoustic peaks, represent the oscillations of the photon–baryon ﬂuid around the time of decoupling. Some features can be closely related to speciﬁc parameters—for instance, the location of the ﬁrst peak probes the spatial geometry, while the relative heights of the peaks probes the baryon density—but many other parameters combine to determine the overall shape. The 2013 data release from the Planck satellite has provided the most powerful results to date on the spectrum of CMB temperature anisotropies, with a precision determination of the temperature power spectrum to beyond ℓ= 2000, shown in Fig. 24.1. The Atacama Cosmology Telescope (ACT) and South Pole Telescope (SPT) experiments extend these results to higher angular resolution, though without full-sky coverage. The most comprehensive measurements of CMB polarization come from the WMAP satellite ﬁnal (9-year) data release , giving the spectrum of E-polarization anisotropies and the correlation spectrum between temperature and polarization (those spectra having ﬁrst been detected by DASI ) . These are consistent with models based on the parameters we have described, and provide accurate determinations of many of those parameters . The data provide an exquisite measurement of the location of the ﬁrst acoustic peak, determining the angular-diameter distance of the last-scattering surface. In combination with other data this strongly constrains the spatial geometry, in a manner consistent with spatial ﬂatness and excluding signiﬁcantly-curved Universes. CMB data also gives a precision measurement of the age of the Universe. It gives a baryon density consistent with, and at higher precision than, that coming from BBN. It aﬃrms the need for both dark matter and dark energy. It shows no evidence for dynamics of the dark energy, being consistent with a pure cosmological constant (w = −1). The density perturbations are consistent with a power-law primordial spectrum, and there is no indication yet of tensor August 21, 2014 13:18 12 24. The Cosmological Parameters Figure 24.1: The angular power spectrum of the CMB temperature anisotropies from Planck, from Ref. 1. Note the x-axis switches from logarithmic to linear at ℓ= 50. The solid line shows the prediction from the best-ﬁtting ΛCDM model and the band indicates the cosmic variance uncertainty. [Figure courtesy ESA/Planck Collaboration.] perturbations. The current best-ﬁt for the reionization optical depth from CMB data, τ = 0.091, is in line with models of how early structure formation induces reionization. Planck has also made the ﬁrst all-sky map of the CMB lensing ﬁeld, which probes the entire matter distribution in the Universe; this detection corresponds to about 25σ and adds some additional constraining power to the CMB-only data-sets. ACT previously announced the ﬁrst detection of gravitational lensing of the CMB from the four-point correlation of temperature variations . These measurements agree with the expected eﬀect in the standard cosmology. 24.3.4. Galaxy clustering : The power spectrum of density perturbations depends on the nature of the dark matter. Within the ΛCDM model, the power spectrum shape depends primarily on the primordial power spectrum and on the combination Ωmh which determines the horizon scale at matter–radiation equality, with a subdominant dependence on the baryon density. The matter distribution is most easily probed by observing the galaxy distribution, but this must be done with care as the galaxies do not perfectly trace the dark matter distribution. Rather, they are a ‘biased’ tracer of the dark matter. The need to allow for such bias is emphasized by the observation that diﬀerent types of galaxies show bias with respect to each other. In particular scale-dependent and stochastic biasing may introduce a systematic eﬀect on the determination of cosmological parameters from redshift surveys. Prior knowledge from simulations of galaxy formation or from gravitational lensing data August 21, 2014 13:18 24. The Cosmological Parameters 13 could help to quantify biasing. Furthermore, the observed 3D galaxy distribution is in redshift space, i.e., the observed redshift is the sum of the Hubble expansion and the line-of-sight peculiar velocity, leading to linear and non-linear dynamical eﬀects which also depend on the cosmological parameters. On the largest length scales, the galaxies are expected to trace the location of the dark matter, except for a constant multiplier b to the power spectrum, known as the linear bias parameter. On scales smaller than 20 h−1 Mpc or so, the clustering pattern is ‘squashed’ in the radial direction due to coherent infall, which depends approximately on the parameter β ≡Ω0.6 m /b (on these shorter scales, more complicated forms of biasing are not excluded by the data). On scales of a few h−1 Mpc, there is an eﬀect of elongation along the line of sight (colloquially known as the ‘ﬁnger of God’ eﬀect) which depends on the galaxy velocity dispersion. 24.3.4.1. Baryonic acoustic oscillations: The power spectra of the 2-degree Field (2dF) Galaxy Redshift Survey and the Sloan Digital Sky Survey (SDSS) are well ﬁt by a ΛCDM model and both surveys showed evidence for BAOs [29,30]. The Baryon Oscillation Spectroscopic Survey (BOSS) of Luminous Red Galaxies (LRGs) in the SDSS found consistency with the dark energy equation of state w = −1 to within ±0.06 . Similar results for w were obtained by the WiggleZ survey . The BAO data from recent galaxy redshift surveys together with SN Ia data are shown in a Hubble diagram in Fig. 24.2. 24.3.4.2. Redshift distortion: There is renewed interest in the ‘redshift distortion’ eﬀect. As the measured redshift of a galaxy is the sum of its redshift due to the Hubble expansion and its peculiar velocity, this distortion depends on cosmological parameters via the perturbation growth rate in linear theory f(z) = d ln δ/d ln a ≈Ωγ(z), where γ ≃0.55 for the ΛCDM model and is diﬀerent for modiﬁed gravity models. Recent observational results show that by measuring f(z) it is feasible to constrain γ and rule out certain modiﬁed gravity models [35,36]. We note the degeneracy of the redshift-distortion pattern and the geometric distortion (the so-called Alcock–Paczynski eﬀect), e.g. as illustrated by the WiggleZ survey . 24.3.4.3. Integrated Sachs–Wolfe eﬀect: The integrated Sachs–Wolfe (ISW) eﬀect, described in the article by Scott and Smoot, is the change in CMB photon energy when propagating through the changing gravitational potential wells of developing cosmic structures. In linear theory, the ISW signal is expected in universes where there is dark energy, curvature, or modiﬁed gravity. Correlating the large-angle CMB anisotropies with very large scale structures, ﬁrst proposed in Ref. 38, has provided results which vary from no detection of this eﬀect to 4σ detection [39,40]. August 21, 2014 13:18 14 24. The Cosmological Parameters Figure 24.2: The cosmic distance scale with redshift. This modern version of the ’Hubble Diagram’ combines data from SN Ia as standard candles and BAO as standard rulers in the LRG SSDS, BOSS, 6dFGRS, and WiggleZ galaxy surveys and from the BOSS Lyman-alpha at high redshift. [Figure courtesy of C. Blake, based on Ref. 33.] 24.3.4.4. Limits on neutrino mass from galaxy surveys and other probes: Large-scale structure data constraints on Ων due to the neutrino free-streaming eﬀect . Presently there is no clear detection, and upper limits on neutrino mass are commonly estimated by comparing the observed galaxy power spectrum with a four-component model of baryons, cold dark matter, a cosmological constant, and massive neutrinos. Such analyses also assume that the primordial power spectrum is adiabatic, scale-invariant, and Gaussian. Potential systematic eﬀects include biasing of the galaxy distribution and non-linearities of the power spectrum. An upper limit can also be derived from CMB anisotropies alone, while additional cosmological data-sets can improve the results. Results using a photometric redshift sample of LRGs combined with WMAP, BAO, Hubble constant and SNe Ia data gave a 95% conﬁdence upper limit on the total neutrino mass of 0.28eV . Recent spectroscopic redshift surveys, with more accurate redshifts but fewer galaxies, yielded similar upper limits for assumed ﬂat ΛCDM model and additional data-sets: 0.34eV from BOSS and 0.29eV from WiggleZ . Planck + WMAP polarization + highL CMB give an upper limit of 0.66eV, and with additional BAO data 0.23eV. The eﬀective number of relativistic degrees of freedom is Neﬀ= 3.30 ± 0.27 in good agreement with the standard value Neﬀ= 3.046. While August 21, 2014 13:18 24. The Cosmological Parameters 15 the latest cosmological data do not yet constrain the sum of neutrino masses to below 0.2eV, as the lower limit on neutrino mass from terrestrial experiments is 0.06eV, it looks promising that future cosmological surveys will detect the neutrino mass. 24.3.5. Clusters of galaxies : A cluster of galaxies is a large collection of galaxies held together by their mutual gravitational attraction. The largest ones are around 1015 Solar masses, and are the largest gravitationally-collapsed structures in the Universe. Even at the present epoch they are relatively rare, with only a few percent of galaxies being in clusters. They provide various ways to study the cosmological parameters. The ﬁrst objects of a given kind form at the rare high peaks of the density distribution, and if the primordial density perturbations are Gaussian distributed, their number density is exponentially sensitive to the size of the perturbations, and hence can strongly constrain it. Clusters are an ideal application in the present Universe. They are usually used to constrain the amplitude σ8, as a sphere of radius 8 h−1 Mpc contains about the right amount of material to form a cluster. One of the most useful observations at present are of X-ray emission from hot gas lying within the cluster, whose temperature is typically a few keV, and which can be used to estimate the mass of the cluster. A theoretical prediction for the mass function of clusters can come either from semi-analytic arguments or from numerical simulations. The same approach can be adopted at high redshift (which for clusters means redshifts of order one) to attempt to measure σ8 at an earlier epoch. The evolution of σ8 is primarily driven by the value of the matter density Ωm, with a sub-dominant dependence on the dark energy properties. The Planck observations were used to produce a sample of 189 clusters selected by the SZ eﬀect. The cluster mass function was constructed using a relation between the SZ signal Y and cluster mass M. For an assumed ﬂat ΛCDM model, the Planck Collaboration found σ8 = 0.77 ± 0.02 and Ωm = 0.29 ± 0.02 . Somewhat larger values of both parameters are preferred by the Planck’s measurements of the primary CMB anisotropies. The discrepancy might be resolved, for example, by using a diﬀerent Y –M calibration. For comparison with other results in the literature see their Fig. 10. 24.3.6. Clustering in the inter-galactic medium : It is commonly assumed, based on hydrodynamic simulations, that the neutral hydrogen in the inter-galactic medium (IGM) can be related to the underlying mass distribution. It is then possible to estimate the matter power spectrum on scales of a few megaparsecs from the absorption observed in quasar spectra, the so-called Lyman-α forest. The usual procedure is to measure the power spectrum of the transmitted ﬂux, and then to infer the mass power spectrum. Photo-ionization heating by the ultraviolet background radiation and adiabatic cooling by the expansion of the Universe combine to give a simple power-law relation between the gas temperature and the baryon density. It also follows that there is a power-law relation between the optical depth τ and ρb. Therefore, the observed ﬂux F = exp(−τ) is strongly correlated with ρb, which itself traces the mass density. The matter and ﬂux power spectra can be related by Pm(k) = b2(k)PF(k) , (24.10) August 21, 2014 13:18 16 24. The Cosmological Parameters where b(k) is a bias function which is calibrated from simulations. The BOSS survey has been used to detect and measure the BAO feature in the Lyman-α forest ﬂuctuation at redshift z = 2.4, with a result impressively consistent with the standard ΛCDM model . The Lyman-α ﬂux power spectrum has also been used to constrain the nature of dark matter, for example constraining the amount of warm dark matter . 24.3.7. Gravitational lensing : Images of background galaxies are distorted by the gravitational eﬀect of mass variations along the line of sight. Deep gravitational potential wells such as galaxy clusters generate ‘strong lensing’, leading to arcs, arclets and multiple images, while more moderate perturbations give rise to ‘weak lensing’. Weak lensing is now widely used to measure the mass power spectrum in selected regions of the sky (see Ref. 48 for reviews). As the signal is weak, the image of deformed galaxy shapes (the ‘shear map’) must be analyzed statistically to measure the power spectrum, higher moments, and cosmological parameters. The shear measurements are mainly sensitive to a combination of Ωm and the amplitude σ8. For example, the weak-lensing signal detected by the CFHTLens Survey (over 154 sq. deg. in 5 optical bands) yields, for a ﬂat ΛCDM model, σ8(Ωm/0.27)0.6 = 0.79 ± 0.03 . Earlier results for comparison are summarized in Ref. 48. There are various systematic eﬀects in the interpretation of weak lensing, e.g., due to atmospheric distortions during observations, the redshift distribution of the background galaxies, the intrinsic correlation of galaxy shapes, and non-linear modeling uncertainties. 24.3.8. Peculiar velocities : Deviations from the Hubble ﬂow directly probe the mass perturbations in the Universe, and hence provide a powerful probe of the dark matter . Peculiar velocities are deduced from the diﬀerence between the redshift and the distance of a galaxy. The observational diﬃculty is in accurately measuring distances to galaxies. Even the best distance indicators (e.g., the Tully–Fisher relation) give an uncertainty of 15% per galaxy, hence limiting the application of the method at large distances. Peculiar velocities are mainly sensitive to Ωm, not to ΩΛ or dark energy. While at present cosmological parameters derived from peculiar velocities are strongly aﬀected by random and systematic errors, a new generation of surveys may improve their accuracy. Three promising approaches are the 6dF near-infrared survey of 15,000 peculiar velocities, peculiar velocities of SNe Ia, and the kinematic Sunyaev–Zel’dovich eﬀect. 24.4. Bringing observations together Although it contains two ingredients—dark matter and dark energy—which have not yet been veriﬁed by laboratory experiments, the ΛCDM model is almost universally accepted by cosmologists as the best description of the present data. The approximate values of some of the key parameters are Ωb ≈0.05, Ωc ≈0.25, ΩΛ ≈0.70, and a Hubble constant h ≈0.70. The spatial geometry is very close to ﬂat (and usually assumed to be precisely ﬂat), and the initial perturbations Gaussian, adiabatic, and nearly scale-invariant. August 21, 2014 13:18 24. The Cosmological Parameters 17 The most powerful data source is the CMB, which on its own supports all these main tenets. Values for some parameters, as given in Ade et al. and Hinshaw et al. , are reproduced in Table 24.1. These particular results presume a ﬂat Universe. The constraints are somewhat strengthened by adding additional data-sets such as BAO, as shown in the Table, though most of the constraining power resides in the CMB data. We see that the Planck and WMAP constraints are similar, though with some shifts within the uncertainties. For these six-parameter ﬁts the parameter uncertainties are also comparable; the additional precision of Planck data versus WMAP is only really apparent when considering signiﬁcantly larger parameter sets. If the assumption of spatial ﬂatness is lifted, it turns out that the CMB on its own only weakly constrains the spatial curvature, due to a parameter degeneracy in the angular-diameter distance. However inclusion of other data readily removes this. For example, inclusion of BAO data, plus the assumption that the dark energy is a cosmological constant, yields a constraint on Ωtot ≡P Ωi +ΩΛ of Ωtot = 1.0005 ± 0.0033 . Results of this type are normally taken as justifying the restriction to ﬂat cosmologies. One parameter which is very robust is the age of the Universe, as there is a useful coincidence that for a ﬂat Universe the position of the ﬁrst peak is strongly correlated with the age. The CMB data give 13.81 ± 0.05 Gyr (assuming ﬂatness). This is in good agreement with the ages of the oldest globular clusters and radioactive dating. The baryon density Ωb is now measured with high accuracy from CMB data alone, and is consistent with the determination from BBN; Fields et al. in this volume quote the range 0.021 ≤Ωbh2 ≤0.025 (95% conﬁdence). While ΩΛ is measured to be non-zero with very high conﬁdence, there is no evidence of evolution of the dark energy density. Mortonson et al. in this volume quote the constraint w = −1.13+0.13 −0.11 on a constant equation of state from a compilation of CMB and BAO data, with the cosmological constant case w = −1 giving an excellent ﬁt to the data. Allowing more complicated forms of dark energy weakens the limits. The data provide strong support for the main predictions of the simplest inﬂation models: spatial ﬂatness and adiabatic, Gaussian, nearly scale-invariant density pertur-bations. But it is disappointing that there is no sign of primordial gravitational waves, with the CMB data compilation providing an upper limit r < 0.11 at 95% conﬁdence (weakening to 0.26 if running is allowed). The spectral index is clearly required to be less than one by this data, though the strength of that conclusion can weaken if additional parameters are included in the model ﬁts. Tests have been made for various types of non-Gaussianity, a particular example being a parameter fNL which measures a quadratic contribution to the perturbations. Various non-gaussianity shapes are possible (see Ref. 51 for details), and current constraints on the popular ‘local’, ‘equilateral’, and ‘orthogonal’ types are flocal NL = 3 ± 6, fequil NL = −42 ± 75, and fortho NL = −25 ± 39 (these look weak, but prominent non-Gaussianity requires the product fNL∆R to be large, and ∆R is of order 10−5). Clearly none of these give any indication of primordial non-gaussianity. August 21, 2014 13:18 18 24. The Cosmological Parameters Table 24.1: Parameter constraints reproduced from Ref. 2 (Table 5) and Ref. 4 (Table 4), with some additional rounding. All columns assume the ΛCDM cosmology with a power-law initial spectrum, no tensors, spatial ﬂatness, and a cosmological constant as dark energy. Planck take the sum of neutrino masses ﬁxed to 0.06eV, while WMAP set it to zero. Above the line are the six parameter combinations actually ﬁt to the data in the Planck analysis (θMC is a measure of the sound horizon at last scattering); those below the line are derived from these. Two diﬀerent data combinations including Planck are shown to highlight the extent to which additional data improve constraints. The ﬁrst column is a combination of CMB data only — Planck temperature plus WMAP polarization data plus high-resolution data from ACT and SPT — while the second column adds BAO data from the SDSS, BOSS, 6dF, and WiggleZ surveys. For comparison the last column shows the ﬁnal nine-year results from the WMAP satellite, combined with the same BAO data and high-resolution CMB data (which they call eCMB). Note WMAP use ΩΛ directly as a ﬁt parameter rather than θMC. The perturbation amplitude ∆2 R is speciﬁed at the scale 0.05 Mpc−1 for Planck, but 0.002 Mpc−1 for WMAP, so the spectral index ns needs to be taken into account in comparing them. Uncertainties are shown at 68% conﬁdence. Planck+WP Planck+WP WMAP9+eCMB +highL +highL+BAO +BAO Ωbh2 0.02207 ± 0.00027 0.02214 ± 0.00024 0.02211 ± 0.00034 Ωch2 0.1198 ± 0.0026 0.1187 ± 0.0017 0.1162 ± 0.0020 100 θMC 1.0413 ± 0.0006 1.0415 ± 0.0006 − ns 0.958 ± 0.007 0.961 ± 0.005 0.958 ± 0.008 τ 0.091+0.013 −0.014 0.092 ± 0.013 0.079+0.011 −0.012 ln(1010∆2 R) 3.090 ± 0.025 3.091 ± 0.025 3.212 ± 0.029 h 0.673 ± 0.012 0.678 ± 0.008 0.688 ± 0.008 σ8 0.828 ± 0.012 0.826 ± 0.012 0.822+0.013 −0.014 Ωm 0.315+0.016 −0.017 0.308 ± 0.010 0.293 ± 0.010 ΩΛ 0.685+0.017 −0.016 0.692 ± 0.010 0.707 ± 0.010 August 21, 2014 13:18 24. The Cosmological Parameters 19 24.5. Outlook for the future The concordance model is now well established, and there seems little room left for any dramatic revision of this paradigm. A measure of the strength of that statement is how diﬃcult it has proven to formulate convincing alternatives. Should there indeed be no major revision of the current paradigm, we can expect future developments to take one of two directions. Either the existing parameter set will continue to prove suﬃcient to explain the data, with the parameters subject to ever-tightening constraints, or it will become necessary to deploy new parameters. The latter outcome would be very much the more interesting, oﬀering a route towards understanding new physical processes relevant to the cosmological evolution. There are many possibilities on oﬀer for striking discoveries, for example: • The cosmological eﬀects of a neutrino mass may be unambiguously detected, shedding light on fundamental neutrino properties; • Detection of primordial non-Gaussianities would indicate that non-linear processes inﬂuence the perturbation generation mechanism; • Detection of variation in the dark-energy density (i.e., w ̸= −1) would provide much-needed experimental input into the nature of the properties of the dark energy. These provide more than enough motivation for continued eﬀorts to test the cosmological model and improve its accuracy. Over the coming years, there are a wide range of new observations which will bring further precision to cosmological studies. Indeed, there are far too many for us to be able to mention them all here, and so we will just highlight a few areas. The CMB observations will improve in several directions. A current frontier is the study of polarization, ﬁrst detected in 2002 by DASI and for which power spectrum measurements have now been made by several experiments. Planck will announce its ﬁrst polarization results in 2014. Future measurements may be able to separately detect the two modes of polarization and a number of projects are underway with this goal. An impressive array of dark energy surveys are already operational, under construction, or proposed, including ground-based imaging surveys the Dark Energy Survey and LSST, spectroscopic surveys such as MS-DESI, and space missions Euclid and WFIRST. An exciting area for the future is radio surveys of the redshifted 21-cm line of hydrogen. Because of the intrinsic narrowness of this line, by tuning the bandpass the emission from narrow redshift slices of the Universe will be measured to extremely high redshift, probing the details of the reionization process at redshifts up to perhaps 20. LOFAR is the ﬁrst instrument able to do this and is beginning its operations. In the longer term, the Square Kilometre Array (SKA) will take these studies to a precision level. The development of the ﬁrst precision cosmological model is a major achievement. However, it is important not to lose sight of the motivation for developing such a model, which is to understand the underlying physical processes at work governing the Universe’s evolution. On that side, progress has been much less dramatic. For instance, there are many proposals for the nature of the dark matter, but no consensus as to which is August 21, 2014 13:18 20 24. The Cosmological Parameters correct. The nature of the dark energy remains a mystery. Even the baryon density, now measured to an accuracy of a percent, lacks an underlying theory able to predict it within orders of magnitude. Precision cosmology may have arrived, but at present many key questions remain to motivate and challenge the cosmology community. References: 1. P.A.R. Ade et al. (Planck Collab. 2013 I), arXiv:1303.5062v1. 2. P.A.R. Ade et al. (Planck Collab. 2013 XVI), arXiv:1303.5076v1. 3. C. Bennett et al., to appear, Astrophys. J. Supp., arXiv:1212.5225v3. 4. G. Hinshaw et al., to appear, Astrophys. J. Supp., arXiv:1212.5226v3. 5. S. Fukuda et al., Phys. Rev. Lett. 85, 3999 (2000); Q.R. Ahmad et al., Phys. Rev. Lett. 87, 071301 (2001). 6. E.W. Kolb and M.S. Turner, The Early Universe, Addison–Wesley (Redwood City, 1990). 7. D.H. Lyth and A.R. Liddle, The Primordial Density Perturbation, Cambridge University Press (2009). 8. A.R. Liddle and D.H. Lyth, Phys. Lett. B291, 391 (1992). 9. A. Lewis, A Challinor, and A. Lasenby, Astrophys. J. 538, 473 (2000); D. Blas, J. Lesgourgues, and T. Tram, JCAP 1107, 034 (2011). 10. D. Fixsen, Astrophys. J. 707, 916 (2009). 11. A. Kosowsky and M.S. Turner, Phys. Rev. D52, 1739 (1995). 12. K.A. Malik and D. Wands, Phys. Reports 475, 1 (2009). 13. P.A.R. Ade et al. (Planck Collab. 2013 XXV), arXiv:1303.5085v1. 14. D.H. Lyth and D. Wands, Phys. Lett. B524, 5 (2002); K. Enqvist and M.S. Sloth, Nucl. Phys. B626, 395 (2002); T. Moroi and T. Takahashi, Phys. Lett. B522, 215 (2001). 15. G. Mangano et al., Nucl. Phys. B729, 221 (2005). 16. S. Riemer-Sørensen, D. Parkinson, and T.M. Davis, PASA 30, e029 (2013). 17. J.K. Webb et al., Phys. Rev. Lett. 107, 191101 (2011); J.A. King et al., MNRAS 422, 3370 (2012); P. Molaro et al., Astron. & Astrophys. 555, 68 (2013). 18. P.A.R. Ade et al., (Planck Collab. 2013 XXVI), arXiv:1303.5086v1. 19. W.L. Freedman et al., Astrophys. J. 553, 47 (2001). 20. A.G. Riess et al., Astrophys. J. 730, 119 (2011). 21. B. Leibundgut, Ann. Rev. Astron. Astrophys. 39, 67 (2001). 22. A.G. Riess et al., Astron. J. 116, 1009 (1998); P. Garnavich et al., Astrophys. J. 509, 74 (1998); S. Perlmutter et al., Astrophys. J. 517, 565 (1999). 23. A. Conley et al., Astrophys. J. Supp. 192, 1 (2011); M. Sullivan et al., Astrophys. J. 737, 102 (2011). 24. M. Kowalski et al., Astrophys. J. 686, 749 (2008). 25. R. Kessler et al., Astrophys. J. Supp. 185, 32 (2009). 26. A. Lewis and S. Bridle, Phys. Rev. D66, 103511 (2002). 27. J. Kovac et al., Nature 420, 772 (2002). 28. S. Das et al., Phys. Rev. Lett. 107, 021301 (2011). August 21, 2014 13:18 24. The Cosmological Parameters 21 29. D. Eisenstein et al., Astrophys. J. 633, 560 (2005). 30. S. Cole et al., MNRAS 362, 505 (2005). 31. A. Sanchez et al., arXiv:1303.4396. 32. D. Parkinson et al., arXiv:1210.2130. 33. C. Blake et al., MNRAS 418, 1707 (2011). 34. N. Kaiser, MNRAS 227, 1 (1987). 35. L. Guzzo et al., Nature 451, 541 (2008). 36. A. Nusser and M. Davis, Astrophys. J. 736, 93 (2011). 37. C. Blake et al., arXiv:1204.3674. 38. R.G. Crittenden and N. Turok, Phys. Rev. Lett. 75, 2642 (1995). 39. T. Giannatonio et al., MNRAS 426, 258 (2012). 40. P.A.R. Ade et al. (Planck Collab. 2013 XIX), arXiv:1303.5076v1. 41. J. Lesgourgues and S. Pastor, Phys. Reports 429, 307 (2006). 42. S. Thomas, F.B. Abdalla, and O. Lahav, Phys. Rev. Lett. 105, 031301 (2010). 43. G.-B. Zhao et al., MNRAS, doi:10.1093/mnras/stt1710 (early on-line). 44. S. Riemer-Sørensen et al., Phys. Rev. D85, 081101 (2012). 45. P.A.R. Ade et al. (Planck Collab. 2013 XX), arXiv:1303.5080v1. 46. A. Slosar et al., arXiv:1301.3459. 47. M. Viel et al., arXiv:1306.2314. 48. A. Refregier, Ann. Rev. Astron. Astrophys. 41, 645 (2003); R. Massey et al., Nature 445, 286 (2007); H. Hoekstra and B. Jain, Ann. Rev. Nucl. and Part. Sci. 58, 99 (2008). 49. M. Kilbinger et al., MNRAS 430, 2200 (2013); C. Heymans et al., MNRAS 427, 146 (2012). 50. A. Dekel, Ann. Rev. Astron. Astrophys. 32, 371 (1994). 51. P.A.R. Ade et al. (Planck Collab. 2013 XXIV), arXiv:1303.5084v1. August 21, 2014 13:18
15050	https://chemistwizards.com/normality-chemistry/	Normality Chemistry Skip to content Chemist Wizards Home Chemistry Organic Chemistry Biochemistry Inorganic chemistry Blog Chemistry Tools Chemistry Book About us Contact US Privacy Policy Software Normality Chemistry What is the definition for Normality? The concentration of the solution expressed in a number of equivalent dissolved per liter of solution is known as the normality chemistry. Normality is represented by N. Table of Contents Toggle Normality Chemistry What is the definition for Normality? Units of normality: Formula of Normality chemistry: Steps for calculation : How normality is calculated in Titration? Relationship between Molarity and Normality Molarity: Normality: Conversion of molarity to normality: Describe the difference between Normality Vs Molarity: What is the difference between Normality and Molality? Units of normality: Its units are eq/L or meg/L. The latter unit is mostly expressed in medical reporting. Formula of Normality chemistry: These are the formulas to calculate normality: Steps for calculation : Here are some steps mentioned which have to be followed to calculate normality: In the first step, students should gather information about the equivalent weight of the reacting substance or the solute. Then search the molecular weight and valence through literature. In the next step, the number of gram equivalents of solute is calculated. Then calculate the volume in liters. In the last step, normality is calculated by using the above formula and replacing the values. How normality is calculated in Titration? Titration is a technique in which a solution of known concentration and volume is gradually added to another solution of unknown concentration until the neutralizing reaction occurs. By using the given formula, the normality of the acid and base titration is calculated. Where, N 1 = Indicates the Normality of the Acidic solution V 1 = Designate the Volume of the Acidic solution N 2 = Indicates the Normality of the basic solution V 3 = Designate the Volume of the basic solution Relationship between Molarity and Normality Both normality and molarity are necessary terms that play crucial roles in chemistry. They are used for the qualitative measurement of a solute or substance. Molarity: It is the number of moles of solute per volume of the solution in liters. It is commonly used for the determination of pH such as dissociation or equilibrium constants etc. Formula of Molarity Its formula is mentioned below: Normality: It is the number of equivalents dissolved per liter of the solution. Formula of normality Normality of bases In the case of bases, the number of H+ are counted which are present in the acid molecule which it can donate. The given formula is used to calculate the normality of bases: Normality of Acids In the case of acids, the number of OH- is counted which are present in the base molecule that it can donate. The given formula is used to calculate the normality of acids: Conversion of molarity to normality: Molarity can be converted into normality by using the given formula: Describe the difference between Normality Vs Molarity: What is the difference between Normality andMolality? By Kaneez FatimaSeptember 12, 2024Chemistry, Organic Chemistry1 Comment on Normality Chemistry Post navigation Definition of molarity-an overview with super tricks Transition Elements 2 Alive and Impactful Types- Outer and Inner transition elements Leave a Reply Cancel reply Your email address will not be published.Required fields are marked Comment Name Email Website [x] Save my name, email, and website in this browser for the next time I comment. Δ Search Search Categories Analytical chemistry Biochemistry Blog Chemistry Inorganic chemistry Organic Chemistry Physical chemistry Recent Posts Chemical Bonding Carbohydrates Is Cream of Wheat Good for you Hemoglobins: Definition, Level, Structure, Functions & Types Industrial Biochemistry and Microbes About Chemistwizards.com, founded in 2023, emerged as a digital haven for chemical lovers and chemistry students. It started with a commitment to provide an opportunity for the sharing of information, analysis, and educational materials while clarifying the field of chemistry. Blog Chemical Bonding Carbohydrates Is Cream of Wheat Good for you Hemoglobins: Definition, Level, Structure, Functions & Types Industrial Biochemistry and Microbes Pages Instructions Terms and Conditions Privacy Policy Theme Mavix Education by Creativ Themes
15051	https://www.studocu.com/en-us/messages/question/7480566/phosphine-ph3-g-decomposes-according-to-the-equation-4-ph3g-p4g-6-h2gthe-kinetics	[Solved] Phosphine PH3 g decomposes according to the equation 4 PH3g P4g 6 - General Chemistry II and Lab (COM) [SGR #6] (CHEM 114) - Studocu Skip to main content Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes Home My Library AI Notes Ask AI AI Quiz New Recent Phosphine, PH3 (g), decomposes according to the equation 4 PH3(g) -> P4(g) + 6 H2(g) The kinetics of the decomposition of phosphine at 950 K was followed by measuring the total pressure in the system as a function of time. The data in the table were obtained in a run where the reaction chamber contained only pure phosphine at the start of the reaction. Time(min) 0.00 P total (Torr) 100 Time(min) 40.0 P total (Torr) 151 Time(min) 80.0 P total (Torr) 168 Time(min) 100 P total (Torr) 171 Choose the rate law that describes this reaction. Calculate the value of the rate constant and determine the correct units.General Chemistry II and Lab (COM) [SGR #6] (CHEM 114) My Library Courses You don't have any courses yet. Add Courses Studylists You don't have any Studylists yet. Create a Studylist South Dakota State University General Chemistry II and Lab (COM) [SGR #6] Question Phosphine PH3 g decomposes according to the equation 4 PH3g P4g 6 South Dakota State University General Chemistry II and Lab (COM) [SGR #6] Question Anonymous Student 1 year ago Phosphine, PH3 (g), decomposes according to the equation 4 PH3(g) -> P4(g) + 6 H2(g) The kinetics of the decomposition of phosphine at 950 K was followed by measuring the total pressure in the system as a function of time. The data in the table were obtained in a run where the reaction chamber contained only pure phosphine at the start of the reaction. Time(min) 0.00 P total (Torr) 100 Time(min) 40.0 P total (Torr) 151 Time(min) 80.0 P total (Torr) 168 Time(min) 100 P total (Torr) 171 Choose the rate law that describes this reaction. Calculate the value of the rate constant and determine the correct units. Like 0 Related documents Experiment 23 - Kinetics Analysis and Exploration General Chemistry II and Lab (COM) [SGR #6]Coursework None Experiment 22 Lab Report: Reduction of Methylene Blue Analysis General Chemistry II and Lab (COM) [SGR #6]Other None Chem 114 Final Exam Notes: Chap 17 - Reaction Rates & Rate Equations General Chemistry II and Lab (COM) [SGR #6]Lecture notes None CHapter 14 Notes General Chemistry II and Lab (COM) [SGR #6]Summaries None CHEM 114 Chapter 14 - Acid/Base Theories & Equilibrium Constants General Chemistry II and Lab (COM) [SGR #6]Lecture notes 100%(1) Chapter 13 notes General Chemistry II and Lab (COM) [SGR #6]Summaries None CHEM 114 Chapter 13: Stable Configurations & Equilibrium Constants General Chemistry II and Lab (COM) [SGR #6]Lecture notes None Answer Created with AI 1 year ago Answer Rate Law The rate law for a reaction describes how the rate of reaction depends on the concentration of the reactants. For the given reaction: js 4 PH3(g) -> P4(g) + 6 H2(g) The rate law can be written as: js Rate = k[PH3]^n Where: Rate is the rate of the reaction k is the rate constant [PH3] is the concentration of PH3 n is the order of the reaction with respect to PH3 Given that the total pressure of the system increases over time, it suggests that the reaction is first order with respect to PH3. This is because in a first order reaction, the rate of reaction is directly proportional to the concentration of one reactant. In this case, as the concentration of PH3 decreases, the total pressure (which is proportional to the total number of moles of gas in the system) increases. Therefore, the rate law for this reaction is: js Rate = k[PH3] Rate Constant The rate constant, k, can be calculated using the integrated rate law for a first order reaction, which is: js ln([PH3]0/[PH3]) = kt Where: [PH3]0 is the initial concentration of PH3 [PH3] is the concentration of PH3 at time t k is the rate constant t is the time Given the data in the table, you can calculate k by rearranging the equation to: js k = ln([PH3]0/[PH3]) / t And substituting the values from the table. Units of the Rate Constant The units of the rate constant for a first order reaction are typically s^-1 or min^-1, depending on the units of time used in the experiment. In this case, since the time is given in minutes, the units of the rate constant would be min^-1. Like 0 AI answers may contain errors. Please double check important information and use responsibly. Ask a new question Discover more from: General Chemistry II and Lab (COM) [SGR #6] CHEM 114 South Dakota State University 46 Documents Go to course 1 Experiment Pre-lab 29 General Chemistry II and Lab (COM) [SGR #6]100%(7) 4 Experiment 27: Determining the pKa of Bromothymol Blue Indicator General Chemistry II and Lab (COM) [SGR #6]100%(3) 5 Experiment 32 Report - FeSCN2+ Formation Constant Lab Analysis General Chemistry II and Lab (COM) [SGR #6]83%(6) 45 CHEM 114 Chapter 14 - Acid/Base Theories & Equilibrium Constants General Chemistry II and Lab (COM) [SGR #6]100%(1) Discover more from: General Chemistry II and Lab (COM) [SGR #6] CHEM 114South Dakota State University46 Documents Go to course 1 Experiment Pre-lab 29 General Chemistry II and Lab (COM) [SGR #6]100%(7) 4 Experiment 27: Determining the pKa of Bromothymol Blue Indicator General Chemistry II and Lab (COM) [SGR #6]100%(3) 5 Experiment 32 Report - FeSCN2+ Formation Constant Lab Analysis General Chemistry II and Lab (COM) [SGR #6]83%(6) 45 CHEM 114 Chapter 14 - Acid/Base Theories & Equilibrium Constants General Chemistry II and Lab (COM) [SGR #6]100%(1) 1 Experiment Pre-lab 30 General Chemistry II and Lab (COM) [SGR #6]67%(3) 4 Comprehensive Guide to Redox Reactions and Balancing Techniques General Chemistry II and Lab (COM) [SGR #6]None Related Answered Questions 4 months ago The oxidation number of phosphorus in H2PO3 General Chemistry II and Lab (COM) [SGR #6] (CHEM 114) 7 months ago What is the equilibrium constant of the following reaction at 50˚C? Cl₂(g) + 2BrF(g) ⇄ 2ClF(g) + Br₂(l) ΔHr°=79 kJ mol⁻¹ ΔSr°=-93.00 J mol⁻¹ K⁻¹ ΔGr°=106 kJ mol⁻¹ R = 8.3145 J mol⁻¹ K⁻¹ 2 × 10⁻¹³ 1 × 10¹⁷ 7 × 10⁻¹⁸ 52 1.9 × 10⁻² General Chemistry II and Lab (COM) [SGR #6] (CHEM 114) 8 months ago the acid and the base were more concentrated than the label said, will there be high, low or no bias General Chemistry II and Lab (COM) [SGR #6] (CHEM 114) 8 months ago a) NO2 - (aq) + Al(s) BASE ⎯⎯ → NH3(g) + AlO2 - (aq) solve this redox reaction General Chemistry II and Lab (COM) [SGR #6] (CHEM 114) 1 year ago Fill in the blank spaces in the table. Draw the structures of these ten unbranched alkanes, using Figure 1.1.1 (hexane) as a model. What do the names of these molecules have in common? What is the relationship between the number of carbon atoms and the number of hydrogen atoms in an unbranched alkane? Provide a general formula (use n for the number of carbon atoms). Draw a graph of boiling points versus number of carbon atoms in unbranched alkanes. How do the boiling points vary with the number of carbons? Propose an explanation for this observation. General Chemistry II and Lab (COM) [SGR #6] (CHEM 114) 1 year ago for a particular redox reaction, Cr is oxidized to CrO4^2- and Fe^3+ is reduced to Fe^2+. complete and balance the equation for this reaction in basic solution General Chemistry II and Lab (COM) [SGR #6] (CHEM 114) Ask AI Home My Library Discovery Discovery Universities High Schools High School Levels Teaching resources Lesson plan generator Test generator Live quiz generator Ask AI English (US) United States Company About us Studocu Premium Academic Integrity Jobs Blog Dutch Website Study Tools All Tools Ask AI AI Notes AI Quiz Generator Notes to Quiz Videos Notes to Audio Infographic Generator Contact & Help F.A.Q. Contact Newsroom Legal Terms Privacy policy Cookie Settings Cookie Statement Copyright & DSA English (US) United States Studocu is not affiliated to or endorsed by any school, college or university. 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15052	https://www.khanacademy.org/math/ap-calculus-ab/ab-diff-contextual-applications-new/ab-4-6/v/linear-approximation-example	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
15053	https://maths.catrust.co.uk/posts/488	Teaching fractions and ratio through bar models \| CAM Maths CAM Maths Blog Log in Teaching fractions and ratio through bar models Sunday, 6 Mar 2022 According to the NCETM, “the bar model is used in teaching for mastery to help children to 'see' mathematical structure. It is not a method for solving problems, but a way of revealing the mathematical structure within a problem and gaining insight and clarity to help solve it” (NCETM, 2015). The bar model has been gaining traction for a number of years, especially in primary schools, but I have previously been resistant to it. There have been various things which had dissuaded me, for example where advocates of bar modelling had demonstrated hopelessly convoluted uses of bar models when a much more sensible algebraic approach existed, or to a lesser extent, my (as yet unresolved) issue with having negative lengths. However, having seen the potential of bar models for teaching ratio and proportion, I decided to teach a series of lessons using bar models on fractions, finding fractions of amounts and sharing in a given ratio to my middle-attaining year 8 class over the course of three weeks. About a month after this series of lessons, I interviewed three pairs of students who had attended the lessons. During the interviews, I asked them to complete five questions (below) and then asked them to reflect on the approach they had used. I did not require them to use a bar model (and indeed not all of them did) but I did prompt two students with the question “could you draw a bar model?” if they were struggling to get started. The questions discussed during the interviews: Find 3 5 3 5 of 40. n 5 n 5 of 45 = 36. Find . Albert and Bella share 20 sweets in the ratio 3:2. What can you work out? Charlie and Delia share some sweets in the ratio 5:3. Delia gets 12 sweets. What can you work out? Eve and Fred’s ages can be expressed as the ratio 3:5. Fred is 6 years older than Eve. What can you work out? One of the difficulties of using a bar model to represent a mathematical problem is making explicit the relationship between the bar model and the underlying mathematics. Making this relationship explicit requires careful consideration of the language to be used, both by me and by students in the class. I started by revisiting representations of fractions, specifically in bar form, with a full bar representing the whole. I chose to use the word bar to refer to the whole, and referred to the parts of the bar as sections. I did consider using only the more abstract part and whole but found that it was useful to have a way of discussing the mechanics of drawing and manipulating a bar. Figure 1: an overview of the language of the bar After the work representing fractions, we moved onto fractions of amounts. At this point, I emphasised the fact that a whole does not necessarily mean you have one of something. For example, in figure 2, a whole is equivalent 15 and so 1 2 3 1 2 3 of the whole is equivalent to 25. This distinction was also important in the work on ratio where, for example, if something is shared in the ratio a:b a:b, a a and b b are parts and a+b a+b represents the whole. Figure 2: a bar model showing 1 2 3 1 2 3 One advantage of using a bar model which was identified by students is the fact that it is a convenient way of laying out your work. One student made the point that “you could do it another way but the bar helps you keep track of what bit you’re on – it’s all there in one place”. It is the perennial battle of maths teachers to persuade students that writing things down is helpful. Fortunately, however, students seem to be more willing to write things down on a bar model than they would usually be. Looking through exercise books, it was relatively common to see students drawing a bar model for each question, even though the strategy for the question was the same (as below). Figure 3: an example of a student using a bar model repeatedly Another student made the point that “otherwise you have to keep everything in your head and that can be quite confusing”. We know that people on average can hold around 4 chunks of information in their working memory at one time (Cowan, 2001) so encouraging students to use strategies which reduce the burden on their working memory seems sensible, especially when some students seem to notice this advantage of bar models themselves. The students I spoke to tended to use bar models to record their thinking throughout the process of answering the question as can be seen below. It’s clear from the image (and was corroborated by follow up questions) that this student has continually referred back to the model. First the student notes the fact that the question is working with fifths, drawing a bar with 5 sections and then labelling them 1 5 1 5. Next the student calculates the value of one of the sections and labels each section accordingly. The student then counts up in nines (shading with a little squiggle as she goes) until she reaches 36. Having done the shading during this process, there is no difficulty in remembering how many nines she has reached and she can immediately see that n=4 n=4. This is by no means a straightforward question: the question must be translated into at least two calculations; those calculations must then be performed; and the results of those calculations interpreted to produce an answer to the original question. The use of the bar model has reduced the load on the working memory allowing this student to effectively answer the question where, based on my experience of similar students in similar groups, I wouldn’t necessarily have expected a complete answer. Figure 4: an illustration of the ongoing use of a bar model. Whilst helping students keep track of their workings is a definite advantage of using bar models, I think the real power comes from the way it supports students to make sense of difficult, non-routine problems and translate them into a method for solving the problem. This advantage was more obvious for the more complicated, later questions. For example, when asked how she knew to divide by 5 on question 1, one student replied “because it’s three fifths” whereas in discussions of later questions she began to refer to the “8 sections” of the bar rather than referring directly to the fractions or ratios involved. Half of the students chose not to use a bar model for the early questions because they could see how to answer the question without doing so. For the later questions, students generally found the bar model helpful, with one student stating that “I’d probably have struggled to do it without it”. This point was also made by another student who said “if I didn’t have the bar, I would have been so confused. I wouldn’t have known where to start.” Often with more complicated problems, students don’t know where to start so they simply don’t. Encouraging the use of bar modelling means students always have a place to start. Moreover, their starting point does not require a full understanding of the question so they can start without having to have a fully formed idea of the path to a solution. In addition, the bar model gives them something which they can use to monitor the progress of their solution: does what I have done fit with the bar model I have drawn? One student illustrated this point by getting an initially incorrect solution to question 2 which he was then able to correct because the bar model made it clear that it couldn’t be correct (an answer of n=6 n=6 which would have been greater than 45). Another student explained that “I did something at first but then I changed it but the bar helped me realise I was right”. Reflecting on the progress of a solution is something which students often find challenging but it seems that the use of a bar model can support students with both of these aspects of problem solving. One final advantage of the bar model in making sense of problems is the fact that students can visualise the question. This point was made by four of the students I spoke to, who made comments about how visualising it helped them. When describing how they answered questions they made comments like “I could see how see how many” or “I could see Delia’s part of the ratio was 3”. The student’s work below illustrates the way students were able to see the key information from the question on the bar model they had drawn. The student below has drawn a line indicating Eve’s three parts, and then sectioned off three of Fred’s parts before labelling the extra two parts which Fred has. The visual nature of the bar model has clearly supported this student to make some correct deductions on what would be, at least with more traditional methods, a very challenging question. Figure 5: visualising the important information from the question As an aside, a number of colleagues and I debated about whether to draw ratio bar models as lines or as stacked bars. The advantage of stacked bars is that it makes questions like “how much more does Fred have than Eve” significantly easier whilst placing the bars in a line enables parallels with finding fractions of amounts to be drawn more easily as the bar model which is drawn is the same in both cases. We settled on drawing the bar models in a line which, as can be seen from figure 3, doesn’t seem to have impacted their usefulness for questions like this. Figure 6: stacked bar models vs lines When introducing a support for students within a particular topic, a question in the back of one’s mind must always be “how will we help students to move away from relying on this support once it has served its purpose?”. One slight concern which has arisen as a result of using bar models is that students tended to count on rather than multiply after finding the value of one unit, perhaps because of the way the bar gives you something you can count. This was illustrated by a student who, having found out that one fifth was worth nine, said he “wrote out the nine times table until I got to 36” whilst another repeatedly added on the unit value rather than multiplying to get the desired number of units. This could potentially cause confusion if there was a question which required them to find, for example, 2 1 2 2 1 2 parts (although I think fractional parts may in itself present some difficulty in bar modelling). However, in general, I think the bar model was used as a way of making sense of the question, rather than as a method in itself. Students tended to describe the process they were carrying out (for example “I divided by 8” or “I worked out how many 9s go into 45”) rather than referring to the bar itself. The bar informed the method rather than being the method. As the questions became more complex, the references to the bar itself did increase (“I drew a box with eight sections”) but generally this was alongside a reference to a calculation (“so I multiplied eight by four”). Encouragingly, there was also at least one student who had initially used the bar model but no longer did so, having made sense of the sorts of questions we had been working on. He said that “when we first started doing this topic, I was using the bar model. It was definitely helpful at first.” This is exactly how I would like students to make use of bar models: use them when they’re helpful and then have the confidence in their understanding of the underlying structure to move onto more concise ways of recording their work once they are no longer useful. The students I spoke to were unanimous in saying that they found the bar models useful. I found that they were hugely powerful in supporting students to interpret questions and then generate a method which they could work through, as well as giving them an accessible way into solving more complicated problems. Getting started can be the most difficult part for some students and bar models allow all students to make a productive start. They also allowed students a way to record what they were doing, making some questions more manageable than they would otherwise be. Furthermore, the connection between ratio and proportion is made much clearer than through more traditional ways of teaching these topics. The link between questions like “share £10 in the ratio 3:7” and “find 3 10 3 10 of £10” is made much more explicit. I also felt that my explanations were much clearer and that my efforts to help students understand why we were doing each step was much less reliant on verbal explanations than it has been previously, and indeed has been when teaching other topics. Often persuading students to write supporting calculations can be a challenge but bar models afforded them a clear and relatively concise way of recording their working. Overall, I am very much persuaded that bar models are an excellent way of helping students make sense of ratio and proportion. The connections between topics that they emphasise is hugely beneficial for students, as well as the way they reveal the underlying structure of a problem, and I look forward to seeing how I can expand their use in other areas. James Baker, Joint Head of Maths at Comberton Village College, Cambridgeshire This article was first published by the Association of Teachers of Mathematics inMT280on 8 February 2022 NCETM, 2015, The Bar Model, viewed 14 June 2021, Cowan, N 2001, The magical number 4 in short-term memory: A reconsideration of mental storage capacity, Behavioural and Brain Sciences, vol. 24, no. 1, pp. 87-114. About Contact Help ✖ We use cookies to enhance your experience. By continuing to visit this site you agree to our use of cookies. Learn more
15054	https://www.commoncoresheets.com/equivalent-fractions-missing-number/446/download	Name: Answer Key Finding Equivalent Fractions Math www.CommonCoreSheets.com 11-10 95 90 85 80 75 70 65 60 55 50 11-20 45 40 35 30 25 20 15 10 50 A n s w e r s Ex. 8 1. 90 2. 81 3. 20 4. 24 5. 32 6. 28 7. 20 8. 10 9. 80 10. 5 11. 42 12. 10 13. 49 14. 12 15. 8 16. 12 17. 63 18. 8 19. 32 20. 24 Find the number that makes an equivalent fraction. Ex) 1 = 83 24 1) 4 = 36 10 90 2) 4 = 36 9 81 3) 1 = 10 2 20 4) 5 = 15 8 24 5) 4 = 32 8 64 6) 4 = 28 6 42 7) 4 = 20 6 30 8) 1 = 52 10 9) 6 = 60 8 80 10) 1 = 53 15 11) 4 = 24 7 42 12) 2 = 10 4 20 13) 7 = 49 8 56 14) 3 = 12 4 16 15) 2 = 89 36 16) 4 = 12 6 18 17) 6 = 42 9 63 18) 2 = 88 32 19) 8 = 32 10 40 20) 2 = 16 3 24 Name: Answer Key Finding Equivalent Fractions Math www.CommonCoreSheets.com 1A n s w e r s Ex. 8 1. 90 2. 81 3. 20 4. 24 5. 32 6. 28 7. 20 8. 10 9. 80 10. 5 11. 42 12. 10 13. 49 14. 12 15. 8 16. 12 17. 63 18. 8 19. 32 20. 24 Find the number that makes an equivalent fraction. Ex) 1 = 83 24 1) 4 = 36 10 90 2) 4 = 36 9 81 3) 1 = 10 2 20 4) 5 = 15 8 24 5) 4 = 32 8 64 6) 4 = 28 6 42 7) 4 = 20 6 30 8) 1 = 52 10 9) 6 = 60 8 80 10) 1 = 53 15 11) 4 = 24 7 42 12) 2 = 10 4 20 13) 7 = 49 8 56 14) 3 = 12 4 16 15) 2 = 89 36 16) 4 = 12 6 18 17) 6 = 42 9 63 18) 2 = 88 32 19) 8 = 32 10 40 20) 2 = 16 3 24 1-10 95 90 85 80 75 70 65 60 55 50 11-20 45 40 35 30 25 20 15 10 50 Name: Answer Key Finding Equivalent Fractions Math www.CommonCoreSheets.com Modified 1Find the number that makes an equivalent fraction. 80 24 8 63 32 8 81 10 32 20 10 42 90 12 49 20 24 5 12 28 Ex) 1 = 83 24 1) 4 = 36 10 90 2) 4 = 36 9 81 3) 1 = 10 2 20 4) 5 = 15 8 24 5) 4 = 32 8 64 6) 4 = 28 6 42 7) 4 = 20 6 30 8) 1 = 52 10 9) 6 = 60 8 80 10) 1 = 53 15 11) 4 = 24 7 42 12) 2 = 10 4 20 13) 7 = 49 8 56 14) 3 = 12 4 16 15) 2 = 89 36 16) 4 = 12 6 18 17) 6 = 42 9 63 18) 2 = 88 32 19) 8 = 32 10 40 20) 2 = 16 3 24 A n s w e r s Ex. 8 1. 90 2. 81 3. 20 4. 24 5. 32 6. 28 7. 20 8. 10 9. 80 10. 5 11. 42 12. 10 13. 49 14. 12 15. 8 16. 12 17. 63 18. 8 19. 32 20. 24
15055	https://www.geeksforgeeks.org/maths/vector-projection-formula/	Vector Projection - Formula, Derivation & Examples Last Updated : 23 Jul, 2025 Suggest changes 2 Likes Vector Projection is the shadow of a vector over another vector. It allows you to determine how one vector influences another in a specific direction. The projection vector is obtained by multiplying the vector with the Cos of the angle between the two vectors. A vector has both magnitude and direction. Vector Projection is essential in solving numerical in physics and mathematics. In this article, we will learn about what is vector projection, vector projection formula, the example, and derivation of the formula, and some other related concepts in detail. What is Vector Projection? Vector Projection is a method of finding components of a vector along the direction of a second vector. By projecting a vector on another vector we obtain a vector which represents the component of the first vector along the direction of the second vector. It represents the length of the shadow of a vector over another vector. The vector projection of a vector is obtained by multiplying the vector with the Cos of the angle between the two vectors. Let's say we have two vectors 'a' and 'b' and we have to find the projection of the vector a on vector b then we will multiply the vector 'a' with cosθ where θ is the angle between vector a and vector b. Table of Content What is Vector Projection? Vector Projection Formula Derivation of Vector Projection Formula Vector Projection Formula Examples Practical Applications and Significance of Vector Projection Examples of Vector Projection in Real-World Problem-Solving Vector Projection Formula If Ais represented as A and Bs represented as B, the Vector Projection of A on B is given as the product of A with Cos θ where θ is the angle between A and B. The other formula for Vector Projection of A on B is given as the product of A and B divided by the magnitude of B. The Projection Vector obtained so is a scalar multiple of A and has a direction in the direction of B. Projection of Vector a on Vector bProj ba=∣b∣a.b a.b = Dot product of a and b \|b\| = magnitude of b Derivation of Vector Projection Formula The vector projection formula derivation is discussed below: Let us assume, OP = A and OQ = B and the angle between OP and OQ is θ. Drawn PN perpendicular to OQ. In the right triangle OPN, Cos θ = ON/OP ⇒ ON = OP Cos θ ⇒ ON = \|A\| Cos θ ON is the projection vector of A on B A.B=∣A∣∣B∣cosθ ⇒ A.B=∣B(∣A∣∣cosθ) ⇒ A.B=∣B∣ON ⇒ ON = ∣B∣A.B Hence, the ON = ∣A∣.B^ Thus the Vector Projection of A on B is given as ∣B∣A.B The Vector Projection of B on A is given as ∣A∣A.B Also Check: Types of Vectors Vector Projection Important Terms To find the vector projection we need to learn to find the angle between two vectors and also to calculate the dot product between two vectors. Angle Between Two Vectors The angle between the two vectors is given as the inverse of the cosine of the dot product of two vectors divided by the product of the magnitude of two vectors. Let's say we have two vectors A and B angle between them is θ ⇒ cos θ = ∣A∣.∣B∣A.B ⇒ θ = cos-1∣A∣.∣B∣A.B Dot Product of Two Vectors Let's say we have two vectors A and Bdefined as A=a1i^+a2j^+a3k^ and B=b1i^+b2j^+b3k^ then the dot product between them is given as A.B=(a1i^+a2j^+a3k^)(b1i^+b2j^+b3k^) ⇒ A.B= a1b1 + a2b2 +a3b3 Related Article: Vector Addition Unit Vector Vector Algebra Linear Algebra Vector Projection Formula Examples Example 1. Find the projection of a vector4i^+2j^+k^ on 5i^−3j^+3k^. Solution: Here, a=4i^+2j^+k^b=5i^−3j^+3k^ . We know, projection of Vector a on Vector b = ∣b∣a.b ∣52+(−3)2+32∣(4.(5)+2(−3)+1.(3))=4317 Example 2. Find the projection of the vector5i^+4j^+k^ on3i^+5j^−2k^ Solution: Here, a=5i^+4j^+k^b=3i^+5j^−2k^. We know, projection of Vector a on Vector b = ∣b∣a.b ∣32+52+(−2)2∣(5.(3)+4(5)+1.(−2))=3833 Example 3. Find the projection of the vector5i^−4j^+k^on3i^−2j^+4k^ Solution: Here, a=5i^−4j^+k^b=3i^−2j^+4k^. We know, projection of Vector a on Vector b = ∣b∣a.b ∣32+(−2)2+(4)2∣(5.(3)+((−4).(−2))+1.(4))=2949 Example 4. Find the projection of the vector 2i^−6j^+k^on8i^−2j^+4k^. Solution: Here, a=2i^−6j^+k^b=8i^−2j^+4k^ We know, projection of Vector a on Vector b = ∣b∣a.b ∣82+(−2)2+(4)2∣(2.(8)+((−6).(−2))+1.(4))=8432 Example 5. Find the projection of the vector 2i^−j^+5k^on4i^−j^+k^. Solution: Here, a=2i^−j^+5k^b=4i^−j^+k^. We know, projection of Vector a on Vector b = ∣b∣a.b ∣42+(−1)2+(1)2∣(2.(4)+((−1).(−1))+5.(1))=1814 Example 6: Find the angle between the vectors a=4i^+3j^−k^b=2i^−j^+2k^ Solution : \vec{a}=4\hat i + 3\hat j - \hat k \\vec{b}=2\hat i - \hat j + 2\hat k cos θ = ∣A∣.∣B∣A.B a.b = (4 x 2) + (3 x (-1)) + (( -1) x 2) = 8 - 3 - 2 = 3 \|a\| = √ (4) 2 + (3) 2 + (- 1) 2 \|a\| = √ 16 + 9 + 1 = √ 26 \|b\| = √ (2) 2 + (- 1) 2 + (2) 2 \|b\| = √ 4 + 1 + 4 =\|b\| = √ 9 = 3 Now computig value cos θ cos θ = a.b/\|a\|\|b\| = 3 /3√26 =1/√26 θ = cos-1 1/√26 θ =78.7 Practical Applications and Significance of Vector Projection Physics Force Decomposition: In physics, the vector projection formula is crucial for decomposing forces into components parallel and perpendicular to surfaces. For example, understanding the force exerted by a rope in a game of tug-of-war requires projecting the force vector onto the direction of the rope. Work Calculation: The work done by a force during displacement is calculated using vector projection. The work is the dot product of the force vector and the displacement vector, essentially projecting one vector onto another to find the component of force in the direction of displacement. Engineering Structural Analysis: Engineers use vector projection to analyze stresses on components. By projecting force vectors onto structural axes, they can determine the stress components in different directions, aiding in the design of safer and more efficient structures. Fluid Dynamics: In fluid dynamics, vector projection helps in analyzing fluid flow around objects. By projecting velocity vectors of fluid onto surfaces, engineers can study flow patterns and forces, crucial for aerodynamic design and hydraulic engineering. Computer Graphics Rendering Techniques: Vector projection is fundamental in computer graphics for rendering shadows and reflections. By projecting light vectors onto surfaces, graphics software calculates the angles and intensities of shadows and reflections, enhancing realism in 3D models. Animation and Game Development: In animation, vector projection is used to simulate movements and interactions. For instance, determining how a character moves over uneven terrain involves projecting motion vectors onto the terrain surface, allowing for realistic animations. Check: Basis Vectors in Linear Algebra Examples of Vector Projection in Real-World Problem-Solving Example 1: GPS Navigation Context: In GPS navigation systems, vector projection is used to calculate the shortest path between two points on the earth’s surface. Application: By projecting the displacement vector between two geographical locations onto the earth's surface vector, GPS algorithms can accurately calculate distances and directions, optimizing travel routes. Example 2: Sports Analytics Context: In sports analytics, particularly in soccer or basketball, vector projection helps in analyzing player movements and ball trajectories. Application: By projecting the movement vectors of players onto the game field or court, analysts can study patterns, speeds, and efficiency of movements, contributing to strategic planning and performance improvement. Example 3: Renewable Energy Engineering Context: In the design of wind turbines, understanding the wind force components is essential for optimizing energy production. Application: Engineers project wind velocity vectors onto the plane of the turbine blades. This analysis helps in determining the optimal angle and orientation of blades to maximize the capture of wind energy. Example 4: Augmented Reality (AR) Context: In augmented reality applications, vector projection is used to accurately place virtual objects in real-world spaces. Application: By projecting vectors from virtual objects onto real-world planes captured by AR devices, developers can ensure that virtual objects interact realistically with the environment, enhancing user experience. Check: Components of Vector Conclusion Vector Projection is a valuable tool that helps break down complex problems into simpler parts. By understanding how one vector influences another, you can solve a wide range of challenges in fields like physics, engineering, and computer graphics. Whether you're working on calculating forces, designing structures, or creating realistic visuals, mastering vector projection makes these tasks much easier. 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15056	https://economics.stackexchange.com/questions/32928/difference-between-opportunity-cost-and-marginal-cost	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. current community your communities more stack exchange communities Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Difference between Opportunity cost and marginal cost I understand that the marginal cost is the cost of producing one additional unit of a good or service. Is it correct to say that the opportunity cost is part of the marginal cost (I'm trying to figure out what is meant by this picture below)? And secondly, is this statement correct "whatever a producer receive aboves its opportunity cost of providing the good/service is its producer surplus?" I read this online. Isn't it suggesting that the opportunity cost is the marginal benefit (because price received minus the points on the supply curve is producer surplus)? 1 Answer 1 Opportunity cost is the price of doing something in terms of something else. For example, cost of taking trip to Prague may be giving up new bike. In this broad sense marginal cost of producing one unit of q would be also it’s opportunity cost because you could use the same resources to produce something else. Opportunity cost of producing 1 widget at 5€ might be giving up possibility to produce 2 pins at 2,5€ each. What the quote says is not that only time has opportunity cost but that you should not forget to add the opportunity cost of the persons time to the costs. Also opportunity cost is not producer surplus. You don’t need to give up producer surplus to produce the particular good or service you are producing. You could include that in opportunity costs if you would be looking at it from a point of view of person deciding to leave the market to go to do some other business or retiring Your Answer Thanks for contributing an answer to Economics Stack Exchange! But avoid … Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. Sign up or log in Post as a guest Required, but never shown Post as a guest Required, but never shown By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answers Find the answer to your question by asking. Explore related questions See similar questions with these tags. Related Hot Network Questions Subscribe to RSS To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Economics Company Stack Exchange Network Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA . rev 2025.9.26.34547
15057	https://www.sciencedirect.com/science/article/pii/S1877065709002905	Skip to article My account Sign in View PDF Annals of Physical and Rehabilitation Medicine Volume 53, Issue 1, February 2010, Pages 34-41 Update article / Mise au point Post-polio syndrome: Pathophysiological hypotheses, diagnosis criteria, medication therapeuticsSyndrome post-poliomyÃ©litique : hypothÃ¨ses physiopathologiques, critÃ¨res diagnostiques, traitements mÃ©dicamenteux Author links open overlay panel, , , , , , , rights and content Under an Elsevier user license Open archive Abstract Post-polio syndrome (PPS) refers to a clinical disorder affecting polio survivors with sequelae years after the initial polio attack. These patients report new musculoskeletal symptoms, loss of muscular strength or endurance. PPS patients are tired, in pain and experience new and unusual muscular deficits, on healthy muscles as well as deficient muscles initially affected by the Poliovirus. Once a clinical diagnosis is established, the therapeutic options can be discussed. Some pathophysiological mechanisms have been validated by research studies on PPS (inflammatory process in cerebrospinal fluid [CSF] and cytokines of the immune system). Several studies have been conducted to validate medications (pyridostigmine, immunoglobulin, coenzyme Q10) or physical exercises protocols. This article focuses on the relevance and efficacy that can be expected from these therapeutics. Very few studies reported some improvements. Medications combined to individual and supervised exercise training programs are promising therapeutic strategies for PPS patients care management. RÃ©sumÃ© Le syndrome post-poliomyÃ©litique (SPP) est un diagnostic clinique. Il touche les personnes avec des sÃ©quelles de poliomyÃ©lite qui se plaignent de nouveaux troubles musculosquelettiques, dune perte de force ou dendurance musculaire. Un patient avec un SPP est fatiguÃ©, douloureux et perÃ§oit de nouvelles dÃ©ficiences musculaires inhabituelles, que ce soit sur des muscles sains ou dÃ©ficients Ã la suite de linfection par le Poliovirus. Une fois le diagnostic clinique Ã©tabli, les alternatives thÃ©rapeutiques peuvent Ãªtre discutÃ©es. Certains mÃ©canismes physiopathologiques ont Ã©tÃ© validÃ©s par des travaux de recherche (dysrÃ©gulation inflammatoire dans le liquide cÃ©phalorachidien et cytokines). Un certain nombre de travaux ont Ã©tÃ© menÃ©s pour valider des procÃ©dures thÃ©rapeutiques mÃ©dicamenteuses ou des protocoles dexercices physiques. Cet article fait le point sur lintÃ©rÃªt et lefficacitÃ© attendue de ces thÃ©rapeutiques. Peu dÃ©tudes thÃ©rapeutiques ont conclu Ã des effets positifs. Les traitements mÃ©dicamenteux combinÃ©s Ã des exercices physiques individualisÃ©s et supervisÃ©s apparaissent comme des stratÃ©gies thÃ©rapeutiques prometteuses pour la prise en charge de personnes atteintes dun SPP. Keywords Post-polio syndrome Therapeutic uses Diagnostic techniques Diagnosis Pathophysiology Mots clÃ©s Syndrome post-poliomyÃ©litique Ãtiologies PathogÃ©nie Traitements mÃ©dicamenteux Cited by (0) Copyright © 2009 Elsevier Masson SAS. All rights reserved.
15058	https://cs.stackexchange.com/questions/87225/maximizing-the-sum-of-adjacent-pairs-of-elements	optimization - Maximizing the sum of adjacent pairs of elements - Computer Science Stack Exchange Join Computer Science By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Maximizing the sum of adjacent pairs of elements Ask Question Asked 7 years, 8 months ago Modified7 years, 8 months ago Viewed 221 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. I encountered the following interesting problem on stackoverflow: Given numbers a(1)<⋯<a(n)a(1)<⋯<a(n), find a permutation π π that maximizes ∑i=1 n−1 a(π(i))a(π(i+1)).∑i=1 n−1 a(π(i))a(π(i+1)). The answers there claim that the answer doesn't depend on the numbers themselves, and is always given by the order …,a(n−5),a(n−3),a(n−1),a(n),a(n−2),a(n−4),……,a(n−5),a(n−3),a(n−1),a(n),a(n−2),a(n−4),… However, the proof (given in the comments) isn't convincing. Is the answer on stackoverflow correct? Can we prove it? optimization permutations Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Improve this question Follow Follow this question to receive notifications asked Jan 24, 2018 at 8:35 Yuval FilmusYuval Filmus 281k 27 27 gold badges 317 317 silver badges 515 515 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. The following answer is due to Reuven Bar-Yehuda. Let us call an order maximal if it maximizes the objective function ∑n−1 i=1 a(π(i))a(π(i+1))∑i=1 n−1 a(π(i))a(π(i+1)). Lemma 1. If π π is maximal then for all 1<i<j<n 1<i<j<n, π(i−1)<π(j+1)iff π(i)<π(j).π(i−1)<π(j+1)iff π(i)<π(j). Proof. Suppose that π(i−1)<π(j+1)π(i−1)<π(j+1) and π(i)>π(j)π(i)>π(j). Form the permutation σ σ by reversing the order all π(i),…,π(j)π(i),…,π(j). Then ∑k=1 n−1 a(σ(k))a(σ(k+1))−∑k=1 n−1 a(π(k))a(π(k+1))=a(σ(i−1))a(σ(i))+a(σ(j))a(σ(j+1))−a(π(i−1))a(π(i))−a(π(j))a(π(j+1))=a(π(i−1))a(π(j))+a(π(i))a(π(j+1))−a(π(i−1))a(π(i))−a(π(j))a(π(j+1))=(a(π(i−1))−a(π(j+1))(a(π(j))−a(π(i))>0.□∑k=1 n−1 a(σ(k))a(σ(k+1))−∑k=1 n−1 a(π(k))a(π(k+1))=a(σ(i−1))a(σ(i))+a(σ(j))a(σ(j+1))−a(π(i−1))a(π(i))−a(π(j))a(π(j+1))=a(π(i−1))a(π(j))+a(π(i))a(π(j+1))−a(π(i−1))a(π(i))−a(π(j))a(π(j+1))=(a(π(i−1))−a(π(j+1))(a(π(j))−a(π(i))>0.◻ Lemma 2. If π π is maximal then π(1)<π(i)π(1)<π(i) for all i≠n i≠n and π(n)<π(j)π(n)<π(j) for all j≠1 j≠1. Proof. Both claims have a similar proof, so we only prove the first. Suppose that π(1)>π(i)π(1)>π(i) for some i≠n i≠n. Form the permutation σ σ by reversing the order of π(1),…,π(i)π(1),…,π(i). Then ∑k=1 n−1 a(σ(k))a(σ(k+1))−∑k=1 n−1 a(π(k))a(π(k+1))=a(σ(i))a(σ(i+1))−a(π(i))a(π(i+1))=(a(π(1))−a(π(i))a(π(i+1))>0.□∑k=1 n−1 a(σ(k))a(σ(k+1))−∑k=1 n−1 a(π(k))a(π(k+1))=a(σ(i))a(σ(i+1))−a(π(i))a(π(i+1))=(a(π(1))−a(π(i))a(π(i+1))>0.◻ Theorem. If π π is maximal then π π is the following permutation or its reverse: 1,3,5,…,6,4,2.1,3,5,…,6,4,2. Proof. Lemma 2 shows that a(π(1))a(π(1)) and a(π(n))a(π(n)) are both smaller than a(π(2)),…,a(π(n−1))a(π(2)),…,a(π(n−1)), and so {π(1),π(n)}={1,2}{π(1),π(n)}={1,2}. Without loss of generality, assume that π(1)=1 π(1)=1 and π(n)=2 π(n)=2. Lemma 1, applied with i=2 i=2 and j=3,…,n−1 j=3,…,n−1, shows that π(2)π(2) is smaller than π(3),…,π(n−1)π(3),…,π(n−1), and so π(2)=3 π(2)=3. Lemma 1, applied with j=n−1 j=n−1 and i=3,…,n−2 i=3,…,n−2, shows that π(n−1)π(n−1) is smaller than π(3),…,π(n−2)π(3),…,π(n−2), and so π(n−1)=4 π(n−1)=4. Continuing in this way, we recover the rest of π π. □◻ Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jan 24, 2018 at 8:35 Yuval FilmusYuval Filmus 281k 27 27 gold badges 317 317 silver badges 515 515 bronze badges Add a comment\| Your Answer Thanks for contributing an answer to Computer Science Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. 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15059	https://www.aimspress.com/aimspress-data/math/2020/6/PDF/math-05-06-364.pdf	AIMS Mathematics, 5(6): 5685–5699. DOI: 10.3934/math.2020364 Received: 20 February 2020 Accepted: 30 June 2020 Published: 06 July 2020 Research article Explicit formulas for the p-adic valuations of Fibonomial coeﬃcients II Phakhinkon Phunphayap and Prapanpong Pongsriiam∗ Department of Mathematics, Faculty of Science, Silpakorn University, Nakhon Pathom, 73000, Thailand Correspondence: Email: prapanpong@gmail.com, pongsriiam p@silpakorn.edu. Abstract: In this article, we give explicit formulas for the p-adic valuations of the Fibonomial coeﬃcients pan n F for all primes p and positive integers a and n. This is a continuation from our previous article extending some results in the literature, which deal only with p = 2, 3, 5, 7 and a = 1. Then we use these formulas to characterize the positive integers n such that pn n F is divisible by p, where p is any prime which is congruent to ±2 (mod 5). Keywords: Fibonacci number; binomial coeﬃcient; Fibonomial coeﬃcient; p-adic valuation; p-adic order; divisibility Mathematics Subject Classiﬁcation: 11B39; 11B65; 11A63 1. Introduction The Fibonacci sequence (Fn)n≥1 is given by the recurrence relation Fn = Fn−1 + Fn−2 for n ≥3 with the initial values F1 = F2 = 1. For each m ≥1 and 1 ≤k ≤m, the Fibonomial coeﬃcients m k F is deﬁned by m k ! F = F1F2F3 · · · Fm (F1F2F3 · · · Fk)(F1F2F3 · · · Fm−k) = Fm−k+1Fm−k+2 · · · Fm F1F2F3 · · · Fk . Similar to the binomial coeﬃcients, we deﬁne m k F = 1 if k = 0 and m k F = 0 if k > m, and it is well-known that m k F is always an integer for every m ≥1 and k ≥0. Recently, there has been an increasing interest in the study of Fibonomial coeﬃcients. Marques and Trojovsk´ y [25, 26] start the investigation on the divisibility of Fibonomial coeﬃcients by determining the integers n ≥1 such that pn n F is divisible by p for p = 2, 3. Marques, Sellers, and Trojovsk´ y show that p divides pa+1 pa F for p ≡±2 (mod 5) and a ≥1. Marques and Trojovsk´ and Trojovsk´ y extend their results further and obtained the p-adic valuation of pa+1 pa F in . Then Ballot [2, Theorem 2] generalizes the Kummer-like theorem of Knuth and 5686 Wilf and uses it to give a generalization of Marques and Trojovsk´ y’s results. In particular, Ballot [2, Theorems 3.6, 5.2, and 5.3] ﬁnds all integers n such that p \| pn n U for any nondegenerate fundamental Lucas sequence U and p = 2, 3 and for p = 5, 7 in the case U = F. Phunphayap and Pongsriiam provide the most general formula for the p-adic valuation of Fibonomial coeﬃcients in the most general form m n F. For other recent results on the divisibility properties of the Fibonacci numbers, the Fibonomial coeﬃcients, and other combinatorial numbers, see for example [3–5, 11–13, 16, 17, 28, 30, 32–34, 37, 38, 41, 43]. For some identities involving Fibonomial coeﬃcients and generalizations, we refer the reader to the work of Kilic and his coauthors [7, 8, 18–21]. For the p-adic valuations of Eulerian, Bernoulli, and Stirling numbers, see [6, 9, 14, 23, 40]. Hence the relation p \| pan n F has been studied only in the case p = 2, 3, 5, 7 and a = 1. In this article, we extend the investigation on pan n F to the case of any prime p and any positive integer a. Replacing n by pa and pa by p, this becomes Marques and Trojovsk´ y’s results [27, 42]. Substituting a = 1, p ∈{2, 3, 5, 7}, and letting n be arbitrary, this reduces to Ballot’s theorems . So our results are indeed an extension of those previously mentioned. To obtain such the general result for all p and a, the calculation is inevitably long but we try to make it as simple as possible. As a reward, we can easily show in Corollaries 9 and 10 that 4n n F is odd if and only if n is a nonnegative power of 2, and 8n n F is odd if and only if n = (1 + 3 · 2k)/7 for some k ≡1 (mod 3). We organize this article as follows. In Section 2, we give some preliminaries and results which are needed in the proof of the main theorems. In Section 3, we calculate the p-adic valuation of pan n F for all a, p, and n, and use it to give a characterization of the positive integers n such that pan n F is divisible by p where p is any prime which is congruent to ±2 (mod 5). Remark that there also is an interesting pattern in the p-adic representation of the integers n such that pn n F is divisible by p. The proof is being prepared but it is a bit too long to include in this paper. We are trying to make it simpler and shorter and will publish it in the future. For more information and some recent articles related to the Fibonacci numbers, we refer the readers to [15,35,36,39] and references therein. 2. Preliminaries and lemmas Throughout this article, unless stated otherwise, x is a real number, p is a prime, a, b, k, m, n, q are integers, m, n ≥1, and q ≥2. The p-adic valuation (or p-adic order) of n, denoted by νp(n), is the exponent of p in the prime factorization of n. In addition, the order (or the rank) of appearance of n in the Fibonacci sequence, denoted by z(n), is the smallest positive integer m such that n \| Fm, ⌊x⌋is the largest integer less than or equal to x, {x} is the fractional part of x given by {x} = x −⌊x⌋, ⌈x⌉is the smallest integer larger than or equal to x, and a mod m is the least nonnegative residue of a modulo m. Furthermore, for a mathematical statement P, the Iverson notation [P] is deﬁned by [P] =        1, if P holds; 0, otherwise. We deﬁne sq(n) to be the sum of digits of n when n is written in base q, that is, if n = (akak−1 . . . a0)q = akqk + ak−1qk−1 + · · · + a0 where 0 ≤ai < q for every i, then sq(n) = ak + ak−1 + · · · + a0. Next, we recall some well-known and useful results for the reader’s convenience. AIMS Mathematics Volume 5, Issue 6, 5685–5699. 5687 Lemma 1. Let p , 5 be a prime. Then the following statements hold. (i) n \| Fm if and only if z(n) \| m (ii) z(p) \| p + 1 if and only if p ≡±2 (mod 5) and z(p) \| p −1, otherwise. (iii) gcd(z(p), p) = 1. Proof. These are well-known. See, for example, in [31, Lemma 1] for more details. □ Lemma 2. (Legendre’s formula) Let n be a positive integer and let p be a prime. Then νp(n!) = ∞ X k=1 $ n pk % = n −sp(n) p −1 . We will deal with a lot of calculations involving the ﬂoor function. So we recall the following results, which will be used throughout this article, sometimes without reference. Lemma 3. For k ∈Z and x ∈R, the following holds (i) ⌊k + x⌋= k + ⌊x⌋, (ii) {k + x} = {x}, (iii) ⌊x⌋+ ⌊−x⌋=        −1, if x < Z; 0, if x ∈Z, (iv) 0 ≤{x} < 1 and {x} = 0 if and only if x ∈Z. (v) ⌊x + y⌋=        ⌊x⌋+ ⌊y⌋, if {x} + {y} < 1; ⌊x⌋+ ⌊y⌋+ 1, if {x} + {y} ≥1, (vi) j ⌊x⌋ k k = j x k k for k ≥1. Proof. These are well-known and can be proved easily. For more details, see in [10, Chapter 3]. We also refer the reader to [1,29] for a nice application of these properties. □ The next three theorems given by Phunphayap and Pongsriiam are important tools for obtaining the main results of this article. Theorem 4. [31, Theorem 7] Let p be a prime, a ≥0, ℓ≥0, and m ≥1. Assume that p ≡±1 (mod m) and δ = [ℓ. 0 (mod m)] is the Iverson notation. Then νp $ℓpa m % ! ! =              ℓ(pa−1) m(p−1) −a n ℓ m o + νp j ℓ m k ! , if p ≡1 (mod m); ℓ(pa−1) m(p−1) −a 2δ + νp j ℓ m k ! , if p ≡−1 (mod m) and a is even; ℓ(pa−1) m(p−1) −a−1 2 δ − n ℓ m o + νp j ℓ m k ! , if p ≡−1 (mod m) and a is odd. Theorem 5. [31, Theorem 11 and Corollary 12] Let 0 ≤k ≤m be integers. Then the following statements hold. AIMS Mathematics Volume 5, Issue 6, 5685–5699. 5688 (i) Let A2 = ν2 j m 6 k ! −ν2 j k 6 k ! −ν2 j m−k 6 k ! . If r = m mod 6 and s = k mod 6, then ν2 m k ! F ! =                                  A2, if r ≥s and (r, s) , (3, 1), (3, 2), (4, 2); A2 + 1, if (r, s) = (3, 1), (3, 2), (4, 2); A2 + 3, if r < s and (r, s) , (0, 3), (1, 3), (2, 3), (1, 4), (2, 4), (2, 5); A2 + 2, if (r, s) = (0, 3), (1, 3), (2, 3), (1, 4), (2, 4), (2, 5). (ii) ν5 m k F = ν5 m k . (iii) Suppose that p is a prime, p , 2, and p , 5. If m′ = j m z(p) k , k′ = j k z(p) k , r = m mod z(p), and s = k mod z(p), then νp m k ! F ! = νp m′ k′ !! + [r < s] νp $m −k + z(p) z(p) %! + νp(Fz(p)) ! . Theorem 6. [31, Theorem 13] Let a, b, ℓ1, and ℓ2 be positive integers and b ≥a. For each p , 5, assume that ℓ1pb > ℓ2pa and let mp = j ℓ1pb−a z(p) k and kp = j ℓ2 z(p) k . Then the following statements hold. (i) If a ≡b (mod 2), then ν2 ℓ12b ℓ22a F is equal to                      ν2 m2 k2 , if ℓ1 ≡ℓ2 (mod 3) or ℓ2 ≡0 (mod 3); a + 2 + ν2 (m2 −k2) + ν2 m2 k2 , if ℓ1 ≡0 (mod 3) and ℓ2 . 0 (mod 3); l a 2 m + 1 + ν2 (m2 −k2) + ν2 m2 k2 , if ℓ1 ≡1 (mod 3) and ℓ2 ≡2 (mod 3); l a+1 2 m + ν2 m2 k2 , if ℓ1 ≡2 (mod 3) and ℓ2 ≡1 (mod 3), and if a . b (mod 2), then ν2 ℓ12b ℓ22a F is equal to                      ν2 m2 k2 , if ℓ1 ≡−ℓ2 (mod 3) or ℓ2 ≡0 (mod 3); a + 2 + ν2 (m2 −k2) + ν2 m2 k2 , if ℓ1 ≡0 (mod 3) and ℓ2 . 0 (mod 3); l a+1 2 m + ν2 m2 k2 , if ℓ1 ≡1 (mod 3) and ℓ2 ≡1 (mod 3); l a 2 m + 1 + ν2 (m2 −k2) + ν2 m2 k2 , if ℓ1 ≡2 (mod 3) and ℓ2 ≡2 (mod 3). (ii) Let p , 5 be an odd prime and let r = ℓ1pb mod z(p) and s = ℓ2pa mod z(p). If p ≡±1 (mod 5), then νp ℓ1pb ℓ2pa ! F ! = [r < s] a + νp mp −kp + νp(Fz(p)) + νp mp kp !! , AIMS Mathematics Volume 5, Issue 6, 5685–5699. 5689 and if p ≡±2 (mod 5), then νp ℓ1pb ℓ2pa F is equal to                                                                              νp mp kp , if r = s or ℓ2 ≡0 (mod z(p)); a + νp(Fz(p)) + νp mp −kp + νp mp kp , if ℓ1 ≡0 (mod z(p)) and ℓ2 . 0 (mod z(p)); a 2 + νp mp kp , if r > s, ℓ1, ℓ2 . 0 (mod z(p)), and a is even; a 2 + νp(Fz(p)) + νp mp −kp + νp mp kp , if r < s, ℓ1, ℓ2 . 0 (mod z(p)), and a is even; a+1 2 + νp mp −kp + νp mp kp , if r > s, ℓ1, ℓ2 . 0 (mod z(p)), and a is odd; a−1 2 + νp(Fz(p)) + νp mp kp , if r < s, ℓ1, ℓ2 . 0 (mod z(p)), and a is odd. In fact, Phunphayap and Pongsriiam obtain other results analogous to Theorems 5 and 6 too but we do not need them in this article. 3. Main results We begin with the calculation of the 2-adic valuation of 2an n F and then use it to determine the integers n such that 2n n F, 4n n F, 8n n F are even. Then we calculate the p-adic valuation of pan n F for all odd primes p. For binomial coeﬃcients, we know that ν2 2n n = s2(n). For Fibonomial coeﬃcients, we have the following result. Theorem 7. Let a and n be positive integers, ε = [n . 0 (mod 3)], and A = j (2a−1)n 3·2ν2(n) k . Then the following statements hold. (i) If a is even, then ν2 2an n ! F ! = δ + A −a 2ε −ν2(A!) = δ + s2(A) −a 2ε, (3.1) where δ = [n mod 6 = 3, 5]. In other words, δ = 1 if n ≡3, 5 (mod 6) and δ = 0 otherwise. (ii) If a is odd, then ν2 2an n ! F ! = δ + A −a −1 2 ε −ν2(A!) = δ + s2(A) −a −1 2 ε, (3.2) where δ = (n mod 6)−1 2 [2 ∤n] + l ν2(n)+3−n mod 3 2 m [n mod 6 = 2, 4]. In other words, δ = (n mod 6)−1 2 if n is odd, δ = 0 if n ≡0 (mod 6), δ = l ν2(n) 2 m + 1 if n ≡4 (mod 6), and δ = l ν2(n)+1 2 m if n ≡2 (mod 6). AIMS Mathematics Volume 5, Issue 6, 5685–5699. 5690 Proof. The second equalities in (3.1) and (3.2) follow from Legendre’s formula. So it remains to prove the ﬁrst equalities in (3.1) and (3.2). To prove (i), we suppose that a is even and divide the consideration into two cases. Case 1. 2 ∤n. Let r = 2an mod 6 and s = n mod 6. Then s ∈{1, 3, 5}, r ≡2an ≡4n ≡4s (mod 6), and therefore (r, s) = (4, 1), (0, 3), (2, 5). In addition, A = j (2a−1)n 3 k = (2a−1)n 3 and δ = [s = 3, 5]. By Theorem 5(i), the left–hand side of (3.1) is A2 if s = 1 and A2 + 2 if s = 3, 5, where A2 = ν2 j 2an 6 k ! −ν2 j n 6 k ! −ν2 j(2a−1)n 6 k ! . We obtain by Theorem 4 that ν2 $2an 6 % ! ! = ν2 $2a−1n 3 % ! ! = (2a−1 −1)n 3 −a −2 2 ε − n 3 + ν2 n 3 ! . By Legendre’s formula and Lemma 3, we have ν2 n 6 ! = ν2 n 3 ! − n 6 , ν2 $(2a −1)n 6 % ! ! = ν2 $(2a −1)n 3 % ! ! − $(2a −1)n 6 % = ν2(A!) − $(2a −1)n 6 % , n 6 + $(2a −1)n 6 % = n −s 6 + 2an −r 6 −n −s 6 + r −s 6 = 2an −r 6 −[s ∈{3, 5}]. From the above observation, we obtain A2 = (2a−1 −1)n 3 −a −2 2 ε − n 3 + 2an −r 6 −[s ∈{3, 5}] −ν2(A!) = A −a −2 2 ε − n 3 −r 6 −[s ∈{3, 5}] −ν2(A!) =              A −a 2 −ν2(A!), if s = 1; A −ν2(A!) −1, if s = 3; A −a 2 −ν2(A!) −1, if s = 5. It is now easy to check that A2 (if s = 1), A2 + 2 (if s = 3, 5) are the same as δ + A −a 2ε −ν2(A!) in (3.1). So (3.1) is veriﬁed. Case 2. 2 \| n. We write n = 2bℓwhere 2 ∤ℓand let m = j 2aℓ 3 k , k = j ℓ 3 k , r = 2aℓmod 3, and s = ℓmod 3. Since a is even, r = s. Then we apply Theorem 6(i) to obtain ν2 2an n ! F ! = ν2 ℓ2a+b ℓ2b ! F ! = ν2 m k !! = ν2(m!) −ν2(k!) −ν2((m −k)!). (3.3) We see that ℓ. 0 (mod 3) if and only if n . 0 (mod 3). In addition, A = (2a−1)ℓ 3 and δ = 0. By Theorem 4, we have ν2(m!) = A −a 2ε + ν2(k!). In addition, m −k = $2aℓ 3 % − $ℓ 3 % = 2aℓ−r 3 −ℓ−s 3 = 2aℓ−ℓ 3 = A. AIMS Mathematics Volume 5, Issue 6, 5685–5699. 5691 So ν2((m −k)!) = ν2(A!). Substituting these in (3.3), we obtain (3.1). This completes the proof of (i). To prove (ii), we suppose that a is odd and divide the proof into two cases. Case 1. 2 ∤n. This case is similar to Case 1 of the previous part. So we let r = 2an mod 6 and s = n mod 6. Then s ∈{1, 3, 5}, r ≡2an ≡2n ≡2s (mod 6), (r, s) = (2, 1), (0, 3), (4, 5), δ = s−1 2 , and the left–hand side of (3.2) is A2 if s = 1, A2 + 2 if s = 3, and A2 + 3 if s = 5, where A2 = ν2 j 2an 6 k ! −ν2 j n 6 k ! −ν2 j (2a−1)n 6 k ! . In addition, we have ν2 $2an 6 % ! ! = (2a−1 −1)n 3 −a −1 2 ε + ν2 n 3 ! , ν2 n 6 ! = ν2 n 3 ! − n 6 , ν2 $(2a −1)n 6 % ! ! = ν2(A!) − $(2a −1)n 6 % , n 6 + $(2a −1)n 6 % = 2an −r 6 −[s ∈{3, 5}]. Therefore A2 = (2a−1 −1)n 3 −a −1 2 ε + 2an −r 6 −[s ∈{3, 5}] −ν2(A!). Furthermore, A = $(2a −1)n 3 % = 2an −r 3 −n −s 3 + r −s 3 =              (2a−1)n 3 −1 3, if s = 1; (2a−1)n 3 , if s = 3; (2a−1)n 3 −2 3, if s = 5, which implies that A = (2a−1)n 3 −r 6. Then A2 = A −a −1 2 ε −[s ∈{3, 5}] −ν2(A!). It is now easy to check that A2 (if s = 1), A2 + 2 (if s = 3), and A2 + 3 (if s = 5), are the same as δ + A −a−1 2 ε −ν2(A!) in (3.2). So (3.2) is veriﬁed. Case 2. 2 \| n. This case is similar to Case 2 of the previous part. So we write n = 2bℓwhere 2 ∤ℓand let m = j 2aℓ 3 k , k = j ℓ 3 k , r = 2aℓmod 3, and s = ℓmod 3. We obtain by Theorem 6 that ν2 2an n F is equal to ν2 ℓ2a+b ℓ2b ! F ! =              ν2 m k , if ℓ≡0 (mod 3); l b+1 2 m + ν2 m k , if ℓ≡1 (mod 3); l b 2 m + 1 + ν2(m −k) + ν2 m k , if ℓ≡2 (mod 3). (3.4) By Theorem 4, we have ν2(m!) = (2a −1)ℓ 3 −a −1 2 ε − (ℓ 3 ) + ν2(k!). AIMS Mathematics Volume 5, Issue 6, 5685–5699. 5692 Since (2a −1)ℓ≡ℓ(mod 3), n(2a−1)ℓ 3 o = n ℓ 3 o . This implies that ν2(m!) = A −a−1 2 ε + ν2(k!). In addition, (r, s) = (0, 0), (2, 1), (1, 2), and m −k = $2aℓ 3 % − $ℓ 3 % = 2aℓ−r 3 −ℓ−s 3 = (2a −1)ℓ−(r −s) 3 = A + [s = 2]. From the above observation, we obtain ν2 m k !! = ν2(m!) −ν2(k!) −ν2((m −k)!) =        A −a−1 2 ε −ν2(A!), if s = 0, 1; A −a−1 2 ε −ν2((A + 1)!), if s = 2. Substituting this in (3.4), we see that ν2 2an n ! F ! =              A −ν2(A!), if ℓ≡0 (mod 3); l b+1 2 m + A −a−1 2 −ν2(A!), if ℓ≡1 (mod 3); l b 2 m + 1 + A −a−1 2 −ν2(A!), if ℓ≡2 (mod 3). (3.5) Recall that n = 2bℓ≡(−1)bℓ(mod 3). So (3.5) implies that ν2 2an n ! F ! =                            A −ν2(A!), if n ≡0 (mod 3); b 2 + 1 + A −a−1 2 −ν2(A!), if n ≡1 (mod 3) and b is even; b+1 2 + 1 + A −a−1 2 −ν2(A!), if n ≡1 (mod 3) and b is odd; b 2 + 1 + A −a−1 2 −ν2(A!), if n ≡2 (mod 3) and b is even; b+1 2 + A −a−1 2 −ν2(A!), if n ≡2 (mod 3) and b is odd, which is the same as (3.2). This completes the proof. □ We can obtain the main result of Maques and Trojovsk´ y as a corollary. Corollary 8. (Marques and Trojovsk´ y ) 2n n F is even for all n ≥2. Proof. Let n ≥2 and apply Theorem 7 with a = 1 to obtain ν2 2n n F = δ + s2(A). If n . 0, 1 (mod 6), then δ > 0. If n ≡0 (mod 6), then n ≥3 · 2ν2(n), and so A ≥1 and s2(A) > 0. If n ≡1 (mod 6), then A = j n 3 k > 1 and so s2(A) > 0. In any case, ν2 2n n F > 0. So 2n n F is even. □ Corollary 9. Let n ≥2. Then 4n n F is even if and only if n is not a power of 2. In other words, for each n ∈N, 4n n F is odd if and only if n = 2k for some k ≥0. Proof. Let δ, ε, and A be as in Theorem 7. If n = 2k for some k ≥1, then we apply Theorem 7 with a = 2, δ = 0, ε = 1, A = 1 leading to ν2 4n n F = 0, which implies that 4n n F is odd. Suppose n is not a power of 2. By Theorem 7, ν2 4n n F = δ + s2(A) −ε ≥s2(A) −1. Since n is not a power of 2, the sum s2(n) ≥2. It is easy to see that s2(m) = s2(2cm) for any c, m ∈N. Therefore s2(A) = s2 n 2ν2(n) = s2 2ν2(n) · n 2ν2(n) = s2(n) ≥2, which implies ν2 4n n F ≥1, as required. □ Observe that 2, 22, 23 are congruent to 2, 4, 1 (mod 7), respectively. This implies that if k ≥1 and k ≡1 (mod 3), then (1 + 3 · 2k)/7 is an integer. We can determine the integers n such that 8n n F is odd as follows. AIMS Mathematics Volume 5, Issue 6, 5685–5699. 5693 Corollary 10. 8n n F is odd if and only if n = 1+3·2k 7 for some k ≡1 (mod 3). Proof. Let a, δ, A, ε be as in Theorem 7. We ﬁrst suppose n = (1 + 3 · 2k)/7 where k ≥1 and k ≡1 (mod 3). Then n ≡7n ≡1+3·2k ≡1 (mod 6). Then a = 3, ε = 1, δ = 0, A = 2k, and so ν2 8n n F = 0. Therefore 8n n F is odd. Next, assume that 8n n F is odd. Observe that A ≥2 and s2(A) > 0. If n ≡0 (mod 3), then ε = 0 and ν2 8n n F = δ + s2(A) > 0, which is not the case. Therefore n ≡1, 2 (mod 3), and so ε = 1. If n ≡0 (mod 2), then δ = l ν2(n)+3−n mod 3 2 m ≥1, and so 8n n F ≥s2(A) > 0, which is a contradiction. So n ≡1 (mod 2). This implies n ≡1, 5 (mod 6). But if n ≡5 (mod 6), then δ ≥2 and ν2 8n n F > 0, a contradiction. Hence n ≡1 (mod 6). Then δ = 0. Since s2(A) −1 = ν2 8n n F = 0, we see that A = 2k for some k ≥1. Then 7n−1 3 = j 7n 3 k = A = 2k, which implies n = 1+3·2k 7 , as required. □ Theorem 11. For each a, n ∈N, ν5 5an n F = ν5 5an n = s5((5a−1)n) 4 . In particular, 5an n F is divisible by 5 for every a, n ∈N. Proof. The ﬁrst equality follows immediately from Theorem 5(ii). By Legendre’s formula, ν5 n k = s5(k)+s5(n−k)−s5(n) 4 for all n ≥k ≥1. So ν5 5an n F is s5(n) + s5(5an −n) −s5(5an) 4 = s5((5a −1)n) 4 . □ Theorem 12. Let p , 2, 5, a, n ∈N, r = pan mod z(p), s = n mod z(p), and A = j n(pa−1) pνp(n)z(p) k . Then the following statements hold. (i) If p ≡±1 (mod 5), then νp pan n F is equal to A p −1 −a ( n pνp(n)z(p) ) −νp(A!) = sp(A) p −1 −a ( n pνp(n)z(p) ) . (3.6) (ii) If p ≡±2 (mod 5) and a is even, then νp pan n F is equal to A p −1 −a 2[s , 0] −νp(A!) = sp(A) p −1 −a 2[s , 0]. (3.7) (iii) If p ≡±2 (mod 5) and a is odd, then νp pan n F is equal to $ A p −1 % −a −1 2 [s , 0] −νp(A!) + δ, (3.8) where δ = j νp(n) 2 k + [2 ∤νp(n)][r > s] + [r < s]νp(Fz(p)) [r , s], or equivalently, δ = 0 if r = s, δ = j νp(n) 2 k + νp(Fz(p)) if r < s, and δ = l νp(n) 2 m if r > s. Proof. We ﬁrst prove (i) and (ii). So we suppose that the hypothesis of (i) or (ii) is true. By writing νp(A!) = A−sp(A) p−1 , we obtain the equalities in (3.6) and (3.7). By Lemma 1(ii), pa ≡1 (mod z(p)). Then r = s. AIMS Mathematics Volume 5, Issue 6, 5685–5699. 5694 Case 1. p ∤n. Let m = j pan z(p) k and k = j n z(p) k . Then we obtain by Theorem 5(iii) that νp pan n ! F ! = νp m k !! = νp(m!) −νp(k!) −νp((m −k)!). (3.9) By Lemma 1(ii) and Theorem 4, we see that if p ≡±1 (mod 5), then p ≡1 (mod z(p)) and νp(m!) = νp $ npa z(p) % ! ! = n(pa −1) z(p)(p −1) −a ( n z(p) ) + νp (k!) , (3.10) and if p ≡±2 (mod 5) and a is even, then p ≡−1 (mod z(p)) and νp(m!) = n(pa −1) z(p)(p −1) −a 2[s , 0] + νp (k!) . (3.11) Since z(p) \| pa −1 and p ∤n, A = n(pa−1) z(p) . Therefore m −k = $ pan z(p) % − $ n z(p) % = pan −r z(p) −n −s z(p) = n(pa −1) z(p) = A. (3.12) Substituting (3.10), (3.11), and (3.12) in (3.9), we obtain (3.6) and (3.7). Case 2. p \| n. Let n = pbℓwhere p ∤ℓ, m = j ℓpa z(p) k , and k = j ℓ z(p) k . Since r = s, we obtain by Theorem 6 that νp pan n F is equal to νp ℓpa+b ℓpb ! F ! = νp m k !! = νp(m!) −νp(k!) −νp((m −k)!). (3.13) Since gcd(p, z(p)) = 1, we see that ℓ≡0 (mod z(p)) ⇔n ≡0 (mod z(p)) ⇔s = 0. Similar to Case 1, we have νp(m!) = ℓ(pa−1) z(p)(p−1) −a n ℓ z(p) o + νp(k!) if p ≡±1 (mod 5), νp(m!) = ℓ(pa−1) z(p)(p−1) −a 2[s , 0] + νp (k!) if p ≡±2 (mod 5) and a is even, ℓpa ≡ℓ(mod z(p)), A = ℓ(pa−1) z(p) , and m −k = A. So (3.13) leads to (3.6) and (3.7). This proves (i) and (ii). To prove (iii), suppose that p ≡±2 (mod 5) and a is odd. By Lemma 1(ii), p ≡−1 (mod z(p)). In addition, pa−1 p−1 = pa−1 + pa−2 + . . . + 1 ≡1 (mod z(p)). We divide the consideration into two cases. Case 1. p ∤n. This case is similar to Case 1 of the previous part. So we apply Theorems 4 and 5(iii). Let m = j pan z(p) k and k = j n z(p) k . Then νp(m!) = n(pa −1) z(p)(p −1) −a −1 2 [s , 0] − ( n z(p) ) + νp(k!), m −k = pan −r z(p) −n −s z(p) = n(pa −1) −(r −s) z(p) , A = $npa −r z(p) −n −s z(p) + r −s z(p) % = m −k + $r −s z(p) % . Since pa−1 p−1 ≡1 (mod z(p)), n(pa−1) p−1 ≡n (mod z(p)). This implies that n n(pa−1) z(p)(p−1) o = n n z(p) o . Therefore νp(m!) = $ n(pa −1) z(p)(p −1) % −a −1 2 [s , 0] + νp(k!) = $ A p −1 % −a −1 2 [s , 0] + νp(k!). AIMS Mathematics Volume 5, Issue 6, 5685–5699. 5695 From the above observation, if r ≥s, then A = m −k and νp pan n ! F ! = νp m k !! = $ A p −1 % −a −1 2 [s , 0] −νp(A!), which leads to (3.8). If r < s, then A = m −k −1, j pan−n+z(p) z(p) k = A + 1, and νp pan n F is equal to $ A p −1 % −a −1 2 [s , 0] −νp((A + 1)!) + νp(A + 1) + νp(Fz(p)) = $ A p −1 % −a −1 2 [s , 0] −νp(A!) + νp(Fz(p)), which is the same as (3.8). Case 2. p \| n. Let n = pbℓwhere p ∤ℓ, m = j ℓpa z(p) k , and k = j ℓ z(p) k . Similar to Case 1, s = 0 ⇔ℓ≡0 (mod z(p)). In addition, ℓ(pa−1) p−1 ≡ℓ(mod z(p)), and so we obtain by Theorem 4 that νp(m!) = j A p−1 k − a−1 2 [s , 0]+νp(k!). The calculation of νp pan n F = νp ℓpa+b ℓpb F is done by the applications of Theorem 6 and is divided into several cases. Suppose r = s. Then pa+bℓ≡pan ≡r ≡s ≡n ≡pbℓ(mod z(p)). Since (p, z(p)) = 1, this implies ℓpa ≡ℓ(mod z(p)). Therefore A = j ℓpa−ℓ z(p) k = ℓpa−ℓ z(p) = m −k and νp pan n ! F ! = νp m k !! = νp(m!) −νp(k!) −νp((m −k)!), which is (3.8). Obviously, if ℓ≡0 (mod z(p)), then r = s, which is already done. So from this point on, we assume that r , s and ℓ. 0 (mod z(p)). Recall that p ≡−1 (mod z(p)) and a is odd. So if b is odd, then r ≡npa ≡−n ≡−pbℓ≡ℓ (mod z(p)), s ≡n ≡pbℓ≡−ℓ≡ℓpa (mod z(p)), and A = $ℓpa −s z(p) −ℓ−r z(p) + s −r z(p) % = ℓpa −s z(p) −ℓ−r z(p) + $s −r z(p) % = m −k + $s −r z(p) % . Similarly, if b is even, then r = ℓpa mod z(p), s = ℓmod z(p), and A = m −k + j r−s z(p) k . Let R = j A p−1 k −a−1 2 [s , 0] −νp(A!) + δ be the quantity in (3.8). From the above observation and the application of Theorem 6, we obtain νp pan n F as follows. If r > s and b is even, then A = m −k and νp pan n ! F ! = b 2 + νp m k !! = b 2 + $ A p −1 % −a −1 2 [s , 0] −νp(A!) = R. If r > s and b is odd, then A = m −k −1 and νp pan n ! F ! = b + 1 2 + νp(A + 1) + νp m k !! = b + 1 2 + $ A p −1 % −a −1 2 [s , 0] −νp(A!) = R. AIMS Mathematics Volume 5, Issue 6, 5685–5699. 5696 If r < s and b is even, then A = m −k −1 and νp pan n ! F ! = b 2 + νp Fz(p) + νp(A + 1) + νp m k !! = b 2 + νp(Fz(p)) + $ A p −1 % −a −1 2 [s , 0] −νp(A!) = R. If r < s and b is odd, then A = m −k and νp pan n ! F ! = b −1 2 + νp Fz(p) + νp m k !! = b −1 2 + νp(Fz(p)) + $ A p −1 % −a −1 2 [s , 0] −νp(A!) = R. This completes the proof. □ In the next two corollaries, we give some characterizations of the integers n such that pan n F is divisible by p. Corollary 13. Let p be a prime and let a and n be positive integers. If n ≡0 (mod z(p)), then p \| pan n F. Proof. We ﬁrst consider the case p , 2, 5. Assume that n ≡0 (mod z(p)) and r, s, A, and δ are as in Theorem 12. Then n pνp(n)z(p), A p−1 ∈Z, r = s = 0, and δ = 0. Every case in Theorem 12 leads to νp pan n F = sp(A) p−1 > 0, which implies p \| pan n F. If p = 5, then the result follows immediately from Theorem 11. If p = 2, then every case of Theorem 7 leads to ν2 2an n F ≥s2(A) > 0, which implies the desired result. □ Corollary 14. Let p , 2, 5 be a prime and let a, n, r, s, and A be as in Theorem 12. Assume that p ≡±2 (mod 5) and n . 0 (mod z(p)). Then the following statements hold. (i) Assume that a is even. Then p \| pan n F if and only if sp(A) > a 2(p −1). (ii) Assume that a is odd and p ∤n. If r < s, then p \| pan n F. If r ≥s, then p \| pan n F if and only if sp(A) ≥a+1 2 (p −1). (iii) Assume that a is odd and p \| n. If r , s, then p \| pan n F. If r = s, then p \| pan n F if and only if sp(A) ≥a+1 2 (p −1). Proof. We use Lemmas 2 and 3 repeatedly without reference. For (i), we obtain by (3.7) that νp pan n ! F ! = sp(A) p −1 −a 2, which is positive if and only if sp(A) > a 2(p −1). This proves (i). To prove (ii) and (iii), we let δ be as in Theorem 12 and divide the consideration into two cases. Case 1. p ∤n. If r < s, then we obtain by Theorem 5(iii) that νp pan n F ≥νp(Fz(p)) ≥1. Suppose r ≥s. Then δ = 0 and (3.8) is $ A p −1 % −a −1 2 −νp(A!) = $ A p −1 % −a −1 2 −A −sp(A) p −1 = sp(A) p −1 − ( A p −1 ) −a −1 2 . AIMS Mathematics Volume 5, Issue 6, 5685–5699. 5697 If sp(A) ≥a+1 2 (p −1), then (3.8) implies that νp pan n ! F ! ≥1 − ( A p −1 ) > 0. Similarly, if sp(A) < a+1 2 (p −1), then νp pan n F < 1 − n A p−1 o ≤1. This proves (ii). Case 2. p \| n. We write n = pbℓwhere p ∤ℓ. Then b ≥1. Recall that νp(Fz(p)) ≥1. If r , s, then Theorem 6 implies that νp pan n ≥ b 2 if b is even and it is ≥ b+1 2 if b is odd. In any case, νp pan n F ≥1. So p \| pan n F. If r = s, then δ = 0 and we obtain as in Case 1 that p \| pan n F if and only if sp(A) ≥a+1 2 (p −1). This proves (iii). □ Corollary 15. Let p , 2, 5 be a prime and let A = n(p−1) pνp(n)z(p). Assume that p ≡±1 (mod 5). Then p \| pn n F if and only if sp(A) ≥p −1. Proof. We remark that by Lemma 1(ii), A is an integer. Let x = n pνp(n)z(p). We apply Theorem 12(i) with a = 1. If sp(A) ≥p −1, then (3.6) implies that νp pn n F ≥1 −{x} > 0. If sp(A) < p −1, then νp pn n F < 1 −{x} ≤1. This completes the proof. □ 4. Conclusions We give exact formulas for the p-adic valuations of Fibonomial coeﬃcients of the form pan n F for all primes p and a, n ∈N. Then we use it to characterize the integers n such that pan n F is divisible by p. Acknowledgments Phakhinkon Phunphayap receives a scholarship from Science Achievement Scholarship of Thailand(SAST). This research was jointly supported by the Thailand Research Fund and the Faculty of Science Silpakorn University, grant number RSA5980040. Conﬂict of interest The authors declare that there is no conﬂict of interests regarding the publication of this article. References 1. S. Aursukaree, T. Khemaratchatakumthorn, P. Pongsriiam, Generalizations of Hermite’s identity and applications, Fibonacci Quart., 57 (2019), 126–133. 2. C. Ballot, Divisibility of Fibonomials and Lucasnomials via a general Kummer rule, Fibonacci Quart., 53 (2015), 194–205. 3. C. Ballot, The congruence of Wolstenholme for generalized binomial coeﬃcients related to Lucas sequences, J. Integer Seq., 18 (2015), Article 15.5.4. AIMS Mathematics Volume 5, Issue 6, 5685–5699. 5698 4. C. Ballot, Lucasnomial Fuss-Catalan numbers and related divisibility questions, J. Integer Seq., 21 (2018), Article 18.6.5. 5. C. Ballot, Divisibility of the middle Lucasnomial coeﬃcient, Fibonacci Quart., 55 (2017), 297–308. 6. F. N. Castro, O. E. Gonz´ alez, L. A. Medina, The p-adic valuation of Eulerian numbers: Trees and Bernoulli numbers, Experiment. Math., 2 (2015), 183–195. 7. W. Chu, E. Kilic, Quadratic sums of Gaussian q-binomial coeﬃcients and Fibonomial coeﬃcients, Ramanujan J., 51 (2020), 229–243. 8. W. Chu, E. Kilic, Cubic sums of q-binomial coeﬃcients and the Fibonomial coeﬃcients, Rocky Mountain J. Math., 49 (2019), 2557–2569. 9. Y. L. Feng, M. Qiu. Some results on p-adic valuations of Stirling numbers of the second kind, AIMS Math., 5 (2020), 4168–4196. 10. R. L. Graham, D. E. Knuth, O. Patashnik, Concrete Mathematics: A Foundation for Computer Science, Second Edition, Addison–Wesley, 1994. 11. V. J. W. Guo, Proof of a generalization of the (B.2) supercongruence of Van Hamme through a q-microscope, Adv. Appl. Math., 116 (2020), Art. 102016. 12. V. J. W. Guo, J. C. Liu, Some congruences related to a congruence of Van Hamme, Integral Transforms Spec. Funct., 31 (2020), 221–231. 13. V. J. W. Guo, W. Zudilin, A common q-analogue of two supercongruences, Results Math., 75(2020), Art. 46. 14. S. F. Hong, M. Qiu, On the p-adic properties of Stirling numbers of the ﬁrst kind, Acta Math. Hungar., 161 (2020), 366–395. 15. M. Jaidee, P. Pongsriiam, Arithmetic functions of Fibonacci and Lucas numbers, Fibonacci Quart., 57 (2019), 246–254. 16. N. Khaochim, P. Pongsriiam, The general case on the order of appearance of product of consecutive Lucas numbers, Acta Math. Univ. Comenian., 87 (2018), 277–289. 17. N. Khaochim, P. Pongsriiam, On the order of appearance of product of Fibonacci numbers, Contrib. Discrete Math., 13 (2018), 45–62. 18. E. Kilic, I. Akkus, On Fibonomial sums identities with special sign functions: Analytically q-calculus approach, Math. Slovaca, 68 (2018), 501–512. 19. E. Kilic, I. Akkus, H. Ohtsuka, Some generalized Fibonomial sums related with the Gaussian q-binomial sums, Bull. Math. Soc. Sci. Math. Roumanie, 55 (2012), 51–61. 20. E. Kilic, H. Prodinger, Closed form evaluation of sums containing squares of Fibonomial coeﬃcients, Math. Slovaca, 66 (2016), 757–767. 21. E. Kilic, H. Prodinger, Evaluation of sums involving products of Gaussian q-binomial coeﬃcients with applications, Math. Slovaca, 69 (2019), 327–338. 22. D. Knuth, H. Wilf, The power of a prime that divides a generalized binomial coeﬃcient, J. Reine Angew. Math., 396 (1989), 212–219. 23. T. Komatsu, P. Young, Exact p-adic valuations of Stirling numbers of the ﬁrst kind, J. Number Theory, 177 (2017), 20–27. 24. D. Marques, J. A. Sellers, P. Trojovsk´ y, On divisibility properties of certain Fibonomial coeﬃcients by a prime, Fibonacci Quart., 51 (2013), 78–83. 25. D. Marques, P. Trojovsk´ y, On parity of Fibonomial coeﬃcients, to appear in Util. Math. AIMS Mathematics Volume 5, Issue 6, 5685–5699. 5699 26. D. Marques, P. Trojovsk´ y, On divisibility of Fibonomial coeﬃcients by 3, J. Integer Seq., 15 (2012), Article 12.6.4. 27. D. Marques, P. Trojovsk´ y, The p-adic order of some Fibonomial coeﬃcients, J. Integer Seq., 18 (2015), Article 15.3.1 28. K. Onphaeng, P. Pongsriiam, Subsequences and divisibility by powers of the Fibonacci numbers, Fibonacci Quart., 52 (2014), 163–171. 29. K. Onphaeng, P. Pongsriiam, Jacobsthal and Jacobsthal-Lucas numbers and sums introduced by Jacobsthal and Tverberg, J. Integer Seq., 20 (2017), Article 17.3.6. 30. K. Onphaeng, P. Pongsriiam, The converse of exact divisibility by powers of the Fibonacci and Lucas numbers, Fibonacci Quart., 56 (2018), 296–302. 31. P. Phunphayap, P. Pongsriiam, Explicit formulas for the p-adic valuations of Fibonomial coeﬃcients, J. Integer Seq., 21 (2018), Article 18.3.1. 32. P. Pongsriiam, Exact divisibility by powers of the Fibonacci and Lucas number, J. Integer Seq., 17 (2014), Article 14.11.2. 33. P. Pongsriiam, A complete formula for the order of appearance of the powers of Lucas numbers, Commun. Korean Math. Soc., 31 (2016), 447–450. 34. P. Pongsriiam, Factorization of Fibonacci numbers into products of Lucas numbers and related results, JP J. Algebra Number Theory Appl., 38 (2016), 363–372. 35. P. Pongsriiam, Fibonacci and Lucas numbers associated with Brocard-Ramanujan equation, Commun. Korean Math. Soc., 32 (2017), 511–522. 36. P. Pongsriiam, Fibonacci and Lucas Numbers which are one away from their products, Fibonacci Quart., 55 (2017), 29–40. 37. P. Pongsriiam, Fibonacci and Lucas numbers which have exactly three prime factors and some unique properties of F18 and L18, Fibonacci Quart., 57 (2019), 130–144. 38. P. Pongsriiam, The order of appearance of factorials in the Fibonacci sequence and certain Diophantine equations, Period. Math. Hungar., 79 (2019), 141–156. 39. B. Prempreesuk, P. Noppakaew, P. Pongsriiam, Zeckendorf representation and multiplicative inverse of Fm mod Fn, Int. J. Math. Comput. Sci., 15(2020), 17–25. 40. M. Qiu, S. F. Hong, 2-Adic valuations of Stirling numbers of the ﬁrst kind, Int. J. Number Theory, 15 (2019), 1827–1855. 41. Z. W. Sun, Fibonacci numbers modulo cubes of prime, Taiwanese J. Math., 17 (2013), 1523–1543. 42. P. Trojovsk´ y, The p-adic order of some Fibonomial coeﬃcients whose entries are powers of p, p-Adic Numbers Ultrametric Anal. Appl., 9 (2017), 228–235. 43. W. Zudilin, Congruences for q-binomial coeﬃcients, Ann. Combin., 23 (2019), 1123–1135. c ⃝2020 the Author(s), licensee AIMS Press. This is an open access article distributed under the terms of the Creative Commons Attribution License ( AIMS Mathematics Volume 5, Issue 6, 5685–5699.
15060	https://forum.wordreference.com/threads/saber-preterite-imperfect.2577902/	Saber (Preterite/Imperfect) \| WordReference Forums WordReference.comLanguage Forums ForumsRules/Help/FAQHelp/FAQ MembersCurrent visitors Interface Language Dictionary search: Log inRegister What's newSearch Search [x] Search titles and first posts only [x] Search titles only By: Search Advanced search… Rules/Help/FAQHelp/FAQ MembersCurrent visitors Interface Language Menu Log in Register Install the app Install How to install the app on iOS Follow along with the video below to see how to install our site as a web app on your home screen. Note: This feature may not be available in some browsers. Spanish-English / Español-Inglés Spanish-English Grammar / Gramática Español-Inglés Saber (Preterite/Imperfect) Thread starterRaidersPorVida Start dateFeb 11, 2013 R RaidersPorVida New Member San Diego English-United States Feb 11, 2013 #1 I have noticed that in certain books they teach that the preterite of SABER (supe, supiste, supo, supimos, supisteis, supieron) means "to find out" rather than to know. I would rather use the verb AVERIGUAR when I want to say "to find out." Personally I use the imperfect of SABER when I knew (or used to know) something. However I have hear people use SABER in the preterite tense to say that they knew something in the past. Can someone please give me some insight regarding this? micafe Senior Member United States Spanish - Colombia Feb 11, 2013 #2 I believe that teaching that the preterite of "Saber" is always "to find out" is a big mistake. That verb has several meanings and the use depends on the context. "Yo supe" can be understood as "I found out" or "I knew".. There's nothing wrong with that. You don't always have to use the imperfect when you mean to say "I knew". It works the same as other verbs as far as the past tenses (preterite and imperfect) is concerned. However, I agree that most of the time, it's translated into English as "find out". Last edited: Feb 11, 2013 C cheesycarrion Member USA, English Feb 11, 2013 #3 I've heard Spanish speakers say sentences like "no supe que encontraras empleo" where it clearly means "know." In cases like these would it be interchangeable with "sabía"? micafe Senior Member United States Spanish - Colombia Feb 11, 2013 #4 That sentence is not correct. You can say "no supe/no sabía que hubieras encontrado empleo" They're not exactly interchangeable because the difference is the same as with other verbs. R RicardoElAbogado Senior Member SF Bay Area, California American English Feb 12, 2013 #5 I suspect that there is a difference between averiguar and the preterit of saber. It seems to me that averiguar implies an effort to find out whereas saber is simply the result of learning something without regard to making an effort. Perhaps a native can comment on this. chileno Senior Member Las Vegas, Nv. USA Castellano - Chile Feb 12, 2013 #6 Like micafe said, it means found out and knew, depending of context. I knew it the minute you said that. I found out you went to the party. R Rasmus1504 Senior Member Dinamarca Danés Feb 12, 2013 #7 On another note, how does the verb enterarse sound to the spanish speakers? Last edited: Feb 12, 2013 chileno Senior Member Las Vegas, Nv. USA Castellano - Chile Feb 12, 2013 #8 Rasmus1504 said: On another note, how does the verb enterarse sound to the spanish speakers? Click to expand... to find out = enterarse -> supe = me enteré (I found out) Cenzontle Senior Member English, U.S. Feb 12, 2013 #9 Sabía > I had the knowledge. Supe > My knowledge began (however you like to say this in idiomatic English). "I didn't believe him, because I knew (sabía) he was a liar." "As soon as he opened his mouth, I knew (supe) he was lying." You must log in or register to reply here. Share: BlueskyLinkedInWhatsAppEmailShareLink Spanish-English / Español-Inglés Spanish-English Grammar / Gramática Español-Inglés WR styleSystemLightDark English (EN-us) Contact us Terms and rules Privacy policy Help RSS Community platform by XenForo®© 2010-2025 XenForo Ltd. Back TopBottom
15061	https://www.quora.com/What-is-the-relationship-between-joules-volts-and-coulombs	Something went wrong. Wait a moment and try again. Volt (SI unit) Units and Measurements Coulomb (SI unit) Electrical Units Unit of Measurement Science Physics 5 What is the relationship between joules, volts, and coulombs? 13 Answers Kenneth D. Oglesby Engineer, Scientist, Founder www.mcphysics.org · Author has 1.5K answers and 1.1M answer views · Updated 6y The googled website link: How to convert volts to joules shows that for higher level matter charges (ie masses with a net charge imbalance), quoted below: The energy E in joules (J) is equal to the voltage V in volts (V), times the electrical charge Q in coulombs (C), or : joule = volt × coulomb or J=VC But let us dig deeper: A charge is the direct or NET strength of the very basic charge that exists in all subatomic/ particle/matter. That charge strength also has a charge type (+ or -). A coulomb C is the measure of that strength. Its unit of measurement is given in electrical terms and units as The googled website link: How to convert volts to joules shows that for higher level matter charges (ie masses with a net charge imbalance), quoted below: The energy E in joules (J) is equal to the voltage V in volts (V), times the electrical charge Q in coulombs (C), or : joule = volt × coulomb or J=VC But let us dig deeper: A charge is the direct or NET strength of the very basic charge that exists in all subatomic/ particle/matter. That charge strength also has a charge type (+ or -). A coulomb C is the measure of that strength. Its unit of measurement is given in electrical terms and units as- the rate that electron charges are transported by a constant current flow rate of one ampere in one second; equivalent to approximately 6.242×10^18 negative electron charges or approximately 6.242×10^18 (1.036×10^−5 mol) net positive charge (not count number) of protons. A volt is defined as the difference in charge strengths between two points (between air:ground for lightning; electric potential on a conducting wire for electrical). Its unit of measurement is based on when a one ampere amount of electric/electron charge current ( ie., electron flow rate) passes one watt of power between those two points. A joule J is the work done in physics is- watts/ second needed to move a body one meter with a one-Newton force; in electrical terms -causing a one ampere current to flow through a material, that has the resistance of one ohm, for one second That same relationship between energy/ work done, volt potential (ie volt differential, even if the same charge type) and coulomb charge current flow that applies to celestial bodies also applies down to the most basic building blocks of the Universe, called ‘mono-charges’, which are quantized electric charges with a singular charge type (+ or -) and given charge strength/ potential (volts). 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I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage. No phone tag. No junk emails. Just a better deal in less time than it takes to make coffee. Here’s the link to two comparison sites - the one I used and an alternative that I also tested. If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Related questions How is a Joule related to a coulomb volt? How many joules are equal to one coulomb? Are coulombs equal to joules? What does 'Voltage=1 joule per coulomb' actually mean? Are joules by coulomb and volts equal? John Bailey My contribution to physics was showing how to correctly calculate the capacitance of semi-conductor diodes · Author has 7.3K answers and 6.3M answer views · Updated 5y Nalin Agrawal gave a physicist's answer. A joule and a watt second are the same thing. watt seconds and joules are units of energy An electrical engineer would say a watt second is the energy represented by one ampere of current thorough a one volt potential for one second. volts are a measure of electrical potential amperes are a rate of flow of electrons. 6.241×1018 electrons are in a coulomb. coulombs are a measure of charge. One coulomb flowing for one second would be an ampere. One electron-volt is equal to 1.602176565×10−19 joules Jim.Moore Physics, Math, Systems Engineer, Educator · Author has 22.3K answers and 17.5M answer views · Jul 3 Coulomb is a “Derived Unit” from 2 base units, amps (A) and seconds (s). Watt (unit of power). The watt is 1 joule/s They’re all kissing cousins Coulomb is a “Derived Unit” from 2 base units, amps (A) and seconds (s). Watt (unit of power). The watt is 1 joule/s They’re all kissing cousins Len Gould Author has 7K answers and 6M answer views · 9y 1 coulomb (6.2415 × 10^18) of unit charge passing through a load delivering (dropping) 1 volt will deliver 1 joule of energy to the load. Related questions What is the relationship between volts and joules? How is joule equal to coulomb volt? What is the relation between a volt and coulomb? What is the difference between a coulomb and a joule? What is the equivalent power of one joule in volts and amperes? S K Aravind Passionate about physics · Author has 51 answers and 469.4K answer views · 10y W (joules)/ Q (coulombs) = V (volts) Voltage or potential is defined as the work done in moving a charge from infinity to a point. Hope this helps! Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. 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Jared Dunn Works at Mott MacDonald · 9y A joule is a measure of energy absorbed or delivered. A volt is a measure of electric potential energy difference between two points. A coulomb is just a number of electrical charges. A joule can describe any type of energy transfer, but in answer to the original question, when related specifically to electric charges, a joule is defined as the energy required to move one coulomb over one volt of electric potential energy difference. Isaac Clark Degree in electrical engineering and physics with 35 years engineering exper. · Author has 6.3K answers and 3.4M answer views · Updated 5y I see a number of answers here already that are both correct and complete. These answers describe the basis for each of the units for charge, voltage, and energy. I don’t see any answers that talk about the relation between one unit and the other, so I’ll give an answer to that question by expressing joules and volts in terms of each other. The units for joules can be expressed as volts coulombs / sec; The unit for volts can be expressed as newtons-m/coulomb or joules/coulomb; Promoted by TruthFinder Carl Madden Finance Security Expert · May 20 Have you ever googled yourself? I got curious one night, and Googled myself. Most of the websites out there gave me only a scrap of information or were wrong entirely, but I did find 1 site that ended up being the HOLY GRAIL of information on anyone (including myself!). When I typed my name into TruthFinder, I was simply CREEPED out by how much they knew! It showed my social media accounts, addresses, contact details and even dating profiles - and it was all accurate! I was truly shocked how much of my information came up! I can't say I loved it either... TruthFinder very cleverly combines databases from dozens of sources and c I got curious one night, and Googled myself. Most of the websites out there gave me only a scrap of information or were wrong entirely, but I did find 1 site that ended up being the HOLY GRAIL of information on anyone (including myself!). When I typed my name into TruthFinder, I was simply CREEPED out by how much they knew! It showed my social media accounts, addresses, contact details and even dating profiles - and it was all accurate! I was truly shocked how much of my information came up! I can't say I loved it either... TruthFinder very cleverly combines databases from dozens of sources and compiles them for a complete report card on anyone you want to know more about! I intially searched myself, but WOW now I can basically knowing anyone's details - which has so many benefits and uses in life. I did see a few other sites doing something similar, but TruthFinder was the easiest and gave the most accurate information. I was able to search for nearly anyone in the United States by name, phone number, address, email address. 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Gerard programmer, webmaster, lexicographer · Author has 10.7K answers and 3.6M answer views · 10y 1 J = 1 C × 1 V Sponsored by Stake Stake: Online Casino games - Play & Win Online. Play the best online casino games, slots & live casino games! Unlock VIP bonuses, bet with crypto & win. Nalin Agrawal Loving Physics since 10 years. · Author has 101 answers and 315.9K answer views · 10y One joule is the work done to move one coulomb of electric charge through an electric field of one volt. Mathematically - 1J = 1C X 1V Victor Etta aspiring physicist · 9y Joules is the unit of energy. Whether electrical, electromagnetic kinetic, potential etc. they all have the same unit. Coulombs is the unit of electrical charge Volts is the unit of electromotive force and potential difference. Nathanaël Dufour Degree in Mathematics and Physics · 9y Joules are an amount of Energy. Volts and Coulombs are an amount of charge (or of difference in charges). Yah Boi An Ametur :) · Author has 192 answers and 129K answer views · Updated 8y Well bassicly everything,The lenz-joule Law states the heat created in a conductor is equal to the current and ressistence Of the conductor,when a volt battery does some work the kinetic turns in to the Joules energy equal to the volts Times ressistence,now coloumbs have also to do something but only with the positive coulon because there the electron does the work to create a positive charge and heat,and coloumbs to volts and opposite is equal because volt is the current Electric potential between two dots in time,and Ep in Its calculation accept wats contains also coulombs charges. Edit:Tnx f Well bassicly everything,The lenz-joule Law states the heat created in a conductor is equal to the current and ressistence Of the conductor,when a volt battery does some work the kinetic turns in to the Joules energy equal to the volts Times ressistence,now coloumbs have also to do something but only with the positive coulon because there the electron does the work to create a positive charge and heat,and coloumbs to volts and opposite is equal because volt is the current Electric potential between two dots in time,and Ep in Its calculation accept wats contains also coulombs charges. Edit:Tnx for so many views!!! Related questions How is a Joule related to a coulomb volt? How many joules are equal to one coulomb? Are coulombs equal to joules? What does 'Voltage=1 joule per coulomb' actually mean? Are joules by coulomb and volts equal? What is the relationship between volts and joules? How is joule equal to coulomb volt? What is the relation between a volt and coulomb? What is the difference between a coulomb and a joule? What is the equivalent power of one joule in volts and amperes? How can joules be calculated using volts, watts, and amps? What is the difference between the joule and the electron volt? What is 'electron volt'? What is its relation with joule? Why is it useful to measure electricity in joules per coulomb (volts) rather than just using joules (energy)? Can you explain the difference between a joule, an electron volt, and a Coulomb? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
15062	https://www.agrader.sg/post/4-common-mistakes-to-avoid-when-solving-linear-inequalities-in-secondary-math	top of page BONUS: Receive the 'EverLoop' after-class resources for free when you sign up your child for the weekly lessons. Learn more. JOIN OUR COMMUNITY FOR FREE study tips, downloads, updates & more! Join Telegram Channel Join WhatsApp Channel Enquire Now Search 4 Common Mistakes to Avoid When Solving Linear Inequalities in Secondary Math AGrader Learning Centre Jun 6, 2023 4 min read As your child delves into the world of linear inequalities in Secondary Math, it's essential to steer clear of common pitfalls that can hinder their progress. In this informative article, we highlight the four most common mistakes students encounter when solving linear inequalities questions, providing practical advice and strategies to help them navigate through them. Say goodbye to confusion and hello to success as we equip your child with the knowledge to tackle these challenges head-on! 1. Not flipping the inequality when dividing or multiplying by negative values When solving linear inequalities, a crucial mistake that many students make is forgetting to flip the inequality sign when dividing or multiplying by negative values. This error can lead to incorrect solutions and may negatively impact your overall understanding of the topic. 2. Not accurately representing solutions on a number line Imagine the number line as a road that leads us to the solutions to our inequality. Just as we need accurate directions to reach our destination, we must accurately represent the solutions on the number line to find the correct answer to the inequality. The number line is a visual tool that helps us understand and visualize the solutions to an inequality. It allows us to see which values satisfy the inequality and which do not. When we fail to accurately represent the solutions on the number line, it can lead to confusion and incorrect answers. Here are some common mistakes to avoid: Misunderstanding the inequality symbols (<, >, ≤, ≥): These symbols indicate whether the solution includes values that are greater than, less than, or equal to a certain number. Students sometimes misinterpret the symbols, leading to an incorrect representation of the solutions on the number line. Understanding the use of circles and solid dots on the number line: Note that a circle O indicates that x cannot take on a particular value (> or <) while a solid 🌑dor indicates that x can take on the particular value (≥ or ≤). Shown below is the correct representation of x < -2 and x ≤ -2 on a number line. To avoid these mistakes, students should review the guidelines for including or excluding endpoints based on the inequality symbol. It's essential to correctly identify when the endpoints are included and when they are excluded, as this directly affects the representation of the solution on the number line. 3. Incorrectly combining like terms Like terms are similar terms with the same variable(s) raised to the same exponent(s). Combining them correctly allows us to simplify the equation and work towards finding the solution. However, incorrectly combining like terms can lead to errors and incorrect answers. Here are some reasons why this mistake occurs and how to avoid it: Misunderstanding the rules for combining like terms: It is essential to recognize that terms must have the same variable(s) raised to the same exponent(s) in order to be combined. Students may overlook this requirement and mistakenly combine terms that don't meet the criteria. Careless mistakes in applying mathematical operations: Sometimes, students make errors applying mathematical operations when combining like terms, leading to incorrect solutions. It's crucial to be careful and accurate during intermediary steps to avoid these mistakes. To avoid this mistake, students should follow these steps: Carefully examine the terms: Before combining like terms, students should examine the variables and exponents to ensure they match. Only terms with identical variables and exponents can be combined. Identify like terms: Once the terms have been examined, students should identify the like terms, which are terms that have the same variable(s) raised to the same exponent(s). For example, terms like 3x and 5x can be combined because they have the same variable (x) raised to the same exponent (1). 4. Using incorrect inequality symbols Inequality symbols (<, >, ≤, ≥) are like road signs that guide us to the correct solution. They indicate the relationship between two values and help us determine which values satisfy the inequality. When we use incorrect inequality symbols, it can lead to confusion and inaccurate answers. Students may sometimes mix up the meanings of the symbols > and <. To avoid these mistakes, students can try to find effective ways to remember the meanings of these symbols. One way to remember this is to observe the inequality symbol itself. Any term on the side of the highlighted portion of the inequality symbol, as shown below, is greater than the term on the other side (not highlighted) of the inequality symbol. By being mindful of these key aspects - understanding the meaning behind each inequality symbol and the inclusion/exclusion of endpoints - students can accurately use the correct inequality symbols in linear inequalities. This skill not only helps them interpret the relationships between values but also reinforces their understanding of the topic. Avoiding these pitfalls is essential to becoming proficient at solving linear inequalities. Remembering to flip the inequality sign when dividing or multiplying by negative values and accurately representing solutions on the number line ensures accurate and precise results. Additionally, combining like terms correctly and using the correct inequality symbols are crucial steps to avoid errors and confusion. By incorporating these strategies into your problem-solving approach, you pave the way for success in understanding and solving linear inequalities. Stay focused, practice diligently, and seek guidance whenever needed to overcome these common mistakes and excel in O-level Mathematics. Is your child seeking an extra boost in conquering O-level Math problems? Look no further than AGrader Learning Centre, where we specialize in providing exceptional Secondary Maths Tuition in Singapore. Our Secondary Math Tuition Programme aligns seamlessly with the latest Ministry of Education standards, equipping students with the necessary tools to master math and achieve remarkable scores. Whether you're interested in O-level Math tuition or Secondary 1 to Secondary 3 Math tuition, our dedicated teachers are committed to delivering the best assistance possible. We provide a supportive learning environment that fosters mathematical confidence and helps students excel in their studies. Enrol your child today and watch their mathematical skills soar to new heights! Recent Posts See All Exploring Angles Concepts and Properties for Primary 5 & PSLE Math How to Write an Effective Argumentative Essay in O Level: 5 Tips and Techniques Understanding Money Math Problems in Primary 3: A Comprehensive Guide bottom of page
15063	https://francisbach.com/chebyshev-polynomials/	Polynomial magic I : Chebyshev polynomials – Machine Learning Research Blog Machine Learning Research Blog Francis Bach Menu Home About Home page Menu Polynomial magic I : Chebyshev polynomials Posted on November 4, 2019 December 1, 2019 by Francis Bach Orthogonal polynomials pop up everywhere in applied mathematics and in particular in numerical analysis. Within machine learning and optimization, typically (a) they provide natural basis functions which are easy to manipulate, or (b) they can be used to model various acceleration mechanisms. In this post, I will describe one class of such polynomials, the Chebyshev polynomials (Tchebychev in French, Чебышёв in Russian), whose extremal properties (beyond being orthogonal) are useful in the analysis of accelerated algorithms. Definition and first properties The k k-th Chebyshev polynomialT k T k is classically defined as the unique polynomial such that ∀θ∈[0,2 π],cos(k θ)=T k(cos θ).∀θ∈[0,2 π],cos⁡(k θ)=T k(cos⁡θ). Recurrence. Summing the two equations cos[(k±1)θ]=cos(k θ)cos θ∓sin(k θ)sin θ cos⁡[(k±1)θ]=cos⁡(k θ)cos⁡θ∓sin⁡(k θ)sin⁡θ, the following recurrence relationship can be deduced: ∀k>0,T k+1(X)=2 X T k(X)–T k−1(X).∀k>0,T k+1(X)=2 X T k(X)–T k−1(X). Together with the first two polynomials T 0(X)=1 T 0(X)=1 and T 1(X)=X T 1(X)=X, this leads to T 2(X)=2 X 2–1 T 2(X)=2 X 2–1, T 3(X)=4 X 3–3 X T 3(X)=4 X 3–3 X, and so on. From the recurrence relationships, one can easily deduce that T k T k has the parity of k k and that T k T k has degree k k, with leading coefficient 2 k−1 2 k−1. Oscillatory behavior. For k θ=j π k θ=j π, for j j integer, we have cos(k θ)=(−1)j cos⁡(k θ)=(−1)j, while when j j goes from k k to 0 0, cos θ=cos(j π k)cos⁡θ=cos(j π k) goes from -1 to 1. Thus, on [−1,1][−1,1], T k(x)T k(x) oscillates between −1−1 and 1 1, with equality for cos(j π k)cos(j π k) for j=0,…,k j=0,…,k. This oscillatory behavior is illustrated below and crucial for the extremal properties below. First 33 Chebyshev polynomials, plotted between -1 and 1. Note the stronger oscillatory behavior between -1 and 1 as k k grows. Orthogonality. Using the orthogonality of the Fourier basis on [0,2 π][0,2 π], we have for k≠ℓ k≠ℓ, ∫π 0 cos(k θ)cos(ℓ θ)d θ=0∫0 π cos⁡(k θ)cos⁡(ℓ θ)d θ=0, and with the change of variable x=cos θ x=cos⁡θ, we obtain ∫1−1 T k(x)T ℓ(x)1−x 2−−−−−√d x=0.∫−1 1 T k(x)T ℓ(x)1−x 2 d x=0. Thus the Chebyshev polynomials inherit from many properties from such orthogonal polynomials (such as the two-term recursion above, but for the Chebyshev polynomials, these can obtained more directly). For further properties, see . Extremal properties Chebyshev polynomials exhibit many “extremal properties”, of the form: among all polynomials of degree k k with some form of normalization (e.g., fixed k k-th order coefficient or value at given point), the one with smallest specific norm is proportional to T k T k. The most classical one is as follows: the polynomial P P of degree k k with k k-th order coefficient equal to one, and with minimum ℓ∞ℓ∞-norm max x∈[−1,1]\|P(x)\|max x∈[−1,1]\|P(x)\| on [−1,1][−1,1], is P=1 2 k−1 T k P=1 2 k−1 T k. The proof is particularly elegant and simple (see here). Since this is not the property that we need for optimization, we will consider another one. Proposition (largest value outside of [−1,1][−1,1]). For any polynomial P P of degree k k such that \|P(x)\|≤1\|P(x)\|≤1 for x∈[−1,1]x∈[−1,1], and any z>1 z>1, \|P(z)\|≤T k(z)\|P(z)\|≤T k(z). Proof. By contradiction, we assume that there exists z>1 z>1 such that P(z)>T k(z)P(z)>T k(z) (the other possibility P(z)<−T k(z)P(z)<−T k(z) is done by replacing P P by −P−P). Without loss of generality, we can assume that max x∈[−1,1]\|P(x)\|<1 max x∈[−1,1]\|P(x)\|<1 (by potentially rescaling P P). Then, the polynomial Q=P–T k Q=P–T k of degree k k has alternatively strictly positive and negative values between −1−1 and z>1 z>1. Indeed, Q(z)>0 Q(z)>0, and (−1)j Q(cos(j π k))<0(−1)j Q(cos(j π k))<0 for all j=0,…,k j=0,…,k. Therefore there are k+2 k+2 alternating signs, and thus k+1 k+1 zeros, which implies that Q=0 Q=0 since Q Q has degree k k. This is a contradiction with the existence of z z. Note that for any θ θ, T k(cosh θ)=cosh(k θ)T k(cosh⁡θ)=cosh⁡(k θ) (which can be shown by induction), and T k T k is thus increasing on [1,+∞)[1,+∞) with values quickly increasing as k k grows. See an illustration below. First 33 Chebyshev polynomials, plotted between 1 and 4 in logarithmic scale. Note the exploding behavior as k k grows. Chebyshev acceleration We consider a recursion in R n R n of the form x k=A x k−1–b x k=A x k−1–b, with A∈R n×n A∈R n×n a symmetric matrix and eigenvalues in [−ρ,ρ][−ρ,ρ] with ρ∈[0,1)ρ∈[0,1). Such recursions are ubiquitous in data science, as (1) gradient descent on a strongly-convex quadratic function, or (b) gossip for distributed averaging (see an example in a section below). The recursion converges to the unique (because ρ∈[0,1)ρ∈[0,1)) fixed point x∗∈R n x∗∈R n such that x∗=A x∗–b x∗=A x∗–b. We have, by unrolling the recursion: x k–x∗=A(x k−1–x∗)=A k(x 0–x∗).x k–x∗=A(x k−1–x∗)=A k(x 0–x∗). This leads to the usual exponential convergence rate ∥x k–x∗∥2≤ρ k∥x 0–x∗∥2‖x k–x∗‖2≤ρ k‖x 0–x∗‖2. In the following, writing ρ=1–(1−ρ)ρ=1–(1−ρ) makes explicit the importance of 1−ρ 1−ρ, which is the gap between 1 1 and the largest eigenvalue of A A. Increasing this gap is equivalent to accelerating the convergence rate. In order to speed-up convergence, a classical idea is to take linear combinations of all past iterates. That is, we consider y k=∑k i=0 ν k i x i y k=∑i=0 k ν i k x i for some weights ν k i ν i k such that ∑k i=0 ν k i=1∑i=0 k ν i k=1 (so that if all iterates are already at x∗x∗, then the weighted average stays there). We have y k–x∗=∑i=0 k ν k i(x i–x∗)=∑i=0 k ν k i A i(x 0−x∗)=P k(A)(x 0−x∗),y k–x∗=∑i=0 k ν i k(x i–x∗)=∑i=0 k ν i k A i(x 0−x∗)=P k(A)(x 0−x∗), where P k(X)=∑k i=0 ν k i X i P k(X)=∑i=0 k ν i k X i is a polynomial such that P k(1)=1 P k(1)=1. Therefore, we have: ∥y k–x∗∥2≤max λ∈[−ρ,ρ]\|P k(λ)\|⋅∥x 0–x∗∥2.‖y k–x∗‖2≤max λ∈[−ρ,ρ]\|P k(λ)\|⋅‖x 0–x∗‖2. In order to select the best polynomial, we are looking for P k P k such that P k(1)=1 P k(1)=1 and max λ∈[−ρ,ρ]\|P k(λ)\|max λ∈[−ρ,ρ]\|P k(λ)\| is as small as possible. Up to mapping [−ρ,ρ][−ρ,ρ] to [−1,1][−1,1], we know from the extremal property above that the optimal polynomial is exactly a rescaled Chebyshev polynomial, that is, P k(X)=T k(X/ρ)T k(1/ρ).P k(X)=T k(X/ρ)T k(1/ρ). The maximal value on [−ρ,ρ][−ρ,ρ] is then (T k(1/ρ))−1(T k(1/ρ))−1. In order to compare to ρ k ρ k (no acceleration), we can provide an equivalent of [T k(1/ρ)]−1/k[T k(1/ρ)]−1/k as ρ 1+1−ρ 2√ρ 1+1−ρ 2 (see end of the post). There is no real acceleration when ρ ρ is bounded away from 1, but as ρ ρ tends to 1 1, this can be shown (see also the end of the post) to be equivalent to 1–2(1−ρ)−−−−−−√1–2(1−ρ), with the usual “square root” acceleration: 1−ρ 1−ρ is essentially replaced by 1−ρ−−−−√1−ρ. We thus get an acceleration, but as is, computing y k y k seems to require to store all values of x 1,…,x k x 1,…,x k, which is not practical. Since there is a second-order recursion for Chebyshev polynomials, one can derive one as well, directly for the sequence (y k)(y k). A somewhat lengthy calculation (see end of the post) leads to the recursion y k+1=ω k+1(A y k–b)+(1−ω k+1)y k−1,y k+1=ω k+1(A y k–b)+(1−ω k+1)y k−1, with a sequence ω k+1 ω k+1 also defined by recursion as ω k+1=(1–ρ 2 4 ω k)−1 ω k+1=(1–ρ 2 4 ω k)−1, initialized with ω 1=2 ω 1=2, y 0=x 0 y 0=x 0, y 1=A x 0–b y 1=A x 0–b. Therefore, on top of the usual computation of A y k−b A y k−b, Chebyshev acceleration comes at no extra computational cost. Simpler stationary recursion. In the recursion above, the parameter ω k ω k varies with k k. A similar (then non totally optimal) acceleration can be obtained by replacing all ω k ω k’s by their limit ω ω when k k tends to infinity, which is characterized by the equation ω=(1−ρ 2 4 ω)−1 ω=(1−ρ 2 4 ω)−1 with smallest solutions 1/ρ–1/ρ 2−1√ρ/2 1/ρ–1/ρ 2−1 ρ/2 (see end of the post for detailed computations). The now stationary recursion then becomes y k+1=ω(A y k–b)+(1−ω)y k−1,y k+1=ω(A y k–b)+(1−ω)y k−1, and is exponentially convergent with rate proportional to ρ ω/2=1 1/ρ+1/ρ 2–1√=ρ 1+1–ρ 2√ρ ω/2=1 1/ρ+1/ρ 2–1=ρ 1+1–ρ 2. Thus, the recursion is simpler and the final speed asymptotically the same as full Chebyshev acceleration. Relationship with other acceleration mechanisms Non-adaptive schemes. As seen above for an affine operator F:R n→R n F:R n→R n (i.e., F(x)=A x−b F(x)=A x−b), Chebyshev acceleration takes a recursion of the form x k+1=F(x k),x k+1=F(x k), and linearly combines iterates; it ends up creating second-order recursions of the form y k+1=ω k+1 F(y k)+(1−ω k+1)y k−1,y k+1=ω k+1 F(y k)+(1−ω k+1)y k−1, with the same fixed points. Other formats (with fixed point preservation) can be considered such as y k+1=F(y k)+δ k+1(y k–y k−1),y k+1=F(y k)+δ k+1(y k–y k−1), or y k+1=F(y k)+δ k+1(F(y k)–F(y k−1)),y k+1=F(y k)+δ k+1(F(y k)–F(y k−1)), for some constants δ k+1 δ k+1. When F F is affine, the format does not matter much (and all end up being essentially equivalent), but for gradient descent algorithms where F(x)=x–γ f′(x)F(x)=x–γ f′(x) for some non-quadratic function f f and γ γ a step-size, there is a difference, the last one corresponding to Nesterov acceleration (see a nice post on it here), and the one before to classical momentum, also known as the heavy-ball method (see ). Adaptive schemes. The methods above need to know the bound on the spectrum ρ ρ. They have to commit to such a value (which is typically only known through upper bounds) and cannot “get lucky”, that is, even if the best value ρ ρ is known, they cannot benefit from additional better properties of the spectrum of A A (e.g., clustered eigenvalues). The conjugate gradient method, which accesses the matrix A A with the slightly stronger oracle of computing A x A x any x x (and not only A x–b A x–b), or Anderson acceleration (which does not need a stronger oracle), are adaptive for similar problems [4, 5]. Again, Chebyshev polynomials are present; probably more on this in future posts! Application to accelerated gossip A interesting linear recursion pops out in distributed optimization, where we assume that computers or processors are placed in n n nodes in a network, and the goal is to minimize an average of function f 1,…,f n f 1,…,f n, each of them only accessible by the corresponding node. The nodes are allowed to communicate messages along each edge of a network. Two-dimensional grid with d=8×8=64 d=8×8=64 nodes. The simplest of such problem is the network averaging problem where f i(θ)=1 2(θ–ξ i)2 f i(θ)=1 2(θ–ξ i)2, for a uni-dimensional parameter θ θ and ξ∈R n ξ∈R n. The solution of this consensus is θ∗=1 n∑n i=1 ξ i θ∗=1 n∑i=1 n ξ i. The gossip algorithm consists in iteratively replacing the value θ i θ i at a given node by a weighted average ∑j∼i W i j θ j∑j∼i W i j θ j of the values at neighboring nodes (and node i i). If all n n nodes communicate simultaneously, then the vector θ θ is replaced by W θ W θ, hence a linear recursion θ k+1=W θ k,θ k+1=W θ k, initialized at θ 0=ξ θ 0=ξ. Assuming that W W is symmetric, with non-negative off-diagonal elements, and such that W 1 n=1 n W 1 n=1 n (where 1 n∈R n 1 n∈R n is the vector of all ones), then all eigenvalues of W W except the largest one are included in the interval [−ρ,ρ][−ρ,ρ], with ρ∈(0,1)ρ∈(0,1) for a connected graph. A simple such matrix W W can be obtained from the adjacency matrix A A of the graph, such that A i j=1 A i j=1 if nodes i i and j j are connected, and zero otherwise, as W=I–α L W=I–α L, with L=D i a g(A 1 n)–A L=D i a g(A 1 n)–A the Laplacian matrix and α α selected so that the eigenvalues are all between −ρ−ρ and ρ ρ, except one, which is equal to 1 (see values of ρ ρ and α α at the end of the post). We will see below that this extra eigenvalue which is equal to one is in fact not a problem for analyzing the convergence of this network averaging procedure. When applying the gossip matrix W W iteratively to θ 0=ξ θ 0=ξ, the projection on the eigensubspace corresponding to the unit eigenvalue is not changed, while all other projections on the other eigensubspaces converge to zero at rate at most ρ k ρ k. Thus θ k θ k converges to the constant vector 1 n 1 n 1⊤n ξ 1 n 1 n 1 n⊤ξ at rate ρ k ρ k, and thus to a constant vector, with the average 1 n∑n i=1 ξ i 1 n∑i=1 n ξ i in all components. Given that we have a linear recursion, we can use Chebyshev acceleration defined above and obtain substantial improvements, as illustrated below. For the use of this acceleration within distributed optimization algorithms, see and references therein. Comparison of gossip algorithms on a two-dimensional grid (each cell correspond to a value ξ i∈[−1,1]ξ i∈[−1,1] to average): (left) regular gossip, (center) accelerated second-order recursion with constant coefficients, (right) Chebyshev acceleration. Convergence is much faster with acceleration, (only) slightly better for Chebyshev acceleration, which is the optimal polynomial acceleration. Conclusion Among classical classes of orthogonal polynomials, Chebyshev polynomials are special, because beyond being orthogonal, they satisfy extremal properties that are particularly useful in numerical analysis. In future posts, I plan to go over Jacobi polynomials (which include Legendre, Gegenbauer and Chebyshev polynomials), Hermite polynomials, and finally Bernoulli polynomials (which are not orthogonal but still very special). For all of these, there are natural applications in machine learning. Acknowledgements. I would like to thank Raphaël Berthier for proofreading this blog post and making good clarifying suggestions. References John C. Mason, and David C. Handscomb. Chebyshev polynomials. Chapman and Hall/CRC, 2002. Stephen Boyd, Arpita Ghosh, Balaji Prabhakar, Devavrat Shah. Randomized gossip algorithms.IEEE/ACM Transactions on Networking, 14:2508-2530, 2006. Euhanna Ghadimi, Hamid Reza Feyzmahdavian, and Mikael Johansson. Global convergence of the heavy-ball method for convex optimization. European Control Conference (ECC), 2015. Homer F. Walker, Peng Ni. Anderson acceleration for fixed-point iterations.SIAM Journal on Numerical Analysis,49(4):1715-1735, 2011. Damien Scieur, Alexandre d’Aspremont, Francis Bach. Regularized Nonlinear Acceleration. Mathematical Programming, 2018. Kevin Scaman, Francis Bach, Sébastien Bubeck, Yin-Tat Lee, Laurent Massoulié. Optimal algorithms for smooth and strongly convex distributed optimization in networks. Proceedings of the International Conference on Machine Learning (ICML), 2017. Mieczysław A. Kłopotek. Spectral Analysis of Laplacians of an Unweighted and Weighted Multidimensional Grid Graph — Combinatorial versus Normalized and Random Walk Laplacians. Technical report, ArXiv:1707.05210, 2019. Detailed Computations Limit of[T k(1/ρ)]−1/k[T k(1/ρ)]−1/k. For z≥1 z≥1, then a well-known property of Chebyshev polynomials is that T k(z)=cosh[k a c o s h(z)]T k(z)=cosh⁡[k a c o s h(z)] (which can be shown by induction using basic hyperbolic trigonometry identities). Moreover, we have a c o s h(z)=log(z+z 2−1−−−−−√)a c o s h(z)=log⁡(z+z 2−1) and thus T k(z)=1 2[(z+z 2−1−−−−−√)k+(z–z 2−1−−−−−√)k].T k(z)=1 2[(z+z 2−1)k+(z–z 2−1)k]. For z=1/ρ z=1/ρ, and taking limits, we get that [T k(z)]1/k[T k(z)]1/k tends to z+z 2−1−−−−−√z+z 2−1, which leads to the limit ρ 1+1−ρ 2√ρ 1+1−ρ 2 for [T k(z)]−1/k[T k(z)]−1/k. Then a classical Taylor expansion in 1−ρ 1−ρ leads to 1–2(1−ρ)−−−−−−−√1–2(1−ρ). Recurrence for Chebyshev acceleration. We have, using the recursion for Chebyshev polynomials y k+1–x∗=2 T k+1(1/ρ)(A/ρ)T k(A/ρ)(x 0–x∗)–1 T k+1(1/ρ)T k−1(A/ρ)(x 0–x∗).y k+1–x∗=2 T k+1(1/ρ)(A/ρ)T k(A/ρ)(x 0–x∗)–1 T k+1(1/ρ)T k−1(A/ρ)(x 0–x∗). Using the equality x∗=A x∗−b x∗=A x∗−b, the terms in x∗x∗ cancel (they have to anyway, because P k(1)=1 P k(1)=1). We then get y k+1=(2/ρ)T k(1/ρ)T k+1(1/ρ)(A y k–b)–T k−1(1/ρ)T k+1(1/ρ)y k−1.y k+1=(2/ρ)T k(1/ρ)T k+1(1/ρ)(A y k–b)–T k−1(1/ρ)T k+1(1/ρ)y k−1. Using T k−1(1/ρ)=(2/ρ)T k(1/ρ)–T k+1(1/ρ)T k−1(1/ρ)=(2/ρ)T k(1/ρ)–T k+1(1/ρ), and denoting ω k+1=(2/ρ)T k(1/ρ)T k+1(1/ρ)ω k+1=(2/ρ)T k(1/ρ)T k+1(1/ρ), we get y k+1=ω k+1(A y k–b)+(1−ω k+1)y k−1.y k+1=ω k+1(A y k–b)+(1−ω k+1)y k−1. Reusing one last time the Chebyshev recursion, we get ω−1 k+1=T k+1(1/ρ)(2/ρ)T k(1/ρ)=1–T k−1(1/ρ)(2/ρ)T k(1/ρ)=1–ρ 2 4 ω k,ω k+1−1=T k+1(1/ρ)(2/ρ)T k(1/ρ)=1–T k−1(1/ρ)(2/ρ)T k(1/ρ)=1–ρ 2 4 ω k, which is the desired recursion. Convergence of stationary recursion. The roots of ω=(1−ρ 2 4 ω)−1 ω=(1−ρ 2 4 ω)−1 are the ones of ρ 2 4 ω 2–ω+1=0 ρ 2 4 ω 2–ω+1=0, with smallest solutions ω=1/ρ–1/ρ 2−1√ρ/2 ω=1/ρ–1/ρ 2−1 ρ/2. In order to study the second-order recursion y k+1=ω(A y k–b)+(1−ω)y k−1,y k+1=ω(A y k–b)+(1−ω)y k−1, with constant coefficient, we need to compute the roots of r 2=ω λ r+(1−ω)r 2=ω λ r+(1−ω), for \|λ\|≤ρ\|λ\|≤ρ. The discriminant of this equation is λ 2 ω 2+4(1−ω)≤ρ 2 ω 2+4(1−ω)=0 λ 2 ω 2+4(1−ω)≤ρ 2 ω 2+4(1−ω)=0, and thus the roots are complex conjugate with squared modulus (ω–1)=1 4 ρ 2 ω 2(ω–1)=1 4 ρ 2 ω 2 independent of λ λ. Thus, as k k tends to infinity, ∥y k–x∗∥1/k 2‖y k–x∗‖2 1/k tends to 1 2 ρ ω=(1/ρ–1/ρ 2–1−−−−−−√)=1 1/ρ+1/ρ 2–1√1 2 ρ ω=(1/ρ–1/ρ 2–1)=1 1/ρ+1/ρ 2–1, which is exactly the rate for Chebyshev acceleration. Eigenvalues of the Laplacian matrix of a square grid. Given a chain of length m m such as depicted below, the m×m m×m Laplacian matrix can be shown (see ) to have eigenvalues 2+2 cos k π m 2+2 cos⁡k π m for k=1,…,m k=1,…,m. One-dimensional grid with m=8 m=8. For a two-dimensional grid of size m×m m×m, then the m 2×m 2 m 2×m 2 Laplacian matrix can be shown (see ) to have eigenvalues 4+2 cos k 1 π m+2 cos k 2 π m 4+2 cos⁡k 1 π m+2 cos⁡k 2 π m for k 1,k 2=1,…,m k 1,k 2=1,…,m. Therefore, the second smallest eigenvalue is λ min=2–2 cos π m λ min=2–2 cos⁡π m and the largest eigenvalue is λ max=4+4 cos π m λ max=4+4 cos⁡π m. We then select α α such that 1−α λ min=ρ 1−α λ min=ρ and 1−α λ max=−ρ 1−α λ max=−ρ, leading to α=2 λ min+λ max=2 6+2 cos π m α=2 λ min+λ max=2 6+2 cos⁡π m and finally ρ=λ max–λ min λ max+λ min=2+6 cos π m 6+2 cos π m∼8–3 π 2 m 2 8–π 2 m 2∼1–π 2 4 m 2 ρ=λ max–λ min λ max+λ min=2+6 cos⁡π m 6+2 cos⁡π m∼8–3 π 2 m 2 8–π 2 m 2∼1–π 2 4 m 2. Thus, as a function of n=m 2 n=m 2, the eigengap is proportional to 1/n 1/n. More generally, for the grid of size m m in dimension d d, then we get λ min=2–2 cos π m λ min=2–2 cos⁡π m and λ max=2 d+2 d cos π m λ max=2 d+2 d cos⁡π m, and ρ=λ max–λ min λ max+λ min=2 d−2+(2 d+2)cos π m 2 d+2+(2 d−2)cos π m∼4 d–(d+1)π 2 m 2 4 d–(d−1)π 2 m 2∼1–π 2 2 d m 2.ρ=λ max–λ min λ max+λ min=2 d−2+(2 d+2)cos⁡π m 2 d+2+(2 d−2)cos⁡π m∼4 d–(d+1)π 2 m 2 4 d–(d−1)π 2 m 2∼1–π 2 2 d m 2. Thus, as a function of n=m d n=m d, the eigengap is proportional to 1/n 2/d 1/n 2/d. Moreover, when m m is large, the normalizing factor α α tends to 1/(2 d)1/(2 d). 1 thought on “Polynomial magic I : Chebyshev polynomials” Ahmed Mazarisays: November 7, 2019 at 10:52 am Hi Francis, Thank you for this nice and concise reminder on the properties of Chebyshev polynomials. Indeed, they are used in versatile applications and recently they draw a lots of attention in geometric deep learning and in graph signal processing. They are used to extend CNNs to non-Euclidean data (graphs and manifolds). Along with spectral graph theory, Chebyshev polynomials allow to design a fast and localized (in space) graph convolutional operator with O(K) its complexity which is linear w.r.t to the filters support’s size K (K as the Chebyshev polynomial order) and the number of edges. As a result, GCN (Graph ConvNet) has the same linear computational complexity as Euclidean CNNs. Reference : Michaël Defferrard, Xavier Bresson and Pierre Vandergheynst. Convolutional Neural Networks on Graphs with Fast Localized Spectral Filtering. NIPS, 2016 Thank you Reply Leave a Reply Cancel reply Your email address will not be published.Required fields are marked Comment Name Email Website Δ Recent Posts Beyond Power Laws: Scaling Laws for Next-Token Prediction Revisiting scaling laws via the z-transform Unraveling spectral properties of kernel matrices – II My book is (at last) out! Scaling laws of optimization About I am Francis Bach, a researcher at INRIA in the Computer Science department of Ecole Normale Supérieure, in Paris, France. I have been working on machine learning since 2000, with a focus on algorithmic and theoretical contributions, in particular in optimization. All of my papers can be downloaded from my web page or my Google Scholar page. I also have a Twitter account. I recently published a book “Learning Theory from First Principles“. Search for: Recent Posts Beyond Power Laws: Scaling Laws for Next-Token Prediction Revisiting scaling laws via the z-transform Unraveling spectral properties of kernel matrices – II My book is (at last) out! Scaling laws of optimization Recent Comments Francis Bach on Sums-of-squares for dummies: a view from the Fourier domain K2V on Sums-of-squares for dummies: a view from the Fourier domain Wok on My book is (at last) out! 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15064	https://www.ixl.com/math/grade-4/solve-one-step-variable-equations-addition-and-subtraction	IXL \| Solve one-step variable equations: addition and subtraction \| 4th grade math SKIP TO CONTENT [x] - [x] IXL Learning Sign in- [x] Remember Sign in nowJoin now IXL Learning Learning Math Skills Lessons Videos Games Fluency Zone New! Language arts Skills Videos Games Science Social studies Spanish Recommendations Recommendations wall Skill plans IXL plans South Carolina state standards Textbooks Test prep Awards Student awards Assessment Analytics Takeoff Inspiration Learning All Learning Math Language arts Science Social studies Spanish Recommendations Skill plans Learning Skill plans IXL plans South Carolina state standards Textbooks Test prep Awards Assessment Analytics Takeoff Inspiration Membership Sign in Math Math Language arts Language arts Science Science Social studies Social studies Spanish Spanish Recommendations Recommendations Skill plans Skill plans IXL plans South Carolina state standards Textbooks Test prep Awards Awards enVisionMATH 2.0 Texas - 4th grade Solve one-step variable equations: addition and subtraction TBG Share skill Copy the link to this skill share to facebook share to twitter Time to get in the zone! Your teacher would like you to focus on skills in . Let's pick a skill from these categories. Let's go! 1 Solve one-step variable equations: addition and subtraction TBG Share video Copy the link to this video share to facebook share to twitter You are watching a video preview. Become a member to get full access! You've reached the end of this video preview, but the learning doesn't have to stop! Join IXL today! Become a memberSign in Incomplete answer You did not finish the question. Do you want to go back to the question? Go back Submit Learn with an example Solve for c. c − 5 = 9 c = Submit Back to practice ref_doc_title. Back to practice Learn with an example Learn with an example question Solve for g. 13 = g − 5 g = key idea To solve for a variable, use inverse operations to undo the operations in the equation. Be sure to do the same operation to both sides of the equation. solution Solve. 13 = g − 5 Add 5 to both sides 13 + 5 = g − 5 + 5 Add 5 to both sides Simplify 18 = g Simplify Learn with an example Excellent! You got that right! Continue Learn with an example Jumping to level 1 of 1 Excellent! Now entering the Challenge Zone—are you ready? Questions answered Questions 0 Time elapsed Time 00 00 03 hr min sec SmartScore out of 100 IXL's SmartScore is a dynamic measure of progress towards mastery, rather than a percentage grade. It tracks your skill level as you tackle progressively more difficult questions. Consistently answer questions correctly to reach excellence (90), or conquer the Challenge Zone to achieve mastery (100)! Learn more 0 You met your goal! Teacher tools Group Jam Live Classroom Leaderboards Work it out Not feeling ready yet? These can help:N.10 Complete the addition sentence - up to two digitsN.10 Complete the addition sentence - up to two digits - Second grade CZKP.11 Complete the subtraction sentence - up to two digitsP.11 Complete the subtraction sentence - up to two digits - Second grade YWC Company \| Membership \| Blog \| Help center \| User guides \| Tell us what you think \| Testimonials \| Careers \| Contact us \| Terms of service \| Privacy policy © 2025 IXL Learning. All rights reserved. Follow us First time here? 1 in 4 students uses IXL for academic help and enrichment. Pre-K through 12th grade Sign up nowKeep exploring
15065	https://www.unionems.net/s/EMS-Protocols-For-Adrenal-Crisis.pdf	MEDICATION SAFETY ALERT (EMS PROTOCOLS FOR ADRENAL CRISIS) Dear EMS Provider: Adrenal crisis is a life-threatening condition caused by inadequate adrenocortical function leading to impaired physiological responses to stressors such as illness and injury. This deficit can lead to hypotension, hypoglycemia, shock, and death. Immediate intervention can be lifesaving. You may be encountering an adrenal crisis if: the patient takes hydrocortisone, prednisone, dexamethasone, or other glucocorticoids on a regular basis; and/or is having signs/symptoms including: nausea, vomiting, hypotension, and/or altered mental status. Patients may carry cards or instructions from their physician for management of adrenal crisis with injectable hydrocortisone; if not, follow dosing recommendations below and contact/inform medical control. The following recommended treatment reflects current consensus-based guidelines.1 • 2 mg/kg INJECTABLE HYDROCORTISONE (as sodium succinate or sodium phosphate) up to a maximum of 100 mg • Can be given: IM/IV/IO/SC • General rule: Err on the higher side of a dose range For suspected or confirmed adrenal crisis, it is imperative to meet the following medication safety conditions: • FIRST-LINE MEDICATION: HYDROCORTISONE • RIGHT TIMING: STAT/IMMEDIATELY • RIGHT DOSE: Weight-based dose and/or Age-based dose / Pediatric vs. Adult Age-Based Dose Recommendations • <3 years = 25 mg • 3 to <10 years = 50 mg • 10 years and older = 100 mg Weight-Based Dose Recommendations Adult: 100 mg Pediatric: 2 mg/kg to a maximum of 100 mg Round up to the nearest 10 mg, e.g., 22.5 kg = 50 mg HYDROCORTISONE Sample pediatric dosing calculations: • 10 kg (22 lb) = 20 mg • 30 kg (66 lb) = 60 mg • 50 kg (110 lb) and up = 100 mg When possible, IV fluids (normal saline with dextrose) should be given: 20 mL/kg bolus (repeat up to 60 mL/kg). Hydrocortisone is the steroid (glucocorticoid) of choice for adrenal crisis due to its rapid onset of action and its mineralocorticoid activity (needed for primary adrenal disease). Dexamethasone, along with adequate saline resuscitation, is an acceptable alternative (second-line) approach and can be used to prevent a delay in emergency therapy when injectable hydrocortisone is not immediately available. Note that dexamethasone, compared to hydrocortisone, has a delayed onset of action and lacks mineralocorticoid activity. Contact medical control if the patient is carrying alternate glucocorticoid medications. We encourage EMS directors to make injectable hydrocortisone available on rigs and enable its administration by EMS personnel.2 Kindly review and revise protocols accordingly.3 Thank you for your prompt consideration of this vital clinical alert. Kindest regards, CARES Foundation Medical and Scientific Advisory Board 1 (NASEMSO) 2 (EMS Training Module) 3 (Endocrine Society Guidelines) Mild injury GIVE INJECTABLE HC,1 THEN CALL 9-1-1 AND ADVISE THAT PATIENT HAS ADRENAL INSUFFICIENCY Last dose >2 hrs ago mom Patient has fever Last dose <2 hrs ago Patient has signs of ADRENAL CRISIS: • Unresponsive • Severe blood loss/shock • Persistent vomiting with dehydration, lethargy Single episode of vomiting If <102 F, give double the usual hydrocortisone dose until no fever x 24 hrs If >102 F, give triple the usual hydrocortisone dose until no fever x 24 hrs Give Tylenol or Motrin as needed Keep well-hydrated Repeat dose Give small sips of Pedialyte or other electrolyte drink Avoid acidic or spicy food Keeps dose down Treat injury and observe for signs of shock Recommended Injectable HC1 DOSES Age Dose 0-3 yrs 25 mg 4-9 yrs 50 mg 10+ yrs 100 mg If patient has both fever and vomiting, follow both paths 1Hydrocortisone sodium succinate for IV/IM use Prepared by Karen Lin Su, M.D. (Medical Director, CARES Foundation) Disclaimer: This document is intended for informational purposes only and should not be used in place of medical advice from the patient’s physician or other health-care provider. PHYSICIAN CONTACT INFORMATION, In case of emergency Name of Endocrinologist: _____M.D. Phone Number: ___ Fax Number: _____ Primary Care Provider/Pediatrician: ____M.D. Phone Number: ___ Fax Number: ______ These are the medications I/my child take(s) daily: _____ _____ _____ _____ ______ These are the medications I have/my child has taken today (includes any stress dosing for the day), including approximate time medication(s) was/were taken. _____ _____ _____ _____ _____ EMERGENCY ROOM INSTRUCTIONS I/my child, ___, have/has a rare, inherited, genetic disorder called Congenital Adrenal Hyperplasia (CAH). I am/my child is adrenally insufficient and steroid dependent. I/my child must be seen by a physician IMMEDIATELY because life threatening electrolyte disturbances/adrenal crises are possible with febrile illnesses, fluid depletion from vomiting and diarrhea, surgery, and serious injuries. Time in a waiting area or triage situation is not appropriate. Signs of adrenal crisis include, but are not limited to: weakness, dizziness, nausea and vomiting, hypotension, hypoglycemia, pallor, and lethargy. Treatment should include: IV fluids-D5 normal saline at 20cc/kg for at least one hour then continuous fluid replacement for dehydration and hypotension. STAT basic metabolic panel (sodium, potassium, chloride, carbon dioxide, glucose, BUN, creatinine, and calcium) Initial hydrocortisone IV bolus can be administered IM if IV access an issue • 25mg for children under age 3 • 50mg for children aged 3-10 • 100mg for children older than 10 years or weighing more than 40kg • 100mg for teens and adults Hydrocortisone as a continuous drip (if necessary) or in 4 divided doses IV bolus • 50mg/day for ages 0-3 • 75mg/day for ages 3-10 • 100mg/day for children older than 10 years or weighing more than 40kg • 100mg/day for teens and adults NOTE FROM PHYSICIAN: Please follow the above treatment instructions and contact me as soon as possible. (See panel to left for contact info). Patient’s health issues include: ____________ ___________ Thank you. Please call if additional information is needed. ________ (Physician signature) EMERGENCY INSTRUCTIONS Treatment for Congenital Adrenal Hyperplasia In times of stress Includes information on: • Stress Dosing • Hydrocortisone Injection • Emergency Room Instructions www.CARESFoundation.org info@caresfoundation.org 1-866-227-3737 Your or your child’s body does not make enough of certain essential hormones: cortisol, and in salt wasting CAH, aldosterone. Cortisol, which is produced by the adrenal glands, has many purposes in the body such as maintaining energy supply, fluid, electrolyte balance, blood pressure, normal blood sugar levels, and controlling the body’s reaction to physical stress. Aldosterone is used by the kidneys to maintain a normal blood sodium and fluid balance (salt and water). When cortisol and aldosterone are not produced by the body they must be replaced by medication. Extra hydrocortisone must be given during times of extreme physical stress such as fever, vomiting and diarrhea, surgery, and traumatic injuries (e.g., broken bones and concussions). The extra hydrocortisone is called a “stress dose.” The Florinef dose does not change. Make sure you discuss stress dosing with your or your child’s physician and you know how to proceed in the event of illness. If ill, call physician to alert him/her of your or your child’s condition. Typically, stress dosing is required when . . . FEVER IS GREATER THAN 100.5°F : DOUBLE the hydro-cortisone dose for the entire day FEVER IS GREATER THAN 102°F : TRIPLE the hydrocortisone dose for the entire day VOMITING: Triple dose with vomiting with or without a fever. If you vomit less than 30 minutes after taking the hydrocortisone stress dose, the medication likely was not absorbed and the dose should be repeated. Wait 10-15 minutes after you/your child vomit(s) and repeat triple stress dose of hydrocortisone. If you/your child vomit(s) again, give the injectable hydrocortisone (brand name Solu-Cortef® in the U.S.) and contact your physician. DO NOT DELAY in giving the injectable hydrocortisone. DIARRHEA: Injection may also be needed in the event of diarrhea due to loss of fluids. If diarrhea, no fever and feeling fine, no need to stress dose. If not feeling well, double dose of hydrocortisone recommended. Try small amounts of clear liquids that contain SUGAR (not artificial sweetener) frequently, at least 1 ounce every 15 minutes. Signs of acute adrenal crisis from cortisol deficiency: • Headache • Nausea • Abdominal pain • Confusion • Pale skin • Listlessness • Dehydration • Dizziness If these occur and continue after stress oral dosing, call your or your child’s physician and go to the nearest emergency room immediately. Again, do NOT wait to give the injectable hydrocortisone. It should be given BEFORE a trip to the emergency room or activating 911 if those actions become necessary. EXAMPLE STRESS DOSING: Normal dose: 1 tab + 1/2 tab + 1 tab (total of 2.5 pills) Double dose: 5 total tablets (divide into 1.5 tablets every 8 hours) Triple dose: 7.5 total tablets (divide into 2.5 tablets every 8 hours) HOW TO GIVE AN INJECTION OF HYDROCORTISONE 1. STAY CALM. Wash your hands and gather equipment: needle, syringe, alcohol pad, and vial of hydrocortisone (Solu-Cortef® Act-O-Vial). 2. Mix the medication by pushing down on top of the vial to release the cork into the vial. 3. Shake the vial to mix medicine, take off the top of the vial, and wipe down the rubber stopper with alcohol. 4. Take the cap off the syringe needle and insert into the vial through the rubber stopper. 5. Draw up the medication and replace the needle cap. 6. Select the site for the intramuscular injection typically the outer portion in the middle of the thigh. 7. Use the alcohol to clean the skin at the injection site. 8. Take off the cap of the needle and hold the syringe like a dart. 9. Using your thumb and first two fingers, spread the skin and push down lightly. 10. Dart the needle into the thigh, going at a 90° angle 11. Hold the syringe in place and pull back the plunger to make sure you don’t see blood (which would mean you are in a blood vessel.If you do (which would be rare), withdraw syringe and discard. Prepare another syringe with medication and inject in a slightly different site. (However, if this is the only dose you have, continue with the same syringe, injecting in a slightly different site). 12. Inject medicine then place tissue or cotton ball near the needle. Pull the needle out quickly. 13. Place the needle and syringe in a hard, unbreakable container. 14. Call doctor/911 or go to hospital, if necessary. Steps 2-3 Steps 4-5 Mix the medication and Insert the needle into shake the vial the vial and draw up the medication Steps 6-7 Steps 8-10 Select the site for injec- Hold the syringe like a tion and use the alcohol dart, spread the skin by to clean the skin. pushing down slightly, and dart the needle into the thigh at a 90° angle. Step 11 Step 12 Pull back the plunger to Inject medicine then check for blood. place tissue/cotton ball near the needle and pull needle out quickly. Pictures reprinted with permission from the National Institute of Health Remember, stress dose with: • Fever of 100.5° F or greater • Vomiting • Diarrhea • Physical Trauma (broken bone, concussion, etc.)
15066	https://math.stackexchange.com/questions/1718733/in-a-triangle-abc-if-cot-a-cot-b-cot-c-sqrt3-then-prove-that-tri	algebra precalculus - In a $\triangle ABC,$ If $\cot A+\cot B+\cot C =\sqrt{3},$ Then prove that $\triangle$ is equilateral. - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more In a △A B C,△A B C, If cot A+cot B+cot C=3–√,cot⁡A+cot⁡B+cot⁡C=3, Then prove that △△ is equilateral. [duplicate] Ask Question Asked 9 years, 6 months ago Modified9 years, 6 months ago Viewed 607 times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. This question already has answers here: How to show that the triangle is equilateral triangle? (2 answers) Closed 9 years ago. In a △A B C,△A B C, If cot A+cot B+cot C=3–√,cot⁡A+cot⁡B+cot⁡C=3, Then prove that △△ is equilateral. M y T r y::M y T r y:: Using Jensen's Inequality, Let f(x)=cot x,f(x)=cot⁡x, Where x∈(0,π),x∈(0,π), Then f′(x)=−csc 2 x f′(x)=−csc 2⁡x and f′′(x)=2 csc 2 x⋅cot x f″(x)=2 csc 2⁡x⋅cot⁡x So we get f′′(x)=2 cos x sin 3 x>0 f″(x)=2 cos⁡x sin 3⁡x>0 in x∈(0,π)−{π 2}x∈(0,π)−{π 2} So cot A+cot B+cot C 3≥cot(A+B+C 3)=1 3–√cot⁡A+cot⁡B+cot⁡C 3≥cot⁡(A+B+C 3)=1 3 So we get cot A+cot B+cot C≥3–√cot⁡A+cot⁡B+cot⁡C≥3 But i did not understand how can we prove that △△ are equilateral Thanks algebra-precalculus inequality Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Mar 29, 2016 at 14:20 SchrodingersCat 24.9k 7 7 gold badges 47 47 silver badges 90 90 bronze badges asked Mar 29, 2016 at 13:20 juantheronjuantheron 56.3k 19 19 gold badges 113 113 silver badges 274 274 bronze badges 4 Maybe this will help : If a triangle is equilateral, each of its angles are equal to 60°Tony Barbé –Tony Barbé 2016-03-29 13:24:33 +00:00 Commented Mar 29, 2016 at 13:24 1 Hint: consider when you get an equality in Jensen's inequality.Wojowu –Wojowu 2016-03-29 13:24:43 +00:00 Commented Mar 29, 2016 at 13:24 Thanks lab bhattacharjee, I have seen that, but i want to solve it using Jenson inequality,but i did not understand why not f′′(x)>0 f″(x)>0 for all x∈(0,π)x∈(0,π) , here i have mention that f′′(x)=0 f″(x)=0 at x=π 2 x=π 2,juantheron –juantheron 2016-03-29 13:27:33 +00:00 Commented Mar 29, 2016 at 13:27 1 Suppose x∈(π 2,π)x∈(π 2,π), then cos x<0,sin x>0⟹f′′(x)<0 cos⁡x<0,sin⁡x>0⟹f″(x)<0 so it's not straightforward Jensen is it?Macavity –Macavity 2016-03-29 14:16:55 +00:00 Commented Mar 29, 2016 at 14:16 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful -1 Save this answer. Show activity on this post. You have completed the sum almost all by yourself. That's a really good try using Jensen's inequality. For a real convex function φ, numbers x 1 x 1, x 2 x 2, ...,x n x n in its domain, and positive weights a i a i, Jensen's inequality can be stated as: φ(∑a i x i∑a i)≤∑a i φ(x i)∑a i(1)φ(∑a i x i∑a i)≤∑a i φ(x i)∑a i(1) and the inequality is reversed if φ is concave, which is φ(∑a i x i∑a i)≥∑a i φ(x i)∑a i.(2)φ(∑a i x i∑a i)≥∑a i φ(x i)∑a i.(2) Equality holds if and only if x 1=x 2=⋯=x n x 1=x 2=⋯=x n or φ φ is linear. It is given here that cot A+cot B+cot C=3–√cot⁡A+cot⁡B+cot⁡C=3 Now, observe that cot x cot⁡x is not linear. So the equality in Jensen's inequality holds only if cot A=cot B=cot C cot⁡A=cot⁡B=cot⁡C or in other words, A=B=C A=B=C It is trivial to conclude now that A=B=C=60∘A=B=C=60∘ and hence the triangle is equilateral. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Mar 29, 2016 at 15:03 answered Mar 29, 2016 at 13:48 SchrodingersCatSchrodingersCat 24.9k 7 7 gold badges 47 47 silver badges 90 90 bronze badges 6 @Macavity Yes, you are right. But the equality condition remains the same in both the cases.SchrodingersCat –SchrodingersCat 2016-03-29 14:21:22 +00:00 Commented Mar 29, 2016 at 14:21 My point is not about the equality condition. My point is Jensen inequality is not applicable at all if the function is convex in part of the interval and concave in another part. So while I am not downvoting, the whole proof is wrong.Macavity –Macavity 2016-03-29 14:23:37 +00:00 Commented Mar 29, 2016 at 14:23 @Macavity Why can't we divide the interval into 2 parts? (0,π 2](0,π 2] and [π 2,π)[π 2,π). And then separately apply the inequality. That should be enough I hope... :-)SchrodingersCat –SchrodingersCat 2016-03-29 14:25:50 +00:00 Commented Mar 29, 2016 at 14:25 2 Because the inequality reverses when the function is concave in part of the intervale. And here you have two points where it is convex and one in which it could be concave. There are workarounds for simple cases, but it clearly is far away from the traditional Jensen inequality.Macavity –Macavity 2016-03-29 14:33:45 +00:00 Commented Mar 29, 2016 at 14:33 @Macavity I understand your point. But please note what I said in my first comment, the inequality might reverse in either intervals, still I can divide the interval into 2 parts, apply the inequality in both separately and yet the e q u a l i t y e q u a l i t y c o n d i t i o n c o n d i t i o n remains the same and hence the conclusion can be drawn.SchrodingersCat –SchrodingersCat 2016-03-29 14:37:11 +00:00 Commented Mar 29, 2016 at 14:37 \|Show 1 more comment Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algebra-precalculus inequality See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 2How to show that the triangle is equilateral triangle? Related 10proving the inequality △≤1 4(a+b+c)⋅a b c−−−−−−−−−−−−−√△≤1 4(a+b+c)⋅a b c 5Area of △A B C△A B C whose sides a,b,c a,b,c satisfy 0≤a≤1;1≤b≤2;2≤c≤3 0≤a≤1;1≤b≤2;2≤c≤3 is 4Prove that ∑c y c cot A+cot B−−−−−−−−−−−√≥2 2–√∑c y c cot⁡A+cot⁡B≥2 2 2In a △A B C,∠B=60 0,△A B C,∠B=60 0, Then range of sin A sin C sin⁡A sin⁡C 20Proving a 2 x+b 2 y+c 2 z≥4[A B C]x y+x z+y z−−−−−−−−−−√a 2 x+b 2 y+c 2 z≥4[A B C]x y+x z+y z, for a a, b b, c c the sides of △A B C△A B C, [A B C][A B C] its area, and x x, y y, z z positive reals 9If 2(b c 2+c a 2+a b 2)=b 2 c+c 2 a+a 2 b+3 a b c,2(b c 2+c a 2+a b 2)=b 2 c+c 2 a+a 2 b+3 a b c, then prove that triangle A B C A B C is equilateral 5In an Acute angled Triangle Δ A B C Δ A B C csc(A 2)+csc(B 2)+csc(C 2)=6 csc⁡(A 2)+csc⁡(B 2)+csc⁡(C 2)=6 4Prove ∑tan A−−−−−√≥∑cot A 2−−−−−√∑tan⁡A≥∑cot⁡A 2 4Prove that in a triangle : cot A+cot B+cot C≥3–√cot⁡A+cot⁡B+cot⁡C≥3 Hot Network Questions Making sense of perturbation theory in many-body physics An odd question Riffle a list of binary functions into list of arguments to produce a result What is the feature between the Attendant Call and Ground Call push buttons on a B737 overhead panel? 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15067	https://www.graphpad.com/guides/the-ultimate-guide-to-linear-regression	Prism Prism Overview Analyze, graph and present your work Comprehensive analysis and statistics Elegant graphing and visualizations Share, view and discuss your projects What’s New Latest product features and releases POPULAR USE CASES Power & Sample Size Survival Analysis Non-Linear Regression Machine Learning Flow Cytometry INTEGRATIONS OMIQ Flow Cytometry FCS Express LabArchives Enterprise Resources Support Pricing Cart Sign In Free Trial HomeResourcesUltimate Guide to Linear Regression Ultimate Guide to Linear Regression Get all your linear regression questions answered here. Get the eBook The Ultimate Guide to Linear Regression What is regression? What is the difference between the variables in regression? What are the purposes of regression analysis? How do I know which model best fits the data? What is linear regression? What are the major advantages of linear regression analysis? Assumptions of linear regression Types of linear regression Simple linear regression Multiple linear regression Model selection - choosing which predictor variables to include When linear regression doesn't work Other types of regression Perform your own Linear Regression The Ultimate Guide to Linear Regression Welcome! When most people think of statistical models, their first thought is linear regression models. What most people don’t realize is that linear regression is a specific type of regression. With that in mind, we’ll start with an overview of regression models as a whole. Then after we understand the purpose, we’ll focus on the linear part, including why it’s so popular and how to calculate regression lines-of-best-fit! (Or, if you already understand regression, you can skip straight down to the linear part). This guide will help you run and understand the intuition behind linear regression models. It’s intended to be a refresher resource for scientists and researchers, as well as to help new students gain better intuition about this useful modeling tool. What is regression? In its simplest form, regression is a type of model that uses one or more variables to estimate the actual values of another. There are plenty of different kinds of regression models, including the most commonly used linear regression, but they all have the basics in common. Usually the researcher has a response variable they are interested in predicting, and an idea of one or more predictor variables that could help in making an educated guess. Some simple examples include: Predicting the progression of a disease such as diabetes using predictors such as age, cholesterol, etc. (linear regression) Predicting survival rates or time-to-failure based on explanatory variables (survival analysis) Predicting political affiliation based on a person’s income level and years of education (logistic regression or some other classifier) Predicting drug inhibition concentration at various dosages (nonlinear regression) There are all sorts of applications, but the point is this: If we have a dataset of observations that links those variables together for each item in the dataset, we can regress the response on the predictors. Furthermore: Fitting a model to your data can tell you how one variable increases or decreases as the value of another variable changes. For example, if we have a dataset of houses that includes both their size and selling price, a regression model can help quantify the relationship between the two. (Not that any model will be perfect for this!) The most noticeable aspect of a regression model is the equation it produces. This model equation gives a line of best fit, which can be used to produce estimates of a response variable based on any value of the predictors (within reason). We call the output of the model a point estimate because it is a point on the continuum of possibilities. Of course, how good that prediction actually depends on everything from the accuracy of the data you’re putting in the model to how hard the question is in the first place. Compare this to other methods like correlation, which can tell you the strength of the relationship between the variables, but is not helpful in estimating point estimates of the actual values for the response. What is the difference between the variables in regression? There are two different kinds of variables in regression: The one which helps predict (predictors), and the one you’re trying to predict (response). Predictors were historically called independent variables in science textbooks. You may also see them referred to as x-variables, regressors, inputs, or covariates. Depending on the type of regression model you can have multiple predictor variables, which is called multiple regression. Predictors can be either continuous (numerical values such as height and weight) or categorical (levels of categories such as truck/SUV/motorcycle). The response variable is often explained in layman’s terms as “the thing you actually want to predict or know more about”. It is usually the focus of the study and can be referred to as the dependent variable, y-variable, outcome, or target. In general, the response variable has a single value for each observation (e.g., predicting the temperature based on some other variables), but there can be multiple values (e.g., predicting the location of an object in latitude and longitude). The latter case is called multivariate regression (not to be confused with multiple regression). What are the purposes of regression analysis? Regression Analysis has two main purposes: Explanatory - A regression analysis explains the relationship between the response and predictor variables. For example, it can answer questions such as, does kidney function increase the severity of symptoms in some particular disease process? Predictive - A regression model can give a point estimate of the response variable based on the value of the predictors. How do I know which model best fits the data? The most common way of determining the best model is by choosing the one that minimizes the squared difference between the actual values and the model’s estimated values. This is called least squares. Note that “least squares regression” is often used as a moniker for linear regression even though least squares is used for linear as well as nonlinear and other types of regression. What is linear regression? The most popular form of regression is linear regression, which is used to predict the value of one numeric (continuous) response variable based on one or more predictor variables (continuous or categorical). Most people think the name “linear regression” comes from a straight line relationship between the variables. For most cases, that’s a fine way to think of it intuitively: As a predictor variable increases, the response either increases or decreases at the same rate (all other things equal). If this relationship holds the same for any values of the variables, a straight line pattern will form in the data when graphed, as in the example below: However, the actual reason that it’s called linear regression is technical and has enough subtlety that it often causes confusion. For example, the graph below is linear regression, too, even though the resulting line is curved. The definition is mathematical and has to do with how the predictor variables relate to the response variable. Suffice it to say that linear regression handles most simple relationships, but can’t do complicated mathematical operations such as raising one predictor variable to the power of another predictor variable. The most common linear regression models use the ordinary least squares algorithm to pick the parameters in the model and form the best line possible to show the relationship (the line-of-best-fit). Though it’s an algorithm shared by many models, linear regression is by far the most common application. If someone is discussing least-squares regression, it is more likely than not that they are talking about linear regression. What are the major advantages of linear regression analysis? Linear regression models are known for being easy to interpret thanks to the applications of the model equation, both for understanding the underlying relationship and in applying the model to predictions. The fact that regression analysis is great for explanatory analysis and often good enough for prediction is rare among modeling techniques. In contrast, most techniques do one or the other. For example, a well-tuned AI-based artificial neural network model may be great at prediction but is a “black box” that offers little to no interpretability. There are some other benefits too: Linear regression is computationally fast, particularly if you’re using statistical software. Though it’s not always a simple task to do by hand, it’s still much faster than the days it would take to calculate many other models. The popularity of regression models is itself an advantage. The fact that it is a tried and tested approach used by so many scientists makes for easy collaboration. Assumptions of linear regression Just because scientists' initial reaction is usually to try a linear regression model, that doesn't mean it is always the right choice. In fact, there are some underlying assumptions that, if ignored, could invalidate the model. Random sample - The observations in your data need to be independent from one another. There are many ways that dependence occurs, for example, one common way is with multiple response data, where a single subject is measured multiple times. The measurements on the same individual are presumably correlated, and you couldn’t use linear regression in this case. Independence between predictors - If you have multiple predictors in your model, in theory, they shouldn’t be correlated with one another. If they are, this can cause instability in your model fit, although this affects the interpretation of your model rather than the predictions. See more about multicollinearity here. Homoscedasticity - Meaning ‘equal scatter,’ this says that your residuals (the difference between the model prediction and the observed values) should be just as variable anywhere along the continuum. This is assessed with residual plots. Residuals are normally distributed - In addition to having equal scatter, in the standard linear regression model, the residuals are assumed to come from a normal distribution. This is commonly assessed using a QQ-plot. Linear relationship between predictors and response - The relationships must be linear as described above, ruling out some more complicated mathematical relationships. You can model some “curves” in your data using, say, variable X and variable X^2 ("X squared") as predictors. No uncertainty in predictor measurements - The model assumes that all of the uncertainty is in the response variable. This is the most nuanced assumption: Even if you’re attempting to make inferences about a model with predictors that are themselves estimates, this would not affect you unless you need to attribute the uncertainty to the predictors. This field of study is called “measurement error.” Other things to keep in mind for valid inference: Representative sample - The dataset you are going to use should be a representative (and random!) sample of the population you’re trying to make inferences about. To use an intuitive example, you should not expect all people to act the same as those in your household. Since we often underestimate our own bias, the best bet is to have a random sample when you start. Sample size - If your dataset only has 5 observations in it, the model will be less effective at finding a real pattern (or if one exists) than if it has 100. There is no one-size-fits-all number for every study, but generally 30 or more is considered the low end of what regression needs. Stay in range - Don’t try to make predictions outside the range of the dataset you used to build the model. For example, let’s say you are predicting home values based on square footage. If your dataset only has homes between 1,000 and 3,000 square feet, the model may not be a good judge of the value of an 800 or 4,000 square-foot house. This is called extrapolating, and is not recommended. Types of linear regression The two most common types of regression are simple linear regression and multiple linear regression, which only differ by the number of predictors in the model. Simple linear regression has a single predictor. Simple linear regression It’s called simple for a reason: If you are testing a linear relationship between exactly two continuous variables (one predictor and one response variable), you’re looking for a simple linear regression model, also called a least squares regression line. Are you looking to use more predictors than that? Try a multiple linear regression model. That is the main difference between the two, but there are other considerations and differences involved too. You can use statistical software such as Prism to calculate simple linear regression coefficients and graph the regression line it produces. For a quick simple linear regression analysis, try our free online linear regression calculator. Interpreting a simple linear regression model Remember the y = mx+b formula for a line from grade school? The slope was m, and the y-intercept was b, and both were necessary to draw a line. That’s what you’re basically building here too, but most textbooks and programs will write out the predictive equation for regression this way: Y is your response variable, and X is your predictor. The two 𝛽 symbols are called “parameters”, the things the model will estimate to create your line of best fit. The first (not connected to X) is the intercept, the other (the coefficient in front of X) is called the slope term. As an example, we will use a sample Prism dataset with diabetes data to model the relationship between a person’s glucose level (predictor) and their glycosylated hemoglobin level (response). Once we run the analysis we get this output: Best-fit parameters and the regression equation The first section in the Prism output for simple linear regression is all about the workings of the model itself. They can be called parameters, estimates, or (as they are above) best-fit values. Keep in mind, parameter estimates could be positive or negative in regression depending on the relationship. There you see the slope (for glucose) and the y-intercept. The values for those help us build the equation the model uses to estimate and make predictions: Glycosylated Hemoglobin = 2.24 + (0.0312Glucose) Notice: That same equation is given later in the output, near the bottom of the page. Using this equation, we can plug in any number in the range of our dataset for glucose and estimate that person’s glycosylated hemoglobin level. For instance, a glucose level of 90 corresponds to an estimate of 5.048 for that person’s glycosylated hemoglobin level. But that’s just the start of how these parameters are used. Interpreting parameter estimates You can also interpret the parameters of simple linear regression on their own, and because there are only two it is pretty straightforward. The slope parameter is often the most helpful: It means that for every 1 unit increase in glucose, the estimated glycosylated hemoglobin level will increase by 0.0312 units. As an aside, if it was negative (perhaps -0.04), we would say a 1 unit increase in glucose would actually decrease the estimated response by -0.04. The intercept parameter is useful for fitting the model, because it shifts the best-fit-line up or down. In this example, the value it shows (2.24) is the predicted glycosylated hemoglobin level for a person with a glucose level of 0. In cases like this, the interpretation of the intercept isn’t very interesting or helpful. Simply put, if there’s no predictor with a value of 0 in the dataset, you should ignore this part of the interpretation and consider the model as a whole and the slope. However, notice that if you plug in 0 for a person’s glucose, 2.24 is exactly what the full model estimates. Confidence intervals and standard error The next couple sections seem technical, but really get back to the core of how no model is perfect. We can give “point estimates” for the best-fit parameters today, but there’s still some uncertainty involved in trying to find the true and exact relationship between the variables. Standard error and confidence intervals work together to give an estimate of that uncertainty. Add and subtract the standard error from the estimate to get a fair range of possible values for that true relationship. With this 95% confidence interval, you can say you believe the true value of that parameter is somewhere between the two endpoints (for the slope of glucose, somewhere between 0.0285 and 0.0340). This method may seem too cautious at first, but is simply giving a range of real possibilities around the point estimate. After all, wouldn’t you like to know if the point estimate you gave was wildly variable? This gives you that missing piece. Goodness of fit Determining how well your model fits can be done graphically and numerically. If you know what to look for, there’s nothing better than plotting your data to assess the fit and how well your data meet the assumptions of the model. These diagnostic graphics plot the residuals, which are the differences between the estimated model and the observed data points. A good plot to use is a residual plot versus the predictor (X) variable. Here you want to look for equal scatter, meaning the points all vary roughly the same above and below the dotted line across all x values. The plot on the left looks great, whereas the plot on the right shows a clear parabolic shaped trend, which would need to be addressed. Another way to assess the goodness of fit is with the R-squared statistic, which is the proportion of the variance in the response that is explained by the model. In this case, the value of 0.561 says that 56% of the variance in glycosylated hemoglobin can be explained by this very simple model equation (effectively, that person’s glucose level). The name R-squared may remind you of a similar statistic: Pearson’s R, which measures the correlation between any two variables. Fun fact: As long as you’re doing simple linear regression, the square-root of R-squared (which is to say, R), is equivalent to the Pearson’s R correlation between the predictor and response variable. The reason is that simple linear regression draws on the same mechanisms of least-squares that Pearson’s R does for correlation. Keep in mind, while regression and correlation are similar they are not the same thing. The differences usually come down to the purpose of the analysis, as correlation does not fit a line through the data points. Significance and F-tests So we have a model, and we know how to use it for predictions. We know R-squared gives an idea of how well the model fits the data… but how do we know if there is actually a significant relationship between the variables? A section at the bottom asks that same question: Is the slope significantly non-zero? This is especially important for this model, where the best-fit value (roughly 0.03) seems very close to 0 to the naked eye. How can we feel confident one way or another? In this case, the slope is significantly non-zero: An F-test gives a p-value of less than 0.0001. F-tests answer this for the model as a whole rather than its individual slopes, but in this case there is only one slope anyway. P-values are always interpreted in comparison to a “significance threshold”: If it’s less than the threshold level, the model is said to show a trend that is significantly different from “no relationship” (or, the null hypothesis). And based on how we set up the regression analysis to use 0.05 as the threshold for significance, it tells us that the model points to a significant relationship. There is evidence that this relationship is real. If it wasn’t, then we are effectively saying there is no evidence that the model gives any new information beyond random guessing. In other words: The model may output a number for a prediction, but if the slope is not significant, it may not be worth actually considering that prediction. Graphing linear regression Since a linear regression model produces an equation for a line, graphing linear regression’s line-of-best-fit in relation to the points themselves is a popular way to see how closely the model fits the eye test. Software like Prism makes the graphing part of regression incredibly easy, because a graph is created automatically alongside the details of the model. Here are some more graphing tips, along with an example from our analysis: Multiple linear regression If you understand the basics of simple linear regression, you understand about 80% of multiple linear regression, too. The inner-workings are the same, it is still based on the least-squares regression algorithm, and it is still a model designed to predict a response. But instead of just one predictor variable, multiple linear regression uses multiple predictors. The model equation is similar to the previous one, the main thing you notice is that it’s longer because of the additional predictors. Let’s say you are using 3 predictor variables, the predictive equation will produce 3 slope estimates (one for each) along with an Intercept term: Prism makes it easy to create a multiple linear regression model, especially calculating regression slope coefficients and generating graphics to diagnose how well the model fits. What do I need to know about multicollinearity? The assumptions for multiple linear regression are discussed here. With multiple predictors, in addition to the interpretation getting more challenging, another added complication is with multicollinearity. Multicollinearity occurs when two or more predictor variables “overlap” in what they measure. In other places you will see this referred to as the variables being dependent of one another. Ideally, the predictors are independent and no one predictor influences the values of another. There are various ways of measuring multicollinearity, but the main thing to know is that multicollinearity won’t affect how well your model predicts point values. However, it garbles inference about how each individual variable affects the response. For example, say that you want to estimate the height of a tree, and you have measured the circumference of the tree at two heights from the ground, one meter and two meter. The circumferences will be highly correlated. If you include both in the model, it’s very possible that you could end up with a negative slope parameter for one of those circumferences. Clearly, a tree doesn't get shorter when the circumference gets larger. Instead, that negative slope coefficient is acting as an adjustment to the other variable. What is the difference between simple linear regression and multiple linear regression? Once you’ve decided that your study is a good fit for a linear model, the choice between the two simply comes down to how many predictor variables you include. Just one? Simple linear. More than that? Multiple linear. Based on that, you may be wondering, “Why would I ever do a simple linear regression when multiple linear regression can account for more variables?” Great question! The answer is that sometimes less is more. A common misconception is that the goal of a model is to be 100% accurate. Scientists know that no model is perfect, it is a simplified version of reality. So the goal isn’t perfection: Rather, the goal is to find as simple a model as possible to describe relationships so you understand the system, reach valid scientific conclusions, and design new experiments. Still not convinced? Let’s say you were able to create a model that was 100% accurate for each point in your dataset. Most of the time if you’ve done this, you’ve done one of two things: Come to an obvious conclusion that isn’t practically useful (100% of winning basketball teams score more points than their opponent) OR You’ve modeled not only the trend in your data, but also the random “noise” that is too variable to count on. This is called “overfitting”: You tried so hard to account for every aspect of the past that the model ignores the differences that will arise in the future. Other differences pop up on the technical side. To give some quick examples of that, using multiple linear regression means that: In addition to the overall interpretation and significance of the model, each slope now has its own interpretation and question of significance. R-squared is not as intuitive as it was for simple linear regression. Graphing the equation is not a single line anymore. You could say that multiple linear regression just does not lend itself to graphing as easily. All in all: simple regression is always more intuitive than multiple linear regression! Interpreting multiple linear regression We’ve said that multiple linear regression is harder to interpret than simple linear regression, and that is true. Taking the math and more technical aspects out of the question, overall interpretation is always harder the more factors are involved. But while there are more things to keep track of, the basic components of the thought process remain the same: parameters, confidence intervals and significance. We even use the model equation the same way. Let’s use the same diabetes dataset to illustrate, but with a new wrinkle: In addition to glucose level, we will also include HDL and the person’s age as predictors of their glycosylated hemoglobin level (response). Here’s the output from Prism: Analysis of variance and F-tests While most scientists’ eyes go straight to the section with parameter estimates, the first section of output is valuable and is the best place to start. Analysis of variance tests the model as a whole (and some individual pieces) to tell you how good your model is before you make sense of the rest. It includes the Sum of Squares table, and the F-test on the far right of that section is of highest interest. The “Regression” as a whole (on the top line of the section) has a p-value of less than 0.0001 and is significant at the 0.05 level we chose to use. Each parameter slope has its own individual F-test too, but it is easier to understand as a t-test. Parameter estimates and T-tests Now for the fun part: The model itself has the same structure and information we used for simple linear regression, and we interpret it very similarly. The key is to remember that you are interpreting each parameter in its own right (not something you have to keep in mind with only one parameter!). Prism puts all of the statistics for each parameters in one table, including (for each parameter): The parameter’s estimate itself Its standard error and confidence interval A P-value from a t-test The estimates themselves are straightforward and are used to make the model equation, just like before. In this case the model’s predictive equation is (when rounding to the nearest thousandth): Glycosylated Hemoglobin = 1.870 + 0.029Glucose - 0.005HDL +0.018Age If you remember back to our simple linear regression model, the slope for glucose has changed slightly. That is because we are now accounting for other factors too. This distinction can sometimes change the interpretation of an individual predictor’s effect dramatically. When interpreting the individual slope estimates for predictor variables, the difference goes back to how Multiple Regression assumes each predictor is independent of the others. For simple regression you can say “a 1 point increase in X usually corresponds to a 5 point increase in Y”. For multiple regression it’s more like “a 1 point increase in X usually corresponds to a 5 point increase in Y, assuming every other factor is equal.” That may not seem like a big jump, but it acknowledges 1) that there are more factors at play and 2) the need for those predictors to not have influence on one another for the model to be helpful. The standard errors and confidence intervals are also shown for each parameter, giving an idea of the variability for each slope/intercept on its own. Interpreting each one of these is done exactly the same way as we mentioned in the simple linear regression example, but remember that if multicollinearity exists, the standard errors and confidence intervals get inflated (often drastically). On the end are p-values, which as you might guess, are interpreted just like we did for the first example. The underlying method behind the p-value here is a T-test. These only tell how significant each of the factors are, to evaluate the model as a whole we would need to use the F-test at the top. Evaluating each on its own though is still helpful: In this case it shows that while the other predictors are all significant, HDL shows no significance since we have already considered the other factors. That is not to say that it has no significance on its own, only that it adds no value to a model of just glucose and age. In fact, now that we know this, we could choose to re-run our model with only glucose and age and dial in better parameter estimates for that simpler model. Another difference in interpretation occurs when you have categorical predictor variables such as sex in our example data. When you add categorical variables to a model, you pick a “reference level.” In this case (image below), we selected female as our reference level. The model below says that males have slightly lower predicted response than females (about 0.15 less). Goodness of fit Assessing how well your model fits with multiple linear regression is more difficult than with simple linear regression, although the ideas remain the same, i.e., there are graphical and numerical diagnoses. At the very least, it’s good to check a residual vs predicted plot to look for trends. In our diabetes model, this plot (included below) looks okay at first, but has some issues. Notice that values tend to miss high on the left and low on the right. However, on further inspection, notice that there are only a few outlying points causing this unequal scatter. If you see outliers like above in your analysis that disrupt equal scatter, you have a few options. As for numerical evaluations of goodness of fit, you have a lot more options with multiple linear regression. R-squared is still a go-to if you just want a measure to describe the proportion of variance in the response variable that is explained by your model. However, a common use of the goodness of fit statistics is to perform model selection, which means deciding on what variables to include in the model. If that’s what you’re using the goodness of fit for, then you’re better off using adjusted R-squared or an information criterion such as AICc. Graphing multiple linear regression Graphs are extremely useful to test how well a multiple linear regression model fits overall. With multiple predictors, it’s not feasible to plot the predictors against the response variable like it is in simple linear regression. A simple solution is to use the predicted response value on the x-axis and the residuals on the y-axis (as shown above). As a reminder, the residuals are the differences between the predicted and the observed response values. There are also several other plots using residuals that can be used to assess other model assumptions such as normally distributed error terms and serial correlation. Model selection - choosing which predictor variables to include How do you know which predictor variables to include in your model? It’s a great question and an active area of research. For most researchers in the sciences, you’re dealing with a few predictor variables, and you have a pretty good hypothesis about the general structure of your model. If this is the case, then you might just try fitting a few different models, and picking the one that looks best based on how the residuals look and using a goodness of fit metric such as adjusted R-square or AICc. Why doesn't my model fit well? There are a lot of reasons that would cause your model to not fit well. One reason is having too much unexplained variance in the response. This could be because there were important predictor variables that you didn’t measure, or the relationship between the predictors and the response is more complicated than a simple linear regression model. In this last case, you can consider using interaction terms or transformations of the predictor variables. If prediction accuracy is all that matters to you, meaning that you only want a good estimate of the response and don’t need to understand how the predictors affect it, then there are a lot of clever, computational tools for building and selecting models. We won’t cover them in this guide, but if you want to know more about this topic, look into cross-validation and LASSO regression to get started. Interactions Interactions and transformations are useful tools to address situations where your model doesn't fit well by just using the unmodified predictor variables. Interaction terms are found by multiplying two predictor variables together to create a new “interaction” variable. They greatly increase the complexity of describing how each variable affects the response. The primary use is to allow for more flexibility so that the effect of one predictor variable depends on the value of another predictor variable. For a specific example using the diabetes data above, perhaps we have reason to believe that the effect of glucose on the response (hemoglobin %) changes depending on the age of the patient. Stats software makes this simple to do, but in effect, we multiply glucose by age, and include that new term in our model. Our new model when rounded is: Glycosylated Hemoglobin = 0.42 + 0.044Glucose - 0.004HDL +0.044Age - .0003GlucoseAge For reference, our model without the interaction term was: Glycosylated Hemoglobin = 1.865 + 0.029Glucose - 0.005HDL +0.018Age Adding the interaction term changed the other estimates by a lot! Interpreting what this means is challenging. At the very least, we can say that the effect of glucose depends on age for this model since the coefficients are statistically significant. We might also want to say that high glucose appears to matter less for older patients due to the negative coefficient estimate of the interaction term (-0.0002). However, there is very high multicollinearity in this model (and in nearly every model with interaction terms), so interpreting the coefficients should be done with caution. Even with this example, if we remove a few outliers, this interaction term is no longer statistically significant, so it is unstable and could simply be a byproduct of noisy data. Transformations In addition to interactions, another strategy to use when your model doesn't fit your data well are transformations of variables. You can transform your response or any of your predictor variables. Transformations on the response variable change the interpretation quite a bit. Instead of the model fitting your response variable, y, it fits the transformed y. A common example where this is appropriate is with predicting height for various ages of an animal species. Log transformations on the response, height in this case, are used because the variability in height at birth is very small, but the variability of height with adult animals is much higher. This violates the assumption of equal scatter. In the plots below, notice the funnel type shape on the left, where the scatter widens as age increases. On the right hand side, the funnel shape disappears and the variability of the residuals looks consistent. The linear model using the log transformed y fits much better, however now the interpretation of the model changes. Using the example data above, the predicted model is: ln(y) = -0.4 + 0.2 x This means that a single unit change in x results in a 0.2 increase in the log of y. That doesn't mean much to most people. Instead, you probably want your interpretation to be on the original y scale. To do that, we need to exponentiate both sides of the equation, which (avoiding the mathematical details) means that a 1 unit increase in x results in a 22% increase in y. All of that is to say that transformations can assist with fitting your model, but they can complicate interpretation. When linear regression doesn't work The ubiquitous nature of linear regression is a positive for collaboration, but sometimes it causes researchers to assume (before doing their due diligence) that a linear regression model is the right model for every situation. Sometimes software even seems to reinforce this attitude and the model that is subsequently chosen, rather than the person remaining in control of their research. Sure, linear regression is great for its simplicity and familiarity, but there are many situations where there are better alternatives. Other types of regression Logistic regression Linear vs logistic regression: linear regression is appropriate when your response variable is continuous, but if your response has only two levels (e.g., presence/absence, yes/no, etc.), then look into simple logistic regression or multiple logistic regression. Poisson regression If instead, your response variable is a count (e.g., number of earthquakes in an area, number of males a female horseshoe crab has nesting nearby, etc.), then consider Poisson regression. Nonlinear regression For more complicated mathematical relationships between the predictors and response variables, such as dose-response curves in pharmacokinetics, check out nonlinear regression. ANOVA If you’ve designed and run an experiment with a continuous response variable and your research factors are categorical (e.g., Diet 1/Diet 2, Treatment 1/Treatment 2, etc.), then you need ANOVA models. These are differentiated by the number of treatments (one-way ANOVA, two-way ANOVA, three-way ANOVA) or other characteristics such as repeated measures ANOVA. Principal component regression Principal component regression is useful when you have as many or more predictor variables than observations in your study. It offers a technique for reducing the “dimension” of your predictors, so that you can still fit a linear regression model. Cox proportional hazards regression Cox proportional hazards regression is the go-to technique for survival analysis, when you have data measuring time until an event. Deming regression Deming regression is useful when there are two variables (x and y), and there is measurement error in both variables. One common situation that this occurs is comparing results from two different methods (e.g., comparing two different machines that measure blood oxygen level or that check for a particular pathogen). Perform your own Linear Regression Are you ready to calculate your own Linear Regression? With a consistently clear, practical, and well-documented interface, learn how Prism can give you the controls you need to fit your data and simplify nonlinear regression. Start your 30 day free trial of Prismand get access to: A step by step guide on how to perform Linear Regression Sample data to save you time More tips on how Prism can help your research With Prism, in a matter of minutes you learn how to go from entering data to performing statistical analyses and generating high-quality graphs. Analyze, graph and present your scientific work easily with GraphPad Prism. No coding required. Try for Free
15068	https://www.wyzant.com/resources/answers/723202/a-9-50kg-hanging-object-physics-help	a 9.50kg hanging object.. PHYSICS HELP \| Wyzant Ask An Expert Log inSign up Find A Tutor Search For Tutors Request A Tutor Online Tutoring How It Works For Students FAQ What Customers Say Resources Ask An Expert Search Questions Ask a Question Wyzant Blog Start Tutoring Apply Now About Tutors Jobs Find Tutoring Jobs How It Works For Tutors FAQ About Us About Us Careers Contact Us All Questions Search for a Question Find an Online Tutor Now Ask a Question for Free Login WYZANT TUTORING Log in Sign up Find A Tutor Search For Tutors Request A Tutor Online Tutoring How It Works For Students FAQ What Customers Say Resources Ask An Expert Search Questions Ask a Question Wyzant Blog Start Tutoring Apply Now About Tutors Jobs Find Tutoring Jobs How It Works For Tutors FAQ About Us About Us Careers Contact Us Subject ZIP Search SearchFind an Online Tutor NowAsk Ask a Question For Free Login MathPhysicsPhysics BPhysics C Li L. asked • 10/20/19 a 9.50kg hanging object.. PHYSICS HELP A 9.50-kg hanging object is connected by a light, inextensible cord over a light, frictionless pulley to a 5.00-kg block that is sliding on a flat table. Taking the coefficient of kinetic friction as 0.187, find the tension in the string. (The block slides to the right in the diagram below.) Follow •2 Add comment More Report 1 Expert Answer Best Newest Oldest By: Ian A.answered • 10/22/19 Tutor 5(30) Experienced High school physics and AP physics teacher See tutors like this See tutors like this From newtons 2nd law of motion, F net=ma. Since you have two objects (table weight we'll call object 1 and the hanging weight we'll call object 2) each object needs its own free body diagram and net force, but the mass will be the combined mass (m 1 + m 2) and the acceleration will be the same value. The equation for the hanging mass is easier to start with since it only has forces in the vertical direction. For the sake of this problem, lets call the direction of motion to be the positive direction. This means the hanging, which is moving down, to have a positive gravitational force (F g = mg) and a negative tension force acting on it. (eq1) m 2 g - T = (m 1 + m 2)a For the table weight, since the acceleration in the vertical direction is 0, the normal force and the gravitational force is equal. Thus we can use (eq2) N 1 = m 1 g In the horizontal direction the table weight has friction acting negatively and tension pulling it forward. (eq3) T - F f = (m 1 + m 2)a Because you are given a coefficient of friction, you can substitute the equation F f = μN 1 and because you already have an equation for N 1 you can substitute equation 2 so that Ff = μ m 1g. Putting this all together you can rewrite equation 3 as (eq4) T - μm 1g = (m 1 + m 2)a Now with equations 1 and 4 you have two equations with only two unknowns: a and T. From here it can be solved using any system of equation method. Personally, I like to add the two equations together, which cancels out the tension and solve for acceleration first, then substitute to solve for T. This is usually easier and faster though less accurate due to rounding errors. By adding the two equations together you get (eq5) m 2g -μm 1g = 2(m 1 + m 2)a This equation itself shows gravity of the hanging weight pulling the system forward and the friction of the table weight slowing it down. By substituting your known masses and coefficient of friction you can solve for acceleration: a = 2.89m/s 2 Now you can substitute your value of acceleration into either equation 1 or equation 4 to solve for tension. T = 51.1 N Upvote • 0Downvote Add comment More Report Still looking for help? Get the right answer, fast. Ask a question for free Get a free answer to a quick problem. Most questions answered within 4 hours. OR Find an Online Tutor Now Choose an expert and meet online. 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Home 2. Bookshelves 3. Precalculus & Trigonometry 4. Precalculus (Tradler and Carley) 5. 19: Inverse Trigonometric Functions 6. 19.1: The functions of arcsin, arccos, and arctan Expand/collapse global location 19.1: The functions of arcsin, arccos, and arctan Last updated May 2, 2022 Save as PDF 19: Inverse Trigonometric Functions 19.2: Exercises Page ID 49073 Thomas Tradler and Holly Carley CUNY New York City College of Technology via New York City College of Technology at CUNY Academic Works ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. The Inverse Tangent Function 1. Definition: Inverse Tangent or Arctangent 2. Warning 3. Observation: Inverse Tangent Function 4. Example 19.1.1 The function y=sin−1⁡(x) Definition: Inverse Sine or Arcsine Observation: Inverse Tangent Function Example 19.1.2 The function y=cos−1⁡(x) Definition: Inverse Cosine or Arccosine Observation: Inverse Cosine Example 19.1.3 The inverse trigonometric functions are the inverse functions of the y=sin⁡x, y=cos⁡x, and y=tan⁡x functions restricted to appropriate domains. In this section we give a precise definition of these functions. The Inverse Tangent Function We start with the inverse to the tangent function y=tan⁡(x). Recall that the graph of y=tan⁡(x) is the following: It has vertical asymptotes at x=±π 2,±3⁢π 2,±5⁢π 2,…. Note, that y=tan⁡(x) is not a one-to-one function in the sense of defintion [DEF:1-to-1] on page . (For example, the horizontal line y=1 intersects the graph at x=π 4, x=π 4±π, x=π 4±2⁢π, etc.) However, when we restrict the function to the domain D=(−π 2,π 2) the restricted function is one-to-one, and, for this restricted function, we may take its inverse function. Definition: Inverse Tangent or Arctangent The inverse of the function y=tan⁡(x) with restricted domain D=(−π 2,π 2) and range R=R is called the inverse tangent or arctangent function. It is denoted by y=tan−1⁡(x)or y=arctan⁡(x)⟺tan⁡(y)=x,y∈(−π 2,π 2) The arctangent reverses the input and output of the tangent function, so that the arctangent has domain D=R and range R=(−π 2,π 2). The graph is displayed below. Warning The notation of tan−1⁡(x) and tan 2⁡(x) is slightly inconsistent, since the exponentiation symbol is used above in two different ways. In fact, tan−1⁡(x)=arctan⁡(x) refers to the inverse function of the tan⁡(x) function. However, when we write tan 2⁡(x), we mean tan 2⁡(x)=(tan⁡(x))2=tan⁡(x)⋅tan⁡(x) Therefore, tan−1⁡(x) is the inverse function of tan⁡(x) with respect to the composition operation, whereas tan 2⁡(x) is the square with respect to the usual product in R. Note also that the inverse function of the tangent with respect to the product in R is y=1 tan⁡(x)=cot⁡(x), which is the cotangent. Observation: Inverse Tangent Function The inverse tangent function is an odd function: tan−1⁡(−x)=−tan−1⁡(x) This can be seen by observing that the tangent y=tan⁡(x) is an odd function (that is tan⁡(−x)=−tan⁡(x)), or directly from the symmetry of the graph with respect to the origin (0,0). The next example calculates function values of the inverse tangent function. Example 19.1.1 Recall the exact values of the tangent function from section 17.1: x 0=0∘π 6=30∘π 4=45∘π 3=60∘π 2=90∘tan⁡(x)0 3 3 1 3 undef. Solution From this, we can deduce function values by reversing inputs and outputs, such as: tan⁡(π 6)=3 3⟹tan−1⁡(3 3)=π 6 tan⁡(π 4)=1⟹tan−1⁡(1)=π 4 Also, since tan−1⁡(−x)=−tan−1⁡(x), we obtain the inverse tangent of negative numbers. tan−1⁡(−3)=−tan−1⁡(3)=−π 3 tan−1⁡(−1)=−tan−1⁡(1)=−π 4 We may calculate the inverse tangent of specific values with the calculator using the 2nd and tan keys. For example, tan−1⁡(4.3)≈1.34. Note, that the answer differs, when changing the mode from radians to degree, since tan−1⁡(4.3)≈76.9∘≈1.34. The function y=sin−1⁡(x) Next, we define the inverse sine function. For this, we again first recall the graph of the y=sin⁡(x) function, and note that it is also not one-to-one. However, when restricting the sine to the domain [−π 2,π 2], the restricted function is one-to-one. Note furthermore, that when restricting the domain to [−π 2,π 2], the range is [−1,1], and therefore we cannot extend this to a larger domain in a way such that the function remains a one-to-one function. We use the domain [−π 2,π 2] to define the inverse sine function. Definition: Inverse Sine or Arcsine The inverse of the function y=sin⁡(x) with restricted domain D=[−π 2,π 2] and range R=[−1,1] is called the inverse sine or arcsine function. It is denoted by y=sin−1⁡(x)or y=arcsin⁡(x)⟺sin⁡(y)=x,y∈[−π 2,π 2] The arcsine reverses the input and output of the sine function, so that the arcsine has domain D=[−1,1] and range R=[−π 2,π 2]. The graph of the arcsine is drawn below. Observation: Inverse Tangent Function The inverse sine function is odd: sin−1⁡(−x)=−sin−1⁡(x) This can again be seen by observing that the sine y=sin⁡(x) is an odd function (that is sin⁡(−x)=−sin⁡(x)), or alternatively directly from the symmetry of the graph with respect to the origin (0,0). We now calculate specific function values of the inverse sine. Example 19.1.2 We first recall the known values of the sine. x 0=0∘π 6=30∘π 4=45∘π 3=60∘π 2=90∘sin⁡(x)0 1 2 2 2 3 2 1 Solution These values together with the fact that the inverse sine is odd, that is sin−1⁡(−x)=−sin−1⁡(x), provides us with examples of its function values. sin−1⁡(2 2)=π 4 sin−1⁡(1)=π 2 sin−1⁡(0)=0 sin−1⁡(−1 2)=−sin−1⁡(1 2)=−π 6 Note, that the domain of y=sin−1⁡(x) is D=[−1,1], so that input numbers that are not in this interval give undefined outputs of the inverse sine: sin−1⁡(3)is undefined Input values that are not in the above table may be found with the calculator via the 2nd sin keys. We point out, that the output values depend on wether the calculator is set to radian or degree mode. (Recall that the mode may be changed via the key). The function y=cos−1⁡(x) Finally, we define the inverse cosine. Recall the graph of y=cos⁡(x), and note again that the function is not one-to-one. In this case, the way to restrict the cosine to a one-to-one function is not as clear as in the previous cases for the sine and tangent. By convention, the cosine is restricted to the domain [0,π]. This provides a function that is one-to-one, which is used to define the inverse cosine. Definition: Inverse Cosine or Arccosine The inverse of the function y=cos⁡(x) with restricted domain D=[0,π] and range R=[−1,1] is called the inverse cosine or arccosine function. It is denoted by y=cos−1⁡(x)or y=arccos⁡(x)⟺cos⁡(y)=x,y∈[0,π] The arccosine reverses the input and output of the cosine function, so that the arccosine has domain D=[−1,1] and range R=[0,π]. The graph of the arccosine is drawn below. Observation: Inverse Cosine The inverse cosine function is neither even nor odd. That is, the function cos−1⁡(−x)cannot be computed by simply taking ±cos−1⁡(x). But it does have some symmetry given algebraically by the more complicated relation cos−1⁡(−x)=π−cos−1⁡(x) Proof We can see that if we shift the graph down by π 2 the resulting function is odd. That is to say the function with the rule cos−1⁡(x)−π 2 is odd: cos−1⁡(−x)−π 2=−(cos−1⁡(x)−π 2) which yields upon distributing and adding π 2. Another, more formal approach is as follows. The bottom right relation of [EQU:basic-trig-eqns-wrt-pi] on page states, that we have the relation cos⁡(π−y)=−cos⁡(y) for all y. Let −1≤x≤1, and denote by y=cos−1⁡(x). That is y is the number 0≤y≤π with cos⁡(y)=x. Then we have −x=−cos⁡(y)=cos⁡(π−y) (by equation 17.1.2) Applying cos−1 to both sides gives: cos−1⁡(−x)=cos−1⁡(cos⁡(π−y))=π−y The last equality follows, since cos and cos−1 are inverse to each other, and 0≤y≤π, so that 0≤π−y≤π are also in the range of the cos−1. Rewriting y=cos−1⁡(x) gives the wanted result: cos−1⁡(−x)=π−cos−1⁡(x) This is the equation which we wanted to prove. We now calculate specific function values of the inverse cosine. Example 19.1.3 We first recall the known values of the cosine. x 0=0∘π 6=30∘π 4=45∘π 3=60∘π 2=90∘cos⁡(x)1 3 2 2 2 1 2 0 Solution Here are some examples for function values of the inverse cosine. cos−1⁡(3 2)=π 6,cos−1⁡(1)=0,cos−1⁡(0)=π 2 Negative inputs to the arccosine can be calculated with equation , that is cos−1⁡(−x)=π−cos−1⁡(x), or by going back to the unit circle definition. cos−1⁡(−1 2)=π−cos−1⁡(1 2)=π−π 3=3⁢π−π 3=2⁢π 3 cos−1⁡(−1)=π−cos−1⁡(1)=π−0=π Furthermore, the domain of y=cos−1⁡(x) is D=[−1,1], so that input numbers not in this interval give undefined outputs of the inverse cosine. cos−1⁡(17)is undefined Other input values can be obtained with the calculator by pressing the 2nd cos keys. For example, we obtain the following function values (here using radian measure). This page titled 19.1: The functions of arcsin, arccos, and arctan is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Thomas Tradler and Holly Carley (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform. Back to top 19: Inverse Trigonometric Functions 19.2: Exercises Was this article helpful? Yes No Recommended articles 5.3: Inverse Trigonometric FunctionsWe have briefly mentioned the inverse trigonometric functions before, but we will now define those inverse functions and determine their graphs. 3.2: Technology and Right Triangle TrigonometryThis section explores how to use technology to work with Right Triangle Trigonometry. It covers computing values of trigonometric functions, convertin... 6.2: Technology and Right Triangle TrigonometryThis section explores how to use technology to work with Right Triangle Trigonometry. It covers computing values of trigonometric functions, convertin... 4.2: Technology and Right Triangle TrigonometryThis section explores how to use technology to work with Right Triangle Trigonometry. It covers computing values of trigonometric functions, convertin... 6.2: Technology and Right Triangle TrigonometryThis section explores how to use technology to work with Right Triangle Trigonometry. It covers computing values of trigonometric functions, convertin... Article typeSection or PageAuthorThomas Tradler and Holly CarleyLicenseCC BY-NC-SALicense Version4.0Show Page TOCno Tags arctangent inverse tangent source@ © Copyright 2025 Mathematics LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. 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15070	https://spdbv.unil.ch/TheMolecularLevel/Goodies/Get2NoHistidine.html	Get To Know Histidine Get To Know Histidine A versatile, important, and confusing amino acid. Gale Rhodes Chemistry Department University of Southern Maine The Active Site of Malate Dehydrogenase(1cme.pdb). In the conversion of malate to oxaloacetate, it is proposed that histidine acts as a general base, abstracting the proton from the hydroxyl of malate (MLT313), and driving the transfer of a hydride ion (not shown) onto the nicotinamide ring of NAD+(below malate). If the proposal is correct, then before the proton transfer, the histidine ring is in the neutral imidazole form, tautomer C in the figure below. Proton transfer converts the ring to the imidazolium ion (resonance hybrid B). Convergent stereo image, prepared with DeepView. Physiologically IMPORTANT Ionization States of the Histidine Side Chain Ionic and Tautomeric States of the Histidine Side Chain The side chain of histidine includes the ionizable imidazole ring. The pK a value for the ring is approximately 7.0, so at physiological pH, both the acid and base forms are present. The acid form, the imidazolium ion B, is a resonance hybrid of two practically equivalent contributing forms (remember that these two contributors represent one structure). Either of the two ring nitrogens can release a proton (H+) to produce the conjugate base form, imidazole(A or C). Release of proton from the upper nitrogen of B produces the imidazole tautomer A, and release of proton from the lower nitrogen produces imidazole tautomer C. The tautomers equilibrate by way of the protonated form B. At pH values near 7, all three forms, A, B, and C, are present in equilibrium. Because the acid form, imidazolium B, is a hybrid of the two contributors, it is plausible to show release of a proton from either nitrogen atom in proposing a reaction mechanism involving the imidazolium form as a general acid. Because the base form, imidazole, exists as two equilibrating tautomers A and C, it is plausible to write either tautomer in proposing a reaction mechanism involving the imidazole form as a general base. A Physiologically RARE Ionization State of Histidine The Imidazolate Ion Abstraction of a proton from imidazole (A or C) results in the imidazolate ion D, a resonance hybrid of two practically equivalent contributors. The pK a of imidazole itself is 14.58, so the pK a of imidazole in a histidine residue of a protein should be near this value. Thus this ionization, producing a free imidazolate ion, would not occur under physiological conditions. In proposing a reaction mechanism for a physioligical process, it is not plausible to propose loss of a proton from an imidazole tautomer (A or C) to form a free imidazolate ion D.Any proposal of free imidazolate ion as an intermediate in an enzymatic reaction is simply incorrect. Imidazolate ion can be produced in organic solvents by the action of strong irreversible bases, such as hydride ion. The ion can also serve as a ligand in transition-metal complexes. Thanks to Professor Daryl Eggers of San Jose State University for pointing out to me that imidazolate ion makes at least one appearance in biology, in the histidine side chain that bridges copper and zinc ions in the enzyme copper-zinc superoxide dismutase. Mutations in this enzyme can lead to amyotrophic lateral sclerosis (ALS), also known as Lou Gehrig's disease. The bridging histidine is shown in the illustration below. Imidazolate Ion in a Biological Setting Metal-bridging imidazolate in human Cu-Zn superoxide dismutase (1hl5.pdb). The side chain of histidine 63 appears to be an imidazolate ion, in which both protons of imidazolium ion are replaced by strong Lewis acids, copper (I or II) and zinc (II) ions. The Chemistry of Heterocycles : Structure, Reactions, Syntheses, and Applications, 2nd edition, Theophil Eicher and Siegfried Hauptmann, New York: John Wiley and Sons, 2003, page 166. Back to Goodies List Back to Biochemistry Resources HOME
15071	http://ramanujan.math.trinity.edu/rdaileda/teach/f20/m3341/lectures/lecture9_slides.pdf	Base b Representations of Integers Ryan C. Daileda Trinity University Number Theory Daileda Base b Introduction We usually represent integers as ﬁnite strings of decimal digits, e.g. 8906. This familiar place-value notation is actually shorthand for a sum involving powers of the base 10: 8906 = 8 · 103 + 9 · 102 + 0 · 101 + 6 · 100. The use of base 10 representations is convenient, but otherwise arbitrary. It is possible to express integers in a similar way using any base b > 1. Daileda Base b Base b Expansions Our ﬁrst goal is to establish the following result on the representation of integers in terms of powers of b. Theorem 1 Let b > 1 be an integer. Every n ∈N can be written uniquely in the form n = ambm + am−1bm−1 + · · · + a2b2 + a1b + a0, (1) with ai ∈{0, 1, 2, . . . , b −1} for all i and am ̸= 0. Remark. The expression (1) is called the base b expansion of n. Proof. To establish the existence of base b expansions we induct on n. If n ∈{0, 1, 2, . . . , b −1}, then setting a0 = n works. Daileda Base b Now suppose n ≥b and we have shown that all positive integers less than n have base b expansions. Use the division algorithm to write n = bq + r with r ∈{0, 1, 2, . . . , b −1}. Because n ≥b we must have q ≥1. And since b > 1, we must have q = n −r b ≤n b < n. So q is a positive integer strictly less than n. By the inductive hypothesis we can write q = a′ ℓbℓ+ a′ ℓ−1bℓ−1 + · · · + a′ 1b + a′ 0 with a′ i ∈{0, 1, 2, . . . , b −1} and a′ ℓ̸= 0. Daileda Base b We therefore have n = bq + r = b(a′ ℓbℓ+ a′ ℓ−1bℓ−1 + · · · + a′ 1b + a′ 0) + r = a′ ℓbℓ+1 + a′ ℓ−1bℓ+ · · · + a′ 1b2 + a′ 0b + r, which is a base b expansion for n with m = ℓ+ 1, a0 = r and ai = a′ i−1 for i ≥1. This completes the induction, and proves that every positive integer has a base b expansion. We now prove uniqueness. Suppose that ambm+am−1bm−1+· · ·+a1b+a0 = cℓbℓ+cℓ−1bℓ−1+· · ·+c1b+c0, with ai, ci ∈{0, 1, . . . , b −1}, am ̸= 0 and cℓ̸= 0. Daileda Base b Our ﬁrst goal is to show that m = ℓ. Notice that ambm+am−1bm−1 + · · · + a1b + a0 ≤(b −1)bm + (b −1)bm−1 + · · · + (b −1)b + (b −1) = (b −1)(bm + bm−1 + · · · + b + 1) = (b −1)bm+1 −1 b −1 = bm+1 −1 < bm+1. Furthermore, since cℓ̸= 0, cℓbℓ+ cℓ−1bℓ−1 + · · · + c1b + c0 ≥bℓ. Since we have assumed the two expansions agree, we conclude that bℓ< bm+1 ⇒ℓ< m + 1 ⇒ℓ≤m. Daileda Base b By a symmetric argument, we ﬁnd that m ≤ℓas well, and hence m = ℓ. Now subtract the second expansion from the ﬁrst: (am−cm)bm+(am−1−cm−1)bm−1+· · ·+(a1−c1)b+(a0−b0) = 0. Since \|ai −ci\| < b for all i, we have \|(am−1 −cm−1) bm−1 + · · · + (a1 −c1)b + (a0 −b0) ≤(b −1)bm−1 + · · · + (b −1)b + (b −1) = bm −1, as above. This means that \|am −cm\|bm ≤bm −1, which is impossible unless am = cm. Daileda Base b Now repeat this argument to obtain am−1 = cm−1, am−2 = cm−2, etc. This completes our proof. Remarks. We will denote the base b expansion ambm + am−1bm−1 + · · · + a1b + a0 by the base b place-value notation (amam−1 · · · a1a0)b. When b = 10 we omit the parentheses. The existence proof above gives a recursive procedure for computing base b expansions through the division algorithm. Daileda Base b Example 1 Find the base 3 expansion of 709. Solution. We have: 709 = 3 · 236 + 1, 236 = 3 · 78 + 2, 78 = 3 · 26 + 0, 26 = 3 · 8 + 2, 8 = 3 · 2 + 2, 2 = 3 · 0 + 2. The remainders give the base 3 expansion: 709 = (222021)3. Daileda Base b Example 2 Find the binary (base 2) expansion of 709. Solution. We have: 709 = 2 · 354 + 1, 354 = 2 · 177 + 0, 177 = 2 · 88 + 1, 88 = 2 · 44 + 0, 44 = 2 · 22 + 0, 22 = 2 · 11 + 0, 11 = 2 · 5 + 1, 5 = 2 · 2 + 1, 2 = 2 · 1 + 0, 1 = 2 · 0 + 1. The remainders give the binary expansion: 709 = (1011000101)2 Daileda Base b Repeated Squaring We can use binary expansions to give an extremely eﬃcient algorithm for modular exponentiation. Example 3 Find the remainder when 5709 is divided by 1234. Solution. The binary expansion 709 = (1011000101)2 expresses 709 as a sum of powers of 2: 709 = 29 + 27 + 26 + 22 + 20. We now compute the ﬁrst 9 squares of 5, modulo 1234: 520 = 5 (mod 1234), 52 = 25 (mod 1234), 522 = (52)2 = 625 (mod 1234), Daileda Base b 523 = (522)2 = 6252 = 390625 ≡681 (mod 1234), 524 = (523)2 ≡6812 = 463761 ≡1011 (mod 1234), 525 = (524)2 ≡10112 = 1022121 ≡369 (mod 1234), 526 = (525)2 ≡3692 = 136161 ≡421 (mod 1234), 527 = (526)2 ≡4212 = 177241 ≡779 (mod 1234), 528 = (527)2 ≡7792 = 606841 ≡947 (mod 1234), 529 = (528)2 ≡9472 = 896809 ≡925 (mod 1234). Therefore 5709 = 529+27+26+22+20 = 529 · 527 · 526 · 522 · 520 ≡925 · 779 · 421 · 625 · 5 ≡147 (mod 1234). Daileda Base b Divisibility Tests Fix a base b > 1 and let d be any positive divisor of b −1. Then b ≡1 (mod d). Let n ∈N have the base b expansion (amam−1 · · · a0)b. Then n = ambm + am−1bm−1 + · · · + a1b + a0 ≡am1m + am−11m−1 + · · · + a1 · 1 + a0 (mod d) ≡am + am−1 + · · · a1 + a0 (mod d). We immediately obtain the following divisibility test. Theorem 2 If n = (amam−1 · · · a0)b and d\|b −1, then d\|n if and only if d\|am + am−1 + · · · + a1 + a0. Daileda Base b The nontrivial positive divisors of 10 −1 = 9 are 3 and 9. We can therefore test for divisibility by 3 or 9 by summing the decimal digits of an integer. For example, if n = 9550684, then 9 + 5 + 5 + 0 + 6 + 8 + 4 = 37 ̸≡0 (mod 3), so that 3 ∤n. On the other hand, if n = 3788058, then 3 + 7 + 8 + 8 + 0 + 5 + 8 = 39 ≡0 (mod 3), so that 3\|n (but 9 ∤n). Daileda Base b Suppose instead that d\|b + 1. Then b ≡−1 (mod d) so that (amam−1 · · · a0)b = ambm + am−1bm−1 + · · · + a1b + a0 ≡am(−1)m + am−1(−1)m−1 + · · · + a1(−1) + a0 (mod d) which is the alternating sum of the base b “digits.” Theorem 3 If n = (amam−1 · · · a0)b and d\|b + 1, then d\|n if and only if d\|(−1)mam + (−1)m−1am−1 + · · · −a1 + a0. If b = 10, then b + 1 = 11, so the only nontrivial choice for d is 11. Taking n = 53084471 we ﬁnd that 5 −3 + 0 −8 + 4 −4 + 7 −1 = 0, and hence 11\|n. Daileda Base b
15072	https://openbooks.lib.msu.edu/collegephysics1/part/work-energy-and-energy-resources-2/	Work, Energy, and Energy Resources – College Physics 1 Skip to content Menu Primary Navigation Home Read Sign in Search in book: Search Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. Book Contents Navigation Contents Structure of the course Preface I. Introduction: The Nature of Science and Physics 1.Introduction to Science and the Realm of Physics, Physical Quantities, and Units 2.Physics: An Introduction Science and the Realm of Physics Applications of Physics in the Health and Life Sciences Models, Theories, and Laws; The Role of Experimentation The Evolution of Natural Philosophy into Modern Physics Summary Conceptual Questions 3.Physical Quantities and Units SI Units: Fundamental and Derived Units Units of Time, Length, and Mass: The Second, Meter, and Kilogram Metric Prefixes Known Ranges of Length, Mass, and Time Unit Conversion and Dimensional Analysis Summary Conceptual Questions Problems & Exercises 4.Accuracy, Precision, and Significant Figures Accuracy and Precision of a Measurement Accuracy vs. Precision: Visualizing the Difference Uncertainty in Measurements Summary Conceptual Questions Problems & Exercises 5.Approximation Summary 6.Problems and Exercises Problems & Exercises II. Kinematics 7.Introduction to One-Dimensional Kinematics 8.Displacement Position: Where is the Object? Displacement Distance Section Summary Conceptual Questions Problems & Exercises 9.Vectors, Scalars, and Coordinate Systems Coordinate Systems for One-Dimensional Motion Section Summary Conceptual Questions 10.Time, Velocity, and Speed Time Average Velocity Instantaneous Velocity Speed vs Velocity Graphical View of Motion Section Summary Conceptual Questions Problems & Exercises 11.Acceleration Average Acceleration Acceleration as a Vector Instantaneous Acceleration Sign and Direction Section Summary Conceptual Questions Problems & Exercises 12.Motion Equations for Constant Acceleration in One Dimension Notation: t, x, v, a Putting Equations Together Section Summary Problems & Exercises 13.Problem-Solving Basics for One-Dimensional Kinematics Problem-Solving Steps Unreasonable Results Section Summary Conceptual Questions 14.Falling Objects Gravity One-Dimensional Motion Involving Gravity Section Summary Conceptual Questions Problems & Exercises 15.Graphical Analysis of One-Dimensional Motion Slopes and General Relationships Graph of Displacement vs. Time (a = 0, so v is constant) Graphs of Motion when [latex]a[/latex] is constant but [latex]a\ne 0[/latex] Graphs of Motion Where Acceleration is Not Constant Section Summary Conceptual Questions Problems & Exercises 16.Introduction to Two-Dimensional Kinematics 17.Kinematics in Two Dimensions: An Introduction Two-Dimensional Motion: Walking in a City Summary 18.Vector Addition and Subtraction: Graphical Methods Vectors in Two Dimensions Vector Addition: Head-to-Tail Method Vector Subtraction Multiplication of Vectors and Scalars Summary Conceptual Questions Problems & Exercises 19.Vector Addition and Subtraction: Analytical Methods Resolving a Vector into Perpendicular Components Calculating a Resultant Vector Adding Vectors Using Analytical Methods Summary Conceptual Questions Problems & Exercises 20.Projectile Motion Summary Conceptual Questions Problems & Exercises 21.Addition of Velocities Relative Velocity Relative Velocities and Classical Relativity Summary Conceptual Questions Problems & Exercises III. Dynamics: Force and Newton's Laws of Motion and Applications: Friction, Drag and Elasticity 22.Introduction to Dynamics: Newton’s Laws of Motion 23.Development of Force Concept Section Summary Conceptual Questions 24.Newton’s First Law of Motion: Inertia [pb_glossary id="5659"]Newton’s First Law of Motion[/pb_glossary] Friction and External Forces Cause and Effect: A Fundamental Scientific Shift Mass and Inertia Section Summary Conceptual Questions 25.Newton’s Second Law of Motion: Concept of a System The Cause of Motion Units of Force Application: Weight as a Force Section Summary Conceptual Questions Problem Exercises 26.Newton’s Third Law of Motion: Symmetry in Forces Forces Always Come in Pairs [pb_glossary id="6405"]Newton’s Third Law of Motion[/pb_glossary] Section Summary Conceptual Questions Problem Exercises 27.Normal, Tension, and Other Examples of Forces Normal Force The Tension Force Applications in Healthcare and Biology Resolving Weight into Components Tension Extended Topic: Real Forces and Inertial Frames Section Summary Conceptual Questions Problem Exercises 28.Problem-Solving Strategies Why Problem Solving Matters Section Summary Problem Exercises 29.Further Applications of Newton’s Laws of Motion Integrating Concepts: Newton’s Laws of Motion and Kinematics Summary Conceptual Questions Problem Exercises 30.Friction Friction Section Summary Conceptual Questions Problems & Exercises 31.Drag Forces Section Summary Conceptual Questions Problems & Exercise 32.Elasticity: Stress and Strain Changes in Length—Tension and Compression: Elastic Modulus Changes in Volume: Bulk Modulus Section Summary Conceptual Questions Problems & Exercises 33.Extended Topic: The Four Basic Forces—An Introduction Summary Conceptual Questions Problem Exercises IV. Uniform Circular Motion and Gravitation 34.Introduction to Uniform Circular Motion and Gravitation 35.Rotation Angle and Angular Velocity Rotation Angle Angular Velocity Section Summary Conceptual Questions Problem Exercises 36.Centripetal Acceleration Section Summary Conceptual Questions Problem Exercises 37.Centripetal Force Section Summary Conceptual Questions Problems Exercise 38.Fictitious Forces and Non-inertial Frames: The Coriolis Force Section Summary Conceptual Questions 39.Newton’s Universal Law of Gravitation Section Summary Conceptual Questions Problem Exercises 40.Satellites and Kepler’s Laws: An Argument for Simplicity Kepler’s Laws of Planetary Motion Section Summary Conceptual Questions Problem Exercises V. Work, Energy, and Energy Resources 41.Introduction to Work, Energy, and Energy Resources 42.Work: The Scientific Definition The Scientific Definition of Work Section Summary Conceptual Questions Problems & Exercises 43.Kinetic Energy and the Work-Energy Theorem Work Transfers Energy Section Summary Conceptual Questions Problems & Exercises 44.Gravitational Potential Energy Work Done Against Gravity Converting Between Potential Energy and Kinetic Energy Using Potential Energy to Simplify Calculations Section Summary Conceptual Questions Problems & Exercises 45.Conservative Forces and Potential Energy What is a Conservative Force? Potential Energy: Stored Energy from Position or Shape Spring Potential Energy and Hooke’s Law Conservation of Mechanical Energy Section Summary Conceptual Questions Problems & Exercises 46.Nonconservative Forces Nonconservative Forces in Real Life How Nonconservative Forces Affect Mechanical Energy The Work-Energy Theorem Revisited Key Equation: Mechanical Energy with Nonconservative Forces Making Connections: Take-Home Investigation—Determining Friction from the Stopping Distance Section Summary Problems & Exercises 47.Conservation of Energy Law of Conservation of Energy Energy in the Human Body and Environment Forms of Energy in Life Sciences Energy Transformation in the Real World Efficiency of Energy Transformations Section Summary Conceptual Questions Problems & Exercises 48.Power What is Power and Its Biological and Mechanical Significance Understanding Power in Everyday Contexts Calculating Power from Energy Examples of Power Power and Energy Consumption Energy Awareness and Practical Cost-Saving Sustainability and Energy Transformation Section Summary Conceptual Questions Problems & Exercises 49.Work, Energy, and Power in Humans The Body as an Energy Conversion System Power Consumed at Rest: Basal [pb_glossary id="5760"]Metabolic Rate[/pb_glossary] Section Summary Conceptual Questions Problems & Exercises 50.World Energy Use Global Energy and Human Well-Being Renewable vs. Nonrenewable Energy Sources The World’s Growing Energy Needs Energy Use and Economic Health Energy Conservation: Two Meanings Section Summary Conceptual Questions Problems & Exercises VI. Linear Momentum and Collisions 51.Introduction to Linear Momentum and Collisions 52.Linear Momentum and Force Understanding Linear Momentum Momentum and Newton’s Second Law Section Summary Conceptual Questions Problems & Exercises 53.Impulse Section Summary Conceptual Questions Problems & Exercises 54.Conservation of Momentum Subatomic Collisions and Conservation Section Summary Conceptual Questions Problems & Exercises 55.Elastic Collisions in One Dimension Elastic Collision Internal Kinetic Energy Example 55.1: Calculating Velocities Following an Elastic Collision Making Connections: Take-Home Investigation—Ice Cubes and Elastic Collisions PhET Explorations: Collision Lab Section Summary Conceptual Questions Problems & Exercises 56.Inelastic Collisions in One Dimension Section Summary Conceptual Questions Problems & Exercises 57.Collisions of Point Masses in Two Dimensions Elastic Collisions of Two Objects with Equal Mass Section Summary Conceptual Questions Problems & Exercises 58.Introduction to Rocket Propulsion Section Summary Conceptual Questions Problems & Exercises VII. Statics and Torque 59.Introduction to Statics and Torque 60.The First Condition for Equilibrium Section Summary Conceptual Questions 61.The Second Condition for Equilibrium Conceptual Questions Problems & Exercises 62.Stability Section Summary Conceptual Questions Problems & Exercises 63.Applications of Statics, Including Problem-Solving Strategies Summary Conceptual Questions Problems & Exercises 64.Simple Machines Section Summary Conceptual Questions Problems & Exercises 65.Forces and Torques in Muscles and Joints Section Summary Conceptual Questions Problems & Exercises VIII. Rotational motion and angular momentum 66.Introduction to Rotational Motion and Angular Momentum 67.Angular Acceleration Section Summary Conceptual Questions Problems & Exercises 68.Gyroscopic Effects: Vector Aspects of Angular Momentum Section Summary Conceptual Questions Problem Exercises 69.Collisions of Extended Bodies in Two Dimensions Conceptual Questions Problems & Exercises 70.Angular Momentum and Its Conservation Section Summary Conceptual Questions Problems & Exercises 71.Rotational Kinetic Energy: Work and Energy Revisited How Thick Is the Soup? Or Why Don’t All Objects Roll Downhill at the Same Rate? Section Summary Conceptual Questions Problems & Exercises 72.Kinematics of Rotational Motion Section Summary Problems & Exercises 73.Dynamics of Rotational Motion: Rotational Inertia Section Summary Conceptual Questions Problems & Exercises IX. Fluid Statics 74.Introduction to Fluid Statics 75.What Is a Fluid? Section Summary Conceptual Questions 76.Density Section Summary Conceptual Questions Problems & Exercises 77.Pressure Section Summary Conceptual Questions Problems & Exercises 78.Variation of Pressure with Depth in a Fluid Section Summary Conceptual Questions Problems & Exercises 79.Pascal’s Principle Section Summary Conceptual Questions Problems & Exercises 80.Gauge Pressure, Absolute Pressure, and Pressure Measurement Section Summary Conceptual Questions Problems & Exercises 81.Archimedes’ Principle Section Summary Conceptual Questions Problem Exercises 82.Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action Surface Tension in the Lungs and Capillary Action Section Summary Conceptual Questions Problems & Exercises 83.Pressures in the Body Section Summary Problems & Exercises X. Fluid Dynamics and Its Biological and Medical Applications 84.Introduction to Fluid Dynamics and Its Biological and Medical Applications 85.Flow Rate and Its Relation to Velocity Section Summary Conceptual Questions Problems & Exercises 86.Bernoulli’s Equation Summary Conceptual Questions Problems & Exercises 87.The Most General Applications of Bernoulli’s Equation Summary Conceptual Questions Problems & Exercises 88.Viscosity and Laminar Flow; Poiseuille’s Law Section Summary Problems & Exercises 89.The Onset of Turbulence Section Summary Conceptual Questions Problems & Exercises 90.Motion of an Object in a Viscous Fluid Section Summary Conceptual Questions 91.Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes Diffusion Section Summary Conceptual Questions Problem Exercises XI. Temperature, Kinetic Theory, and the Gas Laws 92.Introduction to Temperature, Kinetic Theory, and the Gas Laws Introduction to Heat, Temperature, and Thermal Energy 93.Temperature Temperature Ranges in the Universe Section Summary Conceptual Questions 94.Thermal Expansion of Solids and Liquids Thermal Expansion Thermal Expansion in Two and Three Dimensions Thermal Stress Section Summary Conceptual Questions Problems & Exercises 95.The Ideal Gas Law Section Summary Conceptual Questions Problems & Exercises 96.Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature Distribution of Molecular Speeds Section Summary Conceptual Questions Problems & Exercises 97.Phase Changes Section Summary Conceptual Questions 98.Humidity, Evaporation, and Boiling Section Summary Conceptual Questions Problems & Exercises XII. Heat and Heat Transfer Methods 99.Introduction to Heat and Heat Transfer Methods 100.Heat Summary Conceptual Questions 101.Temperature Change and Heat Capacity Section Summary Conceptual Questions Problems & Exercises 102.Phase Change and Latent Heat Section Summary Conceptual Questions Problems & Exercises 103.Heat Transfer Methods Summary Conceptual Questions 104.Conduction Summary Conceptual Questions Problems & Exercises 105.Convection Summary Conceptual Questions Problems & Exercises 106.Radiation Summary Conceptual Questions Problems & Exercises XIII. Thermodynamics 107.Introduction to Thermodynamics 108.The First Law of Thermodynamics Section Summary Conceptual Questions Problems & Exercises 109.The First Law of Thermodynamics and Some Simple Processes Section Summary Conceptual Questions Problem Exercises 110.Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency Section Summary Conceptual Questions Problem Exercises 111.Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated Section Summary Conceptual Questions Problem Exercises 112.Applications of Thermodynamics: Heat Pumps and Refrigerators Section Summary Conceptual Questions Problem Exercises 113.Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy Section Summary Conceptual Questions Problem Exercises 114.Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation Section Summary Conceptual Questions Problem Exercises XIV. Oscillatory Motion and Waves and Physics of Hearing. 115.Introduction to Oscillatory Motion and Waves 116.Hooke’s Law: Stress and Strain Revisited Conceptual Questions Problems & Exercises 117.Period and Frequency in Oscillations Section Summary Problems & Exercises 118.Simple Harmonic Motion: A Special Periodic Motion The Link between Simple Harmonic Motion and Waves Section Summary Conceptual Questions Problems & Exercises 119.The Simple Pendulum Conceptual Questions Problems & Exercises 120.Energy and the Simple Harmonic Oscillator Conceptual Questions Problems & Exercises 121.Uniform Circular Motion and Simple Harmonic Motion Section Summary Problems & Exercises 122.Damped Harmonic Motion Section Summary Conceptual Questions Problems & Exercises 123.Forced Oscillations and Resonance Section Summary Conceptual Questions Problems & Exercises 124.Waves Transverse and Longitudinal Waves Section Summary Conceptual Questions Problems & Exercises 125.Superposition and Interference Section Summary Conceptual Questions Problems & Exercises 126.Energy in Waves: Intensity Section Summary Conceptual Questions Problems & Exercises 127.The Physics of Hearing 128.Sound Section Summary 129.Speed of Sound, Frequency, and Wavelength Conceptual Questions Problems & Exercises 130.Sound Intensity and Sound Level Section Summary Conceptual Questions Problems & Exercises 131.Doppler Effect and Sonic Booms Sonic Booms to Bow Wakes Section Summary Conceptual Questions Problems & Exercises 132.Sound Interference and Resonance: Standing Waves in Air Columns Section Summary Conceptual Questions Problems & Exercises 133.Hearing Section Summary Conceptual Questions Problems & Exercises 134.Ultrasound Ultrasound in Medical Therapy Ultrasound in Medical Diagnostics Section Summary Conceptual Questions Problems & Exercises Atomic Masses Selected Radioactive Isotopes Useful Information Glossary of Key Symbols and Notation College Physics 1 Work, Energy, and Energy Resources This week you will read the introductory chapters listed below for this first part and you will have to solve the associated problems in loncapa. You can also start the problems from the extended set 2 pertaining to this chapter. Introduction to Work, Energy, and Energy Resources Work: The Scientific Definition Kinetic Energy and the Work-Energy Theorem Gravitational Potential Energy Conservative Forces and Potential Energy Nonconservative Forces Conservation of Energy Power Work, Energy, and Power in Humans World Energy Use Previous/next navigation Previous: Satellites and Kepler’s Laws: An Argument for Simplicity Next: Introduction to Work, Energy, and Energy Resources Back to top License College Physics 1 Copyright © 2012 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted. Share This Book Pressbooks Powered by Pressbooks Pressbooks User Guide \|Pressbooks Directory \|Contact Pressbooks on YouTubePressbooks on LinkedIn
15073	https://ocw.mit.edu/courses/5-111sc-principles-of-chemical-science-fall-2014/	Course Info Instructor Prof. Catherine Drennan Departments Chemistry As Taught In Fall 2014 Level Undergraduate Topics Science Chemistry Inorganic Chemistry Organic Chemistry Physical Chemistry Learning Resource Types theaters Lecture Videos notes Lecture Notes assignment_turned_in Problem Sets with Solutions grading Exams with Solutions theaters Other Video co_present Instructor Insights Download Course search GIVE NOW about ocw help & faqs contact us 5.111SC \| Fall 2014 \| Undergraduate Principles of Chemical Science Course Description This course provides an introduction to the chemistry of biological, inorganic, and organic molecules. The emphasis is on basic principles of atomic and molecular electronic structure, thermodynamics, acid-base and redox equilibria, chemical kinetics, and catalysis. One year of high school chemistry is the expected … This course provides an introduction to the chemistry of biological, inorganic, and organic molecules. The emphasis is on basic principles of atomic and molecular electronic structure, thermodynamics, acid-base and redox equilibria, chemical kinetics, and catalysis. One year of high school chemistry is the expected background for this freshman-level course. The aims include developing a unified and intuitive view of how electronic structure controls the three-dimensional shape of molecules, the physical and chemical properties of molecules in gases, liquids and solids, and ultimately the assembly of macromolecules as in polymers and DNA. Relationships between chemistry and other fundamental sciences such as biology and physics are emphasized, as are the relationships between the science of chemistry to its applications in environmental science, atmospheric chemistry and electronic devices. Acknowledgements Professor Drennan would like to acknowledge the contributions of MIT Lecturer Dr. Elizabeth Vogel Taylor, Professor Sylvia Ceyer, and Professor Robert Silbey to the development of this course and its materials. Course Info Instructor Prof. Catherine Drennan Departments Chemistry Topics Science Chemistry Inorganic Chemistry Organic Chemistry Physical Chemistry Learning Resource Types theaters Lecture Videos notes Lecture Notes assignment_turned_in Problem Sets with Solutions grading Exams with Solutions theaters Other Video co_present Instructor Insights The WHO, the WHAT, and the WHY of chemistry. (Adapted from a figure by Mary O’Reilly. O’Reilly Science Art, LLC. Photograph by James Kegley.) Download Course Over 2,500 courses & materials Freely sharing knowledge with learners and educators around the world. Learn more © 2001–2025 Massachusetts Institute of Technology Creative Commons License Terms and Conditions Proud member of: © 2001–2025 Massachusetts Institute of Technology
15074	https://artofproblemsolving.com/wiki/index.php/AM-GM_Inequality?srsltid=AfmBOop4NjIAjqNbzcTnd9GvGEGEYT3KoHDmn5FTEzU-aRi97mcaD3-O	Art of Problem Solving AM-GM Inequality - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki AM-GM Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search AM-GM Inequality In algebra, the AM-GM Inequality, also known formally as the Inequality of Arithmetic and Geometric Means or informally as AM-GM, is an inequality that states that any list of nonnegative reals' arithmetic mean is greater than or equal to its geometric mean. Furthermore, the two means are equal if and only if every number in the list is the same. In symbols, the inequality states that for any real numbers , with equality if and only if . The AM-GM Inequality is among the most famous inequalities in algebra and has cemented itself as ubiquitous across almost all competitions. Applications exist at introductory, intermediate, and olympiad level problems, with AM-GM being particularly crucial in proof-based contests. Contents [hide] 1 Proofs 2 Generalizations 2.1 Weighted AM-GM Inequality 2.2 Mean Inequality Chain 2.3 Power Mean Inequality 3 Problems 3.1 Introductory 3.2 Intermediate 3.3 Olympiad 4 See Also Proofs Main article: Proofs of AM-GM All known proofs of AM-GM use induction or other, more advanced inequalities. Furthermore, they are all more complex than their usage in introductory and most intermediate competitions. AM-GM's most elementary proof utilizes Cauchy Induction, a variant of induction where one proves a result for , uses induction to extend this to all powers of , and then shows that assuming the result for implies it holds for . Generalizations The AM-GM Inequality has been generalized into several other inequalities. In addition to those listed, the Minkowski Inequality and Muirhead's Inequality are also generalizations of AM-GM. Weighted AM-GM Inequality The Weighted AM-GM Inequality relates the weighted arithmetic and geometric means. It states that for any list of weights such that , with equality if and only if . When , the weighted form is reduced to the AM-GM Inequality. Several proofs of the Weighted AM-GM Inequality can be found in the proofs of AM-GM article. Mean Inequality Chain Main article: Mean Inequality Chain The Mean Inequality Chain, also called the RMS-AM-GM-HM Inequality, relates the root mean square, arithmetic mean, geometric mean, and harmonic mean of a list of nonnegative reals. In particular, it states that with equality if and only if . As with AM-GM, there also exists a weighted version of the Mean Inequality Chain. Power Mean Inequality Main article: Power Mean Inequality The Power Mean Inequality relates all the different power means of a list of nonnegative reals. The power mean is defined as follows: The Power Mean inequality then states that if , then , with equality holding if and only if Plugging into this inequality reduces it to AM-GM, and gives the Mean Inequality Chain. As with AM-GM, there also exists a weighted version of the Power Mean Inequality. Problems Introductory For nonnegative real numbers , demonstrate that if then . (Solution) Find the maximum of for all positive . (Solution) Intermediate Find the minimum value of for . (Source) Olympiad Let , , and be positive real numbers. Prove that (Source) See Also Proofs of AM-GM Mean Inequality Chain Power Mean Inequality Cauchy-Schwarz Inequality Inequality Retrieved from " Categories: Algebra Inequalities Definition Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
15075	http://physics.bu.edu/~duffy/Ejs/EP_chapter17/EField_and_Potential_2D_v2_Intro%201.html	Electric Field and Potential in Two Dimensions In this simulation, you can explore the concepts of the electric field and the electric potential, in a two-dimensional situation. You can turn on 1 to 5 charged particles, and move a test charge around the plane near these charged particles to sample both the electric field and the electric potential, produced by the charged particles, at various points. You can also turn on a grid of field vectors, which show the direction and, qualitatively, the magnitude of the field at a grid of equally spaced points in the plane in which the charged particles are located. Here are some facts about the electric field from point charges: the magnitude of the electric field (E) produced by a point charge with a charge of magnitude Q, at a point a distance r away from the point charge, is given by the equation E = kQ/r2, where k is a constant with a value of 8.99 x 109 N m2/C2. the direction of the electric field produced by a point charge is away from the charge if the charge is positive, and toward the charge if the charge is negative. electric field is a vector, so when there are multiple point charges present, the net electric field at any point is the vector sum of the electric fields due to the individual charges. Here are some facts about the electric potential from point charge the electric potential (V) produced by a point charge with a charge of magnitude Q, at a point a distance r away from the point charge, is given by the equation: V = kQ/r, where k is a constant with a value of 8.99 x 109 N m2/. electric potential is a scalar, so when there are multiple point charges present, the net electric potential at any point is the sum of the electric potentials due to the individual charges.
15076	https://thephilosophyforum.com/discussion/6542/the-bijection-problem-the-natural-numbers-and-the-even-numbers	Forum Members HELP The bijection problem the natural numbers and the even numbers 12345Next TheMadFool My math is at high school level so bear with me. One definition of an infinite set I know of is that an infinite set is in bijection with a proper subset of itself. Bijection, as I understand it, means that we can pair one item in a set with exactly one item in another. For example take two sets {a, e, i, o, u} and (3, 9, 2, 7} The bijection would be a3, e9, i2, u7 Now let's take two infinite sets: 1) Natural numbers N = {1, 2, 3,...} 2) Even numbers E = {0, 2, 4,...} E is a proper subset of N and according to the definition of infinite sets E is in bijection with N. For any element y in N there's exactly one element in E that's (2y - 2) I.e. 1 in N is paired with 0 in E, 2 in N is paired with 2, 3 in N is paired with 4 in E, so and so forth. In terms of cardinality or largeness or magnitude N = E. They are both infinite and in an equivalent way. However, let's do something different. We take the same sets N and E. We know that N has the even numbers. So we pair the members of E with the even numbers in N. We can do that perfectly and with each member of E in bijection with the even number members of N. What now of the odd numbers in N? They have no matching counterpart in E. Doesn't this mean N > E? What's wrong with my argument? Thanks Marc Goulet Two sets are said to have the same cardinality if there exists a bijection between them. Your second function is not a bijection. GrandMinnow In terms of cardinality or largeness or magnitude N = E. — TheMadFool That's not how we say it. Rather, we say the cardinality of N equals the cardinality of E. card(N) = card(E) we pair the members of E with the even numbers in N. — TheMadFool The even numbers in N are just the members of E. So your pairing is just pairing the members of E with the members of E. What now of the odd numbers in N? They have no matching counterpart in E. Doesn't this mean N > E? — TheMadFool No. N > E means Both hold: (1) there is a pairing from E to a proper subset of N and (2) there is no pairing between N and E. But there is a pairing between N and E, so it is not the case that N > E. The fact that there is a pairing from E to a proper subset of N (e.g. the pairing misses the odd numbers in N) does not contradict that still there is a pairing between N and E. / I recommend this terminology ('<->' means 'if and only if'): f is an injection from S into T <-> (f is a one-to-one function & the domain of f is S & the range of S is a subset of T) [note: here the range of f might not be a proper subset of T since it is allowed that the range of f is T] f is a surjection from S onto T <-> (f is a function & the domain of f is S & the range of S is T) [note: here f might not be a one-to-one function] f is a bijection from S to T <-> (f is an injection from S into T & f is a surjection from S onto T) S is equinumerous with T <-> there is a bijection from S to T S is dominated by T <-> there is an injection from S into T S is strictly dominated by T (i.e., S < T) <-> (there is an injection from S into T & there is no bijection from S to T) GrandMinnow 169 My math is at high school level so bear with me. — TheMadFool If you like I can recommend a few textbooks in first order predicate logic and then set theory that would provide you with a self-course in this subject. Then you would not be prone to confusions about the subject. GrandMinnow OptionsShare TheMadFool 13.8k ↪Marc Goulet ↪GrandMinnow My argument is like Cantor's diagonal argument regarding the absence of bijection between the real numbers and the natural numbers N = {1, 2, 3, 4, 5, 6, 7, 8, 9,...} E = {0, 2, 4, 6, 8, 10, 12, 14, 16,...} Now pair the even numbers in N to the members of E (2,0), (4,2), (6,4), (n,n-2),... There's a bijection but it leaves the odd numbers in N without a matching pair. In other words card(N) > card(E) TheMadFool OptionsShare GrandMinnow 169 No, your argument is nothing like Cantor's argument. Cantor proved that there does not exist a surjection from the one set onto the other, peforce that there is no bijection between them. (No surjection from N onto R, perforce no bijection between them). Your argument just points out that one particular function is not a bijection from the one set onto the other (it's not a bijection from E onto N) The fact that one particular function is not a bijection from the one set onto the other does not prove that no other function that is a bijection from one set onto the other. And indeed we prove that there is a bijection from the one set onto the other. Cantor proved that there is no surjection from N onto R. But we also know that there is a bijection between N and E. It is a clear logical fallacy to infer that because some particular function is not a bijection between N and E that therefore there is no function that is bijection between N and E. I offered to recommend some starting books on this subject so that you can inform yourself to avoid the fallacies and confusions to which you are prone. GrandMinnow OptionsShare sandman 41 The statement " there are as many even integers as integers", is typically demonstrated using a 'one to one' correspondence as shown. N: 1 2 3 4 5 6 ... E: 2 4 6 8 10 12 ... This is contradicted by: 1. Random sampling of integers results in an average of 50% even E, 50% odd D. Statistics can be verified in the real world, and is useful in applications of probability. 2. In the above example, removing E from N leaves D, removing E from E leaves nothing, so where is the logic? An odd feature of this example is the appearance of the same integers in both sets. The 'bijection' for example 1 defines y=2x, as a mapping from N to E. The results are not about the size of sets, but the definition used for mapping. Representing the first 'one to one' correspondence above in a rectangular form, partitioned into subsets: 1 2 4 8... 3 6 12 24... 5 10 20 40... ... The odd integers, all listed in column 1, are paired with the column of even integers to the right. The remaining even integers in each column are paired with the column to the right. The pairing is independent of direction. The odd are paired 1 to 1 with an even. The first subset of even (col 2) is paired with odd (col 1) and even (col 3). The remaining even are paired with two columns. Not all pairings are 1 to 1, and each integer appears only once. A true example of a 1-1 correspondence, "there are as many even integers as odd integers " D: 1 3 5 7 9 11 ... E: 2 4 6 8 10 12 ... A set without limit (infinite) is not measurable, since boundaries enable measurement. I have a straight stick with one end. Can anyone tell me its length? sandman OptionsShare Deleted User 0 This user has been deleted and all their posts removed. Deleted User OptionsShare ssu 9.5k My argument is like Cantor's diagonal argument regarding the absence of bijection between the real numbers and the natural numbers — TheMadFool No, your not doing that. So we pair the members of E with the even numbers in N. We can do that perfectly and with each member of E in bijection with the even number members of N. What now of the odd numbers in N? — TheMadFool Do understand what you are making with a bijection: every member of E has a pair in N and vice versa. That is a bijection, one to one correspondence. You are now somehow assuming it wouldn't be a bijection, but either a surjection or an injection. The idea is most easily understood like this: Natural number N: 1, 2, 3, 4, 5,... Then multiply EVERY natural number by two. :1x2, 2x2, 3x2, 4x2, 5x2,.... Then you get even numbers. E:2, 4, 6,8, 10,... And you can do with let's say by 10 000. Hence there is a bijection between N: 1,2,3,4,5... and the infinite series :10 000, 20 000, 30 000, 40 000, 50 000,... even if there is 9 999 numbers "in between". ssu OptionsShare Eee 159 My math is at high school level so bear with me. — TheMadFool What's wrong with my argument? — TheMadFool Others have pointed it out. I second their criticisms. This is, however, a fascinating subject, and I also second the notion that you should pick up a book on it. There is room for philosophical criticism of the assumptions involved, but I don't see even room for invalidating relatively trivial results. If one learns the game, then such results are trivially true within the game. Whether this or that game is better in various sense is another question. Eee OptionsShare Eee 159 This is contradicted by: 1. Random sampling of integers results in an average of 50% even E, 50% odd D. Statistics can be verified in the real world, and is useful in applications of probability. 2. In the above example, removing E from N leaves D, removing E from E leaves nothing, so where is the logic? An odd feature of this example is the appearance of the same integers in both sets. The 'bijection' for example 1 defines y=2x, as a mapping from N to E. The results are not about the size of sets, but the definition used for mapping. — sandman No. Sorry. The equivalence is defined as the mathematical existence of a bijection. Mathematicians are well aware that they are different sets. And 'statistics can be verified in the real world' is anything but an obvious truth --assuming that it has a definite meaning in the first place. While some mathematicians have little interest in philosophy and may ignore fascinating questions, they aren't slouches when it comes to actual math. Pure math is about as squeaky clean as it gets. The game has rules. What the game means when held against the 'real world' is a non-mathematical question. And the rules were established so that math wouldn't be a sloppy philosophical conversation. A set without limit (infinite) is not measurable, since boundaries enable measurement. — sandman In mathematical analysis, a measure on a set is a systematic way to assign a number to each suitable subset of that set, intuitively interpreted as its size. In this sense, a measure is a generalization of the concepts of length, area, and volume. A particularly important example is the Lebesgue measure on a Euclidean space, which assigns the conventional length, area, and volume of Euclidean geometry to suitable subsets of the n-dimensional Euclidean space Rn. For instance, the Lebesgue measure of the interval [0, 1] in the real numbers is its length in the everyday sense of the word, specifically, 1. — Wiki The measure of Q is 0. But Q has no boundary on its left or right (on the real line.) It also has infinitely many elements. Eee OptionsShare ssu 9.5k There is room for philosophical criticism of the assumptions involved — Eee Sure. Starting from the fact that we don't know the answer to the Continuum Hypothesis. Which tells us quite plainly that we still don't understand everything about mathematical infinity. ssu OptionsShare Eee 159 Sure. Starting from the fact that we don't know the answer to the Continuum Hypothesis. Which tells us quite plainly that we still don't understand everything about mathematical infinity. — ssu Right. And then one can think about constuctivism, finitism, etc. Or whether math is ultimately justified within a culture by its application. Its 'internal' justification (proofs) aren't necessarily why it is trusted or esteemed, especially as proofs get too complex for non-experts. On forums I see those who appear to have no training approaching it 'metaphysically,' seemingly assuming that mathematicians themselves do it this way. But that misses what's essential, that it's a game with rules. It's agnostic about things outside those rules, even if particular mathematicians philosophize. Eee OptionsShare Eee 159 This is perhaps the easiest diagonal argument. One can still ask philosophical questions about the 'frame' or set rules that makes this argument possible (set theory and classical logic), but the argument itself is beautiful and mind-opening. Because I call it beautiful or mind-opening, I do imply that the argument has intuitive content. But the argument doesn't depend on this intuitive content. Eee OptionsShare ssu 9.5k ↪Eee It truly is mind opening. Even now it's a puzzle for us just what it opens to us. When we show with Cantor's diagonal argument that there isn't a bijection between N and R, it's not anymore such a simple proof that we are used to, but an reductio ad absurdum proof. And this has profound implications, which shows itself with the Continuum Hypothesis. Furthermore, we get using a similar technique as the diagonal argument, Gödel's incompleteness theorem's and Turing's answer to the Entscheidungsproblem. And in another way Russell's paradox. All of them in one way or another use what I'd call a negative self-reference. ssu OptionsShare Eee 159 ↪ssu Indeed. I have also studied those other results, especially in Kleene's Mathematical Logic. When I was first exposed to cardinality issues, I was, frankly, amazed and seduced. And I was myself guilty of raising objections I wasn't qualified to raise, too impatient to do the work required. Though I eventually did the work and saw the futility of merely intuitive/metaphysical approach. A person should probably write (at least) a few hundred proofs before philosophizing much about mathematics. FWIW, I do think self-reference is one of the centers of philosophical opportunity. (Strange loops!) Eee OptionsShare TheMadFool 13.8k ↪Eee ↪ssu I understand Cantor's argument well enough to see that there's a pair (1-to-1 correspondence) between the natural numbers and even numbers. Natural numbers N = {1, 2, 3, 4, n, ...} Even numbers E = {2, 4, 6, 8,, 2n...} 2 = 2 x 1 4 = 2 x 2 6 = 2 x 3 . . . n = 2n This then is used to "prove" that the set E is equivalent to set N I get that and thanks for clarifying. However I can see another pairing (1-to-1 correspondence) of numbers in the two sets as follows: N = {2, 4, 6, 8,...,1, 3, 5,...} As you can see I've divided the set N into two parts viz. the even numbers and the odd numbers but note they're included in the same set N The set of even numbers E = {2, 4, 6, 8,...} As you can I see can form a pairing (1-to-1 correspondence) between E and the even numbers in set N like so: (2,2), (4,4), etc. and that leaves the odd numbers without a corresponding pair in the set E. Basically two different bijections are possible. One agrees with Cantor's "proof" but the other contradicts Cantor. You'll have to show that Cantor's bijection is the correct one and the alternative is nonsensical. Can you do that? NOTE: I've excluded zero from the set of even numbers for simplification purposes TheMadFool OptionsShare Eee 159 Basically two different bijections are possible. — TheMadFool I'm guessing that an infinite number of bijections are possible. We can flip segments of length k and proceed as before. This works for any k >= 2. For instance, let k = 3. Then f = { (1,6),(2,4),(3,2),(4,12),(5,10),(6,8),...}. Note that a function is itself just a set of ordered pairs (satisfying a condition that ensures an unambiguous output). But all we need for equivalence, by definition, is a single bijection. N = {2, 4, 6, 8,...,1, 3, 5,...} — TheMadFool Note that sets are not ordered. {1,2} = {2,1}, for instance. The order is in the bijection. And, as you know, the easiest bijection is f(n)=2n, with inverse f−1(e)=e/2. The alternative bijections mentioned above probably have complicated algebraic forms, despite the simplicity of the idea behind them. As you can I see can form a pairing (1-to-1 correspondence) between E and the even numbers in set N like so: (2,2), (4,4), etc. and that leaves the odd numbers without a corresponding pair in the set E. — TheMadFool What you say is a contradiction. A 1-to-1 correspondence is a bijection. As I understand you, you are presenting g:E→N such that g(e)=e. Because odd numbers are not in the range of g, this function is not a surjection, though it is an injection (1-to-1). A bijection is simultaneously a surjection and an injection. Please consider that there are uncountably many functions from E to N. One can use a diagonal argument to prove this. Assume not. Well clearly fk(e)=ke is a countably infinite family of functions. So the functions from E to N are at least countably infinite. So by our assumption (in pursuit of a contradiction), we can arrange the functions from E to N in an countably infinite list fn. Then we define our diagonal function f(e)=fe(e)+1 where the fn is the nth function in a merely countable list of functions from E to N. Since f(n)=fn(n)+1, it differs from every fn at n and is not on the list, despite being a function from E to N. Contradiction! The point of the list fn was to catch all functions from E to N. Hence our assumption that such functions are countable is incorrect. And they are at least countable, so they must be uncountable. Or you can interpret this as: since any possible listing allows for its own diagonal function not on that list, no listing gets them all (providing a bijection from N to the set of those functions.) Every listing casts a shadow or has a blind spot. We just slash diagonally to get something not on the list. It's not surprising that you can find one such function that isn't a bijection. I'm guessing (someone feel like proving it?) that the probability of picking a bijection randomly is 0. Is the subset of bijections countably infinite? Sounds like a fun question. This then is used to "prove" that the set E is equivalent to set N — TheMadFool Keep in mind that this is just one kind of equivalence. It only has as much metaphysical weight as a philosopher feels like giving it. Mathematicians can all agree that X is proven without having to agree on what that means outside of mathematics. 'There exists' has a formal meaning. If x∈R and x≠0, then ∃y∈R such that xy=1. That's an axiom. It's a rule that allows you to put a new piece on the board. Despite the intuitions and applications that quietly drive the 'game,' the game itself is 'safer' than that. Computers can, in theory, check proofs. In practice this is not often considered necessary or desirable. It's like learning a language. Once you learn it, lots of proofs are easy to judge for their correctness. Eee OptionsShare fishfry 3.4k This then is used to "prove" that the set E is equivalent to set N — TheMadFool No, that's a great source of confusion. If there exists a bijection between two sets, even a single one, even if there are plenty of functions that aren't bijections, then we DEFINE the two sets as being cardinally equivalent. It's a definition, not a proof. Given the naturals N and the evens E, there exists a bijection. So we say, by definition, that the two sets have the same cardinality. The fact that there are many functions between the two sets that aren't bijections is irrelevant. The condition is an existential, not universal, quantification. If out of all the functions from one set to another there happens to be a single one that's a bijection, then the two sets are cardinally equivalent, by definition. As has been noted, this is not the only way to compare the relative sizes of sets. There's the subset relation. Clearly E is a proper subset of N so E is "smaller" than N with respect to the proper subset relation Yet another way that was mentioned a few posts back is natural density, also known as asymptotic density. Using this method we see that the limit as n gets large of the proportion of even numbers among the first n numbers goes to 1/2. So the natural density of E in N is 1/2. Likewise the natural density of the multiples of 3 in N is 1/3; and the natural density of the primes is 0. [Proof by the interested reader for that last assertion]. So there are lots of ways to compare the relative size of sets. Cardinality happens to be one way. It's not a very fine-grained one. For example the intervals [0,1] and [0,2] of real numbers have the same cardinality, since there's a bijection between them: namely f(x) = 2x. But [0,1] has length 1 and [0,2] has length 2. The generalization of the idea of length is called measure, and it's yet another way to compare the relative size of sets. In fact measure theory underlies modern probability theory, so it's very important. But as we saw with the intervals, cardinality does not respect measure. In that sense cardinality is a very crude way of thinking about the relative sizes of sets. Just as we use a hammer for one thing and a spatula for another and a torque wrench for yet something else, mathematicians have a lot of tools in the toolbox. Cardinality, as defined by bijection, is just one tool, to be used as appropriate. Basically two different bijections are possible. One agrees with Cantor's "proof" but the other contradicts Cantor. You'll have to show that Cantor's bijection is the correct one and the alternative is nonsensical. — TheMadFool Cantor's definition of cardinal equivalence via bijection is not right or wrong. It's a definition. In the past 140 years it has proven to be both interesting and useful, which is why it's stuck around. But it's not the only way of comparing sets; and it's neither right nor wrong. It's just one of the tools in the mathematical toolbox. E and N have the same cardinality. The natural density of E in N is 1/2. And E is indeed a proper subset of N. All three points of view are equally valid. One or the other viewpoint might be more useful in a particular context. fishfry OptionsShare TheMadFool 13.8k No, that's a great source of confusion. If there exists a bijection between two sets, even a single one, even if there are plenty of functions that aren't bijections, then we DEFINE the two sets as being cardinally equivalent. It's a definition, not a proof. — fishfry That's where the problem is isn't it? The definition is inadequate for the reason that, on one hand, Cantor's "preferred" bijection leads to an equivalence between the set of even numbers and natural numbers but on the other hand there exists another bijection that shows that the set of natural numbers is not equivalent to the set of natural numbers. TheMadFool OptionsShare fishfry 3.4k That's where the problem is isn't it? The definition is inadequate for the reason that, on one hand, Cantor's "preferred" bijection leads to an equivalence between the set of even numbers and natural numbers but on the other hand there exists another bijection that shows that the set of natural numbers is not equivalent to the set of natural numbers. — TheMadFool It's a definition. It can't be right or wrong. You do agree that there exists at least one bijection between N and E, namely f(n) = 2n. You agree with that, yes? Then by definition the two sets have the same cardinality. That's all it means. If there exists a bijection, then the sets are said to have the same cardinality. It's true that there are many functions between E and N that aren't bijections, but it only takes one to satisfy the definition. Do you agree that even if you don't like the definition, E and N satisfy it by virtue of that one bijection? there exists another bijection that shows that the set of natural numbers is not equivalent to the set of natural numbers. — TheMadFool But no. The existence of a non-bijection doesn't prove anything. As long as there's a single bijection, the definition is satisfied. Not because it's right or wrong, but because that's the definition. fishfry OptionsShare TheMadFool 13.8k It's a definition. It can't be right or wrong. — fishfry If a definition leads to a contradiction? TheMadFool OptionsShare fishfry 3.4k If a definition leads to a contradiction? — TheMadFool It doesn't. Do you agree that there exists at least one bijection from E to N? If you agree, then you must agree that E and N have the same cardinality, because the definition says that there must be at least one bijection between them, and this is manifestly the case. ps -- It's like a guy who cheats on his wife. She says to him, "You're a cheater." He says no! Think about all the times I DIDN"T cheat on you. But that's not the point. If you cheat once, that's the definition of a cheater. If you cheated Monday but not on Tuesday or Wednesday, you can't say you're not a cheater because you didn't cheat on Wednesday. Right? The definition is doing it once. Likewise the definition of cardinal equivalence is that there's at least one bijection. It doesn't matter that some other function isn't a bijection. fishfry OptionsShare ssu 9.5k I understand Cantor's argument well enough to see that there's a pair (1-to-1 correspondence) between the natural numbers and even numbers. — TheMadFool That's not technically Cantor's diagonal argument. That proof comes into use only with the reals R. Basically two different bijections are possible. — TheMadFool Please understand you are not talking of a bijection! It isn't a bijection if one group has more members. It's either an injection or an surjection. It's the most simple math there is an actually one of the most hardest. And, if I understand your thought correctly, you are basically saying that there's an injection from E to N. But when you can prove there's a bijection, it doesn't go that it's only an injection. Yet, understand that you are talking about infinite series. Have you ever heard about the Hilbert Hotel? ssu OptionsShare sandman 41 ↪tim wood I don't presuppose a finite stick. They are the norm. Given the definition of infinite, without limit or boundary, and the fact that infinite is not a number or quantifier, a one ended stick is not measurable/quantifiable. You already described the futility of such an effort. Cantor was an illusionist and fooled many people. sandman OptionsShare Deleted User 0 This user has been deleted and all their posts removed. Deleted User OptionsShare sandman 41 ↪tim wood I read Cantor's biography. It helps to understand his motivation to write. I agree, ignorance is abundant. sandman OptionsShare TheMadFool 13.8k It doesn't. Do you agree that there exists at least one bijection from E to N? If you agree, then you must agree that E and N have the same cardinality, because the definition says that there must be at least one bijection between them, and this is manifestly the case. ps -- It's like a guy who cheats on his wife. She says to him, "You're a cheater." He says no! Think about all the times I DIDN"T cheat on you. But that's not the point. If you cheat once, that's the definition of a cheater. If you cheated Monday but not on Tuesday or Wednesday, you can't say you're not a cheater because you didn't cheat on Wednesday. Right? The definition is doing it once. Likewise the definition of cardinal equivalence is that there's at least one bijection. It doesn't matter that some other function isn't a bijection. — fishfry ↪fishfry Do you agree that there exists at least one bijection from E to N? — fishfry So there was nothing wrong when Socrates defined humans as featherless bipeds and someone came along with a chicken plucked of all its feathers and declared "this is a human"? After all there was/is at least one human that fit the definition. TheMadFool OptionsShare TheMadFool 13.8k ↪ssu Thank you very much for the colorful explanation. Appreciate it. However I am actually talking about there being an injection correspondence between even numbers and natural numbers and so bijection correspondence isn't the only game in town. Ergo, the set of even numbers isn't equivalent to the set of natural numbers. TheMadFool OptionsShare simeonz 310 ↪TheMadFool From a programmer's point of view, cardinality equivalence between two sets is about their expressibility through each other. card(N) = card(E) tells me that if I want to define an encoding of the elements of E through the elements of N, or N through elements of E, I can. Thus, I can write/store n to represent "the n-th even number", and 2n to represent "the ordinal n". In any case, as long as I standardize my representation upfront, I am going to be able to decode it uniquely later. card(N) < card(R) tells me that I can never represent a real number through a natural number, no matter how I try to tip-toe around the issue. I can muster the strangest encodings, but it is logical impossibility. Thus, I will be able to express only some real numbers ordinally (or even out of order), but not all of them. P.S. Thus your dilemma can be rephrased as the ability to encode some set and a strictly smaller set through the same representation set. This may be awkward in some sense, but it is just a fact of life. simeonz OptionsShare SophistiCat 2.3k ↪TheMadFool And everyone has been at pains to explain to you that sets with the same cardinality are just that - sets with the same cardinality. They are not (necessarily) equal. They are not "equivalent" - that's not a set theory term. They have the same number of elements only in the special case of finite sets (cardinality is a generalization of set size). It's like as if someone told you that Sarah and Anil are the same age, and you objected that Sarah is a woman and Anil is a man, Sarah is American and Anil is Indian, Sarah is smaller than Anil, so how could they be equivalent?! Cardinality is just one measure of sets, nothing less, nothing more. Get over this already. SophistiCat OptionsShare 12345Next Write Comment Welcome to The Philosophy Forum! 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15077	https://math.stackexchange.com/questions/4351285/is-circle-the-only-shape-that-can-remain-convex-after-folding	geometry - Is circle the only shape that can remain convex after folding? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Is circle the only shape that can remain convex after folding? Ask Question Asked 3 years, 8 months ago Modified1 year, 4 months ago Viewed 1k times This question shows research effort; it is useful and clear 51 Save this question. Show activity on this post. Here "fold" means "fold a piece of paper (along a straight crease)". The sketch below shows that one can always find a fold by which an ellipse or rectangle loses convexity. But it seems a circle remains convex no matter how the crease is chosen? I am not sure how to deal with a problem like this, where the shape is generic and a convenient coordinate system cannot be defined. Update: to be precise, the circle means "disk", which includes both the border and the interior. Same for all other shapes. Update Jan. 13 I am thrilled by the number of upvotes. Here is an interesting experiment. Starting simple, let's use an upright rectangle centered at the origin, and reflect it w.r.t. all creases passing the origin (the experiment can only use a finite subset of such creases). The area union of all reflected shapes appears to be a disk. This observation seems to allow the following general statement: a 2d shape, convex or not, transforms into a new shape by folding. The union of all new shapes is a disk. To be honest, this statement itself sounds like a question in need of a proof. But I feel it is highly related with the original question. So I update it here along with the Mathematica code for plotting the figure (). Enjoy! ``` foldAlongMiddleAxis[mya_, mytheta_] := Block[{a = mya, cs = Cos[mytheta], ss = Sin[mytheta]}, polPrime1 = Polygon[{{cs, ss}, {cs, ss} + a {-ss, cs}, {-cs, -ss} + a {-ss, cs}, {-cs, -ss}}]; polPrime2 = Polygon[{{1, 0}, {1, a}, {-1, a}, {-1, 0}}]; Graphics[{{Opacity, EdgeForm[Gray], polPrime2}, {Opacity, EdgeForm[Gray], polPrime1}}, AspectRatio -> Automatic]] plotAllFoldsAndTrajO[a_, n_] := Show[{ Graphics[{LightPink, Opacity[0.9], EdgeForm[Gray], Polygon[{{1, -a}, {1, a}, {-1, a}, {-1, -a}}]}], Table[foldAlongMiddleAxis[a, mytheta], {mytheta, Range[0, 2 [Pi], [Pi]/n]}], ParametricPlot[{{Cos, Sin} + a {-Sin, Cos}}, {[Theta], 0, 2 [Pi]}, PlotStyle -> {Black, Thick}] }] plotAllFoldsAndTrajO[5, 50] ``` geometry convex-geometry Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jan 25, 2022 at 15:34 RavenclawPrefect 19k 3 3 gold badges 39 39 silver badges 111 111 bronze badges asked Jan 7, 2022 at 22:06 TaoziTaozi 2,743 15 15 silver badges 21 21 bronze badges 11 2 Seems to be true. You're essentially asking which shapes, when folded along any line, don't have a crease. This seems to be the set of shapes that are symmetrical w.r.t. that line, and that, for any line. The circle is the only one AFAIK.Stefan Lafon –Stefan Lafon 2022-01-07 22:12:08 +00:00 Commented Jan 7, 2022 at 22:12 2 @StefanLafon I am not sure your last sentence is correct. For example, a rectangle can remain convex for many folds (e.g., fold a corner a little bit inside). Those folds do not require any symmetrical properties.Taozi –Taozi 2022-01-07 22:31:38 +00:00 Commented Jan 7, 2022 at 22:31 1 @Maksim You can't remove the folded part and the part it overlaps, they are still part of the original paper.Taozi –Taozi 2022-01-07 22:35:51 +00:00 Commented Jan 7, 2022 at 22:35 1 @Taozi: So you are not talking about circles but rather meaning discs ...Maksim –Maksim 2022-01-07 22:41:09 +00:00 Commented Jan 7, 2022 at 22:41 2 @Sergey Sadly that's not true. Think of a rectangle folded such that you match two corners when folding - you obtain a pentagon.Tanny Sieben –Tanny Sieben 2022-01-07 23:33:25 +00:00 Commented Jan 7, 2022 at 23:33 \|Show 6 more comments 3 Answers 3 Sorted by: Reset to default This answer is useful 11 Save this answer. Show activity on this post. This isn't a complete answer, more like a broad sketch of a proof, but it's too long for a comment. First, we can see that the region must be convex, since if it is not convex, a very tiny fold (think of the fold as being a line segment joining two points on the boundary of the region, so if a fold is "tiny" the length of that segment is very small) will not appreciably alter the shape, thus it will remain non-convex. Second, it is intuitive (but no proof) that a convex polygon cannot work, because in such a shape, there must be an interior angle where two adjacent sides meet and has measure less than π π. Then since a polygon must have at least 3 sides, there is at least one other side not adjacent to this angle, which must form a triangle with the other two sides (the angle cannot point "the wrong way" because the region is convex). Therefore, there is a line of reflection between this side and the vertex of the subtending angle that reflects the vertex across to the other side, and this results in a non-convex shape. Then we can generalize further to exclude any convex shapes that have a polygonal vertex, since a similar argument to the above shows that there must be a line that will reflect that vertex in a way that will make it "poke out" of any smooth curve. So we know the region must be smooth. Then I believe the remaining argument is to show that the boundary must have constant curvature. This strikes me as a previously solved problem but I don't have a literature reference. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jan 8, 2022 at 1:52 heropupheropup 145k 15 15 gold badges 114 114 silver badges 201 201 bronze badges 8 (+1) Another possible idea for the next step after smoothness is established... Start with two parallel tangents alongside an arbitrary direction, so that the shape is between them. Consider reflections about a third parallel drawn at equal distance between the two. The points of contact must be symmetric about the axis for the fold to be convex, then - maybe - it follows that the entire figure must be symmetric about the axis. In that case the problem would reduce to How many non-infinite plane curves with infinite reflectional symmetry?.dxiv –dxiv 2022-01-08 03:14:30 +00:00 Commented Jan 8, 2022 at 3:14 This may work: with a smooth shape, pick a fold direction along which it is not symmetrical (i.e. no axis in this direction makes a symmetry). Mostly likely this means there are two points on the border, P and fold(P), with a non-symmetrical tangent. Then place the folding axis in this direction so that it sends P to fold(P). It creates locally a non-convex shape (like the self-intersection point in each figure in the question).Armin Rigo –Armin Rigo 2022-01-08 09:19:09 +00:00 Commented Jan 8, 2022 at 9:19 Your paragraph claiming "we can generalize further to exclude any convex shapes that have a polygonal vertex" is wrong, because there may be more than one non-smooth point so it is not at all easy to prove that you can make that corner poke out when folded.user21820 –user21820 2022-01-08 16:09:42 +00:00 Commented Jan 8, 2022 at 16:09 @user21820HATESSMOKING-HATS Such a boundary that does not contain any isolated non-smooth points would be pathological. You are welcome to construct such a counterexample and demonstrate it meets the conditions of the problem, as at no point did I claim that my reasoning constitutes a formal proof, merely a possible approach to one.heropup –heropup 2022-01-08 16:42:14 +00:00 Commented Jan 8, 2022 at 16:42 No I'm not saying there is a counter-example; all I said was "it is not at all easy to prove". And yes, maybe a proof can be obtained along the lines you gave, but you need to fix your paragraph because it wrongly says "any smooth curve" as if assuming that there is only one non-smooth point that you are using to poke out the other side.user21820 –user21820 2022-01-08 16:43:02 +00:00 Commented Jan 8, 2022 at 16:43 \|Show 3 more comments This answer is useful 2 Save this answer. Show activity on this post. Sketch Proof The boundary of the original convex shape is differentiable almost everywhere. Let A,B,C A,B,C be three points on the boundary where the boundary is differentiable and let t A,t B,t C t A,t B,t C be the respective tangents. Folding along the perpendicular bisector of A B A B shows that t A t A and t B t B must be equally but oppositely inclined to A B A B. Similarly for t B t B and t C t C and for t C t C and t A t A. Then t A,t B,t C t A,t B,t C are tangents to the circle through A,B,C A,B,C. This circle is therefore determined by A,B A,B alone and so contains every point on the boundary where the boundary is differentiable. It therefore contains every point on the boundary. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jan 24, 2022 at 14:33 user502266 user502266 3 What does it mean for t A t A and t B t B to be oppositely inclined to A B A B? As stated I don't understand how this follows from the starting assumption or why it implies that they are tangents to the circle.RavenclawPrefect –RavenclawPrefect 2022-01-25 15:15:05 +00:00 Commented Jan 25, 2022 at 15:15 The reflection in the perpendicular bisector of A B A B, must map t A t A onto t B t B . Otherwise, the convex hull of points arbitrarily close to the image of t A t A and arbitrarily close to t B t B would contain a disc about B B entirely contained in the new convex region. This new region would therefore contain points on A B A B extended which is impossible.user502266 –user502266 2022-01-25 15:29:35 +00:00 Commented Jan 25, 2022 at 15:29 Ah, I had misread the original folding procedure. Thanks!RavenclawPrefect –RavenclawPrefect 2022-01-25 15:33:18 +00:00 Commented Jan 25, 2022 at 15:33 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. I came here from this question, which I found a simple and concise proof for the following: Consider any simply connected shape on a 2D Euclidean plane. If, for any line that intersects this shape, the reflection of the parts of the boundary that lie on one side of the line do not intersect the parts of the boundary that lie on the other side, then the shape must be a circle. Here, two curves intersect means that there is at least one point that lies on both curves, and at least one point that lies on one curve but not another. (So completely overlapping does not count as intersecting.) (I don't understand why that question got closed. Last time I checked, the consensus seems to be that these are related but different questions. If anyone can re-open it, please do. But anyway, here I'm going to reduce this problem to that one.) If it's true that the reflection of the half can't intersect with the other half here as well, then we're done. Is it possible that the reflection intersects with the other half while the shape remains convex? Unfortunately, it's true. A simple example is matching the vertex of a rectangle to the opposite one on the diagonal. The resulting shape is still convex, although the boundaries intersect. But, notice that, if you move the line slightly above, it'll no longer be convex. So, the only remaining task is, to make this argument rigorous. (Which takes way more effort than proving the previous statement, though.) So, let's suppose that: There is a line, L L, that cuts the boundary into two parts, B u B u and B d B d, the upper and lower parts. The reflection of B d B d, B′d B d′, intersects with B u B u, thus there must be a point p i p i which lies on both B u B u and B′d B d′, and another point p d p d which is on B′d B d′ but not on B u B u. Due to symmetry, w.l.o.g, suppose the line L L is horizontal, and suppose p d p d is to the left of p i p i and under B u B u. Here B′d B d′ and B u B u may intersect at a single point p i p i, or they may overlap on a curve where p i p i lies on. It doesn't matter which case it is. Now, we want to show that, if we move the line L L slightly above, then the reflection of B d B d will also move slightly above, which will cause the shape to be non-convex. The following is just the thought process which is not rigorous, but it shows the direction. To do that, consider the point where the two curves splits. If we zoom in enough, the boundaries should appear linear, like this: Then, if we move the blue dotted line a little bit above, it will intersect the red solid line, which results in the shape being concave. So, is that true that if we zoom in enough, the curve appears linear? Now, we introduce the concept of "left tangent line". First, we prove that, for any point p p on one of the curves, for any angle δ δ, there exists a distance d d and an angle θ θ such that, for any point p 1 p 1 that's to the left of p p which is less than d d from p p horizontally, the line that connect p p and p 1 p 1 must make an angle within [θ,θ+δ][θ,θ+δ] with the −x−x axis. Translation to plane words: if we zoom in enough, the curve should appear to be almost linear (If we connect any two points, the angle of the slope of the line should be within a small interval). The proof is actually very simple. Suppose it's not true. There's an angle δ δ such that there's no distance that satisfies this condition. Then, we choose an arbitrary point, p 1 p 1, to the left of p p. Suppose the angle of the line through p 1 p 1 and p p is θ θ. Because the shape is convex, the angle between −x−x axis and the line that connects p p and any point q q on the curve between p 1 p 1 and p p must be greater than θ θ. Then, there must be a point p 2 p 2 where the line through p 2 p 2 and p p makes an angle that is at least θ+δ θ+δ. Then, we must be able to find another point p 3 p 3, which makes an angle θ+2 δ θ+2 δ. We can keep doing that. Since δ δ is finite, if we do this enough times, the line will be rotated for more than π π, so it's impossible that the point is to the left of p p, contradiction. So, we proved that the left slope, thus the left tangent line, must exist, which does not necessarily equal the right slope or right tangent line, which is obvious considering the vertex of a polygon. Now, we are able to limit the tangent line to an arbitrarily small angle. How do we proceed? Basically, for a small region to the left of the point p s p s where the two curves split, we are able to limit the upper curve to be inside the solid triangle, and the lower curve inside the dashed triangle. If the two triangles don't intersect, then we can move the lower triangle a little bit above, so that they intersect: Suppose the line that connects q′d q d′ and p′s p s′ intersects with the line that connects q u q u and p s p s at A A, and the line p u−p′s p u−p s′ and q u−p s q u−p s intersect at B B. Since p u p u is on the curve, every point below the line p u−p′s p u−p s′ (and above horizontal line L L) must be inside the shape, because it must remain convex. But the upper boundary is inside the solid triangle q u p u p s q u p u p s, and the reflection of the lower boundary is inside the dashed triangle q′d p′d p′s q d′p d′p s′. Thus, every point inside the triangle A B p′s A B p s′ is not inside the shape. Contradiction. Now, we still have one assumption: We can find a small enough distance such that the two triangles don't intersect. Unfortunately, this is not always true. It's possible that the two curves have the same tangent line at p s p s, i.e, they're tangent at p s p s. But notice that: we don't really need to use the point where the two curves split. We can move the reflected curve up and down. If that point doesn't work, we can move the curve and let them intersect at another point, where they are not tangent. So, we want to show that, there exist an x x coordinate x 0 x 0, such that the corresponding points A u A u and A d A d with x x-coordinate x 0 x 0 on the curves B u B u and B′d B d′ respectively, have different left slope. Assume that this is not true: For any x x coordinate, the two points on the two curves with this x x coordinate has the same left slope. Then we can just start from the point where they intersect, and every pair of points to the left of it on the two curves must overlap. Thus the two curves to the left of the intersecting point must overlap, contradicting the fact that there exist a point that is to the left of the intersecting point, and lies on one curve but not the other. (Update: Here it's missing a special case: the intersecting point p s p s lies on a vertical part of the curves. Example: We're folding a rectangle along one of its edges. So, to make the argument work, we must add another condition: If the two curves only intersect at a vertical part of them, then consider the line that's parallel to it which cuts the shape into two equal areas, as I did in the answer to the other question. Then, under this condition, if they still intersect, they must intersect at a different point that's not on a vertical part of the curves, because otherwise, if the two curves still only intersect on the vertical part, the boundary must overlap completely, or the shape is not convex. This implies that the shape must be symmetric w.r.t this line, which reduces to the result of other question.) Now it becomes clear how we construct the proof. There exist a pair of points A u A u, A d A d with the same x x coordinate x 0 x 0 that lies on the two curves and have different left slopes. Suppose the difference in the angle of the tangent lines is θ θ. Next, we want to make sure that the two triangles don't overlap, thus we choose an angle δ<θ δ<θ, and there must exist a distance d u d u such that any point on B u B u that has x x coordinate within [x 0−d u,x 0)[x 0−d u,x 0) must satisfy that the line connecting it and p u p u makes an angle within [α,α+δ][α,α+δ] with the −x−x axis. There must exist a distance d d d d for the reflected lower curve B′d B d′ as well, and a corresponding β=α−θ β=α−θ. We choose the minimum of the two distances, d d. Then, we move the line L L vertically by half of the vertical distance of A u A u and A d A d, so that they overlap afterwards. Let's call the point p s p s. Now the situation is just as before, but we have guaranteed that the two triangles don't overlap. Suppose the vertical distance between p u p u and q d q d is h h. Moving the folding line L L above by h 4 h 4 will result in the reflected lower curve rising by h 2 h 2, thus the two triangles must intersect as shown in the figure above, thus causing the shape to be concave. Since the reflected lower boundary can't intersect with the upper boundary, it reduces to the question that I linked to in the beginning. Thus the boundary must be a circle (the shape is a disk). The proof is complete. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited May 18, 2024 at 7:03 answered May 18, 2024 at 5:36 Yi JiangYi Jiang 726 4 4 silver badges 10 10 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. 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Inicio 2. Estantería 3. Matemáticas 4. Libro: Cálculo (Apex) 5. 8: Secuencias y series 6. 8.5: Serie alterna y convergencia absoluta Expandir/contraer ubicación global 8.5: Serie alterna y convergencia absoluta Última actualización 30 oct 2022 Guardar como PDF 8.4: Pruebas de Ratio y Raíz 8.6: Serie Power Page ID 111816 Gregory Hartman et al. Virginia Military Institute ( \newcommand{\kernel}{\mathrm{null}\,}) Índice 1. Colaboradores y Atribuciones Todas las pruebas de convergencia en serie que hemos utilizado requieren que la secuencia subyacente{a n} sea una secuencia positiva. (Podemos relajar esto con el Teorema 64 y afirmar que debe haber un N>0 tal que a n>0 para todos n>N; es decir,{a n} es positivo para todos menos un número finito de valores de n.) En esta sección exploramos series cuya suma incluye términos negativos. Comenzamos con una forma muy específica de serie, donde los términos de la suma se alternan entre ser positivo y negativo. Definición 34: serie alterna Dejar{a n} ser una secuencia positiva. Una serie alterna es una serie de cualquiera de la forma (8.5.1)∑n=1∞(−1)n⁢a n or∑n=1∞(−1)n+1⁢a n. Recordemos los términos de Serie Armónica provienen de la Secuencia Armónica{a n}={1/n}. Una serie alternante importante es la serie armónica alterna: (8.5.2)∑n=1∞(−1)n+1⁢1 n=1−1 2+1 3−1 4+1 5−1 6+⋯ Las series geométricas también pueden ser series alternas cuando r<0. Por ejemplo, si r=−1/2, la serie geométrica es (8.5.3)∑n=0∞(−1 2)n=1−1 2+1 4−1 8+1 16−1 32+⋯ Teorema 60 afirma que las series geométricas convergen cuando\|r\|<1 y da la suma:∑n=0∞r n=1 1−r. Cuando r=−1/2 como arriba, encontramos (8.5.4)∑n=0∞(−1 2)n=1 1−(−1/2)=1 3/2=2 3. Existe un poderoso teorema de convergencia para otras series alternas que cumplen algunas condiciones. teorema 70: prueba en serie alterna Dejar{a n} ser una secuencia positiva, decreciente donde lim n→∞⁡a n=0. Entonces (8.5.5)∑n=1∞(−1)n⁢a n and∑n=1∞(−1)n+1⁢a n convergen. La idea básica detrás del Teorema 70 se ilustra en la Figura 8.5.1. Se muestra una secuencia{a n} decreciente positiva junto con las sumas parciales (8.5.6)S n=∑i=1 n(−1)i+1⁢a i=a 1−a 2+a 3−a 4+⋯+(−1)n+1⁢a n. Porque{a n} es decreciente, disminuye la cantidad por la que S n rebota arriba/abajo. Además, los términos impares S n forman una secuencia acotada decreciente, mientras que los términos pares S n forman una secuencia creciente, acotada. Desde acotado, las secuencias monótonas convergen (ver Teorema 59) y los términos de{a n} aproximación 0, se pueden mostrar los términos pares e impares de S n convergen al mismo límite común L, la suma de la serie. Figura 8.5.1: Ilustrando la convergencia con la Prueba de Serie Alternante. Ejemplo 8.5.1: Applying the Alternating Series Test Determine si la Prueba de Serie Alternada se aplica a cada una de las siguientes series. ∑n=1∞(−1)n+1⁢1 n ∑n=1∞(−1)n⁢ln⁡n n ∑n=1∞(−1)n+1⁢\|sin⁡n\|n 2 Solución Esta es la Serie Armónica Alternante como se ha visto anteriormente. La secuencia subyacente es{a n}={1/n}, que es positiva, decreciente, y se acerca a 0 como n→∞. Por lo tanto, podemos aplicar la Prueba de Serie Alternante y concluir que esta serie converge. Si bien la prueba no establece a qué converge la serie, veremos más adelante que∑n=1∞(−1)n+1⁢1 n=ln⁡2. 2. La secuencia subyacente es{a n}={ln⁡n/n}. Esto es positivo y se acerca a 0 como n→∞ (use la Regla de L'Hopital). Sin embargo, la secuencia no está disminuyendo para todos n. Es sencillo calcular a 1=0,,a 2≈0.347 a 3≈0.366, y a 4≈0.347: la secuencia está aumentando durante al menos los primeros 3 términos. No concluimos inmediatamente que no podemos aplicar la Prueba de Serie Alternante. Más bien, considere el comportamiento a largo plazo de{a n}. Tratando a n=a⁡(n) como una función continua de n definido en[1,∞), podemos tomar su derivada:(8.5.7)a′⁡(n)=1−ln⁡n n 2. La derivada es negativa para todos n≥3 (en realidad, para todos n>e), el significado a⁡(n)=a n es decreciente en[3,∞). Podemos aplicar la Prueba de Serie Alternante a la serie cuando empezamos n=3 y concluimos que∑n=3∞(−1)n⁢ln⁡n n converge; sumando los términos con n=1 y n=2 no cambian la convergencia (es decir, aplicamos el Teorema 64). La lección importante aquí es que como antes, si una serie no cumple con los criterios de la Prueba de Serie Alternante en solo un número finito de términos, aún podemos aplicar la prueba. 3. La secuencia subyacente es{a n}=\|sin⁡n\|/n. Esta secuencia es positiva y se acerca 0 como n→∞. Sin embargo, no es una secuencia decreciente; el valor de\|sin⁡n\| oscila entre 0 y 1 como n→∞. No podemos eliminar un número finito de términos para hacer{a n} decreciente, por lo tanto no podemos aplicar la Prueba de Serie Alternante. Tenga en cuenta que esto no significa que concluyamos que la serie diverge; de hecho, sí converge. Simplemente no podemos concluir esto con base en el Teorema 70. Key Idea 31 da la suma de algunas series importantes. Dos de estos son (8.5.8)∑n=1∞1 n 2=π 2 6≈1.64493 and∑n=1∞(−1)n+1 n 2=π 2 12≈0.82247. Estas dos series convergen a sus sumas a diferentes ritmos. Para ser exactos a dos lugares después del decimal, necesitamos 202 términos de la primera serie aunque solo 13 de la segunda. Para obtener 3 lugares de precisión, necesitamos 1069 términos de la primera serie aunque solo 33 de la segunda. ¿Por qué es que la segunda serie converge mucho más rápido que la primera? Si bien hay muchos factores involucrados al estudiar las tasas de convergencia, la estructura alterna de una serie alterna nos da una poderosa herramienta a la hora de aproximar la suma de una serie convergente. teorema 71: el teorema de aproximación de series alternas Dejar{a n} ser una secuencia que satisfaga las hipótesis de la Prueba de Serie Alternante, y dejar S n y L ser las sumas n th parciales y la suma, respectivamente, de cualquiera∑n=1∞(−1)n⁢a n o∑n=1∞(−1)n+1⁢a n. Entonces \|S n−L\|<a n+1, y L está entre S n y S n+1. La Parte 1 del Teorema 71 establece que la suma n th parcial de una serie alternante convergente estará dentro a n+1 de su suma total. Consideremos las series alternas que miramos antes de la afirmación del teorema,∑n=1∞(−1)n+1 n 2. Ya que a 14=1/14 2≈0.0051, sabemos que S 13 está dentro 0.0051 de la suma total. Además, la Parte 2 del teorema establece que desde S 13≈0.8252 y S 14≈0.8201, sabemos que la suma L se encuentra entre 0.8201 y 0.8252. Un uso de esto es el conocimiento que S 14 es exacto a dos lugares después del decimal. Algunas series alternas convergen lentamente. En Ejemplo 8.5.1 determinamos la serie∑n=1∞(−1)n+1⁢ln⁡n n convergente. Con n=1001, encontramos ln⁡n/n≈0.0069, lo que significa que S 1000≈0.1633 es exacto a uno, tal vez dos, lugares después del decimal. Ya que S 1001≈0.1564, sabemos que la suma L es 0.1564≤L≤0.1633. Ejemplo 8.5.2: Approximating the sum of convergent alternating series Aproximar la suma de las siguientes series, con precisión de dentro 0.001. ∑n=1∞(−1)n+1⁢1 n 3 2.⁢∑n=1∞(−1)n+1⁢ln⁡n n. Solución Usando el Teorema 71, queremos encontrar n dónde 1/n 3<0.001:1 n 3≤0.001=1 1000 n 3≥1000 n≥1000 3 n≥10. Let L be the sum of this series. Por la Parte 1 del teorema,\|S 9−L\|<a 10=1/1000. Podemos calcular S 9=0.902116, que nuestro teorema afirma está dentro 0.001 de la suma total. Podemos usar la Parte 2 del teorema para obtener un resultado aún más preciso. Como sabemos el 10 th término de la serie es−1/1000, podemos calcular fácilmente S 10=0.901116. La parte 2 del teorema establece que L es entre S 9 y S 10, entonces 0.901116<L<0.902116. 2. Queremos encontrar n dónde ln⁡(n)/n<0.001. Empezamos resolviendo ln⁡(n)/n=0.001 para n. Esto no se puede resolver algebraicamente, por lo que usaremos el Método de Newton para aproximar una solución. Vamos f⁡(x)=ln⁡(x)/x−0.001; queremos saber dónde f⁡(x)=0. Hacemos una conjetura que x debe ser “grande”, así que nuestra suposición inicial será x 1=1000. Recordemos cómo funciona el Método de Newton: dada una solución aproximada x n, nuestra siguiente aproximación la x n+1 da(8.5.9)x n+1=x n−f⁡(x n)f′⁡(x n). We find f′⁡(x)=(1−ln⁡(x))/x 2. Esto da x 2=1000−ln⁡(1000)/1000−0.001(1−ln⁡(1000))/1000 2=2000. Usando una computadora, encontramos que el Método de Newton parece converger a una solución x=9118.01 después de 8 iteraciones. Tomando el siguiente entero más alto, tenemos n=9119, donde ln⁡(9119)/9119=0.000999903<0.001. Nuevamente usando una computadora, nos encontramos S 9118=−0.160369. La parte 1 del teorema establece que esto está dentro 0.001 de la suma real L. Conociendo ya el th término 9,119, podemos computar S 9119=−0.159369, es decir−0.159369<L<−0.160369. Observe cómo la primera serie convergió bastante rápido, donde solo necesitábamos 10 términos para alcanzar la precisión deseada, mientras que la segunda serie tomó más de 9,000 términos. Uno de los famosos resultados de las matemáticas es que la Serie Armónica,∑n=1∞1 n diverge, pero la Serie Armónica Alternante∑n=1∞(−1)n+1⁢1 n,, converge. La noción de que alternar los signos de los términos en una serie puede hacer converger una serie nos lleva a las siguientes definiciones. Definición 35: convergencia absoluta y condicional Una serie∑n=1∞a nconverge absolutamente si∑n=1∞\|a n\| converge. Una serie∑n=1∞a nconverge condicionalmente si∑n=1∞a n converge pero∑n=1∞\|a n\| diverge. Así decimos que la Serie Armónica Alternante converge condicionalmente. Ejemplo 8.5.3: Determining absolute and conditional convergence. Determine si las siguientes series convergen absoluta, condicionalmente o divergen. ∑n=1∞(−1)n⁢n+3 n 2+2⁢n+5 2.⁢∑n=1∞(−1)n⁢n 2+2⁢n+5 2 n 3.⁢∑n=3∞(−1)n⁢3⁢n−3 5⁢n−10 Solución Podemos mostrar las(8.5.10)∑n=1∞\|(−1)n⁢n+3 n 2+2⁢n+5\|=∑n=1∞n+3 n 2+2⁢n+5 divergencias de la serie usando la Prueba de Comparación de Límite, comparando con 1/n. La serie∑n=1∞(−1)n⁢n+3 n 2+2⁢n+5 converge mediante la Prueba de Serie Alternante; concluimos que converge condicionalmente. 2. Podemos mostrar la serie(8.5.11)∑n=1∞\|(−1)n⁢n 2+2⁢n+5 2 n\|=∑n=1∞n 2+2⁢n+5 2 n converge usando la Prueba de Ratio. Por lo tanto concluimos∑n=1∞(−1)n⁢n 2+2⁢n+5 2 n converge absolutamente. 3. La serie(8.5.12)∑n=3∞\|(−1)n⁢3⁢n−3 5⁢n−10\|=∑n=3∞3⁢n−3 5⁢n−10 diverge usando la Prueba de n th Término, por lo que no converge absolutamente. La serie∑n=3∞(−1)n⁢3⁢n−3 5⁢n−10 falla las condiciones de la Prueba de Serie Alternante ya(3⁢n−3)/(5⁢n−10) que no se acerca 0 como n→∞. Podemos afirmar además que esta serie diverge; ya que n→∞, la serie efectivamente suma y resta una y 3/5 otra vez. Esto hace que la secuencia de sumas parciales oscile y no converja. Por lo tanto, la serie∑n=1∞(−1)n⁢3⁢n−3 5⁢n−10 diverge. Saber que una serie converge absolutamente nos permite hacer dos declaraciones importantes, dadas en el siguiente teorema. La primera es que la convergencia absoluta es “más fuerte” que la convergencia regular. Es decir, solo porque∑n=1∞a n converge, no podemos concluir∑n=1∞\|a n\| que converja, pero saber que una serie converge absolutamente nos dice que∑n=1∞a n convergerá. Una razón por la que esto es importante es que todas nuestras pruebas de convergencia requieren que la secuencia subyacente de términos sea positiva. Al tomar el valor absoluto de los términos de una serie donde no todos los términos son positivos, muchas veces podemos aplicar una prueba apropiada y determinar la convergencia absoluta. Esto, a su vez, determina que la serie que nos dan también converja. El segundo enunciado se refiere a reordenamientos de series. Cuando se trata de un conjunto finito de números, la suma de los números no depende del orden en que se sumen. (Entonces 1+2+3=3+1+2.) Uno puede sorprenderse al descubrir que al tratar con un conjunto infinito de números, la misma afirmación no siempre es cierta: algunas listas infinitas de números pueden reorganizarse en diferentes órdenes para lograr diferentes sumas. El teorema afirma que los términos de una serie absolutamente convergente pueden reordenarse de ninguna manera sin afectar a la suma. teorema 72: teorema de convergencia absoluta Déjese∑n=1∞a n ser una serie que converja absolutamente. ∑n=1∞a n converge. {b n}Sea cualquier reordenamiento de la secuencia{a n}. Entonces (8.5.13)∑n=1∞b n=∑n=1∞a n. En el Ejemplo 8.5.3, determinamos que la serie en la parte 2 converge absolutamente. El teorema 72 nos dice que la serie converge (que también podríamos determinar usando la Prueba de Serie Alternante). El teorema afirma que reorganizar los términos de una serie absolutamente convergente no afecta su suma. Esto implica que quizás la suma de una serie condicionalmente convergente pueda cambiar con base en la disposición de los términos. En efecto, puede. El Teorema de Reordenamiento de Riemann (llamado así por Bernhard Riemann) establece que cualquier serie condicionalmente convergente puede tener sus términos reordenados para que la suma sea cualquier valor deseado,∞ ¡incluso! Como ejemplo, considere una vez más la Serie Armónica Alternante. Nosotros hemos afirmado que (8.5.14)∑n=1∞(−1)n+1⁢1 n=1−1 2+1 3−1 4+1 5−1 6+1 7⋯=ln⁡2, (ver Idea Clave 31 o Ejemplo 8.5.1). Considera el reordenamiento donde cada término positivo va seguido de dos términos negativos: (8.5.15)1−1 2−1 4+1 3−1 6−1 8+1 5−1 10−1 12⋯ (Convénzase de que estos son exactamente los mismos números que aparecen en la Serie Armónica Alternante, solo en un orden diferente). Ahora agrupa algunos términos y simplifica: \ [\ comenzar {alinear} \ izquierda (1-\ dfrac12\ derecha) -\ dfrac14+\ izquierda (\ dfrac13-\ dfrac16\ derecha) -\ dfrac18+\ izquierda (\ dfrac15-\ dfrac1 {10}\ derecha) -\ dfrac1 {12} +\ cdots &=\ \ dfrac12-\ dfrac12-\ dfrac14+\ dfrac16-\ dfrac18+\ dfrac1 {10} -\ dfrac {1} {12} +\ cdots &= \ dfrac12\ izquierda (1-\ dfrac12+\ dfrac13-\ dfrac14+\ dfrac15-\ dfrac16+\ cdots\ derecha) & =\ dfrac12\ ln 2. \ end {alinear}] Al reorganizar los términos de la serie, ¡hemos llegado a una suma diferente! (Se podría intentar argumentar que la Serie Armónica Alternante en realidad no converge a ln⁡2, porque reorganizar los términos de la serie no debería cambiar la suma. No obstante, la Prueba de Serie Alternante demuestra que esta serie converge a L L, para algún número, y si el reordenamiento no cambia la suma, entonces L=L/2, implicando L=0. Pero el Teorema de Aproximación de Serie Alternante lo demuestra rápidamente L>0. La única conclusión es que el reordenamiento\ emph {did} cambió la suma.) Este es un resultado increíble. Terminamos aquí nuestro estudio de pruebas para determinar la convergencia. La contraportada de este texto contiene una tabla que resume las pruebas que uno puede encontrar útiles. Si bien las series son dignas de estudio en sí mismas, nuestro objetivo final dentro del cálculo es el estudio de Power Series, que consideraremos en la siguiente sección. Utilizaremos series de potencia para crear funciones donde la salida sea el resultado de una suma infinita. Colaboradores y Atribuciones Template:ContribApexCalc This page titled 8.5: Serie alterna y convergencia absoluta is shared under a CC BY-NC license and was authored, remixed, and/or curated by Gregory Hartman et al.. Volver arriba 8.4: Pruebas de Ratio y Raíz 8.6: Serie Power Artículos recomendados 9.5: Serie alternaEn esta sección presentamos series alternas, aquellas series cuyos términos se alternan en signo. Mostraremos en un capítulo posterior que estas serie... 9.3: Serie alterna 8.4: Pruebas de Ratio y RaízLas pruebas de comparación de la sección anterior determinan la convergencia comparando términos de una serie con términos de otra serie cuya converge... 8.2: Serie InfinitaEsta sección nos introduce a las series y define algunos tipos especiales de series cuyas propiedades de convergencia son bien conocidas: sabemos cuan... Tipo de artículoTemaLicenseCC BY-NCShow TOCno Etiquetas absolute convergence Alternating Harmonic Series alternating series source[translate]-math-4203 © Copyright 2025 LibreTexts Español Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? Contact Support Search the Insight Knowledge Base Check System Status× contents readability resources tools ☰ 8.4: Pruebas de Ratio y Raíz 8.6: Serie Power
15079	https://math.stackexchange.com/questions/162291/modular-arithmetic-for-negative-numbers	Modular arithmetic for negative numbers - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR == Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… 1. 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You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Modular arithmetic for negative numbers Ask Question Asked 13 years, 2 months ago Modified5 years, 10 months ago Viewed 7k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. If I have the congruence m 2≡−1(mod 2 k+1) how do I solve for the solutions to this congruence (given that I know k)? elementary-number-theory modular-arithmetic Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jun 24, 2012 at 7:51 talmid 1,667 9 9 silver badges 11 11 bronze badges asked Jun 24, 2012 at 7:34 MyNameIsKhanMyNameIsKhan 918 4 4 gold badges 12 12 silver badges 24 24 bronze badges 11 You can note that m 2=−1≡(2 k+1)−1=2 k(mod 2 k+1). –talmid Commented Jun 24, 2012 at 7:47 Sorry, I do not understand what that means –MyNameIsKhan Commented Jun 24, 2012 at 7:49 1 With congruences you can always add any multiple of the modulus, since they're all equivalent to 0. In this case, since m 2≡−1(mod 2 k+1), and 2 k+1≡0(mod 2 k+1), we have m 2≡−1=−1+0≡−1+2 k+1=2 k(mod 2 k+1) It's a string of equalities and equivalences, consider each ≡ or =, and try to understand what happens in the passage from the left side to the right side of each ≡ or =. In any case, this is only one thing you could do (to avoid the negative number; although you should be comfortable with having a negative right-hand side ... –talmid Commented Jun 24, 2012 at 7:54 For example let's say k = 6. How do I find solutions to m^2 congruent to -1 mod 13? –MyNameIsKhan Commented Jun 24, 2012 at 7:56 1 @talmid: changing −1 to 2 k seems to me to be moving farther away from a solution rather than closer. The triumvirate of CRT, Hensel's and quadratic reciprocity works on the problem as stated (see the discussion here). –anon Commented Jun 24, 2012 at 12:05 \|Show 6 more comments 5 Answers 5 Sorted by: Reset to default This answer is useful 7 Save this answer. Show activity on this post. As others have pointed out, when dealing with congruences the concept of a negative number is meaningless (as is the concept of a positive number). Judging from the comments you are, in addition to this, asking about the solvability of the congruence m 2≡−1(mod q), where q is some odd integer. Here the theory says that this congruence has solutions, if and only if all the prime divisors of q are congruent to 1(mod 4). For example, when q=7 there are no solutions, because 7≢1(mod 4), but when q=13, there are solutions m≡±5. When q is a prime, m=±((q−1)/2)! are the only non-congruent solutions. The general theory goes via the usual route: solve it for prime number moduli => solve for prime power moduli (by "lifting") => solve for general moduli with the aid of Chinese remainder theorem. The number of non-congruent solutions depends on the number of prime factors of q. [Edit: In response to Thomas' comment.] If m 2≡−1(mod q), then this can be lifted to a solution modulo q k for all positivie integers k. Lifting means that if we have, for some positive integer k, a solution m k such that m k≡m(mod q) and m 2 k≡−1(mod q k), then we can find an integer m k+1 such that m k+1=m k+a q k and m 2 k+1≡−1(mod q k+1). This is not difficult, because we know that m 2 k=−1+b q k for some integer b, and can use this to solve a from the congruence m 2 k+1=m 2 k+2 a m k q k+a 2 q 2 k≡−1+(b+2 a m k)q k(mod q k+1), because this is equivalent to the linear congruence b+2 a m k≡0(mod q). Here gcd(2 m k,q)=1 so the solutions a of this congruence form a unique residue class modulo q. As an example, let us lift the solution m=m 1=5 of the congruence m 2≡−1(mod 13) to a square root of −1 modulo 13 2. Here m 2 1=25=−1+2⋅13, so b=2. We want to solve the linear congruence b+2 a m 1=2+2⋅5 a≡0(mod 13). The usual method for solving a linear congruence yields a≡5(mod 13) as the solution. Therefore m 2=5+13 a=5+5⋅13=70 should be a solution. Indeed, m 2 2=4900=−1+4901=−1+13⋅377=−1+13 2⋅29≡−1(mod 13 2). [/Edit] [Edit^2: An example on using the CRT] Assume that we are given the task to solve the equation m 2≡−1(mod 2873). The process begins by factoring 2873. If you know this an advance, you are lucky, because otherwise this could be tricky. We simply observe that 2873=13 2⋅17. It is our lucky day, because all the prime factors are ≡1(mod 4) (of course I reverse engineered this example a bit). The idea is that we next solve the congruences m 2≡−1(mod 13 2) and m 2≡−1(mod 17) separately. In the previous example I showed how to lift the "guessable" solution 5 2≡1(mod 13) to a solution m=70 of equation (2), so let's reuse that. We observe that −70≡99(mod 13 2) is then the other solution of (2). Finding the solutions of (3) is also easy, because 17 is a relatively small number. There are several methods. Either we simply observe that m=±4 are solutions. Or, as above, we can calculate that (17−1)/2=8 and 8!=40320≡13≡−4(mod 17). We can also use the quicker (for primes a bit larger than 17) but non-deterministic method that for any integer a,0solve for prime power moduli (by "lifting") => solve for general moduli with the aid of Chinese remainder theorem. The number of non-congruent solutions depends on the number of prime factors of q. Lifting: If m^2\equiv-1\pmod q, then this can be lifted to a solution modulo q^k for all positivie integers k. Lifting means that if we have, for some positive integer k, a solution m_k such that m_k^2\equiv -1\pmod{q^k}, then we can find an integer m_{k+1} such that m_{k+1}=m_k+aq^k and m_{k+1}^2\equiv -1\pmod{q^{k+1}}. This is not difficult, because we know that m_k^2=-1+bq^k for some integer b, and can use this to solve a from the congruence m_{k+1}^2=m_k^2+2am_kq^k+a^2q^{2k}\equiv-1+(b+2am_k)q^k\pmod{q^{k+1}}, because this is equivalent to the linear congruence b+2am_k\equiv0\pmod q. Here \gcd(2m_k,q)=1 so the solutions a of this congruence form a unique residue class modulo q. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jun 24, 2012 at 18:37 community wiki 2 revsHenning Makholm \endgroup 2 \begingroup Thanks, Henning. I need to work on my clarity and brevity :-)\endgroup –Jyrki Lahtonen Commented Jun 24, 2012 at 18:42 \begingroup@JyrkiLahtonen: Not sure you're at fault. See recent chat transcript.\endgroup –hmakholm left over Monica Commented Jun 24, 2012 at 18:45 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. \begingroup I discuss the solution method here, involving the Chinese Remainder Theorem, Wilson's Theorem, Hensel's Lemma, and Quadratic Reciprocity, and provide some explicit formulas for calculation: To compute n_p (for each prime p\|m), we use this fact: \begin{array}{c l} (p-1)! & \equiv 1\cdot 2\cdots\frac{p-1}{2}\frac{p+1}{2}\cdots (p-1) \ & \equiv 1\cdot 2\cdots\frac{p-1}{2}\left(p-\frac{p-1}{2}\right)\cdots\big(p-(1)\big) \ & \equiv \left(\frac{p-1}{2}\;!\right)^2(-1)^{(p-1)/2} \mod p. \end{array} \tag{$\bigcirc$} When p\equiv1\mod 4, (-1)^{(p-1)/2}=-1, thus (applying Wilson's theorem to the LHS): n_p\equiv \left(\frac{p-1}{2}\right)!\mod p \tag{$\times$} is a square root of -1 modulo p. Note that computing factorials can be expensive, so you want to do it via repeated _modular_ multiplication, which is more efficient. To compute \hat{n}_p, the solution to f(u):= u^2+1\equiv0\mod p^r, we apply HL. First we evaluate f\,' (the derivative) at the argument n_p, getting 2n_p, and compute the multiplicative inverse of 2n_p modulo p^{r-1} and multiply by the integer -f(n_p)/p; in other words \hat{n}_p\equiv-\left(\frac{n_p^2+1}{p}\right)\left[\frac{1}{2n_p}\bmod p^{r-1}\right] \mod p^r \tag{$\square$} Note that since p\|(n_p^2+1), the division in parentheses is integer division, and the reciprocal in brackets is computed via modular arithmetic \bmod p^{r-1}, but then viewed \bmod p^r afterwards. Furthermore, we may multiply \hat{n}_p by -1\bmod p^r to get a _second_ square root of -1\bmod p^r; call these two roots n_p^+ and n_p^- respectively. An arbitrary choice of \pm is available for each p. Using the general case formula for CRT, we glue all of the "localized" solutions together: n\equiv \sum_{p}n_p^{\pm}\left(\frac{m}{p^r}\right)\left[\left(\frac{m}{p^r}\right)^{-1}\bmod p^r \right] \mod m \tag{$\triangle$} Again, a choice of + or - is made for each p. Counterintuitively, these do not designate the notions of "positive" or "negative"; they designate an initially computed root versus an optional, auxiliary root. Above, division in parentheses is done _in the integers_ (so not in modular arithmetic), whereas the reciprocals in brackets are computed \bmod p^r (for various values of p), and then the results are reinterpreted as integers \bmod m. (None of this was present in the original version of my answer, when I CW'd it for virtually duplicating the discussion in Arturo's answer that preceded mine.) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Apr 13, 2017 at 12:21 CommunityBot 1 answered Jun 24, 2012 at 20:07 anonanon 156k 14 14 gold badges 249 249 silver badges 421 421 bronze badges \endgroup 1 \begingroup The OP's letter 'm' is my letter 'n', and I use m to designate 2k+1. All prime divisors of m are necessarily congruent to 1\bmod 4.\endgroup –anon Commented Jun 24, 2012 at 20:12 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. \begingroup Using a generalization of Wilson's theorem \prod_{\gcd(i,n)=1 \leq i\leq n}i\equiv -1 (\text{mod }n) each i will have a pair in the product n-i and if there are an even number of those pairs (since each pair is congruent to the negative of the other one) each -1 in the product will cancel out with the other one and be a square so there is a solution if \phi(n) is a multiple of 4. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Oct 26, 2019 at 23:22 Varun RajkumarVarun Rajkumar 197 1 1 silver badge 9 9 bronze badges \endgroup Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. 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15080	https://learnmathnow.fandom.com/wiki/Completing_the_Square	Completing the Square \| Learn Math Now! Wiki \| Fandom Join the quest! Share your opinion on the upcoming games! We want to hear from you! Sign In Register Learn Math Now! Wiki Explore Main Page All Pages Community Interactive Maps Recent Blog Posts Wiki Content Recently Changed Pages Daleth function (Martinez) The Four Survivalists Gil's Theorems N-bonacci Numbers Polynomials Sums and differences of two squares Pythagorean Theorem Organization Help Forums Policy Copyright Community Site administration Site maintenance Videos The Sixth Spatial Dimension Community Help Sign In Don't have an account? Register Sign In Menu Explore More History Advertisement Skip to content Learn Math Now! Wiki 40 pages Explore Main Page All Pages Community Interactive Maps Recent Blog Posts Wiki Content Recently Changed Pages Daleth function (Martinez) The Four Survivalists Gil's Theorems N-bonacci Numbers Polynomials Sums and differences of two squares Pythagorean Theorem Organization Help Forums Policy Copyright Community Site administration Site maintenance Videos The Sixth Spatial Dimension Community Help Contents 1 When There's a Coefficient 2 When the Linear Term Is An Odd Number 3 Completing the Square to Solve Quadratic Equations 4 Now, Try This! Completing the Square Sign in to edit History Purge Talk (0) This is one way of looking at completing the square, as represented by algebra tiles. In algebra, completing the square is a mathematical method demonstrated by a man named al-Khwarzimi (780–850), who published a book that gave algebra its very name. Suppose you were given the following mathematical expression: x 2 + 16 x. You have been asked to complete the square of this expression. How do you do that? Here's how it's done: Take the original expression... x 2+16 x{\displaystyle x^2 + 16x} Factor out the x... (x+8)2{\displaystyle (x + 8)^2} Square the constant term (the term with no variable)... 8 2=64{\displaystyle 8^2 = 64} (or 8∗8=64{\displaystyle 8 8 = 64}) And apply it to the original expression. All squared terms remain positive when completing the square of an expression. x 2+16 x+64{\displaystyle x^2 + 16x + 64} Congratulations! You've completed the square! [x] Contents 1 When There's a Coefficient 2 When the Linear Term Is An Odd Number 3 Completing the Square to Solve Quadratic Equations 4 Now, Try This! When There's a Coefficient[] When coefficients are involved, the only major difference is the inclusion of one step: dividing the entire expression by that coefficient. For instance, let's say you were given the following expression: 4 x 2 - 16 x. It's very easy, you know. All you have to do is divide the entire expression by 4. 4 x 2−16 x{\displaystyle 4x^2 - 16x} becomes x 2−4 x{\displaystyle x^2 - 4x}. Then, it's all downhill from here. Complete the square by following the steps above, and your answer should come up as this: x 2−4 x+4{\displaystyle x^2 - 4x + 4} When the Linear Term Is An Odd Number[] Suppose you were given the expression x 2 + 7 x. Here's what you do when the linear term is an odd number: Factor out the x and put 7 over 2... From x 2+7 x{\displaystyle x^2 + 7x} to (x+7 2)2{\displaystyle (x + \frac{7}{2}\,)^2}... Square the fraction (there's an easy way)... (7 2)2=7∗7 2∗2=49 4{\displaystyle (\frac{7}{2}\,)^2 = \frac{7 7}{2 2}\, = \frac{49}{4}\,} And apply it to the expression. If you want to, you can simplify the improper fraction into a mixed number or decimal. x 2+7 x+49 4{\displaystyle x^2 + 7x + \frac{49}{4}\,}, which is also x 2+7 x+12 1 4{\displaystyle x^2 + 7x + 12\frac{1}{4}\,} or x 2+7 x+12.25{\displaystyle x^2 + 7x + 12.25} There you have it! Completing the Square to Solve Quadratic Equations[] Al-Khwarzimi showed how to solve quadratic equations by finding the square by: Rewriting the original problem... x 2−6 x−5=0{\displaystyle x^2 - 6x - 5 = 0} Moving the variables to different sides... −5=−x 2+6 x{\displaystyle -5 = -x^2 + 6x} Factoring the quadratic and linear terms to complete the square... 5+9=x 2−6 x+9{\displaystyle 5 + 9 = x^2 - 6x + 9} 14=(x−3)2{\displaystyle 14 = (x - 3)^2} Taking the square roots of both sides... ±14=x−3{\displaystyle \pm \sqrt{14}\, = x - 3} And, finally, simplifying the whole thing. 3±14=x{\displaystyle 3 \pm \sqrt{14}\, = x} This is a little trickier, though it's easy once you get used to it. Now, Try This![] Complete the square to solve for x. Leave the answer in simplest radical form. 4 x 2+4=32 x+8{\displaystyle 4x^2 + 4 = 32x + 8 } Community content is available under CC-BY-SA unless otherwise noted. Comments Start a conversation Sign in to share your thoughts and get the conversation going. SIGN IN Don't have account? Register now Advertisement Explore properties Fandom Fanatical GameSpot Metacritic TV Guide Honest Entertainment Follow Us Overview What is Fandom? About Careers Press Contact Terms of Use Privacy Policy Digital Services Act Global Sitemap Local Sitemap Do Not Sell My Personal Information Community Community Central Support Help Advertise Media Kit Contact Fandom Apps Take your favorite fandoms with you and never miss a beat. Learn Math Now! Wiki is a Fandom Lifestyle Community. View Mobile Site Do Not Sell My Personal Information When you visit our website, we store cookies on your browser to collect information. 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15081	https://www.sciencedirect.com/science/article/abs/pii/S0960077921009358	Synchronization of a nonlinear oscillator with a sum signal from equivalent oscillators - ScienceDirect Typesetting math: 100% Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Search ScienceDirect Article preview Abstract Introduction Section snippets References (36) Cited by (2) Chaos, Solitons & Fractals Volume 153, Part 1, December 2021, 111581 Frontiers Synchronization of a nonlinear oscillator with a sum signal from equivalent oscillators Author links open overlay panel Robson Vieira, Weliton S.Martins, Sergio Barreiro, Rafael A. de Oliveira, Martine Chevrollier, Marcos Oriá Show more Add to Mendeley Share Cite rights and content Highlights •synchronization is studied in a neuron-like configuration with multiple inputs driving a single nonlinear oscillator output. •partial information from multiple drivers leads a response system to follow the trajectory determined by the sum of all drivers. •the quality of synchronization is stable with the number of drivers in the low coupling regime, but improves and follows a power law with the number of drivers in the forcing regime. Abstract Coupling of chaotic oscillators has evidenced conditions where synchronization is possible, therefore a nonlinear system can be driven to a particular state through input from a similar oscillator. Here we expand this concept of control of the operational state of a nonlinear oscillator by showing that it is possible to induce nonlinear systems to follow a linear superposition of trajectories of two (or more) equivalent systems, using only partial information from them. We show that the trajectory of the receiving oscillator converges to that of the sum of the two (or more) drivers. These results are demonstrated through experimental realization in electronic circuits and numerical simulations, and analysed through calculations of transverse Lyapunov exponents. These behaviors are observed in known nonlinear systems, opening the way for improving description and control of complex systems. Introduction The growing interest in complex interacting systems has resulted in investigations of many aspects of their statistical properties , , , , connectivity , , , or synchronization , , . Ubiquitous behaviors were evidenced in complex systems belonging to fields as diverse as the physical, biological, social sciences and their technological applications. Synchronization, particularly, is a phenomenon long-known to occur in linear coupled systems. Nonlinear chaotic oscillators, on the other hand, were at first not expected to synchronize with one another, because of their high sensitivity to initial conditions. However, the demonstration decades ago that two chaotic systems may follow the same trajectory in phase-space opened the way to a huge number of investigations and applications of synchronization involving two or more chaotic oscillators in increasingly complex configurations. Synchronization is crucial in complex systems such as neural and cardiac , systems, as well as in chaos-based communications , , or extreme events studies , . In neural networks, for instance, synchronization , , has been investigated considering that neural systems are, globally, highly connected, with each cluster of the network receiving and sending signaling , . However, individually each biological neuron has a very asymmetric connection configuration, with a very large number of inputs and a single output, the axon , , making it important to investigate single output configurations in network synchronization studies. Early works on coupled nonlinear systems have shown that a chaotic oscillator can converge to a particular solution , when information about such trajectory is supplied to the system, either by substituting variables , or by adding a linear term to the original system. Here we investigate the behavior of a single nonlinear system that is driven by a combination of signals from multiple equivalent oscillators. In a neuron-like configuration where multiple inputs drive a single oscillator output we observe that even partial information, i.e. only one or two variables, from the drivers leads the response system to follow the trajectory determined by the sum of all drivers. The quality of the synchronization of the oscillator with the drivers sum-solution is shown to improve with increasing number of drivers, as a result of the forcing of the system by the increasing of drive inputs. This paper is organized as follows: we first present results of synchronization of one receiving circuit with two drivers for a Gauthier-Bienfang system (G-B, a modified Chua electronic circuit), showing both experimental and numerical results. Next, we carry out a numerical and analytical study of the quality of synchronization in a regime of forcing, when the sum of driving signals becomes larger than the system flux. We explore this sum synchronization as a function of both the number of drivers and the coupling strength between drivers and receiver, in the G-B as well as in the Lorenz system. Then we conclude. Access through your organization Check access to the full text by signing in through your organization. Access through your organization Section snippets Synchronization with two drivers We prepare three equivalent circuits that can independently oscillate in chaotic regime. Two are drivers, described by X˙1,2=F(X 1,2),where F is the vector field describing the flux of the system. A sum of variables of these two oscillators is then added to the flux of the third one, the receiver X T=(x(1),x(2),x(3),…,x(m)), whose dynamics is described by X˙=F(X)+K(X s−X), with X s=c 1 X 1+c 2 X 2 and K is a m×m matrix allowing for the coupling of the X components with the corresponding sum components X s T=(x Multi-drive synchronization To evidence the behavior of synchronization of a nonlinear system with a sum signal from more than two drivers, we prepare N similar systems, oscillating independently from one another. Each trajectory X n(n=1,2,3,…N) is a particular solution of the free-running system equation X˙n=F(X n),with X n T=(x n,y n,z n) and F the vector field describing the flux of the system. Information from the driver trajectories is sent to the nonlinear receiving oscillator by way of a sum signal (see Fig.5), as Synchronization analysis In order to quanepsy the evolution of the synchronization of the system with the sum signal, we calculated the distance vector, defined as X⊥=(X s−X) , that measures the distance transverse to the synchronization manifold. We show in Fig.7 the behavior of \|X⊥\| as a function of the coupling coefficient in the particular case where all the coupling coefficients c n take the same value c. Fig.7 evidences the improvement of synchronization as the coupling strength increases. Results are Conclusion We showed that it is possible to create a new state for a chaotic system, i.e. induce a non-linear system to follow trajectories that are not direct solutions of the system flux equations but a linear combination of such solutions, the master-slave configuration being a particular case (N=1). Such new synchronization behaviour may have applications in communication using chaos , , , for example, through chaos synchronization and by taking advantage of a multiple input in the CRediT authorship contribution statement Robson Vieira: Data curation, Formal analysis, Writing – review & editing. Weliton S. Martins: Data curation, Formal analysis, Writing – review & editing. Sergio Barreiro: Data curation, Funding acquisition, Formal analysis, Writing – review & editing. Rafael A. de Oliveira: Formal analysis, Writing – review & editing. Martine Chevrollier: Formal analysis, Writing – review & editing. Marcos Oriá: Formal analysis, Writing – original draft, Writing – review & editing. Declaration of Competing Interest The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper. Acknowledgement This work was supported by Universidade Federal Rural de Pernambuco (UFRPE) and the Brazilian agencies CNPq, CAPES, and FACEPE. Recommended articles References (36) S. Boccaletti et al. Complex networks: structure and dynamics Phys Rep (2006) A. Arenas et al. Synchronization in complex networks Phys Rep (2008) H. Yu et al. Chaotic synchronization and control in nonlinear-coupled hindmarsh–rose neural systems Chaos, Solitons & Fractals (2006) Y. Shiogai et al. Nonlinear dynamics of cardiovascular ageing Phys Rep (2010) K.J. Friston et al. Dynamic causal modelling Neuroimage (2003) R. Albert et al. Statistical mechanics of complex networks Rev Mod Phys (2002) Sornette D. Probability distributions in complex systems. arXiv preprint... G. Nicolis et al. Foundations of complex systems: emergence, information and prediction (2012) H.L.d.S. Cavalcante et al. Predictability and suppression of extreme events in a chaotic system Phys Rev Lett (2013) S.H. Strogatz Exploring complex networks Nature (2001) M.E. Newman The structure and function of complex networks SIAM Rev (2003) H. Choi et al. Synchronization dependent on spatial structures of a mesoscopic whole-brain network PLoS Comput Biol (2019) L.M. Pecora et al. Cluster synchronization and isolated desynchronization in complex networks with symmetries Nat Commun (2014) C. Schäfer et al. Heartbeat synchronized with ventilation Nature (1998) K.M. Cuomo et al. Circuit implementation of synchronized chaos with applications to communications Phys Rev Lett (1993) A. Argyris et al. Chaos-based communications at high bit rates using commercial fibre-optic links Nature (2005) L. Keuninckx et al. Encryption key distribution via chaos synchronization Sci Rep (2017) A. Mishra et al. 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15082	https://www.physics.harvard.edu/resource/sol46pdf	Solution Week 46 (7/28/03) The birthday problem (a) Given n people, the probability, Pn, that there is not a common birthday among them is Pn = ( 1 − 1365 ) ( 1 − 2365 ) · · · ( 1 − n − 1365 ) . (1) The first factor is the probability that two given people do not have the same birthday. The second factor is the probability that a third person does not have a birthday in common with either of the first two. This continues until the last factor is the probability that the nth person does not have a birthday in common with any of the other n − 1 people. We want Pn < 1/2. If we simply multiply out the above product with suc-cessive values of n, we find that P22 = .524, and P23 = .493. Therefore, there must be at least 23 people in a room in order for the odds to favor at least two of them having the same birthday. Remark: This answer of n = 23 is much smaller than most people expect, so it provides a nice betting opportunity. For n = 30, the odds of a common birthday increase to 70.6%, and most people still find it hard to believe that among 30 people there are probably two who have the same birthday. The table below lists various values of n and the probabilities, 1 − Pn, that at least two people have a common birthday. n 10 20 23 30 50 60 70 100 1 − Pn 11.7% 41.1% 50.7% 70.6% 97.0% 99.4% 99.92% 99.9994% Even for n = 50, most people would be happy to bet, at even odds, that no two people have the same birthday. If they seem a bit hesitant, however, you can simply offer them the irrefusable odds of 10 to 1. (That is, you put down, for example, $10 on the table, and they put down $1. If there is a common birthday, you take all the money. If there is no common birthday, they take all the money.) Since there is a 97 .0% chance of a common birthday among 50 people, a quick calculation shows that you will gain, on average, 67 cents for every $10 you put down. One reason why many people do not believe the n = 23 answer is correct is that they are asking themselves a different question, namely, “How many people need to be present for there to be a 1/2 chance that someone else has my birthday?” The answer to this question is indeed much larger than 23. The probability that no one out of n people has a birthday on a given day is (1 − 1/365) n. For n = 252, this is just over 1/2. And for n = 253, it is just under 1/2. Therefore, you need to come across 253 other people in order to expect that at least one of them has your birthday. (b) First Solution: Given n people, and given N days in a year, the reasoning in part (a) shows that the probability that no two people have the same birthday is Pn = ( 1 − 1 N ) ( 1 − 2 N ) · · · ( 1 − n − 1 N ) . (2) 1If we take the natural log of this equation and use the expansion, ln(1 − x) = −(x + x2/2 + · · · ), (3) then the requirement Pn ≤ 1/2 becomes − ( 1 N + 2 N + · · · n − 1 N ) − 12 ( 1 N 2 + 4 N 2 + · · · (n − 1) 2 N 2 ) − · · · ≤ − ln 2 . (4) Using the sums, n ∑ 1 k = n(n + 1) 2 , and n ∑ 1 k2 = n(n + 1)(2 n + 1) 6 , (5) we can rewrite eq. (4) as n(n − 1) 2N + n(n − 1)(2 n − 1) 12 N 2 + · · · ≥ ln 2 . (6) For large N , the first term will be of order 1 when n ≈ √N , in which case the second- and higher-order terms are negligible. Therefore, keeping only the first term (which behaves like n2/2N ), we find that Pn is equal to 1/2 when n ≈ √2 ln 2 √N . (7) Let’s look at a few cases: • For N = 365, eq. (7) gives n = 22 .5. Since we must have an integral number of people, this agrees with the exact result, n = 23. • For N = 24 · 365 = 8760 (that is, for births in the same hour), we find n = 110 .2. This agrees with the exact result, n = 111, obtained by multiplying out eq. (2). • For N = 60 · 24 · 365 = 525 , 600 (that is, for births in the same minute), we find n = 853 .6. This agrees with the exact result, n = 854, obtained by multiplying out eq. (2) (not by hand!). This is a very small number compared to the more than half a million minutes in a year. Remarks: (1) If we wish to ask how many people need to be in a room in order for the probability to be at least p that some two have the same birthday, then the above derivation is easily modified to yield n ≈ √ 2 ln( 11−p )√N . (8) (2) The alternative question introduced in part (a), “How many people need to be present for there to be a 1/2 chance that someone else has my birthday?”, may also be answered in the large-N limit. The probability that no one out of n people has a birthday on a given day is ( 1 − 1 N )n = ( 1 − 1 N )N (n/N ) ≈ e−n/N . (9) 2Therefore, if n > N ln 2, you can expect that at least one of the n people has your birthday. For N = 365, we find that N ln 2 is slightly less than 253, so this agrees with the result obtained in part (a). Note that this result is linear in N , whereas the result of the original problem in eq. (7) behaves like √N . The reason for this square-root behavior can be seen in the next solution. Second Solution: Given a large number, n, of people, there are (n 2 ) = n(n − 1) /2 ≈ n2/2 pairs of people. The probability that a given pair has the same birthday is 1 /N , so the probability that they do not have the same birthday is 1 − 1/N .1 Therefore, the probability that no pair has a common birthday is Pn = ( 1 − 1 N )n2/2 ≈ ( 1 − 1 N )N (n2/2N ) ≈ e−n2/(2 N ). (10) This equals 1/2 when n ≈ √2 ln 2 √N . (11) in agreement with eq. (7). Remark: We assumed above that all the pairs are independent, as far as writing down the 1 − 1/N probability goes. Let us now show that this approximately true. We will show that for large N and n, the assumptions on the coincidence of birthdays in some pairs do not significantly affect the probability of coincidence in other pairs. More precisely, the relation Pn ≈ e−n2/(2 N ) is true if n ø N 2/3. The reasoning is as follows. Consider n people and N days in a year. Assumptions on the coincidence of birthdays in some pairs will slightly affect the probability of coincidence in other pairs, because the given assumptions restrict the possible birthdays of these other pairs. For example, if it is given that A and B do not have the same birthday, and also that B and C do not, then the probability that A and C do not have the same birthday is 1 − 1/364 (instead of 1 − 1/365), because they are both restricted from having a birthday on B’s birthday, whatever it may be. The maximal restriction that can be placed on the possible birthdays of a given pair occurs when it is known that neither of them has a birthday in common with any of the other n − 2 people. In this case, the possible birthdays for the last two people are restricted to N − (n − 2) days of the year. Therefore, the probability that the last pair does not have a common birthday is 1 − 1/(N − n + 2) > 1 − 1/(N − n). This is the most that any such probability can differ from the naive 1 − 1/N . Therefore, we may say that ( 1 − 1 N − n )n2/2 ≤ Pn ≤ ( 1 − 1 N )n2/2 =⇒ e−n2/2( N −n) ≤ Pn ≤ e−n2/(2 N ). (12) The ratio of these upper and lower bounds on Pn is exp ( − n2 2N + n2 2( N − n) ) = exp ( n3 2N (N − n) ) . (13) 1This is not quite correct for all the pairs, because two pairs are not independent if, for example, they share a common person. But it is accurate enough for our purposes in the large-Nlimit. See the remark below. 3For large N , this ratio is essentially equal to 1, provided that n ø N 2/3. Therefore, Pn ≈ e−n2/(2 N ) if n ø N 2/3. And since the result in eq. (11) is in this realm, it is therefore valid. Extension : We can also ask the following question: How many people must be in a room in order for the probability to be greater than 1/2 that at least b of them have the same birthday? (Assume that there is a very large number, N , of days in a year, and ignore effects that are of subleading order in N .) We can solve this problem in the manner of the second solution above. Given a large number, n, of people, there are (nb ) groups of b people. This is approx-imately equal to nb/b ! (assuming that b ø n). The probability that a given group of b people all have the same birthday is 1 /N b−1, so the probability that they do not all have the same birthday is 1 − (1 /N b−1). 2 Therefore, the probability, P (b) n , that no group of b people all have the same birthday is P (b) n ≈ ( 1 − 1 N b−1 )nb/b ! ≈ e−nb/b !N b−1 . (14) This equals 1/2 when n ≈ (b! ln 2) 1/b N 1−1/b . (15) Remarks: (1) Eq. (15) holds in the large-N limit. If we wish to make another approximation, that of large b, we see that the quantity ( b! ln 2) 1/b goes like b/e , for large b. (This follows from Sterling’s formula, m! ≈ mme−m√2πm .) Therefore, for large N , n, and b (with b ø n ø N ), we have P (b) n = 1 /2 when n ≈ (b/e )N 1−1/b . (16) (2) The right-hand side of equation eq. (15) scales with N according to N 1−1/b .This means that if we look at the numbers of people needed to have a greater than 1/2 chance that pairs, triplets, etc., have common birthdays, we see that these numbers scale like N 1/2, N 2/3, N 3/4, · · · . (17) For large N , these numbers are multiplicatively far apart. Therefore, there are values of n for which we can say, for example, that we are virtually certain that there are pairs and triplets with common birthdays, but also that we are virtually certain that there are no quadruplets with a common birthday. For example, if n = N 17 /24 (which satisfies N 2/3 < n < N 3/4), then eq. (14) shows that the probability that there is a common birthday triplet is 1 − e− 16 N 1/8 ≈ 1, whereas the probability that there is a common birthday quadruplet is 1 − e− 124 N −1/6 ≈ 124 N −1/6 ≈ 0. 2Again, this is not quite correct, because the groups are not all independent. But I believe it is accurate enough for our purposes in the large-Nlimit. I have a proof which I think is correct, but which is very ugly. If anyone has a nice clean proof, let me know. 4
15083	https://www.basic-mathematics.com/interval-notation.html	Home The Basic math blog Member Login Algebra Pre-Algebra Lessons Pre-Algebra Word Problems Pre-Algebra Calculators Algebra Lessons Algebra Word Problems Algebra Proofs Advanced Algebra Algebra Calculators Geometry Geometry Lessons Geometry Word Problems Geometry Proofs Geometry Calculators Trigonometry Lessons Special Math Topics Math Anxiety Rescue Numeration System Basic Concepts of Set Theory Applied math Consumer Math Baseball Math Math for Nurses Statistics Made Easy High School Physics Test Prep Basic Mathematics Store K-12 Tests Math Vocabulary Quizzes SAT Math Prep Math Skills by Grade Level Glossary Ask an Expert Other Websites Worksheets K-12 Worksheets Algebra Worksheets Geometry Worksheets Fun Online Math Games Pre-Algebra Games Math Puzzles Math Tricks Interval notation We use interval notation to represent subsets of real numbers. Suppose that a and b are real numbers such that a < b. Then, the open interval (a,b) represents the set of all real numbers between a and b, except a and b. { x / a < x < b} is the set-builder notation. a < x < b is the inequality description. (a, b) is the interval notation. The closed interval [a,b] represents the set of all real numbers between a and b, including a and b. { x / a ≤ x ≤ b} is the set-builder notation. a ≤ x ≤ b is the inequality description. [a, b] is the interval notation. The table below lists nine types of intervals used to describe subsets of real numbers. Notice the number line representation of each interval. 114Save Basically, an interval is a set containing all numbers between two given numbers or endpoints. The set may have one, both, or neither of the two given numbers. An interval is open if the interval does not contain its endpoints. The symbols ( or ) called parentheses or round brackets are used to indicate that an endpoint is not included in the interval. An interval is closed if the interval contains its endpoints. The symbols [ or ] called brackets are used to indicate that an endpoint is included in the interval. Sometimes, the interval may contain only one of its endpoints. In this case, the interval is half-open. The interval [2, 8) is half-open. It contains 2 and all numbers between 2 and 8. Notice that 8 is not included since the interval is open at 8. Notice the symbol ∞ which mean infinity. -∞ means minus infinity and +∞ means positive infinity. Both -∞ and +∞ are used to show that an interval is unbounded or extends indefinitely to the left or to the right respectively. Why do we need intervals? They can be used to describe the domains of functions, bounds for estimates, and the solution sets of equations and inequalities. The length of an interval with endpoints a and b with a < b is b - a. Converting an inequality to interval notation Express each inequality in interval notation. Examples Inequality Interval notation -6 ≤ x < 1 [-6, 1) x > 20 (20, ∞) x ≥ -1 or x < -4 (-∞, -4) U [-1,∞) Check this site if you want to learn more about inequalities and interval notation. Compound inequality FacebookPinterestWhatsApp Home The Basic math blog Member Login Algebra Pre-Algebra Lessons Pre-Algebra Word Problems Pre-Algebra Calculators Algebra Lessons Algebra Word Problems Algebra Proofs Advanced Algebra Algebra Calculators Geometry Geometry Lessons Geometry Word Problems Geometry Proofs Geometry Calculators Trigonometry Lessons Special Math Topics Math Anxiety Rescue Numeration System Basic Concepts of Set Theory Applied math Consumer Math Baseball Math Math for Nurses Statistics Made Easy High School Physics Test Prep Basic Mathematics Store K-12 Tests Math Vocabulary Quizzes SAT Math Prep Math Skills by Grade Level Glossary Ask an Expert Other Websites Worksheets K-12 Worksheets Algebra Worksheets Geometry Worksheets Fun Online Math Games Pre-Algebra Games Math Puzzles Math Tricks instagramfacebookpinterest About me :: Privacy policy :: Disclaimer :: Donate Careers in mathematics Copyright © 2008-2021. Basic-mathematics.com. All right reserved This site uses cookies, some of which are required for its operation. Privacy policy.
15084	https://brainly.com/question/38761477	[FREE] Find the coordinates of the point M on the line joining B(-2,5) and C(3,2) such that BM = 2CM . - brainly.com 5 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +81,2k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +23,6k Ace exams faster, with practice that adapts to you Practice Worksheets +7,8k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified Find the coordinates of the point M on the line joining B(−2,5) and C(3,2) such that BM=2 CM. 2 See answers Explain with Learning Companion NEW Asked by mbjj3126 • 09/28/2023 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 13136474 people 13M 4.7 0 Upload your school material for a more relevant answer The coordinates of point M on the line segment BC which divides it in a ratio of 2:1 are (-1,4). The position of point M was calculated using the formula for dividing a line segment in a given ratio. Explanation In this mathematical problem, we need to find the coordinates of the point M on the line joining points B(-2,5) and C(3,2) such that the distance BM is twice the distance of CM. Given that the ratio is 2:1, we can use the formula for dividing a line segment in a given ratio. The formula is ((x1 a+x2 b)/(a+b), (y1 a+y2 b)/(a+b)). Here points B and C are (x1, y1) and (x2, y2), respectively, and a:b is 2:1. Hence, by substituting these values into the formula, we obtain M as ((-2 2+3 1)/(2+1), (5 2+2 1)/(2+1)) which simplifies to (-1, 4). Learn more about Coordinates here: brainly.com/question/32034725 SPJ11 Answered by ishuraj00008 •54.9K answers•13.1M people helped Thanks 0 4.7 (6 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 13136474 people 13M 4.8 0 Non-Euclidean Geometry - Henry Manning Radically Modern Introductory Physics Text I - David J. Raymond Nature of Geographic Information - David DiBiase Upload your school material for a more relevant answer The coordinates of point M, which divides the line segment BC in the ratio 2:1, are M(3 4,3). This was calculated using the section formula for dividing a line segment based on a specific ratio. By substituting the appropriate values into the formula, we were able to find the desired coordinates. Explanation To find the coordinates of point M on the line segment joining points B(-2, 5) and C(3, 2) in such a way that the distance from B to M (BM) is twice the distance from C to M (CM), we first express the ratio of these lengths. Since BM = 2CM, we can write the ratio as BM:CM = 2:1. Now, we need to divide the line segment BC in the ratio 2:1. We can use the section formula to find the coordinates of point M. The formula for finding the coordinates of a point that divides a line segment joining coordinates (x1, y1) and (x2, y2) in the ratio a:b is given as: M(x,y)=(a+b x 1⋅b+x 2⋅a,a+b y 1⋅b+y 2⋅a) where (x 1,y 1) corresponds to point B and (x 2,y 2) corresponds to point C. Here, (x 1,y 1)=(−2,5), (x 2,y 2)=(3,2), a=2, and b=1. Substituting the values into the formula, we get: M(x,y)=(2+1(−2)⋅1+3⋅2,2+1 5⋅1+2⋅2) M(x,y)=(3−2+6,3 5+4) M(x,y)=(3 4,3 9) M(x,y)=(3 4,3) Thus, the coordinates of point M that divides the line segment BC in the ratio 2:1 are M(3 4,3). Examples & Evidence Imagine two points, B and C, on a map. If you need to find a point M along the path connecting B to C that is twice as close to B as it is to C, you would follow a similar method to determine M's coordinates based on the given ratio. The section formula used here is a standard mathematical method taught in coordinate geometry, validated by numerous mathematical textbooks and educational resources. Thanks 0 4.8 (4 votes) Advertisement Community Answer This answer helped 8873253 people 8M 5.0 0 Final answer: The coordinates of point M on the line segment joining B(-2,5) and C(3,2) where BM = 2CM are (3, 3). Explanation: Finding Coordinates of Point M on Line Segment BC To find the coordinates of point M located on the line segment joining B(-2, 5) and C(3, 2) where BM = 2CM, we can use the section formula for divided ratios which states that if M divides BC in the ratio m:n, then the coordinates of M (xm, ym) are given by: xm = (nx1 + mx2)/(m + n) ym = (ny1 + my2)/(m + n) Here, the ratio of BM:CM is 2:1, meaning B is twice as close to M as C is. Applying this to the formula gives: xm = (1(-2) + 2(3))/(2 + 1) ym = (1(5) + 2(2))/(2 + 1) Calculating these we get: xm = (3 + 6)/3 = 3 ym = (5 + 4)/3 = 3 Therefore, the coordinates of point M are (3, 3). Answered by BritMarling •16.3K answers•8.9M people helped Thanks 0 5.0 (3 votes) Advertisement ### Free Mathematics solutions and answers Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. Community Answer 4 Click an Item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Express In simplified exponential notation. 18a^3b^2/ 2ab New questions in Mathematics Given that 2 x+1=4(mod 7), where x is an integer, find the least value of x. Consider the equation, where m is a real number coefficient. 4 3(2−m x)=12 What is the value of x in the equation? Solve for s and simplify your answer. 22=2 3s Curtis wants to build a rectangular enclosure for his animals. One side of the pen will be against the barn, so he needs no fence on that side. The other three sides will be enclosed with wire fencing. If Curtis has 750 feet of fencing, you can find the dimensions that maximize the area of the enclosure. a) Let w be the width of the enclosure (perpendicular to the barn) and let l be the length of the enclosure (parallel to the barn). Write a function for the area A of the enclosure in terms of w. b) What width w would maximize the area? c) What is the maximum area? The monthly cost for renting an apartment at a new apartment complex increases annually. If the monthly cost for rent is 675 w h e n t h eco m pl e x o p e n s,t h e f u n c t i o n g(x) = 675 + 45x c anb e u se d t o d e t er min e t h e m o n t h l ycos t f orre n t x$ years after the apartment complex opens. Which best represents the domain of the function in this context? 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15085	https://web.mae.ufl.edu/nkim/eas6939/constrainedopt.pdf	Constrained Optimization 5 Most problems in structural optimization must be formulated as constrained min-imization problems. In a typical structural design problem the objective function is a fairly simple function of the design variables (e.g., weight), but the design has to satisfy a host of stress, displacement, buckling, and frequency constraints. These constraints are usually complex functions of the design variables available only from an analysis of a ﬁnite element model of the structure. This chapter oﬀers a review of methods that are commonly used to solve such constrained problems. The methods described in this chapter are for use when the computational cost of evaluating the objective function and constraints is small or moderate. In these meth-ods the objective function or constraints these are calculated exactly (e.g., by a ﬁnite element program) whenever they are required by the optimization algorithm. This approach can require hundreds of evaluations of objective function and constraints, and is not practical for problems where a single evaluation is computationally ex-pensive. For these more expensive problems we go through an intermediate stage of constructing approximations for the objective function and constraints, or at least for the more expensive functions. The optimization is then performed on the approx-imate problem. This approximation process is described in the next chapter. The basic problem that we consider in this chapter is the minimization of a function subject to equality and inequality constraints minimize f(x) such that hi(x) = 0, i = 1, . . . , ne , gj(x) ≥0, j = 1, . . . , ng . (5.1) The constraints divide the design space into two domains, the feasible domain where the constraints are satisﬁed, and the infeasible domain where at least one of the constraints is violated. In most practical problems the minimum is found on the boundary between the feasible and infeasible domains, that is at a point where gj(x) = 0 for at least one j. Otherwise, the inequality constraints may be removed without altering the solution. In most structural optimization problems the inequality constraints prescribe limits on sizes, stresses, displacements, etc. These limits have 159 Chapter 5: Constrained Optimization great impact on the design, so that typically several of the inequality constraints are active at the minimum. While the methods described in this section are powerful, they can often per-form poorly when design variables and constraints are scaled improperly. To prevent ill-conditioning, all the design variables should have similar magnitudes, and all con-straints should have similar values when they are at similar levels of criticality. A common practice is to normalize constraints such that g(x) = 0.1 correspond to a ten percent margin in a response quantity. For example, if the constraint is an upper limit σa on a stress measure σ, then the constraint may be written as g = 1 −σ σa ≥0 . (5.2) Some of the numerical techniques oﬀered in this chapter for the solution of con-strained nonlinear optimization problems are not able to handle equality constraints, but are limited to inequality constraints. In such instances it is possible to re-place the equality constraint of the form hi(x) = 0 with two inequality constraints hi(x) ≤0 and hi(x) ≥0. However, it is usually undesirable to increase the number of constraints. For problems with large numbers of inequality constraints, it is possible to construct an equivalent constraint to replace them. One of the ways to replace a family of inequality constraints (gi(x) ≥0, i = 1 . . . m) by an equivalent constraint is to use the Kreisselmeier-Steinhauser function (KS-function) deﬁned as KS[gi(x)] = −1 ρ ln[ X i e−ρgi(x)] , (5.3) where ρ is a parameter which determines the closeness of the KS-function to the smallest inequality min[gi(x)]. For any positive value of the ρ, the KS-function is always more negative than the most negative constraint, forming a lower bound envelope to the inequalities. As the value of ρ is increased the KS-functions conforms with the minimum value of the functions more closely. The value of the KS-function is always bounded by gmin ≤KS[gi(x)] ≤gmin −ln(m) ρ . (5.4) For an equality constraint represented by a pair of inequalities, hi(x) ≤0 and − hi(x) ≤0, the solution is at a point where both inequalities are active, hi(x) = −hi(x) = 0, Figure 5.1 . Sobieski shows that for a KS-function deﬁned by such a positive and negative pair of hi, the gradient of the KS-function at the solution point hi(x) = 0 vanishes regardless of the ρ value, and its value approaches to zero as the value of ρ tends to inﬁnity, Figure 5.1 . Indeed, from Eq. (5.4) at x where hi = 0, the KS-function has the property 0 ≥KS(h, −h) ≥−ln(2) ρ . (5.5) 160 Section 5.1: The Kuhn-Tucker conditions Figure 5.1 Kreisselmeier-Steinhauser function for replacing h(x) = 0. Consequently, an optimization problem minimize f(x) such that hk(x) = 0, k = 1, . . . , ne , (5.6) may be reformulated as minimize f(x) such that KS(h1, −h1, h2, −h2, . . . , hne, −hne) ≥−ϵ . (5.7) where ϵ is a small tolerance. 5.1 The Kuhn-Tucker conditions 5.1.1 General Case In general, problem (5.1) may have several local minima. Only under special circum-stances are sure of the existence of single global minimum. The necessary conditions for a minimum of the constrained problem are obtained by using the Lagrange mul-tiplier method. We start by considering the special case of equality constraints only. Using the Lagrange multiplier technique, we deﬁne the Lagrangian function L(x, λ) = f(x) − ne X j=1 λjhj(x) , (5.1.1) 161 Chapter 5: Constrained Optimization where λj are unknown Lagrange multipliers. The necessary conditions for a stationary point are ∂L ∂xi = ∂f ∂xi − ne X j=1 λj ∂hj ∂xi = 0, i = 1, . . . , n , (5.1.2) ∂L ∂λj = hj(x) = 0, j = 1, . . . , ne . (5.1.3) These conditions, however, apply only at a regular point, that is at a point where the gradients of the constraints are linearly independent. If we have constraint gradients that are linearly dependent, it means that we can remove some constraints without aﬀecting the solution. At a regular point, Eqs. (5.1.2) and (5.1.3) represent n + ne equations for the ne Lagrange multipliers and the n coordinates of the stationary point. The situation is somewhat more complicated when inequality constraints are present. To be able to apply the Lagrange multiplier method we ﬁrst transform the inequality constraints to equality constraints by adding slack variables. That is, the inequality constraints are written as gj(x) −t2 j = 0, j = 1, . . . , ng , (5.1.4) where tj is a slack variable which measures how far the jth constraint is from being critical. We can now form a Lagrangian function L(x, t, λ) = f − ng X j=1 λj(gj −t2 j) . (5.1.5) Diﬀerentiating the Lagrangian function with respect to x, λ and t we obtain ∂L ∂xi = ∂f ∂xi − ng X j=1 λj ∂gj ∂xi = 0, i = 1, . . . , n , (5.1.6) ∂L ∂λj = −gj + t2 j = 0, j = 1, . . . , ng , (5.1.7) ∂L ∂tj = 2λjtj = 0, j = 1, . . . , ng . (5.1.8) Equations (5.1.7) and (5.1.8) imply that when an inequality constraint is not critical (so that the corresponding slack variable is non-zero) then the Lagrange multiplier associated with the constraint is zero. Equations (5.1.6) to (5.1.8) are the necessary conditions for a stationary regular point. Note that for inequality constraints a regular point is one where the gradients of the active constraints are linearly independent. These conditions are modiﬁed slightly to yield the necessary conditions for a minimum and are known as the Kuhn-Tucker conditions. The Kuhn-Tucker conditions may be summarized as follows: 162 Section 5.1: The Kuhn-Tucker conditions A point x is a local minimum of an inequality constrained problem only if a set of nonnegative λj’s may be found such that: 1. Equation (5.1.6) is satisﬁed 2. The corresponding λj is zero if a constraint is not active. Figure 5.1.1 A geometrical interpretation of Kuhn-Tucker condition for the case of two constraints. A geometrical interpretation of the Kuhn-Tucker conditions is illustrated in Fig. (5.1.1) for the case of two constraints. ∇g1 and ∇g2 denote the gradients of the two constraints which are orthogonal to the respective constraint surfaces. The vector s shows a typical feasible direction which does not lead immediately to any constraint violation. For the two-constraint case Eq. (5.1.6) may be written as −∇f = −(λ1∇g1 + λ2∇g2) . (5.1.9) Assume that we want to determine whether point A is a minimum or not. To improve the design we need to proceed from point A in a direction s that is usable and feasible. For the direction to be usable, a small move along this direction should decrease the objective function. To be feasible, s should form an obtuse angle with −∇g1 and −∇g2. To be a direction of decreasing f it must form an acute angle with −∇f. Clearly from Figure (5.1.1), any vector which forms an acute angle with −∇f will also form and acute angle with either −∇g1 or −∇g2. Thus the Kuhn-Tucker conditions mean that no feasible design with reduced objective function is to be found in the neighborhood of A. Mathematically, the condition that a direction s be feasible is written as sT∇gj ≥0, j ∈IA , (5.1.10) 163 Chapter 5: Constrained Optimization where IA is the set of active constraints Equality in Eq. (5.1.10) is permitted only for linear or concave constraints (see Section 5.1.2 for deﬁnition of concavity). The condition for a usable direction (one that decreases the objective function) is sT∇f < 0 . (5.1.11) Multiplying Eq. (5.1.6) by si and summing over i we obtain sT∇f = ng X j=1 λjsT∇gj . (5.1.12) In view of Eqs. (5.1.10) and (5.1.11), Eq. (5.1.12) is impossible if the λj’s are positive. If the Kuhn-Tucker conditions are satisﬁed at a point it is impossible to ﬁnd a direction with a negative slope for the objective function that does not violate the constraints. In some cases, though, it is possible to move in a direction which is tangent to the active constraints and perpendicular to the gradient (that is, has zero slope), that is sT∇f = sT∇gj = 0, j ∈IA . (5.1.13) The eﬀect of such a move on the objective function and constraints can be determined only from higher derivatives. In some cases a move in this direction could reduce the objective function without violating the constraints even though the Kuhn-Tucker conditions are met. Therefore, the Kuhn-Tucker conditions are necessary but not suﬃcient for optimality. The Kuhn-Tucker conditions are suﬃcient when the number of active constraints is equal to the number of design variables. In this case Eq. (5.1.13) cannot be satisﬁed with s ̸= 0 because ∇gj includes n linearly independent directions (in n dimensional space a vector cannot be orthogonal to n linearly independent vectors). When the number of active constraints is not equal to the number of design variables suﬃcient conditions for optimality require the second derivatives of the objective function and constraints. A suﬃcient condition for optimality is that the Hessian matrix of the Lagrangian function is positive deﬁnite in the subspace tangent to the active constraints. If we take, for example, the case of equality constraints, the Hessian matrix of the Lagrangian is ∇2L = ∇2f − ne X j=1 λj∇2hj . (5.1.14) The suﬃcient condition for optimality is that sT(∇2L)s > 0, for all s for which sT∇hj = 0, j = 1 . . . , ne . (5.1.15) When inequality constraints are present, the vector s also needs to be orthogonal to the gradients of the active constraints with positive Lagrange multipliers. For active constraints with zero Lagrange multipliers, s must satisfy sT∇gj ≥0, when gj = 0 and λj = 0 . (5.1.16) 164 Section 5.1: The Kuhn-Tucker conditions Example 5.1.1 Find the minimum of f = −x3 1 −2x2 2 + 10x1 −6 −2x3 2 , subject to g1 = 10 −x1x2 ≥0, g2 = x1 ≥0, g3 = 10 −x2 ≥0 . The Kuhn-Tucker conditions are −3x2 1 + 10 + λ1x2 −λ2 = 0, −4x2 −6x2 2 + λ1x1 + λ3 = 0 . We have to check for all possibilities of active constraints. The simplest case is when no constraints are active, λ1 = λ2 = λ3 = 0. We get x1 = 1.826, x2 = 0, f = 6.17. The Hessian matrix of the Lagrangian, ∇2L = −6x1 λ1 λ1 −4 −12x2 , is clearly negative deﬁnite, so that this point is a maximum. We next assume that the ﬁrst constraint is active, x1x2 = 10, so that x1 ̸= 0 and g2 is inactive and therefore λ2 = 0. We have two possibilities for the third constraint. If it is active we get x1 = 1, x2 = 10, λ1 = −0.7, and λ3 = 639.3, so that this point is neither a minimum nor a maximum. If the third constraint is not active λ3 = 0 and we obtain the following three equations −3x2 1 + 10 + λ1x2 = 0, −4x2 −6x2 2 + λ1x1 = 0, x1x2 = 10 . The only solution for these equations that satisﬁes the constraints on x1 and x2 is x1 = 3.847, x2 = 2.599, λ1 = 13.24, f = −73.08 . This point satisﬁes the Kuhn-Tucker conditions for a minimum. However, the Hessian of the Lagrangian at that point ∇2L = −23.08 13.24 13.24 −35.19 , is negative deﬁnite, so that it cannot satisfy the suﬃciency condition. In fact, an examination of the function f at neighboring points along x1x2 = 10 reveals that the point is not a minimum. 165 Chapter 5: Constrained Optimization Next we consider the possibility that g1 is not active, so that λ1 = 0, and −3x2 1 + 10 −λ2 = 0, −4x2 −6x2 2 + λ3 = 0 . We have already considered the possibility of both λ’s being zero, so we need to consider only three possibilities of one of these Lagrange multipliers being nonzero, or both being nonzero. The ﬁrst case is λ2 ̸= 0, λ3 = 0, then g2 = 0 and we get x1 = 0, x2 = 0, λ2 = 10, and f = −6, or x1 = 0, x2 = −2/3, λ2 = 10, and f = −6.99. Both points satisfy the Kuhn-Tucker conditions for a minimum, but not the suﬃciency condition. In fact, the vectors tangent to the active constraints (x1 = 0 is the only one) have the form sT = (0, a), and it is easy to check that sT∇2Ls < 0. It is also easy to check that these points are indeed no minima by reducing x2 slightly. The next case is λ2 = 0, λ3 ̸= 0, so that g3 = 0. We get x1 = 1.826, x2 = 10, λ3 = 640 and f = −2194. this point satisﬁes the Kuhn-Tucker conditions, but it is not a minimum either. It is easy to check that ∇2L is negative deﬁnite in this case so that the suﬃciency condition could not be satisﬁed. Finally, we consider the case x1 = 0, x2 = 10, λ2 = 10, λ3 = 640, f = −2206. Now the Kuhn-Tucker conditions are satisﬁed, and the number of active constraints is equal to the number of design variables, so that this point is a minimum.• • • 5.1.2 Convex Problems There is a class of problems, namely convex problems, for which the Kuhn-Tucker conditions are not only necessary but also suﬃcient for a global minimum. To deﬁne convex problems we need the notions of convexity for a set of points and for a function. A set of points S is convex whenever the entire line segment connecting two points that are in S is also in S. That is if x1, x2 ∈S, then αx1 + (1 −α)x2 ∈S, 0 < α < 1 . (5.1.17) A function is convex if f[αx2 + (1 −α)x1] ≤αf(x2) + (1 −α)f(x1), 0 < α < 1 . (5.1.18) This is shown pictorially for a function of a single variable in Figure (5.1.2). The straight segment connecting any two points on the curve must lie above the curve. Alternatively we note that the second derivative of f is non-negative f ′′(x) ≥0. It can be shown that a function of n variables is convex if its matrix of second derivatives is positive semi-deﬁnite. A convex optimization problem has a convex objective function and a convex feasible domain. It can be shown that the feasible domain is convex if all the inequality constraints gj are concave (that is, −gj are convex) and the equality constraints are linear. A convex optimization problem has only one minimum, and the Kuhn-Tucker conditions are suﬃcient to establish it. Most optimization problems encountered in practice cannot be shown to be convex. However, the theory of convex programming is still very important in structural optimization, as we often approximate optimization problems by a series of convex approximations (see Chapter 9). The simplest such approximation is a linear approximation for the objective function and constraints– this produces a linear programming problem. 166 Section 5.1: The Kuhn-Tucker conditions Figure 5.1.2 Convex function. Example 5.1.2 Figure 5.1.3 Four bar statically determinate truss. Consider the minimum weight design of the four bar truss shown in Figure (5.1.3). For the sake of simplicity we assume that members 1 through 3 have the same area A1 and member 4 has an area A2. The constraints are limits on the stresses in the members and on the vertical displacement at the right end of the truss. Under the speciﬁed loading the member forces and the vertical displacement δ at the end are found to be f1 = 5p, f2 = −p, f3 = 4p, f4 = −2 √ 3p , δ = 6pl E 3 A1 + √ 3 A2 ! . We assume the allowable stresses in tension and compression to be 8.74 × 10−4E and 4.83 × 10−4E, respectively, and limit the vertical displacement to be no greater than 3×10−3l. The minimum weight design subject to stress and displacement constraints 167 Chapter 5: Constrained Optimization can be formulated in terms of nondimensional design variables x1 = 10−3A1E p , x2 = 10−3A2E p , as minimize f = 3x1 + √ 3x2 subject to g1 = 3 −18 x1 −6 √ 3 x2 ≥0, g2 = x1 −5.73 ≥0, g3 = x2 −7.17 ≥0 . The Kuhn-Tucker conditions are ∂f ∂xi − 3 X j=1 λj ∂gj ∂xi = 0, i = 1, 2 , or 3 −18 x2 1 λ1 −λ2 = 0 , √ 3 −6 √ 3 x2 2 λ1 −λ3 = 0 . Consider ﬁrst the possibility that λ1 = 0. Then clearly λ2 = 3, λ3 = √ 3 so that g2 = 0 and g3 = 0, and then x1 = 5.73, x2 = 7.17, g1 = −1.59, so that this solution is not feasible. We conclude that λ1 ̸= 0, and the ﬁrst constraint must be active at the minimum. Consider now the possibility that λ2 = λ3 = 0. We have the two Kuhn-Tucker equations and the equation g1 = 0 for the unknowns λ1, x1, x2. The solution is x1 = x2 = 9.464, λ1 = 14.93, f = 44.78 . The Kuhn-Tucker conditions for a minimum are satisﬁed. If the problem is convex the Kuhn-Tucker conditions are suﬃcient to guarantee that this point is the global minimum. The objective function and the constraint functions g2 and g3 are linear, so that we need to check only g1. For convexity g1 has to be concave or −g1 convex; this holds if the second derivative matrix −A1 of −g1 is positive semi-deﬁnite −A1 = 36/x3 1 0 0 12 √ 3/x3 2 . Clearly, for x1 > 0 and x2 > 0, −A1 is positive deﬁnite so that the minimum that we found is a global minimum.• • • 168 Section 5.2: Quadratic Programming Problems 5.2 Quadratic Programming Problems One of the simplest form of nonlinear constrained optimization problems is in the form of Quadratic Programming (QP) problem. A general QP problem has a quadratic objective function with linear equality and inequality constraints. For the sake of simplicity we consider only an inequality problem with ng constraints stated as minimize f(x) = cTx + 1 2xTQx such that Ax ≥b , xi ≥0, i = 1, . . . , n . (5.2.1) The linear constraints form a convex feasible domain. If the objective function is also convex, then we have a convex optimization problem in which, as discussed in the previous section, the Kuhn-Tucker conditions become suﬃcient for the optimality of the problem. Hence, having a positive semi-deﬁnite or positive deﬁnite Q matrix assures a global minimum for the solution of the problem, if one exists. For many optimization problems the quadratic form xTQx is either positive deﬁnite or positive semi-deﬁnite. Therefore, one of the methods for solving QP problems relies on solving the Kuhn-Tucker conditions. We start by writing the Lagrange function for the Problem (5.2.1) L(x, λ, µ, t, s) = cTx + 1 2xTQx −λT(Ax −{t2 j} −b) −µT(x −{s2 i }) , (5.2.2) where λ and µ are the vectors of Lagrange multipliers for the inequality constraints and the nonnegativity constraints, respectively, and {t2 j} and {s2 i } are the vectors of positive slack variables for the same. The necessary conditions for a stationary point are obtained by diﬀerentiating the Lagrangian with respect to the x, λ, µ, t, and s, ∂L ∂x = c −Qx −ATλ −µ = 0 , (5.2.3) ∂L ∂λ = Ax −{t2 j} −b = 0 , (5.2.4) ∂L ∂µ = x −{s2 i } = 0 , (5.2.5) ∂L ∂tj = 2λjtj = 0, j = 1, . . . , ng , (5.2.6) ∂L ∂si = 2µisi = 0, i = 1, . . . , n . (5.2.7) where ng is the number of inequality constraints, and n is the number of design variables. We deﬁne a new vector {qj} = {t2 j}, j = 1, . . . , ng (q ≥0). After multiplying Eqs. (5.2.6) and (5.2.7) by {tj} and {si}, respectively, and eliminating 169 Chapter 5: Constrained Optimization {si} from the last equation by using Eq. (5.2.5), we can rewrite the Kuhn-Tucker conditions Qx + ATλ + µ = c , (5.2.8) Ax −q = b , (5.2.9) λjqj = 0, j = 1, . . . , ng , (5.2.10) µixi = 0, i = 1, . . . , n , (5.2.11) x ≥0, λ ≥0, and µ ≥0 . (5.2.12) Equations (5.2.8) and (5.2.9) form a set of n+ng linear equations for the solution of unknowns xi, λj, µi, and qj which also need to satisfy Eqs. (5.2.10) and (5.2.11). Despite the nonlinearity of the Eqs. (5.2.10) and (5.2.11), this problem can be solved as proposed by Wolfe by using the procedure described in 3.6.3 for generating a basic feasible solution through the use of artiﬁcial variables. Introducing a set of artiﬁcial variables, yi, i = 1, . . . , n , we deﬁne an artiﬁcial cost function to be minimized, minimize n X i=1 yi (5.2.13) subject to Qx + ATλ + µ + y = c , (5.2.14) Ax −q = b , (5.2.15) x ≥0, λ ≥0, µ ≥0, and y ≥0 . (5.2.16) Equations (5.2.13) through (5.2.16) can be solved by using the standard simplex method with the additional requirement that (5.2.10) and (5.2.11) be satisﬁed. These requirements can be implemented during the simplex algorithm by simply enforcing that the variables λj and qj (and µi and xi) not be included in the basic solution simultaneously. That is, we restrict a non-basic variable µi from entering the basis if the corresponding xi is already among the basic variables. Other methods for solving the quadratic programming problem are also available, and the reader is referred to Gill et al. (, pp. 177–180) for additional details. 5.3 Computing the Lagrange Multipliers As may be seen from example 5.1.1, trying to ﬁnd the minimum directly from the Kuhn-Tucker conditions may be diﬃcult because we need to consider many com-binations of active and inactive constraints, and this would in general involve the solution of highly nonlinear equations. The Kuhn-Tucker conditions are, however, often used to check whether a candidate minimum point satisﬁes the necessary con-ditions. In such a case we need to calculate the Lagrange multipliers (also called the Kuhn-Tucker multipliers) at a given point x. As we will see in the next section, we 170 Section 5.3: Computing the Lagrange Multipliers may also want to calculate the Lagrange multipliers for the purpose of estimating the sensitivity of the optimum solution to small changes in the problem deﬁnition. To calculate the Lagrange multipliers we start by writing Eq. (5.1.6) in matrix notation as ∇f −Nλ = 0 , (5.3.1) where the matrix N is deﬁned by nij = ∂gj ∂xi , j = 1, . . . , r , and i = 1, . . . , n . (5.3.2) We consider only the active constraints and associated lagrange multipliers, and as-sume that there are r of them. Typically, the number, r, of active constraints is less than n, so that with n equations in terms of r unknowns, Eq. (5.3.1) is an overdetermined system. We assume that the gradients of the constraints are linearly independent so that N has rank r. If the Kuhn-Tucker conditions are satisﬁed the equations are consistent and we have an exact solution. We could therefore use a subset of r equations to solve for the Lagrange multipliers. However, this approach may be susceptible to ampliﬁcation of errors. Instead we can use a least-squares approach to solve the equations. We deﬁne a residual vector u u = Nλ −∇f , (5.3.3) A least squares solution of Eq. (5.3.1) will minimize the square of the Euclidean norm of the residual with respect to λ ∥u∥2 = (Nλ −∇f)T(Nλ −∇f) = λTNTNλ −2λTNT∇f + ∇f T∇f . (5.3.4) To minimize ∥u∥2 we diﬀerentiate it with respect to each one of the Lagrange multi-pliers and get 2NTNλ −2NT∇f = 0 , (5.3.5) or λ = (NTN)−1NT∇f . (5.3.6) This is the best solution in the least square sense. However, if the Kuhn-Tucker conditions are satisﬁed it should be the exact solution of Eq. (5.3.1). Substituting from Eq. (5.3.6) into Eq. (5.3.1) we obtain P∇f = 0 , (5.3.7) where P = I −N(NTN)−1NT . (5.3.8) P is called the projection matrix. It will be shown in Section 5.5 that it projects a vector into the subspace tangent to the active constraints. Equation (5.3.7) implies that for the Kuhn-Tucker conditions to be satisﬁed the gradient of the objective function has to be orthogonal to that subspace. In practice Eq. (5.3.6) is no longer popular for the calculation of the Lagrange multipliers. One reason is that the method is ill-conditioned and another is that it is 171 Chapter 5: Constrained Optimization not eﬃcient. An eﬃcient and better conditioned method for least squares calculations is based on the QR factorization of the matrix N. The QR factorization of the matrix N consists of an r × r upper triangular matrix R and an n × n orthogonal matrix Q such that QN = Q1N Q2N = R 0 . (5.3.9) Here Q1 is a matrix consisting of the ﬁrst r rows of Q, Q2 includes the last n −r rows of Q, and the zero represents an (n −r) × r zero matrix (for details of the QR factorization see most texts on numerical analysis, e.g., Dahlquist and Bjorck ). Because Q is an orthogonal matrix, the Euclidean norm of Qu is the same as that of u, or ∥u∥2 = ∥Qu∥2 = ∥QNλ −Q∇f∥2 = R 0 λ −Q∇f 2 = Rλ −Q1∇f −Q2∇f 2 . (5.3.10) From this form it can be seen that ∥u∥2 is minimized by choosing λ so that Rλ = Q1∇f . (5.3.11) The last n −r rows of the matrix Q denoted Q2 are also important in the following. They are orthogonal vectors which span the null space of NT. That is NT times each one of these vectors is zero. Example 5.3.1 Check whether the point (−2, −2, 4) is a local minimum of the problem f = x1 + x2 + x3, g1 = 8 −x2 1 −x2 2 ≥0, g2 = x3 −4 ≥0, g3 = x2 + 8 ≥0 . Only the ﬁrst two constraints are critical at (−2, −2, 4) ∂g1 ∂x1 = −2x1 = 4, ∂g1 ∂x2 = −2x2 = 4, ∂g1 ∂x3 = 0 , ∂g2 ∂x1 = 0, ∂g2 ∂x2 = 0, ∂g2 ∂x3 = 1 , ∂f ∂x1 = ∂f ∂x2 = ∂f ∂x3 = 1 . So N = " 4 0 4 0 0 1 # , ∇f = ( 1 1 1 ) , 172 Section 5.4: Sensitivity of Optimum Solution to Problem Parameters NTN = 32 0 0 1 , NT∇f = 8 1 , λ = (NTN)−1NT∇f = 1/4 1 , also I −N(NTN)−1NT ∇f = 0 . Equation (5.3.7) is satisﬁed, and all the Lagrange multipliers are positive, so the Kuhn-Tucker conditions for a minimum are satisﬁed. • • • 5.4 Sensitivity of Optimum Solution to Problem Parameters The Lagrange multipliers are not only useful for checking optimality, but they also provide information about the sensitivity of the optimal solution to problem parameters. In this role they are extremely valuable in practical applications. In most engineering design optimization problems we have a host of parameters such as material properties, dimensions and load levels that are ﬁxed during the optimization. We often need the sensitivity of the optimum solution to these problem parameters, either because we do not know them accurately, or because we have some freedom to change them if we ﬁnd that they have a large eﬀect on the optimum design. We assume now that the objective function and constraints depend on a param-eter p so that the optimization problem is deﬁned as minimize f(x, p) such that gj(x, p) ≥0, j = 1, . . . , ng . (5.4.1) The solution of the problem is denoted x∗(p) and the corresponding objective function f ∗(p) = f(x∗(p), p). We want to ﬁnd the derivatives of x∗and f ∗with respect to p. The equations that govern the optimum solution are the Kuhn-Tucker conditions, Eq. (5.3.1), and the set of active constraints ga = 0. (5.4.2) where ga denotes the vector of r active constraint functions. Equations (5.3.1) and (5.4.2) are satisﬁed by x∗(p) for all values of p that do not change the set of active constraints. Therefore, the derivatives of these equations with respect to p are zero, provided we consider the implicit dependence of x and λ on p. Diﬀerentiating Eq. (5.3.1) and (5.4.2) with respect to p we obtain (A −Z)dx∗ dp −Ndλ dp + ∂ ∂p (∇f) − ∂N ∂p λ = 0 , (5.4.3) NT dx∗ dp + ∂ga ∂p = 0 , (5.4.4) 173 Chapter 5: Constrained Optimization where A is the Hessian matrix of the objective function f, aij = ∂2f/∂xi∂xj, and Z is a matrix whose elements are zkl = X j ∂2gj ∂xk∂xl λj . (5.4.5) Equations (5.4.3) and (5.4.4) are a system of simultaneous equations for the deriva-tives of the design variables and of the Lagrange multipliers. Diﬀerent special cases of this system are discussed by Sobieski et al. . Often we do not need the derivatives of the design variables or of the Lagrange multipliers, but only the derivatives of the objective function. In this case the sensi-tivity analysis can be greatly simpliﬁed. We can write d f dp = ∂f ∂p + n X l=1 ∂f ∂xl dx∗ l dp = ∂f ∂p + (∇f)T dx∗ dp . (5.4.6) Using Eq. (5.3.1) and (5.4.4) we get d f dp = ∂f ∂p −λT ∂ga ∂p . (5.4.7) Equation (5.4.7) shows that the Lagrange multipliers are a measure of the eﬀect of a change in the constraints on the objective function. Consider, for example, a constraint of the form gj(x) = Gj(x) −p ≥0. By increasing p we make the constraint more diﬃcult to satisfy. Assume that many constraints are critical, but that p aﬀects only this single constraint. We see that ∂gj/∂p = −1, and from Eq. (5.4.7) d f/dp = λj, that is λj is the ‘marginal price’ that we pay in terms of an increase in the objective function for making gj more diﬃcult to satisfy. The interpretation of Lagrange multipliers as the marginal prices of the con-straints also explains why at the optimum all the Lagrange multipliers have to be non-negative. A negative Lagrange multiplier would indicate that we can reduce the objective function by making a constraint more diﬃcult to satisfy— an absurdity. Example 5.4.1 Consider the optimization problem f = x1 + x2 + x3, g1 = p −x2 1 −x2 2 ≥0, g2 = x3 −4 ≥0, g3 = x2 + p ≥0 . This problem was analyzed for p = 8 in Example 5.3.1, and the optimal solution was found to be (−2, −2, 4). We want to ﬁnd the derivative of this optimal solution with respect to p. At the optimal point we have f = 0 and λT = (0.25, 1.0), with the 174 Section 5.5: Gradient Projection and Reduced Gradient Methods ﬁrst two constraints being critical. We can calculate the derivative of the objective function from Eq. (5.4.7) ∂f ∂p = 0, ∂ga ∂p = 1 0 , so d f dp = −0.25 . To calculate the derivatives of the design variables and constraints we need to set up Eqs. (5.4.3) and (5.4.4). We get A = 0, ∂∇f ∂p = 0, ∂N ∂p = 0 . Only g1 has nonzero second derivatives ∂2g1/∂x2 1 = ∂2g1/∂x2 2 = −2 so from Eq. (5.4.5) z11 = −2λ1 = −0.5, z22 = −2λ1 = −0.5, Z = " −.5 0 0 0 −.5 0 0 0 0 # . With N from Example 5.3.1, Eq. (5.4.3) gives us .5˙ x1 −4˙ λ1 = 0 , .5˙ x2 −4˙ λ1 = 0 , ˙ λ2 = 0 , where a dot denotes derivative with respect to p. From Eq. (5.4.4) we get 4˙ x1 + 4˙ x2 + 1 = 0 , ˙ x3 = 0 . The solution of these ﬁve coupled equations is ˙ x1 = ˙ x2 = −0.125, ˙ x3 = 0, ˙ λ1 = −0.0156, ˙ λ2 = 0 . We can check the derivatives of the objective function and design variables by chang-ing p from 8 to 9 and re-optimizing. It is easy to check that we get x1 = x2 = −2.121, x3 = 4, f = −0.242. These values compare well with linear extrapolation based on the derivatives which gives x1 = x2 = −2.125, x3 = 4, f = −0.25.• • • 175 Chapter 5: Constrained Optimization 5.5 Gradient Projection and Reduced Gradient Methods Rosen’s gradient projection method is based on projecting the search direction into the subspace tangent to the active constraints. Let us ﬁrst examine the method for the case of linear constraints . We deﬁne the constrained problem as minimize f(x) such that gj(x) = n X i=1 ajixi −bj ≥0, j = 1, . . . , ng . (5.5.1) In vector form gj = aT j x −bj ≥0 . (5.5.2) If we select only the r active constraints (j ∈IA), we may write the constraint equations as ga = NTx −b = 0 , (5.5.3) where ga is the vector of active constraints and the columns of the matrix N are the gradients of these constraints. The basic assumption of the gradient projection method is that x lies in the subspace tangent to the active constraints. If xi+1 = xi + αs , (5.5.4) and both xi and xi+1 satisfy Eq. (5.5.3), then NTs = 0 . (5.5.5) If we want the steepest descent direction satisfying Eq. (5.5.5), we can pose the problem as minimize sT∇f such that NTs = 0 , and sTs = 1 . (5.5.6) That is, we want to ﬁnd the direction with the most negative directional deriva-tive which satisﬁes Eq. (5.5.5). We use Lagrange multipliers λ and µ to form the Lagrangian L(s, λ, µ) = sT∇f −sTNλ −µ(sTs −1) . (5.5.7) The condition for L to be stationary is ∂L ∂s = ∇f −Nλ −2µs = 0 . (5.5.8) Premultiplying Eq. (5.5.8) by NT and using Eq. (5.5.5) we obtain NT∇f −NTNλ = 0 , (5.5.9) or λ = (NTN)−1NT∇f . (5.5.10) 176 Section 5.5: Gradient Projection and Reduced Gradient Methods So that from Eq. (5.5.8) s = 1 2µ[I −N(NTN)−1NT]∇f = 1 2µP∇f . (5.5.11) P is the projection matrix deﬁned in Eq. (5.3.8). The factor of 1/2µ is not signiﬁcant because s deﬁnes only the direction of search, so in general we use s = −P∇f. To show that P indeed has the projection property, we need to prove that if w is an arbitrary vector, then Pw is in the subspace tangent to the active constraints, that is Pw satisﬁes NTPw = 0 . (5.5.12) We can easily verify this by using the deﬁnition of P. Equation (5.3.8) which deﬁnes the projection matrix P does not provide the most eﬃcient way for calculating it. Instead it can be shown that P = QT 2 Q2 , (5.5.13) where the matrix Q2 consists of the last n −r rows of the Q factor in the QR factorization of N (see Eq. (5.3.9)). A version of the gradient projection method known as the generalized reduced gradient method was developed by Abadie and Carpentier . As a ﬁrst step we select r linearly independent rows of N, denote their transpose as N1 and partition NT as NT = [N1 N2] . (5.5.14) Next we consider Eq. (5.5.5) for the components si of the direction vector. The r equations corresponding to N1 are then used to eliminate r components of s and obtain a reduced order problem for the direction vector. Once we have identiﬁed N1 we can easily obtain Q2 which is given as QT 2 = −N−1 1 N2 I . (5.5.15) Equation (5.5.15) can be veriﬁed by checking that NTQT 2 = 0, so that Q2N = 0, which is the requirement that Q2 has to satisfy (see discussion following Eq. (5.3.11)). After obtaining s from Eq. (5.5.11) we can continue the search with a one di-mensional minimization, Eq. (5.5.4), unless s = 0. When s = 0 Eq. (5.3.7) indicates that the Kuhn-Tucker conditions may be satisﬁed. We then calculate the Lagrange multipliers from Eq. (5.3.6) or Eq. (5.3.11). If all the components of λ are non-negative, the Kuhn-Tucker conditions are indeed satisﬁed and the optimization can be terminated. If some of the Lagrange multipliers are negative, it is an indication that while no progress is possible with the current set of active constraints, it may be possible to proceed by removing some of the constraints associated with negative Lagrange multipliers. A common strategy is to remove the constraint associated with the most negative Lagrange multiplier and repeat the calculation of P and s. If s 177 Chapter 5: Constrained Optimization is now non-zero, a one-dimensional search may be started. If s remains zero and there are still negative Lagrange multipliers, we remove another constraint until all Lagrange multipliers become positive and we satisfy the Kuhn-Tucker conditions. After a search direction has been determined, a one dimensional search must be carried out to determine the value of α in Eq. (5.5.4). Unlike the unconstrained case, there is an upper limit on α set by the inactive constraints. As α increases, some of them may become active and then violated. Substituting x = xi + αs into Eq. (5.5.2) we obtain gj = aT j (xi + αs) −bj ≥0 , (5.5.16) or α ≤−(aT j xi −bj)/aT j s = −gj(xi)/aT j s . (5.5.17) Equation (5.5.17) is valid if aT j s < 0. Otherwise, there is no upper limit on α due to the jth constraint. From Eq. (5.5.17) we get a diﬀerent α, say αj for each constraint. The upper limit on α is the minimum ¯ α = min αj>0, j∋IA αj . (5.5.18) At the end of the move, new constraints may become active, so that the set of active constraints may need to be updated before the next move is undertaken. The version of the gradient projection method presented so far is an extension of the steepest descent method. Like the steepest descent method, it may have slow convergence. The method may be extended to correspond to Newton or quasi-Newton methods. In the unconstrained case, these methods use a search direction deﬁned as s = −B∇f , (5.5.19) where B is the inverse of the Hessian matrix of f or an approximation thereof. The direction that corresponds to such a method in the subspace tangent to the active constraints can be shown to be s = −QT 2 (QT 2 ALQ2)−1Q2∇f , (5.5.20) where AL is the Hessian of the Lagrangian function or an approximation thereof. The gradient projection method has been generalized by Rosen to nonlinear con-straints . The method is based on linearizing the constraints about xi so that N = [∇g1(xi), ∇g2(xi), . . . , ∇gr(xi)] . (5.5.21) 178 Section 5.5: Gradient Projection and Reduced Gradient Methods Figure 5.5.1 Projection and restoration moves. The main diﬃculty caused by the nonlinearity of the constraints is that the one-dimensional search typically moves away from the constraint boundary. This is because we move in the tangent subspace which no longer follows exactly the constraint boundaries. After the one-dimensional search is over, Rosen prescribes a restoration move to bring x back to the constraint boundaries, see Figure 5.5.1. To obtain the equation for the restoration move, we note that instead of Eq. (5.5.2) we now use the linear approximation gj ≈gj(xi) + ∇gT j (¯ xi −xi) . (5.5.22) We want to ﬁnd a correction ¯ xi −xi in the tangent subspace (i.e. P(¯ xi −xi) = 0) that would reduce gj to zero. It is easy to check that ¯ xi −xi = −N(NTN)−1ga(xi) , (5.5.23) is the desired correction, where ga is the vector of active constraints. Equation (5.5.23) is based on a linear approximation, and may therefore have to be applied repeatedly until ga is small enough. In addition to the need for a restoration move, the nonlinearity of the constraints requires the re-evaluation of N at each point. It also complicates the choice of an upper limit for α which guarantees that we will not violate the presently inactive constraints. Haug and Arora suggest a procedure which is better suited for the nonlinear case. The ﬁrst advantage of their procedure is that it does not require a one-dimensional search. Instead, α in Eq. (5.5.4) is determined by specifying a desired speciﬁed reduction γ in the objective function. That is, we specify f(xi) −f(xi+1) ≈γf(xi) . (5.5.24) Using a linear approximation with Eq. (5.5.4) we get α∗= −γf(xi) sT∇f . (5.5.25) The second feature of Haug and Arora’s procedure is the combination of the projection and the restoration moves as xi+1 = xi + α∗s −N(NTN)−1ga , (5.5.26) where Eqs. (5.5.4), (5.5.23) and (5.5.25) are used. 179 Chapter 5: Constrained Optimization Example 5.5.1 Use the gradient projection method to solve the following problem minimize f = x2 1 + x2 2 + x2 3 + x2 4 −2x1 −3x4 subject to g1 = 2x1 + x2 + x3 + 4x4 −7 ≥0 , g2 = x1 + x2 + x2 3 + x4 −5.1 ≥0 , xi ≥0, i = 1, . . . , 4 . Assume that as a result of previous moves we start at the point xT 0 = (2, 2, 1, 0), f(x0) = 5.0, where the nonlinear constraint g2 is slightly violated. The ﬁrst constraint is active as well as the constraint on x4. We start with a combined projection and restoration move, with a target improvement of 10% in the objective function. At x0 N =    2 1 0 1 1 0 1 2 0 4 1 1   , NTN = " 22 9 4 9 7 1 4 1 1 # , (NTN)−1 = 1 11 " 6 −5 −19 −5 6 14 −19 14 73 # , P = I −N(NTN)−1NT = 1 11    1 −3 1 0 −3 9 −3 0 1 −3 1 0 0 0 0 0   , ∇f =      2 4 2 −3      . The projection move direction is s = −P∇f = [8/11, −24/11, 8/11, 0]T. Since the magnitude of a direction vector is unimportant we scale s to sT = [1, −3, 1, 0]. For a 10% improvement in the objective function γ = 0.1 and from Eq. (5.5.25) α∗= −0.1f sT∇f = −0.1 × 5 −8 = 0.0625 . For the correction move we need the vector ga of constraint values , gT a = (0, −0.1, 0), so the correction is −N(NTN)−1ga = −1 110      4 −1 −7 0      . Combining the projection and restoration moves, Eq. (5.5.26) x1 =      2 2 1 0      + 0.0625      1 −3 1 0      − 1 110      4 −1 −7 0      =      2.026 1.822 1.126 0      , we get f(x1) = 4.64, g1(x1) = 0, g2(x1) = 0.016. Note that instead of 10% reduction we got only 7% due to the nonlinearity of the objective function. However, we did satisfy the nonlinear constraint.• • • 180 Section 5.5: Gradient Projection and Reduced Gradient Methods Example 5.5.2 Consider the four bar truss of Example 5.1.2. The problem of ﬁnding the minimum weight design subject to stress and displacement constraints was formulated as minimize f = 3x1 + √ 3x2 subject to g1 = 3 −18 x1 −6 √ 3 x2 ≥0 , g2 = x1 −5.73 ≥0 , g3 = x2 −7.17 ≥0 , where the xi are non-dimensional areas xi = AiE 1000P , i = 1, 2 . The ﬁrst constraint represents a limit on the vertical displacement, and the other two represent stress constraints. Assume that we start the search at the intersection of g1 = 0 and g3 = 0, where x1 = 11.61, x2 = 7.17, and f = 47.25. The gradients of the objective function and two active constraints are ∇f = 3 √ 3 , ∇g1 = 0.1335 0.2021 , ∇g3 = 0 1 , N = 0.1335 0 0.2021 1 . Because N is nonsingular, Eq. (5.3.8) shows that P = 0. Also since the number of linearly independent active constraints is equal to the number of design variables the tangent subspace is a single point, so that there is no more room for progress. Using Eqs. (5.3.6) or (5.3.11) we obtain λ = 22.47 −2.798 . The negative multiplier associated with g3 indicates that this constraint can be dropped from the active set. Now N = 0.1335 0.2021 . The projection matrix is calculated from Eq. (5.3.8) P = 0.6962 −0.4600 −0.4600 0.3036 , s = −P∇f = −1.29 0.854 . We attempt a 5% reduction in the objective function, and from Eq. (5.5.25) α∗= 0.05 × 47.25 [−1.29 0.854] 3 √ 3 = 0.988 . 181 Chapter 5: Constrained Optimization Since there was no constraint violation at x0 we do not need a combined projection and correction step, and x1 = x0 + α∗s = 11.61 7.17 + 0.988 −1.29 0.854 = 10.34 8.01 . At x1 we have f(x1) = 44.89, g1(x1) = −0.0382. Obviously g2 is not violated. If there were a danger of that we would have to limit α∗using Eq. (5.5.17). The violation of the nonlinear constraint is not surprising, and its size indicates that we should reduce the attempted reduction in f in the next move. At x1, only g1 is active so N = ∇g1 = 0.1684 0.1620 . The projection matrix is calculated to be P = 0.4806 −0.4996 −0.4996 0.5194 , s = −P∇f = −0.5764 0.5991 . Because of the violation we reduce the attempted reduction in f to 2.5%, so α∗= − 0.025 × 44.89 [−0.567 0.599] 3 √ 3 = 1.62 . We need also a correction due to the constraint violation (ga = −0.0382) −N(NTN)−1ga = 0.118 0.113 . Altogether x2 = x1+α∗s−N(NTN)−1ga = 10.34 8.01 −1.62 0.576 −0.599 + 0.118 0.113 = 9.52 9.10 . We obtain f(x2) = 44.32, g1(x2) = −0.0328. The optimum design is actually xT = (9.464, 9.464), f(x) = 44.78, so after two iterations we are quite close to the optimum design.• • • 5.6 The Feasible Directions Method The feasible directions method has the opposite philosophy to that of the gradient projection method. Instead of following the constraint boundaries, we try to stay as far away as possible from them. The typical iteration of the feasible direction method starts at the boundary of the feasible domain (unconstrained minimization techniques are used to generate a direction if no constraint is active). 182 Section 5.6: The Feasible Directions Method Figure 5.6.1 Selection of search direction using the feasible directions method. Consider Figure 5.6.1. As a result of a previous move the design is at point x and we look for a direction s which keeps x in the feasible domain and improves the objective function. A vector s is deﬁned as a feasible direction if at least a small step can be taken along it that does not immediately leave the feasible domain. If the constraints are smooth, this is satisﬁed if sT∇gj > 0, j ∈IA , (5.6.1) where IA is the set of critical constraints at x. The direction s is called a usable direction at the point x if in addition sT∇f = sTg < 0 . (5.6.2) That is, s is a direction which reduces the objective function. Among all possible choices of usable feasible directions we seek the direction which is best in some sense. We have two criteria for selecting a direction. On the one hand we want to reduce the objective function as much as possible. On the other hand we want to keep away from the constraint boundary as much as possible. A compromise is deﬁned by the following maximization problem maximize β such that −sT∇gj + θjβ ≤0, j ∈IA , sT∇f + β ≤0, θj ≥0 , \|si\| ≤1 . (5.6.3) The θj are positive numbers called “push-oﬀ” factors because their magnitude deter-mines how far x will move from the constraint boundaries. A value of θj = 0 will result in a move tangent to the boundary of the the jth constraint, and so may be appropriate for a linear constraint. A large value of θj will result in a large angle between the constraint boundary and the move direction, and so is appropriate for a highly nonlinear constraint. 183 Chapter 5: Constrained Optimization The optimization problem deﬁned by Eq. (5.6.3) is linear and can be solved using the simplex algorithm. If βmax > 0, we have found a usable feasible direction. If we get βmax = 0 it can be shown that the Kuhn-Tucker conditions are satisﬁed. Once a direction of search has been found, the choice of step length is typically based on a prescribed reduction in the objective function (using Eq. (5.5.25)). If at the end of the step no constraints are active, we continue in the same direction as long as sT∇f is negative. We start the next iteration when x hits the constraint boundaries, or use a direction based on unconstrained technique if x is inside the feasible domain. Finally, if some constraints are violated after the initial step we make x retreat based on the value of the violated constraints. The method of feasible directions is implemented in the popular CONMIN program . Example 5.6.1 Consider the four bar truss of Example 5.1.2. The problem of ﬁnding the minimum weight design subject to stress and displacement constraints was formulated as minimize f = 3x1 + √ 3x2 subject to g1 = 3 −18 x1 −6 √ 3 x2 ≥0 , g2 = x1 −5.73 ≥0 , g3 = x2 −7.17 ≥0 , where the xi are non-dimensional areas xi = AiE 1000P , i = 1, 2 . The ﬁrst constraint represents a limit on the vertical displacement, and the other two constraints represent stress constraints. Assume that we start the search at the intersection of g1 = 0 and g3 = 0 where xT 0 = (11.61, 7.17) and f = 47.25. The gradient of the objective function and two active constraints are ∇f = 3 √ 3 , ∇g1 = 0.1335 0.2021 , ∇g3 = 0 1 . Selecting θ1 = θ2 = 1, we ﬁnd that Eq. (5.6.3) becomes maximize β subject to −0.1335s1 −0.2021s2 + β ≤0 , −s2 + β ≤0 , 3s1 + √ 3s2 + β ≤0 , −1 ≤s1 ≤1 , −1 ≤s2 ≤1 . 184 Section 5.7: The Feasible Directions Method The solution of this linear program is s1 = −0.6172, s2 = 1, and we now need to execute the one dimensional search x1 = 11.61 7.17 + α −0.6172 1 . Because the objective function is linear, this direction will remain a descent direction indeﬁnitely, and α will be limited only by the constraints. The requirement that g2 is not violated will lead to α = 9.527, x1 = 5.73, x2 = 16.7 which violates g1. We see that because g1 is nonlinear, even though we start the search by moving away from it we still bump into it again (see Figure 5.6.2). It can be easily checked that for α > 5.385 we violate g1. So we take α = 5.385 and obtain x1 = 8.29, x2 = 12.56, f = 46.62. Figure 5.6.2 Feasible direction solution of 4 bar truss example. For the next iteration we have only one active constraint ∇g1 = 0.2619 0.0659 , ∇f = 3 √ 3 . The linear program for obtaining s is maximize β subject to −0.2619s1 −0.0659s2 + β ≤0 , 3s1 + √ 3s2 + β ≤0 , −1 ≤s1 ≤1 , −1 ≤s2 ≤1 . 185 Chapter 5: Constrained Optimization The solution of the linear program is s1 = 0.5512, s2 = −1, so that the one-dimensional search is x = 8.29 12.56 + α 0.5512 −1 . Again α is limited only by the constraints. The lower limit on x2 dictates α ≤5.35. However, the constraint g1 is again more critical. It can be veriﬁed that for α > 4.957 it is violated, so we take α = 4.957, x1 = 11.02, x2 = 7.60, f = 46.22. The optimum design found in Example 5.1.2 is x1 = x2 = 9.464, f = 44.78. The design space and the two iterations are shown in Figure 5.6.2. • • • 5.7 Penalty Function Methods When the energy crisis erupted in the middle seventies, the United States Congress passed legislation intended to reduce the fuel consumption of American cars. The target was an average fuel consumption of 27.5 miles per gallon for new cars in 1985. Rather than simply legislate this limit Congress took a gradual approach, with a diﬀerent limit set each year to bring up the average from about 14 miles per gallon to the target value. Thus the limit was set at 26 for 1984, 25 for 1983, 24 for 1982, and so on. Furthermore, the limit was not absolute, but there was a ﬁne of $50 per 0.1 miles per gallon violation per car. This approach to constraining the automobile companies to produce fuel eﬃcient cars has two important aspects. First, by legislating a penalty proportional to the violation rather than an absolute limit, the government allowed the auto companies more ﬂexibility. That meant they could follow a time schedule that approximated the government schedule without having to adhere to it rigidly. Second, the gradual approach made enforcement easier politically. Had the government simply set the ul-timate limit for 1985 only, nobody would have paid attention to the law in the 1970’s. Then as 1985 moved closer there would have been a rush to develop fuel eﬃcient cars. The hurried eﬀort could mean both non-optimal car designs and political pressure to delay the enforcement of the law. The fuel eﬃciency law is an example in which constraints on behavior or eco-nomic activities are imposed via penalties whose magnitude depends on the degree of violation of the constraints. It is no wonder that this simple and appealing approach has found application in constrained optimization. Instead of applying constraints we replace them by penalties which depend on the degree of constraint violations. This approach is attractive because it replaces a constrained optimization problem by an unconstrained one. The penalties associated with constraint violation have to be high enough so that the constraints are only slightly violated. However, just as there are political problems associated with imposing abrupt high penalties in real life, so there are numerical diﬃculties associated with such a practice in numerical optimization. For this reason we opt for a gradual approach where we start with small penalties and increase them gradually. 186 Section 5.7: Penalty Function Methods 5.7.1 Exterior Penalty Function The exterior penalty function associates a penalty with a violation of a constraint. The term ‘exterior’ refers to the fact that penalties are applied only in the exterior of the feasible domain. The most common exterior penalty function is one which associates a penalty which is proportional to the square of a violation. That is, the constrained minimization problem, Eq. (5.1) minimize f(x) such that hi(x) = 0, i = 1, . . . , ne , gj(x) ≥0, j = 1, . . . , ng , (5.7.1) is replaced by minimize φ(x, r) = f(x) + r ne X i=1 h2 i (x) + r ng X j=1 < −gj >2 r = r1, r2, . . . , ri →∞, (5.7.2) where < a > denote the positive part of a or max(a, 0). The inequality terms are treated diﬀerently from the equality terms because the penalty applies only for con-straint violation. The positive multiplier r controls the magnitude of the penalty terms. It may seem logical to choose a very high value of r to ensure that no con-straints are violated. However, as noted before, this approach leads to numerical diﬃculties illustrated later in an example. Instead the minimization is started with a relatively small value of r, and then r is gradually increased. A typical value for ri+1/ri is 5. A typical plot of φ(x, r) as a function of r is shown in Figure 5.7.1 for a simple example. Figure 5.7.1 Exterior penalty function for f = 0.5x subject to x −4 ≥0. We see that as r is increased, the minimum of φ moves closer to the constraint boundary. However, the curvature of φ near the minimum also increases. It is 187 Chapter 5: Constrained Optimization the high values of the curvature associated with large values of r which often lead to numerical diﬃculties. By using a sequence of values of r, we use the minima obtained for small values of r as starting points for the search with higher r values. Thus the ill-conditioning associated with the large curvature is counterbalanced by the availability of a good starting point. Based on the type of constraint normalization given by Eq. (5.2) we can select a reasonable starting value for the penalty multiplier r. A rule of thumb is that one should start with the total penalty being about equal to the objective function for typical constraint violation of 50% of the response limits. In most optimization problems the total number of active constraints is about the same as or just slightly lower than the number of design variables. Assuming we start with one quarter of the eventual active constraints being violated by about 50% (or g = −0.5) then we have f(x0) ≈r0 n 4(0.5)2, or r0 = 16f(x0) n . (5.7.3) It is also important to obtain a good starting point for restarting the optimization as r is increased. The minimum of the optimization for the previous value of r is a reasonable starting point, but one can do better. Fiacco and McCormick show that the position of the minimum of φ(x, r) has the asymptotic form x∗(r) = a + b/r, as r →∞. (5.7.4) Once the optimum has been found for two values of r, say ri−1, and ri, the vectors a and b may be estimated, and the value of x∗(r) predicted for subsequent values of r. It is easy to check that in order to satisfy Eq. (5.7.4), a and b are given as a = cx∗(ri−1) −x∗(ri) c −1 , b = [x∗(ri−1) −a] ri−1 , (5.7.5) where c = ri−1/ri . (5.7.6) In addition to predicting a good value of the design variables for restarting the op-timization for the next value of r, Eq. (5.7.4) provides us with a useful convergence criterion, namely ∥x∗−a∥≤ϵ1 , (5.7.7) where a is estimated from the last two values of r, and ϵ1 is a speciﬁed tolerance chosen to be small compared to a typical value of ∥x∥. A second convergence criterion is based on the magnitude of the penalty terms, which, as shown in Example 5.7.1, go to zero as r goes to inﬁnity. Therefore, a reasonable convergence criterion is φ −f f ≤ϵ2 . (5.7.8) 188 Section 5.7: Penalty Function Methods Finally, a criterion based on the change in the value of the objective function at the minimum f ∗is also used f ∗(ri) −f ∗(ri−1) f ∗(ri) ≤0 . (5.7.9) A typical value for ϵ2 or ϵ3 is 0.001. Example 5.7.1 Minimize f = x2 1 + 10x2 2 such that x1 + x2 = 4. We have φ = x2 1 + 10x2 2 + r(4 −x1 −x2)2 . The gradient ∇φ is given as g = 2x1(1 + r) + 2rx2 −8r 2x2(10 + r) + 2rx1 −8r . Setting the gradient to zero we obtain x1 = 40 r 10 + 11r, x2 = 4 r 10 + 11r . The solution as a function of r is shown in Table 5.7.1. Table 5.7.1 Minimization of φ for diﬀerent penalty multipliers. r x1 x2 f φ 1 1.905 0.1905 3.992 7.619 10 3.333 0.3333 12.220 13.333 100 3.604 0.3604 14.288 14.144 1000 3.633 0.3633 14.518 14.532 It can be seen that as r is increased the solution converges to the exact solution of xT = (3.636, 0.3636), f = 14.54. The convergence is indicated by the shrinking diﬀerence between the objective function and the augmented function φ. The Hessian of φ is given as H = 2 + 2r 2r 2r 20 + 2r . As r increases this matrix becomes more and more ill-conditioned, as all four compo-nents become approximately 2r. This ill-conditioning of the Hessian matrix for large values of r often occurs when the exterior penalty function is used, and can cause numerical diﬃculties for large problems. We can use Table 5.7.1 to test the extrapolation procedure, Eq. (5.7.4). For example, with the values of r = 1 and r = 10, Eq. (5.7.5) gives a = 0.1x∗(1) −x∗(10) −0.9 = 3.492 0.3492 , 189 Chapter 5: Constrained Optimization b = x∗(1) −a = −0.159 −0.0159 . We can now use Eq. (5.7.4) to ﬁnd a starting point for the optimization for r = 100 to get a + b/100 = (3.490, 0.3490)T , which is substantially closer to x∗(100) = (3.604, 0.3604)T than to x∗(10) = (3.333, 0.3333)T. • • • 5.7.2 Interior and Extended Interior Penalty Functions With the exterior penalty function, constraints contribute penalty terms only when they are violated. As a result, the design typically moves in the infeasible domain. If the minimization is terminated before r becomes very large (for example, because of shortage of computer resources) the resulting designs may be useless. When only inequality constraints are present, it is possible to deﬁne an interior penalty function that keeps the design in the feasible domain. The common form of the interior penalty method replaces the inequality constrained problem minimize f(x) such that gj(x) ≥0, j = 1, . . . , ng , (5.7.10) by minimize φ(x, r) = f(x) + r ng X j=1 1/gj(x) , r = r1, r2, . . . , ri →0, ri > 0 . (5.7.11) Figure 5.7.2 Interior penalty function for f(x) = 0.5x subject to x −4 ≥0. 190 Section 5.7: Penalty Function Methods The penalty term is proportional to 1/gj and becomes inﬁnitely large at the boundary of the feasible domain creating a barrier there (interior penalty function methods are sometimes called barrier methods). It is assumed that the search is conﬁned to the feasible domain. Otherwise, the penalty becomes negative which does not make any sense. Figure 5.7.2 shows the application of the interior penalty function to the simple example used for the exterior penalty function in Figure 5.7.1. Besides the inverse penalty function deﬁned in Eq. (5.7.11), there has been some use of a logarithmic interior penalty function φ(x, r) = f(x) −r ng X j=1 log(gj(x)) . (5.7.12) While the interior penalty function has the advantage over the exterior one in that it produces a series of feasible designs, it also requires a feasible starting point. Unfortunately, it is often diﬃcult to ﬁnd such a feasible starting design. Also, because of the use of approximation (see Chapter 6), it is quite common for the optimization process to stray occasionally into the infeasible domain. For these reasons it may be advantageous to use a combination of interior and exterior penalty functions called an extended interior penalty function. An example is the quadratic extended interior penalty function of Haftka and Starnes φ(x, r) = f(x) + r ng X j=1 p(gj) , r = r1, r2, . . . , ri →0 , (5.7.13) where p(gj) = 1/gj for gj ≥g0 1/g0[3 −3(gj/g0) + (gj/g0)2] for gi < g0 . (5.7.14) It is easy to check that p(gj) has continuity up to second derivatives. The transi-tion parameter g0 which deﬁnes the boundary between the interior and exterior parts of the penalty terms must be chosen so that the penalty associated with the con-straint, rp(gj), becomes inﬁnite for negative gj as r tends to zero. This results in the requirement that r/g3 0 →∞, as r →0 . (5.7.15) This can be achieved by selecting g0 as g0 = cr1/2 , (5.7.16) where c is a constant. It is also possible to include equality constraints with interior and extended in-terior penalty functions. For example, the interior penalty function Eq. (5.7.11) is augmented as φ(x, r) = f(x) + r ng X j=1 1/gj(x) + r−1/2 ne X i=1 h2 i (x) , r = r1, r2, . . . , ri →0 . (5.7.17) 191 Chapter 5: Constrained Optimization Figure 5.7.3 Extended interior penalty function for f(x) = 0.5x subject to g(x) = x −4 ≥0. The considerations for the choice of an initial value of r are similar to those for the exterior penalty function. A reasonable choice for the interior penalty function would require that n/4 active constraints at g = 0.5 (that is 50% margin for properly normalized constraints) would result in a total penalty equal to the objective function. Using Eq. (5.7.11) we obtain f(x) = n 4 r 0.5, or r = 2f(x)/n . For the extended interior penalty function it is more reasonable to assume that the n/4 constraints are critical (g = 0), so that from Eq. (5.7.13) f(x) = rn 4 3 g0 , or r = 4 3g0f(x)/n . A reasonable starting value for g0 is 0.1. As for the exterior penalty function, it is possible to obtain an expression for the asymptotic (as r →0) coordinates of the minimum of φ as x∗(r) = a + br1/2, as r →0 , (5.7.18) and f ∗(r) = a + br1/2, as r →0 . a, b, a and b may be estimated once the minimization has been carried out for two values of r. For example, the estimates for a and b are a = c1/2x∗(ri−1) −x∗(ri) c1/2 −1 , b = x∗(ri−1) −a r1/2 i−1 , (5.7.19) where c = ri/ri−1. As in the case of exterior penalty function, these expressions may be used for convergence tests and extrapolation. 192 Section 5.7: Penalty Function Methods 5.7.3 Unconstrained Minimization with Penalty Functions Penalty functions convert a constrained minimization problem into an unconstrained one. It may seem that we should now use the best available methods for uncon-strained minimization, such as quasi-Newton methods. This may not necessarily be the case. The penalty terms cause the function φ to have large curvatures near the constraint boundary even if the curvatures of the objective function and constraints are small. This eﬀect permits an inexpensive approximate calculation of the Hessian matrix, so that we can use Newton’s method without incurring the high cost of cal-culating second derivatives of constraints. This may be more attractive than using quasi-Newton methods (where the Hessian is also approximated on the basis of ﬁrst derivatives) because a good approximation is obtained with a single analysis rather than with the n moves typically required for a quasi-Newton method. Consider, for example, an exterior penalty function applied to equality constraints φ(x, r) = f(x) + r ne X i=1 h2 i (x) . (5.7.20) The second derivatives of φ are given as ∂2φ ∂xk∂xl = ∂2f ∂xk∂xl + r ne X i=1 2 ∂hi ∂xk ∂hi ∂xl + hi ∂2hi ∂xk∂xl . (5.7.21) Because of the equality constraint, hi is close to zero, especially for the later stages of the optimization (large r), and we can neglect the last term in Eq. (5.7.21). For large values of r we can also neglect the ﬁrst term, so that we can calculate second derivatives of φ based on ﬁrst derivatives of the constraints. The availability of inexpensive second derivatives permits the use of Newton’s method where the number of iterations is typically independent of the number of design variables. Quasi-Newton and conjugate gradient methods, on the other hand, require a number of iterations proportional to the number of design variables. Thus the use of Newton’s method becomes attractive when the number of design variables is large. The application of Newton’s method with the above approximation of second derivatives is known as the Gauss-Newton method. For the interior penalty function we have a similar situation. The augmented objective function φ is given as φ(x, r) = f(x) + r ng X j=1 1/gj(x) , (5.7.22) and the second derivatives are ∂2φ ∂xk∂xl = ∂2f ∂xk∂xl + r ng X j=1 1 g3 j 2 ∂gj ∂xk ∂gj ∂xl −gj ∂2gj ∂xk∂xl . (5.7.23) 193 Chapter 5: Constrained Optimization Now the argument for neglecting the ﬁrst and last terms in Eq. (5.7.23) is somewhat lengthier. First we observe that because of the 1/g3 j term, the second derivatives are dominated by the critical constraints (gj small). For these constraints the last term in Eq. (5.7.23) is negligible compared to the ﬁrst-derivative term because gj is small. Finally, from Eq. (5.7.18) it can be shown that r/g3 j goes to inﬁnity for active constraints as r goes to zero, so that the ﬁrst term in Eq. (5.7.23) can be neglected compared to the second. The same argument can also be used for extended interior penalty functions . The power of the Gauss-Newton method is shown in for a high- aspect-ratio wing made of composite materials (see Figure 5.7.4) designed subject to stress and displacement constraints. Figure 5.7.4 Aerodynamic planform and structural box for high-aspect ratio wing, from . Table 5.7.2 Results of high-aspect-ratio wing study Number of CDC 6600 Total number of design CPU time unconstrained Total number Final variables sec minimizations of analyses mass, kg 13 142 4 21 887.3 25 217 4 19 869.1 32 293 5 22 661.7 50 460 5 25 658.2 74 777 5 28 648.6 146 1708 5 26 513.0 The structural box of the wing was modeled with a ﬁnite element model with 67 nodes and 290 ﬁnite elements. The number of design variables controlling the thickness of the various elements was varied from 13 to 146. The eﬀect of the number of design variables on the number of iterations (analyses) is shown in Table 5.7.2. It is seen that the number of iterations per unconstrained minimization is almost 194 Section 5.7: Penalty Function Methods constant (about ﬁve). With a quasi-Newton method that number may be expected to be similar to the number of design variables. Because of the sharp curvature of φ near the constraint boundary, it may also be appropriate to use specialized line searches with penalty functions . 5.7.4 Integer Programming with Penalty Functions An extension of the penalty function approach has been implemented by Shin et al. for problems with discrete-valued design variables. The extension is based on introduction of additional penalty terms into the augmented-objective function φ(x, r) to reﬂect the requirement that the design variables take discrete values, xi ∈Xi = {di1, di2, . . . , dil}, i ∈Id , (5.7.24) where Id is the set of design variables that can take only discrete values, and Xi is the set of allowable discrete values. Note that several variables may have the same allowable set of discrete values. In this case the augmented objective function which includes the penalty terms due to constraints and the non-discrete values of the design variables is deﬁned as φ(x, r, s) = f(x) + r ng X j=1 p(gj) + s X i∈Id ψd(xi) , (5.7.25) where s is a penalty multiplier for non-discrete values of the design variables, and ψd(xi) the penalty term for non-discrete values of the ith design variable. Diﬀerent forms for the discrete penalty function are possible. The penalty terms ψd(xi) are assumed to take the following sine-function form in Ref. , ψd(xi) = 1 2 sin 2π[xi −1 4(di(j+1) + 3dij)] di(j+1) −dij + 1 , dij ≤xi ≤di(j+1) . (5.7.26) While penalizing the non-discrete valued design variables, the functions ψd(xi) as-sure the continuity of the ﬁrst derivatives of the augmented function at the discrete values of the design variables. The response surfaces generated by Eq. (5.7.25) are determined according to the values of the penalty multipliers r and s. In contrast to the multiplier r, which initially has a large value and decreases as we move from one iteration to another, the value of the multiplier s is initially zero and increases gradually. One of the important factors in the application of the proposed method is to determine when to activate s, and how fast to increase it to obtain discrete optimum design. Clearly, if the initial value of s is too big and introduced too early in the design process, the design variables will be trapped away from the global minimum, resulting in a sub-optimal solution. To avoid this problem, the multiplier s has to be activated after optimization of several response surfaces which include only constraint penalty terms. In fact, since sometimes the optimum design with discrete values is in the neighborhood of the continuous optimum, it may be desirable not to activate 195 Chapter 5: Constrained Optimization the penalty for the non-discrete design variables until reasonable convergence to the continuous solution is achieved. This is especially true for problems in which the intervals between discrete values are very small. A criterion for the activation of the non-discrete penalty multiplier s is the same as the convergence criterion of Eq. (5.7.6), that is φ −f f ≤ϵc . (5.7.27) A typical value for ϵc is 0.01. The magnitude of the non-discrete penalty multiplier, s, at the ﬁrst discrete iteration is calculated such that the penalty associated with the discrete-valued design variables that are not at their allowed values is of the order of 10 percent of the constraint penalty. s ≈0.1rp(g) . (5.7.28) As the iteration for discrete optimization proceeds, the non-discrete penalty multiplier for the new iteration is increased by a factor of the order of 10. It is also important to decide how to control the penalty multiplier for the constraints, r, during the discrete optimization process. If r is decreased for each discrete optimization iteration as in the continuous optimization process, the design can be stalled due to high penalties for constraint violation. Thus, it is suggested that the penalty multiplier r be frozen at the end of the continuous optimization process. However, the nearest discrete solution at this response surface may not be a feasible design, in which case the design must move away from the continuous optimum by moving back to the previous response surface. This can be achieved by increasing the penalty multiplier, r, by a factor of 10. The solution process for the discrete optimization is terminated if the design variables are suﬃciently close to the prescribed discrete values. The convergence criterion for discrete optimization is max i∈Ii min \|xi −dij\| di(j+1) −dij , \|xi −di(j+1)\| di(j+1) −dij ≤ϵd , (5.7.29) where a typical value of the convergence tolerance ϵd is 0.001. Example 5.7.2 Cross-sectional areas of members of a two-bar truss shown in the Figure 5.7.5 are to be selected from a discrete set of values, Ai ∈{1.0, 1.5, 2.0}, i = 1, 2. Determine the minimum weight structure using the modiﬁed penalty function approach such that the horizontal displacement u at the point of application of the force does not exceed 2/3(Fl/E). Use a tolerance ϵc = 0.1 for the activation of the penalty terms for non-discrete valued design variables, and a convergence tolerance for the design variables ϵd = 0.001. 196 Section 5.7: Penalty Function Methods Figure 5.7.5 Two-bar truss. Upon normalization, the design problem is posed as minimize f = W ρl = x1 + x2 subject to g = uE Fl = 1.5 −1/x1 −1/x2 ≥0 , xi = Ai ∈{1.0, 1.5, 2.0}, i = 1, . . . , 2 . Using an initial design of x1 = x2 = 5 and transition parameter g0 = 0.1, we have g = 1.1 > g0, therefore, from Eq. (5.7.14) the penalty terms for the constraints are in the form of p(g) = 1/g. The augmented function for the extended interior penalty function approach is φ = x1 + x2 + r 1.5 −1/x1 −1/x2 . Setting the gradient to zero, we can show that the minimum of the augmented func-tion as a function of the penalty multiplier r is x1 = x2 = 24 + p 576 −36 (16 −4r) 18 . The initial value of the penalty multiplier r is chosen so that the penalty introduced for the constraint is equal to the objective function value, r 1 g(x0) = f(x0), r = 11 . The minima of the augmented function as functions of the penalty multiplier r are shown in Table 5.7.3 . After four iterations the constraint penalty (φ −f) is within the desired range of the objective function to activate the penalty terms for the non-discrete values of the design variables. From Eq. (5.7.25) the augmented function for the modiﬁed penalty function approach has the form φ =x1 + x2 + r 1.5 −1/x1 −1/x2 + s {1 + sin[4π (x1 −1.125)]} 2 +(s/2) {1 + sin[4π (x2 −1.125)]} . 197 Chapter 5: Constrained Optimization Table 5.7.3 Minimization of φ without the discrete penalty r x1 x2 f g φ -5.000 5.000 10.00 1.100 -11 3.544 3.544 7.089 0.9357 18.844 1.1 2.033 2.033 4.065 0.5160 6.197 0.11 1.554 1.554 3.109 0.2134 3.624 0.011 1.403 1.403 2.807 0.0747 2.954 The minimum of the augmented function can again be obtained by setting the gra-dient to zero 1 − r (1.5 −2/x1)2x12 + 2πs cos[4π (x1 −1.125)] = 0 , which can be solved numerically. The initial value of the penalty multiplier s is calculated from Eq. (5.7.28) s = 0.1 (0.011) 1 0.0747 = 0.0147 . The minima of the augmented function (which includes the penalty for the non-discrete valued variables) are shown in Table 5.7.4 as a function of s. Table 5.7.4 Minimization of φ with the discrete penalty r s x1 x2 f φ 0.011 0.0147 1.406 1.406 2.813 2.963 0.1472 1.432 1.432 2.864 3.021 1.472 1.493 1.493 2.986 3.060 14.72 1.499 1.499 2.999 3.065 147.2 1.500 1.500 3.000 3.066 After four discrete iterations we obtain a minimum at x1 = x2 = 3/2. There are two more minima, x = (2, 1) and x = (1, 2), with the same value of the objective function of f = 3.0. • • • 5.8 Multiplier Methods Multiplier methods combine the use of Lagrange multipliers with penalty functions. When only Lagrange multipliers are employed the optimum is a stationary point rather than a minimum of the Lagrangian function. When only penalty functions are employed we have a minimum but also ill-conditioning. By using both we may hope to get an unconstrained problem where the function to be minimized does not suﬀer from ill-conditioning. A good survey of multiplier methods was conducted by 198 Section 5.8: Multiplier Methods Bertsekas . We study ﬁrst the use of multiplier methods for equality constrained problems. minimize f(x) such that hj(x) = 0, j = 1, . . . , ne . (5.8.1) We deﬁne the augmented Lagrangian function L(x, λ, r) = f(x) − ne X j=1 λjhj(x) + r ne X j=1 h2 j(x) . (5.8.2) If all the Lagrange multipliers are set to zero, we get the usual exterior penalty function. On the other hand, if we use the correct values of the Lagrange multipliers, λ∗ j, it can be shown that we get the correct minimum of problem (5.8.1) for any positive value of r. Then there is no need to use the large value of r required for the exterior penalty function. Of course, we do not know what are the correct values of the Lagrange multipliers. Multiplier methods are based on estimating the Lagrange multipliers. When the estimates are good, it is possible to approach the optimum without using large r values. The value of r needs to be only large enough so that L has a minimum rather than a stationary point at the optimum. To obtain an estimate for the Lagrange multipliers we compare the stationarity conditions for L, ∂L ∂xi = ∂f ∂xi − ne X j=1 (λj −2rhj)∂hj ∂xi = 0 , (5.8.3) with the exact conditions for the Lagrange multipliers ∂f ∂xi − ne X j=1 λ∗ j ∂hj ∂xi = 0 . (5.8.4) Comparing Eqs. (5.8.3) and (5.8.4) we expect that λj −2rhj →λ∗ j , (5.8.5) as the minimum is approached. Based on this relation, Hestenes suggested using Eq. (5.8.5) as an estimate for λ∗ j. That is λ(k+1) j = λ(k) j −2r(k)h(k) j , (5.8.6) where k is an iteration number. 199 Chapter 5: Constrained Optimization Example 5.8.1 We repeat Example 5.7.1 using Hestenes’ multiplier method. f(x) = x2 1 + 10x2 2 , h(x) = x1 + x2 −4 = 0 . The augmented Lagrangian is L = x2 1 + 10x2 2 −λ(x1 + x2 −4) + r(x1 + x2 −4)2 . To ﬁnd the stationary points of the augmented Lagrangian we diﬀerentiate with respect to x1 and x2 to get 2x1 −λ + 2r(x1 + x2 −4) = 0 , 20x2 −λ + 2r(x1 + x2 −4) = 0 , which yield x1 = 10x2 = 5λ + 40r 10 + 11r . We want to compare the results with those of Example 5.7.1, so we start with the same initial r value r0 = 1, the initial estimate of λ = 0 and get x1 = (1.905, 0.1905)T, h = −1.905 . So, using Eq. (5.8.6) we estimate λ(1) as λ(1) = −2 × 1 × (−1.905) = 3.81 . We next repeat the optimization with r(1) = 10, λ(1) = 3.81 and get x2 = (3.492, 0.3492)T, h = −0.1587 . For the same value of r, we obtained in Example 5.7.1 x2 = (3.333, 0.3333)T, so that we are now closer to the exact solution of x = (3.636, 0, 3636)T. Now we estimate a new λ from Eq. (5.8.6) λ(2) = 3.81 −2 × 10 × (−0.1587) = 6.984 . For the next iteration we may, for example, ﬁx the value of r at 10 and change only λ. For λ = 6.984 we obtain x3 = (3.624, 0.3624), h = −0.0136 , which shows that good convergence can be obtained without increasing r.• • • 200 Section 5.9: Projected Lagrangian Methods (Sequential Quadratic Programming) There are several ways to extend the multiplier method to deal with inequality constraints. The formulation below is based on Fletcher’s work . The constrained problem that we examine is minimize f(x) such that gj(x) ≥0, j = 1, . . . , ng . (5.8.7) The augmented Lagrangian function is L(x, λ, r) = f(x) + r ng X j=1 ⟨λj 2r −gj⟩2 , (5.8.8) where < a >= max(a, 0). The condition of stationarity of L is ∂f ∂xi −2r ng X j=1 ⟨λj 2r −gj⟩∂gj ∂xi = 0 . (5.8.9) The exact stationarity condition is ∂f ∂xi − ng X j=1 λ∗ j ∂gj ∂xi = 0 , (5.8.10) where it is also required that λ∗ jgj = 0. Comparing Eqs (5.8.9) and (5.8.10) we expect an estimate for λ∗ j of the form λ∗ j = max(λj −2rgj, 0) . (5.8.11) 5.9 Projected Lagrangian Methods (Sequential Quadratic Programming) The addition of penalty terms to the Lagrangian function by multiplier methods converts the optimum from a stationary point of the Lagrangian function to a min-imum point of the augmented Lagrangian. Projected Lagrangian methods achieve the same result by a diﬀerent method. They are based on a theorem that states that the optimum is a minimum of the Lagrangian function in the subspace of vectors orthogonal to the gradients of the active constraints (the tangent subspace). Pro-jected Lagrangian methods employ a quadratic approximation to the Lagrangian in this subspace. The direction seeking algorithm is more complex than for the methods considered so far. It requires the solution of a quadratic programming problem, that is an optimization problem with a quadratic objective function and linear constraints. Projected Lagrangian methods are part of a class of methods known as sequential quadratic programming (SQP)methods. The extra work associated with the solution of the quadratic programming direction seeking problem is often rewarded by faster convergence. 201 Chapter 5: Constrained Optimization The present discussion is a simpliﬁed version of Powell’s projected Lagrangian method . In particular we consider only the case of inequality constraints minimize f(x) such that gj(x) ≥0, j = 1, . . . , ng . (5.9.1) Assume that at the ith iteration the design is at xi, and we seek a move direction s. The direction s is the solution of the following quadratic programming problem minimize φ(s) = f(xi) + sTg(xi) + 1 2sTA(xi, λi)s such that gj(xi) + sT∇gj(xi) ≥0, j = 1, . . . , ng , (5.9.2) where g is the gradient of f, and A is a positive deﬁnite approximation to the Hessian of the Lagrangian function discussed below. This quadratic programming problem can be solved by a variety of methods which take advantage of its special nature. The solution of the quadratic programming problem yields s and λi+1. We then have xi+1 = xi + αs , (5.9.3) where α is found by minimizing the function ψ(α) = f(x) + ng X j=1 µj\|min(0, gj(x))\| , (5.9.4) and the µj are equal to the absolute values of the Lagrange multipliers for the ﬁrst iteration, i.e. µj = max[\|λ(i) j , 1 2(µ(i−1) j + \|λ(i−1) j \|)] , (5.9.5) with the superscript i denoting iteration number. The matrix A is initialized to some positive deﬁnite matrix (e.g the identity matrix) and then updated using a BFGS type equation (see Chapter 4). Anew = A −A∆x∆xTA ∆xTA∆x + ∆l∆lT ∆xT∆x , (5.9.6) where ∆x = xi+1 −xi , ∆l = ∇xL(xi+1, λi) −∇xL(xi, λi) , (5.9.7) where L is the Lagrangian function and ∇x denotes the gradient of the Lagrangian function with respect to x. To guarantee the positive deﬁniteness of A, ∆l is modiﬁed if ∆xT∆l ≤0.2∆xTA∆x and replaced by ∆l′ = θ∆l + (1 −θ)A∆x , (5.9.8) where θ = 0.8∆xTA∆x ∆xTA∆x −∆xT∆l . (5.9.9) 202 Section 5.9: Projected Lagrangian Methods (Sequential Quadratic Programming) Example 5.9.1 Consider the four bar truss of Example 5.1.2. The problem of ﬁnding the minimum weight design subject to stress and displacement constraints was formulated as minimize f = 3x1 + √ 3x2 subject to g1 = 3 −18 x1 −6 √ 3 x2 ≥0 , g2 = x1 −5.73 ≥0 , g3 = x2 −7.17 ≥0 . Assume that we start the search at the intersection of g1 = 0 and g3 = 0 where x1 = 11.61, x2 = 7.17 and f = 47.25. The gradient of the objective function and two active constraints are ∇f = 3 √ 3 , ∇g1 = 0.1335 0.2021 , ∇g3 = 0 1 , N = 0.1335 0 0.2021 1 . We start with A set to the unit matrix so that φ(s) = 47.25 + 3s1 + √ 3s2 + 0.5s2 1 + 0.5s2 2 , and the linearized constraints are g1(s) = 0.1335s1 + 0.2021s2 ≥0 , g2(s) = 5.88 + s1 ≥0 , g3(s) = s2 ≥0 . We solve this quadratic programming problem directly with the use of the Kuhn-Tucker conditions 3 + s1 −0.1335λ1 −λ2 = 0 , √ 3 + s2 −0.2021λ1 −λ3 = 0 . A consideration of all possibilities for active constraints shows that the optimum is obtained when only g1 is active, so that λ2 = λ3 = 0 and λ1 = 12.8, s1 = −1.29, s2 = 0.855. The next design is x1 = 11.61 7.17 + α −1.29 0.855 , where α is found by minimizing ψ(α) of Eq. (5.9.4). For the ﬁrst iteration µj = \|λj\| so ψ = 3(11.61−1.29α)+ √ 3(7.17+0.855α)+12.8 3 − 18 11.61 −1.29α − 6 √ 3 7.17 + 0.855α . 203 Chapter 5: Constrained Optimization By changing α systematically we ﬁnd that ψ is a minimum near α = 2.2, so that x1 = (8.77, 9.05)T, f(x1) = 41.98, g1(x1) = −0.201 . To update A we need ∆x and ∆l. We have L = 3x1 + √ 3x2 −12.8(3 −18/x1 + 6 √ 3/x2) , so that ∇xL = (3 −230.4/x2 1, √ 3 −133.0/x2 2)T , and ∆x = x1 −x0 = −2.84 1.88 , ∆l = ∇xL(x1) −∇xL(x0) = −1.31 0.963 . With A being the identity matrix we have ∆xTA∆x = 11.6, ∆xT∆l = 5.53. Because ∆xT∆l > 0.2∆xTA∆x we can use Eq. (5.9.5) to update A Anew = I −∆x∆xT ∆xT∆x + ∆l∆lT ∆xT∆x = 0.453 0.352 0.352 0.775 . For the second iteration φ(s) = 41.98 + 3s1 + √ 3s2 + 0.5(0.453s2 1 + 0.775s2 2 + 0.704s1s2) , g1(s) = −0.201 + 0.234s1 + 0.127s2 ≥0 , g2(s) = 3.04 + s1 ≥0 , g3(s) = 1.88 + s2 ≥0 . We can again solve the quadratic programming directly with the use of the Kuhn-Tucker conditions 3 + 0.453s1 + 0.352s2 −0.234λ1 −λ2 = 0 , √ 3 + 0.352s1 + 0.775s2 −0.127λ1 −λ3 = 0 . The solution is λ1 = 14.31, λ2 = λ3 = 0, s1 = 1.059, s2 = −0.376 . The one dimensional search seeks to minimize ψ(α) = f(α) + µ1g1(α) , where µ1 = max(λ1, 1 2(\|λ1\| + µold 1 )) = 14.31 . The one-dimensional search yields approximately α = 0.5, so that x2 = (9.30, 8.86)T, f(x2) = 43.25, g1(x2) = −0.108 , so that we have made good progress towards the optimum x∗= (9.46, 9.46)T. • • • 204 Section 5.11: References 5.10 Exercises 1. Check the nature of the stationary points of the constrained problem minimize f(x) = x2 1 + 4x2 2 + 9x2 3 such that x1 + 2x2 + 3x3 ≥30 , x2x3 ≥2 , x3 ≥4 , x1x2 ≥0 . 2. For the problem minimize f(x) = 3x2 1 −2x1 −5x2 2 + 30x2 such that 2x1 + 3x2 ≥8 , 3x1 + 2x2 ≤15 , x2 ≤5 . Check for a minimum at the following points: (a) (5/3, 5.00) (b) (1/3, 5.00) (c) (3.97,1.55). 3. Calculate the derivative of the solution of Example 5.1.2 with respect to a change in the allowable displacement. First use the Lagrange multiplier to obtain the derivative of the objective function, and then calculate the derivatives of the design variables and Lagrange multipliers and verify the derivative of the objective function. Finally, estimate from the derivatives of the solution how much we can change the allowable displacement without changing the set of active constraints. 4. Solve for the minimum of problem 1 using the gradient projection method from the point (17, 1/2, 4). 5. Complete two additional moves in Example 5.5.2. 6. Find a feasible usable direction for problem 1 at the point (17, 1/2, 4). 7. Use an exterior penalty function to solve Example 5.1.2. 8. Use an interior penalty function to solve Example 5.1.2. 9. Consider the design of a box of maximum volume such that the surface area is equal to S and there is one face with an area of S/4. Use the method of multipliers to solve this problem, employing three design variables. 10. Complete two more iterations in Example 5.9.1. 205 Chapter 5: Constrained Optimization 5.11 References Kreisselmeier, G., and Steinhauser, R., “Systematic Control Design by Optimiz-ing a Vector Performance Index,”Proceedings of IFAC Symposium on Computer Aided Design of Control Systems, Zurich, Switzerland, pp. 113-117, 1979. Sobieszczanski-Sobieski, J., “A Technique for Locating Function Roots and for Satisfying Equality Constraints in Optimization,” NASA TM-104037, NASA LaRC, 1991. Wolfe, P.. “The Simplex Method for Quadratic Programming,” Econometrica, 27 (3), pp. 382–398, 1959. Gill, P.E., Murray, W., and Wright, M.H., Practical Optimization, Academic Press, 1981. Dahlquist, G., and Bjorck, A., Numerical Methods, Prentice Hall, 1974. Sobieszczanski-Sobieski, J., Barthelemy, J.F., and Riley, K.M., “Sensitivity of Optimum Solutions of Problem Parameters”, AIAA Journal, 20 (9), pp. 1291– 1299, 1982. Rosen, J.B., “The Gradient Projection Method for Nonlinear Programming— Part I: Linear Constraints”, The Society for Industrial and Appl. Mech. Journal, 8 (1), pp. 181– 217, 1960. Abadie, J., and Carpentier, J., “Generalization of the Wolfe Reduced Gradient Method for Nonlinear Constraints”, in: Optimization (R. Fletcher, ed.), pp. 37– 49, Academic Press, 1969. Rosen, J.B., “The Gradient Projection Method for Nonlinear Programming—Part II: Nonlinear Constraints”, The Society for Industrial and Appl. Mech. Journal, 9 (4), pp. 514–532, 1961. Haug, E.J., and Arora, J.S., Applied Optimal Design: Mechanical and Structural Systems, John Wiley, New York, 1979. Zoutendijk, G., Methods of Feasible Directions, Elsevier, Amsterdam, 1960. Vanderplaats, G.N., “CONMIN—A Fortran Program for Constrained Function Minimization”, NASA TM X-62282, 1973. Fiacco, V., and McCormick, G.P., Nonlinear Programming: Sequential Uncon-strained Minimization Techniques, John Wiley, New York, 1968. Haftka, R.T., and Starnes, J.H., Jr., “Applications of a Quadratic Extended Interior Penalty Function for Structural Optimization”, AIAA Journal, 14 (6), pp.718–724, 1976. Moe, J., “Penalty Function Methods in Optimum Structural Design—Theory and Applications”, in: Optimum Structural Design (Gallagher and Zienkiewicz, eds.), pp. 143–177, John Wiley, 1973. 206 Section 5.11: References Shin, D.K, G¨ urdal, Z., and Griﬃn, O. H. Jr., “A Penalty Approach for Nonlinear Optimization with Discrete Design Variables,” Engineering Optimization, 16, pp. 29–42, 1990. Bertsekas, D.P., “Multiplier Methods: A Survey,” Automatica, 12, pp. 133–145, 1976. Hestenes, M.R., “Multiplier and Gradient Methods,” Journal of Optimization Theory and Applications, 4 (5), pp. 303–320, 1969. Fletcher, R., “An Ideal Penalty Function for Constrained Optimization,” Journal of the Institute of Mathematics and its Applications, 15, pp.319–342, 1975. Powell, M.J.D., “A Fast Algorithm for Nonlinearly Constrained Optimization Calculations”, Proceedings of the 1977 Dundee Conference on Numerical Analy-sis, Lecture Notes in Mathematics, Vol. 630, pp. 144–157, Springer-Verlag, Berlin, 1978. 207 Chapter 5: Constrained Optimization 208
15086	https://www.linkedin.com/pulse/why-shape-triangle-widely-used-structural-engineering-parishith-jayan	Why the shape "TRIANGLE" is widely used in structural engineering? Agree & Join LinkedIn By clicking Continue to join or sign in, you agree to LinkedIn’s User Agreement, Privacy Policy, and Cookie Policy. Welcome back Email or phone Password Show Forgot password? Sign in or By clicking Continue to join or sign in, you agree to LinkedIn’s User Agreement, Privacy Policy, and Cookie Policy. New to LinkedIn? Join now Skip to main contentLinkedIn Articles People Learning Jobs Games Join nowSign in Why the shape "TRIANGLE" is widely used in structural engineering? Report this article Parishith Jayan Parishith Jayan Engineering Head - Design \| Steel Structures \| Engineering Software Developer Published Aug 9, 2020 + Follow Most of the structures that we know comprise of this particular shape "triangle" in any one of its structural systems. What is so important? or why "TRIANGLES"? Just think of basic 2-dimensional geometrical shapes, what do we have? Squares, Rectangles, Triangles, circles, pentagon, hexagon, and the list go on. Of these shapes, why a triangle is called the most stable shape. Let us compare these shapes and understand them. For our example, let's consider the square and the triangle as shown below. Applying a force on one of the edges and studying the behavior would lead to a better understanding regarding the stability of the structure. First, let’s try with the vertical force. What will happen? The load applied over the square moves directly downward through the vertical member and stands still. The triangle when subjected to a vertical force through one of its edges, distributes that force evenly to either side and stands stable. Now, let’s apply the horizontal force. The square could not withstand this force without deformation, it tries to sway along the direction of application of the force and distort from its original shape. Whereas, applying the horizontal force tries to push one edge of the triangle. It is not possible to move the edge without the elongation of one of its sides, this makes them stay stable for the applied force. This simple comparison applies to all the other shapes too. Also, this can be easily verified by making a simple model with popsicles. I tried it myself, and recommend you to do so. To read the extended version that emphasizes the application of triangles in the structural system, go through the link below. 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15087	https://math.wvu.edu/~hlai2/Teaching/Tip-Pdf/Tip3-27.pdf	Compute partial derivatives with Chain Rule Formulae: These are the most frequently used ones: 1. If w = f(x, y) and x = x(t) and y = y(t) such that f, x, y are all diﬀerentiable. Then dw dt = ∂w ∂x dx dt + ∂w ∂y dy dt . (1) 2. If w = f(x1, x2, · · · , xm) and for each i, (1 ≤i ≤n), xi = xi(t1, t2, · · · , tn) such that f, x1, · · · , xm are all diﬀerentiable. Then ∂w ∂ti = ∂w ∂x1 ∂x1 ∂ti + · · · + ∂w ∂xm ∂xm ∂ti . (2) Example (1) : Given w = ln(u + v + z), with u = cos2 t, v = sin2 t and z = t2, ﬁnd dw/dt both by using the chain rule and by expressing w explicitly as a function of t before diﬀerentiating. Solution: First we apply Chain Rule (1): dw dt = ∂w ∂u du dt + ∂w ∂v dv dt + ∂w ∂z dz dt = 1 u + v + z (−2 cos t sin t + 2 sin t cos t + 2t) = 2t u + v + z . Next, we express w as a function of t by substituting u, v and z into w to get w = ln(cos2 t+ sin2 t + t2) = ln(1 + t2). Therefore, dw/dt = 2t/(1 + t2). Example (2) : Given w = yz + zx + xy, x = s2 −t2, y = s2 + t2 and z = s2t2, ﬁnd ∂w ∂s and ∂w ∂t . Solution: This is a partial derivative problem and so we apply Chain Rule (2). ∂w ∂s = ∂w ∂x ∂x ∂s + ∂w ∂y ∂y ∂s + ∂w ∂z ∂z ∂s = (z + y)(2s) + (z + x)2s + (x + y)2st2 ∂w ∂t = ∂w ∂x ∂x ∂t + ∂w ∂y ∂y ∂t + ∂w ∂z ∂z ∂t = −(z + y)(2t) + (z + x)2t + (x + y)2s2t Example (3) : Given p = f(x, y, z), x = x(u, v), y = y(u, v) and z = z(u, v), write the chain rule formulas giving the partial derivatives of the dependent variable p with respect to each independent variable. Solution: This is a partial derivative problem and so we apply Chain Rule (2). ∂p ∂u = ∂f ∂x ∂x ∂u + ∂f ∂y ∂y ∂u + ∂f ∂z ∂z ∂u ∂p ∂v = ∂f ∂x ∂x ∂v + ∂f ∂y ∂y ∂v + ∂f ∂z ∂z ∂v Example (4) : Given x3 + y3 + z3 = xyz, ﬁnd ∂z ∂x and ∂z ∂y as functions of x, y and z. Solution: As z is an implicit function of x and y, implicit diﬀerentiation must be used. Just view z = z(x, y) everywhere z occurs when we diﬀerentiate both sides of the equation. (Step 1) View z = z(x, y) and diﬀerentiate both sides of the equation with respect to x to get 3x2 + 3z2 ∂z ∂x = yz + xy ∂z ∂x. (Step 2) Solve for ∂z ∂x in the resulting equation above. zx = ∂z ∂x = yz −3x2 3z2 −xy. Do the same for y. (Step 1) View z = z(x, y) and diﬀerentiate both sides of the equation with respect to y to get 3y2 + 3z2 ∂z ∂y = xz + xy ∂z ∂y. (Step 2) Solve for ∂z ∂y in the resulting equation above. zy = ∂z ∂y = xz −3y2 3z2 −xy. Example (5) : Given x3 + y3 + z3 = xyz as the equation of a surface, ﬁnd an equation of the plane tangent to this surface at the point P(1, −1, −1). Solution 1: View z = z(x, y), and so the vector n = (zx, zy, −1) at P will be a normal vector of the tangent plane. From Example (4) above, we have, at P(1, −1, −1), zx = yz −3x2 3z2 −xy (x,y,z)=(1,−1,−1) = (−1)(−1) −3 3(−1)2 −1(−1) = −2 4 = −1 2, zy = xz −3y2 3z2 −xy (x,y,z)=(1,−1,−1) = 1(−1) −3 3(−1)2 −1(−1) = −4 4 = −1. Hence the equation of the tangent plane is −1 2(x −1) −(y + 1) −(z + 1) = 0, or (x −1) + 2(y + 1) + 2(z + 1) = 0. Solution 2: One can also set F(x, y, z) = x3 + y3 + z3 −xyz and view the equation of the surface is F(x, y, z) = 0. In this case, the vector u = (Fx, Fy, Fz) at P(1, −1, −1) can be a normal vector of the tangent plane. We compute the partial derivatives: Fx = ∂F ∂x = 3x2 −yz, Fy = ∂F ∂y = 3y2 −xz, Fz = ∂F ∂z = 3z2 −xy. Therefore, u = (2, 4, 4) and so an equation of the tangent plane is 2(x −1) + 4(y + 1) + 4(z + 1) = 0, or (x −1) + 2(y + 1) + 2(z + 1) = 0. Example (6) : Suppose that w = f(u) and that u = x + y. Show that ∂w ∂x = ∂w ∂y . Solution: Notice that ∂u ∂x = 1 = ∂u ∂y . By Chain Rule (2), ∂w ∂x = ∂w ∂u ∂u ∂x = ∂w ∂u ∂u ∂y = ∂w ∂y . Example (7) : Suppose that w = f(x, y) and that x = r cos θ and y = r sin θ. Show that ∂2w ∂x2 + ∂2w ∂y2 = ∂2w ∂r2 + 1 r ∂w ∂r + 1 r2 ∂2w ∂θ2 . Solution: Apply Chain Rule (2) to compute the ﬁrst order of partial derivatives ∂w ∂r = ∂w ∂x ∂x ∂r + ∂w ∂y ∂y ∂r = wx cos θ + wy sin θ ∂w ∂θ = ∂w ∂x ∂x ∂θ + ∂w ∂y ∂y ∂θ = −wxr sin θ + wyr cos θ. Then, we apply Chain Rule (2) again to compute the second order of partial derivatives (making use of wxy = wyx), ∂2w ∂r2 = wxx cos2 θ + wxy cos θ sin θ + wyx sin θ cos θ + wyy sin2 θ (3) = wxx cos2 θ + (wxy + wyx) cos θ sin θ + wyy sin2 θ, and ∂2w ∂θ2 = wxxr2 sin2 θ −wxyr2 sin θ cos θ −wxr cos θ + wyyr2 cos2 θ + wyxr2 cos θ(−sin θ) −wyr sin θ = wxxr2 sin2 θ + wyyr2 cos2 θ −(wxy + wyx)r2 sin θ cos θ −wxr cos θ −wyr sin θ. It follows that 1 r2 ∂2w ∂θ2 = wxx sin2 θ + wyy cos2 θ −(wxy + wyx) sin θ cos θ −wx cos θ + wy sin θ r (4) 1 r ∂w ∂r = wx cos θ + wy sin θ r . (5) Now add equations (3), (4) and (5) side by side and apply sin2 θ + cos2 θ = 1 to get the conclusion.
15088	https://www.echemi.com/community/what-chemical-do-we-use-to-prove-that-carbon-dioxide-is-produced-during-the-reaction-of-hydrochloric-acid-and-calcium-carbonate_mjart2204108640_987.html	What chemical do we use to prove that carbon dioxide is produced during the reaction of hydrochloric acid and calcium carbonate? - ECHEMI  Community  Sign in \| Join free Home>Community>What chemical do we use to prove that carbon dioxide is produced during the reaction of hydrochloric acid and calcium carbonate? Upvote 16 Downvote Calcium carbonate Acids Chemistry Carbon dioxide Calcium Hydrochloric acid Posted by Michael Flynn What chemical do we use to prove that carbon dioxide is produced during the reaction of hydrochloric acid and calcium carbonate? The reaction between Hydrochloric Acid (HCl) and Calcium Carbonate is: CaCO3+2 HCl=====>CaCl2solution+ CO2gas + H2O The reaction is immediate and characterized by strong effervescent due to the evolving of CO2 gas in the air. There are several ways for detecting CO2 gas. The best way is to use Lime Water ( Ca(OH)2). One has to collect CO2 gas in a test tube or in a conical flask. The addition of Lime Water to the collected gas produces a cloudy ,milky and white precipitate of calcium carbonate indicating the presence of CO2 gas , as shown in e following equation: Ca(OH)2solution+ CO2gas======> CaCO3+H2O Cheating - Spouse/PartnerFollowFollowing The reaction between Hydrochloric Acid (HCl) and Calcium Carbonate is: CaCO3+2 HCl=====>CaCl2solution+ CO2gas + H2O The reaction is immediate and characterized by strong effervescent due to the evolving of CO2 gas in the air. There are several ways for detecting CO2 gas. The best way is to use Lime Water ( Ca(OH)2). One has to collect CO2 gas in a test tube or in a conical flask. The addition of Lime Water to the collected gas produces a cloudy ,milky and white precipitate of calcium carbonate indicating the presence of CO2 gas , as shown in e following equation: Ca(OH)2solution+ CO2gas======> CaCO3+H2O More Upvote VOTE Downvote Usely we use lime water to detect carbon dioxide. When a gas mixture with carbon dioxide vomes in contact with lime water you can see after a while a white clouding in the solution. This with clouding is CaCO3 which is slightly soluble in water. The reaction is Ca2+(aq) + 2 OH-(aq) + CO2(aq) → CaCO3(aq) + H2O(aq) Andrew MartinFollowFollowing Usely we use lime water to detect carbon dioxide. When a gas mixture with carbon dioxide vomes in contact with lime water you can see after a while a white clouding in the solution. This with clouding is CaCO3 which is slightly soluble in water. The reaction is Ca2+(aq) + 2 OH-(aq) + CO2(aq) → CaCO3(aq) + H2O(aq) More Upvote VOTE Downvote Related Posts How many ml are in 1 glass of water? 26 Upvotes · 2 Comments What is the best solution for cleaning a meth pipe? 15 Upvotes · 3 Comments What is dicyanin, and what is it used for? 14 Upvotes · 3 Comments How do you prepare 0.5M of an HCl solution? 14 Upvotes · 8 Comments Is it safe to use expired salt packets from neti pots for nasal irrigation? 19 Upvotes · 2 Comments How do I prepare a 1M KOH solution? 16 Upvotes · 8 Comments How do I make a 1N H2SO4 solution? 15 Upvotes · 10 Comments Wholesale About ECHEMI Encyclopedia Contact Us Local Mall Partners Market Price & Insight Marketing Kit Trade Data ECHEMI Group Join Us News Sitemap For Buyers For Suppliers Copyright@Qingdao ECHEMI Digital Technology Co., Ltd. TOP
15089	https://fiveable.me/organic-chem/unit-27/waxes-fats-oils/study-guide/BKPsQTK67d8COw8p	Waxes, Fats, and Oils \| Organic Chemistry Class Notes \| Fiveable \| Fiveable new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom ap study content toolsprintablespricing my subjectsupgrade 🥼Organic Chemistry Unit 27 Review 27.1 Waxes, Fats, and Oils All Study Guides Organic Chemistry Unit 27 – Biomolecules – Lipids Topic: 27.1 🥼Organic Chemistry Unit 27 Review 27.1 Waxes, Fats, and Oils Written by the Fiveable Content Team • Last updated September 2025 Written by the Fiveable Content Team • Last updated September 2025 print study guide copy citation APA 🥼Organic Chemistry Unit & Topic Study Guides Structure and Bonding Polar Covalent Bonds; Acids and Bases Alkanes: Structure and Stereochemistry Cycloalkanes: Structure and Stereochemistry Stereochemistry at Tetrahedral Centers An Overview of Organic Reactions Alkenes Alkenes Alkynes: Intro to Organic Synthesis Organohalides Alkyl Halide Reactions: Substitutions & Eliminations Mass Spec and IR Spectroscopy in Organic Chem NMR Spectroscopy for Structure Determination Conjugated Systems and UV Spectroscopy Benzene and Aromaticity Benzene: Electrophilic Aromatic Substitution Alcohols and Phenols Ethers, Epoxides, Thiols, and Sulfides Aldehydes & Ketones: Nucleophilic Addition Carboxylic Acids and Nitriles Carboxylic Acid Derivatives: Acyl Substitution Carbonyl Alpha–Substitution Reactions Carbonyl Condensation Reactions Amines and Heterocycles Biomolecules Biomolecules: Amino Acids & Proteins Biomolecules 27.1 Waxes, Fats, and Oils 27.2 Soap 27.3 Phospholipids 27.4 Prostaglandins and Other Eicosanoids 27.5 Terpenoids 27.6 Steroids 27.7 Biosynthesis of Steroids Biomolecules Metabolic Pathways in Organic Chemistry Organic Chemistry: Pericyclic Reactions Synthetic Polymers print guide report error Waxes, fats, and oils are lipids with unique chemical compositions. Waxes are esters of long-chain fatty acids and alcohols, while fats and oils are triesters of glycerol and fatty acids. Their structures determine their properties and behaviors in various applications. Fatty acids can be saturated or unsaturated, affecting their melting points and physical states. Hydrogenation of vegetable oils alters their properties, increasing melting points and stability. Understanding lipid behavior is crucial for industries like food production and cosmetics. Chemical Composition and Properties of Waxes, Fats, and Oils Chemical composition of lipids Waxes composed of esters formed from long-chain fatty acids and long-chain alcohols Carbon chains in waxes range from 12 to 34 carbon atoms in length Common examples include beeswax, carnauba wax, and paraffin wax Fats and oils are triesters of glycerol (a trihydric alcohol) and fatty acids Fatty acids are long-chain carboxylic acids typically containing 12 to 18 carbon atoms Fats are solid at room temperature (butter), while oils are liquid at room temperature (olive oil) Triglyceride structure consists of a glycerol backbone with three fatty acid chains attached via ester linkages Properties of fatty acids Saturated fatty acids contain only single bonds between carbon atoms in the hydrocarbon chain Examples include palmitic acid (C16:0) and stearic acid (C18:0) Exhibit higher melting points due to the ability to pack closely together, resulting in stronger intermolecular forces Typically solid at room temperature (coconut oil) Unsaturated fatty acids contain one or more double bonds between carbon atoms in the hydrocarbon chain Examples include oleic acid (C18:1, one double bond) and linoleic acid (C18:2, two double bonds) Exhibit lower melting points due to the presence of double bonds, which create kinks in the hydrocarbon chain and prevent close packing Typically liquid at room temperature (canola oil) The degree of unsaturation determines the melting point and physical state of a fatty acid The more double bonds present, the lower the melting point and the more likely it is to be liquid at room temperature Hydrogenation of vegetable oils Hydrogenation process involves adding hydrogen atoms to unsaturated fatty acids, converting double bonds to single bonds Carried out in the presence of a metal catalyst (typically nickel) at high temperatures and pressures Partial hydrogenation results in some double bonds remaining, while others are converted to single bonds Full hydrogenation converts all double bonds to single bonds Consequences of hydrogenation include: Increased melting point of the fat, making it more solid at room temperature (margarine) Improved shelf life and stability of the fat, as saturated fats are less susceptible to oxidation and rancidity Formation of trans fatty acids in partially hydrogenated fats, which are associated with negative health effects Trans fatty acids have a linear configuration, similar to saturated fatty acids, despite the presence of double bonds Trans fats are known to increase LDL (bad) cholesterol and decrease HDL (good) cholesterol, increasing the risk of heart disease (fried foods, baked goods) Lipid Behavior and Reactions Saponification: The process of hydrolyzing triglycerides with a strong base to produce soap and glycerol Rancidity: The oxidation of unsaturated fatty acids in fats and oils, leading to unpleasant odors and flavors Emulsification: The process of dispersing one liquid in another immiscible liquid, often stabilized by amphipathic molecules (e.g., lecithin in mayonnaise) Lipid bilayers: Self-assembled structures formed by phospholipids in cell membranes, with hydrophilic heads facing the aqueous environment and hydrophobic tails facing inward BackNext 27.2 Study Content & Tools Study GuidesPractice QuestionsGlossaryScore Calculators Company Get $$ for referralsPricingTestimonialsFAQsEmail us Resources AP ClassesAP Classroom every AP exam is fiveable history 🌎 ap world history🇺🇸 ap us history🇪🇺 ap european history social science ✊🏿 ap african american studies🗳️ ap comparative government🚜 ap human geography💶 ap macroeconomics🤑 ap microeconomics🧠 ap psychology👩🏾‍⚖️ ap us government english & capstone ✍🏽 ap english language📚 ap english literature🔍 ap research💬 ap seminar arts 🎨 ap art & design🖼️ ap art history🎵 ap music theory science 🧬 ap biology🧪 ap chemistry♻️ ap environmental science🎡 ap physics 1🧲 ap physics 2💡 ap physics c: e&m⚙️ ap physics c: mechanics math & computer science 🧮 ap calculus ab♾️ ap calculus bc📊 ap statistics💻 ap computer science a⌨️ ap computer science p world languages 🇨🇳 ap chinese🇫🇷 ap french🇩🇪 ap german🇮🇹 ap italian🇯🇵 ap japanese🏛️ ap latin🇪🇸 ap spanish language💃🏽 ap spanish literature go beyond AP high school exams ✏️ PSAT🎓 Digital SAT🎒 ACT honors classes 🍬 honors algebra II🐇 honors biology👩🏽‍🔬 honors chemistry💲 honors economics⚾️ honors physics📏 honors pre-calculus📊 honors statistics🗳️ honors us government🇺🇸 honors us history🌎 honors world history college classes 👩🏽‍🎤 arts👔 business🎤 communications🏗️ engineering📓 humanities➗ math🧑🏽‍🔬 science💶 social science RefundsTermsPrivacyCCPA © 2025 Fiveable Inc. 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15090	https://goldenratioandphi.weebly.com/phi-and-its-properties.html	\| \| \| \| \| \| --- --- \| The Golden Ratio & Phi \| \| \| \| \| --- \| \| \| \| \| Home The Golden Ratio in History Phi and Its Properties Phi Sightings References Euclid, the famous Greek mathematician and father of Geometry, described the Golden Ratio in this manner: having a line segment with endpoints A and B; divide it at a point C so that AC/BC = AB/AC. This gives an equation that has unique algebraic properties (Livio, 2002). To find what makes the equation unique, one must explore the numerical value that is calculated from it. In order to do so, let us take the shorter portion (BC) and assign a measure of 1 unit as its length. Let us give the longer portion (AC) a length of x. Using Euclid’s the definition of AC/BC = AB/AC, we will substitute the lengths accordingly to get . By using the basic algebraic technique of cross multiplying the equation becomes x2=x+1, which is a simple quadratic equation. Further use of algebra by subtracting x+1from both sides results in the quadratic equation x2-x-1=0. As such, using the quadratic formula leads to two solutions: 1.6180339887… and -0.6180339887...The positive value is the numerical value for the Golden Ratio (Livio, 2002). This value is commonly called the Greek letter “Phi”. The second, negative value; therefore, it is not as useful, but is related to Phi (lengths cannot be negative). Entering the numerical value for the Golden Ratio in a calculator gives a couple of interesting results. If the number is entered and then squared (using the x2 key), the result is 2.6180339887…. If the number is entered to find its reciprocal (using the 1/x key), the result is 0.6180339887…. There is something that should be noticed by comparing the values: x = 1.6180339887…, x2 = 2.6180339887…, and 1/x = 0.6180339887… (Livio, 2002). Each of these values is identical after the decimal place; Phi^2 is one more than Phi and Phi is one more than 1/Phi. In other words, Phi^2 = Phi + 1 and 1/Phi = Phi – 1. Phi is the only number in which adding one will yield its square and subtracting one will yield its inverse (Knott, 2011). The inverse of Phi, 1/Phi, is commonly referred to as the “lowercase phi” (Both symbols for Phi and phi are at top of page). Any division of two quantities that results in Phi or phi yields the Golden Ratio. This can be explained by considering the nature of ratios. Take, for instance, the example of having a room with 2 boys and 3 girls. The ratio of boys to girls would be 2/3, or .666666…; however, if we took the ratio of girls to boys it would be 3/2, or 1.5. Although the numerical values are different and are reciprocals of each other, they describe the relationship between the same two quantities; the ordering of the division is the only difference. Therefore, if one orders the division by using BC/AC instead of AC/BC, the result will be phi (.6180339887…) Incidentally, the negative solution that was mentioned earlier is equal to the negative of 1/phi (Livio, 2002). Another interesting property of Phi is that it can be defined in terms of itself. Remember that subtracting 1 from Phi gives its reciprocal: Phi-1 = 1/Phi. Changing the equation by adding 1 to both sides gives Phi = 1 + 1/Phi. If we substitute the second Phi with its equivalent, 1+1/Phi, we get: Phi = 1 + 1/(1 + 1/Phi). Continually repeating this step begins to look like this: This type of equation containing a fraction within a fraction repeatedly is called a continued fraction (Burger & Starbird, 2005). This equation converges to Phi, but does so very slowly because the fraction consists of only ones. This quality leads Livio to call Phi “the most irrational among irrational” numbers (2002, p. 85). Phi can also be written as a nested radical. Let’s return to x^2=x+1, the equation that describes the Golden Ratio. Manipulating the equation by taking the square root of both sides results in a new equation: x = √(x+1) or x = √(1+x). Here it should be noted that x = √(1+x) is an equation that has the same variable in the definition as it has as the solution; x. Therefore, we can substitute the second x in the equation with its equivalent, √(1+x), yielding x = √1+√(1+x). Repeating this process again gives x = √1+ √1+√(1+x), and again: x =√1+ √1+ √1+√(1+x). The result is an unending chain of √1. This pattern continues into infinity, as shown below (Posamentier & Lehmann, 2012). Fibonacci Phi is also related to the Fibonacci sequence of numbers. This sequence of numbers came about as a solution to a problem that Fibonacci posed about fictional rabbits. The puzzle was to figure out how many pairs of rabbits would be produced in a year if each pair of immortal rabbits produced another pair each month. By writing out the answers to each month the sequence develops as 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, … The pattern that is seen from this list is that the next number in the sequence is achieved by adding the previous two numbers (Bentley, 2008). This sequence is commonly mentioned with the Golden Ratio because taking the ratio of a Fibonacci number and the one before it increasing converges to Phi (Hemenway, 2005). In fact, the larger the numbers become the closer to Phi the ratio is. Phyllotaxis, the distribution or arrangement of plant parts on a stem, and many other appearances of Phi in nature, are a result of the Fibonacci numbers (Adam, 2003). For more information on Phi and the Fibonacci numbers click the middle button below. \| \| \| \| --- \| <Golden Ratio in History \| More about phi! \| Phi Sightings> \|
15091	https://www.countries-ofthe-world.com/time-zones-russia.html	Time zones in Russia, UTC time offsets by region and city 7 Continents World Capitals World Flags World Currencies Time Zones Calling Codes Country Domains Best Travel Places Time zones in Russia There are 11 official standard time zones in Russia. This country has the most number of time zones within a contiguous territory and does not observe the Daylight Saving Time (DST), not changing clocks ahead and back. UTC+02:00 Kaliningrad Time Zone includes the westernmost region called Kaliningrad Oblast. UTC+03:00 Moscow Time Zone (MSK) is the most populous one, encompassing 48 federal subjects in the European part of Russia. The widely known tourist places are located there, like Moscow, Saint Petersburg and the ancient towns of the Golden Ring. UTC+04:00 Samara Time Zone covers 4 regions: Samara Oblast, Astrakhan Oblast, Ulyanovsk Oblast, Udmurt Republic. UTC+05:00 Yekaterinburg Time Zone comprises 9 federal subjects, located to the east of the Ural mountains. UTC+06:00 Omsk Time Zone includes 1 region of southwestern Siberia: Tomsk Oblast. UTC+07:00 Krasnoyarsk Time Zone covers 8 regions in the south of Siberia: Novosibirsk Oblast, Tomsk Oblast, Kemerovo Oblast, Krasnoyarsk Krai, Altai Krai, Altai Republic, Republic of Khakassia, Tuva Republic. UTC+08:00 Irkutsk Time Zone includes 2 federal subjects in the southeastern part of Siberia around the famous Baikal Lake. UTC+09:00 Yakutsk Time Zone stretches along the entire Zabaykalsky Krai, Amur Oblast and the western part of Sakha (Yakutia) Republic. UTC+10:00 Vladivostok Time Zone encompasses 4 regions in the Far East and the central part of Sakha (Yakutia) Republic. UTC+11:00 Magadan Time Zone comprises the western part of the Sakha (Yakutia), Sakhalin Oblast and entire the North Kuril Islands. UTC+12:00 Kamchatka Time Zone covers the most eastern areas of Russia, called Chukotka and Kamchatka Krai, bordering with Alaska (USA) and Japan. The Moscow time (MSK) is now widely used in broadcasting and train schedules across the country. That's why, along with the UTC offset, we also mention the offset from the Moscow time for different locations in Russia. No Daylight Saving Time (DST) in 2025 See, in which time zones are the major cities of Russia located, including the capital, Moscow, the regional capitals (are in bold), large metropolises, tourist places and other locations. Alphabetical list of Russian cities with their standard time zones and MSK offsets \| City or regional capital \| Region \| UTC time offset (2025) \| Moscow time offset (2025) \| --- --- \| \| A \| \| Abakan \| Republic of Khakassia \| +07:00 \| +04:00 \| \| Anadyr \| Chukotka Autonomous Okrug \| +12:00 \| +09:00 \| \| Anapa \| Krasnodar Krai \| +03:00 \| ±00:00 \| \| Angarsk \| Irkutsk Oblast \| +08:00 \| +05:00 \| \| Arkhangelsk \| Arkhangelsk Oblast \| +03:00 \| ±00:00 \| \| Armavir \| Krasnodar Krai \| +03:00 \| ±00:00 \| \| Astrakhan \| Astrakhan Oblast \| +04:00 \| +01:00 \| \| B \| \| Barnaul \| Altai Krai \| +07:00 \| +04:00 \| \| Belgorod \| Belgorod Oblast \| +03:00 \| ±00:00 \| \| Birobidzhan \| Jewish Autonomous Oblast \| +10:00 \| +07:00 \| \| Biysk \| Altai Krai \| +07:00 \| +04:00 \| \| Blagoveshchensk \| Amur Oblast \| +09:00 \| +06:00 \| \| Bratsk \| Irkutsk Oblast \| +08:00 \| +05:00 \| \| Bryansk \| Bryansk Oblast \| +03:00 \| ±00:00 \| \| C \| \| Cheboksary \| Chuvash Republic \| +03:00 \| ±00:00 \| \| Chelyabinsk \| Chelyabinsk Oblast \| +05:00 \| +02:00 \| \| Cherepovets \| Vologda Oblast \| +03:00 \| ±00:00 \| \| Cherkessk \| Karachay-Cherkess Republic \| +03:00 \| ±00:00 \| \| Chita \| Zabaykalsky Krai \| +09:00 \| +06:00 \| \| D \| \| Dmitrov \| Moscow Oblast \| +03:00 \| ±00:00 \| \| E \| \| Elista \| Republic of Kalmykia \| +03:00 \| ±00:00 \| \| G \| \| Gatchina \| Leningrad Oblast \| +03:00 \| ±00:00 \| \| Gorno-Altaysk \| Altai Republic \| +07:00 \| +04:00 \| \| Grozny \| Chechen Republic \| +03:00 \| ±00:00 \| \| I \| \| Irkutsk \| Irkutsk Oblast \| +08:00 \| +05:00 \| \| Ivanovo \| Ivanovo Oblast \| +03:00 \| ±00:00 \| \| Izhevsk \| Udmurt Republic \| +04:00 \| +01:00 \| \| K \| \| Kaliningrad \| Kaliningrad Oblast \| +02:00 \| -01:00 \| \| Kaluga \| Kaluga Oblast \| +03:00 \| ±00:00 \| \| Kazan \| Republic of Tatarstan \| +03:00 \| ±00:00 \| \| Kemerovo \| Kemerovo Oblast \| +07:00 \| +04:00 \| \| Khabarovsk \| Khabarovsk Krai \| +10:00 \| +07:00 \| \| Khanty-Mansiysk \| Khanty-Mansi Autonomous Okrug \| +05:00 \| +02:00 \| \| Kirov \| Kirov Oblast \| +03:00 \| ±00:00 \| \| Kislovodsk \| Stavropol Krai \| +03:00 \| ±00:00 \| \| Kolomna \| Moscow Oblast \| +03:00 \| ±00:00 \| \| Komsomolsk-on-Amur \| Khabarovsk Krai \| +10:00 \| +07:00 \| \| Korolyov \| Moscow Oblast \| +03:00 \| ±00:00 \| \| Kostroma \| Kostroma Oblast \| +03:00 \| ±00:00 \| \| Krasnodar \| Krasnodar Krai \| +03:00 \| ±00:00 \| \| Krasnoyarsk \| Krasnoyarsk Krai \| +07:00 \| +04:00 \| \| Kurgan \| Kurgan Oblast \| +05:00 \| +02:00 \| \| Kursk \| Kursk Oblast \| +03:00 \| ±00:00 \| \| Kyzyl \| Tuva Republic \| +07:00 \| +04:00 \| \| L \| \| Lipetsk \| Lipetsk Oblast \| +03:00 \| ±00:00 \| \| M \| \| Magadan \| Magadan Oblast \| +11:00 \| +08:00 \| \| Magas \| Republic of Ingushetia \| +03:00 \| ±00:00 \| \| Magnitogorsk \| Chelyabinsk Oblast \| +05:00 \| +02:00 \| \| Makhachkala \| Republic of Dagestan \| +03:00 \| ±00:00 \| \| Maykop \| Republic of Adygea \| +03:00 \| ±00:00 \| \| Mineralnye Vody \| Stavropol Krai \| +03:00 \| ±00:00 \| \| Mirny \| Sakha Republic \| +09:00 \| +06:00 \| \| Moscow, capital of Russia \| Federal city \| +03:00 \| ±00:00 \| \| Murmansk \| Murmansk Oblast \| +03:00 \| ±00:00 \| \| Murom \| Vladimir Oblast \| +03:00 \| ±00:00 \| \| N \| \| Naberezhnye Chelny \| Republic of Tatarstan \| +03:00 \| ±00:00 \| \| Nakhodka \| Primorsky Krai \| +10:00 \| +07:00 \| \| Nalchik \| Kabardino-Balkar Republic \| +03:00 \| ±00:00 \| \| Naryan-Mar \| Nenets Autonomous Okrug \| +03:00 \| ±00:00 \| \| Nizhnekamsk \| Republic of Tatarstan \| +03:00 \| ±00:00 \| \| Nizhnevartovsk \| Khanty-Mansi Autonomous Okrug \| +05:00 \| +02:00 \| \| Nizhny Novgorod \| Nizhny Novgorod Oblast \| +03:00 \| ±00:00 \| \| Nizhny Tagil \| Sverdlovsk Oblast \| +05:00 \| +02:00 \| \| Norilsk \| Krasnoyarsk Krai \| +07:00 \| +04:00 \| \| Novokuznetsk \| Kemerovo Oblast \| +07:00 \| +04:00 \| \| Novorossiysk \| Krasnodar Krai \| +03:00 \| ±00:00 \| \| Novosibirsk \| Novosibirsk Oblast \| +07:00 \| +04:00 \| \| Novy Urengoy \| Yamalo-Nenets Autonomous Okrug \| +05:00 \| +02:00 \| \| O \| \| Omsk \| Omsk Oblast \| +06:00 \| +03:00 \| \| Orenburg \| Orenburg Oblast \| +05:00 \| +02:00 \| \| Orsk \| Orenburg Oblast \| +05:00 \| +02:00 \| \| Oryol \| Oryol Oblast \| +03:00 \| ±00:00 \| \| Oymyakon \| Sakha Republic \| +10:00 \| +07:00 \| \| P \| \| Penza \| Penza Oblast \| +03:00 \| ±00:00 \| \| Pereslavl-Zalessky \| Yaroslavl Oblast \| +03:00 \| ±00:00 \| \| Perm \| Perm Krai \| +05:00 \| +02:00 \| \| Petropavlovsk-Kamchatsky \| Kamchatka Krai \| +12:00 \| +09:00 \| \| Petrozavodsk \| Republic of Karelia \| +03:00 \| ±00:00 \| \| Pskov \| Pskov Oblast \| +03:00 \| ±00:00 \| \| Pyatigorsk \| Stavropol Krai \| +03:00 \| ±00:00 \| \| R \| \| Rostov-on-Don \| Rostov Oblast \| +03:00 \| ±00:00 \| \| Rostov the Great \| Yaroslavl Oblast \| +03:00 \| ±00:00 \| \| Ryazan \| Ryazan Oblast \| +03:00 \| ±00:00 \| \| S \| \| Saint Petersburg \| Federal city \| +03:00 \| ±00:00 \| \| Salekhard \| Yamalo-Nenets Autonomous Okrug \| +05:00 \| +02:00 \| \| Samara \| Samara Oblast \| +04:00 \| +01:00 \| \| Saransk \| Republic of Mordovia \| +03:00 \| ±00:00 \| \| Saratov \| Saratov Oblast \| +03:00 \| ±00:00 \| \| Sergiyev Posad \| Moscow Oblast \| +03:00 \| ±00:00 \| \| Sevastopol \| Federal city \| +03:00 \| ±00:00 \| \| Simferopol \| Republic of Crimea \| +03:00 \| ±00:00 \| \| Smolensk \| Smolensk Oblast \| +03:00 \| ±00:00 \| \| Sochi \| Krasnodar Krai \| +03:00 \| ±00:00 \| \| Srednekolymsk \| Sakha Republic \| +11:00 \| +08:00 \| \| Stary Oskol \| Belgorod Oblast \| +03:00 \| ±00:00 \| \| Stavropol \| Stavropol Krai \| +03:00 \| ±00:00 \| \| Sterlitamak \| Republic of Bashkortostan \| +05:00 \| +02:00 \| \| Surgut \| Khanty-Mansi Autonomous Okrug \| +05:00 \| +02:00 \| \| Suzdal \| Vladimir Oblast \| +03:00 \| ±00:00 \| \| Syktyvkar \| Komi Republic \| +03:00 \| ±00:00 \| \| T \| \| Taganrog \| Rostov Oblast \| +03:00 \| ±00:00 \| \| Tambov \| Tambov Oblast \| +03:00 \| ±00:00 \| \| Tobolsk \| Tyumen Oblast \| +05:00 \| +02:00 \| \| Tolyatti \| Samara Oblast \| +04:00 \| +01:00 \| \| Tomsk \| Tomsk Oblast \| +07:00 \| +04:00 \| \| Tula \| Tula Oblast \| +03:00 \| ±00:00 \| \| Tver \| Tver Oblast \| +03:00 \| ±00:00 \| \| Tyumen \| Tyumen Oblast \| +05:00 \| +02:00 \| \| U \| \| Ufa \| Republic of Bashkortostan \| +05:00 \| +02:00 \| \| Uglich \| Yaroslavl Oblast \| +03:00 \| ±00:00 \| \| Ulan-Ude \| Republic of Buryatia \| +08:00 \| +05:00 \| \| Ulyanovsk \| Ulyanovsk Oblast \| +04:00 \| +01:00 \| \| V \| \| Veliky Novgorod \| Novgorod Oblast \| +03:00 \| ±00:00 \| \| Vladimir \| Vladimir Oblast \| +03:00 \| ±00:00 \| \| Vladikavkaz \| Republic of North Ossetia-Alania \| +03:00 \| ±00:00 \| \| Vladivostok \| Primorsky Krai \| +10:00 \| +07:00 \| \| Volgograd \| Volgograd Oblast \| +03:00 \| ±00:00 \| \| Vologda \| Vologda Oblast \| +03:00 \| ±00:00 \| \| Volzhsky \| Volgograd Oblast \| +04:00 \| +01:00 \| \| Vorkuta \| Komi Republic \| +03:00 \| ±00:00 \| \| Voronezh \| Voronezh Oblast \| +03:00 \| ±00:00 \| \| Y \| \| Yakutsk \| Sakha Republic \| +09:00 \| +06:00 \| \| Yaroslavl \| Yaroslavl Oblast \| +03:00 \| ±00:00 \| \| Yekaterinburg \| Sverdlovsk Oblast \| +05:00 \| +02:00 \| \| Yessentuki \| Stavropol Krai \| +03:00 \| ±00:00 \| \| Yoshkar-Ola \| Mari El Republic \| +03:00 \| ±00:00 \| \| Yuzhno-Sakhalinsk \| Sakhalin Oblast \| +11:00 \| +08:00 \| How many countries? 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15092	https://www.youtube.com/watch?v=pkzvCHnW-Ts	How to convert decimal to radical using calculator allaboutintegration 4560 subscribers 727 likes Description 84601 views Posted: 2 Oct 2020 How to convert decimal to radical using calculator This video demonstrates how to derive exact value in surd form from decimals with the help of a calculator. Some decimals, not all, can be expressed in exact surd form with the help of a calculator Do check out my other casio fx570ms videos: 1) Z-Score, Normal Distribution (CASIO fx570ms & fx 991ms) : 2) Find Inverse Matrix using (CASIO fx570ms & fx991ms) : 3) Finding 3x3 Matrix Determinant (CASIO fx570ms & fx 991ms) : 4) Standard Deviation, Mean & Sum of Squares (CASIO fx570ms & fx 991ms) : 5) Vector Product, Cross Product (CASIO fx570ms, fx991ms) : 6) Integration using calculator (casio fx570ms/fx991ms) : 7) Radians to degrees on calculator (Casio fx570ms) : 8) Degrees to radians on calculator (Casio fx570ms) : 71 comments Transcript: hey guys my name is bo and welcome to the excel math channel today i'm going to give you a small tips on how to convert decimals into search forms all right um using your calculator itself because um you know some of the questions requires you to express inserts from or express an exact value and uh especially questions such as trigonometry and it will be very troublesome for you to memorize all the square root or whatsoever square root 3 over 2 square root 2 over 2 and so on and so forth or by drawing the triangles to find out what is the exact value of cosine half or maybe like sine 60 or whatever so here is a small little trick on how to actually get an exact value so uh let's take a look at these three examples all right so if you take a look the first example first thing that you need to do is to type in okay let's say for example if you got this decimal which is zero point eight six six zero two five four zero three okay so it right after you have calculated um you know the problems you said i end up with this all right so the next thing you can do is to indicate is to square this all right so you square the decimal and what you get is around 0.749998 which is around 0.75 so as you know that if the square of 0.86 squared is equal to 0.75 and this is equals to three-quarter okay so therefore all right therefore all right zero point eight six six zero three so and so forth this is approximately is the uh square root of three over four all right so let's just take uh the square root of the the fraction down here then this will be equals to square root three over two all right so this is the exact value okay or the sub form of zero point eight six six zero two five four zero three okay so let's take a look at the next one zero point seven zero seven one zero six seven eight one so let's just keep into the calculator zero point seven zero seven one zero six seven eight one all right and then we have to square it so approximately gives you what if we were to square this right square of the 0.71707106781 that will gives you 0.5 approximately and this is equal to half so which means to say that okay 0.7071 06781 is just basically the square root of half okay so this is actually one over square root two okay so last but not least okay we square this also one point seven three two zero five zero eight zero eight we take the square of it uh and then we realize that this is actually a tree so we take the square of this okay first and then we know that this is three and therefore all right uh 1.732050808 so this is equals to square root 3. all right so what do we do is that first thing first okay we square the decimal okay then after that we will convert it into a fraction and after that we will take the square root of the fraction to get back okay our decimals okay or to get back the exact value of the decimals so here's the small little trip for you and that's about it for this video if you guys want more helps do subscribe to this channel and hopefully i can help you guys in the exam that's called full marks all very well in your papers i'm bob signing off from here i will see you in the next video goodbye guys
15093	https://eprint.iacr.org/2012/644.pdf	Impossibility Results for Indiﬀerentiability with Resets Atul Luykx, Elena Andreeva, Bart Mennink, and Bart Preneel Dept. Electrical Engineering, ESAT/COSIC, KU Leuven, and IBBT, Belgium {atul.luykx, elena.andreeva, bart.mennink, bart.preneel}@esat.kuleuven.be Abstract. The indiﬀerentiability framework of Maurer, Renner, and Holenstein (MRH) has gained immense popularity in recent years and has proved to be a powerful way to argue security of cryptosystems that enjoy proofs in the random oracle model. Recently, however, Ristenpart, Shacham, and Shrimpton (RSS) showed that the composition theorem of MRH has a more limited scope than originally thought, and that extending its scope required the introduction of reset-indiﬀerentiability, a notion which no practical domain extenders satisfy with respect to random oracles. In light of the results of RSS, we set out to rigorously tackle the speciﬁcs of indiﬀerentiability and reset-indiﬀerentiability by viewing the notions as special cases of a more general deﬁnition. Our contributions are twofold. Firstly, we provide the necessary formalism to reﬁne the notion of indiﬀerentiability regarding composition. By formalizing the deﬁnition of stage minimal games we expose new notions lying in between regular indiﬀerentiability (MRH) and reset-indiﬀerentiability (RSS). Secondly, we answer the open problem of RSS by showing that it is impossible to build any domain extender which is reset-indiﬀerentiable from a random oracle. This result formally conﬁrms the intuition that reset-indiﬀerentiability is too strong of a notion to be satisﬁed by any hash function. As a consequence we look at the weaker notion of single-reset-indiﬀerentiability, yet there as well we demonstrate that there are no “meaningful” domain extenders which satisfy this notion. Not all is lost though, as we also view indiﬀerentiability in a more general setting and point out the possibility for diﬀerent variants of indiﬀerentiability. Keywords. indiﬀerentiability, reset-indiﬀerentiability, random oracle, hash functions 1 Introduction The notion of indiﬀerentiability was introduced by Maurer, Renner, and Holenstein (MRH) in as an extension of the classical notion of indistinguishability. The main result of allows one to securely replace a functionality ϕ1 (i.e. a random oracle) in any game G with a diﬀerent functionality ϕ2 (i.e. a hash function), as long as ϕ2 is shown to be indiﬀerentiable from ϕ1. Usually ϕ1 is a functionality which allows for easier security analysis of a game G, thereby reducing the work needed in order to show that ϕ2 is secure with various games. Since the introduction of the random oracle (RO) model by Bellare and Rogaway in , the security for many simple and eﬃcient employed cryptographic schemes, such as OEAP and FDH , has been proven in the RO model. An important implication of the result of MRH is that proving that your hash function is indiﬀerentiable from a random oracle allows one to focus on the modular hash function design and conclude security against generic attacks. Coron et al. developed the idea of using the work of MRH in the case of hash function domain extenders. Both the works of MRH and Coron et al. set a line of research on indiﬀerentiability of hash domain extenders versus random oracles, which resulted in numerous indiﬀerentiability proofs of hash domain extenders, such as [1,2,6,7,10,11]. Several variants of indiﬀerentiability [9, 13, 14, 16] have appeared in the literature, where random oracles are replaced with other functionalities. But despite the wide use of the indif-ferentiability security notion, the seminal indiﬀerentiability framework of MRH has received little further theoretical attention and treatment. Recently, however, Ristenpart, Shacham, and Shrimpton (RSS) revealed a surprising result. They showed that the types of games in which indiﬀerentiable constructions can be securely replaced is limited. To understand their core observation and the underlying intricacies of the indiﬀerentiability framework, we will use a slightly modiﬁed form of the notation of RSS (a translation of terms from RSS to MRH can be found in Table 1 in Appendix A). In , MRH deﬁne a functionality ϕ1 to be at least as secure as a functionality ϕ2, if for any given game G and adversaries A1, there exist adversaries A2 such that the quantity Pr GA1 ϕ1 →1 −Pr GA2 ϕ2 →1 (1) is small. The game G is any procedure with binary output, which may run the adversaries, A1 or A2, and the honest interfaces to either ϕ1 or ϕ2. Here, by the honest interface we simply mean that part of ϕ1 or ϕ2 with which the game may interact, as opposed to the adversarial interface which only the adversaries may access. A useful way of thinking about making (1) small is that the objective of G is to distinguish the interaction of A1 and ϕ1 from that of A2 and ϕ2, whereas the objective of ϕ1 is to mimic ϕ2 as well as possible no matter what adversaries A1 it gets. We would like to explicitly mention that the above deﬁnition is given in terms of all games, which becomes an important point later on. The connection between ϕ1 being at least as secure as ϕ2 and the indiﬀerentiability of ϕ1 with respect to ϕ2 is made in Theorem 1 of (which we refer to as the MRH theorem), where it is stated that these two concepts are equivalent. The proof relies on the fact that the adversaries A2 can be explicitly created by deﬁning it to be a combination of the adversaries A1 and an extra procedure, usually called the simulator. The simulator is introduced via indiﬀerentiability, where it is deﬁned to be the adversary of a distinguisher. The distinguisher in turn can be considered as a convenient way of describing all games (anything a game can do, a distinguisher can do, and vice versa), hence one describes G and A1 in terms of a distinguisher and the MRH theorem is proved. Yet in , RSS point out that there is a hidden assumption used in the proof of the MRH theorem which restricts the type of games to which the MRH theorem is applicable. In partic-ular, the games should be restricted to one-games (single-stage games in RSS terminology), namely games which use one stateful adversary. This is because RSS realized that the dis-tinguishers with which all indiﬀerentiability proofs have been performed are only as powerful as one-games (hence, the distinguishers do not cover all games), therefore the simulators are only designed to work against one-games. As a result, if ϕ1 is indiﬀerentiable from ϕ2, then the quantity in (1) is only guaranteed to be small when G is a one-game. To restore the scope of the MRH theorem, RSS introduced reset-indiﬀerentiability as a generalization, where the simulator must be designed to withstand a distinguisher which can reset the simulator’s state an arbitrary number of times, allowing the MRH theorem to apply to any n-game (n-stage or multi-stage game in RSS terminology). Our Contributions We ﬁrst start by formalizing what exactly an n-game is in Section 2. Although RSS informally describe in what they mean by an n(-stage)-game and a minimal n(-stage)-game, in order to further expand upon the topic of indiﬀerentiability we need a rigorous deﬁnition of an n-game. Then, by looking at indiﬀerentiability from a more general point of view, in Section 3 we put the MRH theorem in perspective and note that there are no faults in the application of indiﬀerentiability. Rather, all existing proofs limit the scope of indiﬀerentiability due to the nature of the distinguishers used. From this viewpoint we are led to a generalization of indif-ferentiability: G-indiﬀerentiability, where G is a class of games in which composition will hold. Indiﬀerentiability, as commonly used, corresponds to the class of one-games, games with one stateful adversary. The reset-indiﬀerentiability of RSS corresponds to the class of all games, thereby restoring the implication for any n-game. There are many more types of indiﬀerentia-bility, as G could be any class of games. One of them is single-reset-indiﬀerentiability, where 2 the distinguisher may reset the state of the simulator only once (as opposed to an arbitrary number of times). As a ﬁrst step we consider the possibilities of reset-indiﬀerentiability and single-reset-indiﬀerentiability. We answer the open problem of RSS in proving that there are no domain extenders which are reset-indiﬀerentiable from a random oracle (Section 4), a generalization of the impossibility result from that no single-pass domain extenders (processing the message only once) can be reset-indiﬀerentiable from a random oracle. The intuition behind this impossibility result is that the distinguisher does not allow the simulator to maintain any state as it resets the simulator after every call to it. Going a step further, as a main contribution we prove that there are no “meaningful” domain extenders which are single-reset-indiﬀerentiable from a random oracle (Section 5). Here, “meaningful” domain extenders are ones where the state size of the domain extender has a ﬁnite upper bound (e.g. it cannot grow with the size of the input message), and where modiﬁcations in the input message lead to diﬀerent results with high probability (also known as the “avalanche eﬀect”). The intuition behind the distinguisher is that it bases its strategy on the type of construction. If the domain extender is roughly single-pass, a generalization of the distinguisher by RSS does the job. On the other hand, if the domain extender processes a signiﬁcant amount of bits more than once, the distinguisher resets at a deliberate time and smartly manipulates the simulator inputs in order to distinguish with high probability. The observation that one cannot hope to ﬁnd any meaningful domain extender that achieves single-reset-indiﬀerentiability (let alone reset-indiﬀerentiability) leads to the following conclusion: either there is no hope of ﬁnding a convenient way of linking the security of domain extenders with that of random oracles for n-games (n > 1), or the notions we are looking at are too strong. The distinguishers derived in this work provide evidence as to why the notions are too strong: if we attempt to prove indiﬀerentiability for the class of all n-games, or the class of games covered by single-reset-indiﬀerentiability, then the games corresponding to our distinguishers, which are rather unnatural, are members of these classes and must be taken into account. Hence, rather than ruling out indiﬀerentiability, the question becomes what classes of games allow for composition and are meaningful to consider. A possible direction would be to consider more restricted, but perhaps more natural, classes of games in which to achieve composition. After all, the alternative of proving security for each game individually already restricts one to games which are considered natural. We elaborate on future work in Section 6. 2 Minimal Games Our notation builds on the terminology used in and introduces small modiﬁcations which facilitate the discussion on how we handle the state. 2.1 Procedures and Functionalities We always talk of sequences of procedures in order to be able to leave procedures unspeciﬁed when deﬁning games. These sequences are always assumed to be ﬁnite and are denoted with calligraphic capital letters (e.g. S as opposed to S). If A and B are two procedure sequences, then we say that they export the same interface if and only if they are of the same length and the ith procedure in A exports the same interface as the ith procedure in B. A functionality is a pair of procedure sequences ϕ := (H, P) (instead of ϕ.hon and ϕ.adv as in ). An unspeciﬁed procedure sequence is a sequence of interfaces where the procedures have not been deﬁned; an unspeciﬁed functionality is deﬁned similarly. If A is a procedure sequence which has access to some unspeciﬁed procedure sequence B, and C is a procedure sequence which exports the same interface as B, then we deﬁne AC to be the procedure sequence where calls to B are executed as calls to C. 3 2.2 Games A game G consists of a main procedure with output in {0, 1}. The main procedure is given access to an unspeciﬁed functionality and an unspeciﬁed procedure sequence called the ad-versaries; we denote this by GA ϕ , where A denotes the adversaries and ϕ the unspeciﬁed functionality. By default the adversaries do not know any of the decisions or knowledge of the other adversaries. The game needs to explicitly specify which adversaries can communicate with each other. To this end, the game speciﬁes for each pair of adversaries a storage pro-cedure which exposes a hash table so that the pair may communicate with each other. The game may specify the amount of storage for each pair of adversaries ranging from none to unlimited. If the storage is limited and an adversary asks to store data exceeding the speciﬁed size, then the data is truncated and the remainder is ignored. These storage procedures are one way of formalizing the principle of resetting a simulator by a distinguisher as was proposed by Ristenpart et al. : a reset corresponds to two simulators which may not communicate any information. Each adversary in A is given access to all procedures in P and all storage procedures involving it as part of the pair. The adversaries may not call other adversaries, nor any procedure in H, nor any storage procedure with which it is not involved. All procedures in ϕ may call all other procedures in ϕ, but not the adversaries. Every procedure used in a game has a distinct state. The communications among the procedures is depicted in Figure 1. GA ϕ A A1 A2 . . . . . . Ak ϕa ϕh 1 Fig. 1: Communications between the adversaries in A and the procedures in ϕ. Arrows denote access, and lines denote a shared storage procedure. If we have a game G and adversaries A with an unspeciﬁed procedure sequence, and B is a procedure sequence which exports the same interface as the unspeciﬁed procedure sequence of A, then we can deﬁne a game HB ϕ := GAB ϕ , where the new main procedure is the main procedure of G along with the A-calls absorbed, and the storage procedures of the new adversaries B are determined by looking at AB and seeing how A allows B to communicate. When we write Pr GA ϕ →1 , we mean that the probability is taken over the random choices made by G, the adversaries in A, and the functionality ϕ in order to determine how likely it is that the result of running the main procedure in GA ϕ results in an output of 1. 2.3 n-Games Versus Minimal n-Games Although RSS describe single-stage-games and multi-stage-games, and they discuss when games are equivalent, their discussion does not uniquely identify a single deﬁnition for the 4 concepts. A formal deﬁnition is however required for a good understanding of generalized notions of indiﬀerentiability, and in this section we aim at such deﬁnition. First, a minor detail, is a change from the use of the term multi-stage-game to n-game, where n is the number of adversaries.1 In order to remove trivialities, we require that each of the n adversaries be called at least once by the n-game, since otherwise a game which only calls one adversary once could be looked at as a 2-game simply by adding an adversary which does nothing. In RSS deﬁne equivalency of games as follows: for any ﬁxed functionality ϕ and adversaries A, two games are equivalent if Pr GA ϕ →1 = Pr HA ϕ →1 . Note that A does not change, hence they do not discuss what happens if the number of adversaries changes. In fact, it is not obvious what is meant when an n-game is equivalent to an m-game for some m < n. For example, if one says that G and H are equivalent if there exist adversaries A and B such that Pr GA ϕ →1 = Pr HB ϕ →1 , then this misses the point of looking at equivalency of games in our case since G and H could be completely diﬀerent, yet happen to output 1 with the same probability. There are other ways of changing n-games into “equivalent” m-games which are not useful for the discussion. For example, if you arbitrarily ﬁx n −m + 1 adversaries in an n-game, then you have an m-game which is equivalent to the original n-game in some sense of the word. To unambiguously identify the amount of adversaries in a game, we explicitly formalize the notion of equivalency among games with respect to the amount of communication possible between the adversaries. Then we refer to a game as a minimal n-game if it is not equivalent to any m-game for m < n. The basic idea behind our deﬁnition of a minimal n-game is that if you have a game with multiple adversaries and you allow some of the adversaries to share unlimited state, and this modiﬁcation does not result in a change to the output of the game, then you can safely replace all of the adversaries with unlimited shared state by a single adversary, resulting in a game with fewer adversaries. Let n > 1, let ϕ be a functionality and let Gϕ be an n-game. Let s be a subset of {1, . . . , n}, then we deﬁne Hϕ to be the same as Gϕ, except all storage procedures related to the pairs contained in s now have unlimited storage. Note that for all adversaries A it is the case that Pr GA ϕ →1 ≤Pr HA ϕ →1 . Letting the pairs of adversaries at the positions speciﬁed in s have unlimited communication is the same as replacing all of them with one adversary, seeing as the only thing that separates adversaries is the amount of communication they may have. Hence if we deﬁne the game Gϕ to be the same as Gϕ except all adversaries with a position in s are replaced by a single adversary, we have that Pr HA ϕ →1 = Pr G B ϕ →1 , where B is a shortened version of A where all adversaries at the positions in s are replaced by a single adversary. Deﬁnition 1. Let 1 ≤m < n. An n-game Gϕ can be written as an m-game under ϕ if there exists a subset s of {1, . . . , n} such that \|s\| = n −m + 1 and for all sequences of adversaries A we have that Pr GA ϕ →1 = Pr G B ϕ →1 , with G and B deﬁned as above. Let m be the smallest integer such that Gϕ can be written as an m-game, then we say that Gϕ is a minimal m-game. 1 The authors believe that the word “stage” does not add anything meaningful to the description of the object. 5 Note that if an n-game Gϕ is a minimal m-game, then Gϕ can be written as an i-game for any m ≤i ≤n: one applies the above trick to n −i + 1 adversaries. Example 1. Below we show an example of a two-game which was introduced in . When we call an adversary, we write in superscript the procedures that the adversary has access to. The storage procedure between A1 and A2 is denoted by stn and is limited to n bits. Let ϕ := (H, P). procedure CRP A1,A2 ϕ (p, n, s) m $ ←{0, 1}p AP,stn 1 (m) c $ ←{0, 1}s z $ ←AP,stn 2 (c) return z = H(m∥c) end procedure We consider the case where H is a domain extender and P is an ideal primitive. Note that when n < p the game is really a minimal two-game. Say that A1 is an adversary which simply stores m and A2 is an adversary that calculates H(m′∥c) using P, where m′ is the value A2 gets from querying stn for m. If n < p then the game outputs 1 with very small probability since the storage procedure truncates the message m, whereas if storage is unlimited, then the game outputs 1 with probability 1. Thus, in terms of Deﬁnition 1, we have Pr GA ϕ →1 < Pr G B ϕ →1 and the game is thus a minimal two-game. If n ≥p then we still cannot say it is a minimal one-game. The reason for this is that a particular choice of A may cause the strict inequality Pr GA ϕ →1 < Pr G B ϕ →1 . For instance, A1 could generate some random message m∗of length n and store m∗∥m so that A2 receives no useful information when the storage is limited, but can always win when the storage is unlimited. 3 A Diﬀerent View of Indiﬀerentiability The goal of this section is to explain indiﬀerentiability as just one way of achieving our original aim of composition. In the process we will prove the MRH theorem from , and indicate where exactly the application of the MRH theorem to the CRP game fails with certain domain extenders. Additionally, we will demonstrate that beyond the notions of regular indiﬀerentiability and reset-indiﬀerentiability, there is a whole spectrum of indiﬀerentiability deﬁnitions that one could consider. A functionality ϕ1 is at least as secure as a functionality ϕ2 if for all games G and adver-saries A, there exists a game H and adversary B such that the quantity Pr GA ϕ1 →1 −Pr HB ϕ2 →1 is small. Note that this deﬁnition diﬀers from the one given by MRH in in that we do not require that H = G. This deﬁnition remains suﬃcient in order to achieve composition, since as long as Pr HB ϕ2 →1 is small, Pr GA ϕ1 →1 will be small. So given G and A, the deﬁnition searches for an H and B. If we deﬁne H := G, create S, deﬁne B := AS, and set D := GA, then we get Pr GA ϕ1 →1 −Pr HB ϕ2 →1 = Pr DL ϕ1 →1 −Pr DS ϕ2 →1 , where L is a procedure which just exposes the adversarial interface to ϕ1. This is exactly what indiﬀerentiability is, and we have proved that indiﬀerentiability gives us that ϕ1 is at least 6 as secure as ϕ2. Note that indiﬀerentiability does not place any requirements on the games or procedures. In fact, in this point of view indiﬀerentiability is mostly a rewording of the deﬁnition of “ϕ1 is at least as secure as ϕ2”. Therefore it is not possible that indiﬀerentiability nor the MRH theorem do not apply to the CRP game. 3.1 G-Indiﬀerentiability What has occurred is that all existing indiﬀerentiability proofs actually limit the types of games that are covered by limiting the powers of the distinguisher. Since the distinguishers are taken to be one-games, composition holds at best for minimal one-games. In fact, the types of games covered will only be as powerful as the distinguishers. For example, a one-game distinguisher cannot model all two-games, yet a minimal two-game distinguisher would be able to do such a thing. One can look at diﬀerent types of indiﬀerentiability where the distinguishers are given varying degrees of power. Let G be a class of games; G could for example be all minimal one-games, all minimal two-games, or all minimal one-games along with CRP. What we are really interested in is when we may reduce functionality ϕ1 to functionality ϕ2 in the class G, or in other words, when is ϕ1 G-indiﬀerentiable from ϕ2? So we pick a class of distinguishers, and then we need to show that every game in G can be written as a distinguisher as we have deﬁned it. For example, if we take distinguishers as they have always been used in regular indiﬀerentiability proofs, then we notice that they are all one-games: the adversaries (simulators) are allowed unlimited communication. Conversely, if we take an arbitrary one-game GAL ϕ (with respect to the adversaries L) and we write it as allowing all adversaries in L to have unlimited communication, then we have that GAL ϕ is a one-game distinguisher. Hence we have shown that G-indiﬀerentiability where G is the class of one-games, corresponds exactly to regular indiﬀerentiability. So proving G-indiﬀerentiability for an arbitrary class G comes down to characterizing what the games in G are allowed to do with the functionalities and adversaries. In fact, without this characterization, it is not immediately clear for what class of games one would be proving indiﬀerentiability. For example, reset-indiﬀerentiability allows the distinguisher to reset as many times as it wants, yet the attack on online computable domain extenders in only uses one reset, so it is interesting to see what games we may cover if we restrict the distinguisher to use only one reset, i.e. single-reset-indiﬀerentiability. We know that single-reset-indiﬀerentiability cannot cover any minimal three-games or higher, and we know that a one-game distinguisher already covers all one-games, hence we focus our attention on minimal two-games. Consider the following minimal two-game, where A1 and A2 may communicate up to n bits: procedure GA1,A2 ϕ (p, n) m1∥m2∥m3 $ ←{0, 1}3p AP,stn 1 (m1) AP,stn 2 (m2) z $ ←AP,stn 1 (m3) return z = H(m1∥m2∥m3) end procedure Here the storage procedure is denoted by stn. Any distinguisher which has to model the above game must be able to reset at least twice if each A1 and A2 call contains an H call. A distinguisher must reset once after A1(m1) and once after A2(m2), otherwise it would violate the fact that A1 and A2 do not have unlimited communication. Hence we have an example of a minimal two-game which cannot be modeled by a single-reset distinguisher, and so single-reset-indiﬀerentiability does not cover all minimal two-games. In fact, using similar reasoning, 7 we see that a single-reset distinguisher is only able to mimic minimal two-games where all calls to the ﬁrst adversary happen before all calls to the second adversary occur. So it is important to understand what types of games a given distinguisher is able to cover. 4 Impossibility Of Reset-Indiﬀerentiability In this section we show that it is impossible to create a domain extender which is reset-indiﬀerentiable from a random oracle. This is not surprising as the existence of a domain extender which is reset-indiﬀerentiable from a random oracle would mean that such a domain extender would have to survive being reset an arbitrary number of times. The result in this section is in fact a generalization of the attack of RSS , and is included as a simple illustration. The attack uses the notions of min-entropy H∞and average min-entropy ˜ H∞. Readers not familiar with these two notions can ﬁnd deﬁnitions and basic facts in Appendix B. Let F : {0, 1}M →{0, 1}H be a domain extender that uses the sequence of ideal primitives π = (π1, . . . , πk); let ϕ1 := (F, π) be the corresponding functionality. For an input message m of length M, we denote the sequence of primitive calls made by F for the evaluation of m with (p1, p2, . . . , pn), where pi : Ui →Vi . Note that n and the sequence in which the πi are called could depend on the input message, and that two diﬀerent primitive calls could possibly be the same primitive. For example, Liskov’s zipper hash uses two primitives, π1 and π2: p1 through pn/2 would be π1-calls whereas pn/2+1 to pn would be π2-calls. Theorem 1. Let ϕ2 := (RO, RO) be a functionality where both the adversarial and honest interfaces expose the random oracle RO with range {0, 1}H. Then there exists an n-game distinguisher D such that for all simulators S = (S1, . . . , Sn), Pr DL ϕ1 →1 −Pr DS ϕ2 →1 ≥1 − n X i=1 qi 2M−log\|Ui\| + 1 2H ! , where Si makes qi queries to RO, Pn i=1 qi ≤q and L is a sequence of procedures where L ∈L returns the value given by η(L). We stress that q here denotes the number of queries made by S; the distinguisher D makes n + 1 queries. In order to illustrate the bound, if p1 = . . . = pn = π : {0, 1}a →{0, 1}b with H ≤b, as is the case with many existing (narrow- and wide-pipe) domain extenders, then we get Pr DL ϕ1 →1 −Pr DS ϕ2 →1 ≥1 − q 2M−a + 1 2H , which can be made arbitrarily close to 1 −1/2H, as M can be increased. Notice that we specify that the distinguisher can be a multi-game and does not have to be a one-game. This is equivalent to saying that the distinguisher may reset an arbitrary number of times. Proof. The distinguisher D requires a functionality ϕ = (H, P), where H exports the same interface as F and P exports the same interface as (π1, . . . , πk). Furthermore D requires a sequence A of n adversaries, with n being the number of primitive calls, and if Ai ∈A then η(Ai) = pi. We use the fact that the distinguisher may be a multi-game by letting all storage procedures have zero storage. 8 First the distinguisher selects a message m uniformly at random from {0, 1}M. Then D copies what F does, except a call to pi is replaced by a call to Ai. The distinguisher outputs its result z which it then compares with H(m): D returns 1 when z equals H(m), and 0 otherwise. Note that Pr DL (F,π) →1 = 1 . Now we look at DS (RO,RO). If H(m) is never called by Si for all i then z is independent of H(m), hence the chance that z equals H(m) is at most 1/2H. Let Ei denote the event that Si queries H(m), then Pr DS (RO,RO) →1 ≤Pr (∪iEi) + Pr DS (RO,RO) →1 \| ¬(∪iEi) ≤ n X i=1 Pr (Ei) + 1 2H . Each input given to Si, ui, can be considered a random variable over Ui. At call i all the information that the simulator knows is ui since no communication is allowed between the Si and each Si is only called once. This means that the probability that Si can compute m is bounded above by 2−˜ H∞(m\|ui). We know that ˜ H∞(m \| ui) ≥H∞(m) −log \|Ui\| = M −log \|Ui\| (cf. Appendix B), therefore Pr (Ei) ≤ qi 2M−log\|Ui\| , and we have our desired result. ⊓ ⊔ 5 Impossibility Of Single-Reset-Indiﬀerentiability We present the main impossibility result of this work, namely that there are no “meaningful” domain extenders which are single-reset-indiﬀerentiable from a random oracle. What we mean by “meaningful” will be explained below. Just as the result of Section 4, the attack of this section uses the notions of min-entropy H∞ and average min-entropy ˜ H∞. Readers not familiar with these two notions can ﬁnd deﬁnitions and basic facts in Appendix B. Let F : {0, 1}M →{0, 1}H be a domain extender which uses some primitives π = (π1, . . . , πk). On a given input m ∈{0, 1}M, the domain extender calls its primitives in a certain order: p1, p2, . . . , pn. Write pi : Ui →Vi. Note that the pi are called primitives, not the primitives themselves, hence diﬀerent pi could be the same primitive. Let ϕ1 denote the functionality representing F and its primitives. We require that the internal state of F does not exceed N bits. Concretely this means that for all j, no more than N bits of the outputs of p1, . . . , pj are used in the inputs to pj+1, . . . , pn. For all j < n we can talk of the bits of m used in the inputs to p1, . . . , pj, represented by lj, and the bits of m used in the inputs to pj+1, . . . , pn, or rj. Let aj be the bits of overlap between the two sides, so that \|lj\| + \|rj\| −\|aj\| = M (here we assume that all of m is used in the inputs, otherwise the domain extender would be easily diﬀerentiable from a random oracle). We additionally restrict F such that modifying a bit of the input message m during computation returns F(m) with low probability. Concretely, let ε be such that for all messages m, and for all j, if a bit chosen at random is ﬂipped from aj, and the result is computed as z, then Pr (F(m) = z)) < ε. Our restriction is that ε is suﬃciently close to 0. Note that this restriction is highly related to the well-known “avalanche eﬀect”, a basic property that practical hash functions are required to satisfy. 9 Theorem 2. Let ϕ2 := (RO, RO) be a functionality where both the adversarial and honest interfaces expose the random oracle RO with range {0, 1}H. Then there exists a single-reset distinguisher D such that for all simulators, Pr DL ϕ1 →1 −Pr DS ϕ2 →1 ≥min 1 −1 2H − q 2M/4−2 , 3 8 −3 M −ε 2 , where q is the total number of queries that S makes to RO and L is a sequence of procedures with L ∈L returning the value given by η(L). Proof. Our distinguisher D is a combination of two distinguishers P and Q which take as input a message m and an index j. Intuitively, D runs P if the domain extender is roughly single-pass, and it uses Q for domain extender which re-use a signiﬁcant part of the message. We will describe the distinguishers in terms of the unspeciﬁed functionality ϕ = (H, P). The functionality ϕ1 = (F, π) represents the so-called “real world”, and the functionality ϕ2 = (RO, RO) the “simulated world”. Distinguisher P . The distinguisher P goes through the process of computing with m just as the domain extender does: it performs all of the primitive calls p1, . . . , pn. The only diﬀerence is that it resets after pj. Eventually it arrives at some ﬁnal result z and it outputs 1 if z is equal to H(m) and 0 otherwise. Note that this distinguisher always outputs one in the real world. In the simulated world we have that z is independent of H(m) unless the simulator queries H(m). Let E be the event that the simulator queries H(m). Then Pr P S ϕ2 →1 ≤Pr P S ϕ2 →1 \| ¬E + Pr (E) ≤1 2H + Pr (E) . By basic properties of entropy (cf. Appendix B), the probability that S queries H(m) before the reset is bounded above by q/2H∞(m)−\|lj\| and the probability that S queries H(m) after the reset is bounded above by q/2H∞(m)−N−\|rj\|, where q is the total number of queries that the simulator makes to RO. Hence we get Pr P S ϕ2 →1 ≤1 2H + q 2H∞(m)−\|lj\| + q 2H∞(m)−N−\|rj\| . Distinguisher Q. Just like P, the distinguisher Q goes through the process of computing m like the domain extender and resets after pj. The main diﬀerence is that if \|aj\| > 0, then with probability one half it will ﬂip one of the overlapping bits of m and continue the computation after the reset with the modiﬁed m′. If \|aj\| = 0, we deﬁne Q to simply return 0 (as becomes clear later, D will by construction run Q only if \|aj\| > 0). More formally, Q will take b $ ←{0, 1}, and if b = 1, it also takes c $ ←{1, . . . , \|aj\|}. If b = 1, m′ equals m with the cth bit ﬂipped, whereas if b = 0, m′ = m. It proceeds the second half of the computation with the modiﬁed m′ and eventually arrives at some ﬁnal result z. Now, Q outputs 1 if b = 0 ∧z = H(m) or if b = 1 ∧z ̸= H(m). Denote by B the event that b = 1 (or equivalently, that a bit ﬂip occurred). In the real world, ¬B, z equals H(m). On the other hand given B, z ̸= H(m) with probability at least 1 −ε (see introduction). We ﬁnd Pr QL ϕ1 →1 ≥1 + 1 −ε 2 = 1 −ε 2 . 10 Let U denote the event that simulator after the reset learns c. Then, given that B does not occur then U occurs with probability zero. If B occurs, U occurs with probability at most N \|aj\|: indeed, the simulator learns the adjusted aj, and at most N bits of information of the original one. We can write Pr QS ϕ2 →1 ≤Pr (U) + Pr QS ϕ2 →1 \| ¬U (1 −Pr (U)) ≤ N 2 \|aj\| + Pr QS ϕ2 →1 \| ¬U 1 − N 2 \|aj\| , where we use that if x ≤z and y ≤1, x + y(1 −x) ≤z + y(1 −z). Let E denote the event that the simulator learns aj. We have Pr QS ϕ2 →1 \| ¬U ≤Pr (E \| ¬U) + Pr QS ϕ2 →1 \| ¬E ∧¬U (1 −Pr (E \| ¬U)) . Given that U does not occur, the probability of E is bounded above by 1 \|aj\|−N : indeed, the simulator learns the adjusted aj, and at most N bits of information of the original one, which due to ¬U rules out N possible values from which c is distributed. If the simulator does not learn aj and U does not occur, then the output of the simulator must be independent of B, therefore Pr QS ϕ2 →1 \| ¬E ∧¬U = 1 2. Putting all of the results together, we get Pr QS ϕ2 →1 ≤1 2 1 + 1 \|aj\| −N + N 4 \|aj\| 1 − 1 \|aj\| −N . Distinguisher D. Now, we deﬁne a distinguisher D which picks a message m uniformly at random from {0, 1}M, where M is taken such that M ≥8N and M ≥8 maxi Ui. If there exists a j ∈{1, . . . , n} such that \|aj\| > 1 4M, then D runs Q on input of j and m. In this case, the advantage of D is lower bounded by 3 8 −3 M −ε 2 . Otherwise, suppose for all j we have \|aj\| ≤1 4M. Let j be maximal such that \|lj\| ≤3 4M. We claim that also \|rj\|+N ≤3 4M. Indeed, as j is maximal, we have 3 4M ≤\|lj+1\| ≤\|lj\|+maxi Ui, and thus \|rj\| + N = M −\|lj\| + \|aj\| + N ≤M −3 4M + max i Ui + 1 4M + N ≤3 4M . Hence the claim. Now, D runs P on input of j and m. Note that the entropy of m could reduce by 1 bit due to the fact that the distinguisher is chosen based on m. By virtue of the distinguishers P, the advantage of D is lower bounded by 1 −1 2H − q 2M/4−2 , which completes the proof. ⊓ ⊔ 6 Conclusions and Future Work As we have seen, the indiﬀerentiability framework comes with a range of subtleties which make it diﬃcult to establish composition for a wide variety of games. This was already clear since the work of Ristenpart et al. , who introduced the generalized deﬁnition of reset-indiﬀerentiability. But, as we have shown, there does not exist any domain extender that is reset-indiﬀerentiable, and even if one opts for single-reset-indiﬀerentiability, there is no hope 11 to ﬁnd a secure domain extender. These dead-ends in terms of domain extenders and random oracles, however, do not lead us to question the usefulness of indiﬀerentiability, rather it points out the disconnect between what distinguishers are and the classes of games one is considering. Take for example, the fact that a game G and its adversaries A1 are merged to form a single distinguisher. This creates the unnatural eﬀect that G and A1 are actually cooperating in diﬀerentiating one functionality from another, whereas usually G and A1 are designed with opposing goals (hence the name “adversary”). If we look at the CRP game deﬁned earlier on, we see that the task of the adversaries is complicated by the fact that they may only communicate a ﬁnite amount of bits, and as a result neither adversary gets to see the full message. Generalizing this, one could consider all games which do not provide their adversaries with a complete message and limited communication, and see where this type of indiﬀerentiability would lead. Ultimately the important part is having an understanding of the natural classes of games that exist and are easy to describe. Finally, an alternative approach is to consider the same classes of games, except with diﬀerent functionalities other than random oracles, as is done in . Acknowledgments. This work has been funded in part by the IAP Program P6/26 BCRYPT of the Belgian State (Belgian Science Policy), and in part by the European Com-mission through the ICT program under contract ICT-2007-216676 ECRYPT II. The second author is supported by a Ph.D. Postdoctoral Fellowship from the Flemish Research Foun-dation (FWO-Vlaanderen). The third author is supported by a Ph.D. Fellowship from the Institute for the Promotion of Innovation through Science and Technology in Flanders (IWT-Vlaanderen). References Andreeva, E., Mennink, B., Preneel, B.: On the indiﬀerentiability of the Grøstl hash function. In: Security and Cryptography for Networks 2010. Lecture Notes in Computer Science, vol. 6280, pp. 88–105. Springer-Verlag, Berlin (2010) Bellare, M., Ristenpart, T.: Multi-property-preserving hash domain extension and the EMD transform. In: Advances in Cryptology - ASIACRYPT 2006. Lecture Notes in Computer Science, vol. 4284, pp. 299–314. Springer-Verlag, Berlin (2006) Bellare, M., Rogaway, P.: Random oracles are practical: A paradigm for designing eﬃcient protocols. In: ACM Conference on Computer and Communications Security. pp. 62–73. ACM, New York (1993) Bellare, M., Rogaway, P.: Optimal asymmetric encryption. In: Advances in Cryptology - EUROCRYPT ’94. Lecture Notes in Computer Science, vol. 839, pp. 92–111. Springer-Verlag, Berlin (1994) Bellare, M., Rogaway, P.: The exact security of digital signatures - how to sign with RSA and Rabin. In: Advances in Cryptology - EUROCRYPT ’96. Lecture Notes in Computer Science, vol. 1109, pp. 399–416. Springer-Verlag, Berlin (1996) Bertoni, G., Daemen, J., Peeters, M., Assche, G.: On the indiﬀerentiability of the sponge construction. In: Advances in Cryptology - EUROCRYPT 2008. Lecture Notes in Computer Science, vol. 4965, pp. 181–197. Springer-Verlag, Berlin (2008) Bhattacharyya, R., Mandal, A., Nandi, M.: Security analysis of the mode of JH hash function. In: Fast Software Encryption 2010. Lecture Notes in Computer Science, vol. 6147, pp. 168–191. Springer-Verlag, Berlin (2010) Coron, J., Dodis, Y., Malinaud, C., Puniya, P.: Merkle-Damg˚ ard revisited: How to construct a hash function. In: Advances in Cryptology - CRYPTO 2005. Lecture Notes in Computer Science, vol. 3621, pp. 430–448. Springer-Verlag, Berlin (2005) Dodis, Y., Ristenpart, T., Shrimpton, T.: Salvaging Merkle-Damg˚ ard for practical applications. In: Ad-vances in Cryptology - EUROCRYPT 2009. Lecture Notes in Computer Science, vol. 5479, pp. 371–388. Springer, Heidelberg (2009) Hirose, S., Park, J., Yun, A.: A simple variant of the Merkle-Damg˚ ard scheme with a permutation. In: Advances in Cryptology - ASIACRYPT 2007. Lecture Notes in Computer Science, vol. 4833, pp. 113–129. Springer-Verlag, Berlin (2007) Liskov, M.: Constructing an ideal hash function from weak ideal compression functions. In: Selected Areas in Cryptography 2006. Lecture Notes in Computer Science, vol. 4356, pp. 358–375. Springer-Verlag, Berlin (2007) 12 Maurer, U., Renner, R., Holenstein, C.: Indiﬀerentiability, impossibility results on reductions, and appli-cations to the random oracle methodology. In: Theory of Cryptography Conference 2004. Lecture Notes in Computer Science, vol. 2951, pp. 21–39. Springer-Verlag, Berlin (2004) Naito, Y.: On the indiﬀerentiable hash functions in the multi-stage security games. Cryptology ePrint Archive, Report 2012/014 (2012) Naito, Y., Yoneyama, K., Wang, L., Ohta, K.: How to conﬁrm cryptosystems security: The original Merkle-Damg˚ ard is still alive! In: Advances in Cryptology - ASIACRYPT 2009. vol. 5912, pp. 382–398. Springer-Verlag, Berlin (2009) Ristenpart, T., Shacham, H., Shrimpton, T.: Careful with composition: Limitations of the indiﬀerentia-bility framework. In: Advances in Cryptology - EUROCRYPT 2011. Lecture Notes in Computer Science, vol. 6632, pp. 487–506. Springer-Verlag, Berlin (2011) Yoneyama, K., Miyagawa, S., Ohta, K.: Leaky random oracle (extended abstract). In: Provable Security 2008. Lecture Notes in Computer Science, vol. 5324, pp. 226–240. Springer-Verlag, Berlin (2008) A Translation Between MRH and RSS Terminology Table 1: Translation between MRH and RSS terminology. MRH RSS Random System Procedure Environment Game Cryptosystem/Resource Functionality Public interface Adversarial interface Private interface Honest interface B Basic Properties of Entropy For the purpose of the impossibility results of Sections 4 and 5, we present some basic prop-erties of entropy. Deﬁnition 2. The min-entropy of a random variable x is H∞(x) = −log(maxx′ Pr (x = x′)). Deﬁnition 3. When x and y are two (possibly correlated) random variables the average min-entropy is given by ˜ H∞(x \| y) = −log  X y′ max x′ Pr x = x′ \| y = y′ Pr y = y′  . The probability that an adversary guesses the value of x given y is bounded above by 2−˜ H∞(x\|y). Furthermore, if a random variable y can take on n possible values, then X y′ max x′ Pr x = x′ \| y = y′ Pr y = y′ = X y′ max x′ Pr x = x′ ∩y = y′ ≤n max x′ Pr x = x′ , and so ˜ H∞(x \| y) ≥H∞(x) −log n. 13
15094	https://artofproblemsolving.com/wiki/index.php/2006_USAMO_Problems/Problem_3?srsltid=AfmBOoq_JQXGsQl96IlTo9x6YTLverVbl9QwuspIvveKWqE21JtTOTdu	Art of Problem Solving 2006 USAMO Problems/Problem 3 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 2006 USAMO Problems/Problem 3 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2006 USAMO Problems/Problem 3 Contents [hide] 1 Problem 2 Solutions 2.1 Solution 1 2.2 Solution 2 3 See also Problem (Titu Andreescu, Gabriel Dospinescu) For integral , let be the greatest prime divisor of . By convention, we set and . Find all polynomials with integer coefficients such that the sequence is bounded above. (In particular, this requires for .) Solutions Solution 1 Let be a non-constant polynomial in of degree with integer coefficients, suppose further that no prime divides all the coefficients of (otherwise consider the polynomial obtained by dividing by the gcd of its coefficients). We further normalize by multiplying by , if necessary, to ensure that the leading coefficient (of ) is positive. Let , then is a polynomial of degree or more and . Let be the factorization of into irreducible factors with positive leading coefficients. Such a factorization is unique. Let denote the degree of . Since the factors are either even functions of or come in pairs with . Let , . For any other integer let be the largest prime factor of . Suppose that for some finite constant and all we have . Since the polynomials divide , the same must be true for each of the irreducible polynomials . A theorem of T.Nagell implies that if the ratio is unbounded for large values of . Since in our case the is asymptotically bounded above by for large , we conclude that all the irreducible factors are linear. Since linear polynomials are not even functions of , they must occur in pairs , . Without loss of generality, . Since the coefficients of are relatively prime, so are and , and since , neither polynomial can have any non-negative integer roots, so and thus . On the other hand, by Dirichlet's theorem, , since otherwise the sequence would yield infinitely many prime values with So and therefore is a positive odd integer. Setting , clearly . Since this holds for each factor , it is true for the product of all the factors with the bound determined by the factor with the largest value of . Therefore, for suitable non-negative integers , is a product of polynomials of the form . Now, since , we conclude that is a product of linear factors of the form . Since we restricted ourselves to non-constant polynomials with relatively prime coefficients, we can now relax this condition and admit a possibly empty list of linear factors as well as an arbitrary non-zero integer multiple . Thus for a suitable non-zero integer and non-negative integers , we have: Solution 2 The polynomial has the required properties if and only if where are odd positive integers and is a nonzero integer. It is straightforward to verify that polynomials given by have the required property. If is a prime divisor of but not of , then or for some . Hence . The prime divisors of form a finite set and do not affect whether or not the given sequence is bounded above. The rest of the proof is devoted to showing that any for which is bounded above is given by . Let denote the set of all polynomials with integer coefficients. Given , let denote the set of those primes that divide at least one of the numbers in the sequence . The solution is based on the following lemma. Lemma. If is a nonconstant polynomial then is infinite. Proof. Repeated use will be made of the following basic fact: if and are distinct integers and , then divides . If , then divides for every prime , so is infinite. If , then every prime divisor of satisfies . Otherwise divides , which in turn divides . This yields , which is false. Hence implies that is infinite. To complete the proof, set and observe that and . The preceding argument shows that is infinite, and it follows that is infinite. Suppose is nonconstant and there exists a number such that for all . Application of the lemma to shows that there is an infinite sequence of distinct primes and a corresponding infinite sequence of nonnegative integers such that for all . Consider the sequence where . Then and . Hence , so for all . It follows that there is an integer such that and for infinitely many . Let . Then and . Consequently, for infinitely many , which shows that is a zero of . Since for , must be odd. Then , where . (See the note below.) Observe that must be bounded above. If is constant, we are done. If is nonconstant, the argument can be repeated to show that is given by . Note. The step that gives where follows immediately using a lemma of Gauss. The use of such an advanced result can be avoided by first writing where is rational and . Then continuation gives where is rational and the are odd. Consideration of the leading coefficient shows that the denominator of is for some and consideration of the constant term shows that the denominator is odd. Hence is an integer. Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page. See also viewtopic.php?t=84553 Discussion on AoPS/MathLinks 2006 USAMO (Problems • Resources) Preceded by Problem 2Followed by Problem 4 1•2•3•4•5•6 All USAMO Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Category: Olympiad Number Theory Problems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
15095	https://www.pearson.com/channels/calculus/learn/patrick/02-intro-to-derivatives/basic-graphing-of-the-derivative	Skip to main content My Courses Chemistry General Chemistry Organic Chemistry Analytical Chemistry GOB Chemistry Biochemistry Intro to Chemistry Biology General Biology Microbiology Anatomy & Physiology Genetics Cell Biology Physics Physics Math College Algebra Trigonometry Precalculus Calculus Business Calculus Statistics Business Statistics Social Sciences Psychology Health Sciences Personal Health Nutrition Business Microeconomics Macroeconomics Financial Accounting Product & Marketing Agile & Product Management Digital Marketing Project Management AI in Marketing Programming Introduction to Python Microsoft Power BI Data Analysis - Excel Introduction to Blockchain HTML, CSS & Layout Introduction to JavaScript R Programming Calculators AI Tools Study Prep Blog Study Prep Home Table of contents Skip topic navigation Skip topic navigation Functions 7h 52m Introduction to Functions 16m + Piecewise Functions 10m + Properties of Functions 9m + Common Functions 1h 8m + 5m + Combining Functions 27m + Exponent rules 32m + Exponential Functions 28m + Logarithmic Functions 24m + Properties of Logarithms 34m + Exponential & Logarithmic Equations 35m + Introduction to Trigonometric Functions 38m + Graphs of Trigonometric Functions 44m + Trigonometric Identities 47m + Inverse Trigonometric Functions 48m 1. Limits and Continuity 2h 2m Introduction to Limits 41m + Finding Limits Algebraically 40m + Continuity 40m 2. Intro to Derivatives 1h 33m Tangent Lines and Derivatives 30m + Derivatives as Functions 14m + Basic Graphing of the Derivative 23m + Differentiability 26m 3. Techniques of Differentiation 3h 18m Basic Rules of Differentiation 39m + Higher Order Derivatives 12m + Product and Quotient Rules 55m + Derivatives of Trig Functions 40m + The Chain Rule 49m 4. Applications of Derivatives 2h 38m Motion Analysis 36m + Implicit Differentiation 27m + Related Rates 59m + Linearization 13m + Differentials 21m 5. Graphical Applications of Derivatives 6h 2m Intro to Extrema 23m + Finding Global Extrema 50m + The First Derivative Test 1h 6m + Concavity 1h 1m + The Second Derivative Test 31m + Curve Sketching 44m + Applied Optimization 1h 25m 6. Derivatives of Inverse, Exponential, & Logarithmic Functions 2h 37m Derivatives of Exponential & Logarithmic Functions 1h 52m + Logarithmic Differentiation 20m + Derivatives of Inverse Trigonometric Functions 24m 7. Antiderivatives & Indefinite Integrals 1h 26m Antiderivatives 17m + Indefinite Integrals 25m + Integrals of Trig Functions 15m + Initial Value Problems 27m 8. Definite Integrals 4h 44m Estimating Area with Finite Sums 42m + Riemann Sums 1h 21m + Introduction to Definite Integrals 22m + Fundamental Theorem of Calculus 37m + Average Value of a Function 21m + Substitution 1h 19m 9. Graphical Applications of Integrals 2h 27m Area Between Curves 1h 18m + Introduction to Volume & Disk Method 1h 8m 10. Physics Applications of Integrals 3h 16m Kinematics 1h 13m + Work 2h 3m 11. Integrals of Inverse, Exponential, & Logarithmic Functions 2h 34m Integrals of Exponential Functions 40m + Integrals Involving Logarithmic Functions 58m + Integrals Involving Inverse Trigonometric Functions 55m 12. Techniques of Integration 7h 39m Integration by Parts 2h 30m + Partial Fractions 4h 29m + Improper Integrals 39m 13. Intro to Differential Equations 2h 55m Basics of Differential Equations 48m + Slope Fields 19m + Euler's Method 22m + Separable Differential Equations 1h 25m 14. Sequences & Series 5h 36m Sequences 1h 22m + Review of Factorials 11m + Series 44m + Convergence Tests 3h 17m 15. Power Series 2h 19m Introduction to Power Series 1h 9m + Taylor Series & Taylor Polynomials 1h 10m 16. Parametric Equations & Polar Coordinates 7h 58m Parametric Equations 1h 6m + Calculus with Parametric Curves 1h 13m + Polar Coordinates 2h 5m + Calculus in Polar Coordinates 55m + Conic Sections 2h 36m Intro to Derivatives Basic Graphing of the Derivative Intro to Derivatives Basic Graphing of the Derivative: Videos & Practice Problems Video Lessons Practice Worksheet Topic summary Understanding the relationship between a function and its derivative is crucial in calculus. The derivative represents the slope of the tangent line at any point on the function's graph. When sketching the derivative, identify intervals where the function is increasing (positive derivative), decreasing (negative derivative), or flat (zero derivative). Special cases, such as sharp corners or discontinuities, indicate where the derivative does not exist. Recognizing these behaviors allows for efficient graphing of derivatives, enhancing comprehension of function dynamics. 1 concept Graphing The Derivative Video duration: 6m Play a video: Graphing The Derivative Video Summary Understanding the relationship between a function and its derivative is crucial in calculus, particularly when analyzing the behavior of graphs. The derivative of a function, denoted as , represents the slope of the tangent line to the graph of the function at any given point. This means that if the function is increasing, the derivative is positive; if the function is decreasing, the derivative is negative; and if the function is flat (horizontal), the derivative is zero. To sketch the derivative based on the graph of a function, one can follow a systematic approach. First, observe the behavior of the function as you move along the x-axis. For example, if you consider an interval where the function is increasing, the corresponding derivative will be above the x-axis, indicating positive values. Conversely, in intervals where the function is decreasing, the derivative will be below the x-axis, indicating negative values. At points where the function has a horizontal tangent line, the derivative will equal zero. These points are critical as they represent local maxima or minima in the function. For instance, if a function has a peak or valley, the derivative will touch the x-axis at those points. To illustrate this process, consider a function that increases from negative infinity to a certain point, then decreases, and finally increases again. The derivative can be sketched by marking the points where the derivative is zero (the peaks and valleys) and determining the intervals of positivity and negativity. For example, if the function increases from negative infinity to a point, the derivative will be positive in that interval. If it then decreases, the derivative will be negative until it reaches another point where it flattens out, indicating a zero derivative. In summary, the key to sketching the derivative of a function lies in understanding the function's behavior across its domain. By identifying where the function is increasing, decreasing, or flat, one can effectively represent the derivative graphically. This method not only simplifies the process but also reinforces the fundamental concepts of calculus regarding the relationship between a function and its derivative. Study Smarter with Worksheets. Follow along with each video using our printable worksheets 2 Problem Based on the graph of f(x), describe the graph of the derivative f′(x) on the interval (0,∞). A Below the x-axis B Above the x-axis C On the x-axis 3 Problem Based on the graph of f(x), describe the graph of the derivative f′(x)at the point x=−1. A Above the x-axis B Below the x-axis C On the x-axis 4 Problem Based on the graph of f(x), describe where the derivative curve f′(x) would be below the x-axis. A (-∞,-1) U (0,1) B (-1,0) U (1,∞) C (-∞, -1) U (1,∞) D (-1,1) 5 example Graphing The Derivative Example 1 Video duration: 3m Play a video: Graphing The Derivative Example 1 Video Summary In this discussion, we explore how to sketch the derivative of a function, specifically focusing on a function that resembles a sine wave. To begin, understanding the behavior of tangent lines is crucial, as the derivative at any point corresponds to the slope of the tangent line at that point. The first step is to identify where the derivative is zero, which occurs at the peaks and valleys of the function where flat tangent lines can be drawn. For a function oscillating like a sine wave, we can locate points where the slope is zero. These points will appear on the x-axis of the derivative graph. For instance, if we find a flat tangent line between -2 and -1, we can mark this point on the x-axis at approximately -1.5. Continuing this process, we identify additional points where the derivative crosses the x-axis. Next, we analyze the behavior of the tangent lines as we move from left to right across the function. On the left side of a flat tangent line, the slopes are positive, indicating that the derivative will be above the x-axis. As we move right past this point, the slopes become negative, suggesting that the derivative will drop below the x-axis. This pattern continues, with the slopes gradually flattening out, indicating a transition back towards the x-axis. As we sketch the derivative, we observe that the curve will rise and fall, reflecting the oscillatory nature of the original function. Notably, the resulting graph of the derivative resembles a cosine wave, which leads us to conclude that the derivative of a sine function is indeed a cosine function. This relationship highlights the interconnectedness of trigonometric functions and their derivatives. In summary, by analyzing the slopes of tangent lines and their behavior across the function, we can effectively sketch the derivative, revealing the underlying relationships between sine and cosine functions. This method not only aids in visualizing derivatives but also reinforces the fundamental concepts of calculus. 6 concept Graphing The Derivative - Special Cases Video duration: 4m Play a video: Graphing The Derivative - Special Cases Video Summary Understanding how to graph the derivative of a function is essential, especially when dealing with special cases that may not present as smooth curves. These special cases often include sharp turns and jumps, which can complicate the process of determining the derivative. However, the approach remains fundamentally similar to that of regular functions, with a few additional rules to keep in mind. When graphing the derivative, visualize a person walking along the graph from left to right. For sections of the graph that are horizontal, the derivative is zero, indicating that the slope is flat. This means that the corresponding portion of the derivative graph will lie along the x-axis. Conversely, if the graph is decreasing, the derivative will be negative, placing it below the x-axis. In cases where the graph is a straight line, the slope remains constant, resulting in a constant value for the derivative. One critical aspect to remember is how to handle sharp corners or discontinuities. At these points, the derivative does not exist, which is represented by a jump or hole in the derivative graph. For example, if there is a sharp turn at a specific x-value, the derivative graph will reflect this by having a gap at that point, indicating that the slope cannot be defined there. As you continue to analyze the graph, observe how the tangent lines behave. If the graph is increasing, the derivative will be positive and above the x-axis. The steepness of the tangent lines will affect the value of the derivative; flatter sections correspond to values closer to zero, while steeper sections yield higher values. Again, if there is a jump or discontinuity, the derivative will not exist at that point, resulting in another gap in the graph. In summary, when sketching the derivative of a function, always consider the behavior of the original function. Remember that horizontal lines yield a derivative of zero, decreasing sections result in negative values, and sharp corners or jumps indicate that the derivative does not exist. By applying these principles, you can effectively graph the derivative even in the presence of special cases. 7 Problem Based on the graph f(x), describe all points where the derivative f′(x)would have a jump. A x=−1.5, x=−1, and x=0.5 B x=−1 and x=0.5 C x − 1.5 D Derivative f′(x) has no jumps 8 example Graphing The Derivative - Special Cases Example 2 Video duration: 2m Play a video: Graphing The Derivative - Special Cases Example 2 Video Summary To sketch the derivative of a function, it is essential to identify where the tangent lines to the curve are flat, indicating a slope of zero. These points occur at peaks and valleys of the original function. For instance, if a peak is located at , the derivative will touch the x-axis at this point. Additionally, if there is a continuous flat section of the graph from to , the derivative will also remain on the x-axis throughout this interval. As you analyze the graph from left to right, observe that in regions where the tangent lines have a positive slope, the derivative will be above the x-axis. Conversely, in sections where the tangent lines slope downwards, the derivative will be below the x-axis. This behavior allows for the construction of a continuous line segment for the derivative in these regions. When encountering sharp corners in the original function, such as at points where the slope changes abruptly, it indicates a jump discontinuity in the derivative. At these points, an open circle should be placed on the derivative graph to signify that the derivative does not exist at that specific value. Finally, if the original function has a linear section with a constant slope, the derivative will also be a constant value, resulting in a horizontal line on the derivative graph. This understanding of how to interpret the original function's features—such as peaks, valleys, and linear segments—enables the accurate sketching of its derivative. Do you want more practice? More sets Basic Graphing of the Derivative Intro to Derivatives 3 problems Topic AllySteele 2. Intro to Derivatives 4 topics 7 problems Chapter JonathanCook Take your learning anywhere! Prep for your exams on the go with video lessons and practice problems in our mobile app. Here’s what students ask on this topic: How do you sketch the derivative of a function from its graph? To sketch the derivative of a function from its graph, follow these steps: First, identify intervals where the function is increasing, decreasing, or flat. The derivative is positive where the function is increasing, negative where it is decreasing, and zero where it is flat. Next, locate points where the function has horizontal tangent lines (peaks and valleys) as these correspond to zero derivative values. Finally, consider any sharp corners or discontinuities, as the derivative does not exist at these points. Plot these observations on a new graph to sketch the derivative. What does it mean when the derivative of a function is zero? When the derivative of a function is zero, it means that the slope of the tangent line at that point is horizontal. In other words, the function is neither increasing nor decreasing at that specific point. This typically occurs at local maxima, minima, or points of inflection. On the graph of the function, these points appear as peaks, valleys, or flat regions. Identifying these points is crucial when sketching the derivative of a function. How do sharp corners or discontinuities affect the derivative of a function? Sharp corners or discontinuities in a function indicate points where the derivative does not exist. At a sharp corner, the function has an abrupt change in direction, making it impossible to define a single tangent line. Similarly, at a discontinuity, the function has a break or jump, preventing the derivative from being defined. When sketching the derivative, these points are represented by holes or jumps in the graph of the derivative, indicating that the derivative is undefined at those locations. What is the relationship between the slope of a tangent line and the derivative of a function? The slope of a tangent line to a function at a given point is the value of the derivative of the function at that point. The derivative represents the rate of change of the function, and the tangent line provides a linear approximation of the function's behavior near that point. A positive slope indicates an increasing function, a negative slope indicates a decreasing function, and a zero slope indicates a flat or stationary point. Understanding this relationship is key to interpreting and sketching the derivative of a function. To determine if the derivative of a function is positive or negative, observe the behavior of the function's graph. If the function is increasing (moving uphill) over an interval, the derivative is positive in that interval. Conversely, if the function is decreasing (moving downhill), the derivative is negative. By analyzing the slope of the tangent lines at various points on the graph, you can identify where the derivative is positive, negative, or zero, which helps in sketching the derivative accurately. Your Calculus tutors Math Instructor Math Instructor
15096	https://www.solidessay.com/our-services/how-to-write-an-exemplification-essay	How to Write an Exemplification Essay Custom WritingHow it WorksPricesF.A.Q. Live ChatLogin Services Home>Our Services> How to Write an Exemplification Essay How to Write an Exemplification Essay Share this article: Tweet What is an Exemplification Essay From the word example, an exemplification essay uses examples to elaborate or support a certain claim. It is a way of expressing a generalization supported with examples wherein you can use a collection of ideas in relevance to your subject. You will need examples to be able to explain, elaborate, and prove your point or argument. It can be brief, comprehensive, or both. By having enough details and using specific examples, you will have a good essay. For this to happen, follow the tips listed below on how to write an exemplification essay. How To: Exemplification Essay Writing Tips Tip #1 Think of how your readers will feel about your subject or topic. And when you write an exemplification essay, you have to find, gather, and provide a lot of examples to support your generalization. Tip #2 When you pick an example, make sure that it is relevant to your topic, otherwise your readers will be confused. Direct to the point. Though you have loads of examples, which are necessary, you should only choose the ones that are significant to what you are writing. In other words, include only the strongest examples. Tip #3 Make a list of all your examples so that it will be easier for you to choose which ones you will describe in your essay. It is the same as making an outline to keep track of every detail. Tip #4 Aside from facts, you can also use short stories and anecdotes, as examples in supporting your generalization, as long as they are of relevance to your topic. Be careful in choosing your examples and make sure that your readers can relate to it and make sure they will get your point. Tip #5 Your examples should also be representative, which means they should reflect the majority and represent what is usually happening to most individuals. Tip #6 For your readers to have a better understanding of what you are trying to exemplify, make sure that they will be able to follow your chain of ideas. To avoid confusion, do not include less common situations when providing examples. Tip #7 You can also do a research on samples of exemplification essay so you will have a reference when doing your own essay. Take note on the subject or the topic, as well as the format in the samples you find. They will help you fully understand what an exemplification essay really is. As stated above, an exemplification essay employs a series of examples with the aim of providing an illustration to a point. Therefore, using a good exemplification essay outline in the provision of these examples will ensure that the paper delivers the point in question at its entirety. Below we will provide the purpose and process of developing an outline that is used in the delivery of an exemplification essay. By sampling the outline of an exemplification essay, it is going to be possible to determine some of the key elements that are associated with such types of papers. Purpose of an exemplification essay outline The purpose of an outline in an exemplification essay is to allow for the delivery of the paper in a manner that expounds different exemplars that ensure that the general message is delivered in an understandable manner. For instance, the introduction of the exemplar is within the outline format with the intention of providing the reader with a rough idea of what to be expected within the paper. The introduction could be broken down into paragraphs with the last paragraph containing the thesis statement that is a summation of the key points to be supported or argued against within the exemplification paper. The purpose of the exemplification essay outline is also to provide the reader with some of the key points to be argued within the paper. This means that each one of the subheadings would contain the main points to be discussed within that section of the paper. This makes it easy for the person reading the exemplification essay to identify different key issues being addressed by simply skimming through the paper. This also allows for the instructor to award the student or individual writing the exemplification paper with appropriate points for ensuring that each of the key points has been pointed out and thus addressed to the full. Process of writing an exemplification essay The process of writing an exemplification essay includes the preparation process. This means going through the instructions so that one can identify each element as pointed out at their entirety. This allows for the identification of the key issues to be addressed in the paper. The researcher should also go through the texts, if provided by the instructor, so that one can gather the ideas needed in the preparation of the exemplification paper. It is also important to identify other sources that could provide the researcher with the appropriate content that would ensure that the exemplification essay is easy to understand. Developing a rough draft should allow the student to identify the key topics that are required for the completion of the exemplification essay. It is also important to identify different sources that are associated with each section of the paper. This is so that the exemplification paper can be found to contain relevant examples within each one of the sections. It is also important to identify different issues that may be contained in the exemplification essay during the development of the rough draft. Upon completion of the essay, it is crucial to develop the final draft and have another person correct the paper by going through it carefully. This ensures that the final paper satisfies all the requirements in the instructions. Get Custom-Made Essay EXEMPLIFICATION ESSAY OUTLINE SAMPLE Below is a sample of the exemplification essay: A. Introduction of the exemplification essay This part of the paper should have a hook. The purpose of the hook is to interest the audience. The introduction also provides a brief explanation of the plot or the theme to be explored in the paper and thus the reason the theme was selected. This part of the exemplification essay provides an understanding of the kind of audience that should benefit from reading the paper. The introduction ends in a well-rounded thesis statement that prepares the reader for the different arguments and examples to be used in the paper. B. Body of the exemplification essay Example 1 Example 2 Example 3 C. Conclusion of the exemplification essay The conclusion starts with a restated thesis statement that provides one with the required information that allows for the development of the required solution. The conclusion can also provide an appropriate action that can result in the resolution of any issues that are identified in the exemplification paper. The exemplification essay also provides details of different implications that one could have identified in the paper. The conclusion also provides the reader with thoughts to think about upon completing the readership. Things to Remember About Writing Exemplification Essay Outline It is important to remember that exemplars are written around a thesis statement. Therefore, the exemplification paper should be written in a format that surrounds the key pointers that have originally been identified in the thesis statement. It is also important to remember that one can use as many examples as possible within the paper. However, it is important to ensure that the examples used in the paper are relevant to the issue being addressed within the paper. Therefore, it is crucial that the author begins the paper with the strongest most relevant examples, and ends with the weaker examples. This should ensure that the basis of all the stated requirements have been addressed in the paper. One can also find that different examples used in the paper may result into a different conclusion and thus one can mold the thesis statement to fit these conclusions as found in the paper. The paper could also be organized such that different examples start with the least controversial and finish off with the most controversial ones. One can also choose to start off with the most important examples and finish off with the least important one. The general idea is to follow a pattern that either creates a better understanding in the reader or provides the reader with details of the most important examples. The exemplar is as good as an effective outline. This is because the outline of the paper should provide the reader with all the details they need to identify different issues that can be used to determine the general conclusion of the paper. It is important that the development of the outline goes hand in hand with the available content so that the subtopics used in the paper align with the thesis statement. However, upon revising the paper, one can change different topics so that the solutions to be identified in the paper result in finding the appropriate solution expected for the problems raised in the paper. EXEMPLIFICATION ESSAY TOPICS Below there’s a list of creative exemplification essay topics to be considered for writing. Discuss the impact of the current level of technologies on our culture. Examine the impact that the growing number of fast food restaurants has had on America. Debate on the impact that hoarding has on families. Deliberate the impact that online education has had on traditional education. Converse the impact that businesses feel from social networking. Discuss the impact of religion in the modern culture. Should everyone receive free medical care? Is a college degree necessary in the modern society? Should schools enforce any religious beliefs on their students? Are gun regulations tough enough? Frequently Asked Questions About Writing an Exemplification Essay What is the Purpose of an Exemplification Essay? An exemplification essay exists to prove a point. In writing the exemplification essay, you must put forward arguments and reasons why you have taken a stance or chosen a particular point of view. To do this, the writer must include detailed information and credible sources. How to Write an Introduction for an Exemplification Essay? To introduce your exemplification essay, you will want to give an outline of your main idea or thesis on the chosen topic. Explain what your point is and give an indication of how you plan to prove your point to the reader. How to Write a Thesis Statement for an Exemplification Essay? In order to write a thesis statement for an exemplification essay, you need a firm understanding of the topic and its main points. Based on this, you will formulate your main point or idea that you want to argue and from this you will draft your thesis statement. What is a Good Hook For an Exemplification Essay? A good hook for an exemplification essay is one that is guaranteed to grab the reader’s attention. This includes, but is not limited to, a relevant anecdote, quote, statistic or question. Is There a Difference Between an Illustration Essay and an Exemplification Essay? A well-written illustration essay explains how to use something and does so with the support of detailed information and evidence on how to use. An exemplification essay is written to prove a point. Based on the definitions, it could be stated that an illustration essay and exemplification essay are similar as both are used to explain and prove a point. What is a Good Exemplification Essay Topic? A good exemplification essay topic is one that is not too narrow or broad so that it is difficult to determine the main point/thesis and to argue it. Good exemplification essay topics allow for exploration of ideas with a variety of sources for information. What are Exemplification Essay Transition Words? Exemplification essay transition words include such as words as: for example, for instance, to illustrate, to demonstrate, etc. How to Write a Conclusion Paragraph for an Exemplification Essay? In the conclusion for an exemplification essay, restate your thesis and summarize your main points. Be sure to leave your reader with at least one main point that you want them to recollect. To sum up, writing an exemplification essay is more about using examples to defend or support an argument, or claim. Brief examples are commonly used in an essay, but using concise details will illustrate the ideas clearer. And make sure that your way of writing is unambiguous. Still having trouble? Go over the tips outlined above on how to write an exemplification essay or let experts help you with it. Psst... 98% of SolidEssay users report better grades! Get your original paper! 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15097	https://proofwiki.org/wiki/Sine_and_Cosine_of_Supplementary_Angles	Sine and Cosine of Supplementary Angles - ProofWiki Sine and Cosine of Supplementary Angles From ProofWiki Jump to navigationJump to search Theorem Sine of Supplementary Angle sin(π−θ)=sin θ sin⁡(π−θ)=sin⁡θ where sin sin denotes sine. That is, the sine of an angle equals its supplement. Cosine of Supplementary Angle cos(π−θ)=−cos θ cos⁡(π−θ)=−cos⁡θ where cos cos denotes cosine. That is, the cosine of an angle is the negative of its supplement. Retrieved from " Categories: Sine Function Cosine Function Supplementary Angles Navigation menu Personal tools Log in Request account Namespaces Page Discussion [x] English Views Read View source View history [x] More Search Navigation Main Page Community discussion Community portal Recent changes Random proof Help FAQ P r∞f W i k i P r∞f W i k i L A T E X L A T E X commands ProofWiki.org Proof Index Definition Index Symbol Index Axiom Index Mathematicians Books Sandbox All Categories Glossary Jokes To Do Proofread Articles Wanted Proofs More Wanted Proofs Help Needed Research Required Stub Articles Tidy Articles Improvements Invited Refactoring Missing Links Maintenance Tools What links here Related changes Special pages Printable version Permanent link Page information This page was last modified on 6 November 2022, at 09:40 and is 256 bytes Content is available under Creative Commons Attribution-ShareAlike License unless otherwise noted. Privacy policy About ProofWiki Disclaimers
15098	https://uodiyala.edu.iq/uploads/PDF%20ELIBRARY%20UODIYALA/EL96/Hayt%20Engineering%20Circuit%20Analysis%208th%20txtbk.pdf	This page intentionally left blank The Resistor Color Code Band color Black Brown Red Orange Yellow Green Blue Violet Gray White Numeric value 0 1 2 3 4 5 6 7 8 9 1st number 2nd number Tolerance band (e.g. gold = 5% silver = 10%, none = 20%) Multiplier 1. Write down the numeric value corresponding to the ﬁrst band on the left. 2. Write down the numeric value corresponding to the second band from the left. 3. Write down the number of zeros indicated by the multiplier band, which represents a power of 10 (black = no extra zeros, brown = 1 zero, etc.). A gold multiplier band indicates that the decimal is shifted one place to the left; a silver multiplier band indicates that the decimal is shifted two places to the left. 4. The tolerance band represents the precision. So, for example, we would not be surprised to ﬁnd a 100 5 percent tolerance resistor that measures anywhere in the range of 95 to 105 . Example Red Red Orange Gold = 22,000 or 22 × 103 = 22 k, 5% tolerance Blue Gray Gold = 6.8 or 68 × 10−1 = 6.8 , 20% tolerance Standard 5 Percent Tolerance Resistor Values 1.0 1.1 1.2 1.3 1.5 1.6 1.8 2.0 2.2 2.4 2.7 3.0 3.3 3.6 3.9 4.3 4.7 5.1 5.6 6.2 6.8 7.5 8.2 9.1 10. 11. 12. 13. 15. 16. 18. 20. 22. 24. 27. 30. 33. 36. 39. 43. 47. 51. 56. 62. 68. 75. 82. 91. 100 110 120 130 150 160 180 200 220 240 270 300 330 360 390 430 470 510 560 620 680 750 820 910 1.0 1.1 1.2 1.3 1.5 1.6 1.8 2.0 2.2 2.4 2.7 3.0 3.3 3.6 3.9 4.3 4.7 5.1 5.6 6.2 6.8 7.5 8.2 9.1 k 10. 11. 12. 13. 15. 16. 18. 20. 22. 24. 27. 30. 33. 36. 39. 43. 47. 51. 56. 62. 68. 75. 82. 91. k 100 110 120 130 150 160 180 200 220 240 270 300 330 360 390 430 470 510 560 620 680 750 820 910 k 1.0 1.1 1.2 1.3 1.5 1.6 1.8 2.0 2.2 2.4 2.7 3.0 3.3 3.6 3.9 4.3 4.7 5.1 5.6 6.2 6.8 7.5 8.2 9.1 M TABLE ●14.1 Laplace Transform Pairs f(t) = −1 {F(s)} F(s) = {f(t)} f(t) = −1 {F(s)} F(s) = {f(t)} δ(t) 1 1 β −α (e−αt −e−βt)u(t) 1 (s + α)(s + β) u(t) 1 s sin ωt u(t) ω s2 + ω2 tu(t) 1 s2 cos ωt u(t) s s2 + ω2 tn−1 (n −1)!u(t), n = 1, 2, . . . 1 sn sin(ωt + θ) u(t) s sin θ + ω cos θ s2 + ω2 e−αtu(t) 1 s + α cos(ωt + θ) u(t) s cos θ −ω sin θ s2 + ω2 te−αtu(t) 1 (s + α)2 e−αt sin ωt u(t) ω (s + α)2 + ω2 tn−1 (n −1)!e−αtu(t), n = 1, 2, . . . 1 (s + α)n e−αt cos ωt u(t) s + α (s + α)2 + ω2 TABLE ●6.1 Summary of Basic Op Amp Circuits Name Circuit Schematic Input-Output Relation Inverting Ampliﬁer vout = −Rf R1 vin Noninverting Ampliﬁer vout = 1 + Rf R1 vin Voltage Follower vout = vin (also known as a Unity Gain Ampliﬁer) Summing Ampliﬁer vout = −Rf R (v1 + v2 + v3) Difference Ampliﬁer vout = v2 −v1 Rf R1 – + i i + – vin vout + – Rf R1 – + + – vin vout + – – + vin + – vout + – – + i vout + – R R RL R v1 va vb Rf i3 i2 i1 + – v2 + – v3 + – – + i vout + – R RL R R v1 va vb R i2 i1 + – v2 + – ENGINEERING CIRCUIT ANALYSIS This page intentionally left blank ENGINEERING CIRCUIT ANALYSIS EIGHTH EDITION William H. Hayt, Jr. (deceased) Purdue University Jack E. Kemmerly (deceased) California State University Steven M. Durbin University at Buffalo The State University of New York ENGINEERING CIRCUIT ANALYSIS, EIGHTH EDITION Published by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New York, NY 10020. Copyright © 2012 by The McGraw-Hill Companies, Inc. All rights reserved. Previous editions © 2007, 2002, and 1993. Printed in the United States of America. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 0 DOW/DOW 1 0 9 8 7 6 5 4 3 2 1 ISBN 978-0-07-352957-8 MHID 0-07-352957-5 Vice President & Editor-in-Chief: Marty Lange Vice President & Director of Specialized Publishing: Janice M. Roerig-Blong Editorial Director: Michael Lange Global Publisher: Raghothaman Srinivasan Senior Marketing Manager: Curt Reynolds Developmental Editor: Darlene M. Schueller Lead Project Manager: Jane Mohr Buyer: Kara Kudronowicz Design Coordinator: Brenda A. Rolwes Senior Photo Research Coordinator: John C. Leland Senior Media Project Manager: Tammy Juran Compositor: MPS Limited, a Macmillan Company Typeface: 10/12 Times Roman Printer: R. R. Donnelley Cover Image: © Getty Images Cover Designer: Studio Montage, St. Louis, Missouri MATLAB is a registered trademark of The MathWorks, Inc. PSpice is a registered trademark of Cadence Design Systems, Inc. The following photos are courtesy of Steve Durbin: Page 5, Fig. 2.22a, 2.24a–c, 5.34, 6.1a, 7.2a–c, 7.11a–b, 13.15, 17.29 Library of Congress Cataloging-in-Publication Data Hayt, William Hart, 1920–1999 Engineering circuit analysis / William H. Hayt, Jr., Jack E. Kemmerly, Steven M. Durbin. — 8th ed. p. cm. Includes index. ISBN 978-0-07-352957-8 1. Electric circuit analysis. 2. Electric network analysis. I. Kemmerly, Jack E. (Jack Ellsworth), 1924–1998 II. Durbin, Steven M. III. Title. TK454.H4 2012 621.319'2—dc22 2011009912 www.mhhe.com To Sean and Kristi. The best part of every day. This page intentionally left blank WILLIAM H. HAYT, Jr., received his B.S. and M.S. at Purdue University and his Ph.D. from the University of Illinois. After spending four years in industry, Professor Hayt joined the faculty of Purdue University, where he served as Professor and Head of the School of Electrical Engineering, and as Professor Emeritus after retiring in 1986. Besides Engineering Circuit Analysis, Professor Hayt authored three other texts, including Engineering Electromagnetics, now in its eighth edition with McGraw-Hill. Professor Hayt’s professional society memberships included Eta Kappa Nu, Tau Beta Pi, Sigma Xi, Sigma Delta Chi, Fellow of IEEE, ASEE, and NAEB. While at Purdue, he received numerous teaching awards, including the univer-sity’s Best Teacher Award. He is also listed in Purdue’s Book of Great Teachers, a permanent wall display in the Purdue Memorial Union, dedi-cated on April 23, 1999. The book bears the names of the inaugural group of 225 faculty members, past and present, who have devoted their lives to excellence in teaching and scholarship. They were chosen by their students and their peers as Purdue’s ﬁnest educators. JACK E. KEMMERLY received his B.S. magna cum laude from The Catholic University of America, M.S. from University of Denver, and Ph.D. from Purdue University. Professor Kemmerly ﬁrst taught at Purdue University and later worked as principal engineer at the Aeronutronic Division of Ford Motor Company. He then joined California State University, Fullerton, where he served as Professor, Chairman of the Faculty of Electrical Engi-neering, Chairman of the Engineering Division, and Professor Emeritus. Professor Kemmerly’s professional society memberships included Eta Kappa Nu, Tau Beta Pi, Sigma Xi, ASEE, and IEEE (Senior Member). His pursuits outside of academe included being an ofﬁcer in the Little League and a scoutmaster in the Boy Scouts. STEVEN M. DURBIN received the B.S., M.S. and Ph.D. degrees in Electrical EngineeringfromPurdueUniversity,WestLafayette,Indiana.Subsequently,he was with the Department of Electrical Engineering at Florida State University and FloridaA&M University before joining the University of Canterbury, New Zealand,in2000.SinceAugust2010,hehasbeenwiththeUniversityatBuffalo, The State University of NewYork, where he holds a joint appointment between the Departments of Electrical Engineering and Physics. His teaching interests include circuits, electronics, electromagnetics, solid-state electronics and nanotechnology. His research interests are primarily concerned with the developmentofnewsemiconductormaterials—inparticularthosebasedonox-ide and nitride compounds—as well as novel optoelectronic device structures. HeisafoundingprincipalinvestigatoroftheMacDiarmidInstituteforAdvanced Materials and Nanotechnology, a New Zealand National Centre of Research Excellence,andcoauthorofover100technicalpublications.Heisaseniormem-ber of the IEEE, and a member of Eta Kappa Nu, the Electron Devices Society, the Materials Research Society, the AVS (formerly the American Vacuum Society),theAmericanPhysicalSociety,andtheRoyalSocietyofNewZealand. ABOUT THE AUTHORS • vii This page intentionally left blank PREFACE xv 1 ● INTRODUCTION 1 2 ● BASIC COMPONENTS AND ELECTRIC CIRCUITS 9 3 ● VOLTAGE AND CURRENT LAWS 39 4 ● BASIC NODAL AND MESH ANALYSIS 79 5 ● HANDY CIRCUIT ANALYSIS TECHNIQUES 123 6 ● THE OPERATIONAL AMPLIFIER 175 7 ● CAPACITORS AND INDUCTORS 217 8 ● BASIC RL AND RC CIRCUITS 261 9 ● THE RLC CIRCUIT 321 10 ● SINUSOIDAL STEADY-STATE ANALYSIS 371 11 ● AC CIRCUIT POWER ANALYSIS 421 12 ● POLYPHASE CIRCUITS 457 13 ● MAGNETICALLY COUPLED CIRCUITS 493 14 ● COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM 533 15 ● CIRCUIT ANALYSIS IN THE s-DOMAIN 571 16 ● FREQUENCY RESPONSE 619 17 ● TWO-PORT NETWORKS 687 18 ● FOURIER CIRCUIT ANALYSIS 733 Appendix 1 AN INTRODUCTION TO NETWORK TOPOLOGY 791 Appendix 2 SOLUTION OF SIMULTANEOUS EQUATIONS 803 Appendix 3 A PROOF OF THÉVENIN’S THEOREM 811 Appendix 4 A PSPICE® TUTORIAL 813 Appendix 5 COMPLEX NUMBERS 817 Appendix 6 A BRIEF MATLAB® TUTORIAL 827 Appendix 7 ADDITIONAL LAPLACE TRANSFORM THEOREMS 833 INDEX 839 BRIEF CONTENTS • ix This page intentionally left blank CONTENTS • xi CHAPTER 1 INTRODUCTION 1 1.1 Overview of Text 2 1.2 Relationship of Circuit Analysis to Engineering 4 1.3 Analysis and Design 5 1.4 Computer-Aided Analysis 6 1.5 Successful Problem-Solving Strategies 7 READING FURTHER 8 CHAPTER 2 BASIC COMPONENTS AND ELECTRIC CIRCUITS 9 2.1 Units and Scales 9 2.2 Charge, Current, Voltage, and Power 11 2.3 Voltage and Current Sources 17 2.4 Ohm’s Law 22 SUMMARY AND REVIEW 28 READING FURTHER 29 EXERCISES 29 CHAPTER 3 VOLTAGE AND CURRENT LAWS 39 3.1 Nodes, Paths, Loops, and Branches 39 3.2 Kirchhoff’s Current Law 40 3.3 Kirchhoff’s Voltage Law 42 3.4 The Single-Loop Circuit 46 3.5 The Single-Node-Pair Circuit 49 3.6 Series and Parallel Connected Sources 51 3.7 Resistors in Series and Parallel 55 3.8 Voltage and Current Division 61 SUMMARY AND REVIEW 66 READING FURTHER 67 EXERCISES 67 CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 79 4.1 Nodal Analysis 80 4.2 The Supernode 89 4.3 Mesh Analysis 92 4.4 The Supermesh 98 4.5 Nodal vs. Mesh Analysis: A Comparison 101 4.6 Computer-Aided Circuit Analysis 103 SUMMARY AND REVIEW 107 READING FURTHER 109 EXERCISES 109 CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 123 5.1 Linearity and Superposition 123 5.2 Source Transformations 133 5.3 Thévenin and Norton Equivalent Circuits 141 5.4 Maximum Power Transfer 152 5.5 Delta-Wye Conversion 154 5.6 Selecting an Approach: A Summary of Various Techniques 157 SUMMARY AND REVIEW 158 READING FURTHER 159 EXERCISES 159 CHAPTER 6 THE OPERATIONAL AMPLIFIER 175 6.1 Background 175 6.2 The Ideal Op Amp: A Cordial Introduction 176 6.3 Cascaded Stages 184 6.4 Circuits for Voltage and Current Sources 188 6.5 Practical Considerations 192 6.6 Comparators and the Instrumentation Ampliﬁer 203 SUMMARY AND REVIEW 206 READING FURTHER 207 EXERCISES 208 CHAPTER 7 CAPACITORS AND INDUCTORS 217 7.1 The Capacitor 217 7.2 The Inductor 225 7.3 Inductance and Capacitance Combinations 235 7.4 Consequences of Linearity 238 7.5 Simple Op Amp Circuits with Capacitors 240 7.6 Duality 242 CONTENTS xii 7.7 Modeling Capacitors and Inductors with PSpice 245 SUMMARY AND REVIEW 247 READING FURTHER 249 EXERCISES 249 CHAPTER 8 BASIC RL AND RC CIRCUITS 261 8.1 The Source-Free RL Circuit 261 8.2 Properties of the Exponential Response 268 8.3 The Source-Free RC Circuit 272 8.4 A More General Perspective 275 8.5 The Unit-Step Function 282 8.6 Driven RL Circuits 286 8.7 Natural and Forced Response 289 8.8 Driven RC Circuits 295 8.9 Predicting the Response of Sequentially Switched Circuits 300 SUMMARY AND REVIEW 306 READING FURTHER 308 EXERCISES 309 CHAPTER 9 THE RLC CIRCUIT 321 9.1 The Source-Free Parallel Circuit 321 9.2 The Overdamped Parallel RLC Circuit 326 9.3 Critical Damping 334 9.4 The Underdamped Parallel RLC Circuit 338 9.5 The Source-Free Series RLC Circuit 345 9.6 The Complete Response of the RLC Circuit 351 9.7 The Lossless LC Circuit 359 SUMMARY AND REVIEW 361 READING FURTHER 363 EXERCISES 363 CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 371 10.1 Characteristics of Sinusoids 371 10.2 Forced Response to Sinusoidal Functions 374 10.3 The Complex Forcing Function 378 10.4 The Phasor 383 10.5 Impedance and Admittance 389 10.6 Nodal and Mesh Analysis 394 10.7 Superposition, Source Transformations and Thévenin’s Theorem 397 10.8 Phasor Diagrams 406 SUMMARY AND REVIEW 409 READING FURTHER 410 EXERCISES 410 CHAPTER 11 AC CIRCUIT POWER ANALYSIS 421 11.1 Instantaneous Power 422 11.2 Average Power 424 11.3 Effective Values of Current and Voltage 433 11.4 Apparent Power and Power Factor 438 11.5 Complex Power 441 SUMMARY AND REVIEW 447 READING FURTHER 449 EXERCISES 449 CHAPTER 12 POLYPHASE CIRCUITS 457 12.1 Polyphase Systems 458 12.2 Single-Phase Three-Wire Systems 460 12.3 Three-Phase Y-Y Connection 464 12.4 The Delta () Connection 470 12.5 Power Measurement in Three-Phase Systems 476 SUMMARY AND REVIEW 484 READING FURTHER 486 EXERCISES 486 CHAPTER 13 MAGNETICALLY COUPLED CIRCUITS 493 13.1 Mutual Inductance 493 13.2 Energy Considerations 501 13.3 The Linear Transformer 505 13.4 The Ideal Transformer 512 SUMMARY AND REVIEW 522 READING FURTHER 523 EXERCISES 523 CHAPTER 14 COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM 533 14.1 Complex Frequency 533 14.2 The Damped Sinusoidal Forcing Function 537 14.3 Deﬁnition of the Laplace Transform 540 14.4 Laplace Transforms of Simple Time Functions 543 14.5 Inverse Transform Techniques 546 14.6 Basic Theorems for the Laplace Transform 553 CONTENTS xiii 14.7 The Initial-Value and Final-Value Theorems 561 SUMMARY AND REVIEW 564 READING FURTHER 565 EXERCISES 565 CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 571 15.1 Z(s) and Y(s) 571 15.2 Nodal and Mesh Analysis in the s-Domain 578 15.3 Additional Circuit Analysis Techniques 585 15.4 Poles, Zeros, and Transfer Functions 588 15.5 Convolution 589 15.6 The Complex-Frequency Plane 598 15.7 Natural Response and the s Plane 602 15.8 A Technique for Synthesizing the Voltage Ratio H(s) = Vout/Vin 606 SUMMARY AND REVIEW 610 READING FURTHER 612 EXERCISES 612 CHAPTER 16 FREQUENCY RESPONSE 619 16.1 Parallel Resonance 619 16.2 Bandwidth and High-Q Circuits 627 16.3 Series Resonance 633 16.4 Other Resonant Forms 637 16.5 Scaling 644 16.6 Bode Diagrams 648 16.7 Basic Filter Design 664 16.8 Advanced Filter Design 672 SUMMARY AND REVIEW 677 READING FURTHER 679 EXERCISES 679 CHAPTER 17 TWO-PORT NETWORKS 687 17.1 One-Port Networks 687 17.2 Admittance Parameters 692 17.3 Some Equivalent Networks 699 17.4 Impedance Parameters 708 17.5 Hybrid Parameters 713 17.6 Transmission Parameters 716 SUMMARY AND REVIEW 720 READING FURTHER 721 EXERCISES 722 CHAPTER 18 FOURIER CIRCUIT ANALYSIS 733 18.1 Trigonometric Form of the Fourier Series 733 18.2 The Use of Symmetry 743 18.3 Complete Response to Periodic Forcing Functions 748 18.4 Complex Form of the Fourier Series 750 18.5 Deﬁnition of the Fourier Transform 757 18.6 Some Properties of the Fourier Transform 761 18.7 Fourier Transform Pairs for Some Simple Time Functions 764 18.8 The Fourier Transform of a General Periodic Time Function 769 18.9 The System Function and Response in the Frequency Domain 770 18.10 The Physical Signiﬁcance of the System Function 777 SUMMARY AND REVIEW 782 READING FURTHER 783 EXERCISES 783 APPENDIX 1 AN INTRODUCTION TO NETWORK TOPOLOGY 791 APPENDIX 2 SOLUTION OF SIMULTANEOUS EQUATIONS 803 APPENDIX 3 A PROOF OF THÉVENIN’S THEOREM 811 APPENDIX 4 A PSPICE® TUTORIAL 813 APPENDIX 5 COMPLEX NUMBERS 817 APPENDIX 6 A BRIEF MATLAB® TUTORIAL 827 APPENDIX 7 ADDITIONAL LAPLACE TRANSFORM THEOREMS 833 INDEX 839 This page intentionally left blank T he target audience colors everything about a book, being a major fac-tor in decisions big and small, particularly both the pace and the overall writing style. Consequently it is important to note that the au-thors have made the conscious decision to write this book to the student, and not to the instructor. Our underlying philosophy is that reading the book should be enjoyable, despite the level of technical detail that it must incor-porate. When we look back to the very ﬁrst edition of Engineering Circuit Analysis, it’s clear that it was developed speciﬁcally to be more of a con-versation than a dry, dull discourse on a prescribed set of fundamental top-ics. To keep it conversational, we’ve had to work hard at updating the book so that it continues to speak to the increasingly diverse group of students using it all over the world. Although in many engineering programs the introductory circuits course is preceded or accompanied by an introductory physics course in which electricity and magnetism are introduced (typically from a ﬁelds perspec-tive), this is not required to use this book. After ﬁnishing the course, many students ﬁnd themselves truly amazed that such a broad set of analytical tools have been derived from only three simple scientiﬁc laws—Ohm’s law and Kirchhoff’s voltage and current laws. The ﬁrst six chapters assume only a familiarity with algebra and simultaneous equations; subsequent chapters assume a ﬁrst course in calculus (derivatives and integrals) is being taken in tandem. Beyond that, we have tried to incorporate sufﬁcient details to allow the book to be read on its own. So, what key features have been designed into this book with the student in mind? First, individual chapters are organized into relatively short sub-sections, each having a single primary topic. The language has been up-dated to remain informal and to ﬂow smoothly. Color is used to highlight important information as opposed to merely improve the aesthetics of the page layout, and white space is provided for jotting down short notes and questions. New terms are deﬁned as they are introduced, and examples are placed strategically to demonstrate not only basic concepts, but problem-solving approaches as well. Practice problems relevant to the examples are placed in proximity so that students can try out the techniques for them-selves before attempting the end-of-chapter exercises. The exercises repre-sent a broad range of difﬁculties, generally ordered from simpler to more complex, and grouped according to the relevant section of each chapter. Answers to selected odd-numbered end-of-chapter exercises are posted on the book’s website at www.mhhe.com/haytdurbin8e. Engineering is an intensive subject to study, and students often ﬁnd them-selves faced with deadlines and serious workloads. This does not mean that textbooks have to be dry and pompous, however, or that coursework should never contain any element of fun. In fact, successfully solving a problem of-ten is fun, and learning how to do that can be fun as well. Determining how PREFACE • xv to best accomplish this within the context of a textbook is an ongoing process. The authors have always relied on the often very candid feedback received from our own students at Purdue University; the California State University, Fullerton; Fort Lewis College in Durango, the joint engineering program at Florida A&M University and Florida State University, the Uni-versity of Canterbury (New Zealand) and the University at Buffalo. We also rely on comments, corrections, and suggestions from instructors and students worldwide, and for this edition, consideration has been given to a new source of comments, namely, semianonymous postings on various websites. The ﬁrst edition of Engineering Circuit Analysis was written by Bill Hayt and Jack Kemmerly, two engineering professors who very much en-joyed teaching, interacting with their students, and training generations of future engineers. It was well received due to its compact structure, “to the point” informal writing style, and logical organization. There is no timidity when it comes to presenting the theory underlying a specific topic, or pulling punches when developing mathematical expressions. Everything, however, was carefully designed to assist students in their learning, present things in a straightforward fashion, and leave theory for theory’s sake to other books. They clearly put a great deal of thought into writing the book, and their enthusiasm for the subject comes across to the reader. KEY FEATURES OF THE EIGHTH EDITION • We have taken great care to retain key features from the seventh edition which were clearly working well. These include the general layout and se-quence of chapters, the basic style of both the text and line drawings, the use of four-color printing where appropriate, numerous worked examples and related practice problems, and grouping of end-of-chapter exercises accord-ing to section. Transformers continue to merit their own chapter, and com-plex frequency is brieﬂy introduced through a student-friendly extension of the phasor technique, instead of indirectly by merely stating the Laplace transform integral. We also have retained the use of icons, an idea ﬁrst in-troduced in the sixth edition: Provides a heads-up to common mistakes; Indicates a point that’s worth noting; Denotes a design problem to which there is no unique answer; Indicates a problem which requires computer-aided analysis. The introduction of engineering-oriented analysis and design software in the book has been done with the mind-set that it should assist, not replace, the learning process. Consequently, the computer icon denotes problems that are typically phrased such that the software is used to verify answers, and not simply provide them. Both MATLAB® and PSpice® are used in this context. PREFACE xvi SPECIFIC CHANGES FOR THE EIGHTH EDITION INCLUDE: • A new section in Chapter 16 on the analysis and design of multistage Butterworth ﬁlters • Over 1000 new and revised end-of-chapter exercises • A new overarching philosophy on end-of-chapter exercises, with each section containing problems similar to those solved in worked examples and practice problems, before proceeding to more complex problems to test the reader’s skills • Introduction of Chapter-Integrating Exercises at the end of each chapter. For the convenience of instructors and students, end-of-chapter exercises are grouped by section. To provide the opportunity for assigning exercises with less emphasis on an explicit solution method (for example, mesh or nodal analysis), as well as to give a broader perspective on key topics within each chapter, a select number of Chapter-Integrating Exercises appear at the end of each chapter. • New photos, many in full color, to provide connection to the real world • Updated screen captures and text descriptions of computer-aided analysis software • New worked examples and practice problems • Updates to the Practical Application feature, introduced to help students connect material in each chapter to broader concepts in engineering. Topics include distortion in ampliﬁers, modeling automotive suspension systems, practical aspects of grounding, the relationship of poles to stability, resistivity, and the memristor, sometimes called “the missing element” • Streamlining of text, especially in the worked examples, to get to the point faster • Answers to selected odd-numbered end-of-chapter exercises are posted on the book’s website at www.mhhe.com/haytdurbin8e. I joined the book in 1999, and sadly never had the opportunity to speak to either Bill or Jack about the revision process, although I count myself lucky to have taken a circuits course from Bill Hayt while I was a student at Purdue. It is a distinct privilege to serve as a coauthor to Engineering Circuit Analysis, and in working on this book I give its fundamental philos-ophy and target audience the highest priority. I greatly appreciate the many people who have already provided feedback—both positive and negative— on aspects of previous editions, and welcome others to do so as well, either through the publishers (McGraw-Hill Higher Education) or to me directly (durbin@ieee.org). Of course, this project has been a team effort, as is the case with every modern textbook. In particular I would like to thank Raghu Srinivasan (Global Publisher), Peter Massar (Sponsoring Editor), Curt Reynolds (Mar-keting Manager), Jane Mohr (Project Manager), Brittney-Corrigan-McElroy (Project Manager), Brenda Rolwes (Designer), Tammy Juran (Media Project Manager), and most importantly, Developmental Editor Darlene Schueller, who helped me with many, many details, issues, deadlines, PREFACE xvii and questions. She is absolutely the best, and I’m very grateful for all the support from the team at McGraw-Hill. I would also like to thank various McGraw-Hill representatives, especially Nazier Hassan, who dropped by whenever on campus to just say hello and ask how things were going. Spe-cial thanks are also due to Catherine Shultz and Michael Hackett, former editors who continue to keep in contact. Cadence® and The MathWorks kindly provided assistance with software-aided analysis software, which was much appreciated. Several colleagues have generously supplied or helped with photographs and technical details, for which I’m very grateful: Prof. Masakazu Kobayashi of Waseda University; Dr. Wade Enright, Prof. Pat Bodger, Prof. Rick Millane, Mr. Gary Turner, and Prof. Richard Blaikie of the University of Canterbury; and Prof. Reginald Perry and Prof. Jim Zheng of Florida A&M University and the Florida State University. For the eighth edition, the following individuals deserve acknowledgment and a debt of gratitude for taking the time to review various versions of the manuscript: Chong Koo An, The University of Ulsan Mark S. Andersland, The University of Iowa Marc Cahay, University of Cincinnati Claudio Canizares, University of Waterloo Teerapon Dachokiatawan, King Mongkut’s University of Technology North Bangkok John Durkin, The University of Akron Lauren M. Fuentes, Durham College Lalit Goel, Nanyang Technological University Rudy Hofer, Conestoga College ITAL Mark Jerabek, West Virginia University Michael Kelley, Cornell University Hua Lee, University of California, Santa Barbara Georges Livanos, Humber College Institute of Technology Ahmad Naﬁsi, Cal Poly State University Arnost Neugroschel, University of Florida Pravin Patel, Durham College Jamie Phillips, The University of Michigan Daryl Reynolds, West Virginia University G.V.K.R. Sastry, Andhra University Michael Scordilis, University of Miami Yu Sun, University of Toronto, Canada Chanchana Tangwongsan, Chulalongkorn University Edward Wheeler, Rose-Hulman Institute of Technology Xiao-Bang Xu, Clemson University Tianyu Yang, Embry-Riddle Aeronautical University Zivan Zabar, Polytechnic Institute of NYU PREFACE xviii I would also like to thank Susan Lord, University of San Diego, Archie L. Holmes, Jr., University of Virginia, Arnost Neugroschel, University of Florida, and Michael Scordilis, University of Miami, for their assistance in accuracy checking answers to selected end-of-chapter exercises. Finally, I would like to brieﬂy thank a number of other people who have contributed both directly and indirectly to the eighth edition. First and fore-most, my wife, Kristi, and our son, Sean, for their patience, understanding, support, welcome distractions, and helpful advice. Throughout the day it has always been a pleasure to talk to friends and colleagues about what should be taught, how it should be taught, and how to measure learning. In particular, Martin Allen, Richard Blaikie, Alex Cartwright, Peter Cottrell, Wade Enright, Jeff Gray, Mike Hayes, Bill Kennedy, Susan Lord, Philippa Martin, Theresa Mayer, Chris McConville, Reginald Perry, Joan Redwing, Roger Reeves, Dick Schwartz, Leonard Tung, Jim Zheng, and many others have provided me with many useful insights, as has my father, Jesse Durbin, an electrical engineering graduate of the Indiana Institute of Technology. Steven M. 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Not only do you get single sign-on with Connect and Create, you also get deep integration of McGraw-Hill content and content engines right in Blackboard. Whether you’re choosing a book for your course or building Connect assignments, all the tools you need are right where you want them—inside of Blackboard. Gradebooks are now seamless. When a student completes an integrated Connect assignment, the grade for that assignment automatically (and in-stantly) feeds your Blackboard grade center. McGraw-Hill and Blackboard can now offer you easy access to industry leading technology and content, whether your campus hosts it or we do. Be sure to ask your local McGraw-Hill representative for details. Electronic Textbook Option This text is offered through CourseSmart for both instructors and students. CourseSmart is an online resource where students can purchase the complete text online at almost half the cost of a traditional text. Purchasing the eText-book allows students to take advantage of CourseSmart’s web tools for learn-ing, which include full text search, notes and highlighting, and email tools for sharing notes between classmates. To learn more about CourseSmart options, contact your sales representative or visit www.CourseSmart.com. PREFACE xxi This page intentionally left blank PREAMBLE Although there are clear specialties within the ﬁeld of engineering, all engineers share a considerable amount of common ground, particularly when it comes to problem solving. In fact, many prac-ticing engineers ﬁnd it is possible to work in a large variety of settings and even outside their traditional specialty, as their skill set is often transferrable to other environments. Today’s engineering graduates are employed in a broad range of jobs, from design of individual components and systems, to assisting in solving socio-economic problems such as air and water pollution, urban planning, communication, mass transportation, power generation and distribu-tion, and efﬁcient use and conservation of natural resources. Circuit analysis has long been a traditional introduction to the art of problem solving from an engineering perspective, even for those whose interests lie outside electrical engineering. There are many reasons for this, but one of the best is that in today’s world it’s extremely unlikely for any engineer to encounter a system that does not in some way include electrical circuitry. As circuits be-come smaller and require less power, and power sources become smaller and cheaper, embedded circuits are seemingly everywhere. Since most engineering situations require a team effort at some stage, having a working knowledge of circuit analysis therefore helps to provide everyone on a project with the background needed for effective communication. Consequently, this book is not just about “circuit analysis” from an engineering perspective, but is also about developing basic problem-solving skills as they apply to situations an engineer is likely to encounter. As part of this, we also ﬁnd that we’re develop-ing an intuitive understanding at a general level, and often we can KEY CONCEPTS Linear versus Nonlinear Circuits Four Main Categories of Circuit Analysis: • DC Analysis • Transient Analysis • Sinusoidal Analysis • Frequency Response Circuit Analysis Beyond Circuits Analysis and Design Use of Engineering Software A Problem-Solving Strategy Introduction C H A P T E R 1 1 CHAPTER 1 INTRODUCTION 2 Not all electrical engineers routinely make use of circuit analysis, but they often bring to bear analytical and problem-solving skills learned early on in their careers. A circuit analysis course is one of the ﬁrst exposures to such concepts. (Solar Mirrors: © Corbis; Skyline: © Getty Images/PhotoLink; Oil Rig: © Getty Images; Dish: © Getty Images/J. Luke/PhotoLink) Television sets include many nonlinear circuits. A great deal of them, however, can be understood and analyzed with the assistance of linear models. (© Sony Electronics, Inc.) understand a complex system by its analogy to an electrical circuit. Before launching into all this, however, we’ll begin with a quick preview of the topics found in the remainder of the book, pausing brieﬂy to ponder the difference between analysis and design, and the evolving role computer tools play in modern engineering. 1.1 • OVERVIEW OF TEXT The fundamental subject of this text is linear circuit analysis, which some-times prompts a few readers to ask, “Is there ever any nonlinear circuit analysis?” Sure! We encounter nonlinear circuits every day: they capture and decode signals for our TVs and radios, perform calculations millions of times a second inside microprocessors, convert speech into electrical signals for transmission over phone lines, and execute many other functions outside our ﬁeld of view. In designing, testing, and implementing such nonlinear circuits, detailed analysis is unavoidable. “Then why study linear circuit analysis?” you might ask. An excellent question. The simple fact of the matter is that no physical system (including electrical circuits) is ever perfectly linear. Fortunately for us, however, a great many systems behave in a reasonably SECTION 1.1 OVERVIEW OF TEXT 3 linear fashion over a limited range—allowing us to model them as linear systems if we keep the range limitations in mind. For example, consider the common function f (x) = ex A linear approximation to this function is f (x) ≈1 + x Let’s test this out. Table 1.1 shows both the exact value and the approx-imate value of f (x) for a range of x. Interestingly, the linear approximation is exceptionally accurate up to about x = 0.1, when the relative error is still less than 1%. Although many engineers are rather quick on a calculator, it’s hard to argue that any approach is faster than just adding 1. TABLE ● 1.1 Comparison of a Linear Model for ex to Exact Value x f(x) 1 + x Relative error 0.0001 1.0001 1.0001 0.0000005% 0.001 1.0010 1.001 0.00005% 0.01 1.0101 1.01 0.005% 0.1 1.1052 1.1 0.5% 1.0 2.7183 2.0 26% Quoted to four signiﬁcant ﬁgures. Relative error 100 × ex −(1 + x) ex Linear problems are inherently more easily solved than their nonlinear counterparts. For this reason, we often seek reasonably accurate linear ap-proximations (or models) to physical situations. Furthermore, the linear models are more easily manipulated and understood—making design a more straightforward process. The circuits we will encounter in subsequent chapters all represent linear approximations to physical electric circuits. Where appropriate, brief discus-sions of potential inaccuracies or limitations to these models are provided, but generally speaking we ﬁnd them to be suitably accurate for most applications. When greater accuracy is required in practice, nonlinear models are em-ployed, but with a considerable increase in solution complexity.Adetailed dis-cussion of what constitutes a linear electric circuit can be found in Chap. 2. Linear circuit analysis can be separated into four broad categories: (1) dc analysis, where the energy sources do not change with time; (2) transient analysis, where things often change quickly; (3) sinusoidal analysis, which applies to both ac power and signals; and (4) frequency response, which is the most general of the four categories, but typically assumes something is changing with time. We begin our journey with the topic of resistive cir-cuits, which may include simple examples such as a ﬂashlight or a toaster. This provides us with a perfect opportunity to learn a number of very pow-erful engineering circuit analysis techniques, such as nodal analysis, mesh analysis, superposition, source transformation, Thévenin’s theorem, Norton’s CHAPTER 1 INTRODUCTION 4 theorem, and several methods for simplifying networks of components con-nected in series or parallel. The single most redeeming feature of resistive circuits is that the time dependence of any quantity of interest does not affect our analysis procedure. In other words, if asked for an electrical quan-tity of a resistive circuit at several speciﬁc instants in time, we do not need to analyze the circuit more than once. As a result, we will spend most of our effort early on considering only dc circuits—those circuits whose electrical parameters do not vary with time. Although dc circuits such as ﬂashlights or automotive rear window de-foggers are undeniably important in everyday life, things are often much more interesting when something happens suddenly. In circuit analysis parlance, we refer to transient analysis as the suite of techniques used to study circuits which are suddenly energized or de-energized. To make such circuits interesting, we need to add elements that respond to the rate of change of electrical quantities, leading to circuit equations which include derivatives and integrals. Fortunately, we can obtain such equations using the simple techniques learned in the ﬁrst part of our study. Still, not all time-varying circuits are turned on and off suddenly. Air conditioners, fans, and ﬂuorescent lights are only a few of the many exam-ples we may see daily. In such situations, a calculus-based approach for every analysis can become tedious and time-consuming. Fortunately, there is a better alternative for situations where equipment has been allowed to run long enough for transient effects to die out, and this is commonly referred to as ac or sinusoidal analysis, or sometimes phasor analysis. The ﬁnal leg of our journey deals with a subject known as frequency response. Working directly with the differential equations obtained in time-domain analysis helps us develop an intuitive understanding of the opera-tion of circuits containing energy storage elements (e.g., capacitors and inductors). As we shall see, however, circuits with even a relatively small number of components can be somewhat onerous to analyze, and so much more straightforward methods have been developed. These methods, which include Laplace and Fourier analysis, allow us to transform differential equations into algebraic equations. Such methods also enable us to design circuits to respond in speciﬁc ways to particular frequencies. We make use of frequency-dependent circuits every day when we dial a telephone, select our favorite radio station, or connect to the Internet. 1.2 • RELATIONSHIP OF CIRCUIT ANALYSIS TO ENGINEERING Whether we intend to pursue further circuit analysis at the completion of this course or not, it is worth noting that there are several layers to the con-cepts under study. Beyond the nuts and bolts of circuit analysis techniques lies the opportunity to develop a methodical approach to problem solving, the ability to determine the goal or goals of a particular problem, skill at collecting the information needed to effect a solution, and, perhaps equally importantly, opportunities for practice at verifying solution accuracy. Students familiar with the study of other engineering topics such as ﬂuid ﬂow, automotive suspension systems, bridge design, supply chain manage-ment, or process control will recognize the general form of many of the Frequency-dependent circuits lie at the heart of many electronic devices, and they can be a great deal of fun to design. (© The McGraw-Hill Companies, Inc.) Modern trains are powered by electric motors. Their electrical systems are best analyzed using ac or phasor analysis techniques. (Used with permission. Image copyright © 2010 M. Kobayashi. All rights reserved.) SECTION 1.3 ANALYSIS AND DESIGN 5 equations we develop to describe the behavior of various circuits. We simply need to learn how to “translate” the relevant variables (for example, replacing voltage with force, charge with distance, resistance with friction coefﬁcient, etc.) to ﬁnd that we already know how to work a new type of problem. Very often, if we have previous experience in solving a similar or related problem, our intuition can guide us through the solution of a totally new problem. What we are about to learn regarding linear circuit analysis forms the basis for many subsequent electrical engineering courses. The study of elec-tronics relies on the analysis of circuits with devices known as diodes and transistors, which are used to construct power supplies, ampliﬁers, and dig-ital circuits. The skills which we will develop are typically applied in a rapid, methodical fashion by electronics engineers, who sometimes can analyze a complicated circuit without even reaching for a pencil! The time-domain and frequency-domain chapters of this text lead directly into discussions of signal processing, power transmission, control theory, and communications. We ﬁnd that frequency-domain analysis in particular is an extremely powerful technique, easily applied to any physical system subjected to time-varying excitation, and particularly helpful in the design of ﬁlters. 1.3 • ANALYSIS AND DESIGN Engineers take a fundamental understanding of scientiﬁc principles, com-bine this with practical knowledge often expressed in mathematical terms, and (frequently with considerable creativity) arrive at a solution to a given problem. Analysis is the process through which we determine the scope of a problem, obtain the information required to understand it, and compute the parameters of interest. Design is the process by which we synthesize something new as part of the solution to a problem. Generally speaking, there is an expectation that a problem requiring design will have no unique solution, whereas the analysis phase typically will. Thus, the last step in designing is always analyzing the result to see if it meets speciﬁcations. A molecular beam epitaxy crystal growth facility. The equations governing its operation closely resemble those used to describe simple linear circuits. An example of a robotic manipulator. The feedback control system can be modeled using linear circuit elements to determine situations in which the operation may become unstable. (NASA Marshall Space Flight Center.) CHAPTER 1 INTRODUCTION 6 This text is focused on developing our ability to analyze and solve problems because it is the starting point in every engineering situation. The philosophy of this book is that we need clear explanations, well-placed ex-amples, and plenty of practice to develop such an ability.Therefore, elements of design are integrated into end-of-chapter problems and later chapters so as to be enjoyable rather than distracting. 1.4 • COMPUTER-AIDED ANALYSIS Solving the types of equations that result from circuit analysis can often be-come notably cumbersome for even moderately complex circuits. This of course introduces an increased probability that errors will be made, in addi-tion to considerable time in performing the calculations. The desire to ﬁnd a tool to help with this process actually predates electronic computers, with purely mechanical computers such as the Analytical Engine designed by Charles Babbage in the 1880s proposed as possible solutions. Perhaps the earliest successful electronic computer designed for solution of differential equations was the 1940s-era ENIAC, whose vacuum tubes ﬁlled a large room. With the advent of low-cost desktop computers, however, computer-aided circuit analysis has developed into an invaluable everyday tool which has become an integral part of not only analysis but design as well. One of the most powerful aspects of computer-aided design is the rela-tively recent integration of multiple programs in a fashion transparent to the user. This allows the circuit to be drawn schematically on the screen, re-duced automatically to the format required by an analysis program (such as SPICE, introduced in Chap. 4), and the resulting output smoothly trans-ferred to a third program capable of plotting various electrical quantities of Charles Babbage’s “Difference Engine Number 2,” as completed by the Science Museum (London) in 1991. (© Science Museum/Science & Society Picture Library.) Two proposed designs for a next-generation space shuttle. Although both contain similar elements, each is unique. (NASA Dryden Flight Research Center.) SECTION 1.5 SUCCESSFUL PROBLEM-SOLVING STRATEGIES 7 interest that describe the operation of the circuit. Once the engineer is satis-ﬁed with the simulated performance of the design, the same software can generate the printed circuit board layout using geometrical parameters in the components library. This level of integration is continually increasing, to the point where soon an engineer will be able to draw a schematic, click a few buttons, and walk to the other side of the table to pick up a manufac-tured version of the circuit, ready to test! The reader should be wary, however, of one thing. Circuit analysis soft-ware, although fun to use, is by no means a replacement for good old-fashioned paper-and-pencil analysis. We need to have a solid understanding of how circuits work in order to develop an ability to design them. Simply going through the motions of running a particular software package is a little like playing the lottery: with user-generated entry errors, hidden default parame-ters in the myriad of menu choices, and the occasional shortcoming of human-written code, there is no substitute for having at least an approximate idea of the expected behavior of a circuit. Then, if the simulation result does not agree with expectations, we can ﬁnd the error early, rather than after it’s too late. Still, computer-aided analysis is a powerful tool. It allows us to vary pa-rameter values and evaluate the change in circuit performance, and to con-sider several variations to a design in a straightforward manner. The result is a reduction of repetitive tasks, and more time to concentrate on engineer-ing details. 1.5 • SUCCESSFUL PROBLEM-SOLVING STRATEGIES As the reader might have picked up, this book is just as much about problem solving as it is about circuit analysis. As a result, the expectation is that during your time as an engineering student, you are learning how to solve problems— so just at this moment, those skills are not yet fully developed. As you proceed An ampliﬁer circuit drawn using a commercial schematic capture software package. through your course of study, you will pick up techniques that work for you, and likely continue to do so as a practicing engineer. At this stage, then, we should spend a few moments discussing some basic points. The ﬁrst point is that by far, the most common difﬁculty encountered by engineering students is not knowing how to start a problem. This improves with experience, but early on that’s of no help. The best advice we can give is to adopt a methodical approach, beginning with reading the problem statement slowly and carefully (and more than once, if needed). Since experience usually gives us some type of insight into how to deal with a speciﬁc problem, worked examples appear throughout the book. Rather than just read them, however, it might be helpful to work through them with a pencil and a piece of paper. Once we’ve read through the problem, and feel we might have some use-ful experience, the next step is to identify the goal of the problem—perhaps to calculate a voltage or a power, or to select a component value. Knowing where we’re going is a big help. The next step is to collect as much infor-mation as we can, and to organize it somehow. At this point we’re still not ready to reach for the calculator. It’s best ﬁrst to devise a plan, perhaps based on experience, perhaps based simply on our intuition. Sometimes plans work, and sometimes they don’t. Starting with our initial plan, it’s time to construct an initial set of equations. If they appear complete, we can solve them. If not, we need to either locate more information, modify our plan, or both. Once we have what appears to be a working solution, we should not stop, even if exhausted and ready for a break. No engineering problem is solved unless the solution is tested somehow. We might do this by per-forming a computer simulation, or solving the problem a different way, or perhaps even just estimating what answer might be reasonable. Since not everyone likes to read to learn, these steps are summarized in the adjacent ﬂowchart. This is just one particular problem-solving strategy, and the reader of course should feel free to modify it as necessary. The real key, however, is to try and learn in a relaxed, low-stress environment free of distractions. Experience is the best teacher, and learning from our own mis-takes will always be part of the process of becoming a skilled engineer. READING FURTHER This relatively inexpensive, best-selling book teaches the reader how to develop winning strategies in the face of seemingly impossible problems: G. Polya, How to Solve It. Princeton, N.J.: Princeton University Press, 1971. CHAPTER 1 INTRODUCTION 8 Read the problem statement slowly and carefully. Identify the goal of the problem. Collect the known information. Devise a plan. Determine if additional information is required. Verify the solution. Is it reasonable or expected? End. Yes No Yes No Construct an appropriate set of equations. Attempt a solution. INTRODUCTION In conducting circuit analysis, we often ﬁnd ourselves seeking spe-ciﬁc currents, voltages, or powers, so here we begin with a brief de-scription of these quantities. In terms of components that can be used to build electrical circuits, we have quite a few from which to choose. We initially focus on the resistor, a simple passive compo-nent, and a range of idealized active sources of voltage and current. As we move forward, new components will be added to the inven-tory to allow more complex (and useful) circuits to be considered. A quick word of advice before we begin: Pay close attention to the role of “+” and “−” signs when labeling voltages, and the sig-niﬁcance of the arrow in deﬁning current; they often make the difference between wrong and right answers. 2.1 • UNITS AND SCALES In order to state the value of some measurable quantity, we must give both a number and a unit, such as “3 meters.” Fortunately, we all use the same number system. This is not true for units, and a lit-tle time must be spent in becoming familiar with a suitable system. We must agree on a standard unit and be assured of its permanence and its general acceptability. The standard unit of length, for exam-ple, should not be deﬁned in terms of the distance between two marks on a certain rubber band; this is not permanent, and further-more everybody else is using another standard. The most frequently used system of units is the one adopted by the National Bureau of Standards in 1964; it is used by all major professional engineering societies and is the language in which to-day’s textbooks are written. This is the International System of Units (abbreviated SI in all languages), adopted by the General KEY CONCEPTS Basic Electrical Quantities and Associated Units: Charge, Current, Voltage, and Power Current Direction and Voltage Polarity The Passive Sign Convention for Calculating Power Ideal Voltage and Current Sources Dependent Sources Resistance and Ohm’s Law Basic Components and Electric Circuits C H A P T E R 2 9 CHAPTER 2 BASIC COMPONENTS AND ELECTRIC CIRCUITS 10 Conference on Weights and Measures in 1960. Modified several times since, the SI is built upon seven basic units: the meter, kilogram, second, ampere, kelvin, mole, and candela (see Table 2.1). This is a “metric system,” some form of which is now in common use in most countries of the world, although it is not yet widely used in the United States. Units for other quan-tities such as volume, force, energy, etc., are derived from these seven base units. The “calorie” used with food, drink, and exercise is really a kilocalorie, 4.187 J. TABLE ●2.1 SI Base Units Base Quantity Name Symbol length meter m mass kilogram kg time second s electric current ampere A thermodynamic temperature kelvin K amount of substance mole mol luminous intensity candela cd TABLE ●2.2 SI Preﬁxes Factor Name Symbol Factor Name Symbol 10−24 yocto y 1024 yotta Y 10−21 zepto z 1021 zetta Z 10−18 atto a 1018 exa E 10−15 femto f 1015 peta P 10−12 pico p 1012 tera T 10−9 nano n 109 giga G 10−6 micro μ 106 mega M 10−3 milli m 103 kilo k 10−2 centi c 102 hecto h 10−1 deci d 101 deka da The fundamental unit of work or energy is the joule (J). One joule (a kg m2 s−2 in SI base units) is equivalent to 0.7376 foot pound-force (ft · lbf). Other energy units include the calorie (cal), equal to 4.187 J; the British thermal unit (Btu), which is 1055 J; and the kilowatthour (kWh), equal to 3.6 × 106 J. Power is deﬁned as the rate at which work is done or energy is expended. The fundamental unit of power is the watt (W), deﬁned as 1 J/s. One watt is equivalent to 0.7376 ft · lbf/s or, equivalently, 1/745.7 horsepower (hp). The SI uses the decimal system to relate larger and smaller units to the basic unit, and employs preﬁxes to signify the various powers of 10. A list of preﬁxes and their symbols is given in Table 2.2; the ones most commonly encountered in engineering are highlighted. There is some inconsistency regarding whether units named after a person should be capitalized. Here, we will adopt the most contemporary convention,1,2 where such units are written out in lowercase (e.g., watt, joule), but abbreviated with an uppercase symbol (e.g., W, J). _________ (1) H. Barrell, Nature 220, 1968, p. 651. (2) V. N. Krutikov, T. K. Kanishcheva, S. A. Kononogov, L. K. Isaev, and N. I. Khanov, Measurement Techniques 51, 2008, p. 1045. 11 These preﬁxes are worth memorizing, for they will appear often both in this text and in other technical work. Combinations of several preﬁxes, such as the millimicrosecond, are unacceptable. It is worth noting that in terms of distance, it is common to see “micron (μm)” as opposed to “microme-ter,” and often the angstrom (Å) is used for 10−10 meter. Also, in circuit analysis and engineering in general, it is fairly common to see numbers ex-pressed in what are frequently termed “engineering units.” In engineering notation, a quantity is represented by a number between 1 and 999 and an appropriate metric unit using a power divisible by 3. So, for example, it is preferable to express the quantity 0.048 W as 48 mW, instead of 4.8 cW, 4.8 × 10−2 W, or 48,000 μW. As seen in Table 2.1, the base units of the SI are not derived from fundamental physical quantities. Instead, they represent historically agreed upon measurements, leading to deﬁnitions which occasionally seem backward. For example, it would make more sense physically to deﬁne the ampere based on electronic charge. ■FIGURE 2.1 The deﬁnition of current illustrated using current ﬂowing through a wire; 1 ampere corresponds to 1 coulomb of charge passing through the arbitrarily chosen cross section in 1 second. Cross section Direction of charge motion Individual charges (1) Although the occasional appearance of smoke may seem to suggest otherwise. . . SECTION 2.2 CHARGE, CURRENT, VOLTAGE, AND POWER PRACTICE ● 2.1 A krypton ﬂuoride laser emits light at a wavelength of 248 nm. This is the same as: (a) 0.0248 mm; (b) 2.48 μm; (c) 0.248 μm; (d) 24,800 Å. 2.2 A single logic gate in a prototype integrated circuit is found to be capable of switching from the “on” state to the “off” state in 12 ps. This corresponds to: (a) 1.2 ns; (b) 120 ns; (c) 1200 ns; (d) 12,000 ns. 2.3 A typical incandescent reading lamp runs at 60 W. If it is left on constantly, how much energy (J) is consumed per day, and what is the weekly cost if energy is charged at a rate of 12.5 cents per kilowatthour? Ans: 2.1 (c); 2.2 (d); 2.3 5.18 MJ, $1.26. 2.2 • CHARGE, CURRENT, VOLTAGE, AND POWER Charge One of the most fundamental concepts in electric circuit analysis is that of charge conservation. We know from basic physics that there are two types of charge: positive (corresponding to a proton) and negative (corresponding to an electron). For the most part, this text is concerned with circuits in which only electron ﬂow is relevant. There are many devices (such as bat-teries, diodes, and transistors) in which positive charge motion is important to understanding internal operation, but external to the device we typically concentrate on the electrons which flow through the connecting wires. Although we continuously transfer charges between different parts of a cir-cuit, we do nothing to change the total amount of charge. In other words, we neither create nor destroy electrons (or protons) when running electric circuits.1 Charge in motion represents a current. In the SI system, the fundamental unit of charge is the coulomb (C). It is defined in terms of the ampere by counting the total charge that passes through an arbitrary cross section of a wire during an interval of one second; one coulomb is measured each second for a wire carrying a current of 1 ampere (Fig. 2.1). In this system of units, a single electron has a charge of −1.602 × 10−19 C and a single proton has a charge of +1.602 × 10−19 C. A quantity of charge that does not change with time is typically repre-sented by Q. The instantaneous amount of charge (which may or may not be time-invariant) is commonly represented by q(t), or simply q. This conven-tion is used throughout the remainder of the text: capital letters are reserved for constant (time-invariant) quantities, whereas lowercase letters represent the more general case. Thus, a constant charge may be represented by either Q or q, but an amount of charge that changes over time must be represented by the lowercase letter q. Current The idea of “transfer of charge” or “charge in motion” is of vital importance to us in studying electric circuits because, in moving a charge from place to place, we may also transfer energy from one point to another. The familiar cross-country power-transmission line is a practical example of a device that transfers energy. Of equal importance is the possibility of varying the rate at which the charge is transferred in order to communicate or transfer information. This process is the basis of communication systems such as radio, television, and telemetry. The current present in a discrete path, such as a metallic wire, has both a numerical value and a direction associated with it; it is a measure of the rate at which charge is moving past a given reference point in a speciﬁed direction. Once we have speciﬁed a reference direction, we may then let q(t) be the total charge that has passed the reference point since an arbitrary time t = 0, moving in the deﬁned direction. A contribution to this total charge will be negative if negative charge is moving in the reference direction, or if posi-tive charge is moving in the opposite direction. As an example, Fig. 2.2 shows a history of the total charge q(t) that has passed a given reference point in a wire (such as the one shown in Fig. 2.1). We deﬁne the current at a speciﬁc point and ﬂowing in a speciﬁed direc-tion as the instantaneous rate at which net positive charge is moving past that point in the speciﬁed direction. This, unfortunately, is the historical de-ﬁnition, which came into popular use before it was appreciated that current in wires is actually due to negative, not positive, charge motion. Current is symbolized by I or i, and so i = dq dt The unit of current is the ampere (A), named afterA. M.Ampère, a French physicist. It is commonly abbreviated as an “amp,” although this is unofﬁcial and somewhat informal. One ampere equals 1 coulomb per second. Using Eq. , we compute the instantaneous current and obtain Fig. 2.3. The use of the lowercase letter i is again to be associated with an instantaneous value; an uppercase I would denote a constant (i.e., time-invariant) quantity. The charge transferred between time t0 and t may be expressed as a deﬁnite integral: q(t) q(t0) dq = t t0 i dt′ The total charge transferred over all time is thus given by q(t) = t t0 i dt′ + q(t0) CHAPTER 2 BASIC COMPONENTS AND ELECTRIC CIRCUITS 12 ■FIGURE 2.2 A graph of the instantaneous value of the total charge q(t) that has passed a given reference point since t 0. 3 2 1 0 6 5 4 –1 –2 1 2 3 4 5 6 7 8 q(t) (C) t(s) ■FIGURE 2.3 The instantaneous current i dq/dt, where q is given in Fig. 2.2. –0.5 –1 –1.5 1.5 1 0.5 0 –2 1 2 3 4 5 6 7 8 i(t) (A) t(s) Several different types of current are illustrated in Fig. 2.4. A current that is constant in time is termed a direct current, or simply dc, and is shown by Fig. 2.4a. We will ﬁnd many practical examples of currents that vary si-nusoidally with time (Fig. 2.4b); currents of this form are present in normal household circuits. Such a current is often referred to as alternating current, or ac. Exponential currents and damped sinusoidal currents (Fig. 2.4c and d) will also be encountered later. We create a graphical symbol for current by placing an arrow next to the conductor. Thus, in Fig. 2.5a the direction of the arrow and the value 3 A in-dicate either that a net positive charge of 3 C/s is moving to the right or that a net negative charge of −3 C/s is moving to the left each second. In Fig. 2.5b there are again two possibilities: either −3 A is ﬂowing to the left or +3 A is ﬂowing to the right. All four statements and both ﬁgures represent currents that are equivalent in their electrical effects, and we say that they are equal. Anonelectrical analogy that may be easier to visualize is to think in terms of a personal savings account: e.g., a deposit can be viewed as either a negative cash ﬂow out of your account or a positive ﬂow into your account. It is convenient to think of current as the motion of positive charge, even though it is known that current ﬂow in metallic conductors results from electron motion. In ionized gases, in electrolytic solutions, and in some semiconductor materials, however, positive charges in motion consti-tute part or all of the current. Thus, any deﬁnition of current can agree with the physical nature of conduction only part of the time. The deﬁnition and symbolism we have adopted are standard. It is essential that we realize that the current arrow does not indicate the “actual” direction of current ﬂow but is simply part of a convention that allows us to talk about “the current in the wire” in an unambiguous manner. The arrow is a fundamental part of the deﬁnition of a current! Thus, to talk about the value of a current i1(t) without specifying the arrow is to discuss an undeﬁned entity. For example, Fig. 2.6a and b are meaningless represen-tations of i1(t), whereas Fig. 2.6c is complete. SECTION 2.2 CHARGE, CURRENT, VOLTAGE, AND POWER 13 ■FIGURE 2.4 Several types of current: (a) Direct current (dc). (b) Sinusoidal current (ac). (c) Exponential current. (d) Damped sinusoidal current. i t (d) t i (c) i t (b) i t (a) ■FIGURE 2.5 Two methods of representation for the exact same current. –3 A (b) 3 A (a) i1(t) i1(t) (a) (b) i1(t) i1(t) (c) ■FIGURE 2.6 (a, b) Incomplete, improper, and incorrect deﬁnitions of a current. (c) The correct deﬁnition of i1(t). I2 I1 ■FIGURE 2.7 PRACTICE ● 2.4 In the wire of Fig. 2.7, electrons are moving left to right to create a current of 1 mA. Determine I1 and I2. Ans: I1 = −1 mA; I2 = +1 mA. Voltage We must now begin to refer to a circuit element, something best deﬁned in general terms to begin with. Such electrical devices as fuses, light bulbs, re-sistors, batteries, capacitors, generators, and spark coils can be represented by combinations of simple circuit elements. We begin by showing a very general circuit element as a shapeless object possessing two terminals at which connections to other elements may be made (Fig. 2.8). There are two paths by which current may enter or leave the element. In subsequent discussions we will deﬁne particular circuit elements by describ-ing the electrical characteristics that may be observed at their terminals. In Fig. 2.8, let us suppose that a dc current is sent into terminal A, through the general element, and back out of terminal B. Let us also assume that pushing charge through the element requires an expenditure of energy. We then say that an electrical voltage (or a potential difference) exists be-tween the two terminals, or that there is a voltage “across” the element. Thus, the voltage across a terminal pair is a measure of the work required to move charge through the element. The unit of voltage is the volt,2 and 1 volt is the same as 1 J/C. Voltage is represented by V or v. Avoltagecanexistbetweenapairofelectricalterminalswhetheracurrent is ﬂowing or not. An automobile battery, for example, has a voltage of 12 V across its terminals even if nothing whatsoever is connected to the terminals. According to the principle of conservation of energy, the energy that is expended in forcing charge through the element must appear somewhere else. When we later meet speciﬁc circuit elements, we will note whether that energy is stored in some form that is readily available as electric energy or whether it changes irreversibly into heat, acoustic energy, or some other nonelectrical form. We must now establish a convention by which we can distinguish be-tween energy supplied to an element and energy that is supplied by the element itself. We do this by our choice of sign for the voltage of terminal A with respect to terminal B. If a positive current is entering terminal A of the element and an external source must expend energy to establish this cur-rent, then terminal A is positive with respect to terminal B. (Alternatively, we may say that terminal B is negative with respect to terminal A.) The sense of the voltage is indicated by a plus-minus pair of algebraic signs. In Fig. 2.9a, for example, the placement of the + sign at terminal A indicates that terminal A is v volts positive with respect to terminal B. If we later ﬁnd that v happens to have a numerical value of −5 V, then we may say either that A is −5 V positive with respect to B or that B is 5 V positive with respect to A. Other cases are shown in Fig. 2.9b, c, and d. Just as we noted in our deﬁnition of current, it is essential to realize that the plus-minus pair of algebraic signs does not indicate the “actual” polarity of the voltage but is simply part of a convention that enables us to talk unam-biguously about “the voltage across the terminal pair.” The deﬁnition of any voltage must include a plus-minus sign pair! Using a quantity v1(t) without specifying the location of the plus-minus sign pair is using an undeﬁned term. Figure 2.10a and b do not serve as deﬁnitions of v1(t); Fig. 2.10c does. CHAPTER 2 BASIC COMPONENTS AND ELECTRIC CIRCUITS 14 ■FIGURE 2.8 A general two-terminal circuit element. A B ■FIGURE 2.9 (a, b) Terminal B is 5 V positive with respect to terminal A; (c, d) terminal A is 5 V positive with respect to terminal B. A v = –5 V B – + (d) A v = 5 V B + – (c) A v = –5 V B + – (a) A v = 5 V B – + (b) ■FIGURE 2.10 (a, b) These are inadequate deﬁnitions of a voltage. (c) A correct deﬁnition includes both a symbol for the variable and a plus-minus symbol pair. v1(t) + – (c) (b) + – v1(t) (a) (2) We are probably fortunate that the full name of the 18th century Italian physicist, Alessandro Giuseppe Antonio Anastasio Volta, is not used for our unit of potential difference! PRACTICE ● 2.5 For the element in Fig. 2.11, v1 = 17 V. Determine v2. Ans: v2 = −17 V. Power We have already deﬁned power, and we will represent it by P or p. If one joule of energy is expended in transferring one coulomb of charge through the device in one second, then the rate of energy transfer is one watt. The absorbed power must be proportional both to the number of coulombs trans-ferred per second (current) and to the energy needed to transfer one coulomb through the element (voltage). Thus, p = vi Dimensionally, the right side of this equation is the product of joules per coulomb and coulombs per second, which produces the expected dimension of joules per second, or watts. The conventions for current, voltage, and power are shown in Fig. 2.12. We now have an expression for the power being absorbed by a circuit element in terms of a voltage across it and current through it. Voltage was deﬁned in terms of an energy expenditure, and power is the rate at which en-ergy is expended. However, no statement can be made concerning energy transfer in any of the four cases shown in Fig. 2.9, for example, until the direction of the current is speciﬁed. Let us imagine that a current arrow is placed alongside each upper lead, directed to the right, and labeled “+2 A.” First, consider the case shown in Fig. 2.9c. Terminal A is 5 V positive with respect to terminal B, which means that 5 J of energy is required to move each coulomb of positive charge into terminal A, through the object, and out terminal B. Since we are injecting +2 A (a current of 2 coulombs of positive charge per second) into terminal A, we are doing (5 J/C) × (2 C/s) 10 J of work per second on the object. In other words, the object is absorbing 10 W of power from whatever is injecting the current. We know from an earlier discussion that there is no difference between Fig. 2.9c and Fig. 2.9d, so we expect the object depicted in Fig. 2.9d to also be absorbing 10 W. We can check this easily enough: we are injecting +2 A into terminal A of the object, so +2 A ﬂows out of terminal B. Another way of saying this is that we are injecting −2 A of current into terminal B. It takes −5 J/C to move charge from terminal B to terminal A, so the object is absorbing (−5 J/C) × (−2 C/s) +10 W as expected. The only difﬁculty in describing this particular case is keeping the minus signs straight, but with a bit of care we see the correct answer can be obtained regardless of our choice of positive reference terminal (terminal A in Fig. 2.9c, and terminal B in Fig. 2.9d). SECTION 2.2 CHARGE, CURRENT, VOLTAGE, AND POWER 15 ■FIGURE 2.12 The power absorbed by the element is given by the product p vi. Alternatively, we can say that the element generates or supplies a power −vi. v + – i v2 – + v1 + – ■FIGURE 2.11 Now let’s look at the situation depicted in Fig. 2.9a, again with +2 A in-jected into terminal A. Since it takes −5 J/C to move charge from terminal A to terminal B, the object is absorbing (−5 J/C) × (2 C/s) −10 W. What does this mean? How can anything absorb negative power? If we think about this in terms of energy transfer, −10 J is transferred to the object each second through the 2 A current ﬂowing into terminal A. The object is actu-ally losing energy—at a rate of 10 J/s. In other words, it is supplying 10 J/s (i.e., 10 W) to some other object not shown in the ﬁgure. Negative absorbed power, then, is equivalent to positive supplied power. Let’s recap. Figure 2.12 shows that if one terminal of the element is v volts positive with respect to the other terminal, and if a current i is entering the element through that terminal, then a power p = vi is being absorbed by the element; it is also correct to say that a power p = vi is being delivered to the element. When the current arrow is directed into the element at the plus-marked terminal, we satisfy the passive sign convention. This conven-tion should be studied carefully, understood, and memorized. In other words, it says that if the current arrow and the voltage polarity signs are placed such that the current enters that end of the element marked with the positive sign, then the power absorbed by the element can be expressed by the product of the speciﬁed current and voltage variables. If the numerical value of the product is negative, then we say that the element is absorbing negative power, or that it is actually generating power and delivering it to some exter-nal element. For example, in Fig. 2.12 with v = 5 V and i = −4 A, the element may be described as either absorbing −20 W or generating 20 W. Conventions are only required when there is more than one way to do something, and confusion may result when two different groups try to communicate. For example, it is rather arbitrary to always place “North” at the top of a map; compass needles don’t point “up,” anyway. Still, if we were talking to people who had secretly chosen the opposite convention of placing “South” at the top of their maps, imagine the confusion that could result! In the same fashion, there is a general convention that always draws the current arrows pointing into the positive voltage terminal, regardless of whether the element supplies or absorbs power. This convention is not in-correct but sometimes results in counterintuitive currents labeled on circuit schematics. The reason for this is that it simply seems more natural to refer to positive current ﬂowing out of a voltage or current source that is supply-ing positive power to one or more circuit elements. CHAPTER 2 BASIC COMPONENTS AND ELECTRIC CIRCUITS 16 If the current arrow is directed into the “+” marked ter-minal of an element, then p vi yields the absorbed power. A negative value indicates that power is actually being generated by the element. If the current arrow is directed out of the “+” terminal of an element, then p vi yields the supplied power. A negative value in this case indicates that power is being absorbed. + – (c) –5 A 4 V – + (b) –3 A –2 V + – (a) 3 A 2 V ■FIGURE 2.13 (a, b, c) Three examples of two-terminal elements. EXAMPLE 2.1 Compute the power absorbed by each part in Fig. 2.13. In Fig. 2.13a, we see that the reference current is deﬁned consistent with the passive sign convention, which assumes that the element is absorbing power. With +3 A ﬂowing into the positive reference termi-nal, we compute P = (2 V)(3 A) = 6 W of power absorbed by the element. Figure 2.13b shows a slightly different picture. Now, we have a cur-rent of −3 A ﬂowing into the positive reference terminal. This gives us an absorbed power P = (−2 V)(−3 A) = 6 W Thus, we see that the two cases are actually equivalent:Acurrent of +3Aﬂowing into the top terminal is the same as a current of +3A ﬂowing out of the bottom terminal, or, equivalently, a current of −3A ﬂowing into the bottom terminal. Referring to Fig. 2.13c, we again apply the passive sign convention rules and compute an absorbed power P = (4 V)(−5 A) = −20 W Since we computed a negative absorbed power, this tells us that the element in Fig. 2.13c is actually supplying +20 W (i.e., it’s a source of energy). PRACTICE ● 2.6 Determine the power being absorbed by the circuit element in Fig. 2.14a. 2.7 Determine the power being generated by the circuit element in Fig. 2.14b. 2.8 Determine the power being delivered to the circuit element in Fig. 2.14c at t = 5 ms. Ans: 880 mW; 6.65 W; −15.53 W. 2.3 • VOLTAGE AND CURRENT SOURCES Using the concepts of current and voltage, it is now possible to be more spe-ciﬁc in deﬁning a circuit element. In so doing, it is important to differentiate between the physical device itself and the mathematical model which we will use to analyze its behavior in a circuit. The model is only an approximation. SECTION 2.3 VOLTAGE AND CURRENT SOURCES 17 + – 3.2 A 8e –100t V (c) –3.8 V –1.75 A + – (b) + – 220 mV 4 A (a) ■FIGURE 2.14 Let us agree that we will use the expression circuit element to refer to the mathematical model. The choice of a particular model for any real device must be made on the basis of experimental data or experience; we will usually assume that this choice has already been made. For simplicity, we initially consider circuits with idealized components represented by simple models. All the simple circuit elements that we will consider can be classiﬁed ac-cording to the relationship of the current through the element to the voltage across the element. For example, if the voltage across the element is linearly proportional to the current through it, we will call the element a resistor. Other types of simple circuit elements have terminal voltages which are proportional to the derivative of the current with respect to time (an induc-tor), or to the integral of the current with respect to time (a capacitor). There are also elements in which the voltage is completely independent of the cur-rent, or the current is completely independent of the voltage; these are termed independent sources. Furthermore, we will need to deﬁne special kinds of sources for which either the source voltage or current depends upon a current or voltage elsewhere in the circuit; such sources are referred to as dependent sources. Dependent sources are used a great deal in electronics to model both dc and ac behavior of transistors, especially in ampliﬁer circuits. Independent Voltage Sources The ﬁrst element we will consider is the independent voltage source. The circuit symbol is shown in Fig. 2.15a; the subscript s merely identiﬁes the voltage as a “source” voltage, and is common but not required. An inde-pendent voltage source is characterized by a terminal voltage which is completely independent of the current through it. Thus, if we are given an independent voltage source and are notiﬁed that the terminal voltage is 12 V, then we always assume this voltage, regardless of the current ﬂowing. The independent voltage source is an ideal source and does not repre-sent exactly any real physical device, because the ideal source could theo-retically deliver an inﬁnite amount of energy from its terminals. This ideal-ized voltage source does, however, furnish a reasonable approximation to several practical voltage sources. An automobile storage battery, for exam-ple, has a 12 V terminal voltage that remains essentially constant as long as the current through it does not exceed a few amperes. A small current may ﬂow in either direction through the battery. If it is positive and ﬂowing out of the positively marked terminal, then the battery is furnishing power to the headlights, for example; if the current is positive and ﬂowing into the posi-tive terminal, then the battery is charging by absorbing energy from the alternator.3 An ordinary household electrical outlet also approximates an independent voltage source, providing a voltage vs = 115 √ 2 cos 2π60t V; this representation is valid for currents less than 20 A or so. A point worth repeating here is that the presence of the plus sign at the upper end of the symbol for the independent voltage source in Fig. 2.15a does not necessarily mean that the upper terminal is numerically positive with respect to the lower terminal. Instead, it means that the upper terminal is vs volts positive with respect to the lower. If at some instant vs happens to be negative, then the upper terminal is actually negative with respect to the lower at that instant. CHAPTER 2 BASIC COMPONENTS AND ELECTRIC CIRCUITS 18 By deﬁnition, a simple circuit element is the mathematical model of a two-terminal electrical device, and it can be completely characterized by its voltage-current relationship; it cannot be subdivided into other two-terminal devices. If you’ve ever noticed the room lights dim when an air conditioner kicks on, it’s because the sudden large current demand temporarily led to a voltage drop. After the motor starts moving, it takes less current to keep it in motion. At that point, the current demand is reduced, the voltage returns to its original value, and the wall outlet again provides a reasonable approximation of an ideal voltage source. ■FIGURE 2.15 Circuit symbol of the independent voltage source. vs + – (a) (b) vs + – i (c) vs + – i (3) Or the battery of a friend’s car, if you accidentally left your headlights on. . . Consider a current arrow labeled “i” placed adjacent to the upper conduc-tor of the source as in Fig. 2.15b.The current i is entering the terminal at which the positive sign is located, the passive sign convention is satisﬁed, and the source thus absorbs power p = vsi. More often than not, a source is expected to deliver power to a network and not to absorb it. Consequently, we might choose to direct the arrow as in Fig. 2.15c so that vsi will represent the power delivered by the source. Technically, either arrow direction may be chosen; whenever possible, we will adopt the convention of Fig. 2.15c in this text for voltageandcurrentsources,whicharenotusuallyconsideredpassivedevices. An independent voltage source with a constant terminal voltage is often termed an independent dc voltage source and can be represented by either of the symbols shown in Fig. 2.16a and b. Note in Fig. 2.16b that when the physical plate structure of the battery is suggested, the longer plate is placed at the positive terminal; the plus and minus signs then represent redundant notation, but they are usually included anyway. For the sake of complete-ness, the symbol for an independent ac voltage source is shown in Fig. 2.16c. Independent Current Sources Another ideal source which we will need is the independent current source. Here, the current through the element is completely independent of the voltage across it. The symbol for an independent current source is shown in Fig. 2.17. If is is constant, we call the source an independent dc current source. An ac current source is often drawn with a tilde through the arrow, similar to the ac voltage source shown in Fig. 2.16c. Like the independent voltage source, the independent current source is at best a reasonable approximation for a physical element. In theory it can deliver inﬁnite power from its terminals because it produces the same ﬁnite current for any voltage across it, no matter how large that voltage may be. It is, however, a good approximation for many practical sources, particularly in electronic circuits. Although most students seem happy enough with an independent volt-age source providing a ﬁxed voltage but essentially any current, it is a com-mon mistake to view an independent current source as having zero voltage across its terminals while providing a ﬁxed current. In fact, we do not know a priori what the voltage across a current source will be—it depends entirely on the circuit to which it is connected. Dependent Sources The two types of ideal sources that we have discussed up to now are called independent sources because the value of the source quantity is not affected in any way by activities in the remainder of the circuit. This is in contrast with yet another kind of ideal source, the dependent, or controlled, source, in which the source quantity is determined by a voltage or current existing at some other location in the system being analyzed. Sources such as these appear in the equivalent electrical models for many electronic devices, such as transistors, operational ampliﬁers, and integrated circuits. To distinguish between dependent and independent sources, we introduce the diamond symbols shown in Fig. 2.18. In Fig. 2.18a and c, K is a dimensionless scaling constant. In Fig.2.18b, g is a scaling factor with units of A/V; in Fig. 2.18d, r is a scaling factor with units of V/A. The controlling current ix and the controlling voltage vx must be deﬁned in the circuit. SECTION 2.3 VOLTAGE AND CURRENT SOURCES 19 Terms like dc voltage source and dc current source are commonly used. Literally, they mean “direct-current voltage source” and “direct-current current source,” respectively. Although these terms may seem a little odd or even redundant, the terminology is so widely used there’s no point in ﬁghting it. ■FIGURE 2.16 (a) DC voltage source symbol; (b) battery symbol; (c) ac voltage source symbol. Vs + – (a) (b) V + – vs (c) + – ■FIGURE 2.17 Circuit symbol for the independent current source. is ■FIGURE 2.18 The four different types of dependent sources: (a) current-controlled current source; (b) voltage-controlled current source; (c) voltage-controlled voltage source; (d) current-controlled voltage source. Kix (a) gvx (b) Kvx (c) + – rix (d) + – It does seem odd at ﬁrst to have a current source whose value depends on a voltage, or a voltage source which is controlled by a current ﬂowing through some other element. Even a voltage source depending on a remote voltage can appear strange. Such sources are invaluable for modeling com-plex systems, however, making the analysis algebraically straightforward. Examples include the drain current of a ﬁeld effect transistor as a function of the gate voltage, or the output voltage of an analog integrated circuit as a function of differential input voltage. When encountered during circuit analysis, we write down the entire controlling expression for the dependent source just as we would if it was a numerical value attached to an indepen-dent source. This often results in the need for an additional equation to complete the analysis, unless the controlling voltage or current is already one of the speciﬁed unknowns in our system of equations. CHAPTER 2 BASIC COMPONENTS AND ELECTRIC CIRCUITS 20 ■FIGURE 2.19 (a) An example circuit containing a voltage-controlled voltage source. (b) The additional information provided is included on the diagram. + – + – vL + – v2 + – 5v2 (a) + – + – vL + – v2 = 3 V + – 5v2 (b) ■FIGURE 2.20 In the circuit of Fig. 2.19a, if υ2 is known to be 3 V, ﬁnd υL. We have been provided with a partially labeled circuit diagram and the additional information that v2 = 3 V. This is probably worth adding to our diagram, as shown in Fig. 2.19b. Next we step back and look at the information collected. In examin-ing the circuit diagram, we notice that the desired voltage vL is the same as the voltage across the dependent source. Thus, vL = 5v2 At this point, we would be done with the problem if only we knew v2! Returning to our diagram, we see that we actually do know v2—it was speciﬁed as 3 V. We therefore write v2 = 3 We now have two (simple) equations in two unknowns, and solve to ﬁnd vL = 15 V. An important lesson at this early stage of the game is that the time it takes to completely label a circuit diagram is always a good invest-ment. As a ﬁnal step, we should go back and check over our work to ensure that the result is correct. PRACTICE ● 2.9 Find the power absorbed by each element in the circuit in Fig. 2.20. Ans: (left to right) −56 W; 16 W; −60 W; 160 W; −60 W. 0.25vx – vx 8 A 2 A 5 A 20 V + + – 8 V + – + – 20 V + – 8 V + – – 7 A 12 V EXAMPLE 2.2 Dependent and independent voltage and current sources are active ele-ments; they are capable of delivering power to some external device. For the present we will think of a passive element as one which is capable only of receiving power. However, we will later see that several passive elements are able to store ﬁnite amounts of energy and then return that energy later to various external devices; since we still wish to call such elements passive, it will be necessary to improve upon our two deﬁnitions a little later. Networks and Circuits The interconnection of two or more simple circuit elements forms an elec-trical network. If the network contains at least one closed path, it is also an electric circuit. Note: Every circuit is a network, but not all networks are circuits (see Fig. 2.21)! SECTION 2.3 VOLTAGE AND CURRENT SOURCES 21 ■FIGURE 2.21 (a) A network that is not a circuit. (b) A network that is a circuit. vs (a) + – (b) + – vs A network that contains at least one active element, such as an indepen-dent voltage or current source, is an active network. A network that does not contain any active elements is a passive network. We have now deﬁned what we mean by the term circuit element, and we have presented the deﬁnitions of several speciﬁc circuit elements, the independent and dependent voltage and current sources. Throughout the remainder of the book we will deﬁne only ﬁve additional circuit elements: the resistor, inductor, capacitor, transformer, and the ideal operational ampli-ﬁer (“op amp,” for short). These are all ideal elements. They are important because we may combine them into networks and circuits that represent real devices as accurately as we require. Thus, the transistor shown in Fig. 2.22a and b may be modeled by the voltage terminals designated vgs and the single dependent current source of Fig. 2.22c. Note that the dependent current source produces a current that depends on a voltage elsewhere in the circuit. The parameter gm, commonly referred to as the transconductance, is calculated using transistor-speciﬁc details as well as the operating point de-termined by the circuit connected to the transistor. It is generally a small number, on the order of 10−2 to perhaps 10A/V.This model works pretty well as long as the frequency of any sinusoidal source is neither very large nor very small; the model can be modiﬁed to account for frequency-dependent CHAPTER 2 BASIC COMPONENTS AND ELECTRIC CIRCUITS 22 ■FIGURE 2.22 The Metal Oxide Semiconductor Field Effect Transistor (MOSFET). (a) An IRF540 N-channel power MOSFET in a TO-220 package, rated at 100 V and 22 A; (b) cross-sectional view of a basic MOSFET (R. Jaeger, Microelectronic Circuit Design, McGraw-Hill, 1997); (c) equivalent circuit model for use in ac circuit analysis. (a) L n+ n+ D G S Metal (or polysilicon) Silicon dioxide (SiO2) W B (b) vgs + – (c) g s d s gmvgs p-type substrate (body) Drain region Source region Channel region effects by including additional ideal circuit elements such as resistors and capacitors. Similar (but much smaller) transistors typically constitute only one small part of an integrated circuit that may be less than 2 mm × 2 mm square and 200 μm thick and yet contains several thousand transistors plus various resistors and capacitors. Thus, we may have a physical device that is about the size of one letter on this page but requires a model composed of ten thousand ideal simple circuit elements. We use this concept of “circuit modeling” in a number of electrical engineering topics covered in other courses, including electronics, energy conversion, and antennas. 2.4 • OHM’S LAW So far, we have been introduced to both dependent and independent voltage and current sources and were cautioned that they were idealized active ele-ments that could only be approximated in a real circuit. We are now ready to meet another idealized element, the linear resistor. The resistor is the sim-plest passive element, and we begin our discussion by considering the work of an obscure German physicist, Georg Simon Ohm, who published a pam-phlet in 1827 that described the results of one of the ﬁrst efforts to measure currents and voltages, and to describe and relate them mathematically. One result was a statement of the fundamental relationship we now call Ohm’s law, even though it has since been shown that this result was discovered 46 years earlier in England by Henry Cavendish, a brilliant semirecluse. Ohm’s law states that the voltage across conducting materials is directly proportional to the current ﬂowing through the material, or v = Ri where the constant of proportionality R is called the resistance. The unit of resistance is the ohm, which is 1 V/A and customarily abbreviated by a capital omega, . When this equation is plotted on i-versus-v axes, the graph is a straight line passing through the origin (Fig. 2.23). Equation is a linear equation, and we will consider it as the deﬁnition of a linear resistor. Resistance is normally considered to be a positive quantity, although negative resistances may be simulated with special circuitry. Again, it must be emphasized that the linear resistor is an idealized circuit element; it is only a mathematical model of a real, physical device. “Resistors” may be easily purchased or manufactured, but it is soon found that the voltage-current ratios of these physical devices are reasonably con-stant only within certain ranges of current, voltage, or power, and depend also on temperature and other environmental factors. We usually refer to a linear resistor as simply a resistor; any resistor that is nonlinear will always be described as such. Nonlinear resistors should not necessarily be consid-ered undesirable elements. Although it is true that their presence compli-cates an analysis, the performance of the device may depend on or be greatly improved by the nonlinearity. For example, fuses for overcurrent protection and Zener diodes for voltage regulation are very nonlinear in nature, a fact that is exploited when using them in circuit design. Power Absorption Figure 2.24 shows several different resistor packages, as well as the most common circuit symbol used for a resistor. In accordance with the voltage, current, and power conventions already adopted, the product of v and i gives the power absorbed by the resistor. That is, v and i are selected to satisfy the passive sign convention. The absorbed power appears physically SECTION 2.4 OHM’S LAW 23 ■FIGURE 2.23 Current-voltage relationship for an example 2 linear resistor. Note the slope of the line is 0.5 A/V, or 500 m1. 1 2 3 4 5 6 7 1 2 3 4 5 6 7 8 9 10 V (volts) I (amperes) ■FIGURE 2.24 (a) Several common resistor packages. (b) A 560 power resistor rated at up to 50 W. (c) A 5% tolerance 10-teraohm (10,000,000,000,000 ) resistor manufactured by Ohmcraft. (d) Circuit symbol for the resistor, applicable to all of the devices in (a) through (c). (a) (c) (d) i R v + – (b) 24 as heat and/or light and is always positive; a (positive) resistor is a passive element that cannot deliver power or store energy. Alternative expressions for the absorbed power are p = vi = i2R = v2/R One of the authors (who shall remain anonymous) had the unfortunate experience of inadvertently connecting a 100 , 2 W carbon resistor across a 110 V source. The ensuing ﬂame, smoke, and fragmentation were rather disconcerting, demonstrating clearly that a practical resistor has deﬁnite limits to its ability to behave like the ideal linear model. In this case, the un-fortunate resistor was called upon to absorb 121 W; since it was designed to handle only 2 W, its reaction was understandably violent. CHAPTER 2 BASIC COMPONENTS AND ELECTRIC CIRCUITS EXAMPLE 2.3 The 560 resistor shown in Fig. 2.24b is connected to a circuit which causes a current of 42.4 mA to ﬂow through it. Calculate the voltage across the resistor and the power it is dissipating. The voltage across the resistor is given by Ohm’s law: v = Ri = (560)(0.0424) = 23.7 V The dissipated power can be calculated in several different ways. For instance, p = vi = (23.7)(0.0424) = 1.005 W Alternatively, p = v2/R = (23.7)2/560 = 1.003 W or p = i2R = (0.0424)2(560) = 1.007 W We note several things. First, we calculated the power in three different ways, and we seem to have obtained three different answers! In reality, however, we rounded our voltage to three signiﬁcant digits, which will impact the accuracy of any subsequent quantity we calculate with it. With this in mind, we see that the answers show rea-sonable agreement (within 1%). The other point worth noting is that the resistor is rated to 50 W— since we are only dissipating approximately 2% of this value, the resis-tor is in no danger of overheating. PRACTICE ● With reference to Fig. 2.25, compute the following: 2.10 R if i = −2 μA and v = −44 V. 2.11 The power absorbed by the resistor if v = 1 V and R = 2 kΩ. 2.12 The power absorbed by the resistor if i = 3 nA and R = 4.7 MΩ. Ans: 22 M; 500 μW; 42.3 pW. ■FIGURE 2.25 i R v + – Technically speaking, any material (except for a super-conductor) will provide resistance to current ﬂow. As in all introductory circuits texts, however, we tacitly as-sume that wires appearing in circuit diagrams have zero resistance. This implies that there is no potential differ-ence between the ends of a wire, and hence no power absorbed or heat generated. Although usually not an unreasonable assumption, it does neglect practical con-siderations when choosing the appropriate wire diameter for a speciﬁc application. Resistance is determined by (1) the inherent resistiv-ity of a material and (2) the device geometry. Resistivity, represented by the symbol ρ, is a measure of the ease with which electrons can travel through a certain mater-ial. Since it is the ratio of the electric ﬁeld (V/m) to the areal density of current ﬂowing in the material (A/m2), the general unit of ρ is an · m, although metric pre-ﬁxes are often employed. Every material has a different inherent resistivity, which depends on temperature. Some examples are shown in Table 2.3; as can be seen, there is a small variation between different types of cop-per (less than 1%) but a very large difference between different metals. In particular, although physically stronger than copper, steel wire is several times more resistive. In some technical discussions, it is more common to see the conductivity (symbolized by σ) of a PRACTICAL APPLICATION Wire Gauge (cm) Direction of current flow Cross-sectional area = A cm2 Resistivity = cm ■FIGURE 2.26 Deﬁnition of geometrical parameters used to compute the resistance of a wire. The resistivity of the material is assumed to be spatially uniform. PRACTICAL APPLICATION material quoted, which is simply the reciprocal of the resistivity. The resistance of a particular object is obtained by multiplying the resistivity by the length ℓof the resistor and dividing by the cross-sectional area (A) as in Eq. ; these parameters are illustrated in Fig. 2.26. R = ρ ℓ A TABLE ●2.3 Common Electrical Wire Materials and Resistivities Resistivity at 20°C ASTM Speciﬁcation Temper and Shape (μ·cm) B33 Copper, tinned soft, round 1.7654 B75 Copper, tube, soft, OF copper 1.7241 B188 Copper, hard bus tube, rectangular or square 1.7521 B189 Copper, lead-coated soft, round 1.7654 B230 Aluminum, hard, round 2.8625 B227 Copper-clad steel, hard, round, 4.3971 grade 40 HS B355 Copper, nickel-coated soft, round 1.9592 Class 10 B415 Aluminum-clad steel, hard, round 8.4805 C. B. Rawlins, “Conductor materials,” Standard Handbook for Electrical Engineering, 13th ed., D. G. Fink and H. W. Beaty, eds. New York: McGraw-Hill, 1993, pp. 4-4 to 4-8. American Society of Testing and Materials. (Continued on next page) We determine the resistivity when we select the material from which to fabricate a wire and measure the temperature of the application environment. Since a ﬁnite amount of power is absorbed by the wire due to its resistance, current ﬂow leads to the production of heat. Thicker wires have lower resistance and also dissipate heat more easily but are heavier, take up a larger volume, and are more expensive. Thus, we are motivated by practical considerations to choose the smallest wire that can safely do the job, rather than simply choosing the largest-diameter wire available in an effort to minimize resistive losses. The American Wire Gauge (AWG) is a standard system of specifying wire size. In selecting a wire gauge, smaller AWG corresponds to a larger wire diameter; an abbreviated table of common gauges is given in Table 2.4. Local ﬁre and electrical safety codes typically dictate the required gauge for speciﬁc wiring applications, based on the maximum current expected as well as where the wires will be located. TABLE ● 2.4 Some Common Wire Gauges and the Resistance of (Soft) Solid Copper Wire Conductor Size (AWG) Cross-Sectional Area (mm2) Ohms per 1000 ft at 20°C 28 0.0804 65.3 24 0.205 25.7 22 0.324 16.2 18 0.823 6.39 14 2.08 2.52 12 3.31 1.59 6 13.3 0.3952 4 21.1 0.2485 2 33.6 0.1563 C. B. Rawlins et al., Standard Handbook for Electrical Engineering, 13th ed., D. G. Fink and H. W. Beaty, eds. New York: McGraw-Hill, 1993, p. 4-47. EXAMPLE 2.4 A dc power link is to be made between two islands separated by a distance of 24 miles. The operating voltage is 500 kV and the sys-tem capacity is 600 MW. Calculate the maximum dc current ﬂow, and estimate the resistivity of the cable, assuming a diameter of 2.5 cm and a solid (not stranded) wire. Dividing the maximum power (600 MW, or 600 × 106 W) by the operating voltage (500 kV, or 500 × 103 V) yields a maximum current of 600 × 106 500 × 103 = 1200 A The cable resistance is simply the ratio of the voltage to the current, or Rcable = 500 × 103 1200 = 417 SECTION 2.4 OHM’S LAW 27 Conductance For a linear resistor the ratio of current to voltage is also a constant i v = 1 R = G where G is called the conductance. The SI unit of conductance is the siemens (S), 1 A/V. An older, unofﬁcial unit for conductance is the mho, which was often abbreviated as and is still occasionally written as −1. You will occasionally see it used on some circuit diagrams, as well as in cat-alogs and texts. The same circuit symbol (Fig. 2.24d) is used to represent both resistance and conductance. The absorbed power is again necessarily positive and may be expressed in terms of the conductance by p = vi = v2G = i2 G Thus a 2 resistor has a conductance of 1 2 S, and if a current of 5 A is ﬂowing through it, then a voltage of 10 V is present across the terminals and a power of 50 W is being absorbed. All the expressions given so far in this section were written in terms of instantaneous current, voltage, and power, such as v = iR and p = vi. We should recall that this is a shorthand notation for v(t) = Ri(t) and p(t) = v(t) i(t). The current through and voltage across a resistor must both vary with time in the same manner. Thus, if R = 10 and v = 2 sin 100t V, then i = 0.2 sin 100t A. Note that the power is given by 0.4 sin2 100t W, and a simple sketch will illustrate the different nature of its variation with time. Although the current and voltage are each negative during certain time intervals, the absorbed power is never negative! Resistance may be used as the basis for deﬁning two commonly used terms, short circuit and open circuit. We deﬁne a short circuit as a resistance of zero ohms; then, since v = iR, the voltage across a short circuit must be zero, although the current may have any value. In an analogous manner, We know the length: ℓ= (24 miles) 5280 ft 1 mile 12 in 1 ft 2.54 cm 1 in = 3,862,426 cm Given that most of our information appears to be valid to only two signif-icant ﬁgures, we round this to 3.9 × 106 cm. With the cable diameter speciﬁed as 2.5 cm, we know its cross-sectional area is 4.9 cm2. Thus, ρcable = Rcable A ℓ= 417 4.9 3.9 × 106 = 520 μ · cm PRACTICE ● 2.13 A 500 ft long 24 AWG soft copper wire is carrying a current of 100 mA. What is the voltage dropped across the wire? Ans: 3.26 V. 28 we deﬁne an open circuit as an inﬁnite resistance. It follows from Ohm’s law that the current must be zero, regardless of the voltage across the open circuit. Although real wires have a small resistance associated with them, we always assume them to have zero resistance unless otherwise speciﬁed. Thus, in all of our circuit schematics, wires are taken to be perfect short circuits. SUMMARY AND REVIEW In this chapter, we introduced the topic of units – speciﬁcally those relevant to electrical circuits—and their relationship to fundamental (SI) units. We also discussed current and current sources, voltage and voltage sources, and the fact that the product of voltage and current yields power (the rate of energy consumption or generation). Since power can be either positive or negative depending on the current direction and voltage polarity, the pas-sive sign convention was described to ensure we always know if an element is absorbing or supplying energy to the rest of the circuit. Four additional sources were introduced, forming a general class known as dependent sources. They are often used to model complex systems and electrical com-ponents, but the actual value of voltage or current supplied is typically unknown until the entire circuit is analyzed. We concluded the chapter with the resistor—by far the most common circuit element—whose voltage and current are linearly related (described by Ohm’s law). Whereas the resistiv-ity of a material is one of its fundamental properties (measured in · cm), resistance describes a device property (measured in ) and hence depends not only on resistivity but on the device geometry (i.e., length and area) as well. We conclude with key points of this chapter to review, along with ap-propriate examples. ❑The system of units most commonly used in electrical engineering is the SI. ❑The direction in which positive charges are moving is the direction of positive current ﬂow; alternatively, positive current ﬂow is in the direction opposite that of moving electrons. ❑To deﬁne a current, both a value and a direction must be given. Currents are typically denoted by the uppercase letter “I” for constant (dc) values, and either i(t) or simply i otherwise. ❑To deﬁne a voltage across an element, it is necessary to label the terminals with “+” and “−” signs as well as to provide a value (either an algebraic symbol or a numerical value). ❑Any element is said to supply positive power if positive current ﬂows out of the positive voltage terminal. Any element absorbs positive power if positive current ﬂows into the positive voltage terminal. (Example 2.1) ❑There are six sources: the independent voltage source, the independent current source, the current-controlled dependent current source, the voltage-controlled dependent current source, the voltage-controlled dependent voltage source, and the current-controlled dependent voltage source. (Example 2.2) CHAPTER 2 BASIC COMPONENTS AND ELECTRIC CIRCUITS Note that a current represented by i or i(t) can be constant (dc) or time-varying, but currents represented by the symbol I must be non-time-varying. ❑Ohm’s law states that the voltage across a linear resistor is directly proportional to the current ﬂowing through it; i.e., v = Ri. (Example 2.3) ❑The power dissipated by a resistor (which leads to the production of heat) is given by p = vi = i2R = v2/R. (Example 2.3) ❑Wires are typically assumed to have zero resistance in circuit analysis. When selecting a wire gauge for a speciﬁc application, however, local electrical and ﬁre codes must be consulted. (Example 2.4) READING FURTHER A good book that discusses the properties and manufacture of resistors in considerable depth: Felix Zandman, Paul-René Simon, and Joseph Szwarc, Resistor Theory and Technology. Raleigh, N.C.: SciTech Publishing, 2002. A good all-purpose electrical engineering handbook: Donald G. Fink and H. Wayne Beaty, Standard Handbook for Electrical Engineers, 13th ed., New York: McGraw-Hill, 1993. In particular, pp. 1-1 to 1-51, 2-8 to 2-10, and 4-2 to 4-207 provide an in-depth treatment of topics related to those discussed in this chapter. A detailed reference for the SI is available on the Web from the National Institute of Standards: Ambler Thompson and Barry N. Taylor, Guide for the Use of the International System of Units (SI), NIST Special Publication 811, 2008 edition, www.nist.gov. EXERCISES 2.1 Units and Scales 1. Convert the following to engineering notation: (a) 0.045 W (b) 2000 pJ (c) 0.1 ns (d) 39,212 as (e) 3 ( f ) 18,000 m (g) 2,500,000,000,000 bits (h) 1015 atoms/cm3 2. Convert the following to engineering notation: (a) 1230 fs (b) 0.0001 decimeter (c) 1400 mK (d) 32 nm (e) 13,560 kHz ( f) 2021 micromoles (g) 13 deciliters (h) 1 hectometer 3. Express the following in engineering units: (a) 1212 mV (b) 1011 pA (c) 1000 yoctoseconds (d) 33.9997 zeptoseconds (e) 13,100 attoseconds (f ) 10−14 zettasecond (g) 10−5 second (h) 10−9 Gs 4. Expand the following distances in simple meters: (a) 1 Zm (b) 1 Em (c) 1 Pm (d) 1 Tm (e) 1 Gm (f ) 1 Mm EXERCISES 29 5. Convert the following to SI units, taking care to employ proper engineering notation: (a) 212°F (b) 0°F (c) 0 K (d) 200 hp (e) 1 yard (f ) 1 mile 6. Convert the following to SI units, taking care to employ proper engineering notation: (a) 100C (b) 0C (c) 4.2 K (d) 150 hp (e) 500 Btu (f ) 100 J/s 7. A certain krypton ﬂuoride laser generates 15 ns long pulses, each of which contains 550 mJ of energy. (a) Calculate the peak instantaneous output power of the laser. (b) If up to 100 pulses can be generated per second, calculate the maximum average power output of the laser. 8. When operated at a wavelength of 750 nm, a certain Ti:sapphire laser is capa-ble of producing pulses as short as 50 fs, each with an energy content of 500 μJ. (a) Calculate the instantaneous output power of the laser. (b) If the laser is capable of a pulse repetition rate of 80 MHz, calculate the maximum average output power that can be achieved. 9. An electric vehicle is driven by a single motor rated at 40 hp. If the motor is run continuously for 3 h at maximum output, calculate the electrical energy consumed. Express your answer in SI units using engineering notation. 10. Under insolation conditions of 500 W/m2 (direct sunlight), and 10% solar cell efﬁciency (deﬁned as the ratio of electrical output power to incident solar power), calculate the area required for a photovoltaic (solar cell) array capable of running the vehicle in Exer. 9 at half power. 11. A certain metal oxide nanowire piezoelectricity generator is capable of producing 100 pW of usable electricity from the type of motion obtained from a person jogging at a moderate pace. (a) How many nanowire devices are required to operate a personal MP3 player which draws 1 W of power? (b) If the nanowires can be produced with a density of 5 devices per square micron directly onto a piece of fabric, what area is required, and would it be practical? 12. A particular electric utility charges customers different rates depending on their daily rate of energy consumption: $0.05/kWh up to 20 kWh, and $0.10/kWh for all energy usage above 20 kWh in any 24 hour period. (a) Calculate how many 100 W light bulbs can be run continuously for less than $10 per week. (b) Calculate the daily energy cost if 2000 kW of power is used continuously. 13. The Tilting Windmill Electrical Cooperative LLC Inc. has instituted a differential pricing scheme aimed at encouraging customers to conserve electricity use during daylight hours, when local business demand is at its highest. If the price per kilowatthour is $0.033 between the hours of 9 p.m. and 6 a.m., and $0.057 for all other times, how much does it cost to run a 2.5 kW portable heater continuously for 30 days? 14. Assuming a global population of 9 billion people, each using approximately 100 W of power continuously throughout the day, calculate the total land area that would have to be set aside for photovoltaic power generation, assuming 800 W/m2 of incident solar power and a conversion efﬁciency (sunlight to electricity) of 10%. 2.2 Charge, Current, Voltage, and Power 15. The total charge ﬂowing out of one end of a small copper wire and into an unknown device is determined to follow the relationship q(t) = 5e−t/2 C, where t is expressed in seconds. Calculate the current ﬂowing into the device, taking note of the sign. 16. The current ﬂowing into the collector lead of a certain bipolar junction transistor (BJT) is measured to be 1 nA. If no charge was transferred in or out of the collector lead prior to t = 0, and the current ﬂows for 1 min, calculate the total charge which crosses into the collector. 30 CHAPTER 2 BASIC COMPONENTS AND ELECTRIC CIRCUITS 17. The total charge stored on a 1 cm diameter insulating plate is −1013 C. (a) How many electrons are on the plate? (b) What is the areal density of electrons (number of electrons per square meter)? (c) If additional electrons are added to the plate from an external source at the rate of 106 electrons per second, what is the magnitude of the current ﬂowing between the source and the plate? 18. A mysterious device found in a forgotten laboratory accumulates charge at a rate speciﬁed by the expression q(t) = 9 −10t C from the moment it is switched on. (a) Calculate the total charge contained in the device at t = 0. (b) Calculate the total charge contained at t = 1 s. (c) Determine the current ﬂowing into the device at t = 1 s, 3 s, and 10 s. 19. A new type of device appears to accumulate charge according to the expression q(t) = 10t2 −22t mC (t in s). (a) In the interval 0 ≤t < 5 s, at what time does the current ﬂowing into the device equal zero? (b) Sketch q(t) and i(t) over the interval 0 ≤t < 5 s. 20. The current ﬂowing through a tungsten-ﬁlament light bulb is determined to follow i(t) = 114 sin(100πt) A. (a) Over the interval deﬁned by t = 0 and t = 2 s, how many times does the current equal zero amperes? (b) How much charge is transported through the light bulb in the ﬁrst second? 21. The current waveform depicted in Fig. 2.27 is characterized by a period of 8 s. (a) What is the average value of the current over a single period? (b) If q(0) = 0, sketch q(t), 0 < t < 20 s. EXERCISES 31 2 4 6 8 10 12 2 4 6 8 10 12 14 1 3 5 7 9 11 13 15 i(t) t (s) 1 –1 – 2 – 3 2 3 4 1 2 3 4 5 6 7 8 i(t) t (s) ■FIGURE 2.27 An example of a time-varying current. 22. The current waveform depicted in Fig. 2.28 is characterized by a period of 4 s. (a) What is the average value of the current over a single period? (b) Compute the average current over the interval 1 < t < 3 s. (c) If q(0) = 1 C, sketch q(t), 0 < t < 4 s. ■FIGURE 2.28 An example of a time-varying current. 23. A path around a certain electric circuit has discrete points labeled A, B, C, and D. To move an electron from points A to C requires 5 pJ. To move an electron from B to C requires 3 pJ. To move an electron from A to D requires 8 pJ. (a) What is the potential difference (in volts) between points B and C, assuming a “+” reference at C? (b) What is the potential difference (in volts) between points B and D, assuming a “+” reference at D? (c) What is the potential difference (in volts) between points A and B (again, in volts), assuming a “+” reference at B? 24. Two metallic terminals protrude from a device. The terminal on the left is the positive reference for a voltage called vx (the other terminal is the negative reference). The terminal on the right is the positive reference for a voltage called vy (the other terminal being the negative reference). If it takes 1 mJ of energy to push a single electron into the left terminal, determine the voltages vx and vy. 25. The convention for voltmeters is to use a black wire for the negative reference terminal and a red wire for the positive reference terminal. (a) Explain why two wires are required to measure a voltage. (b) If it is dark and the wires into the voltmeter are swapped by accident, what will happen during the next measurement? 26. Determine the power absorbed by each of the elements in Fig. 2.29. 27. Determine the power absorbed by each of the elements in Fig. 2.30. 32 CHAPTER 2 BASIC COMPONENTS AND ELECTRIC CIRCUITS (a) (b) (c) 1 V 10 mA – + + – + – 1 pA 6 V 2 A 2 A 10 V ■FIGURE 2.29 Elements for Exer. 26. (a) + – – + 2 V 2 V 1 A (b) (c) –16e–t V (t = 500 ms) 8e–t mA 10–3 i1 (i1 = 100 mA) – + ■FIGURE 2.30 Elements for Exer. 27. 28. A constant current of 1 ampere is measured ﬂowing into the positive reference terminal of a pair of leads whose voltage we’ll call vp. Calculate the absorbed power at t = 1 s if vp(t) equals (a) +1 V; (b) −1 V; (c) 2 + 5 cos(5t) V; (d) 4e−2t V, (e) Explain the signiﬁcance of a negative value for absorbed power. 30. The current-voltage characteristic of a silicon solar cell exposed to direct sunlight at noon in Florida during midsummer is given in Fig. 2.32. It is obtained by placing different-sized resistors across the two terminals of the device and measuring the resulting currents and voltages. (a) What is the value of the short-circuit current? (b) What is the value of the voltage at open circuit? (c) Estimate the maximum power that can be obtained from the device. EXERCISES 33 0.5 1.0 1.5 2.0 2.5 3.0 0.125 0.250 0.375 0.500 Voltage (V) Current (A) ■FIGURE 2.32 29. Determine the power supplied by the leftmost element in the circuit of Fig. 2.31. 8 V – + Vx 5Vx + – VR + – A ■FIGURE 2.33 2.3 Voltage and Current Sources 31. Some of the ideal sources in the circuit of Fig. 2.31 are supplying positive power, and others are absorbing positive power. Determine which are which, and show that the algebraic sum of the power absorbed by each element (taking care to preserve signs) is equal to zero. 32. By careful measurements it is determined that a benchtop argon ion laser is consuming (absorbing) 1.5 kW of electric power from the wall outlet, but only producing 5 W of optical power. Where is the remaining power going? Doesn’t conservation of energy require the two quantities to be equal? 33. Refer to the circuit represented in Fig. 2.33, while noting that the same current ﬂows through each element. The voltage-controlled dependent source provides a current which is 5 times as large as the voltage Vx. (a) For VR = 10 V and Vx = 2 V, determine the power absorbed by each element. (b) Is element A likely a passive or active source? Explain. 2 V + – – + 2 A 5 A + – –3 A 10 V 8 V 10 V –4 A + – + – ■FIGURE 2.31 34. Refer to the circuit represented in Fig. 2.33, while noting that the same current ﬂows through each element. The voltage-controlled dependent source provides a current which is 5 times as large as the voltage Vx. (a) For VR = 100 V and Vx = 92 V, determine the power supplied by each element. (b) Verify that the algebraic sum of the supplied powers is equal to zero. 35. The circuit depicted in Fig. 2.34 contains a dependent current source; the magnitude and direction of the current it supplies are directly determined by the voltage labeled v1. Note that therefore i2 = −3v1. Determine the voltage v1 if v2 = 33i2 and i2 = 100 mA. 34 CHAPTER 2 BASIC COMPONENTS AND ELECTRIC CIRCUITS 36. To protect an expensive circuit component from being delivered too much power, you decide to incorporate a fast-blowing fuse into the design. Knowing that the circuit component is connected to 12 V, its minimum power consumption is 12 W, and the maximum power it can safely dissipate is 100 W, which of the three available fuse ratings should you select: 1 A, 4 A, or 10 A? Explain your answer. 37. The dependent source in the circuit of Fig. 2.35 provides a voltage whose value depends on the current ix. What value of ix is required for the dependent source to be supplying 1 W? + – vS v1 + – 3v1 i2 v2 + – ■FIGURE 2.34 + – v2 + – ix –2ix ■FIGURE 2.35 2.4 Ohm’s Law 38. Determine the magnitude of the current ﬂowing through a 4.7 k resistor if the voltage across it is (a) 1 mV; (b) 10 V; (c) 4e−t V; (d) 100 cos(5t) V; (e) −7 V. 39. Real resistors can only be manufactured to a speciﬁc tolerance, so that in effect the value of the resistance is uncertain. For example, a 1 resistor speciﬁed as 5% tolerance could in practice be found to have a value anywhere in the range of 0.95 to 1.05 . Calculate the voltage across a 2.2 k 10% tolerance resistor if the current ﬂowing through the element is (a) 1 mA; (b) 4 sin 44t mA. 40. (a) Sketch the current-voltage relationship (current on the y-axis) of a 2 k resistor over the voltage range of −10 V ≤Vresistor ≤+10 V. Be sure to label both axes appropriately. (b) What is the numerical value of the slope (express your answer in siemens)? 41. Sketch the voltage across a 33 resistor over the range 0 < t < 2π s, if the current is given by 2.8 cos(t) A. Assume both the current and voltage are deﬁned according to the passive sign convention. 42. Figure 2.36 depicts the current-voltage characteristic of three different resistive elements. Determine the resistance of each, assuming the voltage and current are deﬁned in accordance with the passive sign convention. EXERCISES 35 0.05 0.04 0.03 0.02 0.01 0.00 Current (mA) Voltage (V) –0.01 –0.02 –0.03 –0.04 –0.05 –5 –4 –3 –2 –1 0 1 2 3 4 5 0.05 0.04 0.03 0.02 0.01 0.00 Current (mA) Voltage (V) (a) (b) –0.01 –0.02 –0.03 –0.04 –0.05 –5 –4 –3 –2 –1 0 1 2 3 4 5 0.05 0.04 0.03 0.02 0.01 0.00 Current (mA) Voltage (V) (c) –0.01 –0.02 –0.03 –0.04 –0.05 –5 –4 –3 –2 –1 0 1 2 3 4 5 ■FIGURE 2.36 Voltage (V) Current (mA) −2.0 −0.89 −1.2 −0.47 0.0 0.01 1.0 0.44 1.5 0.70 43. Determine the conductance (in siemens) of the following: (a) 0 ; (b) 100 M; (c) 200 m. 44. Determine the magnitude of the current ﬂowing through a 10 mS conductance if the voltage across it is (a) 2 mV; (b) −1 V; (c) 100e−2t V; (d) 5 sin(5t) V; (e) 0 V. 45. A 1% tolerance 1 k resistor may in reality have a value anywhere in the range of 990 to 1010 . Assuming a voltage of 9 V is applied across it, determine (a) the corresponding range of current and (b) the corresponding range of absorbed power. (c) If the resistor is replaced with a 10% tolerance 1 k resistor, repeat parts (a) and (b). 46. The following experimental data is acquired for an unmarked resistor, using a variable-voltage power supply and a current meter. The current meter readout is somewhat unstable, unfortunately, which introduces error into the measurement. (a) Plot the measured current-versus-voltage characteristic. (b) Using a best-ﬁt line, estimate the value of the resistance. 36 47. Utilize the fact that in the circuit of Fig. 2.37, the total power supplied by the voltage source must equal the total power absorbed by the two resistors to show that VR2 = VS R2 R1 + R2 You may assume the same current ﬂows through each element (a requirement of charge conservation). 48. For each of the circuits in Fig. 2.38, ﬁnd the current I and compute the power absorbed by the resistor. CHAPTER 2 BASIC COMPONENTS AND ELECTRIC CIRCUITS n (cm2/Vs) ND (atoms/cm3) 1014 1015 1016 1017 1018 1019 102 103 104 ■FIGURE 2.39 49. Sketch the power absorbed by a 100 resistor as a function of voltage over the range −2 V ≤Vresistor ≤+2 V. Chapter-Integrating Exercises 50. So-called “n-type” silicon has a resistivity given by ρ = (−qNDμn)−1, where ND is the volume density of phosphorus atoms (atoms/cm3), μn is the electron mobility (cm2/V · s), and q = −1.602 × 10−19 C is the charge of each electron. Conveniently, a relationship exists between mobility and ND, as shown in Fig. 2.39. Assume an 8 inch diameter silicon wafer (disk) having a thickness of 300 μm. Design a 10 resistor by specifying a phosphorus concentration in the range of 2 × 1015 cm−3 ≤ND ≤2 × 1017 cm−3, along with a suitable geometry (the wafer may be cut, but not thinned). R1 R2 + – VS VR2 – + ■FIGURE 2.37 5 V 10 k + – I 5 V 10 k + – I –5 V 10 k + – I –5 V 10 k + – I ■FIGURE 2.38 51. Figure 2.39 depicts the relationship between electron mobility μn and dopant density ND for n-type silicon. With the knowledge that resistivity in this material is given by ρ = NDμn/q, plot resistivity as a function of dopant density over the range 1014 cm−3 ≤ND ≤1019 cm−3. 52. Referring to the data of Table 2.4, design a resistor whose value can be varied mechanically in the range of 100 to 500 (assume operation at 20◦C). 53. A 250 ft long span separates a dc power supply from a lamp which draws 25 A of current. If 14 AWG wire is used (note that two wires are needed for a total of 500 ft), calculate the amount of power wasted in the wire. 54. The resistance values in Table 2.4 are calibrated for operation at 20◦C. They may be corrected for operation at other temperatures using the relationship4 R2 R1 = 234.5 + T2 234.5 + T1 where T1 = reference temperature (20◦C in present case) T2 = desired operating temperature R1 = resistance at T1 R2 = resistance at T2 A piece of equipment relies on an external wire made of 28 AWG soft copper, which has a resistance of 50.0 at 20◦C. Unfortunately, the operating environment has changed, and it is now 110.5◦F. (a) Calculate the length of the original wire. (b) Determine by how much the wire should be shortened so that it is once again 50.0 . 55. Your favorite meter contains a precision (1% tolerance) 10 resistor. Unfortunately, the last person who borrowed this meter somehow blew the resistor, and it needs to be replaced. Design a suitable replacement, assuming at least 1000 ft of each of the wire gauges listed in Table 2.4 is readily available to you. 56. At a new installation, you speciﬁed that all wiring should conform to the ASTM B33 speciﬁcation (see Table 2.3). Unfortunately the subcontractor misread your instructions and installed B415 wiring instead (but the same gauge). Assuming the operating voltage is unchanged, (a) by how much will the current be reduced, and (b) how much additional power will be wasted in the lines? (Express both answers in terms of percentage.) 57. If 1 mA of current is forced through a 1 mm diameter, 2.3 meter long piece of hard, round, aluminum-clad steel (B415) wire, how much power is wasted as a result of resistive losses? If instead wire of the same dimensions but conforming to B75 speciﬁcations is used, by how much will the power wasted due to resistive losses be reduced? 58. The network shown in Fig. 2.40 can be used to accurately model the behavior of a bipolar junction transistor provided that it is operating in the forward active mode. The parameter β is known as the current gain. If for this device EXERCISES 37 0.7 V + – Base Emitter Collector IB IC bIB ■FIGURE 2.40 DC model for a bipolar junction transistor operating in forward active mode. (4) D. G. Fink and H. W. Beaty, Standard Handbook for Electrical Engineers, 13th ed. New York: McGraw-Hill, 1993, p. 2–9. 38 β = 100, and IB is determined to be 100 μA, calculate (a) IC, the current ﬂowing into the collector terminal; and (b) the power dissipated by the base-emitter region. 59. A 100 W tungsten ﬁlament light bulb functions by taking advantage of resistive losses in the ﬁlament, absorbing 100 joules each second of energy from the wall socket. How much optical energy per second do you expect it to produce, and does this violate the principle of energy conservation? 60. Batteries come in a wide variety of types and sizes. Two of the most common are called “AA” and “AAA.” A single battery of either type is rated to produce a terminal voltage of 1.5 V when fully charged. So what are the differences between the two, other than size? (Hint: Think about energy.) CHAPTER 2 BASIC COMPONENTS AND ELECTRIC CIRCUITS INTRODUCTION In Chap. 2 we were introduced to independent voltage and current sources, dependent sources, and resistors. We discovered that dependent sources come in four varieties, and are controlled by a voltage or current which exists elsewhere. Once we know the voltage across a resistor, we know its current (and vice versa); this is not the case for sources, however. In general, circuits must be analyzed to determine a complete set of voltages and currents. This turns out to be reasonably straightforward, and only two simple laws are needed in addition to Ohm’s law. These new laws are Kirchhoff’s current law (KCL) and Kirchhoff’s voltage law (KVL), and they are simply restatements of charge and energy conservation, respectively. They apply to any circuit we will ever encounter, although in later chapters we will learn more efﬁcient techniques for speciﬁc types of situations. 3.1 • NODES, PATHS, LOOPS, AND BRANCHES We now focus our attention on the current-voltage relationships in simple networks of two or more circuit elements. The elements will be connected by wires (sometimes referred to as “leads”), which have zero resistance. Since the network then appears as a number of sim-ple elements and a set of connecting leads, it is called a lumped-parameter network. A more difﬁcult analysis problem arises when we are faced with a distributed-parameter network, which contains an essentially inﬁnite number of vanishingly small elements. We will concentrate on lumped-parameter networks in this text. KEY CONCEPTS New Circuit Terms: Node, Path, Loop, and Branch Kirchhoff’s Current Law (KCL) Kirchhoff’s Voltage Law (KVL) Analysis of Basic Series and Parallel Circuits Combination of Series and Parallel Sources Reduction of Series and Parallel Resistor Combinations Voltage and Current Division Ground Connections Voltage and Current Laws C H A P T E R 3 39 CHAPTER 3 VOLTAGE AND CURRENT LAWS 40 In circuits assembled in the real world, the wires will always have ﬁnite resistance. However, this resistance is typically so small compared to other resistances in the circuit that we can neglect it without introducing signiﬁcant error. In our idealized circuits, we will therefore refer to “zero resistance” wires from now on. ■FIGURE 3.1 (a) A circuit containing three nodes and ﬁve branches. (b) Node 1 is redrawn to look like two nodes; it is still one node. (a) 1 2 3 (b) 1 2 3 A point at which two or more elements have a common connection is called a node. For example, Fig. 3.1a shows a circuit containing three nodes. Sometimes networks are drawn so as to trap an unwary student into believing that there are more nodes present than is actually the case. This occurs when a node, such as node 1 in Fig. 3.1a, is shown as two separate junctions connected by a (zero-resistance) conductor, as in Fig. 3.1b. How-ever, all that has been done is to spread the common point out into a common zero-resistance line. Thus, we must necessarily consider all of the perfectly conducting leads or portions of leads attached to the node as part of the node. Note also that every element has a node at each of its ends. Suppose that we start at one node in a network and move through a sim-ple element to the node at the other end. We then continue from that node through a different element to the next node, and continue this movement until we have gone through as many elements as we wish. If no node was encountered more than once, then the set of nodes and elements that we have passed through is deﬁned as a path. If the node at which we started is the same as the node on which we ended, then the path is, by deﬁnition, a closed path or a loop. For example, in Fig. 3.1a, if we move from node 2 through the current source to node 1, and then through the upper right resistor to node 3, we have established a path; since we have not continued on to node 2 again, we have not made a loop. If we proceeded from node 2 through the current source to node 1, down through the left resistor to node 2, and then up through the central resistor to node 1 again, we do not have a path, since a node (actually two nodes) was encountered more than once; we also do not have a loop, because a loop must be a path. Another term whose use will prove convenient is branch. We deﬁne a branch as a single path in a network, composed of one simple element and the node at each end of that element. Thus, a path is a particular collection of branches. The circuit shown in Fig. 3.1a and b contains ﬁve branches. 3.2 • KIRCHHOFF’S CURRENT LAW We are now ready to consider the ﬁrst of the two laws named for Gustav Robert Kirchhoff (two h’s and two f ’s), a German university professor who was born about the time Ohm was doing his experimental work. This axiomatic law is called Kirchhoff’s current law (abbreviated KCL), and it simply states that The algebraic sum of the currents entering any node is zero. This law represents a mathematical statement of the fact that charge cannot accumulate at a node. A node is not a circuit element, and it certainly cannot store, destroy, or generate charge. Hence, the currents must sum to zero. A hydraulic analogy is sometimes useful here: for example, consider three water pipes joined in the shape of a Y. We deﬁne three “currents” as ﬂowing into each of the three pipes. If we insist that water is always ﬂow-ing, then obviously we cannot have three positive water currents, or the pipes would burst. This is a result of our deﬁning currents independent of SECTION 3.2 KIRCHHOFF’S CURRENT LAW 41 the direction that water is actually ﬂowing. Therefore, the value of either one or two of the currents as deﬁned must be negative. Consider the node shown in Fig. 3.2. The algebraic sum of the four cur-rents entering the node must be zero: iA + iB + (−iC) + (−iD) = 0 However, the law could be equally well applied to the algebraic sum of the currents leaving the node: (−i A) + (−iB) + iC + iD = 0 We might also wish to equate the sum of the currents having reference arrows directed into the node to the sum of those directed out of the node: iA + iB = iC + iD which simply states that the sum of the currents going in must equal the sum of the currents going out. ■FIGURE 3.2 Example node to illustrate the applica-tion of Kirchhoff’s current law. iC iB iA iD For the circuit in Fig. 3.3a, compute the current through resistor R3 if it is known that the voltage source supplies a current of 3 A. Identify the goal of the problem. The current through resistor R3, labeled as i on the circuit diagram. Collect the known information. The node at the top of R3 is connected to four branches. Two of these currents are clearly labeled: 2 A ﬂows out of the node into R2, and 5 A ﬂows into the node from the current source. We are told the current out of the 10 V source is 3 A. Devise a plan. If we label the current through R1 (Fig. 3.3b), we may write a KCL equation at the top node of resistors R2 and R3. Construct an appropriate set of equations. Summing the currents ﬂowing into the node: iR1 −2 −i + 5 = 0 The currents ﬂowing into this node are shown in the expanded dia-gram of Fig. 3.3c for clarity. Determine if additional information is required. We have one equation but two unknowns, which means we need to obtain an additional equation. At this point, the fact that we know the 10 V source is supplying 3 A comes in handy: KCL shows us that this is also the current iR1. Attempt a solution. Substituting, we ﬁnd that i = 3 −2 + 5 = 6 A. Verify the solution. Is it reasonable or expected? It is always worth the effort to recheck our work. Also, we can attempt to evaluate whether at least the magnitude of the solution is EXAMPLE 3.1 (Continued on next page) ■FIGURE 3.3 (a) Simple circuit for which the current through resistor R3 is desired. (b) The current through resistor R1 is labeled so that a KCL equation can be written. (c) The currents into the top node of R3 are redrawn for clarity. (a) + – R2 R3 R1 5 A 10 V i 2 A (c) R2 R3 R1 5 A 5 A i 2 A iR1 (iR1 – 2 A) (b) + – R2 R3 R1 5 A 10 V i 2 A iR1 CHAPTER 3 VOLTAGE AND CURRENT LAWS 42 ■FIGURE 3.5 The potential difference between points A and B is independent of the path selected. v1 + – v3 – + v2 + – A C B 1 2 3 A compact expression for Kirchhoff’s current law is N n=1 in = 0 which is just a shorthand statement for i1 + i2 + i3 + · · · + iN = 0 When Eq. or Eq. is used, it is understood that the N current arrows are either all directed toward the node in question, or are all directed away from it. 3.3 • KIRCHHOFF’S VOLTAGE LAW Current is related to the charge ﬂowing through a circuit element, whereas voltage is a measure of potential energy difference across the element. There is a single unique value for any voltage in circuit theory. Thus, the en-ergy required to move a unit charge from point A to point B in a circuit must have a value independent of the path chosen to get from A to B (there is often more than one such path). We may assert this fact through Kirchhoff’s voltage law (abbreviated KVL): The algebraic sum of the voltages around any closed path is zero. In Fig. 3.5, if we carry a charge of 1 C from A to B through element 1, the reference polarity signs for v1 show that we do v1 joules of work.1 Now (1) Note that we chose a 1 C charge for the sake of numerical convenience: therefore, we did (1 C)(v1 J/C) = v1 joules of work. reasonable. In this case, we have two sources—one supplies 5 A, and the other supplies 3 A. There are no other sources, independent or dependent. Thus, we would not expect to ﬁnd any current in the circuit in excess of 8 A. PRACTICE ● 3.1 Count the number of branches and nodes in the circuit in Fig. 3.4. If ix = 3 A and the 18 V source delivers 8 A of current, what is the value of RA? (Hint: You need Ohm’s law as well as KCL.) Ans: 5 branches, 3 nodes, 1. ■FIGURE 3.4 + – + – vx RA 13 A ix 5 6 18 V SECTION 3.3 KIRCHHOFF’S VOLTAGE LAW 43 if, instead, we choose to proceed from A to B via node C, then we expend (v2 −v3) joules of energy. The work done, however, is independent of the path in a circuit, and so any route must lead to the same value for the volt-age. In other words, v1 = v2 −v3 It follows that if we trace out a closed path, the algebraic sum of the volt-ages across the individual elements around it must be zero. Thus, we may write v1 + v2 + v3 + · · · + vN = 0 or, more compactly, N n=1 vn = 0 We can apply KVL to a circuit in several different ways. One method that leads to fewer equation-writing errors than others consists of moving mentally around the closed path in a clockwise direction and writing down directly the voltage of each element whose (+) terminal is entered, and writing down the negative of every voltage ﬁrst met at the (−) sign. Apply-ing this to the single loop of Fig. 3.5, we have −v1 + v2 −v3 = 0 which agrees with our previous result, Eq. . In the circuit of Fig. 3.6, ﬁnd vx and ix. We know the voltage across two of the three elements in the circuit. Thus, KVL can be applied immediately to obtain vx. Beginning with the bottom node of the 5 V source, we apply KVL clockwise around the loop: −5 −7 + vx = 0 so vx = 12 V. KCL applies to this circuit, but only tells us that the same current (ix) ﬂows through all three elements. We now know the voltage across the 100 resistor, however. Invoking Ohm’s law, ix = vx 100 = 12 100 A = 120 mA PRACTICE ● 3.2 Determine ix and vx in the circuit of Fig. 3.7. Ans: vx = −4 V; ix = −400 mA. EXAMPLE 3.2 ■FIGURE 3.6 A simple circuit with two voltage sources and a single resistor. + – + – 5 V 7 V 100 vx + – ix ■FIGURE 3.7 + – + – 3 V 1 V 10 vx + – ix CHAPTER 3 VOLTAGE AND CURRENT LAWS 44 EXAMPLE 3.3 In the circuit of Fig. 3.8 there are eight circuit elements. Find vR2 (the voltage across R2) and the voltage labeled vx. The best approach for ﬁnding vR2 is to look for a loop to which we can apply KVL. There are several options, but the leftmost loop offers a straightforward route, as two of the voltages are clearly speciﬁed. Thus, we ﬁnd vR2 by writing a KVL equation around the loop on the left, starting at point c: 4 −36 + vR2 = 0 which leads to vR2 = 32 V. To ﬁnd vx, we might think of this as the (algebraic) sum of the volt-ages across the three elements on the right. However, since we do not have values for these quantities, such an approach would not lead to a numerical answer. Instead, we apply KVL beginning at point c, moving up and across the top to a, through vx to b, and through the conducting lead to the starting point: +4 −36 + 12 + 14 + vx = 0 so that vx = 6 V An alternative approach: Knowing vR2, we might have taken the shortcut through R2: −32 + 12 + 14 + vx = 0 yielding vx = 6 V once again. ■FIGURE 3.8 A circuit with eight elements for which we desire vR2 and vx. + – + – 4 V – + vx + – vR2 – + vR1 + – v2 + – + – 12 V + – 14 V R1 R2 vs1 b c a 36 V Points b and c, as well as the wire between them, are all part of the same node. PRACTICE ● 3.3 For the circuit of Fig. 3.9, determine (a) vR2 and (b) v2, if vR1 = 1 V. ■FIGURE 3.9 + – + – –8 V – + vx + – vR2 – + vR1 + – v2 + – + – 7 V – + 9 V 12 V R1 R2 3 V b c a Ans: (a) 4 V; (b) −8 V. SECTION 3.3 KIRCHHOFF’S VOLTAGE LAW 45 Determine vx in the circuit of Fig. 3.10a. EXAMPLE 3.4 + – vx (a) 4 8 2 10 60 V ix + – 5 A + – vx (b) 4 8 2 10 60 V ix + – 5 A v10 + – v4 + – v8 + – i4 i10 i2 ■FIGURE 3.10 (a) A circuit for which vx is to be determined using KVL. (b) Circuit with voltages and currents labeled. We begin by labeling voltages and currents on the rest of the elements in the circuit (Fig. 3.10b). Note that vx appears across the 2 resistor and the source ix as well. If we can obtain the current through the 2 resistor, Ohm’s law will yield vx. Writing the appropriate KCL equation, we see that i2 = i4 + ix Unfortunately, we do not have values for any of these three quanti-ties. Our solution has (temporarily) stalled. Since we were given the current ﬂowing from the 60 V source, per-haps we should consider starting from that side of the circuit. Instead of ﬁnding vx using i2, it might be possible to ﬁnd vx directly using KVL. We can write the following KVL equations: −60 + v8 + v10 = 0 and −v10 + v4 + vx = 0 This is progress: we now have two equations in four unknowns, an improvement over one equation in which all terms were unknown. In fact, we know that v8 = 40V through Ohm’s law, as we were told that 5 Aﬂows through the 8 resistor. Thus, v10 = 0 + 60 −40 = 20 V, (Continued on next page) As we have just seen, the key to correctly analyzing a circuit is to ﬁrst me-thodically label all voltages and currents on the diagram. This way, carefully written KCL or KVL equations will yield correct relationships, and Ohm’s law can be applied as necessary if more unknowns than equations are ob-tained initially. We illustrate these principles with a more detailed example. so Eq. reduces to vx = 20 −v4 If we can determine v4, the problem is solved. The best route to ﬁnding a numerical value for the voltage v4 in this case is to employ Ohm’s law, which requires a value for i4. From KCL, we see that i4 = 5 −i10 = 5 −v10 10 = 5 −20 10 = 3 so that v4 = (4)(3) = 12 V and hence vx = 20 −12 = 8 V. CHAPTER 3 VOLTAGE AND CURRENT LAWS 46 ■FIGURE 3.12 (a) A single-loop circuit with four elements. (b) The circuit model with source voltages and resistance values given. (c) Current and voltage reference signs have been added to the circuit. (a) + – + – + – vs1 vs2 (b) R1 R2 i vR2 + – vR1 + – i i i + – + – vs1 vs2 (c) R1 R2 3.4 • THE SINGLE-LOOP CIRCUIT We have seen that repeated use of KCL and KVL in conjunction with Ohm’s law can be applied to nontrivial circuits containing several loops and a num-ber of different elements. Before proceeding further, this is a good time to focus on the concept of series (and, in the next section, parallel) circuits, as they form the basis of any network we will encounter in the future. All of the elements in a circuit that carry the same current are said to be connected in series. As an example, consider the circuit of Fig. 3.10. The 60 V source is in series with the 8 resistor; they carry the same 5 A cur-rent. However, the 8 resistor is not in series with the 4 resistor; they carry different currents. Note that elements may carry equal currents and not be in series; two 100 W light bulbs in neighboring houses may very well carry equal currents, but they certainly do not carry the same current and are not connected in series. Figure 3.12a shows a simple circuit consisting of two batteries and two resistors. Each terminal, connecting lead, and solder glob is assumed to have zero resistance; together they constitute an individual node of the circuit diagram in Fig. 3.12b. Both batteries are modeled by ideal voltage sources; any internal resistances they may have are assumed to be small enough to neglect. The two resistors are assumed to be ideal (linear) resistors. We seek the current through each element, the voltage across each ele-ment, and the power absorbed by each element. Our ﬁrst step in the analy-sis is the assumption of reference directions for the unknown currents. Arbitrarily, let us select a clockwise current i which ﬂows out of the upper terminal of the voltage source on the left. This choice is indicated by an ar-row labeled i at that point in the circuit, as shown in Fig. 3.12c. A trivial PRACTICE ● 3.4 Determine vx in the circuit of Fig. 3.11. Ans: vx = 12.8 V. ■FIGURE 3.11 + – vx 2 8 2 10 30 V ix + – 2 A SECTION 3.4 THE SINGLE-LOOP CIRCUIT 47 application of Kirchhoff’s current law assures us that this same current must also ﬂow through every other element in the circuit; we emphasize this fact this one time by placing several other current symbols about the circuit. Oursecondstepintheanalysisisachoiceofthevoltagereferenceforeach ofthetworesistors.Thepassivesignconventionrequiresthattheresistorcur-rent and voltage variables be deﬁned so that the current enters the terminal at which the positive voltage reference is located. Since we already (arbitrarily) selected the current direction,vR1 andvR2 are deﬁned as in Fig. 3.12c. The third step is the application of Kirchhoff’s voltage law to the only closed path. Let us decide to move around the circuit in the clockwise direction, beginning at the lower left corner, and to write down directly every voltage ﬁrst met at its positive reference, and to write down the nega-tive of every voltage encountered at the negative terminal. Thus, −vs1 + vR1 + vs2 + vR2 = 0 We then apply Ohm’s law to the resistive elements: vR1 = R1i and vR2 = R2i Substituting into Eq. yields −vs1 + R1i + vs2 + R2i = 0 Since i is the only unknown, we ﬁnd that i = vs1 −vs2 R1 + R2 The voltage or power associated with any element may now be obtained by applying v = Ri, p = vi, or p = i2R. PRACTICE ● 3.5 In the circuit of Fig. 3.12b, vs1 = 120 V, vs2 = 30 V, R1 = 30 , and R2 = 15 . Compute the power absorbed by each element. Ans: p120V = −240 W; p30V = +60 W; p30 = 120 W; p15 = 60 W. EXAMPLE 3.5 Compute the power absorbed in each element for the circuit shown in Fig. 3.13a. (Continued on next page) 2vA 30 15 + – 120 V + – vA – + (a) (b) i 2vA 30 15 v30 + + – – 120 V + – vA – + ■FIGURE 3.13 (a) A single-loop circuit containing a dependent source. (b) The current i and voltage v30 are assigned. CHAPTER 3 VOLTAGE AND CURRENT LAWS 48 We ﬁrst assign a reference direction for the current i and a reference po-larity for the voltage v30 as shown in Fig. 3.13b. There is no need to as-sign a voltage to the 15 resistor, since the controlling voltage vA for the dependent source is already available. (It is worth noting, however, that the reference signs for vA are reversed from those we would have assigned based on the passive sign convention.) This circuit contains a dependent voltage source, the value of which remains unknown until we determine vA. However, its algebraic value 2vA can be used in the same fashion as if a numerical value were avail-able. Thus, applying KVL around the loop: −120 + v30 + 2vA −vA = 0 Using Ohm’s law to introduce the known resistor values: v30 = 30i and vA = −15i Note that the negative sign is required since i ﬂows into the negative terminal of vA. Substituting into Eq. yields −120 + 30i −30i + 15i = 0 and so we ﬁnd that i = 8 A Computing the power absorbed by each element: p120v = (120)(−8) = −960 W p30 = (8)2(30) = 1920 W pdep = (2vA)(8) = 2(−15)(8) = −1920 W p15 = (8)2(15) = 960 W + – 12 V 30 8 7 + – vx + – 4vx ■FIGURE 3.14 A simple loop circuit. In the preceding example and practice problem, we were asked to compute the power absorbed by each element of a circuit. It is difﬁcult to think of a situation, however, in which all of the absorbed power quantities of a circuit would be positive, for the simple reason that the energy must come from somewhere. Thus, from simple conservation of energy, we expect that the sum of the absorbed power for each element of a circuit should be zero. In PRACTICE ● 3.6 In the circuit of Fig. 3.14, ﬁnd the power absorbed by each of the ﬁve elements in the circuit. Ans: (CW from left) 0.768 W, 1.92 W, 0.2048 W, 0.1792 W, −3.072 W. SECTION 3.5 THE SINGLE-NODE-PAIR CIRCUIT 49 other words, at least one of the quantities should be negative (neglecting the trivial case where the circuit is not operating). Stated another way, the sum of the supplied power for each element should be zero. More pragmatically, the sum of the absorbed power equals the sum of the supplied power, which seems reasonable enough at face value. Let’s test this with the circuit of Fig. 3.13 from Example 3.5, which consists of two sources (one dependent and one independent) and two resistors. Adding the power absorbed by each element, we ﬁnd all elements pabsorbed = −960 + 1920 −1920 + 960 = 0 In reality (our indication is the sign associated with the absorbed power) the 120 V source supplies +960 W, and the dependent source supplies +1920 W. Thus, the sources supply a total of 960 + 1920 = 2880 W. The resistors are expected to absorb positive power, which in this case sums to a total of 1920 + 960 = 2880 W. Thus, if we take into account each element of the circuit, pabsorbed = psupplied as we expect. Turning our attention to Practice Problem 3.6, the solution to which the reader might want to verify, we see that the absorbed powers sum to 0.768 + 1.92 + 0.2048 + 0.1792 −3.072 = 0. Interestingly enough, the 12 V independent voltage source is absorbing +1.92 W, which means it is dissipating power, not supplying it. Instead, the dependent voltage source appears to be supplying all the power in this particular circuit. Is such a thing possible? We usually expect a source to supply positive power, but since we are employing idealized sources in our circuits, it is in fact possi-ble to have a net power ﬂow into any source. If the circuit is changed in some way, the same source might then be found to supply positive power. The result is not known until a circuit analysis has been completed. 3.5 • THE SINGLE-NODE-PAIR CIRCUIT The companion of the single-loop circuit discussed in Sec. 3.4 is the single-node-pair circuit, in which any number of simple elements are connected between the same pair of nodes. An example of such a circuit is shown in Fig. 3.15a. KVL forces us to recognize that the voltage across each branch is the same as that across any other branch. Elements in a circuit having a common voltage across them are said to be connected in parallel. (Continued on next page) Find the voltage, current, and power associated with each element in the circuit of Fig. 3.15a. We ﬁrst deﬁne a voltage v and arbitrarily select its polarity as shown in Fig. 3.15b. Two currents, ﬂowing in the resistors, are selected in con-formance with the passive sign convention, as shown in Fig. 3.15b. EXAMPLE 3.6 CHAPTER 3 VOLTAGE AND CURRENT LAWS 50 5 A 6 A 1 A 10 10 v + – ■FIGURE 3.16 PRACTICE ● 3.7 Determine v in the circuit of Fig. 3.16. Ans: 50 V. (a) 120 A 30 A R1 R2 (b) 120 A 30 A R1 R2 v + – i1 i2 1 15 1 30 1 15 1 30 ■FIGURE 3.15 (a) A single-node-pair circuit. (b) A voltage and two currents are assigned. Determining either current i1 or i2 will enable us to obtain a value for v. Thus, our next step is to apply KCL to either of the two nodes in the circuit. Equating the algebraic sum of the currents leaving the upper node to zero: −120 + i1 + 30 + i2 = 0 Writing both currents in terms of the voltage v using Ohm’s law i1 = 30v and i2 = 15v we obtain −120 + 30v + 30 + 15v = 0 Solving this equation for v results in v = 2 V and invoking Ohm’s law then gives i1 = 60 A and i2 = 30 A The absorbed power in each element can now be computed. In the two resistors, pR1 = 30(2)2 = 120 W and pR2 = 15(2)2 = 60 W and for the two sources, p120A = 120(−2) = −240 W and p30A = 30(2) = 60 W Since the 120 A source absorbs negative 240 W, it is actually supplying power to the other elements in the circuit. In a similar fashion, we ﬁnd that the 30 A source is actually absorbing power rather than supplying it. SECTION 3.6 SERIES AND PARALLEL CONNECTED SOURCES 51 Determine the value of v and the power supplied by the independent current source in Fig. 3.17. EXAMPLE 3.7 2 k 2ix 6 k 24 mA v + – ix i6 ■FIGURE 3.17 A voltage v and a current i6 are assigned in a single-node-pair circuit containing a dependent source. By KCL, the sum of the currents leaving the upper node must be zero, so that i6 −2ix −0.024 −ix = 0 Again, note that the value of the dependent source (2ix) is treated the same as any other current would be, even though its exact value is not known until the circuit has been analyzed. We next apply Ohm’s law to each resistor: i6 = v 6000 and ix = −v 2000 Therefore, v 6000 −2 −v 2000 −0.024 − −v 2000 = 0 and so v = (600)(0.024) = 14.4 V. Any other information we may want to ﬁnd for this circuit is now eas-ily obtained, usually in a single step. For example, the power supplied by the independent source is p24 = 14.4(0.024) = 0.3456 W (345.6 mW). Ans: 3 A; −5.4 A; 6 A. 2 A 0.1vx 9 18 5.6 A vx + – iB iA iC ■FIGURE 3.18 PRACTICE ● 3.8 For the single-node-pair circuit of Fig. 3.18, ﬁnd iA, iB, and iC. 3.6 • SERIES AND PARALLEL CONNECTED SOURCES It turns out that some of the equation writing that we have been doing for series and parallel circuits can be avoided by combining sources. Note, however, that all the current, voltage, and power relationships in the remainder of the circuit will be unchanged. For example, several voltage CHAPTER 3 VOLTAGE AND CURRENT LAWS 52 EXAMPLE 3.8 Determine the current i in the circuit of Fig. 3.20a by ﬁrst combin-ing the sources into a single equivalent voltage source. To be able to combine the voltage sources, they must be in series. Since the same current (i) ﬂows through each, this condition is satisﬁed. Starting from the bottom left-hand corner and proceeding clockwise, −3 −9 −5 + 1 = −16 V so we may replace the four voltage sources with a single 16 V source having its negative reference as shown in Fig. 3.20b. KVL combined with Ohm’s law then yields −16 + 100i + 220i = 0 or i = 16 320 = 50 mA We should note that the circuit in Fig. 3.20c is also equivalent, a fact easily veriﬁed by computing i. (a) + – + – 9 V 5 V 1 V 3 V – + 100 220 (b) + – 16 V 100 i i (c) – + 16 V 100 220 220 i + – ■FIGURE 3.20 = (a) v1 v2 + – v1 + v2 – v3 v3 + – + – + – ■FIGURE 3.19 (a) Series-connected voltage sources can be replaced by a single source. (b) Parallel current sources can be replaced by a single source. = (b) i1 i2 i3 i1 – i2 + i3 sources in series may be replaced by an equivalent voltage source having a voltage equal to the algebraic sum of the individual sources (Fig. 3.19a). Parallel current sources may also be combined by algebraically adding the individual currents, and the order of the parallel elements may be rearranged as desired (Fig. 3.19b). SECTION 3.6 SERIES AND PARALLEL CONNECTED SOURCES 53 PRACTICE ● 3.9 Determine the current i in the circuit of Fig. 3.21 after ﬁrst replac-ing the four sources with a single equivalent source. Ans: −54 A. + – 4 V 3 V 5 V – + 1 V – + – + 47 7 i ■FIGURE 3.21 Determine the voltage v in the circuit of Fig. 3.22a by ﬁrst combin-ing the sources into a single equivalent current source. The sources may be combined if the same voltage appears across each one, which we can easily verify is the case. Thus, we create a new source, arrow pointing upward into the top node, by adding the currents that ﬂow into that node: 2.5 −2.5 −3 = −3 A One equivalent circuit is shown in Fig. 3.22b. KCL then allows us to write −3 + v 5 + v 5 = 0 Solving, we ﬁnd v = 7.5 V. Another equivalent circuit is shown in Fig. 3.22c. EXAMPLE 3.9 5 5 v + – 2.5 A 2.5 A 3 A (a) 5 5 v + – 3 A (c) 5 5 v + – –3 A (b) ■FIGURE 3.22 (Continued on next page) CHAPTER 3 VOLTAGE AND CURRENT LAWS 54 PRACTICE ● 3.10 Determine the voltage v in the circuit of Fig. 3.23 after ﬁrst replacing the three sources with a single equivalent source. Ans: 50 V. 10 10 v + – 5 A 6 A 1 A ■FIGURE 3.23 To conclude the discussion of parallel and series source combinations, we should consider the parallel combination of two voltage sources and the series combination of two current sources. For instance, what is the equiva-lent of a 5 V source in parallel with a 10 V source? By the deﬁnition of a voltage source, the voltage across the source cannot change; by Kirchhoff’s voltage law, then, 5 equals 10 and we have hypothesized a physical impos-sibility. Thus, ideal voltage sources in parallel are permissible only when each has the same terminal voltage at every instant. In a similar way, two current sources may not be placed in series unless each has the same cur-rent, including sign, for every instant of time. EXAMPLE 3.10 Determine which of the circuits of Fig. 3.24 are valid. The circuit of Fig. 3.24a consists of two voltage sources in parallel. The value of each source is different, so this circuit violates KVL. For exam-ple, if a resistor is placed in parallel with the 5 V source, it is also in paral-lel with the 10 V source. The actual voltage across it is therefore ambigu-ous, and clearly the circuit cannot be constructed as indicated. If we attempt to build such a circuit in real life, we will ﬁnd it impossible to locate “ideal” voltage sources—all real-world sources have an internal resistance. The presence of such resistance allows a voltage difference between the two real sources.Along these lines, the circuit of Fig. 3.24b is perfectly valid. + – 5 V + – 10 V (a) + – R 2 V + – 14 V (b) R 1 A 1 A (c) ■FIGURE 3.24 (a) to (c) Examples of circuits with multiple sources, some of which violate Kirchhoff’s laws. SECTION 3.7 RESISTORS IN SERIES AND PARALLEL 55 The circuit of Fig. 3.24c violates KCL: it is unclear what current actually ﬂows through the resistor R. PRACTICE ● 3.11 Determine whether the circuit of Fig. 3.25 violates either of Kirchhoff’s laws. Ans: No. If the resistor were removed, however, the resulting circuit would. R 5 A 3 A ■FIGURE 3.25 3.7 • RESISTORS IN SERIES AND PARALLEL It is often possible to replace relatively complicated resistor combinations with a single equivalent resistor. This is useful when we are not speciﬁcally interested in the current, voltage, or power associated with any of the indi-vidual resistors in the combinations. All the current, voltage, and power rela-tionships in the remainder of the circuit will be unchanged. Consider the series combination of N resistors shown in Fig. 3.26a. We want to simplify the circuit with replacing the N resistors with a single resistor Req so that the remainder of the circuit, in this case only the voltage source, does not realize that any change has been made. The current, voltage, and power of the source must be the same before and after the replacement. First, apply KVL: vs = v1 + v2 + · · · + vN and then Ohm’s law: vs = R1i + R2i + · · · + RNi = (R1 + R2 + · · · + RN)i Now compare this result with the simple equation applying to the equiv-alent circuit shown in Fig. 3.26b: vs = Reqi v1 + – v2 + – vN + – (a) R1 R2 RN vs + – i (b) Req vs + – i ■FIGURE 3.26 (a) Series combination of N resistors. (b) Electrically equivalent circuit. Helpful Tip: Inspection of the KVL equation for any series circuit will show that the order in which elements are placed in such a circuit makes no difference. Thus, the value of the equivalent resistance for N series resistors is We are therefore able to replace a two-terminal network consisting of N series resistors with a single two-terminal element Req that has the same v-i relationship. It should be emphasized again that we might be interested in the current, voltage, or power of one of the original elements. For example, the voltage of a dependent voltage source may depend upon the voltage across R3. Once R3 is combined with several series resistors to form an equivalent re-sistance, then it is gone and the voltage across it cannot be determined until R3 is identiﬁed by removing it from the combination. In that case, it would have been better to look ahead and not make R3 a part of the combination initially. Req = R1 + R2 + · · · + RN CHAPTER 3 VOLTAGE AND CURRENT LAWS 56 EXAMPLE 3.11 Use resistance and source combinations to determine the current i in Fig. 3.27a and the power delivered by the 80 V source. We ﬁrst interchange the element positions in the circuit, being careful to preserve the proper sense of the sources, as shown in Fig. 3.27b. The 30 V – + (a) 80 V + – + – i 8 10 7 20 V 5 + – + – i 80 V 10 30 V 20 V – + 8 7 5 (b) (c) 90 V + – i 30 ■FIGURE 3.27 (a) A series circuit with several sources and resistors. (b) The elements are rearranged for the sake of clarity. (c) A simpler equivalent. SECTION 3.7 RESISTORS IN SERIES AND PARALLEL 57 Similar simplifications can be applied to parallel circuits. A circuit containing N resistors in parallel, as in Fig. 3.29a, leads to the KCL equation is = i1 + i2 + · · · + iN or is = v R1 + v R2 + · · · + v RN = v Req Thus, 1 Req = 1 R1 + 1 R2 + · · · + 1 RN next step is to then combine the three voltage sources into an equivalent 90 V source, and the four resistors into an equivalent 30 resistance, as in Fig. 3.27c. Thus, instead of writing −80 + 10i −30 + 7i + 5i + 20 + 8i = 0 we have simply −90 + 30i = 0 and so we ﬁnd that i = 3 A In order to calculate the power delivered to the circuit by the 80 V source appearing in the given circuit, it is necessary to return to Fig. 3.27a with the knowledge that the current is 3A. The desired power is then 80 V × 3A 240 W. It is interesting to note that no element of the original circuit remains in the equivalent circuit. PRACTICE ● 3.12 Determine i in the circuit of Fig. 3.28. Ans: −333 mA. + – + – + – i 5 V 5 V 15 5 V 5 25 ■FIGURE 3.28 R2 R1 is RN v + – i2 i1 iN (a) ... ... is Req v + – (b) ■FIGURE 3.29 (a) A circuit with N resistors in parallel. (b) Equivalent circuit. CHAPTER 3 VOLTAGE AND CURRENT LAWS 58 which can be written as R−1 eq = R−1 1 + R−1 2 + · · · + R−1 N or, in terms of conductances, as Geq = G1 + G2 + · · · + G N The simpliﬁed (equivalent) circuit is shown in Fig. 3.29b. A parallel combination is routinely indicated by the following shorthand notation: Req = R1∥R2∥R3 The special case of only two parallel resistors is encountered fairly of-ten, and is given by Req = R1∥R2 = 1 1 R1 + 1 R2 Or, more simply, The last form is worth memorizing, although it is a common error to attempt to generalize Eq. to more than two resistors, e.g., Req = R1R2R3 R1 + R2 + R3 A quick look at the units of this equation will immediately show that the expression cannot possibly be correct. Req = R1R2 R1 + R2 PRACTICE ● 3.13 Determine v in the circuit of Fig. 3.30 by ﬁrst combining the three current sources, and then the two 10 resistors. Ans: 50 V. 10 10 v + – 5 A 6 A 1 A ■FIGURE 3.30 SECTION 3.7 RESISTORS IN SERIES AND PARALLEL 59 Calculate the powerand voltage of the dependent source in Fig. 3.31a. EXAMPLE 3.12 4 A 0.9i3 15 6 6 3 6 A vx + – i3 9 (a) 0.9i3 3 9 18 (b) 2 A v + – i3 0.9i3 3 2 A 6 v + – i3 (c) ■FIGURE 3.31 (a) A multinode circuit. (b) The two independent current sources are combined into a 2 A source, and the 15 resistor in series with the two parallel 6 resistors are replaced with a single 18 resistor. (c) A simpliﬁed equivalent circuit. We will seek to simplify the circuit before analyzing it, but take care not to include the dependent source since its voltage and power charac-teristics are of interest. Despite not being drawn adjacent to one another, the two indepen-dent current sources are in fact in parallel, so we replace them with a 2 A source. The two 6 resistors are in parallel and can be replaced with a single 3 resistor in series with the 15 resistor. Thus, the two 6 resistors and the 15 resistor are replaced by an 18 resistor (Fig. 3.31b). No matter how tempting, we should not combine the remaining three resistors; the controlling variable i3 depends on the 3 resistor and so that resistor must remain untouched. The only further simpliﬁcation, then, is 9 ∥18 = 6 , as shown in Fig. 3.31c. (Continued on next page) CHAPTER 3 VOLTAGE AND CURRENT LAWS 60 5 5 6 9 3 3 3 3 1 A vx + – i3 ■FIGURE 3.32 Applying KCL at the top node of Fig. 3.31c, we have −0.9i3 −2 + i3 + v 6 = 0 Employing Ohm’s law, v = 3i3 which allows us to compute i3 = 10 3 A Thus, the voltage across the dependent source (which is the same as the voltage across the 3 resistor) is v = 3i3 = 10 V The dependent source therefore furnishes v × 0.9i3 = 10(0.9)(10/3) = 30 W to the remainder of the circuit. Now if we are later asked for the power dissipated in the 15 resis-tor, we must return to the original circuit. This resistor is in series with an equivalent 3 resistor; a voltage of 10 V is across the 18 total; therefore, a current of 5/9 A ﬂows through the 15 resistor and the power absorbed by this element is (5/9)2(15) or 4.63 W. PRACTICE ● 3.14 For the circuit of Fig. 3.32, calculate the voltage vx. Ans: 2.819 V. SECTION 3.8 VOLTAGE AND CURRENT DIVISION 61 Three ﬁnal comments on series and parallel combinations might be helpful. The ﬁrst is illustrated by referring to Fig. 3.33a and asking, “Are vs and R in series or in parallel?” The answer is “Both.” The two elements carry the same current and are therefore in series; they also enjoy the same voltage and consequently are in parallel. The second comment is a word of caution. Circuits can be drawn in such a way as to make series or parallel combinations difficult to spot. In Fig. 3.33b, for example, the only two resistors in parallel are R2 and R3, while the only two in series are R1 and R8. The ﬁnal comment is simply that a simple circuit element need not be in series or parallel with any other simple circuit element in a circuit. For exam-ple, R4 and R5 in Fig. 3.33b are not in series or parallel with any other simple circuit element, and there are no simple circuit elements in Fig. 3.33c that are in series or parallel with any other simple circuit element. In other words, we cannot simplify that circuit further using any of the techniques discussed in this chapter. 3.8 • VOLTAGE AND CURRENT DIVISION By combining resistances and sources, we have found one method of short-ening the work of analyzing a circuit. Another useful shortcut is the appli-cation of the ideas of voltage and current division. Voltage division is used to express the voltage across one of several series resistors in terms of the + – (a) vs R (b) vs + – R3 R2 R1 R7 R5 R4 R6 R8 (c) vs + – RC RB RA RD RE iB iA is ■FIGURE 3.33 These two circuit elements are both in series and in parallel. (b) R2 and R3 are in parallel, and R1 and R8 are in series. (c) There are no circuit elements either in series or in parallel with one another. voltage across the combination. In Fig. 3.34, the voltage across R2 is found via KVL and Ohm’s law: v = v1 + v2 = iR1 + iR2 = i(R1 + R2) so i = v R1 + R2 Thus, v2 = iR2 = v R1 + R2 R2 or v2 = R2 R1 + R2 v and the voltage across R1 is, similarly, v1 = R1 R1 + R2 v If the network of Fig. 3.34 is generalized by removing R2 and replacing it with the series combination of R2, R3, . . . , RN, then we have the general result for voltage division across a string of N series resistors which allows us to compute the voltage vk that appears across an arbitrary resistor Rk of the series. vk = Rk R1 + R2 + · · · + RN v CHAPTER 3 VOLTAGE AND CURRENT LAWS 62 i v2 + – v1 + – v + – R1 R2 ■FIGURE 3.34 An illustration of voltage division. Determine vx in the circuit of Fig. 3.35a. EXAMPLE 3.13 i3 vx + – 4 6 12 sin t V 3 + – 4 Ω 2 (b) (a) 12 sin t V vx + – + – ■FIGURE 3.35 A numerical example illustrating resistance combination and voltage division. (a) Original circuit. (b) Simpliﬁed circuit. We ﬁrst combine the 6 and 3 resistors, replacing them with (6)(3)/(6 + 3) = 2 . Since vx appears across the parallel combination, our simpliﬁcation has not lost this quantity. However, further simpliﬁcation of the circuit by replacing the series combination of the 4 resistor with our new 2 resistor would. SECTION 3.8 VOLTAGE AND CURRENT DIVISION 63 Thus, we proceed by simply applying voltage division to the circuit in Fig. 3.35b: vx = (12 sin t) 2 4 + 2 = 4 sin t volts PRACTICE ● 3.15 Use voltage division to determine vx in the circuit of Fig. 3.36. Ans: 2 V. The dual2 of voltage division is current division. We are now given a total current supplied to several parallel resistors, as shown in the circuit of Fig. 3.37. The current ﬂowing through R2 is i2 = v R2 = i(R1∥R2) R2 = i R2 R1R2 R1 + R2 or and, similarly, Nature has not smiled on us here, for these last two equations have a factor which differs subtly from the factor used with voltage division, and some effort is going to be needed to avoid errors. Many students look on the expression for voltage division as “obvious” and that for current division as being “different.” It helps to realize that the larger of two parallel resistors always carries the smaller current. For a parallel combination of N resistors, the current through resistor Rk is ik = i 1 Rk 1 R1 + 1 R2 + · · · + 1 RN i1 = i R2 R1 + R2 i2 = i R1 R1 + R2 + – 10 V 2 3 10 10 vx + – ■FIGURE 3.36 i v + – R2 R1 i1 i2 ■FIGURE 3.37 An illustration of current division. (2) The principle of duality is encountered often in engineering. We will consider the topic brieﬂy in Chap. 7 when we compare inductors and capacitors. Written in terms of conductances, ik = i Gk G1 + G2 + · · · + G N which strongly resembles Eq. for voltage division. CHAPTER 3 VOLTAGE AND CURRENT LAWS 64 EXAMPLE 3.14 Write an expression for the current through the 3 resistor in the circuit of Fig. 3.38. The total current ﬂowing into the 3 –6 combination is i(t) = 12 sin t 4 + 3∥6 = 12 sin t 4 + 2 = 2 sin t A and thus the desired current is given by current division: i3(t) = (2 sin t) 6 6 + 3 = 4 3 sin t A i3 vx + – 4 6 12 sin t V 3 + – ■FIGURE 3.38 A circuit used as an example of current division. The wavy line in the voltage source symbol indicates a sinusoidal variation with time. Unfortunately, current division is sometimes applied when it is not applicable. As one example, let us consider again the circuit shown in Fig. 3.33c, a circuit that we have already agreed contains no circuit ele-ments that are in series or in parallel. Without parallel resistors, there is no way that current division can be applied. Even so, there are too many stu-dents who take a quick look at resistors RA and RB and try to apply current division, writing an incorrect equation such as iA = iS RB RA + RB Remember, parallel resistors must be branches between the same pair of nodes. PRACTICE ● 3.16 In the circuit of Fig. 3.39, use resistance combination methods and current division to ﬁnd i1, i2, and v3. i1 i2 v3 + – 125 50 240 20 2 40 120 mA ■FIGURE 3.39 Ans: 100 mA; 50 mA; 0.8 V. Up to now, we have been drawing circuit schematics in a fashion similar to that of the one shown in Fig. 3.40, where voltages are deﬁned across two clearly marked terminals. Special care was taken to emphasize the fact that voltage cannot be deﬁned at a single point—it is by deﬁnition the difference in potential between two points. However, many schematics make use of the convention of taking the earth as deﬁning zero volts, so that all other voltages are implicitly referenced to this potential. The concept is often referred to as earth ground, and is fun-damentally tied to safety regulations designed to prevent ﬁres, fatal electrical shocks, and related mayhem. The symbol for earth ground is shown in Fig. 3.41a. Since earth ground is deﬁned as zero volts, it is often convenient to use this as a common terminal in schemat-ics. The circuit of Fig. 3.40 is shown redrawn in this fashion in Fig. 3.42, where the earth ground symbol rep-resents a common node. It is important to note that the two circuits are equivalent in terms of our value for va (4.5 V in either case), but are no longer exactly the same. The circuit in Fig. 3.40 is said to be “ﬂoating” in that it could for all practical purposes be installed on a circuit board of a satellite in geosynchronous orbit (or on its way to Pluto). The circuit in Fig. 3.42, however, is some-how physically connected to the ground through a conducting path. For this reason, there are two other symbols that are occasionally used to denote a common terminal. Figure 3.41b shows what is commonly referred to as signal ground; there can be (and often is) a large voltage between earth ground and any terminal tied to signal ground. The fact that the common terminal of a circuit may or may not be connected by some low-resistance pathway to earth ground can lead to potentially dangerous situa-tions. Consider the diagram of Fig. 3.43a, which depicts an innocent bystander about to touch a piece of equip-ment powered by an ac outlet. Only two terminals have been used from the wall socket; the round ground pin PRACTICAL APPLICATION Not the Earth Ground from Geology PRACTICAL APPLICATION of the receptacle was left unconnected. The common terminal of every circuit in the equipment has been tied together and electrically connected to the conducting equipment chassis; this terminal is often denoted using the chassis ground symbol of Fig. 3.41c. Unfortunately, a wiring fault exists, due to either poor manufacturing or perhaps just wear and tear. At any rate, the chassis is not “grounded,” so there is a very large resistance between chassis ground and earth ground. A pseudo-schematic (some liberty was taken with the person’s equivalent re-sistance symbol) of the situation is shown in Fig. 3.43b. The electrical path between the conducting chassis and ground may in fact be the table, which could represent a resistance of hundreds of megaohms or more. The resis-tance of the person, however, is many orders of magni-tude lower. Once the person taps on the equipment to see why it isn’t working properly . . . well, let’s just say not all stories have happy endings. The fact that “ground” is not always “earth ground” can cause a wide range of safety and electrical noise problems. One example is occasionally encountered in older buildings, where plumbing originally consisted of electrically conducting copper pipes. In such buildings, any water pipe was often treated as a low-resistance path to earth ground, and therefore used in many electrical connections. However, when corroded pipes are replaced with more modern and cost-effective + – 9 V 4.7 k 4.7 k va + – ■FIGURE 3.40 A simple circuit with a voltage va deﬁned between two terminals. + – 9 V 4.7 k 4.7 k va + – ■FIGURE 3.42 The circuit of Fig. 3.40, redrawn using the earth ground symbol. The rightmost ground symbol is redundant; it is only necessary to label the positive terminal of va; the negative reference is then implicitly ground, or zero volts. (a) (b) (c) ■FIGURE 3.41 Three different symbols used to represent a ground or common terminal: (a) earth ground; (b) signal ground; (c) chassis ground. (Continued on next page) nonconducting PVC piping, the low-resistance path to earth ground no longer exists. A related problem occurs when the composition of the earth varies greatly over a particular region. In such situations, it is possible to ac-tually have two separated buildings in which the two “earth grounds” are not equal, and current can ﬂow as a result. Within this text, the earth ground symbol will be used exclusively. It is worth remembering, however, that not all grounds are created equal in practice. (a) Wall outlet Requipment Rto ground (b) 115 V + – ■FIGURE 3.43 (a) A sketch of an innocent person about to touch an improperly grounded piece of equipment. It’s not going to be pretty. (b) A schematic of an equivalent circuit for the situation as it is about to unfold; the person has been represented by an equivalent resistance, as has the equipment. A resistor has been used to represent the nonhuman path to ground. SUMMARY AND REVIEW We began this chapter by discussing connections of circuit elements, and introducing the terms node, path, loop, and branch. The next two topics could be considered the two most important in the entire textbook, namely, Kirchhoff’s current law (KCL) and Kirchhoff’s voltage law. The ﬁrst is derived from conservation of charge, and can be thought of in terms of “what goes in (current) must come out.” The second is based on conservation of energy, and can be viewed as “what goes up (potential) must come down.” These two laws allow us to analyze any circuit, linear or otherwise, provided we have a way of relating the voltage and current associated with passive elements (e.g., Ohm’s law for the resistor). In the case of a single-loop circuit, the elements are connected in series and hence each carries the same current. The single-node-pair circuit, in which elements are connected in parallel with one another, is characterized by a single voltage common to each element. Extending these concepts allowed us to develop a means of simplifying voltage sources connected in series, or current sources in parallel; subsequently we obtained classic expressions for series and parallel connected resistors. The ﬁnal topic, that of voltage and current division, ﬁnds considerable use in the design of circuits where a speciﬁc voltage or current is required but our choice of source is limited. Let’s conclude with key points of this chapter to review, highlighting appropriate examples. EXERCISES 67 67 ❑Kirchhoff’s current law (KCL) states that the algebraic sum of the currents entering any node is zero. (Examples 3.1, 3.4) ❑Kirchhoff’s voltage law (KVL) states that the algebraic sum of the voltages around any closed path in a circuit is zero. (Examples 3.2, 3.3) ❑All elements in a circuit that carry the same current are said to be connected in series. (Example 3.5) ❑Elements in a circuit having a common voltage across them are said to be connected in parallel. (Examples 3.6, 3.7) ❑Voltage sources in series can be replaced by a single source, provided care is taken to note the individual polarity of each source. (Examples 3.8, 3.10) ❑Current sources in parallel can be replaced by a single source, provided care is taken to note the direction of each current arrow. (Examples 3.9, 3.10) ❑A series combination of N resistors can be replaced by a single resistor having the value Req = R1 + R2 + · · · + RN . (Example 3.11) ❑A parallel combination of N resistors can be replaced by a single resistor having the value 1 Req = 1 R1 + 1 R2 + · · · + 1 RN (Example 3.12) ❑Voltage division allows us to calculate what fraction of the total voltage across a series string of resistors is dropped across any one resistor (or group of resistors). (Example 3.13) ❑Current division allows us to calculate what fraction of the total current into a parallel string of resistors ﬂows through any one of the resistors. (Example 3.14) READING FURTHER A discussion of the principles of conservation of energy and conservation of charge, as well as Kirchhoff’s laws, can be found in R. Feynman, R. B. Leighton, and M. L. Sands, The Feynman Lectures on Physics. Reading, Mass.: Addison-Wesley, 1989, pp. 4-1, 4-7, and 25-9. Detailed discussions of numerous aspects of grounding practices consistent with the 2008 National Electrical Code® can be found throughout J. E. McPartland, B. J. McPartland, and F. P. Hartwell, McGraw-Hill’s National Electrical Code® 2008 Handbook, 26th ed. New York, McGraw-Hill, 2008. EXERCISES 3.1 Nodes, Paths, Loops, and Branches 1. Referring to the circuit depicted in Fig. 3.44, count the number of (a) nodes; (b) elements; (c) branches. 4 2 A 5 A 14 1.5 2 5 ■FIGURE 3.44 CHAPTER 3 VOLTAGE AND CURRENT LAWS 68 2. Referring to the circuit depicted in Fig. 3.45, count the number of (a) nodes; (b) elements; (c) branches. + – A B C F G E D ■FIGURE 3.46 C A B E D ■FIGURE 3.47 4 5 A 2 A 4 1.5 2 5 ■FIGURE 3.45 EAT AT RALPH’S EAT AT RALPH’S + – + – ■FIGURE 3.48 3. For the circuit of Fig. 3.46: (a) Count the number of nodes. (b) In moving from A to B, have we formed a path? Have we formed a loop? (c) In moving from C to F to G, have we formed a path? Have we formed a loop? 4. For the circuit of Fig. 3.46: (a) Count the number of circuit elements. (b) If we move from B to C to D, have we formed a path? Have we formed a loop? (c) If we move from E to D to C to B, have we formed a path? Have we formed a loop? 5. Refer to the circuit of Fig. 3.47, and answer the following: (a) How many distinct nodes are contained in the circuit? (b) How many elements are contained in the circuit? (c) How many branches does the circuit have? (d) Determine if each of the following represents a path, a loop, both, or neither: (i) A to B (ii) B to D to C to E (iii) C to E to D to B to A to C (iv) C to D to B to A to C to E 3.2 Kirchhoff’s Current Law 6. A local restaurant has a neon sign constructed from 12 separate bulbs; when a bulb fails, it appears as an inﬁnite resistance and cannot conduct current. In wiring the sign, the manufacturer offers two options (Fig. 3.48). From what you’ve learned about KCL, which one should the restaurant owner select? Explain. EXERCISES 69 69 7. Referring to the single node diagram of Fig. 3.49, compute: (a) iB, if iA = 1 A, iD = 2 A, iC = 3 A, and iE = 0; (b) iE, if iA = 1 A, iB = 1 A, iC = 1 A, and iD = 1 A. iC iB iA iD iE ■FIGURE 3.49 + – 6 A 1.5 V 7 A I I 3 A 3 A I 2 A 9 A (a) (b) (c) 1 1 5 ■FIGURE 3.50 + – R2 R3 R1 3 A 2 V 1 A i2 ■FIGURE 3.51 8. Determine the current labeled I in each of the circuits of Fig. 3.50. 9. In the circuit shown in Fig. 3.51, the resistor values are unknown, but the 2 V source is known to be supplying a current of 7 A to the rest of the circuit. Calculate the current labeled i2. + – R2 R3 R1 7 A – 2 V –3 A i2 ■FIGURE 3.52 RA 6 5 + – + – ix –1.6 A 9 V vx ■FIGURE 3.53 + – + – + – V1 V2 R1 R2 1 k 1 k 150IB IB IE IC ■FIGURE 3.54 10. The voltage source in the circuit of Fig. 3.52 has a current of 1 A ﬂowing out of its positive terminal into resistor R1. Calculate the current labeled i2. 11. In the circuit depicted in Fig. 3.53, ix is determined to be 1.5 A, and the 9 V source supplies a current of 7.6 A (that is, a current of 7.6 A leaves the positive reference terminal of the 9 V source). Determine the value of resistor RA. 12. For the circuit of Fig. 3.54 (which is a model for the dc operation of a bipolar junction transistor biased in forward active region), IB is measured to be 100 μA. Determine IC and IE. CHAPTER 3 VOLTAGE AND CURRENT LAWS 70 Vs + – R R R Vx + – ■FIGURE 3.56 v1 + – v3 – + v2 + – A C B 1 2 3 ■FIGURE 3.57 13. Determine the current labeled I3 in the circuit of Fig. 3.55. 2 mA 4.7 k 3 1 5Vx Vx + – I3 ■FIGURE 3.55 14. Study the circuit depicted in Fig. 3.56, and explain (in terms of KCL) why the voltage labeled Vx must be zero. 15. In many households, multiple electrical outlets within a given room are often all part of the same circuit. Draw the circuit for a four-walled room which has a single electrical outlet per wall, with a lamp (represented by a 1 resistor) connected to each outlet. 3.3 Kirchoff’s Voltage Law 16. For the circuit of Fig. 3.57: (a) Determine the voltage v1 if v2 = 0 and v3 = 17 V. (b) Determine the voltage v1 if v2 = 2 V and v3 = 2 V. (c) Determine the voltage v2 if v1 = 7 V and v3 = 9 V. (d) Determine the voltage v3 if v1 = 2.33 V and v2 = 1.70 V. 17. For each of the circuits in Fig. 3.58, determine the voltage vx and the current ix. – + – + 9 V 4 V 7 vx + – ix – + – + 2 V –7 V 8 vx + – ix (a) (b) ■FIGURE 3.58 EXERCISES 71 + – + – + – 1 V 2 V 5 V 2 10 (a) i + – + – + – + – 10 V 1.5 V + – 1.5 V 2 V 2 2 1 V 2 2 (b) i ■FIGURE 3.59 + – + – 4 V – + vx + – vR – + v1 + – v2 + – + – 12 V + – v3 R1 R2 1.5 V b c a 23 V ■FIGURE 3.60 18. Use KVL to obtain a numerical value for the current labeled i in each circuit depicted in Fig. 3.59. 19. In the circuit of Fig. 3.60, it is determined that v1 = 3 V and v3 = 1.5 V. Calcu-late vR and v2. 20. In the circuit of Fig. 3.60, a voltmeter is used to measure the following: v1 = 2 V and v3 = 1.5 V. Calculate vx. 21. Determine the value of vx as labeled in the circuit of Fig. 3.61. + – vx 2 7.3 2 1 2.3 V ix + – 500 mA ■FIGURE 3.61 22. Consider the simple circuit shown in Fig. 3.62. Using KVL, derive the expressions v1 = vs R1 R1 + R2 and v2 = vs R2 R1 + R2 23. (a) Determine a numerical value for each current and voltage (i1, v1, etc.) in the circuit of Fig. 3.63. (b) Calculate the power absorbed by each element and verify that they sum to zero. + – R2 R1 vs v2 + – v1 + – ■FIGURE 3.62 5i2 5v1 5 6 2 V v1 + – v2 + – v4 + – v5 + – v3 + – i4 i2 i5 + – + – i1 i3 ■FIGURE 3.63 CHAPTER 3 VOLTAGE AND CURRENT LAWS 72 24. The circuit shown in Fig. 3.64 includes a device known as an op amp. This device has two unusual properties in the circuit shown: (1) Vd = 0 V, and (2) no current can ﬂow into either input terminal (marked “−” and “+” inside the symbol), but it can ﬂow through the output terminal (marked “OUT”). This seemingly impossible situation—in direct conﬂict with KCL—is a result of power leads to the device that are not included in the symbol. Based on this information, calculate Vout. (Hint: two KVL equations are required, both involving the 5 V source.) 1 k 2.2 k 500 + – – + 2 V 3vx vx + – ■FIGURE 3.66 + – + – ix X 27 33 19 2 V 12 V v1 + – ■FIGURE 3.67 + – 5 V Vd + – Vout + – 100 470 OP AMP OUT + – ■FIGURE 3.64 3.4 The Single-Loop Circuit 25. The circuit of Fig. 3.12b is constructed with the following: vs1 = 8 V, R1 = 1 , vs2 = 16 V, and R2 = 4.7 . Calculate the power absorbed by each element. Verify that the absorbed powers sum to zero. 26. Obtain a numerical value for the power absorbed by each element in the circuit shown in Fig. 3.65. 8vA 2 5 + – 4.5 V + – vA – + ■FIGURE 3.65 27. Compute the power absorbed by each element of the circuit of Fig. 3.66. 28. Compute the power absorbed by each element in the circuit of Fig. 3.67 if the mysterious element X is (a) a 13 resistor; (b) a dependent voltage source labeled 4v1, “” reference on top; (c) a dependent voltage source labeled 4ix, “” reference on top. 29. Kirchhoff’s laws apply whether or not Ohm’s law applies to a particular element. The I-V characteristic of a diode, for example, is given by ID = IS eVD/VT −1 EXERCISES 73 where VT = 27 mV at room temperature and IS can vary from 10−12 to 10−3 A. In the circuit of Fig. 3.68, use KVL/KCL to obtain VD if IS = 29 pA. (Note: This problem results in a transcendental equation, requiring an iterative approach to obtaining a numerical solution. Most scientiﬁc calculators will perform such a function.) 3.5 The Single-Node-Pair Circuit 30. Referring to the circuit of Fig. 3.69, (a) determine the two currents i1 and i2; (b) compute the power absorbed by each element. + – 100 3 V ID VD + – ■FIGURE 3.68 3 A 7 A 2 R1 R2 v + – i1 i2 4 ■FIGURE 3.69 –2 A 3 A 6 R1 R2 v + – i1 i2 10 ■FIGURE 3.70 1 A 2 A 5 5 A v + – 5 ■FIGURE 3.71 3 3ix 1 2 A v + – ix ■FIGURE 3.72 2.8 k 4.7 k 1 k 5 mA 3 mA ■FIGURE 3.73 31. Determine a value for the voltage v as labeled in the circuit of Fig. 3.70, and compute the power supplied by the two current sources. 32. Referring to the circuit depicted in Fig. 3.71, determine the value of the voltage v. 33. Determine the voltage v as labeled in Fig. 3.72, and calculate the power supplied by each current source. 34. Although drawn so that it may not appear obvious at ﬁrst glance, the circuit of Fig. 3.73 is in fact a single-node-pair circuit. (a) Determine the power absorbed by each resistor. (b) Determine the power supplied by each current source. (c) Show that the sum of the absorbed power calculated in (a) is equal to the sum of the supplied power calculated in (c). 41. (a) Determine the values for IX and VY in the circuit shown in Fig. 3.79. (b) Are those values necessarily unique for that circuit? Explain. (c) Simplify the circuit of Fig. 3.79 as much as possible and still maintain the values for v and i. (Your circuit must contain the 1 resistor.) CHAPTER 3 VOLTAGE AND CURRENT LAWS 74 3.6 Series and Parallel Connected Sources 35. Determine the numerical value for veq in Fig. 3.74a, if (a) v1 = 0, v2 = 3 V, and v3 = 3 V; (b) v1 = v2 = v3 = 1 V; (c) v1 = 9 V, v2 = 4.5 V, v3 = 1 V. 36. Determine the numerical value for ieq in Fig. 3.74b, if (a) i1 = 0, i2 = 3 A, and i3 = 3 A; (b) i1 = i2 = i3 = 1 A; (c) i1 = 9 A, i2 = 4.5 A, i3 = 1 A. 37. For the circuit presented in Fig. 3.75, determine the current labeled i by ﬁrst combining the four sources into a single equivalent source. 38. Determine the value of v1 required to obtain a zero value for the current la-beled i in the circuit of Fig. 3.76. 7 A 8 A 2 5 A v + – 3 ■FIGURE 3.77 = (a) v1 v2 v3 + – + – veq + – + – = (b) i1 i2 i3 ieq ■FIGURE 3.74 – + + – 6 V 2 V 12 V 2 V – + 1 k + – i ■FIGURE 3.75 – + – + 4 V v1 2 V 1 V + – 7 7 + – i ■FIGURE 3.76 1.28 A –2.57 A 1 IS v + – 1 ■FIGURE 3.78 –3 A 3 A IX 3 V –4 A 1 v + – i + – 4 V + – VY + – ■FIGURE 3.79 39. (a) For the circuit of Fig. 3.77, determine the value for the voltage labeled v, after ﬁrst simplifying the circuit to a single current source in parallel with two resistors. (b) Verify that the power supplied by your equivalent source is equal to the sum of the supplied powers of the individual sources in the original circuit. 40. What value of IS in the circuit of Fig. 3.78 will result in a zero voltage v? EXERCISES 75 2 2 3 1 1 2 4 (a) (b) ■FIGURE 3.80 + – + – 2 7 1 5 1 V 3 V i ■FIGURE 3.82 2 1 4 (a) 1 4 3 (b) ■FIGURE 3.81 3.7 Resistors in Series and Parallel 42. Determine the equivalent resistance of each of the networks shown in Fig. 3.80. 43. For each network depicted in Fig. 3.81, determine a single equivalent resistance. 44. (a) Simplify the circuit of Fig. 3.82 as much as possible by using source and resistor combinations. (b) Calculate i, using your simpliﬁed circuit. (c) To what voltage should the 1 V source be changed to reduce i to zero? (d) Calculate the power absorbed by the 5 resistor. 45. (a) Simplify the circuit of Fig. 3.83, using appropriate source and resistor com-binations. (b) Determine the voltage labeled v, using your simpliﬁed circuit. (c) Calculate the power provided by the 2 A source to the rest of the circuit. 2 A 1 A 5 5 A v + – 5 ■FIGURE 3.83 46. Making appropriate use of resistor combination techniques, calculate i3 in the circuit of Fig. 3.84 and the power provided to the circuit by the single current source. 3 5 6 3 1 A 9 3 5 3 vx + – i3 ■FIGURE 3.84 CHAPTER 3 VOLTAGE AND CURRENT LAWS 76 4 A 2i 6 15 3 6 9 A 6 3 A i ■FIGURE 3.86 49. Calculate the equivalent resistance Req of the network shown in Fig. 3.87 if R1 = 2R2 = 3R3 = 4R4 etc. and R11 = 3 . Req R2 R5 R8 R3 R1 R4 R7 R10 R11 R6 R9 ■FIGURE 3.87 i v2 + – v1 + – v + – R1 R2 ■FIGURE 3.88 i v + – R2 R1 i1 i2 ■FIGURE 3.89 3 15 6 6 2 A 4i 9 vx + – i ■FIGURE 3.85 47. Calculate the voltage labeled vx in the circuit of Fig. 3.85 after ﬁrst simplify-ing, using appropriate source and resistor combinations. 48. Determine the power absorbed by the 15 resistor in the circuit of Fig. 3.86. 50. Show how to combine four 100 resistors to obtain an equivalent resistance of (a) 25 ; (b) 60 ; (c) 40 . 3.8 Voltage and Current Division 51. In the voltage divider network of Fig. 3.88, calculate (a) v2 if v = 9.2 V and v1 = 3 V; (b) v1 if v2 = 1 V and v = 2 V; (c) v if v1 = 3 V and v2 = 6 V; (d) R1/R2 if v1 = v2; (e) v2 if v = 3.5 V and R1 = 2R2; (f ) v1 if v = 1.8 V, R1 = 1 k, and R2 = 4.7 k. 52. In the current divider network represented in Fig. 3.89, calculate (a) i1 if i = 8 A and i2 = 1 A; (b) v if R1 = 100 k, R2 = 100 k, and i = 1 mA; (c) i2 if i = 20 mA, R1 = 1 , and R2 = 4 ; (d) i1 if i = 10 A, R1 = R2 = 9 ; (e) i2 if i = 10 A, R1 = 100 M, and R2 =1. EXERCISES 77 + – 3 V 2 3 2 10 vx + – ■FIGURE 3.91 v + – R4 R1 i1 i4 R2 i2 R3 i3 ■FIGURE 3.90 53. Choose a voltage v 2.5 V and values for the resistors R1, R2, R3, and R4 in the circuit of Fig. 3.90 so that i1 =1 A, i2 =1.2 A, i3 =8 A, and i4 = 3.1 A. 54. Employ voltage division to assist in the calculation of the voltage labeled vx in the circuit of Fig. 3.91. 55. A network is constructed from a series connection of ﬁve resistors having val-ues 1 , 3 , 5 , 7 , and 9 . If 9 V is connected across the terminals of the network, employ voltage division to calculate the voltage across the 3 resis-tor, and the voltage across the 7 resistor. 56. Employing resistance combination and current division as appropriate, deter-mine values for i1, i2, and v3 in the circuit of Fig. 3.92. v3 + – 1 2 5 4 4 4 25 A i1 i2 ■FIGURE 3.92 57. In the circuit of Fig. 3.93, only the voltage vx is of interest. Simplify the circuit using appropriate resistor combinations and iteratively employ voltage division to determine vx. 2 k 4 k 3 k 7 k 4 k 3 k 3 V + – 1 k vx + – ■FIGURE 3.93 + – + – + – 10 V 20 V 0.7 V 10 k 10 k 1 k 10i1 i1 ■FIGURE 3.94 Chapter-Integrating Exercises 58. The circuit shown in Fig. 3.94 is a linear model of a bipolar junction transistor biased in the forward active region of operation. Explain why voltage division is not a valid approach for determining the voltage across either 10 k resistor. CHAPTER 3 VOLTAGE AND CURRENT LAWS 78 10 10 50 4 40 20 20 2 V + – ■FIGURE 3.97 1 2 2 5 5 2 A 2 ■FIGURE 3.98 6 cos 2300t V 1 k 3 k 3.3 k 15 k gmv v + – vout + – + – ■FIGURE 3.96 12 cos 1000t mV 30 15 k 1 k 10 k gmv v + – vout + – + – ■FIGURE 3.95 61. With regard to the circuit shown in Fig. 3.97, compute (a) the voltage across the two 10 resistors, assuming the top terminal is the positive reference; (b) the power dissipated by the 4 resistor. 59. A common midfrequency model for a ﬁeld effect–based ampliﬁer circuit is shown in Fig. 3.95. If the controlling parameter gm (known as the transconduc-tance) is equal to 1.2 mS, employ current division to obtain the current through the 1 k resistor, and then calculate the ampliﬁer output voltage vout. 60. The circuit depicted in Fig. 3.96 is routinely employed to model the midfre-quency operation of a bipolar junction transistor–based ampliﬁer. Calculate the ampliﬁer output vout if the transconductance gm is equal to 322 mS. 62. Delete the leftmost 10 resistor in the circuit of Fig. 3.97, and compute (a) the current ﬂowing into the left-hand terminal of the 40 resistor; (b) the power supplied by the 2 V source; (c) the power dissipated by the 4 resistor. 63. Consider the seven-element circuit depicted in Fig. 3.98. (a) How many nodes, loops, and branches does it contain? (b) Calculate the current ﬂowing through each resistor. (c) Determine the voltage across the current source, assuming the top terminal is the positive reference terminal. INTRODUCTION Armed with the trio of Ohm’s and Kirchhoff’s laws, analyzing a simple linear circuit to obtain useful information such as the current, voltage, or power associated with a particular element is perhaps starting to seem a straightforward enough venture. Still, for the moment at least, every circuit seems unique, requiring (to some degree) a measure of creativity in approaching the analysis. In this chapter, we learn two basic circuit analysis techniques— nodal analysis and mesh analysis—both of which allow us to investigate many different circuits with a consistent, methodical approach. The result is a streamlined analysis, a more uniform level of complexity in our equations, fewer errors and, perhaps most importantly, a reduced occurrence of “I don’t know how to even start!” Most of the circuits we have seen up to now have been rather simple and (to be honest) of questionable practical use. Such circuits are valuable, however, in helping us to learn to apply fundamental techniques. Although the more complex circuits appearing in this chapter may represent a variety of electrical systems including control circuits, communication networks, motors, or integrated circuits, as well as electric circuit models of nonelectrical systems, we believe it best not to dwell on such speciﬁcs at this early stage. Rather, it is important to initially focus on the methodology of problem solving that we will continue to develop throughout the book. KEY CONCEPTS Nodal Analysis The Supernode Technique Mesh Analysis The Supermesh Technique Choosing Between Nodal and Mesh Analysis Computer-Aided Analysis, Including PSpice and MATLAB Basic Nodal and Mesh Analysis C H A P T E R 4 79 CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 80 4.1 • NODAL ANALYSIS We begin our study of general methods for methodical circuit analysis by considering a powerful method based on KCL, namely nodal analysis. In Chap. 3 we considered the analysis of a simple circuit containing only two nodes. We found that the major step of the analysis was obtaining a single equation in terms of a single unknown quantity—the voltage between the pair of nodes. We will now let the number of nodes increase and correspondingly pro-vide one additional unknown quantity and one additional equation for each added node. Thus, a three-node circuit should have two unknown voltages and two equations; a 10-node circuit will have nine unknown voltages and nine equations; an N-node circuit will need (N −1) voltages and (N −1) equations. Each equation is a simple KCL equation. To illustrate the basic technique, consider the three-node circuit shown in Fig. 4.1a, redrawn in Fig. 4.1b to emphasize the fact that there are only three nodes, numbered accordingly. Our goal will be to determine the volt-age across each element, and the next step in the analysis is critical. We des-ignate one node as a reference node; it will be the negative terminal of our N −1 = 2 nodal voltages, as shown in Fig. 4.1c. A little simpliﬁcation in the resultant equations is obtained if the node connected to the greatest number of branches is identiﬁed as the reference node. If there is a ground node, it is usually most convenient to select it as the reference node, although many people seem to prefer selecting the bot-tom node of a circuit as the reference, especially if no explicit ground is noted. The voltage of node 1 relative to the reference node is named v1, and v2 is deﬁned as the voltage of node 2 with respect to the reference node. These 3.1 A –1.4 A 2 5 (a) 1 1 2 3 5 (b) 2 1 3.1 A –1.4 A 1 2 5 1 2 (c) Reference node 3.1 A –1.4 A + + – – v1 v2 v1 v2 3.1 A –1.4 A 5 1 2 (d) Ref. ■FIGURE 4.1 (a) A simple three-node circuit. (b) Circuit redrawn to emphasize nodes. (c) Reference node selected and voltages assigned. (d) Shorthand voltage references. If desired, an appropriate ground symbol may be substituted for “Ref.” SECTION 4.1 NODAL ANALYSIS 81 two voltages are all we need, as the voltage between any other pair of nodes may be found in terms of them. For example, the voltage of node 1 with respect to node 2 is v1 −v2. The voltages v1 and v2 and their reference signs are shown in Fig. 4.1c. It is common practice once a reference node has been labeled to omit the reference signs for the sake of clarity; the node labeled with the voltage is taken to be the positive terminal (Fig. 4.1d). This is understood to be a type of shorthand voltage notation. We now apply KCL to nodes 1 and 2. We do this by equating the total current leaving the node through the several resistors to the total source current entering the node. Thus, v1 2 + v1 −v2 5 = 3.1 or 0.7v1 −0.2v2 = 3.1 At node 2 we obtain v2 1 + v2 −v1 5 = −(−1.4) or −0.2v1 + 1.2v2 = 1.4 Equations and are the desired two equations in two unknowns, and they may be solved easily. The results are v1 = 5 V and v2 = 2 V. From this, it is straightforward to determine the voltage across the 5 resistor: v5 = v1 −v2 = 3 V. The currents and absorbed powers may also be computed in one step. We should note at this point that there is more than one way to write the KCL equations for nodal analysis. For example, the reader may prefer to sum all the currents entering a given node and set this quantity to zero. Thus, for node 1 we might have written 3.1 −v1 2 −v1 −v2 5 = 0 or 3.1 + −v1 2 + v2 −v1 5 = 0 either of which is equivalent to Eq. . Is one way better than any other? Every instructor and every student develop a personal preference, and at the end of the day the most important thing is to be consistent. The authors prefer constructing KCL equations for nodal analysis in such a way as to end up with all current source terms on one side and all resistor terms on the other. Speciﬁcally, currents entering the node from current sources = currents leaving the node through resistors There are several advantages to such an approach. First, there is never any confusion regarding whether a term should be “v1 −v2” or “v2 −v1;” the The reference node in a schematic is implicitly deﬁned as zero volts. However, it is important to remember that any terminal can be designated as the reference terminal. Thus, the reference node is at zero volts with respect to the other deﬁned nodal voltages, and not necessarily with respect to earth ground. CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 82 Determine the current ﬂowing left to right through the 15 resistor of Fig. 4.2a. EXAMPLE 4.1 ﬁrst voltage in every resistor current expression corresponds to the node for which a KCL equation is being written, as seen in Eqs. and . Second, it allows a quick check that a term has not been accidentally omitted. Sim-ply count the current sources connected to a node and then the resistors; grouping them in the stated fashion makes the comparison a little easier. ■FIGURE 4.2 (a) A four-node circuit containing two independent current sources. (b) The two resistors in series are replaced with a single 10 resistor, reducing the circuit to three nodes. 3 5 15 7 2 A 4 A Ref. v1 v2 (a) 2 A 4 A 10 15 5 v1 v2 i Ref. (b) Nodal analysis will directly yield numerical values for the nodal volt-ages v1 and v2, and the desired current is given by i = (v1 −v2)/15. Before launching into nodal analysis, however, we ﬁrst note that no details regarding either the 7 resistor or the 3 resistor are of inter-est. Thus, we may replace their series combination with a 10 resistor as in Fig. 4.2b. The result is a reduction in the number of equations to solve. Writing an appropriate KCL equation for node 1, 2 = v1 10 + v1 −v2 15 and for node 2, 4 = v2 5 + v2 −v1 15 Rearranging, we obtain 5v1 −2v2 = 60 and −v1 + 4v2 = 60 Solving, we ﬁnd that v1 = 20 V and v2 = 20 V so that v1 −v2 = 0. In other words, zero current is ﬂowing through the 15 resistor in this circuit! SECTION 4.1 NODAL ANALYSIS 83 PRACTICE ● 4.1 For the circuit of Fig. 4.3, determine the nodal voltages v1 and v2. ■FIGURE 4.3 3 4 15 2 5 A 2 A v1 v2 Ans: v1 = −145/8 V, v2 = 5/2 V. Now let us increase the number of nodes so that we may use this tech-nique to work a slightly more difﬁcult problem. Determine the nodal voltages for the circuit of Fig. 4.4a, as refer-enced to the bottom node. Identify the goal of the problem. There are four nodes in this circuit. With the bottom node as our refer-ence, we label the other three nodes as shown in Fig. 4.4b. The circuit has been redrawn for clarity, taking care to identify the two relevant nodes for the 4 resistor. Collect the known information. We have three unknown voltages, v1, v2, and v3. All current sources and resistors have designated values, which are marked on the schematic. Devise a plan. This problem is well suited to nodal analysis, as three independent KCL equations may be written in terms of the current sources and the current through each resistor. Construct an appropriate set of equations. We begin by writing a KCL equation for node 1: −8 −3 = v1 −v2 3 + v1 −v3 4 or 0.5833v1 −0.3333v2 −0.25v3 = −11 At node 2: −(−3) = v2 −v1 3 + v2 1 + v2 −v3 7 EXAMPLE 4.2 ■FIGURE 4.4 (a) A four-node circuit. (b) Redrawn circuit with reference node chosen and voltages labeled. –3 A –8 A (a) 3 7 4 5 1 –25 A 3 7 4 (b) –3 A 1 –8 A –25 A Reference node 5 v1 v2 v3 (Continued on next page) CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 84 or −0.3333v1 + 1.4762v2 −0.1429v3 = 3 And, at node 3: −(−25) = v3 5 + v3 −v2 7 + v3 −v1 4 or, more simply, −0.25v1 −0.1429v2 + 0.5929v3 = 25 Determine if additional information is required. We have three equations in three unknowns. Provided that they are independent, this is sufﬁcient to determine the three voltages. Attempt a solution. Equations through can be solved using a scientiﬁc calculator (Appendix 5), software packages such as MATLAB, or more tradi-tional “plug-and-chug” techniques such as elimination of variables, matrix methods, or Cramer’s rule. Using the latter method, described in Appendix 2, we have v1 = −11 −0.3333 −0.2500 3 1.4762 −0.1429 25 −0.1429 0.5929 0.5833 −0.3333 −0.2500 −0.3333 1.4762 −0.1429 −0.2500 −0.1429 0.5929 = 1.714 0.3167 = 5.412 V Similarly, v2 = 0.5833 −11 −0.2500 −0.3333 3 −0.1429 −0.2500 25 0.5929 0.3167 = 2.450 0.3167 = 7.736 V and v3 = 0.5833 −0.3333 −11 −0.3333 1.4762 3 −0.2500 −0.1429 25 0.3167 = 14.67 0.3167 = 46.32 V Verify the solution. Is it reasonable or expected? Substituting the nodal voltages into any of our three nodal equations is sufﬁcient to ensure we made no computational errors. Beyond that, is it possible to determine whether these voltages are “reasonable” values? We have a maximum possible current of 3 + 8 + 25 = 36 amperes anywhere in the circuit. The largest resistor is 7 , so we do not expect any voltage magnitude greater than 7 × 36 = 252 V. There are, of course, numerous methods available for the solution of linear systems of equations, and we describe several in Appendix 2 in detail. Prior to the advent of the scientiﬁc calculator, Cramer’s rule as seen in Example 4.2 was very common in circuit analysis, although occasionally tedious to implement. It is, however, straightforward to use on a simple SECTION 4.1 NODAL ANALYSIS 85 four-function calculator, and so an awareness of the technique can be valuable. MATLAB, on the other hand, although not likely to be available during an examination, is a powerful software package that can greatly sim-plify the solution process; a brief tutorial on getting started is provided in Appendix 6. For the situation encountered in Example 4.2, there are several options available through MATLAB. First, we can represent Eqs. to in matrix form: ⎡ ⎣ 0.5833 −0.3333 −0.25 −0.3333 1.4762 −0.1429 −0.25 −0.1429 0.5929 ⎤ ⎦ ⎡ ⎣ v1 v2 v3 ⎤ ⎦= ⎡ ⎣ −11 3 25 ⎤ ⎦ so that ⎡ ⎣ v1 v2 v3 ⎤ ⎦= ⎡ ⎣ 0.5833 −0.3333 −0.25 −0.3333 1.4762 −0.1429 −0.25 −0.1429 0.5929 ⎤ ⎦ −1 ⎡ ⎣ −11 3 25 ⎤ ⎦ In MATLAB, we write >> a = [0.5833 -0.3333 -0.25; -0.3333 1.4762 -0.1429; -0.25 -0.1429 0.5929]; >> c = [-11; 3; 25]; >> b = a^-1 c b = 5.4124 7.7375 46.3127 >> where spaces separate elements along rows, and a semicolon separates rows. The matrix named b, which can also be referred to as a vector as it has only one column, is our solution. Thus, v1 = 5.412 V, v2 = 7.738 V, and v3 = 46.31 V (some rounding error has been incurred). We could also use the KCL equations as we wrote them initially if we employ the symbolic processor of MATLAB. eqn1 = '-8 -3 = (v1 - v2)/ 3 + (v1 - v3)/ 4'; >> eqn2 = '-(-3) = (v2 - v1)/ 3 + v2/ 1 + (v2 - v3)/ 7'; >> eqn3 = '-(-25) = v3/ 5 + (v3 - v2)/ 7 + (v3 - v1)/ 4'; >> answer = solve(eqn1, eqn2, eqn3, 'v1', 'v2', 'v3'); >> answer.v1 ans = 720/133 >> answer.v2 ans = 147/19 >> answer.v3 ans = 880/19 >> which results in exact answers, with no rounding errors. The solve() routine is invoked with the list of symbolic equations we named eqn1, eqn2, and eqn3, but the variables v1, v2 and v3 must also be speciﬁed. If solve() is called with fewer variables than equations, an algebraic solution is returned. The form of the solution is worth a quick comment; it is returned in what is referred to in programming parlance as a structure; in this case, we called our structure “answer.’’ Each component of the structure is accessed sepa-rately by name as shown. CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 86 PRACTICE ● 4.2 For the circuit of Fig. 4.5, compute the voltage across each current source. ■FIGURE 4.5 3 A 7 A Reference node 3 5 4 1 2 Ans: v3A = 5.235 V; v7A = 11.47 V. The previous examples have demonstrated the basic approach to nodal analysis, but it is worth considering what happens if dependent sources are present as well. EXAMPLE 4.3 Determine the power supplied by the dependent source of Fig. 4.6a. ■FIGURE 4.6 (a) A four-node circuit containing a dependent current source. (b) Circuit labeled for nodal analysis. vx + – 15 A 1 3i1 2 3 i1 vx + – 15 A 1 3i1 2 3 i1 v2 v1 (a) Ref. (b) SECTION 4.1 NODAL ANALYSIS 87 We choose the bottom node as our reference, since it has a large number of branch connections, and proceed to label the nodal voltages v1 and v2 as shown in Fig. 4.6b. The quantity labeled vx is actually equal to v2. At node 1, we write 15 = v1 −v2 1 + v1 2 and at node 2 3i1 = v2 −v1 1 + v2 3 Unfortunately, we have only two equations but three unknowns; this is a direct result of the presence of the dependent current source, since it is not controlled by a nodal voltage. Thus, we need an additional equation that relates i1 to one or more nodal voltages. In this case, we ﬁnd that i1 = v1 2 which upon substitution into Eq. yields (with a little rearranging) 3v1 −2v2 = 30 and Eq. simpliﬁes to −15v1 + 8v2 = 0 Solving, we ﬁnd that v1 = −40 V, v2 = −75 V, and i1 = 0.5v1 = −20 A. Thus, the power supplied by the dependent source is equal to (3i1)(v2) = (−60)(−75) = 4.5 kW. We see that the presence of a dependent source will create the need for an additional equation in our analysis if the controlling quantity is not a nodal voltage. Now let’s look at the same circuit, but with the controlling variable of the dependent current source changed to a different quantity— the voltage across the 3 resistor, which is in fact a nodal voltage. We will ﬁnd that only two equations are required to complete the analysis. Determine the power supplied by the dependent source of Fig. 4.7a. We select the bottom node as our reference and label the nodal voltages as shown in Fig. 4.7b. We have labeled the nodal voltage vx explicitly for clarity. Note that our choice of reference node is important in this case; it led to the quantity vx being a nodal voltage. Our KCL equation for node 1 is 15 = v1 −vx 1 + v1 2 EXAMPLE 4.4 (Continued on next page) CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 88 ■FIGURE 4.7 (a) A four-node circuit containing a dependent current source. (b) Circuit labeled for nodal analysis. vx + – 15 A 3vx 3 1 2 i1 vx + – 15 A 3vx 3 1 2 i1 vx v1 (a) Ref. (b) ■FIGURE 4.8 5 A A 2 1 2 i1 v2 v1 Ref. and for node x is 3vx = vx −v1 1 + v2 3 Grouping terms and solving, we ﬁnd that v1 = 50 7 V and vx = −30 7 V. Thus, the dependent source in this circuit generates (3vx)(vx) = 55.1 W. PRACTICE ● 4.3 For the circuit of Fig. 4.8, determine the nodal voltage v1 if A is (a) 2i1; (b) 2v1. Ans: (a) 70 9 V; (b) –10 V. Summary of Basic Nodal Analysis Procedure 1. Count the number of nodes (N). 2. Designate a reference node. The number of terms in your nodal equations can be minimized by selecting the node with the great-est number of branches connected to it. 3. Label the nodal voltages (there are N −1 of them). 4. Write a KCL equation for each of the nonreference nodes. Sum the currents ﬂowing into a node from sources on one side of the equation. On the other side, sum the currents ﬂowing out of the node through resistors. Pay close attention to “−” signs. 5. Express any additional unknowns such as currents or voltages other than nodal voltages in terms of appropriate nodal voltages. This situation can occur if voltage sources or dependent sources appear in our circuit. 6. Organize the equations. Group terms according to nodal voltages. 7. Solve the system of equations for the nodal voltages (there will be N −1 of them). SECTION 4.2 THE SUPERNODE 89 These seven basic steps will work on any circuit we ever encounter, although the presence of voltage sources will require extra care. Such situ-ations are discussed next. 4.2 • THE SUPERNODE As an example of how voltage sources are best handled when performing nodal analysis, consider the circuit shown in Fig. 4.9a. The original four-node circuit of Fig. 4.4 has been changed by replacing the 7 resistor be-tween nodes 2 and 3 with a 22 V voltage source. We still assign the same node-to-reference voltages v1, v2, and v3. Previously, the next step was the application of KCL at each of the three nonreference nodes. If we try to do that once again, we see that we will run into some difﬁculty at both nodes 2 and 3, for we do not know what the current is in the branch with the voltage source. There is no way by which we can express the current as a function of the voltage, for the deﬁnition of a voltage source is exactly that the volt-age is independent of the current. There are two ways out of this dilemma. The more difﬁcult approach is to assign an unknown current to the branch which contains the voltage source, proceed to apply KCL three times, and then apply KVL (v3 −v2 = 22) once between nodes 2 and 3; the result is then four equations in four unknowns. The easier method is to treat node 2, node 3, and the voltage source to-gether as a sort of supernode and apply KCL to both nodes at the same time; the supernode is indicated by the region enclosed by the broken line in Fig. 4.9a. This is okay because if the total current leaving node 2 is zero and the total current leaving node 3 is zero, then the total current leaving the combination of the two nodes is zero. This concept is represented graphi-cally in the expanded view of Fig. 4.9b. ■FIGURE 4.9 (a) The circuit of Example 4.2 with a 22 V source in place of the 7 resistor. (b) Expanded view of the region deﬁned as a supernode; KCL requires that all currents ﬂowing into the region sum to zero, or we would pile up or run out of electrons. 3 22 V 4 –3 A 1 –8 A –25 A 5 v1 v2 v3 + – Reference node 22 V (b) (a) + – Determine the value of the unknown node voltage v1 in the circuit of Fig. 4.9a. The KCL equation at node 1 is unchanged from Example 4.2: −8 −3 = v1 −v2 3 + v1 −v3 4 or 0.5833v1 −0.3333v2 −0.2500v3 = −11 Next we consider the 2-3 supernode. Two current sources are con-nected, and four resistors. Thus, 3 + 25 = v2 −v1 3 + v3 −v1 4 + v3 5 + v2 1 or −0.5833v1 + 1.3333v2 + 0.45v3 = 28 EXAMPLE 4.5 (Continued on next page) CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 90 ■FIGURE 4.10 4 A 5 V 9 A Reference node + – 1 2 1 6 1 3 Since we have three unknowns, we need one additional equation, and it must utilize the fact that there is a 22 V voltage source between nodes 2 and 3: v2 −v3 = −22 Solving Eqs. to , the solution for v1 is 1.071 V. PRACTICE ● 4.4 For the circuit of Fig. 4.10, compute the voltage across each current source. Ans: 5.375 V, 375 mV. The presence of a voltage source thus reduces by 1 the number of nonreference nodes at which we must apply KCL, regardless of whether the voltage source extends between two nonreference nodes or is connected between a node and the reference. We should be careful in analyzing circuits such as that of Practice Problem 4.4. Since both ends of the resistor are part of the supernode, there must technically be two corresponding current terms in the KCL equation, but they cancel each other out. We can summarize the supernode method as follows: Summary of Supernode Analysis Procedure 1. Count the number of nodes (N). 2. Designate a reference node. The number of terms in your nodal equations can be minimized by selecting the node with the greatest number of branches connected to it. 3. Label the nodal voltages (there are N −1 of them). 4. If the circuit contains voltage sources, form a supernode about each one. This is done by enclosing the source, its two terminals, and any other elements connected between the two terminals within a broken-line enclosure. 5. Write a KCL equation for each nonreference node and for each supernode that does not contain the reference node. Sum the currents ﬂowing into a node/supernode from current sources on one side of the equation. On the other side, sum the currents ﬂowing out of the node/supernode through resistors. Pay close attention to “−” signs. 6. Relate the voltage across each voltage source to nodal voltages. This is accomplished by simple application of KVL; one such equation is needed for each supernode deﬁned. 7. Express any additional unknowns (i.e., currents or voltages other than nodal voltages) in terms of appropriate nodal voltages. This situation can occur if dependent sources appear in our circuit. 8. Organize the equations. Group terms according to nodal voltages. 9. Solve the system of equations for the nodal voltages (there will be N −1 of them). SECTION 4.2 THE SUPERNODE 91 We see that we have added two additional steps from our general nodal analysis procedure. In reality, however, application of the supernode tech-nique to a circuit containing voltage sources not connected to the reference node will result in a reduction in the number of KCL equations required. With this in mind, let’s consider the circuit of Fig. 4.11, which contains all four types of sources and has ﬁve nodes. Determine the node-to-reference voltages in the circuit of Fig. 4.11. After establishing a supernode about each voltage source, we see that we need to write KCL equations only at node 2 and at the supernode containing the dependent voltage source. By inspection, it is clear that v1 = −12 V. At node 2, v2 −v1 0.5 + v2 −v3 2 = 14 while at the 3-4 supernode, 0.5vx = v3 −v2 2 + v4 1 + v4 −v1 2.5 We next relate the source voltages to the node voltages: v3 −v4 = 0.2vy and 0.2vy = 0.2(v4 −v1) Finally, we express the dependent current source in terms of the assigned variables: 0.5vx = 0.5(v2 −v1) Five nodes requires four KCL equations in general nodal analysis, but we have reduced this requirement to only two, as we formed two separate supernodes. Each supernode required a KVL equation (Eq. and v1 = −12, the latter written by inspection). Neither dependent source was controlled by a nodal voltage, so two additional equations were needed as a result. With this done, we can now eliminate vx and vy to obtain a set of four equations in the four node voltages: −2v1 + 2.5v2 −0.5v3 = 14 0.1v1 − v2 + 0.5v3 + 1.4v4 = 0 v1 = −12 0.2v1 + v3 −1.2v4 = 0 Solving, v1 = −12 V, v2 = −4 V, v3 = 0 V, and v4 = −2 V. PRACTICE ● 4.5 Determine the nodal voltages in the circuit of Fig. 4.12. Ans: v1 = 3 V, v2 = −2.33 V, v3 = −1.91 V, v4 = 0.945 V. EXAMPLE 4.6 ■FIGURE 4.12 0.15vx 4 2 3 Ref. 4 A – + 3 V v3 v1 v4 v2 1 2 vx – + + – ■FIGURE 4.11 A ﬁve-node circuit with four different types of sources. + – + – 0.5vx 2 1 2.5 Ref. 0.5 14 A 12 V v3 v1 v4 v2 vy vx – + – + 0.2vy 4.3 • MESH ANALYSIS As we have seen, nodal analysis is a straightforward analysis technique when only current sources are present, and voltage sources are easily accommo-dated with the supernode concept. Still, nodal analysis is based on KCL, and the reader might at some point wonder if there isn’t a similar approach based on KVL. There is—it’s known as mesh analysis—and although only strictly speaking applicable to what we will shortly deﬁne as a planar circuit, it can in many cases prove simpler to apply than nodal analysis. If it is possible to draw the diagram of a circuit on a plane surface in such a way that no branch passes over or under any other branch, then that circuit is said to be a planar circuit. Thus, Fig. 4.13a shows a planar network, Fig. 4.13b shows a nonplanar network, and Fig. 4.13c also shows a planar network, although it is drawn in such a way as to make it appear nonplanar at ﬁrst glance. CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 92 ■FIGURE 4.13 Examples of planar and nonplanar networks; crossed wires without a solid dot are not in physical contact with each other. + – (a) + – (b) + – (c) We should mention that mesh-type analysis can be applied to nonplanar circuits, but since it is not possible to deﬁne a complete set of unique meshes for such circuits, assignment of unique mesh currents is not possible. In Sec. 3.1, the terms path, closed path, and loop were deﬁned. Before we deﬁne a mesh, let us consider the sets of branches drawn with heavy lines in Fig. 4.14. The ﬁrst set of branches is not a path, since four branches are connected to the center node, and it is of course also not a loop. The sec-ond set of branches does not constitute a path, since it is traversed only by passing through the central node twice. The remaining four paths are all loops. The circuit contains 11 branches. The mesh is a property of a planar circuit and is undeﬁned for a nonpla-nar circuit. We deﬁne a mesh as a loop that does not contain any other loops within it. Thus, the loops indicated in Fig. 4.14c and d are not meshes, whereas those of parts e and f are meshes. Once a circuit has been drawn neatly in planar form, it often has the appearance of a multipaned window; the boundary of each pane in the window may be considered to be a mesh. If a network is planar, mesh analysis can be used to accomplish the analysis. This technique involves the concept of a mesh current, which we introduce by considering the analysis of the two-mesh circuit of Fig. 4.15a. As we did in the single-loop circuit, we will begin by deﬁning a current through one of the branches. Let us call the current ﬂowing to the right through the 6 resistor i1. We will apply KVL around each of the two meshes, and the two resulting equations are sufﬁcient to determine two un-known currents. We next deﬁne a second current i2 ﬂowing to the right in SECTION 4.3 MESH ANALYSIS 93 the 4 resistor. We might also choose to call the current ﬂowing downward through the central branch i3, but it is evident from KCL that i3 may be ex-pressed in terms of the two previously assumed currents as (i1 −i2). The assumed currents are shown in Fig. 4.15b. Following the method of solution for the single-loop circuit, we now ap-ply KVL to the left-hand mesh, −42 + 6i1 + 3(i1 −i2) = 0 or 9i1 −3i2 = 42 Applying KVL to the right-hand mesh, −3(i1 −i2) + 4i2 −10 = 0 or −3i1 + 7i2 = 10 Equations and are independent equations; one cannot be de-rived from the other. With two equations and two unknowns, the solution is easily obtained: i1 = 6 A i2 = 4 A and (i1 −i2) = 2 A If our circuit contains M meshes, then we expect to have M mesh cur-rents and therefore will be required to write M independent equations. Now let us consider this same problem in a slightly different manner by using mesh currents. We deﬁne a mesh current as a current that ﬂows only around the perimeter of a mesh. One of the greatest advantages in the use of mesh currents is the fact that Kirchhoff's current law is automatically satis-ﬁed. If a mesh current ﬂows into a given node, it ﬂows out of it also. ■FIGURE 4.14 (a) The set of branches identiﬁed by the heavy lines is neither a path nor a loop. (b) The set of branches here is not a path, since it can be traversed only by passing through the central node twice. (c) This path is a loop but not a mesh, since it encloses other loops. (d) This path is also a loop but not a mesh. (e, f) Each of these paths is both a loop and a mesh. (a) (b) (c) (d) (e) ( f) ■FIGURE 4.15 (a, b) A simple circuit for which currents are required. + – + – 42 V 10 V 3 6 4 (a) (b) i1 i2 (i1 – i2) + – + – 42 V 10 V 3 6 4 If we call the left-hand mesh of our problem mesh 1, then we may es-tablish a mesh current i1 ﬂowing in a clockwise direction about this mesh. A mesh current is indicated by a curved arrow that almost closes on itself and is drawn inside the appropriate mesh, as shown in Fig. 4.16. The mesh current i2 is established in the remaining mesh, again in a clockwise direc-tion. Although the directions are arbitrary, we will always choose clockwise mesh currents because a certain error-minimizing symmetry then results in the equations. We no longer have a current or current arrow shown directly on each branch in the circuit. The current through any branch must be determined by considering the mesh currents ﬂowing in every mesh in which that branch appears. This is not difﬁcult, because no branch can appear in more than two meshes. For example, the 3 resistor appears in both meshes, and the cur-rent ﬂowing downward through it is i1 −i2. The 6 resistor appears only in mesh 1, and the current ﬂowing to the right in that branch is equal to the mesh current i1. For the left-hand mesh, −42 + 6i1 + 3(i1 −i2) = 0 while for the right-hand mesh, 3(i2 −i1) + 4i2 −10 = 0 and these two equations are equivalent to Eqs. and . CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 94 A mesh current may often be identiﬁed as a branch current, as i1 and i2 have been identiﬁed in this example. This is not always true, however, for consider-ation of a square nine-mesh network soon shows that the central mesh current cannot be identiﬁed as the current in any branch. EXAMPLE 4.7 Determine the power supplied by the 2 V source of Fig. 4.17a. ■FIGURE 4.17 (a) A two-mesh circuit containing three sources. (b) Circuit labeled for mesh analysis. + – + – + – 5 V 1 V 4 5 2 V 2 i1 i2 + – + – + – 5 V 1 V 4 5 2 V 2 (a) (b) We ﬁrst deﬁne two clockwise mesh currents as shown in Fig. 4.17b. Beginning at the bottom left node of mesh 1, we write the following KVL equation as we proceed clockwise through the branches: −5 + 4i1 + 2(i1 −i2) −2 = 0 Doing the same for mesh 2, we write +2 + 2(i2 −i1) + 5i2 + 1 = 0 ■FIGURE 4.16 The same circuit considered in Fig. 4.15b, but viewed a slightly different way. i1 i2 + – + – 42 V 10 V 3 6 4 SECTION 4.3 MESH ANALYSIS 95 ■FIGURE 4.19 A ﬁve-node, seven-branch, three-mesh circuit. i2 i3 i1 + – + – 7 V 6 V 1 2 1 2 3 Rearranging and grouping terms, 6i1 −2i2 = 7 and −2i1 + 7i2 = −3 Solving, i1 = 43 38 = 1.132 A and i2 = −2 19 = −0.1053 A. The current ﬂowing out of the positive reference terminal of the 2 V source is i1 −i2. Thus, the 2 V source supplies (2)(1.237) = 2.474 W. PRACTICE ● 4.6 Determine i1 and i2 in the circuit in Fig. 4.18. Let us next consider the ﬁve-node, seven-branch, three-mesh circuit shown in Fig. 4.19. This is a slightly more complicated problem because of the additional mesh. ■FIGURE 4.18 + – + – 6 V 5 V 14 10 5 5 i1 i2 Ans: +184.2 mA; −157.9 mA. Use mesh analysis to determine the three mesh currents in the circuit of Fig. 4.19. The three required mesh currents are assigned as indicated in Fig. 4.19, and we methodically apply KVL about each mesh: −7 + 1(i1 −i2) + 6 + 2(i1 −i3) = 0 1(i2 −i1) + 2i2 + 3(i2 −i3) = 0 2(i3 −i1) −6 + 3(i3 −i2) + 1i3 = 0 Simplifying, 3i1 −i2 −2i3 = 1 −i1 + 6i2 −3i3 = 0 −2i1 −3i2 + 6i3 = 6 and solving, we obtain i1 = 3 A, i2 = 2 A, and i3 = 3 A. EXAMPLE 4.8 CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 96 PRACTICE ● 4.7 Determine i1 and i2 in the circuit of Fig 4.20. ■FIGURE 4.20 + – 10 V + – 3 V 5 7 4 1 9 i1 i2 10 The previous examples dealt with circuits powered exclusively by inde-pendent voltage sources. If a current source is included in the circuit, it may either simplify or complicate the analysis, as discussed in Sec. 4.4. As seen in our study of the nodal analysis technique, dependent sources generally require an additional equation besides the M mesh equations, unless the controlling variable is a mesh current (or sum of mesh currents). We explore this in the following example. EXAMPLE 4.9 Determine the current i1 in the circuit of Fig. 4.21a. The current i1 is actually a mesh current, so rather than redeﬁne it we label the rightmost mesh current i1 and deﬁne a clockwise mesh current i2 for the left mesh, as shown in Fig. 4.21b. For the left mesh, KVL yields −5 −4i1 + 4(i2 −i1) + 4i2 = 0 and for the right mesh we ﬁnd 4(i1 −i2) + 2i1 + 3 = 0 Grouping terms, these equations may be written more compactly as −8i1 + 8i2 = 5 and 6i1 −4i2 = –3 Solving, i2 = 375 mA, so i1 = −250 mA. ■FIGURE 4.21 (a) A two-mesh circuit containing a dependent source. (b) Circuit labeled for mesh analysis. 2 4 5 V 3 V + – + – + – i1 4 4i1 (a) 2 4 5 V 3 V + – + – + – 4 4i1 (b) i1 i2 Since the dependent source of Fig. 4.21 is controlled by a mesh current (i1), only two equations—Eqs. and —were required to analyze the two-mesh circuit. In the following example, we explore the situation that arises if the controlling variable is not a mesh current. Ans: 2.220 A, 470.0 mA. SECTION 4.3 MESH ANALYSIS 97 Determine the current i1 in the circuit of Fig. 4.22a. EXAMPLE 4.10 ■FIGURE 4.23 3 2 V 6 V + – – + 4 2 5 – + A vx + – i1 i2 ■FIGURE 4.22 (a) A circuit with a dependent source controlled by a voltage. (b) Circuit labeled for mesh analysis. 2 4 5 V 3 V + – + – + – i1 4 2vx 2 4 5 V 3 V + – + – + – 4 2vx (a) (b) vx + – vx + – i2 i1 In order to draw comparisons to Example 4.9 we use the same mesh current deﬁnitions, as shown in Fig. 4.22b. For the left mesh, KVL now yields −5 −2vx + 4(i2 −i1) + 4i2 = 0 and for the right mesh we ﬁnd the same as before, namely, 4(i1 −i2) + 2i1 + 3 = 0 Since the dependent source is controlled by the unknown voltage vx, we are faced with two equations in three unknowns. The way out of our dilemma is to construct an equation for vx in terms of mesh cur-rents, such as vx = 4(i2 −i1) We simplify our system of equations by substituting Eq. into Eq. , resulting in 4i1 = 5 Solving, we ﬁnd that i1 = 1.25 A. In this particular instance, Eq. is not needed unless a value for i2 is desired. PRACTICE ● 4.8 Determine i1 in the circuit of Fig. 4.23 if the controlling quantity A is equal to (a) 2i2; (b) 2vx. Ans: (a) 1.35 A; (b) 546 mA. The mesh analysis procedure can be summarized by the seven basic steps that follow. It will work on any planar circuit we ever encounter, al-though the presence of current sources will require extra care. Such situa-tions are discussed in Sec. 4.4. CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 98 4.4 • THE SUPERMESH How must we modify this straightforward procedure when a current source is present in the network? Taking our lead from nodal analysis, we should feel that there are two possible methods. First, we could assign an unknown voltage across the current source, apply KVL around each mesh as before, and then relate the source current to the assigned mesh currents. This is gen-erally the more difﬁcult approach. A better technique is one that is quite similar to the supernode approach in nodal analysis. There we formed a supernode, completely enclosing the voltage source inside the supernode and reducing the number of non-reference nodes by 1 for each voltage source. Now we create a kind of “supermesh” from two meshes that have a current source as a common element; the current source is in the interior of the supermesh. We thus reduce the number of meshes by 1 for each current source present. If the current source lies on the perimeter of the circuit, then the single mesh in which it is found is ignored. Kirchhoff’s voltage law is thus applied only to those meshes or supermeshes in the reinterpreted network. EXAMPLE 4.11 Determine the three mesh currents in Fig. 4.24a. We note that a 7 A independent current source is in the common bound-ary of two meshes, which leads us to create a supermesh whose interior Summary of Basic Mesh Analysis Procedure 1. Determine if the circuit is a planar circuit. If not, perform nodal analysis instead. 2. Count the number of meshes (M). Redraw the circuit if necessary. 3. Label each of the M mesh currents. Generally, deﬁning all mesh currents to ﬂow clockwise results in a simpler analysis. 4. Write a KVL equation around each mesh. Begin with a conve-nient node and proceed in the direction of the mesh current. Pay close attention to “−” signs. If a current source lies on the periph-ery of a mesh, no KVL equation is needed and the mesh current is determined by inspection. 5. Express any additional unknowns such as voltages or currents other than mesh currents in terms of appropriate mesh cur-rents. This situation can occur if current sources or dependent sources appear in our circuit. 6. Organize the equations. Group terms according to mesh currents. 7. Solve the system of equations for the mesh currents (there will be M of them). SECTION 4.4 THE SUPERMESH 99 ■FIGURE 4.24 (a) A three-mesh circuit with an independent current source. (b) A supermesh is deﬁned by the colored line. i2 i3 i1 + – 7 V 7 A 1 2 1 2 3 i2 i3 i1 + – 7 V 7 A 1 2 (b) (a) 1 2 3 ■FIGURE 4.25 + – 10 V 3 A 5 7 4 1 9 i1 10 The presence of one or more dependent sources merely requires each of these source quantities and the variable on which it depends to be expressed in terms of the assigned mesh currents. In Fig. 4.26, for example, we note that both a dependent and an independent current source are included in the network. Let’s see how their presence affects the analysis of the circuit and actually simpliﬁes it. Evaluate the three unknown currents in the circuit of Fig. 4.26. The current sources appear in meshes 1 and 3. Since the 15 A source is located on the perimeter of the circuit, we may eliminate mesh 1 from consideration—it is clear that i1 = 15 A. We ﬁnd that because we now know one of the two mesh currents relevant to the dependent current source, there is no need to write a supermesh equation about meshes 1 and 3. Instead, we simply relate i1 and i3 to the current from the dependent source using KCL: vx 9 = i3 −i1 = 3(i3 −i2) 9 EXAMPLE 4.12 (Continued on next page) is that of meshes 1 and 3 as shown in Fig. 4.24b. Applying KVL about this loop, −7 + 1(i1 −i2) + 3(i3 −i2) + 1i3 = 0 or i1 −4i2 + 4i3 = 7 and around mesh 2, 1(i2 −i1) + 2i2 + 3(i2 −i3) = 0 or −i1 + 6i2 −3i3 = 0 Finally, the independent source current is related to the mesh currents, i1 −i3 = 7 Solving Eqs. through , we ﬁnd i1 = 9 A, i2 = 2.5 A, and i3 = 2 A. PRACTICE ● 4.9 Determine the current i1 in the circuit of Fig. 4.25. Ans: −1.93 A. ■FIGURE 4.26 A three-mesh circuit with one dependent and one independent current source. i2 i3 vx vx + – i1 15 A 1 2 1 9 1 2 3 CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 100 ■FIGURE 4.27 i1 + – + – 80 V 30 10 20 40 30 V v3 – + 15i1 Ans: 104.2 V We can now summarize the general approach to writing mesh equations, whether or not dependent sources, voltage sources, and/or current sources are present, provided that the circuit can be drawn as a planar circuit: which can be written more compactly as −i1 + 1 3i2 + 2 3i3 = 0 or 1 3i2 + 2 3i3 = 15 With one equation in two unknowns, all that remains is to write a KVL equation about mesh 2: 1(i2 −i1) + 2i2 + 3(i2 −i3) = 0 or 6i2 −3i3 = 15 Solving Eqs. and , we ﬁnd that i2 = 11 A and i3 = 17 A; we already determined that i1 = 15 A by inspection. PRACTICE ● 4.10 Determine v3 in the circuit of Fig. 4.27. Summary of Supermesh Analysis Procedure 1. Determine if the circuit is a planar circuit. If not, perform nodal analysis instead. 2. Count the number of meshes (M). Redraw the circuit if necessary. 3. Label each of the M mesh currents. Generally, deﬁning all mesh currents to ﬂow clockwise results in a simpler analysis. 4. If the circuit contains current sources shared by two meshes, form a supermesh to enclose both meshes. A highlighted enclo-sure helps when writing KVL equations. 5. Write a KVL equation around each mesh/supermesh. Begin with a convenient node and proceed in the direction of the mesh current. Pay close attention to “−” signs. If a current source lies SECTION 4.5 NODAL VS. MESH ANALYSIS: A COMPARISON 101 4.5 • NODAL VS. MESH ANALYSIS: A COMPARISON Now that we have examined two distinctly different approaches to circuit analysis, it seems logical to ask if there is ever any advantage to using one over the other. If the circuit is nonplanar, then there is no choice: only nodal analysis may be applied. Provided that we are indeed considering the analysis of a planar circuit, however, there are situations where one technique has a small advantage over the other. If we plan to use nodal analysis, then a circuit with N nodes will lead to at most (N −1) KCL equations. Each supernode deﬁned will further reduce this number by 1. If the same circuit has M distinct meshes, then we will obtain at most M KVL equations; each supermesh will reduce this number by 1. Based on these facts, we should select the approach that will result in the smaller number of simultaneous equations. If one or more dependent sources are included in the circuit, then each controlling quantity may inﬂuence our choice of nodal or mesh analysis. For example, a dependent voltage source controlled by a nodal voltage does not require an additional equation when we perform nodal analysis. Like-wise, a dependent current source controlled by a mesh current does not re-quire an additional equation when we perform mesh analysis. What about the situation where a dependent voltage source is controlled by a current? Or the converse, where a dependent current source is controlled by a volt-age? Provided that the controlling quantity can be easily related to mesh currents, we might expect mesh analysis to be the more straightforward option. Likewise, if the controlling quantity can be easily related to nodal voltages, nodal analysis may be preferable. One ﬁnal point in this regard is to keep in mind the location of the source; current sources which lie on the periphery of a mesh, whether dependent or independent, are easily treated in mesh analysis; voltage sources connected to the reference terminal are easily treated in nodal analysis. When either method results in essentially the same number of equations, it may be worthwhile to also consider what quantities are being sought. Nodal analysis results in direct calculation of nodal voltages, whereas mesh analysis provides currents. If we are asked to ﬁnd currents through a set of resistors, for example, after performing nodal analysis, we must still invoke Ohm’s law at each resistor to determine the current. on the periphery of a mesh, no KVL equation is needed and the mesh current is determined by inspection. 6. Relate the current ﬂowing from each current source to mesh currents. This is accomplished by simple application of KCL; one such equation is needed for each supermesh deﬁned. 7. Express any additional unknowns such as voltages or currents other than mesh currents in terms of appropriate mesh cur-rents. This situation can occur if dependent sources appear in our circuit. 8. Organize the equations. Group terms according to nodal voltages. 9. Solve the system of equations for the mesh currents (there will be M of them). As an example, consider the circuit in Fig. 4.28. We wish to determine the current ix. We choose the bottom node as the reference node, and note that there are four nonreference nodes. Although this means that we can write four dis-tinct equations, there is no need to label the node between the 100 V source and the 8 resistor, since that node voltage is clearly 100 V. Thus, we label the remaining node voltages v1, v2, and v3 as in Fig. 4.29. CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 102 ■FIGURE 4.28 A planar circuit with ﬁve nodes and four meshes. + – 100 V 8 A 4 3 5 8 2 10 ix ■FIGURE 4.29 The circuit of Fig. 4.28 with node voltages labeled. Note that an earth ground symbol was chosen to designate the reference terminal. + – 100 V 8 A 4 3 5 8 2 10 v2 v1 v3 ix We write the following three equations: v1 −100 8 + v1 4 + v1 −v2 2 = 0 or 0.875v1 −0.5v2 = 12.5 v2 −v1 2 + v2 3 + v2 −v3 10 −8 = 0 or −0.5v1 −0.9333v2 −0.1v3 = 8 v3 −v2 10 + v3 5 + 8 = 0 or −0.1v2 + 0.3v3 = −8 Solving, we ﬁnd that v1 = 25.89 V and v2 = 20.31 V. We determine the current ix by application of Ohm’s law: ix = v1 −v2 2 = 2.79 A SECTION 4.6 COMPUTER-AIDED CIRCUIT ANALYSIS 103 Next, we consider the same circuit using mesh analysis. We see in Fig. 4.30 that we have four distinct meshes, although it is obvious that i4 = −8 A; we therefore need to write three distinct equations. Writing a KVL equation for meshes 1, 2, and 3: −100 + 8i1 + 4(i1 −i2) = 0 or 12i1 −4i2 = 100 4(i2 −i1) + 2i2 + 3(i2 −i3) = 0 or −4i1 + 9i2 −3i3 0 3(i3 −i2) + 10(i3 + 8) + 5i3 = 0 or −3i2 + 18i3 −80 Solving, we ﬁnd that i2 (= ix) = 2.79 A. For this particular problem, mesh analysis proved to be simpler. Since either method is valid, however, working the same problem both ways can also serve as a means to check our answers. 4.6 • COMPUTER-AIDED CIRCUIT ANALYSIS We have seen that it does not take many components at all to create a cir-cuit of respectable complexity. As we continue to examine even more complex circuits, it will become obvious rather quickly that it is easy to make errors during the analysis, and verifying solutions by hand can be time-consuming. A powerful computer software package known as PSpice is commonly employed for rapid analysis of circuits, and the schematic capture tools are typically integrated with either a printed circuit board or integrated circuit layout tool. Originally developed in the early 1970s at the University of California at Berkeley, SPICE (Simulation Program with Integrated Circuit Emphasis) is now an industry standard. MicroSim Cor-poration introduced PSpice in 1984, which built intuitive graphical inter-faces around the core SPICE program. Depending on the type of circuit application being considered, there are now several companies offering variations of the basic SPICE package. Although computer-aided analysis is a relatively quick means of deter-mining voltages and currents in a circuit, we should be careful not to allow simulation packages to completely replace traditional “paper and pencil” analysis. There are several reasons for this. First, in order to design we must be able to analyze. Overreliance on software tools can inhibit the develop-ment of necessary analytical skills, similar to introducing calculators too early in grade school. Second, it is virtually impossible to use a complicated software package over a long period of time without making some type of data-entry error. If we have no basic intuition as to what type of answer to ■FIGURE 4.30 The circuit of Fig. 4.28 with mesh currents labeled. + – 100 V 8 A 4 3 5 8 2 10 ix i3 i4 i2 i1 CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 104 ■FIGURE 4.31 (a) Circuit of Fig. 4.15a drawn using Orcad schematic capture software. (b) Current, voltage, and power display buttons. (c) Circuit after simulation run, with current display enabled. (a) (b) (c) (1) Refer to Appendix 4 for a brief tutorial on PSpice and schematic capture. expect from a simulation, then there is no way to determine whether or not it is valid. Thus, the generic name really is a fairly accurate description: computer-aided analysis. Human brains are not obsolete. Not yet, anyway. As an example, consider the circuit of Fig. 4.15b, which includes two dc voltage sources and three resistors. We wish to simulate this circuit using PSpice so that we may determine the currents i1 and i2. Figure 4.31a shows the circuit as drawn using a schematic capture program.1 SECTION 4.6 COMPUTER-AIDED CIRCUIT ANALYSIS 105 In order to determine the mesh currents, we need only run a bias point sim-ulation. Under PSpice, select New Simulation Proﬁle, type in a name (such as Example), and click on Create. Under the Analysis type: pull-down menu, select Bias Point, then click on OK. Returning to the original schematic win-dow, under PSpice select Run (or use either of the two shortcuts: pressing the F11 key or clicking on the blue “Play” symbol).To see the currents calculated by PSpice, make sure the current button is selected (Fig. 4.31b). The results of our simulation are shown in Fig. 4.31c. We see that the two currents i1 and i2 are 6 A and 4 A, respectively, as we found previously. As a further example, consider the circuit shown in Fig. 4.32a. It contains a dc voltage source, a dc current source, and a voltage-controlled current source. We are interested in the three nodal voltages, which from either nodal or mesh analysis are found to be 82.91 V, 69.9 V, and 59.9 V, respectively, as we move from left to right across the top of the circuit. Figure 4.32b shows this circuit after the simulation was performed. The three nodal voltages are indicated directly on the schematic. Note that in drawing a dependent source using the schematic capture tool, we must explicitly link two terminals of the source to the controlling voltage or current. ■FIGURE 4.32 (a) Circuit with dependent current source. (b) Circuit drawn using a schematic capture tool, with simulation results presented directly on the schematic. 18 33 20 5 A 0.2 V 2 10 V (b) (a) + − V 2 + − PRACTICAL APPLICATION Node-Based PSpice Schematic Creation The most common method of describing a circuit in con-junction with computer-aided circuit analysis is with some type of graphical schematic drawing package, an example output of which was shown in Fig. 4.32. SPICE, however, was written before the advent of such software, and as such requires circuits to be described in a speciﬁc text-based format. The format has its roots in the syntax used for punch cards, which gives it a some-what distinct appearance. The basis for circuit de-scription is the deﬁnition of elements, each terminal of which is assigned a node number. So, although we have just studied two different generalized circuit analysis methods—the nodal and mesh techniques—it is interest-ing that SPICE and PSpice were written using a clearly deﬁned nodal analysis approach. Even though modern circuit analysis is largely done using graphics-oriented interactive software, when errors are generated (usually due to a mistake in drawing the schematic or in selecting a combination of analysis op-tions), the ability to read the text-based “input deck” generated by the schematic capture tool can be invalu-able in tracking down the speciﬁc problem. The easiest way to develop such an ability is to learn how to run PSpice directly from a user-written input deck. Consider, for example, the sample input deck below (lines beginning with an asterisk are comments, and are skipped by SPICE). Example SPICE input deck for simple voltage divider circuit. .OP (Requests dc operating point) R1 1 2 1k (Locates R1 between nodes 1 and 2; value is 1 k) R2 2 0 1k (Locates R2 between nodes 2 and 0; also 1 k) V1 1 0 DC 5 (Locates 5 V source between nodes 1 and 0) End of input deck. We can create the input deck by using the Notepad pro-gram from Windows or our favorite text editor. Saving the ﬁle under the name example.cir, we next invoke PSpice A/D (see Appendix. 4). Under File, we choose Open, lo-cate the directory in which we saved our ﬁle example.cir, and for Files of type: select Circuit Files (.cir). After se-lecting our ﬁle and clicking Open, we see the PSpice A/D window with our circuit ﬁle loaded (Fig. 4.33a). A netlist such as this, containing instructions for the simulation to be performed, can be created by schematic capture software or created manually as in this example. We run the simulation by either clicking the green “play” symbol at the top right, or selecting Run under Simulation. To view the results, we select Output File from un-der the View menu, which provides the window shown in Fig. 4.33b. Here it is worth noting that the output pro-vides the expected nodal voltages (5 V at node 1, 2.5 V across resistor R2), but the current is quoted using the passive sign convention (i.e., 2.5 mA). Text-based schematic entry is reasonably straightfor-ward, but for complex (large number of elements) cir-cuits, it can quickly become cumbersome. It is also easy to misnumber nodes, an error that can be difﬁcult to iso-late. However, reading the input and output ﬁles is often helpful when running simulations, so some experience with this format is useful. At this point, the real power of computer-aided analysis begins to be apparent: Once you have the circuit drawn in the schematic capture program, it is easy to experiment by simply changing component values and observing the effect on currents and voltages.To gain a little experience at this point, try sim-ulating any of the circuits shown in previous examples and practice problems. (a) (b) ■FIGURE 4.33 (a) PSpice A/D window after the input deck describing our voltage divider is loaded. (b) Output window, showing nodal voltages and current from the source (but quoted using the passive sign convention). Note that the voltage across R1 requires post-simulation subtraction. SUMMARY AND REVIEW Although Chap. 3 introduced KCL and KVL, both of which are sufﬁcient to enable us to analyze any circuit, a more methodical approach proves help-ful in everyday situations. Thus, in this chapter we developed the nodal analysis technique based on KCL, which results in a voltage at each node (with respect to some designated “reference” node). We generally need to solve a system of simultaneous equations, unless voltage sources are con-nected so that they automatically provide nodal voltages. The controlling quantity of a dependent source is written down just as we would write down the numerical value of an “independent” source. Typically an additional equation is then required, unless the dependent source is controlled by a nodal voltage. When a voltage source bridges two nodes, the basic tech-nique can be extended by creating a supernode; KCL dictates that the sum of the currents ﬂowing into a group of connections so deﬁned is equal to the sum of the currents ﬂowing out. As an alternative to nodal analysis, the mesh analysis technique was de-veloped through application of KVL; it yields the complete set of mesh cur-rents, which do not always represent the net current ﬂowing through any particular element (for example, if an element is shared by two meshes). The presence of a current source will simplify the analysis if it lies on the periphery of a mesh; if the source is shared, then the supermesh technique is best. In that case, we write a KVL equation around a path that avoids the shared current source, then algebraically link the two corresponding mesh currents using the source. A common question is: “Which analysis technique should I use?” We dis-cussed some of the issues that might go into choosing a technique for a given circuit. These included whether or not the circuit is planar, what types of sources are present and how they are connected, and also what speciﬁc infor-mation is required (i.e., a voltage, current, or power). For complex circuits, it may take a greater effort than it is worth to determine the “optimum” approach, in which case most people will opt for the method with which they feel most comfortable. We concluded the chapter by introducing PSpice, a common cir-cuit simulation tool, which is very useful for checking our results. At this point we wrap up by identifying key points of this chapter to re-view, along with relevant example(s). ❑Start each analysis with a neat, simple circuit diagram. Indicate all element and source values. (Example 4.1) ❑For nodal analysis, ❑Choose one node as the reference node. Then label the node voltages v1, v2, . . . , vN−1. Each is understood to be measured with respect to the reference node. (Examples 4.1, 4.2) ❑If the circuit contains only current sources, apply KCL at each nonreference node. (Examples 4.1, 4.2) ❑If the circuit contains voltage sources, form a supernode about each one, and then apply KCL at all nonreference nodes and supernodes. (Examples 4.5, 4.6) ❑For mesh analysis, ﬁrst make certain that the network is a planar network. ❑Assign a clockwise mesh current in each mesh: i1, i2, . . . , iM. (Example 4.7) ❑If the circuit contains only voltage sources, apply KVL around each mesh. (Examples 4.7, 4.8, 4.9) ❑If the circuit contains current sources, create a supermesh for each one that is common to two meshes, and then apply KVL around each mesh and supermesh. (Examples 4.11, 4.12) CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 108 EXERCISES 109 ❑Dependent sources will add an additional equation to nodal analysis if the controlling variable is a current, but not if the controlling variable is a nodal voltage. (Conversely, a dependent source will add an additional equation to mesh analysis if the controlling variable is a voltage, but not if the controlling variable is a mesh current). (Examples 4.3, 4.4, 4.6, 4.9, 4.10, 4.12) ❑In deciding whether to use nodal or mesh analysis for a planar circuit, a circuit with fewer nodes/supernodes than meshes/supermeshes will result in fewer equations using nodal analysis. ❑Computer-aided analysis is useful for checking results and analyzing circuits with large numbers of elements. However, common sense must be used to check simulation results. READING FURTHER A detailed treatment of nodal and mesh analysis can be found in: R. A. DeCarlo and P. M. Lin, Linear Circuit Analysis, 2nd ed. New York: Oxford University Press, 2001. A solid guide to SPICE is P. Tuinenga, SPICE: A Guide to Circuit Simulation and Analysis Using PSPICE, 3rd ed. Upper Saddle River, N.J.: Prentice-Hall, 1995. EXERCISES 4.1 Nodal Analysis 1. Solve the following systems of equations: (a) 2v2 – 4v1 = 9 and v1 – 5v2 = –4; (b) –v1 + 2v3 = 8; 2v1 + v2 – 5v3 = –7; 4v1 + 5v2 + 8v3 = 6. 2. Evaluate the following determinants: (a) 2 1 −4 3 (b) 0 2 11 6 4 1 3 −1 5 . 3. Employ Cramer’s rule to solve for v2 in each part of Exercise 1. 4. (a) Solve the following system of equations: 3 = v1 5 −v2 −v1 22 + v1 −v3 3 2 −1 = v2 −v1 22 + v2 −v3 14 0 = v3 10 + v3 −v1 3 + v3 −v2 14 (b) Verify your solution using MATLAB. 5. (a) Solve the following system of equations: 7 = v1 2 −v2 −v1 12 + v1 −v3 19 15 = v2 −v1 12 + v2 −v3 2 4 = v3 7 + v3 −v1 19 + v3 −v2 2 (b) Verify your solution using MATLAB. 6. Correct (and verify by running) the following MATLAB code: >> e1 = ‘3 = v/7 - (v2 - v1)/2 + (v1 - v3)/3; >> e2 = ‘2 = (v2 - v1)/2 + (v2 - v3)/14’; >> e ‘0 = v3/10 + (v3 - v1)/3 + (v3 - v2)/14’; >> >> a = sove(e e2 e3, ‘v1’, v2, ‘v3’) 7. Identify the obvious errors in the following complete set of nodal equations if the last equation is known to be correct: 7 = v1 4 −v2 −v 1 + v1 −v3 9 0 = v2 −v1 2 + v2 −v3 2 4 = v3 7 + v3 −v1 19 + v3 −v2 2 8. In the circuit of Fig. 4.34, determine the current labeled i with the assistance of nodal analysis techniques. CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 110 10. With the assistance of nodal analysis, determine v1 −v2 in the circuit shown in Fig. 4.36. ■FIGURE 4.34 5 A 4 A 1 5 2 v1 i v2 ■FIGURE 4.35 2 3 2 A 3 A 1 ■FIGURE 4.36 4 2 1 5 2 A 15 A v1 v2 9. Calculate the power dissipated in the 1 resistor of Fig. 4.35. EXERCISES 111 ■FIGURE 4.39 1 7 5 3 3 5 A 8 A 4 A ■FIGURE 4.40 6 7 2 A 3 A 5 2 1 4 3 i5 ■FIGURE 4.37 3 1 2 A 6 2 4 A 6 i1 v1 + – ■FIGURE 4.38 vP + – 50 10 40 20 100 200 5 A 10 A 2.5 A 2 A 11. For the circuit of Fig. 4.37, determine the value of the voltage labeled v1 and the current labeled i1. 12. Use nodal analysis to ﬁnd vP in the circuit shown in Fig. 4.38. 13. Using the bottom node as reference, determine the voltage across the 5 resistor in the circuit of Fig. 4.39, and calculate the power dissipated by the 7 resistor. 14. For the circuit of Fig. 4.40, use nodal analysis to determine the current i5. 16. Determine the current i2 as labeled in the circuit of Fig. 4.42, with the assistance of nodal analysis. CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 112 ■FIGURE 4.42 3 10 V 2 5 v1 + – v3 – + 0.2v3 0.02v1 i2 ■FIGURE 4.43 vx – + vx 1 A 5 2 3 i1 ■FIGURE 4.44 5 4 V 1 5 A 3 3 A 8 A 2 v1 v2 v3 – + Ref. ■FIGURE 4.46 + – 5 V + – 6 V 2 A 1 2 4 10 ■FIGURE 4.41 5 10 1 A 2 A 5 2 6 5 2 1 4 6 A 2 A 4 1 4 10 2 v1 v3 v7 v2 v6 v4 v5 v8 15. Determine a numerical value for each nodal voltage in the circuit of Fig. 4.41. ■FIGURE 4.45 3 A 9 V 5 A + – 1 9 5 v1 17. Using nodal analysis as appropriate, determine the current labeled i1 in the circuit of Fig. 4.43. 4.2 The Supernode 18. Determine the nodal voltages as labeled in Fig. 4.44, making use of the supernode technique as appropriate. 19. For the circuit shown in Fig. 4.45, determine a numerical value for the voltage labeled v1. 20. For the circuit of Fig. 4.46, determine all four nodal voltages. EXERCISES 113 21. Employing supernode/nodal analysis techniques as appropriate, determine the power dissipated by the 1 resistor in the circuit of Fig. 4.47. ■FIGURE 4.47 + – + – 7 V 3 A 2 A 4 V 1 – + 4 V 3 2 ■FIGURE 4.48 + – – + 1 V 4 V 4 A 6 A 14 + – 3 V 7 7 2 3 2 23. Determine the voltage labeled v in the circuit of Fig. 4.49. 24. Determine the voltage vx in the circuit of Fig. 4.50, and the power supplied by the 1 A source. 22. Referring to the circuit of Fig. 4.48, obtain a numerical value for the power supplied by the 1 V source. ■FIGURE 4.49 5 A 10 V 10 20 12 2 – + + – 1 A 5 V v + – ■FIGURE 4.50 2vx 8 2 5 1 A 8 A vx + – – + ■FIGURE 4.51 2 4 3 V 4 V + – + – + – i1 2 A 0.5i1 25. Consider the circuit of Fig. 4.51. Determine the current labeled i1. 26. Determine the value of k that will result in vx being equal to zero in the circuit of Fig. 4.52. CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 114 ■FIGURE 4.53 2 2 A 5 3 v1 + – v1 4v1 + – ■FIGURE 4.54 2vx 1 2 3 Ref. 3 A + – 1 V v2 v4 v3 v1 4 1 vx + – ■FIGURE 4.55 + – + – 1 V 2 V 1 4 5 28. For the circuit of Fig. 4.54, determine all four nodal voltages. 4.3 Mesh Analysis 29. Determine the currents ﬂowing out of the positive terminal of each voltage source in the circuit of Fig. 4.55. 27. For the circuit depicted in Fig. 4.53, determine the voltage labeled v1 across the 3 resistor. ■FIGURE 4.52 – + + – 1 4 1 1 A 2 V Ref. 3 kvy vx vy EXERCISES 115 ■FIGURE 4.56 i2 i1 – + + – 5 V 12 V 14 7 3 ■FIGURE 4.57 + – + – – + 15 V 21 V 9 9 11 V 1 i1 i2 ■FIGURE 4.58 i2 i3 i1 + – – + 2 V 3 V 1 5 7 6 9 ■FIGURE 4.59 + – 4.7 k 220 5.7 k 4.7 k 1 k 1 k 2.2 k 5 V iy ■FIGURE 4.60 3 2 7 5 + – + – + – 30. Obtain numerical values for the two mesh currents i1 and i2 in the circuit shown in Fig. 4.56. 31. Use mesh analysis as appropriate to determine the two mesh currents labeled in Fig. 4.57. 32. Determine numerical values for each of the three mesh currents as labeled in the circuit diagram of Fig. 4.58. 33. Calculate the power dissipated by each resistor in the circuit of Fig. 4.58. 34. Employing mesh analysis as appropriate, obtain (a) a value for the current iy and (b) the power dissipated by the 220 resistor in the circuit of Fig. 4.59. 35. Choose nonzero values for the three voltage sources of Fig. 4.60 so that no current ﬂows through any resistor in the circuit. CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 116 36. Calculate the current ix in the circuit of Fig. 4.61. 37. Employing mesh analysis procedures, obtain a value for the current labeled i in the circuit represented by Fig. 4.62. 38. Determine the power dissipated in the 4 resistor of the circuit shown in Fig. 4.63. 39. (a) Employ mesh analysis to determine the power dissipated by the 1 resistor in the circuit represented schematically by Fig. 4.64. (b) Check your answer using nodal analysis. 40. Deﬁne three clockwise mesh currents for the circuit of Fig. 4.65, and employ mesh analysis to obtain a value for each. ■FIGURE 4.61 + – 3 V 10 A 4 8 5 8 12 20 ix ■FIGURE 4.62 2 V 1 4 3 4 1 i + – ■FIGURE 4.64 4 A 5ix 1 A 2 1 5 2 ix ■FIGURE 4.65 + – 2 V 1 V 2 9 10 3 10 – + 5 V + – vx + – 0.5vx ■FIGURE 4.63 5 4 4 V 1 V – + + – + – i1 3 2i1 EXERCISES 117 41. Employ mesh analysis to obtain values for ix and va in the circuit of Fig. 4.66. 4.4 The Supermesh 42. Determine values for the three mesh currents of Fig. 4.67. 43. Through appropriate application of the supermesh technique, obtain a numerical value for the mesh current i3 in the circuit of Fig. 4.68, and calculate the power dissipated by the 1 resistor. 44. For the circuit of Fig. 4.69, determine the mesh current i1 and the power dissipated by the 1 resistor. 45. Calculate the three mesh currents labeled in the circuit diagram of Fig. 4.70. ■FIGURE 4.66 4 4 0.2ix 9 V 1 7 7 + – va + – 0.1va + – + – ix ■FIGURE 4.67 i2 i3 i1 1 V 2 A 7 3 2 1 3 + – ■FIGURE 4.68 i3 i1 + – 3 V 10 5 A 4 5 1 17 ■FIGURE 4.69 i1 – + 7 V 5 9 A 1 3 A 10 11 3 5 ■FIGURE 4.70 + – i3 i2 i1 4.7 k 3.5 k 2.2 k 1.7 k 6.2 k 3 A 7 V 8.1 k 3.1 k 1 A 2 A 5.7 k CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 118 ■FIGURE 4.72 i2 i1 i3 5 A 11 12 12 13 13 vx + – vx 1 – 3 ■FIGURE 4.74 + – + – 3 V 2 1 4 1 5 V v3 – + 1.8v3 ■FIGURE 4.75 5 2ia 3 5 A 4 10 6 A + – 4 V + – ia ■FIGURE 4.73 i1 + – + – 1 V 1 4 3 2 8 V 5i1 47. Through careful application of the supermesh technique, obtain values for all three mesh currents as labeled in Fig. 4.72. 48. Determine the power supplied by the 1 V source in Fig. 4.73. 49. Deﬁne three clockwise mesh currents for the circuit of Fig. 4.74, and employ the supermesh technique to obtain a numerical value for each. 50. Determine the power absorbed by the 10 resistor in Fig. 4.75. 46. Employing the supermesh technique to best advantage, obtain numerical val-ues for each of the mesh currents identiﬁed in the circuit depicted in Fig. 4.71. ■FIGURE 4.71 3 A 8 V 3 6 3 V 1 A –2 A 3 2 V 5 1 4 2 i2 i3 i1 + – + – + – EXERCISES 119 ■FIGURE 4.76 6 7 2 A 3 A 5 2 1 4 3 i5 ■FIGURE 4.77 + – + – 30 6 3 240 V 60 V 10 A 12 v1 + – v2 + – ■FIGURE 4.78 11 A 22 V + – 2 9 vx + – 52. The circuit of Fig. 4.76 is modiﬁed such that the 3 A source is replaced by a 3 V source whose positive reference terminal is connected to the 7 resistor. (a) Determine the number of nodal equations required to determine i5. (b) Al-ternatively, how many mesh equations would be required? (c) Would your pre-ferred analysis method change if only the voltage across the 7 resistor were needed? Explain. 53. The circuit of Fig. 4.77 contains three sources. (a) As presently drawn, would nodal or mesh analysis result in fewer equations to determine the voltages v1 and v2? Explain. (b) If the voltage source were replaced with current sources, and the current source replaced with a voltage source, would your answer to part (a) change? Explain? 54. Solve for the voltage vx as labeled in the circuit of Fig. 4.78 using (a) mesh analysis. (b) Repeat using nodal analysis. (c) Which approach was easier, and why? 4.5 Nodal vs. Mesh Analysis: A Comparison 51. For the circuit represented schematically in Fig. 4.76: (a) How many nodal equations would be required to determine i5? (b) Alternatively, how many mesh equations would be required? (c) Would your preferred analysis method change if only the voltage across the 7 resistor were needed? Explain. CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 120 55. Consider the ﬁve-source circuit of Fig. 4.79. Determine the total number of simultaneous equations that must be solved in order to determine v1 using (a) nodal analysis; (b) mesh analysis. (c) Which method is preferred, and does it depend on which side of the 40 resistor is chosen as the reference node? Explain your answer. 58. From the perspective of determining voltages and currents associated with all components, (a) design a ﬁve-node, four-mesh circuit that is analyzed more easily using nodal techniques. (b) Modify your circuit by replacing only one component such that it is now more easily analyzed using mesh techniques. 4.6 Computer-Aided Circuit Analysis 59. Employ PSpice (or similar CAD tool) to verify the solution of Exercise 8. Submit a printout of a properly labeled schematic with the answer highlighted, along with your hand calculations. 60. Employ PSpice (or similar CAD tool) to verify the solution of Exercise 10. Submit a printout of a properly labeled schematic with the two nodal voltages highlighted, along with your hand calculations solving for the same quantities. 61. Employ PSpice (or similar CAD tool) to verify the voltage across the 5 resistor in the circuit of Exercise 13. Submit a printout of a properly labeled schematic with the answer highlighted, along with your hand calculations. 56. Replace the dependent voltage source in the circuit of Fig. 4.79 with a depen-dent current source oriented such that the arrow points upward. The controlling expression 0.1 v1 remains unchanged. The value V2 is zero. (a) Determine the total number of simultaneous equations required to obtain the power dissipated by the 40 resistor if nodal analysis is employed. (b) Is mesh analysis pre-ferred instead? Explain. 57. After studying the circuit of Fig. 4.80, determine the total number of simulta-neous equations that must be solved to determine voltages v1 and v3 using (a) nodal analysis; (b) mesh analysis. ■FIGURE 4.79 + – v1 + – 10 40 20 96 V 4 A 6 A 0.1v1 V2 ■FIGURE 4.80 30 45 100 V 20 50 v1 + – v3 – + 0.2v3 0.02v1 5i2 + – + – + – i2 62. Verify numerical values for each nodal voltage in Exercise 15 by employing PSpice or a similar CAD tool. Submit a printout of an appropriately labeled schematic with the nodal voltages highlighted, along with your hand calculations. 63. Verify the numerical values for i1 and vx as indicated in the circuit accompany-ing Exercise 17, using PSpice or a similar CAD tool. Submit a printout of a properly labeled schematic with the answers highlighted, along with hand calculations. 64. (a) Generate an input deck for SPICE to determine the voltage v9 as labeled in Fig. 4.81. Submit a printout of the output ﬁle with the solution highlighted. (b) Verify your answer by hand. ■FIGURE 4.81 10 5 4 6 9 40 V 7 + – 11 8 v9 + – 2 3 ■FIGURE 4.82 2 i1 A B 2 C D F E Chapter-Integrating Exercises 65. (a) Design a circuit employing only 9 V batteries and standard 5% tolerance value resistors that provide voltages of 1.5 V, 4.5 V, and 5 V and at least one mesh current of 1 mA. (b) Verify your design using PSpice or similar CAD tool. 66. A decorative string of multicolored outdoor lights is installed on a home in a quiet residential area. After plugging the 12 V ac adapter into the electrical socket, the homeowner immediately notes that two bulbs are burned out. (a) Are the individual lights connected in series or parallel? Explain. (b) Simulate the string by writing a SPICE input deck, assuming 44 lights, 12 V dc power supply, 24 AWG soft solid copper wire, and individual bulbs rated at 10 mW each. Submit a printout of the output ﬁle, with the power supplied by the 12 V supply highlighted. (c) Verify your simulation with hand calculations. 67. Consider the circuit depicted in Fig. 4.82. Employ either nodal or mesh analy-sis as a design tool to obtain a value of 200 mA for i1, if elements A, B, C, D, E, and F must be either current or voltage sources with nonzero values. 68. (a) Under what circumstances does the presence of an independent voltage source greatly simplify nodal analysis? Explain. (b) Under what circumstances does the presence of an independent current source signiﬁcantly simplify mesh analysis? Explain. (c) On which fundamental physical principle do we base nodal analysis? (d) On which fundamental physical principle do we base mesh analysis? EXERCISES 121 69. Referring to Fig. 4.83, (a) determine whether nodal or mesh analysis is more appropriate in determining i2 if element A is replaced with a short circuit, then carry out the analysis. (b) Verify your answer with an appropriate PSpice simu-lation. Submit a properly labeled schematic along with the answer highlighted. ■FIGURE 4.83 i1 i2 A + – + – 80 V 30 10 20 40 30 V v3 – + 70. The element marked A in the circuit of Fig. 4.83 is replaced by a 2.5 V inde-pendent voltage source with the positive reference terminal connected to the common node of the 20 and 30 resistors. (a) Determine whether mesh or nodal analysis is more straightforward for determining the voltage marked v3. (b) Verify your answer using PSpice. (c) Would your conclusion for part (a) change if the current i2 were required as well? Explain. CHAPTER 4 BASIC NODAL AND MESH ANALYSIS 122 INTRODUCTION The techniques of nodal and mesh analysis described in Chap. 4 are reliable and extremely powerful methods. However, both require that we develop a complete set of equations to describe a particular circuit as a general rule, even if only one current, voltage, or power quantity is of interest. In this chapter, we investigate several different techniques for isolating speciﬁc parts of a circuit in order to simplify the analysis. After examining each of these techniques, we focus on how one might go about selecting one method over another. 5.1 • LINEARITY AND SUPERPOSITION All of the circuits which we plan to analyze can be classiﬁed as lin-ear circuits, so this is a good time to be more speciﬁc in deﬁning exactly what we mean by that. Having done this, we can then con-sider the most important consequence of linearity, the principle of superposition. This principle is very basic and will appear repeat-edly in our study of linear circuit analysis. As a matter of fact, the nonapplicability of superposition to nonlinear circuits is the very reason they are so difﬁcult to analyze! The principle of superposition states that the response (a desired current or voltage) in a linear circuit having more than one indepen-dent source can be obtained by adding the responses caused by the separate independent sources acting alone. Linear Elements and Linear Circuits We deﬁne a linear element as a passive element that has a linear voltage-currentrelationship.Bya“linearvoltage-current relationship’’ KEY CONCEPTS Superposition: Determining the Individual Contributions of Different Sources to Any Current or Voltage Source Transformation as a Means of Simplifying Circuits Thévenin’s Theorem Norton’s Theorem Thévenin and Norton Equivalent Networks Maximum Power Transfer ↔Y Transformations for Resistive Networks Selecting a Particular Combination of Analysis Techniques Performing dc Sweep Simulations Using PSpice Handy Circuit Analysis Techniques C H A P T E R 5 123 CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 124 we simply mean that multiplication of the current through the element by a constant K results in the multiplication of the voltage across the element by the same constant K.At this time, only one passive element has been deﬁned (the resistor), and its voltage-current relationship v(t) = Ri(t) is clearly linear. As a matter of fact, if v(t) is plotted as a function of i(t), the result is a straight line. We deﬁne a linear dependent source as a dependent current or voltage source whose output current or voltage is proportional only to the ﬁrst power of a speciﬁed current or voltage variable in the circuit (or to the sum of such quantities). We now deﬁne a linear circuit as a circuit composed entirely of inde-pendent sources, linear dependent sources, and linear elements. From this deﬁnition, it is possible to show1 that “the response is proportional to the source,’’ or that multiplication of all independent source voltages and cur-rents by a constant K increases all the current and voltage responses by the same factor K (including the dependent source voltage or current outputs). The Superposition Principle The most important consequence of linearity is superposition. Let us explore the superposition principle by considering ﬁrst the circuit of Fig. 5.1, which contains two independent sources, the current generators that force the currents ia and ib into the circuit. Sources are often called forc-ing functions for this reason, and the nodal voltages that they produce can be termed response functions, or simply responses. Both the forcing functions and the responses may be functions of time. The two nodal equations for this circuit are 0.7v1 −0.2v2 = ia −0.2v1 + 1.2v2 = ib Now let us perform experiment x. We change the two forcing functions to iax and ibx; the two unknown voltages will now be different, so we will call them v1x and v2x. Thus, 0.7v1x −0.2v2x = iax −0.2v1x + 1.2v2x = ibx We next perform experiment y by changing the source currents to iay and iby and measure the responses v1y and v2y: 0.7v1y −0.2v2y = iay −0.2v1y + 1.2v2y = iby ia v1 v2 ib 2 5 Ref. 1 ■FIGURE 5.1 A circuit with two independent current sources. (1) The proof involves ﬁrst showing that the use of nodal analysis on the linear circuit can produce only linear equations of the form a1v1 + a2v2 + · · · + aN vN = b where the ai are constants (combinations of resistance or conductance values, constants appearing in dependent source expressions, 0, or ±1), the vi are the unknown node voltages (responses), and b is an independent source value or a sum of independent source values. Given a set of such equations, if we multiply all the b’s by K, then it is evident that the solution of this new set of equations will be the node voltages Kv1, Kv2, . . . , KvN . The dependent voltage source vs 0.6i1 14v2 is linear, but vs 0.6i1 2 and vs 0.6i1v2 are not. SECTION 5.1 LINEARITY AND SUPERPOSITION 125 These three sets of equations describe the same circuit with three differ-ent sets of source currents. Let us add or “superpose’’ the last two sets of equations. Adding Eqs. and , (0.7v1x + 0.7v1y) −(0.2v2x + 0.2v2y) = iax + iay 0.7v1 − 0.2v2 ia and adding Eqs. and , −(0.2v1x + 0.2v1y) + (1.2v2x + 1.2v2y) = ibx + iby −0.2v1 + 1.2v2 ib where Eq. has been written immediately below Eq. and Eq. below Eq. for easy comparison. The linearity of all these equations allows us to compare Eq. with Eq. and Eq. with Eq. and draw an interesting conclusion. If we select iax and iay such that their sum is ia and select ibx and iby such that their sum is ib, then the desired responses v1 and v2 may be found by adding v1x to v1y and v2x to v2y, respectively. In other words, we can perform experi-ment x and note the responses, perform experiment y and note the responses, and ﬁnally add the two sets of responses. This leads to the fun-damental concept involved in the superposition principle: to look at each independent source (and the response it generates) one at a time with the other independent sources “turned off’’ or “zeroed out.’’ If we reduce a voltage source to zero volts, we have effectively created a short circuit (Fig. 5.2a). If we reduce a current source to zero amps, we have effectively created an open circuit (Fig. 5.2b). Thus, the superposition theorem can be stated as: In any linear resistive network, the voltage across or the current through any re-sistor or source may be calculated by adding algebraically all the individual voltages or currents caused by the separate independent sources acting alone, with all other independent voltage sources replaced by short circuits and all other independent current sources replaced by open circuits. Thus, if there are N independent sources, we must perform N experi-ments, each having only one of the independent sources active and the others inactive/turned off/zeroed out. Note that dependent sources are in general active in every experiment. There is also no reason that an independent source must assume only its given value or a zero value in the several experiments; it is necessary only for the sum of the several values to be equal to the original value. An inac-tive source almost always leads to the simplest circuit, however. The circuit we have just used as an example should indicate that a much stronger theorem might be written; a group of independent sources may be made active and inactive collectively, if we wish. For example, suppose there are three independent sources. The theorem states that we may ﬁnd a given response by considering each of the three sources acting alone and adding the three results. Alternatively, we may ﬁnd the response due to the ﬁrst and second sources operating with the third inactive, and then add to this the response caused by the third source acting alone. This amounts to treating several sources collectively as a sort of “supersource.” + – 0 V No voltage drop across terminals, but current can flow i i (a) v + – v + – 0 A No current flows, but a voltage can appear across the terminals (b) ■FIGURE 5.2 (a) A voltage source set to zero acts like a short circuit. (b) A current source set to zero acts like an open circuit. CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 126 For the circuit of Fig. 5.3a, use superposition to determine the unknown branch current ix. EXAMPLE 5.1 ■FIGURE 5.3 (a) An example circuit with two independent sources for which the branch current ix is desired; (b) same circuit with current source open-circuited; (c) original circuit with voltage source short-circuited. ix + – vs = 3 V is = 2 A 6 9 (a) + – 3 V 6 9 ix (b) ' 2 A 6 9 ix " (c) First set the current source equal to zero and redraw the circuit as shown in Fig. 5.3b. The portion of ix due to the voltage source has been designated i′ x to avoid confusion and is easily found to be 0.2 A. Next set the voltage source in Fig. 5.3a to zero and again redraw the circuit, as shown in Fig. 5.3c. Current division lets us determine that i′′ x (the portion of ix due to the 2 A current source) is 0.8 A. Now compute the total current ix by adding the two individual components: ix = ix\|3 V + ix\|2 A = i′ x + i′′ x or ix = 3 6 + 9 + 2 6 6 + 9 = 0.2 + 0.8 = 1.0 A Another way of looking at Example 5.1 is that the 3 V source and the 2 A source are each performing work on the circuit, resulting in a total cur-rent ix ﬂowing through the 9 resistor. However, the contribution of the 3 V source to ix does not depend on the contribution of the 2 A source, and vice versa. For example, if we double the output of the 2 A source to 4 A, it will now contribute 1.6 A to the total current ix ﬂowing through the 9 resistor. However, the 3 V source will still contribute only 0.2 A to ix, for a new total current of 0.2 + 1.6 = 1.8 A. SECTION 5.1 LINEARITY AND SUPERPOSITION 127 PRACTICE ● 5.1 For the circuit of Fig. 5.4, use superposition to compute the current ix. ■FIGURE 5.4 + – 3.5 V 2 A 15 7 3 5 ix Ans: 660 mA. As we will see, superposition does not generally reduce our workload when considering a particular circuit, since it leads to the analysis of several new circuits to obtain the desired response. However, it is particularly use-ful in identifying the signiﬁcance of various parts of a more complex circuit. It also forms the basis of phasor analysis, which is introduced in Chap. 10. EXAMPLE 5.2 Referring to the circuit of Fig. 5.5a, determine the maximum positive current to which the source Ix can be set before any resistor exceeds its power rating and overheats. Identify the goal of the problem. Each resistor is rated to a maximum of 250 mW. If the circuit allows this value to be exceeded (by forcing too much current through either resistor), excessive heating will occur—possibly leading to (Continued on next page) + – Ix 6 V 100 64 (a) W 1 4 W 1 4 + – 6 V 64 100 i' 100 i' 64 (b) Ix 100 64 (c) i" 100 i" 64 ■FIGURE 5.5 (a) A circuit with two resistors each rated at 1 4 W. (b) Circuit with only the 6 V source active. (c) Circuit with the source Ix active. CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 128 an accident. The 6 V source cannot be changed, so we are looking for an equation involving Ix and the maximum current through each resistor. Collect the known information. Based on its 250 mW power rating, the maximum current the 100 resistor can tolerate is Pmax R = 0.250 100 = 50 mA and, similarly, the current through the 64 resistor must be less than 62.5 mA. Devise a plan. Either nodal or mesh analysis may be applied to the solution of this problem, but superposition may give us a slight edge, since we are primarily interested in the effect of the current source. Construct an appropriate set of equations. Using superposition, we redraw the circuit as in Fig. 5.5b and ﬁnd that the 6 V source contributes a current i′ 100 = 6 100 + 64 = 36.59 mA to the 100 resistor and, since the 64 resistor is in series, i′ 64 = 36.59 mA as well. Recognizing the current divider in Fig. 5.5c, we note that i′′ 64 will add to i′ 64 , but i′′ 100 is opposite in direction to i′ 100 . Therefore, IX can safely contribute 62.5 −36.59 = 25.91 mA to the 64 resistor current, and 50 −(−36.59) = 86.59 mA to the 100 resistor current. The 100 resistor therefore places the following constraint on Ix: Ix < (86.59 × 10−3) 100 + 64 64 and the 64 resistor requires that Ix < (25.91 × 10−3) 100 + 64 100 Attempt a solution. Considering the 100 resistor ﬁrst, we see that Ix is limited to Ix < 221.9 mA. The 64 resistor limits Ix such that Ix < 42.49 mA. In order to satisfy both constraints, Ix must be less than 42.49 mA. If the value is increased, the 64 resistor will overheat long before the 100 resistor does. Verify the solution. Is it reasonable or expected? One particularly useful way to evaluate our solution is to perform a dc sweep analysis in PSpice as described after the next example. An in-teresting question, however, is whether we would have expected the 64 resistor to overheat ﬁrst. Originally we found that the 100 resistor has a smaller maximum current, so it might be reasonable to expect it to limit Ix. However, because Ix opposes the current sent by the 6 V source through the 100 resistor but adds to the 6 V source’s contribution to the current through the 64 resistor, it turns out to work the other way—it’s the 64 resistor that sets the limit on Ix. SECTION 5.1 LINEARITY AND SUPERPOSITION 129 In the circuit of Fig. 5.6a, use the superposition principle to deter-mine the value of ix. EXAMPLE 5.3 + – 10 V 2ix 2 1 3 A ix + – v + – (a) + – + – 10 V 2 1 2ix ix (b) ' ' 3 A v + – + – 2 1 2ix ix (c) " " " ■FIGURE 5.6 (a) An example circuit with two independent sources and one dependent source for which the branch current ix is desired. (b) Circuit with the 3 A source open-circuited. (c) Original circuit with the 10 V source short-circuited. + – 3 V 2 A 7 15 5 4i v1 v2 i ■FIGURE 5.7 First open-circuit the 3 A source (Fig. 5.6b). The single mesh equation is −10 + 2i′ x + i′ x + 2i′ x = 0 so that i′ x = 2 A Next, short-circuit the 10 V source (Fig. 5.6c) and write the single-node equation v′′ 2 + v′′ −2i′′ x 1 = 3 and relate the dependent-source-controlling quantity to v′′: v′′ = 2(−i′′ x ) Solving, we ﬁnd i′′ x = −0.6 A and, thus, ix = i′ x + i′′ x = 2 + (−0.6) = 1.4 A Note that in redrawing each subcircuit, we are always careful to use some type of notation to indicate that we are not working with the original variables. This prevents the possibility of rather disastrous errors when we add the individual results. PRACTICE ● 5.2 For the circuit of Fig. 5.7, use superposition to obtain the voltage across each current source. Ans: v1\|2A = 9.180 V, v2\|2A = −1.148 V, v1\|3V = 1.967 V, v2\|3V = −0.246 V; v1 = 11.147 V, v2 = −1.394 V. Summary of Basic Superposition Procedure 1. Select one of the independent sources. Set all other indepen-dent sources to zero. This means voltage sources are replaced with short circuits and current sources are replaced with open circuits. Leave dependent sources in the circuit. 2. Relabel voltages and currents using suitable notation (e.g., v′, i′′ 2). Be sure to relabel controlling variables of dependent sources to avoid confusion. 3. Analyze the simpliﬁed circuit to ﬁnd the desired currents and/or voltages. 4. Repeat steps 1 through 3 until each independent source has been considered. 5. Add the partial currents and/or voltages obtained from the separate analyses. Pay careful attention to voltage signs and current directions when summing. 6. Do not add power quantities. If power quantities are required, calculate only after partial voltages and/or currents have been summed. Note that step 1 may be altered in several ways. First, independent sources can be considered in groups as opposed to individually if it simpli-ﬁes the analysis, as long as no independent source is included in more than one subcircuit. Second, it is technically not necessary to set sources to zero, although this is almost always the best route. For example, a 3 V source may appear in two subcircuits as a 1.5 V source, since 1.5 + 1.5 = 3 V just as 0 + 3 = 3 V. Because it is unlikely to simplify our analysis, however, there is little point to such an exercise. CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 130 COMPUTER-AIDED ANALYSIS Although PSpice is extremely useful in verifying that we have analyzed a complete circuit correctly, it can also assist us in determining the contribution of each source to a particular response. To do this, we employ what is known as a dc parameter sweep. Consider the circuit presented in Example 5.2, when we were asked to determine the maximum positive current that could be obtained from the current source without exceeding the power rating of either resistor in the circuit. The circuit is shown redrawn using the Orcad Capture CIS schematic tool in Fig. 5.8. Note that no value has been assigned to the current source. After the schematic has been entered and saved, the next step is to specify the dc sweep parameters. This option allows us to specify a range of values for a voltage or current source (in the present case, the current source Ix), rather than a speciﬁc value. Selecting New Simulation Proﬁle under PSpice, we provide a name for our proﬁle and are then provided with the dialog box shown in Fig. 5.9. SECTION 5.1 LINEARITY AND SUPERPOSITION 131 ■FIGURE 5.8 The circuit from Example 5.2. ■FIGURE 5.9 DC Sweep dialog box shown with Ix selected as the sweep variable. Under Analysis Type, we pull down the DC Sweep option, specify the “sweep variable’’ as Current Source, and then type in Ix in the Name box. There are several options under Sweep Type: Linear, Logarithmic, and Value List. The last option allows us to specify each value to assign to Ix. In order to generate a smooth plot, however, we choose to perform a Linear sweep, with a Start Value of 0 mA, an End Value of 50 mA, and a value of 0.01 mA for the Increment. After we perform the simulation, the graphical output package Probe is automatically launched. When the window appears, the horizontal axis (corresponding to our variable, Ix) is displayed, but the vertical axis variable must be chosen. Selecting Add Trace from the Trace menu, we click on I(R1), then type an asterisk in the Trace Expression box, click on I(R1) once again, insert yet another asterisk, and ﬁnally type in 100. This asks Probe to plot the power absorbed by the 100 resistor. In a similar fashion, we repeat the process to add the power (Continued on next page) absorbed by the 64 resistor, resulting in a plot similar to that shown in Fig. 5.10a. A horizontal reference line at 250 mW was also added to the plot by typing 0.250 in the Trace Expression box after selecting Add Trace from the Trace menu a third time. We see from the plot that the 64 resistor does exceed its 250 mW power rating in the vicinity of Ix = 43 mA. In contrast, however, we also see that regardless of the value of the current source Ix (provided that it is between 0 and 50 mA), the 100 resistor will never dissipate 250 mW; in fact, the absorbed power decreases with increasing current from the current source. If we desire a more precise answer, we can make use of the cursor tool, which is invoked by selecting Trace, Cursor, Display from the menu bar. Figure 5.10b shows the result of dragging cursor 1 to 42.52 A, where the 64 resistor is dissipating just over its maximum rated power of 250 mW. Increased precision can be obtained by decreasing the increment value used in the dc sweep. This technique is very useful in analyzing electronic circuits, where we might need, for example, to determine what input voltage is required CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 132 ■FIGURE 5.10 (a) Probe output with text labels identifying the power absorbed by the two resistors individually. A horizontal line indicating 250 mW has also been included, as well as text labels to improve clarity. (b) Cursor dialog box. (a) (b) SECTION 5.2 SOURCE TRANSFORMATIONS 133 Unfortunately, it usually turns out that little if any time is saved in ana-lyzing a circuit containing one or more dependent sources by use of the superposition principle, for there must always be at least two sources in operation: one independent source and all the dependent sources. We must constantly be aware of the limitations of superposition. It is applicable only to linear responses, and thus the most common nonlinear response—power—is not subject to superposition. For example, consider two 1 V batteries in series with a 1 resistor. The power delivered to the re-sistor is 4 W, but if we mistakenly try to apply superposition, we might say that each battery alone furnished 1 W and thus the calculated power is only 2 W. This is incorrect, but a surprisingly easy mistake to make. 5.2 • SOURCE TRANSFORMATIONS Practical Voltage Sources So far, we’ve only worked with ideal sources—elements whose terminal voltage is independent of the current ﬂowing through them. To see the relevance of this fact, consider a simple independent (“ideal”) 9 V source connected to a 1 resistor. The 9 volt source will force a current of 9 amperes through the 1 resistor (perhaps this seems reasonable enough), but the same source would apparently force 9,000,000 amperes through a 1 m resistor (which hopefully does not seem reasonable). On paper, there's nothing to stop us from reducing the resistor value all the way to 0 … but that would lead to a contradiction, as the source would be “trying” to maintain 9 V across a dead short, which Ohm’s law tells us can’t happen (V = 9 = RI = 0?). Whathappensintherealworldwhenwedothistypeofexperiment?For example, if we try to start a car with the headlights already on, we most likely notice the headlights dim as the battery is asked to supply a large (∼100 A or more) starter current in parallel with the current running to the headlights. If we model the 12 V battery with an ideal 12 V source as in Fig. 5.11a, our observation cannot be explained. Another way of saying this is that our model breaks down when the load draws a large current from the source. To better approximate the behavior of a real device, the ideal voltage source must be modiﬁed to account for the lowering of its terminal voltage when large currents are drawn from it. Let us suppose that we observe ex-perimentally that our car battery has a terminal voltage of 12 V when no current is ﬂowing through it, and a reduced voltage of 11 V when 100 A is ﬂowing. How could we model this behavior? Well, a more accurate model might be an ideal voltage source of 12 V in series with a resistor across which 1 V appears when 100 A ﬂows through it. A quick calculation shows that the resistor must be 1 V/100 A 0.01 , and the ideal voltage source and this series resistor constitute a practical voltage source (Fig. 5.11b). + – 12 V (a) + – 12 V 0.01 (b) ■FIGURE 5.11 (a) An ideal 12 V dc voltage source used to model a car battery. (b) A more accurate model that accounts for the observed reduction in terminal voltage at large currents. to a complicated ampliﬁer circuit in order to obtain a zero output voltage. We also notice that there are several other types of parameter sweeps that we can perform, including a dc voltage sweep. The ability to vary temperature is useful only when dealing with component models that have a temperature parameter built in, such as diodes and transistors. Thus, we are using the series combination of two ideal circuit elements, an independent voltage source and a resistor, to model a real device. We do not expect to ﬁnd such an arrangement of ideal elements inside our car battery, of course. Any real device is characterized by a certain current-voltage relationship at its terminals, and our problem is to develop some combination of ideal elements that can furnish a similar current-voltage characteristic, at least over some useful range of current, voltage, or power. In Fig. 5.12a, we show our two-piece practical model of the car battery now connected to some load resistor RL. The terminal voltage of the practical source is the same as the voltage across RL and is marked2 VL. Figure 5.12b shows a plot of load voltage VL as a function of the load current IL for this practical source. The KVL equation for the circuit of Fig. 5.12a may be written in terms of IL and VL: 12 = 0.01IL + VL and thus VL = −0.01IL + 12 This is a linear equation in IL and VL, and the plot in Fig. 5.12b is a straight line. Each point on the line corresponds to a different value of RL. For exam-ple, the midpoint of the straight line is obtained when the load resistance is equal to the internal resistance of the practical source, or RL = 0.01. Here, the load voltage is exactly one-half the ideal source voltage. When RL = ∞and no current whatsoever is being drawn by the load, the practical source is open-circuited and the terminal voltage, or open-circuit voltage, is VLoc = 12 V. If, on the other hand, RL = 0, thereby short-circuiting the load terminals, then a load current or short-circuit cur-rent, ILsc = 1200 A, would ﬂow. (In practice, such an experiment would probably result in the destruction of the short circuit, the battery, and any measuring instruments incorporated in the circuit!) Since the plot of VL versus IL is a straight line for this practical voltage source, we should note that the values of VLoc and ILsc uniquely determine the entire VL–IL curve. The horizontal broken line of Fig. 5.12b represents the VL–IL plot for an ideal voltage source; the terminal voltage remains constant for any value of load current. For the practical voltage source, the terminal voltage has a value near that of the ideal source only when the load current is relatively small. Let us now consider a general practical voltage source, as shown in Fig. 5.13a. The voltage of the ideal source is vs, and a resistance Rs, called an internal resistance or output resistance, is placed in series with it. Again, we must note that the resistor is not really present as a separate component but merely serves to account for a terminal voltage that decreases as the load current increases. Its presence enables us to model the behavior of a physical voltage source more closely. The linear relationship between vL and iL is vL = vs −RsiL CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 134 VL + – + – IL 0.01 12 V (a) RL (b) 0 4 8 6 2 10 12 0 200 400 600 800 Load current IL (A) Source voltage VL (V) 1000 1200 Practical source Ideal source ■FIGURE 5.12 (a) A practical source, which approximates the behavior of a certain 12 V automobile battery, is shown connected to a load resistor RL. (b) The relationship between IL and VL is linear. vL + – + – iL Rs vs (a) RL Practical source Ideal source vLoc = vs vL 0 (b) iLsc = vs/Rs 0 iL ■FIGURE 5.13 (a) A general practical voltage source connected to a load resistor RL. (b) The terminal voltage of a practical voltage source decreases as iL increases and RL vL/iL decreases. The terminal voltage of an ideal voltage source (also plotted) remains the same for any current delivered to a load. (2) From this point on we will endeavor to adhere to the standard convention of referring to strictly dc quantities using capital letters, whereas lowercase letters denote a quantity that we know to possess some time-varying component. However, in describing general theorems which apply to either dc or ac, we will continue to use lowercase to emphasize the general nature of the concept. SECTION 5.2 SOURCE TRANSFORMATIONS 135 and this is plotted in Fig. 5.13b. The open-circuit voltage (RL = ∞, so iL = 0) is vLoc = vs and the short-circuit current (RL = 0, so vL = 0) is iLsc = vs Rs Once again, these values are the intercepts for the straight line in Fig. 5.13b, and they serve to deﬁne it completely. Practical Current Sources An ideal current source is also nonexistent in the real world; there is no physical device that will deliver a constant current regardless of the load re-sistance to which it is connected or the voltage across its terminals. Certain transistor circuits will deliver a constant current to a wide range of load re-sistances, but the load resistance can always be made sufﬁciently large that the current through it becomes very small. Inﬁnite power is simply never available (unfortunately). A practical current source is deﬁned as an ideal current source in paral-lel with an internal resistance Rp. Such a source is shown in Fig. 5.14a, and the current iL and voltage vL associated with a load resistance RL are indi-cated. Application of KCL yields iL = is −vL Rp which is again a linear relationship. The open-circuit voltage and the short-circuit current are vLoc = Rpis and iLsc = is The variation of load current with changing load voltage may be inves-tigated by changing the value of RL as shown in Fig. 5.14b. The straight line is traversed from the short-circuit, or “northwest,’’ end to the open-circuit termination at the “southeast’’ end by increasing RL from zero to inﬁnite ohms. The midpoint occurs for RL = Rp. The load current iL and the ideal source current are approximately equal only for small values of load volt-age, which are obtained with values of RL that are small compared to Rp. Equivalent Practical Sources It may be no surprise that we can improve upon models to increase their accuracy; at this point we now have a practical voltage source model and also a practical current source model. Before we proceed, however, let’s take a moment to compare Fig. 5.13b and Fig. 5.14b. One is for a circuit with a voltage source and the other, with a current source, but the graphs are indistinguishable! It turns out that this is no coincidence. In fact, we are about to show that a practical voltage source can be electrically equivalent to a practical cur-rent source—meaning that a load resistor RL connected to either will have iL Rp (a) RL is vL + – Practical source Ideal source vLoc = Rpis vL (b) iLsc = is iL ■FIGURE 5.14 (a) A general practical current source connected to a load resistor RL. (b) The load current provided by the practical current source is shown as a function of the load voltage. the same vL and iL. This means we can replace one practical source with the other and the rest of the circuit will not know the difference. Consider the practical voltage source and resistor RL shown in Fig. 5.15a, and the circuit composed of a practical current source and resistor RL shown in Fig. 5.15b. A simple calculation shows that the voltage across the load RL of Fig. 5.15a is vL = vs RL Rs + RL A similar calculation shows that the voltage across the load RL in Fig. 5.15b is vL = is Rp Rp + RL · RL The two practical sources are electrically equivalent, then, if Rs = Rp and vs = Rpis = Rsis where we now let Rs represent the internal resistance of either practical source, which is the conventional notation. Let’s try this with the practical current source shown in Fig. 5.16a. Since its internal resistance is 2 , the internal resistance of the equivalent practi-cal voltage source is also 2 ; the voltage of the ideal voltage source con-tained within the practical voltage source is (2)(3) = 6 V. The equivalent practical voltage source is shown in Fig. 5.16b. To check the equivalence, let us visualize a 4 resistor connected to each source. In both cases a current of 1 A, a voltage of 4 V, and a power of 4 W are associated with the 4 load. However, we should note very care-fully that the ideal current source is delivering a total power of 12 W, while the ideal voltage source is delivering only 6 W. Furthermore, the internal resistance of the practical current source is absorbing 8 W, whereas the in-ternal resistance of the practical voltage source is absorbing only 2 W. Thus we see that the two practical sources are equivalent only with respect to what transpires at the load terminals; they are not equivalent internally! CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 136 + – vs Rs RL vL + – iL (a) RL Rp is vL + – iL (b) ■FIGURE 5.15 (a) A given practical voltage source connected to a load RL. (b) The equivalent practical current source connected to the same load. 3 A 2 (a) 6 V 2 (b) + – ■FIGURE 5.16 (a) A given practical current source. (b) The equivalent practical voltage source. Compute the current through the 4.7 k resistor in Fig. 5.17a after transforming the 9 mA source into an equivalent voltage source. It’s not just the 9 mA source at issue, but also the resistance in parallel with it (5 k). We remove these components, leaving two terminals “dan-gling.” We then replace them with a voltage source in series with a 5 k resistor. The value of the voltage source must be (0.09)(5000) 45 V. Redrawing the circuit as in Fig. 5.17b, we can write a simple KVL equation 45 5000I 4700I 3000I 3 0 which is easily solved to yield I 3.307 mA. We can check our answer of course by analyzing the circuit of Fig. 5.17a using either nodal or mesh techniques. EXAMPLE 5.4 SECTION 5.2 SOURCE TRANSFORMATIONS 137 + – I 9 mA 3 V 4.7 k 5 k 3 k (a) + – + – I 3 V 45 V 4.7 k 5 k 3 k (b) ■FIGURE 5.17 (a) A circuit with both a voltage source and a current source. (b) The circuit after the 9 mA source is transformed into an equivalent voltage source. + – 5 V 1 mA 5 k 47 k IX ■FIGURE 5.18 Calculate the current through the 2 resistor in Fig. 5.19a by making use of source transformations to ﬁrst simplify the circuit. We begin by transforming each current source into a voltage source (Fig. 5.19b), the strategy being to convert the circuit into a simple loop. We must be careful to retain the 2 resistor for two reasons: ﬁrst, the dependent source controlling variable appears across it, and second, we desire the current ﬂowing through it. However, we can combine the 17 and 9 resistors, since they appear in series. We also see that the 3 and 4 resistors may be combined into a single 7 resistor, which can then be used to transform the 15 V source into a 15/7Asource as in Fig. 5.19c. Finally, we note that the two 7 resistors can be combined into a single 3.5 resistor, which may be used to transform the 15/7 A current source into a 7.5 V voltage source. The result is a simple loop circuit, shown in Fig. 5.19d. The current I can now be found using KVL: −7.5 + 3.5I −51Vx + 28I + 9 = 0 where Vx = 2I Thus, I = 21.28 mA EXAMPLE 5.5 Ans: 192 μA. PRACTICE ● 5.3 For the circuit of Fig. 5.18, compute the current IX through the 47 k resistor after performing a source transformation on the voltage source. (Continued on next page) CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 138 I Vx + – 3 7 9 17 4 2 5 A 1 A 3Vx (a) + – + – + – I Vx + – 7 17 2 9 4 3 15 V 9 V 51Vx (b) + – + – I Vx + – 7 7 26 2 9 V 51Vx A 15 7 (c) + – + – + – I Vx + – 26 2 3.5 9 V 7.5 V 51Vx (d) ■FIGURE 5.19 (a) A circuit with two independent current sources and one dependent source. (b) The circuit after each source is transformed into a voltage source. (c) The circuit after further combinations. (d) The ﬁnal circuit. ■FIGURE 5.20 + – V + – 4 M 6 M 1 M 200 k 75 A 40 A 3 V Ans: 27.2 V. PRACTICE ● 5.4 For the circuit of Fig. 5.20, compute the voltage V across the 1 M resistor using repeated source transformations. SECTION 5.2 SOURCE TRANSFORMATIONS 139 Several Key Points We conclude our discussion of practical sources and source transformations with a few observations. First, when we transform a voltage source, we must be sure that the source is in fact in series with the resistor under con-sideration. For example, in the circuit of Fig. 5.21, it is perfectly valid to perform a source transformation on the voltage source using the 10 resis-tor, as they are in series. However, it would be incorrect to attempt a source transformation using the 60 V source and the 30 resistor—a very common type of error. In a similar fashion, when we transform a current source and resistor combination, we must be sure that they are in fact in parallel. Consider the current source shown in Fig. 5.22a. We may perform a source transforma-tion including the 3 resistor, as they are in parallel, but after the transfor-mation there may be some ambiguity as to where to place the resistor. In such circumstances, it is helpful to ﬁrst redraw the components to be trans-formed as in Fig. 5.22b. Then the transformation to a voltage source in series with a resistor may be drawn correctly as shown in Fig. 5.22c; the resistor may in fact be drawn above or below the voltage source. It is also worthwhile to consider the unusual case of a current source in series with a resistor, and its dual, the case of a voltage source in parallel 20 4 A 10 60 V 30 0.4i1 + – i1 ■FIGURE 5.21 An example circuit to illustrate how to determine if a source transformation can be performed. 1 A 2 3 5 V 7 3 V + – + – (a) 2 3 5 V 7 3 V + – + – 1 A (b) 2 3 5 V 7 3 V 3 V + – + – + – (c) ■FIGURE 5.22 (a) A circuit with a current source to be transformed to a voltage source. (b) Circuit redrawn so as to avoid errors. (c) Transformed source/resistor combination. with a resistor. Let’s start with the simple circuit of Fig. 5.23a, where we are interested only in the voltage across the resistor marked R2. We note that re-gardless of the value of resistor R1, VR2 = Ix R2. Although we might be tempted to perform an inappropriate source transformation on such a cir-cuit, in fact we may simply omit resistor R1 (provided that it is of no interest to us itself). A similar situation arises with a voltage source in parallel with a resistor, as depicted in Fig. 5.23b. Again, if we are only interested in some quantity regarding resistor R2, we may ﬁnd ourselves tempted to perform some strange (and incorrect) source transformation on the voltage source and resistor R1. In reality, we may omit resistor R1 from our circuit as far as resistor R2 is concerned—its presence does not alter the voltage across, the current through, or the power dissipated by resistor R2. CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 140 Vx R1 R2 + – Ix R1 R2 VR2 + – (a) (b) ■FIGURE 5.23 (a) Circuit with a resistor R1 in series with a current source. (b) A voltage source in parallel with two resistors. Summary of Source Transformation 1. A common goal in source transformation is to end up with either all current sources or all voltage sources in the circuit. This is especially true if it makes nodal or mesh analysis easier. 2. Repeated source transformations can be used to simplify a circuit by allowing resistors and sources to eventually be combined. 3. The resistor value does not change during a source transfor-mation, but it is not the same resistor. This means that currents or voltages associated with the original resistor are irretrievably lost when we perform a source transformation. 4. If the voltage or current associated with a particular resistor is used as a controlling variable for a dependent source, it should not be included in any source transformation. The original resistor must be retained in the ﬁnal circuit, untouched. 5. If the voltage or current associated with a particular element is of interest, that element should not be included in any source transformation. The original element must be retained in the ﬁnal circuit, untouched. 6. In a source transformation, the head of the current source arrow corresponds to the “+” terminal of the voltage source. 7. A source transformation on a current source and resistor requires that the two elements be in parallel. 8. A source transformation on a voltage source and resistor requires that the two elements be in series. SECTION 5.3 THÉVENIN AND NORTON EQUIVALENT CIRCUITS 141 5.3 • THÉVENIN AND NORTON EQUIVALENT CIRCUITS Now that we have been introduced to source transformations and the super-position principle, it is possible to develop two more techniques that will greatly simplify the analysis of many linear circuits. The ﬁrst of these theo-rems is named after L. C. Thévenin, a French engineer working in telegra-phy who published the theorem in 1883; the second may be considered a corollary of the ﬁrst and is credited to E. L. Norton, a scientist with the Bell Telephone Laboratories. Let us suppose that we need to make only a partial analysis of a circuit. For example, perhaps we need to determine the current, voltage, and power delivered to a single “load” resistor by the remainder of the circuit, which may consist of a sizable number of sources and resistors (Fig. 5.24a). Or, perhaps we wish to ﬁnd the response for different values of the load resis-tance. Thévenin’s theorem tells us that it is possible to replace everything except the load resistor with an independent voltage source in series with a resistor (Fig. 5.24b); the response measured at the load resistor will be un-changed. Using Norton’s theorem, we obtain an equivalent composed of an independent current source in parallel with a resistor (Fig. 5.24c). RL Complex network (a) + – RTH VTH RL (b) IN RL RN (c) ■FIGURE 5.24 (a) A complex network including a load resistor RL. (b) A Thévenin equivalent network connected to the load resistor RL. (c) A Norton equivalent network connected to the load resistor RL. Thus, one of the main uses of Thévenin’s and Norton’s theorems is the replacement of a large part of a circuit, often a complicated and uninter-esting part, with a very simple equivalent. The new, simpler circuit enables us to make rapid calculations of the voltage, current, and power which the original circuit is able to deliver to a load. It also helps us to choose the best value of this load resistance. In a transistor power ampliﬁer, for example, the Thévenin or Norton equivalent enables us to determine the maximum power that can be taken from the ampliﬁer and delivered to the speakers. Consider the circuit shown in Fig. 5.25a. Determine the Thévenin equivalent of network A, and compute the power delivered to the load resistor RL. The dashed regions separate the circuit into networks A and B; our main interest is in network B, which consists only of the load resistor RL. Net-work A may be simpliﬁed by making repeated source transformations. EXAMPLE 5.6 (Continued on next page) CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 142 8 V RL 9 Network A (e) + – + – 12 V Network A Network B 3 7 6 RL (a) 4 A 3 6 RL 7 Network A (b) 4 A 2 RL 7 Network A (c) 8 V RL 2 7 Network A (d) + – ■FIGURE 5.25 (a) A circuit separated into two networks. (b)–(d) Intermediate steps to simplifying network A. (e) The Thévenin equivalent circuit. 2 10 8 RL 5 A ■FIGURE 5.26 We ﬁrst treat the 12 V source and the 3 resistor as a practical volt-age source and replace it with a practical current source consisting of a 4 Asource in parallel with 3 (Fig. 5.25b). The parallel resistances are then combined into 2 (Fig. 5.25c), and the practical current source that results is transformed back into a practical voltage source (Fig. 5.25d). The ﬁnal result is shown in Fig. 5.25e. From the viewpoint of the load resistor RL,this network A (the Thévenin equivalent) is equivalent to the original network A; from our viewpoint, the circuit is much simpler, and we can now easily compute the power delivered to the load: PL = 8 9 + RL 2 RL Furthermore, we can see from the equivalent circuit that the maxi-mum voltage that can be obtained across RL is 8 V and corresponds to RL = ∞. A quick transformation of network A to a practical current source (the Norton equivalent) indicates that the maximum current that may be delivered to the load is 8/9 A, which occurs when RL = 0. Neither of these facts is readily apparent from the original circuit. PRACTICE ● 5.5 Using repeated source transformations, determine the Norton equivalent of the highlighted network in the circuit of Fig. 5.26. Ans: 1 A, 5 . SECTION 5.3 THÉVENIN AND NORTON EQUIVALENT CIRCUITS 143 Thévenin’s Theorem Using the technique of source transformation to ﬁnd a Thévenin or Norton equivalent network worked well enough in Example 5.6, but it can rapidly become impractical in situations where dependent sources are present or the circuit is composed of a large number of elements. An alternative is to employ Thévenin’s theorem (or Norton’s theorem) instead. We will state the theorem3 as a somewhat formal procedure and then consider various ways to make the approach more practical depending on the situation we face. A Statement of Thévenin’s Theorem 1. Given any linear circuit, rearrange it in the form of two networks, A and B, connected by two wires. Network A is the network to be simpliﬁed; B will be left untouched. 2. Disconnect network B. Deﬁne a voltage voc as the voltage now appearing across the terminals of network A. 3. Turn off or “zero out” every independent source in network A to form an inactive network. Leave dependent sources unchanged. 4. Connect an independent voltage source with value voc in series with the inactive network. Do not complete the circuit; leave the two terminals disconnected. 5. Connect network B to the terminals of the new network A. All currents and voltages in B will remain unchanged. Note that if either network contains a dependent source, its control variable must be in the same network. Let us see if we can apply Thévenin’s theorem successfully to the circuit we considered in Fig. 5.25. We have already found the Thévenin equivalent of the circuit to the left of RL in Example 5.6, but we want to see if there is an easier way to obtain the same result. (3) A proof of Thévenin’s theorem in the form in which we have stated it is rather lengthy, and therefore it has been placed in Appendix 3, where the curious may peruse it. Use Thévenin’s theorem to determine the Thévenin equivalent for that part of the circuit in Fig. 5.25a to the left of RL. We begin by disconnecting RL, and note that no current ﬂows through the 7 resistor in the resulting partial circuit shown in Fig. 5.27a. Thus, Voc appears across the 6 resistor (with no current through the 7 resistor there is no voltage drop across it), and voltage division enables us to determine that Voc = 12 6 3 + 6 = 8 V EXAMPLE 5.7 CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 144 + – Voc + – 6 3 12 V 7 (a) 6 3 7 (b) RTH ■FIGURE 5.27 (a) The circuit of Fig. 5.25a with network B (the resistor RL) disconnected and the voltage across the connecting terminals labeled as Voc. (b) The independent source in Fig. 5.25a has been killed, and we look into the terminals where network B was connected to determine the effective resistance of network A. + – 9 V 4 2 4 6 5 I2 ■FIGURE 5.28 Turning off network A (i.e., replacing the 12 V source with a short circuit) and looking back into the dead network, we see a 7 resistor connected in series with the parallel combination of 6 and 3 (Fig. 5.27b). Thus, the inactive network can be represented here by a 9 resistor, referred to as the Thévenin equivalent resistance of network A. The Thévenin equivalent then is Voc in series with a 9 resistor, which agrees with our previous result. PRACTICE ● 5.6 Use Thévenin’s theorem to ﬁnd the current through the 2 resistor in the circuit of Fig. 5.28. (Hint: Designate the 2 resistor as network B.) Ans: VTH = 2.571 V, RTH = 7.857 , I2 = 260.8 mA. A Few Key Points The equivalent circuit we have learned how to obtain is completely inde-pendent of network B: we have been instructed to ﬁrst remove network B and then measure the open-circuit voltage produced by network A, an operation that certainly does not depend on network B in any way. The B network is mentioned only to indicate that an equivalent for A may be obtained no mat-ter what arrangement of elements is connected to the A network; the B net-work represents this general network. There are several points about the theorem which deserve emphasis. • The only restriction that we must impose on A or B is that all dependent sources in A have their control variables in A, and similarly for B. • No restrictions are imposed on the complexity of A or B; either one may contain any combination of independent voltage or current sources, linear dependent voltage or current sources, resistors, or any other circuit elements which are linear. • The dead network A can be represented by a single equivalent resis-tance RTH, which we will call the Thévenin equivalent resistance. SECTION 5.3 THÉVENIN AND NORTON EQUIVALENT CIRCUITS 145 This holds true whether or not dependent sources exist in the inactive A network, an idea we will explore shortly. • A Thévenin equivalent consists of two components: a voltage source in series with a resistance. Either may be zero, although this is not usually the case. Norton’s Theorem Norton’s theorem bears a close resemblance to Thévenin’s theorem and may be stated as follows: A Statement of Norton’s Theorem 1. Given any linear circuit, rearrange it in the form of two networks, A and B, connected by two wires. Network A is the network to be simpliﬁed; B will be left untouched. As before, if either network contains a dependent source, its controlling variable must be in the same network. 2. Disconnect network B, and short the terminals of A. Deﬁne a current isc as the current now ﬂowing through the shorted terminals of network A. 3. Turn off or “zero out” every independent source in network A to form an inactive network. Leave dependent sources unchanged. 4. Connect an independent current source with value isc in parallel with the inactive network. Do not complete the circuit; leave the two terminals disconnected. 5. Connect network B to the terminals of the new network A. All currents and voltages in B will remain unchanged. The Norton equivalent of a linear network is the Norton current source isc in parallel with the Thévenin resistance RTH. Thus, we see that in fact it is possible to obtain the Norton equivalent of a network by performing a source transformation on the Thévenin equivalent. This results in a direct relationship between voc, isc, and RTH: voc = RTHisc In circuits containing dependent sources, we will often ﬁnd it more con-venient to determine either the Thévenin or Norton equivalent by ﬁnding both the open-circuit voltage and the short-circuit current and then deter-mining the value of RTH as their quotient. It is therefore advisable to be-come adept at ﬁnding both open-circuit voltages and short-circuit currents, even in the simple problems that follow. If the Thévenin and Norton equiv-alents are determined independently, Eq. can serve as a useful check. Let’s consider three different examples of the determination of a Thévenin or Norton equivalent circuit. CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 146 ■FIGURE 5.29 (a) A given circuit in which the 1 kresistor is identiﬁed as network B. (b) Network A with all independent sources killed. (c) The Thévenin equivalent is shown for network A. (d) The Norton equivalent is shown for network A. (e) Circuit for determining Isc. + – 4 V 2 mA (a) 1 k 3 k 2 k 2 k 3 k RTH (b) + – 4 V 2 mA 2 k 3 k Isc (e) + – 8 V (c) 1 k 5 k 1.6 mA (d) 1 k 5 k From the wording of the problem statement, network B is the 1 k resistor, so network A is everything else. Choosing to ﬁnd the Thévenin equivalent of network A ﬁrst, we apply superposition, noting that no current ﬂows through the 3 k resistor once network B is disconnected. With the current source set to zero, Voc\|4 V 4 V. With the voltage source set to zero, Voc\|2 mA (0.002)(2000) 4 V. Thus, Voc 4 4 8 V. To ﬁnd RTH, set both sources to zero as in Fig. 5.29b. By inspection, RTH 2 k 3 k 5 k. The complete Thévenin equivalent, with network B reconnected, is shown in Fig. 5.29c. The Norton equivalent is found by a simple source transformation of the Thévenin equivalent, resulting in a current source of 8/5000 1.6 mA in parallel with a 5 k resistor (Fig. 5.29d). Check: Find the Norton equivalent directly from Fig. 5.29a. Re-moving the 1 k resistor and shorting the terminals of network A, we EXAMPLE 5.8 Find the Thévenin and Norton equivalent circuits for the network faced by the 1 k resistor in Fig. 5.29a. SECTION 5.3 THÉVENIN AND NORTON EQUIVALENT CIRCUITS 147 ■FIGURE 5.30 + – 3 V 7 mA 2 k 1 k 5 k ﬁnd Isc as shown in Fig. 5.29e by superposition and current division: Isc = Isc\|4 V + Isc\|2 mA = 4 2 + 3 + (2) 2 2 + 3 = 0.8 + 0.8 = 1.6 mA which completes the check. PRACTICE ● 5.7 Determine the Thévenin and Norton equivalents of the circuit of Fig. 5.30. Ans: −7.857 V, −3.235 mA, 2.429 k. When Dependent Sources Are Present Technically speaking, there does not always have to be a “network B” for us to invoke either Thévenin’s theorem or Norton’s theorem; we could instead be asked to ﬁnd the equivalent of a network with two terminals not yet con-nected to another network. If there is a network B that we do not want to in-volve in the simpliﬁcation procedure, however, we must use a little caution if it contains dependent sources. In such situations, the controlling variable and the associated element(s) must be included in network B and excluded from network A. Otherwise, there will be no way to analyze the ﬁnal circuit because the controlling quantity will be lost. If network A contains a dependent source, then again we must ensure that the controlling variable and its associated element(s) cannot be in net-work B. Up to now, we have only considered circuits with resistors and in-dependent sources. Although technically speaking it is correct to leave a dependent source in the “inactive” network when creating a Thévenin or Norton equivalent, in practice this does not result in any kind of simpliﬁca-tion. What we really want is an independent voltage source in series with a single resistor, or an independent current source in parallel with a single resistor—in other words, a two-component equivalent. In the following examples, we consider various means of reducing networks with dependent sources and resistors into a single resistance. CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 148 EXAMPLE 5.9 Determine the Thévenin equivalent of the circuit in Fig. 5.31a. vx + – vx 4000 (a) + – 2 k 4 V 3 k vx + – vx 4000 (b) 2 k 3 k 8 V + – (c) 10 k + – 8 V ■FIGURE 5.31 (a) A given network whose Thévenin equivalent is desired. (b) A possible, but rather useless, form of the Thévenin equivalent. (c) The best form of the Thévenin equivalent for this linear resistive network. + – 20 k 0.01V1 100 V V1 + – ■FIGURE 5.32 To ﬁnd Voc we note that vx = Voc and that the dependent source current must pass through the 2 k resistor, since no current can ﬂow through the 3 k resistor. Using KVL around the outer loop: −4 + 2 × 103 −vx 4000 + 3 × 103(0) + vx = 0 and vx = 8 V = Voc By Thévenin’s theorem, then, the equivalent circuit could be formed with the inactive A network in series with an 8 V source, as shown in Fig. 5.31b. This is correct, but not very simple and not very helpful; in the case of linear resistive networks, we really want a simpler equivalent for the inactive A network, namely, RTH. The dependent source prevents us from determining RTH directly for the inactive network through resistance combination; we therefore seek Isc. Upon short-circuiting the output terminals in Fig. 5.31a, it is apparent that Vx = 0 and the dependent current source is not active. Hence, Isc = 4/(5 × 103) = 0.8 mA. Thus, RTH = Voc Isc = 8 0.8 × 10−3 = 10 k and the acceptable Thévenin equivalent of Fig. 5.31c is obtained. PRACTICE ● 5.8 Find the Thévenin equivalent for the network of Fig. 5.32. (Hint: a quick source transformation on the dependent source might help.) Ans: −502.5 mV, −100.5 . Note: a negative resistance might seem strange—and it is! Such a thing is physically possible only if, for example, we do a bit of clever electronic circuit design to create something that behaves like the dependent current source we represented in Fig. 5.32. SECTION 5.3 THÉVENIN AND NORTON EQUIVALENT CIRCUITS 149 As another example, let us consider a network having a dependent source but no independent source. Find the Thévenin equivalent of the circuit shown in Fig. 5.33a. The rightmost terminals are already open-circuited, hence i = 0. Consequently, the dependent source is inactive, so voc = 0. We next seek the value of RTH represented by this two-terminal network. However, we cannot ﬁnd voc and isc and take their quotient, for there is no independent source in the network and both voc and isc are zero. Let us, therefore, be a little tricky. We apply a 1 A source externally, measure the voltage vtest that results, and then set RTH = vtest/1. Referring to Fig. 5.33b, we see that i = −1 A. Applying nodal analysis, vtest −1.5(−1) 3 + vtest 2 = 1 so that vtest = 0.6 V and thus RTH = 0.6 The Thévenin equivalent is shown in Fig. 5.33c. EXAMPLE 5.10 (a) 3 2 + – 1.5i i vtest + – (b) 3 2 1 A + – 1.5i i 0.6 (c) ■FIGURE 5.33 (a) A network with no independent sources. (b) A hypothetical measurement to obtain RTH. (c) The Thévenin equivalent to the original circuit. A Quick Recap of Procedures We have now looked at three examples in which we determined a Thévenin or Norton equivalent circuit. The ﬁrst example (Fig. 5.29) contained only independent sources and resistors, and several different methods could have been applied to it. One would involve calculating RTH for the inactive network and then Voc for the live network. We could also have found RTH and Isc, or Voc and Isc. In the second example (Fig. 5.31), both independent and dependent sources were present, and the method we used required us to ﬁnd Voc and Isc. We could not easily ﬁnd RTH for the inactive network because the dependent source could not be made inactive. The last example did not contain any independent sources, and therefore the Thévenin and Norton equivalents do not contain an independent source. We found RTH by applying 1 A and ﬁnding vtest = 1 × RTH . We could also apply 1 V and determine i = 1/RTH. These two related techniques can be applied to any circuit with dependent sources, as long as all independent sources are set to zero ﬁrst. Two other methods have a certain appeal because they can be used for any of the three types of networks considered. In the ﬁrst, simply replace network B with a voltage source vs, deﬁne the current leaving its positive terminal as i, analyze network A to obtain i, and put the equation in the form vs = ai + b. Then, a = RTH and b = voc. One of the most common pieces of electrical test equip-ment is the DMM, or digital multimeter (Fig. 5.34), which is designed to measure voltage, current, and resis-tance values. In a voltage measurement, two leads from the DMM are connected across the appropriate circuit element, as depicted in Fig. 5.35. The positive reference terminal of the meter is typically marked “V/,” and the negative reference terminal—often referred to as the common terminal—is typically designated by “COM.” The convention is to use a red lead for the positive reference terminal and a black lead for the common terminal. From our discussion of Thévenin and Norton equiva-lents, it may now be apparent that the DMM has its own Thévenin equivalent resistance. This Thévenin equivalent resistance will appear in parallel with our circuit, and its value can affect the measurement (Fig. 5.36). The DMM does not supply power to the circuit to measure voltage, so its Thévenin equivalent consists of only a resistance, which we will name RDMM. The input resistance of a good DMM is typically 10 M or more. The measured voltage V thus appears across 1 k∥10 M = 999.9 . Using voltage division, we ﬁnd that V = 4.4998 volts, slightly less than the ex-pected value of 4.5 volts. Thus, the ﬁnite input resistance of the voltmeter introduces a small error in the measured value. PRACTICAL APPLICATION The Digital Multimeter ■FIGURE 5.34 A handheld digital multimeter. 1 k 9 V 1 k V/ COM DMM 4.500 VDC + – ■FIGURE 5.35 A DMM connected to measure voltage. + – + – 9 V 1 k RDMM V 1 k ■FIGURE 5.36 DMM in Fig. 5.35 shown as its Thévenin equivalent resistance, RDMM. 10 5 30 20i1 i1 + – ■FIGURE 5.39 See Practice Problem 5.9. We could also apply a current source is, let its voltage be v, and then de-termine is = cv −d, where c = 1/RTH and d = isc (the minus sign arises from assuming both current source arrows are directed into the same node). Both of these last two procedures are universally applicable, but some other method can usually be found that is easier and more rapid. Although we are devoting our attention almost entirely to the analysis of linear circuits, it is good to know that Thévenin’s and Norton’s theorems are both valid if network B is nonlinear; only network A must be linear. PRACTICE ● 5.9 Find the Thévenin equivalent for the network of Fig. 5.39. (Hint: Try a 1 V test source.) Ans: Itest = 50 mA so RTH = 20 . To measure current, the DMM must be placed in se-ries with a circuit element, generally requiring that we cut a wire (Fig. 5.37). One DMM lead is connected to the common terminal of the meter, and the other lead is placed in a connector usually marked “A” to signify cur-rent measurement. Again, the DMM does not supply power to the circuit in this type of measurement. We see that the Thévenin equivalent resistance (RDMM) of the DMM is in series with our circuit, so its value can affect the measurement. Writing a simple KVL equation around the loop, −9 + 1000I + RDMMI + 1000I = 0 Note that since we have reconﬁgured the meter to perform a current measurement, the Thévenin equivalent resistance is not the same as when the meter is conﬁg-ured to measure voltages. In fact, we would ideally like RDMM to be 0 for current measurements, and ∞for voltage measurements. If RDMM is now 0.1 , we see that the measured current I is 4.4998 mA, which is only slightly different from the expected value of 4.5 mA. De-pending on the number of digits that can be displayed by 1 k 9 V 1 k A COM DMM 4.500 mA + – I ■FIGURE 5.37 A DMM connected to measure current. V + – IN RDMM R ■FIGURE 5.38 DMM in resistance measurement conﬁguration replaced by its Norton equivalent, showing RDMM in parallel with the unknown resistor R to be measured. the meter, we may not even notice the effect of nonzero DMM resistance on our measurement. The same meter can be used to determine resistance, provided no independent sources are active during the measurement. Internally, a known current is passed through the resistor being measured, and the voltmeter circuitry is used to measure the resulting voltage. Re-placing the DMM with its Norton equivalent (which now includes an active independent current source to gener-ate the predetermined current), we see that RDMM ap-pears in parallel with our unknown resistor R (Fig. 5.38). As a result, the DMM actually measures R∥RDMM. If RDMM = 10 M and R = 10 , Rmeasured = 9.99999 , which is more than accurate enough for most purposes. However, if R = 10 M, Rmeasured = 5 M. The input resistance of a DMM therefore places a practical upper limit on the values of resistance that can be measured, and special techniques must be used to measure larger resistances. We should note that if a digital multimeter is programmed with knowledge of RDMM, it is pos-sible to compensate and allow measurement of larger resistances. 5.4 • MAXIMUM POWER TRANSFER A very useful power theorem may be developed with reference to a practi-cal voltage or current source. For the practical voltage source (Fig. 5.40), the power delivered to the load RL is pL = i2 L RL = v2 s RL (Rs + RL)2 To ﬁnd the value of RL that absorbs maximum power from the given practical source, we differentiate with respect to RL: dpL d RL = (Rs + RL)2v2 s −v2 s RL(2)(Rs + RL) (Rs + RL)4 and equate the derivative to zero, obtaining 2RL(Rs + RL) = (Rs + RL)2 or Rs = RL Since the values RL = 0 and RL = ∞both give a minimum (pL = 0), and since we have already developed the equivalence between practical voltage and current sources, we have therefore proved the following maximum power transfer theorem: An independent voltage source in series with a resistance Rs, or an independent current source in parallel with a resistance Rs, delivers maximum power to a load resistance RL such that RL = Rs. An alternative way to view the maximum power theorem is possible in terms of the Thévenin equivalent resistance of a network: A network delivers maximum power to a load resistance RL when RL is equal to a the Thévenin equivalent resistance of the network. Thus, the maximum power transfer theorem tells us that a 2 resistor draws the greatest power (4.5 W) from either practical source of Fig. 5.16, whereas a resistance of 0.01 receives the maximum power (3.6 kW) in Fig. 5.11. There is a distinct difference between drawing maximum power from a source and delivering maximum power to a load. If the load is sized such that its Thévenin resistance is equal to the Thévenin resistance of the net-work to which it is connected, it will receive maximum power from that network. Any change to the load resistance will reduce the power delivered to the load. However, consider just the Thévenin equivalent of the network itself. We draw the maximum possible power from the voltage source by drawing the maximum possible current—which is achieved by shorting the network terminals! However, in this extreme example we deliver zero power to the “load”—a short circuit in this case—as p = i2R, and we just set R = 0 by shorting the network terminals. A minor amount of algebra applied to Eq. coupled with the maxi-mum power transfer requirement that RL = Rs = RTH will provide pmax\|delivered to load = v2 s 4Rs = v2 TH 4RTH CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 152 + – vs Rs RL vL + – iL ■FIGURE 5.40 A practical voltage source connected to a load resistor RL. SECTION 5.4 MAXIMUM POWER TRANSFER 153 where vTH and RTH recognize that the practical voltage source of Fig. 5.40 can also be viewed as a Thévenin equivalent of some speciﬁc source. It is also not uncommon for the maximum power theorem to be misin-terpreted. It is designed to help us select an optimum load in order to maxi-mize power absorption. If the load resistance is already speciﬁed, however, the maximum power theorem is of no assistance. If for some reason we can affect the size of the Thévenin equivalent resistance of the network con-nected to our load, setting it equal to the load does not guarantee maximum power transfer to our predetermined load. A quick consideration of the power lost in the Thévenin resistance will clarify this point. The circuit shown in Fig. 5.41 is a model for the common-emitter bipolar junction transistor ampliﬁer. Choose a load resistance so that maximum power is transferred to it from the ampliﬁer, and calculate the actual power absorbed. EXAMPLE 5.11 v + – + – 300 5 k 17 k 1 k RL 2.5 sin 440t mV 0.03v ■FIGURE 5.41 A small-signal model of the common-emitter ampliﬁer, with the load resistance unspeciﬁed. v + – voc + – + – 300 5 k 17 k 1 k 2.5 sin 440t mV 0.03v RTH v + – 300 5 k 17 k 1 k 0.03v (a) (b) ■FIGURE 5.42 (a) Circuit with RL removed and independent source short-circuited. (b) Circuit for determining vTH. Since it is the load resistance we are asked to determine, the maximum power theorem applies. The ﬁrst step is to ﬁnd the Thévenin equivalent of the rest of the circuit. We ﬁrst determine the Thévenin equivalent resistance, which requires that we remove RL and short-circuit the independent source as in Fig. 5.42a. (Continued on next page) 5.5 • DELTA-WYE CONVERSION We saw previously that identifying parallel and series combinations of re-sistors can often lead to a signiﬁcant reduction in the complexity of a circuit. In situations where such combinations do not exist, we can often make use of source transformations to enable such simpliﬁcations. There is another useful technique, called -Y (delta-wye) conversion, that arises out of net-work theory. Consider the circuits in Fig. 5.44. There are no series or parallel combi-nations that can be made to further simplify any of the circuits (note that 5.44a and 5.44b are identical, as are 5.44c and 5.44d), and without any CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 154 + – + – Rout 30 V 20 V + – 40 V 2 k 2 k ■FIGURE 5.43 Since vπ = 0, the dependent current source is an open circuit, and RTH 1 k. This can be veriﬁed by connecting an independent 1A current source across the 1 k resistor; vπ will still be zero, so the dependent source remains inactive and hence contributes nothing to RTH. In order to obtain maximum power delivered into the load, RL should be set to RTH = 1 k. To ﬁnd vTH we consider the circuit shown in Fig. 5.42b, which is Fig. 5.41 with RL removed. We may write voc = −0.03vπ(1000) = −30vπ where the voltage vπ may be found from simple voltage division: vπ = (2.5 × 10−3 sin 440t) 3864 300 + 3864 so that our Thévenin equivalent is a voltage −69.6 sin 440t mV in series with 1 k. The maximum power is given by pmax = v2 TH 4RTH = 1.211 sin2 440t μW PRACTICE ● 5.10 Consider the circuit of Fig. 5.43. (a) If Rout 3 k, ﬁnd the power delivered to it. (b) What is the maximum power that can be delivered to any Rout? (c) What two different values of Rout will have exactly 20 mW delivered to them? Ans: 230 mW; 306 mW; 59.2 k and 16.88 . SECTION 5.5 DELTA-WYE CONVERSION 155 sources present, no source transformations can be performed. However, it is possible to convert between these two types of networks. We ﬁrst deﬁne two voltages vac and vbc, and three currents i1, i2, and i3 as depicted in Fig. 5.45. If the two networks are equivalent, then the terminal voltages and currents must be equal (there is no current i2 in the T-connected network). A set of relationships between RA, RB, RC and R1, R2, and R3 can now be deﬁned simply by performing mesh analysis. For example, for the network of Fig. 5.45a we may write RAi1 −RAi2 = vac −RAi1 + (RA + RB + RC)i2 −RCi3 = 0 −RCi2 +RCi3 = −vbc and for the network of Fig. 5.45b we have (R1 + R3)i1 −R3i3 = vac −R3i1 + (R2 + R3)i3 = −vbc We next remove i2 from Eqs. and using Eq. , resulting in RA − R2 A RA + RB + RC i1 − RARC RA + RB + RC i3 = vac and − RARC RA + RB + RC i1 + RC − R2 C RA + RB + RC i3 = −vbc Comparing terms between Eq. and Eq. , we see that R3 = RARC RA + RB + RC In a similar fashion, we may ﬁnd expressions for R1 and R2 in terms of RA, RB, and RC, as well as expressions for RA, RB, and RC in terms of R1, R2, and R3; we leave the remainder of the derivations as an exercise for the reader. Thus, to convert from a Y network to a network, the new resistor values are calculated using RA = R1R2 + R2R3 + R3R1 R2 RB = R1R2 + R2R3 + R3R1 R3 RC = R1R2 + R2R3 + R3R1 R1 a c b d RB RA RC (a) RB a c b d RA RC (b) R1 R2 R3 a c b d (c) R1 R2 R3 a c b d (d) ■FIGURE 5.44 (a) network consisting of three resistors and three unique connections. (b) Same network drawn as a network. (c) A T network consisting of three resistors. (d) Same network drawn as a Y network. RB RA RC (a) vbc + – vac + – i3 i2 i1 R1 R2 R3 (b) i1 i3 vbc + – vac + – ■FIGURE 5.45 (a) Labeled network; (b) labeled T network. EXAMPLE 5.12 and to convert from a network to a Y network, Application of these equations is straightforward, although identifying the actual networks sometimes requires a little concentration. R1 = RARB RA + RB + RC R2 = RB RC RA + RB + RC R3 = RC RA RA + RB + RC CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 156 (a) 3 1 4 2 5 1 2 3 8 3 2 (b) 2 5 1 2 13 2 19 8 (c) 159 71 (d) R2 R1 R3 ■FIGURE 5.46 (a) A given resistive network whose input resistance is desired. (b) The upper network is replaced by an equivalent Y network. (c, d) Series and parallel combinations result in a single resistance value. Each R is 10 Ω Rin ■FIGURE 5.47 Use the technique of -Y conversion to ﬁnd the Thévenin equivalent resistance of the circuit in Fig. 5.46a. We see that the network in Fig. 5.46a is composed of two -connected networks that share the 3 resistor. We must be careful at this point not to be too eager, attempting to convert both -connected networks to two Y-connected networks. The reason for this may be more obvious after we convert the top network consisting of the 1 , 4 , and 3 resistors into a Y-connected network (Fig. 5.46b). Note that in converting the upper network to a Y-connected network, we have removed the 3 resistor. As a result, there is no way to con-vert the original -connected network consisting of the 2 , 5 , and 3 resistors into a Y-connected network. We proceed by combining the 3 8 and 2 resistors and the 3 2 and 5 resistors (Fig. 5.46c). We now have a 19 8 resistor in parallel with a 13 2 resistor, and this parallel combination is in series with the 1 2 resistor. Thus, we can replace the original network of Fig. 5.46a with a single 159 71 resistor (Fig. 5.46d). PRACTICE ● 5.11 Use the technique of Y- conversion to ﬁnd the Thévenin equivalent resistance of the circuit of Fig. 5.47. Ans: 11.43 . SECTION 5.6 SELECTING AN APPROACH: A SUMMARY OF VARIOUS TECHNIQUES 157 5.6 • SELECTING AN APPROACH: A SUMMARY OF VARIOUS TECHNIQUES In Chap. 3, we were introduced to Kirchhoff’s current law (KCL) and Kirchhoff’s voltage law (KVL). These two laws apply to any circuit we will ever encounter, provided that we take care to consider the entire system that the circuits represent. The reason for this is that KCL and KVL enforce charge and energy conservation, respectively, which are fundamental prin-ciples. Based on KCL, we developed the very powerful method of nodal analysis. A similar technique based on KVL (unfortunately only applicable to planar circuits) is known as mesh analysis and is also a useful circuit analysis approach. For the most part, this text is concerned with developing analytical skills that apply to linear circuits. If we know a circuit is constructed of only lin-ear components (in other words, all voltages and currents are related by linear functions), then we can often simplify circuits prior to employing either mesh or nodal analysis. Perhaps the most important result that comes from the knowledge that we are dealing with a completely linear system is that the principle of superposition applies: given a number of independent sources acting on our circuit, we can add the contribution of each source independently of the other sources. This technique is extremely pervasive throughout the ﬁeld of engineering, and we will encounter it often. In many real situations, we will ﬁnd that although several “sources’’ are acting simultaneously on our “system,’’typically one of them dominates the system response. Superposition allows us to quickly identify that source, provided that we have a reasonably accurate linear model of the system. However, from a circuit analysis standpoint, unless we are asked to ﬁnd which independent source contributes the most to a particular response, we ﬁnd that rolling up our sleeves and launching straight into either nodal or mesh analysis is often a more straightforward tactic. The reason for this is that applying superposition to a circuit with 12 independent sources will require us to redraw the original circuit 12 times, and often we will have to apply nodal or mesh analysis to each partial circuit, anyway. The technique of source transformations, on the other hand, is often a very useful tool in circuit analysis. Performing source transformations can allow us to consolidate resistors or sources that are not in series or parallel in the original circuit. Source transformations may also allow us to convert all or at least most of the sources in the original circuit to the same type (either all voltage sources or all current sources), so nodal or mesh analysis is more straightforward. Thévenin’s theorem is extremely important for a number of reasons. In working with electronic circuits, we are always aware of the Thévenin equivalent resistance of different parts of our circuit, especially the input and output resistances of ampliﬁer stages. The reason for this is that match-ing of resistances is frequently the best route to optimizing the performance of a given circuit. We have seen a small preview of this in our discussion of maximum power transfer, where the load resistance should be chosen to match the Thévenin equivalent resistance of the network to which the load is connected. In terms of day-to-day circuit analysis, however, we ﬁnd that converting part of a circuit to its Thévenin or Norton equivalent is almost as much work as analyzing the complete circuit. Therefore, as in the case of superposition, Thévenin’s and Norton’s theorems are typically applied only when we require specialized information about part of our circuit. SUMMARY AND REVIEW Although we asserted in Chap. 4 that nodal analysis and mesh analysis are sufﬁcient to analyze any circuit we might encounter (provided we have the means to relate voltage and current for any passive element, such as Ohm’s law for resistors), the simple truth is that often we do not really need all voltages, or all currents. Sometimes, it is simply one element, or a small portion of a larger circuit, that has our attention. Perhaps there is some un-certainty in the ﬁnal value of a particular element, and we’d like to see how the circuit performs over the range of expected values. In such instances, we can exploit the fact that at the moment we have conﬁned ourselves to linear circuits. This allows the development of other tools: superposition, where individual contributions of sources can be identiﬁed; source transforma-tions, where a voltage source in series with a resistor can be replaced with a current source in parallel with a resistor; and the most powerful of all— Thévenin (and Norton) equivalents. An interesting offshoot of these topics is the idea of maximum power transfer. Assuming we can represent our (arbitrarily complex) circuit by two networks, one passive and one active, maximum power transfer to the passive network is achieved when its Thévenin resistance is equal to the Thévenin resistance of the active network. Finally, we introduced the con-cept of delta-wye conversion, a process that allows us to simplify some re-sistive networks which at face value are not reducible using standard series-parallel combination techniques. We are still faced with the perpetual question, “Which tool should I use to analyze this circuit?” The answer typically lies in the type of information required about our circuit. Experience will eventually guide us a bit, but it is not always true that there is one “best” approach. Certainly one issue to focus on is whether one or more components might be changed—this can suggest whether superposition, a Thévenin equivalent, or a partial simpliﬁ-cation such as can be achieved with source or delta-wye transformation is the most practical route. We conclude this chapter by reviewing key points, along with identify-ing relevant example(s). ❑The principle of superposition states that the response in a linear circuit can be obtained by adding the individual responses caused by the separate independent sources acting alone. (Examples 5.1, 5.2, 5.3) ❑Superposition is most often used when it is necessary to determine the individual contribution of each source to a particular response. (Examples 5.2, 5.3) ❑A practical model for a real voltage source is a resistor in series with an independent voltage source. A practical model for a real current source is a resistor in parallel with an independent current source. ❑Source transformations allow us to convert a practical voltage source into a practical current source, and vice versa. (Example 5.4) CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 158 EXERCISES 159 ❑Repeated source transformations can greatly simplify analysis of a circuit by providing the means to combine resistors and sources. (Example 5.5) ❑The Thévenin equivalent of a network is a resistor in series with an independent voltage source. The Norton equivalent is the same resistor in parallel with an independent current source. (Example 5.6) ❑There are several ways to obtain the Thévenin equivalent resistance, depending on whether or not dependent sources are present in the network. (Examples 5.7, 5.8, 5.9, 5.10) ❑Maximum power transfer occurs when the load resistor matches the Thévenin equivalent resistance of the network to which it is connected. (Example 5.11) ❑When faced with a Δ-connected resistor network, it is straightforward to convert it to a Y-connected network. This can be useful in simplify-ing the network prior to analysis. Conversely, a Y-connected resistor network can be converted to a Δ-connected network to assist in simpliﬁcation of the network. (Example 5.12) READING FURTHER A book about battery technology, including characteristics of built-in resistance: D. Linden, Handbook of Batteries, 2nd ed. New York: McGraw-Hill, 1995. An excellent discussion of pathological cases and various circuit analysis theorems can be found in: R. A. DeCarlo and P. M. Lin, Linear Circuit Analysis, 2nd ed. New York: Oxford University Press, 2001. EXERCISES 5.1 Linearity and Superposition 1. Linear systems are so easy to work with that engineers often construct linear models of real (nonlinear) systems to assist in analysis and design. Such mod-els are often surprisingly accurate over a limited range. For example, consider the simple exponential function ex. The Taylor series representation of this function is ex ≈1 + x + x2 2 + x3 6 + · · · (a) Construct a linear model for this function by truncating the Taylor series expansion after the linear term. (b) Evaluate your model function at x = 0.000001, 0.0001, 0.01, 0.1, and 1.0. (c) For which values of x does your model yield a “reasonable” approximation to ex? Explain your reasoning. 2. Construct a linear approximation to the function y(t) = 4 sin 2t. (a) Evaluate your approximation at t = 0, 0.001, 0.01, 0.1, and 1.0. (b) For which values of t does your model provide a “reasonable” approximation to the actual (nonlinear) function y(t)? Explain your reasoning. 3. Considering the circuit of Fig. 5.48, employ superposition to determine the two components of i8 arising from the action of the two independent sources, respectively. 4. (a) Employ superposition to determine the current labeled i in the circuit of Fig. 5.49. (b) Express the contribution the 1 V source makes to the total current i in terms of a percentage. (c) Changing only the value of the 10 A source, adjust the circuit of Fig. 5.49 so that the two sources contribute equally to the current i. CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 160 6 A 2 V 8 3 + – i8 ■FIGURE 5.48 10 A 1 V 4 9 + – i ■FIGURE 5.49 5. (a) Employ superposition to obtain the individual contributions each of the two sources in Fig. 5.50 makes to the current labeled ix. (b) Adjusting only the value of the rightmost current source, alter the circuit so that the two sources contribute equally to ix. 6. (a) Determine the individual contributions of each of the two current sources in the circuit of Fig. 5.51 to the nodal voltage v1. (b) Determine the percentage contribution of each of the two sources to the power dissipated by the 2 resistor. ■FIGURE 5.50 5 A 3 A 5 12 2 5 ix v1 v2 1 5 5 2 7 A 4 A ■FIGURE 5.51 EXERCISES 161 7. (a) Determine the individual contributions of each of the two current sources shown in Fig. 5.52 to the nodal voltage labeled v2. (b) Instead of performing two separate PSpice simulations, verify your answer by using a single dc sweep. Submit a labeled schematic, relevant Probe output, and a short descrip-tion of the results. v1 v2 7 1 4 5 2 7 A 2 A ■FIGURE 5.52 8. After studying the circuit of Fig. 5.53, change both voltage source values such that (a) i1 doubles; (b) the direction of i1 reverses, but its magnitude is un-changed; (c) both sources contribute equally to the power dissipated by the 6 resistor. 4 V 10 V 6 4 3 i1 + – + – ■FIGURE 5.53 9. Consider the three circuits shown in Fig. 5.54.Analyze each circuit, and demon-strate that Vx = V ′ x + V ′′ x (i.e., superposition is most useful when sources are set to zero, but the principle is in fact much more general than that). + – + – Vx + – 12 V –15 V 3 k 1 k 2 k + – + – Vx + – 6 V –10 V 3 k 1 k 2 k + – + – Vx + – 6 V –5 V 3 k 1 k 2 k ' " ■FIGURE 5.54 10. (a) Using superposition, determine the voltage labeled vx in the circuit repre-sented in Fig. 5.55. (b) To what value should the 2 A source be changed to re-duce vx by 10%? (c) Verify your answers by performing three dc sweeps in PSpice (one for each source). Submit a labeled schematic, relevant Probe out-put, and a short description of the results. 11. Employ superposition principles to obtain a value for the current Ix as labeled in Fig. 5.56. CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 162 + – + – 5 1 3 2 A 4 V 4 V 2 vx + – ■FIGURE 5.55 + – – + 7 k 2 k 2 A 5 k 1 V 0.2Ix Ix ■FIGURE 5.56 12. (a) Employ superposition to determine the individual contribution from each independent source to the voltage v as labeled in the circuit shown in Fig. 5.57. (b) Compute the power absorbed by the 2 resistor. 2 3 4 A i1 v + – 0.4i1 1 7 6 A + – ■FIGURE 5.57 6 V 4 10 + – 6 A 4 10 ix 4 1 5 1 + – 2ix ■FIGURE 5.58 5.2 Source Transformations 13. Perform an appropriate source transformation on each of the circuits depicted in Fig. 5.58, taking care to retain the 4 resistor in each ﬁnal circuit. EXERCISES 163 14. For the circuit of Fig. 5.59, plot iL versus vL corresponding to the range of 0 ≤R ≤∞. 15. Determine the current labeled I in the circuit of Fig. 5.60 by ﬁrst performing source transformations and parallel-series combinations as required to reduce the circuit to only two elements. vL + – + – iL 5 k R 3 V ■FIGURE 5.59 16. Verify that the power absorbed by the 7 resistor in Fig. 5.22a remains the same after the source transformation illustrated in Fig. 5.22c. 17. (a) Determine the current labeled i in the circuit of Fig. 5.61 after ﬁrst trans-forming the circuit such that it contains only resistors and voltage sources. (b) Simulate each circuit to verify the same current ﬂows in both cases. + – I 3 A 9 V 7 5 4 ■FIGURE 5.60 2 M 12 V 3 M 13 M 7 V + – + – 5 A i ■FIGURE 5.61 18. (a) Using repeated source transformations, reduce the circuit of Fig. 5.62 to a voltage source in series with a resistor, both of which are in series with the 6 M resistor. (b) Calculate the power dissipated by the 6 M resistor using your simpliﬁed circuit. + – 15 V 1.2 M 750 k 7 M 6 M 3.5 M 27 mA ■FIGURE 5.62 19. (a) Using as many source transformations and element combination techniques as required, simplify the circuit of Fig. 5.63 so that it contains only the 7 V source, a single resistor, and one other voltage source. (b) Verify that the 7 V source delivers the same amount of power in both circuits. – + 5 A 7 V 2 A 3 1 3 ■FIGURE 5.63 20. (a) Making use of repeated source transformations, reduce the circuit of Fig. 5.64 such that it contains a single voltage source, the 17 resistor, and one other resis-tor. (b) Calculate the power dissipated by the 17 resistor. (c) Verify your results by simulating both circuits with PSpice or another suitable CAD tool. CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 164 + – 47 10 7 22 7 9 17 12 V IX 2 ■FIGURE 5.64 21. Make use of source transformations to ﬁrst convert all three sources in Fig. 5.65 to voltage sources, then simplify the circuit as much as possible and calculate the voltage Vx which appears across the 4 resistor. Be sure to draw and label your simpliﬁed circuit. Vx + – 1 7 9 10 2 4 3 A 9 A 5Vx 10 ■FIGURE 5.65 22. (a) With the assistance of source transformations, alter the circuit of Fig. 5.66 such that it contains only current sources. (b) Simplify your new circuit as much as possible, and calculate the power dissipated in the 7 resistor. (c) Verify your solution by simulating both circuits with PSpice or another appropriate CAD tool. – + 7 11 10 9 V 4I1 2 A I1 + – ■FIGURE 5.66 23. Transform the dependent source in Fig. 5.67 to a voltage source, then calculate V0. + – V0 + – V1 + – 0.7 V 6 6 7 2 12V1 ■FIGURE 5.67 EXERCISES 165 24. With regard to the circuit represented in Fig. 5.68, ﬁrst transform both voltage sources to current sources, reduce the number of elements as much as possible, and determine the voltage v3. v3 + – 2 V 4v3 6 2v3 3 2 + – + – ■FIGURE 5.68 + – vL + – RL 9 V 3 1 2 ■FIGURE 5.69 0.8 2 5 RL 1 A 5 iL ■FIGURE 5.70 Voc + – 2.5 k 2.5 k 2.3 k 1.8 k 4.2 V 1.1 k + – ■FIGURE 5.71 31. (a) Employ Thévenin’s theorem to obtain a two-component equivalent for the network shown in Fig. 5.73. (b) Determine the power supplied to a 1 M resistor connected to the network if i1 = 19 μA, R1 = R2 = 1.6 M, R2 = 3 M, and R4 = R5 = 1.2 M. (c) Verify your solution by simulating both circuits with PSpice or another appropriate CAD tool. 75 220 122 45 0.7 V + – 0.3 A ■FIGURE 5.72 R3 R5 R4 R2 R1 i1 ■FIGURE 5.73 5.3 Thévenin and Norton Equivalent Circuits 25. Referring to Fig. 5.69, determine the Thévenin equivalent of the network con-nected to RL. (b) Determine vL for RL = 1 , 3.5 , 6.257 , and 9.8 . 26. (a) With respect to the circuit depicted in Fig. 5.69, obtain the Norton equiva-lent of the network connected to RL. (b) Plot the power dissipated in resistor RL as a function of iL corresponding to the range of 0 < RL < 5 . (c) Using your graph, estimate at what value of RL does the dissipated power reach its maximum value. 27. (a) Obtain the Norton equivalent of the network connected to RL in Fig. 5.70. (b) Obtain the Thévenin equivalent of the same network. (c) Use either to calculate iL for RL = 0 , 1 , 4.923 , and 8.107 . 28. (a) Determine the Thévenin equivalent of the circuit depicted in Fig. 5.71 by ﬁrst ﬁnding Voc and Isc (deﬁned as ﬂowing into the positive reference terminal of Voc). (b) Connect a 4.7 k resistor to the open terminals of your new network and calculate the power it dissipates. 29. Referring to the circuit of Fig. 5.71: (a) Determine the Norton equivalent of the circuit by ﬁrst ﬁnding Voc and Isc (deﬁned as ﬂowing into the positive reference terminal of Voc). (b) Connect a 1.7 k resistor to the open terminals of your new network and calculate the power supplied to that resistor. 30. (a) Employ Thévenin’s theorem to obtain a simple two-component equivalent of the circuit shown in Fig. 5.72. (b) Use your equivalent circuit to determine the power delivered to a 100 resistor connected to the open terminals. (c) Verify your solution by analyzing the original circuit with the same 100 resistor connected across the open terminals. 32. Determine the Thévenin equivalent of the network shown in Fig. 5.74 as seen looking into the two open terminals. CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 166 2 A 1 5 3 2 2 V 4 V + – + – vx + – ■FIGURE 5.74 33. (a) Determine the Norton equivalent of the circuit depicted in Fig. 5.74 as seen looking into the two open terminals. (b) Compute power dissipated in a 5 resistor connected in parallel with the existing 5 resistor. (c) Compute the current ﬂowing through a short circuit connecting the two terminals. 34. For the circuit of Fig. 5.75: (a) Employ Norton’s theorem to reduce the net-work connected to RL to only two components. (b) Calculate the downward-directed current ﬂowing through RL if it is a 3.3 k resistor. (c) Verify your answer by simulating both circuits with PSpice or a comparable CAD tool. – + 2.5 V 1 k 7 k 6 k RL 5 k 300 mA ■FIGURE 5.75 35. (a) Obtain a value for the Thévenin equivalent resistance seen looking into the open terminals of the circuit in Fig. 5.76 by ﬁrst ﬁnding Voc and Isc. (b) Con-nect a 1 A test source to the open terminals of the original circuit after shorting the voltage source, and use this to obtain RTH. (c) Connect a 1 V test source to the open terminals of the original circuit after again zeroing the 2 V source, and use this now to obtain RTH. – + 10 20 30 7 7 2 V ■FIGURE 5.76 36. Refer to the circuit depicted in Fig. 5.77. (a) Obtain a value for the Thévenin equivalent resistance seen looking into the open terminals by ﬁrst ﬁnding Voc and Isc. (b) Connect a 1 A test source to the open terminals of the original 1 3 2 4 1 A ■FIGURE 5.77 EXERCISES 167 circuit after deactivating the other current source, and use this to obtain RTH. (c) Connect a 1 V test source to the open terminals of the original circuit, once again zeroing out the original source, and use this now to obtain RTH. 37. Obtain a value for the Thévenin equivalent resistance seen looking into the open terminals of the circuit in Fig. 5.78 by (a) ﬁnding Voc and Isc, and then taking their ratio; (b) setting all independent sources to zero and using resistor combination techniques; (c) connecting an unknown current source to the terminals, deactivating (zero out) all other sources, ﬁnding an algebraic expres-sion for the voltage that develops across the source, and taking the ratio of the two quantities. + – 17 9 6 4 2 222 A 20 V 33 A ■FIGURE 5.78 38. With regard to the network depicted in Fig. 5.79, determine the Thévenin equivalent as seen by an element connected to terminals (a) a and b; (b) a and c; (c) b and c. (d) Verify your answers using PSpice or other suitable CAD tool. (Hint: Connect a test source to the terminals of interest.) a b c 11 4 10 21 2 12 ■FIGURE 5.79 39. Determine the Thévenin and Norton equivalents of the circuit represented in Fig. 5.80 from the perspective of the open terminals. (There should be no dependent sources in your answer.) 40. Determine the Norton equivalent of the circuit drawn in Fig. 5.81 as seen by terminals a and b. (There should be no dependent sources in your answer.) 41. With regard to the circuit of Fig. 5.82, determine the power dissipated by (a) a 1 k resistor connected between a and b; (b) a 4.7 k resistor connected between a and b; (c) a 10.54 k resistor connected between a and b. Vx + – 21 10Vx ■FIGURE 5.80 ix 700 mA 500 1.5 k a b – + 2ix 2500 ■FIGURE 5.81 1 V 10 k 20 k 0.02v1 a b v1 + – + – ■FIGURE 5.82 42. Determine the Thévenin and Norton equivalents of the circuit shown in Fig. 5.83, as seen by an unspeciﬁed element connected between terminals a and b. 43. Referring to the circuit of Fig. 5.84, determine the Thévenin equivalent resis-tance of the circuit to the right of the dashed line. This circuit is a common-source transistor ampliﬁer, and you are calculating its input resistance. CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 168 11 15 20 + – vab + – 0.5vab 0.11vab a b ■FIGURE 5.83 1 M 3 k RL 0.12vgs vs 300 vgs + – + – ■FIGURE 5.84 44. Referring to the circuit of Fig. 5.85, determine the Thévenin equivalent resistance of the circuit to the right of the dashed line. This circuit is a common-collector transistor ampliﬁer, and you are calculating its input resistance. ■FIGURE 5.85 45. The circuit shown in Fig. 5.86 is a reasonably accurate model of an operational ampliﬁer. In cases where Ri and A are very large and Ro ∼0, a resistive load (such as a speaker) connected between ground and the terminal labeled vout will see a voltage −Rf /R1 times larger than the input signal vin. Find the Thévenin equivalent of the circuit, taking care to label vout. ■FIGURE 5.86 5.4 Maximum Power Transfer 46. (a) For the simple circuit of Fig. 5.87, graph the power dissipated by the resis-tor R as a function of R/RS, if 0 ≤R ≤3000 . (b) Graph the ﬁrst derivative of the power versus R/RS, and verify that maximum power is transferred to R when it is equal to RS. vd – + Avd Rf vout Ro Ri R1 + – + – vin 2 M 1 k 2 k 0.02V 300 r vs v + – + – R RS 12 V 1 k + – ■FIGURE 5.87 EXERCISES 169 47. For the circuit drawn in Fig. 5.88, (a) determine the Thévenin equivalent con-nected to Rout. (b) Choose Rout such that maximum power is delivered to it. 48. Study the circuit of Fig. 5.89. (a) Determine the Norton equivalent connected to resistor Rout. (b) Select a value for Rout such that maximum power will be delivered to it. + – + – Rout 2 V 4 V 3 2 ■FIGURE 5.88 2 V 4 A 1 k 2 k 3 V + – Rout + – ■FIGURE 5.89 49. Assuming that we can determine the Thévenin equivalent resistance of our wall socket, why don’t toaster, microwave oven, and TV manufacturers match each appliance’s Thévenin equivalent resistance to this value? Wouldn’t it per-mit maximum power transfer from the utility company to our household appliances? 50. For the circuit of Fig. 5.90, what value of RL will ensure it absorbs the maxi-mum possible amount of power? – + 1 A 3 V 3 2 5 RL ■FIGURE 5.90 9 A 2 A 9 5 3 ■FIGURE 5.91 + – V2 + – 5 V 3.3 7 2 0.1V2 + – ■FIGURE 5.92 51. With reference to the circuit of Fig. 5.91, (a) calculate the power absorbed by the 9 resistor; (b) adjust the size of the 5 resistor so that the new network delivers maximum power to the 9 resistor. 52. Referring to the circuit of Fig. 5.92, (a) determine the power absorbed by the 3.3 resistor; (b) replace the 3.3 resistor with another resistor such that it absorbs maximum power from the rest of the circuit. 53. Select a value for RL in Fig. 5.93 such that it is ensured to absorb maximum power from the circuit. CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 170 10 8 RL 5 0.2v1 + – 4 V v1 + – ■FIGURE 5.93 54. Determine what value of resistance would absorb maximum power from the circuit of Fig. 5.94 when connected across terminals a and b. a b 20 50 10 100 900 mA 0.1vab vab 2vab + – + – ■FIGURE 5.94 5.5 Delta-Wye Conversion 55. Derive the equations required to convert from a Y-connected network to a -connected network. 56. Convert the - (or “-”) connected networks in Fig. 5.95 to Y-connected networks. a c b d a c b d 33 21 17 1.1 k 2.1 k 4.7 k ■FIGURE 5.95 57. Convert the Y- (or “T-”) connected networks in Fig. 5.96 to -connected networks. 33 21 17 a c b d 1.3 k 2.1 k 4.7 k a c b d ■FIGURE 5.96 58. For the network of Fig. 5.97, select a value of R such that the network has an equivalent resistance of 9 . Round your answer to two signiﬁcant ﬁgures. R 30 10 2 3 ■FIGURE 5.97 EXERCISES 171 59. For the network of Fig. 5.98, select a value of R such that the network has an equivalent resistance of 70.6 . 200 100 R 42 68 ■FIGURE 5.98 Each R is 2.2 k Rin ■FIGURE 5.99 61 25 55 46 23 11 31 31 110 63 Rin ■FIGURE 5.100 60. Determine the effective resistance Rin of the network exhibited in Fig. 5.99. 61. Calculate Rin as indicated in Fig. 5.100. 62. Employ /Y conversion techniques as appropriate to determine Rin as labeled in Fig. 5.101. 6 7 6 3 9 20 12 10 4 5 Rin ■FIGURE 5.101 63. (a) Determine the two-component Thévenin equivalent of the network in Fig. 5.102. (b) Calculate the power dissipated by a 1 resistor connected between the open terminals. CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 172 – + 9 V 2 1 11 12 22 10 ■FIGURE 5.102 65. (a) Replace the network in Fig. 5.104 with an equivalent three-resistor network. (b) Perform a PSpice analysis to verify that your answer is in fact equivalent. (Hint: Try adding a load resistor.) 2 2 3 1 1 1 1 2 2 ■FIGURE 5.104 5.6 Selecting an Approach: A Summary of Various Techniques 66. Determine the power absorbed by a resistor connected between the open termi-nal of the circuit shown in Fig. 5.105 if it has a value of (a) 1 ; (b) 100 ; (c) 2.65 k; (d) 1.13 M. 4 mA 7 k 1 k 2.2 k 10 k 10 k 4 k 5 k ■FIGURE 5.105 67. It is known that a load resistor of some type will be connected between termi-nals a and b of the network of Fig. 5.106. (a) Change the value of the 25 V source such that both voltage sources contribute equally to the power delivered to the load resistor, assuming its value is chosen such that it absorbs maximum power. (b) Calculate the value of the load resistor. 10 15 + – 5 + – 10 V 25 V a b ■FIGURE 5.106 64. (a) Use appropriate techniques to obtain both the Thévenin and Norton equiva-lents of the network drawn in Fig. 5.103. (b) Verify your answers by simulating each of the three circuits connected to a 1 resistor. 8 A 2 3 6 4 ■FIGURE 5.103 EXERCISES 173 68. A 2.57 load is connected between terminals a and b of the network drawn in Fig. 5.106. Unfortunately, the power delivered to the load is only 50% of the required amount. Altering only voltage sources, modify the circuit so that the required power is delivered and both sources contribute equally. 69. A load resistor is connected across the open terminals of the circuit shown in Fig. 5.107, and its value was chosen carefully to ensure maximum power trans-fer from the rest of the circuit. (a) What is the value of the resistor? (b) If the power absorbed by the load resistor is three times as large as required, modify the circuit so that it performs as desired, without losing the maximum power transfer condition already enjoyed. 5.4 1.8 3 5 0.8 A 0.1 A 1.2 A ■FIGURE 5.107 70. A backup is required for the circuit depicted in Fig. 5.107. It is unknown what will be connected to the open terminals, or whether it will be purely linear. If a simple battery is to be used, what no-load (“open circuit”) voltage should it have, and what is the maximum tolerable internal resistance? Chapter-Integrating Exercises 71. Three 45 W light bulbs originally wired in a Y network conﬁguration with a 120 V ac source connected across each port are rewired as a network. The neutral, or center, connection is not used. If the intensity of each light is pro-portional to the power it draws, design a new 120 V ac power circuit so that the three lights have the same intensity in the conﬁguration as they did when connected in a Y conﬁguration. Verify your design using PSpice by comparing the power drawn by each light in your circuit (modeled as an appropriately chosen resistor value) with the power each would draw in the original Y-connected circuit. 72. (a) Explain in general terms how source transformation can be used to simplify a circuit prior to analysis. (b) Even if source transformations can greatly sim-plify a particular circuit, when might it not be worth the effort? (c) Multiplying all the independent sources in a circuit by the same scaling factor results in all other voltages and currents being scaled by the same amount. Explain why we don’t scale the dependent sources as well. (d) In a general circuit, if we set an independent voltage source to zero, what current can ﬂow through it? (e) In a general circuit, if we set an independent current source to zero, what voltage can be sustained across its terminals? 73. The load resistor in Fig. 5.108 can safely dissipate up to 1 W before overheating and bursting into ﬂame. The lamp can be treated as a 10.6 resistor if less than 1 A ﬂows through it and a 15 resistor if more than 1 A ﬂows through it. What is the maximum permissible value of Is? Verify your answer with PSpice. ■FIGURE 5.108 Vx Indicator lamp 200 200 1 k Is RL = Load Resistor 5Vx + – + – 74. A certain red LED has a maximum current rating of 35 mA, and if this value is exceeded, overheating and catastrophic failure will result. The resistance of the LED is a nonlinear function of its current, but the manufacturer war-rants a minimum resistance of 47 and a maximum resistance of 117 . Only 9 V batteries are available to power the LED. Design a suitable circuit to deliver the maximum power possible to the LED without damaging it. Use only combinations of the standard resistor values given in the inside front cover. 75. As part of a security system, a very thin 100 wire is attached to a window using nonconducting epoxy. Given only a box of 12 rechargeable 1.5 V AAA batteries, one thousand 1 resistors, and a 2900 Hz piezo buzzer that draws 15 mA at 6 V, design a circuit with no moving parts that will set off the buzzer if the window is broken (and hence the thin wire as well). Note that the buzzer requires a dc voltage of at least 6 V (maximum 28 V) to operate. CHAPTER 5 HANDY CIRCUIT ANALYSIS TECHNIQUES 174 INTRODUCTION At this point we have a good set of circuit analysis tools at our dis-posal, but have focused primarily on somewhat general circuits composed of only sources and resistors. In this chapter, we intro-duce a new component which, although technically nonlinear, can be treated effectively with linear models. This element, known as the operational ampliﬁer or op amp for short, ﬁnds daily usage in a large variety of electronic applications. It also provides us a new element to use in building circuits, and another opportunity to test out our developing analytical skills. 6.1 • BACKGROUND The origins of the operational ampliﬁer date to the 1940s, when basic circuits were constructed using vacuum tubes to perform mathematical operations such as addition, subtraction, multiplica-tion, division, differentiation, and integration. This enabled the con-struction of analog (as opposed to digital) computers tasked with the solution of complex differential equations. The ﬁrst commercially available op amp device is generally considered to be the K2-W, manufactured by Philbrick Researches, Inc. of Boston from about 1952 through the early 1970s (Fig. 6.1a). These early vacuum tube devices weighed 3 oz (85 g), measured 133/ 64 in × 29/ 64 in × 47/ 64 in (3.8 cm × 5.4 cm × 10.4 cm), and sold for about US$22. In contrast, integrated circuit (IC) op amps such as the Fairchild KA741 weigh less than 500 mg, measure 5.7 mm × 4.9 mm × 1.8 mm, and sell for approximately US$0.22. Compared to op amps based on vacuum tubes, modern IC op amps are constructed using perhaps 25 or more transistors all on the same silicon “chip,” as well as resistors and capacitors needed to ob-tain the desired performance characteristics. As a result, they run at KEY CONCEPTS Characteristics of Ideal Op Amps Inverting and Noninverting Ampliﬁers Summing and Difference Ampliﬁer Circuits Cascaded Op Amp Stages Using Op Amps to Build Voltage and Current Sources Nonideal Characteristics of Op Amps Voltage Gain and Feedback Basic Comparator and Instrumentation Ampliﬁer Circuits The Operational Ampliﬁer C H A P T E R 6 175 CHAPTER 6 THE OPERATIONAL AMPLIFIER 176 much lower dc supply voltages (±18 V, for example, as opposed to ±300 V for the K2-W), are more reliable, and considerably smaller (Fig. 6.1b,c). In some cases, the IC may contain several op amps. In addition to the output pin and the two inputs, other pins enable power to be supplied to run the transis-tors, and for external adjustments to be made to balance and compensate the op amp. The symbol commonly used for an op amp is shown in Fig. 6.2a. At this point, we are not concerned with the internal circuitry of the op amp or the IC, but only with the voltage and current relationships that exist between the input and output terminals. Thus, for the time being we will use a simpler elec-trical symbol, shown in Fig. 6.2b. Two input terminals are shown on the left, and a single output terminal appears at the right. The terminal marked by a “+” is referred to as the noninverting input, and the “−” marked terminal is called the inverting input. 6.2 • THE IDEAL OP AMP: A CORDIAL INTRODUCTION In practice, we ﬁnd that most op amps perform so well that we can often make the assumption that we are dealing with an “ideal” op amp. The char-acteristics of an ideal op amp form the basis for two fundamental rules that at ﬁrst may seem somewhat unusual: Ideal Op Amp Rules 1. No current ever ﬂows into either input terminal. 2. There is no voltage difference between the two input terminals. In a real op amp, a very small leakage current will ﬂow into the input (sometimes as low as 40 femtoamperes). It is also possible to obtain a very small voltage across the two input terminals. However, compared to other voltages and currents in most circuits, such values are so small that includ-ing them in the analysis does not typically affect our calculations. When analyzing op amp circuits, we should keep one other point in mind. As opposed to the circuits that we have studied so far, an op amp circuit al-ways has an output that depends on some type of input. Therefore, we will an-alyze op amp circuits with the goal of obtaining an expression for the output in terms of the input quantities. We will ﬁnd that it is usually a good idea to begin the analysis of an op amp circuit at the input, and proceed from there. (b) (a) (c) ■FIGURE 6.1 (a) A Philbrick K2-W op amp, based on a matched pair of 12AX7A vacuum tubes. (b) LMV321 op amp, used in a variety of phone and game applications. (c) LMC6035 operational ampliﬁer, which packs 114 transistors into a package so small that it ﬁts on the head of a pin. (b–c) Copyright © 2011 National Semiconductor Corporation (www.national.com). All rights reserved. Used with permission. – + Offset null V – V + Offset null Input (a) Output – + (b) ■FIGURE 6.2 (a) Electrical symbol for the op amp. (b) Minimum required connections to be shown on a circuit schematic. SECTION 6.2 THE IDEAL OP AMP: A CORDIAL INTRODUCTION 177 The circuit shown in Fig. 6.3 is known as an inverting ampliﬁer. We choose to analyze this circuit using KVL, beginning with the input voltage source. The current labeled i ﬂows only through the two resistors R1 and Rf; ideal op amp rule 1 states that no current ﬂows into the inverting input terminal. Thus, we can write −vin + R1i + Rf i + vout = 0 which can be rearranged to obtain an equation that relates the output to the input: vout = vin −(R1 + Rf )i Given vin = 5 sin 3t mV, R1 4.7 k, and Rf 47 k, we require one additional equation that expresses i only in terms of vout, vin, R1, and/or Rf. This is a good time to mention that we have not yet made use of ideal op amp rule 2. Since the noninverting input is grounded, it is at zero volts. By ideal op amp rule 2, the inverting input is therefore also at zero volts! This does not mean that the two inputs are physically shorted together, and we should be careful not to make such an assumption. Rather, the two input voltages simply track each other: if we try to change the voltage at one pin, the other pin will be driven by internal circuitry to the same value. Thus, we can write one more KVL equation: −vin + R1i + 0 = 0 or i = vin R1 Combining Eq. with Eq. , we obtain an expression for vout in terms of vin: vout = −Rf R1 vin Substituting vin = 5 sin 3t mV, R1 = 4.7 k, and Rf = 47 k, vout = −50 sin 3t mV Since Rf > R1, this circuit ampliﬁes the input voltage signal vin. If we choose Rf < R1, the signal will be attenuated instead. We also note that the output voltage has the opposite sign of the input voltage,1 hence the name “inverting ampliﬁer.” The output is sketched in Fig. 6.4, along with the in-put waveform for comparison. At this point, it is worth mentioning that the ideal op amp seems to be violating KCL. Speciﬁcally, in the above circuit no current ﬂows into or out of either input terminal, but somehow current is able to ﬂow into the output pin! This would imply that the op amp is somehow able to either create elec-trons out of nowhere or store them forever (depending on the direction of current ﬂow). Obviously, this is not possible. The conﬂict arises because we have been treating the op amp the same way we treated passive elements – + i i vout + – R1 Rf + – vin ■FIGURE 6.3 An op amp used to construct an inverting ampliﬁer circuit. The current i ﬂows to ground through the output pin of the op amp. The fact that the inverting input terminal ﬁnds itself at zero volts in this type of circuit conﬁguration leads to what is often referred to as a “virtual ground.” This does not mean that the pin is actually grounded, which is sometimes a source of confusion for students. The op amp makes whatever internal adjustments are necessary to prevent a voltage difference between the input terminals. The input terminals are not shorted together. (1) Or, “the output is 180◦out of phase with the input,” which sounds more impressive. –60 –40 –20 0 20 40 60 vin vin vin vout vout Voltage (mV) 1 2 3 4 5 6 7 t (s) ■FIGURE 6.4 Input and output waveforms of the inverting ampliﬁer circuit. such as the resistor. In reality, however, the op amp cannot function unless it is connected to external power sources. It is through those power sources that we can direct current ﬂow through the output terminal. Although we have shown that the inverting ampliﬁer circuit of Fig. 6.3 can amplify an ac signal (a sine wave in this case having a frequency of 3 rad/s and an amplitude of 5 mV), it works just as well with dc inputs. We consider this type of situation in Fig. 6.5, where values for R1 and Rf are to be selected to obtain an output voltage of −10 V. This is the same circuit as shown in Fig. 6.3, but with a 2.5 V dc input. Since no other change has been made, the expression we presented as Eq. is valid for this circuit as well. To obtain the desired output, we seek a ratio of Rf to R1 of 10/2.5, or 4. Since it is only the ratio that is important here, we simply need to pick a convenient value for one resistor, and the other resis-tor value is then ﬁxed at the same time. For example, we could choose R1 = 100 (so Rf = 400 ), or even Rf = 8 M (so R1 = 2 M). In practice, other constraints (such as bias current) may limit our choices. This circuit conﬁguration therefore acts as a convenient type of voltage ampliﬁer (or attenuator, if the ratio of Rf to R1 is less than 1), but does have the sometimes inconvenient property of inverting the sign of the input. There is an alternative, however, which is analyzed just as easily—the non-inverting amplifier shown in Fig. 6.6. We examine such a circuit in the following example. CHAPTER 6 THE OPERATIONAL AMPLIFIER 178 – + vout + – 2.5 V R1 Rf + – ■FIGURE 6.5 An inverting ampliﬁer circuit with a 2.5 V input. EXAMPLE 6.1 Sketch the output waveform of the noninverting ampliﬁer circuit in Fig. 6.6a. Use vin 5 sin 3t mV, R1 4.7 kΩ, and Rf 47 kΩ. Identify the goal of the problem. We require an expression for vout that only depends on the known quantities vin, R1, and Rf. Collect the known information. Since values have been speciﬁed for the resistors and the input waveform, we begin by labeling the current i and the two input voltages as shown in Fig. 6.6b. We will assume that the op amp is an ideal op amp. Devise a plan. Although mesh analysis is a favorite technique of students, it turns out to be more practical in most op amp circuits to apply nodal analysis, since there is no direct way to determine the current ﬂowing out of the op amp output. Construct an appropriate set of equations. Note that we are using ideal op amp rule 1 implicitly by deﬁning the same current through both resistors: no current ﬂows into the invert-ing input terminal. Employing nodal analysis to obtain our expression for vout in terms of vin, we thus ﬁnd that – + vout + – R1 Rf vin + – (a) – + i i vout + – R1 Rf vin va vb + – (b) ■FIGURE 6.6 (a) An op amp used to construct a noninverting ampliﬁer circuit. (b) Circuit with the current through R1 and Rf deﬁned, as well as both input voltages labeled. SECTION 6.2 THE IDEAL OP AMP: A CORDIAL INTRODUCTION 179 Just like the inverting ampliﬁer, the noninverting ampliﬁer works with dc as well as ac inputs, but has a voltage gain of vout/vin = 1 + (Rf /R1). Thus, if we set Rf = 9 and R1 = 1 , we obtain an output vout which is 10 times larger than the input voltage vin. In contrast to the inverting ampli-ﬁer, the output and input of the noninverting ampliﬁer always have the same sign, and the output voltage cannot be less than the input; the minimum gain is 1. Which ampliﬁer we choose depends on the application we are consid-ering. In the special case of the voltage follower circuit shown in Fig. 6.8, –60 –40 –20 0 20 40 60 vin vin vin vout vout Voltage (mV) 1 2 3 4 5 6 7 t (s) ■FIGURE 6.7 Input and output waveforms for the noninverting ampliﬁer circuit. ■FIGURE 6.8 At node a: 0 = va R1 + va −vout Rf At node b: vb = vin Determine if additional information is required. Our goal is to obtain a single expression that relates the input and output voltages, although neither Eq. nor Eq. appears to do so. However, we have not yet employed ideal op amp rule 2, and we will ﬁnd that in almost every op amp circuit both rules need to be invoked in order to obtain such an expression. Thus, we recognize that va = vb = vin, and Eq. becomes 0 = vin R1 + vin −vout Rf Attempt a solution. Rearranging, we obtain an expression for the output voltage in terms of the input voltage vin: vout = 1 + Rf R1 vin = 11vin = 55 sin 3t mV Verify the solution. Is it reasonable or expected? The output waveform is sketched in Fig. 6.7, along with the input waveform for comparison. In contrast to the output waveform of the inverting ampliﬁer circuit, we note that the input and output are in phase for the noninverting ampliﬁer. This should not be entirely unexpected: it is implicit in the name “noninverting ampliﬁer.” PRACTICE ● 6.1 Derive an expression for vout in terms of vin for the circuit shown in Fig. 6.8. Ans: vout = vin. The circuit is known as a “voltage follower,” since the output voltage tracks or “follows” the input voltage. – + vout + – vin + – RL Rin which represents a noninverting ampliﬁer with R1 set to ∞and Rf set to zero, the output is identical to the input in both sign and magnitude. This may seem rather pointless as a general type of circuit, but we should keep in mind that the voltage follower draws no current from the input (in the ideal case)—it therefore can act as a buffer between the voltage vin and some re-sistive load RL connected to the output of the op amp. We mentioned earlier that the name “operational ampliﬁer” originates from using such devices to perform arithmetical operations on analog (i.e., nondigitized, real-time, real-world) signals. As we see in the following two circuits, this includes both addition and subtraction of input voltage signals. CHAPTER 6 THE OPERATIONAL AMPLIFIER 180 Obtain an expression for vout in terms of v1, v2, and v3 for the op amp circuit in Fig. 6.9, also known as a summing ampliﬁer. EXAMPLE 6.2 We ﬁrst note that this circuit is similar to the inverting ampliﬁer circuit of Fig. 6.3. Again, the goal is to obtain an expression for vout (which in this case appears across a load resistor RL) in terms of the inputs (v1, v2, and v3). Since no current can ﬂow into the inverting input terminal, we can write i = i1 + i2 + i3 Therefore, we can write the following equation at the node labeled va: 0 = va −vout Rf + va −v1 R + va −v2 R + va −v3 R This equation contains both vout and the input voltages, but unfortu-nately it also contains the nodal voltage va. To remove this unknown quantity from our expression, we need to write an additional equation that relates va to vout, the input voltages, Rf , and/or R. At this point, we remember that we have not yet used ideal op amp rule 2, and that we will almost certainly require the use of both rules when analyzing an op amp circuit. Thus, since va = vb = 0, we can write the following: 0 = vout Rf + v1 R + v2 R + v3 R – + i vout + – R R RL R v1 va vb Rf i3 i2 i1 + – v2 + – v3 + – ■FIGURE 6.9 Basic summing ampliﬁer circuit with three inputs. SECTION 6.2 THE IDEAL OP AMP: A CORDIAL INTRODUCTION 181 – + i vout + – R RL R R v1 va vb R i2 i1 + – v2 + – ■FIGURE 6.10 Ans: vout = v2 −v1. Hint: Use voltage division to obtain vb. Rearranging, we obtain the following expression for vout: vout = −Rf R (v1 + v2 + v3) In the special case where v2 = v3 = 0, we see that our result agrees with Eq. , which was derived for essentially the same circuit. There are several interesting features about the result we have just de-rived. First, if we select Rf so that it is equal to R, then the output is the (neg-ative of the) sum of the three input signalsv1, v2, and v3. Further, we can select the ratio of Rf to R to multiply this sum by a ﬁxed constant. So, for exam-ple, if the three voltages represented signals from three separate scales cal-ibrated so that −1 V 1 lb, we could set Rf = R/2.205 to obtain a voltage signal that represented the combined weight in kilograms (to within about 1 percent accuracy due to our conversion factor). Also, we notice that RL did not appear in our ﬁnal expression.As long as its value is not too low, the operation of the circuit will not be affected; at present, we have not considered a detailed enough model of an op amp to predict such anoccurrence.ThisresistorrepresentstheThéveninequivalentofwhateverwe use to monitor the ampliﬁer output. If our output device is a simple voltmeter, then RL represents the Thévenin equivalent resistance seen looking into the voltmeter terminals (typically 10 M or more). Or, our output device might be a speaker (typically 8 ), in which case we hear the sum of the three sepa-rate sources of sound; v1, v2, and v3 might represent microphones in that case. One word of caution: It is frequently tempting to assume that the current labeled i in Fig. 6.9 ﬂows not only through Rf but through RL also. Not true! It is very possible that current is ﬂowing through the output terminal of the op amp as well, so that the currents through the two resistors are not the same. It is for this reason that we almost universally avoid writing KCL equations at the output pin of an op amp, which leads to the preference of nodal over mesh analysis when working with most op amp circuits. For convenience, we summarize the most common op amp circuits in Table 1. PRACTICE ● 6.2 Derive an expression for vout in terms of v1 and v2 for the circuit shown in Fig. 6.10, also known as a difference ampliﬁer. CHAPTER 6 THE OPERATIONAL AMPLIFIER 182 TABLE ●6.1 Summary of Basic Op Amp Circuits Name Circuit Schematic Input-Output Relation Inverting Ampliﬁer vout = −Rf R1 vin Noninverting Ampliﬁer vout = 1 + Rf R1 vin Voltage Follower vout = vin (also known as a Unity Gain Ampliﬁer) Summing Ampliﬁer vout = −Rf R (v1 + v2 + v3) Difference Ampliﬁer vout = v2 −v1 Rf R1 – + i i + – vin vout + – Rf R1 – + + – vin vout + – – + vin + – vout + – – + i vout + – R R RL R v1 va vb Rf i3 i2 i1 + – v2 + – v3 + – – + i vout + – R RL R R v1 va vb R i2 i1 + – v2 + – A point-to-point intercom system can be constructed using a number of different approaches, depending on the intended application environment. Low-power radio frequency (RF) systems work very well and are gener-ally cost-effective, but are subject to interference from other RF sources and are also prone to eavesdropping. Use of a simple wire to connect the two intercom sys-tems instead can eliminate a great deal of the RF inter-ference as well as increase privacy. However, wires are subject to corrosion and short circuits when the plastic insulation wears, and their weight can be a concern in aircraft and related applications (Fig. 6.11). An alternative design would be to convert the electri-cal signal from the microphone to an optical signal, which could then be transmitted through a thin (∼50 μm diameter) optical ﬁber. The optical signal is then con-verted back to an electrical signal, which is ampliﬁed and delivered to a speaker. A schematic diagram of such a system is shown in Fig. 6.12; two such systems would be needed for two-way communication. We can consider the design of the transmission and reception circuits separately, since the two circuits are in fact electrically independent. Figure 6.13 shows a simple PRACTICAL APPLICATION A Fiber Optic Intercom – + LED R1 Rf + – Microphone ■FIGURE 6.13 Circuit used to convert the electrical microphone signal into an optical signal for transmission through a ﬁber. – + R2 R3 Speaker Photodetector ■FIGURE 6.14 Receiver circuit used to convert the optical signal into an audio signal. (Continued on next page) Speaker Photodetector Light source Microphone Amplifier Optical fiber Amplifier ■FIGURE 6.12 Schematic diagram of one-half of a simple ﬁber optic intercom. ■FIGURE 6.11 The application environment often dictates design constraints. (© Michael Melford/Riser/Getty Images.) PRACTICAL APPLICATION signal generation circuit consisting of a microphone, a light-emitting diode (LED), and an op amp used in a noninverting amplifier circuit to drive the LED; not shown are the power connections required for the op amp itself. The light output of the LED is roughly pro-portional to its current, although less so for very small and very large values of current. We know the gain of the ampliﬁer is given by vout vin = 1 + Rf R1 which is independent of the resistance of the LED. In or-der to select values for Rf and R1, we need to know the input voltage from the microphone and the necessary output voltage to power the LED. A quick measurement indicates that the typical voltage output of the micro-phone peaks at 40 mV when someone is using a normal speaking voice. The LED manufacturer recommends op-erating at approximately 1.6 V, so we design for a gain of 1.6/0.04 = 40. Arbitrarily choosing R1 = 1 k leads to a required value of 39 k for Rf. The circuit of Fig. 6.14 is the receiver part of our one-way intercom system. It converts the optical signal from the ﬁber into an electrical signal, amplifying it so that an audible sound emanates from the speaker. 6.3 • CASCADED STAGES Although the op amp is an extremely versatile device, there are numerous applications in which a single op amp will not sufﬁce. In such instances, it is often possible to meet application requirements by cascading several in-dividual op amps together in the same circuit. An example of this is shown in Fig. 6.15, which consists of the summing ampliﬁer circuit of Fig. 6.9 with only two input sources, and the output fed into a simple inverting ampliﬁer. The result is a two-stage op amp circuit. After coupling the LED output of the transmitting circuit to the optical fiber, a signal of approximately 10 mV is measured from the photodetector. The speaker is rated for a maximum of 100 mW and has an equivalent resistance of 8 . This equates to a maximum speaker voltage of 894 mV, so we need to select values of R2 and R3 to obtain a gain of 894/10 = 89.4. With the arbitrary selection of R2 = 10 k, we ﬁnd that a value of 884 k completes our design. This circuit will work in practice, although the non-linear characteristics of the LED lead to a noticeable dis-tortion of the audio signal. We leave improved designs for more advanced texts. – + i vout + – R R R1 R2 v1 va vb vx vc Rf i2 i1 + – v2 + – – + ■FIGURE 6.15 A two-stage op amp circuit consisting of a summing ampliﬁer cascaded with an inverting ampliﬁer circuit. We have already analyzed each of these op amp circuits separately. Based on our previous experience, if the two op amp circuits were discon-nected, we would expect vx = −Rf R (v1 + v2) and vout = −R2 R1 vx In fact, since the two circuits are connected at a single point and the voltage vx is not inﬂuenced by the connection, we can combine Eqs. and to obtain vout = R2 R1 Rf R (v1 + v2) SECTION 6.3 CASCADED STAGES 185 which describes the input-output characteristics of the circuit shown in Fig. 6.15. We may not always be able to reduce such a circuit to familiar stages, however, so it is worth seeing how the two-stage circuit of Fig. 6.15 can be analyzed as a whole. When analyzing cascaded circuits, it is sometimes helpful to begin with the last stage and work backward toward the input stage. Referring to ideal op amp rule 1, the same current ﬂows through R1 and R2. Writing the appro-priate nodal equation at the node labeled vc yields 0 = vc −vx R1 + vc −vout R2 Applying ideal op amp rule 2, we can set vc = 0 in Eq. , resulting in 0 = vx R1 + vout R2 Since our goal is an expression for vout in terms of v1 and v2, we proceed to the ﬁrst op amp in order to obtain an expression for vx in terms of the two input quantities. Applying ideal op amp rule 1 at the inverting input of the ﬁrst op amp, 0 = va −vx Rf + va −v1 R + va −v2 R Ideal op amp rule 2 allows us to replace va in Eq. with zero, since va = vb = 0. Thus, Eq. becomes 0 = vx Rf + v1 R + v2 R We now have an equation for vout in terms of vx (Eq. ), and an equa-tion for vx in terms of v1 and v2 (Eq. ). These equations are identical to Eqs. and , respectively, which means that cascading the two separate circuits as in Fig. 6.15 did not affect the input-output relationship of either stage. Combining Eqs. and , we ﬁnd that the input-output relation-ship for the cascaded op amp circuit is vout = R2 R1 Rf R (v1 + v2) which is identical to Eq. . Thus, the cascaded circuit acts as a summing ampliﬁer, but without a phase reversal between the input and output. By choosing the resistor val-ues carefully, we can either amplify or attenuate the sum of the two input voltages. If we select R2 = R1 and Rf = R, we can also obtain an ampliﬁer circuit where vout = v1 + v2, if desired. CHAPTER 6 THE OPERATIONAL AMPLIFIER 186 Amultiple-tank gas propellant fuel system is installed in a small lunar orbit runabout. The amount of fuel in any tank is monitored by measuring the tank pressure (in psia).2 Technical details for tank capacity as well as sensor pressure and voltage range are given in Table 6.2. Design a circuit which provides a positive dc voltage signal proportional to the total fuel remaining, such that 1 V 100 percent. EXAMPLE 6.3 We see from Table 6.2 that the system has three separate gas tanks, requiring three separate sensors. Each sensor is rated up to 12,500 psia, with a corresponding output of 5 V. Thus, when tank 1 is full, its sensor will provide a voltage signal of 5 × (10,000/12,500) = 4 V; the same is true for the sensor monitoring tank 2. The sensor connected to tank 3, however, will only provide a maximum voltage signal of 5 × (200012,500) 800 mV. One possible solution is the circuit shown in Fig. 6.16a, which em-ploys a summing ampliﬁer stage with v1, v2, and v3 representing the three sensor outputs, followed by an inverting ampliﬁer to adjust the voltage sign and magnitude. Since we are not told the output resistance of the sensor, we employ a buffer for each one as shown in Fig. 6.16b; the result is (in the ideal case) no current ﬂow from the sensor. To keep the design as simple as possible, we begin by choosing R1, R2, R3, and R4 to be 1 k; any value will do as long as all four resistors are equal. Thus, the output of the summing stage is vx = −(v1 + v2 + v3) The ﬁnal stage must invert this voltage and scale it such that the output voltage is 1 V when all three tanks are full. The full condition results in vx = −(4 + 4 + 0.8) = −8.8 V. Thus, the ﬁnal stage needs a voltage ratio of R6/R5 = 1/8.8. Arbitrarily choosing R6 = 1 k, we ﬁnd that a value of 8.8 k for R5 completes the design. TABLE ● 6.2 Technical Data for Tank Pressure Monitoring System Tank 1 Capacity 10,000 psia Tank 2 Capacity 10,000 psia Tank 3 Capacity 2000 psia Sensor Pressure Range 0 to 12,500 psia Sensor Voltage Output 0 to 5 Vdc (2) Pounds per square inch, absolute. This is a differential pressure measurement relative to a vacuum reference. © Corbis SECTION 6.3 CASCADED STAGES 187 ■FIGURE 6.16 (a) A proposed circuit to provide a total fuel remaining readout. (b) Buffer design to avoid errors associated with the internal resistance of the sensor and limitations on its ability to provide current. One such buffer is used for each sensor, providing the inputs v1, v2, and v3 to the summing ampliﬁer stage. – + vout + – R2 R1 R3 R5 R6 v1 vx R4 + – + – v2 v3 + – – + (a) v1 – + Sensor 1 (b) ■FIGURE 6.17 One possible solution to Practice Problem 6.3; all resistors are 10 kΩ (although any value will do as long as they are all equal). Input voltages v1, v2, v3, and v4 represent the voltage signals from the four wheel pad sensors, and vout is the output signal to be connected to the positive input terminal of the DMM. All ﬁve voltages are referenced to ground, and the common terminal of the DMM should be connected to ground as well. – + vout + – v1 + – + – v2 v3 v4 + – – + + – Ans: See Fig. 6.17. PRACTICE ● 6.3 An historic bridge is showing signs of deterioration. Until renova-tions can be performed, it is decided that only cars weighing less than 1600 kg will be allowed across. To monitor this, a four-pad weighing system is designed. There are four independent voltage signals, one from each wheel pad, with 1 mV 1 kg. Design a circuit to provide a positive voltage signal to be displayed on a DMM (digital multimeter) that repre-sents the total weight of a vehicle, such that 1 mV 1 kg. You may as-sume there is no need to buffer the wheel pad voltage signals. 6.4 • CIRCUITS FOR VOLTAGE AND CURRENT SOURCES In this and previous chapters we have often made use of ideal current and volt-age sources, which we assume provide the same value of current or voltage, re-spectively, regardless of how they are connected in a circuit. Our assumption of independence has its limits, of course, as mentioned in Sec. 5.2 when we discussed practical sources which included a “built-in” or inherent resistance. The effect of such a resistance was a reduction of the voltage output of a volt-age source as more current was demanded, or a diminished current output as more voltage was required from a current source.As discussed in this section, it ispossibletoconstructcircuitswithmorereliablecharacteristicsusingopamps. A Reliable Voltage Source One of the most common means of providing a stable and consistent refer-ence voltage is to make use of a nonlinear device known as a Zener diode. Its symbol is a triangle with a Z-like line across the top of the triangle, as shown for a 1N750 in the circuit of Fig. 6.18a. Diodes are characterized by CHAPTER 6 THE OPERATIONAL AMPLIFIER 188 ■FIGURE 6.18 (a) PSpice schematic of a simple voltage reference circuit based on the 1N750 Zener diode. (b) Simulation of the circuit showing the diode voltage Vref as a function of the driving voltage V1. (c) Simulation of the diode current, showing that its maximum rating is exceeded when V1 exceeds 12.3 V. (Note that performing this calculation assuming an ideal Zener diode yields 12.2 V.) (a) + – R1 100 0 0 D1 D1N750 V1 DC = 0 Vref + – (b) (c) SECTION 6.4 CIRCUITS FOR VOLTAGE AND CURRENT SOURCES 189 a strongly asymmetric current-voltage relationship. For small voltages, they either conduct essentially zero current—or experience an exponentially in-creasing current—depending on the voltage polarity. In this way, they dis-tinguish themselves from the simple resistor, where the magnitude of the current is the same for either voltage polarity and hence the resistor current-voltage relationship is symmetric. Consequently, the terminals of a diode are not interchangeable, and have unique names: the anode (the ﬂat part of the triangle) and the cathode (the point of the triangle). AZener diode is a special type of diode designed to be used with a positive voltage at the cathode with respect to the anode; when connected this way, the diode is said to be reverse biased. For low voltages, the diode acts like a resis-tor with a small linear increase in current ﬂow as the voltage is increased. Once a certain voltage (VBR) is reached, however—known as the reverse breakdown voltage or Zener voltage of the diode—the voltage does not signiﬁcantly in-crease further, but essentially any current can ﬂow up to the maximum rating of the diode (75 mA for a 1N750, whose Zener voltage is 4.7 V). Let’s consider the simulation result presented in Fig. 6.18b, which shows the voltage Vref across the diode as the voltage source V1 is swept from 0 to 20 V. Provided V1 remains above 5 V, the voltage across our diode is es-sentially constant. Thus, we could replace V1 with a 9 V battery, and not be too concerned with changes in our voltage reference as the battery voltage begins to drop as it discharges. The purpose of R1 in this circuit is simply to provide the necessary voltage drop between the battery and the diode; its value should be chosen to ensure that the diode is operating at its Zener volt-age but below its maximum rated current. For example, Fig. 6.18c shows that the 75 mA rating is exceeded in our circuit if the source voltage V1 is much greater than 12 V. Thus, the value of resistor R1 should be sized cor-responding to the source voltage available, as we explore in Example 6.4. Design a circuit based on the 1N750 Zener diode that runs on a single 9 V battery and provides a reference voltage of 4.7 V. EXAMPLE 6.4 The 1N750 has a maximum current rating of 75 mA, and a Zener volt-age of 4.7 V. The voltage of a 9 V battery can vary slightly depending on its state of charge, but we neglect this for the present design. A simple circuit such as the one shown in Fig. 6.19a is adequate for our purposes; the only issue is determining a suitable value for the resistor Rref. If 4.7 V is dropped across the diode, then 9 −4.7 = 4.3 V must be dropped across Rref. Thus, Rref = 9 −Vref Iref = 4.3 Iref We determine Rref by specifying a current value. We know that Iref should not be allowed to exceed 75 mA for this diode, and large currents will discharge the battery more quickly. However, as seen in Fig. 6.19b, we cannot simply select Iref arbitrarily; very low currents do not allow (Continued on next page) CHAPTER 6 THE OPERATIONAL AMPLIFIER 190 (b) (a) 1N750 Vref + – Iref 9 V + – Rref 0V (c) DC = 9 D1 D1N750 37.10mA + – 37.10mA 4.733V 37.10mA R1 115 V1 0 ■FIGURE 6.19 (a) A voltage reference circuit based on the 1N750 Zener diode. (b) Diode I-V relationship. (c) PSpice simulation of the ﬁnal design. vout + – Vbat + – R1 Rref Rf – + ■FIGURE 6.20 An op amp–based voltage source using on a Zener voltage reference. the diode to operate in the Zener breakdown region. In the absence of a detailed equation for the diode’s current-voltage relationship (which is clearly nonlinear), we design for 50 percent of the maximum rated cur-rent as a rule of thumb. Thus, Rref = 4.3 0.0375 = 115 Detailed “tweaking” can be obtained by performing a PSpice simu-lation of the ﬁnal circuit, although we see from Fig. 6.19c that our ﬁrst pass is reasonably close (within 1 percent) to our target value. PRACTICE ● 6.4 Design a circuit to provide a reference voltage of 6 V using a 1N750 Zener diode and a noninverting ampliﬁer. Ans: Using the circuit topology shown in Fig. 6.20, choose Vbat = 9 V, Rref = 115 , R1 = 1 k, and Rf = 268 . A Reliable Current Source Consider the circuit shown in Fig. 6.21a, where Vref is provided by a regu-lated voltage source such as the one shown in Fig. 6.19a. The reader may recognize this circuit as a simple inverting ampliﬁer conﬁguration, assum-ing we tap the output pin of the op amp. We can also use this circuit as a cur-rent source, however, where RL represents a resistive load. The input voltage Vref appears across reference resistor Rref, since the noninverting input of the op amp is connected to ground. With no current The basic Zener diode voltage reference circuit of Fig. 6.18a works very well in many situations, but we are limited somewhat in the value of the volt-age depending on which Zener diodes are available. Also, we often ﬁnd that the circuit shown is not well suited to applications requiring more than a few milliamperes of current. In such instances, we may use the Zener reference circuit in conjunction with a simple ampliﬁer stage, as shown in Fig. 6.20. The result is a stable voltage that can be controlled by adjusting the value of either R1 or Rf, without having to switch to a different Zener diode. SECTION 6.4 CIRCUITS FOR VOLTAGE AND CURRENT SOURCES 191 RL IS RL – + OUT IS Vref Rref RL – + OUT IS Vref Rref (a) (b) (c) ■FIGURE 6.21 (a) An op amp–based current source, controlled by the reference voltage Vref. (b) Circuit redrawn to highlight load. (c) Circuit model. Resistor RL represents the Norton equivalent of an unknown passive load circuit. ﬂowing into the inverting input, the current ﬂowing through the load resistor RL is simply Is = Vref Rref In other words, the current supplied to RL does not depend on its resistance—the primary attribute of an ideal current source. It is also worth noting that we are not tapping the output voltage of the op amp here as a quan-tity of interest. Instead, we may view the load resistor RL as the Norton (or Thévenin) equivalent of some unknown passive load circuit, which receives powerfromtheopampcircuit.RedrawingthecircuitslightlyasinFig.6.21b, we see that it has a great deal in common with the more familiar circuit of Fig. 6.21c. In other words, we may use this op amp circuit as an independent current source with essentially ideal characteristics, up to the maximum rated output current of the op amp selected. Design a current source that will deliver 1 mA to an arbitrary resistive load. EXAMPLE 6.5 Basing our design on the circuits of Fig. 6.20 and Fig. 6.21a, we know that the current through our load RL will be given by Is = Vref Rref where values for Vref and Rref must be selected, and a circuit to provide Vref must also be designed. If we use a 1N750 Zener diode in series with a 9 V (Continued on next page) 6.5 • PRACTICAL CONSIDERATIONS A More Detailed Op Amp Model Reduced to its essentials, the op amp can be thought of as a voltage-controlled dependent voltage source. The dependent voltage source provides the output of the op amp, and the voltage on which it depends is applied to the input terminals. A schematic diagram of a reasonable model for a practical op amp is shown in Fig. 6.24; it includes a dependent voltage source with voltage gain A, an output resistance Ro, and an input resistance Ri. Table 6.3 gives typical values for these parameters for several types of commercially available op amps. TheparameterAisreferredtoastheopen-loopvoltagegainoftheopamp, and is typically in the range of 105 to 106. We notice that all of the op amps listed in Table 6.3 have extremely large open-loop voltage gain, especially compared to the voltage gain of 11 that characterized the noninverting ampliﬁer circuit of Example 6.1. It is important to remember the distinction CHAPTER 6 THE OPERATIONAL AMPLIFIER 192 9 V + – 4.9 k 100 RL – + 1N750 IS ■FIGURE 6.22 One possible design for the desired current source. Note the change in current direction from Fig. 6.21b. 9 V + – 9.8 k 100 RL – + 1N750 ■FIGURE 6.23 One possible solution to Practice Problem 6.5. battery and a 100 Ω resistor, we know from Fig. 6.18b that a voltage of 4.9 V will exist across the diode. Thus, Vref = 4.9V, dictating a value of 4.9/10−3 = 4.9 k for Rref. The complete circuit is shown in Fig. 6.22. Note that if we had assumed a diode voltage of 4.7 V instead, the error in our designed current would only be a few percent, well within the typical 5 to 10 percent tolerance in resistor values we might expect. The only issue remaining is whether 1 mA can in fact be provided to any value of RL. For the case of RL 0, the output of the op amp will be 4.9 V, which is not unreasonable. As the load resistor is increased, however, the op amp output voltage increases. Eventually we must reach some type of limit, as discussed in Sec. 6.5. PRACTICE ● 6.5 Design a current source capable of providing 500 μA to a resistive load. Ans: See Fig. 6.23 for one possible solution. Ro Ri Avd vd + – iout iin + – + – vout ■FIGURE 6.24 A more detailed model for the op amp. SECTION 6.5 PRACTICAL CONSIDERATIONS 193 between the open-loop voltage gain of the op amp itself, and the closed-loop voltage gain that characterizes a particular op amp circuit. The “loop” in this case refers to an external path between the output pin and the invert-ing input pin; it can be a wire, a resistor, or another type of element, de-pending on the application. The μA741 is a very common op amp, originally produced by Fairchild Corporation in 1968. It is characterized by an open-loop voltage gain of 200,000, an input resistance of 2 M, and an output resistance of 75 . In order to evaluate how well the ideal op amp model approximates the be-havior of this particular device, let’s revisit the inverting ampliﬁer circuit of Fig. 6.3. TABLE ●6.3 Typical Parameter Values for Several Types of Op Amps Part Number μA741 LM324 LF411 AD549K OPA690 Description General Low-power Low-offset, low-Ultralow input Wideband video purpose quad drift JFET input bias current frequency op amp Open loop gain A 2 × 105 V/V 105 V/V 2 × 105 V/V 106 V/V 2800 V/V Input resistance 2 M 1 T 10 TΩ 190 k Output resistance 75 ∼1 ∼15 Input bias current 80 nA 45 nA 50 pA 75 fA 3 μA Input offset voltage 1.0 mV 2.0 mV 0.8 mV 0.150 mV ±1.0 mV CMRR 90 dB 85 dB 100 dB 100 dB 65 dB Slew rate 0.5 V/μs 15 V/μs 3 V/μs 1800 V/μs PSpice Model ✓ ✓ ✓ Not provided by manufacturer. ✓Indicates that a PSpice model is included in Orcad Capture CIS Lite Edition 16.3. Using the appropriate values for the μA741 op amp in the model of Fig. 6.24, reanalyze the inverting ampliﬁer circuit of Fig. 6.3. We begin by replacing the ideal op amp symbol of Fig. 6.3 with the detailed model, resulting in the circuit shown in Fig. 6.25. Note that we can no longer invoke the ideal op amp rules, since we are not using the ideal op amp model. Thus, we write two nodal equations: 0 = −vd −vin R1 + −vd −vout Rf + −vd Ri 0 = vout + vd Rf + vout −Avd Ro Performing some straightforward but rather lengthy algebra, we elimi-nate vd and combine these two equations to obtain the following EXAMPLE 6.6 (Continued on next page) Derivation of the Ideal Op Amp Rules We have seen that the ideal op amp can be a reasonably accurate model for the behavior of practical devices. However, using our more detailed model which includes a ﬁnite open-loop gain, ﬁnite input resistance, and nonzero output resistance, it is actually straightforward to derive the two ideal op amp rules. Referring to Fig. 6.24, we see that the open circuit output voltage of a practical op amp can be expressed as vout = Avd CHAPTER 6 THE OPERATIONAL AMPLIFIER 194 expression for vout in terms of vin: vout = Ro + Rf Ro −ARf 1 R1 + 1 Rf + 1 Ri −1 Rf −1 vin R1 Substitutingvin = 5 sin 3tmV, R1 = 4.7 k, Rf = 47 k,Ro = 75 , Ri = 2 M, and A = 2 × 105,we obtain vout = −9.999448vin = −49.99724 sin 3t mV Upon comparing this to the expression found assuming an ideal op amp (vout = −10vin = −50 sin 3t mV), we see that the ideal op amp is indeed a reasonably accurate model. Further, assuming an ideal op amp leads to a signiﬁcant reduction in the algebra required to perform the circuit analysis. Note that if we allow A →∞, Ro →0, and Ri →∞, Eq. reduces to vout = −Rf R1 vin which is what we derived earlier for the inverting ampliﬁer when assuming the op amp was ideal. PRACTICE ● 6.6 Assuming a ﬁnite open-loop gain (A), a ﬁnite input resistance (Ri), and zero output resistance (Ro), derive an expression for vout in terms of vin for the op amp circuit of Fig. 6.3. Ans: vout/vin = −ARf Ri/[(1 + A)R1Ri + R1Rf + Rf Ri]. Ro R1 + – Ri Rf Avd vd + – vin vout + – + – ■FIGURE 6.25 Inverting ampliﬁer circuit drawn using detailed op amp model. SECTION 6.5 PRACTICAL CONSIDERATIONS 195 Rearranging this equation, we ﬁnd that vd, sometimes referred to as the differential input voltage, can be written as vd = vout A As we might expect, there are practical limits to the output voltage vout that can be obtained from a real op amp.As described in the next section, we must connect our op amp to external dc voltage supplies in order to power the internal circuitry. These external voltage supplies represent the maximum value of vout, and are typically in the range of 5 to 24 V. If we divide 24 V by the open-loop gain of the μA741 (2 × 105), we obtain vd = 120 μV. Although this is not the same as zero volts, such a small value compared to the output voltage of 24Vis practically zero.An ideal op amp would have in-ﬁnite open-loop gain, resulting in vd = 0regardless of vout; this leads to ideal op amp rule 2. Ideal op amp rule 1 states that “No current ever ﬂows into either input terminal.” Referring to Fig. 6.23, the input current of an op amp is simply iin = vd Ri We have just determined that vd is typically a very small voltage. As we can see from Table 6.3, the input resistance of an op amp is very large, rang-ing from the megaohms to the teraohms! Using the value of vd = 120 μV from above and Ri = 2 M, we compute an input current of 60 pA. This is an extremely small current, and we would require a specialized ammeter (known as a picoammeter) to measure it. We see from Table 6.3 that the typ-ical input current (more accurately termed the input bias current) of a μA741 is 80 nA, three orders of magnitude larger than our estimate. This is a shortcoming of the op amp model we are using, which is not designed to provide accurate values for input bias current. Compared to the other cur-rents ﬂowing in a typical op amp circuit, however, either value is essentially zero. More modern op amps (such as the AD549) have even lower input bias currents. Thus, we conclude that ideal op amp rule 1 is a fairly reason-able assumption. From our discussion, it is clear that an ideal op amp has inﬁnite open-loop voltage gain, and inﬁnite input resistance. However, we have not yet considered the output resistance of the op amp and its possible effects on our circuit. Referring to Fig. 6.24, we see that vout = Avd −Roiout where iout ﬂows from the output pin of the op amp. Thus, a nonzero value of Ro acts to reduce the output voltage, an effect which becomes more pro-nounced as the output current increases. For this reason, an ideal op amp has an output resistance of zero ohms. The μA741 has a maximum output resistance of 75 , and more modern devices such as the AD549 have even lower output resistance. Common-Mode Rejection The op amp is occasionally referred to as a difference ampliﬁer, since the output is proportional to the voltage difference between the two input terminals. This means that if we apply identical voltages to both input terminals, we expect the output voltage to be zero. This ability of the op amp is one of its most attractive qualities, and is known as common-mode rejection. The circuit shown in Fig. 6.26 is connected to provide an output voltage vout = v2 −v1 If v1 = 2 + 3 sin 3t volts and v2 = 2 volts, we would expect the output to be −3 sin 3t volts; the 2 V component common to v1 and v2 would not be ampliﬁed, nor does it appear in the output. For practical op amps, we do in fact ﬁnd a small contribution to the out-put in response to common-mode signals. In order to compare one op amp type to another, it is often helpful to express the ability of an op amp to reject common-mode signals through a parameter known as the common-mode rejection ratio, or CMRR. Deﬁning voCM as the output obtained when both inputs are equal (v1 = v2 = vCM), we can determine ACM, the common-mode gain of the op amp ACM = voCM vCM We then deﬁne CMRR in terms of the ratio of differential-mode gain A to the common-mode gain ACM, or CMRR ≡ A ACM although this is often expressed in decibels (dB), a logarithmic scale: CMRR(dB) ≡20 log10 A ACM dB Typical values for several different op amps are provided in Table 6.3; a value of 100 dB corresponds to an absolute ratio of 105 for A to ACM. Negative Feedback We have seen that the open-loop gain of an op amp is very large, ideally in-ﬁnite. In practical situations, however, its exact value can vary from the value speciﬁed by the manufacturer as typical. Temperature, for example, can have a number of signiﬁcant effects on the performance of an op amp, so that the operating behavior in −20◦C weather may be signiﬁcantly dif-ferent from the behavior observed on a warm sunny day. Also, there are typically small variations between devices fabricated at different times. If we design a circuit in which the output voltage is the open-loop gain times the voltage at one of the input terminals, the output voltage could therefore be difﬁcult to predict with a reasonable degree of precision, and might be expected to change depending on the ambient temperature. A solution to such potential problems is to employ the technique of negative feedback, which is the process of subtracting a small portion of the output from the input. If some event changes the characteristics of the ampliﬁer such that the output tries to increase, the input is decreasing at the same time. Too much negative feedback will prevent any useful ampliﬁca-tion, but a small amount provides stability. An example of negative CHAPTER 6 THE OPERATIONAL AMPLIFIER 196 – + vout + – R R R v1 va vb R + – v2 + – ■FIGURE 6.26 An op amp connected as a difference ampliﬁer. SECTION 6.5 PRACTICAL CONSIDERATIONS 197 feedback is the unpleasant sensation we feel as our hand draws near a ﬂame. The closer we move toward the ﬂame, the larger the negative signal sent from our hand. Overdoing the proportion of negative feedback, how-ever, might cause us to abhor heat, and eventually freeze to death. Positive feedback is the process where some fraction of the output signal is added back to the input. A common example is when a microphone is directed to-ward a speaker—a very soft sound is rapidly ampliﬁed over and over until the system “screams.” Positive feedback generally leads to an unstable system. All of the circuits considered in this chapter incorporate negative feed-back through the presence of a resistor between the output pin and the inverting input. The resulting loop between the output and the input reduces the dependency of the output voltage on the actual value of the open-loop gain (as seen in Example 6.6). This obviates the need to measure the precise open-loop gain of each op amp we use, as small variations in A will not sig-niﬁcantly impact the operation of the circuit. Negative feedback also pro-vides increased stability in situations where A is sensitive to the op amp’s surroundings. For example, if A suddenly increases in response to a change in the ambient temperature, a larger feedback voltage is added to the inverting input. This acts to reduce the differential input voltage vd, and therefore the change in output voltage Avd is smaller. We should note that the closed-loop circuit gain is always less than the open-loop device gain; this is the price we pay for stability and reduced sensitivity to parameter variations. Saturation So far, we have treated the op amp as a purely linear device, assuming that its characteristics are independent of the way in which it is connected in a cir-cuit. In reality, it is necessary to supply power to an op amp in order to run the internal circuitry, as shown in Fig. 6.27. Apositive supply, typically in the range of 5 to 24 V dc, is connected to the terminal marked V+, and a nega-tive supply of equal magnitude is connected to the terminal marked V −. There are also a number of applications where a single voltage supply is acceptable, as well as situations where the two voltage magnitudes may be unequal. The op amp manufacturer will usually specify a maximum power supply voltage, beyond which damage to the internal transistors will occur. The power supply voltages are a critical choice when designing an op amp circuit, because they represent the maximum possible output voltage of the op amp.3 For example, consider the op amp circuit shown in Fig. 6.26, now connected as a noninverting ampliﬁer having a gain of 10. As shown in the PSpice simulation in Fig. 6.28, we do in fact observe linear behavior from the op amp, but only in the range of ±1.71 V for the input voltage. Outside of this range, the output voltage is no longer proportional to the input, reaching a peak magnitude of 17.6 V. This important nonlinear effect is known as saturation, which refers to the fact that further increases in the input voltage do not result in a change in the output voltage. This phenom-enon refers to the fact that the output of a real op amp cannot exceed its – + Offset null V – V + Offset null 18 V 18 V + – + – ■FIGURE 6.27 Op amp with positive and negative voltage supplies connected. Two 18 V supplies are used as an example; note the polarity of each source. (3) In practice, we ﬁnd the maximum output voltage is slightly less than the supply voltage by as much as a volt or so. supply voltages. For example, if we choose to run the op amp with a +9 V supply and a −5 V supply, then our output voltage will be limited to the range of −5 to +9 V. The output of the op amp is a linear response bounded by the positive and negative saturation regions, and as a general rule, we try to design our op amp circuits so that we do not accidentally enter the satu-ration region. This requires us to select the operating voltage carefully based on the closed-loop gain and maximum expected input voltage. Input Offset Voltage As we are discovering, there are a number of practical considerations to keep in mind when working with op amps. One particular nonideality worth mentioning is the tendency for real op amps to have a nonzero output even when the two input terminals are shorted together. The value of the output under such conditions is known as the offset voltage, and the input voltage required to reduce the output to zero is referred to as the input offset voltage. Referring to Table 6.3, we see that typical values for the input offset voltage are on the order of a few millivolts or less. Most op amps are provided with two pins marked either “offset null” or “balance.” These terminals can be used to adjust the output voltage by con-necting them to a variable resistor. A variable resistor is a three-terminal de-vice commonly used for such applications as volume controls on radios. The device comes with a knob that can be rotated to select the actual value of resistance, and has three terminals. Measured between the two extreme terminals, its resistance is ﬁxed regardless of the position of the knob. Us-ing the middle terminal and one of the end terminals creates a resistor whose value depends on the knob position. Figure 6.29 shows a typical cir-cuit used to adjust the output voltage of an op amp; the manufacturer’s data sheet may suggest alternative circuitry for a particular device. CHAPTER 6 THE OPERATIONAL AMPLIFIER 198 ■FIGURE 6.28 Simulated input-output characteristics of a μA741 connected as a noninverting ampliﬁer with a gain of 10, and powered by ±18 V supplies. – + Offset null V – V + Offset null Output –10 V +10 V + – + – ■FIGURE 6.29 Suggested external circuitry for obtaining a zero output voltage. The ±10 V supplies are shown as an example; the actual supply voltages used in the ﬁnal circuit would be chosen in practice. SECTION 6.5 PRACTICAL CONSIDERATIONS 199 Slew Rate Up to now, we have tacitly assumed that the op amp will respond equally well to signals of any frequency, although perhaps we would not be sur-prised to ﬁnd that in practice there is some type of limitation in this regard. Since we know that op amp circuits work well at dc, which is essentially zero frequency, it is the performance as the signal frequency is increased that we must consider. One measure of the frequency performance of an op amp is its slew rate, which is the rate at which the output voltage can respond to changes in the input; it is most often expressed in V/μs. The typical slew rate speciﬁcation for several commercially available devices is provided in Table 6.3, showing values on the order of a few volts per microsecond. One notable exception is the OPA690, which is designed as a high-speed op amp for video applications requiring operation at several hundred MHz. As can be seen, a respectable slew rate of 1800 V/μs is not unrealistic for this device, although its other parameters, particularly input bias current and CMRR, suffer somewhat as a result. The PSpice simulations shown in Fig. 6.30 illustrate the degradation in performance of an op amp due to slew rate limitations. The circuit simu-lated is an LF411 conﬁgured as a noninverting ampliﬁer with a gain of 2 and powered by ±15 V supplies. The input waveform is shown in green, and has (a) (b) (c) ■FIGURE 6.30 Simulated performance of an LF411 op amp connected as a noninverting ampliﬁer having a gain of 2, with ±15 V supplies and a pulsed input waveform. (a) Rise and fall times 1 µs, pulse width 5 µs; (b) rise and fall times 100 ns, pulse width 500 ns; (c) rise and fall times 50 ns, pulse width 250 ns. a peak voltage of 1 V; the output voltage is shown in red. The simulation of Fig. 6.30a corresponds to a rise and fall time of 1 μs which, although a short time span for humans, is easily coped with by the LF411. As the rise and fall times are decreased by a factor of 10 to 100 ns (Fig. 6.30b), we begin to see that the LF411 is having a small difﬁculty in tracking the input. In the case of a 50 ns rise and fall time (Fig. 6.30c), we see that not only is there a signiﬁcant delay between the output and the input, but the waveform is noticeably distorted as well—not a good feature of an ampliﬁer. This ob-served behavior is consistent with the typical slew rate of 15 V/μs speciﬁed in Table 6.3, which indicates that the output might be expected to require roughly 130 ns to change from 0 to 2 V (or 2 V to 0 V). Packaging Modern op amps are available in a number of different types of packages. Some styles are better suited to high temperatures, and there are a variety of different ways to mount ICs on printed-circuit boards. Figure 6.31 shows several different styles of the LM741, manufactured by National Semiconductor. The label “NC” next to a pin means “no connection.” The package styles shown in the ﬁgure are standard conﬁgurations, and are used for a large number of different integrated circuits; occasionally there are more pins available on a package than required. CHAPTER 6 THE OPERATIONAL AMPLIFIER 200 (c) (a) (b) ■FIGURE 6.31 Several different package styles for the LM741 op amp: (a) metal can; (b) dual-in-line package; (c) ceramic ﬂatpak. (Copyright © 2011 National Semiconductor Corporation (www.national.com). All rights reserved. Used with permission.) COMPUTER-AIDED ANALYSIS As we have just seen, PSpice can be enormously helpful in predicting the output of an op amp circuit, especially in the case of time-varying inputs. We will ﬁnd, however, that our ideal op amp model agrees fairly well with PSpice simulations as a general rule. When performing a PSpice simulation of an op amp circuit, we must be careful to remember that positive and negative dc supplies must be connected to the device (with the exception of the LM324, which is de-signed to be a single-supply op amp). Although the model shows the offset null pins used to zero the output voltage, PSpice does not build in any offset, so these pins are typically left ﬂoating (unconnected). Table 6.3 shows the different op amp part numbers available in the Evaluation version of PSpice; other models are available in the com-mercial version of the software and from some manufacturers. SECTION 6.5 PRACTICAL CONSIDERATIONS 201 EXAMPLE 6.7 Our previous analysis using an ideal op amp model predicted a gain of −10. With an input of 5 sin 3t mV, this led to an output voltage of −50 sin 3t mV. However, an implicit assumption in the analysis was that any voltage input would be ampliﬁed by a factor of −10. Based on practical considerations, we expect this to be true for small input volt-ages, but the output will eventually saturate to a value comparable to the corresponding power supply voltage. We perform a dc sweep from −2 to +2 volts, as shown in Fig. 6.33; this is a slightly larger range than the supply voltage divided by the gain, so we expect our results to include the positive and negative saturation regions. Using the cursor tool on the simulation results shown in Fig. 6.34a, the input-output characteristic of the ampliﬁer is indeed linear over a wide input range, corresponding approximately to −1.45 < Vs < +1.45 V (Fig. 6.34b): This range is slightly less than the range deﬁned by dividing the positive and negative supply voltages by the gain. Outside this range, the output of the op amp saturates, with only a slight dependence on the input voltage. In the two saturation re-gions, then, the circuit does not perform as a linear ampliﬁer. Increasing the number of cursor digits (Tools, Options, Number of Cursor Digits) to 10, we ﬁnd that at an input voltage of Vs = 1.0 V, the We begin by drawing the inverting ampliﬁer circuit of Fig. 6.3 using the schematic capture tool as shown in Fig. 6.32. Note that two separate 15 V dc supplies are required to power the op amp. Simulate the circuit of Fig. 6.3 using PSpice. Determine the point(s) at which saturation begins if ±15 V dc supplies are used to power the device. Compare the gain calculated by PSpice to what was predicted using the ideal op amp model. ■FIGURE 6.32 The inverting ampliﬁer of Fig. 6.3 drawn using a μA741 op amp. (Continued on next page) CHAPTER 6 THE OPERATIONAL AMPLIFIER 202 ■FIGURE 6.33 DC sweep setup window. (a) (b) ■FIGURE 6.34 (a) Output voltage of the inverting amplifer circuit, with the onset of saturation identiﬁed with the cursor tool. (b) Close-up of the cursor window. output voltage is −9.99548340, slightly less than the value of −10 predicted from the ideal op amp model, and slightly different from the value of −9.999448 obtained in Example 6.6 using an analytical model. Still, the results predicted by the PSpice μA741 model are within a few SECTION 6.6 COMPARATORS AND THE INSTRUMENTATION AMPLIFIER 203 6.6 • COMPARATORS AND THE INSTRUMENTATION AMPLIFIER The Comparator Every op amp circuit we have discussed up to now has featured an electri-cal connection between the output pin and the inverting input pin. This is known as closed-loop operation, and is used to provide negative feedback as discussed previously. Closed loop is the preferred method of using an op amp as an ampliﬁer, as it serves to isolate the circuit performance from variations in the open-loop gain that arise from changes in temperature or manufacturing differences. There are a number of applications, however, where it is advantageous to use an op amp in an open-loop conﬁguration. Devices intended for such applications are frequently referred to as comparators, as they are designed slightly differently from regular op amps in order to improve their speed in open-loop operation. Figure 6.35a shows a simple comparator circuit where a 2.5 V refer-ence voltage is connected to the noninverting input, and the voltage being compared (vin) is connected to the inverting input. Since the op amp has a very large open-loop gain A (105 or greater, typically, as seen in Table 6.3), it does not take a large voltage difference between the input terminals to drive it into saturation. In fact, a differential input voltage as small as the supply voltage divided by A is required—approximately ±120 μV in the case of the circuit in Fig. 6.35a and A = 105. The distinctive output of the comparator circuit is shown in Fig. 6.35b, where the response swings vout vin – + + – + – + – – + 2.5 V 12 V –12 V V+ V – Offset null Offset null (a) (b) vin (V) vout (V) ■FIGURE 6.35 (a) An example comparator circuit with a 2.5 V reference voltage. (b) Graph of input-output characteristic. hundredths of a percent of either analytical model, demonstrating that the ideal op amp model is indeed a remarkably accurate approximation for modern operational ampliﬁer integrated circuits. PRACTICE ● 6.7 Simulate the remaining op amp circuits described in this chapter, and compare the results to those predicted using the ideal op amp model. between positive and negative saturation, with essentially no linear “ampliﬁcation” region. Thus, a positive 12 V output from the comparator indicates that the input voltage is less than the reference voltage, and a negative 12 V output indicates an input voltage greater than the reference. Opposite behavior is obtained if we connect the reference voltage to the inverting input instead. CHAPTER 6 THE OPERATIONAL AMPLIFIER 204 Design a circuit that provides a “logic 1” 5 V output if a certain voltage signal drops below 3 V, and zero volts otherwise. EXAMPLE 6.8 ■FIGURE 6.36 One possible design for the required circuit. ■FIGURE 6.37 One possible solution to Practice Problem 6.8. The Instrumentation Ampliﬁer The basic comparator circuit acts on the voltage difference between the two input terminals to the device, although it does not technically amplify sig-nals as the output is not proportional to the input. The difference ampliﬁer of Fig. 6.10 also acts on the voltage difference between the inverting and noninverting inputs, and as long as care is taken to avoid saturation, does provide an output directly proportional to this difference. When dealing with a very small input voltage, however, a better alternative is a device vout vsignal – + + – + – – + 12 V –2 V V+ V– Since we want the output of our comparator to swing between 0 and 5 V, we will use an op amp with a single-ended +5 V supply, connected as shown in Fig. 6.36. We connect a +3 V reference voltage to the noninverting input, which may be provided by two 1.5 V batteries in series, or a suitable Zener diode reference circuit. The input voltage signal (designated vsignal) is then connected to the inverting input. In reality, the saturation voltage range of a comparator circuit will be slightly less than that of the supply voltages, so that some adjustment may be required in conjunction with simulation or testing. PRACTICE ● 6.8 Design a circuit that provides a 12 V output if a certain voltage (vsignal) exceeds 0 V, and a −2 V output otherwise. Ans: One possible solution is shown in Fig. 6.37. vout vsignal – + + – + – + – 3 V 5 V V+ V– SECTION 6.6 COMPARATORS AND THE INSTRUMENTATION AMPLIFIER 205 known as an instrumentation ampliﬁer, which is actually three op amp devices in a single package. An example of the common instrumentation ampliﬁer conﬁguration is shown in Fig. 6.38a, and its symbol is shown in Fig. 6.38b. Each input is fed directly into a voltage follower stage, and the output of both voltage fol-lowers is fed into a difference ampliﬁer stage. It is particularly well suited to applications where the input voltage signal is very small (for example, on the order of millivolts), such as that produced by thermocouples or strain gauges, and where a signiﬁcant common-mode noise signal of several volts may be present. ■FIGURE 6.38 (a) The basic instrumentation ampliﬁer. (b) Commonly used symbol. vout vx vx R1 R3 R4 R2 + – v+ v– – + – + – + – + vd RG + – (a) (b) If components of the instrumentation ampliﬁer are fabricated all on the same silicon “chip,” then it is possible to obtain well-matched device char-acteristics and to achieve precise ratios for the two sets of resistors. In order to maximize the CMRR of the instrumentation amplifier, we expect R4/R3 = R2/R1, so that equal ampliﬁcation of common-mode components of the input signals is obtained. To explore this further, we identify the volt-age at the output of the top voltage follower as “v−,” and the voltage at the output of the bottom voltage follower as “v+.” Assuming all three op amps are ideal and naming the voltage at either input of the difference stage vx, we may write the following nodal equations: vx −v− R1 + vx −vout R2 = 0 and vx −v+ R3 + vx R4 = 0 Solving Eq. for vx, we ﬁnd that vx = v+ 1 + R3/R4 and upon substituting into Eq. , obtain an expression for vout in terms of the input: vout = R4 R3 1 + R2/R1 1 + R4/R3 v+ −R2 R1 v− From Eq. it is clear that the general case allows ampliﬁcation of common-mode components to the two inputs. In the speciﬁc case where R4/R3 = R2/R1 = K, however, Eq. reduces to K(v+ −v−) = Kvd, so that (asssuming ideal op amps) only the difference is ampliﬁed and the gain is set by the resistor ratio. Since these resistors are internal to the instru-mentation ampliﬁer and not accessible to the user, devices such as the AD622 allow the gain to be set anywhere in the range of 1 to 1000 by con-necting an external resistor between two pins (shown as RG in Fig. 6.38b). SUMMARY AND REVIEW In this chapter we introduced a new circuit element—a three-terminal device—called the operational ampliﬁer (or more commonly, the op amp). In many circuit analysis situations it is approximated as an ideal device, which leads to two rules that are applied. We studied several op amp circuits in detail, including the inverting ampliﬁer with gain RfR1, the noninverting amplifer with gain 1 RfR1, and the summing ampliﬁer. We were also introduced to the voltage follower and the difference ampliﬁer, although the analysis of these two circuits was left for the reader. The concept of cas-caded stages was found to be particularly useful, as it allows a design to be broken down into distinct units, each of which has a speciﬁc function. We took a slight detour and introduced brieﬂy a two-terminal nonlinear circuit element, the Zener diode, as it provides a practical and straightforward volt-age reference. We then used this element to contruct practical voltage and current sources using op amps, removing some of the mystery as to their origins. Modern op amps have nearly ideal characteristics, as we found when we opted for a more detailed model based on a dependent source. Still, nonide-alities are encountered occasionally, so we considered the role of negative feedback in reducing the effect of temperature and manufacturing-related variations in various parameters, common-mode rejection, and saturation. One of the most interesting nonideal characteristics of any op amp is slew rate. By simulating three different cases, we were able to see how the out-put voltage can struggle to follow the form of the input voltage signal once its frequency becomes high enough. We concluded the chapter with two special cases: the comparator, which intentionally makes use of our ability to saturate a practical (nonideal) op amp, and the instrumentation ampliﬁer, which is routinely used to amplify very small voltages. This is a good point to pause, take a breath, and recap some of the key points. At the same time, we will highlight relevant examples as an aid to the reader. ❑There are two fundamental rules that must be applied when analyzing ideal op amp circuits: 1. No current ever ﬂows into either input terminal. (Example 6.1) 2. No voltage ever exists between the input terminals. ❑Op amp circuits are usually analyzed for an output voltage in terms of some input quantity or quantities. (Examples 6.1, 6.2) ❑Nodal analysis is typically the best choice in analyzing op amp circuits, and it is usually better to begin at the input, and work toward the output. (Examples 6.1, 6.2) CHAPTER 6 THE OPERATIONAL AMPLIFIER 206 READING FURTHER 207 ❑The output current of an op amp cannot be assumed; it must be found after the output voltage has been determined independently. (Example 6.2) ❑The gain of an inverting op amp circuit is given by the equation vout = −Rf R1 vin ❑The gain of a noninverting op amp circuit is given by the equation vout = 1 + Rf R1 vin (Example 6.1) ❑Cascaded stages may be analyzed one stage at a time to relate the output to the input. (Example 6.3) ❑Zener diodes provide a convenient voltage reference. They are not symmetric, however, meaning the two terminals are not interchangeable. (Example 6.4) ❑Op amps can be used to construct current sources which are independent of the load resistance over a speciﬁc current range. (Example 6.5) ❑A resistor is almost always connected from the output pin of an op amp to its inverting input pin, which incorporates negative feedback into the circuit for increased stability. ❑The ideal op amp model is based on the approximation of inﬁnite open-loop gain A, inﬁnite input resistance Ri, and zero output resistance Ro. (Example 6.6) ❑In practice, the output voltage range of an op amp is limited by the supply voltages used to power the device. (Example 6.7) ❑Comparators are op amps designed to be driven into saturation. These circuits operate in open loop, and hence have no external feedback resistor. (Example 6.8) READING FURTHER Two very readable books which deal with a variety of op amp applications are: R. Mancini (ed.), Op Amps Are For Everyone, 2nd ed.Amsterdam: Newnes, 2003.Also available on the Texas Instruments website (www.ti.com). W. G. Jung, Op Amp Cookbook, 3rd ed. Upper Saddle River, N.J.: Prentice-Hall, 1997. Characteristics of Zener and other types of diodes are covered in Chapter 1 of W. H. Hayt, Jr., and G. W. Neudeck, Electronic Circuit Analysis and Design, 2nd ed. New York: Wiley, 1995. One of the ﬁrst reports of the implementation of an “operational ampliﬁer” can be found in J. R. Ragazzini, R. M. Randall, and F. A. Russell, “Analysis of problems in dynamics by electronic circuits,” Proceedings of the IRE 35(5), 1947, pp. 444–452. And an early applications guide for the op amp can be found on the Analog Devices, Inc. website (www.analog.com): George A. Philbrick Researches, Inc., Applications Manual for Computing Ampliﬁers for Modelling, Measuring, Manipulating & Much Else. Norwood, Mass.: Analog Devices, 1998. EXERCISES 6.2 The Ideal Op Amp 1. For the op amp circuit shown in Fig. 6.39, calculate vout if (a) R1 = R2 = 100 and vin = 5 V; (b) R2 = 200R1 and vin = 1 V; (c) R1 = 4.7 k, R2 = 47 k, and vin = 20 sin 5t V. CHAPTER 6 THE OPERATIONAL AMPLIFIER 208 R1 R2 – + vout vin + – ■FIGURE 6.39 2. Determine the power dissipated by a 100 resistor connected between ground and the output pin of the op amp of Fig. 6.39 if vin = 4 V and (a) R1 = 2R2; (b) R1 = 1 k and R2 = 22 k; (c) R1 = 100 and R2 = 101 . 3. Connect a 1 resistor between ground and the output terminal of the op amp of Fig. 6.39, and sketch vout(t) if (a) R1 = R2 = 10 and vin = 5 sin 10t V; (b) R1 = 0.2R2 = 1 k, and vin = 5 cos 10t V; (c) R1 = 10 , R2 = 200 , and vin = 1.5 + 5e−t V. 4. For the circuit of Fig. 6.40, calculate vout if (a) R1 = R2 = 100 k, RL = 100 , and vin = 5 V; (b) R1 = 0.1R2, RL = ∞, and vin = 2 V; (c) R1 = 1 k, R2 = 0, RL = 1 , and vin = 43.5 V. 5. (a) Design a circuit which converts a voltage v1(t) = 9 cos 5t V into 9 sin 5t V. (b) Verify your design by analyzing the ﬁnal circuit. 6. A certain load resistor requires a constant 5 V dc supply. Unfortunately, its resistance value changes with temperature. Design a circuit which supplies the requisite voltage if only 9 V batteries and standard 10% tolerance resistor values are available. 7. For the circuit of Fig. 6.40, R1 = RL = 50 . Calculate the value of R2 required to deliver 5 W to RL if Vin equals (a) 5 V; (b) 1.5 V. (c) Repeat parts (a) and (b) if RL is reduced to 22 . 8. Calculate vout as labeled in the schematic of Fig. 6.41 if (a) iin = 1 mA, Rp = 2.2 k, and R3 = 1 k; (b) iin = 2 A, Rp = 1.1 , and R3 = 8.5 . (c) For each case, state whether the circuit is wired as a noninverting or an inverting ampli-ﬁer. Explain your reasoning. 9. (a) Design a circuit using only a single op amp which adds two voltages v1 and v2 and provides an output voltage twice their sum (i.e., vout = 2v1 + 2v2). (b) Verify your design by analyzing the ﬁnal circuit. 10. (a) Design a circuit that provides a current i which is equal in magnitude to the sum of three input voltages v1, v2, and v3. (Compare volts to amperes.) (b) Verify your design by analyzing the ﬁnal circuit. R1 R2 RL – + vout vin + – ■FIGURE 6.40 – + vout R3 Rp iin ■FIGURE 6.41 EXERCISES 209 11. (a) Design a circuit that provides a voltage vout which is equal to the difference between two voltages v2 and v1 (i.e., vout = v2 −v1), if you have only the following resistors from which to choose: two 1.5 k resistors, four 6 k resistors, or three 500 resistors. (b) Verify your design by analyzing the ﬁnal circuit. 12. Analyze the circuit of Fig. 6.42 and determine a value for V1, which is refer-enced to ground. – + 850 1 M 100 250 10 k 850 9 V V1 + – 1 mA ■FIGURE 6.42 13. Derive an expression for vout as a function of v1 and v2 for the circuit repre-sented in Fig. 6.43. – + R1 R2 R3 v1 Rf + – v2 + – vout ■FIGURE 6.43 14. Explain what is wrong with each diagram in Fig. 6.44 if the two op amps are known to be perfectly ideal. – + vout 1 mA 10 k – + + – vout 1 k + – 5 V + – 10 V (a) (b) ■FIGURE 6.44 15. For the circuit depicted in Fig. 6.45, calculate vout if Is = 2 mA, RY = 4.7 k, RX = 1 k, and Rf = 500 . 16. Consider the ampliﬁer circuit shown in Fig. 6.45. What value of Rf will yield vout = 2 V when Is = 10 mA and RY = 2RX = 500 ? – + Rf RY RX IS vout ■FIGURE 6.45 17. With respect to the circuit shown in Fig. 6.46, calculate vout if vs equals (a) 2 cos 100t mV; (b) 2 sin(4t + 19◦) V. CHAPTER 6 THE OPERATIONAL AMPLIFIER 210 – + + – vs vout 1 k 100 3 k 1 k 10–3v v + – ■FIGURE 6.46 – + vout – + 10 2 V 10 5 2 k + – Rx ■FIGURE 6.47 + – vout – + 10 15 5 k vin + – R4 ■FIGURE 6.48 – + vout – + 500 1.5 V 5 k 1.5 k 5 k 5 k 5 k + – + – v1 ■FIGURE 6.49 6.3 Cascaded Stages 18. Calculate vout as labeled in the circuit of Fig. 6.47 if Rx = 1 k. 19. For the circuit of Fig. 6.47, determine the value of Rx that will result in a value of vout = 10 V. 20. Referring to Fig. 6.48, sketch vout as a function of (a) vin over the range of −2 V ≤vin ≤+ 2 V, if R4 = 2 k; (b) R4 over the range of 1 k ≤R4 ≤10 k, if vin = 300 mV. 21. Obtain an expression for vout as labeled in the circuit of Fig. 6.49 if v1 equals (a) 0 V; (b) 1 V; (c) −5 V; (d) 2 sin 100t V. 22. The 1.5 V source of Fig. 6.49 is disconnected, and the output of the circuit shown in Fig. 6.48 is connected to the left-hand terminal of the 500 resistor instead. Calculate vout if R4 = 2 k and (a) vin = 2 V, v1 = 1 V; (b) vin = 1 V, v1 = 0; (c) vin = 1 V, v1 = −1 V. 23. For the circuit shown in Fig. 6.50, compute vout if (a) v1 = 2v2 = 0.5v3 = 2.2 V and R1 = R2 = R3 = 50 k; (b) v1 = 0, v2 = −8 V, v3 = 9 V, and R1 = 0.5R2 = 0.4R3 = 100 k. EXERCISES 211 – + R1 R2 R3 200 k – + vout + – v3 + – v2 + – v1 ■FIGURE 6.50 24. (a) Design a circuit which will add the voltages produced by three separate pressure sensors, each in the range of 0 ≤vsensor ≤5 V, and produce a positive voltage vout linearly correlated to the voltage sum such that vout = 0 when all three voltages are zero, and vout = 2 V when all three voltages are at their max-imum. (b) Verify your design by analyzing the ﬁnal circuit. 25. (a) Design a circuit which produces an output voltage vout proportional to the difference of two positive voltages v1 and v2 such that vout = 0 when both volt-ages are equal, and vout = 10 V when v1 − v2 = 1 V. (b) Verify your design by analyzing the ﬁnal circuit. 26. (a) Three pressure-sensitive sensors are used to double-check the weight read-ings obtained from the suspension systems of a long-range jet airplane. Each sensor is calibrated such that 10 μV corresponds to 1 kg. Design a circuit which adds the three voltage signals to produce an output voltage calibrated such that 10 V corresponds to 400,000 kg, the maximum takeoff weight of the aircraft. (b) Verify your design by analyzing the ﬁnal circuit. 27. (a) The oxygen supply to a particular bathysphere consists of four separate tanks, each equipped with a pressure sensor capable of measuring between 0 (corresponding to 0 V output) and 500 bar (corresponding to 5 V output). Design a circuit which produces a voltage proportional to the total pressure in all tanks, such that 1.5 V corresponds to 0 bar and 3 V corresponds to 2000 bar. (b) Verify your design by analyzing the ﬁnal circuit. 28. For the circuit shown in Fig. 6.51, let vin = 8 V, and select values for R1, R2, and R3 to ensure an output voltage vout = 4 V. – + vout – + – + R1 R2 R3 vin 50 k 200 k + – ■FIGURE 6.51 29. For the circuit of Fig. 6.52, derive an expression for vout in terms of vin. CHAPTER 6 THE OPERATIONAL AMPLIFIER 212 – + R1 1 V R2 R3 R4 – + R5 R6 vout vin + – + – ■FIGURE 6.52 V3 Vbatt + – 890 1.1 k 400 1N750 – + ■FIGURE 6.53 6.4 Circuits for Voltage and Current Sources 30. Construct a circuit based on the 1N4740 diode which provides a reference volt-age of 10 V if only 9 V batteries are available. Note that the breakdown voltage of this diode is equal to 10 V at a current of 25 mA. 31. Employ a 1N4733 Zener diode to construct a circuit which provides a 4 V ref-erence voltage to a 1 k load, if only 9 V batteries are available as sources. Note that the Zener breakdown voltage of this diode is 5.1 V at a current of 76 mA. 32. (a) Design a circuit which provides a 5 V dc reference voltage to an unknown (nonzero resistance) load, if only a 9 V battery is available as a supply. (b) Verify your design with an appropriate simulation. As part of this, deter-mine the acceptable range for the load resistor. 33. A particular passive network can be represented by a Thévenin equivalent re-sistance between 10 and 125 depending on the operating temperature. (a) Design a circuit which provides a constant 2.2 V to this network regardless of temperature. (b) Verify your design with an appropriate simulation (resis-tance can be varied from within a single simulation, as described in Chap. 8). 34. Calculate the voltage V1 as labeled in the circuit of Fig. 6.53 if the battery is rated at Vbatt equal to (a) 9 V; (b) 12 V. (c) Verify your solutions with appropri-ate simulations, commenting on the possible origin of any discrepancies. 35. (a) Design a current source based on the 1N750 diode which is capable of providing a dc current of 750 μA to a load RL, such that 1 k < RL < 50 k. (b) Verify your design with an appropriate simulation (note that resistance can be varied within a single simulation, as described in Chap. 8). 36. (a) Design a current source able to provide a dc current of 50 mA to an unspeciﬁed load. Use a 1N4733 diode (Vbr = 5.1 V at 76 mA). (b) Use an appropriate simulation to determine the permissible range of load resistance for your design. EXERCISES 213 37. (a) Design a current source able to provide a dc current of 10 mA to an unspec-iﬁed load. Use a 1N4747 diode (Vbr = 20 V at 12.5 mA). (b) Use an appropri-ate simulation to determine the permissible range of load resistance for your design. 38. The circuit depicted in Fig. 6.54 is known as a Howland current source. Derive expressions for vout and IL, respectively as a function of V1 and V2. 39. For the circuit depicted in Fig. 6.54, known as a Howland current source, set V2 = 0, R1 = R3, and R2 = R4; then solve for the current IL when R1 = 2R2 = 1 k and RL = 100 . 6.5 Practical Considerations 40. (a) Employ the parameters listed in Table 6.3 for the μA741 op amp to analyze the circuit of Fig. 6.55 and compute a value for vout. (b) Compare your result to what is predicted using the ideal op amp model. – + R1 V2 R3 RL R4 V1 R2 vout IL ■FIGURE 6.54 – + 250 vout + – 450 mV 1.4 k ■FIGURE 6.55 – + 4.7 k vout + – 2 V 470 ■FIGURE 6.56 41. (a) Employ the parameters listed in Table 6.3 for the μA741 op amp to analyze the circuit of Fig. 6.10 if R = 1.5 k, v1 = 2 V, and v2 = 5 V. (b) Compare your solution to what is predicted using the ideal op amp model. 42. Deﬁne the following terms, and explain when and how each can impact the performance of an op amp circuit: (a) common-mode rejection ratio; (b) slew rate; (c) saturation; (d) feedback. 43. For the circuit of Fig. 6.56, replace the 470 resistor with a short circuit, and compute vout using (a) the ideal op amp model; (b) the parameters listed in Table 6.3 for the μA741 op amp; (c) an appropriate PSpice simulation. (d) Compare the values obtained in parts (a) to (c) and comment on the possi-ble origin of any discrepancies. 44. If the circuit of Fig. 6.55 is analyzed using the detailed model of an op amp (as opposed to the ideal op amp model), calculate the value of open-loop gain A required to achieve a closed-loop gain within 2% of its ideal value. 45. Replace the 2 V source in Fig. 6.56 with a sinusoidal voltage source having a magnitude of 3 V and radian frequency ω = 2πf. (a) Which device, a μA741 op amp or an LF411 op amp, will track the source frequency better over the range 1 Hz < f < 10 MHz? Explain. (b) Compare the frequency performance of the circuit over the range 1 Hz < f < 10 MHz using appropriate PSpice simulations, and compare the results to your prediction in part (a). 46. (a) For the circuit of Fig. 6.56, if the op amp (assume LF411) is powered by matched 9 V supplies, estimate the maximum value to which the 470 resis-tor can be increased before saturation effects become apparent. (b) Verify your prediction with an appropriate simulation. 47. For the circuit of Fig. 6.55, calculate the differential input voltage and the input bias current if the op amp is a(n) (a) μA741; (b) LF411; (c) AD549K; (d) OPA690. 48. Calculate the common-mode gain for each device listed in Table 6.3. Express your answer in units of V/V, not dB. 6.6 Comparators and the Instrumentation Ampliﬁer 49. Human skin, especially when damp, is a reasonable conductor of electricity. If we assume a resistance of less than 10 M for a ﬁngertip pressed across two terminals, design a circuit which provides a +1 V output if this nonmechanical switch is “closed” and −1 V if it is “open.” 50. Design a circuit which provides an output voltage vout based on the behavior of another voltage vin, such that vout = +2.5 V vin > 1 V 1.2 V otherwise 51. For the instrumentation ampliﬁer shown in Fig. 6.38a, assume that the three internal op amps are ideal, and determine the CMRR of the circuit if (a) R1 = R3 and R2 = R4; (b) all four resistors have different values. 52. For the circuit depicted in Fig. 6.57, sketch the expected output voltage vout as a function of vactive for −5 V ≤vactive ≤+5 V, if vref is equal to (a) −3 V; (b) +3 V. 53. For the circuit depicted in Fig. 6.58, (a) sketch the expected output voltage vout as a function of v1 for −5 V ≤v1 ≤+5 V, if v2 = +2 V; (b) sketch the expected output voltage vout as a function of v2 for −5 V ≤v2 ≤+5 V, if v1 = +2 V. 54. For the circuit depicted in Fig. 6.59, sketch the expected output voltage vout as a function of vactive, if −2 V ≤vactive ≤+2 V. Verify your solution using a μA741 (although it is slow compared to op amps designed speciﬁcally for use as comparators, its PSpice model works well, and as this is a dc appli-cation speed is not an issue). Submit a properly labeled schematic with your results. CHAPTER 6 THE OPERATIONAL AMPLIFIER 214 vout vactive vref – + + – + – + – 18 V V+ V – ■FIGURE 6.57 vout v1 v2 – + + – + – + – 5 V –5 V V + V – – + ■FIGURE 6.58 vout vactive – + + – + – – + 12 V –12 V V+ V – ■FIGURE 6.59 55. In digital logic applications, a +5 V signal represents a logic “1” state, and a 0 V signal represents a logic “0” state. In order to process real-world informa-tion using a digital computer, some type of interface is required, which typi-cally includes an analog-to-digital (A/D) converter—a device that converts analog signals into digital signals. Design a circuit that acts as a simple 1-bit A/D, with any signal less than 1.5 V resulting in a logic “0” and any signal greater than 1.5 V resulting in a logic “1.” 56. A common application for instrumentation ampliﬁers is to measure voltages in resistive strain gauge circuits. These strain sensors work by exploiting the changes in resistance that result from geometric distortions, as in Eq. of Chap. 2. They are often part of a bridge circuit, as shown in Fig. 6.60a, where the strain gauge is identiﬁed as RG. (a) Show that Vout = Vin R2 R1+R2 − R3 R3+RGauge . (b) Verify that Vout = 0 when the three ﬁxed-value resistors R1, R2, and R3 are all chosen to be equal to the unstrained gauge resistance RGauge. (c) For the intended application, the gauge selected has an unstrained resistance of 5 k, and a maximum resistance increase of EXERCISES 215 Chapter-Integrating Exercises 57. (a) You’re given an electronic switch which requires 5 V at 1 mA in order to close; it is open with no voltage present at its input. If the only microphone available produces a peak voltage of 250 mV, design a circuit which will ener-gize the switch when someone speaks into the microphone. Note that the audio level of a general voice may not correspond to the peak voltage of the micro-phone. (b) Discuss any issues that may need to be addressed if your circuit were to be implemented. 58. You’ve formed a band, despite advice to the contrary. Actually, the band is pretty good except for the fact that the lead singer (who owns the drum set, the microphones, and the garage where you practice) is a bit tone-deaf. Design a circuit that takes the output from each of the ﬁve microphones your band uses, and adds the voltages to create a single voltage signal which is fed to the am-plifer. Except not all voltages should be equally ampliﬁed. One microphone output should be attenuated such that its peak voltage is 10% of any other microphone’s peak voltage. 59. Cadmium sulﬁde (CdS) is commonly used to fabricate resistors whose value depends on the intensity of light shining on the surface. In Fig. 6.61 a CdS “photocell” is used as the feedback resistor Rf. In total darkness, it has a resistance of 100 k, and a resistance of 10 k under a light intensity of 6 candela. RL represents a circuit that is activated when a voltage of 1.5 V or less is applied to its terminals. Choose R1 and Vs so that the circuit represented by RL is activated by a light of 2 candela or brighter. + – Vref RGauge R2 R1 R3 Vout + – (a) – + 1 2 3 4 RG –IN +IN –VS RG +VS OUTPUT REF 8 7 6 5 AD622 (b) ■FIGURE 6.60 AD622 Speciﬁcations Ampliﬁer gain G can be varied from 2 to 1000 by connecting a resistor between pins 1 and 8 with a value calculated by R = 50.5 G −1 k. © Analog Devices. 50 m is expected. Only ±12 V supplies are available. Using the instrumenta-tion ampliﬁer of Fig. 6.60b, design a circuit that will provide a voltage signal of +1 V when the strain gauge is at its maximum loading. – + + – CdS R1 RL Vs ■FIGURE 6.61 CHAPTER 6 THE OPERATIONAL AMPLIFIER 216 60. A fountain outside a certain ofﬁce building is designed to reach a maximum height of 5 meters at a ﬂow rate of 100 l/s. A variable position valve in line with the water supply to the fountain can be controlled electrically, such that 0 V applied results in the valve being fully open, and 5 V results in the valve being closed. In adverse wind conditions the maximum height of the fountain needs to be adjusted; if the wind velocity exceeds 50 km/h, the height cannot exceed 2 meters. A wind velocity sensor is available which provides a voltage calibrated such that 1 V corresponds to a wind velocity of 25 km/h. Design a circuit which uses the velocity sensor to control the fountain according to speciﬁcations. 61. For the circuit of Fig. 6.43, let all resistor values equal 5 k. Sketch vout as a function of time if (a) v1 = 5 sin 5t V and v2 = 5 cos 5t V; (b) v1 = 4e−t V and v2 = 5e−2t V; (c) v1 = 2 V and v2 = e−t V. INTRODUCTION In this chapter we introduce two new passive circuit elements, the capacitor and the inductor, each of which has the ability to both store and deliver ﬁnite amounts of energy. They differ from ideal sources in this respect, since they cannot sustain a ﬁnite average power ﬂow over an inﬁnite time interval.Although they are classed as linear elements, the current-voltage relationships for these new elements are time-dependent, leading to many interesting circuits. The range of capacitance and inductance values we might encounter can be huge, so that at times they may dominate circuit behavior, and at other times be essentially insigniﬁcant. Such issues continue to be relevant in modern circuit applications, particularly as com-puter and communication systems move to increasingly higher operating frequencies and component densities. 7.1 • THE CAPACITOR Ideal Capacitor Model Previously, we termed independent and dependent sources active elements, and the linear resistor a passive element, although our de-ﬁnitions of active and passive are still slightly fuzzy and need to be brought into sharper focus. We now deﬁne an active element as an element that is capable of furnishing an average power greater than zero to some external device, where the average is taken over an inﬁnite time interval. Ideal sources are active elements, and the operational ampliﬁer is also an active device. A passive element, however, is deﬁned as an element that cannot supply an average power that is greater than zero over an inﬁnite time interval. The resistor falls into this category; the energy it receives is usually transformed into heat, and it never supplies energy. KEY CONCEPTS The Voltage-Current Relationship of an Ideal Capacitor The Current-Voltage Relationship of an Ideal Inductor Calculating Energy Stored in Capacitors and Inductors Response of Capacitors and Inductors to Time-Varying Waveforms Series and Parallel Combinations Op Amp Circuits with Capacitors PSpice Modeling of Energy Storage Elements Capacitors and Inductors C H A P T E R 7 217 We now introduce a new passive circuit element, the capacitor. We deﬁne capacitance C by the voltage-current relationship where v and i satisfy the conventions for a passive element, as shown in Fig. 7.1. We should bear in mind that v and i are functions of time; if needed, we can emphasize this fact by writing v(t) and i(t) instead. From Eq. , we may determine the unit of capacitance as an ampere-second per volt, or coulomb per volt. We will now deﬁne the farad1 (F) as one coulomb per volt, and use this as our unit of capacitance. The ideal capacitor deﬁned by Eq. is only a mathematical model of a real device. A capacitor consists of two conducting surfaces on which charge may be stored, separated by a thin insulating layer that has a very large resistance. If we assume that this resistance is sufﬁciently large that it may be considered inﬁnite, then equal and opposite charges placed on the capacitor “plates” can never recombine, at least by any path within the ele-ment. The construction of the physical device is suggested by the circuit symbol shown in Fig. 7.1. Let us visualize some external device connected to this capacitor and causing a positive current to ﬂow into one plate of the capacitor and out of the other plate. Equal currents are entering and leaving the two terminals, and this is no more than we expect for any circuit element. Now let us ex-amine the interior of the capacitor. The positive current entering one plate represents positive charge moving toward that plate through its terminal lead; this charge cannot pass through the interior of the capacitor, and it therefore accumulates on the plate. As a matter of fact, the current and the increasing charge are related by the familiar equation i = dq dt Now let us consider this plate as an overgrown node and apply Kirchhoff’s current law. It apparently does not hold; current is approaching the plate from the external circuit, but it is not ﬂowing out of the plate into the “internal circuit.” This dilemma bothered a famous Scottish scientist, James Clerk Maxwell, more than a century ago. The uniﬁed electromagnetic the-ory that he subsequently developed hypothesizes a “displacement current” that is present wherever an electric ﬁeld or a voltage is varying with time. The displacement current ﬂowing internally between the capacitor plates is exactly equal to the conduction current flowing in the capacitor leads; Kirchhoff’s current law is therefore satisﬁed if we include both conduction and displacement currents. However, circuit analysis is not concerned with this internal displacement current, and since it is fortunately equal to the conduction current, we may consider Maxwell’s hypothesis as relating the conduction current to the changing voltage across the capacitor. A capacitor constructed of two parallel conducting plates of area A, separated by a distance d, has a capacitanceC = εA/d, where εis the permit-tivity, a constant of the insulating material between the plates; this assumes i = C dv dt i v + – C ■FIGURE 7.1 Electrical symbol and current-voltage conventions for a capacitor. (1) Named in honor of Michael Faraday. CHAPTER 7 CAPACITORS AND INDUCTORS 218 (a) (b) (c) ■FIGURE 7.2 Several examples of commercially available capacitors. (a) Left to right: 270 pF ceramic, 20 μF tantalum, 15 nF polyester, 150 nF polyester. (b) Left: 2000 μF 40 VDC rated electrolytic, 25,000 μF 35 VDC rated electrolytic. (c) Clockwise from smallest: 100 μF 63 VDC rated electrolytic, 2200 μF 50 VDC rated electrolytic, 55 F 2.5 VDC rated electrolytic, and 4800 μF 50 VDC rated electrolytic. Note that generally speaking larger capacitance values require larger packages, with one notable exception above. What was the tradeoff in that case? Determine the current i ﬂowing through the capacitor of Fig. 7.1 for the two voltage waveforms of Fig. 7.3 if C 2 F. EXAMPLE 7.1 (Continued on next page) (b) v (V) –6 –4 –2 0 2 4 6 –1 0 1 t (s) 2 3 4 5 (a) –2 –1 0 1 t (s) v (V) 2 3 4 5 –1 0 1 2 3 4 5 6 7 8 ■FIGURE 7.3 (a) A dc voltage applied to the terminals of the capacitor. (b) A sinusoidal voltage waveform applied to the capacitor terminals. SECTION 7.1 THE CAPACITOR 219 the linear dimensions of the conducting plates are all very much greater than d. For air or vacuum, ε = ε0 = 8.854 pF/m. Most capacitors employ a thin dielectric layer with a larger permittivity than air in order to minimize the device size. Examples of various types of commercially available capacitors are shown in Fig. 7.2, although we should remember that any two conducting surfaces not in direct contact with each other may be characterized by a nonzero (although probably small) capacitance. We should also note that a capacitance of several hundred microfarads (μF) is considered “large.” Several important characteristics of our new mathematical model can be discovered from the deﬁning equation, Eq. . A constant voltage across a capacitor results in zero current passing through it; a capacitor is thus an “open circuit to dc.” This fact is pictorially represented by the capacitor symbol. It is also apparent that a sudden jump in the voltage requires an in-ﬁnite current. Since this is physically impossible, we will therefore prohibit the voltage across a capacitor to change in zero time. (a) –2 –1 0 1 t (s) i (A) 2 3 4 5 –1.5 –1 –0.5 0 0.5 1 1.5 2 (b) i (A) –10 –5 0 5 10 –1 0 1 t (s) 2 3 4 5 ■FIGURE 7.4 (a) i 0 as the voltage applied is dc. (b) The current has a cosine form in response to a sine wave voltage. PRACTICE ● 7.1 Determine the current ﬂowing through a 5 mF capacitor in response to a voltage v equal to: (a) −20 V; (b) 2e−5t V. Ans: 0 A; −50e−5t mA. (2) Note that we are employing the mathematically correct procedure of deﬁning a dummy variable t′ in situations where the integration variable t is also a limit. The current i is related to the voltage v across the capacitor by Eq. : i = C dv dt For the voltage waveform depicted in Fig. 7.3a, dv/dt = 0, so i = 0; the result is plotted in Fig. 7.4a. For the case of the sinusoidal waveform of Fig. 7.3b, we expect a cosine current waveform to ﬂow in response, having the same frequency and twice the magnitude (since C 2 F). The result is plotted in Fig. 7.4b. Integral Voltage-Current Relationships The capacitor voltage may be expressed in terms of the current by integrat-ing Eq. . We ﬁrst obtain dv = 1 C i(t) dt and then integrate2 between the times t0 and t and between the correspond-ing voltages v(t0) and v(t): Equation may also be written as an indeﬁnite integral plus a constant of integration: v(t) = 1 C i dt + k v(t) = 1 C t t0 i(t′) dt′ + v(t0) CHAPTER 7 CAPACITORS AND INDUCTORS 220 Find the capacitor voltage that is associated with the current shown graphically in Fig. 7.5a. The value of the capacitance is 5 μF. EXAMPLE 7.2 20 1 0 2 3 4 –1 (a) i(t) (mA) t (ms) 8 1 0 2 3 4 –1 (b) v(t) (V) t (ms) ■FIGURE 7.5 (a) The current waveform applied to a 5 μF capacitor. (b) The resultant voltage waveform obtained by graphical integration. Equation is the appropriate expression here: v(t) = 1 C t t0 i(t′) dt′ + v(t0) but now it needs to be interpreted graphically. To do this, we note that the difference in voltage between times t and t0 is proportional to the area under the current curve deﬁned by the same two times. The con-stant of proportionality is 1/C. From Fig. 7.5a, we see three separate intervals: t ≤0, 0 ≤t ≤2 ms, and t ≥2 ms. Deﬁning the ﬁrst interval more speciﬁcally as between −∞and 0, so that t0 = −∞, we note two things, both a consequence of the fact that the current has always been zero up to t = 0: First, v(t0) = v(−∞) = 0 Second, the integral of the current between t0 = −∞and 0 is simply zero, since i = 0 in that interval. Thus, v(t) = 0 + v(−∞) −∞≤t ≤0 or v(t) = 0 t ≤0 (Continued on next page) SECTION 7.1 THE CAPACITOR 221 Finally, in many situations we will ﬁnd that v(t0), the voltage initially across the capacitor, is not able to be discerned. In such instances it is mathematically convenient to set t0 = −∞and v(−∞) = 0, so that v(t) = 1 C t −∞ i dt′ Since the integral of the current over any time interval is the correspond-ing charge accumulated on the capacitor plate into which the current is ﬂowing, we may also deﬁne capacitance as q(t) = Cv(t) where q(t) and v(t) represent instantaneous values of the charge on either plate and the voltage between the plates, respectively. 2 1 0 2 3 4 –1 v(t) (V) t (ms) ■FIGURE 7.6 If we now consider the time interval represented by the rectangular pulse, we obtain v(t) = 1 5 × 10−6 t 0 20 × 10−3 dt′ + v(0) Since v(0) = 0, v(t) = 4000t 0 ≤t ≤2 ms For the semi-inﬁnite interval following the pulse, the integral of i(t) is once again zero, so that v(t) = 8 t ≥2 ms The results are expressed much more simply in a sketch than by these analytical expressions, as shown in Fig. 7.5b. PRACTICE ● 7.2 Determine the current through a 100 pF capacitor if its voltage as a function of time is given by Fig. 7.6. Ans: 0 A, −∞≤t ≤1 ms; 200 nA, 1 ms ≤t ≤2 ms; 0 A, t ≥2 ms. Energy Storage To determine the energy stored in a capacitor, we begin with the power delivered to it: p = vi = Cv dv dt The change in energy stored in its electric ﬁeld is simply t t0 p dt′ = C t t0 v dv dt′ dt′ = C v(t) v(t0) v′ dv′ = 1 2C [v(t)]2 −[v(t0)]2 and thus wC(t) −wC(t0) = 1 2C [v(t)]2 −[v(t0)]2 where the stored energy is wC(t0) in joules (J) and the voltage at t0 is v(t0). If we select a zero-energy reference at t0, implying that the capacitor volt-age is also zero at that instant, then Let us consider a simple numerical example. As sketched in Fig. 7.7, a sinusoidal voltage source is in parallel with a 1 M resistor and a 20 μF capacitor. The parallel resistor may be assumed to represent the ﬁnite resis-tance of the dielectric between the plates of the physical capacitor (an ideal capacitor has inﬁnite resistance). wC(t) = 1 2Cv2 CHAPTER 7 CAPACITORS AND INDUCTORS 222 EXAMPLE 7.3 Find the maximum energy stored in the capacitor of Fig. 7.7 and the energy dissipated in the resistor over the interval 0 < t < 0.5 s. Identify the goal of the problem. The energy stored in the capacitor varies with time; we are asked for the maximum value over a speciﬁc time interval. We are also asked to ﬁnd the total amount of energy dissipated by the resistor over this in-terval. These are actually two completely different questions. Collect the known information. The only source of energy in the circuit is the independent voltage source, which has a value of 100 sin 2πt V. We are only interested in the time interval of 0 < t < 0.5 s. The circuit is properly labeled. Devise a plan. Determine the energy in the capacitor by evaluating the voltage. To ﬁnd the energy dissipated in the resistor during the same time interval, integrate the dissipated power, pR = i2 R · R. Construct an appropriate set of equations. The energy stored in the capacitor is simply wC(t) = 1 2Cv2 = 0.1 sin2 2πt J We obtain an expression for the power dissipated by the resistor in terms of the current iR: iR = v R = 10−4 sin 2πt A and so pR = i2 R R = (10−4)(106) sin2 2πt so that the energy dissipated in the resistor between 0 and 0.5 s is wR = 0.5 0 pR dt = 0.5 0 10−2 sin2 2πt dt J Determine if additional information is required. We have an expression for the energy stored in the capacitor; a sketch is shown in Fig. 7.8. The expression derived for the energy dissipated by the resistor does not involve any unknown quantities, and so may also be readily evaluated. Attempt a solution. From our sketch of the expression for the energy stored in the capaci-tor, we see that it increases from zero at t = 0 to a maximum of 100 mJ at t = 1 4 s, and then decreases to zero in another 1 4 s. Thus, wCmax = 100 mJ. Evaluating our integral expression for the energy dissipated in the resistor, we ﬁnd that wR = 2.5 mJ. (Continued on next page) iC iR 20 F 1 M 100 sin 2t V v + – + – ■FIGURE 7.7 A sinusoidal voltage source is applied to a parallel RC network. The 1 M resistor might represent the ﬁnite resistance of the “real” capacitor’s dielectric layer. 0 0.02 0.04 0.06 0.08 0.10 0.1 0.2 0.3 0.4 0.5 0 wC(t) = 0.1 sin2 2t (J) t (s) ■FIGURE 7.8 A sketch of the energy stored in the capacitor as a function of time. SECTION 7.1 THE CAPACITOR 223 –0.015 –0.010 –0.005 0 0.005 0.010 0.015 0.10 0.08 Current (mA) Current (A) 0.06 0.04 0.02 0 0 0.05 0.1 0.15 0.2 0.25 t (s) 0.3 0.35 0.4 0.45 0.5 iC iC iS iS iR iR ■FIGURE 7.9 Plot of the resistor, capacitor, and source currents during the interval of 0 to 500 ms. PRACTICE ● 7.3 Calculate the energy stored in a 1000 μF capacitor at t = 50 μs if the voltage across it is 1.5 cos 105t volts. Ans: 90.52 μJ. Verify the solution. Is it reasonable or expected? We do not expect to calculate a negative stored energy, which is borne out in our sketch. Further, since the maximum value of sin 2πt is 1, the maxi-mum energy expected anywhere would be (1/2)(20 × 10−6)(100)2 = 100 mJ. The resistor dissipated 2.5 mJ in the period of 0 to 500 ms, although the capacitor stored a maximum of 100 mJ at one point during that interval. What happened to the “other” 97.5 mJ? To answer this, we compute the capacitor current iC = 20 × 10−6 dv dt = 0.004π cos 2πt and the current is deﬁned as ﬂowing into the voltage source is = −iC −iR both of which are plotted in Fig. 7.9. We observe that the current ﬂow-ing through the resistor is a small fraction of the source current, not entirely surprising as 1 M is a relatively large resistance value. As current ﬂows from the source, a small amount is diverted to the resistor, with the rest ﬂowing into the capacitor as it charges. After t = 250 ms, the source current is seen to change sign; current is now ﬂowing from the capacitor back into the source. Most of the energy stored in the capacitor is being returned to the ideal voltage source, except for the small fraction dissipated in the resistor. CHAPTER 7 CAPACITORS AND INDUCTORS 224 SECTION 7.2 THE INDUCTOR 225 7.2 • THE INDUCTOR Ideal Inductor Model In the early 1800s the Danish scientist Oersted showed that a current-carrying conductor produced a magnetic field (compass needles were affected in the presence of a wire when current was ﬂowing). Shortly there-after, Ampère made some careful measurements which demonstrated that this magnetic ﬁeld was linearly related to the current which produced it. The next step occurred some 20 years later when the English experimental-ist Michael Faraday and the American inventor Joseph Henry discovered almost simultaneously3 that a changing magnetic ﬁeld could induce a volt-age in a neighboring circuit. They showed that this voltage was proportional to the time rate of change of the current producing the magnetic ﬁeld. The constant of proportionality is what we now call the inductance, symbolized by L, and therefore where we must realize that v and i are both functions of time. When we wish to emphasize this, we may do so by using the symbols v(t) and i(t). The circuit symbol for the inductor is shown in Fig. 7.10, and it should be noted that the passive sign convention is used, just as it was with the re-sistor and the capacitor. The unit in which inductance is measured is the henry (H), and the deﬁning equation shows that the henry is just a shorter expression for a volt-second per ampere. v = L di dt (3) Faraday won. Important Characteristics of an Ideal Capacitor 1. There is no current through a capacitor if the voltage across it is not changing with time.Acapacitor is therefore an open circuit to dc. 2. A ﬁnite amount of energy can be stored in a capacitor even if the current through the capacitor is zero, such as when the voltage across it is constant. 3. It is impossible to change the voltage across a capacitor by a ﬁnite amount in zero time, as this requires an inﬁnite current through the capacitor. (A capacitor resists an abrupt change in the voltage across it in a manner analogous to the way a spring resists an abrupt change in its displacement.) 4. A capacitor never dissipates energy, but only stores it. Although this is true for the mathematical model, it is not true for a physical capacitor due to ﬁnite resistances associated with the dielectric as well as the packaging. iL L vL + – ■FIGURE 7.10 Electrical symbol and current-voltage conventions for an inductor. (a) (b) ■FIGURE 7.11 (a) Several different types of commercially available inductors, sometimes also referred to as “chokes.” Clockwise, starting from far left: 287 μH ferrite core toroidal inductor, 266 μH ferrite core cylindrical inductor, 215 μH ferrite core inductor designed for VHF frequencies, 85 μH iron powder core toroidal inductor, 10 μH bobbin-style inductor, 100 μH axial lead inductor, and 7 μH lossy-core inductor used for RF suppression. (b) An 11 H inductor, measuring 10 cm (tall) × 8 cm (wide) × 8 cm (deep). CHAPTER 7 CAPACITORS AND INDUCTORS 226 The inductor whose inductance is deﬁned by Eq. is a mathematical model; it is an ideal element which we may use to approximate the behavior of a real device. A physical inductor may be constructed by winding a length of wire into a coil. This serves effectively to increase the current that is causing the magnetic ﬁeld and also to increase the “number” of neigh-boring circuits into which Faraday’s voltage may be induced. The result of this twofold effect is that the inductance of a coil is approximately propor-tional to the square of the number of complete turns made by the conduc-tor out of which it is formed. For example, an inductor or “coil” that has the form of a long helix of very small pitch is found to have an inductance of μN 2A/s, where A is the cross-sectional area, s is the axial length of the he-lix, N is the number of complete turns of wire, and μ (mu) is a constant of the material inside the helix, called the permeability. For free space (and very closely for air), μ = μ0 = 4π × 10−7 H/m = 4π nH/cm. Several ex-amples of commercially available inductors are shown in Fig. 7.11. Let us now scrutinize Eq. to determine some of the electrical charac-teristics of the mathematical model. This equation shows that the voltage across an inductor is proportional to the time rate of change of the current through it. In particular, it shows that there is no voltage across an inductor carrying a constant current, regardless of the magnitude of this current. Accordingly, we may view an inductor as a short circuit to dc. Another fact that can be obtained from Eq. is that a sudden or dis-continuous change in the current must be associated with an inﬁnite voltage across the inductor. In other words, if we wish to produce an abrupt change in an inductor current, we must apply an inﬁnite voltage. Although an infinite-voltage forcing function might be amusing theoretically, it can never be a part of the phenomena displayed by a real physical device. As we SECTION 7.2 THE INDUCTOR 227 Given the waveform of the current in a 3 H inductor as shown in Fig. 7.12a, determine the inductor voltage and sketch it. EXAMPLE 7.4 ■FIGURE 7.12 (a) The current waveform in a 3 H inductor. (b) The corresponding voltage waveform, v = 3 di/dt. (a) 1 –1 1 0 2 3 i(t) (A) t (s) (b) 3 –3 –1 1 0 2 3 v(t) (V) t (s) Deﬁning the voltage v and the current i to satisfy the passive sign con-vention, we may obtain v from Fig. 7.12a using Eq. : v = 3 di dt Since the current is zero for t < −1 s, the voltage is zero in this inter-val. The current then begins to increase at the linear rate of 1 A/s, and thus a constant voltage of L di/dt = 3 V is produced. During the following 2 s interval, the current is constant and the voltage is there-fore zero. The ﬁnal decrease of the current results in di/dt = −1 A/s, yielding v = −3 V. For t > 3 s, i(t) is a constant (zero), so that v(t) = 0 for that interval. The complete voltage waveform is sketched in Fig. 7.12b. shall see shortly, an abrupt change in the inductor current also requires an abrupt change in the energy stored in the inductor, and this sudden change in energy requires inﬁnite power at that instant; inﬁnite power is again not a part of the real physical world. In order to avoid inﬁnite voltage and inﬁnite power, an inductor current must not be allowed to jump instantaneously from one value to another. If an attempt is made to open-circuit a physical inductor through which a ﬁnite current is ﬂowing, an arc may appear across the switch. This is use-ful in the ignition system of some automobiles, where the current through the spark coil is interrupted by the distributor and the arc appears across the spark plug. Although this does not occur instantaneously, it happens in a very short timespan, leading to the creation of a large voltage. The pres-ence of a large voltage across a short distance equates to a very large electric ﬁeld; the stored energy is dissipated in ionizing the air in the path of the arc. Equation may also be interpreted (and solved, if necessary) by graphical methods, as seen in Example 7.4. ■FIGURE 7.13 4 2 –2 –4 –6 1 2 3 4 –3 –2 –1 5 6 7 iL (mA) t (ms) Find the inductor voltage that results from applying the current waveform shown in Fig. 7.14a to the inductor of Example 7.4. EXAMPLE 7.5 1 –1 1 0 2 2.1 –0.1 3 (a) i(t) (A) t (s) v(t) (V) 30 –30 –1 1 0 2 2.1 –0.1 3 (b) t (s) ■FIGURE 7.14 (a) The time required for the current of Fig. 7.12a to change from 0 to 1 and from 1 to 0 is decreased by a factor of 10. (b) The resultant voltage waveform. The pulse widths are exaggerated for clarity. Note that the intervals for the rise and fall have decreased to 0.1 s. Thus, the magnitude of each derivative will be 10 times larger; this condition is shown in the current and voltage sketches of Fig. 7.14a and b. In the voltage waveforms of Fig. 7.13b and 7.14b, it is interest-ing to note that the area under each voltage pulse is 3 V· s. CHAPTER 7 CAPACITORS AND INDUCTORS 228 Just for curiosity’s sake, let’s continue in the same vein for a moment. A further decrease in the rise and fall times of the current waveform will pro-duce a proportionally larger voltage magnitude, but only within the interval PRACTICE ● 7.4 The current through a 200 mH inductor is shown in Fig. 7.13. Assume the passive sign convention, and ﬁnd vL at t equal to (a) 0; (b) 2 ms; (c) 6 ms. Ans: 0.4 V; 0.2 V; −0.267 V. Let us now investigate the effect of a more rapid rise and decay of the current between the 0 and l A values. ■FIGURE 7.15 (a) The time required for the current of Fig. 7.14a to change from 0 to 1 and from 1 to 0 is decreased to zero; the rise and fall are abrupt. (b) The resultant voltage across the 3 H inductor consists of a positive and a negative inﬁnite spike. 1 –1 1 0 2 3 (a) i(t) (A) t (s) (to ) –1 1 0 2 (to – ) 3 (b) v(t) (V) t (s) in which the current is increasing or decreasing. An abrupt change in the current will cause the inﬁnite voltage “spikes” (each having an area of 3 V· s) that are suggested by the waveforms of Fig. 7.15a and b; or, from the equally valid but opposite point of view, these inﬁnite voltage spikes are required to produce the abrupt changes in the current. PRACTICE ● 7.5 The current waveform of Fig. 7.14a has equal rise and fall times of duration 0.1 s (100 ms). Calculate the maximum positive and negative voltages across the same inductor if the rise and fall times, respectively, are changed to (a) 1 ms, 1 ms; (b) 12 μs, 64 μs; (c) 1 s, 1 ns. Ans: 3 kV, −3 kV; 250 kV, −46.88 kV; 3 V, −3 GV. Integral Voltage-Current Relationships We have deﬁned inductance by a simple differential equation, v = L di dt and we have been able to draw several conclusions about the characteristics of an inductor from this relationship. For example, we have found that we may consider an inductor to be a short circuit to direct current, and we have agreed that we cannot permit an inductor current to change abruptly from one value to another, because this would require that an inﬁnite voltage and power be associated with the inductor. The simple deﬁning equation for inductance contains still more information, however. Rewritten in a slightly different form, di = 1 L v dt it invites integration. Let us ﬁrst consider the limits to be placed on the two integrals. We desire the current i at time t, and this pair of quantities there-fore provides the upper limits on the integrals appearing on the left and right sides of the equation, respectively; the lower limits may also be kept general by merely assuming that the current is i(t0) at time t0. Thus, i(t) i(t0) di′ = 1 L t t0 v(t′) dt′ which leads to the equation i(t) −i(t0) = 1 L t t0 v dt′ or Equation expresses the inductor voltage in terms of the current, whereas Eq. gives the current in terms of the voltage. Other forms are i(t) = 1 L t t0 v dt′ + i(t0) SECTION 7.2 THE INDUCTOR 229 The voltage across a 2 H inductor is known to be 6 cos 5t V. Determine the resulting inductor current if i(t −π/2) 1 A. From Eq. , i(t) = 1 2 t t0 6 cos 5t′ dt′ + i(t0) or i(t) = 1 2 6 5 sin 5t −1 2 6 5 sin 5t0 + i(t0) = 0.6 sin 5t −0.6 sin 5t0 + i(t0) The ﬁrst term indicates that the inductor current varies sinusoidally; the second and third terms together represent a constant which becomes known when the current is numerically speciﬁed at some instant of time. Using the fact that the current is 1 A at t = −π/2 s, we identify t0 as −π/2 with i(t0) = 1, and ﬁnd that i(t) = 0.6 sin 5t −0.6 sin(−2.5π) + 1 or i(t) = 0.6 sin 5t + 1.6 Alternatively, from Eq. , i(t) = 0.6 sin 5t + k and we establish the numerical value of k by forcing the current to be 1 A at t = −π/2: 1 = 0.6 sin(−2.5π) + k or k = 1 + 0.6 = 1.6 and so, as before, i(t) = 0.6 sin 5t + 1.6 EXAMPLE 7.6 CHAPTER 7 CAPACITORS AND INDUCTORS 230 also possible for the latter equation. We may write the integral as an indeﬁ-nite integral and include a constant of integration k: i(t) = 1 L v dt + k We also may assume that we are solving a realistic problem in which the selection of t0 as −∞ensures no current or energy in the inductor. Thus, if i(t0) = i(−∞) = 0, then i(t) = 1 L t −∞ v dt′ Let us investigate the use of these several integrals by working a simple example where the voltage across an inductor is speciﬁed. SECTION 7.2 THE INDUCTOR 231 We should not make any snap judgments, however, as to which single form of Eqs. , , and we are going to use forever after; each has its advantages, depending on the problem and the application. Equation represents a long, general method, but it shows clearly that the constant of integration is a current. Equation is a somewhat more concise expression of Eq. , but the nature of the integration constant is suppressed. Finally, Eq. is an excellent expression, since no constant is necessary; however, it applies only when the current is zero at t = −∞and when the analytical expression for the current is not indeterminate there. Energy Storage Let us now turn our attention to power and energy. The absorbed power is given by the current-voltage product p = vi = Li di dt The energy wL accepted by the inductor is stored in the magnetic ﬁeld around the coil. The change in this energy is expressed by the integral of the power over the desired time interval: t t0 p dt′ = L t t0 i di dt′ dt′ = L i(t) i(t0) i′ di′ = 1 2 L [i(t)]2 −[i(t0)]2 Thus, wL(t) −wL(t0) = 1 2 L [i(t)]2 −[i(t0)]2 Equation is going to cause trouble with this particular voltage. We based the equation on the assumption that the current was zero when t = −∞. To be sure, this must be true in the real, physical world, but we are working in the land of the mathematical model; our ele-ments and forcing functions are all idealized. The difﬁculty arises after we integrate, obtaining i(t) = 0.6sin 5t′ t −∞ and attempt to evaluate the integral at the lower limit: i(t) = 0.6 sin 5t −0.6 sin(−∞) The sine of ±∞is indeterminate, and therefore we cannot evaluate our expression. Equation is only useful if we are evaluating func-tions which approach zero as t →−∞. PRACTICE ● 7.6 A 100 mH inductor has voltage vL = 2e−3t V across its terminals. Determine the resulting inductor current if iL(−0.5) = 1 A. Ans: −20 3 e−3t + 30.9 A. ■FIGURE 7.16 A sinusoidal current is applied as a forcing function to a series RL circuit. The 0.1 represents the inherent resistance of the wire from which the inductor is fabricated. i vL + – vR + – 0.1 3 H 12 sin A t 6 Find the maximum energy stored in the inductor of Fig. 7.16, and calculate how much energy is dissipated in the resistor in the time during which the energy is being stored in, and then recovered from, the inductor. EXAMPLE 7.7 The energy stored in the inductor is wL = 1 2 Li2 = 216 sin2 πt 6 J and this energy increases from zero at t = 0 to 216 J at t = 3 s. Thus, the maximum energy stored in the inductor is 216 J. After reaching its peak value at t = 3 s, the energy has completely left the inductor 3 s later. Let us see what price we have paid in this coil for the privilege of storing and removing 216 J in these 6 seconds. The power dissipated in the resistor is easily found as pR = i2R = 14.4 sin2 πt 6 W and the energy converted into heat in the resistor within this 6 s interval is therefore wR = 6 0 pR dt = 6 0 14.4 sin2 π 6 t dt where we have again assumed that the current is i(t0) at time t0. In using the energy expression, it is customary to assume that a value of t0 is selected at which the current is zero; it is also customary to assume that the energy is zero at this time. We then have simply where we now understand that our reference for zero energy is any time at which the inductor current is zero. At any subsequent time at which the cur-rent is zero, we also ﬁnd no energy stored in the coil. Whenever the current is not zero, and regardless of its direction or sign, energy is stored in the in-ductor. It follows, therefore, that energy may be delivered to the inductor for a part of the time and recovered from the inductor later. All the stored energy may be recovered from an ideal inductor; there are no storage charges or agent’s commissions in the mathematical model. A physical coil, however, must be constructed out of real wire and thus will always have a resistance associated with it. Energy can no longer be stored and recovered without loss. These ideas may be illustrated by a simple example. In Fig. 7.16, a 3 H inductor is shown in series with a 0.1 resistor and a sinusoidal current source, is = 12 sin πt 6 A. The resistor should be interpreted as the resistance of the wire which must be associated with the physical coil. wL(t) = 1 2 Li2 CHAPTER 7 CAPACITORS AND INDUCTORS 232 We summarize by listing four key characteristics of an inductor which result from its deﬁning equation v = L di/dt: SECTION 7.2 THE INDUCTOR 233 or wR = 6 0 14.4 1 2 1 −cos π 3 t dt = 43.2 J Thus, we have expended 43.2 J in the process of storing and then re-covering 216 J in a 6 s interval. This represents 20 percent of the maxi-mum stored energy, but it is a reasonable value for many coils having this large an inductance. For coils having an inductance of about 100 μH, we might expect a ﬁgure closer to 2 or 3 percent. PRACTICE ● 7.7 Let L = 25 mH for the inductor of Fig. 7.10. (a) Find vL at t = 12 ms if iL = 10te−100t A. (b) Find iL at t = 0.1 s if vL = 6e−12t V and iL(0) = 10 A. If iL = 8(1 −e−40t) mA, ﬁnd (c) the power being delivered to the inductor at t = 50 ms and (d) the energy stored in the inductor at t = 40 ms. Ans: −15.06 mV; 24.0 A; 7.49 μW; 0.510 μJ. Important Characteristics of an Ideal Inductor 1. There is no voltage across an inductor if the current through it is not changing with time. An inductor is therefore a short circuit to dc. 2. A ﬁnite amount of energy can be stored in an inductor even if the voltage across the inductor is zero, such as when the current through it is constant. 3. It is impossible to change the current through an inductor by a ﬁnite amount in zero time, for this requires an inﬁnite voltage across the inductor. (An inductor resists an abrupt change in the current through it in a manner analogous to the way a mass resists an abrupt change in its velocity.) 4. The inductor never dissipates energy, but only stores it. Although this is true for the mathematical model, it is not true for a physical inductor due to series resistances. It is interesting to anticipate our discussion of duality in Sec. 7.6 by rereading the previous four statements with certain words replaced by their “duals.” If capacitor and inductor, capacitance and inductance, voltage and current, across and through, open circuit and short circuit, spring and mass, and displacement and velocity are interchanged (in either direction), the four statements previously given for capacitors are obtained. So far, we have been introduced to three different two-terminal passive elements: the resistor, the capacitor, and the inductor. Each has been defined in terms of its current-voltage relationship (v = Ri, i = C dv/dt, and v = L di/dt, respectively). From a more fundamen-tal perspective, however, we can view these three ele-ments as part of a larger picture relating four basic quantities, namely, charge q, current i, voltage v, and ﬂux linkage ϕ. Charge, current, and voltage are discussed in Chap. 2. Flux linkage is the product of magnetic ﬂux and the number of turns of conducting wire linked by the ﬂux, and it can be expressed in terms of the voltage v across the coil as ϕ = v dt or v = dϕ/dt. Figure 7.17 graphically represents how these four quantities are interrelated. First, apart from any circuit elements and their characteristics, we have dq = i dt (Chap. 2) and now dϕ = v dt. Charge is related to volt-age in the context of a capacitor, since C = dq/dv or dq = C dv. The element we call a resistor provides a direct relationship between voltage and current, which for consistency can be expressed as dv = R di. Continuing our counterclockwise journey around the perimeter of Fig. 7.17, we note that our original expression connecting the voltage and current associated with an inductor can be written in terms of current i and ﬂux linkage ϕ, since rearranging yields v dt = L di, and we know dϕ = v dt. Thus, for the inductor, we can write dϕ = L di. So far, we have traveled from q to v with the aid of a capacitor, v to i using the resistor, and i to ϕ using the inductor. However, we have not yet used any element to connect ϕ and q, although symmetry suggests such a thing should be possible. In the early 1970s, Leon Chua found himself thinking along these lines, and postulated a new device—a “missing” two-terminal circuit element— and named it the memristor.1 He went on to demonstrate that the electrical characteristics of a memristor should be nonlinear, and depend on its history—in other words, a memristor might be characterized by having a memory (hence its name). Independent of his work, others had proposed a similar device, not so much for its practical use in real circuits, but for its potential in device model-ing and signal processing. Not a great deal was heard of this hypothetical element afterward, at least until Dmitri Strukov and coworkers at HP Labs in Palo Alto published a short paper in 2008 claiming to have “found” the memristor.2 They offer several reasons why it took almost four decades to realize the general type of device Chua hypothesized in 1971, but one of the most interesting has to do with size. In making their prototype memristor, nanotechnology (the art of fabricating devices with at least one dimension less than 1000 nm, which is approx-imately 1% of the diameter of human hair) played a key role. A 5 nm thick oxide layer sandwiched between platinum electrodes comprises the entire device. The nonlinear electrical characteristics of the prototype immediately generated considerable excitement, most notably for its potential applications in integrated cir-cuits, where devices are already approaching their small-est realistic size; and many believe new types of devices will be required to further extend integrated circuit density and functionality. Whether the memristor is the circuit element that will allow this remains to be seen— despite the report of a prototype, there remains much work to be done before it becomes practical. PRACTICAL APPLICATION In Search of the Missing Element PRACTICAL APPLICATION Capacitor dq Cdv Resistor dv Rdi Memristor d Mdq Inductor d Ldi v i q d vdt dq idt ■FIGURE 7.17 A graphical representation of the four basic two-terminal passive elements (resistor, capacitor, inductor, and memristor) and their interrelationships. Note that ﬂux linkage is more commonly represented by the Greek letter λ to distinguish it from ﬂux: then λ = Nϕ where N is the number of turns and ϕ is the ﬂux. (Reprinted by permission from Macmillan Publishers Ltd. Nature Publishing Group, “Electronics: The Fourth Element,” Volume 453, pg. 42, 2008.) (1) L. O. Chua, “Memristor—The missing circuit element,” IEEE Transactions on Circuit Theory CT-18 (5), 1971, p. 507. (2) D. B. Strukov, G. S. Snider, D. R. Stewart, and R. S. Williams, “The missing memristor found,” Nature 453, 2008, p. 80. ■FIGURE 7.18 (a) A circuit containing N inductors in series. (b) The desired equivalent circuit, in which Leq = L1 + L2 + · · · + LN. + – i vN LN + – vs (a) v2 + – v1 + – L1 L2 + – i (b) Leq vs SECTION 7.3 INDUCTANCE AND CAPACITANCE COMBINATIONS 235 7.3 • INDUCTANCE AND CAPACITANCE COMBINATIONS Now that we have added the inductor and capacitor to our list of passive cir-cuit elements, we need to decide whether or not the methods we have de-veloped for resistive circuit analysis are still valid. It will also be convenient to learn how to replace series and parallel combinations of either of these elements with simpler equivalents, just as we did with resistors in Chap. 3. We look ﬁrst at Kirchhoff’s two laws, both of which are axiomatic. However, when we hypothesized these two laws, we did so with no restric-tions as to the types of elements constituting the network. Both, therefore, remain valid. Inductors in Series Now we may extend the procedures we have derived for reducing various combinations of resistors into one equivalent resistor to the analogous cases of inductors and capacitors. We shall ﬁrst consider an ideal voltage source applied to the series combination of N inductors, as shown in Fig. 7.18a. We desire a single equivalent inductor, with inductance Leq, which may re-place the series combination so that the source current i(t) is unchanged. The equivalent circuit is sketched in Fig. 7.18b. Applying KVL to the orig-inal circuit, vs = v1 + v2 + · · · + vN = L1 di dt + L2 di dt + · · · + L N di dt = (L1 + L2 + · · · + L N) di dt or, written more concisely, vs = N n=1 vn = N n=1 Ln di dt = di dt N n=1 Ln But for the equivalent circuit we have vs = Leq di dt and thus the equivalent inductance is Leq = L1 + L2 + · · · + L N or Leq = N n=1 Ln The inductor which is equivalent to several inductors connected in series is one whose inductance is the sum of the inductances in the original circuit. This is exactly the same result we obtained for resistors in series. Inductors in Parallel The combination of a number of parallel inductors is accomplished by writ-ing the single nodal equation for the original circuit, shown in Fig. 7.19a, is = N n=1 in = N n=1 1 Ln t t0 v dt′ + in(t0) = N n=1 1 Ln t t0 v dt′ + N n=1 in(t0) and comparing it with the result for the equivalent circuit of Fig. 7.19b, is = 1 Leq t t0 v dt′ + is(t0) Since Kirchhoff’s current law demands that is(t0) be equal to the sum of the branch currents at t0, the two integral terms must also be equal; hence, Leq = 1 1/L1 + 1/L2 + · · · + 1/L N For the special case of two inductors in parallel, Leq = L1L2 L1 + L2 and we note that inductors in parallel combine exactly as do resistors in parallel. Capacitors in Series In order to ﬁnd a capacitor that is equivalent to N capacitors in series, we use the circuit of Fig. 7.20a and its equivalent in Fig. 7.20b to write vs = N n=1 vn = N n=1 1 Cn t t0 i dt′ + vn(t0) = N n=1 1 Cn t t0 i dt′ + N n=1 vn(t0) and vs = 1 Ceq t t0 i dt′ + vs(t0) However, Kirchhoff’s voltage law establishes the equality of vs(t0) and the sum of the capacitor voltages at t0; thus Ceq = 1 1/C1 + 1/C2 + · · · + 1/CN CHAPTER 7 CAPACITORS AND INDUCTORS 236 + – i vN + – vs CN (a) v2 + – v1 + – C2 C1 + – i (b) Ceq vs ■FIGURE 7.20 (a) A circuit containing N capacitors in series. (b) The desired equivalent circuit, where Ceq [1/C1 + 1/C2 + · · · + 1/CN]−1. ■FIGURE 7.19 (a) The parallel combination of N inductors. (b) The equivalent circuit, where Leq [1/L1 + 1/L2 + · · · + 1/LN]−1. (a) is LN L1 L2 iN i2 i1 v + – (b) Leq is v + – SECTION 7.3 INDUCTANCE AND CAPACITANCE COMBINATIONS 237 and capacitors in series combine as do conductances in series, or resistors in parallel. The special case of two capacitors in series, of course, yields Ceq = C1C2 C1 + C2 Capacitors in Parallel Finally, the circuits of Fig. 7.21 enable us to establish the value of the ca-pacitor which is equivalent to N parallel capacitors as Ceq = C1 + C2 + · · · + CN and it is no great source of amazement to note that capacitors in parallel combine in the same manner in which we combine resistors in series, that is, by simply adding all the individual capacitances. These formulas are well worth memorizing. The formulas applying to se-ries and parallel combinations of inductors are identical to those for resistors, so they typically seem “obvious.” Care should be exercised, however, in the case of the corresponding expressions for series and parallel combinations of capacitors, as they are opposite those of resistors and inductors, frequently leading to errors when calculations are made too hastily. ■FIGURE 7.21 (a) The parallel combination of N capacitors. (b) The equivalent circuit, where Ceq C1 + C2 + · · · + CN. (a) is CN C1 C2 iN i2 i1 v + – (b) Ceq is v + – Simplify the network of Fig. 7.22a using series-parallel combinations. The 6 μF and 3 μF series capacitors are ﬁrst combined into a 2 μF equivalent, and this capacitor is then combined with the 1 μF element with which it is in parallel to yield an equivalent capacitance of 3 μF. In addition, the 3 H and 2 H inductors are replaced by an equivalent 1.2 H inductor, which is then added to the 0.8 H element to give a total equivalent inductance of 2 H. The much simpler (and probably less expensive) equivalent network is shown in Fig. 7.22b. PRACTICE ● 7.8 Find Ceq for the network of Fig. 7.23. EXAMPLE 7.8 (a) 2 H 3 H 0.8 H 1 F 6 F 3 F (b) 2 H 3 F ■FIGURE 7.22 (a) A given LC network. (b) A simpler equivalent circuit. 0.4 F 2 F 1 F 0.8 F 7 F 5 F 12 F 5 F Ceq ■FIGURE 7.23 Ans: 3.18 μF. The network shown in Fig. 7.24 contains three inductors and three capacitors, but no series or parallel combinations of either the inductors or the capacitors can be achieved. Simpliﬁcation of this network cannot be accomplished using the techniques presented here. CHAPTER 7 CAPACITORS AND INDUCTORS 238 3 H 5 H 4 F 6 F 1 H 2 F ■FIGURE 7.24 An LC network in which no series or parallel combinations of either the inductors or the capacitors can be made. 7.4 • CONSEQUENCES OF LINEARITY Next let us turn to nodal and mesh analysis. Since we already know that we may safely apply Kirchhoff’s laws, we can apply them in writing a set of equations that are both sufﬁcient and independent. They will be constant-coefﬁcient linear integrodifferential equations, however, which are hard enough to pronounce, let alone solve. Consequently, we shall write them now to gain familiarity with the use of Kirchhoff’s laws in RLC circuits and discuss the solution of the simpler cases in subsequent chapters. Write appropriate nodal equations for the circuit of Fig. 7.25. EXAMPLE 7.9 ■FIGURE 7.25 A four-node RLC circuit with node voltages assigned. R L C1 vs vs v2 v1 is iL + – C2 Node voltages are already chosen, so we sum currents leaving the cen-tral node: 1 L t t0 (v1 −vs) dt′ + iL(t0) + v1 −v2 R + C2 dv1 dt = 0 where iL(t0) is the value of the inductor current at the time the integra-tion begins. At the right-hand node, C1 d(v2 −vs) dt + v2 −v1 R −is = 0 Rewriting these two equations, we have v1 R + C2 dv1 dt + 1 L t t0 v1 dt′ −v2 R = 1 L t t0 vs dt′ −iL(t0) −v1 R + v2 R + C1 dv2 dt = C1 dvs dt + is These are the promised integrodifferential equations, and we note several interesting points about them. First, the source voltage vs happens to enter the equations as an integral and as a derivative, but not simply as vs. Since both sources are speciﬁed for all time, we should be able to evaluate the derivative or integral. Second, the initial value of the induc-tor current, iL(t0), acts as a (constant) source current at the center node. SECTION 7.4 CONSEQUENCES OF LINEARITY 239 We will not attempt the solution of integrodifferential equations here. It is worthwhile pointing out, however, that when the voltage forcing functions are sinusoidal functions of time, it will be possible to deﬁne a voltage-current ratio (called impedance) or a current-voltage ratio (called admittance) for each of the three passive elements. The factors operating on the two node voltages in the preceding equations will then become simple multiplying fac-tors, and the equations will be linear algebraic equations once again.These we may solve by determinants or a simple elimination of variables as before. We may also show that the beneﬁts of linearity apply to RLC circuits as well. In accordance with our previous deﬁnition of a linear circuit, these circuits are also linear because the voltage-current relationships for the inductor and capacitor are linear relationships. For the inductor, we have v = L di dt and multiplication of the current by some constant K leads to a voltage that is also greater by a factor K. In the integral formulation, i(t) = 1 L t t0 v dt′ + i(t0) it can be seen that, if each term is to increase by a factor of K, then the ini-tial value of the current must also increase by this same factor. A corresponding investigation of the capacitor shows that it, too, is lin-ear. Thus, a circuit composed of independent sources, linear dependent sources, and linear resistors, inductors, and capacitors is a linear circuit. In this linear circuit the response is again proportional to the forcing function. The proof of this statement is accomplished by ﬁrst writing a general system of integrodifferential equations. Let us place all the terms having the form of Ri, L di/dt, and 1/C i dt on the left side of each equation, and keep the independent source voltages on the right side. As a simple example, one of the equations might have the form Ri + L di dt + 1 C t t0 i dt′ + vC(t0) = vs If every independent source is now increased by a factor K, then the right side of each equation is greater by the factor K. Now each term on the left side is either a linear term involving some loop current or an initial capacitor voltage. In order to cause all the responses (loop currents) to in-crease by a factor K, it is apparent that we must also increase the initial capacitor voltages by a factor K. That is, we must treat the initial capacitor voltage as an independent source voltage and increase it also by a factor K. In a similar manner, initial inductor currents appear as independent source currents in nodal analysis. The principle of proportionality between source and response can thus be extended to the general RLC circuit, and it follows that the principle of superposition also applies. It should be emphasized that initial inductor PRACTICE ● 7.9 If vC(t) = 4 cos 105t V in the circuit in Fig. 7.26, ﬁnd vs(t). Ans: −2.4 cos 105t V. ■FIGURE 7.26 + – vs(t) vC + – 2 mH 80 nF currents and capacitor voltages must be treated as independent sources in applying the superposition principle; each initial value must take its turn in being rendered inactive. In Chap. 5 we learned that the principle of superposition is a natural consequence of the linear nature of resistive circuits. The resistive circuits are linear because the voltage-current rela-tionship for the resistor is linear and Kirchhoff’s laws are linear. Before we can apply the superposition principle to RLC circuits, how-ever, it is first necessary to develop methods of solving the equations describing these circuits when only one independent source is present. At this time we should feel convinced that a linear circuit will possess a re-sponse whose amplitude is proportional to the amplitude of the source. We should be prepared to apply superposition later, considering an inductor current or capacitor voltage speciﬁed at t = t0 as a source that must be deactivated when its turn comes. Thévenin’s and Norton’s theorems are based on the linearity of the ini-tial circuit, the applicability of Kirchhoff’s laws, and the superposition prin-ciple. The general RLC circuit conforms perfectly to these requirements, and it follows, therefore, that all linear circuits that contain any combina-tions of independent voltage and current sources, linear dependent voltage and current sources, and linear resistors, inductors, and capacitors may be analyzed with the use of these two theorems, if we wish. 7.5 • SIMPLE OP AMP CIRCUITS WITH CAPACITORS In Chap. 6 we were introduced to several different types of ampliﬁer circuits based on the ideal op amp. In almost every case, we found that the output was related to the input voltage by some combination of resistance ratios. If we replace one or more of these resistors with a capacitor, it is possible to obtain some interesting circuits in which the output is proportional to either the derivative or integral of the input voltage. Such circuits ﬁnd widespread use in practice. For example, a velocity sensor can be connected to an op amp cir-cuit that provides a signal proportional to the acceleration, or an output signal can be obtained that represents the total charge incident on a metal electrode during a speciﬁc period of time by simply integrating the measured current. To create an integrator using an ideal op amp, we ground the noninvert-ing input, install an ideal capacitor as a feedback element from the output back to the inverting input, and connect a signal source vs to the inverting input through an ideal resistor as shown in Fig. 7.27. Performing nodal analysis at the inverting input, 0 = va −vs R1 + i We can relate the current i to the voltage across the capacitor, i = Cf dvCf dt resulting in 0 = va −vs R1 + Cf dvCf dt Invoking ideal op amp rule 2, we know that va = vb = 0, so 0 = −vs R1 + Cf dvCf dt CHAPTER 7 CAPACITORS AND INDUCTORS 240 ■FIGURE 7.27 An ideal op amp connected as an integrator. + – vout – + vs va vb R1 Cf i i vCf + – + – SECTION 7.5 SIMPLE OP AMP CIRCUITS WITH CAPACITORS 241 Integrating and solving for vout, we obtain vCf = va −vout = 0 −vout = 1 R1Cf t 0 vs dt′ + vCf (0) or vout = − 1 R1Cf t 0 vs dt′ −vCf (0) We therefore have combined a resistor, a capacitor, and an op amp to form an integrator. Note that the ﬁrst term of the output is 1/RC times the negative of the integral of the input from t′ = 0 to t, and the second term is the nega-tive of the initial value of vCf . The value of (RC)−1 can be made equal to unity, if we wish, by choosing R = 1 M and C = 1 μF, for example; other selections may be made that will increase or decrease the output voltage. Before we leave the integrator circuit, we might anticipate a question from an inquisitive reader, “Could we use an inductor in place of the capaci-tor and obtain a differentiator?” Indeed we could, but circuit designers usu-ally avoid the use of inductors whenever possible because of their size, weight, cost, and associated resistance and capacitance. Instead, it is possi-ble to interchange the positions of the resistor and capacitor in Fig. 7.27 and obtain a differentiator. Derive an expression for the output voltage of the op amp circuit shown in Fig. 7.28. EXAMPLE 7.10 + – vout – + vs va vb Rf C1 i vRf + – + – ■FIGURE 7.28 An ideal op amp connected as a differentiator. + – vout – + vs va vb Lf R1 i i vLf + – + – ■FIGURE 7.29 We begin by writing a nodal equation at the inverting input pin, with vC1 △ va −vs: 0 = C1 dvC1 dt + va −vout Rf Invoking ideal op amp rule 2, va = vb = 0. Thus, C1 dvC1 dt = vout Rf Solving for vout, vout = Rf C1 dvC1 dt Since vC1 = va −vs = −vs, vout = −Rf C1 dvs dt So, simply by swapping the resistor and capacitor in the circuit of Fig. 7.27, we obtain a differentiator instead of an integrator. PRACTICE ● 7.10 Derive an expression for vout in terms of vs for the circuit shown in Fig. 7.29. Ans: vout = −L f /R1 dvs/dt. vC + – 5 3 2 cos 6t V 4 H 8 F i1 i2 + – ■FIGURE 7.30 A given circuit to which the deﬁnition of duality may be applied to determine the dual circuit. Note that vc(0) = 10 V. iL 3 S 5 S Ref. 8 H 4 F v2 v1 2 cos 6t A ■FIGURE 7.31 The exact dual of the circuit of Fig. 7.30. 7.6 • DUALITY The concept of duality applies to many fundamental engineering concepts. In this section, we shall deﬁne duality in terms of the circuit equations. Two circuits are “duals” if the mesh equations that characterize one of them have the same mathematical form as the nodal equations that characterize the other. They are said to be exact duals if each mesh equation of one circuit is numerically identical with the corresponding nodal equation of the other; the current and voltage variables themselves cannot be identical, of course. Duality itself merely refers to any of the properties exhibited by dual circuits. Let us use the deﬁnition to construct an exact dual circuit by writing the two mesh equations for the circuit shown in Fig. 7.30. Two mesh currents i1 and i2 are assigned, and the mesh equations are 3i1 + 4 di1 dt −4 di2 dt = 2 cos 6t −4 di1 dt + 4 di2 dt + 1 8 t 0 i2 dt′ + 5i2 = −10 We may now construct the two equations that describe the exact dual of our circuit. We wish these to be nodal equations, and thus begin by re-placing the mesh currents i1 and i2 in Eqs. and by the two nodal voltages v1 and v2, respectively. We obtain 3v1 + 4 dv1 dt −4 dv2 dt = 2 cos 6t −4 dv1 dt + 4 dv2 dt + 1 8 t 0 v2 dt′ + 5v2 = −10 and we now seek the circuit represented by these two nodal equations. Let us ﬁrst draw a line to represent the reference node, and then we may establish two nodes at which the positive references for v1 and v2 are lo-cated. Equation indicates that a current source of 2 cos 6t A is con-nected between node 1 and the reference node, oriented to provide a current entering node 1. This equation also shows that a 3 S conductance appears between node 1 and the reference node. Turning to Eq. , we ﬁrst con-sider the nonmutual terms, i.e., those terms which do not appear in Eq. , and they instruct us to connect an 8 H inductor and a 5 S conductance (in parallel) between node 2 and the reference. The two similar terms in Eqs. and represent a 4 F capacitor present mutually at nodes 1 and 2; the circuit is completed by connecting this capacitor between the two nodes. The constant term on the right side of Eq. is the value of the inductor current at t = 0; in other words, iL(0) = 10 A. The dual circuit is shown in Fig. 7.31; since the two sets of equations are numerically identi-cal, the circuits are exact duals. Dual circuits may be obtained more readily than by this method, for the equations need not be written. In order to construct the dual of a given cir-cuit, we think of the circuit in terms of its mesh equations. With each mesh we must associate a nonreference node, and, in addition, we must supply the reference node. On a diagram of the given circuit we therefore place a CHAPTER 7 CAPACITORS AND INDUCTORS 242 5 5 8 H 4 F 3 3 2 cos 6t V 2 cos 6t A 4 H Ref. 8 F + – ■FIGURE 7.32 The dual of the circuit of Fig. 7.30 is constructed directly from the circuit diagram. SECTION 7.6 DUALITY 243 node in the center of each mesh and supply the reference node as a line near the diagram or a loop enclosing the diagram. Each element that appears jointly in two meshes is a mutual element and gives rise to identical terms, except for sign, in the two corresponding mesh equations. It must be re-placed by an element that supplies the dual term in the two corresponding nodal equations. This dual element must therefore be connected directly be-tween the two nonreference nodes that are within the meshes in which the given mutual element appears. The nature of the dual element itself is easily determined; the mathe-matical form of the equations will be the same only if inductance is replaced by capacitance, capacitance by inductance, conductance by resistance, and resistance by conductance. Thus, the 4 H inductor which is common to meshes 1 and 2 in the circuit of Fig. 7.30 appears as a 4 F capacitor con-nected directly between nodes 1 and 2 in the dual circuit. Elements that appear only in one mesh must have duals that appear be-tween the corresponding node and the reference node. Referring again to Fig. 7.30, the voltage source 2 cos 6t V appears only in mesh 1; its dual is a current source 2 cos 6t A, which is connected only to node 1 and the refer-ence node. Since the voltage source is clockwise-sensed, the current source must be into-the-nonreference-node-sensed. Finally, provision must be made for the dual of the initial voltage present across the 8 F capacitor in the given circuit. The equations have shown us that the dual of this initial volt-age across the capacitor is an initial current through the inductor in the dual circuit; the numerical values are the same, and the correct sign of the initial current may be determined most readily by considering both the initial volt-age in the given circuit and the initial current in the dual circuit as sources. Thus, if vC in the given circuit is treated as a source, it would appear as −vC on the right side of the mesh equation; in the dual circuit, treating the cur-rent iL as a source would yield a term −iL on the right side of the nodal equa-tion. Since each has the same sign when treated as a source, then, if vC(0) = 10 V, iL(0) must be 10 A. The circuit of Fig. 7.30 is repeated in Fig. 7.32, and its exact dual is con-structed on the circuit diagram itself by merely drawing the dual of each given element between the two nodes that are inside the two meshes that are common to the given element. A reference node that surrounds the given circuit may be helpful. After the dual circuit is redrawn in more standard form, it appears as shown in Fig. 7.31. (a) + – (b) ■FIGURE 7.33 (a) The dual (in gray) of a given circuit (in black) is constructed on the given circuit. (b) The dual circuit is drawn in more conventional form for comparison to the original. (4) Someone suggested, “The voltage is across all over the parallel circuit.” An additional example of the construction of a dual circuit is shown in Fig. 7.33a and b. Since no particular element values are speciﬁed, these two circuits are duals, but not necessarily exact duals. The original circuit may be recovered from the dual by placing a node in the center of each of the ﬁve meshes of Fig. 7.33b and proceeding as before. The concept of duality may also be carried over into the language by which we describe circuit analysis or operation. For example, if we are given a volt-age source in series with a capacitor, we might wish to make the important statement, “The voltage source causes a current to ﬂow through the capacitor.” The dual statement is, “The current source causes a voltage to exist across the inductor.” The dual of a less carefully worded statement, such as “The current goes round and round the series circuit,” may require a little inventiveness.4 Practice in using dual language can be obtained by reading Thévenin’s theorem in this sense; Norton’s theorem should result. We have spoken of dual elements, dual language, and dual circuits. What about a dual network? Consider a resistor R and an inductor L in series. The dual of this two-terminal network exists and is most readily obtained by connecting some ideal source to the given network. The dual circuit is then obtained as the dual source in parallel with a conductance G with the same magnitude as R, and a capacitance C having the same magni-tude as L. We consider the dual network as the two-terminal network that is connected to the dual source; it is thus a pair of terminals between which G and C are connected in parallel. Before leaving the deﬁnition of duality, we should point out that duality is defined on the basis of mesh and nodal equations. Since nonplanar circuits cannot be described by a system of mesh equations, a circuit that cannot be drawn in planar form does not possess a dual. We shall use duality principally to reduce the work that we must do to analyze the simple standard circuits. After we have analyzed the series RL circuit, the parallel RC circuit requires less attention, not because it is less important, but because the analysis of the dual network is already known. Since the analysis of some complicated circuit is not apt to be well known, duality will usually not provide us with any quick solution. CHAPTER 7 CAPACITORS AND INDUCTORS 244 (a) 8e–106t mA 10 0.2 F v + – (b) v2 + – i 0.1 0.2 H + – 8e–106t mV v1 + – ■FIGURE 7.34 (a) (b) ■FIGURE 7.35 (a) Capacitor property editor window. (b) Display Properties dialog box, obtained by right-clicking in the IC box. Ans: −8e−106t mV; 16e−106t mV; −80e−106t mA. 7.7 • MODELING CAPACITORS AND INDUCTORS WITH PSPICE When using PSpice to analyze circuits containing inductors and capacitors, it is frequently necessary to be able to specify the initial condition of each element [i.e., vC(0) and iL(0)]. This is achieved by double-clicking on the element symbol, resulting in the dialog box shown in Fig. 7.35a. At the far right (not shown), we ﬁnd the value of the capacitance, which defaults to 1 nF. We can also specify the initial condition (IC), set to 2 V in Fig. 7.35a. Click-ing on the right mouse button and selecting Display results in the dialog box shown in Fig. 7.35b, which allows the initial condition to be displayed on the schematic. The procedure for setting the initial condition of an inductor is es-sentially the same. We should also note that when a capacitor is ﬁrst placed in the schematic, it appears horizontally; the positive reference terminal for the initial voltage is the left terminal. SECTION 7.7 MODELING CAPACITORS AND INDUCTORS WITH PSPICE 245 PRACTICE ● 7.11 Write the single nodal equation for the circuit of Fig. 7.34a, and show, by direct substitution, that v = −80e−106t mV is a solution. Knowing this, ﬁnd (a) v1; (b) v2; and (c) i for the circuit of Fig. 7.34b. Simulate the output voltage waveform of the circuit in Fig. 7.36 if vs = 1.5 sin 100t V, R1 = 10 k, Cf = 4.7 μF, and vC(0) = 2 V. EXAMPLE 7.11 We begin by drawing the circuit schematic, making sure to set the ini-tial voltage across the capacitor (Fig. 7.37). Note that we had to convert the frequency from 100 rad/s to 100/2π = 15.92 Hz. ■FIGURE 7.36 An integrating op amp circuit. + – vout – + vs R1 Cf vC + – + – ■FIGURE 7.37 The schematic representation of the circuit shown in Fig. 7.36, with the initial capacitor voltage set to 2 V. ■FIGURE 7.38 Dialog box for setting up a transient analysis. We choose a ﬁnal time of 0.5 s to obtain several periods of the output waveform (1/15.92 ≈0.06 s). In order to obtain time-varying voltages and currents, we need to per-form what is referred to as a transient analysis. Under the PSpice menu, we create a New Simulation Proﬁle named op amp integrator, which leads to the dialog box recreated in Fig. 7.38. Run to time represents the CHAPTER 7 CAPACITORS AND INDUCTORS 246 time at which the simulation is terminated; PSpice will select its own discrete times at which to calculate the various voltages and currents. Occasionally we obtain an error message stating that the transient solu-tion could not converge, or the output waveform does not appear as smooth as we would like. In such situations, it is useful to set a value for Maximum step size, which has been set to 0.5 ms in this example. From our earlier analysis and Eq. , we expect the output to be proportional to the negative integral of the input waveform, i.e., vout = 0.319 cos 100t −2.319 V, as shown in Fig. 7.39. The initial condition of 2 V across the capacitor has combined with a constant term from the integration to result in a nonzero average value for the output, unlike the input which has an average value of zero. ■FIGURE 7.39 Probe output for the simulated integrator circuit along with the input waveform for comparison. SUMMARY AND REVIEW 247 SUMMARY AND REVIEW A large number of practical circuits can be effectively modeled using only resistors and voltage/current sources. However, most interesting everyday occurrences somehow involve something changing with time, and in such cases intrinsic capacitances and/or inductances can become important. We employ such energy storage elements consciously as well, for example, in the design of frequency-selective ﬁlters, capacitor banks, and electric vehicle mo-tors. An ideal capacitor is modeled as having inﬁnite shunt resistance, and a current which depends on the time rate of change of the terminal voltage. Capacitance is measured in units of farads (F). Conversely, an ideal inductor is modeled as having zero series resistance, and a terminal voltage which de-pends on the time rate of change of the current. Inductance is measured in units of henrys (H). Either element can store energy; the amount of energy present in a capacitor (stored in its electric ﬁeld) is proportional to the square of the terminal voltage, and the amount of energy present in an inductor (stored in its magnetic ﬁeld) is proportional to the square of its current. As we found for resistors, we can simplify some connections of capaci-tors (or inductors) using series/parallel combinations. The validity of such equivalents arises from KCL and KVL. Once we have simpliﬁed a circuit as much as possible (taking care not to “combine away” a component which is used to deﬁne a current or voltage of interest to us), nodal and mesh analy-sis can be applied to circuits with capacitors and inductors. However, the resulting integrodifferential equations are often nontrivial to solve, and so we will consider some practical approaches in the next two chapters. Simple circuits, however, such as those which involve a single operational ampli-ﬁer, can be analyzed easily. We found (to our surprise) that such circuits can be used as signal integrators or differentiators. Consequently, they provide an output signal that tells us how some input quantity (accumulating charge during ion implantation into a silicon wafer, for example) varies with time. As a ﬁnal note, capacitors and inductors provide a particularly strong example of the concept known as duality. KCL and KVL, mesh and nodal analysis are other examples. Circuits are rarely analyzed using this idea, but it is nevertheless important, since the implication is that we only need to learn roughly “half” of the complete set of concepts, and then determine how to translate the remainder. Some people ﬁnd this helpful; others don’t. Regardless, capacitors and inductors are straightforward to model in PSpice and other circuit simulation tools, allowing us to check our answers. The difference between those elements and resistors in such software packages is that we must take care to set the initial condition properly. As an additional review aid, here we list some key points from the chapter, and identify relevant example(s). ❑The current through a capacitor is given by i = C dv/dt. (Example 7.1) ❑The voltage across a capacitor is related to its current by v(t) = 1 C t t0 i(t′) dt′ + v(t0) (Example 7.2) ❑A capacitor is an open circuit to dc voltages. (Example 7.1) ❑The voltage across an inductor is given by v = L di/dt. (Examples 7.4, 7.5) ❑The current through an inductor is related to its voltage by i(t) = 1 L t t0 v dt′ + i(t0) (Example 7.6) ❑An inductor is a short circuit to dc currents. (Examples 7.4, 7.5) ❑The energy presently stored in a capacitor is given by 1 2Cv2, whereas the energy presently stored in an inductor is given by 1 2 Li2; both are referenced to a time at which no energy was stored. (Examples 7.3, 7.7) ❑Series and parallel combinations of inductors can be combined using the same equations as for resistors. (Example 7.8) CHAPTER 7 CAPACITORS AND INDUCTORS 248 (b) v (V) –4 –2 0 2 4 –1 0 1 t (s) 2 3 4 5 (a) –1 0 1 t (s) v (V) 2 3 4 5 1 2 3 4 5 6 7 ■FIGURE 7.40 EXERCISES 249 ❑Series and parallel combinations of capacitors work the opposite way as they do for resistors. (Example 7.8) ❑Since capacitors and inductors are linear elements, KVL, KCL, superposition, Thévenin’s and Norton’s theorems, and nodal and mesh analysis apply to their circuits as well. (Example 7.9) ❑A capacitor as the feedback element in an inverting op amp leads to an output voltage proportional to the integral of the input voltage. Swapping the input resistor and the feedback capacitor leads to an output voltage proportional to the derivative of the input voltage. (Example 7.10) ❑PSpice allows us to set the initial voltage across a capacitor, and the initial current through an inductor. A transient analysis provides details of the time-dependent response of circuits containing these types of elements. (Example 7.11) READING FURTHER A detailed guide to characteristics and selection of various capacitor and inductor types can be found in: H. B. Drexler, Passive Electronic Component Handbook, 2nd ed., C. A. Harper, ed. New York: McGraw-Hill, 2003, pp. 69–203. C. J. Kaiser, The Inductor Handbook, 2nd ed. Olathe, Kans.: C.J. Publish-ing, 1996. Two books that describe capacitor-based op amp circuits are: R. Mancini (ed.), Op Amps Are For Everyone, 2nd ed. Amsterdam: Newnes, 2003. W. G. Jung, Op Amp Cookbook, 3rd ed. Upper Saddle River, N.J.: Prentice-Hall, 1997. EXERCISES 7.1 The Capacitor 1. Making use of the passive sign convention, determine the current ﬂowing through a 220 nF capacitor for t ≥0 if its voltage vC(t) is given by (a) −3.35 V; (b) 16.2e−9t V; (c) 8 cos 0.01t mV; (d) 5 + 9 sin 0.08t V. 2. Sketch the current ﬂowing through a 13 pF capacitor for t ≥0 as a result of the waveforms shown in Fig. 7.40. Assume the passive sign convention. 1 2 1 2 3 4 3 4 5 6 v (V) t (s) ■FIGURE 7.41 3. (a) If the voltage waveform depicted in Fig. 7.41 is applied across the termi-nals of a 1 μF electrolytic capacitor, graph the resulting current, assuming the passive sign convention. (b) Repeat part (a) if the capacitor is replaced with a 17.5 pF capacitor. CHAPTER 7 CAPACITORS AND INDUCTORS 250 4. A capacitor is constructed from two copper plates, each measuring 1 mm × 2.5 mm and 155 μm thick. The two plates are placed such that they face each other and are separated by a 1 μm gap. Calculate the resulting capacitance if (a) the intervening dielectric has a permittivity of 1.35ε0; (b) the intervening dielectric has a permittivity of 3.5ε0; (c) the plate separation is increased by 3.5 μm and the gap is ﬁlled with air; (d) the plate area is doubled and the 1 μm gap is ﬁlled with air. 5. Two pieces of gadolinium, each measuring 100 μm × 750 μm and 604 nm thick, are used to construct a capacitor. The two plates are arranged such that they face each other and are separated by a 100 nm gap. Calculate the resulting capacitance if (a) the intervening dielectric has a permittivity of 13.8ε0; (b) the intervening dielectric has a permittivity of 500ε0; (c) the plate separation is increased by 100 nm and the gap is ﬁlled with air; (d) the plate area is quadru-pled and the 100 nm gap is ﬁlled with air. 6. Design a 100 nF capacitor constructed from 1 μm thick gold foil, and which ﬁts entirely within a volume equal to that of a standard AAA battery, if the only dielectric available has a permittivity of 3.1ε0. 7. Design a capacitor whose capacitance can be varied mechanically with a simple vertical motion, between the values of 100 nF and 300 nF. 8. Design a capacitor whose capacitance can be varied mechanically over the range of 50 nF and 100 nF by rotating a knob 90◦. 9. Asilicon pn junction diode is characterized by a junction capacitance deﬁned as Cj = Ksε0 A W where Ks = 11.8 for silicon, ε0 is the vacuum permittivity, A = the cross-sectional area of the junction, and W is known as the depletion width of the junction. Width W depends not only on how the diode is fabricated, but also on the voltage applied to its two terminals. It can be computed using W = 2Ksε0 qN (Vbi −VA) Thus, diodes are frequently used in electronic circuits, since they can be thought of as voltage-controlled capacitors. Assuming parameter values of N = 1018 cm−3, Vbi = 0.57 V, and using q = 1.6 × 10−19 C, calculate the capacitance of a diode with cross-sectional area A 1 μm × 1 μm at applied voltages of VA = −1, −5, and −10 volts. 10. Assuming the passive sign convention, sketch the voltage which develops across the terminals of a 2.5 F capacitor in response to the current waveforms shown in Fig. 7.42. EXERCISES 251 11. The current ﬂowing through a 33 mF capacitor is shown graphically in Fig. 7.43. (a) Assuming the passive sign convention, sketch the resulting voltage waveform across the device. (b) Compute the voltage at 300 ms, 600 ms, and 1.1 s. (c) 3 –1 1 0 2 3 i(t) (A) t (s) (a) (b) 2 –2 –1 1 0 2 3 i(t) (A) t (s) 2 –2 –1 1 0 2 3 i(t) (A) t (s) ■FIGURE 7.42 4 8 0 0.4 0.2 0.8 0.6 1.2 1.0 1.4 i (A) t (s) ■FIGURE 7.43 + – 1.2 V 22 40 9.8 mF 9.8 mF vC + – 1.2 V 22 40 vC + – (a) (b) + – ■FIGURE 7.44 12. Calculate the energy stored in a capacitor at time t = 1 s if (a) C = 1.4 F and vC = 8 V, t > 0; (b) C = 23.5 pF and vC = 0.8 V, t > 0; (c) C = 17 nF, vC(1) = 12 V, vC(0) = 2 V, and wC(0) = 295 nJ. 13. A 137 pF capacitor is connected to a voltage source such that vC(t) = 12e−2tV, t ≥0 and vC(t) = 12 V, t < 0. Calculate the energy stored in the capacitor at t equal to (a) 0; (b) 200 ms; (c) 500 ms; (d) 1 s. 14. Calculate the power dissipated in the 40 resistor and the voltage labeled vC in each of the circuits depicted in Fig. 7.44. 15. For each circuit shown in Fig. 7.45, calculate the voltage labeled vC. CHAPTER 7 CAPACITORS AND INDUCTORS 252 3 mF 4.5 nA 7 5 10 13 vC – + 3 mF 4.5 nA 13 7 5 10 vC – + (a) (b) ■FIGURE 7.45 7.2 The Inductor 16. Design a 30 nH inductor using 29 AWG solid soft copper wire. Include a sketch of your design and label geometrical parameters as necessary for clarity. Assume the coil is ﬁlled with air only. 17. If the current ﬂowing through a 75 mH inductor has the waveform shown in Fig. 7.46, (a) sketch the voltage which develops across the inductor terminals for t ≥0, assuming the passive sign convention; and (b) calculate the voltage at t = 1 s, 2.9 s, and 3.1 s. 2 1 –1 1 0 2 3 i(t) (A) t (s) ■FIGURE 7.46 2 2 3 4 5 3 4 5 6 7 i(t) (A) t (s) ■FIGURE 7.47 (a) 4.7 k 12 nH 14 k vL + – vL + – vL + – vL + – is (c) 4.7 k 12 nH + – iL iL vs (b) is (d) 12 nH + – iL vs 4.7 k 4.7 k 14 k 12 nH iL ■FIGURE 7.48 18. The current through a 17 nH aluminum inductor is shown in Fig. 7.47. Sketch the resulting voltage waveform for t ≥0, assuming the passive sign convention. 19. Determine the voltage for t ≥0 which develops across the terminals of a 4.2 mH inductor, if the current (deﬁned consistent with the passive sign convention) is (a) −10 mA; (b) 3 sin 6t A; (c) 11 + 115 √ 2 cos(100πt −9◦) A; (d) 13e−t nA; (e) 3 + te−14t A. 20. Determine the voltage for t ≥0 which develops across the terminals of an 8 pH inductor, if the current (deﬁned consistent with the passive sign convention) is (a) 8 mA; (b) 800 mA; (c) 8 A; (d) 4e−t A; (e) −3 + te−t A. 21. Calculate vL and iL for each of the circuits depicted in Fig. 7.48, if is = 1 mA and vs = 2.1 V. 3 2 1 –1 –2 1 2 3 4 –3 –2 –1 5 6 7 iL (mA) t (s) ■FIGURE 7.49 4.7 k 16 k 7 k 10 V 4.7 k ix 2 H 6 H 8 H 2 k 4 k 5 k 2 A 1 k ix 10 A 4 nH 3 F (a) (b) + – ■FIGURE 7.50 24. Determine the current ﬂowing through a 6 mH inductor if the voltage (deﬁned such that it is consistent with the passive sign convention) is given by (a) 5 V; (b) 100 sin 120πt, t ≥0 and 0, t < 0. 25. The voltage across a 2 H inductor is given by vL = 4.3t, 0 ≤t ≤50 ms. With the knowledge that iL(−0.1) = 100 μA, calculate the current (assuming it is deﬁned consistent with the passive sign convention) at t equal to (a) 0; (b) 1.5 ms; (c) 45 ms. 26. Calculate the energy stored in a 1 nH inductor if the current ﬂowing through it is (a) 0 mA; (b) 1 mA; (c) 20 A; (d) 5 sin 6t mA, t > 0. 27. Determine the amount of energy stored in a 33 mH inductor at t = 1 ms as a result of a current iL given by (a) 7 A; (b) 3 −9e−103t mA. 28. Making the assumption that the circuits in Fig. 7.50 have been connected for a very long time, determine the value for each current labeled ix. EXERCISES 253 22. The current waveform shown in Fig. 7.14 has a rise time of 0.1 (100 ms) and a fall time of the same duration. If the current is applied to the “+” voltage refer-ence terminal of a 200 nH inductor, sketch the expected voltage waveform if the rise and fall times are changed, respectively, to (a) 200 ms, 200 ms; (b) 10 ms, 50 ms; (c) 10 ns, 20 ns. 23. Determine the inductor voltage which results from the current waveform shown in Fig. 7.49 (assuming the passive sign convention) at t equal to (a) −1 s; (b) 0 s; (c) 1.5 s; (d) 2.5 s; (e) 4 s; (f) 5 s. + – vx + – 20 15 12 20 1 V 5 F 5 A x y 2 H 3 H 20 F 5 H ■FIGURE 7.51 + – 50 mH 4 V 10 k 47 k ■FIGURE 7.52 ■FIGURE 7.53 30. For the circuit shown in Fig. 7.52, (a) compute the Thévenin equivalent seen by the inductor; (b) determine the power being dissipated by both resistors; (c) calculate the energy stored in the inductor. 7.3 Inductance and Capacitance Combinations 31. If each capacitor has a value of 1 F, determine the equivalent capacitance of the network shown in Fig. 7.53. 32. Determine an equivalent inductance for the network shown in Fig. 7.54 if each inductor has value L. 29. Calculate the voltage labeled vx in Fig. 7.51, assuming the circuit has been running a very long time, if (a) a 10 resistor is connected between terminals x and y; (b) a 1 H inductor is connected between terminals x and y; (c) a 1 F capacitor is connected between terminals x and y; (d) a 4 H inductor in parallel with a 1 resistor is connected between terminals x and y. CHAPTER 7 CAPACITORS AND INDUCTORS 254 ■FIGURE 7.55 4 F 2 F 2 F 1 F 5 F 12 F Ceq 8 F 5 F 7 F ■FIGURE 7.54 33. Using as many 1 nH inductors as you like, design two networks, each of which has an equivalent inductance of 1.25 nH. 34. Compute the equivalent capacitance Ceq as labeled in Fig. 7.55. a b 5 F 1 F 2 F 12 F 4 F 10 F 12 F 7 F Ceq ■FIGURE 7.56 a b 5 H 1 H 2 H 12 H 4 H 10 H 12 H 7 H Leq ■FIGURE 7.57 + – 2 V R R R R vx + – R L L C C C ■FIGURE 7.58 b a 1 nH 2 nH 2 nH 4 nH 1 nH 7 nH ■FIGURE 7.60 ■FIGURE 7.59 37. Reduce the circuit depicted in Fig. 7.58 to as few components as possible. 38. Refer to the network shown in Fig. 7.59 and ﬁnd (a) Req if each element is a 10 resistor; (b) Leq if each element is a 10 H inductor; and (c) Ceq if each element is a 10 F capacitor. 39. Determine the equivalent inductance seen looking into the terminals marked a and b of the network represented in Fig. 7.60. EXERCISES 255 36. Apply combinatorial techniques as appropriate to obtain a value for the equiva-lent inductance Leq as labeled on the network of Fig. 7.57. 35. Determine the equivalent capacitance Ceq of the network shown in Fig. 7.56. 40. Reduce the circuit represented in Fig. 7.61 to the smallest possible number of components. CHAPTER 7 CAPACITORS AND INDUCTORS 256 is R L L L L R C C C L C ■FIGURE 7.61 ■FIGURE 7.62 L4 L5 L6 L2 L1 L3 ■FIGURE 7.63 R L C1 vs vs v2 v1 is iL + – C2 i1 i2 ■FIGURE 7.65 41. Reduce the network of Fig. 7.62 to the smallest possible number of compo-nents if each inductor is 1 nH and each capacitor is 1 mF. 42. For the network of Fig. 7.63, L1 = 1 H, L2 = L3 = 2 H, L4 = L5 = L6 = 3 H. (a) Find the equivalent inductance. (b) Derive an expression for a general network of this type having N stages, assuming stage N is composed of N inductors, each having inductance N henrys. 43. Extend the concept of -Y transformations to simplify the network of Fig. 7.64 if each element is a 2 pF capacitor. 44. Extend the concept of -Y transformations to simplify the network of Fig. 7.64 if each element is a 1 nH inductor. 7.4 Consequences of Linearity 45. With regard to the circuit represented in Fig. 7.65, (a) write a complete set of nodal equations and (b) write a complete set of mesh equations. ■FIGURE 7.64 vL + – 20 60 mH 9 V 5 F 30 mA 20 mA 40 cos 103t mA vC + – ■FIGURE 7.69 + – + – 50 mH 50 100 0.2vx 1 F vx + – 40e–20t V 20e–20t mA ■FIGURE 7.70 vs + – i(t) 2 F 1 F 4 F v1 + – v2 + – ■FIGURE 7.68 EXERCISES 257 46. (a) Write nodal equations for the circuit of Fig. 7.66. (b) Write mesh equations for the same circuit. 47. In the circuit shown in Fig. 7.67, let is = 60e−200t mA with i1(0) = 20 mA. (a) Find v(t) for all t. (b) Find i1(t) for t ≥0. (c) Find i2(t) for t ≥0. 48. Let vs = 100e−80t V and v1(0) = 20 V in the circuit of Fig. 7.68. (a) Find i(t) for all t. (b) Find v1(t) for t ≥0. (c) Find v2(t) for t ≥0. 8 mH iL vL + – vC + – + – vs vC (0) = 12 V, iL(0) = 2 A 20 10 5 F i20 ■FIGURE 7.66 6 H 4 H i1 i2 is v(t) + – 3 H ■FIGURE 7.67 49. If it is assumed that all the sources in the circuit of Fig. 7.69 have been con-nected and operating for a very long time, use the superposition principle to ﬁnd vC(t) and vL(t). 50. For the circuit of Fig. 7.70, assume no energy is stored at t = 0, and write a complete set of nodal equations. 7.5 Simple Op Amp Circuits with Capacitors 51. Interchange the location of R1 and Cf in the circuit of Fig. 7.27, and assume that Ri = ∞, Ro = 0, and A = ∞for the op amp. (a) Find vout(t) as a function of vs(t). (b) Obtain an equation relating vo(t) and vs(t) if A is not assumed to be inﬁnite. 52. For the integrating ampliﬁer circuit of Fig. 7.27, R1 = 100 k, Cf = 500 μF, and vs = 20 sin 540t mV. Calculate vout if (a) A = ∞, Ri = ∞, and Ro = 0; (b) A = 5000, Ri = 1 M, and Ro = 3 . + – vout – + vs + – Rf L1 ■FIGURE 7.71 + – vout – + vs + – Rf Cf R1 ■FIGURE 7.72 53. Derive an expression for vout in terms of vs for the ampliﬁer circuit shown in Fig. 7.71. 54. In practice, circuits such as those depicted in Fig. 7.27 may not function correctly unless there is a conducting pathway between the output and input terminals of the op amp. (a) Analyze the modiﬁed integrating ampliﬁer circuit shown in Fig. 7.72 to obtain an expression for vout in terms of vs, and (b) compare this expression to Eq. . 55. A new piece of equipment designed to make crystals from molten constituents is experiencing too many failures (cracked products). The production manager wants to monitor the cooling rate to see if this is related to the problem. The system has two output terminals available, where the voltage across them is linearly proportional to the crucible temperature such that 30 mV corresponds to 30◦C and 1 V corresponds to 1000◦C. Design a circuit whose voltage output represents the cooling rate, calibrated such that 1 V = 1◦C/s. 56. A confectionary company has decided to increase the production rate of its milk chocolate bars to compensate for a recent increase in the cost of raw materials. However, the wrapping unit cannot accept more than 1 bar per sec-ond, or it drops bars. A 200 mV peak-to-peak sinusoidal voltage signal is avail-able from the bar-making system which feeds into the wrapping unit, such that its frequency matches the bar production frequency (i.e., 1 Hz = 1 bar/s). Design a circuit that provides a voltage output sufﬁcient to power a 12 V audi-ble alarm when the production rate exceeds the capacity of the wrapping unit. 57. One problem satellites face is exposure to high-energy particles, which can cause damage to sensitive electronics as well as solar arrays used to provide power. A new communications satellite is equipped with a high-energy proton detector measuring 1 cm × 1 cm. It provides a current directly equal to the number of protons impinging the surface per second. Design a circuit whose output voltage provides a running total of the number of proton hits, calibrated such that 1 V = 1 million hits. 58. The output of a velocity sensor attached to a sensitive piece of mobile equip-ment is calibrated to provide a signal such that 10 mV corresponds to linear motion at 1 m/s. If the equipment is subjected to sudden shock, it can be dam-aged. Since force = mass × acceleration, monitoring of the rate of change of velocity can be used to determine if the equipment is transported improperly. (a) Design a circuit to provide a voltage proportional to the linear acceleration such that 10 mV = 1 m/s2. (b) How many sensor-circuit combinations does this application require? 59. A ﬂoating sensor in a certain fuel tank is connected to a variable resistor (often called a potentiometer) such that a full tank (100 liters) corresponds to 1 and an empty tank corresponds to 10 . (a) Design a circuit that provides an output voltage which indicates the amount of fuel remaining, so that 1 V = empty and 5 V = full. (b) Design a circuit to indicate the rate of fuel consumption by providing a voltage output calibrated to yield 1 V = 1 l/s. 7.6 Duality 60. (a) Draw the exact dual of the circuit depicted in Fig. 7.73. (b) Label the new (dual) variables. (c) Write nodal equations for both circuits. 61. (a) Draw the exact dual of the simple circuit shown in Fig. 7.74. (b) Label the new (dual) variables. (c) Write mesh equations for both circuits. CHAPTER 7 CAPACITORS AND INDUCTORS 258 6 H 4 H i1 i2 is v(t) + – 3 H ■FIGURE 7.73 + – 4 H 2 V 10 7 ■FIGURE 7.74 EXERCISES 259 62. (a) Draw the exact dual of the simple circuit shown in Fig. 7.75. (b) Label the new (dual) variables. (c) Write mesh equations for both circuits. 10 H 100 10 F iL + – vs ■FIGURE 7.75 20 2 80 100 V 16 ix 1 H 2 H 3 H ■FIGURE 7.76 2 H 3 4 5 8 F 7 F 6 F + – 10e–2t V 1 H ■FIGURE 7.77 63. (a) Draw the exact dual of the simple circuit shown in Fig. 7.76. (b) Label the new (dual) variables. (c) Write nodal and mesh equations for both circuits. 64. Draw the exact dual of the circuit shown in Fig. 7.77. Keep it neat! 7.7 Modeling Capacitors and Inductors with PSpice 65. Taking the bottom node in the circuit of Fig. 7.78 as the reference terminal, calculate (a) the current through the inductor and (b) the power dissipated by the 7 resistor. (c) Verify your answers with an appropriate PSpice simulation. 66. For the four-element circuit shown in Fig. 7.79, (a) calculate the power absorbed in each resistor; (b) determine the voltage across the capacitor; (c) compute the energy stored in the capacitor; and (d) verify your answers with an appropriate PSpice simulation. (Recall that calculations can be performed in Probe.) 80 k 46 k + – 6 mH 7 V ■FIGURE 7.78 80 k 46 k + – 10 F 7 V ■FIGURE 7.79 67. (a) Compute iL and vx as indicated in the circuit of Fig. 7.80. (b) Determine the energy stored in the inductor and in the capacitor. (c) Verify your answers with an appropriate PSpice simulation. CHAPTER 7 CAPACITORS AND INDUCTORS 260 810 1 F iL 120 440 k 6 mA 2 H + – vx ■FIGURE 7.80 68. For the circuit depicted in Fig. 7.81, the value of iL(0) = 1 mA. (a) Compute the energy stored in the element at t = 0. (b) Perform a transient simulation of the circuit over the range of 0 ≤t ≤500 ns. Determine the value of iL at t = 0, 130 ns, 260 ns, and 500 ns. (c) What fraction of the initial energy remains in the inductor at t = 130 ns? At t = 500 ns? 69. Assume an initial voltage of 9 V across the 10 μF capacitor shown in Fig. 7.82 (i.e., v(0) = 9 V). (a) Compute the initial energy stored in the capacitor. (b) For t > 0, do you expect the energy to remain in the capacitor? Explain. (c) Perform a transient simulation of the circuit over the range of 0 ≤t ≤2.5 s and determine v(t) at t = 460 ms, 920 ms, and 2.3 s. (c) What fraction of the initial energy remains stored in the capacitor at t = 460 ms? At t = 2.3 s? 70. Referring to the circuit of Fig. 7.83, (a) calculate the energy stored in each energy storage element; (b) verify your answers with an appropriate PSpice simulation. Chapter-Integrating Exercises 71. For the circuit of Fig. 7.28, (a) sketch vout over the range of 0 ≤t ≤5 ms if Rf = 1 k, C1 = 100 mF, and vs is a 1 kHz sinusoidal source having a peak voltage of 2 V. (b) Verify your answer with an appropriate transient simulation, plotting both vs and vout in Probe. (Hint: Between plotting traces, add a second y axis using Plot, Add Y Axis. This allows both traces to be seen clearly.) 72. (a) Sketch the output function vout of the ampliﬁer circuit in Fig. 7.29 over the range of 0 ≤t ≤100 ms if vs is a 60 Hz sinusoidal source having a peak voltage of 400 mV, R1 is 1 k, and L f is 80 nH. (b) Verify your answer with an appropriate transient simulation, plotting both vs and vout in Probe. (Hint: Between plotting traces, add a second y axis using Plot, Add Y Axis. This allows both traces to be seen clearly.) 73. For the circuit of Fig. 7.71, (a) sketch vout over the range of 0 ≤t ≤2.5 ms if Rf = 100 k, L1 = 100 mH, and vs is a 2 kHz sinusoidal source having a peak voltage of 5 V. (b) Verify your answer with an appropriate transient simu-lation, plotting both vs and vout in Probe. (Hint: Between plotting traces, add a second y axis using Plot, Add Y Axis. This allows both traces to be seen clearly.) 74. Consider the modiﬁed integrator depicted in Fig. 7.72. Take R1 = 100 , Rf = 10 M, and C1 = 10 mF. The source vs provides a 10 Hz sinusoidal voltage having a peak amplitude of 0.5 V. (a) Sketch vout over the range of 0 ≤t ≤500 ms. (b) Verify your answer with an appropriate transient simula-tion, plotting both vs and vout in Probe. (Hint: Between plotting traces, add a second y axis using Plot, Add Y Axis. This allows both traces to be seen clearly.) iL 46 k 6 mH ■FIGURE 7.81 46 k 10 F v(t) + – ■FIGURE 7.82 1 2 2 mH 4 V 2 mA + – 1 F + – vx 5vx + – ■FIGURE 7.83 INTRODUCTION In Chap. 7 we wrote equations for the response of several circuits containing both inductance and capacitance, but we did not solve any of them. Now we are ready to proceed with the solution of the simpler circuits, namely, those which contain only resistors and inductors, or only resistors and capacitors. Although the circuits we are about to consider have a very elementary appearance, they are also of practical importance. Networks of this form ﬁnd use in electronic ampliﬁers, automatic control systems, operational ampliﬁers, communications equip-ment, and many other applications. Familiarity with these simple circuits will enable us to predict the accuracy with which the output of an ampliﬁer can follow an input that is changing rapidly with time, or to predict how quickly the speed of a motor will change in response to a change in its ﬁeld current. Our understand-ing of simple RL and RC circuits will also enable us to suggest modiﬁcations to the ampliﬁer or motor in order to obtain a more desirable response. 8.1 • THE SOURCE-FREE RL CIRCUIT The analysis of circuits containing inductors and/or capacitors is de-pendent upon the formulation and solution of the integrodifferential equations that characterize the circuits. We will call the special type of equation we obtain a homogeneous linear differential equation, which is simply a differential equation in which every term is of the ﬁrst degree in the dependent variable or one of its derivatives. A solution is obtained when we have found an expression for the KEY CONCEPTS RL and RC Time Constants Natural and Forced Response Calculating the Time-Dependent Response to DC Excitation How to Determine Initial Conditions and Their Effect on the Circuit Response Analyzing Circuits with Step Function Input and with Switches Construction of Pulse Waveforms Using Unit-Step Functions The Response of Sequentially Switched Circuits Basic RLand RC Circuits C H A P T E R 8 261 CHAPTER 8 BASIC RL AND RC CIRCUITS 262 dependent variable that satisﬁes both the differential equation and also the prescribed energy distribution in the inductors or capacitors at a prescribed instant of time, usually t = 0. The solution of the differential equation represents a response of the circuit, and it is known by many names. Since this response depends upon the general “nature’’ of the circuit (the types of elements, their sizes, the interconnection of the elements), it is often called a natural response. However, any real circuit we construct cannot store energy forever; the resistances intrinsically associated with inductors and capacitors will even-tually convert all stored energy into heat. The response must eventually die out, and for this reason it is frequently referred to as the transient response. Finally, we should also be familiar with the mathematicians’contribution to the nomenclature; they call the solution of a homogeneous linear differen-tial equation a complementary function. When we consider independent sources acting on a circuit, part of the re-sponse will resemble the nature of the particular source (or forcing function) used; this part of the response, called the particular solution, the steady-state response, or the forced response, will be “complemented’’ by the comple-mentary response produced in the source-free circuit. The complete response of the circuit will then be given by the sum of the complementary function and the particular solution. In other words, the complete response is the sum of the naturalresponseandtheforcedresponse.Thesource-freeresponsemay be called the natural response, the transient response, the free response, or the complementary function, but because of its more descriptive nature, we will most often call it the natural response. We will consider several different methods of solving these differential equations. The mathematical manipulation, however, is not circuit analysis. Our greatest interest lies in the solutions themselves, their meaning, and their interpretation, and we will try to become sufﬁciently familiar with the form of the response that we are able to write down answers for new circuits by just plain thinking. Although complicated analytical methods are needed when simpler methods fail, a well-developed intuition is an invaluable resource in such situations. We begin our study of transient analysis by considering the simple series RL circuit shown in Fig. 8.1. Let us designate the time-varying current asi(t); we will represent the value of i(t)at t = 0as I0; in other words, i(0) = I0. We therefore have Ri + vL = Ri + L di dt = 0 or di dt + R L i = 0 Our goal is an expression for i(t) which satisﬁes this equation and also has the value I0 at t = 0. The solution may be obtained by several different methods. A Direct Approach One very direct method of solving a differential equation consists of writ-ing the equation in such a way that the variables are separated, and then i(t) vR + – vL + – R L ■FIGURE 8.1 A series RL circuit for which i(t) is to be determined, subject to the initial condition that i(0) I0. It may seem pretty strange to discuss a time-varying current ﬂowing in a circuit with no sources! Keep in mind that we only know the current at the time speci-ﬁed as t 0; we don’t know the current prior to that time. In the same vein, we don’t know what the circuit looked like prior to t 0, either. In order for a current to be ﬂowing, a source had to have been present at some point, but we are not privy to this information. Fortunately, it is not required in order to analyze the circuit we are given. SECTION 8.1 THE SOURCE-FREE RL CIRCUIT 263 integrating each side of the equation. The variables in Eq. are i and t, and it is apparent that the equation may be multiplied by dt, divided by i, and arranged with the variables separated: di i = −R L dt Since the current is I0 at t = 0 and i(t) at time t, we may equate the two deﬁnite integrals which are obtained by integrating each side between the corresponding limits: i(t) I0 di′ i′ = t 0 −R L dt′ Performing the indicated integration, ln i′ i I0 = −R L t′ t 0 which results in ln i −ln I0 = −R L (t −0) After a little manipulation, we ﬁnd that the current i(t) is given by i(t) = I0e−Rt/L We check our solution by ﬁrst showing that substitution of Eq. in Eq. yields the identity 0 = 0, and then showing that substitution of t = 0 in Eq. produces i(0) = I0. Both steps are necessary; the solution must satisfy the differential equation which characterizes the circuit, and it must also satisfy the initial condition. ■FIGURE 8.2 A simple RL circuit in which energy is stored in the inductor at t 0. 200 50 mH ■FIGURE 8.3 Circuit for Practice Problem 8.1. 1 k 500 nH iR If the inductor of Fig. 8.2 has a current iL 2 A at t 0, ﬁnd an expression for iL(t) valid for t > 0, and its value at t 200 μs. This is the identical type of circuit just considered, so we expect an inductor current of the form iL = I0e−Rt/L where R = 200 , L = 50 mH and I0 is the initial current ﬂowing through the inductor at t = 0. Thus, iL(t) = 2e−4000t Substituting t = 200 × 10−6 s, we ﬁnd that iL(t) = 898.7 mA, less than half the initial value. PRACTICE ● 8.1 Determine the current iR through the resistor of Fig. 8.3 at t = 1 ns if iR(0) = 6 A. Ans: 812 mA. EXAMPLE 8.1 An Alternative Approach The solution may also be obtained by a slight variation of the method we just described. After separating the variables, we now also include a con-stant of integration. Thus, di i = − R L dt + K and integration gives us ln i = −R L t + K The constant K cannot be evaluated by substitution of Eq. in the orig-inal differential equation ; the identity 0 = 0 will result, because Eq. is a solution of Eq. for any value of K (try it out on your own). The con-stant of integration must be selected to satisfy the initial condition i(0) = I0. Thus, at t = 0, Eq. becomes ln I0 = K and we use this value for K in Eq. to obtain the desired response ln i = −R L t + ln I0 or i(t) = I0e−Rt/L as before. A More General Solution Approach Either of these methods can be used when the variables are separable, but this is not always the situation. In the remaining cases we will rely on a very powerful method, the success of which will depend upon our intuition or experience. We simply guess or assume a form for the solution and then test our assumptions, ﬁrst by substitution in the differential equation, and then by applying the given initial conditions. Since we cannot be expected to guess the exact numerical expression for the solution, we will assume a so-lution containing several unknown constants and select the values for these constants in order to satisfy the differential equation and the initial condi-tions. Many of the differential equations encountered in circuit analysis have a solution which may be represented by the exponential function or by the sum of several exponential functions. Let us assume a solution of Eq. in exponential form, i(t) = Aes1t where A and s1 are constants to be determined. After substituting this as-sumed solution in Eq. , we have As1es1t + A R L es1t = 0 CHAPTER 8 BASIC RL AND RC CIRCUITS 264 SECTION 8.1 THE SOURCE-FREE RL CIRCUIT 265 or s1 + R L Aes1t = 0 In order to satisfy this equation for all values of time, it is necessary that A = 0, or s1 = −∞, or s1 = −R/L. But if A = 0 or s1 = −∞, then every response is zero; neither can be a solution to our problem. Therefore, we must choose s1 = −R L and our assumed solution takes on the form i(t) = Ae−Rt/L The remaining constant must be evaluated by applying the initial condi-tion i(0) = I0. Thus, A = I0, and the ﬁnal form of the assumed solution is (again) i(t) = I0e−Rt/L A summary of the basic approach is outlined in Fig. 8.4. A Direct Route: The Characteristic Equation In fact, there is a more direct route that we can take. In obtaining Eq. , we solved s1 + R L = 0 which is known as the characteristic equation. We can obtain the charac-teristic equation directly from the differential equation, without the need for substitution of our trial solution. Consider the general ﬁrst-order differential equation a d f dt + bf = 0 where a and b are constants. We substitute s1 for d f/dt and s0 for f, result-ing in a d f dt + bf = (as + b) f = 0 From this we may directly obtain the characteristic equation as + b = 0 which has the single root s = −b/a. The solution to our differential equa-tion is then f = Ae−bt/a This basic procedure is easily extended to second-order differential equa-tions, as we will explore in Chap. 9. ■FIGURE 8.4 Flowchart for the general approach to solution of ﬁrst-order differential equations where, based on experience, we can guess the form of the solution. Assume a general solution with appropriate constants. Substitute the trial solution into the differential equation and simplify the result. Determine the value for one constant that does not result in a trivial solution. Invoke the initial condition(s) to determine values for the remaining constant(s). End. CHAPTER 8 BASIC RL AND RC CIRCUITS 266 EXAMPLE 8.2 For the circuit of Fig. 8.5a, ﬁnd the voltage labeled v at t 200 ms. Identify the goal of the problem. The schematic of Fig. 8.5a actually represents two different circuits: one with the switch closed (Fig. 8.5b) and one with the switch open (Fig. 8.5c). We are asked to ﬁnd v(0.2)for the circuit shown in Fig. 8.5c. Collect the known information. Both new circuits are drawn and labeled correctly. We next make the assumption that the circuit in Fig. 8.5b has been connected for a long time, so that any transients have dissipated. We may make such an as-sumption as a general rule unless instructed otherwise. This circuit de-termines iL(0). Devise a plan. The circuit of Fig. 8.5c may be analyzed by writing a KVL equation. Ultimately we want a differential equation with only v and t as vari-ables; we will then solve the differential equation for v(t). Construct an appropriate set of equations. Referring to Fig. 8.5c, we write −v + 10iL + 5 diL dt = 0 Substituting iL = −v/40, we ﬁnd that 5 40 dv dt + 10 40 + 1 v = 0 or, more simply, dv dt + 10v = 0 Determine if additional information is required. From previous experience, we know that a complete expression for v will require knowledge of v at a speciﬁc instant of time, with t = 0 being the most convenient. We might be tempted to look at Fig. 8.5b and write v(0) = 24 V, but this is only true just before the switch opens. The resistor voltage can change to any value in the instant that the switch is thrown; only the inductor current must remain unchanged. In the circuit of Fig. 8.5b, iL = 24/10 = 2.4A since the inductor acts like a short circuit to a dc current. Therefore, iL(0) = 2.4 Ain the circuit of Fig. 8.5c, as well—a key point in analyzing this type of cir-cuit. Therefore, in the circuit of Fig. 8.5c, v(0) = (40)(−2.4) = −96 V. Attempt a solution. Any of the three basic solution techniques can be brought to bear; let’s start by writing the characteristic equation corresponding to Eq. : s + 10 = 0 24 V – + v 5 H 40 10 t = 0 iL 24 V – + v 5 H (b) (a) (c) t 0 40 10 iL – + v 5 H t 0 40 10 iL + – + – ■FIGURE 8.5 (a) A simple RL circuit with a switch thrown at time t 0. (b) The circuit as it exists prior to t 0. (c) The circuit after the switch is thrown, and the 24 V source is removed. SECTION 8.1 THE SOURCE-FREE RL CIRCUIT 267 10 V – + v 5 H 6 4 t = 0 iL ■FIGURE 8.6 Circuit for Practice Problem 8.2. Solving, we ﬁnd that s = −10, so v(t) = Ae−10t (which, upon substitution into the left-hand side of Eq. , results in −10Ae−10t + 10Ae−10t = 0 as expected.) We ﬁnd A by setting t = 0 in Eq. and employing the fact that v(0) = −96 V. Thus, v(t) = −96e−10t and so v(0.2) = −12.99 V, down from a maximum of −96 V. Verify the solution. Is it reasonable or expected? Instead of writing a differential equation in v, we could have written our differential equation in terms of iL: 40iL + 10iL + 5 diL dt = 0 or diL dt + 10iL = 0 which has the solution iL = Be−10t. With iL(0) = 2.4, we ﬁnd that iL(t) = 2.4e−10t. Since v = −40iL, we once again obtain Eq. . We should note: it is no coincidence that the inductor current and the resistor voltage have the same exponential dependence! Accounting for the Energy Before we turn our attention to the interpretation of the response, let us re-turn to the circuit of Fig. 8.1, and check the power and energy relationships. The power being dissipated in the resistor is pR = i2R = I 2 0 Re−2Rt/L and the total energy turned into heat in the resistor is found by integrating the instantaneous power from zero time to inﬁnite time: wR = ∞ 0 pR dt = I 2 0 R ∞ 0 e−2Rt/L dt = I 2 0 R −L 2R e−2Rt/L ∞ 0 = 1 2 L I 2 0 This is the result we expect, because the total energy stored initially in the inductor is 1 2 LI 2 0 , and there is no longer any energy stored in the induc-tor at inﬁnite time since its current eventually drops to zero. All the initial energy therefore is accounted for by dissipation in the resistor. PRACTICE ● 8.2 Determine the inductor voltage v in the circuit of Fig. 8.6 for t > 0. Ans: −25e−2tV. 8.2 • PROPERTIES OF THE EXPONENTIAL RESPONSE Let us now consider the nature of the response in the series RL circuit. We have found that the inductor current is represented by i(t) = I0e−Rt/L At t = 0, the current has value I0, but as time increases, the current decreases and approaches zero. The shape of this decaying exponential is seen by the plot of i(t)/I0 versus t shown in Fig. 8.7. Since the function we are plotting is e−Rt/L, the curve will not change if R/L remains unchanged. Thus, the same curve must be obtained for every series RL circuit having the same L/R ratio. Let us see how this ratio affects the shape of the curve. CHAPTER 8 BASIC RL AND RC CIRCUITS 268 0 1 t i I0 ■FIGURE 8.7 A plot of e−Rt/L versus t. If we double the ratio of L to R, the exponent will be unchanged only if t is also doubled. In other words, the original response will occur at a later time, and the new curve is obtained by moving each point on the original curve twice as far to the right. With this larger L/R ratio, the current takes longer to decay to any given fraction of its original value. We might have a tendency to say that the “width’’ of the curve is doubled, or that the width is proportional to L/R. However, we ﬁnd it difﬁcult to deﬁne our term width, because each curve extends from t = 0 to ∞! Instead, let us consider the time that would be required for the current to drop to zero if it continued to drop at its initial rate. The initial rate of decay is found by evaluating the derivative at zero time: d dt i I0 t=0 = −R L e−Rt/L t=0 = −R L We designate the value of time it takes for i/I0 to drop from unity to zero, as-suming a constant rate of decay, by the Greek letter τ (tau). Thus, R L τ = 1 or τ = L R SECTION 8.2 PROPERTIES OF THE EXPONENTIAL RESPONSE 269 The ratio L/R has the units of seconds, since the exponent −Rt/L must be dimensionless. This value of time τ is called the time constant and is shown pictorially in Fig. 8.8. The time constant of a series RL circuit may be found graphically from the response curve; it is necessary only to draw the tangent to the curve at t = 0 and determine the intercept of this tangent line with the time axis. This is often a convenient way of approximating the time constant from the display on an oscilloscope. 0 1 t i I0 ■FIGURE 8.8 The time constant τ is L/R for a series RL circuit. It is the time required for the response curve to drop to zero if it decays at a constant rate equal to its initial rate of decay. 0 2 3 1 0.37 0.14 0.05 t i I0 ■FIGURE 8.9 The current in a series RL circuit is reduced to 37 percent of its initial value at t = τ, 14 percent at t 2τ, and 5 percent at t 3τ. An equally important interpretation of the time constant τ is obtained by determining the value of i(t)/I0 at t = τ . We have i(τ) I0 = e−1 = 0.3679 or i(τ) = 0.3679I0 Thus, in one time constant the response has dropped to 36.8 percent of its initial value; the value of τ may also be determined graphically from this fact, as indicated by Fig. 8.9. It is convenient to measure the decay of the current at intervals of one time constant, and recourse to a hand calculator shows that i(t)/I0 is 0.3679 at t = τ, 0.1353 at t = 2τ, 0.04979 at t = 3τ, 0.01832 at t = 4τ, and 0.006738 at t = 5τ. At some point three to ﬁve time constants after zero time, most of us would agree that the current is a CHAPTER 8 BASIC RL AND RC CIRCUITS 270 negligible fraction of its former self. Thus, if we are asked, “How long does it take for the current to decay to zero?’’ our answer might be, “About ﬁve time constants.’’ At that point, the current is less than 1 percent of its original value! The transient analysis capability of PSpice is very useful when consider-ing the response of source-free circuits. In this example, we make use of a special feature that allows us to vary a component parameter, similar to the way we varied the dc voltage in other simulations. We do this by adding the component PARAM to our schematic; it may be placed any-where, as we will not wire it into the circuit. Our complete RL circuit is shown in Fig. 8.10, which includes an initial inductor current of 1 mA. In order to relate our resistor value to the proposed parameter sweep, we must perform three tasks. First, we provide a name for our parameter, which we choose to call Resistance for the sake of simplic-ity. This is accomplished by double-clicking on the PARAMETERS: label in the schematic, which opens the Property Editor for this pseudo-component. Clicking on New Column results in the dialog box shown in Fig. 8.11a, in which we enter Resistance under Name and a place-holder value of 1 under Value. Our second task consists of linking the COMPUTER-AIDED ANALYSIS ■FIGURE 8.10 Simple RL circuit drawn using the schematic capture tool. PRACTICE ● 8.3 In a source-free series RL circuit, ﬁnd the numerical value of the ratio: (a) i(2τ)/i(τ); (b) i(0.5τ)/i(0); (c) t/τ if i(t)/i(0) = 0.2; (d) t/τ if i(0) −i(t) = i(0) ln 2. Ans: 0.368; 0.607; 1.609; 1.181. SECTION 8.2 PROPERTIES OF THE EXPONENTIAL RESPONSE 271 value of R1 to our parameter sweep, which we accomplish by double-clicking on the default value of R1 on the schematic, resulting in the dialog box of Fig. 8.11b. Under Value, we simply enter {Resistance}. (Note the curly brackets are required.) (a) (b) ■FIGURE 8.11 (a) Add New Column dialog box in the Property Editor for PARAM. (b) Resistor value dialog box. Our third task consists of setting up the simulation, which includes setting transient analysis parameters as well as the values we desire for R1. Under PSpice we select New Simulation Proﬁle (Fig. 8.12a), in which we select Time Domain (Transient) for Analysis type, 300 ns for Run to time, and tick the Parametric Sweep box under Options. This last action results in the dialog box shown in Fig. 8.12b, in which we select Global parameter for Sweep variable and enter Resistance for Parameter name. The ﬁnal setup step required is to select Logarithmic under Sweep type, a Start value of 10, an End value of 1000, and 1 Points/Decade; alternatively we could list the desired resistor values using Value list. After running the simulation, the notiﬁcation box shown in Fig. 8.13 appears, listing the available data sets for plotting (Resistance = 10, 100, and 1000 in this case). A particular data set is selected by high-lighting it; we select all three for this example. Upon selecting the (a) (b) ■FIGURE 8.12 (a) Simulation dialog box. (b) Parameter sweep dialog box. ■FIGURE 8.13 Available data sections dialog box. (Continued on next page) 8.3 • THE SOURCE-FREE RC CIRCUIT Circuits based on resistor-capacitor combinations are more common than their resistor-inductor analogs. The principal reasons for this are the smaller losses present in a physical capacitor, lower cost, better agreement between the simple mathematical model and the actual device behavior, and also smaller size and lighter weight, both of which are particularly important for integrated-circuit applications. CHAPTER 8 BASIC RL AND RC CIRCUITS 272 Why does a larger value of the time constant L/R produce a response curve that decays more slowly? Let us consider the effect of each element. In terms of the time constant τ, the response of the series RL circuit may be written simply as i(t) = I0e−t/τ An increase in L allows a greater energy storage for the same initial current, and this larger energy requires a longer time to be dissipated in the resistor. We may also increase L/R by reducing R. In this case, the power ﬂowing into the resistor is less for the same initial current; again, a greater time is required to dissipate the stored energy. This effect is seen clearly in our simulation result of Fig. 8.14. ■FIGURE 8.14 Probe output for the three resistances. inductor current from our Trace variable choices in Probe, we obtain three graphs at once, as shown (after labeling by hand) in Fig. 8.14. SECTION 8.3 THE SOURCE-FREE RC CIRCUIT 273 i v + – C R ■FIGURE 8.15 A parallel RC circuit for which v(t) is to be determined, subject to the initial condition that v(0) V0. Let us see how closely the analysis of the parallel (or is it series?) RC circuit shown in Fig. 8.15 corresponds to that of the RL circuit. We will as-sume an initial stored energy in the capacitor by selecting v(0) = V0 The total current leaving the node at the top of the circuit diagram must be zero, so we may write C dv dt + v R = 0 Division by C gives us dv dt + v RC = 0 Equation has a familiar form; comparison with Eq. di dt + R L i = 0 shows that the replacement of i by v and L/R by RC produces the identical equation we considered previously. It should, for the RC circuit we are now analyzing is the dual of the RL circuit we considered ﬁrst. This duality forces v(t) for the RC circuit and i(t) for the RL circuit to have identical expressions if the resistance of one circuit is equal to the reciprocal of the resistance of the other circuit, and if L is numerically equal to C. Thus, the response of the RL circuit i(t) = i(0)e−Rt/L = I0e−Rt/L enables us to immediately write v(t) = v(0)e−t/RC = V0e−t/RC for the RC circuit. Suppose instead that we had selected the current i as our variable in the RC circuit, rather than the voltage v. Applying Kirchhoff’s voltage law, 1 C t t0 i dt′ −v0(t0) + Ri = 0 we obtain an integral equation as opposed to a differential equation. How-ever, taking the time derivative of both sides of this equation, i C + R di dt = 0 and replacing i with v/R, we obtain Eq. again: v RC + dv dt = 0 Equation could have been used as our starting point, but the application of duality principles would not have been as natural. Let us discuss the physical nature of the voltage response of the RC circuit as expressed by Eq. .At t = 0 we obtain the correct initial condition, and as t becomes inﬁnite, the voltage approaches zero. This latter result agrees with our thinking that if there were any voltage remaining across the capaci-tor, then energy would continue to ﬂow into the resistor and be dissipated as heat. Thus, a ﬁnal voltage of zero is necessary. The time constant of the RC circuit may be found by using the duality relationships on the expression for the time constant of the RL circuit, or it may be found by simply noting the time at which the response has dropped to 37 percent of its initial value: τ RC = 1 so that Our familiarity with the negative exponential and the signiﬁcance of the time constant τ enables us to sketch the response curve readily (Fig. 8.16). Larger values of R or C provide larger time constants and slower dissipation of the stored energy. A larger resistance will dissipate a smaller power with a given voltage across it, thus requiring a greater time to convert the stored energy into heat; a larger capacitance stores a larger energy with a given voltage across it, again requiring a greater time to lose this initial energy. τ = RC CHAPTER 8 BASIC RL AND RC CIRCUITS 274 0 0.368V0 V0 t v ■FIGURE 8.16 The capacitor voltage v(t) in the parallel RC circuit is plotted as a function of time. The initial value of v(t) is V0. For the circuit of Fig. 8.17a, ﬁnd the voltage labeled v at t 200 μs. To ﬁnd the requested voltage, we will need to draw and analyze two separate circuits: one corresponding to before the switch is thrown (Fig. 8.17b), and one corresponding to after the switch is thrown (Fig. 8.17c). The sole purpose of analyzing the circuit of Fig. 8.17b is to obtain an initial capacitor voltage; we assume any transients in that circuit died out long ago, leaving a purely dc circuit. With no current through either the capacitor or the 4 resistor, then, v(0) = 9 V We next turn our attention to the circuit of Fig. 8.17c, recognizing that τ = RC = (2 + 4)(10 × 10−6) = 60 × 10−6 s Thus, from Eq. , v(t) = v(0)e−t/RC = v(0)e−t/60×10−6 EXAMPLE 8.3 SECTION 8.4 A MORE GENERAL PERSPECTIVE 275 8.4 • A MORE GENERAL PERSPECTIVE As seen indirectly from Examples 8.2 and 8.3, regardless of how many re-sistors we have in the circuit, we obtain a single time constant (either τ = L/R or τ = RC) when only one energy storage element is present. We can formalize this by realizing that the value needed for R is in fact the Thévenin equivalent resistance seen by our energy storage element. (Strange as it may seem, it is even possible to compute a time constant for a circuit containing dependent sources!) General RL Circuits As an example, consider the circuit shown in Fig. 8.19. The equivalent re-sistance the inductor faces is Req = R3 + R4 + R1R2 R1 + R2 9 V 9 V – + v – + v 10 F 10 F 10 F 2 4 t = 0 (b) (a) (c) t 0 2 4 – + v t 0 2 4 + – + – ■FIGURE 8.17 (a) A simple RC circuit with a switch thrown at time t 0. (b) The circuit as it exists prior to t 0. (c) The circuit after the switch is thrown, and the 9 V source is removed. v + – 2 F t = 0 70 50 V 80 + – ■FIGURE 8.18 ■FIGURE 8.19 A source-free circuit containing one inductor and several resistors is analyzed by determining the time constant τ = L/Req. iL i1 i2 R1 R2 R4 R3 L The capacitor voltage must be the same in both circuits at t = 0; no such restriction is placed on any other voltage or current. Substituting Eq. into Eq. , v(t) = 9e−t/60×10−6 V so that v(200 × 10−6) = 321.1 mV (less than 4 percent of its maximum value). PRACTICE ● 8.4 Noting carefully how the circuit changes once the switch in the circuit of Fig. 8.18 is thrown, determine v(t) at t = 0 and at t = 160 μs. Ans: 50 V, 18.39 V. and the time constant is therefore τ = L Req If several inductors are present in a circuit and can be combined using series and/or parallel combination, then Eq. can be further generalized to τ = Leq Req where Leq represents the equivalent inductance. Slicing Thinly: The Distinction Between 0+ and 0− Let’s return to the circuit of Fig. 8.19, and assume that some ﬁnite amount of energy is stored in the inductor at t = 0, so that iL(0) ̸= 0. The inductor current iL is iL = iL(0)e−t/τ and this represents what we might call the basic solution to the problem. It is quite possible that some current or voltage other than iL is needed, such as the current i2 in R2. We can always apply Kirchhoff’s laws and Ohm’s law to the resistive portion of the circuit without any difﬁculty, but current divi-sion provides the quickest answer in this circuit: i2 = − R1 R1 + R2 [iL(0)e−t/τ] It may also happen that we know the initial value of some current other than the inductor current. Since the current in a resistor may change in-stantaneously, we will indicate the instant after any change that might have occurred at t = 0 by the use of the symbol 0+; in more mathematical lan-guage, i1(0+) is the limit from the right of i1(t) as t approaches zero.1 Thus, if we are given the initial value of i1 as i1(0+), then the initial value of i2 is i2(0+) = i1(0+) R1 R2 From these values, we obtain the necessary initial value of iL: iL(0+) = −[i1(0+) + i2(0+)] = −R1 + R2 R2 i1(0+) and the expression for i2 becomes i2 = i1(0+) R1 R2 e−t/τ Let us see if we can obtain this last expression more directly. Since the in-ductor current decays exponentially as e−t/τ, every current throughout the circuit must follow the same functional behavior. This is made clear by con-sidering the inductor current as a source current that is being applied to a re-sistive network. Every current and voltage in the resistive network must have the same time dependence. Using these ideas, we therefore express i2 as i2 = Ae−t/τ where τ = L Req CHAPTER 8 BASIC RL AND RC CIRCUITS 276 We could also write τ = L RTH , where RTH is the Thévenin equivalent resistance “seen” by the inductor L. Note that iL(0+) is always equal to iL(0−). This is not necessarily true for the inductor voltage or any resistor voltage or current, since they may change in zero time. (1) Note that this is a notational convenience only. When faced with t 0+ or its companion t 0−in an equation, we simply use the value zero. This notation allows us to clearly differentiate between the time be-fore and after an event, such as a switch opening or closing, or a power supply being turned on or off. SECTION 8.4 A MORE GENERAL PERSPECTIVE 277 and A must be determined from a knowledge of the initial value of i2. Since i1(0+) is known, the voltage across R1 and R2 is known, and R2i2(0+) = R1i1(0+) leads to i2(0+) = i1(0+) R1 R2 Therefore, i2(t) = i1(0+) R1 R2 e−t/τ A similar sequence of steps will provide a rapid solution to a large num-ber of problems. We ﬁrst recognize the time dependence of the response as an exponential decay, determine the appropriate time constant by combin-ing resistances, write the solution with an unknown amplitude, and then de-termine the amplitude from a given initial condition. This same technique can be applied to any circuit with one inductor and any number of resistors, as well as to those special circuits containing two or more inductors and also two or more resistors that may be simpliﬁed by resistance or inductance combination to one inductor and one resistor. Determine both i1 and iL in the circuit shown in Fig. 8.20a for t > 0. EXAMPLE 8.4 iL i1 18 V t = 0 90 3 mH 2 mH 1 mH 50 60 120 + – (a) iL i1 90 3 mH 2 mH 1 mH (b) 50 60 120 ■FIGURE 8.20 (a) A circuit with multiple resistors and inductors. (b) After t 0, the circuit simpliﬁes to an equivalent resistance of 110 in series with Leq 2.2 mH. (Continued on next page) After t = 0, when the voltage source is disconnected as shown in Fig. 8.20b, we easily calculate an equivalent inductance, Leq = 2 × 3 2 + 3 + 1 = 2.2 mH an equivalent resistance, in series with the equivalent inductance, Req = 90(60 + 120) 90 + 180 + 50 = 110 CHAPTER 8 BASIC RL AND RC CIRCUITS 278 ■FIGURE 8.21 2 A 2 8 t = 0 0.4 H i2 i1 iL WecanverifyouranalysisusingPSpiceandtheswitchmodelSw_tOpen, although it should be remembered that this part is actually just two resis-tance values: one corresponding to before the switch opens at the speciﬁed time (the default value is 10 m), and one for after the switch opens (the default value is 1 M). If the equivalent resistance of the remainder of the circuit is comparable to either value, the values should be edited by double-clicking on the switch symbol in the circuit schematic. Note that there is also a switch model that closes at a speciﬁed time: Sw_tClose. PRACTICE ● 8.5 At t 0.15 s in the circuit of Fig. 8.21, ﬁnd the value of (a) iL; (b) i1; (c) i2. Ans: 0.756 A; 0; 1.244 A. We have now considered the task of ﬁnding the natural response of any circuit which can be represented by an equivalent inductor in series with an equivalent resistor. A circuit containing several resistors and several induc-tors does not always possess a form which allows either the resistors or the inductors to be combined into single equivalent elements. In such instances, there is no single negative exponential term or single time constant associ-ated with the circuit. Rather, there will, in general, be several negative and the time constant, τ = Leq Req = 2.2 × 10−3 110 = 20 μs Thus, the form of the natural response is Ke−50,000t, where K is an unknown constant. Considering the circuit just prior to the switch open-ing (t = 0−), iL = 18/50 A. Since iL(0+) = iL(0−), we know that iL = 18/50 A or 360 mA at t = 0+ and so iL = 360 mA t < 0 360e−50,000t mA t ≥0 There is no restriction on i1 changing instantaneously at t = 0, so its value at t = 0−(18/90 A or 200 mA) is not relevant to ﬁnding i1 for t > 0. Instead, we must ﬁnd i1(0+) through our knowledge of iL(0+). Using current division, i1(0+) = −iL(0+) 120 + 60 120 + 60 + 90 = −240 mA Hence, i1 = 200 mA t < 0 −240e−50,000t mA t ≥0 SECTION 8.4 A MORE GENERAL PERSPECTIVE 279 exponential terms, the number of terms being equal to the number of induc-tors that remain after all possible inductor combinations have been made. General RC Circuits Many of the RC circuits for which we would like to ﬁnd the natural response contain more than a single resistor and capacitor. Just as we did for the RL cir-cuits, we ﬁrst consider those cases in which the given circuit may be reduced to an equivalent circuit consisting of only one resistor and one capacitor. Let us suppose ﬁrst that we are faced with a circuit containing a single capacitor, but any number of resistors. It is possible to replace the two-terminal resistive network which is across the capacitor terminals with an equivalent resistor, and we may then write down the expression for the ca-pacitor voltage immediately. In such instances, the circuit has an effective time constant given by τ = ReqC where Req is the equivalent resistance of the network. An alternative per-spective is that Req is in fact the Thévenin equivalent resistance “seen” by the capacitor. If the circuit has more than one capacitor, but they may be replaced somehow using series and/or parallel combinations with an equivalent capacitance Ceq, then the circuit has an effective time constant given by τ = RCeq with the general case expressed as τ = ReqCeq It is worth noting, however, that parallel capacitors replaced by an equiva-lent capacitance would have to have identical initial conditions. Find v(0+) and i1(0+) for the circuit shown in Fig. 8.22a if v(0−) V0. EXAMPLE 8.5 R1 R2 R3 C (a) v + – i1 Req C (b) v + – ■FIGURE 8.22 (a) A given circuit containing one capacitor and several resistors. (b) The resistors have been replaced by a single equivalent resistor; the time constant is simply τ = ReqC. (Continued on next page) We ﬁrst simplify the circuit of Fig. 8.22a to that of Fig. 8.22b, enabling us to write v = V0e−t/ReqC Our method can be applied to circuits with one energy storage element and one or more dependent sources as well. In such instances, we may write an appropriate KCL or KVL equation along with any necessary supporting equations, distill this down into a single differential equation, and extract the characteristic equation to ﬁnd the time constant. Alternatively, we may begin by ﬁnding the Thévenin equivalent resistance of the network con-nected to the capacitor or inductor, and use this in calculating the appropri-ate RL or RC time constant—unless the dependent source is controlled by a voltage or current associated with the energy storage element, in which case the Thévenin approach cannot be used. CHAPTER 8 BASIC RL AND RC CIRCUITS 280 + – 120 V 4 F 250 600 100 2 k 400 1250 vo + – vC + – t = 0 ■FIGURE 8.23 where v(0+) = v(0−) = V0 and Req = R2 + R1R3 R1 + R3 Every current and voltage in the resistive portion of the network must have the form Ae−t/ReqC, where A is the initial value of that current or voltage. Thus, the current in R1, for example, may be expressed as i1 = i1(0+)e−t/τ where τ = R2 + R1R3 R1 + R3 C and i1(0+) remains to be determined from the initial condition.Any cur-rent ﬂowing in the circuit at t = 0+ must come from the capacitor. There-fore, since v cannot change instantaneously, v(0+) = v(0−) = V0 and i1(0+) = V0 R2 + R1R3/(R1 + R3) R3 R1 + R3 PRACTICE ● 8.6 Find values of vC and vo in the circuit of Fig. 8.23 at t equal to (a) 0−; (b) 0+; (c) 1.3 ms. Ans: 100 V, 38.4 V; 100 V, 25.6 V; 59.5 V, 15.22 V. SECTION 8.4 A MORE GENERAL PERSPECTIVE 281 For the circuit of Fig. 8.24a, ﬁnd the voltage labeled vC for t > 0 if vC(0−) 2 V. EXAMPLE 8.6 20 20 1.5i1 1.5i1 1 A i1 vC 1 F 10 10 + – i1 Vx + – (a) (b) ■FIGURE 8.24 (a) A simple RC circuit containing a dependent source not controlled by a capacitor voltage or current. (b) Circuit for ﬁnding the Thévenin equivalent of the network connected to the capacitor. The dependent source is not controlled by a capacitor voltage or current, so we can start by ﬁnding the Thévenin equivalent of the network to the left of the capacitor. Connecting a 1Atest source as in Fig. 8.24b, Vx = (1 + 1.5i1)(30) where i1 = 1 20 20 10 + 20Vx = Vx 30 Performing a little algebra, we ﬁnd that Vx = −60 V, so the network has a Thévenin equivalent resistance of −60 (unusual, but not im-possible when dealing with a dependent source). Our circuit therefore has a negative time constant τ = −60(1 × 10−6) = −60 μs The capacitor voltage is therefore vC(t) = Aet/60×10−6 V where A = vC(0+) = vC(0−) = 2 V. Thus, vC(t) = 2et/60×10−6 V which, interestingly enough is unstable: it grows exponentially with time. This cannot continue indeﬁnitely; one or more elements in the circuit will eventually fail. Alternatively, we could write a simple KCL equation for the top node of Fig. 8.24a vC = 30 1.5i1 −10−6 dvC dt where i1 = vC 30 (Continued on next page) Some circuits containing a number of both resistors and capacitors may be replaced by an equivalent circuit containing only one resistor and one ca-pacitor; it is necessary that the original circuit be one which can be broken into two parts, one containing all resistors and the other containing all ca-pacitors, such that the two parts are connected by only two ideal conductors. Otherwise, multiple time constants and multiple exponential terms will be required to describe the behavior of the circuit (one time constant for each energy storage element remaining in the circuit after it is reduced as much as possible). As a parting comment, we should be wary of certain situations involving only ideal elements which are suddenly connected together. For example, we may imagine connecting two ideal capacitors in series having unequal voltages prior to t = 0. This poses a problem using our mathematical model of an ideal capacitor; however, real capacitors have resistances associated with them through which energy can be dissipated. 8.5 • THE UNIT-STEP FUNCTION We have been studying the response of RL and RC circuits when no sources or forcing functions were present. We termed this response the natural response, because its form depends only on the nature of the circuit. The reason that any response at all is obtained arises from the presence of initial CHAPTER 8 BASIC RL AND RC CIRCUITS 282 ■FIGURE 8.25 Circuit for Practice Problem 8.7. 2 1.5v1 2 mF 1 v1 + – vC + – Substituting Eq. into Eq. and performing some algebra, we obtain dvC dt − 1 60 × 10−6 vC = 0 which has the characteristic equation s − 1 60 × 10−6 = 0 Thus, s = 1 60 × 10−6 and so vC(t) = Aet/60×10−6 V as we found before. Substitution of A = vC(0+) = 2 results in Eq. , our expression for the capacitor voltage for t > 0. PRACTICE ● 8.7 (a) Regarding the circuit of Fig. 8.25, determine the voltage vC(t) for t > 0 if vC(0−) = 11 V. (b) Is the circuit “stable”? Ans: (a) vC(t) = 11e−2×103t/3 V, t > 0. (b) Yes; it decays (exponentially) rather than grows with time. SECTION 8.5 THE UNIT-STEP FUNCTION 283 energy storage within the inductive or capacitive elements in the circuit. In some cases we were confronted with circuits containing sources and switches; we were informed that certain switching operations were per-formed at t = 0 in order to remove all the sources from the circuit, while leaving known amounts of energy stored here and there. In other words, we have been solving problems in which energy sources are suddenly removed from the circuit; now we must consider that type of response which results when energy sources are suddenly applied to a circuit. We will focus on the response which occurs when the energy sources suddenly applied are dc sources. Since every electrical device is intended to be energized at least once, and since most devices are turned on and off many times in the course of their lifetimes, our study applies to many prac-tical cases. Even though we are now restricting ourselves to dc sources, there are still many cases in which these simpler examples correspond to the operation of physical devices. For example, the ﬁrst circuit we will analyze could represent the buildup of the current when a dc motor is started. The generation and use of the rectangular voltage pulses needed to represent a number or a command in a microprocessor provide many examples in the ﬁeld of electronic or transistor circuitry. Similar circuits are found in the synchronization and sweep circuits of television receivers, in communica-tion systems using pulse modulation, and in radar systems, to name but a few examples. We have been speaking of the “sudden application’’of an energy source, and by this phrase we imply its application in zero time.2 The operation of a switch in series with a battery is thus equivalent to a forcing function which is zero up to the instant that the switch is closed and is equal to the battery voltage thereafter. The forcing function has a break, or discontinuity, at the instant the switch is closed. Certain special forcing functions which are discontinuous or have discontinuous derivatives are called singularity functions, the two most important of these singularity functions being the unit-step function and the unit-impulse function. We deﬁne the unit-step forcing function as a function of time which is zero for all values of its argument less than zero and which is unity for all positive values of its argument. If we let (t −t0) be the argument and rep-resent the unit-step function by u, then u(t −t0) must be zero for all values of t less than t0, and it must be unity for all values of t greater than t0. At t = t0, u(t −t0) changes abruptly from 0 to 1. Its value at t = t0 is not de-ﬁned, but its value is known for all instants of time that are arbitrarily close to t = t0. We often indicate this by writing u(t− 0 ) = 0 and u(t+ 0 ) = 1. The concise mathematical deﬁnition of the unit-step forcing function is u(t −t0) = 0 t < t0 1 t > t0 and the function is shown graphically in Fig. 8.26. Note that a vertical line of unit length is shown at t = t0. Although this “riser’’ is not strictly a part of the deﬁnition of the unit step, it is usually shown in each drawing. (2) Of course, this is not physically possible. However, if the time scale over which such an event occurs is very short compared to all other relevant time scales that describe the operation of a circuit, this is approxi-mately true, and mathematically convenient. 0 t t0 1 u(t – t0) ■FIGURE 8.26 The unit-step forcing function, u(t −t0). We also note that the unit step need not be a time function. For example, u(x −x0) could be used to denote a unit-step function where x might be a distance in meters, for example, or a frequency. Very often in circuit analysis a discontinuity or a switching action takes place at an instant that is deﬁned as t = 0. In that case t0 = 0, and we then represent the corresponding unit-step forcing function by u(t −0), or more simply u(t). This is shown in Fig. 8.27. Thus u(t) = 0 t < 0 1 t > 0 The unit-step forcing function is in itself dimensionless. If we wish it to represent a voltage, it is necessary to multiply u(t −t0) by some constant voltage, such as 5 V. Thus, v(t) = 5u(t −0.2) V is an ideal voltage source which is zero before t = 0.2 s and a constant 5 V after t = 0.2 s. This forc-ing function is shown connected to a general network in Fig. 8.28a. Physical Sources and the Unit-Step Function Perhaps we should ask what physical source is the equivalent of this discontinuous forcing function. By equivalent, we mean simply that the voltage-current characteristics of the two networks are identical. For the step-voltage source of Fig. 8.28a, the voltage-current characteristic is simple: the voltage is zero prior to t = 0.2 s, it is 5 V after t = 0.2 s, and the current may be any (ﬁnite) value in either time interval. Our ﬁrst thoughts might produce the attempt at an equivalent shown in Fig. 8.28b, a 5 V dc source in series with a switch which closes at t = 0.2 s. This network is not equiva-lent for t < 0.2 s, however, because the voltage across the battery and switch is completely unspeciﬁed in this time interval. The “equivalent’’ source is an open circuit, and the voltage across it may be anything. After t = 0.2 s, the networks are equivalent, and if this is the only time interval in which we are interested, and if the initial currents which ﬂow from the two networks are identical at t = 0.2 s, then Fig. 8.28b becomes a useful equiv-alent of Fig. 8.28a. In order to obtain an exact equivalent for the voltage-step forcing func-tion, we may provide a single-pole double-throw switch. Before t = 0.2 s, the switch serves to ensure zero voltage across the input terminals of the general network. After t = 0.2 s, the switch is thrown to provide a constant input voltage of 5 V. At t = 0.2 s, the voltage is indeterminate (as is the step forcing function), and the battery is momentarily short-circuited (it is CHAPTER 8 BASIC RL AND RC CIRCUITS 284 u(t) 0 t 1 ■FIGURE 8.27 The unit-step forcing function u(t) is shown as a function of t. + – General network (a) 5u(t – 0.2) V + – 5 V t = 0.2 s General network (b) + – 5 V General network (c) t = 0.2 s ■FIGURE 8.28 (a) A voltage-step forcing function is shown as the source driving a general network. (b) A simple circuit which, although not the exact equivalent of part (a), may be used as its equivalent in many cases. (c) An exact equivalent of part (a). SECTION 8.5 THE UNIT-STEP FUNCTION 285 fortunate that we are dealing with mathematical models!). This exact equiv-alent of Fig. 8.28a is shown in Fig. 8.28c. Figure 8.29a shows a current-step forcing function driving a general network. If we attempt to replace this circuit by a dc source in parallel with a switch (which opens at t = t0), we must realize that the circuits are equiva-lent after t = t0 but that the responses after t = t0 are alike only if the initial conditions are the same. The circuit in Fig. 8.29b implies no voltage exists across the current source terminals for t < t0. This is not the case for the cir-cuit of Fig. 8.29a. However, we may often use the circuits of Fig. 8.29a and b interchangeably. The exact equivalent of Fig. 8.29a is the dual of the cir-cuit of Fig. 8.28c; the exact equivalent of Fig. 8.29b cannot be constructed with current- and voltage-step forcing functions alone.3 The Rectangular Pulse Function Some very useful forcing functions may be obtained by manipulating the unit-step forcing function. Let us deﬁne a rectangular voltage pulse by the following conditions: v(t) = ⎧ ⎨ ⎩ 0 t < t0 V0 t0 < t < t1 0 t > t1 The pulse is drawn in Fig. 8.30. Can this pulse be represented in terms of the unit-step forcing function? Let us consider the difference of the two unit steps,u(t −t0)−u(t −t1).ThetwostepfunctionsareshowninFig.8.31a, and their difference is a rectangular pulse. The source V0u(t −t0)−V0u(t −t1) which provides us with the desired voltage is indicated in Fig. 8.31b. (3) The equivalent can be drawn if the current through the switch prior to t = t0 is known. (4) Apparently, we’re pretty good at the controls of this car. A reaction time of 70 ns? General network (a) I0u(t – t0) I0 General network (b) t = t0 ■FIGURE 8.29 (a) A current-step forcing function is applied to a general network. (b) A simple circuit which, although not the exact equivalent of part (a), may be used as its equivalent in many cases. V0 0 v (t) t0 t1 t ■FIGURE 8.30 A useful forcing function, the rectangular voltage pulse. ■FIGURE 8.31 (a) The unit steps u(t −t0) and −u(t −t1). (b) A source which yields the rectangular voltage pulse of Fig. 8.30. If we have a sinusoidal voltage source Vm sin ωt which is suddenly con-nected to a network at t = t0, then an appropriate voltage forcing function would be v(t) = Vmu(t −t0) sin ωt. If we wish to represent one burst of en-ergy from the transmitter for a radio-controlled car operating at 47 MHz (295 Mrad/s), we may turn the sinusoidal source off 70 ns later by a second unit-step forcing function.4 The voltage pulse is thus v(t) = Vm[u(t −t0) −u(t −t0 −7 × 10−8)] sin(295 × 106t) This forcing function is sketched in Fig. 8.32. 0 –1 1 t0 t1 u(t – t0) –u(t – t1) (a) t + – + – (b) V0u(t – t1) V0u(t – t0) v(t) + – PRACTICE ● 8.8 Evaluate each of the following at t = 0.8: (a) 3u(t) −2u(−t) + 0.8u(1 −t); (b) [4u(t)]u(−t); (c) 2u(t) sin πt. Ans: 3.8; 0; 1.176. 8.6 • DRIVEN RL CIRCUITS We are now ready to subject a simple network to the sudden application of a dc source. The circuit consists of a battery whose voltage is V0 in series with a switch, a resistor R, and an inductor L. The switch is closed at t = 0, as indicated on the circuit diagram of Fig. 8.33a. It is evident that the cur-rent i(t) is zero before t = 0, and we are therefore able to replace the battery and switch by a voltage-step forcing function V0u(t), which also produces no response prior to t = 0. After t = 0, the two circuits are clearly identical. Hence, we seek the current i(t) either in the given circuit of Fig. 8.33a or in the equivalent circuit of Fig. 8.33b. We will ﬁnd i(t) at this time by writing the appropriate circuit equation and then solving it by separation of the variables and integration. After we obtain the answer and investigate the two parts of which it is composed, we will see that there is physical signiﬁcance to each of these two terms. With a more intuitive understanding of how each term originates, we will be able to produce more rapid and more meaningful solutions to every problem in-volving the sudden application of any source. Applying Kirchhoff’s voltage law to the circuit of Fig. 8.33b, we have Ri + L di dt = V0u(t) Since the unit-step forcing function is discontinuous at t = 0, we will ﬁrst consider the solution for t < 0 and then for t > 0. The application of zero voltage since t = −∞forces a zero response, so that i(t) = 0 t < 0 For positive time, however, u(t) is unity and we must solve the equation Ri + L di dt = V0 t > 0 CHAPTER 8 BASIC RL AND RC CIRCUITS 286 0 t0 + 7 10–8 t (s) v (t) Vm –Vm t0 ■FIGURE 8.32 A 47 MHz radio-frequency pulse, described by v(t) Vm[u(t −t0) −u(t −t0 −7 × 10−8)] sin(259 ×106t). + – V0 L R (a) i(t) t = 0 + – V0 u(t) L (b) R i(t) ■FIGURE 8.33 (a) The given circuit. (b) An equivalent circuit, possessing the same response i(t) for all time. SECTION 8.6 DRIVEN RL CIRCUITS 287 The variables may be separated in several simple algebraic steps, yielding L di V0 −Ri = dt and each side may be integrated directly: −L R ln(V0 −Ri) = t + k In order to evaluate k, an initial condition must be invoked. Prior to t = 0, i(t) is zero, and thus i(0−) = 0. Since the current in an inductor cannot change by a ﬁnite amount in zero time without being associated with an inﬁnite voltage, we thus have i(0+) = 0. Setting i = 0 at t = 0, we obtain −L R ln V0 = k and, hence, −L R [ln(V0 −Ri) −ln V0] = t Rearranging, V0 −Ri V0 = e−Rt/L or i = V0 R −V0 R e−Rt/L t > 0 Thus, an expression for the response valid for all t would be i = V0 R −V0 R e−Rt/L u(t) A More Direct Procedure This is the desired solution, but it has not been obtained in the simplest man-ner. In order to establish a more direct procedure, let us try to interpret the two terms appearing in Eq. . The exponential term has the functional form of the natural response of the RL circuit; it is a negative exponential, it ap-proaches zero as time increases, and it is characterized by the time constant LR. The functional form of this part of the response is thus identical with that which is obtained in the source-free circuit. However, the amplitude of this exponential term depends on the source voltage V0. We might gener-alize, then, that the response will be the sum of two terms, where one term has a functional form identical to that of the source-free response, but has an am-plitude that depends on the forcing function. But what of the other term? Equation also contains a constant term, V0/R. Why is it present? The answer is simple: the natural response approaches zero as the energy is gradually dissipated, but the total response must not approach zero. Eventu-ally the circuit behaves as a resistor and an inductor in series with a battery. Since the inductor looks like a short circuit to dc, the only current now ﬂowing is V0/R. This current is a part of the response that is directly attrib-utable to the forcing function, and we call it the forced response. It is the response that is present a long time after the switch is closed. The complete response is composed of two parts, the natural response and the forced response. The natural response is a characteristic of the cir-cuit and not of the sources. Its form may be found by considering the source-free circuit, and it has an amplitude that depends on both the initial amplitude of the source and the initial energy storage. The forced response has the characteristics of the forcing function; it is found by pretending that all switches were thrown a long time ago. Since we are presently concerned only with switches and dc sources, the forced response is merely the solu-tion of a simple dc circuit problem. CHAPTER 8 BASIC RL AND RC CIRCUITS 288 For the circuit of Fig. 8.34, ﬁnd i(t) for t ∞, 3−, 3+, and 100 μs after the source changes value. Long after any transients have died out (t →∞), the circuit is a simple dc circuit driven by a 12 V voltage source. The inductor appears as a short circuit, so i(∞) = 12 1000 = 12 mA What is meant by i(3−)? This is simply a notational convenience to indicate the instant before the voltage source changes value. For t < 3, u(t −3) = 0. Thus, i(3−) = 0 as well. At t = 3+, the forcing function 12u(t −3) = 12 V. However, since the inductor current cannot change in zero time, i(3+) = i(3−) = 0. The most straightforward approach to analyzing the circuit for t > 3 s is to rewrite Eq. as i(t′) = V0 R −V0 R e−Rt′/L u(t′) and note that this equation applies to our circuit as well if we shift the time axis such that t′ = t −3 Therefore, with V0/R = 12 mA and R/L = 20,000 s−1, i(t −3) = 12 −12e−20,000(t−3) u(t −3) mA which can be written more simply as i(t) = 12 −12e−20,000(t−3) u(t −3) mA since the unit-step function forces a zero value for t < 3, as required. Substituting t = 3.0001 s into Eq. or , we ﬁnd that i = 10.38 mA at a time 100 μs after the source changes value. EXAMPLE 8.7 ■FIGURE 8.34 A simple RL circuit driven by a voltage-step forcing function. + – i(t) 1 k 50 mH 12u(t – 3) V SECTION 8.7 NATURAL AND FORCED RESPONSE 289 Developing an Intuitive Understanding The reason for the two responses, forced and natural, may be seen from physical arguments. We know that our circuit will eventually assume the forced response. However, at the instant the switches are thrown, the initial inductor currents (or, in RC circuits, the voltages across the capacitors) will have values that depend only on the energy stored in these elements. These currents or voltages cannot be expected to be the same as the currents and voltages demanded by the forced response. Hence, there must be a transient period during which the currents and voltages change from their given ini-tial values to their required ﬁnal values. The portion of the response that provides the transition from initial to ﬁnal values is the natural response (often called the transient response, as we found earlier). If we describe the response of the simple source-free RL circuit in these terms, then we should say that the forced response is zero and that the natural response serves to connect the initial response dictated by the stored energy with the zero value of the forced response. This description is appropriate only for those circuits in which the nat-ural response eventually dies out. This always occurs in physical circuits where some resistance is associated with every element, but there are a number of “pathologic’’ circuits in which the natural response is nonvan-ishing as time becomes inﬁnite. Those circuits in which trapped currents circulate around inductive loops, or voltages are trapped in series strings of capacitors, are examples. 8.7 • NATURAL AND FORCED RESPONSE There is also an excellent mathematical reason for considering the complete response to be composed of two parts—the forced response and the natural response. The reason is based on the fact that the solution of any linear differential equation may be expressed as the sum of two parts: the comple-mentary solution (natural response) and the particular solution (forced response). Without delving into the general theory of differential equations, let us consider a general equation of the type met in the previous section: di dt + Pi = Q or di + Pi dt = Q dt We may identify Q as a forcing function and express it as Q(t) to em-phasize its general time dependence. Let us simplify the discussion by PRACTICE ● 8.9 The voltage source 60 −40u(t) V is in series with a 10 resistor and a 50 mH inductor. Find the magnitudes of the inductor current and voltage at t equal to (a) 0−; (b) 0+; (c) ∞; (d) 3 ms. Ans: 6 A, 0 V; 6 A, 40 V; 2 A, 0 V; 4.20 A, 22.0 V. assuming that P is a positive constant. Later, we will also assume that Q is constant, thus restricting ourselves to dc forcing functions. In any standard text on elementary differential equations, it is shown that if both sides of Eq. are multiplied by a suitable “integrating factor,” then each side becomes an exact differential that can be integrated directly to obtain the solution. We are not separating the variables, but merely ar-ranging them in such a way that integration is possible. For this equation, the integrating factor is e P dt or simply ePt, since P is a constant. We mul-tiply each side of the equation by this integrating factor and obtain ePt di + iPePt dt = QePt dt The form of the left side may be simpliﬁed by recognizing it as the exact differential of iePt: d(iePt) = ePt di + iPePt dt so that Eq. becomes d(iePt) = QePt dt Integrating each side, iePt = QePt dt + A where A is a constant of integration. Multiplication by e−Pt produces the so-lution for i(t), i = e−Pt QePt dt + Ae−Pt If our forcing function Q(t) is known, then we can obtain the functional form of i(t) by evaluating the integral. We will not evaluate such an integral for each problem, however; instead, we are interested in using Eq. to draw several very general conclusions. The Natural Response We note ﬁrst that, for a source-free circuit, Q must be zero, and the solution is the natural response in = Ae−Pt We will ﬁnd that the constant P is never negative for a circuit with only resistors, inductors, and capacitors; its value depends only on the passive circuit elements5 and their interconnection in the circuit. The natural re-sponse therefore approaches zero as time increases without limit. This must be the case for the simple RL circuit, because the initial energy is gradually dissipated in the resistor, leaving the circuit in the form of heat. There are also idealized circuits in which P is zero; in these circuits the natural re-sponse does not die out. We therefore ﬁnd that one of the two terms making up the complete re-sponse has the form of the natural response; it has an amplitude which will CHAPTER 8 BASIC RL AND RC CIRCUITS 290 (5) If the circuit contains a dependent source or a negative resistance, P may be negative. SECTION 8.7 NATURAL AND FORCED RESPONSE 291 depend on (but not always be equal to) the initial value of the complete response and thus on the initial value of the forcing function also. The Forced Response We next observe that the ﬁrst term of Eq. depends on the functional form of Q(t), the forcing function. Whenever we have a circuit in which the natural response dies out as t becomes inﬁnite, this ﬁrst term must com-pletely describe the form of the response after the natural response has dis-appeared. This term is typically called the forced response; it is also called the steady-state response, the particular solution, or the particular integral. For the present, we have elected to consider only those problems in-volving the sudden application of dc sources, and Q(t) will therefore be a constant for all values of time. If we wish, we can now evaluate the integral in Eq. , obtaining the forced response i f = Q P and the complete response i(t) = Q P + Ae−Pt For the RL series circuit, Q/P is the constant current V0/R and 1/P is the time constant τ. We should see that the forced response might have been obtained without evaluating the integral, because it must be the complete response at inﬁnite time; it is merely the source voltage divided by the series resistance. The forced response is thus obtained by inspection of the ﬁnal circuit. Determination of the Complete Response Let us use the simple RL series circuit to illustrate how to determine the complete response by the addition of the natural and forced responses. The circuit shown in Fig. 8.35 was analyzed earlier, but by a longer method. The desired response is the current i(t), and we ﬁrst express this current as the sum of the natural and the forced current, i = in + i f The functional form of the natural response must be the same as that ob-tained without any sources. We therefore replace the step-voltage source by a short circuit and recognize the old RL series loop. Thus, in = Ae−Rt/L where the amplitude A is yet to be determined; since the initial condition applies to the complete response, we cannot simply assume A = i(0). We next consider the forced response. In this particular problem the forced response must be constant, because the source is a constant V0 for all positive values of time. After the natural response has died out, there can be no voltage across the inductor; hence, a voltage V0 appears across R, and the forced response is simply i f = V0 R ■FIGURE 8.35 A series RL circuit that is used to illustrate the method by which the complete response is obtained as the sum of the natural and forced responses. + – V0 u(t) L R i(t) Note that the forced response is determined completely; there is no un-known amplitude. We next combine the two responses to obtain i = Ae−Rt/L + V0 R and apply the initial condition to evaluate A. The current is zero prior to t = 0, and it cannot change value instantaneously since it is the current ﬂowing through an inductor. Thus, the current is zero immediately after t = 0, and 0 = A + V0 R so i = V0 R (1 −e−Rt/L) Note carefully that A is not the initial value of i, since A = −V0/R, while i(0) = 0. In considering source-free circuits, we found that A was the initial value of the response. When forcing functions are present, however, we must ﬁrst ﬁnd the initial value of the response and then substitute this in the equation for the complete response to ﬁnd A. This response is plotted in Fig. 8.36, and we can see the manner in which the current builds up from its initial value of zero to its ﬁnal value of V0/R. The transition is effectively accomplished in a time 3τ. If our circuit repre-sents the field coil of a large dc motor, we might have L = 10 H and R = 20 , obtaining τ = 0.5 s. The ﬁeld current is thus established in about 1.5 s. In one time constant, the current has attained 63.2 percent of its ﬁnal value. CHAPTER 8 BASIC RL AND RC CIRCUITS 292 ■FIGURE 8.36 The current ﬂowing through the inductor of Fig. 8.35 is shown graphically. A line extending the initial slope meets the constant forced response at t τ. 0 0.632V0/R V0/R 2 3 i t Determine i(t) for all values of time in the circuit of Fig. 8.37. EXAMPLE 8.8 ■FIGURE 8.37 The circuit of Example 8.8. + – 50 V 2 50u(t) V 6 3 H i(t) + – The circuit contains a dc voltage source as well as a step-voltage source. We might choose to replace everything to the left of the inductor by the Thévenin equivalent, but instead let us merely recog-nize the form of that equivalent as a resistor in series with some voltage source. The circuit contains only one energy storage element, the inductor. We ﬁrst note that τ = L Req = 3 1.5 = 2 s SECTION 8.7 NATURAL AND FORCED RESPONSE 293 As a ﬁnal example of this method by which the complete response of any circuit subjected to a transient may be written down almost by inspection, let us again consider the simple RL series circuit, but subjected to a voltage pulse. ■FIGURE 8.38 The response i(t) of the circuit shown in Fig. 8.37 is sketched for values of times less and greater than zero. 25 50 0 –2 2 4 6 Forced response begins around t > 3 i(t) (A) t (s) and recall that i = i f + in The natural response is therefore a negative exponential as before: in = Ke−t/2 A t > 0 Since the forcing function is a dc source, the forced response will be a constant current. The inductor acts like a short circuit to dc, so that i f = 100 2 = 50 A Thus, i = 50 + Ke−0.5t A t > 0 In order to evaluate K, we must establish the initial value of the inductor current. Prior to t = 0, this current is 25 A, and it cannot change instantaneously. Thus, 25 = 50 + K or K = −25 Hence, i = 50 −25e−0.5t A t > 0 We complete the solution by also stating i = 25 A t < 0 or by writing a single expression valid for all t, i = 25 + 25(1 −e−0.5t)u(t) A The complete response is sketched in Fig. 8.38. Note how the natural response serves to connect the response for t < 0 with the constant forced response. PRACTICE ● 8.10 A voltage source, vs = 20u(t) V, is in series with a 200 resistor and a 4 H inductor. Find the magnitude of the inductor current at t equal to (a) 0−; (b) 0+; (c) 8 ms; (d) 15 ms. Ans: 0; 0; 33.0 mA; 52.8 mA. CHAPTER 8 BASIC RL AND RC CIRCUITS 294 Find the current response in a simple series RL circuit when the forcing function is a rectangular voltage pulse of amplitude V0 and duration t0. We represent the forcing function as the sum of two step-voltage sources V0u(t) and −V0u(t −t0), as indicated in Fig. 8.39a and b, and we plan to obtain the response by using superposition. Let i1(t) designate that part of i(t) which is due to the upper source V0u(t) acting alone, and let i2(t) represent that part due to −V0u(t −t0) acting alone. Then, i(t) = i1(t) + i2(t) Our object is now to write each of the partial responses i1 and i2 as the sum of a natural and a forced response. The response i1(t) is familiar; this problem was solved in Eq. : i1(t) = V0 R (1 −e−Rt/L) t > 0 Note that this solution is only valid for t > 0 as indicated; i1 = 0 for t < 0. We now turn our attention to the other source and its response i2(t). Only the polarity of the source and the time of its application are differ-ent. There is no need therefore to determine the form of the natural response and the forced response; the solution for i1(t) enables us to write i2(t) = −V0 R [1 −e−R(t−t0)/L] t > t0 where the applicable range of t, t > t0, must again be indicated; and i2 = 0 for t < t0. We now add the two solutions, but do so carefully, since each is valid over a different interval of time. Thus, i(t) = 0 t < 0 i(t) = V0 R (1 −e−Rt/L) 0 < t < t0 and i(t) = V0 R (1 −e−Rt/L) −V0 R (1 −e−R(t−t0)/L) t > t0 or more compactly, i(t) = V0 R e−Rt/L(eRt0/L −1) t > t0 EXAMPLE 8.9 ■FIGURE 8.39 (a) A rectangular voltage pulse which is to be used as the forcing function in a simple series RL circuit. (b) The series RL circuit, showing the representation of the forcing function by the series combination of two independent voltage-step sources. The current i(t) is desired. 0 v (t) (a) t t0 V0 + – + – –V0u(t – t0) V0u(t) (b) R L v (t) + – i(t) ■FIGURE 8.40 Two possible response curves are shown for the circuit of Fig. 8.39b. (a) τ is selected as t0/2. (b) τ is selected as 2t0. t0 1 2 0 t0 2t0 t i(t) (a) V0/R 0 t0 2t0 3t0 t i(t) (b) V0/R Although Eqs. through completely describe the response of the circuit in Fig. 8.39b to the pulse waveform of Fig. 8.39a, the current waveform itself is sensitive to both the circuit time constant τ and the volt-age pulse duration t0. Two possible curves are shown in Fig. 8.40. SECTION 8.8 DRIVEN RC CIRCUITS 295 The left curve is drawn for the case where the time constant is only one-half as large as the length of the applied pulse; the rising portion of the exponential has therefore almost reached V0/R before the decaying expo-nential begins. The opposite situation is shown to the right; there, the time constant is twice t0 and the response never has a chance to reach the larger amplitudes. The procedure we have been using to ﬁnd the response of an RL circuit after dc sources have been switched on or off (or in or out of the circuit) at some instant of time is summarized in the following. We assume that the circuit is reducible to a single equivalent resistance Req in series with a sin-gle equivalent inductance Leq when all independent sources are set equal to zero. The response we seek is represented by f (t). PRACTICE ● 8.11 The circuit shown in Fig. 8.41 has been in the form shown for a very long time. The switch opens at t = 0. Find iR at t equal to (a) 0−; (b) 0+; (c) ∞; (d) 1.5 ms. Ans: 0; 10 mA; 4 mA; 5.34 mA. 8.8 • DRIVEN RC CIRCUITS The complete response of any RC circuit may also be obtained as the sum of the natural and the forced response. Since the procedure is virtually identical to what we have already discussed in detail for RL circuits, the best approach at this stage is to illustrate it by working a relevant example com-pletely, where the goal is not just a capacitor-related quantity but the current associated with a resistor as well. iR 0.1 H t = 0 60 40 10 mA ■FIGURE 8.41 1. With all independent sources zeroed out, simplify the circuit to determine Req, Leq, and the time constant τ = Leq/Req. 2. Viewing Leq as a short circuit, use dc analysis methods to ﬁnd iL(0−), the inductor current just prior to the discontinuity. 3. Again viewing Leq as a short circuit, use dc analysis methods to ﬁnd the forced response. This is the value approached by f (t) as t →∞; we represent it by f (∞). 4. Write the total response as the sum of the forced and natural responses: f (t) = f (∞) + Ae−t/τ . 5. Find f (0+) by using the condition that iL(0+) = iL(0−). If desired, Leq may be replaced by a current source iL(0+) [an open circuit if iL(0+) = 0] for this calculation. With the exception of inductor currents (and capacitor voltages), other currents and volt-ages in the circuit may change abruptly. 6. f (0+) = f (∞) + A and f (t) = f (∞) + [ f (0+) −f (∞)] e−t/τ, or total response = ﬁnal value + (initial value −ﬁnal value) e−t/τ. CHAPTER 8 BASIC RL AND RC CIRCUITS 296 Find the capacitor voltage vC(t) and the current i(t) in the 200 resistor of Fig. 8.42 for all time. EXAMPLE 8.10 We begin by considering the state of the circuit at t < 0, corresponding to the switch at position a as represented in Fig. 8.42b. As usual, we assume no transients are present, so that only a forced response due to the 120 V source is relevant to ﬁnding vC(0−). Simple voltage division then gives us the initial voltage, vC(0) = 50 50 + 10(120) = 100 V Since the capacitor voltage cannot change instantaneously, this voltage is equally valid at t = 0−and t = 0+. The switch is now thrown to b, and the complete response is vC = vC f + vCn a b 10 60 200 50 120 V 50 V 50 mF vC(t) + – i(t) t = 0 t 0 50 V + – 120 V + – 200 50 10 60 50 mF vC + – i(t) (b) (a) t 0 50 V + – + – + – 50 200 60 50 mF vC + – i(t) (c) ■FIGURE 8.42 (a) An RC circuit in which the complete responses vC and i are obtained by adding a forced response and a natural response. (b) Circuit for t ≤0. (c) Circuit for t ≥0. SECTION 8.8 DRIVEN RC CIRCUITS 297 The corresponding circuit has been redrawn in Fig. 8.42c for conve-nience. The form of the natural response is obtained by replacing the 50 V source by a short circuit and evaluating the equivalent resistance to ﬁnd the time constant (in other words, we are ﬁnding the Thévenin equivalent resistance “seen’’ by the capacitor): Req = 1 1 50 + 1 200 + 1 60 = 24 Thus, vCn = Ae−t/ReqC = Ae−t/1.2 In order to evaluate the forced response with the switch at b, we wait until all the voltages and currents have stopped changing, thus treating the capacitor as an open circuit, and use voltage division once more: vCf = 50 200 ∥50 60 + 200 ∥50 = 50 (50)(200)/250 60 + (50)(200)/250 = 20 V Consequently, vC = 20 + Ae−t/1.2 V and from the initial condition already obtained, 100 = 20 + A or vC = 20 + 80e−t/1.2 V t ≥0 and vC = 100 V t < 0 This response is sketched in Fig. 8.43a; again the natural response is seen to form a transition from the initial to the ﬁnal response. Next we attack i(t). This response need not remain constant during the instant of switching. With the contact at a, it is evident that i = 50/260 = 192.3 milliamperes. When the switch moves to position b, the forced response for this current becomes i f = 50 60 + (50)(200)/(50 + 200) 50 50 + 200 = 0.1 ampere The form of the natural response is the same as that which we already determined for the capacitor voltage: in = Ae−t/1.2 Combining the forced and natural responses, we obtain i = 0.1 + Ae−t/1.2 amperes 20 100 0 –1 1 2 (a) 3 vC (V) t (s) 0.1 0.192 0.5 0 –1 1 2 (b) 3 i(t) (A) t (s) ■FIGURE 8.43 The responses (a) vC and (b) i are plotted as functions of time for the circuit of Fig. 8.42. (Continued on next page) We conclude by listing the duals of the statements given at the end of Sec. 8.7. The procedure we have been using to ﬁnd the response of an RC circuit after dc sources have been switched on or off, or in or out of the circuit, at some instant of time, say t = 0, is summarized in the following. We assume that the circuit is reducible to a single equivalent resistance Req in parallel with a single equivalent capacitance Ceq when all independent sources are set equal to zero. The response we seek is represented by f (t). CHAPTER 8 BASIC RL AND RC CIRCUITS 298 ■FIGURE 8.44 To evaluate A, we need to know i(0+). This is found by ﬁxing our attention on the energy-storage element (the capacitor). The fact that vC must remain 100 V during the switching interval is the governing condition which establishes the other currents and voltages at t = 0+. Since vC(0+) = 100 V, and since the capacitor is in parallel with the 200 resistor, we ﬁnd i(0+) = 0.5 ampere, A = 0.4 ampere, and thus i(t) = 0.1923 ampere t < 0 i(t) = 0.1 + 0.4e−t/1.2 ampere t > 0 or i(t) = 0.1923 + (−0.0923 + 0.4e−t/1.2)u(t) amperes where the last expression is correct for all t. The complete response for all t may also be written concisely by using u(−t), which is unity for t < 0 and 0 for t > 0. Thus, i(t) = 0.1923u(−t) + (0.1 + 0.4e−t/1.2)u(t) amperes This response is sketched in Fig. 8.43b. Note that only four numbers are needed to write the functional form of the response for this single-energy-storage-element circuit, or to prepare the sketch: the constant value prior to switching (0.1923 ampere), the instantaneous value just after switching (0.5 ampere), the constant forced response (0.1 ampere), and the time constant (1.2 s). The appropriate negative exponential function is then easily written or drawn. PRACTICE ● 8.12 For the circuit of Fig. 8.44, ﬁnd vC(t) at t equal to (a) 0−; (b) 0+; (c) ∞; (d) 0.08 s. Ans: 20 V; 20 V; 28 V; 24.4 V. + – 20 k 25 k 80 k 1 mA 5 F vC(t) + – 10u(t) V iR SECTION 8.8 DRIVEN RC CIRCUITS 299 EXAMPLE 8.11 10 22 F 4.7 v + – 5e–2000t u(t) A ■FIGURE 8.45 A simple RC circuit driven by an exponentially decaying forcing function. Determine an expression for v(t) in the circuit of Fig. 8.45 valid for t > 0. Based on experience, we expect a complete response of the form v(t) = vf + vn where vf will likely resemble our forcing function and vn will have the form Ae−t/τ. What is the circuit time constant τ? We replace our source with an open circuit and ﬁnd the Thévenin equivalent resistance in parallel with the capacitor: Req = 4.7 + 10 = 14.7 Thus, our time constant is τ = ReqC = 323.4 μs, or equivalently 1/τ = 3.092 ×103 s−1. There are several ways to proceed, although perhaps the most straightforward is to perform a source transformation, resulting in a voltage source 23.5e−2000tu(t) V in series with 14.7 and 22 μF. (Note this does not change the time constant.) Writing a simple KVL equation for t > 0, we ﬁnd that 23.5e−2000t = (14.7)(22 × 10−6) dv dt + v (Continued on next page) As we have just seen, the same basic steps that apply to the analysis of RL circuits can be applied to RC circuits as well. Up to now, we have con-ﬁned ourselves to the analysis of circuits with dc forcing functions only, de-spite the fact that Eq. holds for more general functions such as Q(t) = 9 cos(5t −7o) or Q(t) = 2e−5t. Before concluding this section, we explore one such non-dc scenario. 1. With all independent sources zeroed out, simplify the circuit to determine Req, Ceq, and the time constant τ = ReqCeq. 2. Viewing Ceq as an open circuit, use dc analysis methods to ﬁnd vC(0−), the capacitor voltage just prior to the discontinuity. 3. Again viewing Ceq as an open circuit, use dc analysis methods to ﬁnd the forced response. This is the value approached by f (t) as t →∞; we represent it by f (∞). 4. Write the total response as the sum of the forced and natural responses: f (t) = f (∞) + Ae−t/τ . 5. Find f (0+) by using the condition that vC(0+) = vC(0−). If desired, Ceq may be replaced by a voltage source vC(0+) [a short circuit if vC(0+) = 0] for this calculation. With the exception of capacitor voltages (and inductor currents), other voltages and cur-rents in the circuit may change abruptly. 6. f (0+) = f (∞) + A and f (t) = f (∞) + [ f (0+) −f (∞)]e−t/τ , or total response = ﬁnal value + (initial value −ﬁnal value) e−t/τ. 8.9 • PREDICTING THE RESPONSE OF SEQUENTIALLY SWITCHED CIRCUITS In Example 8.9 we brieﬂy considered the response of an RL circuit to a pulse waveform, in which a source was effectively switched into and sub-sequently switched out of the circuit. This type of situation is common in practice, as few circuits are designed to be energized only once (passenger vehicle airbag triggering circuits, for example). In predicting the response of simple RL and RC circuits subjected to pulses and series of pulses— sometimes referred to as sequentially switched circuits—the key is the rel-ative size of the circuit time constant to the various times that deﬁne the pulse sequence.The underlying principle behind the analysis will be whether the energy storage element has time to fully charge before the pulse ends, and whether it has time to fully discharge before the next pulse begins. Consider the circuit shown in Fig. 8.47a, which is connected to a pulsed voltage source described by seven separate parameters defined in Fig. 8.47b. The waveform is bounded by two values, V1 and V2. The time tr required to change from V1 to V2 is called the rise time (TR), and the time tf required to change from V2 to V1 is called the fall time (TF). The duration Wp of the pulse is referred to as the pulse width (PW), and the period T of the waveform (PER) is the time it takes for the pulse to repeat. Note also that SPICE allows a time delay (TD) before the pulse train CHAPTER 8 BASIC RL AND RC CIRCUITS 300 ■FIGURE 8.46 A simple RC circuit driven by a sinusoidal forcing function. 10 22 F 4.7 v + – 5 cos 3t u(t) A A little rearranging results in dv dt + 3.092 × 103v = 72.67 × 103 e−2000t which, upon comparison with Eqs. and , allows us to write the complete response as v(t) = e−Pt QePtdt + Ae−Pt where in our case P = 1/τ = 3.092 × 103 and Q(t) = 72.67 × 103e−2000t. We therefore ﬁnd that v(t) = e−3092t 72.67 × 103e−2000te3092tdt + Ae−3092t V Performing the indicated integration, v(t) = 66.55e−2000t + Ae−3092t V Our only source is controlled by a step function with zero value for t < 0, so we know that v(0−) = 0. Since v is a capacitor voltage, v(0+) = v(0−), and we therefore ﬁnd our initial condition v(0) = 0 eas-ily enough. Substituting this into Eq. , we ﬁnd A = −66.55 V and so v(t) = 66.55(e−2000t −e−3092t) V t > 0 PRACTICE ● 8.13 Determine the capacitor voltage v in the circuit of Fig. 8.46 for t > 0. Ans: 23.5 cos 3t + 22.8 × 10−3 sin 3t −23.5e−3092t V. SECTION 8.9 PREDICTING THE RESPONSE OF SEQUENTIALLY SWITCHED CIRCUITS 301 begins, which can be useful in allowing initial transient responses to decay for some circuit conﬁgurations. For the purposes of this discussion, we set a zero time delay, V1 = 0, and V2 = 9 V. The circuit time constant is τ = RC = 1 ms, so we set the rise and fall times to be 1 ns. Although SPICE will not allow a voltage to change in zero time since it solves the differential equations using discrete time intervals, compared to our circuit time constant 1 ns is a reasonable ap-proximation to “instantaneous.” We will consider four basic cases, summarized in Table 8.1. In the ﬁrst two cases, the pulse width Wp is much longer than the circuit time constant τ, so we expect the transients resulting from the beginning of the pulse to die out before the pulse is over. In the latter two cases, the opposite is true: the pulse width is so short that the capacitor does not have time to fully charge before the pulse ends. A similar issue arises when we consider the re-sponse of the circuit when the time between pulses (T −Wp) is either short (Case II) or long (Case III) compared to the circuit time constant. V2 V1 t TD PER PW TR TF (a) (b) ■FIGURE 8.47 (a) Schematic of a simple RC circuit connected to a pulsed voltage waveform. (b) Diagram of the SPICE VPULSE parameter deﬁnitions. TABLE ● 8.1 Four Separate Cases of Pulse Width and Period Relative to the Circuit Time Constant of 1 ms Case Pulse Width Wp Period T I 10 ms (τ ≪Wp) 20 ms (τ ≪T −Wp) II 10 ms (τ ≪Wp) 10.1 ms (τ ≫T −Wp) III 0.1 ms (τ ≫Wp) 10.1 ms (τ ≪T −Wp) IV 0.1 ms (τ ≫Wp) 0.2 ms (τ ≫T −Wp) We qualitatively sketch the circuit response for each of the four cases in Fig. 8.48, arbitrarily selecting the capacitor voltage as the quantity of inter-est as any voltage or current is expected to have the same time dependence. In Case I, the capacitor has time to both fully charge and fully discharge (Fig. 8.48a), whereas in Case II (Fig. 8.48b), when the time between pulses is reduced, it no longer has time to fully discharge. In contrast, the capaci-tor does not have time to fully charge in either Case III (Fig. 8.48c) or Case IV (Fig. 8.48d). Case I: Time Enough to Fully Charge and Fully Discharge We can obtain exact values for the response in each case, of course, by per-forming a series of analyses. We consider Case I ﬁrst. Since the capacitor has time to fully charge, the forced response will correspond to the 9 V dc driving voltage. The complete response to the ﬁrst pulse is therefore vC(t) = 9 + Ae−1000t V With vC(0) = 0, A = −9 V and so vC(t) = 9(1 −e−1000t) V in the interval of 0 < t < 10 ms. At t = 10 ms, the source drops suddenly to 0 V, and the capacitor begins to discharge through the resistor. In this time interval we are faced with a simple “source-free” RC circuit, and we can write the response as vC(t) = Be−1000(t−0.01) 10 < t < 20 ms where B = 8.99959 V is found by substituting t = 10 ms in Eq. ; we will be pragmatic here and round this to 9 V, noting that the value calculated is consistent with our assumption that the initial transient dissipates before the pulse ends. At t = 20 ms, the voltage source jumps immediately back to 9 V. The capacitor voltage just prior to this event is given by substituting t = 20 ms in Eq. , leading to vC(20 ms) = 408.6 μV, essentially zero compared to the peak value of 9 V. CHAPTER 8 BASIC RL AND RC CIRCUITS 302 (a) (b) (c) (d) ■FIGURE 8.48 Capacitor voltage for the RC circuit, with pulse width and period as in (a) Case I; (b) Case II; (c) Case III; and (d) Case IV. SECTION 8.9 PREDICTING THE RESPONSE OF SEQUENTIALLY SWITCHED CIRCUITS 303 If we keep to our convention of rounding to four signiﬁcant digits, the capacitor voltage at the beginning of the second pulse is zero, which is the same as our starting point. Thus, Eqs. and form the basis of the re-sponse for all subsequent pulses, and we may write vC(t) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 9(1 −e−1000t) V 0 ≤t ≤10 ms 9e−1000(t−0.01) V 10 < t ≤20 ms 9(1 −e−1000(t−0.02)) V 20 < t ≤30 ms 9e−1000(t−0.03) V 30 < t ≤40 ms and so on. Case II: Time Enough to Fully Charge But Not Fully Discharge Next we consider what happens if the capacitor is not allowed to completely discharge (Case II). Equation still describes the situation in the interval of 0 < t < 10 ms, and Eq. describes the capacitor voltage in the inter-val between pulses, which has been reduced to 10 < t < 10.1 ms. Just prior to the onset of the second pulse at t = 10.1 ms, vC is now 8.144 V; the capacitor has only had 0.1 ms to discharge, and therefore still retains 82 percent of its maximum energy when the next pulse begins. Thus, in the next interval, vC(t) = 9 + Ce−1000(t−10.1×10−3) V 10.1 < t < 20.1 ms where vC(10.1 ms) = 9 + C = 8.144 V, so C = −0.856 V and vC(t) = 9 −0.856e−1000(t−10.1×10−3) V 10.1 < t < 20.1 ms which reaches the peak value of 9 V much more quickly than for the previ-ous pulse. Case III: No Time to Fully Charge But Time to Fully Discharge What if it isn’t clear that the transient will dissipate before the end of the volt-age pulse? In fact, this situation arises in Case III. Just as we wrote for Case I, vC(t) = 9 + Ae−1000t V still applies to this situation, but now only in the interval 0 < t < 0.1 ms. Our initial condition has not changed, so A = −9 V as before. Now, how-ever, just before this first pulse ends at t = 0.1 ms, we find that vC = 0.8565 V. This is a far cry from the maximum of 9 V possible if we allow the capacitor time to fully charge, and is a direct result of the pulse lasting only one-tenth of the circuit time constant. The capacitor now begins to discharge, so that vC(t) = Be−1000(t−1×10−4) V 0.1 < t < 10.1 ms We have already determined that vC(0.1−ms) = 0.8565 V, so vC(0.1+ ms) = 0.8565 V and substitution into Eq. yields B = 0.8565 V. Just prior to the onset of the second pulse at t = 10.1 ms, the capacitor volt-age has decayed to essentially 0 V; this is the initial condition at the start of the second pulse and so Eq. can be rewritten as vC(t) = 9 −9e−1000(t−10.1×10−3) V 10.1 < t < 10.2 ms to describe the corresponding response. CHAPTER 8 BASIC RL AND RC CIRCUITS 304 (a) (b) (c) (d) ■FIGURE 8.49 PSpice simulation results corresponding to (a) Case I; (b) Case II; (c) Case III; (d) Case IV. Case IV: No Time to Fully Charge or Even Fully Discharge In the last case, we consider the situation where the pulse width and period are so short that the capacitor can neither fully charge nor fully discharge in any one period. Based on experience, we can write vC(t) = 9 −9e−1000t V 0 < t < 0.1 ms vC(t) = 0.8565e−1000(t−1×10−4) V 0.1 < t < 0.2 ms vC(t) = 9 + Ce−1000(t−2×10−4) V 0.2 < t < 0.3 ms vC(t) = De−1000(t−3×10−4) V 0.3 < t < 0.4 ms Just prior to the onset of the second pulse at t = 0.2 ms, the capacitor voltage has decayed to vC = 0.7750 V; with insufﬁcient time to fully discharge, it retains a large fraction of the little energy it had time to store initially. For the interval of 0.2 < t < 0.3 ms, substitution of vC(0.2+) = vC(0.2−) = 0.7750 V into Eq. yields C = −8.225 V. Continuing, we evaluate Eq. at t = 0.3 ms and calculate vC = 1.558 V just prior to the end of the second pulse. Thus, D = 1.558 V and our SECTION 8.9 PREDICTING THE RESPONSE OF SEQUENTIALLY SWITCHED CIRCUITS 305 capacitor is slowly charging to ever increase voltage levels over several pulses. At this stage it might be useful if we plot the detailed responses, so we show the PSpice simulation results of Cases I through IV in Fig. 8.49. Note in particular that in Fig. 8.49d, the small charge/discharge transient response similar in shape to that shown in Fig. 8.49a–c is superimposed on a charging-type response of the form (1 −e−t/τ). Thus, it takes about 3 to 5 circuit time constants for the capacitor to charge to its maximum value in situations where a single period does not allow it to fully charge or discharge! What we have not yet done is predict the behavior of the response for t ≫5τ, although we would be interested in doing so, especially if it was not necessary to consider a very long sequence of pulses one at a time. We note that the response of Fig. 8.49d has an average value of 4.50 V from about 4 ms onward. This is exactly half the value we would expect if the voltage source pulse width allowed the capacitor to fully charge. In fact, this long-term average value can be computed by multiplying the dc capacitor volt-age by the ratio of the pulse width to the period. PRACTICE ● 8.14 With regard to Fig. 8.50a, sketch iL(t) in the range of 0 < t < 6 s for (a) vS(t) = 3u(t) −3u(t −2) + 3u(t −4) −3u(t −6) + · · ·; (b) vS(t) = 3u(t) −3u(t −2) + 3u(t −2.1) −3u(t −4.1) + · · ·. Ans: See Fig. 8.50b; see Fig. 8.50c. ■FIGURE 8.50 (a) Circuit for Practice Problem 8.14; (b) solution to part (a); (c) solution to part (b). iL 100 mH vS (t) 1 (a) + – 0 0 2 4 1 2 3 4 5 6 iL (A) t (s) (b) 0 0 2 4 1 2 3 4 5 6 iL (A) t (s) (c) Modern digital integrated circuits such as programmable array logic (PALs) and microprocessors (Fig. 8.51) are composed of interconnected transistor circuits known as gates. Digital signals are represented symbolically by com-binations of ones and zeros, and can be either data or in-structions (such as “add” or “subtract”). Electrically, we represent a logic “1” by a “high” voltage, and a logic “0” by a “low” voltage. In practice, there is a range of voltages that correspond to each; for example, in the 7400 series of TTL logic integrated circuits, any voltage between 2 and 5 V will be interpreted as a logic “1,” and any voltage between 0 and 0.8 V will be interpreted as a logic “0.” Voltages between 0.8 and 2 V do not corre-spond to either logic state, as shown in Fig. 8.52. A key parameter in digital circuits is the speed at which we can effectively use them. In this sense, “speed” refers to how quickly we can switch a gate from one logic state to another (either logic “0” to logic “1” or vice PRACTICAL APPLICATION Frequency Limits in Digital Integrated Circuits 0 0 1 2 3 4 5 6 100 400 300 200 500 600 700 1000 900 800 vin (V) Time (s) Logic ‘0’ Logic ‘1’ ■FIGURE 8.52 Charge/discharge characteristic of a pathway capacitance identifying the TTL voltage ranges for logic “1” and logic “0,” respectively. ■FIGURE 8.51 A silicon wafer with multiple, identical integrated circuit dies. Each die is smaller than a US 1 cent coin. Reprinted with permission of Intel Corporation. SUMMARY AND REVIEW In this chapter we learned that circuits containing a single energy storage element (either an inductor or a capacitor) can be described by a character-istic time scale, namely, the circuit time constant (τ = L/R, or τ = RC, respectively). If we attempt to change the amount of energy stored in the element (either charging or discharging), every voltage and current in the cir-cuit will include an exponential term of the form e−t/τ. After approximately 5 time constants from the moment we attempted to alter the amount of stored energy, the transient response has essentially disappeared and we are left simply with a forced response which arises from the independent sources driving the circuit at times t > 0. When determining the forced response in a purely dc circuit, we may treat inductors as short circuits and capacitors as open circuits. versa), and the time delay required to convey the output of one gate to the input of the next gate. Although tran-sistors contain “built-in” capacitances that affect their switching speed, it is the interconnect pathways that presently limit the speed of the fastest digital integrated circuits. We can model the interconnect pathway between two logic gates using a simple RC circuit (although as feature sizes continue to decrease in modern designs, more detailed models are required to accurately predict circuit performance). For example, consider a 2000 μm long pathway 2 μm wide. We can model this pathway in a typical silicon-based integrated circuit as having a capacitance of 0.5 pF and a resistance of 100 , shown schematically in Fig. 8.53. Let’s assume the voltage vout represents the output voltage of a gate that is changing from a logic “0” state to a logic “1” state. The voltage vin appears across the in-put of a second gate, and we are interested in how long it takes vin to reach the same value as vout. Assuming the 0.5 pF capacitance that characterizes the interconnect pathway is initially discharged [i.e., vin(0) = 0], calculating the RC time constant for our pathway as τ = RC = 50 ps, and defining t = 0 as when vout changes, we obtain the expression vin(t) = Ae−t/τ + vout(0) Setting vin(0) = 0, we ﬁnd that A = −vout(0) so that vin(t) = vout(0)[1 −e−t/τ] Upon examining this equation, we see that vin will reach the value vout(0) after ∼5τ or 250 ps. If the volt-age vout changes again before this transient time period is over, then the capacitance does not have sufﬁcient time to fully charge. In such situations, vin will be less than vout(0). Assuming that vout(0) equals the minimum logic “1” voltage, for example, this means that vin will not correspond to a logic “1.” If vout now suddenly changes to 0 V (logic “0”), the capacitance will begin to discharge so that vin decreases further. Thus, by switch-ing our logic states too quickly, we are unable to trans-fer the information from one gate to another. The fastest speed at which we can change logic states is therefore (5τ)−1. This can be expressed in terms of the maximum operating frequency: fmax = 1 2(5τ) = 2 GHz where the factor of 2 represents a charge/discharge pe-riod. If we desire to operate our integrated circuit at a higher frequency so that calculations can be performed faster, we need to reduce the interconnect capacitance and/or the interconnect resistance. 100 0.5 pF vout + – vin + – ■FIGURE 8.53 Circuit model for an integrated circuit pathway. We started our analysis with so-called source-free circuits to introduce the idea of time constants without unnecessary distractions; such circuits have zero forced response and a transient response derived entirely from the energy stored at t = 0. We reasoned that a capacitor cannot change its volt-age in zero time (or an inﬁnite current results), and indicated this by intro-ducing the notation vC(0+) = vC(0−). Similarly, the current through an in-ductor cannot change in zero time, or iL(0+) = iL(0−). The complete response is always the sum of the transient response and the forced re-sponse. Applying the initial condition to the complete response allows us to determine the unknown constant which multiplies the transient term. We spent a little time discussing modeling switches, both analytically and within the context of PSpice. A common mathematical representation makes use of the unit-step function u(t −t0), which has zero value for t < t0, unity value for t > t0, and is indeterminate for t = t0. Unit-step functions can “activate” a circuit (connecting sources so current can ﬂow) for values of t preceding a speciﬁc time as well as after. Combinations of step functions can be used to create pulses and more complex waveforms. In the case of sequentially switched circuits, where sources are connected and dis-connected repeatedly, we found the behavior of the circuits to depend strongly on both period and pulse width as they compare to the circuit time constant. This is a good time to highlight some key points worth reviewing, along with relevant example(s). ❑The response of a circuit having sources suddenly switched in or out of a circuit containing capacitors and inductors will always be composed of two parts: a natural response and a forced response. ❑The form of the natural response (also referred to as the transient response) depends only on the component values and the way they are wired together. (Examples 8.1, 8.2) ❑A circuit reduced to a single equivalent capacitance C and a single equivalent resistance R will have a natural response given by v(t) = V0e−t/τ , where τ = RC is the circuit time constant. (Examples 8.3, 8.5) ❑A circuit reduced to a single equivalent inductance L and a single equivalent resistance R will have a natural response given by i(t) = I0e−t/τ , where τ = L/R is the circuit time constant. (Example 8.4) ❑Circuits with dependent sources can be represented by a resistance using Thévenin procedures. ❑The unit-step function is a useful way to model the closing or opening of a switch, provided we are careful to keep an eye on the initial conditions. (Examples 8.7, 8.9) ❑The form of the forced response mirrors the form of the forcing function. Therefore, a dc forcing function always leads to a constant forced response. (Examples 8.7, 8.8) ❑The complete response of an RL or RC circuit excited by a dc source will have the form f (0+) = f (∞) + A and f (t) = f (∞) + [ f (0+) −f (∞)]e−t/τ , or total response = ﬁnal value + (initial value −ﬁnal value)e−t/τ. (Examples 8.9, 8.10, 8.11) ❑The complete response for an RL or RC circuit may also be determined by writing a single differential equation for the quantity of interest and solving. (Examples 8.2, 8.11) ❑When dealing with sequentially switched circuits, or circuits connected to pulsed waveforms, the relevant issue is whether the energy storage element has sufﬁcient time to fully charge and to fully discharge, as measured relative to the circuit time constant. READING FURTHER A guide to solution techniques for differential equations can be found in: W. E. Boyce and R. C. DiPrima, Elementary Differential Equations and Boundary Value Problems, 7th ed. New York: Wiley, 2002. CHAPTER 8 BASIC RL AND RC CIRCUITS 308 EXERCISES 309 A detailed description of transients in electric circuits is given in: E. Weber, Linear Transient Analysis Volume I. New York: Wiley, 1954. (Out of print, but in many university libraries.) EXERCISES 8.1 The Source-Free RL Circuit 1. Setting R = 1 k and L = 1 nH for the circuit represented in Fig. 8.1, and with the knowledge that i(0) = −3 mA, (a) write an expression for i(t) valid for all t ≥0; (b) compute i(t) at t = 0, t = 1 ps, 2 ps, and 5 ps; and (c) calcu-late the energy stored in the inductor at t = 0, t = 1 ps, and t = 5 ps. 2. If i(0) = 1 A and R = 100 for the circuit of Fig. 8.1, (a) select L such that i(50 ms) = 368 mA; (b) compute the energy stored in the inductor at t = 0, 50 ms, 100 ms, and 150 ms. 3. Referring to the circuit shown in Fig. 8.1, select values for both elements such that L/R = 1 and (a) calculate vR(t) at t = 0, 1, 2, 3, 4, and 5 s; (b) compute the power dissipated in the resistor at t = 0, 1 s, and 5 s. (c) At t = 5 s, what is the percentage of the initial energy still stored in the inductor? 4. The circuit depicted in Fig. 8.1 is constructed from components whose value is unknown. If a current i(0) of 6 μA initially ﬂows through the inductor, and it is determined that i(1 ms) = 2.207 μA, calculate the ratio of R to L. 5. Determine the characteristic equation of each of the following differential equations: (a) 5v + 14 dv dt = 0; (b) −9 di dt −18i = 0; (c) di dt + 18i + R B i = 0; (d) d2f dt2 + 8 d f dt + 2 f = 0. 6. For the following characteristic equations, write corresponding differential equations and ﬁnd all roots, whether real, imaginary, or complex: (a) 4s + 9 = 0; (b) 2s −4 = 0; (c) s2 + 7s + 1 = 0; (d) 5s2 + 8s + 18 = 0. 7. With the assumption that the switch in the circuit of Fig. 8.54 has been closed a long, long, long time, calculate iL(t) at (a) the instant just before the switch opens; (b) the instant just after the switch opens; (c) t = 15.8 μs; (d) t = 31.5 μs; (e) t = 78.8 μs. 300 4 mA 2 mH t = 0 220 iL v + – ■FIGURE 8.54 8. The switch in Fig. 8.54 has been closed since Catﬁsh Hunter last pitched for the New York Yankees. Calculate the voltage labeled v as well as the energy stored in the inductor at (a) the instant just prior to the switch being thrown open; (b) the instant just after the switch is opened; (c) t = 8 μs; (d) t = 80 μs. 9. The switch in the circuit of Fig. 8.55 has been closed a ridiculously long time before suddenly being thrown open at t = 0. (a) Obtain expressions for iL and v in the circuit of Fig. 8.55 which are valid for all t ≥0. (b) Calculate iL(t) and v(t) at the instant just prior to the switch opening, at the instant just after the switch opening, and at t = 470 μs. CHAPTER 8 BASIC RL AND RC CIRCUITS 310 10 V – + v 40 mH 25 10 t = 0 iL + – 50 ■FIGURE 8.55 10. Assuming the switch initially has been open for a really, really long time, (a) obtain an expression for iW in the circuit of Fig. 8.56 which is valid for all t ≥0; (b) calculate iW at t = 0 and t = 1.3 ns. 1.5 V 20 H 5 k 10 k iL iW t = 0 + – ■FIGURE 8.56 8.2 Properties of the Exponential Response 11. (a) Graph the function f(t) = 10e−2t over the range of 0 ≤t ≤2.5 s using linear scales for both y and x axes. (b) Replot with a logarithmic scale for the y axis. [Hint: the function semilogy() can be helpful here.] (c) What are the units of the 2 in the argument of the exponential? (d) At what time does the function reach a value of 9? 8? 1? 12. The current i(t) ﬂowing through a 1 resistor is given by i(t) = 5e−10t mA, t ≥0. (a) Determine the values of t for which the resistor voltage magnitude is equal to 5 V, 2.5 V, 0.5 V, and 5 mV. (b) Graph the function over the range of 0 ≤t ≤1 s using linear scales for both axes. (c) Draw a tangent to your curve at t = 100 ms, and determine where the tangent intersects the time axis. 13. The thickness of a solar cell must be chosen carefully to ensure photons are properly absorbed; even metals can be partly transparent when rolled out into very thin foils. If the incident light ﬂux (number of photons per unit area per unit time) at the solar cell surface (x = 0) is given by 0, and the intensity of light a distance x inside the solar cell is given by (x), the behavior of (x) is described by the equation d/dx + α = 0. Here, α, known as the absorption coefﬁcient, is a constant speciﬁc to a given semiconductor material. (a) What is the SI unit for α? (b) Obtain an expression for (x) in terms of 0, α, and x. (c) How thick should the solar cell be made in order to absorb at least 38% of the incident light? Express your answer in terms of α. (d) What happens to the light which enters the solar cell at x = 0 but is not absorbed? 14. For the circuit of Fig. 8.5, compute the time constant if the 10 resistor is replaced with (a) a short circuit; (b) a 1 resistor; (c) a series connection of two 5 resistors; (d) a 100 resistor. (e) Verify your answers with a suitable parameter sweep simulation. (Hint: the cursor tool might come in handy, and the answer does not depend on the initial current you choose for the inductor.) 15. Design a circuit which will produce a voltage of 1 V at some initial time, and a voltage of 368 mV at a time 5 seconds later. You may specify an initial induc-tor current without showing how it arises. EXERCISES 311 8.3 The Source-Free RC Circuit 16. The resistor in the circuit of Fig. 8.57 has been included to model the dielectric layer separating the plates of the 3.1 nF capacitor, and has a value of 55 M. The capacitor is storing 200 mJ of energy just prior to t = 0. (a) Write an expression for v(t) valid for t ≥0. (b) Compute the energy remaining in the capacitor at t = 170 ms. (c) Graph v(t) over the range of 0 < t < 850 ms, and identify the value of v(t) when t = 2τ. 17. The resistor in the circuit of Fig. 8.57 has a value of 1 and is connected to a 22 mF capacitor. The capacitor dielectric has inﬁnite resistance, and the device is storing 891 mJ of energy just prior to t = 0. (a) Write an expression for v(t) valid for t ≥0. (b) Compute the energy remaining in the capacitor at t = 11 ms and 33 ms. (c) If it is determined that the capacitor dielectric is much leakier than expected, having a resistance as low as 100 k, repeat parts (a) and (b). 18. Calculate the time constant of the circuit depicted in Fig. 8.57 if C = 10 mF and R is equal to (a) 1 ; (b) 10 ; (c) 100 . (d) Verify your answers with an appropriate parameter sweep simulation. (Hint: the cursor tool might come in handy, and the time constant does not depend on the initial voltage across the capacitor.) 19. Design a capacitor-based circuit that will provide (a) a voltage of 9 V at some time t = 0, and a voltage of 1.2 V at a time 4 ms later; (b) a current of 1 mA at some time t = 0, and a reduced current of 50 μA at a time 100 ns later. (You can choose to design two separate circuits if desired, and do not need to show how the initial capacitor voltage is set.) 20. It is safe to assume that the switch drawn in the circuit of Fig. 8.58 has been closed such a long time that any transients which might have arisen from ﬁrst connecting the voltage source have disappeared. (a) Determine the circuit time constant. (b) Calculate the voltage v(t) at t = τ, 2τ, and 5τ. R C i v + – ■FIGURE 8.57 4 V – + v 2 nF 200 100 150 t = 0 + – ■FIGURE 8.58 8 V – + v 50 mF 12 9 t = 0 + – i1 ■FIGURE 8.59 10 k 5 k 1 k 3 k 20 k t = 0 + – 12 V 5 F v + – ■FIGURE 8.60 21. We can safely assume the switch in the circuit of Fig. 8.59 was closed a very long time prior to being thrown open at t = 0. (a) Determine the circuit time constant. (b) Obtain an expression for i1(t) which is valid for t > 0. (c) Deter-mine the power dissipated by the 12 resistor at t = 500 ms. 22. The switch above the 12 V source in the circuit of Fig. 8.60 has been closed since just after the wheel was invented. It is ﬁnally thrown open at t = 0. (a) Compute the circuit time constant. (b) Obtain an expression for v(t) valid for t > 0. (c) Calculate the energy stored in the capacitor 170 ms after the switch is opened. 23. For the circuit represented schematically in Fig. 8.61, (a) calculate v(t) at t = 0, t = 984 s, and t = 1236 s; (b) determine the energy still stored in the capacitor at t = 100 s. CHAPTER 8 BASIC RL AND RC CIRCUITS 312 v + – 12 mF t = 0 21 k 20 V 82 k + – ■FIGURE 8.61 t = 0 10 k 150 nF 2 mA 10 k v + – i ■FIGURE 8.62 i4 R4 R3 R1 L R2 iL v + – ■FIGURE 8.63 5 10 3 iL 6 nH t = 0 ix – + vL + – 4 V ■FIGURE 8.64 24. For the circuit depicted in Fig. 8.62, (a) compute the circuit time constant; (b) determine v in the instant just before the switch is closed; (c) obtain an expression for v(t) valid for t > 0; (d) calculate v(3 ms). 25. The switch drawn in Fig. 8.62 has been open a ponderously long time. (a) De-termine the value of the current labeled i just prior to the switch being closed. (b) Obtain the value of i just after the switch is closed. (c) Compute the power dissipated in each resistor over the range of 0 < t < 15 ms. (d) Graph your answer to part (c). 8.4 A More General Perspective 26. (a) Obtain an expression for v(t), the voltage which appears across resistor R3 in the circuit of Fig. 8.63, which is valid for t > 0. (b) If R1 = 2R2 = 3R3 = 4R4 = 1.2 k, L = 1 mH, and iL(0−) = 3 mA, calculate v(t = 500 ns). 27. For the circuit of Fig. 8.64, determine ix, iL, and vL at t equal to (a) 0−; (b) 0+. EXERCISES 313 28. The switch shown in Fig. 8.65 has been closed for 6 years prior to being ﬂipped open at t = 0. Determine iL, vL, and vR at t equal to (a) 0−; (b) 0+; (c) 1 μs; (d) 10 μs. + – 1.2 V 1 k vR + – t = 0 1 k 2 k 30 mH iL vL + – ■FIGURE 8.65 iL i1 5 A t = 0 3 2 H 1 H 3 H 2 8 ■FIGURE 8.66 9 mA 4 1 t = 0 5 mH i2 i1 iL ■FIGURE 8.67 + – 2 V 10 mH 3 1 5 vx + – t = 0 ■FIGURE 8.68 29. Obtain expressions for both i1(t) and iL(t) as labeled in Fig. 8.66, which are valid for t > 0. 30. The voltage across the resistor in a simple source-free RL circuit is given by 5e−90t V, t > 0. The inductor value is not known. (a) At what time will the inductor voltage be exactly one-half of its maximum value? (b) At what time will the inductor current reach 10% of its maximum value? 31. Referring to Fig. 8.67, calculate the currents i1 and i2 at t equal to (a) 1 ms; (b) 3 ms. 32. (a) Obtain an expression for vx as labeled in the circuit of Fig. 8.68. (b) Evalu-ate vx at t = 5 ms. (c) Verify your answer with an appropriate PSpice simula-tion. (Hint: employ the part named Sw_tClose.) 33. Design a complete circuit which provides a voltage vab across two terminals labeled a and b, respectively, such that vab = 5 V at t = 0−, 2 V at t = 1 s, and less than 60 mV at t = 5. Verify the operation of your circuit using an appropri-ate PSpice simulation. (Hint: employ the part named Sw_tOpen or Sw_tClose as appropriate.) 34. For the part Sw_tOpen, PSpice actually employs a sequence of simulations where the part is ﬁrst replaced with a resistor having value 1 M, and then replaced with a resistor having value 10 m corresponding to when the switch opens. (a) Evaluate the reliability of these default values by simulating the circuit of Fig. 8.55, and evaluating iL at t = 1 ns. (b) Repeat part (a) with RCLOSED changed to 1 . Did this change your answer? (c) Repeat part (a) with ROPEN changed to 100 k and RCLOSED reset to its default value. Did this change your answer? (Hint: double-click on the part to access its attributes.) 35. Select values for the resistors R0 and R1 in the circuit of Fig. 8.69 such that vC(0.65) = 5.22 V and vC(2.21) = 1 V. CHAPTER 8 BASIC RL AND RC CIRCUITS 314 R1 R0 10 mF 60 10 vC + – 12.5 V t = 2 s t = 0 + – ■FIGURE 8.69 + – 1 V 1 F 6 k 2 k 2 k 4 k 1 k 5 k vo + – vC + – t = 0 ■FIGURE 8.71 1.5i1 i1 vC 1 F 3 k 6 k 5 k + – t = 0 + – 10 V ■FIGURE 8.72 C vC + – v + – i1 10 8 20 ■FIGURE 8.70 100 V 5 F 20 F 5 k 20 k B A vR + – v2 + – v1 + – i(t) t = 0 ■FIGURE 8.73 36. A quick measurement determines that the capacitor voltage vC in the circuit of Fig. 8.70 is 2.5 V at t = 0−. (a) Determine vC(0+), i1(0+), and v(0+). (b) Select a value of C so that the circuit time constant is equal to 14 s. 37. Determine vC(t) and vo(t) as labeled in the circuit represented by Fig. 8.71 for t equal to (a) 0−; (b) 0+; (c) 10 ms; (d) 12 ms. 38. For the circuit shown in Fig. 8.72, determine (a) vC(0−); (b) vC(0+); (c) the circuit time constant; (d) vC(3 ms). 39. The switch in Fig. 8.73 is moved from A to B at t = 0 after being at A for a long time. This places the two capacitors in series, thus allowing equal and opposite dc voltages to be trapped on the capacitors. (a) Determine v1(0−), v2(0−), and vR(0−). (b) Find v1(0+), v2(0+), and vR(0+). (c) Determine the time constant of vR(t). (d) Find vR(t), t > 0. (e) Find i(t). ( f ) Find v1(t) and v2(t) from i(t) and the initial values. (g) Show that the stored energy at t = ∞ plus the total energy dissipated in the 20 k resistor is equal to the energy stored in the capacitors at t = 0. EXERCISES 315 40. The inductor in Fig. 8.74 is storing 54 nJ at t = 0−. Compute the energy remaining at t equal to (a) 0+; (b) 1 ms; (c) 5 ms. 8.5 The Unit-Step Function 41. Evaluate the following functions at t = −2, 0, and +2: (a) f(t) = 3u(t); (b) g(t) = 5u(−t) + 3; (c) h(t) = 5u(t −3); (d) z(t) = 7u(1 −t) + 4u(t + 3). 42. Evaluate the following functions at t = −1, 0, and +3: (a) f(t) = tu(1 −t); (b) g(t) = 8 + 2u(2 −t); (c) h(t) = u(t + 1) −u(t −1) + u(t + 2) −u(t −4); (d) z(t) = 1 + u(3 −t) + u(t −2). 43. Sketch the following functions over the range −3 ≤t ≤3: (a) v(t) = 3 −u(2 −t) −2u(t) V; (b) i(t) = u(t) −u(t −0.5) + u(t −1) −u(t −1.5) + u(t −2) −u(t −2.5) A; (c) q(t) = 8u(−t) C. 44. Use step functions to construct an equation that describes the waveform sketched in Fig. 8.75. 48 mH 10 40 iL iL 5 ■FIGURE 8.74 0 1 2 3 t –1 –2 f (t) ■FIGURE 8.75 1 2 3 2 –4 4 4 5 t (s) v(t) V –2 ■FIGURE 8.76 + – i(t) 1 H 3 k 9u(t – 1) V ■FIGURE 8.77 45. Employing step functions as appropriate, describe the voltage waveform graphed in Fig. 8.76. 8.6 Driven RL Circuits 46. With reference to the simple circuit depicted in Fig. 8.77, compute i(t) for (a) t = 0−; (b) t = 0+; (c) t = 1−; (d) t = 1+; (e) t = 2 ms. 47. For the circuit given in Fig. 8.78, (a) determine vL(0−), vL(0+), iL(0−), and iL(0+); (b) calculate iL(150 ns). (c) Verify your answer to part (b) with an appropriate PSpice simulation. CHAPTER 8 BASIC RL AND RC CIRCUITS 316 3 mH 100 25 vL + – iL 2u(t) mA ■FIGURE 8.78 1.2 V 1 k 1 k 2 k 10 mH 50u(t) mA + – iL ■FIGURE 8.79 45 mH 20 i(t) + – 2u(t) V ■FIGURE 8.80 30 5 5 H vR + – 12u(t) V + – i(t) ■FIGURE 8.81 50 mH 20 60 4.5 V iL t = 0 4.5 V + – + – ■FIGURE 8.83 5 H 100 400 5 V + – 6u(t) V + – i(t) ■FIGURE 8.82 48. The circuit depicted in Fig. 8.79 contains two independent sources, one of which is only active for t > 0. (a) Obtain an expression for iL(t) valid for all t; (b) calculate iL(t) at t = 10 μs, 20 μs, and 50 μs. 49. The circuit shown in Fig. 8.80 is powered by a source which is inactive for t < 0. (a) Obtain an expression for i(t) valid for all t. (b) Graph your answer over the range of −1 ms ≤t ≤10 ms. 50. For the circuit shown in Fig. 8.81, (a) obtain an expression for i(t) valid for all time; (b) obtain an expression for vR(t) valid for all time; and (c) graph both i(t) and vR(t) over the range of −1 s ≤t ≤6 s. 8.7 Natural and Forced Response 51. For the two-source circuit of Fig. 8.82, note that one source is always on. (a) Obtain an expression for i(t) valid for all t; (b) determine at what time the energy stored in the inductor reaches 99% of its maximum value. 52. (a) Obtain an expression for iL as labeled in Fig. 8.83 which is valid for all val-ues of t. (b) Sketch your result over the range −1 ms ≤t ≤3 ms. EXERCISES 317 53. Obtain an expression for i(t) as labeled in the circuit diagram of Fig. 8.84, and determine the power being dissipated in the 40 resistor at t = 2.5 ms. 40 30 m 30 i(t) t = 0 100 mA 200 mA ■FIGURE 8.84 + – + – 50 nH 5 5 3i1 i1 2u(t) V ■FIGURE 8.85 1 nF 2 k 1 k 3u(t) V vC + – + – ■FIGURE 8.87 300 nF 10 V 1 k 3 k iA t = 0 + – ■FIGURE 8.88 + – + – 9u(t) V –9u(t – 1) V R i(t) 4 H ■FIGURE 8.86 2 mF 1 mA 10 20 15 30 ix t = 0 ■FIGURE 8.89 54. Obtain an expression for i1 as indicated in Fig. 8.85 that is valid for all values of t. 55. Plot the current i(t) in Fig. 8.86 if (a) R = 10 ; (b) R = 1 . In which case does the inductor (temporarily) store the most energy? Explain. 8.8 Driven RC Circuits 56. (a) Obtain an expression for vC in the circuit of Fig. 8.87 valid for all values of t. (b) Sketch vC(t) over the range 0 ≤t ≤4 μs. 57. Obtain an equation which describes the behavior of iA as labeled in Fig. 8.88 over the range of −1 ms ≤t ≤5 ms. 58. The switch in the circuit of Fig. 8.89 has been closed an incredibly long time, before being thrown open at t = 0. (a) Evaluate the current labeled ix at t = 70 ms. (b) Verify your answer with an appropriate PSpice simulation. 59. The switch in the circuit of Fig. 8.89 has been open a really, really incredibly long time, before being closed without further fanfare at t = 0. (a) Evaluate the current labeled ix at t = 70 ms. (b) Verify your answer with an appropriate PSpice simulation. 60. The “make-before-break” switch shown in Fig. 8.90 has been in position a since the ﬁrst episode of “Jonny Quest” aired on television. It is moved to position b, ﬁnally, at time t = 0. (a) Obtain expressions for i(t) and vC(t) valid for all values of t. (b) Determine the energy remaining in the capacitor at t = 33 μs. CHAPTER 8 BASIC RL AND RC CIRCUITS 318 a b 5 k 20 k 10 50 10 V 2 F vC (t) + – i(t) t = 0 + – 6 V + – ■FIGURE 8.90 a b 5 3 1 1 2 10 mA 1 mF vC (t) + – i(t) t = 0 4 V + – ■FIGURE 8.91 20 F 3 V 10 5 0.5vx vC + – vx + – t = 0 + – ■FIGURE 8.92 61. The switch in the circuit of Fig. 8.91, often called a make-before-break switch (since during switching it brieﬂy makes contact to both parts of the circuit to ensure a smooth electrical transition), moves to position b at t = 0 only after being in position a long enough to ensure all initial transients arising from turning on the sources have long since decayed. (a) Determine the power dissipated by the 5 resistor at t = 0−. (b) Determine the power dissipated in the 3 resistor at t = 2 ms. 62. Referring to the circuit represented in Fig. 8.92, (a) obtain an equation which describes vC valid for all values of t; (b) determine the energy remaining in the capacitor at t = 0+, t = 25 μs, and t = 150 μs. EXERCISES 319 63. The dependent source shown in Fig. 8.92 is unfortunately installed upside down during manufacturing, so that the terminal corresponding to the arrowhead is actually wired to the negative reference terminal of the voltage source. This is not detected by the quality assurance team so the unit ships out wired improperly. The capacitor is initially discharged. If the 5 resistor is only rated to 2 W, after what time t is the circuit likely to fail? 64. For the circuit represented in Fig. 8.93, (a) obtain an expression for v which is valid for all values of t; (b) sketch your result for 0 ≤t ≤3 s. 1 1 F 1 v + – 12e–2t u(t) V + – ■FIGURE 8.93 – + 8 mF 50 2 vx + – 9u(t) V + – ■FIGURE 8.94 R C vS + – vC + – ■FIGURE 8.95 65. Obtain an expression for the voltage vx as labeled in the op amp circuit of Fig. 8.94. 8.9 Predicting the Response of Sequentially Switched Circuits 66. Sketch the current iL of the circuit in Fig. 8.50a if the 100 mH inductor is replaced by a 1 nH inductor, and is subjected to the waveform vs(t) equal to (a) 5u(t) −5u(t −10−9) + 5u(t −2 × 10−9) V, 0 ≤t ≤4 ns; (b) 9u(t) −5u(t −10−8) + 5u(t −2 × 10−8) V, 0 ≤t ≤40 ns. 67. The 100 mH inductor in the circuit of Fig. 8.50a is replaced with a 1 H inductor. Sketch the inductor current iL if the source vs(t) is equal to (a) 5u(t) −5u(t −0.01) + 5u(t −0.02) V, 0 ≤t ≤40 ms; (b) 5u(t) −5u(t −10) + 5u(t −10.1) V, 0 ≤t ≤11 s. 68. Sketch the voltage vC across the capacitor of Fig. 8.95 for at least 3 periods if R = 1 , C = 1 F, and vs(t) is a pulsed waveform having (a) minimum of 0 V, maximum of 2 V, rise and fall times of 1 ms, pulse width of 10 s, and period of 10 s; (b) minimum of 0 V, maximum of 2 V, rise and fall times of 1 ms, pulse width of 10 ms, and period of 10 ms. (c) Verify your answers with appropriate PSpice simulations. 69. Sketch the voltage vC across the capacitor of Fig. 8.95 for at least 3 periods if R = 1 , C = 1 F, and vs(t) is a pulsed waveform having (a) minimum of 0 V, maximum of 2 V, rise and fall times of 1 ms, pulse width of 10 s, and period of 10 ms; (b) minimum of 0 V, maximum of 2 V, rise and fall times of 1 ms, pulse width of 10 ms, and period of 10 s. (c) Verify your answers with appropriate PSpice simulations. Chapter-Integrating Exercises 70. The circuit in Fig. 8.96 contains two switches that always move in perfect syn-chronization. However, when switch A opens, switch B closes, and vice versa. Switch A is initially open, while switch B is initially closed; they change positions every 40 ms. Using the bottom node as the reference node, determine the voltage across the capacitor at t equal to (a) 0−; (b) 0+; (c) 40−ms; (d) 40+ ms; (e) 50 ms. CHAPTER 8 BASIC RL AND RC CIRCUITS 320 10 7 3 A 9 A 10 mF A B ■FIGURE 8.96 – + vo + – vC + – + – vs 300 nF 10 15 ■FIGURE 8.98 + – 3 mH 10 4 0.1vx vx + – 2u(t) mA ■FIGURE 8.97 71. In the circuit of Fig. 8.96, when switch A opens, switch B closes, and vice versa. Switch A is initially open, while switch B is initially closed; they change positions every 400 ms. Determine the energy in the capacitor at t equal to (a) 0−; (b) 0+; (c) 200 ms; (d) 400−ms; (e) 400+ ms; (f ) 700 ms. 72. Refer to the circuit of Fig. 8.97, which contains a voltage-controlled dependent voltage source in addition to two resistors. (a) Compute the circuit time constant. (b) Obtain an expression for vx valid for all t. (c) Plot the power dissipated in the resistor over the range of 6 time constants. (d) Repeat parts (a) to (c) if the dependent source is installed in the circuit upside down. (e) Are both circuit conﬁgurations “stable”? Explain. 73. In the circuit of Fig. 8.97, a 3 mF capacitor is accidentally installed instead of the inductor. Unfortunately, that’s not the end of the problems, as it’s later determined that the real capacitor is not really well modeled by an ideal capaci-tor, and the dielectric has a resistance of 10 k (which should be viewed as connected in parallel to the ideal capacitor). (a) Compute the circuit time con-stant with and without taking the dielectric resistance into account. By how much does the dielectric change your answer? (b) Calculate vx at t = 200 ms. Does the dielectric resistance affect your answer signiﬁcantly? Explain. 74. For the circuit of Fig. 8.98, assuming an ideal op amp, derive an expression for vo(t) if vs is equal to (a) 4u(t) V; (b) 4e−130,000tu(t) V. INTRODUCTION In Chap. 8 we studied circuits which contained only one energy storage element, combined with a passive network which partly determined how long it took either the capacitor or the inductor to charge/discharge. The differential equations which resulted from analysis were always ﬁrst-order. In this chapter, we consider more complex circuits which contain both an inductor and a capacitor. The result is a second-order differential equation for any voltage or current of interest. What we learned in Chap. 8 is easily extended to the study of these so-called RLC circuits, although now we need two initial conditions to solve each differential equation. Such circuits occur routinely in a wide variety of applications, including oscillators and frequency ﬁlters. They are also very useful in mod-eling a number of practical situations, such as automobile suspen-sion systems, temperature controllers, and even the response of an airplane to changes in elevator and aileron positions. 9.1• THE SOURCE-FREE PARALLEL CIRCUIT There are two basic types of RLC circuits: parallel connected, and series connected. We could start with either, but somewhat arbitrar-ily choose to begin by considering parallel RLC circuits. This particular combination of ideal elements is a reasonable model for portions of many communication networks. It represents, for exam-ple, an important part of certain electronic amplifiers found in radios, and enables the ampliﬁers to produce a large voltage ampli-ﬁcation over a narrow band of signal frequencies (with almost zero ampliﬁcation outside this band). Just as we did with RL and RC circuits, we ﬁrst consider the nat-ural response of a parallel RLC circuit, where one or both of the KEY CONCEPTS Resonant Frequency and Damping Factor of Series and Parallel RLC Circuits Overdamped Response Critically Damped Response Underdamped Response Making Use of Two Initial Conditions Complete (Natural + Forced) Response of RLC Circuits Representing Differential Equations Using Op Amp Circuits The RLC Circuit C H A P T E R 9 321 energy storage elements have some nonzero initial energy (the origin of which for now is unimportant). This is represented by the inductor current and the capacitor voltage, both speciﬁed at t = 0+. Once we’re comfortable with this part of RLC circuit analysis, we can easily include dc sources, switches, or step sources in the circuit. Then we ﬁnd the total response, which will be the sum of the natural response and the forced response. Frequency selectivity of this kind enables us to listen to the transmission of one station while rejecting the transmission of any other station. Other ap-plications include the use of parallel RLC circuits in frequency multiplexing andharmonic-suppressionﬁlters.However,evenasimplediscussionofthese principles requires an understanding of such terms as resonance, frequency response, and impedance, which we have not yet discussed. Let it sufﬁce to say, therefore, that an understanding of the natural behavior of the parallel RLC circuit is fundamentally important to future studies of communications networks and ﬁlter design, as well as many other applications. When a physical capacitor is connected in parallel with an inductor and the capacitor has associated with it a ﬁnite resistance, the resulting network can be shown to have an equivalent circuit model like that shown in Fig. 9.1. The presence of this resistance can be used to model energy loss in the capacitor; over time, all real capacitors will eventually discharge, even if dis-connected from a circuit. Energy losses in the physical inductor can also be taken into account by adding an ideal resistor (in series with the ideal inductor). For simplicity, however, we restrict our discussion to the case of an essentially ideal inductor in parallel with a “leaky” capacitor. Obtaining the Differential Equation for a Parallel RLC Circuit In the following analysis we will assume that energy may be stored initially in both the inductor and the capacitor; in other words, nonzero initial values of both inductor current and capacitor voltage may be present.With reference to the circuit of Fig. 9.1, we may then write the single nodal equation v R + 1 L t t0 v dt′ −i(t0) + C dv dt = 0 Note that the minus sign is a consequence of the assumed direction for i. We must solve Eq. subject to the initial conditions i(0+) = I0 and v(0+) = V0 When both sides of Eq. are differentiated once with respect to time, the re-sult is the linear second-order homogeneous differential equation C d2v dt2 + 1 R dv dt + 1 L v = 0 whose solution v(t) is the desired natural response. Solution of the Differential Equation ThereareanumberofinterestingwaystosolveEq..Mostofthesemethods we will leave to a course in differential equations, selecting only the quickest and simplest method to use now. We will assume a solution, relying upon our CHAPTER 9 THE RLC CIRCUIT 322 ■FIGURE 9.1 The source-free parallel RLC circuit. R L C v Ref. i SECTION 9.1 THE SOURCE-FREE PARALLEL CIRCUIT 323 intuitionandmodestexperiencetoselectoneoftheseveralpossibleformsthat are suitable. Our experience with ﬁrst-order equations might suggest that we at least try the exponential form once more. Thus, we assume v = Aest being as general as possible by allowing A and s to be complex numbers if necessary. Substituting Eq. in Eq. , we obtain CAs2est + 1 R Asest + 1 L Aest = 0 or Aest Cs2 + 1 R s + 1 L = 0 In order for this equation to be satisﬁed for all time, at least one of the three factors must be zero. If either of the ﬁrst two factors is set equal to zero, then v(t) = 0. This is a trivial solution of the differential equation which cannot satisfy our given initial conditions.We therefore equate the remaining factor to zero: Cs2 + 1 R s + 1 L = 0 This equation is usually called the auxiliary equation or the characteristic equation, as we discussed in Sec. 8.1. If it can be satisﬁed, then our assumed solution is correct. Since Eq. is a quadratic equation, there are two solu-tions, identiﬁed as s1 and s2: s1 = − 1 2RC + 1 2RC 2 − 1 LC and s2 = − 1 2RC − 1 2RC 2 − 1 LC If either of these two values is used for s in the assumed solution, then that solution satisﬁes the given differential equation; it thus becomes a valid solution of the differential equation. Let us assume that we replace s by s1 in Eq. , obtaining v1 = A1es1t and, similarly, v2 = A2es2t The former satisﬁes the differential equation C d2v1 dt2 + 1 R dv1 dt + 1 L v1 = 0 and the latter satisﬁes C d2v2 dt2 + 1 R dv2 dt + 1 L v2 = 0 Addingthesetwodifferentialequationsandcombiningsimilarterms,wehave C d2(v1 + v2) dt2 + 1 R d(v1 + v2) dt + 1 L (v1 + v2) = 0 Linearity triumphs, and it is seen that the sum of the two solutions is also a solution. We thus have the general form of the natural response v(t) = A1es1t + A2es2t where s1 and s2 are given by Eqs. and ; A1 and A2 are two arbi-trary constants which are to be selected to satisfy the two speciﬁed initial conditions. Deﬁnition of Frequency Terms The form of the natural response as given in Eq. offers little insight into the nature of the curve we might obtain if v(t) were plotted as a function of time. The relative amplitudes of A1 and A2, for example, will certainly be important in determining the shape of the response curve. Furthermore, the constants s1 and s2 can be real numbers or conjugate complex numbers, depending upon the values of R, L, and C in the given network. These two cases will produce fundamentally different response forms. Therefore, it will be helpful to make some simplifying substitutions in Eq. . Since the exponents s1t and s2t must be dimensionless, s1 and s2 must have the unit of some dimensionless quantity “per second.” From Eqs. and we therefore see that the units of 1/2RC and 1/ √ LC must also be s−1 (i.e., seconds−1). Units of this type are called frequencies. Let us deﬁne a new term, ω0 (omega-sub-zero, or just omega-zero): ω0 = 1 √ LC and reserve the term resonant frequency for it. On the other hand, we will call 1/2RC the neper frequency, or the exponential damping coefﬁcient, and represent it by the symbol α (alpha): α = 1 2RC This latter descriptive expression is used because α is a measure of how rapidly the natural response decays or damps out to its steady, ﬁnal value (usually zero). Finally, s, s1, and s2, which are quantities that will form the ba-sis for some of our later work, are called complex frequencies. We should note that s1, s2, α, and ω0 are merely symbols used to simplify the discussion of RLC circuits; they are not mysterious new properties of any kind. It is easier, for example, to say “alpha” than it is to say “the reciprocal of 2RC.” Let us collect these results. The natural response of the parallel RLC circuit is v(t) = A1es1t + A2es2t where s1 = −α + α2 −ω2 0 s2 = −α − α2 −ω2 0 CHAPTER 9 THE RLC CIRCUIT 324 SECTION 9.1 THE SOURCE-FREE PARALLEL CIRCUIT 325 α = 1 2RC ω0 = 1 √ LC and A1 and A2 must be found by applying the given initial conditions. We note two basic scenarios possible with Eqs. and depending on the relative sizes of α and ω0 (dictated by the values of R, L, and C). If α > ω0, s1 and s2 will both be real numbers, leading to what is referred to as an overdamped response. In the opposite case, where α < ω0, both s1 and s2 will have nonzero imaginary components, leading to what is known as an underdamped response. Both of these situations are considered separately in the following sections, along with the special case of α = ω0, which leads to what is called a critically damped response.We should also note that the gen-eral response comprised by Eqs. through describes not only the volt-age but all three branch currents in the parallel RLC circuit; the constants A1 and A2 will be different for each, of course. EXAMPLE 9.1 Consider a parallel RLC circuit having an inductance of 10 mH and a capacitance of 100 μF. Determine the resistor values that would lead to overdamped and underdamped responses. We ﬁrst calculate the resonant frequency of the circuit: ω0 = 1 LC = 1 (10 × 10−3)(100 × 10−6) = 103 rad/s An overdamped response will result if α > ω0; an underdamped response will result if α < ω0. Thus, 1 2RC > 103 and so R < 1 (2000)(100 × 10−6) or R < 5 leads to an overdamped response; R > 5 leads to an underdamped response. PRACTICE ● 9.1 A parallel RLC circuit contains a 100 resistor and has the parameter values α = 1000 s−1 and ω0 = 800 rad/s. Find (a) C; (b) L; (c) s1; (d) s2. Ans: 5 μF; 312.5 mH; −400 s−1; −1600 s−1. The ratio of α to ω0 is called the damping ratio by control system engineers and is designated by ζ (zeta). Overdamped: α > ω0 Critically damped: α = ω0 Underdamped: α < ω0 9.2• THE OVERDAMPED PARALLEL RLC CIRCUIT A comparison of Eqs. and shows that α will be greater than ω0 if LC > 4R2C2. In this case the radical used in calculating s1 and s2 will be real, and both s1 and s2 will be real. Moreover, the following inequalities α2 −ω2 0 < α −α − α2 −ω2 0 < −α + α2 −ω2 0 < 0 may be applied to Eqs. and to show that both s1 and s2 are negative real numbers.Thus, the response v(t)can be expressed as the (algebraic) sum of two decreasing exponential terms, both of which approach zero as time in-creases. In fact, since the absolute value of s2 is larger than that of s1, the term containings2 hasthemorerapidrateofdecrease,and,forlargevaluesoftime, we may write the limiting expression v(t) →A1es1t →0 as t →∞ The next step is to determine the arbitrary constants A1 and A2 in confor-mance with the initial conditions. We select a parallel RLC circuit with R = 6 , L = 7 H, and, for ease of computation, C = 1 42 F. The initial energy storage is speciﬁed by choosing an initial voltage across the circuit v(0) = 0 and an initial inductor currenti(0) = 10 A, where v and iare deﬁned in Fig. 9.2. We may easily determine the values of the several parameters α = 3.5 s1 = −1 ω0 = √ 6 s2 = −6 (all s−1) and immediately write the general form of the natural response v(t) = A1e−t + A2e−6t Finding Values for A1 and A2 Only the evaluation of the two constants A1 and A2 remains. If we knew the response v(t) at two different values of time, these two values could be sub-stituted in Eq. and A1 and A2 easily found. However, we know only one instantaneous value of v(t), v(0) = 0 and, therefore, 0 = A1 + A2 We can obtain a second equation relating A1 and A2 by taking the deriva-tive of v(t) with respect to time in Eq. , determining the initial value of this derivative through the use of the remaining initial condition i(0) = 10, and equating the results. So, taking the derivative of both sides of Eq. , dv dt = −A1e−t −6A2e−6t and evaluating the derivative at t = 0, dv dt t=0 = −A1 −6A2 CHAPTER 9 THE RLC CIRCUIT 326 ■FIGURE 9.2 A parallel RLC circuit used as a numer-ical example. The circuit is overdamped. v i iC iR 7 H F 1 42 6 SECTION 9.2 THE OVERDAMPED PARALLEL RLC CIRCUIT 327 we obtain a second equation. Although this may appear to be helpful, we do not have a numerical value for the initial value of the derivative, so we do not yet have two equations in two unknowns . . . Or do we? The expression dv/dt suggests a capacitor current, since iC = C dv dt Kirchhoff’s current law must hold at any instant in time, as it is based on conservation of electrons. Thus, we may write −iC(0) + i(0) + iR(0) = 0 Substituting our expression for capacitor current and dividing by C, dv dt t=0 = iC(0) C = i(0) + iR(0) C = i(0) C = 420 V/s since zero initial voltage across the resistor requires zero initial current through it. We thus have our second equation, 420 = −A1 −6A2 and simultaneous solution of Eqs. and provides the two amplitudes A1 = 84 and A2 = −84. Therefore, the ﬁnal numerical solution for the nat-ural response of this circuit is v(t) = 84(e−t −e−6t) V For the remainder of our discussions concerning RLC circuits, we will always require two initial conditions in order to completely specify the response. One condition will usually be very easy to apply—either a voltage or current at t = 0. It is the second condition that usually requires a little effort. Although we will often have both an initial current and an initial voltage at our disposal, one of these will need to be applied indirectly through the derivative of our assumed solution. Find an expression for vC(t) valid for t > 0 in the circuit of Fig. 9.3a. EXAMPLE 9.2 (Continued on next page) 200 300 20 nF 5 mH 150 V vC + – iL iR iC iC t = 0 200 20 nF 5 mH iL iR iC (a) (b) + – ■FIGURE 9.3 (a) An RLC circuit that becomes source-free at t = 0. (b) The circuit for t > 0, in which the 150 V source and the 300 resistor have been shorted out by the switch, and so are of no further relevance to vC. Identify the goal of the problem. We are asked to ﬁnd the capacitor voltage after the switch is thrown. This action leads to no sources remaining connected to either the inductor or the capacitor. CHAPTER 9 THE RLC CIRCUIT 328 Collect the known information. After the switch is thrown, the capacitor is left in parallel with a 200 resistor and a 5 mH inductor (Fig. 9.3b). Thus, α = 1/2RC = 125,000 s−1, ω0 = 1/ √ LC = 100,000 rad/s, s1 = −α + α2 −ω2 0 = −50,000 s−1 and s2 = −α − α2 −ω2 0 = −200,000 s−1. Devise a plan. Since α > ω0, the circuit is overdamped and so we expect a capacitor voltage of the form vC(t) = A1es1t + A2es2t We know s1 and s2; we need to obtain and invoke two initial condi-tions to determine A1 and A2. To do this, we will analyze the circuit at t = 0−(Fig. 9.4a) to ﬁnd iL(0−) and vC(0−). We will then analyze the circuit at t = 0+ with the assumption that neither value changes. Construct an appropriate set of equations. From Fig. 9.4a, in which the inductor has been replaced with a short circuit and the capacitor with an open circuit, we see that iL(0−) = − 150 200 + 300 = −300 mA and vC(0−) = 150 200 200 + 300 = 60 V ■FIGURE 9.4 (a) The equivalent circuit at t = 0−; (b) equivalent circuit at t = 0+, drawn using ideal sources to represent the initial inductor current and initial capacitor voltage. 200 300 150 V iL(0–) (a) vC(0–) + – 200 iR(0+) iC(0+) vC(0+) = vC(0–) = 60 V iL(0+) = iL(0–) = –0.3 A (b) + – + – SECTION 9.2 THE OVERDAMPED PARALLEL RLC CIRCUIT 329 ■FIGURE 9.5 10 H 24 48 vC + – iC iR iL t = 0 3u(–t) A F 1 240 Ans: 1 A; 48 V; 2 A; −3 A; −17.54 V. In Fig. 9.4b, we draw the circuit at t = 0+, representing the inductor current and capacitor voltage by ideal sources for simplicity. Since neither can change in zero time, we know that vC(0+) = 60 V. Determine if additional information is required. We have an equation for the capacitor voltage: vC(t) = A1e−50,000t + A2e−200,000t. We now know vC(0) = 60 V, but a third equation is still required. Differentiating our capacitor voltage equation, we ﬁnd dvC dt = −50,000A1e−50,000t −200,000A2e−200,000t which can be related to the capacitor current as iC = C(dvC/dt). Returning to Fig. 9.4b, KCL yields iC(0+) = −iL(0+) −iR(0+) = 0.3 −[vC(0+)/200] = 0 Attempt a solution. Application of our ﬁrst initial condition yields vC(0) = A1 + A2 = 60 and application of our second initial condition yields iC(0) = −20 × 10−9(50,000A1 + 200,000A2) = 0 Solving, A1 = 80 V and A2 = −20 V, so that vC(t) = 80e−50,000t −20e−200,000t V, t > 0 Verify the solution. Is it reasonable or expected? At the very least, we can check our solution at t = 0, verifying that vC(0) = 60 V. Differentiating and multiplying by 20 × 10−9, we can also verify that iC(0) = 0. Also, since we have a source-free circuit for t > 0, we expect that vC(t) must eventually decay to zero as t approaches ∞, which our solution does. PRACTICE ● 9.2 After being open for a long time, the switch in Fig. 9.5 closes at t = 0. Find (a) iL(0−); (b) vC(0−); (c) iR(0+); (d) iC(0+); (e) vC(0.2). As noted previously, the form of the overdamped response applies to any voltage or current quantity, as we explore in the following example. CHAPTER 9 THE RLC CIRCUIT 330 The circuit of Fig. 9.6a reduces to a simple parallel RLC circuit after t = 0. Determine an expression for the resistor current iR valid for all time. EXAMPLE 9.3 ■FIGURE 9.6 (a) Circuit for which iR is required. (b) Equivalent circuit for t = 0−. (c) Equivalent circuit for t = 0+. (c) 30 k iR(0+) iC(0+) vC(0+) = 3.75 V iL(0+) = 125 A + – (b) 2 k 4 V vC(0–) + – iR(0–) iL(0–) + – 30 k (a) 2 pF iR 4 V 2 k 12 mH 30 k t = 0 + – For t > 0, we have a parallel RLC circuit with R = 30 k, L = 12 mH, and C = 2 pF. Thus, α = 8.333 × 106 s−1 and ω0 = 6.455 × 106 rad/s. We therefore expect an overdamped response, with s1 = −3.063 × 106 s−1 and s2 = −13.60 × 106 s−1, so that iR(t) = A1es1t + A2es2t, t > 0 To determine numerical values for A1 and A2, we ﬁrst analyze the circuit at t = 0−, as drawn in Fig. 9.6b. We see that iL(0−) = iR(0−) = 4/32 × 103 = 125 μA, and vC(0−) = 4 × 30/32 = 3.75 V. In drawing the circuit at t = 0+ (Fig. 9.6c), we only know that iL(0+) = 125 μA and vC(0+) = 3.75 V. However, by Ohm’s law we can calculate that iR(0+) = 3.75/30 × 103 = 125 μA, our ﬁrst initial condition. Thus, iR(0) = A1 + A2 = 125 × 10−6 SECTION 9.2 THE OVERDAMPED PARALLEL RLC CIRCUIT 331 Graphical Representation of the Overdamped Response Now let us return to Eq. and see what additional information we can de-termine about this circuit. We may interpret the ﬁrst exponential term as hav-ing a time constant of 1 s and the other exponential, a time constant of 1 6 s. Each starts with unity amplitude, but the latter decays more rapidly; v(t) is never negative.As time becomes inﬁnite, each term approaches zero, and the response itself dies out as it should. We therefore have a response curve which is zero at t = 0, is zero at t = ∞, and is never negative; since it is not everywhere zero, it must possess at least one maximum, and this is not a difﬁcult point to determine exactly. We differentiate the response dv dt = 84(−e−t + 6e−6t) set the derivative equal to zero to determine the time tm at which the voltage becomes maximum, 0 = −e−tm + 6e−6tm manipulate once, e5tm = 6 and obtain tm = 0.358 s ■FIGURE 9.7 Circuit for Practice Problem 9.3. 625 pH iR iL 3 4 pF How do we obtain a second initial condition? If we multiply Eq. by 30 × 103, we obtain an expression for vC(t). Taking the derivative and multiplying by 2 pF yield an expression for iC(t): iC = C dvC dt = (2 × 10−12)(30 × 103)(A1s1es1t + A2s2es2t) By KCL, iC(0+) = iL(0+) −iR(0+) = 0 Thus, −(2 × 10−12)(30 × 103)(3.063 × 106A1 + 13.60 × 106A2) = 0 Solving Eqs. and , we ﬁnd that A1 = 161.3 μA and A2 = −36.34 μA. Thus, iR = 125 μA t < 0 161.3e−3.063×106t −36.34e−13.6×106t μA t > 0 PRACTICE ● 9.3 Determine the current iR through the resistor of Fig. 9.7 for t > 0 if iL(0−) = 6 A and vC(0+) = 0 V. The conﬁguration of the circuit prior to t = 0 is not known. Ans: iR(t) = 2.437(e−7.823 ×1010t −e−0.511×1010t) A. and v(tm) = 48.9 V A reasonable sketch of the response may be made by plotting the two exponential terms 84e−t and 84e−6t and then taking their difference. This technique is illustrated by the curves of Fig. 9.8; the two exponentials are shown lightly, and their difference, the total response v(t), is drawn as a col-ored line. The curves also verify our previous prediction that the functional behavior of v(t) for very large t is 84e−t, the exponential term containing the smaller magnitude of s1 and s2. CHAPTER 9 THE RLC CIRCUIT 332 ■FIGURE 9.8 The response v(t) = 84(e−t −e−6t) of the network shown in Fig. 9.2. 20 40 60 80 0 –20 1 2 3 4 i(0) = 10 A v(0) = 0 = 3.5 0 = 6 Overdamped v(t) (V) t (s) 7 H F 1 42 6 v i A frequently asked question is the length of time it actually takes for the transient part of the response to disappear (or “damp out”). In practice, it is often desirable to have this transient response approach zero as rapidly as possible, that is, to minimize the settling time ts. Theoretically, of course, ts is inﬁnite, because v(t) never settles to zero in a ﬁnite time. However, a neg-ligible response is present after the magnitude of v(t) has settled to values that remain less than 1 percent of its maximum absolute value \|vm\|. The time that is required for this to occur we deﬁne as the settling time. Since \|vm\| = vm = 48.9 V for our example, the settling time is the time required for the response to drop to 0.489 V. Substituting this value for v(t) in Eq. and neglecting the second exponential term, known to be negligi-ble here, the settling time is found to be 5.15 s. For t > 0, the capacitor current of a certain source-free parallel RLC circuit is given by iC(t) 2e−2t −4e−t A. Sketch the current in the range 0 < t < 5 s, and determine the settling time. We ﬁrst sketch the two terms as shown in Fig. 9.9, then subtract them to ﬁnd iC(t). The maximum value is clearly \|−2\| = 2 A. We therefore need to ﬁnd the time at which \|iC\| has decreased to 20 mA, or 2e−2ts −4e−ts = −0.02 EXAMPLE 9.4 SECTION 9.2 THE OVERDAMPED PARALLEL RLC CIRCUIT 333 ■FIGURE 9.9 The current response iC(t) = 2e−2t −4e−t A, sketched alongside its two components. 1 0 2 4 3 –1 –2 1 iC(t) (A) t (s) 5 4 4e–t 2e–2t iC(t) 2 3 This equation can be solved using an iterative solver routine on a sci-entiﬁc calculator, which returns the solution ts = 5.296 s. If such an op-tion is not available, however, we can approximate Eq. for t ≥ts as −4e−ts = −0.02 Solving, ts = −ln 0.02 4 = 5.298 s which is reasonably close (better than 0.1% accuracy) to the exact solution. PRACTICE ● 9.4 (a) Sketch the voltage vR(t) = 2e−t −4e−3t V in the range 0 < t < 5 s. (b) Estimate the settling time. (c) Calculate the maximum positive value and the time at which it occurs. Ans: See Fig. 9.10; 5.9 s; 544 mV, 896 ms. ■FIGURE 9.10 Response sketched for Practice Problem 9.4a. –0.5 –1.0 0 1.0 0.5 –1.5 –2.0 vR(t) (V) t (s) 5.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 9.3• CRITICAL DAMPING The overdamped case is characterized by α > ω0 or LC > 4R2C2 and leads to negative real values for s1 and s2 and to a response expressed as the algebraic sum of two negative exponentials. Now let us adjust the element values until α and ω0 are equal. This is a very special case which is termed critical damping. If we were to attempt to construct a parallel RLC circuit that is critically damped, we would be at-tempting an essentially impossible task, for we could never make α exactly equal to ω0. For completeness, however, we will discuss the critically damped circuit here, because it shows an interesting transition between overdamping and underdamping. Critical damping is achieved when or α = ω0 LC = 4R2C2 L = 4R2C critical damping We can produce critical damping by changing the value of any of the three elements in the numerical example discussed at the end of Sec. 9.1. We will select R, increasing its value until critical damping is obtained, and thus leave ω0 unchanged. The necessary value of R is 7 √ 6/2 ; L is still 7 H, and C re-mains 1 42 F. We thus ﬁnd α = ω0 = √ 6 s−1 s1 = s2 = − √ 6 s−1 and recall the initial conditions that were speciﬁed, v(0) = 0 and i(0) = 10A. Form of a Critically Damped Response We proceed to attempt to construct a response as the sum of two exponentials, v(t) ? = A1e− √ 6t + A2e− √ 6t which may be written as v(t) ? = A3e− √ 6t At this point, some of us might be feeling that something’s wrong. We have a response that contains only one arbitrary constant, but there are two initial conditions, v(0) = 0 and i(0) = 10 amperes, both of which must be satisﬁed by this single constant. If we select A3 = 0, then v(t) = 0, which is consistent with our initial capacitor voltage. However, although there is no energy stored in the capacitor at t = 0+, we have 350 J of energy initially stored in the inductor. This energy will lead to a transient current ﬂowing out of the inductor, giving rise to a nonzero voltage across all three elements.This seems to be in direct conﬂict with our proposed solution. CHAPTER 9 THE RLC CIRCUIT 334 “Impossible” is a pretty strong term. We make this statement because in practice it is unusual to obtain components that are closer than 1 percent of their speciﬁed values. Thus, obtaining L precisely equal to 4R2C is theoretically possible, but not very likely, even if we’re willing to measure a drawer full of components until we ﬁnd the right ones. SECTION 9.3 CRITICAL DAMPING 335 If a mistake has not led to our difﬁculties, we must have begun with an incorrect assumption, and only one assumption has been made.We originally hypothesized that the differential equation could be solved by assuming an exponential solution, and this turns out to be incorrect for this single special case of critical damping. When α = ω0, the differential equation, Eq. , becomes d2v dt2 + 2α dv dt + α2v = 0 The solution of this equation is not a tremendously difﬁcult process, but we will avoid developing it here, since the equation is a standard type found in the usual differential-equation texts. The solution is v = e−αt(A1t + A2) It should be noted that the solution is still expressed as the sum of two terms, where one term is the familiar negative exponential and the second is t times a negative exponential. We should also note that the solution contains the two expected arbitrary constants. Finding Values for A1 and A2 Let us now complete our numerical example. After we substitute the known value of α in Eq. , obtaining v = A1te− √ 6t + A2e− √ 6t we establish the values of A1 and A2 by ﬁrst imposing the initial condition on v(t)itself, v(0) = 0. Thus, A2 = 0. This simple result occurs because the ini-tial value of the response v(t)was selected as zero; the more general case will require the solution of two equations simultaneously. The second initial con-dition must be applied to the derivative dv/dt just as in the overdamped case. We therefore differentiate, remembering that A2 = 0: dv dt = A1t(− √ 6)e− √ 6t + A1e− √ 6t evaluate at t = 0: dv dt t=0 = A1 and express the derivative in terms of the initial capacitor current: dv dt t=0 = iC(0) C = iR(0) C + i(0) C where reference directions for iC, iR, and i are deﬁned in Fig. 9.2. Thus, A1 = 420 V The response is, therefore, v(t) = 420te−2.45t V Graphical Representation of the Critically Damped Response Before plotting this response in detail, let us again try to anticipate its form by qualitative reasoning. The speciﬁed initial value is zero, and Eq. con-curs. It is not immediately apparent that the response also approaches zero as t becomes inﬁnitely large, because te−2.45t is an indeterminate form. How-ever, this obstacle is easily overcome by use of L’Hôspital’s rule, which yields lim t→∞v(t) = 420 lim t→∞ t e2.45t = 420 lim t→∞ 1 2.45e2.45t = 0 and once again we have a response that begins and ends at zero and has posi-tive values at all other times.Amaximum value vm again occurs at time tm; for our example, tm = 0.408 s and vm = 63.1 V This maximum is larger than that obtained in the overdamped case, and is a result of the smaller losses that occur in the larger resistor; the time of the maximum response is slightly later than it was with overdamping. The set-tling time may also be determined by solving vm 100 = 420tse−2.45ts for ts (by trial-and-error methods or a calculator’s SOLVE routine): ts = 3.12 s which is a considerably smaller value than that which arose in the over-damped case (5.15 s).As a matter of fact, it can be shown that, for given val-ues of L and C, the selection of that value of R which provides critical damp-ing will always give a shorter settling time than any choice of R that produces an overdamped response. However, a slight improvement (reduction) in set-tling time may be obtained by a further slight increase in resistance; a slightly underdamped response that will undershoot the zero axis before it dies out will yield the shortest settling time. CHAPTER 9 THE RLC CIRCUIT 336 ■FIGURE 9.11 The response v(t) = 420te−2.45t of the network shown in Fig. 9.2 with R changed to provide critical damping. 20 40 60 80 0 –20 1 2 3 4 v(t) (V) t (s) 7 H F 1 42 8.57 v + – i SECTION 9.3 CRITICAL DAMPING 337 The response curve for critical damping is drawn in Fig. 9.11; it may be compared with the overdamped (and underdamped) case by reference to Fig. 9.16. Select a value for R1 such that the circuit of Fig. 9.12 will be char-acterized by a critically damped response for t > 0, and a value for R2 such that v(0) 2 V. EXAMPLE 9.5 R2 v + – t = 0 5u(–t) A 1 nF 4 H R1 ■FIGURE 9.12 A circuit that reduces to a parallel RLC circuit after the switch is thrown. We note that at t = 0−, the current source is on, and the inductor can be treated as a short circuit. Thus, v(0−) appears across R2, and is given by v(0−) = 5R2 and a value of 400 m should be selected for R2 to obtain v(0) = 2 V. After the switch is thrown, the current source has turned itself off and R2 is shorted. We are left with a parallel RLC circuit comprised of R1, a 4 H inductor, and a 1nF capacitor. We may now calculate (for t > 0) α = 1 2RC = 1 2 × 10−9R1 and ω0 = 1 √ LC = 1 √ 4 × 10−9 = 15,810 rad/s Therefore, to establish a critically damped response in the circuit for t > 0, we need to set R1 = 31.63 k. (Note: since we have rounded to four signiﬁcant ﬁgures, the pedantic can rightly argue that this is still not exactly a critically damped response—a difﬁcult situation to create.) 9.4• THE UNDERDAMPED PARALLEL RLC CIRCUIT Let us continue the process begun in Sec. 9.3 by increasing R once more to obtain what we will refer to as an underdamped response. Thus, the damping coefﬁcient α decreases while ω0 remains constant, α2 becomes smaller than ω2 0, and the radicand appearing in the expressions for s1 and s2 becomes neg-ative. This causes the response to take on a much different character, but it is fortunately not necessary to return to the basic differential equation again. By using complex numbers, the exponential response turns into a damped sinu-soidal response; this response is composed entirely of real quantities, the complex quantities being necessary only for the derivation.1 The Form of the Underdamped Response We begin with the exponential form v(t) = A1es1t + A2es2t where s1,2 = −α ± α2 −ω2 0 and then let α2 −ω2 0 = √ −1 ω2 0 −α2 = j ω2 0 −α2 where j ≡√−1. We now take the new radical, which is real for the underdamped case, and call it ωd, the natural resonant frequency: ωd = ω2 0 −α2 The response may now be written as v(t) = e−αt(A1e jωdt + A2e−jωdt) CHAPTER 9 THE RLC CIRCUIT 338 ■FIGURE 9.13 4 H R2 R1 v + – t = 0 0.5u(–t) A 1 F Electrical engineers use “j” instead of “i” to represent √−1 to avoid confusion with currents. (1) A review of complex numbers is presented in Appendix 5. PRACTICE ● 9.5 (a) Choose R1 in the circuit of Fig. 9.13 so that the response after t = 0 will be critically damped. (b) Now select R2 to obtain v(0) = 100 V. (c) Find v(t) at t = 1 ms. Ans: 1 k; 250 ; −212 V. SECTION 9.4 THE UNDERDAMPED PARALLEL RLC CIRCUIT 339 or, in the longer but equivalent form, v(t) = e−αt (A1 + A2) e jωdt + e−jωdt 2 + j(A1 −A2) e jωdt −e−jωdt j2 Applying identities described in Appendix 5, the term in the ﬁrst square brackets in the preceding equation is identically equal to cos ωdt, and the sec-ond is identically sin ωdt. Hence, v(t) = e−αt[(A1 + A2) cos ωdt + j (A1 −A2) sin ωdt] and the multiplying factors may be assigned new symbols: v(t) = e−αt(B1 cos ωdt + B2 sin ωdt) where Eqs. and are identical. It may seem a little odd that our expression originally appeared to have a complex component, and now is purely real. However, we should remember that we originally allowed for A1 and A2 to be complex as well as s1 and s2. In any event, if we are dealing with the underdamped case, we have now left complex numbers behind. This must be true since α, ωd, and t are real quan-tities, so that v(t) itself must be a real quantity (which might be presented on an oscilloscope, a voltmeter, or a sheet of graph paper). Equation is the desired functional form for the underdamped response, and its validity may be checked by direct substitution in the original differential equation; this exercise is left to the doubters. The two real constants B1 and B2 are again selected to ﬁt the given initial conditions. We return to our simple parallel RLC circuit of Fig. 9.2 with R = 6 , C = 1/42 F, and L = 7 H, but now increase the resistance further to 10.5 . Thus, α = 1 2RC = 2 s−1 ω0 = 1 √ LC = √ 6 s−1 and ωd = ω2 0 −α2 = √ 2 rad/s Except for the evaluation of the arbitrary constants, the response is now known: v(t) = e−2t(B1 cos √ 2t + B2 sin √ 2t) Finding Values for B1 and B2 The determination of the two constants proceeds as before. If we still assume that v(0) = 0 and i(0) = 10, then B1 must be zero. Hence v(t) = B2e−2t sin √ 2t The derivative is dv dt = √ 2B2e−2t cos √ 2t −2B2e−2t sin √ 2t and at t = 0 it becomes dv dt t=0 = √ 2B2 = iC(0) C = 420 where iC is deﬁned in Fig. 9.2. Therefore, v(t) = 210 √ 2e−2t sin √ 2t Graphical Representation of the Underdamped Response Notice that, as before, this response function has an initial value of zero because of the initial voltage condition we imposed, and a ﬁnal value of zero because the exponential term vanishes for large values of t. As t increases from zero through small positive values, v(t) increases as 210 √ 2 sin √ 2t, because the exponential term remains essentially equal to unity. But, at some time tm, the exponential function begins to decrease more rapidly than sin √ 2t is increasing; thus v(t) reaches a maximum vm and begins to de-crease. We should note that tm is not the value of t for which sin √ 2t is a maximum, but must occur somewhat before sin √ 2t reaches its maximum. When t = π/ √ 2, v(t) is zero. Thus, in the interval π/ √ 2 < t < √ 2π, the response is negative, becoming zero again at t = √ 2π. Hence, v(t) is an oscillatory function of time and crosses the time axis an inﬁnite number of times at t = nπ/ √ 2, where n is any positive integer. In our example, how-ever, the response is only slightly underdamped, and the exponential term causes the function to die out so rapidly that most of the zero crossings will not be evident in a sketch. The oscillatory nature of the response becomes more noticeable as α de-creases. If α is zero, which corresponds to an inﬁnitely large resistance, then v(t) is an undamped sinusoid that oscillates with constant amplitude. There is never a time at which v(t)drops and stays below 1 percent of its maximum value; the settling time is therefore inﬁnite. This is not perpetual motion; we have merely assumed an initial energy in the circuit and have not provided any means to dissipate this energy. It is transferred from its initial location in the inductor to the capacitor, then returns to the inductor, and so on, forever. The Role of Finite Resistance Aﬁnite R in the parallel RLC circuit acts as a kind of electrical transfer agent. Every time energy is transferred from L to C or from C to L, the agent exacts a commission. Before long, the agent has taken all the energy, wantonly dissi-pating every last joule.The L and C are left without a joule of their own, with-out voltage and without current. Actual parallel RLC circuits can be made to have effective values of R so large that a natural undamped sinusoidal re-sponse can be maintained for years without supplying any additional energy. Returning to our speciﬁc numerical problem, differentiation locates the ﬁrst maximum of v(t), vm1 = 71.8 V at tm1 = 0.435 s the succeeding minimum, vm2 = −0.845 V at tm2 = 2.66 s CHAPTER 9 THE RLC CIRCUIT 340 SECTION 9.4 THE UNDERDAMPED PARALLEL RLC CIRCUIT 341 and so on. The response curve is shown in Fig. 9.14. Additional response curves for increasingly more underdamped circuits are shown in Fig. 9.15. ■FIGURE 9.14 The response v(t) 210 √ 2e−2t sin √ 2t of the network shown in Fig. 9.2 with R increased to produce an underdamped response. 20 40 60 80 0 –20 1 2 3 4 v(t) (V) t (s) 7 H F 1 42 10.5 v + – i vm1 vm2 ■FIGURE 9.15 Simulated underdamped voltage response of the network for three different resistance values, showing an increase in the oscillatory behavior as R is increased. The settling time may be obtained by a trial-and-error solution, and for R = 10.5 , it turns out to be 2.92 s, somewhat smaller than for critical damping. Note that ts is greater than tm2 because the magnitude of vm2 is greater than 1 percent of the magnitude of vm1. This suggests that a slight de-crease in R would reduce the magnitude of the undershoot and permit ts to be less than tm2. The overdamped, critically damped, and underdamped responses for this network as simulated by PSpice are shown on the same graph in Fig. 9.16. A comparison of the three curves makes the following general CHAPTER 9 THE RLC CIRCUIT 342 ■FIGURE 9.16 Simulated overdamped, critically damped, and underdamped voltage response for the example network, obtained by varying the value of the parallel resistance R. Determine iL(t) for the circuit of Fig. 9.17a, and plot the waveform. At t = 0, both the 3 A source and the 48 resistor are removed, leaving the circuit shown in Fig. 9.17b. Thus, α = 1.2 s−1 and ω0 = 4.899 rad/s. Since α < ω0, the circuit is underdamped, and we therefore expect a response of the form iL(t) = e−αt(B1 cos ωdt + B2 sin ωdt) where ωd = ω2 0 −α2 = 4.750 rad/s. The only remaining step is to ﬁnd B1 and B2. Figure 9.17c shows the circuit as it exists at t = 0−. We may replace the inductor with a short circuit and the capacitor with an open circuit; the result is vC(0−) = 97.30 V and iL(0−) = 2.027 A. Since neither quantity can change in zero time, vC(0+) = 97.30 V and iL(0+) = 2.027 A. Substituting iL(0) = 2.027 into Eq. yields B1 = 2.027 A. To determine the other constant, we ﬁrst differentiate Eq. : diL dt = e−αt(−B1ωd sin ωdt + B2ωd cos ωdt) −αe−at(B1 cos ωdt + B2 sin ωdt) EXAMPLE 9.6 conclusions plausible: • When the damping is changed by increasing the size of the parallel resistance, the maximum magnitude of the response is greater and the amount of damping is smaller. • The response becomes oscillatory when underdamping is present, and the minimum settling time is obtained for slight underdamping. SECTION 9.4 THE UNDERDAMPED PARALLEL RLC CIRCUIT 343 ■FIGURE 9.17 (a) A parallel RLC circuit for which the current iL(t) is desired. (b) Circuit for t ≥0. (c) Circuit for determining the initial conditions. (a) 10 H 100 48 vC + – iC iR iL t = 0 3u(–t) A F 1 240 (b) (c) 100 10 H iC iL iR F 1 240 vC + – 3 A 100 48 10 H vC + – iL iC iR F 1 240 and note that vL(t) = L(diL/dt). Referring to the circuit of Fig. 9.17b, we see that vL(0+) = vC(0+) = 97.3 V. Thus, multiplying Eq. by L = 10 H and setting t = 0, we ﬁnd that vL(0) = 10(B2ωd) −10αB1 = 97.3 Solving, B2 2.561 A, so that iL = e−1.2t(2.027 cos 4.75t + 2.561 sin 4.75t) A which we have plotted in Fig. 9.18. ■FIGURE 9.18 Plot of iL(t), showing obvious signs of being an underdamped response. 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 –1.5 –1.0 –0.5 0 0.5 1.0 1.5 2.0 2.5 3.0 t (s) iL(t) (A) CHAPTER 9 THE RLC CIRCUIT 344 ■FIGURE 9.19 + – 3 V 2 H 10 F 50 k 100 k 500 v + – t = 0 5u(–t) V PRACTICE ● 9.6 The switch in the circuit of Fig. 9.19 has been in the left position for a long time; it is moved to the right at t = 0. Find (a) dv/dt at t = 0+; (b) v at t = 1 ms; (c) t0, the ﬁrst value of t greater than zero at which v = 0. Ans: −1400 V/s; 0.695 V; 1.609 ms. COMPUTER-AIDED ANALYSIS One useful feature in Probe is the ability to perform mathematical opera-tions on the voltages and currents that result from a simulation. In this example, we will make use of that ability to show the transfer of energy in a parallel RLC circuit from a capacitor that initially stores a speciﬁc amount of energy (1.25 μJ) to an inductor that initially stores no energy. We choose a 100 nF capacitor and a 7 μH inductor, which immedi-ately enables us to calculate ω0 = 1.195 × 106 s−1. In order to consider overdamped, critically damped, and underdamped cases, we need to select the parallel resistance in such a way as to obtain α > ω0 (over-damped), α = ω0 (critically damped), and α < ω0 (underdamped). From our previous discussions, we know that for a parallel RLC circuit α = (2RC)−1. We select R = 4.1833 as a close approximation to the critically damped case; obtaining α precisely equal to ω0 is effectively impossible. If we increase the resistance, the energy stored in the other two elements is dissipated more slowly, resulting in an underdamped response. We select R = 100 so that we are well into this regime, and use R = 1 (a very small resistance) to obtain an overdamped response. We therefore plan to run three separate simulations, varying only the resistance R between them. The 1.25 μJ of energy initially stored in the capacitor equates to an initial voltage of 5 V, and so we set the initial condition of our capacitor accordingly. Once Probe is launched, we select Add under the Trace menu. We wish to plot the energy stored in both the inductor and the capacitor as a function of time. For the capacitor, w = 1 2Cv2, so we click in the Trace Expression window, type in “0.5100E-9” (without the quotes), click on V(C1:1), return to the Trace Expression window and enter “”, click on V(C1:1) once again, and then select Ok. We repeat the sequence to obtain the energy stored in the inductor, using 7E-6 instead of 100E-9, and clicking on I(L1:1) instead of V(C1:1). SECTION 9.5 THE SOURCE-FREE SERIES RLC CIRCUIT 345 9.5• THE SOURCE-FREE SERIES RLC CIRCUIT We now wish to determine the natural response of a circuit model composed of an ideal resistor, an ideal inductor, and an ideal capacitor connected in series. The ideal resistor may represent a physical resistor connected into a series LC or RLC circuit; it may represent the ohmic losses and the losses in the ferromagnetic core of the inductor; or it may be used to represent all these and other energy-absorbing devices. ■FIGURE 9.20 Energy transfer in a parallel RLC circuit with (a) R = 100 (underdamped); (b) R = 4.1833 (critically damped); and (c) R = 1 (overdamped). (a) (b) (c) The Probe output plots for three separate simulations are provided in Fig. 9.20. In Fig. 9.20a, we see that the energy remaining in the circuit is continuously transferred back and forth between the capacitor and the inductor until it is (eventually) completely dissipated by the resistor. Decreasing the resistance to 4.1833 yields a critically damped circuit, resulting in the energy plot of Fig. 9.20b. The oscillatory energy transfer between the capacitor and the inductor has been dramatically reduced. We see that the energy transferred to the inductor peaks at approximately 0.8 μs, and then drops to zero. The overdamped response is plotted in Fig. 9.20c. We note that the energy is dissipated much more quickly in the case of the overdamped response, and that very little energy is transferred to the inductor, since most of it is now quickly dissipated in the resistor. The series RLC circuit is the dual of the parallel RLC circuit, and this sin-gle fact is sufﬁcient to make its analysis a trivial affair. Figure 9.21a shows the series circuit. The fundamental integrodifferential equation is L di dt + Ri + 1 C t t0 i dt′ −vC(t0) = 0 and should be compared with the analogous equation for the parallel RLC cir-cuit, drawn again in Fig. 9.21b, C dv dt + 1 R v + 1 L t t0 v dt′ −iL(t0) = 0 The respective second-order equations obtained by differentiating these two equations with respect to time are also duals: L d2i dt2 + R di dt + 1 C i = 0 C d2v dt2 + 1 R dv dt + 1 L v = 0 Our complete discussion of the parallel RLC circuit is directly applicable to the series RLC circuit; the initial conditions on capacitor voltage and in-ductor current are equivalent to the initial conditions on inductor current and capacitor voltage; the voltage response becomes a current response. It is therefore possible to reread the previous four sections using dual language and thereby obtain a complete description of the series RLC circuit. This process, however, is apt to induce a mild neurosis after the ﬁrst few para-graphs and does not really seem to be necessary. A Brief Résumé of the Series Circuit Response In terms of the circuit shown in Fig. 9.21a, the overdamped response is i(t) = A1es1t + A2es2t where s1,2 = −R 2L ± R 2L 2 − 1 LC = −α ± α2 −ω2 0 and thus α = R 2L ω0 = 1 √ LC The form of the critically damped response is i(t) = e−αt(A1t + A2) and the underdamped response may be written i(t) = e−αt(B1 cos ωdt + B2 sin ωdt) CHAPTER 9 THE RLC CIRCUIT 346 ■FIGURE 9.21 (a) The series RLC circuit which is the dual of (b) a parallel RLC circuit. Element values are, of course, not identical in the two circuits. i L C R (a) vL + – vC + – L R C (b) v + – iC iL SECTION 9.5 THE SOURCE-FREE SERIES RLC CIRCUIT 347 where ωd = ω2 0 −α2 It is evident that if we work in terms of the parameters α, ω0, and ωd, the mathematical forms of the responses for the dual situations are identical. An increase in α in either the series or parallel circuit, while keeping ω0 constant, tends toward an overdamped response.The only caution that we need exert is in the computation of α, which is 1/2RC for the parallel circuit and R/2L for the series circuit; thus, α is increased by increasing the series resistance or de-creasing the parallel resistance.The key equations for parallel and series RLC circuits are summarized in Table 9.1 for convenience. TABLE ●9.1 Summary of Relevant Equations for Source-Free RLC Circuits Type Condition Criteria α ω0 Response Parallel Series Parallel Series Parallel Series Overdamped Critically damped Underdamped α > ω0 α = ω0 α < ω0 1 √ LC 1 √ LC 1 √ LC A1es1t + A2es2t , where s1,2 = −α ± √ α2 −ω2 e−αt (A1t + A2) e−αt(B1 cos ωdt + B2 sin ωdt), where ωd = ω2 0 −α2 EXAMPLE 9.7 Given the series RLC circuit of Fig. 9.22 in which L = 1 H, R = 2 k, C = 1/401 μF, i(0) = 2 mA, and vC(0) = 2 V, ﬁnd and sketch i(t), t > 0. We ﬁnd that α = R/2L = 1000 s−1 and ω0 = 1/ √ LC = 20,025 rad/s. This indicates an underdamped response; we therefore calculate the value of ωd and obtain 20,000 rad/s. Except for the evaluation of the two arbitrary constants, the response is now known: i(t) = e−1000t(B1 cos 20,000t + B2 sin 20,000t) Since we know that i(0) = 2 mA, we may substitute this value into our equation for i(t) to obtain B1 = 0.002 A and thus i(t) = e−1000t(0.002 cos 20,000t + B2 sin 20,000t) A (Continued on next page) ■FIGURE 9.22 A simple source-free RLC circuit with energy stored in both the inductor and the capacitor at t = 0. i L C R vL + – vC + – 1 2RC R 2L 1 2RC R 2L 1 2RC R 2L CHAPTER 9 THE RLC CIRCUIT 348 The remaining initial condition must be applied to the derivative; thus, di dt = e−1000t(−40 sin 20,000t + 20,000B2 cos 20,000t −2 cos 20,000t −1000B2 sin 20,000t) and di dt t=0 = 20,000B2 −2 = vL(0) L = vC(0) −Ri(0) L = 2 −2000(0.002) 1 = −2 A/s so that B2 = 0 The desired response is therefore i(t) = 2e−1000t cos 20,000t mA A good sketch may be made by ﬁrst drawing in the two portions of the exponential envelope, 2e−1000t and −2e−1000t mA, as shown by the broken lines in Fig. 9.23. The location of the quarter-cycle points of the sinusoidal wave at 20,000t = 0, π/2, π, etc., or t = 0.07854k ms, k = 0, 1, 2, . . ., by light marks on the time axis then permits the oscilla-tory curve to be sketched in quickly. ■FIGURE 9.23 The current response in an underdamped series RLC circuit for which α = 1000 s−1, ω0 = 20,000 s−1, i(0) = 2 mA, and vC(0) = 2 V. The graphical construction is simpliﬁed by drawing in the envelope, shown as a pair of broken lines. 0 1 2 –1 –2 0.2 0.4 0.6 0.8 1.0 i(t) (mA) t (ms) The settling time can be determined easily here by using the upper portion of the envelope. That is, we set 2e−1000ts mA equal to 1 percent of its maximum value, 2 mA. Thus, e−1000ts = 0.01, and ts = 4.61 ms is the approximate value that is usually used. SECTION 9.5 THE SOURCE-FREE SERIES RLC CIRCUIT 349 As a ﬁnal example, we pause to consider situations where the circuit in-cludes a dependent source. If no controlling current or voltage associated with the dependent source is of interest, we may simply ﬁnd the Thévenin equiva-lent connected to the inductor and capacitor. Otherwise, we are likely faced with having to write an appropriate integrodifferential equation, take the indi-cated derivative, and solve the resulting differential equation as best we can. ■FIGURE 9.24 0.5 H 40 F 100 i u(–t) A Ans: 100 s−1; 224 rad/s; 1 A; 0; −0.1204 A. Find an expression for vC(t) in the circuit of Fig. 9.25a, valid for t > 0. EXAMPLE 9.8 (Continued on next page) ■FIGURE 9.25 (a) An RLC circuit containing a dependent source. (b) Circuit for ﬁnding Req. 10 V 2 9 t = 0 5 H 2 mF i + – 3i + – vC – + (a) 2 1 A 9 i + – 3i (b) vtest + – PRACTICE ● 9.7 With reference to the circuit shown in Fig. 9.24, ﬁnd (a) α; (b) ω0; (c) i(0+); (d) di/dt\|t=0+; (e) i(12 ms). As we are interested only in vC(t), it is perfectly acceptable to begin by ﬁnding the Thévenin equivalent resistance connected in series with the CHAPTER 9 THE RLC CIRCUIT 350 inductor and capacitor at t = 0+. We do this by connecting a 1 A source as shown in Fig. 9.25b, from which we deduce that vtest = 11i −3i = 8i = 8(1) = 8 V Thus, Req = 8 , so α = R/2L = 0.8 s−1 and ω0 = 1/ √ LC = 10 rad/s, meaning that we expect an underdamped response with ωd = 9.968 rad/s and the form vC(t) = e−0.8t(B1 cos 9.968t + B2 sin 9.968t) In considering the circuit at t = 0−, we note that iL(0−) = 0 due to the presence of the capacitor. By Ohm’s law, i(0−) = 5 A, so vC(0+) = vC(0−) = 10 −3i = 10 −15 = −5 V This last condition substituted into Eq. yields B1 = −5 V. Taking the derivative of Eq. and evaluating at t = 0 yield dvC dt t=0 = −0.8B1 + 9.968B2 = 4 + 9.968B2 We see from Fig. 9.25a that i = −C dvC dt Thus, making use of the fact that i(0+) = iL(0−) = 0 in Eq. yields B2 = −0.4013 V, and we may write vC(t) = −e−0.8t(5 cos 9.968t + 0.4013 sin 9.968t) V t > 0 The PSpice simulation of this circuit, shown in Fig. 9.26, conﬁrms our analysis. ■FIGURE 9.26 PSpice simulation of the circuit shown in Fig. 9.25a. The analytical result is plotted using a dashed red line. SECTION 9.6 THE COMPLETE RESPONSE OF THE RLC CIRCUIT 351 9.6• THE COMPLETE RESPONSE OF THE RLC CIRCUIT We now consider those RLC circuits in which dc sources are switched into the network and produce forced responses that do not necessarily vanish as time becomes inﬁnite. The general solution is obtained by the same procedure that was fol-lowed for RL and RC circuits. The basic steps are (not necessarily in this order) as follows: 1. Determine the initial conditions. 2. Obtain a numerical value for the forced response. 3. Write the appropriate form of the natural response with the necessary number of arbitrary constants. 4. Add the forced response and natural response to form the complete response. 5. Evaluate the response and its derivative at t = 0, and employ the ini-tial conditions to solve for the values of the unknown constants. We note that it is generally this last step that causes the most trouble for stu-dents, as the circuit must be carefully evaluated at t = 0 to make full use of the initial conditions. Consequently, although the determination of the initial conditions is basically no different for a circuit containing dc sources from what it is for the source-free circuits that we have already covered in some de-tail, this topic will receive particular emphasis in the examples that follow. Most of the confusion in determining and applying the initial conditions arises for the simple reason that we do not have a rigorous set of rules laid down for us to follow. At some point in each analysis, a situation usually arises in which some thinking is involved that is more or less unique to that particular problem. This is almost always the source of the difﬁculty. The Easy Part The complete response (arbitrarily assumed to be a voltage response) of a second-order system consists of a forced response, vf (t) = Vf PRACTICE ● 9.8 Find an expression for iL(t) in the circuit of Fig. 9.27, valid for t > 0, if vC(0−) = 10 V and iL(0−) = 0. Note that although it is not helpful to apply Thévenin techniques in this instance, the action of the dependent source links vC and iL such that a ﬁrst-order linear differential equation results. Ans: iL(t) = −30e−300t A, t > 0. ■FIGURE 9.27 Circuit for Practice Problem 9.8. 2 5 H 2 iL 3vC 10 mF vC – + which is a constant for dc excitation, and a natural response, vn(t) = Aes1t + Bes2t Thus, v(t) = Vf + Aes1t + Bes2t We assume that s1, s2, and Vf have already been determined from the cir-cuit and the given forcing functions; A and B remain to be found. The last equation shows the functional interdependence of A, B, v, and t; and substitu-tion of the known value of v at t = 0+ thus provides us with a single equation relating A and B, v(0+) = Vf + A + B. This is the easy part. The Other Part Another relationship between A and B is necessary, unfortunately, and this is normally obtained by taking the derivative of the response, dv dt = 0 + s1Aes1t + s2Bes2t and inserting the known value of dv/dt at t = 0+. We thus have two equa-tions relating A and B, and these may be solved simultaneously to evaluate the two constants. The only remaining problem is that of determining the values of v and dv/dt at t = 0+. Let us suppose that v is a capacitor voltage, vC. Since iC = C dvC/dt, we should recognize the relationship between the initial value of dv/dt and the initial value of some capacitor current. If we can es-tablish a value for this initial capacitor current, then we will automatically es-tablish the value of dv/dt. Students are usually able to get v(0+) very easily, but are inclined to stumble a bit in ﬁnding the initial value of dv/dt. If we had selected an inductor current iL as our response, then the initial value ofdiL/dt would be intimately related to the initial value of some inductor voltage.Vari-ables other than capacitor voltages and inductor currents are determined by expressing their initial values and the initial values of their derivatives in terms of the corresponding values for vC and iL. We will illustrate the procedure and ﬁnd all these values by the careful analysis of the circuit shown in Fig. 9.28.To simplify the analysis, an unusual value of capacitance is used again. CHAPTER 9 THE RLC CIRCUIT 352 ■FIGURE 9.28 (a) An RLC circuit that is used to illustrate several procedures by which the initial conditions may be obtained. The desired response is nominally taken to be vC(t). (b) t 0−. (c) t > 0. 3 H 5 A 30 vC + – vL + – vR + – iR iL iC 4u(t) A F 1 27 (a) vR + – iC iL vL + – vC + – iR 5 A 3 H 30 (b) F 1 27 vR + – iC iL vL + – vC + – iR 5 A 4 A 3 H 30 (c) F 1 27 SECTION 9.6 THE COMPLETE RESPONSE OF THE RLC CIRCUIT 353 There are three passive elements in the circuit shown in Fig. 9.28a, and a voltage and a current are deﬁned for each. Find the values of these six quantities at both t 0−and t 0+. EXAMPLE 9.9 Our object is to ﬁnd the value of each current and voltage at both t = 0−and t = 0+. Once these quantities are known, the initial values of the derivatives may be found easily. 1. t = 0−At t = 0−, only the right-hand current source is active as depicted in Fig. 9.28b. The circuit is assumed to have been in this state forever, so all currents and voltages are constant. Thus, a dc current through the inductor requires zero voltage across it: vL(0−) = 0 and a dc voltage across the capacitor (−vR) requires zero current through it: iC(0−) = 0 We next apply Kirchhoff’s current law to the right-hand node to obtain iR(0−) = −5 A which also yields vR(0−) = −150 V We may now use Kirchhoff’s voltage law around the left-hand mesh, ﬁnding vC(0−) = 150 V while KCL enables us to ﬁnd the inductor current, iL(0−) = 5 A 2. t = 0+ During the interval from t = 0−to t = 0+, the left-hand current source becomes active and many of the voltage and current values at t = 0−will change abruptly. The corresponding circuit is shown in Fig. 9.28c. However, we should begin by focusing our attention on those quantities which cannot change, namely, the inductor current and the capacitor voltage. Both of these must remain constant during the switching interval. Thus, iL(0+) = 5 A and vC(0+) = 150 V Since two currents are now known at the left node, we next obtain iR(0+) = −1 A and vR(0+) = −30 V so that iC(0+) = 4 A and vL(0+) = 120 V and we have our six initial values at t = 0−and six more at t = 0+. Among these last six values, only the capacitor voltage and the inductor current are unchanged from the t = 0−values. CHAPTER 9 THE RLC CIRCUIT 354 We could have employed a slightly different method to evaluate these currents and voltages at t = 0−and t = 0+. Prior to the switching operation, only direct currents and voltages exist in the circuit. The inductor may therefore be replaced by a short circuit, its dc equivalent, while the capacitor is replaced by an open circuit. Redrawn in this manner, the cir-cuit of Fig. 9.28a appears as shown in Fig. 9.29a. Only the current source at the right is active, and its 5 A ﬂow through the resistor and the inductor. We therefore have iR(0−) = −5 A and vR(0−) = −150 V, iL(0−) = 5 A and vL(0−) = 0, and iC(0−) = 0 and vC(0−) = 150 V, as before. ■FIGURE 9.29 (a) A simple circuit equivalent to the circuit of Fig. 9.28a for t = 0−. (b) Equivalent circuit with labeled voltages and currents valid at the instant deﬁned by t = 0+. + – 5 A 5 A 30 (b) vC + – vL + – vR + – iR iL iC 4 A 150 V 5 A 30 (a) vC + – vL + – vR + – iR iL iC 0 A We now turn to the problem of drawing an equivalent circuit that will assist us in determining the several voltages and currents at t = 0+. Each capacitor voltage and each inductor current must remain constant during the switching interval. These conditions are ensured by replacing the induc-tor with a current source and the capacitor with a voltage source. Each source serves to maintain a constant response during the discontinuity. The equivalent circuit of Fig. 9.29b results. It should be noted that the circuit shown in Fig. 9.29b is valid only for the interval between 0−and 0+. The voltages and currents at t = 0+ are obtained by analyzing this dc circuit. The solution is not difﬁcult, but the relatively large number of sources present in the network does produce a somewhat strange sight. However, problems of this type were solved in Chap. 3, and nothing new is involved. Attacking the currents ﬁrst, we begin at the upper left node and see that iR(0+) = 4 −5 = −1 A. Moving to the upper right node, we ﬁnd that iC(0+) = −1 + 5 = 4 A. And, of course, iL(0+) = 5 A. Next we consider the voltages. Using Ohm’s law, we see that vR(0+) = 30(−1) = −30 V. For the inductor, KVL gives us vL(0+) = −30 + 150 = 120 V. Finally, including vC(0+) = 150 V, we have all the values at t = 0+. SECTION 9.6 THE COMPLETE RESPONSE OF THE RLC CIRCUIT 355 ■FIGURE 9.30 10 F 1 mH 20 is vR + – vC + – iL PRACTICE ● 9.9 Let is = 10u(−t) −20u(t) A in Fig. 9.30. Find (a) iL(0−); (b) vC(0+); (c) vR(0+); (d) iL(∞); (e) iL(0.1 ms). Complete the determination of the initial conditions in the circuit of Fig. 9.28, repeated in Fig. 9.31, by ﬁnding values at t = 0+ for the ﬁrst derivatives of the three voltage and three current variables deﬁned on the circuit diagram. EXAMPLE 9.10 We begin with the two energy storage elements. For the inductor, vL = L diL dt and, speciﬁcally, vL(0+) = L diL dt t=0+ Thus, diL dt t=0+ = vL(0+) L = 120 3 = 40 A/s Similarly, dvC dt t=0+ = iC(0+) C = 4 1/27 = 108 V/s ■FIGURE 9.31 Circuit of Fig. 9.28, repeated for Example 9.10. 3 H 5 A 30 vC + – vL + – vR + – iR iL iC 4u(t) A F 1 27 Ans: 10 A; 200 V; 200 V; −20 A; 2.07 A. (Continued on next page) CHAPTER 9 THE RLC CIRCUIT 356 The other four derivatives may be determined by realizing that KCL and KVL are both satisﬁed by the derivatives also. For example, at the left-hand node in Fig. 9.31, 4 −iL −iR = 0 t > 0 and thus, 0 −diL dt −diR dt = 0 t > 0 and therefore, diR dt t=0+ = −40 A/s The three remaining initial values of the derivatives are found to be dvR dt t=0+ = −1200 V/s dvL dt t=0+ = −1092 V/s and diC dt t=0+ = −40 A/s Before leaving this problem of the determination of the necessary initial values, we should point out that at least one other powerful method of deter-mining them has been omitted: we could have written general nodal or loop equations for the original circuit. Then the substitution of the known zero values of inductor voltage and capacitor current at t = 0−would uncover several other response values at t = 0−and enable the remainder to be found easily. A similar analysis at t = 0+ must then be made. This is an important method, and it becomes a necessary one in more complicated circuits which cannot be analyzed by our simpler step-by-step procedures. Now let us brieﬂy complete the determination of the response vC(t) for the original circuit of Fig. 9.31.With both sources dead, the circuit appears as a series RLC circuit and s1 and s2 are easily found to be −1 and −9, respec-tively. The forced response may be found by inspection or, if necessary, by drawing the dc equivalent, which is similar to Fig. 9.29a, with the addition of a 4Acurrent source. The forced response is 150 V. Thus, vC(t) = 150 + Ae−t + Be−9t and vC(0+) = 150 = 150 + A + B or A + B = 0 Then, dvC dt = −Ae−t −9Be−9t SECTION 9.6 THE COMPLETE RESPONSE OF THE RLC CIRCUIT 357 and dvC dt t=0+ = 108 = −A −9B Finally, A = 13.5 B = −13.5 and vC(t) = 150 + 13.5(e−t −e−9t) V A Quick Summary of the Solution Process In summary, then, whenever we wish to determine the transient behavior of a simple three-element RLC circuit, we must ﬁrst decide whether we are con-fronted with a series or a parallel circuit, so that we may use the correct rela-tionship for α. The two equations are α = 1 2RC (parallel RLC) α = R 2L (series RLC) Our second decision is made after comparing α with ω0, which is given for either circuit by ω0 = 1 √ LC If α > ω0, the circuit is overdamped, and the natural response has the form fn(t) = A1es1t + A2es2t where s1,2 = −α ± α2 −ω2 0 If α = ω0, then the circuit is critically damped and fn(t) = e−αt(A1t + A2) And ﬁnally, if α < ω0, then we are faced with the underdamped response, fn(t) = e−αt(A1 cos ωdt + A2 sin ωdt) where ωd = ω2 0 −α2 Our last decision depends on the independent sources. If there are none acting in the circuit after the switching or discontinuity is completed, then the circuit is source-free and the natural response accounts for the complete re-sponse. If independent sources are still present, then the circuit is driven and a forced response must be determined. The complete response is then the sum f (t) = f f (t) + fn(t) This is applicable to any current or voltage in the circuit. Our ﬁnal step is to solve for unknown constants given the initial conditions. Earlier, we alluded to the fact that the concepts investi-gated in this chapter actually extend beyond the analysis of electric circuits. In fact, the general form of the differ-ential equations we have been working with appears in many ﬁelds—we need only learn how to “translate’’new parameter terminology. For example, consider a simple automotive suspension, as shown in Fig. 9.32. The pis-ton is not attached to the cylinder, but is attached to both the spring and the wheel. The moving parts therefore are the spring, the piston, and the wheel. We will model this physical system by ﬁrst determin-ing the forces in play. Deﬁning a position function p(t) which describes where the piston lies within the cylin-der, we may write FS, the force on the spring, as FS = K p(t) where K is known as the spring constant and has units of lb/ft. The force on the wheel FW is equal to the mass of the wheel times its acceleration, or FW = m d2 p(t) dt2 where m is measured in lb · s2/ft. Last but not least is the force of friction Ff acting on the piston Ff = μf dp(t) dt where μf is the coefficient of friction, with units of lb · s/ft. From our basic physics courses we know that all forces acting in our system must sum to zero, so that m d2 p(t) dt2 + μf dp(t) dt + Kp(t) = 0 This equation most likely had the potential to give us nightmares at one point in our academic career, but no longer. We compare Eq. to Eqs. and and immediately see a distinct resemblance, at least in the general form. Choosing Eq. , the differential equa-tion describing the inductor current of a series-connected RLC circuit, we observe the following correspondences: Mass m → inductance L Coefﬁcient of friction μf → resistance R Spring constant K → inverse of the C−1 capacitance Position variable p(t) → current variable i(t) So, if we are willing to talk about feet instead of am-peres, lb · s2/ft instead of H, ft/lb instead of F, and lb · s/ft instead of , we can apply our newly found skills at modeling RLC circuits to the task of evaluating automo-tive shock absorbers. Take a typical car wheel of 70 lb. The mass is found by dividing the weight by the earth’s gravitational accel-eration (32.17 ft/s2), resulting in m = 2.176 lb · s2/ft. The curb weight of our car is 1985 lb, and the static displace-ment of the spring is 4 inches (no passengers). The spring constant is obtained by dividing the weight on each shock absorber by the static displacement, so that we have K = ( 1 4)(1985)(3 ft−1) = 1489 lb/ft. We are also told that the coefﬁcient of friction for our piston-cylinder assem-bly is 65 lb · s/ft. Thus, we can simulate our shock ab-sorber by modeling it with a series RLC circuit having R = 65 , L = 2.176 H, and C = K −1 = 671.6 μF. The resonant frequency of our shock absorber is ω0 = (LC)−1/2 = 26.16 rad/s, and the damping coefﬁ-cient is α = R/2L = 14.94 s−1. Since α < ω0, our shock absorber represents an underdamped system; this means that we expect a bounce or two after we run over a pot-hole. A stiffer shock (larger coefﬁcient of friction, or a larger resistance in our circuit model) is typically desir-able when curves are taken at high speeds—at some point this corresponds to an overdamped response. How-ever, if most of our driving is over unpaved roads, a slightly underdamped response is preferable. PRACTICAL APPLICATION Modeling Automotive Suspension Systems PRACTICAL APPLICATION ■FIGURE 9.32 Typical automotive suspension system. © Transtock Inc./Alamy. SECTION 9.7 THE LOSSLESS LC CIRCUIT 359 9.7• THE LOSSLESS LC CIRCUIT When we considered the source-free RLC circuit, it became apparent that the resistor served to dissipate any initial energy stored in the circuit. At some point it might occur to us to ask what would happen if we could remove the resistor. IfthevalueoftheresistanceinaparallelRLCcircuitbecomesinﬁnite, or zero in the case of a series RLC circuit, we have a simple LC loop in which an oscillatory response can be maintained forever. Let us look brieﬂy at an ex-ample of such a circuit, and then discuss another means of obtaining an iden-tical response without the need of supplying any inductance. Consider the source-free circuit of Fig. 9.34, in which the large values L = 4 H and C = 1 36 F are used so that the calculations will be simple. We let i(0) = −1 6 A and v(0) = 0. We ﬁnd that α = 0 and ω2 0 = 9 s−2, so that ωd = 3 rad/s. In the absence of exponential damping, the voltage v is simply v = A cos 3t + B sin 3t Since v(0) = 0, we see that A = 0. Next, dv dt t=0 = 3B = −i(0) 1/36 But i(0) = −1 6 ampere, and therefore dv/dt = 6V/s at t = 0. We must have B = 2V and so v = 2 sin 3t V which is an undamped sinusoidal response; in other words, our voltage response does not decay. Now let us see how we might obtain this voltage without using an LC cir-cuit. Our intentions are to write the differential equation that v satisﬁes and then to develop a conﬁguration of op amps that will yield the solution of the equation.Although we are working with a speciﬁc example, the technique is a general one that can be used to solve any linear homogeneous differential equation. For the LC circuit of Fig. 9.34, we select v as our variable and set the sum of the downward inductor and capacitor currents equal to zero: 1 4 t t0 v dt′ −1 6 + 1 36 dv dt = 0 Differentiating once, we have 1 4v + 1 36 d2v dt2 = 0 or d2v dt2 = −9v ■FIGURE 9.34 This circuit is lossless, and it provides the undamped response v = 2 sin 3t V, if v(0) = 0 and i(0) = −1 6 A. 4 H v + – i F 1 36 PRACTICE ● 9.10 Let vs = 10 + 20u(t) V in the circuit of Fig. 9.33. Find (a) iL(0); (b) vC(0); (c) iL, f ; (d) iL(0.1 s). Ans: 0.2 A; 10 V; 0.6 A; 0.319 A. ■FIGURE 9.33 + – 1 mF 15.625 H 50 vC + – iL vs CHAPTER 9 THE RLC CIRCUIT 360 ■FIGURE 9.35 The inverting operational ampliﬁer provides a gain vo/vs = −Rf/R1, assuming an ideal op amp. + – – + R1 Rf vo + – vs ■FIGURE 9.36 Two integrators and an inverting ampliﬁer are connected to provide the solution of the differential equation d2v/dt2 = −9v. – + 6 V 1 F 1 F 1 M 1 M 10 k Rf = 90 k A d2v dt2 –9v – v = 2 sin 3t V t = 0 t = 0 – + t = 0 – + dv dt Inordertosolvethisequation,weplantomakeuseoftheoperationalampliﬁer as an integrator. We assume that the highest-order derivative appearing in the differential equation here, d2v/dt2, is available in our conﬁguration of op amps at an arbitrary point A.We now make use of the integrator, with RC = 1, as discussed in Sec. 7.5.The input isd2v/dt2, and the output must be−dv/dt, where the sign change results from using an inverting op amp conﬁguration for the integrator. The initial value of dv/dt is 6 V/s, as we showed when we ﬁrst analyzed the circuit, and thus an initial value of−6Vmust be set in the in-tegrator. The negative of the ﬁrst derivative now forms the input to a second integrator. Its output is therefore v(t), and the initial value is v(0) = 0. Now it only remains to multiply v by −9 to obtain the second derivative we assumed at point A. This is ampliﬁcation by 9 with a sign change, and it is easily accomplished by using the op amp as an inverting ampliﬁer. Figure 9.35 shows the circuit of an inverting ampliﬁer. For an ideal op amp, both the input current and the input voltage are zero. Thus, the current going “east’’ through R1 is vs/R1, while that traveling west through Rf is vo/Rf . Since their sum is zero, we have vo vs = −Rf R1 Thus, we can design for a gain of −9by setting Rf = 90kand R1 = 10k, for example. If we let R be 1 M and C be 1 μF in each of the integrators, then vo = − t 0 vs dt′ + vo(0) in each case. The output of the inverting ampliﬁer now forms the assumed in-put at point A, leading to the conﬁguration of op amps shown in Fig. 9.36. If the left switch is closed at t = 0 while the two initial-condition switches are opened at the same time, the output of the second integrator will be the un-damped sine wave v = 2 sin 3t V. Note that both the LC circuit of Fig. 9.34 and the op amp circuit of Fig. 9.36 have the same output, but the op amp circuit does not contain a SUMMARY AND REVIEW 361 single inductor. It simply acts as though it contained an inductor, providing the appropriate sinusoidal voltage between its output terminal and ground. This can be a considerable practical or economic advantage in circuit design, as inductors are typically bulky, more costly than capacitors, and have more losses associated with them (and therefore are not as well approximated by the “ideal’’model). PRACTICE ● 9.11 Give new values for Rf and the two initial voltages in the circuit of Fig. 9.36 if the output represents the voltage v(t) in the circuit of Fig. 9.37. SUMMARY AND REVIEW The simple RL and RC circuits examined in Chap. 8 essentially did one of two things as the result of throwing a switch: charge or discharge. Which one happened was determined by the initial charge state of the energy storage element. In this chapter, we considered circuits that had two energy storage elements (a capacitor and an inductor), and found that things could get pretty interesting. There are two basic configurations of such RLC circuits: parallel connected and series connected. Analysis of such a circuit yields a second-order partial differential equation, consistent with the number of distinct energy storage elements (if we construct a circuit using only resistors and capacitors such that the capacitors cannot be combined using series/parallel techniques, we also obtain—eventually—a second-order partial differential equation). Depending on the value of the resistance connected to our energy storage elements, we found the transient response of an RLC circuit could be either overdamped (decaying exponentially) or underdamped (decaying, but oscillatory), with a “special case” of critically damped which is difﬁcult to achieve in practice. Oscillations can be useful (for example, in transmitting information over a wireless network) and not so useful (for example, in accidental feedback situations between an amplifier and a microphone at a concert). Although the oscillations are not sustained in the circuits we examined, we have at least seen one way to create them at will, ■FIGURE 9.37 8 H 12 V 5 mF 1 5 v(t) + – t = 0 Ans: 250 k; 400 V; 10 V. CHAPTER 9 THE RLC CIRCUIT 362 and design for a speciﬁc frequency of operation if so desired. We didn’t end up spending a great deal of time with the series connected RLC circuit because with the exception of α, the equations are the same; only a minor adjustment in how we employ initial conditions to ﬁnd the two unknown constants characterizing the transient response is needed. Along those lines, there were two “tricks,” if you will, that we encountered. One is that to employ the second initial condition, we need to take the derivative of our response equation. The second is that whether we’re employing KCL or KVL to make use of that initial condition, we’re doing so at the instant that t = 0; appreciating this fact can simplify equations dramatically by setting t = 0 early. We wrapped up the chapter by considering the complete response, and our approach to this did not differ signiﬁcantly from what we did in Chap. 8. We closed with a brief section on a topic that might have occurred to us at some point—what happens when we remove the resistive losses completely (by setting parallel resistance to ∞, or series resistance to 0)? We end up with an LC circuit, and we saw that we can approximate such an animal with an op amp circuit. By now the reader is likely ready to ﬁnish reviewing key concepts of the chapter, so we’ll stop here and list them, along with corresponding exam-ples in the text. ❑Circuits that contain two energy storage devices that cannot be com-bined using series-parallel combination techniques are described by a second-order differential equation. ❑Series and parallel RLC circuits fall into one of three categories, depend-ing on the relative values of R, L, and C: Overdamped α > ω0 Critically damped α = ω0 Underdamped α < ω0 (Example 9.1) ❑For series RLC circuits, α = R/2L and ω0 = 1/ √ LC. (Example 9.7) ❑For parallel RLC circuits, α = 1/2RC and ω0 = 1/ √ LC. (Example 9.1) ❑The typical form of an overdamped response is the sum of two exponen-tial terms, one of which decays more quickly than the other: e.g., A1e−t + A2e−6t. (Examples 9.2, 9.3, 9.4) ❑The typical form of a critically damped response is e−αt(A1t + A2). (Example 9.5) ❑The typical form of an underdamped response is an exponentially damped sinusoid: e−αt(B1 cos ωdt + B2 sin ωdt). (Examples 9.6, 9.7, 9.8) ❑During the transient response of an RLC circuit, energy is transferred between energy storage elements to the extent allowed by the resistive component of the circuit, which acts to dissipate the energy initially stored. (See Computer-Aided Analysis section.) ❑The complete response is the sum of the forced and natural responses. In this case the total response must be determined before solving for the constants. (Examples 9.9, 9.10) EXERCISES 363 READING FURTHER An excellent discussion of employing PSpice in the modeling of automotive suspension systems can be found in R.W. Goody, MicroSim PSpice for Windows, vol. I, 2nd ed. Englewood Cliffs, N.J.: Prentice-Hall, 1998. Many detailed descriptions of analogous networks can be found in Chap. 3 of E. Weber, Linear Transient Analysis Volume I. New York: Wiley, 1954. (Out of print, but in many university libraries.) EXERCISES 9.1 The Source-Free Parallel Circuit 1. For a certain source-free parallel RLC circuit, R = 1 k, C = 3 μF, and L is such that the circuit response is overdamped. (a) Determine the value of L. (b) Write the equation for the voltage v across the resistor if it is known that v(0−) = 9 V and dv/dt\|t=0+ = 2 V/s. 2. Element values of 10 mF and 2 nH are employed in the construction of a sim-ple source-free parallel RLC circuit. (a) Select R so that the circuit is just barely overdamped. (b) Write the equation for the resistor current if its initial value is iR(0+) = 13 pA and diE/dt\|t=0+ = 1 nA/s. 3. If a parallel RLC circuit is constructed from component values C = 16 mF and L = 1 mH, choose R such that the circuit is (a) just barely overdamped; (b) just barely underdamped; (c) critically damped. (d) Does your answer for part (a) change if the resistor tolerance is 1%? 10%? (e) Increase the exponential damping coefﬁcient for part (c) by 20%. Is the circuit now underdamped, over-damped, or still critically damped? Explain. 4. Calculate α, ω0, s1, and s2 for a source-free parallel RLC circuit if (a) R = 4 , L = 2.22 H, and C = 12.5 mF; (b) L = 1 nH, C = 1 pF, and R is 1% of the value required to make the circuit underdamped. (c) Calculate the damping ratio for the circuits of parts (a) and (b). 5. You go to construct the circuit in Exercise 1, only to ﬁnd no 1 k resistors. In fact, all you are able to locate in addition to the capacitor and inductor is a 1 meter long piece of 24 AWG soft solid copper wire. Connecting it in parallel to the two components you did ﬁnd, compute the value of α, ω0, s1, and s2, and verify that the circuit is still overdamped. 6. Consider a source-free parallel RLC circuit having α = 108 s−1, ω0 = 103 rad/s, and ω0L = 5 . (a) Show that the stated units of ω0L are correct. (b) Compute s1 and s2. (c) Write the general form of the natural response for the capacitor voltage. (d) By appropriate substitution, verify that your answer to part (c) is indeed a solution to Eq. if the inductor and capacitor each initially store 1 mJ of energy, respectively. 7. A parallel RLC circuit is constructed with R = 500 , C = 10 μF, and L such that it is critically damped. (a) Determine L. Is this value large or small for a printed-circuit board mounted component? (b) Add a resistor in parallel to the existing components such that the damping ratio is equal to 10. (c) Does in-creasing the damping ratio further lead to an overdamped, critically damped, or underdamped circuit? Explain. 9.2 The Overdamped Parallel RLC Circuit 8. The circuit of Fig. 9.2 is modiﬁed substantially, with the resistor being re-placed with a 1 k resistor, the inductor swapped out for a smaller 7 mH ver-sion, the capacitor replaced with a 1 nF alternative, and now the inductor is ini-tially discharged while the capacitor is storing 7.2 mJ. (a) Compute α, ω0, s1, and s2, and verify that the circuit is still overdamped. (b) Obtain an expression CHAPTER 9 THE RLC CIRCUIT 364 for the current ﬂowing through the resistor which is valid for t > 0. (c) Calcu-late the magnitude of the resistor current at t = 10 μs. 9. The voltage across a capacitor is found to be given by vC(t) = 10e−10t −5e−4t V. (a) Sketch each of the two components over the range of 0 ≤t ≤1.5 s. (b) Graph the capacitor voltage over the same time range. 10. The current ﬂowing through a certain inductor is found to be given by iL(t) = 0.20e−2t −0.6e−3t V. (a) Sketch each of the two components over the range of 0 ≤t ≤1.5 s. (b) Graph the inductor current over the same time range. (c) Graph the energy remaining in the inductor over 0 ≤t ≤1.5 s. 11. The current ﬂowing through a 5 resistor in a source-free parallel RLC circuit is determined to be iR(t) = 2e−t −3e−8t V, t > 0. Determine (a) the maximum current and the time at which it occurs; (b) the settling time; (c) the time t corresponding to the resistor absorbing 2.5 W of power. 12. For the circuit of Fig. 9.38, obtain an expression for vC(t) valid for all t > 0. 13. Consider the circuit depicted in Fig. 9.38. (a) Obtain an expression for iL(t) valid for all t > 0. (b) Obtain an expression for iR(t) valid for all t > 0. (c) Determine the settling time for both iL and iR. 14. With regard to the circuit represented in Fig. 9.39, determine (a) iC(0−); (b) iL(0−); (c) iR(0−); (d) vC(0−); (e) iC(0+); ( f ) iL(0+); (g) iR(0+); (h) vC(0+). 250 mF 20 k 0.1 H 6 V vC + – iL iC iR t = 0 + – 2 13 ■FIGURE 9.38 15. (a) Assuming the passive sign convention, obtain an expression for the voltage across the 1 resistor in the circuit of Fig. 9.39 which is valid for all t > 0. (b) Determine the settling time of the resistor voltage. 16. With regard to the circuit presented in Fig. 9.40, (a) obtain an expression for v(t) which is valid for all t > 0; (b) calculate the maximum inductor current and identify the time at which it occurs; (c) determine the settling time. 250 mH 1 48 vC + – iC iR iL t = 0 10u(–t) mA 2 mF ■FIGURE 9.39 0.2 4 mF v + – 5u(–t) A iC t = 0 1 mH ■FIGURE 9.40 1 H 310 mA 14 v(t) + – i(t) t = 0 360 F ■FIGURE 9.41 17. Obtain expressions for the current i(t) and voltage v(t) as labeled in the circuit of Fig. 9.41 which are valid for all t > 0. EXERCISES 365 18. Replace the 14 resistor in the circuit of Fig. 9.41 with a 1 resistor. (a) Obtain an expression for the energy stored in the capacitor as a function of time, valid for t > 0. (b) Determine the time at which the energy in the capacitor has been reduced to one-half its maximum value. (c) Verify your answer with an appropriate PSpice simulation. 19. Design a complete source-free parallel RLC circuit which exhibits an overdamped response, has a settling time of 1 s, and has a damping ratio of 15. 20. For the circuit represented by Fig. 9.42, the two resistor values are R1 = 0.752 and R2 = 1.268 , respectively. (a) Obtain an expression for the energy stored in the capacitor, valid for all t > 0; (b) determine the settling time of the cur-rent labeled iA. + – vC + – iA 1.5 V R1 5 F R2 2iA t = 0 2 H + – ■FIGURE 9.42 9.3 Critical Damping 21. A motor coil having an inductance of 8 H is in parallel with a 2 μF capacitor and a resistor of unknown value. The response of the parallel combination is determined to be critically damped. (a) Determine the value of the resistor. (b) Compute α. (c) Write the equation for the current ﬂowing into the resistor if the top node is labeled v, the bottom node is grounded, and v = Rir. (d) Verify that your equation is a solution to the circuit differential equation, dir dt + 2α dir dt + α2ir = 0 22. The condition for critical damping in an RLC circuit is that the resonant fre-quency ω0 and the exponential damping factor α are equal. This leads to the relationship L = 4R2C, which implies that 1 H = 1 2 · F. Verify this equiva-lence by breaking down each of the three units to fundamental SI units (see Chap. 2). 23. A critically damped parallel RLC circuit is constructed from component values 40 , 8 nF, and 51.2 μH, respectively. (a) Verify that the circuit is indeed criti-cally damped. (b) Explain why, in practice, the circuit once fabricated is un-likely to be truly critically damped. (c) The inductor initially stores 1 mJ of en-ergy while the capacitor is initially discharged. Determine the magnitude of the capacitor voltage at t = 500 ns, the maximum absolute capacitor voltage, and the settling time. 24. Design a complete (i.e., with all necessary switches or step function sources) parallel RLC circuit which has a critically damped response such that the capacitor voltage at t = 1 s is equal to 9 V and the circuit is source-free for all t > 0. 25. A critically damped parallel RLC circuit is constructed from component values 40 and 2 pF. (a) Determine the value of L, taking care not to overround. (b) Explain why, in practice, the circuit once fabricated is unlikely to be truly critically damped. (c) The inductor initially stores no energy while the capacitor is initially storing 10 pJ. Determine the power absorbed by the resistor at t = 2 ns, the maximum absolute inductor current \|iL\|, and the settling time. 26. For the circuit of Fig. 9.43, is(t) = 30u(−t) mA. (a) Select R1 so that v(0+) = 6 V. (b) Compute v(2 ms). (c) Determine the settling time of the capacitor voltage. (d) Is the inductor current settling time the same as your answer to part (c)? 27. The current source in Fig. 9.43 is is(t) = 10u(1 −t) μA. (a) Select R1 such that iL(0+) = 2 μA. Compute iL at t = 500 ms and t = 1.002 ms. 28. The inductor in the circuit of Fig. 9.41 is changed such that the circuit response is now critically damped. (a) Determine the new inductor value. (b) Calculate the energy stored in both the inductor and the capacitor at t = 10 ms. 29. The circuit of Fig. 9.42 is rebuilt such that the quantity controlling the dependent source is now 100iA, a 2 μF capacitor is used instead, and R1 = R2 = 10 . (a) Calculate the inductor value required to obtain a critically damped response. (b) Determine the power being absorbed by R2 at t = 300 μs. 9.4 The Underdamped Parallel RLC Circuit 30. (a) With respect to the parallel RLC circuit, derive an expression for R in terms of C and L to ensure the response is underdamped. (b) If C = 1 nF and L = 10 mH, select R such that an underdamped response is (just barely) achieved. (c) If the damping ratio is increased, does the circuit become more or less underdamped? Explain. (d) Compute α and ωd for the value of R you selected in part (b). 31. The circuit of Fig. 9.1 is constructed using component values 10 k, 72 μH, and 18 pF. (a) Compute α, ωd, and ω0. Is the circuit overdamped, critically damped, or underdamped? (b) Write the form of the natural capacitor voltage response v(t). (c) If the capacitor initially stores 1 nJ of energy, compute v at t = 300 ns. 32. The source-free circuit depicted in Fig. 9.1 is constructed using a 10 mH in-ductor, a 1 mF capacitor, and a 1.5 k resistor. (a) Calculate α, ωd, and ω0. (b) Write the equation which describes the current i for t > 0. (c) Determine the maximum value of i, and the time at which it occurs, if the inductor initially stores no energy and v(0−) = 9 V. 33. (a) Graph the current i for the circuit described in Exercise 32 for resistor values 1.5 k, 15 k, and 150 k. Make three separate graphs and be sure to extend the corresponding time axis to 6π/ωd in each case. (b) Determine the corresponding settling times. 34. Analyze the circuit described in Exercise 32 to ﬁnd v(t), t > 0, if R is equal to (a) 2 k; (b) 2 . (c) Graph both responses over the range of 0 ≤t ≤60 ms. (d) Verify your answers with appropriate PSpice simulations. 35. For the circuit of Fig. 9.44, determine (a) iC(0−); (b) iL(0−); (c) iR(0−); (d) vC(0−); (e) iC(0+); ( f ) iL(0+); (g) iR(0+); (h) vC(0+). CHAPTER 9 THE RLC CIRCUIT 366 R1 v + – t = 0 is 200 F 20 mH 5 iL ■FIGURE 9.43 20 mH 50 2 vC + – vL + – iC iR iL t = 0 3u(–t) A 2.5 F ■FIGURE 9.44 36. Obtain an expression for vL(t), t > 0, for the circuit shown in Fig. 9.44. Plot the waveform for at least two periods of oscillation. EXERCISES 367 37. For the circuit of Fig. 9.45, determine (a) the ﬁrst time t > 0 when v(t) = 0; (b) the settling time. + – 2 V 20 mH 2 mF 5 5 2 v + – t = 0 5u(–t) V + – ■FIGURE 9.45 500 m vC + – iL 2.5u(–t) A 250 mF 160 mH ■FIGURE 9.46 38. (a) Design a parallel RLC circuit that provides a capacitor voltage which oscil-lates with a frequency of 100 rad/s, with a maximum value of 10 V occurring at t = 0, and the second and third maxima both in excess of 6 V. (b) Verify your design with an appropriate PSpice simulation. 39. The circuit depicted in Fig. 9.46 is just barely underdamped. (a) Compute α and ωd. (b) Obtain an expression for iL(t) valid for t > 0. (c) Determine how much energy is stored in the capacitor, and in the inductor, at t = 200 ms. 40. When constructing the circuit of Fig. 9.46, you inadvertently install a 500 M resistor by mistake. (a) Compute α and ωd. (b) Obtain an expression for iL(t) valid for t > 0. (c) Determine how long it takes for the energy stored in the inductor to reach 10% of its maximum value. 9.5 The Source-Free Series RLC Circuit 41. The circuit of Fig. 9.21a is constructed with a 160 mF capacitor and a 250 mH inductor. Determine the value of R needed to obtain (a) a critically damped response; (b) a “just barely” underdamped response. (c) Compare your answers to parts (a) and (b) if the circuit was a parallel RLC circuit. 42. Component values of R = 2 , C = 1 mF, and L = 2 mH are used to construct the circuit represented in Fig. 9.21a. If vC(0−) = 1 V and no current initially ﬂows through the inductor, calculate i(t) at t = 1 ms, 2 ms, and 3 ms. 43. The series RLC circuit described in Exercise 42 is modiﬁed slightly by adding a 2 resistor in parallel to the existing resistor. The initial capacitor voltage remains 1 V, and there is still no current ﬂowing in the inductor prior to t = 0. (a) Calculate vC(t) at 4 ms. (b) Sketch vC(t) over the interval 0 ≤t ≤10 s. 44. The simple three-element series RLC circuit of Exercise 42 is constructed with the same component values, but the initial capacitor voltage vC(0−) = 2 V and the initial inductor current i(0−) = 1 mA. (a) Obtain an expression for i(t) valid for all t > 0. (b) Verify your solution with an appropriate simulation. 45. The series RLC circuit of Fig. 9.22 is constructed using R = 1 k, C = 2 mF, and L = 1 mH. The initial capacitor voltage vC is −4 V at t = 0−. There is no current initially ﬂowing through the inductor. (a) Obtain an expression for vC(t) valid for t > 0. (b) Sketch over 0 ≤t ≤6 μs. 46. With reference to the circuit depicted in Fig. 9.47, calculate (a) α; (b) ω0; (c) i(0+); (d) di/dt\|0+; (e) i(t) at t = 6 s. CHAPTER 9 THE RLC CIRCUIT 368 12 H 0.5 F 140 i 0.5u(–t) A ■FIGURE 9.47 9 V 30 100 t = 0 90 mH i + – 2i + – 40 F vC – + ■FIGURE 9.48 5 500 mH 1 mF iL i1 80 20i1 5u(–t) mA + – vC + – ■FIGURE 9.49 20 mF 10 H R is vR + – vC + – iL ■FIGURE 9.50 47. Obtain an equation for vC as labeled in the circuit of Fig. 9.48 valid for all t > 0. 48. With reference to the series RLC circuit of Fig. 9.48, (a) obtain an expression for i, valid for t > 0; (b) calculate i(0.8 ms) and i(4 ms); (c) verify your answers to part (b) with an appropriate PSpice simulation. 49. Obtain an expression for i1 as labeled in Fig. 9.49 which is valid for all t > 0. 9.6 The Complete Response of the RLC Circuit 50. In the series circuit of Fig. 9.50, set R = 1 . (a) Compute α and ω0. (b) If is = 3u(−t) + 2u(t) mA, determine vR(0−), iL(0−), vC(0−), vR(0+), iL(0+), vC(0+), iL(∞), and vC(∞). EXERCISES 369 51. Evaluate the derivative of each current and voltage variable labeled in Fig. 9.51 at t = 0+. 0.6 H 10 mA 20 k vC + – vL + – vR + – iR iL iC 15u(t) mA 5 nF ■FIGURE 9.51 + – 5 mF 6 mH 15 vC + – iL vs ■FIGURE 9.52 2 mH 20 nF 10 iL i1 ■FIGURE 9.53 3 2 mH 4 F 10 vC + – is ■FIGURE 9.55 0.01 H 0.5 F 6 V 1 5 vC + – t = 0 + – ■FIGURE 9.54 52. Consider the circuit depicted in Fig. 9.52. If vs(t) = −8 + 2u(t) V, determine (a) vC(0+); (b) iL(0+); (c) vC(∞); (d) vC(t = 150 ms). 53. The 15 resistor in the circuit of Fig. 9.52 is replaced with a 500 m alterna-tive. If the source voltage is given by vs = 1 −2u(t) V, determine (a) iL(0+); (b) vC(0+); (c) iL(∞); (d) vC(4 ms). 54. In the circuit shown in Fig. 9.53, obtain an expression for iL valid for all t > 0 if i1 = 8 −10u(t) mA. 55. The 10 resistor in the series RLC circuit of Fig. 9.53 is replaced with a 1 k resistor. The source i1 = 5u(t) −4 mA. Obtain an expression for iL valid for all t > 0. 56. For the circuit represented in Fig. 9.54, (a) obtain an expression for vC(t) valid for all t > 0. (b) Determine vC at t = 10 ms and t = 600 ms. (c) Verify your answers to part (b) with an appropriate PSpice simulation. 57. Replace the 1 resistor in Fig. 9.54 with a 100 m resistor, and the 5 resis-tor with a 200 m resistor. Assuming the passive sign convention, obtain an expression for the capacitor current which is valid for t > 0. 58. With regard to the circuit of Fig. 9.55, obtain an expression for vC valid for t ≥0 if is(t) = 3u(−t) + 5u(t) mA. 59. (a) Adjust the value of the 3 resistor in the circuit represented in Fig. 9.55 to obtain a “just barely” overdamped response. (b) Determine the ﬁrst instant (t > 0) at which an equal (and nonzero) amount of energy is stored in the capacitor and the inductor if is(t) = 2u(t) A. (c) Calculate the corresponding energy. (d) At what subsequent time will the energy stored in the inductor be twice that stored in the capacitor at the same instant? 9.7 The Lossless LC Circuit 60. Design an op amp circuit to model the voltage response of the LC circuit shown in Fig. 9.56. Verify your design by simulating both the circuit of Fig. 9.56 and your circuit using an LF 411 op amp, assuming v(0) = 0 and i(0) = 1 mA. 10 pH 2 nF v + – i ■FIGURE 9.56 61. Refer to Fig. 9.57, and design an op amp circuit whose output will be i(t) for t > 0. 62. Replace the capacitor in the circuit of Fig. 9.56 with a 20 H inductor in parallel with a 5 μF capacitor. Design an op amp circuit whose output will be i(t) for t > 0. Verify your design by simulating both the capacitor-inductor circuit and your op amp circuit. Use an LM111 op amp in the PSpice simulation. 63. A source-free RC circuit is constructed using a 1 k resistor and a 3.3 mF capacitor. The initial voltage across the capacitor is 1.2 V. (a) Write the differ-ential equation for v, the voltage across the capacitor, for t > 0. (b) Design an op amp circuit that provides v(t) as the output. 64. A source-free RL circuit contains a 20 resistor and a 5 H inductor. If the initial value of the inductor current is 2 A: (a) write the differential equation for i for t > 0, and (b) design an op amp integrator to provide i(t) as the output, using R1 = 1 M and Cf = 1 μF. Chapter-Integrating Exercises 65. The capacitor in the circuit of Fig. 9.58 is set to 1 F. Determine vC(t) at (a) t = −1 s; (b) t = 0+; (c) t = 20 s. CHAPTER 9 THE RLC CIRCUIT 370 67. Obtain an expression for the current labeled i1 in the circuit of Fig. 9.58 which is valid for t > 0, if the current source is replaced with a source 5u(t + 1) A. 68. Design a parallel RLC circuit which produces an exponentially damped sinu-soidal pulse with a peak voltage of 1.5 V and at least two additional peaks with voltage magnitude greater than 0.8 V. Verify your design with an appropriate PSpice simulation. 69. Design a series RLC circuit which produces an exponentially damped sinu-soidal pulse with a peak voltage of 1.5 V and at least two additional peaks with voltage magnitude greater than 0.8 V. Verify your design with an appropriate PSpice simulation. 10 H 1 1 i1 iL C + – vC + – –2i1 3u(–t) A ■FIGURE 9.58 10 H i1 1 –2i1 3u(–t) A + – iL C vC + – ■FIGURE 9.59 20 H 1 mF i(t) 2u(–t) A ■FIGURE 9.57 66. (a) What value of C for the circuit of Fig. 9.59 will result in an overdamped response? (b) Set C = 1 F and obtain an expression for iL(t) valid for t > 0. INTRODUCTION The complete response of a linear electric circuit is composed of two parts, the natural response and the forced response. The natural response is the short-lived transient response of a circuit to a sudden change in its condition. The forced response is the long-term steady-state response of a circuit to any independent sources present. Up to this point, the only forced response we have consid-ered is that due to dc sources. Another very common forcing function is the sinusoidal waveform. This function describes the voltage available at household electrical sockets as well as the volt-age of power lines connected to residential and industrial areas. In this chapter, we assume that the transient response is of little interest, and the steady-state response of a circuit (a television set, a toaster, or a power distribution network) to a sinusoidal voltage or current is needed. We will analyze such circuits using a powerful technique that transforms integrodifferential equations into algebraic equations. Before we see how that works, it’s useful to quickly review a few important attributes of general sinusoids, which will describe pretty much all currents and voltages through-out the chapter. 10.1 • CHARACTERISTICS OF SINUSOIDS Consider a sinusoidally varying voltage v(t) = Vm sin ωt shown graphically in Figs. 10.1a and b. The amplitude of the sine wave is Vm,and the argument is ωt. The radian frequency, or angular frequency, is ω. In Fig. 10.1a, Vm sin ωt is plotted as a function of the argument ωt, and the periodic nature of the sine wave is evident. KEY CONCEPTS Characteristics of Sinusoidal Functions Phasor Representation of Sinusoids Converting Between the Time and Frequency Domains Impedance and Admittance Reactance and Susceptance Parallel and Series Combinations in the Frequency Domain Determination of Forced Response Using Phasors Application of Circuit Analysis Techniques in the Frequency Domain Sinusoidal Steady-State Analysis C H A P T E R 10 371 The function repeats itself every 2π radians, and its period is therefore 2π radians. In Fig. 10.1b, Vm sin ωt is plotted as a function of t and the period is now T.Asine wave having a period T must execute 1/T periods each second; its frequency f is 1/T hertz, abbreviated Hz. Thus, f = 1 T and since ωT = 2π we obtain the common relationship between frequency and radian frequency, Lagging and Leading A more general form of the sinusoid, v(t) = Vm sin(ωt + θ) includes a phase angle θ in its argument. Equation is plotted in Fig. 10.2 as a function of ωt, and the phase angle appears as the number of radians by which the original sine wave (shown in green color in the sketch) is shifted to the left, or earlier in time. Since corresponding points on the sinusoid Vm sin(ωt + θ) occur θ rad, or θ/ω seconds, earlier, we say that Vm sin(ωt + θ) leads Vm sin ωt by θ rad. Therefore, it is correct to describe ω = 2πf CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 372 Vm –Vm 0 2 v(t) t (rad) (a) 2 3 2 2 – Vm –Vm 0 T v(t) t (s) (b) T 4 T 2 3T 4 T 4 – ■FIGURE 10.1 The sinusoidal function v(t) = Vm sin ωt is plotted (a) versus ωt and (b) versus t. Vm –Vm Vm sin (t + ) Vm sin t 2 v t ■FIGURE 10.2 The sine wave Vm sin(ωt + θ) leads Vm sin ωt by θ rad. SECTION 10.1 CHARACTERISTICS OF SINUSOIDS 373 sin ωt as lagging sin(ωt + θ) by θ rad, as leading sin(ωt + θ) by −θ rad, or as leading sin(ωt −θ) by θ rad. In either case, leading or lagging, we say that the sinusoids are out of phase. If the phase angles are equal, the sinusoids are said to be in phase. In electrical engineering, the phase angle is commonly given in degrees, rather than radians; to avoid confusion we should be sure to always use the degree symbol. Thus, instead of writing v = 100 sin 2π1000t −π 6 we customarily use v = 100 sin(2π1000t −30◦) In evaluating this expression at a speciﬁc instant of time, e.g., t = 10−4 s, 2π1000t becomes 0.2π radian, and this should be expressed as 36° before 30° is subtracted from it. Don’t confuse your apples with your oranges. Converting Sines to Cosines The sine and cosine are essentially the same function, but with a 90° phase difference. Thus, sin ωt = cos(ωt −90◦). Multiples of 360° may be added to or subtracted from the argument of any sinusoidal function without changing the value of the function. Hence, we may say that v1 = Vm1 cos(5t + 10◦) = Vm1 sin(5t + 90◦+ 10◦) = Vm1 sin(5t + 100◦) leads v2 = Vm2 sin(5t −30◦) by 130°. It is also correct to say that v1 lags v2 by 230°, since v1 may be written as v1 = Vm1 sin(5t −260◦) We assume that Vm1 and Vm2 are both positive quantities. A graphical representation is provided in Fig. 10.3; note that the frequency of both sinu-soids (5 rad/s in this case) must be the same, or the comparison is meaning-less. Normally, the difference in phase between two sinusoids is expressed by that angle which is less than or equal to 180° in magnitude. The concept of a leading or lagging relationship between two sinusoids will be used extensively, and the relationship is recognizable both mathe-matically and graphically. ■FIGURE 10.3 A graphical representation of the two sinusoids v1 and v2. The magnitude of each sine function is represented by the length of the corresponding arrow, and the phase angle by the orientation with respect to the positive x axis. In this diagram, v1 leads v2 by 100° + 30° = 130°, although it could also be argued that v2 leads v1 by 230°. It is customary, however, to express the phase difference by an angle less than or equal to 180° in magnitude. 100 –30 0 –260 v1 v2 Note that: −sin ωt = sin(ωt ± 180◦) −cos ωt = cos(ωt ± 180◦) ∓sin ωt = cos(ωt ± 90◦) ±cos ωt = sin(ωt ± 90◦) Two sinusoidal waves whose phases are to be compared must: 1. Both be written as sine waves, or both as cosine waves. 2. Both be written with positive amplitudes. 3. Each have the same frequency. Recall that to convert radians to degrees, we simply multiply the angle by 180/π. CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 374 ■FIGURE 10.4 A series RL circuit for which the forced response is desired. i L R vs (t) = Vm cos t + – PRACTICE ● 10.1 Find the angle by which i1 lags v1 if v1 = 120 cos(120πt −40◦) V and i1 equals (a) 2.5 cos(120πt + 20◦) A; (b) 1.4 sin(120πt −70◦) A; (c) −0.8 cos(120πt −110◦) A. 10.2 Find A, B, C, and φ if 40 cos(100t −40◦) −20 sin(100t + 170◦) = A cos 100t + B sin 100t = C cos(100t + φ). Ans: 10.1: −60◦; 120°; −110◦. 10.2: 27.2; 45.4; 52.9; −59.1◦. 10.2 • FORCED RESPONSE TO SINUSOIDAL FUNCTIONS Now that we are familiar with the mathematical characteristics of sinusoids, we are ready to apply a sinusoidal forcing function to a simple circuit and obtain the forced response. We will ﬁrst write the differential equation that applies to the given circuit. The complete solution of this equation is com-posed of two parts, the complementary solution (which we call the natural response) and the particular integral (or forced response). The methods we plan to develop in this chapter assume that we are not interested in the short-lived transient or natural response of our circuit, but only in the long-term or “steady-state” response. The Steady-State Response The term steady-state response is used synonymously with forced response, and the circuits we are about to analyze are commonly said to be in the “sinusoidal steady state.” Unfortunately, steady state carries the connota-tion of “not changing with time” in the minds of many students. This is true for dc forcing functions, but the sinusoidal steady-state response is deﬁ-nitely changing with time. The steady state simply refers to the condition that is reached after the transient or natural response has died out. The forced response has the mathematical form of the forcing function, plus all its derivatives and its ﬁrst integral. With this knowledge, one of the methods by which the forced response may be found is to assume a solution composed of a sum of such functions, where each function has an unknown amplitude to be determined by direct substitution in the differential equa-tion. As we are about to see, this can be a lengthy process, so we will be sufﬁciently motivated to seek out a simpler alternative. Consider the series RL circuit shown in Fig. 10.4. The sinusoidal source voltage vs = Vm cos ωt has been switched into the circuit at some remote time in the past, and the natural response has died out completely. We seek the forced (or “steady-state”) response, which must satisfy the differential equation L di dt + Ri = Vm cos ωt obtained by applying KVL around the simple loop. At any instant where the derivative is equal to zero, we see that the current must have the form i ∝cos ωt. Similarly, at an instant where the current is equal to zero, the SECTION 10.2 FORCED RESPONSE TO SINUSOIDAL FUNCTIONS 375 derivative must be proportional to cos ωt, implying a current of the form sin ωt. We might expect, therefore, that the forced response will have the general form i(t) = I1 cos ωt + I2 sin ωt where I1 and I2 are real constants whose values depend upon Vm, R, L, and ω. No constant or exponential function can be present. Substituting the assumed form for the solution in the differential equation yields L(−I1ω sin ωt + I2ω cos ωt) + R(I1 cos ωt + I2 sin ωt) = Vm cos ωt If we collect the cosine and sine terms, we obtain (−LI1ω + RI2) sin ωt + (LI2ω + RI1 −Vm) cos ωt = 0 This equation must be true for all values of t, which can be achieved only if the factors multiplying cos ωt and sin ωt are each zero. Thus, −ωLI1 + RI2 = 0 and ωLI2 + RI1 −Vm = 0 and simultaneous solution for I1 and I2 leads to I1 = RVm R2 + ω2L2 I2 = ωLVm R2 + ω2L2 Thus, the forced response is obtained: i(t) = RVm R2 + ω2L2 cos ωt + ωLVm R2 + ω2L2 sin ωt A More Compact and User-Friendly Form Although accurate, this expression is slightly cumbersome; a clearer picture of the response can be obtained by expressing it as a single sinusoid or cosinusoid with a phase angle. We choose to express the response as a cosine function, i(t) = A cos(ωt −θ) At least two methods of obtaining the values of A and θ suggest them-selves. We might substitute Eq. directly in the original differential equation, or we could simply equate the two solutions, Eqs. and . Selecting the latter method, and expanding the function cos(ωt −θ): A cos θ cos ωt + A sin θ sin ωt = RVm R2 + ω2L2 cos ωt + ωLVm R2 + ω2L2 sin ωt All that remains is to collect terms and perform a bit of algebra, an exercise left to the reader. The result is θ = tan−1 ωL R and A = Vm √ R2 + ω2L2 and so the alternative form of the forced response therefore becomes i(t) = Vm √ R2 + ω2L2 cos ωt −tan−1 ωL R Several useful trigonometric identities are provided on the inside cover of the book. With this form, it is easy to see that the amplitude of the response is proportional to the amplitude of the forcing function; if not, the linearity concept would have to be discarded. The current is seen to lag the applied voltage by tan−1 (ωL/R), an angle between 0 and 90°. When ω = 0 or L = 0, the current must be in phase with the voltage; since the former situ-ation is direct current and the latter provides a resistive circuit, the result agrees with our previous experience. If R = 0, the current lags the voltage by 90°. In an inductor, then, if the passive sign convention is satisﬁed, the current lags the voltage by exactly 90°. In a similar manner we can show that the current through a capacitor leads the voltage across it by 90°. The phase difference between the current and voltage depends upon the ratio of the quantity ωL to R. We call ωL the inductive reactance of the in-ductor; it is measured in ohms, and it is a measure of the opposition that is offered by the inductor to the passage of a sinusoidal current. CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 376 Find the current iL in the circuit shown in Fig. 10.5a, if the transients have already died out. EXAMPLE 10.1 iL (a) 100 30 mH 25 10 cos 103t V + – (b) b a 100 25 10 cos 103t V + – voc + – 20 (c) 8 cos 103t V 30 mH iL + – ■FIGURE 10.5 (a) The circuit for Example 10.1, in which the current iL is desired. (b) The Thévenin equivalent is desired at terminals a and b. (c) The simpliﬁed circuit. Once upon a time, the symbol E (for electromotive force) was used to designate voltages. Then every stu-dent learned the phase “ELI the ICE man” as a reminder that voltage leads current in an inductive circuit, while current leads voltage in a capacitive circuit. Now that we use V instead, it just isn’t the same. Although this circuit has a sinusoidal source and a single inductor, it contains two resistors and is not a single loop. In order to apply the results of the preceding analysis, we need to seek the Thévenin equivalent as viewed from terminals a and b in Fig. 10.5b. The open-circuit voltage voc is voc = (10 cos 103t) 100 100 + 25 = 8 cos 103t V SECTION 10.2 FORCED RESPONSE TO SINUSOIDAL FUNCTIONS 377 Since there are no dependent sources in sight, we ﬁnd Rth by shorting out the independent source and calculating the resistance of the passive network, so Rth = (25 × 100)/(25 + 100) = 20 . Now we do have a series RL circuit, with L = 30 mH, Rth = 20 , and a source voltage of 8 cos 103t V, as shown in Fig. 10.5c. Thus, applying Eq. , which was derived for a general RL series circuit, iL = 8 202 + (103 × 30 × 10−3)2 cos 103t −tan−1 30 20 = 222 cos(103t −56.3◦) mA The voltage and current waveforms are plotted in Fig. 10.6. ■FIGURE 10.6 Voltage and current waveforms on a dual axis plot, generated using MATLAB: EDU» t = linspace(0,8e-3,1000); EDU» v = 8cos(1000t); EDU» i = 0.222cos(1000t −56.3pi/180); EDU» plotyy(t,v,t,i); EDU» xlabel(‘time (s)’); ■FIGURE 10.7 iR 3 k 100 mH 1 k vs + – is iL vL + – Note that there is not a 90° phase difference between the current and voltage waveforms of the plot. This is because we are not plotting the inductor voltage, which is left as an exercise for the reader. PRACTICE ● 10.3 Let vs = 40 cos 8000t V in the circuit of Fig. 10.7. Use Thévenin’s theorem where it will do the most good, and ﬁnd the value at t = 0 for (a) iL; (b) vL; (c) iR; (d) is. Ans: 18.71 mA; 15.97 V; 5.32 mA; 24.0 mA. 10.3 • THE COMPLEX FORCING FUNCTION The method we just employed works—the correct answer is obtained in a straightforward manner. However, it isn’t particularly graceful, and after being applied to a few circuits, it remains as clunky and cumbersome as the ﬁrst time we use it. The real problem isn’t the time-varying source—it’s the inductor (or capacitor), since a purely resistive circuit is no more difﬁcult to analyze with sinusoidal sources than with dc sources, as only algebraic equations result. It turns out that if the transient response is of no interest to us, there is an alternative approach for obtaining the sinusoidal steady-state response of any linear circuit. The distinct advantage of this alternative is that it allows us to relate the current and voltage associated with any element using a simple algebraic expression. The basic idea is that sinusoids and exponentials are related through complex numbers. Euler’s identity, for example, tells us that e jθ = cos θ + j sin θ Whereas taking the derivative of a cosine function yields a (negative) sine function, the derivative of an exponential is simply a scaled version of the same exponential. If at this point the reader is thinking, “All this is great, but there are no imaginary numbers in any circuit I ever plan to build!” that may be true. What we’re about to see, however, is that adding imaginary sources to our circuits leads to complex sources which (surprisingly) simplify the analysis process. It might seem like a strange idea at ﬁrst, but a moment’s reﬂection should remind us that superposition requires any imaginary sources we might add to cause only imaginary responses, and real sources can only lead to real responses. Thus, at any point, we should be able to separate the two by simply taking the real part of any complex voltage or current. In Fig. 10.8, a sinusoidal source Vm cos(ωt + θ) is connected to a general network, which we will assume to contain only passive elements (i.e., no independent sources) in order to keep things simple. A current response in some other branch of the network is to be determined, and the parameters appearing in Eq. are all real quantities. CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 378 ■FIGURE 10.8 The sinusoidal forcing function Vm cos(ωt + θ) produces the steady-state sinusoidal response Im cos(ωt + φ). Im cos (t + ) Vm cos (t + ) N + – Appendix 5 deﬁnes the complex number and related terms, reviews complex arithmetic, and develops Euler’s identity and the relationship between exponential and polar forms. We have shown that we may represent the response by the general co-sine function Im cos(ωt + φ) A sinusoidal forcing function always produces a sinusoidal forced response of the same frequency in a linear circuit. SECTION 10.3 THE COMPLEX FORCING FUNCTION 379 Now let us change our time reference by shifting the phase of the forc-ing function by 90°, or changing the instant that we call t = 0. Thus, the forcing function Vm cos(ωt + θ −90◦) = Vm sin(ωt + θ) when applied to the same network will produce a corresponding response Im cos(ωt + φ −90◦) = Im sin(ωt + φ) We next depart from physical reality by applying an imaginary forcing function, one that cannot be applied in the laboratory but can be applied mathematically. Imaginary Sources Lead to . . . Imaginary Responses We construct an imaginary source very simply; it is only necessary to mul-tiply Eq. by j, the imaginary operator. We thus apply jVm sin(ωt + θ) What is the response? If we had doubled the source, then the principle of linearity would require that we double the response; multiplication of the forcing function by a constant k would result in the multiplication of the response by the same constant k. The fact that our constant is √−1 does not destroy this relationship. The response to the imaginary source of Eq. is thus jIm sin(ωt + φ) The imaginary source and response are indicated in Fig. 10.9. ■FIGURE 10.9 The imaginary sinusoidal forcing function jVm sin(ωt + θ) produces the imaginary sinusoidal response jIm sin(ωt + φ) in the network of Fig. 10.8. jIm sin (t + ) jVm sin (t + ) N + – Applying a Complex Forcing Function We have applied a real source and obtained a real response; we have also applied an imaginary source and obtained an imaginary response. Since we are dealing with a linear circuit, we may use the superposition theorem to ﬁnd the response to a complex forcing function which is the sum of the real and imaginary forcing functions. Thus, the sum of the forcing functions of Eqs. and , Vm cos(ωt + θ) + jVm sin(ωt + θ) must produce a response that is the sum of Eqs. and , Im cos(ωt + φ) + jIm sin(ωt + φ) Electrical engineers use “j” instead of “i” to represent √ −1 to avoid confusion with currents. The complex source and response may be represented more simply by ap-plying Euler’s identity, i.e., cos(ωt + θ) + j sin(ωt + θ) = e j(ωt+θ). Thus, the source of Eq. may be written as Vme j(ωt+θ) and the response of Eq. is Ime j(ωt+φ) The complex source and response are illustrated in Fig. 10.10. CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 380 ■FIGURE 10.10 The complex forcing function V me j (ωt+θ) produces the complex response I me j (ωt+θ) in the network of Fig. 10.8. Ime j (t + ) Vme j (t + ) N + – ■FIGURE 10.11 A simple circuit in the sinusoidal steady state is to be analyzed by the application of a complex forcing function. i L R vs = Vm cos t + – Again, linearity assures us that the real part of the complex response is produced by the real part of the complex forcing function, while the imagi-nary part of the response is caused by the imaginary part of the complex forcing function. Our plan is that instead of applying a real forcing function to obtain the desired real response, we will substitute a complex forcing function whose real part is the given real forcing function; we expect to obtain a complex response whose real part is the desired real response. The advantage of this procedure is that the integrodifferential equations describ-ing the steady-state response of a circuit will now become simple algebraic equations. An Algebraic Alternative to Differential Equations Let’s try out this idea on the simple RL series circuit shown in Fig. 10.11. The real source Vm cos ωt is applied; the real response i(t) is desired. Since Vm cos ωt = Re{Vm cos ωt + jVm sin ωt} = Re{Vme jωt} the necessary complex source is Vme jωt We express the complex response that results in terms of an unknown amplitude Im and an unknown phase angle φ: Ime j(ωt+φ) Writing the differential equation for this particular circuit, Ri + L di dt = vs SECTION 10.3 THE COMPLEX FORCING FUNCTION 381 we insert our complex expressions for vs and i: RIme j(ωt+φ) + L d dt (Ime j(ωt+φ)) = Vme jωt take the indicated derivative: RIme j(ωt+φ) + jωLIme j(ωt+φ) = Vme jωt and obtain an algebraic equation. In order to determine the values of Im and φ, we divide throughout by the common factor e jωt: RIme jφ + jωLIme jφ = Vm factor the left side: Ime jφ(R + jωL) = Vm rearrange: Ime jφ = Vm R + jωL and identify Im and φ by expressing the right side of the equation in expo-nential or polar form: Ime jφ = Vm √ R2 + ω2L2 e j[−tan−1(ωL/R)] Thus, Im = Vm √ R2 + ω2L2 and φ = −tan−1 ωL R In polar notation, this may be written as Im φ or Vm/ R2 + ω2L2 −tan−1(ωL/R) The complex response is given by Eq. . Since Im and φ are readily iden-tiﬁed, we can write the expression for i(t) immediately. However, if we feel like using a more rigorous approach, we may obtain the real response i(t) by reinserting the e jωt factor on both sides of Eq. and taking the real part. Either way, we ﬁnd that i(t) = Im cos(ωt + φ) = Vm √ R2 + ω2L2 cos ωt −tan−1 ωL R which agrees with the response obtained in Eq. for the same circuit. CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 382 EXAMPLE 10.2 (If you have trouble working this practice problem, turn to Appendix 5.) For the simple RC circuit of Fig. 10.12a, substitute an appropriate complex source and use it to solve for the steady-state capacitor voltage. Since the real source is 3 cos 5t, we “replace” it with a complex source 3e j5t V. We’ll call the new capacitor voltage vC2 and deﬁne a capacitor current iC2 consistent with the passive sign convention (Fig. 10.12b). The differential equation can be now obtained by simple application of KVL, −3e j5t + 1iC2 + vC2 = 0 or −3e j5t + 2 dvC2 dt + vC2 = 0 We anticipate a steady-state response of the same form as our source; in other words, vC2 = Vme j5t Substituting this into our differential equation and rearranging terms yields j10Vme j5t + Vme j5t = 3e j5t Canceling the exponential term, we ﬁnd that Vm = 3 1 + j10 = 3 √ 1 + 102 −tan−1(10/1) V and our steady-state capacitor voltage is given by Re{vC2} = Re{29.85e– j84.3◦e j5t mV} = 29.85 cos(5t −84.3◦) mV PRACTICE ● 10.4 Evaluate and express the result in rectangular form: (a) (2/30◦)(5/−110◦); (b) (5/−200◦) + 4/20◦. Evaluate and express the result in polar form: (c) (2 −j7)/(3 −j); (d) 8 −j4 + [(5/80◦)/(2/20◦)]. 10.5 If the use of the passive sign convention is speciﬁed, ﬁnd the (a) complex voltage that results when the complex current 4e j800t A is applied to the series combination of a 1 mF capacitor and a 2 resistor; (b) complex current that results when the complex voltage 100e j2000t V is applied to the parallel combination of a 10 mH inductor and a 50 resistor. Ans: 10.4: 21.4 −j6.38; −0.940 + j3.08; 2.30/−55.6◦; 9.43/−11.22◦. 10.5: 9.43e j(800t−32.0◦) V; 5.39e j(2000t−68.2◦) A. ■FIGURE 10.12 (a) An RC circuit for which the sinusoidal steady-state capacitor voltage is required. (b) Modiﬁed circuit, with the real source replaced with a complex source. + – vC + – 1 2 F 3 cos 5t V (a) + – vC2 + – 1 2 F 3ej5t V (b) iC2 SECTION 10.4 THE PHASOR 383 10.4 • THE PHASOR In the last section, we saw that the addition of an imaginary sinusoidal source led to algebraic equations which describe the sinusoidal steady-state re-sponse of a circuit. An intermediate step of our analysis was the “canceling” of the complex exponential term—once its derivative was taken, we appar-ently had no further use for it until the real form of the response was desired. Even then, it was possible to read the magnitude and phase angle directly from our analysis, and hence skip the step where we overtly take the real part. Another way of looking at this is that every voltage and current in our circuit contain the same factor e jωt, and the frequency, although relevant to our analysis, does not change as we move through the circuit. Dragging it around, then, is a bit of a waste of time. Looking back at Example 10.2, then, we could represent our source as 3e j0◦V (or even just 3 V) and our capacitor voltage as Vme jφ, which we ultimately found was 0.02985e−j84.3◦V. Knowledge of the source frequency is implicit here; without it, we are unable to reconstruct any voltage or current. These complex quantities are usually written in polar form rather than exponential form in order to achieve a slight additional saving of time and effort. For example, a source voltage v(t) = Vm cos ωt = Vm cos(ωt + 0◦) we now represent in complex form as Vm/0◦ and its current response i(t) = Im cos(ωt + φ) becomes Im φ This abbreviated complex representation is called a phasor.1 Let us review the steps by which a real sinusoidal voltage or current is transformed into a phasor, and then we will be able to deﬁne a phasor more meaningfully and to assign a symbol to represent it. A real sinusoidal current i(t) = Im cos(ωt + φ) is expressed as the real part of a complex quantity by invoking Euler’s identity i(t) = Re Ime j(ωt+φ) We then represent the current as a complex quantity by dropping the in-struction Re{}, thus adding an imaginary component to the current without affecting the real component; further simpliﬁcation is achieved by sup-pressing the factor e jωt: I = Ime jφ and writing the result in polar form: I = Im φ ej0 = cos 0 + j sin 0 = 1 Remember that none of the steady-state circuits we are considering will respond at a frequency other than that of the excitation source, so that the value of ω is always known. (1) Not to be confused with the phaser, an interesting device featured in a popular television series. . . . The process by which we change i(t) into I is called a phasor transformation from the time domain to the frequency domain. Several useful trigonometric identities are provided on the inside cover for convenience. EXAMPLE 10.3 Transform the time-domain voltage v(t) 100 cos(400t −30°) volts into the frequency domain. The time-domain expression is already in the form of a cosine wave with a phase angle. Thus, suppressing ω = 400 rad/s, V = 100/−30◦volts Note that we skipped several steps in writing this representation di-rectly. Occasionally, this is a source of confusion for students, as they may forget that the phasor representation is not equal to the time-domain voltage v(t). Rather, it is a simpliﬁed form of a complex function formed by adding an imaginary component to the real function v(t). PRACTICE ● 10.7 Transform each of the following functions of time into phasor form: (a) −5 sin(580t −110◦); (b) 3 cos 600t −5 sin(600t + 110◦); (c) 8 cos(4t −30◦) + 4 sin(4t −100◦). Hint: First convert each into a single cosine function with a positive magnitude. Ans: 5/−20◦; 2.41/−134.8◦; 4.46/−47.9◦. This abbreviated complex representation is the phasor representation; phasors are complex quantities and hence are printed in boldface type. Cap-ital letters are used for the phasor representation of an electrical quantity because the phasor is not an instantaneous function of time; it contains only amplitude and phase information. We recognize this difference in view-point by referring to i(t) as a time-domain representation and terming the phasor I a frequency-domain representation. It should be noted that the frequency-domain expression of a current or voltage does not explicitly in-clude the frequency. The process of returning to the time domain from the frequency domain is exactly the reverse of the previous sequence. Thus, given the phasor voltage V = 115/−45◦volts and the knowledge that ω = 500 rad/s, we can write the time-domain equivalent directly: v(t) = 115 cos(500t −45◦) volts If desired as a sine wave, v(t) could also be written v(t) = 115 sin(500t + 45◦) volts CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 384 PRACTICE ● 10.6 Let ω = 2000 rad/s and t = 1 ms. Find the instantaneous value of each of the currents given here in phasor form: (a) j10 A; (b) 20 + j10 A; (c) 20 + j(10/20◦) A. Ans: −9.09 A; −17.42 A; −15.44 A. i(t) = Im cos (t + ) i(t) = Re{Imej(t + )} I = Ime j I = Im ■FIGURE 10.13 A resistor and its associated voltage and current in (a) the time domain, v = Ri; and (b) the frequency domain, V = RI. i v = Ri + – (a) R I V = RI + – (b) R SECTION 10.4 THE PHASOR 385 Ohm’s law holds true both in the time domain and in the frequency domain. In other words, the voltage across a resistor is always given by the resistance times the current ﬂowing through the element. The real power of the phasor-based analysis technique lies in the fact that it is possible to deﬁne algebraic relationships between the voltage and current for inductors and capacitors, just as we have always been able to do in the case of resistors. Now that we are able to transform into and out of the frequency domain, we can proceed to our simplification of sinusoidal steady-state analysis by establishing the relationship between the phasor voltage and phasor current for each of the three passive elements. The Resistor The resistor provides the simplest case. In the time domain, as indicated by Fig. 10.13a, the deﬁning equation is v(t) = Ri(t) Now let us apply the complex voltage v(t) = Vme j(ωt+θ) = Vm cos(ωt + θ) + jVm sin(ωt + θ) and assume the complex current response i(t) = Ime j(ωt+φ) = Im cos(ωt + φ) + jIm sin(ωt + φ) so that Vme j(ωt+θ) = Ri(t) = RIme j(ωt+φ) Dividing throughout by e jωt, we ﬁnd Vme jθ = RIme jφ or, in polar form, Vm/θ = RIm φ But Vm/θ and Im φ merely represent the general voltage and current pha-sors V and I. Thus, V = RI The voltage-current relationship in phasor form for a resistor has the same form as the relationship between the time-domain voltage and cur-rent. The deﬁning equation in phasor form is illustrated in Fig. 10.13b. The angles θ and φ are equal, so that the current and voltage are always in phase. As an example of the use of both the time-domain and frequency-domain relationships, let us assume that a voltage of 8 cos(100t −50◦) V is across a 4 resistor. Working in the time domain, we ﬁnd that the current must be i(t) = v(t) R = 2 cos(100t −50◦) A The phasor form of the same voltage is 8/−50◦V, and therefore I = V R = 2/−50◦ A If we transform this answer back to the time domain, it is evident that the same expression for the current is obtained. We conclude that there is no saving in time or effort when a resistive circuit is analyzed in the frequency domain. The Inductor Let us now turn to the inductor. The time-domain representation is shown in Fig. 10.14a, and the deﬁning equation, a time-domain expression, is v(t) = L di(t) dt After substituting the complex voltage equation and complex current equation in Eq. , we have Vme j(ωt+θ) = L d dt Ime j(ωt+φ) Taking the indicated derivative: Vme j(ωt+θ) = jωLIme j(ωt+φ) and dividing through by e jωt: Vme jθ = jωLIme jφ we obtain the desired phasor relationship V = jωLI The time-domain differential equation has become the algebraic equation in the frequency domain. The phasor relationship is indicated in Fig. 10.14b. Note that the angle of the factor jωL is exactly +90◦and that I must therefore lag V by 90° in an inductor. CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 386 ■FIGURE 10.14 An inductor and its associated voltage and current in (a) the time domain, v = Ldi/dt; and (b) the frequency domain, V = jωLI. di dt i v = L + – (a) L I V = jLI + – (b) L Apply the voltage 8/−50° V at a frequency ω 100 rad/s to a 4 H inductor, and determine the phasor current and the time-domain current. We make use of the expression we just obtained for the inductor, I = V jωL = 8/−50◦ j100(4) = −j0.02/−50◦= (1/−90◦)(0.02/−50◦) or I = 0.02/−140◦A If we express this current in the time domain, it becomes i(t) = 0.02 cos(100t −140◦) A = 20 cos(100t −140◦) mA EXAMPLE 10.4 SECTION 10.4 THE PHASOR 387 The Capacitor The ﬁnal element to consider is the capacitor. The time-domain current-voltage relationship is i(t) = C dv(t) dt The equivalent expression in the frequency domain is obtained once more by letting v(t) and i(t) be the complex quantities of Eqs. and , tak-ing the indicated derivative, suppressing e jωt, and recognizing the phasors V and I. Doing this, we ﬁnd I = jωCV Thus, I leads V by 90° in a capacitor. This, of course, does not mean that a current response is present one-quarter of a period earlier than the voltage that caused it! We are studying steady-state response, and we ﬁnd that the current maximum is caused by the increasing voltage that occurs 90° earlier than the voltage maximum. The time-domain and frequency-domain representations are compared in Fig. 10.15a and b. We have now obtained the V-I relationships for the three passive elements. These results are summarized in Table 10.1, where the time-domain v-i expressions and the frequency-domain V-I relation-ships are shown in adjacent columns for the three circuit elements. All the phasor equations are algebraic. Each is also linear, and the equations relat-ing to inductance and capacitance bear a great similarity to Ohm’s law. In fact, we will indeed use them as we use Ohm’s law. ■FIGURE 10.15 (a) The time-domain and (b) the frequency-domain relationships between capacitor current and voltage. i = C v + – (a) C dv dt I = jCV V + – (b) C R i v + – L i v + – C i v + – R I V + – V + – V + – I I jL 1/jC TABLE 10.1 Comparison of Time-Domain and Frequency-Domain ● Voltage-Current Expressions Time Domain Frequency Domain v = Ri V = RI v = L di dt V = jωLI v = 1 C i dt V = 1 jωC I Kirchhoff’s Laws Using Phasors Kirchhoff’s voltage law in the time domain is v1(t) + v2(t) + · · · + vN(t) = 0 We now use Euler’s identity to replace each real voltage vi by a complex voltage having the same real part, suppress e jωt throughout, and obtain V1 + V2 + · · · + VN = 0 Thus, we see that Kirchhoff’s voltage law applies to phasor voltages just as it did in the time domain. Kirchhoff’s current law can be shown to hold for phasor currents by a similar argument. CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 388 ■FIGURE 10.17 A three-mesh circuit. Each source operates at the same frequency ω. IC IR2 IL 2 2 H 1 F 1 Vs Is + – IR1 ■FIGURE 10.16 The series RL circuit with a phasor voltage applied. VL + – VR + – + – Vs L R I Now let us look brieﬂy at the series RL circuit that we have considered several times before. The circuit is shown in Fig. 10.16, and a phasor cur-rent and several phasor voltages are indicated. We may obtain the desired response, a time-domain current, by ﬁrst ﬁnding the phasor current. From Kirchhoff’s voltage law, VR + VL = Vs and using the recently obtained V-I relationships for the elements, we have RI + jωLI = Vs The phasor current is then found in terms of the source voltage Vs: I = Vs R + jωL Let us select a source-voltage amplitude of Vm and phase angle of 0°. Thus, I = Vm/0◦ R + jωL The current may be transformed to the time domain by ﬁrst writing it in polar form: I = Vm √ R2 + ω2L2 [−tan−1(ωL/R)] and then following the familiar sequence of steps to obtain in a very simple manner the same result we obtained the “hard way’’ earlier in this chapter. EXAMPLE 10.5 For the RLC circuit of Fig. 10.17, determine Is and is(t) if both sources operate at ω = 2 rad/s, and IC = 2̸ 28 A. The fact that we are given IC and asked for Is is all the prompting we need to consider applying KCL. If we label the capacitor voltage VC consistent with the passive sign convention, then VC = 1 jωC IC = −j 2 IC = −j 2 (2/28◦) = (0.5/−90◦)(2/28◦) = 1/−62◦V This voltage also appears across the 2 resistor, so that the current IR2 ﬂowing downward through that branch is IR2 = 1 2VC = 1 2/−62◦A KCL then yields Is = IR2 + IC = 1/−62◦+ 1 2/−62◦= (3/2)/−62◦A. (We should note the addition of these polar quantities was trivial since the resistor and capacitor currents have the same angle, i.e., are in phase.) Thus Is and knowledge of ω permit us to write is(t) directly: is(t) = 1.5 cos (2t −62◦) A PRACTICE ● 10.8 In the circuit of Fig. 10.17, both sources operate at ω = 1 rad/s. If IC = 2/28◦A and IL = 3/53◦A, calculate (a) Is; (b) Vs; (c) iR1(t). Ans: 3/−62◦A; (b) 3.71/−4.5◦V; (c) 3.22 cos (t −4.5◦) A. SECTION 10.5 IMPEDANCE AND ADMITTANCE 389 10.5 • IMPEDANCE AND ADMITTANCE The current-voltage relationships for the three passive elements in the fre-quency domain are (assuming that the passive sign convention is satisﬁed) V = RI V = jωLI V = I jωC If these equations are written as phasor voltage/phasor current ratios V I = R V I = jωL V I = 1 jωC we ﬁnd that these ratios are simple quantities that depend on element values (and frequency also, in the case of inductance and capacitance). We treat these ratios in the same manner that we treat resistances, provided we remember that they are complex quantities. Let us deﬁne the ratio of the phasor voltage to the phasor current as impedance, symbolized by the letter Z. The impedance is a complex quan-tity having the dimensions of ohms. Impedance is not a phasor and cannot be transformed to the time domain by multiplying by e jωt and taking the real part. Instead, we think of an inductor as being represented in the time domain by its inductance L and in the frequency domain by its impedance jωL. A capacitor in the time domain has a capacitance C; in the frequency domain, it has an impedance 1/jωC. Impedance is a part of the frequency domain and not a concept that is a part of the time domain. Series Impedance Combinations The validity of Kirchhoff’s two laws in the frequency domain leads to the fact that impedances may be combined in series and parallel by the same rules we established for resistances. For example, at ω = 10 × 103 rad/s, a 5 mH inductor in series with a 100 μF capacitor may be replaced by the sum of the individual impedances. The impedance of the inductor is ZL = jωL = j50 and the impedance of the capacitor is ZC = 1 jωC = −j ωC = −j1 The impedance of the series combination is therefore Zeq = ZL + ZC = j50 −j1 = j49 The impedance of inductors and capacitors is a function of frequency, and this equivalent impedance is thus applicable only at the single frequency at which it was calculated, ω = 10,000 rad/s. If we change the frequency to ω = 5000 rad/s, for example, Zeq = j23 . Parallel Impedance Combinations The parallel combination of the 5 mH inductor and the 100 μF capacitor at ω = 10,000 rad/s is calculated in exactly the same fashion in which we Note that 1 j −j. ZR = R ZL = j ωL ZC = 1 j ωC calculated parallel resistances: Zeq = ( j50)(−j1) j50 −j1 = 50 j49 = −j1.020 At ω = 5000 rad/s, the parallel equivalent is −j2.17 . Reactance Of course, we may choose to express impedance in either rectangular (Z = R + jX) or polar (Z = \|Z\|/θ) form. In rectangular form, we can see clearly the real part which arises only from real resistances, and an imagi-nary component, termed the reactance, which arises from the energy storage elements. Both resistance and reactance have units of ohms, but reactance will always depend upon frequency. An ideal resistor has zero reactance; an ideal inductor or capacitor is purely reactive (i.e., character-ized by zero resistance). Can a series or parallel combination include both a capacitor and an inductor, and yet have zero reactance? Sure! Consider the series connection of a 1 resistor, a 1 F capacitor, and a 1 H inductor driven at ω = 1 rad/s. Zeq = 1 −j(1)(1) + j(1)(1) = 1 . At that partic-ular frequency, the equivalent is a simple 1 resistor. However, even small deviations from ω = 1 rad/s lead to nonzero reactance. CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 390 EXAMPLE 10.6 Determine the equivalent impedance of the network shown in Fig. 10.18a, given an operating frequency of 5 rad/s. ■FIGURE 10.18 (a) A network that is to be replaced by a single equivalent impedance. (b) The elements are replaced by their impedances at ω = 5 rad/s. 10 6 (a) 500 mF 2 H 200 mF 10 6 (b) –j0.4 j10 –j We begin by converting the resistors, capacitors, and inductor into the corresponding impedances as shown in Fig. 10.18b. SECTION 10.5 IMPEDANCE AND ADMITTANCE 391 It is important to note that the resistive component of the impedance is not necessarily equal to the resistance of the resistor that is present in the net-work. For example, a 10 resistor and a 5 H inductor in series at ω = 4 rad/s have an equivalent impedance Z = 10 + j20 , or, in polar form, 22.4/63.4◦. In this case, the resistive component of the impedance is equal to the resistance of the series resistor because the network is a simple series network. However, if these same two elements are placed in parallel, the equivalent impedance is 10( j20)/(10 + j20) , or 8 + j4 . The resistive component of the impedance is now 8 . Upon examining the resulting network, we observe that the 6 im-pedance is in parallel with −j0.4 . This combination is equivalent to (6)(−j0.4) 6 −j0.4 = 0.02655 −j0.3982 which is in series with both the −j and j10 impedances, so that we have 0.0265 −j0.3982 −j + j10 = 0.02655 + j8.602 This new impedance is in parallel with 10 , so that the equivalent impedance of the network is 10 ∥(0.02655 + j8.602) = 10(0.02655 + j8.602) 10 + 0.02655 + j8.602 = 4.255 + j4.929 Alternatively, we can express the impedance in polar form as 6.511/49.20◦. PRACTICE ● 10.9 With reference to the network shown in Fig. 10.19, ﬁnd the input impedance Zin that would be measured between terminals: (a) a and g; (b) b and g; (c) a and b. ■FIGURE 10.19 10 20 mH 5 mH = 1000 rad/s 100 F 200 F a b g g Ans: 2.81 + j4.49 ; 1.798 −j1.124 ; 0.1124 −j3.82 . CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 392 EXAMPLE 10.7 Find the current i(t) in the circuit shown in Fig. 10.20a. ■FIGURE 10.20 (a) An RLC circuit for which the sinusoidal forced response i(t) is desired. (b) The frequency-domain equivalent of the given circuit at ω 3000 rad/s. 1.5 k 1 k H + – (a) vs(t) = 40 sin 3000t V i(t) 1 3 F 1 6 1.5 k 1 k j1 k –j2 k + – (b) I V s = 40 –90 V Identify the goal of the problem. We need to ﬁnd the sinusoidal steady-state current ﬂowing through the 1.5 k resistor due to the 3000 rad/s voltage source. Collect the known information. We begin by drawing a frequency-domain circuit. The source is trans-formed to the frequency-domain representation 40/−90◦V, the fre-quency domain response is represented as I, and the impedances of the inductor and capacitor, determined at ω = 3000 rad/s, are j k and −j2 k, respectively. The corresponding frequency-domain circuit is shown in Fig. 10.20b. Devise a plan. We will analyze the circuit of Fig. 10.20b to obtain I; combining im-pedances and invoking Ohm’s law is one possible approach. We will then make use of the fact that we know ω = 3000 rad/s to convert I into a time-domain expression. Construct an appropriate set of equations. Zeq = 1.5 + ( j)(1 −2 j) j + 1 −2 j = 1.5 + 2 + j 1 −j = 1.5 + 2 + j 1 −j 1 + j 1 + j = 1.5 + 1 + j3 2 = 2 + j1.5 = 2.5/36.87◦k SECTION 10.5 IMPEDANCE AND ADMITTANCE 393 Before we begin to write great numbers of equations in the time domain or in the frequency domain, it is very important that we shun the construc-tion of equations that are partly in the time domain, partly in the frequency domain, and wholly incorrect. One clue that a faux pas of this type has been committed is the sight of both a complex number and a t in the same equa-tion, except in the factor e jωt. And, since e jωt plays a much bigger role in derivations than in applications, it is pretty safe to say that students who ﬁnd they have just created an equation containing j and t, or / and t, have cre-ated a monster that the world would be better off without. The phasor current is then simply I = Vs Zeq Determine if additional information is required. Substituting known values, we ﬁnd that I = 40/−90◦ 2.5/36.87◦mA which, along with the knowledge that ω = 3000 rad/s, is sufﬁcient to solve for i(t). Attempt a solution. This complex expression is easily simpliﬁed to a single complex number in polar form: I = 40 2.5/−90◦−36.87◦mA = 16.00/−126.9◦mA Upon transforming the current to the time domain, the desired response is obtained: i(t) = 16 cos(3000t −126.9◦) mA Verify the solution. Is it reasonable or expected? The effective impedance connected to the source has an angle of +36.87°, indicating that it has a net inductive character, or that the current will lag the voltage. Since the voltage source has a phase angle of −90◦(once converted to a cosine source), we see that our answer is consistent. PRACTICE ● 10.10 In the frequency-domain circuit of Fig. 10.21, ﬁnd (a) I1; (b) I2; (c) I3. Ans: 28.3/45◦A; 20/90◦A; 20/0◦A. ■FIGURE 10.21 I2 I3 5 + – 100 0 V –j5 j5 I1 For example, a few equations back we saw I = Vs Zeq = 40/−90◦ 2.5/36.9◦= 16/−126.9◦mA Please do not try anything like the following: i(t) = 40 sin 3000t 2.5/36.9◦ or i(t) = 40 sin 3000t 2 + j1.5 Admittance Although the concept of impedance is very useful, and familiar in a way based on our experience with resistors, the reciprocal is often just as valu-able. We deﬁne this quantity as the admittance Y of a circuit element or passive network, and it is simply the ratio of current to voltage: The real part of the admittance is the conductance G, and the imaginary part is the susceptance B. All three quantities (Y, G, and B) are measured in siemens. The real part of the admittance is the conductance G, and the imaginary part of the admittance is the susceptance B. Thus, Y = G + jB = 1 Z = 1 R + jX Equation should be scrutinized carefully; it does not state that the real part of the admittance is equal to the reciprocal of the real part of the impedance, or that the imaginary part of the admittance is equal to the reci-procal of the imaginary part of the impedance! PRACTICE ● 10.11 Determine the admittance (in rectangular form) of (a) an impedance Z = 1000 + j400 ; (b) a network consisting of the parallel combination of an 800 resistor, a 1 mH inductor, and a 2 nF capacitor, if ω = 1 Mrad/s; (c) a network consisting of the series com-bination of an 800 resistor, a 1 mH inductor, and a 2 nF capacitor, if ω = 1 Mrad/s. Ans: 0.862 −j0.345 mS; 1.25 + j1 mS; 0.899 −j0.562 mS. 10.6 • NODAL AND MESH ANALYSIS We previously achieved a great deal with nodal and mesh analysis techniques, and it’s reasonable to ask if a similar procedure might be valid in terms of pha-sors and impedances for the sinusoidal steady state. We already know that both of Kirchhoff’s laws are valid for phasors; also, we have an Ohm-like law for the passive elements V = ZI. We may therefore analyze circuits by nodal techniques in the sinusoidal steady state. Using similar arguments, we can establish that mesh analysis methods are valid (and often useful) as well. CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 394 YR = 1 R YL = 1 j ωL YC = j ωC There is a general (unitless) term for both impedance and admittance—immitance—which is sometimes used, but not very often. ■FIGURE 10.22 A frequency-domain circuit for which node voltages V1 and V2 are identiﬁed. 0.5 A –90 10 j5 5 A 0 1 – j5 V1 V2 j10 –j10 Find the time-domain node voltages v1(t) and v2(t) in the circuit shown in Fig. 10.22. EXAMPLE 10.8 (Continued on next page) SECTION 10.6 NODAL AND MESH ANALYSIS 395 Two current sources are given as phasors, and phasor node voltages V1 and V2 are indicated. At the left node we apply KCL, yielding: V1 5 + V1 −j10 + V1 −V2 −j5 + V1 −V2 j10 = 1/0◦= 1 + j0 At the right node, V2 −V1 −j5 + V2 −V1 j10 + V2 j5 + V2 10 = −(0.5/−90◦) = j0.5 Combining terms, we have (0.2 + j0.2)V1 −j0.1V2 = 1 and −j0.1V1 + (0.1 −j0.1)V2 = j0.5 These equations are easily solved on most scientiﬁc calculators, result-ing in V1 = 1 −j2 V and V2 = −2 + j4 V. The time-domain solutions are obtained by expressing V1 and V2 in polar form: V1 = 2.24/−63.4◦ V2 = 4.47/116.6◦ and passing to the time domain: v1(t) = 2.24 cos(ωt −63.4◦) V v2(t) = 4.47 cos(ωt + 116.6◦) V Note that the value of ω would have to be known in order to compute the impedance values given on the circuit diagram. Also, both sources must be operating at the same frequency. CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 396 ■FIGURE 10.23 V2 V1 50 –90 mA 20 0 mA j50 mS – j25 mS 40 mS Obtain expressions for the time-domain currents i1 and i2 in the circuit given as Fig. 10.24a. EXAMPLE 10.9 ■FIGURE 10.24 (a) A time-domain circuit containing a dependent source. (b) The corresponding frequency-domain circuit. 3 + – + – (a) 10 cos 103t V 2i1 4 mH 500 F i2 i1 3 I1 I2 + – + – (b) 10 0 V 2I1 j4 –j2 PRACTICE ● 10.12 Use nodal analysis on the circuit of Fig. 10.23 to ﬁnd V1 and V2. Ans: 1.062/23.3◦V; 1.593/−50.0◦V. Now let us look at an example of mesh analysis, keeping in mind again that all sources must be operating at the same frequency. Otherwise, it is im-possible to deﬁne a numerical value for any reactance in the circuit. As we see in the next section, the only way out of such a dilemma is to apply superposition. SECTION 10.7 SUPERPOSITION, SOURCE TRANSFORMATIONS, AND THÉVENIN’S THEOREM 397 ■FIGURE 10.25 10 0 V I1 I2 + – 15 90 V 20 0 V j5 – j4 3 + – + – Noting from the left source that ω = 103 rad/s, we draw the frequency-domain circuit of Fig. 10.24b and assign mesh currents I1 and I2. Around mesh 1, 3I1 + j4(I1 −I2) = 10/0◦ or (3 + j4)I1 −j4I2 = 10 while mesh 2 leads to j4(I2 −I1) −j2I2 + 2I1 = 0 or (2 −j4)I1 + j2I2 = 0 Solving, I1 = 14 + j8 13 = 1.24/29.7◦A I2 = 20 + j30 13 = 2.77/56.3◦A Hence, i1(t) = 1.24 cos(103t + 29.7◦) A i2(t) = 2.77 cos(103t + 56.3◦) A PRACTICE ● 10.13 Use mesh analysis on the circuit of Fig. 10.25 to ﬁnd I1 and I2. Ans: 4.87/−164.6◦A; 7.17/−144.9◦A. 10.7 • SUPERPOSITION, SOURCE TRANSFORMATIONS, AND THÉVENIN’S THEOREM After inductors and capacitors were introduced in Chap. 7, we found that circuits containing these elements were still linear, and that the beneﬁts of linearity were again available. Included among these were the superposition principle, Thévenin’s and Norton’s theorems, and source transformations. Thus, we know that these methods may be used on the circuits we are now considering; the fact that we happen to be applying sinusoidal sources and are seeking only the forced response is immaterial. The fact that we are an-alyzing the circuits in terms of phasors is also immaterial; they are still lin-ear circuits. We might also remember that linearity and superposition were invoked when we combined real and imaginary sources to obtain a complex source. Transistor-based ampliﬁer circuits are an integral part of many modern electronic instruments. One common application is in mobile telephones (Fig. 10.26), where audio signals are superimposed on high-frequency car-rier waves. Unfortunately, transistors have built-in capac-itances that lead to limitations in the frequencies at which they can be used, and this fact must be considered when choosing a transistor for a particular application. Figure 10.27a shows what is commonly referred to as a high-frequency hybrid-π model for a bipolar junc-tion transistor. In practice, although transistors are non-linear devices, we ﬁnd that this simple linear circuit does a reasonably accurate job of modeling the actual device behavior. The two capacitors Cπ and Cμ are used to represent internal capacitances that characterize the particular transistor being used; additional capacitors as well as resistors can be added to increase the accuracy of the model as needed. Figure 10.27b shows the transistor model inserted into an ampliﬁer circuit known as a com-mon emitter ampliﬁer. Assuming a sinusoidal steady-state signal repre-sented by its Thévenin equivalent Vs and Rs, we are in-terested in the ratio of the output voltage Vout to the input voltage Vin. The presence of the internal transistor ca-pacitances leads to a reduction in ampliﬁcation as the frequency of Vs is increased; this ultimately limits the frequencies at which the circuit will operate properly. Writing a single nodal equation at the output yields −gmVπ = Vout −Vin 1/jωCμ + Vout RC ∥RL PRACTICAL APPLICATION Cutoff Frequency of a Transistor Ampliﬁer ■FIGURE 10.26 Transistor ampliﬁers are used in many devices, including mobile phones. Linear circuit models are often used to analyze their performance as a function of frequency. © PNC/Getty Images/RF. ■FIGURE 10.27 (a) High-frequency hybrid-π transistor model. (b) Common-emitter ampliﬁer circuit using the hybrid-π transistor model. r C (a) gmV C Collector Emitter Base V + – r C (b) gmV C Rs Vin Vout RB RC RL + – V + – Vs One ﬁnal comment is in order. Up to this point, we have restricted our-selves to considering either single-source circuits or multiple-source circuits in which every source operates at the exact same frequency. This is necessary in order to deﬁne speciﬁc impedance values for inductive and capacitive elements. However, the concept of phasor analysis can be easily extended to circuits with multiple sources operating at different frequencies. In such Solving for Vout in terms of Vin, and noting that Vπ = Vin, we obtain an expression for the ampliﬁer gain Vout Vin = −gm(RC∥RL)(1/jωCμ) + (RC∥RL) (RC∥RL) + (1/jωCμ) = −gm(RC∥RL) + jω(RC∥RL)Cμ 1 + jω(RC∥RL)Cμ Given the typical values gm = 30 mS, RC = RL = 2 k, and Cμ = 5pF, we can plot the magnitude of the gain as a function of frequency (recalling that ω = 2πf ). The semilogarithmic plot is shown in Fig. 10.28a, and the MATLAB script used to generate the ﬁgure is given in Fig. 10.28b. It is interesting, but maybe not totally sur-prising, to see that a characteristic such as the ampliﬁer gain is dependent on frequency. In fact, we might be able to contemplate using such a circuit as a means of ﬁltering out frequencies we aren’t interested in. However, at least for relatively low frequencies, we see that the gain is essentially independent of the frequency of our input source. When characterizing ampliﬁers, it is common to ref-erence the frequency at which the gain is reduced to 1/ √ 2 times its maximum value. From Fig. 10.28a, we see that the maximum gain magnitude is 30, and the gain magnitude is reduced to 30/ √ 2 = 21 at a fre-quency of approximately 30 MHz. This frequency is often called the cutoff or corner frequency of the ampli-ﬁer. If operation at a higher frequency is required, either the internal capacitances must be reduced (i.e., a differ-ent transistor must be used) or the circuit must be re-designed in some way. We should note at this point that deﬁning the gain relative to Vin does not present a complete picture of the frequency-dependent behavior of the ampliﬁer.This may be apparent if we brieﬂy consider the capacitance Cπ: as ω →∞, ZCπ →0, so Vin →0. This effect does not manifest itself in the simple equation we derived.Amore comprehensive approach is to develop an equation for Vout in terms of Vs, in which case both capacitances will appear in the expression; this requires a little bit more algebra. ■FIGURE 10.28 (a) Ampliﬁer gain as a function of frequency. (b) MATLAB script used to create plot. No longer amplifying effectively (a) EDU» frequency = logspace(3,9,100); EDU» numerator = -30e-31000 + ifrequency10005e-12; EDU» denominator = 1 + ifrequency10005e-12; EDU» for k = 1:100 gain(k) = abs(numerator(k)/denominator(k)); end EDU» semilogx(frequency/2/pi,gain); EDU» xlabel(‘Frequency (Hz)’); EDU» ylabel(‘Gain’); EDU» axis([100 1e8 0 35]); instances, we simply employ superposition to determine the voltages and currents due to each source, and then add the results in the time domain. If several sources are operating at the same frequency, superposition will also allow us to consider those sources at the same time, and add the resulting response to the response(s) of any other source(s) operating at a different frequency. (b) CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 400 EXAMPLE 10.10 Use superposition to ﬁnd V1 for the circuit of Fig. 10.22, repeated as Fig. 10.29a for convenience. ■FIGURE 10.29 (a) Circuit of Fig. 10.22 for which V1 is desired, (b) V1 may be found by using superposition of the separate phasor responses. 0.5 A –90 10 j5 5 A 0 1 – j5 V1 V2 j10 –j10 (a) Ref. (b) 0.5 A –90 1 0 A 2 + j4 4 – j2 V1 V2 –j10 First we redraw the circuit as Fig. 10.29b, where each pair of parallel impedances is replaced by a single equivalent impedance. That is, 5∥−j10 is 4 −j2 ; j10∥−j5 is −j10 ; and 10 ∥j5 is equal to 2 + j4 . To ﬁnd V1, we ﬁrst activate only the left source and ﬁnd the partial response, V1L. The 1/0◦source is in parallel with an impedance of (4 −j2) ∥(−j10 + 2 + j4) so that V1L = 1/0◦(4 −j2)(−j10 + 2 + j4) 4 −j2 −j10 + 2 + j4 = −4 −j28 6 −j8 = 2 −j2 V With only the right source active, current division and Ohm’s law yield V1R = (−0.5/−90◦) 2 + j4 4 −j2 −j10 + 2 + j4 (4 −j2) = −1 V Summing, then, V1 = V1L + V1R = 2 −j2 −1 = 1 −j2 V which agrees with our previous result from Example 10.8. As we will see, superposition is also extremely useful when dealing with a circuit in which not all sources operate at the same frequency. Determine the Thévenin equivalent seen by the −j10 impedance of Fig. 10.31a, and use this to compute V1. EXAMPLE 10.11 SECTION 10.7 SUPERPOSITION, SOURCE TRANSFORMATIONS, AND THÉVENIN’S THEOREM 401 ■FIGURE 10.30 V2 V1 50 –90 mA 20 0 mA j50 mS –j25 mS 40 mS ■FIGURE 10.31 (a) Circuit of Fig. 10.29b. The Thévenin equivalent seen by the −j10 impedance is desired. (b) Voc is deﬁned. (c) Zth is deﬁned. (d) The circuit is redrawn using the Thévenin equivalent. Ref. (a) 0.5 A –90 1 0 A 2 + j4 4 – j2 V1 V2 –j10 0 A 1 –90 A 0.5 4 – j2 2 + j4 (b) V oc + – 4 – j2 2 + j4 Zth (c) 6 + j2 1 2 –j10 (d) V th + – I1 PRACTICE ● 10.14 If superposition is used on the circuit of Fig. 10.30, ﬁnd V1 with (a) only the 20/0◦mA source operating; (b) only the 50/−90◦mA source operating. Ans: 0.1951 −j0.556 V; 0.780 + j0.976 V. (Continued on next page) CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 402 ■FIGURE 10.32 10 3 30 A – j5 j5 a b The open-circuit voltage, deﬁned in Fig. 10.31b, is Voc = (1/0◦)(4 −j2) −(−0.5/−90◦)(2 + j4) = 4 −j2 + 2 −j1 = 6 −j3 V The impedance of the inactive circuit of Fig. 10.31c as viewed from the load terminals is simply the sum of the two remaining impedances. Hence, Zth = 6 + j2 Thus, when we reconnect the circuit as in Fig. 10.31d, the current di-rected from node 1 toward node 2 through the −j10 load is I12 = 6 −j3 6 + j2 −j10 = 0.6 + j0.3 A We now know the current ﬂowing through the −j10 impedance of Fig. 10.31a. Note that we are unable to compute V1 using the circuit of Fig. 10.31d as the reference node no longer exists. Returning to the original circuit, then, and subtracting the 0.6 + j0.3 A current from the left source current, the downward current through the (4 −j2) branch is found: I1 = 1 −0.6 −j0.3 = 0.4 −j0.3 A and, thus, V1 = (0.4 −j0.3)(4 −j2) = 1 −j2 V as before. We might have been clever and used Norton’s theorem on the three elements on the right of Fig. 10.31a, assuming that our chief interest is in V1. Source transformations can also be used repeatedly to simplify the circuit. Thus, all the shortcuts and tricks that arose in Chaps. 4 and 5 are available for circuit analysis in the frequency domain. The slight additional complexity that is apparent now arises from the necessity of using complex numbers and not from any more involved theoretical considerations. PRACTICE ● 10.15 For the circuit of Fig. 10.32, ﬁnd the (a) open-circuit voltage Vab; (b) downward current in a short circuit between a and b; (c) Thévenin equivalent impedance Zab in parallel with the current source. Ans: 16.77/−33.4◦V; 2.60 + j1.500 A; 2.5 −j5 . SECTION 10.7 SUPERPOSITION, SOURCE TRANSFORMATIONS, AND THÉVENIN’S THEOREM 403 Determine the power dissipated by the 10 resistor in the circuit of Fig. 10.33a. EXAMPLE 10.12 After glancing at the circuit, we might be tempted to write two quick nodal equations, or perhaps perform two sets of source transformations and launch immediately into ﬁnding the voltage across the 10 resistor. Unfortunately, this is impossible, since we have two sources operat-ing at different frequencies. In such a situation, there is no way to com-pute the impedance of any capacitor or inductor in the circuit—which ω would we use? The only way out of this dilemma is to employ superposition, group-ing all sources with the same frequency in the same subcircuit, as shown in Fig. 10.33b and c. In the subcircuit of Fig. 10.33b, we quickly compute the current I′ using current division: I′ = 2/0◦ −j0.4 10 −j −j0.4 = 79.23/−82.03◦mA ■FIGURE 10.33 (a) A simple circuit having sources operating at different frequencies. (b) Circuit with the left source killed. (c) Circuit with the right source killed. (a) 10 5 cos 3t A 2 cos 5t A 0.2 F 0.5 F (b) 10 –j –j0.4 0 A 2 I –j0.6667 –j1.667 (c) 10 0 A 5 I (Continued on next page) In future studies of signal processing, we will also be introduced to the method of Jean-Baptiste Joseph Fourier, a French mathematician who developed a technique for representing almost any arbitrary function by a combination of sinusoids. When working with lin-ear circuits, once we know the response of a particular circuit to a general sinusoidal forcing function, we can easily predict the response of the circuit to an arbitrary waveform represented by a Fourier series function, simply by using superposition. CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 404 ■FIGURE 10.34 3 H 1 H 4 4 cos 5t V 3 cos 2t V + – + – i so that i′ = 79.23 cos(5t −82.03◦) mA Likewise, we ﬁnd that I′′ = 5/0◦ −j1.667 10 −j0.6667 −j1.667 = 811.7/−76.86◦mA so that i′′ = 811.7 cos(3t −76.86◦) mA It should be noted at this point that no matter how tempted we might be to add the two phasor currents I′ and I′′, in Fig. 10.33b and c, this would be incorrect. Our next step is to add the two time-domain currents, square the result, and multiply by 10 to obtain the power absorbed by the 10 resistor in Fig. 10.33a: p10 = (i′ + i′′)2 × 10 = 10[79.23 cos(5t −82.03◦) + 811.7 cos(3t −76.86◦)]2 μW PRACTICE ● 10.16 Determine the current i through the 4 resistor of Fig. 10.34. Ans: i = 175.6 cos(2t −20.55◦) + 547.1 cos(5t −43.16◦) mA. COMPUTER-AIDED ANALYSIS We have several options in PSpice for the analysis of circuits in the sinu-soidal steady state. Perhaps the most straightforward approach is to make use of two specially designed sources: VAC and IAC. The magnitude and phase of either source is selected by double-clicking on the part. Let’s simulate the circuit of Fig. 10.20a, shown redrawn in Fig. 10.35. The frequency of the source is not selected through the Property Editor, but rather through the ac sweep analysis dialog box. This is accomplished by choosing AC Sweep/Noise for Analysis when pre-sented with the Simulation Settings window. We select a Linear sweep SECTION 10.7 SUPERPOSITION, SOURCE TRANSFORMATIONS, AND THÉVENIN’S THEOREM 405 and set Total Points to 1. Since we are only interested in the frequency of 3000 rad/s (477.5 Hz), we set both Start Frequency and End Frequency to 477.5 as shown in Fig. 10.36. ■FIGURE 10.35 The circuit of Fig. 10.20a, operating at ω = 3000 rad/s. The current through the 1.5 k resistor is desired. ■FIGURE 10.36 Dialog box for setting source frequency. Note that an additional “component’’ appears in the schematic. This component is called IPRINT, and allows a variety of current parameters to be printed. In this simulation, we are interested in the AC, MAG, and PHASE attributes. In order for PSpice to print these quantities, double-click on the IPRINT symbol in the schematic, and enter yes in each of the appropriate ﬁelds. The simulation results are obtained by choosing View Output File under PSpice in the Capture CIS window. FREQ IM(V_PRINT1) IP(V—PRINT1) 4.775E+02 1.600E-02 -1.269E+02 Thus, the current magnitude is 16 mA, and the phase angle is −126.9◦, so that the current through the 1.5 k resistor is i = 16 cos(3000t −126.9◦) mA = 16 sin(3000t −36.9◦) mA 10.8 • PHASOR DIAGRAMS The phasor diagram is a name given to a sketch in the complex plane show-ing the relationships of the phasor voltages and phasor currents throughout a speciﬁc circuit. We are already familiar with the use of the complex plane in the graphical identiﬁcation of complex numbers and in their addition and subtraction. Since phasor voltages and currents are complex numbers, they may also be identiﬁed as points in a complex plane. For example, the phasor voltage V1 = 6 + j8 = 10/53.1◦V is identified on the complex voltage plane shown in Fig. 10.37. The x axis is the real voltage axis, and the y axis is the imaginary voltage axis; the voltage V1 is located by an arrow drawn from the origin. Since addition and subtraction are particularly easy to perform and display on a complex plane, phasors may be easily added and subtracted in a phasor diagram. Multiplication and division result in the addition and subtraction of angles and a change of amplitude. Figure 10.38a shows the sum of V1 and a second phasor voltage V2 = 3 −j4 = 5/−53.1◦V, and Fig. 10.38b shows the current I1, which is the product of V1 and the admit-tance Y = 1 + j1 S. This last phasor diagram shows both current and voltage phasors on the same complex plane; it is understood that each will have its own amplitude scale, but a common angle scale. For example, a phasor voltage 1 cm long might represent 100 V, while a phasor current 1 cm long could indicate 3 mA. Plotting both phasors on the same diagram enables us to easily determine which waveform is leading and which is lagging. The phasor diagram also offers an interesting interpretation of the time-domain to frequency-domain transformation, since the diagram may be in-terpreted from either the time- or the frequency-domain viewpoint. Up to this point, we have been using the frequency-domain interpretation, as we have been showing phasors directly on the phasor diagram. However, let us proceed to a time-domain viewpoint by ﬁrst showing the phasor voltage V = Vm/α as sketched in Fig. 10.39a. In order to transform V to the time domain, the next necessary step is the multiplication of the phasor by e jωt; thus we now have the complex voltage Vme jαe jωt = Vm/ωt + α. This voltage may also be interpreted as a phasor, one which possesses a phase angle that increases linearly with time. On a phasor diagram it therefore represents a rotating line segment, the instantaneous position being ωt radi-ans ahead (counterclockwise) of Vm/α. Both Vm/α and Vm/ωt + α are shown on the phasor diagram of Fig. 10.39b. The passage to the time CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 406 ■FIGURE 10.38 (a) A phasor diagram showing the sum of V1 = 6 + j8 V and V2 = 3 −j4 V, V1 + V2 = 9 + j4 V 9.85/24.0◦V. (b) The phasor diagram shows V1 and I1, where I1 = YV1 and Y = (1 + j1) S 2/45◦S. The current and voltage amplitude scales are different. I1 = (1 + j1)V 1 = (√2 45)V 1 V1 V2 V1 + V2 (a) V1 (b) 45 ■FIGURE 10.37 A simple phasor diagram shows the single voltage phasor V1 = 6 + j8 = 10/53.1◦V. Imaginary axis (V) 6 j8 Real axis (V) 10 53.1 V1 ■FIGURE 10.39 (a) The phasor voltage V m/α. (b) The complex voltage V m/ωt + α is shown as a phasor at a particular instant of time. This phasor leads V m/α by ωt radians. (a) Vm Vm (b) Vm t + t + Vm t SECTION 10.8 PHASOR DIAGRAMS 407 domain is now completed by taking the real part of Vm/ωt + α. The real part of this complex quantity is the projection of Vm/ωt + α on the real axis: Vm cos(ωt + α). In summary, then, the frequency-domain phasor appears on the phasor diagram, and the transformation to the time domain is accomplished by al-lowing the phasor to rotate in a counterclockwise direction at an angular ve-locity of ω rad/s and then visualizing the projection on the real axis. It is helpful to think of the arrow representing the phasor V on the phasor dia-gram as the photographic snapshot, taken at ωt = 0, of a rotating arrow whose projection on the real axis is the instantaneous voltage v(t). Let us now construct the phasor diagrams for several simple circuits.The series RLC circuit shown in Fig. 10.40a has several different voltages asso-ciated with it, but only a single current. The phasor diagram is constructed most easily by employing the single current as the reference phasor. Let us arbitrarily select I = Im/0◦and place it along the real axis of the phasor dia-gram, Fig. 10.40b.The resistor, capacitor, and inductor voltages may then be calculated and placed on the diagram, where the 90° phase relationships stand out clearly. The sum of these three voltages is the source voltage, and for this circuit, which is in what we will deﬁne in a subsequent chapter as the “resonant condition’’ since ZC = −ZL, the source voltage and resistor voltage are equal. The total voltage across the resistor and inductor or resis-tor and capacitor is obtained from the diagram by adding the appropriate phasors as shown. Figure 10.41a is a simple parallel circuit in which it is logical to use the single voltage between the two nodes as a reference phasor. Suppose that V = 1/0◦V. The resistor current, IR = 0.2/0◦A, is in phase with this volt-age, and the capacitor current, IC = j0.1 A, leads the reference voltage by 90°. After these two currents are added to the phasor diagram, shown as Fig. 10.41b, they may be summed to obtain the source current. The result is Is = 0.2 + j0.1 A. If the source current is speciﬁed initially as the convenient value of 1/0◦A and the node voltage is not initially known, it is still convenient to begin construction of the phasor diagram by assuming a node voltage (for example, V = 1/0◦V once again) and using it as the reference phasor. The diagram is then completed as before, and the source current that ﬂows as a result of the assumed node voltage is again found to be 0.2 + j0.1 A. The true source current is 1/0◦A, however, and thus the true node voltage is obtained by multiplying the assumed node voltage by 1/0◦/(0.2 + j0.1); ■FIGURE 10.40 (a) A series RLC circuit. (b) The phasor diagram for this circuit; the current I is used as a convenient reference phasor. 10 VC + – VL + – VR + – + – Vs –j50 j50 (a) I VL VC VR = Vs VR + VL VR + VC (b) I ■FIGURE 10.41 (a) A parallel RC circuit. (b) The phasor diagram for this circuit; the node voltage V is used as a convenient reference phasor. IR IC 5 50 F Is V + – = 2000 rad/s (a) (b) Is = 0.2 + j0.1 A IC = j0.1 A IR = 0.2 A V = 1 0 V the true node voltage is therefore 4 −j2 V = √ 20/−26.6◦V. The as-sumed voltage leads to a phasor diagram which differs from the true phasor diagram by a change of scale (the assumed diagram is smaller by a factor of 1/ √ 20) and an angular rotation (the assumed diagram is rotated coun-terclockwise through 26.6°). CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 408 ■FIGURE 10.42 A simple circuit for which several currents are required. V + – j0.3 S –j0.1 S IL IC 0.2 S Is Ix IR Construct a phasor diagram showing IR, IL, and IC for the circuit of Fig. 10.42. Combining these currents, determine the angle by which Is leads IR, IC, and Ix. We begin by choosing a suitable reference phasor. Upon examining the circuit and the variables to be determined, we see that once V is known, IR, IL, and IC can be computed by simple application of Ohm’s law. Thus, we select V = 1/0◦V for simplicity’s sake, and subsequently compute IR = (0.2)1/0◦ = 0.2/0◦A IL = (−j0.1)1/0◦= 0.1/−90◦A IC = ( j0.3)1/0◦ = 0.3/90◦A The corresponding phasor diagram is shown in Fig. 10.43a. We also need to ﬁnd the phasor currents Is and Ix. Figure 10.43b shows the de-termination of Ix = IL + IR = 0.2 −j0.1 = 0.224/−26.6◦A, and Fig. 10.43c shows the determination of Is = IC + Ix = 0.283/45◦A. From Fig. 10.43c, we ascertain that Is leads IR by 45°, IC by −45◦, and Ix by 45° + 26.6° = 71.6°. These angles are only relative, however; the exact numerical values will depend on Is, upon which the actual value of V (assumed here to be 1/0◦V for convenience) also depends. EXAMPLE 10.13 PRACTICE ● 10.17 Select some convenient reference value for IC in the circuit of Fig. 10.44; draw a phasor diagram showing VR, V2, V1, and Vs; and measure the ratio of the lengths of (a) Vs to V1; (b) V1 to V2; (c) Vs to VR. Ans: 1.90; 1.00; 2.12 ■FIGURE 10.43 (a) Phasor diagram constructed using a reference value of V = 1/0◦. (b) Graphical determination of Ix = IL + IR. (c) Graphical determination of Is = IC + Ix. IL IC IR (a) IL IC IR Ix = IL + IR (b) IL Ix IC IR Is = IC + Ix (c) ■FIGURE 10.44 2 2 VR + – V2 + – V1 + – IC + – Vs j2 –j1 SUMMARY AND REVIEW 409 SUMMARY AND REVIEW This chapter dealt with the steady-state response of circuits to sinusoidal excitation. This is a limited analysis of a circuit in some respects, as the transient behavior is completely ignored. In many situations, such an approach is more than adequate, and reducing the amount of information we seek about a circuit speeds up the analysis considerably. The fundamental idea behind what we did was that an imaginary source was added to every real sinusoidal source; then Euler’s identity converted the source to a com-plex exponential. Since the derivative of an exponential is simply another exponential, what would otherwise be integrodifferential equations arising from mesh or nodal analysis become algebraic equations. A few new terms were introduced: lagging, leading, impedance, ad-mittance, and a particularly important one phasor. Phasor relationships between current and voltage gave rise to the concept of impedance, where resistors are represented by a real number (resistance, as before), and inductors are represented by Z = jωL while capacitors are represented by −j/ωC(ω being the operating frequency of our sources). From that point forward, all the circuit analysis techniques learned in Chaps. 3 to 5 apply. It might seem odd to have an imaginary number as part of our solution, but we found that recovering the time-domain solution to our analysis is straightforward once the voltage or current is expressed in polar form. The magnitude of our quantity of interest is the magnitude of the cosine func-tion, the phase angle is the phase of the cosine term, and the frequency is obtained from the original circuit (it disappears from view during the analy-sis, but the circuits we are analyzing do not change it in any way). We con-cluded the chapter with an introduction to the concept of phasor diagrams. Prior to inexpensive scientiﬁc calculators such tools were invaluable in an-alyzing many sinusoidal circuits. They still ﬁnd use in analysis of ac power systems, as we see in subsequent chapters. A concise list of key concepts of the chapter is presented below for the convenience of the reader, along with the corresponding example numbers. ❑If two sine waves (or two cosine waves) both have positive magnitudes and the same frequency, it is possible to determine which waveform is leading and which is lagging by comparing their phase angles. ❑The forced response of a linear circuit to a sinusoidal voltage or current source can always be written as a single sinusoid having the same frequency as the sinusoidal source. (Example 10.1) ❑A phasor has both a magnitude and a phase angle; the frequency is understood to be that of the sinusoidal source driving the circuit. (Example 10.2) ❑A phasor transform may be performed on any sinusoidal function, and vice versa: Vm cos(ωt + φ) ↔Vm φ. (Example 10.3) ❑When transforming a time-domain circuit into the corresponding frequency-domain circuit, resistors, capacitors, and inductors are replaced by impedances (or, occasionally, by admittances). (Examples 10.4, 10.6) • The impedance of a resistor is simply its resistance. • The impedance of a capacitor is 1/jωC . • The impedance of an inductor is jωL . ❑Impedances combine both in series and in parallel combinations in the same manner as resistors. (Example 10.6) ❑All analysis techniques previously used on resistive circuits apply to circuits with capacitors and/or inductors once all elements are replaced by their frequency-domain equivalents. (Examples 10.5, 10.7, 10.8, 10.9, 10.10, 10.11) ❑Phasor analysis can only be performed on single-frequency circuits. Otherwise, superposition must be invoked, and the time-domain partial responses added to obtain the complete response. (Example 10.12) ❑The power behind phasor diagrams is evident when a convenient forcing function is used initially, and the ﬁnal result scaled appropriately. (Example 10.13) READING FURTHER A good reference to phasor-based analysis techniques can be found in: R. A. DeCarlo and P. M. Lin, Linear Circuit Analysis, 2nd ed. New York: Oxford University Press, 2001. Frequency-dependent transistor models are discussed from a phasor perspec-tive in Chap. 7 of: W. H. Hayt, Jr., and G. W. Neudeck, Electronic Circuit Analysis and Design, 2nd ed. New York: Wiley, 1995. EXERCISES 10.1 Characteristics of Sinusoids 1. Evaluate the following: (a) 5 sin (5t −9◦) at t = 0, 0.01, and 0.1 s; (b) 4 cos 2t and 4 sin (2t + 90◦) at t = 0, 1, and 1.5 s; (c) 3.2 cos (6t + 15◦) and 3.2 sin (6t + 105◦) at t = 0, 0.01, and 0.1 s. 2. (a) Express each of the following as a single cosine function: 5 sin 300t, 1.95 sin (πt −92◦), 2.7 sin (50t + 5◦) 10 cos 50t. (b) Express each of the following as a single sine function: 66 cos (9t −10◦), 4.15 cos 10t, 10 cos (100t −9◦) + 10 sin (100t + 19◦). 3. Determine the angle by which v1 leads i1 if v1 = 10 cos (10t −45◦) and i1 is equal to (a) 5 cos 10t; (b) 5 cos (10t −80◦); (c) 5 cos (10t −40◦); (d) 5 cos (10t + 40◦); (e) 5 sin (10t −19◦). 4. Determine the angle by which v1 lags i1 if v1 = 34 cos (10t + 125◦) and i1 is equal to (a) 5 cos 10t; (b) 5 cos (10t −80◦); (c) 5 cos (10t −40◦); (d) 5 cos (10t + 40◦); (e) 5 sin (10t −19◦). 5. Determine which waveform in each of the following pairs is lagging: (a) cos 4t, sin 4t; (b) cos (4t −80◦), cos (4t); (c) cos (4t + 80◦), cos 4t; (d) −sin 5t, cos (5t + 2◦); (e) sin 5t + cos 5t, cos (5t −45◦). 6. Calculate the ﬁrst three instants in time (t > 0) for which the following functions are zero, by ﬁrst converting to a single sinusoid: (a) cos 3t −7 sin 3t; (b) cos (10t + 45◦); (c) cos 5t−sin 5t; (d) cos 2t + sin 2t−cos 5t + sin 5t. 7. (a) Determine the ﬁrst two instants in time (t > 0) for which each of the functions in Exercise 6 are equal to unity, by ﬁrst converting to a single sinusoid. (b) Verify your answers by plotting each waveform using an appropriate software package. CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 410 EXERCISES 411 8. The concept of Fourier series is a powerful means of analyzing periodic waveforms in terms of sinusoids. For example, the triangle wave in Fig. 10.45 can be represented by the inﬁnite sum v(t) = 8 π2 sin πt −1 32 sin 3πt + 1 52 sin 5πt −1 72 sin 7πt + · · · where in practice perhaps the ﬁrst several terms provide an accurate enough approximation. (a) Compute the exact value of v(t) at t = 0.25 s by ﬁrst obtaining an equation for the corresponding segment of the waveform. (b) Compute the approximate value at t = 0.25 s using the ﬁrst term of the Fourier series only. (c) Repeat part (b) using the ﬁrst three terms. (d) Plot v(t) using the ﬁrst term only. (e) Plot v(t) using the ﬁrst two terms only. ( f ) Plot v(t) using the ﬁrst three terms only. ■FIGURE 10.46 iR 1 3 mH 10 vs + – is iL vL + – 9. Household electrical voltages are typically quoted as either 110 V, 115 V, or 120 V. However, these values do not represent the peak ac voltage. Rather, they represent what is known as the root mean square of the voltage, deﬁned as Vrms = 1 T T 0 V 2 m cos2(ωt) dt where T = the period of the waveform, Vm is the peak voltage, and ω = the waveform frequency ( f = 60 Hz in North America). (a) Perform the indicated integration, and show that for a sinusoidal voltage, Vrms = Vm √ 2 (b) Compute the peak voltages corresponding to the rms voltages of 110, 115, and 120 V. 10.2 Forced Response to Sinusoidal Functions 10. If the source vs in Fig. 10.46 is equal to 4.53 cos (0.333 × 10−3t + 30◦) V, (a) obtain is, iL, and iR at t = 0 assuming no transients are present; (b) obtain an expression for vL (t) in terms of a single sinusoid, valid for t > 0, again assuming no transients are present. ■FIGURE 10.45 1 –1 1 2 3 v(t) (V) t (s) 11. Assuming there are no longer any transients present, determine the current labeled iL in the circuit of Fig. 10.47. Express your answer as a single sinusoid. CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 412 ■FIGURE 10.48 + – vC + – 15 2 mF 3 cos 40t V ■FIGURE 10.47 iL 1 2 10 mH 1 25 cos 100t A 12. Calculate the power dissipated in the 2 resistor of Fig. 10.47 assuming there are no transients present. Express your answer in terms of a single sinusoidal function. 13. Obtain an expression for vC as labeled in Fig. 10.48, in terms of a single sinu-soidal function. You may assume all transients have died out long before t = 0. 14. Calculate the energy stored in the capacitor of the circuit depicted in Fig. 10.48 at t = 10 ms and t = 40 ms. 15. Obtain an expression for the power dissipated in the 10 resistor of Fig. 10.49, assuming no transients present. ■FIGURE 10.49 iL 10 cos 6t A 0.5 H 0.2iL 10.3 The Complex Forcing Function 16. Express the following complex numbers in rectangular form: (a) 50/−75◦; (b) 19e j30◦; 2.5/−30◦+ 0.5/45◦. Convert the following to polar form: (c) (2 + j2)(2 −j2); (d) (2 + j2)(5/22◦). 17. Express the following in polar form: (a) 2 + e j35◦;(b) (j)(j)(j); (c) 1. Express the following in rectangular form: (d) 2 + e j35◦; (e) −j9 + 5/55◦. 18. Evaluate the following, and express your answer in polar form: (a) 4(8 −j8); (b) 4/5◦−2/15◦; (c) (2 + j9) −5/0◦; (d) −j 10 + 5 j −3/40◦+ 2; (e) (10 + j5) (10 −j5) (3/40◦) + 2. 19. Evaluate the following, and express your answer in rectangular form: (a) 3 (3/30◦); (b) 2/25◦+ 5/−10◦; (c) (12 + j90) −5/30◦; (d) 10 + 5 j 8 −j + 2/60◦+ 1; (e) (10 + 5 j) (10 −5 j) (3/40◦) + 2. 20. Perform the indicated operations, and express the answer in both rectangular and polar forms: (a) 2 + j3 1 + 8/90◦−4; (b) 10/25◦ 5/−10◦+ 3/15◦ 3 −j5 j2; (c) (1 −j)(1 + j) + 1/0◦ −j (3/−90◦) + j 5/−45◦. 21. Insert an appropriate complex source into the circuit represented in Fig. 10.50, and use it to determine steady-state expressions for iC(t) and vC(t). ■FIGURE 10.50 + – vC + – 5 130 mF 5 sin 20t V iC EXERCISES 413 22. For the circuit of Fig. 10.51, if is = 5 cos 10t A, use a suitable complex source replacement to obtain a steady-state expression for iL(t). 23. In the circuit depicted in Fig. 10.51, is is modiﬁed such that the 2 resistor is replaced by a 20 resistor. If iL(t) = 62.5 /31.3◦mA, determine is. 24. Employ a suitable complex source to determine the steady-state current iL in the circuit of Fig. 10.52. ■FIGURE 10.51 iL 0.4 H is 2 ■FIGURE 10.52 iL 6 0.01 F 0.4 H 5 sin (35t – 10) V 10.4 The Phasor 25. Transform each of the following into phasor form: (a) 75.928 cos (110.1t); (b) 5 cos (55t −42◦); (c) −sin(8000t + 14◦); (d) 3 cos 10t −8 cos(10t + 80◦). 26. Transform each of the following into phasor form: (a) 11 sin 100t; (b) 11 cos 100t; (c) 11 cos(100t −90◦); (d) 3 cos 100t – 3 sin 100t. 27. Assuming an operating frequency of 1 kHz, transform the following phasor expressions into a single cosine function in the time domain: (a) 9/65◦V; (b) 2/31◦ 4/25◦A; (c) 22/14◦−8/33◦V. 28. The following complex voltages are written in a combination of rectangular and polar form. Rewrite each, using conventional phasor notation (i.e., a magnitude and angle): (a) 2 −j 5/45◦V; (b) 6/20◦ 1000 −jV; (c) ( j)(52.5/−90◦) V. 29. Assuming an operating frequency of 50 Hz, compute the instantaneous voltage at t = 10 ms and t = 25 ms for each of the quantities represented in Exercise 26. 30. Assuming an operating frequency of 50 Hz, compute the instantaneous voltage at t = 10 ms and t = 25 ms for each of the quantities represented in Exercise 27. 31. Assuming the passive sign convention and an operating frequency of 5 rad/s, calcualte the phasor voltage which develops across the following when driven by the phasor current I = 2/0◦mA: (a) a 1 k resistor; (b) a 1 mF capacitor; (c) a 1 nH inductor. 32. (a) A series connection is formed between a 1 resistor, a 1 F capacitor, and a 1 H inductor, in that order. Assuming operation at ω = 1 rad/s, what are the magnitude and phase angle of the phasor current which yields a voltage of 1/30◦V across the resistor (assume the passive sign convention)? (b) Compute the ratio of the phasor voltage across the resistor to the phasor voltage which appears across the capacitor-inductor combination. (c) The frequency is doubled. Calculate the new ratio of the phasor voltage across the resistor to the phasor voltage across the capacitor-inductor combination. 33. Assuming the passive sign convention and an operating frequency of 314 rad/s, calculate the phasor voltage V which appears across each of the following when driven by the phasor current I = 10/0◦mA: (a) a 2 resistor; (b) a 1 F capacitor; (c) a 1 H inductor; (d) a 2 resistor in series with a 1 F capacitor; (e) a 2 resistor in series with a 1 H inductor. (f) Calculate the instantaneous value of each voltage determined in parts (a) to (e) at t = 0. 34. In the circuit of Fig. 10.53, which is shown in the phasor (frequency) domain, I10 is determined to be 2/42◦mA. If V = 40/132◦mV: (a) what is the likely type of element connected to the right of the 10 resistor and (b) what is its value, assuming the voltage source operates at a frequency of 1000 rad/s? 35. The circuit of Fig. 10.53 is shown represented in the phasor (frequency) do-main. If I10 = 4/35◦A, V = 10/35◦, and I = 2/35◦A, (a) across what type of element does V appear, and what is its value? (b) Determine the value of Vs. CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 414 10.5 Impedance and Admittance 36. (a) Obtain an expression for the equivalent impedance Zeq of a 1 resistor in series with a 10 mH inductor as a function of ω. (b) Plot the magnitude of Zeq as a function of ω over the range 1 < ω < 100 krad/s (use a logarithmic scale for the frequency axis). (c) Plot the angle (in degrees) of Zeq as a function of ω over the range 1 < ω < 100 krad/s (use a logarithmic scale for the frequency axis). [Hint: semilogx() in MATLAB is a useful plotting function.] 37. Determine the equivalent impedance of the following, assuming an operating frequency of 20 rad/s: (a) 1 k in series with 1 mF; (b) 1 k in parallel with 1 mH; (c) 1 k in parallel with the series combination of 1 F and 1 H. 38. (a) Obtain an expression for the equivalent impedance Zeq of a 1 resistor in series with a 10 mF capacitor as a function of ω. (b) Plot the magnitude of Zeq as a function of ω over the range 1 < ω < 100 krad/s (use a logarithmic scale for the frequency axis). (c) Plot the angle (in degrees) of Zeq as a function of ω over the range 1 < ω < 100 krad/s (use a logarithmic scale for the frequency axis). [Hint: semilogx() in MATLAB is a useful plotting function.] 39. Determine the equivalent admittance of the following, assuming an operating frequency of 1000 rad/s: (a) 25 in series with 20 mH; (b) 25 in parallel with 20 mH; (c) 25 in parallel with 20 mH in parallel with 20 mF; (d) 1 in series with 1 F in series with 1 H; (e) 1 in parallel with 1 F in parallel with 1 H. 40. Consider the network depicted in Fig. 10.54, and determine the equivalent impedance seen looking into the open terminals if (a) ω = 1 rad/s; (b) ω = 10 rad/s; (c) ω = 100 rad/s. 41. Exchange the capacitor and inductor in the network shown in Fig. 10.54, and calculate the equivalent impedance looking into the open terminals if ω = 25 rad/s. 42. Find V in Fig. 10.55 if the box contains (a) 3 in series with 2 mH; (b) 3 in series with 125 μF; (c) 3 , 2 mH, and 125 μF in series; (d) 3 , 2 mH, and 125 μF in series, but ω = 4 krad/s. 43. Calculate the equivalent impedance seen at the open terminals of the network shown in Fig. 10.56 if f is equal to (a) 1 Hz; (b) 1 kHz; (c) 1 MHz; (d) 1 GHz; (e) 1 THz. ■FIGURE 10.53 I 25 10 + – Vs V + – I10 ■FIGURE 10.54 25 55 20 mH 20 10 mF ■FIGURE 10.55 3 –20 A = 2 krad/s V + – ■FIGURE 10.56 60 60 60 10 mH 30 F a b EXERCISES 415 44. Employ phasor-based analysis to obtain an expression for i(t) in the circuit of Fig. 10.57. 45. Design a suitable combination of resistors, capacitors, and/or inductors which has an equivalent impedance at ω = 100 rad/s of (a) 1 using at least one inductor; (b) 7/10◦; (c) 3 – j4 . 46. Design a suitable combination of resistors, capacitors, and/or inductors which has an equivalent admittance at ω = 10 rad/s of (a) 1 S using at least one capacitor; (b) 12/−18◦S ; (c) 2 + j mS. 10.6 Nodal and Mesh Analysis 47. For the circuit depicted in Fig. 10.58, (a) redraw with appropriate phasors and impedances labeled; (b) employ nodal analysis to determine the two nodal voltages v1(t) and v2 (t). ■FIGURE 10.57 1 mF 20 mH 4 cos (100t – 20) A i(t) 5 2 ■FIGURE 10.58 2 10 mH 5 3 2.2 mF 4.7 mF v1(t) v2(t) 3 cos (100t + 62) A 2 cos 80t A 48. For the circuit illustrated in Fig. 10.59, (a) redraw, labeling appropriate phasor and impedance quantities; (b) determine expressions for the three time-domain mesh currents. ■FIGURE 10.59 1.5 cos (10t – 42) mA 2.5 cos 10t mA 100 mH 220 mF 2 v1(t) v2(t) 49. Referring to the circuit of Fig. 10.59, employ phasor-based analysis techniques to determine the two nodal voltages. 50. In the phasor-domain circuit represented by Fig. 10.60, let V1 = 10/−80◦V, V2 = 4/−0◦V, and V3 = 2/−23◦V. Calculate I1 and I2. ■FIGURE 10.60 I1 V1 I2 + – V2 + – V3 + – j30 55 – j20 51. With regard to the two-mesh phasor-domain circuit depicted in Fig. 10.60, calculate the ratio of I1 to I2 if V1 = 3/0◦V, V2 = 5.5/−130◦V, and V3 = 1.5/17◦V. 52. Employ phasor analysis techniques to obtain expressions for the two mesh cur-rents i1 and i2 as shown in Fig. 10.61. CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 416 ■FIGURE 10.61 2 + – + – 2.5 cos (10t + 9) V 5i1 1 H 330 mF i2 i1 ■FIGURE 10.62 j2 j3.8 –j4 2 1 V2 I2 I1 IB 53. Determine IB in the circuit of Fig. 10.62 if I1 = 5/−18◦A and I2 = 2/5◦A. 54. Determine V2 in the circuit of Fig. 10.62 if I1 = 15/0◦A and I2 = 25/131◦A. 55. Employ phasor analysis to obtain an expression for vx as labeled in the circuit of Fig. 10.63. ■FIGURE 10.63 vx + – ix + – 2 4.7 2 1 4 cos 20t V 100 mH 890 mF 56. Determine the current ix in the circuit of Fig. 10.63. 57. Obtain an expression for each of the four (clockwise) mesh currents for the cir-cuit of Fig. 10.64 if v1 = 133 cos(14t + 77◦) V and v2 = 55 cos(14t + 22◦) V. ■FIGURE 10.64 28 mH 32 mH 100 mF Ref. 0.4 0.8 0.6 v1 v2 + – + – EXERCISES 417 58. Determine the nodal voltages for the circuit of Fig. 10.64, using the bottom node as the reference node, if v1 = 0.009 cos (500t + 0.5ο) V and v2 = 0.004 cos (500t + 1.5ο) V. 59. The op amp shown in Fig. 10.65 has an inﬁnite input impedance, zero output impedance, and a large but ﬁnite (positive, real) gain, A = −Vo/Vi .(a) Con-struct a basic differentiator by letting Zf = Rf , ﬁnd Vo/Vs, and then show that Vo/Vs →−jωC1Rf as A →∞. (b) Let Zf represent Cf and Rf in parallel, ﬁnd Vo/Vs, and then show that Vo/Vs →−jωC1Rf /(1 + jωCf Rf ) as A →∞. 60. Obtain an expression for each of the four mesh currents labeled in the circuit of Fig. 10.66. 10.7 Superposition, Source Transformations, and Thévenin’s Theorem 61. Determine the individual contribution each current source makes to the two nodal voltages V1 and V2 as represented in Fig. 10.67. ■FIGURE 10.66 + – 0.005i1 70 mH 250 mF 250 mF 9 cos 20t V 9 sin 20t V 3 5 + – + – i1 i2 i3 i4 ■FIGURE 10.68 V2 I2 I1 V1 j3 – j5 2 ■FIGURE 10.67 3 k j2 k j8 k –j5 k –j3 k 3 k V1 V2 3 –41 mA 5 13 mA ■FIGURE 10.69 Ref. j2 V1 V2 (2 – j) 1.5 A 24 2 A 38 4 10 I1 ■FIGURE 10.65 + – Vo + – Vi – + + – C1 Zf Vs 62. Determine V1 and V2 in Fig. 10.68 if I1 = 33/3◦mA and I2 = 51/−91◦mA. 63. The phasor domain circuit of Fig. 10.68 was drawn assuming an operating frequency of 2.5 rad/s. Unfortunately, the manufacturing unit installed the wrong sources, each operating at a different frequency. If i1(t) = 4 cos 40t mA and i2(t) = 4 sin 30t mA, calculate v1(t) and v2(t). 64. Obtain the Thévenin equivalent seen by the (2 j) impedance of Fig. 10.69, and employ it to determine the current I1. 65. The (2 −j) impedance in the circuit of Fig. 10.69 is replaced with a (1 + j) impedance. Perform a source transformation on each source, simplify the resulting circuit as much as possible, and calculate the current ﬂowing through the (1 + j) impedance. 66. With regard to the circuit depicted in Fig. 10.70, (a) calculate the Thévenin equivalent seen looking into the terminals marked a and b; (b) determine the Norton equivalent seen looking into the terminals marked a and b; (c) compute the current ﬂowing from a to b if a (7 – j2) impedance is connected across them. CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 418 ■FIGURE 10.70 12 22 30 A j10 –j34 a b ■FIGURE 10.73 15 mF 25 mF 1 5 + – 5 cos 20t A 110 cos 20t V ■FIGURE 10.71 vL + – + – 1 5 mH vs3 is2 is1 ■FIGURE 10.72 i1 + – 1 2 50 mF 50 mH 2.1 cos 20t V 3 sin 20t A v1 + – 69. Determine the power dissipated by the 1 resistor in the circuit of Fig. 10.73. Verify your solution with an appropriate PSpice simulation. 67. In the circuit of Fig. 10.71, is1 = 8 cos (4t −9◦) mA, is2 = 5 cos 4t and vs3 = 2 sin 4t. (a) Redraw the circuit in the phasor domain; (b) reduce the circuit to a single current source with the assistance of a source transformation; (c) calcu-late vL(t). (d) Verify your solution with an appropriate PSpice simulation. 68. Determine the individual contribution of each source in Fig. 10.72 to the volt-age v1(t). 70. Use ω = 1 rad/s, and ﬁnd the Norton equivalent of the network shown in Fig. 10.74. Construct the Norton equivalent as a current source IN in parallel with a resistance RN and either an inductance LN or a capacitance CN. ■FIGURE 10.74 a b VL + – + – 2 H 1 F 0.25VL 1 0 V EXERCISES 419 10.8 Phasor Diagrams 71. The source Is in the circuit of Fig. 10.75 is selected such that V = 5/120◦V. (a) Construct a phasor diagram showing IR, IL, and IC. (b) Use the diagram to determine the angle by which Is leads IR, IC, and Is. ■FIGURE 10.75 j10 S –j2 S 1 S IC IL Is V + – Ix IR 74. In the circuit of Fig. 10.77, (a) ﬁnd values for I1, I2, and I3. (b) Show Vs, I1, I2, and I3 on a phasor diagram (scales of 50 V/in and 2 A/in work ﬁne). (c) Find Is graphically and give its amplitude and phase angle. ■FIGURE 10.76 IC IR 2 VR + – VC + – Vs = 100 0 V IL + – j2.5 –j1 VL + – ■FIGURE 10.77 I1 I2 Is I3 30 50 j40 – j30 40 30 V s = 120 0 V + – 72. Let V1 = 100/0◦V, \|V2\| = 140 V, and \|V1 + V2\| = 120 V. Use graphical methods to ﬁnd two possible values for the angle of V2. 73. (a) Calculate values for IL, IR, IC, VL, VR, and VC for the circuit shown in Fig. 10.76. (b) Using scales of 50 V to 1 in and 25 A to 1 in, show all seven quantities on a phasor diagram, and indicate that IL = IR + IC and Vs = VL + VR. ■FIGURE 10.78 5 3 V2 + – VR + – V1 + – IC + – Vs j2 – j4 75. The voltage source Vs in Fig. 10.78 is chosen such that IC = 1/0◦A. (a) Draw a phasor diagram showing V1, V2, Vs, and VR. (b) Use the diagram to deter-mine the ratio of V2 to V1. Chapter-Integrating Exercises 76. For the circuit shown in Fig. 10.79, (a) draw the phasor representation of the circuit; (b) determine the Thévenin equivalent seen by the capacitor, and use it to calculate vC(t). (c) Determine the current ﬂowing out of the positive refer-ence terminal of the voltage source. (d) Verify your solution with an appropri-ate PSpice simulation. CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS 420 77. The circuit of Fig. 10.79 is unfortunately operating differently than speciﬁed; the frequency of the current source is only 19 rad/s. Calculate the actual capac-itor voltage, and compare it to the expected voltage had the circuit been operat-ing correctly. 78. For the circuit shown in Fig. 10.80, (a) draw the corresponding phasor repre-sentation; (b) obtain an expression for Vo/Vs; (c) Plot \|Vo/Vs\|, the magnitude of the phasor voltage ratio, as a function of frequency ω over the range 0.01 ≤ ω ≤100 rad/s (use a logarithmic x axis). (d) Does the circuit transfer low fre-quencies or high frequencies more effectively to the output? 79. (a) Replace the inductor in the circuit of Fig. 10.80 with a 1 F capacitor and repeat Exercise. 10.78. (b) If we design the “corner frequency” of the circuit as the frequency at which the output is reduced to 1/ √ 2 times its maximum value, redesign the circuit to achieve a corner frequency of 2 kHz. 80. Design a purely passive network (containing only resistors, capacitors, and inductors) which has an impedance of (22 −j7)/5/8◦ at a frequency of f = 100 MHz. ■FIGURE 10.79 100 mH 15 mF 150 mH 2 1 2 sin (20t + 45) A 5 sin (20t + 12) V + – vC + – ■FIGURE 10.80 + – vo(t) vs(t) 1 1 H + – INTRODUCTION Often an integral part of circuit analysis is the determination of either power delivered or power absorbed (or both). In the context of ac power, we ﬁnd that the rather simple approach we have taken previously does not provide a convenient picture of how a particu-lar system is operating, so we introduce several different power-related quantities in this chapter. We begin by considering instantaneous power, the product of the time-domain voltage and time-domain current associated with the element or network of interest. The instantaneous power is sometimes quite useful in its own right, because its maximum value might have to be limited to avoid exceeding the safe operating range of a physical device. For example, transistor and vacuum-tube power ampliﬁers both produce a distorted output, and speakers give a distorted sound, when the peak power exceeds a certain limiting value. However, we are mainly interested in instantaneous power for the simple reason that it provides us with the means to calculate a more important quantity, the average power. In a similar way, the progress of a cross-country road trip is best described by the average velocity; our interest in the instan-taneous velocity is limited to the avoidance of maximum velocities that will endanger our safety or arouse the highway patrol. In practical problems we will deal with values of average power which range from the small fraction of a picowatt available in a telemetry signal from outer space, to the few watts of audio power supplied to the speakers in a good stereo system, to the sev-eral hundred watts required to run the morning coffeepot, or to the 10 billion watts generated at the Grand Coulee Dam. Still, we will KEY CONCEPTS Calculating Instantaneous Power Average Power Supplied by a Sinusoidal Source Root-Mean-Square (RMS) Values Reactive Power The Relationship Between Complex, Average, and Reactive Power Power Factor of a Load AC Circuit Power Analysis C H A P T E R 11 421 see that even the concept of average power has its limitations, especially when dealing with the energy exchange between reactive loads and power sources. This is easily handled by introducing the concepts of reactive power, complex power, and the power factor—all very common terms in the power industry. 11.1 • INSTANTANEOUS POWER The instantaneous power delivered to any device is given by the product of the instantaneous voltage across the device and the instantaneous current through it (the passive sign convention is assumed). Thus,1 p(t) = v(t)i(t) If the device in question is a resistor of resistance R, then the power may be expressed solely in terms of either the current or the voltage: p(t) = v(t)i(t) = i2(t)R = v2(t) R If the voltage and current are associated with a device that is entirely induc-tive, then p(t) = v(t)i(t) = Li(t) di(t) dt = 1 L v(t) t −∞ v(t′) dt′ where we will arbitrarily assume that the voltage is zero at t = −∞. In the case of a capacitor, p(t) = v(t)i(t) = Cv(t) dv(t) dt = 1 C i(t) t −∞ i(t′) dt′ where a similar assumption about the current is made. For example, consider the series RL circuit as shown in Fig. 11.1, excited by a step-voltage source. The familiar current response is i(t) = V0 R (1 −e−Rt/L)u(t) and thus the total power delivered by the source or absorbed by the passive network is p(t) = v(t)i(t) = V 2 0 R (1 −e−Rt/L)u(t) since the square of the unit-step function is simply the unit-step function itself. The power delivered to the resistor is pR(t) = i2(t)R = V 2 0 R (1 −e−Rt/L)2u(t) CHAPTER 11 AC CIRCUIT POWER ANALYSIS 422 ■FIGURE 11.1 The instantaneous power that is delivered to R is pR (t) = i 2(t)R = (V 2 0 /R )(1 −e−R t/L )2u(t). + – i vL + – V0u(t) L R (1) Earlier, we agreed that lowercase variables in italics were understood to be functions of time, and we have carried on in this spirit up to now. However, in order to emphasize the fact that these quantities must be evaluated at a speciﬁc instant in time, we will explicitly denote the time dependence throughout this chapter. SECTION 11.1 INSTANTANEOUS POWER 423 In order to determine the power absorbed by the inductor, we ﬁrst obtain the inductor voltage: vL(t) = L di(t) dt = V0e−Rt/Lu(t) + LV0 R (1 −e−Rt/L) du(t) dt = V0e−Rt/Lu(t) since du(t)/dt is zero for t > 0 and (1 −e−Rt/L) is zero at t = 0. The power absorbed by the inductor is therefore pL(t) = vL(t)i(t) = V 2 0 R e−Rt/L(1 −e−Rt/L)u(t) Only a few algebraic manipulations are required to show that p(t) = pR(t) + pL(t) which serves to check the accuracy of our work; the results are sketched in Fig. 11.2. Power Due to Sinusoidal Excitation Let us change the voltage source in the circuit of Fig. 11.1 to the sinusoidal source Vm cos ωt. The familiar time-domain steady-state response is i(t) = Im cos(ωt + φ) where Im = Vm √ R2 + ω2L2 and φ = −tan−1 ωL R The instantaneous power delivered to the entire circuit in the sinusoidal steady state is, therefore, p(t) = v(t)i(t) = Vm Im cos(ωt + φ) cos ωt which we will ﬁnd convenient to rewrite in a form obtained by using the trigonometric identity for the product of two cosine functions. Thus, p(t) = Vm Im 2 [cos(2ωt + φ) + cos φ] = Vm Im 2 cos φ + Vm Im 2 cos(2ωt + φ) The last equation possesses several characteristics that are true in general for circuits in the sinusoidal steady state. One term, the ﬁrst, is not a function of time; and a second term is included which has a cyclic variation at twice the applied frequency. Since this term is a cosine wave, and since sine waves and cosine waves have average values which are zero (when averaged over an integral number of periods), this example suggests that the average power is 1 2Vm Im cos φ; as we will see shortly, this is indeed the case. ■FIGURE 11.2 Sketch of p(t), pR(t), and pL(t). As the transient dies out, the circuit returns to steady-state operation. Since the only source remaining in the circuit is dc, the inductor eventually acts as a short circuit absorbing zero power. 0 1 2 3 4 5 6 7 8 9 10 Power supplied by source Power absorbed by resistor Power absorbed by inductor t Power 11.2 • AVERAGE POWER When we speak of an average value for the instantaneous power, the time interval over which the averaging process takes place must be clearly de-ﬁned. Let us ﬁrst select a general interval of time from t1 to t2. We may then obtain the average value by integrating p(t) from t1 to t2 and dividing the result by the time interval t2 −t1. Thus, P = 1 t2 −t1 t2 t1 p(t) dt The average value is denoted by the capital letter P, since it is not a function of time, and it usually appears without any speciﬁc subscripts that identify it as an average value. Although P is not a function of time, it is a function CHAPTER 11 AC CIRCUIT POWER ANALYSIS 424 EXAMPLE 11.1 A voltage source, 40 + 60u(t) V, a 5 μF capacitor, and a 200 resistor form a series circuit. Find the power being absorbed by the capacitor and by the resistor at t 1.2 ms. At t = 0−, no current is ﬂowing and so 40 V appears across the capaci-tor. At t = 0+, the voltage across the capacitor-resistor series combina-tion jumps to 100 V. Since vC cannot change in zero time, the resistor voltage at t = 0+ is 60 V. The current ﬂowing through all three elements at t = 0+ is therefore 60/200 = 300 mA and for t > 0 is given by i(t) = 300e−t/τ mA where τ = RC = 1 ms. Thus, the current ﬂowing at t = 1.2 ms is 90.36 mA, and the power being absorbed by the resistor at that instant is simply i2(t)R = 1.633 W The instantaneous power absorbed by the capacitor is i(t)vC(t). Recog-nizing that the total voltage across both elements for t > 0 will always be 100 V, and that the resistor voltage is given by 60e−t/τ , vC(t) = 100 −60e−t/τ and we ﬁnd that vC(1.2 ms) = 100 −60e−1.2 = 81.93 V. Thus, the power being absorbed by the capacitor at t = 1.2 ms is (90.36 mA) (81.93 V) = 7.403 W. PRACTICE ● 11.1 A current source of 12 cos 2000t A, a 200 resistor, and a 0.2 H inductor are in parallel. Assume steady-state conditions exist. At t = 1 ms, ﬁnd the power being absorbed by the (a) resistor; (b) inductor; (c) sinusoidal source. Ans: 13.98 kW; −5.63 kW; −8.35 kW. SECTION 11.2 AVERAGE POWER 425 of t1 and t2, the two instants of time which deﬁne the interval of integration. This dependence of P on a speciﬁc time interval may be expressed in a simpler manner if p(t) is a periodic function. We consider this important case ﬁrst. Average Power for Periodic Waveforms Let us assume that our forcing function and the circuit responses are all pe-riodic; a steady-state condition has been reached, although not necessarily the sinusoidal steady state. We may deﬁne a periodic function f (t) mathe-matically by requiring that f (t) = f (t + T) where T is the period. We now show that the average value of the instanta-neous power as expressed by Eq. may be computed over an interval of one period having an arbitrary beginning. A general periodic waveform is shown in Fig. 11.3 and identiﬁed as p(t). We ﬁrst compute the average power by integrating from t1 to a time t2 which is one period later, t2 = t1 + T: P1 = 1 T t1+T t1 p(t) dt and then by integrating from some other time tx to tx + T : Px = 1 T tx+T tx p(t) dt The equality of P1 and Px should be evident from the graphical interpre-tation of the integrals; the periodic nature of the curve requires the two areas to be equal. Thus, the average power may be computed by integrating the instantaneous power over any interval that is one period in length and then dividing by the period: P = 1 T tx+T tx p(t) dt It is important to note that we may also integrate over any integral number of periods, provided that we divide by the same integral number of periods. Thus, P = 1 nT tx+nT tx p(t) dt n = 1, 2, 3, . . . If we carry this concept to the extreme by integrating over all time, another useful result is obtained. We ﬁrst provide ourselves with symmetrical limits on the integral P = 1 nT nT/2 −nT/2 p(t) dt and then take the limit as n becomes inﬁnite, P = lim n→∞ 1 nT nT/2 −nT/2 p(t) dt ■FIGURE 11.3 The average value P of a periodic function p(t) is the same over any period T. t1 t1 + T tx + T tx p(t) t As long as p(t) is a mathematically well-behaved function, as all physi-cal forcing functions and responses are, it is apparent that if a large integer n is replaced by a slightly larger number which is not an integer, then the value of the integral and of P is changed by a negligible amount; moreover, the error decreases as n increases. Without justifying this step rigorously, we therefore replace the discrete variable nT with the continuous variable τ: P = lim τ→∞ 1 τ τ/2 −τ/2 p(t) dt We will ﬁnd it convenient on several occasions to integrate periodic func-tions over this “inﬁnite period.” Average Power in the Sinusoidal Steady State Now let us obtain the general result for the sinusoidal steady state. We assume the general sinusoidal voltage v(t) = Vm cos(ωt + θ) and current i(t) = Im cos(ωt + φ) associated with the device in question. The instantaneous power is p(t) = Vm Im cos(ωt + θ) cos(ωt + φ) Again expressing the product of two cosine functions as one-half the sum of the cosine of the difference angle and the cosine of the sum angle, p(t) = 1 2Vm Im cos(θ −φ) + 1 2Vm Im cos(2ωt + θ + φ) we may save ourselves some integration by an inspection of the result. The ﬁrst term is a constant, independent of t. The remaining term is a cosine function; p(t) is therefore periodic, and its period is 1 2T . Note that the pe-riod T is associated with the given current and voltage, and not with the power; the power function has a period 1 2T . However, we may integrate over an interval of T to determine the average value if we wish; it is neces-sary only that we also divide by T. Our familiarity with cosine and sine waves, however, shows that the average value of either over a period is zero. There is thus no need to integrate Eq. formally; by inspection, the av-erage value of the second term is zero over a period T (or 1 2T ), and the av-erage value of the ﬁrst term, a constant, must be that constant itself. Thus, This important result, introduced in the previous section for a specific circuit, is therefore quite general for the sinusoidal steady state. The average power is one-half the product of the crest amplitude of the voltage, the crest amplitude of the current, and the cosine of the phase-angle difference be-tween the current and the voltage; the sense of the difference is immaterial. Two special cases are worth isolating for consideration: the average power delivered to an ideal resistor and that to an ideal reactor (any combi-nation of only capacitors and inductors). P = 1 2Vm Im cos(θ −φ) CHAPTER 11 AC CIRCUIT POWER ANALYSIS 426 Recall that T = 1 f = 2π ω . SECTION 11.2 AVERAGE POWER 427 Given the time-domain voltage v 4 cos(πt/6) V, ﬁnd both the average power and an expression for the instantaneous power that result when the corresponding phasor voltage V 4/0° V is applied across an impedance Z 2/60° . The phasor current is V/Z = 2/−60◦A, and so the average power is P = 1 2(4)(2) cos 60◦= 2 W We can write the time-domain voltage, v(t) = 4 cos πt 6 V and the time-domain current, i(t) = 2 cos πt 6 −60◦ A The instantaneous power, therefore, is given by their product: p(t) = 8 cos πt 6 cos πt 6 −60◦ = 2 + 4 cos πt 3 −60◦ W All three quantities are sketched on the same time axis in Fig. 11.4. Both the 2 W average value of the power and its period of 6 s, one-half the period of either the current or the voltage, are evident. The zero value of the instantaneous power at each instant when either the voltage or current is zero is also apparent. EXAMPLE 11.2 ■FIGURE 11.4 Curves of v(t), i(t), and p(t) are plotted as functions of time for a simple circuit in which the phasor voltage V = 4/0◦V is applied to the impedance Z = 2/60◦ at ω = π/6 rad/s. 6 p v i 4 –2 –4 1 –1 6 4 –3 2 3 5 8 12 p, v, i (W , V , A) t(s) PRACTICE ● 11.2 Given the phasor voltage V = 115 √ 2/45◦V across an impedance Z = 16.26/19.3◦, obtain an expression for the instantaneous power, and compute the average power if ω = 50 rad/s. Ans: 767.5 + 813.2 cos(100t + 70.7◦) W; 767.5 W. Average Power Absorbed by an Ideal Resistor The phase-angle difference between the current through and the voltage across a pure resistor is zero. Thus, PR = 1 2Vm Im cos 0 = 1 2Vm Im or PR = 1 2 I 2 m R or PR = V 2 m 2R The last two formulas, enabling us to determine the average power de-livered to a pure resistance from a knowledge of either the sinusoidal cur-rent or voltage, are simple and important. Unfortunately, they are often mis-used. The most common error is made in trying to apply them in cases where the voltage included in Eq. is not the voltage across the resistor. If care is taken to use the current through the resistor in Eq. and the voltage across the resistor in Eq. , satisfactory operation is guaranteed. Also, do not forget the factor of 1 2! Average Power Absorbed by Purely Reactive Elements The average power delivered to any device which is purely reactive (i.e., contains no resistors) must be zero. This is a direct result of the 90◦phase dif-ference which must exist between current and voltage; hence, cos(θ −φ) = cos ± 90◦= 0 and PX = 0 The average power delivered to any network composed entirely of ideal inductors and capacitors is zero; the instantaneous power is zero only at spe-ciﬁc instants. Thus, power ﬂows into the network for a part of the cycle and out of the network during another portion of the cycle, with no power lost. CHAPTER 11 AC CIRCUIT POWER ANALYSIS 428 Keep in mind that we are computing the average power delivered to a resistor by a sinusoidal source; take care not to confuse this quantity with the instantaneous power, which has a similar form. Find the average power being delivered to an impedance ZL 8 −j11 by a current I 5/20◦A. We may ﬁnd the solution quite rapidly by using Eq. . Only the 8 resistance enters the average-power calculation, since the j11 compo-nent will not absorb any average power. Thus, P = 1 2(52)8 = 100 W PRACTICE ● 11.3 Calculate the average power delivered to the impedance 6/25◦ by the current I = 2 + j5 A. Ans: 78.85 W. EXAMPLE 11.3 SECTION 11.2 AVERAGE POWER 429 EXAMPLE 11.4 Find the average power absorbed by each of the three passive elements in Fig. 11.5, as well as the average power supplied by each source. ■FIGURE 11.5 The average power delivered to each reactive element is zero in the sinusoidal steady state. I1 I2 2 j2 –j2 0 V 10 0 V 20 + – + – Without even analyzing the circuit, we already know that the average power absorbed by the two reactive elements is zero. The values of I1 and I2 are found by any of several methods, such as mesh analysis, nodal analysis, or superposition. They are I1 = 5 −j10 = 11.18/−63.43◦A I2 = 5 −j5 = 7.071/−45◦A The downward current through the 2 resistor is I1 −I2 = −j5 = 5/−90◦A so that Im = 5 A, and the average power absorbed by the resistor is found most easily by Eq. : PR = 1 2 I 2 m R = 1 2(52)2 = 25 W This result may be checked by using Eq. or Eq. . We next turn to the left source. The voltage 20/0◦V and associated current I1 = 11.18/−63.43◦A satisfy the active sign convention, and thus the power delivered by this source is Pleft = 1 2(20)(11.18) cos[0◦−(−63.43◦)] = 50 W In a similar manner, we ﬁnd the power absorbed by the right source using the passive sign convention, Pright = 1 2(10)(7.071) cos(0◦+ 45◦) = 25 W Since 50 = 25 + 25, the power relations check. ■FIGURE 11.6 –j100 2 j45 I1 I2 0 V 5 50 V 10 + – + – Ans: 0, 37.6 mW, 0, 42.0 mW, −4.4 mW. PRACTICE ● 11.4 For the circuit of Fig. 11.6, compute the average power delivered to each of the passive elements. Verify your answer by computing the power delivered by the two sources. Maximum Power Transfer We previously considered the maximum power transfer theorem as it applied to resistive loads and resistive source impedances. For a Thévenin source Vth and impedance Zth = Rth + jXth connected to a load ZL = RL + j X L, it may be shown that the average power delivered to the load is a maxi-mum when RL = Rth and X L = −Xth, that is, when ZL = Z∗ th. This result is often digniﬁed by calling it the maximum power transfer theorem for the sinusoidal steady state: CHAPTER 11 AC CIRCUIT POWER ANALYSIS 430 ■FIGURE 11.7 A simple loop circuit used to illustrate the derivation of the maximum power transfer theorem as it applies to circuits operating in the sinusoidal steady state. VL + – Vth IL + – Zth ZL The notation Z denotes the complex conjugate of the complex number Z. It is formed by replacing all “j”s with “−j ”s. See Appendix 5 for more details. An independent voltage source in series with an impedance Zth or an independent current source in parallel with an impedance Zth delivers a maximum average power to that load impedance ZL which is the conjugate of Zth, or ZL = Z∗ th. The details of the proof are left to the reader, but the basic approach can be understood by considering the simple loop circuit of Fig. 11.7. The Thévenin equivalent impedance Zth may be written as the sum of two com-ponents, Rth + jXth, and in a similar fashion the load impedance ZL may be written as RL + jX L. The current ﬂowing through the loop is IL = Vth Zth + ZL = Vth Rth + jXth + RL + j X L = Vth Rth + RL + j(Xth + X L) and VL = Vth ZL Zth + ZL = Vth RL + jX L Rth + jXth + RL + jX L = Vth RL + jX L Rth + RL + j(Xth + X L) The magnitude of IL is \|Vth\| (Rth + RL)2 + (Xth + X L)2 and the phase angle is Vth −tan−1 Xth + X L Rth + RL Similarly, the magnitude of VL is \|Vth\| R2 L + X2 L (Rth + RL)2 + (Xth + X L)2 and its phase angle is Vth + tan−1 X L RL −tan−1 Xth + X L Rth + RL SECTION 11.2 AVERAGE POWER 431 Referring to Eq. , then, we ﬁnd an expression for the average power P delivered to the load impedance ZL: P = 1 2\|Vth\|2 R2 L + X2 L (Rth + RL)2 + (Xth + X L)2 cos tan−1 X L RL In order to prove that maximum average power is indeed delivered to the load when ZL = Z∗ th, we must perform two separate steps. First, the derivative of Eq. with respect to RL must be set to zero. Second, the derivative of Eq. with respect to XL must be set to zero. The remaining details are left as an exercise for the avid reader. EXAMPLE 11.5 A particular circuit is composed of the series combination of a sinusoidal voltage source 3 cos(100t −3°) V, a 500 resistor, a 30 mH inductor, and an unknown impedance. If we are assured that the voltage source is delivering maximum average power to the unknown impedance, what is its value? The phasor representation of the circuit is sketched in Fig. 11.8. The circuit is easily seen as an unknown impedance Z? in series with a Thévenin equivalent consisting of the 3/−3◦V source and a Thévenin impedance 500 + j3 . Since the circuit of Fig. 11.8 is already in the form required to employ the maximum average power transfer theorem, we know that maximum average power will be transferred to an impedance equal to the complex conjugate of Zth, or Z? = Z∗ th = 500 −j3 This impedance can be constructed in several ways, the simplest being a 500 resistor in series with a capacitor having impedance −j3 . Since the operating frequency of the circuit is 100 rad/s, this corre-sponds to a capacitance of 3.333 mF. PRACTICE ● 11.5 If the 30 mH inductor of Example 11.5 is replaced with a 10 μF capacitor, what is the value of the inductive component of the unknown impedance Z? if it is known that Z? is absorbing maximum power? Ans: 10 H. ■FIGURE 11.8 The phasor representation of a simple series circuit composed of a sinusoidal voltage source, a resistor, an inductor, and an unknown impedance. 500 j3 –3 V 3 + – Z? Average Power for Nonperiodic Functions We should pay some attention to nonperiodic functions. One practical ex-ample of a nonperiodic power function for which an average power value is desired is the power output of a radio telescope directed toward a “radio star.” Another is the sum of a number of periodic functions, each function having a different period, such that no greater common period can be found for the combination. For example, the current i(t) = sin t + sin πt CHAPTER 11 AC CIRCUIT POWER ANALYSIS 432 is nonperiodic because the ratio of the periods of the two sine waves is an irrational number. At t = 0, both terms are zero and increasing. But the ﬁrst term is zero and increasing only when t = 2πn, where n is an integer, and thus periodicity demands that πt or π(2πn) must equal 2πm, where m is also an integer. No solution (integral values for both m and n) for this equation is possible. It may be illuminating to compare the nonperiodic expression in Eq. with the periodic function i(t) = sin t + sin 3.14t where 3.14 is an exact decimal expression and is not to be interpreted as 3.141592. . . . With a little effort,2 it can be shown that the period of this cur-rent wave is 100π seconds. The average value of the power delivered to a 1 resistor by either a pe-riodic current such as Eq. or a nonperiodic current such as Eq. may be found by integrating over an inﬁnite interval. Much of the actual inte-gration can be avoided because of our thorough knowledge of the average values of simple functions. We therefore obtain the average power delivered by the current in Eq. by applying Eq. : P = lim τ→∞ 1 τ τ/2 −τ/2 (sin2 t + sin2 πt + 2 sin t sin πt) dt We now consider P as the sum of three average values. The average value of sin2 t over an inﬁnite interval is found by replacing sin2 t with ( 1 2 −1 2 cos 2t); the average is simply 1 2. Similarly, the average value of sin2 πt is also 1 2. And the last term can be expressed as the sum of two cosine func-tions, each of which must certainly have an average value of zero. Thus, P = 1 2 + 1 2 = 1 W An identical result is obtained for the periodic current of Eq. . Applying this same method to a current function which is the sum of several sinusoids of different periods and arbitrary amplitudes, i(t) = Im1 cos ω1t + Im2 cos ω2t + · · · + ImN cos ωNt we ﬁnd the average power delivered to a resistance R, P = 1 2 I 2 m1 + I 2 m2 + · · · + I 2 mN R The result is unchanged if an arbitrary phase angle is assigned to each component of the current. This important result is surprisingly simple when we think of the steps required for its derivation: squaring the current function, integrating, and taking the limit. The result is also just plain (2) T1 = 2π and T2 = 2π/3.14. Therefore, we seek integral values of m and n such that 2πn = 2πm/3.14, or 3.14n = m, or 314 100 n = m or 157n = 50m. Thus, the smallest integral values for n and m are n = 50 and m = 157. The period is therefore T = 2πn = 100π, or T = 2π(157/3.14) = 100π s. Find the average power delivered to a 4 resistor by the current i1 2 cos 10t −3 cos 20t A. Since the two cosine terms are at different frequencies, the two average-power values may be calculated separately and added. Thus, this current delivers 1 2(22)4 + 1 2(32)4 = 8 + 18 = 26 W to a 4 resistor. EXAMPLE 11.6 Find the average power delivered to a 4 resistor by the current i2 2 cos 10t −3 cos 10t A. Here, the two components of the current are at the same frequency, and they must therefore be combined into a single sinusoid at that frequency. Thus, i2 = 2 cos 10t −3 cos 10t = −cos 10t delivers only 1 2(12)4 = 2 W of average power to a 4 resistor. PRACTICE ● 11.6 A voltage source vs is connected across a 4 resistor. Find the average power absorbed by the resistor if vs equals (a) 8 sin 200t V; (b) 8 sin 200t −6 cos(200t −45◦) V; (c) 8 sin 200t −4 sin 100t V; (d) 8 sin 200t −6 cos(200t −45◦) −5 sin 100t + 4 V. Ans: 8.00 W; 4.01 W; 10.00 W; 11.14 W. SECTION 11.3 EFFECTIVE VALUES OF CURRENT AND VOLTAGE 433 surprising, because it shows that, in this special case of a current such as Eq. , where each term has a unique frequency, superposition is applic-able to power. Superposition is not applicable for a current which is the sum of two direct currents, nor is it applicable for a current which is the sum of two sinusoids of the same frequency. 11.3 • EFFECTIVE VALUES OF CURRENT AND VOLTAGE In North America, most power outlets deliver a sinusoidal voltage having a frequency of 60 Hz and a “voltage” of 115 V (elsewhere, 50 Hz and 240 V are typically encountered). But what is meant by “115 volts”? This is cer-tainly not the instantaneous value of the voltage, for the voltage is not a con-stant. The value of 115 V is also not the amplitude which we have been symbolizing as Vm; if we displayed the voltage waveform on a calibrated oscilloscope, we would ﬁnd that the amplitude of this voltage at one of our ac outlets is 115 √ 2, or 162.6, volts. We also cannot ﬁt the concept of an av-erage value to the 115 V, because the average value of the sine wave is zero. We might come a little closer by trying the magnitude of the average over a positive or negative half cycle; by using a rectiﬁer-type voltmeter at the out-let, we should measure 103.5 V. As it turns out, however, 115 V is the effective value of this sinusoidal voltage; it is a measure of the effectiveness of a voltage source in delivering power to a resistive load. Effective Value of a Periodic Waveform Let us arbitrarily deﬁne effective value in terms of a current waveform, al-though a voltage could equally well be selected. The effective value of any periodic current is equal to the value of the direct current which, ﬂowing through an R ohm resistor, delivers the same average power to the resistor as does the periodic current. In other words, we allow the given periodic current to ﬂow through the resistor, determine the instantaneous power i2R, and then ﬁnd the average value of i2R over a period; this is the average power. We then cause a direct EXAMPLE 11.7 current to ﬂow through this same resistor and adjust the value of the direct current until the same value of average power is obtained. The resulting magnitude of the direct current is equal to the effective value of the given periodic current. These ideas are illustrated in Fig. 11.9. The general mathematical expression for the effective value of i(t) is now easily obtained. The average power delivered to the resistor by the periodic current i(t) is P = 1 T T 0 i2R dt = R T T 0 i2 dt where the period of i(t) is T. The power delivered by the direct current is P = I 2 effR Equating the power expressions and solving for Ieff, we get The result is independent of the resistance R, as it must be to provide us with a worthwhile concept. A similar expression is obtained for the effective value of a periodic voltage by replacing i and Ieff by v and Veff, respectively. Notice that the effective value is obtained by ﬁrst squaring the time func-tion, then taking the average value of the squared function over a period, and ﬁnally taking the square root of the average of the squared function. In short, the operation involved in ﬁnding an effective value is the (square) root of the mean of the square; for this reason, the effective value is often called the root-mean-square value, or simply the rms value. Effective (RMS) Value of a Sinusoidal Waveform The most important special case is that of the sinusoidal waveform. Let us select the sinusoidal current i(t) = Im cos(ωt + φ) which has a period T = 2π ω and substitute in Eq. to obtain the effective value Ieff = 1 T T 0 I 2 m cos2(ωt + φ) dt = Im ω 2π 2π/ω 0 1 2 + 1 2 cos(2ωt + 2φ) dt = Im ω 4π [t]2π/ω 0 = Im √ 2 Ieff = 1 T T 0 i2 dt CHAPTER 11 AC CIRCUIT POWER ANALYSIS 434 ■FIGURE 11.9 If the resistor receives the same average power in parts a and b, then the effective value of i(t) is equal to Ieff, and the effective value of v(t) is equal to Veff. + – i(t) R v(t) (a) R Ieff Veff (b) SECTION 11.3 EFFECTIVE VALUES OF CURRENT AND VOLTAGE 435 Thus the effective value of a sinusoidal current is a real quantity which is in-dependent of the phase angle and numerically equal to 1/ √ 2 = 0.707 times the amplitude of the current. A current √ 2 cos(ωt + φ) A, therefore, has an effective value of 1 A and will deliver the same average power to any resis-tor as will a direct current of 1 A. It should be noted carefully that the √ 2 factor that we obtained as the ratio of the amplitude of the periodic current to the effective value is applicable only when the periodic function is sinusoidal. For a sawtooth waveform, for example, the effective value is equal to the maximum value divided by √ 3. The factor by which the maximum value must be divided to obtain the effec-tive value depends on the mathematical form of the given periodic function; it may be either rational or irrational, depending on the nature of the function. Use of RMS Values to Compute Average Power The use of the effective value also simpliﬁes slightly the expression for the average power delivered by a sinusoidal current or voltage by avoiding use of the factor 1 2. For example, the average power delivered to an R ohm resistor by a sinusoidal current is P = 1 2 I 2 m R Since Ieff = Im/ √ 2, the average power may be written as P = I 2 effR The other power expressions may also be written in terms of effective values: P = VeffIeff cos(θ −φ) P = V 2 eff R Although we have succeeded in eliminating the factor 1 2 from our average-power relationships, we must now take care to determine whether a sinusoidal quantity is expressed in terms of its amplitude or its effective value. In practice, the effective value is usually used in the ﬁelds of power transmission or distribution and of rotating machinery; in the areas of elec-tronics and communications, the amplitude is more often used. We will as-sume that the amplitude is speciﬁed unless the term “rms” is explicitly used, or we are otherwise instructed. In the sinusoidal steady state, phasor voltages and currents may be given either as effective values or as amplitudes; the two expressions dif-fer only by a factor of √ 2. The voltage 50/30◦V is expressed in terms of an amplitude; as an rms voltage, we should describe the same voltage as 35.4/30◦V rms. Effective Value with Multiple-Frequency Circuits In order to determine the effective value of a periodic or nonperiodic waveform which is composed of the sum of a number of sinusoids of dif-ferent frequencies, we may use the appropriate average-power relation-ship of Eq. , developed in Sec. 11.2, rewritten in terms of the effective values of the several components: P = I 2 1eff + I 2 2eff + · · · + I 2 Neff R The fact that the effective value is deﬁned in terms of an equivalent dc quantity provides us with average-power formulas for resistive circuits which are identical with those used in dc analysis. Note that the effective value of a dc quantity K is simply K, not K √ 2 . From this we see that the effective value of a current which is composed of any number of sinusoidal currents of different frequencies can be expressed as Ieff = I 2 1eff + I 2 2eff + · · · + I 2 Neff These results indicate that if a sinusoidal current of 5 A rms at 60 Hz ﬂows through a 2 resistor, an average power of 52(2) = 50 W is absorbed by the resistor; if a second current—perhaps 3 A rms at 120 Hz, for example— is also present, the absorbed power is 32(2) + 50 = 68 W. Using Eq. instead, we ﬁnd that the effective value of the sum of the 60 and 120 Hz currents is 5.831 A. Thus, P = 5.8312(2) = 68 W as before. However, if the second current is also at 60 Hz, the effective value of the sum of the two 60 Hz currents may have any value between 2 and 8 A. Thus, the ab-sorbed power may have any value between 8 W and 128 W, depending on the relative phase of the two current components. PRACTICE ● 11.7 Calculate the effective value of each of the periodic voltages: (a) 6 cos 25t; (b) 6 cos 25t + 4 sin(25t + 30◦); (c) 6 cos 25t + 5 cos2(25t); (d) 6 cos 25t + 5 sin 30t + 4 V. Ans: 4.24 V; 6.16 V; 5.23 V; 6.82 V. CHAPTER 11 AC CIRCUIT POWER ANALYSIS 436 Several useful techniques are available through PSpice for calculation of power quantities. In particular, the built-in functions of Probe allow us to both plot the instantaneous power and compute the average power. For example, consider the simple voltage divider circuit of Fig. 11.10, which is being driven by a 60 Hz sine wave with an ■FIGURE 11.10 A simple voltage divider circuit driven by a 115 V rms source operating at 60 Hz. COMPUTER-AIDED ANALYSIS SECTION 11.3 EFFECTIVE VALUES OF CURRENT AND VOLTAGE 437 ■FIGURE 11.11 Current and instantaneous power associated with resistor R1 of Fig. 11.10. amplitude of 115 √ 2 V. We begin by performing a transient simulation over one period of the voltage waveform, 1 60 s. The current along with the instantaneous power dissipated in resistor R1 is plotted in Fig. 11.11 by employing the Add Plot to Window option under Plot. The instantaneous power is periodic, with a nonzero average value and a peak of 6.61 W. The easiest means of using Probe to obtain the average power, which we expect to be 1 2 162.6 1000 1000+1000 (81.3 × 10−3) = 3.305 W, is to make use of the built-in “running average” function. Once the Add Traces dialog box appears (Trace, Add Trace . . .), type AVG(I(R1) ∗I(R1) ∗1000) in the Trace Expression window. As can be seen in Fig. 11.12, the average value of the power over either one or two periods is 3.305 W, in agreement with the hand calculation. Note that since PSpice only calculates at speciﬁc times, the circuit was not simulated at precisely 8.333 ms and hence Cursor 1 in-dicates a slightly higher average power. Probe also allows us to compute the average over a speciﬁc interval using the built-in function avgx. For example, to use this function to compute the average power over a single period, which in this case is 1/120 = 8.33 ms, we would enter AVGX(I(R1) ∗I(R1) ∗1000, 8.33 m) Either approach will result in a value of 3.305 W at the endpoint of the plot. 0 1 (Continued on next page) CHAPTER 11 AC CIRCUIT POWER ANALYSIS 438 ■FIGURE 11.12 Calculated running average of the power dissipated by resistor R1. 11.4 • APPARENT POWER AND POWER FACTOR Historically, the introduction of the concepts of apparent power and power factor can be traced to the electric power industry, where large amounts of electric energy must be transferred from one point to another; the efﬁciency with which this transfer is effected is related directly to the cost of the elec-tric energy, which is eventually paid by the consumer. Customers who pro-vide loads which result in a relatively poor transmission efﬁciency must pay a greater price for each kilowatthour (kWh) of electric energy they actually receive and use. In a similar way, customers who require a costlier invest-ment in transmission and distribution equipment by the power company will also pay more for each kilowatthour unless the company is benevolent and enjoys losing money. Let us ﬁrst deﬁne apparent power and power factor and then show brieﬂy how these terms are related to practical economic situations. We assume that the sinusoidal voltage v = Vm cos(ωt + θ) is applied to a network and the resultant sinusoidal current is i = Im cos(ωt + φ) The phase angle by which the voltage leads the current is therefore (θ −φ). The average power delivered to the network, assuming a passive sign SECTION 11.4 APPARENT POWER AND POWER FACTOR 439 convention at its input terminals, may be expressed either in terms of the maximum values: P = 1 2Vm Im cos(θ −φ) or in terms of the effective values: P = VeffIeff cos(θ −φ) If our applied voltage and current responses had been dc quantities, the average power delivered to the network would have been given simply by the product of the voltage and the current. Applying this dc technique to the sinusoidal problem, we should obtain a value for the absorbed power which is “apparently’’ given by the familiar product Veff Ieff. However, this product of the effective values of the voltage and cur-rent is not the average power; we define it as the apparent power. Di-mensionally, apparent power must be measured in the same units as real power, since cos(θ −φ) is dimensionless; but in order to avoid confu-sion, the term volt-amperes, or VA, is applied to the apparent power. Since cos(θ −φ) cannot have a magnitude greater than unity, the magni-tude of the real power can never be greater than the magnitude of the apparent power. The ratio of the real or average power to the apparent power is called the power factor, symbolized by PF. Hence, PF = average power apparent power = P VeffIeff In the sinusoidal case, the power factor is simply cos(θ −φ), where (θ −φ) is the angle by which the voltage leads the current. This relation-ship is the reason why the angle (θ −φ) is often referred to as the PF angle. For a purely resistive load, the voltage and current are in phase, (θ −φ) is zero, and the PF is unity. In other words, the apparent power and the average power are equal. Unity PF, however, may also be achieved for loads that contain both inductance and capacitance if the element values and the operating frequency are carefully selected to provide an input imped-ance having a zero phase angle. A purely reactive load, that is, one contain-ing no resistance, will cause a phase difference between the voltage and current of either plus or minus 90°, and the PF is therefore zero. Between these two extreme cases there are the general networks for which the PF can range from zero to unity. A PF of 0.5, for example, indi-cates a load having an input impedance with a phase angle of either 60° or −60◦; the former describes an inductive load, since the voltage leads the current by 60°, while the latter refers to a capacitive load. The ambiguity in the exact nature of the load is resolved by referring to a leading PF or a lag-ging PF, the terms leading or lagging referring to the phase of the current with respect to the voltage. Thus, an inductive load will have a lagging PF and a capacitive load a leading PF. Apparent power is not a concept which is limited to sinusoidal forcing functions and responses. It may be determined for any current and voltage waveshapes by simply taking the product of the effective values of the current and voltage. CHAPTER 11 AC CIRCUIT POWER ANALYSIS 440 EXAMPLE 11.8 Calculate values for the average power delivered to each of the two loads shown in Fig. 11.13, the apparent power supplied by the source, and the power factor of the combined loads. Identify the goal of the problem. The average power refers to the power drawn by the resistive compo-nents of the load elements; the apparent power is the product of the effective voltage and the effective current of the load combination. Collect the known information. The effective voltage is 60 V rms, which appears across a combined load of 2 −j + 1 + j5 = 3 + j4 . Devise a plan. Simple phasor analysis will provide the current. Knowing voltage and current will enable us to calculate average power and apparent power; these two quantities can be used to obtain the power factor. Construct an appropriate set of equations. The average power P supplied to a load is given by P = I 2 effR where R is the real part of the load impedance. The apparent power supplied by the source is VeffIeff, where Veff = 60 V rms. The power factor is calculated as the ratio of these two quantities: PF = average power apparent power = P VeffIeff Determine if additional information is required. We require Ieff: I = 60/0◦ 3 + j4 = 12/−53.13◦A rms so Ieff = 12 A rms, and ang I = −53.13◦. Attempt a solution. The average power delivered to the top load is given by Pupper = I 2 effRtop = (12)2(2) = 288 W and the average power delivered to the right load is given by Plower = I 2 effRright = (12)2(1) = 144 W The source itself supplies an apparent power of VeffIeff = (60)(12) = 720 VA. Finally, the power factor of the combined loads is found by consid-ering the voltage and current associated with the combined loads. This ■FIGURE 11.13 A circuit in which we seek the average power delivered to each element, the apparent power supplied by the source, and the power factor of the combined load. I 2 – j1 1 + j5 0 V rms 60 + – 11.5 • COMPLEX POWER As we saw in Chap. 10, “complex” numbers do not actually “complicate” analysis. By allowing us to carry two pieces of information together through a series of calculations via the “real” and “imaginary” compo-nents, they often greatly simplify what might otherwise be tedious calcu-lations. This is particularly true with power, since we have resistive as well as inductive and capacitive elements in a general load. In this sec-tion, we define complex power to allow us to calculate the various con-tributions to the total power in a clean, efficient fashion. The magnitude of the complex power is simply the apparent power. The real part is the average power and—as we are about to see—the imaginary part is a new quantity, termed the reactive power, which describes the rate of energy transfer into and out of reactive load components (e.g., inductors and capacitors). We deﬁne complex power with reference to a general sinusoidal voltage Veff = Veff/θ across a pair of terminals and a general sinusoidal current Ieff = Ieff φ ﬂowing into one of the terminals in such a way as to satisfy the passive sign convention. The average power P absorbed by the two-terminal network is thus P = VeffIeff cos(θ −φ) Complex nomenclature is next introduced by making use of Euler’s formula in the same way as we did in introducing phasors. We express P as P = VeffIeff Re{e j(θ−φ)} SECTION 11.5 COMPLEX POWER 441 ■FIGURE 11.14 Is 0 V rms 60 + – 2 – j1 ZL power factor is, of course, identical to the power factor for the source. Thus PF = P VeffIeff = 432 60(12) = 0.6 lagging since the combined load is inductive. Verify the solution. Is it reasonable or expected? The total average power delivered to the source is 288 + 144 = 432 W. The average power supplied by the source is P = VeffIeff cos(ang V −ang I) = (60)(12) cos(0 + 53.13◦) = 432 W so we see the power balance is correct. We might also write the combined load impedance as 5/53.1◦, identify 53.1° as the PF angle, and thus have a PF of cos 53.1° = 0.6 lagging. PRACTICE ● 11.8 For the circuit of Fig. 11.14, determine the power factor of the combined loads if ZL = 10 . Ans: 0.9966 leading. CHAPTER 11 AC CIRCUIT POWER ANALYSIS 442 or P = Re{Veffe jθ Ieffe−jφ} The phasor voltage may now be recognized as the ﬁrst two factors within the brackets in the preceding equation, but the second two factors do not quite correspond to the phasor current, because the angle includes a minus sign, which is not present in the expression for the phasor current. That is, the phasor current is Ieff = Ieffe jφ and we therefore must make use of conjugate notation: I∗ eff = Ieffe−jφ Hence P = Re{VeffI∗ eff} and we may now let power become complex by deﬁning the complex power S as S = VeffI∗ eff If we ﬁrst inspect the polar or exponential form of the complex power, S = VeffIeff e j(θ−φ) we see that the magnitude of S, VeffIeff, is the apparent power. The angle of S, (θ −φ), is the PF angle (i.e., the angle by which the voltage leads the current). In rectangular form, we have S = P + jQ where P is the average power, as before. The imaginary part of the complex power is symbolized as Q and is termed the reactive power. The dimensions of Q are the same as those of the real power P, the complex power S, and the apparent power \|S\|. In order to avoid confusion with these other quanti-ties, the unit of Q is deﬁned as the volt-ampere-reactive (abbreviated VAR). From Eqs. and , it is seen that Q = VeffIeff sin(θ −φ) The physical interpretation of reactive power is the time rate of energy ﬂow back and forth between the source (i.e., the utility company) and the reac-tive components of the load (i.e., inductances and capacitances). These components alternately charge and discharge, which leads to current ﬂow from and to the source, respectively. The relevant quantities are summarized in Table 11.1 for convenience. The Power Triangle A commonly employed graphical representation of complex power is known as the power triangle, and is illustrated in Fig. 11.15. The diagram shows that only two of the three power quantities are required, as the third may be obtained by trigonometric relationships. If the power triangle lies in the ﬁrst quadrant (θ −φ > 0); the power factor is lagging (corresponding to an inductive load); and if the power triangle lies in the fourth quadrant (θ −φ < 0), the power factor is leading (corresponding to a capacitive ■FIGURE 11.15 The power triangle representation of complex power. S P Q − Re Im The sign of the reactive power characterizes the nature of a passive load at which Veff and Ieff are speciﬁed. If the load is inductive, then (θ −φ) is an angle between 0 and 90°, the sine of this angle is positive, and the reactive power is positive. A capacitive load results in a negative reactive power. SECTION 11.5 COMPLEX POWER 443 load). A great deal of qualitative information concerning our load is there-fore available at a glance. Another interpretation of reactive power may be seen by constructing a phasor diagram containing Veff and Ieff as shown in Fig. 11.16. If the phasor current is resolved into two components, one in phase with the voltage, hav-ing a magnitude Ieff cos(θ −φ), and one 90° out of phase with the voltage, with magnitude equal to Ieff sin \|θ −φ\|, then it is clear that the real power is given by the product of the magnitude of the voltage phasor and the component of the phasor current which is in phase with the voltage. More-over, the product of the magnitude of the voltage phasor and the component of the phasor current which is 90° out of phase with the voltage is the reac-tive power Q. It is common to speak of the component of a phasor which is 90° out of phase with some other phasor as a quadrature component. Thus Q is simply Veff times the quadrature component of Ieff. Q is also known as the quadrature power. Power Measurement Strictly speaking, a wattmeter measures average real power P drawn by a load, and a varmeter reads the average reactive power Q drawn by a load. However, it is common to ﬁnd both features in the same meter, which is often also capable of measuring apparent power and power factor (Fig. 11.17). TABLE ●11.1 Summary of Quantities Related to Complex Power Quantity Symbol Formula Units Average power P VeffIeff cos(θ −φ) watt (W) Reactive power Q VeffIeff sin(θ −φ) volt-ampere-reactive (VAR) Complex power S P + jQ VeffIeff/θ −φ volt-ampere (VA) VeffI∗ eff Apparent power \|S\| VeffIeff volt-ampere (VA) ■FIGURE 11.17 A clamp-on digital powermeter manufactured by Amprobe, capable of measuring ac currents up to 400 A and voltages up to 600 V. Copyright AMPROBE. ■FIGURE 11.16 The current phasor Ieff is resolved into two components, one in phase with the voltage phasor Veff and the other 90° out of phase with the voltage phasor. This latter component is called a quadrature component. Real Ieff cos ( – ) Ieff sin \| – \| Imaginary Ieff V eff – PRACTICAL APPLICATION Power Factor Correction When electric power is being supplied to large industrial consumers by a power company, the company will fre-quently include a PF clause in its rate schedules. Under this clause, an additional charge is made to the consumer whenever the PF drops below a certain speciﬁed value, usually about 0.85 lagging. Very little industrial power is consumed at leading PFs, because of the nature of typi-cal industrial loads. There are several reasons that force the power company to make this additional charge for low PFs. In the first place, it is evident that larger current-carrying capacity must be built into its genera-tors in order to provide the larger currents that go with lower-PF operation at constant power and constant volt-age. Another reason is found in the increased losses in its transmission and distribution system. In an effort to recoup losses and encourage its cus-tomers to operate at high PF, a certain utility charges a penalty of $0.22/kVAR for each kVAR above a bench-mark value computed as 0.62 times the average power demand: S = P + j Q = P + j0.62P = P(1 + j0.62) = P(1.177/31.8◦) This benchmark targets a PF of 0.85 lagging, as cos 31.8° = 0.85 and Q is positive; this is represented graphically in Fig. 11.18. Customers with a PF smaller thanthebenchmarkvaluearesubjecttoﬁnancialpenalties. The reactive power requirement is commonly ad-justed through the installation of compensation capaci-tors placed in parallel with the load (typically at the substation outside the customer’s facility). The value of the required capacitance can be shown to be C = P(tan θold −tan θnew) ωV 2 rms where ω is the frequency, θold is the present PF angle, ■FIGURE 11.18 Plot showing acceptable ratio of reactive power to average power for power factor benchmark of 0.85 lagging. PF greater than 0.85 PF less than 0.85 Corresponds to PF 0.85 lagging 0 0 1000 2000 3000 4000 5000 Average Power (kW) 6000 7000 8000 900010000 1000 2000 3000 4000 5000 6000 7000 Reactive Power (kVAR) It is easy to show that the complex power delivered to several intercon-nected loads is the sum of the complex powers delivered to each of the individual loads, no matter how the loads are interconnected. For example, consider the two loads shown connected in parallel in Fig. 11.20. If rms values are assumed, the complex power drawn by the combined load is S = VI∗= V(I1 + I2)∗= V(I∗ 1 + I∗ 2) and thus S = VI∗ 1 + VI∗ 2 as stated. ■FIGURE 11.20 A circuit used to show that the complex power drawn by two parallel loads is the sum of the complex powers drawn by the individual loads. V + – I2 I1 I S2 S1 and θnew is the target PF angle. For convenience, however, compensation capacitor banks are manufac-tured in specific increments rated in units of kVAR capacity. An example of such an installation is shown in Fig. 11.19. Now let us consider a speciﬁc example. A particular industrial machine plant has a monthly peak demand of 5000 kW and a monthly reactive requirement of 6000 kVAR. Using the rate schedule above, what is the annual cost to this utility customer associated with PF penalties? If compensation is available through the utility company at a cost of $2390 per 1000 kVAR in-crement and $3130 per 2000 kVAR increment, what is the most cost-effective solution for the customer? The PF of the installation is the angle of the complex power S, which in this case is 5000 + j6000 kVA. Thus, the angle is tan−1(6000/5000) = 50.19◦and the PF is 0.64 lagging.The benchmark reactive power value, com-puted as 0.62 times the peak demand, is 0.62(5000) = 3100 kVAR. So, the plant is drawing 6000 −3100 = 2900 kV ARmore reactive power than the utility company is willing to allow without penalty. This represents an annual assessment of 12(2900)(0.22) = $7656 in addi-tion to regular electricity costs. If the customer chooses to have a single 1000 kVAR increment installed (at a cost of $2390), the excess reactive power draw is reduced to 2900 −1000 = 1900 kVAR, so that the annual penalty is now 12(1900)(0.22) = $5016. The total cost this year is then $5016 + $2390 = $7406, for a savings of $250. If the customer chooses to have a single 2000 kVAR incre-ment installed (at a cost of $3130), the excess reactive power draw is reduced to 2900 −2000 = 900 kVAR, so that the annual penalty is now 12(900)(0.22) = $2376. The total cost this year is then $2376 + $3130 = $5506, for a ﬁrst-year savings of $2150. If, however, the customer goes overboard and installs 3000 kVAR of compensation capacitors so that no penalty is assessed, it will actually cost $14 more in the ﬁrst year than if only 2000 kVAR were installed. ■FIGURE 11.19 A compensation capacitor installation. (Courtesy of Nokian Capacitors Ltd.) EXAMPLE 11.9 An industrial consumer is operating a 50 kW (67.1 hp) induction motor at a lagging PF of 0.8. The source voltage is 230 V rms. In order to obtain lower electrical rates, the customer wishes to raise the PF to 0.95 lagging. Specify a suitable solution. Although the PF might be raised by increasing the real power and maintaining the reactive power constant, this would not result in a lower bill and is not a cure that interests the consumer. A purely reac-tive load must be added to the system, and it is clear that it must be added in parallel, since the supply voltage to the induction motor must not change. The circuit of Fig. 11.21 is thus applicable if we interpret S1 as the induction motor’s complex power and S2 as the complex power drawn by the corrective device. ■FIGURE 11.21 V + – I2 I1 I S2 S1 (motor) (corrective device) (Continued on next page) CHAPTER 11 AC CIRCUIT POWER ANALYSIS 446 ■FIGURE 11.22 0 V rms 120 + – 5 1 j10 –j10 The complex power supplied to the induction motor must have a real part of 50 kW and an angle of cos−1(0.8), or 36.9°. Hence, S1 = 50/36.9◦ 0.8 = 50 + j37.5 kVA In order to achieve a PF of 0.95, the total complex power must become S = S1 + S2 = 50 0.95 cos−1(0.95) = 50 + j16.43 kVA Thus, the complex power drawn by the corrective load is S2 = −j21.07 kVA The necessary load impedance Z2 may be found in several simple steps. We select a phase angle of 0° for the voltage source, and therefore the current drawn by Z2 is I∗ 2 = S2 V = −j21,070 230 = −j91.6 A or I2 = j91.6 A Therefore, Z2 = V I2 = 230 j91.6 = −j2.51 If the operating frequency is 60 Hz, this load can be provided by a 1056 μF capacitor connected in parallel with the motor. However, its initial cost, maintenance, and depreciation must be covered by the reduction in the electric bill. PRACTICE ● 11.9 For the circuit shown in Fig. 11.22, ﬁnd the complex power absorbed by the (a) 1 resistor; (b) −j10 capacitor; (c) 5 + j10 impedance; (d) source. Ans: 26.6 + j0 VA; 0 −j1331 VA; 532 + j1065 VA; −559 + j266 VA. SUMMARY AND REVIEW 447 TABLE ●11.2 A Summary of AC Power Terms Term Symbol Unit Description Instantaneous power p(t) W p(t) = v(t)i(t). It is the value of the power at a speciﬁc instant in time. It is not the product of the voltage and current phasors! Average power P W In the sinusoidal steady state, P = 1 2 Vm Im cos(θ −φ), where θ is the angle of the voltage and φ is the angle of the current. Reactances do not contribute to P. Effective or rms value Vrms or Irms V or A Deﬁned, e.g., as Ieff = 1 T T 0 i2 dt ; if i(t) is sinusoidal, then Ieff = Im/ √ 2. Apparent power \|S\| VA \|S\| = VeffIeff, and is the maximum value the average power can be; P = \|S\| only for purely resistive loads. Power factor PF None Ratio of the average power to the apparent power. The PF is unity for a purely resistive load, and zero for a purely reactive load. Reactive power Q VAR A means of measuring the energy ﬂow rate to and from reactive loads. Complex power S VA A convenient complex quantity that contains both the average power P and the reactive power Q: S = P + jQ. PRACTICE ● 11.10 A 440 V rms source supplies power to a load ZL = 10 + j2 through a transmission line having a total resistance of 1.5 . Find (a) the average and apparent power supplied to the load; (b) the average and apparent power lost in the transmission line; (c) the average and apparent power supplied by the source; (d) the power factor at which the source operates. Ans: 14.21 kW, 14.49 kVA; 2.131 kW, 2.131 kVA; 16.34 kW, 16.59 kVA; 0.985 lag. SUMMARY AND REVIEW In this chapter, we introduced a fair number of new power-related terms (summarized in Table 11.2), which might have come as a bit of a surprise af-ter watts did so well for us up to this point. The new terminology is largely relevant to ac power systems, where voltages and currents are generally as-sumed to be sinusoidal (the prevalence of switched-mode power supplies in many computer systems can alter this situation, a topic of more advanced power engineering texts). After clarifying what is meant by instantaneous power, we discussed the concept of average power P. This quantity is not a function of time, but is a strong function of the phase difference between sinusoidal voltage and current waveforms. Purely reactive elements such as ideal inductors and capacitors absorb zero average power. Since such ele-ments do increase the magnitude of the current ﬂowing between the source and load, however, two new terms ﬁnd common usage: apparent power and power factor. The average power and apparent power are identical when voltage and current are in phase (i.e., associated with a purely resistive load). The power factor gives us a numerical guage of how reactive a CHAPTER 11 AC CIRCUIT POWER ANALYSIS 448 particular load is: a unity power factor (PF) corresponds to a purely resistive load (if inductors are present, they are being “canceled” by an appropriate capacitance); a zero PF indicates a purely reactive load, and the sign of the angle indicates whether the load is capacitive or inductive. Putting all of these concepts together allowed us to create a more compact representation known as complex power, S. The magnitude of S is the apparent power, P is the real part of S, and Q, the reactive power (zero for resistive loads), is the imaginary part of S. Along the way, we paused to introduce the notion of effective values of current and voltage, often referred to as rms values. Care must be taken from this point forward to establish whether a particular voltage or current value is being quoted as a magnitude or its corresponding rms value, as an ∼40% error can be introduced. Interestingly, we also discovered an exten-sion of the maximum power theorem encountered in Chap. 5, namely, that maximum average power is delivered to a load whose impedance ZL is the complex conjugate of the Thévenin equivalent impedance of the network to which it is connected. For convenience, key points of the chapter are summarized below, along with corresponding example numbers. ❑The instantaneous power absorbed by an element is given by the expression p(t) = v(t)i(t). (Examples 11.1, 11.2) ❑The average power delivered to an impedance by a sinusoidal source is 1 2Vm Im cos(θ −φ), where θ = the voltage phase angle and φ = the phase angle of the current. (Example 11.2) ❑Only the resistive component of a load draws nonzero average power. The average power delivered to the reactive component of a load is zero. (Examples 11.3, 11.4) ❑Maximum average power transfer occurs when the condition ZL = Z∗ th is satisﬁed. (Example 11.5) ❑When multiple sources are present, each operating at a different frequency, the individual contributions to average power may be summed. This is not true for sources operating at the same frequency. (Examples 11.6, 11.7) ❑The effective or rms value of a sinusoidal waveform is obtained by dividing its amplitude by √ 2. ❑The power factor (PF) of a load is the ratio of its average dissipated power to the apparent power. (Example 11.8) ❑A purely resistive load will have a unity power factor. A purely reactive load will have a zero power factor. (Example 11.8) ❑Complex power is deﬁned as S = P + jQ, or S = VeffI∗ eff. It is measured in units of volt-amperes (VA). (Example 11.9) ❑Reactive power Q is the imaginary component of the complex power, and is a measure of the energy ﬂow rate into or out of the reactive components of a load. Its unit is the volt-ampere-reactive (VAR). (Example 11.9) ❑Capacitors are commonly used to improve the PF of industrial loads to minimize the reactive power required from the utility company. (Example 11.9) EXERCISES 449 READING FURTHER A good overview of ac power concepts can be found in Chap. 2 of: B. M. Weedy, Electric Power Systems, 3rd ed. Chichester, England: Wiley, 1984. Contemporary issues pertaining to ac power systems can be found in: International Journal of Electrical Power & Energy Systems. Guildford, England: IPC Science and Technology Press, 1979–. ISSN: 0142-0615. EXERCISES 11.1 Instantaneous Power 1. Determine the instantaneous power delivered to the 1 resistor of Fig. 11.23 at t = 0, t = 1 s, and t = 2 s if vs is equl to (a) 9 V; (b) 9 sin 2t V; (c) 9 sin (2t + 13◦) V; (d ) 9e−t V. 2. Determine the power absorbed at t = 1.5 ms by each of the three elements of the circuit shown in Fig. 11.24 if vs is equal to (a) 30u(−t) V; (b) 10 + 20u(t) V. 3. Calculate the power absorbed at t = 0−, t = 0+, and t = 200 ms by each of the elements in the circuit of Fig. 11.25 if vs is equal to (a) −10u(−t) V; (b) 20 + 5u(t) V. i vs 4 1 + – ■FIGURE 11.23 6 10 mF is ■FIGURE 11.26 i vC vs + – 500 4 F + – ■FIGURE 11.24 250 mH 1 i(t) vs + – ■FIGURE 11.25 2.5 cos 10t A 1 4 4 F ■FIGURE 11.27 4. Three elements are connected in parallel: a 1 k resistor, a 15 mH inductor, and a 100 cos (2 × 105t) mA sinusoidal source. All transients have long since died out, so the circuit is operating in steady state. Determine the power being absorbed by each element at t = 10 μs. 5. Let is = 4u(−t) A in the circuit of Fig. 11.26. (a) Show that, for all t > 0, the instantaneous power absorbed by the resistor is equal in magnitude but opposite in sign to the instantaneous power absorbed by the capacitor. (b) Determine the power absorbed by the resistor at t = 60 ms. 6. The current source in the circuit of Fig. 11.26 is given by is = 8 −7u(t) A. Compute the power absorbed by all three elements at t = 0−, t = 0+, and t = 75 ms. 7. Assuming no transients are present, calcualte the power absorbed by each element shown in the circuit of Fig. 11.27 at t = 0, 10, and 20 ms. 8. Calculate in Fig. 11.28 the power absorbed by the inductor at t = 0 and t = 1 s if vs = 10u(t) V. 9. A 100 mF capacitor is storing 100 mJ of energy up until the point when a con-ductor of resistance 1.2 falls across its terminals. What is the instantaneous power dissipated in the conductor at t = 120 ms? If the speciﬁc heat capacity3 of the conductor is 0.9 kJ/kg · K and its mass is 1 g, estimate the increase in temperature of the conductor in the ﬁrst second of the capacitor discharge, assuming both elements are initially at 23°C. 10. If we take a typical cloud-to-ground lightning stroke to represent a current of 30 kA over an interval of 150 μs, calculate (a) the instantaneous power deliv-ered to a copper rod having resistance 1.2 m during the stroke; (b) the total energy delivered to the rod. 11.2 Average Power 11. The phasor current I = 9/15◦mA (corresponding to a sinusoid operating at 45 rad/s) is applied to the series combination of a 18 k resistor and a 1 μF capacitor. Obtain an expression for (a) the instantaneous power and (b) the average power absorbed by the combined load. 12. A phasor voltage V = 100/45◦V (the sinusoid operates at 155 rad/s) is applied to the parallel combination of a 1 resistor and a 1 mH inductor. (a) Obtain an expression for the average power absorbed by each passive element. (b) Graph the instantaneous power supplied to the parallel combination, along with the instantaneous power absorbed by each element separately. (Use a single graph.) 13. Calculate the average power delivered by the current 4 −j2 A to (a) Z = 9 ; (b) Z = −j1000 ; (c) Z = 1 −j2 + j3 ; (d) Z = 6/32◦; (e) Z = 1.5/−19◦ 2 + j k. 14. With regard to the two-mesh circuit depicted in Fig. 11.29, determine the aver-age power absorbed by each passive element, and the average power supplied by each source, and verify that the total supplied average power = the total absorbed average power. CHAPTER 11 AC CIRCUIT POWER ANALYSIS 450 1 10 2 F 0.5 H + – vs ■FIGURE 11.28 –j3.1 j7 5 I1 I2 0 V 152 –40 V 79 + – + – ■FIGURE 11.29 (3) Assume the speciﬁc heat capacity c is given by c = Q/m · T , where Q = the energy delivered to the conductor, m is its mass, and T is the increase in temperature. –j1.5 3 1 j2.8 3 V 194 + – ■FIGURE 11.30 15. (a) Calculate the average power absorbed by each passive element in the circuit of Fig. 11.30, and verify that it equals the average power supplied by the source. (b) Check your solution with an appropriate PSpice simulation. EXERCISES 451 16. (a) What load impedance ZL will draw the maximum average power from the source shown in Fig. 11.31? (b) Calculate the maximum average power supplied to the load. 17. The inductance of Fig. 11.31 is replaced by the impedance 9 −j8 k. Repeat Exercise 16. 18. Determine the average power supplied by the dependent source in the circuit of Fig. 11.32. j700 225 60 V 15 + – ZL ■FIGURE 11.31 VC + – 2 –j2 3 + – 0 V 20 2V C ■FIGURE 11.32 Ix j1.92 –j2 A 4.8 8 1.6Ix ■FIGURE 11.33 20. (a) Compute the average value of each waveform shown in Fig. 11.34. (b) Square each waveform, and determine the average value of each new periodic waveform. 1 –2 0 2 4 8 6 10 –2 (a) i (A) t (s) 5 0 –1 –2 1 2 3 4 5 6 –3 (b) i (A) t (ms) ■FIGURE 11.34 19. (a) Calculate the average power supplied to each passive element in the circuit of Fig. 11.33. (b) Determine the power supplied by each source. (c) Replace the 8 resistive load with an impedance capable of drawing maximum aver-age power from the remainder of the circuit. (d) Verify your solution with a PSpice simulation. CHAPTER 11 AC CIRCUIT POWER ANALYSIS 452 21. Calculate the average power delivered to a 2.2 load by the voltage vs equal to (a) 5 V; (b) 4 cos 80t −8 sin 80t V; (c) 10 cos 100t + 12.5 cos (100t + 19◦) V. 11.3 Effective Values of Current and Voltage 22. Calculate the effective value of the following waveforms: (a) 7 sin 30t V; (b) 100 cos 80t mA; (c) 120 √ 2 cos (5000t −45◦) V; (d) 100 √ 2 sin (2t + 72◦) A. 23. Determine the effective value of the following waveforms: (a) 62.5 cos 100t mV; (b) 1.95 cos 2t A; (c) 208 √ 2 cos (100πt + 29◦) V; (d) 400 √ 2 sin (2000t −14◦) A. 24. Compute the effective value of (a) i(t) = 3 sin 4t A; (b) v(t) = 4 sin 20t cos 10t; (c) i(t) = 2 −sin 10t mA; (d) the waveform plotted in Fig. 11.35. 2.82 1 2 3 4 5 6 7 i(t) (mA) t (s) ■FIGURE 11.35 25. For each waveform plotted in Fig. 11.34, determine its frequency, period, and rms value. 26. Determine both the average and rms value of each waveform depicted in Fig. 11.36. 9 1 2 3 4 5 6 7 i(t) (mA) t (s) 1 –1 –0.1 0.1 0.2 0.3 0.4 0.5 f (t) t (ms) (a) (b) ■FIGURE 11.36 27. The series combination of a 1 k resistor and a 2 H inductor must not dissipate more than 250 mW of power at any instant. Assuming a sinusoidal current with ω = 500 rad/s, what is the largest rms current that can be tolerated? 28. For each of the following waveforms, determine its period, frequency and effec-tive value: (a) 5 V; (b) 2 sin 80t −7 cos 20t + 5 V; (c) 5 cos 50t + 3 sin 50t V; (d) 8 cos2 90t mA. (e) Verify your answers with an appropriate simulation. 29. With regard to the circuit of Fig. 11.37, determine whether a purely real value of R can result in equal rms voltages across the 14 mH inductor and the resistor R. If so, calculate R and the rms voltage across it; if not, explain why not. EXERCISES 453 30. (a) Calculate both the average and rms values of the waveform plotted in Fig. 11.38. (b) Verify your solutions with appropriate PSpice simulations (Hint: you may want to employ two pulse waveforms added together). 208 cos 40t V R 14 mH 28 mH + – ■FIGURE 11.37 1 2 –1 0 1 2 3 4 5 6 7 v(t) t (s) ■FIGURE 11.38 11.4 Apparent Power and Power Factor 31. For the circuit of Fig. 11.39, compute the average power delivered to each load, the apparent power supplied by the source, and the power factor of the combined loads if (a) Z1 = 14/32◦ and Z2 = 22 ; (b) Z1 = 2/0◦ and Z2 = 6 −j ; (c) Z1 = 100/70◦ and Z2 = 75/90◦. 32. Calculate the power factor of the combined loads of the circuit depicted in Fig. 11.39 if (a) both loads are purely resistive; (b) both loads are purely induc-tive and ω = 100 rad/s; (c) both loads are purely capacitive and ω = 200 rad/s; (d) Z1 = 2Z2 = 5 −j8 . 33. A given load is connected to an ac power system. If it is known that the load is characterized by resistive losses and either capacitors, inductors, or neither (but not both), which type of reactive element is part of the load if the power factor is measured to be (a) unity; (b) 0.85 lagging; (c) 0.221 leading; (d) cos (−90◦)? 34. An unknown load is connected to a standard European household outlet (240 V rms, 50 Hz). Determine the phase angle difference between the voltage and current, and whether the voltage leads or lags the current, if (a) V = 240/243◦V rms and I = 3/9◦A rms; (b) the power factor of the load is 0.55 lagging; (c) the power factor of the load is 0.685 leading; (d) the capac-itive load draws 100 W average power and 500 volt-amperes apparent power. 35. (a) Design a load which draws an average power of 25 W at a leading PF of 0.88 from a standard North American household outlet (120 V rms, 60 Hz). (b) Design a capacitor-free load which draws an average power of 150 W and an apparent power of 25 W from a household outlet in eastern Japan (110 V rms, 50 Hz). 36. Assuming an operating frequency of 40 rad/s for the circuit shown in Fig. 11.40, and a load impedance of 50/−100◦, calculate (a) the instanta-neous power separately delivered to the load and to the 1 k shunt resistance at t = 20 ms; (b) the average power delivered to both passive elements; (c) the apparent power delivered to the load; (d) the power factor at which the source is operating. 37. Calculate the power factor at which the source in Fig. 11.40 is operating if the load is (a) purely resistive; (b) 1000 + j900 ; (c) 500/−5◦. I 3 V rms 119 + – Z1 Z2 ■FIGURE 11.39 I 1 k 20 mA 275 Load ■FIGURE 11.40 CHAPTER 11 AC CIRCUIT POWER ANALYSIS 454 43. Referring to the network represented in Fig. 11.21, if the motor draws complex power 150/24◦VA, (a) determine the PF at which the source is operating; (b) determine the impedance of the corrective device required to change the PF of the source to 0.98 lagging. (c) Is it physically possible to obtain a leading PF for the source? Explain. 44. Determine the complex power absorbed by each passive component in the circuit of Fig. 11.43, and the power factor at which the source is operating. 38. Determine the load impedance for the circuit depicted in Fig. 11.40 if the source is operating at a PF of (a) 0.95 leading; (b) unity; (c) 0.45 lagging. 39. For the circuit of Fig. 11.41, ﬁnd the apparent power delivered to each load, and the power factor at which the source operates, if (a) ZA = 5 −j2 , ZB = 3 , ZC = 8 + j4 , and ZD = 15/−30◦; (b) ZA = 2/−15◦, ZB = 1 , ZC = 2 + j , and ZD = 4/45◦. 0 V rms 200 + – ZA ZB ZD ZC ■FIGURE 11.41 1.5 S S 1.0 0.5 1 (a) 2 3 Re (W) Re (kW) 1 2 3 4 1 (b) 2 Im (VAR) Im (kVAR) ■FIGURE 11.42 11.5 Complex Power 40. Compute the complex power S (in polar form) drawn by a certain load if it is known that (a) it draws 100 W average power at a lagging PF of 0.75; (b) it draws a current I = 9 + j5 A rms when connected to the voltage 120/32◦V rms; (c) it draws 1000 W average power and 10 VAR reactive power at a leading PF; (d) it draws an apparent power of 450 W at a lagging PF of 0.65. 41. Calculate the apparent power, power factor, and reactive power associated with a load if it draws complex power S equal to (a) 1 + j0.5 kVA; (b) 400 VA; (c) 150/−21◦VA; (d ) 75/25◦VA. 42. For each power triangle depicted in Fig. 11.42, determine S (in polar form) and the PF. 45 V rms 240 + – 18 18 j10 1000 –j5 ■FIGURE 11.43 EXERCISES 455 j20 –j10 20 10 + – 0 V rms 100 ■FIGURE 11.44 45. What value of capacitance must be added in parallel to the 10 resistor of Fig. 11.44 to increase the PF of the source to 0.95 at 50 Hz? 46. The kiln operation of a local lumberyard has a monthly average power demand of 175 kW, but associated with that is an average monthly reactive power draw of 205 kVAR. If the lumberyard’s utility company charges $0.15 per kVAR for each kVAR above the benchmark value (0.7 times the peak average power demand), (a) estimate the annual cost to the lumberyard from PF penalties; (b) calculate the money saved in the ﬁrst and second years, respectively, if 100 kVAR compensation capacitors are available for purchase at $75 each (installed). 47. Calculate the complex power delivered to each passive component of the cir-cuit shown in Fig. 11.45, and determine the power factor of the source. 10 15 + – –17 V rms 50 –j25 j30 ■FIGURE 11.45 48. Replace the 10 resistor in the circuit of Fig. 11.45 with a 200 mH inductor, assume an operating frequency of 10 rad/s, and calculate (a) the PF of the source; (b) the apparent power supplied by the source; (c) the reactive power delivered by the source. 49. Instead of including a capacitor as indicated in Fig. 11.45, the circuit is erro-neously constructed using two identical inductors, each having an impedance of j30 W at the operating frequency of 50 Hz. (a) Compute the complex power delivered to each passive component. (b) Verify your solution by calculating the complex power supplied by the source. (c) At what power factor is the source operating? 50. Making use of the general strategy employed in Example 11.9, derive Eq. , which enables the corrective value of capacitance to be calculated for a general operating frequency. Chapter-Integrating Exercises 51. A load is drawing 10 A rms when connected to a 1200 V rms supply running at 50 Hz. If the source is operating at a lagging PF of 0.9, calculate (a) the peak voltage magnitude; (b) the instantaneous power absorbed by the load at t = 1 ms; (c) the apparent power supplied by the source; (d) the reactive power supplied to the load; (e) the load impedance; and ( f ) the complex power supplied by the source (in polar form). 52. For the circuit of Fig. 11.46, assume the source operates at a frequency of 100 rad/s. (a) Determine the PF at which the source is operating. (b) Calculate the apparent power absorbed by each of the three passive elements. (c) Compute the average power supplied by the source. (d ) Determine the CHAPTER 11 AC CIRCUIT POWER ANALYSIS 456 53. Remove the 50 resistor in Fig. 11.46, assume an operating frequency of 50 Hz, and (a) determine the power factor at which the load is operating; (b) compute the average power delivered by the source; (c) compute the instantaneous power absorbed by the inductance at t = 2 ms; (d) determine the capacitance that must be connected between the terminals marked a and b to increase the PF of the source to 0.95. 54. A source 45 sin 32t V is connected in series with a 5 resistor and a 20 mH inductor. Calculate (a) the reactive power delivered by the source; (b) the apparent power absorbed by each of the three elements; (c) the complex power S absorbed by each element; (d) the power factor at which the source is operating. 55. For the circuit of Fig. 11.37, (a) derive an expression for the complex power delivered by the source in terms of the unknown resistance R; (b) compute the necessary capacitance that must be added in parallel to the 28 mH inductor to achieve a unity PF. 50 a b 0 A 5 j60 80 ■FIGURE 11.46 Thévenin equivalent seen looking into the terminals marked a and b, and calculate the average power delivered to a 100 resistor connected between the same terminals. INTRODUCTION The vast majority of power is supplied to consumers in the form of sinusoidal voltages and currents, typically referred to as alternating current or simply ac. Although there are exceptions, for example, some types of train motors, most equipment is designed to run on either 50 or 60 Hz. Most 60 Hz systems are now standardized to run on 120 V, whereas 50 Hz systems typically correspond to 240 V (both voltages being quoted in rms units). The actual voltage delivered to an appliance can vary somewhat from these values, and distribution systems employ signiﬁcantly higher voltages to minimize the current and hence cable size. Originally Thomas Edison advocated a purely dc power distribution network, purport-edly due to his preference for the simple algebra required to analyze such circuits. Nikola Tesla and George Westinghouse, two other pioneers in the ﬁeld of electricity, proposed ac distribution systems as the achievable losses were signiﬁcantly lower. Ulti-mately they were more persuasive, despite some rather theatrical demonstrations on the part of Edison. The transient response of ac power systems is of interest when determining the peak power demand, since most equipment requires more current to start up than it does to run continuously. Often, however, it is the steady-state operation that is of primary interest, so our experience with phasor-based analysis will prove to be handy. In this chapter we introduce a new type of voltage source, the three-phase source, which can be connected in either a three- or four-wire Y conﬁguration or a three-wire conﬁguration. Loads can also be either Y- or -connected, depending on the application. KEY CONCEPTS Single-Phase Power Systems Three-Phase Power Systems Three-Phase Sources Line Versus Phase Voltage Line Versus Phase Current Y-Connected Networks -Connected Networks Balanced Loads Per-Phase Analysis Power Measurement in Three-Phase Systems Polyphase Circuits C H A P T E R 12 457 12.1 • POLYPHASE SYSTEMS So far, whenever we used the term “sinusoidal source’’ we pictured a single sinusoidal voltage or current having a particular amplitude, frequency, and phase. In this chapter, we introduce the concept of polyphase sources, focus-ing on three-phase systems in particular. There are distinct advantages in us-ing rotating machinery to generate three-phase power rather than single-phase power, and there are economical advantages in favor of the transmission of power in a three-phase system.Although most of the electri-cal equipment we have encountered so far is single-phase, three-phase equipment is not uncommon, especially in manufacturing environments. In particular, motors used in large refrigeration systems and in machining facil-ities are often wired for three-phase power. For the remaining applications, once we have become familiar with the basics of polyphase systems, we will ﬁnd that it is simple to obtain single-phase power by just connecting to a sin-gle “leg’’of a polyphase system. Let us look brieﬂy at the most common polyphase system, a balanced three-phase system. The source has three terminals (not counting a neutral or ground connection), and voltmeter measurements will show that sinu-soidal voltages of equal amplitude are present between any two terminals. However, these voltages are not in phase; each of the three voltages is 120° out of phase with each of the other two, the sign of the phase angle depend-ing on the sense of the voltages. One possible set of voltages is shown in Fig. 12.1. A balanced load draws power equally from all three phases. At no instant does the instantaneous power drawn by the total load reach zero; in fact, the total instantaneous power is constant. This is an advantage in rotating machinery, for it keeps the torque on the rotor much more constant than it would be if a single-phase source were used. As a result, there is less vibration. CHAPTER 12 POLYPHASE CIRCUITS 458 ■FIGURE 12.1 An example set of three voltages, each of which is 120◦out of phase with the other two. As can be seen, only one of the voltages is zero at any particular instant. –0.4 –0.6 –0.2 0.2 0 0.4 0.6 0.8 1.0 –0.8 –1.0 Volts t (s) 0.04 0.035 0.03 0.025 0.02 0.015 0.01 0.005 0 SECTION 12.1 POLYPHASE SYSTEMS 459 The use of a higher number of phases, such as 6- and 12-phase systems, is limited almost entirely to the supply of power to large rectiﬁers. Recti-ﬁers convert alternating current to direct current by only allowing current to ﬂow to the load in one direction, so that the sign of the voltage across the load remains the same. The rectiﬁer output is a direct current plus a smaller pulsating component, or ripple, which decreases as the number of phases increases. Almost without exception, polyphase systems in practice contain sources which may be closely approximated by ideal voltage sources or by ideal voltage sources in series with small internal impedances. Three-phase current sources are extremely rare. Double-Subscript Notation It is convenient to describe polyphase voltages and currents using double-subscript notation. With this notation, a voltage or current, such as Vab or IaA, has more meaning than if it were indicated simply as V3 or Ix. By def-inition, the voltage of point a with respect to point b is Vab. Thus, the plus sign is located at a, as indicated in Fig. 12.2a. We therefore consider the double subscripts to be equivalent to a plus-minus sign pair; the use of both would be redundant. With reference to Fig. 12.2b, for example, we see that Vad = Vab + Vcd. The advantage of the double-subscript notation lies in the fact that Kirchhoff’s voltage law requires the voltage between two points to be the same, regardless of the path chosen between the points; thus Vad = Vab + Vbd = Vac + Vcd = Vab + Vbc + Vcd , and so forth. The beneﬁt of this is that KVL may be satisﬁed without reference to the circuit diagram; correct equations may be written even though a point, or subscript letter, is included which is not marked on the diagram. For example, we might have written Vad = Vax + Vxd, where x identiﬁes the location of any interesting point of our choice. One possible representation of a three-phase system of voltages1 is shown in Fig. 12.3. Let us assume that the voltages Van, Vbn, and Vcn are known: Van = 100/0◦V Vbn = 100/−120◦V Vcn = 100/−240◦V The voltage Vab may be found, with an eye on the subscripts, as Vab = Van + Vnb = Van −Vbn = 100/0◦−100/−120◦V = 100 −(−50 −j86.6) V = 173.2/30◦V The three given voltages and the construction of the phasor Vab are shown on the phasor diagram of Fig. 12.4. A double-subscript notation may also be applied to currents. We deﬁne the current Iab as the current ﬂowing from a to b by the most direct path. In ■FIGURE 12.2 (a) The deﬁnition of the voltage Vab. (b) Vad = Vab + Vbc + Vcd = Vab + Vcd. Vab + – + – a b (a) (b) a c b d ■FIGURE 12.3 A network used as a numerical example of double-subscript voltage notation. 120° V + – + – + – c b a n 100 –120° V 100 0° V 100 ■FIGURE 12.4 This phasor diagram illustrates the graphical use of the double-subscript voltage convention to obtain Vab for the network of Fig. 12.3. Vcn Vbn Van Vnb Vab = Van + Vnb 30° 120° 120° (1) In keeping with power industry convention, rms values for currents and voltages will be used implicitly throughout this chapter. CHAPTER 12 POLYPHASE CIRCUITS 460 ■FIGURE 12.5 An illustration of the use and misuse of the double-subscript convention for current notation. Icd? Icd? Iab + – a c d b ■FIGURE 12.7 (a) A single-phase three-wire source. (b) The representation of a single-phase three-wire source by two identical voltage sources. Single-phase 3-wire source a n b (a) + – + – V1 V1 a n b (b) ■FIGURE 12.6 5 A 2 A 8 A 4 A 10 A c g k h i d e l a b j f Ans: 12.1: 114.0/20.2◦V; 41.8/145.0◦V; 44.0/20.6◦V. 12.2: −3 A; 7 A; 7 A. 12.2 • SINGLE-PHASE THREE-WIRE SYSTEMS Before studying polyphase systems in detail, it can be helpful ﬁrst to look at a simple single-phase three-wire system. A single-phase three-wire source is deﬁned as a source having three output terminals, such as a, n, and b in Fig. 12.7a, at which the phasor voltages Van and Vnb are equal. The source may therefore be represented by the combination of two iden-tical voltage sources; in Fig. 12.7b, Van = Vnb = V1. It is apparent that Vab = 2Van = 2Vnb, and we therefore have a source to which loads operat-ing at either of two voltages may be connected. The normal North American household system is single-phase three-wire, permitting the operation of both 110 V and 220 V appliances. The higher-voltage appliances are nor-mally those drawing larger amounts of power; operation at higher voltage results in a smaller current draw for the same power. Smaller-diameter wire may consequently be used safely in the appliance, the household distribution every complete circuit we consider, there must of course be at least two pos-sible paths between the points a and b, and we agree that we will not use double-subscript notation unless it is obvious that one path is much shorter, or much more direct. Usually this path is through a single element. Thus, the current Iab is correctly indicated in Fig. 12.5. In fact, we do not even need the direction arrow when talking about this current; the subscripts tell us the direction. However, the identiﬁcation of a current as Icd for the circuit of Fig. 12.5 would cause confusion. PRACTICE ● 12.1 Let Vab = 100/0◦V, Vbd = 40/80◦V, and Vca = 70/200◦V. Find (a) Vad; (b) Vbc; (c) Vcd. 12.2 Refer to the circuit of Fig. 12.6 and let If j = 3 A, Ide = 2 A, and Ihd = −6 A. Find (a) Icd; (b) Ief ; (c) Ii j. SECTION 12.2 SINGLE-PHASE THREE-WIRE SYSTEMS 461 system, and the distribution system of the utility company, as larger-diameter wire must be used with higher currents to reduce the heat produced due to the resistance of the wire. The name single-phase arises because the voltages Van and Vnb, being equal, must have the same phase angle. From another viewpoint, however, the voltages between the outer wires and the central wire, which is usually re-ferred to as the neutral, are exactly 180◦out of phase. That is, Van = −Vbn, and Van + Vbn = 0. Later, we will see that balanced polyphase systems are characterized by a set of voltages of equal amplitude whose (phasor) sum is zero. From this viewpoint, then, the single-phase three-wire system is really a balanced two-phase system. Two-phase, however, is a term that is tradi-tionally reserved for a relatively unimportant unbalanced system utilizing two voltage sources 90° out of phase. Let us now consider a single-phase three-wire system that contains identical loads Zp between each outer wire and the neutral (Fig. 12.8). We ﬁrst assume that the wires connecting the source to the load are perfect conductors. Since Van = Vnb then, IaA = Van Zp = IBb = Vnb Zp and therefore InN = IBb + IAa = IBb −IaA = 0 Thus there is no current in the neutral wire, and it could be removed with-out changing any current or voltage in the system. This result is achieved through the equality of the two loads and of the two sources. Effect of Finite Wire Impedance We next consider the effect of a ﬁnite impedance in each of the wires. If lines aA and bB each have the same impedance, this impedance may be added to Zp, resulting in two equal loads once more, and zero neutral cur-rent. Now let us allow the neutral wire to possess some impedance Zn. Without carrying out any detailed analysis, superposition should show us that the symmetry of the circuit will still cause zero neutral current. More-over, the addition of any impedance connected directly from one of the outer lines to the other outer line also yields a symmetrical circuit and zero neutral current. Thus, zero neutral current is a consequence of a balanced, or symmetrical, load; nonzero impedance in the neutral wire does not destroy the symmetry. The most general single-phase three-wire system will contain unequal loads between each outside line and the neutral and another load directly between the two outer lines; the impedances of the two outer lines may be expected to be approximately equal, but the neutral impedance is often slightly larger. Let us consider an example of such a system, with particular interest in the current that may ﬂow now through the neutral wire, as well as the overall efﬁciency with which our system is transmitting power to the unbalanced load. ■FIGURE 12.8 A simple single-phase three-wire system. The two loads are identical, and the neutral current is zero. + – + – a A N B b Vnb Van Zp Zp n CHAPTER 12 POLYPHASE CIRCUITS 462 EXAMPLE 12.1 Analyze the system shown in Fig. 12.9 and determine the power delivered to each of the three loads as well as the power lost in the neutral wire and each of the two lines. Identify the goal of the problem. The three loads in the circuit are the 50 resistor, the 100 resistor, and a 20 + j10 impedance. Each of the two lines has a resistance of 1 , and the neutral wire has a resistance of 3 . We need the cur-rent through each of these in order to determine power. Collect the known information. We have a single-phase three-wire system; the circuit diagram of Fig. 12.9 is completely labeled. The computed currents will be in rms units. Devise a plan. The circuit is conducive to mesh analysis, having three clearly deﬁned meshes. The result of the analysis will be a set of mesh currents, which can then be used to compute absorbed power. Construct an appropriate set of equations. The three mesh equations are: −115/0◦+ I1 + 50(I1 −I2) + 3(I1 −I3) = 0 (20 + j10)I2 + 100(I2 −I3) + 50(I2 −I1) = 0 −115/0◦+ 3(I3 −I1) + 100(I3 −I2) + I3 = 0 which can be rearranged to obtain the following three equations 54I1 −50I2 −3I3 = 115/0◦ −50I1 + (170 + j10)I2 −100I3 = 0 −3I1 −100I2 +104I3 = 115/0◦ Determine if additional information is required. We have a set of three equations in three unknowns, so it is possible to attempt a solution at this point. ■FIGURE 12.9 A typical single-phase three-wire system. I1 I3 I2 0° V rms 115 0° V rms 115 + – + – a b n 1 50 20 j10 100 3 1 A B N SECTION 12.2 SINGLE-PHASE THREE-WIRE SYSTEMS 463 Attempt a solution. Solving for the phasor currents I1, I2, and I3 using a scientiﬁc calcula-tor, we ﬁnd I1 = 11.24/−19.83◦A I2 = 9.389/−24.47◦A I3 = 10.37/−21.80◦A The currents in the outer lines are thus IaA = I1 = 11.24/−19.83◦A and IbB = −I3 = 10.37/158.20◦A and the smaller neutral current is InN = I3 −I1 = 0.9459/−177.7◦A The average power drawn by each load may thus be determined: P50 = \|I1 −I2\|2 (50) = 206 W P100 = \|I3 −I2\|2 (100) = 117 W P20+ j10 = \|I2\|2 (20) = 1763 W The total load power is 2086 W. The loss in each of the wires is next found: PaA = \|I1\|2 (1) = 126 W PbB = \|I3\|2 (1) = 108 W PnN = \|InN\|2 (3) = 3 W giving a total line loss of 237 W. The wires are evidently quite long; otherwise, the relatively high power loss in the two outer lines would cause a dangerous temperature rise. Verify the solution. Is it reasonable or expected? The total absorbed power is 206 + 117 + 1763 + 237, or 2323 W, which may be checked by ﬁnding the power delivered by each voltage source: Pan = 115(11.24) cos 19.83◦= 1216 W Pbn = 115(10.37) cos 21.80◦= 1107 W or a total of 2323 W. The transmission efﬁciency for the system is η = total power delivered to load total power generated = 2086 2086 + 237 = 89.8% This value would be unbelievable for a steam engine or an internal combustion engine, but it is too low for a well-designed distribution system. Larger-diameter wires should be used if the source and the load cannot be placed closer to each other. Note that we do not need to include a factor of 1 2 since we are working with rms current values. (Continued on next page) Imagine the heat produced by two 100 W light bulbs! These outer wires must dissipate the same amount of power. In order to keep their temperature down, a large surface area is required. 12.3 • THREE-PHASE Y-Y CONNECTION Three-phase sources have three terminals, called the line terminals, and they may or may not have a fourth terminal, the neutral connection. We will begin by discussing a three-phase source that does have a neutral connec-tion. It may be represented by three ideal voltage sources connected in a Y, as shown in Fig. 12.11; terminals a, b, c, and n are available. We will con-sider only balanced three-phase sources, which may be deﬁned as having \|Van\| = \|Vbn\| = \|Vcn\| and Van + Vbn + Vcn = 0 These three voltages, each existing between one line and the neutral, are called phase voltages. If we arbitrarily choose Van as the reference, or deﬁne Van = Vp/0◦ where we will consistently use Vp to represent the rms amplitude of any of the phase voltages, then the deﬁnition of the three-phase source indicates that either Vbn = Vp/−120◦ and Vcn = Vp/−240◦ or Vbn = Vp/120◦ and Vcn = Vp/240◦ The former is called positive phase sequence, or abc phase sequence, and is shown in Fig. 12.12a; the latter is termed negative phase sequence, or cba phase sequence, and is indicated by the phasor diagram of Fig. 12.12b. CHAPTER 12 POLYPHASE CIRCUITS 464 A phasor diagram showing the two source voltages, the currents in the outer lines, and the current in the neutral is constructed in Fig. 12.10. The fact that IaA + IbB + InN = 0 is indicated on the diagram. ■FIGURE 12.11 A Y-connected three-phase four-wire source. + – + – + – A B N C a n c b Van Vbn Vcn ■FIGURE 12.10 The source voltages and three of the currents in the circuit of Fig. 12.9 are shown on a phasor diagram. Note that Ia A + IbB + In N = 0. IbB IaA + IbB InN IaA Vbn Van PRACTICE ● 12.3 Modify Fig. 12.9 by adding a 1.5 resistance to each of the two outer lines, and a 2.5 resistance to the neutral wire. Find the average power delivered to each of the three loads. Ans: 153.1 W; 95.8 W; 1374 W. SECTION 12.3 THREE-PHASE Y-Y CONNECTION 465 The actual phase sequence of a physical three-phase source depends on the arbitrary choice of the three terminals to be lettered a, b, and c. They may always be chosen to provide positive phase sequence, and we will assume that this has been done in most of the systems we consider. Line-to-Line Voltages Let us next find the line-to-line voltages (often simply called the line voltages) which are present when the phase voltages are those of Fig. 12.12a. It is easiest to do this with the help of a phasor diagram, since the angles are all multiples of 30◦. The necessary construction is shown in Fig. 12.13; the results are Vab = √ 3Vp/30◦ Vbc = √ 3Vp/−90◦ Vca = √ 3Vp/−210◦ Kirchhoff’s voltage law requires the sum of these three voltages to be zero; the reader is encouraged to verify this as an exercise. If the rms amplitude of any of the line voltages is denoted by VL, then one of the important characteristics of the Y-connected three-phase source may be expressed as VL = √ 3Vp Note that with positive phase sequence, Van leads Vbn and Vbn leads Vcn, in each case by 120◦, and also that Vab leads Vbc and Vbc leads Vca, again by 120◦. The statement is true for negative phase sequence if “lags’’ is substituted for “leads.’’ Now let us connect a balanced Y-connected three-phase load to our source, using three lines and a neutral, as drawn in Fig. 12.14. The load is 0° Van = Vp –120° Vbn = Vp –240° V Vcn = Vp (+) sequence (a) 0° Van = Vp 240° Vcn = Vp 120° Vbn = Vp (–) sequence (b) ■FIGURE 12.12 (a) Positive, or abc, phase sequence. (b) Negative, or cba, phase sequence. Vcn Vca Vbn Vbc Van Vab 30 ■FIGURE 12.13 A phasor diagram which is used to determine the line voltages from the given phase voltages. Or, algebraically, Vab = Van −Vbn = V p/0◦−V p/−120◦= V p −V p cos(−120◦) − j V p sin(−120◦) = V p(1 + 1 2 + j 3/2) = 3V p/30◦. + – + – + – a c n N C A B b Zp Zp Zp ■FIGURE 12.14 A balanced three-phase system, connected Y-Y and including a neutral. represented by an impedance Zp between each line and the neutral. The three line currents are found very easily, since we really have three single-phase circuits that possess one common lead:2 IaA = Van Zp IbB = Vbn Zp = Van/−120◦ Zp = IaA/−120◦ IcC = IaA/−240◦ and therefore INn = IaA + IbB + IcC = 0 Thus, the neutral carries no current if the source and load are both bal-anced and if the four wires have zero impedance. How will this change if an impedance ZL is inserted in series with each of the three lines and an imped-ance Zn isinsertedintheneutral?Thelineimpedancesmaybecombinedwith the three load impedances; this effective load is still balanced, and a perfectly conducting neutral wire could be removed. Thus, if no change is produced in the system with a short circuit or an open circuit between n and N, any imped-ance may be inserted in the neutral and the neutral current will remain zero. It follows that, if we have balanced sources, balanced loads, and bal-anced line impedances, a neutral wire of any impedance may be replaced by any other impedance, including a short circuit or an open circuit; the replacement will not affect the system’s voltages or currents. It is often helpful to visualize a short circuit between the two neutral points, whether a neutral wire is actually present or not; the problem is then reduced to three single-phase problems, all identical except for the consistent difference in phase angle. We say that we thus work the problem on a “per-phase’’ basis. CHAPTER 12 POLYPHASE CIRCUITS 466 (2) This can be seen to be true by applying superposition and looking at each phase one at a time. EXAMPLE 12.2 For the circuit of Fig. 12.15, ﬁnd both the phase and line currents, and the phase and line voltages throughout the circuit; then calcu-late the total power dissipated in the load. + – + – + – 60° 100 0° V rms 200 Balanced (+) sequence B C A N n b c a ■FIGURE 12.15 A balanced three-phase three-wire Y-Y connected system. SECTION 12.3 THREE-PHASE Y-Y CONNECTION 467 Vca Vcn Vab Van Vbc Vbn IbB IaA IcC 30 60 ■FIGURE 12.16 The phasor diagram that applies to the circuit of Fig. 12.15. Since one of the source phase voltages is given and we are told to use the positive phase sequence, the three phase voltages are: Van = 200/0◦V Vbn = 200/−120◦V Vcn = 200/−240◦V The line voltage is 200 √ 3 = 346 V; the phase angle of each line volt-age can be determined by constructing a phasor diagram, as we did in Fig. 12.13 (as a matter of fact, the phasor diagram of Fig. 12.13 is applicable), subtracting the phase voltages using a scientiﬁc calculator, or by invoking Eqs. to . We ﬁnd that Vab is 346/30◦V, Vbc = 346/−90◦V, and Vca = 346/−210◦V. The line current for phase A is IaA = Van Zp = 200/0◦ 100/60◦= 2/−60◦A Since we know this is a balanced three-phase system, we may write the remaining line currents based on IaA: IbB = 2 (−60◦−120◦) = 2/−180◦A IcC = 2 (−60◦−240◦) = 2/−300◦A Finally, the average power absorbed by phase A is Re{VanI∗ aA}, or PAN = 200(2) cos(0◦+ 60◦) = 200 W Thus, the total average power drawn by the three-phase load is 600 W. The phasor diagram for this circuit is shown in Fig. 12.16. Once we knew any of the line voltage magnitudes and any of the line current magnitudes, the angles for all three voltages and all three currents could have been obtained by simply reading the diagram. PRACTICE ● 12.4 A balanced three-phase three-wire system has a Y-connected load. Each phase contains three loads in parallel: −j100 , 100 , and 50 + j50 . Assume positive phase sequence with Vab = 400/0◦V. Find (a) Van; (b) IaA; (c) the total power drawn by the load. Ans: 231/−30◦V; 4.62/−30◦A; 3200 W. Before working another example, this would be a good opportunity to quickly explore a statement made in Sec. 12.1, i.e., that even though phase voltages and currents have zero value at speciﬁc instants in time (every 1/120 s in North America), the instantaneous power delivered to the total load is never zero. Consider phase A of Example 12.2 once more, with the phase voltage and current written in the time domain: vAN = 200 √ 2 cos(120πt + 0◦) V and i AN = 2 √ 2 cos(120πt −60◦) A The factor of 2 is required to convert from rms units. Thus, the instantaneous power absorbed by phase A is pA(t) = vANiAN = 800 cos(120πt) cos(120πt −60◦) = 400[cos(−60◦) + cos(240πt −60◦)] = 200 + 400 cos(240πt −60◦) W in a similar fashion, pB(t) = 200 + 400 cos(240πt −300◦) W and pC(t) = 200 + 400 cos(240πt −180◦) W The instantaneous power absorbed by the total load is therefore p(t) = pA(t) + pB(t) + pC(t) = 600 W independent of time, and the same value as the average power computed in Example 12.2. CHAPTER 12 POLYPHASE CIRCUITS 468 EXAMPLE 12.3 A balanced three-phase system with a line voltage of 300 V is sup-plying a balanced Y-connected load with 1200 W at a leading PF of 0.8. Find the line current and the per-phase load impedance. The phase voltage is 300/ √ 3 V and the per-phase power is 1200/3 = 400 W. Thus the line current may be found from the power relationship 400 = 300 √ 3 (IL)(0.8) and the line current is therefore 2.89A. The phase impedance magnitude is given by Zp = Vp IL = 300/ √ 3 2.89 = 60 Since the PF is 0.8 leading, the impedance phase angle is −36.9◦; thus Zp = 60/−36.9◦. PRACTICE ● 12.5 A balanced three-phase three-wire system has a line voltage of 500 V. Two balanced Y-connected loads are present. One is a capacitive load with 7 −j2 per phase, and the other is an inductive load with 4 + j2 per phase. Find (a) the phase voltage; (b) the line current; (c) the total power drawn by the load; (d) the power factor at which the source is operating. Ans: 289 V; 97.5 A; 83.0 kW; 0.983 lagging. SECTION 12.3 THREE-PHASE Y-Y CONNECTION 469 EXAMPLE 12.4 I2 IL I1 + – 300 3 V rms 200 W 400 W 0.8 PF leading ■FIGURE 12.17 The per-phase circuit that is used to analyze a balanced three-phase example. A balanced 600 W lighting load is added (in parallel) to the system of Example 12.3. Determine the new line current. We ﬁrst sketch a suitable per-phase circuit, as shown in Fig. 12.17. The 600 W load is assumed to be a balanced load evenly distributed among the three phases, resulting in an additional 200 W consumed by each phase. The amplitude of the lighting current (labeled I1) is determined by 200 = 300 √ 3 \|I1\| cos 0◦ so that \|I1\| = 1.155 A In a similar way, the amplitude of the capacitive load current (labeled I2) is found to be unchanged from its previous value, since the voltage across it has remained the same: \|I2\| = 2.89 A If we assume that the phase with which we are working has a phase voltage with an angle of 0◦, then since cos−1(0.8) = 36.9◦, I1 = 1.155/0◦A I2 = 2.89/+36.9◦A and the line current is IL = I1 + I2 = 3.87/+26.6◦A We can check our results by computing the power generated by this phase of the source Pp = 300 √ 3 3.87 cos(+26.6◦) = 600 W which agrees with the fact that the individual phase is known to be sup-plying 200 W to the new lighting load, as well as 400 W to the original load. PRACTICE ● 12.6 Three balanced Y-connected loads are installed on a balanced three-phase four-wire system. Load 1 draws a total power of 6 kW at unity PF, load 2 pulls 10 kVA at PF 0.96 lagging, and load 3 demands 7 kW at 0.85 lagging. If the phase voltage at the loads is 135 V, if each line has a resistance of 0.1 , and if the neutral has a resistance of 1 , ﬁnd (a) the total power drawn by the loads; (b) the combined PF of the loads; (c) the total power lost in the four lines; (d) the phase voltage at the source; (e) the power factor at which the source is operating. Ans: 22.6 kW; 0.954 lag; 1027 W; 140.6 V; 0.957 lagging. If an unbalanced Y-connected load is present in an otherwise balanced three-phase system, the circuit may still be analyzed on a per-phase basis if the neutral wire is present and if it has zero impedance. If either of these conditions is not met, other methods must be used, such as mesh or nodal analysis. However, engineers who spend most of their time with unbalanced three-phase systems will ﬁnd the use of symmetrical components a great time saver. We leave this topic for more advanced texts. 12.4 • THE DELTA () CONNECTION An alternative to the Y-connected load is the -connected conﬁguration, as shown in Fig. 12.18. This type of conﬁguration is very common, and does not possess a neutral connection. CHAPTER 12 POLYPHASE CIRCUITS 470 + – + – + – a b c A C B n Zp Zp Zp ■FIGURE 12.18 A balanced -connected load is present on a three-wire three-phase system. The source happens to be Y-connected. Let us consider a balanced -connected load which consists of an imped-ance Zp inserted between each pair of lines. With reference to Fig. 12.18, let us assume known line voltages VL = \|Vab\| = \|Vbc\| = \|Vca\| or known phase voltages Vp = \|Van\| = \|Vbn\| = \|Vcn\| where VL = √ 3Vp and Vab = √ 3Vp/30◦ as we found previously. Because the voltage across each branch of the is known, the phase currents are easily found: IAB = Vab Zp IBC = Vbc Zp IC A = Vca Zp and their differences provide us with the line currents, such as IaA = IAB −IC A Since we are working with a balanced system, the three phase currents are of equal amplitude: Ip = \|IAB\| = \|IBC\| = \|IC A\| The line currents are also equal in amplitude; the symmetry is apparent from the phasor diagram of Fig. 12.19. We thus have IL = \|IaA\| = \|IbB\| = \|IcC\| and IL = √ 3Ip SECTION 12.4 THE DELTA () CONNECTION 471 Let us disregard the source for the moment and consider only the bal-anced load. If the load is -connected, then the phase voltage and the line voltage are indistinguishable, but the line current is larger than the phase current by a factor of √ 3; with a Y-connected load, however, the phase cur-rent and the line current refer to the same current, and the line voltage is greater than the phase voltage by a factor of √ 3. VCA ICA IAB IaA IcC Vcn VAB Van VBC Vbn IBC IbB ■FIGURE 12.19 A phasor diagram that could apply to the circuit of Fig. 12.18 if Zp were an inductive impedance. EXAMPLE 12.5 Again, keep in mind that we are assuming all voltages and currents are quoted as rms values. Determine the amplitude of the line current in a three-phase system with a line voltage of 300 V that supplies 1200 W to a -connected load at a lagging PF of 0.8; then ﬁnd the phase impedance. Let us again consider a single phase. It draws 400 W, 0.8 lagging PF, at a 300 V line voltage. Thus, 400 = 300(Ip)(0.8) and Ip = 1.667 A and the relationship between phase currents and line currents yields IL = √ 3(1.667) = 2.89 A Next, the phase angle of the load is cos−1(0.8) = 36.9◦, and therefore the impedance in each phase must be Zp = 300 1.667/36.9◦= 180/36.9◦ PRACTICE ● 12.7 Each phase of a balanced three-phase -connected load consists of a 200 mH inductor in series with the parallel combination of a 5 μF capacitor and a 200 resistance. Assume zero line resistance and a phase voltage of 200 V at ω = 400 rad/s. Find (a) the phase current; (b) the line current; (c) the total power absorbed by the load. Ans: 1.158 A; 2.01 A; 693 W. The √ 3 factor not only relates phase and line quantities but also appears in a useful expression for the total power drawn by any balanced three-phase load. If we assume a Y-connected load with a power-factor angle θ, the power taken by any phase is Pp = VpIp cos θ = VpIL cos θ = VL √ 3 IL cos θ and the total power is P = 3Pp = √ 3VL IL cos θ In a similar way, the power delivered to each phase of a -connected load is Pp = VpIp cos θ = VL Ip cos θ = VL IL √ 3 cos θ giving a total power P = 3Pp = √ 3VL IL cos θ Thus Eq. enables us to calculate the total power delivered to a balanced load from a knowledge of the magnitude of the line voltage, of the line cur-rent, and of the phase angle of the load impedance (or admittance), regard-less of whether the load is Y-connected or -connected. The line current in CHAPTER 12 POLYPHASE CIRCUITS 472 EXAMPLE 12.6 Determine the amplitude of the line current in a three-phase system with a 300 V line voltage that supplies 1200 W to a Y-connected load at a lagging PF of 0.8. (This is the same circuit as in Example 12.5, but with a Y-connected load instead.) On a per-phase basis, we now have a phase voltage of 300/ √ 3 V, a power of 400 W, and a lagging PF of 0.8. Thus, 400 = 300 √ 3 (Ip)(0.8) and Ip = 2.89 (and so IL = 2.89 A) The phase angle of the load is again 36.9◦, and thus the impedance in each phase of the Y is Zp = 300/ √ 3 2.89 /36.9◦= 60/36.9◦ PRACTICE ● 12.8 Abalanced three-phase three-wire system is terminated with two -connected loads in parallel. Load 1 draws 40 kVAat a lagging PF of 0.8, while load 2 absorbs 24 kW at a leading PF of 0.9.Assume no line resistance, and let Vab = 440/30◦V. Find (a) the total power drawn by the loads; (b) the phase current IAB1 for the lagging load; (c) IAB2; (d) IaA. Ans: 56.0 kW; 30.3/−6.87◦A; 20.2/55.8◦A; 75.3/−12.46◦A. SECTION 12.4 THE DELTA () CONNECTION 473 TABLE 12.1 Comparison of Y- and -Connected Three-Phase Loads. Vp Is the Voltage ● Magnitude of Each Y-Connected Source Phase Load Phase Voltage Line Voltage Phase Current Line Current Power per Phase VAN = Vp/0◦ VBN = Vp/−120◦ VC N = Vp/−240◦ VAB = Vab = √ 3Vp/30◦ VBC = Vbc = √ 3Vp/−90◦ VC A = Vca = √ 3Vp/−210◦ Y VAB = Vab = ( √ 3/30◦)VAN = √ 3Vp/30◦ VBC = Vbc = ( √ 3/30◦)VBN = √ 3Vp/−90◦ VC A = Vca = ( √ 3/30◦)VC N = √ 3Vp/−210◦ VAB = Vab = √ 3Vp/30◦ VBC = Vbc = √ 3Vp/−90◦ VC A = Vca = √ 3Vp/−210◦ IaA = IAN = VAN Zp IbB = IBN = VBN Zp IcC = IC N = VC N Zp IAB = VAB Zp IBC = VBC Zp IC A = VC A Zp IaA = IAN = VAN Zp IbB = IBN = VBN Zp IcC = IC N = VC N Zp IaA = ( √ 3/−30◦)VAB Zp IbB = ( √ 3/−30◦)VBC Zp IcC = ( √ 3/−30◦)VC A Zp √ 3VL IL cos θ where cos θ = power factor of the load √ 3VL IL cos θ where cos θ = power factor of the load Examples 12.5 and 12.6 can now be obtained in two simple steps: 1200 = √ 3(300)(IL)(0.8) Therefore, IL = 5 √ 3 = 2.89 A A brief comparison of phase and line voltages as well as phase and line currents is presented in Table 12.1 for both Y- and -connected loads powered by a Y-connected three-phase source. -Connected Sources The source may also be connected in a conﬁguration. This is not typical, however, for a slight unbalance in the source phases can lead to large cur-rents circulating in the loop. For example, let us call the three single-phase sources Vab, Vbc, and Vcd. Before closing the by connecting d to a, let us determine the unbalance by measuring the sum Vab + Vbc + Vca. Suppose that the amplitude of the result is only 1 percent of the line voltage. The circulating current is thus approximately 1 3 percent of the line voltage divided by the internal impedance of any source. How large is this impedance apt to be? It must depend on the current that the source is expected to deliver with a negligible drop in terminal voltage. If we assume that this maximum current causes a 1 percent drop in the terminal voltage, then the circulating current is one-third of the maximum current! This reduces the useful current capacity of the source and also increases the losses in the system. PRACTICAL APPLICATION Power-Generating Systems ■FIGURE 12.20 Wind-energy harvesting installation at Altamont Pass, California, which consists of over 7000 individual windmills. (© Digital Vision/PunchStock) Today, electrical power is generated by a rather wide variety of techniques. For example, direct conversion of solar energy into electricity using photovoltaic (solar cell) technology results in the production of dc power. Despite representing a very environmentally friendly technology, however, photovoltaic-based installations are presently more expensive than other means of producing electric-ity, and require the use of inverters to convert the dc power into ac. Other technologies such as wind turbine, geo-thermal, hydrodynamic, nuclear, and fossil fuel–based generators are often more economical by comparison. In these systems, a shaft is rotated through the action of a prime mover, such as wind on a propeller, or water or steam on turbine blades (Fig. 12.20). Once a prime mover has been harnessed to generate rotational movement of a shaft, there are several means of converting this mechanical energy into electrical energy. One example is a synchronous generator (Fig. 12.21). These machines are composed of two main sections: a stationary part, called the stator, and a rotating part, termed the rotor. DC current is supplied to coils of wire wound about the rotor to generate a magnetic field, which is rotated through the action of the prime mover. A set of three-phase voltages is then induced at a sec-ond set of windings around the stator. Synchronous generators get their name from the fact that the fre-quency of the ac voltage produced is synchronized with the mechanical rotation of the rotor. The actual demand on a stand-alone generator can vary greatly as various loads are added or removed, such as when air conditioning units kick on, lighting is turned on or off, etc. The voltage output of a generator should ideally be independent of the load, but this is not the case in practice. The voltage EA induced in any given stator phase, often referred to as the internal generated voltage, has a magnitude given by E A = Kφω where K is a constant dependent on the way the machine is constructed, φ is the magnetic ﬂux produced by the ﬁeld windings on the rotor (and hence is independent of the load), and ω is the speed of rotation, which depends only on the prime mover and not the load attached to the generator. Thus, changing the load does not affect the magnitude of EA. The internal generated voltage can be related to the phase voltage Vφ and the phase current IA by EA = Vφ + j XSIA where XS is the synchronous reactance of the generator. The voltage regulation of a generator is deﬁned as % regulation = Vno load −Vfull load Vfull load × 100 Ideally, the regulation should be as close to zero as pos-sible, but this can only be accomplished if the dc current used to control the ﬂux φ around the ﬁeld winding is var-ied in order to compensate for changing load conditions; this can quickly become rather cumbersome. Thus, when designing a power generation facility several smaller generators connected in parallel are usually preferable to one large generator capable of handling the peak load. Each generator can be operated at or near full load, so that the voltage output is essentially constant; individual generators can be added or removed from the system depending on the demand. V jXSIA EA IA (a) EA V jXSIA IA IA E A jXSI A I A V (b) ■FIGURE 12.22 Phasor diagrams describing the effect of loading on a stand-alone synchronous generator. (a) Generator connected to a load having a lagging power factor of cos θ. (b) An additional load is added without changing the power factor. The magnitude of the internal generated voltage EA remains the same while the output current increases. Consequently, the output voltage Vφ is reduced. ■FIGURE 12.21 The 24-pole rotor of a synchronous generator as it is being lowered into position. (Photo courtesy Dr. Wade Enright, Te Kura Pukaha Vira O Te Whare Wananga O Waitaha, Aotearoa.) If the load is increased, then a larger current I′ A will be drawn from the generator. If the power factor is not changed (i.e., the angle between Vφ and IA remains con-stant), Vφ will be reduced since EA cannot change. For example, consider the phasor diagram of Fig. 12.22a, which depicts the voltage-current output of a single phase of a generator connected to a load with a lagging power factor of cos θ. The internal generated voltage EA is also shown. If an additional load is added without changing the power factor, as represented in Fig. 12.22b, the supplied current IA increases to I′ A. However, the magnitude of the internal generated volt-age, formed by the sum of the phasors j X SI′ A and V′ φ, must remain unchanged. Thus, E′ A = E A, and so the voltage output (V′ φ) of the generator will be slightly reduced, as depicted in Fig. 12.22b. We should also note that balanced three-phase sources may be trans-formed from Y to , or vice versa, without affecting the load currents or voltages. The necessary relationships between the line and phase voltages are shown in Fig. 12.13 for the case where Van has a reference phase angle of 0◦. This transformation enables us to use whichever source connection we prefer, and all the load relationships will be correct. Of course, we can-not specify any currents or voltages within the source until we know how it is actually connected. Balanced three-phase loads may be transformed between Y- and -connected conﬁgurations using the relation ZY = Z 3 which is probably worth remembering. 12.5 • POWER MEASUREMENT IN THREE-PHASE SYSTEMS Use of the Wattmeter In large electrical systems, not only are voltage and current important to know, but power is quoted so often that measuring it directly proves highly valuable. This is typically performed using a device known as a wattmeter, which must have the ability to establish both the voltage and the current associated with either the source, the load, or both. Modern devices are very similar to the digital multimeter, providing a numerical display of the quan-tity being measured. These devices frequently make use of the fact that current gives rise to a magnetic ﬁeld, which can be measured without break-ing the circuit. However, in the ﬁeld we still encounter analog versions of the multimeter, and they continue to have some advantages over digital versions, such as the ability to function without a separate power source (e.g., battery), and secondary information that comes from watching a needle move as opposed to numbers seemingly jumping around randomly on a display. Thus, in this section, we focus on power measurement using a traditional analog meter, as switching to a digital device is straightforward if one is available. Before embarking on a discussion of the specialized tech-niques used to measure power in three-phase systems, it is to our advantage to brieﬂy consider how a wattmeter is used in a single-phase circuit. Power measurement is most often accomplished at frequencies below a few hundred hertz through the use of a wattmeter that contains two separate coils. One of these coils is made of heavy wire, having a very low resis-tance, and is called the current coil; the second coil is composed of a much greater number of turns of ﬁne wire, with relatively high resistance, and is termed the potential coil, or voltage coil. Additional resistance may also be inserted internally or externally in series with the potential coil. The torque applied to the moving system and the pointer is proportional to the instanta-neous product of the currents ﬂowing in the two coils. The mechanical iner-tia of the moving system, however, causes a deﬂection that is proportional to the average value of this torque. The wattmeter is used by connecting it into a network in such a way that the current ﬂowing in the current coil is the current ﬂowing into the network CHAPTER 12 POLYPHASE CIRCUITS 476 SECTION 12.5 POWER MEASUREMENT IN THREE-PHASE SYSTEMS 477 (a) Potential coil Current coil + + Passive network 10 V1 = 10 j5 I (b) + + 90° 100 V V2 = 0° 100 V + – + – ■FIGURE 12.23 (a) A wattmeter connection that will ensure an upscale reading for the power absorbed by the passive network. (b) An example in which the wattmeter is installed to give an upscale indication of the power absorbed by the right source. and the voltage across the potential coil is the voltage across the two terminals of the network. The current in the potential coil is thus the input voltage di-vided by the resistance of the potential coil. It is apparent that the wattmeter has four available terminals, and correct connections must be made to these terminals in order to obtain an upscale reading on the meter. To be speciﬁc, let us assume that we are measuring the power absorbed by a passive network. The current coil is inserted in series with one of the two conductors connected to the load, and the potential coil is installed between the two conductors, usually on the “load side’’ of the current coil. The potential coil terminals are often indicated by arrows, as shown in Fig. 12.23a. Each coil has two terminals, and the proper relation-ship between the sense of the current and voltage must be observed. One end of each coil is usually marked (+), and an upscale reading is obtained if a positive current is ﬂowing into the (+) end of the current coil while the (+) terminal of the potential coil is positive with respect to the unmarked end. The wattmeter shown in the network of Fig. 12.23a therefore gives an upscale deﬂection when the network to the right is absorbing power. A re-versal of either coil, but not both, will cause the meter to try to deﬂect down-scale; a reversal of both coils will never affect the reading. As an example of the use of such a wattmeter in measuring average power, let us consider the circuit shown in Fig. 12.23b. The connection of the wattmeter is such that an upscale reading corresponds to a positive ab-sorbed power for the network to the right of the meter, that is, the right source. The power absorbed by this source is given by P = \|V2\| \|I\| cos(ang V2 −ang I) CHAPTER 12 POLYPHASE CIRCUITS 478 Using superposition or mesh analysis, we ﬁnd the current is I = 11.18/153.4◦A and thus the absorbed power is P = (100)(11.18) cos(0◦−153.4◦) = −1000 W The pointer therefore rests against the downscale stop. In practice, the potential coil can be reversed more quickly than the current coil, and this reversal provides an upscale reading of 1000 W. 4 150 + j130 V 6 –j12 j2 z y x + + 0° V 100 + – + – ■FIGURE 12.24 PRACTICE ● 12.9 Determine the wattmeter reading in Fig. 12.24, state whether or not the potential coil had to be reversed in order to obtain an upscale reading, and identify the device or devices absorbing or generating this power. The (+) terminal of the wattmeter is connected to (a) x; (b) y; (c) z. Ans: 1200 W, as is, P6 (absorbed); 2200 W, as is, P4 + P6 (absorbed); 500 W, reversed, absorbed by 100 V. The Wattmeter in a Three-Phase System At ﬁrst glance, measurement of the power drawn by a three-phase load seems to be a simple problem. We need place only one wattmeter in each of the three phases and add the results. For example, the proper connections for a Y-connected load are shown in Fig. 12.25a. Each wattmeter has its cur-rent coil inserted in one phase of the load and its potential coil connected between the line side of that load and the neutral. In a similar way, three wattmeters may be connected as shown in Fig. 12.25b to measure the total power taken by a -connected load. The methods are theoretically correct, but they may be useless in practice because the neutral of the Y is not al-ways accessible and the phases of the are not available. A three-phase ro-tating machine, for example, has only three accessible terminals, those which we have been calling A, B, and C. Clearly, we have a need for a method of measuring the total power drawn by a three-phase load having only three accessible terminals; mea-surements may be made on the “line’’ side of these terminals, but not on the “load’’side. Such a method is available, and is capable of measuring the power taken by an unbalanced load from an unbalanced source. Let us SECTION 12.5 POWER MEASUREMENT IN THREE-PHASE SYSTEMS 479 + + A A B x C N C B a b c + + + + ZC ZB ZA ■FIGURE 12.26 A method of connecting three wattmeters to measure the total power taken by a three-phase load. Only the three terminals of the load are accessible. (a) + + + + + + n C B A + + + + C A B (b) + + ■FIGURE 12.25 Three wattmeters are connected in such a way that each reads the power taken by one phase of a three-phase load, and the sum of the readings is the total power. (a) A Y-connected load. (b) A -connected load. Neither the loads nor the source need be balanced. connect three wattmeters in such a way that each has its current coil in one line and its voltage coil between that line and some common point x, as shown in Fig. 12.26. Although a system with a Y-connected load is illus-trated, the arguments we present are equally valid for a -connected load. The point x may be some unspeciﬁed point in the three-phase system, or it may be merely a point in space at which the three potential coils have a common node. The average power indicated by wattmeter A must be PA = 1 T T 0 vAxiaA dt where T is the period of all the source voltages. The readings of the other two wattmeters are given by similar expressions, and the total average power drawn by the load is therefore P = PA + PB + PC = 1 T T 0 (vAxiaA + vBxibB + vCxicC) dt Each of the three voltages in the preceding expression may be written in terms of a phase voltage and the voltage between point x and the neutral, vAx = vAN + vNx vBx = vBN + vNx vCx = vC N + vNx and, therefore, P = 1 T T 0 (vANiaA + vBNibB + vC NicC) dt + 1 T T 0 vNx(iaA + ibB + icC) dt However, the entire three-phase load may be considered to be a supernode, and Kirchhoff’s current law requires iaA + ibB + icC = 0 Thus P = 1 T T 0 (vANiaA + vBNibB + vC NicC) dt Reference to the circuit diagram shows that this sum is indeed the sum of the average powers taken by each phase of the load, and the sum of the read-ings of the three wattmeters therefore represents the total average power drawn by the entire load! Let us illustrate this procedure with a numerical example before we dis-cover that one of these three wattmeters is really superﬂuous. We will as-sume a balanced source, Vab = 100/0◦V Vbc = 100/−120◦V Vca = 100/−240◦V or Van = 100 √ 3 /−30◦V Vbn = 100 √ 3 /−150◦V Vcn = 100 √ 3 /−270◦V and an unbalanced load, ZA = −j10 ZB = j10 ZC = 10 CHAPTER 12 POLYPHASE CIRCUITS 480 SECTION 12.5 POWER MEASUREMENT IN THREE-PHASE SYSTEMS 481 Let us assume ideal wattmeters, connected as illustrated in Fig. 12.26, with point x located on the neutral of the source n. The three line currents may be obtained by mesh analysis, IaA = 19.32/15◦A IbB = 19.32/165◦A IcC = 10/−90◦A The voltage between the neutrals is VnN = Vnb + VBN = Vnb + IbB( j10) = 157.7/−90◦ The average power indicated by each wattmeter may be calculated, PA = VpIaA cos(angVan −ang IaA) = 100 √ 3 19.32 cos(−30◦−15◦) = 788.7 W PB = 100 √ 3 19.32 cos(−150◦−165◦) = 788.7 W PC = 100 √ 3 10 cos(−270◦+ 90◦) = −577.4 W or a total power of 1 kW. Since an rms current of 10 A ﬂows through the resistive load, the total power drawn by the load is P = 102(10) = 1 kW and the two methods agree. The Two-Wattmeter Method We have proved that point x, the common connection of the three potential coils, may be located any place we wish without affecting the algebraic sum of the three wattmeter readings. Let us now consider the effect of placing point x, this common connection of the three wattmeters, directly on one of the lines. If, for example, one end of each potential coil is returned to B, then there is no voltage across the potential coil of wattmeter B and this meter must read zero. It may therefore be removed, and the algebraic sum of the remaining two wattmeter readings is still the total power drawn by the load. When the location of x is selected in this way, we describe the method of power measurement as the two-wattmeter method. The sum of the readings indicates the total power, regardless of (1) load unbalance, (2) source un-balance, (3) differences in the two wattmeters, and (4) the waveform of the periodic source. The only assumption we have made is that wattmeter cor-rections are sufﬁciently small that we can ignore them. In Fig. 12.26, for example, the current coil of each meter has passing through it the line current drawn by the load plus the current taken by the potential coil. Since the latter current is usually quite small, its effect may be estimated from a knowledge of the resistance of the potential coil and voltage across it. These two quantities enable a close estimate to be made of the power dissipated in the potential coil. In the numerical example described previously, let us now assume that two wattmeters are used, one with current coil in line A and potential coil between lines A and B, the other with current coil in line C and potential coil Note that the reading of one of the wattmeters is negative. Our previous discussion on the basic use of a wattmeter indicates that an upscale reading on that meter can only be obtained after either the potential coil or the current coil is reversed. between C and B. The ﬁrst meter reads P1 = VAB IaA cos(ang VAB −ang IaA) = 100(19.32) cos(0◦−15◦) = 1866 W and the second P2 = VCB IcC cos(ang VCB −ang IcC) = 100(10) cos(60◦+ 90◦) = −866 W and, therefore, P = P1 + P2 = 1866 −866 = 1000 W as we expect from recent experience with the circuit. In the case of a balanced load, the two-wattmeter method enables the PF angle to be determined, as well as the total power drawn by the load. Let us assume a load impedance with a phase angle θ; either a Y or connection may be used and we will assume the connection shown in Fig. 12.27. The construction of a standard phasor diagram, such as that of Fig. 12.19, en-ables us to determine the proper phase angle between the several line volt-ages and line currents. We therefore determine the readings P1 = \|VAB\| \|IaA\| cos(ang VAB −ang IaA) = VL IL cos(30◦+ θ) and P2 = \|VC B\| \|IcC\| cos(ang VC B −ang IcC) = VL IL cos(30◦−θ) The ratio of the two readings is P1 P2 = cos(30◦+ θ) cos(30◦−θ) If we expand the cosine terms, this equation is easily solved for tan θ, tan θ = √ 3 P2 −P1 P2 + P1 CHAPTER 12 POLYPHASE CIRCUITS 482 + + + A B C 1 a + 2 c b ⏐Zp⏐ ⏐Zp⏐ ⏐Zp⏐ ■FIGURE 12.27 Two wattmeters connected to read the total power drawn by a balanced three-phase load. SECTION 12.5 POWER MEASUREMENT IN THREE-PHASE SYSTEMS 483 Thus, equal wattmeter readings indicate a unity PF load; equal and op-posite readings indicate a purely reactive load; a reading of P2 which is (algebraically) greater than P1 indicates an inductive impedance; and a reading of P2 which is less than P1 signiﬁes a capacitive load. How can we tell which wattmeter reads P1 and which reads P2? It is true that P1 is in line A, and P2 is in line C, and our positive phase-sequence system forces Van to lag Vcn. This is enough information to differentiate between two watt-meters, but it is confusing to apply in practice. Even if we were unable to distinguish between the two, we know the magnitude of the phase angle, but not its sign. This is often sufﬁcient information; if the load is an induction motor, the angle must be positive and we do not need to make any tests to determine which reading is which. If no previous knowledge of the load is assumed, then there are several methods of resolving the ambiguity. Perhaps the simplest method is that which involves adding a high-impedance reac-tive load, say, a three-phase capacitor, across the unknown load. The load must become more capacitive. Thus, if the magnitude of tan θ (or the mag-nitude of θ) decreases, then the load was inductive, whereas an increase in the magnitude of tan θ signiﬁes an original capacitive impedance. EXAMPLE 12.7 The balanced load in Fig. 12.28 is fed by a balanced three-phase system having Vab = 230/0◦V rms and positive phase sequence. Find the reading of each wattmeter and the total power drawn by the load. The potential coil of wattmeter #1 is connected to measure the voltage Vac, and its current coil is measuring the phase current IaA. Since we know to use the positive phase sequence, the line voltages are Vab = 230/0◦V Vbc = 230/−120◦V Vca = 230/120◦V Note that Vac = −Vca = 230/−60◦V. + + + A 4 j15 1 a + B N C 2 b c ■FIGURE 12.28 A balanced three-phase system connected to a balanced three-phase load, the power of which is being measured using the two-wattmeter technique. (Continued on next page) SUMMARY AND REVIEW Polyphase circuits are not encountered directly by everyone, but are part of almost every large building installation. In this chapter we studied how three voltages, each 120◦out of phase with the others, can be supplied by a single generator (and hence have the same frequency) and connected to a three-component load. For the sake of convenience we introduced the double-subscript notation, which is commonly employed. A three-phase system will have at least three terminals; a neutral wire connection is not mandatory but is common at least for the source. If a -connected load is employed, then there is no neutral connection to it. When a neutral wire is present, we can deﬁne phase voltages Van, Vbn, and Vcn between each phase (a, b, or c) and neutral. Kirchhoff’s voltage law requires that these three phase voltages sum to zero, regardless of whether positive or negative phase sequence relates their angles. Line voltages (i.e., between phases) can be related directly to the phase voltages; for a -connected load they are equal. In a similar fashion, line currents and phase currents can be directly related to one another; in a Y-connected load, they are equal. At ﬁrst glance such systems can appear to be somewhat complicated, but symmetry often allows us to perform the analysis on a per-phase basis, simplifying our calculations considerably. CHAPTER 12 POLYPHASE CIRCUITS 484 Since this measurement would result in the meter pegged at downscale, one of the coils would need to be reversed in order to take the reading. The phase current IaA is given by the phase voltage Van divided by the phase impedance 4 + j15 , IaA = Van 4 + j15 = (230/ √ 3)/−30◦ 4 + j15 A = 8.554/−105.1◦A We may now compute the power measured by wattmeter #1 as P1 = \|Vac\| \|IaA\| cos(ang Vac −ang IaA) = (230)(8.554) cos(−60◦+ 105.1◦) W = 1389 W In a similar fashion, we determine that P2 = \|Vbc\| \|IbB\| cos(ang Vbc −ang IbB) = (230)(8.554) cos(−120◦−134.9◦) W = −512.5 W Thus, the total average power absorbed by the load is P = P1 + P2 = 876.5 W PRACTICE ● 12.10 For the circuit of Fig. 12.26, let the loads be ZA = 25/60◦, ZB = 50/−60◦, ZC = 50/60◦, VAB = 600/0◦V rms with (+) phase sequence, and locate point x at C. Find (a) PA; (b) PB; (c) PC. Ans: 0; 7200 W; 0. SUMMARY AND REVIEW 485 A concise list of key concepts of the chapter is presented below for the convenience of the reader, along with the corresponding example numbers. ❑The majority of electricity production is in the form of three-phase power. ❑Most residential electricity in North America is in the form of single-phase alternating current at a frequency of 60 Hz and an rms voltage of 115 V. Elsewhere, 50 Hz at 240 V rms is most common. ❑Double-subscript notation is commonly employed in power systems for both voltages and currents. (Example 12.1) ❑Three-phase sources can be either Y- or -connected. Both types of sources have three terminals, one for each phase; Y-connected sources have a neutral connection as well. (Example 12.2) ❑In a balanced three-phase system, each phase voltage has the same magnitude, but is 120◦out of phase with the other two. (Example 12.2) ❑Loads in a three-phase system may be either Y- or -connected. ❑In a balanced Y-connected source with positive (“abc’’) phase sequence, the line voltages are Vab = √ 3Vp/30◦ Vbc = √ 3Vp/−90◦ Vca = √ 3Vp/−210◦ where the phase voltages are Van = Vp/0◦ Vbn = Vp/−120◦ Vcn = Vp/−240◦ (Example 12.2) ❑In a system with a Y-connected load, the line currents are equal to the phase currents. (Examples 12.3, 12.4, 12.6) ❑In a -connected load, the line voltages are equal to the phase voltages. (Example 12.5) ❑In a balanced system with positive phase sequence and a balanced -connected load, the line currents are Ia = IAB √ 3/−30◦ Ib = IBC √ 3/−150◦ Ic = IC A √ 3/+90◦ where the phase currents are IAB = VAB Z = Vab Z IBC = VBC Z = Vbc Z IC A = VC A Z = Vca Z (Example 12.5) ❑Most power calculations are performed on a per-phase basis, assuming a balanced system; otherwise, nodal/mesh analysis is always a valid approach. (Examples 12.3, 12.4, 12.5) ❑The power in a three-phase system (balanced or unbalanced) can be measured with only two wattmeters. (Example 12.7) ❑The instantaneous power in any balanced three-phase system is constant. READING FURTHER A good overview of ac power concepts can be found in Chap. 2 of: B. M. Weedy, Electric Power Systems, 3rd ed. Chichester, England: Wiley, 1984. A comprehensive book on generation of electrical power from wind is: T. Burton, D. Sharpe, N. Jenkins, and E. Bossanyi, Wind Energy Handbook. Chichester, England: Wiley, 2001. EXERCISES 12.1 Polyphase Systems 1. An unknown three terminal device has leads named b, c, and e. When installed in one particular circuit, measurements indicated that Vec = –9 V and Veb = –0.65 V. (a) Calculate Vcb. (b) Determine the power dissipated in the b-e junction if the current Ib ﬂowing into the terminal marked b is equal to 1 μA. 2. A common type of transistor is known as the MESFET, which is an acronym for metal-semiconductor ﬁeld effect transistor. It has three terminals, named the gate (g), the source (s), and the drain (d). As an example, consider one particular MESFET operating in a circuit such that Vsg = 0.2 V and Vds = 3 V. (a) Calculate Vgs and Vdg. (b) If a gate current Ig = 100 pA is found to be ﬂowing into the gate terminal, compute the power lost at the gate-source junction. 3. For a certain Y-connected three-phase source, Van = 400/33◦V, Vbn = 400/153◦V, and Vcx = 160/208◦V. Determine (a) Vcn; (b) Van −Vbn; (c) Vax; (d) Vbx. 4. Describe what is meant by a “polyphase” source, state one possible advantage of such sources that might outweigh their additional complexity over single-phase sources of power, and explain the difference between “balanced” and “unbalanced” sources. 5. Several of the voltages associated with a certain circuit are given by V12 = 9/30◦V, V32 = 3/130◦V, and V14 = 2/10◦V. Determine V21, V13, V34, and V24. 6. The nodal voltages which describe a particular circuit can be expressed as V14 = 9 −j V, V24 = 3 + j3 V, and V34 = 8 V. Calculate V12, V32, and V13. Express your answers in phasor form. 7. In the circuit of Fig. 12.29, the resistor markings unfortunately have been omitted, but several of the currents are known. Speciﬁcally, Iad = 1 A. (a) Compute Iab, Icd, Ide, If e, and Ibe. (b) If Vba = 125 V, determine the value of the resistor linking nodes a and b. CHAPTER 12 POLYPHASE CIRCUITS 486 8 A 10 A c d e a b f ■FIGURE 12.29 8. For the circuit shown in Fig. 12.30, (a) determine Igh, Icd, and Idh. (b) Calcu-late Ied, Iei, and Ij f . (c) If all resistors in the circuit each have a value of 1 , determine the three clockwise-ﬂowing mesh currents. EXERCISES 487 9. Additional resistors are added in parallel to the resistors between terminals d and e, and terminals f and j, respectively, of the circuit in Fig. 12.30. (a) Which voltages may still be described using double-subscript notation? (b) Which line currents may still be described by double-subscript notation? 12.2 Single-Phase Three-Wire Systems 10. Most consumer electronics are powered by 110 V outlets, but several types of appliances (such as clothes dryers) are powered from 220 V outlets. Lower voltages are generally safer. What, then, motivates manufactures of some pieces of equipment to design them to run on 220 V? 11. The single-phase three-wire system of Fig. 12.31 has three separate load impedances. If the source is balanced and Van = 110 + j0 V rms, (a) express Van and Vbn in phasor notation. (b) Determine the phasor voltage which appears across the impedance Z3. (c) Determine the average power delivered by the two sources if Z1 = 50 + j0 , Z2 = 100 + j45 , and Z3 = 100 −j90 . (d) Represent load Z3 by a series connection of two elements, and state their respective values if the sources operate at 60 Hz. 12. For the system represented in Fig. 12.32, the ohmic losses in the neutral wire are so small they can be neglected and it can be adequately modeled as a short circuit. (a) Calculate the power lost in the two lines as a result of their nonzero resistance. (b) Compute the average power delivered to the load. (c) Determine the power factor of the total load. 5 A 4 A c g h i d e j f ■FIGURE 12.30 + – + – a A N B b Vnb Van Z1 Z3 Z2 n ■FIGURE 12.31 10 + j2 A N C B 10 + j2 ■FIGURE 12.33 I1 I3 I2 0° V rms 115 0° V rms 115 + – + – a b n 1 200 j500 10 50 1 A B N ■FIGURE 12.32 13. Referring to the balanced load represented in Fig. 12.33, if it is connected to a three-wire balanced source operating at 50 Hz such that VAN = 115 V, (a) determine the power factor of the load if the capacitor is omitted; (b) deter-mine the value of capacitance C that will achieve a unity power factor for the total load. 14. In the three-wire system of Fig. 12.32, (a) replace the 50 resistor with a 200 resistor, and calculate the current ﬂowing through the neutral wire. (b) Determine a new value for the 50 resistor such that the neutral wire current magnitude is 25% that of line current IaA. 12.3 Three-Phase Y-Y Connection 15. (a) Show that if Van = 400/33◦V, Vbn = 400/−87◦V, and Vbn = 400/−207◦V, that Van + Vbn + Vcn = 0. (b) Do the voltages in part (a) represent positive or negative phase sequence? Explain. 16. Consider a simple positive phase sequence, three-phase, three-wire system operated at 50 Hz and with a balance load. Each phase voltage of 240 V is connected across a load composed of a series-connected 50 and 500 mH combination. Calculate (a) each line current; (b) the power factor of the load; (c) the total power supplied by the three-phase source. 17. Assume the system shown in Fig. 12.34 is balanced, Rw = 0, Van = 208/0◦V, and a positive phase sequence applies. Calculate all phase and line currents, and all phase and line voltages, if Zp is equal to (a) 1 k; (b) 100 + j48 ; (c) 100 −j48. CHAPTER 12 POLYPHASE CIRCUITS 488 + – + – + – c b a B A n N C Zp Vcn Vbn Van Rw Rw Rw Zp Zp ■FIGURE 12.34 18. Repeat Exercise 17 with Rw = 10 , and verify your answers with appropriate PSpice simulations if the operating frequency is 60 Hz. 19. Each impedance Zp in the balanced three-phase system of Fig. 12.34 is con-structed using the parallel combination of a 1 mF capacitance, a 100 mH inductance, and a 10 resistance. The sources have positive phase sequence and operate at 50 Hz. If Vab = 208/0◦V, and Rw = 0, calculate (a) all phase voltages; (b) all line voltages; (c) all three line currents; (d) the total power drawn by the load. 20. With the assumption that the three-phase system pictured in Fig. 12.34 is balanced with a line voltage of 100 V, calculate the line current and per-phase load impedance if Rw = 0 and the load draws (a) 1 kW at a PF of 0.85 lagging; (b) 300 W per phase at a PF of 0.92 leading. 21. The balanced three-phase system of Fig. 12.34 is characterized by a positive phase sequence and a line voltage of 300 V. And Zp is given by the parallel combination of a 5 −j3 capacitive load and a 9 + j2 inductive load. If Rw = 0, calculate (a) the power factor of the source; (b) the total power supplied by the source. (c) Repeat parts (a) and (b) if Rw = 1 . 22. A balanced Y-connected load of 100 + j50 is connected to a balanced three-phase source. If the line current is 42 A and the source supplies 12 kW, determine (a) the line voltage; (b) the phase voltage. 23. A three-phase system is constructed from a balanced Y-connected source operating at 50 Hz and having a line voltage of 210 V, and each phase of the balanced load draws 130 W at a leading power factor of 0.75. (a) Calculate the line current and the total power supplied to the load. (b) If a purely resistive load of 1 is connected in parallel with each existing load, calculate the new line current and total power supplied to the load. (c) Verify your answers using appropriate PSpice simulations. EXERCISES 489 24. Returning to the balanced three-phase system described in Exercise 21, deter-mine the complex power delivered to the load for both Rw = 0 and Rw = 1 . 25. Each load in the circuit of Fig. 12.34 is composed of a 1.5 H inductor in parallel with a 100 μF capacitor and a 1 k resistor. The resistance is labeled Rw = 0 .Using positive phase sequence with Vab = 115/0◦V at f = 60 Hz, determine the rms line current and the total power delivered to the load. Verify your answers with an appropriate PSpice simulation. 12.4 The Delta () Connection 26. A particular balanced three-phase system is supplying a -connected load with 10 kW at a leading power factor of 0.7. If the phase voltage is 208 V and the source operates at 50 V, (a) compute the line current; (b) determine the phase impedance; (c) calculate the new power factor and new total power delivered to the load if a 2.5 H inductor is connected in parallel with each phase of the load. 27. If each of the three phases in a balanced -connected load is composed of a 10 mF capacitor in parallel with a series-connected 470 resistor and 4 mH inductor combination, assume a phase voltage of 400 V at 50 Hz. (a) Calculate the phase current; (b) the line current; (c) the line voltage; (d) the power factor at which the source operates; (e) the total power delivered to the load. 28. A three-phase load is to be powered by a three-wire three-phase Y-connected source having phase voltage of 400 V and operating at 50 Hz. Each phase of the load consists of a parallel combination of a 500 resistor, 10 mH inductor, and 1 mF capacitor. (a) Compute the line current, line voltage, phase current, and power factor of the load if the load is also Y-connected. (b) Rewire the load so that it is -connected and ﬁnd the same quantities requested in part (a). 29. For the two situations described in Exercise 28, compute the total power delivered to each of the two loads. 30. Two -connected loads are connected in parallel and powered by a balanced Y-connected system. The smaller of the two loads draws 10 kVA at a lagging PF of 0.75, and the larger draws 25 kVA at a leading PF of 0.80. The line voltage is 400 V. Calculate (a) the power factor at which the source is operating; (b) the total power drawn by the two loads; (c) the phase current of each load. 31. For the balanced three-phase system shown in Fig. 12.35, it is determined that 100 W is lost in each wire. If the phase voltage of the source is 400 V, and the load draws 12 kW at a lagging PF of 0.83, determine the wire resistance Rw. + – + – + – c b a B A n C Vcn Vbn Van Rw Rw Rw Zp Zp Zp ■FIGURE 12.35 32. The balanced -connected load in Fig. 12.35 is demanding 10 kVA at a lagging PF of 0.91. If line losses are negligible, calculate IbB and Van if Vca = 160/30◦V and the source voltages are described using positive phase sequence. 33. Repeat Exercise 32 if Rw = 1 . Verify your solution using an appropriate PSpice simulation. 34. Compute IaA, IAB, and Van if the -connected load of Fig. 12.35 draws a total complex power of 1800 + j700 W, Rw = 1.2 , and the source generates a complex power of 1850 + j700 W. 35. A balanced three-phase system having line voltage of 240 V rms contains a -connected load of 12 + j k per phase and also a Y-connected load of 5 + j3 k per phase. Find the line current, the power taken by the combined load, and the power factor of the load. 12.5 Power Measurement in Three-Phase Systems 36. Determine the wattmeter reading (stating whether or not the leads had to be reversed to obtain it) in the circuit of Fig. 12.36 if terminals A and B, respectively, are connected to (a) x and y; (b) x and z; (c) y and z. CHAPTER 12 POLYPHASE CIRCUITS 490 A x y z B + + 200 V rms + – ■FIGURE 12.36 V2 + – I1 + – 440 V rms 60 Hz 1 H ■FIGURE 12.37 + + 200 cos 500t V 20iC 2.5 sin 500t A + – iC + – ■FIGURE 12.38 37. A wattmeter is connected into the circuit of Fig. 12.37 so that I1 enters the (+) terminal of the current coil, while V2 is the voltage across the potential coil. Find the wattmeter reading, and verify your solution with an appropriate PSpice simulation. 38. Find the reading of the wattmeter connected in the circuit of Fig. 12.38. EXERCISES 491 39. (a) Find both wattmeter readings in Fig. 12.39 if VA = 100/0◦V rms, VB = 50/90◦V rms, ZA = 10 −j10 , ZB = 8 + j6 , and ZC = 30 + j10 . (b) Is the sum of these readings equal to the total power taken by the three loads? Verify your answer with an appropriate PSpice simulation. + + A B + + ZC + – + – VB VA ZB ZA ■FIGURE 12.39 40. Circuit values for Fig. 12.40 are Vab = 200/0◦, Vbc = 200/120◦, Vca = 200/240◦V rms, Z4 = Z5 = Z6 = 25/30◦, Z1 = Z2 = Z3 = 50/−60◦. Find the reading for each wattmeter. + + A a b c Z6 Z4 + + B + + C Z2 Z1 Z3 Z5 ■FIGURE 12.40 Chapter-Integrating Exercises 41. Explain under what circumstances a -connected load might be preferred over a Y-connected load which draws the same average and complex powers. 42. A certain 208 V, 60 Hz, three-phase source is connected in a Y conﬁguration and exhibits positive phase sequence. Each phase of the balanced load consists of a coil best modeled as a 0.2 resistance in series with a 580 mH inductance. (a) Determine the line voltages and the phase currents if the load is -connected. (b) Repeat part (a) if the load is Y-connected instead. 44. The computer equipment in a small manufacturing plant all runs on standard 120 VAC, but only 208 VAC three-phase power is available. Explain how the computer equipment can be connected to the existing power wiring. 43. (a) Is the load represented in Fig. 12.41 considered a three-phase load? Explain. (b) If ZAN = 1 −j7 , ZBN = 3/22◦ and ZAB = 2 + j , calculate all phase (and line) currents and voltages assuming a phase to neutral voltage of 120 VAC (the two phases are 180◦out of phase). (c) Under what circumstances does current ﬂow in the neutral wire? CHAPTER 12 POLYPHASE CIRCUITS 492 a n b A N B ZAN ZBN ZAB ■FIGURE 12.41 INTRODUCTION Whenever current flows through a conductor, whether as ac or dc, a magnetic field is generated about that conductor. In the context of circuits, we often refer to the magnetic flux through a loop of wire. This is the average normal component of the mag-netic field density emanating from the loop multiplied by the surface area of the loop. When a time-varying magnetic field generated by one loop penetrates a second loop, a voltage is induced between the ends of the second wire. In order to distin-guish this phenomenon from the “inductance’’ we defined earlier, more properly termed “self-inductance,’’ we will define a new term, mutual inductance. There is no such device as a “mutual inductor,’’ but the principle forms the basis for an extremely important device—the transformer. A transformer consists of two coils of wire separated by a small distance, and is commonly used to convert ac voltages to higher or lower values depending on the application. Every electrical appliance that requires dc current to operate but plugs into an ac wall outlet makes use of a transformer to adjust voltage levels prior to rectiﬁcation, a function typically performed by diodes and described in every introductory electronics text. 13.1 • MUTUAL INDUCTANCE When we deﬁned inductance in Chap. 7, we did so by specifying the relationship between the terminal voltage and current, v(t) = L di(t) dt KEY CONCEPTS Mutual Inductance Self-Inductance The Dot Convention Reﬂected Impedance T and Equivalent Networks The Ideal Transformer Turns Ratio of an Ideal Transformer Impedance Matching Voltage Level Adjustment PSpice Analysis of Circuits with Transformers Magnetically Coupled Circuits C H A P T E R 13 493 where the passive sign convention is assumed. The physical basis for such a current-voltage characteristic rests upon two things: 1. The production of a magnetic ﬂux by a current, the ﬂux being proportional to the current in linear inductors. 2. The production of a voltage by the time-varying magnetic ﬁeld, the voltage being proportional to the time rate of change of the magnetic ﬁeld or the magnetic ﬂux. Coefﬁcient of Mutual Inductance Mutual inductance results from a slight extension of this same argument. A current ﬂowing in one coil establishes a magnetic ﬂux about that coil and also about a second coil nearby.The time-varying ﬂux surrounding the second coil produces a voltage across the terminals of the second coil; this voltage is pro-portional to the time rate of change of the current ﬂowing through the ﬁrst coil. Figure 13.1a shows a simple model of two coils L1 and L2, sufﬁciently close togetherthattheﬂuxproducedbyacurrenti1(t)ﬂowingthroughL1establishes an open-circuit voltage v2(t) across the terminals of L2. Without considering the proper algebraic sign for the relationship at this point, we deﬁne the coef-ﬁcient of mutual inductance, or simply mutual inductance, M21, as v2(t) = M21 di1(t) dt CHAPTER 13 MAGNETICALLY COUPLED CIRCUITS 494 ■FIGURE 13.1 (a) A current i1 through L1 produces an open-circuit voltage v2 across L2. (b) A current i2 through L2 produces an open-circuit voltage v1 across L1. (a) L2 M v2 L1 i1 + – (b) L2 v1 L1 i2 + – M (1) Mutual inductance is not universally assumed to be positive. It is particularly convenient to allow it to “carry its own sign’’ when three or more coils are involved and each coil interacts with each other coil. We will restrict our attention to the more important simple case of two coils. The order of the subscripts on M21 indicates that a voltage response is pro-duced at L2 by a current source at L1. If the system is reversed, as indicated in Fig. 13.1b, then we have v1(t) = M12 di2(t) dt Two coefficients of mutual inductance are not necessary, however; we will use energy relationships a little later to prove that M12 and M21 are equal. Thus, M12 = M21 = M. The existence of mutual coupling between two coils is indicated by a double-headed arrow, as shown in Fig. 13.1a and b. Mutual inductance is measured in henrys and, like resistance, induc-tance, and capacitance, is a positive quantity.1 The voltage M di/dt, however, may appear as either a positive or a negative quantity depending on whether the current is increasing or decreasing at a particular instant in time. SECTION 13.1 MUTUAL INDUCTANCE 495 ■FIGURE 13.2 Current entering the dotted terminal of one coil produces a voltage that is sensed positively at the dotted terminal of the second coil. Current entering the undotted terminal of one coil produces a voltage that is sensed positively at the undotted terminal of the second coil. i1 di1 dt (a) L2 v2 = M L1 + – M i1 di1 dt (d) L2 v2 = M L1 + – M i1 di1 dt (b) L2 v2 = –M L1 + – M i1 di1 dt (c) L2 v2 = –M L1 + – M Dot Convention The inductor is a two-terminal element, and we are able to use the passive sign convention in order to select the correct sign for the voltage L di/ dt or jωLI. If the current enters the terminal at which the positive voltage reference is located, then the positive sign is used. Mutual inductance, however, cannot be treated in exactly the same way because four terminals are involved. The choice of a correct sign is established by use of one of sev-eral possibilities that include the “dot convention,” or by an examination of the particular way in which each coil is wound. We will use the dot conven-tion and merely look brieﬂy at the physical construction of the coils; the use of other special symbols is not necessary when only two coils are coupled. The dot convention makes use of a large dot placed at one end of each of the two coils which are mutually coupled. We determine the sign of the mutual voltage as follows: A current entering the dotted terminal of one coil produces an open-circuit voltage with a positive voltage reference at the dotted terminal of the second coil. Thus, in Fig. 13.2a, i1 enters the dotted terminal of L1, v2 is sensed posi-tively at the dotted terminal of L2, and v2 = M di1/dt. We have found pre-viously that it is often not possible to select voltages or currents throughout a circuit so that the passive sign convention is everywhere satisﬁed; the same situation arises with mutual coupling. For example, it may be more convenient to represent v2 by a positive voltage reference at the undotted terminal, as shown in Fig. 13.2b; then v2 = −M di1/dt. Currents that enter the dotted terminal are also not always available, as indicated by Fig. 13.2c and d. We note then that: A current entering the undotted terminal of one coil provides a voltage that is positively sensed at the undotted terminal of the second coil. Note that the preceding discussion does not include any contribution to the voltage from self-induction, which would occur if i2 were nonzero. We will consider this important situation in detail, but a quick example first is appropriate. EXAMPLE 13.1 For the circuit shown in Fig. 13.3, (a) determine v1 if i2 = 5 sin 45tA and i1 = 0; (b) determine v2 if i1 = −8e−t A and i2 = 0. (a) Since the current i2 is entering the undotted terminal of the right coil, the positive reference for the voltage induced across the left coil is the undotted terminal. Thus, we have an open-circuit voltage v1 = −(2)(45)(5 cos 45t) = −450 cos 45t V appearing across the terminals of the left coil as a result of the time-varying magnetic ﬂux generated by i2 ﬂowing into the right coil. ■FIGURE 13.3 The dot convention provides a rela-tionship between the terminal at which a current enters one coil, and the positive voltage reference for the other coil. v2 + – v1 + – i1 i2 L2 L1 M = 2 H (Continued on next page) CHAPTER 13 MAGNETICALLY COUPLED CIRCUITS 496 Combined Mutual and Self-Induction Voltage So far, we have considered only a mutual voltage present across an open-circuited coil. In general, a nonzero current will be ﬂowing in each of the two coils, and a mutual voltage will be produced in each coil because of the current ﬂowing in the other coil. This mutual voltage is present indepen-dently of and in addition to any voltage of self-induction. In other words, the voltage across the terminals of L1 will be composed of two terms, L1 di1/dt and M di2/dt, each carrying a sign depending on the current directions, the assumed voltage sense, and the placement of the two dots. In the portion of a circuit drawn in Fig. 13.4, currents i1 and i2 are shown, each entering a dotted terminal. The voltage across L1 is thus composed of two parts, v1 = L1 di1 dt + M di2 dt as is the voltage across L2, v2 = L2 di2 dt + M di1 dt In Fig. 13.5 the currents and voltages are not selected with the object of obtaining all positive terms for v1 and v2. By inspecting only the reference symbols for i1 and v1, it is apparent that the passive sign convention is not satisﬁed and the sign of L1 di1/dt must therefore be negative. An identical conclusion is reached for the term L2 di2/dt. The mutual term of v2 is signed by inspecting the direction of i1 and v2; since i1 enters the dotted terminal and v2 is sensed positive at the dotted terminal, the sign of M di1/dt must be positive. Finally, i2 enters the undotted terminal of L2, and v1 is sensed positive at the undotted terminal of L1; hence, the mutual portion of v1, M di2/dt, must also be positive. Thus, we have v1 = −L1 di1 dt + M di2 dt v2 = −L2 di2 dt + M di1 dt The same considerations lead to identical choices of signs for excitation by a sinusoidal source operating at frequency ω V1 = −jωL1I1 + jωMI2 V2 = −jωL2I2 + jωMI1 Since no current ﬂows through the coil on the left, there is no contribu-tion to v1 from self-induction. (b) We now have a current entering a dotted terminal, but v2 has its positive reference at the undotted terminal. Thus, v2 = −(2)(−1)(−8e−t) = −16e−t V PRACTICE ● 13.1 Assuming M = 10 H, coil L2 is open-circuited, and i1 = −2e−5t A, ﬁnd the voltage v2 for (a) Fig. 13.2a; (b) Fig. 13.2b. Ans: 100e−5t V; −100e−5t V. i1 i2 L2 v2 L1 + – v1 + – M ■FIGURE 13.4 Since the pairs v1, i1 and v2, i2 each satisfy the passive sign convention, the voltages of self-induction are both positive; since i1 and i2 each enter dotted terminals, and since v1 and v2 are both positively sensed at the dotted terminals, the voltages of mutual induction are also both positive. i1 i2 L2 v2 L1 + – v1 _ + M ■FIGURE 13.5 Since the pairs v1, i1 and v2, i2 are not sensed according to the passive sign convention, the voltages of self-induction are both negative; since i1 enters the dotted terminal and v2 is positively sensed at the dotted terminal, the mutual term of v2 is positive; and since i2 enters the undotted terminal and v1 is positively sensed at the undotted terminal, the mutual term of v1 is also positive. SECTION 13.1 MUTUAL INDUCTANCE 497 Physical Basis of the Dot Convention We can gain a more complete understanding of the dot symbolism by looking at the physical basis for the convention; the meaning of the dots is now inter-preted in terms of magnetic ﬂux. Two coils are shown wound on a cylindrical form in Fig. 13.6, and the direction of each winding is evident. Let us assume that the current i1 is positive and increasing with time. The magnetic ﬂux that i1 produces within the form has a direction which may be found by the right-hand rule: when the right hand is wrapped around the coil with the ﬁngers pointing in the direction of current ﬂow, the thumb indicates the direction of the ﬂux within the coil. Thus i1 produces a ﬂux which is directed downward; since i1 is increasing with time, the ﬂux, which is proportional to i1, is also increasing with time. Turning now to the second coil, let us also think of i2 as positive and increasing; the application of the right-hand rule shows that i2 also produces a magnetic ﬂux which is directed downward and is increasing. In other words, the assumed currents i1 and i2 produce additive ﬂuxes. The voltage across the terminals of any coil results from the time rate of change of the ﬂux within that coil. The voltage across the terminals of the ﬁrst coil is therefore greater with i2 ﬂowing than it would be if i2 were zero; i2 in-duces a voltage in the ﬁrst coil which has the same sense as the self-induced voltage in that coil. The sign of the self-induced voltage is known from the passive sign convention, and the sign of the mutual voltage is thus obtained. The dot convention enables us to suppress the physical construction of the coils by placing a dot at one terminal of each coil such that currents entering dot-marked terminals produce additive ﬂuxes. It is apparent that there are always two possible locations for the dots, because both dots may always be moved to the other ends of the coils and additive ﬂuxes will still result. i1 i2 ■FIGURE 13.6 The physical construction of two mutually coupled coils. From a consideration of the direction of magnetic ﬂux produced by each coil, it is shown that dots may be placed either on the upper terminal of each coil or on the lower terminal of each coil. EXAMPLE 13.2 For the circuit shown in Fig. 13.7a, ﬁnd the ratio of the output voltage across the 400 resistor to the source voltage, expressed using phasor notation. (Continued on next page) I1 I2 0° V 1 j90 j10 400 j k V 1 = 10 = 10 rad/s + – V 2 + – (b) v1 = 10 cos 10t V M = 9 H v2 + – + – 400 1 100 H (a) 1 H i2 i1 ■FIGURE 13.7 (a) A circuit containing mutual inductance in which the voltage ratio V2/V1 is desired. (b) Self- and mutual inductances are replaced by the corresponding impedances. Identify the goal of the problem. We need a numerical value for V2. We will then divide by 10/0◦V. Collect the known information. We begin by replacing the 1 H and 100 H inductances by their corre-sponding impedances, j10 and j k, respectively (Fig. 13.7b). We also replace the 9 H mutual inductance by jωM = j90 . CHAPTER 13 MAGNETICALLY COUPLED CIRCUITS 498 Devise a plan. Mesh analysis is likely to be a good approach, as we have a circuit with two clearly deﬁned meshes. Once we ﬁnd I2, V2 is simply 400I2. Construct an appropriate set of equations. In the left mesh, the sign of the mutual term is determined by applying the dot convention. Since I2 enters the undotted terminal of L2, the mutual voltage across L1 must have the positive reference at the undotted terminal. Thus, (1 + j10)I1 −j90I2 = 10/0◦ Since I1 enters the dot-marked terminal, the mutual term in the right mesh has its (+) reference at the dotted terminal of the 100 H induc-tor. Therefore, we may write (400 + j1000)I2 −j90I1 = 0 Determine if additional information is required. We have two equations in two unknowns, I1 and I2. Once we solve for the two currents, the output voltage V2 may be obtained by multiplying I2 by 400 . Attempt a solution. Upon solving these two equations with a scientiﬁc calculator, we ﬁnd that I2 = 0.172/−16.70◦A Thus, V2 V1 = 400(0.172/−16.70◦) 10/0◦ = 6.880/−16.70◦ Verify the solution. Is it reasonable or expected? We note that the output voltage V2 is actually larger in magnitude than the input voltage V1. Should we always expect this result? The answer is no. As we will see in later sections, the transformer can be constructed to achieve either a reduction or an increase in the voltage. We can perform a quick estimate, however, and at least ﬁnd an upper and lower bound for our answer. If the 400 resistor is replaced with a short circuit, V2 = 0. If instead we replace the 400 resistor with an open circuit, I2 = 0 and hence V1 = (1 + jωL1)I1 and V2 = jωMI1 Solving, we ﬁnd that the maximum value we could expect for V2/V1 is 8.955/5.711◦. Thus, our answer at least appears reasonable. SECTION 13.1 MUTUAL INDUCTANCE 499 The output voltage of the circuit in Fig. 13.7a is greater in magnitude than the input voltage, so that a voltage gain is possible with this type of circuit. It is also interesting to consider this voltage ratio as a function of ω. To ﬁnd I2( jω) for this particular circuit, we write the mesh equations in terms of an unspeciﬁed angular frequency ω: (1 + jω)I1 −jω9I2 = 10/0◦ and −jω9I1 + (400 + jω100)I2 = 0 Solving by substitution, we ﬁnd that I2 = j90ω 400 + j500ω −19ω2 Thus, we obtain the ratio of output voltage to input voltage as a function of frequency ω V2 V1 = 400I2 10 = jω3600 400 + j500ω −19ω2 The magnitude of this ratio, sometimes referred to as the circuit transfer function, is plotted in Fig. 13.8 and has a peak magnitude of approximately ■FIGURE 13.8 The voltage gain \|V2/V1\| of the circuit shown in Fig. 13.7a is plotted as a function of ω using the following MATLAB script: >> w = linspace(0,30,1000); >> num = jw3600; >> for indx = 1:1000 den = 400 + j500w(indx) −19w(indx)w(indx); gain(indx) = num(indx)/den; end >> plot(w, abs(gain)); >> xlabel('Frequency (rad/s)'); >> ylabel('Magnitude of Voltage Gain'); CHAPTER 13 MAGNETICALLY COUPLED CIRCUITS 500 Ans: 20e−1000t = 3i1 + 0.002 di1/dt −0.003 di2/dt ; 10i2 + 0.005 di2/dt − 0.003 di1/dt = 0. 3 mH 10 5 mH 2 mH i2 i1 3 + – vs ■FIGURE 13.9 PRACTICE ● 13.2 For the circuit of Fig. 13.9, write appropriate mesh equations for the left mesh and the right mesh if vs = 20e−1000t V. EXAMPLE 13.3 Write a complete set of phasor mesh equations for the circuit of Fig. 13.10a. 3 6 H 7 H v1 i2 i3 i1 M = 2 H + – 5 1 F 5 (b) (a) j2 j6 j 1 j7 3 V1 I1 I2 I3 + – ■FIGURE 13.10 (a) A three-mesh circuit with mutual coupling. (b) The 1 F capacitance as well as the self- and mutual inductances are replaced by their corresponding impedances. 7 near a frequency of 4.6 rad/s. However, for very small or very large fre-quencies, the magnitude of the transfer function is less than unity. The circuit is still passive, except for the voltage source, and the voltage gain must not be mistakenly interpreted as a power gain. At ω = 10 rad/s, the voltage gain is 6.88, but the ideal voltage source, having a terminal volt-age of 10 V, delivers a total power of 8.07 W, of which only 5.94 W reaches the 400 resistor. The ratio of the output power to the source power, which we may deﬁne as the power gain, is thus 0.736. SECTION 13.2 ENERGY CONSIDERATIONS 501 Ans: Vs = (3 + j10)I1 −j15I2; 0 = −j15I1 + (10 + j25)I2 . 3 mH 10 5 mH 2 mH i2 i1 3 + – vs ■FIGURE 13.11 13.2 • ENERGY CONSIDERATIONS Let us now consider the energy stored in a pair of mutually coupled induc-tors. The results will be useful in several different ways. We will ﬁrst justify our assumption that M12 = M21, and we may then determine the maximum possible value of the mutual inductance between two given inductors. Once again, our ﬁrst step is to replace both the mutual inductance and the two self-inductances with their corresponding impedances as shown in Fig. 13.10b. Applying Kirchhoff’s voltage law to the ﬁrst mesh, a positive sign for the mutual term is ensured by selecting (I3 −I2) as the current through the second coil. Thus, 5I1 + 7 jω(I1 −I2) + 2 jω(I3 −I2) = V1 or (5 + 7 jω)I1 −9 jωI2 + 2 jωI3 = V1 The second mesh requires two self-inductance terms and two mutual inductance terms. Paying close attention to dots, we obtain 7 jω(I2 −I1) + 2 jω(I2 −I3) + 1 jωI2 + 6 jω(I2 −I3) + 2 jω(I2 −I1) = 0 or −9 jωI1 + 17 jω + 1 jω I2 −8 jωI3 = 0 Finally, for the third mesh, 6 jω(I3 −I2) + 2 jω(I1 −I2) + 3I3 = 0 or 2 jωI1 −8 jωI2 + (3 + 6 jω)I3 = 0 Equations to may be solved by any of the conventional methods. PRACTICE ● 13.3 For the circuit of Fig. 13.11, write an appropriate mesh equation in terms of the phasor currents I1 and I2 for the (a) left mesh; (b) right mesh. Equality of M12 and M21 The pair of coupled coils shown in Fig. 13.12 has currents, voltages, and po-larity dots indicated. In order to show that M12 = M21 we begin by letting all currents and voltages be zero, thus establishing zero initial energy stor-age in the network. We then open-circuit the right-hand terminal pair and in-crease i1 from zero to some constant (dc) value I1 at time t = t1. The power entering the network from the left at any instant is v1i1 = L1 di1 dt i1 and the power entering from the right is v2i2 = 0 since i2 = 0. The energy stored within the network when i1 = I1 is thus t1 0 v1i1 dt = I1 0 L1i1 di1 = 1 2 L1I 2 1 We now hold i1 constant (i1 = I1), and we let i2 change from zero at t = t1 to some constant value I2 at t = t2. The energy delivered from the right-hand source is thus t2 t1 v2i2 dt = I2 0 L2i2 di2 = 1 2 L2I 2 2 However, even though the value of i1 remains constant, the left-hand source also delivers energy to the network during this time interval: t2 t1 v1i1 dt = t2 t1 M12 di2 dt i1 dt = M12I1 I2 0 di2 = M12I1I2 The total energy stored in the network when both i1 and i2 have reached constant values is Wtotal = 1 2 L1I 2 1 + 1 2 L2I 2 2 + M12I1I2 Now, we may establish the same ﬁnal currents in this network by allow-ing the currents to reach their ﬁnal values in the reverse order, that is, ﬁrst increasing i2 from zero to I2 and then holding i2 constant while i1 increases from zero to I1. If the total energy stored is calculated for this experiment, the result is found to be Wtotal = 1 2 L1I 2 1 + 1 2 L2I 2 2 + M21I1I2 The only difference is the interchange of the mutual inductances M21 and M12. The initial and ﬁnal conditions in the network are the same, however, and so the two values of the stored energy must be identical. Thus, M12 = M21 = M and W = 1 2 L1I 2 1 + 1 2 L2I 2 2 + M I 1I2 CHAPTER 13 MAGNETICALLY COUPLED CIRCUITS 502 i1 i2 L2 v2 L1 + – v1 + – M ■FIGURE 13.12 A pair of coupled coils with a mutual inductance of M12 M21 M. SECTION 13.2 ENERGY CONSIDERATIONS 503 If one current enters a dot-marked terminal while the other leaves a dot-marked terminal, the sign of the mutual energy term is reversed: W = 1 2 L1I 2 1 + 1 2 L2I 2 2 −MI1I2 Although Eqs. and were derived by treating the ﬁnal values of the two currents as constants, these “constants’’ can have any value, and the energy expressions correctly represent the energy stored when the instanta-neous values of i1 and i2 are I1 and I2, respectively. In other words, lower-case symbols might just as well be used: w(t) = 1 2 L1 [i1(t)]2 + 1 2 L2 [i2(t)]2 ± M [i1(t)] [i2(t)] The only assumption upon which Eq. is based is the logical establish-ment of a zero-energy reference level when both currents are zero. Establishing an Upper Limit for M Equation may now be used to establish an upper limit for the value of M. Since w(t) represents the energy stored within a passive network, it cannot be negative for any values of i1, i2, L1, L2, or M. Let us assume ﬁrst that i1 and i2 are either both positive or both negative; their product is therefore positive. From Eq. , the only case in which the energy could possibly be negative is w = 1 2 L1i2 1 + 1 2 L2i2 2 −Mi1i2 which we may write, by completing the square, as w = 1 2 √L1i1 −√L2i2 2 + √L1L2i1i2 −Mi1i2 Since in reality the energy cannot be negative, the right-hand side of this equation cannot be negative. The ﬁrst term, however, may be as small as zero, so we have the restriction that the sum of the last two terms cannot be negative. Hence, L1L2 ≥M or M ≤ L1L2 There is, therefore, an upper limit to the possible magnitude of the mutual inductance; it can be no larger than the geometric mean of the inductances of the two coils between which the mutual inductance exists. Although we have derived this inequality on the assumption that i1 and i2 carried the same algebraic sign, a similar development is possible if the signs are opposite; it is necessary only to select the positive sign in Eq. . We might also have demonstrated the truth of inequality from a physical consideration of the magnetic coupling; if we think of i2 as being zero and the current i1 as establishing the magnetic ﬂux linking both L1 and L2, it is apparent that the ﬂux within L2 cannot be greater than the ﬂux within L1, which represents the total ﬂux. Qualitatively, then, there is an upper limit to the magnitude of the mutual inductance possible between two given inductors. The Coupling Coefﬁcient The degree to which M approaches its maximum value is described by the coupling coefﬁcient, deﬁned as k = M √L1L2 Since M ≤√L1L2, 0 ≤k ≤1 The larger values of the coefﬁcient of coupling are obtained with coils which are physically closer, which are wound or oriented to provide a larger common magnetic ﬂux, or which are provided with a common path through a material which serves to concentrate and localize the magnetic flux (a high-permeability material). Coils having a coefﬁcient of coupling close to unity are said to be tightly coupled. CHAPTER 13 MAGNETICALLY COUPLED CIRCUITS 504 EXAMPLE 13.4 i1 i2 L2 v2 L1 + – v1 + – M ■FIGURE 13.13 Two coils with a coupling coefﬁcient of 0.6, L1 = 0.4 H and L2 = 2.5 H. 2.5 H 0.4 H 3 is x M y ■FIGURE 13.14 In Fig. 13.13, let L1 = 0.4 H, L2 = 2.5 H, k = 0.6, and i1 = 4i2 = 20 cos(500t −20◦) mA. Determine both v1(0) and the total energy stored in the system at t = 0. In order to determine the value of v1, we need to include the contribu-tions from both the self-inductance of coil 1 and the mutual inductance. Thus, paying attention to the dot convention, v1(t) = L1 di1 dt + M di2 dt To evaluate this quantity, we require a value for M. This is obtained from Eq. , M = k L1L2 = 0.6 (0.4)(2.5) = 0.6 H Thus, v1(0) = 0.4[−10 sin(−20◦)] + 0.6[−2.5 sin(−20◦)] = 1.881V. The total energy is found by summing the energy stored in each in-ductor, which has three separate components since the two coils are known to be magnetically coupled. Since both currents enter a “dotted’’ terminal, w(t) = 1 2 L1[i1(t)]2 + 1 2 L2[i2(t)]2 + M[i1(t)] [i2(t)] Since i1(0) = 20 cos (–20o) = 18.79 mA and i2(0) = 0.25i1(0) = 4.698 mA, we ﬁnd that the total energy stored in the two coils at t = 0 is 151.2 μJ. PRACTICE ● 13.4 Let is = 2 cos 10t A in the circuit of Fig. 13.14, and ﬁnd the total energy stored in the passive network at t = 0 if k = 0.6 and terminals x and y are (a) left open-circuited; (b) short-circuited. Ans: 0.8 J; 0.512 J. SECTION 13.3 THE LINEAR TRANSFORMER 505 13.3 • THE LINEAR TRANSFORMER We are now ready to apply our knowledge of magnetic coupling to the description of two speciﬁc practical devices, each of which may be repre-sented by a model containing mutual inductance. Both of the devices are transformers, a term which we deﬁne as a network containing two or more coils which are deliberately coupled magnetically (Fig. 13.15). In this sec-tion we consider the linear transformer, which happens to be an excellent model for devices used at radio frequencies, or higher frequencies. In Sec. 13.4 we will consider the ideal transformer, which is an idealized unity-coupled model of a physical transformer that has a core made of some magnetic material, usually an iron alloy. ■FIGURE 13.15 A selection of small transformers for use in electronic applications; the AA battery is shown for scale only. In Fig. 13.16 a transformer is shown with two mesh currents identiﬁed. The ﬁrst mesh, usually containing the source, is called the primary, while the second mesh, usually containing the load, is known as the secondary. The inductors labeled L1 and L2 are also referred to as the primary and secondary, respectively, of the transformer. We will assume that the transformer is linear. This implies that no magnetic material (which may cause a nonlin-ear ﬂux-versus-current relationship) is employed. Without such material, however, it is difﬁcult to achieve a coupling coefﬁcient greater than a few tenths. The two resistors serve to account for the resistance of the wire out of which the primary and secondary coils are wound, and any other losses. Reﬂected Impedance Consider the input impedance offered at the terminals of the primary circuit. The two mesh equations are Vs = (R1 + jωL1)I1 −jωMI2 M VL + – + – Vs I2 I1 R1 R2 L2 L1 ZL ■FIGURE 13.16 A linear transformer containing a source in the primary circuit and a load in the secondary circuit. Resistance is also included in both the primary and the secondary. and 0 = −jωMI1 + (R2 + jωL2 + ZL)I2 We may simplify by deﬁning Z11 = R1 + jωL1 and Z22 = R2 + jωL2 + ZL so that Vs = Z11I1 −jωMI2 0 = −jωMI1 + Z22I2 Solving the second equation for I2 and inserting the result in the ﬁrst equa-tion enable us to ﬁnd the input impedance, Zin = Vs I1 = Z11 −( jω)2M2 Z22 Before manipulating this expression any further, we can draw several exciting conclusions. In the ﬁrst place, this result is independent of the lo-cation of the dots on either winding, for if either dot is moved to the other end of the coil, the result is a change in sign of each term involving M in Eqs. to . This same effect could be obtained by replacing M by (−M), and such a change cannot affect the input impedance, as Eq. demonstrates. We also may note in Eq. that the input impedance is sim-ply Z11 if the coupling is reduced to zero. As the coupling is increased from zero, the input impedance differs from Z11 by an amount ω2M2/Z22, termed the reﬂected impedance. The nature of this change is more evident if we expand this expression Zin = Z11 + ω2M2 R22 + j X22 and rationalize the reﬂected impedance, Zin = Z11 + ω2M2R22 R2 22 + X2 22 + −jω2M2X22 R2 22 + X2 22 Since ω2M2R22/(R2 22 + X2 22) must be positive, it is evident that the presence of the secondary increases the losses in the primary circuit. In other words, the presence of the secondary might be accounted for in the primary circuit by increasing the value of R1. Moreover, the reactance which the secondary reﬂects into the primary circuit has a sign which is opposite to that of X22, the net reactance around the secondary loop. This reactance X22 is the sum of ωL2 and XL; it is necessarily positive for induc-tive loads and either positive or negative for capacitive loads, depending on the magnitude of the load reactance. CHAPTER 13 MAGNETICALLY COUPLED CIRCUITS 506 Zin is the impedance seen looking into the primary coil of the transformer. PRACTICE ● 13.5 Element values for a certain linear transformer are R1 = 3 , R2 = 6 , L1 = 2 mH, L2 = 10 mH, and M = 4 mH. If ω = 5000 rad/s, ﬁnd Zin for ZL equal to (a) 10 ; (b) j20 ; (c) 10 + j20 ; (d) −j20 . Ans: 5.32 + j2.74 ; 3.49 + j4.33 ; 4.24 + j4.57 ; 5.56 −j2.82 . SECTION 13.3 THE LINEAR TRANSFORMER 507 T and Equivalent Networks It is often convenient to replace a transformer with an equivalent network in the form of a T or . If we separate the primary and secondary resistances from the transformer, only the pair of mutually coupled inductors remains, as shown in Fig. 13.17. Note that the two lower terminals of the transformer are connected together to form a three-terminal network. We do this because both of our equivalent networks are also three-terminal networks. The differential equations describing this circuit are, once again, v1 = L1 di1 dt + M di2 dt and v2 = M di1 dt + L2 di2 dt The form of these two equations is familiar and may be easily inter-preted in terms of mesh analysis. Let us select a clockwise i1 and a counter-clockwise i2 so that i1 and i2 are exactly identiﬁable with the currents in Fig. 13.17. The terms M di2/dt in Eq. and M di1/dt in Eq. indi-cate that the two meshes must then have a common self-inductance M. Since the total inductance around the left-hand mesh is L1, a self-inductance of L1 −M must be inserted in the ﬁrst mesh, but not in the second mesh. Similarly, a self-inductance of L2 −M is required in the second mesh, but not in the first mesh. The resultant equivalent network is shown in Fig. 13.18. The equivalence is guaranteed by the identical pairs of equations relating v1, i1, v2, and i2 for the two networks. i1 i2 L2 v2 L1 + – v1 + – M ■FIGURE 13.17 A given transformer which is to be replaced by an equivalent or T network. i1 i2 v2 + – v1 + – M L1 – M L2 – M ■FIGURE 13.18 The T equivalent of the transformer shown in Fig. 13.17. If either of the dots on the windings of the given transformer is placed on the opposite end of its coil, the sign of the mutual terms in Eqs. and will be negative. This is analogous to replacing M with −M, and such a replacement in the network of Fig. 13.18 leads to the correct equivalent for this case. (The three self-inductance values would then be L1 + M, −M, and L2 + M.) The inductances in the T equivalent are all self-inductances; no mutual inductance is present. It is possible that negative values of inductance may be obtained for the equivalent circuit, but this is immaterial if our only de-sire is a mathematical analysis. There are times when procedures for syn-thesizing networks to provide a desired transfer function lead to circuits containing a T network having a negative inductance; this network may then be realized by use of an appropriate linear transformer. CHAPTER 13 MAGNETICALLY COUPLED CIRCUITS 508 i1 i2 (a) 40 mH 60 mH 30 mH A C B D i2 i1 (b) C D A B 40 mH 20 mH –10 mH ■FIGURE 13.19 (a) A linear transformer used as an example. (b) The T-equivalent network of the transformer. The equivalent network is not obtained as easily. It is more complicated, and it is not used as much. We develop it by solving Eq. for di2/dt and ■FIGURE 13.20 (a) C D A B Lz Ly Lx (b) 3.5 H 6 H 2 H A C B D Ans: −1.5, 2.5, 3.5 H; 5.5, 9.5, −3.5 H. EXAMPLE 13.5 Find the T equivalent of the linear transformer shown in Fig. 13.19a. We identify L1 = 30 mH, L2 = 60 mH, and M = 40 mH, and note that the dots are both at the upper terminals, as they are in the basic circuit of Fig. 13.17. Hence, L1 −M = −10 mH is in the upper left arm, L2 −M = 20 mH is at the upper right, and the center stem contains M = 40 mH. The complete equivalent T is shown in Fig. 13.19b. To demonstrate the equivalence, let us leave terminals C and D open-circuited and apply vAB = 10 cos 100t V to the input in Fig. 13.19a. Thus, i1 = 1 30 × 10−3 10 cos(100t) dt = 3.33 sin 100t A and vCD = M di1 dt = 40 × 10−3 × 3.33 × 100 cos 100t = 13.33 cos 100t V Applying the same voltage in the T equivalent, we ﬁnd that i1 = 1 (−10 + 40) × 10−3 10 cos(100t) dt = 3.33 sin 100t A once again. Also, the voltage at C and D is equal to the voltage across the 40 mH inductor. Thus, vCD = 40 × 10−3 × 3.33 × 100 cos 100t = 13.33 cos 100t V and the two networks yield equal results. PRACTICE ● 13.6 (a) If the two networks shown in Fig. 13.20 are equivalent, specify values for Lx, Ly, and Lz. (b) Repeat if the dot on the secondary in Fig. 13.20b is located at the bottom of the coil. SECTION 13.3 THE LINEAR TRANSFORMER 509 substituting the result in Eq. : v1 = L1 di1 dt + M L2 v2 −M2 L2 di1 dt or di1 dt = L2 L1L2 −M2 v1 − M L1L2 −M2 v2 If we now integrate from 0 to t, we obtain i1 −i1(0)u(t) = L2 L1L2 −M2 t 0 v1 dt′ − M L1L2 −M2 t 0 v2 dt′ In a similar fashion, we also have i2 −i2(0)u(t) = −M L1L2 −M2 t 0 v1 dt′ + L1 L1L2 −M2 t 0 v2 dt′ Equations and may be interpreted as a pair of nodal equations; a step-current source must be installed at each node in order to provide the proper initial conditions. The factors multiplying each integral have the general form of inverses of certain equivalent inductances. Thus, the second coefﬁcient in Eq. , M/(L1L2 −M2), is 1/L B, or the reciprocal of the inductance extending between nodes 1 and 2, as shown on the equivalent network, Fig. 13.21. So L B = L1L2 −M2 M i1 L1L2 – M2 L2 – M L1L2 – M2 L1 – M L1L2 – M2 M i1(0)u(t) i2(0)u(t) LA LC LB v1 + – v2 + – i2 ■FIGURE 13.21 The network which is equivalent to the transformer shown in Fig. 13.17. The ﬁrst coefﬁcient in Eq. , L2/(L1L2 −M2), is 1/L A + 1/L B. Thus, 1 L A = L2 L1L2 −M2 − M L1L2 −M2 or L A = L1L2 −M2 L2 −M Finally, LC = L1L2 −M2 L1 −M No magnetic coupling is present among the inductors in the equivalent , and the initial currents in the three self-inductances are zero. We may compensate for a reversal of either dot in the given transformer by merely changing the sign of M in the equivalent network. Also, just as we found in the equivalent T, negative self-inductances may appear in the equivalent network. CHAPTER 13 MAGNETICALLY COUPLED CIRCUITS 510 i2 i1 C D A B –20 mH 5 mH 10 mH ■FIGURE 13.22 The equivalent of the linear transformer shown in Fig. 13.19a. It is assumed that i 1(0) = 0 and i 2(0) = 0. (b) 3.4 H 6 H 2 H A C B D (a) i2 i1 C D A B LA LC LB ■FIGURE 13.23 COMPUTER-AIDED ANALYSIS The ability to simulate circuits that contain magnetically coupled inductances is a useful skill, especially with circuit dimensions continuing to decrease. As various loops and partial loops of conductors are brought closer in new designs, various circuits and subcircuits that are intended to be isolated from one another inadvertently become coupled through stray magnetic ﬁelds, and interact with one another. PSpice allows us to incorporate this effect through the component K_Linear, which links a pair of inductors in the schematic by a coupling coefﬁcient k in the range of 0 ≤k ≤1. For example, consider the circuit of Fig. 13.19a, which consists of two coils whose coupling is described by a mutual inductance of M = 40 mH, corresponding to a coupling coefﬁcient of k = 0.9428. The basic circuit schematic is shown in Fig. 13.24a. Note that when ﬁrst EXAMPLE 13.6 Find the equivalent network of the transformer in Fig. 13.19a, assuming zero initial currents. Since the term L1L2 −M2 is common to LA, LB, and LC, we begin by evaluating this quantity, obtaining 30 × 10−3 × 60 × 10−3 −(40 × 10−3)2 = 2 × 10−4 H2 Thus, L A = L1L2 −M2 L2 −M = 2 × 10−4 20 × 10−3 = 10 mH LC = L1L2 −M2 L1 −M = −20 mH and L B = L1L2 −M2 M = 5 mH The equivalent network is shown in Fig. 13.22. If we again check our result by letting vAB = 10 cos 100t V with terminals C-D open-circuited, the output voltage is quickly obtained by voltage division: vCD = −20 × 10−3 5 × 10−3 −20 × 10−3 10 cos 100t = 13.33 cos 100t V as before. Thus, the network in Fig. 13.22 is electrically equivalent to the networks in Fig. 13.19a and b. PRACTICE ● 13.7 If the networks in Fig. 13.23 are equivalent, specify values (in mH) for LA, LB, and LC. Ans: L A = 169.2 mH, L B = 129.4 mH, LC = −314.3 mH. SECTION 13.3 THE LINEAR TRANSFORMER 511 placed horizontally in the schematic, the dotted terminal is on the left, and this is the pin about which the symbol is rotated.Also note that the K_Linear component is not “wired’’into the schematic anywhere; its loca-tion is arbitrary. The two coupled inductors, L1 and L2, are speciﬁed along with the coupling coefﬁcient through the Property Editor (Fig 13.24b). The circuit is connected to a 100 rad/s (15.92 Hz) sinusoidal voltage source, a fact which is incorporated by performing a single-frequency ac sweep. It is also necessary to add two resistors to the schematic in order for PSpice to perform the simulation without generating an error message. First, a small series resistance has been inserted between the voltage source and L1; a value of 1 p was selected to minimize its effects. Second, a 1000 M resistor (essentially inﬁnite) was connected to L2. The output of the simulation is a voltage magnitude of 13.33 V ■FIGURE 13.24 (a) The circuit of Fig. 13.19a, modiﬁed to meet simulation requirements. (b) Property Editor dialog box showing how various inductors to be linked are named. (a) (b) CHAPTER 13 MAGNETICALLY COUPLED CIRCUITS 512 I2 I1 L2 L1 V1 + – V2 + – 1 : a k = 1 ZL ■FIGURE 13.25 An ideal transformer is connected to a general load impedance. and a phase angle of −3.819 × 10−8 degrees (essentially zero), in agreement with the values calculated by hand in Example 13.5. PSpice also provides two different transformer models, a linear transformer, XFRM_LINEAR, and an ideal transformer XFRM_NONLINEAR, a circuit element which is the subject of the following section. The linear transformer requires that values be speci-ﬁed for the coupling coefﬁcient and both coil inductances. The ideal transformer also requires a coupling coefﬁcient, but, as we shall see, an ideal transformer has inﬁnite or nearly inﬁnite inductance values. Thus, the remaining parameters required for the part XFRM_NONLINEAR are the number of turns of wire that form each coil. 13.4• THE IDEAL TRANSFORMER An ideal transformer is a useful approximation of a very tightly coupled trans-former in which the coupling coefﬁcient is essentially unity and both the pri-mary and secondary inductive reactances are extremely large in comparison with the terminating impedances. These characteristics are closely approached by most well-designed iron-core transformers over a reasonable range of fre-quencies for a reasonable range of terminal impedances. The approximate analysis of a circuit containing an iron-core transformer may be achieved very simply by replacing that transformer with an ideal transformer; the ideal trans-former may be thought of as a ﬁrst-order model of an iron-core transformer. Turns Ratio of an Ideal Transformer One new concept arises with the ideal transformer: the turns ratio a. The self-inductance of a coil is proportional to the square of the number of turns of wire forming the coil. This relationship is valid only if all the ﬂux estab-lished by the current ﬂowing in the coil links all the turns. In order to develop this result quantitatively it is necessary to utilize magnetic ﬁeld concepts, a subject that is not included in our discussion of circuit analysis. However, a qualitative argument may sufﬁce. If a current i ﬂows through a coil of N turns, then N times the magnetic ﬂux of a single-turn coil will be produced. If we think of the N turns as being coincident, then all the ﬂux certainly links all the turns. As the current and ﬂux change with time, a volt-age is then induced in each turn which is N times larger than that caused by a single-turn coil. Thus, the voltage induced in the N-turn coil must be N 2 times the single-turn voltage. From this, the proportionality between induc-tance and the square of the numbers of turns arises. It follows that L2 L1 = N 2 2 N 2 1 = a2 or Figure 13.25 shows an ideal transformer to which a secondary load is connected. The ideal nature of the transformer is established by several a = N2 N1 SECTION 13.4 THE IDEAL TRANSFORMER 513 conventions: the use of the vertical lines between the two coils to indicate the iron laminations present in many iron-core transformers, the unity value of the coupling coefﬁcient, and the presence of the symbol 1:a, suggesting a turns ratio of N1 to N2. Let us analyze this transformer in the sinusoidal steady state. The two mesh equations are V1 = jωL1I1 −jωMI2 and 0 = −jωMI1 + (ZL + jωL2)I2 First, consider the input impedance of an ideal transformer. By solving Eq. for I2 and substituting in Eq. , we obtain V1 = I1 jωL1 + I1 ω2M2 ZL + jωL2 and Zin = V1 I1 = jωL1 + ω2M2 ZL + jωL2 Since k = 1, M2 = L1L2 so Zin = jωL1 + ω2L1L2 ZL + jωL2 Besides a unity coupling coefﬁcient, another characteristic of an ideal transformer is an extremely large impedance for both the primary and sec-ondary coils, regardless of the operating frequency. This suggests that the ideal case would be for both L1 and L2 to tend to inﬁnity. Their ratio, how-ever, must remain ﬁnite, as speciﬁed by the turns ratio. Thus, L2 = a2L1 leads to Zin = jωL1 + ω2a2L2 1 ZL + jωa2L1 Now if we let L1 become inﬁnite, both of the terms on the right-hand side of the preceding equation become inﬁnite, and the result is indeterminate. Thus, it is necessary to ﬁrst combine these two terms: Zin = jωL1ZL −ω2a2L2 1 + ω2a2L2 1 ZL + jωa2L1 or Zin = jωL1ZL ZL + jωa2L1 = ZL ZL/jωL1 + a2 Now as L1 →∞, we see that Zin becomes Zin = ZL a2 for ﬁnite ZL. This result has some interesting implications, and at least one of them appears to contradict one of the characteristics of the linear transformer. The input impedance of an ideal transformer is proportional to the load impedance, the proportionality constant being the reciprocal of the square of the turns ratio. In other words, if the load impedance is a capacitive imped-ance, then the input impedance is a capacitive impedance. In the linear trans-former, however, the reﬂected impedance suffered a sign change in its reac-tive part; a capacitive load led to an inductive contribution to the input impedance. The explanation of this occurrence is achieved by ﬁrst realiz-ing that ZL/a2 is not the reﬂected impedance, although it is often loosely called by that name. The true reﬂected impedance is inﬁnite in the ideal transformer; otherwise it could not “cancel’’ the inﬁnite impedance of the primary inductance. This cancellation occurs in the numerator of Eq. . The impedance ZL/a2 represents a small term which is the amount by which an exact cancellation does not occur. The true reﬂected impedance in the ideal transformer does change sign in its reactive part; as the primary and secondary inductances become inﬁnite, however, the effect of the inﬁnite primary-coil reactance and the inﬁnite, but negative, reﬂected reactance of the secondary coil is one of cancellation. The ﬁrst important characteristic of the ideal transformer is therefore its ability to change the magnitude of an impedance, or to change impedance level. An ideal transformer having 100 primary turns and 10,000 secondary turns has a turns ratio of 10,000/100, or 100. Any impedance placed across the secondary then appears at the primary terminals reduced in magnitude by a factor of 1002, or 10,000. A 20,000 resistor looks like 2 , a 200 mH inductor looks like 20 μH, and a 100 pF capacitor looks like 1 μF. If the pri-mary and secondary windings are interchanged, then a = 0.01 and the load impedance is apparently increased in magnitude. In practice, this exact change in magnitude does not always occur, for we must remember that as we took the last step in our derivation and allowed L1 to become inﬁnite in Eq. , it was necessary to neglect ZL in comparison with jωL2. Since L2 can never be inﬁnite, it is evident that the ideal transformer model will be-come invalid if the load impedances are very large. Use of Transformers for Impedance Matching A practical example of the use of an iron-core transformer as a device for changing impedance level is in the coupling of an ampliﬁer to a speaker sys-tem. In order to achieve maximum power transfer, we know that the resis-tance of the load should be equal to the internal resistance of the source; the speaker usually has an impedance magnitude (often assumed to be a resis-tance) of only a few ohms, while an ampliﬁer may possess an internal resistance of several thousand ohms. Thus, an ideal transformer is required in which N2 < N1. For example, if the ampliﬁer internal impedance is 4000 and the speaker impedance is 8 , then we desire that Zg = 4000 = ZL a2 = 8 a2 or a = 1 22.4 and thus N1 N2 = 22.4 CHAPTER 13 MAGNETICALLY COUPLED CIRCUITS 514 SECTION 13.4 THE IDEAL TRANSFORMER 515 Use of Transformers for Current Adjustment There is a simple relationship between the primary and secondary currents I1 and I2 in an ideal transformer. From Eq. , I2 I1 = jωM ZL + jωL2 Once again we allow L2 to become inﬁnite, and it follows that I2 I1 = jωM jωL2 = L1 L2 or Thus, the ratio of the primary and secondary currents is the turns ratio. If we have N2 > N1, then a > 1, and it is apparent that the larger current ﬂows in the winding with the smaller number of turns. In other words, N1I1 = N2I2 It should also be noted that the current ratio is the negative of the turns ratio if either current is reversed or if either dot location is changed. In our example in which an ideal transformer was used to change the im-pedance level to efﬁciently match a speaker to an ampliﬁer, an rms current of 50 mA at 1000 Hz in the primary causes an rms current of 1.12 A at 1000 Hz in the secondary. The power delivered to the speaker is (1.12)2(8), or 10 W, and the power delivered to the transformer by the power amplifier is (0.05)24000, or 10 W. The result is comforting, since the ideal transformer contains neither an active device which can generate power nor any resistor which can absorb power. Use of Transformers for Voltage Level Adjustment Since the power delivered to the ideal transformer is identical with that de-livered to the load, whereas the primary and secondary currents are related by the turns ratio, it should seem reasonable that the primary and secondary voltages must also be related to the turns ratio. If we deﬁne the secondary voltage, or load voltage, as V2 = I2ZL and the primary voltage as the voltage across L1, then V1 = I1Zin = I1 ZL a2 The ratio of the two voltages then becomes V2 V1 = a2 I2 I1 I2 I1 = 1 a or The ratio of the secondary to the primary voltage is equal to the turns ra-tio. We should take care to note that this equation is opposite that of Eq. , and this is a common source of error for students. This ratio may also be neg-ative if either voltage is reversed or either dot location is changed. Simply by choosing the turns ratio, therefore, we now have the ability to change any ac voltage to any other ac voltage. If a > 1, the secondary volt-age will be greater than the primary voltage, and we have what is commonly referred to as a step-up transformer. If a < 1, the secondary voltage will be less than the primary voltage, and we have a step-down transformer. Utility companies typically generate power at a voltage in the range of 12 to 25 kV. Although this is a rather large voltage, transmission losses over long distances can be reduced by increasing the level to several hundred thousand volts using a step-up transformer (Fig. 13.26a). This voltage is then reduced to several tens of kilovolts at substations for local power distribution using step-down transformers (Fig. 13.26b). Additional step-down transformers are located outside buildings to reduce the voltage from the transmission voltage to the 110 or 220 V level required to operate machinery (Fig. 13.26c). Combining the voltage and current ratios, Eqs. and , V2I2 = V1I1 and we see that the primary and secondary complex voltamperes are equal. The magnitude of this product is usually specified as a maximum allowable value on power transformers. If the load has a phase angle θ, or ZL = \|ZL\|/θ then V2 leads I2 by an angle θ. Moreover, the input impedance is ZL/a2, and thus V1 also leads I1 by the same angle θ. If we let the voltage and cur-rent represent rms values, then \|V2\| \|I2\| cos θ must equal \|V1\| \|I1\| cos θ, and all the power delivered to the primary terminals reaches the load; none is absorbed by or delivered to the ideal transformer. The characteristics of the ideal transformer that we have obtained have all been determined by phasor analysis. They are certainly true in the sinusoidal steady state, but we have no reason to believe that they are cor-rect for the complete response. Actually, they are applicable in general, and the demonstration that this statement is true is much simpler than the phasor-based analysis we have just completed. Our analysis, however, has served to point out the speciﬁc approximations that must be made on a more exact model of an actual transformer in order to obtain an ideal transformer. For example, we have seen that the reactance of the secondary winding must be much greater in magnitude than the impedance of any load that is connected to the secondary. Some feeling for those operating conditions under which a transformer ceases to behave as an ideal transformer is thus achieved. V2 V1 = a = N2 N1 CHAPTER 13 MAGNETICALLY COUPLED CIRCUITS 516 (a) (b) (c) ■FIGURE 13.26 (a) A step-up transformer used to increase the generator output voltage for transmission. (b) Substation transformer used to reduce the voltage from the 220 kV transmission level to several tens of kilovolts for local distribution. (c) Step-down transformer used to reduce the distribution voltage level to 240 V for power consumption. (Photos courtesy of Dr. Wade Enright, Te Kura Pukaha Vira O Te Whare Wananga O Waitaha, Aotearoa.) SECTION 13.4 THE IDEAL TRANSFORMER 517 EXAMPLE 13.7 For the circuit given in Fig. 13.27, determine the average power dissipated in the 10 k resistor. V2 + – V1 + – 1 : 10 I1 I2 + – 100 10 k 50 V rms ■FIGURE 13.27 A simple ideal transformer circuit. The average power dissipated by the 10 k resistor is simply P = 10,000\|I2\|2 The 50 V rms source “sees’’ a transformer input impedance of ZL/a2 or 100 . Thus, we obtain I1 = 50 100 + 100 = 250 mA rms From Eq. , I2 = (1/a)I1 = 25 mA rms, so we ﬁnd that the 10 k resistor dissipates 6.25 W. PRACTICE ● 13.8 Repeat Example 13.7 using voltages to compute the dissipated power. Ans: 6.25 W. Voltage Relationship in the Time Domain Let us now determine how the time-domain quantities v1 and v2 are related in the ideal transformer. Returning to the circuit shown in Fig. 13.17 and the two equations, and , describing it, we may solve the second equa-tion for di2/dt and substitute in the ﬁrst equation: v1 = L1 di1 dt + M L2 v2 −M2 L2 di1 dt However, for unity coupling, M2 = L1L2, and so v1 = M L2 v2 = L1 L2 v2 = 1 a v2 The relationship between primary and secondary voltage therefore does apply to the complete time-domain response. The phase angles can be ignored in this example as they do not impact the calculation of average power dissipated by a purely resistive load. PRACTICAL APPLICATION Superconducting Transformers For the most part, we have neglected the various types of losses that may be present in a particular transformer. When dealing with large power transformers, however, close attention must be paid to such nonidealities, de-spite overall efﬁciencies of typically 97 percent or more. Although such a high efﬁciency may seem nearly ideal, it can represent a great deal of wasted energy when the transformer is handling several thousand amperes. So-called i2R (pronounced “eye-squared-R”) losses rep-resent power dissipated as heat, which can increase the temperature of the transformer coils. Wire resistance increases with temperature, so heating only leads to greater losses. High temperatures can also lead to degradation of the wire insulation, resulting in shorter transformer life. As a result, many modern power trans-formers employ a liquid oil bath to remove excess heat from the transformer coils. Such an approach has its drawbacks, however, including environmental impact and ﬁre danger from leaking oil as a result of corrosion over time (Fig. 13.28). One possible means of improving the performance of such transformers is to make use of superconducting wire to replace the resistive coils of a standard trans-former design. Superconductors are materials that are resistive at high temperature, but suddenly show no re-sistance to the ﬂow of current below a critical tempera-ture. Most elements are superconducting only near ■FIGURE 13.28 Fire that broke out in 2004 at the 340,000 V American Electric Power Substation near Mishawaka, Indiana. (© AP/Wide World Photos) An expression relating primary and secondary current in the time domain is most quickly obtained by dividing Eq. throughout by L1, v1 L1 = di1 dt + M L1 di2 dt = di1 dt + a di2 dt and then invoking one of the hypotheses underlying the ideal transformer: L1 must be inﬁnite. If we assume that v1 is not inﬁnite, then di1 dt = −a di2 dt Integrating, i1 = −ai2 + A where A is a constant of integration that does not vary with time. Thus, if we neglect any direct currents in the two windings and ﬁx our attention only on the initial cost is much higher than for a traditional resis-tive transformer. At present, many companies (including utilities) are driven by short-term cost considerations and are not always eager to make large capital invest-ments with only long-term cost beneﬁts. absolute zero, requiring expensive liquid helium–based cryogenic cooling. With the discovery in the 1980s of ce-ramic superconductors having critical temperatures of 90 K (−183°C) and higher, it became possible to replace helium–based equipment with significantly cheaper liquid nitrogen systems. Figure 13.29 shows a prototype partial-core super-conducting transformer being developed at the Univer-sity of Canterbury. This design employs environmentally benign liquid nitrogen in place of an oil bath, and is also signiﬁcantly smaller than a comparably rated conven-tional transformer. The result is a measurable improve-ment in overall transformer efﬁciency, which translates into operational cost savings for the owner. Still, all designs have disadvantages that must be weighed against their potential advantages, and super-conducting transformers are no exception. The most sig-niﬁcant obstacle at present is the relatively high cost of fabricating superconducting wire several kilometers in length compared to copper wire. Part of this is due to the challenge of fabricating long wires from ceramic materi-als, but part of it is also due to the silver tubing used to surround the superconductor to provide a low-resistance current path in the event of a cooling system failure (al-though less expensive than silver, copper reacts with the ceramic and is therefore not a viable alternative). The net result is that although a superconducting transformer is likely to save a utility money over a long period of time—many transformers see over 30 years of service— ■FIGURE 13.29 Prototype 15 kVA partial core superconducting power transformer. (Photo courtesy of Department of Electrical and Computer Engineering, University of Canterbury.) the time-varying portion of the response, then i1 = −ai2 The minus sign arises from the placement of the dots and selection of the current directions in Fig. 13.17. The same current and voltage relationships are therefore obtained in the time domain as were obtained previously in the frequency domain, pro-vided that dc components are ignored. The time-domain results are more general, but they have been obtained by a less informative process. Equivalent Circuits The characteristics of the ideal transformer which we have established may be utilized to simplify circuits in which ideal transformers appear. Let us as-sume, for purposes of illustration, that everything to the left of the primary terminals has been replaced by its Thévenin equivalent, as has the network to the right of the secondary terminals. We thus consider the circuit shown in Fig. 13.30. Excitation at any frequency ω is assumed. CHAPTER 13 MAGNETICALLY COUPLED CIRCUITS 520 V2 + – V1 + – I2 I1 Vs2 Vs1 1 : a k = 1 + – + – Zg1 Zg2 ■FIGURE 13.30 The networks connected to the primary and secondary terminals of an ideal transformer are represented by their Thévenin equivalents. V2 + – I2 Vs2 aVs1 + – + – a2Zg1 Zg2 ■FIGURE 13.31 The Thévenin equivalent of the network to the left of the secondary terminals in Fig. 13.30 is used to simplify that circuit. Each primary voltage may therefore be multiplied by the turns ratio, each primary current divided by the turns ratio, and each primary imped-ance multiplied by the square of the turns ratio; and then these modiﬁed voltages, currents, and impedances replace the given voltages, currents, and impedances plus the transformer. If either dot is interchanged, the equiva-lent may be obtained by using the negative of the turns ratio. Thévenin’s or Norton’s theorem may be used to achieve an equivalent circuit that does not contain a transformer. For example, let us determine the Thévenin equivalent of the network to the left of the secondary terminals. Open-circuiting the secondary, I2 = 0 and therefore I1 = 0 (remember that L1 is infinite). No voltage appears across Zg1, and thus V1 = Vs1 and V2oc = aVs1. The Thévenin impedance is obtained by setting Vs1 to zero and utilizing the square of the turns ratio, being careful to use the reciprocal turns ratio, since we are looking in at the secondary terminals. Thus, Zth2 = Zg1a2. As a check on our equivalent, let us also determine the short-circuit sec-ondary current I2sc. With the secondary short-circuited, the primary genera-tor faces an impedance of Zg1, and, thus, I1 = Vs1/Zg1. Therefore, I2sc = Vs1/aZg1. The ratio of the open-circuit voltage to the short-circuit current is a2Zg1, as it should be. The Thévenin equivalent of the transformer and primary circuit is shown in the circuit of Fig. 13.31. SECTION 13.4 THE IDEAL TRANSFORMER 521 Note that this equivalence, as illustrated by Fig. 13.31, is possible only if the network connected to the two primary terminals, and that connected to the two secondary terminals, can be replaced by their Thévenin equiva-lents. That is, each must be a two-terminal network. For example, if we cut the two primary leads at the transformer, the circuit must be divided into two separate networks; there can be no element or network bridging across the transformer between primary and secondary. A similar analysis of the transformer and the secondary network shows that everything to the right of the primary terminals may be replaced by an identical network without the transformer, each voltage being divided by a, each current being multiplied by a, and each impedance being divided by a2. A reversal of either winding requires the use of a turns ratio of −a. EXAMPLE 13.8 For the circuit given in Fig. 13.32, determine the equivalent circuit in which the transformer and the secondary circuit are replaced, and also that in which the transformer and the primary circuit are replaced. V2 + – V1 + – 1 : 10 I1 I2 + – 100 10 k 50 V rms ■FIGURE 13.32 A simple circuit in which a resistive load is matched to the source impedance by means of an ideal transformer. This is the same circuit we analyzed in Example 13.7. As before, the input impedance is 10,000/(10)2, or 100 and so \|I1\| = 250 mA rms. We can also compute the voltage across the primary coil \|V1\| = \|50 −100I1\| = 25 V rms and thus ﬁnd that the source delivers (25 × 10−3)(50) = 12.5 W, of which (25 × 10−3)2(100) = 6.25 W is dissipated in the internal resis-tance of the source and 12.5 −6.25 = 6.25 W is delivered to the load. This is the condition for maximum power transfer to the load. If the secondary circuit and the ideal transformer are removed by the use of the Thévenin equivalent, the 50 V source and 100 resistor sim-ply see a 100 impedance, and the simpliﬁed circuit of Fig. 13.33a is obtained. The primary current and voltage are now immediately evident. If, instead, the network to the left of the secondary terminals is replaced by its Thévenin equivalent, we ﬁnd (keeping in mind the location of the dots) Vth = −10(50) = −500 V rms, and Zth = (−10)2(100) = 10 k; the resulting circuit is shown in Fig. 13.33b. I1 + – 100 100 50 V rms V1 + – (a) I2 + – 10 k –500 V rms V2 + – (b) 10 k ■FIGURE 13.33 The circuit of Fig. 13.32 is simpliﬁed by replacing (a) the transformer and secondary circuit by the Thévenin equivalent or (b) the transformer and primary circuit by the Thévenin equivalent. CHAPTER 13 MAGNETICALLY COUPLED CIRCUITS 522 PRACTICE ● 13.9 Let N1 = 1000 turns and N2 = 5000 turns in the ideal transformer shown in Fig. 13.34. If ZL = 500 −j400 , ﬁnd the average power delivered to ZL for (a) I2 = 1.4/20◦A rms; (b) V2 = 900/40◦V rms; (c) V1 = 80/100◦V rms; (d) I1 = 6/45◦A rms; (e) Vs = 200/0◦V rms. V2 + – V1 + – N1 : N2 + – 10 Vs I1 I2 ZL ■FIGURE 13.34 SUMMARY AND REVIEW Transformers play a critical role in the power industry, allowing voltages to be stepped up for transmission, and stepped down to the level required for individual pieces of equipment. In this chapter, we studied transformers in the broader context of magnetically coupled circuits, where the magnetic ﬂux associated with current can link two or more elements in a circuit (or even neighboring circuits). This is most easily understood by extending the concept of inductance studied in Chap. 7 to introduce the idea of mutual inductance (also having units of henrys). We saw that the coefﬁcient M of mutual inductance is limited to less than the geometric mean of the two inductances being coupled (i.e., M ≤√L1L2), and made use of the dot convention to determine the polarity of the voltage induced across one inductance as a result of current ﬂowing through the other. When the two inductances are not particularly close, M might be rather small. However, in the case of a well-designed transformer, it might approach its maximum value. To describe such situations, we introduced the concept of the cou-pling coefﬁcient k. When dealing with a linear transformer, analysis may be assisted by representing the element with an equivalent T (or, less com-monly, ) network, but a great deal of circuit analysis is performed assum-ing an ideal transformer. In such instances we no longer concern ourselves with M or k, but rather the turns ratio a. We saw that the voltages across the primary and secondary coils, as well as their individual currents, are related by this parameter. This approximation is very useful for both analysis and design. We concluded the chapter with a brief discussion of how Thévenin’s theorem can be applied to circuits with ideal transformers. We could continue, as the study of inductively coupled circuits is an in-teresting and important topic, but at this point it might be appropriate to list some of the key concepts we have already discussed, along with corre-sponding example numbers. ❑Mutual inductance describes the voltage induced at the ends of a coil due to the magnetic ﬁeld generated by a second coil. (Example 13.1) Ans: 980 W; 988 W; 195.1 W; 720 W; 692 W. EXERCISES 523 ❑The dot convention allows a sign to be assigned to the mutual inductance term. (Example 13.1) ❑According to the dot convention, a current entering the dotted terminal of one coil produces an open-circuit voltage with a positive voltage reference at the dotted terminal of the second coil. (Examples 13.1, 13.2, 13.3) ❑The total energy stored in a pair of coupled coils has three separate terms: the energy stored in each self-inductance ( 1 2 Li2), and the energy stored in the mutual inductance (Mi1i2). (Example 13.4) ❑The coupling coefﬁcient is given by k = M/√L1L2, and is restricted to values between 0 and 1. (Example 13.4) ❑A linear transformer consists of two coupled coils: the primary winding and the secondary winding. (Examples 13.5, 13.6) ❑An ideal transformer is a useful approximation for practical iron-core transformers. The coupling coefﬁcient is taken to be unity, and the in-ductance values are assumed to be inﬁnite. (Examples 13.7, 13.8) ❑The turns ratio a = N2/N1 of an ideal transformer relates the primary and secondary coil voltages: V2 = aV1. (Example 13.8) ❑The turns ratio a also relates the currents in the primary and secondary coils: I1 = aI2. (Examples 13.7, 13.8) READING FURTHER Almost everything you ever wanted to know about transformers can be found in: M. Heathcote, J&P Transformer Book, 12th ed. Oxford: Reed Educational and Professional Publishing Ltd., 1998. Another comprehensive transformer title is: W. T. McLyman, Transformer and Inductor Design Handbook, 3rd ed. New York: Marcel Dekker, 2004. A good transformer book with a strong economic focus is: B. K. Kennedy, Energy Efﬁcient Transformers. New York: McGraw-Hill, 1998. EXERCISES 13.1 Mutual Inductance 1. Consider the two inductances depicted in Fig. 13.35. Set L1 = 10 mH, L2 = 5 mH, and M = 1 mH. Determine the steady-state expression for (a) v1 if i1 = 0 and i2 = 5 cos 8t A; (b) v2 if i1 = 3 sin 100t A and i2 = 0; (c) v2 if i1 = 5 cos (8t – 40◦) A and i2 = 4 sin 8t A. i1 i2 L2 L1 M v2 + – v1 + – ■FIGURE 13.35 2. With respect to Fig. 13.36, assume L1 = 400 mH, L2 = 230 mH, and M = 10 mH. Determine the steady-state expression for (a) v1 if i1 = 0 and i2 = 2 cos 40t A; (b) v2 if i1 = 5 cos (40t 15◦) A and i2 = 0. (c) Repeat parts (a) and (b) if M is increased to 300 mH. 3. In Fig. 13.37, set L1 = 1 μH, L2 = 2 μH, and M = 150 nH. Obtain a steady-state expression for (a) v1 if i2 = –cos 70t mA and i1 = 0; (b) v2 if i1 = 55 cos (5t – 30◦) A; (c) v2 if i1 = 6 sin 5t A and i2 = 3 sin 5t. CHAPTER 13 MAGNETICALLY COUPLED CIRCUITS 524 i1 i2 L2 L1 M v2 + – v1 + – ■FIGURE 13.36 v2 + – M L1 L2 i1 ■FIGURE 13.40 i1 i2 L2 L1 M v2 + – v1 – + ■FIGURE 13.37 i1 i2 L2 L1 M v2 + – v1 – + ■FIGURE 13.38 1 2 3 4 (a) 1 2 3 4 (b) 2 1 3 4 (c) ■FIGURE 13.39 4. For the conﬁguration of Fig. 13.38, L1 = 0.5L2 = 1 mH and M = 0.85√L1L2. Calculate v2(t) if (a) i2 = 0 and i1 = 5e–t mA; (b) i2 = 0 and i1 = 5 cos 10t mA; (c) i2 = 5 cos 70t mA and i1 = 0.5i2. 5. The physical construction of three pairs of coupled coils is shown in Fig. 13.39. Show the two different possible locations for the two dots on each pair of coils. 6. In the circuit of Fig. 13.40, i1 = 5 sin (100t – 80◦) mA, L1 = 1 H, and L2 = 2 H. If v2 = 250 sin (100t – 80◦) mV, calculate M. 7. In the circuit represented in Fig. 13.40, determine i1 if v2(t) = 4 cos 5t V, L1 = 1 mH, L2 = 4 mH, and M = 1.5 mH. 8. Calculate v1 and v2 if i1 = 5 sin 40t mA and i2 = 5 cos 40t mA, L1 = 1 mH, L2 = 3 mH, and M = 0.5 mH, for the coupled inductances shown in (a) Fig. 13.37; (b) Fig. 13.38. EXERCISES 525 9. Calculate v1 and v2 if i1 = 3 cos (2000t + 13◦) mA and i2 = 5 sin 400t mA, L1 = 1 mH, L2 = 3 mH, and M = 200 nH, for the coupled inductances shown in (a) Fig. 13.35; (b) Fig. 13.36. 10. For the circuit of Fig. 13.41, calculate I1, I2, V2V1, and I2I1. I1 I2 4.7 k 500 j750 j2 k 870 j1.8 k 0° V V 1 = 40 + – V 2 + – ■FIGURE 13.41 I1 I2 1 1 j2 j4 1 j6 0° V V 1 = 40 + – V 2 + – ■FIGURE 13.42 2 2 mH 1 mH v1 i2 i3 i1 M = 500 nH + – 1.8 1 mF ■FIGURE 13.43 1 H + – 5 10 3 H 5 H 2 cos 10t V ■FIGURE 13.44 15 mH 3 mH 8 (speaker) M + – 15 cos 200t V ■FIGURE 13.45 11. For the circuit of Fig. 13.42, plot the magnitude of V2V1 as a function of frequency ω, over the range 0 ≤ω ≤2 rad/s. 12. For the circuit of Fig. 13.43, (a) draw the phasor representation; (b) write a complete set of mesh equations; (c) calculate i2(t) if v1(t) = 8 sin 720t V. 13. In the circuit of Fig. 13.43, M is reduced by an order of magnitude. Calculate i3 if v1 = 10 cos (800t 20◦) V. 14. In the circuit shown in Fig. 13.44, ﬁnd the average power absorbed by (a) the source; (b) each of the two resistors; (c) each of the two inductances; (d) the mutual inductance. 15. The circuit of Fig. 13.45 is designed to drive a simple 8 speaker. What value of M results in 1 W of average power being delivered to the speaker? 16. Consider the circuit of Fig. 13.46. The two sources are is1 = 2 cos t mA and is2 = 1.5 sin t mA. If M1 = 2 H, M2 = 0 H, and M3 = 10 H, calculate vAG(t). CHAPTER 13 MAGNETICALLY COUPLED CIRCUITS 526 A B C G 3 H 5 H M3 M2 20 H is1 is2 M1 ■FIGURE 13.46 I3 I1 3 H 2 sin 3t V I2 + – 5 4 H 12 2 10 H ■FIGURE 13.47 + – vs + – vx + – iC 100vx 1 F 15 mH 10 mH 40 mH ■FIGURE 13.48 V1 + – V2 + – IA IB M L1 L2 vA + – L2 L1 (a) M i2 vB + – i1 (b) ■FIGURE 13.49 20. For the coupled inductor network of Fig. 13.49a, set L1 = 20 mH, L2 = 30 mH, M = 10 mH, and obtain equations for vA and vB if (a) i1 = 0 and i2 = 5 sin 10t; (b) i1 = 5 cos 20t and i2 = 2 cos (20t – 100◦) mA. (c) Express V1 and V2 as functions of IA and IB for the network shown in Fig. 13.49b. 19. Determine an expression for iC(t) valid for t > 0 in the circuit of Fig. 13.48, if vs(t) = 10t2u(t)/(t2 + 0.01) V. 17. For the circuit of Fig. 13.46, M1 = 1 H, M2 = 1.5 H, and M3 = 2 H. If is1 = 8 cos 2t A and is2 = 7 sin 2t A, calculate (a) VAB; (b) VAG; (c) VCG. 18. For the circuit of Fig. 13.47, ﬁnd the currents i1(t), i2(t), and i3(t) if f = 60 Hz. EXERCISES 527 22. Find V1( jω) and V2( jω) in terms of I1( jω) and I2( jω) for each circuit of Fig. 13.51. I1 100 0 V 4 5 6 H 0 H 3 H 4 H 5 H 2 H 6 + – I3 I2 ■FIGURE 13.50 V2 + – V 1 + – (a) L1 L2 R1 R2 M I1 I2 V2 – + V 1 + – (b) L1 L2 R1 R2 M I1 I2 ■FIGURE 13.51 Zin 5 0.2 H 0.5 H 0.1 H 2 ■FIGURE 13.52 i1 i2 L2 v2 L1 + – v1 + – M ■FIGURE 13.53 + – k + – v1 v2 ■FIGURE 13.54 23. (a) Find Zin( jω) for the network of Fig. 13.52. (b) Plot Zin over the frequency range of 0 ≤ω ≤1000 rad/s. (c) Find Zin( jω) for ω = 50 rad/s. 13.2 Energy Considerations 24. For the coupled coils of Fig. 13.53, L1 = L2 = 10 H, and M is equal to its maximum possible value. (a) Compute the coupling coefﬁcient k. (b) Calculate the energy stored in the magnetic ﬁeld linking the two coils at t = 200 ms if i1 = 10 cos 4t mA and i2 = 2 cos 4t mA. 25. With regard to the coupled inductors shown in Fig. 13.53, L1 = 10 mH, L2 = 5 mH, and k = 0.75. (a) Compute M. (b) If i1 = 100 sin 40t mA, and i2 = 0, compute the energy stored in each coil and in the magnetic ﬁeld coupling the two inductors at t = 2 ms. (c) Repeat part (b) if i2 is set to 75 cos 40t mA. 26. For the circuit of Fig. 13.54, L1 = 2 mH, L2 = 8 mH, and v1 = cos 8t V. (a) Obtain an equation for v2(t). (b) Plot V2 as a function of k. (c) Plot the phase angle (in degrees) of V2 as a function of k. 21. Note that there is no mutual coupling between the 5 H and 6 H inductors in the circuit of Fig. 13.50. (a) Write a set of equations in terms of I1( jω), I2( jω), and I3( jω). (b) Find I3( jω) if ω = 2 rad/s. 13.3 The Linear Transformer 30. Assume the following values for the circuit depicted schematically in Fig. 13.16: R1 = 10 , R2 = 1 , L1 = 2 μH, L2 = 1 μH, and M = 500 nH. Calculate the input impedance for ω = 10 rad/s if ZL is equal to (a) 1 ; (b) j ; (c) –j ; (d) 5/33◦. 31. Determine the T equivalent of the linear transformer represented in Fig. 13.57 (draw and label an appropriate diagram). 27. Connect a load ZL = 5/33◦ to the right-hand terminals of Fig. 13.53. Derive an expression for the input impedance at f = 100 Hz, seen looking into the left-hand terminals, if L1 = 1.5 mH, L2 = 3 mH, and k = 0.55. 28. Consider the circuit represented in Fig. 13.55. The coupling coefﬁcient k = 0.75. If is = 5 cos 200t mA, calculate the total energy stored at t = 0 and t = 5 ms if (a) a-b is open-circuited (as shown); (b) a-b is short-circuited. CHAPTER 13 MAGNETICALLY COUPLED CIRCUITS 528 is 1.2 mH 12 mH 100 m M 3 mH a b ■FIGURE 13.55 1 H 1 H 1 H 1 H 4 cos 5t V 1 2 k2 = 0.82 k1 = 0.64 + – – + v1 – + v2 ■FIGURE 13.56 i1 i2 25 nH 13 nH 130 nH A C B D ■FIGURE 13.57 i1 i2 1 H 2 H 25 H A C B D ■FIGURE 13.58 29. Compute v1, v2, and the average power delivered to each resistor in the circuit of Fig. 13.56. 32. (a) Draw and label an appropriate diagram of a T equivalent network for the linear transformer shown in Fig. 13.58. (b) Verify the two are equivalent by connecting a voltage vAB = 5 sin 45t V and calculating the open-circuit voltage vCD. EXERCISES 529 i2 i1 C D A B Lz Ly Lx ■FIGURE 13.59 i2 i1 C D A B LA LC LB ■FIGURE 13.60 2 1.5 2 H 8 H 4 H vs + – iL v1 + – ■FIGURE 13.61 k = 1 k = 0.9 50 mH 5 mH 25 mH 10 mH 10 vo + – 2 + – vs ■FIGURE 13.62 33. Represent the T network shown in Fig. 13.59 as an equivalent linear trans-former if (a) Lx = 1 H, Ly = 2 H, and Lz = 4 H; (b) Lx = 10 mH, Ly = 50 mH, and Lz = 22 mH. 38. (a) For the circuit of Fig. 13.62, if vs = 8 cos 1000t V, calculate vo. (b) Verify your solution with an appropriate PSpice simulation. 37. For the circuit of Fig. 13.61, determine an expression for (a) ILVs; (b) V1Vs. 34. Assuming zero initial currents, obtain an equivalent network of the trans-former depicted in Fig. 13.57. 35. (a) Draw and label a suitable equivalent network of the linear transformer shown in Fig. 13.58, assuming zero initial currents. (b) Verify their equivalence with an appropriate simulation. 36. Represent the network of Fig. 13.60 as an equivalent linear transformer with zero initial currents if (a) LA = 1 H, LB = 2 H, and LC = 4 H; (b) LA = 10 mH, LB = 50 mH, and LC = 22 mH. 39. With respect to the network shown in Fig. 13.63, derive an expression for Z( jω) if M1 and M2 are set to their respective maximum values. 13.4 The Ideal Transformer 40. Calculate I2 and V2 for the ideal transformer circuit of Fig. 13.64 if (a) V1 = 4/32◦V and ZL = 1 – j ; (b) V1 = 4/32◦V and ZL = 0; (c) V1 = 2/118◦V and ZL = 1.5/10◦. CHAPTER 13 MAGNETICALLY COUPLED CIRCUITS 530 41. With respect to the ideal transformer circuit depicted in Fig. 13.64, calculate I2 and V2 if (a) I1 = 244/0◦mA and ZL = 5 – j2 ; (b) I1 = 100/10◦mA and ZL = j2 . 42. Calculate the average power delivered to the 400 m and 21 resistors, respectively, in the circuit of Fig. 13.65. 5 H M1 M2 500 mH 1 H 250 mH 1 3 H Z( j) ■FIGURE 13.63 I2 I1 V1 + – V2 + – 1 : 6 ZL ■FIGURE 13.64 v2 + – v1 + – 1 : 100 i1 i2 + – 3 400 m 21 2 cos 280t V ■FIGURE 13.65 50 38 1.5 1:4 1:9 5 cos 120t A 9 ■FIGURE 13.66 2 4 2.7 k 2:15 5:1 25 cos 120t mA + – v1 v2 + – 100 ■FIGURE 13.67 43. With regard to the ideal transformer circuit represented in Fig. 13.65, deter-mine an equivalent circuit in which (a) the transformer and primary circuit are replaced, so that V2 and I2 are unchanged; (b) the transformer and secondary circuit are replaced, so that V1 and I1 are unchanged. 44. Calculate the average power delivered to each resistor shown in Fig. 13.66. 45. With respect to the circuit depicted in Fig. 13.67, calculate (a) the voltages v1 and v2; (b) the average power delivered to each resistor. EXERCISES 531 + – 4 3 10 140 1:5 188 30 V V2 + – Ix ■FIGURE 13.68 v2 + – 1 4 vs 30:1 + – a b ■FIGURE 13.69 2 cos 10t V 1:1 a:b + – 1.85 R1 R2 iL RL vx + – ■FIGURE 13.70 46. Calculate Ix and V2 as labeled in Fig. 13.68. 47. The ideal transformer of the circuit in Fig. 13.68 is removed, ﬂipped across its vertical axis, and reconnected such that the same terminals remain connected to the negative terminal of the source. (a) Calculate Ix and V2. (b) Repeat part (a) if both dots are placed at the bottom terminals of the transformer. 48. For the circuit of Fig. 13.69, vs = 117 sin 500t V. Calculate v2 if the terminals marked a and b are (a) left open-circuited; (b) short-circuited; (c) bridged by a 2 resistor. 49. The turns ratio of the ideal transformer in Fig. 13.69 is changed from 30:1 to 1:3. Take vs = 720 cos 120πt V, and calculate v2 if terminals a and b are (a) short-circuited; (b) bridged by a 10 resistor; (c) bridged by a 1 M resistor. 50. For the circuit of Fig. 13.70, R1 = 1 , R2 = 4 , and RL = 1 . Select a and b to achieve a peak voltage of 200 V magnitude across RL. 51. Calculate vx for the circuit of Fig. 13.70 if a = 0.01b = 1, R1 = 300 , R2 = 14 , and RL = 1 k. 52. (a) Referring to the ideal transformer circuit in Fig. 13.70, determine the load current iL if b = 0.25a = 1, R1 = 2.2 , R2 = 3.1 , and RL = 200 . (b) Verify your solution with an appropriate PSpice simulation. 53. Determine the Thévenin equivalent of the network in Fig. 13.71 as seen look-ing into terminals a and b. 54. Calculate V2 and the average power delivered to the 8 resistor of Fig. 13.72 if Vs = 10/15◦V, and the control parameter c is equal to (a) 0; (b) 1 mS. 3 1 3:2 10Ix + – Ix a b ■FIGURE 13.71 2 8 Vs a:b + – V2 + – cV2 ■FIGURE 13.72 CHAPTER 13 MAGNETICALLY COUPLED CIRCUITS 532 5 Vs + – V2 + – 40 L2 L1 ■FIGURE 13.73 55. (a) For the circuit of Fig. 13.72, take c = –2.5 mS and select values of a and b such that 100 W average power is delivered to the 8 load when Vs = 5/−35◦V, (b) Verify your solution with an appropriate PSpice simulation. Chapter-Integrating Exercises 56. A transformer whose nameplate reads operates with primary and secondary voltages of 2300 V and 230 V rms, respectively, and can supply 25 kVA from its secondary winding. If this transformer is supplied with 2300 V rms and is connected to secondary loads requiring 8 kW at unity PF and 15 kVA at 0.8 PF lagging, (a) what is the primary current? (b) How many kilowatts can the transformer still supply to a load operating at 0.95 PF lagging? (c) Verify your answers with PSpice. 57. A friend brings a vintage stereo system back from a recent trip to Warnemünde, unaware that it was designed to operate on twice the supply voltage (240 VAC) available at American household outlets. Design a circuit to allow your friend to listen to the stereo in the United States, assuming the operating frequency (50 Hz in Germany, 60 Hz in the United States) difference can be neglected. 58. The friend referred to in Exercise 57 attempts to justify the erroneous assump-tion made regarding the stereo by pointing out that the wall outlet in the W.C. (bathroom) had a socket for his U.S. electric razor, clearly marked 120 VAC. He failed to notice that the small sign below the outlet clearly stated “Razors only.” With the knowledge that all power lines running into the room operated at 240 VAC, draw the likely circuit built into the bathroom wall outlet, and explain why it is limited to “razors only.” 59. Obtain an expression for V2Vs in the circuit of Fig. 13.73 if (a) L1 = 100 mH, L2 = 500 mH, and M is its maximum possible value; (b) L1 = 5L2 = 1.4 H and k = 87% of its maximum possible value; (c) the two coils can be treated as an ideal transformer, the left-hand coil having 500 turns and the right-hand coil having 10,000 turns. 2300/230 V, 25 kVA 60. You notice your neighbor has installed a large coil of wire in close proximity to the power line coming into your house (underground cables are not available in your neighborhood). (a) What is the likely intention of your neighbor? (b) Is the plan likely to succeed? Explain. (c) When confronted, your neighbor sim-ply shrugs and claims there’s no way it can cost you anything, anyway, since nothing of his is touching anything on your property. True or not? Explain. INTRODUCTION When faced with time-varying sources, or a circuit with switches installed, we have several choices with respect to the analysis approach. Chapters 7 through 9 detail direct differential equation– based analysis, which is particularly useful when examining turn-on or turn-off transients. In contrast, Chaps. 10 to 13 describe analysis situations where sinusoidal excitation is assumed, and transients are of little or no interest. Unfortunately, not all sources are sinu-soidal, and there are times when both transient and steady-state responses are required. In such instances, the Laplace transform proves to be a highly valuable tool. Many textbooks simply launch straight into the Laplace trans-form integral, but this approach conveys no intuitive understand-ing. For this reason, we have chosen to ﬁrst introduce what may strike the reader at ﬁrst as a somewhat odd concept—the notion of a “complex” frequency. Simply a mathematical convenience, com-plex frequency allows us to manipulate both periodic and nonperi-odic time-varying quantities in parallel, greatly simplifying the analysis. After getting a feel for the basic technique, we develop it as a speciﬁc circuit analysis tool in Chap. 15. 14.1 • COMPLEX FREQUENCY We introduce the notion of complex frequency by considering a (purely real) exponentially damped sinusoidal function, such as the voltage v(t) = Vmeσt cos(ωt + θ) KEY CONCEPTS Complex Frequency Laplace Transform Inverse Transform Use of Transform Tables Method of Residuals Using MATLAB to Manipulate Polynomials Using MATLAB to Determine Residues of Rational Fractions Initial-Value Theorem Final-Value Theorem Complex Frequency and the Laplace Transform C H A P T E R 14 533 where σ (sigma) is a real quantity and is usually negative. Although we often refer to this function as being “damped,’’ it is conceivable that we might occasionally encounter a situation where σ > 0 and hence the sinusoidal amplitude is increasing. (In Chap. 9, our study of the natural response of the RLC circuit also indicates that σ is the negative of the exponential damping coefﬁcient.) Note that we may construct a constant voltage from Eq. by letting σ = ω = 0: v(t) = Vm cos θ = V0 If we set only σ equal to zero, then we obtain a general sinusoidal voltage v(t) = Vm cos(ωt + θ) and if ω = 0, we have the exponential voltage v(t) = Vm cos θ eσt = V0eσt Thus, the damped sinusoid of Eq. includes as special cases the dc Eq. , sinusoidal Eq. , and exponential Eq. functions. Some additional insight into the signiﬁcance of σ can be obtained by comparing the exponential function of Eq. with the complex representa-tion of a sinusoidal function with a zero-degree phase angle, v(t) = V0e jωt It is apparent that the two functions, Eqs. and , have much in common. The only difference is that the exponent in Eq. is real and the one in Eq. is imaginary. The similarity between the two functions is emphasized by describing σ as a “frequency.’’ This choice of terminology will be discussed in detail in the following sections, but for now we need merely note that σ is specifically termed the real part of the complex frequency. It should not be called the “real frequency,’’ however, for this is a term that is more suitable for f (or, loosely, for ω). We will also refer to σ as the neper frequency, the name arising from the dimensionless unit of the exponent of e. Thus, given e7t, the dimensions of 7t are nepers (Np), and 7 is the neper frequency in nepers per second. The General Form The forced response of a network to a general forcing function of the form of Eq. can be found very simply by using a method almost identical with that used in phasor-based analysis. Once we are able to ﬁnd the forced response to this damped sinusoid, we will also have found the forced response to a dc voltage, an exponential voltage, and a sinusoidal voltage. First we consider σ and ω as the real and imaginary parts of a complex frequency. We suggest that any function that may be written in the form f (t) = Kest where K and s are complex constants (independent of time) is characterized by the complex frequency s. The complex frequency s is therefore simply CHAPTER 14 COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM 534 The neper itself was named after the Scottish philoso-pher and mathematician John Napier (1550–1617) and his napierian logarithm system; the spelling of his name is historically uncertain (see, for example, H. A. Wheeler, IRE Transactions on Circuit Theory 2, 1955, p. 219). SECTION 14.1 COMPLEX FREQUENCY 535 the factor that multiplies t in this complex exponential representation. Until we are able to determine the complex frequency of a given function by inspection, it is necessary to write the function in the form of Eq. . The DC Case We apply this deﬁnition ﬁrst to the more familiar forcing functions. For example, a constant voltage v(t) = V0 may be written in the form v(t) = V0e(0)t Therefore, we conclude that the complex frequency of a dc voltage or cur-rent is zero (i.e., s = 0). The Exponential Case The next simple case is the exponential function v(t) = V0eσt which is already in the required form. The complex frequency of this volt-age is therefore σ (i.e., s = σ + j0). The Sinusoidal Case Now let us consider a sinusoidal voltage, one that may provide a slight surprise. Given v(t) = Vm cos(ωt + θ) we desire to ﬁnd an equivalent expression in terms of the complex expo-nential. From our past experience, we therefore use the formula we derived from Euler’s identity, cos(ωt + θ) = 1 2[e j(ωt+θ) + e−j(ωt+θ)] and obtain v(t) = 1 2Vm[e j(ωt+θ) + e−j(ωt+θ)] = 1 2Vme jθ e jωt + 1 2Vme−jθ e−jωt or v(t) = K1es1t + K2es2t We have the sum of two complex exponentials, and two complex frequencies are therefore present, one for each term. The complex frequency of the ﬁrst term is s = s1 = jω, and that of the second term is s = s2 = −jω. These two values of s are conjugates, or s2 = s∗ 1 and the two values of K are also conjugates: K1 = 1 2Vme jθ and K2 = K∗ 1 = 1 2Vme−jθ. The entire ﬁrst term and the entire second term are therefore conjugates, which we might have expected inasmuch as their sum must be a real quantity, v(t). The complex conjugate of any number can be obtained by simply replacing all occurrences of ‘’j’’ with ‘’j.’’ The concept arises from our arbitrary choice of j = + −1. However, the negative root is just as valid, which leads us to the deﬁnition of a complex conjugate. The Exponentially Damped Sinusoidal Case Finally, let us determine the complex frequency or frequencies associated with the exponentially damped sinusoidal function, Eq. . We again use Euler’s formula to obtain a complex exponential representation: v(t) = Vmeσt cos(ωt + θ) = 1 2Vmeσt[e j(ωt+θ) + e−j(ωt+θ)] and thus v(t) = 1 2Vme jθe j(σ+ jω)t + 1 2Vme−jθe j(σ−jω)t We find that a conjugate complex pair of frequencies, s1 = σ + jω and s2 = s∗ 1 = σ −jω, is also required to describe the exponentially damped sinusoid. In general, neither σ nor ω is zero, and the exponentially varying sinusoidal waveform is the general case; the constant, sinusoidal, and exponential waveforms are special cases. The Relationship of s to Reality A positive real value of s, e.g., s = 5 + j0, identiﬁes an exponentially in-creasing function Ke+5t, where K must be real if the function is to be a physical one. A negative real value for s, such as s = −5 + j0, refers to an exponentially decreasing function Ke−5t. A purely imaginary value of s, such as j10, can never be associated with a purely real quantity. The functional form is Ke j10t, which can also be writ-ten as K(cos 10t + j sin 10t); it obviously possesses both a real and an imag-inary part, each of which is sinusoidal. In order to construct a real function, it is necessary to consider conjugate values of s, such as s1,2 = ± j10, with which must be associated conjugate values of K. Loosely speaking, however, we may identify either of the complex frequencies s1 = + j10 or s2 = −j10 with a sinusoidal voltage at the radian frequency of 10 rad/s; the presence of the conjugate complex frequency is understood.The amplitude and phase an-gle of the sinusoidal voltage will depend on the choice of K for each of the two frequencies. Thus, selecting s1 = j10 and K1 = 6 −j8, where v(t) = K1es1t + K2es2t s2 = s∗ 1 and K2 = K∗ 1 we obtain the real sinusoid 20 cos(10t −53.1◦). In a similar manner, a general value for s, such as 3 −j5, can be associ-ated with a real quantity only if it is accompanied by its conjugate, 3 + j5. Speaking loosely again, we may think of either of these two conjugate fre-quencies as describing an exponentially increasing sinusoidal function, e3t cos 5t; the speciﬁc amplitude and phase angle will again depend on the values of the conjugate complex K’s. By now we should have achieved some appreciation of the physical na-ture of the complex frequency s; in general, it describes an exponentially varying sinusoid. The real part of s is associated with the exponential varia-tion; if it is negative, the function decays as t increases; if it is positive, the function increases; and if it is zero, the sinusoidal amplitude is constant. The larger the magnitude of the real part of s, the greater is the rate of exponen-tial increase or decrease. The imaginary part of s describes the sinusoidal variation; it is speciﬁcally the radian frequency. A large magnitude for the imaginary part of s indicates a more rapidly changing function of time. CHAPTER 14 COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM 536 Note that \|6 −j8\| = 10, so that V m = 2\|K\| = 20. Also, ang(6 −j8) = −53.13◦. Large magnitudes for the real part of s, the imaginary part of s, or the magnitude of s indicate a rapidly varying function. SECTION 14.2 THE DAMPED SINUSOIDAL FORCING FUNCTION 537 It is customary to use the letter σ to designate the real part of s, and ω (not jω) to designate the imaginary part: s = σ + jω The radian frequency is sometimes referred to as the “real frequency,’’ but this terminology can be very confusing when we ﬁnd that we must then say that “the real frequency is the imaginary part of the complex frequency’’! When we need to be speciﬁc, we will call s the complex frequency, σ the neper frequency, ω the radian frequency, and f = ω/2π the cyclic fre-quency; when no confusion seems likely, it is permissible to use “frequency’’ to refer to any of these four quantities. The neper frequency is measured in nepers per second, radian frequency is measured in radians per second, and complex frequency s is measured in units which are variously termed complex nepers per second or complex radians per second. 14.2 • THE DAMPED SINUSOIDAL FORCING FUNCTION It is time to put this concept of complex frequency to work. The general exponentially varying sinusoid, which we may represent with the voltage function v(t) = Vmeσt cos(ωt + θ) can be expressed in terms of the complex frequency s by making use of Euler’s identity as before: v(t) = Re{Vmeσte j(ωt+θ)} or v(t) = Re{Vmeσte j(−ωt−θ)} Either representation is suitable, and the two expressions should remind us that a pair of conjugate complex frequencies is associated with a sinusoid or an exponentially damped sinusoid. Equation is more directly related to the given damped sinusoid, and we will concern ourselves principally with it. PRACTICE ● 14.1 Identify all the complex frequencies present in these real functions: (a) (2e−100t + e−200t) sin 2000t; (b) (2 −e−10t) cos(4t + φ); (c) e−10t cos 10t sin 40t. 14.2 Use real constants A, B, C, φ, and so forth, to construct the general form of the real function of time for a current having components at these frequencies: (a) 0, 10, −10 s−1; (b) −5, j8, −5 −j8 s−1; (c) −20, 20, −20 + j20, 20 −j20 s−1. Ans: 14.1: −100 + j2000, −100 −j2000, −200 + j2000, −200 −j2000 s−1; j4, −j4, −10 + j4, −10 −j4 s−1; −10 + j30, −10 −j30, −10 + j50, −10 −j50 s−1; 14.2: A + Be10t + Ce−10t ; Ae−5t + B cos(8t + φ1) + Ce−5t × cos(8t + φ2); Ae−20t + Be20t + Ce−20t cos(20t + φ1) + De20t cos(20t + φ2). Collecting factors, we now substitute s = σ + jω into v(t) = Re{Vme jθe(σ+ jω)t} and obtain v(t) = Re{Vme jθest} Before we apply a forcing function of this form to any circuit, we note the resemblance of this last representation of the damped sinusoid to the corre-sponding representation of the undamped sinusoid in Chap. 10, Re{Vme jθe jωt} The only difference is that we now have s where we previously had jω. In-stead of restricting ourselves to sinusoidal forcing functions and their radian frequencies, we have now extended our notation to include the damped sinusoidal forcing function at a complex frequency. It should be no surprise at all to see later in this section that we will develop a frequency-domain description of the exponentially damped sinusoid in exactly the same way that we did for the sinusoid; we will simply omit the Re{ } nota-tion and suppress est. We are now ready to apply the exponentially damped sinusoid, as given by Eq. , , , or , to an electrical network, where the forced response—perhaps a current in some branch of the network—is desired. Since the forced response has the form of the forcing function, its integral, and its derivatives, the response may be assumed to be i(t) = Imeσt cos(ωt + φ) or i(t) = Re{Ime jφest} where the complex frequency of both the source and the response must be identical. If we now recall that the real part of a complex forcing function pro-duces the real part of the response while the imaginary part of the complex forcing function causes the imaginary part of the response, then we are again led to the application of a complex forcing function to our network. We will obtain a complex response whose real part is the desired real re-sponse. Actually, we will work with the Re{ } notation omitted, but we should realize that it may be reinserted at any time and that it must be rein-serted whenever we desire the time-domain response. Thus, given the real forcing function v(t) = Re{Vme jθest} we apply the complex forcing function Vme jθest; the resultant forced re-sponse Ime jφest is complex, and it must have as its real part the desired time-domain forced response i(t) = Re{Ime jφest} The solution of our circuit analysis problem consists of the determination of the unknown response amplitude Im and phase angle φ. CHAPTER 14 COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM 538 SECTION 14.2 THE DAMPED SINUSOIDAL FORCING FUNCTION 539 Before we actually carry out the details of an analysis problem and see how the procedure resembles what we used in sinusoidal analysis, it is worthwhile to outline the steps of the basic method. 1. We ﬁrst characterize the circuit with a set of loop or nodal integrodifferential equations. 2. The given forcing functions, in complex form, and the assumed forced responses, also in complex form, are substituted in the equations and the indicated integrations and differentiations are performed. 3. Each term in every equation will then contain the same factor est. We divide throughout by this factor, or “suppress est,’’ understanding that it must be reinserted if a time-domain description of any response function is desired. With the Re{ } notation and the est factor gone, we have converted all the voltages and currents from the time domain to the frequency domain. The integrodifferential equations become algebraic equations, and their solution is obtained just as easily as in the sinusoidal steady state. Let us illustrate the basic method by a numerical example. EXAMPLE 14.1 Apply the forcing function v(t) 60e−2t cos(4t + 10°) V to the series RLC circuit shown in Fig. 14.1, and specify the forced response by ﬁnding values for Im and φ in the time-domain expression i(t) Im e−2t cos(4t + φ). We ﬁrst express the forcing function in Re{ } notation: v(t) = 60e−2t cos(4t + 10◦) = Re{60e−2te j(4t+10◦)} = Re{60e j10◦e(−2+ j4)t} or v(t) = Re{Vest} where V = 60/10◦ and s = −2 + j4 After dropping Re{ }, we are left with the complex forcing function 60/10◦est Similarly, we represent the unknown response by the complex quantity Iest, where I = Im φ. Our next step must be the integrodifferential equation for the circuit. From Kirchhoff’s voltage law, we obtain v(t) = Ri + L di dt + 1 C i dt = 2i + 3 di dt + 10 i dt ■FIGURE 14.1 A series RLC circuit to which a damped sinusoidal forcing function is applied. A frequency-domain solution for i(t) is desired. i(t) v (t) + – 2 3 H 0.1 F (Continued on next page) If the notation here seems unfamiliar, the reader may wish to pause and read Appendix 5, particularly section 4, which deals with the polar form of complex number representation. 14.3 • DEFINITION OF THE LAPLACE TRANSFORM Our constant goal has been one of analysis: given some forcing function at one point in a linear circuit, determine the response at some other point. For the first several chapters, we played only with dc forcing functions and responses of the form V0e0. However, after the introduction of inductance and capacitance, the sudden dc excitation of simple RL and RC circuits produced responses varying exponentially with time: V0eσt. When we considered the RLC circuit, the responses took on the form of the exponen-tiallyvaryingsinusoid, V0eσt cos(ωt + θ).Allthisworkwasaccomplishedin the time domain, and the dc forcing function was the only one we considered. As we advanced to the use of the sinusoidal forcing function, the tedium and complexity of solving the integrodifferential equations caused us to CHAPTER 14 COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM 540 and we substitute the given complex forcing function and the assumed complex forced response in this equation: 60/10◦est = 2Iest + 3sIest + 10 s Iest The common factor est is next suppressed: 60/10◦= 2I + 3sI + 10 s I and thus I = 60/10◦ 2 + 3s + 10/s We now let s = −2 + j4 and solve for the complex current I: I = 60/10◦ 2 + 3(−2 + j4) + 10/(−2 + j4) After manipulating the complex numbers, we ﬁnd I = 5.37/−106.6◦ Thus, Im is 5.37 A, φ is −106.6◦, and the forced response can be writ-ten directly (recalling that s = −2 + j4) as i(t) = 5.37e−2t cos(4t −106.6◦) A We have thus solved the problem by reducing a calculus-based expression to an algebraic expression. This is only a small indication of the power of the technique we are about to study. PRACTICE ● 14.3 Give the phasor current that is equivalent to the time-domain current: (a) 24 sin(90t + 60◦) A; (b) 24e−10t cos(90t + 60◦) A; (c) 24e−10t cos 60◦× cos 90t A. If V = 12/35◦V, ﬁnd v(t) for s equal to (d) 0; (e) −20 s−1; ( f ) −20 + j5 s−1. Ans: 24/−30◦A; 24/60◦A; 12/0◦A; 9.83 V; 9.83e−20t V; 12e−20t cos(5t + 35◦) V. SECTION 14.3 DEFINITION OF THE LAPLACE TRANSFORM 541 begin casting about for an easier way to work problems. The phasor trans-form was the result, and we might remember that we were led to it through consideration of a complex forcing function of the form V0e jθe jωt. As soon as we concluded that we did not need the factor containing t, we were left with the phasor V0e jθ; we had arrived at the frequency domain. Now a little ﬂexing of our cerebral cortex has caused us to apply a forcing function of the form V0e jθe(σ+ jω)t, leading to the invention of the complex frequency s, and thereby relegating all our previous functional forms to special cases: dc (s = 0), exponential (s = σ), sinusoidal (s = jω), and ex-ponential sinusoid (s = σ + jω). By analogy to our previous experience with phasors, we saw that in these cases we may omit the factor containing t, and once again obtain a solution by working in the frequency domain. The Two-Sided Laplace Transform We know that sinusoidal forcing functions lead to sinusoidal responses, and also that exponential forcing functions lead to exponential responses. However, as practicing engineers we will encounter many waveforms that are neither sinusoidal nor exponential, such as square waves, sawtooth waveforms, and pulses beginning at arbitrary instants of time. When such forcing functions are applied to a linear circuit, we will see that the response is neither similar to the form of the excitation waveform nor exponential. As a result, we are not able to eliminate the terms containing t to form a frequency-domain response. This is rather unfortunate, as working in the frequency domain has proved to be rather useful. There is a solution, however, which makes use of a technique that allows us to expand any function into a sum of exponential waveforms, each with its own complex frequency. Since we are considering linear circuits, we know that the total response of our circuit can be obtained by simply adding the individual response to each exponential waveform. And, in dealing with each exponential waveform, we may once again neglect any terms contain-ing t, and work instead in the frequency domain. It unfortunately takes an inﬁnite number of exponential terms to accurately represent a general time function, so that taking a brute-force approach and applying superposition to the exponential series might be somewhat insane. Instead, we will sum these terms by performing an integration, leading to a frequency-domain function. We formalize this approach using what is known as a Laplace transform, deﬁned for a general function f(t) as F(s) = ∞ −∞ e−st f(t) dt The mathematical derivation of this integral operation requires an under-standing of Fourier series and the Fourier transform, which are discussed in Chap. 18. The fundamental concept behind the Laplace transform, how-ever, can be understood based on our discussion of complex frequency and our prior experience with phasors and converting back and forth between the time domain and the frequency domain. In fact, that is precisely what the Laplace transform does: it converts the general time-domain function f(t) into a corresponding frequency-domain representation, F(s). The Two-Sided Inverse Laplace Transform Equation deﬁnes the two-sided, or bilateral, Laplace transform of f(t). The term two-sided or bilateral is used to emphasize the fact that both pos-itive and negative values of t are included in the range of integration. The inverse operation, often referred to as the inverse Laplace transform, is also deﬁned as an integral expression1 f (t) = 1 2π j σ0+ j∞ σ0−j∞ estF(s) ds where the real constant σ0 is included in the limits to ensure convergence of this improper integral; the two equations and constitute the two-sided Laplace transform pair. The good news is that Eq. need never be invoked in the study of circuit analysis: there is a quick and easy alternative to look forward to learning. The One-Sided Laplace Transform In many of our circuit analysis problems, the forcing and response functions do not exist forever in time, but rather they are initiated at some speciﬁc instant that we usually select as t = 0. Thus, for time functions that do not exist for t < 0, or for those time functions whose behavior for t < 0 is of no interest,thetime-domaindescriptioncanbethoughtofasv(t)u(t).Thedeﬁn-ing integral for the Laplace transform is taken with the lower limit at t = 0− in order to include the effect of any discontinuity at t = 0, such as an impulse or a higher-order singularity. The corresponding Laplace transform is then F(s) = ∞ −∞ e−st f (t)u(t) dt = ∞ 0−e−st f (t) dt This deﬁnes the one-sided Laplace transform of f (t), or simply the Laplace transform of f (t), one-sided being understood. The inverse transform ex-pression remains unchanged, but when evaluated, it is understood to be valid only for t > 0. Here then is the deﬁnition of the Laplace transform pair that we will use from now on: The script may also be used to indicate the direct or inverse Laplace transform operation: F(s) = { f (t)} and f (t) = −1{F(s)} f (t) = 1 2π j σ0+ j∞ σ0−j∞ estF(s) ds f (t) ⇔F(s) F(s) = ∞ 0−e−st f (t) dt CHAPTER 14 COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM 542 (1) If we ignore the distracting factor of 1/2π j and view the integral as a summation over all frequencies such that f (t) ∝[F(s) ds]est , this reinforces the notion that f (t) is indeed a sum of complex frequency terms having a magnitude proportional to F(s). SECTION 14.4 LAPLACE TRANSFORMS OF SIMPLE TIME FUNCTIONS 543 EXAMPLE 14.2 14.4 • LAPLACE TRANSFORMS OF SIMPLE TIME FUNCTIONS In this section we will begin to build up a catalog of Laplace transforms for those time functions most frequently encountered in circuit analysis; we will assume for now that the function of interest is a voltage, although such a choice is strictly arbitrary. We will create this catalog, at least initially, by utilizing the deﬁnition, V(s) = ∞ 0−e−stv(t) dt = {v(t)} which, along with the expression for the inverse transform, v(t) = 1 2π j σ0+ j∞ σ0−j∞ estV(s) ds = −1{V(s)} establishes a one-to-one correspondence between v(t) and V(s). That is, for every v(t) for which V(s) exists, there is a unique V(s). At this point, we may be looking with some trepidation at the rather ominous form given for the inverse transform. Fear not! As we will see shortly, an introductory study of Laplace transform theory does not require actual evaluation of this integral. By going from the time domain to the frequency domain and tak-ing advantage of the uniqueness just mentioned, we will be able to generate a catalog of transform pairs that will already contain the corresponding time function for nearly every transform that we wish to invert. PRACTICE ● 14.4 Let f (t) = −6e−2t[u(t + 3) −u(t −2)]. Find the (a) two-sided F(s); (b) one-sided F(s). Ans: 6 2+s[e−4−2s −e6+3s]; 6 2+s[e−4−2s −1]. Compute the Laplace transform of the function f(t) = 2u(t −3). In order to ﬁnd the one-sided Laplace transform of f (t) = 2u(t −3), we must evaluate the integral F(s) = ∞ 0−e−st f(t) dt = ∞ 0−e−st2u(t −3) dt = 2 ∞ 3 e−st dt Simplifying, we ﬁnd F(s) = −2 s e−st ∞ 3 = −2 s (0 −e−3s) = 2 s e−3s Before we continue, however, we should pause to consider whether there is any chance that the transform may not even exist for some v(t) that concerns us. A set of conditions sufﬁcient to ensure the absolute conver-gence of the Laplace integral for Re{s} > σ0 is 1. The function v(t) is integrable in every ﬁnite interval t1 < t < t2, where 0 ≤t1 < t2 < ∞. 2. lim t→∞e−σ0t\|v(t)\| exists for some value of σ0. Time functions that do not satisfy these conditions are seldom encountered by the circuit analyst.2 The Unit-Step Function u(t) Now let us look at some speciﬁc transforms. We ﬁrst examine the Laplace transform of the unit-step function u(t). From the deﬁning equation, we may write {u(t)} = ∞ 0−e−stu(t) dt = ∞ 0 e−st dt = −1 s e−st ∞ 0 = 1 s for Re{s} > 0, to satisfy condition 2. Thus, u(t) ⇔1 s and our ﬁrst Laplace transform pair has been established with great ease. The Unit-Impulse Function δ(t −t0) A singularity function whose transform is of considerable interest is the unit-impulse function δ(t −t0). This function, plotted in Fig. 14.2, seems rather strange at ﬁrst but is enormously useful in practice. The unit-impulse function is deﬁned to have an area of unity, so that δ(t −t0) = 0 t ̸= t0 t0+ε t0−ε δ(t −t0) dt = 1 where ε is a small constant. Thus, this “function’’ (a naming that makes many purist mathematicians cringe) has a nonzero value only at the point t0. For t0 > 0−, we therefore ﬁnd the Laplace transform to be {δ(t −t0)} = ∞ 0−e−stδ(t −t0) dt = e−st0 δ(t −t0) ⇔e−st0 CHAPTER 14 COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM 544 (2) Examples of such functions are et2 and eet, but not tn or nt. For a somewhat more detailed discussion of the Laplace transform and its applications, refer to Clare D. McGillem and George R. Cooper, Continuous and Discrete Signal and System Analysis, 3d ed. Oxford University Press, North Carolina: 1991, Chap. 5. The double arrow notation is commonly used to indicate Laplace transform pairs. ■FIGURE 14.2 The unit-impulse function δ(t t0). This function is often used to approximate a signal pulse whose duration is very short compared to circuit time constants. t0 t SECTION 14.4 LAPLACE TRANSFORMS OF SIMPLE TIME FUNCTIONS 545 In particular, note that we obtain δ(t) ⇔1 for t0 = 0. An interesting feature of the unit-impulse function is known as the sifting property. Consider the integral of the impulse function multiplied by an arbitrary function f (t): ∞ −∞ f (t)δ(t −t0) dt Since the function δ(t −t0) is zero everywhere except at t = t0, the value of this integral is simply f (t0). The property turns out to be very useful in sim-plifying integral expressions containing the unit-impulse function. The Exponential Function e−αt Recalling our past interest in the exponential function, we examine its transform, {e−αtu(t)} = ∞ 0−e−αte−st dt = − 1 s + α e−(s+α)t ∞ 0 = 1 s + α and therefore, e−αtu(t) ⇐ ⇒ 1 s + α It is understood that Re{s} > −α. The Ramp Function tu(t) As a ﬁnal example, for the moment, let us consider the ramp function tu(t). We obtain {tu(t)} = ∞ 0−te−st dt = 1 s2 tu(t) ⇔1 s2 either by a straightforward integration by parts or from a table of integrals. So what of the function te−αtu(t)? We leave it to the reader to show that te−αtu(t) ⇔ 1 (s + α)2 There are, of course, quite a few additional time functions worth con-sidering, but it may be best if we pause for the moment to consider the reverse of the process—the inverse Laplace transform—before returning to add to our list. 14.5 • INVERSE TRANSFORM TECHNIQUES The Linearity Theorem Although we mentioned that Eq. can be applied to convert an s-domain expression into a time-domain expression, we also alluded to the fact that this is more work than required—if we’re willing to exploit the uniqueness of any Laplace transform pair. In order to fully capitalize on this fact, we must ﬁrst introduce one of several helpful and well-known Laplace transform theorems—the linearity theorem. This theorem states that the Laplace trans-form of the sum of two or more time functions is equal to the sum of the transforms of the individual time functions. For two time functions we have { f1(t) + f2(t)} = ∞ 0−e−st[ f1(t) + f2(t)] dt = ∞ 0−e−st f1(t) dt + ∞ 0−e−st f2(t) dt = F1(s) + F2(s) As an example of the use of this theorem, suppose that we have a Laplace transform V(s) and want to know the corresponding time function v(t). It will often be possible to decompose V(s) into the sum of two or more functions, e.g., V1(s) and V2(s), whose inverse transforms, v1(t) and v2(t), are already tabulated. It then becomes a simple matter to apply the linearity theorem and write v(t) = −1{V(s)} = −1{V1(s) + V2(s)} = −1{V1(s)} + −1{V2(s)} = v1(t) + v2(t) Another important consequence of the linearity theorem is evident by studying the deﬁnition of the Laplace transform. Since we are working with an integral, the Laplace transform of a constant times a function is equal to the constant times the Laplace transform of the function. In other words, {kv(t)} = k{v(t)} or kv(t) ⇔kV(s) where k is a constant of proportionality. This result is extremely handy in many situations that arise from circuit analysis, as we will see. CHAPTER 14 COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM 546 This is known as the “additive property’’ of the Laplace transform. This is known as the “homogeneity property’’ of the Laplace transform. PRACTICE ● 14.5 Determine V(s) if v(t) equals (a) 4δ(t) −3u(t); (b) 4δ(t −2) −3tu(t); (c) [u(t)] [u(t −2)]. 14.6 Determine v(t) if V(s) equals (a) 10; (b) 10/s; (c) 10/s2; (d) 10/[s(s + 10)]; (e) 10s/(s + 10). Ans: 14.5: (4s −3)/s; 4e−2s −(3/s2); e−2s/s. 14.6: 10δ(t); 10u(t); 10tu(t); u(t) −e−10tu(t); 10δ(t) −100e−10tu(t). SECTION 14.5 INVERSE TRANSFORM TECHNIQUES 547 Inverse Transform Techniques for Rational Functions In analyzing circuits with multiple energy storage elements, we will often encounter s-domain expressions that are ratios of s-polynomials. We thus expect to routinely encounter expressions of the form V(s) = N(s) D(s) where N(s) and D(s) are polynomials in s. The values of s which lead to N(s) = 0 are referred to as zeros of V(s), and those values of s which lead to D(s) = 0 are referred to as poles of V(s). Rather than rolling up our sleeves and invoking Eq. each time we need to ﬁnd an inverse transform, it is often possible to decompose these ex-pressions using the method of residues into simpler terms whose inverse transforms are already known. The criterion for this is that V(s) must be a rational function for which the degree of the numerator N(s) must be less than that of the denominator D(s). If it is not, we must ﬁrst perform a sim-ple division step, as shown in the following example. The result will include an impulse function (assuming the degree of the numerator is the same as that of the denominator) and a rational function. The inverse transform of the ﬁrst is simple; the straightforward method of residues applies to the ra-tional function if its inverse transform is not already known. EXAMPLE 14.3 In practice, it is seldom necessary to ever invoke Eq. for functions encountered in circuit analysis, provided that we are clever in using the various techniques pre-sented in this chapter. Given the function G(s) = (7/s) −31/(s + 17), obtain g(t). This s-domain function is composed of the sum of two terms, 7/s and −31/(s + 17). Through the linearity theorem we know that g(t) will be composed of two terms as well, each the inverse Laplace transform of one of the two s-domain terms: g(t) = −1 7 s −−1 31 s + 17 Let’s begin with the ﬁrst term. The homogeneity property of the Laplace transform allows us to write that −1 7 s = 7−1 1 s = 7u(t) Thus, we have made use of the known transform pair u(t) ⇔1/s and the homogeneity property to ﬁnd this ﬁrst component of g(t). In a similar fashion, we ﬁnd that −1 31 s + 17 = 31e−17tu(t). Putting these two terms together, g(t) = [7 −31e−17t]u(t) PRACTICE ● 14.7 Given H(s) = 2 s −4 s2 + 3.5 (s + 10)(s + 10) , obtain h(t). Ans: h(t) = [2 −4t + 3.5te−10t]u(t). CHAPTER 14 COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM 548 In employing the method of residues, essentially performing a partial fraction expansion of V(s), we focus our attention on the roots of the denominator. Thus, it is ﬁrst necessary to factor the s-polynomial that com-prises D(s) into a product of binomial terms. The roots of D(s) may be any combination of distinct or repeated roots, and may be real or complex. It is worth noting, however, that complex roots always occur as conjugate pairs, provided that the coefﬁcients of D(s) are real. Distinct Poles and the Method of Residues As a speciﬁc example, let us determine the inverse Laplace transform of V(s) = 1 (s + α)(s + β) The denominator has been factored into two distinct roots, −α and −β. Although it is possible to substitute this expression in the deﬁning equation for the inverse transform, it is much easier to utilize the linearity theorem. Using partial-fraction expansion, we can split the given transform into the sum of two simpler transforms, V(s) = A s + α + B s + β where A and B may be found by any of several methods. Perhaps the quick-est solution is obtained by recognizing that A = lim s→−α (s + α)V(s) −(s + α) (s + β) B = lim s→−α 1 s + β −0 = 1 β −α In this equation, we use the single-fraction (i.e., nonexpanded) version of V(s). EXAMPLE 14.4 Calculate the inverse transform of F(s) 2s + 2 s . Since the degree of the numerator is equal to the degree of the denominator, F(s) is not a rational function. Thus, we begin by performing long division: 2 F(s) = s 2s + 4 2s 4 so that F(s) = 2 + (4/s). By the linearity theorem, −1{F(s)} = −1{2} + −1 4 s = 2δ(t) + 4u(t) (It should be noted that this particular function can be simpliﬁed without the process of long division; such a route was chosen to provide an example of the basic process.) PRACTICE ● 14.8 Given the function Q(s) = 3s2 −4 s2 , ﬁnd q(t). Ans: q(t) = 3δ(t) −4tu(t). SECTION 14.5 INVERSE TRANSFORM TECHNIQUES 549 Recognizing that the second term is always zero, in practice we would sim-ply write A = (s + α)V(s)\|s=−α Similarly, B = (s + β)V(s)\|s=−β = 1 α −β and therefore, V(s) = 1/(β −α) s + α + 1/(α −β) s + β We have already evaluated inverse transforms of this form, and so v(t) = 1 β −α e−αtu(t) + 1 α −β e−βtu(t) = 1 β −α (e−αt −e−βt)u(t) If we wished, we could now include this as a new entry in our catalog of Laplace pairs, 1 β −α (e−αt −e−βt)u(t) ⇔ 1 (s + α)(s + β) This approach is easily extended to functions whose denominators are higher-order s-polynomials, although the operations can become somewhat tedious. It should also be noted that we did not specify that the constants A and B must be real. However, in situations where α and β are complex, we will ﬁnd that α and β are also complex conjugates (this is not required mathematically, but is required for physical circuits). In such instances, we will also ﬁnd that A = B∗; in other words, the coefﬁcients will be complex conjugates as well. EXAMPLE 14.5 Determine the inverse transform of P(s) = 7s + 5 s2 + s We see that P(s) is a rational function (the degree of the numerator is one, whereas the degree of the denominator is two), so we begin by factoring the denominator and write: P(s) = 7s + 5 s(s + 1) = a s + b s + 1 where our next step is to determine values for a and b. Applying the method of residues, a = 7s + 5 s + 1 s=0 = 5 and b = 7s + 5 s s=−1 = 2 (Continued on next page) CHAPTER 14 COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM 550 EXAMPLE 14.6 Compute the inverse transform of the function V(s) = 2 s3 + 12s2 + 36s We note that the denominator can be easily factored, leading to V(s) = 2 s(s + 6)(s + 6) = 2 s(s + 6)2 We may now write P(s) as P(s) = 5 s + 2 s + 1 the inverse transform of which is simply p(t) = [5 + 2e−t]u(t). PRACTICE ● 14.9 Given the function Q(s) = 11s + 30 s2 + 3s , ﬁnd q(t). Ans: q(t) = [10 + e−3t]u(t). Repeated Poles A closely related situation is that of repeated poles. Consider the function V(s) = N(s) (s −p)n which we want to expand into V(s) = an (s −p)n + an−1 (s −p)n−1 + · · · + a1 (s −p) To determine each constant, we ﬁrst multiply the nonexpanded version of V(s) by (s −p)n. The constant an is found by evaluating the resulting ex-pression at s = p. The remaining constants are found by differentiating the expression (s −p)nV(s) the appropriate number of times prior to evaluat-ing at s = p, and dividing by a factorial term. The differentiation procedure removes the constants previously found, and evaluating at s = p removes the remaining constants. For example, an−2 is found by evaluating 1 2! d2 ds2 [(s −p)nV(s)]s= p and the term an−k is found by evaluating 1 k! dk dsk [(s −p)nV(s)]s= p To illustrate the basic procedure, let’s ﬁnd the inverse Laplace transform of a function having a combination of both situations: one pole at s = 0 and two poles at s = −6. SECTION 14.5 INVERSE TRANSFORM TECHNIQUES 551 As promised, we see that there are indeed three poles, one at s = 0, and two at s = −6. Next, we expand the function into V(s) = a1 (s + 6)2 + a2 (s + 6) + a3 s and apply our new procedure to obtain the unknown constants a1 and a2; we will ﬁnd a3 using the previous procedure. Thus, a1 = (s + 6)2 2 s(s + 6)2 s=−6 = 2 s s=−6 = −1 3 and a2 = d ds (s + 6)2 2 s(s + 6)2 s=−6 = d ds 2 s s=−6 = −2 s2 s=−6 = −1 18 The remaining constant a3 is found using the procedure for distinct poles a3 = s 2 s(s + 6)2 s=0 = 2 62 = 1 18 Thus, we may now write V(s) as V(s) = −1 3 (s + 6)2 + −1 18 (s + 6) + 1 18 s Using the linearity theorem, the inverse transform of V(s) can now be found by simply determining the inverse transform of each term. We see that the ﬁrst term on the right is of the form K (s + α)2 and making use of Eq. , we ﬁnd that its inverse transform is −1 3te−6tu(t). In a similar fashion, we ﬁnd that the inverse transform of the second term is −1 18e−6tu(t), and that of the third term is 1 18u(t). Thus, v(t) = −1 3te−6tu(t) −1 18e−6tu(t) + 1 18u(t) or, more compactly, v(t) = 1 18[1 −(1 + 6t)e−6t]u(t) PRACTICE ● 14.10 Determine g(t) if G(s) = 3 s3 + 5s2 + 8s + 4 . Ans: g(t) = 3[e−t −te−2t −e−2t]u(t). COMPUTER-AIDED ANALYSIS MATLAB, a very powerful numerical analysis package, can be used to assist in the solution of equations arising from the analysis of circuits with time-varying excitation in several different ways. The most straightforward technique makes use of ordinary differential equation CHAPTER 14 COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM 552 (ODE) solver routines ode23() and ode45(). These two routines are based on numerical methods of solving differential equations, with ode45() having greater accuracy. The solution is determined only at discrete points, however, and therefore is not known for all values of time. For many applications this is adequate, provided a sufﬁcient density of points is used. The Laplace transform technique provides us with the means of obtaining an exact expression for the solution of differential equations, and as such has many advantages over the use of numerical ODE solution techniques. Another signiﬁcant advantage to the Laplace transform technique will become apparent in subsequent chapters when we study the signiﬁcance of the form of s-domain expressions, particularly once we factor the denominator polynomials. As we have already seen, lookup tables can be handy when working with Laplace transforms, although the method of residues can become somewhat tedious for functions with higher-order polynomials in their denominators. In these instances MATLAB can also be of assistance, as it contains several useful functions for the manipulation of polynomial expressions. In MATLAB, the polynomial p(x) = anxn + an−1xn−1 + · · · + a1x + a0 is stored as the vector [an an−1 . . . a1 a0]. Thus, to deﬁne the polynomi-als N(s) = 2 and D(s) = s3 + 12s2 + 36s we write EDU» N = ; EDU» D = [1 12 36 0]; The roots of either polynomial can be obtained by invoking the function roots(p), where p is a vector containing the coefﬁcients of the polyno-mial. For example, EDU» q = [1 8 16]; EDU» roots(q) yields ans = −4 −4 MATLAB also enables us to determine the residues of the rational function N(s)/D(s) using the function residue(). For example, EDU» [r p y] = residue(N, D); returns three vectors r, p, and y, such that N(s) D(s) = r1 x −p1 + r2 x −p2 + · · · + rn x −pn + y(s) in the case of no multiple poles; in the case of n multiple poles, N(s) D(s) = r1 (x −p) + r2 (x −p)2 + · · · + rn (x −p)n + y(s) SECTION 14.6 BASIC THEOREMS FOR THE LAPLACE TRANSFORM 553 Note that as long as the order of the numerator polynomial is less than the order of the denominator polynomial, the vector y(s) will always be empty. Executing the command without the semicolon results in the output r 0.0556 0.3333 0.0556 p 6 6 0 y [ ] which agrees with the answer found in Example 14.6. 14.6 • BASIC THEOREMS FOR THE LAPLACE TRANSFORM We are now able to consider two theorems that might be considered collec-tively the raison d’être for Laplace transforms in circuit analysis—the time differentiation and integration theorems. These will help us transform the derivatives and integrals appearing in the time-domain circuit equations. Time Differentiation Theorem Let us look at time differentiation ﬁrst by considering a time function v(t) whose Laplace transform V(s) is known to exist. We want the transform of the ﬁrst derivative of v(t), dv dt = ∞ 0−e−st dv dt dt This can be integrated by parts: U = e−st dV = dv dt dt with the result dv dt = v(t)e−st ∞ 0−+ s ∞ 0−e−stv(t) dt The ﬁrst term on the right must approach zero as t increases without limit; otherwise V(s) would not exist. Hence, dv dt = 0 −v(0−) + sV(s) and dv dt ⇔sV(s) −v(0−) Similar relationships may be developed for higher-order derivatives: d2v dt2 ⇔s2V(s) −sv(0−) −v′(0−) d3v dt3 ⇔s3V(s) −s2v(0−) −sv′(0−) −v′′(0−) where v′(0−) is the value of the ﬁrst derivative of v(t) evaluated at t = 0−, v′′(0−) is the initial value of the second derivative of v(t), and so forth. When all initial conditions are zero, we see that differentiating once with re-spect to t in the time domain corresponds to multiplication by s in the fre-quency domain; differentiating twice in the time domain corresponds to multiplication by s2 in the frequency domain, and so on. Thus, differentia-tion in the time domain is equivalent to multiplication in the frequency do-main. This is a substantial simpliﬁcation! We also notice that, when the ini-tial conditions are not zero, their presence is still accounted for. CHAPTER 14 COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM 554 EXAMPLE 14.7 Given the series RL circuit shown in Fig. 14.3, calculate the current through the 4 resistor. Identify the goal of the problem. We need to ﬁnd an expression for the current labeled i(t). Collect the known information. The network is driven by a step voltage, and we are given an initial value of the current (at t = 0−) of 5 A. Devise a plan. Applying KVL to the circuit will result in a differential equation with i(t) as the unknown. Taking the Laplace transform of both sides of this equation will convert it to the s-domain. Solving the resulting algebraic equation for I(s), the inverse Laplace transform will yield i(t). Construct an appropriate set of equations. Using KVLto write the single-loop equation in the time domain, 2 di dt + 4i = 3u(t) Now, we take the Laplace transform of each term, so that 2[sI(s) −i(0−)] + 4I(s) = 3 s Determine if additional information is required. We have an equation that may be solved for the frequency-domain representation I(s) of our goal, i(t). Attempt a solution. We next solve for I(s), substituting i(0−) = 5: (2s + 4)I(s) = 3 s + 10 i(t) + – 2 H 4 i(0–) = 5 A 3u(t) V ■FIGURE 14.3 A circuit that is analyzed by trans-forming the differential equation 2 di /dt + 4i = 3u(t) into 2[sI(s) −i (0−)] + 4I(s) = 3/s. SECTION 14.6 BASIC THEOREMS FOR THE LAPLACE TRANSFORM 555 Time-Integration Theorem The same kind of simpliﬁcation can be accomplished when we meet the operation of integration with respect to time in our circuit equations. Let us determine the Laplace transform of the time function described by t 0−v(x) dx, t 0−v(x) dx = ∞ 0−e−st t 0−v(x) dx dt Integrating by parts, we let u = t 0−v(x) dx dv = e−st dt du = v(t) dt v = −1 s e−st and I(s) = 1.5 s(s + 2) + 5 s + 2 Applying the method of residues to the ﬁrst term, 1.5 s + 2 s=0 = 0.75 and 1.5 s s=−2 = −0.75 so that I(s) = 0.75 s + 4.25 s + 2 We then use our known transform pairs to invert: i(t) = 0.75u(t) + 4.25e−2tu(t) = (0.75 + 4.25e−2t)u(t) A Verify the solution. Is it reasonable or expected? Based on our previous experience with this type of circuit, we expected a dc forced response plus an exponentially decaying natural response. At t = 0, we obtain i(0) = 5 A, as required, and as t →∞, i(t) →3 4 A as we would expect. Our solution for i(t) is therefore complete. Both the forced re-sponse 0.75u(t) and the natural response 4.25e−2tu(t) are present, and the initial condition was automatically incorporated into the solution. The method illustrates a very painless way of obtaining the complete solution of many differential equations. PRACTICE ● 14.11 Use Laplace transform methods to ﬁnd i(t) in the circuit of Fig. 14.4. Ans: (0.25 + 4.75e−20t)u(t) A. i + – 4 (t) + u(t) V 0.2 H ■FIGURE 14.4 Then t 0−v(x) dx = t 0−v(x) dx −1 s e−st t=∞ t=0−− ∞ 0−−1 s e−stv(t) dt = −1 s e−st t 0−v(x) dx ∞ 0−+ 1 s V(s) But, since e−st →0 as t →∞, the ﬁrst term on the right vanishes at the up-per limit, and when t →0−, the integral in this term likewise vanishes. This leaves only the V(s)/s term, so that t 0−v(x) dx ⇔V(s) s and thus integration in the time domain corresponds to division by s in the frequency domain. Once more, a relatively complicated calculus operation in the time domain simpliﬁes to an algebraic operation in the frequency domain. CHAPTER 14 COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM 556 EXAMPLE 14.8 Determine i(t) for t > 0 in the series RC circuit shown in Fig. 14.5. We ﬁrst write the single-loop equation, u(t) = 4i(t) + 16 t −∞ i(t′) dt′ In order to apply the time-integration theorem, we must arrange for the lower limit of integration to be 0−. Thus, we set 16 t −∞ i(t′) dt′ = 16 0− −∞ i(t′) dt′ + 16 t 0−i(t′) dt′ = v(0−) + 16 t 0−i(t′) dt′ Therefore, u(t) = 4i(t) + v(0−) + 16 t 0−i(t′) dt′ We next take the Laplace transform of both sides of this equation. Since we are utilizing the one-sided transform, L{v(0−)} is simply L{v(0−)u(t)}, and thus 1 s = 4I(s) + 9 s + 16 s I(s) Solving for I(s), I(s) = − 2 s + 4 the desired result is immediately obtained, i(t) = −2e−4tu(t) A i(t) 4 v(0–) = 9 V + – u(t) 1 F 16 v(t) + – ■FIGURE 14.5 A circuit illustrating the use of the Laplace transform pair t 0−i (t ′) dt ′ ⇔1 s I(s). SECTION 14.6 BASIC THEOREMS FOR THE LAPLACE TRANSFORM 557 EXAMPLE 14.9 Find v(t) for the same circuit, repeated as Fig. 14.6 for convenience. This time we write a single nodal equation, v(t) −u(t) 4 + 1 16 dv dt = 0 Taking the Laplace transform, we obtain V(s) 4 −1 4s + 1 16sV(s) −v(0−) 16 = 0 or V(s) 1 + s 4 = 1 s + 9 4 Thus, V(s) = 4 s(s + 4) + 9 s + 4 = 1 s − 1 s + 4 + 9 s + 4 = 1 s + 8 s + 4 and taking the inverse transform, v(t) = (1 + 8e−4t)u(t) V To check this result, we note that ( 1 16)dv/dt should yield the previous expression for i(t). For t > 0, 1 16 dv dt = 1 16(−32)e−4t = −2e−4t which is in agreement with what was found in Example 14.8. PRACTICE ● 14.12 Find v(t) at t = 800 ms for the circuit of Fig. 14.7. i(t) + – 4 1 F 16 v(0–) = 9 V u(t) v(t) + – ■FIGURE 14.6 The circuit of Fig. 14.5 repeated, in which the voltage v(t) is required. 5 + – 2tu(t) V 0.1 F v(t) + – ■FIGURE 14.7 Ans: 802 mV. Laplace Transforms of Sinusoids To illustrate the use of both the linearity theorem and the time-differentiation theorem, not to mention the addition of a most important pair to our forth-coming Laplace transform table, let us establish the Laplace transform of sin ωt u(t). We could use the deﬁning integral expression with integration by parts, but this is needlessly difﬁcult. Instead, we use the relationship sin ωt = 1 2 j (e jωt −e−jωt) The transform of the sum of these two terms is just the sum of the trans-forms, and each term is an exponential function for which we already have the transform. We may immediately write {sin ωt u(t)} = 1 2 j 1 s −jω − 1 s + jω = ω s2 + ω2 sin ωt u(t) ⇔ ω s2 + ω2 We next use the time-differentiation theorem to determine the transform of cos ωt u(t), which is proportional to the derivative of sin ωt. That is, {cos ωt u(t)} = 1 ω d dt [sin ωt u(t)] = 1 ωs ω s2 + ω2 cos ωt u(t) ⇔ s s2 + ω2 The Time-Shift Theorem As we have seen in some of our earlier transient problems, not all forcing functions begin at t = 0. What happens to the transform of a time function if that function is simply shifted in time by some known amount? In partic-ular, if the transform of f (t)u(t) is the known function F(s), then what is the transform of f (t −a)u(t −a), the original time function delayed by a seconds (and not existing for t < a)? Working directly from the deﬁnition of the Laplace transform, we get { f (t −a)u(t −a)} = ∞ 0−e−stf (t −a)u(t −a) dt = ∞ a−e−stf (t −a) dt for t ≥a−. Choosing a new variable of integration, τ = t −a, we obtain { f (t −a)u(t −a)} = ∞ 0−e−s(τ+a) f (τ) dτ = e−asF(s) Therefore, f (t −a)u(t −a) ⇔e−asF(s) (a ≥0) This result is known as the time-shift theorem, and it simply states that if a time function is delayed by a time a in the time domain, the result in the fre-quency domain is a multiplication by e−as. CHAPTER 14 COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM 558 Note that we have made use of the fact that sin ωt t =0 = 0. SECTION 14.6 BASIC THEOREMS FOR THE LAPLACE TRANSFORM 559 1 2 3 4 6 5 u(t – 2) –u(t – 5) t ■FIGURE 14.8 Plot of u(t −2) −u(t −5). At this point we have obtained a number of entries for the catalog of Laplace transform pairs that we agreed to construct earlier. Included are the transforms of the impulse function, the step function, the exponential function, the ramp function, the sine and cosine functions, and the sum of two exponentials. In addition, we have noted the consequences in the s domain of the time-domain operations of addition, multiplication by a constant, differentiation, and inte-gration. These results are collected in Tables 14.1 and 14.2; several others which are derived in Appendix 7 are also included. 10 5 0 1 2 3 4 5 6 f (t) t (s) ■FIGURE 14.9 TABLE ●14.1 Laplace Transform Pairs f(t) = −1 {F(s)} F(s) = {f(t)} f(t) = −1 {F(s)} F(s) = {f(t)} δ(t) 1 1 β −α (e−αt −e−βt)u(t) 1 (s + α)(s + β) u(t) 1 s sin ωt u(t) ω s2 + ω2 tu(t) 1 s2 cos ωt u(t) s s2 + ω2 tn−1 (n −1)!u(t), n = 1, 2, . . . 1 sn sin(ωt + θ) u(t) s sin θ + ω cos θ s2 + ω2 e−αtu(t) 1 s + α cos(ωt + θ) u(t) s cos θ −ω sin θ s2 + ω2 te−αtu(t) 1 (s + α)2 e−αt sin ωt u(t) ω (s + α)2 + ω2 tn−1 (n −1)!e−αtu(t), n = 1, 2, . . . 1 (s + α)n e−αt cos ωt u(t) s + α (s + α)2 + ω2 EXAMPLE 14.10 Determine the transform of the rectangular pulse v(t) = u(t −2) − u(t −5). This pulse, shown plotted in Fig. 14.8, has unit value for the time inter-val 2 < t < 5, and zero value elsewhere. We know that the transform of u(t) is just 1/s, and since u(t −2) is simply u(t) delayed by 2 s, the transform of this delayed function is e−2s/s. Similarly, the transform of u(t −5) is e−5s/s. It follows, then, that the desired transform is V(s) = e−2s s −e−5s s = e−2s −e−5s s It was not necessary to revert to the deﬁnition of the Laplace transform in order to determine V(s). PRACTICE ● 14.13 Obtain the Laplace transform of the time function shown in Fig. 14.9. Ans: (5/s)(2e−2s −e−4s −e−5s). PRACTICAL APPLICATION Stability of a System PRACTICAL APPLICATION Many years ago (or so it seems), one of the authors was driving along a country road and attempting to make use of the electronic speed control (“cruise control”) feature of the automobile. After turning the system on and man-ually setting the vehicular speed at precisely the posted speed limit,3 the “set” button was depressed and the ac-celerator pedal released; at this point the system was ex-pected to maintain the set speed by regulating the fuel ﬂow as needed. Unfortunately, something else happened instead. The vehicle speed immediately dropped by about 10 percent, to which the cruise control electronics responded by increasing the fuel ﬂow. The two events were not well matched, so that a few moments later the vehicle speed overshot the set point—resulting in a sudden (and signif-icant) drop in fuel ﬂow—which led to a reduction in the vehicle speed. The cycle continued to the consternation of the driver, who eventually gave up and turned off the system. Clearly the system response was not optimized—in fact, as built the system was unstable. System stability is a major engineering concern across a wide variety of problems (cruise controls, temperature regulators, and tracking systems, to name but a few), and the techniques developed in this chapter are invaluable in allowing the stability of a particular system to be examined. One of the powerful aspects of working in the s-domain as made possible by the Laplace transform is that instead of describing the response of a particular system through an integrodifferential equation, we can obtain a system transfer function represented by the ratio of two s-polynomials. The issue of stability is then easily addressed by studying the denominator of the transfer function: no pole should have a positive real component. There are quite a few techniques that can be applied to the problem of determining the stability of a particular system. One simple technique is known as the Routh test. Consider the s-domain system function (a concept developed further in Chap. 15) H(s) = N(s) D(s) The s-polynomial represented by D(s) can be written as ansn + an−1sn−1 + · · · + a1s + a0. Without factoring the polynomial, not much about the poles can be determined at a glance. If all the coefﬁcients an . . . a0 are positive and nonzero, the Routh procedure has us arrange them in the following pattern: We next create a third row by cross-multiplying the two rows: an−1an−2 −anan−3 an−1 an−1an−4 −anan−5 an−1 and a fourth row by cross-multiplying the second and third rows. This process continues until we have n + 1 rows of numerical values. All that remains is to scan down the leftmost column for changes in sign. The num-ber of sign changes indicates the number of poles with a positive real component; any sign changes indicate an unstable system. For example, let’s assume the automobile cruise con-trol system behind the author’s vexation has a system transfer function with denominator D(s) = 7s4 + 4s3 + s2 + 13s + 2 All of the coefﬁcients of this fourth-order s-polynomial are positive and nonzero, so we construct the corre-sponding Routh table: 7 1 2 4 13 0 −21.75 2 13.37 2 from which we see two sign changes in the leftmost col-umn. Thus, the system is in fact unstable (which explains its failure to perform) as two of its poles have positive real components. (3) Since no cameras were present, nobody can prove otherwise. an an2 an4 . . . an1 an3 an5 . . . © Donovan Reese/Getty Images SECTION 14.7 THE INITIAL-VALUE AND FINAL-VALUE THEOREMS 561 14.7 • THE INITIAL-VALUE AND FINAL-VALUE THEOREMS The last two fundamental theorems that we will discuss are known as the initial-value and final-value theorems. They will enable us to evaluate f (0+) and f (∞) by examining the limiting values of sF(s). Such an ability can be invaluable; if only the initial and ﬁnal values are needed for a partic-ular function of interest, there is no need to take the time to perform an in-verse transform operation. The Initial-Value Theorem To derive the initial-value theorem, we consider the Laplace transform of the derivative once again, d f dt = sF(s) −f (0−) = ∞ 0−e−st d f dt dt TABLE ●14.2 Laplace Transform Operations Operation f(t) F(s) Addition f1(t) ± f2(t) F1(s) ± F2(s) Scalar multiplication k f (t) kF(s) Time differentiation d f dt sF(s) −f (0−) d2f dt2 s2F(s) −s f (0−) −f ′(0−) d3f dt3 s3F(s) −s2 f (0−) −s f ′(0−) −f ′′(0−) Time integration t 0−f (t) dt 1 s F(s) t −∞ f (t) dt 1 s F(s) + 1 s 0− −∞ f (t) dt Convolution f1(t) ∗f2(t) F1(s)F2(s) Time shift f (t −a)u(t −a), a ≥0 e−asF(s) Frequency shift f (t)e−at F(s + a) Frequency differentiation t f (t) −dF(s) ds Frequency integration f (t) t ∞ s F(s) ds Scaling f (at), a ≥0 1 a F s a Initial value f (0+) lim s→∞sF(s) Final value f (∞) lim s→0 sF(s), all poles of sF(s) in LHP Time periodicity f (t) = f (t + nT), 1 1 −e−Ts F1(s), n = 1, 2, . . . where F1(s) = T 0−f (t)e−st dt We now let s approach inﬁnity. By breaking the integral into two parts, lim s→∞[sF(s) −f (0−)] = lim s→∞ 0+ 0−e0 d f dt dt + ∞ 0+ e−st d f dt dt we see that the second integral must approach zero in the limit, since the in-tegrand itself approaches zero. Also, f (0−) is not a function of s, and it may be removed from the left limit: −f (0−) + lim s→∞[sF(s)] = lim s→∞ 0+ 0− d f = lim s→∞[ f (0+) −f (0−)] = f (0+) −f (0−) and ﬁnally, f (0+) = lim s→∞[sF(s)] or lim t→0+ f (t) = lim s→∞[sF(s)] This is the mathematical statement of the initial-value theorem. It states that the initial value of the time function f (t) can be obtained from its Laplace transform F(s) by ﬁrst multiplying the transform by s and then let-ting s approach inﬁnity. Note that the initial value of f (t) we obtain is the limit from the right. The initial-value theorem, along with the ﬁnal-value theorem that we will consider in a moment, is useful in checking the results of a transforma-tion or an inverse transformation. For example, when we ﬁrst calculated the transform of cos(ω0t)u(t), we obtained s/(s2 + ω2 0). After noting that f (0+) = 1, we can make a partial check on the validity of this result by applying the initial-value theorem: lim s→∞ s s s2 + ω2 0 = 1 and the check is accomplished. The Final-Value Theorem The ﬁnal-value theorem is not quite as useful as the initial-value theorem, for it can be used only with a certain class of transforms. In order to deter-mine whether a transform ﬁts into this class, the denominator of F(s) must be evaluated to ﬁnd all values of s for which it is zero, i.e., the poles of F(s). Only those transforms F(s)whose poles lie entirely within the left half of the s plane (i.e., σ < 0), except for a simple pole at s = 0, are suitable for use with the ﬁnal-value theorem. We again consider the Laplace trans-form of df/dt, ∞ 0−e−st df dt dt = sF(s) −f (0−) this time in the limit as s approaches zero, lim s→0 ∞ 0−e−st df dt dt = lim s→0[sF(s) −f (0−)] = ∞ 0− d f dt dt CHAPTER 14 COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM 562 SECTION 14.7 THE INITIAL-VALUE AND FINAL-VALUE THEOREMS 563 We assume that both f (t) and its ﬁrst derivative are transformable. Now, the last term of this equation is readily expressed as a limit, ∞ 0− df dt dt = lim t→∞ t 0− d f dt dt = lim t→∞[ f (t) −f (0−)] By recognizing that f (0−) is a constant, a comparison of the last two equa-tions shows us that lim t→∞f (t) = lim s→0[sF(s)] which is the ﬁnal-value theorem. In applying this theorem, it is necessary to know that f (∞), the limit of f (t) as t becomes inﬁnite, exists or—what amounts to the same thing—that the poles of F(s) all lie within the left half of the s plane except for (possibly) a simple pole at the origin. The product sF(s) thus has all of its poles lying within the left half plane. EXAMPLE 14.11 Use the ﬁnal-value theorem to determine f(∞) for the function (1 −e−at)u(t), where a > 0. Without even using the ﬁnal-value theorem, we see immediately that f (∞) = 1. The transform of f (t) is F(s) = 1 s − 1 s + a = a s(s + a) The poles of F(s) are s = 0 and s = −a. Thus, the nonzero pole of F(s) lies in the left-hand s-plane, as we were assured that a > 0; we ﬁnd that we may indeed apply the ﬁnal-value theorem to this function. Multiply-ing by s and letting s approach zero, we obtain lim s→0[sF(s)] = lim s→0 a s + a = 1 which agrees with f (∞). If f (t) is a sinusoid, however, so that F(s) has poles on the jω axis, then a blind use of the ﬁnal-value theorem might lead us to conclude that the ﬁnal value is zero. We know, however, that the ﬁnal value of either sin ω0t or cos ω0t is indeterminate. So, beware of jω-axis poles! PRACTICE ● 14.14 Without ﬁnding f (t) ﬁrst, determine f (0+) and f (∞) for each of the following transforms: (a) 4e−2s(s + 50)/s; (b) (s2 + 6)/(s2 + 7); (c) (5s2 + 10)/[2s(s2 + 3s + 5)]. Ans: 0, 200; ∞, indeterminate (poles lie on the jω axis); 2.5, 1. CHAPTER 14 COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM 564 SUMMARY AND REVIEW The primary topic of this chapter was the Laplace transform, a mathemati-cal tool for converting well-behaved time-domain functions into frequency-domain expressions. Before introducing the transform, we ﬁrst considered the notion of a complex frequency, which we referred to as s. This conve-nient term has both a real (σ) and imaginary (ω) component, so can be writ-ten as s = σ + jω. In reality, this is shorthand for an exponentially damped sinusoid, and we noted that several common functions are actually special cases of this function. Limited circuit analysis can be performed with this generalized function, but its real purpose was to simply acquaint the reader with the idea of a so-called complex frequency. One of the most surprising things is that day-to-day circuit analysis does not require direct implementation of either the Laplace transform integral or its corresponding inverse integral! Instead, look-up tables are routinely em-ployed, and the s-polynomials which result from analyzing circuits in the s domain are factored into smaller, easily recognizable terms. This works because each Laplace transform pair is unique. There are several theorems associated with Laplace transforms which do see daily usage, however. These include the linearity theorem, the time-differentiation theorem, and the time-integration theorem. The time-shift as well as initial-value and ﬁnal-value theorems are also commonly employed. The Laplace technique is not restricted to circuit analysis, or even electrical engineering for that matter. Any system which is described by integrodifferential equations can make use of the concepts studied in this chapter. At this stage, however, it is probably best to review the key con-cepts already discussed, highlighting appropriate examples. ❑The concept of complex frequency allows us to consider the exponentially damped and oscillatory components of a function simultaneously. (Example 14.1) ❑The complex frequency s = σ + jω is the general case; dc (s = 0), exponential (ω = 0), and sinusoidal (σ = 0) functions are special cases. ❑Analyzing circuits in the s domain results in the conversion of time-domain integrodifferential equations into frequency-domain algebraic equations. (Example 14.1) ❑In circuit analysis problems, we convert time-domain functions into the frequency domain using the one-sided Laplace transform: F(s) = ∞ 0−e−st f (t) dt. (Example 14.2) ❑The inverse Laplace transform converts frequency-domain expressions into the time domain. However, it is seldom needed due to the existence of tables listing Laplace transform pairs. (Example 14.3) ❑The unit-impulse function is a common approximation to pulses with very narrow widths compared to circuit time constants. It is nonzero only at a single point, and has unity area. ❑{ f1(t) + f2(t)} = { f1(t)} + { f2(t)} (additive property) ❑{k f (t)} = k{ f (t)}, k = constant (homogeneity property) EXERCISES 565 ❑Inverse transforms are typically found using a combination of partial-fraction expansion techniques and various operations (Table 14.2) to simplify s-domain quantities into expressions that can be found in trans-form tables (such as Table 14.1). (Examples 14.4, 14.5, 14.6, 14.10) ❑The differentiation and integration theorems allow us to convert inte-grodifferential equations in the time domain into simple algebraic equations in the frequency domain. (Examples 14.7, 14.8, 14.9) ❑The initial-value and ﬁnal-value theorems are useful when only the speciﬁc values f (t = 0+) or f (t →∞) are desired. (Example 14.11) READING FURTHER An easily readable development of the Laplace transform and some of its key properties can be found in Chap. 4 of: A. Pinkus and S. Zafrany, Fourier Series and Integral Transforms. Cambridge, United Kingdom: Cambridge University Press, 1997. A much more detailed treatment of integral transforms and their application to science and engineering problems can be found in: B. Davies, Integral Transforms and Their Applications, 3rd ed. New York: Springer-Verlag, 2002. Stability and the Routh test are discussed in Chap. 5 of: K. Ogata, Modern Control Engineering, 4th ed. Englewood Cliffs, N.J.: Prentice-Hall, 2002. EXERCISES 14.1 Complex Frequency 1. Determine the conjugate of each of the following: (a) 8 −j; (b) 8e−9t; (c) 22.5; (d) 4e j9; (e) j2e−j11. 2. Compute the complex conjugate of each of the following expressions: (a) −1; (b) −j 5/20◦; (c) 5e−j5 + 2e j3; (d) (2 + j)(8/30◦)e j2t . 3. Several real voltages are written down on a piece of paper, but coffee spills across half of each one. Complete the voltage expression if the legible part is (a) 5e−j50t; (b) (2 + j)e j9t; (c) (1 −j)e j78t; (d) −je−5t. Assume the units of each voltage are volts (V). 4. State the complex frequency or frequencies associated with each function: (a) f (t) = sin 100t; (b) f(t) = 10; (c) g(t) = 5e−7t cos 80t; (d) f(t) = 5e8t ; (e) g(t) = (4e−2t −e−t) cos(4t −95◦). 5. For each of the following functions, determine both the complex frequency s as well as s: (a) 7e−9t sin (100t + 9◦); (b) cos 9t; (c) 2 sin 45t; (d) e7t cos 7t. 6. Use real constants A, B, θ, φ, etc. to construct the general form of a real time function characterized by the following frequency components: (a) 10 −j3 s−1; (b) 0.25 s−1; (c) 0, 1, −j, 1 + j (all s−1). 7. The following voltage sources AeBt cos(Ct + θ) are connected (one at a time) to a 280 resistor. Calculate the resulting current at t = 0, 0.1, and 0.5 s, assuming the passive sign convention: (a) A = 1 V, B = 0.2 Hz, C = 0, θ = 45◦; (b) A = 285 mV, B = −1 Hz, C = 2 rad/s, θ = −45◦. 8. Your neighbor’s cell phone interferes with your laptop speaker system when-ever the phone is connecting to the local network. Connecting an oscilloscope to the output jack of your computer, you observe a voltage waveform that can be described by a complex frequency s = −1 + j200π s−1. (a) What can you deduce about your neighbor’s movements? (b) The imaginary part of the complex frequency starts to decrease suddenly. Alter your deduction as appropriate. 9. Compute the real part of each of the following complex functions: (a) v(t) = 9e−j4t V; (b) v(t) = 12 −j9 V; (c) 5 cos 100t −j43 sin 100t V; (d) (2 + j)e j3t V. 10. Your new assistant has measured the signal coming from a piece of test equipment, writing v(t) = Vxe(−2+ j60)t , where Vx = 8 −j100 V. (a) There is a missing term. What is it, and how can you tell it’s missing? (b) What is the complex frequency of the signal? (c) What is the signiﬁcance of the fact that Im{Vx} > Re{Vx}? (d) What is the signiﬁcance of the fact that \|Re{s}\| < \|Im{s}\|? 14.2 The Damped Sinusoidal Forcing Function 11. State the time-domain voltage v(t) which corresponds to the voltage V = 19/84◦V if s is equal to (a) 5 s−1; (b) 0; (c) −4 + j s−1. 12. For the circuit of Fig. 14.10, the voltage source is chosen such that it can be represented by the complex frequency domain function Vest, with V = 2.5/−20◦V and s = −1 + j100 s−1. Calculate (a) s; (b) v(t), the time- domain representation of the voltage source; (c) the current i(t). 13. With regard to the circuit depicted in Fig. 14.10, determine the time-domain voltage v(t) which corresponds to a frequency-domain current i(t) = 5/30◦A for a complex frequency of (a) s = −2 + j2 s−1; (b) s = −3 + j s−1. 14. For the circuit depicted in Fig. 14.11, take s = −200 + j150 s−1. Determine the ratio of the frequency-domain voltages V2 and V1, which correspond to v2(t) and v1(t), respectively. 15. If the complex frequency describing the circuit of Fig. 14.11 is s = −150 + j100 s−1, determine the time-domain voltage which corresponds to a frequency-domain voltage V2 = 5/−25◦V. 16. Calculate the time-domain voltage v in the circuit of Fig. 14.12 if the frequency-domain representation of the current source is 2.3/5◦A at a complex frequency of s = −1 + j2 s−1. 17. The circuit of Fig. 14.12 is operated for an extended period of time without interruption. The frequency-domain voltage which develops across the three elements can be represented as 1.8/75◦V at a complex frequency of s = −2 + j1.5 s−1. Determine the time-domain current is. 18. The circuit of Fig. 14.13 is driven by vS(t) = 10 cos 5t V. (a) Determine the complex frequency of the source. (b) Determine the frequency-domain repre-sentation of the source: (c) Compute the frequency-domain representation of ix. (d) Obtain the time-domain expression for ix. 19. The frequency-domain current Ix which ﬂows through the 2.2 resistor of Fig. 14.13 can be represented as 2/10◦A at a complex frequency of s = −1 + j0.5 s−1. Determine the time-domain voltage vs. 20. Let is1 = 20e−3t cos 4t A and is2 = 30e−3t sin 4t A in the circuit of Fig. 14.14. (a) Work in the frequency domain to ﬁnd Vx. (b) Find vx(t). CHAPTER 14 COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM 566 5 2 H 0.1 F vx + – is1 is2 ■FIGURE 14.14 i(t) v (t) + – 1.5 250 mF 1.8 H ■FIGURE 14.10 i(t) + – 21 100 mH vs v2 + – v1 + – ■FIGURE 14.11 250 mF is v + – 2 ■FIGURE 14.12 1 400 mF 2.2 1.5 H + – vS ix ■FIGURE 14.13 EXERCISES 567 14.3 Deﬁnition of the Laplace Transform 21. Calculate, with the assistance of Eq. (and showing intermediate steps), the Laplace transform of the following: (a) 2.1u(t); (b) 2u(t −1); (c) 5u(t −2) − 2u(t); (d) 3u(t −b), where b > 0. 22. Employ the one-sided Laplace transform integral (with intermediate steps explicitly included) to compute the s-domain expressions which correspond to the following: (a) 5u(t −6); (b) 2e−tu(t); (c) 2e−tu(t −1); (d) e−2t sin 5tu(t). 23. With the assistance of Eq. and showing appropriate intermediate steps, compute the one-sided Laplace transform of the following: (a) (t −1)u(t −1); (b) t2u(t); (c) sin 2tu(t); (d) cos 100t u(t). 24. The Laplace transform of tf(t), assuming { f (t)} = F(s), is given by −d ds F(s). Test this by comparing the predicted result to what is found by directly employing Eq. for (a) tu(t); (b) t2u(t); (c) t3u(t); (d) te−tu(t). 14.4 Laplace Transforms of Simple Time Functions 25. For the following functions, specify the range of σ0 for which the one-sided Laplace transform exists: (a) t + 4; (b) (t + 1)(t −2); (c) e−t/2u(t); (d) sin 10t u(t + 1). 26. Show, with the assistance of Eq. , that { f (t) + g(t) + h(t)} = { f (t)} + {g(t)}+ {h(t)}. 27. Determine F(s) if f(t) is equal to (a) 3u(t −2); (b) 3e−2tu(t) + 5u(t); (c) δ(t) + u(t) −tu(t); (d) 5δ(t). 28. Obtain an expression for G(s) if g(t) is given by (a) [5u(t)]2 −u(t); (b) 2u(t) −2u(t −2); (c) tu(2t); (d) 2e−tu(t) + 3u(t). 29. Without recourse to Eq. , obtain an expression for f(t) if F(s) is given by (a) 1 s ; (b) 1.55 −2 s ; (c) 1 s + 1.5; (d) 5 s2 + 5 s + 5. (Provide some brief explanation of how you arrived at your solution.) 30. Obtain an expression for g(t) without employing the inverse Laplace transform integral, if G(s) is known to be (a) 1.5 (s + 9)2 ; (b) 2 s −0; (c) π; (d) a (s + 1)2 −a, a > 0. (Provide some brief explanation of your solution process for each.) 31. Evaluate the following: (a) δ(t) at t = 1; (b) 5δ(t + 1) + u(t + 1) at t = 0; (c) 2 −1 δ(t) dt; (d) 3 − 2 −1 2δ(t) dt. 32. Evaluate the following: (a) [δ(2t)]2 at t = 1; (b) 2δ(t −1) + u(−t + 1) at t = 0; (c) 1 3 0.003 −0.001 δ(t) dt; (d) 1 1 2 2 0 δ(t −1) dt\| 2 . 33. Evaluate the following expressions at t = 0: (a) +∞ −∞2δ(t −1) dt; (b) +∞ −∞δ(t + 1) dt u(t + 1) ; (c) 3 +∞ −∞δ(t −2) dt [u(1 −t)]3 − u(t + 2); (d) +∞ −∞δ(t −1) dt +∞ −∞δ(t + 1) dt 2 . 34. Evaluate the following: (a) +∞ −∞e−100δ t −1 5 dt; (b) +∞ −∞4tδ(t −2) dt; (c) +∞ −∞4t2δ(t −1.5) dt; (d) +∞ −∞(4 −t)δ(t −1) dt +∞ −∞(4 −t)δ(t + 1) dt . 14.5 Inverse Transform Techniques 35. Determine the inverse transform of F(s) equal to (a) 5 + 5 s2 − 5 (s + 1); (b) 1 s + 5 (0.1s + 4) −3; (c) −1 2s + 1 (0.5s)2 + 4 (s + 5)(s + 5) + 2; (d) 4 (s + 5)(s + 5) + 2 s + 1 + 1 s + 3 . 36. Obtain an expression for g(t) if G(s) is given by (a) 3(s + 1) (s + 1)2 + 2s s2 − 1 (s + 2)2; (b) − 10 (s + 3)3 ; (c) 19 − 8 (s + 3)2 + 18 s2 + 6s + 9 . 37. Reconstruct the time-domain function if its transform is (a) s s(s + 2); (b) 1; (c) 3 s + 2 (s2 + 2s + 4) ; (d) 4 s 2s + 3 . 38. Determine the inverse transform of V(s) equal to (a) s2 + 2 s + 1; (b) s + 8 s + 2 s2 ; (c) s + 1 s(s + 2) + 2s2 −1 s2 ; (d) s2 + 4s + 4 s . 39. Obtain the time-domain expression which corresponds to each of the following s-domain functions: (a) 2 3s + 1 2 s2 + 3s; (b) 7 − s + 1 s s2 + 3s + 1 ; (c) 2 s2 + 1 s + s + 2 s 2 2 + 4s + 6 ; (d) 2 (s + 1)(s + 1); (e) 14 (s + 1)2(s + 4)(s + 5). 40. Find the inverse Laplace transform of the following: (a) 1 s2 + 9s + 20; (b) 4 s3 + 18s2 + 17s + 1 s ; (c) (0.25) 1 s 2 2 + 1.75s + 2.5 ; (d) 3 s(s + 1)(s + 4)(s + 5)(s + 2). (e) Verify your answers with MATLAB. 41. Determine the inverse Laplace transform of each of the following s-domain expressions: (a) 1 (s + 2)2(s + 1); (b) s (s2 + 4s + 4)(s + 2); (c) 8 s3 + 8s2 + 21s + 18. (d) Verify your answers with MATLAB. 42. Given the following expressions in the s-domain, determine the corresponding time-domain functions: (a) 1 3s − 1 2s + 1 + 3 s3 + 8s2 + 16s −1; (b) 1 3s + 5 + 3 s3/8 + 0.25s2 ; (c) 2s (s + a)2 . 43. Compute −1 {G(s)} if G(s) is given by (a) 3s (s/2 + 2)2(s + 2); (b) 3 −3 s (2s2 + 24s + 70)(s + 5) ; (c) 2 − 1 s + 100 + s s2 + 100 ; (d) {tu(2t)}. 44. Obtain the time-domain expression which corresponds to the following s-domain functions: (a) s (s + 2)3 ; (b) 4 (s + 1)4(s + 1)2 ; (c) 1 s2(s + 4)2(s + 6)3 −2s2 s + 9. (d) Verify your solutions with MATLAB. 14.6 Basic Theorems for the Laplace Transform 45. Take the Laplace transform of the following equations: (a) 5 di/dt −7 d2i/dt2 + 9i = 4; (b) m d2p dt2 + μf dp dt + kp(t) = 0, the equation that describes the “force-free’’ response of a simple shock CHAPTER 14 COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM 568 EXERCISES 569 absorber system; (c) dnp dt = −np τ + GL , with τ = constant, which describes the recombination rate of excess electrons (np) in p-type silicon under optical illumination (GL is a constant proportional to the light intensity). 46. With regard to the circuit depicted in Fig. 14.15, the initial voltage across the capacitor is v(0−) = 1.5 V and the current source is is = 700u(t) mA. (a) Write the differential equation which arises from KCL, in terms of the nodal voltage v(t). (b) Take the Laplace transform of the differential equation. (c) Determine the frequency-domain representation of the nodal voltage. (d) Solve for the time-domain voltage v(t). 47. For the circuit of Fig. 14.15, if Is = 2 s + 1 mA, (a) write the time-domain nodal equation in terms of v(t); (b) solve for V(s); (c) determine the time-domain voltage v(t). 48. The voltage source in the circuit of Fig. 14.4 is replaced with one whose s-domain equivalent is 2 s − 1 s + 1 V. The initial condition is unchanged. (a) Write the s-domain KVL equation in terms of I(s). (b) Solve for i(t). 49. For the circuit of Fig. 14.16, vs(t) = 2u(t) V and the capacitor initially stores zero energy. (a) Write the time-domain loop equation in terms of the current i(t). (b) Obtain the s-domain representation of this integral equation. (c) Solve for i(t). 50. The s-domain representation of the voltage source in Fig. 14.16 is Vs(s) = 2 s + 1 V. The initial voltage across the capacitor, deﬁned using the passive sign convention in terms of the current i, is 4.5 V. (a) Write the time-domain integral equation that arises from application of KVL. (b) By ﬁrst solving for I(s), determine the time-domain current i(t). 51. If the current source of Fig. 14.17 is given by 450u(t) mA, and ix(0) = 150 mA, work initially in the s-domain to obtain an expression for v(t) valid for t > 0. 52. Obtain, through purely legitimate means, an s-domain expression which corre-sponds to the time-domain wavefrom plotted in Fig. 14.18. 53. Apply the Routh test to the following system functions, and state whether the system is stable or unstable: (a) H(s) = s −500 s3 + 13s2 + 47s + 35 ; (b) H(s) = s −500 s3 + 13s2 + s + 35 . 54. Apply the Routh test to the following system functions, and state whether the system is stable or unstable; then factor each denominator to identify the poles of H(s) and verify the accuracy of the Routh test for these functions: (a) H(s) = 4s s2 + 3s + 8 ; (b) H(s) = s −9 s2 + 2s + 1 . 14.7 The Initial-Value and Final-Value Theorems 55. Employ the initial-value theorem to determine the initial value of each of the following time-domain functions: (a) 2u(t); (b) 2e−tu(t); (c) u(t −6); (d) cos 5t u(t). 56. Employ the initial-value theorem to determine the initial value of each of the following time-domain functions: (a) u(t −3); (b) 2e−(t−2)u(t −2); (c) u(t −2) + [u(t)]2 2 ; (d) sin 5t e−2tu(t). 57. Make use of the ﬁnal-value theorem (if appropriate) to ascertain f (∞) for (a) 1 s + 2 −2 s ; (b) 2s (s + 2)(s + 1); (c) 1 (s + 2)(s + 4) + 2 s ; (d) 1 (s2 + s −6)(s + 9). is 500 mF v + – 2 ■FIGURE 14.15 i(t) 200 mF + – vs v(t) + – 5 ■FIGURE 14.16 ix is 1.5 H v + – 1 ■FIGURE 14.17 12 6 0 1 2 3 4 5 6 f(t) t (s) ■FIGURE 14.18 CHAPTER 14 COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM 570 + – iC 100 mF 2 1 is vs ■FIGURE 14.19 64. Referring to the circuit depicted in Fig. 14.19 and working in the s-domain to develop an expression for IC(s), determine iC(t) for t > 0 if is(t) = 2u(t + 2) A and vs(t) is equal to (a) 2u(t) V; (b) te−tu(t) V. 65. For the circuit of Fig. 14.20, I(s) = 5 s + 1 (s + 1)2 + 104 A. (a) Determine the initial value of the inductor current. (b) Determine the ﬁnal value of the induc-tor voltage, assuming it is deﬁned consistent with the passive sign convention. 58. Without recourse to f (t), determine f (0+) and f (∞) (or show they do not exist) for each of the following s-domain expressions: (a) 1 s + 18; (b) 10 1 s2 + 3 s ; (c) s2 −4 s3 + 8s2 + 4s; (d) s2 + 2 s3 + 3s2 + 5s. 59. Apply the initial- or ﬁnal-value theorems as appropriate to determine f (0+) and f (∞) for the following functions: (a) s + 2 s2 + 8s + 4; (b) 1 s2(s + 4)2(s + 6)3 −2s2 s + 9; (c) 4s2 + 1 (s + 1)2(s + 2)2 . 60. Determine which of the following functions are appropriate for the ﬁnal-value theorem: (a) 1 (s −1); (b) 10 s2 −4s + 4; (c) 13 s3 −5s2 + 8s −6; (d) 3 2s3 −10s2 + 16s −12. Chapter-Integrating Exercises 61. The voltage v(t) = 8e−2tu(t) V is applied to an unlabeled two-terminal device. Your assistant misunderstands you and only records the s-domain current which results. Determine what type of element it is and its value if I(s) is equal to (a) 1 s + 2 A; (b) 4 s(s + 2) A. 62. (a) Create an s-domain function F(s) which corresponds to an initial value f (0−) = 16 yet has an indeterminate ﬁnal value. (b) Obtain an expression for f (t). (c) If this waveform represents the voltage across a 2 F capacitor, determine the current ﬂowing through the device (assume the passive sign convention). 63. For the circuit of Fig. 14.19, let is(t) = 5u(t) A and vs(t) = e−4tu(t + 1) V. Working initially in the s-domain, obtain an expression for iC(t) valid for t > 0. + – 5 H i ■FIGURE 14.20 INTRODUCTION Having been introduced to the concept of complex frequency and to the Laplace transform technique, we now are ready to see the details of how circuit analysis in the s-domain actually works. As the reader might suspect, particularly if Chap. 10 has already been studied, in fact several shortcuts are routinely applied. The ﬁrst of these is to create a new way of viewing capacitors and inductors, so that s-domain nodal and mesh equations can be written directly. As part of this method, we will learn how to take care to account for initial conditions. Another “shortcut” is the concept of a circuit transfer function. This general function can be exploited to predict the response of a circuit to various inputs, its stability, and even its frequency-selective response. 15.1 • Z(s) AND Y(s) The key concept that makes phasors so useful in the analysis of sinusoidal steady-state circuits is the transformation of resistors, capacitors, and inductors into impedances. Circuit analysis then proceeds using the basic techniques of nodal or mesh analysis, su-perposition, source transformation, as well as Thévenin or Norton equivalents. This concept can be extended to the s-domain, since the sinusoidal steady state is included in s-domain analysis as a special case (where σ = 0). Resistors in the Frequency Domain Let’s begin with the simplest situation: a resistor connected to a voltage source v(t). Ohm’s law speciﬁes that v(t) = Ri(t) Taking the Laplace transform of both sides, V(s) = RI(s) KEY CONCEPTS Extending the Concept of Impedance to the s-Domain Modeling Initial Conditions with Ideal Sources Applying Nodal, Mesh, Superposition, and Source Transformation in the s-Domain Thévenin and Norton’s Theorems Applied to s-Domain Circuits Manipulating s-Domain Algebraic Expressions with MATLAB Identifying Poles and Zeros in Circuit Transfer Functions Impulse Response of a Circuit Use of Convolution to Determine System Response Response as a Function of σ and ω Using Pole-Zero Plots to Predict the Natural Response of Circuits Synthesizing Speciﬁc Voltage Transfer Functions Using Op Amps Circuit Analysis in the s-Domain C H A P T E R 15 571 Thus, the ratio of the frequency-domain representation of the voltage to the frequency-domain representation of the current is simply the resistance, R. Since we are working in the frequency domain, we refer to this quantity as an impedance for the sake of clarity, but still assign it the unit ohms (): Z(s) ≡V(s) I(s) = R Just as we found in working with phasors in the sinusoidal steady state, the impedance of a resistor does not depend on frequency. The admittance Y(s) of a resistor, deﬁned as the ratio of I(s) to V(s), is simply 1/R; the unit of admittance is the siemen (S). Inductors in the Frequency Domain Next, we consider an inductor connected to some time-varying voltage source v(t), as shown in Fig. 15.1a. Since v(t) = L di dt taking the Laplace transform of both sides of this equation yields V(s) = L[sI(s) −i(0−)] We now have two terms: sLI(s) and Li(0−). In situations where we have zero initial energy stored in the inductor (i.e., i(0−) = 0), then V(s) = sLI(s) so that Z(s) ≡V(s) I(s) = sL Equation may be further simpliﬁed if we are only interested in the sinusoidal steady-state response. It is permissible to neglect the initial con-ditions in such instances as they only affect the nature of the transient response. Thus, we substitute s = jω and ﬁnd Z( jω) = jωL as was obtained previously in Chap. 10. Modeling Inductors in the s-Domain Although we refer to the quantity in Eq. as the impedance of an induc-tor, we must remember that it was obtained by assuming zero initial current. In the more general situation where energy is stored in the element at t = 0−, this quantity is not sufﬁcient to represent the inductor in the fre-quency domain. Fortunately, it is possible to include the initial condition by modeling an inductor as an impedance in combination with either a voltage or current source. To do this, we begin by rearranging Eq. as V(s) = sLI(s) −Li(0−) The second term on the right will be a constant: the inductance L in henrys multiplied by the initial current i(0−) in amperes. The result is a constant CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 572 ■FIGURE 15.1 (a) Inductor in the time domain. (b) The complete model for an inductor in the frequency domain, consisting of an impedance sL and a voltage source −Li(0−) that incorporates the effects of nonzero initial conditions on the element. v(t) + – i(t) (a) L (b) –Li(0–) Z(s) = sL V(s) + – I(s) + – SECTION 15.1 Z(s) AND Y(s) 573 voltage term that is subtracted from the frequency-dependent term sLI(s). A small leap of intuition at this point leads us to the realization that we can model a single inductor L as a two-component frequency-domain element, as shown in Fig. 15.1b. The frequency-domain inductor model shown in Fig. 15.1b consists of an impedance sL and a voltage source Li(0−). The voltage across the impedance sL is given by Ohm’s law as sLI(s). Since the two-element com-bination in Fig. 15.1b is linear, every circuit analysis technique previously explored can be brought to bear in the s-domain as well. For example, it is possible to perform a source transformation on the model in order to obtain an impedance sL in parallel with a current source [−Li(0−)]/sL = −i(0−)/s. This can be veriﬁed by taking Eq. and solving for I(s): I(s) = V(s) + Li(0−) sL = V(s) sL + i(0−) s We are once again left with two terms. The ﬁrst term on the right is simply an admittance 1/sL times the voltage V(s). The second term on the right is a current, although it has units of ampere · seconds. Thus, we can model this equation with two separate components: an admittance 1/sL in parallel with a current source i(0−)/s; the resulting model is shown in Fig. 15.2. The choice of whether to use the model of Fig. 15.1b or that shown in Fig. 15.2 is usually made depending on which one will result in simpler equations. Note that although Fig. 15.2 shows the inductor symbol labeled with an admit-tance Y(s) = 1/sL, it can also be viewed as an impedance Z(s) = sL; again, the choice of which to use is generally based on personal preference. ■FIGURE 15.2 An alternative frequency-domain model for the inductor, consisting of an admittance 1/sL and a current source i (0−)/s. V(s) + – I(s) Y(s) = i(0–) s 1 sL A brief comment on units is in order. When we take the Laplace trans-form of a current i(t), we are integrating over time. Thus, the units of I(s) are technically ampere·seconds; in a similar fashion, the units of V(s) are volt·seconds. However, it is the convention to drop the seconds and assign I(s) the units of amperes, and to measure V(s) in volts. This convention does not present any problems until we scrutinize an equation such as Eq. , and see a term like i(0−)/s seemingly in conﬂict with the units of I(s) on the left-hand side. Although we will continue to measure these phasor quantities in “amperes” and “volts,” when checking the units of an equation to verify algebra, we must remember the seconds! CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 574 EXAMPLE 15.1 Calculate the voltage v(t) shown in Fig. 15.3a, given an initial current i(0−) 1 A. ■FIGURE 15.3 (a) A simple resistor-inductor circuit for which the voltage v (t) is desired. (b) The equivalent frequency-domain circuit, including the initial current in the inductor through the use of a series voltage source −Li(0−). We begin by ﬁrst converting the circuit in Fig. 15.3a to its frequency-domain equivalent, shown in Fig. 15.3b; the inductor has been replaced with a two-component model: an impedance sL = 2s , and an inde-pendent voltage source −Li(0−) = −2 V. We seek the quantity labeled V(s), as its inverse transform will result in v(t). Note that V(s) appears across the entire inductor model, and not just the impedance component. Taking a straightforward route, we write I(s) = 3 s + 8 + 2 1 + 2s = s + 9.5 (s + 8)(s + 0.5) and V(s) = 2s I(s) −2 so that V(s) = 2s(s + 9.5) (s + 8)(s + 0.5) −2 Before attempting to take the inverse Laplace transform of this ex-pression, it is well worth a little time and effort to simplify it ﬁrst. Thus, V(s) = 2s −8 (s + 8)(s + 0.5) Employing the technique of partial-fraction expansion (on paper or with the assistance of MATLAB), we ﬁnd that V(s) = 3.2 s + 8 − 1.2 s + 0.5 + – 1 2 H 3e–8tu(t) volts v(t) + – i(t) (a) (b) + – + – V(s) + – 1 I(s) V –2 V 2s 3 s + 8 SECTION 15.1 Z(s) AND Y(s) 575 Modeling Capacitors in the s-Domain The same concepts apply to capacitors in the s-domain as well. Following the passive sign convention as illustrated in Fig. 15.5a, the governing equa-tion for capacitors is i = C dv dt ■FIGURE 15.4 + – 12 4u(t) V 3 H i(0–) = –4 A i(t) ■FIGURE 15.5 (a) Capacitor in the time domain, with v (t) and i (t) labeled. (b) Frequency-domain model of a capacitor with initial voltage v (0−). (c) An equivalent model obtained by performing a source transformation. v(t) + – i(t) (a) C V(s) + – I(s) (b) Cv(0–) Y(s) = sC V(s) + – I(s) (c) v(0–) s Z(s) = 1 sC + – Referring to Table 14.1, then, the inverse transform is found to be v(t) = [3.2e−8t −1.2e−0.5t]u(t) volts PRACTICE ● 15.1 Determine the current i(t) in the circuit of Fig. 15.4. Ans: 1 3[1 −13e−4t]u(t) A. Taking the Laplace transform of both sides results in I(s) = C[sV(s) −v(0−)] or I(s) = sCV(s) −Cv(0−) which can be modeled as an admittance sC in parallel with a current source Cv(0−) as shown in Fig. 15.5b. Performing a source transformation on this circuit (taking care to follow the passive sign convention) results in an equivalent model for the capacitor consisting of an impedance 1/sC in series with a voltage source v(0−)/s, as shown in Fig. 15.5c. In working with these s-domain equivalents, we should be careful not to be confused with the independent sources being used to include initial conditions. The initial condition for an inductor is given as i(0−); this term may appear as part of either a voltage source or a current source, depending on which model is chosen. The initial condition for a capacitor is given as v(0−); this term may thus appear as part of either a voltage source or a current source.Avery common mistake for students working with s-domain analysis for the ﬁrst time is to always use v(0−) for the voltage source component of the model, even when dealing with an inductor. CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 576 EXAMPLE 15.2 Determine vC(t) in the circuit of Fig. 15.6a, given an initial voltage vC(0−) −2 V. Identify the goal of the problem. An expression for the capacitor voltage, vC(t). Collect the known information. The problem speciﬁes an initial capacitor voltage of −2 V. Devise a plan. Our ﬁrst step is to draw the equivalent s-domain circuit; in doing so, we must choose between the two possible capacitor models. With no clear beneﬁt in choosing one over the other, we select the current-source-based model, as in Fig. 15.6b. Construct an appropriate set of equations. We proceed with the analysis by writing a single nodal equation: −1 = VC 2/s + VC −9/s 3 Determine if additional information is required. We have one equation in one unknown, the frequency-domain representation of the desired capacitor voltage. Attempt a solution. Solving for VC, we ﬁnd that VC = 18/s −6 3s + 2 = −2 s −3 s(s + 2/3) Partial fraction expansion yields VC = 9 s − 11 s + 2/3 We obtain vC(t) by taking the inverse Laplace transform of this expression, resulting in vC(t) = 9u(t) −11e−2t/3u(t) V or, more compactly, vC(t) = [9 −11e−2t/3]u(t) V Verify the solution. Is it reasonable or expected? Aquick check for t = 0 yields vC(t) = −2 V, as it should based on our knowledge of the initial condition.Also, as t →∞, vC(t) →9 V, as we would expect from Fig. 15.6a once the transient has died out. ■FIGURE 15.6 (a) A circuit for which the current vC (t) is required. (b) Frequency-domain equivalent circuit, employing the current-source based model to account for the initial condition of the capacitor. + – 3 –1 A V VC(s) + – (b) 2 s 9 s + – 3 9u(t) V 0.5 F vC(t) + – (a) SECTION 15.1 Z(s) AND Y(s) 577 The results of this section are summarized in Table 15.1. Note that in each case, we have assumed the passive sign convention. PRACTICE ● 15.2 Repeat Example 15.2 using the voltage-source-based capacitor model. Ans: [9 −11e−2t/3]u(t) V. TABLE 15.1 Summary of Element Representations in ● the Time and Frequency Domains Time Domain Frequency Domain V(s) + – I(s) Z(s) = R V(s) = RI(s) V(s) + – I(s) Y(s) = I(s) = V(s) 1 R 1 R v(t) + – i(t) L v(t) = L di dt V(s) + – I(s) Z(s) = sL V(s) = sLI(s) –Li(0–) + – –Li(0–) I(s) = + V(s) + – I(s) Y(s) = i(0–) s 1 sL V(s) sL i(0–) s V(s) + – I(s) Z(s) = V(s) = + + – I(s) sC v(0–) s v(0–) s 1 sC Resistor Inductor Capacitor v(t) + – i(t) R v(t) = Ri(t) I(s) = sCV(s) – Cv(0–) V(s) + – I(s) Y(s) = sC Cv(0–) v(t) + – i(t) i(t) = C dv dt C 15.2 • NODAL AND MESH ANALYSIS IN THE s-DOMAIN In Chap. 10, we learned how to transform time-domain circuits driven by sinusoidal sources into their frequency-domain equivalents. The beneﬁts of this transformation were immediately evident, as we were no longer required to solve integrodifferential equations. Nodal and mesh analysis of such circuits (restricted to determining only the steady-state response) resulted in algebraic expressions in terms of jω, ω being the frequency of the sources. We have now seen that the concept of impedance can be extended to the more general case of complex frequency (s = σ + jω). Once we transform circuits from the time domain into the frequency domain, performing nodal or mesh analysis will once again result in purely algebraic expressions, this time in terms of the complex frequency s. Solution of the resulting equa-tions requires the use of variable substitution, Cramer’s rule, or software capable of symbolic algebra manipulation (e.g., MATLAB). In this section, we present two examples of reasonable complexity so that we may examine these issues in greater detail. First, however, we consider how MATLAB can be used to assist us in such endeavors. CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 578 COMPUTER-AIDED ANALYSIS In Chap. 14, we saw that MATLAB can be used to determine the residues of rational functions in the s-domain, making the inverse Laplace transform process signiﬁcantly easier. However, the software package is actually much more powerful, having numerous built-in routines for manipulation of algebraic expressions. In fact, as we will see in this example, MATLAB is even capable of performing inverse Laplace transforms directly from the rational functions we obtain through circuit analysis. Let’s begin by seeing how MATLAB can be used to work with alge-braic expressions. These expressions are stored as character strings, with the apostrophe symbol (‘) used in the deﬁning expression. For ex-ample, we previously represented the polynomial p(s) = s3 −12s + 6 as a vector: EDU» p [1 0 −12 6]; However, we can also represent it symbolically: EDU» p ‘sˆ3 −12s + 6’; These two representations are not equal in MATLAB; they are two distinct concepts. When we wish to manipulate an algebraic expression symbolically, the second representation is necessary. This ability is especially useful in working with simultaneous equations. Consider the set of equations (3s + 10)I1 −10I2 = 4 s + 2 −10I1 + (4s + 10)I2 = − 2 s + 1 SECTION 15.2 NODAL AND MESH ANALYSIS IN THE s-DOMAIN 579 Using MATLAB’s symbolic notation, we deﬁne two string variables: EDU» eqn1 ‘(3s+10)I1 −10I2 4/(s+2)’; EDU» eqn2 ‘−10I1 + (4s+10)I2 −2/(s+1)’; Note that the entire equation has been included in each string; our goal is to solve the two equations for the variables I1 and I2. MATLAB provides a special routine, solve(), that can manipulate the equations for us. It is invoked by listing the separate equations (deﬁned as strings), followed by a list of the unknowns (also deﬁned as strings): EDU» solution solve(eqn1, eqn2, ‘I1’, ‘I2’); The answer is stored in the variable solution, although in a somewhat unexpected form. MATLAB returns the answer in what is termed a struc-ture, a construct that will be familiar to C programmers.At this stage, however, all we need to know is how to extract our answer. If we type EDU» I1 solution.I1 we obtain the response I1 2(4s+9)/(s+1)/(6sˆ2+47s+70) indicating that an s-polynomial expression has been assigned to the variable I1; a similar operation is used for the variable I2. We can now proceed directly to determining the inverse Laplace transform using the function ilaplace(): EDU» i1 ilaplace(I1) i1 10/29exp(−t)−172/667exp(−35/6t)−2/23exp(−2t) In this manner, we can quickly obtain the solution to simultaneous equations resulting from nodal or mesh analysis, and also obtain the inverse Laplace transforms. The command ezplot(i1) allows us to see what the solution looks like, if we’re so inclined. It should be noted that complicated expressions sometimes may confuse MATLAB; in such situations, ilaplace() may not return a useful answer. It is worth mentioning a few related functions, as they can also be used to quickly check answers obtained by hand. The function numden() converts a rational function into two separate variables: one containing the numerator, and the other containing the denominator. For example, EDU» [N,D] numden(I1) returns two algebraic expressions stored in N and D, respectively: N 8s+18 D (s+1)(6sˆ2+47s+70) (Continued on next page) CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 580 In order to apply our previous experience with the function residue(), we need to convert each symbolic (string) expression into a vector containing the coefﬁcients of the polynomial. This is achieved using the command sym2poly(): EDU» n sym2poly(N); and EDU» d sym2poly(D) d 6 53 117 70 after which we can determine the residues: EDU» [r p y] residue(n,d) r p y −0.2579 −5.8333 [ ] −0.0870 −2.0000 0.3448 −1.0000 which is in agreement with what we obtained using ilaplace(). With these new MATLAB skills, (or a deep-seated desire to try an alter-native approach such as Cramer’s rule or direct substitution), we are ready to proceed to analyze a few circuits. EXAMPLE 15.3 Determine the two mesh currents i1 and i2 in the circuit of Fig. 15.7a. There is no energy initially stored in the circuit. + – + – 4e–2t u(t) V 2e–t u(t) V 4 H 10 (a) i2(t) i1(t) F 1 3 + – + – V 3/s 4s 10 (b) I2(s) I1(s) 2 s + 1 V 4 s + 2 ■FIGURE 15.7 (a) A two-mesh circuit for which the individual mesh currents are desired. (b) The frequency-domain equivalent circuit. As always, our ﬁrst step is to draw the appropriate frequency-domain equivalent circuit. Since we have zero energy stored in the circuit at SECTION 15.2 NODAL AND MESH ANALYSIS IN THE s-DOMAIN 581 + – + – 3u(t) V 2u(t) V F 1 H 3 i2(t) i1(t) 1 4 ■FIGURE 15.8 Ans: i1 = e−2t/3 cos 4 3 √ 2t + √ 2/8 e−2t/3 sin 4 3 √ 2t A; i2 = −2 3 + 2 3e−2t/3 cos 4 3 √ 2t + 13 √ 2/24 e−2t/3 sin 4 3 √ 2t A. We were (indirectly) told that no current ﬂows through the inductor at t = 0−. Therefore, i 2(0−) = 0, and consequently i 2(0+) must be zero as well. Does this result hold true for our answer? t = 0−, we replace the 1 3 F capacitor with a 3/s impedance, and the 4 H inductor with a 4s impedance, as shown in Fig. 15.7b. Next, we write two mesh equations just as we have before: − 4 s + 2 + 3 s I1 + 10I1 −10I2 = 0 or 3 s + 10 I1 −10I2 = 4 s + 2 (mesh 1) and − 2 s + 1 + 10I2 −10I1 + 4sI2 = 0 or −10I1 + (4s + 10)I2 = 2 s + 1 (mesh 2) Solving for I1 and I2, we ﬁnd that I1 = 2s(4s2 + 19s + 20) 20s4 + 66s3 + 73s2 + 57s + 30 A and I2 = 30s2 + 43s + 6 (s + 2)(20s3 + 26s2 + 21s + 15) A All that remains is for us to take the inverse Laplace transform of each function, after which we ﬁnd that i1(t) = −96.39e−2t −344.8e−t + 841.2e−0.15t cos 0.8529t + 197.7e−0.15t sin 0.8529t mA and i2(t) = −481.9e−2t −241.4e−t + 723.3e−0.15t cos 0.8529t + 472.8e−0.15t sin 0.8529t mA PRACTICE ● 15.3 Find the mesh currents i1 and i2 in the circuit of Fig. 15.8. You may assume no energy is stored in the circuit at t = 0−. CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 582 EXAMPLE 15.4 Calculate the voltage vx in the circuit of Fig. 15.9 using nodal analysis techniques. 4 H + – + – 2 + 5u(t) V 4u(t) V 1 F 1 2 vx Ref ■FIGURE 15.9 A simple four-node circuit containing two energy storage elements. + – + – 4s 1 1 A Ref V 7 s 2 s V 4 s Vx ■FIGURE 15.10 The s-domain equivalent circuit of Fig. 15.9. The ﬁrst step is to draw the corresponding s-domain circuit. We see that the 1 2 F capacitor has an initial voltage of 2 V across it at t = 0−, requiring that we employ one of the two models of Fig. 15.5. Since we are to use nodal analysis, perhaps the model of Fig. 15.5b is the better route. The resulting circuit is shown in Fig. 15.10. With two of the three nodal voltages speciﬁed, we have only one nodal equation to write: −1 = Vx −7/s 2/s + Vx + Vx −4/s 4s so that Vx = 10s2 + 4 s(2s2 + 4s + 1) = 5s2 + 2 s s + 1 + √ 2 2 s + 1 − √ 2 2 The nodal voltage vx is found by taking the inverse Laplace transform, and we ﬁnd that vx = [4 + 6.864e−1.707t −5.864e−0.2929t]u(t) or vx = 4 −e−t 9 √ 2 sinh √ 2 2 t −cosh √ 2 2 t u(t) Is our answer correct? One way to check is to evaluate the capacitor voltage at t = 0, since we know it to be 2 V. Thus, VC = 7 s −Vx = 4s2 + 28s + 3 s(2s2 + 4s + 1) Multiplying VC by s and taking the limit as s →∞, we ﬁnd that vc(0+) = lim s→∞ 4s2 + 28s + 3 2s2 + 4s + 1 = 2 V as expected. SECTION 15.2 NODAL AND MESH ANALYSIS IN THE s-DOMAIN 583 + – + – 1 + 4u(t) V 1 + 4u(t) V Ref 4 H 1 F 1 2 vx ■FIGURE 15.11 For Practice Problem 15.4. PRACTICE ● 15.4 Employnodalanalysistocalculatevx(t)forthecircuitof Fig.15.11. Ans: [5 + 5.657(e−1.707t −e−0.2929t)]u(t). EXAMPLE 15.5 Use nodal analysis to determine the voltages v1, v2, and v3 in the circuit of Fig. 15.12a. No energy is stored in the circuit at t 0−. 100 0.1e–3t u(t) amperes 6 H F 0.2v2(t) v1(t) v2(t) v3(t) (a) F 1 7 1 2 100 A 6s 7/s 2/s 0.2V2 V1 V2 V3 (b) 0.1 s + 3 ■FIGURE 15.12 (a) A four-node circuit containing two capacitors and one inductor, none of which are storing energy at t = 0−. (b) The frequency-domain equivalent circuit. This circuit consists of three separate energy storage elements, none of which is storing any energy at t = 0−. Thus, each may be replaced by their corresponding impedance as shown in Fig. 15.12b. We also note the presence of a dependent current source controlled by the nodal voltage v2(t). Beginning at node 1, we write the following equation: 0.1 s + 3 = V1 −V2 100 (Continued on next page) CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 584 8 H 10u(t) A 3u(t) A 2 v1(t) v2(t) v3(t) 3 H ■FIGURE 15.13 or 10 s + 3 = V1 −V2 (node 1) and at node 2, 0 = V2 −V1 100 + V2 7/s + V2 −V3 6s or −42sV1 + (600s2 + 42s + 700)V2 −700V3 = 0 (node 2) and ﬁnally, at node 3, −0.2V2 = V3 −V2 6s + V3 2/s or (1.2s −1)V2 + (3s2 + 1)V3 = 0 Solving this set of equations for the nodal voltages, we obtain V1 = 3 100s3 + 7s2 + 150s + 49 (s + 3)(30s3 + 45s + 14) V2 = 7 3s2 + 1 (s + 3)(30s3 + 45s + 14) V3 = −1.4 6s −5 (s + 3)(30s3 + 45s + 14) The only remaining step is to take the inverse Laplace transform of each voltage, so that, for t > 0, v1(t) = 9.789e−3t + 0.06173e−0.2941t + 0.1488e0.1471t cos(1.251t) + 0.05172e0.1471t sin(1.251t) V v2(t) = −0.2105e−3t + 0.06173e−0.2941t + 0.1488e0.1471t cos(1.251t) + 0.05172e0.1471t sin(1.251t) V v3(t) = −0.03459e−3t + 0.06631e−0.2941t −0.03172e0.1471t cos(1.251t) −0.06362e0.1471t sin(1.251t) V Note that the response grows exponentially as a result of the action of the dependent current source. In essence, the circuit is “running away,” indicating that at some point a component will melt, explode, or fail in some related fashion. Although analyzing such circuits can evidently entail a great deal of work, the advantages to s-domain techniques are clear once we contemplate performing the analysis in the time domain! PRACTICE ● 15.5 Use nodal analysis to determine the voltages v1, v2, and v3 in the circuit of Fig. 15.13. Assume there is zero energy stored in the inductors at t = 0−. Ans: v1(t) = −30δ(t) −14u(t) V; v2(t) = −14u(t) V; v3(t) = 24δ(t) −14u(t) V. SECTION 15.3 ADDITIONAL CIRCUIT ANALYSIS TECHNIQUES 585 15.3 • ADDITIONAL CIRCUIT ANALYSIS TECHNIQUES Depending on the speciﬁc goal in analyzing a particular circuit, we often find that we can simplify our task by carefully choosing our analysis technique. For example, it is seldom desirable to apply superposition to a circuit containing 215 independent sources, as such an approach requires analysis of 215 separate circuits! By treating passive elements such as ca-pacitors and inductors as impedances, however, we are free to employ any of the circuit analysis techniques studied in Chaps. 3, 4, and 5 to circuits that have been transformed to their s-domain equivalents. Thus, superposition, source transformations, Thévenin’s theorem, and Norton’s theorem all apply in the s-domain. EXAMPLE 15.6 Simplify the circuit of Fig. 15.14a using source transformations, and determine an expression for the voltage v(t). With no initial currents or voltages speciﬁed, and a u(t) multiplying the voltage source, we conclude that there is no energy initially stored in the circuit. Thus, we draw the frequency-domain circuit as shown in Fig. 15.14b. Our strategy will be to perform several source transformations in succession in order to combine the two 2/s impedances and the 10 resistor; we must leave the 9s impedance alone as the desired quan-tity V(s) appears across its terminals. We may now transform the voltage source and the leftmost 2/s impedance into a current source I(s) = 2s s2 + 9 s 2 = s2 s2 + 9 A in parallel with a 2/s impedance. As depicted in Fig. 15.15a, after this transformation, we have Z1 ≡ (2/s)∥10 = 20/(10s + 2) facing the current source. Performing an-other source transformation, we obtain a voltage source V2(s)such that V2(s) = s2 s2 + 9 20 10s + 2 (a) 10 2 cos 3t u(t) volts 0.5 F 0.5 F v(t) + – + – 9 H (b) 10 V 2/s 2/s V(s) + – 9s 2s s2 + 9 + – ■FIGURE 15.14 (a) Circuit to be simpliﬁed using source transformations. (b) Frequency-domain representation. (Continued on next page) V2 + – (b) Z2 V(s) + – 9s A V(s) + – (a) 9s s2 s2 + 9 Z1 2 s ■FIGURE 15.15 (a) Circuit after ﬁrst source transformation. (b) Final circuit to be analyzed for V(s). This voltage source is in series with Z1 and also with the remaining 2/s impedance; combining Z1 and 2/s into a new impedance Z2 yields Z2 = 20 10s + 2 + 2 s = 40s + 4 s(10s + 2) CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 586 + – 7 5u(t) V 3 H 6 H 0.25 F A B ■FIGURE 15.16 The resulting circuit is shown in Fig. 15.15b. At this stage, we are now ready to obtain an expression for the voltage V(s) using simple voltage division: V(s) = s2 s2 + 9 20 10s + 2 9s 9s + 40s + 4 s(10s + 2) = 180s4 (s2 + 9)(90s3 + 18s2 + 40s + 4) Both terms in the denominator possess complex roots. Employing MATLAB to expand the denominator and then determine the residues, EDU» d1 ’sˆ2 + 9’; EDU» d2 ’90sˆ3 + 18sˆ2 + 40s + 4’; EDU» d symmul(d1,d2); EDU» denominator expand(d); EDU» den sym2poly(denominator); EDU» num [180 0 0 0 0]; EDU» [r p y] residue(num,den); we ﬁnd V(s) =1.047 + j0.0716 s −j3 + 1.047 −j0.0716 s + j3 − 0.0471 + j0.0191 s + 0.04885 −j0.6573 − 0.0471 −j0.0191 s + 0.04885 + j0.6573 + 5.590 × 10−5 s + 0.1023 Taking the inverse transform of each term, writing 1.047 + j0.0716 as 1.049e j3.912◦and 0.0471 + j0.0191 as 0.05083e j157.9◦results in v(t) =1.049e j3.912◦e j3tu(t) + 1.049e−j3.912◦e−j3tu(t) + 0.05083e−j157.9◦e−0.04885te−j0.6573tu(t) + 0.05083e+ j157.9◦e−0.04885te+ j0.6573tu(t) + 5.590 × 10−5e−0.1023tu(t) Converting the complex exponentials to sinusoids then allows us to write a slightly simpliﬁed expression for our voltage v(t) =[5.590 × 10−5e−0.1023t + 2.098 cos(3t + 3.912◦) + 0.1017e−0.04885t cos(0.6573t + 157.9◦)]u(t) V PRACTICE ● 15.6 Using the method of source transformation, reduce the circuit of Fig. 15.16 to a single s-domain current source in parallel with a single impedance. Ans: Is = 35 s2(18s + 63) A, Zs = 72s2 + 252s 18s3 + 63s2 + 12s + 28 . Note that each term having a complex pole has a companion term that is its complex conjugate. For any physical system, complex poles will always occur in conjugate pairs. SECTION 15.3 ADDITIONAL CIRCUIT ANALYSIS TECHNIQUES 587 EXAMPLE 15.7 Find the frequency-domain Thévenin equivalent of the highlighted network shown in Fig. 15.17a. RE RL RC gv r C C vs Rs v – + vo + – (a) + – RE 1 A RL RC gV r 1/sC 1/sC V + – Vo + – (b) Vin + – ■FIGURE 15.17 (a) An equivalent circuit for the “common base” transistor ampliﬁer. (b) The frequency-domain equivalent circuit with a 1 A test source substituted for the input source represented by vs and R s . We are being asked to determine the Thévenin equivalent of the circuit connected to the input device; this quantity is often referred to as the input impedance of the ampliﬁer circuit. After converting the circuit to its frequency-domain equivalent, we replace the input device (vs and Rs) with a 1 A “test” source, as shown in Fig. 15.17b. The input imped-ance Zin is then Zin = Vin 1 or simply Vin. We must ﬁnd an expression for this quantity in terms of the 1 A source, resistors and capacitors, and/or the dependent source parameter g. Writing a single nodal equation at the input, then, we ﬁnd that 1 + gVπ = Vin Zeq where Zeq ≡RE 1 sCπ rπ = RErπ rπ + RE + sRErπCπ Since Vπ = −Vin, we ﬁnd that Zin = Vin = RErπ rπ + RE + sRErπCπ + gRErπ This particular circuit is known as the “hybrid π” model for a special type of single-transistor circuit known as the common base ampliﬁer. The two capacitors, Cπ and Cμ, represent capacitances internal to the transistor, and are typically on the order of a few pF. The resistor RL in the circuit represents the Thévenin equivalent resistance of the output device, which could be a speaker or even a semiconductor laser. The voltage source vs and the resistor R s together represent the Thévenin equivalent of the input device, which may be a microphone, a light-sensitive resistor, or possibly a radio antenna. 15.4 • POLES, ZEROS, AND TRANSFER FUNCTIONS In this section, we revisit terminology ﬁrst introduced in Chap. 14, namely, poles, zeros, and transfer functions. Consider the simple circuit in Fig. 15.19a. The s-domain equivalent is given in Fig. 15.19b, and nodal analysis yields 0 = Vout 1/sC + Vout −Vin R Rearranging and solving for Vout, we ﬁnd Vout = Vin 1 + sRC or H(s) ≡Vout Vin = 1 1 + sRC where H(s) is the transfer function of the circuit, deﬁned as the ratio of the output to the input. We could just as easily specify a particular current as either the input or output quantity, leading to a different transfer function for the same circuit. Circuit schematics are typically read from left to right, so designers often place the input of a circuit on the left of the schematic and the output terminals on the right, at least to the extent where it is possible. The concept of a transfer function is very important, both in terms of cir-cuit analysis as well as in other areas of engineering. There are two reasons for this. First, once we know the transfer function of a particular circuit, we can easily ﬁnd the output that results from any input. All we need to do is multiply H(s) by the input quantity, and take the inverse transform of the resulting expression. Second, the form of the transfer function contains a great deal of information about the behavior we might expect from a partic-ular circuit (or system). As alluded to in the Practical Application of Chap. 14, in order to eval-uate the stability of a system it is necessary to determine the poles and zeros of the transfer function H(s); we will explore this issue in detail shortly. Writing Eq. as H(s) = 1/RC s + 1/RC we see that the magnitude of this function approaches zero as s →∞. Thus, we say that H(s) has a zero at s = ∞. The function approaches inﬁnity at CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 588 PRACTICE ● 15.7 Working in the s-domain, ﬁnd the Norton equivalent connected to the 1 resistor in the circuit of Fig. 15.18. Ans: Isc = 3(s + 1)/4s A; Zth = 4/(s + 1) . + – 1 4 3u(t) V 0.25 F ■FIGURE 15.18 vin(t) + – vout(t) + – (a) R C Vin(s) + – Vout(s) + – (b) R sC 1 ■FIGURE 15.19 (a) A simple resistor-capacitor circuit, with an input voltage and output voltage speciﬁed. (b) The s-domain equivalent circuit. When computing magnitude, it is customary to consider +∞and −∞as being the same frequency. The phase angle of the response at very large positive and negative values of ω need not be the same, however. SECTION 15.5 CONVOLUTION 589 s = −1/RC; we therefore say that H(s) has a pole at s = −1/RC. These frequencies are termed critical frequencies, and their early identiﬁcation sim-pliﬁes the construction of the response curves we will develop in Sec. 15.7. 15.5 • CONVOLUTION The s-domain techniques we have developed up to this point are very use-ful in determining the current and voltage response of a particular circuit. However, in practice we are often faced with circuits to which arbitrary sources can be connected, and require an efﬁcient means of determining the new output each time. This is easily accomplished if we can characterize the basic circuit by a transfer function called the system function. The analysis can proceed in either the time domain or the frequency do-main, although it is generally more useful to work in the frequency domain. In such situations, we have a simple four-step process: By these means some relatively complicated integral expressions will be reduced to simple functions of s, and the mathematical operations of inte-gration and differentiation will be replaced by the simpler operations of algebraic multiplication and division. The Impulse Response Consider a linear electrical network N, without initial stored energy, to which a forcing function x(t) is applied. At some point in this circuit, a response function y(t) is present. We show this in block diagram form in Fig. 15.20a along with sketches of generic time functions. The forcing function is shown to exist only in the interval a < t < b. Thus, y(t) exists only for t > a. The question that we now wish to answer is this: “If we know the form of x(t), how is y(t) described?” To answer this question, we need to know something about N, such as its response when the forcing function is a unit impulse δ(t). That is, we are assuming that we know h(t), the response function resulting when a unit impulse is supplied as the forcing function at t = 0, as shown in Fig. 15.20b. The function h(t) is commonly called the unit-impulse response function, or the impulse response. Based on our knowledge of Laplace transforms, we can view this from a slightly different perspective. Transforming x(t) into X(s) and y(t) into Y(s), we deﬁne the system transfer function H(s) as H(s) ≡Y(s) X(s) 1. Determine the circuit system function (if not already known); 2. Obtain the Laplace transform of the forcing function to be applied; 3. Multiply this transform and the system function; and ﬁnally 4. Perform an inverse transform operation on the product to ﬁnd the output. If x(t) = δ(t), then according to Table 14.1, X(s) = 1. Thus, H(s) = Y(s) and so in this instance h(t) = y(t). Instead of applying the unit impulse at time t = 0, let us now suppose that it is applied at time t = λ (lambda). We see that the only change in the output is a time delay. Thus, the output becomes h(t −λ) when the input is δ(t −λ), as shown in Fig. 15.20c. Next, suppose that the input impulse has some strength other than unity. Speciﬁcally, let the strength of the im-pulse be numerically equal to the value of x(t) when t = λ. This value x(λ) is a constant; we know that the multiplication of a single forcing function in a linear circuit by a constant simply causes the response to change propor-tionately. Thus, if the input is changed to x(λ)δ(t −λ), then the response becomes x(λ)h(t −λ), as shown in Fig. 15.20d. Now let us sum this latest input over all possible values of λ and use the result as a forcing function for N. Linearity decrees that the output must be equal to the sum of the responses resulting from the use of all possible val-ues of λ. Loosely speaking, the integral of the input produces the integral of the output, as shown in Fig. 15.20e. But what is the input now? Given the sifting property1 of the unit impulse, we see that the input is simply x(t), the original input. Thus, Fig. 15.20e may be represented as in Fig. 15.20f. CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 590 N y(t) x(t) b a t (a) x(t) b a t y(t) N h(t) (t) t (b) (1) x(t) = (t) t y(t) = h(t) N (c) h(t – ) (t – ) N (d) x() h(t – ) x() (t – ) N (e) x() h(t – ) d x() (t – ) d ∫ – ∫ – N ( f ) x() h(t – ) d = y(t) x(t) ∫ – ■FIGURE 15.20 A conceptual development of the convolution integral. (1) The sifting property of the impulse function, described in Section 14.5, states that ∞ −∞f (t)δ(t −t0) dt = f (t0). The Convolution Integral If the input to our system N is the forcing function x(t), we know the output must be the function y(t) as depicted in Fig. 15.20a. Thus, from Fig. 15.20f we conclude that y(t) = ∞ −∞ x(λ)h(t −λ) dλ where h(t) is the impulse response of N. This important relationship is known far and wide as the convolution integral. In words, this last equation states that the output is equal to the input convolved with the impulse response. It is often abbreviated by means of y(t) = x(t) ∗h(t) where the asterisk is read “convolved with.” Equation sometimes appears in a slightly different but equivalent form. If we let z = t −λ, then dλ = −dz, and the expression for y(t) becomes y(t) = −∞ ∞ −x(t −z)h(z) dz = ∞ −∞ x(t −z)h(z) dz and since the symbol that we use for the variable of integration is unimpor-tant, we can modify Eq. to write y(t) = x(t) ∗h(t) = ∞ −∞ x(z)h(t −z) dz = ∞ −∞ x(t −z)h(z) dz Convolution and Realizable Systems The result that we have in Eq. is very general; it applies to any linear system. However, we are usually interested in physically realizable systems, those that do exist or could exist, and such systems have a property that modiﬁes the convolution integral slightly. That is, the response of the system cannot begin before the forcing function is applied. In particular, h(t) is the response of the system resulting from the application of a unit impulse at t = 0. Therefore, h(t) cannot exist for t < 0. It follows that, in the second integral of Eq. , the integrand is zero when z < 0; in the ﬁrst integral, the integrand is zero when (t −z) is negative, or when z > t. Therefore, for realizable systems the limits of integration change in the convolution integrals: y(t) = x(t) ∗h(t) = t −∞ x(z)h(t −z) dz = ∞ 0 x(t −z)h(z) dz SECTION 15.5 CONVOLUTION 591 Be careful not to confuse this new notation with multiplication! Equations and are both valid, but the latter is more speciﬁc when we are speaking of realizable linear systems, and well worth memorizing. Graphical Method of Convolution Before discussing the signiﬁcance of the impulse response of a circuit any further, let us consider a numerical example that will give us some insight into just how the convolution integral can be evaluated. Although the ex-pression itself is simple enough, the evaluation is sometimes troublesome, especially with regard to the values used as the limits of integration. Suppose that the input is a rectangular voltage pulse that starts at t = 0, has a duration of 1 second, and is 1 V in amplitude: x(t) = vi(t) = u(t) −u(t −1) Suppose also that this voltage pulse is applied to a circuit whose impulse response is known to be an exponential function of the form: h(t) = 2e−tu(t) We wish to evaluate the output voltage vo(t), and we can write the answer immediately in integral form, y(t) = vo(t) = vi(t) ∗h(t) = ∞ 0 vi(t −z)h(z) dz = ∞ 0 [u(t −z) −u(t −z −1)][2e−zu(z)] dz Obtaining this expression for vo(t) is simple enough, but the presence of the many unit-step functions tends to make its evaluation confusing and possi-bly even a little obnoxious. Careful attention must be paid to the determina-tion of those portions of the range of integration in which the integrand is zero. Let us use some graphical assistance to help us understand what the con-volution integral says. We begin by drawing several z axes lined up one above the other, as shown in Fig. 15.21. We know what vi(t) looks like, and so we know what vi(z) looks like also; this is plotted as Fig. 15.21a. The function vi(−z) is simply vi(z) run backward with respect to z, or rotated about the ordinate axis; it is shown in Fig. 15.21b. Next we wish to repre-sent vi(t −z), which is vi(−z) after it is shifted to the right by an amount z = t as shown in Fig. 15.21c. On the next z axis, in Fig. 15.21d, our impulse response h(z) = 2e−zu(z) is plotted. The next step is to multiply the two functions vi(t −z) and h(z); the re-sult for an arbitrary value of t < 1 is shown in Fig. 15.21e. We are after a value for the output vo(t), which is given by the area under the product curve (shown shaded in the ﬁgure). First consider t < 0. There is no overlap between vi(t −z) and h(z), so vo = 0. As we increase t, we slide the pulse shown in Fig. 15.21c to the right, leading to an overlap with h(z) once t > 0. The area under the corre-sponding curve of Fig. 15.21e continues to increase as we increase the value of t until we reach t = 1. As t increases above this value, a gap opens up between z = 0 and the leading edge of the pulse, as shown in Fig. 15.21f. As a result, the overlap with h(z) decreases. CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 592 1 1 z (a) vi (z) 1 –1 z (b) vi(–z) 1 t – 1 t z vi(t – z) (c) 2 1 z (d) h(z) 2 1 1 2 3 z (e) vi(t – z)h(z) t Area = vi(t – z) h(z) dz = vo(t) ∫ 0 t – 1 t vi(t – z) 0 2 1 1 z ( f ) ■FIGURE 15.21 Graphical concepts in evaluating a convolution integral. SECTION 15.5 CONVOLUTION 593 In other words, for values of t that lie between zero and unity, we must integrate from z = 0 to z = t; for values of t that exceed unity, the range of integration is t −1 < z < t. Thus, we may write vo(t) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ 0 t < 0 t 0 2e−z dz = 2(1 −e−t) 0 ≤t ≤1 t t−1 2e−z dz = 2(e −1)e−t t > 1 This function is shown plotted versus the time variable t in Fig. 15.22, and our solution is completed. 2 1 t vo(t) 1 0 2 3 ■FIGURE 15.22 The output function vo obtained by graphical convolution. EXAMPLE 15.8 Apply a unit-step function, x(t) u(t), as the input to a system whose impulse response is h(t) u(t) −2u(t −1) + u(t −2), and determine the corresponding output y(t) x(t) h(t). Our ﬁrst step is to plot both x(t) and h(t), as shown in Fig. 15.23. 1 0 x(t) t (a) (b) 0 1 2 1 h(t) t –1 ■FIGURE 15.23 Sketches of (a) the input signal x(t) = u(t) and (b) the unit-impulse response h(t) = u(t) −2u(t −1) + u(t −2), for a linear system. We arbitrarily choose to evaluate the ﬁrst integral of Eq. , y(t) = t −∞ x(z)h(t −z) dz and prepare a sequence of sketches to help select the correct limits of integration. Figure 15.24 shows these functions in order: the input x(z) as a function of z; the impulse response h(z); the curve of h(−z), (Continued on next page) CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 594 1 0 x(z) z (a) 1 0 1 2 h(z) z (b) –1 1 0 –1 –2 h(–z) z (c) –1 1 t – 2 t – 1 h(t – z) z (d) –1 t ■FIGURE 15.24 (a) The input signal and (b) the impulse response are plotted as functions of z. (c) h(−z) is obtained by ﬂipping h(z) about the vertical axis, and (d) h(t −z) results when h(−z) is slid t units to the right. It is now possible to visualize the product of the ﬁrst graph, x(z), and the last, h(t −z), for the various ranges of t. When t is less than zero, there is no overlap, and y(t) = 0 t < 0 For the case sketched in Fig. 15.24d, h(t −z) has a nonzero overlap with x(z) from z = 0 to z = t, and each is unity in value. Thus, y(t) = t 0 (1 × 1) dz = t 0 < t < 1 When t lies between 1 and 2, h(t −z) has slid far enough to the right to bring under the step function that part of the negative square wave extending from z = 0 to z = t −1. We then have y(t) = t−1 0 [1 × (−1)] dz + t t−1 (1 × 1) dz = −z z=t−1 z=0 + z z=t z=t−1 Therefore, y(t) = −(t −1) + t −(t −1) = 2 −t 1 < t < 2 Finally, when t is greater than 2, h(t −z) has slid far enough to the right that it lies entirely to the right of z = 0. The intersection with the unit step is complete, and y(t) = t−1 t−2 [1 × (−1)] dz + t t−1 (1 × 1) dz = −z z=t−1 z=t−2 + z z=t z=t−1 which is just h(z) rotated about the vertical axis; and h(t −z), obtained by sliding h(−z) to the right t units. For this sketch, we have selected t in the range 0 < t < 1. SECTION 15.5 CONVOLUTION 595 Convolution and the Laplace Transform Convolution has applications in a wide variety of disciplines beyond linear circuit analysis, including image processing, communications, and semi-conductor transport theory. It is often helpful therefore to have a graphical intuition of the basic process, even if the integral expressions of Eqs. and are not always the best solution route. One powerful alternative approach makes use of properties of the Laplace transform—hence our in-troduction to convolution in this chapter. Let F1(s) and F2(s) be the Laplace transforms of f1(t) and f2(t), respec-tively, and consider the Laplace transform of f1(t) ∗f2(t), { f1(t) ∗f2(t)} = ∞ −∞ f1(λ) f2(t −λ) dλ One of these time functions will typically be the forcing function that is applied at the input terminals of a linear circuit, and the other will be the unit-impulse response of the circuit. Since we are now dealing with time functions that do not exist prior to t = 0−(the deﬁnition of the Laplace transform forces us to assume this), the lower limit of integration can be changed to 0−. Then, using the deﬁnition of the Laplace transform, we get { f1(t) ∗f2(t)} = ∞ 0−e−st ∞ 0− f1(λ) f2(t −λ) dλ dt Since e−st does not depend upon λ, we can move this factor inside the inner integral. If we do this and also reverse the order of integration, the result is { f1(t) ∗f2(t)} = ∞ 0− ∞ 0−e−st f1(λ) f2(t −λ) dt dλ Continuing with the same type of trickery, we note that f1(λ) does not depend upon t, and so it can be moved outside the inner integral: { f1(t) ∗f2(t)} = ∞ 0− f1(λ) ∞ 0−e−st f2(t −λ) dt dλ 1 0 1 2 y(t) t ■FIGURE 15.25 The result of convolving x (t) and h(t) as shown in Fig. 15.23. or y(t) = −(t −1) + (t −2) + t −(t −1) = 0 t > 2 These four segments of y(t) are collected as a continuous curve in Fig. 15.25. PRACTICE ● 15.8 Repeat Example 15.8 using the second integral of Eq. . 15.9 The impulse response of a network is given by h(t) = 5u(t −1). If an input signal x(t) = 2[u(t) −u(t −3)] is applied, determine the output y(t) at t equal to (a) −0.5; (b) 0.5; (c) 2.5; (d) 3.5. Ans: 15.9: 0, 0, 15, 25. We then make the substitution x = t −λ in the bracketed integral (where we may treat λ as a constant): { f1(t) ∗f2(t)} = ∞ 0− f1(λ) ∞ −λ e−s(x+λ) f2(x) dx dλ = ∞ 0− f1(λ)e−sλ ∞ −λ e−sx f2(x) dx dλ = ∞ 0− f1(λ)e−sλ[F2(s)] dλ = F2(s) ∞ 0− f1(λ)e−sλdλ Since the remaining integral is simply F1(s), we ﬁnd that { f1(t) ∗f2(t)} = F1(s) · F2(s) Stated slightly differently, we may conclude that the inverse transform of the product of two transforms is the convolution of the individual inverse transforms, a result that is sometimes useful in obtaining inverse transforms. CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 596 EXAMPLE 15.9 Find v(t) by applying convolution techniques, given that V(s) = 1/[(s + α) (s + β)]. We obtained the inverse transform of this particular function in Sec. 14.5 using a partial-fraction expansion. We now identify V(s) as the product of two transforms, V1(s) = 1 s + α and V2(s) = 1 s + β where v1(t) = e−αtu(t) and v2(t) = e−βtu(t) The desired v(t) can be expressed as v(t) = −1{V1(s)V2(s)} = v1(t) ∗v2(t) = ∞ 0−v1(λ)v2(t −λ) dλ = ∞ 0−e−αλu(λ)e−β(t−λ)u(t −λ) dλ = t 0−e−αλe−βteβλ dλ = e−βt t 0−e(β−α)λ dλ = e−βt e(β−α)t −1 β −α u(t) SECTION 15.5 CONVOLUTION 597 Further Comments on Transfer Functions As we have noted several times before, the output vo(t) at some point in a linear circuit can be obtained by convolving the input vi(t) with the unit-impulse response h(t). However, we must remember that the impulse response results from the application of a unit impulse at t = 0 with all initial conditions zero. Under these conditions, the Laplace transform of vo(t) is {vo(t)} = Vo(s) = {vi(t) ∗h(t)} = Vi(s)[{h(t)}] Thus, the ratio Vo(s)/Vi(s) is equal to the transform of the impulse re-sponse, which we shall denote by H(s), {h(t)} = H(s) = Vo(s) Vi(s) From Eq. we see that the impulse response and the transfer function make up a Laplace transform pair, h(t) ⇔H(s) This is an important fact that we shall explore further in Sec. 15.7, after becoming familiar with the concept of pole-zero plots and the complex-frequency plane. At this point, however, we are already able to exploit this new concept of convolution for circuit analysis. or, more compactly, v(t) = 1 β −α (e−αt −e−βt)u(t) which is the same result that we obtained before using partial-fraction expansion. Note that it is necessary to insert the unit step u(t) in the result because all (one-sided) Laplace transforms are valid only for nonnegative time. PRACTICE ● 15.10 Repeat Example 15.8, performing the convolution in the s-domain. Was the result easier to obtain by this method? Not unless one is in love with convolution integrals! The partial-fraction-expansion method is usually simpler, assuming that the expansion itself is not too cumber-some. However, the operation of convolution is easier to perform in the s-domain, since it only requires multiplication. EXAMPLE 15.10 Determine the impulse response of the circuit in Fig. 15.26a, and use this to compute the forced response vo(t) if the input vin(t) 6e−tu(t) V. (Continued on next page) vin + – vo + – (a) 1 (b) 2 500 mF vo + – 1 2 500 mF (t) + – ■FIGURE 15.26 (a) A simple circuit to which an exponential input is applied at t = 0. (b) Circuit used to determine h(t). 15.6• THE COMPLEX-FREQUENCY PLANE As evident in the last several examples, circuits with even a comparatively small number of elements can lead to rather unwieldy s-domain expres-sions. In such instances, a graphical representation of a particular circuit response or transfer function can provide useful insights. In this section, we introduce one such technique, based on the idea of the complex-frequency plane (Fig. 15.28). Complex frequency has two components (σ and ω), so we are naturally drawn to representing our functions using a three-dimensional model. Since ω represents an oscillating function, there is no physical distinc-tion between a positive and negative frequency. In the case of σ, however, which can be identiﬁed with an exponential term, positive values are in-creasing in magnitude, whereas negative values are decaying. The origin of the s plane corresponds to dc (no time variation). A pictorial summary of these ideas is presented in Fig. 15.29. CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 598 Vo + – 1 2 1 V + – 2 s ■FIGURE 15.27 Circuit used to ﬁnd H(s). We ﬁrst connect an impulse voltage pulse δ(t) V to the circuit as shown in Fig. 15.26b. Although we may work in either the time domain with h(t) or the s-domain with H(s), we choose the latter, so we next consider the s-domain representation of Fig. 15.26b as depicted in Fig. 15.27. The impulse response H(s) is given by H(s) = Vo 1 so our immediate goal is to ﬁnd Vo—a task easily performed by simple voltage division: Vo vin=∂(t) = 2 2/s + 2 = s s + 1 = H(s) We may now ﬁnd vo(t) when vin = 6e−tu(t) using convolution, as vin = −1{Vin(s) · H(s)} Since Vin(s) = 6/(s + 1), Vo = 6s (s + 1)2 = 6 s + 1 − 6 (s + 1)2 Taking the inverse Laplace transform, we ﬁnd that vo(t) = 6e−t(1 −t)u(t) V PRACTICE ● 15.11 Referring to the circuit of Fig. 15.26a, use convolution to obtain vo(t) if vin = tu(t) V. Ans: vo(t) = (1 −e−t)u(t) V. j s plane ■FIGURE 15.28 The complex-frequency plane, also referred to as the s plane. SECTION 15.6 THE COMPLEX-FREQUENCY PLANE 599 To construct an appropriate three-dimensional representation of some function F(s), we ﬁrst note that we have its magnitude in mind, although the phase will have a strong complex-frequency dependence as well and may be graphed in a similar fashion. Thus, we will begin by substituting σ + jω for s in our function F(s), then proceed to determine an expression for \|F(s)\|. We next draw an axis perpendicular to the s plane, and use this to plot \|F(s)\| for each value of σ and ω. The basic process is illustrated in the following example. j t t t t t t t t t ■FIGURE 15.29 An illustration of the physical meaning of positive and negative values for σ and ω, as would be represented on the complex-frequency plane. When ω = 0, a function will have no oscillatory component; when σ = 0, the function is purely sinusoidal except when ω is also zero. EXAMPLE 15.11 Sketch the admittance of the series combination of a 1 H inductor and a 3 resistor as a function of both jω and σ. The admittance of these two series elements is given by Y(s) = 1 s + 3 Substituting s = σ + jω, we ﬁnd the magnitude of the function is \|Y(s)\| = 1 (σ + 3)2 + ω2 (Continued on next page) Pole-Zero Constellations This approach works well for relatively simple functions, but a more prac-tical method is needed in general. Let us visualize the s plane once again as CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 600 j (a) \|Y\| –3 \|Y\| (b) \|Y\| (c) –3 ■FIGURE 15.30 (a) A cutaway view of a clay model whose top surface represents \|Y(s)\| for the series combination of a 1 H inductor and a 3 resistor. (b) \|Y(s)\| as a function of ω. (c) \|Y(s)\| as a function of σ. When s = −3 + j0, the response magnitude is inﬁnite; when s is inﬁnite, the magnitude of Y(s) is zero. Thus our model must have inﬁ-nite height over the point (−3 + j0), and it must have zero height at all points inﬁnitely far away from the origin. A cutaway view of such a model is shown in Fig. 15.30a. Once the model is constructed, it is simple to visualize the variation of \|Y\| as a function of ω (with σ = 0) by cutting the model with a perpendicular plane containing the jω axis. The model shown in Fig. 15.30a happens to be cut along this plane, and the desired plot of \|Y\| versus ω can be seen; the curve is also drawn in Fig. 15.30b. In a similar manner, a vertical plane containing the σ axis enables us to obtain \|Y\| versus σ (with ω = 0), as shown in Fig. 15.30c. PRACTICE ● 15.12 Sketch the magnitude of the impedance Z(s) = 2 + 5s as a function of σ and jω. Ans: See Fig. 15.31. ■FIGURE 15.31 Solution for Practice Problem 15.12, generated with the following code: EDU» sigma = linspace(−10,10,21); EDU» omega = linspace(−10,10,21); EDU» [X, Y] = meshgrid(sigma,omega); EDU» Z = abs(2 + 5X + j5Y); EDU» colormap(hsv); EDU» s = [−5 3 8]; EDU» surﬂ(X,Y,Z,s); EDU» view (−20,5) SECTION 15.6 THE COMPLEX-FREQUENCY PLANE 601 the ﬂoor and then imagine a large elastic sheet laid on it. We now ﬁx our attention on all the poles and zeros of the response. At each zero, the re-sponse is zero, the height of the sheet must be zero, and we therefore tack the sheet to the ﬂoor. At the value of s corresponding to each pole, we may prop up the sheet with a thin vertical rod. Zeros and poles at inﬁnity must be treated by using a large-radius clamping ring or a high circular fence, re-spectively. If we have used an inﬁnitely large, weightless, perfectly elastic sheet, tacked down with vanishingly small tacks, and propped up with inﬁ-nitely long, zero-diameter rods, then the elastic sheet assumes a height that is exactly proportional to the magnitude of the response. These comments may be illustrated by considering the conﬁguration of the poles and zeros, sometimes called a pole-zero constellation, that locates all the critical frequencies of a frequency-domain quantity, for example, an impedance Z(s). A pole-zero constellation for an example impedance is shown in Fig. 15.32; in such a diagram, poles are denoted by crosses and zeros by circles. If we visualize an elastic-sheet model, tacked down at s = −2 + j0 and propped up at s = −1 + j5 and at s = −1 −j5, we should see a terrain whose distinguishing features are two mountains and one conical crater or depression. The portion of the model for the upper LHP is shown in Fig. 15.32b. j s plane –2 –1 + j5 –1 – j5 (a) j –1 + j5 \|Z(j)\| \|Z()\| (b) ■FIGURE 15.32 (a) The pole-zero constellation of some impedance Z(s). (b) A portion of the elastic-sheet model of the magnitude of Z(s). Let us now build up the expression for Z(s) that leads to this pole-zero conﬁguration. The zero requires a factor of (s + 2) in the numerator, and the two poles require the factors (s + 1 −j5) and (s + 1 + j5) in the denominator. Except for a multiplying constant k, we now know the form of Z(s): Z(s) = k s + 2 (s + 1 −j5)(s + 1 + j5) or Z(s) = k s + 2 s2 + 2s + 26 In order to determine k, we require a value for Z(s) at some s other than a critical frequency. For this function, let us suppose we are told Z(0) 1. By direct substitution in Eq. , we ﬁnd that k is 13, and therefore Z(s) = 13 s + 2 s2 + 2s + 26 The plots \|Z(σ)\| versus σ and \|Z( jω)\| versus ω may be obtained ex-actly from Eq. , but the general form of the function is apparent from the pole-zero conﬁguration and the elastic-sheet analogy. Portions of these two curves appear at the sides of the model shown in Fig. 15.32b. CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 602 PRACTICE ● 15.13 The parallel combination of 0.25 mH and 5 is in series with the parallel combination of 40 μF and 5 . (a) Find Zin(s), the input impedance of the series combination. (b) Specify all the zeros of Zin(s). (c) Specify all the poles of Zin(s). (d) Draw the pole-zero conﬁguration. Ans: 5(s2 + 10,000s + 108)/(s2 + 25,000s + 108) ; −5 ± j8.66 krad/s; −5, −20 krad/s. 15.7 • NATURAL RESPONSE AND THE s PLANE At the beginning of this chapter, we explored how working in the frequency domain through the Laplace transform allows us to consider a broad range of time-varying circuits, discarding integrodifferential equations and working algebraically instead. This approach is very powerful, but it does suffer from not being a very visual process. In contrast, there is a tremendous amount of information contained in the pole-zero plot of a forced response. In this section, we consider how such plots can be used to obtain the complete response of a circuit—natural plus forced—provided the initial conditions are known. The advantage of such an approach is a more intuitive linkage between the location of the critical frequencies, easily visualized through the pole-zero plot, and the desired response. Let us introduce the method by considering the simplest example, a se-ries RL circuit as shown in Fig. 15.33. A general voltage source vs(t) causes the current i(t) to ﬂow after closure of the switch at t = 0. The com-plete response i(t) for t > 0 is composed of a natural response and a forced response: i(t) = in(t) + i f (t) We may ﬁnd the forced response by working in the frequency domain, as-suming, of course, that vs(t) has a functional form that we can transform to the frequency domain; if vs(t) = 1/(1 + t2), for example, we must proceed as best we can from the basic differential equation for the circuit. For the circuit of Fig. 15.33, we have If (s) = Vs R + sL or If (s) = 1 L Vs s + R/L + – i(t) vs(t) L R t = 0 ■FIGURE 15.33 An example that illustrates the determination of the complete response through a knowledge of the critical frequencies of the impedance faced by the source. SECTION 15.7 NATURAL RESPONSE AND THE s PLANE 603 What does it mean to “operate” at a complex frequency? How could we possibly accomplish such a thing in a real laboratory? In this instance, it is important to remember how we invented complex frequency to begin with: it is a means of describing a sinusoidal function of frequency ω multiplied by an exponential function eσt . Such types of signals are very easy to generate with real (i.e., nonimaginary) laboratory equip-ment. Thus, we need only set the value for σ and the value for ω in order to “operate” at s = σ + jω. Next we consider the natural response. From previous experience, we know that the form will be a decaying exponential with the time constant L/R, but let’s pretend that we are ﬁnding it for the ﬁrst time. The form of the natural (source-free) response is, by deﬁnition, independent of the forc-ing function; the forcing function contributes only to the magnitude of the natural response. To find the proper form, we turn off all independent sources; here, vs(t) is replaced by a short circuit. Next, we try to obtain the natural response as a limiting case of the forced response. Returning to the frequency-domain expression of Eq. , we obediently set Vs = 0. On the surface, it appears that I(s) must also be zero, but this is not necessarily true if we are operating at a complex frequency that is a simple pole of I(s). That is, the denominator and the numerator may both be zero so that I(s) need not be zero. Let us inspect this new idea from a slightly different vantage point. We ﬁx our attention on the ratio of the desired forced response to the forcing function. We designate this ratio H(s) and deﬁne it to be the circuit transfer function. Then, If (s) Vs = H(s) = 1 L(s + R/L) In this example, the transfer function is the input admittance faced by Vs. We seek the natural (source-free) response by setting Vs = 0. However, If (s) = VsH(s), and if Vs = 0, a nonzero value for the current can be obtained only by operating at a pole of H(s). The poles of the transfer func-tion therefore assume a special signiﬁcance. In this particular example, we see that the pole of the transfer function occurs at s = −R/L + j0, as shown in Fig. 15.34. If we choose to operate at this particular complex frequency, the only ﬁnite current that could result must be a constant in the s-domain (i.e., frequency-independent). We thus obtain the natural response I s = −R L + j0 = A where A is an unknown constant. We next desire to transform this natural response to the time domain. Our knee-jerk reaction might be to attempt to apply inverse Laplace transform techniques in this situation. However, we have already speciﬁed a value of s, so that such an approach is not valid. Instead, we look to the real part of our general function est, such that in(t) = Re{Aest} = Re{Ae−Rt/L} In this case we ﬁnd in(t) = Ae−Rt/L so that the total response is then i(t) = Ae−Rt/L + i f (t) –R/L j ■FIGURE 15.34 Pole-zero constellation of the transfer function H(s) showing the single pole at s = −R/L . and A may be determined once the initial conditions are speciﬁed for this circuit. The forced response i f (t) is obtained by ﬁnding the inverse Laplace transform of If (s). A More General Perspective Figure 15.35 shows single sources connected to networks containing no independent sources. The desired response, which might be some current I1(s) or some voltage V2(s), may be expressed by a transfer function that displays all the critical frequencies. To be speciﬁc, we select the response V2(s) in Fig. 15.35a: V2(s) Vs = H(s) = k (s −s1)(s −s3) · · · (s −s2)(s −s4) · · · The poles of H(s) occur at s = s2, s4, . . . , and so a ﬁnite voltage V2(s) at each of these frequencies must be a possible functional form for the natural response. Thus, we think of a zero-volt source (which is just a short-circuit) applied to the input terminals; the natural response that occurs when the input terminals are short-circuited must therefore have the form v2n(t) = A2es2t + A4es4t + · · · where each A must be evaluated in terms of the initial conditions (including the initial value of any voltage source applied at the input terminals). To ﬁnd the form of the natural response i1n(t) in Fig. 15.35a, we should determine the poles of the transfer function, H(s) = I1(s)/Vs. The transfer functions applying to the situations depicted in Fig. 15.35b would be I1(s)/Is and V2(s)/Is, and their poles then determine the natural responses i1n(t) and v2n(t), respectively. If the natural response is desired for a network that does not contain any independent sources, then a source Vs or Is may be inserted at any conve-nient point, restricted only by the condition that the original network is obtained when the source is set to zero. The corresponding transfer function is then determined and its poles specify the natural frequencies. Note that the same frequencies must be obtained for any of the many source locations possible. If the network already contains a source, that source may be set equal to zero and another source inserted at a more convenient point. CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 604 V2(s) + – I1(s) Vs Network without independent sources + – (a) V2(s) + – I1(s) Is Network without independent sources (b) ■FIGURE 15.35 The poles of the response, I1(s) or V2(s), produced by (a) a voltage source Vs or (b) a current source Is . The poles determine the form of the natural response, i 1n(t) or v2n(t), that occurs when Vs is replaced by a short circuit or Is by an open circuit and some initial energy is available. SECTION 15.7 NATURAL RESPONSE AND THE s PLANE 605 A Special Case Before we illustrate this method with an example, completeness requires us to acknowledge a special case that might arise. This occurs when the net-work in Fig. 15.35a or b contains two or more parts that are isolated from each other. For example, we might have the parallel combination of three networks: R1 in series with C, R2 in series with L, and a short circuit. Clearly, a voltage source in series with R1 and C cannot produce any current in R2 and L; that transfer function would be zero. To ﬁnd the form of the nat-ural response of the inductor voltage, for example, the voltage source must be installed in the R2L network. A case of this type can often be recognized by an inspection of the network before a source is installed; but if it is not, then a transfer function equal to zero will be obtained. When H(s) = 0, we obtain no information about the frequencies characterizing the natural response, and a more suitable location for the source must be used. EXAMPLE 15.12 For the source-free circuit of Fig. 15.36, determine expressions for i1 and i2 for t > 0, given the initial conditions i1(0) = i2(0) = 11 A. Let us install a voltage source Vs between points x and y and ﬁnd the transfer function H(s) = I1(s)/Vs, which also happens to be the input admittance seen by the voltage source. We have I1(s) = Vs 2s + 1 + 6s/(3s + 2) = (3s + 2)Vs 6s2 + 13s + 2 or H(s) = I1(s) Vs = 1 2 s + 2 3 (s + 2) s + 1 6 From recent experience, we know at a glance that i1 must be of the form i1(t) = Ae−2t + Be−t/6 The solution is completed by using the given initial conditions to estab-lish the values of A and B. Since i1(0) is given as 11 amperes, 11 = A + B The necessary additional equation is obtained by writing the KVL equation around the perimeter of our circuit: 1i1 + 2 di1 dt + 2i2 = 0 and solving for the derivative: di1 dt t=0 = −1 2[2i2(0) + 1i1(0)] = −22 + 11 2 = −2A −1 6 B i1 i2 x y 2 H 3 H 1 2 ■FIGURE 15.36 A circuit for which the natural responses i 1 and i 2 are desired. (Continued on next page) The process that we must pursue to evaluate the amplitude coefﬁcients of the natural response is a detailed one, except in those cases where the ini-tial values of the desired response and its derivatives are obvious. However, we should not lose sight of the ease and rapidity with which the form of the natural response can be obtained. 15.8 • A TECHNIQUE FOR SYNTHESIZING THE VOLTAGE RATIO H(s) = Vout/Vin Much of the discussion in this chapter has been related to the poles and zeros of a transfer function. We have located them on the complex-frequency plane, we have used them to express transfer functions as ratios of factors or polynomials in s, we have calculated forced responses from them, and in Sec. 15.7 we have used their poles to establish the form of the natural response. Now let us see how we might determine a network that can provide a desired transfer function. We consider only a small part of the general prob-lem, working with a transfer function of the form H(s) = Vout(s)/Vin(s), as indicated in Fig. 15.41. For simplicity, we restrict H(s) to critical frequen-cies on the negative σ axis (including the origin). Thus, we will consider transfer functions such as H1(s) = 10(s + 2) s + 5 CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 606 Thus, A = 8 and B = 3, and so the desired solution is i1(t) = 8e−2t + 3e−t/6 amperes The natural frequencies constituting i2 are the same as those of i1, and a similar procedure used to evaluate the arbitrary constants leads to i2(t) = 12e−2t −e−t/6 amperes PRACTICE ● 15.16 If a current source i1(t) = u(t) A is present at a-b in Fig. 15.37 with the arrow entering a, ﬁnd H(s) = Vcd/I1, and specify the natural frequencies present in vcd(t). a b c d 300 200 0.1 F ■FIGURE 15.37 Ans: 120s/(s + 20,000) , −20,000 s−1. At several points throughout this book, we have investi-gated the behavior of various circuits responding to sinu-soidal excitation. The creation of sinusoidal waveforms, however, is an interesting topic in itself. Generation of large sinusoidal voltages and currents is straightforward using magnets and rotating coils of wire, for example, but such an approach is not easily scaled down for creation of small signals. Instead, low-current applications typically make use of what is known as an oscillator, which ex-ploits the concept of positive feedback using an appropri-ate ampliﬁer circuit. Oscillator circuits are an integral component of many consumer products, such as the global positioning satellite (GPS) receiver of Fig. 15.38. R + 1/sC and Zp = R∥(1/sC). The values we choose for R and C allow us to design an oscillator having a speciﬁc frequency (the internal capacitances of the op amp itself will limit the maximum frequency that can be obtained). In order to determine the relationship between R, C, and the oscillation frequency, we seek an expres-sion for the ampliﬁer gain, Vo/Vi. Recalling the two ideal op amp rules as discussed in Chap. 6 and examining the circuit in Fig. 15.39 closely, we recognize that Zp and Zs form a voltage divider such that Vi = Vo Zp Zp + Zs Simplifying the expressions for Zp = R∥(1/sC) = R/(1 + sRC), and Zs = R + 1/sC = (1 + sRC)/sC, we ﬁnd that Vi Vo = R 1 + sRC 1 + sRC sC + R 1 + sRC = sRC 1 + 3sRC + s2R2C2 Since we are interested in the sinusoidal steady-state op-eration of the ampliﬁer, we replace s with jω, so that Vi Vo = jωRC 1 + 3 jωRC + ( jω)2R2C2 = jωRC 1 −ω2R2C2 + 3 jωRC This expression for the gain is real only when ω = 1/RC. Thus, we can design an ampliﬁer to operate at a particular frequency f = ω/2π = 1/2π RC by select-ing values for R and C. ■FIGURE 15.38 Many consumer electronic products, such as this GPS receiver, rely on oscillator circuits to provide a reference frequency. © Nick Koudis/Photodisc/Getty Images/RF. One straightforward but useful oscillator circuit is known as the Wien-bridge oscillator, shown in Fig. 15.39. The circuit resembles a noninverting op amp circuit, with a resistor R1 connected between the inverting input pin and ground, and a resistor Rf connected between the output and the inverting input pin. The resistor Rf supplies what is referred to as a negative feedback path, since it connects the output of the ampliﬁer to the invert-ing input. Any increase Vo in the output then leads to a reduction of the input, which in turn leads to a smaller output; this process increases the stability of the output voltage Vo. The gain of the op amp, deﬁned as the ratio of Vo to Vi, is determined by the relative sizes of R1 and Rf . The positive feedback loop consists of two sepa-rate resistor-capacitor combinations, defined as Zs = R1 Vi R R 1/sC 1/sC Vo Rf – + ■FIGURE 15.39 A Wien-bridge oscillator circuit. (Continued on next page) PRACTICAL APPLICATION Design of Oscillator Circuits PRACTICAL APPLICATION As an example, let’s design a Wien-bridge oscillator to generate a sinusoidal signal at a frequency of 20 Hz, the commonly accepted lower frequency of the audio range. We require a frequency ω = 2π f = (6.28)(20) = 125.6 rad/s. Once we specify a value for R, the necessary value for C is known (and vice versa). Assuming that we happen to have a 1 μF capacitor handy, we thus compute a required resistance of R = 7962 . Since this is not a standard resistor value, we will likely have to use several resistors in series and/or parallel combinations to obtain the necessary value. Referring to Fig. 15.39 in prepara-tion for simulating the circuit using PSpice, however, we notice that no values for Rf or R1 have been speciﬁed. Although Eq. correctly speciﬁes the relationship between Vo and Vi, we may also write another equation relating these quantities: 0 = Vi R1 + Vi −Vo Rf which can be rearranged to obtain Vo Vi = 1 + Rf R1 Setting ω = 1/RC in Eq. results in Vi Vo = 1 3 Therefore, we need to select values of R1 and Rf such that Rf /R1 = 2. Unfortunately, if we proceed to perform a transient PSpice analysis on the circuit selecting Rf = 2 k and R1 = 1 k, for example, we will likely be disappointed in the outcome. In order to ensure that the circuit is indeed unstable (a necessary condition for oscillations to begin), it is necessary to have Rf /R1 slightly greater than 2. The simulated output of our ﬁnal design (R = 7962 , C = 1 μF, Rf = 2.01 k, and R1 = 1 k) is shown in Fig. 15.40. Note that the magni-tude of the oscillations is increasing in the plot; in practice, nonlinear circuit elements are required to stabi-lize the voltage magnitude of the oscillator circuit. ■FIGURE 15.40 Simulated output of the Wien-bridge oscillator designed for operation at 20 Hz. or H2(s) = −5s (s + 8)2 or H3(s) = 0.1s(s + 2) Let us begin by ﬁnding the voltage gain of the network of Fig. 15.42, which contains an ideal op amp. The voltage between the two input terminals of the op amp is essentially zero, and the input impedance of the op amp is essentially inﬁnite. We therefore may set the sum of the currents entering the inverting input terminal equal to zero: Vin Z1 + Vout Zf = 0 or Vout Vin = −Zf Z1 – + Vin + – Vout + – Z1 Zf ■FIGURE 15.42 For an ideal op amp, H(s) = Vout/Vin = −Zf /Z1. Vout + – Network + – Vin ■FIGURE 15.41 Given H(s) = Vout/Vin, we seek a network having a speciﬁed H(s). SECTION 15.8 A TECHNIQUE FOR SYNTHESIZING THE VOLTAGE RATIO H(s) = Vout/Vin 609 If Zf and Z1 are both resistances, the circuit acts as an inverting ampliﬁer, or possibly an attenuator (if the ratio is less than unity). Our present inter-est, however, lies with those cases in which one of these impedances is a re-sistance while the other is an RC network. In Fig. 15.43a, we let Z1 = R1, while Zf is the parallel combination of Rf and Cf . Therefore, Zf = Rf /sCf Rf + (1/sCf ) = Rf 1 + sCf Rf = 1/Cf s + (1/Rf Cf ) and H(s) = Vout Vin = −Zf Z1 = − 1/R1Cf s + (1/Rf Cf ) We have a transfer function with a single (ﬁnite) critical frequency, a pole at s = −1/Rf Cf . Moving on to Fig. 15.43b, we now let Zf be resistive while Z1 is an RC parallel combination: Z1 = 1/C1 s + (1/R1C1) and H(s) = Vout Vin = −Zf Z1 = −Rf C1 s + 1 R1C1 The only ﬁnite critical frequency is a zero at s = −1/R1C1. For our ideal op amps, the output or Thévenin impedance is zero and therefore Vout and Vout/Vin are not functions of any load ZL that may be placed across the output terminals. This includes the input to another op amp as well, and therefore we may connect circuits having poles and zeros at specified locations in cascade, where the output of one op amp is connected directly to the input of the next, and thus generates any desired transfer function. – + Vin + – Vout + – R1 (a) Rf Cf – + Vin + – Vout + – (b) R1 C1 Rf ■FIGURE 15.43 (a) The transfer function H(s) = Vout/Vin has a pole at s = −1/R f Cf . (b) Here, there is a zero at s = −1/R1C1. EXAMPLE 15.13 Synthesize a circuit that will yield the transfer function H(s) Vout/Vin 10(s + 2)/(s + 5 ). The pole at s = −5 may be obtained by a network of the form of Fig. 15.43a. Calling this network A, we have 1/RfACfA = 5. We arbi-trarily select RfA = 100 k; therefore, CfA = 2 μF. For this portion of the complete circuit, HA(s) = − 1/R1ACfA s + (1/RfACfA) = −5 × 105/R1A s + 5 Next, we consider the zero at s = −2. From Fig. 15.43b, 1/R1BC1B = 2, and, with R1B = 100 k, we have C1B = 5 μF. Thus HB(s) = −RfBC1B s + 1 R1BC1B = −5 × 10−6RfB(s + 2) (Continued on next page) CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 610 SUMMARY AND REVIEW After having been exposed to the concept of complex frequency in Chap. 14, we applied that concept to circuit analysis in this chapter. The ﬁrst topic was impedance—perhaps familiar to those who have already read Chap. 10. The concept of impedance (or admittance) allows us to directly construct s-domain equations which describe nodal voltages, mesh currents, etc., without having to rely on taking the Laplace transform of each term of an integrodifferential equation. Somewhat surprisingly, we found that the impedance of inductors and capacitors includes the initial condition of that element. From that point forward, all of our familiar circuit analysis techniques apply. The only difﬁculty encountered is in fac-toring higher-order polynomials in order to perform an inverse transform. We also introduced the notion of a system transfer function, which allows the input to a network to be changed easily, and the new output predicted. Working in the s-domain proved very natural here, and we saw that convo-lution of two time-domain functions is easily performed by multiplying their s-domain equivalents. – + Vin + – 25 k 100 k 2 F – + Vout + – 100 k 5 F 100 k ■FIGURE 15.44 This network contains two ideal op amps and gives the voltage transfer function H(s) = Vout/Vin = 10(s + 2)/(s + 5). and H(s) = HA(s)HB(s) = 2.5 RfB R1A s + 2 s + 5 We complete the design by letting RfB = 100 k and R1A = 25 k. The result is shown in Fig. 15.44. The capacitors in this circuit are fairly large, but this is a direct consequence of the low frequencies selected for the pole and zero of H(s). If H(s) were changed to 10(s + 2000)/(s + 5000), we could use 2 and 5 nF values. PRACTICE ● 15.17 Specify suitable element values for Z1 and Zf in each of three cascaded stages to realize the transfer function H(s) = −20s2/(s + 1000). Ans: 1 μF ∥∞, 1 M; 1 μF ∥∞, 1 M; 100 k ∥10 nF,5 M. SUMMARY AND REVIEW 611 The third major topic of the chapter was the complex-frequency plane, which allows us to create a graphical representation of any s-domain ex-pression. In particular, it provides a tidy means for readily identifying poles and zeros. Since the sources connected to a circuit only determine the mag-nitude of the transient response, and not the form of the transisent response itself, we found that s-domain analysis can reveal details about the natural as well as forced response of a network. We concluded the chapter with a description of how op amps can be employed to synthesize a desired trans-fer function, placing poles and zeros where they are needed through cas-caded stages. This topic will be revisited in future studies of signal analysis, and the concept of convolution in particular is applicable to a broad range of appli-cations. At this stage, however, perhaps we should pause and allow the reader to focus on key issues and identify relevant examples as a start to re-viewing what we have discussed. ❑Resistors may be represented in the frequency domain by an impedance having the same magnitude. (Example 15.1) ❑Inductors may be represented in the frequency domain by an impedance sL. If the initial current is nonzero, then the impedance must be placed in series with a voltage source −Li(0−) or in parallel with a current source i(0−)/s. (Example 15.1) ❑Capacitors may be represented in the frequency domain by an impedance 1/sC. If the initial voltage is nonzero, then the impedance must be placed in series with a voltage source v(0−)/s or in parallel with a current source Cv(0−). (Example 15.2) ❑Nodal and mesh analysis in the s-domain lead to simultaneous equations in terms of s-polynomials. MATLAB is a particularly useful tool for solving such systems of equations. (Examples 15.3, 15.4, 15.5) ❑Superposition, source transformation, and the Thévenin and Norton theorems all apply in the s-domain. (Examples 15.6, 15.7) ❑A circuit transfer function H(s) is deﬁned as the ratio of the s-domain output to the s-domain input. Either quantity may be a voltage or current. (Example 15.8) ❑The zeros of H(s) are those values that result in zero magnitude. The poles of H(s) are those values that result in inﬁnite magnitude. ❑Convolution provides us with both an analytic and a graphical means of determining the output of a circuit from its impulse response h(t). (Examples 15.8, 15.9, 15.10) ❑There are several graphical approaches to representing s-domain expressions in terms of poles and zeros. Such plots can be used to synthesize a circuit to obtain a desired response. (Example 15.11) ❑A source-free circuit can be analyzed using s-domain techniques to determine its transient response. ❑Single op amp stages can be used to synthesize transfer functions having either a zero or a pole. More complex functions can be synthesized by cascading multiple stages. These models are summarized in Table 15.1. CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 612 READING FURTHER More details regarding s-domain analysis of systems, use of Laplace transforms, and properties of transfer functions can be found in: K. Ogata, Modern Control Engineering, 4th ed. Englewood Cliffs, N.J.: Prentice-Hall, 2002. A good discussion of various types of oscillator circuits can be found in: R. Mancini, Op Amps for Everyone, 2nd ed. Amsterdam: Newnes, 2003. and G. Clayton and S. Winder, Operational Ampliﬁers, 5th ed. Amsterdam: Newnes, 2003. EXERCISES 15.1 Z(s) and Y(s) 1. Draw an s-domain equivalent of the circuit depicited in Fig. 15.45, if the only quantity of interest is v(t). (Hint: omit the source, but don’t ignore it.) 2. For the circuit of Fig. 15.46, the only quantity of interest is the voltage v(t). Draw an appropriate s-domain equivalent circuit. (Hint: omit the source, but don’t ignore it.) 3. For the circuit represented in Fig. 15.47, draw an s-domain equivalent and analyze it to obtain a value for i(t) if i(0) is equal to (a) 0; (b) 2 A. ■FIGURE 15.49 + – 73 1 30 mH 2000 F 7.2 V v(t) + – t = 0 ■FIGURE 15.50 Zin 500 mH 3.3 250 mF ■FIGURE 15.45 4 1.5 H 2 A v(t) + – t = 0 ■FIGURE 15.46 10 500 mF 3 A v(t) + – t = 0 ■FIGURE 15.47 + – 2.7 1.1 H 1.5e–tu(t) + 2u(t) V v(t) + – i(t) ■FIGURE 15.48 + – 5 2 A V VC(s) + – 1.5 s 2 s 4. For the circuit of Fig. 15.47, draw an s-domain equivalent and analyze it to obtain a value for v(t) if i(0) is equal to (a) 0; (b) 3 A. 5. With respect to the s-domain circuit drawn in Fig. 15.48, (a) calculate VC(s); (b) determine vC(t), t > 0; (c) draw the time-domain representation of the circuit. 6. Draw all possible s-domain equivalents (t > 0) of the circuit shown in Fig. 15.49. 7. Determine the input impedance Zin(s) seen looking into the terminals of the network depicted in Fig. 15.50. Express your answer as a ratio of two s-polynomials. EXERCISES 613 8. With respect to the network of Fig. 15.51, obtain an expression for the input admittance Y(s) as labeled. Express your answer as a ratio of two s-polynomials. 9. For the circuit of Fig. 15.52, (a) draw both s-domain equivalent circuits; (b) choose one and solve for V(s); (c) determine v(t). ■FIGURE 15.51 Y(s) 1.5 333 mF 4.7 1.7 H ■FIGURE 15.52 + – 200 mH 2e–2tu(t) V i(0–) = 0.5 A 1 k v(t) + – i(t) ■FIGURE 15.53 2e–tu(t) V –4u(t) V 1 i3(t) i2(t) i1(t) + – + – 500 mF 500 mF 3 ■FIGURE 15.54 250 mF – + + – 2u(t) V 3u(t) V 5 800 mH vx Ref 10. Determine the input impedance 1/Y(s) of the network represented in Fig. 15.51 if the 1.5 resistor is replaced with the parallel combination of a 100 mF capacitor and a 1 resistor, and the initial current through the inductor (deﬁned as ﬂowing downward) is 540 mA. 15.2 Nodal and Mesh Analysis in the s-Domain 11. For the circuit given in Fig. 15.53, (a) draw the s-domain equivalent; (b) write the three s-domain mesh equations; (c) determine i1, i2, and i3. 12. Replace the −4u(t) label in the circuit of Fig. 15.53 with 4e−tu(t) V. Calculate i1, i2, and i3 if it transpires that the initial current through the inductor, i2−i3, is equal to 50 mA. 13. For the circuit shown in Fig. 15.54, (a) write an s-domain nodal equation for Vx(s); (b) solve for vx(t). CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 614 15. The 2u(t) A source in Fig. 15.55 is replaced with a 4e−tu(t) A source. Employ s-domain analysis to determine the power dissipated by the 1 resistor. 16. Calculate the power dissipated in the 3 resistor of Fig. 15.56, if v1(0−) = 2 V. ■FIGURE 15.55 1 2 2 H 2u(t) A 5u(t) A v1 v2 200 mF ■FIGURE 15.56 3 1.5u(t) V 2u(t) A v1 v2 5 2 + – 400 mF ■FIGURE 15.57 5 2 H 0.1 F vx + – is1 is2 ■FIGURE 15.58 2 4 250 mF 450 mF 0.1v2(t) v1(t) v2(t) v3(t) 2u(t) A ■FIGURE 15.59 + – 2 1 mH 0.005i1 1000 F 750 F 6 cos (2t –13°) u(t) V 6 cos 2t u(t) V i4 i2 i3 i1 + – + – 17. For the circuit shown in Fig. 15.57, let is1 = 3u(t) A and is2 = 5 sin 2t A. Working initially in the s-domain, obtain an expression for vx(t). 18. For the circuit of Fig. 15.58, (a) draw the corresponding s-domain circuit; (b) solve for v1(t), v2(t), and v3(t); (c) verify your solution with an appropriate PSpice simulation. 19. Determine the mesh currents i1(t) and i2(t) in Fig. 15.59 if the current through the 1 mH inductor (i2 −i4) is 1 A at t = 0−. Verify that your answer approaches the answer obtained using phasor analysis as the circuit response eventually reaches steady state. 14. Determine v1 and v2 for the circuit of Fig. 15.55 using nodal analysis in the s-domain. EXERCISES 615 20. Assuming no energy initially stored in the circuit of Fig. 15.60, determine the value of v2 at t equal to (a) 1 ms; (b) 100 ms; (c) 10 s. ■FIGURE 15.61 + – 8s 20 14 Z 12 s I(s) V (s + 1)(s + 2) s ■FIGURE 15.62 4 2 3 H iC(t) i 2u(t) A 1.2 F ■FIGURE 15.63 3.5 is 5 H 3 10 ix iL ■FIGURE 15.64 2 5 5 H I2 5I2 a b ■FIGURE 15.60 3v2 5v2 100 2 mH 500 F 600 F 14tu(t) V + – v2 15.3 Additional Circuit Analysis Techniques 21. Using repeated source transformations, obtain an s-domain expression for the Thévenin equivalent seen by the element labeled Z in the circuit of Fig. 15.61. 22. Calculate I(s) as labeled in the circuit of Fig. 15.61 if the element Z has impedance of (a) 2 ; (b) 1 2s ; (c) s + 1 2s + 3 . 23. For the circuit shown in Fig. 15.62, determine the s-domain Thévenin equivalent seen by the (a) 2 resistor; (b) 4 resistor; (c) 1.2 F capacitor; (d) current source. 24. Calculate both currents labeled in the circuit of Fig. 15.62. 25. For the circuit of Fig. 15.63, take is(t) = 5u(t) A and determine (a) the Thévenin equivalent impedance seen by the 10 resistor; (b) the inductor current iL(t). 26. If the current source of Fig. 15.63 is 1.5e−2tu(t) A, and iL(0−) = 1 A, deter-mine ix(t). 27. For the s-domain circuit of Fig. 15.64, determine the Thévenin equivalent seen looking into the terminals marked a and b. CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 616 28. (a) Use superposition in the s-domain to ﬁnd an expression for V1(s) as labeled in Fig. 15.65. (b) Find v1(t). ■FIGURE 15.65 2 5 3 1 1 F 3 H 1 H 2 F cos 2t u(t) V v1(t) 4ix 4 cos 4t u(t) V ix + – + – + – a b d c 29. If the top right voltage source of Fig. 15.65 is open-circuited, determine the Thévenin equivalent seen looking into the terminals marked a and b. 30. If the bottom left voltage source of Fig. 15.65 is open-circuited, determine the Thévenin equivalent seen looking into the terminals marked c and d. 15.4 Poles, Zeros, and Transfer Functions 31. Determine the poles and zeros of the following s-domain functions: (a) s s + 12.5; (b) s(s + 1) (s + 5)(s + 3); (c) s + 4 s2 + 8s + 7; (d) s2 −s −2 3s3 + 24s2 + 21s. 32. Use appropriate means to ascertain the poles and zeros of (a) s + 4; (b) 2s s2 −8s + 16; (c) 4 s3 + 8s + 7; (d) s −5 s3 −7s + 6. 33. Consider the following expressions and determine the critical frequencies of each: (a) 5 + s−1; (b) s(s + 1)(s + 4) (s + 5)(s + 3)2 ; (c) 1 s2 + 4; (d) 0.5s2 −18 s2 + 1 . 34. For the network represented schematically in Fig. 15.66, (a) write the transfer function H(s) ≡Vout(s)/Vin(s); (b) determine the poles and zeros of H(s). 35. For each of the two networks represented schematically in Fig. 15.67, (a) write the transfer function H(s) ≡Vout(s)/Vin(s); (b) determine the poles and zeros of H(s). ■FIGURE 15.66 vin(t) + – vout(t) + – C R ■FIGURE 15.67 vin(t) + – vout(t) + – (a) L R vin(t) + – vout(t) + – (b) R L 36. Determine the criticial frequencies of Zin as deﬁned in Fig. 15.50. 37. Specify the poles and zeros of Y(s) as deﬁned by Fig. 15.51. 38. If a network is found to have the transfer function H(s) = s s2 + 8s + 7 , determine the s-domain output voltage for vin(t) equal to (a) 3u(t) V; (b) 25e−2tu(t) V; (c) 4u(t + 1) V; (d) 2 sin 5t u(t) V. EXERCISES 617 39. A particular network is known to be characterized by the transfer function H(s) = s + 1/(s2 + 23s + 60). Determine the critical frequencies of the output if the input is (a) 2u(t) + 4δ(t); (b) −5e−tu(t); (c) 4te−2tu(t); (d) 5 √ 2e−10t cos 5t u(t) V. 40. For the network represented in Fig. 15.68, determine the critical frequencies of Zin(s). 15.5 Convolution 41. Referring to Fig. 15.69, employ Eq. to obtain x(t) ∗y(t). ■FIGURE 15.68 Zin(s) 100 75 50s 25 s ■FIGURE 15.69 1 0 x(t) t 3 1 0 y(t) t ■FIGURE 15.70 1 H 4 5 vin(t) + – vo(t) + – ■FIGURE 15.71 j2 –2 j –1 15.6 The Complex-Frequency Plane 46. A 2 resistor is placed in series with a 250 mF capacitor. Sketch the magni-tude of the equivalent impedance as a function of (a) σ; (b) ω; (c) σ and ω, using an elastic-sheet type approach. (d) Verify your solutions using MATLAB. 47. Sketch the magnitude of Z(s) = s2 + s as a function of (a) σ; (b) ω; (c) σ, and ω, using an elastic-sheet type approach. (d) Verify your solutions using MATLAB. 48. Sketch the pole-zero constellation of each of the following: (a) s(s + 4) (s + 5)(s + 2); (b) s −1 s2 + 8s + 7; (c) s2 + 1 s(s2 + 10s + 16; (d) 5 s2 + 2s + 5 . 49. The partially labeled pole-zero constellation of a particular transfer function H(s) is shown in Fig. 15.71. Obtain an expression for H(s) if H(0) is equal to (a) 1; (b) −5. (c) Is the system H(s) represents expected to be stable or unstable? Explain. 42. With respect to the functions x(t) and y(t) as plotted in Fig. 15.69, use Eq. to obtain (a) x(t) ∗x(t); (b) y(t) ∗δ(t). 43. Employ graphical convolution techniques to determine f ∗g if f (t) = 5u(t) and g(t) = 2u(t) −2u(t −2) + 2u(t −4) −2u(t −6). 44. Let h(t) = 2e−3tu(t) and x(t) = u(t) −δ(t). Find y(t) = h(t) ∗x(t) by (a) using convolution in the time domain; (b) ﬁnding H(s) and X(s) and then obtaining −1{H(s)X(s)}. 45. (a) Determine the impulse response h(t) of the network shown in Fig. 15.70. (b) Use convolution to determine vo(t) if vin(t) = 8u(t) V. CHAPTER 15 CIRCUIT ANALYSIS IN THE s-DOMAIN 618 50. The three-element network shown in Fig. 15.72 has an input impedance ZA(s) that has a zero at s = −10 + j0. If a 20 resistor is placed in series with the network, the zero of the new impedance shifts to s = −3.6 + j0. Calculate R and C. 51. Let H(s) = 100(s + 2)/(s2 + 2s + 5) and (a) show the pole-zero plot for H(s); (b) ﬁnd H( jω); (c) ﬁnd \|H( jω)\|; (d) sketch \|H( jω)\| versus ω; (e) ﬁnd ωmax, the frequency at which \|H( jω)\| is a maximum. 15.7 Natural Response and the s Plane 52. Determine expressions for i1(t) and i2(t) for the circuit of Fig. 15.73, assum-ing v1(0−) = 2 V and v2(0−) = 0 V. 53. The 250 mF capacitor in the circuit of Fig. 15.73 is replaced with a 2 H induc-tor. If v1(t) = 0 V and i1(0−) −i2(0−) = 1 A, obtain an expression for i2(t). 54. In the network of Fig. 15.74, a current source ix(t) = 2u(t) A is connected between terminals c and d such that the arrow of the source points upward. Determine the natural frequencies present in the voltage vab(t) which results. 55. With regard to the circuit shown in Fig. 15.75, let i1(0−) = 1 A and i2(0−) = 0. (a) Determine the poles of Iin(s)/Vin(s); (b) use this information to obtain expressions for i1(t) and i2(t). ■FIGURE 15.72 ZA 5 R C ■FIGURE 15.73 i1 i2 x y 250 mF 500 mF 4 2 v1 + – v1 + – ■FIGURE 15.74 a b c d C R1 R2 ■FIGURE 15.75 iin i2 i1 + – 500 mH 1 H 1.5 2 2u(t) V 15.8 A Technique for Synthesizing the Voltage Ratio H(s) = Vout/Vin 56. Design a circuit which produces the transfer function H(s) = Vout/Vin equal to (a) 5(s + 1); (b) 5 (s + 1); (c) 5s + 1 s + 2 . 57. Design a circuit which produces the transfer function H(s) = Vout/Vin equal to (a) 2(s + 1)2; (b) 3 (s + 500)(s + 100). 58. Design a circuit which produces the transfer function H(s) = Vout Vin = 5 s + 104 s + 2 × 105 . 59. Design a circuit which produces the transfer function H(s) = Vout Vin = 3 s + 50 (s + 75)2 . 60. Find H(s) = Vout/Vin as a ratio of polynomials in s for the op-amp circuit of Fig. 15.42, given the impedance values (in ): (a) Z1(s) = 103 + (108/s), Zf (s) = 5000; (b) Z1(s) = 5000, Zf (s) = 103 + (108/s); (c) Z1(s) = 103 + (108/s), Zf (s) = 104 + (108/s). Chapter-Integrating Exercises 61. Design a circuit that provides a frequency of 16 Hz, which is near the lower end of the human hearing range. Verify your design with an appropriate simulation. 62. Design a circuit which provides a dual-tone multifrequency (DTMF) signal corresponding to the number 9, which is a voltage output composed of a 1477 Hz signal and an 852 Hz signal. 63. (a) Design a circuit which provides a signal at 261.6 Hz, which is approxi-mately middle “C”. Use only standard 5% tolerance resistance values. (b) Estimate the likely frequency range of your signal generator based on the range of possible resistor values which can be used in construction. 64. (a) Many people with partial hearing loss, especially the elderly, have difﬁculty in detecting standard smoke detectors. An alternative is to lower the frequency to approximately 500 Hz. Design a circuit which provides such a signal, using only standard 10% tolerance resistor and capacitor values. (b) Estimate the actual frequency range expected from your design if it is manufactured based on the possible range of component values. 65. Design a circuit which provides either a 200 Hz signal or a 400 Hz signal by closing appropriate switches. INTRODUCTION We have already been introduced to the concept of frequency response, meaning that the behavior of our circuit can change dramatically depending on the frequency (or frequencies) of operation—a radical departure from our ﬁrst experiences with simple dc circuits. In this chapter we take the topic to a more re-ﬁned level, as even simple circuits designed for speciﬁc frequency response can be enormously useful in a wide variety of everyday applications. In fact, we make use of frequency-selective circuits throughout the day, probably without even realizing it. For exam-ple, switching to our favorite radio station is in fact tuning our radio to selectively amplify a narrow band of signal frequencies; heating microwave popcorn is possible while watching television or talking on a cell phone because the frequencies of each device can be isolated from one another. In addition, studying frequency response and ﬁlters can be particularly enjoyable as it provides us with a vehicle for pushing past analysis of existing circuits, and enabling the design of complex circuits from scratch to meet some-times stringent speciﬁcations. We’ll start this journey with a short discussion of resonance, loss, quality factor, and bandwidth— important concepts for ﬁlters as well as any circuit (or system, for that matter) containing energy storage elements. 16.1 • PARALLEL RESONANCE Suppose that a certain forcing function is found to contain sinusoidal componentshavingfrequencieswithintherangeof10to100Hz.Now let us imagine that this forcing function is applied to a network that KEY CONCEPTS Resonant Frequency of Circuits with Inductors and Capacitors Quality Factor Bandwidth Frequency and Magnitude Scaling Bode Diagram Techniques Low- and High-Pass Filters Bandpass Filter Design Active Filters Butterworth Filter Design Frequency Response C H A P T E R 16 619 CHAPTER 16 FREQUENCY RESPONSE 620 has the property that all sinusoidal voltages with frequencies from zero to 200 Hz applied at the input terminals appear doubled in magnitude at the output terminals, with no change in phase angle. The output function is therefore an undistorted facsimile of the input function, but with twice the amplitude. If, however, the network has a frequency response such that the magnitudes of input sinusoids between 10 and 50 Hz are multiplied by a dif-ferent factor than are those between 50 and 100 Hz, then the output would in general be distorted; it would no longer be a magniﬁed version of the in-put. This distorted output might be desirable in some cases and undesirable in others. That is, the network frequency response might be chosen deliber-ately to reject some frequency components of the forcing function, or to emphasize others. Such behavior is characteristic of tuned circuits or resonant circuits, as we will see in this chapter. In discussing resonance we will be able to apply all the methods we have discussed in presenting frequency response. Resonance In this section we will begin the study of a very important phenomenon that may occur in circuits that contain both inductors and capacitors. The phe-nomenon is called resonance, and it may be loosely described as the condi-tion existing in any physical system when a ﬁxed-amplitude sinusoidal forcing function produces a response of maximum amplitude. However, we often speak of resonance as occurring even when the forcing function is not sinusoidal. The resonant system may be electrical, mechanical, hydraulic, acoustic, or some other kind, but we will restrict our attention, for the most part, to electrical systems. Resonance is a familiar phenomenon. Jumping up and down on the bumper of an automobile, for example, can put the vehicle into rather large oscillatory motion if the jumping is done at the proper frequency (about one jump per second), and if the shock absorbers are somewhat decrepit. How-ever, if the jumping frequency is increased or decreased, the vibrational response of the automobile will be considerably less than it was before. A further illustration is furnished in the case of an opera singer who is able to shatter crystal goblets by means of a well-formed note at the proper fre-quency. In each of these examples, we are thinking of frequency as being adjusted until resonance occurs; it is also possible to adjust the size, shape, and material of the mechanical object being vibrated, but this may not be so easily accomplished physically. The condition of resonance may or may not be desirable, depending upon the purpose which the physical system is to serve. In the automotive example, a large amplitude of vibration may help to separate locked bumpers, but it would be somewhat disagreeable at 65 mi/h (105 km/h). Let us now deﬁne resonance more carefully. In a two-terminal electrical network containing at least one inductor and one capacitor, we deﬁne reso-nance as the condition which exists when the input impedance of the net-work is purely resistive. Thus, a network is in resonance (or resonant) when the voltage and current at the network input terminals are in phase. SECTION 16.1 PARALLEL RESONANCE 621 We will also ﬁnd that a maximum-amplitude response is produced in the network when it is in the resonant condition. We ﬁrst apply the deﬁnition of resonance to a parallel RLC network dri-ven by a sinusoidal current source as shown in Fig. 16.1. In many practical situations, this circuit is a very good approximation to the circuit we might build in the laboratory by connecting a physical inductor in parallel with a physical capacitor, where the parallel combination is driven by an energy source having a very high output impedance. The steady-state admittance offered to the ideal current source is Y = 1 R + j ωC −1 ωL Resonance occurs when the voltage and current at the input terminals are in phase. This corresponds to a purely real admittance, so that the nec-essary condition is given by ωC −1 ωL = 0 The resonant condition may be achieved by adjusting L, C, or ω; we will devote our attention to the case for which ω is the variable. Hence, the res-onant frequency ω0 is ω0 = 1 √ LC rad/s or f0 = 1 2π √ LC Hz This resonant frequency ω0 is identical to the resonant frequency deﬁned in Eq. , Chap. 9. The pole-zero conﬁguration of the admittance function can also be used to considerable advantage here. Given Y(s), Y(s) = 1 R + 1 sL + sC or Y(s) = C s2 + s/RC + 1/LC s we may display the zeros of Y(s) by factoring the numerator: Y(s) = C (s + α −jωd)(s + α + jωd) s where α and ωd represent the same quantities that they did when we dis-cussed the natural response of the parallel RLC circuit in Sec. 9.4. That is, α is the exponential damping coefﬁcient, α = 1 2RC V + – ILC IC IL I R L C ■FIGURE 16.1 The parallel combination of a resistor, an inductor, and a capacitor, often referred to as a parallel resonant circuit. CHAPTER 16 FREQUENCY RESPONSE 622 and ωd is the natural resonant frequency (not the resonant frequency ω0), ωd = ω2 0 −α2 The pole-zero constellation shown in Fig. 16.2a follows directly from the factored form. In view of the relationship among α, ωd, and ω0, it is apparent that the distance from the origin of the s plane to one of the admittance zeros is nu-merically equal to ω0. Given the pole-zero conﬁguration, the resonant fre-quency may therefore be obtained by purely graphical methods. We merely swing an arc, using the origin of the s plane as a center, through one of the zeros. The intersection of this arc and the positive jω axis locates the point s = jω0. It is evident that ω0 is slightly greater than the natural resonant frequency ωd, but their ratio approaches unity as the ratio of ωd to α in-creases. Resonance and the Voltage Response Next let us examine the magnitude of the response, the voltage V(s)indicated in Fig. 16.1, as the frequency ωof the forcing function is varied. If we assume a constant-amplitude sinusoidal current source, the voltage response is pro-portional to the input impedance. This response can be obtained from the pole-zero plot of the impedance Z(s) = s/C (s + α −jωd)(s + α + jωd) shown in Fig. 16.2b. The response of course starts at zero, reaches a maxi-mum value in the vicinity of the natural resonant frequency, and then drops again to zero as ω becomes inﬁnite. The frequency response is sketched in Fig. 16.3. The maximum value of the response is indicated as R times the amplitude of the source current, implying that the maximum magnitude of the circuit impedance is R; moreover, the response maximum is shown to occur exactly at the resonant frequency ω0. Two additional frequencies, ω1 and ω2, which we will later use as a measure of the width of the response – s-plane Y(s) j0 jd –jd 0 j (a) jd j –jd – (b) ■FIGURE 16.2 (a) The pole-zero constellation of the input admittance of a parallel resonant circuit is shown on the s-plane; ω2 0 = α2 + ω2 d . (b) The pole-zero constellation of the input impedance. 1 0 2 \|V(j)\| \|I\|R 0.707\|I\|R ■FIGURE 16.3 The magnitude of the voltage response of a parallel resonant circuit is shown as a function of frequency. SECTION 16.1 PARALLEL RESONANCE 623 curve, are also identiﬁed. Let us ﬁrst show that the maximum impedance magnitude is R and that this maximum occurs at resonance. The admittance, as speciﬁed by Eq. , possesses a constant conduc-tance and a susceptance which has a minimum magnitude (zero) at reso-nance. The minimum admittance magnitude therefore occurs at resonance, and it is 1/R. Hence, the maximum impedance magnitude is R, and it occurs at resonance. At the resonant frequency, therefore, the voltage across the parallel reso-nant circuit of Fig. 16.1 is simply IR, and the entire source current I ﬂows through the resistor. However, current is also present in L and C. For the inductor, IL,0 = VL,0/jω0L = IR/jω0L, and the capacitor current at reso-nance is IC,0 = ( jω0C)VC,0 = jω0CRI. Since 1/ω0C = ω0L at resonance, we ﬁnd that IC,0 = −IL,0 = jω0CRI and IC,0 + IL,0 = ILC = 0 Thus, the net current ﬂowing into the LC combination is zero. The maxi-mum value of the response magnitude and the frequency at which it occurs are not always found so easily. In less standard resonant circuits, we may ﬁnd it necessary to express the magnitude of the response in analytical form, usually as the square root of the sum of the real part squared and the imaginary part squared; then we should differentiate this expression with respect to frequency, equate the derivative to zero, solve for the frequency of maximum response, and ﬁnally substitute this frequency in the magni-tude expression to obtain the maximum-amplitude response. The procedure may be carried out for this simple case merely as a corroborative exercise; but, as we have seen, it is not necessary. Quality Factor It should be emphasized that, although the height of the response curve of Fig. 16.3 depends only upon the value of R for constant-amplitude excita-tion, the width of the curve or the steepness of the sides depends upon the other two element values also. We will shortly relate the “width of the response curve” to a more carefully deﬁned quantity, the bandwidth, but it is helpful to express this relationship in terms of a very important parameter, the quality factor Q. We will ﬁnd that the sharpness of the response curve of any resonant cir-cuit is determined by the maximum amount of energy that can be stored in the circuit, compared with the energy that is lost during one complete period of the response. We deﬁne Q as The proportionality constant 2π is included in the deﬁnition in order to sim-plify the more useful expressions for Q which we will now obtain. Since Q = quality factor ≡2π maximum energy stored total energy lost per period We should be very careful not to confuse the quality factor with charge or reactive power, all of which unfortunately are represented by the letter Q. CHAPTER 16 FREQUENCY RESPONSE 624 energy can be stored only in the inductor and the capacitor, and can be lost only in the resistor, we may express Q in terms of the instantaneous energy associated with each of the reactive elements and the average power PR dissipated in the resistor: Q = 2π [wL(t) + wC(t)]max PRT where T is the period of the sinusoidal frequency at which Q is evaluated. Now let us apply this deﬁnition to the parallel RLC circuit of Fig. 16.1 and determine the value of Q at the resonant frequency; this value of Q is denoted by Q0. We select the current forcing function i(t) = Im cos ω0t and obtain the corresponding voltage response at resonance, v(t) = Ri(t) = RIm cos ω0t The energy stored in the capacitor is then wC(t) = 1 2Cv2 = I2 m R2C 2 cos2 ω0t and the instantaneous energy stored in the inductor is given by wL(t) = 1 2 Li2 L = 1 2 L 1 L v dt 2 = 1 2L RIm ω0 sin ω0t 2 so that wL(t) = I2 m R2C 2 sin2 ω0t The total instantaneous stored energy is therefore constant: w(t) = wL(t) + wC(t) = I2 m R2C 2 and this constant value must also be the maximum value. In order to ﬁnd the energy lost in the resistor in one period, we take the average power absorbed by the resistor (see Sec. 11.2), PR = 1 2I2 m R and multiply by one period, obtaining PRT = 1 2 f0 I2 m R We thus ﬁnd the quality factor at resonance: Q0 = 2π I2 m R2C/2 I2 m R/2 f0 or Q0 = 2π f0RC = ω0RC This equation (as well as any expression in Eq. ) holds only for the sim-ple parallel RLC circuit of Fig. 16.1. Equivalent expressions for Q0 which SECTION 16.1 PARALLEL RESONANCE 625 are often quite useful may be obtained by simple substitution: Q0 = R C L = R \|XC,0\| = R \|X L,0\| So we see that for this speciﬁc circuit, decreasing the resistance de-creases Q0; the lower the resistance, the greater the amount of energy lost in the element. Intriguingly, increasing the capacitance increases Q0, but increasing the inductance leads to a reduction in Q0. These statements, of course, apply to operation of the circuit at the resonant frequency. Other Interpretations of Q Another useful interpretation of Q is obtained when we inspect the inductor and capacitor currents at resonance, as given by Eq. , IC,0 = −IL,0 = jω0CRI = jQ0I Note that each is Q0 times the source current in amplitude and that each is 180◦out of phase with the other. Thus, if we apply 2 mA at the resonant fre-quency to a parallel resonant circuit with a Q0 of 50, we ﬁnd 2 mA in the resistor, and 100 mA in both the inductor and the capacitor. A parallel reso-nant circuit can therefore act as a current ampliﬁer, but not, of course, as a power ampliﬁer, since it is a passive network. Resonance, by deﬁnition, is fundamentally associated with the forced response, since it is deﬁned in terms of a (purely resistive) input impedance, a sinusoidal steady-state concept. The two most important parameters of a resonant circuit are perhaps the resonant frequency ω0 and the quality fac-tor Q0. Both the exponential damping coefﬁcient and the natural resonant frequency may be expressed in terms of ω0 and Q0: α = 1 2RC = 1 2(Q0/ω0C)C or α = ω0 2Q0 and ωd = ω2 0 −α2 or ωd = ω0 1 − 1 2Q0 2 Damping Factor For future reference it may be helpful to note one additional relationship involving ω0 and Q0. The quadratic factor appearing in the numerator of Eq. , s2 + 1 RC s + 1 LC CHAPTER 16 FREQUENCY RESPONSE 626 may be written in terms of α and ω0: s2 + 2αs + ω2 0 In the ﬁeld of system theory or automatic control theory, it is traditional to write this factor in a slightly different form that utilizes the dimensionless parameter ζ (zeta), called the damping factor: s2 + 2ζω0s + ω2 0 Comparison of these expressions allows us to relate ζ to other parameters: ζ = α ω0 = 1 2Q0 EXAMPLE 16.1 Consider a parallel RLC circuit such that L 2 mH, Q0 5, and C 10 nF. Determine the value of R and the magnitude of the steady-state admittance at 0.1ω0, ω0, and 1.1ω0. We derived several expressions for Q0, a parameter directly related to energy loss, and hence the resistance in our circuit. Rearranging the expression in Eq. , we calculate R = Q0 L C = 2.236 k Next, we compute ω0, a term we may recall from Chap. 9, ω0 = 1 √ LC = 223.6 krad/s or, alternatively, we may exploit Eq. and obtain the same answer, ω0 = Q0 RC = 223.6 krad/s The admittance of any parallel RLC network is simply Y = 1 R + jωC + 1 jωL and hence \|Y\| = 1 R + jωC + 1 jωL evaluated at the three designated frequencies is equal to \|Y(0.9ω0)\| = 6.504 × 10−4 S \|Y(ω0)\| = 4.472 × 10−4 S \|Y(1.1ω0)\| = 6.182 × 10−4 S We thus obtain a minimum impedance at the resonant frequency, or a maximum voltage response to a particular input current. If we quickly compute the reactance at these three frequencies, we ﬁnd X(0.9ω0) = −4.72 × 10−4 S X(1.1ω0) = 4.72 × 10−4 S X(ω0) = −1.36 × 10−7 We leave it to the reader to show that our value for X(ω0) is nonzero only as a result of rounding error. SECTION 16.2 BANDWIDTH AND HIGH-Q CIRCUITS 627 Now let us interpret Q0 in terms of the pole-zero locations of the ad-mittance Y(s) of the parallel RLC circuit. We will keep ω0 constant; this may be done, for example, by changing R while holding L and C con-stant. As Q0 is increased, the relationships relating α, Q0, and ω0 indicate that the two zeros must move closer to the jω axis. These relationships also show that the zeros must simultaneously move away from the σ axis. The exact nature of the movement becomes clearer when we remember that the point at which s = jω0 could be located on the jω axis by swing-ing an arc, centered at the origin, through one of the zeros and over to the positive jω axis; since ω0 is to be held constant, the radius must be con-stant, and the zeros must therefore move along this arc toward the positive jω axis as Q0 increases. The two zeros are indicated in Fig. 16.4, and the arrows show the path they take as R increases. When R is inﬁnite, Q0 is also inﬁnite, and the two zeros are found at s = ± jω0 on the jω axis. As R decreases, the zeros move toward the σ axis along the circular locus, joining to form a double zero on the σ axis at s = −ω0 when R = 1 2 L/C or Q0 = 1 2. This condi-tion may be recalled as that for critical damping, so that ωd = 0 and α = ω0. Lower values of R and lower values of Q0 cause the zeros to sep-arate and move in opposite directions on the negative σ axis, but these low values of Q0 are not really typical of resonant circuits and we need not track them any further. Later, we will use the criterion Q0 ≥5 to describe a high-Q circuit. When Q0 = 5, the zeros are located at s = −0.1ω0 ± j0.995ω0, and thus ω0 and ωd differ by only one-half of 1 percent. 16.2 • BANDWIDTH AND HIGH-Q CIRCUITS We continue our discussion of parallel resonance by deﬁning half-power frequencies and bandwidth, and then we will make good use of these new concepts in obtaining approximate response data for high-Q circuits. The “width” of a resonance response curve, such as the one shown in Fig. 16.3, may now be deﬁned more carefully and related to Q0. Let us ﬁrst deﬁne the two half-power frequencies ω1 and ω2 as those frequencies at which the magnitude of the input admittance of a parallel resonant circuit is greater than the magnitude at resonance by a factor of √ 2. Since the response curve PRACTICE ● 16.1 A parallel resonant circuit is composed of the elements R = 8 k, L = 50 mH, and C = 80 nF. Compute (a) ω0; (b) Q0; (c) ωd; (d) α; (e) ζ. 16.2 Determine the values of R, L, and C in a parallel resonant circuit for which ω0 = 1000 rad/s, ωd = 998 rad/s, and Yin = 1 mS at resonance. Ans: 16.1: 15.811 krad/s; 10.12; 15.792 krad/s; 781 Np/s; 0.0494. 16.2: 1000 ; 126.4 mH; 7.91 μF. j0 jd –j0 –0 0 0 – –jd j Q0 = 1 Y(s) 2 R = Q0 = ∞ R = ∞ 1 2 L C ■FIGURE 16.4 The two zeros of the admittance Y(s), located at s = −α ± jωd , provide a semicircular locus as R increases from 1 2 L/C to ∞. CHAPTER 16 FREQUENCY RESPONSE 628 of Fig. 16.3 displays the voltage produced across the parallel circuit by a sinusoidal current source as a function of frequency, the half-power fre-quencies also locate those points at which the voltage response is 1/ √ 2, or 0.707, times its maximum value. A similar relationship holds for the imped-ance magnitude. We will designate ω1 as the lower half-power frequency and ω2 as the upper half-power frequency. Bandwidth The (half-power) bandwidth of a resonant circuit is deﬁned as the differ-ence of these two half-power frequencies. B ≡ω2 −ω1 We tend to think of bandwidth as the “width” of the response curve, even though the curve actually extends from ω = 0 to ω = ∞. More exactly, the half-power bandwidth is measured by that portion of the response curve which is equal to or greater than 70.7 percent of the maximum value, as illustrated in Fig. 16.5. These names arise from the fact that a voltage which is 1 / 2 times the resonant voltage is equivalent to a squared voltage which is one-half the squared voltage at resonance. Thus, at the half-power frequencies, the resistor absorbs one-half the power that it does at resonance. 1 0 2 \|V(j)\| \|I\|R 0.707\|I\|R ■FIGURE 16.5 The bandwidth of the circuit response is highlighted in green; it corresponds to the portion of the response curve greater than or equal to 70.7% of the maximum value. We can express the bandwidth in terms of Q0 and the resonant fre-quency. In order to do so, we ﬁrst express the admittance of the parallel RLC circuit, Y = 1 R + j ωC −1 ωL in terms of Q0: Y = 1 R + j 1 R ωω0CR ω0 −ω0R ωω0L or Y = 1 R 1 + jQ0 ω ω0 −ω0 ω SECTION 16.2 BANDWIDTH AND HIGH-Q CIRCUITS 629 We note again that the magnitude of the admittance at resonance is 1/R, and then realize that an admittance magnitude of √ 2/R can occur only when a frequency is selected such that the imaginary part of the bracketed quantity has a magnitude of unity. Thus Q0 ω2 ω0 −ω0 ω2 = 1 and Q0 ω1 ω0 −ω0 ω1 = −1 Solving, we have ω1 = ω0 ⎡ ⎣ 1 + 1 2Q0 2 − 1 2Q0 ⎤ ⎦ ω2 = ω0 ⎡ ⎣ 1 + 1 2Q0 2 + 1 2Q0 ⎤ ⎦ Although these expressions are somewhat unwieldy, their difference provides a very simple formula for the bandwidth: B = ω2 −ω1 = ω0 Q0 Equations and may be multiplied by each other to show that ω0 is exactly equal to the geometric mean of the half-power frequencies: ω2 0 = ω1ω2 or ω0 = √ω1ω2 Circuits possessing a higher Q0 have a narrower bandwidth, or a sharper response curve; they have greater frequency selectivity, or higher quality (factor). Approximations for High-Q Circuits Many resonant circuits are deliberately designed to have a large Q0 in order to take advantage of the narrow bandwidth and high frequency selectivity associated with such circuits. When Q0 is larger than about 5, it is possible to make some useful approximations in the expressions for the upper and lower half-power frequencies and in the general expressions for the re-sponse in the neighborhood of resonance. Let us arbitrarily refer to a “high-Q circuit” as one for which Q0 is equal to or greater than 5. The pole-zero conﬁguration of Y(s) for a parallel RLC circuit having a Q0 of about 5 is shown in Fig. 16.6. Since α = ω0 2Q0 then α = 1 2B Keep in mind that ω2 > ω0, while ωl < ω0. j2 = j(0 + B) jd = j0 j 1 2 1 2 j1 = j(0 – B) s plane s1 Y(s) 1 2 B – 1 2 B s2 . . . ■FIGURE 16.6 The pole-zero constellation of Y(s) for a parallel RLC circuit. The two zeros are exactly 1 2B Np/s (or rad/s) to the left of the jω axis and approximately jω0 rad/s (or Np/s) from the σ axis. The upper and lower half-power frequencies are separated exactly B rad/s, and each is approximately 1 2B rad/s away from the resonant frequency and the natural resonant frequency. CHAPTER 16 FREQUENCY RESPONSE 630 and the locations of the two zeros s1 and s2 may be approximated: s1,2 = −α ± jωd ≈−1 2B ± jω0 Moreover, the locations of the two half-power frequencies (on the positive jω axis) may also be determined in a concise approximate form: ω1,2 = ω0 ⎡ ⎣ 1 + 1 2Q0 2 ∓ 1 2Q0 ⎤ ⎦≈ω0 1 ∓ 1 2Q0 or ω1,2 ≈ω0 ∓1 2B In a high-Q circuit, therefore, each half-power frequency is located ap-proximately one-half bandwidth from the resonant frequency; this is indi-cated in Fig. 16.6. The approximate relationships for ω1 and ω2 in Eq. may be added to show that ω0 is approximately equal to the arithmetic mean of ω1 and ω2 in high-Q circuits: ω0 ≈1 2(ω1 + ω2) Now let us visualize a test point slightly above jω0 on the jω axis. In order to determine the admittance offered by the parallel RLC network at this fre-quency, we construct the three vectors from the critical frequencies to the test point. If the test point is close to jω0, then the vector from the pole is approximately jω0 and that from the lower zero is nearly j2ω0. The admit-tance is therefore given approximately by Y(s) ≈C ( j2ω0)(s −s1) jω0 ≈2C(s −s1) where C is the capacitance, as shown in Eq. . In order to determine a use-ful approximation for the vector (s −s1), let us consider an enlarged view of that portion of the s plane in the neighborhood of the zero s1 (Fig. 16.7). In terms of its cartesian components, we see that s −s1 ≈1 2B + j(ω −ω0) where this expression would be exact if ω0 were replaced by ωd. We now substitute this equation in the approximation for Y(s), Eq. , and factor out 1 2B: Y(s) ≈2C 1 2B 1 + j ω −ω0 1 2B or Y(s) ≈1 R 1 + j ω −ω0 1 2B j0 (approx.) s – s1 B j s = j 1 2 s1 ■FIGURE 16.7 An enlarged portion of the pole-zero constellation for Y(s) of a high-Q 0 parallel RLC circuit. SECTION 16.2 BANDWIDTH AND HIGH-Q CIRCUITS 631 The fraction (ω −ω0)/( 1 2B) may be interpreted as the “number of half-bandwidths off resonance” and abbreviated by N. Thus, Y(s) ≈1 R (1 + jN) where N = ω −ω0 1 2B At the upper half-power frequency, ω2 ≈ω0 + 1 2B, N = +1, and we are one half-bandwidth above resonance. For the lower half-power frequency, ω1 ≈ω0 −1 2B, so that N = −1, locating us one half-bandwidth below resonance. Equation is much easier to use than the exact relationships we have had up to now. It shows that the magnitude of the admittance is \|Y( jω)\| ≈1 R 1 + N 2 while the angle of Y( jω) is given by the inverse tangent of N: ang Y( jω) ≈tan−1 N EXAMPLE 16.2 Estimate the location of the two half-power frequencies of the voltage response of a parallel RLC network for which R 40 k, L 1 H, and C 1 64 μF, and determine the approximate value of the admittance for an operating frequency of 8200 rad/s. Identify the goal of the problem. We seek the lower and upper half-power frequencies of the voltage response as well as Y(ω0). Since we are asked to “estimate” and “approximate,” the implication is that this is a high-Q circuit, an assumption we should verify. Collect the known information. Given R, L, and C, we are able to compute ω0 and Q0. If Q0 ≥5, we may invoke approximate expressions for half-power frequencies and admittance near resonance, but regardless could compute these quan-tities exactly if required. Devise a plan. To use approximate expressions, we must ﬁrst determine Q0, the quality factor at resonance, as well as the bandwidth. The resonant frequency ω0 is given by Eq. as 1/ √ LC = 8 krad/s. Thus, Q0 = ω0RC = 5, and the bandwidth is ω0/Q0 = 1.6 krad/s. The value of Q0 for this circuit is sufﬁcient to employ “high-Q” approximations. (Continued on next page) CHAPTER 16 FREQUENCY RESPONSE 632 Construct an appropriate set of equations. The bandwidth is simply B = ω0 Q0 = 1600 rad/s and so ω1 ≈ω0 −B 2 = 7200 rad/s ω1 ≈ω0 + B 2 = 8800 rad/s Equation states that Y(s) ≈1 R (1 + jN) so \|Y( jω)\| ≈1 R 1 + N 2 and ang Y( jω) ≈tan−1 N Determine if additional information is required. We still require N, which tells us how many half-bandwidths ω is from the resonant frequency ω0: N = (8.2 −8)/0.8 = 0.25 Attempt a solution. Now we are ready to employ our approximate relationships for the magnitude and angle of the network admittance, ang Y ≈tan−1 0.25 = 14.04◦ and \|Y\| ≈25 1 + (0.25)2 = 25.77 μS Verify the solution. Is it reasonable or expected? An exact calculation of the admittance using Eq. shows that Y( j8200) = 25.75/13.87◦μS The approximate method therefore leads to values of admittance mag-nitude and angle that are reasonably accurate (better than 2 percent) for this frequency. We leave it to the reader to judge the accuracy of our prediction for ω1 and ω2. PRACTICE ● 16.3 Amarginally high-Q parallel resonant circuit has f0 = 440 Hz with Q0 = 6. Use Eqs. and to obtain accurate values for (a) f1; (b) f2. Now use Eq. to calculate approximate values for (c) f1; (d) f2. Ans: 404.9 Hz; 478.2 Hz; 403.3 Hz; 476.7 Hz. SECTION 16.3 SERIES RESONANCE 633 We conclude our coverage of the parallel resonant circuit by reviewing some key conclusions we have reached: • The resonant frequency ω0 is the frequency at which the imaginary part of the input admittance becomes zero, or the angle of the admittance becomes zero. For this circuit, ω0 = 1/ √ LC. • The circuit’s ﬁgure of merit Q0 is deﬁned as 2π times the ratio of the maximum energy stored in the circuit to the energy lost each period in the circuit. For this circuit, Q0 = ω0RC. • We deﬁned two half-power frequencies, ω1 and ω2, as the frequencies at which the admittance magnitude is √ 2 times the minimum admittance magnitude. (These are also the frequencies at which the voltage response is 70.7 percent of the maximum response.) • The exact expressions for ω1 and ω2 are ω1,2 = ω0 ⎡ ⎣ 1 + 1 2Q0 2 ∓ 1 2Q0 ⎤ ⎦ • The approximate (high-Q0) expressions for ω1 and ω2 are ω1,2 ≈ω0 ∓1 2B • The half-power bandwidth B is given by B = ω2 −ω1 = ω0 Q0 • The input admittance may also be expressed in approximate form for high-Q circuits: Y ≈1 R (1 + jN) = 1 R 1 + N 2/tan−1 N where N is deﬁned as the number of half-bandwidths off resonance, or N = ω −ω0 1 2B This approximation is valid for 0.9ω0 ≤ω ≤1.1ω0. 16.3 • SERIES RESONANCE Although we probably ﬁnd less use for the series RLC circuit than we do for the parallel RLC circuit, it is still worthy of our attention. We will consider the circuit shown in Fig. 16.8. It should be noted that the various circuit + – Is Vs Cs Ls Rs ■FIGURE 16.8 A series resonant circuit. CHAPTER 16 FREQUENCY RESPONSE 634 elements are given the subscript s (for series) for the time being in order to avoid confusing them with the parallel elements when the circuits are compared. Our discussion of parallel resonance occupied two sections of consider-able length. We could now give the series RLC circuit the same kind of treatment, but it is much cleverer to avoid such needless repetition and use the concept of duality. For simplicity, let us concentrate on the conclusions presented in the last paragraph of the preceding section on parallel reso-nance. The important results are contained there, and the use of dual lan-guage enables us to transcribe this paragraph to present the important results for the series RLC circuit. “We conclude our coverage of the series resonant circuit by reviewing some key conclusions we have reached: • The resonant frequency ω0 is the frequency at which the imaginary part of the input impedance becomes zero, or the angle of the impedance becomes zero. For this circuit, ω0 = 1/√CsLs. • The circuit’s ﬁgure of merit Q0 is deﬁned as 2π times the ratio of the maximum energy stored in the circuit to the energy lost each period in the circuit. For this circuit, Q0 = ω0L S/RS. • We deﬁned two half-power frequencies, ω1 and ω2, as the frequencies at which the impedance magnitude is √ 2 times the minimum impedance magnitude. (These are also the frequencies at which the current response is 70.7 percent of the maximum response.) • The exact expressions for ω1 and ω2 are ω1,2 = ω0 1 + 1 2Q0 2 ∓ 1 2Q0 • The approximate (high-Q0) expressions for ω1 and ω2 are ω1,2 ≈ω0 ∓1 2B • The half-power bandwidth B is given by B = ω2 −ω1 = ω0 Q0 • The input admittance may also be expressed in approximate form for high-Q circuits: Y ≈1 R (1 + jN) = 1 R 1 + N 2/tan−1 N where N is deﬁned as the number of half-bandwidths off resonance, or N = ω −ω0 1 2B This approximation is valid for 0.9ω0 ≤ω ≤1.1ω0. From this point on, we will no longer identify series resonant circuits by use of the subscript s, unless clarity requires it. Again, this paragraph is the same as the last paragraph of Sec. 16.2, with the parallel RLC language converted to series RLC language using duality (hence the quotation marks). SECTION 16.3 SERIES RESONANCE 635 EXAMPLE 16.3 The voltage vs 100 cos ωt mV is applied to a series resonant circuit composed of a 10 resistance, a 200 nF capacitance, and a 2 mH inductance. Use both exact and approximate methods to calculate the current amplitude if ω 48 krad/s. The resonant frequency of the circuit is given by ω0 = 1 √ LC = 1 (2 × 10−3)(200 × 10−9) = 50 krad/s Since we are operating at ω = 48 krad/s, which is within 10 percent of the resonant frequency, it is reasonable to apply our approximate relationships to estimate the equivalent impedance of the network provided that we ﬁnd that we are working with a high-Q circuit: Zeq ≈R 1 + N 2/tan−1 N where N is computed once we determine Q0. This is a series circuit, so Q0 = ω0L R = (50 × 103)(2 × 10−3) 10 = 10 which qualiﬁes as a high-Q circuit. Thus, B = ω0 Q0 = 50 × 103 10 = 5 krad/s The number of half-bandwidths off resonance (N) is therefore N = ω −ω0 B/2 = 48 −50 2.5 = −0.8 Thus, Zeq ≈R 1 + N 2/tan−1 N = 12.81/−38.66◦ The approximate current magnitude is then \|Vs\| \|Zeq\| = 100 12.81 = 7.806 mA Using the exact expressions, we ﬁnd that I = 7.746/39.24◦mA and thus \|I\| = 7.746 mA PRACTICE ● 16.4 A series resonant circuit has a bandwidth of 100 Hz and contains a 20 mH inductance and a 2 μF capacitance. Determine (a) f0; (b) Q0; (c) Zin at resonance; (d) f2. Ans: 796 Hz; 7.96; 12.57 + j0 ; 846 Hz (approx.). CHAPTER 16 FREQUENCY RESPONSE 636 The series resonant circuit is characterized by a minimum impedance at resonance, whereas the parallel resonant circuit produces a maximum reso-nant impedance. The latter circuit provides inductor currents and capacitor currents at resonance which have amplitudes Q0 times as great as the source current; the series resonant circuit provides inductor voltages and capacitor voltages which are greater than the source voltage by the factor Q0s. The series circuit thus provides voltage ampliﬁcation at resonance. A comparison of our results for series and parallel resonance, as well as the exact and approximate expressions we have developed, appears in Table 16.1. TABLE ●16.1 A Short Summary of Resonance Yp I IL IC R L C Zs V R L C VC + – VL + – + – Q0 = ω0RC α = 1 2RC Q0 = ω0L R α = R 2L \|IL( jω0)\| = \|IC( jω0)\| = Q0\|I( jω0)\| \|VL( jω0)\| = \|VC( jω0)\| = Q0\|V( jω0)\| Yp = 1 R 1 + jQ0 ω ω0 −ω0 ω Zs = R 1 + jQ0 ω ω0 −ω0 ω Exact expressions ω0 = 1 √ LC = √ω1ω2 ωd = ω2 0 −α2 = ω0 1 − 1 2Q0 2 ω1,2 = ω0 ⎡ ⎣ 1 + 1 2Q0 2 ∓ 1 2Q0 ⎤ ⎦ N = ω −ω0 1 2B B = ω2 −ω1 = ω0 Q0 = 2α Approximate expressions (Q0 ≥5 0.9ω0 ≤ω ≤1.1ω0) ωd ≈ω0 ω1,2 ≈ω0 ∓1 2B ω0 ≈1 2(ω1 + ω2) Yp ≈ √ 1 + N 2 R /tan−1 N Zs ≈R √ 1 + N 2/tan−1 N SECTION 16.4 OTHER RESONANT FORMS 637 16.4 • OTHER RESONANT FORMS The parallel and series RLC circuits of the previous two sections represent idealized resonant circuits. The degree of accuracy with which the idealized model ﬁts an actual circuit depends on the operating frequency range, the Q of the circuit, the materials present in the physical elements, the element sizes, and many other factors. We are not studying the techniques for deter-mining the best model of a given physical circuit, for this requires some knowledge of electromagnetic ﬁeld theory and the properties of materials; we are, however, concerned with the problem of reducing a more compli-cated model to one of the two simpler models with which we are more familiar. The network shown in Fig. 16.9a is a reasonably accurate model for the parallel combination of a physical inductor, capacitor, and resistor. The resistor labeled R1 is a hypothetical resistor that is included to account for the ohmic, core, and radiation losses of the physical coil. The losses in the dielectric within the physical capacitor, as well as the resistance of the phys-ical resistor in the given RLC circuit, are accounted for by the resistor la-beled R2. In this model, there is no way to combine elements and produce a simpler model which is equivalent to the original model for all frequencies. We will show, however, that a simpler equivalent may be constructed which is valid over a frequency band that is usually large enough to include all fre-quencies of interest. The equivalent will take the form of the network shown in Fig. 16.9b. Before we learn how to develop such an equivalent circuit, let us ﬁrst consider the given circuit, Fig. 16.9a. The resonant radian frequency for this network is not 1/ √ LC, although if R1 is sufﬁciently small it may be very close to this value. The deﬁnition of resonance is unchanged, and we may determine the resonant frequency by setting the imaginary part of the input admittance equal to zero: Im{Y( jω)} = Im 1 R2 + jωC + 1 R1 + jωL = 0 or Im 1 R2 + jωC + 1 R1 + jωL R1 −jωL R1 −jωL = Im 1 R2 + jωC + R1 −jωL R2 1 + ω2L2 = 0 Thus, we have the resonance condition that C = L R2 1 + ω2L2 and so ω0 = 1 LC − R1 L 2 (a) Y R1 L C R2 (b) Re Le Ce ■FIGURE 16.9 (a) A useful model of a physical network which consists of a physical inductor, capacitor, and resistor in parallel. (b) A network which can be equivalent to part (a) over a narrow frequency band. CHAPTER 16 FREQUENCY RESPONSE 638 We note that ω0 is less than 1/ √ LC, but sufﬁciently small values of the ra-tio R1/L may result in a negligible difference between ω0 and 1/ √ LC. The maximum magnitude of the input impedance also deserves consideration. It is not R2, and it does not occur at ω0 (or at ω = 1/ √ LC). The proof of these statements will not be shown, because the expressions soon become algebraically cumbersome; the theory, however, is straightfor-ward. Let us be content with a numerical example. Frequency (rad/s) Impedance magnitude (ohms) ■FIGURE 16.10 Plot of \|Z\| vs. ω, generated using the following MATLAB script: EDU» omega = linspace(0,10,100); EDU» for i = 1:100 Y(i) = 1/3 + jomega(i)/8 + 1/(2 + jomega(i)); Z(i) = 1/Y(i); end EDU» plot(omega,abs(Z)); EDU» xlabel(‘frequency (rad/s)’); EDU» ylabel(‘impedance magnitude (ohms)’); EXAMPLE 16.4 Using the values R1 2 , L 1 H, C 125 mF, and R2 3 for Fig. 16.9a, determine the resonant frequency and the impedance at resonance. Substituting the appropriate values in Eq. , we ﬁnd ω0 = 8 −22 = 2 rad/s and this enables us to calculate the input admittance, Y = 1 3 + j2 1 8 + 1 2 + j(2)(1) = 1 3 + 1 4 = 0.583 S and then the input impedance at resonance: Z( j2) = 1 0.583 = 1.714 At the frequency which would be the resonant frequency if R1 were zero, 1 √ LC = 2.83 rad/s the input impedance would be Z( j2.83) = 1.947/−13.26◦ As can be seen in Fig. 16.10, however, the frequency at which the maximum impedance magnitude occurs, indicated by ωm, can be deter-mined to be ωm = 3.26 rad/s, and the maximum impedance magnitude is Z( j3.26) = 1.980/−21.4◦ The impedance magnitude at resonance and the maximum magnitude differ by about 16 percent. Although it is true that such an error may be neglected occasionally in practice, it is too large to neglect on an exam. (The later work in this section will show that the Q of the inductor-resistor combination at 2 rad/s is unity; this low value accounts for the 16 percent discrepancy.) PRACTICE ● 16.5 Referring to the circuit of Fig. 16.9a, let R1 = 1 k and C = 2.533 pF. Determine the inductance necessary to select a resonant frequency of 1 MHz. (Hint: Recall that ω = 2π f .) Ans: 10 mH. SECTION 16.4 OTHER RESONANT FORMS 639 Equivalent Series and Parallel Combinations In order to transform the given circuit of Fig. 16.9a into an equivalent of the form shown in Fig. 16.9b, we must discuss the Q of a simple series or par-allel combination of a resistor and a reactor (inductor or capacitor). We ﬁrst consider the series circuit shown in Fig. 16.11a. The Q of this network is again deﬁned as 2π times the ratio of the maximum stored energy to the en-ergy lost each period, but the Q may be evaluated at any frequency we choose. In other words, Q is a function of ω. It is true that we will choose to evaluate it at a frequency which is, or apparently is, the resonant frequency of some network of which the series arm is a part. This frequency, however, is not known until a more complete circuit is available. The reader is encouraged to show that the Q of this series arm is \|Xs\|/Rs, whereas the Q of the parallel network of Fig. 16.11b is Rp/\|X p\|. Let us now carry out the details necessary to ﬁnd values for Rp and X p so that the parallel network of Fig. 16.11b is equivalent to the series network of Fig. 16.11a at some single speciﬁc frequency. We equate Ys and Yp, Ys = 1 Rs + jXs = Rs −jXs R2 s + X2 s = Yp = 1 Rp −j 1 X p and obtain Rp = R2 s + X2 s Rs X p = R2 s + X2 s Xs Dividing these two expressions, we ﬁnd Rp X p = Xs Rs It follows that the Q’s of the series and parallel networks must be equal: Qp = Qs = Q The transformation equations may therefore be simpliﬁed: Rp = Rs(1 + Q2) X p = Xs 1 + 1 Q2 Also Rs and Xs may be found if Rp and X p are the given values; the trans-formation in either direction may be performed. If Q ≥5, little error is introduced by using the approximate rela-tionships Rp ≈Q2Rs X p ≈Xs (Cp ≈Cs or L p ≈Ls) ■FIGURE 16.11 (a) A series network which consists of a resistance R s and an inductive or capacitive reactance Xs may be transformed into (b) a parallel network such that Ys = Yp at one speciﬁc frequency. The reverse transformation is equally possible. (a) Y s Rs jXs Rp (b) Yp jXp CHAPTER 16 FREQUENCY RESPONSE 640 As a further example of the replacement of a more complicated resonant circuit by an equivalent series or parallel RLC circuit, let us consider a prob-lem in electronic instrumentation. The simple series RLC network in Fig. 16.13a is excited by a sinusoidal voltage source at the network’s reso-nant frequency. The effective (rms) value of the source voltage is 0.5 V, and we wish to measure the effective value of the voltage across the capacitor with an electronic voltmeter (VM) having an internal resistance of 100,000 . That is, an equivalent representation of the voltmeter is an ideal voltmeter in parallel with a 100 k resistor. EXAMPLE 16.5 Find the parallel equivalent of the series combination of a 100 mH inductor and a 5 resistor at a frequency of 1000 rad/s. Details of the network to which this series combination is connected are unavailable. At ω = 1000 rad/s, Xs = 1000(100 × 10−3) = 100 . The Q of this series combination is Q = Xs Rs = 100 5 = 20 Since the Q is sufﬁciently high (20 is much greater than 5), we use Eqs. and to obtain Rp ≈Q2Rs = 2000 and L p ≈Ls = 100 mH Our assertion here is that a 100 mH inductor in series with a 5 resistor provides essentially the same input impedance as does a 100 mH induc-tor in parallel with a 2000 resistor at the frequency 1000 rad/s. To check the accuracy of the equivalence, let us evaluate the input impedance for each network at 1000 rad/s. We ﬁnd Zs( j1000) = 5 + j100 = 100.1/87.1◦ Zp( j1000) = 2000( j100) 2000 + j100 = 99.9/87.1◦ and conclude that the accuracy of our approximation at the transforma-tion frequency is pretty impressive. The accuracy at 900 rad/s is also reasonably good, because Zs( j900) = 90.1/86.8◦ Zp( j900) = 89.9/87.4◦ PRACTICE ● 16.6 At ω = 1000 rad/s, ﬁnd a parallel network that is equivalent to the series combination in Fig. 16.12a. 16.7 Find a series equivalent for the parallel network shown in Fig. 16.12b, assuming ω = 1000 rad/s. Ans: 16.6: 8 H, 640 k; 16.7: 5 H, 250 . 8 H 100 (a) 100 k 5 H (b) ■FIGURE 16.12 (a) A series network for which an equivalent parallel network (at ω = 1000 rad/s) is needed. (b) A parallel network for which an equivalent series network (at ω = 1000 rad/s) is needed. An “ideal” meter is an instrument that measures a particular quantity of interest without disturbing the circuit being tested. Although this is impossible, modern instruments can come very close to being ideal in this respect. SECTION 16.4 OTHER RESONANT FORMS 641 Before the voltmeter is connected, we compute that the resonant fre-quency is 105 rad/s, Q0 = 50, the current is 25 mA, and the rms capacitor voltage is 25 V. (As indicated at the end of Sec. 16.3, this voltage is Q0 times the applied voltage.) Thus, if the voltmeter were ideal, it would read 25 V when connected across the capacitor. However, when the actual voltmeter is connected, the circuit shown in Fig. 16.13b results. In order to obtain a series RLC circuit, it is now neces-sary to replace the parallel RC network with a series RC network. Let us assume that the Q of this RC network is sufﬁciently high that the equivalent series capacitor will be the same as the given parallel capacitor. We do this in order to approximate the resonant frequency of the ﬁnal series RLC circuit. Thus, if the series RLC circuit also contains a 0.01 μF capacitor, the resonant frequency remains 105 rad/s. We need to know this estimated res-onant frequency in order to calculate the Q of the parallel RC network; it is Q = Rp \|X p\| = ωRpCp = 105(105)(10−8) = 100 Since this value is greater than 5, our vicious circle of assumptions is justi-ﬁed, and the equivalent series RC network consists of the capacitor Cs = 0.01 μF and the resistor Rs ≈Rp Q2 = 10 Hence, the equivalent circuit of Fig. 16.13c is obtained. The resonant Q of this circuit is now only 33.3, and thus the voltage across the capacitor in the circuit of Fig. 16.13c is 16 2 3 V. But we need to ﬁnd \|V′ C\|, the voltage across the series RC combination; we obtain \|V′ C\| = 0.5 30 \|10 −j1000\| = 16.67 V VC + – 20 0.01 F 100 k 0.5 V rms = 0 10 mH (a) + – VM VC + – 20 0.01 F 10 0.5 V rms = 0 = 105 10 mH (c) + – . ' VC + – 20 0.01 F 100 k 0.5 V rms = 0 = 105 10 mH (b) + – ' . ■FIGURE 16.13 (a) A given series resonant circuit in which the capacitor voltage is to be measured by a nonideal electronic voltmeter. (b) The effect of the voltmeter is included in the circuit; it reads V ′ c . (c) A series resonant circuit is obtained when the parallel RC network in part (b) is replaced by the series RC network which is equivalent at 105 rad/s. CHAPTER 16 FREQUENCY RESPONSE 642 In the laboratory one afternoon, Dr. Abel gave Sean three practical cir-cuit devices: a resistor, an inductor, and a capacitor, having nominal element values of 20 , 20 mH, and 1μF. The student was asked to connect a variable-frequency voltage source to the series combination of these three elements, to measure the resultant voltage across the resistor as a function of frequency, and then to calculate numerical values for the resonant fre-quency, the Q at resonance, and the half-power bandwidth. The student was also asked to predict the results of the experiment before making the measurements. Sean, ﬁrst drew an equivalent circuit for this problem that was like the circuit of Fig. 16.14, and then calculated f0 = 1 2π √ LC = 1 2π √ 20 × 10−3 × 10−6 = 1125 Hz Q0 = ω0L R = 7.07 B = f0 Q0 = 159 Hz Next, Sean made the measurements that Dr. Abel requested, compared them with the predicted values, and then felt a strong urge to transfer to the busi-ness school. The results were f0 = 1000 Hz Q0 = 0.625 B = 1600 Hz Sean knew that discrepancies of this magnitude could not be characterized as being “within engineering accuracy” or “due to meter errors.” Sadly, the results were handed to the professor. Remembering many past errors in judgment, some of which were even (possibly) self-made, Dr. Abel smiled kindly and called Sean’s attention to the Q-meter (or impedance bridge) which is present in most well-equipped laboratories, and suggested that it might be used to ﬁnd out what these prac-tical circuit elements really looked like at some convenient frequency near resonance, 1000 Hz, for example. Upon doing so, Sean discovered that the resistor had a measured value of 18 and the inductor was 21.4 mH with a Q of 1.2, while the capacitor had a capacitance of 1.41 μF and a dissipation factor (the reciprocal of Q) equal to 0.123. So, with the hope that springs eternal within the heart of every engi-neering undergraduate, Sean reasoned that a better model for the practical inductor would be 21.4 mH in series with ωL/Q = 112 , while a more vs vo + – 1 F 20 20 mH + – ■FIGURE 16.14 A ﬁrst model for a 20 mH inductor, a 1 μF capacitor, and a 20 resistor in series with a voltage generator. The capacitor voltage and \|V′ C\| are essentially equal, since the voltage across the 10 resistor is quite small. The ﬁnal conclusion must be that an apparently good voltmeter may still produce a severe effect on the response of a high-Q resonant circuit. A similar effect may occur when a nonideal ammeter is inserted in the circuit. We wrap up this section with a technical fable. O nce upon a time there was a student named Sean, who had a professor identiﬁed simply as Dr. Abel. SECTION 16.4 OTHER RESONANT FORMS 643 appropriate model for the capacitor would be 1.41 μF in series with 1/ωC Q = 13.9 . Using these data, Sean prepared the modiﬁed circuit model shown as Fig. 16.15 and calculated a new set of predicted values: f0 = 1 2π √ 21.4 × 10−3 × 1.41 × 10−6 = 916 Hz Q0 = 2π × 916 × 21.4 × 10−3 143.9 = 0.856 B = 916/0.856 = 1070 Hz Since these results were much closer to the measured values, Sean was much happier. Dr. Abel, however, being a stickler for detail, pondered the differences in the predicted and measured values for both Q0 and the band-width. “Have you,” Dr. Abel asked, “given any consideration to the output impedance of the voltage source?” “Not yet,” said Sean, trotting back to the laboratory bench. It turned out that the output impedance in question was 50 and so Sean added this value to the circuit diagram, as shown in Fig. 16.16. Using the new equivalent resistance value of 193.9 , improved values for Q0 and B were then obtained: Q0 = 0.635 B = 1442 Hz vs vo + – 1.41 F 18 112 13.9 21.4 mH + – ■FIGURE 16.15 An improved model in which more accurate values are used and the losses in the inductor and capacitor are acknowledged. vo + – vs + – 1.41 F 18 50 13.9 21.4 mH + – 112 ■FIGURE 16.16 The ﬁnal model also contains the output resistance of the voltage source. Since all the theoretical and experimental values now agreed within 10 per-cent, Sean was once again an enthusiastic, conﬁdent engineering student, motivated to start homework early and read the textbook prior to class.1 Dr. Abel simply nodded her head agreeably as she moralized: When using real devices, Watch the models that you choose; Think well before you calculate, And mind your Z’s and Q’s! (1) Okay, this last part is a bit much. Sorry about that. CHAPTER 16 FREQUENCY RESPONSE 644 16.5 • SCALING Some of the examples and problems that we have been solving have in-volved circuits containing passive element values ranging around a few ohms, a few henrys, and a few farads. The applied frequencies were a few radians per second. These particular numerical values were used not be-cause they are those commonly met in practice, but because arithmetic ma-nipulations are so much easier than they would be if it were necessary to carry along various powers of 10 throughout the calculations. The scaling procedures that will be discussed in this section enable us to analyze net-works composed of practical-sized elements by scaling the element values to permit more convenient numerical calculations. We will consider both magnitude scaling and frequency scaling. Let us select the parallel resonant circuit shown in Fig. 16.17a as our example. The impractical element values lead to the unlikely response curve drawn as Fig. 16.17b; the maximum impedance is 2.5 , the resonant fre-quencyis1rad/s, Q0 is5,andthebandwidthis0.2rad/s.Thesenumericalval-ues are much more characteristic of the electrical analog of some mechanical system than they are of any basically electrical device. We have convenient numbers with which to calculate, but an impractical circuit to construct. Z (a) 2.5 2 F 1 H 2 (b) 0 0.5 1 1.5 2 0.5 1 1.5 2 2.5 (rad/s) Z () ■FIGURE 16.17 (a) A parallel resonant circuit used as an example to illustrate magnitude and frequency scaling. (b) The magnitude of the input impedance is shown as a function of frequency. Recall that “ordinate” refers to the vertical axis and “abscissa” refers to the horizontal axis. Our goal is to scale this network in such a way as to provide an impedance maximum of 5000 at a resonant frequency of 5 × 106 rad/s, or 796 kHz. In other words, we may use the same response curve shown in Fig. 16.17b if PRACTICE ● 16.8 The series combination of 10 and 10 nF is in parallel with the series combination of 20 and 10 mH. (a) Find the approximate resonant frequency of the parallel network. (b) Find the Q of the RC branch. (c) Find the Q of the RL branch. (d) Find the three-element equivalent of the original network. Ans: 105 rad/s; 100; 50; 10 nF ∥10 mH ∥33.3 k. SECTION 16.5 SCALING 645 every number on the ordinate scale is increased by a factor of 2000 and every number on the abscissa scale is increased by a factor of 5 × 106.We will treat this as two problems: (1) scaling in magnitude by a factor of 2000 and (2) scaling in frequency by a factor of 5 × 106. Magnitude scaling is deﬁned as the process by which the impedance of a two-terminal network is increased by a factor of Km, the frequency re-maining constant. The factor Km is real and positive; it may be greater or smaller than unity. We will understand that the shorter statement ‘‘the net-work is scaled in magnitude by a factor of 2” indicates that the impedance of the new network is to be twice that of the old network at any frequency. Let us now determine how we must scale each type of passive element. To increase the input impedance of a network by a factor of Km, it is sufﬁcient to increase the impedance of each element in the network by this same factor. Thus, a resistance R must be replaced by a resistance KmR. Each inductance must also exhibit an impedance which is Km times as great at any frequency. In order to increase an impedance sL by a factor of Km when s remains constant, the inductance L must be replaced by an inductance KmL. In a similar manner, each capacitance C must be replaced by a capac-itance C/Km. In summary, these changes will produce a network which is scaled in magnitude by a factor of Km: R →Km R L →KmL C →C Km ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭ magnitude scaling When each element in the network of Fig. 16.17a is scaled in magnitude by a factor of 2000, the network shown in Fig. 16.18a results. The response curve shown in Fig. 16.18b indicates that no change in the previously drawn response curve need be made other than a change in the scale of the ordinate. Let us now take this new network and scale it in frequency. We deﬁne frequency scaling as the process by which the frequency at which any im-pedance occurs is increased by a factor of K f . Again, we will make use of the shorter expression “the network is scaled in frequency by a factor of 2” to indicate that the same impedance is now obtained at a frequency twice as great. Frequency scaling is accomplished by scaling each passive element in frequency. It is apparent that no resistor is affected. The impedance of any inductor is sL, and if this same impedance is to be obtained at a frequency K f times as great, then the inductance L must be replaced by an inductance of L/K f . Similarly, a capacitance C is to be replaced by a capacitance C/K f . Thus, if a network is to be scaled in frequency by a factor of K f , then the changes necessary in each passive element are R →R L →L K f C →C K f ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭ frequency scaling Z' (a) 5 k 10–3 F 1000 H (b) 0 0.5 1 1.5 2 1 2 3 4 5 (rad/s) ⏐Z'⏐ (k) ■FIGURE 16.18 (a) The network of Fig. 16.17a after being scaled in magnitude by a factor K m = 2000. (b) The corresponding response curve. CHAPTER 16 FREQUENCY RESPONSE 646 When each element of the magnitude-scaled network of Fig. 16.18a is scaled in frequency by a factor of 5 × 106 , the network of Fig. 16.19a is obtained. The corresponding response curve is shown in Fig. 16.19b. The circuit elements in this last network have values which are easily achieved in physical circuits; the network can actually be built and tested. It follows that, if the original network of Fig. 16.17a were actually an analog of some mechanical resonant system, we could have scaled this analog in both magnitude and frequency in order to achieve a network which we might construct in the laboratory; tests that are expensive or inconvenient to run on the mechanical system could then be made on the scaled electrical system, and the results should then be “unscaled” and converted into me-chanical units to complete the analysis. An impedance that is given as a function of s may also be scaled in mag-nitude or frequency, and this may be done without any knowledge of the speciﬁc elements of which the two-terminal network is composed. In order to scale Z(s) in magnitude, the deﬁnition of magnitude scaling shows that it is necessary only to multiply Z(s) by Km in order to obtain the magnitude-scaled impedance. Hence, the impedance Z′(s) of the magnitude-scaled network is Z′(s) = KmZ(s) If Z′(s) is now to be scaled in frequency by a factor of 5 × 106, then Z′′(s) and Z′(s) are to provide identical values of impedance if Z′′(s) is evaluated at a frequency K f times that at which Z′(s) is evaluated, or Z′′(s) = Z′ s K f Although scaling is a process normally applied to passive elements, de-pendent sources may also be scaled in magnitude and frequency. We assume that the output of any source is given as kxvx or kyiy, where kx has the dimensions of an admittance for a dependent current source and is dimen-sionless for a dependent voltage source, while ky has the dimensions of ohms for a dependent voltage source and is dimensionless for a dependent ■FIGURE 16.19 (a) The network of Fig. 16.18a after being scaled in frequency by a factor K f = 5 × 106. (b) The corresponding response curve. Z'' (a) 5 k 200 H 200 pF (b) 0 2.5 5 7.5 10 1 2 3 4 5 (Mrad/s) ⏐Z''⏐(k) SECTION 16.5 SCALING 647 current source. If the network containing the dependent source is scaled in magnitude by Km, then it is necessary only to treat kx or ky as if it were the type of element consistent with its dimensions. That is, if kx (or ky) is di-mensionless, it is left unchanged; if it is an admittance, it is divided by Km; and if it is an impedance, it is multiplied by Km. Frequency scaling does not affect the dependent sources. EXAMPLE 16.6 Scale the network shown in Fig. 16.20 by Km 20 and Kf 50, and then ﬁnd Zin(s) for the scaled network. ■FIGURE 16.20 (a) A network to be magnitude-scaled by a factor of 20, and frequency-scaled by a factor of 50. (b) The scaled network. (c) A 1 A test source is applied to the input terminals in order to obtain the impedance of the unscaled network in part (a). 0.5s 0.2V1 20/s 1 A Vin + – V1 + – (c) Zin V1 + – 0.2V1 0.05 F 0.5 H (a) Zin 200 mH 0.01V1 50 F V1 + – (b) Magnitude scaling of the capacitor is accomplished by dividing 0.05 F by the scaling factor Km = 20, and frequency scaling is accomplished by dividing by K f = 50. Carrying out both operations simultaneously, Cscaled = 0.05 (20)(50) = 50 μF The inductor is also scaled: Lscaled = (20)(0.5) 50 = 200 mH In scaling the dependent source, only magnitude scaling need be con-sidered, as frequency scaling does not affect dependent sources. Since (Continued on next page) CHAPTER 16 FREQUENCY RESPONSE 648 16.6 • BODE DIAGRAMS In this section we will discover a quick method of obtaining an approximate picture of the amplitude and phase variation of a given transfer function as functions of ω. Accurate curves may, of course, be plotted after calculating values with a programmable calculator or a computer; curves may also be produced directly on the computer. Our object here, however, is to obtain a better picture of the response than we could visualize from a pole-zero plot, but yet not mount an all-out computational offensive. this is a voltage-controlled current source, the multiplying constant 0.2 has units of A/V, or S. Since the factor has units of admittance, we divide by Km, so that the new term is 0.01V1. The resulting (scaled) network is shown in Fig. 16.20b. To ﬁnd the impedance of the new network, we need to apply a 1 A test source at the input terminals. We may work with either circuit; however, let’s proceed by ﬁrst ﬁnding the impedance of the unscaled network in Fig. 16.20a, and then scaling the result. Referring to Fig. 16.20c, Vin = V1 + 0.5s(1 −0.2V1) Also, V1 = 20 s (1) Performing the indicated substitution followed by a little algebraic manipulation yields Zin = Vin 1 = s2 −4s + 40 2s To scale this quantity to correspond to the circuit of Fig. 16.20b we multiply by Km = 20, and replace s with s/Kf = s/50. Thus, Zinscaled = 0.2s2 −40s + 20,000 s PRACTICE ● 16.9 Aparallel resonant circuit is deﬁned by C = 0.01 F, B = 2.5 rad/s, and ω0 = 20 rad/s. Find the values of R and L if the network is scaled in (a) magnitude by a factor of 800; (b) frequency by a factor of 104; (c) magnitude by a factor of 800 and frequency by a factor of 104. Ans: 32 k, 200 H; 40 , 25 μH; 32 k, 20 mH. SECTION 16.6 BODE DIAGRAMS 649 The decibel is named in honor of Alexander Graham Bell. (2) Note that we are being slightly dishonest here by using 20 log 2 = 6 dB rather than 6.02 dB. It is customary, however, to represent √ 2 as 3 dB; since the dB scale is inherently logarithmic, the small inaccuracy is seldom signiﬁcant. The Decibel (dB) Scale The approximate response curve we construct is called an asymptotic plot, or a Bode plot, or a Bode diagram, after its developer, Hendrik W. Bode, who was an electrical engineer and mathematician with the Bell Telephone Laboratories. Both the magnitude and phase curves are shown using a logarithmic frequency scale for the abscissa, and the magnitude itself is also shown in logarithmic units called decibels (dB). We deﬁne the value of \|H( jω)\| in dB as follows: HdB = 20 log \|H( jω)\| where the common logarithm (base 10) is used. (A multiplier of 10 instead of 20 is used for power transfer functions, but we will not need it here.) The inverse operation is \|H( jω)\| = 10HdB/20 Before we actually begin a detailed discussion of the technique for drawing Bode diagrams, it will help to gain some feeling for the size of the decibel unit, to learn a few of its important values, and to recall some of the properties of the logarithm. Since log1 = 0, log 2 = 0.30103, and log 10 = 1, we note the correspondences: \|H( jω)\| = 1 ⇔HdB = 0 \|H( jω)\| = 2 ⇔HdB ≈6 dB \|H( jω)\| = 10 ⇔HdB = 20 dB An increase of \|H( jω)\| by a factor of 10 corresponds to an increase in HdB by 20 dB. Moreover, log 10n = n, and thus 10n ⇔20n dB, so that 1000 corresponds to 60 dB, while 0.01 is represented as −40 dB. Using only the values already given, we may also note that 20 log 5 = 20 log 10 2 = 20 log 10 −20 log 2 = 20 −6 = 14 dB, and thus 5 ⇔14 dB. Also, log √x = 1 2 log x, and therefore √ 2 ⇔3 dB and 1/ √ 2 ⇔−3 dB.2 We will write our transfer functions in terms of s, substituting s = jω when we are ready to ﬁnd the magnitude or phase angle. If desired, the mag-nitude may be written in terms of dB at that point. PRACTICE ● 16.10 Calculate HdB at ω = 146 rad/s if H(s) equals (a) 20/(s + 100); (b) 20(s + 100); (c) 20s. Calculate \|H( jω)\| if HdB equals (d) 29.2 dB; (e) −15.6 dB; ( f ) −0.318 dB. Ans: −18.94 dB; 71.0 dB; 69.3 dB; 28.8; 0.1660; 0.964. CHAPTER 16 FREQUENCY RESPONSE 650 Determination of Asymptotes Our next step is to factor H(s) to display its poles and zeros. We ﬁrst con-sider a zero at s = −a, written in a standardized form as H(s) = 1 + s a The Bode diagram for this function consists of the two asymptotic curves approached by HdB for very large and very small values of ω. Thus, we begin by ﬁnding \|H( jω)\| = 1 + jω a = 1 + ω2 a2 and thus HdB = 20 log 1 + jω a = 20 log 1 + ω2 a2 When ω ≪a, HdB ≈20 log 1 = 0 (ω ≪a) This simple asymptote is shown in Fig. 16.21. It is drawn as a solid line for ω < a, and as a dashed line for ω > a. –20 20 40 0 0.1a 0.01a 10a 100a a HdB (log) ■FIGURE 16.21 The Bode amplitude plot for H(s) = 1 + s/a consists of the low- and high-frequency asymptotes, shown as dashed lines. They intersect on the abscissa at the corner frequency. The Bode plot represents the response in terms of two asymptotes, both straight lines and both easily drawn. When ω ≫a, HdB ≈20 log ω a (ω ≫a) At ω = a, HdB = 0; at ω = 10a, HdB = 20 dB; and at ω = 100a, HdB = 40 dB. Thus, the value of HdB increases 20 dB for every 10-fold increase in fre-quency. The asymptote therefore has a slope of 20 dB/decade. Since HdB increases by 6 dB when ω doubles, an alternate value for the slope is 6 dB/ octave. The high-frequency asymptote is also shown in Fig. 16.21, a solid A decade refers to a range of frequencies deﬁned by a factor of 10, such as 3 Hz to 30 Hz, or 12.5 MHz to 125 MHz. An octave refers to a range of frequencies deﬁned by a factor of 2, such as 7 GHz to 14 GHz. SECTION 16.6 BODE DIAGRAMS 651 line for ω > a, and a broken line for ω < a. Note that the two asymptotes intersect at ω = a, the frequency of the zero. This frequency is also desc-ribed as the corner, break, 3 dB, or half-power frequency. Smoothing Bode Plots Let us see how much error is embodied in our asymptotic response curve. At the corner frequency (ω = a), HdB = 20 log 1 + a2 a2 = 3 dB as compared with an asymptotic value of 0 dB. At ω = 0.5a, we have HdB = 20 log √ 1.25 ≈1 dB Thus, the exact response is represented by a smooth curve that lies 3 dB above the asymptotic response at ω = a, and 1 dB above it at ω = 0.5a (and also at ω = 2a). This information can always be used to smooth out the corner if a more exact result is desired. Multiple Terms Most transfer functions will consist of more than a simple zero (or simple pole). This, however, is easily handled by the Bode method, since we are in fact working with logarithms. For example, consider a function H(s) = K 1 + s s1 1 + s s2 where K = constant, and −s1 and −s2 represent the two zeros of our function H(s). For this function HdB may be written as HdB = 20 log K 1 + jω s1 1 + jω s2 = 20 log K 1 + ω s1 2 1 + ω s2 2 or HdB = 20 log K + 20 log 1 + ω s1 2 + 20 log 1 + ω s2 2 which is simply the sum of a constant (frequency-independent) term 20 log K and two simple zero terms of the form previously considered. In other words, we may construct a sketch of HdB by simply graphically adding the plots of the separate terms. We explore this in the following example. Note that we continue to abide by the convention of taking 2 as corresponding to 3 dB. CHAPTER 16 FREQUENCY RESPONSE 652 Phase Response Returning to the transfer function of Eq. , we would now like to deter-mine the phase response for the simple zero, ang H( jω) = ang 1 + jω a = tan−1 ω a This expression is also represented by its asymptotes, although three straight-line segments are required. For ω ≪a, ang H( jω) ≈0◦, and we use this as our asymptote when ω < 0.1a: ang H( jω) = 0◦ (ω < 0.1a) At the high end, ω ≫a, we have ang H( jω) ≈90◦, and we use this above ω = 10a: ang H( jω) = 90◦ (ω > 10a) Since the angle is 45◦at ω = a, we now construct the straight-line asymp-tote extending from 0◦at ω = 0.1a, through 45◦at ω = a, to 90◦at ω = 10a. This straight line has a slope of 45◦/decade. It is shown as a solid curve in Fig. 16.24, while the exact angle response is shown as a broken line. The maximum differences between the asymptotic and true responses are ±5.71◦at ω = 0.1a and 10a. Errors of ∓5.29◦occur at ω = 0.394a and ■FIGURE 16.23 (a) The Bode plots for the factors of H(s) = 20(1 + s/100) are sketched individually. (b) The composite Bode plot is shown as the sum of the plots of part (a). (a) 1 10 100 + 20 dB/dec. 20 log 20 = 26 dB 1000 40 20 (log) (rad/s) HdB (b) 1 10 100 26 dB 1000 40 20 (log) (rad/s) HdB + 20 dB/dec. EXAMPLE 16.7 H(s) = Zin (s) 0.2 H 20 ■FIGURE 16.22 If H(s) is selected as Zin(s) for this network, then the Bode plot for HdB is as shown in Fig. 16.23b. Obtain the Bode plot of the input impedance of the network shown in Fig. 16.22. We have the input impedance, Zin(s) = H(s) = 20 + 0.2s Putting this in standard form, we obtain H(s) = 20 1 + s 100 The two factors constituting H(s) are a zero at s = −100, leading to a break frequency of ω = 100 rad/s, and a constant equivalent to 20 log 20 = 26 dB. Each of these is sketched lightly in Fig. 16.23a. Since we are working with the logarithm of \|H( jω)\|, we next add together the Bode plots corresponding to the individual factors. The resultant magnitude plot appears as Fig. 16.23b. No attempt has been made to smooth out the corner with a +3 dB correction at ω = 100 rad/s; this is left to the reader as a quick exercise. PRACTICE ● 16.11 Construct a Bode magnitude plot for H(s) = 50 + s. Ans: 34 dB, ω < 50 rad/s; slope = +20 dB/decade ω > 50 rad/s. SECTION 16.6 BODE DIAGRAMS 653 135° 90° 45° 0° 0.1a 0.01a 10a 100a a ang H(j) (log) ■FIGURE 16.24 The asymptotic angle response for H(s) = 1 + s/a is shown as the three straight-line segments in solid color. The endpoints of the ramp are 0◦at 0.1a and 90◦at 10a. The dashed line represents a more accurate (smoothed) response. 2.54a; the error is zero at ω = 0.159a, a, and 6.31a. The phase angle plot is typically left as a straight-line approximation, although smooth curves can also be drawn in a manner similar to that depicted in Fig. 16.24 by the dashed line. PRACTICE ● 16.12 Draw the Bode phase plot for the transfer function of Example 16.7. Ans: 0◦, ω ≤10; 90◦, ω ≥1000; 45◦, ω = 100; 45◦/dec slope, 10 < ω < 1000. (ω in rad/s). Additional Considerations in Creating Bode Plots We next consider a simple pole, H(s) = 1 1 + s/a Since this is the reciprocal of a zero, the logarithmic operation leads to a Bode plot which is the negative of that obtained previously. The amplitude is 0 dB up to ω = a, and then the slope is −20 dB/decade for ω > a. The angle plot is 0◦for ω < 0.1a, −90◦for ω > 10a, and −45◦at ω = a, and it has a slope of −45◦/decade when 0.1a < ω < 10a.The reader is encouraged to generate the Bode plot for this function by working directly with Eq. . It is worth pausing brieﬂy here to consider what the phase plot is telling us. In the case of a simple zero at s = a, we see that for frequencies much less than the corner frequency, the phase of the response function is 0◦. For high frequencies, however (ω ≫a), the phase is 90◦. In the vicinity of the corner frequency, the phase of the transfer function varies somewhat rapidly. The actual phase angle imparted to the response can therefore be selected through the design of the circuit (which determines a). CHAPTER 16 FREQUENCY RESPONSE 654 EXAMPLE 16.8 Obtain the Bode plot for the gain of the circuit shown in Fig. 16.26. Vx + – Vout + – + – 4 kV 10 nF 5 kV Vin Vx 200 20 F 1 kV ■FIGURE 16.26 If H(s) = Vout/Vin, this ampliﬁer is found to have the Bode amplitude plot shown in Fig. 16.27b, and the phase plot shown in Fig. 16.28. We work from left to right through the circuit and write the expression for the voltage gain, H(s) = Vout Vin = 4000 5000 + 106/20s −1 200 5000(108/s) 5000 + 108/s which simpliﬁes (mercifully) to H(s) = −2s (1 + s/10)(1 + s/20,000) Another term that can appear in H(s) is a factor of s in the numerator or denominator. If H(s) = s, then HdB = 20 log \|ω\| Thus, we have an inﬁnite straight line passing through 0 dB at ω = 1 and having a slope everywhere of 20 dB/decade. This is shown in Fig. 16.25a. If the s factor occurs in the denominator, a straight line is obtained having a slope of −20 dB/decade and passing through 0 dB at ω = 1, as shown in Fig. 16.25b. Another simple term found in H(s) is the multiplying constant K. This yields a Bode plot which is a horizontal straight line lying 20 log \|K\| dB above the abscissa. It will actually be below the abscissa if \|K\| < 1. 20 –20 0.1 1 10 (a) 100 Slope of + 20 dB per decade HdB (log) (rad/s) 20 –20 0.1 1 10 (b) 100 HdB (log) (rad/s) Slope of – 20 dB per decade ■FIGURE 16.25 The asymptotic diagrams are shown for (a) H(s) = s and (b) H(s) = 1/s. Both are inﬁnitely long straight lines passing through 0 dB at ω = 1 and having slopes of ±20 dB/decade. SECTION 16.6 BODE DIAGRAMS 655 Before we construct the phase plot for the ampliﬁer of Fig. 16.26, let us take a few moments to investigate several of the details of the magnitude plot. First, it is wise not to rely too heavily on graphical addition of the indi-vidual magnitude plots. Instead, the exact value of the combined magnitude plot may be found easily at selected points by considering the asymptotic 40 20 –20 0.1 1 10 100 6 dB 103 104 105 106 (a) (log) (rad/s) Vout Vin dB 40 20 –20 0.1 1 10 100 103 104 105 106 (b) (log) (rad/s) Vout Vin dB ■FIGURE 16.27 (a) Individual Bode magnitude sketches are made for the factors (−2), (s), (1 + s/10)−1, and (1 + s/20,000)−1. (b) The four separate plots of part (a) are added to give the Bode magnitude plots for the ampliﬁer of Fig. 16.26. PRACTICE ● 16.13 Construct a Bode magnitude plot for H(s) equal to (a) 50/(s + 100); (b) (s + 10)/(s + 100); (c) (s + 10)/s. Ans: (a) −6 dB, ω < 100; −20 dB/decade, ω > 100; (b) −20 dB, ω < 10; +20 dB/decade, 10 < ω < 100; 0 dB, ω > 100; (c) 0 dB, ω > 10; −20 dB/decade, ω < 10. We see a constant, 20 log \| −2\| = 6 dB, break points at ω = 10 rad/s and ω = 20,000 rad/s, and a linear factor s. Each of these is sketched in Fig. 16.27a, and the four sketches are added to give the Bode magni-tude plot in Fig. 16.27b. CHAPTER 16 FREQUENCY RESPONSE 656 value of each factor of H(s) at the point in question. For example, in the ﬂat region of Fig. 16.27a between ω = 10 and ω = 20,000, we are below the corner at ω = 20,000, and so we represent (1 + s/20,000) by 1; but we are above ω = 10, so (1 + s/10) is represented as ω/10. Hence, HdB = 20 log −2ω (ω/10)(1) = 20 log 20 = 26 dB (10 < ω < 20,000) We might also wish to know the frequency at which the asymptotic re-sponse crosses the abscissa at the high end. The two factors are expressed here as ω/10 and ω/20,000; thus HdB = 20 log −2ω (ω/10)(ω/20,000) = 20 log 400,000 ω Since HdB = 0 at the abscissa crossing, 400,000/ω = 1, and therefore ω = 400,000 rad/s. Many times we do not need an accurate Bode plot drawn on printed semilog paper. Instead we construct a rough logarithmic frequency axis on simple lined paper. After selecting the interval for a decade—say, a dis-tance L extending from ω = ω1 to ω = 10ω1 (where ω1 is usually an integral power of 10)—we let x locate the distance that ω lies to the right of ω1, so that x/L = log(ω/ω1). Of particular help is the knowledge that x = 0.3L when ω = 2ω1, x = 0.6L at ω = 4ω1, and x = 0.7L at ω = 5ω1. EXAMPLE 16.9 Draw the phase plot for the transfer function given by Eq. , H(s) −2s/[(1 + s/10)(1 + s/20,000)]. We begin by inspecting H( jω): H( jω) = −j2ω (1 + jω/10)(1 + jω/20,000) The angle of the numerator is a constant, −90◦. The remaining factors are represented as the sum of the angles con-tributed by break points at ω = 10 and ω = 20,000. These three terms appear as broken-line asymptotic curves in Fig. 16.28, and their sum is shown as the solid curve. An equivalent representation is obtained if the curve is shifted upward by 360◦. Exact values can also be obtained for the asymptotic phase response. For example, at ω = 104 rad/s, the angle in Fig. 16.28 is obtained from the numerator and denominator terms in Eq. . The numerator angle is −90◦. The angle for the pole at ω = 10 is −90◦, since ω is greater than 10 times the corner frequency. Between 0.1 and 10 times the corner fre-quency, we recall that the slope is −45◦per decade for a simple pole. For the break point at 20,000 rad/s, we therefore calculate the angle, −45◦log(ω/0.1a) = −45◦log[10,000/(0.1 × 20,000)] = −31.5◦. SECTION 16.6 BODE DIAGRAMS 657 Higher-Order Terms The zeros and poles that we have been considering are all ﬁrst-order terms, such as s±1, (1 + 0.2s)±1, and so forth. We may extend our analysis to higher-order poles and zeros very easily, however. A term s±n yields a magnitude response that passes through ω = 1 with a slope of ±20n dB/decade; the phase response is a constant angle of ±90n◦. Also, a multiple zero, (1 + s/a)n, must represent the sum of n of the magnitude-response curves, or n of the phase-response curves of the simple zero. We therefore obtain an asymptotic magnitude plot that is 0 dB for ω < a and has a slope of 20n dB/decade when ω > a; the error is −3n dB at ω = a, and −n dB at ω = 0.5a and 2a. The phase plot is 0◦for ω < 0.1a, 90n◦for ω > 10a , 45n◦at ω = a , and a straight line with a slope of 45n◦/ decade for 0.1a < ω < 10a, and it has errors as large as ±5.71n◦at two frequencies. The asymptotic magnitude and phase curves associated with a factor such as (1 + s/20)−3 may be drawn quickly, but the relatively large errors associated with the higher powers should be kept in mind. 90° –90° –180° –270° 1 0° 10 100 103 2 105 2 103 104 105 106 107 ang H(j) (log) (rad/s) ■FIGURE 16.28 The solid curve displays the asymptotic phase response of the ampliﬁer shown in Fig. 16.26. The algebraic sum of these three contributions is −90◦−90◦ −31.5◦= −211.5◦, a value that appears to be moderately near the asymptotic phase curve of Fig. 16.28. PRACTICE ● 16.14 Draw the Bode phase plot for H(s) equal to (a) 50/(s + 100); (b) (s + 10)/(s + 100); (c) (s + 10)/s. Ans: (a) 0◦, ω < 10; −45◦/decade, 10 < ω < 1000; −90◦, ω > 1000; (b) 0◦, ω < 1; +45◦/decade, 1 < ω < 10; 45◦, 10 < ω < 100; −45◦/decade, 100 < ω < 1000; 0◦, ω > 1000; (c) −90◦, ω < 1; +45◦/decade, 1 < ω < 100; 0◦, ω > 100. CHAPTER 16 FREQUENCY RESPONSE 658 Complex Conjugate Pairs The last type of factor we should consider represents a conjugate complex pair of poles or zeros. We adopt the following as the standard form for a pair of zeros: H(s) = 1 + 2ζ s ω0 + s ω0 2 The quantity ζ is the damping factor introduced in Sec. 16.1, and we will see shortly that ω0 is the corner frequency of the asymptotic response. If ζ = 1, we see that H(s) = 1 + 2(s/ω0) + (s/ω0)2 = (1 + s/ω0)2, a second-order zero such as we have just considered. If ζ > 1, then H(s) may be factored to show two simple zeros. Thus, if ζ = 1.25, then H(s) = 1 + 2.5(s/ω0) + (s/ω0)2 = (1 + s/2ω0)(1 + s/0.5ω0), and we again have a familiar situation. A new case arises when 0 ≤ζ ≤1. There is no need to ﬁnd values for the conjugate complex pair of roots. Instead, we determine the low- and high-frequency asymptotic values for both the magnitude and phase re-sponse, and then apply a correction that depends on the value of ζ. For the magnitude response, we have HdB = 20 log \|H( jω)\| = 20 log 1 + j2ζ ω ω0 − ω ω0 2 When ω ≪ω0, HdB = 20 log \|1\| = 0 dB. This is the low-frequency asymp-tote. Next, if ω ≫ω0, only the squared term is important, and HdB = 20 log \|−(ω/ω0)2\| = 40 log(ω/ω0). We have a slope of +40 dB/decade. This is the high-frequency asymptote, and the two asymptotes intersect at 0 dB, ω = ω0. The solid curve in Fig. 16.29 shows this asymptotic repre-sentation of the magnitude response. However, a correction must be applied –10 10 20 ζ = 1 ζ = 0.5 ζ = 0.25 ζ = 0.1 30 0.10 +40 dB/dec. 0 100 0.010 HdB (log) (rad/s) ■FIGURE 16.29 Bode amplitude plots are shown for H(s) = 1 + 2ζ(s/ω0) + (s/ω0)2 for several values of the damping factor ζ. SECTION 16.6 BODE DIAGRAMS 659 in the neighborhood of the corner frequency. We let ω = ω0 in Eq. and have HdB = 20 log j2ζ ω ω0 = 20 log(2ζ) If ζ = 1, a limiting case, the correction is +6 dB; for ζ = 0.5, no correction is required; and if ζ = 0.1, the correction is −14 dB. Knowing this one cor-rection value is often sufﬁcient to draw a satisfactory asymptotic magnitude response. Figure 16.29 shows more accurate curves for ζ = 1, 0.5, 0.25, and 0.1, as calculated from Eq. . For example, if ζ = 0.25, then the exact value of HdB at ω = 0.5ω0 is HdB = 20 log \|1 + j0.25 −0.25\| = 20 log 0.752 + 0.252 = −2.0 dB The negative peaks do not show a minimum value exactly at ω = ω0, as we can see by the curve for ζ = 0.5. The valley is always found at a slightly lower frequency. If ζ = 0, then H( jω0) = 0 and HdB = −∞. Bode plots are not usually drawn for this situation. Our last task is to draw the asymptotic phase response for H( jω) = 1 + j2ζ(ω/ω0) −(ω/ω0)2. Below ω = 0.1ω0, we let ang H( jω) = 0◦; above ω = 10ω0, we have ang H( jω) = ang [−(ω/ω0)2] = 180◦. At the corner frequency, ang H( jω0) = ang ( j2ζ) = 90◦. In the interval 0.1ω0 < ω < 10ω0, we begin with the straight line shown as a solid curve in Fig. 16.30. It extends from (0.1ω0, 0◦), through (ω0, 90◦), and terminates at (10ω0, 180◦); it has a slope of 90◦/decade. 30° 60° 90° +90°/dec ζ = 0.1 ζ = 0.25 ζ = 0.5 ζ = 1 120° 150° 180° 0.10 100 0 0.010 ang H(j) (log) (rad/s) ■FIGURE 16.30 The straight-line approximation to the phase characteristic for H( jω) = 1 + j2ζ(ω/ω0) −(ω/ω0)2 is shown as a solid curve, and the true phase response is shown for ζ = 1, 0.5, 0.25, and 0.1 as broken lines. We must now provide some correction to this basic curve for various values of ζ. From Eq. , we have ang H( jω) = tan−1 2ζ(ω/ω0) 1 −(ω/ω0)2 One accurate value above and one below ω = ω0 may be sufﬁcient to give an approximate shape to the curve. If we take ω = 0.5ω0, we find ang H( j0.5ω0) = tan−1(4ζ/3), while the angle is 180◦−tan−1(4ζ/3) at CHAPTER 16 FREQUENCY RESPONSE 660 ω = 2ω0. Phase curves are shown as broken lines in Fig. 16.30 for ζ = 1, 0.5, 0.25, and 0.1; heavy dots identify accurate values at ω = 0.5ω0 and ω = 2ω0. If the quadratic factor appears in the denominator, both the magnitude and phase curves are the negatives of those just discussed. We conclude with an example that contains both linear and quadratic factors. EXAMPLE 16.10 Construct the Bode plot for the transfer function H(s) 100,000s/[(s + 1)(10,000 + 20s + s2)]. Let’s consider the quadratic factor ﬁrst and arrange it in a form such that we can see the value of ζ. We begin by dividing the second-order factor by its constant term, 10,000: H(s) = 10s (1 + s)(1 + 0.002s + 0.0001s2) An inspection of the s2 term next shows that ω0 = √1/0.0001 = 100. Then the linear term of the quadratic is written to display the factor 2, the factor (s/ω0), and ﬁnally the factor ζ: H(s) = 10s (1 + s)[1 + (2)(0.1)(s/100) + (s/100)2] We see that ζ = 0.1. The asymptotes of the magnitude-response curve are sketched in lightly in Fig. 16.31: 20 dB for the factor of 10, an inﬁnite straight line through ω = 1 with a +20 dB/decade slope for the s factor, a corner at ω = 1 for the simple pole, and a corner at ω = 100 with a slope of −40 dB/decade for the second-order term in the denominator. Adding these four curves and supplying a correction of +14 dB for the qua-dratic factor lead to the heavy curve of Fig. 16.31. –20 20 0.1 1 10 100 HdB (log) (rad/s) ■FIGURE 16.31 The Bode magnitude plot of the transfer function H(s) = 100,000s (s + 1)(10,000 + 20s + s2) . The phase response contains three components: +90◦for the factor s; 0◦for ω < 0.1, −90◦for ω > 10, and −45◦/decade for the simple pole; and 0◦for ω < 10, −180◦for ω > 1000, and −90◦per decade for SECTION 16.6 BODE DIAGRAMS 661 the quadratic factor. The addition of these three asymptotes plus some improvement for ζ = 0.1 is shown as the solid curve in Fig. 16.32. 90° –90° –180° 1 0.1 10 100 1000 ang H(j) (log) (rad/s) ■FIGURE 16.32 The Bode phase plot of the transfer function H(s) = 100,000s (s + 1)(10,000 + 20s + s2) . PRACTICE ● 16.15 If H(s) = 1000s2/(s2 + 5s + 100), sketch the Bode amplitude plot and calculate a value for (a) ω when HdB = 0; (b) HdB at ω = 1; (c) HdB as ω →∞. Ans: 0.316 rad/s; 20 dB; 60 dB. COMPUTER-AIDED ANALYSIS The technique of generating Bode plots is a valuable one. There are many situations in which an approximate diagram is needed quickly (such as on exams, or when evaluating a particular circuit topology for a speciﬁc application), and simply knowing the general shape of the re-sponse is adequate. Further, Bode plots can be invaluable when design-ing ﬁlters in terms of enabling us to select factors and coefﬁcient values. In situations where exact response curves are required (such as when verifying a ﬁnal circuit design), there are several computer-assisted op-tions available to the engineer. The ﬁrst technique we will consider here is the use of MATLAB to generate a frequency response curve. In order to accomplish this, the circuit must ﬁrst be analyzed to obtain the cor-rect transfer function. However, it is not necessary to factor or simplify the expression. Consider the circuit in Fig. 16.26. We previously determined that the transfer function for this circuit can be expressed as H(s) = −2s (1 + s/10)(1 + s/20,000) (Continued on next page) CHAPTER 16 FREQUENCY RESPONSE 662 We seek a detailed graph of this function over the frequency range of 100 mrad/s to 1 Mrad/s. Since the ﬁnal graph will be plotted on a logarithmic scale, there is no need to uniformly space our discrete frequencies. Instead, we use the MATLAB function logspace() to gen-erate a frequency vector, where the ﬁrst two arguments represent the power of 10 for starting and ending frequencies, respectively (−1 and 6 in the present example), and the third argument is the total number of points desired. Thus, our MATLAB script is EDU» w = logspace(−1,6,100); EDU» denom = (1+jw/10) . (1+jw/20000); EDU» H = −2jw ./ denom; EDU» Hdb = 20log10(abs(H)); EDU» semilogx(w,Hdb) EDU» xlabel(’frequency (rad/s)’) EDU» ylabel (’\|H(jw)\| (dB)’) which yields the graph depicted in Fig. 16.33. ■FIGURE 16.33 Plot of H dB generated using MATLAB. A few comments about the MATLAB code are warranted. First, note that we have substituted s = jω in our expression for H(s). Also, MATLAB treats the variable w as a vector, or one-dimensional matrix. As such, this variable can cause difﬁculties in the denominator of an expression as MATLAB will attempt to apply matrix algebra rules to any expression. Thus, the denominator of H( jω) is computed in a separate line, and the operator “.” is required instead of “” to multiply the two terms. This new operator is equivalent to the following MATLAB code: EDU» for k = 1:100 denom = (1 + jw(k)/10) (1 + jw(k)/20000); end In a similar fashion, the new operator “./” is used in the subsequent line of code. The results are desired in dB, so the function log10() is SECTION 16.6 BODE DIAGRAMS 663 invoked; log() represents the natural logarithm in MATLAB. Finally, the new plot command semilogx() is used to generate a graph with the x axis having a logarithmic scale. The reader is encouraged at this point to return to previous examples, and use these techniques to generate exact curves for comparison to the corresponding Bode plots. PSpice is also commonly used to generate frequency response curves, especially to evaluate a ﬁnal design. Figure 16.34a depicts the circuit of Fig. 16.26, where the voltage across the resistor R3 represents the desired output voltage. The source component VAC has been (a) (b) ■FIGURE 16.34 (a) The circuit of Fig. 16.26. (b) Frequency response of the circuit plotted in dB scale. (Continued on next page) CHAPTER 16 FREQUENCY RESPONSE 664 16.7 • BASIC FILTER DESIGN The design of ﬁlters is a very practical (and interesting) subject, worthy of a separate textbook in its own right. In this section, we introduce some of the basic concepts of ﬁltering, and explore both passive and active ﬁlter circuits. These circuits may be very simple, consisting of a single capacitor or induc-tor whose addition to a given network leads to improved performance. They may also be fairly sophisticated, consisting of many resistors, capacitors, in-ductors, and op amps in order to obtain the precise response curve required employed with a ﬁxed voltage of 1 V for convenience. An ac sweep simulation is required to determine the frequency response of our circuit; Fig. 16.34b was generated using 10 points per decade (with Decade selected under Logarithmic AC Sweep Type) from 10 mHz to 1 MHz. Note the simulation has been performed in Hz, not rad/s, so the cursor tool is indicating a bandwidth of 3.137 kHz. Again, the reader is encouraged to simulate example circuits and compare the results to the Bode plots generated previously. ⏐H⏐(dB) Frequency (Hz) (a) 101 102 103 Passband 104 105 106 107 –60 –50 –40 –30 –20 –10 0 10 Stopband ⏐H⏐(dB) Frequency (Hz) (d) 100 105 1010 Stopband –60 –80 –100 –120 –140 –40 –20 0 High-frequency passband Low-frequency passband ⏐H⏐(dB) Frequency (Hz) (c) 102 103 Passband 104 105 106 107 –60 –50 –40 –30 –20 –10 0 10 Low-frequency stopband High-frequency stopband ⏐H⏐(dB) Frequency (Hz) (b) 101 102 103 Passband 104 105 106 107 –60 –50 –40 –30 –20 –10 0 10 Stopband ■FIGURE 16.35 Frequency response curves for (a) a low-pass ﬁlter; (b) a high-pass ﬁlter; (c) a bandpass ﬁlter; (d) a bandstop ﬁlter. In each diagram, a solid dot corresponds to −3 dB. SECTION 16.7 BASIC FILTER DESIGN 665 for a given application. Filters are used in modern electronics to obtain dc voltages in power supplies, eliminate noise in communication channels, separate radio and television channels from the multiplexed signal provided by antennas, and boost the bass signal in a car stereo, to name just a few applications. The underlying concept of a ﬁlter is that it selects the frequencies that may pass through a network. There are several varieties, depending on the needs of a particular application. A low-pass ﬁlter, the response of which is illustrated in Fig. 16.35a, passes frequencies below a cutoff frequency, while signiﬁcantly damping frequencies above that cutoff. A high-pass ﬁlter, on the other hand, does just the opposite, as shown in Fig. 16.35b. The chief ﬁgure of merit of a ﬁlter is the sharpness of the cutoff, or the steepness of the curve in the vicinity of the corner frequency. In general, steeper response curves require more complex circuits. Combining a low-pass and a high-pass ﬁlter can lead to what is known as a bandpass ﬁlter, as illustrated by the response curve shown in Fig. 16.35c. In this type of ﬁlter, the region between the two corner frequencies is referred to as the passband; the region outside the passband is referred to as the stop-band. These terms may also be applied to the low- and high-pass ﬁlters, as indicated in Fig. 16.35a and b. We can also create a bandstop ﬁlter, which allows both high and low frequencies to pass but attenuates any signal with a frequency between the two corner frequencies (Fig. 16.35d). The notch ﬁlter is a specialized bandstop ﬁlter, designed with a narrow response characteristic that blocks a single frequency component of a signal. Multiband ﬁlters are also possible; these are ﬁlter circuits which have multiple passbands and stopbands. The design of such filters is straightforward, but beyond the range of this book. Passive Low-Pass and High-Pass Filters A ﬁlter can be constructed by simply using a single capacitor and a single resistor, as shown in Fig. 16.36a. The transfer function for this low-pass ﬁl-ter circuit is H(s) ≡Vout Vin = 1 1 + RCs H(s) has a single corner frequency, which occurs at ω = 1/RC, and a zero at s = ∞, leading to its “low-pass” ﬁltering behavior. Low frequencies (s →0) result in \|H(s)\| near its maximum value (unity, or 0 dB), and high frequencies (s →∞) result in \|H(s)\| →0. This behavior can be understood qualitatively by considering the impedance of the capacitor: as the fre-quency increases, the capacitor begins to act like a short-circuit to ac signals, leading to a reduction in the output voltage. An example response curve for such a ﬁlter with R = 500 and C = 2 nF is shown in Fig. 16.36b; the cor-ner frequency of 159 kHz (1 Mrad/s) can be found by moving the cursor to −3 dB. The sharpness of the response curve in the vicinity of the cutoff frequency can be improved by moving to a circuit containing additional re-active (i.e., capacitive and/or inductive) elements. A high-pass ﬁlter can be constructed by simply swapping the locations of the resistor and capacitor in Fig. 16.36a, as we see in the next example. (a) (b) Vin + – Vout + – R C ■FIGURE 16.36 (a) A simple low-pass RC ﬁlter. (b) Simulated frequency response for R = 500 and C = 2 nf, showing a corner frequency at 159 kHz. CHAPTER 16 FREQUENCY RESPONSE 666 EXAMPLE 16.11 Design a high-pass ﬁlter with a corner frequency of 3 kHz. We begin by selecting a circuit topology. Since no requirements as to the sharpness of the response are given, we choose the simple circuit of Fig. 16.37. The transfer function of this circuit is easily found to be H(s) ≡Vout Vin = RCs 1 + RCs which has a zero at s = 0 and a pole at s = −1/RC, leading to “high-pass” ﬁlter behavior (i.e., \|H\| →0 as ω →∞). The corner frequency of the ﬁlter circuit is ωc = 1/RC, and we seek a value of ωc = 2π fc = 2π(3000) = 18.85 krad/s. Again, we must select a value for either R or C. In practice, our decision would most likely be based on the values of resistors and capacitors at hand, but since no such information has been provided here, we are free to make arbitrary choices. We therefore choose the standard resistor value 4.7 k for R, leading to a requirement of C = 11.29 nF. The only remaining step is to verify our design with a PSpice simu-lation; the predicted frequency response curve is shown in Fig. 16.38. + – Vin R C Vout + – ■FIGURE 16.37 A simple high-pass ﬁlter circuit, for which values for R and C must be selected to obtain a cutoff frequency of 3 kHz. ■FIGURE 16.38 Simulated frequency response of the ﬁnal ﬁlter design, showing a cutoff (3 dB) frequency of 3 kHz as expected. PRACTICE ● 16.16 Design a high-pass ﬁlter with a cutoff frequency of 13.56 MHz, a common RF power supply frequency. Verify your design using PSpice. SECTION 16.7 BASIC FILTER DESIGN 667 Bandpass Filters We have already seen several circuits earlier in this chapter which could be classiﬁed as “bandpass” ﬁlters (e.g., Figs. 16.1 and 16.8). Consider the simple circuit of Fig. 16.39, in which the output is taken across the resistor. The transfer function of this circuit is easily found to be AV = sRC LCs2 + RCs + 1 The magnitude of this function is (after a few algebraic maneuvers) \|AV \| = ωRC (1 −ω2LC)2 + ω2R2C2 which, in the limit of ω →0, becomes \|AV \| ≈ωRC →0 and in the limit of ω →∞becomes \|AV \| ≈R ωL →0 We know from our experience with Bode plots that Eq. represents three critical frequencies: one zero and two poles. In order to obtain a band-pass ﬁlter response with a peak value of unity (0 dB), both pole frequencies must be greater than 1 rad/s, the 0 dB crossover frequency of the zero term. These two critical frequencies can be obtained by factoring Eq. or de-termining the values of ω at which Eq. is equal to 1/ √ 2. The center fre-quency of this ﬁlter then occurs at ω = 1/ √ LC. Thus, applying a minor amount of algebraic manipulation after setting Eq. equal to 1/ √ 2, we ﬁnd that 1 −LCω2 c 2 = ω2 c R2C2 Taking the square root of both sides yields LCω2 c + RCωc −1 = 0 Applying the quadratic equation, we ﬁnd that ωc = −R 2L ± √ R2C2 + 4LC 2LC Negative frequency is a nonphysical solution to our original equation, and so only the positive radicand of Eq. is applicable. However, we may have been a little too hasty in taking the positive square root of both sides of Eq. . Considering the negative square root as well, which is equally valid, we also obtain ωc = R 2L ± √ R2C2 + 4LC 2LC from which it can be shown that only the positive radicand is physical. Thus, we obtain ωL from Eq. and ωH from Eq. ; since ωH −ωL = B, simple algebra shows that B = R/L. Vo + – + – Vi R C L ■FIGURE 16.39 A simple bandpass ﬁlter, constructed using a series RLC circuit. CHAPTER 16 FREQUENCY RESPONSE 668 EXAMPLE 16.12 Design a bandpass ﬁlter characterized by a bandwidth of 1 MHz and a high-frequency cutoff of 1.1 MHz. We choose the circuit topology of Fig. 16.39, and begin by determining the corner frequencies required. The bandwidth is given by fH −fL, so fL = 1.1 × 106 −1 × 106 = 100 kHz and ωL = 2πfL = 628.3 krad/s The high-frequency cutoff (ωH) is simply 6.912 Mrad/s. In order to proceed to design a circuit with these characteristics, it is necessary to obtain an expression for each frequency in terms of the variables R, L, and C. Setting Eq. equal to 2π(1.1 × 106) allows us to solve for 1/LC, as we already know that B = 2π( fH −fL) = 6.283 × 106. 1 2B + 1 4B2 + 1 LC 1/2 = 2π(1.1 × 106) Solving, we ﬁnd that 1/LC = 4.343 × 1012. Arbitrarily selecting L = 50 mH, we obtain R = 314 k and C = 4.6 pF. It should be noted that there is no unique solution for this “design” problem—R, L, or C can be selected as a starting point. PSpice veriﬁcation of our design is shown in Fig. 16.40. ■FIGURE 16.40 Simulated response of the bandpass ﬁlter design showing a bandwidth of 1 MHz and a high-frequency cutoff of 1.1 MHz as desired. SECTION 16.7 BASIC FILTER DESIGN 669 The type of circuit we have been considering is known as a passive ﬁlter, as it is constructed of only passive components (i.e., no transistors, op amps, or other “active” elements). Although passive ﬁlters are relatively common, they are not well suited to all applications. The gain (deﬁned as the output voltage divided by the input voltage) of a passive ﬁlter can be dif-ﬁcult to set, and ampliﬁcation is often desirable in ﬁlter circuits. Active Filters The use of an active element such as the op amp in ﬁlter design can overcome many of the shortcomings of passive ﬁlters. As we saw in Chap. 6, op amp circuits can easily be designed to provide gain. Op amp circuits can also exhibit inductorlike behavior through the strategic location of capacitors. The internal circuitry of an op amp contains very small capacitances (typically on the order of 100 pF), and these limit the maximum frequency at which the op amp will function properly. Thus, any op amp circuit will behave as a low-pass ﬁlter, with a cutoff frequency for modern devices of perhaps 20 MHz or more (depending on the circuit gain). PRACTICE ● 16.17 Design a bandpass ﬁlter with a low-frequency cutoff of 100 rad/s and a high-frequency cutoff of 10 krad/s. Ans: One possible answer of many: R = 990 , L = 100 mH, and C = 10 μF. EXAMPLE 16.13 Design an active low-pass ﬁlter with a cutoff frequency of 10 kHz and a voltage gain of 40 dB. For frequencies much less than 10 kHz, we require an ampliﬁer circuit capable of providing a gain of 40 dB, or 100 V/V. This can be accom-plished by simply using a noninverting ampliﬁer (such as the one shown in Fig. 16.41a) with Rf R1 + 1 = 100 (Continued on next page) – + Vo R1 Rf V1 (a) – + Vo R1 C Rf V1 V+ (b) R2 + – ■FIGURE 16.41 (a) A simple noninverting op amp circuit. (b)A low-pass ﬁlter consisting of a resistor R2 and a capacitor C has been added to the input. CHAPTER 16 FREQUENCY RESPONSE 670 (a) (b) ■FIGURE 16.42 (a) Frequency response for ﬁlter circuit using a μA741 op amp, showing a corner frequency of 6.4 kHz. (b) Frequency response of the same ﬁlter circuit, but using an LF411 op amp instead. The cutoff frequency for this circuit is 10 kHz, the desired value. To provide a high-frequency corner at 10 kHz, we require a low-pass ﬁlter at the input to the op amp (as in Fig. 16.41b). To derive the transfer function, we begin at the noninverting input, V+ = Vi 1/sC R2 + 1/sC = Vi 1 1 + sR2C At the inverting input we have Vo −V+ Rf = V+ R1 Combining these two equations and solving for Vo, we ﬁnd that Vo = Vi 1 1 + sR2C 1 + Rf R1 The maximum value of the gain AV = Vo/Vi is 1 + Rf /R1, so we set this quantity equal to 100. Since neither resistor appears in the expres-sion for the corner frequency (R2C)−1, either may be selected ﬁrst. We thus choose R1 = 1 k, so Rf = 99 k. Arbitrarily selecting C = 1 μF, we ﬁnd that R2 = 1 2π(10 × 103)C = 15.9 At this point, our design is complete. Or is it? The simulated frequency response of this circuit is shown in Fig. 16.42a. It is readily apparent that our design does not in fact meet the 10 kHz cutoff speciﬁcation. What did we do wrong? A careful check of our algebra does not yield any errors, so an erroneous assumption must have been made somewhere. The simulation was performed using a μA741 op amp, as opposed to the ideal op amp assumed in the deriva-tions. It turns out that this is the source of our discomfort—the same circuit with an LF411 op amp substituted for the μA741 results in a cutoff frequency of 10 kHz as desired; the corresponding simulation result is shown in Fig. 16.42b. Unfortunately, the μA741 op amp with a gain of 40 dB has a corner frequency in the vicinity of 10 kHz, which cannot be neglected in this instance. The LF411, however, does not reach its ﬁrst corner frequency until approximately 75 kHz, which is far enough away from 10 kHz that it does not affect our design. PRACTICE ● 16.18 Design a low-pass ﬁlter circuit with a gain of 30 dB and a cutoff frequency of 1 kHz. Ans: One possible answer of many: R1 = 100 k, Rf = 3.062 M, R2 = 79.58 , and C = 2 μF. RACTICAL APPLICATION Bass, Treble, and Midrange Adjustment PRACTICAL APPLICATION ■FIGURE 16.44 Block diagram of a simple graphic equalizer circuit. This signal must be sent to three different op amp cir-cuits, each with a different ﬁlter at the input. The bass ad-justment circuit will require a low-pass ﬁlter, the treble adjustment circuit will require a high-pass ﬁlter, and the midrange adjustment circuit requires a bandpass ﬁlter. The output of each op amp circuit is then fed into a sum-ming ampliﬁer circuit; a block diagram of the complete circuit is shown in Fig. 16.44. The ability to independently adjust the bass, treble, and midrange settings on a sound system is commonly desir-able, even in the case of inexpensive equipment. The au-dio frequency range (at least for the human ear) is commonly accepted to be 20 Hz to 20 kHz, with bass corresponding to lower frequencies (< 500 Hz or so) and treble corresponding to higher frequencies (> 5 kHz or thereabouts). Designing a simple graphic equalizer is a relatively straightforward endeavor, although a system such as that shown in Fig. 16.43 requires a bit more effort. In the bass, midrange, treble type equalizer common on many portable radios, the main signal (provided by the radio receiver circuit, or perhaps a CD player) consists of a wide spectrum of frequencies having a bandwidth of ap-proximately 20 kHz. would call this resistor the volume control. The low-pass ﬁlter network restricts the frequencies that will enter the op amp and hence be ampliﬁed; the corner frequency is simply (R2C)−1. If the circuit designer needs to allow the user to also select the break frequency for the ﬁlter, R2 may be replaced by a potentiometer, or, alternatively, C could be replaced by a variable capacitor. The remaining stages are constructed in essentially the same way, but with a different ﬁlter network at the input. In order to keep the resistors, capacitors, and op amps separate, we should add an appropriate subscript to each as an indication of the stage to which it belongs (t, m, b). Beginning with the treble stage, we have already en-countered problems in using the μA741 in the 10 to 20 kHz range at high gain, so perhaps the LF411 is a bet-ter choice here as well. Selecting a treble cutoff fre-quency of 5 kHz (there is some variation among values selected by different audio circuit designers), we require 1 R2tCt = 2π(5 × 103) = 3.142 × 104 Arbitrarily selecting Ct = 1 μF results in a required value of 31.83 for R2t. Selecting Cb = 1 μF as well (perhaps we can negotiate a quantity discount), we need R2b = 318.3 for a bass cutoff frequency of 500 Hz. We leave the design of a suitable bandpass ﬁlter for the reader. The next part of our design is to choose suitable val-ues for R1t and R1b, as well as the corresponding feed-back resistors. Without any instructions to the contrary, it is probably simplest to make both stages identical. Therefore, we arbitrarily select both R1t and R1b as 1 k, and Rf t and Rf b as 10 k potentiometers (meaning that the range will be from 0 to 10 k). This allows the volume of one signal to be up to 11 times louder than the other. In case we need our design to be portable, we se-lect ±9 V supply voltages, although this can be easily changed if needed. – + Vo R1 C Rf Vin R2 + – ■FIGURE 16.45 The bass adjustment section of the ampliﬁer circuit. ■FIGURE 16.43 An example of a graphic equalizer. Courtesy of Alesis. Low-pass filter Bandpass filter High-pass filter Amplifier Amplifier Amplifier Summing amplifier Speaker Vin Our basic building block is shown in Fig. 16.45. This circuit consists of a noninverting op amp circuit charac-terized by a voltage gain of 1 + Rf /R1, and a simple low-pass ﬁlter composed of a resistor R2 and a capacitor C. The feedback resistor Rf is a variable resistor (some-times referred to as a potentiometer), and allows the gain to be varied through the rotation of a knob; the layperson (Continued on next page) 16.8 • ADVANCED FILTER DESIGN Although the basic ﬁlters we have encountered so far function adequately for a number of applications, their characteristics are far from an ideal “step-function-like” magnitude response. Fortunately, we have alternatives— known as higher-order ﬁlters—with improved behavior, at the cost of increased complexity and more components. For example, the general low-pass ﬁlter transfer function of order n may be written as N(s) = Ka0 sn + an−1sn−1 + · · · + a1s + a0 and that of the general high-pass ﬁlter (of order n) is only subtly different: N(s) = Ksn sn + an−1sn−1 + · · · + a1s + a0 Now that the design of the ﬁlter stage is complete, we are ready to consider the design of the summing stage. For the sake of simplicity, we should power this op amp stage with the same voltage sources as the other stages, which limits the maximum output voltage magnitude to less than 9 V. We use an inverting op amp conﬁguration, with the output of each of the ﬁlter op amp stages fed di-rectly into its own 1 k resistor. The other terminal of ■FIGURE 16.46 Simulated frequency response of the equalizer design. each 1 k resistor is then connected to the inverting input of the summing ampliﬁer stage. The appropriate potentiometer for the summing ampliﬁer stage must be selected in order to prevent saturation, so that knowledge of both the input voltage range and the output speaker wattage is required. A limited simulation of the ﬁnal design is shown in Fig. 16.46. SECTION 16.8 ADVANCED FILTER DESIGN 673 To represent a bandpass ﬁlter, we need only alter the numerator to Ksn/2, and the band-reject ﬁlter (shown in Fig. 16.35d) has the transfer function N(s) = K s2 + ω2 0 n/2 sn + an−1sn−1 + · · · + a1s + a0 Design of a speciﬁc ﬁlter, then, requires selecting the appropriate transfer function, and choosing a class of polynomials which specify the coefﬁcients a1, a2, etc. In this section, we introduce ﬁlters based on Butterworth and Chebyshev polynomials, two of the most commonly employed in ﬁlter design. The low-pass Butterworth ﬁlter is one of the best-known ﬁlters. It is characterized by an amplitude magnitude \|H( jω)\| = K 1 + (ω/ωc)2n n = 1, 2, 3, . . . which is sketched in Fig. 16.47a for n = 1, 2, 3, and 5; K is a real constant, and ωc represents the critical frequency. As can be seen, the magnitude approaches a step-function-like shape as the order n increases. In contrast, the low-pass Chebyshev ﬁlter is characterized by rather prominent ripples in the passband, the number of which depends upon the order of the ﬁlter as illustrated in Fig. 16.47b. Its magnitude response is described by \|H( jω)\| = K 1 + β2C2 n(ω/ωc) n = 1, 2, 3, . . . where β is a real constant known as the ripple factor and Cn(ω/ωc) denotes the Chebyshev polynomial of the ﬁrst kind of degree n. For convenience, selected coefﬁcients of both polynomial types are listed in Table 16.2. The Sallen-Key Ampliﬁer As seen in Sec. 15.8, we may create an op-amp-based ﬁlter circuit having a double pole simply by cascading two circuits such as the one shown in Fig. 15.49a, in which case we obtain a transfer function H(s) = (1/R1Cf )2 s2 + 2/Rf Cf + (1/Rf Cf )2 (b) (a) ■FIGURE 16.47 Plot of \|H( jω)\| for ﬁrst-, second-, and third-order low-pass (a) Butterworth ﬁlters and (b) Chebyshev ﬁlters. All ﬁlters were normalized to a corner frequency of 1 rad/s. CHAPTER 16 FREQUENCY RESPONSE 674 If we wish to improve upon this basic approach, a circuit worth considering is known as the Sallen-Key ampliﬁer, shown in Fig. 16.48, conﬁgured as a low-pass ﬁlter. Analysis of this circuit by nodal analysis is straightforward. We ﬁrst deﬁne the gain G of the noninverting ampliﬁer as G ≡RA + RB RB Then voltage division yields Vy = Vx 1 1 + R2C2s and we may write a single nodal equation 0 = Vx −Vi R1 + Vx −Vy R2 + Vx −Vo 1/sC1 Subtituting Eqs. and into Eq. and performing a few algebraic maneuvers, we arrive at an expression for the transfer function of the ampliﬁer, Vo Vi = G/R1R2C1C2 s2 + 1 R1C1 + 1 R2C1 + 1 −G R2C2 s + 1 R1R2C1C2 Noting that the coefﬁcients in Table 16.2 represent ﬁlters with a cutoff fre-quency of 1 rad/s, so that when ﬁnished we will need to make use of the simple scaling techniques described in Sec. 16.5, we are now ready to explore the design of a second-order Butterworth low-pass ﬁlter. + – Vx Vy Vi Vo R2 RA RB R1 C2 C1 ■FIGURE 16.48 General low-pass Sallen-Key ﬁlter circuit. TABLE ● 16.2 Coefﬁcients for Low-Pass Butterworth and Chebyshev (β = 0.9976, or 3 dB) Filter Functions, Normalized to ωc = 1 Butterworth n a0 a1 a2 a3 a4 1 1.0000 2 1.0000 1.4142 3 1.0000 2.0000 2.0000 4 1.0000 2.6131 3.4142 2.6131 5 1.0000 3.2361 5.2361 5.2361 3.2361 Chebyshev (β = 0.9976) n a0 a1 a2 a3 a4 1 1.0024 2 0.7080 0.6449 3 0.2506 0.9284 0.5972 4 0.1770 0.4048 1.1691 0.5816 5 0.0626 0.4080 0.5489 1.4150 0.5744 SECTION 16.8 ADVANCED FILTER DESIGN 675 EXAMPLE 16.14 Design a second-order low-pass Butterworth ﬁlter having a gain of 4 and a corner frequency at 1400 rad/s. We begin by selecting the Sallen-Key prototype shown in Fig. 16.48, and opt for the simpliﬁcation which arises when we set R1 = R2 = R and C1 = C2 = C. With a second-order Butterworth ﬁlter we expect from Table 16.2 to have a denominator polynomial s2 + 1.4142s + 1 and comparing to Eq. RC = 1 and 2 RC + 1 −G RC = 1.414 hence G = RA + RB RB = 1.586 We ﬁrst set values for the two resistors in our gain network (which do not need to undergo scaling) by arbitrarily choosing RB = 1 k, so that RA = 586 . Next, we note that if C = 1 F, then R = 1 , neither of which is a particularly conventional value. Instead we select C′ = 1 μF; this requires scaling the resistor by 106. We also need to frequency-scale to 1400 rad/s. Thus, 10−6 F = 1 F kmkf = 1 F 1400km and km = 714 . Consequently, R′ = km R = 714 . Unfortunately, our design is not quite done. We were constrained to an ampliﬁer gain of 1.586, or 4 dB, but the speciﬁcations called for a gain of 4, or 12 dB. The only option available to us is to feed the output of our circuit into a noninverting ampliﬁer such as the one in Fig. 6.6a. Choosing 1 k and 1.52 k for R1 (output stage) and Rf completes the design. PRACTICE ● 16.19 Design a second-order Butterworth low-pass ﬁlter having a gain of 10 dB and a cutoff frequency of 1000 Hz. Ans: A two-stage circuit, with the output of the circuit of Fig. 16.48 fed into the input of a noninverting ampliﬁer, with component values C1 = C2 = 1 μF, R1 = R2 = 159 , RA = 586 , RB = 1 k (stage 1) and R1 = 1 k, Rf = 994 (stage 2). CHAPTER 16 FREQUENCY RESPONSE 676 Design of high-pass ﬁlters based on the Sallen-Key model is similarly straightforward; the only modiﬁcation required is to replace capacitors C1 and C2 with resistors, and resistors R1 and R2 with capacitors. The remain-der of the circuit remains unchanged. Nodal analysis of the resulting circuit with C1 = C2 = C and R1 = R2 = R yields a0 = 1 R2C2 and a1 = 3 −G RC as we found for the low-pass ﬁlter. Higher-order ﬁlters can be realized by cascading appropriate op amp stages. For example, Butterworth ﬁlters of odd order (e.g., 3, 5, . . .) require an additional pole at s = −1. Thus, a third-order Butterworth filter is constructed using a Sallen-Key stage which provides a transfer function denominator D(s) of (s + 1) s2 + s + 1 s3 + 2s2 + 2s + 1 or D(s) = s2 + s + 1 with an additional op amp stage such as the one in Fig. 15.49a to provide the term (s + 1). Design a third-order low-pass Butterworth ﬁlter having a voltage gain magnitude of 4 and a corner frequency at 2000 rad/s. We begin by again selecting the Sallen-Key prototype shown in Fig. 16.48, and opt for the simpliﬁcation which arises when we set R1 = R2 = R and C1 = C2 = C. We will also add an input stage of the form shown in Fig. 15.49a to add the necessary pole. The basic design is shown in Fig. 16.49. EXAMPLE 16.15 + – – + Vx Vy Vo Vi R R R1 RA RB C C Rf Cf ■FIGURE 16.49 Basic structure of the proposed third-order low-pass Butterworth ﬁlter, with component values still to be chosen. SUMMARY AND REVIEW 677 Comparing Eqs. , , and , we determine that our design must ensure that 1 = 1 R2C2 and 1 = 3 −G RC Consequently, RC = 1 and G = 4. If we choose RA = 3 k, it follows that RB = 1 k. We may scale these values later if we choose when adjusting for operation at 2000 rad/s, but this is unnecessary as the dc gain is set by the ratio of the two resistors. Initially we design for R = 1 and C = 1 F as this automatically satiﬁes the RC = 1 requirement. Neither value being easy to locate, however, we select a more reasonable capacitor value of 0.1 μF, which combined with our frequency scaling factor kf = 2000, results in a resistor scaling factor km = 5000. Thus, R = 5 k in our ﬁnal design. All that remains is to select values for R1, Rf , and Cf in our front-end stage. Recall that the transfer function of this stage is − 1/R1Cf s + (1/Rf Cf ) Setting Rf = 1 and Cf = 1 F initially allows the pole to be properly located prior to scaling operations, which dictate that we build the circuit with Rf = 5 k and Cf = 0.1 μF. Our only remaining choice, then, is to ensure that R1 allows us to meet our gain requirement of 4. Since we have already achieved this with our Sallen-Key stage, R1 must be equal to Rf , or 5 k. Design of Chebyshev ﬁlters proceeds along the same lines as that of Butterworth ﬁlters, except we have more choices now with the ripple factor. Also, for ﬁlters not having a 3 dB ripple factor, the critical frequency is where the ripple channel in the passband terminates, which is slightly dif-ferent than what we have speciﬁed previously. Filters with order n > 2 are constructed by cascading stages, either multiple Sallen-Key stages for even orders, or a simple stage such as Fig. 15.49a in conjunction with the appro-priate number of Sallen-Key stages for odd orders. For ﬁlters with a speciﬁc gain requirement, an op amp stage containing only resistors is typically required at the output. SUMMARY AND REVIEW We began this chapter with a short discussion of resonance. Of course the reader was likely to already have an intuitive understanding of the basic concept—timing when to kick our legs on a swing as a child; watching videos of crystal glasses shattering under the power of a trained soprano’s voice; instinctively slowing down when driving over a corrugated surface. In CHAPTER 16 FREQUENCY RESPONSE 678 the context of linear circuit analysis, we found (perhaps surprisingly) that a frequency can be chosen even for networks with capacitors and inductors such that the voltage and current are in phase (hence the network appears purely resistive at that particular frequency). How quickly our circuit response changes as we move “off resonance” was related to a new term— the quality factor (Q) of our circuit. After deﬁning what is meant by critical frequencies for our circuit response, we introduced the concept of bandwidth, and discovered that our expressions may be simpliﬁed rather dramatically for high-Q (Q > 5) circuits. We brieﬂy extended this discussion to consider the differences between series and parallel circuits near resonance, along with more practical networks which cannot be classed as either. The remainder of this chapter dealt with the analysis and design of ﬁlter circuits. Prior to launching into that discussion, the topic of “scaled” circuit components dealt with both frequency and magnitude scaling as a conve-nient design tool. We also introduced the handy method of Bode plots, which allows us to quickly sketch a reasonable approximation to the response of a ﬁlter circuit as a function of frequency. We next considered both passive and active ﬁlters, starting with simple designs using a single capacitor to achieve either low-pass or high-pass behavior. Shortly there-after, bandpass ﬁlter design was studied. Although they are straightforward to work with, the response of such simple circuits is not particularly abrupt. As an alternative, ﬁlter designs based on either Butterworth or Chebyshev polynomials were examined, with higher-order ﬁlters yielding sharper mag-nitude response at the expense of increased complexity. ❑Resonance is the condition in which a ﬁxed-amplitude sinusoidal forcing function produces a response of maximum amplitude. (Example 16.1) ❑An electrical network is in resonance when the voltage and current at the network input terminals are in phase. (Example 16.1) ❑The quality factor is proportional to the maximum energy stored in a network divided by the total energy lost per period. ❑A half-power frequency is deﬁned as the frequency at which the mag-nitude of a circuit response function is reduced to 1/ √ 2 times its maxi-mum value. ❑A high-Q circuit is a resonant circuit in which the quality factor is ≥5. (Example 16.2) ❑The bandwidth of a resonant circuit is deﬁned as the difference between the upper and lower half-power frequencies. ❑In a high-Q circuit, each half-power frequency is located approximately one-half bandwidth from the resonant frequency. (Example 16.2) ❑A series resonant circuit is characterized by a low impedance at resonance, whereas a parallel resonant circuit is characterized by a high impedance at resonance. (Examples 16.1 and 16.3) ❑A series resonant circuit and a parallel resonant circuit are equivalent if Rp = Rs(1 + Q2) and X p = Xs(1 + Q−2). (Examples 16.4, 16.5) ❑Impractical values for components often make design easier. The trans-fer function of a network may be scaled in magnitude or frequency using appropriate replacement values for components. (Example 16.6) EXERCISES 679 ❑Bode diagrams allow the rough shape of a transfer function to be plotted quickly from the poles and zeros. (Examples 16.7, 16.8, 16.9, 16.10) ❑The four basic types of ﬁlters are low-pass, high-pass, bandpass, and bandstop. (Examples 16.11, 16.12) ❑Passive ﬁlters use only resistors, capacitors, and inductors; active ﬁlters are based on op amps or other active elements. (Example 16.13) ❑Butterworth and Chebyshev ﬁlters can be designed based on the simple Sallen-Key ampliﬁer. Filter gain typically must be adjusted by adding a purely resistor-based ampliﬁer circuit at the output. READING FURTHER A good discussion of a large variety of ﬁlters can be found in: J. T. Taylor and Q. Huang, eds., CRC Handbook of Electrical Filters. Boca Raton, Fla.: CRC Press, 1997. A comprehensive compilation of various active ﬁlter circuits and design proce-dures is given in: D. Lancaster, Lancaster’s Active Filter Cookbook, 2nd ed. Burlington, Mass.: Newnes, 1996. Additional ﬁlter references which the reader might ﬁnd useful include: D. E. Johnson and J. L. Hilburn, Rapid Practical Design of Active Filters. New York: John Wiley & Sons, Inc., 1975. J. V. Wait, L. P. Huelsman, and G. A. Korn, Introduction to Operational Ampliﬁer Theory and Applications, 2nd ed. New York: McGraw-Hill, 1992. EXERCISES 16.1 Parallel Resonance 1. Compute Q0 and ζ for a simple parallel RLC network if (a) R = 1 k, C = 10 mF, and L = 1 H; (b) R = 1 , C = 10 mF, and L = 1 H; (c) R = 1 k, C = 1 F, and L = 1 H; (d) R = 1 , C = 1 F, and L = 1 H. 2. For the circuit shown in Fig. 16.1, let R = 1 k, C = 22 mF, and L = 12 mH. (a) Calculate α, ω0, ζ, f0, and ωd for the circuit. (b) If I = 1/0◦A, plot V, ILC, IL, and IC as a function of frequency, and verify that I and V are in phase at ω0. (c) What is the relationship of IL to IC at ω0? 3. A certain parallel RLC circuit is built using component values L = 50 mH and C = 33 mF. If Q0 = 10, determine the value of R, and the sketch the magni-tude of the steady-state impedance over the range of 2 < ω < 40 rad/s. 4. A parallel RLC network is constructed using R = 5 , L = 100 mH, and C = 1 mF. (a) Compute Q0. (b) Determine at which frequencies the imped-ance magnitude drops to 90% of its maximum value. 5. For the network of Fig. 16.50, derive an expression for the steady-state input impedance and determine the frequency at which it has maximum amplitude. 6. Plot the input admittance of the network depicted in Fig. 16.50 using a loga-rithmic frequency scale over the range 0.01ω0 < ω0 < 100ω0, and determine the resonant frequency and the bandwidth of the network. 7. Delete the 2 resistor in the network of Fig. 16.50 and determine (a) the mag-nitude of the input impedance at resonance; (b) the resonant frequency; (c) Q0. 1 100 k 2 200 mH 10 F ■FIGURE 16.50 CHAPTER 16 FREQUENCY RESPONSE 680 8. Delete the 1 resistor in the network of Fig. 16.50 and determine (a) the mag-nitude of the input impedance at resonance; (b) the resonant frequency; (c) Q0. 9. A varactor is a semiconductor device whose reactance may be varied by apply-ing a bias voltage. The quality factor can be expressed3 as Q ≈ ωCJ RP 1 + ω2C2 J RP RS where CJ is the junction capacitance (which depends on the voltage applied to the device), RS is the series resistance of the device, and RP is an equivalent parallel resistance term. (a) If CJ = 3.77 pF at 1.5 V, RP = 1.5 M, and RS = 2.8 , plot the quality factor as a function of frequency ω. (b) Differenti-ate the expression for Q to obtain both ω0 and Qmax. 16.2 Bandwidth and High-Q Circuits 10. The circuit of Fig. 16.1 is built using component values L = 1 mH and C = 100 μF. If Q0 = 15, determine the bandwidth and estimate the magnitude and angle of the input impedance for operation at (a) 3162 rad/s; (b) 3000 rad/s; (c) 3200 rad/s; (d) 2000 rad/s. (e) Verify your estimates using an exact expres-sion for Y( jω). 11. A parallel RLC network is constructed with a 5 mH inductor, and the remain-ing component values are chosen such that Q0 = 6.5 and ω0 = 1000 rad/s. Determine the approximate value of the input impedance magnitude for opera-tion at (a) 500 rad/s; (b) 750 rad/s; (c) 900 rad/s; (d) 1100 rad/s. (e) Plot your estimates along with the exact result using a linear frequency (rad/s) axis. 12. A parallel RLC network is constructed with a 200 μH inductor, and the remain-ing component values are chosen such that Q0 = 8 and ω0 = 5000 rad/s. Use approximate expressions to estimate the input impedance angle for operation at (a) 2000 rad/s; (b) 3000 rad/s; (c) 4000 rad/s; (d) 4500 rad/s. (e) Plot your estimates along with the exact result using a linear frequency (rad/s) axis. 13. Find the bandwidth of each of the response curves shown in Fig. 16.51. (3) S. M. Sze, Physics of Semiconductor Devices, 2d ed. New York: Wiley, 1981, p. 116. 0 0.2 0.4 0.6 0.8 1 0 1 2 3 4 5 6 (a) f (kHz) 1.5 1.0 0.5 102 103 104 105 106 107 108 (b) f (Hz) ■FIGURE 16.51 14. A parallel RLC circuit is constructed such that it has the impedance magnitude characteristic plotted in Fig. 16.52. (a) Determine the resistor value. (b) Deter-mine the capacitor value if a 1 H inductor was used. (c) Obtain values for the bandwidth, Q0, and both the low half-power frequency and the high half-power frequency. EXERCISES 681 16.3 Series Resonance 15. A series RLC circuit is constructed employing component values R = 100 and L = 1.5 mH along with a sinusoidal voltage source vs. If Q0 = 7, deter-mine (a) the magnitude of the impedance at 500 Mrad/s; (b) the current which ﬂows in response to a voltage vs = 2.5 cos(425 × 106t) V. 16. With regard to the series RLC circuit described in Exercise 15, adjust the resis-tor value such that Q0 is reduced to 5, and (a) estimate the angle of the imped-ance at 90 krad/s, 100 krad/s, and 110 krad/s. (b) Determine the percent error in the estimated values, compared to the exact expression. 17. An RLC circuit is constructed using R = 5 , L = 20 mH, and C = 1 mF. Calculate Q0, the bandwidth, and the magnitude of the impedance at 0.95ω0 if the circuit is (a) parallel-connected; (b) series-connected. (c) Verify your solu-tions using appropriate PSpice simulations. (Hint: a large resistor in parallel with the capacitor will avoid error messages associated with no dc path to ground, and a small resistance in series with the VAC source will avoid short-ing by the inductor during dc bias point determination.) 18. Inspect the circuit of Fig. 16.53, noting the amplitude of the source voltage. Now decide whether you would be willing to put your bare hands across the capacitor if the circuit were actually built in the lab. Plot \|VC\| versus ω to justify your answer. 0 0.5 1 1.5 2 0.5 1.0 1.5 2.0 2.5 (rad/s) ⏐Z⏐ () ■FIGURE 16.52 125 1.5 V 10 0.105V1 4 H VC + – + – F 1 4 V1 + – ■FIGURE 16.53 50 nF 0.5VR 10 1 mH VR + – Zin ■FIGURE 16.54 19. After deriving Zin(s) in Fig. 16.54, ﬁnd (a) ω0; (b) Q0. CHAPTER 16 FREQUENCY RESPONSE 682 16.4 Other Resonant Forms 20. For the network of Fig. 16.9a, R1 = 100 , R2 = 150 , L = 30 mH, and C is chosen so that ω0 = 750 rad/s. Calculate the impedance magnitude at (a) the fre-quency corresponding to resonance when R1 = 0; (b) 700 rad/s; (c) 800 rad/s. 21. Assuming an operating frequency of 200 rad/s, ﬁnd a series equivalent of the parallel combination of a 500 resistor and (a) a 1.5 μF capacitor; (b) a 200 mH inductor. 22. If the frequency of operation is either 40 rad/s or 80 rad/s, ﬁnd a parallel equiv-alent of the series combination of a 2 resistor and (a) a 100 mF capacitor; (b) a 3 mH inductor. 23. For the network represented in Fig. 16.55, determine the resonant frequency and the corresponding value of \|Zin\|. 22 10 100 H 15 75 mH 50 F 100 mH Zin ■FIGURE 16.55 24. For the circuit shown in Fig. 16.56, the voltage source has magnitude 1 V and phase angle 0◦. Determine the resonant frequency ω0 and the value of Vx at 0.95ω0. 5 mH 12 mH 1.8 5 35 mF + – Vx + – ■FIGURE 16.56 16.5 Scaling 25. A parallel RLC circuit is constructed using component values R = 1 , C = 3 F, and L = 1 3 H. Determine the required component values if the network is to have (a) a resonant frequency of 200 kHz; (b) a peak impedance of 500 k; (c) a resonant frequency of 750 kHz and an impedance magnitude at resonance of 25 . 26. Aseries RLC circuit is constructed using component values R = 1 , C = 5 F, and L = 1 5 H. Determine the required component values if the network is to have (a) a resonant frequency of 430 Hz; (b) a peak impedance of 100 ; (c) a resonant frequency of 75 kHz and an impedance magnitude at resonance of 15 k. 27. Scale the network shown in Fig. 16.57 by Km = 200 and K f = 700, and obtain an expression for the new impedance Zin(s). 5 1 H 500 mF 0.2I1 I1 Zin (s) ■FIGURE 16.57 EXERCISES 683 28. The ﬁlter shown in Fig. 16.58a has the response curve shown in Fig. 16.58b. (a) Scale the ﬁlter so that it operates between a 50 source and a 50 load and has a cutoff frequency of 20 kHz. (b) Draw the new response curve. 31.8 H 9.82 H 9.82 H 100 2.57 nF 2.57 nF 100 Vout + – 0° V 100 + – (a) (b) 1 2 3 50 f (MHz) ⏐Vout⏐ (V) ■FIGURE 16.58 29. (a) Draw the new conﬁguration for Fig. 16.59 after the network is scaled by Km = 250 and K f = 400. (b) Determine the Thévenin equivalent of the scaled network at ω = 1 krad/s. 0.1 F 2 H a b 5 4Ix Ix + – ■FIGURE 16.59 16.6 Bode Diagrams 30. Sketch the Bode magnitude and phase plots for the following functions: (a) 3 + 4s; (b) 1 3 + 4s. 31. For the following functions, sketch the Bode magnitude and phase plots: (a) 25 1 + s 3 (5 + s); (b) 0.1 (1 + 5s)(2 + s) . 32. Use the Bode approach to sketch the magnitude of each of the following responses, then verify your solutions with appropriate MATLAB simulations: (a) 3 s s2 + 7s + 10 ; (b) 4 s3 + 7s2 + 12s. 33. If a particular network is described by transfer function H(s), plot the magnitude of H(s) as a function of frequency for H(s) equal to (a) s + 300 s(5s + 8); (b) s(s2 + 7s + 7) s(2s + 4)2 . 34. Sketch the phase plot of each of the following transfer functions: (a) s + 1 s(s + 2)2 ; (b) 5s2 + s s + 2 . CHAPTER 16 FREQUENCY RESPONSE 684 35. Determine the Bode magnitude plot for the following transfer functions, and compare to what is predicted using MATLAB: (a) s2 + 0.2s + 1; (b) s 4 2 + 0.1 s 4 + 1 . 36. Determine the phase plot corresponding to each of the transfer functions in Exercises 33 and 35, and compare your sketches to what is predicted using MATLAB. 37. Determine the Bode magnitude plot for each of the following: (a) 3 + 0.1s + s2/3 s2 + 1 ; (b) 2s2 + 9s + 20 s2(s + 1)3 . 38. For the circuit of Fig. 16.60, (a) derive an expression for the transfer function H(s) = Vout/Vin. (b) Sketch the corresponding Bode magnitude and phase plots. – + Vin + – Vout + – 50 100 – + 200 250 mF 250 mF ■FIGURE 16.60 39. (a) Modify the circuit shown in Fig. 16.60 to add a double pole at 0.05 rad/s and a zero at 0.01 rad/s. (b) Sketch the corresponding Bode magnitude and phase plots. 16.7 Basic Filter Design 40. (a) Design a high-pass ﬁlter with a corner frequency of 100 rad/s. (b) Verify your design with an appropriate PSpice simulation. 41. (a) Design a low-pass ﬁlter with a break frequency of 1450 rad/s. (b) Sketch the Bode magnitude and phase plots for your design. (c) Verify your ﬁlter performance with an appropriate simulation. 42. (a) Design a bandpass ﬁlter characterized by a bandwidth of 1000 rad/s and a low-frequency corner of 250 Hz. (b) Verify your design with an appropriate PSpice simulation. 43. Design a bandpass ﬁlter having a low-frequency cutoff of 500 Hz and a high-frequency cutoff of 1580 Hz. 44. Design a notch ﬁlter which removes 60 Hz “noise” from power line inﬂuences on a particular signal by taking the output across the inductor-capacitor series connection in the circuit of Fig. 16.39. 45. Design a low-pass ﬁlter characterized by a voltage gain of 25 dB and a corner frequency of 5000 rad/s. 46. Design a high-pass ﬁlter characterized by a voltage gain of 30 dB and a corner frequency of 50 rad/s. 47. (a) Design a two-stage op amp ﬁlter circuit with a bandwidth of 1000 rad/s, a low-frequency cutoff of 100 rad/s, and a voltage gain of 20 dB. (b) Verify your design with an appropriate PSpice simulation. 48. Design a circuit which removes the entire audio frequency range (approxi-mately 20 Hz to 20 kHz, for human hearing), but ampliﬁes the signal voltage of all other frequencies by a factor of 15. EXERCISES 685 49. Depending on which song you’re listening to, your MP3 player sometimes provides too little bass, even when the appropriate setting is maximized. Design a ﬁlter which allows you to vary the gain real time of all signals less than 500 Hz prior to reaching your earphones. Include a diagram of the overall system. 16.8 Advanced Filter Design 50. Show that the circuit represented by Eq. cannot be implemented as either a Butterworth or a Chebyshev low-pass ﬁlter. 51. Design a second-order low-pass ﬁlter having a voltage gain of 5 dB and a cutoff frequency of 1700 kHz based on (a) Butterworth polynomials; (b) Chebyshev polynomials for 3 dB ripple factor. 52. If a high-pass ﬁlter is required having gain of 6 dB and a cutoff frequency of 350 Hz, design a suitable second-order Butterworth-based solution. 53. (a) Design a second-order low-pass Butterworth ﬁlter with a cutoff frequency of 890 rad/s and a voltage gain of 8 dB. (b) Verify your design with an appro-priate PSpice simulation. 54. (a) Design a second-order high-pass Butterworth ﬁlter with a cutoff frequency of 2000 Hz and a voltage gain of 4.5 dB. (b) Verify your design with an appro-priate PSpice simulation. 55. A third-order low-pass Butterworth ﬁlter is required having a cutoff frequency of 1200 Hz and a voltage gain of at least 3 dB. Design a suitable circuit. 56. (a) Design a third-order low-pass Butterworth ﬁlter having a gain of 13 dB and a corner frequency at 1800 Hz. (b) Compare your ﬁlter response to that of a Chebyshev ﬁlter with the same speciﬁcations. 57. Design a fourth-order high-pass Butterworth ﬁlter having a minimum gain of 15 dB and a corner frequency of 1100 rad/s. 58. Choose parameters for the circuit described by Eq. such that it has a cutoff frequency at 450 rad/s, and compare its performance to a comparable second-order Butterworth ﬁlter. Chapter-Integrating Exercises 59. Design a parallel resonant circuit for an AM radio so that a variable inductor can adjust the resonant frequency over the AM broadcast band, 535 to 1605 kHz, with Q0 = 45 at one end of the band and Q0 ≤45 throughout the band. Let R = 20 k, and specify values for C, Lmin, and Lmax. 60. Derive an expression for the transfer function Vout/Vin which describes the cir-cuit shown in Fig. 16.61, and sketch its magnitude as a function of frequency. + – Vout Vin R3 R1 R2 C2 C1 ■FIGURE 16.61 61. The network of Fig. 16.36 was implemented as a low-pass ﬁlter designed with a corner frequency of 1250 rad/s. Its performance is inadequate in two respects: (1) a voltage gain of at least 2 dB is required, and (2) the magnitude of the output voltage does not decrease quickly enough in the stopband. CHAPTER 16 FREQUENCY RESPONSE 686 Design a better alternative if only one op amp is available and only two 1 μF capacitors can be located. 62. Determine the effect of component tolerance on the circuit designed in Example 16.14 if each component is speciﬁed to be only within 10% of its stated value. 63. Derive an expression for the transfer function Vout/Vin which describes the cir-cuit shown in Fig. 16.62, and sketch its magnitude as a function of frequency. 64. For the circuit shown in Fig. 16.62, select component values to design for corner frequencies at 500 rad/s and 1500 rad/s. Verify your design. + – C2 RA RB R1 R2 C1 R3 ■FIGURE 16.62 65. Design a bandpass ﬁlter that spans the portion of the audio spectrum from 200 Hz to 2 kHz which has a minimum gain of 5 dB, and a steeper magnitude characteristic on the high-frequency side than on the low-frequency side. Verify your design using a suitable simulation. INTRODUCTION A general network having two pairs of terminals, one often labeled the “input terminals’’ and the other the “output terminals,’’ is a very important building block in electronic systems, communica-tion systems, automatic control systems, transmission and distribu-tion systems, or other systems in which an electrical signal or elec-tric energy enters the input terminals, is acted upon by the network, and leaves via the output terminals. The output terminal pair may very well connect with the input terminal pair of another network. When we studied the concept of Thévenin and Norton equivalent networks in Chap. 5, we were introduced to the idea that it is not always necessary to know the detailed workings of part of a circuit. This chapter extends such concepts to linear networks, resulting in parameters that allow us to predict how any network will interact with other networks. 17.1 • ONE-PORT NETWORKS A pair of terminals at which a signal may enter or leave a network is called a port, and a network having only one such pair of terminals is called a one-port network, or simply a one-port. No connections may be made to any other nodes internal to the one-port, and it is therefore evident that ia must equal ib in the one-port shown in Fig. 17.1a. When more than one pair of terminals is present, the network is known as a multiport network. The two-port network to which this chapter is principally devoted is shown in Fig. 17.1b. The currents in the two leads making up each port must be equal, and so it follows that ia = ib and ic = id in the two-port shown in Fig. 17.1b. Sources and loads must be connected directly across the two termi-nals of a port if the methods of this chapter are to be used. In other KEY CONCEPTS The Distinction Between One-Port and Two-Port Networks Admittance (y) Parameters Impedance (z) Parameters Hybrid (h) Parameters Transmission (t) Parameters Transformation Methods Between y, z, h, and t Parameters Circuit Analysis Techniques Using Network Parameters Two-Port Networks C H A P T E R 17 687 words, each port can be connected only to a one-port network or to a port of another multiport network. For example, no device may be connected between terminals a and c of the two-port network in Fig. 17.1b. If such a cir-cuit must be analyzed, general loop or nodal equations should be written. Some of the introductory study of one- and two-port networks is ac-complished best by using a generalized network notation and the abbrevi-ated nomenclature for determinants introduced in Appendix 2. Thus, if we write a set of loop equations for a passive network, Z11I1 + Z12I2 + Z13I3 + · · · + Z1NIN = V1 Z21I1 + Z22I2 + Z23I3 + · · · + Z2NIN = V2 Z31I1 + Z32I2 + Z33I3 + · · · + Z3NIN = V3 ZN1I1 + ZN2I2 + ZN3I3 + · · · + ZN NIN = VN · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · then the coefﬁcient of each current will be an impedance Zi j(s), and the circuit determinant, or determinant of the coefﬁcients, is Z = Z11 Z12 Z13 · · · Z1N Z21 Z22 Z23 · · · Z2N Z31 Z32 Z33 · · · Z3N · · · · · · · · · · · · · · · ZN1 ZN2 ZN3 · · · ZN N Here N loops have been assumed, the currents appear in subscript order in each equation, and the order of the equations is the same as that of the cur-rents. We also assume that KVL is applied so that the sign of each Zii term (Z11, Z22, . . . , ZN N) is positive; the sign of any Zi j(i ̸= j) or mutual term may be either positive or negative, depending on the reference directions assigned to Ii and Ij. If there are dependent sources within the network, then it is possible that not all the coefﬁcients in the loop equations must be resistances or impedances. Even so, we will continue to refer to the circuit determinant as Z. The use of minor notation (Appendix 2) allows for the input or driving-point impedance at the terminals of a one-port network to be expressed very concisely. The result is also applicable to a two-port network if one of the two ports is terminated in a passive impedance, including an open or a short circuit. Let us suppose that the one-port network shown in Fig. 17.2 is com-posed entirely of passive elements and dependent sources; linearity is also assumed. An ideal voltage source V1 is connected to the port, and the source current is identiﬁed as the current in loop 1. Employing Cramer’s rule, then, I1 = V1 Z12 Z13 · · · Z1N 0 Z22 Z23 · · · Z2N 0 Z32 Z33 · · · Z3N · · · · · · · · · · · · · · · 0 ZN2 ZN3 · · · ZN N Z11 Z12 Z13 · · · Z1N Z21 Z22 Z23 · · · Z2N Z31 Z32 Z33 · · · Z3N · · · · · · · · · · · · · · · ZN1 ZN2 ZN3 · · · ZN N CHAPTER 17 TWO-PORT NETWORKS 688 ia ib (a) ia ic id ib (b) a b c d ■FIGURE 17.1 (a) A one-port network. (b) A two-port network. ■FIGURE 17.2 An ideal voltage source V1 is connected to the single port of a linear one-port network containing no independent sources; Zin = z/11. I1 + – V1 Linear network Cramer’s rule is reviewed in Appendix 2. SECTION 17.1 ONE-PORT NETWORKS 689 or, more concisely, I1 = V111 Z Thus, Zin = V1 I1 = Z 11 EXAMPLE 17.1 Calculate the input impedance for the one-port resistive network shown in Fig. 17.3. ■FIGURE 17.3 An example one-port network containing only resistive elements. 2 Ω 4 Ω 10 Ω 20 Ω 1 Ω 5 Ω V1 + – I3 I2 I4 I1 We ﬁrst assign the four mesh currents as shown and write the corre-sponding mesh equations by inspection: V1 = 10I1 −10I2 0 = −10I1 + 17I2 −2I3 −5I4 0 = −2I2 + 7I3 − I4 0 = −5I2 −I3 + 26I4 The circuit determinant is then given by Z = 10 −10 0 0 −10 17 −2 −5 0 −2 7 −1 0 −5 −1 26 and has the value 9680 4. Eliminating the ﬁrst row and ﬁrst column, we have 11 = 17 −2 −5 −2 7 −1 −5 −1 26 = 2778 3 (Continued on next page) CHAPTER 17 TWO-PORT NETWORKS 690 Thus, Eq. provides the value of the input impedance, Zin = 9680 2778 = 3.485 PRACTICE ● 17.1 Find the input impedance of the network shown in Fig. 17.4 if it is formed into a one-port network by breaking it at terminals (a) a and a′; (b) b and b′; (c) c and c′. ■FIGURE 17.4 5 Ω 6 Ω 7 Ω 3 Ω 4 Ω 2 Ω a c c' a' b b' Ans: 9.47 ; 10.63 ; 7.58 . EXAMPLE 17.2 Find the input impedance of the network shown in Fig. 17.5. ■FIGURE 17.5 A one-port network containing a dependent source. 2 Ω 4 Ω 10 Ω 0.5Ia 1 Ω 5 Ω V1 + – Ia I3 I2 I4 I1 The four mesh equations are written in terms of the four assigned mesh currents: 10I1 −10I2 = V1 −10I1 + 17I2 −2I3 −5I4 = 0 −2I2 + 7I3 −I4 = 0 SECTION 17.1 ONE-PORT NETWORKS 691 and I4 = −0.5Ia = −0.5(I4 −I3) or −0.5I3 + 1.5I4 = 0 Thus we can write Z = 10 −10 0 0 −10 17 −2 −5 0 −2 7 −1 0 0 −0.5 1.5 = 590 4 while 11 = 17 −2 −5 −2 7 −1 0 −0.5 1.5 = 159 3 giving Zin = 590 159 = 3.711 ■FIGURE 17.6 + – 5 S 10 S 3V2 0.2V3 V1 V2 V3 20 S 2 S Ans: 10.68 S; 13.16 S. We may also select a similar procedure using nodal equations, yielding the input admittance: Yin = 1 Zin = Y 11 where 11 now refers to the minor of Y. PRACTICE ● 17.2 Write a set of nodal equations for the circuit of Fig. 17.6, cal-culate Y, and then ﬁnd the input admittance seen between (a) node 1 and the reference node; (b) node 2 and the reference. Exercises 9 and 10 at the end of the chapter give one-ports that can be built using operational ampliﬁers. These exercises illustrate that negative resistances may be obtained from networks whose only passive circuit elements are resistors, and that inductors may be simulated with only resis-tors and capacitors. 17.2 • ADMITTANCE PARAMETERS Let us now turn our attention to two-port networks. We will assume in all that follows that the network is composed of linear elements and contains no independent sources; dependent sources are permissible, however. Fur-ther conditions will also be placed on the network in some special cases. CHAPTER 17 TWO-PORT NETWORKS 692 EXAMPLE 17.3 Use Eq. to again determine the input impedance of the network shown in Fig. 17.3, repeated here as Fig. 17.7. ■FIGURE 17.7 The circuit from Example 17.1, repeated for convenience. 2 Ω 4 Ω 10 Ω 20 Ω 1 Ω 5 Ω V1 + – I3 I2 I4 I1 We ﬁrst order the node voltages V1, V2, and V3 from left to right, select the reference at the bottom node, and then write the system admittance matrix by inspection: Y = 0.35 −0.2 −0.05 −0.2 1.7 −1 −0.05 −1 1.3 = 0.3473 S3 11 = 1.7 −1 −1 1.3 = 1.21 S2 so that Yin = 0.3473 1.21 = 0.2870 S which corresponds to Zin = 1 0.287 = 3.484 which agrees with our previous answer to within expected rounding error (we only retained four digits throughout the calculations). SECTION 17.2 ADMITTANCE PARAMETERS 693 We will consider the two-port as it is shown in Fig. 17.8; the voltage and current at the input terminals are V1 and I1, and V2 and I2 are speciﬁed at the output port. The directions of I1 and I2 are both customarily selected as into the network at the upper conductors (and out at the lower conductors). Since the network is linear and contains no independent sources within it, I1 may be considered to be the superposition of two components, one caused by V1 and the other by V2. When the same argument is applied to I2, we may begin with the set of equations I1 = y11V1 + y12V2 I2 = y21V1 + y22V2 where the y’s are no more than proportionality constants, or unknown coef-ﬁcients, for the present. However, it should be clear that their dimensions must be A/V, or S. They are therefore called the y (or admittance) param-eters, and are deﬁned by Eqs. and . The y parameters, as well as other sets of parameters we will deﬁne later in the chapter, are represented concisely as matrices. Here, we deﬁne the (2 × 1) column matrix I, I = I1 I2 the (2 × 2) square matrix of the y parameters, y = y11 y12 y21 y22 and the (2 × 1) column matrix V, V = V1 V2 Thus, we may write the matrix equation I = yV, or I1 I2 = y11 y12 y21 y22 V1 V2 and matrix multiplication of the right-hand side gives us the equality I1 I2 = y11V1 + y12V2 y21V1 + y22V2 These (2 × 1) matrices must be equal, element by element, and thus we are led to the deﬁning equations, and . The most useful and informative way to attach a physical meaning to the y parameters is through a direct inspection of Eqs. and . Consider Eq. , for example; if we let V2 be zero, then we see that y11 must be given by the ratio of I1 to V1. We therefore describe y11 as the admittance measured at the input terminals with the output terminals short-circuited (V2 = 0). Since there can be no question which terminals are short-circuited, y11 is best described as the short-circuit input admittance. Alternatively, we might describe y11 as the reciprocal of the input impedance measured with the out-put terminals short-circuited, but a description as an admittance is obviously more direct. It is not the name of the parameter that is important. Rather, it ■FIGURE 17.8 A general two-port with terminal voltages and currents speciﬁed. The two-port is composed of linear elements, possibly including dependent sources, but not containing any independent sources. V2 + – V1 + – I2 I1 Linear network The notation adopted in this text to represent a matrix is standard, but also can be easily confused with our previous notation for phasors or general complex quantities. The nature of any such symbol should be clear from the context in which it is used. is the conditions which must be applied to Eq. or , and hence to the network, that are most meaningful; when the conditions are determined, the parameter can be found directly from an analysis of the circuit (or by exper-iment on the physical circuit). Each of the y parameters may be described as a current-voltage ratio with either V1 0 (the input terminals short-circuited) or V2 0 (the output terminals short-circuited): Because each parameter is an admittance which is obtained by short-circuiting either the output or the input port, the y parameters are known as the short-circuit admittance parameters. The speciﬁc name of y11 is the short-circuit input admittance, y22 is the short-circuit output admittance, and y12 and y21 are the short-circuit transfer admittances. y11 = I1 V1 V2=0 y12 = I1 V2 V1=0 y21 = I2 V1 V2=0 y22 = I2 V2 V1=0 CHAPTER 17 TWO-PORT NETWORKS 694 EXAMPLE 17.4 Find the four short-circuit admittance parameters for the resistive two-port shown in Fig. 17.9. The values of the parameters may be easily established by applying Eqs. to , which we obtained directly from the deﬁning equa-tions, and . To determine y11, we short-circuit the output and ﬁnd the ratio of I1 to V1. This may be done by letting V1 1 V, for then y11 = I1. By inspection of Fig. 17.9, it is apparent that 1 V applied at the input with the output short-circuited will cause an input current of ( 1 5 + 1 10), or 0.3 A. Hence, y11 = 0.3 S In order to ﬁnd y12, we short-circuit the input terminals and apply 1 V at the output terminals. The input current ﬂows through the short circuit and is I1 = −1 10 A. Thus y12 = −0.1 S By similar methods, y21 = −0.1 S y22 = 0.15 S The describing equations for this two-port in terms of the admittance parameters are, therefore, I1 = 0.3V1 −0.1V2 I2 = −0.1V1 + 0.15V2 ■FIGURE 17.9 A resistive two-port. 10 Ω 5 Ω 20 Ω V2 + – V1 + – I2 I1 SECTION 17.2 ADMITTANCE PARAMETERS 695 In general, it is easier to use Eq. , , , or when only one parameter is desired. If we need all of them, however, it is usually easier to assign V1 and V2 to the input and output nodes, to assign other node-to-reference voltages at any interior nodes, and then to carry through with the general solution. In order to see what use might be made of such a system of equations, let us now terminate each port with some specific one-port network. and y = 0.3 −0.1 −0.1 0.15 (all S) It is not necessary to ﬁnd these parameters one at a time by using Eqs. to , however. We may ﬁnd them all at once—as shown in the next example. EXAMPLE 17.5 Assign node voltages V1 and V2 in the two-port of Fig. 17.9 and write the expressions for I1 and I2 in terms of them. We have I1 = V1 5 + V1 −V2 10 = 0.3V1 −0.1V2 and I2 = V2 −V1 10 + V2 20 = −0.1V1 + 0.15V2 These equations are identical with Eqs. and , and the four y parameters may be read from them directly. PRACTICE ● 17.3 By applying the appropriate 1 V sources and short circuits to the circuit shown in Fig. 17.10, ﬁnd (a) y11; (b) y21; (c) y22; (d) y12. Ans: 0.1192 S; −0.1115 S; 0.1269 S; −0.1115 S. ■FIGURE 17.10 20 Ω 10 Ω 40 Ω 5 Ω V2 + – V1 + – I2 I1 Consider the simple two-port network of Example 17.4, shown in Fig. 17.11 with a practical current source connected to the input port and a resistive load connected to the output port. A relationship must now exist between V1 and I1 that is independent of the two-port network. This rela-tionship may be determined solely from this external circuit. If we apply KCL (or write a single nodal equation) at the input, I1 = 15 −0.1V1 For the output, Ohm’s law yields I2 = −0.25V2 Substituting these expressions for I1 and I2 in Eqs. and , we have 15 = 0.4V1 −0.1V2 0 = −0.1V1 + 0.4V2 from which are obtained V1 = 40 V V2 = 10 V The input and output currents are also easily found: I1 = 11 A I2 = −2.5 A and the complete terminal characteristics of this resistive two-port are then known. The advantages of two-port analysis do not show up very strongly for such a simple example, but it should be apparent that once the y parameters are determined for a more complicated two-port, the performance of the two-port for different terminal conditions is easily determined; it is neces-sary only to relate V1 to I1 at the input and V2 to I2 at the output. In the example just concluded, y12 and y21 were both found to be −0.1 S. It is not difﬁcult to show that this equality is also obtained if three general impedances ZA, ZB, and ZC are contained in this network. It is somewhat more difﬁcult to determine the speciﬁc conditions which are necessary in order that y12 = y21, but the use of determinant notation is of some help. Let us see if the relationships of Eqs. to can be expressed in terms of the impedance determinant and its minors. Since our concern is with the two-port and not with the speciﬁc net-works with which it is terminated, we will let V1 and V2 be represented by two ideal voltage sources. Equation is applied by letting V2 0 (thus short-circuiting the output) and ﬁnding the input admittance. The network now, however, is simply a one-port, and the input impedance of a one-port was found in Sec. 17.1. We select loop 1 to include the input terminals, and let I1 be that loop’s current; we identify (−I2) as the loop current in loop 2 CHAPTER 17 TWO-PORT NETWORKS 696 ■FIGURE 17.11 The resistive two-port network of Fig. 17.9, terminated with speciﬁc one-port networks. 10 Ω 5 Ω 10 Ω 15 A 20 Ω 4 Ω V2 + – V1 + – I2 I1 SECTION 17.2 ADMITTANCE PARAMETERS 697 and assign the remaining loop currents in any convenient manner. Thus, Zin\|V2=0 = Z 11 and, therefore, y11 = 11 Z Similarly, y22 = 22 Z In order to ﬁnd y12, we let V1 0 and ﬁnd I1 as a function of V2. We ﬁnd that I1 is given by the ratio I1 = 0 Z12 · · · Z1N −V2 Z22 · · · Z2N 0 Z32 · · · Z3N · · · · · · · · · · · · 0 ZN2 · · · ZN N Z11 Z12 · · · Z1N Z21 Z22 · · · Z2N Z31 Z32 · · · Z3N · · · · · · · · · · · · ZN1 ZN2 · · · ZN N Thus, I1 = −(−V2)21 Z and y12 = 21 Z In a similar manner, we may show that y21 = 12 Z The equality of y12 and y21 is thus contingent on the equality of the two minors of Z—12 and 21. These two minors are 21 = Z12 Z13 Z14 · · · Z1N Z32 Z33 Z34 · · · Z3N Z42 Z43 Z44 · · · Z4N · · · · · · · · · · · · · · · ZN2 ZN3 ZN4 · · · ZN N and 12 = Z21 Z23 Z24 · · · Z2N Z31 Z33 Z34 · · · Z3N Z41 Z43 Z44 · · · Z4N · · · · · · · · · · · · · · · ZN1 ZN3 ZN4 · · · ZN N Their equality is shown by ﬁrst interchanging the rows and columns of one minor (for example, 21), an operation which any college algebra book proves is valid, and then letting every mutual impedance Zij be replaced by Zji. Thus, we set Z12 = Z21 Z23 = Z32 etc. This equality of Zij and Zji is evident for the three familiar passive elements—the resistor, capacitor, and inductor—and it is also true for mutual inductance. However, it is not true for every type of device which we may wish to include inside a two-port network. Speciﬁcally, it is not true in general for a dependent source, and it is not true for the gyrator, a useful model for Hall-effect devices and for waveguide sections containing fer-rites. Over a narrow range of radian frequencies, the gyrator provides an additional phase shift of 180° for a signal passing from the output to the input over that for a signal in the forward direction, and thus y12 = −y21. A common type of passive element leading to the inequality of Zij and Zji, however, is a nonlinear element. Any device for which Zij = Zji is called a bilateral element, and a circuit which contains only bilateral elements is called a bilateral circuit. We have therefore shown that an important property of a bilateral two-port is y12 = y21 and this property is gloriﬁed by stating it as the reciprocity theorem: If we had been working with the admittance determinant of the circuit and had proved that the minors 21 and 12 of the admittance determinant Y were equal, then we should have obtained the reciprocity theorem in its dual form: In any passive linear bilateral network, if the single current source Ix between nodes x and x′ produces the voltage response Vy between nodes y and y′, then the removal of the current source from nodes x and x′ and its insertion between nodes y and y′ will produce the voltage re-sponse Vy between nodes x and x′. In any passive linear bilateral network, if the single voltage source Vx in branch x produces the current response Iy in branch y, then the removal of the voltage source from branch x and its insertion in branch y will produce the current response Iy in branch x. CHAPTER 17 TWO-PORT NETWORKS 698 PRACTICE ● 17.4 In the circuit of Fig. 17.10, let I1 and I2 represent ideal current sources. Assign the node voltage V1 at the input, V2 at the output, and Vx from the central node to the reference node. Write three nodal equations, eliminate Vx to obtain two equations, and then rearrange these equations into the form of Eqs. and so that all four y parameters may be read directly from the equations. Another way of stating the theorem is to say that the interchange of an ideal voltage source and an ideal ammeter in any passive, linear, bilateral circuit will not change the ammeter reading. In other words, the interchange of an ideal current source and an ideal voltmeter in any passive linear bilateral circuit will not change the voltmeter reading. SECTION 17.3 SOME EQUIVALENT NETWORKS 699 17.3 • SOME EQUIVALENT NETWORKS When analyzing electronic circuits, it is often necessary to replace the active device (and perhaps some of its associated passive circuitry) with an equivalent two-port containing only three or four impedances. The validity of the equivalent may be restricted to small signal amplitudes and a single frequency, or perhaps a limited range of frequencies. The equivalent is also a linear approximation of a nonlinear circuit. However, if we are faced with a network containing a number of resistors, capacitors, and inductors, plus a transistor labeled 2N3823, then we cannot analyze the circuit by any of the techniques we have studied previously; the transistor must ﬁrst be re-placed by a linear model, just as we replaced the op amp by a linear model in Chap. 6. The y parameters provide one such model in the form of a two-port network that is often used at high frequencies. Another common linear model for a transistor appears in Sec. 17.5. The two equations that determine the short-circuit admittance parameters, I1 = y11V1 + y12V2 I2 = y21V1 + y22V2 have the form of a pair of nodal equations written for a circuit containing two nonreference nodes. The determination of an equivalent circuit that leads to Eqs. and is made more difﬁcult by the inequality, in general, of y12 and y21; it helps to resort to a little trickery in order to obtain a pair of equa-tions that possess equal mutual coefﬁcients. Let us both add and subtract y12V1 (the term we would like to see present on the right side of Eq. ): I2 = y12V1 + y22V2 + (y21 −y12)V1 or I2 −(y21 −y12)V1 = y12V1 + y22V2 The right-hand sides of Eqs. and now show the proper symmetry for a bilateral circuit; the left-hand side of Eq. may be interpreted as the algebraic sum of two current sources, one an independent source I2 entering node 2, and the other a dependent source (y21 −y12)V1 leaving node 2. Let us now “read’’ the equivalent network from Eqs. and . We ﬁrst provide a reference node, and then a node labeled V1 and one labeled V2. 17.5 Find y for the two-port shown in Fig. 17.12. ■FIGURE 17.12 10 Ω 0.2V2 0.5I1 5 Ω V2 + – V1 + – I2 I1 Ans: 17.4: 0.1192 −0.1115 −0.1115 0.1269 (all S). 17.5: 0.6 0 −0.2 0.2 (all S). From Eq. , we establish the current I1 ﬂowing into node 1, we supply a mutual admittance (−y12) between nodes 1 and 2, and we supply an admit-tance of (y11 + y12) between node 1 and the reference node. With V2 0, the ratio of I1 to V1 is then y11, as it should be. Now consider Eq. ; we cause the current I2 to ﬂow into the second node, we cause the current (y21 −y12)V1 to leave the node, we note that the proper admittance (−y12) exists between the nodes, and we complete the circuit by installing the ad-mittance (y22 + y12) from node 2 to the reference node. The completed cir-cuit is shown in Fig. 17.13a. Another form of equivalent network is obtained by subtracting and adding y21V2 in Eq. ; this equivalent circuit is shown in Fig. 17.13b. If the two-port is bilateral, then y12 = y21, and either of the equivalents reduces to a simple passive network. The dependent source disappears. This equivalent of the bilateral two-port is shown in Fig. 17.13c. There are several uses to which these equivalent circuits may be put. In the ﬁrst place, we have succeeded in showing that an equivalent of any complicated linear two-port exists. It does not matter how many nodes or loops are contained within the network; the equivalent is no more complex than the circuits of Fig. 17.13. One of these may be much simpler to use than the given circuit if we are interested only in the terminal characteristics of the given network. The three-terminal network shown in Fig. 17.14a is often referred to as a of impedances, while that in Fig. 17.14b is called a Y. One network may be replaced by the other if certain speciﬁc relationships between the imped-ances are satisﬁed, and these interrelationships may be established by use of the y parameters. We ﬁnd that y11 = 1 ZA + 1 ZB = 1 Z1 + Z2Z3/(Z2 + Z3) y12 = y21 = −1 ZB = −Z3 Z1Z2 + Z2Z3 + Z3Z1 y22 = 1 ZC + 1 ZB = 1 Z2 + Z1Z3/(Z1 + Z3) CHAPTER 17 TWO-PORT NETWORKS 700 (y21 – y12) V1 V2 + – V1 + – I2 I1 y11 + y12 y22 + y12 –y12 (a) V2 + – V1 + – I2 I1 (c) y11 + y12 y22 + y12 –y12 (y12 – y21) V2 V2 + – V1 + – I2 I1 (b) y11 + y21 y22 + y21 –y21 ■FIGURE 17.13 (a, b) Two-ports which are equivalent to any general linear two-port. The dependent source in part (a) depends on V1, and that in part (b) depends on V2. (c) An equivalent for a bilateral network. ■FIGURE 17.14 The three-terminal network (a) and the three-terminal Y network (b) are equivalent if the six impedances satisfy the conditions of the Y- (or -T) transformation, Eqs. to . (a) ZB ZA ZC (b) Z3 Z2 Z1 SECTION 17.3 SOME EQUIVALENT NETWORKS 701 These equations may be solved for ZA, ZB, and ZC in terms of Z1, Z2, and Z3: or, for the inverse relationships: These equations enable us to transform easily between the equivalent Y and networks, a process known as the Y- transformation (or -T transfor-mation if the networks are drawn in the forms of those letters). In going from Y to , Eqs. to , ﬁrst ﬁnd the value of the common numera-tor as the sum of the products of the impedances in the Y taken two at a time. Each impedance in the is then found by dividing the numerator by the impedance of that element in the Y which has no common node with the desired element. Conversely, given the , ﬁrst take the sum of the three impedances around the ; then divide the product of the two impedances having a common node with the desired Y element by that sum. These transformations are often useful in simplifying passive networks, particularly resistive ones, thus avoiding the need for any mesh or nodal analysis. Z1 = ZAZB ZA + ZB + ZC Z2 = ZBZC ZA + ZB + ZC Z3 = ZCZA ZA + ZB + ZC ZA = Z1Z2 + Z2Z3 + Z3Z1 Z2 ZB = Z1Z2 + Z2Z3 + Z3Z1 Z3 ZC = Z1Z2 + Z2Z3 + Z3Z1 Z1 EXAMPLE 17.6 Find the input resistance of the circuit shown in Fig. 17.15a. The reader may recall these useful relationships from Chap. 5, where their derivation was described. ■FIGURE 17.15 (a) A resistive network whose input resistance is desired. This example is repeated from Chap. 5. (b) The upper is replaced by an equivalent Y. (c, d) Series and parallel combinations give the equivalent input resistance 159 71 . (a) 1 Ω 4 Ω 3 Ω 2 Ω 5 Ω 159 71 (d) Ω 1 2 (c) Ω 13 2 Ω 19 8 Ω 1 2 (b) 2 Ω 5 Ω Ω 3 2 Ω 3 8 Ω (Continued on next page) Now let us tackle a slightly more complicated example, shown as Fig. 17.16. We note that the circuit contains a dependent source, and thus the Y- transformation is not applicable. CHAPTER 17 TWO-PORT NETWORKS 702 We ﬁrst make a -Y transformation on the upper appearing in Fig. 17.15a. The sum of the three resistances forming this is 1 + 4 + 3 = 8 . The product of the two resistors connected to the top node is 1 × 4 = 4 2. Thus, the upper resistor of the Y is 4 8, or 1 2 . Repeating this procedure for the other two resistors, we obtain the network shown in Fig. 17.15b. We next make the series and parallel combinations indicated, obtaining in succession Fig. 17.15c and d. Thus, the input resistance of the circuit in Fig. 17.15a is found to be 159 71 , or 2.24 . EXAMPLE 17.7 The circuit shown in Fig. 17.16 is an approximate linear equivalent of a transistor ampliﬁer in which the emitter terminal is the bottom node, the base terminal is the upper input node, and the collector terminal is the upper output node. A 2000 resistor is connected between collector and base for some special application and makes the analysis of the circuit more difﬁcult. Determine the y param-eters for this circuit. Identify the goal of the problem. Cutting through the problem-speciﬁc jargon, we realize that we have been presented with a two-port network and require the y parameters. Collect the known information. Figure 17.16 shows a two-port network with V1, I1, V2, and I2 already indicated, and a value for each component has been provided. Devise a plan. There are several ways we might think about this circuit. If we recognize it as being in the form of the equivalent circuit shown in Fig. 17.13a, then we may immediately determine the values of the y parameters. If recognition is not immediate, then the y parameters ■FIGURE 17.16 The linear equivalent circuit of a transistor in common-emitter conﬁguration with resistive feedback between collector and base. V2 + – V1 + – I1 I2 500 Ω 0.0395V1 10 kΩ 2000 Ω SECTION 17.3 SOME EQUIVALENT NETWORKS 703 may be determined for the two-port by applying the relationships of Eqs. to . We also might avoid any use of two-port analysis methods and write equations directly for the circuit as it stands. The ﬁrst option seems best in this case. Construct an appropriate set of equations. By inspection, we ﬁnd that −y21 corresponds to the admittance of our 2 k resistor, that y11 + y12 corresponds to the admittance of the 500 resistor, the gain of the dependent current source corresponds to y21 −y12, and ﬁnally that y22 + y12 corresponds to the admittance of the 10 k resistor. Hence we may write y12 = − 1 2000 y11 = 1 500 −y12 y21 = 0.0395 + y12 y22 = 1 10,000 −y12 Determine if additional information is required. With the equations written as they are, we see that once y12 is com-puted, the remaining y parameters may also be obtained. Attempt a solution. Plugging the numbers into a calculator, we ﬁnd that y12 = − 1 2000 = −0.5 mS y11 = 1 500 − − 1 2000 = 2.5 mS y22 = 1 10,000 − − 1 2000 = 0.6 mS and y21 = 0.0395 + − 1 2000 = 39 mS The following equations must then apply: I1 = 2.5V1 −0.5V2 I2 = 39V1 + 0.6V2 where we are now using units of mA, V, and mS or k. Verify the solution. Is it reasonable or expected? Writing two nodal equations directly from the circuit, we ﬁnd I1 = V1 −V2 2 + V1 0.5 or I1 = 2.5V1 −0.5V2 and −39.5V1 + I2 = V2 −V1 2 + V2 10 or I2 = 39V1 + 0.6V2 which agree with Eqs. and obtained directly from the y parameters. Now let us make use of Eqs. and by analyzing the performance of the two-port in Fig. 17.16 under several different operating conditions. We ﬁrst provide a current source of 1/0◦mA at the input and connect a 0.5 k (2 mS) load to the output. The terminating networks are therefore both one-ports and give us the following speciﬁc information relating I1 to V1 and I2 to V2: I1 = 1 (for any V1) I2 = −2V2 We now have four equations in the four variables, V1, V2, I1, and I2. Sub-stituting the two one-port relationships in Eqs. and , we obtain two equations relating V1 and V2: 1 = 2.5V1 −0.5V2 0 = 39V1 + 2.6V2 Solving, we ﬁnd that V1 = 0.1 V V2 = −1.5 V I1 = 1 mA I2 = 3 mA These four values apply to the two-port operating with a prescribed input (I1 1 mA) and a speciﬁed load (RL = 0.5 k). The performance of an ampliﬁer is often described by giving a few spe-ciﬁc values. Let us calculate four of these values for this two-port with its terminations. We will deﬁne and evaluate the voltage gain, the current gain, the power gain, and the input impedance. The voltage gain GV is GV = V2 V1 From the numerical results, it is easy to see that GV = −15. The current gain GI is deﬁned as GI = I2 I1 and we have GI = 3 Let us deﬁne and calculate the power gain GP for an assumed sinusoidal excitation. We have G P = Pout Pin = Re −1 2V2I∗ 2 Re 1 2V1I∗ 1 = 45 The device might be termed either a voltage, a current, or a power ampliﬁer, since all the gains are greater than unity. If the 2 k resistor were removed, the power gain would rise to 354. The input and output impedances of the ampliﬁer are often desired in or-der that maximum power transfer may be achieved to or from an adjacent two-port. We deﬁne the input impedance Zin as the ratio of input voltage to current: Zin = V1 I1 = 0.1 k CHAPTER 17 TWO-PORT NETWORKS 704 SECTION 17.3 SOME EQUIVALENT NETWORKS 705 This is the impedance offered to the current source when the 500 load is connected to the output. (With the output short-circuited, the input imped-ance is necessarily 1/y11, or 400 .) It should be noted that the input impedance cannot be determined by replacing every source with its internal impedance and then combining resistances or conductances. In the given circuit, such a procedure would yield a value of 416 . The error, of course, comes from treating the dependent source as an independent source. If we think of the input imped-ance as being numerically equal to the input voltage produced by an input current of 1 A, the application of the 1 A source produces some input volt-age V1, and the strength of the dependent source (0.0395V1) cannot be zero. We should recall that when we obtain the Thévenin equivalent impedance of a circuit containing a dependent source along with one or more indepen-dent sources, we must replace the independent sources with short circuits or open circuits, but a dependent source must not be deactivated. Of course, if the voltage or current on which the dependent source depends is zero, then the dependent source will itself be inactive; occasionally a circuit may be simpliﬁed by recognizing such an occurrence. Besides GV, GI, GP, and Zin, there is one other performance parameter that is quite useful. This is the output impedance Zout, and it is determined for a different circuit conﬁguration. The output impedance is just another term for the Thévenin impedance appearing in the Thévenin equivalent circuit of that portion of the network faced by the load. In our circuit, which we have assumed is driven by a 1/0◦mA current source, we therefore replace this independent source with an open circuit, leave the dependent source alone, and seek the input im-pedance seen looking to the left from the output terminals (with the load removed). Thus, we deﬁne Zout = V2\|I2=1 A with all other independent sources deactivated and RL removed We therefore remove the load resistor, apply 1/0◦mA (since we are work-ing in V, mA, and k) at the output terminals, and determine V2. We place these requirements on Eqs. and , and obtain 0 = 2.5V1 −0.5V2 1 = 39V1 + 0.6V2 Solving, V2 = 0.1190 V and thus Zout = 0.1190 k An alternative procedure might be to ﬁnd the open-circuit output volt-age and the short-circuit output current. That is, the Thévenin impedance is the output impedance: Zout = Zth = −V2oc I2sc Carrying out this procedure, we ﬁrst rekindle the independent source so that I1 1 mA, and then open-circuit the load so that I2 0. We have 1 = 2.5V1 −0.5V2 0 = 39V1 + 0.6V2 and thus V2oc = −1.857 V Next, we apply short-circuit conditions by setting V2 0 and again let I1 1 mA. We ﬁnd that I1 = 1 = 2.5V1 −0 I2 = 39V1 + 0 and thus I2sc = 15.6 mA The assumed directions of V2 and I2 therefore result in a Thévenin or out-put impedance Zout = −V2oc I2sc = −−1.857 15.6 = 0.1190 k as before. We now have enough information to enable us to draw the Thévenin or Norton equivalent of the two-port of Fig. 17.16 when it is driven by a 1/0◦mA current source and terminated in a 500 load. Thus, the Norton equivalent presented to the load must contain a current source equal to the short-circuit current I2sc in parallel with the output impedance; this equiva-lent is shown in Fig. 17.17a. Also, the Thévenin equivalent offered to the 1/0◦mA input source must consist solely of the input impedance, as drawn in Fig. 17.17b. Before leaving the y parameters, we should recognize their usefulness in describing the parallel connection of two-ports, as indicated in Fig. 17.18. When we ﬁrst deﬁned a port in Sec. 17.1, we noted that the currents enter-ing and leaving the two terminals of a port had to be equal, and there could be no external connections made that bridged between ports. Apparently the parallel connection shown in Fig. 17.18 violates this condition. However, if each two-port has a reference node that is common to its input and output port, and if the two-ports are connected in parallel so that they have a com-mon reference node, then all ports remain ports after the connection. Thus, for the A network, IA = yAVA CHAPTER 17 TWO-PORT NETWORKS 706 ■FIGURE 17.17 (a) The Norton equivalent of the network in Fig. 17.16 to the left of the output terminal, I1 = 1/0◦mA. (b) The Thévenin equivalent of that portion of the network to the right of the input terminals, if I2 = −2V2 mA. V2 + – I2 119 Ω 15.6 mA (a) V1 + – I1 100 Ω (b) ■FIGURE 17.18 The parallel connection of two two-port networks. If both inputs and outputs have the same reference node, then the admittance matrix y = yA + yB. VA2 + – VA1 + – I1 I2 IA2 IA1 IB1 IB2 Network A Network B SECTION 17.3 SOME EQUIVALENT NETWORKS 707 where IA = IA1 IA2 and VA = VA1 VA2 and for the B network IB = yBVB But VA = VB = V and I = IA + IB Thus, I = (yA + yB)V and we see that each y parameter of the parallel network is given as the sum of the corresponding parameters of the individual networks, y = yA + yB This may be extended to any number of two-ports connected in parallel. PRACTICE ● 17.6 Find y and Zout for the terminated two-port shown in Fig. 17.19. 17.7 Use -Y and Y- transformations to determine Rin for the network shown in (a) Fig. 17.20a; (b) Fig. 17.20b. ■FIGURE 17.19 + – V1 + – V2 + – I1 I2 200 Ω 1 kΩ 3 kΩ 5 kΩ 20I1 10–3V2 Vs ■FIGURE 17.20 Each R is 47 Ω (a) Rin 4 Ω 2 Ω 1 Ω 18 Ω 12 Ω 2 Ω 6 Ω 3 Ω (b) Rin Ans: 17.6: 2 × 10−4 −10−3 −4 × 10−3 20.3 × 10−3 (S); 51.1 . 17.7: 53.71 , 1.311 . 17.4 • IMPEDANCE PARAMETERS The concept of two-port parameters has been introduced in terms of the short-circuit admittance parameters. There are other sets of parameters, however, and each set is associated with a particular class of networks for which its use provides the simplest analysis. We will consider three other types of parameters, the open-circuit impedance parameters, which are the subject of this section; and the hybrid and the transmission parameters, which are discussed in following sections. We begin again with a general linear two-port that does not contain any independent sources; the currents and voltages are assigned as before (Fig. 17.8). Now let us consider the voltage V1 as the response produced by two current sources I1 and I2. We thus write for V1 V1 = z11I1 + z12I2 and for V2 V2 = z21I1 + z22I2 or V = V1 V2 = zI = z11 z12 z21 z22 I1 I2 Of course, in using these equations it is not necessary that I1 and I2 be current sources; nor is it necessary that V1 and V2 be voltage sources. In general, we may have any networks terminating the two-port at either end. As the equations are written, we probably think of V1 and V2 as given quan-tities, or independent variables, and I1 and I2 as unknowns, or dependent variables. The six ways in which two equations may be written to relate these four quantities deﬁne the different systems of parameters. We study the four most important of these six systems of parameters. The most informative description of the z parameters, defined in Eqs. and , is obtained by setting each of the currents equal to zero. Thus Since zero current results from an open-circuit termination, the z param-eters are known as the open-circuit impedance parameters. They are easily related to the short-circuit admittance parameters by solving Eqs. and z11 = V1 I1 I2=0 z12 = V1 I2 I1=0 z21 = V2 I1 I2=0 z22 = V2 I2 I1=0 CHAPTER 17 TWO-PORT NETWORKS 708 SECTION 17.4 IMPEDANCE PARAMETERS 709 for I1 and I2: I1 = V1 z12 V2 z22 z11 z12 z21 z22 or I1 = z22 z11z22 −z12z21 V1 − z12 z11z22 −z12z21 V2 Using determinant notation, and being careful that the subscript is a lower-case z, we assume that z ̸= 0 and obtain y11 = 11 z = z22 z y12 = −21 z = −z12 z and from solving for I2, y21 = −12 z = −z21 z y22 = 22 z = z11 z In a similar manner, the z parameters may be expressed in terms of the admittance parameters. Transformations of this nature are possible between any of the various parameter systems, and quite a collection of occasionally useful formulas may be obtained. Transformations between the y and z pa-rameters (as well as the h and t parameters which we will consider in the following sections) are given in Table 17.1 as a helpful reference. TABLE ●17.1 Transformations Between y, z, h, and t Parameters y z h t y y11 y12 z22 z −z12 z 1 h11 −h12 h11 t22 t12 −t t12 y21 y22 −z21 z z11 z h21 h11 h h11 −1 t12 t11 t12 z y22 y −y12 y z11 z12 h h22 h12 h22 t11 t21 t t21 −y21 y y11 y -z21 z22 −h21 h22 1 h22 1 t21 t22 t21 h 1 y11 −y12 y11 z z22 z12 z22 h11 h12 t12 t22 t t22 y21 y11 y y11 −z21 z22 1 z22 h21 h22 −1 t22 t21 t22 t −y22 y21 −1 y21 z11 z21 z z21 −h h21 −h11 h21 t11 t12 −y y21 −y11 y21 1 z21 z22 z21 −h22 h21 −1 h21 t21 t22 For all parameter sets: p = p11p22 −p12p21. If the two-port is a bilateral network, reciprocity is present; it is easy to show that this results in the equality of z12 and z21. Equivalent circuits may again be obtained from an inspection of Eqs. and ; their construction is facilitated by adding and subtract-ing either z12I1 in Eq. or z21I2 in Eq. . Each of these equivalent circuits contains a dependent voltage source. Let us leave the derivation of such an equivalent to some leisure moment, and consider next an example of a rather general nature. Can we construct a general Thévenin equivalent of the two-port, as viewed from the output terminals? It is necessary ﬁrst to assume a speciﬁc input circuit conﬁguration, and we will select an independent voltage source Vs (positive sign at top) in series with a generator impedance Zg. Thus Vs = V1 + I1Zg Combining this result with Eqs. and , we may eliminate V1 and I1 and obtain V2 = z21 z11 + Zg Vs + z22 − z12z21 z11 + Zg I2 The Thévenin equivalent circuit may be drawn directly from this equation; it is shown in Fig. 17.21. The output impedance, expressed in terms of the z parameters, is Zout = z22 − z12z21 z11 + Zg If the generator impedance is zero, the simpler expression Zout = z11z22 −z12z21 z11 = z 22 = 1 y22 Zg = 0 is obtained. For this special case, the output admittance is identical to y22, as indicated by the basic relationship of Eq. . CHAPTER 17 TWO-PORT NETWORKS 710 ■FIGURE 17.21 The Thévenin equivalent of a general two-port, as viewed from the output terminals, expressed in terms of the open-circuit impedance parameters. + – V2 + – I2 Vs z21 z11 + Zg z22 – z12 z21 z11 + Zg EXAMPLE 17.8 Given the set of impedance parameters z = 103 10 −106 104 (all ) which is representative of a bipolar junction transistor operating in the common-emitter conﬁguration, determine the voltage, current, and power gains, as well as the input and output impedances. The two-port is driven by an ideal sinusoidal voltage source Vs in series with a 500 resistor, and terminated in a 10 k load resistor. The two describing equations for the two-port are V1 = 103I1 + 10I2 V2 = −106I1 + 104I2 SECTION 17.4 IMPEDANCE PARAMETERS 711 and the characterizing equations of the input and output networks are Vs = 500I1 + V1 V2 = −104I2 From these last four equations, we may easily obtain expressions for V1, I1, V2, and I2 in terms of Vs: V1 = 0.75Vs I1 = Vs 2000 V2 = −250Vs I2 = Vs 40 From this information, it is simple to determine the voltage gain, GV = V2 V1 = −333 the current gain, GI = I2 I1 = 50 the power gain, G P = Re −1 2V2I∗ 2 Re 1 2V1I∗ 1 = 16,670 and the input impedance, Zin = V1 I1 = 1500 The output impedance may be obtained by referring to Fig. 17.21: Zout = z22 − z12z21 z11 + Zg = 16.67 k In accordance with the predictions of the maximum power transfer theorem, the power gain reaches a maximum value when ZL = Z∗ out = 16.67 k; that maximum value is 17,045. The y parameters are useful when two-ports are interconnected in paral-lel, and, in a dual manner, the z parameters simplify the problem of a series connection of networks, shown in Fig. 17.22. Note that the series connection is not the same as the cascade connection that we will discuss later in con-nection with the transmission parameters. If each two-port has a common reference node for its input and output, and if the references are connected together as indicated in Fig. 17.22, then I1 ﬂows through the input ports of the two networks in series. A similar statement holds for I2. Thus, ports remain ports after the interconnection. It follows that I = IA = IB and V = VA + VB = zAIA + zBIB = (zA + zB)I = zI CHAPTER 17 TWO-PORT NETWORKS 712 where z = zA + zB so that z11 = z11A + z11B, and so forth. ■FIGURE 17.22 The series connection of two two-port networks is made by connecting the four common reference nodes together; then the matrix z = zA + zB . V1 + – V2A + – V2B + – I2 = I2A I1 = I1B I1 = I1A V1A + – I1 V1B + – Network A Network B PRACTICE ● 17.8 Find z for the two-port shown in (a) Fig. 17.23a; (b) Fig. 17.23b. 17.9 Find z for the two-port shown in Fig. 17.23c. ■FIGURE 17.23 + – V2 + – V1 + – 25 Ω 20 Ω 50 Ω 0.5V2 (c) V1 + – V2 + – 25 Ω 20 Ω 50 Ω (a) V1 + – V2 + – 25 Ω 40 Ω 20 Ω 50 Ω (b) Ans: 17.8: 45 25 25 75 (), 21.2 11.76 11.76 67.6 (). 17.9: 70 100 50 150 (). SECTION 17.5 HYBRID PARAMETERS 713 17.5 • HYBRID PARAMETERS The difﬁculty in measuring quantities such as the open-circuit impedance parameters arises when a parameter such as z21 must be measured. A known sinusoidal current is easily supplied at the input terminals, but because of the exceedingly high output impedance of the transistor circuit, it is difﬁcult to open-circuit the output terminals and yet supply the necessary dc biasing voltages and measure the sinusoidal output voltage. A short-circuit current measurement at the output terminals is much simpler to implement. The hybrid parameters are deﬁned by writing the pair of equations relat-ing V1, I1, V2, and I2 as if V1 and I2 were the independent variables: V1 = h11I1 + h12V2 I2 = h21I1 + h22V2 or V1 I2 = h I1 V2 The nature of the parameters is made clear by ﬁrst setting V2 = 0. Thus, h11 = V1 I1 V2=0 = short-circuit input impedance h21 = I2 I1 V2=0 = short-circuit forward current gain Letting I1 = 0, we obtain h12 = V1 V2 I1=0 = open-circuit reverse voltage gain h22 = I2 V2 I1=0 = open-circuit output admittance Since the parameters represent an impedance, an admittance, a voltage gain, and a current gain, they are called the “hybrid’’ parameters. The subscript designations for these parameters are often simpliﬁed when they are applied to transistors. Thus, h11, h12, h21, and h22 become hi, hr, hf, and ho, respectively, where the subscripts denote input, reverse, for-ward, and output. EXAMPLE 17.9 V1 + – V2 + – I1 I2 1 Ω 6 Ω 4 Ω ■FIGURE 17.24 A bilateral network for which the h parameters are found: h12 = −h21. Find h for the bilateral resistive circuit drawn in Fig. 17.24. With the output short-circuited (V2 = 0), the application of a 1 A source at the input (I1 = 1 A) produces an input voltage of 3.4 V (V1 = 3.4 V); hence, h11 = 3.4 . Under these same conditions, the output current is easily obtained by current division: I2 = −0.4 A; thus, h21 = −0.4. The remaining two parameters are obtained with the input open-circuited (I1 = 0). We apply 1 V to the output terminals (V2 = 1 V). (Continued on next page) CHAPTER 17 TWO-PORT NETWORKS 714 17.11 If h 5 2 −0.5 0.1 S , ﬁnd (a) y; (b) z. Ans: 17.10: 20 1 −1 25 ms , 8 0.8 −0.8 20 ms . 17.11: 0.2 −0.4 −0.1 0.3 (S), 15 20 5 10 (). V1 + – V2 + – 40 Ω 20 Ω (a) V1 + – V2 + – 40 Ω 10 Ω (b) ■FIGURE 17.25 The response at the input terminals is 0.4 V (V1 = 0.4 V), and thus h12 0.4. The current delivered by this source at the output terminals is 0.1 A (I2 = 0.1 A), and therefore h22 0.1 S. We therefore have h = 3.4 0.4 −0.4 0.1 S . It is a consequence of the reciprocity theorem that h12 = −h21 for a bilateral network. PRACTICE ● 17.10 Find h for the two-port shown in (a) Fig. 17.25a; (b) Fig. 17.25b. The circuit shown in Fig. 17.26 is a direct translation of the two deﬁning equations, and . The ﬁrst represents KVL about the input loop, while the second is obtained from KCL at the upper output node. This circuit is also a popular transistor equivalent circuit. Let us assume some reasonable values for the common-emitter conﬁguration: h11 = 1200 , h12 = 2 × 10−4, h21 = 50, h22 = 50 × 10−6 S, a voltage generator of 1/0◦mV in series with 800 , and a 5 k load. For the input, 10−3 = (1200 + 800)I1 + 2 × 10−4V2 and at the output, I2 = −2 × 10−4V2 = 50I1 + 50 × 10−6V2 + – V1 + – V2 + – I1 I2 h11 (Ω) h12V2 h21I1 h22 ( ) Ω ■FIGURE 17.26 The four h parameters are referred to a two-port. The pertinent equations are V1 = h11I1 + h12V2 and I2 = h21I1 + h22V2. Parameter values for bipolar junction transistors are commonly quoted in terms of h parameters. Invented in the late 1940s by researchers at Bell Laboratories (Fig. 17.27), the transistor is a nonlinear three-terminal passive semiconductor device that forms the basis for al-most all ampliﬁers and digital logic circuits. are valid for all voltages and currents is not possible. Therefore, it is common practice to quote h parameters at a speciﬁc value of collector current IC and collector-emitter voltage VCE. Another consequence of the nonlin-earity of the device is that ac h parameters and dc h parameters are often quite different in value. There are many types of instruments which may be employed to obtain the h parameters for a particular transistor. One example is a semiconductor parameter analyzer, shown in Fig. 17.29. This instrument sweeps the desired current (plotted on the vertical axis) against a specified voltage (plotted on the horizontal axis). A “family” of curves is produced by varying a third param-eter, often the base current, in discrete steps. As an example, the manufacturer of the 2N3904 NPN silicon transistor quotes h parameters as indicated in Table 17.2; note that the speciﬁc parameters are given alternative designations (hie, hre, etc.) by transistor engineers. The measurements were made at IC = 1.0 mA, VCE = 10 V dc, and f = 1.0 kHz. (Continued on next page) PRACTICAL APPLICATION Characterizing Transistors ■FIGURE 17.27 Photograph of the ﬁrst demonstrated bipolar junction transistor (“bjt”). Lucent Technologies Inc./Bell Labs The three terminals of a transistor are labeled the base (b), collector (c), and emitter (e) as shown in Fig. 17.28, and are named after their roles in the trans-port of charge within the device. The h parameters of a bipolar junction transistor are typically measured with the emitter terminal grounded, also known as the common-emitter conﬁguration; the base is then desig-nated as the input and the collector as the output. As mentioned previously, however, the transistor is a non-linear device, and so deﬁnition of h parameters which ■FIGURE 17.29 Display snapshot of an HP 4155A Semiconductor Parameter Analyzer used to measure the h parameters of a 2N3904 bipolar junction transistor (bjt). VCE VCB VBE + – + – + – IC IE IB Collector Emitter Base ■FIGURE 17.28 Schematic of a bjt showing currents and voltages deﬁned using the IEEE convention. Just for fun, one of the authors and a friend decided to measure these parameters for themselves. Grabbing an inexpensive device off the shelf and using the instru-ment in Fig. 17.29, they found hoe = 3.3 μmhos h f e = 109 hie = 3.02 k hre = 4 × 10−3 measurement, as we were sweeping below and above IC = 1 mA. Transistors, unfortunately, can change their properties rather dramatically as a function of tempera-ture; the manufacturer values were speciﬁcally for 25◦C. Once the sweep was changed to minimize device heat-ing, we obtained a value of 2.0 × 10−4 for hre. Linear cir-cuits are by far much easier to work with, but nonlinear circuits can be much more interesting! the ﬁrst three of which were all well within the manufac-turer’s published tolerances, although much closer to the minimum values than to the maximum values. The value for hre, however, was an order of magnitude larger than the maximum value specified by the manufacturer’s datasheet! This was rather disconcerting, as we thought we were doing pretty well up to that point. Upon further reﬂection, we realized that the experi-mental setup allowed the device to heat up during the TABLE ●17.2 Summary of 2N3904 AC Parameters Parameter Name Speciﬁcation Units hie (h11) Input impedance 1.0–10 k hre (h12) Voltage feedback ratio 0.5–8.0 × 10−4 – hfe (h21) Small-signal current gain 100–400 – hoe (h22) Output admittance 1.0–40 μmhos Solving, I1 = 0.510 μA V1 = 0.592 mV I2 = 20.4 μA V2 = −102 mV Through the transistor we have a current gain of 40, a voltage gain of −172, and a power gain of 6880. The input impedance to the transistor is 1160 , and a few more calculations show that the output impedance is 22.2 k. Hybrid parameters may be added directly when two-ports are connected in series at the input and in parallel at the output. This is called a series-parallel interconnection, and it is not used very often. 17.6 • TRANSMISSION PARAMETERS The last two-port parameters that we will consider are called the t param-eters, the ABCD parameters, or simply the transmission parameters. They are deﬁned by V1 = t11V2 −t12I2 and I1 = t21V2 −t22I2 or V1 I1 = t V2 −I2 where V1, V2, I1, and I2 are deﬁned as usual (Fig. 17.8). The minus signs that appear in Eqs. and should be associated with the output SECTION 17.6 TRANSMISSION PARAMETERS 717 current, as (−I2). Thus, both I1 and −I2 are directed to the right, the direc-tion of energy or signal transmission. Other widely used nomenclature for this set of parameters is t11 t12 t21 t22 = A B C D Note that there are no minus signs in the t or ABCD matrices. Looking again at Eqs. to , we see that the quantities on the left, often thought of as the given or independent variables, are the input voltage and current, V1 and I1; the dependent variables, V2 and I2, are the output quantities. Thus, the transmission parameters provide a direct relationship between input and output. Their major use arises in transmission-line analy-sis and in cascaded networks. Let us find the t parameters for the bilateral resistive two-port of Fig. 17.30a. To illustrate one possible procedure for ﬁnding a single param-eter, consider t12 = V1 −I2 V2=0 We therefore short-circuit the output (V2 = 0) and set V1 1 V, as shown in Fig. 17.30b. Note that we cannot set the denominator equal to unity by plac-ing a 1 A current source at the output; we already have a short circuit there. The equivalent resistance offered to the 1 V source is Req = 2 + (4∥10) , and we then use current division to get −I2 = 1 2 + (4∥10) × 10 10 + 4 = 5 34 A Hence, t12 = 1 −I2 = 34 5 = 6.8 If it is necessary to ﬁnd all four parameters, we write any convenient pair of equations using all four terminal quantities, V1, V2, I1, and I2. From Fig. 17.30a, we have two mesh equations: V1 = 12I1 + 10I2 V2 = 10I1 + 14I2 Solving Eq. for I1, we get I1 = 0.1V2 −1.4I2 so that t21 = 0.1 S and t22 = 1.4. Substituting the expression for I1 in Eq. , we ﬁnd V1 = 12(0.1V2 −1.4I2) + 10I2 = 1.2V2 −6.8I2 and t11 1.2 and t12 = 6.8 , once again. For reciprocal networks, the determinant of the t matrix is equal to unity: t = t11t22 −t12t21 = 1 V1 + – V2 + – I1 I2 2 Ω 4 Ω 10 Ω (a) –I2 2 Ω 1 V 4 Ω 10 Ω (b) + – ■FIGURE 17.30 (a) A two-port resistive network for which the t parameters are to be found. (b) To ﬁnd t12, set V1 = 1 V with V2 = 0; then t12 = 1/(−I2) = 6.8 . In the resistive example of Fig. 17.30, t = 1.2 × 1.4 −6.8 × 0.1 = 1. Good! We conclude our two-port discussion by connecting two two-ports in cascade, as illustrated for two networks in Fig. 17.31. Terminal voltages and currents are indicated for each two-port, and the corresponding t parameter relationships are, for network A, CHAPTER 17 TWO-PORT NETWORKS 718 EXAMPLE 17.10 Find the t parameters for the cascaded networks shown in Fig. 17.32. V1 + – V4 + – V2 V3 + – I1 –I4 I3 –I2 Network A Network B ■FIGURE 17.31 When two-port networks A and B are cascaded, the t parameter matrix for the combined network is given by the matrix product t = tA tB. V1 I1 = tA V2 −I2 = tA V3 I3 and for network B, V3 I3 = tB V4 −I4 Combining these results, we have V1 I1 = tAtB V4 −I4 Therefore, the t parameters for the cascaded networks are found by the ma-trix product, t = tAtB This product is not obtained by multiplying corresponding elements in the two matrices. If necessary, review the correct procedure for matrix multi-plication in Appendix 2. 2 Ω 4 Ω 10 Ω Network A 4 Ω 8 Ω 20 Ω Network B ■FIGURE 17.32 A cascaded connection. SECTION 17.6 TRANSMISSION PARAMETERS 719 Network A is the two-port of Fig. 17.32, and, therefore tA = 1.2 6.8 0.1 S 1.4 while network B has resistance values twice as large, so that tB = 1.2 13.6 0.05 S 1.4 For the combined network, t = tAtB = 1.2 6.8 0.1 1.4 1.2 13.6 0.05 1.4 = 1.2 × 1.2 + 6.8 × 0.05 1.2 × 13.6 + 6.8 × 1.4 0.1 × 1.2 + 1.4 × 0.05 0.1 × 13.6 + 1.4 × 1.4 and t = 1.78 25.84 0.19 S 3.32 PRACTICE ● 17.12 Given t = 3.2 8 0.2 S 4 , ﬁnd (a) z; (b) t for two identical networks in cascade; (c) z for two identical networks in cascade. Ans: 16 56 5 20 (); 11.84 57.6 1.44 S 17.6 ; 8.22 87.1 0.694 12.22 (). COMPUTER-AIDED ANALYSIS The characterization of two-port networks using t parameters creates the opportunity for vastly simpliﬁed analysis of cascaded two-port network circuits. As seen in this section, where, for example, tA = 1.2 6.8 0.1 S 1.4 and tB = 1.2 13.6 0.05 S 1.4 we found that the t parameters characterizing the cascaded network can be found by simply multiplying tA and tB: t tA tB Such matrix operations are easily carried out using scientiﬁc calculators or software packages such as MATLAB. The MATLAB script, (Continued on next page) SUMMARY AND REVIEW In this chapter we encountered a somewhat abstract way to represent net-works. This new approach is especially useful if the network is passive, and will either be connected somehow to other networks at some point, or perhaps component values will frequently be changed. We introduced the concept through the idea of a one-port network, where all we really did was determine the Thévenin equivalent resistance (or impedance, more generally speaking). Our ﬁrst exposure to the idea of a two-port network (perhaps one port is an input, the other an output?) was through admittance parameters, also called y parameters. The result is a matrix which, when multiplied by the vector containing the terminal voltages, yields a vector with the currents into each port. Alittle manipulation yielded what we called -Y equivalents in Chap. 5. The direct counterpart to y parameters are z parameters, where each matrix element is the ratio of a voltage to a current. Occasionally y and z parameters are not particularly convenient, so we also introduced “hybrid” or h parameters, as well as “transmission” or t parameters, also referred to as ABCD parameters. Table 17.1 summarizes the conversion process between y, z, h, and t parameters; having one set of parameters which completely describes a network is sufﬁcent regardless of what type of matrix we prefer for a partic-ular analysis. CHAPTER 17 TWO-PORT NETWORKS 720 for example, would be EDU» tA = [1.2 6.8; 0.1 1.4]; EDU» tB = [1.2 13.6; 0.05 1.4]; EDU» t = tAtB t = 1.7800 25.8400 0.1900 3.3200 as we found in Example 17.10. In terms of entering matrices in MATLAB, each has a case-sensitive variable name (tA, tB, and t in this example). Matrix elements are en-tered a row at a time, beginning with the top row; rows are separated by a semicolon. Again, the reader should always be careful to remember that the order to operations is critical when performing matrix algebra. For example, tBtA results in a totally different matrix than the one we sought: tB tA 2.8 27.2 0.2 2.3 For simple matrices such as seen in this example, a scientiﬁc calculator is just as handy (if not more so). However, larger cascaded networks are more easily handled on a computer, where it is more convenient to see all arrays on the screen simultaneously. READING FURTHER 721 As a convenience to the reader, we will now proceed directly to a list of key concepts in the chapter, along with correponding examples. ❑In order to employ the analysis methods described in this chapter, it is critical to remember that each port can only be connected to either a one-port network or a port of another multiport network. ❑The input impedance of a one-port (passive) linear network can be obtained using either nodal or mesh analysis; in some instances the set of coefﬁcients can be written directly by inspection. (Examples 17.1, 17.2, 17.3) ❑The deﬁning equations for analyzing a two-port network in terms of its admittance (y) parameters are: I1 = y11V1 + y12V2 and I2 = y21V1 + y22V2 where y11 = I1 V1 V2=0 y12 = I1 V2 V1=0 y21 = I2 V1 V2=0 and y22 = I2 V2 V1=0 (Examples 17.4, 17.5, 17.7) ❑The deﬁning equations for analyzing a two-port network in terms of its impedance (z) parameters are: V1 = z11I1 + z12I2 and V2 = z21I1 + z22I2 (Example 17.8) ❑The deﬁning equations for analyzing a two-port network in terms of its hybrid (h) parameters are: V1 = h11I1 + h12V2 and I2 = h21I1 + h22V2 (Example 17.9) ❑The deﬁning equations for analyzing a two-port network in terms of its transmission (t) parameters (also called the ABCD parameters) are: V1 = t11V2 −t12I2 and I1 = t21V2 −t22I2 (Example 17.10) ❑It is straightforward to convert between h, z, t, and y parameters, depending on circuit analysis needs; the transformations are summarized in Table 17.1. (Example 17.6) READING FURTHER Further details of matrix methods for circuit analysis can be found in: R. A. DeCarlo and P. M. Lin, Linear Circuit Analysis, 2nd ed. New York: Oxford University Press, 2001. Analysis of transistor circuits using network parameters is described in: W. H. Hayt, Jr., and G. W. Neudeck, Electronic Circuit Analysis and Design, 2nd ed. New York: Wiley, 1995. CHAPTER 17 TWO-PORT NETWORKS 722 EXERCISES 17.1 One-Port Networks 1. Consider the following system of equations: −2I1 + 4I2 = 3 5I1 + I2 −9I3 = 0 2I1 −5I2 + 4I3 = −1 (a) Write the set of equations in matrix form. (b) Determine Z and 11. (c) Calculate I1. 2. For the following system of equations, 100V1 −45V2 + 30V3 = 0.2 75V1 + 80V3 = −0.1 48V1 + 200V2 + 42V3 = 0.5 (a) Write the set of equations in matrix form. (b) Use Y to calculate V2 only. 3. With regard to the passive network depicted in Fig. 17.33, (a) obtain the four mesh equations; (b) compute Z; and (c) calculate the input impedance. CHAPTER 17 TWO-PORT NETWORKS 722 10 kΩ 1 kΩ 470 Ω 4.7 kΩ 2.2 kΩ 2.2 kΩ V1 + – I3 I2 I4 I1 ■FIGURE 17.33 4. Determine the input impedance of the network shown in Fig. 17.34 after ﬁrst calculating Z. 1 kΩ 100 Ω 870 Ω 220 Ω 100 Ω 870 Ω V1 + – I3 I2 I4 I5 I1 ■FIGURE 17.34 5. For the one-port network represented schematically in Fig. 17.35, choose the bottom node as the reference; name the junction between the 3, 10, and 20 S conductances V2 and the remaining node V3. (a) Write the three nodal equations. (b) Compute Y. (c) Calculate the input admittance. V1 + – 3 S 10 S 20 S 5 S 2 S ■FIGURE 17.35 EXERCISES 723 6. Calculate Z and Zin for the network of Fig. 17.36 if ω is equal to (a) 1 rad/s; (b) 320 krad/s. 7. Set ω = 100π rad/s in the one-port of Fig. 17.36. (a) Calculate Y and the input admittance at ω, Yin(ω). (b) A sinusoidal current source having magnitude 100 A, frequency 100π rad/s, and 0◦phase is connected to the network. Calculate the voltage across the current source (express answer as a phasor). 8. With reference to the one-port of Fig. 17.37, which contains a dependent cur-rent source controlled by a resistor voltage, (a) calculate Z; (b) compute Zin. 100 mH 100 mH Zin 50 mH 6 Ω 20 nF ■FIGURE 17.36 V1 + – 4 Ω 0.2V1 10 Ω 5 Ω 10 Ω Zin ■FIGURE 17.37 9. For the ideal op amp circuit represented in Fig. 17.38, the input resistance is deﬁned by looking between the positive input terminal of the op amp and ground. (a) Write the appropriate nodal equations for the one-port. (b) Obtain an expression for Rin. Is your answer somewhat unexpected? Explain. 10. (a) If both the op amps shown in the circuit of Fig. 17.39 are assumed to be ideal (Ri = ∞, Ro = 0, and A = ∞), ﬁnd Zin. (b) R1 = 4 k, R2 = 10 k, R3 = 10 k, R4 = 1 k, and C = 200 pF; show that Zin = jωLin, where Lin = 0.8 mH. – + 28 Ω 28 Ω Rx Rin ■FIGURE 17.38 10 kΩ 1 kΩ 8 kΩ V2 + – V1 + – I2 I1 ■FIGURE 17.40 – + – + Zin R1 R2 R3 C R4 ■FIGURE 17.39 17.2 Admittance Parameters 11. Obtain a complete set of y parameters which describe the two-port shown in Fig. 17.40. CHAPTER 17 TWO-PORT NETWORKS 724 12. (a) Determine the short-circuit admittance parameters which completely describe the two-port network of Fig. 17.41. (b) If V1 = 3 V and V2 = –2 V, use your answer in part (a) to compute I1 and I2. 13. (a) Determine the y parameters for the two-port of Fig. 17.42. (b) Deﬁne the bottom node of Fig. 17.42 as the reference node, and apply nodal analysis to obtain expressions for I1 and I2 in terms of V1 and V2. Use these expressions to write down the admittance matrix. (c) If V1 = 2V2 = 10 V, calculate the power dissipated in the 100 mS conductance. 14. Obtain an complete set of y parameters to describe the two-port network depicted in Fig. 17.43. 0.05 S 0.15 S 0.1 S 0.25 S V2 + – V1 + – I2 I1 ■FIGURE 17.42 10 kΩ 10 Ω 1 kΩ 8 kΩ 4 Ω V2 + – V1 + – I2 I1 + – 15 V ■FIGURE 17.44 1 Ω 2 Ω V2 + – V1 + – I2 I1 5I1 ■FIGURE 17.45 540 Ω 200 Ω 510 Ω 400 Ω V2 + – V1 + – I2 I1 ■FIGURE 17.43 15. The circuit of Fig. 17.44 is simply the two-port of Fig. 17.40 terminated by a passive one-port and a separate one-port consisting of a voltage source in series with a resistor. (a) Determine the complete set of admittance parameters which describe the two-port network. (Hint: draw the two-port by itself, properly labeled with a voltage and current at each port.) (b) Calculate the power dissi-pated in the passive one-port, using your answer to part (a). 16. Replace the 10 resistor of Fig. 17.44 with a 1 k resistor, the 15 V source with a 9 V source, and the 4 resistor with a 4 k resistor. (a) Determine the complete set of admittance parameters which describe the two-port network consisting of the 1 k, 10 k, and 8 k resistors. (Hint: draw the two port by itself, properly labeled with a voltage and current at each port.) (b) Calculate the power dissipated in the passive one-port, using your answer to part (a). 17. Determine the admittance parameters which describe the two-port shown in Fig. 17.45. 10 Ω 8 Ω 11 Ω 20 Ω V2 + – V1 + – I2 I1 ■FIGURE 17.41 EXERCISES 725 19. Employ an appropriate method to obtain y for the network of Fig. 17.47. 5 kΩ 0.6V2 0.1I1 10 kΩ 20 kΩ V2 + – V1 + – I2 I1 ■FIGURE 17.46 V1 + – V2 + – I2 I1 5 Ω 1 Ω 0.3I1 2 Ω ■FIGURE 17.47 G S D S Cgd Cgs rd gmv Cds v + – ■FIGURE 17.48 (a) For the conﬁguration stated above, which transistor terminal is used as the input, and which terminal is used as the output? (b) Derive expressions for the parameters yis, yrs, yf s, and yos deﬁned in Eqs. and , in terms of the model parameters Cgs, Cgd, gm,rd, and Cds of Fig. 17.48. (c) Compute yis, yrs, yf s, and yos if gm = 4.7 mS, Cgs = 3.4 pF, Cgd = 1.4 pF, Cds = 0.4 pF, and rd = 10 k. 17.3 Some Equivalent Networks 21. For the two-port displayed in Fig. 17.49, (a) determine the input resistance; (b) compute the power dissipated by the network if connected in parallel with a 2 A current source; (c) compute the power dissipated by the network if connected in parallel with a 9 V voltage source. 5 kΩ 20 kΩ 12 kΩ 2.2 kΩ 4.7 kΩ ■FIGURE 17.49 20. The metal-oxide-semiconductor ﬁeld effect transistor (MOSFET), a three-terminal nonlinear element used in many electronics applications, is often speciﬁed in terms of its y parameters. The ac parameters are strongly dependent on the measurement conditions, and commonly named yis, yrs, yf s, and yos, as in Ig = yisVgs + yrsVds Id = yf sVgs + yosVds where Ig is the transistor gate current, Id is the transistor drain current, and the third terminal (the source) is common to the input and output during the mea-surement. Thus, Vgs is the voltage between the gate and the source, and Vds is the voltage between the drain and the source. The typical high-frequency model used to approximate the behavior of a MOSFET is shown in Fig. 17.48. 18. Obtain the y parameter for the network shown in Fig. 17.46 and use it to determine I1 and I2 if (a) V1 = 0, V2 = 1 V; (b) V1 = −8 V, V2 = 3 V; (c) V1 = V2 = 5 V. CHAPTER 17 TWO-PORT NETWORKS 726 24. Determine the input impedance Zin of the one-port shown in Fig. 17.52 if ω is equal to (a) 50 rad/s; (b) 1000 rad/s. CHAPTER 17 TWO-PORT NETWORKS 726 5 H 1 H 5 H 3 H 2 H Zin 50 mF 0.02 F ■FIGURE 17.51 6 Ω 2 H 5 H 3 H 5 mF 3 mF 4 Ω Zin ■FIGURE 17.52 4 MΩ 600 kΩ 2 MΩ 500 kΩ 1 MΩ 3 MΩ 700 kΩ 220 kΩ 400 kΩ Rin ■FIGURE 17.53 23. Determine the input impedance Zin of the one-port shown in Fig. 17.51 if ω is equal to (a) 50 rad/s; (b) 1000 rad/s. 25. Employ -Y conversion techniques as appropriate to determine the input resis-tance Rin of the one-port shown in Fig. 17.53. 6 Ω 3 Ω 2 Ω a c b d 470 Ω a c b d 220 Ω 100 Ω ■FIGURE 17.50 22. With reference to the two networks in Fig. 17.50, convert the -connected network to a Y-connected network, and vice versa. EXERCISES 727 27. (a) Determine the parameter values required to model the network of Fig. 17.43 with the alternative network shown in Fig. 17.13a. (b) Verify that the two networks are in fact equivalent by computing the power dissipated in a 2 resistor connected to the right of each network and connecting a 1 A current source to the left-hand terminals. 28. (a) The network of Fig. 17.13b is equivalent to the network of Fig. 17.43 assuming the appropriate parameter values are chosen. (a) Compute the neces-sary parameter values. (b) Verify the equivalence of the two networks by termi-nating each with a 1 resistor (across their V2 terminals), connecting a 10 mA source to the other terminals, and showing that I1, V1, I2, and V2 are equal for both networks. 29. Compute the three parameter values necessary to construct an equivalent network for Fig. 17.43 modeled after the network of Fig. 17.13c. Verify their equivalence with an appropriate PSpice simulation. (Hint: connect some type of source(s) and load(s).) 30. It is possible to construct an alternative two-port to the one shown in Fig. 17.47, by selecting appropriate parameter values as labeled on the diagram in Fig. 17.13. (a) Construct such an equivalent network. (b) Verify their equiva-lence with an appropriate PSpice simulation. (Hint: connect some type of source(s) and load(s).) 31. Let y = 0.1 −0.05 −0.5 0.2 (S) for the two-port of Fig. 17.55. Find (a) GV; (b) GI; (c) GP; (d) Zin; (e) Zout. ( f ) If the reverse voltage gain GV,rev is deﬁned as V1/V2 with Vs = 0 and RL removed, calculate GV,rev. (g) If the insertion power gain Gins is deﬁned as the ratio of P5 with the two-port in place to P5 with the two-port replaced by jumpers connecting each input terminal to the corresponding output terminal, calculate Gins. 5 Ω 9 Ω 3 Ω 6 Ω 7 Ω 2 Ω 12 Ω 10 Ω 12 Ω 2 Ω 4 Ω 6 Ω 4 Ω ■FIGURE 17.54 26. Employ appropriate techniques to ﬁnd a value for the input resistance of the one-port network represented by the schematic of Fig. 17.54. V2 + – y V1 + – 10 Ω 5 Ω Vs = 1 V + – ■FIGURE 17.55 17.4 Impedance Parameters 32. Convert the following z parameters to y parameters, or vice versa, as appropriate: z = 2 3 5 1 z = 1000 470 2500 900 y = 0.001 0.005 0.006 0.03 S y = 1 2 −1 3 S 33. By employing Eqs. to , obtain a complete set of z parameters for the network given in Fig. 17.56. 100 Ω 50 Ω a c b d 25 Ω ■FIGURE 17.56 34. The network of Fig. 17.56 is terminated with a 10 resistor across terminals b and d, and a 6 mA sinusoidal current source operating at 100 Hz in parallel with a 50 resistor is connected across terminals a and c. Calculate the volt-age, current, and power gains, respectively, as well as the input and output impedance. 35. The two-port networks of Fig. 17.50 are connected in series. (a) Determine the impedance parameters for the series connection by ﬁrst ﬁnding the z parame-ters of the individual networks. (b) If the two networks are instead connected in parallel, determine the admittance parameters of the combination by ﬁrst ﬁnding the y parameters of the individual networks. (c) Verify your answer to part (b) by using Table 17.1 in conjunction with your answer to part (a). 36. (a) Use an appropriate method to obtain the impedance parameters which describe the network illustrated in Fig. 17.57. (b) If a 1 V source in series with a 1 k resistor is connected to the left-hand port such that the negative refer-ence terminal of the source is connected to the common terminal of the net-work, and a 5 k load is connected across the right-hand terminals, compute the current, voltage, and power gain. 37. Determine the impedance parameters for the two-port exhibited in Fig. 17.58. CHAPTER 17 TWO-PORT NETWORKS 728 CHAPTER 17 TWO-PORT NETWORKS 728 + – V2 + – V1 + – 4 kΩ 12 kΩ 10 kΩ 3 kΩ 0.2V2 ■FIGURE 17.57 V1 + – V2 + – 2 Ω 5 Ω 0.1V1 0.8V2 + – ■FIGURE 17.58 V1 + – V2 + – I1 I2 30 Ω 50 Ω 100 Ω 0.08V1 0.2V2 ■FIGURE 17.59 V1 + – V2 + – 10 kΩ 100 kΩ 0.01V1 1 pF 5 pF ■FIGURE 17.60 38. Obtain both the impedance and admittance parameters for the two-port network of Fig. 17.59. 39. Find the four z parameters at ω = 108 rad/s for the transistor high-frequency equivalent circuit shown in Fig. 17.60. 17.5 Hybrid Parameters 40. Determine the h parameters which describe the purely resistive network shown in Fig. 17.56 by connecting appropriate 1 V, 1 A, and short circuits to terminals as required. EXERCISES 729 729 42. If h for some particular two-port is given by h = 2 k −3 5 0.01 S , calculate (a) z; (b) y. 43. A certain two-port network is described by hybrid parameters h = 100 −2 5 0.1 S . Determine the new h parameters if a 25 resistor is connected in parallel with (a) the input; (b) the output. 44. A bipolar junction transistor is connected in common-emitter conﬁguration, and found to have h parameters h11 = 5 k, h12 = 0.55 × 10−4, h21 = 300, and h22 = 39 μS. (a) Write h in matrix form. (b) Determine the small-signal current gain. (c) Determine the output resistance in k. (d) If a sinusoidal voltage source having frequency 100 rad/s and amplitude 5 mV in series with a 100 resistor is connected to the input terminals, calculate the peak voltage which appears across the output terminals. 45. The two-port which plays a central role in the circuit of Fig. 17.62 can be characterized by hybrid parameters h = 1 −1 2 0.5 S . Determine I1, I2, V1, and V2. V1 + – V2 + – 25 Ω 50 Ω V1 + – V2 + – 25 Ω 50 Ω ■FIGURE 17.61 41. Obtain the h parameters of the two-ports of Fig. 17.61. + – V2 + – V1 + – 5 Ω 2 Ω 1 V I1 I2 ■FIGURE 17.62 46. The two networks of Fig. 17.61 are connected in series by connecting the terminals as illustrated in Fig. 17.22 (assume the left-hand network of Fig. 17.61 is network A). Determine the new set of h parameters which describe the series connection. 47. The two networks of Fig. 17.61 are connected in parallel by tying the corre-sponding input terminals together, and then tying the corresponding output terminals together. Determine the new set of h parameters which describe the parallel connection. 48. Find y, z, and h for both of the two-ports shown in Fig. 17.63. If any parameter is inﬁnite, skip that parameter set. (a) R (b) R ■FIGURE 17.63 CHAPTER 17 TWO-PORT NETWORKS 730 6 Ω 4 Ω 3 Ω 5 Ω 1 Ω 2 Ω A B C V1 + – V2 + – I2 I1 ■FIGURE 17.65 55. Consider the two separate two-ports of Fig. 17.61. Determine the ABCD matrix which characterizes the cascaded network resulting from connecting (a) the output of the left-hand network to the input of the right-hand network; (b) the output of the right-hand network to the input of the left-hand network. 56. (a) Determine the t parameters which describe the two-port of Fig. 17.58. (b) Compute Zout if a practical voltage source having a 100 series resistance is connected to the input terminals of the network. 17.6 Transmission Parameters 50. (a) With the assistance of appropriate mesh equations, determine the ABCD matrix which represents the two-port shown in Fig. 17.9. (b) Convert your answer to h. 51. (a) Employ suitably written mesh equations to obtain the t parameters which characterize the network of Fig. 17.57. (b) If currents I1 and I2 are deﬁned as ﬂowing into the (+) reference terminals of V1 and V2, respectively, compute the voltages if I1 = 2I2 = 3 mA. 52. Consider the following matrices: a = 5 2 4 1 b = 1.5 1 1 0.5 c = −4 2 Calculate (a) a · b; (b) b · a; (c) a · c; (d) b · c; (e) b · a · c; (f) a · a. 53. Two networks are represented by the following impedance matrices: z1 = 4.7 0.5 0.87 1.8 k and z2 = 1.1 2.2 0.89 1.8 k, respectively. (a) Determine the t matrix which characterizes the cascaded network resulting from connecting network 2 to the output of network 1. (b) Reverse the order of the networks and compute the new t matrix which results. 54. The two-port of Fig. 17.65 can be viewed as three separate cascaded two-ports A, B, and C. (a) Compute t for each network. (b) Obtain t for the cas-caded network. (c) Verify your answer by naming the two middle nodes Vx and Vy, respectively, writing nodal equations, obtaining the admittance parameters from your nodal equations, and converting to t parameters using Table 17.1. + – V1 + – V2 + – 10 kΩ 1 kΩ 10–5V2 100V1 ■FIGURE 17.64 49. (a) Find h for the two-port of Fig. 17.64. (b) Find Zout if the input contains Vs in series with Rs = 200 . EXERCISES 731 Chapter-Integrating Exercises 59. (a) Obtain y, z, h, and t parameters for the network shown in Fig. 17.67 using either the deﬁning equations or mesh/nodal equations. (b) Verify your answers, using the relationships in Table 17.1. R (a) V1 + – V2 + – 2 Ω 20 Ω 1:4 10 Ω 50 Ω (d) 1:a (c) R (b) ■FIGURE 17.66 10 Ω 10 Ω a c b d 5 Ω ■FIGURE 17.67 60. Four networks, each identical to the one depicted in Fig. 17.67, are connected in parallel such that all terminals labeled a are tied together, all terminals designated b are tied together, and all terminals labeled c and d are connected. Obtain the y, z, h, and t parameters which describe the parallel-connected network. 61. A cascaded 12-element network is formed using four two-ports identical to the one shown in Fig. 17.67. Determine the y, z, h, and t parameters which describe the result. 62. The concept of ABCD matrices extends to systems beyond electrical circuits. For example, they are commonly employed for ray-tracing calculations in opti-cal systems. In that case, we envision parallel input and output planes in xy, skewered by an optical axis z. An inbound ray crosses the input plane a distance x = rin from the optical axis, making an angle θin. The corresponding 57. Three identical networks as the one depicted in Fig. 17.56 are cascaded together. Determine the t parameters which fully represent the result. 58. (a) Find ta, tb, and tc for the networks shown in Fig. 17.66a, b, and c. (b) By using the rules for interconnecting two-ports in cascade, ﬁnd t for the network of Fig. 17.66d. parameters rout, θout for the outbound ray crossing the output plane are then given by the ABCD matrix such that rout θout = A B C D rin θin Each type of optical element (e.g., mirror, lens, or even propagation through free space) has its own ABCD matrix. If the ray passes through several ele-ments, the net effect can be predicted by simply cascading the individual ABCD matrices (in the proper order). (a) Obtain expressions for A, B, C, and D similar to Eqs. to . (b) If the ABCD matrix for a perfectly reﬂecting ﬂat mirror is given by 1 0 0 1 , sketch the system along with the inbound and outbound rays, taking care to note the orientation of the mirror. 63. Continuing from Exercise 62, the behavior of a ray propating through free space a distance d can be modeled with the ABCD matrix 1 d 0 1 . (a) Show that the same result is obtained (rout, θout) whether a single ABCD matrix is used with d, or two cascaded matrices are used, each with d/2. (b) What are the units of A, B, C, and D, respectively? CHAPTER 17 TWO-PORT NETWORKS 732 INTRODUCTION In this chapter we continue our introduction to circuit analysis by studying periodic functions in both the time and frequency domains. Speciﬁcally, we will consider forcing functions which are periodic and have functional natures which satisfy certain mathematical restrictions that are characteristic of any function which we can generate in the laboratory. Such functions may be represented as the sum of an inﬁnite number of sine and cosine functions which are harmonically related. Therefore, since the forced response to each sinusoidal component can be determined easily by sinusoidal steady-state analysis, the response of the linear network to the general periodic forcing function may be obtained by superposing the partial responses. The topic of Fourier series is of vital importance in a number of ﬁelds, particularly communications. The use of Fourier-based tech-niques to assist in circuit analysis, however, had been slowly falling out of fashion for a number of years. Now as we face an increas-ingly larger fraction of global power usage coming from equipment employing pulse-modulated power supplies (e.g., computers), the subject of harmonics in power systems and power electronics is rapidly becoming a serious problem in even large-scale generation plants. It is only with Fourier-based analysis that the underlying problems and possible solutions can be understood. 18.1 • TRIGONOMETRIC FORM OF THE FOURIER SERIES We know that the complete response of a linear circuit to an arbi-trary forcing function is composed of the sum of a forced response and a natural response. The natural response has been considered KEY CONCEPTS Representing Periodic Functions as a Sum of Sines and Cosines Harmonic Frequencies Even and Odd Symmetry Half-Wave Symmetry Complex Form of the Fourier Series Discrete Line Spectra Fourier Transform Using Fourier Series and Fourier Transform Techniques in Circuit Analysis System Response and Convolution in the Frequency Domain Fourier Circuit Analysis C H A P T E R 18 733 both in the time domain (Chaps. 7, 8, and 9) and in the frequency domain (Chaps. 14 and 15). The forced response has also been considered fromsev-eral perspectives, including the phasor-based techniques of Chap. 10. As we have discovered, in some cases we need both components of the total re-sponse of a particular circuit, while in others we need only the natural or the forced response. In this section, we refocus our attention on forcing func-tions that are sinusoidal in nature, and discover how to write a general peri-odic function as a sum of such functions—leading us into a discussion of a new set of circuit analysis procedures. Harmonics Some feeling for the validity of representing a general periodic function by an inﬁnite sum of sine and cosine functions may be gained by considering a sim-ple example. Let us ﬁrst assume a cosine function of radian frequency ω0, v1(t) = 2 cos ω0t where ω0 = 2πf0 and the period T is T = 1 f0 = 2π ω0 Although T does not usually carry a zero subscript, it is the period of the fundamental frequency. The harmonics of this sinusoid have frequencies nω0, where ω0 is the fundamental frequency and n = 1, 2, 3, . . . . The frequency of the ﬁrst harmonic is the fundamental frequency. Next let us select a third-harmonic voltage v3a(t) = cos 3ω0t The fundamental v1(t), the third harmonic v3a(t), and the sum of these two waves are shown as functions of time in Fig. 18.1a. Note that the sum is also periodic, with period T = 2π/ω0. The form of the resultant periodic function changes as the phase and amplitude of the third-harmonic component change. Thus, Fig. 18.1b shows theeffectofcombiningv1(t)andathirdharmonicofslightlylargeramplitude, v3b(t) = 1.5 cos 3ω0t By shifting the phase of the third harmonic by 90 degrees to give v3c(t) = sin 3ω0t the sum, shown in Fig. 18.1c, takes on a still different character. In all cases, the period of the resultant waveform is the same as the period of the funda-mental waveform. The nature of the waveform depends on the amplitude and phase of every possible harmonic component, and we will ﬁnd that we are able to generate waveforms which have extremely nonsinusoidal characteristics by an appropriate combination of sinusoidal functions. After we have become familiar with the use of the sum of an inﬁnite number of sine and cosine functions to represent a periodic waveform, we will consider the frequency-domain representation of a general nonperiodic waveform in a manner similar to the Laplace transform. CHAPTER 18 FOURIER CIRCUIT ANALYSIS 734 SECTION 18.1 TRIGONOMETRIC FORM OF THE FOURIER SERIES 735 ■FIGURE 18.1 Several of the inﬁnite number of different waveforms which may be obtained by combining a fundamental and a third harmonic. The fundamental is v1 = 2 cos ω0t, and the third harmonic is (a) v3a = cos 3ω0t ; (b) v3b = 1.5 cos 3ω0t ; (c) v3c = sin 3ω0t . (a) (c) 7.85 6.28 4.71 3.14 1.57 0t –3 –2 –1 0 1 2 3 v1(t) v(t) v3a(t) 7.85 6.28 4.71 3.14 1.57 0t –3 –2 –1 0 1 2 3 v1(t) v(t) v3c(t) 7.85 6.28 4.71 3.14 1.57 0t –3 –2 –1 0 1 2 3 v1(t) v(t) v3b(t) (b) 4 – 4 PRACTICE ● 18.1 Let a third-harmonic voltage be added to the fundamental to yield v = 2 cos ω0t + Vm3 sin 3ω0t, the waveform shown in Fig. 18.1c for Vm3 = 1. (a) Find the value of Vm3 so that v(t) will have zero slope at ω0t = 2π/3. (b) Evaluate v(t) at ω0t = 2π/3. Ans: 0.577; −1.000. The Fourier Series We ﬁrst consider a periodic function f (t), deﬁned in Sec. 11.2 by the func-tional relationship f (t) = f (t + T) where T is the period. We further assume that the function f (t) satisﬁes the following properties: CHAPTER 18 FOURIER CIRCUIT ANALYSIS 736 We will take f(t) to represent either a voltage or a current waveform, and any such waveform which we can actually produce must satisfy these four conditions; perhaps it should be noted, however, that certain mathematical functions do exist for which these four conditions are not satisﬁed. 1. f (t) is single-valued everywhere; that is, f (t) satisﬁes the mathe-matical deﬁnition of a function. 2. The integral t0+T t0 \| f (t)\| dt exists (i.e., is not inﬁnite) for any choice of t0. 3. f (t) has a ﬁnite number of discontinuities in any one period. 4. f (t) has a ﬁnite number of maxima and minima in any one period. Given such a periodic function f (t), the Fourier theorem states that f (t) may be represented by the inﬁnite series f (t) = a0 + a1 cos ω0t + a2 cos 2ω0t + · · · + b1 sin ω0t + b2 sin 2ω0t + · · · = a0 + ∞ n=1 (an cos nω0t + bn sin nω0t) where the fundamental frequency ω0 is related to the period T by ω0 = 2π T and where a0, an, and bn are constants that depend upon n and f (t). Equa-tion is the trigonometric form of the Fourier series for f (t), and the process of determining the values of the constants a0, an, and bn is called Fourier analysis. Our object is not the proof of this theorem, but only a sim-ple development of the procedures of Fourier analysis and a feeling that the theorem is plausible. Some Useful Trigonometric Integrals Before we discuss the evaluation of the constants appearing in the Fourier series, let us collect a set of useful trigonometric integrals. We let both n and k represent any element of the set of integers 1, 2, 3, . . . . In the following integrals, 0 and T are used as the integration limits, but it is understood that any interval of one period is equally correct. T 0 sin nω0t dt = 0 T 0 cos nω0t dt = 0 T 0 sin kω0t cos nω0t dt = 0 SECTION 18.1 TRIGONOMETRIC FORM OF THE FOURIER SERIES 737 T 0 sin kω0t sin nω0t dt = 0 (k ̸= n) T 0 cos kω0t cos nω0t dt = 0 (k ̸= n) Those cases which are excepted in Eqs. and are also easily evaluated; we obtain T 0 sin2 nω0t dt = T 2 T 0 cos2 nω0t dt = T 2 Evaluation of the Fourier Coefﬁcients The evaluation of the unknown constants in the Fourier series may now be accomplished readily. We ﬁrst attack a0. If we integrate each side of Eq. over a full period, we obtain T 0 f (t) dt = T 0 a0 dt + T 0 ∞ n=1 (an cos nω0t + bn sin nω0t) dt But every term in the summation is of the form of Eq. or , and thus T 0 f (t) dt = a0T or a0 = 1 T T 0 f (t) dt This constant a0 is simply the average value of f (t) over a period, and we therefore describe it as the dc component of f (t). To evaluate one of the cosine coefﬁcients, for example, ak, the coefﬁcient of cos kω0t, we ﬁrst multiply each side of Eq. by cos kω0t and then integrate both sides of the equation over a full period: T 0 f (t) cos kω0t dt = T 0 a0 cos kω0t dt + T 0 ∞ n=1 an cos kω0t cos nω0t dt + T 0 ∞ n=1 bn cos kω0t sin nω0t dt From Eqs. , , and we note that every term on the right-hand side of this equation is zero except for the single an term where k = n. We evaluate that term using Eq. , and in so doing we ﬁnd ak, or an: an = 2 T T 0 f (t) cos nω0t dt CHAPTER 18 FOURIER CIRCUIT ANALYSIS 738 This result is twice the average value of the product f (t) cos nω0t over a period. In a similar way, we obtain bk by multiplying by sin kω0t, integrating over a period, noting that all but one of the terms on the right-hand side are zero, and performing that single integration by Eq. . The result is bn = 2 T T 0 f (t) sin nω0t dt which is twice the average value of f (t) sin nω0t over a period. Equations to now enable us to determine values for a0 and all the an and bn in the Fourier series, Eq. , as summarized below: f (t) = a0 + ∞ n=1 (an cos nω0t + bn sin nω0t) ω0 = 2π T = 2πf0 a0 = 1 T T 0 f (t) dt an = 2 T T 0 f (t) cos nω0t dt bn = 2 T T 0 f (t) sin nω0t dt EXAMPLE 18.1 The “half-sinusoidal” waveform shown in Fig. 18.2 represents the voltage response obtained at the output of a half-wave rectiﬁer circuit, a nonlinear circuit whose purpose is to convert a sinusoidal input voltage to a (pulsating) approximation to dc. Find the Fourier series representation of this waveform. Identify the goal of the problem. We are presented with a periodic function and are asked to ﬁnd the Fourier series representation. If not for the removal of all negative voltages, the problem would be trivial, as only one sinusoid would be required. ■FIGURE 18.2 The output of a half-wave rectiﬁer to which a sinusoidal input is applied. 0 –0.2 –0.4 0.4 0.2 Vm v (t) t (s) SECTION 18.1 TRIGONOMETRIC FORM OF THE FOURIER SERIES 739 Collect the known information. In order to represent this voltage as a Fourier series, we must ﬁrst determine the period and then express the graphical voltage as an analytical function of time. From the graph, the period is seen to be T = 0.4 s and thus f0 = 2.5 Hz and ω0 = 5π rad/s Devise a plan. The most straightforward approach is to apply Eqs. to to calculate the set of coefﬁcients a0, an, and bn. To do this, we need a functional expression for v(t), the most straightforward being deﬁned over the interval t = 0 to t = 0.4 as v(t) = ⎧ ⎨ ⎩ Vm cos 5πt 0 ≤t ≤0.1 0 0.1 ≤t ≤0.3 Vm cos 5πt 0.3 ≤t ≤0.4 However, choosing the period to extend from t = −0.1 to t = 0.3 will result in fewer equations and, hence, fewer integrals: v(t) = Vm cos 5πt −0.1 ≤t ≤0.1 0 0.1 ≤t ≤0.3 This form is preferable, although either description will yield the correct results. Construct an appropriate set of equations. The zero-frequency component is easily obtained: a0 = 1 0.4 0.3 −0.1 v(t) dt = 1 0.4 0.1 −0.1 Vm cos 5πt dt + 0.3 0.1 0 dt The amplitude of a general cosine term is an = 2 0.4 0.1 −0.1 Vm cos 5πt cos 5πnt dt and the amplitude of a general sine term is bn = 2 0.4 0.1 −0.1 Vm cos 5πt sin 5πnt dt which, in fact, is always zero, and hence will not be considered further. Determine if additional information is required. The form of the function we obtain upon integrating is different when n is unity than it is for any other choice of n. If n = 1, we have a1 = 5Vm 0.1 −0.1 cos2 5πt dt = Vm 2 Notice that integration over an entire period must be broken up into subintervals of the period, in each of which the functional form of v(t) is known. (Continued on next page) CHAPTER 18 FOURIER CIRCUIT ANALYSIS 740 whereas if n is not equal to unity, we ﬁnd an = 5Vm 0.1 −0.1 cos 5πt cos 5πnt dt Attempt a solution. Solving, we ﬁnd that a0 = Vm π an = 5Vm 0.1 −0.1 1 2[cos 5π(1 + n)t + cos 5π(1 −n)t] dt or an = 2Vm π cos(πn/2) 1 −n2 (n ̸= 1) (A similar integration shows that bn = 0 for any value of n, and the Fourier series thus contains no sine terms.) The Fourier series is therefore obtained from Eqs. , , , and : v(t) = Vm π + Vm 2 cos 5πt + 2Vm 3π cos 10πt −2Vm 15π cos 20πt + 2Vm 35π cos 30πt −· · · Verify the solution. Is it reasonable or expected? Our solution can be checked by plugging values into Eq. and truncating after a speciﬁc number of terms. Another approach, how-ever, is to plot the function as shown in Fig. 18.3 for n = 1, 2, and 6. ■FIGURE 18.3 Equation truncated after n = 1 term, n = 2 term, and n = 6 term, showing convergence to the half-sinusoid v (t). A magnitude of V m = 1 has been chosen for convenience. 1.2 1.0 0.8 0.6 v(t) (volts) 0.4 0.2 0 –0.2 –0.8 –1 –0.6 –0.4 –0.2 0 Time (seconds) 0.2 0.4 0.6 0.8 1 n = 1 n = 6 n = 2 It should be pointed out, incidentally, that the expres-sion for an when n ̸= 1 will yield the correct result for n = 1 in the limit as n →1. SECTION 18.1 TRIGONOMETRIC FORM OF THE FOURIER SERIES 741 ■FIGURE 18.4 –1 0 –1 +1 1 2 3 4 v (V) t (s) (a) –1 –1 0 1 1 2 3 4 v (V) t (s) (c) –1 –1 0 1 1 2 3 4 v (V) t (s) (b) As can be seen, as more terms are included, the more the plot resem-bles that of Fig. 18.2. PRACTICE ● 18.2 A periodic waveform f (t) is described as follows: f (t) = −4, 0 < t < 0.3; f (t) = 6, 0.3 < t < 0.4; f (t) = 0, 0.4 < t < 0.5; T = 0.5. Evaluate (a) a0; (b) a3; (c) b1. 18.3 Write the Fourier series for the three voltage waveforms shown in Fig. 18.4. Line and Phase Spectra We depicted the function v(t) of Example 18.1 graphically in Fig. 18.2, and analytically in Eq. —both representations being in the time domain. The Fourier series representation of v(t) given in Eq. is also a time-domain expression, but may be transformed into a frequency-domain representation as well. For example, Fig. 18.5 shows the amplitude of each frequency component of v(t), a type of plot known as a line spectrum. Here, the mag-nitude of each frequency component (i.e., \|a0\|, \|a1\|, etc.) is indicated by the length of the vertical line at the corresponding frequency (f0, f1, etc.); for the sake of convenience, we have taken Vm = 1. Given a different value of Vm, we simply scale the y axis values by the new value. Ans:18.2:−1.200;1.383;−4.44.18.3:(4/π)(sin πt + 1 3 sin 3πt + 1 5 sin 5πt + · · ·) V; (4/π)(cos πt −1 3 cos 3πt + 1 5 cos 5πt −· · ·)V; (8/π2)(sin πt −1 9 sin 3πt + 1 25 sin 5πt −· · ·). Such a plot, sometimes referred to as a discrete spectrum, gives a great deal of information at a glance. In particular, we can see how many terms of the series are required to obtain a reasonable approximation of the original waveform. In the line spectrum of Fig. 18.5, we note that the 8th and 10th harmonics (20 and 25 Hz, respectively) add only a small correction. Trun-cating the series after the 6th harmonic therefore should lead to a reasonable approximation; the reader can judge this for herself/himself by considering Fig. 18.3. One note of caution must be injected. The example we have considered contains no sine terms, and the amplitude of the nth harmonic is therefore \|an\|. If bn is not zero, then the amplitude of the component at a frequency nω0 must be a2 n + b2 n. This is the general quantity which we must show in a line spectrum. When we discuss the complex form of the Fourier series, we will see that this amplitude is obtained more directly. In addition to the amplitude spectrum, we may construct a discrete phase spectrum. At any frequency nω0, we combine the cosine and sine terms to determine the phase angle φn: an cos nω0t + bn sin nω0t = a2 n + b2 n cos nω0t + tan−1 −bn an = a2 n + b2 n cos(nω0t + φn) or φn = tan−1 −bn an In Eq. , φn = 0◦or 180◦for every n. The Fourier series obtained for this example includes no sine terms and no odd harmonics (except the fundamental) among the cosine terms. It is CHAPTER 18 FOURIER CIRCUIT ANALYSIS 742 ■FIGURE 18.5 The discrete line spectrum of v(t) as represented in Eq. , showing the ﬁrst seven frequency components. A magnitude of V m = 1 has been chosen for convenience. 30 25 20 10 15 Frequency (Hz) 5 0 –5 0.5 0.4 0.3 0.2 0.1 0 Harmonic Amplitude SECTION 18.2 THE USE OF SYMMETRY 743 possibletoanticipatetheabsenceofcertaintermsinaFourierseries,beforeany integrations are performed, by an inspection of the symmetry of the given time function. We will investigate the use of symmetry in the following section. 18.2 • THE USE OF SYMMETRY Even and Odd Symmetry The two types of symmetry which are most readily recognized are even-function symmetry and odd-function symmetry, or simply even symmetry and odd symmetry. We say that f (t) possesses the property of even symmetry if f (t) = f (−t) Such functions as t2, cos 3t, ln(cos t), sin2 7t, and a constant C all possess even symmetry; the replacement of t by (−t) does not change the value of any of these functions. This type of symmetry may also be recognized graphically, for if f (t) = f (−t), then mirror symmetry exists about the f (t) axis. The function shown in Fig. 18.6a possesses even symmetry; if the ﬁgure were to be folded along the f (t) axis, then the portions of the graph for positive and negative time would ﬁt exactly, one on top of the other. We deﬁne odd symmetry by stating that if odd symmetry is a property of f (t), then f (t) = −f (−t) In other words, if t is replaced by (−t), then the negative of the given func-tion is obtained; for example, t, sin t, t cos 70t, t √ 1 + t2, and the function sketched in Fig. 18.6b are all odd functions and possess odd symmetry. The graphical characteristics of odd symmetry are apparent if the portion of f (t) for t > 0 is rotated about the positive t axis and the resultant ﬁgure is then rotated about the f (t) axis; the two curves will ﬁt exactly, one on top of the other. That is, we now have symmetry about the origin, rather than about the f (t) axis as we did for even functions. Having deﬁnitions for even and odd symmetry, we should note that the product of two functions with even symmetry, or of two functions with odd symmetry, yields a function with even symmetry. Furthermore, the product of an even and an odd function gives a function with odd symmetry. Symmetry and Fourier Series Terms Now let us investigate the effect that even symmetry produces in a Fourier series. If we think of the expression which equates an even function f (t) and the sum of an inﬁnite number of sine and cosine functions, then it is apparent that the sum must also be an even function. A sine wave, however, is an odd function, and no sum of sine waves can produce any even function other than zero (which is both even and odd). It is thus plausible that the Fourier series of any even function is composed of only a constant and cosine functions. Let us now show carefully that bn = 0. We have bn = 2 T T/2 −T/2 f (t) sin nω0t dt = 2 T 0 −T/2 f (t) sin nω0t dt + T/2 0 f (t) sin nω0t dt ■FIGURE 18.6 (a) A waveform showing even symmetry. (b) A waveform showing odd symmetry. T –T f (t) t (a) 0 T 0 –T f (t) t (b) We replace the variable t in the ﬁrst integral by −τ, or τ = −t, and make use of the fact that f (t) = f (−t) = f (τ): bn = 2 T 0 T/2 f (−τ) sin(−nω0τ)(−dτ) + T/2 0 f (t) sin nω0t dt = 2 T − T/2 0 f (τ) sin nω0τ dτ + T/2 0 f (t) sin nω0t dt but the symbol we use to identify the variable of integration cannot affect the value of the integral. Thus, T/2 0 f (τ) sin nω0τ dτ = T/2 0 f (t) sin nω0t dt and bn = 0 (even sym.) No sine terms are present. Therefore, if f (t) shows even symmetry, then bn = 0; conversely, if bn = 0, then f (t) must have even symmetry. A similar examination of the expression for an leads to an integral over the half period extending from t = 0 to t = 1 2T : an = 4 T T/2 0 f (t) cos nω0t dt (even sym.) The fact that an may be obtained for an even function by taking “twice the integral over half the range’’ should seem logical. A function having odd symmetry can contain no constant term or cosine terms in its Fourier expansion. Let us prove the second part of this state-ment. We have an = 2 T T/2 −T/2 f (t) cos nω0t dt = 2 T 0 −T/2 f (t) cos nω0t dt + T/2 0 f (t) cos nω0t dt and we now let t = −τ in the ﬁrst integral: an = 2 T 0 T/2 f (−τ) cos(−nω0τ) (−dτ) + T/2 0 f (t) cos nω0t dt = 2 T T/2 0 f (−τ) cos nω0τ dτ + T/2 0 f (t) cos nω0t dt But f (−τ) = −f (τ), and therefore an = 0 (odd sym.) CHAPTER 18 FOURIER CIRCUIT ANALYSIS 744 SECTION 18.2 THE USE OF SYMMETRY 745 A similar, but simpler, proof shows that a0 = 0 (odd sym.) With odd symmetry, therefore, an = 0 and a0 = 0; conversely, if an = 0 and a0 = 0, odd symmetry is present. The values of bn may again be obtained by integrating over half the range: bn = 4 T T/2 0 f (t) sin nω0t dt (odd sym.) Half-Wave Symmetry The Fourier series for both of these square waves have one other interesting characteristic: neither contains any even harmonics.1 That is, the only fre-quency components present in the series have frequencies which are odd multiples of the fundamental frequency; an and bn are zero for even values of n. This result is caused by another type of symmetry, called half-wave symmetry. We will say that f (t) possesses half-wave symmetry if f (t) = −f t −1 2T or the equivalent expression, f (t) = −f t + 1 2T Except for a change of sign, each half cycle is like the adjacent half cycles. Half-wave symmetry, unlike even and odd symmetry, is not a function of the choice of the point t = 0. Thus, we can state that the square wave (Fig. 18.4a or b) shows half-wave symmetry. Neither waveform shown in Fig. 18.6 has half-wave symmetry, but the two somewhat similar functions plotted in Fig. 18.7 do possess half-wave symmetry. It may be shown that the Fourier series of any function which has half-wave symmetry contains only odd harmonics. Let us consider the coefﬁ-cients an. We have again an = 2 T T/2 −T/2 f (t) cos nω0t dt = 2 T 0 −T/2 f (t) cos nω0t dt + T/2 0 f (t) cos nω0t dt which we may represent as an = 2 T (I1 + I2) Now we substitute the new variable τ = t + 1 2T in the integral I1: I1 = T/2 0 f τ −1 2T cos nω0 τ −1 2T dτ = T/2 0 −f (τ) cos nω0τ cos nω0T 2 + sin nω0τ sin nω0T 2 dτ (1) Constant vigilance is required to avoid confusion between an even function and an even harmonic, or between an odd function and an odd harmonic. For example, b10 is the coefﬁcient of an even harmonic, and it is zero if f (t) is an even function. ■FIGURE 18.7 (a) A waveform somewhat similar to the one shown in Fig. 18.6a but possessing half-wave symmetry. (b) A waveform somewhat similar to the one shown in Fig. 18.6b but possessing half-wave symmetry. T 0 –T f (t) t (a) T 0 –T f (t) t (b) But ω0T is 2π, and thus sin nω0T 2 = sin nπ = 0 Hence I1 = −cos nπ T/2 0 f (τ) cos nω0τ dτ After noting the form of I2, we therefore may write an = 2 T (1 −cos nπ) T/2 0 f (t) cos nω0t dt The factor (1 −cos nπ) indicates that an is zero if n is even. Thus, an = ⎧ ⎨ ⎩ 4 T T/2 0 f (t) cos nω0t dt n odd 0 n even 1 2-wave sym. A similar investigation shows that bn is also zero for all even n, and therefore bn = ⎧ ⎨ ⎩ 4 T T/2 0 f (t) sin nω0t dt n odd 0 n even 1 2-wave sym. It should be noted that half-wave symmetry may be present in a wave-form which also shows odd symmetry or even symmetry. The waveform sketched in Fig. 18.7a, for example, possesses both even symmetry and half-wave symmetry. When a waveform possesses half-wave symmetry and either even or odd symmetry, then it is possible to reconstruct the waveform if the function is known over any quarter-period interval. The value of an or bn may also be found by integrating over any quarter period. Thus, an = 8 T T/4 0 f (t) cos nω0t dt n odd an = 0 n even bn = 0 all n ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ 1 2-wave and even sym. an = 0 all n bn = 8 T T/4 0 f (t) sin ω0t dt n odd bn = 0 n even ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ 1 2-wave and odd sym. Table 18.1 provides a short summary of the simpliﬁcations arising from the various types of symmetry discussed. CHAPTER 18 FOURIER CIRCUIT ANALYSIS 746 It is always worthwhile to spend a few moments investigating the symmetry of a function for which a Fourier series is to be determined. SECTION 18.2 THE USE OF SYMMETRY 747 TABLE ●18.1 Summary of Symmetry-Based Simpliﬁcations in Fourier Series Symmetry Type Characteristic Simpliﬁcation bn = 0 an = 0 an = ⎧ ⎨ ⎩ 4 T T/2 0 f (t) cos nω0t dt n odd 0 n even bn = ⎧ ⎨ ⎩ 4 T T/2 0 f (t) sin nω0t dt n odd 0 n even an = ⎧ ⎨ ⎩ 8 T T/4 0 f (t1) cos nω0t dt n odd 0 n even bn = 0 all n an = 0 all n bn = ⎧ ⎨ ⎩ 8 T T/4 0 f (t) sin nω0t dt n odd 0 n even Even Odd Half-Wave Half-Wave and Even Half-Wave and Odd f (t) = −f (t) f (t) = −f (−t) f (t) = −f t −T 2 or f (t) = −f t + T 2 f (t) = −f t −T 2 and f (t) = −f (t) or f (t) = −f t + T 2 and f (t) = −f (t) f (t) = −f t −T 2 and f (t) = −f (−t) or f (t) = −f t + T 2 and f (t) = −f (−t) PRACTICE ● 18.4 Sketch each of the functions described; state whether or not even symmetry, odd symmetry, and half-wave symmetry are present; and give the period: (a) v = 0, −2 < t < 0 and 2 < t < 4; v = 5, 0 < t < 2; v = −5, 4 < t < 6; repeats; (b) v = 10, 1 < t < 3; v = 0, 3 < t < 7; v = −10, 7 < t < 9; repeats; (c) v = 8t, −1 < t < 1; v = 0, 1 < t < 3; repeats. 18.5 Determine the Fourier series for the waveforms of Practice Problem 18.4a and b. Ans: 18.4: No, no, yes, 8; no, no, no, 8; no, yes, no, 4. 18.5: ∞ n=1(odd) 10 nπ sin nπ 2 cos nπt 4 + sin nπt 4 ; ∞ n=1 10 nπ sin 3nπ 4 −3 sin nπ 4 cos nπt 4 + cos nπ 4 −cos 3nπ 4 sin nπt 4 . 18.3 • COMPLETE RESPONSE TO PERIODIC FORCING FUNCTIONS Through the use of the Fourier series, we may now express an arbitrary periodic forcing function as the sum of an inﬁnite number of sinusoidal forcing functions. The forced response to each of these functions may be determined by conventional steady-state analysis, and the form of the nat-ural response may be determined from the poles of an appropriate network transfer function. The initial conditions existing throughout the network, including the initial value of the forced response, enable the amplitude of the natural response to be selected; the complete response is then obtained as the sum of the forced and natural responses. CHAPTER 18 FOURIER CIRCUIT ANALYSIS 748 EXAMPLE 18.2 For the circuit of Fig. 18.8a, determine the periodic response i(t) corresponding to the forcing function shown in Fig. 18.8b if i(0) 0. The forcing function has a fundamental frequency ω0 = 2 rad/s, and its Fourier series may be written down by comparison with the Fourier series developed for the waveform of Fig. 18.4b in the solution of Practice Problem 18.3, vs(t) = 5 + 20 π ∞ n=1(odd) sin 2nt n We will ﬁnd the forced response for the nth harmonic by working in the frequency domain. Thus, vsn(t) = 20 nπ sin 2nt and Vsn = 20 nπ /−90◦= −j 20 nπ The impedance offered by the RL circuit at this frequency is Zn = 4 + j(2n)2 = 4 + j4n and thus the component of the forced response at this frequency is Ifn = Vsn Zn = −j5 nπ(1 + jn) Transforming to the time domain, we have i fn = 5 nπ 1 √ 1 + n2 cos(2nt −90◦−tan−1 n) = 5 π(1 + n2) sin 2nt n −cos 2nt ■FIGURE 18.8 (a) A simple series RL circuit subjected to a periodic forcing function vs(t). (b) The form of the forcing function. 10 0 – 2 vs(t) (V) t (s) 2 2 3 2 (b) + – (a) 4 vs(t) t = 0 2 H i(t) Recall that V m sin ωt is equal to V m cos(ωt −90◦), corresponding to V m/−90◦= −jVm . SECTION 18.3 COMPLETE RESPONSE TO PERIODIC FORCING FUNCTIONS 749 Since the response to the dc component is simply 5 V/4 = 1.25 A, the forced response may be expressed as the summation i f (t) = 1.25 + 5 π ∞ n=1(odd) sin 2nt n(1 + n2) −cos 2nt 1 + n2 The familiar natural response of this simple circuit is the single expo-nential term [characterizing the single pole of the transfer function, If /Vs = 1/(4 + 2s)] in(t) = Ae−2t The complete response is therefore the sum i(t) = i f (t) + in(t) Letting t = 0, we ﬁnd A using i(0) = 0: A = −1.25 + 5 π ∞ n=1(odd) 1 1 + n2 Although correct, it is more convenient to use the numerical value of the summation. The sum of the ﬁrst 5 terms of 1/(1 + n2) is 0.671, the sum of the ﬁrst 10 terms is 0.695, the sum of the ﬁrst 20 terms is 0.708, and the exact sum is 0.720 to three signiﬁcant ﬁgures. Thus A = −1.25 + 5 π (0.720) = −0.104 and i(t) = −0.104e−2t + 1.25 + 5 π ∞ n=1(odd) sin 2nt n(1 + n2) −cos 2nt 1 + n2 amperes In obtaining this solution, we have had to use many of the most general concepts introduced in this and the preceding 17 chapters. Some we did not have to use because of the simple nature of this particular circuit, but their places in the general analysis were indicated. In this sense, we may look upon the solution of this problem as a signiﬁcant achievement in our intro-ductory study of circuit analysis. In spite of this glorious feeling of accom-plishment, however, it must be pointed out that the complete response, as obtained in Example 18.2 in analytical form, is not of much value as it stands; it furnishes no clear picture of the nature of the response. What we really need is a sketch of i(t) as a function of time. This may be obtained by a laborious calculation at a sufﬁcient number of instants of time; a desktop computer or a programmable calculator can be of great assistance here. The sketch may be approximated by the graphical addition of the natural re-sponse, the dc term, and the ﬁrst few harmonics; this is an unrewarding task. When all is said and done, the most informative solution of this problem is probably obtained by making a repeated transient analysis. That is, the form of the response can certainly be calculated in the interval from t = 0 to t = π/2 s; it is an exponential rising toward 2.5 A. After determining the value at the end of this ﬁrst interval, we have an initial condition for the next (π/2)-second interval. The process is repeated until the response assumes a generally periodic nature. The method is eminently suitable to this example, for there is negligible change in the current waveform in the successive periods π/2 < t < 3π/2 and 3π/2 < t < 5π/2. The complete current response is sketched in Fig. 18.9. CHAPTER 18 FOURIER CIRCUIT ANALYSIS 750 ■FIGURE 18.9 The initial portion of the complete response of the circuit of Fig. 18.8a to the forcing function of Fig. 18.8b. 0 2 0.5 1.0 1.5 2.0 2.5 t (s) i(t) (A) 2 3 2 PRACTICE ● 18.6 Use the methods of Chap. 8 to determine the value of the current sketched in Fig. 18.9 at t equal to (a) π/2; (b) π; (c) 3π/2. Ans: 2.392 A; 0.1034 A; 2.396 A. 18.4 • COMPLEX FORM OF THE FOURIER SERIES In obtaining a frequency spectrum, we have seen that the amplitude of each frequency component depends on both an and bn; that is, the sine term and the cosine term both contribute to the amplitude. The exact expression for this amplitude is a2 n + b2 n. It is also possible to obtain the amplitude directly by using a form of Fourier series in which each term is a cosine function with a phase angle; the amplitude and phase angle are functions of f (t) and n. An even more convenient and concise form of the Fourier series is obtained if the sines and cosines are expressed as exponential functions with complex multiplying constants. Let us ﬁrst take the trigonometric form of the Fourier series: f (t) = a0 + ∞ n=1 (an cos nω0t + bn sin nω0t) and then substitute the exponential forms for the sine and cosine. After rearranging, f (t) = a0 + ∞ n=1 e jnω0t an −jbn 2 + e−jnω0t an + jbn 2 The reader may recall the identities sin α = e jα −e−jα j 2 and cos α = e jα + e−jα 2 SECTION 18.4 COMPLEX FORM OF THE FOURIER SERIES 751 We now deﬁne a complex constant cn: cn = 1 2(an −jbn) (n = 1, 2, 3, . . .) The values of an, bn, and cn all depend on n and f (t). Suppose we now replace n with (−n); how do the values of the constants change? The coef-ﬁcients an and bn are deﬁned by Eqs. and , and it is evident that a−n = an but b−n = −bn From Eq. , then, c−n = 1 2(an + jbn) (n = 1, 2, 3, . . .) Thus, cn = c∗ −n We also let c0 = a0 We may therefore express f (t) as f (t) = c0 + ∞ n=1 cne jnω0t + ∞ n=1 c−ne−jnω0t or f (t) = ∞ n=0 cne jnω0t + ∞ n=1 c−ne−jnω0t Finally, instead of summing the second series over the positive integers from 1 to ∞, let us sum over the negative integers from −1 to −∞: f (t) = ∞ n=0 cne jnω0t + −∞ n=−1 cne jnω0t or f (t) = ∞ n=−∞ cne jnω0t By agreement, a summation from −∞to ∞is understood to include a term for n = 0. Equation is the complex form of the Fourier series for f (t); its con-ciseness is one of the most important reasons for its use. In order to obtain the expression by which a particular complex coefﬁcient cn may be evalu-ated, we substitute Eqs. and in Eq. : cn = 1 T T/2 −T/2 f (t) cos nω0t dt −j 1 T T/2 −T/2 f (t) sin nω0t dt CHAPTER 18 FOURIER CIRCUIT ANALYSIS 752 and then we use the exponential equivalents of the sine and cosine and simplify: cn = 1 T T/2 −T/2 f (t)e−jnω0t dt Thus, a single concise equation serves to replace the two equations required for the trigonometric form of the Fourier series. Instead of evaluating two integrals to ﬁnd the Fourier coefﬁcients, only one integration is required; moreover, it is almost always a simpler integration. It should be noted that the integral of Eq. contains the multiplying factor 1/T , whereas the in-tegrals for an and bn both contain the factor 2/T . Collecting the two basic relationships for the exponential form of the Fourier series, we have f (t) = ∞ n=−∞ cne jnω0t cn = 1 T T/2 −T/2 f (t)e−jnω0t dt where ω0 = 2π/T as usual. The amplitude of the component of the exponential Fourier series at ω = nω0, where n = 0, ±1, ±2, . . . , is \|cn\|. We may plot a discrete frequency spectrum giving \|cn\| versus nω0 or nf0, using an abscissa that shows both positive and negative values; and when we do this, the graph is symmetrical about the origin, since Eqs. and show that \|cn\| = \|c−n\|. We note also from Eqs. and that the amplitude of the sinusoidal component at ω = nω0, where n = 1, 2, 3, . . . , is a2 n + b2 n = 2\|cn\| = 2\|c−n\| = \|cn\| + \|c−n\|. For the dc component, a0 = c0. The exponential Fourier coefﬁcients, given by Eq. , are also affected by the presence of certain symmetries in f (t). Thus, appropriate expres-sions for cn are cn = 2 T T/2 0 f (t) cos nω0t dt (even sym.) cn = −j2 T T/2 0 f (t) sin nω0t dt (odd sym.) cn = ⎧ ⎨ ⎩ 2 T T/2 0 f (t)e−jnω0t dt n odd, 1 2-wave sym. 0 n even, 1 2-wave sym. [33a] [33b] cn = ⎧ ⎨ ⎩ 4 T T/4 0 f (t) cos nω0t dt n odd, 1 2-wave and even sym. 0 n even, 1 2-wave and even sym. [34a] [34b] cn = ⎧ ⎨ ⎩ −j4 T T/4 0 f (t) sin nω0t dt n odd, 1 2-wave and odd sym. 0 n even, 1 2-wave and odd sym. [35a] [35b] SECTION 18.4 COMPLEX FORM OF THE FOURIER SERIES 753 EXAMPLE 18.3 Determine cn for the square wave of Fig. 18.10. This square wave possesses both even and half-wave symmetry. If we ignore the symmetry and use our general equation , with T = 2 and ω0 = 2π/2 = π, we have cn = 1 T T/2 −T/2 f (t)e−jnω0t dt = 1 2 −0.5 −1 −e−jnπt dt + 0.5 −0.5 e−jnπt dt − 1 0.5 e−jnπt dt = 1 2 −1 −jnπ (e−jnπt) −0.5 −1 + 1 −jnπ (e−jnπt) 0.5 −0.5 + −1 −jnπ (e−jnπt) 1 0.5 = 1 j2nπ (e jnπ/2 −e jnπ −e−jnπ/2 + e jnπ/2 + e−jnπ −e−jnπ/2) = 2 e jnπ/2 −e−jnπ/2 j2nπ −e jnπ −e−jnπ j2nπ = 1 nπ 2 sin nπ 2 −sin nπ We thus ﬁnd that c0 = 0, c1 = 2/π, c2 = 0, c3 = −2/3π, c4 = 0, c5 = 2/5π, and so forth. These values agree with the trigonometric Fourier series given as the answer we obtained in Practice Problem 18.3 for the same waveform shown in Fig. 18.4b if we remember that an = 2cn when bn = 0. Utilizing the symmetry of the waveform (even and half-wave), there is less work when we apply Eqs. [34a] and [34b], leading to cn = 4 T T/4 0 f (t) cos nω0t dt = 4 2 0.5 0 cos nπt dt = 2 nπ (sin nπt) 0.5 0 = 2 nπ sin nπ 2 (n odd) 0 (n even) These results are the same as those we just obtained when we did not take the symmetry of the waveform into account. ■FIGURE 18.10 A square wave function possessing both even and half-wave symmetry. –1 0 –1 1 1 2 3 4 v (V) t (s) Now let us consider a more difﬁcult, more interesting example. CHAPTER 18 FOURIER CIRCUIT ANALYSIS 754 EXAMPLE 18.4 A certain function f(t) is a train of rectangular pulses of amplitude V0 and duration τ, recurring periodically every T seconds, as shown in Fig. 18.11. Find the exponential Fourier series for f(t). The fundamental frequency is f0 = 1/T . No symmetry is present, and the value of a general complex coefﬁcient is found from Eq. : cn = 1 T T/2 −T/2 f (t)e−jnω0t dt = V0 T t0+τ t0 e−jnω0t dt = V0 −jnω0T (e−jnω0(t0+τ) −e−jnω0t0) = 2V0 nω0T e−jnω0(t0+τ/2) sin 1 2nω0τ = V0τ T sin 1 2nω0τ 1 2nω0τ e−jnω0(t0+τ/2) The magnitude of cn is therefore \|cn\| = V0τ T sin 1 2nω0τ 1 2nω0τ and the angle of cn is ang cn = −nω0 t0 + τ 2 (possibly plus 180◦) Equations and represent our solution to this exponential Fourier series problem. The Sampling Function The trigonometric factor in Eq. occurs frequently in modern communi-cation theory, and it is called the sampling function. The “sampling” refers to the time function of Fig. 18.11 from which the sampling function is derived. The product of this sequence of pulses and any other function f (t) ■FIGURE 18.11 A periodic sequence of rectangular pulses. 0 V0 T t t0 2T –T v(t) SECTION 18.4 COMPLEX FORM OF THE FOURIER SERIES 755 represents samples of f (t) every T seconds if τ is small and V0 = 1. We deﬁne Sa(x) = sin x x Because of the way in which it helps to determine the amplitude of the var-ious frequency components in f (t), it is worth our while to discover the im-portant characteristics of this function. First, we note that Sa(x) is zero whenever x is an integral multiple of π; that is, Sa(nπ) = 0 n = 1, 2, 3, . . . When x is zero, the function is indeterminate, but it is easy to show that its value is unity: Sa(0) = 1 The magnitude of Sa(x) therefore decreases from unity at x = 0 to zero at x = π. As x increases from π to 2π, \|Sa(x)\| increases from zero to a max-imum less than unity, and then decreases to zero once again. As x continues to increase, the successive maxima continually become smaller because the numerator of Sa(x) cannot exceed unity and the denominator is continually increasing. Also, Sa(x) shows even symmetry. Now let us construct the line spectrum. We ﬁrst consider \|cn\|, writing Eq. in terms of the fundamental cyclic frequency f0: \|cn\| = V0τ T sin(nπf0τ) nπf0τ The amplitude of any cn is obtained from Eq. by using the known values τ and T = 1/f0 and selecting the desired value of n, n = 0, ±1, ±2, . . . . Instead of evaluating Eq. at these discrete frequencies, let us sketch the envelope of \|cn\| by considering the frequency nf0 to be a contin-uous variable. That is, f, which is nf0, can actually take on only the discrete values of the harmonic frequencies 0, ± f0, ±2 f0, ±3 f0, and so forth, but we may think of n for the moment as a continuous variable. When f is zero, \|cn\| is evidently V0τ/T , and when f has increased to 1/τ, \|cn\| is zero. The resultant envelope is first sketched as in Fig. 18.12a. The line spectrum is then obtained by simply erecting a vertical line at each harmonic frequency, as shown in the sketch. The amplitudes shown are those of the cn. The particular case sketched applies to the case where τ/T = 1/(1.5π) = 0.212. In this example, it happens that there is no har-monic exactly at that frequency at which the envelope amplitude is zero; another choice of τ or T could produce such an occurrence, however. In Fig. 18.12b, the amplitude of the sinusoidal component is plotted as a function of frequency. Note again that a0 = c0 and a2 n + b2 n = \|cn\| + \|c−n\|. There are several observations and conclusions which we may make about the line spectrum of a periodic sequence of rectangular pulses, as given in Fig. 18.12b. With respect to the envelope of the discrete spectrum, it is evident that the “width” of the envelope depends upon τ, and not upon T. As a matter of fact, the shape of the envelope is not a function of T. It follows that the bandwidth of a ﬁlter which is designed to pass the periodic pulses is a function of the pulse width τ, but not of the pulse period T; an CHAPTER 18 FOURIER CIRCUIT ANALYSIS 756 14f0 10f0 12f0 8f0 6f0 4f0 2f0 0 0 f (Hz) –2f0 –4f0 –6f0 –8f0 –10f0 –12f0 –14f0 V0 T 1 0.5V0 T \|cn\| (a) (b) 14f0 10f0 12f0 8f0 6f0 4f0 2f0 0 0 f (Hz) 2V0 T 1 V0 T √a2 n + b2 n ■FIGURE 18.12 (a) The discrete line spectrum of \|cn\| versus f = nf0, n = 0, ±1, ±2, . . . , corresponding to the pulse train shown in Fig. 18.11. (b) a2 + b2 versus f = nf0, n = 0, 1, 2, . . . , for the same pulse train. SECTION 18.5 DEFINITION OF THE FOURIER TRANSFORM 757 inspection of Fig. 18.12b indicates that the required bandwidth is about 1/τ Hz. If the pulse period T is increased (or the pulse repetition frequency f0 is decreased), the bandwidth 1/τ does not change, but the number of spectral lines between zero frequency and 1/τ Hz increases, albeit discon-tinuously; the amplitude of each line is inversely proportional to T. Finally, a shift in the time origin does not change the line spectrum; that is, \|cn\| is not a function of t0. The relative phases of the frequency components do change with the choice of t0. PRACTICE ● 18.7 Determine the general coefﬁcient cn in the complex Fourier series for the waveform shown in (a) Fig. 18.4a; (b) Fig. 18.4c. Ans: −j2/(nπ) for n odd, 0 for n even; −j[4/(n2π2)] sin nπ/2 for all n. 18.5 • DEFINITION OF THE FOURIER TRANSFORM Now that we are familiar with the basic concepts of the Fourier series representation of periodic functions, let us proceed to deﬁne the Fourier transform by ﬁrst recalling the spectrum of the periodic train of rectangular pulses we obtained in Sec. 18.4. That was a discrete line spectrum, which is the type that we must always obtain for periodic functions of time. The spectrum was discrete in the sense that it was not a smooth or continuous function of frequency; instead, it had nonzero values only at specific frequencies. There are many important forcing functions, however, that are not peri-odic functions of time, such as a single rectangular pulse, a step function, a ramp function, or the somewhat strange type of function called the impulse function deﬁned in Chap. 14. Frequency spectra may be obtained for such nonperiodic functions, but they will be continuous spectra in which some energy, in general, may be found in any nonzero frequency interval, no matter how small. We will develop this concept by beginning with a periodic function and then letting the period become infinite. Our experience with periodic rectangular pulses should indicate that the envelope will decrease in ampli-tude without otherwise changing shape, and that more and more frequency components will be found in any given frequency interval. In the limit, we should expect an envelope of vanishingly small amplitude, ﬁlled with an inﬁnite number of frequency components separated by vanishingly small frequency intervals. The number of frequency components between 0 and 100 Hz, for example, becomes infinite, but the amplitude of each one approaches zero. At ﬁrst thought, a spectrum of zero amplitude is a puzzling concept. We know that the line spectrum of a periodic forcing function shows the amplitude of each frequency component. But what does the zero-amplitude continuous spectrum of a nonperiodic forcing function signify? That question will be answered in the following section; now we proceed to carry out the limiting procedure just suggested. We begin with the exponential form of the Fourier series: f (t) = ∞ n=−∞ cne jnω0t where cn = 1 T T/2 −T/2 f (t)e−jnω0t dt and ω0 = 2π T We now let T →∞ and thus, from Eq. , ω0 must become vanishingly small. We represent this limit by a differential: ω0 →dω Thus 1 T = ω0 2π →dω 2π Finally, the frequency of any “harmonic” nω0 must now correspond to the general frequency variable which describes the continuous spectrum. In other words, n must tend to inﬁnity as ω0 approaches zero, so that the prod-uct is ﬁnite: nω0 →ω When these four limiting operations are applied to Eq. , we ﬁnd that cn must approach zero, as we had previously presumed. If we multiply each side of Eq. by the period T and then undertake the limiting process, a nontrivial result is obtained: cnT → ∞ −∞ f (t)e−jωtdt The right-hand side of this expression is a function of ω (and not of t), and we represent it by F( jω): F( jω) = ∞ −∞ f (t)e−jωtdt Now let us apply the limiting process to Eq. . We begin by multi-plying and dividing the summation by T, f (t) = ∞ n=−∞ cnTe jnω0t 1 T next replacing cnT with the new quantity F( jω), and then making use of expressions and . In the limit, the summation becomes an integral, and f (t) = 1 2π ∞ −∞ F( jω)e jωtdω CHAPTER 18 FOURIER CIRCUIT ANALYSIS 758 SECTION 18.5 DEFINITION OF THE FOURIER TRANSFORM 759 Equations and are collectively called the Fourier transform pair. The function F( jω) is the Fourier transform of f (t), and f (t) is the inverse Fourier transform of F( jω). This transform-pair relationship is very important! We should memorize it, draw arrows pointing to it, and mentally keep it on the conscious level. We emphasize the importance of these relations by repeating them in boxed form: F( jω) = ∞ −∞ e−jωt f(t) dt [46a] f (t) = 1 2π ∞ −∞ e jωtF( jω) dω [46b] The exponential terms in these two equations carry opposite signs for the exponents. To keep them straight, it may help to note that the positive sign is associated with the expression for f (t), as it is with the complex Fourier series, Eq. . It is appropriate to raise one question at this time. For the Fourier trans-form relationships of Eq. , can we obtain the Fourier transform of any arbitrarily chosen f (t)? It turns out that the answer is afﬁrmative for almost any voltage or current that we can actually produce. A sufﬁcient condition for the existence of F( jω) is that ∞ −∞ \| f (t)\| dt < ∞ This condition is not necessary, however, because some functions that do not meet it still have a Fourier transform; the step function is one such ex-ample. Furthermore, we will see later that f (t) does not even need to be nonperiodic in order to have a Fourier transform; the Fourier series repre-sentation for a periodic time function is just a special case of the more gen-eral Fourier transform representation. As we indicated earlier, the Fourier transform-pair relationship is unique. For a given f (t) there is one speciﬁc F( jω); and for a given F( jω) there is one speciﬁc f (t). The reader may have already noticed a few similarities between the Fourier transform and the Laplace trans-form. Key differences between the two include the fact that initial energy storage is not easily incorporated in circuit analysis using Fourier transforms while it is very easily incorporated in the case of Laplace transforms. Also, there are several time functions (e.g., the increasing exponential) for which a Fourier transform does not exist. However, if it is spectral information as opposed to transient response in which we are primarily concerned, the Fourier transform is the ticket. EXAMPLE 18.5 Use the Fourier transform to obtain the continuous spectrum of the single rectangular pulse Fig. 18.13a. The pulse is a truncated version of the sequence considered previously in Fig. 18.11, and is described by f (t) = V0 t0 < t < t0 + τ 0 t < t0 and t > t0 + τ (Continued on next page) CHAPTER 18 FOURIER CIRCUIT ANALYSIS 760 ■FIGURE 18.13 (a) A single rectangular pulse identical to those of the sequence in Fig. 18.11. (b) A plot of \|F( jω)\| corresponding to the pulse, with V 0 = 1, τ = 1, and t0 = 0. The frequency axis has been normalized to the value of f0 = 1/1.5 π corresponding to Fig. 18.12a to allow comparison; note that f0 has no meaning or relevance in the context of F( jω). 15 10 5 0 –5 –10 –15 ff0 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 \|F( j)\| (b) 0 2T T (a) –T t0 V0 v (t) t The Fourier transform of f (t) is found from Eq. [46a]: F( jω) = t0+τ t0 V0e−jωtdt and this may be easily integrated and simpliﬁed: F( jω) = V0τ sin 1 2ωτ 1 2ωτ e−jω(t0+τ/2) The magnitude of F( jω) yields the continuous frequency spectrum, and it is of the form of the sampling function. The value of F(0) is V0τ. The shape of the spectrum is identical with the envelope in Fig. 18.12b. A plot of \|F( jω)\| as a function of ω does not indicate the magnitude of the voltage present at any given frequency. What is it, then? Examination of SECTION 18.6 SOME PROPERTIES OF THE FOURIER TRANSFORM 761 Eq. shows that, if f (t) is a voltage waveform, then F( jω) is dimensionally “volts per unit frequency,” a concept that was introduced in Sec. 15.1. PRACTICE ● 18.8 If f (t) = −10 V, −0.2 < t < −0.1 s, f (t) = 10 V, 0.1 < t < 0.2 s, and f (t) = 0 for all other t, evaluate F( jω) for ω equal to (a) 0; (b) 10π rad/s; (c) −10π rad/s; (d) 15π rad/s; (e) −20π rad/s. 18.9 If F( jω) = −10 V/(rad/s) for −4 < ω < −2 rad/s, +10 V/(rad/s) for 2 < ω < 4 rad/s, and 0 for all other ω, ﬁnd the numerical value of f (t) at t equal to (a) 10−4 s; (b) 10−2 s; (c) π/4 s; (d) π/2 s; (e) π s. Ans: 18.8: 0; j1.273 V/(rad/s); −j1.273 V/(rad/s); −j0.424 V/(rad/s); 0. 18.9: j1.9099 × 10−3 V; j0.1910 V; j4.05 V; −j4.05 V; 0. 18.6 • SOME PROPERTIES OF THE FOURIER TRANSFORM Our object in this section is to establish several of the mathematical proper-ties of the Fourier transform and, even more important, to understand its physical signiﬁcance. We begin by using Euler’s identity to replace e−jωt in Eq. [46a]: F( jω) = ∞ −∞ f (t) cos ωt dt −j ∞ −∞ f (t) sin ωt dt Since f (t), cos ωt, and sin ωt are all real functions of time, both the integrals in Eq. are real functions of ω. Thus, by letting F( jω) = A(ω) + jB(ω) = \|F( jω)\|e jφ(ω) we have A(ω) = ∞ −∞ f (t) cos ωt dt B(ω) = − ∞ −∞ f (t) sin ωt dt \|F( jω)\| = A2(ω) + B2(ω) and φ(ω) = tan−1 B(ω) A(ω) Replacing ω by −ω shows that A(ω) and \|F( jω)\| are both even functions of ω, while B(ω) and φ(ω) are both odd functions of ω. Now, if f (t) is an even function of t, then the integrand of Eq. is an odd function of t, and the symmetrical limits force B(ω) to be zero; thus, if f (t) is even, its Fourier transform F( jω) is a real, even function of ω, and the phase function φ(ω) is zero or π for all ω. However, if f (t) is an odd function of t, then A(ω) = 0 and F( jω) is both odd and a pure imagi-nary function of ω; φ(ω) is ±π/2. In general, however, F( jω) is a complex function of ω. Finally, we note that the replacement of ω by −ω in Eq. forms the conjugate of F( jω). Thus, F(−jω) = A(ω) −jB(ω) = F∗( jω) and we have F( jω)F(−jω) = F( jω)F∗( jω) = A2(ω) + B2(ω) = \|F( jω)\|2 Physical Signiﬁcance of the Fourier Transform With these basic mathematical properties of the Fourier transform in mind, we are now ready to consider its physical signiﬁcance. Let us suppose that f (t) is either the voltage across or the current through a 1 resistor, so that f 2(t) is the instantaneous power delivered to the 1 resistor by f (t). Inte-grating this power over all time, we obtain the total energy delivered by f (t) to the 1 resistor, W1 = ∞ −∞ f 2(t) dt Now let us resort to a little trickery. Thinking of the integrand in Eq. as f (t) times itself, we replace one of those functions with Eq. [46b]: W1 = ∞ −∞ f (t) 1 2π ∞ −∞ e jωtF( jω) dω dt Since f (t) is not a function of the variable of integration ω, we may move it inside the bracketed integral and then interchange the order of integration: W1 = 1 2π ∞ −∞ ∞ −∞ F( jω)e jωt f (t) dt dω Next we shift F( jω) outside the inner integral, causing that integral to be-come F(−jω): W1 = 1 2π ∞ −∞ F( jω)F(−jω) dω = 1 2π ∞ −∞ \|F( jω)\|2 dω Collecting these results, ∞ −∞ f 2(t) dt = 1 2π ∞ −∞ \|F( jω)\|2 dω Equation is a very useful expression known as Parseval’s theorem. This theorem, along with Eq. , tells us that the energy associated with f (t) can be obtained either from an integration over all time in the time domain or by 1/(2π) times an integration over all (radian) frequency in the fre-quency domain. Parseval’s theorem also leads us to a greater understanding and interpre-tation of the meaning of the Fourier transform. Consider a voltage v(t) with Fourier transform Fv( jω) and 1 energy W1: W1 = 1 2π ∞ −∞ \|Fv( jω)\|2 dω = 1 π ∞ 0 \|Fv( jω)\|2 dω CHAPTER 18 FOURIER CIRCUIT ANALYSIS 762 Marc Antoine Parseval-Deschenes was a rather obscure French mathematician, geographer, and occasional poet who published these results in 1805, seventeen years before Fourier published his theorem. SECTION 18.6 SOME PROPERTIES OF THE FOURIER TRANSFORM 763 where the rightmost equality follows from the fact that \|Fv( jω)\|2 is an even function of ω. Then, since ω = 2πf, we can write W1 = ∞ −∞ \|Fv( jω)\|2 df = 2 ∞ 0 \|Fv( jω)\|2 df Figure 18.14 illustrates a typical plot of \|Fv( jω)\|2 as a function of both ω and f. If we divide the frequency scale up into vanishingly small increments df , Eq. shows us that the area of a differential slice under the\|Fv( jω)\|2 curve, having a width df , is \|Fv( jω)\|2 df . This area is shown shaded. The sum of all such areas, as f ranges from minus to plus inﬁnity, is the total 1 energy con-tained in v(t). Thus, \|Fv( jω)\|2 is the (1 ) energy density or energy per unit bandwidth (J/Hz) of v(t), and this energy density is always a real, even, non-negative function of ω. By integrating \|Fv( jω)\|2 over an appropriate fre-quency interval, we are able to calculate that portion of the total energy lying within the chosen interval. Note that the energy density is not a function of the phase of Fv( jω), and thus there are an inﬁnite number of time functions and Fourier transforms that possess identical energy-density functions. ■FIGURE 18.14 The area of the slice \|Fv ( j ω)\|2 is the 1 energy associated with v(t) lying in the bandwidth df. 0 0 ⏐Fv ( j)2⏐ f d df EXAMPLE 18.6 The one-sided [i.e., v(t) = 0 for t < 0] exponential pulse v(t) = 4e−3tu(t) V is applied to the input of an ideal bandpass ﬁlter. If the ﬁlter pass-band is deﬁned by 1 < \| f \| < 2 Hz, calculate the total output energy. We call the ﬁlter output voltage vo(t). The energy in vo(t) will there-fore be equal to the energy of that part of v(t) having frequency com-ponents in the intervals 1 < f < 2 and −2 < f < −1. We determine the Fourier transform of v(t), Fv( jω) = 4 ∞ −∞ e−jωte−3tu(t) dt = 4 ∞ 0 e−(3+ jω)tdt = 4 3 + jω (Continued on next page) CHAPTER 18 FOURIER CIRCUIT ANALYSIS 764 and then we may calculate the total 1 energy in the input signal by either W1 = 1 2π ∞ −∞ \|Fv( jω)\|2 dω = 8 π ∞ −∞ dω 9 + ω2 = 16 π ∞ 0 dω 9 + ω2 = 8 3 J or W1 = ∞ −∞ v2(t) dt = 16 ∞ 0 e−6t dt = 8 3 J The total energy in vo(t), however, is smaller: Wo1 = 1 2π −2π −4π 16 dω 9 + ω2 + 1 2π 4π 2π 16 dω 9 + ω2 = 16 π 4π 2π dω 9 + ω2 = 16 3π tan−1 4π 3 −tan−1 2π 3 = 358 mJ In general, we see that an ideal bandpass ﬁlter enables us to remove energy from prescribed frequency ranges while still retaining the energy contained in other frequency ranges. The Fourier transform helps us to describe the ﬁltering action quantitatively without actually evaluating vo(t), although we will see later that the Fourier transform can also be used to obtain the expression for vo(t) if we wish to do so. PRACTICE ● 18.10 If i(t) = 10e20t[u(t + 0.1) −u(t −0.1)] A, ﬁnd (a) Fi( j0); (b) Fi( j10); (c) Ai(10); (d) Bi(10); (e) φi(10). 18.11 Find the 1 energy associated with the current i(t) = 20e−10tu(t) A in the interval (a) −0.1 < t < 0.1 s; (b) −10 < ω < 10 rad/s; (c) 10 < ω < ∞rad/s. Ans: 18.10: 3.63 A/(rad/s); 3.33/−31.7◦A/(rad/s); 2.83 A/(rad/s); −1.749 A/(rad/s); −31.7◦. 18.11: 17.29 J; 10 J; 5 J. 18.7 • FOURIER TRANSFORM PAIRS FOR SOME SIMPLE TIME FUNCTIONS The Unit-Impulse Function We now seek the Fourier transform of the unit impulse δ(t −t0), a function we introduced in Sec. 14.4. That is, we are interested in the spectral proper-ties or frequency-domain description of this singularity function. If we use the notation F{ } to symbolize “Fourier transform of {},” then F {δ(t −t0)} = ∞ −∞ e−jωtδ(t −t0) dt From our earlier discussion of this type of integral, we have F {δ(t −t0)} = e−jωt0 = cos ωt0 −j sin ωt0 SECTION 18.7 FOURIER TRANSFORM PAIRS FOR SOME SIMPLE TIME FUNCTIONS 765 This complex function of ω leads to the 1 energy-density function, \|F {δ(t −t0)}\|2 = cos2 ωt0 + sin2 ωt0 = 1 This remarkable result says that the (1 ) energy per unit bandwidth is unity at all frequencies, and that the total energy in the unit impulse is inﬁnitely large. No wonder, then, that we must conclude that the unit impulse is “impractical” in the sense that it cannot be generated in the laboratory. More-over, even if one were available to us, it must appear distorted after being subjected to the ﬁnite bandwidth of any practical laboratory instrument. Since there is a unique one-to-one correspondence between a time func-tion and its Fourier transform, we can say that the inverse Fourier transform of e−jωt0 is δ(t −t0). Utilizing the symbol F −1{ } for the inverse transform, we have F −1{e−jωt0} = δ(t −t0) Thus, we now know that 1 2π ∞ −∞ e jωte−jωt0 dω = δ(t −t0) even though we would fail in an attempt at the direct evaluation of this improper integral. Symbolically, we may write δ(t −t0) ⇔e−jωt0 where ⇔indicates that the two functions constitute a Fourier transform pair. Continuing with our consideration of the unit-impulse function, let us consider a Fourier transform in that form, F( jω) = δ(ω −ω0) which is a unit impulse in the frequency domain located at ω = ω0. Then f (t) must be f (t) = F −1{F( jω)} = 1 2π ∞ −∞ e jωtδ(ω −ω0) dω = 1 2π e jω0t where we have used the sifting property of the unit impulse. Thus we may now write 1 2π e jω0t ⇔δ(ω −ω0) or e jω0t ⇔2πδ(ω −ω0) Also, by a simple sign change we obtain e−jω0t ⇔2πδ(ω + ω0) Clearly, the time function is complex in both expressions and , and does not exist in the real world of the laboratory. However, we know that cos ω0t = 1 2e jω0t + 1 2e−jω0t and it is easily seen from the deﬁnition of the Fourier transform that F { f1(t)} + F { f2(t)} = F { f1(t) + f2(t)} Therefore, F {cos ω0t} = F 1 2e jω0t + F 1 2e−jω0t = πδ(ω −ω0) + πδ(ω + ω0) which indicates that the frequency-domain description of cos ω0t shows a pair of impulses, located at ω = ±ω0. This should not be a great surprise, for in our ﬁrst discussion of complex frequency in Chap. 14, we noted that a sinusoidal function of time was always represented by a pair of imaginary frequencies located at s = ± jω0. We have, therefore, cos ω0t ⇔π[δ(ω + ω0) + δ(ω −ω0)] The Constant Forcing Function To ﬁnd the Fourier transform of a constant function of time, f (t) = K, our ﬁrst inclination might be to substitute this constant in the deﬁning equation for the Fourier transform and evaluate the resulting integral. If we did, we would find ourselves with an indeterminate expression on our hands. Fortunately, however, we have already solved this problem, for from expression , e−jω0t ⇔2πδ(ω + ω0) We see that if we simply let ω0 = 0, then the resulting transform pair is 1 ⇔2πδ(ω) from which it follows that K ⇔2πKδ(ω) and our problem is solved. The frequency spectrum of a constant function of time consists only of a component at ω = 0, which we knew all along. The Signum Function As another example, let us obtain the Fourier transform of a singularity function known as the signum function, sgn(t), deﬁned by sgn(t) = −1 t < 0 1 t > 0 or sgn(t) = u(t) −u(−t) Again, if we should try to substitute this time function in the deﬁning equa-tion for the Fourier transform, we would face an indeterminate expression upon substitution of the limits of integration. This same problem will arise every time we attempt to obtain the Fourier transform of a time function that does not approach zero as \|t\| approaches inﬁnity. Fortunately, we can avoid this situation by using the Laplace transform, as it contains a built-in con-vergence factor that cures many of the inconvenient ills associated with the evaluation of certain Fourier transforms. CHAPTER 18 FOURIER CIRCUIT ANALYSIS 766 SECTION 18.7 FOURIER TRANSFORM PAIRS FOR SOME SIMPLE TIME FUNCTIONS 767 Along those lines, the signum function under consideration can be writ-ten as sgn(t) = lim a→0[e−atu(t) −eatu(−t)] Notice that the expression within the brackets does approach zero as \|t\| gets very large. Using the deﬁnition of the Fourier transform, we obtain F {sgn(t)} = lim a→0 ∞ 0 e−jωte−atdt − 0 −∞ e−jωteatdt = lim a→0 −j2ω ω2 + a2 = 2 jω The real component is zero, since sgn(t) is an odd function of t. Thus, sgn(t) ⇔2 jω The Unit-Step Function As a ﬁnal example in this section, let us look at the familiar unit-step func-tion, u(t). Making use of our work on the signum function in the preceding paragraphs, we represent the unit step by u(t) = 1 2 + 1 2sgn(t) and obtain the Fourier transform pair u(t) ⇔ πδ(ω) + 1 jω Table 18.2 presents the conclusions drawn from the examples discussed in this section, along with a few others that have not been detailed here. EXAMPLE 18.7 Use Table 18.2 to ﬁnd the Fourier transform of the time function 3e−t cos 4t u(t). From the next to the last entry in the table, we have e−αt cos ωdt u(t) ⇔ α + jω (α + jω)2 + ω2 d We therefore identify α as 1 and ωd as 4, and have F( jω) = 3 1 + jω (1 + jω)2 + 16 PRACTICE ● 18.12 Evaluate the Fourier transform at ω = 12 for the time function (a) 4u(t) −10δ(t); (b) 5e−8tu(t); (c) 4 cos 8tu(t); (d ) −4 sgn(t). 18.13 Find f (t) at t = 2 if F( jω) is equal to (a) 5e−j3ω −j(4/ω); (b) 8[δ(ω −3) + δ(ω + 3)]; (c) (8/ω) sin 5ω. Ans: 18.12:10.01/−178.1◦;0.347/−56.3◦;−j0.6; j0.667. 18.13:2.00;2.45;4.00. CHAPTER 18 FOURIER CIRCUIT ANALYSIS 768 TABLE ●18.2 A Summary of Some Fourier Transform Pairs f(t) f(t) F {f(t)} = F(jω) \|F(jω)\| 2 T – 2 T d –d 1 () (2) () 0 –0 () (2) 0 1 t 1 T 2 – T 2 t 1 t 1 t 1 t –1 1 t 1 t –1 Complex t (1) t0 t δ(t −t0) e−jωt0 e jω0t 2πδ(ω −ω0) cos ω0t π[δ(ω + ω0) + δ(ω −ω0)] 1 2πδ(ω) sgn(t) 2 jω u(t) πδ(ω) + 1 jω e−αtu(t) 1 α + jω [e−αt cos ωdt]u(t) α + jω (α + jω)2 + ω2 d u(t + 1 2 T) −u(t −1 2 T) T sin ωT 2 ωT 2 SECTION 18.8 THE FOURIER TRANSFORM OF A GENERAL PERIODIC TIME FUNCTION 769 18.8 • THE FOURIER TRANSFORM OF A GENERAL PERIODIC TIME FUNCTION In Sec. 18.5 we remarked that we would be able to show that periodic time functions, as well as nonperiodic functions, possess Fourier transforms. Let us now establish this fact on a rigorous basis. Consider a periodic time func-tion f (t) with period T and Fourier series expansion, as outlined by Eqs. , , and , repeated here for convenience: f (t) = ∞ n=−∞ cne jnω0t cn = 1 T T/2 −T/2 f (t)e−jnω0tdt and ω0 = 2π T Bearing in mind that the Fourier transform of a sum is just the sum of the transforms of the terms in the sum, and that cn is not a function of time, we can write F { f (t)} = F ∞ n=−∞ cne jnω0t ! = ∞ n=−∞ cnF {e jnω0t} After obtaining the transform of e jnω0t from expression , we have f (t) ⇔2π ∞ n=−∞ cnδ(ω −nω0) This shows that f (t) has a discrete spectrum consisting of impulses located at points on the ω axis given by ω = nω0, n = . . . , −2, −1, 0, 1, . . . . The strength of each impulse is 2π times the value of the corresponding Fourier coefﬁcient appearing in the complex form of the Fourier series expansion for f (t). As a check on our work, let us see whether the inverse Fourier transform of the right side of expression is once again f (t). This inverse trans-form can be written as F −1{F( jω)} = 1 2π ∞ −∞ e jωt " 2π ∞ n=−∞ cnδ(ω −nω0) # dω ? = f (t) Since the exponential term does not contain the index of summation n, we can interchange the order of the integration and summation operations: F −1{F( jω)} = ∞ n=−∞ ∞ −∞ cne jωtδ(ω −nω0) dω ? = f (t) Because it is not a function of the variable of integration, cn can be treated as a constant. Then, using the sifting property of the impulse, we obtain F −1{F( jω)} = ∞ n=−∞ cne jnω0t ? = f (t) CHAPTER 18 FOURIER CIRCUIT ANALYSIS 770 which is exactly the same as Eq. , the complex Fourier series expan-sion for f (t). The question marks in the preceding equations can now be removed, and the existence of the Fourier transform for a periodic time function is established. This should come as no great surprise, however. In the last section we evaluated the Fourier transform of a cosine func-tion, which is certainly periodic, although we made no direct reference to its periodicity. However, we did use a backhanded approach in getting the transform. But now we have a mathematical tool by which the transform can be obtained more directly. To demonstrate this procedure, consider f (t) = cos ω0t once more. First we evaluate the Fourier coefﬁcients cn: cn = 1 T T/2 −T/2 cos ω0t e−jnω0tdt = 1 2 n = ±1 0 otherwise Then F { f (t)} = 2π ∞ n=−∞ cnδ(ω −nω0) This expression has values that are nonzero only when n = ±1, and it fol-lows, therefore, that the entire summation reduces to F {cos ω0t} = π[δ(ω −ω0) + δ(ω + ω0)] which is precisely the expression that we obtained before. What a relief! PRACTICE ● 18.14 Find (a) F{5 sin2 3t}; (b) F{A sin ω0t}; (c) F{6 cos(8t + 0.1π)}. Ans: 2.5π[2δ(ω) −δ(ω + 6) −δ(ω −6)]; jπ A[δ(ω + ω0) −δ(ω −ω0)]; [18.85/18◦] δ(ω −8) + [18.85/−18◦] δ(ω + 8). 18.9• THE SYSTEM FUNCTION AND RESPONSE IN THE FREQUENCY DOMAIN In Sec. 15.5, the problem of determining the output of a physical system in terms of the input and the impulse response was solved by using the convolution integral and initially working in the time domain. The input, the output, and the impulse response are all time functions. Subsequently, we found that it was often more convenient to perform such operations in the frequency domain, as the Laplace transform of the convolution of two functions is simply the product of each function in the frequency domain. Along the same lines, we ﬁnd the same is true when working with Fourier transforms. To do this we examine the Fourier transform of the system output. As-suming arbitrarily that the input and output are voltages, we apply the basic 771 SECTION 18.9 THE SYSTEM FUNCTION AND RESPONSE IN THE FREQUENCY DOMAIN deﬁnition of the Fourier transform and express the output by the convolution integral: F {v0(t)} = F0( jω) = ∞ −∞ e−jωt ∞ −∞ vi(t −z)h(z) dz dt where we again assume no initial energy storage. At first glance this expression may seem rather formidable, but it can be reduced to a result that is surprisingly simple. We may move the exponential term inside the inner integral because it does not contain the variable of integration z. Next we reverse the order of integration, obtaining F0( jω) = ∞ −∞ ∞ −∞ e−jωtvi(t −z)h(z) dt dz Since it is not a function of t, we can extract h(z) from the inner integral and simplify the integration with respect to t by a change of variable, t −z = x: F0( jω) = ∞ −∞ h(z) ∞ −∞ e−jω(x+z)vi(x) dx dz = ∞ −∞ e−jωzh(z) ∞ −∞ e−jωxvi(x) dx dz But now the sun is starting to break through, for the inner integral is merely the Fourier transform of vi(t). Furthermore, it contains no z terms and can be treated as a constant in any integration involving z. Thus, we can move this transform, Fi( jω), completely outside all the integral signs: F0( jω) = Fi( jω) ∞ −∞ e−jωzh(z) dz Finally, the remaining integral exhibits our old friend once more, another Fourier transform! This one is the Fourier transform of the impulse re-sponse, which we will designate by the notation H( jω). Therefore, all our work has boiled down to the simple result: F0( jω) = Fi( jω)H( jω) = Fi( jω)F {h(t)} This is another important result: it deﬁnes the system function H( jω) as the ratio of the Fourier transform of the response function to the Fourier trans-form of the forcing function. Moreover, the system function and the impulse response constitute a Fourier transform pair: h(t) ⇔H( jω) The development in the preceding paragraph also serves to prove the general statement that the Fourier transform of the convolution of two time functions is the product of their Fourier transforms, F { f (t) ∗g(t)} = Ff ( jω)Fg( jω) CHAPTER 18 FOURIER CIRCUIT ANALYSIS The foregoing comments might make us wonder once again why we would ever choose to work in the time domain at all, but we must always remember that we seldom get something for nothing. A poet once said, “Our sincerest laughter/with some pain is fraught.”2 The pain herein is the occasional dif-ﬁculty in obtaining the inverse Fourier transform of a response function, for reasons of mathematical complexity. On the other hand, a simple desktop computer can convolve two time functions with magniﬁcent celerity. For that matter, it can also obtain an FFT (fast Fourier transform) quite rapidly. Consequently there is no clear-cut advantage between working in the time domain and in the frequency domain. A decision must be made each time a new problem arises; it should be based on the information available and on the computational facilities at hand. Consider a forcing function of the form vi(t) = u(t) −u(t −1) and a unit-impulse response deﬁned by h(t) = 2e−tu(t) We ﬁrst obtain the corresponding Fourier transforms. The forcing function is the difference between two unit-step functions. These two functions are identical, except that one is initiated 1 s after the other. We will evaluate the response due to u(t); the response due to u(t −1) is the same, but delayed in time by 1 s. The difference between these two partial responses will be the total response due to vi(t). The Fourier transform of u(t) was obtained in Sec. 18.7: F {u(t)} = πδ(ω) + 1 jω The system function is obtained by taking the Fourier transform of h(t), listed in Table 18.2, F {h(t)} = H( jω) = F {2e−tu(t)} = 2 1 + jω The inverse transform of the product of these two functions yields that component of vo(t) caused by u(t), vo1(t) = F −1 2πδ(ω) 1 + jω + 2 jω(1 + jω) $ Using the sifting property of the unit impulse, the inverse transform of the ﬁrst term is just a constant equal to unity. Thus, vo1(t) = 1 + F −1 2 jω(1 + jω) $ The second term contains a product of terms in the denominator, each of the form (α + jω), and its inverse transform is found most easily by making use of the partial-fraction expansion that we developed in Sec. 14.5. Let us 772 (2) P. B. Shelley, “To a Skylark,” 1821. To recapitulate, if we know the Fourier transforms of the forcing function and the impulse response, then the Fourier transform of the response function can be ob-tained as their product. The result is a description of the response function in the frequency domain; the time-domain description of the response function is obtained by simply taking the inverse Fourier transform. Thus we see that the process of convolution in the time domain is equivalent to the relatively simple operation of multi-plication in the frequency domain. SECTION 18.9 THE SYSTEM FUNCTION AND RESPONSE IN THE FREQUENCY DOMAIN 773 select a technique for obtaining a partial-fraction expansion that has one big advantage—it always works, although faster methods are usually available for most situations. We assign an unknown quantity in the numerator of each fraction, here two in number, 2 jω(1 + jω) = A jω + B 1 + jω and then substitute a corresponding number of simple values for jω. Here we let jω = 1: 1 = A + B 2 and then let jω = −2: 1 = −A 2 −B This leads to A = 2 and B = −2. Thus, F −1 2 jω(1 + jω) $ = F −1 2 jω − 2 1 + jω $ = sgn(t) −2e−tu(t) so that vo1(t) = 1 + sgn(t) −2e−tu(t) = 2u(t) −2e−tu(t) = 2(1 −e−t)u(t) It follows that vo2(t), the component of vo(t) produced by u(t −1), is vo2(t) = 2(1 −e−(t−1))u(t −1) Therefore, vo(t) = vo1(t) −vo2(t) = 2(1 −e−t)u(t) −2(1 −e−t+1)u(t −1) The discontinuities at t = 0 and t = 1 dictate a separation into three time intervals: vo(t) = ⎧ ⎨ ⎩ 0 t < 0 2(1 −e−t) 0 < t < 1 2(e −1)e−t t > 1 PRACTICE ● 18.15 The impulse response of a certain linear network is h(t) = 6e−20tu(t). The input signal is 3e−6tu(t) V. Find (a) H( jω); (b) Vi( jω); (c) Vo( jω); (d ) vo(0.1); (e) vo(0.3); ( f ) vo,max. Ans: 6/(20 + jω); 3/(6 + jω); 18/[(20 + jω)(6 + jω)]; 0.532 V; 0.209 V; 0.5372. CHAPTER 18 FOURIER CIRCUIT ANALYSIS 774 COMPUTER-AIDED ANALYSIS The material presented in this chapter forms the foundation for many advanced ﬁelds of study, including signal processing, communica-tions, and controls. We are only able to introduce some of the more fundamental concepts within the context of an introductory circuits text, but even at this point some of the power of Fourier-based analy-sis can be brought to bear. As a ﬁrst example, consider the op amp circuit of Fig. 18.15, constructed in PSpice using a μA741 opera-tional ampliﬁer. ■FIGURE 18.16 Simulated output voltage of the ampliﬁer circuit shown in Fig. 18.15. The circuit has a voltage gain of −10, and so we would expect a sinusoidal output of 10 V amplitude. This is indeed what we obtain from a transient analysis of the circuit, as shown in Fig. 18.16. ■FIGURE 18.15 An inverting ampliﬁer circuit with a voltage gain of −10, driven by a sinusoidal input operating at 100 Hz. SECTION 18.9 THE SYSTEM FUNCTION AND RESPONSE IN THE FREQUENCY DOMAIN 775 PSpice allows us to determine the frequency spectrum of the output voltage through what is known as a fast Fourier transform (FFT), a discrete-time approximation to the exact Fourier transform of the sig-nal. From within Probe, we select Fourier under the Trace menu; the result is the plot shown in Fig. 18.17. As expected, the line spectrum for the output voltage of this ampliﬁer circuit consists of a single feature at a frequency of 100 Hz. ■FIGURE 18.17 Discrete approximation to the Fourier transform of Fig. 18.16. As the input voltage magnitude is increased, the output of the ampli-ﬁer approaches the saturation condition determined by the positive and negative dc supply voltages (±15 V in this example). This behavior is evident in the simulation result of Fig. 18.18, which corresponds to an input voltage magnitude of 1.8 V. A key feature of interest is that the output voltage waveform is no longer a pure sinusoid. As a result, we expect nonzero values at harmonic frequencies to appear in the ■FIGURE 18.18 Transient analysis simulation results for the ampliﬁer circuit when the input voltage magnitude is increased to 1.8 V. Saturation effects manifest them-selves in the plot as clipped waveform extrema. (Continued on next page) CHAPTER 18 FOURIER CIRCUIT ANALYSIS 776 ■FIGURE 18.19 Frequency spectrum of the waveform depicted in Fig. 18.18, showing the pres-ence of several harmonic components in addition to the fundamental frequency. The ﬁnite width of the features is an artifact of the numerical discretization (a set of discrete time values was used). (a) (b) ■FIGURE 18.20 (a) Severe effects of ampliﬁer saturation are observed in the simulated response to a 15 V sinusoidal input. (b) An FFT of the waveform shows a signiﬁcant increase in the fraction of energy present in harmonics as opposed to the fundamental frequency of 100 Hz. SECTION 18.10 THE PHYSICAL SIGNIFICANCE OF THE SYSTEM FUNCTION 777 18.10 • THE PHYSICAL SIGNIFICANCE OF THE SYSTEM FUNCTION In this section we will try to connect several aspects of the Fourier trans-form with work we completed in earlier chapters. Given a general linear two-port network N without any initial energy storage, we assume sinusoidal forcing and response functions, arbitrarily taken to be voltages, as shown in Fig. 18.21. We let the input voltage be simply A cos(ωxt + θ), and the output can be described in general terms as vo(t) = B cos(ωxt + φ), where the amplitude B and phase angle φ are functions of ωx. In phasor form, we can write the forcing and response func-tions as Vi = Ae jθ and Vo = Be jφ. The ratio of the phasor response to the phasor forcing function is a complex number that is a function of ωx: Vo Vi = G(ωx) = B Ae j(φ−θ) where B/A is the amplitude of G and φ −θ is its phase angle. This transfer function G(ωx) could be obtained in the laboratory by varying ωx over a large range of values and measuring the amplitude B/A and phase φ −θ for each value of ωx. If we then plotted each of these parameters as a func-tion of frequency, the resultant pair of curves would completely describe the transfer function. frequency spectrum of the function, as is the case in Fig. 18.19. The ef-fect of reaching saturation in the ampliﬁer circuit is a distortion of the signal; if connected to a speaker, we do not hear a “clean” 100 Hz waveform. Instead, we now hear a superposition of waveforms which include not only the 100 Hz fundamental frequency, but signiﬁcant har-monic components at 300 and 500 Hz as well. Further distortion of the waveform would increase the amount of energy in harmonic frequen-cies, so that contributions from higher-frequency harmonics would become more signiﬁcant. This is evident in the simulation results of Fig. 18.20a and b, which show the output voltage in the time and frequency domains, respectively. vo(t) = B cos (x t + ) vi (t) = A cos (x t + ) N + – + – ■FIGURE 18.21 Sinusoidal analysis can be used to determine the transfer function H( jωx ) = (B/A)e j (φ−θ) , where B and φ are functions of ωx . Now let us hold these comments in the backs of our minds for a moment as we consider a slightly different aspect of the same analysis problem. For the circuit with sinusoidal input and output shown in Fig. 18.21, what is the system function H( jω)? To answer this question, we begin with the deﬁnition of H( jω) as the ratio of the Fourier transforms of the output and the input. Both of these time functions involve the functional form cos(ωxt + β), whose Fourier transform we have not evaluated as yet, although we can handle cos ωxt. The transform we need is F {cos(ωxt + β)} = ∞ −∞ e−jωt cos(ωxt + β) dt If we make the substitution ωxt + β = ωxτ , then F {cos(ωxt + β)} = ∞ −∞ e−jωτ+ jωβ/ωx cos ωxτ dτ = e jωβ/ωxF {cos ωxt} = πe jωβ/ωx[δ(ω −ωx) + δ(ω + ωx)] This is a new Fourier transform pair, cos(ωxt + β) ⇔πe jωβ/ωx[δ(ω −ωx) + δ(ω + ωx)] which we can now use to evaluate the desired system function, H( jω) = F {B cos(ωxt + φ)} F {A cos(ωxt + θ)} = πBe jωφ/ωx[δ(ω −ωx) + δ(ω + ωx)] πAe jωθ/ωx[δ(ω −ωx) + δ(ω + ωx)] = B Ae jω(φ−θ)/ωx Now we recall the expression for G(ωx), G(ωx) = B Ae j(φ−θ) where B and φ were evaluated at ω = ωx, and we see that evaluating H( jω) at ω = ωx gives H(ωx) = G(ωx) = B Ae j(φ−θ) Since there is nothing special about the x subscript, we conclude that the system function and the transfer function are identical: H( jω) = G(ω) The fact that one argument is ω while the other is indicated by jω is imma-terial and arbitrary; the j merely makes possible a more direct comparison between the Fourier and Laplace transforms. Equation represents a direct connection between Fourier transform techniques and sinusoidal steady-state analysis. Our previous work on steady-state sinusoidal analysis using phasors was but a special case of the more general techniques of Fourier transform analysis. It was ‘‘special’’ in the sense that the inputs and outputs were sinusoids, whereas the use of Fourier transforms and system functions enables us to handle nonsinusoidal forcing functions and responses. Thus, to ﬁnd the system function H( jω) for a network, all we need to do is to determine the corresponding sinusoidal transfer function as a function of ω (or jω). CHAPTER 18 FOURIER CIRCUIT ANALYSIS 778 SECTION 18.10 THE PHYSICAL SIGNIFICANCE OF THE SYSTEM FUNCTION 779 EXAMPLE 18.8 Find the voltage across the inductor of the circuit shown in Fig. 18.22a when the input voltage is a simple exponentially decay-ing pulse, as indicated. We need the system function; but it is not necessary to apply an impulse, ﬁnd the impulse response, and then determine its inverse transform. Instead we use Eq. to obtain the system function H( jω) by assuming that the input and output voltages are both sinusoids described by their corresponding phasors, as shown in Fig. 18.22b. Using voltage division, we have H( jω) = Vo Vi = j2ω 4 + j2ω The transform of the forcing function is F {vi(t)} = 5 3 + jω and thus the transform of vo(t) is given as F {vo(t)} = H( jω)F{vi(t)} = j2ω 4 + j2ω 5 3 + jω = 15 3 + jω − 10 2 + jω where the partial fractions appearing in the last step help to determine the inverse Fourier transform vo(t) = F −1 15 3 + jω − 10 2 + jω $ = 15e−3tu(t) −10e−2tu(t) = 5(3e−3t −2e−2t)u(t) Our problem is completed without fuss, convolution, or differential equations. PRACTICE ● 18.16 Use Fourier transform techniques on the circuit of Fig. 18.23 to ﬁnd i1(t) at t = 1.5 ms if is equals (a) δ(t) A; (b) u(t) A; (c) cos 500t A. + – vo(t) + – (a) 4 2 H vi (t) = 5e–3tu(t) + – (b) 4 j2 Vi Vo + – ■FIGURE 18.22 (a) The response vo(t) caused by vi (t) is desired. (b) The system function H( jω) may be determined by sinusoidal steady-state analysis: H( jω) = Vo/Vi . 4 6 20 mH is i1 ■FIGURE 18.23 Ans: −141.7 A; 0.683 A; 0.308 A. Although a great deal of progress has been made toward developing a complete understanding of the function of muscle, there remain many open questions. A great deal of research in this ﬁeld has been carried out using verte-brate skeletal muscle, in particular the sartorius or leg muscle of the frog (Fig. 18.24). shows an electron micrograph of frog sartorius muscle tissue, sectioned in such a fashion as to highlight the reg-ular arrangement of myosin, a ﬁlamentary type of con-tractile protein. Of interest to structural biologists are the periodicity and disorder of these proteins over a large area of muscle tissue. In order to develop a model for these characteristics, a numerical approach is preferable, where the analysis of such images can be automated. As can be seen in the ﬁgure, however, the image produced by the electron microscope can be contaminated by a high level of background noise, making automated iden-tiﬁcation of the myosin ﬁlaments prone to error. Introduced with the intent of ultimately assisting us in the analysis of time-varying linear circuits, the Fourier-based techniques of this chapter are in fact very powerful general methods which ﬁnd application in many other situations. Among these, the ﬁeld of image processing makes frequent use of Fourier techniques, especially through the fast Fourier transform and related numerical methods. The image of Fig. 18.25 can be described by a spatial function f (x, y) where f (x, y) = 0 corresponds to white, f (x, y) = 1 corresponds to red, and (x, y) denotes a pixel location in the image. Deﬁning a ﬁlter function h(x, y) that has the appearance of Fig. 18.26a, the convolution operation g(x, y) = f (x, y) ∗h(x, y) results in the image of Fig. 18.26b in which the myosin filaments (viewed on end) are more clearly identiﬁable. In practice, this image processing is performed in the frequency domain, where the FFT of both f and h are cal-culated, and the resulting matrices multiplied together. PRACTICAL APPLICATION Image Processing ■FIGURE 18.24 Close-up of a frog against an orange background. © IT Stock/PunchStock/RF. ■FIGURE 18.25 Electron micrograph of a region of frog sartorius muscle tissue. False color has been employed for clarity. Courtesy Professor John M. Squire, Imperial College, London. Of the many analytical techniques scientists use, one of the most common is electron microscopy. Figure 18.25 Epilogue Returning again to Eq. , the identity between the system function H( jω) and the sinusoidal steady-state transfer function G(ω), we may now consider the system function as the ratio of the output phasor to the input phasor. Sup-pose that we hold the input-phasor amplitude at unity and the phase angle at zero. Then the output phasor is H( jω). Under these conditions, if we record the output amplitude and phase as functions of ω, for all ω, we have recorded the system function H( jω) as a function of ω, for all ω. We thus have exam-ined the system response under the condition that an inﬁnite number of sinu-soids, all with unity amplitude and zero phase, were successively applied at the input. Now suppose that our input is a single unit impulse, and look at the impulse response h(t). Is the information we examine really any different from what we just obtained? The Fourier transform of the unit impulse is a constant equal to unity, indicating that all frequency components are present, all with the same magnitude, and all with zero phase. Our system response is the sum of the responses to all these components. The result might be viewed at the output on a cathode-ray oscilloscope. It is evident that the system func-tion and the impulse-response function contain equivalent information re-garding the response of the system. We therefore have two different methods of describing the response of a system to a general forcing function; one is a time-domain description, and the other a frequency-domain description. Working in the time domain, we An inverse FFT operation then produces the ﬁltered im-age of Fig. 18.26b. Why does this convolution equate to a ﬁltering operation? The myosin ﬁlament arrangement possesses hexagonal symmetry, as does the ﬁlter func-tion h(x, y)—in a sense, both the myosin filament arrangement and the filter function possess the same spatial frequencies. The convolution of f with h results in a reinforcement of the hexagonal pattern within the orig-inal image, and the removal of noise pixels (which do not possess hexagonal symmetry). This can be under-stood qualitatively if we model a horizontal row of Fig. 18.25 as a sinusoidal function f (x) = cos ω0t, which has the Fourier transform shown in Fig. 18.27a—a matched pair of impulse functions separated by 2ω0. If we convolve this function with a filter function h(x) = cos ω1t, the Fourier transform of which is de-picted in Fig. 18.27b, we get zero if ω1 ̸= ω0; the fre-quencies (periodicities) of the two functions do not match. If, instead, we choose a ﬁlter function with the same frequency as f (x), the convolution has a nonzero value at ω = ±ω0. (a) 0 0 F (b) 1 1 F ■FIGURE 18.27 (a) Fourier transform of f (x ) = cos ω0t . (b) Fourier transform of h(x ) = cos ω1t . (a) ■FIGURE 18.26 (a) Spatial ﬁlter having hexagonal symmetry. (b) Image after convolution and inverse discrete Fourier transform are performed, showing a reduction in background noise. Courtesy Professor John M. Squire, Imperial College, London. (b) CHAPTER 18 FOURIER CIRCUIT ANALYSIS 782 convolve the forcing function with the impulse response of the system to obtain the response function. As we saw when we ﬁrst considered convolu-tion, this procedure may be interpreted by thinking of the input as a contin-uum of impulses of different strengths and times of application; the output which results is a continuum of impulse responses. In the frequency domain, however, we determine the response by multiplying the Fourier transform of the forcing function by the system function. In this case we interpret the transform of the forcing function as a frequency spectrum, or a continuum of sinusoids. Multiplying this by the system function, we obtain the response function, also as a continuum of sinusoids. SUMMARY AND REVIEW Whether we choose to think of the output as a continuum of impulse responses or as a continuum of sinusoidal responses, the linearity of the net-work and the superposition principle enable us to determine the total output as a time function by summing over all frequencies (the inverse Fourier transform), or as a frequency function by summing over all time (the Fourier transform). Unfortunately, both of these techniques have some difﬁculties or limi-tations associated with their use. In using convolution, the integral itself can often be rather difﬁcult to evaluate when complicated forcing func-tions or impulse response functions are present. Furthermore, from the experimental point of view, we cannot really measure the impulse re-sponse of a system because we cannot actually generate an impulse. Even if we approximated the impulse by a narrow high-amplitude pulse, we would probably drive our system into saturation and out of its linear oper-ating range. With regard to the frequency domain, we encounter one absolute limita-tion in that we may easily hypothesize forcing functions that we would like to apply theoretically that do not possess Fourier transforms. Moreover, if we wish to ﬁnd the time-domain description of the response function, we must evaluate an inverse Fourier transform, and some of these inversions can be extremely difﬁcult. Finally, neither of these techniques offers a very convenient method of handling initial conditions. For this, the Laplace transform is clearly superior. The greatest beneﬁts derived from the use of the Fourier transform arise through the abundance of useful information it provides about the spectral properties of a signal, particularly the energy or power per unit bandwidth. Some of this information is also easily obtained through the Laplace trans-form; we must leave a detailed discussion of the relative merits of each to more advanced signals and systems courses. So, why has this all been withheld until now? The best answer is proba-bly that these powerful techniques can overcomplicate the solution of simple problems and tend to obscure the physical interpretation of the per-formance of the simpler networks. For example, if we are interested only in the forced response, then there is little point in using the Laplace transform EXERCISES 783 ❑The harmonic frequencies of a sinusoid having the fundamental frequency ω0 are nω0, where n is an integer. (Examples 18.1, 18.2) ❑The Fourier theorem states that provided a function f (t) satisﬁes certain key properties, it may be represented by the inﬁnite series a0 + ∞ n=1(an cos nω0t + bn sin nω0t), where a0 = (1/T) T 0 f (t) dt, an = (2/T) T 0 f (t) cos nω0t dt, and bn = (2/T) T 0 f (t) sin nω0t dt. (Example 18.1) ❑A function f (t) possesses even symmetry if f (t) = f (−t). ❑A function f (t) possesses odd symmetry if f (t) = −f (−t). ❑A function f (t) possesses half-wave symmetry if f (t) = −f (t −1 2T). ❑The Fourier series of an even function is composed of only a constant and cosine functions. ❑The Fourier series of an odd function is composed of only sine functions. ❑The Fourier series of any function possessing half-wave symmetry contains only odd harmonics. ❑The Fourier series of a function may also be expressed in complex or exponential form, where f (t) = ∞ n=−∞cne jnω0t and cn = (1/T) T/2 −T/2 f (t)e−jnω0t dt. (Examples 18.3, 18.4) ❑The Fourier transform allows us to represent time-varying functions in the frequency domain, in a manner similar to that of the Laplace transform. The deﬁning equations are F( jω) = ∞ −∞e−jωt f (t) dt and f (t) = (1/2π) ∞ −∞e jωtF( jω) dω. (Examples 18.5, 18.6, 18.7) ❑Fourier transform analysis can be implemented to analyze circuits containing resistors, inductors, and/or capacitors in a manner similar to what is done using Laplace transforms. (Example 18.8) READING FURTHER A very readable treatment of Fourier analysis can be found in: A. Pinkus and S. Zafrany, Fourier Series and Integral Transforms. Cambridge: Cambridge University Press, 1997. Finally, for those interested in learning more about muscle research, including electron microscopy of tissue, an excellent treatment can be found in: J. Squire, The Structural Basis of Muscular Contraction. New York: Plenum Press, 1981. EXERCISES 18.1 Trigonometric Form of the Fourier Series 1. Determine the fundamental frequency, fundamental radian frequency, and period of the following: (a) 5 sin 9t; (b) 200 cos 70t; (c) 4 sin(4t −10◦); (d) 4 sin(4t + 10◦). 2. Plot multiple periods of the ﬁrst, third, and ﬁfth harmonics on the same graph of each of the following periodic waveforms (three separate graphs in total are desired): (a) 3 sin t; (b) 40 cos 100t; (c) 2 cos(10t −90◦). and obtaining both the forced and natural response after laboring through a difﬁcult inverse transform operation. Well, we could go on, but all good things must come to an end. Best of luck to you in your future studies. CHAPTER 18 FOURIER CIRCUIT ANALYSIS 784 3. Calculate a0 for the following: (a) 4 sin 4t; (b) 4 cos 4t; (c) 4 + cos 4t; (d) 4 cos(4t + 40◦). 4. Compute a0, a1, and b1 for the following functions: (a) 2 cos 3t; (b) 3 −cos 3t; (c) 4 sin(4t −35◦). 5. (a) Calculate the Fourier coefﬁcients a0, a1, a2, a3, b1, b2, and b3 for the peri-odic function f (t) = 2u(t) −2u(t + 1) + 2u(t + 2) −2u(t + 3) + · · · . (b) Sketch f(t) and the Fourier series truncated after n = 3 over 3 periods. 6. (a) Compute the Fourier coefﬁcients a0, a1, a2, a3, a4, b1, b2, b3, and b4 for the periodic function g(t) partially sketched in Fig. 18.28. (b) Plot g(t) along with the Fourier series representation truncated after n = 4. 5 10 –5 0 5 10 f (t) t (s) ■FIGURE 18.29 7. For the periodic waveform f(t) represented in Fig. 18.29, calculate a1, a2, a3 and b1, b2, b3. 8. With respect to the periodic waveform sketched in Fig. 18.29, let gn(t) repre-sent the Fourier series representation of f(t) truncated at n. [For example, if n = 1, g1(t) has three terms, deﬁned through a0, a1 and b1.] (a) Sketch g2(t), g3(t), and g5(t), along with f(t). (b) Calculate f(2.5), g2(2.5), g3(2.5), and g5(2.5). 9. With respect to the periodic waveform g(t) sketched in Fig. 18.28, deﬁne yn(t) which represents the Fourier series representation truncated at n. (For example, y2(t) has ﬁve terms, deﬁned through a0, a1, a2, b1, and b2.) (a) Sketch y3(t) and y5(t) along with g(t). (b) Compute y1(0.5), y2(0.5), y3(0.5), and g(0.5). 10. Determine expressions for an and bn for g(t −1) if the periodic waveform g(t) is deﬁned as sketched in Fig. 18.28. 11. Plot the line spectrum (limited to the six largest terms) for the waveform shown in Fig. 18.4a. 12. Plot the line spectrum (limited to the ﬁve largest terms) for the waveform of Fig. 18.4b. 13. Plot the line spectrum (limited to the ﬁve largest terms) for the waveform rep-resented by the graph of Fig. 18.4c. 2 –2 2 1 4 3 g (t) t –2 ■FIGURE 18.28 EXERCISES 785 18.2 The Use of Symmetry 14. State whether the following exhibit odd symmetry, even symmetry, and/or half-wave symmetry: (a) 4 sin 100t; (b) 4 cos 100t; (c) 4 cos(4t + 70◦); (d) 4 cos 100t + 4; (e) each waveform in Fig. 18.4. 15. Determine whether the following exhibit odd symmetry, even symmetry, and/or half-wave symmetry: (a) the waveform in Fig. 18.28; (b) g(t −1), if g(t) is represented in Fig. 18.28; (c) g(t + 1), if g(t) is represented in Fig. 18.28; (d) the waveform of Fig. 18.29. 16. The nonperiodic waveform g(t) is deﬁned in Fig. 18.30. Use it to create a new function y(t) such that y(t) is identical to g(t) over the range of 0 < t < 4 and also is characterized by a period T = 8 and has (a) odd symmetry; (b) even symmetry; (c) both even and half-wave symmetry; (d) both odd and half-wave symmetry. 2 0 1 3 4 5 t g (t) ■FIGURE 18.30 t (s) 1 2 3 –3 –2 –1 –1 1 v(t) (V) ■FIGURE 18.31 17. Calculate a0, a1, a2, a3 and b1, b2, b3 for the periodic waveform v(t) repre-sented in Fig. 18.31. 18. The waveform of Fig. 18.31 is shifted to create a new waveform such that vnew(t) = v(t + 1). Calculate a0, a1, a2, a3 and b1, b2, b3. 19. Design a triangular waveform having a peak magnitude of 3, a period of 2 s, and characterized by (a) half-wave and even symmetry; (b) half-wave and odd symmetry. 20. Make use of symmetry as much as possible to obtain numerical values for a0, an, and bn, 1 ≤n ≤10, for the waveform shown in Fig. 18.32. t (ms) f (t) 2 –10 4 –4 –2 6 8 12 14 10 ■FIGURE 18.32 CHAPTER 18 FOURIER CIRCUIT ANALYSIS 786 18.3 Complete Response to Periodic Forcing Functions 21. For the circuit of Fig. 18.33a, calculate v(t) if is(t) is given by Fig. 18.33b and v(0) = 0. 1 2 F is (a) v + – t = 0 10 0 – 2 is (mA) t (s) 2 2 3 2 (b) ■FIGURE 18.33 t (s) vs(t) (V) 12 0 – 10 10 5 ■FIGURE 18.34 22. If the waveform shown in Fig. 18.34 is applied to the circuit of Fig. 18.8a, calculate i(t). 23. The circuit of Fig. 18.35a is subjected to the waveform depicted in Fig. 18.35b. Determine the steady-state voltage v(t). (a) 10 is 5 mH 10 iL v + – (b) –0.1 0 –2 +2 0.1 0.2 0.3 0.4 iS (A) t (s) ■FIGURE 18.35 EXERCISES 787 24. Apply the waveform of Fig. 18.36 to the circuit of Fig. 18.35b, and calculate the steady-state current iL(t). –2 –3 0 3 2 4 6 8 iS (A) t (s) ■FIGURE 18.36 25. If the current waveform of Fig. 18.36 is applied to the circuit of Fig. 18.33a, calculate the steady-state voltage v(t). 18.4 Complex Form of the Fourier Series 26. Let the function v(t) be deﬁned as indicated in Fig. 18.10. Determine cn for (a) v(t + 0.5); (b) v(t–0.5). 27. Calculate c0, c±1, and c±2 for the waveform of Fig. 18.36. 28. Determine the ﬁrst ﬁve terms of the exponential Fourier series representation of the waveform graphed in Fig. 18.33b. 29. For the periodic waveform shown in Fig. 18.37, determine (a) the period T; (b) c0, c±1, c±2, and c±3. t (s) –10 –3 –2 –1 1 3 4 –10 f (t) 2 ■FIGURE 18.37 100 1 2 3 4 5 6 f (t) t (ms) ■FIGURE 18.38 30. For the periodic waveform represented in Fig. 18.38, calculate (a) the period T; (b) c0, c±1, c±2, and c±3. CHAPTER 18 FOURIER CIRCUIT ANALYSIS 788 31. A pulse sequence has a period of 5 μs, an amplitude of unity for −0.6 < t < −0.4 μs and for 0.4 < t < 0.6 μs, and zero amplitude elsewhere in the period interval. This series of pulses might represent the decimal number 3 being transmitted in binary form by a digital computer. (a) Find cn. (b) Evaluate c4. (c) Evaluate c0. (d) Find \|cn\|max. (e) Find N so that \|cn\| ≤0.1\|cn\|max for all n > N. ( f ) What bandwidth is required to transmit this portion of the spectrum? 32. Let a periodic voltage vs(t) is equal to 40 V for 0 < t < 1 96 s, and to 0 for 1 96 < t < 1 16 s. If T = 1 16 s, ﬁnd (a) c3; (b) the power delivered to the load in the circuit of Fig. 18.39. f (Hz) 0 15 30 45 60 Load 1 v0 vs + – v0 + – 5 Ω 10 mH vs ■FIGURE 18.39 18.5 Deﬁnition of the Fourier Transform 33. Given g(t) = 5 −1 < t < 1 0 elsewhere $ sketch (a) g(t); (b) G( jω). 34. For the function v(t) = 2u(t) −2u(t + 2) + 2u(t + 4) −2u(t + 6) V, sketch (a) v(t); (b) V(jω). 35. Employ Eq. [46a] to calculate G(jω) if g(t) is (a) 5e−tu(t); (b) 5te−tu(t). 36. Obtain the Fourier transform F( jω) of the single triangle pulse plotted in Fig. 18.40. 37. Determine the Fourier transform F(jω) of the single sinusoidal pulse wave-form shown in Fig. 18.41. 15 0 3 – 3 f (t) t ■FIGURE 18.40 5 –5 2 – f (t) t 2 ■FIGURE 18.41 18.6 Some Properties of the Fourier Transform 38. For g(t) = 3e−tu(t), calculate (a) G( jω); (b) Ag(1); (c) Bg(1); (d) φ(ω). 39. The voltage pulse 2e−tu(t) V is applied to the input of an ideal bandpass ﬁlter. The passband of the ﬁlter is deﬁned by 100 < \| f \| < 500 Hz. Calculate the total output energy. EXERCISES 789 40. Given that v(t) = 4e−\|t\| V, calculate the frequency range in which 85% of the 1 energy lies. 41. Calculate the 1 energy associated with the function f (t) = 4te−3tu(t). 42. Use the deﬁnition of the Fourier transform to prove the following results, where F{ f (t)} = F( jω): (a) F{ f (t −t0)} = e−jωt0F{ f (t)}; (b) F{d f (t)/dt} = jωF{ f (t)}; (c) F{ f (kt)} = (1/\|k\|)F( jω/k); (d) F{ f (−t)} = F(−jω); (e) F{t f (t)} = j d[F( jω)]/dω. 18.7 Fourier Transform Pairs for Some Simple Time Functions 43. Determine the Fourier transform of the following: (a); 5u(t) −2 sgn(t); (b) 2 cos 3t −2; (c) 4e−j3t + 4e j3t + 5u(t). 44. Find the Fourier transform of each of the following: (a) 85u(t + 2) −50u(t −2); (b) 5δ(t) −2 cos 4t. 45. Sketch f(t) and \|F( jω)\| if f(t) is given by (a) 2 cos 10t; (b) e−4tu(t); (c) 5sgn(t). 46. Determine f(t) if F(jω) is given by (a) 4δ(ω); (b) 2/(5000 + jω); (c) e−j120ω. 47. Obtain an expression for f(t) if F( jω) is given by (a) −j 231 ω ; (b) 1 + j2 1 + j4; (c) 5δ(ω) + 1 2 + j10 . 18.8 The Fourier Transform of a General Periodic Time Function 48. Calculate the Fourier transform of the following functions: (a) 2 cos2 5t; (b) 7 sin 4t cos 3t; (c) 3 sin(4t −40◦). 49. Determine the Fourier transform of the periodic function g(t), which is deﬁned over the range 0 < t < 10 s by g(t) = 2u(t) −3u(t −4) + 2u(t −8). 50. If F( jω) = 20 ∞ n=1[1/(\|n\|! + 1)]δ(ω −20n), ﬁnd the value of f (0.05). 51. Given the periodic waveform shown in Fig. 18.42, determine its Fourier transform. 5 –5 1 2 3 4 5 6 7 – 6 –5 – 4 –3 –2 –1 f (t) t (s) ■FIGURE 18.42 18.9 The System Function and Response in the Frequency Domain 52. If a system is described by transfer function h(t) = 2u(t) + 2u(t −1), use convolution to calculate the output (time domain) if the input is (a) 2u(t); (b) 2te−2tu(t). 53. Given the input function x(t) = 5e−5tu(t), employ convolution to obtain a time-domain output if the system transfer function h(t) is given by (a) 3u(t + 1); (b) 10te−tu(t). 54. (a) Design a noninverting ampliﬁer having a gain of 10. If the circuit is con-structed using a μA741 op amp powered by ±15 V supplies, determine the FFT of the output through appropriate simulations if the input voltage operates at 1 kHz and has magnitude (b) 10 mV; (c) 1 V; (d) 2 V. CHAPTER 18 FOURIER CIRCUIT ANALYSIS 790 55. (a) Design an inverting ampliﬁer having a gain of 5. If the circuit is con-structed using a μA741 op amp powered by ±10 V supplies, perform appro-priate simulations to determine the FFT of the output voltage if the input volt-age has a frequency of 10 kHz and magnitude (b) 500 mV; (c) 1.8 V; (d) 3 V. 18.10 The Physical Signiﬁcance of the System Function 56. With respect to the circuit of Fig. 18.43, calculate vo(t) using Fourier tech-niques if vi(t) = 2te−tu(t) V. 57. After the inductor of Fig. 18.43 is surreptitiously replaced with a 2 F capacitor, calculate vo(t) using Fourier techniques if vi(t) is equal to (a) 5u(t) V; (b) 3e−4tu(t) V. 58. Employ Fourier-based techniques to calculate vC(t) as labeled in Fig. 18.44 if vi(t) is equal to (a) 2u(t) V; (b) 2δ(t) V. 59. Employ Fourier-based techniques to calculate vo(t) as labeled in Fig. 18.45 if vi(t) is equal to (a) 5u(t) V; (b) 3δ(t) V. 60. Employ Fourier-based techniques to calculate vo(t) as labeled in Fig. 18.45 if vi(t) is equal to (a) 5u(t −1) V; (b) 2 + 8e−tu(t) V. Chapter-Integrating Exercises 61. Apply the pulse waveform of Fig. 18.46a as the voltage input vi(t) to the cir-cuit shown in Fig. 18.44, and calculate vC(t). 2 1 –2 0 2 4 6 h(t) t (a) 10 5 –2 0 2 4 6 x (t) t (b) ■FIGURE 18.46 62. Apply the pulse waveform of Fig. 18.46b as the voltage input vi(t) to the circuit shown in Fig. 18.44, and calculate vC(t). 63. Apply the pulse waveform of Fig. 18.46a as the voltage input vi(t) to the circuit shown in Fig. 18.44, and calculate iC(t), deﬁned consistent with the passive sign convention. 64. Apply the pulse waveform of Fig. 18.46b as the voltage input vi(t) to the circuit shown in Fig. 18.45, and calculate vo(t). 65. Apply the pulse waveform of Fig. 18.46b as the voltage input vi(t) to the circuit shown in Fig. 18.45, and calculate vo(t). + – vo(t) + – 2.5 5 H vi(t) ■FIGURE 18.43 500 mF + – 2 vC + – 200 mH vi(t) ■FIGURE 18.44 1 500 mF + – 800 mH vo(t) + – vi(t) ■FIGURE 18.45 APPENDIX 1 • 791 AN INTRODUCTION TO NETWORK TOPOLOGY After working many circuits problems, it slowly becomes evident that many of the circuits we see have quite a bit in common, at least in terms of the arrangement of components. From this realization, it is possible to create a more abstract view of circuits which we call network topology, a subject we introduce in this appendix. A1.1 • TREES AND GENERAL NODAL ANALYSIS We now plan to generalize the method of nodal analysis that we have come to know and love. Since nodal analysis is applicable to any network, we cannot promise that we will be able to solve a wider class of circuit prob-lems. We can, however, look forward to being able to select a general nodal analysis method for any particular problem that may result in fewer equa-tions and less work. We must ﬁrst extend our list of deﬁnitions relating to network topology. We begin by deﬁning topology itself as a branch of geometry which is con-cerned with those properties of a geometrical ﬁgure which are unchanged when the ﬁgure is twisted, bent, folded, stretched, squeezed, or tied in knots, with the provision that no parts of the ﬁgure are to be cut apart or to be joined together. A sphere and a tetrahedron are topologically identical, as are a square and a circle. In terms of electric circuits, then, we are not now concerned with the particular types of elements appearing in the circuit, but only with the way in which branches and nodes are arranged. As a matter of fact, we usually suppress the nature of the elements and simplify the draw-ing of the circuit by showing the elements as lines. The resultant drawing is called a linear graph, or simply a graph. A circuit and its graph are shown in Fig. A1.1. Note that all nodes are identiﬁed by heavy dots in the graph. Since the topological properties of the circuit or its graph are unchanged when it is distorted, the three graphs shown in Fig. A1.2 are all topologi-cally identical with the circuit and graph of Fig. A1.1. Topological terms that we already know and have been using correctly are Node: A point at which two or more elements have a common connection. Path: A set of elements that may be traversed in order without passing through the same node twice. Branch: A single path, containing one simple element, which connects one node to any other node. + – + – (a) (b) ■FIGURE A1.1 (a) A given circuit. (b) The linear graph of this circuit. Loop: A closed path. Mesh: A loop which does not contain any other loops within it. Planar circuit: A circuit which may be drawn on a plane surface in such a way that no branch passes over or under any other branch. Nonplanar circuit: Any circuit which is not planar. The graphs of Fig. A1.2 each contain 12 branches and 7 nodes. Three new properties of a linear graph must now be deﬁned—a tree, a cotree, and a link. We deﬁne a tree as any set of branches which does not contain any loops and yet connects every node to every other node, not nec-essarily directly. There are usually a number of different trees which may be drawn for a network, and the number increases rapidly as the complex-ity of the network increases. The simple graph shown in Fig. A1.3a has eight possible trees, four of which are shown by heavy lines in Fig. A1.3b, c, d, and e. APPENDIX 1 AN INTRODUCTION TO NETWORK TOPOLOGY 792 (a) (e) (d) (c) (b) ■FIGURE A1.3 (a) The linear graph of a three-node network. (b, c, d, e) Four of the eight different trees which may be drawn for this graph are shown by the black lines. In Fig. A1.4a a more complex graph is shown. Figure A1.4b shows one possible tree, and Fig. A1.4c and d show sets of branches which are not trees because neither set satisﬁes the deﬁnition. After a tree has been speciﬁed, those branches that are not part of the tree form the cotree, or complement of the tree. The lightly drawn branches in Fig. A1.3b to e show the cotrees that correspond to the heavier trees. Once we understand the construction of a tree and its cotree, the concept of the link is very simple, for a link is any branch belonging to the cotree. It is evident that any particular branch may or may not be a link, depending on the particular tree which is selected. The number of links in a graph may easily be related to the number of branches and nodes. If the graph has N nodes, then exactly (N −1) branches are required to construct a tree because the ﬁrst branch chosen connects two nodes and each additional branch includes one more node. (a) (c) (b) ■FIGURE A1.2 (a, b, c) Alternative linear graphs of the circuit of Fig. A1.1. (a) (d) (c) (b) ■FIGURE A1.4 (a) A linear graph. (b) A possible tree for this graph. (c, d) These sets of branches do not satisfy the deﬁnition of a tree. APPENDIX 1 AN INTRODUCTION TO NETWORK TOPOLOGY 793 Thus, given B branches, the number of links L must be L = B −(N −1) or L = B −N + 1 There are L branches in the cotree and (N −1) branches in the tree. In any of the graphs shown in Fig. A1.3, we note that 3 = 5 −3 + 1, and in the graph of Fig. A1.4b, 6 = 10 −5 + 1. A network may be in several disconnected parts, and Eq. may be made more general by replacing +1 with +S, where S is the number of separate parts. However, it is also possi-ble to connect two separate parts by a single conductor, thus causing two nodes to form one node; no current can ﬂow through this single conductor. This process may be used to join any number of separate parts, and thus we will not suffer any loss of generality if we restrict our attention to circuits for which S = 1. We are now ready to discuss a method by which we may write a set of nodal equations that are independent and sufﬁcient. The method will enable us to obtain many different sets of equations for the same network, and all the sets will be valid. However, the method does not provide us with every possible set of equations. Let us ﬁrst describe the procedure, illustrate it by three examples, and then point out the reason that the equations are inde-pendent and sufﬁcient. Given a network, we should: 1. Draw a graph and then identify a tree. 2. Place all voltage sources in the tree. 3. Place all current sources in the cotree. 4. Place all control-voltage branches for voltage-controlled dependent sources in the tree, if possible. 5. Place all control-current branches for current-controlled dependent sources in the cotree, if possible. These last four steps effectively associate voltages with the tree and currents with the cotree. We now assign a voltage variable (with its plus-minus pair) across each of the (N −1) branches in the tree. A branch containing a voltage source (dependent or independent) should be assigned that source voltage, and a branch containing a controlling voltage should be assigned that controlling voltage. The number of new variables that we have introduced is therefore equal to the number of branches in the tree (N −1), reduced by the number of voltage sources in the tree, and reduced also by the number of control voltages we were able to locate in the tree. In Example A1.3, we will ﬁnd that the number of new variables required may be zero. Having a set of variables, we now need to write a set of equations that are sufficient to determine these variables. The equations are obtained through the application of KCL. Voltage sources are handled in the same way that they were in our earlier attack on nodal analysis; each voltage source and the two nodes at its terminals constitute a supernode or a part of a supernode. Kirchhoff’s current law is then applied at all but one of the re-maining nodes and supernodes. We set the sum of the currents leaving the node in all of the branches connected to it equal to zero. Each current is ex-pressed in terms of the voltage variables we just assigned. One node may be ignored, just as was the case earlier for the reference node. Finally, in case there are current-controlled dependent sources, we must write an equation for each control current that relates it to the voltage variables; this also is no different from the procedure used before with nodal analysis. Let us try out this process on the circuit shown in Fig. A1.5a. It contains four nodes and ﬁve branches, and its graph is shown in Fig. A1.5b. APPENDIX 1 AN INTRODUCTION TO NETWORK TOPOLOGY 794 EXAMPLE A1.1 Find the value of vx in the circuit of Fig. A1.5a. ■FIGURE A1.5 (a) A circuit used as an example for general nodal analysis. (b) The graph of the given circuit. (c) The voltage source and the control voltage are placed in the tree, while the current source goes in the cotree. (d)The tree is completed and a voltage is assigned across each tree branch. vx 14 (a) 4 15 8 100 V + – vx + – (d) vx + – v1 + – 100 V + – (c) (b) In accordance with steps 2 and 3 of the tree-drawing procedure, we place the voltage source in the tree and the current source in the cotree. Following step 4, we see that the vx branch may also be placed in the tree, since it does not form any loop which would violate the deﬁnition of a tree. We have now arrived at the two tree branches and the single link shown in Fig. A1.5c, and we see that we do not yet have a tree, since the right node is not connected to the others by a path through tree branches. The only possible way to complete the tree is shown in Fig. A1.5d. The 100 V source voltage, the control voltage vx, and a new voltage variable v1 are next assigned to the three tree branches as shown. We therefore have two unknowns, vx and v1, and we need to obtain two equations in terms of them. There are four nodes, but the presence APPENDIX 1 AN INTRODUCTION TO NETWORK TOPOLOGY 795 EXAMPLE A1.2 Find the values of vx and vy in the circuit of Fig. A1.6a. ■FIGURE A1.6 (a) A circuit with ﬁve nodes. (b) A tree is chosen such that both voltage sources and both control voltages are tree branches. of the voltage source causes two of them to form a single supernode. Kirchhoff’s current law may be applied at any two of the three remaining nodes or supernodes. Let’s attack the right node ﬁrst. The current leaving to the left is −v1/15, while that leaving downward is −vx/14. Thus, our ﬁrst equation is −v1 15 + −vx 14 = 0 The central node at the top looks easier than the supernode, and so we set the sum of the current to the left (−vx/8), the current to the right (v1/15), and the downward current through the 4- resistor equal to zero. This latter current is given by the voltage across the resistor divided by 4 , but there is no voltage labeled on that link. However, when a tree is constructed according to the deﬁnition, there is a path through it from any node to any other node. Then, since every branch in the tree is assigned a voltage, we may express the voltage across any link in terms of the tree-branch voltages. This downward current is therefore (−vx + 100)/4, and we have the second equation, −vx 8 + v1 15 + −vx + 100 4 = 0 The simultaneous solution of these two nodal equations gives v1 = −60 V vx = 56 V + – (a) vx + – vy + – 1 V 1 S 1 S 2 S 2 S 2 A 4vy 2vx + – (b) vx + – vy + – + – 1 V + – 4vy We draw a tree so that both voltage sources and both control voltages appear as tree-branch voltages and, hence, as assigned variables. As it happens, these four branches constitute a tree, Fig. A1.6b, and tree-branch voltages vx, 1, vy, and 4vy are chosen, as shown. (Continued on next page) APPENDIX 1 AN INTRODUCTION TO NETWORK TOPOLOGY 796 Both voltage sources deﬁne supernodes, and we apply KCL twice, once to the top node, 2vx + 1(vx −vy −4vy) = 2 and once to the supernode consisting of the right node, the bottom node, and the dependent voltage source, 1vy + 2(vy −1) + 1(4vy + vy −vx) = 2vx Instead of the four equations we would expect using previously studied techniques, we have only two, and we ﬁnd easily that vx = 26 9 V and vy = 4 3 V. Find the value of vx in the circuit of Fig. A1.7a. The two voltage sources and the control voltage establish the three-branch tree shown in Fig. A1.7b. Since the two upper nodes and the lower right node all join to form one supernode, we need write only one KCL equation. Selecting the lower left node, we have −1 −vx 4 + 3 + −vx + 30 + 6vx 5 = 0 and it follows that vx = −32 3 V. In spite of the apparent complexity of this circuit, the use of general nodal analysis has led to an easy solution. Employing mesh currents or node-to-reference voltages would require more equations and more effort. EXAMPLE A1.3 + – vx + – (a) 5 4 30 V 1 A 2 A 3 A 6vx + – vx + – (b) 6vx + – 30 V – + ■FIGURE A1.7 (a) A circuit for which only one general nodal equation need be written. (b) The tree and the tree-branch voltages used. We will discuss the problem of ﬁnding the best analysis scheme in the following section. If we needed to know some other voltage, current, or power in the pre-vious example, one additional step would give the answer. For example, the power provided by the 3 A source is 3 −30 −32 3 = −122 W Let us conclude by discussing the sufﬁciency of the assumed set of tree-branch voltages and the independence of the nodal equations. If these tree-branch voltages are sufﬁcient, then the voltage of every branch in either the tree or the cotree must be obtainable from a knowledge of the values of all the tree-branch voltages. This is certainly true for those branches in the tree. For the links we know that each link extends between two nodes, and, by deﬁnition, the tree must also connect those two nodes. Hence, every link voltage may also be established in terms of the tree-branch voltages. Once the voltage across every branch in the circuit is known, then all the currents may be found by using either the given value of the current if the branch consists of a current source, by using Ohm’s law if it is a resistive branch, or by using KCL and these current values if the branch happens to be a voltage source. Thus, all the voltages and currents are determined and sufﬁciency is demonstrated. APPENDIX 1 AN INTRODUCTION TO NETWORK TOPOLOGY 797 To demonstrate independency, let us satisfy ourselves by assuming the situation where the only sources in the network are independent current sources. As we have noticed earlier, independent voltage sources in the cir-cuit result in fewer equations, while dependent sources usually necessitate a greater number of equations. With independent current sources only, there will then be precisely (N −1) nodal equations written in terms of (N −1) tree-branch voltages. To show that these (N −1) equations are independ-ent, visualize the application of KCL to the (N −1) different nodes. Each time we write the KCL equation, there is a new tree branch involved—the one which connects that node to the remainder of the tree. Since that circuit element has not appeared in any previous equation, we must obtain an in-dependent equation. This is true for each of the (N −1) nodes in turn, and hence we have (N −1) independent equations. PRACTICE ● A1.1 (a) How many trees may be constructed for the circuit of Fig. A1.8 that follow all ﬁve of the tree-drawing suggestions listed earlier? (b) Draw a suitable tree, write two equations in two unknowns, and ﬁnd i3. (c) What power is supplied by the dependent source? ■FIGURE A1.8 Ans: 1; 7.2 A; 547 W. + – + – i3 12 8 5 9 A 25 V A1.2 • LINKS AND LOOP ANALYSIS Now we will consider the use of a tree to obtain a suitable set of loop equations. In some respects this is the dual of the method of writing nodal equations. Again it should be pointed out that, although we are able to guar-antee that any set of equations we write will be both sufﬁcient and inde-pendent, we should not expect that the method will lead directly to every possible set of equations. We again begin by constructing a tree, and we use the same set of rules as we did for general nodal analysis. The objective for either nodal or loop analysis is to place voltages in the tree and currents in the cotree; this is a mandatory rule for sources and a desirable rule for controlling quantities. Now, however, instead of assigning a voltage to each branch in the tree, we assign a current (including reference arrow, of course) to each element in the cotree or to each link. If there were 10 links, we would assign exactly 10 link currents. Any link that contains a current source is assigned that source current as the link current. Note that each link current may also be APPENDIX 1 AN INTRODUCTION TO NETWORK TOPOLOGY 798 thought of as a loop current, for the link must extend between two speciﬁc nodes, and there must also be a path between those same two nodes through the tree. Thus, with each link there is associated a single speciﬁc loop that includes that one link and a unique path through the tree. It is evident that the assigned current may be thought of either as a loop current or as a link current. The link connotation is most helpful at the time the currents are be-ing deﬁned, for one must be established for each link; the loop interpreta-tion is more convenient at equation-writing time, because we will apply KVL around each loop. Let us try out this process of deﬁning link currents by considering the circuit shown in Fig. A1.9a. The tree selected is one of several that might be constructed for which the voltage source is in a tree branch and the current source is in a link. Let us ﬁrst consider the link containing the current source. The loop associated with this link is the left-hand mesh, and so we show our link current ﬂowing about the perimeter of this mesh (Fig. A1.9b). An obvi-ous choice for the symbol for this link current is ‘‘7 A.’’ Remember that no other current can ﬂow through this particular link, and thus its value must be exactly the strength of the current source. + – (a) 2 1 1 2 3 7 V 7 A (b) 7 A iB iA ■FIGURE A1.9 (a) A simple circuit. (b) A tree is chosen such that the current source is in a link and the voltage source is in a tree branch. We next turn our attention to the 3 resistor link. The loop associated with it is the upper right-hand mesh, and this loop (or mesh) current is deﬁned as iA and also shown in Fig. A1.9b. The last link is the lower 1 resistor, and the only path between its terminals through the tree is around the perimeter of the circuit. That link current is called iB, and the arrow indicating its path and reference direction appears in Fig. A1.9b. It is not a mesh current. Note that each link has only one current present in it, but a tree branch may have any number from 1 to the total number of link currents assigned. The use of long, almost closed, arrows to indicate the loops helps to indicate which loop currents ﬂow through which tree branch and what their refer-ence directions are. A KVL equation must now be written around each of these loops. The variables used are the assigned link currents. Since the voltage across a cur-rent source cannot be expressed in terms of the source current, and since we have already used the value of the source current as the link current, we dis-card any loop containing a current source. APPENDIX 1 AN INTRODUCTION TO NETWORK TOPOLOGY 799 EXAMPLE A1.4 For the example of Fig. A1.9, ﬁnd the values of iA and iB. We ﬁrst traverse the iA loop, proceeding clockwise from its lower left corner. The current going our way in the 1 resistor is (i A −7), in the 2 element it is (iA + iB), and in the link it is simply iB. Thus 1(iA −7) + 2(i A + iB) + 3iA = 0 For the iB link, clockwise travel from the lower left corner leads to −7 + 2(i A + iB) + 1iB = 0 Traversal of the loop deﬁned by the 7 A link is not required. Solving, we have iA = 0.5 A, iB = 2 A, once again. The solution has been achieved with one less equation than before! EXAMPLE A1.5 Evaluate i1 in the circuit shown in Fig. A1.10a. + – + – + – i1 (a) 5 2 4 19 V 30 V 25 V 4 A 1.5i1 4 A i1 (c) 1.5i1 (b) ■FIGURE A1.10 (a) A circuit for which i 1 may be found with one equation using general loop analysis. (b) The only tree that satisﬁes the rules outlined in Sec. A1.1. (c) The three link currents are shown with their loops. This circuit contains six nodes, and its tree therefore must have ﬁve branches. Since there are eight elements in the network, there are three links in the cotree. If we place the three voltage sources in the tree and the two current sources and the current control in the cotree, we are led to the tree shown in Fig. A1.10b. The source current of 4 A deﬁnes a (Continued on next page) APPENDIX 1 AN INTRODUCTION TO NETWORK TOPOLOGY 800 How may we demonstrate sufﬁciency? Let us visualize a tree. It con-tains no loops and therefore contains at least two nodes to each of which only one tree branch is connected. The current in each of these two branches is easily found from the known link currents by applying KCL. If there are other nodes at which only one tree branch is connected, these tree-branch currents may also be immediately obtained. In the tree shown in Fig. A1.11, we thus have found the currents in branches a, b, c, and d. Now we move along the branches of the tree, ﬁnding the currents in tree branches e and f; the process may be continued until all the branch currents are determined. The link currents are therefore sufﬁcient to determine all branch currents. It is helpful to look at the situation where an incorrect ‘‘tree’’ has been drawn which contains a loop. Even if all the link currents were zero, a current might still circulate about this ‘‘tree loop.’’ Hence, the link currents could not determine this current, and they would not represent a sufﬁcient set. Such a ‘‘tree’’ is by deﬁnition impossible. To demonstrate independence, let us satisfy ourselves by assuming the situation where the only sources in the network are independent voltage sources. As we have noticed earlier, independent current sources in the cir-cuit result in fewer equations, while dependent sources usually necessitate a greater number of equations. If only independent voltage sources are pres-ent, there will then be precisely (B −N + 1) loop equations written in terms of the (B −N + 1) link currents. To show that these (B −N + 1) loop equations are independent, it is necessary only to point out that each represents the application of KVL around a loop which contains one link not appearing in any other equation. We might visualize a different resis-tance R1, R2, . . . , RB−N+1 in each of these links, and it is then apparent that one equation can never be obtained from the others, since each contains one coefﬁcient not appearing in any other equation. Hence, the link currents are sufﬁcient to enable a complete solution to be obtained, and the set of loop equations which we use to ﬁnd the link currents is a set of independent equations. ■FIGURE A1.11 A tree that is used as an example to illustrate the sufﬁciency of the link currents. loop as shown in Fig. A1.10c. The dependent source establishes the loop current 1.5i1 around the right mesh, and the control current i1 gives us the remaining loop current about the perimeter of the circuit. Note that all three currents ﬂow through the 4 resistor. We have only one unknown quantity, i1, and after discarding the loops deﬁned by the two current sources, we apply KVL around the outside of the circuit: −30 + 5(−i1) + 19 + 2(−i1 −4) + 4(−i1 −4 + 1.5i1) −25 = 0 Besides the three voltage sources, there are three resistors in this loop. The 5 resistor has one loop current in it, since it is also a link; the 2 resistor contains two loop currents; and the 4 resistor has three. A carefully drawn set of loop currents is a necessity if errors in skipping currents, utilizing extra ones, or erring in choosing the correct direction are to be avoided. The foregoing equation is guaranteed, however, and it leads to i1 = −12 A. a e d g c f b APPENDIX 1 AN INTRODUCTION TO NETWORK TOPOLOGY 801 Having looked at both general nodal analysis and loop analysis, we should now consider the advantages and disadvantages of each method so that an intelligent choice of a plan of attack can be made on a given analy-sis problem. The nodal method in general requires (N −1) equations, but this num-ber is reduced by 1 for each independent or dependent voltage source in a tree branch, and increased by 1 for each dependent source that is voltage-controlled by a link voltage, or current-controlled. The loop method basically involves (B −N + 1) equations. However, each independent or dependent current source in a link reduces this number by 1, while each dependent source that is current-controlled by a tree-branch current, or is voltage-controlled, increases the number by 1. As a grand ﬁnale for this discussion, let us inspect the T-equivalent circuit model for a transistor shown in Fig. A1.12, to which is connected a sinusoidal source, 4 sin 1000t mV, and a 10 k load. EXAMPLE A1.6 Find the input (emitter) current ie and the load voltage vL in the circuit of Fig. A1.12, assuming typical values for the emitter resistance re = 50 ; the base resistance rb = 500 ; the collector resistance rc = 20 k; and the common-base forward-current-transfer ratio α = 0.99. vL + – vs = 4 sin 1000t mV ie ie Collector Emitter Base 10 k rb re rc + – ■FIGURE A1.12 A sinusoidal voltage source and a 10 k load are connected to the T-equivalent circuit of a transistor. The common connection between the input and output is at the base terminal of the transistor, and the arrangement is called the common-base conﬁguration. Although the details are requested in the practice problems that follow, we should see readily that the analysis of this circuit might be accomplished by drawing trees requiring three general nodal equations (N −1 −1 + 1) or two loop equations (B −N + 1 −1). We might also note that three equations are required in terms of node-to-reference voltages, as are three mesh equations. No matter which method we choose, these results are obtained for this speciﬁc circuit: ie = 18.42 sin 1000t μA vL = 122.6 sin 1000t mV (Continued on next page) APPENDIX 1 AN INTRODUCTION TO NETWORK TOPOLOGY 802 and we therefore ﬁnd that this transistor circuit provides a voltage gain (vL/vs) of 30.6, a current gain (vL/10,000ie) of 0.666, and a power gain equal to the product 30.6(0.666) = 20.4. Higher gains could be secured by operating this transistor in a common-emitter conﬁguration. PRACTICE ● A1.2 Draw a suitable tree and use general loop analysis to ﬁnd i10 in the circuit of (a) Fig. A1.13a by writing just one equation with i10 as the variable; (b) Fig. A1.13b by writing just two equations with i10 and i3 as the variables. i10 (a) 2 k 10 k 20 k 5 k 5 mA 0.4i10 + – i10 i3 20 4 6 24 10 2 A 100 V (b) 3i3 ■FIGURE A1.13 A1.3 For the transistor ampliﬁer equivalent circuit shown in Fig. A1.12, let re = 50 , rb = 500 , rc = 20 k, and α = 0.99, and ﬁnd both ie and vL by drawing a suitable tree and using (a) two loop equations; (b) three nodal equations with a common reference node for the voltage; (c) three nodal equations without a common reference node. A1.4 Determine the Thévenin and Norton equivalent circuits presented to the 10 k load in Fig. A1.12 by ﬁnding (a) the open-circuit value of vL; (b) the (downward) short-circuit current; (c) the Thévenin equivalent resistance.All circuit values are given in Practice ProblemA1.3. Ans: A1.2: −4.00 mA; 4.69 A. A1.3: 18.42 sin 1000t μA; 122.6 sin 1000t mV. A1.4: 147.6 sin 1000t mV; 72.2 sin 1000t μA; 2.05 k. APPENDIX 2 • 803 SOLUTION OF SIMULTANEOUS EQUATIONS Consider the simple system of equations 7v1 −3v2 −4v3 = −11 −3v1 + 6v2 −2v3 = 3 −4v1 −2v2 + 11v3 = 25 This set of equations could be solved by a systematic elimination of the variables. Such a procedure is lengthy, however, and may never yield an-swers if done unsystematically for a greater number of simultaneous equa-tions. Fortunately, there are many more options available to us, some of which we will explore in this appendix. The Scientiﬁc Calculator Perhaps the most straightforward approach when faced with a system of equations such as Eqs. to , in which we have numerical coefﬁcients and are only interested in the speciﬁc values of our unknowns (as opposed to algebraic relationships), is to employ any of the various scientiﬁc calcu-lators presently on the market. For example, on a Texas Instruments TI-84, we can employ the Polynomial Root Finder and Simultaneous Equation Solver (you may need to install the application using TI ConnectTM). Pressing the APPS key and scrolling down, ﬁnd the application named PLYSmlt2. Running and continuing past the welcome screen shows the Main Menu of Fig. A2.1a. Selecting the second menu item results in the screen shown in Fig. A2.1b, where we have chosen three equations in three unknowns. After (a) (b) (c) (d) ■FIGURE A2.1 Screen sequence for solving Eqs. to as seen on a TI-84 running the Simultaneous Equation Solver application. pressing NEXT, we are presented with a screen similar to that shown in Fig. A2.1c. After we have ﬁnished entering all coefﬁcients, pressing the SOLVE button yields the Solution screen depicted in Fig. A2.1d. Since we did not name the variables, a slight mental conversion is required to realize X1 = v1, X2 = v2, etc. It should be noted that each calculator capable of solving simultaneous equations has its own procedure for entering the required information— therefore, it is a good idea not to throw away anything marked ‘‘Owner’s Manual’’or ‘‘Instructions,’’no matter how tempting such an action might be. Matrices Another powerful approach to the solution of a system of equations is based on the concept of matrices. Consider Eqs. , , and . The array of the constant coefﬁcients of the equations, G = ⎡ ⎣ 7 −3 −4 −3 6 −2 −4 −2 11 ⎤ ⎦ is called a matrix; the symbol G has been selected since each element of the matrix is a conductance value. A matrix itself has no ‘‘value’’; it is merely an ordered array of elements. We use a letter in boldface type to represent a matrix, and we enclose the array itself by square brackets. Amatrix having m rows and n columns is called an (m × n) (pronounced ‘‘m by n’’) matrix. Thus, A = 2 0 5 −1 6 3 is a (2 × 3) matrix, and the G matrix of our example is a (3 × 3) matrix. An (n × n) matrix is a square matrix of order n. An (m × 1) matrix is called a column matrix, or a vector. Thus, V = V1 V2 is a (2 × 1) column matrix of phasor voltages, and I = I1 I2 is a (2 × 1) phasor-current vector.A(1 × n) matrix is known as a row vector. Two (m × n) matrices are equal if their corresponding elements are equal. Thus, if ajk is that element of A located in row j and column k and bjk is the element at row j and column k in matrix B, then A = B if and only if ajk = bjk for all 1 ≤j ≤m and 1 ≤k ≤n. Thus, if V1 V2 = z11I1 + z12I2 z21I1 + z22I2 then V1 = z11I1 + z12I2 and V2 = z21I1 + z22I2. APPENDIX 2 SOLUTION OF SIMULTANEOUS EQUATIONS 804 APPENDIX 2 SOLUTION OF SIMULTANEOUS EQUATIONS 805 Two (m × n) matrices may be added by adding corresponding elements. Thus, 2 0 5 −1 6 3 + 1 2 3 −3 −2 −1 = 3 2 8 −4 4 2 Next let us consider the matrix product AB, where A is an (m × n) matrix and B is a (p × q) matrix. If n = p, the matrices are said to be conformal, and their product exists. That is, matrix multiplication is deﬁned only for the case where the number of columns of the ﬁrst matrix in the product is equal to the number of rows in the second matrix. The formal deﬁnition of matrix multiplication states that the product of the (m × n) matrix A and the (n × q) matrix B is an (m × q) matrix having elements cjk, 1 ≤j ≤m and 1 ≤k ≤q, where cjk = aj1b1k + aj2b2k + · · · + ajnbnk That is, to ﬁnd the element in the second row and third column of the prod-uct, we multiply each of the elements in the second row of A by the corre-sponding element in the third column of B and then add the n results. For example, given the (2 × 3) matrix A and the (3 × 2) matrix B, a11 a12 a13 a21 a22 a23 ⎡ ⎣ b11 b12 b21 b22 b31 b32 ⎤ ⎦= (a11b11 + a12b21 + a13b31) (a11b12 + a12b22 + a13b32) (a21b11 + a22b21 + a23b31) (a21b12 + a22b22 + a23b32) The result is a (2 × 2) matrix. As a numerical example of matrix multiplication, we take 3 2 1 −2 −2 4 ⎡ ⎣ 2 3 −2 −1 4 −3 ⎤ ⎦= 6 4 16 −16 where 6 = (3)(2) + (2)(−2) + (1)(4), 4 = (3)(3) + (2)(−1) + (1)(−3), and so forth. Matrix multiplication is not commutative. For example, given the (3 × 2) matrix C and the (2 × 1) matrix D, it is evident that the product CD may be calculated, but the product DC is not even deﬁned. As a ﬁnal example, let tA = 2 3 −1 4 and tB = 3 1 5 0 so that both tAtB and tBtA are deﬁned. However, tAtB = 21 2 17 −1 Matrix Inversion If we write our system of equations using matrix notation, ⎡ ⎣ 7 −3 −4 −3 6 −2 −4 −2 11 ⎤ ⎦ ⎡ ⎣ v1 v2 v3 ⎤ ⎦= ⎡ ⎣ −11 3 25 ⎤ ⎦ we may solve for the voltage vector by multiplying both sides of Eq. by the inverse of our matrix G: G−1 ⎡ ⎣ 7 −3 −4 −3 6 −2 −4 −2 11 ⎤ ⎦ ⎡ ⎣ v1 v2 v3 ⎤ ⎦= G−1 ⎡ ⎣ −11 3 25 ⎤ ⎦ This procedure makes use of the identity G−1G = I, where I is the identity matrix, a square matrix of the same size as G, with zeros everywhere except on the diagonal. Each element on the diagonal of an identity matrix is unity. Thus, Eq. becomes ⎡ ⎣ 1 0 0 0 1 0 0 0 1 ⎤ ⎦ ⎡ ⎣ v1 v2 v3 ⎤ ⎦= G−1 ⎡ ⎣ −11 3 25 ⎤ ⎦ which may be simpliﬁed to ⎡ ⎣ v1 v2 v3 ⎤ ⎦= G−1 ⎡ ⎣ −11 3 25 ⎤ ⎦ since the identity matrix times any vector is simply equal to that vector (the proof is left to the reader as a 30-second exercise). The solution of our system of equations has therefore been transformed into the problem of obtaining the inverse matrix of G. Many scientiﬁc calculators provide the means of per-forming matrix algebra. APPENDIX 2 SOLUTION OF SIMULTANEOUS EQUATIONS 806 PRACTICE ● A2.1 Given A = 1 −3 3 5 , B = 4 −1 −2 3 , C = 50 30 , and V = V1 V2 , ﬁnd (a) A + B; (b) AB; (c) BA; (d) AV + BC; (e) A2 = AA. Ans: 5 −4 1 8 ; 10 −10 2 12 ; 1 −17 7 21 ; V1 −3V2 + 170 3V1 + 5V2 −10 ; −8 −18 18 16 . while tBtA = 5 13 10 15 APPENDIX 2 SOLUTION OF SIMULTANEOUS EQUATIONS 807 Returning to the TI-84, we press 2ND and MATRIX to obtain the screen shown in Fig. A2.2a. Scrolling horizontally to EDIT, we press the ENTER key, and select a 3 × 3 matrix, resulting in a screen similar to that shown in Fig. A2.2b. Once we have ﬁnished entering the matrix, we press 2ND and QUIT. Returning to the MATRIX editor, we create a 3 1 vec-tor named B, as shown in Fig. A2.2c. We are now (ﬁnally) ready to solve for the solution vector. Pressing 2ND and MATRIX, under NAMES we select [A] and press ENTER, followed by the x−1 key. Next, we select [B] the same way (we could have pressed the multiplication key in between but it is not necessary). The result of our calculation is shown in Fig. A2.2d, and agrees with our earlier exercise. Determinants Although a matrix itself has no ‘‘value,’’ the determinant of a square ma-trix does have a value. To be precise, we should say that the determinant of a matrix is a value, but common usage enables us to speak of both the array itself and its value as the determinant. We shall symbolize a determinant by , and employ a suitable subscript to denote the matrix to which the determinant refers. Thus, G = 7 −3 −4 −3 6 −2 −4 −2 11 Note that simple vertical lines are used to enclose the determinant. The value of any determinant is obtained by expanding it in terms of its minors. To do this, we select any row j or any column k, multiply each el-ement in that row or column by its minor and by (−1) j+k, and then add the products. The minor of the element appearing in both row j and column k (a) (b) (c) (d) ■FIGURE A2.2 Sequence of screens for matrix solution. (a) Matrix editor screen; (b) entering terms; (c) creating right-hand side vector; (d) solving matrix equation. APPENDIX 2 SOLUTION OF SIMULTANEOUS EQUATIONS 808 is the determinant obtained when row j and column k are removed; it is indicated by jk. As an example, let us expand the determinant G along column 3. We ﬁrst multiply the (−4) at the top of this column by (−1)1+3 = 1 and then by its minor: (−4)(−1)1+3 −3 6 −4 −2 and then repeat for the other two elements in column 3, adding the results: −4 −3 6 −4 −2 + 2 7 −3 −4 −2 + 11 7 −3 −3 6 The minors contain only two rows and two columns. They are of order 2, and their values are easily determined by expanding in terms of minors again, here a trivial operation. Thus, for the ﬁrst determinant, we expand along the ﬁrst column by multiplying (−3) by (−1)1+1 and its minor, which is merely the element (−2), and then multiplying (−4) by (−1)2+1 and by 6. Thus, −3 6 −4 −2 = (−3)(−2) −4(−6) = 30 It is usually easier to remember the result for a second-order determinant as ‘‘upper left times lower right minus upper right times lower left.’’ Finally, G = −4[(−3)(−2) −6(−4)] + 2[(7)(−2) −(−3)(−4)] + 11[(7)(6) −(−3)(−3)] = −4(30) + 2(−26) + 11(33) = 191 For practice, let us expand this same determinant along the ﬁrst row: G = 7 6 −2 −2 11 −(−3) −3 −2 −4 11 + (−4) −3 6 −4 −2 = 7(62) + 3(−41) −4(30) = 191 The expansion by minors is valid for a determinant of any order. Repeating these rules for evaluating a determinant in more general terms, we would say, given a matrix a, a = ⎡ ⎢ ⎣ a11 a12 . . . a1N a21 a22 . . . a2N aN1 aN2 . . . aN N ⎤ ⎥ ⎦ . . . . . . . . . . . . . . . . . . . . . that a may be obtained by expansion in terms of minors along any row j: a = aj1(−1) j+1j1 + aj2(−1) j+2j2 + · · · + aj N(−1) j+Nj N = N n=1 ajn(−1) j+njn or along any column k: a = a1k(−1)1+k1k + a2k(−1)2+k2k + · · · + aNk(−1)N+kNk = N n=1 ank(−1)n+knk The cofactor Cjk of the element appearing in both row j and column k is simply (−1) j+k times the minor jk. Thus, C11 = 11, but C12 = −12. We may now write a = N n=1 ajnCjn = N n=1 ankCnk As an example, let us consider this fourth-order determinant: = 2 −1 −2 0 −1 4 2 −3 −2 −1 5 −1 0 −3 3 2 We ﬁnd 11 = 4 2 −3 −1 5 −1 −3 3 2 = 4(10 + 3) + 1(4 + 9) −3(−2 + 15) = 26 12 = −1 2 −3 −2 5 −1 0 3 2 = −1(10 + 3) + 2(4 + 9) + 0 = 13 and C11 = 26, whereas C12 = −13. Finding the value of for practice, we have = 2C11 + (−1)C12 + (−2)C13 + 0 = 2(26) + (−1)(−13) + (−2)(3) + 0 = 59 Cramer’s Rule We next consider Cramer’s rule, which enables us to ﬁnd the values of the unknown variables. It is also useful in solving systems of equations where numerical coefﬁcients have not yet been speciﬁed, thus confounding our calculators. Let us again consider Eqs. , , and ; we deﬁne the deter-minant 1 as that determinant which is obtained when the ﬁrst column of G is replaced by the three constants on the right-hand sides of the three equations. Thus, 1 = −11 −3 −4 3 6 −2 25 −2 11 We expand along the ﬁrst column: 1 = −11 6 −2 −2 11 −3 −3 −4 −2 11 + 25 −3 −4 6 −2 = −682 + 123 + 750 = 191 APPENDIX 2 SOLUTION OF SIMULTANEOUS EQUATIONS 809 Cramer’s rule then states that v1 = 1 G = 191 191 = 1 V and v2 = 2 G = 7 −11 −4 −3 3 −2 −4 25 11 = 581 −63 −136 191 = 2 V and ﬁnally, v3 = 3 G = 7 −3 −11 −3 6 3 −4 −2 25 = 1092 −291 −228 191 = 3 V Cramer’s rule is applicable to a system of N simultaneous linear equations in N unknowns; for the ith variable vi: vi = i G APPENDIX 2 SOLUTION OF SIMULTANEOUS EQUATIONS 810 PRACTICE ● A2.2 Evaluate (a) 2 −3 −2 5 ; (b) 1 −1 0 4 2 −3 3 −2 5 ; (c) 2 −3 1 5 −3 1 −1 0 0 4 2 −3 6 3 −2 5 ; (d) Find i2 if 5i1 −2i2 −i3 = 100, −2i1 + 6i2 −3i3 −i4 = 0, −i1 −3i2 + 4i3 −i4 = 0, and −i2 −i3 = 0. Ans: 4; 33; −411; 1.266. APPENDIX 3 • 811 A PROOF OF THÉVENIN’S THEOREM Here we prove Thévenin’s theorem in the same form in which it is stated in Sec. 5.4 of Chap. 5: A B (no inde-pendent sources) i (a) A (dead) B (no inde-pendent sources) i (c) + – vx = voc voc + – A B (no inde-pendent sources) 0 (b) + – vx ■FIGURE A3.1 (a) A general linear network A and a network B that contains no independent sources. Controls for dependent sources must appear in the same part of the network. (b) The Thévenin source is inserted in the circuit and adjusted until i = 0. No voltage appears across network B and thus v x = voc . The Thévenin source thus produces a current −i while network A provides i . (c) The Thévenin source is reversed and network A is killed. The current is therefore i . Given any linear circuit, rearrange it in the form of two networks A and B connected by two wires. Deﬁne a voltage voc as the open-circuit volt-age which appears across the terminals of A when B is disconnected. Then all currents and voltages in B will remain unchanged if all inde-pendent voltage and current sources in A are ‘‘killed’’ or ‘‘zeroed out,’’ and an independent voltage source voc is connected, with proper polar-ity, in series with the dead (inactive) A network. We will effect our proof by showing that the original A network and the Thévenin equivalent of the A network both cause the same current to ﬂow into the terminals of the B network. If the currents are the same, then the voltages must be the same; in other words, if we apply a certain current, which we might think of as a current source, to the B network, then the cur-rent source and the B network constitute a circuit that has a speciﬁc input voltage as a response. Thus, the current determines the voltage. Alterna-tively we could, if we wished, show that the terminal voltage at B is unchanged, because the voltage also determines the current uniquely. If the input voltage and current to the B network are unchanged, then it follows that the currents and voltages throughout the B network are also unchanged. Let us ﬁrst prove the theorem for the case where the B network is inactive (no independent sources).After this step has been accomplished, we may then use the superposition principle to extend the theorem to include B networks that contain independent sources. Each network may contain dependent sources, provided that their control variables are in the same network. The current i, ﬂowing in the upper conductor from the A network to the B network in Fig. A3.1a, is therefore caused entirely by the independent APPENDIX 3 A PROOF OF THÉVENIN’S THEOREM 812 sources present in the A network. Suppose now that we add an additional voltage source vx, which we shall call the Thévenin source, in the conduc-tor in which i is measured, as shown in Fig. A3.1b, and then adjust the mag-nitude and time variation of vx until the current is reduced to zero. By our deﬁnition of voc, then, the voltage across the terminals of A must be voc, since i = 0. Network B contains no independent sources, and no current is entering its terminals; therefore, there is no voltage across the terminals of the B network, and by Kirchhoff’s voltage law the voltage of the Thévenin source is voc volts, vx = voc. Moreover, since the Thévenin source and the A network jointly deliver no current to B, and since the A network by itself delivers a current i, superposition requires that the Thévenin source acting by itself must deliver a current of −i to B. The source acting alone in a reversed direction, as shown in Fig. A3.1c, therefore produces a current i in the upper lead. This situation, however, is the same as the conclusion reached by Thévenin’s theorem: the Thévenin source voc acting in series with the inactive A network is equivalent to the given network. Now let us consider the case where the B network may be an active net-work. We now think of the current i, ﬂowing from the A network to the B network in the upper conductor, as being composed of two parts, i A and iB, where i A is the current produced by A acting alone and the current iB is due to B acting alone. Our ability to divide the current into these two compo-nents is a direct consequence of the applicability of the superposition principle to these two linear networks; the complete response and the two partial responses are indicated by the diagrams of Fig. A3.2. A If: B (dead) iA (a) A then: B i = iA + iB (c) A and: B (dead) iB (b) ■FIGURE A3.2 Superposition enables the current i to be considered as the sum of two partial responses. The partial response iA has already been considered; if network B is in-active, we know that network A may be replaced by the Thévenin source and the inactive A network. In other words, of the three sources which we must keep in mind—those in A, those in B, and the Thévenin source—the partial response i A occurs when A and B are dead and the Thévenin source is active. Preparing for the use of superposition, we now let A remain inactive, but turn on B and turn off the Thévenin source; by deﬁnition, the partial re-sponse iB is obtained. Superimposing the results, the response when A is dead and both the Thévenin source and B are active is iA + iB. This sum is the original current i, and the situation wherein the Thévenin source and B are active but A is dead is the desired Thévenin equivalent circuit. Thus the active network A may be replaced by its Thévenin source, the open-circuit voltage, in series with the inactive A network, regardless of the status of the B network; it may be either active or inactive. APPENDIX 4 • 813 A PSpice® TUTORIAL SPICE is an acronym for Simulation Program with Integrated Circuit Emphasis. A powerful program, it is an industry standard and used through-out the world for a variety of circuit analysis applications. SPICE was origi-nally developed in the early 1970s by Donald O. Peterson and coworkers at the University of California at Berkeley. Interestingly, Peterson advocated free and unhindered distribution of knowledge created in university labs, choosing to make an impact as opposed to proﬁting ﬁnancially. In 1984, MicroSim Corporation introduced a PC version of SPICE called PSpice®, which built intuitive graphical interfaces around the core SPICE software routines. There are now several variations of SPICE available commercially, as well as competing software products. The goal of this appendix is to simply introduce the basics of computer-aided circuit analysis; more details are presented in the main text as well as in the references listed under Reading Further. Advanced topics covered in the references include how to determine the sensitivity of an output variable to changes in a speciﬁc component value; how to obtain plots of the output versus a source value; determining ac output as a function of source fre-quency; methods for performing noise and distortion analyses; nonlinear component models; and how to model temperature effects on speciﬁc types of circuits. The acquisition of MicroSim by OrCAD, and the subsequent acquisition of OrCAD by Cadence, has led to quite a few changes in this popular cir-cuit simulation package. At the time of this writing, OrCAD Capture CIS-Demo 16.3, is the current professional release; a scaled-back version is available for free download (www.cadence.com). This new version replaces the popular PSpice Student Release 9.1, and although slightly dif-ferent, particularly in terms of the schematic editing, should seem generally familiar to users of previous PSpice releases. The documentation which accompanies the demo version of OrCAD 16.3 lists several restrictions that do not apply to the professional (commercially available) version. The most signiﬁcant is that only circuits having 60 or fewer parts may be saved and simulated; larger circuits can be drawn and viewed, however. We have chosen to work with the OrCAD Capture schematics editor, as the current version is very similar fundamentally to the PSpice A/D Schematic Capture editor. Although at present Cadence also provides PSpice A/D for download, it is no longer supported. Getting Started A computer-aided circuit analysis consists of three separate steps: (1) draw-ing the schematic, (2) simulating the circuit, and (3) extracting the desired information from the simulation output. The OrCAD Capture schematic APPENDIX 4 A PSpice® TUTORIAL 814 editor is launched through the Windows programs list found under the menu; upon selecting OrCAD Capture CIS-Demo, the schematics editor opens, as shown in Fig. A4.1. ■FIGURE A4.1 Capture CIS-Demo window. Under the File menu, select New, then Project; the window of Fig. A4.2a appears. After providing a simulation ﬁlename and a directory path, the window of Fig. A4.2b appears (simply select the “blank project” option). We are now presented with the main schematics editor window, as in Fig. A4.3. (a) (b) ■FIGURE A4.2 (a) New Project window. ( b) Create PSpice Project window. APPENDIX 4 A PSpice® TUTORIAL 815 815 ■FIGURE A4.3 Main Capture CIS-Demo schematics page. At this point, we’re ready to draw a circuit, so let’s try a simple voltage divider for purposes of illustration. We will ﬁrst place the necessary com-ponents on the grid, and then wire them together. Pulling down the Place menu, we choose Part. Typing a lowercase ‘‘r’’ as shown, we click OK and are now able to move a resistor symbol around the schematic window using the mouse. A single left click places a resistor (named R1) at the mouse location; a second left click places a second resis-tor on our schematic (named R2). A single right click and selecting End Mode cancel further resistor placements. The second resistor does not have the appropriate orientation, but is easily manipulated by highlighting it with a single right click, then selecting Rotate. If we do not know the name of the desired part, we can browse through the library of parts provided. If 1 k resistors are not desired—for example, perhaps two 500 resistors were called for—we change the default values simply by double-clicking the ‘‘1k’’ next to the appropriate symbol. No voltage divider circuit is complete of course without a voltage source (vsrc). Double-clicking DC=, we choose a value of 9 V for our source. One further component is required: SPICE requires a reference (or ground) node to be speciﬁed. Clicking the GND symbol to the far right of the schematic win-dow, we choose 0/Source from the options. Our progress so far is shown in Fig. A4.4a; all that remains is to wire the components together. This is accomplished by selecting the Place wire (W) icon. The left and right mouse keys control each wire (some experimenting is called for here—afterward, se-lect any unwanted wire segments and hit the Delete key). Our ﬁnal circuit is shown in Fig.A4.4b. It is worth noting that the editor will allow the user to wire through a resistor (thus shorting it out), which can be difﬁcult to see. Generally a warning symbol appears before wiring to an inappropriate location. Prior to simulating our circuit, we save it by clicking the save icon or selecting Save from the File menu. From the PSpice menu, we select New Simulation Proﬁle, and type Voltage Divider in the dialog box that appears. The Simulation Settings dialog box that appears allows us to set parameters for a variety of types of simulations; for the present example we need to select Bias Point under the Analysis type menu. Once again pulling down the PSpice menu, we select Run. The simulation results are shown in Fig. A4.5. APPENDIX 4 A PSpice® TUTORIAL 816 ■FIGURE A4.5 Simulation results. Fortunately, our simulation yields the expected result—an even split of our source voltage across the two equal-valued resistors. We can also view the simulation results by selecting View Output File under the PSpice menu. Scrolling down to the end of this ﬁle, we see the following lines: NODE VOLTAGE NODE VOLTAGE (N00109) 9.0000 (N00116) 4.5000 where node 109 is the positive reference of our voltage source, and node 116 is the junction between the two resistors. This information is available at the top of the ﬁle. READING FURTHER Two very good books devoted to SPICE and PSpice simulation are: P. W. Tuinenga, SPICE: A Guide to Circuit Simulation and Analysis Using PSpice. Englewood Cliffs, N.J.: Prentice-Hall, 1995. R. W. Goody, OrCAD PSpice for Windows Volume 1: DC and AC Circuits, 3rd ed. Englewood Cliffs, N.J.: Prentice-Hall, 2001. An interesting history of circuit simulators, as well as Donald Peterson’s contri-butions to the ﬁeld, can be found in: T. Perry, “Donald O. Peterson [electronic engineering biography],” IEEE Spectrum 35 (1998) 22–27. (a) (b) ■FIGURE A4.4 (a) Parts placed on the grid. (b) Fully wired circuit, ready to simulate. APPENDIX 5 • 817 COMPLEX NUMBERS This appendix includes sections covering the deﬁnition of a complex num-ber, the basic arithmetic operations for complex numbers, Euler’s identity, and the exponential and polar forms of the complex number. We ﬁrst intro-duce the concept of a complex number. A5.1• THE COMPLEX NUMBER Our early training in mathematics dealt exclusively with real numbers, such as 4, −2 7, and π. Soon, however, we began to encounter algebraic equations, such as x2 = −3, which could not be satisﬁed by any real number. Such an equation can be solved only through the introduction of the imaginary unit, or the imaginary operator, which we shall designate by the symbol j. By deﬁnition, j2 = −1, and thus j = √−1, j3 = −j, j4 = 1, and so forth. The product of a real number and the imaginary operator is called an imag-inary number, and the sum of a real number and an imaginary number is called a complex number. Thus, a number having the form a + jb, where a and b are real numbers, is a complex number. We shall designate a complex number by means of a special single sym-bol; thus, A = a + jb. The complex nature of the number is indicated by the use of boldface type; in handwritten material, a bar over the letter is cus-tomary. The complex number A just shown is described as having a real component or real part a and an imaginary component or imaginary part b. This is also expressed as Re{A} = a Im{A} = b The imaginary component of A is not jb. By deﬁnition, the imaginary com-ponent is a real number. It should be noted that all real numbers may be regarded as complex numbers having imaginary parts equal to zero. The real numbers are there-fore included in the system of complex numbers, and we may now consider them as a special case. When we deﬁne the fundamental arithmetic opera-tions for complex numbers, we should therefore expect them to reduce to the corresponding deﬁnitions for real numbers if the imaginary part of every complex number is set equal to zero. Since any complex number is completely characterized by a pair of real numbers, such as a and b in the previous example, we can obtain some visual assistance by representing a complex number graphically on a rec-tangular, or Cartesian, coordinate system. By providing ourselves with a real axis and an imaginary axis, as shown in Fig. A5.1, we form a complex plane, or Argand diagram, on which any complex number can be repre-sented as a single point. The complex numbers M = 3 + j1 and N = 2 −j2 Mathematicians designate the imaginary operator by the symbol i , but it is customary to use j in electrical engineering in order to avoid confusion with the symbol for current. The choice of the words imaginary and complex is unfortunate. They are used here and in the mathematical literature as technical terms to designate a class of numbers. To interpret imaginary as ‘‘not pertaining to the physical world’’ or complex as ‘‘complicated’’ is neither justiﬁed nor intended. We deﬁne two complex numbers as being equal if, and only if, their real parts are equal and their imaginary parts are equal. Graphically, then, to each point in the complex plane there corresponds only one complex num-ber, and conversely, to each complex number there corresponds only one point in the complex plane. Thus, suppose we are given the two complex numbers A = a + jb and B = c + jd Then, if A = B it is necessary that a = c and b = d A complex number expressed as the sum of a real number and an imaginary number, such as A = a + jb, is said to be in rectangular or cartesian form. Other forms for a complex number will appear shortly. Let us now deﬁne the fundamental operations of addition, subtraction, multiplication, and division for complex numbers. The sum of two complex numbers is deﬁned as the complex number whose real part is the sum of the real parts of the two complex numbers and whose imaginary part is the sum of the imaginary parts of the two complex numbers. Thus, (a + jb) + (c + jd) = (a + c) + j(b + d) For example, (3 + j4) + (4 −j2) = 7 + j2 The difference of two complex numbers is taken in a similar manner; for example, (3 + j4) −(4 −j2) = −1 + j6 APPENDIX 5 COMPLEX NUMBERS 818 ■FIGURE A5.1 The complex numbers M = 3 + j1 and N = 2 −j2 are shown on the complex plane. are indicated. It is important to understand that this complex plane is only a visual aid; it is not at all essential to the mathematical statements which follow. j3 j2 j1 –j1 –j2 –1 1 2 3 M N 4 5 0 Real Imaginary APPENDIX 5 COMPLEX NUMBERS 819 819 Addition and subtraction of complex numbers may also be accomplished graphically on the complex plane. Each complex number is represented as a vector, or directed line segment, and the sum is obtained by completing the parallelogram, illustrated by Fig. A5.2a, or by connecting the vectors in a head-to-tail manner, as shown in Fig. A5.2b. A graphical sketch is often useful as a check for a more exact numerical solution. The product of two complex numbers is deﬁned by (a + jb)(c + jd) = (ac −bd) + j(bc + ad) This result may be easily obtained by a direct multiplication of the two binomial terms, using the rules of the algebra of real numbers, and then simplifying the result by letting j2 = −1. For example, (3 + j4)(4 −j2) = 12 −j6 + j16 −8 j2 = 12 + j10 + 8 = 20 + j10 It is easier to multiply the complex numbers by this method, particularly if we immediately replace j2 by −1, than it is to substitute in the general for-mula that deﬁnes the multiplication. Before deﬁning the operation of division for complex numbers, we should deﬁne the conjugate of a complex number. The conjugate of the complex number A = a + jb is a −jb and is represented as A∗. The con-jugate of any complex number is therefore easily obtained by merely chang-ing the sign of the imaginary part of the complex number. Thus, if A = 5 + j3 then A∗= 5 −j3 It is evident that the conjugate of any complicated complex expression may be found by replacing every complex term in the expression by its conju-gate, which may be obtained by replacing every j in the expression by −j. The deﬁnitions of addition, subtraction, and multiplication show that the following statements are true: the sum of a complex number and its conju-gate is a real number; the difference of a complex number and its conjugate is an imaginary number; and the product of a complex number and its conjugate is a real number. It is also evident that if A∗is the conjugate of A, then A is the conjugate of A∗; in other words, A = (A∗)∗. A complex number and its conjugate are said to form a conjugate complex pair of numbers. We now deﬁne the quotient of two complex numbers: A B = (A)(B∗) (B)(B∗) and thus a + jb c + jd = (ac + bd) + j(bc −ad) c2 + d2 ■FIGURE A5.2 (a) The sum of the complex numbers M = 3 + j 1 and N = 2 −j 2 is obtained by constructing a parallelogram. (b) The sum of the same two complex numbers is found by a head-to-tail combination. Inevitably in a physical problem a complex number is somehow accompanied by its conjugate. M N M + N = 5 – j1 Real Imaginary (a) M M + N = 5 – j1 N Real Imaginary (b) We multiply numerator and denominator by the conjugate of the de-nominator in order to obtain a denominator which is real; this process is called rationalizing the denominator. As a numerical example, 3 + j4 4 −j2 = (3 + j4)(4 + j2) (4 −j2)(4 + j2) = 4 + j22 16 + 4 = 0.2 + j1.1 The addition or subtraction of two complex numbers which are each ex-pressed in rectangular form is a relatively simple operation; multiplication or division of two complex numbers in rectangular form, however, is a rather unwieldy process. These latter two operations will be found to be much simpler when the complex numbers are given in either exponential or polar form. These forms will be introduced in Secs. A5.3 and A5.4. APPENDIX 5 COMPLEX NUMBERS 820 PRACTICE ● A5.1 Let A = −4 + j5, B = 3 −j2, and C = −6 −j5, and ﬁnd (a)C −B;(b)2A −3B + 5C;(c) j5C2(A + B);(d)B Re{A} + A Re{B}. A5.2 Using the same values for A, B, and C as in the previous problem, ﬁnd (a) [(A −A∗)(B + B∗)∗]∗; (b) (1/C) −(1/B)∗; (c) (B + C)/(2BC). Ans: A5.1: −9 −j3; −47 −j9; 27 −j191; −24 + j23. A5.2: −j60; −0.329 + j0.236; 0.0662 + j0.1179. A5.2• EULER’S IDENTITY In Chap. 9 we encounter functions of time which contain complex numbers, and we are concerned with the differentiation and integration of these func-tions with respect to the real variable t. We differentiate and integrate such functions with respect to t by exactly the same procedures we use for real functions of time. That is, the complex constants are treated just as though they were real constants when performing the operations of differentiation or integration. If f(t) is a complex function of time, such as f(t) = a cos ct + jb sin ct then df(t) dt = −ac sin ct + jbc cos ct and f(t) dt = a c sin ct −j b c cos ct + C where the constant of integration C is a complex number in general. It is sometimes necessary to differentiate or integrate a function of a complex variable with respect to that complex variable. In general, the successful accomplishment of either of these operations requires that the function which is to be differentiated or integrated satisfy certain condi-tions. All our functions do meet these conditions, and integration or differ-entiation with respect to a complex variable is achieved by using methods identical to those used for real variables. At this time we must make use of a very important fundamental rela-tionship known as Euler’s identity (pronounced ‘‘oilers’’). We shall prove this identity, for it is extremely useful in representing a complex number in a form other than rectangular form. The proof is based on the power series expansions of cos θ, sin θ, and ez, given toward the back of your favorite college calculus text: cos θ = 1 −θ2 2! + θ4 4! −θ6 6! + · · · sin θ = θ −θ3 3! + θ5 5! −θ7 7! + · · · or cos θ + j sin θ = 1 + jθ −θ2 2! −j θ3 3! + θ4 4! + j θ5 5! −· · · and ez = 1 + z + z2 2! + z3 3! + z4 4! + z5 5! + · · · so that e jθ = 1 + jθ −θ2 2! −j θ3 3! + θ4 4! + · · · We conclude that e jθ = cos θ + j sin θ or, if we let z = −jθ, we ﬁnd that e−jθ = cos θ −j sin θ By adding and subtracting Eqs. and , we obtain the two expressions which we use without proof in our study of the underdamped natural response of the parallel and series RLC circuits, cos θ = 1 2(e jθ + e−jθ) sin θ = −j 1 2(e jθ −e−jθ) APPENDIX 5 COMPLEX NUMBERS 821 PRACTICE ● A5.3 Use Eqs. through to evaluate (a) e−j1; (b) e1−j1; (c) cos(−j1); (d) sin(−j1). A5.4 Evaluateatt = 0.5:(a)(d/dt)(3 cos 2t −j2 sin 2t); (b) t 0(3 cos 2t −j2 sin 2t) dt; Evaluate at s = 1 + j2: (c) ∞ s s−3 ds; (d) (d/ds)[3/(s + 2)]. Ans: A5.3: 0.540 −j0.841; 1.469 −j2.29; 1.543; −j1.175. A5.4: −5.05 −j2.16; 1.262 −j0.460; −0.06 −j0.08; −0.0888 + j0.213. APPENDIX 5 COMPLEX NUMBERS 822 A5.3• THE EXPONENTIAL FORM Let us now take Euler’s identity e jθ = cos θ + j sin θ and multiply each side by the real positive number C: Ce jθ = C cos θ + jC sin θ The right-hand side of Eq. consists of the sum of a real number and an imaginary number and thus represents a complex number in rectangular form; let us call this complex number A, where A = a + jb. By equating the real parts a = C cos θ and the imaginary parts b = C sin θ then squaring and adding Eqs. and , a2 + b2 = C2 or C = + a2 + b2 and dividing Eq. by Eq. : b a = tan θ or θ = tan−1 b a we obtain the relationships of Eqs. and , which enable us to determine C and θ from a knowledge of a and b. For example, if A = 4 + j2, then we identify a as 4 and b as 2 and ﬁnd C and θ: C = 42 + 22 = 4.47 θ = tan−1 2 4 = 26.6◦ We could use this new information to write A in the form A = 4.47 cos 26.6◦+ j4.47 sin 26.6◦ but it is the form of the left-hand side of Eq. which will prove to be the more useful: A = Ce jθ = 4.47e j26.6◦ A complex number expressed in this manner is said to be in exponential form. The real positive multiplying factor C is known as the amplitude or magnitude, and the real quantity θ appearing in the exponent is called the argument or angle. A mathematician would always express θ in radians and would write A = 4.47e j0.464 but engineers customarily work in terms of degrees. The use of the degree symbol (◦) in the exponent should make confusion impossible. To recapitulate, if we have a complex number which is given in rectan-gular form, A = a + jb and wish to express it in exponential form, A = Ce jθ we may ﬁnd C and θ by Eqs. and . If we are given the complex num-ber in exponential form, then we may ﬁnd a and b by Eqs. and . When A is expressed in terms of numerical values, the transformation between exponential (or polar) and rectangular forms is available as a built-in operation on most handheld scientiﬁc calculators. One question will be found to arise in the determination of the angle θ by using the arctangent relationship of Eq. . This function is multivalued, and an appropriate angle must be selected from various possibilities. One method by which the choice may be made is to select an angle for which the sine and cosine have the proper signs to produce the required values of a and b from Eqs. and . For example, let us convert V = 4 −j3 to exponential form. The amplitude is C = 42 + (−3)2 = 5 and the angle is θ = tan−1 −3 4 Avalue of θ has to be selected which leads to a positive value for cos θ, since 4 = 5 cos θ, and a negative value for sin θ, since −3 = 5 sin θ. We therefore obtain θ = −36.9◦, 323.1◦, −396.9◦, and so forth. Any of these angles is correct, and we usually select that one which is the simplest, here, −36.9◦. We should note that the alternative solution of Eq. , θ = 143.1◦, is not correct, because cos θ is negative and sin θ is positive. A simpler method of selecting the correct angle is available if we repre-sent the complex number graphically in the complex plane. Let us ﬁrst se-lect a complex number, given in rectangular form, A = a + jb, which lies in the ﬁrst quadrant of the complex plane, as illustrated in Fig. A5.3. If we draw a line from the origin to the point which represents the complex num-ber, we shall have constructed a right triangle whose hypotenuse is evi-dently the amplitude of the exponential representation of the complex num-ber. In other words, C = √ a2 + b2. Moreover, the counterclockwise angle which the line makes with the positive real axis is seen to be the angle θ of the exponential representation, because a = C cos θ and b = C sin θ. Now if we are given the rectangular form of a complex number which lies in an-other quadrant, such as V = 4 −j3, which is depicted in Fig. A5.4, the cor-rect angle is graphically evident, either −36.9◦or 323.1◦for this example. The sketch may often be visualized and need not be drawn. APPENDIX 5 COMPLEX NUMBERS 823 Imaginary a = C cos C = a2 + b2 b = C sin Real ■FIGURE A5.3 A complex number may be represented by a point in the complex plane through choosing the correct real and imaginary parts from the rectangular form, or by selecting the magnitude and angle from the exponential form. –j3 C = 5 V 4 Imaginary Real = 323.1° = –36.9° ■FIGURE A5.4 The complex number V = 4 −j 3 = 5e−j 36.9◦is represented in the complex plane. If the rectangular form of the complex number has a negative real part, it is often easier to work with the negative of the complex number, thus avoiding angles greater than 90◦in magnitude. For example, given I = −5 + j2 we write I = −(5 −j2) and then transform (5 −j2) to exponential form: I = −Ce jθ where C = √ 29 = 5.39 and θ = tan−1 −2 5 = −21.8◦ We therefore have I = −5.39e−j21.8◦ The negative sign may be removed from the complex number by increasing or decreasing the angle by 180◦, as shown by reference to a sketch in the complex plane. Thus, the result may be expressed in exponential form as I = 5.39e j158.2◦ or I = 5.39e−j201.8◦ Note that use of an electronic calculator in the inverse tangent mode always yields angles having magnitudes less than 90◦. Thus, both tan−1[(−3) /4] and tan−1[3/(−4)] come out as −36.9◦. Calculators that provide rectangular-to-polar conversion, however, give the correct angle in all cases. One last remark about the exponential representation of a complex num-ber should be made. Two complex numbers, both written in exponential form, are equal if, and only if, their amplitudes are equal and their angles are equivalent. Equivalent angles are those which differ by multiples of 360◦. For example, if A = Ce jθ and B = De jφ, then if A = B, it is necessary that C = D and θ = φ ± (360◦)n, where n = 0, 1, 2, 3, . . . . APPENDIX 5 COMPLEX NUMBERS 824 PRACTICE ● A5.5 Express each of the following complex numbers in exponential form, using an angle lying in the range −180◦< θ ≤180◦: (a) −18.5 −j26.1; (b) 17.9 −j12.2; (c) −21.6 + j31.2. A5.6 Express each of these complex numbers in rectangular form: (a) 61.2e−j111.1◦; (b) −36.2e j108◦; (c) 5e−j2.5. Ans: A5.5: 32.0e−j125.3◦; 21.7e−j34.3◦; 37.9e j124.7◦. A5.6: −22.0 −j57.1; 11.19 −j34.4; −4.01 −j2.99. A5.4• THE POLAR FORM The third (and last) form in which we may represent a complex number is essentially the same as the exponential form, except for a slight difference in symbolism. We use an angle sign (/ ) to replace the combination e j. Thus, the exponential representation of a complex number A, A = Ce jθ may be written somewhat more concisely as A = C/θ The complex number is now said to be expressed in polar form, a name which suggests the representation of a point in a (complex) plane through the use of polar coordinates. It is apparent that transformation from rectangular to polar form or from polar form to rectangular form is basically the same as transformation between rectangular and exponential form. The same relationships exist between C, θ, a, and b. The complex number A = −2 + j5 is thus written in exponential form as A = 5.39e j111.8◦ and in polar form as A = 5.39/111.8◦ In order to appreciate the utility of the exponential and polar forms, let us consider the multiplication and division of two complex numbers repre-sented in exponential or polar form. If we are given A = 5/53.1◦ and B = 15/−36.9◦ then the expression of these two complex numbers in exponential form A = 5e j53.1◦ and B = 15e−j36.9◦ enables us to write the product as a complex number in exponential form whose amplitude is the product of the amplitudes and whose angle is the algebraic sum of the angles, in accordance with the normal rules for multi-plying two exponential quantities: (A)(B) = (5)(15)e j(53.1◦−36.9◦) or AB = 75e j16.2◦= 75/16.2◦ From the deﬁnition of the polar form, it is evident that A B = 0.333/90◦ Addition and subtraction of complex numbers are accomplished most eas-ily by operating on complex numbers in rectangular form, and the addition or subtraction of two complex numbers given in exponential or polar form should begin with the conversion of the two complex numbers to rectangu-lar form. The reverse situation applies to multiplication and division; two APPENDIX 5 COMPLEX NUMBERS 825 numbers given in rectangular form should be transformed to polar form, un-less the numbers happen to be small integers. For example, if we wish to multiply (1 −j3) by (2 + j1), it is easier to multiply them directly as they stand and obtain (5 −j5). If the numbers can be multiplied mentally, then time is wasted in transforming them to polar form. We should now endeavor to become familiar with the three different forms in which complex numbers may be expressed and with the rapid con-version from one form to another. The relationships among the three forms seem almost endless, and the following lengthy equation summarizes the various interrelationships A = a + jb = Re{A} + jIm{A} = Ce jθ = a2 + b2e j tan−1(b/a) = a2 + b2/tan−1(b/a) Most of the conversions from one form to another can be done quickly with the help of a calculator, and many calculators are equipped to solve linear equations with complex numbers. We shall ﬁnd that complex numbers are a convenient mathematical arti-ﬁce which facilitates the analysis of real physical situations. APPENDIX 5 COMPLEX NUMBERS 826 PRACTICE ● A5.7 Express the result of each of these complex-number manipula-tions in polar form, using six signiﬁcant ﬁgures just for the pure joy of calculating (a) [2 −(1/−41◦)]/(0.3/41◦); (b) 50/(2.87/83.6◦+ 5.16/63.2◦); (c) 4/18◦−6/−75◦+ 5/28◦. A5.8 Find Z in rectangular form if (a) Z + j2 = 3/Z; (b) Z = 2 ln(2 −j3); (c) sin Z = 3. Ans: A5.7: 4.69179/−13.2183◦; 6.318 33/−70.4626◦; 11.5066/54.5969◦. A5.8: ±1.414 −j1; 2.56 −j1.966; 1.571 ± j1.763. APPENDIX 6 • 827 A BRIEF MATLAB® TUTORIAL The intention of this tutorial is to provide a very brief introduction to some basic concepts required to use a powerful software package known as MATLAB. The use of MATLAB is a completely optional part of the mate-rial in this textbook, but as it is becoming an increasingly common tool in all areas of electrical engineering, we felt that it was worthwhile to provide students with the opportunity to begin exploring some of its features, par-ticularly in plotting 2D and 3D functions, performing matrix operations, solving simultaneous equations, and manipulating algebraic expressions. Many institutions now provide the full version of MATLAB for their stu-dents, but at the time of this writing, a student version is available at reduced cost from The MathWorks, Inc. ( academia/student_version/). Getting Started MATLAB is launched by clicking on the program icon; the typical opening window is shown in Fig. A6.1. Programs may be run from ﬁles or by directly entering commands in the window. MATLAB also has extensive online help resources, useful for both beginners and advanced users alike. Typical MATLAB programs very much resemble programs written in C, although familiarity with this language is by no means required. ■FIGURE A6.1 MATLAB command window upon start-up. APPENDIX 6 A BRIEF MATLAB® TUTORIAL 828 Second color has been used to differentiate program-generated text from user-generated text for the convenience of the reader only. We should note that the most recent versions of MATLAB use color to separate different types of text (functions, variables, etc.) and to highlight potential typographical errors. Variables and Mathematical Operations MATLAB makes a great deal more sense once the user realizes that all vari-ables are matrices, even if simply 1 × 1 matrices. Variable names can be up to 19 characters in length, which is extremely useful in constructing pro-grams with adequate readability. The ﬁrst character must be a letter, but all remaining characters can be any letter or number; the underscore (_) character may also be used. Variable names in MATLAB are case-sensitive. MATLAB includes several predeﬁned variables. Relevant predeﬁned vari-ables for the material presented in this text include: eps The machine accuracy realmin The smallest (positive) ﬂoating-point number handled by the computer realmax The largest ﬂoating-point number handled by the computer inf Inﬁnity (deﬁned as 10) NaN Literally, ‘‘Not a Number.’’This includes situations such as 0/0 pi π (3.14159 . . .) i, j Both are initially deﬁned as √−1. They may be assigned other values by the user A complete list of currently deﬁned variables can be obtained with the com-mand who. Variables are assigned by using an equal sign (=). If the state-ment is terminated with a semicolon (;), then another prompt appears. If the statement is simply terminated by a carriage return (i.e., by pressing the En-ter key), then the variable is repeated. For example, EDU» input_voltage = 5; EDU» input_current = 1e−3 input_current = 1.0000e−003 EDU» Complex variables are easy to deﬁne in MATLAB: for example, EDU» s = 9 + j5; creates a complex variable s with value 9 + j5. Amatrix other than a 1 × 1 matrix is deﬁned using brackets. For example, we would express the matrix t = 2 −1 3 0 in MATLAB as EDU» t = [2 −1; 3 0]; Note that the matrix elements are entered a row at a time; row elements are separated by a space, and rows are separated by a semicolon (;). The same arithmetic operations are available for matrices; so, for example, we may ﬁnd t + t as EDU» t + t ans = 4 −2 6 0 APPENDIX 6 A BRIEF MATLAB® TUTORIAL 829 829 ∧ power \ left division ∗ multiplication + addition / right (ordinary) division − subtraction Arithmetic operators include: The order of operations is important. The order of precedence is power, then multiplication and division, then addition and subtraction. EDU» x 1 + 5 ^ 2 3 x = 76 The concept of left division may seem strange at ﬁrst, but is very useful in matrix algebra. For example, EDU» 1/5 ans = 0.2000 EDU» 1\5 ans = 5 EDU» 5\1 ans = 0.2000 And, in the case of the matrix equation Ax = B, where A = 2 4 1 6 and B = −1 2 we ﬁnd x with EDU» A = [2 4; 1 6]; EDU» B = [−1; 2]; EDU» x = A\B x = −1.7500 0.6250 Alternatively, we can also write EDU» x = A^−1B x = −1.7500 0.6250 or EDU» inv(A)B ans = −1.7500 0.6250 When in doubt, parentheses can help a great deal. Some Useful Functions Space requirements prevent us from listing every function contained in MATLAB. Some of the more basic ones include: APPENDIX 6 A BRIEF MATLAB® TUTORIAL 830 830 abs(x) \|x\| log 10(x) log10 x exp(x) ex sin(x) sin x asin(x) sin−1 x sqrt(x) √x cos(x) cos x acos(x) cos−1 x log(x) ln x tan(x) tan x atan(x) tan−1 x Functions useful for manipulating complex variables include: real(s) Re{s} imag(s) Im{s} abs(s) √ a2 + b2, where s ≡a + jb angle(s) tan−1(b/a), where s ≡a + jb conj(s) complex conjugate of s Another extremely useful command, often forgotten, is simply help. Occasionally we require a vector, such as when we plan to create a plot. The command linspace(min, max, number of points) is invaluable in such instances: EDU» frequency = linspace(0,10,5) frequency = 0 2.5000 5.0000 7.5000 10.0000 A useful cousin is the command logspace(). Generating Plots Plotting with MATLAB is extremely easy. For example, Fig. A6.2 shows the result of executing the following MATLAB program: EDU» x = linspace(0,2pi,100); EDU» y = sin(x); EDU» plot(x,y); EDU» xlabel('Angle (radians)'); EDU» ylabel('f(x)'); APPENDIX 6 A BRIEF MATLAB® TUTORIAL 831 Writing Programs Although the MATLAB examples in this text are presented as lines typed into the Command Window, it is possible (and often prudent, if repetition is an issue) to write a program so that calculations are more convenient. This is accomplished in MATLAB by writing what is termed an m-ﬁle. This is simply a text file saved with a ‘‘.m’’ extension (for example, ﬁrst_program.m). In a nod to Kernighan and Ritchie, we pull down New Script under the File menu, which opens up the editor. (Note that you can use another editor, for example, WordPad, if you prefer.) We type in r = input('Hello, World') as shown in Fig. A6.3. ■FIGURE A6.3 Example m-ﬁle created in m-ﬁle editor. ■FIGURE A6.2 An example plot of sin(x), 0 < x< 2π, generated using MATLAB. The variable x is a vector comprised of 100 equally spaced elements. APPENDIX 6 A BRIEF MATLAB® TUTORIAL 832 We leave it to the reader to choose when to write a program/m-ﬁle and when to simply use the Command Window directly. READING FURTHER There are a large number of excellent MATLAB references available, with new titles appearing regularly. Two worth looking at are: D. C. Hanselman and B. L. Littleﬁeld, Mastering MATLAB 7. Upper Sad-dle River, N.J.: Prentice-Hall, 2005. W. J. Palm, III, Introduction to MATLAB 7 for Engineers, 2nd ed. New York: McGraw-Hill, 2005. ■FIGURE A6.4 Example m-ﬁle named example1.m for generating sine wave plot. We next save it as ﬁrst_program in a convenient directory, taking care to select MATLAB Files (.m) under File Type. Under the File menu, we select Open, and ﬁnd ﬁrst_program.m. This reopens the editor (so we could have skipped closing it earlier). We run our program by hitting f5 or selecting Run under the Debug menu. In the Command Window, we see our greeting; MATLAB is waiting for a keyboard response, so just hit the Enter key. Let’s expand a previous example to allow the magnitude to be user-selected as in Fig. A6.4. We are now allowed to enter an arbitrary amplitude for our plot. APPENDIX 7 • 833 ADDITIONAL LAPLACE TRANSFORM THEOREMS In this appendix, we brieﬂy present several Laplace transform theorems typically used in more advanced situations in addition to those described in Chap. 14. Transforms of Periodic Time Functions The time-shift theorem is very useful in evaluating the transform of periodic time functions. Suppose that f (t) is periodic with a period T for positive values of t. The behavior of f (t) for t < 0 has no effect on the (one-sided) Laplace transform, as we know. Thus, f (t) can be written as f (t) = f (t −nT) n = 0, 1, 2, . . . If we now deﬁne a new time function which is nonzero only in the ﬁrst period of f (t), f1(t) = [u(t) −u(t −T)] f (t) then the original f (t) can be represented as the sum of an inﬁnite number of such functions, delayed by integral multiples of T . That is, f (t) = [u(t) −u(t −T)] f (t) + [u(t −T) −u(t −2T)] f (t) + [u(t −2T) −u(t −3T)] f (t) + · · · = f1(t) + f1(t −T) + f1(t −2T) + · · · or f (t) = ∞ n=0 f1(t −nT) The Laplace transform of this sum is just the sum of the transforms, F(s) = ∞ n=0 { f1(t −nT)} so that the time-shift theorem leads to F(s) = ∞ n=0 e−nTsF1(s) where F1(s) = { f1(t)} = T 0−e−st f (t) dt Since F1(s) is not a function of n, it can be removed from the summation, and F(s) becomes F(s) = F1(s)[1 + e−Ts + e−2Ts + · · ·] When we apply the binomial theorem to the bracketed expression, it sim-pliﬁes to 1/(1 −e−Ts). Thus, we conclude that the periodic function f (t), with period T, has a Laplace transform expressed by F(s) = F1(s) 1 −e−Ts where F1(s) = {[u(t) −u(t −T)] f (t)} is the transform of the ﬁrst period of the time function. To illustrate the use of this transform theorem for periodic functions, let us apply it to the familiar rectangular pulse train, Fig. A7.1. We may de-scribe this periodic function analytically: v(t) = ∞ n=0 V0[u(t −nT) −u(t −nT −τ)] t > 0 APPENDIX 7 ADDITIONAL LAPLACE TRANSFORM THEOREMS 834 ■FIGURE A7.1 A periodic train of rectangular pulses for which F(s) = (V 0/s)(1 −e−sτ )/(1 −e−sT ). 0 v(t) V0 t T T + 2T 2T + The function V1(s) is simple to calculate: V1(s) = V0 τ 0−e−st dt = V0 s (1 −e−sτ) Now, to obtain the desired transform, we just divide by (1 −e−sT ): V(s) = V0(1 −e−sτ) s(1 −e−sT ) We should note how several different theorems show up in the transform in Eq. . The (1 −e−sT ) factor in the denominator accounts for the periodic-ity of the function, the e−sτ term in the numerator arises from the time de-lay of the negative square wave that turns off the pulse, and the V0/s factor is, of course, the transform of the step functions involved in v(t). APPENDIX 7 ADDITIONAL LAPLACE TRANSFORM THEOREMS 835 835 Frequency Shifting The next new theorem establishes a relationship between F(s) = { f (t)} and F(s + a). We consider the Laplace transform of e−at f (t), {e−at f (t)} = ∞ 0−e−ste−at f (t) dt = ∞ 0−e−(s+a)t f (t) dt Looking carefully at this result, we note that the integral on the right is iden-tical to that deﬁning F(s) with one exception: (s + a) appears in place of s. Thus, e−at f (t) ⇔F(s + a) EXAMPLE A7.1 Determine the transform of the periodic function of Fig. A7.2. We begin by writing an equation which describes f (t), a function composed of alternating positive and negative impulse functions. f (t) = 2δ(t −1) −2δ(t −3) + 2δ(t −5) −2δ(t −7) + · · · Deﬁning a new function f1 and recognizing a period T = 4 s, f1(t) = 2[δ(t −1) −δ(t −3)] we can make use of the time periodicity operation as listed in Table 14.2 to ﬁnd F(s) F(s) = 1 1 −e−Ts F1(s) where F1(s) = T 0−f (t)e−st dt = 4 0−f1(t)e−st dt There are several ways to evaluate this integral. The easiest is to recognize that its value will remain the same if the upper limit is increased to ∞, allowing us to make use of the time-shift theorem. Thus, F1(s) = 2[e−s −e−3s] Our example is completed by multiplying Eq. by the factor indicated in Eq. , so that F(s) = 2 1 −e−4s (e−s −e−3s) = 2e−s 1 + e−2s PRACTICE ● A7.1 Determine the Laplace transform of the periodic function shown in Fig. A7.3. Ans: 8 s2 + π2/4 s + (π/2)e−s + (π/2)e−3s −se−4s 1 −e−4s . 0 1 2 3 4 5 (2) (2) 6 7 (–2) (–2) 8 f(t) t (s) ■FIGURE A7.2 A periodic function based on unit-impulse functions. 0 1 2 3 4 (cosine) 5 6 7 t (s) 8 f (t) ■FIGURE A7.3 We conclude that replacing sby (s + a)in the frequency domain corresponds to multiplication by e−at in the time domain. This is known as the frequency-shift theorem. It can be put to immediate use in evaluating the transform of the exponentially damped cosine function that we used extensively in previ-ous work. Beginning with the known transform of the cosine function, {cos ω0t} = F(s) = s s2 + ω2 0 then the transform of e−at cos ω0t must be F(s + a): {e−at cos ω0t} = F(s + a) = s + a (s + a)2 + ω2 0 APPENDIX 7 ADDITIONAL LAPLACE TRANSFORM THEOREMS 836 Differentiation in the Frequency Domain Next let us examine the consequences of differentiating F(s) with respect to s. The result is d dsF(s) = d ds ∞ 0−e−st f (t) dt = ∞ 0−−te−st f (t) dt = ∞ 0−e−st[−t f (t)] dt which is simply the Laplace transform of [−t f (t)]. We therefore conclude that differentiation with respect to s in the frequency domain results in mul-tiplication by −t in the time domain, or −t f (t) ⇔d dsF(s) Suppose now that f (t) is the unit-ramp function tu(t), whose transform we know is 1/s2. We can use our newly acquired frequency-differentiation the-orem to determine the inverse transform of 1/s3 as follows: d ds 1 s2 = −2 s3 ⇔−t−1 1 s2 = −t2u(t) and t2u(t) 2 ⇔1 s3 Continuing with the same procedure, we ﬁnd t3 3!u(t) ⇔1 s4 and in general t(n−1) (n −1)!u(t) ⇔1 sn PRACTICE ● A7.2 Find {e−2t sin(5t + 0.2π)u(t)}. Ans: (0.588s + 4.05)/(s2 + 4s + 29). Integration in the Frequency Domain The effect on f (t) of integrating F(s) with respect to s may be shown by be-ginning with the deﬁnition once more, F(s) = ∞ 0−e−st f (t) dt performing the frequency integration from s to ∞, ∞ s F(s) ds = ∞ s ∞ 0−e−st f (t) dt ds interchanging the order of integration, ∞ s F(s) ds = ∞ 0− ∞ s e−st ds f (t) dt and performing the inner integration, ∞ s F(s) ds = ∞ 0− −1 t e−st ∞ s f (t) dt = ∞ 0− f (t) t e−st dt Thus, f (t) t ⇔ ∞ s F(s) ds For example, we have already established the transform pair sin ω0t u(t) ⇔ ω0 s2 + ω2 0 Therefore, sin ω0t u(t) t = ∞ s ω0 ds s2 + ω2 0 = tan−1 s ω0 ∞ s and we have sin ω0t u(t) t ⇔π 2 −tan−1 s ω0 APPENDIX 7 ADDITIONAL LAPLACE TRANSFORM THEOREMS 837 PRACTICE ● A7.3 Find {t sin(5t + 0.2π)u(t)}. Ans: (0.588s2 + 8.09s −14.69)/(s2 + 25)2. PRACTICE ● A7.4 Find {sin2 5tu(t)/t}. Ans: 1 4 ln[(s2 + 100)/s2]. The Time-Scaling Theorem We next develop the time-scaling theorem of Laplace transform theory by evaluating the transform of f (at), assuming that { f (t)} is known. The procedure is very simple: { f (at)} = ∞ 0−e−st f (at) dt = 1 a ∞ 0−e−(s/a)λf (λ) dλ where the change of variable at = λ has been employed. The last integral is recognizable as 1/a times the Laplace transform of f (t), except that s is re-placed by s/a in the transform. It follows that f (at) ⇔1 a F s a As an elementary example of the use of this time-scaling theorem, consider the determination of the transform of a 1 kHz cosine wave. Assuming we know the transform of a 1 rad/s cosine wave, cos t u(t) ⇔ s s2 + 1 the result is {cos 2000πt u(t)} = 1 2000π s/2000π (s/2000π)2 + 1 = s s2 + (2000π)2 APPENDIX 7 ADDITIONAL LAPLACE TRANSFORM THEOREMS 838 PRACTICE ● A7.5 Find {sin2 5t u(t)}. Ans: 50/[s(s2 + 100)]. A A1 and A2 values critical damping and, 335 overdamped parallel RLC circuit, 326–327 μA741 op amp, 193–194, 195, 198 ABCD parameters, two-port networks, 716–720, 730–731 abc phase sequence, 464–465 Absorbed power, 16, 19, 48–49 by element, 48–49 in resistors, 23–27 ac circuit analysis. See ac circuit power analysis; Circuit analysis ac circuit power analysis, 421–456. See also Complex power apparent power/power factor, 438–441, 453–454 average power. See Average power instantaneous power, 422–424, 447, 450–451 maximum average power, 431 RMS values of current/voltage, 433–438, 447 average power computations, 435 multiple-frequency circuits, 435–436 periodic waveform values, 433–434 sinusoidal waveform values, 434–435 sinusoidal excitation, instantaneous power, 423, 450–451 sinusoidal steady state theorem, 430–431 Active element, 217 Active ﬁlters, 669–670 Active network, 21 AD549K op amp, 193, 195 AD622 op amp, 206 Addition, Laplace transform operation, 561 Additive ﬂuxes, 497 Additive property, of the Laplace transform, 546 Admittance, 239, 572 parameters. See Two-port networks in sinusoidal steady-state, 394 Algebraic alternatives, complex forcing functions, 380–381 American Wire Gauge (AWG), 26 Ampère, A.M., 12 Amperes, 10, 11, 12 Ampliﬁers, equivalent networks and, 704–706 Amplitude exponential form of complex number, 822–824 of response, proportional forcing function, 376 of sinusoids, 371 Amprobe, 443 Analysis of circuits. See Circuit analysis computer-aided. See Computer-aided analysis deﬁned, 5–6 Fourier circuit. See Fourier circuit analysis mesh. See Nodal and mesh analysis nodal. See Nodal and mesh analysis power. See ac circuit power analysis PSpice Type command, 105 sinusoidal steady-state. See Sinusoidal steady-state analysis transient, 3, 4, 270–272 Analytical Engine, 6 Angles, exponential complex numbers, 822–824 Angular frequency, of sinusoids, 371 Anode, 189 Apparent power, 439, 443, 447 power factor and, 438–441, 453–454 Argand diagram, 817–818 Argument exponential form of complex number, 822–824 of sinusoids, 371 Arrows, for current, 9, 13 Asymptotes, Bode diagrams and, 650–651 Attenuator, 178, 609 Automotive suspensions, modeling, 358 Auxiliary equation, 323 Average power, 443, 447 ac circuits, 424–433, 447, 450–452 ideal resistor absorption of, 428 maximum, 431 maximum transfer of, 430–432 nonperiodic functions, 431–433 periodic waveforms, 425–426 reactive element absorption of, 428–429 RMS value and, 435 in the sinusoidal steady state, 426–427 superposition and, 433 12AX7A vacuum tube, 176 B B1 and B2 values, 339–340 Babbage, Charles, 6 Balanced load, 458 Balanced three-phase system, 458 Bandpass ﬁlters, 665, 667–669 Band-reject ﬁlters, 673 Bandstop ﬁlters, 665 Bandwidth, and high-Q circuits, 628–633, 680–681 Base, of transistors, 715 Basic components and electric circuits, 9–38 charge, 11–12, 30–33 current. See Current Ohm’s law. See Ohm’s law power. See Power units and scales, 9–11, 29–30 voltage. See Voltage INDEX • 839 Note: Page numbers followed by an “n” refer to footnotes. INDEX 840 Bass, treble, and midrange ﬁlters, 671–672 Beaty, H. Wayne, 29 Bias Point command (PSpice), 105 Bilateral circuit, 698 Bilateral element, 698 Bode, Hendrik W., 649 Bode diagrams/plots, 648–664, 683–684 additional considerations, 653–657 asymptotes, determining, 650–651 complex conjugate pairs, 658–661 computer-aided analysis for, 661–664 decibel (dB) scale, 649 higher-order terms and, 657 multiple terms in, 651 phase response and, 652–653 smoothing of, 651 Bossanyi, E., 486 Boyce, W.E., 308 Branch current, 94 Branches, deﬁned, 791 Break frequency, 651 Buffer design, 180 Burton, T., 486 Butterworth ﬁlters, 673–674 Butterworth polynomials, 673 C Candela, 10 Capacitors, 217–225 deﬁned, 218 duality. See Duality energy storage, 222–224 ideal, 217–220, 225 integral voltage-current relationships, 220–222, 249–252 linearity, consequences of, 238–240, 254–257 modeling of ideal capacitors, 217–220 with PSpice, 245–247, 259–260 in the s-domain, 575–576 op amp circuits with, 240–241, 257–258 in parallel, 237–238 phasor relationships for, 387–388 s-domain circuits and, 575–577 in series, 236–237 Cartesian form, complex numbers, 818 Cascaded op amps, 184–187, 210–212, 609 Cathode, 189 Cavendish, Henry, 22 cba phase sequence, 464–465 Characteristic equation, 265–267, 323 Charge, 11–12, 30–33 conservation of, 11, 157 distance and, 5 Chassis ground, 65–66 Chebyshev ﬁlters, 673–674 Chebyshev polynomials, 673 Chua, L.O., 234 Circuit analysis. See also Circuit analysis techniques engineering and, 4–5 linear. See Linear circuits nonlinear. See Nonlinear circuit analysis in the s-domain. See s-domain circuit analysis software, 7. See also Computer-aided analysis Circuit analysis techniques, 123–174 delta-wye conversion, 154–156, 170–172 linearity and superposition, 123–133, 159–162 maximum power transfer, 152–154, 168–170 Norton equivalent circuits. See Thévenin/Norton equivalent circuits selection process for, 157–158, 172–173 source transformations. See Source transformations superposition. See Superposition Thévenin equivalent circuits. See Thévenin/Norton equivalent circuits Circuits analysis of. See Circuit analysis components of. See Basic components and electric circuits elements of, 17–18, 21 networks and, 21–22 response résumé, source-free series RLC, 346–347 transfer functions for, 499 Clayton, G., 612 Closed-loop operation, op amps, 203 Closed-loop voltage gain, 193 Closed paths, 43, 92 Coefﬁcient of mutual inductance, 494 Coils, in wattmeters, 476–477 Collectors, 715 Column matrix, 804 Common-emitter conﬁguration, 715 Common mode rejection ratio (CMRR), op amps, 195–196 Comparators, 203–204, 214–215 Complementary function, source-free RL circuits, 262 Complementary solution. See Natural responses Complete response, 733–734 driven RL circuits, 291–295, 317–319 to periodic forcing functions, 748–750 of RLC circuits. See RLC circuits Complex conjugate pairs, Bode diagrams and, 658–661 Complex forcing function. See Sinusoidal steady-state analysis Complex form, of Fourier series, 750–757 Complex frequency, 324 dc case, 535 deﬁned, 533–537 exponential case, 535 exponentially damped sinusoids, 536 general form, 534–535, 565–566 neper frequency, 534, 537 radian frequency, 537 s-domain circuit analysis and, 598–606 at complex frequencies, 603 graphing and, 599, 617–618 natural response and, 602–606, 618 general perspective, 604 special case, 605 operating at complex frequencies, 603 pole-zero constellations, 600–602 response as a function of σ, 598–599 s in relation to reality, 536–537 sinusoidal case, 535 Complex numbers, 819–828 arithmetic operations for, 818–820 described, 817–818 Euler’s identity, 820–821 exponential form of, 822–824 INDEX 841 imaginary unit (operator), 817 polar form of, 824–826 rectangular (cartesian) form of, 818 Complex plane, 817–818 s-domain circuit analysis and. See Complex frequency Complex power, 441–447, 454–455 apparent power, 439, 443, 447 and power factor, 438–441, 453–454 average power, 443 complex power, 441, 443 formula, 441–442 measuring, 443–444 power factor, 438–441, 453–454 correction, 444–445 power factor (PF) lagging, 439 leading, 439 power triangle, 442–443 quadrature component, 443 quadrature power, 443 reactive power, 441, 442–443, 447 terminology, 447 volt-ampere (VA), 439 volt-ampere-reactive (VAR) units, 442 watt (W), 447 Complex representation, phasor as abbreviation for, 383 Components. See Basic components and electric circuits Computer-aided analysis, 6–7, 130–133. See also MATLAB; PSpice Bode diagrams and, 661–664 fast Fourier Transform, 774–777 Laplace transforms and, 551–553 magnetically coupled circuits, 510–512 nodal and mesh analysis, 103–107, 120–121, 578–580 op amps, 200–203 s-domain nodal and mesh analysis, 578–580 sinusoidal steady-state analysis, 404–405 source-free parallel RLC circuits, 344–345 source-free RL circuits, 270–272 system function, 774–777 for two port networks, 719–720 Conductance, 27–28, 394 Conformal matrices, 805 Conservation of charge, 11, 157 Conservation of energy, 14, 48, 157 Constant charge, 12 Controlled sources, of voltage/current, 18, 19–21 Convolution Laplace transform operation, 561, 595–596 s-domain circuit analysis and, 589–598 convolution integral, 591 four-step process for analysis, 589 graphical methods of, 592–593 impulse response, 589–590, 617 Laplace transform and, 595–596 realizable systems and, 591–592 transfer function comments, 597 Cooper, George R., 544n Corner frequency, 651 Cosines, sines converted to, 373 Cotree, 792–793 Coulomb, 11 Coupling coefﬁcient, 504 Cramer’s rule, 84, 809–810 Create command (PSpice), 105 Critical frequencies, s-domain circuit analysis, 589 Critically damped response, RLC circuits form of, 334–335 graphical representation, 336–337 source-free circuits parallel, 325, 347 series, 346–347 Current, 9, 11, 12–13, 30–33 actual direction vs. convention, 13 branch current, 94 capacitor voltage-current relationships, 220–222, 249–252 coil, 476 current-controlled current source, 18, 19–21 current-controlled voltage source, 18, 19–21 effective values of, 433–438, 452–453 gain, ampliﬁers, 704 graphical symbols for, 13 laws. See Voltage and current laws mesh, 92, 93–95, 505 response, resonance and, 622 sources controlled, 18, 19–21 practical, 135, 139–140 reliable, op amps, 190–192, 212–213 series/parallel connections, 51–55, 74 and voltage. See Voltage superposition applicable to, 433 types of, 13 and voltage division, 61–64, 76–77 Current level adjustment, ideal transformers for, 517 Cutoff frequency, transistor ampliﬁer, 398–399 D Damped sinusoidal forcing function, 537–540, 566 Damped sinusoidal response, 338 Damping factor, parallel resonance and, 625–627 Damping out, of transients, 332 Davies, B., 565 3 dB frequency, 651 dc (direct current) analysis, 3 case, complex frequency, 535 current source, 19 parameter sweep, 130–133 short circuits to, 226 sources, 19, 175 Dead network, 144, 147 Decade (of frequencies), 650 DeCarlo, R.A., 109, 159, 410, 721 Decibel (dB) scale, Bode diagrams, 649 Delivered power, 19 Delta () connection, 470–476, 489–490 connected sources, 473–476 Y-connected loads vs., 473 Delta () of impedances, equivalent networks, 700–702 Delta-wye conversion, 154–156, 170–172 Dependent sources linear, 124 Thévenin/Norton equivalent circuits, 147–149 of voltage/current, 18, 19–21 Derivative of the current voltage, 18 Design, deﬁned, 5–6 Determinants, 807–809 Difference ampliﬁer, 181–184, 195–196 summary, 182 Difference Engine, 6 Differential equations algebraic alternative, sinusoidal steady-state, 380–381 for source-free parallel RLC circuits, 322–324 Differential input voltage, 195 Digital integrated circuits, frequency limits in, 306–307 Digital multimeter (DMM), 150–151 DiPrima, R.C., 308 Direct approach, source-free RL circuits, 262–263 Direction of travel, current, 12 Direct procedure, driven RL circuits, 287–289 Discrete spectrum, 742 Dissipation of power, 49 Distance, charge and, 5 Distinct poles, method of residues and, 548–549 Distributed-parameter networks, 39 Dot convention circuit transfer function, 499 mutual inductance, 495–499, 523–527 physical basis of, 497–500 power gain, 499 Double-subscript notation, polyphase circuits, 459–460 Drexler, H.B., 249 Driven RC circuits, 295–300 Driven RL circuits, 286–289, 315–316 complete response determination, 291–295, 317–319 direct procedure, 287–289 intuitive understanding of, 289 natural and forced response, 288, 289–295, 316–317 Duality, 233, 242–245, 258–259 E Earth ground, 65–66 Edison, Thomas, 457 Effective (RMS) value. See RMS value Electric circuits. See Circuits Emitters, 715 Energy, 14 accounting, source-free RL circuits, 267 conservation of, 14, 48, 157 density, 763 instantaneous, stored, 624 magnetically coupled circuits. See Magnetically coupled circuits storage capacitors, 222–224 storage inductors, 231–233 work units, 10 Engineering, circuit analysis and, 4–5 Engineering units, 11 ENIAC, 6 Equivalent circuits, ideal transformers, 519–521 Equivalent combinations, frequency response and, 639–644 Equivalent networks, two-port. See Two-port networks Equivalent practical sources, 135–138 Equivalent resistance, 56, 144 Equivalent voltage sources, 133 Euler’s identity, 380, 383, 441 Even functions, 745n Even harmonics, 745, 745n Even symmetry, Fourier series analysis, 743, 747 Exponential case, complex frequency, 535 Exponential damping coefﬁcient, 324, 621 Exponential form, complex numbers, 822–824 Exponential function eαt, 545 Exponentially damped sinusoids, 536 Exponential response, RL circuits, 268–272, 310 F Fairchild Corp., 175, 193 Fall time, of wave forms, 300 farad (F), 218 Faraday, Michael, 218n, 225, 226 Fast Fourier transform (FFT), 772, 774–777 image processing example, 780 Feedback control, 5 Feynman. R., 67 Fiber optic intercom, 183–184 Filters (frequency), 664–672, 684–685 active, 669–670 bandpass, 665, 667–669 band-reject, 673 bandstop, 665 bass/treble/midrange adjustment, 671–672 Butterworth, 673–674 Chebyshev, 673–674 higher order, 672–677, 685 high-pass, 665–666, 676 low-pass, 665–666, 674 multiband, 665 notch, 665 passive deﬁned, 669 low-pass and high-pass, 665–666 practical application, 671–672 Final-value, Laplace transforms, 562–563 Finite resistance, underdamped source-free parallel RLC, 340–342 Finite wire impedance, 461 Fink, Donald G., 29 Flowchart, for problem-solving, 8 Force, voltage and, 5 Forced responses, 371, 733–734 driven RL circuits, 288, 316–317 to sinusoids. See Sinusoidal steady-state analysis source-free RL circuits, 262 Forcing functions, 124 sinusoidal waveform as, 371 source-free RL circuits, 262 Forms of responses critically damped RLC circuits, 334–335 underdamped source-free parallel RLC circuits, 338–339 Fourier circuit analysis, 4, 733–790. See also Fourier series; Fourier transform complete response to periodic forcing functions, 748–750 image processing, 780–781 practical application, 780–781 Fourier series coefﬁcients, 737–738 complex form, 750–757 sampling function, 754–757 symmetry, use of, 743–747 INDEX 842 INDEX 843 even and odd symmetry, 743, 747 Fourier terms and, 743–745 half-wave symmetry, 745–746, 747 for simpliﬁcation purposes, 747 trigonometric form of, 733–743 coefﬁcients, evaluating, 737–738 derived, 735–736 equation for, 736 harmonics, 734–735 integrals, useful, 736–737 line spectra, 741–742 phase spectra, 742–743 Fourier transform. See also Fourier transform pairs deﬁned, 757–761 fast Fourier transform (FFT), 772, 774–777 image processing example, 780 of general periodic time function, 769–770 physical signiﬁcance of, 762–763 properties of, 761–764 system function, frequency domain. See System function Fourier transform pairs, 759 for constant forcing function, 766 for signum function, 766–767 summary of, 768 for unit-impulse function, 764–766 for unit step function, 767 Free response, source-free RL circuits, 262 Frequency angular, of sinusoids, 371 complex. See Complex frequency cutoff, transistor ampliﬁer, 398–399 differentiation, Laplace transforms, 561, 836–837 domain. See Frequency domain fundamental frequency, 734 integration, Laplace transforms, 561, 837 limits, digital integrated circuits, 306–307 multiple, RMS value with, 435–436 natural resonant, 338–339 op amps and, 199–200 radian, of sinusoids, 371 response. See Frequency response scaling, 644–648, 682–683 selectivity, parallel resonance and, 629 shift, Laplace transforms, 561, 835–836 of sinusoids, 372–373 source-free parallel RLC circuits, 324–325 unit deﬁnitions for, 324 Frequency domain phasor representation, 384 system function and, 770–777 time domain converted to, 539 V-I expressions, phasor relationships and, 387 Frequency response, 3, 4, 619–686 Bode diagrams. See Bode diagram/plots equivalent series/parallel combinations, 639–644 ﬁlters. See Filters (frequency) parallel resonance. See Parallel resonance resonant forms, other, 637–644, 682 scaling, 644–648, 682–683 series resonance, 633–636, 681 Friction coefﬁcient, 5 Fundamental frequency, 734 G Gain, of op amps, 607 General Conference on Weights and Measures, 9–10 General form, complex frequency, 534–535, 565–566 General practical voltage source, 134 General RC circuits, 279–282 General RL circuits, 275–276, 312–315 General solution, source-free RL circuits, 264–265 George A. Philbrick Researches, Inc., 208 Global positioning systems (GPS), 607 Goody, R.W., 363, 816 Graphics/Graphical on complex-frequency (s) plane, 599, 617–618 of convolution, s-domain analysis, 592–593 of critically damped response, RLC circuits, 336–337 of current, symbols for, 13 overdamped response, RLC circuits, 331–332 underdamped response, RLC circuits, 340 Ground (neutral) connection, 65–66, 458 Groups, of independent sources, 125 H Half-power frequency, 651 Half-wave symmetry, Fourier, 745–746, 747 Hanselman, D.C., 832 Harmonics, Fourier, 734–735 Harper, C.A., 249 Hartwell, F.P., 67 Hayt, W.H., Jr., 207, 410, 721 Heathcote, M., 523 henry (H), 225 Henry, Joseph, 225 Higher order ﬁlters, 672–677, 685 Higher-order terms, Bode diagrams, 657 High-pass ﬁlters, 665, 676 passive, 665–666 High-Q circuits approximations for, 629–633 bandwidth and, 629–633, 680–681 Hilburn, J.L., 679 Homogeneity property, Laplace transforms, 546 Homogeneous linear differential equations, 261–262 H(s) Vout/Vin, synthesizing, 606–610, 618 Huang, Q., 679 Huelsman, L.P., 679 Hybrid parameters, two-port networks, 713–716, 729–730 I Ideal capacitor model, 217–220 Ideal inductor model, 225–229 Ideal operational ampliﬁers. See Operational ampliﬁers Ideal resistor, average power, absorption, 428 Ideal sources, of voltage, 18 Ideal transformers, 512–522 for current level adjustment, 517 equivalent circuits, 519–521 for impedance matching, 514 step-down transformers, 516 step-up transformers, 516 turns ratio of, 512–514 for voltage level adjustment, 515–516 voltage relationship in the time domain, 517–521, 530–532 Ideal voltage sources, 133–135 Image processing, Fourier analysis and, 780–781 Imaginary sources →imaginary responses, 379–380 Imaginary unit (operator)/component, 817 of complex forcing function, 378 of complex power, 441 imaginary sources →imaginary responses, 379–380 Immittance, 394 Impedance, 239, 571–572 input, 587 matching, 514 sinusoidal steady-state, 389–394, 414–415 deﬁned, 389 parallel impedance combinations, 388 reactance and, 357 resistance and, 390 series impedance combinations, 389 Impulse response, convolution and, 589–590, 617 Inactive network, 147 Independent current sources, 18, 19 Independent voltage sources, 18–19 Inductors/Inductance, 225–234, 252–254, 493 characteristics, ideal, 233 deﬁned, 225 duality. See Duality energy storage, 231–233 in the frequency domain, 572, 577 ideal inductor model, 225–229 inductive reactance, 376 inﬁnite voltage spikes, 229 integral voltage-current relationships, 229–231 linearity, consequences of, 238–240, 254–257 modeled, 245–247, 259–260, 572–575 in parallel, 236 phasor relationships for, 386, 413–414 in series, 235–236 in the time domain, 577 Inﬁnite voltage spikes, inductors and, 229 Initial value, Laplace transforms, 561–562 In-phase sinusoids, 372–373 Input bias, 195 Input impedance, 587 ampliﬁers, 704–706 one-port networks, 688–692 Input offset voltage, op amps, 198 Instantaneous charge, 12 Instantaneous power, 422–424, 447, 450–451 Instantaneous stored energy, parallel resonance and, 624 Instrumentation ampliﬁer, 204–206, 214–215 Integral of the current voltage, 18 Integral voltage-current relationships capacitors, 220–222, 249–252 inductors, 229–231 Internal generated voltage, 474 Internal resistance, 134 International System of Units (SI), 9–10 Intuitive understanding, driven RL circuits, 289 Inverse transforms. See Laplace transform(s) Inversion, of matrices, 806–807 Inverting ampliﬁer, 177, 182 Inverting input, 176 J Jenkins, N., 486 Johnson, D.E., 679 Joules, 10 Jung, W.G., 207, 249 K K2-W op amp, 176 Kaiser, C.J., 249 kelvin, 10 Kennedy, B.K., 523 Kilograms, 10 kilowatthour (kWh), 438 Kirchhoff, Gustav Robert, 40 Kirchhoff’s laws current law (KCL), 39, 40–42, 68–70 nodal analysis and, 80, 157 phasors and, 387–388 voltage law (KVL), 39, 42–46, 70–72 circuit analysis and, 157 in mesh analysis, 98 order of elements and, 55 Korn, G.A., 679 L Lagging power factor, 439 Lagging sinusoids, 372–373 Lancaster, D., 679 Laplace analysis, 4 Laplace transform(s), 533–570 computer-aided analysis, 551–553 convolution and, 595–596 damped sinusoidal forcing function, 537–540, 566 deﬁned, 540–543, 567 for exponential function eατ, 545 frequency-differentiation theorem, 836–837 frequency-integration theorem, 837 frequency-shift theorem, 835–836 initial-value/ﬁnal-value theorems, 561–563, 569–570 inverse transform techniques, 546–551, 568 distinct poles/method of residues, 548–549 linearity theorem, 546–547 for rational functions, 547–548 repeated poles, 550 one-sided, 542–543 operations, table of, 561 pairs, 559 of periodic time functions, 833–835 for ramp function tu(t), 545 sifting property, 545 of simple time functions, 543–546, 567 sinusoid theorem, 558 system stability theorem, 560 theorems for, 553–561, 568–569 time differentiation theorem, 553–554 time-integration theorem, 555–556 time-scaling theorem, 838 time-shift theorem, 558, 833–835 two-sided inverse Laplace transform, 542 two-sided Laplace transform, 541 for unit-impulse function α(t t0), 544–545 for unit-step function u(t), 544 LC circuit, lossless, 359–361, 369–370 INDEX 844 INDEX 845 Leading sinusoids, 372–373 Leighton, R.B., 67 LF411 op amp, 193, 200 Lin, P.M., 109, 159, 410, 721 Linden, D., 159 Linear circuits, 2–4 complex forcing functions, 379–380 conservation laws, 157 dc analysis, 3 frequency response analysis, 3, 4 linear voltage-current relationships, 123–124 transient analysis, 3, 4 Linear dependent source, 124 Linear elements, 123–124 Linear homogeneous differential equations, 261–262 Linearity, 123–124 consequences, capacitors/inductors, 238–240, 254–257 inverse transform theorem, 546–547 Linear resistor, 23 Linear transformers, 505–512, 528–530 primary mesh current, 505 reﬂected impedance, 505–506 secondary mesh current, 505 T and equivalent networks, 507–510 Linear voltage-current relationship, 123–124 Line spectra, Fourier series analysis, 741–742 Line terminals, 464 Line-to-line voltages, three-phase Y-Y connection, 465–466 Links, 792–793 loop analysis and, 797–802 Littleﬁeld, B.L., 832 LM324 op amp, 193 LM741 op amp, 200 LMC6035 op amp, 176 LMV321 dual op amp, 176 Loop analysis, links and, 797–802 deﬁned, 792 mesh analysis and, 92 Lossless LC circuit, 359–361, 369–370 Lower half-power frequency, 628 Low-pass ﬁlters, 665, 674 passive, 665–666 Lumped-parameter networks, 39 M M, upper limit for, 503 M12/M21 equality, magnetically coupled circuits, 502–503 Magnetically coupled circuits, 493–532. See also Transformers computer-aided analysis, 510–512 coupling coefﬁcient, 504 energy considerations, 501–504, 527–528 equality of M12 and M21, 502–503 ideal transformers. See Ideal transformers linear transformers, 505–512, 528–530 magnetic ﬂux, 493, 494, 497 mutual inductance. See Mutual inductance upper limit for M, establishing, 503 Magnetic ﬂux, 493, 494, 497 Magnitude exponential form of complex number, 822–824 scaling, 644–648, 682–683 Mancini, R., 207, 249, 612 MATLAB, 85, 551–553 tutorial, 827–832 Matrices determinants of, 807–809 inversion of, 806–807 matrix form of equations, 85 simultaneous equations, solving, 804–810 Maximum average power, 431 Maximum power transfer, 152–154, 168–170, 430–432 Maxwell, James Clerk, 218 McGillem, Clare D., 544n McLyman, W.T., 523 McPartland, B.J., 67 McPartland, J.P., 67 Memristor, 234 Mesh. See Nodal and mesh analysis Meters, 10 Method of residues, 548–549 Metric system of units, 10 microfarads (μF), 219 MicroSim Corporation, 103 Midrange ﬁlters, 671–672 Models/Modeling, 3 of automotive suspension systems, 358 of ideal capacitors, 217–220 of inductors ideal inductors, 225–229 with PSpice, 245–247, 259–260 in the s-domain, 572–575 of op amps, detailed, 192–194 Moles, 10 MOSFET, 22 Multiband ﬁlters, 665 Multiple-frequency circuits, RMS value with, 435–436 Multiple terms, in Bode diagrams, 651 Multiport network, 687. See also Two-port networks Mutual inductance, 493–501 additive ﬂuxes, 497 coefﬁcient of, 494 dot convention, 495–499, 523–527 circuit transfer function, 499 physical basis of, 497–500 power gain, 499 magnetic ﬂux, 493, 494, 497 self-inductance added to, 496 N 1N750 Zener diode, 189–190 2N3904, ac parameters, 716 Nanotechnology, 234 Napier, John, 534 NASA Dryden Space Flight Center, 6 National Bureau of Standards, 9 National Semiconductor Corp., 176, 200 Natural resonant frequency, 338–339, 622 Natural responses, 282, 371, 374, 733–734 and the complex-frequency (s) plane, 602–606, 618 driven RL circuits, 288, 289–295, 316–317 source-free RL circuits, 262 Negative charge, 11 Negative feedback op amps, 196–197 path, 607 Negative phase sequence, 464–465 Negative (absorbed) power, 16, 19 Negative resistances, 692 Neper frequency, 537 deﬁned, 324 Nepers (Np), 534 Networks, 21–22 active, 21 passive, 21 topology. See Network topology two-port. See Two-port networks Network topology, 791–802 links and loop analysis, 797–802 trees and general nodal analysis, 791–797 Neudeck, G.W., 207, 410, 721 Neutral (ground) connection, 458, 464 New Simulation Proﬁle command (PSpice), 105 Nodal and mesh analysis, 3, 79–122 compared, 101–103, 119–120 computer-aided, 103–107, 120–121, 578–580 location of sources and, 101 mesh analysis, 92–98, 114–117, 157 Kirchhoff’s voltage law applied to, 98 mesh current, 92, 93–95, 505 mesh deﬁned, 792 procedure, summarized, 98 supermesh, 98, 100–101, 117–118 nodal analysis, 3, 80–89, 109–112, 157 basic procedure, summary, 88–89 Kirchhoff s current law and, 80 nodes deﬁned, 40, 791 procedure, summarized, 98 reference node, 80 sinusoidal steady-state analysis, 394–397, 415–417 supermesh, 98, 100–101, 117–118 supernodes, 89–91, 112–114 trees and, 791–797 voltage source effects, 89–91, 112–114 node-base PSpice schematics, 106–107 s-domain circuit analysis and, 578–584, 613–615 computer-aided, 578–580 of sinusoidal steady-states, 394–397, 415–417 Noninverting ampliﬁer circuit, 182 output waveform, 178–179 Noninverting input, 176 Nonlinear circuit analysis, 2 Nonperiodic functions, average power for, 431–433 Nonplanar circuit, deﬁned, 792 Norton, E.L., 141 Norton equivalents. See Thévenin/Norton equivalent circuits Notch ﬁlters, 665 Number systems, units and scales, 9 Numerical value, of current, 12 O Octave (of frequencies), 650 Odd functions, 745n Odd harmonics, 745n Odd symmetry, Fourier series analysis, 743, 747 Øersted, Hans Christian, 225 Ogata, K., 565, 612 Ohm, Georg Simon, 22 Ohms (), 22 Ohm’s law, 22–28, 34–36 conductance, 27–28 deﬁned, 22 power absorption in resistors, 23–27 practical application, 25–26 resistance units deﬁned, 22 One-port networks, 687–692, 722–723 input impedance calculations for, 688–692 One-sided Laplace transform, 542–543 OPA690 op amp, 193, 199 Op amps. See Operational ampliﬁers Open circuit, 27–28 to dc, 219 impedance parameters, 708–709 Open-loop conﬁguration, op amps, 203 voltage gain, 192–193 Operating at complex frequencies, 603 Operational ampliﬁers, 175–216 μA741 op amp, 193–194, 195, 198 AD549K op amp, 193, 195 AD622 op amp, 206 capacitors with, 240–241, 257–258 cascaded stages, 184–187, 210–212 common mode rejection, 195–196 comparators, 203–204, 214–215 computer-aided analysis, 200–203 frequency and, 199–200 ideal, 176–184, 208–210 derivation of, 194–195 difference ampliﬁer, 181–184, 195–196 inverting ampliﬁer, 177, 182 noninverting ampliﬁer circuit, 178–179, 182 rules, 176 summary, 182 summing ampliﬁer, 180–181, 182 voltage follower circuit, 179, 182 input offset voltage, 198 instrumentation ampliﬁer, 204–206, 214–215 LF411 op amp, 193, 200 LM324 op amp, 193 LM741 op amp, 200 LMC6035 op amp, 176 LMV321 dual op amp, 176 modeling, 192–194 negative feedback, 196–197 OPA690 op amp, 193, 199 outputs depending on inputs, 176 packaging, 200 parameter values, typical, 193 Philbrick K2-W op amp, 176 positive feedback, 197 practical considerations, 192–203, 213 reliable current sources, 190–192, 212–213 reliable voltage sources, 188–190, 212–213 saturation, 197–198 slew rate, 199–200 tank pressure monitoring system, 186–187 Operations, Laplace transform, table of, 561 Order of elements, KVL and, 55 Oscillator, 607 circuit design, 607–608 function, 340 Out-of-phase sinusoids, 372–373 Output impedance, ampliﬁers, 705 Output resistance, 134 Overdamped response source-free parallel RLC circuits, 325, 326–333, 347, 363–365 A1 and A2 values, ﬁnding, 326–327 graphical representation of, 331–332 source-free series RLC circuits, 346–347 INDEX 846 INDEX 847 P Packages, op amp, 200 Pairs, Laplace transform, 559 Palm, W.J., III, 832 and T equivalent networks, 507–510 Parallel element combinations, 49 capacitors, 237–238 impedance combinations, 389–390 inductors, 236 series/parallel combination equivalents, 639–644 Parallel resonance, 619–627, 636, 679–680 bandwidth and high-Q circuits, 628–633, 680–681 current response and, 622 damping exponential coefﬁcient, 621 factor, 625–627 deﬁned, 620–622 frequency selectivity, 629 instantaneous stored energy, 624 key conclusions on, 633 natural resonant frequency, 622 quality factor (Q), 623–627 bandwidth and, 628–633, 680–681 damping factor and, 625–627 other interpretations of Q, 625 summary of, 636 voltage response and, 622–623 Parameter values, op amps, 193 Parseval-Deschenes, Marc Antione, 762 Particular integral, 291 Particular solution, 291 source-free RL circuits, 262 Passband, 665 Passive element, 217 Passive ﬁlters deﬁned, 669 low-pass and high-pass, 665–666 Passive network, 21 Passive sign convention, 16 Path deﬁned, 791 mesh analysis, 92 voltage, 14 Periodic functions/waveforms, 432. See also Sinusoidal steady-state analysis; Sinusoidal waveforms ac average power of, 425–426 complete response to, 748–750 fall time of, 300 as forcing functions, 371 Laplace transforms of, 833–835 as output, noninverting ampliﬁers, 178–179 period T of, 300, 372 pulse width of, 300 rise time of, 300 RMS values for, 433–434 time delay of, 300 Perry, T., 816 Peterson, Donald O., 813 Phase angle θ, 372 Phase comparison, sinusoidal waves, 373 Phase response, Bode diagrams and, 652–653 Phase spectra, Fourier series analysis, 742–743 Phase voltages, 464 Phasor(s), 4, 384, 413–414, 571. See also Phasor relationships for R, L, and C diagrams, sinusoidal steady-states, 406–408, 419 Phasor relationships for R, L, and C as abbreviated complex representation, 383 capacitors, 387–388 frequency-domain representation, 384 frequency-domain V-I expressions, 387 impedance deﬁned from. See Sinusoidal steady-state analysis inductors, 386, 413–414 Kirchhoff’s laws using, 387–388 phasor representation, 384 resistors, 385–386 time-domain representation, 384 time-domain V-I expressions, 387 Philbrick, George A., 208 Philbrick K2-W op amp, 176 Philbrick Researches, Inc., 175 Physically realizable systems, 591–592 Physical signiﬁcance, of Fourier transforms, 762–763 Physical sources, unit-step function and, 284–285 Pinkus, A., 565, 783 Planar circuit, 92, 101 deﬁned, 792 Polar form, of complex numbers, 824–826 Poles, 547 method of residues and, 548–549 pole-zero constellations, 600–602 repeated, inverse transforms, 550 zeros, and transfer functions, 588–589, 616–617 Polya, G., 8 Polyphase circuits, 457–492 delta () connection, 470–476, 489–490 of sources, 473–476 Y-connected loads vs., 473 double-subscript notation, 459–460 polyphase systems, 458–460, 486–487 single-phase three-wire systems, 460–464, 487 three-phase Y-Y connection. See Three-phase Y-Y connection Port, 687 Positive charge, 11 Positive feedback, 197, 607 Positive phase sequence, 464–465 Positive power, 16, 18 Potential coil, 476 Potential difference, 14 Potentiometer, 671 Power, 9, 15–17, 30–33. See also ac circuit power analysis absorbed. See Absorbed power average. See Average power dissipation, 49 expression for, 15 factor. See Power factor gain, 499, 704 generating systems, 474–475 maximum transfer of, 152–154, 168–170 measuring. See Power measurement negative. See Absorbed power positive, 16, 18 reactive, 442, 447 superposition applicable to, 433 terminology recap, 447 triangle, 442–443 units, 10 Power factor (PF), 447 apparent power and, 438–441, 453–454 complex power, 438–441, 453–454 correction, 444–445 lagging, 439 leading, 439 Power factor (PF) angle, 439 Power measurement, 443–444 three-phase systems, 476–484, 490–491 two-wattmeter method, 481–483 wattmeters, use of, 476–478 wattmeter theory and formulas, 478–481 Practical current sources, 135, 139–140 Practical voltage sources, 133–135, 139–140 Preﬁxes, SI, 10–11 Primary mesh current, 505 Prime mover, 474 Probe software, 344–345 Problem-solving strategies, 1, 7–8 PSpice, 103, 105–107, 130–133 Bias Point command, 105 capacitors modeled with, 245–247, 259–260 Create command, 105 inductors modeled with, 245–247, 259–260 New Simulation Proﬁle command, 105 node-base schematics, 106–107 Run command, 105 for sinusoidal steady-state analysis, 404–405 for transient analysis, 270–272 tutorial, 813–816 Type command, 105 Pulse width (PW), of waveforms, 300 Purely reactive elements, average power absorption, 428–429 Q Quadrature power, 443 Quality factor (Q). See Parallel resonance R Radian frequency, 371, 537 Ragazzini, J.R., 207 Ramp function tu(t), Laplace transform for, 545 Randall, R.M., 207 Rational functions, inverse transforms for, 547–548 Rawlins, C.B., 25n, 26n RC circuits driven, 295–300 general, 279–282 sequentially switched, 300–305, 319 I: time to fully charge/fully discharge, 302–303, 304 II: time to fully charge but not fully discharge, 303, 304 III: no time to fully charge but time to fully discharge, 303, 304 IV: no time to fully charge or fully discharge, 304–305 source-free, 272–275, 311–312 time constant (τ), 274 unit-step function, 282–286, 315 Reactance impedance and, 390 inductive, 376 synchronous, 474 Reactive elements, average power absorption, 428–429 Reactive power, 442–443, 447 Realizable systems, s-domain analysis, 591–592 Real portion, of complex forcing function, 378 Real sources →real responses, complex forcing functions, 379–380 Reciprocity theorem, 698 Rectangular form, complex numbers, 818 Rectangular pulse function, 285–286 Rectiﬁers/Rectiﬁcation, 459, 493 Reference node, 80 Reﬂected impedance, 505–506 Reliable current sources, op amps, 190–192, 212–213 Reliable voltage sources, op amps, 188–190, 212–213 Repeated poles, inverse transform techniques, 550 Resistance/Resistors/Resistivity, 9, 25. See also Ohm’s law equivalent, 56 in the frequency domain, 571–572 ideal, average power absorption, 428 impedance and, 390, 391 internal, 134 linear, 23 output, 134 phasor relationships for, 385–386 in s-domain circuit analysis, 571–572, 577 in series and parallel, 55–61, 75–76 in the time domain, 577 variable. See Potentiometer Resonance, 324 current response and, 622 parallel. See Parallel resonance series, 633–636, 681 summary table for, 636 voltage response and, 622–623 Resonant frequency, 324 Response, 123 in the frequency domain, 770–777 as a function of the σ s-domain, 598–599 functions, 124 source-free series RLC circuits, 346–347 Ripple factor, 673, 677 Rise time (TR), of waveforms, 300 RLC circuits, 321–370 automotive suspensions modeled, 358 complete response of, 351–359, 368–369 complicated part, 352–357 uncomplicated part, 351–352 lossless LC circuit, 359–361, 369–370 phasor relationships for. See Phasor relationships for R, L, and C solution process summary, 357–359 source-free critical damping, 334–338, 365–366 A1 and A2 values, 335 form of critically damped response, 334–335 graphical representation of, 336–337 source-free parallel circuits, 321–325, 363 computer-aided analysis, 344–345 critically damped response, 325, 347 differential equation for, 322–324 equations summary, 347 frequency terms deﬁned, 324–325 INDEX 848 overdamped response, 325, 326–333, 347, 363–365 A1 and A2 values, 326–327 graphical representation, 331–332 underdamped response, 325, 338–345, 347, 366–367 B1 and B2 values, 339–340 ﬁnite resistance, role of, 340–342 form of, 338–339 graphical representation, 340 source-free series circuits, 345–351, 367–368 circuit response résumé, 346–347 critically damped response, 346–347 equations summary, 347 overdamped response, 346–347 underdamped response, 346–347 RL circuits driven. See Driven RL circuits exponential response properties, 268–272, 310 exponential response time constant (τ), 268–269 general, 275–276, 312–315 natural response. See Natural responses sequentially switched, 300–305, 319 I: time to fully charge/fully discharge, 302–303, 304 II: time to fully charge but not fully discharge, 303, 304 III: no time to fully charge but time to fully discharge, 303, 304 IV: no time to fully charge or fully discharge, 304–305 slicing thinly: 0 vs. 0, 276–279 source-free, 261–268, 309–310 alternative approach, 264 complementary function, 262 computer-aided analysis, 270–272 direct approach, 262–263 energy, accounting for, 267 forced response, 262 forcing function, 262 free response, 262 general solution approach, 264–265 natural response, 262 the particular solution, 262 the steady-state response, 262 transient response, 262 unit-step function, 282–286, 315 RMS value for average power, 435 for current and voltage, 433–438, 447 with multiple-frequency circuits, 435–436 for periodic waveforms, 433–434 for sinusoidal waveforms, 434–435 Robotic manipulator, 5 Root-mean-square (RMS) value. See RMS value Rotor, 474 Row vector, 804 Run command (PSpice), 105 Russell, F.A., 207 S s, deﬁned, 536–537 Sallen-key ampliﬁer, 673–677 Sampling function, Fourier series, 754–757 Sands, M.L., 67 Saturation, op amp, 197–198 Scalar multiplication, 561 Scales, units and, 9–11, 29–30 Scaling and frequency response, 644–648, 682–683 Laplace transform operation, 561 Scientiﬁc calculators, 803–804 s-domain circuit analysis, 571–618 additional techniques, 585–589, 615–616 complex frequency and. See Complex frequency convolution and. See Convolution H(s) Vout/Vin voltage ratio, synthesized, 606–610, 618 nodal and mesh analysis in, 578–584, 613–615 computer-aided analysis, 578–580 poles, zeros, and transfer functions, 588–589, 616–617 Thévenin equivalent technique, 587–588 Z(s) and Y(s), 571–577, 612–613 capacitors in frequency domain, 577 modeled in the s domain, 575–576 in time domain, 577 inductors in frequency domain, 572, 577 modeled in the s domain, 572–575 in time domain, 577 resistors in frequency domain, 571–572, 577 in time domain, 577 summary of element representations, 577 Secondary mesh current, 505 Seconds, 10 Self-inductance, 493 added to mutual inductance, 496 Sequentially switched RL or RC circuits. See RC circuits; RL circuits Series connections, 46 capacitors, 236–237 impedance combinations, 389 inductors in, 235–236 and parallel combinations. See also Source transformations connected sources, 51–55, 74, 139–140 other resonant forms, 639–644 Series resonance, 633–636, 681 Settling time, 332 Sharpe, D., 486 Short circuit(s), 27–28 admittance and, 708–709 for equivalent networks, 699–700 input admittance, 693–694 output admittance, 694 transfer admittance, 694 two-port networks, 694 to dc, 226 SI base units, 10 siemen (S), 572 Sifting property, 545 Signal ground, 65–66 Signs passive convention, 16 for voltages, 9, 14 Simon, Paul-René, 29 INDEX 849 INDEX 850 Simple time functions, Laplace transforms of, 543–546, 567 Simulation Program with Integrated Circuit Emphasis, 103 Simultaneous equations, solving, 803–810 Cramer’s rule, 809–810 determinants and, 807–809 matrices, 804–810 scientiﬁc calculators and, 803–804 Sines, converted to cosines, 373 Single-loop circuit, 46–49, 72–73 Single-node-pair circuit, 49–51, 73 Single-phase three-wire systems, 460–464, 487 Singularity functions, 283 Sinusoids complex frequency case, 535 as forcing functions, 619–620 Laplace transforms of, 558 Sinusoidal steady-state analysis, 3, 371–420 ac circuit average power, 426–427 admittance, 394 amplitude, 371 angular frequency, 371 argument, 371 characteristics of sinusoids, 371–374, 410–411 complex forcing function, 378–382, 412–413 algebraic alternative to differential equations, 380–381 applying, 379–380 imaginary part, 378 imaginary sources →imaginary responses, 379–380 real part, 378 real sources →real responses, 379–380 superposition theorem, 379–380 computer-aided analysis, 404–405 conductance, 394 cutoff frequency, transistor ampliﬁer, 398–399 forced responses to sinusoids, 371, 374–377, 411–412 alternative form of, 375–376 amplitude, response vs. forcing function, 376 steady-state, 374–375 frequency, 372–373 immittance, 394 impedance. See Impedance lagging and leading, 372–373 natural response, 371 nodal and mesh analysis, 394–397, 415–417 out-of-phase, 372–373 period, 372 in phase, 372–373 phase comparison requirements, 373 phasor diagrams, 406–408, 419 phasor relationships and. See Phasor relationships for R, L and C radian frequency, 371 sines converted to cosines, 373 sinusoidal waveform forcing function, 371 superposition, source transformations, and, 397–405, 417–418 susceptance, 394 Sinusoidal waveforms as forcing functions, 371 oscillator circuit design and, 607–608 phase comparison, 373 RMS values of current/voltage, 434–435 SI preﬁxes, 10–11 Slew rate, op amps, 199–200 Slicing thinly: 0 vs. 0, RL circuits, 276–279 Smoothing, of Bode diagrams, 651 Snider, G.S., 234n Solve() routine, 86 Source-free RC circuits, 272–275, 311–312 Source-free RLC circuits. See RLC circuits Source-free RL circuits. See RL circuits Source transformations, 3, 133–140, 157, 162–165 equivalent practical sources, 135–138 key concept requirements, 139–140 practical current sources, 135, 139–140 practical voltage sources, 133–135, 139–140 and sinusoidal steady-state analysis, 397–405, 417–418 summary, 140 SPICE, 6, 103. See also PSpice Square matrix, 804 Squire, J., 783 Stability, of a system, 560 Stator, 474 Steady-state analysis/response, 291. See also Sinusoidal steady-state analysis source-free RL circuits, 262 Step-down transformers, 516 Step-up transformers, 516 Stewart, D.R., 234n Stopband, 665 Structure (programming), 86 Strukov, D.B., 234n Summing ampliﬁer, 180–181, 182 Superconducting transformers, 518–519 Supermesh, 98, 100–101, 117–118 Supernodes, 89–91, 112–114 Superposition, 3, 123–133, 158, 159–162, 379–380 applicable to current, 433 applicable to power, 433 basic procedure, 130 limitations of, 133 sinusoidal steady-state analysis, 397–405, 417–418 superposition theorem, 125 Supplied power, 16 equaling absorbed power, 49 Susceptance, 394 Suspension systems, automotive, modeling of, 358 Symmetrical components, 470 Symmetry, use of, Fourier series analysis, 743–747 Synchronous generator, 474 Synchronous reactance, 474 System function, 589 computer-aided analysis, 774–777 fast Fourier transform (FFT), 772, 774–777 image processing example, 780 physical signiﬁcance of, 777–779 response, in frequency domain, 770–777 Systems, stability of, 560 Szwarc, Joseph, 29 INDEX 851 T T and equivalent networks, 507–510 Tank pressure monitoring system., 186–187 Taylor, Barry N., 29 Taylor, J.T., 679 Tesla, Nikola, 457 Thévenin, L.C., 141 Thévenin/Norton equivalent circuits, 3–4, 141–151, 157–158, 165–168, 172–173 Norton’s theorem, 3–4, 145–147, 157–158, 172–173 linearity for capacitors/ inductors, 240 resistance, 144, 157–158, 172–173 s-domain circuit analysis, 587–588 Thévenin’s theorem, 3, 141, 143–145, 157–158, 172–173 linearity for capacitors/ inductors, 240 proof of, 811–812 and sinusoidal steady-state analysis, 397–405, 417–418 two-port networks, 705–706 when dependent sources are present, 147–149 Thompson, Ambler, 29 Three-phase system, balanced, 458 Three-phase Y-Y connection, 464–470, 488–489 abc phase sequence, 464–465 cba phase sequence, 464–465 Delta () connection vs., 473 line-to-line voltages, 465–466 negative phase sequence, 464–465 positive phase sequence, 464–465 power measurement in. See Power measurement total instantaneous power, 467–468 with unbalanced load, 470 Tightly coupled coils, 504 Time constant (τ) exponential response of RL circuits, 268–269 RC circuits, 274 Time delay (TD) of waveforms, 300 Time differentiation, Laplace transforms and, 553–554, 561 Time domain capacitors in, 577 converted to frequency domain, 539 ideal transformer voltage relationships in, 517–521, 530–532 inductors in, 577 representation, phasors, 384 resistors in, 577 V-I expressions, phasor relationships and, 387 Time functions, simple, Laplace transforms of, 543–546, 567 Time integration, Laplace transforms and, 555–556, 561 Time periodicity, Laplace transforms and, 561, 833–835 Time-scaling theorem, Laplace transforms and, 838 Time shift, Laplace transforms and, 558, 561, 833–835 Topology, 791. See also Network topology Total instantaneous power, three-phase, 458, 467–468 T parameters, two-port networks, 716–720, 730–731 Transconductance, 21 Transfer functions, 499, 588, 597 Transfer of charge, 12 Transformations source. See Source transformations between y, z, h, and t parameters, 709 Transformers, 493. See also Magnetically coupled circuits ideal. See Ideal transformers linear. See Linear transformers superconducting, 518–519 Transient analysis, 3, 4 PSpice capability for, 270–272 Transient response, 289 source-free RL circuits, 262 Transistors, 22, 398–399, 715–716 Transmission parameters, two-port networks, 716–720, 730–731 Treble ﬁlters, 671–672 Trees, 791–797 Trigonometric form, of Fourier series. See Fourier series Trigonometric integrals, Fourier series analysis, 736–737 Tuinenga, P., 109, 816 Turns ratio, ideal transformers, 512–514 Two-port networks, 687–732 ABCD parameters, 716–720, 730–731 admittance parameters, 692–699, 723–725 bilateral circuit, 698 bilateral element, 698 reciprocity theorem, 698 short-circuit admittance parameters, 694 short-circuit input admittance, 693–694 short-circuit output admittance, 694 short-circuit transfer admittance, 694 y parameters, 694–695, 706–707 computer-aided analysis for, 719–720 equivalent networks, 699–707, 725–727 ampliﬁers, 704–706 of impedances method, 700–702 Norton equivalent method, 705–706 short-circuit admittance method, 699–700 Thevenin equivalent method, 705–706 Y- not applicable, 702 yV subtraction/addition method, 699 hybrid parameters, 713–716, 729–730 impedance parameters, 708–712, 727–728 one-port networks. See One-port networks t parameters, 716–720, 730–731 transistors, characterizing, 715–716 transmission parameters, 716–720, 730–731 Two-sided inverse Laplace transform, 542 Two-sided Laplace transform, 541 U Unbalanced Y-connected loads, 470 Underdamped response source-free parallel RLC circuits. See RLC circuits source-free series RLC circuits, 346–347 Unit-impulse function, 283 Laplace transform for, 544–545 Units and scales, 9–11, 29–30 Unit-step function u(t), 282–286, 315 Fourier transform pairs for, 767 Laplace transforms for, 544 and physical sources, 284–285 RC circuits, 282–286, 315 rectangular, 285–286 RL circuits, 282–286, 315 Unity gain ampliﬁer, 182 Upper half-power frequency, 628 V Vectors, 85, 804 Volta, Alessandro Giuseppe Antonio Anastasio, 14n Voltage, 9, 14–15, 30–33 actual polarity vs. convention, 14 current sources and, 17–22, 33–34, 51–55, 74 active elements, 21 circuit element, 21 dependent sources of voltage/current, 18, 19–21 derivative of the current voltage, 18 independent current sources, 19 independent voltage sources, 18–19 integral of the current voltage, 18 networks and circuits, 21–22 passive elements, 21 effective values of, 433–438, 452–453 force and, 5 input offset, op amps, 198 integral voltage-current relationships, for capacitors, 220–222, 249–252 internally generated, 474 laws. See Voltage and current laws sources. See Voltage sources voltage and current division, 61–64, 76–77 Voltage ampliﬁer, 178 Voltage and current division, 61–64, 76–77 Voltage and current laws, 39–78 branches, 39–40, 67–68 equivalent resistance, 55 Kirchhoff’s current law (KCL), 39, 40–42, 68–70 Kirchhoff’s voltage law (KVL), 39, 42–46, 70–72 order of elements and, 55 loops, 39–40, 67–68 nodes, 39–40, 67–68 paths, 39–40, 67–68 resistors in series and parallel, 55–61, 75–76 series and parallel connected sources, 51–55, 74 single-loop circuit, 46–49, 72–73 single-node-pair circuit, 49–51, 73 voltage and current division, 61–64, 76–77 Voltage coil, 476 Voltage-controlled current source, 19 Voltage-controlled voltage source, 19–20 Voltage follower circuit, 179, 182 Voltage gain, ampliﬁers, 704 Voltage level adjustment, ideal transformers for, 515–516 Voltage ratio H(s) Vout/Vin, synthesizing, 606–610, 618 Voltage regulation, 475 Voltage relationship, ideal transformers, time domain, 517–521, 530–532 Voltage response, resonance and, 622–623 Voltage sources ideal, 133–135 practical, 133–135 reliable, op amps, 188–190, 212–213 series and parallel connected sources, 51–55, 74 source effects, nodal and mesh analysis, 89–91, 112–114 Volt-ampere-reactive (VAR) units, 442 complex power, 441 Volt-amperes (VA), 439 W Wait, J.V., 679 Wattmeters, for three-phase systems theory and formulas, 478–481 two wattmeter method, 481–483 use, 476–478 Watts (W), 10, 447 Weber, E., 308, 363 Weedy, B.M., 449, 486 Westinghouse, George, 457 Wheeler, H.A., 534 Wien-bridge oscillator, 607 Williams, R.S., 234n Winder, S., 612 Wire gauges, 25–26 Work (energy) units, 10 Y Y parameters, two-port networks, 694–695, 706–707 Y(s) and Z(s). See s-domain circuit analysis YV method, for equivalent networks, 699 Z Zafrany, S., 565, 783 Zandman, Felix, 29 Zener diode, 188–190, 212–213 Zener voltage, 189 Zeros, 547 s-domain circuit analysis pole-zero constellations, 600–602 zeros, poles, and transfer functions, 588–589 Zero vs. Zero, slicing thinly: RL circuits, 276–279 Zeta (ζ) damping factor, 626 Z parameters, 708–712, 727–728 Z(s), Y(s) and. See s-domain circuit analysis INDEX 852 A Short Table of Integrals sin2 ax dx = x 2 −sin 2ax 4a cos2 ax dx = x 2 + sin 2ax 4a x sin ax dx = 1 a2 (sin ax −ax cos ax) x2 sin ax dx = 1 a3 (2ax sin ax + 2 cos ax −a2x2 cos ax) x cos ax dx = 1 a2 (cos ax + ax sin ax) x2 cos ax dx = 1 a3 (2ax cos ax −2 sin ax + a2x2 sin ax) sin ax sin bx dx = sin(a −b)x 2(a −b) −sin(a + b)x 2(a + b) ; a2 ̸= b2 sin ax cos bx dx = −cos(a −b)x 2(a −b) −cos(a + b)x 2(a + b) ; a2 ̸= b2 cos ax cos bx dx = sin(a −b)x 2(a −b) + sin(a + b)x 2(a + b) ; a2 ̸= b2 xeaxdx = eax a2 (ax −1) x2eaxdx = eax a3 (a2x2 −2ax + 2) eax sin bx dx = eax a2 + b2 (a sin bx −b cos bx) eax cos bx dx = eax a2 + b2 (a cos bx + b sin bx) dx a2 + x2 = 1 a tan−1 x a ∞ 0 sin ax x dx = ⎧ ⎪ ⎨ ⎪ ⎩ 1 2π a > 0 0 a = 0 −1 2π a < 0 π 0 sin2 x dx = π 0 cos2 x dx = π 2 π 0 sin mx sin nx dx = π 0 cos mx cos nx dx = 0; m ̸= n, m and n integers π 0 sin mx cos nx dx = ⎧ ⎨ ⎩ 0 m −n even 2m m2 −n2 m −n odd A Short Table of Trigonometric Identities sin(α ± β) = sin α cos β ± cos α sin β cos(α ± β) = cos α cos β ∓sin α sin β cos(α ± 90◦) = ∓sin α sin(α ± 90◦) = ± cos α cos α cos β = 1 2 cos(α + β) + 1 2 cos(α −β) sin α sin β = 1 2 cos(α −β) −1 2 cos(α + β) sin α cos β = 1 2 sin(α + β) + 1 2 sin(α −β) sin 2α = 2 sin α cos α cos 2α = 2 cos2 α −1 = 1 −2 sin2 α = cos2 α −sin2 α sin2 α = 1 2(1 −cos 2α) cos2 α = 1 2(1 + cos 2α) sin α = ejα −e−jα j2 cos α = ejα + e−jα 2 e±jα = cos α ± j sin α A cos α + B sin α = A2 + B2 cos α + tan−1 −B A
15099	https://flexbooks.ck12.org/cbook/ck-12-chemistry-flexbook-2.0/section/13.4/primary/lesson/pressure-units-and-conversions-chem/	Skip to content Math Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up 13.4 Pressure Units and Conversions Written by:Ck12 Science Fact-checked by:The CK-12 Editorial Team Last Modified: Just Now Lesson Can you guess how old this car is? There are several benefits to maintaining the proper air pressure in a car tire. The ride is smoother and safer than with lowered pressure. The car gets better gas mileage and the tires don’t wear out as fast. The recommended pressure for that model of car (usually somewhere between 32-35 psi) is usually listed in the owner’s manual or stamped somewhere inside the door. The pressure on the tire is the maximum pressure for that tire, not the recommended one. Tire pressure is best measured when the tire is cold since driving the car for a while will heat up the air in the tire and increase the pressure. Pressure Units and Conversion A barometer measures gas pressure by the height of the column of mercury. One unit of gas pressure is the millimeter of mercury (mmHg). An equivalent unit to the mmHg is called the torr, in honor of the inventor of the barometer, Evangelista Torricelli. The pascal (Pa) is the standard unit of pressure. A pascal is a very small amount of pressure, so the more useful unit for everyday gas pressures is the kilopascal (kPa). A kilopascal is equal to 1000 pascals. Another commonly used unit of pressure is the atmosphere (atm). Standard atmospheric pressure is called 1 atm of pressure and is equal to 760 mmHg and 101.3 kPa. Atmospheric pressure is also often stated as pounds/square inch (psi). The atmospheric pressure at sea level is 14.7 psi. @$$\begin{align}1 \ \text{atm}=760 \ \text{mmHg} = 760 \ \text{torr} = 101.3 \ \text{KPa} = 14.7 \ \text{psi}\end{align}@$$ It is important to be able to convert between different units of pressure. To do so, we will use the equivalent standard pressures shown above. Sample Problem: Pressure Unit Conversions The atmospheric pressure in a mountainous location is measured to be 613 mmHg. What is this pressure in atm and in kPa? Step 1: List the known quantities and plan the problem. Known given: 613 mmHg 1 atm = 760 mmHg 101.3 kPa = 760 mmHg Unknown pressure = ? atm pressure = ? kPa Use conversion factors from the equivalent pressure units to convert from mmHg to atm and from mmHg to kPa. Step 2: Solve. @$$\begin{align}613 \ \text{mmHg}\times\frac{1\ \text{atm}}{760 \ \text{mmHg}} &=0.807 \ \text{atm}\ 613 \ \text{mmHg}\times\frac{101.3 \ \text{kPa}}{760 \ \text{mmHg}} &=81.7 \ \text{kPa}\end{align}@$$ Step 3: Think about your result. The air pressure is about 80% that of standard atmospheric pressure at sea level. For significant figure purposes, the standard pressure of 760 mmHg has three significant figures. Summary Calculations are described for converting between different pressure units. Review What instrument measures gas pressure by the height of the column of mercury? 1 atm = ___ torr 32.02 atm = ___ kPa 542 mmHg = ___ psi The pressure in a car tire is 35 psi. How many atmospheres is that? Asked by Students Here are the top questions that students are asking Flexi for this concept: \| Image \| Reference \| Attributions \| --- \| \| \| Credit: User:IFCAR/Wikimedia Commons Source: License: Public Domain \| Student Sign Up Are you a teacher? Having issues? Click here By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? Adaptive Practice Save this section to your Library in order to add a Practice or Quiz to it. 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