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9500 | https://www.shmoop.com/common-core-standards/ccss-hs-a-ced-2.html | More on Common Core Standards
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High School: Functions
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High School: Algebra
High School: Algebra
Creating Equations HSA-CED.A.2
2. Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.
This standard has two significant components. The first is translating word problems into equations with two or more variables. The more the merrier. Well, maybe not in this case. Translating word problems to create simple equations with two or more variables is not that different conceptually from creating equations with one variable.
The main difference is that more complicated mathematical relationships such as systems of equations, functions, and proportions may develop (along with nausea, headaches, and spontaneous yodeling). In any case, this aspect of this standard should be taught with the previous one.
The second component is creating graphs of equations on coordinate axes, which incorporates multiple skills such as visual perception, interpreting data, and synthesizing information. Such graphs relate to equations with multiple equations by relating one variable to another.
Take lines, for example. In the form y = mx + b, we can look at either x or y and any defined value for x will give us a defined value for y, and vice versa. Graphs can help visualize these relationships between variables and facilitate the connection of equations to the graphs that represent them.
Yearnin' for more graphin'? Don't worry. There'll be more down the line.
Drills
The ratio of nut chocolates to cordials in an assortment box is 2:1. Which equality describes the contents of an assortment box?
Correct Answer:
Answer Explanation:
The relationship 2:1 is already expressed as a ratio; so, this is just a simple equality of ratio of nut chocolates to cordials = 2:1.
The ratio of nut chocolates to cordials in an assortment box is 2:1. Which relation describes the contents of an assortment box?
Correct Answer:
Answer Explanation:
The key word "ratio" indicates division. The two ratios involved are nut chocolates to cordials and 2:1. Dividing in each case gives .
The ratio of nut chocolates to cordials in an assortment box is 2:1. Which relation describes the contents of an assortment box? (Let n = number of nut chocolates, and c = number of cordials.)
Correct Answer:
Answer Explanation:
Substituting in variables for the nut chocolates (n) and the cordials (c) give . The simplifies to 2. This gives the equation , where n = number of nut chocolates, and c = number of cordials.
The ratio of nut chocolates to cordials in an assortment box is 2:1. If there are 5 cordials in an assortment box, how many nut chocolates are there?
Correct Answer:
Answer Explanation:
The equation describes the ratio of nuts to cordials in an assortment box. The assortment box has 5 cordials so, c = 5. Substituting in, this gives . Multiplying both sides by 5 gives n = 2 × 5 = 10. There are 10 nut chocolates in the box.
The Fuzzlegump School for Gifted and Not-So-Gifted Children requires that three chaperones go on any field trip. More chaperones are required if there are more than 30 students on the trip. For every 12 additional students, another chaperone is required. Which of these inequalities describes the number of chaperones required?
Correct Answer:
Answer Explanation:
There are two different requirements for chaperones. Three are required for any field trip with up to 30 students. Then, for every 12 students over the 30-student mark, an additional chaperone is required. These two requirements need to be combined to calculate the minimum number of chaperones for a trip. This gives the inequality chaperones required ≥ three chaperones for 30 students and another chaperone for every 12 additional students.
The Fuzzlegump School for Gifted and Not-So-Gifted Children requires that three chaperones go on any field trip. More chaperones are required if there are more than 30 students on the trip. For every 12 additional students, another chaperone is required. Which relation describes the minimum number of chaperones required?
Correct Answer:
Answer Explanation:
The inequality is chaperones required ≥ three chaperones for 30 students and another chaperone for every 12 additional students. The "and" indicates addition to link three chaperones with the additional chaperone for every 12 students over the initial 30. The key is then recognizing that the, "every 12 additional students," holds only for more than 30 students. The key phrase, "more than" indicates subtraction to get how many more than 30 students there are. And the key words, "for every" indicates division. These give the relation chaperones required ≥ three chaperones + .
The Fuzzlegump School for Gifted and Not-So-Gifted Children requires that three chaperones go on any field trip. More chaperones are required if there are more than 30 students on the trip. For every 12 additional students, another chaperone is required. Which equation describes the minimum number of chaperones required? (Let c = chaperones and s = students.)
Correct Answer:
Answer Explanation:
The relation forming the basis for the equation is chaperones required ≥ three chaperones + . Translating in the variables c = chaperones and s = students gives the equation .
The Fuzzlegump School for Gifted and Not-So-Gifted Children requires that three chaperones go on any field trip. More chaperones are required if there are more than 30 students on the trip. For every 12 additional students, another chaperone is required. If 60 students attend the field trip, how many chaperones are required?
Correct Answer:
Answer Explanation:
The equation describes the relationship between number of chaperones (c) and number of students (s). Substituting in 60 students with s = 60 gives . Simplifying the calculation gives and then c ≥ 3 + 2.5 so c ≥ 5.5. Since there can't be half a chaperone (unless you count robots), they need 6 chaperones on the trip.
Which graph represents x = 1?
Correct Answer:
Answer Explanation:
We need our x value to be 1 regardless of the y value. For any and every y value we choose, we should have x = 1. No matter where we travel along the y-axis, we'll always end up moving one unit to the right of it and drawing a point there. This creates a perfectly vertical line that intersects with the x-axis at 1.
Which graph represents y = 2x + 3?
Correct Answer:
Answer Explanation:
We can think of the equation as being a machine that gives us an output value for every number we input. If we input 0 for x, we should get 3 for y; if we input 1 for x, we should get 5 for y. Those two points alone are enough to graph the line on the coordinate plane and match it with (D).
Aligned Resources
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9501 | https://www.youtube.com/playlist?list=PLPLlcVV9fXj_nIwm8uKW0T5nOzDlNsCQ6 | Chapter 4: Engineering Statics by Meriam and Kraige - YouTube
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9502 | https://math.libretexts.org/Bookshelves/Algebra/College_Algebra_1e_(OpenStax)/03%3A_Functions/3.07%3A_Absolute_Value_Functions | 3.7: Absolute Value Functions - Mathematics LibreTexts
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3: Functions
College Algebra 1e (OpenStax)
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Wed, 06 Oct 2021 15:16:50 GMT
3.7: Absolute Value Functions
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3.7: Absolute Value Functions
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3.6: Transformation of Functions
3.8: Inverse Functions
Page ID 15058
OpenStax
OpenStax
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Understanding Absolute Value
2. Graphing an Absolute Value Function
3. Solving an Absolute Value Equation
4. Solving an Absolute Value Inequality
5. Key Concepts
6. Glossary
Learning Objectives
Graph an absolute value function.
Solve an absolute value equation.
Until the 1920s, the so-called spiral nebulae were believed to be clouds of dust and gas in our own galaxy, some tens of thousands of light years away. Then, astronomer Edwin Hubble proved that these objects are galaxies in their own right, at distances of millions of light years. Today, astronomers can detect galaxies that are billions of light years away. Distances in the universe can be measured in all directions. As such, it is useful to consider distance as an absolute value function. In this section, we will investigate absolute value functions.
Figure 3.7.1: Distances in deep space can be measured in all directions. As such, it is useful to consider distance in terms of absolute values. (credit: "s58y"/Flickr)
Understanding Absolute Value
Recall that in its basic form f(x)=|x|, the absolute value function, is one of our toolkit functions. The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. Algebraically, for whatever the input value is, the output is the value without regard to sign.
Absolute Value Function
The absolute value function can be defined as a piecewise function
(3.7.1)f(x)=|x|={x if x≥0−x if x<0
Example 3.7.1: Determine a Number within a Prescribed Distance
Describe all values x within or including a distance of 4 from the number 5.
Solution
We want the distance between x and 5 to be less than or equal to 4. We can draw a number line, such as the one in , to represent the condition to be satisfied.
Figure 3.7.2: Number line describing the difference of the distance of 4 away from 5.
The distance from x to 5 can be represented using the absolute value as |x−5|. We want the values of x that satisfy the condition |x−5|≤4.
Analysis
Note that
−4≤x−5 x−5≤4 1≤x x≤9
So |x−5|≤4 is equivalent to 1≤x≤9.
However, mathematicians generally prefer absolute value notation.
Exercise 3.7.1
Describe all values x within a distance of 3 from the number 2.
Answer
|x−2|≤3
Example 3.7.2: Resistance of a Resistor
Electrical parts, such as resistors and capacitors, come with specified values of their operating parameters: resistance, capacitance, etc. However, due to imprecision in manufacturing, the actual values of these parameters vary somewhat from piece to piece, even when they are supposed to be the same. The best that manufacturers can do is to try to guarantee that the variations will stay within a specified range, often ±1%, ±5%, or ±10%.
Suppose we have a resistor rated at 680 ohms, ±5%. Use the absolute value function to express the range of possible values of the actual resistance.
Solution
5% of 680 ohms is 34 ohms. The absolute value of the difference between the actual and nominal resistance should not exceed the stated variability, so, with the resistance R in ohms,
|R−680|≤34
Exercise 3.7.2
Students who score within 20 points of 80 will pass a test. Write this as a distance from 80 using absolute value notation.
Answer
Using the variable p for passing, |p−80|≤20
Graphing an Absolute Value Function
The most significant feature of the absolute value graph is the corner point at which the graph changes direction. This point is shown at the origin in Figure 3.7.3.
Figure 3.7.3: Graph of an absolute function.
Figure 3.7.3 shows the graph of y=2|x–3|+4. The graph of y=|x| has been shifted right 3 units, vertically stretched by a factor of 2, and shifted up 4 units. This means that the corner point is located at (3,4) for this transformed function.
Figure 3.7.4: Graph of the different types of transformations for an absolute function.
Example 3.7.3: Writing an Equation for an Absolute Value Function
Write an equation for the function graphed in Figure 3.7.5.
Figure 3.7.5: Graph of an absolute function.
Solution
The basic absolute value function changes direction at the origin, so this graph has been shifted to the right 3 units and down 2 units from the basic toolkit function. See Figure 3.7.6.
Figure 3.7.6: Graph of two transformations for an absolute function at (3,−2).
We also notice that the graph appears vertically stretched, because the width of the final graph on a horizontal line is not equal to 2 times the vertical distance from the corner to this line, as it would be for an unstretched absolute value function. Instead, the width is equal to 1 times the vertical distance as shown in Figure 3.7.7.
Figure 3.7.7: Graph of two transformations for an absolute function at (3,−2) and the ratios between the two different transformations.
From this information we can write the equation
f(x)=2|x−3|−2,treating the stretch as a vertial stretch, or f(x)=|2(x−3)|−2,treating the stretch as a horizontal compression.
Analysis
Note that these equations are algebraically equivalent—the stretch for an absolute value function can be written interchangeably as a vertical or horizontal stretch or compression.
Q & A
If we couldn’t observe the stretch of the function from the graphs, could we algebraically determine it?
Answer
Yes. If we are unable to determine the stretch based on the width of the graph, we can solve for the stretch factor by putting in a known pair of values for x and f(x).
f(x)=a|x−3|−2
Now substituting in the point (1,2)
2=a|1−3|−2 4=2a a=2
Exercise 3.7.3
Write the equation for the absolute value function that is horizontally shifted left 2 units, is vertically flipped, and vertically shifted up 3 units.
Answer
f(x)=−|x+2|+3
Q & A
Do the graphs of absolute value functions always intersect the vertical axis? The horizontal axis?
Answer
Yes, they always intersect the vertical axis. The graph of an absolute value function will intersect the vertical axis when the input is zero.
No, they do not always intersect the horizontal axis. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. It is possible for the absolute value function to intersect the horizontal axis at zero, one, or two points (Figure 3.7.8).
Figure 3.7.8: (a) The absolute value function does not intersect the horizontal axis. (b) The absolute value function intersects the horizontal axis at one point. (c) The absolute value function intersects the horizontal axis at two points.
Solving an Absolute Value Equation
Now that we can graph an absolute value function, we will learn how to solve an absolute value equation. To solve an equation such as 8=|2x−6|, we notice that the absolute value will be equal to 8 if the quantity inside the absolute value is 8 or -8. This leads to two different equations we can solve independently.
2x−6=8 or 2x−6=−8
2x=14 2x=−2 x=7 x=−1
Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point.
An absolute value equation is an equation in which the unknown variable appears in absolute value bars. For example,
|x|=4,|2x−1|=3,|5x+2|−4=9.
Solutions to Absolute Value Equations
For real numbers A and B, an equation of the form |A|=B, with B≥0, will have solutions when A=B or A=−B. If B<0, the equation |A|=B has no solution.
How To ...
Given the formula for an absolute value function, find the horizontal intercepts of its graph.
Isolate the absolute value term.
Use |A|=B to write A=B or −A=B, assuming B>0.
Solve for x.
Example 3.7.4: Finding the Zeros of an Absolute Value Function
For the function f(x)=|4x+1|−7, find the values of x such that f(x)=0.
Solution
0=|4x+1|−7 Substitute 0 for f(x).7=|4x+1|Isolate the absolute value on one side of the equation.7=4x+1 or−7=4x+1 Break into two separate equations and solve.6=4x−8=4x x=6 4=1.5 x=−8 4=−2
The function outputs 0 when x=1.5 or x=−2 (Figure 3.7.9).
Figure 3.7.9: Graph of an absolute function with x-intercepts at -2 and 1.5.
Exercise 3.7.4
For the function f(x)=|2x−1|−3, find the values of x such that f(x)=0.
Solution
x=−1 or x=2
Q & A
Should we always expect two answers when solving |A|=B?
Answer
No. We may find one, two, or even no answers. For example, there is no solution to 2+|3x−5|=1.
How To ...
Given an absolute value equation, solve it.
Isolate the absolute value term.
Use |A|=B to write A=B or A=−B.
Solve for x.
Example 3.7.5: Solving an Absolute Value Equation
Solve 1=4|x−2|+2.
Solution
Isolating the absolute value on one side of the equation gives the following.
1=4|x−2|+2−1=4|x−2|−1 4=|x−2|
The absolute value always returns a positive value, so it is impossible for the absolute value to equal a negative value. At this point, we notice that this equation has no solutions.
Q & A
In Example 3.7.5, if f(x)=1 and g(x)=4|x−2|+2 were graphed on the same set of axes, would the graphs intersect?
Answer
No. The graphs of f and g would not intersect, as shown in Figure 3.7.10. This confirms, graphically, that the equation 1=4|x−2|+2 has no solution.
Figure 3.7.10: Graph of g(x)=4|x−2|+2 and f(x)=1.
Find where the graph of the function f(x)=−|x+2|+3 intersects the horizontal and vertical axes.
f(0)=1, so the graph intersects the vertical axis at (0,1). f(x)=0 when x=−5 and x=1 so the graph intersects the horizontal axis at (−5,0) and (1,0).
Solving an Absolute Value Inequality
Absolute value equations may not always involve equalities. Instead, we may need to solve an equation within a range of values. We would use anabsolute value inequalityto solve such an equation. An absolute value inequality is an equation of the form
|A|B,
or
|A|≥B,
where an expression A (and possibly but not usually B) depends on a variable x. Solving the inequality means finding the set of all x that satisfy the inequality. Usually this set will be an interval or the union of two intervals.
There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two functions. The advantage of the algebraic approach is it yields solutions that may be difficult to read from the graph.
For example, we know that all numbers within 200 units of 0 may be expressed as
|x|<200
or
−200<x<200
Suppose we want to know all possible returns on an investment if we could earn some amount of money within 200of 600. We can solve algebraically for the set of values x such that the distance between x and 600 is less than 200. We represent the distance between x and 600 as |x−600|.
(3.7.2)|x−600|<200
or
(3.7.3)−200<x−600<200
−200+600<x−600+600<200+600 400<x<800
This means our returns would be between 400and 800.
Sometimes an absolute value inequality problem will be presented to us in terms of a shifted and/or stretched or compressed absolute value function, where we must determine for which values of the input the function’s output will be negative or positive.
How To ...
Given an absolute value inequality of the form |x−A|≤B for real numbers a and b where b is positive, solve the absolute value inequality algebraically.
Find boundary points by solving |x−A|=B.
Test intervals created by the boundary points to determine where |x−A|≤B.
Write the interval or union of intervals satisfying the inequality in interval, inequality, or set-builder notation.
Example 3.7.6: Solving an Absolute Value Inequality
Solve |x−5|≤4.
Solution
With both approaches, we will need to know first where the corresponding equality is true. In this case we first will find where |x−5|=4. We do this because the absolute value is a function with no breaks, so the only way the function values can switch from being less than 4 to being greater than 4 is by passing through where the values equal 4. Solve |x−5|=4.
x−5=4 or x=9 x−5=−4 x=1
After determining that the absolute value is equal to 4 at x=1 and x=9, we know the graph can change only from being less than 4 to greater than 4 at these values. This divides the number line up into three intervals:
x<1,1<x<9,and x>9.
To determine when the function is less than 4, we could choose a value in each interval and see if the output is less than or greater than 4, as shown in Table 3.7.1.
Table 3.7.1| Interval test x | f(x) | <4 or >4 |
---
| x<1 | 0 | |0−5|=5 | Greater than |
| 1<x<9 | 6 | |6−5|=1 | Less than |
| x>9 | 11 | |11−5|=6 | Greater than |
Because 1≤x≤9 is the only interval in which the output at the test value is less than 4, we can conclude that the solution to |x−5|≤4 is 1≤x≤9, or [1,9].
To use a graph, we can sketch the function f(x)=|x−5|. To help us see where the outputs are 4, the line g(x)=4 could also be sketched as in Figure 3.7.11.
Figure 3.7.11: Graph to find the points satisfying an absolute value inequality.
We can see the following:
The output values of the absolute value are equal to 4 at x=1 and x=9.
The graph of f is below the graph of g on 1<x<9. This means the output values of f(x) are less than the output values of g(x).
The absolute value is less than or equal to 4 between these two points, when 1≤x≤9. In interval notation, this would be the interval [1,9].
Analysis
For absolute value inequalities,
|x−A|C,−C<x−A<C,x−A<−C or x−A>C.
The < or > symbol may be replaced by ≤ or ≥.
So, for this example, we could use this alternative approach.
|x−5|≤4−4≤x−5≤4 Rewrite by removing the absolute value bars.−4+5≤x−5+5≤4+5 Isolate the x.1≤x≤9
Exercise 3.7.5
Solve |x+2|≤6.
Answer
−8≤x≤4
How To ...
Given an absolute value function, solve for the set of inputs where the output is positive (or negative).
Set the function equal to zero, and solve for the boundary points of the solution set.
Use test points or a graph to determine where the function’s output is positive or negative.
Example 3.7.7: Using a Graphical Approach to Solve Absolute Value Inequalities
Given the function f(x)=−1 2|4x−5|+3, determine the x-values for which the function values are negative.
Solution
We are trying to determine where f(x)<0, which is when −1 2|4x−5|+3<0. We begin by isolating the absolute value.
−1 2|4x−5|<−3 Multiply both sides by –2, and reverse the inequality.|4x−5|>6
Next we solve for the equality |4x−5|=6.
4x−5=6 4x−5=−6 4x−6=6
or
4x=−1 x=11 4 x=−1 4
Now, we can examine the graph of f to observe where the output is negative. We will observe where the branches are below the x-axis. Notice that it is not even important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at x=−1 4 and x=11 4 and that the graph has been reflected vertically. See Figure 3.7.12.
Figure 3.7.12: Graph of an absolute function with x-intercepts at -0.25 and 2.75.]
We observe that the graph of the function is below the x-axis left of x=−1 4 and right of x=11 4. This means the function values are negative to the left of the first horizontal intercept at x=−1 4, and negative to the right of the second intercept at x=11 4. This gives us the solution to the inequality.
x<−1 4 or x>1 1 4
In interval notation, this would be (−∞,−0.25)∪(2.75,∞).
Exercise 3.7.6
Solve −2|k−4|≤−6.
Answer
k≤1 or k≥7; in interval notation, this would be (−∞,1]∪[7,∞)
Key Concepts
The absolute value function is commonly used to measure distances between points.
Applied problems, such as ranges of possible values, can also be solved using the absolute value function.
The graph of the absolute value function resembles a letter V. It has a corner point at which the graph changes direction.
In an absolute value equation, an unknown variable is the input of an absolute value function.
If the absolute value of an expression is set equal to a positive number, expect two solutions for the unknown variable.
An absolute value equation may have one solution, two solutions, or no solutions.
An absolute value inequality is similar to an absolute value equation but takes the form | A |B, or | A |≥B.It can be solved by determining the boundaries of the solution set and then testing which segments are in the set.
Absolute value inequalities can also be solved graphically.
Glossary
absolute value equation
an equation of the form |A|=B, with B≥0; it will have solutions when A=B or A=−B
absolute value inequality
a relationship in the form |A|B, or |A|≥B
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3.8: Inverse Functions
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9503 | https://www.youtube.com/watch?v=y9gxcWEhqws | EJEMPLO DE DIAGRAMAS DE VIGAS CON EL USO DE LAS FUNCIONES DE SINGULARIDAD
PROFE JN El canal del ingeniero
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Posted: 22 Mar 2019
Este video muestra como usar las funciones de singularidad para construir diagramas de vigas
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Transcript:
Introducción [Música] INTRODUCIÓN muy buenos días para todos y todas en este nuevo vide tutorial vamos a mirar cómo sacamos las ecuaciones para hacer los diagramas de esta Viga esta Viga ya la tenemos desarrollada en otro video el cual les dejo en la descripción el link lo dejo en la descripción para que con comparen los dos métodos allá lo sacamos por el método de las ecuaciones que es haciendo Cortes y ir buscando las ecuaciones en esta ocasión lo vamos a hacer por funciones de singularidad que es un método mucho más fácil y es equivalente para desarrollarlo por este TABLA DE FUNCIONES DE SINGULARIDAD método Pues necesitamos esta tablita Esta es la tabla de funciones de singularidad en un video anterior pues ya habíamos explicado esta tablita en el video anterior a este ustedes pueden encontrar la explicación de esta tablita ahora vamos a hacer la aplicación Entonces lo primero que debemos hacer REACCIONES EN LOS APOYOS acá es Hallar las reacciones en los apoyos Hallar el valor de la reacción en a y la reacción en B bueno Y aquí coloqué las dos reacciones la reacción en B la coloqué subiendo la reacción en a la coloqué bajando ustedes las pueden colocar en la dirección que quieran yo ya sé que van en estas direcciones por qué Porque piensen que toda esta carga genera una rotación en este sentido entonces para contrarestar esa rotación la reacción en a hace esto tiene que ir bajando para hacer eso y y ahí se contrarrestan las dos rotaciones como lo sabía el video anterior Pues el hecho lo apliqué pero repito no importa hacia donde usted las coloque cuando las calcule el signo le va a decir si está en lo correcto o no Bueno ahora hago sumatoria de momentos respecto a b respecto a una de las incógnitas no puede ser en a o en B yo escogí hacerlo sumatorio de momento respecto a b igual a c porque estamos en equilibrio pero recuerden que todas estas cargas esta carga esta sección de carga triangular y esta sección de carga rectangular deben ser cambiadas por las cargas equivalentes y que a la carga equivalente la hallo como el área de la figura de carga en este caso la primera carga es el Triángulo entonces todo esta carga la voy a cambiar por una equivalente que es la carga triangular que la hallo como el área de este triangulito no el área del triangulito de carga exactamente el área de esta figura de esa figura triangular y esta figura triangular pues tiene de base aquí la base son estos 3 m y la altura son estos 300 n porque arranque en C altura 300 entonces diríamos base por altura sobre dos entonces son 3 M 300 n sobre Met y todo esto sobre 2 pues 300 3 me da 900 Divo en 2 450 cierto metros y metros se cancelan y nos quedan 450 n entonces voy a decir que el equivalente son 450 n a la carga repito la carga triangular el la otra figura de carga pues es un rectángulo Ah Bueno pero antes de pasar al rectángulo Me faltó decir que esta se ubica en el centroide del triángulo el centroide del ángulo del triángulo con respecto a este que es el ángulo recto Acá está el ángulo recto queda a un tercio de él Entonces como todo esto mide 3 m cierto la tercera parte aquí a la tercera parte desde el ángulo recto la tercera parte de tres pues es uno quiere decir que esto está puesto a 1 m ahí está ahí va a estar la carga equivalente esta carga la carga rectangular que es la verde el área son estos 300 que tiene de altura cierto Esta es la altura 300 nm por la base que son 4 m 4 3 12 quiere decir que son 10000 la carga equivalente 10000 n y ya vemos que está puesta en el centroide del rectángulo y el centroide de un rectángulo pues está en la mitad de él quiere decir que si eso mide 4 de base esa carga está puesta a 2 m de uno de los extremos y ahí están mis cargas equivalentes Entonces ya sabemos que nos olvidamos de esta carga triangular y esta rectangular y que trabajamos con la de 450 y con la de 1200 para los momentos Entonces qué tenemos sumatoria de momento respecto al punto B empiezo como es respecto a B ojo ubicados aquí en B la primera carga con la que voy a trabajar es ra entonces voy a multiplicar la carga ra por la distancia hasta el punto B entonces de acá desde ra hasta el punto B Pues hay 3 m entonces acá por 3 m Listo ya sabemos que esta rotación queda en este sentido cierto respecto al punto b y que eso es antihorario positiva No porque las rotaciones que van en este sentido que se el sentido antihorario las estamos asumiendo positivas bueno la siguiente carga que nos genera rotación es la carga de 450 como va bajando alrededor de B genera esta rotación también positiva no antihorario entonces Cuánta rotación genera Pues el valor de la carga que son los 450 n por la distancia al punto B Entonces desde la carga al punto B hay 1 m aquí por 1 m cierto ya dijimos que positiva bueno la siguiente pues rb que es la siguiente carga está sobre el punto de la rotación lo que no genera rotación y la siguiente será la de 100 no entonces aquí 100 Newton Por qué distancia hay desde la de 1200 hasta el punto B pues 2 m estos 2 m ent aquí por 2 m y esta rotación que genera la carga de 1200 respecto aquí al punto B no ahí será en este sentido o sea sentido horario ya sabemos que el sentido horario es negativa entonces aquí colocamos menos Listo y como no hay más cargas Entonces ya cerramos igualando a cero obviamente de acá la única incógnita es ra entonces hacen el proceso matemático y Al despejar pues obtengo que ra son 650 n y esto nos da positivo bueno y lo siguiente Recuerden que es la sumatoria de fuerzas en y = a 0 Porque si yo hago sumatoria de fuerzas en Y entonces miremos voy a asmir ya sabemos Bueno ya sabemos que se arba es positivo cierto entonces acá qué tengo la única que va hacia arriba fíjese es rb luego es la que coloqué aquí positiva todas las demás van bajando ra que ya tengo el valor 650 entonces aquí -650 esta de 450 y esta de 10000 ambas bajando o sea ambas negativas y como no hay más cargas cierro igualando a cer0 o sea tenemos 1 2 3 4 cargas ahí están una 2 3 4 bueno de acá despejo de r pasando estos valores al otro lado y me queda que rb son 2300 n Bueno una vez que ya se tienen las reacciones ahora por ese método el paso a seguir es este entonces CÓMO UTILIZAR LAS FUNCIONES DE SINGULARIDAD Este es mi Viga cierto como voy a utilizar las funciones de singularidad entonces dice que tenemos un triángulo de carga y un rectángulo de carga entonces voy a jugar solo con triángulos y rectángulos Pero si yo vengo aquí a mi tablita Pues yo no tengo una carga como esta un triángulo y un rectángulo seguido fíjese Solo tengo o triángulo o rectángulo pero no tengo una situación así como de la la de la figura Entonces qué me toca hacer me toca trabajar con una sola y hacer combinaciones Entonces qué hice yo Esta RESOLUCIÓN DEL PROBLEMA es mi situación de carga que debo resolver Cómo la voy a resolver Pues voy a decir que esto es igual a la suma de estas dos cosas entonces lo que voy a hacer es voy a considerar toda una carga triangular toda esta miren entonces aquí está voy a asumir que todo esto es un solo triángulo de carga desde la punta hasta el final para utilizar esta y a este triángulo de carga Voy a quitarle este triángulo también de carga que va desde esta reacción en adelante fíjese que estas van subiendo y estas van bajando si tomamos esta Bueno aquí y la saqué Aparte esta es esta misma miren al sumarla con esta Pues voy a terminar anulando cada una de estas porciones miren ahí voy anulando cada una de estas porciones o sea esta me anula esta Y si yo anulo cada una de estas si yo anulo este pedazo este pedazo este pedazo este pedazo o sea lo haría el Triángulo si yo anulo el Triángulo al anular el Triángulo simplemente me queda este rectángulo hasta acá no que es la situación que yo tengo en el dibujo miren o sea yo podría decir que finalmente esto es tener todo un triángulo así y quitar tener un triángulo de arriba hasta abajo no de arriba hasta abajo tener todo este triángulo y a todo este triángulo a toda esta sección triangular le quito este pedazo le quito este triangulito entonces al quitar este triangulito cierto al quitar este triangulito a todo el Triángulo amarillo todo este triángulo al quitarle al borrarle al borrarle este triangulito pues ahí ya me queda lo que yo necesito pero Cómo quito aquí matemáticamente este triangulito pues matemáticamente simplemente sumo este pedazo a esto para que me la anule bueno Y una vez que ya he hecho eso ya puedo plantear mi función de singularidad para estas dos cargas toda toda la carga triangular que está sobre la viga todo este que estamos aquí delineando y esta carga de abajo también triangular como si tuviera dos triángulos de carga Y entonces qué voy a decir voy a empezar por la función de carga wx entonces la función de carga para el caso triangular fíjese Entonces voy aquí al caso triangular Ahí está función de carga es la constante de proporcionalidad que dice la pendiente la recta cierto del triángulo de carga la pendiente de esta recta factor de X Men a donde a es esta distancia la distancia del extremo libre a donde inicia la carga y esto a la un quiere decir que si lo aplico acá pues necesito la pendiente de esta recta recordemos que la pendiente es la altura sobre la base cierto o la constante de proporcionalidad es la altura sobre la base para el caso de un triángulo o para el caso mejor de una línea cuánto tenemos de altura Entonces si ustedes se fijan la altura acá Bueno entonces para esta constante Aquí está mi triángulo tomé el Triángulo porque recordemos que acá mírenla es este triángulo amarillo que teníamos hasta acá no este acá cuánto tenemos de altura Recuerden que aquí tenemos 300 que ya se fracciona acá hasta que hay 300 Esa es la altura del triángulo 300 y la base del triángulo pues son estos 3 m entonces aquí 300 y 3 m bueno estos 300 listo eh entonces la altura entre la base quedaría 300 que estos son newm cierto dividido en 3 m entonces la constante de proporcionalidad Pues nos da 100 bueno y las unidades realmente aquí No necesitamos unidades para la constante de proporcionalidad simplemente es el valor que vamos a colocar acá 100 luego como estamos utilizando esta cierto entonces Entonces wx igual al valor de la constante de proporcionalidad factor de X - a a la 1 entonces aquí wx = la constante de proporcionalidad 100 función de singularidad X - a miren que a es la distancia del extremo libre aquí está cierto la distancia del extremo libre a donde empieza la carga Pero nuestra carga empieza justo aquí en la orilla justo en el punto a Entonces eso es entonces aquí me va a quedar x - 0 cierto y esto dará 1 Esta es la función de singularidad esta es la función de singularidad para todo este triángulo para todo el Triángulo encima Pero a esta función de singularidad debo también sumarle la de este triángulo de abajo ahora lo que voy a tener en cuenta es la constante de proporcionalidad acá resulta que este triangulito es exactamente Este triangulito pero fesen que esta recta está en el otro sentido entonces la constante de proporcionalidad Sencillamente es -1 si aquí esta pendiente es positiva miren las pendientes que van así son positivas las pendientes que van opuestas a ellas como esta es negativa Entonces es la única diferencia pero es la misma pendiente entonces quiere decir que aquí vamos a colocar -1 cierto que es la constante de proporcionalidad y ahora viene la función de singularidad x- de acá la distancia que hay de donde inicia el Triángulo al extremo libre que es el punto a o bueno donde empieza la viga en a son 3 m entonces aquí vamos a escribir x - 3 y cerramos y ya tenemos la función de carga para esta Viga que es la equivalente esta Viga es la misma situación derivo ya sabemos que derivamos y obtenemos la función de entonces decimos que la función de cante es 100 100 función de singularidad ahora al cuadrado y esto sobre 2 menos los mismos 100 x - 3 la función de singularidad al cuadrado y esto sobre 2 bueno pero hay que hacer aquí varias cositas primero 100 sobre 2 Pues nos da 50 cierto Entonces esto nos da 50 este otro también 100 -2 Pues también nos va a dar 50 Aquí nos va a dar 50 listo por un lado por el otro lado recordemos que si aquí arrancamos con signo eh más arriba cierto ese 100 es positivo quiere decir que aquí es Negativo si aquí es negativo aquí es positivo y me falta sumar las reacciones aquí las dos reacciones porque ya en el cortante debo tener en cuenta esas dos reacciones Entonces como son cargas puntuales si yo vengo acá Aquí está cara punal cierto Cómo la sumo en el cortante Aquí está el cortante Entonces me dice menos la carga P O sea si la carga p va bajando entonces men p x - a a la 0 donde a es la distancia de aquí a la carga luego vamos con la primera con la de 650 entonces Ahá -650 entonces quitemos de una vez este más coloquemos -650 que es la carga Newton que es la carga p x - a en en este caso a es la distancia de donde está la carga al origen pero pues ese es cer0 y Esto va a cero cierto eso dice la función de singularidad miren para el caso del cortante - p x - a a la 0 donde a es la distancia que hay de acá del extremo libre a donde está ubicada la carga para la siguiente como está tamb bien carga puntual de 2,300 entonces aquí tendremos pero esta Va subiendo Entonces será más entonces aquí coloquemos 2300 factor o bueno función de singularidad X - a en este caso la 2300 está a 3 cierto aquí a 3 de distancia de acá Acá hay 3 a la 2300 entonces aquí -3 a la 0 y listo Esta es la función de carga ya Bueno A partir de la función de carga Recuerden que construimos la función de momento mdx Y entonces aquí si no se cambian los signos no se dejan igual lo único que se hace es integrar entonces me quedará -50 x - 0 cierto a la 3 sobre 3 + 50 fact de x - 3 como estaba aquí arriba a la 2 entonces recuerde que asumimos aquí sumamos 1 a la 3 sobre 3 y aquí - 650 x - 0 entonces aquí pasará a la 1 No aumenta en un grado + 2300 factor de x - 3 Aquí también subimos en un grado estaba la c y queda la 1 y ahí están las funciones de singularidad para mi Viga estas funciones estas funciones son las de esa Viga que estamos trí pero como esa Viga es equivalente a la viga que me dieron aquí mente quiere decir que estas son las funciones de singularidad de esta viga y ya con estas funciones vamos a hacer el diagrama entonces Los invito que en el otro video veamos la simplificación de estas funciones y el uso de estas funciones Cómo se utilizan ya a la hora de hacer el diagrama y Cómo saco de estas las equivalentes a las de corte a la forma a la forma de las funciones matemáticas algebraicas que que obtenía aquí por corte |
9504 | https://link.springer.com/book/10.1007/978-0-387-54109-9 | Skip to main content
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Quadratic Diophantine Equations
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© 2015
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Authors:
: Titu Andreescu
0,
Dorin Andrica
1
Titu Andreescu
School of Natural Sciences and Mathematics, University of Texas at Dallas, Richardson, USA
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2. Dorin Andrica
1. Faculty of Mathematics & Computer Science, "Babeş-Bolyai" University, Cluj-Napoca, Romania
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Includes both theoretical and computational examples
Explores new computational techniques for quadratic diophantine equations
Techniques presented will shed light on important open problems
Includes supplementary material: sn.pub/extras
Part of the book series:
Developments in Mathematics (DEVM, volume 40)
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This monograph treats the classical theory of quadratic Diophantine equations and guides the reader through the last two decades of computational techniques and progress in the area. These new techniques combined with the latest increases in computational power shed new light on important open problems. The authors motivate the study of quadratic Diophantine equations with excellent examples, open problems and applications. Moreover, the exposition aptly demonstrates many applications of results and techniques from the study of Pell-type equations to other problems in number theory.
The book is intended for advanced undergraduate and graduate students as well as researchers. It challenges the reader to apply not only specific techniques and strategies, but also to employ methods and tools from other areas of mathematics, such as algebra and analysis.
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Keywords
Pell's equation
algebra
diophantine equations
number theory
Table of contents (7 chapters)
Front Matter
Pages i-xviii
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2. ### Why Quadratic Diophantine Equations?
Titu Andreescu, Dorin Andrica
Pages 1-8
3. ### Continued Fractions, Diophantine Approximation, and Quadratic Rings
Titu Andreescu, Dorin Andrica
Pages 9-29
4. ### Pell’s Equation
Titu Andreescu, Dorin Andrica
Pages 31-53
5. ### General Pell’s Equation
Titu Andreescu, Dorin Andrica
Pages 55-105
6. ### Equations Reducible to Pell’s Type Equations
Titu Andreescu, Dorin Andrica
Pages 107-143
7. ### Diophantine Representations of Some Sequences
Titu Andreescu, Dorin Andrica
Pages 145-167
8. ### Other Applications
Titu Andreescu, Dorin Andrica
Pages 169-199
9. ### Back Matter
Pages 201-211
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Reviews
“The book under review is an excellent book on the interesting subject of quadratic Diophantine equations. It is well written, well organized, and contains a wealth of material that one does not expect to find in a book of its size, with full proofs of scores of theorems. … This reviewer does not know any book that covers similar material, and sees it as a very valuable and much needed addition to the literature on number theory.” (Mowaffaq Hajja, zbMATH 1376.11001, 2018)
“Diophantine analysis aims to solve equations (usually polynomial) in integers (or rationals). … this work settles the classical foundation, then develops state-of-the-art issues, especially concerning computation. … Summing Up: Recommended. Lower-division undergraduates through professionals/practitioners.” (D. V. Feldman, Choice, Vol. 53 (9), May, 2016)
“The primary focus of this book under review is the integer solutions of Pell equations, their generalisations and related diophantine equations, along with applications of these equations. … The book is suitable for readers from the level of a motivated undergraduate upwards, who are interested in the classical techniques for solving such equations. … There is also an extensive bibliography.” (Paul M. Voutier, Mathematical Reviews, March, 2016)
Authors and Affiliations
School of Natural Sciences and Mathematics, University of Texas at Dallas, Richardson, USA
Titu Andreescu
### Faculty of Mathematics & Computer Science, "Babeş-Bolyai" University, Cluj-Napoca, Romania
Dorin Andrica
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Bibliographic Information
Book Title: Quadratic Diophantine Equations
Authors: Titu Andreescu, Dorin Andrica
Series Title:
Developments in Mathematics
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Publisher: Springer New York, NY
eBook Packages:
Mathematics and Statistics, Mathematics and Statistics (R0)
Copyright Information: Springer Science+Business Media New York 2015
Hardcover ISBN: 978-0-387-35156-8Published: 30 June 2015
Softcover ISBN: 978-1-4939-3880-3Published: 09 October 2016
eBook ISBN: 978-0-387-54109-9Published: 29 June 2015
Series ISSN:
1389-2177
Series E-ISSN:
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Edition Number: 1
Number of Pages: XVIII, 211
Topics:
Number Theory,
Algebra
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9505 | https://www.youtube.com/watch?v=S6BHQMk8C_A | Points inside/outside/on a circle | Mathematics I | High School Math | Khan Academy
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Sal uses the distance formula to determine whether the point (-6,-6) is inside, outside, or on the circle centered at (-1,-3) with radius 6.
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Transcript:
a circle is centered at the point C which has the coordinates -1 comma -3 and has a radius of six where does the point P which is has the coordinates -6 comma -6 lie and we have three options inside the circle on the circle or outside the circle and the key realization here is just what a circle is all about if we have the point C which is the center of a circle a circle of radius six so let me draw that radius so let's say that is its radius it is six units the circle will look something like this remember the circle is a set of all points that are exactly six units away from that Center so that's that's the definition of a circle it's the set of all points that are exactly six units away from the center so if for example p is less than six units away it's going to be inside the circle if it's exactly six units away it's going to be on the circle and if it's more than six units away it's going to be outside of the circle so the key is is let's find the distance between these two points if the distance is less than six inside distance equals six we're on the circle distance more than six we are outside of the circle so let's do that so if we wanted to find and there's different notations for the distance well I'll just write D or I could write the distance between C and P is going to be equal to and the distance formula comes straight out of the Pythagorean theorem but it's going to be the square root of our change in x^2 plus our change in y^2 so what is our change in X so our change in X if we view c as our starting point and P as our end point but we could do it either way our change in X our change in X is -6 minus1 so6 minus -1 and we're going to square it so what we have inside here that is change in X so we're taking our change in X SAR and then plus our change in y^2 so we are going we're going from -3 to -6 so our change in y is -6 minus -3 -6- -3 and we're going to square everything so that is our change in y inside the parenthesis and we're going to square it so this is equal to this is equal to -66 plus positive 1 is one way to think about it so this is -5^ SAR and then this is -6 + 3 so+ -3 2ar and once again you can see our change in x isg five we go five lower in X and we're going F three lower in y so our change in y is3 so this is equal to the square OT of 25 square > of 25 + 9 < TK of 25 + 9 which is equal to the square < TK of 34 now the key is is the square OT of 34 less than 6 greater than six or equal to six well we know that 6 is equal to the square < TK of 36 so the Square < t of 34 is less than theare < of 36 so I could write theun of 34 is less than theare < TK of 36 and so the < TK of 34 is less than 6 these or squ 36 is 6 and so since the distance between C and P is less than six we are going to be on the inside of the circle if I somehow got square of 36 here then we'd be on the circle and if I somehow got square root of 37 here or something larger we would have been outside the circle |
9506 | https://www.orpha.net/en/disease/detail/444490 | Share
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Familial chylomicronemia syndrome
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Disease definition
A rare genetic hyperlipidemia characterized by excessive increase in plasma triglyceride levels due to the accumulation of chylomicrons, which manifests biochemically as severe hypertriglyceridemia. Clinical manifestations include recurrent episodes of acute pancreatitis, abdominal pain, nausea, fatigue, diarrhea, hepatosplenomegaly, eruptive xanthomas, lipemia retinalis and failure to thrive. Children may be asymptomatic with later onset of symptoms. The condition is not associated with severe atherosclerosis.
ORPHA:444490
Classification level: Disorder
Prevalence: 1-9 / 1 000 000
Inheritance: Autosomal recessive
Age of onset: Adolescent, Adult, Childhood, Infancy
ICD-10: E78.3
ICD-11: 5C80.1
OMIM: 145750 118830 207750 238600 615947 144650
UMLS: C5442313
GARD: 6414
Summary
Epidemiology
Familial chylomicronemia syndrome (FCS) has an estimated prevalence of 1/300,000 (ranging from 1/100,000 to 1/1,000,000 in Europe and North America). Founder effects are seen in Quebec and the Cayman Islands.
Clinical description
Presentation in infancy includes failure to thrive, abdominal pain, nausea and vomiting progressing to acute pancreatitis. Symptoms and signs include fatigue, irritability, lipemia retinalis, eruptive xanthomas on trunk, back and gluteal region, and hepatosplenomegaly. Lipemic plasma indicates severely elevated plasma triglyceride levels due to pathological presence of chylomicrons (>10 mmol/L or >875 mg/dL).
Etiology
Most FCS patients have bi-allelic loss-of-function variants in LPL encoding lipoprotein lipase. Other minor causal genes encoding factors that interact with LPL, including apolipoprotein (apo) A-V (APOA5), apo C-II (APOC2), glycosylphosphatidylinositol-anchored high-density lipoprotein binding protein 1 (GPIHBP1) and lipase maturation factor 1 (LMF1).
Diagnostic methods
Definitive diagnosis is made by DNA sequencing. Biochemical methods are non-specific with technical performance challenges.
Differential diagnosis
In young adults, multifactorial (polygenic) chylomicronemia is 50 to 100 times as common as FCS and is associated with many secondary factors, hepatosteatosis, renal impairment or proteinuria. Patients with autoimmune diseases can develop antibodies against GPIHBP1 and develop chylomicronemia secondarily. Patients with undiagnosed partial lipodystrophy may present with chylomicronemia and pancreatitis.
Antenatal diagnosis
Antenatal diagnosis is not commonly offered, since FCS is manageable with dietary intervention.
Genetic counseling
Transmission is autosomal recessive. Genetic counseling should be offered to at-risk couples (i.e. parents of a known FCS child or where both individuals are known carriers of a disease-causing mutation) informing them that there is a 25% risk of having an affected child at each pregnancy.
Management and treatment
The mainstay of treatment is fat restriction, reducing total fat to <10% of calories or <20 grams per day, in combination with weight maintenance, exercise, and avoidance of processed foods, alcohol and smoking. Standard lipid-lowering medications and plasmapheresis are ineffective. Volanesorsen (apo C-III antisense oligonucleotide) is available in the EU but not North America; similar agents are in development.
Prognosis
Some patients have lived for six decades after the initial FCS diagnosis with minimal sequelae reportedly by adhering to a strict diet. Chylomicrons are not atherogenic, so atherosclerosis is uncommon in FCS patients, unlike other severe hypertriglyceridemia patients.
Last update: March 2023 - Expert reviewer(s): Dr Robert HEGELE
A summary on this disease is available in Français, Español, Deutsch, Italiano, Português, Nederlands,
Detailed information
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Diagnostic tests (78)
Patient organisation(s) (130)
Federation/alliance(s) (44)
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The documents contained in this website are presented for information purposes only. The material is in no way intended to replace professional medical care by a qualified specialist and should not be used as a basis for diagnosis or treatment. |
9507 | https://physics.info/motion-equations/practice.shtml | The Physics
Hypertextbook
Opus in profectus
Equations of Motion
Practice
practice problem 1
The speed limit of a particular section of freeway is 25 m/s. The right travel lane is connected to an exit ramp with a short auxiliary lane. Cars would have a comfortable deceleration of −2.0 m/s2 for 3.0 s in the auxiliary lane if they were driving at the speed limit.
What speed will cars have when they are done decelerating in this way? (This is also the speed limit of the exit ramp.)
What minimum length should the auxiliary lane be to allow for this deceleration?
Drivers don't always drive at the speed limit, and highway engineers take this into consideration.
Assume a car could decelerate at four times the "comfortable" rate without losing control. At what maximum speed could a car enter an auxiliary lane with the length you calculated in part b. and still exit at the intended speed?
Assume a driver was traveling on the freeway at the speed you calculated in part c. What distance is needed to decelerate this car to the speed limit of the exit ramp at the "comfortable" rate ?
solution
State the givens and the unknown. Use the first equation of motion — the one where speed is a function of time.
| | |
--- |
| v0 = | 25 m/s |
| a = | −2.0 m/s2 |
| ∆t = | 3.0 s |
| v = | ? |
| |
| v = v0 + at v = (25 m/s) + (−2.0 m/s2)(3.0 s) v = 19 m/s |
2. Restate the givens from the previous part, since they're all still valid, but now is distance is the unknown. Use the second equation of motion — the one where distance is a function of time.
| | |
--- |
| v0 = | 25 m/s |
| a = | −2.0 m/s2 |
| ∆t = | 3.0 s |
| ∆s = | ? |
| |
| ∆s = v0t + ½at2 ∆s = (25 m/s)(3.0 s) + ½(−2.0 m/s2)(3.0 s)2 ∆s = 66 m |
3. The final speed calculated in part a. is still the final speed. (All cars need to exit at the same speed.) The distance calculated in part b. is still the distance. (All cars get the same amount of space to slow down.) The new acceleration is four times the old one. The new initial speed is the new unknown. No time is given or can be inferred. None is needed. Use the third equation of motion — the one where speed is a function of distance and time is not a part of the equation.
| | |
--- |
| v = | 19 m/s |
| a = | −8.0 m/s2 |
| ∆s = | 66 m |
| v0 = | ? |
| |
| v2 = v02 + 2a∆s v02 = v2 − 2a∆s v02 = (19 m/s)2 − 2(−8.0 m/s2)(66 m) 2v0 = 37.6 m/s |
4. The final speed calculated in part a. is still the final speed. (All cars still need to exit at the same speed.) The acceleration is the value given at the start of the problem. (Let's allow the speed demons to accelerate comfortably, too.) The new initial speed is the value we just calculated. The new distance is the new unknown. Again no time is given or can be inferred, so again use the third equation of motion.
| | |
--- |
| v = | 19 m/s |
| a = | −2.0 m/s2 |
| v0 = | 37.6 m/s |
| ∆s = | ? |
| |
| ∆v2 = v02 + 2a∆s 2∆s = (v2 − v02)/2a 2∆s = [(19 m/s)2 − (37.6 m/s)2]/[2(−2.0 m/s2)] 2∆s = 264 m |
practice problem 2
A car with an initial velocity of 60 mph needs 144 feet to come to a complete stop. Determine the stopping distance of this same car with an initial velocity of…
30 mph
20 mph
10 mph
Note: The rate of change of velocity is not affected by inital velocity in this problem. Fast cars and slow cars slow down at the same rate.
solution
First method.
The hard way to solve this problem is to do it the way that many students think is the easy way — "numbers in, answers out" or "plug and chug". This method appears easy since it requires little thought, but it turns out to be quite demanding.
First, convert to SI units.
| | | | | | |
--- --- --- |
| 60 mile | | 1609 m | | 1 hour | = 26.8 m/s |
| 1 hour | 1 mile | 3600 s |
| 30 mile | | 1609 m | | 1 hour | = 13.4 m/s |
| 1 hour | 1 mile | 3600 s |
| 20 mile | | 1609 m | | 1 hour | = 8.94 m/s |
| 1 hour | 1 mile | 3600 s |
| 10 mile | | 1609 m | | 1 hour | = 4.47 m/s |
| 1 hour | 1 mile | 3600 s |
| | | | | | |
| 144 feet | | 1 mile | | 1609 m | = 43.9 m |
| 1 | 5280 feet | 1 mile |
Then calculate the deceleration from 60 mph.
| | |
--- |
| v0 = | 26.8 m/s |
| v = | 0 m/s |
| ∆s = | 43.9 m |
| a = | ? |
| | |
| | | |
---
| v2 = | v02 + 2a∆s | |
| |
| a = | | | | v2 − v02 | | 2∆s | | |
| |
| a = | −(26.8 m/s)2 | |
| 2(43.9 m) | |
| a = −8.18 m/s2 | | |
| |
Then use this number to calculate the distances for the other speeds.
v2 = v02 + 2a∆s
Eliminate the zero term and solve for displacement.
| | |
--- |
| ∆s = | − v02 |
| 2a |
Numbers in. Answers out.
| | | |
---
| ∆s = | −(13.4 m/s)2 | = 11.0 m |
| 2(−8.18 m/s2) |
| ∆s = | −(8.94 m/s)2 | = 4.89 m |
| 2(−8.18 m/s2) |
| ∆s = | −(4.47 m/s)2 | = 1.22 m |
| 2(−8.18 m/s2) |
And finally, convert back into English units.
| | | | | | |
--- --- --- |
| 11.0 m | | 1 mile | | 5280 feet | = 36 feet |
| 1 | 1609 m | 1 mile |
| 4.89 m | | 1 mile | | 5280 feet | = 16 feet |
| 1 | 1609 m | 1 mile |
| 1.22 m | | 1 mile | | 5280 feet | = 04 feet |
| 1 | 1609 m | 1 mile |
Second method.
Standard problem solving techniques work, but they're a monumental waste of time for this problem. Any small error would destroy the answers and waste personal mental energy, which is something we'd all like to avoid. The easy way to solve this problem does not involve any trickery. It requires that you identify and understand the key concepts needed to solve the problem. Halfway through the mass of equations, an important assumption was made. It was assumed that the braking acceleration of the car would remain constant for all initial velocities. This problem is then one of determining the relationship between displacement and velocity. The equation that does this is…
v2 = v02 + 2a∆s
which shows that displacement is proportional to velocity squared (when acceleration is constant and either the the initial or final velocity is zero).
∆s ∝ v2
In this problem we're comparing stopping distances at 30, 20, and 10 mph to those at 60 mph. The square of the ratio of the new velocity to the original velocity will be the ratio of the new stopping distance to the original stopping distance.
| | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- ---
| v2 | | | | | | | | ∝ | ∆s | | | | | | | |
| | | | | | | | | | | | | | | | | |
| ⎛ ⎜ ⎝ | 30 mph | ⎞2 ⎟ ⎠ | = | ⎛ ⎜ ⎝ | 1 | ⎞2 ⎟ ⎠ | | = | | 1 | = | 36 feet | | | | |
| 60 mph | 2 | | | 4 | 144 feet | | | | |
| ⎛ ⎜ ⎝ | 20 mph | ⎞2 ⎟ ⎠ | = | ⎛ ⎜ ⎝ | 1 | ⎞2 ⎟ ⎠ | | = | | 1 | = | 16 feet | | | | |
| 60 mph | 3 | | | 9 | 144 feet | | | | |
| ⎛ ⎜ ⎝ | 10 mph | ⎞2 ⎟ ⎠ | = | ⎛ ⎜ ⎝ | 1 | ⎞2 ⎟ ⎠ | | = | | 1 | = | 04 feet | | | | |
| 60 mph | 6 | | | 36 | 144 feet | | | | |
These are the same answers we got using the "plug and chug" method.
practice problem 3
A typical commercial jet airliner needs to reach a speed of 180 knots before it can take off. (A knot is a nautical mile per hour and is nearly equal to half a meter per second.) If such a plane spends 30 s on the runway estimate…
its acceleration.
the minimum runway length.
solution
To determine acceleration, I recommend using the definition of acceleration.
List the givens and unknown first.
| | |
--- |
| v0 = | 0 m/s |
| v = | 180 kts ≈ 90 m/s |
| t = | 30 s |
| a = | ? |
Then use the equation.
| | | | |
--- --- |
| | | | --- | | a = | ∆v | | ∆t | |
| | | | --- | | a = | 90 m/s | | 30 s | |
| | | | --- | | a = 3 m/s2 ≈ ⅓ g | | | | |
2. There are two ways to determine the runway length. Either method yields the same solution. Let's list the givens and unknown first.
| | |
--- |
| v0 = | 0 m/s |
| v = | 180 kts ≈ 90 m/s |
| t = | 30 s |
| a = | 3 m/s2 |
| ∆s = | ? |
Here's the solution using the distance-time equation.
| | |
--- |
| s = | s0 + v0t + ½at2 |
| ∆s = | ½(3 m/s2)(30 s)2 |
| ∆s = 1350 m | |
And here's the solution using the average velocity equation.
| |
| ∆s = vt = ½(v + v0)t ∆s = ½(90 m/s + 0 m/s)(30 m/s) ∆s = 1350 m |
practice problem 4
A 10 car subway train is sitting in a station. It reaches its cruising speed after accelerating at 0.75 m/s2 for distance equivalent to the length of the station (184 m). It then travels at a constant speed towards the next station 18 blocks away (1425 m).
Determine the train's cruising speed.
Determine the time it took for the train to accelerate from rest to its cruising speed.
How long does it take the train to travel the 18 blocks to the next station?
The driver stops the train in the second station in half the distance it took to start it at the first station.
What is deceleration of the train in the second station?
solution
List the givens and the unknown for the train as it departs the station. Pick an appropriate equation of motion. I suggest the velocity-displacement equation, a.k.a. the third equation of motion. Almost no algebra is needed. Put numbers in. Get answer out.
| | |
--- |
| v0 = | 0 m/s |
| a = | 0.75 m/s2 |
| ∆s = | 184 m |
| v = | ? |
| | |
--- |
| v2 = | v02 + 2a∆s |
| v2 = | 2(0.75 m/s2)(184 m) |
| v = | 16.6 m/s |
This is about 60 km/h or 37 mph
2. There are several ways to determine the time it took to reach cruising speed.
List only the quantities given in the problem and state the new unknown. Pick a new equation. I suggest the displacement-time equation, a.k.a. the second equation of motion. Some algebra is needed. This is followed by the usual numbers in, answer out.
| | |
--- |
| v0 = | 0 m/s |
| a = | 0.75 m/s2 |
| ∆s = | 184 m |
| t = | ? |
| | |
--- |
| ∆s = | v0t + ½at2 |
| t2 = | 2∆s/a |
| t2 = | 2(184 m)/(0.75 m/s2) |
| t = | 22.2 s |
We could also use the results of our first calculation and add it to the list of known quantities. Adding the final speed to this list means we could use the velocity-time equation, a.k.a. the first equation of motion. We wouldn't need the displacement of the subway if we did this. We would still need a little bit of algebra, however. (Also, the answer may be slightly different depending on how many digits you saved from your calculation for part a.)
| | |
--- |
| v0 = | 0 m/s |
| v = | 16.6 m/s |
| a = | 0.75 m/s2 |
| t = | ? |
| | |
--- |
| v = | v0 + at |
| t = | v/a |
| t = | (16.6 m/s)/(0.75 m/s2) |
| t = | 22.2 s |
We could also use every known and calculated quantity plus the two equations for average speed and some algebra. I don't recommend this method, but it works.
| | |
--- |
| v0 = | 0 m/s |
| v = | 16.6 m/s |
| a = | 0.75 m/s2 |
| ∆s = | 184 m |
| t = | ? |
| | | | |
--- --- |
| v = | ∆s | = | v0 + v |
| ∆t | 2 |
| t = | | | | 2∆s | | v | | | |
| t = | | | | 2(184 m) | | 16.6 m/s | | | |
| t = | 22.2 s | | |
| |
3. After the train leaves the station, it travels with a constant speed. That makes for an easy problem. No need to distinguish between initial speed, final speed, and average speed anymore. The speed is just the speed.
| | |
--- |
| v = | 16.6 m/s |
| ∆s = | 1425 m |
| ∆t = | ? |
| | |
--- |
| v = | ∆s/∆t |
| ∆t = | v/∆a |
| ∆t = | (1425 m)/(16.6 m/s) |
| ∆t = | 85.8 s |
4. This could be done as a GUESS problem (given, unknown, equation, substitute, solve), but given the way it's worded this is a proportion problem.
The initial and final velocities get switched, which means the sign will change but the absolute value of the change is unchanged. These are our constants. We know nothing about the time. We could make an inference about it, but let's not. Distance is halved and acceleration is the goal. Find an equation with initial and final velocities, acceleration, and distance — but not time.
v2 = v02 + 2a∆s
The two changing quantities are in the same term. Everything else is essentially constant, but with a sign switch. That means that the product of acceleration and distance is also constant. They're inversely proportional.
a ∝ 1∆s
Half the distance means twice the acceleration. That plus a sign switch gives us…
a = −2(0.75 m/s2)
a = −1.50 m/s2
| | |
--- |
| | | |
9508 | https://www.khanacademy.org/math/cc-sixth-grade-math/x0267d782:coordinate-plane/cc-6th-coordinate-plane/v/graphing-points-and-naming-quadrants-exercise | Points and quadrants example (video) | Khan Academy
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6th grade math
Course: 6th grade math>Unit 9
Lesson 1: Four quadrants
Points on the coordinate plane examples
Plotting a point (ordered pair)
Finding the point not graphed
Points on the coordinate plane
Points on the coordinate plane
Quadrants of the coordinate plane
Points and quadrants example
Quadrants on the coordinate plane
Coordinate plane parts review
Graphing coordinates review
Math>
6th grade math>
Coordinate plane>
Four quadrants
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Points and quadrants example
GA.Math: 5.PAR.6.2, 6.PAR.8.1, 6.PAR.8.2, 6.PAR.8.3
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About About this video Transcript
Plotting points on a graph involves moving along the x and y axes. Positive x-values move right, negative x-values move left. Positive y-values move up, negative y-values move down. The graph is divided into four quadrants, helping us locate points easily.Created by Sal Khan.
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scott stevenson 12 years ago Posted 12 years ago. Direct link to scott stevenson's post “quadrants are numbered in...” more quadrants are numbered in counterclockwise order starting at the upper right quadrant, how would octants be numbered? starting where and what order?
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Beabfekad Zikie 5 years ago Posted 5 years ago. Direct link to Beabfekad Zikie's post “In which quadrant does th...” more In which quadrant does the point (0,1) lie?
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TJ 5 years ago Posted 5 years ago. Direct link to TJ's post “Points that lie directly ...” more Points that lie directly on the x and y axis aren't considered to be in a quadrant. In this case, you would just say that the point (0, 1) lies on the y-axis
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jirehrhipolito 4 years ago Posted 4 years ago. Direct link to jirehrhipolito's post “Why are the quadrants des...” more Why are the quadrants designated as such? Is there any particular reason? Thank You.
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Jasper Wang 4 years ago Posted 4 years ago. Direct link to Jasper Wang's post “It's just a convention, I...” more It's just a convention, I believe. Some other places have different orders for their quadrants, though I think the regular order (shown by Sal) is in the order you encounter them as directional angles increase. Hope this helps :D
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sarah 11 years ago Posted 11 years ago. Direct link to sarah's post “Why do we need to learn h...” more Why do we need to learn how to plot?
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Elias Jay Hudson 11 years ago Posted 11 years ago. Direct link to Elias Jay Hudson's post “For programming purposes ...” more For programming purposes (to map images), and also to plot real areas of space.
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laroshc29 2 years ago Posted 2 years ago. Direct link to laroshc29's post “Just to ask, is it a nece...” more Just to ask, is it a necessity to say "comma" in between numbers? because I think it's just [7,7] is read "7 by 7."
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Werdna 2 years ago Posted 2 years ago. Direct link to Werdna's post “or"seven seven".” more or"seven seven".
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Brian Buckley 2 years ago Posted 2 years ago. Direct link to Brian Buckley's post “who made up the naming co...” more who made up the naming convention
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Julia C 9 years ago Posted 9 years ago. Direct link to Julia C's post “So the quadrants always s...” more So the quadrants always start in the top right corner, and it proceeds in a counterclockwise manner? I'm just making sure I'm fully understanding which quadrant is which, because I'm pretty sure I'm going to forget which is which.
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p.m. 9 years ago Posted 9 years ago. Direct link to p.m.'s post “That's right! It is kind...” more That's right! It is kind of hard to remember.
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Max Chow =) 2 years ago Posted 2 years ago. Direct link to Max Chow =)'s post “why have 4 boxs?” more why have 4 boxs?
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🫧𓇼𓏲ੈ✩‧₊˚🐋🐚 2 years ago Posted 2 years ago. Direct link to 🫧𓇼𓏲ੈ✩‧₊˚🐋🐚's post “Quadrant one (QI) is the ...” more Quadrant one (QI) is the top right fourth of the coordinate plane, where there are only positive coordinates. Quadrant two (QII) is the top left fourth of the coordinate plane. Quadrant three (QIII) is the bottom left fourth. Quadrant four (QIV) is the bottom right fourth.
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Demetrios V 9 years ago Posted 9 years ago. Direct link to Demetrios V's post “What does Q stand for on ...” more What does Q stand for on the answer key?
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Opal 2 years ago Posted 2 years ago. Direct link to Opal's post “Quadrant” more Quadrant
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Snoopy 10 years ago Posted 10 years ago. Direct link to Snoopy's post “Why do we need to do x ax...” more Why do we need to do x axis first and then y axis? Thanks, have a good day!
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Sophia 7 years ago Posted 7 years ago. Direct link to Sophia's post “I'm pretty sure this is n...” more I'm pretty sure this is not in the video, but x is the independant variable and y is the dependant variable. In a line, for example, y=x+1, whatever x is, y will change with it. It does not always make sense, but y changes accordingly to x.
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Video transcript
Plot 4 comma negative 1, and select the quadrant in which the point lies. So 4, the first number in our ordered pair, that's our x-coordinate. That says how far to move in the horizontal or the x-direction. It's a positive 4. So I'm going to go 4 to the right. And then the second coordinate says, what do we do in the vertical direction or in the y-direction? It's a negative 1. Since it's negative, we're going to go down. And it's a negative 1, so we're going to go down 1. So that right over there is the point 4 comma negative 1. So I've plotted it, but now I have to select which quadrant the point lies in. And this is just a naming convention. This is the first quadrant. This is the second quadrant. This is the third quadrant. And this is the fourth quadrant. So the point lies in the fourth quadrant, quadrant IV. And I guess you have to know your Roman numerals a little bit to know that's representing quadrant IV. Let's do a couple more of these. Plot 8 comma negative 4, and select the quadrant in which the point lies. Well, my x-coordinate is 8 so I go 8 in the positive x-direction. And then my y-coordinate is negative 4, so I go 4 down. And this is sitting, again, in not the first, not the second, not the third, but the fourth quadrant, in quadrant IV. Let's do one more of these. Hopefully we get a different quadrant. So we want to plot the point negative 5 comma 5. So now my x-coordinate is negative. It's negative 5. So I'm going to move to the left in the x-direction. So I go to negative 5. And my y-coordinate is positive so I go up 5, so negative 5 comma 5. And this is sitting not in the first quadrant, but the second quadrant. And of course, this is the third and the fourth. So this is sitting in the second quadrant. Check answer, and we got it right.
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9509 | http://brennen.caltech.edu/fluidbook/externalflows/vortexshedding/introduction.pdf | An Internet Book on Fluid Dynamics Introduction to Vortex Shedding As we discussed in section (Dbc), the wake behind a cylinder in a uniform stream grows as the Reynolds number is increased above about Re = 5. Between 5 < Re < 50 the steady attached vortices that form the wake behind the cylinder grow with Re. However about Re = 50 the width of the wake approaches the diameter of the cylinder and the wake becomes unstable in that it begins to oscillate back and forth behind the cylinder, veering from one side to the other. About Re = 60 this process leads to vortices breaking off and being convected downstream, with vortices alternately being shed from one side and then the other as illustrated in Figure 1. This phenomena occurs with a wide range of objects of many different sizes and Figure 1: Karman vortex street behind a cylinder. Reproduced with permission of T.Colonius.
shapes. Figure 2 presents another example, the large cavitating vortices behind a cavitating hydrofoil.
Figure 2: A cavitating vortex street behind a hydrofoil with a flap.
The pattern of alternate vortex shedding is known as a Karman vortex street and it continues as the Reynolds number increases from 60 to 5000 when a steadier laminar wake forms behind the cylinder.
However, vortices continue to be shed further downstream at the rear of the laminar wake as indicated by the flow visualized in Figure 3.
Figure 3: Formation of vortices in the wake of a circular cylinder.
The frequency of vortex shedding, f (Hz), is non-dimensionalized in the Strouhal number, St = 2fR/U, where U is the velocity of the impinging uniform stream and R is the radius of the cylinder. As illustrated by the data presented in Figure 4, experiments show that the measured Strouhal numbers remain remarkably constant at a value around 0.2 as the Reynolds number is increased from about 60 to 107 though the shedding begins to lose coherence around Re = 105 as turbulence invades the flow from the wake.
Because it occurs over such a wide range of Reynolds numbers (and object shapes), vortex shedding is a ubiquitous phenomenon. It occurs with scenarios as diverse as an oscillating violin string and large underwater structures (particularly when they are smooth or cylindrical). Figure 5 represents a particular meteorological example in which clouds in the atmosphere reflect the vortices shed by the island of Tristan de Cunha. Vortex shedding has also caused destruction of many man-made structures because the lateral forces generated during the shedding of each vortex can be very substantial.
Early in the industrial revolution tall brick chimney stacks would sometimes sway and even break under the oscillatory lateral loads generated during vortex shedding. In modern times projections or ribs are often attached to large circular structures in order to fix the location of flow separation and thus stabilize the flow by minimizing the vortex shedding; Figure 6 shows a section of modern chimney with such fins attached. Also, in the early Figure 4: Measured Strouhal numbers for a circular cylinder as a function of Reynolds number. Adapted from Roshko (1954), Jones (1968) and White (1999).
Figure 5: Karman vortex street reflected in the clouds above the island of Tristan de Cunha.
days of suspension bridges the vibration induced by vortex shedding sometimes caused the destruction of those bridges. Perhaps the most famous case, shown in Figure 7 was the self-destruction of the Tacoma Narrows suspension bridge.
Figure 6: Fins on large industrial chimney.
Figure 7: Failure of the Tacoma Narrows suspension bridge as a result of vortex shedding. |
9510 | http://thejavamathematician.blogspot.com/2015/03/horners-rule-for-polynomial-computation.html | Monday, March 9, 2015
Horner's Rule for Polynomial Computation
Suppose I had a polynomial anxn + an-1xn-1 + … + a1x + a0, and a point x at which I wanted to evaluate that polynomial. Now our immediate inclination is to just substitute the point straight into the polynomial to get an(x)n + an-1(x)n-1 + … + a1(x) + a0 and then work it through in the obvious way: we raise x to the nth power and multiply by an, then we raise x to the (n-1)th power and multiply by an-1, and so on, adding them all together at the end. That seems like a lot of work though. We could be more efficient by working it the other way around: that is, starting with x and caching the intermediate powers of x as we work our way up. That would definitely cut down on some of the multiplications. But is there actually an even better way to do this polynomial evaluation? As it turns out, there is – we can use Horner's Rule! Not only is Horner's Rule more efficient than either of the above approaches, it is actually an optimal algorithm for polynomial computation: that is, we can prove mathematically that any other rule for polynomial computation can do no better than Horner's Rule. So let's investigate how this rule works, and write up some Java code to test it out! [We're back with the SICP exercise blog posts! This one is inspired by Exercise 2.34.]
Polynomial Computation – The Obvious Way
For the purposes of comparison, let's quickly review the ways that we would have otherwise calculated a polynomial. We'll start with the most obvious way:
Here we use result as a variable to hold the intermediate answers each time. By the end, the number in result will be the value of the polynomial at that point. So for example, let's say we had the polynomial 5x4 + 2x3 – 3x2 + x – 7, and we wanted the value of the polynomial at the point x=3. We'd work it through like this:
I've written it in a more "math-like" form here, but notice that we're essentially working through the above set of steps. We compute each of the terms (and each of the powers of x=3) separately, and add them up as we go across. We could easily write some Java code to do this, right? We have a class Polynomial1 with a method computePolynomial() that takes in a polynomial (represented here as an array of doubles, where the ith element of the array is the coefficient ai of the polynomial), and a value x at which we are to evaluate the polynomial. In the computePolynomial() method, we start from the nth term of the polynomial, and determine the value of that term. Then as we iterate in the for loop we work our way down, accumulating the terms in result. So that we can compare each of the polynomial evaluation methods that we'll look at, we've also defined some static variables totalAdditions and totalMultiplications to keep track of those operations when the program runs. These get updated in the loop, and we'll display the values of those variables in the output at the end. Running our little program on that polynomial example we had above, we get the following: java Polynomial1 //> Polynomial: 5x^4 + 2x^3 - 3x^2 + x - 7 //> x=3: 428.0 //> Total additions: 5 //> Total multiplications: 10 We got the right answer at least. But we're doing quite a few multiplications for such a small polynomial! In general, computing an nth-degree polynomial this way means that we end up doing n additions and (0 + 1 + 2 + … + n) multiplications (since to get the aixi term we multiply x by itself (i-1) times, and then multiply on the ai coefficient to give us i multiplications for that step.) This can be seen in the number of additions and multiplications we had for both the worked solution, and our program output (we've got an extra addition showing up in the program since we added the very first term to result, which is initially 0.0. We could always rewrite the program to make that first addition an assignment instead if we wanted.) So we can summarise the efficiency of this polynomial computation scheme as follows:
We can easily see here that the number of multiplications is going to grow quadratically as the degree of the polynomial increases. We can do better than this though, of course.
Polynomial Computation – Caching Powers of x
As we mentioned earlier, we can already improve the efficiency of the computation by noticing that we're calculating many of the same powers of x for each term. Consider that example with the polynomial 5x4 + 2x3 – 3x2 + x – 7. Here we end up calculating x2 three separate times, since we would have calculated x2 as we were calculating both x3 and x4! So one way to get an improvement is to cache those intermediate powers of x. Instead of starting with the anxn term and working our way down, we'll start from the other end of the polynomial and work our way up – this means that we'll have the previous powers of x on hand that we can use to calculate the higher-order terms. Our new set of steps looks like this:
So you can see that we keep x2 around so we can use it to get x3, which we keep around to help us get x4, and so on. If we went through our example by hand using this method, we would get the following working:
And of course, this easily translates into another small program that we can use to compute polynomials: This Polynomial2 class is very similar to the class we had before, and our computePolynomial() method takes the same input and returns the same output. But instead of starting at the anxn term and working our way down the polynomial, this time around we start at the a0 term (which we can make an assignment), and work our way up. We also use the variable powerOfX to keep the previous x term around, so that we can use it to compute the next term of the polynomial. This should work to reduce the number of multiplications that we end up doing. When we run the code, the output is as follows: java Polynomial2 //> Polynomial: 5x^4 + 2x^3 - 3x^2 + x - 7 //> x=3: 428.0 //> Total additions: 4 //> Total multiplications: 8 So we've managed to reduce the number of multiplications with this new method! And since we assigned the a0 coefficient instead of adding it to 0, we've only had 4 additions this time too.
Computing an nth-degree polynomial this way will always take n additions and
2 + 2 + … + 2 = 2n multiplications:
This is better – at least the number of multiplications we need to do now grows linearly with the degree of the polynomial.
Horner's Rule
We can reduce the number of multiplications even further. Consider that we can always rewrite a polynomial in the following equivalent way:
This is the main idea of Horner's Rule, which is named after the mathematician William George Horner. Basically, we start with the an coefficient, multiply it by x and add on the an-1 coefficient. Then we multiply that whole result by x and add on the an-2 coefficient. Then we multiply everything by x again, and add on the an-3 coefficient, and so on for the rest of the polynomial terms. You can see that by working through it that way, we should get the right powers of x coming out: an ends up multiplied with x n times, an-1 ends up multiplied by x n-1 times, and similarly right down to the a1 term, which is multiplied by x once. The computation algorithm captures this process exactly:
And that's the entire algorithm! (It even looks prettier than the other algorithms we had above.) As an example, if we were to work through our polynomial example by hand, we'd do the following:
And this algorithm is easy enough to translate into working Java code, too. As you can see, in our computePolynomial() method we assign the an coefficient to result to start with, and then for each iteration of the loop, we multiply the current value of result by x and add on the next coefficient. Running the code gives us the following output for our example: java HornersRule //> Polynomial: 5x^4 + 2x^3 - 3x^2 + x - 7 //> x=3: 428.0 //> Total additions: 4 //> Total multiplications: 4 So we've now got 4 additions and 4 multiplications, which is nice! As the final point of this blog post, we can notice that computing an nth-degree polynomial with Horner's Rule will always take only n additions and multiplications:
And in fact, this is the best we can do. The mathematician Alexander Ostrowski proved in 1954 that Horner's Rule uses the optimal number of additions, and another mathematician Victor Pan proved in 1966 that Horner's Rule uses the optimal number of multiplications. So Horner's Rule is an optimal algorithm for polynomial computation!
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9511 | https://en.wikipedia.org/wiki/Acceleration_due_to_gravity | Jump to content
Acceleration due to gravity
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Acceleration due to gravity, acceleration of gravity or gravitational acceleration may refer to:
Gravitational acceleration, the acceleration caused by the gravitational attraction of massive bodies in general
Gravity of Earth, the acceleration caused by the combination of gravitational attraction and centrifugal force of the Earth
Standard gravity, or g, the standard value of gravitational acceleration at sea level on Earth
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9512 | https://www.chegg.com/homework-help/questions-and-answers/define-function-1-2-2-g-x-x-1-2-x-2-also-define-derivative-g-x--plot-function-2-25x-2-25-r-q85866685 | Your solution’s ready to go!
Our expert help has broken down your problem into an easy-to-learn solution you can count on.
Question: Define the function 𝑔(𝑥)=(𝑥+1)2(𝑥−2)g(x)=(x+1)2(x−2), and also define its derivative 𝑔′(𝑥)g′(x). Plot the function from 𝑥=−2…2.5x=−2…2.5 In [ ]: Run the FindRootNewtonRaphson command with the final (optional) argument set to True to see the steps the algorithm takes. Run it with: the initial value of 𝑥0=−1.5x0=−1.5 the initial value of
Define the function 𝑔(𝑥)=(𝑥+1)2(𝑥−2)g(x)=(x+1)2(x−2), and
also define its derivative 𝑔′(𝑥)g′(x). Plot the function
from 𝑥=−2…2.5x=−2…2.5
In [ ]:
Run the FindRootNewtonRaphson command with the final (optional)
argument set to True to see the steps the algorithm takes. Run it
with:
In [ ]:
What do you notice about the convergence rates in these two
cases? Let's examine it more carefully. The function below returns
the individual steps of the Newton-Raphson algorithm.
In :
Using the above function:
In :
Plot the absolute difference between the steps and the relevant
root, i.e., |𝑥𝑛−𝑥∗||xn−x∗|, on a log scale.
In [ ]:
You should find the repeated difference for the repeated root
looks linear on the log-scale, which is the same as we saw for the
bisection method. This (correctly) suggests that for repeated roots
the Newton-Raphson converges linearly (rather than
quadratically)
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9513 | https://www.mathsisfun.com/time-add-subtract.html | Adding and Subtracting Time
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Adding and Subtracting Time
Add or subtract the hours and minutes separately.
But you may need to do some adjusting if the minutes end up 60 or more, or less than zero!
Adding Times
Follow these steps:
Add the hours
Add the minutes
If the minutes are 60 or more, subtract 60 from the minutes and add 1 to hours
Like this:
Easy example: What is 2:45 + 1:10 ?
Add the Hours: 2+1 = 3
Add the Minutes: 45+10 = 55
The minutes are OK, so the answer is 3:55
Hard example: What is 2:45 + 1:20 ?
Add the Hours: 2+1 = 3
Add the Minutes: 45+20 = 65
The minutes are 60 or more, so
subtract 60 from minutes (65−60 =5 Minutes)
and add 1 to Hours (3+1 = 4 Hours)
The answer is 4:05
Subtracting Times
Follow these steps:
Subtract the hours
Subtract the minutes
If the minutes are negative, add 60 to the minutes and subtract 1 from hours
Like this:
Easy example: What is 4:10 − 1:05 ?
Subtract the Hours: 4−1 = 3
Subtract the Minutes: 10−5 = 5
The minutes are OK, so the answer is 3:05
Hard example: What is 4:10 − 1:35 ?
Subtract the Hours: 4−1 = 3
Subtract the Minutes: 10−35 = −25
The minutes are less than 0, so:
add 60 to Minutes (−25+60 = 60−25 =35 Minutes)
and subtract 1 from Hours (3−1 =2 Hours)
The answer is 2:35
(Note: did you see how we changed "−25+60" to "60−25" ... that is perfectly fine and makes the calculation easier.)
Adjusting The Hours
The hours can also end up too large or small!
To fix that first we must know if it is 24-Hour Clock or AM/PM:
24-Hour Clock
When Hours end up less than zero: add 24
When hours end up more than 23: subtract 24
Example: What is 16:20 + 9:35 ?
Add the Hours: 16+9 = 25
Add the Minutes: 20+35 = 55
The minutes are OK
The hours are more than 23, so subtract 24: 25−24 = 1
The answer is 1:55 of the next day.
Example: What is 4:10 − 6:15 ?
Subtract the Hours: 4−6 = −2
Subtract the Minutes: 10−15 = −5
The minutes are less than 0, so:
add 60 to Minutes: −5+60 = 60−5 =55 Minutes
and subtract 1 from Hours: −2−1 =−3 Hours
The hours are less than 0, so add 24: −3+24 = 24−3 = 21
The answer is 21:55 of the previous day.
12-Hour Clock (AM/PM)
For 12-Hour Clock it is best to convert to 24-Hour Clock, do the calculations, then convert back.
Example: What is 3:20 PM − 16:05 ?
3:20 PM is 15:20
Subtract the Hours: 15−16 = −1
Subtract the Minutes: 20−5 = 15
The minutes are OK
The hours are less than 1, so add 24: −1+24 = 24−1 = 23
The answer is 23:15, which is 11:15 PM of the previous day.
Mathopolis:Q1)Q2)Q3)Q4)Q5)Q6)Q7)Q8)Q9)Q10)
Time: AM/PM and 24 Hour ClockSun ClockWorld Time ZonesEarth's Orbit and Daylight
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9514 | https://brians.wsu.edu/2016/05/19/joint-possessives/ | Skip to main content
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Common Errors in English Usage and More
Posts
joint possessives
yanira.vargas
When writing about jointly owned objects, people often fret about where to place apostrophes. The standard pattern is to treat the two partners as a single unit—a couple—and put an apostrophe only after the last name: “John and Jane’s villa,” “Ben & Jerry’s ice cream.” Add more owners and you still use only one apostrophe: “Bob and Carol and Ted and Alice’s party.”
If each person owns his or her own item, then each owner gets an apostrophe: “John’s and Jane’s cars“ (each of them separately owns a car).
But when you begin to introduce pronouns the situation becomes much murkier. “Jane and his villa” doesn’t sound right because it sounds like Jane and the villa make a pair. The most common solution—“Jane’s and his villa”—violates the rule about using the possessive form only on the last partner in the ownership. However, most people don’t care and using this form won’t raise too many eyebrows.
How about when you have two pronouns? “She and his villa” definitely won’t work. “Her and his villa” might get by, but if you say “his and her villa” you inevitably remind people of the common phrase “his and hers” with a very different meaning: male and female, as in a sale on “his and hers scarves.”
If you have time to think ahead, especially when writing, the best solution is to avoid this sort of construction altogether by rewording: “Jane and John have a villa outside Florence. Their villa is beautiful.” “The villa owned by Jane and him is beautiful.” “The villa is Jane’s and his.” “The villa that he and she own is beautiful.”
Things get tricky when using personal pronouns instead of names. Note that “I’s” is not an acceptable substitute for “my.” It’s not “directions to my wife and I’s house,” but if you say “directions to my wife and my house” it sounds as if you were providing directions to your wife plus directions to your house. Stick with simpler constructions like “our house.”
Other awkward examples you might want to avoid: “your and my shares” (better: “your share and mine”), “their and our shares” (better: their share and ours”), and “his and her shares” (not too bad, but “his share and hers” is better).
Back to list of errors
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9515 | https://math.stackexchange.com/questions/129849/linear-homogeneous-recurrence-relations-with-repeated-roots-motivation-behind-l | Stack Exchange Network
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Linear homogeneous recurrence relations with repeated roots; motivation behind looking for solutions of the form $nx^n$?
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$\begingroup$
If we have a linear homogeneous recurrence relation, such as $t_{k+1}=4t_k-4t_{k-1}$, and attempt to find solutions of the form $t_n=x^n$ for some $x \in \mathbb{R} \setminus {0}$, we obtain the characteristic polynomial $(x-2)^2=0$. The characteristic polynomial has repeated roots, so the standard technique is to look for solutions of the form $t_n=nx^n$ too.
It's routine to check that $t_n=n x^n$ is indeed another solution to the recurrence. But, if I didn't already know that $t_n=n x^n$ is a solution, why would I think to look at solutions of this form?
Question: What motivation is there to consider solutions such as $t_n=n x^n$ in the case of repeated roots?
I find it easy to motivate looking for solutions of the form $t_n=x^n$, since e.g. for the Fibonacci numbers $F(n)$, we can prove exponential upper and lower bounds on $F(n)$, and it seems reasonable to suspect that similar inequalities hold more generally.
recurrence-relations
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asked Apr 10, 2012 at 0:38
Douglas S. StonesDouglas S. Stones
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$\begingroup$ It'd be clearer to not use $x$ in both the characteristic polynomial and in $x^n$, e.g. use $c^n$. $\endgroup$
Bill Dubuque
– Bill Dubuque
2012-04-10 00:46:00 +00:00
Commented Apr 10, 2012 at 0:46
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2 Answers 2
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$\begingroup$
Linear Algebra:
To set the stage, let $V$ be the vector space of sequences $a_0, a_1, ... $ (say of elements of an algebraically closed field) indexed by non-negative integers. On this vector space define the shift operator $S : V \to V$, which sends a sequence $a_i$ to the sequence $(Sa)_i = a_{i+1}$. To say that $a$ satisfies a linear recurrence with characteristic polynomial $p$ is then precisely to say that $$p(S) a = 0.$$
Since we're working over an algebraically closed field, we can factor $$p(S) = \prod_{i=1}^n (S - \lambda_i I)^{m_i}$$
and then write our recurrence as $$\prod_{i=1}^n (S - \lambda_i I)^{m_i} a = 0.$$
In particular, any sequence satisfying $(S - \lambda_i I)^{m_i} a = 0$ (a generalized eigenvector of $S$ with eigenvector $\lambda_i$) satisfies the recurrence, and standard results in linear algebra imply that generalized eigenvectors span the entire space of solutions.
So it suffices to solve the generalized eigenvector equation. First, I'm not sure if you know this, so it's worth saying: if $m_i = 1$, then the space of solutions to $$(S - \lambda_i I) a = 0$$
is spanned by $a_n = \lambda_i^n$ and this is the clearest reason I can think of (eigenvector decomposition) to look for exponential solutions.
Before we tackle the general case, another special case: $\lambda_i = 1$. In this case $S - I$ is just the forward difference operator, and the sequences satisfying $$(S - I)^n a = 0$$
are precisely the polynomial sequences of degree less than $n$. This should be a familiar fact about the forward difference operator, and it turns out to imply the general result. Indeed, suppose $a$ satisfies $$(S - \lambda_i I)^n a = 0$$
and write $a_n = \lambda_i^n b_n$ for some sequence $b_n$.
Lemma: $((S - \lambda_i I)a)_n = \lambda_i^{n+1} ((S - I)b)_n$.
Proof. Straightforward computation.
By induction, it follows that $$(S - \lambda_i I)^n a = 0 \Leftrightarrow (S - I)^n b = 0$$
and the conclusion follows. (I haven't discussed the case $\lambda_i = 0$ but the solutions are easy to describe here.)
A Limiting Argument:
We'll work through a simple representative example since the goal here is motivation. Consider a sequence $a$ satisfying $(S - \lambda_1 I)(S - \lambda_2 I) a = 0$ where $\lambda_1 \neq \lambda_2$, and let's see what happens if we let $\lambda_1 \to \lambda_2$. To take this limit sensibly we'll fix initial conditions $a_0 = 0, a_1 = 1$. This will give $$a_n = \frac{\lambda_1^n - \lambda_2^n}{\lambda_1 - \lambda_2}$$
Now taking the limit $\lambda_2 \to \lambda_1$ and applying, for example, l'Hopital's rule, we obtain $$a_n = n \lambda_1^{n-1}.$$
Differential Equations:
The shift operator $S$ defined above has another incarnation as differentiation $D$: indeed, we just need to associate to a sequence $a_0, a_1, ...$ the exponential generating function $$A(t) = \sum_{n=0}^{\infty} a_n \frac{t^n}{n!}.$$
Solutions to $p(S) a = 0$ then become solutions to differential equations $p(D) A = 0$. But now that we can work in the formalism of differential equations we can appeal to physical intuition: in physical terms, the relevant phenomenon is resonance. More concretely, let's work with a forced harmonic oscillator $$m \frac{d^2 x}{dt^2} + kx = \sin \omega t.$$
(Think of periodically pushing a swing.) Note that $x$ satisfies $(m D^2 + k)(D^2 + \omega^2) x = 0$. This polynomial has repeated roots when $\omega^2 = \frac{m}{k}$, that is, when the frequency of the forcing function matches the "natural frequency" of the system (if physicists have better terms for these I've forgotten them). In that case one can observe experimentally in the relevant physical systems (e.g. the Tacoma Narrows bridge) that the amplitude of the oscillations increase linearly with time, and one can observe mathematically that the solution involves terms like $t \sin \omega t$ (and one simple way to see this is to use the limiting argument above). Computing the Taylor series of such solutions then gives the familiar extra factor of $n$.
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edited Apr 10, 2012 at 23:26
answered Apr 10, 2012 at 1:02
Qiaochu YuanQiaochu Yuan
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$\begingroup$ Fantastic answer! Only the Tacome Narrows bridge example seems imprecise since recent research suggests that not ressonance, but a non-linear phenomenum was the cause of the bridge collapse: "Large-Amplitude Periodic Oscillations in Suspension Bridges: Some New Connections with Nonlinear Analysis" Lazer, McKenna, SIAM Review, vol.32, no.4. (1990) or, for a simpler model, "The Collapse of the Tacoma Narrows Suspension Bridge: A Modern Viewpoint", Lewis, ODE book $\endgroup$
Lucas Seco
– Lucas Seco
2013-10-03 03:40:51 +00:00
Commented Oct 3, 2013 at 3:40
$\begingroup$ I'm sorry, but what is $m_i$ when you change to product notation? $\endgroup$
Kenny
– Kenny
2017-02-23 02:09:54 +00:00
Commented Feb 23, 2017 at 2:09
$\begingroup$ @Kenneth: $m_i$ is the multiplicity of $\lambda_i$ as a root of the polynomial $p$. $\endgroup$
Qiaochu Yuan
– Qiaochu Yuan
2017-02-23 03:14:20 +00:00
Commented Feb 23, 2017 at 3:14
Add a comment |
5
$\begingroup$
One way to look at this would be using generating functions.
The generating function comes out to
$$F(x) = \frac{Q(x)}{P(x)}$$
where the polynomial $P(x)$ is got from the characteristic polynomial, by replacing $x$ by $\frac{1}{x}$ and multiplying by an appropriate power of $x$.
Now using the partial fraction decomposition
$$ \frac{Q(x)}{(x-a_1)^{n_1} \dots (x-a_k)^{n_k}} = \sum_{i=1}^{k} \sum_{j=1}^{n_i} \frac{A_{ij}}{(x-a_1)^j}$$
tells us that we should get a polynomial in $n$.
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answered Apr 10, 2012 at 0:54
AryabhataAryabhata
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$\endgroup$
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$\begingroup$ This answer uses ordinary generating functions whereas my answer above uses exponential generating functions; the connection between the two is given (essentially) by the Laplace transform. $\endgroup$
Qiaochu Yuan
– Qiaochu Yuan
2012-04-10 01:43:39 +00:00
Commented Apr 10, 2012 at 1:43
$\begingroup$ The polynomial in $n$ comes from $\binom{-j}{n} = (-1)^n \binom{j + n - 1}{j - 1}$, a polynomial of degree $j - 1$ in $n$. $\endgroup$
vonbrand
– vonbrand
2014-04-21 17:51:38 +00:00
Commented Apr 21, 2014 at 17:51
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Mass points
Mass points is a technique in Euclidean geometry that can greatly simplify the proofs of many theorems concerning polygons, and is helpful in solving complex geometry problems involving lengths. In essence, it involves using a local coordinate system to identify points by the ratios into which they divide line segments. Mass points are generalized by barycentric coordinates.
Mass point geometry was invented by Franz Mobius in 1827 along with his theory of homogeneous coordinates. The technique did not catch on until the 1960s when New York high school students made it popular. The technique greatly simplifies certain problems.
Contents
[hide]
1 Uses
2 Examples
3 Example 1
3.1 Solution
4 Example 2
4.1 Solution
5 Problems
6 Videos
Uses
Mass point geometry involves systematically assigning 'weights' to points using ratios of lengths relating vertices, which can then be used to deduce other lengths, using the fact that the lengths must be inversely proportional to their weight (just like a balanced lever). Additionally, the point dividing the line has a mass equal to the sum of the weights on either end of the line (like the fulcrum of a lever). The way to systematically assign weights to the points involves first choosing a point for the entire figure to balance around. From there, WLOG a first weight can be assigned. From the first weight, others can be derived using a few simple rules. Given a line with point on it, and the mass put on a point P is denoted as ,
If two points balance, the product of the mass and distance from a line of balance of one point will equal the product of the mass and distance from the same line of balance of the other point. In other words, .
If two points are balanced, the point on the balancing line used to balance them has a mass equal to the sum of the masses of the two points. That is, .
Examples
Example 1
Consider a triangle with its three medians drawn, with the intersection points being corresponding to and respectively. Let the centroid of triangle be . Prove is and the corresponding identities for medians from and .
Solution
Thus, if we label point with a weight of , must also have a weight of since and are equidistant from . By the same process, we find must also have a weight of 1. Now, since and both have a weight of , must have a weight of (as is true for and ). Thus, if we label the centroid , we can deduce that is - the inverse ratio of their weights.
Example 2
has point on , point on , and point on . , , and intersect at point . The ratio is and the ratio is . Find the ratio of
Solution
Throughout this solution, let denote the weight at point . Since , let , which makes . Now, look at . Since (this is a general property commonly used in many mass points problems, in fact it is the same property we used above to determine ), we have . Then, (another property of mass points). Finally, we have .
Problems
2019 AMC 8 Problems/Problem 24
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Area of Equilateral Triangle
Last Updated : 23 Jul, 2025
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The area of an equilateral triangle is the amount of space enclosed within its three equal sides. For an equilateral triangle, where all three sides and all three internal angles are equal (each angle measuring 60 degrees), the area can be calculated using the formula 3 4×a 2\frac{\sqrt{3}}{4}\times a^{2}4 3×a 2 where a is length of a side of Equilateral Triangle. An equilateral triangle is a triangle whose all side is equal to 60°.
Area of Equilateral Triangle
In simple words, Area of an equilateral triangle is the space occupied by an equilateral triangle. Let's know more about Area of Equilateral Triangle with formula, proof and examples.
Table of Content
Area of Equilateral Triangle
Area of Equilateral Triangle Formula
Area of Equilateral Triangle Formula Proof
Derivation of Area of Equilateral Triangle using Trigonometry
Properties of Equilateral Triangle
Solved Examples on Area of Equilateral triangle
Area of Equilateral Triangle
The area of an equilateral triangle is the amount of space enclosed within its three equal sides of an Equilateral Triangle. It is measured in square units. Area of an equilateral depends upon the length of a side of an equilateral triangle. Let's learn the formula used to define the area of an equilateral triangle.
Examples of Equilateral Triangle
Area of Equilateral Triangle Formula
Area of an equilateral triangle is the space occupied between the sides of the equilateral triangle in a plane. Below is the formula for finding the area of a triangle whose base and height are given is
Area = 1 2×b a s e×h e i g h t\frac{1}{2}\times base \times height 2 1×ba se×h e i g h t
If only sides of the triangle are given. Let an equilateral triangle of side 'a' be given then area of Equilateral Triangle is
3 4×a 2\frac{\sqrt{3}}{4} \times a^2 4 3×a 2
Area of Equilateral Triangle Formula Proof
Let's calculate the area for a given equilateral triangle of side a. It is known that the area of a triangle is given as 1/2 × Base × Height.
Here the base is a. Let's find the height of this triangle in order to find the area. It can clearly be seen that the height can be found using the Pythagoras theorem since it is one of the sides of the right-angled triangle.
Applying Pythagoras' theorem,
h 2 + (a/2)2 = a 2
⇒ h 2 = (3a 2/4)
⇒ h = √3a/2
Now the height of this equilateral triangle is known. Now, substitute this value of height into our formula,
Area =1/2 × Base × Height
⇒ Area = 1/2 × a × √3a/2 =√3a 2/4
Area = √3a2/4
Derivation of Area of Equilateral Triangle using Trigonometry
Suppose the sides of a triangle are given, then the height can be calculated using the sine formula. Let the sides of a triangle ABC be a, b, and the angle corresponding to them be A, B, and C. Now, the height of a triangle is
h = a × Sin B = b × Sin C = c × Sin A
Now, area of ABC = ½ × a × (b × sin C)
⇒ area of ABC = ½ × b × (c × sin A)
⇒ area of ABC = ½ × c (a × sin B)
Since it is an equilateral triangle, A = B = C = 60° and a = b = c
⇒ Area = ½ × a × (a × Sin 60°)
⇒ Area = ½ × a 2 × Sin 60°
⇒ Area = ½ × a 2 × √3/2 = √3a 2/4
Area of Equilateral Triangle = (√3/4)a2
Read More
Trigonometry
Perimeter of the Equilateral Triangle
An equilateral triangle is a triangle with all three sides and the perimeter of any figure is the sum of all its sides. So, the perimeter of an equilateral triangle of side of length "a" is given by
Must Read
Perimeter of an equilateral triangle
Properties of Equilateral Triangle
An equilateral triangle is one triangle in which all three sides are equal. For an equilateral triangle PQR, PQ = QR = RP. A few important properties of an equilateral triangle are:
All three sides are equal in an Equilateral Triangle.
In an equilateral triangle, all three internal angles are equal to each other and their value is 60°.
For an equilateral triangle, the median, angle bisector, and perpendicular all are the same.
Ortho-centre and centroid of an equilateral triangle are the same points.
In an equilateral triangle, there are three lines of symmetry and also 3rd order rotational symmetry as well.
Area of an equilateral triangle is √3 a 2/ 4.
Perimeter of an equilateral triangle is 3a.
Must Read
Area of Triangle
Area of Square
Area of Rhombus
Area of Rectangle
Area of Parallelogram
Area of Circle
Solved Examples on Area of Equilateral triangle
Example 1: Find the area of the triangle whose all sides measure 4 units.
Solution:
As given all sides are of equal length hence, we can say that it is an equilateral triangle.
So we can apply the formula to directly find the area of this triangle.
Area = √3a 2/4 = √3 × 4 2/4 = 4√3 units 2
Example 2: Find the perimeter of the triangle whose sides are given as 3 cm, 4 cm, and 5 cm.
Solution:
Sum of all the sides of any triangle is the perimeter of triangle
Hence, the perimeter of this given triangle is (3 + 4 + 5) cm
i.e. Perimeter is 12 cm
Example 3: Find the height of the equilateral triangle whose side is 4 cm.
Solution:
The formula for the height is given by:h = √3a/2
h = (√3 × 4)/2= 2√3 cm
Hence the height of the triangle is 2√3 cm
Example 4: Find the perimeter and area of the equilateral triangle whose side is given as 4 cm.
Solution:
Side (s) = 4 cm
For any equilateral triangle the perimeter is calculated as 3 × s
Primeter(P) = 3 × 4 = 12 cm
Area = √3a 2/4
= √3(4)2/4
= √3(16) / 4 cm 2
Area = 4√3 cm 2
Example 5: Find the area of an equilateral triangle when the perimeter is 18 cm.
Solution:
Perimeter of an equilateral triangle = 18 cm
Perimeter of the equilateral triangle = 3a
3a = 18, a = 6
The length of side is 6 cm.
Area, A = √3 a 2/ 4 sq units
= √3 (6)2/ 4 cm 2 ⇒ 36 √3 / 4
= 9√3 cm2
Then area of the equilateral triangle is9√3 cm2
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科普中国-充分必要条件
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充分必要条件，又称充要条件，是数学和逻辑学中的一种重要概念。具体来说，如果能从命题p推出命题q，而且也能从命题q推出命题p ，则称p是q的充分必要条件，且q也是p的充分必要条件。这种关系在数学中通常用“当且仅当”来表示。
充分必要条件是一种双向推理的关系，即一个命题的成立必然导致另一个命题成立，反之亦然。这一概念在逻辑推理和数学证明中具有重要作用，能够帮助人们更准确地分析和解决问题。
定义
1.1从数学命题角度理解
命题p、q，如果p推出q且q推出p，则p是q的充分必要条件，简称充要条件。
p推出q，p是q的充分条件，同时q是p的必要条件。
例如：a、b一正一负推出ab<0，ab<0推出a、b一正一负，则a、b一正一负和ab<0互为充要条件。
简单地说就是在证p与q时,前面那个推出后面那个就是充分条件；后面那个推出前面那个就是必要条件；前面能推出后面、后面也能推出前面就是充要条件。
如果既有p推出q，又有q推出p，则记作p=q，就说p是q的充要条件，也可以说q是p的充要条件，或者若p推出q，但q推不出p，则p是q的充分不必要条件，q是p的必要不充分条件。
例如“两个三角形全等”是“两个三角形面积相等”的充分不必要条件，|x|=|y|是“x²=y²”的充要条件。
1.2从集合角度理解
假设A是条件，B是结论，设C、D分别为A、B所描述对象的集合，则有下列定义和推论：
（1）由A可以推出B，由B可以推出A，则A是B的充分必要条件（此时）
（2）由A可以推出B，由B不可以推出A，则A是B的充分不必要条件（此时）；
（3）由A不可以推出B，由B可以推出A，则A是B的必要不充分条件（此时）；
（4）由A不可以推出B，由B不可以推出A，则A是B的既不充分也不必要条件（此时）1
1.3从逻辑学角度理解
如果有事物情况A，则必然有事物情况B；如果没有事物情况A，则必然没有事物情况B，A就是B的充分必要条件。
充分必要条件是逻辑学在研究假言命题及假言推理时引出的。陈述某一事物情况是另一件事物情况的充分必要条件的假言命题叫作充分必要条件假言命题。充分必要条件假言命题的一般形式是：p当且仅当q。符号为：p←→q(读作“p等值q”) 。
例如：“三角形等边当且仅当三角形等角。”是一个充分必要条件假言命题。根据充分必要条件假言命题的逻辑性质进行的推理叫充分必要条件假言推理。
相关概念
2.1充分条件
充分条件是指如果条件A成立，那么条件B一定成立。换句话说，条件A是条件B成立的一个保证。
例如：假设有两个命题：
命题A：今天下雨了。
命题B：地面是湿的。
在这个例子中，命题A（今天下雨了）是命题B（地面是湿的）的充分条件。因为如果今天下雨了，那么地面一定会湿。但是，地面湿并不一定意味着今天下雨了，因为地面也可能因为其他原因（如洒水）而湿。
2.2必要条件
必要条件是指如果没有条件A，那么结果B一定不会发生。换句话说，如果发生了结果B，那么条件A一定是已经满足的。
例如：假设有两个命题：
命题A：植物得到充足的水分。
命题B：植物能够生长良好。
在这个例子中，命题A（植物得到充足的水分）是命题B（植物能够生长良好）的必要条件。因为如果植物要生长良好，它必须得到充足的水分。然而，即使植物得到了充足的水分，也可能因为其他因素（如光照不足或土壤营养不足）而无法生长良好。因此，充足的水分是植物生长良好的必要条件，但不一定是充分条件。
2.3充分不必要条件
充分不必要条件是指如果条件A成立，那么结果B一定成立，但是结果B成立并不一定需要条件A。换句话说，条件A是结果B成立的一个充分条件，但不是必要条件。
2.4必要不充分条件
必要不充分条件指的是，如果结果B发生，那么条件A一定已经满足（即A是B的必要条件），但是即使条件A满足，结果B也不一定会发生（即A不是B的充分条件）。换句话说，条件A的存在对于结果B的发生是必要的，但单独的条件A不足以保证结果B的发生。
假言推理
假言推理是根据假言命题的逻辑性质进行的推理，分为充分条件假言推理、必要条件假言推理和充分必要条件假言推理三种形式。
3.1 充分条件假言推理
定义：充分条件假言命题通常用“如果……那么……”的形式表示，例如“如果今天下雨，那么地面就会湿”。
推理规则：肯定前件就要肯定后件；否定后件就要否定前件。
正确形式：
肯定前件式：如果p，那么q；p，所以，q
否定后件式：如果p，那么q；非q，所以，非p
3.2 必要条件假言推理
定义：必要条件假言命题通常用“只有……才……”的形式表示，例如“只有年满十八岁，才有选举权”。
推理规则：否定前件就要否定后件；肯定后件就要肯定前件。
正确形式：
否定前件式：只有p，才q；非p，所以，非q
肯定后件式：只有p，才q；q，所以，p
3.3. 充分必要条件假言推理
定义：充分必要条件假言命题通常用“当且仅当……”的形式表示，例如“一个数是偶数当且仅当它能被2整除”
推理规则：肯定前件就要肯定后件；肯定后件就要肯定前件；否定前件就要否定后件；否定后件就要否定前件。
正确形式：
肯定前件式：p当且仅当q；p，所以，q
肯定后件式：p当且仅当q；q，所以，p
否定前件式：p当且仅当q；非p，所以，非q
否定后件式：p当且仅当q；非q，所以，非p2
简史
在公元前6世纪，古希腊思想家们已经开始研究逻辑推理的概念，活跃的国家政治生活鼓励人们开展讨论和发展辩论的技巧。例如：古希腊思想家巴门尼德（Parmenniddes，公元前6世纪后期）及其弟子芝诺（Zeno，公元前5世纪）就在“归谬法”中提出了逻辑论证的基本原则——假定要证明的命题不成立从而引出矛盾；否定后件律——先证明若A正确，则B也正确，然后证明B不正确，结论是A也不正确。”
受到这一思想的影响，古希腊哲学家和科学家亚里士多德在他的著作《形而上学》中，提到了“必需的”相关定义和概念，亚里士多德认为，逻辑论证应建立在三段论（syllogism）的基础上，三段论指的是由所陈述的事情，必定可得出另外的某些结论的论证过程。古希腊数学家欧几里得在《几何原本》中，提到了一些充分必要条件的理论。3
举例
数学问题
三角形等边与三角形等角：一个三角形是等边的当且仅当它是等角的。这意味着如果一个三角形的三条边都相等，那么它的三个角也一定相等；反之，如果一个三角形的三个角都相等，那么它的三条边也一定相等。
工程问题
电路设计：在电路设计中，如果一个电路能够正常工作，那么它必须满足基尔霍夫定律；反之，如果一个电路满足基尔霍夫定律，那么它能够正常工作。
生活问题
健康与饮食：一个人如果摄入足够的营养，那么他的身体就会健康；反之，如果一个人的身体是健康的，那么他一定摄入了足够的营养。
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9519 | https://www.vcalc.com/wiki/combinations | {::}\_n"C"\_k = ( n ! ) / ( k !( n - k )! )
Enter a value for all fields
The COMBINATIONS (nCk) calculator computes the number of combination possible in a set of k elements containing exactly one of each unique element from a finite set of n objects where order does not matter.
INSTRUCTIONS: Enter the following:
(n) Total Number of Objects (e.g. 52 cards).
(k) Number in Subset (e.g. 5 card poker hand)
Combinations (nCk): The calculator computes and return the number of combinations. Combinations also plays the core role in the Binomial Coefficientand in the construction of Pascal's Triangle.
The Math / Science
The combination equation computes the number of possible combinations of k elements (unique sets) from within the set of n objects. Unlike permutations, the order of the distinct selected items are not considered in the set of combinations. We denote the number of combinations as C(n,k). The number C(n,k) or C\_"(n,k)" is the the probability of choosing one specific set of k elements from a set of n-objects.
The formula for the number of combinations is as follows:
nCk = (n!)/(k!(n-k)!)
where:
nCk = combinations of k objects from a set of n objects
n = total size of set
k = size of subset
EXAMPLES
This equations can be used to answer a question like: How many different hands can be dealt in a 5 card Poker deal.
The answer is: 52! / ( 5! ( 52-5)! ) = 2598960
So, from a set of 52 cards, there are 2,598,960 possible different hands that could be dealt.
Another example would be the number of combinations of players on the field from a 50 player football team where every player can play any position on both offense or defense. Since each player is a unique person and the number of players on the field is eleven, the number of possible player combinations would be:
C(50,11) = 50! / ( 11! (50-11)! ) = 37,353,738,800
Combinatorics Calculators
n! - factorial
n! = sqrt(2 pi n) (n/e)^n - Stirling's factorial approximation
nCk - combinations
Cr(n,k) - combinations with repetitions
nPk - permutations
Pr(n) - permutations with repetitions
Pcomp(n,2) - Comparison permutations
Pc(n,r) - Circular Permutation |
9520 | https://www.youtube.com/watch?v=PjtLtKk8dXA | Solve Exponential Equation using Substitution: Quadratic form
Maths with Jay
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Solve exponential equation by factoring
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Transcript:
hello welcome to maths with Jay the most important thing with an equation like this is to know how start to deal with it and what's important here is that you know that if you've got a number raised to the power of two numbers like a to the power of M times n then we know that we can think of that as either a to the M or to the N or a to the N or to the M so in this particular question what we want to do is to rewrite it so that the unknown only appears in one place in a sense what we're thinking about is at the moment we've got three to the 2x and 3 to the X so the unknown is appearing in two places and in two different ways but if we could write both of them in the same sort of way then we'd have a quadratic so what we're going to do in this question is we're going to write the 3 to the 2x as 3 to the X all squared so in fact we're using the fact that we can write this sort of thing as a to the N to the M aren't we right so once we realize that then it becomes straightforward so we're going to write 3 to the X all squared and then the rest of the equation will stay the same and now you can see that what we've got is something squared minus 10 times that something plus 9 is syrup so we've got a quadratic and to make it even more obvious let's actually make a substitution so we're going to let y equal that's something Y is 3 to the X so now this will look even more like a straightforward quadratic so all we're trying to do first of all is to find Y so let's factorize this so we're looking for two numbers that multiply to nine and that sum add up to negative ten so we've got minus 1 and minus 9 so we know that y must be 1 or 9 and remember that Y was actually 3 to the X so now I think 3 to the X again so 3 to the X is 1 or it's nine now if these were difficult numbers you could use your calculator to work out what X is you would take log to both sides but very straightforward here because we know that really we've effectively got powers 3 only so anything to the power of 0 is 1 so X is 0 3 to the X is 1 and we know that if 3 to the power of something is neither that something must be too because 3 squared is 9 and then of course we want to check our answer answers I should say so first of all let's check x equals 0 so that would give off substituting back into the original equation 3 to the power 2 times 0 minus 10 times 3 to the 0 plus 9 so 3 to 0 is 1 so we've got 1 minus 10 plus 9 so that is equal to 0 so x equals 0 is a solution and then let's check x equals 2 and that will give us 3 to the power of 2 times 2 so 3 to the 4 minus 10 times 3 squared plus 9 and that's going to be 81 minus 10 9 minus 90 plus 9 and that again is 0 so we know that our answers are correct you you |
9521 | https://courses.lumenlearning.com/suny-physics/chapter/15-7-statistical-interpretation-of-entropy-and-the-second-law-of-thermodynamics-the-underlying-explanation/ | Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation
Learning Objectives
By the end of this section, you will be able to:
Identify probabilities in entropy.
Analyze statistical probabilities in entropic systems.
Figure 1. When you toss a coin a large number of times, heads and tails tend to come up in roughly equal numbers. Why doesn’t heads come up 100, 90, or even 80% of the time? (credit: Jon Sullivan, PDPhoto.org)
The various ways of formulating the second law of thermodynamics tell what happens rather than why it happens. Why should heat transfer occur only from hot to cold? Why should energy become ever less available to do work? Why should the universe become increasingly disorderly? The answer is that it is a matter of overwhelming probability. Disorder is simply vastly more likely than order.
When you watch an emerging rain storm begin to wet the ground, you will notice that the drops fall in a disorganized manner both in time and in space. Some fall close together, some far apart, but they never fall in straight, orderly rows. It is not impossible for rain to fall in an orderly pattern, just highly unlikely, because there are many more disorderly ways than orderly ones. To illustrate this fact, we will examine some random processes, starting with coin tosses.
Coin Tosses
What are the possible outcomes of tossing 5 coins? Each coin can land either heads or tails. On the large scale, we are concerned only with the total heads and tails and not with the order in which heads and tails appear. The following possibilities exist:
5 heads,0 tails4 heads,1 tail3 heads,2 tails2 heads,3 tails1 head,4 tails0 head,5 tails
These are what we call macrostates. A macrostate is an overall property of a system. It does not specify the details of the system, such as the order in which heads and tails occur or which coins are heads or tails.
Using this nomenclature, a system of 5 coins has the 6 possible macrostates just listed. Some macrostates are more likely to occur than others. For instance, there is only one way to get 5 heads, but there are several ways to get 3 heads and 2 tails, making the latter macrostate more probable. Table 1 lists of all the ways in which 5 coins can be tossed, taking into account the order in which heads and tails occur. Each sequence is called a microstate—a detailed description of every element of a system.
| Table 1. 5-Coin Toss | | |
---
| | Individual microstates | Number of microstates |
| 5 heads, 0 tails | HHHHH | 1 |
| 4 heads, 1 tail | HHHHT, HHHTH, HHTHH, HTHHH, THHHH | 5 |
| 3 heads, 2 tails | HTHTH, THTHH, HTHHT, THHTH, THHHT HTHTH, THTHH, HTHHT, THHTH, THHHT | 10 |
| 2 heads, 3 tails | TTTHH, TTHHT, THHTT, HHTTT, TTHTH, THTHT, HTHTT, THTTH, HTTHT, HTTTH | 10 |
| 1 head, 4 tails | TTTTH, TTTHT, TTHTT, THTTT, HTTTT | 5 |
| 0 heads, 5 tails | TTTTT | 1 |
| | | Total: 32 |
The macrostate of 3 heads and 2 tails can be achieved in 10 ways and is thus 10 times more probable than the one having 5 heads. Not surprisingly, it is equally probable to have the reverse, 2 heads and 3 tails. Similarly, it is equally probable to get 5 tails as it is to get 5 heads. Note that all of these conclusions are based on the crucial assumption that each microstate is equally probable. With coin tosses, this requires that the coins not be asymmetric in a way that favors one side over the other, as with loaded dice. With any system, the assumption that all microstates are equally probable must be valid, or the analysis will be erroneous.
The two most orderly possibilities are 5 heads or 5 tails. (They are more structured than the others.) They are also the least likely, only 2 out of 32 possibilities. The most disorderly possibilities are 3 heads and 2 tails and its reverse. (They are the least structured.) The most disorderly possibilities are also the most likely, with 20 out of 32 possibilities for the 3 heads and 2 tails and its reverse. If we start with an orderly array like 5 heads and toss the coins, it is very likely that we will get a less orderly array as a result, since 30 out of the 32 possibilities are less orderly. So even if you start with an orderly state, there is a strong tendency to go from order to disorder, from low entropy to high entropy. The reverse can happen, but it is unlikely.
| Table 2. 100-Coin Toss | | |
---
| Macrostate | | Number of microstates |
| Heads | Tails | (W) |
| 100 | 0 | 1 |
| 99 | 1 | 1.0 × 102 |
| 95 | 5 | 7.5 × 107 |
| 90 | 10 | 1.7 × 1013 |
| 75 | 25 | 2.4 × 1023 |
| 60 | 40 | 1.4 × 1028 |
| 55 | 45 | 6.1 × 1028 |
| 51 | 49 | 9.9 × 1028 |
| 50 | 50 | 1.0 × 1029 |
| 49 | 51 | 9.9 × 1028 |
| 45 | 55 | 6.1 × 1028 |
| 40 | 60 | 1.4 × 1028 |
| 25 | 75 | 2.4 × 1023 |
| 10 | 90 | 1.7 × 1013 |
| 5 | 95 | 7.5 × 107 |
| 1 | 99 | 1.0 × 102 |
| 0 | 100 | 1 |
| | | Total: 1.27 ×1030 |
This result becomes dramatic for larger systems. Consider what happens if you have 100 coins instead of just 5. The most orderly arrangements (most structured) are 100 heads or 100 tails. The least orderly (least structured) is that of 50 heads and 50 tails. There is only 1 way (1 microstate) to get the most orderly arrangement of 100 heads. There are 100 ways (100 microstates) to get the next most orderly arrangement of 99 heads and 1 tail (also 100 to get its reverse). And there are 1.0 × 1029 ways to get 50 heads and 50 tails, the least orderly arrangement. Table 2 is an abbreviated list of the various macrostates and the number of microstates for each macrostate. The total number of microstates—the total number of different ways 100 coins can be tossed—is an impressively large 1.27 × 1030. Now, if we start with an orderly macrostate like 100 heads and toss the coins, there is a virtual certainty that we will get a less orderly macrostate. If we keep tossing the coins, it is possible, but exceedingly unlikely, that we will ever get back to the most orderly macrostate. If you tossed the coins once each second, you could expect to get either 100 heads or 100 tails once in 2 × 1022 years! This period is 1 trillion (1012) times longer than the age of the universe, and so the chances are essentially zero. In contrast, there is an 8% chance of getting 50 heads, a 73% chance of getting from 45 to 55 heads, and a 96% chance of getting from 40 to 60 heads. Disorder is highly likely.
Disorder in a Gas
The fantastic growth in the odds favoring disorder that we see in going from 5 to 100 coins continues as the number of entities in the system increases. Let us now imagine applying this approach to perhaps a small sample of gas. Because counting microstates and macrostates involves statistics, this is called statistical analysis. The macrostates of a gas correspond to its macroscopic properties, such as volume, temperature, and pressure; and its microstates correspond to the detailed description of the positions and velocities of its atoms. Even a small amount of gas has a huge number of atoms: 1.0 cm3 of an ideal gas at 1.0 atm and 0º C has 2.7 × 1019 atoms. So each macrostate has an immense number of microstates. In plain language, this means that there are an immense number of ways in which the atoms in a gas can be arranged, while still having the same pressure, temperature, and so on.
The most likely conditions (or macrostates) for a gas are those we see all the time—a random distribution of atoms in space with a Maxwell-Boltzmann distribution of speeds in random directions, as predicted by kinetic theory. This is the most disorderly and least structured condition we can imagine. In contrast, one type of very orderly and structured macrostate has all of the atoms in one corner of a container with identical velocities. There are very few ways to accomplish this (very few microstates corresponding to it), and so it is exceedingly unlikely ever to occur. (See Figure 2b.) Indeed, it is so unlikely that we have a law saying that it is impossible, which has never been observed to be violated—the second law of thermodynamics.
Figure 2. (a) The ordinary state of gas in a container is a disorderly, random distribution of atoms or molecules with a Maxwell-Boltzmann distribution of speeds. It is so unlikely that these atoms or molecules would ever end up in one corner of the container that it might as well be impossible. (b) With energy transfer, the gas can be forced into one corner and its entropy greatly reduced. But left alone, it will spontaneously increase its entropy and return to the normal conditions, because they are immensely more likely.
The disordered condition is one of high entropy, and the ordered one has low entropy. With a transfer of energy from another system, we could force all of the atoms into one corner and have a local decrease in entropy, but at the cost of an overall increase in entropy of the universe. If the atoms start out in one corner, they will quickly disperse and become uniformly distributed and will never return to the orderly original state (Figure 2b). Entropy will increase. With such a large sample of atoms, it is possible—but unimaginably unlikely—for entropy to decrease. Disorder is vastly more likely than order.
The arguments that disorder and high entropy are the most probable states are quite convincing. The great Austrian physicist Ludwig Boltzmann (1844–1906)—who, along with Maxwell, made so many contributions to kinetic theory—proved that the entropy of a system in a given state (a macrostate) can be written as S= k lnW, where k = 1.38 × 10−23 J/K is Boltzmann’s constant, and lnW is the natural logarithm of the number of microstates W corresponding to the given macrostate. W is proportional to the probability that the macrostate will occur. Thus entropy is directly related to the probability of a state—the more likely the state, the greater its entropy. Boltzmann proved that this expression for S is equivalent to the definition ΔS=QT, which we have used extensively.
Thus the second law of thermodynamics is explained on a very basic level: entropy either remains the same or increases in every process. This phenomenon is due to the extraordinarily small probability of a decrease, based on the extraordinarily larger number of microstates in systems with greater entropy. Entropy can decrease, but for any macroscopic system, this outcome is so unlikely that it will never be observed.
Example 1. Entropy Increases in a Coin Toss
Suppose you toss 100 coins starting with 60 heads and 40 tails, and you get the most likely result, 50 heads and 50 tails. What is the change in entropy?
Strategy
Noting that the number of microstates is labeled W in Table 2 for the 100-coin toss, we can use ΔS= Sf − Si = k lnWf – klnWi to calculate the change in entropy.
Solution
The change in entropy is ΔS= Sf – Si = k lnWf – klnWi,
where the subscript i stands for the initial 60 heads and 40 tails state, and the subscript f for the final 50 heads and 50 tails state. Substituting the values for W from Table 2 gives
ΔS=(1.38×10−23 J/K)[ln(1.0×1029)−ln(1.4×1029)] =2.7×10−23 J/K
Discussion
This increase in entropy means we have moved to a less orderly situation. It is not impossible for further tosses to produce the initial state of 60 heads and 40 tails, but it is less likely. There is about a 1 in 90 chance for that decrease in entropy (−2.7 × 10−23 J/K) to occur. If we calculate the decrease in entropy to move to the most orderly state, we get ΔS = −92 × 10−23 J/K. There is about a 1 in 1030 chance of this change occurring. So while very small decreases in entropy are unlikely, slightly greater decreases are impossibly unlikely. These probabilities imply, again, that for a macroscopic system, a decrease in entropy is impossible. For example, for heat transfer to occur spontaneously from 1.00 kg of 0ºC ice to its 0ºC environment, there would be a decrease in entropy of 1.22 × 103 J/K. Given that a ΔS 10−21 J/K corresponds to about a 1 in 1030 chance, a decrease of this size (103 J/K) is an utter impossibility. Even for a milligram of melted ice to spontaneously refreeze is impossible.
Problem-Solving Strategies for Entropy
Examine the situation to determine if entropy is involved.
Identify the system of interest and draw a labeled diagram of the system showing energy flow.
Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful.
Make a list of what is given or can be inferred from the problem as stated (identify the knowns). You must carefully identify the heat transfer, if any, and the temperature at which the process takes place. It is also important to identify the initial and final states.
Solve the appropriate equation for the quantity to be determined (the unknown). Note that the change in entropy can be determined between any states by calculating it for a reversible process.
Substitute the known value along with their units into the appropriate equation, and obtain numerical solutions complete with units.
To see if it is reasonable: Does it make sense? For example, total entropy should increase for any real process or be constant for a reversible process. Disordered states should be more probable and have greater entropy than ordered states.
Section Summary
Disorder is far more likely than order, which can be seen statistically.
The entropy of a system in a given state (a macrostate) can be written as S= klnW,where k= 1.38 × 10−23 J/K is Boltzmann’s constant, and lnW is the natural logarithm of the number of microstates W corresponding to the given macrostate.
Conceptual Questions
Explain why a building made of bricks has smaller entropy than the same bricks in a disorganized pile. Do this by considering the number of ways that each could be formed (the number of microstates in each macrostate).
Problems & Exercises
Using Table 2, verify the contention that if you toss 100 coins each second, you can expect to get 100 heads or 100 tails once in 2 × 1022 years; calculate the time to two-digit accuracy.
What percent of the time will you get something in the range from 60 heads and 40 tails through 40 heads and 60 tails when tossing 100 coins? The total number of microstates in that range is 1.22 × 1030. (Consult Table 2.)
(a) If tossing 100 coins, how many ways (microstates) are there to get the three most likely macrostates of 49 heads and 51 tails, 50 heads and 50 tails, and 51 heads and 49 tails? (b) What percent of the total possibilities is this? (Consult Table 2.)
(a) What is the change in entropy if you start with 100 coins in the 45 heads and 55 tails macrostate, toss them, and get 51 heads and 49 tails? (b) What if you get 75 heads and 25 tails? (c) How much more likely is 51 heads and 49 tails than 75 heads and 25 tails? (d) Does either outcome violate the second law of thermodynamics?
(a) What is the change in entropy if you start with 10 coins in the 5 heads and 5 tails macrostate, toss them, and get 2 heads and 8 tails? (b) How much more likely is 5 heads and 5 tails than 2 heads and 8 tails? (Take the ratio of the number of microstates to find out.) (c) If you were betting on 2 heads and 8 tails would you accept odds of 252 to 45? Explain why or why not.
| Table 3. 10-Coin Toss | | |
---
| Macrostate | | Number of Microstates |
| Heads | Tails | (W) |
| 10 | 0 | 1 |
| 9 | 1 | 10 |
| 8 | 2 | 45 |
| 7 | 3 | 120 |
| 6 | 4 | 210 |
| 5 | 5 | 252 |
| 4 | 6 | 210 |
| 3 | 7 | 120 |
| 2 | 8 | 45 |
| 1 | 9 | 10 |
| 0 | 10 | 1 |
| | | Total: 1024 |
6. (a) If you toss 10 coins, what percent of the time will you get the three most likely macrostates (6 heads and 4 tails, 5 heads and 5 tails, 4 heads and 6 tails)? (b) You can realistically toss 10 coins and count the number of heads and tails about twice a minute. At that rate, how long will it take on average to get either 10 heads and 0 tails or 0 heads and 10 tails?
7. (a) Construct a table showing the macrostates and all of the individual microstates for tossing 6 coins. (Use Table 3 as a guide.) (b) How many macrostates are there? (c) What is the total number of microstates? (d) What percent chance is there of tossing 5 heads and 1 tail? (e) How much more likely are you to toss 3 heads and 3 tails than 5 heads and 1 tail? (Take the ratio of the number of microstates to find out.)
8. In an air conditioner, 12.65 MJ of heat transfer occurs from a cold environment in 1.00 h. (a) What mass of ice melting would involve the same heat transfer? (b) How many hours of operation would be equivalent to melting 900 kg of ice? (c) If ice costs 20 cents per kg, do you think the air conditioner could be operated more cheaply than by simply using ice? Describe in detail how you evaluate the relative costs.
Glossary
macrostate:an overall property of a system
microstate: each sequence within a larger macrostate
statistical analysis: using statistics to examine data, such as counting microstates and macrostates
Selected Solutions to Problems & Exercises
1. It should happen twice in every 1.27 × 1030 s or once in every 6.35 × 1029 s
(6.35×1029s)(1 h3600 s)(1 d24 h)(1 y365.25 d)=2.0×1022y
3. (a) 3.0 × 1029; (b) 24%
5. (a) −2.38 × 10−23 J/K; (b) 5.6 times more likely; (c) If you were betting on two heads and 8 tails, the odds of breaking even are 252 to 45, so on average you would break even. So, no, you wouldn’t bet on odds of 252 to 45.
7. (b) 7; (c) 64; (d) 9.38%; (e) 3.33 times more likely (20 to 6)
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9522 | http://dynamicmathematicslearning.com/Centroids-of-Quadrilaterals-Hans-Humenberger-2023.pdf | 1 Azim Premji University At Right Angles, November 2023 Problem Corner Dedicated to Arnold Kirsch (Germany, 1922-2013) on the 10th anniversary of his death.
Abstract: We analyze briefly different kinds of centroids of quadrilaterals and give geometrical and elementary proofs that in the world of quadrilaterals, only parallelograms have the property that their laminar centroid coincides with the vertex centroid. This paper is based on short papers (in German) by Arnold Kirsch (Kassel, Germany, 1922-2013) published between 1987 and 1995. We think these deserve to be better known – published proofs in mathematical journals in English language are usually rather complex (e.g., Kim 2016, 2020).
Definitions. A lamina is a flat object of uniform thickness.
The laminar centroid of a flat region is the centre of gravity of the region when it is regarded as a thin lamina. It is also called the geometric centroid. The vertex centroid of a polygon is the centre of gravity of a system of unit masses placed at the vertices of the polygon.
In the world of triangles, the laminar centroid always coincides with the vertex centroid (intersection point of the medians). This is an elementary and well-known fact. We proceed to prove this using the principle of levers from elementary physics.
Centroids of Quadrilaterals and a Peculiarity of Parallelograms HANS HUMENBERGER Keywords: Geometry, centroids, parallelograms, laminar centroid, vertex centroid 2 Azim Premji University At Right Angles, November 2023 Lemma 1: The vertex centroid G of a pair of point masses (weights w1and w2) lies on the connecting line and the corresponding distances l1, l2 have the ratio l1 l2 = w2 w1. For mechanical purposes, one can imagine that at the centroid G, a combined weight w1 + w2 is concentrated. In terms of analytical geometry, the point G is the weighted arithmetic mean of the points G1 and G2: G = w1 w1 + w2 · G1 + w2 w1 + w2 · G2.
Figure 1. Law of levers.
Drawing on this fact, one can give a physically motivated proof that the medians of a triangle concur, intersecting each other in the ratio 2: 1. Assume that at the vertices of a triangle we have unit point masses, and we want to determine the centroid of these three point masses (we call this the ‘vertex centroid’).
Lemma 1 tells us that the centroid of the pair of unit masses at A and B is at the midpoint MAB of AB, where we then have mass 2. Using Lemma 1 again, we see that the centroid of the three unit-masses lies on the median mc = CMAB, at the point G such that CG: GMAB = 2: 1. We may imagine all three unit-masses to be concentrated at G (with total mass 3).
Figure 2. The triangle centroid as the vertex centroid.
Since the same must hold for the other two medians, and the centroid is unique, we have proven two things: (1) The medians concur at a point that trisects all three of them; and (2) the point of concurrence is the vertex centroid. Note that this approach does not explain why the laminar centroid of the triangle lies at the same point. Here is one approach which explains why. We divide the triangle into infinitesimally thin stripes parallel to AB. Each of these stripes has its center of mass at its midpoint, so the center of mass of the whole lamina must lie somewhere on the line consisting of all these midpoints, which is the median CMAB. By a symmetric argument, it must also lie on the other two medians, hence the intersection point of the three medians is also the laminar centroid.
Let us denote the laminar centroid of a polygon by GL and its vertex centroid by GV.
3 Azim Premji University At Right Angles, November 2023 The following must be noted. The property GL = GV is a peculiarity of triangles (in the sense that it is true for all triangles), but for other polygons this is not necessarily true. Of course, for regular polygons GL = GV still holds (by symmetry, both must lie at the centre of the polygon), but for general polygons it is of great interest to ask: For which polygons is it true that GL = GV? We will restrict our exploration in this only article to quadrilaterals and ask: Which quadrilaterals have the property GL=GV? (We are not aware if there are any results of this kind for polygons with more than 4 vertices.) It is easy to see that parallelograms have the property that GL = GV (= intersection point of the two diagonals). Assume we have a parallelogram ABCD with unit mass at each vertex. Then according to Lemma 1, the intersection point of the diagonals (where they bisect each other) is the centroid of the two masses at A and C (mass 2 units), and also of the two masses at B and D (again mass 2 units). Hence the point of intersection of the diagonals is the vertex centroid GV.
To see that this is GL too, we consider the laminar centroids of triangles ABC and ADC, namely, GL(ABC) and GL(ADC). (These coincide with respective vertex centroids.) Both lie on the diagonal BC and lie at equal distance from the point of intersection of the diagonals (note the half-turn symmetry of a parallelogram with this point as centre). For mechanical purposes we can imagine the whole masses of triangles ABC and ADC being concentrated at these two triangle laminar centroids. And since these masses (areas) are equal, it follows that the laminar centroid of the whole parallelogram lies at the intersection point of the diagonals (Figure 3). So, all parallelograms have the property that GL = GV.
Figure 3. The intersection point of the diagonals of a parallelogram are GV and GL.
What has been proved above is well known. Now for the not so well-known part: parallelograms are the only quadrilaterals with the property GL=GV. Here Arnold Kirsch (of Germany, a very deserving professor for mathematics and mathematics education at the University of Kassel) came up with a very elementary proof which students in grades 9 or 10 can follow and which not only answers the question (verification) but explains why it is true (explanation). Verification and explanation are two important functions of proof (but there are also others, see De Villiers 2012).
We did not find an elementary proof in the English literature (if somebody happens to know one, please inform the author), hence we wanted to share his ingenious ideas, formulated in German (Kirsch 1987, 1995), with potentially more readers in the English language.
Theorem: A quadrilateral has the property that GL=GV if and only if it is a parallelogram.
4 Azim Premji University At Right Angles, November 2023 For this topic, we can omit crossed quadrilaterals because it is not so clear what is the interior of such a quadrilateral, and we restrict to convex or concave quadrilaterals (Figure 4a, 4b). In both cases there is an interior diagonal (AC in Fig. 4) which itself or its extension meets the other diagonal (BD in Fig. 4).
Figure 4a. Convex quadrilateral.
Figure 4b. Concave quadrilateral.
The part “if” of the Theorem (the easy and well-known part) has been dealt with above. For the “only if” part we use another lemma. We lay the groundwork for this by stating two facts: (a) If the vertex D of ∆DAC is moved parallel to AC by u, then the centroid G2 of ∆DAC is moved by 1 3u (see Figure 5; here M denotes the midpoint of AC). This should be clear since G2 = D+A+C 3 .
Figure 5. If D ′ = D + u, then G ′ 2 = G2 + 1 3u.
(b) If vertices A and C of ∆ABC are moved along the straight line AC by v, then the centroid G1 of ∆ABCis moved by 2 3v (see Figure 6; M is the midpoint of AC; M ′ is the midpoint of A′C ′). This is so since G1 = A+B+C 3 .
5 Azim Premji University At Right Angles, November 2023 Figure 6. If A′ = A + v and C ′ = C + v, then G ′ 1 = G1 + 2 3v.
These two facts and the following lemma may also be formulated using idea of shear mappings, but this is probably not very well-known at school. Knowledge of shear mappings is not necessary; we can do without it (the intercept theorem or homothety suffice).
Lemma 2: Let ABCD be a quadrilateral with interior diagonal AC. Let the points A, C be translated along AC by a vector v to A′, C ′; let the points B, D be translated by −v to B ′, D ′. Then quadrilateral A ′B ′C ′D ′ has the same vertex centroid as quadrilateral ABCD, and the laminar centroid GL of ABCD maps to the laminar centroid G ′ L of A ′B ′C ′D ′ via the translation 1 3v (see Figure 7).
Figure 7. To Lemma 2 Proof of Lemma 2: The first claim in Lemma 2 is immediately clear because the total shift of all four points together is 0.
From the facts a) and b) presented earlier, the shifts G1 − →G ′ 1 and G2 − →G ′ 2 of the centroids of the triangles are given by 1 3v. Now by Lemma 1, the laminar centroid GL of ABCD divides line segment G1G2 in the same ratio as the laminar centroid G ′ L of A ′B ′C ′D ′ divides line segment G′ 1G′ 2. This ratio is given by the areas of the two triangles; these are the weights. But since the areas of the triangles do not change 6 Azim Premji University At Right Angles, November 2023 (same base and same altitude), the ratios are the same! It follows that the shift GL − →G ′ L is given by 1 3v, too. ■ Now we are ready to prove the ‘only if’ part of the Theorem. Let ABCD be a quadrilateral with interior diagonal AC and the property GL = GV. We want to prove that it must be a parallelogram. We apply the operation of Lemma 2 to ABCD, choosing the vector v along AC in such a way that the diagonal A′C ′ of the new quadrilateral is bisected by the other diagonal B ′D ′ at points M ′ = N ′ (for this, we choose v = 1 2MN where N denotes the intersection point of AC and BD; see Figure 8).
Figure 8. The operation of Lemma 2 with v = 1 2MN.
After this operation B ′D ′ will surely be an interior diagonal of the quadrilateral A′B ′C ′D ′, and again we apply the operation of Lemma 2, this second time with vector w parallel to B ′D ′, and we choose w in such a way that the diagonal B ′′D ′′ of the image quadrilateral A ′′B ′′C ′′D ′′is bisected by the other diagonal A ′′C ′′. Now both diagonals bisect each other, which means that A ′′B ′′C ′′D ′′ must be a parallelogram. We know that in parallelograms GL = GV holds, so GL (A′′B ′′C ′′D ′′) = GV (A′′B ′′C ′′D ′′) .
Since the vertex centroid did not change when applying the two operations, we know that GV (A′′B ′′C ′′D ′′) = GV (ABCD) , hence GL (A′′B ′′C ′′D ′′) = GV (ABCD) .
On the other hand, we know that GL (A′′B ′′C ′′D ′′) = GL (ABCD) + 1 3v + w, and since GV (ABCD) = GL(ABCD), this means that this added shift vector 1 3v + w must vanish. This vanishes only for v = 0 = w, which means that ABCD is a parallelogram. ■ This was, roughly spoken (we made some additional sketches and did not translate literally), the version of Kirsch 1987. Then K. Seebach (Munich) came up with another purely geometric proof (Seebach 1994) using the principle of homothety. And it was again A. Kirsch, 1995, who made this proof still easier and 7 Azim Premji University At Right Angles, November 2023 shorter, and he used the words (translated from German) “hereby probably the ideal geometric proof of the statement is found!” This proof needs the knowledge how to construct the laminar centroid of a quadrilateral ABCD. First, draw the diagonal AC and the laminar centroids of ∆ABC and ∆ADC, i.e., GL(ABC) and GL(ADC). The laminar centroid GL(ABCD) of the whole quadrilateral must lie on the line segment connecting GL(ABC) and GL(ADC) (we even know where, but now that is not important).
Doing the same with the other diagonal BD, we know that GL(ABCD) must be the point of intersection of the segments GL (ABC) GL(ADC and GL (ABD) GL(CBD). One also must know how to construct the vertex centroid of a quadrilateral ABCD: it is the midpoint of the line segment joining the midpoints of the two diagonals. These two principles were already used in Figure 3.
Assume that ABCD is not a parallelogram. Let S be the intersection point of the diagonals (Figure 9); then S is not simultaneously the midpoint of both the diagonals.
Figure 9. Very short and purely geometric proof – convex case Then, using the well-known properties of triangle centroids, the intercept theorem, and its converse, one can see immediately (note the parallelogram with opposite vertices S and GL(ABCD): SGL (ABCD) = 2 3SM 1 + 2 3SM 2 = 2 3 (SM 1 + SM 2) ̸= 1 2 (SM 1 + SM 2) = SGV (ABCD) , (∗) hence GL (ABCD) ̸= GV(ABCD).
Remarks • Note that our precondition is that while we do not have M1 = S = M2(because ABCD is not a parallelogram), the case M1 = S ̸= M2 is covered by the above.
• The use of vector notation in () is just for abbreviation; one could easily avoid it and describe with more words the resulting parallelogram with opposite vertices S and GL(ABCD). Thus, this proof can be seen as purely geometric, and not analytic, although we used vectors in ().
8 Azim Premji University At Right Angles, November 2023 In the concave case nearly nothing changes (Figure 10), the only difference is that S lies in the exterior of ABCD and the two triangles ∆ABD and ∆CBD are not “addeD ” (for getting the quadrilateral ABCD) but “subtracteD ”.
Figure 10. Short and purely geometric proof – concave case Here is another short proof of the Theorem with coordinates, vectors, and an oblique coordinate system.
We use an oblique coordinate system. The origin lies in the intersection point of the diagonals of the quadrilateral. The first axis is the straight line AC and the second BD. Then the vertices are: A = (a, 0); B = (0, b), b < 0; C = (c, 0), c > a; D = (0, d) , d > 0.
Then the vertex centroid is given by GV = (a + c 4 , b + d 4 ) .
The centroid of the ∆ABC is G1 = (a + c 3 , b 3 ) , and the centroid of ∆ADC is G2 = (a + c 3 , d 3 ) .
According to Lemma 1, the laminar centroid GL of the quadrilateral ABCD is the weighted mean of the points G1 and G2, where the weights are the triangle areas or weights proportional to these areas, namely, −b and d: GL = −b (−b) + d · (a + c 3 , b 3 ) + d (−b) + d · (a + c 3 , d 3 ) = (a + c 3 , b + d 3 ) .
Hence, GL = GV holds if and only if a = −c and b = −d, i.e., ABCD is a parallelogram. ■ 9 Azim Premji University At Right Angles, November 2023 Conclusion In many cases school students get wrong impressions concerning centroids (e.g., that there is only one kind of centroid, or that if distinguished at all, the laminar centroid necessarily coincides with the vertex centroid, as with triangles). Dealing with that topic in case of quadrilaterals (how to determine the laminar centroid of a quadrilateral, parallelograms have the property GL = GV, and only they have this property, and so on) provides a possible way to prevent this misconception. Many proofs for “only parallelograms have this property” are too complicated to be treated at school but Kirsch’s proofs are elementary, purely geometric and students can easily follow every single step. Of course, it cannot be expected that students find these steps on their own; this was an ingenious idea of Arnold Kirsch. And using the alternative proof using analytic geometry provides a good opportunity to make use of oblique coordinate systems.
References 1. De Villiers, M. (2012). Rethinking proof with the Geometer’s Sketchpad. Key Curriculum Press, Emeryville 2. Kim, D. S. et al. (2016). Centroids and some characterizations of parallelograms. Commun. Korean Math. Soc. 31(3), 637-645 3. Kim, D. S., Kim, I. (2020). Various Centroids of Quadrilaterals without symmetry. Journal of the Chungcheong Mathematical Society 33(4), 429-444, available at 4. Kirsch, A. (1987). Bemerkung zum Vierecksschwerpunkt. Didaktik der Mathematik 15(1), 34-36 5. Kirsch, A. (1995). Vierecksschwerpunkte – Eine Ergänzung zum Beitrag von Karl Seebach über Vierecksschwerpunkte. Didaktik der Mathematik 23(3), 328 6. Seebach, K. (1994). Nochmals Vierecksschwerpunkte. Didaktik der Mathematik 22(4), 309-315 HANS HUMENBERGER received his PhD at the University of Vienna and later his habilitation in the field of mathematics education. He works at his alma mater as a professor for mathematics with special emphasis on mathematics education. He also heads the working group didactics of mathematics and school mathematics and he, along with his team, is responsible for the education of teachers at secondary and high school level. He has written many papers in German and English in mathematics education and mathematics, and also several books, most of them in German. His main fields of interest are mathematics as a process, applications of mathematics, problem solving, geometry and stochastics. For more details see his homepage: hans.humenberger/ Email: hans.humenberger@univie.ac.at University of Vienna, Faculty for Mathematics, Oskar-Morgenstern-Platz 1, 1090 Vienna, Austria |
9523 | https://www.themathdoctors.org/polynomials-a-matter-of-degree/ | Typesetting math: 100%
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Polynomials: A Matter of Degrees
Last time we examined why polynomials are defined as they are. This time, let’s look at some tricky aspects of the concept of “degree”, mostly involving something being zero.
Recall that the degree of a polynomial is the highest degree of any of its terms (that is, the degree of the leading term); and the degree of a term is the exponent (a non-negative integer) on the variable (or, when there are more than one variable, the sum of the exponents in the term).
But what is the degree when there is no exponent, namely a constant term? This will be uncertain, then obvious, and then uncertain again!
Degree of a constant term
Here is a question from 2002:
```
Degree of a Constant
Why is the degree of a constant zero?
We have just finished studying that any number to the zero power is one.
It makes sense that the degree of a monomial like 2x is one, but it doesn't make sense to me that in a constant like 5, the degree is zero.
```
Joy is referring to the degree of a constant term, such as the 5 in x2+2x+5. The first term has degree 2, the second 1 because 2x=2x1; but where does the degree of zero come from?
Doctor Ian answered:
```
Hi Joy,
It looks like you mostly figured it out yourself. Each term in a polynomial is the product of a constant (which may be 1 or 0) and a variable raised to an exponent, e.g.,
3x^4 + 0x^3 + 1x^2 + 5x^1 + 7x^0
For convenience, we leave out terms where the coefficient is zero,
3x^4 + 1x^2 + 5x^1 + 7x^0
and we leave multiplication by 1 implied,
3x^4 + x^2 + 5x^1 + 7x^0
We abbreviate x^1 as just x,
3x^4 + x^2 + 5x + 7x^0
and we simplify x^0 to 1... which is left implied (see above):
3x^4 + x^2 + 5x + 7
```
So coefficients, exponents, and even entire terms can be omitted when they do nothing, but we can still think of them as existing. The constant term is one whose exponent (when written out) is zero. And that’s the degree.
```
So it's convenient to write a polynomial this way, but it's much cleaner for the definition to require no special cases:
n
\ i
p(x) = / a x
--- i
i=0
```
If you have seen summation notation, this is just another way to write a sum like he started with:
∑i=0naixi=a0x0+a1x1+a2x2+⋯+anxn
This says to add up terms aixi while i varies from 0 through n, which is the degree of the polynomial (as long as an≠0).
```
When n=0, we get
0
a x = a
0 0
When you look at it this way, a constant is just a very short polynomial. This is nice, because you can just have one set of rules for dealing with polynomials of any degree, and constants are covered by it.
```
This, by the way, is a common theme of math: When there is a special case, we make definitions so that we can include it with the others. When we say that each term is a product of a coefficient and some number of variables, that includes the coefficient being 1, or the number of variables being 0 (the degree), just as we saw that a sum of terms can be only one term.
But isn’t that indeterminate?
The previous year, though, we had a question from a teacher who questioned this, for a good reason:
```
Degree of Constant Function
Dear Dr Math,
I teach at Galileo High School in San Francisco and we teachers are stumped on this question:
First, we agree that F(x) = 1 is a polynomial function of degree 0. We tell the kids that this is true because this function is equivalent to the function F(x) = 1x^0.
My students recently pointed out that these functions are not equivalent. This is because F(x) = 1 has a domain over all the reals, but F(x) = 1x^0 has a discontinuity at x = 0.
Now we think F(x) = 1x^0 is not a polynomial function (because polynomials shouldn't have discontinuities), but F(x) = 1 is a polynomial. And F(x) = 1 still has degree 0 but for reasons we can't explain well.
```
The issue is one we briefly touched on in Zero Divided By Zero: Undefined and Indeterminate. We call the form 00 indeterminate, meaning that a limit limx→af(x)g(x) where both functions approach 0 may take different values depending on how each of them does so. So normally, and particularly in calculus, 00 is considered undefined, just as 00 is; we need to use special techniques, such as L’Hôpital’s rule, rather than just substituting and calling it 1. So isn’t x0 undefined for x=0?
I answered:
```
Hi, Masha.
Have you seen the Dr. Math FAQ on 0^0?
I've never been quite happy with what that says, but you've convinced me that it's right!
```
Here is a quote from that FAQ:
Other than the times when we want it to be indeterminate, 0^0 = 1 seems to be the most useful choice for 0^0 . This convention allows us to extend definitions in different areas of mathematics that would otherwise require treating 0 as a special case. Notice that 0^0 is a discontinuity of the function f(x,y) = x^y, because no matter what number you assign to 0^0, you can’t make x^y continuous at (0,0), since the limit along the line x=0 is 0, and the limit along the line y=0 is 1.
This means that depending on the context where 0^0 occurs, you might wish to substitute it with 1, indeterminate or undefined/nonexistent.
I was, until writing this, a little unsatisfied with saying that 00=1. Context makes a difference.
Although 0^0 is properly considered indeterminate, it is for many purposes taken to be equal to 1. What you describe is a very good reason for doing so: it makes x^0 a continuous function, _always_ equal to 1. Having made that definition, there's no more trouble!
So we define 00=1 so that polynomials make sense!
It's important to distinguish between a discontinuity like that of x^-1 at x = 0, where the function is actually undefined and the discontinuity is not removable, and one like x^0, where it is only indeterminate and the discontinuity can be removed by a proper definition. Even if there were no consensus on doing this with 0^0 in general, there would be no problem with calling a constant function a degree-zero polynomial, because the difficulty is so easily dealt with.
In other words, we can ignore the problem in the context of polynomials, by making an appropriate definition. So that’s what we do!
And what if the constant is zero?
There’s always one more level to take it to, as this 2003 question did:
```
Degree of Zero
Dear Dr. Math,
I just wanted to know what the degree of zero was.
Thank you.
```
Now we’re moving from the degree of a constant term, to the degree of a specific constant term, namely 0. Doctor Rick took the challenge, referring first to those two answers above:
```
Hi, Kristi.
That's an interesting question. I assume you're asking for the degree of the function f(x) = 0 regarded as a polynomial.
In general, a constant function is regarded as a polynomial of degree zero, as we discuss here:
Degree of a Constant
Degree of Constant Function
This is true because a constant such as 2 can be regarded as 2x^0, and the degree of a polynomial is the highest power of the variable that has a non-zero coefficient.
```
Are you starting to see the problem?
```
I presume this is why you asked the question - there is no term in f(x) = 0 that has a non-zero coefficient. So do we say that this function has no degree?
To me, yes, this is the most sensible answer. Think about degree another way: a polynomial has (at most) as many zeros as its degree. (It has exactly as many zeros as its degree if we count the multiplicities of the zeros, for instance, x^3-10x^2+33x-36 = (x-3)^2(x-4) has 3 zeros: 3, of multiplicity 2, and 4, of multiplicity 1.)
A non-zero constant function like f(x) = 5 has zero zeros, in keeping with its degree of zero. But f(x) = 0 has an infinite number of zeros: every value of x is a zero of the function. Thus it makes sense to say that the degree of f(x) = 0 is undefined.
```
Another way to look at this would be that, if you were given 0 as a term (rather than, as discussed here, the entire function), you would want to ask, does this term represent 0 as a constant term, or 0x, a linear term, or 0x2, a quadratic term, or what? It could equally well be any of those. So we can’t decide what its degree is; it is, in a way, indeterminate.
The zero polynomial is so special that we can’t assign it a degree. If we could, it would be expected to have that many zeros. At best, then, we could call its degree infinite. But it seems better to say “undefined”.
A polynomial with degree 1, but 3 terms?
A recent question involves a different sort of issue with terms and degrees, so I’ll close with it. This came from Za in February:
I’ve learnt that a linear polynomial is a polynomial with degree as 1 and that it only consists of 2 terms.
But “√2 + 2 + y” is a polynomial with degree 1 even though it consists of 3 terms. So, is it a linear polynomial? If not, why?
A linear polynomial has degree 1, and typically looks like
ax+b,
where a and b are the coefficients.
It can have at most two terms, whose degrees will be 1 and 0. So what is happening in Za’s example?
2–√+2+y
Terms are the things we add; and we’re adding three things, not just two!
Doctor Fenton answered:
Hi Za,
Yes, that is a linear polynomial, because the sum 2+√2 is one number (approximately 3.414213562…), so it counts as a single term. A linear polynomial (in the variable y) has the form
a1y + a0 ,
where a1 and a0 are numbers. Those numbers might be complicated combinations of other numbers, but they are still a single number, so a1y is a single term, and the constant a0 is a single term.
This is something I would never have thought to teach: When we talk about coefficients in a polynomial, and say they can be any kind of number, that includes numbers that can only be written (exactly) as a sum of simpler numbers. In Za’s example, this happens because the constant term is a surd (radical sum); but it also happens when coefficients are complex numbers, like this:
(1+i)x2−2ix+(11−3i)
We just don’t happen to do that very often!
Note my use of parentheses to write the coefficients. To display Za’s linear polynomial as a sum of two distinct terms, we can similarly write
y+(2+2–√)
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9524 | https://germanna.edu/sites/default/files/2022-03/Complex%20Numbers.pdf | Working With Complex Numbers Provided by the Academic Center for Excellence 1 Reviewed August 2014 Working with Complex Numbers Real Numbers “Real” numbers are the numbers with which you are already familiar. They can be positive, negative, or zero. They can be rational or irrational. 1, 42, 8 3 −, Pi (π), and 2 are all examples of “real” numbers. Imaginary Numbers Imaginary numbers are defined as numbers which when squared produce a negative number. This seems counterintuitive at first since squaring any real number, positive or negative, produces a positive number result; ex: 5² = 25 and -5² = 25. Imaginary numbers were created in order to allow the square of a number to be negative. They are called imaginary because they do not really exist, but are needed to solve some algebraic equations. Some examples of imaginary numbers include: 8i, 8.2i, 1 −, and 4 3 i . Note that all of these examples have either an i in them or a negative number in an even root. The constant i and its powers Imaginary numbers (and complex numbers) can be recognized by the presence of “ i ” in the number. The symbol i represents 1 −. Therefore the square of any negative number will result in i multiplied by the square root of the number; ex: i i 5 25 25 = = − The powers of i are in a cycle of 4 - if you multiply i by itself four times, the result is 1 since i 2 is -1, and (-1)2 is 1. Therefore, the cycle repeats itself, successive powers of i result in i0=1, i 1=i, i 2=-1, i 3=-i, i 4=1, i5=i, … Complex Numbers Complex numbers are numbers which have both a “real” part and an “imaginary” part. An example of a complex number is 5 + 6i. In this example, 5 is the real part and 6i is the imaginary part of the complex number. Addition and subtraction: When adding and subtracting complex numbers, the real part is added or subtracted independently, and the imaginary part is added or subtracted independently. Remember not to combine unlike terms. You can not add or subtract together the real parts and the imaginary parts! • For example, to add (2+3i) and (4+5i), you would add 2 and 4 to get 6 for the real part, and add 3i and 5i to get 8i for the imaginary part. (2 + 3i) + (4 + 5i) = (6 + 8i) Working With Complex Numbers Provided by the Academic Center for Excellence 2 Reviewed August 2014 • Subtracting works similarly; subtract the real part and the imaginary part separately… o For example, (6 + 7i) – (3 + 2i) = (3 + 5i) Multiplication: To multiply two complex numbers together, use the FOIL method (First, Outer, Inner, Last). Remember that since the definition of i is 1 −, i2 is equal to -1. (2 + 3i) (4 + 5i) = (2· 4 + 2· 5i + 3i· 4 + 3i· 5i) = (8 + 10i + 12i + 15i2) = [8 + 22i + 15· (-1)] = (8 + 22i – 15) = (-7 + 22i) The “Complex Conjugate”: The complex conjugate of a complex number is given by changing the sign of the imaginary part. For instance, the complex conjugate of 3 + 2i is 3 – 2i. The sign of the real part always remains the same. This is very similar to the special factoring rule, factoring the difference of two perfect squares (a² + b²) = (a + b)(a – b). The complex conjugate is used to create a completely real number (with no imaginary part) by multiplying the complex number and complex conjugate together. It is commonly used to obtain a real-number denominator when the original denominator was a complex number. For example, (3 + 2i)(3 – 2i) = 9 – 6i + 6i – 4i² = 9 – 4(-1) = 9 + 4 = 13. Division: One way to divide complex numbers is to write the division problem in fraction form. Then multiply both the top and bottom by the complex conjugate (explained above) of the denominator. This will turn the denominator into a pure real number, and the new numerator can then be divided by this number to provide the answer. Example 1: Example 2: 41 2 23 25 20 20 16 15 12 10 8 1 25 20 20 16 1 15 12 10 8 2 25 20 20 16 2 15 12 10 8 5 4 5 4 5 4 3 2 5 4 3 2 i i i i i i i i i i i i i i i i i i i i i + − = + − + − + − − = − − − + − + + − − = − − + + + − − = + + − + − = − + − = − − + + = + + Di C Di C Di C Bi A Di C Bi A = + + + − 2 2 D C BD BCi ADi AC i D C AD BC D C BD AC + − + + + 2 2 2 2 Provided by the Academic Center for Excellence 3 Working With Complex Numbers Available related handouts: GCC’s Academic Center for Excellence (ACE) has other math handouts available on our website and in the handout rack in front of the ACE Centers at each campus. The following handouts could be especially useful in working with types of problems that involve complex numbers: • Exponents, Radicals, and Scientific Notation • Factoring methods • Polynomial fractions • The Quadratic Formula and the Discriminant Provided by the Academic Center for Excellence 4 Working With Complex Numbers Practice problems: Try the following problems; the solutions are on the back of this page. If you have any questions, stop by the ACE Center at either campus, or call us for an appointment. 1. (2 + 3i) + (3 + 2i) 2. (2 + 3i) – (3 + 2i) 3. (2 + 3i)(3 + 2i) 4. i i 2 3 3 2 + + 5. (6 + 7i)2 6. (1 + 2i)(2 + 3i)(3 + 4i) 7. (6 + 7i)2 – (7 + 6i)2 8. [(6 + 7i) - (7 + 6i)]2 9. i 2 10. i 3 11. i 100 12. i 50 13. ) 7 6 ( ) 5 4 ( ) 4 3 )( 3 2 ( ) 2 1 ( i i i i i − + − + − + + 14. (1 + 2i) 7 15. i + i2 + i3 + i 4 + i5 + i6 + i7 + i8 + i9 + i10 + i11 + i12 Provided by the Academic Center for Excellence 5 Working With Complex Numbers Solutions to practice exercises: 1. (2 + 3i) + (3 + 2i) = 5 + 5i 2. (2 + 3i) – (3 + 2i) = -1 + i 3. (2 + 3i) (3 + 2i) = (2 · 3 + 2 · 2i + 3i · 3 + 3i · 2i) = 6 + 4i + 9i – 6 = 13i 4. i i 2 3 3 2 + + = − − + + i i i i 2 3 2 3 2 3 3 2 = 13 5 12 i + 5. (6 + 7i)2 = (6 + 7i)(6 + 7i) = 36 + 42i + 42i – 49 = -13 + 84i 6. (1 + 2i)(2 + 3i)(3 + 4i) =(2 + 3i + 4i - 6)(3 + 4i) = (-4 + 7i)(3+4i) = -12 - 16i + 21i - 28 = -40 + 5i 7. (6 + 7i)2 - (7 + 6i)2 = (-13 + 84i) - (13 + 84i) = -26 8. [(6+7i)-(7+6i)]2 = [-1 + i ]2 = (1 -i - i + i2) = -2i 9. i2 = 1 − 2 = -1 10. i3 = -i 11. i100 =1 (all powers of i which are divisible by 4 are equal to 1) 12. i50 = -1 (all powers of i which are divisible by 2, but not 4, are equal to -1) 13. = − − + + = − + − − + + = − + − + − + + i i i i i i i i i i i i 12 10 18 ) 2 1 ( ) 7 6 ( ) 5 4 ( 18 ) 2 1 ( ) 7 6 ( ) 5 4 ( ) 4 3 )( 3 2 ( ) 2 1 ( i i i i i i i i i i i i i 122 347 61 109 122 103 61 48 ) 2 1 ( 244 206 192 ) 2 1 ( 144 100 12 10 216 180 ) 2 1 ( 12 10 12 10 12 10 18 ) 2 1 ( + = + + + = + + + = + + − + + + = + + − − + + 14. (1 + 2i)7 = (1 + 2i)4 (1 + 2i)2 (1 + 2i) = (1 + 2i)2 (1 + 2i)2 (1 + 2i)2 (1 + 2i) = (1 + 4i - 4)(1 + 4i - 4)(1 + 4i - 4)(1 + 2i) = (1 + 4i – 4 + 4i – 16 – 16i – 4 – 16i + 16)(1 + 2i + 4i – 8 – 4 – 2i) = (-7 - 24i)(-11 - 2i) = (77 + 14i + 264i – 48) = 29 + 278i 15. i + i 2 + i3 + i4 + i5 + i6 + i7 + i8 + i9 + i10 + i11 + i12 = i + -1 + -i + 1 + i + -1 + -i + 1 + i + -1 + -i + 1 = 0 |
9525 | https://arxiv.org/pdf/2407.08121 | On the characteristic and diameter of planar integral point sets ∗
N.N. Avdeev †
, E.A. Lushina
Voronezh State University
July 12, 2024
Abstract. A point set M in Euclidean plane is called an integral point set in semi-general position if all the distances between the elements of M are integers, and M
does not contain collinear triples. We improve the lower bound for diameter of such sets in the particular case when the characteristic of the set is of the form 4k + 1 or
4k + 2 . To achieve that, we combine hyperbolae-based and grid-based toolsets.
1 Introduction
A planar integral point set (IPS) is a set of points in the plane, such that the distance between any pair of its points is an integer, and at least one triple of its points is non-collinear. The latter condition is essential to avoid subsets of a straight line; those are de-facto equivalent to subsets of integer numbers and form a completely different combinatorial object. In 1945, Erd¨ os gave an elegant proof [1, 8] that every IPS is finite. In any IPS he chose a non-collinear triple {M1, M 2, M 3} ∈ M so that any other point M0 ∈ M lies on either the straight line M1M2, the perpendicular bisector to the segment M1M2,or one of |M1M2| − 1 hyperbolae, where |M1M2| stands for length of line segment
M1M2. Applying the same argument to the line segment M1M3, Erd¨ os concluded that #M ≤ 4 · | M1M2| · | M1M3|, where #M stands for cardinality of M .Thus, we can easily infer the lower bound for diameter of an IPS M :
diam M ≥√#M
2 .
∗The work of the first author was partially supported by the Theoretical Physics and Mathematics Advancement Foundation “BASIS”.
†nickkolok@mail.ru, avdeev@math.vsu.ru
1
arXiv:2407.08121v1 [math.CO] 11 Jul 2024
That was the first lower bound for the diameter ever; however, Erd¨ os did not only estimate the diameter, but also established the toolset for further investigation, the core of which is a system of cofocal hyperbolae. Using this toolset, in 2003 Solymosi proved that
diam M ≥ c · #M.
Although Solymosi did not give the constant c explicitly, it can be inferred from his proof that c = 124 .In the constant (for n ≥ 4) was improved to 0.3457 employing Point Packing in a Square Problem [7, 17]. In the approach has been further developed, and the constant has been tightened to 511 . Finally, in for IPS M in semi-general position (that is an IPS with no collinear triples) it was proved that
diam M ≥
n
5
5/4
(1) (in assumption that the set has at east 4 points). All these lower bounds are based on the Erdos’s framework — cofocal hyperbolae. Meanwhile in 1988 Kemnitz introduced a characteristic of an IPS, that is a squarefree integer q such as the area of a triangle formed by any triple of points from the IPS is comparable with √q. (Indeed, Kemnitz proved this fact for any point set with rational distances — in contrast with IPS, those can be infinite even if they contain non-collinear triples, see for an example construction and for some known limitations.) In 2000s, Kurz introduced the function d(2 , n ) that evaluates to the minimal possible diameter of planar IPS of cardinality n. Then Kurz employed Kemnitz’s results and found [15, Subsection 4.2] exact values of d(2 , n ) up to n = 122
by exhaustive computer search. (Taking the characteristic into consideration allows to boost such search significantly — basically because all triples of an IPS form triangles with equal characteristic.) For generalization in higher dimensions, we refer the reader to . In the present paper, we put the power of Erd¨ os’s and Kemnitz’s approaches to-gether. Due to squarefree nature of characteristic, it can be of the form 4k + 1 , 4k + 2
2or 4k + 3 , where k is integer. We prove that for a particular case of characteristic
4k + 1 and 4k + 2 the bound (1) can be improved to
diam M ≥
25 36 n
5/4
.
In Section 2, we give all the required notions and known results. In Section 3, we discuss some examples of integral point sets in order to demonstrate that none of the classes 4k + 1 , 4k + 2 , 4k + 3 is too exotic nor pathological. Section 4 is devoted to the connection between Erd¨ os curves and characteristic. In Section 5, we prove some auxiliary results, and we proceed to the main one in Section 6.
2 Basic Notions and Results
In this Section, we provide rigorous definitions and list some known results. For the sake of brevity, the following notations will be used for sets of positive integers, non-negative integers and all the integers resp.:
N = {1, 2, 3, 4, ... }, N0 = N ∪ { 0}, N± = {0, ±1, ±2, ±3, ±4, ... }
For a finite set M , we will denote its cardinality by #M .
Definition 2.1. A planar integral point set (IPS) is a set M of non-collinear points in the plane R2 such that for any pair of points M1, M 2 ∈ M the Euclidean distance
|M1M2| between points M1 and M2 is integral. Notation: M ∈ M, and also M ∈ Mn
for n = # M .When we say that a set in non-collinear, we mean that it has at least one non-collinear triple. If we tighten the condition and require all the triples to be non-collinear, we get the next definition.
Definition 2.2. A planar IPS M is said to be in semi-general position if no three points of M are collinear. Notation: M ∈ ˙M, and also M ∈ ˙Mn for n = # M .In the present paper, we mostly focus on IPS in semi-general position. However, the next restriction step can be done as the following. 3Definition 2.3. A planar IPS M is said to be in general position if no three points of M are collinear and no four points of M are concircular. Notation: M ∈ M, and also M ∈ Mn for n = # M .
Definition 2.4. The diameter of an integral point set M is defined by setting
diam M = max
M1,M 2∈M
|M1M2|.
Definition 2.5. Two numbers a and b are commensurable if their ratio a/b is a rational number. For example, the pair (7 , 2/3) is commensurable, the pair (√3/3, √12) is also com-mensurable, but the pair (√2, √3) is not.
Definition 2.6. A number is called squarefree if its only perfect square divisor is 1. The first squarefree numbers are: 1, 2, 3, 5, 6, 7, 10, 11, 13, ...
Definition 2.7. The characteristic of a planar IPS M is a squarefree number q such that the area of any triangle M1M2M3, {M1, M 2, M 3} ⊂ M , is commensurable with
√q. Notation: char M = q.
Definition 2.8. If points M1, M 2 ∈ M ∈ M, then the line segment M1M2 is said to be an edge of M .The following result is due to Kemnitz :
Theorem 2.9 (the Grid Theorem) . A set M ∈ Mn with characteristic p can be placed on the grid ai
2m; bi
√p
2m
,
where ai, b i ∈ N±, and m can be taken as the length of any edge of the set M .
For the sake of completeness, we sketch the proof for the Grid Theorem below.
Proof. Let M1, M 2 ∈ M ∈ M and |M1M2| = m. Set M1 = ( −m/ 2, 0) , M2 = ( m/ 2, 0) .Then for any Mi ∈ M , Mi = ( x, y ) one has |MiM1| = k ∈ N0, |MiM2| = n ∈ N0. The 4point Mi belongs to the intersection of two circles, whose equations are
−m
2 − x
2
y2 = k2,
m
2 − x
2
y2 = n2,
where k + n ≥ m.The solution is
x = k2 − n2
2m = ai
2m,y = ±
r
k2 −
m
2 + x
2
= ±
r
k2 −
m
2 + ai
2m
2
= ±bi
√q
2m ,
and the claim follows.
Definition 2.10. Let M1M2 be an edge of M ∈ M. For N ∈ N±, |n| < |M1M2| we will say that the set of points
{M0 : |M0M1| − | M0M2| = n}
is called the n-th Erdos curve .Obviously, the 0-th Erdos curve is the perpendicular bisector of M1M2, and all the other Erdos curves are branches of cofocal hyperbolae. On Figure 1, the Erdos curves are shown for an edge of length 3. Thus, an edge M1M2 generates 2|M1M2| − 1 Erdos curves. For the sake of brevity, Erdos curves with odd numbers are named odd Erdos curves, and the ones with even numbers are named even Erdos curves. Two following definitions are used to classify integral point sets with many collinear triples :
Definition 2.11. A planar integral point sets of n points with n − 1 points on a straight line is called a facher set. For 9 ≤ n ≤ 122 , the minimal possible diameter is attained at a facher set .
Definition 2.12. A planar integral point set situated on two parallel straight lines is called a rails set. 5Figure 1: Erdos curves
Definition 2.13. The part of a plane between two parallel straight lines with distance
ρ between them is called a strip of width ρ.
Lemma 2.14. If a triangle T with minimal height ρ is situated in a strip, then the width of the strip is at least ρ.
Lemma 2.15. [3, Lemma 4]; [6, Lemma 2.4] Let M ∈ M(2 , n ), diam M = d. Then
M is situated in a square of side length d.
Definition 2.16. [6, Definition 2.5] A cross for points M1 and M2, denoted by
cr (M1, M 2), is the union of two straight lines: the line through M1 and M2, and the perpendicular bisector of line segment M1M2.
Lemma 2.17. [6, Theorem 3.10] Each set M ∈ Mn such that for some M1, M 2 ∈ M
equality |M1M2| = 1 holds, consists of n−1 points, including M1 and M2, on a straight line, and one point out of the line, on the perpendicular bisector of line segment M1M2.
Our attention will be mostly restricted to planar integral point sets with character-6istic of the form 4k + 1 and 4k + 2 in semi-general position. Thus, if an IPS M satisfies these conditions, we will write that M ∈ M′, and also M ∈ M′
n
for n = # M .
3 Integral Point Sets with Various Characteristics
Let us demonstrate that the classes 4k + 1 , 4k + 2 nor 4k + 3 are neither too exotic nor pathological. In order to do this, we will provide important examples of IPS for each class. For convenience, we use the notation [22–24]: √p/q ∗ { (x1, y 1), ..., (xn, y n)}, which means that each abscissa is multiplied by 1/q and each ordinate is multiplied by √p/q ,i.e.
√p/q ∗ { (x1, y 1), ..., (xn, y n)} =
x1
q , y1
√pq
, ...,
xn
q , yn
√pq
,
where q is the characteristic of the IPS. Such notation is made possible by the Grid Theorem. When we think about IPS with characteristic 4k + 1 , the very first example that comes to our mind is Egyptian triangle, that is the triangle with sides (3; 4; 5) . Egyp-tian triangle is obviously an IPS with characteristic 1 = 4 · 0 + 1 .Facher sets with characteristic 1, that are called semi-crabs , are investigated in . There are also much more complex IPS with characteristic 1; for those from ˙M7, we refer the reader to . Figure 2 shows a rails IPS of characteristic 385 = 4 · 96 + 1 presented in . Its coordinates are
√385 /2 ∗ { (±1105 , 48) , (±2189; 0) , (±1587; 0) , (±1269; 0) ,
(±763; 0) , (±623; 0) , (±529; 0) , (±339; 0) }
In , the first ever known planar integral point set M7 ∈ ˙M7 is given, see Figure 3. Its coordinates are
√2002 /2227 ∗ { (0; 0) , (2227 2 · 10; 0) , (26127018; 932064) , (32142553; 411864) ,
(17615968; 238464) , (7344908; 411864) , (19079044; 54168) },
7Figure 2: IPS of cardinality 16 and diameter 2189 and char M7 = 2002 = 4 · 500 + 2 .It is noticeable that the second example of an IPS from ˙M7 given in the same article has the same characteristic. Also the largest known rails IPS presented in with 104 points on one straight line and the rest 2 on another (that gives the cardinality of 106) has the characteristic of 154 = 4 · 38 + 2 . (We do not list the coordinates of that set here due to its diameter which is 2745754098774581800288844387372160.) As for characteristic 4k+3 , we should mention that the upper bound for the minimal diameter of planar integral point set given in employs IPS with characteristic 3.Moreover, all the IPS of minimal possible diameter provided in [9, §5, Figure 1] (for cardinalities from 3 to 9) have characteristic of form 4k + 3 .For the sake of completeness, on Figure 4 we give an example of rails set with 3 points on one line and 8 points on the other (first presented in [4, Figure 1]) whose characteristic is
255255 = 3 · 5 · 7 · 11 · 13 · 17 = 4 · 63813 + 3
and whose coordinates are
P3,8 = √255255 /2 ∗ { (1767; −3); (2791; −3); (4071; −3); (−306; 0); (0; 0); (1798; 0); (2304; 0); (2760; 0); (3534; 0); (4040; 0); (4558; 0) }
8Figure 3: The heptagon in general position
Figure 4: The rails set with characteristic 255255 94 Erd¨ os Curves and Characteristic Lemma 4.1. Any set M = {A, B, C } ∈ M3 with edge |AB | = m, where |AC | − |BC | = m − s, m, s ∈ N, s is an odd number and s ≤ m, has a characteristic of the form p = 4 k + 3 , k ∈ N0.Proof. Let |BC | = n. Then, from the triangle inequality |AC | < |AB | + |BC |, we have that |AC | = m + n − s, s ∈ N and s ≤ m. By the Grid Theorem, we can assume that A = ( −m
2
; 0) , B = ( m
2
; 0) , and C = ( a
2m
; b√p
2m
). We find the distance between points A and C, and points B and C in coordinates, and form a system of equations
√(a+m2)2+pb 2
2m
= m + n − s.
√(a−m2)2+pb 2
2m
= n
(2) Multiply each equation by 2m and square both sides of the equations in system (2)
(a + m2)2 + pb 2 = 4 m2(m + n − s)2,
(a − m2)2 + pb 2 = 4 m2n2
(3) and then subtract the second equation from the first one:
(a + m2)2 − (a − m2)2 = 4 m2(m + n − s)2 − 4m2n2,
4m2a = 4 m2(( m + n − s)2 − n2),
so
a = ( m − s)( m + 2 n − s). (4) Substitute the obtained expression (4) into the second equation of the system (3):
(( m − s)( m + 2 n − s) − m2)2 + pb 2 = 4 m2n2,
that immediately gives
pb 2 = s(2 m + 2 n − s)(2 n − s)(2 m − s). (5) Since s is assumed to be odd, let s = 2 t + 1 , where t ∈ N0. Then equation (5) can be rewritten as:
pb 2 = (2 t + 1)(2 m + 2 n − 2t − 1)(2 n − 2t − 1)(2 m − 2t − 1)
10 or
pb 2 + 1 = 4( −4t4 + 8 t3m + 8 t3n − 8t3 − 4t2m2 − 12 t2mn + 12 t2m −−4t2n2 + 12 t2n − 6t2 + 4 tm 2n − 4tmn 2 − 12 tmn + 6 tm −−4tn 2 + 6 tn − 2t + 2 m2n − m2 + 2 mn 2 − 3mn + m − n2 + n).
If equation (6) has integer solutions, then both p and b are odd. The right-hand side is divisible by 4, which implies that pb 2 ≡ 3 mod 4 . Since b2 ≡ 1(mod 4) , it follows that p = 4 k + 3 , k ∈ N0.The following theorem follows immediately.
Theorem 4.2 (The Weeding Theorem I) . Let M = {M1, M 2, M 3} ∈ M3 be an IPS of semi-general position with a characteristic different from 4k + 3 , k ∈ N0. Then, the length of the edge M1M2 and its n-th Erd˝ os curve, which contains the point M3,have the same parity. Proof. Let |M1M2| = m, m ∈ N, and let the n-th Erd˝ os curve satisfy |M1M3| − |M2M3| = n. If m is an odd number, then by Lemma 4.1, the difference n = |M1M3|− |M2M3| equals m − s, where s is an even number, is also an odd number. Let m be an even number; then n = |M1M3| − | M2M3| = m − s is an even number, as a difference of two even numbers. The case when the length of an edge is even is slightly more specific and allows a more precise statement.
Lemma 4.3. Every set M = {A, B, C } ∈ M3 with an even edge |AB | = 2 q, where
|AC | − | BC | = 2 q − s, q, s ∈ N, s is an odd number and s ≤ 2q, has a characteristic of the form p = 8 k + 7 , k ∈ N0.Proof. Let us set m = 2 q in the Lemma 4.1. Then equation (5) turns into
pb 2 = s(4 q + 2 n − s)(2 n − s)(4 q − s). (7) Taking into account that s = 2 t + 1 , t ∈ N0, we bring equation (7) to the form:
pb 2 = (2 t + 1)(4 q + 2 n − 2t − 1)(2 n − 2t − 1)(4 q − 2t − 1)
11 or, after expansion,
pb 2 + 1 = 8( −2t4 + 8 t3q + 4 t3n − 4t3 − 8t2q2 − 12 t2qn + 12 t2q −−2t2n2 + 6 t2n − 3t2 + 8 tq 2n − 4tqn 2 − 12 tqn + 6 tq −−2tn 2 + 3 tn − t + 4 q2n − 2q2 + 2 qn 2 − 3qn + q) −−4n(n − 1) .
Among two consecutive natural numbers, exactly one is even, so 4n(n − 1) ≡ 0(mod 8) . Thus, the right-hand side of equation (8) is a multiple of 8. Then pb 2 ≡ 7(mod 8) . A necessary condition for the existence of an integer solution to equation (8) is that numbers p and b must be odd. Since b2 ≡ 1 (mod 8) , it follows that p = 8 k +7 ,
k ∈ N0.
Theorem 4.4 (The Weeding Theorem II) . Let an IPS in semi-general position M =
{M1, M 2, M 3} ∈ M3 have an even edge length |M1M2|, and let point M3 lie on an odd Erd˝ os curve of edge M1M2. Then the set M has characteristic p = 8 k + 7 , k ∈ N0.Proof. Indeed, let |M1M2| = 2 q, q ∈ N. For the n-th Erd˝ os curve we have |M1M3| − |M2M3| = n. According to Lemma 4.3, the difference n = |M1M3| − | M2M3| = 2 q − s,where s is an odd number, is also an odd number and char M = 8 k + 7 , k ∈ N0.
5 Auxiliary Results Lemma 5.1. Every set M ∈ Mn
′
with an edge |M1M2| = 2 has a cardinality n = 3 .Proof. Let M ∈ Mn
′
and M1, M 2 ∈ M . Then all points in the set M lie on cr (M1, M 2).Otherwise, according to Lemma 4.3, char M = 8 k + 7 , k ∈ N0.Suppose, to the contrary, that M = M1, M 2, M 3, M 4 ∈ M4
′
. Employ the Grid Theorem and set M1(−1; 0) , M 2(1; 0) , O (0; 0) . Then M3 and M4 lie on the perpendic-ular bisector to M1M2. Since the distance |M3M4| is an integer, the area of triangle
M1M3M4 is a rational number. By Definition 2.7, this means that char M = 1 . Then by the Grid Theorem we have M3(0; t/ 4) . Without loss of generality, let us assume that t ∈ N. Let |M1M3| = s. Applying the Pythagorean theorem to triangle OM 1M3,12 we get
1 + t2
16 = s2,
or equivalently,
16 s2 − t2 = 16 . (9) Now, we find the solutions to equation (9) in positive integers. To do this, we introduce the substitution: t = 4 k, k ∈ N. Then equation (9) takes the form:
s2 − k2 = 1 . (10) Break down the left-hand side of equation (10) into factors:
(s − k)( s + k) = 1 .
Since s, k ∈ N, the number 1 can be represented as the product of two integers in two ways: 1 · 1 and −1 · (−1) . In the first case, (s − k) = 1 and (s + k) = 1 , which gives
s = 1 , k = 0 , which contradicts that k ∈ N. In the second case, (s − k) = −1 and
(s + k) = −1, which gives s = −1, k = 0 , which also contradicts that s, k ∈ N.Thus, equation (9) has no positive integer solutions. The obtained contradiction completes the proof. The following result is a key element for bounds presented in and :
Lemma 5.2. [20, Observation 1] If a triangle T has integer side lengths a ≤ b ≤ c,then its minimal height m is at least a − 14
1/2.
If we exclude characteristics of the form 4k + 3 , this Solymosi’s result can be sharp-ened slightly.
Lemma 5.3. Any triangle with sides a ≤ b ≤ c, where c = a + b − 1 and a, b, c ∈ N,has the characteristic of the form p = 4 k + 3 , k ∈ N0.Proof. Indeed, consider the triangle ABC and let |BC | = a, |AC | = b. Using the triangle inequality |AB | < |BC | + |AC |, we can represent the length of side AB as
|AB | = a + b − s, s ∈ N and s ≤ a. For s = 1 , the conditions of Lemma 4.1 are satisfied: s is an odd number and s ≤ a. Therefore, the triangle with sides a, b, and
c = a + b − 1 has a characteristic of the form p = 4 k + 3 , k ∈ N0.13 Lemma 5.4. Let a triangle ABC have the characteristic different from 4k+3 , k ∈ N0,with a ≤ b ≤ c. Then the smallest height of the triangle ABC is at least (2 a − 1) 1/2.Proof. By Lemma 5.3, a triangle with a side c = a + b − 1 cannot have a characteristic different from 4k + 3 , k ∈ N0. Therefore, in a triangle with integer sides, we have
a + b ≥ c + 2 .
The height h of the triangle ABC , dropped onto side c, can be found from the formula for its area: S = hc/ 2, which gives h = 2 S/c . To find the area of the triangle, we use the Heron’s formula in the following form:
S = 14
q
4a2c2 − (c2 + a2 − b2)2.
Then
h2 =
24c ·
q
4a2c2 − (c2 + a2 − b2)2
2
= 14c2 · 4a2c2 − (c2 + a2 − b2)2
= a2 −
c2 + a2 − b2
2c
2
(11) From three heights of the triangle, the smallest one is that dropped onto its largest side. For fixed sides a and b, let us set c = a + b − 2. Then
c2 + a2 − b2
c = c + a + bc (a − b) = c + c + 2
c (a − b) =
c +
1 + 2
c
(a − b) ≤ c + a − b = 2 a − 1
and this equality is possible when a = b. We rewrite expression (11) as follows:
h2 ≥ a2 −
2a − 12
2
= a2 − (a − 1) 2 = a2 − a2 + 2 a − 1 = 2 a − 1.
Thus, the smallest height of the triangle with characteristic different from 4k + 3 is at least (2 a − 1) 1/2.
Corollary 5.5. For the height of a triangle ABC , where A, B, C ⊂ M ∈ M4
′
, with sides 3 ≤ a ≤ b ≤ c, the following estimate holds:
(2 a − 1) 1/2 ≥√5
√3a1/2,
where the difference between the left and right-hand sides increases with increasing a.
14 6 The Main Bound
The following lemma is a classical Erdos-style intersection-enumeration one, em-powered by our Weeding Theorem I.
Lemma 6.1. Let M1, M 2, M 3, M 4 ⊂ M ∈ Mn (points M2 and M3 may coincide, while the others are distinct), where n ≥ 4. Then #M ≤ | M1M2| · | M3M4| − 2.Proof. We will distinguish three cases: two even edges, two odd edges and two edges with different parity. Let’s consider the first case. Suppose both edges M1M2 and M3M4 have even lengths. Then, for each point N ∈ M , one of the following conditions is satisfied: a) N belongs to cr (M1, M 2), which implies that there are no more than 4 points (no more than 2 on each of the lines); b) N belongs to cr (M3, M 4), which implies that there are no more than 4 points (no more than 2 on each of the lines); c) N belongs to the intersection of one of the (|M1M2|/2 − 1) hyperbolae with one of the (|M3M4|/2 − 1) hyperbolae, which implies that there are no more than
4( |M1M2|/2 − 1)( |M3M4|/2 − 1) points. Assuming that the edges M1M2 and M3M4 have even lengths, we infer that
|M1M2| ≥ 4 and |M3M4| ≥ 4. Then,
4
|M1M2|
2 − 1
|M3M4|
2 − 1
4 + 4 = = 4
|M1M2| · | M3M4|
4 − |M1M2|
2 − |M3M4|
2 + 1
4 + 4 = = |M1M2| · | M3M4| − 2|M1M2| − 2|M3M4| + 4 + 4 + 4 ≤ | M1M2| · | M3M4| − 4
< |M1M2| · | M3M4| − 2.
Let’s consider the second case. Suppose both edges M1M2 and M3M4 have odd lengths. Then, for each point N ∈ M , one of the following conditions is satisfied: a) N belongs to cr (M1, M 2), which implies that there are no more than 2 points (otherwise, char M = 4 k + 3 ); 15 b) N belongs to cr (M3, M 4), which implies that there are no more than 2 points; c) N belongs to the intersection of one of (|M1M2| − 1) /2 hyperbolae with one of
(|M3M4| − 1) /2 hyperbolae, which implies that there are no more than (|M1M2| −
1)( |M3M4| − 1) points. Assuming that the edges M1M2 and M3M4 have odd lengths, we infer that
|M1M2| ≥ 3 and |M3M4| ≥ 3. Then,
(|M1M2| − 1)( |M3M4| − 1) + 2 + 2 = |M1M2| · | M3M4| − | M1M2| − | M3M4| + 4 ≤≤ | M1M2| · | M3M4| − 2.
Let’s consider the third case. Suppose the lengths of edges M1M2 and M3M4 are different. Without loss of generality suppose the edge M1M2 has even length while edge
M3M4 has odd length. Then, for each point N ∈ M , one of the following conditions is satisfied: a) N belongs to cr (M1, M 2), which implies that there are no more than 4 points (no more than 2 on each of the lines); b) N belongs to cr (M3, M 4), which implies that there are no more than 2 points (otherwise char M = 4 k + 3 ); c) N belongs to the intersection of one of (|M1M2|/2 − 1) hyperbolae with one of
(|M3M4| − 1) /2 hyperbolae, which implies that there are no more than 4( |M1M2|/2 −
1)( |M3M4| − 1) /2 points. From the parity assumption for edges M1M2 and M3M4 we can infer that |M1M2| ≥
4 and |M3M4| ≥ 3. Then,
4
|M1M2|
2 − 1
|M3M4| − 12
2 + 4 = = ( |M1M2| − 2)( |M3M4| − 1) + 6 = = |M1M2| · | M3M4| − | M1M2| − 2|M3M4| + 2 + 6 ≤ | M1M2| · | M3M4| − 2.
This proves the assertion. The following function was introduced in
d(2 , n ) = min
M∈Mn
diam M,
16 and some values were given:
d(2 , 3) = 1,d(2 , 4) = 4,d(2 , 5) = d(2 , 6) = 8 ,d(2 , 7) = 33 ,d(2 , 8) = d(2 , 9) = 56 ,d(2 , 10) = ... = d(2 , 12) = 105 ,d(2 , 13) = d(2 , 14) = 532 ,d(2 , 15) = ... = d(2 , 18) = 735 ,d(2 , 19) = ... = d(2 , 24) = 1995 ,d(2 , 25) = ... = d(2 , 27) = 9555 ,d(2 , 28) = 10672 ,d(2 , 29) = ... = d(2 , 36) = 13975 ,d(2 , 37) > 20 000 .
Now we are finally ready to proof our main result. It improves the bound (1) in our special case.
Theorem 6.2. Let M ∈ M′
n
, i.e., M is a set in semi-general position with charac-teristic different from 4k + 3 , k ∈ N0. Then, for every integer n ≥ 3, the inequality
diam M ≥
25 n
36
5/4
holds. Proof. For n = 3 , we have diam M ≥ 3 (achieved by an isosceles triangle with sides 2, 3, 3), and the assertion is obvious. Let us consider M ∈ Mn
′
, n ≥ 4, diam M = p.Choose points M1, M 2, M 3, M 4 ∈ M (points M2 and M3 may coincide, while the others must be pairwise distinct) such that 17 min
A,B ∈M
|AB | = |M1M2|
min
A,B ∈M\M1
|AB | = |M3M4| = m
If m ≤ 65 p2/5, then by Lemma 6.1,
n ≤ | M1M2| · | M3M4| − 2 ≤ 36 25 p4/5 − 2
or equivalently
p ≥
25( n + 2) 36
5/4
≥
25 n
36
5/4
which is exactly the claim of the theorem. Now we have to consider m > 65 p2/5. Then for any points A, B ∈ M \ M1, we have
|AB | > 65 p2/5. By Corollary 5.5 and Lemma 2.14, no three points from M \ M1 lie in a strip of width √5
√3 ·
s
6p2/5
5 = √2 · p1/5.
By Lemma 2.15, the set M lies in a square with side length p. We cover this square by q strips, p4/5
√2
≤ q < p4/5
√2
1 , such that the width of each strip does not exceed
√2 · p1/5. Each of the obtained strips contains no more than two points from M \ M1,so
n ≤ 2 p4/5
√2 + 1
!
1 = 2p4/5
√2 + 3 = √2p4/5 + 3 . (13) From inequality (13), we obtain
p ≥
n − 3
√2
5/4
. (14) According to the results (12), for 3 ≤ n ≤ 36 the theorem is true. Moreover, it is known that for all 37 ≤ n ≤ 74, we have d(2 , n ) > 20 000 and our estimate on the diameter is also true. Indeed, d(2 , 74) ≥ 25 36 · 745/4 ≈ 10 , 655 . This estimate is weaker than the available numerical results. Therefore, from now on, we can assume that n ≥ 74.18 For estimation (14) and n ≥ 74, we have
p ≥
n − 3
√2
5/4
≥
25 n
36
5/4
.
Thus, for any n ≥ 3 the inequality diam M ≥ 25 n
36
5/4 holds.
7 Conclusion
The bound proved above (as well as (1)) may appear to be far from precise values of d(2 , n ). However, it’s easy to see that values of d(2; n) tend to repeat often; thus, it is rather not unrealistic that the bounds may converge to the precise values. Also, we should notice that our approach and specifically Weeding Theorems do not require semi-general position and can be applied to tighten the bound from in the special case of characteristic 4k + 1 or 4k + 2 . However, we are not ready to accept this tedious challenge yet. Another research area that is able to utilize our results fruitfully is the maximization of IPS. The Weeding Theorems I and II can sometimes make the exhaustive search up to 4 times faster.
8 Acknowledgements
The author thanks Dr. Prof. E.M. Semenov for the fruitful discussion and ideas, Dr. A.S. Chervinskaia for the idea of using the term “facher” and proofreading, Dr. A.S. Us-achev for proofreading, and E.A. Momot for the “SciLexic” project that helped the author with some word usage issues.
References
N. H. Anning and P. Erd˝ os, Integral distances, Bull. Amer. Math. Soc. 51. 8 (1945), 598–600, doi : 10.1090/S0002-9904-1945-08407-9 . A. R. Antonov and S. Kurz, Maximal integral point sets over Z2, Int. J. Com-put. Math. 87. 12 (2008), 2653–2676, doi : 10.1080/00207160902993636 , arXiv:
0804.1280 .19 N. N. Avdeev and E. M. Semenov, Integral point sets in the plane and Euclidean space (Множества точек с целочисленными расстояниями на плоскости и в евклидовом пространстве), Matematicheskij forum (Itogi nauki. Yug Rossii)
(2018), 217–236. N. N. Avdeev, R. E. Zvolinsky, and E. A. Momot, On particular diameter bounds for integral point sets in higher dimensions (2019), arXiv: 1909.10386 , url :
. N. Avdeev, On diameter bounds for planar integral point sets in semi-general position (2019), arXiv: 1907.09331 , url : abs/2019arXiv190709331A/abstract . N. Avdeev, On existence of integral point sets and their diameter bounds, Aus-tralas. J. Combin. 77. 1 (2020), 100–116, arXiv: 1906.11926 , url : https: //ui.adsabs.harvard.edu/abs/2019arXiv190611926A/abstract . A. Costa, Valid constraints for the point packing in a square problem, Discrete Appl. Math. 161. 18 (2013), 2901–2909. P. Erd˝ os, Integral distances, Bull. Amer. Math. Soc. 51. 12 (1945), 996, doi : 10. 1090/S0002-9904-1945-08490-0 . H. Harborth, A. Kemnitz, and M. M¨ oller, An upper bound for the minimum diameter of integral point sets, Discrete Comput. Geom. 9. 4 (1993), 427–432,
doi : 10.1007/bf02189331 . G. B. Huff, Diophantine problems in geometry and elliptic ternary forms, Duke Math. J. 15. 2 (1948), 443–453, doi : 10.1215/S0012-7094-48-01543-9 . A. Kemnitz, Punktmengen mit ganzzahligen Abst¨ anden, 1988. T. Kreisel and S. Kurz, There are integral heptagons, no three points on a line, no four on a circle, Discrete Comput. Geom. 39. 4 (2008), 786–790, doi : 10.1007/ s00454-007-9038-6 .20 S. Kurz, On the characteristic of integral point sets in Em, Australas. J. Combin.
36 (2006), 241–248, arXiv: math/0511704 . S. Kurz and R. Laue, Bounds for the minimum diameter of integral point sets,
Australas. J. Combin. 39 (2007), 233–240, arXiv: 0804.1296 . S. Kurz and A. Wassermann, On the minimum diameter of plane integral point sets, Ars Combin. 101 (2011), 265–287, arXiv: 0804.1307 . S. Kurz et al., Constructing 7-clusters, Serdica J. Comput. 8. 1 (2014), 47–70, arXiv: 1312.2318 . M. C. Mark´ ot and T. Csendes, A new verified optimization technique for the "packing circles in a unit square" problems, SIAM J. Optim. 16. 1 (2005), 193–219. E. A. Momot, R. E. Zvolinskiy, and A. E. Zvolinskiy, An example of rails integral point set with cardinality 106, Прикладная математика и фундаментальная информатика, 2022, 114–115. H. Nozaki, Lower bounds for the minimum diameter of integral point sets, Aus-tralas. J. Combin. 56 (2013), 139–143. J. Solymosi, Note on integral distances, Discrete Comput. Geom. 30. 2 (2003), 337–342, doi : 10.1007/s00454-003-0014-7 . J. Solymosi and F. De Zeeuw, On a question of Erd˝ os and Ulam, Discrete Comput. Geom. 43. 2 (2010), 393–401, arXiv: 0806.3095 . Н. Н. Авдеев, On integral point sets in special position, Некоторые вопросы анализа, алгебры, геометрии и математического образования: материалы международной молодежной научной школы «Актуальные направления математического анализа и смежные вопросы» 8 (2018), 5–6. Н. Н. Авдеев, Об отыскании целоудалённых множеств специального вида, Актуальные проблемы прикладной математики, информатики имеханики - сборник трудов Международной научной конференции. Научно-исследовательские публикации, 2018, 492–498. 21 Н. Н. Авдеев and Е. М. Семёнов, Множества точек с целочисленными расстояниями на плоскости и в евклидовом пространстве, Russian,
Математический форум (Итоги науки. Юг России) (2018), 217–236. М. Смуров and А. Спивак, Покрытия полосками, Квант 5 (1998), 6–12. 22 |
9526 | https://byjus.com/maths/pemdas/ | In Mathematics, we do operations like addition, subtraction, multiplication and division. These operations are performed by a certain rule or say there is an order of operation. PEMDAS rule is one of the rules which is exactly equal to BODMAS rule. The full form of PEMDAS is given below:
P – Parentheses [{()}]
E – Exponents (Powers and Roots)
MD- Multiplication and Division (left to right) (× and ÷)
AS – Addition and Subtraction (left to right) (+ and -)
whereas the full form of BODMAS is – Brackets Order Division Multiplication Addition and Subtraction.
PEMDAS term is used mainly in the US but in India and the UK, we call it as BODMAS. But there is no difference between them. The order of operations for brackets, orders, addition, subtraction, multiplication and division is the same for both the rule. The PEMDAS rule can be remembered using the acronym “Please Excuse My Dear Aunt Sally”.
PEMDAS formula is nothing but the order of calculations by means of which we calculate difficult equations step by step. Let us discuss it with some examples.
| |
| Arithmetic Operations Exponents And Powers Addition And Subtraction Of Integers Multiplication And Division Of Integers |
PEMDAS Rule
PEMDAS rule states that the order of operation starts with the parentheses first or the calculation which is enclosed in brackets. Then the operation is performed on exponents(degree or square roots) and later we do operations on multiplication & division and at last addition and subtraction. Let us discuss in brief.
PEMDAS: Order of Operations
P: Solve the calculation or equation which are present in the parentheses or brackets like small brackets( ), curly brackets{ } or big brackets[ ]. Priority is given to brackets first.
E: Exponential expressions should be calculated first before the operations of multiplication, division, addition and subtraction. Usually, they are expressed in power or roots, like 22 or √4.
MD: Then perform multiplication or division from left to right, whichever comes first in the equation.
AS: At last, perform addition or subtraction whichever comes first while moving from left to right.
PEMDAS Vs BODMAS
There is only an abbreviation difference between them.
| | |
--- |
| P – Parentheses First | B – Brackets First |
| E – Exponents | O – Orders |
| M – Multiplication | D – Division |
| D – Division | M – Multiplication |
| A – Addition | A – Addition |
| S – Subtraction | S – Subtraction |
In Canada, this order of operation is also mentioned as BEDMAS(Brackets, exponents, division, multiplication, addition and subtraction). Though the order of operation has given different names in different countries, the meaning for all is the same.
PEMDAS Examples with Answers
Let us see how to solve different problems using PEMDAS rule in maths.
Example 1: Solve 58÷ (4 x 5) + 32
Solution:
58 ÷ (4 x 5) + 32
As per the PEMDAS rule, first, we have to perform the operation which is in the parentheses.
= 58 ÷ 20 + 32
Now perform the exponent/power operation
= 58 ÷ 20 + 9
The division should be performed.
= 2.9 + 9
And the last, addition.
= 11.9Therefore, 58 ÷ (4 x 5) + 32 = 11.9
Example 2:
(\begin{array}{l}Simplify\ the\ expression:\ \sqrt{1+8}+12\end{array} )
Solution:
As per the PEMDAS rule, first we need to perform the operation of exponent, i.e. square root
For this first we need to add the numbers under the square root.
(\begin{array}{l}=\sqrt{1+8}+12\ =\sqrt{9}+12\ =3+12\=15\end{array} )
Example 3:
(\begin{array}{l}Simplify:\frac{5+4}{1+2}-3\end{array} )
Solution:
A horizontal fractional line also acts as a symbol of grouping:
(\begin{array}{l}=\frac{5+4}{1+2}-3\=\frac{9}{3}-3\=3-3\=0\end{array} )
Example 4: Calculate: [25 + {14 – (3 x 6)}]
Solution:
Given,
[25 + {14 – (3 x 6)}]
As per PEMDAS, here we have perform the operations within the parentheses, first (), second {} and finally []
=[25 + {14 – 18}]
= [25 +{-4}]
Here, we have to perform multiplication for the signs
= 25 – 4
= 21
Also, learn more Maths topics and download BYJU’S – The learning app for interactive videos.
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9527 | https://github.com/SUSTech-data/LLM-MATH | GitHub - SUSTech-data/LLM-MATH
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LLM-MATH
基于自然语言处理的生成式大模型已经在很多领域，包括知识，写作，理解上达到甚至超过了人类水平。但与大模型在许多领域上超越人类的表现不同, 在需要复杂推理的数学领域上，大模型与人类、甚至大模型与大模型之间的差异依然显著, GPT-4的数学能力只相当于一个成绩中等的数学系学生(Frieder et al. 2023), 而在普遍被用于衡量数学能力的GSM8K（小学水平的数学应用题）, MATH数据集（数学竞赛及本科水平）上, GPT-4要优于目前最先进的开源大模型2-4倍, 因此模型的数学能力也可以被间接用于评估模型的性能。 另一方面, 随着大语言模型的能力增长，对其在特定领域，如数学的应用也引发了广泛关注。不仅仅是解决基础数学应用题，更高级的自动定理证明也逐渐被看作是大模型的潜在应用领域。这使得如何提高模型的推理和数学能力，利用大模型解决更复杂的数学问题, 成为日益受到关注的问题。
数学作为一个极度依赖多步逻辑的学科，解决数学问题的能力客观地反映了模型的逻辑水平和对上下文之间隐含知识的充分理解，因此增强大模型的数学能力是真正意义上提高大模型智能水平的重要标志。 对于数学相关的题目和任务，我们按照解决问题所需要的模型能力层级，将其分为三级, 并维护相应的排行榜。
算术推理
这类题目的问题和答案都主要基于自然语言描述，答案附带含有四则运算的算术式对题目中出现的数量关系进行转换和理解，并借由步骤来分隔解题逻辑，最终通过一步一步地推理和运算，得到最终答案，其难度主要是小学数学题，小学奥数题和部分初中题目，代表性的任务数据集是GSM8k。解决这类问题需要模型具备逐步推理能力和简单的计算能力。
数学应用题（Math word problem)
这类题目同样含有自然语言描述，但题目包含代数式和方程，往往没有充足的自然语言描述的实体对象，是相较于应用题更加抽象的数学题目。模型需要经过代数符号与代数式的推导和方程求解得到最终答案，其难度类似于中学数学竞赛以及本科数学的题目，代表性的数据集是MATH。解决这类问题需要模型具备较为完整的数学知识体系，理解抽象概念和符号的能力同时根据这些概念进行创造性推理的能力。
定理证明
这类问题包括复杂的数学证明和所有我们真正关心的数学问题，由于其严谨的逻辑性和抽象性，其难度远远超过了前两类问题。由于其自然语言数据的相对匮乏和命题的多样性，我们无法通过简单分析模型的输出评判模型回答的对错，对于这类问题，我们需要将问题转化为机器证明语言，通过训练和提示让模型输出合法的计算机语句来完成命题的证明。代表性的数据集和评价指标包括miniF2F、ProofNet和IMO Grand Challenge。解决这类问题需要模型的数学能力与代码能力的深度对齐，甚至需要外挂知识库和评估模式来指导和优化模型产生正确的输出。
Frieder, Simon, Luca Pinchetti, Ryan-Rhys Griffiths, Tommaso Salvatori, Thomas Lukasiewicz, Philipp Christian Petersen, Alexis Chevalier, and Julius Berner. 2023. “Mathematical Capabilities of ChatGPT.” arXiv.
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9528 | https://ericrowland.github.io/investigations/sumsofpowers.html | Sums of Consecutive Powers
Sums of Consecutive Powers
Contents I. The Formula Skeleton II. Bernoulli Polynomials ReferencesSupplements and Links Sums of Consecutive Powers Project
I. The Formula Skeleton We would like to arrive at a computationally simpler formula for the function (1) that computes the sum of the first n consecutive like p th powers. Clearly for zeroth powers, For p = 1, we desire a formula for the sum of the first n natural numbers. The first few sums are These numbers are known as the triangular numbers because they are the number of dots in the following isosceles right "triangles": The explicit formula can be obtained by a simple geometric interpetation: The numbers of dots in the triangles are half of the numbers of dots in the following rectangles: So the n th triangular number is n (n + 1) / 2, and we have the identity For second powers, there is no obvious geometric interpretation. The sequence begins: It's not too difficult to guess the correct formula, especially if we suspect that it is given by a polynomial in n, as in the p = 1 case: In general, we may suspect that the sum of the first natural numbers raised to the p th power is a polynomial in n of degree p + 1. If this is true, then on a case by case basis it is possible to find this polynomial for a given p (for example, by finding an interpolating polynomial to the first p + 2 points). However, it would be more enlightening to find a closed-form expression for the sum f p(n). Without foresight, one approach to this is the following: Record the expressions for several low values of p and look for patterns. By interpolation, we find: There are no obvious patterns in the larger factors, and the irreducible quartic factor in the formula for p = 6 is somewhat foreboding. Let us change our view of these slightly and look instead at the fully expanded polynomials: Immediately we notice some patterns: The leading coefficient appears to be 1 / (p + 1), and the coefficient of n p seems to always be 1 / 2. By clearing away the variables and putting the coefficients in a table to line them up (ignoring the zero constant term 0 n 0 in each case), we see some more structure: Several columns are simply 0. The coefficient of n p – 1 is evidently linear: p / 12. It turns out that the coefficient of n p – 3 is given by a cubic in p: –p (p – 1) (p – 2) / 720. In fact, for even k the coefficients of n p + 1 – k are given by a fairly simple polynomial of degree k – 1: The product can be written more succinctly as , suggesting that we rewrite the above polynomials as binomial coefficients: Let B k be the (signed) constant appearing in the expression for the coefficient of n p + 1 – k. Note that the zero columns in the table of coefficients imply that for odd k> 1, B k = 0. With the exception of understanding B k, we have arrived at the following conjectural formula for f p(n): It turns out that the coefficients B k are simply the Bernoulli numbers: Furthermore, we can even bring the first two terms into the sum if we include a factor of (–1)k to take care of the fact that B 1 = – 1 / 2 while the coefficient of n p is 1 / 2: (2)
II. Bernoulli Polynomials To write the expression (2) more concisely, we now define Bernoulli polynomials and explore some of their properties. For p ≥ 1, let B p(n) be a polynomial in n of degree p satisfying (3) This defines B p(n) up to its constant term; we define its constant term B p(0) to be the Bernoulli number B p. We call B p(n) the p th Bernoulli polynomial, after Jakob Bernoulli (1654–1705), who discovered them in his attempt to find the general formula for sums of consecutive powers. The first few Bernoulli polynomials are One property of the Bernoulli polynomials is that where "B k" is to be interpreted as B k. (We do not prove this here but simply take it from the literature.) Then considering f p(n – 1) rather than f p(n) puts (2) in this form: or Bernoulli's form, (4) Of course, this wasn't Bernoulli's approach. Bernoulli noted that since the Bernoulli polynomials are defined by (3) up to a constant and then we must have for some constant C p if f p(n) is a polynomial of degree p + 1. For p ≠ 0 we may redefine f p(n) as so that f p(0) = 0. Putting n = 0 in (3) gives B p = B p(1) for p> 1, so for p> 1 This again gives (4). At last we will actually prove that (4) holds. We do this for every p ≥ 2 by induction on n. (The cases p = 0 and p = 1 were proven in Section I.) Clearly for n = 0 Assume that (4) holds for n. Then by the definition (3) of B p(n), so (4) holds for n + 1, completing the induction.
References
Relevant links:
Power Sum on MathWorld
Bernoulli Number on MathWorld
Bernoulli Polynomial on MathWorld
Introduction on Bernoulli's Numbers: Bernoulli numbers in sums of powers, power series, and the Riemann zeta function
The Bernoulli Number Page: a program that computes Bernoulli numbers and some advanced topics
ProjectExpositions
Pascal's Simplices•Pythagorean Triples•Regular Polygons
Regular Polyhedra•Regular Polytopes•Sums of Consecutive Powers
Mathematical Induction•Modular Arithmetic•Polynomial Equations
Investigations Home•Calculators•Popular Books
Eric Rowland |
9529 | https://www.sciencedirect.com/science/article/pii/S0092867421001665 | The basic immunology of asthma - ScienceDirect
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Summary
Keywords
Pathophysiology of asthma
Type 2-high asthma is controlled by airway epithelial cells
Adaptive type 2 immunity in asthma
Contribution of B cells and IgE to asthma
Eosinophils: More than a marker of type 2-high asthma
ILC2s: Does their function overlap with Th2 cells in type 2-high asthma?
Non- (or low-) type 2 asthma: All down to neutrophils?
Virus-induced asthma exacerbations
Biologicals in the treatment of asthma
Conclusions
References
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Volume 184, Issue 6, 18 March 2021, Pages 1469-1485
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The basic immunology of asthma
Author links open overlay panelHamida Hammad 1 2, Bart N.Lambrecht 1 2 3
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The basic immunology of asthma
Cell, Volume 184, Issue 9, 29 April 2021, Pages 2521-2522
Hamida Hammad, Bart N. Lambrecht
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Summary
In many asthmatics, chronic airway inflammation is driven by IL-4-, IL-5-, and IL-13-producing Th2 cells or ILC2s. Type 2 cytokines promote hallmark features of the disease such as eosinophilia, mucus hypersecretion, bronchial hyperresponsiveness (BHR), IgE production, and susceptibility to exacerbations. However, only half the asthmatics have this “type 2-high” signature, and “type 2-low” asthma is more associated with obesity, presence of neutrophils, and unresponsiveness to corticosteroids, the mainstay asthma therapy. Here, we review the underlying immunological basis of various asthma endotypes by discussing results obtained from animal studies as well as results generated in clinical studies targeting specific immune pathways.
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Keywords
asthma
lung
immune cells
endotypes
biologics
Asthma is a chronic inflammatory disease of the airways leading to cough, wheeze, shortness of breath, and chest tightness. Asthma symptoms are driven by the inflammation of the airways, which triggers processes such as mucus production, remodeling of the airway wall, and bronchial hyperresponsiveness (BHR, which is the tendency of smooth muscle cells to react to non-specific stimuli such as cold air. Asthma often starts at a young age (childhood-onset asthma), but some patients can develop asthma later in life (late-onset asthma). Childhood-onset and late-onset asthma differ in many ways. Late-onset asthma is more severe and less associated with allergy than childhood-onset asthma. In children, atopy, lower-lung function, and respiratory-tract infections especially with rhinovirus, represent major risk factors for the persistence of asthma. Whether the underlying inflammation present in asthmatic children drives the pathogenicity of respiratory viruses or whether frequent viral infections at young age set the stage for asthma to develop is still not understood. Because inflammation is so central in asthma pathogenesis, it is not surprising that the primary goal of asthma treatment has been to achieve the control of symptoms and underlying inflammation to avoid future disease exacerbations.
Unsupervised clustering methodology to look at the disease course and clinical features of asthma has revealed that this is a heterogeneous disease, reflected by major patient-specific differences in age of onset, associated risk factors, and degrees of severity, comorbidity, and response to treatment (Wu et al., 2019). We also realize now that not all asthma patients have reversible airway obstruction, and persistent airway obstruction caused by airway-wall remodeling and mucus plugging of the airways is now seen as one of the biggest unmet medical needs of clinical asthma care (Dunican et al., 2018). Historically, only two main forms of asthma have been identified: allergic and nonallergic asthma, but this has turned out to be an oversimplification.
Allergic asthma tends to begin in childhood and is associated with T helper 2 (Th2) cell responses, which are also seen in other allergic conditions such as atopic dermatitis or allergic rhinitis. This form of asthma is induced by early life encounters with environmental allergens such as house dust mite (HDM), pollen, cockroach, or animal dander but can also be induced later in life when a new, e.g., occupational allergen, is encountered. Upon recognizing allergens, allergen-specific Th2 cells produce type 2 cytokines (interleukin [IL]-4, IL-5, IL-9, and IL-13), that lead to the accumulation of high numbers of eosinophils in the airway wall, mucus overproduction, and synthesis of immunoglobulin E (IgE) by allergen-specific B cells, which can be detected in the serum or through a positive skin-prick test. Although the mechanisms and environmental risk and protective factors behind allergic sensitization in childhood are well understood and extensively modeled in mice (reviewed in Lambrecht and Hammad, 2017), why the disease localizes to the airways, and persists into adulthood, is less clear. Disease onset coincides, however, with a very crucial period of immune system and lung structural development in early childhood. Lifelong homeostasis and susceptibility to immune-mediated diseases such as asthma are shaped during the neonatal period. Therefore, alterations in the lung environment during this “window of opportunity” could lead to changes in immune cell and organ behavior, that persist long after the initial trigger is gone (de Kleer et al., 2016a; Saglani et al., 2018). In contrast to allergic asthma, non-allergic asthma is usually late onset, is more common in females and in obese patients (Pakkasela et al., 2020), and can sometimes be very difficult to treat. Late-onset asthma phenotypes were classified into Th2 and non-Th2 late-onset asthma. The non-Th2 form is often associated with obesity, aging, and smoking. The Th2-associated form is often accompanied by recurrent and chronic rhinosinusitis with nasal polyps (CRSwNP) and with sensitivity to aspirin and can be associated with high eosinophil numbers in the airways (Bachert et al., 2020).
As with many chronic inflammatory diseases, clinicians now realize that the division of asthma into just two clinical forms has been an oversimplification. In the last years, the classification of asthma phenotypes has evolved into asthma endotypes such as the type 2-high or -ultra-high (essentially eosinophilic; see Figure 1) and type 2-low (non-eosinophilic, sometimes neutrophilic, and metabolic; see Figure 2; Peters et al., 2019). Endotypes are defined by underlying pathophysiological mechanisms, which might lead to direct differences in responsiveness to common therapies such as inhaled corticosteroids or specific biologicals (Fahy, 2015; Lötvall et al., 2011; Lambrecht et al., 2019). As such, the type 2-high endotype is orchestrated by Th2-associated cytokines such as IL-4, IL-5, and IL-13, with ultra-type 2-high asthma reflecting a more severe form of the disease (Figure 1). The type 2-low endotype is more complex and no biomarkers have been identified so far. Therefore, type 2-low asthma generally includes all asthmatic patients with no type 2-high inflammation (Figure 2).
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Figure 1. Schematic of our current understanding of type 2-high asthma
Type 2 asthma is characterized by high production of IL-4, IL-5, and IL-13, which leads to asthma features with the accumulation of type 2-associated cells such as eosinophils and mast cells in lung tissue and to mucus production. Four asthma phenotypes have been identified, ranging from mild allergic early-onset to severe late-onset Th2 high to ultra-high asthma.
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Figure 2. Schematic of our current understanding of type 2-low asthma
Type 2-low asthma is characterized by lack of type 2 biomarkers, presence of neutrophils, obesity, and/or unresponsiveness to corticosteroids. Different phenotypes with a broad range of severity have been identified.
In this review, we will focus on the underlying immunological basis of the various asthma endotypes, highlighting articles where authors imply causality through rigorous experimentation. We will discuss results obtained from animal studies in which molecular pathways have been unraveled in great molecular detail and results generated in clinical studies targeting particular pathways using molecule-specific biologics.
Pathophysiology of asthma
The main feature of asthma is airway obstruction, which is caused by a reduction in the diameter of the airways. The narrowing of the airways is mediated by chronic inflammation of the airway wall, characterized by the infiltration and activation of immune cells such as dendritic cells (DCs), eosinophils, neutrophils, lymphocytes, innate lymphoid cells (ILCs), and mast cells. A complex interplay between these immune cell types and with neighboring structural cells such as epithelial cells leads to the development of asthma features such as BHR which, in most cases, is reversible by the use of bronchodilators. However, in more severe forms of asthma, airway obstruction does not always normalize following therapy. In such patients, the development of long-lasting mucus plugs in smaller airways might explain the fixed airway obstruction (Dunican et al., 2018). In addition, in patients with severe asthma, other mechanisms, including airway remodeling (which consists of airway smooth muscle hyperplasia, goblet cellmetaplasia, and exaggerated subepithelial collagen deposition), might also contribute greatly to the pathogenesis of asthma.
Type 2-high asthma is controlled by airway epithelial cells
Our understanding on the role of epithelial cells (ECs) in asthma has advanced greatly in the last years. Nowadays, ECs are considered as more than just a physical barrier between the body and the outside world and are recognized as immunologically active cells that can orchestrate inflammatory responses to external triggers. In asthma, the epithelial cell barrier is altered by the loss of tight junction proteins that keep them together, leading to increased permeability. Allergens such as HDM or cockroach, endowed with enzymatic activities, are able to cleave intercellular junctions leading to the loss of cell-cell contacts (Wan et al., 1999). Epithelial cell damage is actually a feature present in all phenotypes of asthma and correlates with disease severity (Papi et al., 2018). Moreover, changes in EC functions are seen at very young age (Carsin et al., 2016), leading to the idea that ECs may contribute in some way to the initiation of asthma early in life.
Epithelial-derived cytokines contribute to type 2-high asthma
Lung ECs express a myriad of pattern recognition receptors such as toll-like receptors (TLRs), nucleotide-binding oligomerization do-main (NOD)-like receptors (NLRs), C-type lectin receptors (CLRs), retinoic acid-inducible gene (RIG)-I-like receptors (RLRs), protease-activated receptors, and purinergic receptors. These receptors allow ECs to respond to a variety of external triggers by producing chemokines and cytokines (Figure 3). In mice, allergen exposure is able to trigger the production of IL-1α, granulocyte-macrophage colony-stimulating factor (GM-CSF) (Willart et al., 2012), macrophage colony stimulating factor (M-CSF) (Moon et al., 2018), and transforming growth factor β (TGF-β) (Denney et al., 2015) by ECs, but the best-studied epithelial-derived cytokines in the context of asthma are IL-33, thymic stromal lymphopoietin (TSLP), and IL-25. These “alarmins” are produced by ECs in response to the allergens HDM, Aspergillus fumigatus, and the cat dander protein Feld1, mainly in a TLR4/Myd88-dependent manner (Cayrol et al., 2018; Hammad et al., 2009; Willart et al., 2012). All these epithelial-derived cytokines can contribute to the establishment of Th2 responses in murine models of eosinophilic asthma (Lambrecht et al., 2019).
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Figure 3. Early innate immune response driving asthma
Most allergens and many air pollutants trigger production by epithelial cell of cytokines and chemokines, through activation of toll-like receptors (TLRs) or protease-activated receptors (PARs). This then leads to activation of dendritic cells (DCs) that migrate to the draining nodes to promote Th2 development, and to activation of innate immune cells that get recruited to the airways, and subsequently produce many mediators that contribute to airway inflammation. Cytokines such as thymic stromal lymphopoietin (TSLP), interleukin-33 (IL-33) and granulocyte-macrophage colony stimulating factor (GM-CSF), and chemokines such as monocyte chemokine proteins (MCPs), eotaxins, and C-C chemokines (CCL17 and CCL22) are all potential drug targets for new biologicals. Some environmental exposures such as chronic exposure to farm dust can suppress the airway epithelial response to allergens because these induce expression of a negative regulator of nuclear factor kB (NF-kB) activation called TNF-a induced protein-3 (TNFAIP3). In this way, chronic farm dust exposure can suppress manifestations of asthma.
In humans, the exact contribution of these epithelial-derived alarmins to asthma is difficult to assess, but certainly they are also present in patients with different phenotypes of asthma. IL-33 and IL1RL1, encoding the IL-33 receptor ST2, are among the most highly replicated susceptibility loci for asthma (Moffatt et al., 2010). IL-33 is increased in epithelial cells and bronchoalveolar lavages from asthmatic patients and correlates with disease severity (Li et al., 2018). Patients with allergic eosinophilic asthma have increased serum IL-33 levels and higher ST2 expression on blood and suptum eosinophils compared to non-allergic and non-eosinophilic phenotypes (Gasiuniene et al., 2019).
IL-25 transcripts and proteins are elevated in type 2-high asthmatic patients, and patients with higher levels of IL-25 had greater airway hyperresponsiveness, more airway and blood eosinophils, higher serum IgE, more subepithelial thickening, and higher expression of Th2 signature genes (Cheng et al., 2014).
TSLP mRNA expression is also increased in the airway mucosa of asthmatic patients and correlates with disease severity (Ying et al., 2005). It was shown that TSLP could contribute to ILC2 longevity and be responsible for their resistance to corticoidtreatment (Kabata et al., 2013). Although classically considered as a Th2-associated cytokine, TSLP expression in bronchoalveolar lavages has been shown to also correlate with neutrophilic inflammation (Mitchell et al., 2018).
Epithelial cell-dendritic cell interactions are required in asthma
Our understanding of the mode of action of these epithelial-derived cytokines mainly comes from murine models. Indeed, the absence of epithelial-derived cytokines or cytokine signaling in these models failed to lead to Th2 immunity to allergens. It was shown that the way epithelial-derived cytokines participate in Th2 immunity in the lung comes from their ability to activate a number of immune cells involved in asthma, such as eosinophils, Th2 cells, mast cells, basophils (Chan et al., 2019), but also ILC2s and DCs (Roan et al., 2019; Figure 3).
The initiation of Th2 responses in the lungs and in other organs has been attributed to a subset of conventional DCs (conventional type 2 or cDC2s) that require IRF4 for their development (Deckers et al., 2017). Several epithelial-derived cytokines, such as IL-33, TSLP, IL-1α, or GM-CSF, have the ability to target SIRPα+CD11b+ cDC2s involved in T helper 2 cell differentiation (Moon et al., 2020; Willart et al., 2012). On the contrary, bona fide lung XCR1+CD103+ cDC1s and monocyte-derived DCs are not able to induce Th2 responses and were even shown to protect from asthma development through the production of the Th1-associated cytokine IL-12 that can inhibit Th2 responses (Conejero et al., 2017).
To be able to induce proper Th2 immunity, cDC2s that have been exposed to epithelial-derived cytokines need to migrate from the lung tissue to the draining lymph nodes, a process potentially controlled by ILC2-derived IL-13 and by type I interferon (IFN) (Halim et al., 2014; Webb et al., 2017). The question of why type I IFN, normally associated with anti-microbial responses, has evolved to Th2 immune responses remains unclear. Results from our lab have shown that, following exposure to Th2-inducing triggers such as allergen or helminths, type I IFN induces a very peculiar subset of inflammatory cDC2s that has hybrid characteristics of cDC1s (IRF8 expression) and monocyte-derived cells (CD64 expression) (Bosteels et al., 2020). The exact contribution of this subset to the initiation of Th2 immunity needs to be clarified, since experimental data clearly identify CD64– cDC2s and not inflammatory CD64+ cDC2s as the driver of Th2 cell differentiation (Bosteels et al., 2020).
DCs have an important role not only in sensitization to allergens but also in ongoing asthma both in humans and in mice. The number of DCs is increased in the airways of patients with asthma. In the lung, most conventional CD1c-expressing DCs found in type 2-high patients express the high-affinity receptor FcεRI, suggesting a role for IgE and these DCs in Th2 airway inflammation, and suggesting they are highly related to cDC2s (Dutertre et al., 2019; Naessens et al., 2020). The difficulty of obtaining lung tissue samples from patients with ongoing asthma has made functional studies on DCs difficult. However, in a recent study, sputum was used as a proxy for the lung compartment, and the Siebold group has identified inflammatory CD11b+IRF4+CD103– DCs expressing thymic stromal lymphopoietin (TSLPR) and type 2-associated chemokines in patients with severe type 2-high asthma (Peters et al., 2019). This finding therefore provides a rationale for targeting CD11b+ cDC2s in patients with type 2-high asthma. However, whether these cells require type I IFN to maintain type 2 immunity and are related to an inflammatory cDC2 subset identified in murine models remains to be addressed. In mice, early experiments have shown that the depletion of CD11c+ DCs during ongoing allergen challenge suppressed all features of asthma (van Rijt et al., 2005). Despite yielding important results, one caveat with this strategy of CD11c+ cell depletion is that it affects all subsets of lung DCs. Therefore, the exact contribution of DC subsets to inflammation in ongoing asthma still needs to be addressed using more cell-specific depletion strategies, or by purifying DC subsets to look at their gene-expression profiles. Using the latter strategy, our group has found that monocyte-derived DCs, which accumulate after an allergen challenge, were producing chemokines involved in the attraction of Th2 cells and eosinophils to the lungs but were not good at promoting Th2 cell differentiation (Bosteels et al., 2020; Plantinga et al., 2013). However, a recent paper proposed that cDC1s were the drivers of eosinophil accumulation in the lung following repeated exposure to allergens (Yi et al., 2018). How each subset exactly contributes to the maintenance of asthma features still needs to be addressed both in patients and in murine models of the disease.
Adaptive type 2 immunity in asthma
All asthma patients with airway eosinophilia, reflected by high sputum and/or blood eosinophilia, are now said to be type 2-high patients, and some endotypes of asthma even have an ultra-type 2-high immune response (Peters et al., 2019). Within the type 2-high subtype, the major pathways involved in airway inflammation (i.e., IL-4, IL-13, and IL-5 signaling) are driven by T helper type 2 (Th2) cells, which are induced or restimulated in the lung by DCs under the influence of epithelial-derived cytokines. Such effector Th2 cells have been shown to arise from IL-4-committed T follicular helper cells upon allergen challenge in mice (Ballesteros-Tato et al., 2016). Whether such a differentiation path of Th2 cells also exist in humans is not known.
Th2 cells and their cytokines control type 2-high asthma
The idea that eosinophilic asthma is a Th2-driven disease comes from several observations. First, allergen-specific Th2 cells and their associated cytokines are clearly present in the bronchoalveolar lavage of asthmatic patients and mice with allergic eosinophilic asthma (Coquet et al., 2015; Robinson et al., 1992; Tibbitt et al., 2019). Then, circulating allergen-specific Th2 cells producing IL-4, IL-5, IL-13, and even IL-9 can be found in the blood of allergic asthmatic patients (Seumois et al., 2020). In addition, early studies performed in murine models of type 2-high asthma induced by the inhaled antigen ovalbumin (OVA) have demonstrated that the depletion of CD4+ T cells prevented asthma development and that, on the opposite side, the adoptive transfer of in vitro-polarized Th2 cells from mice with transgenic expression of an OVA peptide-specific T cell antigen receptor (TCR) lead to the induction of asthma features (Cohn et al., 1997). Interestingly, Th2 cells obtained from humans with type 2-high asthma and from murine models of allergic eosinophilic asthma show similar patterns of gene expression. These cells produce high levels of Th2 cytokines, express specific receptors recognizing epithelial-derived signals, and, unexpectedly, are characterized by the expression of the nuclear receptor PPAR-γ (Tibbitt et al., 2019). PPAR-γ in Th2 cells seems to be crucial to drive the pathogenicity of these cells since it contributes to Th2 cytokine production, by upregulating the expression of the IL-33 receptor (Chen et al., 2017).
Although ovalbumin-induced models are not very physiological (sensitization is achieved by intraperitoneal administration of the protein emulsified in alum), they are characterized by a pure Th2 response (IL-4, IL-5, and IL-13 production and presence of antigen-specific IgE), and they have allowed us to obtain a detailed picture of Th2 cell effector functions. Studies in these models have led to the finding that, by acting on the IL-4Rα, IL-4 was shown to induce IgE immunoglobulin class switching by B cell, to promote bronchial hyperresponsiveness, and induce the expression of adhesion molecules such as ICAM-1 and VCAM1 that allow the extravasation of inflammatory cells such as eosinophils into inflamed lungs (Godar et al., 2018). Many of these effects are also induced by IL-13 that also uses the IL-4Rα for signaling (Godar et al., 2018). Whereas IL-13 was originally thought to not be involved in stimulating IgE class-switching, recently an IL-13-producing T follicular helper (T FH) cell subset boosting anaphylactic IgE was identified in mice and humans (Gowthaman et al., 2019). Studies in mice also showed that IL-5 drives the recruitment and the activation of eosinophils, but its effect on driving BHR has been controversial (Corry et al., 1996).
IL-9: A real player in type 2-high asthma?
IL-9 is the most enigmatic type 2 effector cytokine in asthma. It is produced by a multitude of cells including highly differentiated Th2 cells in allergic asthmatic patients (Seumois et al., 2020), Th9 cells (Veldhoen et al., 2008), and human eosinophils and neutrophils (Gounni et al., 2000; Sun et al., 2018). In humans, studies have shown increased IL-9 expression in bronchoalveolar lavages from patients with asthma (Erpenbeck et al., 2003) and by PBMCs from patients with allergic eosinophilic asthma (Seumois et al., 2020). However, the function of IL-9 in type 2-high asthma is unknown. In mice, the role of IL-9 has been studied more effectively. In murine models of type 2-high asthma, IL-9 seems to be a critical player in allergic airway inflammation since the administration of blocking antibodies in the airways or the use of IL-9-deficient mice reduced all the features of asthma, including BHR (Du et al., 2020). On the contrary, the administration of Th9 cells to mice induced asthma features (bronchial hyperresponsiveness, eosinophilia, and mucus production) similar to a transfer of Th2 cells (Staudt et al., 2010). Experiments in mice have shown a clear role for IL-9 in driving several pathological features of asthma. More evidence on its role in patients with type 2-high asthma is still needed.
Resident memory T cells: The new players in asthma
In mice and in humans, memory lymphocytes consist of circulating (central memory and effector memory) and non-circulating cells (resident memory located in peripheral tissues). Mainly CD4+ resident memory T (Trm) cells have been studied in the context of eosinophilic asthma. These cells reside in the lung very long after cessation of allergen exposure and can be found in the lungs of mice subjected to experimental asthma and in patients with type 2-high asthma (Hondowicz et al., 2016; Vieira Braga et al., 2019). Trm cells produce more Th2 cytokines that circulating Th2 cells and get reactivated rapidly in situ after re-exposure to allergens, although the mechanism of allergen-specific Trm cell reactivation in tissues is still not very clear. A possible hypothesis is that this could be mediated by CD11b+ dendritic cells (DCs) presenting the antigen locally without migration to the lymph nodes. Given the heterogeneity of lung DCs (Plantinga et al., 2013), it is not clear whether these CD11b+ DCs are conventional DCs or monocyte-derived cells.
One outstanding question is whether effector Th2 cells and CD4+ Trm cells perform overlapping functions or whether each cell population can contribute differently to asthma pathogenesis. An interesting feature of allergen-specific lung Trm cells is that, at least in murine models, they are sufficient to induce features of asthma. Indeed, experiments performed in mice showed that reducing the number of circulating memory cells while keeping the pool of lung Trm cells intact was sufficient to induce several features of the disease (Hondowicz et al., 2016; Vieira Braga et al., 2019). This was mainly explained by the difference in localization of circulating and resident-memory T cells in the lung. Circulating memory Th2 cells were found to preferentially localize in the parenchyma and were responsible for eosinophil and T cell recruitment to the lung. Trm cells were mainly found near the airways and induced mucus production, eosinophil activation, and bronchial hyperresponsiveness (Rahimi et al., 2020). These data are in line with the fact that expression of the IL-33 receptor, ST2, was found on lung Trm cells in mice (Bošnjak et al., 2019) and suggest that Trm cells could be directly affected by epithelial-derived IL-33 to perform at least some of their functions. Moreover, ST2+ memory Th2 have also been shown to persist in lymphoid structures under the influence of IL-33-producing lymphatic endothelial cell and immunofibroblasts, which could also produce a favorable niche for Trm cell survival (Shinoda et al., 2016). The field of Trm cell biology is still in its infancy and the use of specific mouse models allowing to target Trm cells would help to pinpoint their exact contribution to asthma pathogenesis. Moreover, more data on Trm cells in human asthma are needed to confirm whether or not they share, at least, some similarities with murine Trm cells regarding their localization and functions.
Contribution of B cells and IgE to asthma
After Th2 cells have been generated in lung-draining lymph nodes, part of them will interact locally with B cells, which will mature into plasma cells and antibody-producing cells. Under the influence of the Th2 cytokines IL-4 and IL-13, B cells will preferentially produce IgE. Although IgE production mainly takes place in secondary lymphoid organs, there is evidence that this event can also happen in the lung mucosa of patients with asthma (so-called local IgE production) (Manise et al., 2013).
The role of IgE is very complex and is linked to its ability to affect several immune and structural cells involved in allergic asthma. IgE can bind to the high-affinity FcεRI and the low-affinity CD23 receptors. FcεRI is expressed on basophils, mast cells, eosinophils, and DCs but also on airway smooth muscle cells, endothelial cells, and epithelial cells (Redhu and Gounni, 2013). In asthma, IgE is mainly produced from a pool of memory IgG-positive B cells that class switch to IgE and become long-lived plasma cells in response to IL-4 and/or IL-13 derived from T FH cells (Hoof et al., 2020).
Effects of IgE on mast cells and basophils
The interaction between IgE and FcεRI on mast cells and basophils is an important part of the allergic cascade. The crosslinking of two adjacent IgE molecules by the allergen activates mast cells and basophils to release biologically active preformed mediators such as histamine and neutral proteases such as tryptase and chymase. They also produce large amounts of lipid mediators (cysteinyl leukotriens [CysLT] or prostaglandin D2 [PGD2]) as well as Th2-associated cytokines (IL-4, IL-5, IL-13, IL-9), all reinforcing the pro-inflammatory Th2 environment present (Figure 4). The interaction of IgE with mast cells and basophils is responsible for the rapid phase of the allergic response, which is characterized by an increased vascular permeability and increased cellular recruitment in the lung. In addition, in the airways, mast cells localize near the submucosal mucous glands, and the release of PGD2, LTC4, IL-4, and IL-13 can trigger mucus hyperproduction by goblet cells (Carter and Bradding, 2011). Finally, the localization of mast cells in airway smooth muscles is a key feature in the pathogenesis of asthma and contributes to smooth muscle hypertrophy and hyperplasia, and to the establishment of BHR (Elieh Ali Komi and Bjermer, 2019).
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Figure 4. IgE effector functions in asthma
When allergens crosslink allergen-specific IgE on mast cells and basophils, this leads to immediate degranulation and release of many mediators such as prostaglandin D2 (PGD2) that can act on the type 1 prostaglandin D2 receptor (DP1) receptor or the CRTH2 (chemokine receptor expressed by Th2 cells, a.k.a.DP2) receptor or serotonine acting on the HT5R to cause inflammation and immediate bronchoconstriction, in the so-called early asthmatic response. Histamine will also cause bronchoconstriction and will cause plasma leakage from nearby vessels. Cytokines released with PGD2 will prepare the vessel wall for later extravasation of leukocytes, and in this way mast cells and basophils can contribute to the late-phase response as well. Crosslinkining of IgE on antigen-presenting type 2 conventional DCs (cDC2s) leads to improved antigen uptake and better presentation of Th2 cells that further contribute to the late allergic response by producing cytokines and activating the recruited cell types. When allergens crosslink IgE on plasmacytoid DCs (pDCs), however, this leads to suppressed capacity of these cells to produce type I interferons (IFN), thus contributing to the increased susceptibility of asthmatics to respiratory viral infections that cause exacerbations of disease.
IgE: More than basophil and mast cell activation
Over the years, it has become obvious that IgE mediates more than just the early phase of the allergic response. The two receptors for IgE are expressed by airway smooth muscle cells. As such, airway smooth muscle cells can also respond directly to IgE by producing cytokines and chemokines but also by proliferating and contracting, therefore contributing to airway hyperreactivity (Ferreira et al., 2018; Figure 4).
In addition, IgE can also bind to FcεRI on DCs and facilitate allergen presentation to memory Th2 lymphocytes. IgE-mediated allergen presentation actually decreases the threshold to mount allergen-specific Th2 cell responses (Khan and Grayson, 2010; Figure 4). This, in turn, leads to an increased production of allergen-specific IgE by B cells, and this vicious cycle contributes to pathogenic mechanisms in asthma. Another way IgE contributes to asthma pathogenesis through DCs is through its ability to hamper plasmacytoid DC functions when binding FcεRI. Plasmacytoid DCs are cells specialized in type I IFN production and play an important role in anti-viral responses. By binding on plasmacytoid DCs, IgE reduces intracellular type 1 IFN signaling (Schroeder et al., 2005), which could have detrimental effects upon encounter with respiratory viruses during which type I IFNs are crucial for viral clearance (Figure 4).
Eosinophils: More than a marker of type 2-high asthma
Development and recruitment of eosinophils
Eosinophils develop in the bone marrow from CD34+ stem cells. The commitment of these stem cells to the eosinophil lineage requires expression of specific transcription factors, namely, GATA-1, PU.1, and CCAAT/enhancing binding protein (a.k.a C/EBP). Eosinophil progenitors mature in the bone marrow and acquire IL-5R expression before being released into blood vessels. After entering the circulation, and in the absence of inflammation, eosinophils either stay in the bone marrow or enter various peripheral tissues where they will exert homeostatic functions including maintenance of plasma cells or metabolic homeostasis (Wu et al., 2011).
IL-5 represents the key cytokine for mediating maturation of eosinophils in the bone marrow. IL-5 is not a chemoattractant but mobilizes eosinophils from the bone marrow during allergic inflammation. IL-5 is also driving activation, proliferation, and survival of eosinophils in peripheral tissues. In general, CCR3, the receptor for eotaxin, along with expression of several adhesion molecules such as VCAM1, allow the recruitment of eosinophils from the bloodstream into inflamed tissues.
Different eosinophil populations in tissues
In humans, two distinct population of eosinophils, normodense and hypodense, have been identified based on their respective density. The hypodense population is more activated and is increased in the blood and BAL of asthmatic patients. Moreover, hypodense eosinophils are sensitive to corticosteroid treatments (Kuo et al., 1994).
In mice, similar as in humans, two subsets of eosinophils have been described based on the inflammatory status of the organ. In the absence of inflammation, IL-5-independent resident eosinophils (Siglec F int CD101 low) are present in the parenchyma. Following HDM exposure, IL-5-dependent inflammatory eosinophils (Siglec F int CD101 low) are found around the airways. Resident eosinophils are related to human normodense eosinophils, whereas inflammatory eosinophils are related to human hyposense eosinophils (Mesnil et al., 2016).
Role of eosinophils in asthma pathogenesis
It is now clear from the literature that eosinophils exert both immunomodulatory and pro-inflammatory functions. In type 2-high asthma, eosinophils are recruited from the bloodstream to the lung where they get activated by IL-5 released by Th2 cells. Activated eosinophils might exert their biological effects in the lung through a myriad of factors including cytotoxic proteins (e.g., major binding protein [MBP], eosinophil peroxidase [EPO], eosinophil cationic protein [ECP], and eosinophil-derived neurotoxin [EDN]), Th2 cytokines (IL-4, IL-5, IL-9, IL-13, and IL-25), acute proinflammatory cytokines (e.g., tumor necrosis factor [TNF]-α, IL-1β, IL-6, and IL-8), chemokines, and lipid mediators (e.g., leukotriene C4). All these factors contribute to several features of asthma including BHR and goblet cell metaplasia. Moreover, persistent airway inflammation associated with eosinophils leads to continued damage to lung structural cells induced by the release of cytotoxic granule proteins (MBP, EPO, EDN) (Cañas et al., 2018). Eosinophil-associated fibrogenic factors (such as TGF-β) will, in turn, lead to airway remodeling characterized by smooth muscle thickening, goblet cell metaplasia, and extracellular matrix protein deposition (Kanda et al., 2020). Moreover, the localization of activated eosinophils and the release of granule content near airway nerves can also change the tone of parasympathetic and sensory nerves and promote BHR (Drake et al., 2018).
Role of extracellular traps and eosinophil-derived crystals in asthma
Upon appropriate stimulation, eosinophils can form extracellular traps (EETs), which are extracellular DNA fibers (Ueki et al., 2016). Although they have an important function in immunity against extracellular pathogens, EETs and extracellular histone-rich DNA are also contributing to the pathogenesis of asthma. Indeed, peripheral EET-forming eosinophils are elevated in severe asthmatics and can activate lung epithelial cells to produce IL-33 and TSLP (Choi et al., 2020). In eosinophils, EET formation is also strongly linked with the production of Charcot-Leyden crystals (CLCs) (Persson et al., 2019). These are bipyramidal crystals made of the protein Galectin-10, one of the most abundant cytoplasmic proteins of eosinophils (Su, 2018). CLC crystals and soluble galectin-10 are found in increased amounts in asthmatic patients (Nyenhuis et al., 2019) and CLCs have been mostly described inside mucus plugs (Persson et al., 2019). It is believed that the presence of EET together with CLCs could profoundly alter the rheology of the mucus, turning it into a more elastic gel that is harder to cough up. In mouse models, CLCs were found to be immunogenic and promoted the development of Th2 immunity and were shown to represent an interesting therapeutic target since their dissolution using a specific antibody prevented Th2 immunity, IgE synthesis and mucus production in a humanized mouse model (Persson et al., 2019). In CRSwNP, CLCs can also be abundantly found. Here, crystals seem to promote neutrophil recruitment and induction of NETosis in neutrophils (Gevaert et al., 2020). The crystals therefore seem to lodge in a mucus hydrogel that is stuck in the airways, which could be a very localized trigger for creating a favorable niche for type 2 immune cells to persist for long times.
Despite the increased attention these eosinophil-derived crystals have received these last years, many questions still remain as to their precise function and evolutionary development. Do these crystals contribute to asthma or are they just a marker for the presence of eosinophils? Do these crystals elicit any type of response from immune or neighboring structural cells in the lung? Are they present in all patients with type 2-high asthma or only in more severe cases who are refractory to treatment? More research on the contribution of these CLCs to asthma is warranted to address these aspects.
ILC2s: Does their function overlap with Th2 cells in type 2-high asthma?
Characterization and activation of ILC2s
Natural ILCs mainly reside in mucosal tissues such as the gastrointestinal tract, the reproductive tract, and the respiratory tract, where they contribute to homeostasis, immunosurveillance, immunoregulation, and tissue repair (Ricardo-Gonzalez et al., 2018). In contrast to these resident ILCs, inflammatory ILCs can respond to microbes, helminths, and allergens by cytokine production within a few hours and migrate from one site of exposure to another. ILCs have a phenotype close to the one of lymphocytes but lack the expression of TCR, BCR, and any myeloid lineage markers. Different groups of ILCs (ILC1, ILC2, and ILC3) with unique phenotypes and different functions have been identified (Meininger et al., 2020). Recent studies have shown that the specific gene signature of ILCs depends on the tissue microenvironment in humans (Yudanin et al., 2019) and in mice (Zeis et al., 2020). As such, ILCs perform tissue-specific functions that depend on the signals these cells receive. ILC2s resemble Th2 cells since they also express the prototypical Th2-associated transcription factor GATA-3 and produce IL-5, IL-9, and IL-13 (Figure 5). Like effector Th2 cells, ILC2s also heavily rely on PPAR-γ regulation (Xiao et al., 2020). ILCs store fatty acid nutrients in lipid droplets and use these droplets as an emergency energy source to proliferate when stimulated by cytokines (Karagiannis et al., 2020).
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Figure 5. Central role of Th2 and innate lymphocytes in controlling key features of type 2-high asthma
Type 2-high asthma is defined by presence of airway eosinophils and by the biomarker iNOS in exhaled air. It is often accompanied by goblet cell metaplasia of the airway epithelium and high-level production of the airway mucin MUC5AC. Th2 memory lymphocytes get restimulated by dendritic cells (DCs) upon allergen recognition, and subsequently produce IL-4, IL-5, and IL-13. This scenario likely occurs in allergic asthmatics. In some patients with type 2-high asthma, innate lymphocytes type 2 (ILC2) get activated directly by epithelial cytokines such as TSLP and IL-33 to produce the cytokines IL-5 and IL-13. Th2 cells and ILC2s share many features such as expression of cytokine receptors, transcription factor GATA3, and CRTH2 (chemokine receptor expressed by Th2 cells). IL-5 drives the development and activation of airway eosinophils. These eosinophils are the source of galectin-10 which forms Charcot-Leyden proteins in the airways. Eosinophil activation also leads to their degranulation and release of toxic substances such as eosinophil peroxidase (EPO) that can crosslink airway mucus, and eosinophil cationic protein (ECP) and eosinophil-derived neurotoxin (EDN) that are very damaging to lung structural cells and also activate innate immune cells. IL-13 drives iNOS production in airway epithelial cells, goblet cell metaplasia and bronchial hyperreactivity. IL-4 promotes IgE synthesis (allergen-specific in the case of allergic asthma) and primes the vessel wall for extravasation of eosinophils through induction of vascular cell adhesion molecule (VCAM-1) and intercellular adhesion molecule (ICAM-1), by acting on IL-4R. When IL-4R is blocked, this can explain increased blood eosinophilia, with reduced tissue eosinophilia.
ILC2s mainly respond to the epithelial-derived cytokines IL-33, IL-25, and TSLP, which induce not only their proliferation but also activation (Guo et al., 2015; Huang et al., 2015). ILC2s in the lung can also get activated by the neuropeptideneuromedin U (NMU) in the presence of high levels of epithelial-derived cytokines, showing that neuronal activation following allergen exposure can enhance the effect of epithelial-derived cytokines on ILC2s (Wallrapp et al., 2017).
Role of ILC2s in asthma
In humans, ILC2s were found to be an important player in asthma. The number of ILC2s is increased in the blood and in the bronchoalveolar lavage of patients with asthma (Winkler et al., 2019). In human asthmatics, allergen challenge also led to expression of TL1A, a known ligand for DR3 expressed by ILC2s. Allergen-induced TL1A thus might contribute to the expansion and activation of human ILC2s (Machida et al., 2020). The ex vivo co-culture of ILC2s with human bronchial epithelial cells showed that IL-13 derived from ILC2s could disturb lung epithelial cell barrier function (Sugita et al., 2018). ILC2s also seem to be more active in females and in patients with CRSwNP, potentially explaining the gender bias in non-allergic type 2-high asthmatics (Wang et al., 2020).
The mechanism by which ILC2s contribute to type 2 responses in the lung have started to be slowly unraveled with the use of murine models (Figure 5). However, the results obtained in these models are difficult to translate to patients with asthma because immunologists have mainly used models of allergic eosinophilic asthma, which do not reflect the human non-allergic eosinophilic asthma. In mice, the accumulation of ILC2s in allergic lung can be explained by the fact that HDM exposure is sufficient to induce the production of IL-33 by lung epithelial cells, a process that is greatly facilitated in early life (de Kleer et al., 2016a; Hammad et al., 2009). Under the influence of IL-33, ILC2s produce IL-13 and IL-5, thus promoting key asthma features such as tissue eosinophilia and BHR (Klein Wolterink et al., 2012; Figure 5). However, ILC2s can also act very early on in the process of Th2 sensitization, by promoting DC migration to the draining nodes via release of IL-13 (Halim et al., 2014), thus facilitating induction of Th2 immunity to allergens. Moreover, in parasitic infections, ILC2s have been shown to have antigen-presenting capacities through their expression of major histocompatibility complex class II (MHCII) and costimulatory molecules (Angkasekwinai et al., 2017). Whether they perform similar functions following allergen exposure in the lung remains to be addressed.
The main issue that remains is whether ILC2 and Th2 cells overlap in their functions in asthma, given that they produce similar cytokine patterns. One way to address this question would be to have models where ILC2s can be specifically deleted. Moreover, it would be interesting to assess their real contribution to disease features in allergic versus non allergic asthma. Although murine models are available for the former, they are still missing for the latter.
Non- (or low-) type 2 asthma: All down to neutrophils?
Non- or low-type 2 phenotypes of asthma are defined by the absence of Th2 cytokine signatures (such as blood and sputum eosinophilia, increased FeNO) rather than by the existence of alternative immune signatures. Non-type 2 asthma has been associated with later age at onset of the disease, use of high dose of corticosteroids, and obesity (Tliba and Panettieri, 2019; Figure 2). What these patients have in common is that they respond poorly to corticosteroids. Despite this, these patients are often treated with these drugs for long periods of time, which contributes to the worsening of their obesity. Obese patients with a type 2-low signature are often female gender with late-onset asthma. In these patients, studies have identified increases in the IL-6 pathway in epithelial brushings and sputum cells (Li et al., 2020).
The most consistent pathways identified in type 2-low asthma patients are those related to the inflammasome, especially those related to the IL-1β pathway. By discarding patients with sputum eosinophils, the U-BIOPRED consortium has identified a cluster of patients with inflammasome expression, neutrophilic inflammation, and high corticosteroids usage (Rossios et al., 2018). Interestingly, a study from the Severe Asthma Research Program has identified patients with similar characteristics, but, in addition, these patients had extracellular DNA in their sputum, which was identified as neutrophil-derived extracellular traps (Lachowicz-Scroggins et al., 2019). This group of patients also had sputum neutrophils, caspase-1 activation, and high IL-1β levels, consistent with the activation of the inflammasome pathway (Figure 6). Neutrophilic asthma has been proposed as a phenotype of asthma. However, clinical trials, performed in patients with neutrophilic asthma using CXCR2 antagonists to block the recruitment of neutrophils by CXCL8, failed to change any of the asthma outcomes (O’Byrne et al., 2016). Therefore, whether neutrophils have a function in asthma or are simply bystander cells is still an open question (Figure 6). A recent study using children and adolescents with neutrophilic asthma has shown that children with high neutrophil counts had increased concentrations of granule proteins (myeloperoxidase and elastase) suggesting that these cells might contribute to lung tissue damage. In addition, exposure of primary neutrophils to BAL fluids from children with high neutrophil counts increased their phagocytic capacities and formation of NETs, which are cytotoxic for epithelial cells (Grunwell et al., 2019). These data indicate that neutrophils can be real effector cells in severe non-type 2 asthma and that the environment in the lung of these patients drive neutrophil pathogenic functions.
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Figure 6. Proposed role of neutrophils in non-type 2 asthma
The airway epithelium and alveolar macrophages can get activated by environmental triggers such as microbes or pollutants. In response to these stimuli, they produce pro-inflammatory cytokines (IL-1β, IL-6) and the epithelium also produces CXCL8 (also known as IL-8), a potent neutrophil attractant. Under the influence of the cytokine environment, Th1 and Th17 cells are induced and further contribute to neutrophil recruitment and activation. Activated neutrophils release factors such as neutrophil elastase, myeloperoxidase, or ROS that will induce epithelial cell damage and contribute to increased mucus production.
In patients with moderate to severe forms of asthma, the levels of IL-17A, IL-17F, and IL-22 are increased in their bronchoalveolar lavage and correlates with disease severity (Al-Ramli et al., 2009). The presence of neutrophils in some asthma patients has generated interest in the cytokine IL-17. However, even using murine models of experimental asthma, its role is still unclear mainly because animal models have yielded confusing results showing that IL-17 can sometimes protect and sometimes favor asthma depending on the timing of IL-17 blockade. In these studies, IL-17 was protective only during the challenge phase (Schnyder-Candrian et al., 2006). Despite the promising results of IL-17 blockade in animal models, the main issue here is that the models used did not reflect human type 2-low asthma. It has been very difficult for immunologists to model neutrophilic inflammation associated with asthma features in mice. To do this, groups had to resort to the use of adjuvants (e.g., high dose of LPS) endowed with pro-Th1 and -Th17 properties. Using one of such adjuvants called c-di-GMP, researchers have managed to develop a model of asthma with high numbers of lung neutrophils, low Th2 cytokines, high IFN-γ and IL-17 levels, and presence of BHR. When analyzing the contribution of cytokines to asthma features, they found out that IFN-γ and not IL-17 could drive asthma features including BHR (Raundhal et al., 2015). Whether such mouse model could represent human type 2-low asthma is not certain, but sheds light on the fact that other cytokines than IL-17 should be considered to explain neutrophilic asthma. In view of these findings in mice, it is not surprising that clinical trials aiming at blocking the IL-17 pathway in patients failed to show efficacy (Busse et al., 2013). One caveat in the latter study is that the number of neutrophils was not an inclusion criterion, and it is likely that the patients selected were heterogeneous regarding airway neutrophil numbers. Moreover, the IL-17 pathway expression has been identified in some studies but is rather inconsistent, and one study has even shown the presence of tissue eosinophils within a group of asthma patients with a Th17-high signature, showing the possible coexistence of type 2-high and Th17-high signatures in a subset of patients (Choy et al., 2015). Similar findings were recently also reported in mouse models of asthma (Tibbitt et al., 2019; Tortola et al., 2020). To make things even more complex, neutrophils have been shown to influence type 2-high asthma. Indeed, studies have reported that neutrophil extracellular traps induced by rhinovirus were involved in type 2-high asthma exacerbations (Toussaint et al., 2017). A few years later, the same group showed that neutrophil extracellular traps induced by low-dose microbial exposure facilitated the development of type 2-high asthma in mice (Radermecker et al., 2019). These data show that the relationship between Th17-associated neutrophilia and type 2-high asthma is very complex. Therefore, it has been proposed that targeting both Th17 and Th2 pathways might be beneficial in some patients. The dual blockade of IL-13 and IL-17 already showed promising results in mice (Choy et al., 2015) but still needs to be studied in humans, especially in view of the negative trials using IL-17 blockade only.
Virus-induced asthma exacerbations
Asthma exacerbations can be induced by different stimuli, including allergens, pollution, cold air, and microbes. Such agents induced an enhanced inflammation in the lung that worsens the symptoms of the existing disease. Among these agents, respiratory viruses, especially respiratory syncytial virus (RSV) and rhinovirus (RV), are the major drivers of asthma exacerbations in children and adults, respectively (Jartti and Gern, 2017). Respiratory viruses primarily infect lung epithelial cells. In response to viral replication, healthy epithelial cells produce antiviral factors that ultimately lead to viral clearance. In the chronically inflamed airways of asthmatic patients, the antiviral response of epithelial cells might be altered, causing sustained inflammation and exacerbation of symptoms (Busse et al., 2010).
Role of type I interferons
Once the link between viral infections and asthma exacerbation was established, many groups have attempted to identify mechanisms of the underlying virus-induced asthma exacerbation. Following infection, the host cells generally induce an inflammatory response as a way to counteract the virus. Generally, in response to viruses, infected cells produce type I/III IFN. Interestingly, exacerbations are mainly induced in patients with high body mass index, with high eosinophil numbers (type 2-high patients), and in patients with decreased production of type I interferons (especially IFN-β) and type III interferons (IFN-λ) (Denlinger et al., 2017; Wark et al., 2005; Zhu et al., 2019). A recent single-cell RNA sequencing analysis, performed on upper-airway cells of patients with viral infection progressing to asthma exacerbation, showed that higher expression of Th2- and lower expression of type I IFN-related genes were associated with a shorter time to exacerbation (Altman et al., 2019). It is also believed that the levels of IgE may render patients more vulnerable to infections (Denlinger et al., 2017). In this case, the binding of IgE to FcεRI on plasmacytoid DCs was shown to block their capacity to produce type I interferons (Schroeder et al., 2005), allowing the virus to spread easier, and induce more damage and inflammation, leading to asthma exacerbation. In addition, murine models of infection with PVM (pneumovirus of mice), the counterpart of RSV, have highlighted that a defect in type I interferon-producing plasmacytoid DCs predisposes to viral bronchiolitis and asthma. However, the authors have attributed this to TLR7 and not to IgE/FcεRI (Kaiko et al., 2013). Therefore, whether other pathways than IgE contribute to the link between viral infections and asthma in humans remains to be fully addressed.
Epithelial-derived cytokines contribute to virus-induced asthma exacerbations
Because asthma exacerbations are more frequent in patients with blood eosinophils and higher IgE levels, suggestive of the type 2-high endotype, researchers have begun looking at different Th2-associated cell populations and pro-Th2 cytokines to understand the possible reasons underlying virus-induced exacerbations. Recent studies have shown that, during virus-induced asthma exacerbation in mice, IL-33 was produced in high amounts by lung epithelial cells, could suppress antiviral responses by dampening type I IFN production, and could favor asthma exacerbations (Lynch et al., 2016). Single-cell RNA sequencing (RNA-seq) analysis, performed in patients with virus-induced asthma exacerbation, identified a gene core associated with IL-33 and epithelial cell repair. This module also contained CDHR3, a susceptibility locus in childhood asthma encoding for a protein involved in the entrance of Rhinovirus-C in epithelial cells, suggesting that IL-33 in the context of viral infection may leave the epithelium more susceptible to repeated infections (Altman et al., 2019). One caveat in this study is that upper-airway cells were used as a proxy for lower-airway cells. Although some studies have shown that nasal cells are transcriptionally similar to lung cells (Poole et al., 2014), other studies have observed differences in immune responses, including interferon responses (Mihaylova et al., 2018). Future studies using lung cells are needed to confirm the involvement of the identified pathways.
A role for innate type 2 cells in virus-induced asthma exacerbations?
Because exposure to respiratory viruses increases the production of IL-33 and TSLP by lung epithelial cells, it is not really surprising to observe that infants hospitalized with severe respiratory virus infection had increased numbers of ILC2s in their nasal aspirates compared to children with mild infection (Vu et al., 2019). Similarly, in mice, influenza infection increased ILC2 numbers in the lung (Li et al., 2019). However, during influenza-induced asthma exacerbation, lung ILC2-derived Th2 cytokines were found to be initially lower than those derived from Th2 cells, but ILC2-derived Th2 cytokines peaked at the time of virus clearance (Li et al., 2019). Although influenza virus is not frequently involved in asthma exacerbations, these data nevertheless suggest that, during virus-induced asthma exacerbation, there is probably a division of labor between Th2 cells that contribute early on to increased inflammation and ILC2s that contribute at a later time point to the repair of the lung. In line with this, once the virus was cleared from the airways, ILC2s expressed Amphiregulin, a protein involved in tissue repair (Li et al., 2019). Whether the observations made with influenza also apply for other viruses more relevant to asthma exacerbations remains to be fully addressed.
Viral infections are also always accompanied by high levels of type I (IFN-α/β) and type II (IFN-γ), which are known to inhibit ILC2 functions (Duerr et al., 2016; Moro et al., 2016) and to affect the biology of DCs, which have important functions in Th2 responses to allergens (Webb et al., 2017) as well as in antiviral responses (Bosteels et al., 2020). During viral infections, type I IFN induces a specific population of SIRPα+IFNAR+ conventional DC2 with high levels of Fc receptors and with strong capacities to activate CD4+ and CD8+ T cell-specific antiviral responses (Bosteels et al., 2020). Such DC population also appears in the lung upon exposure to allergens. The link between type I IFN, IFNAR+ cDC2, and virus-induced asthma exacerbation is very complex and still unclear. It is, however, tempting to speculate that, in a Th2-prone environment as in type 2-high asthma, the low levels of type I IFN present in the lung have an impact on these IFNAR+ cDC2s, leading to less virus-induced migration to the draining lymph nodes, and as a consequence to less virus-specific CD4 and CD8 T cell responses.
With less potent antiviral responses, increased levels of IL-33 induced by respiratory viruses would pave the way for a stronger activation of Th2 cells and ILC2s, both contributing to enhanced asthma features, including bronchial hyperresponsiveness and eosinophilia. However, to make things even more complex, eosinophils themselves have been suggested to have antiviral capacities. Indeed, extracellular eosinophil-derived neurotoxin (EDN) displays RNase activity and can enter viral capsids to degrade RNA from RSV (Samarasinghe et al., 2017). Moreover, mice overexpressing both Eotaxin2 and IL-5 were protected against a lethal pneumovirus infection (Percopo et al., 2014). On the other hand, the depletion of eosinophils with anti-IL-5 antibodies increased the viral load in HDM-sensitized mice (Ravanetti et al., 2017). Moreover, eosinophils from mice and healthy donors are able to capture several respiratory viruses and reduce their infectivity. However, this is not the case for eosinophils obtained from patients with asthma (Sabogal Piñeros et al., 2019). In patients with asthma, the inability of eosinophils to control respiratory viruses may lead to increased viral loads, and, with less type I IFN being produced and less DC-mediated antiviral responses, an ideal scenario would be set to allow frequent virus-induced asthma exacerbations.
Biologicals in the treatment of asthma
The acknowledgment that asthma is a heterogeneous disease has prompted pharmaceutical companies to develop new drugs to specifically target effector cells, cytokines, or molecules involved in different types of asthma.
Inhibition of eosinophils and IL-5
It is now clear that eosinophils represent a proinflammatory granulocyte that plays a major role in type 2-high asthma. Eosinophilic inflammation in the airways is associated with asthma disease severity (Figure 1), and there is a direct correlation between tissue and blood eosinophil count and the frequency of asthma exacerbations, and the occurrence of irreversible airway obstruction (Graff et al., 2020).
The differentiation, activation, and survival of eosinophils is driven by IL-5, which is secreted by Th2 cells, mast cells, and innate lymphoid cells. Therefore, inhibiting IL-5 or IL-5 signaling seemed like a good option in asthma. The first biologicals targeting the IL-5 pathway was Mepolizumab, an IgG1 antibody directed against IL-5. The first trials using all-comers have proved disappointing (Leckie et al., 2000), but later on trials using biomarker-identified patients showed efficacy of Mepolizumab in reducing corticosteroid use and even improving lung function (Nair et al., 2009). The success of Mepolizumab was also seen with Reslizumab, a similar anti-IL-5 IgG4 antibody (Castro et al., 2015). The third biologics targeting the IL-5 pathway to show clinical efficacy was Benralizumab, a cytotoxic antibody that kills cells expressing the IL-5 receptor alpha chain such as eosinophils and basophils. The depletion of eosinophils with Benralizumab in severe forms of asthma lead to reduced exacerbation rate and reduced need for oral corticosteroids, strengthening the importance of eosinophils in asthma pathobiology (Busse et al., 2019).
Inhibition of IL-4RA
IL-4RA binds both IL-4 and IL-13. Given the importance of both cytokines in driving IgE production by B cells, BHR, goblet cell metaplasia, mucus production, basement membrane thickening, and fibrosis, IL-4RA was considered as an attractive target in asthma.
The use of Dupilumab, a humanized monoclonal antibody that blocks IL-4Rα, was first used in patients with eosinophilic asthma. Patients using Dupilumab showed improved lung function and asthma symptoms (Wenzel et al., 2013). Dupilumab was also found to be efficacious in patients with non-eosinophilic asthma (Wenzel et al., 2016), but the effects were larger for patients with eosinophilic asthma with regards to improved lung function and reduced exacerbation frequency. In some 15% of patients, however, this treatment is associated with a temporary rise in blood eosinophils, most likely because of reduced extravasation from the bloodstream (Castro et al., 2018). Given that it also reduced severity and recurrence of CRSwNP, and features of atopic dermatitis, Dupilumab could be a good choice in asthmatics with comorbid allergic disease (Agache et al., 2020).
Inhibiting IL-9
IL-9 is believed to play a role in asthma although the extent of its involvement in asthma features still remains to be fully addressed, especially in humans. A clinical trial using a humanized anti-IL-9 monoclonal antibody in patients with moderate to severe asthma failed to show efficacy (Oh et al., 2013). It was suggested that the phenotypic heterogeneity of the patients used in this study was a reason for failure. Future studies to identify correlations between Th9/IL-9 and asthma phenotypes should provide a better understanding of the role of IL-9 in asthma and allow identification of subgroups of patients that would benefit from blocking IL-9 therapies, certainly given the recent observations that one of the major discriminators between HDM allergic donors without asthma and HDM allergic asthmatics seems to be T cell production of IL-9 in the latter (Seumois et al., 2020).
Inhibiting IgE
Although the biologics directed against the Th2 cytokines IL-4, IL-13, and IL-5 show exciting results, the first-ever approved biologicals in asthma was in fact directed against IgE. Considering that IgE is involved in the onset of the allergic asthma as well as in the chronic phase of the disease, it is not surprising that omalizumab (a monoclonal antibody that reduces the binding of IgE to its high-affinity receptor leading to the downregulation of FcεRI expression and to an impaired function of several FcεRI+ cells; MacGlashan et al., 1997) demonstrated efficacy in clinical trials by reducing exacerbation rates, reducing the doses of corticosteroids needed to control the disease, and by improving lung function (Casale et al., 2019; Holgate et al., 2004; Humbert et al., 2005). Interestingly, the levels of IgE in patients seem to be a poor predictor of response to omalizumab (Bousquet et al., 2004). One retrospective study showed that patients with high eosinophil counts had the greatest benefit (Casale et al., 2019), although this has been refuted (Humbert et al., 2018). It also appears that omalizumab improves lung function in adult and pediatric patients with eosinophilic nonallergic asthma (Bourgoin-Heck et al., 2018; Pillai et al., 2016). This is not surprising since, in this phenotype of asthma, patients show a Th2 cell signature with tissue eosinophilia and can have high total serum IgE (Beeh et al., 2000). Therefore, blocking IgE represents a safe and effective treatment of type 2-high but not type 2-low inflammation (Hanania et al., 2013). This being said, anti-IgE treatment is not the panacea for all type 2-high asthma patients. Indeed, only about 70% of children and adults treated with omalizumab show good to excellent responses to the treatment (Humbert et al., 2018). Deciding which patients should be best treated with which therapy remains a challenging task for physicians.
Inhibiting epithelial-derived cytokines
Given the inability of available anti-type 2 biologicals to prevent all asthma exacerbations and given the complexity in mechanisms involved in asthma, there was a need to investigate other means of curing asthma. Because of their central and very upstream role, epithelial-derived alarmins (IL-1, IL-33, IL-25, and TSLP) have been considered as interesting targets. Biologicals targeting TSLP and IL-33 are the most advanced and have entered clinical trials, although results of the clinical trials have only been published for TSLP. Topline results from the phase 3 Navigator clinical trial using the blocking TSLP antibody Tezepelumab were reported. Very strikingly, the drug improved exacerbation frequency not only in type 2-high patients but also in those with type 2-low disease, and low blood eosinophil counts, making it the first biological on the horizon that may benefit a broad category of asthma patients (Corren et al., 2017).
Anti-type 2 biologics are not a panacea for all patients with type 2-high asthma
Although the use of the new anti-type 2 biologics has changed the life of severe asthmatic patients, clinical trials have shown that not all patients with type 2-high asthma respond well to anti-type 2 biologics. This means that, even if eosinophil count in the blood is sometimes a good biomarker, more is needed to ensure that patients respond to one or another biologics. Other type 2-high biomarkers such as FeNO or periostin can be used together with eosinophil counts to identify asthma subgroups. However, it is very likely that the further identification of novel biomarkers might be necessary to further improve the response to specific biologics. The more extensive use of omics techniques might help identify such new biomarkers in patients who failed to respond to certain agents.
Another point is that, even when considering eosinophilic asthma, some patients will present long-lasting, difficult-to-cough-up, mucus impactions in the airways, caused by eosinophil activation (Dunican et al., 2018). Such more-severe patients also show the presence of eosinophil-derived Charcot-Leyden crystals in these mucus plugs (Persson et al., 2019), which would contribute to fixed-airway obstruction. It remains to be seen whether such patients with such crystals in elastic mucus would benefit from an eosinophil-targeted therapy and show improved lung function or whether an add-on therapy that would be more directed to the crystals would be beneficial.
Finally, the fact that not all type 2-high patients respond to anti-type 2 biologics raises the possibility that non-type 2 pathways might also be at play in such refractory patients. Indeed, inducible nitric oxide synthase (iNOS) is induced not only by Th2 cytokines but also by the Th1 cytokine IFN-γ (Guo et al., 1997), suggesting that elevated FeNO might not solely be explained by type 2 inflammation. In addition, one study has found tissue eosinophils in asthma patients with a Th17-high signature, showing the possible coexistence of type 2-high and Th17-high signatures in a subset of patients (Choy et al., 2015). Therefore, strategies targeting multiple pathways might represent an alternative way to treat subsets of asthmatic patients.
Conclusions
Clinicians now realize that asthma is a very heterogeneous disease with different endotypes. This is best reflected by the results of the different clinical trials using different biologics in patients with asthma. Asthma is the consequence of a complex interaction between structural cells, such as epithelial cells, and immune cells occurring in the context of exposure to specific environmental triggers in very specific periods of life. Despite the fact that we gradually understand more of the immunology of type 2-high asthma, more investigative work is clearly needed for type 2-low asthma before we can rationally design therapies based on a well-understood endotype.
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2023, Lancet Show abstract Asthma is one of the most common chronic non-communicable diseases worldwide and is characterised by variable airflow obstruction, causing dyspnoea and wheezing. Highly effective therapies are available; asthma morbidity and mortality have vastly improved in the past 15 years, and most patients can attain good asthma control. However, undertreatment is still common, and improving patient and health-care provider understanding of when and how to adjust treatment is crucial. Asthma management consists of a cycle of assessment of asthma control and risk factors and adjustment of medications accordingly. With the introduction of biological therapies, management of severe asthma has entered the precision medicine era—a shift that is driving clinical ambitions towards disease remission. Patients with severe asthma often have co-existing conditions contributing to their symptoms, mandating a multidimensional management approach. In this Seminar, we provide a clinically focused overview of asthma; epidemiology, pathophysiology, diagnosis, and management in children and adults.
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2022, Molecular Aspects of Medicine Show abstract Asthma is a chronic disease of the airways, which affects more than 350 million people worldwide. It is the most common chronic disease in children, affecting at least 30 million children and young adults in Europe. Asthma is a complex, partially heritable disease with a marked heterogeneity. Its development is influenced both by genetic and environmental factors. The most common, as well as the most well characterized subtype of asthma is allergic eosinophilic asthma, which is characterized by a type 2 airway inflammation. The prevalence of asthma has substantially increased in industrialized countries during the last 60 years. The mechanisms underpinning this phenomenon are incompletely understood, however increased exposure to various environmental pollutants probably plays a role. Disease inception is thought to be enabled by a disadvantageous shift in the balance between protective and harmful lifestyle and environmental factors, including exposure to protective commensal microbes versus infection with pathogens, collectively leading to airway epithelial cell damage and disrupted barrier integrity. Epithelial cell-derived cytokines are one of the main drivers of the type 2 immune response against innocuous allergens, ultimately leading to infiltration of lung tissue with type 2 T helper (T H 2) cells, type 2 innate lymphoid cells (ILC2s), M2 macrophages and eosinophils. This review outlines the mechanisms responsible for the orchestration of type 2 inflammation and summarizes the novel findings, including but not limited to dysregulated epithelial barrier integrity, alarmin release and innate lymphoid cell stimulation.
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9530 | https://math.stackexchange.com/questions/449041/counting-the-number-of-elements-in-a-conjugacy-class-of-s-n | Skip to main content
counting the number of elements in a conjugacy class of Sn
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I want to know if there is some systematic way (using some combinatorial argument)
to find the number of elements of conjugacy classes of Sn for some given n.
For example, let's consider S5. If the representative for the conjugacy class
is an m-cycle then Dummit and Foote gives a formula on how to compute the number of elements in the conjugacy class. This is not a problem. But what about when the representative is not an m-cycle. As an example we can consider the conjugacy class that gives rise by the partition 2+3 of 5. A representative for the conjugacy class would be (12)(345). How can I find the number of such elements?.
Question?:
Does (52)⋅(33)⋅2 give me what I want?
Reasoning: For the first parenthesis I need to choose 2 elements out of 5 and for the second set of parenthesis I need to choose 3 out of the remaining 3 (noting that they can't be repeats). Finally we can permute these two parenthesis in two ways, thus giving me the above number.
Is this reasoning correct?. If not how does one find the number of elements of such conjugacy classes.
As always, any help is greatly appreciated.
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asked Jul 21, 2013 at 21:40
CousinCousin
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This is exercise 4.3.33 of D&F Abstract Algebra.
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More systematically, you have n! choices to arrange 1,…,n. Place them into the parentheses pattern in this order to obtain an element of the conjugacy class.
For each r-cycle, you divide by r as only the cyclic order within a cycle plays a role, not which element we start with. Then, if there are nr cycles of length r, divide by nr! as the order in which the cycles are listed is not important. Note that this must also be done for the cycles of length 1!
This gives us
n!∏rrnrnr!
Thus in S5, there are 5!2⋅3 conjugates of (12)(345). Similarly, there are 7!2⋅2⋅2!⋅3! conjugates of (12)(34) in S7.
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answered Jul 21, 2013 at 21:49
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Nice explanation. Thanks. This helps a lot :-)
– Cousin
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Jul 22, 2013 at 3:00
@Hagen, where does the 3! come in your second example? Thanks
– messi
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Jul 22, 2013 at 14:47
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@messi: The explicit cycle decomposition of (12)(34) in S7 is (12)(34)(5)(6)(7), that is, it has 3 one-cycles.
– Cihan
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Jul 24, 2013 at 17:12
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Cycles of length 1! or cycles of length 1? (Yes, I know it's the same thing ;-)
– Marc van Leeuwen
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9531 | https://en.wikipedia.org/wiki/Parity_problem_(sieve_theory) | Jump to content
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1 Statement
2 Example
3 Parity-sensitive sieves
4 Karatsuba phenomenon
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Parity problem (sieve theory)
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From Wikipedia, the free encyclopedia
In number theory, a limitation of sieve theory
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In number theory, the parity problem refers to a limitation in sieve theory that prevents sieves from giving good estimates in many kinds of prime-counting problems. The problem was identified and named by Atle Selberg in 1949. Beginning around 1996, John Friedlander and Henryk Iwaniec developed some parity-sensitive sieves that make the parity problem less of an obstacle.
Statement
[edit]
Terence Tao gave this "rough" statement of the problem:
Parity problem. If A is a set whose elements are all products of an odd number of primes (or are all products of an even number of primes), then (without injecting additional ingredients), sieve theory is unable to provide non-trivial lower bounds on the size of A. Also, any upper bounds must be off from the truth by a factor of 2 or more.
This problem is significant because it may explain why it is difficult for sieves to "detect primes," in other words to give a non-trivial lower bound for the number of primes with some property. For example, in a sense Chen's theorem is very close to a solution of the twin prime conjecture, since it says that there are infinitely many primes p such that p + 2 is either prime or the product of two primes (semiprime). The parity problem suggests that, because the case of interest has an odd number of prime factors (namely 1), it won't be possible to separate out the two cases using sieves.
Example
[edit]
This example is due to Selberg and is given as an exercise with hints by Cojocaru & Murty.: 133–134
The problem is to estimate separately the number of numbers ≤ x with no prime divisors ≤ x1/2, that have an even (or an odd) number of prime factors. It can be shown that, no matter what the choice of weights in a Brun- or Selberg-type sieve, the upper bound obtained will be at least (2 + o(1)) x / ln x for both problems. But in fact the set with an even number of factors is empty and so has size 0. The set with an odd number of factors is just the primes between x1/2 and x, so by the prime number theorem its size is (1 + o(1)) x / ln x. Thus these sieve methods are unable to give a useful upper bound for the first set, and overestimate the upper bound on the second set by a factor of 2.
Parity-sensitive sieves
[edit]
Beginning around 1996 John Friedlander and Henryk Iwaniec developed some new sieve techniques to "break" the parity problem. One of the triumphs of these new methods is the Friedlander–Iwaniec theorem, which states that there are infinitely many primes of the form a2 + b4.
Glyn Harman relates the parity problem to the distinction between Type I and Type II information in a sieve.
Karatsuba phenomenon
[edit]
In 2007 Anatolii Alexeevitch Karatsuba discovered an imbalance between the numbers in an arithmetic progression with given parities of the number of prime factors. His papers were published after his death.
Let be a set of natural numbers (positive integers) that is, the numbers . The set of primes, that is, such integers , , that have just two distinct divisors (namely, and ), is denoted by , . Every natural number , , can be represented as a product of primes (not necessarily distinct), that is where , and such representation is unique up to the order of factors.
If we form two sets, the first consisting of positive integers having even number of prime factors, the second consisting of positive integers having an odd number of prime factors, in their canonical representation, then the two sets are approximately the same size.
If, however, we limit our two sets to those positive integers whose canonical representation contains no primes in arithmetic progression, for example , or the progression , , , , then of these positive integers, those with an even number of prime factors will tend to be fewer than those with odd number of prime factors. Karatsuba discovered this property. He found also a formula for this phenomenon, a formula for the difference in cardinalities of sets of natural numbers with odd and even amount of prime factors, when these factors are complied with certain restrictions. In all cases, since the sets involved are infinite, by "larger" and "smaller" we mean the limit of the ratio of the sets as an upper bound on the primes goes to infinity. In the case of primes containing an arithmetic progression, Karatsuba proved that this limit is infinite.
We restate the Karatsuba phenomenon using mathematical terminology.
Let and be subsets of , such that , if contains an even number of prime factors, and , if contains an odd number of prime factors. Intuitively, the sizes of the two sets and are approximately the same. More precisely, for all , we define and , where is the cardinality of the set of all numbers from such that , and is the cardinality of the set of all numbers from such that . The asymptotic behavior of and was derived by E. Landau:
This shows that
that is and are asymptotically equal.
Further,
so that the difference between the cardinalities of the two sets is small.
On the other hand, if we let be a natural number, and be a sequence of natural numbers, , such that ; ; every are different modulo ; Let be a set of primes belonging to the progressions ; . ( is the set of all primes not dividing ).
We denote as a set of natural numbers, which do not contain prime factors from , and as a subset of numbers from with even number of prime factors, as a subset of numbers from with odd number of prime factors. We define the functions
Karatsuba proved that for , the asymptotic formula
is valid, where is a positive constant.
He also showed that it is possible to prove the analogous theorems for other sets of natural numbers, for example, for numbers which are representable in the form of the sum of two squares, and that sets of natural numbers, all factors of which do belong to , will display analogous asymptotic behavior.
The Karatsuba theorem was generalized for the case when is a certain unlimited set of primes.
The Karatsuba phenomenon is illustrated by the following example. We consider the natural numbers whose canonical representation does not include primes belonging to the progression , . Then this phenomenon is expressed by the formula:
Notes
[edit]
^ Tao, Terence (2007-06-05). "Open question: The parity problem in sieve theory". Retrieved 2008-08-11.
^ Cojocaru, Alina Carmen; M. Ram Murty (2005). An introduction to sieve methods and their applications. London Mathematical Society Student Texts. Vol. 66. Cambridge University Press. ISBN 0-521-61275-6.
^ Friedlander, John; Henryk Iwaniec (1997-02-18). "Using a parity-sensitive sieve to count prime values of a polynomial". Proceedings of the National Academy of Sciences. 94 (4): 1054–1058. Bibcode:1997PNAS...94.1054F. doi:10.1073/pnas.94.4.1054. PMC 19742. PMID 11038598. 1054–1058.
^ Friedlander, John; Henryk Iwaniec (1998). "Asymptotic sieve for primes". Annals of Mathematics. 148 (3): 1041–1065. arXiv:math/9811186. Bibcode:1998math.....11186F. doi:10.2307/121035. JSTOR 121035. S2CID 11574656.
^ Harman, Glyn (2007). Prime-detecting sieves. London Mathematical Society Monographs. Vol. 33. Princeton University Press. pp. 45, 335. ISBN 978-0-691-12437-7. Zbl 1220.11118.
^ Karatsuba, A. A. (2011). "A property of the set of prime numbers". Russian Mathematical Surveys. 66 (2): 209–220. Bibcode:2011RuMaS..66..209K. doi:10.1070/RM2011v066n02ABEH004739.
^ Karatsuba, A. A. (2011). "A property of the Set of Primes as a Multiplicative Basis of Natural Numbers". Doklady Mathematics (84:1): 1–4.
^ Landau, E. (1912). "Über die Anzahl der Gitter punkte in gewissen Bereichen". Gött. Nachricht.: 687–771.
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What are the three stop codons?
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A codon consists of a sequence of three consecutive nucleotide bases that together form a unit of genetic code in a DNA or RNA molecule. Codons are named according to the order of their bases.
The three stop codons in mRNA are UAA, UAG, and UGA, where U stands for uracil, A for adenine, and G for guanine. Stop codons provide an end point for protein synthesis. They do not encode any amino acid. The ribosome pauses and falls off the mRNA. Stop codons are also called nonsense or termination codons.
There is only one start codon known as AUG. This start codon marks the beginning of a protein. The stretch of codons between the start codon and a stop codon is called an open reading frame.
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How can I prove that a polynomial with degree $n$ is continuous everywhere in $\mathbb{R}$ using definitions?
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How can I prove that a polynomial with degree $n$ is continuous everywhere in $\mathbb{R}$ using definitions?
With induction. I can show that this polynomial is continuous at $x_0$ but I do not know how to prove that it is continuous everywhere in $\mathbb{R}$. I know that a function is continuous on an interval if it is continuous at every point on the interval but this implies that I need to show for every $x_0\in\mathbb{R}$, the polynomial is continuous.
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edited Aug 26, 2017 at 20:27
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asked Mar 30, 2014 at 4:51
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$\begingroup$ Related: math.stackexchange.com/questions/314552/… $\endgroup$
Sujaan Kunalan
– Sujaan Kunalan
2014-03-30 04:55:39 +00:00
Commented Mar 30, 2014 at 4:55
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$\begingroup$ Your question is not clear. Saying "I can show that this polynomial is continuous at $x_0$", which $x_0$ do you mean, it is not mentioned in the statement you want to prove. If it is an arbitrary real number, your proof is complete. Did you maybe mean instead that you proved continuity at $x=0$? $\endgroup$
Marc van Leeuwen
– Marc van Leeuwen
2014-03-30 05:38:47 +00:00
Commented Mar 30, 2014 at 5:38
$\begingroup$ No, I meant arbitrary $x_0$. $\endgroup$
user48876
– user48876
2014-03-30 07:34:45 +00:00
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$f(x)=x$ is continuous everywhere
If $f(x)$ and $g(x)$ are continuous in $D$ then $f(x)g(x)$ in continuous on $D$.
Using 2 and 1 show that $x^n$ is continuous for every $n\in \mathbb{N}$
If $f(x)$ and $g(x)$ are continuous on $D$ then $f(x)+g(x)$ is continous on $D$
Now use 3 and 4.
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answered Mar 30, 2014 at 5:08
hrkrshnnhrkrshnn
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$\begingroup$ Hi, I've follow your steps and would like to check if my proof is okay 1)elementary proof 2)algebraic property of coninuous functions proof 3) Proof: If f1(x)=x and f2(x)=x are both continuous on D then f1(x)f2(x)=x^2 is continuous. Hence we can conclude that f1(x)....fm(x)=x^n, for n=m(m+1)/2 is continuous on D. 4)algebraic property of continuous functions 5)proof: from 3) we know that x^n is contionuous on D and hence if we add a polynomial of degree n-1 or smaller then by 4) we can conclude that x^n+x^(n-1)+...+1 is continuous. Hence all polynomials are continuous. $\endgroup$
user634512
– user634512
2019-03-29 18:15:01 +00:00
Commented Mar 29, 2019 at 18:15
$\begingroup$ @ThePoorJew. Yes looks okay $\endgroup$
hrkrshnn
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2019-04-01 13:21:02 +00:00
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Using the definition of a limit it is easy to prove (and you should prove as a simple exercise) that $$\lim_{x \to a}k = k, \lim_{x \to a}x = a\tag{1}$$ Using these results and induction and laws of algebra of limits it is an easy matter to prove that any polynomial function $f$ with real coefficients is continuous everywhere.
We will use induction on the degree of a polynomial. Polynomials of degree $0$ are constants and by first result in $(1)$ a constant is continuous everywhere. Now we assume that any polynomial of degree less than $n$ is continuous everywhere and let $f$ be a polynomial of degree $n$. Let $a$ be any arbitrary real number. Let $$f(x) = a_{0}x^{n} + a_{1}x^{n - 1} + \cdots + a_{n - 1}x + a_{n}$$ and then we have \begin{align} \lim_{x \to a}f(x) &= \lim_{x \to a}a_{0}x^{n} + a_{1}x^{n - 1} + \cdots + a_{n - 1}x + a_{n}\notag\ &= \lim_{x \to a}x(a_{0}x^{n - 1} + a_{1}x^{n - 2} + \cdots + a_{n - 1}) + a_{n}\notag\ &= \lim_{x \to a}xg(x) + a_{n}\notag\ &\,\,\,\,\,\,\,\,\text{(note that }g(x)\text{ is a polynomial of degree }(n - 1))\notag\ &= \lim_{x \to a}x\cdot\lim_{x \to a}g(x) + \lim_{x \to a}a_{n}\text{ (laws of algebra of limits)}\notag\ &= ag(a) + a_{n}\text{ (}g(x)\text{ is continuous and using (1))}\notag\ &= a(a_{0}a^{n - 1} + a_{1}a^{n - 2} + \cdots + a_{n - 1}) + a_{n}\notag\ &= a_{0}a^{n} + a_{1}a^{n - 1} + \cdots + a_{n - 1}a + a_{n}\notag\ &= f(a)\notag \end{align} Hence polynomial $f(x)$ is continuous at $a$. Since $a$ was an arbitrary real number it follows that $f(x)$ is continuous everywhere. The proof is now complete by principle of mathematical induction and every polynomial with real coefficients is continuous everywhere.
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Lemma 1: Limit of power function is the corresponding value of the function itself. $$ \begin{align} \lim_{x\to a}a_ix^i&=a_i\lim_{x\to a}x^i\ &=a_i\prod_{j=1}^{i}\lim_{x\to a}x\tag{product rule of limit}\ &=a_ia^i \end{align} $$
Every polynomial function is continuous Let the polynomial be $P(x)=a_0+a_1x+a_2x^2+a_3x^3+...+a_nx^n=\sum_{i=0}^n a_ix^i$ $$ \begin{align} \lim_{x\to a}P(x)&=\lim_{x\to a}\sum_{i=0}^{n}a_ix^i\ &=\sum_{i=0}^{n}\lim_{x\to a}a_ix^i\tag{summation rule of limit}\ &=\sum_{i=0}^{n}a_ia^i\tag{lemma 1}\ &=P(a)\ &Q.E.D \end{align} $$
In addition, Every rational function is also continuous Let the rational function be $R(x)=\frac{P_1(x)}{P_2(x)}$, given that $P_2(x)\neq 0$ $$ \begin{align} \lim_{x\to a}R(x)&=\lim_{x\to a}\frac{P_1(x)}{P_2(x)}\ &=\frac{\lim_{x\to a}P_1(x)}{\lim_{x\to a}P_2(x)}\tag{reciprocal rule of limit}\ &=\frac{P_1(a)}{P_2(a)}\tag{see proof of polynomial}\ &=R(a)\ &Q.E.D \end{align} $$
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7.G. Grade 7 - Geometry
7.G.A. Draw, construct, and describe geometrical figures and describe the relationships between them.
7.G.A.1. Solve problems involving scale drawings of geometric figures, including computing actual lengths and areas from a scale drawing and reproducing a scale drawing at a different scale.
Approximating the area of a circle
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9536 | http://230nsc1.phy-astr.gsu.edu/hbase/thermo/temperp.html | Temperature
Temperature =========== A convenient operational definition of temperature is that it is a measure of the average translational kinetic energy associated with the disordered microscopic motion of atoms and molecules. The flow of heat is from a high temperature region toward a lower temperature region. The details of the relationship to molecular motion are described in kinetic theory.The temperature defined from kinetic theory is called the kinetic temperature. Temperature is not directly proportional to internal energy since temperature measures only the kinetic energy part of the internal energy, so two objects with the same temperature do not in general have the same internal energy (see water-metal example). Temperatures are measured in one of the three standard temperature scales (Celsius, Kelvin, and Fahrenheit). More generalized view of temperatureIndex Temperature concepts
HyperPhysics ThermodynamicsR NaveGo Back
Temperature Scales ================== The Celsius, Kelvin, and Fahrenheit temperature scales are shown in relation to the phase change temperatures of water. The Kelvin scale is called absolute temperature and the Kelvin is the SI unit for temperature. The triple point of water is 273.16 K, and that is an international standard temperature point. The freezing point of water at one atmosphere pressure, 0.00°C, is 0.01K below that at 273.15 K. If you want to be really precise about it, the boiling point is 373.125 K, or 99.975 °C relative to the standard pressure freezing point. But for general purposes, just 0 °C and 100 °C are precise enough. Standard temperature pointsIndex Temperature concepts Internal energy concepts
HyperPhysics ThermodynamicsR NaveGo Back
Temperature Standard Points =========================== While the typical treatment of temperature scales takes the freezing point of water to be 0C and the boiling point at standard pressure to be 100C, there are more precise treatments of standard points for defining temperatures. By international agreement, one standard point is the triple point of water which has been defined to be 273.16K. The freezing point of water at atmospheric pressure is .01K below this at 273.15K. In order to obtain a second standard point by means of a thermometer which doesn't depend on the particular substance used to make it, a constant-volume gas thermometer was chosen to measure the boiling point of water. This method is based upon the ideal gas law, i.e., the assumption that if the volume is fixed, the temperature is directly proportional to the pressure. This measurement leads to a boiling point of 373.125K or 99.975 C above freezing at standard pressure. This measurement is independent of the gas used to make the thermometer. Ordinary gases do not behave exactly as ideal gases and are better described by the van der Waalsequation of state, but as they are extrapolated to zero pressure, they all project to the same value for the zero of the Kelvin scale. Table of standard temperature pointsIndex Temperature concepts Internal energy concepts
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9537 | https://brainly.com/question/39488894 | [FREE] The prime factorization of a perfect cube (other than 0 or 1) will only have exponents that are multiples - brainly.com
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The prime factorization of a perfect cube (other than 0 or 1) will only have exponents that are multiples of 3.
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In mathematics, a number is a perfect cube if it can be expressed as the product of three equal integers. The exponents of the prime factorization of a perfect cube will always be multiples of 3.
Explanation
In mathematics, exponents are a way of expressing repeated multiplication. This concept is used extensively when dealing with squares and cubes. Here, moving on to the concept of a perfect cube, a number is said to be a perfect cube if it can be expressed as the product of three equal integers. For instance, 27 (3 3 3) is a perfect cube.
When you write the prime factorization of a number, you are essentially breaking the number down into its base prime factors. For a perfect cube, the exponents of the prime factorization will always be multiples of 3. This is because to get a perfect cube, each prime factor has to be present three times.
Let's illustrate this with an example. Consider the perfect cube 64. The prime factorization of 64 is 26 (i.e., 2 2 2 2 2 2), and the exponent is indeed a multiple of 3. This rule holds for all perfect cubes and is a helpful way to determine if a given number is indeed a perfect cube.
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A perfect cube can be expressed as the product of three equal integers. Its prime factorization will have exponents that are multiples of 3, as each prime factor must appear three times. This property holds for all perfect cubes, aiding in their identification.
Explanation
In mathematics, a number is defined as a perfect cube if it can be represented as the product of three identical factors. For example, the number 27 can be expressed as 3×3×3, which makes it a perfect cube.
When discussing prime factorization, we break down numbers into their prime building blocks. A crucial property of perfect cubes is that when you perform the prime factorization of such a number, all the exponents of the prime factors will be multiples of 3. This is due to the requirement that in order to form a cube, each prime must appear three times.
Let’s illustrate this with some examples:
Consider the perfect cube 8. The prime factorization of 8 is 2 3 (i.e., 2×2×2). Here, the exponent 3 is a multiple of 3.
Another example is 125. Its prime factorization is 5 3 (i.e., 5×5×5), where again the exponent is a multiple of 3.
This property of having exponents as multiples of 3 holds true for all perfect cubes, making it a useful rule when determining if a number is a perfect cube or not.
To summarize, a perfect cube's prime factorization will only have exponents that are multiples of 3, reaffirming the definition that a perfect cube must consist of factors multiplied by themselves three times.
Examples & Evidence
Examples include 8, which factors to 2^3, and 27, which factors to 3^3, both demonstrating that exponents are multiples of 3.
The definition of perfect cubes and their properties can be found in algebra textbooks and essential mathematics resources.
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What is the difference between the signs “≤”, “<”, and “>” in mathematics?
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Raphael
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They all describe something similar: how values compare to each other.
Specifically, the following statements are true.
< means less than. The item on the left has a lower value than that on the right.
≤ means less than or equal to. The item on the left has either the same or a lower value than that on the right.
means greater than. The item on the left has a higher value than that on the right.
≥ means greater than or equal to. The item on the left has either the same or a higher value than that on the right.
Basically, think of it like this. The “open” end of the symbol faces the larger object,
They all describe something similar: how values compare to each other.
Specifically, the following statements are true.
< means less than. The item on the left has a lower value than that on the right.
≤ means less than or equal to. The item on the left has either the same or a lower value than that on the right.
means greater than. The item on the left has a higher value than that on the right.
≥ means greater than or equal to. The item on the left has either the same or a higher value than that on the right.
Basically, think of it like this. The “open” end of the symbol faces the larger object, and the “closed” end the smaller object. If a line is present on the bottom of the symbol, the two objects can also have the same value.
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Related questions
In mathematics, what is the difference between = and: =?
What does the "=:" mean in mathematics?
What is the name of the math symbol that looks like an equal sign, but the top line is wavy? What does it symbolize?
What is the symbol ≤? What is it called and how does it work (in maths)?
Mathematics:
What's the difference between an equal sign with two bars and an equal sign with three bars?
Fredrick Chris Engallado
2y
< means “is less than"
For example 3<4, 3 is less than 4
≤ means “is less than or equal to"
Example X ≤ Y
means “is greater than"
Example 4>3
≥ means “is greater than or equal to"
Example Y ≥ X
Tim Sabin
BS in Computer Science & Mathematics, Pace University, Pleasantville, NY (Graduated 1975)
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Author has 9.2K answers and 5.6M answer views
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4y
Related
What is the symbol ≤? What is it called and how does it work (in maths)?
The < symbol is “less than”, and = is “equal to”. Put them together, you have ≤ which is “less than or equal to”. It's a comparison operator.
Similar with > = ≥
Example: False: 0 < 0. True: 0 ≤ 0
Allen Knutson
Math professor at Cornell University
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In mathematics, what is the difference between = and: =?
Research mathematician here (just back from the joint meetings of the American Mathematical Society / Mathematical Association of America), and I’m going to pull rank and say that yes, among researchers “:=” is (becoming) standard notation for “is defined as” rather than “these two previously defined quantities are in fact equal”. When I say “becoming”, I mean I’ve seen it in use for decades, trickling down from blackboard use to actual publications. (For example “iff” for “if and only if” was commonly used in talks before it became commonly used in papers.) Not everybody uses it yet but they
Research mathematician here (just back from the joint meetings of the American Mathematical Society / Mathematical Association of America), and I’m going to pull rank and say that yes, among researchers “:=” is (becoming) standard notation for “is defined as” rather than “these two previously defined quantities are in fact equal”. When I say “becoming”, I mean I’ve seen it in use for decades, trickling down from blackboard use to actual publications. (For example “iff” for “if and only if” was commonly used in talks before it became commonly used in papers.) Not everybody uses it yet but they should!
I’m pretty sure that it derives from Pascal (as mentioned by other posters). One funny thing is that mathematicians will also use “=:” from time to time, to mean “let’s assign a short name to this complicated expression we came up with”, as in “4 + x^2 - y =: D”, and obviously you can’t do that in Pascal.
Related questions
In mathematics, what is the difference between = and: =?
What does the "=:" mean in mathematics?
What is the name of the math symbol that looks like an equal sign, but the top line is wavy? What does it symbolize?
What is the symbol ≤? What is it called and how does it work (in maths)?
Mathematics:
What's the difference between an equal sign with two bars and an equal sign with three bars?
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Polygon area calculator
The calculator below will find the area of any polygon if you know the coordinates of each vertex.
This will work for
triangles,
regular
and
irregular polygons,
convex
or
concave polygons.
It uses the same method as in
Area of a polygon but does the arithmetic for you.
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Instructions
Enter the vertices in order, either clockwise or counter-clockwise starting at any vertex.
Enter the x,y coordinates of each vertex into the table. Empty rows will be ignored.
Click on "Calculate".
Unlike the
manual method, you do not need to enter the first vertex again at the end,
and you can go in either direction around the polygon.
The internal programming of the calculator takes care of it all for you.
There are other, often easier ways to calculate the area of triangles and regular polygons. See
Area of a regular polygon
Area of a triangle (Given base and altitude)
Area of a triangle (Heron's formula - given lengths of the three sides)
Area of a triangle (By formula, given coordinates of vertices)
Area of a triangle (Box method, given coordinates of vertices)
Limitations
The calculator will produce the wrong answer for crossed polygons,
where one side crosses over another, as shown below.
Other Coordinate Geometry topics
Introduction to coordinate geometry
The coordinate plane
The origin of the plane
Axis definition
Coordinates of a point
Distance between two points
Introduction to Lines
in Coordinate Geometry
Line (Coordinate Geometry)
Ray (Coordinate Geometry)
Segment (Coordinate Geometry)
Midpoint Theorem
Distance from a point to a line
- When line is horizontal or vertical
- Using two line equations
- Using trigonometry
- Using a formula
Intersecting lines
Cirumscribed rectangle (bounding box)
Area of a triangle (formula method)
Area of a triangle (box method)
Centroid of a triangle
Incenter of a triangle
Area of a polygon
Algorithm to find the area of a polygon
Area of a polygon (calculator)
Rectangle
Definition and properties, diagonals
Area and perimeter
Square
Definition and properties, diagonals
Area and perimeter
Trapezoid
Definition and properties, altitude, median
Area and perimeter
Parallelogram
Definition and properties, altitude, diagonals
Print blank graph paper
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9540 | https://obrc.ouhsc.edu/Portals/1308a/Assets/documents/WebinarSlides/Maintaining%20Lactation%206-2015.pdf | MAINTAINING LACTATION WHEN SEPARATED: STEP 5 MEDICAL INDICATION FOR SUPPLEMENTATION: STEP 6 Keri Hale, RD/LD, IBCLC Becky Mannel, BS, IBCLC DISCLOSURE Both speakers are employed by the nonprofit Oklahoma Mothers’ Milk Bank CALL TO ACTION TO SUPPORT BREASTFEEDING, JANUARY 2011 US SURGEON GENERAL “One of the most highly effective preventive measures a mother can take to protect the health of her infant and herself is to breastfeed.” U.S. SURGEON GENERAL’S CALL TO ACTION TO SUPPORT BREASTFEEDING 2011 Action 12: Identify and address obstacles to greater availability of safe banked donor milk for fragile infants.
MAINTAINING LACTATION WHEN SEPARATED (STEP 5) BFHI Step 5: “Show mothers how to breastfeed and how to maintain lactation, even if they are separated from their infants.” How soon to start milk expression Importance of hand expression for colostrum Hands on pumping Skin to skin asap Pumping/hand expression needs to be started within 4-6 hours of birth or separation, ideally right away HOW DOES MOM PUMP/EXPRESS FOR BEST RESULTS?
START WITH HAND EXPRESSION PRACTICE SKIN TO SKIN / KANGAROO CARE TO OBTAIN AND MAINTAIN MILK SUPPLY RELAX AND STIMULATE MILK RELEASE • Look at Babies Picture • Baby Item • Listen to Music • Watch a video • Power Nap • Prop Feet Up • Comfy Chair • Quiet Area USE HANDS ON PUMPING PAINFUL PUMPING NEEDS TO BE EVALUATED • Ensure that flanges are fitting properly • Flanges that are too small may irritate mom’s nipple • Pain during pumping can impair your milk release • Ensure suction is not set too strong • Time pumping sessions MEDICAL INDICATIONS FOR SUPPLEMENTATION BFHI Step 6: Give infants no food or drink other than breast-milk, unless medically indicated. RESOURCES ON MEDICAL INDICATIONS FOR SUPPLEMENTATION OF A BREASTFED BABY • American Academy of Family Physicians • American Academy of Pediatrics • Baby Friendly Hospital Initiative • The Joint Commission • Academy of Breastfeeding Medicine AAFP: HOSPITAL USE OF INFANT FORMULA IN BREASTFEEDING INFANTS “RESPECT THE DECISION OF THE MOTHER WHO CHOOSES TO BREASTFEED EXCLUSIVELY BY NOT OFFERING FORMULA, WATER OR PACIFIERS TO AN INFANT UNLESS THERE IS A SPECIFIC PHYSICIAN ORDER.” Main Breastfeeding Policy: Supplementation may be done with expressed mother’s milk, pasteurized human milk from a donor, or infant formula. AAP POLICY STATEMENT: BREASTFEEDING AND THE USE OF HUMAN MILK, 2012 • Ensure 8 to 12 feedings at the breast every 24 h • Ensure formal evaluation and documentation of breastfeeding by trained caregivers (including position, latch, milk transfer, examination) at least for each nursing shift • Give no supplements (water, glucose water, commercial infant formula, or other fluids) to breastfeeding newborn infants unless medically indicated using standard evidence-based guidelines for the management of hyperbilirubinemia and hypoglycemia • Avoid routine pacifier use in the postpartum period BABY-FRIENDLY HOSPITAL INITIATIVE • Breastfed infants will only be supplemented with a physicians order and only for clinical evidence of a medically justifiable reason. • Developing a list of current, evidence-based medical indications that is communicated to all staff and care providers is recommended. THE JOINT COMMISSION PERINATAL CARE CORE MEASURE: EXCLUSIVE BREAST MILK FEEDING • Exclusive breast milk feeding is defined as a newborn receiving only breast milk and no other liquids or solids except for drops or syrups consisting of vitamins, minerals, or medicines.
• Use of donor breast milk is allowable ACADEMY OF BREASTFEEDING MEDICINE CLINICAL PROTOCOL #3: HOSPITAL GUIDELINES FOR THE USE OF SUPPLEMENTARY FEEDINGS IN THE HEALTHY, TERM BREASTFED NEONATE REVISED 2009 NOTE: ABM Clinical Protocols are now readily available through the National Guideline Clearinghouse website.
Visit www.guideline.gov SUPPLEMENTARY FEEDINGS: FEEDINGS PROVIDED IN PLACE OF BREASTFEEDING. THIS MAY INCLUDE EXPRESSED OR BANKED BREASTMILK AND/OR BREASTMILK SUBSTITUTES/FORMULA.
“ANY FOODS GIVEN PRIOR TO 6 MONTHS” Complementary feedings: Feedings provided in addition to breastfeeding when breastmilk alone is no longer sufficient.
“foods or liquids given in addition to breastfeeding AFTER 6 months” ABM: SUPPLEMENTATION IS NOT INDICATED (EVALUATION AND BREASTFEEDING MANAGEMENT MAY BE NEEDED) • Sleepy infant, <8 feedings in 1st 24-48 hrs, <7% weight loss, no s/s illness • Healthy, term, AGA infant, bilirubin <18 mg/dl after 72 hrs, feeding/stooling/weight loss WNL • Infant fussy at night • Infant constantly feeding for several hours • Tired/sleeping mother ABM: INDICATIONS FOR SUPPLEMENTAL FEEDINGS IN TERM, HEALTHY INFANTS • Separation • Unable to feed at breast (congenital malformation, illness) • Inborn error of metabolism (eg, galactosemia) • Contraindicated maternal medications ABM: POSSIBLE INDICATIONS FOR SUPPLEMENTATION IN TERM, HEALTHY INFANTS • INFANT: • Asymptomatic hypoglycemia, unresolved by frequent breastfeeding • Hyperbilirubinemia • Poor milk transfer • Significant dehydration • Meconium stools on day 5 • Weight loss of 8-10% AND delayed lactogenesis II day 5 or later ABM: POSSIBLE INDICATIONS FOR SUPPLEMENTATION IN TERM, HEALTHY INFANTS • MOTHER: • Delayed lactogenesis II (day 3-5 or later) AND inadequate intake • Sheehan’s syndrome • Retained placenta • Primary glandular insufficiency • Poor milk production r/t breast pathology/surgery • Intolerable pain during feedings unrelieved by interventions AVERAGE REPORTED INTAKES OF COLOSTRUM BY HEALTHY BREASTFED INFANTS Infants fed artificial milk ad lib typically have higher intakes than breastfed infants. There is no definitive research available on supplemental feedings thus supplements should reflect the normal amounts of colostrum, the size of the infant’s stomach, and the age and size of the infant. Feeding volumes should be by infant satiation cues.
1st 24hrs 2-10mL/feed 24-48hrs 5-15mL/feed (~25mL/Kg) 48-72hrs 15-30mL/feed (~45mL/kg) 72-96hrs 30-60mL/feed (~100mL/Kg) HOW TO GIVE A SUPPLEMENTAL FEED BEFORE BREASTFEEDING IS ESTABLISHED There is little evidence about the safety or efficacy of most alternative feeding methods and their effect on breastfeeding Cup feeding has been shown safe for both term and preterm infants and may help preserve breastfeeding duration among those who require multiple supplemental feedings REDUCING THE NEED FOR SUPPLEMENTATION Skin to skin contact immediately after birth Early initiation of breastfeeding Rooming in Help with position and latch Early, regular assessment of feedings Teach to feed on cue ACCREDITED BY AND MEMBER OF THE HUMAN MILK BANKING ASSOCIATION OF NORTH AMERICA. 13th milk bank in the U.S.!
PASTEURIZED DONOR MILK ON A MOTHER/BABY UNIT Store in a freezer separated from medications/foods Monitor temperature daily Can use a countertop refrigerator/freezer PASTEURIZED DONOR MILK ON A MOTHER/BABY UNIT Stock several bottles to have immediately available Milk bank can provide long expiration dates Document PDM in medical record Document “batch #” in medical record EACH BOTTLE OF PDM IS LABELED: BATCH NUMBER CALORIES/OUNCE EXPIRATION DATE PROTEIN CONTENT HANDLING PDM ON A MOTHER/BABY UNIT Can thaw just enough to draw off a small feeding (10 ml) Can use same bottle for more than one baby Can refreeze if still has ice crystals Thawed PDM can be refrigerated for at least 48 hours Pasteurized Donor Human Milk Maintains Microbiological Purity for 4 Days at 4°C, JHL 2014 CURRENT GOALS Increase production capacity Provide milk to level II NICUs and mother/baby units Provide milk for outpatients Financial stabilityThank you to Sarkeys Foundation!! www.okmilkbank.org 405-297-LOVE info@okmilkbank.org keri@okmilkbank.org becky@okmilkbank.org |
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与导数的定义有关的题目
原创 已于 2024-10-13 14:27:48 修改·1.2k 阅读
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文章标签：
#概率论#抽象代数#密码学
于 2024-10-13 14:27:11 首次发布
题目一：证明一个关于导数的不等式
设函数 f(x)f(x)f(x) 在区间 [a,b][a, b][a,b] 上二阶可导，且 f′′(x)≥0f''(x) \ge 0 f′′(x)≥0 (即 f(x)f(x)f(x) 是凸函数)。 证明：对于任意 x,y∈[a,b]x, y \in [a, b]x,y∈[a,b]，都有：
f(x)+f(y)≥2f(x+y2)f(x) + f(y) \ge 2f\left(\frac{x+y}{2}\right)f(x)+f(y)≥2 f(2 x+y)
证明思路：
利用导数定义和中值定理: 首先考虑函数 g(t)=f(x+t(y−x))g(t) = f(x+t(y-x))g(t)=f(x+t(y−x))。 g(0)=f(x)g(0) = f(x)g(0)=f(x)，g(1)=f(y)g(1) = f(y)g(1)=f(y)。 根据中值定理，存在 θ1∈(0,1)\theta_1 \in (0, 1)θ 1∈(0,1) 使得 g′(θ1)=f(y)−f(x)g'( \theta_1 ) = f(y) - f(x)g′(θ 1)=f(y)−f(x).
利用泰勒展开: 对 f(x)f(x)f(x) 在 x+y2\frac{x+y}{2}2 x+y 处进行泰勒展开：
f(x)=f(x+y2)+f′(x+y2)(x−x+y2)+f′′(ξ1)2(x−x+y2)2f(x) = f\left(\frac{x+y}{2}\right) + f'\left(\frac{x+y}{2}\right)\left(x - \frac{x+y}{2}\right) + \frac{f''(\xi_1)}{2}\left(x - \frac{x+y}{2}\right)^2 f(x)=f(2 x+y)+f′(2 x+y)(x−2 x+y)+2 f′′(ξ 1)(x−2 x+y)2
f(y)=f(x+y2)+f′(x+y2)(y−x+y2)+f′′(ξ2)2(y−x+y2)2f(y) = f\left(\frac{x+y}{2}\right) + f'\left(\frac{x+y}{2}\right)\left(y - \frac{x+y}{2}\right) + \frac{f''(\xi_2)}{2}\left(y - \frac{x+y}{2}\right)^2 f(y)=f(2 x+y)+f′(2 x+y)(y−2 x+y)+2 f′′(ξ 2)(y−2 x+y)2
其中 ξ1\xi_1 ξ 1 和 ξ2\xi_2 ξ 2 是介于 xx x 和 x+y2\frac{x+y}{2}2 x+y，以及 yy y 和 x+y2\frac{x+y}{2}2 x+y 之间的数。
结合条件 f′′(x)≥0f''(x) \ge 0 f′′(x)≥0： 由于 f′′(x)≥0f''(x) \ge 0 f′′(x)≥0，且 (x−x+y2)2=(y−x+y2)2\left(x - \frac{x+y}{2}\right)^2 = \left(y - \frac{x+y}{2}\right)^2(x−2 x+y)2=(y−2 x+y)2，可以得到 f(x)+f(y)≥2f(x+y2)f(x) + f(y) \ge 2f\left(\frac{x+y}{2}\right)f(x)+f(y)≥2 f(2 x+y).
题目二：证明一个关于导数极限的结论
设函数 f(x)f(x)f(x) 在 x=0x=0 x=0 处可导，且 f(0)=0f(0) = 0 f(0)=0。 证明：
limx→0f(x)x=f′(0)\lim_{x \to 0} \frac{f(x)}{x} = f'(0)x→0 limx f(x)=f′(0)
证明思路：
这似乎是一个显而易见的结论，但需要严格地从导数的定义出发证明。
导数定义:f′(0)=limx→0f(x)−f(0)x−0=limx→0f(x)xf'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0} \frac{f(x)}{x}f′(0)=lim x→0x−0 f(x)−f(0)=lim x→0x f(x) (因为 f(0)=0f(0) = 0 f(0)=0)
极限的性质: 直接应用极限的代数性质即可证明。 因为 f(x)f(x)f(x) 在 x=0x=0 x=0 可导，所以极限存在且等于 f′(0)f'(0)f′(0)。
题目三：构造一个反例
试构造一个函数 f(x)f(x)f(x)，它满足 limx→0f(x)x=0\lim_{x \to 0} \frac{f(x)}{x} = 0 lim x→0x f(x)=0，但 f(x)f(x)f(x) 在 x=0x=0 x=0 处不可导。
证明思路：
这道题考察对导数定义的深刻理解，以及极限与导数之间的联系。 需要找到一个函数，其在 x=0x=0 x=0 处的极限为 0，但是其导数在 x=0x=0 x=0 处不存在。
一个例子是：
f(x)={x2sin(1x),x≠00,x=0f(x) =
{x 2 sin(1 x),0,x≠0 x=0{x 2 sin(1 x),x≠0 0,x=0
f(x)={x 2 sin(x 1),0,x=0 x=0
可以验证 limx→0f(x)x=0\lim_{x \to 0} \frac{f(x)}{x} = 0 lim x→0x f(x)=0，但是 f′(0)f'(0)f′(0) 不存在。
题目四：关于一致收敛与可导性
设 {fn(x)}{f_n(x)}{f n(x)} 是定义在 [a,b][a, b][a,b] 上的一列函数，且每个 fn(x)f_n(x)f n(x) 都在 [a,b][a, b][a,b] 上可导。 假设 {fn(x)}{f_n(x)}{f n(x)} 在 [a,b][a, b][a,b] 上一致收敛于 f(x)f(x)f(x)，且 {fn′(x)}{f_n'(x)}{f n′(x)} 在 [a,b][a, b][a,b] 上一致收敛于 g(x)g(x)g(x)。 证明：f(x)f(x)f(x) 在 [a,b][a, b][a,b] 上可导，且 f′(x)=g(x)f'(x) = g(x)f′(x)=g(x)。
解题思路提示:
这道题需要利用一致收敛的性质和微积分基本定理。 关键步骤是利用积分中值定理：
从 fn′(x)f_n'(x)f n′(x) 的一致收敛性出发，考虑积分 ∫axfn′(t)dt\int_a^x f_n'(t) dt∫a xf n′(t)d t。
利用微积分基本定理，将积分与 fn(x)f_n(x)f n(x) 联系起来。
利用一致收敛的性质，证明极限与积分可以交换顺序。
最终得到 f′(x)=g(x)f'(x) = g(x)f′(x)=g(x)。 这需要非常细致的 ϵ−δ\epsilon-\delta ϵ−δ 分析。
题目五：关于隐函数的导数
设 F(x,y)F(x, y)F(x,y) 是一个连续可微函数，且 F(x0,y0)=0F(x_0, y_0) = 0 F(x 0,y 0)=0。 假设 ∂F∂y(x0,y0)≠0\frac{\partial F}{\partial y}(x_0, y_0) \ne 0∂y∂F(x 0,y 0)=0。 根据隐函数定理，存在 xx x 的一个邻域，使得方程 F(x,y)=0F(x, y) = 0 F(x,y)=0 定义了一个隐函数 y=f(x)y = f(x)y=f(x)。 证明：
f′(x0)=−∂F∂x(x0,y0)∂F∂y(x0,y0)f'(x_0) = -\frac{\frac{\partial F}{\partial x}(x_0, y_0)}{\frac{\partial F}{\partial y}(x_0, y_0)}f′(x 0)=−∂y∂F(x 0,y 0)∂x∂F(x 0,y 0)
解题思路提示:
这道题需要直接从隐函数定理的证明出发。 隐函数定理的证明通常利用反函数定理和微分方程的解的存在唯一性定理，非常复杂。 本题要求从导数定义出发，通过构造极限证明上述结果。
考虑极限 limx→x0f(x)−f(x0)x−x0\lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0}lim x→x 0x−x 0f(x)−f(x 0)。
利用 F(x,f(x))=0F(x, f(x)) = 0 F(x,f(x))=0 以及全微分公式。
谨慎地进行极限运算，最终得到结论。
题目六：一个涉及到瑕积分的题目
设 f(x)f(x)f(x) 在 [0,1][0, 1][0,1] 上连续，且 ∫01f(x)dx=0\int_0^1 f(x) dx = 0∫0 1f(x)d x=0。 证明：存在 ξ∈(0,1)\xi \in (0, 1)ξ∈(0,1) 使得 ∫0ξf(x)dx=0\int_0^\xi f(x) dx = 0∫0 ξf(x)d x=0.
解题思路提示:
这道题看起来比较简单，但需要仔细分析。 不能直接使用积分中值定理，因为题目没有假设 f(x)f(x)f(x) 的正负性。
定义函数 F(x)=∫0xf(t)dtF(x) = \int_0^x f(t) dt F(x)=∫0 xf(t)d t。 分析 F(x)F(x)F(x) 的性质。
利用 F(0)=0F(0) = 0 F(0)=0 和 F(1)=0F(1) = 0 F(1)=0，结合连续函数的性质，证明存在 ξ∈(0,1)\xi \in (0, 1)ξ∈(0,1) 使得 F(ξ)=0F(\xi) = 0 F(ξ)=0.
需要用到介值定理或罗尔定理。
题目七：关于泛函分析的应用
设 HH H 是一个希尔伯特空间，{xn}{x_n}{x n} 是 HH H 中的一个序列，满足 limn→∞∥xn∥=0\lim_{n \to \infty} \|x_n\| = 0 lim n→∞∥x n∥=0。 证明：存在一个子序列 {xnk}{x_{n_k}}{x n k} 使得对任意 x∈Hx \in H x∈H，都有 limk→∞⟨xnk,x⟩=0\lim_{k \to \infty} \langle x_{n_k}, x \rangle = 0 lim k→∞⟨x n k,x⟩=0。
解题思路提示:
这道题目需要运用泛函分析中的弱收敛概念。
首先，理解希尔伯特空间中弱收敛的定义。
利用巴拿赫-阿劳格鲁定理或其他相关定理，证明存在一个弱收敛的子序列。
仔细分析弱收敛的性质，证明极限为 0。
题目八：关于分部积分的巧妙应用
计算积分：∫0∞sinxxdx\int_0^\infty \frac{\sin x}{x} dx∫0∞x s i n xd x。
解题思路提示:
这道题目需要运用分部积分，以及一些技巧处理无穷积分。 这不是一个简单的分部积分，需要考虑如何构造合适的函数并处理极限。 可以考虑以下步骤：
构造一个合适的函数，使其与 sinxx\frac{\sin x}{x}x s i n x 相关。
运用分部积分法，多次积分。
利用狄利克雷判别法或者其他方法处理无穷积分的收敛性。
题目九：关于测度论的应用
设 ff f 是一个可测函数，定义在 [0,1][0, 1][0,1] 上，且 0≤f(x)≤10 \le f(x) \le 1 0≤f(x)≤1 几乎处处成立。 证明：存在一个可测集 E⊂[0,1]E \subset [0, 1]E⊂[0,1] 使得 ∫Ef(x)dx=12∫01f(x)dx\int_E f(x) dx = \frac{1}{2} \int_0^1 f(x) dx∫Ef(x)d x=2 1∫0 1f(x)d x.
解题思路提示:
这道题需要用到测度论中的基本概念和定理。
考虑函数 F(x)=∫0xf(t)dtF(x) = \int_0^x f(t) dt F(x)=∫0 xf(t)d t。 分析 F(x)F(x)F(x) 的性质。
利用测度论中的介值定理的推广形式。
证明存在一个可测集 EE E 满足条件。 注意需要利用可测函数的性质。
题目十：高维空间中的微分形式
计算在 R3\mathbb{R}^3 R 3 上的三维微分形式 ω=x dy∧dz+y dz∧dx+z dx∧dy\omega = x \, dy \wedge dz + y \, dz \wedge dx + z \, dx \wedge dy ω=x d y∧d z+y d z∧d x+z d x∧d y 的外微分 dωd\omega d ω。
解题思路提示:
这道题需要对微分形式的外微分运算有深入的理解。
回顾三维微分形式的外微分运算规则。
小心地计算 dωd\omega d ω，注意运算顺序。
简化结果，并得到最终答案。
题目十一：关于Sobolev空间中的嵌入定理
设 Ω⊂Rn\Omega \subset \mathbb{R}^n Ω⊂R n 是一个有界开集，且边界光滑。 考虑Sobolev空间 W1,p(Ω)W^{1,p}(\Omega)W 1,p(Ω)。 证明或反驳：对于 1≤p<n1 \le p < n 1≤p<n，W1,p(Ω)W^{1,p}(\Omega)W 1,p(Ω) 连续嵌入到 Lq(Ω)L^q(\Omega)L q(Ω) 中，其中 q=npn−pq = \frac{np}{n-p}q=n−p n p。 并讨论边界光滑性的影响。
解题思路提示: 这需要熟练掌握Sobolev嵌入定理及其证明。 需要理解Sobolev空间的定义、弱导数的概念以及相关的积分不等式（例如，Sobolev不等式和Gagliardo-Nirenberg不等式）。 边界光滑性会影响嵌入定理的成立条件，需要深入分析。
题目十二：关于非线性偏微分方程的解的存在性
证明或反驳：对于非线性抛物型方程 ut=Δu+u3u_t = \Delta u + u^3 u t=Δ u+u 3 在有界区域 Ω\Omega Ω 上，在狄利克雷边界条件下，存在一个全局光滑解。
解题思路提示: 这需要对非线性偏微分方程理论有深入的了解。 需要考虑能量估计、正则性理论以及潜在的解的爆破现象。 需要用到高级的分析技巧，例如，利用最大值原理、能量泛函方法等。 这通常是一个研究性问题，而非简单的习题。
题目十三：关于遍历理论中的一个问题
设 (X,B,μ,T)(X, \mathcal{B}, \mu, T)(X,B,μ,T) 是一个保测变换，其中 T:X→XT: X \to X T:X→X 是保测的。 假设存在一个集合 A∈BA \in \mathcal{B}A∈B 使得 μ(A)>0\mu(A) > 0 μ(A)>0，且对任意 n≥1n \ge 1 n≥1，μ(A∩T−nA)>0\mu(A \cap T^{-n}A) > 0 μ(A∩T−n A)>0。 证明或反驳：TT T 是遍历的。
解题思路提示: 这需要对遍历理论有相当深入的了解，包括遍历性、混合性、以及相关的定理，例如Poincaré复现定理。 需要构造反例或利用遍历理论中的高级工具进行证明。
题目十四：关于黎曼几何中的一个问题
考虑一个黎曼流形 (M,g)(M, g)(M,g)。 假设 MM M 是紧致的，且Ricci曲率处处为正。 证明或反驳：MM M 的基本群是有限的。
解题思路提示: 这需要对黎曼几何有深入的了解，包括Ricci曲率、基本群、以及相关的定理，例如Myers定理。 这需要应用黎曼几何中的许多高级概念和工具。
题目十五：关于复分析中的一个积分
计算积分：∮∣z∣=1ezz2(z−2)dz \oint_{|z|=1} \frac{e^z}{z^2(z-2)} dz ∮∣z∣=1z 2(z−2)e zd z 其中积分路径为单位圆 ∣z∣=1|z| = 1∣z∣=1，逆时针方向。
解题思路:
这道题可以用柯西积分公式及其推广来解决。
奇点分析: 被积函数在 z=0z=0 z=0 处有二阶极点，在 z=2z=2 z=2 处有一阶极点。
柯西积分公式的应用: 由于积分路径是单位圆，只有 z=0z=0 z=0 在积分路径内部。 我们需要计算 z=0z=0 z=0 处的留数。
留数计算: 对于二阶极点 z=0z=0 z=0，留数的计算公式为：Res(f,0)=limz→0ddz(z2ezz2(z−2))=limz→0ddzezz−2=limz→0(z−2)ez−ez(z−2)2=−14 Res(f, 0) = \lim_{z \to 0} \frac{d}{dz} (z^2 \frac{e^z}{z^2(z-2)}) = \lim_{z \to 0} \frac{d}{dz} \frac{e^z}{z-2} = \lim_{z \to 0} \frac{(z-2)e^z - e^z}{(z-2)^2} = \frac{-1}{4} R es(f,0)=z→0 limd z d(z 2 z 2(z−2)e z)=z→0 limd z dz−2 e z=z→0 lim(z−2)2(z−2)e z−e z=4−1
积分计算: 根据留数定理，积分值为 2πi2\pi i 2 πi 乘以所有在积分路径内部奇点的留数之和。 因此，∮∣z∣=1ezz2(z−2)dz=2πi×(−14)=−πi2 \oint_{|z|=1} \frac{e^z}{z^2(z-2)} dz = 2\pi i \times (-\frac{1}{4}) = -\frac{\pi i}{2} ∮∣z∣=1z 2(z−2)e zd z=2 πi×(−4 1)=−2 πi
题目十六：关于概率论中的一个问题
设 X1,X2,…X_1, X_2, \dots X 1,X 2,… 是独立同分布的随机变量，E[Xi]=0E[X_i] = 0 E[X i]=0, Var(Xi)=1Var(X_i) = 1 Va r(X i)=1。 设 Sn=∑i=1nXiS_n = \sum_{i=1}^n X_i S n=∑i=1 nX i。 证明：对任意 ϵ>0\epsilon > 0 ϵ>0，limn→∞P(∣Sn∣>ϵn)=0\lim_{n \to \infty} P(|S_n| > \epsilon \sqrt{n}) = 0 lim n→∞P(∣S n∣>ϵ n)=0。
解题思路:
这道题可以用切比雪夫不等式来解决。
方差计算:Var(Sn)=Var(∑i=1nXi)=∑i=1nVar(Xi)=nVar(S_n) = Var(\sum_{i=1}^n X_i) = \sum_{i=1}^n Var(X_i) = n Va r(S n)=Va r(∑i=1 nX i)=∑i=1 nVa r(X i)=n (因为 XiX_i X i 独立)。
切比雪夫不等式: 切比雪夫不等式指出，对任意随机变量 YY Y 和 ϵ>0\epsilon > 0 ϵ>0，P(∣Y−E[Y]∣≥ϵ)≤Var(Y)ϵ2P(|Y - E[Y]| \ge \epsilon) \le \frac{Var(Y)}{\epsilon^2}P(∣Y−E[Y]∣≥ϵ)≤ϵ 2 Va r(Y)。
应用切比雪夫不等式: 令 Y=SnY = S_n Y=S n，E[Y]=0E[Y] = 0 E[Y]=0。 则有：P(∣Sn∣≥ϵn)≤Var(Sn)(ϵn)2=nϵ2n=1ϵ2 P(|S_n| \ge \epsilon \sqrt{n}) \le \frac{Var(S_n)}{(\epsilon \sqrt{n})^2} = \frac{n}{\epsilon^2 n} = \frac{1}{\epsilon^2} P(∣S n∣≥ϵ n)≤(ϵ n)2 Va r(S n)=ϵ 2 n n=ϵ 2 1
弱大数定律: 虽然上述结果不直接证明极限为 0，但它说明概率有界。 结合弱大数定律，因为 E[Xi]=0E[X_i]=0 E[X i]=0， Snn→0\frac{S_n}{n} \to 0 n S n→0 以概率 1，这暗示着 P(∣Sn∣>ϵn)→0P(|S_n| > \epsilon \sqrt{n}) \to 0 P(∣S n∣>ϵ n)→0 当 n→∞n \to \infty n→∞。 更严格的证明需要用到中心极限定理。
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9542 | https://www.mathsisfun.com/prime-factorization.html | Prime Factorization
Prime Numbers
A Prime Number is:
a whole number above 1 that cannot be made by multiplying other whole numbers
When it can be made by multiplying other whole numbers it is a Composite Number, like this:
2 is Prime, 3 is Prime, 4 is Composite (=2×2), 5 is Prime, and so on...
(The first few prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19 and 23, and we have a prime number chart if you need more.)
Factors
"Factors" are the numbers you multiply together to get another number:
Prime Factorization
"Prime Factorization" is finding which prime numbers multiply together to make the original number.
Here are some examples:
Example: What are the prime factors of 12 ?
It is best to start working from the smallest prime number, which is 2, so let's check:
12 ÷ 2 = 6
Yes, it divided exactly by 2. We have taken the first step!
But 6 is not a prime number, so we need to go further. Let's try 2 again:
6 ÷ 2 = 3
Yes, that worked also. And 3 is a prime number, so we have the answer:
12 = 2 × 2 × 3
As you can see, every factor is a prime number, so the answer is right.
It is neater to show repeated numbers using exponents:
Without exponents: 2 × 2 × 3
With exponents: 22 × 3
Example: What is the prime factorization of 147 ?
Can we divide 147 exactly by 2?
147 ÷ 2 = 73½
No we can't. The answer should be a whole number, and 73½ is not.
Let's try the next prime number, 3:
147 ÷ 3 = 49
That worked, now try factoring 49.
The next prime, 5, does not work. But 7 does, so we get:
49 ÷ 7 = 7
And that is as far as we need to go, because all the factors are prime numbers.
147 = 3 × 7 × 7 = 3 × 72
Example: What is the prime factorization of 17 ?
Hang on ... 17 is a Prime Number.
So that is as far as we can go.
17 = 17
Another Method
We just did factorization by starting at the smallest prime and working upwards.
But sometimes it is easier to break a number down into any factors we can ... then work those factor down to primes.
Example: What are the prime factors of 90 ?
Break 90 into 9 × 10
The prime factors of 9 are 3 and 3
The prime factors of 10 are 2 and 5
So the prime factors of 90 are 3, 3, 2 and 5
90 = 2 × 32 × 5
Factor Tree
A "Factor Tree" can help: find any factors of the number, then the factors of those numbers, etc, until we can't factor any more.
Example: 48
48 = 8 × 6, so we write down "8" and "6" below 48
Now we continue and factor 8 into 4 × 2
Then 4 into 2 × 2
And lastly 6 into 3 × 2
We can't factor any more, so we have found the prime factors.
Which reveals that 48 = 2 × 2 × 2 × 2 × 3
48 = 24 × 3
Why find Prime Factors?
A prime number can only be divided by 1 or itself, so it cannot be factored any further!
Every other whole number can be broken down into prime number factors.
| | |
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| | It is like the Prime Numbers are the basic building blocks of all numbers. |
This idea can be very useful when working with big numbers, such as in Cryptography.
Cryptography
Cryptography is the study of secret codes. Prime Factorization is important to people who try to make (or break) secret codes based on numbers.
That is because factoring very large numbers is very hard, and can take computers a long time to do.
Unique
And here is another thing:
There is only one (unique!) set of prime factors for any number.
Example: the prime factors of 330 are 2, 3, 5 and 11
330 = 2 × 3 × 5 × 11
There is no other possible set of prime numbers that can be multiplied to make 330.
This idea is so important it is called the Fundamental Theorem of Arithmetic.
Prime Factorization Tool
OK, we have one more method ... use our Prime Factorization Tool that can work out the prime factors for numbers up to 9007199254740991.
370, 1055, 1694, 1695, 1696, 1697
Prime and Composite Numbers Prime Numbers Chart Prime Color Chart Prime Factorization Tool Divisibility Rules
Copyright © 2024 Rod Pierce |
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Comparison of a Combined Nontreponemal (VDRL) and Treponemal Immunoblot to Traditional Nontreponemal and Treponemal Assays
Alicia A Franken
Joyce H Oliver
Christine M Litwin
Correspondence to: Christine M. Litwin, Department of Pathology, Medical College of Georgia, Georgia Regents University, 1120 15th Street, Augusta, GA 30912. E‐mail: clitwin@gru.edu
Corresponding author.
Received 2012 Nov 15; Accepted 2013 Nov 12; Collection date 2015 Jan.
Abstract
Background
Serology is the mainstay for the diagnosis and management of patients with syphilis. Newer technologies such as immunoblotting are now available for the diagnosis of syphilis.
Methods
A commercial IgM/IgG immunoblot assay that detects both nontreponemal (VDRL‐Venereal Disease Research Laboratory) and treponemal antibodies was compared with standard nontreponemal and treponemal assays. The immunoblot and T. pallidum particle agglutination assay (TP‐PA) were performed on 198 samples. Ninety‐seven samples were Rapid plasma reagin (RPR)‐positive and one hundred one were RPR‐negative. Positive RPR samples were titered by VDRL.
Results
The agreement, sensitivity, and specificity of the IgM/IgG VDRL results of the immunoblot compared to RPR were 74.2% (95% CI: 67.2–80.2), 77.3% (95% CI: 70.2–83.4), and 71.3% (95% CI: 64.4–77.1), respectively. The agreement, sensitivity, and specificity of the IgM/IgG treponemal immunoblot compared to TP‐PA were 100% for all parameters, if the ten equivocal results were not used in the calculation.
Conclusion
The treponemal portion of the ViraBlot IgM/IgG immunoblot compared well with the treponemal confirmation assay and could be a useful supplemental method to fluorescent treponemal antibody or TP‐PA for the confirmation of syphilis. The addition of the detection of nontreponemal antibodies to the immunoblot assay, however, may not be of added benefit to the overall assay, due to decreased sensitivity and specificity compared to standard assays.
Keywords: treponemal, nontreponemal, syphilis, immunoblot, RPR
INTRODUCTION
Syphilis, a chronic infection caused by Treponema pallidum, is classified into multiple, distinct disease stages: primary, secondary, latent (early and late), and tertiary 1, 2. Serologic laboratory testing for syphilis continues to be the standard tool for the diagnosis of all stages of syphilis infection. The classic diagnostic algorithm includes a nontreponema‐based screening test (RPR‐Rapid plasma reagin, VDRL‐Venereal Disease Research Laboratory) and a treponemal‐based confirmatory assay 3, 4. Confirmatory assays include the fluorescent treponemal antibody absorption test (FTA‐ABS) and the T. pallidum particle agglutination assay (TP‐PA) 2. These serum‐based assays have been shown to have close to 100% sensitivity and specificity in secondary disease; however, they have diagnostic limitations in cases of early and late disease states 4. New diagnostic assays such as Western blot testing have been developed commercially in an effort to eliminate these and other issues seen with conventional testing. Western blot assays have shown promise in previous studies; however, they still have not received FDA approval 5.
A number of different Western blot and Immunoblot assays have been developed for the diagnosis of syphilis. Several studies have identified antibodies to immunodeterminants with molecular masses of 15, 17, 44.5, and 47 kDa that appear to confirm a diagnosis of syphilis 6. Western blot assays have also been created to specifically measure the IgM and IgG response to their immunodeterminants. IgG‐specific assays appear to be at least as sensitive and specific as the conventional FTA‐ABS test and TP‐PA 5, 7. The IgM Western immunoblots have shown promise with the diagnosis of acute disease and may prove to be a useful tool for the diagnosis of congenital infection. There are now commercially available immunoblots that use both nontreponemal (VDRL) and cloned recombinant treponemal antigens to help in the diagnosis of acute versus past syphilis infection.
In this study, we compared the results of a commercial IgM and IgG Western immunoblot for the detection of VDRL and treponemal antibodies to the results of traditional nontreponemal (RPR and VDRL) and treponemal (TP‐PA) serum assays. These results were subsequently compared to the patient's clinical history.
METHODS
Human Sera
A total of 198 human serum samples sent to Georgia Health Sciences University (GHSU) immunology laboratory for syphilis testing were collected from January 2011 to June 2012. Ninety‐seven RPR‐positive and 101 RPR‐negative samples were collected consecutively. Procedures were followed in accordance with ethical standards established by GHSU in accordance with the Helsinki Declaration of 1975. The protocol used was approved by the GHSU Institutional Review Board (no. 549) to meet the Health Information Portability and Accountability Act guidelines. Specimens were stored at −20°C until testing and then stored at 2 to 8°C.
Nontreponema‐Based Testing
All samples were tested by RPR with the 18 mm circle quantitative card test according to the manufacturer's protocol (Becton Dickinson & Co, Sparks, MD). Any samples reactive by RPR were titered by VDRL. The VDRL assay was run according to the manufacturer's protocol. Samples were reported as nonreactive, weakly reactive, or with the highest titer dilution that produced a reactive result.
Treponema‐Based Testing
All samples were tested by Serodia TP‐PA (Fujiribio Diagnostics, Inc, Seguin, TX) a gelatin particle agglutination assay. This assay was performed according to the manufacturers’ protocol. The results were read as nonreactive or reactive.
Nontreponemal and Treponemal Immunoblot Testing
All samples were tested using the Treponema ViraBlot IgG test kit and Treponema ViraBlot IgM test kit (ViraMed Biotech AG, Planegg, Germany). Each assay was performed according to the manufacturer's protocol. In brief, 20 μl of the sample diluted in 1.5 ml of diluent/wash buffer was incubated with one test strip for 30 min at room temperature. The strips were then washed three times for 5 min. Each strip was then incubated with 1.5 ml of diluted conjugate (IgG‐ or IgM‐specific alkaline phosphatase anti‐human conjugate) at room temperature for 15 min. The strips were then washed three times for 5 min with diluent/wash buffer and subsequently washed for 1 min with 1.5 ml of distilled water at room temperature. Then, the strips were incubated for at least 5 min with 1.5 ml of chromogen/substrate solution by rocking at room temperature. The reaction was, then, stopped after the control band appeared by decanting and washing three times with distilled water. The strips were dried and interpreted. The test was deemed valid if the function control band and the appropriate conjugate control band (IgM or IgG) were clearly visible (Fig. 1).
Figure 1.
VDRL and Treponemal immunoblot. The test was deemed valid if the serum control band and the appropriate conjugate control band (IgM or IgG) were clearly visible. Each of five VDRL bands were assessed for intensity compared to the cutoff control and assigned a number of units. For the Treponema bands (p47, p44.5, p17, p15), band intensity less than the cutoff control band was considered a weak band; band intensity equal to or greater than the cutoff control band was considered a reactive band.
For both the IgG and IgM VDRL interpretation, each of five VDRL bands were assessed for intensity and assigned a number of units (Fig. 1). If the VDRL band was absent or if the intensity was lower than the cutoff control band, it was assigned a value of 0 units. If the VDRL band intensity was equal to or greater than the cutoff control band, then it was assigned a value of 1 unit. If the VDRL band intensity was much greater than the cutoff control, then it was assigned a value of 2 units. All of the units for each of the five VDRL bands were added together to calculate the total VDRL ViraBlot units. The ViraBlot units were then interpreted as negative, low‐reactive, medium reactive, or high‐reactive (Table 1).
Table 1.
Interpretation of VDRL1‐VDRL5 Intensities in VDRL ViraBlot Units
| VDRL ViraBlot units | Interpretation |
--- |
| 0 | VDRL‐negative, no lipoid antibody activity. |
| 1–2 | VDRL‐reactive, low lipoid antibody activity. |
| 3–6 | VDRL‐reactive, medium lipoid antibody activity. |
| 7–10 | VDRL‐reactive, high lipoid antibody activity. |
For the interpretation of the treponema‐specific portion of the blot, band intensity lower than the cutoff control band was considered a weak band. Band intensity equal to or greater than the cutoff control band was considered reactive. Interpretation of the IgM and IgG Western immunoblot reactivity as negative, equivocal, or positive is presented in Tables 2 and 3, respectively.
Table 2.
Interpretation of IgM Treponema‐Specific Bands
| IgM bands | Result | Interpretation |
---
| No bands or one reactive band from p44.5 or one to four weak bands. | Negative | No specific antibodies against Treponema pallidum detectable. |
| One reactive band from p47, p17, or p15 or one reactive band p44.5 together with one to three weak bands. | Equivocal | T. pallidum infection is suspected. Check a second sample after 2–3 weeks. |
| At least one reactive band from p47, p17, p15 together with one to three weak bands. | Positive | IgM antibodies against T. pallidum are detectable. An infection with T. pallidum is very likely. |
Table 3.
Interpretation of IgG Treponema‐Specific Bands
| IgG bands | Result | Interpretation |
---
| No bands or one to three weak bands or one reactive band or one reactive band together with one weak band. | Negative | No specific antibodies against Treponema pallidum detectable. |
| One reactive band together with two to three weak bands or four weak bands. | Equivocal | T. pallidum infection is suspected. Check a second sample after 2–3 weeks control. |
| At least two reactive bands. | Positive | IgG‐antibodies against T. pallidum are detectable. An infection with T. pallidum is very likely. |
Statistical Analysis
Comparison of the RPR, TP‐PA, and syphilis history results with the nontreponemal and treponemal immunoblot were analyzed using a Yates’ corrected Chi‐square test to determine the agreement, clinical sensitivity, clinical specificity, and 95% confidence intervals for sensitivity and specificity. Spreadsheets and additional calculations were performed using an Excel spreadsheet (Microsoft Corp., Redmond, WA).
RESULTS
The IgM/IgG VDRL Immunoblot Compared to RPR With Titer to VDRL
A total of 198 specimens were tested by RPR and IgM/IgG immunoblot. Ninety‐seven specimens were RPR‐positive and one hundred one were nonreactive. The total agreement between RPR and the IgM/IgG VDRL immunoblot was 74.2% (95% CI: 67.2–80.2). Seventy‐five RPR reactive samples (70 TP‐PA reactive, 5 TP‐PA nonreactive) were positive on the IgM /IgG VDRL for a sensitivity of 77.3% (95% CI: 70.2–83.4). Of the 101 nonreactive RPR specimens, 29 were positive on the IgM/IgG Immunoblot VDRL for a specificity of 71.3% (95% CI: 64.4–77.1%). Of the 22 RPR‐positive/VDRL immunoblot negative discrepant samples, 17 were TP‐PA‐positive and 5 were TP‐PA‐negative. Of the 29 RPR‐negative/VDRL immunoblot positive discrepant samples, 3 were TP‐PA‐positive and 26 were TP‐PA‐negative.
When the IgM/IgG VDRL immunoblot was compared to standard VDRL testing, there was an increase in sensitivity but a decrease in specificity. The total agreement was 73.2% (95% CI: 66.7–78.0). The sensitivity was 84.7% (95% CI: 75.8–91.3%) and specificity was 66.7% (95% CI: 61.6–70.4%).
The units of reactivity for both the IgM and IgG VDRL immunoblots were added together and plotted against the inverse VDRL titer on a semilogarithmic scale (Fig. 2). A weak correlation was seen with an r
2 value of 0.160. In samples with VDRL titers of less than 64, the correlation was stronger with an r
2 value of 0.523.
Figure 2.
Semilogarithmic plot of the inverse VDRL titer value versus the sum of the IgM and IgG Units on VDRL immunoblot.
The IgM/IgG Treponemal Immunoblot Compared to TP‐PA
Of the 93 samples that were TP‐PA‐reactive, 85 samples were positive on either the IgM or the IgG treponemal immunoblot. Eight TP‐PA‐positive samples were equivocal on either or both the IgM and IgG immunoblot. There were 105 samples that were TP‐PA nonreactive. All except two samples were negative on the IgM/IgG treponemal immunoblots. The two samples were equivocal on the IgG treponemal immunoblot and negative for IgM. If equivocal results are not used in the calculation, the sensitivity, specificity, and agreement would all be 100% (95% CI: 97–100) for the combined IgM/IgG immunoblot. If the IgG immunoblot alone is compared with the TP‐PA, then the sensitivity, specificity, and agreement would be 98.8% (95% CI: 95.3–98.8), 98.1% (95% CI: 97.1–100), and 99.5% (95% CI: 96.2–99.5), respectively. If equivocal samples are counted as discrepants (eight TP‐PA‐positive, two TP‐PA‐negative) then the sensitivity, specificity, and agreement would be 91.4% (95% CI:
86.7–93.2), 98.1% (95% CI: 93.9–99.7), and 94.9% (95% CI: 90.6–96.6).
There were 21 (22.6%) positive and 5 (5.3%) equivocal IgM treponemal immunoblots of the 93 TP‐PA‐reactive samples. Twenty of the twenty‐one positive IgM immunoblots were also positive on IgG. One sample was positive on IgM and equivocal on IgG and another was equivocal on IgM immunoblot, but negative on IgG immunoblot. Otherwise, no other negative IgG immunoblot result was observed for all 93 samples.
IgM VDRL Immunoblot Results in New Infection
We compared the results of the IgM VDRL portion of the immunoblot to laboratory and clinical patient history of syphilis infection for the ability to discriminate between patients with a new infection and patients with a history of past infection. Past infection was defined as a diagnosis of syphilis at least 6 months prior to testing. Of the 42 patients with no known prior history of syphilis but reactive on TP‐PA, 26 were positive on the IgM VDRL immunoblot, for a sensitivity of 61.9% (95% CI: 49.7–73.6) (Table 4). Twenty‐three of fifty‐three with a past history syphilis were negative by the IgM immunoblot for a specificity of 43.4% (95% CI: 33.7–52.6). The total agreement was 51.6% (95% CI: 40.8–61.9).
Table 4.
VDRL and Treponemal IgM Results in New Syphilis Infection
| VDRL IgM+ | | | | VDRL IgM− | | |
--- --- ---
| History of syphilis (total) | Trep. IgM+ or +/− | Trep. IgM – | Total VDRL IgM+ | Trep. IgM+ | Trep. IgM− | Total VDRL IgM− |
| No (42) | 15 | 11 | 26 | 2 | 14 | 16 |
| Yes (53) | 7 | 23 | 30 | 2 | 21 | 23 |
+, positive; −, negative; +/−, equivocal; Trep., treponemal.
IgM Treponemal Immunoblot Results in New Infection
We compared the results of the IgM treponemal portion of the immunoblot to laboratory and clinical patient history of syphilis infection for the ability to detect new infection. Of the 42 total patients with no known prior history of syphilis, 17 were positive or equivocal on the IgM immunoblot (13 VDRL IgM‐positive and 2 VDRL IgM‐negative) for a sensitivity of 40.5% (95% CI: 29.2–50.0). Forty‐four of fifty‐three patients with a past history of syphilis were negative for IgM and positive or equivocal for IgG, for a specificity of 83.0% (95% CI: 74.1–90.5%). The total agreement was 64.2% (95% CI: 54.2–72.6).
DISCUSSION
We previously studied three IgG T. pallidum Western blots and immunoblots compared with FTA‐ABS/TP‐PA results, including the ViraBlot immunoblot assay, Virotech immunoblot assay (Genzyme Virotech GmbH), and the Marblot traditional Western blot strip test system (MarDx Diagnostics) 5. The ViraBlot assay, however, also contains a VDRL (nontreponemal) portion on both the IgM and the IgG immunoblot strips, which was not analyzed in that previous study. To completely evaluate the entire ViroBlot immunoblot assay system, this study also includes an analysis of the IgM treponemal antibody results and the IgM and IgG VDRL antibody response in comparison with standard RPR, VDRL, and TP‐PA. Clinical information regarding previous history of syphilis infection and lab results was also available in this study.
In our previous study, the overall agreement, sensitivity, and specificity of the IgG ViraBlot treponemal assay were 97.0%, 95.5% (95% CI: 90.4–97.9), and 97.8% (95% CI: 95.2–99.0%), respectively. This study showed 100% (95% CI: 97–100) agreement, sensitivity, and specificity when compared to TP‐PA with the combination of the IgM/IgG treponemal immunoblot results, when samples with only equivocal results were excluded. The IgG ViraBlot treponemal assay alone compared to TP‐PA had an overall agreement, sensitivity, and specificity of 99.5% (95% CI: 96.2–99.5), 98.8% (95% CI: 95.3–98.8), and 100% (95% CI: 97.1–100), respectively.
In a recent study by Binnicker et al. 8 of seven different treponema‐specific tests that included an evaluation of the IgG ViraBlot assay, the agreement, sensitivity, and specificity of the IgG ViraBlot were 98.0%, 96.8% (95% CI: 90.6–99.3), and 98.6% (95% CI 95.7–99.7) when compared to a consensus panel. There were four discrepant samples identified in that study. Two out of ninety‐four consensus positive samples were IgG immunoblot negative and two out of two hundred eight consensus negative samples were IgG immunoblot positive.
Using 95% confidence intervals, the sensitivity and specificity of the IgG ViraBlot assay in this study are similar to those found with the other two studies. Since the addition of the IgM immunoblot results to the IgG results did not significantly change the sensitivity and specificity of the assay, performing only the IgG ViraBlot assay for the confirmation of treponemal infection would probably be sufficient under most clinical circumstances. The sensitivity, however, may potentially be increased if the IgM immunoblot is also performed, especially if early primary syphilis is suspected.
In our previous study, there were no equivocal results seen with the ViraBlot assay, but equivocal results were observed for the Virotech and Marblot assays in 2.5 and 12.5%, respectively. In this study, equivocal results were observed in ten IgG immunoblots (5.5%) and five (2.5%) IgM immunoblots. In the study by Binnicker et al. 8, the equivocal rate for the IgG ViraBlot was 1.3%. The difference in the reporting of equivocal results between our two studies may be due to a change in the laboratory personnel performing the test between the two different laboratories or the type of population sampled. The previous study was performed at a major reference laboratory, whereas the present study was performed in a university hospital laboratory.
The IgM and IgG antibody response to T. pallidum has been studied at length in humans and experimentally infected animals 9. Antitreponemal IgM antibodies are produced at about 2 weeks after infection, while IgG antibodies are produced at about 1 month after exposure. Theoretically, in very early syphilis infection, there is the potential that a patient may be positive on the IgM treponemal immunoblot but negative on the IgG immunoblot. Yet, with the exception of one sample, all of the positive TP‐PA samples were at least equivocally reactive with the IgG treponemal immunoblot.
After therapy of the primary and secondary stages of syphilis, T. pallidum IgM antibodies decrease quickly, and have been found to be absent within 6–12 months. A number of studies have suggested that decreasing treponemal IgM levels indicate adequate treatment 10. The absence of IgM has been demonstrated in 84% of patients with syphilis that were treated in the past 11. In our study, we obtained similar results, with 83.0% of patients with a past history of syphilis being negative for IgM but positive or equivocal for IgG. The sensitivity of the IgM immunoblot to discriminate between a recent infection and a past infection was 40.5%. The low sensitivity may be the result of a delay in clinical and laboratory detection of greater than 6 months post initial infection.
When the combined IgM/IgG results of the VDRL immunoblot were compared to RPR reactivity, the sensitivity was low at 77.3% with 17 of the 22 discrepant samples confirming positive on TP‐PA. When the combined IgM/IgG results were compared to VDRL reactivity, the sensitivity was 84.7%. In general, the VDRL and RPR quantitative results cannot be compared directly because RPR titers are often slightly higher than VDRL titers 4. A similar conclusion can be made with the VDRL IgM/IgG immunoblot. The VDRL IgM/IgG immunoblot did not compare well with the VDRL titer except in samples with higher titers.
The VDRL IgM immunoblot was positive in 61.9% of new infections. In patients with a past infection, 56.6% were also positive with the VDRL IgM immunoblot, indicating that an IgM nontreponemal response will occur during reinfection with T. pallidum. It is interesting to note that only two samples that were negative on IgM VDRL immunoblot were positive on the IgM treponemal immunoblot.
Overall, the treponemal portion of the ViraBlot IgM/IgG immunoblot compares well with other treponemal confirmation assays and could be a useful supplemental method to FTA or TP‐PA for the confirmation of syphilis. The IgM/IgG VDRL portion of the immunoblot assay, however, did not compare well either to the RPR or VDRL standard assays and may not be sensitive enough to use as a monitor for syphilis treatment response. The lack of specificity of the VDRL portion of the immunoblot was also problematic. In conclusion, the addition of the detection of nontreponemal antibodies to the immunoblot assay may not be of added benefit to the overall assay, due to decreased sensitivity and specificity compared to standard assays.
Grant sponsor: Department of Pathology, Georgia Health Sciences University.
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9544 | https://byjus.com/physics/derivation-of-terminal-velocity/ | What is Terminal Velocity?
Terminal velocity is defined as the highest velocity attained by an object falling through a fluid. It is observed when the sum of drag force and buoyancy is equal to the downward gravity force acting on the object. The acceleration of the object is zero as the net force acting on the object is zero.
How to find Terminal Velocity?
In fluid mechanics, for an object to attain its terminal velocity, it should have a constant speed against the force exerted by the fluid through which it is moving.
The mathematical representation of terminal velocity is:
vt is the terminal velocity, m is the mass of the falling object, g is the acceleration due to gravity, Cd is the drag coefficient, 𝜌 density of the fluid through which the object is falling, and A is the area projected by the object.
Terminal Velocity Derivation
Deriving terminal velocity using mathematical terms according to the drag equation as follows:
Where b is the constant depending on the type of drag
(integrating the equations)
Where,
After integration,
After substituting for vt
Therefore, above is the derivation of terminal velocity.
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Related Physics Articles:
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| Acceleration Due to Gravity | Universal Law Of Gravitation |
| What is a Frictional force? | Fluid Friction |
Frequently Asked Questions – FAQs
Q1
When does terminal velocity exist?
When the speed of a moving object is no longer increasing or decreasing; the object’s acceleration (or deceleration) is zero.
Q2
What is terminal velocity?
Terminal velocity is defined as the highest velocity attained by an object that is falling through a fluid.
Q3
Who discovered terminal velocity?
Galileo discovered terminal velocity.
Q4
Does terminal velocity exist in a vacuum?
In vacuum since there is no drag force, the terminal velocity does not exist.
Q5
How does terminal velocity work?
Terminal velocity, steady speed achieved by an object freely falling through a gas or liquid. An object dropped from rest will increase its speed until it reaches terminal velocity; an object forced to move faster than its terminal velocity will, upon release, slow down to this constant velocity.
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9545 | https://haas.berkeley.edu/wp-content/uploads/Understanding-IL-Playbook-3.pdf | UNDERSTANDING INCLUSIVE LANGUAGE: A FRAMEWORK CENTER FOR EQUITY, GENDER & LEADERSHIP THIS GUIDE WAS WRITTEN BY JULIA NEE AND GENEVIEVE MACFARLANE SMITH, WITH VALUABLE INPUT FROM ISHITA RUSTAGI, KELLIE MCELHANEY, AND LEANN QUASTHOFF. INVALUABLE SUPPORT - PARTICULARLY FOR SECTION 3 (INCLUSIVE LANGUAGE FRAMEWORK) - WAS PROVIDED BY A WORKING GROUP (SEE APPENDIX A. METHODOLOGY). WHILE WE SOUGHT TO INCLUDE A VARIETY OF PERSPECTIVES, THIS WORK MAY OVERREPRESENT CERTAIN VIEWPOINTS, AS ALL OF THE PEOPLE INVOLVED IN THE PROJECT WERE LIVING IN THE UNITED STATES OR UNITED KINGDOM AT THE TIME OF THE RESEARCH, AND ARE ALL SPEAKERS OF ENGLISH. ALL SHORTCOMINGS AND ERRORS ARE THE AUTHORS’ ALONE.
Table of Contents 1 4 7 7 9 13 14 14 15 18 18 19 29 29 31 32 EXECUTIVE SUMMARY WHY INCLUSIVE LANGUAGE MATTERS INCLUSIVE LANGUAGE FRAMEWORK Principles for inclusive language Key questions in assessing word choice Maintaining a growth mindset RECOMMENDATIONS Use metaphorical and other figurative language thoughtfully Learn to distinguish borrowing and appropriation Avoid language with harmful associations Honor how people identify themselves Understand inclusive language across different identities Consider intersectionality Work with others to adopt inclusive language together APPENDIX REFERENCES Using inclusive language is a key aspect of creating an environment where people from a diversity of backgrounds and life experiences feel comfortable contributing and feel they belong. Choosing words that are more inclusive over harmful alternatives can be an important step toward building a more inclusive environment, and can help support effective communication to diverse audiences.² Our inclusive language framework includes four guiding principles and five key questions that allow us to consider whether a term is inclusive in a given context (see table 1) –. The framework includes guiding questions to help us continually learn so that we can communicate more effectively and inclusively across a diversity of audiences. Note that each term should be evaluated in context: some terms may be inclusive in one context but not another. Center for Equity, Gender & Leadership 1 UNDERSTANDING INCLUSIVE LANGUAGE: A FRAMEWORK 1. EXECUTIVE SUMMARY Inclusive language is not about memorizing a fixed list of "good" and "bad" words. Rather, it’s about centering compassion in how we communicate and recognizing that inclusive language and communication is an ongoing process.
Because language is always changing and context-dependent, it is not possible to create a fixed list of words that “should” or “shouldn’t” be used. Instead of focusing on policing language use, we invite you to approach language with a mindset of being open to learning and unlearning, recognizing that context matters and language changes, and being willing to change behaviors and strive to do better. Ultimately, inclusive language is a choice that each of us can make: a choice to act in a more compassionate and thoughtful way towards our colleagues, friends, neighbors, and community members. GUIDING PRINCIPLES KEY QUESTIONS Inclusive language conveys respect to all people.
Inclusive language communicates a message effectively through precise language.
Inclusive language recognizes diversity.
Inclusive language involves continual improvement.
1.
2.
3.
4.
Does this term center dominant³ groups as the default and/or perpetuate harmful stereotypes? If yes: look for a more inclusive term.
Does this term have a harmful history or association? If yes: look for a more inclusive term Does this term contribute to communication in which everyone feels respected and seen? If no: look for a more inclusive term Does this term convey the intended meaning to all people precisely and effectively? If no: look for a more inclusive term.
Is this term the best we can find to communicate the message? If no: consider using a more inclusive term.
1.
2.
3.
4.
5. Center for Equity, Gender & Leadership 2 In addition to the framework, we share seven recommendations for inclusive language: Table 1. Inclusive Language Framework Center for Equity, Gender & Leadership 3 Use metaphorical and other figurative language thoughtfully This paper explains why inclusive language matters (section 2), outlines our inclusive language framework, and illustrates how it can be applied (section 3). In section 4, we delve into the seven specific recommendations for inclusive language. Our framework and recommendations are based on a systematic literature review, with further information on our methodology in Appendix A.
Learn to distinguish borrowing terms from other languages from appropriating terms Avoid language with harmful associations Honor how people identify themselves Understand inclusive language across different identities 1 2 3 4 4 5 4 6 Consider intersectionality 4 7 Work with others to adopt inclusive language Language can impact how we think about and approach problems.
Language and reality are mutually reinforcing : our language reflects the world around us and influences how we think and what we do. If we use harmful language (see box 1 for a definition), we may be encouraged to think in those same harmful terms. For example, using the term “illegal aliens,” which dehumanizes undocumented migrants, has been linked to increased violence against (perceived) immigrants, while use of more positive terms like “DREAMers” results in more positive views of undocumented migrants . Center for Equity, Gender & Leadership 4 2. Why Inclusive Language Matters BOX 1. WHAT IS HARMFUL LANGUAGE?
There are many types of harmful language, making “harmful language” difficult to define. We consider harmful language to be any unnecessary language that results in harms such as exploitation; mental, emotional, or physical distress; violation of privacy; damage to personal reputation; economic harm; decreased participation in civic or public life (i.e. silencing); increased tendency to hate, fear, discriminate against, or endorse violence against other people (including through dehumanization); or promoting falsehoods [10, p. 257]. We include the caveat, “unnecessary,” as some language can cause mental or emotional distress, but nevertheless is necessary. For example, learning that systemic racism continues to impact members of minoritized groups can be distressing for White people who may feel negative emotions such as guilt or shame when talking about racism. However, it is necessary to talk about and understand systemic racism in order to address it.
Use of imprecise language, like some metaphors, can also influence how we reason about or approach an issue. For example, in one study, when crime was described as a “virus,” respondents were more likely to recommend policies that addressed the root cause of crime (following the metaphor of removing a virus from an infected person), but when crime was described as a “beast,” they were more likely to recommend policies that focused on enforcement (following the metaphor of caging an animal) . As this study suggests, metaphors can impact our thinking and responses.
Metaphors may bias us from pursuing potentially beneficial solutions if those solutions don’t fit with the metaphor –. For example, if crime is viewed metaphorically as being a “beast,” we are more likely to pursue approaches that combat crime in ways that follow the metaphor and parallel combat against a beast (such as greater police enforcement). Even if other approaches (such as addressing root issues like poverty and police violence) could be more effective, they may be seen as less attractive options if they clash with the dominant narrative.
Language can affect wellbeing and life outcomes. Overtly harmful language such as slurs and hate speech (while not a focus of this brief) may be the most recognizable examples of harmful language, with immense impacts on individuals and communities. But oftentimes, harmful language is more subtle.
Verbal microaggressions,⁴ for example, can lead to negative impacts for individuals who are affected by them. Microaggressions — including verbal microaggressions — have been tied to lower physical and psychological health outcomes, including lower self-esteem, greater anxiety and depression, difficulty sleeping, tension headaches and backaches, extreme fatigue, elevated blood pressure, and diminished cognitive function –. They also negatively impact productivity , , particularly for people of color, LGBTQ, disabled, and other marginalized and minoritized populations , , . On the other hand, using inclusive language can result in profoundly greater life outcomes.
For example, using an individual’s pronouns correctly is associated with a 50% decrease in suicide among trans youth . Individuals have commented time and again that they experience negative effects when they encounter harmful language. For example, security analyst Brian Anderson noted how terms like “black hat,” which rely on metaphors Center for Equity, Gender & Leadership 5 linking blackness with badness , have impacted him and his family. He wrote on Twitter, “Black people have struggled for lifetimes with negative imagery associated with blackness. The studies are clear that children see goodness and beauty in whiteness, ugliness and bad in black. MY children have to deal with this just like our parents did before us” . Because words can cause harm, and because there are more inclusive alternatives available, we can use those more inclusive terms to acknowledge that we hear Anderson and others’ concerns, and to show that we value the inclusion and well-being of all individuals.
Using inclusive language can support personal and organizational growth.
Using inclusive language (which we define in section 3 below) presents an opportunity to improve communication by crafting a message that is easier for a broader audience to take up and understand. It is an opportunity for growth and caring for others through learning how they may interpret what is said .
Moreover, many companies have publicly committed to furthering diversity, equity, and inclusion within their teams and products. Using language that promotes equity and inclusion supports the company’s efforts. On the other hand, the harmful effects of non-inclusive language may be amplified by hierarchical employer/employee or teacher/student relationships in the workplace or academic setting . The presence of harmful language in workplace environments — including within datasets and codes — can contribute to alienation and may ultimately lead to inequitable workplace representation of people from different identity groups –. In the face of the potential harm that non-inclusive language can cause, we can instead strive to use more inclusive language to build an environment in which individuals holding any identities feel a sense of safety and belonging that allows everyone to participate equitably. Center for Equity, Gender & Leadership 6 What does it mean to use inclusive language and how might we identify harmful terms? Given that we each have our own experiences and perspectives, inclusive language might not mean using the exact same set of words in all contexts and for all people. This set of four guiding principles will help us to consider whether the words we are using might be harmful within a given context, and, if so, identify more inclusive alternatives.
Principle 1: Inclusive language conveys respect to all people .
Inclusive language recognizes the humanity of all people and allows for self-determination of how individuals wish to be identified . It invites people to participate in the conversation by respecting their identities , . Using words that affirm individuals’ identities is crucial. As Thomas writes, “The act of being seen and being named is empowering; it confirms the basic right to exist.” When we talk about people, we have the potential to impact how they are seen by others, and as a result, we take on responsibility when we represent others with our words . Inclusive language also promotes equal Center for Equity, Gender & Leadership 7 3. Inclusive Language Framework 3.1 Principles for Inclusive Language Language is a powerful tool that can be used to bring people together or to foment divisions. In all environments — including technical environments — the way we communicate can have powerful effects on others, as described above. Inclusive language is an important step in building inclusive environments. In this section, we present an overview of our inclusive language framework, designed to help identify potentially un-inclusive language as well as more inclusive alternatives. opportunities by avoiding stereotypes and limiting conceptualizations of people’s potentials , . Stereotypes and limiting conceptualizations of people’s potential can be conveyed through direct uses, such as gender-exclusive terminology, and indirect uses, such as through metaphors that draw harmful connections. Inclusive language is also context-dependent. While it may be important to use a particular word or phrase in one context, that term may be considered harmful in another. As an example, it may be crucial to use one set of pronouns with trans individuals in private spaces where they feel comfortable, and avoid those pronouns in public spaces where they do not feel safe if their trans identity is known.
Principle 2: Inclusive language communicates a message effectively through precise language , , . Using precise terms that distinguish between multiple experiences is more inclusive than using vague terms. For example, with respect to age, terms like “elderly” imprecisely categorize the diverse experiences of older adults. Instead, we could use more precise terms that better describe the demographic of focus, whether retirees, people with grandchildren, or people aged 60+ , . We can also reflect on the use of metaphors and other figures of speech to ensure that the comparison drawn by that language is accurate and does not form harmful or misleading connections –, , . Using precise, inclusive language also helps improve the efficacy of the message, as it may reduce the strain placed on individuals who are harmed by non-inclusive language and instead allow them to focus on the content of the message rather than its wording .
Principle 3: Inclusive language acknowledges diversity , . Instead of minimizing differences and homogenizing the experiences of a diversity of individuals, we can be sensitive to the differences each of us bring as a result of our identities and life experiences . For example, instead of using imprecise terms like “non-White” (through which many different groups are homogenized into a unit that contrasts with the dominance of Whiteness) or “non-native Center for Equity, Gender & Leadership 8 English speaker” (through which many people with different linguistic backgrounds are grouped together) we can use more precise terms that recognize the unique experiences of individuals and groups we are talking about.
We can refer to specific identity groups (Black, Indigenous, Asian American, Pacific Islander, etc.) or use more precise descriptors of linguistic backgrounds (Spanish-speaker, adult English-learner, etc.). Acknowledging diversity also requires us to suspend our assumptions: we should learn about an individual's unique experiences rather than forming assumptions based on their identities . Principle 4: Inclusive language involves continual improvement. Language and language preferences will continue to change and shift over time, and we must be ready to learn and shift with those changes. Approaching the topic with a sense of humility can help . While we can continue to learn more and practice using inclusive language, there will always be more to learn, and we are likely to make mistakes along the way. We should be ready to be wrong and be willing to make changes to improve. What is important is that we acknowledge any harm that has been done and learn from those mistakes. Moreover, it can be helpful to approach adoption of inclusive language from a place of empathy and generosity. Instead of focusing on the effort required to adopt new terms, we can seek to understand why terms may be harmful for others and focus on the benefits of creating more inclusive spaces through our efforts to use more inclusive language. Inclusive language is about taking advantage of opportunities to use more effective words and phrases in our communication.
Center for Equity, Gender & Leadership 9 3.2 Key Questions in Assessing Terms Our guiding questions can help determine whether a term is inclusive. These questions can be applied to language used across contexts, whether in the workplace, in code or datasets, or in interpersonal communication. In considering these questions, think not just about the specific term, but also the context in which it is used. In some cases, terms may be inclusive in some contexts, but not in others.
Instead of using terms like “non-White,” which centers whiteness as the default and groups a diverse set individuals and experiences into a single, imprecise category, use specific terms for specific groups or experiences.
This acknowledges the unique and intersectional experiences of different individuals and groups and pushes back against the hegemony of dominant groups.
Dominant narratives often reflect the perspectives of people with power, but they may be misleading. For example, describing people who owned slaves as “masters” rather than “enslavers” may minimize the harm of what they did. Other terms may reflect harmful stereotypes. Referring to a “chairman” of the board (rather than a “chair” or “chairperson”), for example, reflects a stereotype that this is a role for a man, and not for people of other genders. Instead of replicating the dominant perspective, we can use more inclusive language that makes other perspectives visible.
Using terminology that has racist origins or is associated with negative stereotypes — whether used in coding or in casual conversation — can lead to people who encounter such harmful language feeling disrespected or unincluded. We consider any language which has an unnecessary negative impact on individuals to be harmful.
Language has an impact on society. Language that contains harmful biases or stereotypes can perpetuate them and result in negative outcomes for individuals who are described with biased or stereotypical language. Avoid using language that contains harmful stereotypes or metaphors.
Use of non-inclusive terminology or categories when collecting or 1. Does this term center dominant groups as the default and/or perpetuate harmful stereotypes?
2. Does this term have a harmful history or association?
3. Does this term contribute to communication in which everyone feels respected and seen?
Center for Equity, Gender & Leadership 10 Center for Equity, Gender & Leadership 11 If you are using a term other than self-reported identity, be clear about how and why you are using that term. For example, use of perceived identity categories (rather than self-reported identity categories) may be important in studies examining bias. If you report perceived identity, note that in your work and explain why. In some cases, members of the same group may have different preferences in how to describe that group.
Again, be transparent in explaining your decision to use one term or another. Using terms that have ties to specific identity groups to describe broader or more general experiences can also be harmful, as it minimizes the diversity of individuals’ different experiences. For example, referring to a group of friends as a “tribe” can minimize the distinct histories of Indigenous tribes and can contribute to an environment where members of Indigenous tribes may feel unseen.
Precise terms eliminate ambiguity. Terms and phrases with multiple meanings are imprecise because their meaning within a given context may be ambiguous between those multiple meanings (this has been argued, for example, with respect to the terms master/slave in coding, which have been used to describe several different dependency relationships within coding). They also avoid overgeneralizations. For example, stating that “Women are socialized to be caregivers” may be an overgeneralization; “Many individuals who are raised as girls are socialized to be caregivers” may be more precise. By clarifying that many (but not displaying data (such as presenting only a gender binary or using racial categories that don’t reflect the terms preferred by members of those groups) may result in audiences feeling unseen and disrespected. Replace outdated and harmful terms with terms that are supported by community members. 4. Does this term convey the intended meaning to all people precisely and effectively?
Center for Equity, Gender & Leadership 12 Effective communication means that the audience is able to understand the intended meaning of the author, speaker, or signer. In some cases, the language used may cause certain audience members to feel harm or distress; this may decrease the efficacy of the communication, as those affected may have their attention diverted from the message in order to consider and attend to the impacts of the harmful language used. Using more inclusive language can reduce this communicative barrier.
If there is a term that is more precise or effective, or which generates respect, honors self-identification, promotes equal opportunities, and acknowledges diversity, we can use that term. We can also create new terms to replace problematic terms if no suitable replacements exist.
If you are feeling hesitant to use a new term, reflect on your reasons. The motivation behind shifting to more inclusive language use is to create a more inclusive society. Change can be uncomfortable, but it is necessary if we wish to create a more inclusive future.
all) individuals have been raised this way, we are more inclusive of individuals whose experiences do not fit the description. Moreover, we can specify when we are referring to individuals who were raised as girls so that we are inclusive of people who were raised as girls but who do not identify as women.
5. Can we find an alternate term that does a better job?
As you ask yourself the five questions above, follow the flowchart in Figure 1. FIGURE 1: INCLUSIVE LANGUAGE FRAMEWORK Does this term center dominant groups as the default, and/or perpetuate harmful stereotypes?
Consider Context Does this term contribute to communication in which everyone feels respected and seen?
Does this term have a harmful history or association?
Does this term convey the intended meaning to all people precisely and effectively?
No Yes Continue using the term.
No Center for Equity, Gender & Leadership 13 Set Intentions Start from a place where you feel ready and open to learning. Then ask the following questions: Yes Center Inclusion Can we find an alternate term that does a better job at communicating the message?
No Yes Implement Positive Change Use the term with caution or decide not to use it. Consider explaining your choice to your audience.
Use the more inclusive term.
We want to acknowledge that this process may cause some discomfort. On the one hand, you may feel harmed by some of the terms that come up in these discussions. On the other hand, you may find that you have used terms that others perceive to be harmful — even if it was not your intention. Realizing that some terms that you use often are harmful may feel upsetting, and changing your everyday language patterns may feel uncomfortable. We encourage you to approach this process with a growth mindset. Instead of focusing on reasons to maintain the status quo, think of the opportunities for increased inclusivity that may come with changes in how we use language. 3.3 Maintaining a Growth Mindset While using metaphorical and other figurative language can help convey a message in a way that can be easily understood by the audience, use such language thoughtfully –. When metaphors are invoked, the full context of the comparison is brought to bear on the situation, and the audience’s understanding of both elements being compared may be changed as a result . For example, using a phrase like “infant mortality” to describe the rate of defective, newly-assembled electronics draws a comparison between human infants and inanimate machines, and in the process both personifies machines and dehumanizes infants . As mentioned above in the example of different policies being endorsed whether crime is described as a “virus” or a “beast,” metaphors have been shown to influence the audience’s reactions to the same phenomenon . This shows that metaphors can shape how we think about issues and events, and as a result, it is important to Center for Equity, Gender & Leadership 14 4.
4.1. In this section, we briefly lay out some recommendations for using inclusive language, including guidance on metaphorical and appropriative language, language with harmful associations, and language to talk about different identities and their intersections. We close with some suggestions for building an inclusive language community that can help support inclusive language adoption within an organization.
Recommendations for Inclusive Language Use metaphorical and other figurative language thoughtfully Center for Equity, Gender & Leadership 15 critically examine metaphors within our language. This is particularly true given that certain types of metaphors — particularly militaristic and technological metaphors — are more prevalent (at least within US discourse) and reflect racial and gender stereotypes that may be perpetuated through their use . Even if a metaphor is intended to be invoked in a specific context, the speaker, writer, or signer does not have complete control over how that metaphor is perceived by the audience ; the audience may bring their own experiences and prejudices to the metaphor and draw additional conclusions. Moreover, for audiences with different cultural and linguistic backgrounds, metaphorical, idiomatic, or other non-literal phrases may be unnecessarily confusing or isolating if they are not part of audience members’ backgrounds or understandings . While metaphorical and figurative language can play a role in illuminating complex topics for non-specialist audiences, it is important to think carefully about how all aspects of potential metaphorical comparisons might play out and what impacts those associations could have.
4.2. Learn to distinguish borrowing and appropriation Language is constantly changing and evolving, and this includes adopting and using words from other languages. For example, many words describing new technologies come from the languages spoken by the people who invented them, such as karaoke and anime (both borrowed from Japanese). While borrowing is a regular part of language change, in some situations a dominant group takes linguistic elements from a group with less power and uses them to their own benefit. This results not in borrowing but in appropriation. Often, the dominant group also simultaneously devalues members of the minoritized group who use the exact terms that the dominant group has appropriated and benefits from [38, p. 28]. For example, White people often adopt phrases that were coined by speakers of African American English — including terms such as on fleek, bae, and squad — in order to portray themselves as “cool,” but when Black speakers use the same phrases, they are often judged as “uneducated” or “unprofessional” –. Whether or not a term is being borrowed or appropriated is highly context specific, and terms might be considered Center for Equity, Gender & Leadership 16 harmful and appropriative in one context but supportive and inclusive in another. Remember that language is dynamic and context dependent, so it can be difficult or impossible to determine whether using a word involves appropriation without taking into consideration the context of use. Moreover, in seeking to be more inclusive in language use, focus on impact: whether or not we know the true origin of a term, if its use reinforces harmful power dynamics, it would be more inclusive to use an alternate term.
There can be harmful material consequences to linguistic appropriation.
Appropriated terms show up in advertisements, music, and the other media that result in financial gain for members of the dominant group , . This can be especially problematic when appropriated terms are used to reference stereotypical features of a marginalized group. For example, African American English terms have been used in advertising to evoke stereotypes of hypermasculinity that are associated with Black men; use of those terms then reinforces harmful stereotypes .
Similar patterns can be seen in the appropriation of what Jane Hill calls mock Spanish: use of phrases like no problemo that are loosely based on the Spanish language. English speakers who use mock Spanish may gain status through being perceived as humorous, cosmopolitan individuals who have access to a second language, but Spanish speakers who use similar phrases may be perceived as “lazy” individuals who are “incapable” of using English .
Speakers of Indigenous languages in the United States face similar problems.
Words taken from Indigenous languages, such as the Massachusett word “mugwump” meaning “war leader,” have been appropriated into English and assigned new, derogatory meanings that in turn make them unappealing for use within their original contexts [38, p. 164]. This type of “language theft” has negative impacts on speakers of minoritized languages and language varieties who may no longer feel comfortable using their own languages.
While borrowing words and phrases from different languages and communities Center for Equity, Gender & Leadership Is this a term that originated in a community that I’m a part of? If not, would members of the community that coined this term be treated similarly to me if they used the term in this context?
If the term is not from a community that you’re a part of, and you’ll gain some benefit from using the term that members of the community where the term originated would not, this is likely appropriation. Think about whether using the term is useful and necessary, and consider using an alternate term. What is the response of the community where this term originated when I use it in this context? Are they upset by my use of the term?
If community members are upset by your use of the term, consider replacing it with an alternative that does not cause a negative reaction.
In some contexts, members of a community may want you to use a particular term while others may feel upset. There are no absolute rules about when to use or avoid such terms. In many cases, there will be no clear answer, and it may be useful to open a discussion with community members, learn more about how they react to your use of a term, and consider why you use it in that context. is common, it is important to consider not only the words or phrases being borrowed, but also the larger power dynamics at play within a specific context.
When thinking about whether a term is appropriative in a particular context, you might consider: 1.
2.
If the use of a word or phrase is well-regarded only if used by the dominant group (but not by the minoritized group from which it was taken), it is likely appropriation. In that case, consider using a different term and also think critically about why the term may be highly valued in one context but not another. Changing our language to be more inclusive is one important step toward greater equity, but so is shifting the balance of social and economic power between different groups.
17 Center for Equity, Gender & Leadership 18 4.3. Avoid language with harmful associations Identity-based slurs and phrases that have harmful or derogatory associations should be avoided. The use of such harmful language has been shown to contribute to negative self-images among members of the group affected, as well as other disparities in healthcare access and outcomes, economic status, and legal support . Crucially, what is most important is not the true meaning or origin of a harmful term, but rather the way that the term is received by an individual . For example, attention has been drawn to the harmfulness of the term nitty gritty. This is due to a proposed link between that term and its possible use by enslavers who used it to describe detritus that would collect at the bottom of ships used to transport enslaved people across the Atlantic .
Whether or not this term was used by enslavers in this way –, the association has been made and is now salient ; this association can cause harm, and as a result, using a more inclusive term, such as details, is advised.
The potential for language to do harm is also context dependent. Some community members, for example, have worked to reclaim harmful slurs and redefine them in positive ways. Respect the self-identification of individuals while understanding that the same term may be differently received in different contexts. If you are not a member of an identity group, your use of a reclaimed term may still be perceived as harmful, so it may be best not to use such terms .
4.4. Honor how people identify themselves When describing individuals and their identities, what is most important is to respect the self-identification of individuals whenever possible and to represent people accurately. Moreover, consider whether the identities you reference are contextually important and include them only if they are. If you’re not sure how a person identifies, it’s best to ask them in a respectful way, being mindful that some people may wish to keep their identities private. It may help to explain the reason why knowing a person’s identity is important. For example, if you are introducing a guest speaker on a panel and their identity may be relevant to the Center for Equity, Gender & Leadership 19 relevant to the conversation, you might ask, “Are there aspects of your identity you’d like me to mention when I introduce you?” See below for information on honoring self-identification across different identities. The overview provided is not comprehensive. Self-identification may vary from person to person, over time, and across contexts, and as a result the suggestions may not always apply. This guidance is meant to serve as a starting point to orient you toward possible ways that people may self-identify, as well as highlight some areas in which there is a lack of consensus about which terms are most inclusive. 4.5. Understand inclusive language across different identities Take care to present narratives that do not reinforce negative or harmful stereotypes, and work to make sure that the voices and perspectives of the people being discussed are represented. Center the humanity of individuals by referring them as people and using identity categories as adjectives, not nouns (i.e. Black people and not Blacks, people with Autism/Austistic people and not Autistics, gay people and not gays, etc.) . More specific points for consideration and commonly used harmful terms are presented below. However, keep in mind that language is constantly changing and is dependent on context.
The recommendations provided below are a good starting point, but remember to honor self-identification and use terms that fit the context. For example, while the majority of members of a racial group might prefer one term (such as Black), specific individuals might have a different preference (such as African American) that should be respected. While some terms may be harmful if used by people who do not hold that identity, they may be important tools of empowerment when used by group members. 4.5.1. Gender, sex, and sexual orientation Gender representation in language matters. Research shows that the use of masculine terms such as “chairman,” even when intended to reference individuals of all genders, evokes and reinforces stereotypes of men within those roles –. Center for Equity, Gender & Leadership 20 To address this imbalance, some have advocated for using feminine terms as the default (i.e. “mailwoman”), while others have advocated for using masculine and feminine terms together (i.e. “mailman/woman”). However, these solutions still rely on a conceptualization of gender as a male/female binary, and as a result may be exclusive of individuals with other gender identities, such as nonbinary people . Using gender-neutral or gender-inclusive terms, such as “postal worker,” is maximally inclusive . Using gender-inclusive language has been shown to shape stereotypes about who might occupy a role, resulting in greater inclusivity . Note, however, that gender-neutral terms have still been shown to invoke some stereotypes of men occupying positions, particularly when in reality those positions show an overrepresentation of men . This highlights the importance of changing both language and social reality in pursuit of gender equity. Gender diversity in examples is also important , so take care to create examples, case studies, and other tools that reflect gender diversity — in the identities of the characters referenced and in the activities that they’re depicted performing to ensure that examples are not reinforcing gendered stereotypes.
Shifting to gender-inclusive language use entails different challenges for different languages. Most languages have some gendered terms such as mother and father, but some languages also have grammatical gender, meaning that all nouns are assigned an (often arbitrary) “gender.” For example, in Spanish, the word casa ‘house’ is feminine while the word hogar ‘home’ is masculine. When using these nouns, other parts of the sentence (such as the endings of adjectives) also change to reflect the gender of the noun. As a result of linguistic differences, different strategies for implementing gender-inclusive language may be better suited to different language contexts , . However, gender-inclusive language is possible and preferable even for languages with grammatical gender .
Respecting individuals’ gender identity is important . Using terms — including pronouns — that align with an individual’s identity can lead to more positive life outcomes, including improved mental health . The differences in life outcomes based on respecting or disrespecting pronouns are stark: trans Center for Equity, Gender & Leadership 21 youth who reported having their pronouns used by most or all of the people in their lives attempted suicide 50% less than those whose pronouns were not used . Pay attention to how people introduce themselves or ask for their name and pronouns if appropriate and follow their lead. In some situations, not everyone may feel safe or comfortable sharing their pronouns, particularly trans and nonbinary people who may face violence or other negative consequences if their identities become known. One way to invite those who feel comfortable to share their pronouns is to lead by example: introduce yourself with your name and pronouns as a way of normalizing the practice. However, avoid requiring people to identify their pronouns in case this may cause harm. If you’re not sure what pronouns a person uses and it’s not appropriate to ask, use the person’s name instead. Not everyone uses he or she as their pronouns. Many nonbinary people use singular they. While singular “they” is common in casual speech (consider, for example, what you would say if you only saw someone’s shadow: “I couldn’t tell who they were'') and was even used by Shakespeare , there has been resistance to using this term. However, it is becoming increasingly common and follows communication guidelines set out by the Linguistic Society of America , the American Psychological Society , the Associated Press , and Chicago Manual of Style , among other sources. For more tips on adopting inclusive pronoun use, both personally and within groups that you lead, consult resources like “Pronouns 101: Introduction to your Loved One’s New Pronouns” by Kirby Conrod or “Gender friendly teaching in higher education: Guidelines for affirmative and inclusive pronoun practices” by Sofia Melendez and Archie Crowley . There are differences between gender, sex, and sexual orientation. Gender is a social construct and identity; sex refers to biological sex assignment (i.e. sex assigned at birth). It is important to be precise in referring to the gender or sex of individuals rather than assuming cisgender identities (i.e. when an individual’s gender identity is the same as the sex they were assigned at birth) that overlook transgender identities (i.e. when an individual’s gender identity is not the same as the sex they were assigned at birth). Terms like “man/male” and “woman/female” have been casually used to refer to different aspects of both Center for Equity, Gender & Leadership 22 Women grow up being taught to accommodate others’ needs.
Women face negative assumptions about their professional capabilities.
All women need access to cervical cancer screenings. (p. 98) gender and sex. However, such uses may be imprecise and overlook areas of variation between individuals . An individual may, for example, be assigned male (sex) at birth but identify as female (gender). As Linguist Lal Zimman notes, not all casual uses of a term like “women” describe the same demographic: 1.
2.
3.
In the sentences above, “women” refers to three distinct groups of people: (1) individuals socialized as girls (including trans men who were raised as girls); (2) individuals perceived as women (including male-identifying individuals who are perceived by others as female); and (3) individuals who have a cervix (regardless of their gender identity). Instead of simply using “women” to describe these distinct groups, we can be more specific (i.e. people raised as girls / people perceived as women / people with a cervix), and as a result, more inclusive of individuals who may otherwise be incorrectly included or excluded from the group being mentioned. Being deliberate in deciding whether gender and sex information is important to report — and if so, which aspects of gender and sex are being referenced in the context — can help ensure that descriptions are precise and inclusive.
Sexual orientation is another aspect of identity that is independent of sex and gender. Sexual orientation refers to the “sexual and emotional attraction” that one person may feel for another . This can include the degree to which an individual feels sexual or emotional attraction, including identities as sexual, demisexual, asexual, and others . It may also include whether individuals are attracted to people with similar or different gender identities . People who are attracted to individuals of another gender may be referred to as “straight” or “heterosexual”; assuming this sexual orientation (as opposed to other orientations such as lesbian, gay, asexual, bisexual, queer, polysexual or pansexual) is known as heterosexism. When referring to multiple sexual orientations and gender identities, define the terms that you use and be sure to Center for Equity, Gender & Leadership 23 use them precisely. For example, terms such as LGBTQ+ (i.e. lesbian, gay, bisexual, transgender, queer, and more) are common. However, they should not be used when other terms would be more precise; for example, if reporting on trans youth, say trans youth rather than LGBTQ community. Race & ethnicity can be considered overlapping categories. They share some features, such as relying on physical traits such as facial features, skin color, and hair type . However, racial classification is often imposed while ethnic identity often emerges from the group itself . Moreover, race is associated with a power hierarchy, with certain racial groups deemed inferior by other racial groups, while ethnicity centers on common origins, ancestry, or heritage .
Some scholars argue that race and ethnicity are distinct but overlapping categories while others argue that race is a subset of ethnicity , , . 4.5.2. Race & Ethnicity Race is a social contract. While differences related to language, nationality, and culture have always existed, race emerged as a socially significant category during the Trans-Atlantic slave trade as a means of justifying slavery by promoting White superiority . “Scientific” arguments for differences between races relied on biological arguments that have been debunked as pseudo-science. In fact, there are much greater genetic differences within populations than between populations or races . Moreover, many “scientific” arguments for racial differences overlook the role of systemic forces like housing segregation and economic discrimination that result in disparate life outcomes between racial groups . Respecting self-identification is important. No matter how we conceptualize race and ethnicity as categories, it is essential to respect individuals’ self-identification to the extent possible. Different individuals who may consider themselves members of the same racial or ethnic group may have different preferences in the labels they use to describe their identities, and we should honor those individual preferences whenever possible. At the same time, it is Center for Equity, Gender & Leadership 24 important to be precise. In some instances, we may be talking not about self-identification but perceived identification. For example, in reporting on how former police officer Eric Parker treated Sureshbhai Patel, a man from India who police targeted after receiving a call about a “suspicious black man,” Parker’s treatment of Patel hinges not on Patel’s self-identification, but rather on Parker’s perception of Patel .
It is also important to use terms that precisely and accurately describe the group you wish to refer to. For example, African American refers to Americans with African ancestry, while Black is a larger category that also includes non-American citizens who are racially categorized as Black. While some Black Americans identify as either/both Black and/or African American, these terms are not interchangeable. Similarly, BIPOC (standing for Black, Indigenous, and people of color) serves to include all people of color (i.e. non-White people) while highlighting the unique challenges faced by Black and Indigenous people that may not be shared by other people of color. When talking about race and ethnicity, choose terms that best describe the group and context under discussion.
Avoid assuming a particular race/ethnicity is the “default” or “unmarked.” Everyone has a race — including White people. When talking about race/ethnicity, be sure not only to mention the race/ethnicities of minoritized groups, but also of the dominant group. Moreover, avoid using catch-all terms like “other” or “non-White” that push diverse groups of people into the same homogenous category. Using terms like “non-White” also fails to recognize the diversity of experiences of people of color in their own right and instead positions them only in opposition to Whiteness. Be precise in referring to an individual’s nationality or immigration status.
Remember that nationality and ethnicity are distinct: ethnicity reflects membership in a cultural group, while nationality indicates membership within a nation-state. To avoid invoking potentially harmful stereotypes, avoid imprecise 4.5.3. Nationality & religion Center for Equity, Gender & Leadership 25 phrases like “the Mexicans” or “the French,” and instead use more precise terms such as “Mexican citizens” or “the French people interviewed for this story” .
Remember, however, that not all residents are citizens; a more precise term could be “people” or “the public” if citizenship is not the focus of the discussion .
Do not use the term “illegal” to describe human beings — regardless of immigration status. This follows guidance from the Associated Press , Los Angeles Times , Human Rights Watch , and other organizations and activists . If an individual has entered a country without following immigration protocols, best practice is to describe their particular immigration status if relevant (i.e. “overstayed a student visa” or “was brought to the United States by parents without a visa”) . Some suggest using terms such as “undocumented,” “unauthorized,” or “irregular” migrants/immigrants . While some have argued that the term “illegal immigrants” is precise, concise, and less dehumanizing than other terms (such as “illegals” that reduce immigrants to their immigration status), others argue the opposite . The term “illegal immigrant” is misleading legally, as it implies criminality, while presence in the United States without proper documents is a civil offense, not criminal .
Moreover, it is similar to calling a criminal defendant “guilty” rather than “accused,” as the term is often used to describe individuals whose cases have yet to be presented in court . It is important to remember that all humans, regardless of their nation of origin, immigration status, or residency, are first and foremost human beings. Using terms like “illegals” dehumanizes those who are described with that term and can lead to discrimination and violence against members of minoritized groups .
Instead of assuming someone's religious identity, ask them. Avoid making assumptions about religion based on race, ethnicity, geography or other factors, and instead allow individuals to self-identify . Even if a region has an “official” religion or language, individuals may resist those policies and identify with others . Similarly, when describing the extent to which an individual practices a religion, terms like “practicing” or “observant” are preferred to more subjective terms such as “devout” or “pious,” though it is best to ask an individual how they Center for Equity, Gender & Leadership 26 self-identify . Be cautious about using terms and phrases that rely on religious metaphors or traditions that may be unknown to individuals from other religious traditions, as this may add to a feeling of exclusion and make your message less clear .
Avoid negative or condescending framings. Instead of phrases like “afflicted by ADHD” or “confined to a wheelchair” that imply something negative about the disability, use more neutral phrases like “has ADHD” or “uses a wheelchair” instead . Similarly, refrain from using euphemisms like “handi-capable” or Person-first and identity-first language can each be inclusive. Person-first language requires that personhood of the individual comes first: person with a disability (rather than disabled person) or individual with autism (rather than autistic person), for example. Such language pushes back against the dehumanization of people with disabilities as well as homogenization of the unique experiences of people with disabilities , . Referring to “people with diabetes,” for example, may help to highlight that each person with diabetes may have a unique experience, while “diabetics” may downplay such uniqueness . But not all disability communities agree with this point of view.
Others advocate for identity-first language, such as “disabled person” or “autistic person,” as being disabled or autistic is an important part of their identities that cannot be separated from their person –. Moreover, this pattern fits with the placement of other positive adjectives in English, i.e.
“intelligent person” rather than “person with intelligence,” and thereby reduces the implication that there is something negative about having a disability .
Moreover, some disability communities advance the use of capitalized identity-first terms, like Deaf, to indicate membership in the community of Deaf people . Because preferences differ between individuals, it is best to listen to how people describe themselves or ask how they prefer to be referred to , .
When it is not possible to learn an individual’s preference, using person-first or alternating between person-first and identity-first phrasing is recommended , , . 4.5.4. Ability Use precise terms that avoid negative stereotypes about aging. When referring to older adults, it is important to use inclusive language that doesn’t advance negative perceptions of the aging process, as this can result in decreased functional health of individuals described in negative terms . Being precise may help. Common terms such as “elderly” or “senior citizens” can be imprecise and fail to describe a meaningful population. For example, not all “seniors” may be “citizens” and not all “elderly” people may be at similar life stages: some may be experiencing health problems, some may not; some may be retired, some may not; some may have grandchildren, some may not . Instead of using an imprecise term like “elderly,” consider using more specific criteria that are relevant to the demographic in question: retired adults, people over 65, or grandparents, for instance. Moreover, “elderly” is rarely a term used by individuals to self-identify, while other terms (i.e. grandparent, retiree, etc.) are.
If use of a more specific, self-identified term is not possible, terms such as “older Center for Equity, Gender & Leadership 27 “special,” as these may be perceived as patronizing . Think critically about depictions of people with disabilities and ensure that they are balanced, depicting multiple aspects of disabled people’s identities and experiences, and that they don’t reinforce negative stereotypes or serve only to entertain or inspire others . Avoid using disability-related terms in derogatory ways. Using these terms — whether they are outdated (i.e. lame, insane, crazy, etc.) or current possible diagnoses (i.e. blind, ADHD) — casually to refer to people without those disabilities is harmful. It can reinforce negative stereotypes about people with disabilities. It can also minimize the perceived impact of having a disability by drawing comparisons between personality traits (i.e. detail-oriented) and diagnosable disorders (i.e. OCD). Finally, avoid using phrases that rely on metaphors linked to disabilities, such as “deaf to our concerns” or “crippled by debt,” and use more precise terms instead, such as “ignored our concerns” or “with large debts.” 4.5.5. Older Adults Be precise. Socioeconomic status — sometimes referred to as class — may refer to the income, educational attainment, occupational prestige, or subjective perceptions of social status of individuals and groups . When mentioning socioeconomic status, it is useful to be as precise as possible: if distinguishing between “high” and “low” income individuals, name the criteria for determining those categories, such as contextual or environmental indicators, i.e. median neighborhood household income, percentage of (un)employed people, or proportion of children who qualify for free or reduced-price lunch at school .
Use person-first language like “people who receive TANF (Temporary Assistance for Needy Families)” or “people experiencing homelessness” rather than “welfare recipients” or “homeless people.” Moreover, ensure that socioeconomic status is not being used imprecisely to implicitly reference race or other identities. If race, for example, is an important factor in the discussion, name it. Center for Equity, Gender & Leadership 28 adults” or “older people” are preferred as they center the humanity of older individuals and highlight that aging is a process: one doesn’t simply wake up one day in a new category of “old” but instead becomes gradually older .
Using “older adults” is not entirely unproblematic, as it is often used in contrast to “adults” (rather than “younger adults”) and thus promotes youth as the default. However, it is preferred to other terms. As Elana Buch notes, it is difficult to find a term to refer to older people that is viewed positively, as she states, “I’d argue that the reason there isn’t consensus about a preferred term has everything to do with ageism rather than that the terms themselves are problematic” . Thus, in addition to using more inclusive terms like “older adult” it is also important to work against negative stereotypes and conceptions about aging.
4.5.6. Socioeconomic Status Think critically about how socioeconomic disparities are presented and strive to avoid deficit-based language. For example, instead of writing about “high school dropouts” (which focuses on individuals’ shortcomings), write about Center for Equity, Gender & Leadership 29 “people with grade school education” (which focuses on individuals’ achievements). Similarly, be sure to visualize systemic barriers to opportunity and achievement rather than placing blame on individuals who face those barriers. This may include shifting from a description of “achievement gaps,” for example, to “systemic inequities” and avoiding negative phrasing like “poverty stricken” or “afflicted by poverty” in favor of more neutral phrases like “with income below the national average” , . Avoid terms that have become associated with negative stereotypes, particularly if those stereotypes are also associated with race such as “ghetto,” “inner-city,” or “the projects.” Use alternatives such as naming the specific neighborhood or subsidized housing initiative under discussion.
Individuals belong to multiple, often overlapping identity categories.
Membership in multiple marginalized groups may result in unique experiences that are distinct from experiences of people who are members of only one group or the other. As Kimberlé Crenshaw, who coined the term, notes, the experiences of people with overlapping identities, like Black women, “is greater than the sum of racism and sexism” [90, p. 140], cited in, . Keeping in mind that the preferences and experiences of individual people will differ depending on their unique and intersecting identities can be helpful as we reflect on how to use language in a way that is inclusive for everyone.
4.6. Consider intersectionality Adopting more inclusive language takes practice, and one way to support that practice is to form a community where inclusive language is promoted and valued , . It’s important to have a growth mindset, remembering that through making mistakes, we learn and become more adept at using new inclusive language practices. Members of a supportive inclusive language community can point out instances where more inclusive language could be used and can help others remember to use inclusive alternatives.
4.7. Work with others to adopt inclusive language together Center for Equity, Gender & Leadership 30 Acknowledge the mistake you made and the harm or offense it caused. Let the person who was harmed know that it was not acceptable. Explain what happened without excusing it. Let the other person know that you’ve identified what went wrong and the negative impact it had. Express your remorse. Offer to make amends.
As you adopt more inclusive language, you will also make mistakes; that is part of the learning process and should be expected. When you make a mistake, it can be helpful to apologize sincerely. You may want to: 1.
2.
3.
4.
For example, imagine that you’ve just described a successful presentation as “a cakewalk” and your colleague points out that term’s harmful history. You might say, “You’re right, using that phrase is harmful. I’m sorry, and I’ll do my best to use a better term in the future.” While using inclusive language is important, it’s also important to build support for other social equity initiatives. While changing our language practices can help support more inclusive and equitable environments, it is not sufficient on its own. Instead, working to build personal, social, and institutional practices that support minoritized voices is also crucial.
Center for Equity, Gender & Leadership 31 This paper was informed by a systematic literature review carried out by a multidisciplinary research team May to September 2021. This work was undertaken by the Center for Equity, Gender & Leadership at the UC Berkeley Haas School of Business with support from Google. EGAL also formed and collaborated with a working group of academic and community leaders with representation across different identities and expertise. Through the literature review, we sought to (1) define and understand ‘inclusive language’; (2) clarify the impacts of harmful versus inclusive language use; and (3) explore how these issues may vary across different identities. We performed searches of academic publications, articles, reports, blogs, and websites using key terms related to language equity and inclusion (language, inclusive/inclusion, equity, harmful, coding, race, impacts). We then formed and collaborated with a working group of academic and community leaders with representation across different identities and expertise. The working group members, including Dr. April Baker-Bell, Karen Bouris, Stuart Getty, Dr. Wesley Y. Leonard, and Dr. Kellie McElhaney, helped refine and inform the inclusive language framework (section 3). Appendix A: Methodology Center for Equity, Gender & Leadership 32 C. Kramsch, “Language, Thought, and Culture,” in The Handbook of Applied Linguistics, A. Davies and C. Elder, Eds. Malden, MA: Blackwell Publishing Ltd, 2004, pp. 235–261.
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Acids and Bases
How to Find the Most Acidic Proton
A very common exam question in the chapter of Acids and Bases is to ask for the most acidic proton in a molecule or when comparing two molecules.
So, how do we do this?
If you have already covered and learned the key principles of acids and bases, such as the p K a value, its relationship to acids, and the effects of hybridization and resonance on acidity, you know that it’s going to be a long discussion touching on ARIO and stuff like that.
However, as always, let’s try to make it simple, at least for starters, and go from there.
The first thing you should do when choosing the most acidicprotons is identifythose that are connected toheteroatoms(any atoms other than carbon). The good news is that it’s going to be either N, O, or S, as halogens make only one bond in organic chemistry, and there is not much left that is so common in organic chemistry.
Let’s look at these examples – they should be(come) very easy for you in the course of your organic chemistry class:
You should be like, “Well, obviously all the colored protons are more acidic.” And yes, a lot of the time, this is going to be correct because hydrogens on electronegative atoms such as O and N are usually more acidic.
Let’s recall why that is the case. Remember, a strong acid is one that “has no problem losing a proton” becauseit can successfully stabilize the negative charge formed as a result of this loss.
Now, what is anegative charge? A negative charge is an excess of electron density such that it exceeds the standard bonding number of the atom. We know from general chemistry that electronegative atomslike electrons and therefore, they are generally good at stabilizing the negative charge.
This is essentially the “A” of “ARIO” – the atom that handles the negative charge of the conjugate base.
For example, if we compare the acidity of an alcohol with an amine, that would be a less obvious comparison; however, the idea is the same:
Oxygen is more electronegative than nitrogen, and it stabilizes the negative charge better; thus, alcohols are stronger acids than amines.
So, as a general trend, let’s remember thatthe acidity of H–A increases as the electronegativity of A increases, going from left to right in the periodic table:
Charge and Acidity
Most or a significant portion of the examples we discuss for understanding the acidity of protons are on neutral molecules. However, there are common protonated species in organic chemistry that play important roles in many chemical and biological processes. For example, the protonated forms of amines in amino acids and elsewhere, alcohols, and last but not least, water. We all know the hydronium ion (H 3 O+) and how scientifically more acidic it is than water:
So, the general trend is that protonated species tend to be a lot more acidic than their neutral counterparts. This is intuitive, isn’t it? Why would a molecule want to be protonated? Remember, when discussing major and minor resonance contributors, we said that neutral species are more stable.
Because of this, the term CARIO, where C stands for charge, instead of ARIO, is used.
Now, just because “Charge” stands before “Atom” in this abbreviation, it does not mean any protonated atom is always going to be more acidic than a neutral atom. To share a secret with you, you’re going to see that anytime your professor says “always,” there’s going to be an exception (or a few). Hopefully in the lecture, and not on the test 😉.
Here is an example where the proton on a neutral atom is more acidic than the one on a protonated atom.
Atomic Size and Acidity
We mentioned in the previous sentence that the acidity increases for the atoms going from left to right in the periodic table because that’s how electronegativity increases, too.
A trickysituation arises when we godown a group in the periodic table. Remember, the atomic radius gets larger, and helps handle the negative charge better despite the atom being less electronegative. As an example, compare the acidity of alcohols and thiols.
The p K a values suggest that thiols are about 1000 times more acidic than alcohols.
The reason for this is the ability of larger atoms to better stabilize the negative charge:
You can think about it this way: the negative charge is spread around a larger surface/volume, thus it is better stabilized.
Therefore,down the periodic table, the atomic size determines the acidity and not the electronegativity.
Inductive Effect and Acid Strength
Inductive effect is anelectronic effect that occurs through sigma (σ) bonds.If the atom or the groupdonates electron density,it is said to have apositive inductive effect (+I).Atoms or groups thatpull electron densityfrom another atom or a group through sigma bond(s), areelectron-withdrawingand denoted as–Igroups.
Perhaps the most important feature of the inductive effect is that itworks through several sigma bonds.
Let’s see how the inductive effect influences the acidic and basic properties of molecules. For example, 2-fluoroethanol is a stronger acid than ethanol, as seen from their p K a values:
What makes 2-fluoroethanol a stronger acid? The acid strength increases with better stabilization of the negative charge of its conjugate base, so what happens here is that fluorine pulls the electron density from the negatively charged oxygen, thus stabilizing the conjugate base of the alcohol.
The way it works is that the F-C bond reduces the electron density of the carbon, which in turn does the same to the other carbon, and finally, the carbon connected to the oxygen helps it handle the negative charge by reducing its electron density:
This pattern is very common and very well-studied for carboxylic acids. The presence of a halogen increases their acidity through the electron-withdrawing inductive effect. Thegreater the electronegativityof the halogen, the greater the inductive effect and thus thestronger the acid.
Fluoroacetic acid is stronger than chloroacetic acid, which is stronger than bromoacetic acid, which, in turn, is stronger than acetic acid by itself because the electronegativity of these halogens decreases in this order: F > Cl > Br.
Interestingly, it is not only the nature of the electron-withdrawing atom that affects the charge stabilization but also itsdistancefrom the carboxylate group. This is expected because theinductive effect gets weakerwith the increasingnumber of sigma bondsbetween the two centers:
For example, compare the acidity of butanoic acid with 2, 3, and 4-chlorobuatanoic acid:
2-chlorobutanoic acid has the lowest p K a because the chlorine is the closest to the carboxylic group, and thus the inductive effect is the most profound.
So, if you have two protons connected to the same atom, the one closer to an electron-withdrawing group is going to be more acidic:
Acidity and Resonance
Let’s compare the acidity of ethanol and acetic acid. We can see, from the p K a values, that acetic acid is a much stronger acid than ethanol, even though upon dissociation, the negative charge ends up on the same element, which is oxygen:
This is because the electrons on the oxygens of carboxylic acid aredelocalized(in resonance with the other oxygen) and the negative charge is handled by both atoms, while the oxygen in the alcohol handles the negative charge alone:
Another example of theresonancestabilizationof the conjugate base, thus making the acid stronger, isphenol.Comparedtocyclohexanol, it is again abouta million timesmore acidic, which isagainbecause ofresonancestabilization of the conjugate base.Althoughthis is a bit more complicated, as we also say that thelonepairs on phenol are notdelocalizedover the benzenering,as that would disturb thearomaticity, but that is another topic, and a lot of you have probably not talked aboutaromatic compoundsyet.
To keep this fair,another factorcontributing to the increased acidity of phenol is thehybridization of the carbon atomsin the aromatic ring. These are sp 2 hybridized and sp 2 hybridized carbon atoms are more electronegative than sp 3 carbons because they have more s character.
As a result, they have an electron-withdrawinginductive effectwhile the aliphatic carbons are electron donors. So, the sp 2 carbon is more helpful in stabilizing the negative charge on the oxygen via an electron-withdrawing inductive effect (Recall theARIO).
In general, themore s character, themore electronegativethe atom is. These are the percentages of the s orbital (s-character) in each hybrid orbital:
To have all theseparameters identicaland see what would be theeffect ofresonanceelectron withdrawing, we can also compare theacidityofphenolwithnitrophenols. The nitro group contains aN=Opi bond, and it is aresonance-withdrawinggroup like, for example, different carbonyl derivatives, such as aldehydes, ketones, carboxylic acids,etc. This pi bond allows for a resonance stabilization of the negative charge that is formed upon the deprotonation of the OH group.
Both 3-nitrophenol (m-nitrophenol) and 4-nitrophenol (p-nitrophenol) are more acidic than phenol.
Notice thatp-nitrophenolis alsomore acidic than m-nitrophenol, and that has to do with the fact it the nitro group in themeta position can only stabilize the negative charge viainductive electron-withdrawing effect, whereas inp-nitrophenol, the negative charge is stabilized by bothinductive and resonance effects.
You do not need to worry about the terms “meta”, para, “aromatic etc., if you have not reached the chapter ofaromatic chemistry, which is an organic 2 topic.
The take-home message for you here is thatresonance-withdrawing groups can stabilize the conjugate baseby delocalizing its lone pairs, thusincreasing the acidityof the compound. Once again,recallthat the weaker/more stable the conjugate base, the stronger the acid.
Watch out for Resonance Rules
Do not fall for resonance-stabilization anytime you see a double bond near the proton. Not every lone pair (negative charge) can be resonance delocalized. You need to follow the rules for drawing resonance structures. This is mainly about not exceeding the octeton carbon and other second-row elements, andnot breaking or making single bondswhen drawing resonance forms. Here are some examples of correct and incorrect resonance transformations, and you can check the linked article for more details.
Resonance vs Electronegativity
The resonance effect can sometimes make a proton connected to carbon, or, in general, a less electronegative atom, more acidic. For example, the red proton in molecule A is between two resonance-withdrawing carbonyl groups, and it is more acidic than the OH proton. This is despite the fact that it is connected to a carbon, which is less electronegative than oxygen.
So, we have seen how the resonance stabilization by two carbonyl groups makes proton A more acidic than proton B. Notice that proton C is connected to a carbon, but it only has one adjacent carbonyl group to stabilize the conjugate base. It turns out this is not enough to make it more acidic than the OH proton. In fact, it is about 1000 times weaker than typical alcohols.
At the end, I’ll include a simplified table of p K a values for common functional groups, so you can use it to practice. You’ll “always” be allowed to use a p K a table on the test, but make it a goal to identify acidic protonsusing the principles we discussed.
You don’t need to memorize everyp K avalue, but being able to recognize when one proton is significantly more acidic than another is a valuable skill to have.
Practice
1.
Determine if the blue- or red-colored proton is more acidic in each of the following compounds:
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Identify the most acidic proton in each molecule and then, based on that, the more acidic compound for each pair shown below. Consider the electronegativity, hybridization, and possible resonance stabilization of the conjugate bases.
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For each pair of compounds below, highlight the most acidic proton and identify the more acidic compound:
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Synthesis of Aromatic Compounds From Benzene
Arenediazonium Salts in Electrophilic Aromatic Substitution
Reactions at the Benzylic Position
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Meta Substitution on Activated Aromatic Ring
Electrophilic Aromatic Substitution Practice Problems
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How to Name a Compound with Multiple Functional Groups
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Reactions of Aldehydes and Ketones with Alcohols: Acetals and Hemiacetals
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Aldehydes and Ketones to Carboxylic Acids
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Fischer Esterification
Ester Hydrolysis by Acid and Base-Catalyzed Hydrolysis
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Esters Reaction with Amines – The Aminolysis Mechanism
Ester Reactions Summary and Practice Problems
Preparation of Acyl (Acid) Chlorides (ROCl)
Reactions of Acid Chlorides (ROCl) with Nucleophiles
R 2 CuLi Organocuprates - Gilman Reagent
Reaction of Acyl Chlorides with Grignard and Gilman (Organocuprate) Reagents
Reduction of Acyl Chlorides by LiAlH 4, NaBH4, and LiAl(OtBu)3 H
Reduction of Carboxylic Acids and Their Derivatives
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Amides - Structure and Reactivity
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Amides Hydrolysis: Acid and Base-Catalyzed Mechanism
Amide Dehydration Mechanism by SOCl 2, POCl 3, and P 2 O 5
Amide Reduction Mechanism by LiAlH4
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Amides from Carboxylic Acids-DCC and EDC Coupling
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The Reactions of Nitriles
Converting Nitriles to Amides
Carboxylic Acids to Ketones
Esters to Ketones
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9548 | https://www.youtube.com/playlist?list=PLU5aQXLWR3_yYS0ZYRA-5g5YSSYLNZ6Mc | Descriptive statistics | Probability and Statistics | Khan Academy - YouTube
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9549 | https://resources.finalsite.net/images/v1727122674/reedschoolsorg/aufbz6dtups0iktpgszi/89_parent_newsletter.pdf | Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 9: Solving Quadratic Equations Students will… • Solve quadratic equations by graphing. • Solve quadratic equations by taking square roots. • Solve quadratic equations by completing the square. • Solve quadratic equations by the quadratic formula. • Use discriminants to determine the number of real solutions of quadratic equations. • Choose a method to solve quadratic equations. • Solve systems of linear and quadratic equations. • Solve real-life problems. Standards California Common Core: A.REI.4a: Use the method of completing the square to transform any quadratic equation in x into an equation of the form (x - p)2 = q that has the same solutions. Derive the quadratic formula from this form. A.REI.4b: Solve quadratic equations by inspection (e.g., for x2 = 49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a ± bi for real numbers a and b. A.REI.7: Solve a simple system consisting of a linear equation and a quadratic equation in two variables algebraically and graphically. A.REI.11: Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the solutions approximately, e.g., using technology to graph the functions, make tables of values, or find successive approximations. Include cases where f(x) and/or g(x) are linear, polynomial, rational, absolute value, exponential, and logarithmic functions. A.SSE.3b: Complete the square in a quadratic expression to reveal the maximum or minimum value of the function it defines. F.IF.8a: Use the process of factoring and completing the square in a quadratic function to show zeros, extreme values, and symmetry of the graph, and interpret these in terms of a context. Key Terms A quadratic equation is a nonlinear equation that can be written in the standard form where Another method for solving quadratic equations is completing the square. In this method, a constant c is added to the expression so that is a perfect square trinomial. Quadratic formula: where and The expression in the quadratic formula is the discriminant. Games • Quadratic Quandry • Transform Me These are available online in the Game Closet at www.bigideasmath.com. Essential Questions • How can you use a graph to solve a quadratic equation in one variable? • How can you determine the number of solutions of a quadratic equation of the form • How can you use “completing the square” to solve a quadratic equation? • How can you use the discriminant to determine the number of solutions of a quadratic equation? • How can you solve a system of two equations when one is linear and the other is quadratic? Copyright © Big Ideas Learning, LLC All rights reserved. Quick Review • The solutions of a quadratic equation are also called roots. • A zero of a function y = f(x) is an x-value for which the value of the function is zero. • The square of a real number cannot be negative. • Before completing the square, make sure the leading coefficient is 1. • You can use the roots of a quadratic equation to factor the related expression. • The solutions of a quadratic equation may be real numbers or imaginary numbers. What’s the Point? Solving quadratic equations is very useful in real-life for finding what amount of goods being supplied is exactly equal to the amount of goods demanded for the economy to be in equilibrium The STEM Videos available online show ways to use mathematics in real-life situations. The Chapter 9: Wolf Population Growth STEM Video is available online at www.bigideasmath.com. Solving Quadratic Equations Using Square Roots You can solve by taking the square root of each side. • When has two real solutions, • When has one real solution, • When has no real solutions. Completing the Square To complete the square for an expression of the form follow these steps. Step 1: Find one-half of b, the coefficient of x. Step 2: Square the result from Step 1. Step 3: Add the result from Step 2 to Factor the resulting expression as the square of a binomial. Quadratic Formula The real solutions of the quadratic equation are where and Interpreting the Discriminant • two real solutions two x-intercepts • one real solution one x-intercept • no real solutions no x-intercepts Reference Tools An Information Wheel can be used to organize information about a concept. Students write the concept in the middle of the “wheel.” Then students write information related to the concept on the “spokes” of the wheel. This type of organizer serves as a good summary tool because any information related to a concept can be included. Methods for Solving Quadratic Equations Method Advantages Disadvantages Factoring • Straightforward when equation can be factored easily • Some equations are not factorable. Graphing • Can easily see the number of solutions • Use when approximate solutions are sufficient. • Can use a graphing calculator • May not give exact solutions Using Square Roots • Use to solve equations of the form 2 .
x d = • Can only be used for certain equations Completing the Square • Best used when a = 1 and b is even • May involve difficult calculations Quadratic Formula • Can be used for any quadratic equations • Gives exact solutions • Takes time to do calculations |
9550 | https://www.willseye.org/wp-content/uploads/2018/04/DiabeticRet_PatientBro_FNL.pdf | 840 Walnut Street, Philadelphia PA 19107 www.willseye.org A Patient’s Guide to Diabetic Retinopathy Diabetic Retinopathy 1 Diabetic Retinopathy Figure 1: Cross section of the eye showing the center of the retina (Macula), retinal blood vessels and optic nerve.
1. Definition Diabetic retinopathy is a complication of diabetes mellitus. Diabetes mellitus is a condition in which the blood sugar level is elevated because the body is unable to use and store sugar. This high sugar content damages blood vessels in the body over time and can affect a variety of body organs such as the eyes, heart, and kidneys. Diabetes affects the eyes by causing deterioration of blood vessels in the retina. Breakdown of retinal blood vessels may result in fluid leaking into the center of the retina (macular edema) or abnormal blood vessels that grow on the surface of the retina (neovascularization) which can bleed and scar. This can lead to loss of central and possibly peripheral vision.
3 2 2. Causes and Associations The longer someone has diabetes mellitus, the more likely they will develop diabetic retinopathy. After 25 years, nearly all people with diabetes mellitus will show some signs of diabetic retinopathy. The severity of diabetic retinopathy is also related to blood glucose (sugar) control. Keeping blood glucose levels down to as normal as possible reduces the degree and rate of progression of diabetic retinopathy and other diabetic complications in the body. The hemoglobin A1C level reflects how well blood sugar control has been achieved over the past several months. The goal in managing diabetes is to keep the hemoglobin A1C level less than 7. Patients should discuss the appropriate target A1C with doctor that helps them manage their diabetes. Maintaining blood pressure and blood lipid levels in normal ranges can also have a favorable impact on the course of diabetic retinopathy 3. Symptoms Symptoms of diabetic retinopathy include gradual, progressive blurring of vision, sudden, severe vision loss, floaters or fluctuating vision. It is important to recognize that people with diabetic retinopathy may not necessarily have visual changes even in more advanced stages. It is important and mandatory that people with diabetes mellitus have their eyes examined at least annually.
4. Examination A complete ophthalmic examination is important in the assessment of diabetic retinopathy and this includes vision testing, drops to dilate pupils, and a complete examination of the front and back of the eye.
What the Doctor Sees There are two major types of diabetic retinopathy: non-proliferative retinopathy and proliferative retinopathy. Non-proliferative diabetic retinopathy is the earlier stage and is characterized by visible damage to small retinal blood vessels. These blood vessels may develop balloon-like swelling called microaneurysms. Microaneurysms and other areas of abnormal retinal blood vessels may leak fluid, causing the retina to swell or bleed. This may lead to vision loss. Leakage in the center of the retina (macula), known as macular edema, is the most common mechanism of vision loss in people with diabetic retinopathy. Non-proliferative diabetic retinopathy is the most common form of diabetic retinopathy, accounting for approximately 80% of all cases.
Figure 2: Non-proliferative diabetic retinopathy with macular edema (arrow) 5 4 Some people progress to the more advanced proliferative diabetic retinopathy stage.
Proliferative diabetic retinopathy is characterized by such severe small retinal vessel damage and reduced oxygenization of the retina that the retina reacts by growing abnormal blood vessels (neovascularization.) These abnormal blood vessels are fragile and can bleed and pull on the retina as they grow. Bleeding into the vitreous cavity of the eye (vitreous hemorrhage) can result in sudden and sometimes severe loss of vision. This type of hemorrhage is painless and, early on, may be seen as cobweb-like floaters in one’s vision. Symptoms of new floaters and any sudden vision change in a person with diabetic retinopathy should be evaluated promptly by an ophthalmologist.
Proliferative diabetic retinopathy can also lead to traction retinal detachments. The retinal neovascularization can grow to be large and then contract, pull, and lift the retina. Retinal detachment can lead to loss of vision if it involves the macula.
Testing People with diabetic retinopathy may have several types of tests in evaluating their condition. Fundus Photography: People may undergo photographs of their retinas to document the stage and findings of diabetic retinopathy. There are no risks associated with this simple test.
Fluorescein Angiography: Fluorescein angiography may be used to determine the extent of diabetic retinopathy or to detect areas of leakage or bleeding that may lead to vision loss. The test is performed by injecting sodium fluorescein dye into a peripheral vein with a small needle. This dye then goes through the body and eyes, as well as the retina to show blood flow and various features of retinopathy such as leakage. It is considered a routine and safe test, but people should expect some temporary, mild yellowish tinting of skin and orange colored urine. Most people have no difficulty with this test, although a low percentage of people will experience some transient, mild nausea after the injection. Very rarely, allergic or even more severe reactions can occur.
Figure 4: Fluorescein Angiography Figure 3: Proliferative diabetic retinopathy (Dark arrow shows leakage of new vessals on the optic nerve. Orange arrow shows hemmorrhage in front of the retina). 7 6 Figure 6 Optical Coherence Tomography (OCT) OCT imaging is a fast, non-invasive test that uses low energy laser to scan the macula and determine whether there is swelling or distortion of the macula in the setting of diabetic retinopathy. The test is also useful to assess the response to diabetic macular edema treatment.
5. Prognosis People who maintain healthy, active lifestyles and who optimize their blood sugar control have the best chance of slowing progression of diabetic retinopathy and preserving good vision. It is very important that people with diabetes mellitus undergo at least an annual eye exam, whether or not they have any vision symptoms. It is important to remember that diabetic retinopathy may progress and not cause any symptoms. It is also very important for people to understand that their blood glucose (sugar) control should be as good as possible with the goal of keeping the hemoglobin A1C level less than 7. 6. Treatment Injections: Medicines injected into the eye such as anti-VEGF drugs (eg. Lucentis, Avastin, and Eylea) and steroids (Triamcinolone or Triesence and Ozurdex) are commonly used to treat diabetic macular edema and sometimes also the proliferative manifestations of the condition such as when there is vitreous hemorrhaging. Both classes of medicines have been proven in large-scale studies to be highly effective Figure 5: Optical Coherence Tomography B Scan Ultrasonography: This test utilizes standard, non-invasive ultrasound technology and is used in the office to view the retina when the retina cannot be directly visualized by the doctor with standard examination techniques such as in the setting of a severe vitreous hemorrhage.
Typically the doctor will recommend the test to rule of retinal tears and any pulling on or detachment of the retina. Certain types of retinal detachment may need relatively urgent surgery.
9 8 edema. It is performed in one session, is generally painless, and can take up to 2 to 3 months to see the desired “drying” effect. The other type of laser treatment is called panretinal (or “scatter”) photocoagulation. These are longer, more extensive treatments that are used to shrink abnormal vessels and reduce the number or severity of vitreous hemorrhages. Panretinal laser treatments may be divided into several sessions and can be associated with some ache in the eye during or after the treatment. The desired effects may take 4 to 6 weeks or more. Both focal and panretinal laser treatments may need to be repeated to control the diabetic retinopathy problems.
In general, laser treatments are intended to stabilize or prevent progression of various diabetic retinopathy complications and may or may not result in noticeable vision improvement. The best results with the best chances of preserving a good level of vision are achieved when diabetic retinopathy-related problems are detected early.
Lastly, laser treatments may not work in everyone and other treatments may be needed.
Figure 7: Laser Photocoagulation in reducing macular edema and improving vision.
Repeat injections may be necessary for long-term control of the problems.
The injections are performed in the office using topical drop anesthesia. They are very well tolerated and complications are rare. There is a very small risk of infection with any eye injection. Steroid injections also may be associated with elevating the pressure of the eye or causing progression of cataract. People should discuss risks and benefits of all treatments, including injection therapies, with their eye specialist.
Laser: Laser photocoagulation is a well-established, standard treatment for diabetic retinopathy. A laser delivers a split-second burst of intense light energy to treat leaky retinal blood vessels or promote shrinkage of abnormal blood vessels (neovascularization) that can cause bleeding. Laser photocoagulation has been proven in large clinical trials to reduce the risk of both moderate and severe vision loss in people with diabetic retinopathy.
Laser photocoagulation is performed in the office setting with the patient seated in front of the laser unit. The eye is anesthetized with drops, and a contact lens in placed on the eye to focus the laser-aiming beam for some of the laser procedures.
People will experience brief, bright flashes of lights.
Some people may experience discomfort during or after laser photocoagulation, especially with the longer (panretinal) treatments, but generally it is a well-tolerated office procedure.
There are two main types of laser treatments in diabetic retinopathy. One is called focal laser and this is the technique used to control diabetic macular 11 10 Vitrectomy: People with diabetic retinopathy may require vitrectomy surgery in an operating room setting. A vitrectomy is performed when there is bleeding or retinal traction that is causing loss of vision in people with advanced diabetic retinopathy. In this surgical procedure, small instruments are inserted into the eye under microscopic visualization, and both the vitreous hemorrhage and any scar tissue are removed. The vitreous gel typically is replaced with clear fluids at the end of the case. Laser photocoagulation may be performed at the time of surgery, and in some cases, a gas bubble or silicone oil may be placed to hold the retina in position if there are retinal holes or detachment. The prognosis for people who require vitrectomy surgery depends upon the status of the underlying diabetic-related effects on the retina.
Figure 8: Pars plana vitreous for hemorrhage Notes About Us Wills Eye Hospital is a global leader in ophthalmology.
Established in 1832 as the nation’s first hospital specializing in eye care, we now are known worldwide for our clinical expertise. Today, we continue to shape the field of ophthalmology thanks to our talented, skilled physicians and staff who are dedicated to improving and pre serving sight.
Wills Eye’s core strengths include: Research: We maintain a close connection between innovative research and advanced care as our physicians pursue research that can be translated quickly into clinical care.
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9551 | https://pmc.ncbi.nlm.nih.gov/articles/PMC6274679/ | The Pathophysiology of Gestational Diabetes Mellitus - PMC
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Int J Mol Sci
. 2018 Oct 26;19(11):3342. doi: 10.3390/ijms19113342
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The Pathophysiology of Gestational Diabetes Mellitus
Jasmine F Plows
Jasmine F Plows
1 Department of Preventive Medicine, University of Southern California, Los Angeles, CA 90033, USA; plows@usc.edu
Find articles by Jasmine F Plows
1, Joanna L Stanley
Joanna L Stanley
2 Liggins Institute, University of Auckland, Auckland 1023, New Zealand; jostnly@gmail.com (J.L.S.); c.reynolds@auckland.ac.nz (C.M.R.)
Find articles by Joanna L Stanley
2, Philip N Baker
Philip N Baker
3 University of Leicester, University Road, Leicester LE1 7RH, UK; philip.baker@leicester.ac.uk
Find articles by Philip N Baker
3, Clare M Reynolds
Clare M Reynolds
2 Liggins Institute, University of Auckland, Auckland 1023, New Zealand; jostnly@gmail.com (J.L.S.); c.reynolds@auckland.ac.nz (C.M.R.)
Find articles by Clare M Reynolds
2, Mark H Vickers
Mark H Vickers
2 Liggins Institute, University of Auckland, Auckland 1023, New Zealand; jostnly@gmail.com (J.L.S.); c.reynolds@auckland.ac.nz (C.M.R.)
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2,
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1 Department of Preventive Medicine, University of Southern California, Los Angeles, CA 90033, USA; plows@usc.edu
2 Liggins Institute, University of Auckland, Auckland 1023, New Zealand; jostnly@gmail.com (J.L.S.); c.reynolds@auckland.ac.nz (C.M.R.)
3 University of Leicester, University Road, Leicester LE1 7RH, UK; philip.baker@leicester.ac.uk
Correspondence: m.vickers@auckland.ac.nz; Tel.: +64-9-9236687
Received 2018 Sep 27; Accepted 2018 Oct 21; Collection date 2018 Nov.
© 2018 by the authors.
Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (
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PMCID: PMC6274679 PMID: 30373146
Abstract
Gestational diabetes mellitus (GDM) is a serious pregnancy complication, in which women without previously diagnosed diabetes develop chronic hyperglycemia during gestation. In most cases, this hyperglycemia is the result of impaired glucose tolerance due to pancreatic β-cell dysfunction on a background of chronic insulin resistance. Risk factors for GDM include overweight and obesity, advanced maternal age, and a family history or any form of diabetes. Consequences of GDM include increased risk of maternal cardiovascular disease and type 2 diabetes and macrosomia and birth complications in the infant. There is also a longer-term risk of obesity, type 2 diabetes, and cardiovascular disease in the child. GDM affects approximately 16.5% of pregnancies worldwide, and this number is set to increase with the escalating obesity epidemic. While several management strategies exist—including insulin and lifestyle interventions—there is not yet a cure or an efficacious prevention strategy. One reason for this is that the molecular mechanisms underlying GDM are poorly defined. This review discusses what is known about the pathophysiology of GDM, and where there are gaps in the literature that warrant further exploration.
Keywords: gestational diabetes, pregnancy, pathophysiology, physiology, pathology, molecular
1. Introduction
Gestational diabetes mellitus (GDM) is a common pregnancy complication, in which spontaneous hyperglycemia develops during pregnancy . According to the most recent (2017) International Diabetes Federation (IDF) estimates, GDM affects approximately 14% of pregnancies worldwide, representing approximately 18 million births annually . Risk factors include overweight/obesity, westernized diet and micronutrient deficiencies, advanced maternal age, and a family history of insulin resistance and/or diabetes. While GDM usually resolves following delivery, it can have long-lasting health consequences, including increased risk for type 2 diabetes (T2DM) and cardiovascular disease (CVD) in the mother, and future obesity, CVD, T2DM, and/or GDM in the child. This contributes to a vicious intergenerational cycle of obesity and diabetes that impacts the health of the population as a whole. Unfortunately, there is currently no widely-accepted treatment or prevention strategy for GDM, except lifestyle intervention (diet and exercise) and occasionally insulin therapy—which is only of limited effectiveness due to the insulin resistance that is often present. While emerging oral antidiabetics, such as glyburide and metformin, are promising, concerns remain about their long-term safety for the mother and the child [3,4]. Therefore, safe, effective, and easy-to-administer new treatments are sought. In order to develop such treatments, a thorough understanding of the pathophysiology of GDM is required. This review will discuss what is known about the pathophysiology of GDM and what has yet to be elucidated. In order to do so, a contextual summary of glucose regulation during normal pregnancy, classification of GDM, forms of GDM, risk factors for GDM, and consequences of GDM is first required.
1.1. Glucose Regulation during Healthy Pregnancy
During healthy pregnancy, the mother’s body undergoes a series of physiological changes in order to support the demands of the growing fetus. These include adaptations to the cardiovascular, renal, hematologic, respiratory, and metabolic systems. One important metabolic adaptation is in insulin sensitivity. Over the course of gestation, insulin sensitivity shifts depending on the requirements of pregnancy. During early gestation, insulin sensitivity increases, promoting the uptake of glucose into adipose stores in preparation for the energy demands of later pregnancy . However, as pregnancy progresses, a surge of local and placental hormones, including estrogen, progesterone, leptin, cortisol, placental lactogen, and placental growth hormone together promote a state of insulin resistance . As a result, blood glucose is slightly elevated, and this glucose is readily transported across the placenta to fuel the growth of the fetus. This mild state of insulin resistance also promotes endogenous glucose production and the breakdown of fat stores, resulting in a further increase in blood glucose and free fatty acid (FFA) concentrations . Evidence in animals suggests that, in order to maintain glucose homeostasis, pregnant women compensate for these changes through hypertrophy and hyperplasia of pancreatic β-cells, as well as increased glucose-stimulated insulin secretion (GSIS) . The importance of placental hormones in this process is exemplified by the fact that maternal insulin sensitivity returns to pre-pregnancy levels within a few days of delivery . For reasons that will be explored in this review, the normal metabolic adaptations to pregnancy do not adequately occur in all pregnancies, resulting in GDM.
1.2. Classification and Prevalence of Gestational Diabetes
The American Diabetes Association (ADA) formally classifies GDM as “diabetes first diagnosed in the second or third trimester of pregnancy that is not clearly either preexisting type 1 or type 2 diabetes” . However, the exact threshold for a diagnosis of GDM depends on the criteria used, and so far, there has been a lack of consensus amongst health professionals. It is now advised by the ADA, the World Health Organization (WHO), the International Federation of Gynaecology and Obstetrics, and the Endocrine Society, that the International Association of Diabetes and Pregnancy Study Group (IADPSG) criteria be used in the diagnosis of GDM . The IADPSG criteria was developed based on the results of the Hyperglycemia and Adverse Pregnancy Outcomes (HAPO) Study—a large multinational and multicenter study of 23,000 pregnant women . One major finding of the HAPO Study was a continuous risk of adverse maternal and fetal outcomes with increasing maternal glycaemia—even below the diagnostic threshold for GDM—suggesting the that criteria for intervention needed to be adjusted. The IADPSG therefore recommends that all women undergo a fasting plasma glucose (FPG) test at their first prenatal visit (where a reading ≥92 mg/dL is indicative of GDM), and that women with FPG <92 mg/dL undergo a 2-h 75 g oral glucose tolerance test (OGTT) between 24 and 28 weeks’ gestation. These glycemic cut-offs are lower than other guidelines, and only one abnormal glucose reading is required for diagnosis, which has resulted in a drastic increase in the number of cases of GDM and associated healthcare costs . For this reason, there has been much discussion amongst experts as to whether the IADPSG criteria should be modified to only screen at-risk women (i.e., women of advanced maternal age, those who are overweight/obese, who are in high-risk ethnic groups, or with a family history of diabetes). However, some studies suggest that such efforts would miss a substantial number of GDM cases without significantly reducing the cost [13,14,15]. Therefore, the IADPSG criteria are the most widely recommended guideline today, although alternate criteria remain in some centers and countries (Table 1).
Table 1.
Various criteria for gestational diabetes mellitus (GDM) diagnosis using oral glucose tolerance test (OGTT).
| Criteria | Pregnancies | Timing of OGTT | Steps | Glucose Load (g) | Glucose Threshold (mmol/L) |
:---: :---: :---: |
| Fasting | 1 h | 2 h | 3 h |
| O’Sullivan, 1964 | All | 24–28 weeks | 2 | 100 | 5.0 | 9.2 | 8.1 | 6.9 |
| WHO, 1999 | All | 24–28 weeks | 1 | 75 | 7.0 | — | 7.8 | — |
| American Diabetes Association (ADA), 2004 | High and medium risk | 14–18 weeks for high risk, 28–32 weeks for medium risk | 2 | 100 | 5.3 | 10.0 | 8.6 | 7.8 |
| National Institute for Health and Care Excellence (NICE), 2015 | High risk | As early as possible | 1 | 75 | 5.6 | — | 7.8 | — |
| IADPSG, 2010 | All | 24–28 weeks | 1 | 75 | 5.1 | 10.0 | 8.5 | — |
| WHO, 2013 |
| ADA, 2016 |
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The inconsistencies in screening and diagnosis of GDM make worldwide estimates difficult. Using the IADPSG’s criteria, the International Diabetes Federation (IDF) estimated that 18 million live births worldwide (14%) were affected by gestational diabetes in 2017 . South-East Asia had the highest prevalence of GDM at 24.2%, while the lowest prevalence was seen in Africa at 10.5%. Almost 90% of cases of hyperglycemia in pregnancy occurred in low- and middle-income countries, where access to maternal healthcare is limited. Even within-countries, GDM prevalence varies depending on race/ethnicity and socioeconomic status. Aboriginal Australians, Middle Easterners, and Pacific Islanders are the most at-risk groups for GDM . Within the United States, Native Americans, Hispanics, Asians, and African-American women are at a higher risk of GDM than Caucasian women . There is also some evidence that GDM prevalence varies by season, with more diagnoses of GDM in summer than winter .
1.3. Forms of Gestational Diabetes
Outside of pregnancy, three distinct forms of diabetes mellitus are described: autoimmune diabetes (type 1), diabetes occurring on a background of insulin resistance (type 2), and diabetes as a result of other causes, including genetic mutation, diseases of the exocrine pancreas (e.g., pancreatitis), and drug- or chemical-induced diabetes (such as after organ transplantation or in the treatment of human immunodeficiency virus infection and acquired immune deficiency syndrome (HIV/AIDS)) [1,19]. While there is evidence that GDM can occur in all three settings [20,21], the vast majority (~80%) of GDM cases present as β-cell dysfunction on a background of chronic insulin resistance, to which the normal insulin resistance of pregnancy is partially additive . Thus, affected women tend to have an even greater degree of insulin resistance than healthy pregnant women, and therefore have further reductions in glucose utilization and increased glucose production and FFA concentrations . It is thought that β-cells deteriorate due to excessive insulin production in response to excess energy consumption and insulin resistance, exhausting the cells over time. The fact that this pathology closely resembles that of T2DM has spurred much debate about whether the two diseases should be considered to be etiologically indistinct [24,25]. As this form of GDM is by far the most common, it will be the focus of this review.
1.4. Risk Factors for Gestational Diabetes
Epidemiological studies of risk factors for GDM are limited and are typically afflicted by confounding factors [26,27]. In addition, inconsistencies in diagnostic criteria for GDM and measurements of risk factors make it difficult to compare findings across studies. Despite these concerns, several risk factors for GDM emerge consistently. These include overweight/obesity , excessive gestational weight gain , westernized diet , ethnicity , genetic polymorphisms , advanced maternal age , intrauterine environment (low or high birthweight ), family and personal history of GDM , and other diseases of insulin resistance, such as polycystic ovarian syndrome (PCOS) .
Each of these risk factors are either directly or indirectly associated with impaired β-cell function and/or insulin sensitivity. For example, overweight and obesity are intrinsically linked with prolonged, excessive calorie intake, which overwhelms β-cell insulin production and insulin signaling pathways. Even independently of body mass index (BMI) and overall caloric intake, diet and nutrition are associated with GDM. Diets that are high in saturated fats, refined sugars, and red and processed meats are consistently associated with an increased risk of GDM [36,37], while diets high in fiber, micronutrients, and polyunsaturated fats are consistently associated with a reduced risk of GDM [38,39,40]. Saturated fats directly interfere with insulin signaling , and they can also induce inflammation and endothelial dysfunction—both pathogenic factors in GDM . On the other hand, n-3 polyunsaturated fatty acids, including those derived from fish and seafood, have anti-inflammatory properties . The relationship between processed meat and GDM remains strong, even after adjustment for fatty acids, cholesterol, heme iron, and protein content . It has been suggested that by-products related to the processing of meat could be responsible—such as nitrates (a common preservative in processed meats), or advanced glycation end products (AGEs), which have both been implicated in β-cell toxicity [44,45]. Interestingly, even independently of meat consumption, high protein diets are associated with GDM [46,47,48]. One theory for this is the role of amino acids as substrates for hepatic glucose production , and in hepatic lipotoxicity . The inverse association between dietary fiber and GDM may be the result of reduced appetite or slowed glucose absorption, reducing demand on β-cells and insulin signaling mediators .
Low and high birthweight are likely risk factors for GDM because of their association with insulin resistance. Low birthweight is often the result of undernutrition in the womb, either as a result of maternal undernutrition or placental insufficiency. It is believed that the fetus compensates for undernutrition in the womb by epigenetically altering the expression of genes that are involved in fat storage, energy utilisation, and appetite regulation. Further, animal studies suggest that undernutrition in utero is associated with reduced β-cell number . These alterations persist after birth—a phenomenon referred to as “developmental programming” . While potentially beneficial in times of famine, a mismatch between nutritional status in the womb and nutritional status once born can contribute to the development of obesity and metabolic disease [53,54]. On the opposite end of the spectrum, overnutrition in the womb—such as can occur in GDM—can result in fetal overgrowth. These individuals are more likely to have experienced hyperglycemia and β-cell fatigue even before birth, predisposing them to hyperglycemia during times of later metabolic stress, such as during pregnancy .
1.5. Consequences of Gestational Diabetes
The importance of aiming to understand and effectively treat or prevent GDM is illustrated by the wide-ranging consequences of GDM for both the mother and the fetus.
Mother—GDM increases the risk of a number of short-term and long-term maternal health issues. In addition to the stress of normal pregnancy, GDM is associated with antenatal depression . There is also an increased risk of additional pregnancy complications, including preterm birth and preeclampsia, and, in many cases, surgical delivery of the baby is required . Approximately 60% of women with a past history of GDM develop T2DM later in life . Each additional pregnancy also confers a threefold increase in the risk of T2DM in women with a history of GDM. Further, women with a previous case of GDM have a yearly risk of conversion to T2DM of ~2 to 3% . Emerging evidence also suggests that the vasculature of women with a prior case of GDM is permanently altered, predisposing them to cardiovascular disease (CVD). A recent study reported a 63% increased risk of CVD amongst women with a history of GDM, which was partly, but not fully, explained by BMI . This is of major concern, as CVD is the number one cause of death in the world .
Child—GDM also poses short- and long-term consequences for the infant. The aforementioned increase in placental transport of glucose, amino acids, and fatty acids stimulate the fetus’s endogenous production of insulin and insulin-like growth factor 1 (IGF-1). Together, these can cause fetal overgrowth, often resulting in macrosomia at birth . As previously mentioned, excess fetal insulin production can stress the developing pancreatic β-cells, contributing to β-cell dysfunction and insulin resistance, even prenatally . Macrosomia is also a risk factor for shoulder dystocia—a form of obstructed labor. Thus, babies of GDM pregnancies are usually delivered by caesarean section [63,64]. Once delivered, these babies are at increased risk of hypoglycemia, which is likely due to formed dependence on maternal hyperglycemia (fetal hyperinsulinemia), which can contribute to brain injury if not properly managed . There is also evidence that GDM increases the risk of stillbirth . In the long term, babies that are born of GDM pregnancies are at increased risk of obesity, T2DM, CVD, and associated metabolic diseases. Children born to mothers with GDM have almost double the risk of developing childhood obesity when compared with nondiabetic mothers, even after adjusting for confounders such as maternal BMI [67,68], and impaired glucose tolerance can be detected as young as five years old . Females are therefore more likely to experience GDM in their own pregnancies, contributing to a vicious intergenerational cycle of GDM .
2. Pathophysiology of Gestational Diabetes
The remainder of this review will discuss molecular processes underlying the pathophysiology of GDM. GDM is usually the result of β-cell dysfunction on a background of chronic insulin resistance during pregnancy and thus both β-cell impairment and tissue insulin resistance represent critical components of the pathophysiology of GDM. In most cases, these impairments exist prior to pregnancy and can be progressive—representing an increased risk of T2DM post-pregnancy . A number of additional organs and systems contribute to, or are affected by, GDM. These include the brain, adipose tissue, liver, muscle, and placenta.
2.1. β-Cell Dysfunction
The primary function of β-cells is to store and secrete insulin in response to glucose load. When β-cells lose the ability to adequately sense blood glucose concentration, or to release sufficient insulin in response, this is classified as β-cell dysfunction. β-cell dysfunction is thought to be the result of prolonged, excessive insulin production in response to chronic fuel excess . However, the exact mechanisms underlying β-cell dysfunction can be varied and complex [73,74]. Defects can occur at any stage of the process: pro-insulin synthesis, post-translational modifications, granule storage, sensing of blood glucose concentrations, or the complex machinery underlying exocytosis of granules. Indeed, the majority of susceptibility genes that are associated with GDM are related to β-cell function, including potassium voltage-gated channel KQT-like 1 (Kcnq1) and glucokinase (Gck). Minor deficiencies in the β-cell machinery may only be exposed in times of metabolic stress, such as pregnancy .
β-cell dysfunction is exacerbated by insulin resistance. Reduced insulin-stimulated glucose uptake further contributes to hyperglycemia, overburdening the β-cells, which have to produce additional insulin in response. The direct contribution of glucose to β-cell failure is described as glucotoxicity . Thus, once β-cell dysfunction begins, a vicious cycle of hyperglycemia, insulin resistance, and further β-cell dysfunction is set in motion.
Animal studies suggest that β-cell number is also an important determinant of glucose homeostasis. For example, Zucker fatty (ZF) rats that were subjected to 60% pancreatectomy mostly recover β-cell mass by one week post-surgery, but still develop hyperglycemia. In these cases, the short-term but dramatic reduction in β-cell mass overburdens the remaining β-cells, resulting in severely reduced glucose-stimulated insulin secretion and the depletion of internal insulin granule stores . Sprague Dawley rats, which are usually very resistant to the development of diabetes, experience substantial loss of β-cell mass (50% reduction) by 15-weeks old when growth-restricted in utero via bilateral uterine artery ligation . This loss of β-cell mass has been linked to epigenetic downregulation of pancreatic homeobox transcription factor (Pdx1), which is essential for normal β-cell differentiation in the embryo . Prolactin is also essential for adequate β-cell proliferation, as demonstrated in mouse knockouts of the prolactin receptor (PrlR−/−) . In addition, glucotoxicity is also thought to result in β-cell apoptosis over time . Pancreatic samples from T2DM patients can show a reduction of β-cell mass by 40–60% , but less than 24% loss after five years of disease has also been reported . Reduced β-cell hyperplasia may also play a role in GDM, based on animal studies and limited post-mortem human studies . Therefore, reduced β-cell mass, reduced β-cell number, β-cell dysfunction, or a mix of all three contribute to GDM, depending on the individual.
2.2. Chronic Insulin Resistance
Insulin resistance occurs when cells no longer adequately respond to insulin. At the molecular level, insulin resistance is usually a failure of insulin signaling, resulting in inadequate plasma membrane translocation of glucose transporter 4 (GLUT4)—the primary transporter that is responsible for bringing glucose into the cell to use as energy (Figure 1). The rate of insulin-stimulated glucose uptake is reduced by 54% in GDM when compared with normal pregnancy . While insulin receptor abundance is usually unaffected, reduced tyrosine or increased serine/threonine phosphorylation of the insulin receptor dampens insulin signaling . In addition, altered expression and/or phosphorylation of downstream regulators of insulin signaling, including insulin receptor substrate (IRS)-1, phosphatidylinositol 3-kinase (PI3K), and GLUT4, has been described in GDM . Many of these molecular changes persist beyond pregnancy .
Figure 1.
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Simplified diagram of insulin signaling. Binding of insulin to the insulin receptor (IR) activates IRS-1. Adiponectin promotes IRS-1 activation through AMP-activated protein kinase (AMPK), while pro-inflammatory cytokines activate protein kinase C (PKC) via IκB kinase (IKK), which inhibits IRS-1. IRS-1 activates phosphatidylinositol-3-kinase (PI3K), which phosphorylates phosphatidylinositol-4, 5-bisphosphate (PIP2) to phosphatidylinositol-3, 4, 5-phosphate (PIP3). PIP3 activates Akt2, which promotes GLUT4 translocation and glucose uptake into the cell.
Several of the previously discussed risk factors for GDM are thought to exert their effects by interfering with insulin signaling. For example, saturated fatty acids increase intracellular concentrations of diacylglycerol within myocytes, activating protein kinase C (PKC) and inhibiting tyrosine kinase, IRS-1 and PI3K . Pro-inflammatory cytokines and adiponectin also modify this process, as discussed below.
A diagram of the relationship between β-cell dysfunction, insulin resistance, and GDM is provided in Figure 2.
Figure 2.
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β-cell, blood glucose, and insulin sensitivity during normal pregnancy and GDM. During normal pregnancy, β-cells undergo hyperplasia and hypertrophy in order to meet the metabolic demands of pregnancy. Blood glucose rises as insulin sensitivity falls. Following pregnancy, β-cells, blood glucose, and insulin sensitivity return to normal. During gestational diabetes, β-cells fail to compensate for the demands of pregnancy, and, when combined with reduced insulin sensitivity, this results in hyperglycemia. Following pregnancy, β-cells, blood glucose, and insulin sensitivity may return to normal or may remain impaired on a pathway toward GDM in future pregnancy or T2DM. Pancreas image obtained from The Noun Project under the terms and conditions of the Creative Commons Attribution (CC BY) license ( by artist Arif Fajar Vulianto.
2.3. Neurohormonal Networks
Neurohormonal dysfunction has been implicated in the pathogenesis of diseases of insulin resistance, such as that present in GDM. This network regulates appetite, active energy expenditure, and basal metabolic rate, and it is made up of a complex network of central (e.g., cortical centers that control cognitive, visual, and “reward” cues) and peripheral (e.g., satiety and hunger hormones) signals [87,88]. These contribute to GDM by influencing adiposity and glucose utilization. This network is highly regulated by the circadian clock, which may explain why pathological sleep disorders or those individuals undertaking shift work are correlated with GDM rates [89,90]. Neural networks controlling body weight are most likely set in early life, as demonstrated in animal studies. For example, rats that are both under- and over-fed in early life experience epigenetic alteration of the regulatory set-point of hypothalamic neurons [91,92]. This adds to the previously mentioned suggestion that predisposition to GDM may be set in the womb.
Some of the most important regulators of neurohormonal metabolic control are adipokines—cell signaling proteins that are secreted primarily by adipose tissue. These include leptin and adiponectin:
2.3.1. Leptin
Leptin is a satiety hormone secreted primarily by adipocytes in response to adequate fuel stores. It primarily acts on neurons within the arcuate nucleus of the hypothalamus to decrease appetite and increase energy expenditure. Specifically, leptin inhibits appetite-stimulators neuropeptide Y (NPY) and agouti-related peptide (AgRP), and it activates the anorexigenic polypeptide pro-opiomelanocortin (POMC) . When leptin was first discovered, it was lauded as a potential treatment for obesity . However, it was soon revealed that the majority of obese individuals do not respond to leptin, and instead demonstrate leptin resistance. While leptin treatment is effective in obesity that is caused by leptin and leptin receptor genetic polymorphisms, these are rare (<5% of obese individuals) . Therefore, obesity is associated with excessive plasma leptin concentration (hyperleptinemia) as a result of leptin resistance, and plasma leptin concentrations are generally proportional to the degree of adiposity . Leptin resistance can occur either as a defect in blood-brain barrier leptin transport, or through intracellular mechanisms that are similar to insulin resistance . Like insulin resistance, a degree of leptin resistance occurs in normal pregnancy, presumably to bolster fat stores beyond what would usually be required in the non-pregnant state. Leptin resistance is further increased in GDM, resulting in hyperleptinemia . However, pre-pregnancy BMI is a stronger predictor of circulating leptin than GDM per se .
The placenta also secretes leptin during human pregnancy. In fact, the placenta is responsible for the majority of plasma leptin during pregnancy . Placental leptin production is increased in GDM, probably as a result of placental insulin resistance, and this further contributes to hyperleptinemia. This is also thought to facilitate amino acid transport across the placenta, contributing to fetal macrosomia .
2.3.2. Adiponectin
Similar to leptin, adiponectin is a hormone that is primarily secreted by adipocytes. However, plasma adiponectin concentrations are inversely proportional to adipose tissue mass, with low concentrations in obese individuals. GDM is similarly associated with decreased adiponectin . In contrast to leptin, there is a stronger association of adiponectin with insulin resistance than with adiposity . This suggests that adiponectin plays an important role in the pathogenesis of GDM, independent of obesity. Adiponectin enhances insulin signaling and fatty acid oxidation, and it inhibits gluconeogenesis . It does so by activating AMP-activated protein kinase (AMPK) within insulin-sensitive cells, which facilitates the action of IRS-1 (Figure 1), and by activating the transcription factor peroxisome proliferator-activated receptor alpha (PPARα) in the liver. Furthermore, adiponectin stimulates insulin secretion, by upregulating insulin gene expression and exocytosis of insulin granules from β-cells .
Adiponectin is also expressed at low concentration from the syncytiotrophoblast of the placenta where it is regulated by cytokines, such as tumor necrosis factor alpha (TNF-α), interleukin (IL)-6, interferon gamma (IFN-γ), and leptin . The role of placental adiponectin in normal and GDM pregnancy is unclear . However, emerging evidence suggests adiponectin impairs insulin signaling and amino acid transport across the placenta, limiting fetal growth. Therefore, adiponectin gene methylation in the placenta is associated with maternal glucose intolerance and fetal macrosomia .
2.4. Adipose Tissue
Originally believed to exist only as a passive depot of energy, the discovery of leptin in 1994 established adipose tissue as an essential endocrine organ. Adipose tissue both ensures that energy is partitioned safely and it actively secretes circulatory factors, including adipokines (the aforementioned leptin and adiponectin) and cytokines (such as TNF-α, IL-6, and IL-1β), which have wide-ranging metabolic effects.
2.4.1. Energy Storage
The storage capability of adipose tissue is essential for metabolic health. This is exemplified through two extremes: rare disorders in which white adipose tissue is absent lead to severe metabolic syndrome, whereas some obese individuals (with excessive white adipose tissue) do not develop metabolic syndrome at all . Therefore, the ability to partition excess calories into adipose tissue rather than ectopically in the liver, muscle, or pancreas, appears to serve as a protective measure. Non-diabetic obese individuals exhibit adequate adipose tissue expansion in response to fuel surfeit, and therefore maintain healthy blood glucose concentrations, sufficient β-cell compensation, and avoid chronic insulin resistance [110,111]. In this way, key organs avoid glucose and fatty acid-induced tissue damage. As previously mentioned, early pregnancy is marked by an increase in adipose tissue mass, while later pregnancy promotes the mobilization of fats from adipose tissue in order to fuel fetal growth. Both of these processes are thought to be limited in GDM . GDM is associated with reduced adipocyte differentiation and increased adipocyte size (hypertrophy), accompanied by downregulated gene expression of insulin signaling regulators, fatty acid transporters, and key adipogenic transcription factors, such as PPARγ . The combination of insulin resistance and reduced adipocyte differentiation hinders the tissue’s ability to safely dispose of excess energy, contributing to gluco- and lipo-toxicity in other peripheral organs. Indeed, both T2DM and GDM are associated with lipid deposition in muscle and liver [114,115].
2.4.2. Adipose Tissue Inflammation
Obesity, T2DM and GDM are associated with an increased number of resident adipose tissue macrophages (ATM) that secrete pro-inflammatory cytokines, including TNF-α, IL-6, and IL-1β. The importance of a low-grade inflammatory state in the pathogenesis of insulin resistance has recently become apparent. Pro-inflammatory cytokines have been discovered to both impair insulin signaling and inhibit insulin release from β-cells. These factors induce insulin resistance either by diminishing insulin receptor (IR) tyrosine kinase activity, increasing serine phosphorylation of IRS-1, or through the STAT3-SOCS3 pathway, which degrades IRS-1 [85,116]. Circulating concentrations of pro-inflammatory cytokines are increased in GDM [107,117]. Plasma TNF-α, in particular, is strongly correlated with insulin resistance . Similarly, placental gene expression of TNF-α, IL-1β and their receptors has been reported to be increased in GDM [118,119]. However, the relationship between pregnancy and inflammation is complex. For example, Lappas et al. (2010) reported that GDM placentae secrete fewer pro-inflammatory cytokines (3 of 16 studied: IL-1β, TNF-α and M1P1B) than healthy placentae (13 out of 16 studied) . This suggests that, while chronic low-grade inflammation appears to be important in the pathogenesis of GDM, the relationship may not be straightforward.
2.5. Liver
GDM is associated with upregulated hepatic glucose production (gluconeogenesis). Gluconeogenesis is increased in the fasted state, and not adequately suppressed in the fed state . This is not believed to be entirely the result of inaccurate glucose sensing due to insulin resistance, as the majority of glucose uptake by the liver (~70%) is not insulin dependent. Common factors between the insulin signaling pathway and the pathways controlling gluconeogenesis, such as PI3K, might contribute to these effects . Increased protein intake and muscle breakdown may also stimulate the process by providing excess gluconeogenesis substrate . Despite this, the liver does not seem to be a primary pathogenic driver of T2DM or GDM .
2.6. Skeletal and Cardiac Muscle
Traditionally, skeletal muscle insulin resistance was believed to play a causal role in T2DM. However, skeletal muscle insulin resistance now appears to be a consequence of hyperglycemia—a protective measure to prevent metabolic stress and steatosis . Even following a short period of overfeeding, cardiac and skeletal muscle develop insulin resistance in order to divert the excess energy into adipose tissue . This is an important distinction when considering potential treatments for GDM: attempts to directly reverse skeletal muscle insulin resistance, without reducing plasma glucose concentrations, could be detrimental .
Separate to insulin sensitivity, T2DM and GDM are associated with a reduced number and function of mitochondria within skeletal muscle cells . This could be the result of genetics, early-life programming, or chronic inactivity. Therefore, decreased number and function of mitochondria is likely an additional contributor to reduced glucose utilization in GDM.
2.7. Gut Microbiome
There is emerging evidence that microbial organisms within the gut—the “gut microbiome”—might contribute to metabolic diseases, including GDM. The gut microbiome can be influenced by early-life events, such as preterm delivery and breastfeeding, and by events in later life, such as diet composition and antibiotic use. The gut microbiome has been consistently reported to differ between metabolically healthy and obese individuals, including during pregnancy . Furthermore, a study of stool bacteria in women with a past case of GDM reported a lower proportion of the phylum Firmicutes and higher proportion of the family Prevotellaceae as compared with normoglycemic pregnancy . Similar associations have been observed in obesity , T2DM , fatty liver disease , and elevated total plasma cholesterol . Firmicutes metabolize dietary plant polysaccharides. This may explain some of the dietary risk factors for GDM that are discussed earlier. Both red meat and animal protein decrease levels of Firmicutes, while high dietary fiber increase them . However, the findings by Fugmann et al. (2015) remained after adjustment for dietary habits . Therefore, Firmicutes appear to be relevant to pathogenesis of GDM independent of diet, although the mechanisms underlying this are unknown. Prevotellaceae are mucin-degrading bacteria that may contribute to increased gut permeability. Gut permeability is regulated by tight junction proteins, such as zonulin (ZO-1). Increased “free” plasma/serum ZO-1 is associated with type 1 diabetes (T1DM), T2DM , and GDM . Increased gut permeability is thought to facilitate the movement of inflammatory mediators from the gut into the circulation, promoting systemic insulin resistance [134,136].
2.8. Oxidative Stress
Oxidative stress describes an imbalance between pro-oxidants and antioxidants in cells. Oxidative stress can lead to cellular damage by interfering with the state of proteins, lipids and DNA, and has been implicated in the pathogenesis of many diseases, including GDM . Reactive oxygen species (ROS) are described as free radical and nonradical derivatives of oxygen, and include superoxide anion (O 2−), hydroxyl radical (•OH) and hydrogen peroxide (H 2 O 2) . A hyperglycemic environment is associated with oxidative stress, and GDM women have been reported to overproduce free radicals and have impaired free-radical scavenging mechanisms . ROS inhibit insulin-stimulated glucose uptake by interfering with both IRS-1 and GLUT4 . ROS also slow glycogen synthesis in the liver and muscle. Pro-inflammatory cytokines, such as TNF-α, may also contribute to oxidative stress by increasing the expression and the activation of ROS precursors, like NADPH oxidase 4 (NOX4) .
Interestingly, iron supplementation in women already replete in iron is associated with GDM . Several studies suggest that this relationship is the result of increased oxidative stress. Iron is a transitional metal and it can catalyze the reaction from O 2− and H 2 O 2 to the extremely reactive •OH within mitochondria . On the contrary, selenium and zinc are transitional metals that are necessary for the activity of some antioxidant enzymes, which may explain their inverse association with GDM .
Homocysteine—a non-protein α-amino acid that is formed by the demethylation of methionine—is also thought to contribute to GDM via oxidative stress. Exposure of β-cells to even small amounts of homocysteine results in dysfunction and impaired insulin secretion . A recent meta-analysis examined the relationship between serum homocysteine concentration and GDM in ten eligible studies. The authors reported significantly higher homocysteine concentrations among women with GDM as compared with those without GDM . B vitamins, including folic acid, B2, B6, and B12 are essential for homocysteine homeostasis, and this may be one reason why deficiencies and imbalances of these micronutrients are associated with GDM .
2.9. Placental Transport
The placenta contributes to insulin resistance during pregnancy via its secretion of hormones and cytokines. As the barrier between the maternal and fetal environments, the placenta itself is also exposed to hyperglycemia and its consequences during GDM. This can impact transport of glucose, amino acids, and lipids across the placenta:
Glucose— Glucose is the primary energy source for the fetus and the placenta, and therefore must be readily available at all times. For this reason, insulin is not required for the placental transport of glucose. Instead, glucose transport occurs via GLUT1, by carrier-mediated sodium-independent diffusion . However, the placenta still expresses the insulin receptor, and insulin signaling can influence placental metabolism of glucose . The receptiveness of the placenta to glucose uptake means that it is particularly sensitive to maternal hyperglycemia, and this directly contributes to increased fetal growth and macrosomia.
Protein— Amino acid transport across the placenta is also an important determinant of fetal growth. GDM is associated with increased System A and L activity . These can also be modulated by pro-inflammatory cytokines, such as TNF-α and IL-6 . Altered amino acid transport may also be one mechanism by which excess protein intake contributes to GDM.
Lipids— Finally, while GDM has traditionally been described as a disease of hyperglycemia, the rise in obesity-associated GDM has prompted a greater focus on the role of hyperlipidemia in GDM. The majority of placental gene expression alterations in GDM occur in lipid pathways (67%), as compared with glucose pathways (9%) . Preferential activation of placental lipid genes is also associated with GDM compared with T1DM . These data correlate with the results of the HAPO Study, which revealed independent effects of maternal obesity and glucose on excessive fetal growth . Therefore, it appears that GDM influences the placental transport of glucose, amino acids, and fatty acids, and that all three must be considered when discussing the impact of GDM on placental function and fetal growth.
In addition to these alterations in placental transport, GDM has been associated with other changes in the placenta. Some recent studies have reported that GDM is associated with placenta global DNA hypermethylation . Similarly, studies of the placental proteome have identified differences in the expression of proteins between GDM and non-GDM placentas . However, more research is required before the role of placental epigenetic and proteomic modifications in GDM is fully understood . There has also been recent interest in small noncoding single-stranded segments of RNA, called microRNAs (miRNAs), expressed in placental trophoblast cells. miRNAs are involved in a number of cellular processes, including proliferation, differentiation, and apoptosis. Emerging evidence suggests that exosomes containing miRNAs are shed from the placenta during gestation and released into the maternal circulation, which can in turn influence the functioning of other cells, potentially contributing to the pathogenesis of GDM [157,158]. Interestingly, exposure to endocrine disrupting chemicals (EDCs), including bisphenol A (BPA—found in food packaging materials and consumer products) has been associated with GDM, and it has been suggested that this could be because EDCs induce exosome signaling from the placenta . Interestingly, EDCs including BPA have also been associated with alterations in methylation, perhaps linking the two mechanisms . A summary diagram of the pathophysiology of GDM is presented in Figure 3.
Figure 3.
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Organs involved in the pathophysiology of GDM (Images in this figure were obtained from The Noun Project under the terms and conditions of the Creative Commons Attribution (CC BY) license ( Brain and Gut by Hunotika; Liver by Lavmik; Pancreas by Arif Fajar Vulianto; Placenta by Charmeleon Design; Muscle by Misha Petrishchev).
3. Opportunities and Considerations for Future Study
Uncovering the intricate molecular mechanisms underlying GDM is challenging, but necessary for our greater understanding of the disease and how these could assist in the design of new treatments. As β-cell dysfunction and insulin resistance are the hallmarks of GDM, the greatest emphasis should be placed on further understanding the mechanisms underlying these processes. For example, why do β-cells exhibit proper hyperplasia and hypertrophy in some pregnancies, but not others? How could we modify these processes to bolster pancreatic function and prevent hyperglycemia in at-risk individuals? As already mentioned, increasing insulin sensitivity could have unintended consequences by promoting uptake of glucose into tissue where energy should not be stored, such as the liver and skeletal muscle. Instead, investment into adipose-specific insulin sensitivity should be examined. While improving adipose capacity (and in theory increasing adipose tissue mass) might seem counterintuitive, in actuality, it should reduce hyperglycemia while ensuring that excess energy is stored safely. Of course, many of the mechanisms underlying GDM are not unique to GDM, encompassing other common disorders of insulin sensitivity, such as T2DM, prediabetes, and PCOS. Therefore, determination of pathways influencing development of these metabolic disorders may also shed light on GDM, and potentially accelerate opportunities for prevention and/or treatment. This is an important consideration, as the study of GDM (as a disorder of pregnancy) is limited for ethical reasons. Finally, the ability to study large amounts of data through computer technology is rapidly advancing the fields of genomics, epigenetics, proteomics, metagenomics (the microbiome), and metabolomics (the study of the small-molecule intermediates and products of metabolism). It is hopeful that the advancement of these large-scale techniques may assist in our understanding of the pathogenesis of GDM in the future.
4. Conclusions
Pregnancy is a state of high metabolic activity, in which maintaining glucose homeostasis is of upmost importance. When hyperglycemia is detected in the pregnant mother, this is referred to as GDM, although controversy remains over diagnostic criteria. It is likely that genetic, epigenetic, and environmental factors all contribute to the development of GDM, and that the mechanisms involved are complex and advance over a substantial period of time. However, in the majority of cases, pancreatic β-cells fail to compensate for a chronic fuel surfeit, leading to eventual insulin resistance, hyperglycemia, and an increased supply of glucose to the growing fetus. There is also evidence that adipose expandability, low-grade chronic inflammation, gluconeogenesis, oxidative stress, and placental factors contribute to the pathology of GDM. Greater understanding of these processes and their contribution to GDM is required in order to develop effective treatments and prevention strategies.
Author Contributions
J.F.P. primarily wrote the manuscript. J.L.S., C.M.R., P.N.B. and M.H.V. provided supervisory and editorial assistance.
Funding
This research received no external funding.
Conflicts of Interest
The authors declare no conflict of interest.
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9552 | https://stats.stackexchange.com/questions/270428/how-do-i-show-that-the-sample-median-minimizes-the-sum-of-absolute-deviations | self study - How do I show that the sample median minimizes the sum of absolute deviations? - Cross Validated
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How do I show that the sample median minimizes the sum of absolute deviations? [duplicate]
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Mean and Median properties (4 answers)
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I want to show that the sample median x~x~ minimizes the sum of absolute deviations,
i.e., x~=a r g m i n a∑n i=1|x i−a|x~=a r g m i n a∑i=1 n|x i−a|
To show this, so far I have:
Sum of absolute deviation about a is
=∑n i=1|x i−a|=∑i=1 n|x i−a|
Assume that x 1≤x 2≤...≤x n x 1≤x 2≤...≤x n
CASE 1: When n is ODD (n = 2m + 1)
We get,
x 1≤x 2≤...≤x m≤x m+1≤...≤x 2 m≤x 2 m+1 x 1≤x 2≤...≤x m≤x m+1≤...≤x 2 m≤x 2 m+1
From this we can see that
|x 1−a|+|x 2 m+1−a||x 1−a|+|x 2 m+1−a| is least when x 1≤a≤x 2 m+1 x 1≤a≤x 2 m+1
|x 2−a|+|x 2 m−a||x 2−a|+|x 2 m−a| is least when x 2≤a≤x 2 m x 2≤a≤x 2 m
...
Thus,
|x m+1−a||x m+1−a| is least when a=x m+1=m e d i a n a=x m+1=m e d i a n
CASE 2: When n is EVEN (n = 2m) We get,
x 1≤x 2≤...≤x m≤x m+1≤...≤x 2 m−1≤x 2 m x 1≤x 2≤...≤x m≤x m+1≤...≤x 2 m−1≤x 2 m
From this we can see that
|x 1−a|+|x 2 m−a||x 1−a|+|x 2 m−a| is least when x 1≤a≤x 2 m x 1≤a≤x 2 m
...
Thus,
|x m−a|+|x m+1−a||x m−a|+|x m+1−a| is least when x m≤a≤x m+1 x m≤a≤x m+1
Is this enough to show that sample median minimizes the sum of absolute deviations?
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asked Mar 28, 2017 at 21:41
user155019user155019
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1 The proof seems to be absolutely correct to me. I like the fact that you knew that the self-study tag was required.Michael R. Chernick –Michael R. Chernick 2017-03-28 21:47:41 +00:00 Commented Mar 28, 2017 at 21:47
1 The idea of a proof is explained under "Absolute (L 1 L 1) Loss" in my answer at stats.stackexchange.com/a/114363/919. The generalization to any percentile (the median is the 50th percentile) is addressed at stats.stackexchange.com/questions/251600. Neither one requires splitting into odd and even cases: they can be handled in the same manner.whuber –whuber♦ 2017-03-28 22:56:54 +00:00 Commented Mar 28, 2017 at 22:56
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When all of the x i x i are distinct, this is easy. If
f(m)=∑i=1 n|x i−m|f(m)=∑i=1 n|x i−m|
then
f′(m)=∑i=1 n s i g n(x i−m)f′(m)=∑i=1 n s i g n(x i−m)
which equals zero when there are an equal number of elements of x 1,...,x m x 1,...,x m that are above and below m m, which is the definition of the median, m⋆m⋆. As a function of m m, this is decreasing on (−∞,m⋆)(−∞,m⋆) and increasing on (m⋆,∞)(m⋆,∞), so m⋆m⋆ is a minimizer.
Note: If n n is even, the sample median is not necessarily uniquely defined. In that case, what is the "right" point estimate of the median is debatable. So, m⋆m⋆ could be called any value over the open interval where f′f′ is zero (a common convention is taking the midpoint... e.g. see what happens when you type median(1:4) into R....you will get 2.5), but the basic logic of what I wrote above proves it would still minimize the MAD....But so would any value between 2 and 3...
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edited Mar 29, 2017 at 3:05
answered Mar 29, 2017 at 2:43
gammergammer
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1 The problem with this proof is that f′f′ is not generally zero on an open interval, or anywhere. Example, f(x)=|x−1|+|x−2|+|x−3|f(x)=|x−1|+|x−2|+|x−3|. Or even f(x)=|x|f(x)=|x|.Flounderer –Flounderer 2017-03-29 02:55:03 +00:00 Commented Mar 29, 2017 at 2:55
Not sure if this was your point but I can see this proof won't always work if there are ties in the data set, e.g. if the x i x i are 1,1,1,3. I modified the answer to indicate this applies when the x i x i are all distinct, e.g. the median of a sample from a continuous distribution.gammer –gammer 2017-03-29 03:09:17 +00:00 Commented Mar 29, 2017 at 3:09
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9553 | https://www.sjsu.edu/faculty/watkins/Digitsum0.htm | | applet-magic.com Thayer Watkins Silicon Valley, Tornado Alley & BB Island USA |
| |
| The Sum of Digits for Multiples of Numbers |
It is well known that the digits of multiples of nine sum to nine; i.e.,
9→9, 18→1+8=9, 27→2+7=9, . . ., 99→9+9=18→1+8=9, 108→1+0+8=9,
etc. Less well known is that the sum of digits of multiples of other numbers have simple
patterns although not so simple as the case of nine. These are shown below:
| | |
--- |
| Number | Repeating Cycle of Sum of Digits of Multiples |
| 2 | {2,4,6,8,1,3,5,7,9} |
| 3 | {3,6,9,3,6,9,3,6,9} |
| 4 | {4,8,3,7,2,6,1,5,9} |
| 5 | {5,1,6,2,7,3,8,4,9} |
| 6 | {6,3,9,6,3,9,6,3,9} |
| 7 | {7,5,3,1,8,6,4,2,9} |
| 8 | {8,7,6,5,4,3,2,1,9} |
| 9 | {9,9,9,9,9,9,9,9,9} |
| 10 | {1,2,3,4,5,6,7,8,9} |
| 11 | {2,4,6,8,1,3,5,7,9} |
| 12 | {3,6,9,3,6,9,3,6,9} |
| 13 | {4,8,3,7,2,6,1,5,9} |
It is asserted that the sum of digits follows a repeating sequence of length 9. This is
so because for any decimal representation x, the number 10x (ten times x) will have the same sum of digits.
Multiplication by the base number 10 simply moves the
digits one place to the left and puts a zero in the units place. More generally,
DigitSum(10kn) = DigitSum(n).
This operation, multiplication by the k-th power of 10, moves the digits k places to the
left and places k zeroes on the right. The sum of digits is the same.
Less obviously adding a nine at any place in a decimal representation reduces the digit by one and adds one to the
digit in the next higher place, and thus the sum of the digits is not altered. That is to say,
DigitSum(n+910k) = DigitSum(n).
Going back to the above table, what is immediately suggested by the information presented there is that
the sequence for a multiple digit number m is the sequence for the sum of the digits of m,
the digit sum of m.
For example, the sequence for 12 is the same as the sequence for 1+2=3. Likewise the
sequence for 13 is the same as the sequence for 1+3=4. This of course also holds true for
10.
The sequences for 3 and 6 are composed of the subsequences {3,6,9} and {6,3,9}, respectively.
Thus the sequences for these digits have three copies of the subsequences of length 3. The
sequence for 9 is nine copies of a subsequence of length 1. Note that {3,6,9}=3{1,2,3} and that
{6,3,9}=3{2,1,3}. It is also worth noting that for all digits except 3, 6 and 9 the length of the sequence is
nine and that for 3, 6 and 9 the lengths of their sequences, 3 and 1, are factors of 9.
The structures of the length 9 sequences are noteworthy. Here is a naive perception of the structure
of the sequences. For 2 the sequence is the even
digits in ascending order then the odd digits also in ascending order. For 4 the sequence
is two generally descending sequences interleaved; i.e., 4,3,2,1,9 with 8,7,6,5 interleaved.
For 5 the sequences is also two interleaved sequences but ascending rather than descending; i.e.,
5,6,7,8,9 with 1,2,3,4 interleaved. For 7 the sequence is a descending sequence of the odd
digits starting with 7 then the even digits in descending order with 9 the odd digit left
out coming last. For 8 the sequence is a single descending sequence of digits starting
with 8 and finishing with the digit 9 which was left out of the descending sequence.
Later a more mathematical explanation of the structure of the sequences will be given.
The DigitSum Function
Let n be a number and let Dec10(n) be the decimal representation of n.
Let DigitSum(z) be the ultimate sum of digits of a decimal representation z; i.e., if the
sum of digits of a decimal representation is greater than nine then the sum of that number's
digits
is computed until a single digit is ultimately obtained.
This function, DigitSum( ), has several interesting properties; i.e.,
DigitSum(x+y) = DigitSum(DigitSum(x) + DigitSum(y))
DigitSum(x−y) = DigitSum(DigitSum(x) − DigitSum(y))
DigitSum(xy)=DigitSum(DigitSum(x)y))
DigitSum(xy)=DigitSum(xDigitSum(y))
DigitSum(xy)=DigitSum(DigitSum(x)DigitSum(y))
See Digit Sums for an analysis and proof of these propositions.
The proposition that DigitSum(xy)=DigitSum(DigitSum(x)y)) establishes that the sequences for the
multiples of 12 and of 13 are the same as the sequences for 3 (1+2) and 4 (1+3),
respectively.
Decimal Representation of Numbers
In order for the following to make sense one must stop thinking of a number in terms of
its decimal representation and think of a number in terms of an appropriate number of tally
marks so three would be (|||) and eleven (||||||||||||).
Consider how one obtains the decimal representation of a number. To get the last digit one
divides the number by ten and takes the remainder as the last digit. That last digit is
subtracted from the number and the result divided by ten. Then the decimal representation of
that quotient is sought. The process is repeated and the next the last digit is obtained.
An alternate characterization of the process of finding the decimal representation of a number is
that the k-th power digit for a whole number n is:
ck = (trunc[n/10k])%10
where trunc[] means the fractional part is thrown away and m%10 means the remainder after
division by 10. In terms of the terminology from the programming language Pascal the
formula is
ck(n) = (n div 10k) mod 10
The sum of the digits for a number n is then
Sum = Σk ck(n),
but this is not necessarily the digit sum for the number. The process has to be repeated
iteratively on the sum of the digits.
The Explanation of the Patterns of the Sequences
Consider two digits, a and b. If their sum is less than ten then
DigitSum(a+b) = DigitSum(a)+DigitSum(b) and hence DigitSum(a+b) = DigitSum(DigitSum(a)+DigitSum(b)).
But, if the sum of a and b is ten or more then the decimal representation of their sum is a one in
the ten's place and (a+b−10) in the unit's place. Thus
DigitSum(a+b) = 1 + (a+b−10) = a+b−(10−1) = a+b−9
For digits the DigitSum(a)=a and DigitSum(b)=b so
DigitSum(DigitSum(a)+DigitSum(b)) = 1 + (a+b−10).
Therefore for any two digits a and b
DigitSum(a+b) = DigitSum(DigitSum(a)+DigitSum(b)).
This applies as well to the digits in the k-th place. Thus the general proposition for
any two decimal representations of numbers, x and y
DigitSum(a+b) = DigitSum(DigitSum(a)+DigitSum(b)).
For differences, if a and b are digits and a>b then
DigitSum(a−b) = DigitSum(DigitSum(a)−DigitSum(b)).
On the other hand if a<b then (a−b)=−(b−a) and,
since DigitSum(−c) = DigitSum(c), it also holds that
DigitSum(a−b) = DigitSum(DigitSum(a)−DigitSum(b)).
Again this extends to the digits in any place in a decimal representation of a number.
For any decimal numbers x and y then
DigitSum(x±y) = DigitSum(DigitSum(x)±DigitSum(y)).
Since multiplication is simply repeated addition it also follows that
DigitSum(xy) = DigitSum(DigitSum(x)y) DigitSum(xy) = DigitSum(xDigitSum(y)) and finally DigitSum(xy) = DigitSum(DigitSum(x)DigitSum(y)).
It was previously noted that for any two digits whose sum is greater than ten,
DigitSum(a+b) = 1 + (a+b−10) = a+b−(10−1) = a+b−9
In general then for any decimal representation x
DigitSum(x) = Sumofdigits(x) − m9 where m is such that DigitSum(x) is reduced to a single digit.
Another way of expressing this is that the DigitSum for a number n is simply the
remainder after division by 9; i.e., DigitSum(n)=(n%9). DigitSum arithmetic is simply
arithmetic modulo 9.
For comparison the multiplication table for modulo 9 arithmetic is:
<
| Multiplication Table for Modulo 9 Arithmetic | | | | | | | | |
--- --- --- ---
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 0 | 2 | 4 | 6 | 8 | 1 | 3 | 5 | 7 |
| 0 | 3 | 6 | 0 | 3 | 6 | 0 | 3 | 6 |
| 0 | 4 | 8 | 3 | 7 | 2 | 6 | 1 | 5 |
| 0 | 5 | 1 | 6 | 2 | 7 | 3 | 8 | 4 |
| 0 | 6 | 3 | 0 | 6 | 3 | 0 | 6 | 3 |
| 0 | 7 | 5 | 3 | 1 | 2 | 6 | 4 | 2 |
| 0 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |
If the 0's were replaced by 9's and the table rearranged so the first column becomes the last column and the first row
becomes the last row the result would be
identical to the table for the sequences of digit sums.
| The Rearrangement of the Multiplication Table for Modulo 9 Arithmetic With 9 Substituted for 0 | | | | | | | | |
--- --- --- ---
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 2 | 4 | 6 | 8 | 1 | 3 | 5 | 7 | 9 |
| 3 | 6 | 9 | 3 | 6 | 9 | 3 | 6 | 9 |
| 4 | 8 | 3 | 7 | 2 | 6 | 1 | 5 | 9 |
| 5 | 1 | 6 | 2 | 7 | 3 | 8 | 4 | 9 |
| 6 | 3 | 9 | 6 | 3 | 9 | 6 | 3 | 9 |
| 7 | 5 | 3 | 1 | 2 | 6 | 4 | 2 | 9 |
| 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 9 |
| 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 |
Some Proofs
The proof of the property that the DigitSum of any multiple of nine is equal to nine starts with the obvious proposition
DigitSum(10n) = DigitSum(n).
From this it follows from the proposition that
DigitSum(x−y)=DigitSum(x)−DigitSum(y) that
DigitSum((10−1)n) = 0 and thus DigitSum(9n) = 0 but in modulo 9 arithmetic 0 is the same as 9, so DigitSum(9n) = 9
The way to understand the sequence for 8 is that when 8 is added to any digit except 0 or 1
in any place the digit is reduced by 2 and one added to the digit of the next place. This
results in a net decrease in the sum of digits of one. Thus when 8 is added to 8 the digit
becomes 6, a reduction of 2, and 1 is added to the next higher digit, a net decrease in the
sum of the digits of 1. Thus added 8 to 8 results in a sum of digits of 7. Adding 8 to 7
results in a sum of digits of 6, and so on down to adding 8 to 1 which gives 9. Even this
fits into the rule in the sense that if 1 were reduce by 1 the result would be 0 which is
equivalent to 9 modulo 9. Likewise adding 7 to a digit reduces it by 3 and adds 1 to the
digit in the next place, a net reduction in the sum of digits of 2. Thus when 7 is added to 7
the sum of digits is reduced to 5. When 7 is added to 5 the sum of digits is reduced to 3.
When 7 is added to 3 the sum of digits is reduced to 1. If 7 is added to 1 the result is 8,
but that 8 could be consider as a reduction of 1 by 2 modulo 9. The addition of 7 to 8 results
in a sum of digits of 6 and so on down to a sum of digits of 2. The addition of 7 to 2
results in a sum of digits of 9, but that 9 can be considered 0 modulo 9 and thus it is
a reduction of 2.
Generalization to Other Number Bases
There is nothing special about 9; it is simply the number base ten less one. The digitsum sequences for multiples in the
hexadecimal (base 16) number system is:
| | |
--- |
| Number | Repeating Cycle of Sum of Digits of Multiples |
| 2 | {2,4,6,8,a,c,e,1,3,5,7,9,b,d,f} |
| 3 | {3,6,9,c,f,3,6,9,c,f,3,6,9,c,f} |
| 4 | {4,8,c,1,5,9,d,2,6,a,e,3,7,b,f} |
| 5 | {5,a,f,5,a,f,5,a,f,5,a,f,5,a,f} |
| 6 | {6,c,3,9,f,6,c,3,9,f,6,c,3,9,f} |
| 7 | {7,e,6,d,5,c,4,b,3,a,2,9,1,8,f} |
| 8 | {8,1,9,2,a,3,b,4,c,5,d,6,e,7,f} |
| 9 | {9,3,c,6,f,9,3,c,6,f,9,3,c,6,f} |
| a | {a,5,f,a,5,f,a,5,f,a,5,f,a,5,f} |
| b | {b,7,3,e,a,6,2,d,9,5,1,c,8,4,f} |
| c | {c,9,6,3,f,c,9,6,3,f,c,9,6,3,f} |
| d | {d,b,9,7,5,3,1,e,c,a,8,6,4,2,f} |
| e | {e,d,c,b,a,9,8,7,6,5,4,3,2,1,f} |
| f | {f,f,f,f,f,f,f,f,f,f,f,f,f,f,f,f} |
In general, for numbers of base b the digit sums are equivalent to arithmetic
modulo (b−1).
The sequences which contain subsequences are the multiples of the factors of (b−1)
other than unity.
In the case of b=ten these factors are 3 and 9. Therefore there are subsequences for
{3,6,9}. The lengths of the subsequences are (b−1) divided by the factors; i.e.,
in the case of b=ten 3 and 1.
For b=sixteen the factors are three, five
and fifteen and so subsequences occur for {3,6,9,c,f,5,a} and the lengths of the
subsequences are 3, 5 and 1.
| |
| HOME PAGE OF applet-magic HOME PAGE OF Thayer Watkins | |
9554 | https://mathcentre.ac.uk/resources/uploaded/half-lives.pdf | Calculating the value after a specified time period, or the time taken to reach a specified value. Step 1: Tabulate the time and value for each half-life 2hr = 1 half −life = 1200 ÷ 2 = 600𝑚𝑚/𝐿 20 AT A GLANCE/ PHARMACY CALCULATIONS HALF-LIVES 17 Half life The half-life of a drug is is the period of time required for its concentration or amount in the body to be reduced by exactly one-half. The symbol for half-life is T1/2. Example 1 Drug A has a half-life of 2 hours. If the initial plasma level of the drug, given as a single dose, is 1200mg/L, what will its plasma level be after 8 hours? Method 4hr = 2 half −life = 600 ÷ 2 = 300𝑚𝑚/𝐿 6hr = 3 half −life = 300 ÷ 2 = 150𝑚𝑚/𝐿 8hr = 4 half −life = 150 ÷ 2 = 𝟕𝟕𝒎𝒎/𝑳 0 20 40 60 80 100 0 1 2 3 4 5 6 7 8 VALUE (%) T1/2 ?
Student Learning Advisory Service Contact us Please come and see us if you need any academic advice or guidance. Canterbury Our offices are next to Santander Bank Open Monday to Friday, 09.00 – 17.00 E: learning@kent.ac.uk T: 01227 824016 Medway We are based in room G0-09, in the Gillingham Building and in room DB034, in the Drill Hall Library. Open Monday to Friday, 09.00 – 17.00 E: learningmedway@kent.ac.uk T: 01634 888884 The Student Learning Advisory Service (SLAS) is part of the Unit for the Enhancement of Learning and Teaching (UELT) www.kent.ac.uk/learning kent.slas @unikentSLAS Acknowledgments All materials checked by Dr Scott Wildman, Dr Cleopatra Branch, Jerome Durodie and Andrew Lea, Medway School of Pharmacy, Anson Building, Central Avenue, Chatham Maritime, Chatham, Kent. ME4 4TB. This leaflet has been produced in conjunction with sigma Network for Excellence in Mathematics and Statistics Support Step 3: Example 2 Drug B has a half-life of 3 hours. If the initial plasma level of the drug, given as a single dose, is 3600mg/L, what will its plasma level be after 10 hours? Note: In this case the time/value does not coincide with an exact half-life interval. Method Step 1: Tabulate the time and value for each half-life, to the next higher time/value interval. 3hr = 1 half −life = 3600 ÷ 2 = 1800𝑚𝑚/𝐿 6hr = 2 half −life = 1800 ÷ 2 = 900𝑚𝑚/𝐿 9hr = 3 half −life = 900 ÷ 2 = 450𝑚𝑚/𝐿 12hr = 4 half −life = 450 ÷ 2 = 225𝑚𝑚/𝐿 Step 2: Tabulate the times and values between 9hr and 12 hr. Since 10hr equals 9hr + 1/3 of the interval to 12hr, the value will equal that at 9hr – 1/3 of the difference, time and value being inversely proportional. 450 −225 = 225 225 × 1 3 ൗ= 75 450 −75 = 𝟑𝟑𝟑𝟑𝟑/𝑳 ⓑ Multiply the difference: ⓐ Calculate the difference: ⓒ Subtract from upper value Example 2 Drug C has a half-life of 8 hours. If the initial plasma level of the drug is, given as a single dose , is 4800mg/L, how long will it take for the plasma level to fall to 180mg/L? Note: Here we are solving for time rather than value. Method Step 1: Tabulate the time and value for each half-life, to the next higher time/value interval. 8hr = 1 half −life = 4800 ÷ 2 = 2400𝑚𝑚/𝐿 16hr = 2 half −life = 2400 ÷ 2 = 1200𝑚𝑚/𝐿 24hr = 3 half −life = 1200 ÷ 2 = 600𝑚𝑚/𝐿 32hr = 4 half −life = 600 ÷ 2 = 300𝑚𝑚/𝐿 40hr = 4 half −life = 600 ÷ 2 = 150𝑚𝑚/𝐿 Step 2: Tabulate the values and times between 300mg/l and 150mg/l. Since 180mg/l equals 300mg/l – 0.8 x 150mg/l, the time will equal 32hr + 0.8 x 8hr, value and time being inversely proportional. Step 3: 40 −32 = 8ℎ𝑟= 480 𝑚𝑚𝑚 480 × 0.8 = 384 𝑚𝑚𝑚= 6ℎ𝑟 24𝑚𝑚𝑚 32ℎ𝑟+ 6ℎ𝑟 24𝑚𝑚𝑚= 𝟑𝟑𝟑𝟑 𝟐𝟐𝟐𝟐𝟐 ⓑ Multiply the difference: ⓐ Calculate the difference: ⓒ Add to lower value Q1 Drug D has a half-life of 90 min. If the initial plasma level of the drug, given as a single dose, is 2688mg/L, what will its plasma level be after 8hr? Q2 Drug E has a half-life of 16 hours. If the initial plasma level of the drug, given as a single dose, is 512mg/L, how long will it take for the plasma level to fall to 24mg/L? Answers: Q1 = 70mg/L. Q2 = 72hr. 9hr 12hr 450mg/L 225mg/L 10hr 11hr ?
32hr 40hr 300mg/L 150mg/L 180mg/L 120mg/L ? |
9555 | https://www.litcharts.com/literary-devices-and-terms/connotation | Download
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Definition
Examples
Function
Resources
Connotation Definition
What is connotation? Here’s a quick and simple definition:
Connotation is the array of emotions and ideas suggested by a word in addition to its dictionary definition. Most words carry meanings, impressions, or associations apart from or beyond their literal meaning. For example, the words "child" and "kid" mean the same thing, but to call someone a "kid" has a slightly informal and therefore disrespectful connotation. The nature of connotations is that they are not explicitly defined, so they can be used in both purposeful and accidental ways to convey subtle meaning or subtext: you might call someone a "kid" on purpose to imply some disrespect, or you may do so without realizing the connotation of your words.
Some additional key details about connotation:
Words can have positive, negative or neutral connotations. For instance, the word "peace" has a positive connotation, "coffin" has a negative one, and "table" is neutral. Connotation is an important aspect of diction, a term that refers to the choice and usage of words by writers.
The connotations a word carries are often subjective, meaning that they might change depending on an individual's experience, geographical location, or time period. In other words, connotation is deeply dependent on context.
Writers may use connotation to evoke specific emotions in their readers without explicitly telling them what to feel. Connotation is vital to the arts, but is also extremely useful in business, advertising, and politics.
Connotation Pronunciation
Here's how to pronounce connotation: kon-oh-tay-shun
Connotation Explained
Every word has a literal definition that you can look up in the dictionary, but most of the words people use on a daily basis carry associations that aren't written down as part of their strict definition. This "hidden" layer of meaning includes the array of emotions, cultural associations, and ideas that a given word invokes whenever it's used—all of which is determined by the history and context of the word's usage.
For example, the word "Hollywood" refers to a specific area in the city of Los Angeles, but the word is used so often as a metonym for the American film industry as a whole that it has come to evoke ideas of glamor, artifice, and fame. These associations are the connotations the word carries, and they're separate from the literal meaning of the word, which is known as its denotation.
Connotation vs. Denotation
Connotation is commonly defined in contrast to denotation, a literary term that refers to the "dictionary definition," or the explicit and literal definition of a word or phrase. The word "swan," for example, denotes a swan. But what "swan" connotes is grace, beauty, love, and purity.
Connotation and Context
The connotations a word carries may be different for different people, depending on a wide variety of factors, including their life experiences, where they live, and when. For instance, not everybody is aware that swans can actually be very aggressive, but for those who are, the word "swan" might actually have a connotation of violence, viciousness, and brutality.
Most words don't carry two such opposite connotations, but this example underscores that depending on where you are, what you know, and who you're speaking to, a certain word or phrase could have such different connotations that it could offend one person and delight another.
Positive, Negative and Neutral Connotation
Connotations can be positive, negative, or neutral, depending on the associations evoked by a given word. When a writer is choosing the right words to express an idea, considering whether a given word has positive, negative, or neutral connotations is vital to getting across the right idea to a reader. There may be three words that all denote the same thing (i.e., the words are synonyms), but perhaps only one of them carries the correct connotation to accurately express a certain idea. For instance, imagine a writer choosing between the words "thin," "skinny," and "lithe" to describe a character. All are adjectives that denote slenderness, but each has a different connotation:
"Skinny" has a somewhat negative connotation, implying that a person may be slightly too thin.
"Lithe" has a positive connotation, implying a thinness that is elegant and attractive.
"Thin" is neutral, casting no judgement on the person's weight at all.
Depending on how the writer wants the reader to think of their character—underweight, attractive, or simply thin—a different word should be selected.
Connotation vs. Symbolism
Connotation is in some ways similar to, and at times confused with, symbolism, a literary device in which one thing suggests or alludes to something else. Someone might mistakenly say, for instance, that a word connotes something that it actually symbolizes. So it's worthwhile to understand how these two devices overlap, and what makes them different from one another.
Symbolism is the use of one thing—usually a physical object or phenomenon—to represent something more abstract. For instance, a rose is often used as a symbol of love.
Connotation is specific to words and phrases, not physical objects. So the word "lithe," for example, has a connotation of attractiveness and elegance, but it doesn't symbolize attractiveness and elegance.
A famous example of a symbol in literature occurs inTo Kill a Mockingbird, when Atticus tells his children Jem and Scout that it's a sin to kill a mockingbird because mockingbirds cause no harm to anyone; they just sing. Because of these traits, mockingbirds in the novel symbolize innocence and beauty, while killing a mockingbird symbolizes an act of senseless cruelty. However, the word "mockingbird" doesn't actually carry a connotation of innocence or beauty.
Connotation and Figurative Language
Figurative language refers to any language that uses words or phrases that have meanings that are different from their literal interpretation. Metaphors, similes, and hyperboles are all examples of figurative language. The simile, "Mary's eyes are like the sea," for instance, is not trying to say that Mary's eyes are literally the same as a large body of salt-water. Instead, it is using the comparison to say something that is at once non-literal and more powerful about Mary's eyes: that they are deep, unknowable, powerful, can shift from calm to playfulness to rage.
There is another way to describe what this simile is doing: through the comparison that it is making, it connects Mary's eyes not to the denotation of the word "sea," but to the connotations of "sea." Put more broadly: figurative language often functions at the connotative level, and involve writers mixing connotations across words, or using the connotations of one word or image to create a new understanding of a different word or image.
Connotation Examples
Since connotation simply refers to the additional, sometimes hidden meaning of a word, examples of it are essentially infinite. The examples included here come from poetry, fiction, advertising, and painting to illustrate a few different ways connotation can be used to evoke specific ideas or emotions in the reader or viewer.
Example of Connotation in Poetry
In his poem "I Hear America Singing," Walt Whitman carefully chooses his words in order to imbue the poem with a sense of ruggedness and virility.
I hear America singing, the varied carols I hear,
Those of mechanics, each one singing his as it should be blithe and strong,
The carpenter singing his as he measures his plank or beam,
The mason singing his as he makes ready for work, or leaves off work,
...
Each singing what belongs to him or her and to none else,
The day what belongs to the day—at night the party of young fellows, robust, friendly,
Singing with open mouths their strong melodious songs.
The poem is brimming with words like "mechanic," "carpenter," and "mason," all of which describe traditional working-class American jobs that conjure up images of physical labor and hard work. Rather than speaking directly to his readers about the American common man, Whitman makes use of connotation to spin an image of the strength and rugged individualism of Americans. Just imagine if Whitman had instead chosen to describe the soft singing of whittlers and tinkers, or bankers and legal clerks—and the different connotations those occupations would have brought to the poem.
Example of Connotation in Fiction
George Orwell's novel Animal Farm is a dystopian retelling of the events surrounding the Russian revolution, and is commonly described as an allegory, or a highly symbolic narrative that conveys a hidden meaning. Orwell uses connotation to make his characters representative of figures and forces in real life. He casts pigs as the oppressive and dogmatic ruling class because the word "pig" carries a strong connotation of corruption and greed. The horse characters, such as Boxer, populate the laboring class of the farm, because "workhorses" are associated with strenuous physical labor.
Orwell went even further with connotation in Animal Farm, and considered connotation when naming his characters. A great example is Mr. Whymper: the untrustworthy and money-grubbing human the animals choose as their interpreter. Even if you haven't read Animal Farm, or don't know who Mr. Whymper is, his name has an immediately negative connotation, as it sounds similar to "whimper," a word that connotes weakness and cowardice. Choosing this word for his character cements the idea in readers' heads from the outset that Mr. Whymper should be treated with suspicion or disdain—and indeed, Mr. Whymper turns out to be a cowardly, two-faced character.
Example of Connotation in Advertising and Branding
Connotation is used in advertisements and branding to get emotional responses out of the consumer. Companies can take advantage of the array of associations people naturally make with a certain word or phrase and, if they are successful, sell consumers on an idea or a feeling rather than just a product.
If you take a look at the following list of cars from the 1960's, there's an obvious theme:
Thunderbird
Falcon
Charger
Comet
Mustang
Barracuda
Each name has been chosen specifically for its connotation, and in every case the desired connotation is speed, wildness, and power.The car manufacturers of the Mustang, for example, chose a word that denotes wild horses and connotes beauty, freedom, and power.
Why Writers Use Connotation
Writers use connotation to transmit meaning without explicitly telling a reader what to think feel. When a writer chooses one word over another that means the same thing, they are giving preference to the set of associations carried by that word. These associations (thoughts, feelings, memories, words, images, or ideas) help to deepen and elaborate on the meaning of a word or phrase beyond its literal definition. Furthermore, writers may choose words for their positive, negative, or neutral connotations in order to evoke the right response in the reader. On a larger scale, writers might also choose to create or use particular characters, environments, images, and events in order to create specific associations in the reader's mind. In this way, connotations are used to indirectly influence the way people receive and process language on a daily basis.
Other Helpful Connotation Resources
Wikipedia Page on Connotation: A brief outline of connotation.
Dictionary Definition of Connotation: A concise and helpful definition of connotation that includes further examples. Put another way, it's the denotation of connotation.
YouTube Video Explaining the Difference between Connotation and Denotation: A helpful video with animations that outline the differences between these two literary terms.
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9556 | https://math.stackexchange.com/questions/1095235/equivalence-of-controllability-and-reachability-in-discrete-time-systems | Skip to main content
Equivalence of controllability and reachability in discrete time systems
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I am trying to prove that the statements;
Σd is controllable,
Σd is reachable,
The pair (A,B) is controllable (in other words =X).
are equivalent for discrete time systems.
I defined controllability as being able to get from an initial state x1 to any other state x2, reachability to get from the origin 0 to any other state x2 and null-controllability to get from an initial state x1 to the origin 0.
I see how being able to get from any state x1 to any state x2 (thus controllability) implies reachability as you can take the origin as x1 and being able to reach any state x2.
However i can't figure out how to proof it the other way around (that reachability implies conrollability).
Also i don't really understand how the last statement is different from the first statement.
discrete-mathematics
control-theory
systems-theory
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asked Jan 7, 2015 at 13:54
GWielinkGWielink
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Reaching from x1 to x2 in k steps is equivalent to reaching from 0 to x2−Akx1 in k steps, which can be immediately seen from the solution of the difference equation. Since all the states are reachable from 0, this implies controllability.
Null-controllability is equivalent to reachability if the system does not have finite modes, i.e. 0 eigenvalues. If the system has an uncontrollable finite mode, this particular node will be 0 in a finite time (more precisely at most n steps, where n is the system order), regardless of the input. Therefore null-controllability is equivalent to "all uncontrollable modes are finite".
Note: Not to be confused with stabilizability, which is "all uncontrollable modes are stable". This doesn't imply null-controllability since uncontrollable stable modes may be infinite, i.e. they cannot reach 0 in a finite time. However, null-controllability implies stabilizability.
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answered Jan 7, 2015 at 18:56
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9557 | https://www.chicagomanualofstyle.org/book/ed17/part2/ch07/toc.html | Chapter 7 Contents
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7:Spelling, Distinctive Treatment of Words, and Compounds
Overview
7.1Recommended dictionaries
7.2Spellings peculiar to particular disciplines
7.3Non-US spelling
7.4Supplementing the dictionary
Plurals
7.5Standard plural forms
7.6Alternative plural forms
7.7Plurals of compound nouns
7.8Plurals for centuries
7.9Plurals of proper nouns
7.10Plural form for Native American group names
7.11Singular form used for the plural
7.12Plural form of italicized words
7.13Plural form for words in quotation marks
7.14Plurals of noun coinages
7.15Plurals for letters, abbreviations, and numerals
Possessives
The General Rule
7.16Possessive form of most nouns
7.17Possessive of proper nouns, abbreviations, and numbers
7.18Possessive of words and names ending in unpronounced “s”
7.19Possessive of names like “Euripides”
Exceptions to the General Rule
7.20Possessive of nouns plural in form, singular in meaning
7.21“For... sake” expressions
7.22An alternative practice for words ending in “s”
Particularities of the Possessive
7.23Joint versus separate possession
7.24Compound possessives
7.25Possessive to mean “of”
7.26Double possessive
7.27Possessive versus attributive forms for groups
7.28Possessive with gerund
7.29Possessive with italicized or quoted terms
Contractions and Interjections
7.30Contractions
7.31Interjections
“A” and “An”
7.32“A” and “an” before “h”
7.33“A” and “an” before abbreviations, symbols, and numerals
Ligatures
7.34When not to use ligatures
7.35When ligatures should be used
Word Division
7.36Dictionary word division
7.37Word divisions that should be avoided
7.38Dividing according to pronunciation
7.39Dividing after a vowel
7.40Dividing compounds, prefixes, and suffixes
7.41Dividing words ending in “ing”
7.42Dividing proper nouns and personal names
7.43Dividing numerals
7.44Dividing numerals with abbreviated units of measure
7.45Division in run-in lists
7.46Dividing URLs and email addresses
7.47Hyphenation and appearance
Italics, Capitals, and Quotation Marks
7.48Setting off proper names and titles of works
7.49Italics and markup
Emphasis
7.50Italics for emphasis
7.51Boldface or underscore for emphasis
7.52Capitals for emphasis
Words from Other Languages
7.53Unfamiliar words and phrases from other languages
7.54Roman for familiar words from other languages
7.55Roman for Latin words and abbreviations
Highlighting Key Terms and Expressions
7.56Italics or boldface for key terms
7.57“Scare quotes”
7.58Mixing single and double quotation marks
7.59“So-called”
7.60Common expressions and figures of speech
7.61Signs and notices
7.62Mottoes
Words as Words and Letters as Letters
7.63Words and phrases used as words
7.64Letters as letters
7.65Scholastic grades
7.66Letters standing for names
7.67Letters as shapes
7.68Names of letters
7.69Rhyme schemes
Music: Some Typographic Conventions
7.70Suggested references for music publishing
7.71Musical pitches
7.72Octaves
7.73Chords
7.74“Major” and “minor”
7.75Dynamics
Computer Terms
7.76Application-specific versus generic usage
7.77Capitalization for keys, menu items, and file formats
7.78Keyboard combinations and shortcuts
7.79Setting off file names and words to be typed or selected
7.80Terms like “web” and “internet”
Compounds and Hyphenation
7.81To hyphenate or not to hyphenate
7.82Compounds defined
7.83The trend toward closed compounds
7.84Hyphens and readability
7.85Compound modifiers before or after a noun
7.86Adverbs ending in “ly”
7.87Multiple hyphens
7.88Suspended hyphens
7.89Hyphenation guide
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9558 | https://groupsmadesimple.wordpress.com/2020/06/06/parity-in-practice-the-15-puzzle/ | Skip to content
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Parity in practice: The 15-Puzzle
February 1880 in The New York Times’ editorial: “… just now the chief amusement of the New York mind, … a mental epidemic …. In a month from now, the whole population of North America will be at it, and when the 15 puzzle crosses the seas, it is sure to become an English mania.” This mania, fuelled by the $1,000 prize offered by Sam Loyd for the first solution, is to solve the 15-Puzzle. Pictured in Loyd’s promotional poster below, the puzzle consists of a four by four grid holding 15 numbered tiles and a space to slide the tiles around, with the 14 and 15 tiles the wrong way around. The challenge is to slide the tiles around to bring all the tiles back into order. In Loyd’s own words: Nobody wanted to give up as everyone was confident of imminent success. It was said that navigators allowed their ships to run aground, engine drivers took their trains past stations, and farmers neglected their ploughs.
But the $1,000 remained unclaimed, and always will be, as the puzzle is impossible to solve. Loyd (a recreational mathematician and prolific puzzle writer) would have known this, and we can imagine he delighted in watching the fruitless efforts of the American public, whilst he continued to sell them copies of the puzzle. Perhaps though we might forgive him his unscrupulous behaviour for his greater service of bringing the mathematical notion of parity to attention of the general public. Many mathematical arguments turn on the idea of keeping track of odd and even, and the 15-Puzzle is a great example.
Moves as transpositions of the blank tile
You can buy copies of the puzzle today or play online to get a feel for it and these typically come with the numbers in the right order (the solved position, below left). Like you would with a Rubik’s cube, you can jumble up the tiles and see if you can get to some other arrangement, or just try and get back again to where you started (there are videos online with strategies). So the question becomes what rearrangements can or can’t you get to from jumbling up the base position, and why isn’t Loyd’s set up (below right) amongst them?
Again like a Rubik’s cube, the 15-puzzle is based on taking a set of disordered objects and permuting them into a desired order. This means that we can apply what we know mathematically about permutations to help us understand the puzzle. We start by thinking of each legal move as a permutation. Each move slides a tile adjacent to the empty space into that the empty space, which has the effect of swapping or interchanging the empty space with that tile. If we think of the empty space as a tile like any other (the blank tile, or the 16th tile, as purposes suit) then every legal move swaps two tiles. Permutations that just swap two items have a special name, the transpositions, and this means every legal move can be thought of as a transposition of tile 16 with one of the other tiles neighbouring it.
Tracking the parity of the number of moves
What we are going to do is find two different metrics for tracking the parity of the number of moves from the base position (ie whether this is even or odd) to a given jumbled up position. This will tell us if the number of moves were odd or even, even if we don’t know what the moves where. We will take as our example the nine moves shown in the figure below, where the blank tile has swapped position nine times (shown by the arrows) to end up in the position of tile 7. This example shows us our two metrics, which I talk about after.
Our first metric to keep track of on each move is distance the blank tile ends up from its starting position at the bottom right hand corner. Distance here doesn’t mean diagonally, it means the Manhattan distance following the columns and rows of the puzzle: so called because if you where trying to find the shortest distance to walk between A to B in downtown Manhattan you would have to follow the grid layout of the streets. In the example above the distance is 3, because one must travel 1 across and 2 down back to the bottom right corner (or equivalently, 2 down and 1 across).
This distance will always have the same parity as the number of moves: so after the 9 moves above the distance is 3, and both 9 and 3 are odd. To see why this is true, see the picture on the right with the distance on each position labelled. You start at distance 0, then each move takes you alternatively to an odd numbered distance (shaded), then to an even (unshaded), and so on. Wherever the blank tile ends up, we know at a glance if the number of moves carried out were odd or even.
The next metric is the parity of the resulting permutation of the tiles. The resulting permutation is written out in two line notation where each column tells us where each tile is mapped to, ie the first column tells us that tile 2 is permuted to the position of tile 6, the second that tile 3 is permuted to position of tile 2,… and generally tile i is permuted to position . In a way similar to the integers, permutations have parity depending on whether they can be written as an odd or even number of transpositions: if this is unfamiliar see my final post for an explanation. Because each move is a successive transposition, the resulting permutation must have the same parity as the number of moves: so the permutation in the example is generated by nine transpositions and is odd.
Similarly to before if we lose track of the moves carried out, we can still tell if they were odd or even in number by checking the parity of the permutation. This isn’t too tricky but it’s not quite so “at a glance” as the previous metric. It’s fair game to use any way that you know of to check the parity of a permutation (you could count the inversions, or write it in cycle notation), but I quite like a fast and elementary method: draw it out and count the number of crossings to see if that is odd or even (read about why this works in my previous post). On the picture above, we draw a line down from 2 to 6, from 3 down to 2, and so on to give 11 crossings, which as expected is odd. We might get a different number of crossings if we ordered the tiles differently on each row, but it will always be odd.
Why Loyd’s set up is impossible and the general case
So we have two metrics for checking if an odd or even number of moves have been carried out from the base position. These two metrics better agree, or we are in trouble. If presented with an arbitrary arrangement, we don’t know if it has been obtain from shuffling the genuine base case or a trick board with the tiles in a different order—but if our two metrics don’t stack up we definitely didn’t start from the genuine base case.
This is what happens with Loyd’s board with 14 and 15 the wrong way around. The blank has returned to the starting position, indicating an even number of moves, but the overall permutation has one transpostion, 14 and 15, indicating an odd number of moves where performed, which is a contradiction. This shows we can’t get from the base case to Loyd’s set up!
It’s worth a quick comment on the parity of the permutation in this case. You could check it is odd by drawing it out: the line from 14 to 15 and 15 to 14 will cross once, indicating it is odd. Or you can just see that swapping the 14 and 15 tile is a single transposition (imagine prising them out of the puzzle, swapping them around and putting them back in), and so any other expression as transpositions must also be odd in length. The laws of mathematics are ambivalent about whether your transpositions follow the legal rules of the puzzle, or are just any transpositions.
So in general, if the two metrics are inconsistent for a given board, then that board is not solvable from the base position. What if the metrics are consistent, is that board definitely solvable? The answer, it turns out, is yes although it is more involved so I won’t prove it, (see links at bottom of this post). What’s more, those that aren’t solvable from the base case are solvable from Loyd’s set up with the 14 and 15 inverted. Hence, there are essentially only two different versions of this puzzle, in the sense that the 16! possible configurations of the tiles partitions into two halves: rearrangements of the base puzzle, and rearrangements of Loyd’s set up. There is no overlap, else you would be able to rearrange from the base case, to that arrangement, and then onto Loyd’s set up, which we know is impossible.
Spot the odd one out
Time to try out our new-found understanding on some real boards. There is a great YouTube channel called Numberphile on a variety of mathematical topics, and they take a look at the 15 puzzle here. The presenter owns a copy of the puzzle (with the usual base set up) that came with an instruction card with six configurations to try and solve, shown below. But it turns out only five of them are possible, although Numberphile didn’t know if this was deliberate mischief or accidental on the part of the puzzle makers! Here are the configurations—can you spot the rogue entry?
We’ll go through them now and work it out which ones are in and which one is out. By default the ‘1 to 15’ configuration is solvable, being the starting configuration.
Vertical
The Vertical configuration is the simplest so we will start with this one. The rows are switched around with the columns, so the numbers going down each column. The blank space is still in the bottom right (zero Manhattan distance), so for this configuration to be solvable we need the permutation of the tiles to be even. It’s easy to see that it is, as recall that an even permutation can be written as an even number of transpositions, and this 6 transpositions, swapping the 6 tiles above the diagonal each with one below the diagonal. So 2 swaps with 5, 3 with 9 and so on.
Drawn out above, each pair of tiles that are transposed generate a crossing, giving an even number of crossings. So yes, we can do this one. Unfortunately none of this work tells us how we would actually do it in practice!
Vertical Odd / Even
Next up is the Vertical Odd / Even configuration: working down the columns we write out all the odd numbers in increasing order, then the even numbers. Again the blank tile remains in place so we need the permutation to be even for this configuration to be solvable.
Unlike the last case, it is trickier to see how we would write this permutation as transpositions: the tiles don’t conveniently pair up for swaps. So instead we can rely on drawing out the picture: which makes a really nice pattern. The number of crossings is the 7th triangular number, 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28, the , and as this is even, the permutation is even and this configuration is solvable too.
Skip Odd / Even
In the Skip odd / even configuration, the numbers 1 to 8 follow a snakelike path at the top, and then the remaining numbers a similar snakelike path at the bottom. The blank space has travelled 3 squares from the bottom right to bottom left, so this configuration is solvable just if the permutation is odd. Drawing it out reveals 15 crossings, which is odd, so this configuration too is solvable.
Spiral
The Spiral configuration has the tiles looping round the edge in increasing order in, as the name suggests, a spiral. The blank space moves three spaces, two to the left and one up, and since three is odd we need the permutation to be odd if the permutation is to be solvable.
The problem with this one is if you draw out this permutation with the tiles in the usual increasing order, the picture is a mess, see right! Counting the number of crossings (taking care to count between just two lines at a time) becomes fiddly. We can take advantage of a couple of tricks to get to the simplified picture above. First note the top row with tiles 1 to 4 stays the same, so we can safely leave them out the drawing. Second is to remember that it doesn’t matter what order you write the numbers out, as long as they are in the same order in the top and bottom rows. The number of crossings will vary depending on this order, but the parity is invariant: it will stay odd or even however you draw it.
I’ve drawn it out to follow the cycle of the permutation: start at 5 (the smallest tile to move), tile 5 moves to the position of tile 8, so write out 8 next, then tile 8 moves to the position of tile 15, and so on. The fact you can keep going with this right up until the last tile, tile 12, shows you that you can think of this permutation as a single cycle, where the cycle advances each tile to the position of the next one. If you are familiar with cycle notation, then this permutation is given as:
The fact you can draw a picture like the above, with just one line crossing all the others, tells you that any cycle is odd just if n is even, and vice versa.
With 11 crossings, the permutation is odd so this configuration is possible. This doesn’t give us any indication how many moves it takes to solve, although it turns out to be quite tricky, requiring 61 moves!
15 to 1
As you might have guessed, this last configuration is the culprit, and is impossible to get to—although you could get to it from Loyd’s board with 14 and 15 the wrong way around. The 15 to 1 configuration reverses the order of the tiles, swapping tile 15 with tile 1, tile 14 with tile 2, … and tile 9 and tile 7. Tile 8 stays in place in the middle. So this is another configuration that arises as transpositions, 7 swaps in total, indicating an odd permutation. But because the blank tile stays in place it would require an even number of moves, so we can’t do it.
Wrap Up
After publishing his 14-15 puzzle, Loyd continued to publish boards for the American public to solve, all of which were impossible. We now know how to spot them by checking the parity of the blank tile against the parity of the permutation, using nothing more complicated than drawing a picture.
All 16! possible configurations of the board come in two varieties, half of them can be solved from the base position, and the other half can be solved by first adding in a transposition (such as swapping the 14 and 15 tile) and then solving.
Want to know more? You might find these links helpful.
You can play the 15-puzzle online here;
And you can watch Professor Steven Bradlow on Numberphile discuss the 15-puzzle (same link as mentioned above);
Cut the Knot has a nice short discussion of the 15-Puzzle and other similar sliding puzzles;
And Keith Conrad has a longer discussion on the 15-Puzzle which clearly makes the link with the even permutations and the alternating group;
We claimed that all boards where the parities agree are possible to solve, to prove it, work through Archer’s technical paper A Modern Treatment of the 15-Puzzle; take care as Archer treats a board as a permutation of the non-blank tiles 1 – 15 (which must be even), unlike our approach based on all tiles 1-16 (which may be even or odd);
British mathematician J J Slyvester, writing the in the American Journal of Mathematics in 1879, said whoever has made himself master of [this puzzle] may fairly be said to have taken his first lesson in the theory of determinants; Anyone keen enough to want to continue the lesson would do well to read this three page paper from Clark University;
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I’m enjoying these posts; they’re well written and interesting. Thanks Tim!
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9559 | https://math.stackexchange.com/questions/1710750/how-to-integrate-int-frac-sqrt-x-sqrt-1-xdx | real analysis - How to integrate $\int\frac{\sqrt x}{\sqrt {1-x}}dx$? - Mathematics Stack Exchange
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How to integrate ∫x√1−x√d x∫x 1−x d x?
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I have
∫x−−√1−x−−−−−√d x∫x 1−x d x
I place x−−√=t x=t so I have
∫2 t 2 1−t 2−−−−−√d t∫2 t 2 1−t 2 d t
then by parts I have
2 t 2 arcsin t−4∫t arcsin t d t 2 t 2 arcsint−4∫t arcsint d t
Now, how can I proceed? If I still use by parts, I go back.
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edited Mar 23, 2016 at 21:53
Daniel R
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asked Mar 23, 2016 at 20:39
L.G.A.G.L.G.A.G.
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Putting x=sin 2 t x=sin 2t, we get:
∫x−−√1−x−−−−−√d x∫x 1−x d x
=∫2 sin 2 t cos t d t cos t=∫2 sin 2t cost d t cost
=2∫sin 2 t d t=2∫sin 2t d t
=∫(1−cos 2 t)d t=∫(1−cos2 t)d t
=t−1 2 sin 2 t+C=t−1 2 sin2 t+C
You can substitute t t back
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edited Mar 23, 2016 at 20:54
answered Mar 23, 2016 at 20:45
NikunjNikunj
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Looks like a good substitution, but your d x d x is incorrect.Mike –Mike 2016-03-23 20:50:38 +00:00 Commented Mar 23, 2016 at 20:50
@Mike, thanks, edited.Nikunj –Nikunj 2016-03-23 20:52:11 +00:00 Commented Mar 23, 2016 at 20:52
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Hint:
The standard substitution for integrals of functions of the form f(x,a x+b c x+d−−−−√)f(x,a x+b c x+d), where f(x,y)f(x,y) is a rational function of two variables , is to use the substitution t=a x+b c x+d−−−−√t=a x+b c x+d. So here, one sets
t=x 1−x−−−−−√⟺t 2=x 1−x,t≥0.t=x 1−x⟺t 2=x 1−x,t≥0.
The substitution results in
∫x−−√1−x−−−−−√d x=∫2 t 2(1+t 2)2 d t=2(∫1 1+t 2 d t−∫1(1+t 2)2 d t)∫x 1−x d x=∫2 t 2(1+t 2)2 d t=2(∫1 1+t 2 d t−∫1(1+t 2)2 d t)
Now I 1=∫1 1+t 2 d t=arctan t I 1=∫1 1+t 2 d t=arctant and I 2=∫1(1+t 2)2 d t I 2=∫1(1+t 2)2 d t is classically obtained from I 1 I 1 by an integration by parts.
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answered Mar 23, 2016 at 21:06
BernardBernard
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For I 2 I 2 by parts I have t(1+t 2)2+4∫t 2(1+t 2)3 d t t(1+t 2)2+4∫t 2(1+t 2)3 d t and how can I proceed?L.G.A.G. –L.G.A.G. 2016-03-23 21:24:32 +00:00 Commented Mar 23, 2016 at 21:24
He means to apply by parts in I 1 I 1 Nikunj –Nikunj 2016-03-23 21:25:10 +00:00 Commented Mar 23, 2016 at 21:25
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Hint: Write 2 t 2 2 t 2 as 2 t 2−2+2 2 t 2−2+2 before the step where you apply by parts
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answered Mar 23, 2016 at 20:45
user2277550user2277550
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9560 | https://www.whitman.edu/mathematics/cgt_online/book/section06.03.html | Suppose we are interested in a more detailed inventory of the
colorings of an object, namely, instead of the total number of
colorings we seek the number of colorings with a given number of each
color. The method presented here was first published by J. Howard
Redfield in 1927. In 1937 it was independently rediscovered by George
Pólya, who then greatly popularized the result by applying it to many
counting problems, in particular to the enumeration of chemical
compounds.
Example 6.3.1 How many distinct ways are there to color the vertices of a regular
pentagon modulo D5D5 so that one vertex is red, two are blue, and two
are green?
We can approach this as before, that is, the answer is
1|D5|∑σ∈D5|fix(σ)|,
1|D5|∑σ∈D5|fix(σ)|,
where fix(σ)fix(σ) now means the colorings with one red,
two blues, and two greens that are fixed by σσ.
No longer can we use the simple expression of
corollary 6.2.8.
The identity permutation fixes all colorings, so we need to know how
many colorings of the pentagon use one red, two blues, and two
greens. This is an easy counting problem: the number is
(52)(32)=30(52)(32)=30.
If σσ is a non-trivial rotation, |fix(σ)|=0|fix(σ)|=0, since the
only colorings fixed by a rotation have all vertices the same color.
If σσ is a reflection, the single vertex fixed by σσ must
be red, and then the remaining 2-cycles are colored blue and green in
one of two ways, so |fix(σ)|=2|fix(σ)|=2.
Thus, the number of distinct colorings is
110(30+0+0+0+0+2+2+2+2+2)=4.
110(30+0+0+0+0+2+2+2+2+2)=4.
◻
What we seek is a way to streamline this process, since in general the
computations of |fix(σ)| can be tedious. We begin by recasting
the formula of corollary 6.2.8.
Definition 6.3.2 The type of a permutation σ∈Sn is
τ(σ)=(τ1(σ),τ2(σ),…,τn(σ)),
where τi(σ) is the number of
i-cycles in the cycle form of σ.
◻
Note that ∑ni=1τi(σ)=#σ.
Now instead of the simple
1|G|∑σ∈Gk#σ
let us write
1|G|∑σ∈Gxτ1(σ)1xτ2(σ)2⋯xτn(σ)n.If we substitute xi=k for every i, we get the original form of
the sum, but the new version carries more information about each
σ.
Suppose we want to know the number of colorings fixed by some σ
that use i reds and j blues, where of course i+j=n. Using ideas
familiar from generating functions, consider the following expression:
(r+b)τ1(σ)(r2+b2)τ2(σ)⋯(rn+bn)τn(σ).
If we multiply out, we get a sum of terms of the form
rpbq, each representing some particular way of coloring the vertices of
cycles red and blue so that the total number of red vertices is p
and the number of blue vertices is q, and moreover this coloring
will be fixed by σ. When we collect like terms, the coefficient
of ribj is the number of colorings fixed by σ that use i
reds and j blues. This means that the coefficient of
ribj in
∑σ∈G(r+b)τ1(σ)(r2+b2)τ2(σ)⋯(rn+bn)τn(σ)is
∑σ∈G|fix(σ)|where fix(σ) is the set of colorings using i reds and j
blues that are fixed by σ. Finally, then, the number of distinct
colorings using i reds and j
blues is this coefficient divided by |G|.
This means that by multiplying out
1|G|∑σ∈G(r+b)τ1(σ)(r2+b2)τ2(σ)⋯(rn+bn)τn(σ)and collecting like terms,
we get a list of the number of distinct colorings using any
combination of reds and blues, each the coefficient of a different
term; we call this the inventory of colorings.
If we substitute r=1 and b=1, we get the sum of the
coefficients, namely, the total number of distinct colorings with two colors.
Definition 6.3.3 The cycle index of G is
PG=1|G|∑σ∈Gn∏i=1xτi(σ)i.
◻
Example 6.3.4 Consider again example 6.2.6,
in which we found the number of colorings of a square with two colors.
The cycle index of D4 is
18(x41+x14+x22+x14+x22+x22+x21x2+x21x2)=18x41+14x21x2+38x22+14x4.
Substituting as above gives
18(r+b)4+14(r+b)2(r2+b2)+38(r2+b2)2+14(r4+b4)=r4+r3b+2r2b2+rb3+b4.Thus there is one all red coloring, one with three reds and one blue,
and so on, as shown in figure 6.2.4.
◻
There is nothing special about the use of two colors. If we want
to use three colors, we substitute ri+bi+gi for xi in the
cycle index, and for k colors we substitute something like
ci1+ci2+ci3+⋯+cik.
Example 6.3.5
Let's do the number of 3-colorings of the square. Since we
already have the cycle index, we need only substitute
xi=ri+bi+gi and expand. We get
18(r+b+g)4+14(r+b+g)2(r2+b2+g2)+38(r2+b2+g2)2+14(r4+b4+g4)=b4+b3g+b3r+2b2g2+2b2gr+2b2r2+bg3+2bg2r+2bgr2+br3+g4+g3r+2g2r2+gr3+r4.
So, for example, there are two squares with two blue vertices, one
green, and one red, from the b2gr term.
◻
Example 6.3.6
Consider again example 6.2.7, in which we counted the number of four-vertex
graphs. Following that example, we get
PG=124(x61+6x2x4+8x23+3x21x22+6x21x22),
and substituting for the variables xi gives
r6+r5b+2r4b2+3r3b3+2r2b4+rb5+b6.Recall that the "colors'' of the edges in this example are
"included'' and "excluded''. If we set b=1 and r=i (for
"included'') we get
i6+i5+2i4+3i3+2i2+i+1,interpreted as one graph with 6 edges, one with 5, two with 4,
three with 3, two with 2, one with 1, and one with zero edges, since
1=i0. (Of course, if we are interested in the inventory in the
final form, we can skip the step using r and b by substituting
xj=ij+1 in the cycle index.)
◻
It is possible, though a bit difficult, to see that for n vertices
the cycle index is
PG=∑jn∏k=11kjkjk!⌊n/2⌋∏k=1(xkxk−12k)j2k⌊(n−1)/2⌋∏k=1xkj2k+12k+1⌊n/2⌋∏k=1xkC(jk,2)k∏1≤r<s≤n−1xgcd(r,s)jrjslcm(r,s),
where the sums are over all partitions j=(j1,j2,…,jn)
of n, that is, over all j such that
j1+2j2+3j3+⋯+njn=n, and C(m,2)=(m2).
This is where the formula 6.2.1
comes from, substituting xi=2 for all i.
With this formula and a computer it is easy to compute the inventory
of n-vertex graphs when n is not too large. When n=5, the
inventory is
i10+i9+2i8+4i7+6i6+6i5+6i4+4i3+2i2+i+1.
We can use
Sage
to do the computations involved in finding the cycle
index for graphs.
Exercises 6.3
Ex 6.3.1
Find the cycle index PG for the group of permutations of
the vertices of a regular tetrahedron induced by the rigid
motions. (See exercise 1 in
section 6.2.)
Ex 6.3.2
Using the previous exercise, write out a full inventory of colorings
of the vertices of a regular tetrahedron induced by the rigid
motions, with three colors, as in
example 6.3.5.
You may use Sage or some other computer algebra system.
Ex 6.3.3
Find the cycle index PG for the group of permutations of
the edges of K5.
(See exercise 5 in
section 6.2. Don't use the general formula above.)
Ex 6.3.4
Using the previous exercise, write out a full inventory of the graphs
on five vertices, as in example 6.3.6.
You may use Sage or some other computer algebra system. |
9561 | https://en.wikipedia.org/wiki/Catalan_number | Jump to content
Search
Contents
1 Properties
2 Applications in combinatorics
3 Proof of the formula
3.1 First proof
3.2 Second proof
3.3 Third proof
3.4 Fourth proof
3.5 Fifth proof
3.6 Sixth proof
4 Hankel matrix
5 History
6 Generalizations
7 Catalan k-fold convolution
8 See also
9 Notes
10 References
11 External links
Catalan number
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From Wikipedia, the free encyclopedia
Recursive integer sequence
Not to be confused with Catalan's constant.
The Catalan numbers are a sequence of natural numbers that occur in various counting problems, often involving recursively defined objects. They are named after Eugène Catalan, though they were previously discovered in the 1730s by Minggatu.
The n-th Catalan number can be expressed directly in terms of the central binomial coefficients by
The first Catalan numbers for n = 0, 1, 2, 3, ... are
: 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, ... (sequence A000108 in the OEIS).
Properties
[edit]
An alternative expression for Cn is
: for
which is equivalent to the expression given above because . This expression shows that Cn is an integer, which is not immediately obvious from the first formula given. This expression forms the basis for a proof of the correctness of the formula.
Another alternative expression is
which can be directly interpreted in terms of the cycle lemma; see below.
The Catalan numbers satisfy the recurrence relations
and
Asymptotically, the Catalan numbers grow as in the sense that the quotient of the n-th Catalan number and the expression on the right tends towards 1 as n approaches infinity.
This can be proved by using the asymptotic growth of the central binomial coefficients, by Stirling's approximation for , or via generating functions.
The only Catalan numbers Cn that are odd are those for which n = 2k − 1; all others are even. The only prime Catalan numbers are C2 = 2 and C3 = 5. More generally, the multiplicity with which a prime p divides Cn can be determined by first expressing n + 1 in base p. For p = 2, the multiplicity is the number of 1 bits, minus 1. For p an odd prime, count all digits greater than (p + 1) / 2; also count digits equal to (p + 1) / 2 unless final; and count digits equal to (p − 1) / 2 if not final and the next digit is counted. The only known odd Catalan numbers that do not have last digit 5 are C0 = 1, C1 = 1, C7 = 429, C31, C127 and C255. The odd Catalan numbers, Cn for n = 2k − 1, do not have last digit 5 if n + 1 has a base 5 representation containing 0, 1 and 2 only, except in the least significant place, which could also be a 3.
The Catalan numbers have the integral representations
which immediately yields .
This has a simple probabilistic interpretation. Consider a random walk on the integer line, starting at 0. Let −1 be a "trap" state, such that if the walker arrives at −1, it will remain there. The walker can arrive at the trap state at times 1, 3, 5, 7..., and the number of ways the walker can arrive at the trap state at time is . Since the 1D random walk is recurrent, the probability that the walker eventually arrives at −1 is .
Applications in combinatorics
[edit]
There are many counting problems in combinatorics whose solution is given by the Catalan numbers. The book Enumerative Combinatorics: Volume 2 by combinatorialist Richard P. Stanley contains a set of exercises which describe 66 different interpretations of the Catalan numbers. Following are some examples, with illustrations of the cases C3 = 5 and C4 = 14.
Cn is the number of Dyck words of length 2n. A Dyck word is a string consisting of n X's and n Y's such that no initial segment of the string has more Y's than X's. For example, the following are the Dyck words up to length 6:
XY
XXYY XYXY
XXXYYY XYXXYY XYXYXY XXYYXY XXYXYY
Re-interpreting the symbol X as an open parenthesis and Y as a close parenthesis, Cn counts the number of expressions containing n pairs of parentheses which are correctly matched:
((())) (()()) (())() ()(()) ()()()
Cn is the number of different ways n + 1 factors can be completely parenthesized (or the number of ways of associating n applications of a binary operator, as in the matrix chain multiplication problem). For n = 3, for example, we have the following five different parenthesizations of four factors:
((ab)c)d (a(bc))d (ab)(cd) a((bc)d) a(b(cd))
Successive applications of a binary operator can be represented in terms of a full binary tree, by labeling each leaf a, b, c, d. It follows that Cn is the number of full binary trees with n + 1 leaves, or, equivalently, with a total of n internal nodes:
Cn is the number of non-isomorphic ordered (or plane) trees with n + 1 vertices. See encoding ordered trees as binary trees. For example, Cn is the number of possible parse trees for a sentence (assuming binary branching), in natural language processing.
Cn is the number of monotonic lattice paths along the edges of a grid with n × n square cells, which do not pass above the diagonal. A monotonic path is one which starts in the lower left corner, finishes in the upper right corner, and consists entirely of edges pointing rightwards or upwards. Counting such paths is equivalent to counting Dyck words: X stands for "move right" and Y stands for "move up".
The following diagrams show the case n = 4:
This can be represented by listing the Catalan elements by column height:
[0,0,0,0] [0,0,0,1] [0,0,0,2] [0,0,1,1]
[0,1,1,1] [0,0,1,2] [0,0,0,3] [0,1,1,2] [0,0,2,2] [0,0,1,3]
[0,0,2,3] [0,1,1,3] [0,1,2,2] [0,1,2,3]
A convex polygon with n + 2 sides can be cut into triangles by connecting vertices with non-crossing line segments (a form of polygon triangulation). The number of triangles formed is n and the number of different ways that this can be achieved is Cn. The following hexagons illustrate the case n = 4:
Cn is the number of stack-sortable permutations of {1, ..., n}. A permutation w is called stack-sortable if S(w) = (1, ..., n), where S(w) is defined recursively as follows: write w = unv where n is the largest element in w and u and v are shorter sequences, and set S(w) = S(u)S(v)n, with S being the identity for one-element sequences.
Cn is the number of permutations of {1, ..., n} that avoid the permutation pattern 123 (or, alternatively, any of the other patterns of length 3); that is, the number of permutations with no three-term increasing subsequence. For n = 3, these permutations are 132, 213, 231, 312 and 321. For n = 4, they are 1432, 2143, 2413, 2431, 3142, 3214, 3241, 3412, 3421, 4132, 4213, 4231, 4312 and 4321.
Cn is the number of noncrossing partitions of the set {1, ..., n}. A fortiori, Cn never exceeds the n-th Bell number. Cn is also the number of noncrossing partitions of the set {1, ..., 2n} in which every block is of size 2.
Cn is the number of ways to tile a stairstep shape of height n with n rectangles. Cutting across the anti-diagonal and looking at only the edges gives full binary trees. The following figure illustrates the case n = 4:
Cn is the number of ways to form a "mountain range" with n upstrokes and n downstrokes that all stay above a horizontal line. The mountain range interpretation is that the mountains will never go below the horizon.
Mountain Ranges
| | | 1 way |
| | /\ | 1 way |
| | 0000000/\/\/\,0/00\ | 2 ways |
| | 0000000000000000000000000000000000/\00000000000/\0000/\000000/\/\0000/00\/\/\/\,0/\/00\,0/00\/\,0/0000\,0/0000\ | 5 ways |
Cn is the number of standard Young tableaux whose diagram is a 2-by-n rectangle. In other words, it is the number of ways the numbers 1, 2, ..., 2n can be arranged in a 2-by-n rectangle so that each row and each column is increasing. As such, the formula can be derived as a special case of the hook-length formula.
123 124 125 134 135 456 356 346 256 246
is the number of length n sequences that start with , and can increase by either or , or decrease by any number (to at least ). For these are . From a Dyck path, start a counter at 0. An X increases the counter by 1 and a Y decreases it by 1. Record the values at only the X's. Compared to the similar representation of the Bell numbers, only is missing.
Proof of the formula
[edit]
There are several ways of explaining why the formula
solves the combinatorial problems listed above. The first proof below uses a generating function. The other proofs are examples of bijective proofs; they involve literally counting a collection of some kind of object to arrive at the correct formula.
First proof
[edit]
We first observe that all of the combinatorial problems listed above satisfy Segner's recurrence relation
For example, every Dyck word w of length ≥ 2 can be written in a unique way in the form
: w = Xw1Yw2
with (possibly empty) Dyck words w1 and w2.
The generating function for the Catalan numbers is defined by
The recurrence relation given above can then be summarized in generating function form by the relation
in other words, this equation follows from the recurrence relation by expanding both sides into power series. On the one hand, the recurrence relation uniquely determines the Catalan numbers; on the other hand, interpreting xc2 − c + 1 = 0 as a quadratic equation of c and using the quadratic formula, the generating function relation can be algebraically solved to yield two solution possibilities
: or .
From the two possibilities, the second must be chosen because only the second gives
: .
The square root term can be expanded as a power series using the binomial series
Thus,
Second proof
[edit]
See also: Method of images § Mathematics for discrete cases
We count the number of paths which start and end on the diagonal of an n × n grid. All such paths have n right and n up steps. Since we can choose which of the 2n steps are up or right, there are in total monotonic paths of this type. We define a bad path as one that crosses the main diagonal and touches the next higher diagonal (red in the illustration).
The part of the path after the higher diagonal is then flipped about that diagonal, as illustrated with the red dotted line. This swaps all the right steps to up steps and vice versa. In the section of the path that is not reflected, there is one more up step than right steps, so therefore the remaining section of the bad path has one more right step than up steps. When this portion of the path is reflected, it will have one more up step than right steps. Therefore the whole path after reflection has two more up step than right steps.
Since there are still 2n steps, there are now n + 1 up steps and n − 1 right steps. So, instead of reaching (n, n), all bad paths after reflection end at (n − 1, n + 1). Because every monotonic path in the (n − 1) × (n + 1) grid meets the higher diagonal, and because the reflection process is reversible, the reflection is therefore a bijection between bad paths in the original grid and monotonic paths in the new grid.
The number of bad paths is therefore:
and the number of Catalan paths (i.e. good paths) is obtained by removing the number of bad paths from the total number of monotonic paths of the original grid,
In terms of Dyck words, we start with a (non-Dyck) sequence of n X's and n Y's and interchange all X's and Y's after the first Y that violates the Dyck condition. After this Y, note that there is exactly one more Y than there are Xs.
Third proof
[edit]
This bijective proof provides a natural explanation for the term n + 1 appearing in the denominator of the formula for Cn. A generalized version of this proof can be found in a paper of Rukavicka Josef (2011).
Given a monotonic path, the exceedance of the path is defined to be the number of vertical edges above the diagonal. For example, in Figure 2, the edges above the diagonal are marked in red, so the exceedance of this path is 5.
Given a monotonic path whose exceedance is not zero, we apply the following algorithm to construct a new path whose exceedance is 1 less than the one we started with.
Starting from the bottom left, follow the path until it first travels above the diagonal.
Continue to follow the path until it touches the diagonal again. Denote by X the first such edge that is reached.
Swap the portion of the path occurring before X with the portion occurring after X.
In Figure 3, the black dot indicates the point where the path first crosses the diagonal. The black edge is X, and we place the last lattice point of the red portion in the top-right corner, and the first lattice point of the green portion in the bottom-left corner, and place X accordingly, to make a new path, shown in the second diagram.
The exceedance has dropped from 3 to 2. In fact, the algorithm causes the exceedance to decrease by 1 for any path that we feed it, because the first vertical step starting on the diagonal (at the point marked with a black dot) is the only vertical edge that changes from being above the diagonal to being below it when we apply the algorithm - all the other vertical edges stay on the same side of the diagonal.
It can be seen that this process is reversible: given any path P whose exceedance is less than n, there is exactly one path which yields P when the algorithm is applied to it. Indeed, the (black) edge X, which originally was the first horizontal step ending on the diagonal, has become the last horizontal step starting on the diagonal. Alternatively, reverse the original algorithm to look for the first edge that passes below the diagonal.
This implies that the number of paths of exceedance n is equal to the number of paths of exceedance n − 1, which is equal to the number of paths of exceedance n − 2, and so on, down to zero. In other words, we have split up the set of all monotonic paths into n + 1 equally sized classes, corresponding to the possible exceedances between 0 and n. Since there are monotonic paths, we obtain the desired formula
Figure 4 illustrates the situation for n = 3. Each of the 20 possible monotonic paths appears somewhere in the table. The first column shows all paths of exceedance three, which lie entirely above the diagonal. The columns to the right show the result of successive applications of the algorithm, with the exceedance decreasing one unit at a time. There are five rows, that is C3 = 5, and the last column displays all paths no higher than the diagonal.
Using Dyck words, start with a sequence from . Let be the first X that brings an initial subsequence to equality, and configure the sequence as . The new sequence is .
Fourth proof
[edit]
This proof uses the triangulation definition of Catalan numbers to establish a relation between Cn and Cn+1.
Given a polygon P with n + 2 sides and a triangulation, mark one of its sides as the base, and also orient one of its 2n + 1 total edges. There are (4n + 2)Cn such marked triangulations for a given base.
Given a polygon Q with n + 3 sides and a (different) triangulation, again mark one of its sides as the base. Mark one of the sides other than the base side (and not an inner triangle edge). There are (n + 2)Cn + 1 such marked triangulations for a given base.
There is a simple bijection between these two marked triangulations: We can either collapse the triangle in Q whose side is marked (in two ways, and subtract the two that cannot collapse the base), or, in reverse, expand the oriented edge in P to a triangle and mark its new side.
Thus
: .
Write
Because
we have
Applying the recursion with gives the result.
Fifth proof
[edit]
This proof is based on the Dyck words interpretation of the Catalan numbers, so is the number of ways to correctly match n pairs of brackets. We denote a (possibly empty) correct string with c and its inverse with c'. Since any c can be uniquely decomposed into , summing over the possible lengths of immediately gives the recursive definition
: .
Let b be a balanced string of length 2n, i.e. b contains an equal number of and , so . A balanced string can also be uniquely decomposed into either or , so
Any incorrect (non-Catalan) balanced string starts with , and the remaining string has one more than , so
Also, from the definitions, we have:
Therefore, as this is true for all n,
Sixth proof
[edit]
This proof is based on the Dyck words interpretation of the Catalan numbers and uses the cycle lemma of Dvoretzky and Motzkin.
We call a sequence of X's and Y's dominating if, reading from left to right, the number of X's is always strictly greater than the number of Y's. The cycle lemma states that any sequence of X's and Y's, where , has precisely dominating circular shifts. To see this, arrange the given sequence of X's and Y's in a circle. Repeatedly removing XY pairs leaves exactly X's. Each of these X's was the start of a dominating circular shift before anything was removed. For example, consider . This sequence is dominating, but none of its circular shifts , , and are.
A string is a Dyck word of X's and Y's if and only if prepending an X to the Dyck word gives a dominating sequence with X's and Y's, so we can count the former by instead counting the latter. In particular, when , there is exactly one dominating circular shift. There are sequences with exactly X's and Y's. For each of these, only one of the circular shifts is dominating. Therefore there are distinct sequences of X's and Y's that are dominating, each of which corresponds to exactly one Dyck word.
Hankel matrix
[edit]
The n × n Hankel matrix whose (i, j) entry is the Catalan number Ci+j−2 has determinant 1, regardless of the value of n. For example, for n = 4 we have
Moreover, if the indexing is "shifted" so that the (i, j) entry is filled with the Catalan number Ci+j−1 then the determinant is still 1, regardless of the value of n. For example, for n = 4 we have
Taken together, these two conditions uniquely define the Catalan numbers.
Another feature unique to the Catalan–Hankel matrix is that the n × n submatrix starting at 2 has determinant n + 1.
et cetera.
History
[edit]
The Catalan sequence was described in 1751 by Leonhard Euler, who was interested in the number of different ways of dividing a polygon into triangles. The sequence is named after Eugène Charles Catalan, who discovered the connection to parenthesized expressions during his exploration of the Towers of Hanoi puzzle. The reflection counting trick (second proof) for Dyck words was found by Désiré André in 1887.
The name “Catalan numbers” originated from John Riordan.
In 1988, it came to light that the Catalan number sequence had been used in China by the Mongolian mathematician Mingantu by 1730. That is when he started to write his book Ge Yuan Mi Lu Jie Fa [The Quick Method for Obtaining the Precise Ratio of Division of a Circle], which was completed by his student Chen Jixin in 1774 but published sixty years later. Peter J. Larcombe (1999) sketched some of the features of the work of Mingantu, including the stimulus of Pierre Jartoux, who brought three infinite series to China early in the 1700s.
For instance, Ming used the Catalan sequence to express series expansions of and in terms of .
Generalizations
[edit]
The Catalan numbers can be interpreted as a special case of the Bertrand's ballot theorem. Specifically, is the number of ways for a candidate A with n + 1 votes to lead candidate B with n votes.
The two-parameter sequence of non-negative integers is a generalization of the Catalan numbers. These are named super-Catalan numbers, per Ira Gessel. These should not confused with the Schröder–Hipparchus numbers, which sometimes are also called super-Catalan numbers.
For , this is just two times the ordinary Catalan numbers, and for , the numbers have an easy combinatorial description. However, other combinatorial descriptions are only known for and , and it is an open problem to find a general combinatorial interpretation.
Sergey Fomin and Nathan Reading have given a generalized Catalan number associated to any finite crystallographic Coxeter group, namely the number of fully commutative elements of the group; in terms of the associated root system, it is the number of anti-chains (or order ideals) in the poset of positive roots. The classical Catalan number corresponds to the root system of type . The classical recurrence relation generalizes: the Catalan number of a Coxeter diagram is equal to the sum of the Catalan numbers of all its maximal proper sub-diagrams.
The Catalan numbers are a solution of a version of the Hausdorff moment problem.
For coprime positive integers r and s, the rational Catalan numbers count the number of lattice paths with steps of unit length rightwards and upwards from (0,0) to (r,s) that never go above the line ry = sx.
Catalan k-fold convolution
[edit]
The Catalan k-fold convolution, where k = m, is:
See also
[edit]
Mathematics portal
Associahedron
Bertrand's ballot theorem
Binomial transform
Catalan's triangle
Catalan–Mersenne number
Delannoy number
Fuss–Catalan number
List of factorial and binomial topics
Lobb numbers
Motzkin number
Narayana number
Narayana polynomials
Schröder number
Schröder–Hipparchus number
Semiorder
Tamari lattice
Wedderburn–Etherington number
Wigner's semicircle law
Notes
[edit]
^ Koshy, Thomas; Salmassi, Mohammad (2006). "Parity and primality of Catalan numbers" (PDF). The College Mathematics Journal. 37 (1): 52–53. doi:10.2307/27646275. JSTOR 27646275.
^ Sloane, N. J. A. (ed.). "Sequence A000108 (Catalan numbers)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation.
^ "Catalan Number".
^ Choi, Hayoung; Yeh, Yeong-Nan; Yoo, Seonguk (2020), "Catalan-like number sequences and Hausdorff moment sequences", Discrete Mathematics, 343 (5): 111808, 11, arXiv:1809.07523, doi:10.1016/j.disc.2019.111808, MR 4052255, S2CID 214165563, Example 3.1
^ Feng, Qi; Bai-Ni, Guo (2017), "Integral Representations of the Catalan Numbers and Their Applications", Mathematics, 5 (3): 40, doi:10.3390/math5030040,Theorem 1
^ Dyck paths
^ Stanley p.221 example (e)
^ Črepinšek, Matej; Mernik, Luka (2009). "An efficient representation for solving Catalan number related problems" (PDF). International Journal of Pure and Applied Mathematics. 56 (4): 589–604.
^ A. de Segner, Enumeratio modorum, quibus figurae planae rectilineae per diagonales dividuntur in triangula. Novi commentarii academiae scientiarum Petropolitanae 7 (1758/59) 203–209.
^ Rukavicka Josef (2011), On Generalized Dyck Paths, Electronic Journal of Combinatorics online
^ Dershowitz, Nachum; Zaks, Shmuel (1980), "Enumerations of ordered trees", Discrete Mathematics, 31: 9–28, doi:10.1016/0012-365x(80)90168-5, hdl:2027/uiuo.ark:/13960/t3kw6z60d
^ Dvoretzky, Aryeh; Motzkin, Theodore (1947), "A problem of arrangements", Duke Mathematical Journal, 14 (2): 305–313, doi:10.1215/s0012-7094-47-01423-3
^ Dershowitz, Nachum; Zaks, Shmuel (January 1990). "The Cycle Lemma and Some Applications" (PDF). European Journal of Combinatorics. 11 (1): 35–40. doi:10.1016/S0195-6698(13)80053-4.
^ Stanley, Richard P. (2021). "Enumerative and Algebraic Combinatorics in the 1960's and 1970's". arXiv:2105.07884 [math.HO].
^ Larcombe, Peter J. "The 18th century Chinese discovery of the Catalan numbers" (PDF).
^ "Ming Antu, the First Inventor of Catalan Numbers in the World". Archived from the original on 2020-01-31. Retrieved 2014-06-24.
^ Chen, Xin; Wang, Jane (2012). "The super Catalan numbers S(m, m + s) for s ≤ 4". arXiv:1208.4196 [math.CO].
^ Gheorghiciuc, Irina; Orelowitz, Gidon (2020). "Super-Catalan Numbers of the Third and Fourth Kind". arXiv:2008.00133 [math.CO].
^ Sergey Fomin and Nathan Reading, "Root systems and generalized associahedra", Geometric combinatorics, IAS/Park City Math. Ser. 13, American Mathematical Society, Providence, RI, 2007, pp 63–131. arXiv:math/0505518
^ Choi, Hayoung; Yeh, Yeong-Nan; Yoo, Seonguk (2020), "Catalan-like number sequences and Hausdorff moment sequences", Discrete Mathematics, 343 (5): 111808, 11, arXiv:1809.07523, doi:10.1016/j.disc.2019.111808, MR 4052255, S2CID 214165563
^ Krattenthaler, Christian (2015). "Lattice Path Enumeration" (PDF). In Bóna, Miklós (ed.). Handbook of Enumerative Combinatorics. Discrete Mathematics and Its Applications (1 ed.). CRC Press. p. 598. ISBN 9780429170317.
^ Bowman, D.; Regev, Alon (2014). "Counting symmetry: classes of dissections of a convex regular polygon". Adv. Appl. Math. 56: 35–55. arXiv:1209.6270. doi:10.1016/j.aam.2014.01.004. S2CID 15430707.
References
[edit]
Stanley, Richard P. (2015), Catalan numbers. Cambridge University Press, ISBN 978-1-107-42774-7.
Conway and Guy (1996) The Book of Numbers. New York: Copernicus, pp. 96–106.
Gardner, Martin (1988), Time Travel and Other Mathematical Bewilderments, New York: W.H. Freeman and Company, pp. 253–266 (Ch. 20), Bibcode:1988ttom.book.....G, ISBN 0-7167-1924-X
Koshy, Thomas (2008), Catalan Numbers with Applications, Oxford University Press, ISBN 978-0-19-533454-8
Koshy, Thomas & Zhenguang Gao (2011) "Some divisibility properties of Catalan numbers", Mathematical Gazette 95:96–102.
Larcombe, P.J. (1999). "The 18th century Chinese discovery of the Catalan numbers" (PDF). Mathematical Spectrum. 32: 5–7.
Stanley, Richard P. (1999), Enumerative combinatorics. Vol. 2, Cambridge Studies in Advanced Mathematics, vol. 62, Cambridge University Press, ISBN 978-0-521-56069-6, MR 1676282
Egecioglu, Omer (2009), A Catalan–Hankel Determinant Evaluation (PDF)
Gheorghiciuc, Irina; Orelowitz, Gidon (2020), Super-Catalan Numbers of the Third and Fourth Kind, arXiv:2008.00133
External links
[edit]
Stanley, Richard P. (1998), Catalan addendum to Enumerative Combinatorics, Volume 2 (PDF)
Weisstein, Eric W. "Catalan Number". MathWorld.
Davis, Tom: Catalan numbers. Still more examples.
"Equivalence of Three Catalan Number Interpretations" from The Wolfram Demonstrations Project
Learning materials related to Partition related number triangles at Wikiversity
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The Pauli Exclusion Principle
The Pauli Exclusion Principle
52,451 Views 03:06 min September 3, 2020
Overview
The arrangement of electrons in the orbitals of an atom is called its electron configuration. We describe an electron configuration with a symbol that contains three pieces of information:
The number of the principal quantum shell, n,
The letter that designates the orbital type (the subshell, l), and
A superscript number that designates the number of electrons in that particular subshell.
For example, the notation 2p4 indicates four electrons in a p subshell (l = 1) with a principal quantum number (n) of 2. The notation 3d 8 indicates eight electrons in the d subshell ( l = 2) of the principal shell for which n = 3.
While the three quantum numbers work well for describing electron orbitals, some experiments showed that they were not sufficient to explain all observed results. It was demonstrated in the 1920s that when hydrogen-line spectra are examined at extremely high resolution, some lines are actually not single peaks but, rather, pairs of closely spaced lines. This is the so-called fine structure of the spectrum, and it implies that there are additional small differences in energies of electrons even when they are located in the same orbital. These observations led Samuel Goudsmit and George Uhlenbeck to propose that electrons have a fourth quantum number. They called this the spin quantum number or ms.
In an applied magnetic field, an electron has two possible orientations with different energies, one with spin up, aligned with the magnetic field, and one with spin down, aligned against it.
The fourth quantum number, spin quantum number (ms) describes these two different spin states of an electron. A spin quantum number has two possible values, −1/2 (spin down) and +1/2 (spin-up).
Electron spin describes an intrinsic electron "rotation" or "spinning." Each electron acts as a tiny magnet or a tiny rotating object with angular momentum, or as a loop with an electric current, even though this rotation or current cannot be observed in terms of spatial coordinates.
The magnitude of the overall electron spin can only have one value, and an electron can only “spin” in one of two quantized states. One is termed the α state, with the z component of the spin being in the positive direction of the z-axis. This corresponds to the spin quantum number ms = +1/2. The other is called the β state, with the z component of the spin being negative and ms = −1/2.
Any electron, regardless of the atomic orbital it is located in, can only have one of those two values of the spin quantum number. The energies of electrons having different spins are different if an external magnetic field is applied.
An electron in an atom is completely described by four quantum numbers: n, l, ml, and ms. The first three quantum numbers define the orbital, and are interdependent, while the fourth quantum number is independent of other quantum numbers as it describes an intrinsic electron property called spin. An Austrian physicist Wolfgang Pauli (Nobel Prize in Physics: 1945) formulated a general principle that gives the last piece of information that we need to understand the general behavior of electrons in atoms. The Pauli exclusion principle can be formulated as follows: No two electrons in the same atom can have exactly the same set of all the four quantum numbers. What this means is that two electrons can share the same orbital (the same set of the quantum numbers n, l, and ml) only if their spin quantum numbers have different values. Since the spin quantum number (ms) can only have two values +1/2 and -1/2, no more than two electrons can occupy the same orbital (and if two electrons are located in the same orbital, they must have opposite spins). Therefore, any atomic orbital can be populated by only zero, one, or two electrons.
The orbital diagram style of the electron configuration represents each orbital within an occupied subshell as a box or line and each electron as an arrow. The orbital diagram of hydrogen, which an electron configuration of 1s1, is:
An upward arrow indicates a plus-half spin, or spin-up, and a downward arrow signifies a minus-half spin, or spin-down. The orbital diagram of hydrogen, therefore, has one upward arrow.
The electron configuration of helium is 1s2. The two electrons have three identical quantum numbers, as they belong to the same shell and subshell. Their spin quantum numbers are different, in accordance with the Pauli exclusion principle. Electrons with opposite spins are called paired if they occupy the same orbital.
This text is adapted from Openstax, Chemistry 2e, Section 6.3: Development of Quantum Theory.
Transcript
Atomic orbitals are the regions where an atom’s electrons are most likely to be found. But how many electrons can each orbital hold?
The Pauli exclusion principle answers this question, as it means that no two electrons in an atom can have the same set of four quantum numbers.
Every orbital corresponds to fixed principal; angular momentum, or azimuthal; and magnetic quantum number values. For example, an electron in the 1s orbital always has a principal quantum number of one and azimuthal and magnetic quantum numbers of zero.
Accordingly, electrons must have different spin quantum number values, or spins, to reside in the same atomic orbital. Recall that the spin quantum number has only two possible values: +1/2 and −1/2. Therefore, only two electrons can occupy the same orbital.
Hence, each s subshell, which has one orbital, can accommodate only two electrons, and each p subshell, which has three orbitals, can hold six electrons. Each of the d and f subshells has a maximum capacity of ten and fourteen electrons, respectively.
The distribution of electrons among an atom’s atomic orbitals is represented by its electron configuration in text or diagram form.
Consider a ground-state hydrogen atom, where one electron occupies the lowest-energy orbital: 1s. The written electron configuration denotes each occupied subshell with the number of the corresponding shell, the letter of the subshell, and a superscript number specifying the number of electrons in the subshell.
The orbital diagram style of the electron configuration denotes each orbital within an occupied subshell as a box or line and each electron as an arrow. An upward arrow indicates a +1/2 spin, or spin-up, and a downward arrow signifies a −1/2 spin, or spin-down. The orbital diagram of hydrogen, therefore, has one upward arrow.
The electron configuration of helium is 1s2. The two electrons have three identical quantum numbers, as they belong to the same shell and subshell. Their spin quantum numbers are different, in accordance with the Pauli exclusion principle.
Electrons with opposite spins are called “paired” if they occupy the same orbital.
For lithium, which has an atomic number of three, the two electrons in the 1s orbital are paired and the electron in the 2s orbital is unpaired. Conventionally, unpaired electrons are represented as spin-up.
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9563 | https://bmcresnotes.biomedcentral.com/articles/10.1186/1756-0500-7-544 | Advertisement
Factors associated with genital human papillomavirus infection among adult females in the United States, NHANES 2007–2010
BMC Research Notes
volume 7, Article number: 544 (2014)
Cite this article
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Abstract
Background
Patients with human papillomavirus (HPV) infection are at risk of developing cancer later in their life. Current research estimates the prevalence of genital HPV infection and explores the factors that are associated with the infection.
Findings
The National Health and Nutrition Examination Survey 2007–2010 was used in this research study. The study population included females in the United States aged 18–59 years. The weighted prevalence of HPV infection was 41.9%. An estimated 59.4% of non-Hispanic black females had HPV infection. In a multivariate analysis, number of sexual partners, race, age, education level, marital status, income, smoking, and insurance status were associated with HPV infection. HPV infection was 5.77 times more likely for women with >11 sexual partners compared to women with 0–1 partners. Non-Hispanic black females were 1.87 times more likely to have HPV infection compared to non-Hispanic white females. Participants with only a high school degree had a 58% increased prevalence compared to college-educated women. Uninsured women had a 39% increased prevalence compared to those with insurance.
Conclusion
This study found that 41.9% of U.S. females aged 18–59 years tested positive for genital HPV infection. We determined that individuals with more sexual partners, with a lower education level, with non-Hispanic black race, and with no insurance were the populations at greatest risk. It is necessary to continue monitoring the prevalence of this infection in the general population to provide a basis for effective treatment and prevention in the target populations.
Findings
Genital human papillomavirus (HPV) is the most common sexually transmitted infection in the United States. HPV includes a group of more than 150 related viruses. Many of these viruses can be easily spread through direct skin contact during sexual intercourse [1e
]")]. More than 50% of individuals engaging in sexual activities are infected with at least one type of HPV in their lifetime. An estimated 42.5% of U.S. females aged 14–59 years had a genital HPV infection in 2003–2006 [2: 566-573.")]. Some patients with HPV infection can be restored back to health, while infection progresses to cancer in others [3: 630-632.")].
More than 40 HPV types can infect the genital areas of men and women . These types can be classified as high-risk, probable high-risk, low-risk, and undetermined risk for the development of cervical cancer. Low-risk types (HPV 6, 11) are mostly associated with genital warts. High-risk types (HPV 16, 18) can contribute to precancerous lesions, low-grade cervical intraepithelial lesions and high-grade cervical intraepithelial lesions, as well as anogenital cancers [5: S16-24.")]. HPV infection is a major cause of cervical cancer [6: 3036-3046.")], and 11,818 women in the U.S. were diagnosed with cervical cancer in 2010 .
The primary goals of our current research are to identify the potential factors associated with HPV infection and to estimate the prevalence of infection for U.S. females from 2007 to 2010. Eventually, we aim to help create programs targeting high-prevalence populations to prevent HPV infection and lower their risk of getting cervical cancer.
Methods and materials
Data was obtained from the National Health and Nutrition Examination Survey (NHANES) 2007–2010 . The survey data includes information about demographic and socioeconomic status, mental health, and dental health, as well as physiological and laboratory measurements. The survey interviews about 12,000 people biannually.
Survey design and population
The NHANES used a multistage probability sample design to select the participants. Consenting participants completed a household interview followed by a physical examination and interviews at a Mobile Examination Center (MEC). Non-Hispanic (NH) black and low income groups were oversampled in the NHANES to allow for an accurate statistical estimation in these population groups . The protocol was approved by the National Center for Health Statistics (NCHS) institutional review board. Further information regarding study design and methods for oversampling is available on the NHANES website .
From 2007 to 2010, 10,010 females of all ages were interviewed. The combined unweighted household interview response rate for that period was 78.9%; the examination response rate was 76.3%. All females aged 18–59 years (n = 4242) who visited the MEC were asked to self-collect a cervicovaginal swab sample. Out of all samples collected, 3738 (88.1%) were reported as positive or negative and were used in the final analysis. Combining the data for years 2007 through 2010 was justified. There was no significant difference in HPV prevalence between 2007–2008 and 2009–2010 for all but 5 of 37 HPV types, as detected by the Linear Array assay (data not shown).
Demographic and behavioral data
Demographic information, including gender, age, race, education, marital status, and country of birth, was obtained from all participants during the household interviews. The poverty index was calculated according to the U.S. Census definition. This method divides total family income by the poverty threshold after adjusting for family size at the time of the interview.
Sexual history information, including if the participant had ever had sex and the age of first sexual experience, was self-reported by participants using an audio computer-assisted self-interview. Respondents who reported having sex (described as vaginal, oral, or anal) were asked additional questions about their lifetime sexual history and about any sexual encounters in the prior 12 months. These additional questions addressed number of sexual partners, sexual orientation, and condom usage.
Specimen collection and processing, laboratory methods
As described by Dunne et al. , self-collected cervicovaginal swab samples were obtained from female participants aged 18–59 who had an examination in the MEC. Swabs were given to the NHANES personnel, stored at room temperature, and mailed within 1 week to the Centers for Disease Control and Prevention (CDC) laboratory. There, they were kept at 4°C and extracted within 1 month of collection.
Multiplex polymerase chain reaction (PCR) was used for detection of 37 HPV types within the Alphapapillomavirus genera. Samples were reported as HPV positive if any of the 37 HPV deoxyribonucleic acid (DNA) types were detected, including high-risk (16, 18, 26, 31, 33, 35, 39, 45, 51, 52, 53, 56, 58, 59, 66, 68, 73, 82) and low-risk (6, 11, 40, 42, 54, 55, 61, 62, 64, 67, 69, 70, 71, 72, 81, 82 subtype IS39, 83, 84, 89) types. If the strips were positive for any of the types, the sample was coded as positive. If the strips were negative for all of the types and beta-globin was detected, the sample was coded as negative. If there was no beta-globin present in the sample and no HPV type was detected, the sample was coded as inadequate [8, 11: 412-414.")–14: 430-441.")].
Statistical analysis
We estimated the overall prevalence of infection for any HPV type with respect to sociodemographic and sexual behavioral characteristics. Due to the complexity of the design, all estimates were measured using 4-year MEC sample weights provided by NCHS to account for the unequal probabilities of selection and adjustment for nonresponse. The weighting methodology has been described previously . Taylor series linearization was used to estimate variance in a complex cluster survey design . Confidence intervals (CIs) were calculated using a logit transformation with the standard error of the logit prevalence based on the delta method and applying SUDAAN estimated standard errors .
The Wald χ2 statistic was used to assess bivariate association between HPV and sociodemographic or behavioral characteristics. No adjustments were made to the p- values for multiple comparisons. An unconditional logistic regression model was used to determine associations between genital HPV infection and these factors. Variables for adjustment in multivariate logistic regression models were selected based on bivariate associations with p-values <0.2. Goodness of fit for the final step of the model was assessed using the Hosmer-Lemeshow Satterthwaite adjusted F test.
SAS software version 9.4 for Windows (SAS Institute Inc. Gary, North Carolina) and SAS callable SUDAAN 11.0.1 were used for the statistical analyses. Two-sided p-values less than 0.05 were considered statistically significant.
Results
HPV types and species by oncogenic risk category are presented in Figure 1. In the NHANES 2007–2010, there were 3738 females aged 18 to 59 years with an HPV evaluation result in the final analysis. The 37 HPV types were grouped as high-risk (Figure 1 Left Panel) or low-risk (Figure 1 Right Panel). The prevalence of high-risk HPV types 33 and 58 was less than 2%; HPV strains 16 and 18 had a prevalence of 4.92% and 1.77%, respectively. HPV strain 53 had the highest prevalence of 6.26% in the high-risk HPV category, and HPV strain 62 had the highest prevalence of 5.93% in the low-risk category.
HPV types and species by oncogenic risk category. Left panel, High-Risk HPV Types. Right panel, Low-Risk HPV Types.
In a univariate analysis (Table 1), the overall weighted prevalence of HPV infection was 41.9%. An estimated 59.4% NH black females and 38.7% of NH white females, the lowest prevalence, were positive for HPV. There was a statistically significant difference between subjects of different races. The Prevalence Ratio for NH black females was 2.3 times that of NH white females.
There was a lower trend of HPV infection with increased age, higher education level, and decreased family poverty level. There was a bimodal pattern in the prevalence of HPV for age, with a high HPV prevalence of 56.1% in the 18–24 age group and with a second peak in the 40–44 age group. HPV prevalence was 48% for women with less than a high school education, whereas it was 31.4% for those with a college education. Participants living in a family with an income-to-poverty ratio (PIR) of <130% showed a 55.3% prevalence of HPV infection. In terms of marital status, prevalence was lowest (29.4%) for married women and highest (>47%) for women classified as single, divorced, or never married.
The association of HPV infection with behavioral factors is presented in Table 2. Females who smoked <100 cigarettes in their lifetime (Prevalence Ratio [PR] = 0.52) were less likely to have HPV infection. Females who did not have health insurance (PR = 1.9) or a routine place to receive health care (PR =1.39) were more likely to have HPV infection.Females who reported never having sex were 71% less likely to have HPV infection compared to females who reported having sex. The prevalence of HPV infection increased 89% for females having sex before 16 years of age compared to after 16. There was a trend of increasing HPV infection in women with a higher number of sexual partners in their lifetime (or yearly). The prevalence of infection was 8.9 times higher for women with >11 sexual partners in their lifetime compared to women with 0–1 partners. This trend was noticed among all races. Overall, NH black females showed the highest prevalence of HPV infection regardless of their number of partners (Figure 2). Compared to women who used a condom during sex, those who did not use a condom had twice the prevalence of HPV infection. Lastly, women who identified themselves as being ‘lesbian’ or ‘bisexual’ had a 72% increased prevalence of HPV infection compared to those who identified as ‘straight.’
Weighted prevalence of genital HPV among female adult participants aged 18–59 by lifetime sexual partners and race.
In a multivariate logistic regression analysis, adjusting for other factors, the number of lifetime sexual partners had a significant association with HPV infection. HPV infection was 5.77 times more likely for women with >11 partners compared to women with 0–1 partners. NH black females were 1.87 times more likely to have an HPV infection compared to NH white females. Age was negatively associated with the prevalence of HPV, except for an increased peak in the 44–49 age group. Having a college degree and being married were also associated with lower HPV prevalence. Family income-to-poverty ratio, a habit of smoking cigarettes, and insurance status remained significant factors associated with HPV infection.
When the results were adjusted for other factors, country of birth, recreational drug usage, ever having sex, age at first sexual encounter, condom usage during sex, and sexual orientation were no longer associated with HPV infection.
Discussion and conclusion
From the NHANES 2007–2010 data set, we found that the overall prevalence of genital HPV infection was 41.9% (95% CI: 39.6-44.2%) in females aged 18–59 in the United States. We compared this result with two previous NHANES analyses observing data for females in the United States. Our estimate was similar to the result of the first study, which found a 42.5% (95% CI: 40.3-44.7%) prevalence for females aged 14–59 in the 2003–2006 data set . The confidence intervals overlapped, indicating no statistical difference between these two estimations. Our estimate was higher than the result of the second study, which found a prevalence of 26.8% among females, aged 14–59 in the 2003–2004 data set. This discrepancy could be due to a change in the laboratory methods to detect HPV DNA resulting in increased sensitivity of the test [2, 18, 19].
In our study, there was a bimodal prevalence of genital HPV for age, consistent with the same two previous NHANES studies [2, 10]. This may indicate increased sexual activity in the 18–24 age group [20, 21] causing the transmission of HPV by sexual contact. The reason for the increased prevalence in the 45–49 age group is unclear. We speculate that it may be due to an increased incidence , a difference in sexual behaviors across birth cohorts , or a change in marital status.
HPV infection was associated with certain racial and ethnic groups. NH black females had an increased prevalence even when controlling for factors such as number of sexual partners in their lifetime (Figure 2) or other factors (Table 3). Interestingly, even in the group with 0–1 sexual partners, NH black females had twice the prevalence of HPV infection compared to other racial and ethnic groups. This indicates that number of sexual partners is not the only factor that impacts the prevalence of HPV infection.
Some studies have shown an inconsistent association between education level and prevalence [24, 25]. However, in others, education level was negatively associated with HPV infection [2, 10]. In our study, with increased education, the prevalence of HPV infection decreased. We speculate that this trend could be connected to factors such as increased awareness of HPV or adoption of safe sexual practices. Safe practices might include, but are not limited to, regular usage of condoms and limiting the number of sexual partners. Therefore, health education interventions could be introduced to reduce the HPV prevalence by increasing awareness, encouraging safe sex , teaching about the routes of HPV transmission, and promoting the use of the HPV vaccine .
The number of sexual partners played a statistically significant role in HPV infection. This finding mirrors those seen in other studies [2, 10]. We also found an association of health insurance status with HPV infection. A meta-analysis showed that cost of vaccination and lack of insurance coverage are barriers that prevent women from obtaining the vaccine . Individuals with private health insurance are more likely to hear about the HPV vaccine and three times more likely to get the vaccine compared to uninsured patients or those with public insurance plans . In addition, poverty and having smoked <100 cigarettes in their lifetime were associated with HPV infection.
In conclusion, 41.9% of U.S. females aged 18–59 years tested positive for genital HPV infection. This study found that an increased number of sexual partners, a lower level of education, non-Hispanic black race, and a lack of insurance were factors of concern with HPV infection. Continuing to monitor the prevalence of HPV in the general population can establish a basis for possible interventions focusing on at-risk groups.
Availability of supporting data
All data used in this research can be downloaded from the following website:
Ethics statement
Use of data from the NHANES 2007–2010 is approved by the National Center for Health Statistics (NCHS) Research Ethics Review Board (ERB) Approval for NHANES 2009–2010 (Continuation of Protocol #2005-06), NHANES 2007–2008 (Continuation of Protocol #2005-06) and NHANES 2005–2006 (Protocol #2005-06).
Abbreviations
Human papillomavirus
Non-hispanic
Centers for Disease Control and Prevention
National Health and Nutrition Examination Survey
Mobile Examination Center
National Center for Health Statistics
Polymerase chain reaction
Deoxyribonucleic acid
Income-to-poverty ratio
Prevalence ratio
Statistical analysis system
Survey data analysis.
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Acknowledgements
The authors wish to acknowledge the NHANES 2007–2010 participants and investigators as well as the National Center for Health Statistics and its Research Data Center for making the public data available. However, the authors take full and sole responsibility for the integrity of the data analysis and the contents of this article. We also wish to thank Talicia A. Tarver and Shenika McCary for their assistance in preparing this manuscript and the reviewers for their critical comments and suggestions.
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Authors and Affiliations
Department of Medicine, Feist-Weiller Cancer Center, LSU Health Shreveport, Shreveport, LA, USA
Runhua Shi, Srinivas Devarakonda, Lihong Liu, Hannah Taylor & Glenn Mills
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Correspondence to
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The authors have no financial or non-financial competing interests to disclose.
Authors’ contributions
RS carried out design of the study, data management, statistical analysis, and writing of the manuscript. HT and SD helped to write and finalize the manuscript. LL conceived the study, participated in its design, and helped to write the manuscript. GM helped in the study’s design and in writing the manuscript. All authors read and approved the final manuscript.
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Shi, R., Devarakonda, S., Liu, L. et al. Factors associated with genital human papillomavirus infection among adult females in the United States, NHANES 2007–2010.
BMC Res Notes 7, 544 (2014).
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9564 | https://libraryguides.centennialcollege.ca/c.php?g=717032&p=5124576 | Volumes of Revolution - Calculus - Library Guides at Centennial College
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Volumes of Solids of Revolution
A solid of revolution is when a function is rotated about an axis. Using integration, we are able to determine the volume of solids of revolution. Consider a portion of the graph of f(x)\=x2:
Figure 1 Figure 2
Vertical Disk Method
The vertical disk method is used when rotating a graph around a horizontal axis. Figure 2 from above is a good example of when we wish to use this method. Before integrating, it is helpful to draw a diagram; we will use f(x)\=x2 once again. After drawing the solid of revolution we will proceed to draw a "slice" of it:
We can view the slice as a disk that has the x-axis going through the middle of it. Notice that the radius of the disk is simply f(x) at that particular x-value. Taking the formula for the area of a circle, we get that this disk has area π(f(x))2 and thus has volume Vi\=π(f(x))2Δx (where Δx is the width). It follows then that the volume of this shape is simply the addition of the volumes of all of the disks_:_ V≅∑i\=0nπf(xi)2Δxi
Recall our definition of the Definite Integral: ∫abf(x)dx\=limn→∞∑i\=0n−1f(xi)Δx
If we take the limit of our V as n→∞ we get that the volume of our solid is V\=∫abπ(f(x))2dx
EXAMPLE
Draw the graph of f(x)\=x and determine the volume resulting from rotating f(x) about the x-axis in the interval of [1,5].
See the video below for the solution:
Rotating between Two Curves
When dealing with volumes of solids of revolution, in many instances the section that we are rotating exists between two functions. Consider f(x)\=x and g(x)\=(x−2)4+1
If we rotate the region between the functions on the interval of [2,3], we see that the radius of the disk is f(x)−g(x).
Rewriting our above formula, we obtain a new formula for the volume of a solid obtained by rotating a region between two functions:
V\=∫abπ(f(x)−g(x))2dx
Horizontal Disk Method
The horizontal disk method is used when rotating a graph around a vertical axis. We will use the region between the graphs of x\=(y−1)2 and x\=y+1 for this example (note that these are functions in terms of y). Similarly to the vertical disk method, after drawing the solid of revolution we will proceed to draw a "slice" of it:
The radius of the "slice"/disk from x\=y+1 to the y-axis is simply R1\=x. The radius of the "slice"/disk from x\=(y−1)2 to the y-axis is R2\=x. Similarly to the vertical disk method, the radius of the disk formed by rotating the region between the two curves is found by subtracting the distance that is outlined by the grey dashes in the diagram above. We now see that R\=R1−R2.
Recall our formula for volume when dealing with two curves:
V\=∫abπ(f(x)−g(x))2dx
We can substitute out R, but we still need to determine our a and b values. From the diagram we can see that the points of intersection of the two functions are at x\=1,4
DO NOT SUBSTITUTE x\=1 AND x\=4 AS THE ENDPOINTS!
Although it may be tempting, since we are dealing with functions in terms of y, our a and b values need to be in terms of y as well. We can use either function to determine what our y-values are:
⟹1\=y+1
⟹0\=y
⟹ The first point of intersection is (1,0)
⟹4\=y+1
⟹3\=y
⟹ The second point of intersection is (4,3)
We can now substitute all of our values and proceed to use integration to solve for the volume:
⟹V\=∫03π((y+1)−(y−1)2)2dx
⟹ \=π(∫03(y+1−y2+2y−1)2dx)
⟹ \=π(∫03(−y2+3y)2dx)
⟹ \=π(∫03y4−6y3+9y2dx)
⟹ \=π(15y5−32y4+3y3∣03)
⟹ \=π(15(3)5−32(3)4+3(3)3−0)
⟹ \=81π10 OR 8.1π units cubed
Cylindrical Shells
The cylindrical shell method is prominently used when rotating a graph about a vertical axis, but it can be used when rotating about a horizontal axis as well. Consider f(x)\=2x2−x3 on the interval of [0,2].
If we rotate the region between x\=0 and x\=2 about the y−axis, we obtain a cylindrical shell as shown below.
We see that the height of the shell is f(x) and the radius is x. Using our formula for calculating the volume of a cylinder, we obtain the following general formula for determining the volume of a solid of revolution:
V\=∫ab2πx2f(x)dx
See the video below for the solution to this example:
Try changing the function and rotating it on different axis. Use the slider to adjust the angle of rotation.
Volumes of Revolution Tip Chart
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9565 | https://www.kristakingmath.com/blog/measures-of-parallelograms | Measures of parallelograms, including angles, sides, and diagonals — Krista King Math | Online math help
Measures of parallelograms, including angles, sides, and diagonals
Defining all the measures of a parallelogram
A parallelogram is a quadrilateral that has opposite sides that are parallel.
The parallel sides let you know a lot about a parallelogram.
Here are the special properties of parallelograms:
Parallelogram
Two pairs of opposite parallel sides
Opposite sides are equal lengths
Opposite angles are congruent
m∠1=m∠3
m∠2=m∠4
Consecutive angles are supplementary
m∠1+m∠2=180∘
m∠2+m∠3=180∘
m∠3+m∠4=180∘
m∠4+m∠1=180∘
Diagonals bisect each other (cut each other in half)
How to solve for every measure of a parallelogram, including angles, side lengths, and the lengths of diagonals
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Finding the measure of a interior angle of a parallelogram
Example
Find the measure of angle y, given JKLM is a parallelogram.
Opposite angles of parallelograms are congruent, so
m∠JML=m∠JKL=57∘
Now we can use the fact that opposite sides of a parallelogram are parallel to state that JK∥ML. This means that the diagonal JL of the parallelogram is also a transversal of these two parallel lines. This means that ∠KLJ and ∠MJL are alternate interior angles. Alternate interior angle pairs are congruent, so m∠KLJ=m∠MJL=y.
The measures of the three interior angles of a triangle add up to 180∘, so we can set up an equation for the sum of the interior angles of △JML and solve for y.
y+57∘+64∘=180∘
y=59∘
Example
If STUV is a parallelogram, and if VT=4n+34 and VE=7n−3, what is the length of ET?
We know that the diagonals of a parallelogram bisect each other. Let’s add this information into the diagram.
Now we can see the relationships we need. Because the diagonals bisect, VE=ET and VE=(1/2)VT. We can use what we know to find the length of VE and then we’ll know the length of ET as well.
VE=21VT
7n−3=21(4n+34)
7n−3=2n+17
5n=20
n=4
Now we can substitute back in to find the length of VE, which is equal to the length of ET.
VE=ET=7n−3
VE=ET=7(4)−3
VE=ET=25
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9566 | https://math.libretexts.org/Courses/Lumen_Learning/Beginning_Algebra_(Lumen)/01%3A_Solving_Equations_and_Inequalities/1.03%3A_Multi-Step_Linear_Equations | Skip to main content
1.3: Multi-Step Linear Equations
Last updated
: Sep 27, 2020
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1.2: Solving Linear Equations
1.4: Problem Solving
Page ID
: 51441
( \newcommand{\kernel}{\mathrm{null}\,})
Learning Objectives
Solve multi-step equations
Use properties of equality to isolate variables and solve algebraic equations
Solve equations containing absolute values
Use the distributive property
Use the properties of equality and the distributive property to solve equations containing parentheses
Clear fractions and decimals from equations to make them easier to solve
Classify solutions to linear equations
Solve equations that have one solution, no solution, or an infinite number of solutions
Recognize when a linear equation that contains absolute value does not have a solution
Use properties of equality to isolate variables and solve algebraic equations
Steps With an End In Sight
There are some equations that you can solve in your head quickly. For example—what is the value of y in the equation y=3y=3. You only needed to do one thing to get the answer: divide 6 by 2.
Other equations are more complicated. Solving 4(13t+12)=64(13t+12)=6 without writing anything down is difficult! That’s because this equation contains not just a variable but also fractions and terms inside parentheses. This is a multi-step equation, one that takes several steps to solve. Although multi-step equations take more time and more operations, they can still be simplified and solved by applying basic algebraic rules.
Remember that you can think of an equation as a balance scale, with the goal being to rewrite the equation so that it is easier to solve but still balanced. The addition property of equality and the multiplication property of equality explain how you can keep the scale, or the equation, balanced. Whenever you perform an operation to one side of the equation, if you perform the same exact operation to the other side, you’ll keep both sides of the equation equal.
If the equation is in the form ax+b=cax+b=c, where x is the variable, you can solve the equation as before. First “undo” the addition and subtraction, and then “undo” the multiplication and division.
Example
Solve 3y+2=113y+2=11.
[reveal-answer q=”843520″]Show Solution[/reveal-answer]
[hidden-answer a=”843520″]
Subtract 2 from both sides of the equation to get the term with the variable by itself.
3y+2=11−2−2_3y=93y+2=11−2−2–––––––––––––3y=9
Divide both sides of the equation by 3 to get a coefficient of 1 for the variable.
3y_=9_39y=33y–––=9–39y=3
Answer
y=3y=3[/hidden-answer]
In the following video we show examples of solving two step linear equations.
A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=48
Example
Solve 3x+5x+4−x+7=883x+5x+4−x+7=88.
[reveal-answer q=”455516″]Show Solution[/reveal-answer]
[hidden-answer a=”455516″]
There are three like terms 5x5x, and –x–x involving a variable. Combine these like terms. 4 and 7 are also like terms and can be added.
3x+5x+4−x+7=887x+4+7=883x+5x+4−x+7=887x+4+7=88
The equation is now in the form ax+b=cax+b=c, so we can solve as before.
7x+11=887x+11=88
Subtract 11 from both sides.
7x+11=88−11−11_7x=777x+11=88−11−11––––––––––––7x=77
Divide both sides by 7.
7x_=77_77x=117x–––=77–––77x=11
Answer
x=11x=11[/hidden-answer]
In the following video, we show an example of solving a linear equation that requires combining like terms.
A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=48
Some equations may have the variable on both sides of the equal sign, as in this equation: 4x−6=2x+104x−6=2x+10.
To solve this equation, we need to “move” one of the variable terms. This can make it difficult to decide which side to work with. It doesn’t matter which term gets moved, 2x2x, however, to avoid negative coefficients, you can move the smaller term.
Examples
Solve: 4x−6=2x+104x−6=2x+10
[reveal-answer q=”457216″]Show Solution[/reveal-answer]
[hidden-answer a=”457216″]
Choose the variable term to move—to avoid negative terms choose 2x2x
4x−6=2x+10−2x−2x_4x−6=104x−6=2x+10−2x−2x–––––––––––––4x−6=10
Now add 6 to both sides to isolate the term with the variable.
4x−6=10+6+6_4x=164x−6=10+6+6––––––––––4x=16
Now divide each side by 4 to isolate the variable x.
4x4=164x=44x4=164x=4
[/hidden-answer]
In this video, we show an example of solving equations that have variables on both sides of hte equal sign.
A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=48
Solving Multi-Step Equations With Absolute Value
We can apply the same techniques we used for solving a one-step equation which contains absolute value to an equation that will take more than one step to solve. Let’s start with an example where the first step is to write two equations, one equal to positive 26 and one equal to negative 26.
Example
Solve for p. |2p–4|=26|2p–4|=26
[reveal-answer q=”371950″]Show Solution[/reveal-answer]
[hidden-answer a=”371950″]
Write the two equations that will give an absolute value of 26.
2p−4=26or2p−4=−262p−4=26or2p−4=−26
Solve each equation for p by isolating the variable.
2p−4=262p−4=−26+4+4_+4+4_2p_=30_2p_=−22_2=22=2p=15orp=−112p−4=262p−4=−26+4+4–––––––––––+4+4––––––––––––2p–––=30–––2p–––=−22––––2=22=2p=15orp=−11
Check the solutions in the original equation.
|2p−4|=26|2p−4|=26|2(15)−4|=26|2(−11)−4|=26|30−4|=26|−22−4|=26|26|=26|−26|=26|2p−4|=26|2p−4|=26|2(15)−4|=26|2(−11)−4|=26|30−4|=26|−22−4|=26|26|=26|−26|=26
Both solutions check!
Answer
p=−11p=−11[/hidden-answer]
In the next video, we show more examples of solving a simple absolute value equation.
A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=48
Now let’s look at an example where you need to do an algebraic step or two before you can write your two equations. The goal here is to get the absolute value on one side of the equation by itself. Then we can proceed as we did in the previous example.
Example
Solve for w. 3|4w–1|–5=103|4w–1|–5=10
[reveal-answer q=”303228″]Show Solution[/reveal-answer]
[hidden-answer a=”303228″]
Isolate the term with the absolute value by adding 5 to both sides.
3|4w−1|−5=10+5+5_3|4w−1|=153|4w−1|−5=10+5+5–––––––––––––3|4w−1|=15
Divide both sides by 3. Now the absolute value is isolated.
3|4w−1|_=15_33|4w−1|=53|4w−1|––––––––––=15–––33|4w−1|=5
Write the two equations that will give an absolute value of 5 and solve them.
4w−1=5or4w−1=−5+1+1_+1+1_4w_=6_4w_=−4_4444w=32w=−1w=32or−14w−1=5or4w−1=−5+1+1––––––––––+1+1––––––––––––4w–––=6–4w–––=−4–––4444w=32w=−1w=32or−1
Check the solutions in the original equation.
3|4w−1|−5=103|4w−1|−5=103|4(32)−1|−5=103|4w−1|−5=103|122−1|−5=103|4(−1)−1|−5=103|6−1|−5=103|−4−1|−5=103(5)−5=103|−5|−5=1015−5=1015−5=1010=1010=103|4w−1|−5=103|4w−1|−5=103∣∣4(32)−1∣∣−5=103|4w−1|−5=103∣∣122−1∣∣−5=103|4(−1)−1|−5=103|6−1|−5=103|−4−1|−5=103(5)−5=103|−5|−5=1015−5=1015−5=1010=1010=10
Both solutions check
Answer
w=−1orw=32w=−1orw=32[/hidden-answer]
In the two videos that follow, we show examples of how to solve an absolute value equation that requires you to isolate the absolute value first using mathematical operations.
A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=48
A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=48
The Distributive Property
As we solve linear equations, we often need to do some work to write the linear equations in a form we are familiar with solving. This section will focus on manipulating an equation we are asked to solve in such a way that we can use the skills we learned for solving multi-step equations to ultimately arrive at the solution.
Parentheses can make solving a problem difficult, if not impossible. To get rid of these unwanted parentheses we have the distributive property. Using this property we multiply the number in front of the parentheses by each term inside of the parentheses.
The Distributive Property of Multiplication
For all real numbers a, b, and c, a(b+c)=ab+ac.
What this means is that when a number multiplies an expression inside parentheses, you can distribute the multiplication to each term of the expression individually. Then, you can follow the steps we have already practiced to isolate the variable and solve the equation.
Example
Solve for 4(2a+3)=28
[reveal-answer q=”372387″]Show Solution[/reveal-answer]
[hidden-answer a=”372387″]
Apply the distributive property to expand 8a+12
4(2a+3)=288a+12=28
Subtract 12 from both sides to isolate the variable term.
8a+12=28−12−12_8a=16
Divide both terms by 8 to get a coefficient of 1.
8a_=16_88a=2
Answer
a=2[/hidden-answer]
In the video that follows, we show another example of how to use the distributive property to solve a multi-step linear equation.
A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=48
In the next example, you will see that there are parentheses on both sides of the equal sign, so you will need to use the distributive property twice. Notice that you are going to need to distribute a negative number, so be careful with negative signs!
Example
Solve for 2(4t−5)=−3(2t+1)
[reveal-answer q=”302387″]Show Solution[/reveal-answer]
[hidden-answer a=”302387″]
Apply the distributive property to expand 8t−10 and −6t−3. Be careful in this step—you are distributing a negative number, so keep track of the sign of each number after you multiply.
2(4t−5)=−3(2t+1)8t−10=−6t−3
Add −6t to both sides to begin combining like terms.
8t−10=−6t−3+6t+6t_14t−10=−3
Add 10 to both sides of the equation to isolate t.
14t−10=−3+10+10_14t=7
The last step is to divide both sides by 14 to completely isolate t.
14t=714t14=714
Answer
t=12
We simplified the fraction 12[/hidden-answer]
In the following video, we solve another multi-step equation with two sets of parentheses.
A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=48
Sometimes, you will encounter a multi-step equation with fractions. If you prefer not working with fractions, you can use the multiplication property of equality to multiply both sides of the equation by a common denominator of all of the fractions in the equation. This will clear all the fractions out of the equation. See the example below.
Example
Solve 12x−3=2−34x by clearing the fractions in the equation first.
[reveal-answer q=”129951″]Show Solution[/reveal-answer]
[hidden-answer a=”129951″]
Multiply both sides of the equation by 4, the common denominator of the fractional coefficients.
12x−3=2−34x4(12x−3)=4(2−34x)
Use the distributive property to expand the expressions on both sides. Multiply.
4(12x)−4(3)=4(2)−4(−34x)42x−12=8−124x2x−12=8−3x
Add 3x to both sides to move the variable terms to only one side. Add 12 to both sides to move the variable terms to only one side.
2x−12=8−3x+3x+3x_5x−12=8
Add 12 to both sides to move the constant terms to the other side.
5x−12=8+12+12_5x=20
Divide to isolate the variable.
5x_=5_55x=4
Answer
x=4[/hidden-answer]
Of course, if you like to work with fractions, you can just apply your knowledge of operations with fractions and solve.
In the following video, we show how to solve a multi-step equation with fractions.
A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=48
Regardless of which method you use to solve equations containing variables, you will get the same answer. You can choose the method you find the easiest! Remember to check your answer by substituting your solution into the original equation.
Sometimes, you will encounter a multi-step equation with decimals. If you prefer not working with decimals, you can use the multiplication property of equality to multiply both sides of the equation by a a factor of 10 that will help clear the decimals. See the example below.
Example
Solve 3y+10.5=6.5+2.5y by clearing the decimals in the equation first.
[reveal-answer q=”159951″]Show Solution[/reveal-answer]
[hidden-answer a=”159951″]
Since the smallest decimal place represented in the equation is 0.10, we want to multiply by 10 to make 1.0 and clear the decimals from the equation.
3y+10.5=6.5+2.5y10(3y+10.5)=10(6.5+2.5y)
Use the distributive property to expand the expressions on both sides.
10(3y)+10(10.5)=10(6.5)+10(2.5y)
Multiply.
30y+105=65+25y
Move the smaller variable term, 25y, by subtracting it from both sides.
30y+105=65+25y−25y−25y_5y+105=65
Subtract 105 from both sides to isolate the term with the variable.
5y+105=65−105−105_5y=−40
Divide both sides by 5 to isolate the y.
5y_=−40_55x=−8
Answer
x=−8[/hidden-answer]
In the following video, we show another example of clearing decimals first to solve a multi-step linear equation.
A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=48
Here are some steps to follow when you solve multi-step equations.
Solving Multi-Step Equations
(Optional) Multiply to clear any fractions or decimals.
Simplify each side by clearing parentheses and combining like terms.
Add or subtract to isolate the variable term—you may have to move a term with the variable.
Multiply or divide to isolate the variable.
Check the solution.
Classify Solutions to Linear Equations
There are three cases that can come up as we are solving linear equations. We have already seen one, where an equation has one solution. Sometimes we come across equations that don’t have any solutions, and even some that have an infinite number of solutions. The case where an equation has no solution is illustrated in the next examples.
Equations with no solutions
Example
Solve for x. 12+2x–8=7x+5–5x
[reveal-answer q=”790409″]Show Solution[/reveal-answer]
[hidden-answer a=”790409″]
Combine like terms on both sides of the equation.
12+2x−8=7x+5−5x2x+4=2x+5
Isolate the x term by subtracting 2x from both sides.
2x+4=2x+5−2x−2x_4=5
This false statement implies there are no solutions to this equation. Sometimes, we say the solution does not exist, or DNE for short.
[/hidden-answer]
This is not a solution! You did not find a value for x. Solving for x the way you know how, you arrive at the false statement 4=5. Surely 4 cannot be equal to 5!
This may make sense when you consider the second line in the solution where like terms were combined. If you multiply a number by 2 and add 4 you would never get the same answer as when you multiply that same number by 2 and add 5. Since there is no value of x that will ever make this a true statement, the solution to the equation above is “no solution.”
Be careful that you do not confuse the solution x=0 means that the value 0 satisfies the equation, so there is a solution. “No solution” means that there is no value, not even 0, which would satisfy the equation.
Also, be careful not to make the mistake of thinking that the equation 12+2x–8=7x+5–5x true.
Think About It
Try solving these equations. How many steps do you need to take before you can tell whether the equation has no solution or one solution?
a) Solve 8y=3(y+4)+y
Use the textbox below to record how many steps you think it will take before you can tell whether there is no solution or one solution.
[practice-area rows=”1″][/practice-area]
[reveal-answer q=”933839″]Show Solution[/reveal-answer]
[hidden-answer a=”933839″]
Solve 8y=3(y+4)+y
First, distribute the 3 into the parentheses on the right-hand side.
8y=3(y+4)+y=8y=3y+12+y
Next, begin combining like terms.
8y=3y+12+y=8y=4y+12
Now move the variable terms to one side. Moving the 4y will help avoid a negative sign.
8y=4y+12−4y−4y_4y=12
Now, divide each side by 4y.
4y4=124y=3
Because we were able to isolate y on one side and a number on the other side, we have one solution to this equation.
[/hidden-answer]
b) Solve 2(3x−5)−4x=2x+7
Use the textbox below to record how many steps you think it will take before you can tell whether there is no solution or one solution.
[practice-area rows=”1″][/practice-area]
[reveal-answer q=”937839″]Show Solution[/reveal-answer]
[hidden-answer a=”937839″]
Solve 2(3x−5)−4x=2x+7.
First, distribute the 2 into the parentheses on the left-hand side.
2(3x−5)−4x=2x+76x−10−4x=2x+7
Now begin simplifying. You can combine the x terms on the left-hand side.
6x−10−4x=2x+72x−10=2x+7
Now, take a moment to ponder this equation. It says that 2x+7. Can some number times two minus 10 be equal to that same number times two plus seven?
Let’s pretend x=3.
Is it true that 2(3)+7=13. NO! We don’t even really need to continue solving the equation, but we can just to be thorough.
Add 10 to both sides.
2x−10=2x+7+10+10_2x=2x+17
Now move 2x from the right hand side to combine like terms.
2x=2x+17−2x−2x_0=17
We know that 0 and 17 are not equal, so there is no number that x could be to make this equation true.
This false statement implies there are no solutions to this equation, or DNE (does not exist) for short.
[/hidden-answer]
Algebraic Equations with an Infinite Number of Solutions
You have seen that if an equation has no solution, you end up with a false statement instead of a value for x. It is possible to have an equation where any value for x will provide a solution to the equation. In the example below, notice how combining the terms −4x on the left leaves us with an equation with exactly the same terms on both sides of the equal sign.
Example
Solve for x. 5x+3–4x=3+x
[reveal-answer q=”773733″]Show Solution[/reveal-answer]
[hidden-answer a=”773733″]Combine like terms on both sides of the equation.
5x+3−4x=3+xx+3=3+x
Isolate the x term by subtracting x from both sides.
x+3=3+x−x−x_3=3
This true statement implies there are an infinite number of solutions to this equation, or we can also write the solution as “All Real Numbers”
[/hidden-answer]
You arrive at the true statement “x=0 into the original equation—you will get a true statement! Try x=−34, and it also will check!
This equation happens to have an infinite number of solutions. Any value for x that you can think of will make this equation true. When you think about the context of the problem, this makes sense—the equation x+3=3+x means “some number plus 3 is equal to 3 plus that same number.” We know that this is always true—it’s the commutative property of addition!
In the following video, we show more examples of attempting to solve a linear equation with either no solution or many solutions.
A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=48
Example
Solve for x. 3(2x−5)=6x−15
[reveal-answer q=”973733″]Show Solution[/reveal-answer]
[hidden-answer a=”973733″]
Distribute the 3 through the parentheses on the left-hand side.
3(2x−5)=6x−156x−15=6x−15
Wait! This looks just like the previous example. You have the same expression on both sides of an equal sign. No matter what number you choose for x, you will have a true statement. We can finish the algebra:
6x−15=6x−15−6x−6x_−15=−15
This true statement implies there are an infinite number of solutions to this equation.
[/hidden-answer]
In this video, we show more examples of solving linear equations with either no solutions or many solutions.
A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=48
In the following video, we show more examples of solving linear equations with parentheses that have either no solution or many solutions.
A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=48
Absolute value equations with no solutions
As we are solving absolute value equations it is important to be aware of special cases. An absolute value is defined as the distance from 0 on a number line, so it must be a positive number. When an absolute value expression is equal to a negative number, we say the equation has no solution, or DNE. Notice how this happens in the next two examples.
Example
Solve for x. 7+|2x−5|=4
[reveal-answer q=”173733″]Show Solution[/reveal-answer]
[hidden-answer a=”173733″]Notice absolute value is not alone. Subtract 7 from each side to isolate the absolute value.
7+|2x−5|=4−7−7_|2x−5|=−3
Result of absolute value is negative! The result of an absolute value must always be positive, so we say there is no solution to this equation, or DNE.
[/hidden-answer]
Example
Solve for x. −12|x+3|=6
[reveal-answer q=”173738″]Show Solution[/reveal-answer]
[hidden-answer a=”173738″]Notice absolute value is not alone, multiply both sides by the reciprocal of −2.
−12|x+3|=6(−2)−12|x+3|=(−2)6|x+3|=−12
Again, we have a result where an absolute value is negative!
There is no solution to this equation, or DNE.
[/hidden-answer]
In this last video, we show more examples of absolute value equations that have no solutions.
A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=48
Summary
Equations are mathematical statements that combine two expressions of equal value. An algebraic equation can be solved by isolating the variable on one side of the equation using the properties of equality. To check the solution of an algebraic equation, substitute the value of the variable into the original equation.
Complex, multi-step equations often require multi-step solutions. Before you can begin to isolate a variable, you may need to simplify the equation first. This may mean using the distributive property to remove parentheses or multiplying both sides of an equation by a common denominator to get rid of fractions. Sometimes it requires both techniques. If your multi-step equation has an absolute value, you will need to solve two equations, sometimes isolating the absolute value expression first.
We have also seen that solutions to equations can fall into three categories:
One solution
No solution, DNE (does not exist)
Many solutions, also called infinitely many solutions or All Real Numbers
And sometimes, we don’t need to do much algebra to see what the outcome will be.
CC licensed content, Original
Revision and Adaptation. Provided by: Lumen Learning. License: CC BY: Attribution
Graphic: Equation, term, expression. Provided by: Lumen Learning. License: CC BY: Attribution
Solving One Step Equations Using Multiplication and Division (Basic). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution
Solving Absolute Value Equation Using Multiplication and Division. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution
Solving Two Step Equations (Basic). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution
Solving an Equation that Requires Combining Like Terms. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution
Solve an Equation with Variable on Both Sides. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution
Solving an Equation with One Set of Parentheses. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution
Solving an Equation with Parentheses on Both Sides. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution
Solving an Equation with Fractions (Clear Fractions). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution
Solving an Equation with Decimals (Clear Decimals). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution
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9567 | https://en.wikipedia.org/wiki/Logarithmic_mean_temperature_difference | Jump to content
Logarithmic mean temperature difference
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Method of calculating heat transfer in flow systems
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| | This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. Find sources: "Logarithmic mean temperature difference" – news · newspapers · books · scholar · JSTOR (May 2020) (Learn how and when to remove this message) |
In thermal engineering, the logarithmic mean temperature difference (LMTD) is used to determine the temperature driving force for heat transfer in flow systems, most notably in heat exchangers. The LMTD is a logarithmic average of the temperature difference between the hot and cold feeds at each end of the double pipe exchanger. For a given heat exchanger with constant area and heat transfer coefficient, the larger the LMTD, the more heat is transferred. The use of the LMTD arises straightforwardly from the analysis of a heat exchanger with constant flow rate and fluid thermal properties.
Definition
[edit]
We assume that a generic heat exchanger has two ends (which we call "A" and "B") at which the hot and cold streams enter or exit on either side; then, the LMTD is defined by the logarithmic mean as follows:
where ΔTA is the temperature difference between the two streams at end A, and ΔTB is the temperature difference between the two streams at end B. When the two temperature differences are equal, this formula does not directly resolve, so the LMTD is conventionally taken to equal its limit value, which is in this case trivially equal to the two differences.
With this definition, the LMTD can be used to find the exchanged heat in a heat exchanger:
where (in SI units):
Q is the exchanged heat duty (watts),
U is the heat transfer coefficient (watts per kelvin per square meter),
A is the exchange area.
Note that estimating the heat transfer coefficient may be quite complicated.
This holds both for cocurrent flow, where the streams enter from the same end, and for countercurrent flow, where they enter from different ends.
In a cross-flow, in which one system, usually the heat sink, has the same nominal temperature at all points on the heat transfer surface, a similar relation between exchanged heat and LMTD holds, but with a correction factor. A correction factor is also required for other more complex geometries, such as a shell and tube exchanger with baffles.
Derivation
[edit]
Assume heat transfer is occurring in a heat exchanger along an axis z, from generic coordinate A to B, between two fluids, identified as 1 and 2, whose temperatures along z are T1(z) and T2(z).
The local exchanged heat flux at z is proportional to the temperature difference:
The heat that leaves the fluids causes a temperature gradient according to Fourier's law:
where ka, kb are the thermal conductivities of the intervening material at points A and B respectively. Summed together, this becomes
| | | |
---
| | | 1 |
where K = ka + kb.
The total exchanged energy is found by integrating the local heat transfer q from A to B:
Notice that B − A is clearly the pipe length, which is distance along z, and D is the circumference. Multiplying those gives Ar the heat exchanger area of the pipe, and use this fact:
In both integrals, make a change of variables from z to ΔT:
With the relation for ΔT (equation 1), this becomes
Integration at this point is trivial, and finally gives:
: ,
from which the definition of LMTD follows.
Assumptions and limitations
[edit]
It has been assumed that the rate of change for the temperature of both fluids is proportional to the temperature difference; this assumption is valid for fluids with a constant specific heat, which is a good description of fluids changing temperature over a relatively small range. However, if the specific heat changes, the LMTD approach will no longer be accurate.
A particular case for the LMTD are condensers and reboilers, where the latent heat associated to phase change is a special case of the hypothesis. For a condenser, the hot fluid inlet temperature is then equivalent to the hot fluid exit temperature.
It has also been assumed that the heat transfer coefficient (U) is constant, and not a function of temperature. If this is not the case, the LMTD approach will again be less valid
The LMTD is a steady-state concept, and cannot be used in dynamic analyses. In particular, if the LMTD were to be applied on a transient in which, for a brief time, the temperature difference had different signs on the two sides of the exchanger, the argument to the logarithm function would be negative, which is not allowable.
No phase change during heat transfer
Changes in kinetic energy and potential energy are neglected
Logarithmic Mean Pressure Difference
[edit]
A related quantity, the logarithmic mean pressure difference or LMPD, is often used in mass transfer for stagnant solvents with dilute solutes to simplify the bulk flow problem.
References
[edit]
^ "Basic Heat Transfer". www.swep.net. Retrieved 2020-05-12.
^ "MIT web course on Heat Exchangers". [MIT].
Kay J M & Nedderman R M (1985) Fluid Mechanics and Transfer Processes, Cambridge University Press
Retrieved from "
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9568 | https://math.stackexchange.com/questions/3706818/proving-that-a-map-is-a-quotient-map | general topology - Proving that a map is a quotient map - Mathematics Stack Exchange
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Proving that a map is a quotient map
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I am trying assignments in topology and I got stuck on this question:
Prove that the map f:R 2→R f:R 2→R defined by
f(x,y)=y 3+x y 2+x+y f(x,y)=y 3+x y 2+x+y
is a quotient map.
I have done a course on topology but quotient maps were not covered in it. So, I am unable to do it.
I am using this definition of quotient map: A map p p is called a quotient map if p:X→Y p:X→Y is such that
(a) p p is surjective,
(b) p p is continuous,
(c) U U belonging to Y Y, p−1(U)p−1(U) open in X X implies U U is open in Y Y.
For my function, the first 2 conditions are satisfied but I am not able to prove that it satisfies the 3rd condition.
Kindly tell.
general-topology
quotient-spaces
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edited Jun 5, 2020 at 12:18
Aryaman Maithani
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asked Jun 5, 2020 at 11:29
user775699 user775699
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I haven't worked this idea through, but for the condition (c) I think you can try to prove the contrapositive: if U U is a subset of Y Y which is not open, then p−1(U)p−1(U) is not open either. I feel like there should be a more clever way of proving this condition, but right now I can't find it Louis Hainaut –Louis Hainaut 2020-06-05 11:42:28 +00:00 Commented Jun 5, 2020 at 11:42
A continuous surjection that is open (or closed) is a quotient map. [This is a sufficient but not necessary condition, there are quotient maps that are neither open nor closed.] Try showing that f f is open in 2.(b).Daniel Fischer –Daniel Fischer 2020-06-05 11:50:15 +00:00 Commented Jun 5, 2020 at 11:50
@DanielFischer Can you please give a hint for proving it open?user775699 –user775699 2020-06-05 12:10:06 +00:00 Commented Jun 5, 2020 at 12:10
2 If a function g:U→R m g:U→R m, where U⊂R n U⊂R n is open, is differentiable, then g g is open at all points where the derivative D g D g is surjective. (Looking up the implicit function theorem and/or the inverse function theorem can help if you don't remember how that goes.) Here we have m=1 m=1, so the surjectivity of D f D f is equivalent to the non-vanishing of the gradient of f f.Daniel Fischer –Daniel Fischer 2020-06-05 12:23:04 +00:00 Commented Jun 5, 2020 at 12:23
@OP what is the context of this exercise? Are you supposed to work with the standard topology on R R?Zest –Zest 2020-06-06 04:52:33 +00:00 Commented Jun 6, 2020 at 4:52
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Use the final topology on R R.
Let T R×R=T 1 T R×R=T 1 denote the standard topology on R 2 R 2. Now consider the final topology on R R induced by f f, i.e.
T 2={V⊂R∣f−1(V)∈T 1}.T 2={V⊂R∣f−1(V)∈T 1}.
Now by construction, f f is obviously T 1−T 2 T 1−T 2 continous. Since you've already proven f f to be surjective it follows that f f is a quotient map.
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edited Jun 5, 2020 at 18:38
answered Jun 5, 2020 at 13:50
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Is it not true with the standard topology on R R? This seems to trivialize the problem.Keshav –Keshav 2020-06-05 18:45:01 +00:00 Commented Jun 5, 2020 at 18:45
I haven't checked whether the given map is a quotient map on the standard topology of R R. I do not see any indication that this has to be proven explicitly for the standard topology. Usually questions like these always explicitly mention the given topologies. However, since the final topology is the finest topology on R R for which f f is continous and quotient maps usually appear in the context of quotient spaces, i found this approach most natural. Note that (c) is exactly the condition the final topology satisfies.Zest –Zest 2020-06-06 01:40:17 +00:00 Commented Jun 6, 2020 at 1:40
Hm, I disagree, I think usually people are interested in R R with the usual topology, and ask questions on top of it. I agree that the context of the question would clear this up though. It seems to be true with the usual topologies, looking at the gradient of f f.Keshav –Keshav 2020-06-06 04:49:04 +00:00 Commented Jun 6, 2020 at 4:49
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9569 | https://pdg.lbl.gov/2014/reviews/rpp2014-rev-passage-particles-matter.pdf | Passage of particles through matter 1 32. PASSAGE OF PARTICLES THROUGH MATTER . . . . . . . . . . . . . . . . . . . . . .
2 32.1. Notation . . . . . . . . . . . . . . . . . . . .
2 32.2. Electronic energy loss by heavy particles . . . . . .
2 32.2.1. Moments and cross sections . . . . . . . . . .
2 32.2.2. Maximum energy transfer in a single collision . . . . . . . . . . . . . . . . . . . . .
4 32.2.3. Stopping power at intermediate ener-gies . . . . . . . . . . . . . . . . . . . . . .
5 32.2.4. Mean excitation energy . . . . . . . . . . . .
6 32.2.5. Density effect . . . . . . . . . . . . . . . .
7 32.2.6. Energy loss at low energies . . . . . . . . . .
9 32.2.7. Energetic knock-on electrons (δ rays) . . . . .
10 32.2.8. Restricted energy loss rates for rela-tivistic ionizing particles . . . . . . . . . . . . .
10 32.2.9. Fluctuations in energy loss . . . . . . . . . .
11 32.2.10. Energy loss in mixtures and com-pounds . . . . . . . . . . . . . . . . . . . . .
14 32.2.11. Ionization yields . . . . . . . . . . . . . .
14 32.3. Multiple scattering through small angles . . . . . .
15 32.4. Photon and electron interactions in mat-ter . . . . . . . . . . . . . . . . . . . . . . . .
17 32.4.1. Collision energy losses by e± . . . . . . . . .
17 32.4.2. Radiation length . . . . . . . . . . . . . .
18 32.4.3. Bremsstrahlung energy loss by e± . . . . . . .
19 32.4.4. Critical energy . . . . . . . . . . . . . . .
21 32.4.5. Energy loss by photons . . . . . . . . . . . .
21 32.4.6. Bremsstrahlung and pair production at very high energies . . . . . . . . . . . . . . .
25 32.4.7. Photonuclear and electronuclear in-teractions at still higher energies . . . . . . . . . .
26 32.5. Electromagnetic cascades . . . . . . . . . . . . .
27 32.6. Muon energy loss at high energy . . . . . . . . .
30 32.7. Cherenkov and transition radiation . . . . . . . .
33 32.7.1. Optical Cherenkov radiation . . . . . . . . .
33 32.7.2. Coherent radio Cherenkov radiation . . . . . .
34 32.7.3. Transition radiation . . . . . . . . . . . . .
35 K.A. Olive et al. (PDG), Chin. Phys. C38, 090001 (2014) ( August 21, 2014 13:18 2 32. Passage of particles through matter 32. PASSAGE OF PARTICLES THROUGH MATTER Revised September 2013 by H. Bichsel (University of Washington), D.E. Groom (LBNL), and S.R. Klein (LBNL).
This review covers the interactions of photons and electrically charged particles in matter, concentrating on energies of interest for high-energy physics and astrophysics and processes of interest for particle detectors (ionization, Cherenkov radiation, transition radiation). Much of the focus is on particles heavier than electrons (π±, p, etc.). Although the charge number z of the projectile is included in the equations, only z = 1 is discussed in detail. Muon radiative losses are discussed, as are photon/electron interactions at high to ultrahigh energies. Neutrons are not discussed. The notation and important numerical values are shown in Table 32.1.
32.1.
Notation 32.2.
Electronic energy loss by heavy particles [1–33] 32.2.1.
Moments and cross sections : The electronic interactions of fast charged particles with speed v = βc occur in single collisions with energy losses W , leading to ionization, atomic, or collective excitation.
Most frequently the energy losses are small (for 90% of all collisions the energy losses are less than 100 eV). In thin absorbers few collisions will take place and the total energy loss will show a large variance ; also see Sec. 32.2.9 below. For particles with charge ze more massive than electrons (“heavy” particles), scattering from free electrons is adequately described by the Rutherford differential cross section , dσR(W; β) dW = 2πr2 emec2z2 β2 (1 −β2W/Wmax) W 2 , (32.1) where Wmax is the maximum energy transfer possible in a single collision. But in matter electrons are not free. W must be finite and depends on atomic and bulk structure. For electrons bound in atoms Bethe used “Born Theorie” to obtain the differential cross section dσB(W; β) dW = dσR(W, β) dW B(W) .
(32.2) Electronic binding is accounted for by the correction factor B(W). Examples of B(W) and dσB/dW can be seen in Figs. 5 and 6 of Ref. 1.
Bethe’s theory extends only to some energy above which atomic effects are not important. The free-electron cross section (Eq. (32.1)) can be used to extend the cross section to Wmax. At high energies σB is further modified by polarization of the medium, and this “density effect,” discussed in Sec. 32.2.5, must also be included. Less important corrections are discussed below.
The mean number of collisions with energy loss between W and W + dW occurring in a distance δx is Neδx (dσ/dW)dW, where dσ(W; β)/dW contains all contributions. It is convenient to define the moments Mj(β) = Ne δx Z W j dσ(W; β) dW dW , (32.3) August 21, 2014 13:18 32. Passage of particles through matter 3 Table 32.1: Summary of variables used in this section. The kinematic variables β and γ have their usual relativistic meanings.
Symbol Definition Value or (usual) units α fine structure constant e2/4πǫ0ℏc 1/137.035 999 074(44) M incident particle mass MeV/c2 E incident part. energy γMc2 MeV T kinetic energy, (γ −1)Mc2 MeV W energy transfer to an electron MeV in a single collision k bremsstrahlung photon energy MeV mec2 electron mass × c2 0.510 998 928(11) MeV re classical electron radius e2/4πǫ0mec2 2.817 940 3267(27) fm NA Avogadro’s number 6.022 141 29(27) × 1023 mol−1 z charge number of incident particle Z atomic number of absorber A atomic mass of absorber g mol−1 K 4πNAr2 emec2 0.307 075 MeV mol−1 cm2 I mean excitation energy eV (Nota bene!) δ(βγ) density effect correction to ionization energy loss ℏωp plasma energy p ρ ⟨Z/A⟩× 28.816 eV p 4πNer3 e mec2/α | − →ρ in g cm−3 Ne electron density (units of re)−3 wj weight fraction of the jth element in a compound or mixture nj ∝number of jth kind of atoms in a compound or mixture X0 radiation length g cm−2 Ec critical energy for electrons MeV Eµc critical energy for muons GeV Es scale energy p 4π/α mec2 21.2052 MeV RM Moliere radius g cm−2 so that M0 is the mean number of collisions in δx, M1 is the mean energy loss in δx, (M2 −M1)2 is the variance, etc. The number of collisions is Poisson-distributed with mean M0. Ne is either measured in electrons/g (Ne = NAZ/A) or electrons/cm3 (Ne = NA ρZ/A). The former is used throughout this chapter, since quantities of interest (dE/dx, X0, etc.) vary smoothly with composition when there is no density dependence.
August 21, 2014 13:18 4 32. Passage of particles through matter Muon momentum 1 10 100 Stopping power [MeV cm2/g] Lindhard-Scharff Bethe Radiative Radiative effects reach 1% Without δ Radiative losses βγ 0.001 0.01 0.1 1 10 100 100 10 1 0.1 1000 104 105 [MeV/c] 100 10 1 [GeV/c] 100 10 1 [TeV/c] Minimum ionization Eµc Nuclear losses µ− µ+ on Cu Anderson-Ziegler Fig. 32.1: Stopping power (= ⟨−dE/dx⟩) for positive muons in copper as a function of βγ = p/Mc over nine orders of magnitude in momentum (12 orders of magnitude in kinetic energy). Solid curves indicate the total stopping power. Data below the break at βγ ≈0.1 are taken from ICRU 49 , and data at higher energies are from Ref. 5. Vertical bands indicate boundaries between different approximations discussed in the text. The short dotted lines labeled “µ−” illustrate the “Barkas effect,” the dependence of stopping power on projectile charge at very low energies .
dE/dx in the radiative region is not simply a function of β.
32.2.2.
Maximum energy transfer in a single collision : For a particle with mass M, Wmax = 2mec2 β2γ2 1 + 2γme/M + (me/M)2 .
(32.4) In older references [2,8] the “low-energy” approximation Wmax = 2mec2 β2γ2, valid for 2γme ≪M, is often implicit. For a pion in copper, the error thus introduced into dE/dx is greater than 6% at 100 GeV. For 2γme ≫M, Wmax = Mc2 β2γ.
At energies of order 100 GeV, the maximum 4-momentum transfer to the electron can exceed 1 GeV/c, where hadronic structure effects significantly modify the cross sections.
This problem has been investigated by J.D. Jackson , who concluded that for hadrons (but not for large nuclei) corrections to dE/dx are negligible below energies where radiative effects dominate. While the cross section for rare hard collisions is modified, the average stopping power, dominated by many softer collisions, is almost unchanged.
August 21, 2014 13:18 32. Passage of particles through matter 5 32.2.3.
Stopping power at intermediate energies : The mean rate of energy loss by moderately relativistic charged heavy particles, M1/δx, is well-described by the “Bethe equation,” ¿ −dE dx À = Kz2 Z A 1 β2 ·1 2 ln 2mec2β2γ2Wmax I2 −β2 −δ(βγ) 2 ¸ .
(32.5) It describes the mean rate of energy loss in the region 0.1 < ∼βγ < ∼1000 for intermediate-Z materials with an accuracy of a few %. With the symbol definitions and values given in Table 32.1, the units are MeV g−1cm2. Wmax is defined in Sec. 32.2.2. At the lower limit the projectile velocity becomes comparable to atomic electron “velocities” (Sec. 32.2.6), and at the upper limit radiative effects begin to be important (Sec. 32.6). Both limits are Z dependent. A minor dependence on M at the highest energies is introduced through Wmax, but for all practical purposes ⟨dE/dx⟩in a given material is a function of β alone.
Few concepts in high-energy physics are as misused as ⟨dE/dx⟩. The main problem is that the mean is weighted by very rare events with large single-collision energy deposits.
Even with samples of hundreds of events a dependable value for the mean energy loss cannot be obtained. Far better and more easily measured is the most probable energy loss, discussed in Sec. 32.2.9. The most probable energy loss in a detector is considerably below the mean given by the Bethe equation.
In a TPC (Sec. 33.6.5), the mean of 50%–70% of the samples with the smallest signals is often used as an estimator.
Although it must be used with cautions and caveats, ⟨dE/dx⟩as described in Eq. (32.5) still forms the basis of much of our understanding of energy loss by charged particles.
Extensive tables are available[4,5, pdg.lbl.gov/AtomicNuclearProperties/].
For heavy projectiles, like ions, additional terms are required to account for higher-order photon coupling to the target, and to account for the finite size of the target radius.
These can change dE/dx by a factor of two or more for the heaviest nuclei in certain kinematic regimes .
The function as computed for muons on copper is shown as the “Bethe” region of Fig. 32.1. Mean energy loss behavior below this region is discussed in Sec. 32.2.6, and the radiative effects at high energy are discussed in Sec. 32.6. Only in the Bethe region is it a function of β alone; the mass dependence is more complicated elsewhere. The stopping power in several other materials is shown in Fig. 32.2. Except in hydrogen, particles with the same velocity have similar rates of energy loss in different materials, although there is a slow decrease in the rate of energy loss with increasing Z. The qualitative behavior difference at high energies between a gas (He in the figure) and the other materials shown in the figure is due to the density-effect correction, δ(βγ), discussed in Sec. 32.2.5. The stopping power functions are characterized by broad minima whose position drops from βγ = 3.5 to 3.0 as Z goes from 7 to 100. The values of minimum ionization as a function of atomic number are shown in Fig. 32.3.
In practical cases, most relativistic particles (e.g., cosmic-ray muons) have mean energy loss rates close to the minimum; they are “minimum-ionizing particles,” or mip’s.
Eq. (32.5) may be integrated to find the total (or partial) “continuous slowing-down approximation” (CSDA) range R for a particle which loses energy only through ionization and atomic excitation. Since dE/dx depends only on β, R/M is a function of E/M or August 21, 2014 13:18 6 32. Passage of particles through matter 1 2 3 4 5 6 8 10 1.0 10 100 1000 10 000 0.1 Pion momentum (GeV/c) Proton momentum (GeV/c) 1.0 10 100 1000 0.1 1.0 10 100 1000 0.1 βγ = p/Mc Muon momentum (GeV/c) H2 liquid He gas C Al Fe Sn Pb 〈–dE/dx〉 (MeV g—1cm2) 1.0 10 100 1000 10 000 0.1 Figure 32.2: Mean energy loss rate in liquid (bubble chamber) hydrogen, gaseous helium, carbon, aluminum, iron, tin, and lead. Radiative effects, relevant for muons and pions, are not included. These become significant for muons in iron for βγ > ∼1000, and at lower momenta for muons in higher-Z absorbers. See Fig. 32.23.
pc/M. In practice, range is a useful concept only for low-energy hadrons (R < ∼λI, where λI is the nuclear interaction length), and for muons below a few hundred GeV (above which radiative effects dominate). R/M as a function of βγ = p/Mc is shown for a variety of materials in Fig. 32.4.
The mass scaling of dE/dx and range is valid for the electronic losses described by the Bethe equation, but not for radiative losses, relevant only for muons and pions.
32.2.4.
Mean excitation energy : “The determination of the mean excitation energy is the principal non-trivial task in the evaluation of the Bethe stopping-power formula” .
Recommended values have varied substantially with time. Estimates based on experimental stopping-power measurements for protons, deuterons, and alpha particles and on oscillator-strength distributions and dielectric-response functions were given in ICRU 49 .
See also ICRU 37 .
These values, shown in Fig. 32.5, have since been widely used. Machine-readable versions can also be found .
August 21, 2014 13:18 32. Passage of particles through matter 7 0.5 1.0 1.5 2.0 2.5 1 2 5 10 20 50 100 Z H He Li Be B C NO Ne Sn Fe Solids Gases H2 gas: 4.10 H2 liquid: 3.97 2.35 — 0.28 ln(Z) 〈–dE/dx〉 (MeV g—1cm2) Figure 32.3: Stopping power at minimum ionization for the chemical elements.
The straight line is fitted for Z > 6. A simple functional dependence on Z is not to be expected, since ⟨−dE/dx⟩also depends on other variables.
32.2.5.
Density effect : As the particle energy increases, its electric field flattens and extends, so that the distant-collision contribution to Eq. (32.5) increases as ln βγ.
However, real media become polarized, limiting the field extension and effectively truncating this part of the logarithmic rise [2–8,15–16]. At very high energies, δ/2 →ln(ℏωp/I) + ln βγ −1/2 , (32.6) where δ(βγ)/2 is the density effect correction introduced in Eq. (32.5) and ℏωp is the plasma energy defined in Table 32.1. A comparison with Eq. (32.5) shows that |dE/dx| then grows as ln βγ rather than ln β2γ2, and that the mean excitation energy I is replaced by the plasma energy ℏωp. The ionization stopping power as calculated with and without the density effect correction is shown in Fig. 32.1. Since the plasma frequency scales as the square root of the electron density, the correction is much larger for a liquid or solid than for a gas, as is illustrated by the examples in Fig. 32.2.
The density effect correction is usually computed using Sternheimer’s parameteriza-tion : δ(βγ) = 2(ln 10)x −C if x ≥x1; 2(ln 10)x −C + a(x1 −x)k if x0 ≤x < x1; 0 if x < x0 (nonconductors); δ0102(x−x0) if x < x0 (conductors) (32.7) Here x = log10 η = log10(p/Mc). C (the negative of the C used in Ref. 15) is obtained by equating the high-energy case of Eq. (32.7) with the limit given in Eq. (32.6). The other parameters are adjusted to give a best fit to the results of detailed calculations for momenta below Mc exp(x1). Parameters for elements and nearly 200 compounds and mixtures of interest are published in a variety of places, notably in Ref. 16. A recipe for August 21, 2014 13:18 8 32. Passage of particles through matter 0.05 0.1 0.02 0.5 0.2 1.0 5.0 2.0 10.0 Pion momentum (GeV/c) 0.1 0.5 0.2 1.0 5.0 2.0 10.0 50.0 20.0 Proton momentum (GeV/c) 0.05 0.02 0.1 0.5 0.2 1.0 5.0 2.0 10.0 Muon momentum (GeV/c) βγ = p/Mc 1 2 5 10 20 50 100 200 500 1000 2000 5000 10000 20000 50000 R/M (g cm−2 GeV−1) 0.1 2 5 1.0 2 5 10.0 2 5 100.0 H2 liquid He gas Pb Fe C Figure 32.4: Range of heavy charged particles in liquid (bubble chamber) hydrogen, helium gas, carbon, iron, and lead. For example: For a K+ whose momentum is 700 MeV/c, βγ = 1.42. For lead we read R/M ≈396, and so the range is 195 g cm−2 (17 cm).
finding the coefficients for nontabulated materials is given by Sternheimer and Peierls , and is summarized in Ref. 5.
The remaining relativistic rise comes from the β2γ growth of Wmax, which in turn is due to (rare) large energy transfers to a few electrons. When these events are excluded, the energy deposit in an absorbing layer approaches a constant value, the Fermi plateau (see Sec. 32.2.8 below).
At even higher energies (e.g., > 332 GeV for muons in iron, and at a considerably higher energy for protons in iron), radiative effects are more important than ionization losses. These are especially relevant for high-energy muons, as discussed in Sec. 32.6.
August 21, 2014 13:18 32. Passage of particles through matter 9 0 10 20 30 40 50 60 70 80 90 100 8 10 12 14 16 18 20 22 Iadj/Z (eV) Z Barkas & Berger 1964 Bichsel 1992 ICRU 37 (1984) (interpolated values are not marked with points) Figure 32.5: Mean excitation energies (divided by Z) as adopted by the ICRU .
Those based on experimental measurements are shown by symbols with error flags; the interpolated values are simply joined. The grey point is for liquid H2; the black point at 19.2 eV is for H2 gas. The open circles show more recent determinations by Bichsel .
The dash-dotted curve is from the approximate formula of Barkas used in early editions of this Review.
32.2.6.
Energy loss at low energies : Shell corrections C/Z must be included in the square brackets of of Eq. (32.5) [4,11,13,14] to correct for atomic binding having been neglected in calculating some of the contributions to Eq. (32.5). The Barkas form was used in generating Fig. 32.1. For copper it contributes about 1% at βγ = 0.3 (kinetic energy 6 MeV for a pion), and the correction decreases very rapidly with increasing energy.
Equation 32.2, and therefore Eq. (32.5), are based on a first-order Born approximation.
Higher-order corrections, again important only at lower energies, are normally included by adding the “Bloch correction” z2L2(β) inside the square brackets (Eq.(2.5) in ) .
An additional “Barkas correction” zL1(β) reduces the stopping power for a negative particle below that for a positive particle with the same mass and velocity. In a 1956 paper, Barkas et al. noted that negative pions had a longer range than positive pions .
The effect has been measured for a number of negative/positive particle pairs, including a detailed study with antiprotons .
A detailed discussion of low-energy corrections to the Bethe formula is given in ICRU 49 .
When the corrections are properly included, the Bethe treatment is accurate to about 1% down to β ≈0.05, or about 1 MeV for protons.
For 0.01 < β < 0.05, there is no satisfactory theory. For protons, one usually relies on the phenomenological fitting formulae developed by Andersen and Ziegler [4,19].
As tabulated in ICRU 49 , the nuclear plus electronic proton stopping power in copper is 113 MeV cm2 g−1 at T = 10 keV (βγ = 0.005), rises to a maximum of 210 MeV cm2 g−1 at T ≈120 keV (βγ = 0.016), then falls to 118 MeV cm2 g−1 at T = 1 MeV (βγ = 0.046).
August 21, 2014 13:18 10 32. Passage of particles through matter Above 0.5–1.0 MeV the corrected Bethe theory is adequate.
For particles moving more slowly than ≈0.01c (more or less the velocity of the outer atomic electrons), Lindhard has been quite successful in describing electronic stopping power, which is proportional to β .
Finally, we note that at even lower energies, e.g., for protons of less than several hundred eV, non-ionizing nuclear recoil energy loss dominates the total energy loss [4,20,21].
32.2.7.
Energetic knock-on electrons (δ rays) : The distribution of secondary electrons with kinetic energies T ≫I is d2N dTdx = 1 2 Kz2 Z A 1 β2 F(T) T 2 (32.8) for I ≪T ≤Wmax, where Wmax is given by Eq. (32.4). Here β is the velocity of the primary particle. The factor F is spin-dependent, but is about unity for T ≪Wmax.
For spin-0 particles F(T) = (1 −β2T/Wmax); forms for spins 1/2 and 1 are also given by Rossi 2. Additional formulae are given in Ref. 22.
Equation (32.8) is inaccurate for T close to I .
δ rays of even modest energy are rare. For a β ≈1 particle, for example, on average only one collision with Te > 10 keV will occur along a path length of 90 cm of Ar gas .
A δ ray with kinetic energy Te and corresponding momentum pe is produced at an angle θ given by cos θ = (Te/pe)(pmax/Wmax) , (32.9) where pmax is the momentum of an electron with the maximum possible energy transfer Wmax.
32.2.8.
Restricted energy loss rates for relativistic ionizing particles : Further insight can be obtained by examining the mean energy deposit by an ionizing particle when energy transfers are restricted to T ≤Wcut ≤Wmax. The restricted energy loss rate is −dE dx ¯ ¯ ¯ ¯ T<Wcut = Kz2 Z A 1 β2 ·1 2 ln 2mec2β2γ2Wcut I2 −β2 2 µ 1 + Wcut Wmax ¶ −δ 2 ¸ .
(32.10) This form approaches the normal Bethe function (Eq. (32.5)) as Wcut →Wmax. It can be verified that the difference between Eq. (32.5) and Eq. (32.10) is equal to R Wmax Wcut T(d2N/dTdx)dT, where d2N/dTdx is given by Eq. (32.8).
Since Wcut replaces Wmax in the argument of the logarithmic term of Eq. (32.5), the βγ term producing the relativistic rise in the close-collision part of dE/dx is replaced by a constant, and |dE/dx|T<Wcut approaches the constant “Fermi plateau.” (The density effect correction δ eliminates the explicit βγ dependence produced by the distant-collision contribution.) This behavior is illustrated in Fig. 32.6, where restricted loss rates for two examples of Wcut are shown in comparison with the full Bethe dE/dx and the Landau-Vavilov most probable energy loss (to be discussed in Sec. 32.2.9 below).
August 21, 2014 13:18 32. Passage of particles through matter 11 Landau/Vavilov/Bichsel ∆p/x for: Bethe Tcut = 10 dE/dx|min Tcut = 2 dE/dx|min Restricted energy loss for: 0.1 1.0 10.0 100.0 1000.0 1.0 1.5 0.5 2.0 2.5 3.0 MeV g−1 cm2 (Electonic loses only) Muon kinetic energy (GeV) Silicon x/ρ = 1600 µm 320 µm 80 µm Figure 32.6: Bethe dE/dx, two examples of restricted energy loss, and the Landau most probable energy per unit thickness in silicon. The change of ∆p/x with thickness x illustrates its a ln x + b dependence. Minimum ionization (dE/dx|min) is 1.664 MeV g−1 cm2. Radiative losses are excluded. The incident particles are muons.
“Restricted energy loss” is cut at the total mean energy, not the single-collision energy above Wcut It is of limited use. The most probable energy loss, discussed in the next Section, is far more useful in situations where single-particle energy loss is observed.
32.2.9.
Fluctuations in energy loss : For detectors of moderate thickness x (e.g.
scintillators or LAr cells), the energy loss probability distribution f(∆; βγ, x) is ade-quately described by the highly-skewed Landau (or Landau-Vavilov) distribution [24,25].
The most probable energy loss is † ∆p = ξ · ln 2mc2β2γ2 I + ln ξ I + j −β2 −δ(βγ) ¸ , (32.11) where ξ = (K/2) ⟨Z/A⟩(x/β2) MeV for a detector with a thickness x in g cm−2, and j = 0.200 . ‡ While dE/dx is independent of thickness, ∆p/x scales as a ln x + b. The density correction δ(βγ) was not included in Landau’s or Vavilov’s work, but it was later G < ∼0.05–0.1, where G is given by Rossi [Ref. 2, Eq. 2.7(10)]. It is Vavilov’s κ .
It is proportional to the absorber’s thickness, and as such parameterizes the constants describing the Landau distribution. These are fairly insensitive to thickness for G < ∼0.1, the case for most detectors.
† Practical calculations can be expedited by using the tables of δ and β from the text ver-sions of the muon energy loss tables to be found at pdg.lbl.gov/AtomicNuclearProperties.
‡ Rossi , Talman , and others give somewhat different values for j. The most probable loss is not sensitive to its value.
August 21, 2014 13:18 12 32. Passage of particles through matter included by Bichsel .
The high-energy behavior of δ(βγ) (Eq. (32.6)) is such that ∆p − → βγ> ∼100 ξ · ln 2mc2ξ (ℏωp)2 + j ¸ .
(32.12) Thus the Landau-Vavilov most probable energy loss, like the restricted energy loss, reaches a Fermi plateau. The Bethe dE/dx and Landau-Vavilov-Bichsel ∆p/x in silicon are shown as a function of muon energy in Fig. 32.6. The energy deposit in the 1600 µm case is roughly the same as in a 3 mm thick plastic scintillator.
f(Δ) [MeV−1] Electronic energy loss Δ [MeV] Energy loss [MeV cm2/g] 150 100 50 0 0.4 0.5 0.6 0.7 0.8 1.0 0.9 0.8 1.0 0.6 0.4 0.2 0.0 Mj(Δ)/Mj(∞) Landau-Vavilov Bichsel (Bethe-Fano theory) Δp Δ fwhm M0(Δ)/M0(∞) Μ1(Δ)/Μ1(∞) 10 GeV muon 1.7 mm Si 1.2 1.4 1.6 1.8 2.0 2.2 2.4 < > Figure 32.7: Electronic energy deposit distribution for a 10 GeV muon traversing 1.7 mm of silicon, the stopping power equivalent of about 0.3 cm of PVC scintillator [1,13,28].
The Landau-Vavilov function (dot-dashed) uses a Rutherford cross section without atomic binding corrections but with a kinetic energy transfer limit of Wmax. The solid curve was calculated using Bethe-Fano theory. M0(∆) and M1(∆) are the cumulative 0th moment (mean number of collisions) and 1st moment (mean energy loss) in crossing the silicon. (See Sec. 32.2.1. The fwhm of the Landau-Vavilov function is about 4ξ for detectors of moderate thickness. ∆p is the most probable energy loss, and ⟨∆⟩divided by the thickness is the Bethe ⟨dE/dx⟩.
The distribution function for the energy deposit by a 10 GeV muon going through a detector of about this thickness is shown in Fig. 32.7. In this case the most probable energy loss is 62% of the mean (M1(⟨∆⟩)/M1(∞)). Folding in experimental resolution displaces the peak of the distribution, usually toward a higher value. 90% of the collisions (M1(⟨∆⟩)/M1(∞)) contribute to energy deposits below the mean. It is the very rare high-energy-transfer collisions, extending to Wmax at several GeV, that drives the mean into the tail of the distribution. The large weight of these rare events makes the mean of an experimental distribution consisting of a few hundred events subject to large fluctuations and sensitive to cuts. The mean of the energy loss given by the Bethe August 21, 2014 13:18 32. Passage of particles through matter 13 100 200 300 400 500 600 0.0 0.2 0.4 0.6 0.8 1.0 0.50 1.00 1.50 2.00 2.50 640 µm (149 mg/cm2) 320 µm (74.7 mg/cm2) 160 µm (37.4 mg/cm2) 80 µm (18.7 mg/cm2) 500 MeV pion in silicon Mean energy loss rate w f (∆/x) ∆/x (eV/µm) ∆p/x ∆/x (MeV g−1 cm2) Figure 32.8: Straggling functions in silicon for 500 MeV pions, normalized to unity at the most probable value δp/x. The width w is the full width at half maximum.
equation, Eq. (32.5), is thus ill-defined experimentally and is not useful for describing energy loss by single particles.♮It rises as ln γ because Wmax increases as γ at high energies. The most probable energy loss should be used.
A practical example: For muons traversing 0.25 inches of PVT plastic scintillator, the ratio of the most probable E loss rate to the mean loss rate via the Bethe equation is [0.69, 0.57, 0.49, 0.42, 0.38] for Tµ = [0.01, 0.1, 1, 10, 100] GeV. Radiative losses add less than 0.5% to the total mean energy deposit at 10 GeV, but add 7% at 100 GeV. The most probable E loss rate rises slightly beyond the minimum ionization energy, then is essentially constant.
The Landau distribution fails to describe energy loss in thin absorbers such as gas TPC cells and Si detectors , as shown clearly in Fig. 1 of Ref. 1 for an argon-filled TPC cell. Also see Talman .
While ∆p/x may be calculated adequately with Eq. (32.11), the distributions are significantly wider than the Landau width w = 4ξ [Ref. 26, Fig. 15].
Examples for 500 MeV pions incident on thin silicon detectors are shown in Fig. 32.8.
For very thick absorbers the distribution is less skewed but never approaches a Gaussian.
The most probable energy loss, scaled to the mean loss at minimum ionization, is shown in Fig. 32.9 for several silicon detector thicknesses.
♮It does find application in dosimetry, where only bulk deposit is relevant.
August 21, 2014 13:18 14 32. Passage of particles through matter 1 3 0.3 30 300 10 100 1000 βγ (= p/m) 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 (∆p/x) / dE/dxmin 80 µm (18.7 mg/cm2) 160 µm (37.4 mg/cm2) x = 640 µm (149 mg/cm2) 320 µm (74.7 mg/cm2) Figure 32.9: Most probable energy loss in silicon, scaled to the mean loss of a minimum ionizing particle, 388 eV/µm (1.66 MeV g−1cm2).
32.2.10.
Energy loss in mixtures and compounds : A mixture or compound can be thought of as made up of thin layers of pure elements in the right proportion (Bragg additivity). In this case, ¿dE dx À = X wj ¿dE dx À j , (32.13) where dE/dx|j is the mean rate of energy loss (in MeV g cm−2) in the jth element.
Eq. (32.5) can be inserted into Eq. (32.13) to find expressions for ⟨Z/A⟩, ⟨I ⟩, and ⟨δ⟩; for example, ⟨Z/A⟩= P wjZj/Aj = P njZj/ P njAj. However, ⟨I ⟩as defined this way is an underestimate, because in a compound electrons are more tightly bound than in the free elements, and ⟨δ⟩as calculated this way has little relevance, because it is the electron density that matters. If possible, one uses the tables given in Refs. 16 and 29, that include effective excitation energies and interpolation coefficients for calculating the density effect correction for the chemical elements and nearly 200 mixtures and compounds. Otherwise, use the recipe for δ given in Ref. 5 and 17, and calculate ⟨I⟩following the discussion in Ref. 10. (Note the “13%” rule!) 32.2.11.
Ionization yields : Physicists frequently relate total energy loss to the number of ion pairs produced near the particle’s track. This relation becomes complicated for relativistic particles due to the wandering of energetic knock-on electrons whose ranges exceed the dimensions of the fiducial volume. For a qualitative appraisal of the nonlocality of energy deposition in various media by such modestly energetic knock-on electrons, see Ref. 30. The mean local energy dissipation per local ion pair produced, W, while essentially constant for relativistic particles, increases at slow particle speeds .
For gases, W can be surprisingly sensitive to trace amounts of various contaminants .
Furthermore, ionization yields in practical cases may be greatly influenced by such factors August 21, 2014 13:18 32. Passage of particles through matter 15 as subsequent recombination .
32.3.
Multiple scattering through small angles A charged particle traversing a medium is deflected by many small-angle scatters.
Most of this deflection is due to Coulomb scattering from nuclei as described by the Rutherford cross section. (However, for hadronic projectiles, the strong interactions also contribute to multiple scattering.) For many small-angle scatters the net scattering and displacement distributions are Gaussian via the central limit theorem. Less frequent “hard” scatters produce non-Gaussian tails. These Coulomb scattering distributions are well-represented by the theory of Moli ere .
Accessible discussions are given by Rossi and Jackson , and exhaustive reviews have been published by Scott and Motz et al. .
Experimental measurements have been published by Bichsel 37 and by Shen et al. 38. If we define θ0 = θ rms plane = 1 √ 2 θrms space , (32.14) then it is sufficient for many applications to use a Gaussian approximation for the central 98% of the projected angular distribution, with an rms width given by [39,40] θ0 = 13.6 MeV βcp z p x/X0 h 1 + 0.038 ln(x/X0) i .
(32.15) Here p, βc, and z are the momentum, velocity, and charge number of the incident particle, and x/X0 is the thickness of the scattering medium in radiation lengths (defined below).
This value of θ0 is from a fit to Moliere distribution for singly charged particles with β = 1 for all Z, and is accurate to 11% or better for 10−3 < x/X0 < 100.
Eq. (32.15) describes scattering from a single material, while the usual problem involves the multiple scattering of a particle traversing many different layers and mixtures. Since it is from a fit to a Moli ere distribution, it is incorrect to add the individual θ0 contributions in quadrature; the result is systematically too small. It is much more accurate to apply Eq. (32.15) once, after finding x and X0 for the combined scatterer.
The nonprojected (space) and projected (plane) angular distributions are given approximately by 1 2π θ2 0 exp −θ2 space 2θ2 0 dΩ, (32.16) 1 √ 2π θ0 exp − θ2 plane 2θ2 0 dθplane , (32.17) where θ is the deflection angle. In this approximation, θ2 space ≈(θ2 plane,x +θ2 plane,y), where the x and y axes are orthogonal to the direction of motion, and dΩ≈dθplane,x dθplane,y.
Deflections into θplane,x and θplane,y are independent and identically distributed.
Shen et al.’s measurements show that Bethe’s simpler methods of including atomic electron effects agrees better with experiment than does Scott’s treatment.
August 21, 2014 13:18 16 32. Passage of particles through matter x splane yplane Ψplane θplane x/2 Figure 32.10: Quantities used to describe multiple Coulomb scattering. The particle is incident in the plane of the figure.
Fig. 32.10 shows these and other quantities sometimes used to describe multiple Coulomb scattering. They are ψ rms plane = 1 √ 3 θ rms plane = 1 √ 3 θ0 , (32.18) y rms plane = 1 √ 3 x θ rms plane = 1 √ 3 x θ0 , (32.19) s rms plane = 1 4 √ 3 x θ rms plane = 1 4 √ 3 x θ0 .
(32.20) All the quantitative estimates in this section apply only in the limit of small θ rms plane and in the absence of large-angle scatters. The random variables s, ψ, y, and θ in a given plane are correlated. Obviously, y ≈xψ. In addition, y and θ have the correlation coefficient ρyθ = √ 3/2 ≈0.87. For Monte Carlo generation of a joint (y plane, θplane) distribution, or for other calculations, it may be most convenient to work with independent Gaussian random variables (z1, z2) with mean zero and variance one, and then set yplane =z1 x θ0(1 −ρ2 yθ)1/2/ √ 3 + z2 ρyθx θ0/ √ 3 (32.21) =z1 x θ0/ √ 12 + z2 x θ0/2 ; (32.22) θplane =z2 θ0 .
(32.23) Note that the second term for y plane equals x θplane/2 and represents the displacement that would have occurred had the deflection θplane all occurred at the single point x/2.
For heavy ions the multiple Coulomb scattering has been measured and compared with various theoretical distributions .
August 21, 2014 13:18 32. Passage of particles through matter 17 32.4.
Photon and electron interactions in matter At low energies electrons and positrons primarily lose energy by ionization, although other processes (Møller scattering, Bhabha scattering, e+ annihilation) contribute, as shown in Fig. 32.11. While ionization loss rates rise logarithmically with energy, bremsstrahlung losses rise nearly linearly (fractional loss is nearly independent of energy), and dominates above the critical energy (Sec. 32.4.4 below), a few tens of MeV in most materials 32.4.1.
Collision energy losses by e± : Stopping power differs somewhat for electrons and positrons, and both differ from stopping power for heavy particles because of the kinematics, spin, charge, and the identity of the incident electron with the electrons that it ionizes. Complete discussions and tables can be found in Refs. 10, 11, and 29.
For electrons, large energy transfers to atomic electrons (taken as free) are described by the Møller cross section. From Eq. (32.4), the maximum energy transfer in a single collision should be the entire kinetic energy, Wmax = mec2(γ −1), but because the particles are identical, the maximum is half this, Wmax/2. (The results are the same if the transferred energy is ǫ or if the transferred energy is Wmax −ǫ. The stopping power is by convention calculated for the faster of the two emerging electrons.) The first moment of the Møller cross section 22 is the stopping power: ¿ −dE dx À =1 2K Z A 1 β2 · ln mec2β2γ2{mec2(γ −1)/2} I2 +(1 −β2) −2γ −1 γ2 ln 2 + 1 8 µγ −1 γ ¶2 −δ # (32.24) The logarithmic term can be compared with the logarithmic term in the Bethe equation (Eq. (32.2)) by substituting Wmax = mec2(γ −1)/2. The two forms differ by ln 2.
Electron-positron scattering is described by the fairly complicated Bhabha cross section .
There is no identical particle problem, so Wmax = mec2(γ −1). The first moment of the Bhabha equation yields ¿ −dE dx À =1 2K Z A 1 β2 · ln mec2β2γ2{mec2(γ −1)} 2I2 +2 ln 2 −β2 12 µ 23 + 14 γ + 1 + 10 (γ + 1)2 + 4 (γ + 1)3 ¶ −δ ¸ .
(32.25) Following ICRU 37 , the density effect correction δ has been added to Uehling’s equations in both cases.
For heavy particles, shell corrections were developed assuming that the projectile is equivalent to a perturbing potential whose center moves with constant velocity. This assumption has no sound theoretical basis for electrons. The authors of ICRU 37 estimated the possible error in omitting it by assuming the correction was twice as great as for a proton of the same velocity. At T = 10 keV, the error was estimated to be ≈2% for water, ≈9% for Cu, and ≈21% for Au.
As shown in Fig. 32.11, stopping powers for e−, e+, and heavy particles are not dramatically different. In silicon, the minimum value for electrons is 1.50 MeV cm2/g (at August 21, 2014 13:18 18 32. Passage of particles through matter γ = 3.3); for positrons, 1.46 MeV cm2/g (at γ = 3.7), and for muons, 1.66 MeV cm2/g (at γ = 3.58).
32.4.2.
Radiation length : High-energy electrons predominantly lose energy in matter by bremsstrahlung, and high-energy photons by e+e−pair production. The characteristic amount of matter traversed for these related interactions is called the radiation length X0, usually measured in g cm−2. It is both (a) the mean distance over which a high-energy electron loses all but 1/e of its energy by bremsstrahlung, and (b) 7 9 of the mean free path for pair production by a high-energy photon .
It is also the appropriate scale length for describing high-energy electromagnetic cascades. X0 has been calculated and tabulated by Y.S. Tsai : 1 X0 = 4αr2 e NA A n Z2£ Lrad −f(Z) ¤ + Z L′ rad o .
(32.26) For A = 1 g mol−1, 4αr2 eNA/A = (716.408 g cm−2)−1. Lrad and L′ rad are given in Table 32.2. The function f(Z) is an infinite sum, but for elements up to uranium can be represented to 4-place accuracy by f(Z) =a2 · (1 + a2)−1 + 0.20206 −0.0369 a2 + 0.0083 a4 −0.002 a6 ¸ , (32.27) where a = αZ .
Table 32.2: Tsai’s Lrad and L′ rad, for use in calculating the radiation length in an element using Eq. (32.26).
Element Z Lrad L′ rad H 1 5.31 6.144 He 2 4.79 5.621 Li 3 4.74 5.805 Be 4 4.71 5.924 Others > 4 ln(184.15 Z−1/3) ln(1194 Z−2/3) The radiation length in a mixture or compound may be approximated by 1/X0 = X wj/Xj , (32.28) where wj and Xj are the fraction by weight and the radiation length for the jth element.
August 21, 2014 13:18 32. Passage of particles through matter 19 Figure 32.11: Fractional energy loss per radiation length in lead as a function of electron or positron energy. Electron (positron) scattering is considered as ionization when the energy loss per collision is below 0.255 MeV, and as Møller (Bhabha) scattering when it is above. Adapted from Fig. 3.2 from Messel and Crawford, Electron-Photon Shower Distribution Function Tables for Lead, Copper, and Air Absorbers, Pergamon Press, 1970. Messel and Crawford use X0(Pb) = 5.82 g/cm2, but we have modified the figures to reflect the value given in the Table of Atomic and Nuclear Properties of Materials (X0(Pb) = 6.37 g/cm2).
32.4.3.
Bremsstrahlung energy loss by e± : At very high energies and except at the high-energy tip of the bremsstrahlung spectrum, the cross section can be approximated in the “complete screening case” as dσ/dk = (1/k)4αr2 e © ( 4 3 −4 3y + y2)[Z2(Lrad −f(Z)) + Z L′ rad] + 1 9(1 −y)(Z2 + Z) ª , (32.29) where y = k/E is the fraction of the electron’s energy transferred to the radiated photon.
At small y (the “infrared limit”) the term on the second line ranges from 1.7% (low Z) to 2.5% (high Z) of the total. If it is ignored and the first line simplified with the definition of X0 given in Eq. (32.26), we have dσ dk = A X0NAk ¡ 4 3 −4 3y + y2¢ .
(32.30) This cross section (times k) is shown by the top curve in Fig. 32.12.
This formula is accurate except in near y = 1, where screening may become incomplete, and near y = 0, where the infrared divergence is removed by the interference of bremsstrahlung amplitudes from nearby scattering centers (the LPM effect) [45,46] and dielectric suppression [47,48].
These and other suppression effects in bulk media are discussed in Sec. 32.4.6.
August 21, 2014 13:18 20 32. Passage of particles through matter 0 0.4 0.8 1.2 0 0.25 0.5 0.75 1 y = k/E Bremsstrahlung (X0NA/A) ydσLPM/dy 10 GeV 1 TeV 10 TeV 100 TeV 1 PeV 10 PeV 100 GeV Figure 32.12: The normalized bremsstrahlung cross section k dσLP M/dk in lead versus the fractional photon energy y = k/E. The vertical axis has units of photons per radiation length.
With decreasing energy (E < ∼10 GeV) the high-y cross section drops and the curves become rounded as y →1. Curves of this familar shape can be seen in Rossi (Figs. 2.11.2,3); see also the review by Koch & Motz .
2 5 10 20 50 100 200 Copper X0 = 12.86 g cm−2 Ec = 19.63 MeV dE/dx × X0 (MeV) Electron energy (MeV) 10 20 30 50 70 100 200 40 Brems = ionization Ionization Rossi: Ionization per X0 = electron energy Total Brems ≈E Exact bremsstrahlung Figure 32.13: Two definitions of the critical energy Ec.
Except at these extremes, and still in the complete-screening approximation, the number of photons with energies between kmin and kmax emitted by an electron travelling a distance d ≪X0 is Nγ = d X0 " 4 3 ln µkmax kmin ¶ −4(kmax −kmin) 3E + k2 max −k2 min 2E2 # .
(32.31) August 21, 2014 13:18 32. Passage of particles through matter 21 Ec (MeV) Z 1 2 5 10 20 50 100 5 10 20 50 100 200 400 610 MeV _ Z + 1.24 710 MeV _ Z + 0.92 Solids Gases H He Li Be B C NO Ne Sn Fe Figure 32.14: Electron critical energy for the chemical elements, using Rossi’s definition .
The fits shown are for solids and liquids (solid line) and gases (dashed line). The rms deviation is 2.2% for the solids and 4.0% for the gases.
(Computed with code supplied by A. Fass´ o.) 32.4.4.
Critical energy : An electron loses energy by bremsstrahlung at a rate nearly proportional to its energy, while the ionization loss rate varies only logarithmically with the electron energy. The critical energy Ec is sometimes defined as the energy at which the two loss rates are equal .
Among alternate definitions is that of Rossi , who defines the critical energy as the energy at which the ionization loss per radiation length is equal to the electron energy. Equivalently, it is the same as the first definition with the approximation |dE/dx|brems ≈E/X0. This form has been found to describe transverse electromagnetic shower development more accurately (see below). These definitions are illustrated in the case of copper in Fig. 32.13.
The accuracy of approximate forms for Ec has been limited by the failure to distinguish between gases and solid or liquids, where there is a substantial difference in ionization at the relevant energy because of the density effect. We distinguish these two cases in Fig. 32.14. Fits were also made with functions of the form a/(Z + b)α, but α was found to be essentially unity. Since Ec also depends on A, I, and other factors, such forms are at best approximate.
Values of Ec for both electrons and positrons in more than 300 materials can be found at pdg.lbl.gov/AtomicNuclearProperties.
32.4.5.
Energy loss by photons : Contributions to the photon cross section in a light element (carbon) and a heavy element (lead) are shown in Fig. 32.15. At low energies it is seen that the photoelectric effect dominates, although Compton scattering, Rayleigh scattering, and photonuclear absorption also contribute. The photoelectric cross section is characterized by discontinuities (absorption edges) as thresholds for photoionization of various atomic levels are reached. Photon attenuation lengths for a variety of elements are shown in Fig. 32.16, and data for 30 eV< k <100 GeV for all elements are available from the web pages given in the caption. Here k is the photon energy.
August 21, 2014 13:18 22 32. Passage of particles through matter Photon Energy 1 Mb 1 kb 1 b 10 mb 10 eV 1 keV 1 MeV 1 GeV 100 GeV (b) Lead (Z = 82) - experimental σtot σp.e.
κe Cross section (barns/atom) Cross section (barns/atom) 10 mb 1 b 1 kb 1 Mb (a) Carbon (Z = 6) σRayleigh σg.d.r.
σCompton σCompton σRayleigh κnuc κnuc κe σp.e.
experimental σtot Figure 32.15: Photon total cross sections as a function of energy in carbon and lead, showing the contributions of different processes : σp.e. = Atomic photoelectric effect (electron ejection, photon absorption) σRayleigh = Rayleigh (coherent) scattering–atom neither ionized nor excited σCompton = Incoherent scattering (Compton scattering offan electron) κnuc = Pair production, nuclear field κe = Pair production, electron field σg.d.r. = Photonuclear interactions, most notably the Giant Dipole Resonance .
In these interactions, the target nucleus is broken up.
Original figures through the courtesy of John H. Hubbell (NIST).
August 21, 2014 13:18 32. Passage of particles through matter 23 Photon energy 100 10 10–4 10–5 10–6 1 0.1 0.01 0.001 10 eV 100 eV 1 keV 10 keV 100 keV 1 MeV 10 MeV 100 MeV 1 GeV 10 GeV 100 GeV Absorption length λ (g/cm2) Si C Fe Pb H Sn Figure 32.16: The photon mass attenuation length (or mean free path) λ = 1/(µ/ρ) for various elemental absorbers as a function of photon energy. The mass attenuation coefficient is µ/ρ, where ρ is the density. The intensity I remaining after traversal of thickness t (in mass/unit area) is given by I = I0 exp(−t/λ). The accuracy is a few percent. For a chemical compound or mixture, 1/λeff≈P elements wZ/λZ, where wZ is the proportion by weight of the element with atomic number Z. The processes responsible for attenuation are given in Fig. 32.11. Since coherent processes are included, not all these processes result in energy deposition. The data for 30 eV < E < 1 keV are obtained from constants (courtesy of Eric M. Gullikson, LBNL). The data for 1 keV < E < 100 GeV are from through the courtesy of John H.
Hubbell (NIST).
August 21, 2014 13:18 24 32. Passage of particles through matter Figure 32.17: Probability P that a photon interaction will result in conversion to an e+e−pair. Except for a few-percent contribution from photonuclear absorption around 10 or 20 MeV, essentially all other interactions in this energy range result in Compton scattering offan atomic electron. For a photon attenuation length λ (Fig. 32.16), the probability that a given photon will produce an electron pair (without first Compton scattering) in thickness t of absorber is P[1 −exp(−t/λ)].
0 0.25 0.5 0.75 1 0 0.25 0.50 0.75 1.00 x = E/k Pair production (X0NA/A) dσLPM/dx 1 TeV 10 TeV 100 TeV 1 PeV 10 PeV 1 EeV 100 PeV Figure 32.18: The normalized pair production cross section dσLP M/dy, versus fractional electron energy x = E/k.
August 21, 2014 13:18 32. Passage of particles through matter 25 The increasing domination of pair production as the energy increases is shown in Fig. 32.17. Using approximations similar to those used to obtain Eq. (32.30), Tsai’s formula for the differential cross section reduces to dσ dx = A X0NA £ 1 −4 3x(1 −x) ¤ (32.32) in the complete-screening limit valid at high energies. Here x = E/k is the fractional energy transfer to the pair-produced electron (or positron), and k is the incident photon energy. The cross section is very closely related to that for bremsstrahlung, since the Feynman diagrams are variants of one another. The cross section is of necessity symmetric between x and 1 −x, as can be seen by the solid curve in Fig. 32.18. See the review by Motz, Olsen, & Koch for a more detailed treatment .
Eq. (32.32) may be integrated to find the high-energy limit for the total e+e− pair-production cross section: σ = 7 9(A/X0NA) .
(32.33) Equation Eq. (32.33) is accurate to within a few percent down to energies as low as 1 GeV, particularly for high-Z materials.
32.4.6.
Bremsstrahlung and pair production at very high energies : At ultra-high energies, Eqns. 32.29–32.33 will fail because of quantum mechanical interference between amplitudes from different scattering centers. Since the longitudinal momentum transfer to a given center is small (∝k/E(E −k), in the case of bremsstrahlung), the interaction is spread over a comparatively long distance called the formation length (∝E(E −k)/k) via the uncertainty principle. In alternate language, the formation length is the distance over which the highly relativistic electron and the photon “split apart.” The interference is usually destructive. Calculations of the “Landau-Pomeranchuk-Migdal” (LPM) effect may be made semi-classically based on the average multiple scattering, or more rigorously using a quantum transport approach [45,46].
In amorphous media, bremsstrahlung is suppressed if the photon energy k is less than E2/(E + ELP M) , where ELP M = (mec2)2αX0 4πℏcρ = (7.7 TeV/cm) × X0 ρ .
(32.34) Since physical distances are involved, X0/ρ, in cm, appears. The energy-weighted bremsstrahlung spectrum for lead, k dσLP M/dk, is shown in Fig. 32.12. With appropriate scaling by X0/ρ, other materials behave similarly.
For photons, pair production is reduced for E(k −E) > k ELP M. The pair-production cross sections for different photon energies are shown in Fig. 32.18.
If k ≪E, several additional mechanisms can also produce suppression. When the formation length is long, even weak factors can perturb the interaction. For example, the emitted photon can coherently forward scatter offof the electrons in the media.
This definition differs from that of Ref. 54 by a factor of two. ELP M scales as the 4th power of the mass of the incident particle, so that ELP M = (1.4 × 1010 TeV/cm) × X0/ρ for a muon.
August 21, 2014 13:18 26 32. Passage of particles through matter Because of this, for k < ωpE/me ∼10−4, bremsstrahlung is suppressed by a factor (kme/ωpE)2 .
Magnetic fields can also suppress bremsstrahlung.
In crystalline media, the situation is more complicated, with coherent enhancement or suppression possible. The cross section depends on the electron and photon energies and the angles between the particle direction and the crystalline axes .
32.4.7.
Photonuclear and electronuclear interactions at still higher energies : At still higher photon and electron energies, where the bremsstrahlung and pair production cross-sections are heavily suppressed by the LPM effect, photonuclear and electronuclear interactions predominate over electromagnetic interactions.
At photon energies above about 1020 eV, for example, photons usually interact hadronically. The exact cross-over energy depends on the model used for the photonuclear interactions. These processes are illustrated in Fig. 32.19. At still higher energies (> ∼1023 eV), photonuclear interactions can become coherent, with the photon interaction spread over multiple nuclei. Essentially, the photon coherently converts to a ρ0, in a process that is somewhat similar to kaon regeneration .
k [eV] 10 log 10 12 14 16 18 20 22 24 26 (Interaction Length) [m] 10 log −1 0 1 2 3 4 5 BH σ Mig σ A γ σ A γ σ + Mig σ Figure 32.19: Interaction length for a photon in ice as a function of photon energy for the Bethe-Heitler (BH), LPM (Mig) and photonuclear (γA) cross sections .
The Bethe-Heitler interaction length is 9X0/7, and X0 is 0.393 m in ice.
Similar processes occur for electrons. As electron energies increase and the LPM effect suppresses bremsstrahlung, electronuclear interactions become more important.
At energies above 1021eV, these electronuclear interactions dominate electron energy loss .
August 21, 2014 13:18 32. Passage of particles through matter 27 32.5.
Electromagnetic cascades When a high-energy electron or photon is incident on a thick absorber, it initiates an electromagnetic cascade as pair production and bremsstrahlung generate more electrons and photons with lower energy. The longitudinal development is governed by the high-energy part of the cascade, and therefore scales as the radiation length in the material. Electron energies eventually fall below the critical energy, and then dissipate their energy by ionization and excitation rather than by the generation of more shower particles. In describing shower behavior, it is therefore convenient to introduce the scale variables t = x/X0 , y = E/Ec , (32.35) so that distance is measured in units of radiation length and energy in units of critical energy.
0.000 0.025 0.050 0.075 0.100 0.125 0 20 40 60 80 100 (1/E0)dE/dt t = depth in radiation lengths Number crossing plane 30 GeV electron incident on iron Energy Photons × 1/6.8 Electrons 0 5 10 15 20 Figure 32.20: An EGS4 simulation of a 30 GeV electron-induced cascade in iron.
The histogram shows fractional energy deposition per radiation length, and the curve is a gamma-function fit to the distribution. Circles indicate the number of electrons with total energy greater than 1.5 MeV crossing planes at X0/2 intervals (scale on right) and the squares the number of photons with E ≥1.5 MeV crossing the planes (scaled down to have same area as the electron distribution).
Longitudinal profiles from an EGS4 simulation of a 30 GeV electron-induced cascade in iron are shown in Fig. 32.20. The number of particles crossing a plane (very close to Rossi’s Π function ) is sensitive to the cutoffenergy, here chosen as a total energy of 1.5 MeV for both electrons and photons. The electron number falls offmore quickly than energy deposition. This is because, with increasing depth, a larger fraction of the cascade energy is carried by photons. Exactly what a calorimeter measures depends on the device, but it is not likely to be exactly any of the profiles shown. In gas counters it may be very close to the electron number, but in glass Cherenkov detectors and other devices with “thick” sensitive regions it is closer to the energy deposition (total track August 21, 2014 13:18 28 32. Passage of particles through matter length). In such detectors the signal is proportional to the “detectable” track length Td, which is in general less than the total track length T. Practical devices are sensitive to electrons with energy above some detection threshold Ed, and Td = T F(Ed/Ec). An analytic form for F(Ed/Ec) obtained by Rossi is given by Fabjan in Ref. 58; see also Amaldi .
The mean longitudinal profile of the energy deposition in an electromagnetic cascade is reasonably well described by a gamma distribution : dE dt = E0 b (bt)a−1e−bt Γ(a) (32.36) The maximum tmax occurs at (a −1)/b. We have made fits to shower profiles in elements ranging from carbon to uranium, at energies from 1 GeV to 100 GeV. The energy deposition profiles are well described by Eq. (32.36) with tmax = (a −1)/b = 1.0 × (ln y + Cj) , j = e, γ , (32.37) where Ce = −0.5 for electron-induced cascades and Cγ = +0.5 for photon-induced cascades. To use Eq. (32.36), one finds (a −1)/b from Eq. (32.37) and Eq. (32.35), then finds a either by assuming b ≈0.5 or by finding a more accurate value from Fig. 32.21.
The results are very similar for the electron number profiles, but there is some dependence on the atomic number of the medium. A similar form for the electron number maximum was obtained by Rossi in the context of his “Approximation B,” (see Fabjan’s review in Ref. 58), but with Ce = −1.0 and Cγ = −0.5; we regard this as superseded by the EGS4 result.
Carbon Aluminum Iron Uranium 0.3 0.4 0.5 0.6 0.7 0.8 10 100 1000 10 000 b y = E/Ec Figure 32.21: Fitted values of the scale factor b for energy deposition profiles obtained with EGS4 for a variety of elements for incident electrons with 1 ≤E0 ≤100 GeV. Values obtained for incident photons are essentially the same.
The “shower length” Xs = X0/b is less conveniently parameterized, since b depends upon both Z and incident energy, as shown in Fig. 32.21. As a corollary of this August 21, 2014 13:18 32. Passage of particles through matter 29 Z dependence, the number of electrons crossing a plane near shower maximum is underestimated using Rossi’s approximation for carbon and seriously overestimated for uranium. Essentially the same b values are obtained for incident electrons and photons.
For many purposes it is sufficient to take b ≈0.5.
The length of showers initiated by ultra-high energy photons and electrons is somewhat greater than at lower energies since the first or first few interaction lengths are increased via the mechanisms discussed above.
The gamma function distribution is very flat near the origin, while the EGS4 cascade (or a real cascade) increases more rapidly. As a result Eq. (32.36) fails badly for about the first two radiation lengths; it was necessary to exclude this region in making fits.
Because fluctuations are important, Eq. (32.36) should be used only in applications where average behavior is adequate. Grindhammer et al. have developed fast simulation algorithms in which the variance and correlation of a and b are obtained by fitting Eq. (32.36) to individually simulated cascades, then generating profiles for cascades using a and b chosen from the correlated distributions .
The transverse development of electromagnetic showers in different materials scales fairly accurately with the Moliere radius RM, given by [62,63] RM = X0 Es/Ec , (32.38) where Es ≈21 MeV (Table 32.1), and the Rossi definition of Ec is used.
In a material containing a weight fraction wj of the element with critical energy Ecj and radiation length Xj, the Moli ere radius is given by 1 RM = 1 Es X wj Ecj Xj .
(32.39) Measurements of the lateral distribution in electromagnetic cascades are shown in Refs. 62 and 63. On the average, only 10% of the energy lies outside the cylinder with radius RM. About 99% is contained inside of 3.5RM, but at this radius and beyond composition effects become important and the scaling with RM fails. The distributions are characterized by a narrow core, and broaden as the shower develops. They are often represented as the sum of two Gaussians, and Grindhammer describes them with the function f(r) = 2r R2 (r2 + R2)2 , (32.40) where R is a phenomenological function of x/X0 and ln E.
At high enough energies, the LPM effect (Sec. 32.4.6) reduces the cross sections for bremsstrahlung and pair production, and hence can cause significant elongation of electromagnetic cascades .
August 21, 2014 13:18 30 32. Passage of particles through matter 32.6.
Muon energy loss at high energy At sufficiently high energies, radiative processes become more important than ionization for all charged particles. For muons and pions in materials such as iron, this “critical energy” occurs at several hundred GeV. (There is no simple scaling with particle mass, but for protons the “critical energy” is much, much higher.) Radiative effects dominate the energy loss of energetic muons found in cosmic rays or produced at the newest accelerators. These processes are characterized by small cross sections, hard spectra, large energy fluctuations, and the associated generation of electromagnetic and (in the case of photonuclear interactions) hadronic showers [64–72].
As a consequence, at these energies the treatment of energy loss as a uniform and continuous process is for many purposes inadequate.
It is convenient to write the average rate of muon energy loss as −dE/dx = a(E) + b(E) E .
(32.41) Here a(E) is the ionization energy loss given by Eq. (32.5), and b(E) is the sum of e+e− pair production, bremsstrahlung, and photonuclear contributions. To the approximation that these slowly-varying functions are constant, the mean range x0 of a muon with initial energy E0 is given by x0 ≈(1/b) ln(1 + E0/Eµc) , (32.42) where Eµc = a/b. Fig. 32.22 shows contributions to b(E) for iron. Since a(E) ≈0.002 GeV g−1 cm2, b(E)E dominates the energy loss above several hundred GeV, where b(E) is nearly constant. The rates of energy loss for muons in hydrogen, uranium, and iron are shown in Fig. 32.23 .
Muon energy (GeV) 0 1 2 3 4 5 6 7 8 9 106 b(E) (g−1cm2) Iron btotal bpair bbremsstrahlung bnuclear 102 10 1 103 104 105 Figure 32.22: Contributions to the fractional energy loss by muons in iron due to e+e−pair production, bremsstrahlung, and photonuclear interactions, as obtained from Groom et al. except for post-Born corrections to the cross section for direct pair production from atomic electrons.
August 21, 2014 13:18 32. Passage of particles through matter 31 Figure 32.23: The average energy loss of a muon in hydrogen, iron, and uranium as a function of muon energy. Contributions to dE/dx in iron from ionization and the processes shown in Fig. 32.22 are also shown.
The “muon critical energy” Eµc can be defined more exactly as the energy at which radiative and ionization losses are equal, and can be found by solving Eµc = a(Eµc)/b(Eµc). This definition corresponds to the solid-line intersection in Fig. 32.13, and is different from the Rossi definition we used for electrons. It serves the same function: below Eµc ionization losses dominate, and above Eµc radiative effects dominate. The dependence of Eµc on atomic number Z is shown in Fig. 32.24.
The radiative cross sections are expressed as functions of the fractional energy loss ν. The bremsstrahlung cross section goes roughly as 1/ν over most of the range, while for the pair production case the distribution goes as ν−3 to ν−2 .
“Hard” losses are therefore more probable in bremsstrahlung, and in fact energy losses due to pair production may very nearly be treated as continuous. The simulated momentum distribution of an incident 1 TeV/c muon beam after it crosses 3 m of iron is shown in Fig. 32.25. The most probable loss is 8 GeV, or 3.4 MeV g−1cm2. The full width at half maximum is 9 GeV/c, or 0.9%. The radiative tail is almost entirely due to bremsstrahlung, although most of the events in which more than 10% of the incident energy lost experienced relatively hard photonuclear interactions. The latter can exceed detector resolution , necessitating the reconstruction of lost energy. Tables in Ref. 5 list the stopping power as 9.82 MeV g−1cm2 for a 1 TeV muon, so that the mean loss should be 23 GeV (≈23 GeV/c), for a final momentum of 977 GeV/c, far below the peak.
This agrees with the indicated mean calculated from the simulation. Electromagnetic and hadronic cascades in detector materials can obscure muon tracks in detector planes and reduce tracking efficiency .
August 21, 2014 13:18 32 32. Passage of particles through matter __ (Z + 2.03)0.879 __ (Z + 1.47)0.838 100 200 400 700 1000 2000 4000 Eµc (GeV) 1 2 5 10 20 50 100 Z 7980 GeV 5700 GeV H He Li Be B CNO Ne Sn Fe Solids Gases Figure 32.24: Muon critical energy for the chemical elements, defined as the energy at which radiative and ionization energy loss rates are equal .
The equality comes at a higher energy for gases than for solids or liquids with the same atomic number because of a smaller density effect reduction of the ionization losses.
The fits shown in the figure exclude hydrogen. Alkali metals fall 3–4% above the fitted function, while most other solids are within 2% of the function. Among the gases the worst fit is for radon (2.7% high).
950 960 970 980 990 1000 Final momentum p [GeV/c] 0.00 0.02 0.04 0.06 0.08 0.10 1 TeV muons on 3 m Fe Mean 977 GeV/c Median 987 GeV/c dN/dp [1/(GeV/c)] FWHM 9 GeV/c Figure 32.25: The momentum distribution of 1 TeV/c muons after traversing 3 m of iron as calculated with the MARS15 Monte Carlo code by S.I. Striganov .
August 21, 2014 13:18 32. Passage of particles through matter 33 32.7.
Cherenkov and transition radiation [33,77,78] A charged particle radiates if its velocity is greater than the local phase velocity of light (Cherenkov radiation) or if it crosses suddenly from one medium to another with different optical properties (transition radiation). Neither process is important for energy loss, but both are used in high-energy and cosmic-ray physics detectors.
θc γc η Cherenkov wavefront Particle velocity v = βc v = vg Figure 32.26: Cherenkov light emission and wavefront angles. In a dispersive medium, θc + η ̸= 900.
32.7.1.
Optical Cherenkov radiation : The angle θc of Cherenkov radiation, relative to the particle’s direction, for a particle with velocity βc in a medium with index of refraction n is cos θc = (1/nβ) or tan θc = p β2n2 −1 ≈ p 2(1 −1/nβ) for small θc, e.g. in gases.
(32.43) The threshold velocity βt is 1/n, and γt = 1/(1−β2 t )1/2. Therefore, βtγt = 1/(2δ +δ2)1/2, where δ = n −1. Values of δ for various commonly used gases are given as a function of pressure and wavelength in Ref. 79. For values at atmospheric pressure, see Table 6.1.
Data for other commonly used materials are given in Ref. 80.
Practical Cherenkov radiator materials are dispersive. Let ω be the photon’s frequency, and let k = 2π/λ be its wavenumber. The photons propage at the group velocity vg = dω/dk = c/[n(ω) + ω(dn/dω)]. In a non-dispersive medium, this simplies to vg = c/n.
In his classical paper, Tamm showed that for dispersive media the radiation is concentrated in a thin conical shell whose vertex is at the moving charge, and whose opening half-angle η is given by cot η = · d dω (ω tan θc) ¸ ω0 = · tan θc + β2ω n(ω) dn dω cot θc ¸ ω0 , (32.44) where ω0 is the central value of the small frequency range under consideration.
(See Fig. 32.26.) This cone has a opening half-angle η, and, unless the medium is non-dispersive (dn/dω = 0), θc + η ̸= 900. The Cherenkov wavefront ‘sideslips’ along August 21, 2014 13:18 34 32. Passage of particles through matter with the particle .
This effect has timing implications for ring imaging Cherenkov counters , but it is probably unimportant for most applications.
The number of photons produced per unit path length of a particle with charge ze and per unit energy interval of the photons is d2N dEdx = αz2 ℏc sin2 θc = α2z2 re mec2 µ 1 − 1 β2n2(E) ¶ ≈370 sin2 θc(E) eV−1cm−1 (z = 1) , (32.45) or, equivalently, d2N dxdλ = 2παz2 λ2 µ 1 − 1 β2n2(λ) ¶ .
(32.46) The index of refraction n is a function of photon energy E = ℏω, as is the sensitivity of the transducer used to detect the light. For practical use, Eq. (32.45) must be multiplied by the the transducer response function and integrated over the region for which β n(ω) > 1. Further details are given in the discussion of Cherenkov detectors in the Particle Detectors section (Sec. 33 of this Review).
When two particles are close together (lateral separation < ∼1 wavelength), the electromagnetic fields from the particles may add coherently, affecting the Cherenkov radiation. Because of their opposite charges, the radiation from an e+e−pair at close separation is suppressed compared to two independent leptons .
32.7.2.
Coherent radio Cherenkov radiation : Coherent Cherenkov radiation is produced by many charged particles with a non-zero net charge moving through matter on an approximately common “wavefront”—for example, the electrons and positrons in a high-energy electromagnetic cascade. The signals can be visible above backgrounds for shower energies as low as 1017 eV; see Sec. 34.3.3 for more details. The phenomenon is called the Askaryan effect .
Near the end of a shower, when typical particle energies are below Ec (but still relativistic), a charge imbalance develops. Photons can Compton-scatter atomic electrons, and positrons can annihilate with atomic electrons to contribute even more photons which can in turn Compton scatter. These processes result in a roughly 20% excess of electrons over positrons in a shower. The net negative charge leads to coherent radio Cherenkov emission. The radiation includes a component from the decelerating charges (as in bremsstrahlung). Because the emission is coherent, the electric field strength is proportional to the shower energy, and the signal power increases as its square. The electric field strength also increases linearly with frequency, up to a maximum frequency determined by the lateral spread of the shower. This cutoffoccurs at about 1 GHz in ice, and scales inversely with the Moliere radius. At low frequencies, the radiation is roughly isotropic, but, as the frequency rises toward the cutofffrequency, the radiation becomes increasingly peaked around the Cherenkov angle. The radiation is linearly polarized in the plane containing the shower axis and the photon direction. A measurement of the signal polarization can be used to help determine the shower direction. The characteristics of this radiation have been nicely demonstrated in a series of experiments at SLAC . A detailed discussion of the radiation can be found in Ref. 87.
August 21, 2014 13:18 32. Passage of particles through matter 35 32.7.3.
Transition radiation : The energy radiated when a particle with charge ze crosses the boundary between vacuum and a medium with plasma frequency ωp is I = αz2γℏωp/3 , (32.47) where ℏωp = q 4πNer3 e mec2/α = q ρ (in g/cm3) ⟨Z/A⟩× 28.81 eV .
(32.48) For styrene and similar materials, ℏωp ≈20 eV; for air it is 0.7 eV.
The number spectrum dNγ/d(ℏω diverges logarithmically at low energies and decreases rapidly for ℏω/γℏωp > 1. About half the energy is emitted in the range 0.1 ≤ℏω/γℏωp ≤1. Inevitable absorption in a practical detector removes the divergence.
For a particle with γ = 103, the radiated photons are in the soft x-ray range 2 to 40 keV.
The γ dependence of the emitted energy thus comes from the hardening of the spectrum rather than from an increased quantum yield.
10−3 10−2 10−4 10−5 10 1 100 1000 25 µm Mylar/1.5 mm air γ = 2 ×104 Without absorption With absorption 200 foils Single interface x-ray energy ω (keV) dS/d( ω), differential yield per interface (keV/keV) Figure 32.27: X-ray photon energy spectra for a radiator consisting of 200 25 µm thick foils of Mylar with 1.5 mm spacing in air (solid lines) and for a single surface (dashed line). Curves are shown with and without absorption. Adapted from Ref. 88.
The number of photons with energy ℏω > ℏω0 is given by the answer to problem 13.15 August 21, 2014 13:18 36 32. Passage of particles through matter in Ref. 33, Nγ(ℏω > ℏω0) = αz2 π "µ ln γℏωp ℏω0 −1 ¶2 + π2 12 # , (32.49) within corrections of order (ℏω0/γℏωp)2. The number of photons above a fixed energy ℏω0 ≪γℏωp thus grows as (ln γ)2, but the number above a fixed fraction of γℏωp (as in the example above) is constant. For example, for ℏω > γℏωp/10, Nγ = 2.519 αz2/π = 0.59% × z2.
The particle stays “in phase” with the x ray over a distance called the formation length, d(ω) = (2c/ω)(1/γ2 + θ2 + ω2 p/ω2)−1. Most of the radiation is produced in this distance. Here θ is the x-ray emission angle, characteristically 1/γ. For θ = 1/γ the formation length has a maximum at d(γωp/ √ 2) = γc/ √ 2 ωp. In practical situations it is tens of µm.
Since the useful x-ray yield from a single interface is low, in practical detectors it is enhanced by using a stack of N foil radiators—foils L thick, where L is typically several formation lengths—separated by gas-filled gaps. The amplitudes at successive interfaces interfere to cause oscillations about the single-interface spectrum. At increasing frequencies above the position of the last interference maximum (L/d(w) = π/2), the formation zones, which have opposite phase, overlap more and more and the spectrum saturates, dI/dω approaching zero as L/d(ω) →0. This is illustrated in Fig. 32.27 for a realistic detector configuration.
For regular spacing of the layers fairly complicated analytic solutions for the intensity have been obtained [88,89].
Although one might expect the intensity of coherent radiation from the stack of foils to be proportional to N2, the angular dependence of the formation length conspires to make the intensity ∝N.
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9571 | https://pmc.ncbi.nlm.nih.gov/articles/PMC3122495/ | Histopathologic Diagnosis of Fungal Infections in the 21st Century - PMC
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Clin Microbiol Rev
. 2011 Apr;24(2):247–280. doi: 10.1128/CMR.00053-10
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Histopathologic Diagnosis of Fungal Infections in the 21st Century
Jeannette Guarner
Jeannette Guarner
1 Department of Pathology and Laboratory Medicine, Emory University, Atlanta, Georgia
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1,, Mary E Brandt
Mary E Brandt
2 Mycotic Diseases Branch, Centers for Disease Control and Prevention, Atlanta, Georgia
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1 Department of Pathology and Laboratory Medicine, Emory University, Atlanta, Georgia
2 Mycotic Diseases Branch, Centers for Disease Control and Prevention, Atlanta, Georgia
Corresponding author. Mailing address: Department of Pathology and Laboratory Medicine, Emory University Hospital, 1364 Clifton Rd., Atlanta, GA 30322. Phone: (404) 712-2631. Fax: (404) 728-7995. E-mail: jguarne@emory.edu.
Copyright © 2011, American Society for Microbiology
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PMCID: PMC3122495 PMID: 21482725
Abstract
Summary: Fungal infections are becoming more frequent because of expansion of at-risk populations and the use of treatment modalities that permit longer survival of these patients. Because histopathologic examination of tissues detects fungal invasion of tissues and vessels as well as the host reaction to the fungus, it is and will remain an important tool to define the diagnostic significance of positive culture isolates or results from PCR testing. However, there are very few instances where the morphological characteristics of fungi are specific. Therefore, histopathologic diagnosis should be primarily descriptive of the fungus and should include the presence or absence of tissue invasion and the host reaction to the infection. The pathology report should also include a comment stating the most frequent fungi associated with that morphology as well as other possible fungi and parasites that should be considered in the differential diagnosis. Alternate techniques have been used to determine the specific agent present in the histopathologic specimen, including immunohistochemistry, in situ hybridization, and PCR. In addition, techniques such as laser microdissection will be useful to detect the now more frequently recognized dual fungal infections and the local environment in which this phenomenon occurs.
INTRODUCTION
Fungal infections are becoming more frequent because of expansion of at-risk populations and use of treatment modalities that permit longer survival of these patients (109). Some of the changes in endemic fungal infections can be attributed to climate changes, extension of human habitats, ease of travel, and shifting populations. At-risk populations for opportunistic fungal infections or disseminated endemic fungal infections include patients who have received transplants, those prescribed immunosuppressive and chemotherapeutic agents, HIV-infected patients, premature infants, the elderly, and patients undergoing major surgery. Thus, a shift in the mycoses encountered in the health care setting has occurred. Prior to the 21st century, bloodstream infections were more frequently caused by Candida spp., and agents of invasive pulmonary infections included primarily endemic mycoses and Aspergillus spp. Today, fungi previously considered nonpathogenic, including mucoraceous genera (formerly called zygomycetes) and a variety of both hyaline and dematiaceous molds, are commonly seen in immunocompromised patients. In addition, diagnosis of infection versus colonization with these fungi is a frequent problem that has important treatment implications for these patients. Furthermore, advances in diagnostic radiology and in patient support (such as platelet transfusions, etc) have allowed greater ability to pursue specific diagnoses by collecting tissue biopsy specimens from body sites formerly not available for histopathologic examination.
The advantages of obtaining these specimens have created a series of diagnostic challenges because of the limited amount of tissue obtained and the architectural distortion produced by these new procedures. In addition, the therapeutic armamentarium now available and the presence of resistance of these fungi to different drugs have compounded the diagnostic challenges. Histopathology continues to be a rapid and cost-effective means of providing a presumptive or definitive diagnosis of an invasive fungal infection. However, the use of fungal silver impregnation stains (Grocott or Gomori methenamine silver [GMS]) cannot alone solve these challenges, and newer diagnostic techniques may be required. Microbiologists, pathologists, and clinicians need to be aware of the limitations of tissue diagnosis, the pitfalls of morphological diagnosis, and the tests that can be performed with tissue and other samples to make organism-specific diagnoses. In this review we present epidemiologic, clinical, and morphological findings and interpretation pitfalls regarding the most frequently encountered yeasts and molds, as well as alternative testing that can be performed with other samples. Table 1 summarizes the clinical presentations and host reactions produced by the mycoses discussed in this review, Fig. 1 to 3 summarize the morphologies of these fungi and the differential diagnoses for each, and Table 2 summarizes the alternate testing that can be performed with specimens that were not sent to the pathology laboratory. Additionally, we present methods that can be used for diagnosis of specific yeasts and molds in formalin-fixed, paraffin-embedded tissue submitted for histopathologic diagnosis, as well as a series of scenarios that should help guide the diagnosis and treatment of patients with mycoses.
Table 1.
Clinical presentation of and host reaction to the more common mycoses
| Fungus(i) | Clinical presentation | Host responsea | Comment(s) |
:---: :---: |
| Blastomyces dermatitidis | Asymptomatic | Tissue descriptions unavailable | Epidemiologic evidence |
| Acute pneumonia | Mixed suppurativeb inflammation | Seen in areas of endemicity |
| Chronic pneumonia | Mixed suppurative and granulomatousc inflammation | Most frequently diagnosed |
| ARDS | Diffuse alveolar damage | Can be fatal |
| Disseminated | Various inflammatory responses depending on immune status | Skin, soft tissue, bone, GU, or CNS is primarily involved |
| Cutaneous | Mixed suppurative and granulomatous inflammation | Rare, due to direct cutaneous inoculation |
| Cryptococcus spp. | Asymptomatic | Minimal reaction | Rare, epidemiologic evidence |
| Pneumonia | Predominantly granulomatous inflammation, can have abundant fibrosis | More frequent in immunocompetent individuals |
| Cryptococcoma | Granuloma with various degrees of necrosis and fibrosis | More frequent with C. gattii |
| Pleural effusion | Various inflammatory responses depending on immune status | More frequent in immunosuppressed patients |
| Disseminated | Various inflammatory responses depending on immune status, abundant extracellular yeasts may efface tissue architecture, necrosis may be present | Frequent in immunosuppressed patients; involves the CNS (producing meningitis or cryptococcomas), skin, bones, or other tissues |
| Histoplasma capsulatum | Asymptomatic | Tissue descriptions not available | Occurs when low numbers of microconidia are inhaled |
| Acute pneumonia | Nodules showing vascular necrosis associated with lympho-histiocytic vasculitis and rare granulomatous inflammation | Occurs when high numbers of microconidia are inhaled |
| ARDS | Diffuse alveolar damage | Can be fatal |
| Mediastinitis | Granulomatous inflammation | Occurs upon initial inhalation of the microconidia |
| Chronic pneumonia | Granulomas with various degrees of necrosis and calcification | Can present as a nodule or cavity |
| Disseminated | Various inflammatory responses depending on immune status, abundant intracellular yeasts may efface tissue architecture, necrosis may be present | Seen upon initial infection or as reactivation of latent disease in patients with T-cell deficiencies; can involve skin, GI tract, liver, spleen, and bone marrow |
| Coccidioides immitis/posadasii | Asymptomatic | Tissue descriptions unavailable | Epidemiologic evidence |
| Acute pneumonia | Suppurative and granulomatous inflammation | Presents as lobar infiltrates and adenopathy |
| Chronic pneumonia | Mixed suppurative (including eosinophils) and granulomatous inflammation with a rim of lymphocytes, Splendore-Höeppli phenomenon likely | Can present as a nodule or cavity |
| Disseminated | Various inflammatory responses depending on immune status | Occurs in certain risk groups (those with diabetes, use of steroids, and others); can involve skin, lymph nodes, bones, joints, and CNS |
| Cutaneous | Mixed suppurative and granulomatous inflammation | Rare, due to direct cutaneous inoculation |
| Candida spp. | Superficial infections | Minimal to suppurative inflammation depending on immune status of individual | Skin and mucous membranes of GI and GU tracts in immunocompetent and immunosuppressed individuals |
| Invasive disease | Various inflammatory responses depending on immune status, primarily suppurative inflammation with rare granulomas, invasion of blood vessels, necrotizing vasculitis | Occurs primarily as a health care-associated infection (patients with vascular access devices, with recent surgeries, receiving broad-spectrum antibiotics, or immunosuppressed), can involve all organs |
| Pneumocystis jirovecii | Asymptomatic | Minimal reaction | Has been found in the lungs of children |
| Pneumonia | Minimal reaction; rarely atypical reactions such as fibrosis, granulomas, and others | Affects immunosuppressed patients |
| Disseminated | Minimal reaction; rarely atypical reactions such as fibrosis, granulomas, and others | Rare, affects immunosuppressed patients |
| Sporothrix schenckii | Cutaneous | Mixed suppurative (including eosinophils) and granulomatous inflammation, Splendore-Höeppli phenomenon frequent, presence of asteroid bodies, epidermis with pseudoepitheliomatous hyperplasia | Associated with handling contaminated soil or animals, draining lymph nodes are frequently affected |
| Disseminated | Various inflammatory responses depending on immune status | Affects bone, joints, meninges, and other organs |
| Penicillium marneffei | Cutaneous | Mixed suppurative and granulomatous inflammation with various degrees of necrosis | Seen mostly in Southeast Asia |
| Disseminated | Various inflammatory responses depending on immune status, may consist only of necrosis and infected macrophages | Seen mostly in Southeast Asia; P. marneffei infection represents the most frequent AIDS-defining illness |
| Paracoccidioides brasiliensis | Acute pneumonia | Mixed suppurative and granulomatous inflammation | Correlates with hormonal, genetic, immunologic, and nutritional status |
| Chronic pneumonia | Mixed suppurative and granulomatous inflammation surrounded by fibrosis | Correlates with hormonal, genetic, immunologic, and nutritional status; if organism is swallowed, can cause GI disease |
| Disseminated | Mixed suppurative and granulomatous inflammation, bone may show osteonecrosis | Involvement of bone marrow, adrenal glands, CNS, and other tissues |
| Cutaneous | Mixed suppurative and granulomatous inflammation, epidermis with epitheliomatous hyperplasia | Rare, due to direct cutaneous inoculation |
| Rhinosporidium seeberid | Nose, nasopharynx, ocular areas | Granulomatous inflammation with fibrosis and granulation tissue | Presents as a mass or polyp |
| Disseminated | Chronic and granulomatous inflammation | Rare, can involve other mucous membranes or cutaneous sites and internal organs |
| Aspergillus spp. | Allergic bronchopulmonary aspergillosis | Allergic mucous with eosinophils, Curshmann's spirals, Charcot-Leyden crystals; mucosa with suppurative and granulomatous inflammation, vasculitis, and fibrosis | Hypersensitivity reaction to fungi, most frequently A. fumigatus; seen frequently in patients with cystic fibrosis or steroid-dependent asthma |
| Allergic fungal rhinosinusitis | Similar to that for allergic bronchopulmonary aspergillosis | Hypersensitivity reaction to fungi similar to that for allergic bronchopulmonary aspergillosis |
| Chronic pulmonary aspergillosis | The wall surrounding the fungus ball consists of fibrosis | Occurs in immunocompetent individuals with a variety of lung conditions (tuberculosis, emphysema, and others) in which the cavity or lesion is colonized and then a “fungus ball” or aspergilloma forms |
| Chronic necrotizing pulmonary aspergillosis | The wall surrounding the fungus ball consists of a layer of necrosis, granulation tissue, granulomatous inflammation, and fibrosis | Occurs in immunosuppressed individuals with chronic pulmonary aspergillosis, where the fungus is invading the tissues locally |
| Invasive disease | Angioinvasion by hyphae with consequent necrosis or hemorrhage of surrounding tissue | Seen in severely immunosuppressed patients, involves the lungs, CNS, and other tissues |
| Mucorales genera | Cutaneous | Angioinvasion by hyphae with consequent necrosis or hemorrhage of surrounding tissue; inflammation, if present, is frequently suppurative, less commonly granulomatous, but varies depending on immune status | Necrotic (black) skin lesion in immunosuppressed patients |
| Rhinocerebral | Similar to that for cutaneous disease | Particularly frequent in diabetic patients but can occur in any immunosuppressed patient |
| Pulmonary | Similar to that for cutaneous disease | Multiple pulmonary nodules and pleural effusions in immunosuppressed patients |
| Invasive disease | Similar to that for cutaneous disease | Risk factors include cancer chemotherapy and stem cell transplantation |
| Entomophthorales | Mucocutaneous | Fibrosis, granulation tissue, mixed eosinophilic and granulomatous inflammation; Splendore-Höeppli phenomenon present | Presents as a mass in immunocompetent individuals, the lesion can be in the GI tract |
| Hyaline septated molds (Fusarium spp., Scedosporium spp., Trichoderma spp., Paecilomyces spp., and others) | Superficial infections | Mild inflammation | Occur in skin, cornea, and nails; Fusarium spp. are the most common of these organisms in causing superficial infections |
| Range of diseases similar to that for Aspergillus: allergic, chronic pulmonary, and invasive | Similar to that for Aspergillus | Some organisms have some peculiarities (for example, Scedosporium spp. are associated with pneumonia after near drowning, and Trichoderma spp. have been observed in patients undergoing dialysis) |
| Dematiaceous fungi (Madurella spp., Fonsecaea spp., Cladophialophora spp., Exophiala spp., Curvularia spp., Bipolaris spp., and others) | Superficial infections | Mild inflammation | Occur in skin, cornea, and nails |
| Deep skin infections | Mixed suppurative and granulomatous inflammation with reactive epidermal changes, including pseudoepitheliomatous hyperplasia and draining sinuses | See Table 3 |
| Range of diseases similar to that for Aspergillus: allergic, chronic pulmonary, and invasive | Similar to that for Aspergillus | Bipolaris and Curvularia are most frequently associated with eosinophilia and allergic sinusitis or allergic bronchopulmonary mycosis, Cladophialophora bantiana is most frequently associated with brain abscesses |
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a
Fungal morphology is presented in Fig. 1, 2, and 3.
b
Suppurative inflammation refers to presence of congestion, edema, necrosis, and an inflammatory infiltrate with a predominance of neutrophils.
c
Granulomatous inflammation refers to presence of epithelioid macrophages including multinucleated giant cells, lymphocytes, and necrosis.
d
R. seeberi is not a fungus but is included for purposes of contrast with fungi that have similar morphology.
Fig. 1.
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Morphology, description, diagnosis, and comment for endemic fungal infections that present as yeasts in tissues. All photographs are of Grocott methenamine silver (GMS)-stained specimens except for the inset in the second row, which is a mucicarmine stain. For each type of infection, alternative testing and correlation with culture, epidemiologic, and clinical features are necessary. (The photographs of Cryptococcus spp., Histoplasma capsulatum, and Coccidioides spp. are reprinted from the CDC Public Health Image Library collection.)
Fig. 3.
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Morphology, description, diagnosis, and comment for fungal infections that show characteristic yeast morphology in tissues. Except for the last row, which shows an H&E-stained asteroid body, all photographs are of Grocott methenamine silver (GMS)-stained specimens (including the inset of S. schenckii, which is counterstained with H&E). For each type of infection, alternative testing and correlation with culture, epidemiologic, and clinical features are necessary. (The photographs of Pneumocystis, Penicillium marneffei, and Paracoccidioides brasiliensis are reprinted from the CDC Public Health Image Library collection.)
Table 2.
Alternative testing that can be performed with nontissue specimens based on the morphology present in tissue and the suspected fungi
| Morphology using H&E, GMS, and PAS staining | Suspect fungus(i) | Alternative test with nontissue samples | Comment(s) |
:---: :---: |
| Broad-based budding yeasts (10–15 μm) | Blastomyces dermatitidis | Antigens in urine or serum | EIA antigen assay for urine, serum, BAL fluid, or CSF is available at MiraVista Diagnostics (Indianapolis, IN) (sensitivity has been reported to be around 90%, but this test should be performed simultaneously with Histoplasma testing because of cross-reactivity between the two antigens); EIA can be used to follow response to treatment |
| Serology | There is poor sensitivity (9 to 28%) and specificity if using complement fixation and immunodiffusion, radioimmunodiffusion and EIAs have better sensitivity (77 to 83%) and specificity (95%) but are not commercially available |
| Cultures of sputum or BAL fluid | Diagnostic yields of 86 to 92%, respectively; however, delays in diagnosis are inherent to the technique |
| Narrow-based budding yeasts (4–10 μm) with a thick capsule | Cryptococcus spp. | Cryptococcal antigen | Sensitivity of >90% but may have up to 7% false-positive results; latex agglutination and EIA with serum and CSF are FDA approved, but testing in urine is not; laboratories can perform the test with urine samples if they validate for the sample type |
| Cultures of blood, CSF, and other fluids | Use of canavanine-glycine-bromthymol blue medium is helpful to distinguish Cryptococcus gattii from C. neoformans |
| Small yeasts (2–4 μm) with narrow-based budding grouped in clusters inside macrophages | Histoplasma capsulatum | Antigens in urine or serum | Several commercial assays are available; sensitivity is about 90% in cases of disseminated disease and 75% during acute pulmonary disease; false-positive results occur in cases with other endemic mycoses, with blastomycosis being the most important overlap; Aspergillus galactomannan also cross-reacts |
| Serology | Accomplished by the use of complement fixation or immunodiffusion with about 80% sensitivity, useful for immunocompetent individuals but may not be useful when the patient is immunosuppressed |
| Cultures of blood and other fluids | It is important to use lysis-centrifugation to release the organisms from phagocytic cells; although specific, culture may take up to 6 weeks |
| Spherules with multiple endospores | Coccidioides immitis/posadasii | Antigens in urine or serum | An EIA for Coccidioides antigen in urine, serum, and CSF is commercially available (MiraVista Diagnostics, Indianapolis, IN); studies using rabbit antibodies against a Coccidioides galactomannan in an EIA showed that 71% of patients with coccidioidomycosis have antigenuria; however, EIA cross-reacts in 10% of patients with other endemic mycoses |
| Serology | Measurement of combined IgM and IgG by EIA shows a sensitivity and specificity of >95%, immunodiffusion can also be used to measure IgM and IgG |
| Cultures of any sample | Coccidioides is a select agent with stringent culture regulations |
| DNA detection in sputum | PCR detection of ITS2 or Ag2/PRA targets has been published, showing good sensitivity, but assays are not commercially available |
| Small yeasts (3–5 μm) intermingled with pseudohyphae and/or hyphae | Candida spp. | Detection of beta-d-glucan in serum | Sensitivity will vary between 57 to 90% depending on the patient population and cutoff values used for the assay, specificity ranges from 44 to 92% |
| Cultures of blood and other fluids | Blood cultures occur in 50 to 70% of cases, and when positive, they indicate invasive disease; peptide nucleic acid fluorescent in situ hybridization assay (Yeast “Traffic Light” PNA-FISH assay; AdvanDx Inc., Woburn, MA) of positive blood culture smears can identify the most frequent Candida spp. without the need for subcultures |
| DNA detection in whole blood, serum, or plasma | Several PCRs have been developed and validated; they seem to have increased sensitivity compared to blood cultures, but the clinical significance of these findings is still being researched |
| Nonpigmented (hyaline), septated hyphae with acute-angle branching | Aspergillus spp., Fusarium spp., Scedosporium spp., Trichoderma spp., and Paecilomyces spp. | Detection of galactomannan in serum or BAL fluid | Galactomannan antigen detection using commercially available Aspergillus EIA (Platelia Aspergillus test; Bio-Rad, Hercules, CA) has sensitivities that range from 40 to 100% and specificities from 56 to 100%, depending on the patient population; false-positive results occur in 50% of patients taking certain antibiotics or plasmalyte or having other fungal infections; other genera in this morphological category do not have specific alternative tests, although galactomannan has been shown to be positive in patients with Paecilomyces infections |
| Detection of beta-d-glucan in serum | Measured using commercially available EIA (Fungitell kit; Associates of Cape Cod, East Falmouth, MA) with sensitivities that range from 50 to 100% and specificities of from 44 to 98%, the presence of beta-glucan is not specific for particular fungal genera |
| Cultures of blood | Fusarium spp. may have positive blood cultures, other genera are rarely cultured from the blood, even in the presence of invasive disease |
| Nonpigmented (hyaline), pauciseptate ribbon-like hyphae with right-angle branching | Mucorales genera | Serology | Serologic tests for invasive disease have been attempted but have not been clinically useful |
| Culture of blood | Mucorales genera are rarely recovered from blood, even with invasive disease |
| Pigmented irregular hyphae and yeast-like structures both with septations | Dematiaceous fungi, including Madurella spp., Fonsecaea spp., Cladophialophora spp., Exophiala spp., Curvularia spp., Bipolaris spp., and others | Serology | Not available |
| Culture of blood or other fluids | These genera are rarely cultured from the blood, even in the presence of invasive disease |
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MORPHOLOGIES OF ORGANISMS IN TISSUES IN VARIOUS DISEASES
Diseases Where Yeasts or Yeast-Like Structures Are Usually Seen in Tissues
Blastomycosis.
(i) Epidemiologic and clinical situations when blastomycosis should be considered in the differential diagnosis.
Blastomycosis was first described by Gilchrist at the end of the 19th century as a dermatologic infection caused by a protozoan (58). Today we know it is caused by the dimorphic fungus Blastomyces dermatitidis, and the majority of cases involve primarily lung disease. The organism has been isolated from moist soil rich in organic debris in the Mississippi and Ohio River valleys and around the Great Lakes and the Saint Lawrence River, which include multiple states in the United States (southeastern, south central, and upper midwestern states) and several Canadian provinces (99). Blastomycosis was so prevalent in Chicago that it has been referred to as “Chicago disease” (118); however, sporadic cases have been reported from areas where the disease is not endemic, such as Colorado, Texas, Kansas, and Nebraska in North America, and from other countries around the world (150).
Inhalation of the B. dermatitidis conidia is the usual route of infection, and a variety of responses can occur in the lung, including asymptomatic infections, acute and chronic pneumonia, and fatal acute respiratory distress syndrome (ARDS) (99, 118). In general, acute pneumonia is rarely identified outside outbreaks. More frequently, pulmonary blastomycosis is diagnosed in cases of chronic pneumonia or when a lung neoplasia is suspected. In 20 to 40% of cases the disease has become systemic at the time of diagnosis and there is skin, soft tissue, bone, genitourinary (GU), or central nervous system involvement (9, 87, 99, 118). In rare cases skin lesions without lung involvement have been described, suggesting direct cutaneous inoculation. The skin lesions most frequently described are painless ulcers or verrucous lesions. Blastomycosis in immunocompromised patients appears to be more severe and more frequently fatal (99). It should be noted that in patients with central nervous system involvement, diabetes mellitus is an important predisposing factor (9).
(ii) Morphological characteristics that set blastomycosis apart.
B. dermatitidis in tissue appears as yeasts that measure 8 to 15 μm in diameter, have thick refractile cell walls, and may show a single, broad-based bud (Fig. 1). The yeasts can be observed in a variety of specimens, including sputum, bronchoalveolar lavage (BAL) fluid, fine-needle aspirates from lung, skin, or other lesions, cerebrospinal fluid (CSF), and surgical resections (9, 87). The thick refractile cell wall of this organism gives the appearance of a space between the fungal cell contents and the surrounding tissue when hematoxylin and eosin (H&E) stain is used. Inside the cell wall, the multiple nuclei of the yeast stain with hematoxylin. Occasionally, B. dermatitidis can show smaller yeast forms, the so-called microforms. In addition, B. dermatitidis can be seen with a variety of routinely used preparations and stains such as KOH and Papanicolaou stain. The contour of the yeast is best highlighted by staining the cell wall with fungal silver stains such as GMS or periodic acid-Schiff (PAS) stain (87, 118). The inflammatory reaction accompanying the yeasts is primarily granulomatous with various degrees of neutrophilic infiltrate; thus, it has been described as pyogranulomatous inflammation (118). It needs to be remembered that blastomycosis can be concomitantly present with neoplasias and tuberculosis.
(iii) Pitfalls in morphological diagnosis.
Few studies have systematically compared the presence of broad-based budding yeasts in histopathologic or cytologic specimens with culture or other diagnostic methods that would confirm the diagnosis of blastomycosis. A retrospective study of 53 patients showed that Coccidioides immitis, Candida albicans, or Aspergillus was recovered from 4 pathologic specimens (10%) demonstrating broad-based budding yeasts in direct histopathologic examination (118). An earlier study of patients with blastomycosis commented that a high percentage of their cultures were overgrown with Candida (87). This suggests that not all broad based-budding yeasts in the 8- to 15-μm size range are Blastomyces. Since histopathologic or cytologic results can usually be provided before the culture is available, there is pressure to use these results to guide treatment, particularly because B. dermatitidis can take up to 3 weeks to grow or may not grow at all from these specimens. The sensitivity of culture varies depending on the sample that was obtained and may range from 62% to 100% (87, 96). The diagnostic yield of histopathology will depend on the expertise present in the center where the patient is seen (99). Because of the possibility of histopathologic false-positive results, pathologists should describe the yeast and budding pattern that are observed in the tissue specimen and should list the yeasts that can have this morphology in the report comment field (Fig. 1). In addition, alternative tests are necessary to determine that the patient truly has blastomycosis, especially in cases from areas where the disease is not endemic or when the clinical picture is not typical.
(iv) Alternative testing.
Blastomycosis antigens can be detected in the urine and serum by using an enzyme immunoassay (EIA). The sensitivity and specificity for antigen detection have been reported to be above 90%; however, cross-reactivity occurs in patients with histoplasmosis, paracoccidioidomycosis, and penicilliosis due to Penicillium marneffei (50). Because of the cross-reactivity, it is important that antigen tests for both blastomycosis and histoplasmosis be ordered. Detection of antibodies to B. dermatitidis in serum using traditional complement fixation and immunodiffusion has poor sensitivity and specificity; however, as antigens have been better purified and used in radioimmunodiffusion and EIAs, the sensitivity and specificity of serology have significantly improved (99). Cultures of the lesion should be encouraged, particularly in regions where the disease is not endemic.
Cryptococcosis.
(i) Epidemiologic and clinical situations when cryptococcosis should be considered in the differential diagnosis.
Human cryptococcosis is caused by several Cryptococcus species, including C. neoformans and C. gattii. C. neoformans is responsible for the majority of infections found in immunocompromised individuals, while C. gattii causes infections in immunocompetent and compromised hosts (72). C. neoformans var. grubii (serotype A) and C. neoformans var. neoformans (serotype D) have a worldwide distribution and are found primarily associated with pigeon guano. HIV infection is the most frequent predisposing factor for cryptococcal disease; however, other conditions are associated with cryptococcosis, including underlying lung, liver, or renal disease, immunosuppressant use, malignancies, and autoimmune diseases. C. gattii (serotypes B and C) has been found in eucalyptus and other tropical and subtropical trees in a more limited geographic distribution, including Australia, Papua New Guinea, parts of Africa, Mexico, and southern California. In 1999, C. gattii was first reported as a human pathogen in Vancouver Island, Canada, and it has now been found to cause disease in other areas in British Columbia, Canada, and in the northwest United States (Washington and Oregon) (53).
Independent of the species, humans inhale cryptococcal yeasts or basidiospores, and thus the lung is the primary infection site (53, 72, 90). Few exposed individuals remain asymptomatic, while the majority display pneumonia, cryptococcomas, or pleural effusions. From the lung, cryptococci can disseminate to the central nervous system (producing meningitis or cryptococcomas), skin, bones, or other tissues. C. gattii is associated with a higher incidence of solid lesions in the lungs and brain than C. neoformans. The frequency of disseminated disease is dependent on the immune status of the patient: in immunocompetent individuals the most frequent presentation is pulmonary, while immunosuppressed patients commonly present with central nervous system involvement.
(ii) Morphological characteristics that set cryptococcosis apart.
Cryptococci are encapsulated, spherical to oval yeast that measure 5 to 10 μm in diameter and have narrow-based budding (Fig. 1) (31, 56). A thick polysaccharide capsule gives these organisms the characteristic appearance of having a clear space around them that can be seen in tissue sections with H&E stains. When testing CSF, India ink can be used as a negative stain to highlight the capsule. Because of the capsule, the buds appear separate from the mother cells. The polysaccharide capsule stains with Alcian blue and Mayer's or Southgate's mucicarmine stain. As with all other yeasts, the wall of the organism stains with GMS and PAS stains. In addition, cryptococci stain with Fontana-Masson stain because they contain melanin.
The inflammatory reaction against cryptococci seen in histopathology varies from well-formed granulomas where the organisms are found inside macrophages and giant cells to minimal inflammation with abundant extracellular organisms that efface the tissue architecture (31, 56). The granulomatous inflammatory reactions also show a spectrum from abundant necrosis to fibrous granulomas. In some instances the fibrosis is intense, with the fibroblast having plump spindle cells in a storiform pattern accompanied by a background of lymphocytes and plasma cells, giving the appearance of an inflammatory pseudotumor (151). Some authors have correlated the inflammatory reaction to the immune status of the patient and the presence or absence of capsule.
(iii) Pitfalls in morphological diagnosis.
In some patients with cryptococcosis, the yeast may produce lesser amount of the characteristic polysaccharide capsule; thus, these organisms may resemble other yeasts of similar size, such as Candida spp. or Histoplasma. Staining these specimens with Fontana-Masson stain may prove that the yeast produces melanin, which is characteristic of cryptococci. The use of cryptococcal antigen tests with serum and CSF may not be helpful for patients with poorly encapsulated cryptococci, because most of the serologic tests detect antigens present in the capsule (56).
(iv) Alternative testing.
Cryptococcal antigen testing using latex agglutination or EIA can be performed with serum and CSF. These techniques show a sensitivity and specificity of above 90%; however, false-negative results can occur due to low fungal burden or a prozone phenomenon, while false-positive results have been seen in patients infected with Trichosporon spp. or Klebsiella pneumoniae, in those with positive rheumatoid factor, or if the reagent was incubated with the specimen beyond the recommended time. The presence of antigens or yeast forms in the various fluids and tissues is important for diagnosis and for determining the amount of cryptococci present, but continued infection can be determined only by cultures (72). Cultures, particularly using canavanine-glycine-bromthymol blue medium, which turns blue in the presence of C. gattii, are helpful to determine the species of Cryptococcus and are indispensable for determining antifungal susceptibility when indicated. However, it should be noted that breakpoints for antifungal drugs have not been established for Cryptococcus spp.
Histoplasmosis.
(i) Epidemiologic and clinical situations when histoplasmosis should be considered in the differential diagnosis.
Histoplasma capsulatum is found worldwide in old buildings, caves, and soil rich in bird and bat droppings; nevertheless, there are areas of endemicity, including the Ohio and Mississippi River valleys in the United States, Central and South America, southern Europe, parts of Africa, and southeastern Asia (78). In most areas of the world human histoplasmosis is caused by H. capsulatum var. capsulatum; however, in western and central regions of sub-Saharan Africa the African clade of Histoplasma capsulatum, formerly named H. capsulatum var. duboisii, can be found (93). Although histoplasmosis can occur in outbreaks when old buildings are renovated/demolished or groups of tourists visit caves, most cases are sporadic.
Histoplasmosis is acquired by inhaling microconidia, and depending on the amount of fungus inhaled and immune status, the host may show no symptoms or may have acute or chronic pulmonary disease or disseminated infection (22, 78). Once inhaled, the conidia are ingested by lung alveolar macrophages, where the organisms convert into the yeast phase. The phagocytized organisms survive inside macrophages for weeks and can disseminate as the macrophages travel in the lymphatic system. H. capsulatum is an intracellular pathogen that can remain viable inside macrophages until specific cell-mediated immunity kills the organisms. Disseminated disease occurs upon initial infection or as reactivation of latent disease in individuals who have T-cell immunodeficiencies such as AIDS, hematologic malignancies, or solid organ transplants or who use corticosteroids or tumor necrosis factor antagonists (63).
Patients with high exposure loads or those who are immunosuppressed can present with acute pneumonia or ARDS (32, 78). As macrophages travel to mediastinal lymph nodes, patients may present with mediastinitis, and as macrophages disseminate through the body, the organisms spread to other organs, forming nodules. Usual sites of dissemination include the skin, gastrointestinal (GI) tract, liver, spleen, and bone marrow. Although rare, central nervous system infection can occur. Chronic cavitary histoplasmosis that may be clinically indistinguishable from tuberculosis is a frequent presentation of older patients with emphysema or immunocompetent individuals exposed to lower fungal loads.
(ii) Morphological characteristics that set histoplasmosis apart.
H. capsulatum var. capsulatum in tissue is an oval 2- to 4-μm yeast that may show narrow-based buds (Fig. 1) (32). With H&E stain, the basophilic yeast cytoplasm is separated from the surrounding tissue by a clear zone corresponding to the cell wall. The cell wall is highlighted with GMS and PAS stains. Because the yeasts are initially ingested by macrophages, they appear to be clustered, and some authors have suggested that this is an important diagnostic feature. This clustering within histiocytes and occasionally within neutrophils is the presentation of Histoplasma in fluids stained with Papanicolaou stain or blood smears stained with Giemsa stain (60). African histoplasmosis shows similar clustering inside phagocytic cells (particularly large multinucleated giant cells), but the yeasts are larger (8 to 15 μm in diameter) than with H. capsulatum and may be pigmented.
Few lung tissue samples from patients with acute pulmonary histoplasmosis have been studied, and these have shown nodular areas of parenchymal and vascular necrosis associated with lympho-histiocytic vasculitis (105). The histopathologic picture resembles that for lymphomatoid granulomatosis, but scattered small granulomas with small yeasts in the parenchyma should suggest the diagnosis of histoplasmosis. Chronic lung infections that radiographically appear as coin lesions show typical granulomatous inflammation with central necrosis and calcified material (32). Yeasts are usually found in this necrotic calcified material, which can be lost during processing and cutting of the tissue. In immunosuppressed patients, sheets of macrophages filled with yeasts characterize disseminated disease. The collections of macrophages distort the organ architecture and produce necrotic areas. Because the morphology of H. capsulatum is not specific, it is important to perform clinico-epidemiologic correlation.
(iii) Pitfalls in morphological diagnosis.
Several fungi can be confused with H. capsulatum var. capsulatum when studying tissue sections (22, 60): the small variant of B. dermatitidis (in these cases the presence of broad-based budding and seeking larger forms can be helpful in making the diagnosis of blastomycosis), capsule-deficient cryptococci (in these cases size variation and looking for weakly positive mucicarmine-staining yeasts may suggest the diagnosis of cryptococcosis), endospores of Coccidioides spp. (looking for remnants of a ruptured spherule or an intact spherule is paramount for diagnosis of coccidioidomycosis), Pneumocystis jirovecii (this organism lacks budding and has an intracystic focus), Penicillium marneffei (this organism shows formation of a transverse septum rather than budding), and Candida glabrata (this organism may show more size variability than in histoplasmosis, and the inflammation is primarily neutrophilic).
In addition, several protozoa can also show intracellular organisms of similar size, including the agents of leishmaniasis, toxoplasmosis, and Chagas' disease, which should be differentiated from histoplasmosis (32, 60). The histopathologic difference between these organisms and Histoplasma is that H&E stains the entire protozoan and none shows the halo produced by the fungal cell wall. Kinetoplasts (a distinct hematoxylin-stained bar to the side of the nucleus) should be observed if the patient has leishmaniasis or Chagas' disease. Infected cells in toxoplasmosis and Chagas' disease are somatic (cardiomyocytes or neurons) rather than histiocytes.
In summary, definitive diagnosis of histoplasmosis can be difficult with tissue sections, and if a portion of the tissue specimen was not sent for cultures, alternative testing should be considered.
(iv) Alternative testing.
Cultures for Histoplasma using blood samples can aid in the diagnosis of disseminated disease. However, since these are primarily intracellular organisms, lysis-centrifugation methods should be used to release the yeasts from histiocytes. Furthermore, the organism grows slowly, so cultures require incubation for 4 to 6 weeks before they are called negative (77). Although testing for antibodies can be performed using complement fixation or immunodiffusion, production of antibodies may not occur in immunodeficient patients (22, 78). False-positive serology can occur in patients with lymphoma, tuberculosis, and other fungal infections, particularly blastomycosis. Detection of antigen in urine and serum can be performed using EIA with various results. Antigen is concentrated in the urine, making Histoplasma antigen detection in this specimen more reliable. Similarly to antibody testing, there are false-positive results with antigen testing. The cross-reactivity with blastomycosis is particularly problematic because histoplasmosis and blastomycosis have overlapping endemicity and histopathologically these yeasts can sometimes be confused. Nonetheless, in patients with nondisseminated histoplasmosis the antigen burden is lower, and thus sensitivity is lower. Combining the results of detection of antigen in urine and serum may increase the sensitivity in patients with pulmonary histoplasmosis (157). Diagnosis of exposure to Histoplasma has been performed using intradermal reaction to histoplasmin, but this reagent is not available in the United States (55).
Coccidioidomycosis.
(i) Epidemiologic and clinical situations when coccidioidomycosis should be considered in the differential diagnosis.
Coccidioides immitis is endemic in California desertic areas, in particular the San Joaquin valley (1, 117). Coccidioides posadasii is present in desertic regions of northwest Mexico, Arizona, Utah, New Mexico, and West Texas and in desertic areas in Argentina, Paraguay, and parts of Central America (5). However, very little difference in morphology or clinical presentation has been found between the two species. There is a clear correlation between the incidence of disease and environmental factors: coccidioidomycosis increases when there are rainy summers followed by dry winters, following earthquakes, or when humans settle and develop the previously mentioned desertic areas. In any of these instances, Coccidioides arthrospores are released in higher numbers than the usual baseline (117).
Humans inhale the arthroconidia, which in the lung are transformed into multinucleated spherical structures that contain hundreds of endospores (117). It is estimated that up to 60% of individuals exposed have no symptoms, while the remainder may present with what appears to be acute community-acquired pneumonia (1, 5). In those patients with acute pneumonia, the chest X rays show lobar infiltrates and adenopathy. Several erythematous rashes (multiforme, nodosum, or toxic) are reflections of immune response to the acute infection. The majority of acute pulmonary infections known as “valley fever” resolve; however, in a minority of patients the infection may become chronic progressive, showing either a cavity or nodule. Extrapulmonary disease can occur in members of certain risk groups, including African Americans, Asians, pregnant women, patients with diabetes, or patients receiving corticosteroids for a variety of conditions. The most common sites of dissemination include skin, lymph nodes, bones, and joints; nonetheless, the most feared complication is extension to the central nervous system. Besides acquisition of infection through the respiratory route, there are rare reports of direct inoculation of skin, giving rise to primary cutaneous lesions, or acquisition through transplanted organs (21, 102, 116).
(ii) Morphological characteristics that set coccidioidomycosis apart.
Spherules of various sizes (10 to 100 μm) with multiple endospores (2 to 5 μm) are characteristic of coccidioidomycosis and can be seen with routine H&E staining (116, 142). The walls of some of the spherules may appear to be ruptured, and the endospores spill into surrounding tissues. Active lesions contain multiple organisms, while resolving or residual lesions usually show lower number of organisms. GMS highlights spherule walls and endospores (Fig. 1). In contrast, reactions with PAS stain vary with age of the structures: young endospores and spherules stain strongly, while staining fades as the organisms mature. Occasionally, mycelia can be observed in cavitary lung or skin lesions (142). The sensitivity for histopathologic detection of Coccidioides is 84%, and that for cytology is 75% (1).
The inflammatory reaction to endospores is predominantly neutrophilic, while reaction to spherules is granulomatous. Thus, early in the infection the lesions tend to show pyogranulomas because the concentration of spherules and endospores is high. Lymphocytic clusters of B and T cells around well-formed granulomas with necrosis have been described and appear to be an important response to coccidioidomycosis (88). Eosinophils can be abundant, releasing eosinophilic major basic protein, and can create the Splendore-Höeppli phenomenon (an intense rim of eosinophilic material around the fungal elements) (127).
(iii) Pitfalls in morphological diagnosis.
Rhinosporidium seeberi, a mesomycetozoan parasite that causes palate and nasopharyngeal polyps, produces large sporangia (some can be seen with the naked eye) with multiple internal endospores. R. seeberi has very similar morphology, but its sporangia and endospores are larger than Coccidioides spherules, and its inner sporangial wall stains with mucicarmine. When coccidioidomycosis is suspected, it is important to look for spherules; endospores outside spherules or young spherules without endospores can be confused with Blastomyces, Histoplasma, Emmonsia, Candida, Pneumocystis, and other yeasts (142). It also needs to be remembered that in immunosuppressed patients, more than one infection may coexist; thus, in areas of endemicity, Pneumocystis and Coccidioides could be found in the same specimen.
(iv) Alternative testing.
In the United States, Coccidioides spp. are select agents that are governed by specific rules related to their possession, use, and transfer. These fungi grow easily in the laboratory (93% sensitivity), and the arthroconidia can be easily aerosolized; thus, it is mandatory that all work with mycology cultures is performed using a class II biological safety cabinet (1, 142). Because of the culture constraints, laboratories have studied genomic assays targeting either internal transcribed spacer 2 (ITS2) or proline-rich antigen (Ag2/PRA) for diagnosis of coccidioidomycosis with up to 98% sensitivity (5).
Detection of antibodies to Coccidioides can be an important diagnostic tool. Today, IgM and IgG are generally measured using EIA and/or immunodiffusion; however, some laboratories continue to use tube precipitin to measure IgM and complement fixation to measure IgG antibodies (142). False-negative serology has been seen in up to 38% of patients with hematogenous infection and 46% of fatal cases (1). Detection of antigens in the urine using EIA has been shown in 71% of patients with coccidioidomycosis but shows cross-reaction in 10% of patients with other endemic mycoses (49).
Candidiasis.
(i) Epidemiologic and clinical situations when candidiasis should be considered in the differential diagnosis.
Candida albicans colonizes the human oropharynx and vagina, and a small number of viable organisms can be cultured from these surfaces (38, 153). Superficial infections in the gastrointestinal or genitourinary tract occur when there are microbial imbalances caused by fluctuations in reproductive hormones, antibiotic use, and immunosuppression that may have causes ranging from HIV infection to diabetes (38, 52). Invasive candidiasis occurs primarily as a health care-associated infection, and patients at risk include those receiving broad-spectrum antibiotics, immunosuppressants, or total parenteral nutrition and those with vascular access devices, recent surgeries, malignancies, and neutropenia (40). C. albicans is the most frequent species isolated from the blood, accounting for one-third to two-thirds of bloodstream isolates (38). Certain species are associated with particular risk factors; for example, Candida parapsilosis is seen in patients with hyperalimentation or indwelling devices or in neonates, while Candida glabrata is seen in patients who are receiving azoles. Once Candida organisms gain access to the blood, secondary seeding into all organs can occur. Common localizations include the liver and spleen in neutropenic hemato-oncologic patients, the endocardium in those with prosthetic heart valves or other intravascular prosthetic devices, and the eye in those with longstanding candidemia (160). Candida organisms are capable of forming biofilms in catheters and devices, permitting their growth even when achievable doses of appropriate antifungal agents are given (40, 160).
(ii) Morphological characteristics that set candidiasis apart.
In tissues Candida organisms appear as mats of yeasts measuring 3 to 5 μm in diameter intermingled with pseudohyphae (also referred to as filaments) (Fig. 2) (94). The filaments may show periodic constrictions. The organisms can be seen with H&E, GMS, and PAS stains. The predominant Candida species that does not produce filaments is C. glabrata. Histopathologic examination of specimens is very important to define invasion of tissues and vessels, since growth from skin, lung, and the gastrointestinal or genitourinary tract is only indicative of colonization (40).
Fig. 2.
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Morphology, description, diagnosis, and comment for fungal infections that present with hyphae or pseudohyphae in tissues. All photographs are of Grocott methenamine silver (GMS)-stained specimens except for the inset in the fourth row, which is a Fontana-Masson stain. The brown color observed in the fungal element is melanin. For each type of infection, alternative testing and correlation with culture, epidemiologic, and clinical features are necessary. (The photograph in the fourth row is reprinted from the CDC Public Health Image Library collection.)
The usual host reaction, whether in superficial or invasive candidiasis, consists primarily of neutrophilic inflammation with some lymphocytes and macrophages, fibrin, and coagulative necrosis (94). Giant cells and granulomas can be seen but are sparse. As Candida organisms invade blood vessels, they can cause mycotic aneurysms or thrombophlebitis. Necrotizing vasculitis has been described in candidemia, but organisms are not observed in the necrotic vessels, supporting the concept that Candida soluble fractions cause the necrotizing lesions (141). In patients with neutropenia the necrosis is usually accompanied by hemorrhage, and few lymphocytes and macrophages can be observed. In patients with endocarditis and vegetations, platelets are an important component (20). In gynecologic Pap smears, superficial Candida infections are associated with enlarged, hyperchromatic nuclei with perinuclear halos; these changes can be confused with low-grade, squamous intraepithelial lesions (104).
(iii) Pitfalls in morphological diagnosis.
Candida spp. are yeasts that can produce pseudohyphae, and they thus require differentiation from other yeasts and molds that produce true hyphae in tissue. The most frequent differential diagnosis is with Aspergillus spp. and Trichosporon spp. (94). Elongated Candida pseudohyphae can appear to be branching but are differentiated because pseudohyphae are slender and do not have septations. Germinating Candida blastospores can also appear to be branching but can be distinguished by the absence of a constriction between the base of the blastospore and the germ tube. Histoplasma is the differential diagnosis for C. glabrata, since pseudohyphae will not be produced by these species.
(iv) Alternative testing.
Blood cultures are indispensable for diagnosis in invasive disease, although positive blood cultures are estimated to occur in 50 to 70% of cases. The peptide nucleic acid fluorescent in situ hybridization assay can be used to identify the most common species of Candida in smears made from positive blood culture bottles. Multiplex-tandem PCR detection of Candida spp. in whole blood, serum, or plasma has yielded better and faster results than blood cultures; however, this method is still for research use only and will need to be validated if used for diagnostic purposes (85). Detection of β-d-glucan in serum has been used, but the sensitivity and specificity will vary with the type of patient and the cutoff value used for the test (167). Sensitivity is lowest (57%) when using single specimens from patients with candidemia in the intensive care unit or monitoring once weekly with liver transplant patients, while it is highest (97%) when using a low cutoff value (60 pg/ml). Specificity ranges from 44 to 92% with single specimens from patients with candidemia using high cutoff values (80 pg/ml).
Diseases caused by other fungi and organisms resembling fungi that display yeasts or yeast-like organisms in tissue.
(i) Epidemiologic and clinical situations when yeasts or yeast-like organisms should be considered in the differential diagnosis.
Pneumocystis pneumonia was an unusual infection until 1982, when AIDS appeared in the western world (28). Until the advent and widespread use of highly active antiretroviral treatment (HAART), Pneumocystis pneumonia was frequently diagnosed in patients with AIDS and was associated with high mortality. Although the prevalence of Pneumocystis pneumonia in the United States has dropped, it continues to be a problem in patients with AIDS who have suboptimal access to HIV testing and health care, patients receiving immunosuppressant medications chronically, or those with altered immune status (79, 83). Pneumocystis is an interesting organism that has protozoan and fungal characteristics and has been classified as both at different times in history (112). In addition, the species that is pathogenic to humans (previously named P. carinii) has recently been renamed P. jirovecii. Although the mode of transmission of Pneumocystis is not known, an airborne mechanism is suspected since the majority of infected patients present with pneumonia. Extrapulmonary disease has been documented but is very rare (112).
Sporothrix schenckii is the cause of lymphocutaneous and fixed cutaneous lesions traditionally associated with injuries that occur during handling of contaminated soil, plants, and wood. Transmission in persons who handle sick animals but do not recall having had a skin injury has also been documented, particularly in cases involving children and cats (10). Most of the lesions occur in exposed areas, predominantly in hands, arms, and face, with rare instances of systemic disease affecting bone, joints, meninges, and other internal organs. Erythema nodosum and arthralgias (without infection of the joints) have been described in patients with cutaneous and lymphocutaneous disease. Although S. schenckii is found worldwide, infections are more common in tropical and temperate climates. Outbreaks of this disease in humans and animals in South Africa and Brazil have been described (144).
Of all the Penicillium species, P. marneffei is the most frequent cause of pathology in humans (98). Disseminated disease with this organism is seen primarily in immunosuppressed patients, particularly those with AIDS, who have resided in or visited Southeast Asia (Thailand, Vietnam, and the southern part of China) (48). In some areas it is the third most common opportunistic infection in patients with AIDS, after tuberculosis and cryptococcosis. The disease presents with fever that may be accompanied with chills, weight loss, anemia, cutaneous or subcutaneous lesions, lymphadenopathy, hepatosplenomegaly, respiratory symptoms, and osteoarticular lesions.
Paracoccidioides brasiliensis is the agent of paracoccidioidomycosis, which occurs mostly in Brazil, Colombia, Venezuela, Ecuador, and Argentina, with occasional cases in countries in Central and South America (126). Rare cases have been documented in other countries decades after the patient resided in a country where the disease is endemic, suggesting reactivation of paracoccidioidomycosis that had been acquired at an earlier age. Exposure to P. brasiliensis does not guarantee the development of active disease. Hormonal, genetic, immunologic, and nutritional statuses appear to determine if the patient will present with acute or chronic or with localized or disseminated disease. In adults the disease predominates in males, which has been supported by in vitro studies showing that 17β-estradiol blocks or delays the transition from the mycelial or conidial forms into the pathogenic yeast forms (139). The majority of patients present with lung disease, which can disseminate hematogenously to bone marrow, adrenal glands, brain, and other tissues, or the organisms are expectorated and lesions can form in the oropharyngeal tract. In the lung the disease can present as interstitial infiltrates or cavities. Less frequently, P. brasiliensis is inoculated directly into the skin or oral mucosa, causing lesions in these areas. If there is lymphadenopathy, the necrotic contents may drain into the skin.
R. seeberi (described above) is another organism that cannot be cultured and for which animal models do not exist. It has been placed phylogenetically at the divergence between fungal and animal organisms (6). The organism causes rhinosporidiosis, a disease characterized by polyploid tumors that occur primarily in the nose, nasopharynx, and ocular areas (27). Rarely the disease can disseminate to other mucous membranes or cutaneous sites as well as viscera. Rhinosporidiosis occurs in hot tropical climates in all continents, and cases have been reported from Uganda, Sri Lanka, India, Brazil, Argentina, Texas, and other locations.
Another unusual fungus that can be present in histopathologic specimens is Emmonsia crescens. This dimorphic fungus is inhaled or inoculated into the skin when humans are in close contact with rodent burrows and aerosolized conidia. The amount of conidia inoculated will determine the presentation: small inocula are asymptomatic and present as walled-off granulomas, while larger inocula can result in acute severe pulmonary disease. The disease is known as adiaspiromycosis or haplomycosis, due to the presence in tissue of adiaspores, which are large, thick-walled structures formed by transformation of hyphae. Emmonsia pasteuriana has also been reported in a patient with AIDS (47); this species does not produce adiaspores in tissue.
(ii) Morphological characteristics.
In tissue sections stained with H&E, Pneumocystis pneumonia presents as foamy intra-alveolar eosinophilic exudates with minimal inflammatory infiltrate. In Papanicolaou-stained respiratory cytology specimens, the organisms blend into the mucous blue-green background. GMS staining demonstrates that the foamy material in tissue sections or cytologic specimens corresponds to multiple organisms which are thin-walled spheres of 2 to 5 μm that have an intracystic focus (capsular dot) (Fig. 3) (64). Collapsed organisms are usually found intermingled with intact organisms. Atypical inflammatory reactions to Pneumocystis have been documented, including interstitial pulmonary fibrosis, granulomas, hyaline membranes, and interstitial lung infiltrates (64). In these instances, it is difficult to suspect that Pneumocystis is present in the lesion.
S. schenckii in tissues appears as round, oval, or cigar-shaped yeasts of 2 to 6 μm or larger in diameter that may show narrow-based or tube-like budding (34). S. schenckii yeasts are not easy to identify with H&E stains, and thus GMS and PAS stains should be used to highlight their contour. In cases of sporotrichosis, star-like, eosinophilic material (Splendore-Höeppli phenomenon) surrounding yeasts can be observed in 40 to 92% of cases (24, 138). These structures have been called asteroid bodies, and Sporothrix has been demonstrated in the centers of these structures using immunohistochemistry (138). The infection is usually present in a background of granulomatous inflammation with neutrophils and microabscesses and various degrees of fibrosis. The epidermis in cutaneous lesions shows pseudoepitheliomatous hyperplasia and microabscesses. Yeasts are most abundant in the microabscesses. In epidermal microabscesses, the yeasts may be mixed with hyphae. S. schenckii has also been identified using fine-needle aspiration of lesions (57).
P. marneffei yeasts are small (2 to 5 μm in diameter) and do not bud but divide by fission, thus resembling a sausage with a transverse septum (Fig. 3) (98, 168). These transverse septa will appear to be thicker than the wall of the yeast when using the GMS stain. P. marneffei yeasts are usually found inside macrophages and thus appear in clusters with occasional single extracellular organisms. Immunosuppressed patients show an anergic necrotizing host response with abundant infected macrophages accompanied by various degrees of necrosis; however, in patients where immunity is less compromised, granulomatous and abscess formation have been described. P. marneffei has also been identified using fine-needle aspiration of a lymph node (89).
P. brasiliensis in tissue sections stained with H&E are spherical yeasts that vary in size from 4 to 60 μm and have an optically clear space between the fungus and surrounding tissue (92). The pathognomonic “pilot wheel” created by multiple buds surrounding the parent cell is highlighted with GMS but may be difficult to observe with H&E stains (Fig. 3). Not all yeasts show buds around the entire circumference. The yeasts are usually found inside multinucleated giant cells. The host reaction ranges from mixed granulomatous and neutrophilic inflammation to necrotic granulomas surrounded by fibrosis. The epidermis in skin lesions shows pseudoepitheliomatous hyperplasia. In the bone marrow there may be coagulative necrosis with osteonecrosis (129).
Diagnosis of rhinosporidiosis can be performed only using histopathology. R. seeberi presents as large (50- to 100-mm) round structures that can be seen with the naked eye as yellowish pinhead-sized spots in the polyp (27). Microscopically, these structures vary in size, corresponding to different stages in the development of the organism, and have a densely eosinophilic wall that either encloses smaller round structures or can be empty and containing amorphous eosinophilic material. Microscopic features of this organism are enhanced by using GMS, PAS, and mucicarmine stains (164). The host response is predominantly granulomatous inflammation admixed with fibrosis and granulation tissue. The epidermis and mucosal epithelium show hyperplastic features, and R. seeberi can be seen in these cellular layers.
On H&E staining, adiaspores are the primary feature of pulmonary adiaspiromycosis. Adiaspores are thick, double-walled spherules that measure 20 to 400 μm or more and are empty or contain eosinophilic hyaline globules (162). When stained with GMS, the entire wall thickness stains and shows fenestrations. Adiaspores usually elicit a granulomatous inflammatory reaction.
(iii) Pitfalls in morphological diagnosis.
The differential diagnosis of Pneumocystis includes Histoplasma. In the usual clinico-pathologic setting, the intracystic body of Pneumocystis is the key to differentiate this organism from Histoplasma. However, Pneumocystis may be difficult to differentiate from Histoplasma when present inside granulomas or when there is extrapulmonary disease.
S. schenckii should be differentiated from Candida (the presence of pseudohyphae may be useful unless it is C. glabrata), Histoplasma (finding clusters of organisms may indicate histoplasmosis), and Leishmania (by finding kinetoplasts). In addition, Hamazaki-Wesenberg bodies (pigmented elliptical structures seen in sarcoidosis) have been confused with S. schenckii (34). The presence of the Splendore-Höeppli phenomenon is an important distinctive feature of sporotrichosis.
P. marneffei should also be differentiated from Candida, Histoplasma, Toxoplasma gondii, and Leishmania, and this can be achieved by finding the characteristic P. marneffei transverse septum that occurs during fission of the organism.
Paracoccidioidomycosis may not show the characteristic multiple budding but may show only one or two buds. Since C. neoformans, S. schenckii and Lacazia loboi may have two and three buds, these fungi should be considered in the differential diagnosis. L. loboi is the cause of a nodular skin infection that histopathologically forms chains of cells connected to one another by a tubular structure in a granulomatous background (121). The diagnosis of paracoccidioidomycosis should be considered only when yeasts are variable in size and multiple teardrop and tubular buds are seen coming from a parent cell. Because of the variability in size, many yeasts can be confused with P. brasiliensis if organisms are present in low numbers and the characteristic “pilot wheel” is not seen.
Morphologically, coccidioidomycosis and rhinosporidiosis show either a spherule (Coccidioides) or a sporangium (Rhinosporidium) containing distinct endospore-like structures; however, R. seeberi sporangia are larger and mucicarmine positive, and the clinical setting is different from that of coccidioidomycosis (27).
The differential diagnosis for adiaspiromycosis in tissues has usually been helminthic infections, since there are no other fungi with these histologic characteristics (42). Helminths will show musculature and internal organs that are not present in adiaspiromycosis. Adiaspores may collapse and form various shapes that resemble other fungi, helminths, or pollen grains. Adiaspores are distinguished from Coccidioides spherules in that spherules contain endospores while adiaspores are empty.
(iv) Alternative testing.
Pneumocystis cannot be recovered in culture, and thus alternative testing is usually performed using direct immunofluorescence. Although β-d-glucan in serum cross-reacts with Candida and Aspergillus organisms, it has been shown to have good sensitivity for detection of Pneumocystis (83). In addition to cultures for detection of S. schenckii, a skin test (sporothricin) has been used in areas of endemicity outside the United States. P. marneffei can be cultured from blood, skin lesions, or aspirates from lymph nodes. Serologic testing specific for P. marneffei has been studied in Southeast Asia (48). Serologic tests detecting IgG antibodies against P. brasiliensis are available in countries where it is endemic and are used for diagnosis and to follow up treatment(123). Skin tests using paracoccidioidin also have been used for epidemiologic studies (75). Alternative testing for rhinosporidiosis does not exist. Molecular diagnosis with fresh tissues has been used for several of these infections, using either specific primers for these fungi or panfungal primers (23, 45, 83).
Diseases Where Hyphae Are Usually Seen in Tissue
Aspergillosis.
(i) Epidemiologic and clinical situations when aspergillosis should be considered in the differential diagnosis.
The genus Aspergillus encompasses a large number of molds that reproduce asexually by producing unbranched chains of conidia from a bulbous structure called a vesicle (13). Aspergillus spp. are ubiquitous in the environment and have been used for centuries to ferment rice to produce sake or soybeans to produce soy sauce. In industry, Aspergillus niger is used to produce citric acid and numerous commercial enzymes. The genus has evolved to survive in a variety of habitats from damp basements and decaying vegetation to colonization of mammals. Aspergillus fumigatus is the species most frequently associated with human disease, but other species, including those that are important for industry (A. niger), can cause disease in immunosuppressed hosts. Besides the host immune status, genetic predisposition to disease with Aspergillus spp. is an important component.
Aspergillosis encompasses three entities: allergic bronchopulmonary aspergillosis (ABPA), chronic pulmonary aspergillosis/aspergilloma, and invasive or systemic aspergillosis (13, 136). ABPA is an exaggerated hypersensitivity reaction to a variety of fungi, most frequently A. fumigatus, in cystic fibrosis and steroid-dependent asthma patients. These patients may be asymptomatic with infiltrates on chest X rays or have symptoms of bronchiectasis with copious production of sputum that contains brown specks or hemoptysis in addition to wheezing, fatigue, weight loss, and chest pain. Diagnostic criteria for the different stages have been established for the disease (2, 59, 119). However, all have in common the recognition of specific immunity toward Aspergillus antigens, which are not standardized tests (136). Once the patient has reached the fibrotic stage, the disease is not reversible; thus, diagnosis at earlier stages is important. Related to ABPA is allergic fungal rhinosinusitis, a hypersensitivity reaction to noninvasive fungal elements, particularly of the Aspergillus spp. (29, 145). Both ABPA and allergic fungal rhinosinusitis show allergic mucin with noninvasive hyphae, respiratory atopy, positive skin tests to the etiologic fungal organism, elevated total IgE, peripheral eosinophilia, association with major histocompatibility complex (MHC) class II antigens, and a favorable response to steroids. In addition, ABPA shows elevated fungus-specific IgE and IgG antibodies.
Chronic pulmonary aspergillosis usually occurs in nonimmunocompromised individuals with a variety of predisposing conditions (152). Lung cavities caused by tuberculosis are the most frequently described predisposing condition; however, emphysema, sarcoidosis, bronchiectasis, ankylosing spondylitis, and other infections have been described. As the lung tissue is destroyed by the fungus, a cavity forms or enlarges and a fungus ball or aspergilloma can be produced. From the pathologic perspective, two different entities with similar signs and symptoms have been described: thin-walled aspergillomas and chronic cavitary/necrotizing pulmonary aspergillosis (136). In patients with chronic necrotizing pulmonary aspergillosis, further immunosuppressive conditions can be encountered, such as diabetes, HIV infection, advanced age, chronic steroid use, malnutrition, or alcohol abuse (155). These entities evolve slowly over several months to years. Hemoptysis is the most common sign and the one associated with life-threatening events. Some aspergillomas are asymptomatic and found only when an X ray is performed for other reasons, or the patient may have cough, fever, weight loss, and malaise. Patients with chronic pulmonary aspergillosis have serologic or microbiologic evidence that an Aspergillus sp. is involved in the process.
Invasive pulmonary aspergillosis is a disease that occurs in severely immunocompromised patients, including patients with prolonged neutropenia, hematopoietic stem cell and solid organ transplant recipients, patients with AIDS, premature newborns, and patients with chronic granulomatous disease (65, 148). The disease most frequently involves the respiratory tract, and the signs and symptoms include fever, cough, dyspnea, and hemoptysis. Pleuritic chest pain may be present if there is lung infarction. In a computed tomography (CT) scan, there is a lung nodule with the halo sign (ground glass opacity surrounding the nodule) or with the air crescent sign that occurs when there is fungal vascular invasion and hemorrhage. Neurological signs of stroke or seizures indicate that the fungus has reached the central nervous system. An international consensus defining proven, probable, and possible invasive fungal infections for research purposes was published in 2002 and revised in 2008 (7, 43). It is important to notice that the highest level of certainty for proven invasive fungal infections includes demonstration of fungal elements in diseased tissue obtained by either a biopsy sample or fine-needle aspiration of the lesion; however, the procedures to obtain these specimens may not be possible because of the host's underlying condition. In those cases, clinicians will have to rely on alternative testing.
(ii) Morphological characteristics that set aspergillosis apart.
Aspergillus spp. are usually described as thin (3- to 12-μm), septate, acute-angle (45°) or dichotomous branching hyphae (Fig. 2). Vesicles with conidia can be observed when the fungi are present in cavitary lesions or sinuses. However, it needs to be remembered that multiple Aspergillus species may cause aspergillosis, and this description may not reflect the characteristics of all species or changes that may occur after antifungal treatment. For example, with A. niger infection, calcium oxalate crystals may be found in respiratory specimens. A. terreus is the only Aspergillus species known to produce round or pear-shaped aleurioconidia directly along the lateral hyphal walls.
In ABPA, the lung appears to have thick, tenacious mucous material in normal, fibrotic, or bronchioectatic airways. Microscopically, the mucus contains inflammatory cells (mostly eosinophils, lymphocytes, and macrophages), Curshmann's spirals (desquamated epithelium with eosinophils), Charcot-Leyden crystals, and scant hyphae. The bronchial wall may show a spectrum of changes that include inflammation with eosinophils, neutrophils, and macrophages, granulomas, vasculitis, interstitial fibrosis, and microabscesses (136). The presence of microabscesses with hyphae could represent an early stage of invasive aspergillosis. The diagnosis of allergic fungal rhinosinusitis is one of exclusion and is based on histopathologic demonstration in mucosa obtained during surgery of scattered hyphae in the inspissated allergic mucus (mucus showing laminated concretions of pyknotic and degranulated eosinophils and Charcot-Leyden crystals) with no fungal invasion or necrosis (145). Some authors describe scattered, mild granulomatous inflammation in the mucosa of the sinuses (41).
The fungus ball in chronic pulmonary aspergillosis consists of hyphae enmeshed in necrotic material. In thin-walled aspergillomas the reaction surrounding the fungus ball consists of fibrosis, while in chronic cavitary/necrotizing pulmonary aspergillosis there is a necrotic tissue layer with abundant hyphae surrounded by granulation tissue and an outer layer of fibrosis (155). Additional reactions in the cavity wall may include granulomas, eosinophils with formation of the Splendore-Höeppli phenomenon around hyphae, calcium oxalate crystalloids, and hemorrhage or hemosiderin-laden macrophages (170). Differentiation of cases with chronic cavitary/necrotizing pulmonary aspergillosis from those with invasive aspergillosis may be difficult, since vessels in the wall of the cavity may show hyphal invasion and various degrees of thrombosis. Aspergillomas are usually parenchymal lesions, but if the aspergilloma arises in a bronchiectasis, it will be bronchocentric.
In tissue from neutropenic patients with invasive pulmonary aspergillosis, angioinvasion is demonstrated by the presence of septate, right-angle-branching hyphae in the vessel wall and hemorrhage (154). The lesions result in either wedge-shaped pulmonary infarctions or a well-circumscribed spherical nodule with a vessel in the center (65). Because of the patient's neutropenia, these lesions show little inflammatory reaction.
(iii) Pitfalls in morphological diagnosis.
A study of 122 specimens showed concordance in 83% of cases with septated, acute-angle-branching hyphae in histology and the presence of Aspergillus spp. in culture, while Scedosporium spp., Fusarium spp., Pseudallescheria spp., Phialophora verrucosa, and Trichophyton spp. were recovered in culture from discordant cases (86). Cultures of the allergic inspissated mucus of patients with allergic fungal rhinosinusitis have yielded multiple species of Aspergillus, Bipolaris spicifera, and Curvularia lunata as well as Staphylococcus aureus. For these patients nasal swabs are not useful for diagnosis (145). Fontana-Masson stains may be useful to demonstrate pigments in dematiaceous organisms (Bipolaris and Curvularia).
Cases with dual infection with Aspergillus spp. and Candida or mucoraceous genera have been described and pose important diagnostic dilemmas. To be able to identify dual infections, it is crucial to have alternative diagnostic testing of tissues such as immunohistochemistry, in situ hybridization, or PCR, which at this time are available only for research use (70, 147). In addition, in invasive pulmonary aspergillosis, cultures are positive in only 50% of bronchoalveolar lavage fluid specimens (148), and organisms recovered from BAL fluid samples may reflect colonization rather than the actual pathogen. Lastly, detection of Aspergillus spp. in blood cultures in cases with invasive disease is approximately 5%.
(iv) Alternative testing.
Galactomannan and (1→3)-β-d-glucan are components of the Aspergillus cell wall and can be measured using commercially available EIAs. Galactomannan can be measured in serum and bronchoalveolar lavage fluid specimens using the Platelia Aspergillus test (Bio-Rad, Hercules, CA). Different studies of this test have shown sensitivities that range from 40 to 100% and specificities of from 56 to 100%, depending on the population of patients tested and the cutoff values used (62, 97, 167). The most important problem with galactomannan testing is that false-positive results occur in approximately 50% of patients taking antibiotics (piperacillin, amoxicillin, or ticarcillin), 100% of patients receiving substances that contain products of A. niger fermentation (plasmalyte), and various percentages of patients with infections with other fungi, including Penicillium, Paecilomyces, Alternaria, and Histoplasma. Serial galactomannan testing of sera from patients at risk is one strategy to define whether the patient is colonized, has invasive disease, or is responding appropriately to treatment.
The (1→3)-β-d-glucan is a characteristic fungal cell wall constituent common to a broad range of fungal pathogens. The commercially available assay is the Fungitell kit (Associates of Cape Cod, East Falmouth, MA) (114). Circulating (1→3)-β-d-glucan was detected in patients with systemic fungal infections, including invasive aspergillosis, candidemia, and Pneumocystis pneumonia (3, 120). False-positive reactions are known to occur in some patients who are receiving piperacillin/tazobactam (101).
Mucormycosis (zygomycosis).
(i) Epidemiologic and clinical situations when mucormycosis should be considered in the differential diagnosis.
Even though diseases caused by ribbon-like, pauciseptate, hyaline molds were originally described in the 1800s, the nomenclature of these molds has not been completely settled (132). The names that have been given to these molds in the medical literature include Zygomycota (currently regarded as invalid) and Mucorales. It is currently accepted that the broader subphylum Mucoromycotina has two orders, the Mucorales and the Entomophthorales. Fungi classified as Entomophthorales were originally identified as parasites or pathogens of insects that occasionally cause mucocutaneous disease in immunocompetent human hosts. Conversely, those fungi classified as Mucorales cause a spectrum of predominantly angioinvasive disease in immunosuppressed patients (109). Of the Mucorales, Rhizopus is the genus most frequently causing human disease, while Mucor spp. cause disease in fewer than 20% of cases.
Mucorales genera are ubiquitous in the environment and can be found in soil and decomposing matter (132). Their spores are easily airborne, which can cause contamination of laboratory media; thus, finding these molds in clinical cultures has to be correlated with the patient's history and clinical findings to define whether the cultured mold should be considered the cause of disease. Inhaled spores cause disease in the upper and lower respiratory tracts of immunosuppressed persons. In immunocompetent or immunosuppressed hosts, spores can be inoculated into the skin and subcutaneous tissues by trauma, needle exposure, or insect bites, and they can be ingested, causing gastrointestinal disease. In immunosuppressed patients, cutaneous lesions have been linked to adhesive hospital products such as the tape used for maintaining intravenous devices or tubing in place.
Host defense against the Mucorales genera is primarily through macrophages that inhibit germination of spores and neutrophils that use the oxidative burst to kill proliferating hyphal elements; thus, patients who have diseases affecting the function of these two cell types will be at risk for infection (132). Diabetic ketoacidosis causes dysfunction of macrophages and is the most frequent risk factor for sinusitis and rhinocerebral infection. As expected, cancer chemotherapy and stem cell transplantation have emerged in the past 2 decades as major risk factors for invasive pulmonary mucormycoses. In addition, these fungi thrive when iron is present in the host, and those patients receiving iron-chelating agents such as deferoxamine to reduce iron overload are also at risk. Other risk factors include prematurity and injection drug use. If host defenses are poor, the spores germinate in the original inoculation site and invade tissues, including blood vessels. During the initial phases of the infection there is edema, but as the hyphae invade blood vessels the tissue undergoes necrosis and has a characteristic black color. In contrast, immunocompetent individuals infected with Entomophthorales produce an intense inflammatory response and present with a mass in the skin, respiratory sinuses, or gastrointestinal tract.
The three most frequent primary clinical manifestations of mucormycosis are rhinocerebral, pulmonary, and cutaneous infections. Any of these primary manifestations can give rise to disseminated disease. Patients with rhinocerebral mucormycosis have fever and nasal discharge. Those with pulmonary mucormycosis have fever, multiple pulmonary nodules, and pleural effusions. Mortality due to disseminated disease is extremely high but varies depending on the associated risk factor and clinical presentation (137). Early detection at the primary site is imperative to institute surgical and antifungal treatment. For diagnosis, tissue should be obtained for both culture and histopathology.
(ii) Morphological characteristics that set mucormycosis apart.
Mucorales genera can be seen in biopsy and cytologic (needle aspirate and bronchoalveolar lavage fluid) specimens. Tissue identification of these molds is a very important diagnostic tool, since it distinguishes the presence of the fungus as a pathogen in the specimen from a culture contaminant and is indispensable to define whether there is blood vessel invasion. Mucorales genera produce nonpigmented, wide (5- to 20-μm), thin-walled, ribbon-like hyphae with few septations (pauciseptate) and right-angle branching (132). The hyphae may vary in width, appear folded or crinkled, and be sparse or fragmented. In lesions exposed to air, thick-walled spherical structures can form at the ends of the hyphae (35). Routine H&E stains may show only the cell wall with no structure inside. In cytologic specimens the hyphae will be highlighted with Papanicolaou and calcofluor white stains. On occasion the hyphae are very degenerate, and many of the characteristics may not be appreciated in the specimen; however, the pathologist should mention that degenerate hyphal elements are observed in the specimen, since this identifies the source where the fungus is found and will rule out contamination if there are questions about contamination in the culture. Stains that can help highlight the fungal wall include GMS and PAS stains, although fragmentation and necrosis of the fungal elements may cause these stains, in particular GMS, to be either faintly positive or negative (Fig. 2).
In immunosuppressed hosts, the hyphal elements will be found with abundant necrosis, hemorrhage, and blood vessel thrombosis (12). Important diagnostic features include identification of fungal elements invading the blood vessel wall or inside their lumen. Sparse neutrophilic inflammation can be found in the periphery of the lesion. Entomophthorales infections in immunocompetent hosts are accompanied by intense granulomatous inflammation with abundant neutrophils and eosinophils, fibrosis, and granulation tissue (111). The Splendore-Höeppli phenomenon in these cases can be prominent. Sporulation can occasionally be observed.
(iii) Pitfalls in morphological diagnosis.
The major morphological differentiation between Mucorales genera and other molds is with other fungi that produce nonpigmented hyphae in tissue, including Aspergillus spp., other hyaline septated molds (such as Fusarium and Scedosporium), and Candida spp. (132). The presence of abundant septation and acute-angle branching should suggest the diagnosis of Aspergillus spp. or another hyaline septate mold, while yeasts with pseudohyphae should suggest Candida spp. Poor staining of hyphae with GMS should suggest mucormycosis. To be able to specifically identify Mucorales in tissues or to detect dual infections by Mucorales genera and other fungi, it is important to use immunohistochemistry, in situ hybridization, or PCR (70, 147).
(iv) Alternative testing.
Culture of the tissue specimen is indispensable for organism-specific diagnosis. Attention to using gentle processing, is important since aggressive grinding of the tissue may render the fragile fungal elements nonviable (132). Mucorales genera are fast-growing fungi, but unfortunately, the yield of cultures is low. Although serologic tests have been attempted, they are not recommended.
Diseases caused by Fusarium, Scedosporium, and other hyaline septated molds.
(i) Epidemiologic and clinical situations when hyaline septated molds should be considered in the differential diagnosis.
Infections caused by colorless septate fungi are referred to as hyalohyphomycoses (109). The major organisms included in this category are Fusarium spp., Scedosporium spp., Trichoderma spp., Paecilomyces spp., Scopulariopsis spp., Acremonium spp., Schizophyllum spp., Phaeoacremonium spp., and Trichosporon spp. These organisms are ubiquitous in nature, leading to frequent mucocutaneous and inhalational exposure, and the distinction between colonization and infection may be difficult to assess in immunosuppressed patients with positive cultures. Although histopathology has limitations, it remains crucial because it provides information regarding tissue invasion. Neutropenia is an important risk factor for invasive disease caused by all of the hyaline molds (39, 113).
Fusarium spp. can cause superficial infections such as keratitis, onychomycosis, sinusitis, and cutaneous infections in normal hosts (113). It has been recovered frequently from wounds and skin of burned patients and has been described in association with contaminated water supplies. Invasive infection, particularly in the lung, has been seen in patients with hematologic malignancies, prolonged neutropenia, and graft-versus-host disease and is second in frequency to infections by Aspergillus spp. in these patients. Skin lesions that result from disseminated disease are painful, pruritic nodules that evolve to have central necrosis, giving a characteristic “bull's eye” appearance, and eventually ulcerate to produce ecthyma gangrenosum-like lesions. Dissemination can occur to all organs, including the heart (26). Fusarium spp. are frequently recovered from blood cultures in patients with disseminated disease, and Fusarium solani is the most common species encountered. In addition, Fusarium has the ability to produce toxins, and mycotoxicosis can occur in animals and humans when they ingest food contaminated with a toxin-producing species (113).
Scedosporium spp., and for some species their sexual state Pseudallescheria spp., are also ubiquitous in soil and water and can cause disease through inhalation of conidia or direct inoculation of mucous membranes and skin (39). These organisms can cause a range of diseases similar to those observed with Aspergillus, from allergic responses in sinuses and lungs and colonization of lung cavities with formation of fungus balls to invasive pulmonary and disseminated organ involvement. Disseminated disease caused by Scedosporium spp. is seen primarily in patients with AIDS, primary immunodeficiencies, and hematologic malignancies and patients who have received transplants and/or corticosteroids. However, pneumonia or brain abscesses can be observed in normal hosts and should be particularly considered after near drowning in polluted water. In addition, individuals who participate in agricultural activities may present with white-grain mycetomas caused by Scedosporium spp. that involve skin, soft tissues, and bone.
Lastly, Trichoderma spp. have been observed in patients undergoing dialysis in addition to those with the typical causes of immunosuppression (109). Paecilomyces spp. are a cause of keratitis and endophthalmitis (109). Organisms of both of these genera as well as Scopulariopsis spp., Acremonium spp., Schizophyllum spp., Phaeoacremonium spp., and Trichosporon spp. can cause invasive diseases with clinical presentations indistinguishable from those caused by Aspergillus.
(ii) Morphological characteristics that set hyaline molds apart.
In tissue the features of hyaline septated molds are similar to those seen with aspergillosis (113). The fungi invade vessels and cause thrombosis and necrosis of the surrounding tissues. Microscopically the hyphae are septated, show acute-angle branching, and are not pigmented (Fig. 2). The contours of the hyphae and presence of septa are highlighted by using PAS and GMS stains. Occasionally, the hyphae of hyaline fungi swell and appear globose.
Fusarium and Scedosporium may show prominently constricted hyphae with varicosities and intercalated chlamydoconidia (163). Fusarium spp. can sporulate in tissues, and thus there may be a combination of yeast-like structures and hyphae in histopathologic preparations (113). Scedosporium spp. rarely sporulate in tissues but can do so in patients with fungus ball cavities; in this case the ovoid conidia are pigmented, correlating with the melanin-like pigment of the colonies (39).
(iii) Pitfalls in morphological diagnosis.
These organisms cannot be distinguished from each other or from Aspergillus spp. Thus, the presence of nonpigmented, septate hyphae with acute-angle branching in tissue should not be signed out as “compatible or suggestive of Aspergillus,” since there are important treatment implications (67). These organisms should be signed out descriptively, and a comment stating that the fungal elements could correspond to Aspergillus spp., Fusarium spp., Scedosporium spp., Trichoderma spp., Paecilomyces spp., Scopulariopsis spp., Acremonium spp., Schizophyllum spp., Phaeoacremonium spp., or Trichosporon spp. should be added (Fig. 2). When the hyphae of these fungi appear globose, the differential diagnosis is with fungi in the Mucorales genera. In these instances, the width of the hyphae can be assessed at the areas of septation, since these areas are less affected by the swelling.
(iv) Alternative testing.
Blood cultures for Fusarium have a high yield compared to those for Aspergillus because of adventitious sporulation in tissue. However, caution has to be exercised when these fungi are cultured, since these molds are ubiquitous in the environment and frequently cause contamination or colonization. The (1→3)-β-d-glucan test is positive in these infections and cannot help distinguish them from other invasive fungal infections, including candidiasis (113). If culture is not available, galactomannan testing should be performed to try to differentiate hyaline molds from Aspergillus spp. It is important to make organism-specific diagnoses of these molds because some Fusarium species and Scedosporium frequently demonstrate resistance to azoles and echinocandins, thus requiring combination therapy (36, 113).
Diseases caused by Bipolaris/Curvularia and other dematiaceous fungi.
(i) Epidemiologic and clinical situations when dematiaceous fungi should be considered in the differential diagnosis.
Dematiaceous fungi are naturally pigmented molds whose hyphae and conidia contain melanin. Dematiaceous fungi are ubiquitous in the environment and can be found in soil, plants, and organic debris worldwide. More than 60 genera have been implicated in human disease (130). Melanin is considered an important virulence factor, since disruption of pigment production by Exophiala dermatitidis leads to reduced virulence in animal models (44).
Dematiaceous molds primarily cause skin and soft tissue infections that are preceded by trauma or environmental exposure, and they can give rise to three clinical entities: eumycetoma, chromoblastomycosis, and phaeohyphomycosis (54, 109). The name “chromo” comes from the Greek root meaning color, while “phaeo” also comes from the Greek and means dusky, with both names indicating the color of the lesion or the pigment production by the causative organism. Table 3 presents the differences among these three clinical entities. In addition, dematiaceous fungi can cause a variety of other clinical syndromes, including onychomycosis, keratitis, allergic disease, pneumonia, brain abscesses, and disseminated disease. In patients with respiratory syndromes, the infection is acquired by inhalation of conidia. The dematiaceous fungi most frequently associated with eosinophilia and allergic sinusitis or allergic bronchopulmonary mycosis are Bipolaris and Curvularia (130). The clinico-pathologic characteristics of allergic diseases are similar to those described for Aspergillus. Pneumonia, brain abscesses, and disseminated disease are frequently seen in immunocompromised patients such as HIV-infected patients, those who have received a transplant or are receiving corticosteroids, and patients with cancer, diabetes, or chronic diseases (11). However, brain abscesses have also been described in immunocompetent hosts (131). It is postulated that brain abscess and disseminated disease are secondary to a primary pneumonic focus rather than skin disease. Cladophialophora bantiana is the dematiaceous fungus most frequently associated with brain abscesses (130). Mortality is low for patients with superficial diseases, but once the disease becomes disseminated, mortality increases to approximately 80%.
Table 3.
Clinical presentation of dematiaceous molds
| Condition | Clinical presentation | Tissues involved | General agents that can cause the entity | Molds most frequently associated with clinical presentation | Other molds associated |
:---: :---: :---: |
| Eumycetoma | Painless mass in extremities with multiple sinuses that drain pus and grains | Skin, soft tissues, fascia, bone | Bacteria, hyaline and pigmented molds | Madurella spp. | Scedosporium, Acremonium, Exophiala, Aspergillus, and others |
| Chromoblastomycosis | Tumor-like lesions of skin (nodules, verruca, scar, plaques) | Skin, underlying soft tissues | Pigmented molds | Fonsecaea pedrosoi, Cladophialophora carrionii | Phialophora verrucosa, Rhinocladiella spp., Exophiala spp., and others |
| Phaeohyphomycosis | Immunocompetent individuals present with single lesion (cyst, plaque) in the area of inoculation; in immunosuppressed patients the inoculation site may be a nodule, eschar, ulcer, or other and may disseminate to other organs | Skin, soft tissue, sinuses, lungs, brain | Pigmented molds | Exophiala jeanselmei, Exophiala dermatitidis, Bipolaris spp. | Alternaria, Curvularia, Exserohilum |
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(ii) Morphological characteristics that set dematiaceous fungi apart.
Diagnosis of these entities requires histopathologic examination of the tissues and culture. Although biopsy specimens are best for histology and culture, scrapings of the lesions can also be used. The biopsy specimens should be obtained from areas with pigment, as these are sites where fungi can be found. All dematiaceous fungi show pigmented hyphae; however, the degree of pigmentation will vary. The pigmented round cells characteristic of chromoblastomycosis, which are also called “copper penny lesions” or sclerotic bodies, may show internal septations. In cases where pigmentation is not evident with H&E stains, Fontana-Masson staining may be necessary to demonstrate the melanin pigment (54). The hyphae tend to be thin (2 to 6 μm wide) but are irregularly swollen (also referred to as toruloid or moniliform) with prominent septa that show constrictions (Fig. 2) (33, 135). The frequency of branching will depend on the fungus. The hyphae may show terminal or intercalated vesicular swellings with thick walls resembling chlamydoconidia. Pigmented yeast-like cells can also be seen and may show septation and budding. In addition, GMS and PAS stains can be used to highlight the fungal wall. The fungal elements are usually seen in multinucleated giant cells or in the extracellular necrotic material. Specific morphological features have been described for the three distinct skin and soft tissue entities.
Mycetomas are characterized by the production of grains. Dematiaceous fungi produce black grains while other fungi (usually Scedosporium and Acremonium) or bacteria produce white grains. Histologically, grains are interwoven mycelial aggregates that are lined with intense eosinophilic material (Splendore-Höeppli phenomenon) (30). The grains are localized in dermal abscesses and sinus tracts that connect the abscess to the skin surface. The sinus tracts are surrounded by granulation tissue and granulomatous inflammation.
Chromoblastomycosis forms characteristic sclerotic bodies in tissue or muriform cells. These are darkly pigmented, thick-walled, round to polyhedral fungal elements that have undergone septation in multiple planes (124). They are usually found in the dermis or epidermis, including the keratin layer, and are surrounded by connective tissue and neutrophilic inflammation. Fungal elements are found singly or in clusters. In chromoblastomycosis, the epidermis usually shows an intense reaction with pseudoepitheliomatous hyperplasia and hyperkeratosis. The inflammatory infiltrate is primarily granulomatous with microabscesses.
The phaeohyphomycotic cyst presents histopathologically as a single dermal lesion with minimal changes in the epidermis (33). The cyst wall consists of dense collagenous tissue and granulomatous inflammation with abundant giant cells. In the center of the cyst there are foci of geographic necrosis and foreign bodies, presumed splinters that carried the infection. Fungal elements (yeast-like structures and septated hyphae) can be found throughout the lesion. When phaeohyphomycosis becomes disseminated in patients with cancer, fungal elements can be observed in 83% of specimens (11). The inflammatory reaction in disseminated disease is usually nonspecific (neutrophils and mononuclear cells), with only one-fourth of the specimens showing necrosis and even less commonly granulomas.
(iii) Pitfalls in morphological diagnosis.
The different dematiaceous fungi cannot be distinguished from one another by histology. Some dematiaceous fungi may show very little melanin and may appear as hyaline hyphae; thus, a Fontana-Masson stain is needed to highlight the pigment. However, caution should be exercised when interpreting Fontana-Masson staining, since many Aspergillus spp., some Mucorales genera, and Trichosporon can also show positive staining (80).
(iv) Alternative testing.
Cultures should be obtained to have an organism-specific diagnosis. Because dematiaceous fungi are ubiquitous in the environment, contamination of cultures occurs frequently. A study in a cancer center showed that only 11% of 348 specimens from which dematiaceous fungi were cultured could be associated with cases of proven or probable invasive fungal disease (11). Thus, finding the fungal elements by histopathology is indispensable for diagnosis. Serodiagnostic tests are not available for these infections.
Dermatophyte disease.
(i) Epidemiologic and clinical situations when dermatophytes should be considered in the differential diagnosis.
Superficial mycoses are believed to affect 20 to 25% of the world population (4). Dermatophytes are the predominant cause, with three main anamorphic (asexual or imperfect) genera infecting humans: Epidermophyton, Trichophyton, and Microsporum (165). Dermatophytes are found worldwide, usually associated with keratinous material that serves as the source for human and animal infections. Some species have more restricted geographic locations; for example, Trichophyton soudanense is primarily found in central Africa, while Trichophyton rubrum, Trichophyton mentagrophytes, Microsporum canis, and Epidermophyton floccosum are distributed worldwide. Migration of populations has resulted in changes in the distribution of the different dermatophytes. In addition, there has been a decline of these infections with improvements in living conditions and the use of antifungal agents.
Dermatophytes are acquired through contact of an individual with conidia present in soil, animals, or objects (e.g., combs, shoes, or clothing) (100). Certain conditions that allow longer contact between conidia and skin (spaces between nails) and have the appropriate humidity and temperature (interdigital areas or hair) are more prone to harbor these infections. These infections are designated tinea or ringworm, followed by the affected body site. Examples include tinea unguium or onychomycosis (in nails), tinea cruris (in the groin), tinea corporis, tinea circinata, or tinea glabrosa (in the extremities), tinea pedis (athlete's foot), tinea capitis or tonsurans (in the scalp) and tinea barbae (in the beard). All tineas have a similar clinical pattern, showing a ring of scaling inflammatory skin that is accompanied by burning and itching. If there is an intense reaction in hair follicles, the disease is known as Majocchi's granuloma.
Dermatophytes are fungi that invade the keratin of immunocompetent and immunosuppressed hosts by virtue of their keratinolytic proteases (161). These fungi express carbohydrate-specific adhesins that allow attachment to epithelial cells, and they produce multiple serine and metallo-endoproteases that allow digestion of the keratin network. Dermatophytes cause chronic skin disease since they adapt and are not rapidly eliminated by the host immune response. Trichophyton cell wall mannans are capable of inhibiting lymphoproliferative responses in vitro and may also be responsible for inhibiting turnover of the stratum corneum. The host response is primarily delayed-type hypersensitivity and varies in degree depending on the host immune status and the infecting dermatophyte. In addition, there is an association between dermatophyte infections and allergy, particularly asthma.
Besides dermatophytes, other fungi can cause superficial skin and hair infections (8). The clinico-pathologic presentations and the responsible fungi are presented in Table 4. The two most important groups are Malassezia spp. and Candida spp. Malassezia infections cause tinea versicolor and are particularly associated with application of oils and lotions, use of corticosteroids, exposure to sunlight, hydrosis, and possibly seborrheic dermatitis (100). In patients with AIDS, superficial Candida infections are common and include cheilitis (painful fissures in the lip commissures) and a variety of mucocutaneous lesions, including atrophic plaques, pseudomembranous lesions, or leukoplasia (125).
Table 4.
Clinical presentation of dermatophytes
| Name of superficial infection | Clinical presentation | Extension to hair follicle | Fungus(i) | Systemic disease | KOH preparations | Morphology in tissue sections |
:---: :---: :---:
| Tinea or ringworm, followed by the location in the body | Round lesions with scaly border, accompanied by pruritus and burning | Yes; when suppurative known as kerion, when chronic known as Majocchi's granuloma | Dermatophytes (Epidermophyton spp., Trichophytum spp., Microsporum spp.) | Very rare but can invade the dermis and soft tissues, causing mycetomas | Hyphae with or without septations | Hyphae cannot be visualized in the keratin with H&E, special stains are needed |
| Tinea versicolor | Hypo and hyperpigmentation in patients with oily and sweaty skin, fine scales when scratching | Yes, known as Pityrosporum folliculits | Malassezia spp. | Systemic infections may occur in premature neonates receiving parenteral nutrition and in other immunosuppressed hosts | Yeasts and hyphae (“spaghetti and meat balls”) | Faintly basophilic hyphae in the stratum corneum |
| Tinea nigra | Brown to black macule, usually in palms, with some scaling | No | Phaeoannellomyces werneckii | Not described | Darkly pigmented, septated, and branching hyphae | Pigmented hyphae in the stratum corneum |
| White piedra | Creamy-white, small, soft nodules in hair shafts | No | Trichosporon spp. | Immunosuppressed patients may have lung infiltrates, renal involvement, and fungemia | Septate hyphae perpendicular to hair shaft | Not used for diagnosis |
| Black piedra | Hard dark nodules in hair shafts | No | Piedraia hortae | Not described | Collections of crescent ascospores surrounded by pigmented hyphae | Not used for diagnosis |
| Superficial candidiasis | Intertrigo, chronic paronychia, onychodystrophy, cheilitis | Yes | Candida spp. | Yes, particularly in patients with AIDS and depending on the level of immunosuppression | Yeasts, pseudohyphae may be observed | Fungal elements may be seen through the biopsy, vascular invasion must be determined |
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(ii) Morphological characteristics that set dermatomycosis apart.
To detect dermatophytes in the keratin skin layer, it is necessary to use GMS or PAS stain since these are hyaline fungi that are difficult to observe in the keratin layer using H&E (66). Hyphae and aleurioconidia can be visualized and are particularly prominent in hair follicles. The host reaction to the fungus is very variable. In the keratin layer there may be mild hyperkeratosis with focal parakeratosis. In acute lesions the epidermis shows spongiosis and neutrophilic microabscesses. Lastly, the dermis shows various degrees of perivascular lymphocytes and plasma cells, and in some cases there may be prominent papillary dermal edema (71). When a dermatophyte causes severe inflammation of hair follicles and shafts, it is called kerion if the infiltrate is primarily neutrophilic or Majocchi's granuloma if chronic with prominent mononuclear inflammation. In both instances giant cells can be present and frequently contain fragments of hyphal elements. On rare occasions dermatophytes can invade the epidermis and dermis, producing nodular lesions that resemble mycetomas.
(iii) Pitfalls in morphological diagnosis.
Superficial fungal infections caused by Candida spp. and Malassezia spp. should be considered in the differential diagnosis of dermatophytes. In comparison with dermatophytes, these organisms tend to stain basophilic with H&E. In the case of Candida spp., it is indispensable to make sure that the infection is superficial and no vascular invasion is present. If dermal or vascular invasion is observed, it is important to rule out skin manifestations of systemic fungemias.
(iv) Alternative testing.
Lesion scrapings can be clarified with potassium hydroxide (KOH), and then either viewed under a light microscope or stained with calcofluor white and viewed under a fluorescence microscope. This will identify the presence of fungal elements but will not determine which fungus is present. Cultures are important to differentiate between dermatophytic disease and superficial skin infections caused by other fungi or yeasts. Dermatophyte test medium (DTM) changes color with production of alkaline metabolites produced by dermatophytes, potentially allowing faster identification than when using traditional fungal culture media. The diagnosis should be established before starting treatment because of the length, cost, and potential side effects of the drugs used. In addition, knowing the dermatophyte species may help establish preventive measures with pets or other possible sources of reinfection. PCR testing of skin scrapings and other specimens has been used to identify dermatophytes and other fungi that can cause onychomycosis and superficial skin infections; however, PCR testing has not been approved in the United States for diagnostic purposes (76).
Diseases caused by other molds (coelomycetes).
(i) Epidemiologic and clinical situations when coelomycetes should be considered in the differential diagnosis.
Fungi that in the asexual stage produce conidia inside a fruiting body-containing cavity lined by either host or fungal tissue are known as coelomycetes. The fruiting body can have a variety of shapes (spherical or disc-like), and the cavity opens to the environment through an ostiole or pore (156). In general these fungi infect plants and thus are found ubiquitously in the environment. Humans acquire these organisms during contact with infected plants or soil through cuts or punctures in the skin. Coelomycetes cause superficial skin infections, onychomycosis, and keratitis/endophthalmitis and can occasionally become systemic in immunocompromised hosts (51). All these fungi are pigmented and cause cutaneous and subcutaneous phaeohyphomycosis and black-grain mycetomas. A number of fungi are considered to be in this group; Phoma spp., Colletotrichum spp., and Nattrassia mangiferae are some examples.
(ii) Morphological characteristics that set coelomycetes apart.
It is important to perform histopathologic studies in these cases, since coelomycetes are ubiquitous and culture contamination may occur. The fungal elements are quite pleomorphic and can be moniliform, bead-like yeasts to short branched or unbranched hyphae. Some tissues can display yeast-like elements, depending on the plane in which the tissue was cut. The fungal elements show pigment.
(iii) Alternative testing.
These fungi can be difficult to identify because they do not produce reproductive structures in tissue although they show moderate to rapid growth. Thus, panfungal PCR amplification with sequencing of the product has been used to identify cultured isolates (51).
TECHNIQUES THAT ARE USED FOR IDENTIFICATION IN HISTOPATHOLOGIC MATERIAL
Histochemical Staining
Methods.
Histopathologic examination of tissues to detect fungi is and will remain an important tool to define the diagnostic significance of positive culture isolates, including fungal invasion of tissues and vessels as well as the host reaction to the fungus. Histopathology can also provide rapid presumptive diagnosis of the fungus while waiting for fungal culture results, or it may provide the only available material when no culture growth occurs or cultures were not ordered. Histopathologic examination of biopsy specimens, surgical resection specimens, and autopsy material should always start with H&E staining of the tissue. GMS and PAS staining should be performed if a fungus is suspected after review of tissue sections because of presence of an inflammatory tissue response or when there is high clinical suspicion even if the H&E stain is unrevealing. In addition, mucin (mucicarmine) or melanin (Fontana-Masson) stains can be extremely useful for identification of Cryptococcus (mucin and melanin) and dematiaceous fungi that may not produce abundant pigment (melanin). Table 5 presents the different stains and alternative methods for fungi that can be used in tissue sections. It is important to remember that some fungal infections, particularly during the acute phases, give rise to neutrophilic or suppurative inflammation, so epidemiologic circumstances and clinical history may also prompt the need to use these special stains. In immunosuppressed patients there may be no inflammation but necrosis may be present because of fungal invasion of blood vessels and associated thrombosis. Fungal invasion of blood vessels weakens the wall and can result in hemorrhage. Communication between clinicians and pathologists is invaluable to define the cases in which GMS, PAS, Fontana-Masson, or mucicarmine stains may be useful to highlight fungi. As the use of less invasive procedures has become more prevalent in medicine, cytologic specimens have become common samples. Bronchoalveolar lavage fluid is a frequent specimen used to diagnose pulmonary infections (81). Other cytologic specimens include sputum, cerebrospinal fluid, and fine-needle aspirates of lesions. Most fungi can be visualized with the routine stains used for cytologic preparations, including Papanicolaou and Giemsa stains; however, the wall of the fungi is not highlighted with either stain. In order to stain the fungal wall, GMS and PAS stains can be used in cytologic specimens. Calcofluor white, an alternative stain that highlights the fungal elements, can be used with fresh cytologic specimens if a fluorescence microscope is available; alternatively, potassium hydroxide wet preparations can be studied using bright-field or phase-contrast microscopy. Bronchoalveolar lavage and cerebrospinal fluid specimens can also be used for detection of fungal antigens.
Table 5.
Stains and alternative methods that can be used with tissue sections
| Stain(s) | Application(s) | Color and fungi stained |
:---:
| Hematoxylin and eosin (H&E) | Used routinely in pathology to demonstrate tissue morphology; in the case of fungal infections, this stain helps identify the inflammatory host reaction, such as multinucleated giant cells, necrotic material, hemorrhage, and the Splendore-Höeppli phenomenon; most fungi can be observed with this stain, particularly the nuclei of yeast-like cells or if the fungus is naturally pigmented (however, the fungal elements may be difficult to distinguish from the background) | All fungi show pink cytoplasm, blue nuclei, no color for the wall |
| Fungal silver stains (Grocott and Gomori methenamine silver [GMS]) | They highlight the wall of the fungus and thus are useful for screening the tissue sample, can be combined with H&E in such a way that the fungus and the host reaction can be clearly observed | The fungal cell wall appears black or dark brown for all fungi, the surrounding tissue is usually green, the Mucorales may stain very pale |
| Periodic acid-Schiff (PAS) | Detects glycogen in tissues, fungal walls contain large amounts of glycogen and thus PAS can be used for screening for fungal organisms | The fungal cell wall appears pink to red purple; depending on the counterstain used, the nuclei can be blue |
| Gridley fungus | Stains the walls of most fungi | The fungal cell wall appears purplish red, the background is usually yellow |
| Mucine stains (Mayer's or Southgate mucicarmine, Alcian blue) | Stains mucopolysaccharides, including the capsules of a variety of organisms; also stains mucus, which can be present in a variety of human cells | Useful to highlight the capsules of cryptococci, which appear red or blue depending on the stain used |
| Melanin stains (Fontana-Masson) | Stains melanin present in some fungi; also stains melanin present in human tissues, such as in the skin epidermis | Cryptococcus and the dematiaceous fungi will take a black to dark brown color |
| Bacterial stains (tissue Gram stains or acid-fast stains) | Many fungi take bacterial stains; additionally, some filamentous bacteria (actinomycetes and Nocardia) have to be differentiated from fungi | Candida spp. stain purple/blue (Gram positive) with the Gram stain; Blastomyces and Histoplasma can be acid-fast staining (take a red color) |
| Immunohistochemistry (IHC) | Uses antibodies against the different fungi, antibodies can be monoclonal or polyclonal, assays are not FDA approved and require validation by each laboratory | Assays for Blastomyces, Cryptococcus, Histoplasma, Coccidioides, Pneumocystis, Sporothrix, Paracoccidioides, Penicillium, Candida, Aspergillus, and Mucorales have been published in the literature; depending on the colorimetric developer, the fungi can stain dark brown or red; the counterstain is usually hematoxylin; cross-reactivity of the antibodies is a concern |
| In situ hybridization (ISH) | Uses molecular probes for different fungi, probes are generally ribosomal since there are multiple copies of ribosomal genes in each fungal cell, assays are not FDA approved and require validation by each laboratory | Probes published in the literature include those for Blastomyces, Cryptococcus, Histoplasma, Coccidioides, Pneumocystis, Sporothrix, Candida, Aspergillus, Mucorales, Fusarium, and Pseudallescheria; depending on the colorimetric developer, the fungi can stain dark brown or red; sensitivity and specificity of the probes vary |
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Retrospective studies that correlate culture results with histopathology and cytology showed that the overall accuracy for microscopic morphological techniques can vary from 20 to 80% (140, 143, 158). The lowest correlation has been reported for invasive septate molds (158). Even though GMS and PAS stains were used more frequently in the cases correctly diagnosed than those misclassified, special stains did not significantly improve pathologists' diagnostic capabilities (140). Misclassification of cases occurs when the pathologist has a false sense of his or her ability to categorize fungal organisms by genus based on microscopic morphology alone, when inappropriate terminology is used such that other potential molds within a particular category are not included in the differential diagnosis, or when there is a lack of knowledge of morphological mimics of yeasts and hyphal forms. The misclassifications with greater potential for adverse consequences occurred when there were few, folded, fragmented, and/or necrotic fungal elements in the specimen and the structures could not be adequately categorized as septate versus pauciseptate hyaline molds. Therefore, clinicians need to be aware that misclassifications in histopathologic examination occur in at least 20% of cases, and pathologists need to give as much information as possible without overextending their diagnostic capabilities (140, 143).
In order to avoid misclassifications, pathologists should describe the fungal elements observed in the sample and refrain from trying to offer a specific diagnosis. Pathologists need to remember that there are very few instances where morphological characteristics are specific. Some groups have suggested the use of templates or synoptic reporting for the diagnosis of fungal infections. In Fig. 1, 2, and 3 we have suggested templates for reporting on histopathologic specimens according to the morphological characteristics found. It is important to recognize that the diagnosis is primarily descriptive of the fungus and should include whether or not there is invasion of the tissues and vessels, the amount of fungal elements observed, and the host reaction to the infection (inflammation, necrosis, or hemorrhage). The comment section of the report should clearly state the fungi most frequently associated with that morphology as well as other possible organisms (fungi and parasites) that should be considered in the differential diagnosis. All pathology reports should also include a statement in the comment section regarding the importance of correlating clinico-epidemiologic features and results of cultures and other laboratory tests.
Advantages.
Histopathology is indispensable in some instances to define whether an organism recovered in culture represents contamination, colonization or true infection. Tissue and vascular invasion and necrosis are important histopathologic features that can help make the distinction.
Fungal cultures may not always grow; thus, histopathologic presumptive diagnosis may be the only evidence of a fungal infection. For example, Mucorales genera are molds that usually grow within 24 to 48 h; however, if the original specimen is ground too aggressively, the hyphal elements may be destroyed and the culture may not grow (132). In immunocompetent individuals with a chronic solitary nodule caused by an endemic mycosis, yeasts may be observed in tissue sections but the culture may not grow because the yeasts are nonviable (166). Poor recovery of Aspergillus spp. and other septate hyaline molds in cultures has been attributed to previous treatment, use of prophylactic antifungal medications, or possible differences in the physiologic states of the mold in vivo versus in vitro (158). Another situation that can result in loss of ability to recover an organism in culture even when a viable organism was initially present in the clinical material is when a small quantity of biopsy material is sent to the microbiology laboratory with a request for multiple cultures, such as for aerobic bacteria, anaerobic bacteria, acid-fast bacilli (AFB), and fungi, all of which require inoculation of multiple media and slide preparations. Sometimes this small amount of material must be even further split between microbiology and histopathology. Lastly, certain fungi, such as Pneumocystis, do not grow using current microbiology practices and require detection using histopathologic or cytologic techniques (140).
In some instances, fungal cultures may take weeks to be completed, and thus a preliminary histopathologic diagnosis of a fungal infection may be ready before the culture and may provide sufficient information for clinicians to start treatment. For example, Scedosporium, Sporothrix, Blastomyces, and Coccidioides may take up to 3 to 4 days to grow and Histoplasma and Paracoccidioides may take more than 2 weeks; thus, the histopathologic diagnosis could be available earlier than culture results (140).
Disadvantages.
In the previous sections we have discussed the pitfalls for each fungus in the comparison of its morphology with those of other fungi or parasites. In addition to these pitfalls, pathologists must differentiate possible fungal structures from stained normal human tissue structures. Particularly when using GMS stains, normal tissue structures that can be confused with yeasts include neurosecretory granules and melanin, while hyphae require differentiation from collagen fibers, basement membranes, and other silver-staining filamentous structures. Figure 4 shows neurosecretory granules, which compared to any of the yeasts tend to be irregular, smaller, and located inside neurosecretory cells, while basement membranes and collagen fibers tend to show lower staining intensity and less sharpness than hyphae. To aid in the recognition of these different tissue structures and fungi, it is important to colocalize the GMS-stained structures in tissue sections (usually from a consecutive level) stained with H&E or PAS stain. The combination of GMS counterstained with H&E, instead of Light Green, is a way to achieve the colocalization in one slide. In addition, pathologists should assess the presence or absence of internal structures that can be observed in fungi (nuclei and cytoplasm) which stain with H&E but not with GMS.
Fig. 4.
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Diagnostic pitfalls. When using GMS stains, normal tissue structures can appear as yeasts or hyphae. (A) Neurosecretory granules (arrow). (B) Collagen fibers, with one even showing a “pseudoseptum” (arrow). (C) In specimens with few organisms, hyphae cut transversally can appear as yeasts that may have “pseudobudding” (arrows).
Another disadvantageous occurrence that can happen during interpretation of special stains is when the sample shows some transversally cut hyphae, which then appear as yeasts that may even appear to be budding (Fig. 4). In these cases it is important to go deeper into the block to see if more fungal elements cut longitudinally are present in the sample.
Histopathology usually cannot provide the fungal genus and species, which are very important for treatment. For example, a case with hyaline septate hyphae includes in the differential Aspergillus spp., Fusarium, Scedosporium, and others. Thus, treatment should be with voriconazole, which is effective for all these fungi; however, treating with or adding amphotericin B should be done only for A. fumigatus and F. solani. Itraconazole or echinocandins could be used for A. fumigatus but would not be effective for F. solani or Scedosporium (140). Detection of yeasts with pseudohyphae suggests Candida spp. with a differential diagnosis that includes septate hyaline molds; thus, fluconazole could be used to cover most Candida spp. but would lack activity against C. krusei, C. glabrata, or the other septated hyaline molds.
Infections with more than one fungus have been reported in burn and immunosuppressed patients (70, 143). Even though histopathology might be expected to be an adequate method to identify these double infections, the morphological diversity may be subtle and not appreciated. Thus, other tests should be used to determine if more than one organism is present.
In immunosuppressed patients, isolation of different fungi during short periods of time may raise the question of whether there are either different fungal infections in different locations, subsequent infections by different fungi, or an infection with two different fungi in which one of the organisms had intrinsic resistance to the treatment given. In Fig. 5 we present an example of a neutropenic patient with chronic lymphocytic leukemia who was diagnosed first as having an invasive mold (by culture A. fumigatus) in the lung and then 3 days later with invasive fungal rhinosinusitis (by culture Fusarium spp.) and died 22 days later with disseminated fungal infection (suspected mucormycosis by immunohistochemistry). The sequence of photographs stained with GMS shows the usual difficulty in distinguishing hyaline septated molds from those that are pauciseptated by using histopathology, particularly in the second and third samples, where morphological characteristics could have been altered because of previous treatment of this patient with amphotericin B and voriconazole.
Fig. 5.
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Sequential specimens stained with GMS (magnification, ×120) showing mold infections in a neutropenic patient with chronic lymphocytic leukemia. (A and B) Hyaline septated hyphae in the lung. The culture was positive for Aspergillus fumigatus. (C) Hyphae (by culture a Fusarium sp.) in a nasal debridement sample obtained in the same patient 3 days after the lung biopsy. The morphology of fungal elements could be confused with mucormycosis, since there are few septations and the hyphae twist and turn. (D) Hyaline pauciseptated hyphae in a lung specimen obtained at autopsy 22 days after the nasal debridement. The specimen stained positive using immunohistochemistry for mucormycosis. The morphology of the hyphae is distorted, probably due to previous antifungal treatment.
Immunohistochemistry
Methods.
Immunohistochemistry refers to the use of antibodies to detect targets (fungal antigens) in tissue sections so that the morphology of the target as well as the surrounding tissues can be seen. For these assays it is necessary to cut a very thin section of the tissue (similar to that used for histopathology) and place it on a glass slide. The tissue can be fresh frozen or paraffin embedded. Tissue that has been embedded is usually formalin fixed, which can cause distortion of antigens, particularly if fixation has been lengthy. In general, pathologists prefer to use formalin-fixed, paraffin-embedded (FFPE) tissues since this is the routine histopathologic procedure that renders the material noninfectious and easily stored at room temperature. For performing the assay, paraffin needs to be removed from the tissue, which is usually done with reagents that dehydrate the tissue. The tissue needs to be rehydrated and either treated with enzymes or subjected to other antigen retrieval techniques that will make fungal antigens available for the antibodies, and then the antifungal antibodies are applied. In some methods the primary antibody has been enzyme or fluorescence labeled, while other procedures call for a secondary labeled antibody whose target is usually the Fc portion of the primary antibody. Once the labeled antibodies have been incubated for an appropriate amount of time, the tissue sections are washed to remove unbound antibodies. If fluorescent labels are used, the tissue section can be visualized with a fluorescence microscope. Although fluorescent antibodies have been used for fungi in specialized centers, this labeling method does not permit visualization of surrounding tissues and the preparations are not permanent. In the case of enzyme labels such as peroxidase, the color needs to be developed with the appropriate substrates and the tissue then counterstained, enabling visualization of the fungus and the surrounding tissues.
The choice of the primary antibody is very important to define the best targets. Antibodies used in other methods such as immunodiffusion or complement fixation have been tested in immunohistochemical assays, since there are few commercially available antibodies specifically validated for immunohistochemistry in FFPE tissues (128). Immunohistochemical reagents that detect Aspergillus spp. and mucormycetes in tissue are commercially available (AbD Serotec). It is very important to validate these assays, starting with extensive assessment of cross-reactivities of the primary antibodies with cultures and tissues that have been treated in a manner similar to that for the unknown specimens that will be tested (73). The widespread presence of common antigens in fungi has resulted in very few clinically useful specific primary antibodies. For example, several studies using a variety of monoclonal and polyclonal anti-Aspergillus antibodies have shown a broad range of cross-reactivities with other hyaline septated molds, mucormycetes, and some yeasts (122, 128, 146).
Studies of immunohistochemical assays for fungi have reported uniformly distributed staining of the fungal organism (128), although we have noted that nonviable hyphae may not show staining when these assays are used, particularly in tissues with mucormycosis. In addition, antigen staining has been noted outside the fungal organisms, similar to what has been well documented in immunohistochemical assays for bacterial infections (149). To be able to adequately interpret immunohistochemical assays, it is imperative to use appropriate antibody controls (an irrelevant antibody of the same type [monoclonal versus polyclonal]) in a sequential patient tissue slide so as to define the amount of nonspecific staining present for each case (46).
Advantages.
In theory, immunohistochemistry for fungi has many advantages, including the combination of morphology (the fungal element itself, its localization in the tissue, and the inflammatory reaction) with specific detection of the organism using specimens that are routinely processed in pathology laboratories to render them noninfectious. Multiple automation platforms are commercially available for immunohistochemical assays, reducing the cost and turnaround time. Lastly, enzyme-labeled antibodies result in a permanent record of the reaction. As new non-cross-reactive antibodies are developed and tested using this technique, immunohistochemistry may provide an inexpensive and rapid alternative to more costly assays that do not combine morphology with detection of the specific fungus. In addition, double-staining immunohistochemical assays will permit the simultaneous detection of more than one fungus in tissue sections (70, 73, 143).
Disadvantages.
At this time, many of the available antibodies cross-react with multiple fungi and cannot be used for detection of specific organisms. Evaluation, verification, and validation of the antibodies and immunohistochemical assays must be performed in the laboratory before results can be used for patient care purposes. Since antibodies are analyte-specific reagents (ASR), verification and validation in the United States must follow federal regulations (25).
In Situ Hybridization
Methods.
In situ hybridization refers to the use of probes to detect the presence of specific fungal nucleic acids while preserving the tissue morphology so that the morphology of the fungus and the tissue reaction to the organism can be visualized. For these assays, a thin tissue section is placed on a slide and the hybridization is performed directly on the slide. Similarly to immunohistochemistry, the tissue can be either frozen or FFPE. If the tissue is paraffin embedded, preparation of the tissue sections before hybridization includes deparaffinization and rehydration. After rehydration, proteins bound to nucleic acids should be digested using enzymes (pepsin or proteinase K). Tissues are then prehybridized with a solution that contains formamide, salmon sperm DNA, and yeast RNA to decrease the amount of nonspecific binding of the probe to DNA and RNA in the tissues. The tissue DNA is then denatured and hybridized with the probe of choice. After the hybridization, excess or unbound probe is washed using different concentrations of standard saline citrate. The probe is then detected in a variety of ways, depending on how the probe was labeled. Lastly, the tissues are counterstained.
Most probes used to detect fungi using in situ hybridization assays have been unique, organism-specific rRNA (18S, 28S, or 5.8S) targets for various molds and yeasts (67–69, 82, 103). rRNA is distributed through the organism in large amounts, providing ample opportunity for hybridization. rRNA probes have shown strong signals when hybridized to their homologous target. For Aspergillus, another target that has also been used is alkaline proteinase (108). Choosing the size of the probe is important for penetration of tissues and can have a significant impact on signal intensity (61). Probes of between 400 and 750 bp seem to have the best penetration. Most frequently probes have been labeled with digoxigenin, and an immunohistochemical assay using antidigoxigenin is used for detection. Probes with over 200 bp have more digoxigenin labels, and thus their signal is stronger. Probes have also been designed with locked nucleic acids, which are modified nucleotides linked through a methylene unit, or peptide nucleic acid-labeled probes that provide uncharged, neutral backbones have been used (103, 110). Both the locked and peptide nucleic acid probes improve hybridization and help reduce the time of detection, making in situ hybridization assays much more rapid to complete (3- to 4-h assays). Various systems, such as catalyzed reported deposition, have been used to enhance detection (68, 69).
The probes that have been used are for the most part genus specific. Probes for Blastomyces, Coccidioides, Cryptococcus, Sporothrix, Pneumocystis, Candida, Fusarium, and Pseudallescheria have shown strong signals with good analytic specificity (67–69, 82). On the other hand, probes for Histoplasma showed very low sensitivity when the organism was demonstrated in an area of necrosis (69). Also, some Aspergillus probes have shown cross-reactivity with other septate hyaline molds, particularly Fusarium (67). In some assays, probes for Mucorales genera have been difficult to interpret because of high background and low signal intensity, in addition to uneven distribution of the signal in the hyphal elements (68).
In most published reports, in situ hybridization is performed in cases where fungal elements can be first demonstrated with routine histopathologic stains, since GMS and PAS stains have shown better sensitivity than hybridization assays as a screening method for detection of fungal elements (67–69). In situ hybridization is used to confirm and identify the specific organism present in the tissue. Thus, each case usually requires in situ hybridization with an array of probes designed to identify the fungi in the differential diagnosis. However, hybridization may be uneven, and nonstaining hyphal elements may represent either infection by another fungus for which there is no probe or nonviable portions of the organism. The use of multiple probes in one case can potentially demonstrate the presence of dual infections.
Advantages.
In situ hybridization offers the highest degree of specificity compared to histochemical stains and immunohistochemistry (the other morphology-based techniques). As with histochemistry and immunohistochemistry, in situ hybridization assays performed with FFPE tissues have the advantage of using noninfectious specimens that are routinely processed in pathology laboratories. Automated platforms for in situ hybridization already exist; however, costs are higher and the turnaround time is usually longer than for immunohistochemical assays. At this time, differentiation of Fusarium from other septate hyaline molds is very important clinically and has been demonstrated using in situ hybridization (67, 103). Although probes for all fungi are not yet commercially available, probes that are used in other settings, such as fluorescent in situ hybridization for Cryptococcus and Candida spp. in blood cultures and CSF, could potentially be validated for in situ hybridization platforms for FFPE tissue (91, 95).
Disadvantages.
In situ hybridization is not a screening method, since GMS and PAS stains are more sensitive for detection of fungi (67–69). Once a yeast or mold has been detected in tissue using histochemistry, a panel of probes should be used to define the genus present. The panels should be constructed based on the fungi in the differential diagnosis for that tissue and that patient population. Although more expensive, the use of panels will allow for detection of single and dual infections. At this time, probes for in situ hybridization assays in tissues are not commercially available. Laboratory-developed assays should be evaluated, verified, and validated in the laboratory before results are used for clinical diagnosis and patient care as indicated by federal regulations (25).
PCR-Based Methods
Methods.
The PCR has been used to detect fungal DNA in FFPE tissues. FFPE tissue is not the sample of choice. Fresh nonembedded tissues have shown a sensitivity for PCR detection of fungi of 97%, while the sensitivity of paraffin-embedded material is only 68% (84). The fungal DNA extracted from FFPE specimens can be degraded and in low concentration, and it often contains substances that inhibit protein digestion or DNA amplification. However, when fungal elements are detected in FFPE tissue sections and fungus culture is not available, PCR can in some cases determine the organism that is causing the infection. Depending on the DNA extraction method, the nucleic acid quality, determined as the percentage of samples in which a human housekeeping gene control is recovered, can vary from 60 to 90%, and consequently the PCR efficiency is between 57 and 93% (106). Each tissue sample needs to be tested with a human control DNA to determine the quality of the nucleic acid extracted.
Either of two approaches can be used for analyzing extracted DNA preparations: one is to target sequences specific to a particular organism; the second is to amplify a gene that is present in all fungi (i.e., panfungal) and then sequence the fungal DNA. Genus-specific PCR probes that have been used with FFPE samples include those for Aspergillus, Rhizopus, B. dermatitidis, Coccidioides, H. capsulatum, and P. brasiliensis (14–16, 18, 19, 23). The majority of the published assays target specific rRNA genes (18S, 28S, and 5.8S), the intervening internal transcribed spacer (ITS1 and ITS2), or particular portions of the fungal genome and use nested or seminested PCR methods. The second approach of amplifying a panfungal gene and then sequencing the product is very appealing because this approach is expected to detect a large variety of fungi that can infect humans (84, 133). For this approach, primers should include sequences present in multiple copies within the fungal genome but contain highly variable regions that allow species identification. Primers for the ITS1 and ITS 2 regions have been used for the amplification reaction (106). The PCR product(s) obtained is then visualized, purified, and sequenced. The identification is made by comparing the sequence to those in a sequence database such as GenBank (37).
There is no side-by-side comparison of culture with PCR from FFPE tissues. A prospective study comparing culture to PCR from fresh frozen tissue showed that cultures were positive in 63% of cases while PCR was positive in 96% (134). However, results from retrospective studies have shown that once tissues are formalin fixed, PCR positivity can be as low as 60% (106). Studies that compare all detection methods for fungi, including histopathology, PCR, in situ hybridization, and immunohistochemistry performed with FFPE tissues, are not available. Histopathology still appears to be the best screening tool to define the presence of fungal elements and verify that the fungus is causing disease in the tissue (19); however, to define the specific fungus or fungi present, nucleic acid detection (PCR or in situ hybridization) is probably more specific than immunohistochemistry at this time.
Advantages.
Although morphology is not preserved, the great advantage of PCR in FFPE tissue samples is determination of the specific agent that has been observed by histopathology. With the use of PCR it has become evident that infections with multiple agents are more frequent than previously suspected (70, 84, 106, 133). To detect double infections, it is necessary to use panels of primers for more than one organism.
Disadvantages.
Nucleic acids obtained from FFPE material are frequently damaged (cross-linked) or may contain PCR inhibitors and thus may be unable to generate an adequate PCR product, may not display homology to the primer used, or may generate a product that cannot be sequenced (84). Selection of the DNA extraction method is crucial to obtain the best yield from this material (106). In addition, when fungal elements are scant in tissues, the amount of DNA obtained may be insufficient to perform a PCR assay.
The choice of PCR primer is important. There is insufficient variation in the ITS1 region to differentiate certain species, including C. neoformans, some Candida spp., and Fusarium spp.; thus, analysis of other regions should be considered (37, 84). In addition, false-positive results with specific H. capsulatum primers and difficulties in identifying Coccidioides in formalin-fixed tissues have been reported (15, 17). A Mayo Clinic study of 147 FFPE samples showed that histology found more coccidioidomycosis cases than PCR (19). Lastly, it is estimated that 10 to 20% of the sequences in GenBank are misidentified (106).
PCR assays continue to be labor-intensive and costly. The turnaround time for paraffin material is still approximately 4 to 5 days (for deparaffinization, DNA extraction, PCR, and sequencing). At this time, PCR assays for tissues are not commercially available, so laboratory-developed assays should be evaluated, verified, and validated by the laboratory before results are used for clinical diagnosis and patient care as indicated by federal regulations (25).
Laser Microdissection
Methods.
Laser microdissection combines microscopy with laser technology to enable the study of specific cell types. Once the cells of interest are isolated, a variety of studies/tests can be performed using these cells specifically, and the results are not masked or diluted by surrounding cells or tissue constituents within the tissue sample. Two microdissection technologies are available: laser capture microdissection and laser cutting (107).
Tissue sections (usually thicker than the ones used for histopathology) are placed in special slides that allow for easy separation from the rest of the specimen. The tissue can be fresh frozen or FFPE and can be stained with H&E or by other methods that permit visualization of the desired cells in a bright-field microscope. A narrow-beam laser is used to focus on (for laser capture) or cut around (for laser cutting) the cells of interest. Capture microdissection uses an infrared laser, while cutting microdissection uses a UV laser. The cells that have been targeted or cut out are collected in a plastic cap or tube.
An important aspect of microdissection is that the tissue preparation before the tissue is placed on the slides must be tailored to the secondary test. Preparation includes fixation (or lack thereof) and staining. Exposing the tissues to the smallest amounts of chemicals ensures the least alterations to the sample for subsequent tests. Possible secondary tests include PCR for nucleic acids, electron microscopy for cellular organelles, and mass spectrometry or two-dimensional polyacrylamide gel electrophoresis for proteins (115, 159). If prepared correctly, the microdissected cells can even be cultured.
Until now laser microdissection has been used primarily for diagnosis and research on neoplastic diseases, but several researchers have reported using this technique to study infectious diseases (74, 169). Use of this tool in the ecology of dual fungal infections would enable us to better understand the pathophysiology in these cases. Laser microdissection has been used for identification of a single hypha of A. fumigatus from fresh frozen bird tissues (115). Although DNA sequences were obtained from 60% of the animals, a PCR product was not obtained from the remainder, suggesting that the amount of DNA will depend on the number of hyphae and the number of intact nuclei selected.
Advantages.
The greatest advantage of laser microdissection is that the elements to be tested can be specifically selected. Thus, dual infections and the local environment in which this occurs can be studied in detail. Another advantage is that the material obtained will not be contaminated with nonfungal tissues.
Disadvantages.
Laser capture microdissection instruments are expensive, and many laboratories may not have one or be able to afford them. From the technical perspective, there is the possibility that certain components may not be preserved because the heat in the laser may destroy the elements selected for secondary testing (107). Contamination may occur if a microdissection system requires using particular devices to select the cells, compared to placing the cells in the tube where the second test will take place. The disadvantages that occur for the secondary testing will be the same as those observed in these tests, with the added disadvantage of potentially obtaining very small amounts of material. Another disadvantage that has been reported is that intact fungal nuclei must be obtained for DNA extraction. Particularly with mucormycete organisms that are pauciseptate, hyphal material may be successfully collected but may not contain nuclei.
INTERPRETATION IN DIFFERENT SITUATIONS
Fungal Elements in Tissue but All of the Specimen in Formalin
The situation where fungal elements are in tissue but all of the specimen is in formalin is relatively frequent for several reasons: (i) cancer is the most frequent diagnosis of lesions that are resected or biopsied, and the routine procedure for tissues obtained in operating and endoscopy rooms is to place them in formalin so that morphology for cancer diagnosis is preserved; (ii) not all lesions are studied using frozen sections (iii) when a frozen section is obtained and shows fungal elements, the pathologist may not think it necessary to remind the surgeon that a sterile portion of the lesion should be sent to the microbiology laboratory for cultures; and (iv) communication between infectious disease physicians and surgeons is not always established before the tissue is obtained. Training and better communication among surgeons, interventional radiologists, pathologists, and infectious disease physicians needs to take place so that this situation decreases to a minimum.
However, when fungal elements are seen in tissue but the entire specimen is in formalin, the pathologist should describe the fungal elements, define as part of the diagnosis line whether tissue invasion by the fungi is occurring, and provide a comment that enumerates the fungi that can display the morphological features described (see “Histochemical Staining” above) (Fig. 1, 2, and 3). Identifying the pathology caused by the fungal elements (inflammatory reaction, invasion of vessels, necrosis, or hemorrhage) tells the clinician that the fungi are not contaminating or colonizing the tissue. Descriptive diagnoses of the fungal elements together with a comment listing the fungi with consistent morphology are important to guide treatment (140). For example, the presence of septated, nonpigmented hyphae will result in treatment with voriconazole plus amphotericin B since Fusarium, Aspergillus, and Scedosporium could have that morphology, while a patient whose tissue displays pauciseptate hyphae will receive amphotericin B or posaconazole since this morphology most probably corresponds to the Mucorales genera. If the pathologist observes more than one type of fungal element in the tissue (for example, pauciseptate hyphae and yeasts with pseudohyphae), this should be noted, particularly in view of the increasing numbers of dual infections that are being reported.
In addition, alternative testing that is relevant to the clinico-epidemiologic presentation of the patient and can be performed with nontissue specimens (serology, antigen detection, skin testing, and blood cultures) can be suggested to clinicians by the microbiology laboratory and infectious disease specialist (as presented for each fungal infection in this review). Lastly, immunohistochemistry, in situ hybridization, and PCR with FFPE samples have already been validated for research purposes in some centers, enabling detection of specific fungi when the entire specimen is placed in formalin.
Positive Cultures but Tissue without Fungal Elements
The situation where there are positive cultures but tissue without fungal elements can occur in four instances: (i) the fungus present in the cultures is a colonizer in the patient; (ii) the fungus present in the culture is a contaminant in the laboratory; (iii) tissue is sampled from two different areas, with one sample sent to microbiology and the other to pathology; or (iv) the pathologic specimen has not been extensively studied with adequate special stains. In this situation it is important to know the pathology that was observed in the specimen and the location and number of colonies found in the plates. When only one or two colonies are observed at the area of inoculation of the plate and there is minimal pathology in the tissue sections, it is reasonable to think that the fungus present in the culture may be a colonizer. When only one or two colonies are observed away from the area of primary inoculation of the plate and there is minimal pathology in the tissue sections, it is reasonable to think that there was contamination of the fungal culture. It is also possible that the surgeon sampled two different areas of the tissue and that the one containing viable fungi was sent to microbiology while the second sample, not containing the fungal elements, was sent to pathology. Thus, there will be positive cultures, but the tissue will not contain fungal elements. Lastly, the specimen sent to pathology may not have shown the pathology usually associated with fungal infections and thus the pathologist may not have asked for special stains, or the fungal elements may be so sparse that they could have been missed. In any of these instances it is indispensable to correlate the laboratory findings (pathology and microbiology) with the clinical presentation of the patient and determine if alternative testing is required. For example, if a Fusarium colony was found in the middle of the plate and there was necrosis or hemorrhage in the tissue sample from a severely neutropenic patient, obtaining blood cultures may be a reasonable alternative test since invasive infections with Fusarium can be found in the blood. However, in the same scenario, if the organism cultured had been an Aspergillus sp. or a member of the Mucorales genera, obtaining blood cultures would not be helpful.
In instances where the tissue sections show pathology (inflammation, necrosis, or hemorrhage), it is important that the pathologist, clinician, and microbiology laboratory communicate to define the need to examine deeper in the tissue block and verify that slides were stained with GMS and PAS stains. Any fungus in low abundance may be difficult to identify, and deeper sections stained with GMS and PAS stains may be necessary. In addition, hyphae in patients infected with Mucorales genera may be quite distorted and negative or faintly GMS positive; thus, using PAS stains is indispensable. If the pathology present in the tissues is consistent with allergic fungal disease, finding the fungal elements may be difficult. Appropriate alternative testing may be required, depending on the culture findings, the pathology encountered, and the clinical presentation.
Fungal Elements in Tissue but Cultures without Growth
The situation where there are fungal elements in tissue but cultures without growth can occur in three instances: (i) when the tissue in the microbiology laboratory is ground too aggressively and the fungal cells are destroyed; (ii) when the fungus in the tissue is not viable; or (iii) when tissue is sampled from two different areas, with one sample sent to microbiology and the other to pathology. Mucorales genera are particularly prone to be destroyed with aggressive processing of the tissues. Thus, it has been recommended that the sample not be ground or homogenized but instead that direct plating of larger tissue pieces on fungal media should be routinely performed. If the pathology specimen shows hyphae with few septations compatible with Mucorales genera, positive growth should occur within 3 days, and the microbiology laboratory can then give the clinician a presumptive diagnosis. If growth has not occurred in the expected time, it is possible that there will be no growth. However, because Histoplasma and Paracoccidioides grow slowly in vitro, fungal cultures are usually incubated for at least 4 weeks in most laboratories.
Nonviable fungi in tissue are frequent in chronic, walled-off infections with endemic yeasts, such as in cryptococcosis, histoplasmosis, or coccidioidomycosis. This could also happen if the patient has received antifungal medications. It is also possible that the surgeon sampled two different areas of the tissue and that the one containing fungi was sent to pathology while the second sample, not containing the fungal elements or containing nonviable fungi, was sent to microbiology. Thus, there will be negative cultures but the tissue will contain fungal elements. Regardless of the reason for the lack of growth, fungal elements in tissue causing pathology should be treated as described in the first situation, i.e., where there are fungal elements in tissue but all of the specimen is in formalin.
Discrepancy between Culture Results and Histopathologic Findings
A discrepancy between culture results and histopathologic findings can occur because either (i) the characteristic morphology of the fungus has been altered due to use of antifungal medications or host responses or (ii) there is a dual infection and only one fungus is growing in culture. With more cases being studied using PCR, it has become evident that dual infections are more frequent than previously realized. Although the selection of cases from the CDC study may be biased (due to difficult cases being sent for diagnosis to a reference facility), dual infections may occur in up to 20% of cases (106). In these situations, the pathologic description may refer only to the most abundant fungus present in the specimen, and the second fungus may not have been mentioned. If a second fungus is not mentioned but grows in the culture, the results may appear to be discrepant. Reasons for the recovery of only one fungus in instances of dual infections include too-aggressive processing where larger fungal cells are destroyed but smaller ones survive, the growth of one fungus inhibiting the growth of the second, or inappropriate culture conditions such that only one fungus can be recovered.
The true frequency of dual fungal infections is not well established, and the technology that exists can now permit studies that address this important question. Dual-color immunohistochemistry and in situ hybridization as well as laser microdissection followed by panfungal or specific PCR could be used to address this issue.
Discrepant results can appear problematic but can be solved by reviewing the patient's clinical history to see if the patient has received antifungals or how long the infection has been present and by reviewing with the pathologist the morphological characteristics of the fungal elements present in the slide. Providing descriptions of the fungal elements in the tissue sections for the diagnosis with a comment listing possible organisms that can show that particular morphology should help decrease the number of discrepancies between culture results and histopathology.
Biographies
Jeannette Guarner was brought up in Mexico City, where she obtained her medical degree from LaSalle University. She did her residency and surgical pathology fellowship at Emory University in Atlanta, GA. After residency she returned to Mexico City, where she was the Director of the Clinical Laboratory at the National Cancer Institute. In 1997 she returned to Atlanta and worked at the Centers for Disease Control and Prevention (CDC) in the Infectious Disease Pathology Branch. During her 10-year tenure at CDC, she was involved in the histopathologic study of high-profile outbreaks, including the U.S. anthrax bioterrorism attack, the introduction of West Nile virus to the Americas, and the discovery of severe acute respiratory syndrome (SARS) coronavirus. In 2007 she joined the faculty at Emory University as Associate Professor of Pathology and Laboratory Medicine. She continues to do research on the diagnosis and pathogenesis of infections using histopathologic material as a guest researcher at CDC.
Mary E. Brandt, a native of Philadelphia, PA, received her B.S. at Chestnut Hill College, an M.S. in clinical microbiology at Thomas Jefferson University, and a Ph.D. in microbiology and immunology at Temple University. She joined the Centers for Disease Control and Prevention in 1991 and is now the Chief of the Mycotic Diseases Branch. She directs the CDC Fungus Reference Laboratory and the CDC mold identification training course. She is a mycology section editor of the Manual of Clinical Microbiology as well as an associate editor or editorial board member of several clinical microbiology journals. Her research interests include the detection, identification, and molecular epidemiology of medically important fungi.
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9572 | https://www.sciencebuddies.org/blog/scientific-method-examples-for-teachers | Jump to main content
Blog Posts
Four Ways to Teach the Scientific Method
By Amy Cowen
on July 16, 2025 8:00 AM
Teaching students about the scientific method helps prepare them to do science experiments and to design and execute their own science fair projects to explore science questions. These four free plug-and-play resources help you approach the scientific method with tools and learning aids that suit the needs of your classroom.
Teaching Steps of the Scientific Method
Whether preparing students to do classroom science experiments and science fair projects or reviewing the way science experiments work and how scientists approach testing science questions, educators teach and use the scientific method with K-12 students at all grade levels.
The scientific method offers a set of steps that help students understand the process of asking a question, doing research, forming a hypothesis, and then putting the hypothesis to the test with an experiment. The steps of the scientific method include:
Ask a question.
Do background research.
Construct a hypothesis.
Do an experiment.
Analyze data from the experiment.
Draw conclusions.
Communicate results.
In learning about the scientific method, students learn about independent, dependent, and controlled variables; the importance of doing multiple trials; what to do when an experiment doesn't go as planned; and how to analyze data and evaluate whether or not the hypothesis was supported.
While the steps of the scientific method do not work for testing every question, and scientists may not follow the steps exactly every time, the scientific method does help organize the flow of a science experiment and helps reinforce the relationship between the science question, the hypothesis, the variables, and experimentation. Students can use these steps as a repeatable sequence for organizing and conducting a science fair project or experiment. With a hypothesis statement formulated to indicate the variable that will be changed (independent) in order to see what happens to another variable (dependent), students learn to design and conduct experiments that can lead to meaningful data.
Four Free Resources to Teach the Scientific Method
As you get ready to introduce or review the scientific method, see how these free resources can help you reach students in different ways:
Scientific Method Video
Scientific Method Lesson Plans
Scientific Method Video Lesson
Scientific Method Project Guide
1. Scientific Method Video
The Scientific Method: Steps and Examples video walks students through the scientific method using a sample science project involving making popping boba. In the video, students follow along as a testable science question is selected, background research is conducted, a hypothesis is formed, a procedure is identified, and an experiment is performed.
You might watch the video with students in the classroom and discuss the steps or assign watching the video as homework. (Note: Use the "Assign" button to make assignments for your LMS!)
2. Scientific Method Lesson Plans
Video also available en Español
With our NGSS-aligned lesson plans, educators can teach the scientific method in the classroom using a hands-on activity that helps students put the steps in action as they work through the steps of a project.
There are two Lesson Plans designed for teaching the scientific method to students:
Teaching the Scientific Method with Paper Rockets, grades 3-5
Paper Rockets to Learn the Scientific Method, grades 6-8
Designed for use in elementary and middle school, these lessons help educators lead an interactive, guided activity to teach the scientific method. In both lessons, students explore steps of the scientific method while building and testing paper rockets. Each lesson plan contains grade-level NGSS alignment information, preparatory information for educators, a hands-on activity, discussion prompts, a worksheet, and an assessment tool. To learn more about Science Buddies Lesson Plans, see How to Use a Science Buddies Lesson Plan.
3. Scientific Method Video Lesson
In the Scientific Method: How Do Scientists Make Discoveries? video lesson, Science Buddies teaches students about the scientific method. The video lesson introduces the scientific method and leads students in a hands-on activity to explore the role of animal camouflage in evolution. During the activity, students practice each step of the scientific method including doing background research, making a hypothesis, conducting an experiment, analyzing data, and drawing conclusions. The self-paced video lesson format allows students to follow along, pausing to do steps of the experiment and to fill in the worksheet.
4. Scientific Method Project Guide
The Steps of the Scientific Method guide can be used alone or in combination with the video and lesson plans described above. This resource summarizes the scientific method and contains an interactive diagram that allows students to click on individual steps to review information about the different steps.
This project guide can be assigned using the Google Classroom button. An assessment tool is also provided for Google Classroom educators.
What about the Engineering Design Process?
Like the scientific method, the engineering design process offers a framework that can help guide the design and execution of an engineering project. These methods are similar in some ways, but they differ in significant ways as well.
This video helps explain the differences between the methods:
See also: Comparing the Engineering Design Process and the Scientific Method.
For resources to teach the engineering design process, see 4 Ways to Teach Engineering Design.
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9573 | https://teachy.ai/en/project/high-school/10th-grade/mathematics-en/the-art-of-ap-a-creative-journey-through-arithmetic-progression | Activities of The Art of A.P - A Creative Journey through Arithmetic Progression
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Arithmetic Progression: Sum
Project: The Art of A.P - A Creative Journey through Arithmetic Progression
Lara from Teachy
Subject Mathematics
Mathematics
Source Teachy Original
Teachy Original
Topic Arithmetic Progression: Sum
Arithmetic Progression: Sum
Contextualization
In this project, we will delve into the world of the most basic and widely used numerical sequence in mathematics: Arithmetic Progression (A.P). This sequence is characterized by a common difference between consecutive terms, making its study very intuitive and straightforward.
The simplicity of A.P. compared to other numerical sequences does not diminish its importance. In fact, A.P. is so fundamental that it can be considered the basis upon which many other mathematical concepts are built. For example, A.P. is used to solve series sum problems, in number theory, in statistics, and even in physics, where it can represent, for instance, the speed of an object in uniform motion.
Furthermore, A.P. has practical applications in many aspects of everyday life. If we think of any situation where a value increases or decreases constantly, we are dealing with an A.P. This can include, for example, saving money with a fixed rate, calculating simple interest, predicting the population of a city growing at a constant rate, and so on.
In this project, we will specifically focus on A.P. sum, which is the process of adding all the terms of an A.P. The sum of A.P. is a fundamental skill in mathematics that has practical applications in many fields, from engineering to social sciences. Therefore, understanding A.P. sum and knowing how to calculate it is a valuable skill that we should all master.
To delve deeper into the topic, we suggest the following resources:
Arithmetic Progression (A.P) - Mundo Educação
Arithmetic Progression - Brasil Escola
Arithmetic Progression (A.P) - Khan Academy
Practical Activity
Activity Title: "The Art of A.P - A Creative Journey through Arithmetic Progression"
Project Objective:
The objective of this project is to develop students' understanding of Arithmetic Progression (A.P) and, specifically, A.P. sum. They will use A.P. sum to create an individual or collective work of art, and later, write a detailed report on the process and concepts used.
Detailed Project Description:
Students will be divided into groups of 3 to 5 people. Each group will create a work of art using A.P. sum. This work of art can be anything the students choose (a drawing, a sculpture, a song, etc.), as long as it uses A.P. sum in a meaningful way and is visible in the artwork.
Required Materials:
Paper and pencil for initial sketches and calculations.
Materials for creating the artwork (e.g., colored pencils, clay, musical instruments, etc.).
Internet access for research.
Computer with word processor for writing the report.
Detailed Step-by-Step for Activity Execution:
Group formation and initial discussion on A.P. sum and its possible applications in art.
Research and study on A.P. sum. We recommend using the sources mentioned in the Introduction.
Choosing the type of art to be produced and planning how to incorporate A.P. sum into it.
Creating the artwork. Remember to document the entire creation process, including the challenges faced and the solutions found.
Writing the report. The report should contain four main sections: Introduction, Development, Conclusions, and Bibliography used.
Project Deliverables:
The artwork: the main deliverable is the artwork created by the students. It should illustrate the use of A.P. sum and its practical application.
Report: after completing the practical part, students must produce a written report on the work they have done. The report should include:
Introduction: the relevance of the topic, the application of A.P. sum in the real world and in the specific project.
Development: the theory of A.P. sum, a detailed description of the activity, the methodology used, and the results.
Conclusion: recap of the main points, lessons learned, and conclusions about the project.
Bibliography: the sources used to carry out the project.
Project presentation: each group must present their artwork to the class and explain how A.P. sum was used in its creation. They are encouraged to share any lessons learned or problems encountered during the project.
Students have one week to complete the project. By the end, they should have acquired technical skills in mathematics and also socio-emotional skills, such as time management, communication, problem-solving, creative thinking, and proactivity.
Need materials to present the project topic in class?
On the Teachy platform, you can find a variety of ready-to-use materials on this topic! Games, slides, activities, videos, lesson plans, and much more...
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9574 | https://math.stackexchange.com/questions/2503266/show-n1-n-is-decreasing | sequences and series - Show $n^{1/n}$ is decreasing. - Mathematics Stack Exchange
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Show n 1/n n 1/n is decreasing.
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Asked 7 years, 11 months ago
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This is getting to be a habit! I have done a fair amount of looking around however I cannot comment on other peoples questions yet in order to get further help when I don't understand something. My question is essentially this one Show n 1 n n 1 n is decreasing for n≥3 n≥3.
The answer chosen as the accepted solution helps me see where to begin, but I am unable to see where to go from there. Another thing is that in my question I am asked to show the sequence is decreasing for n≥3 n≥3 starting with
(1+1 n)n≤n(1+1 n)n≤n
for all n≥3 n≥3. So although the linked question provides some help, it doesn't work from this point which is why I'm not sure how to proceed.
I really appreciate the help!
Just want to say a massive thank you to Jorge, ab123 and Luiz for all your help, I don't even want to imagine how frustrating that must have been for you!
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edited Nov 3, 2017 at 18:47
MathsIsFunMathsIsFun
asked Nov 3, 2017 at 18:09
MathsIsFunMathsIsFun
193 2 2 silver badges 9 9 bronze badges
8
you can prove (1+1 n)n<e(1+1 n)n<e Asinomás –Asinomás 2017-11-03 18:11:17 +00:00 Commented Nov 3, 2017 at 18:11
1 See that's all well and good but I dont know how MathsIsFun –MathsIsFun 2017-11-03 18:11:56 +00:00 Commented Nov 3, 2017 at 18:11
@Emily They used the binomial series expansion of (1+x)n(1+x)n, where x=1/n x=1/n ab123 –ab123 2017-11-03 18:14:56 +00:00 Commented Nov 3, 2017 at 18:14
The linked post did or the comment above sorry?MathsIsFun –MathsIsFun 2017-11-03 18:15:41 +00:00 Commented Nov 3, 2017 at 18:15
@Emily the top answer in the linked post ab123 –ab123 2017-11-03 18:16:07 +00:00 Commented Nov 3, 2017 at 18:16
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We want to prove n 1/n≥(n+1)1/(n+1)n 1/n≥(n+1)1/(n+1) for n≥3 n≥3.
This is equivalent to n n+1≥(n+1)n n n+1≥(n+1)n
Which is equivalent to (n+1 n)n≤n(n+1 n)n≤n
Notice (n+1 n)n=(1+1 n)n(n+1 n)n=(1+1 n)n.
So we have to prove (1+1 n)n≤n(1+1 n)n≤n for n≥3 n≥3.
Notice that (1+1 n)n=∑i=0 n(n i)1 n i(1+1 n)n=∑i=0 n(n i)1 n i by newton's theorem.
∑i=0 n(n i)1 n i=∑i=0 n n!i!(n−i)!n i=∑i=0 n n(n−1)…(n−i+1)i!n i≤∑i=0 n 1 i!≤1+∑i=1 n 1 2 i−1≤1+2≤n∑i=0 n(n i)1 n i=∑i=0 n n!i!(n−i)!n i=∑i=0 n n(n−1)…(n−i+1)i!n i≤∑i=0 n 1 i!≤1+∑i=1 n 1 2 i−1≤1+2≤n
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edited Nov 3, 2017 at 18:42
answered Nov 3, 2017 at 18:18
AsinomásAsinomás
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I can follow the logic, however I don't understand how the first three lines follow from one another. Basically I dont understand why the first line is equivalent to the second etc MathsIsFun –MathsIsFun 2017-11-03 18:22:56 +00:00 Commented Nov 3, 2017 at 18:22
Just take n(n+1)t h n(n+1)t h power on both sides of inequation ab123 –ab123 2017-11-03 18:23:48 +00:00 Commented Nov 3, 2017 at 18:23
1 if x x and y y are positive we have x≤y x≤y if and only if x n(n+1)≤y n(n+1)x n(n+1)≤y n(n+1)Asinomás –Asinomás 2017-11-03 18:23:56 +00:00 Commented Nov 3, 2017 at 18:23
1 (n 1/n)n(n+1)=n n(n+1)n=n n+1(n 1/n)n(n+1)=n n(n+1)n=n n+1 Asinomás –Asinomás 2017-11-03 18:27:09 +00:00 Commented Nov 3, 2017 at 18:27
1 @Emily (n 1/n)n(n+1)=n n+1(n 1/n)n(n+1)=n n+1 ab123 –ab123 2017-11-03 18:27:34 +00:00 Commented Nov 3, 2017 at 18:27
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I am not sure whether you are looking for a solution containing more powerful tools like derivatives. I try to show you this approach anyway.
Consider the function f(x)=x 1 x,x>0 f(x)=x 1 x,x>0. When x x is a natural number this gives exactly your sequence.
Take log log of both sides: log(f(x))=log(x 1 x)=1 x log(x)log(f(x))=log(x 1 x)=1 x log(x) and differentiate them:
f′(x)f(x)=−1 x 2 log(x)+1 x 2 f′(x)f(x)=−1 x 2 log(x)+1 x 2
hence
f′(x)=x 1 x−2(1−log(x)).f′(x)=x 1 x−2(1−log(x)).
Since x 1 x−2>0 x 1 x−2>0, this is negative when
1−log(x)<0→x>e.1−log(x)<0→x>e.
So f f is decreasing when x>e x>e. If you restrict to natural numbers, then n 1 n n 1 n is decreasing for n≥3 n≥3.
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answered Nov 3, 2017 at 18:35
GibbsGibbs
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Let x=y x,x=y x, for x≥1.x≥1. Then we can show that
d y d x=y x 2(1−ln x).d y d x=y x 2(1−lnx).
It is not difficult to see that this implicit function gives a maximum for y y at x=e x=e then decreasing so that y=1 y=1 is a horizontal asymptote for the graph.
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answered Nov 3, 2017 at 18:34
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From the derivative it is not clear that a horizontal asymptote for large positive y y is y=1 y=1. But we can use a L'hopital-esqe argument, e.g. lim x→∞x 1/x=lim x→∞e ln(x 1/x)=lim x→∞e ln(x)x=e lim x→∞ln(x)x=e lim x→∞1/x 1=e 0 lim x→∞x 1/x=lim x→∞e ln(x 1/x)=lim x→∞e ln(x)x=e lim x→∞ln(x)x=e lim x→∞1/x 1=e 0 john –john 2021-03-22 03:50:41 +00:00 Commented Mar 22, 2021 at 3:50
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In this case, we have to show that a sequence (x n)n∈N(x n)n∈N given by x n:=n 1/n x n:=n 1/n is decreasing for n≥3 n≥3. To do this, we should check that
x n+1≤x n x n+1≤x n this is (n+1)1/(n+1)≤n 1/n(n+1)1/(n+1)≤n 1/n.
raise both members of this inequality by n(n+1)n(n+1) we have
(n+1)n≤n(n+1)(n+1)n≤n(n+1)
⇔⇔
(n+1)n/n n≤n(n+1)n/n n≤n
⇔⇔
(1+1/n)n≤n(1+1/n)n≤n.
if you already know how to prove this last inequality you can use these equivalences to prove that the sequence is decreasing.
another way of proving that {(1+1/n)n}n∈N{(1+1/n)n}n∈N is decreasing and converges to the number e e:
the inequality (b n+1−a n+1)/(b−a)<(n+1)b(n+1)(b n+1−a n+1)/(b−a)<(n+1)b(n+1) for all 0≤a<b 0≤a<b.
To prove this use the binomial expansion
b n+1−a n+1=(b−a)(b n+a b(n−1)+a 2 b(n−2)+...+a(n−1)b+a n)b n+1−a n+1=(b−a)(b n+a b(n−1)+a 2 b(n−2)+...+a(n−1)b+a n)
Then
(b n+1−a n+1)/(b−a)<b n+b b(n−1)+b 2 b(n−2)+...+b(n−1)b+b n=(n+1)b(n+1)(b n+1−a n+1)/(b−a)<b n+b b(n−1)+b 2 b(n−2)+...+b(n−1)b+b n=(n+1)b(n+1).
Now with this inequality you can get
b n[b−(n+1)(b−a)]<a n+1 b n[b−(n+1)(b−a)]<a n+1
if we set a=1+1/(n+1)a=1+1/(n+1) and b=1+1/n b=1+1/n, we have 0≤a<b 0≤a<b and the term in brackets reduces to 1 1 and we have
(1+1/n)n<(1+1/(n+1))n+1(1+1/n)n<(1+1/(n+1))n+1
so we prove that the sequence is decreasing. to show that the sequence converges, it suffices to show that it is bounded. We will again use the inequality above.
Set a=1 a=1 and b=1+1/(2 n)b=1+1/(2 n). This time the term in the brackets reduces to 1/2 1/2, and we have
(1+1/(2 n))n<2(1+1/(2 n))n<2
thus (1+1/(2 n))2 n<4(1+1/(2 n))2 n<4
but {(1+1/n)n}n∈N{(1+1/n)n}n∈N is decreasing, then
(1+1/n)n<(1+1/(2 n))2 n<4(1+1/n)n<(1+1/(2 n))2 n<4 for all n∈N n∈N.
Therefore the sequence {(1+1/n)n}n∈N{(1+1/n)n}n∈N is decreasing and bounded. With this we conclude that {(1+1/n)n}n∈N{(1+1/n)n}n∈N is is a convergent sequence and its limit is denoted by e e.
This answer complements the answer given by Jorge Fernández.
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edited Nov 3, 2017 at 19:32
answered Nov 3, 2017 at 18:38
Luiz ColloviniLuiz Collovini
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n 1/n=exp((1/n)log(n)).n 1/n=exp((1/n)log(n)).
Since exp(x)exp(x) is strictly increasing it suffices to show that
for m>n m>n:
(1/n)log(n)>(1/m)log(m).(1/n)log(n)>(1/m)log(m).
f(x):=(1/x)log(x),x≥3.f(x):=(1/x)log(x),x≥3.
f′(x)=−(1/x 2)log(x)+1/x 2.f′(x)=−(1/x 2)log(x)+1/x 2.
f′(x)=(1/x 2)[1−log(x)].f′(x)=(1/x 2)[1−log(x)].
f′(x)<0 f′(x)<0 for x>3.x>3.
f(x)f(x) strictly decreasing for x>3,x>3,
→→:
For m>n:m>n:
n 1/n>m 1/m n 1/n>m 1/m, I.e. strictly decreasing.
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edited Nov 3, 2017 at 20:14
answered Nov 3, 2017 at 20:07
Peter SzilasPeter Szilas
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I had written the first answer below, which only uses Bernoulli's Inequality, because it was close to the approach in the question, looking at (1+1 n)n+1 n+1(1+1 n)n+1 n+1 instead of (1+1 n)n n(1+1 n)n n (they are equal). Then I saw that (n+1)n n n+1(n+1)n n n+1 was decreasing, which prompted the second, simpler answer.
First Answer
Note that (1+1 n)n+1(1+1 n)n+1 is decreasing:
(1+1 n−1)n(1+1 n)n+1=n−1 n(1+1(n−1)(n+1))n+1≥n−1 n(1+1 n−1)=1(1a)(1b)(1c)(1a)(1+1 n−1)n(1+1 n)n+1=n−1 n(1+1(n−1)(n+1))n+1(1b)≥n−1 n(1+1 n−1)(1c)=1
Explanation:
(1a):(1a): rewrite the fraction
(1b):(1b): Bernoulli's Inequality
(1c):(1c): simplify
If f(n)f(n) is decreasing, then for n≥1 n≥1, f(n)≤f(1)f(n)≤f(1); that is,
(1+1 n)n+1≤4(2)(2)(1+1 n)n+1≤4
Therefore, for n≥3 n≥3.
(n+1)n n n+1=(1+1 n)n+1 n+1≤4 n+1≤1(3a)(3b)(3c)(3a)(n+1)n n n+1=(1+1 n)n+1 n+1(3b)≤4 n+1(3c)≤1
Explanation:
(3a):(3a): rewrite the fraction
(3b):(3b): apply (2)(2)
(3c):(3c):n≥3 n≥3
Inequality (3)(3) says that, for n≥3 n≥3, n 1/n n 1/n is decreasing; that is,
(n+1)1/(n+1)≤n 1/n(4)(4)(n+1)1/(n+1)≤n 1/n
Second Answer
Note that (n+1)n n n+1(n+1)n n n+1 is decreasing:
n n−1(n−1)n(n+1)n n n+1=(n 2 n 2−1)n>1(5a)(5b)(5a)n n−1(n−1)n(n+1)n n n+1=(n 2 n 2−1)n(5b)>1
Thus, for n≥3 n≥3,
(n+1)n n n+1≤4 3 3 4=64 81<1(6a)(6b)(6c)(6a)(n+1)n n n+1≤4 3 3 4(6b)=64 81(6c)<1
Explanation:
(6a):(6a): if f(n)f(n) is decreasing, then for n≥3 n≥3, f(n)≤f(3)f(n)≤f(3)
(6b):(6b): evaluate
(6c):(6c): compare
Inequality (6)(6) says that, for n≥3 n≥3, n 1/n n 1/n is decreasing; that is,
(n+1)1/(n+1)<n 1/n(7)(7)(n+1)1/(n+1)<n 1/n
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edited Oct 30, 2023 at 12:52
answered Oct 30, 2023 at 2:06
robjohn♦robjohn
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9575 | https://www.imaios.com/en/e-anatomy/anatomical-structures/ophthalmic-nerve-1557858964 | View the module
Ophthalmic nerve
Nervus ophthalmicus
Latin synonym:
Nervus cranialis V1
Synonym:
Cranial nerve V1; Ophthalmic division of trigeminal nerve
Related terms:
Ophthalmic nerve; Ophthalmic division [Va; V1]
Definition
IMAIOS
The Ophthalmic Nerve [Va; V1], or first division of the trigeminal, is a sensory nerve. It supplies branches to the cornea, ciliary body, and iris; to the lacrimal gland and conjunctiva; to the part of the mucous membrane of the nasal cavity; and to the skin of the eyelids, eyebrow, forehead, and nose. It is the smallest of the three divisions of the trigeminal nerve, and arises from the upper part of the semilunar ganglion as a short, flattened band, about 2.5 cm. long, which passes forward along the lateral wall of the cavernous sinus, below the oculomotor and trochlear nerves; just before entering the orbit, through the superior orbital fissure, it divides into three branches, lacrimal, frontal, and nasociliary.
The ophthalmic nerve is joined by filaments from the cavernous plexus of the sympathetic, and communicates with the oculomotor, trochlear, and abducent nerves; it gives off a recurrent filament which passes between the layers of the tentorium.
References
This definition incorporates text from a public domain edition of Gray's Anatomy (20th U.S. edition of Gray's Anatomy of the Human Body, published in 1918 – from
Gallery |
9576 | https://engineering.purdue.edu/~wassgren/teaching/ME30900/Examples/statics_14.pdf | statics_14 Page 1 of 2 The rigid, L-shaped gate shown in the figure can rotate about the hinge and rests against the rigid support at point A. What is the minimum horizontal force, F required to hold the gate closed if its width is w = 3 m and the lengths are h = 4 m and l = 2 m? The height of the free surface above the hinge is H = 3 m. You may neglect the weight of the gate and the friction in the hinge. Note that the back of the gate is exposed to the atmosphere. water h l F H hinge gate A statics_14 Page 2 of 2 SOLUTION: Balance moments about the hinge, , (1) , (2) , (3) , (4) . (5) Using the given data, r = 1000 kg/m3 = 9.81 m/s2 w = 3 m H = 3 m h = 4 m l = 2 m Þ F = 437 kN M hinge ∑ = 0 = y −H ( ) moment arm length ρgy pressure wdy ( ) area y=H y=H+h ∫ + x moment arm length ρg H + h ( ) pressure wdx ( ) area x=0 x=l ∫ − hF moment due to applied force ρgw y −H ( )ydy y=H y=H+h ∫ + ρg H + h ( )w xdx x=0 x=l ∫ −hF = 0 hF = ρgw 1 3 y3 −1 2 Hy2 ( )y=H y=H+h + 1 2 ρg H + h ( )wl2 hF = ρgw 1 3 H + h ( ) 3 −H 3 ⎡ ⎣ ⎤ ⎦−1 2 H H + h ( ) 2 −H 2 ⎡ ⎣ ⎤ ⎦ ( ) { }+ 1 2 ρg H + h ( )wl2 F = ρgw 1 2 Hh + 1 3 h2 + 1 2 H h +1 ( )l2 ⎡ ⎣ ⎤ ⎦ water h l F H hinge x y |
9577 | https://helpfulprofessor.com/negative-correlation-examples/ | Published Time: 2023-03-23T23:38:10+00:00
10 Negative Correlation Examples
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10 Negative Correlation Examples
Written by Viktoriya Sus (MA)
Viktoriya Sus (MA)
Viktoriya Sus is an academic writer specializing mainly in economics and business from Ukraine. She holds a Master’s degree in International Business from Lviv National University and has more than 6 years of experience writing for different clients. Viktoriya is passionate about researching the latest trends in economics and business. However, she also loves to explore different topics such as psychology, philosophy, and more.
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| August 25, 2023
Reviewed by Chris Drew (PhD)
Chris Drew (PhD)
This article was peer-reviewed and edited by Chris Drew (PhD). The review process on Helpful Professor involves having a PhD level expert fact check, edit, and contribute to articles. Reviewers ensure all content reflects expert academic consensus and is backed up with reference to academic studies. Dr. Drew has published over 20 academic articles in scholarly journals. He is the former editor of the Journal of Learning Development in Higher Education and holds a PhD in Education from ACU.
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A negative correlation is a relationship between two variables in which one variable decreases as the other increases.
As a negative correlation example from psychology, one might observe a negative correlation between happiness and the number of hours worked; that is, as working time increases, contentment diminishes.
From poverty and life expectancy to crime rates and education levels, as well as employment rates and inflation – negative correlations can be found in many areas.
Researchers can quantify the strength of a negative correlation between two variables and measure its effect on one another by utilizing techniques such as regression analysis.
This process makes it possible to assess how changes in one variable may influence the other.
Negative correlations can be a powerful tool for researchers, allowing them to uncover and reveal cause-and-effect relationships between various aspects of our world.
By studying these connections, we can gain much-needed insights into the environment around us.
Contentsshow
Definition of Negative Correlation
Negative Correlation Examples
Types of Correlations
How to Determine Negative Correlation
Negative Correlation Strength
Importance of Negative Correlation
Negative Correlation in Portfolio Diversification
Conclusion
References
Definition of Negative Correlation
Economics,math, statistics,psychology, and philosophy all study a phenomenon known as negative correlation: the relationship between two variables where an increase in one causes a decrease in the other.
The relationship between variables is inverse: as one increases, the other decreases. For instance, when temperatures rise, snowfall diminishes accordingly.
According to psychologists Hinote and Wasserman (2020),
“…negative (or inverse) correlation describes a situation when one variable increases the other systematically decreases, or vice versa” (p. 56).
Moreover, a negative correlation is an indication of the strength of the association between two variables and carries heavy implications for understanding how diverse phenomena interact with one another.
Simply, negative correlations can be used to explain why certain phenomena occur, such as why an increase in one variable causes a decrease in the other.
Negative Correlation Examples
Negative Correlation in Psychology:When interpersonal conflict increases, satisfaction in relationships decreases. This is because conflicts can erode trust, communication, and intimacy between individuals.
Negative Correlation in Sociology: As poverty rises, life expectancy decreases. This is because poorer people have less access to nutrition, healthcare, and other goods and services that help achieve a healthy lifestyle.
Negative Correlation in Education: Higher student-teacher ratios are correlated with lower student achievement scores. When there are too many students per teacher, it becomes difficult for individual attention to be given, which can ultimately impact student performance negatively.
Negative Correlation in the Environment: A study of air pollution and asthma rates reveals a clear correlation: the higher the levels of air contamination, the poorer out health becomes. We could also state this as a positive correlation: when there is more smog in our atmosphere, we experience more asthma.
Psychology:When a student’s procrastination increases, their academic performance decreases. This is because putting off tasks can lead to incomplete work, or increase stress levels, or cause missed deadlines.
Medicine:Excessive sugar intake has been linked to decreased oral hygiene, as it not only stimulates bacterial growth but also weakens tooth enamel.
Consumer Economics: As interest rates go up, house prices go down. This is because high interest rates make it harder to get a mortgage. With less demand in the market, the less people will be inclined to buy a house.
Consumer Economics: When mortgage interest rates surge, consumers feel the pinch of extra housing costs and decrease their spending in other areas. Consequently, an upswing in these rates leads to a drop in discretionary spending.
Biology: When consumption of ‘adult drinks’ rises, mental acuity and physical coordination decrease sharply. It is for this reason that it is illegal to drive when under the influence.
Consumer Economics: Higher unemployment rates coincide with reduced consumer confidence levels – when people don’t feel secure about their jobs, they tend to hold back from spending money or taking risks financially, suppressing economic growth over time if left unchecked.
Consumer Economics: As the weather increases, consumption of ice cream goes down. This is because our bodies crave cold foods more during hot weather than cold weather.
Sports Sciences: As exercise levels increase, risk of cancer decreases. As a result, most government health agencies recommend a minimum of 30 minutes of cardiac exercise per day.
Types of Correlations
While a negative correlation describes a relationship between two variables that decreases when one increases, a positive correlation is an opposite. A zero correlation implies no relationship between the two variables at all.
Negative correlation is a phenomenon in which two variables are related so that an increase in one of the variables leads to a decrease in the other and vice versa (DePoy & Gilson, 2016).
Positive correlation, on the other hand, is when two variables are related such that an increase in one variable results in an increase in the other (DePoy & Gilson, 2016). For example, an increase in sugar consumption will lead to decreased dental health (negative correlation). In contrast, an increase in exercise leads to increased physical fitness (positive correlation). In both cases, there is a direct relationship between the two variables but with different outcomes – one negative and one positive.
Zero correlation describes two variables that are completely unrelated to each other. It means that changes in one variable will not affect the other in any way (DePoy & Gilson, 2016).
Illusory correlation happen when two variables (people, events, or behaviors), are perceived to have a relationship, when in fact, there is no logical reason for them to be correlated. For example, if you see that a coin has flipped heads six out of six times, you may think that heads is likely to turn up again the next time. This is an illusion. There is still a 50/50 chance that tails will turn up.
How to Determine Negative Correlation
The Pearson Correlation Coefficient formula is one of the most commonly used methods for determining the strength of negative correlations between two variables.
This formula considers the mean and standard deviation of both variables, as well as the covariance between them.
The formula for calculating Pearson Correlation Coefficient is as follows:
Correlation(X,Y) = Covariance(X,Y) / (StdDev(X)StdDev(Y)),
where covariance measures how two variables change together, StdDev is the standard deviation of each variable, and X and Y are your two variables (Sharma, 2019).
If you plug in your data points and calculate a result that is anywhere between -1 to 0, then you have established a negative correlation between the two variables.
Negative Correlation Strength
Negative correlation strength is measured on a scale from -1 to 0, with -1 being the strongest possible level of correlation and 0 indicating no correlation at all.
The closer the correlation coefficient gets to -1, the stronger the negative relationship between two different variables (McMillan, 2008).
For example, let’s say we’re looking at temperature and precipitation data for a specific area over time. A data set that showed temperatures were decreasing when rain was increasing would indicate a strong negative relationship.
If we ran our analysis and got back a correlation coefficient of -0.9, it would be safe to assume that there was indeed a strong negative connection between these two variables.
Alternatively, if we got back a coefficient of only -0.5 when running our analysis, this would indicate that while there is still some connection between these two variables (temperature decreasing when rain increases), it isn’t as strong as before (−0.9).
It could mean that other factors influence this relationship besides just temperatures and rainfall.
Importance of Negative Correlation
Negative correlation is an important concept in data analysis, as it helps better understand the relationships between two variables.
For example, if studying how temperature affects a species’ growth rate, a negative correlation might indicate that as temperatures increase, the growth rate of the species decreases.
This knowledge can be used to make informed decisions about what actions should be taken to preserve or improve long-term growth rates for the species.
In predictive analytics, understanding negative correlations can help us anticipate and respond to changes in trends.
For instance, analyzing correlations can reveal connections between stock prices and other macroeconomic indicators when working with financial data sets.
By recognizing those connections, better predictions can be made about future market movements, and investment opportunities can be identified.
Negative correlation may also provide insight into cause-and-effect relationships between otherwise unrelated variables that are not easy to identify.
A classic example of this is the strong negative correlation between smoking tobacco and life expectancy: as more people smoke cigarettes over time, life expectancy tends to decline.
Understanding this correlation allows public health professionals to target interventions and create campaigns that reduce smoking prevalence.
Negative Correlation in Portfolio Diversification
Negative correlations in portfolio diversificationarisewhen two investments have an inverse relationship. For example, as one asset goes up, the other goes down, and vice versa(Pula et al., 2012).
This type of correlation is used to reduce risk in a portfolio, as it spreads out losses over different assets and can help protect investors from large market swings.
Negative correlations are usually found between stocks and bonds or stocks and commodities such as gold or oil.
For example, bonds tend to increase in value when stocks go down due to their safe-haven status. Similarly, when oil prices drop, gold prices tend to rise as investors flock toward the safety of precious metals.
Another way negative correlations can be used in portfolio diversification is by investing in different geographic regions or market sectors (Lossen, 2007).
Investing in emerging markets like India or China may provide some protection against investments in established markets like the US or UK because rises and falls may not necessarily occur simultaneously across countries or regions.
Negative correlation can also be exploited through short selling, where a trader will sell borrowed shares with the expectation that they will fall in price so that they can be bought back at a lower cost later on.
This strategy helps protect against losses if a trader is wrong about their bet and provides an additional layer of protection for an investor’s portfolio.
Conclusion
Negative correlation is an important concept in data analysis, predictive analytics, and portfolio diversification.
It can provide insight into cause-and-effect relationships between variables that are not obvious and can be used to identify investment opportunities or protect against losses.
When analyzing correlations, it is important to remember that correlation does not imply causation, and further research is needed to understand the underlying relationships between variables.
However, properly understanding negative correlations can be valuable for making better data-driven decisions.
References
DePoy, E., & Gilson, S. (2016).Social work research and evaluation. SAGE Publications.
Hinote, B. P., & Wasserman, J. A. (2020).Social and behavioral science for health professionals. Rowman & Littlefield Publishers.
Lossen, U. (2007).Portfolio strategies of private equity firms. Springer Science & Business Media.
Mcmillan, J. H. (2008).Assessment essentials for standards-based education. Corwin Press.
Pula, J. S., Berisha, G., & Ahmeti, S. (2012). The impact of portfolio diversification in the performance and the risk of investments of kosovo pension savings trust.Journal of Business and Economics,3(3), 198–211.
Sharma, J. K. (2019).Business statistics. Vikas Publishing.
Viktoriya Sus (MA)
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Viktoriya Sus is an academic writer specializing mainly in economics and business from Ukraine. She holds a Master’s degree in International Business from Lviv National University and has more than 6 years of experience writing for different clients. Viktoriya is passionate about researching the latest trends in economics and business. However, she also loves to explore different topics such as psychology, philosophy, and more.
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9578 | https://www.youtube.com/watch?v=X2yLePs_0zY | A variable line passes through a fixed point P. The algebraic sum of the perpendiculars drawn fr...
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A variable line passes through a fixed point P. The algebraic sum of the perpendiculars drawn from the points (2,0), (0,2) and (1,1) on the line is zero. Find the coordinate of the point P.
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Transcript:
foreign okay so let's move to the question the algebraic system algebraic storm of perpendicular drawn from the point this on the line is zero so then find the coordinates of point B so one thing you can keep in mind if in any question fixed point is there then there may be concept of what family of a straight line because in family of straight line there is a fixed Point through which all the infinite lines path so so let's move to the question and suppose the fixed Point p is a comma B suppose the fixed Point p is a common B okay and the slope is n okay so equation of line will be y minus b equals to M times of x minus B and we can write y minus m a working 30 x minus a y minus MX plus a m minus b equals to 0 and this is equivalent of line so from point two comma 0 D1 I am going to D1 D1 is will keep y 0 and x minus 2 so 0 minus 2 m Plus a m minus B upon 2 tender 1 plus M Square 1 plus M square and D2 into also denominator will be root 1 plus M square and 46.2 and for this point D3 denominator will be 1 plus n Square so when we'll add this all A2 is equal to 1 plus M Square and in numerator X will be 0 and Y will be 2 so 2 plus a m minus B and D3 will be equals to square root 1 plus M square and we'll put 1 comma 1 so value will be 1 minus M plus a n minus t and as every sum of all three distances T1 plus D2 plus T3 is 0 so it will add all the three denominator will be 10 1 plus M Square okay and it will become 2 plus 1 this is all 2 plus 1 will become 3 okay minus 2 m minus M will be minus 3 m again am plus am plus am plus 3 AM okay n minus B minus B minus 3 B and this addition is equals to 0 so this is in form of numerator upon denominator numerator upon denominator is zero so numerator must be 0 so we can write 3 minus 3M plus 3 AM minus 3B equals to zero okay so I am going to take 3 minus 3B in one side and I'm going to take M minus M common so this will be 3 minus 3 a equals to 0. okay so this always satisfies the equation okay so put this the value of B if value of B is 1 and value of a is one then this always satisfied that's equals to 0 so this equation always satisfies when the value of a and when the value of B is positive one so this line is always passes through 1 comma 1. thank you student s is super |
9579 | https://www.chegg.com/homework-help/questions-and-answers/use-diagram-answer-question-follows-15-ft-10-ft-hi-30-ft-part-art-installation-students-er-q73778171 | Solved Use the diagram below to answer the question that | Chegg.com
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Use the diagram below to answer the question that follows. 15 ft. 10 ft hi 30 ft As part of an art installation, students erect a 15-foot-high post and a 10-foot-high post 30 feet apart. They attach wire from the top of each post to the bottom of the opposite post as shown in the diagram. How high above the ground is the point where the two wires intersect?
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Question: Use the diagram below to answer the question that follows. 15 ft. 10 ft hi 30 ft As part of an art installation, students erect a 15-foot-high post and a 10-foot-high post 30 feet apart. They attach wire from the top of each post to the bottom of the opposite post as shown in the diagram. How high above the ground is the point where the two wires intersect?
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Consider the similarity between triangles Δ A B C and Δ A N M and set up the proportion A N A B=h 30 to express A N in terms of h. Therefor…
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Transcribed image text:
Use the diagram below to answer the question that follows. 15 ft. 10 ft hi 30 ft As part of an art installation, students erect a 15-foot-high post and a 10-foot-high post 30 feet apart. They attach wire from the top of each post to the bottom of the opposite post as shown in the diagram. How high above the ground is the point where the two wires intersect? OA. 5 feet OB. 5 5 24 feet OC. 6 feet OD. 6 feet
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9580 | https://www.youtube.com/watch?v=HvpkKnDUdWI | Related Rates - Growth of Circle Area as Radius Grows and Growth of Cone Height as Volume Grows
Dr. Pierce's Physics & Math
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We look at two common examples of related rates, in calculus. 0:00 Growth of a Circle's Area with Radius Growth and 4:15 Growth of a Cone's Height with Volume Growth. We use the following steps: (1) Determine how two variables (A and B) are related - write A as a function of B, or A(B). (2) Determine how the rates of change of those variables are related, using the chain rule; dA/dt = dA/dB dB/dt.
Transcript:
all right my friends we're gonna take a little look at related rates for those of you who are beginning your study of calculus we'll start with a simpler problem and then we'll ramp this up and look at something a little tougher um so first we'll start with if the radius of a circle is growing at two centimeters per second which is a constant rate of growth let's find the rate at which the area is growing when the circle has well first a radius of one centimeter and then a radius of three centimeters now why you might get different answers for say part A and Part B that's uh evident if you draw a little picture what's happening so imagine the circle that starts real small and then starts to grow in radius at a constant rate well when the circle is say one centimeter in radius and we let a little wiggle in time go by and we add some radius to it you would add this sort of magenta colored ring of say additional material when the circle is has a radius of three centimeters when you add that same little bit of radius say over a given amount of time you would add this bluish area and you can see that the bluish area is larger than the um the magenta area and so what happens is as the radius of the circle grows the rate of change of the area is going to get larger and larger so let's see that play out well so if we want to relate the rate of growth of the area to the rate of growth of the radius we first just need to write down how the area and radius are related well of course they're related as a equals pi r squared so this relates the area and the radius well now we need to relate the rate of change of the area to the rate of change of the radius well this is where the chain rule comes in if we want the rate of change of the area we have to take the derivative of this expression with respect to R and then multiply by the time rate of change of the radius so this is just nothing other than the chain rule well so let's go for it we'll use the chain rule on this expression so we have d80t equals well we have to take the derivative of this expression with respect to R so you bring the exponent down in front and then reduce it by one so you'd get 2 pi r to the first so that's here and then you just multiply by the rate of change of the radius which is actually given in the problem be two centimeters per second um and so we now have an expression that relates the rate of change of the area to the rate of change of the radius and so we can just start plugging in some numbers now so let's do part a so d a DT is going to be 2 pi times the radius of one centimeter for part A and then times Dr DT which is just going to be 2 throughout this whole problem and you can see when you multiply all this together you'll get 4 Pi centimeters squared per second of additional area right that's the rate at which we're accumulating area when the circle only has a radius of one centimeter well to do Part B we just need to plug in some different numbers now so dadt again is just 2 pi r where now the r is equal to three times the rate of change of the radius which is still two two centimeters per second and you notice when you multiply all this together 2 times 3 would be 6 times 2 is 12. so now it's 12 Pi centimeters squared per second so we're actually accumulating area now at three times the rate that we were before um which actually makes sense because if you look at the rate of change of area it does just go like the radius it's proportional to the radius um and so if we're out at 3 times the radius we'll have three times uh greater um rate of change of area and so that's a pretty standard related rate problem you'll see something like that pretty often we'll look at a slightly more difficult case with a related rate problem pretty popular one um and this says if water is being added to a 60 degree cone shaped cup at a rate of five centimeter cubed per second find the rate at which the water level is rising when the water depth is well these two values by a 60 degree cone I mean that's the angle at the base of the cone so let's draw a little picture of this so we can figure out what's happening so the idea is you have this cone um the bottom angle here is 60 degrees and you just start filling this thing with water so it starts with no water in it and then the water is going to go up and up and up and up what we're doing though is we're adding a constant volume of uh a water at a constant rate at a constant rate such that the volume that you're adding is per second let's say is constant well the thing is when you are down here in the lower section of the cone a given volume added volume of water is going to make the level rise quite a bit where as you get higher and higher as you add that same volume the level is not going to rise quite so much just because this cross-sectional area is is growing and growing and growing as you go up um and so we're supposed to find the rate at which the water level rises well from the picture you can anticipate that you're going to get a much faster water level rise when you're near the bottom than you will when you get a little higher um so let's go for it again we're trying to relate the um the rate of additional the volume per time to the rate at which the um the height increases here so DV DT and dhdt we're trying to relate well first then what we need to do is relate the volume and the height that we are well we know that in general for a cone the volume of a cone is one-third pi r squared times the height of the cone um but where this expression is not quite good enough for us yet is we really just need a direct relationship between V and H but unfortunately we have this kind of confounding variable R in here well so what we need is just an expression where V is purely a function of H not with a changing R in there also well so we can do that because we know the angle at the bottom of this cone um so as it turns out this radius is just a h tan tangent of 30. now why 30 even though this is a 60 degree cone base if you look at the relationship between R here and H here um you can see that the relevant angle if you have a little right triangle here you the angle that we need to worry about is actually half of the cone so if you imagine running a vertical line from the base of the cone right here to the end of this arrow that represents the radius we have this little triangle that's here and the angle sort of from the apex of the of the cone here from the vertical to the edge of the cone that would be 30 degrees if this is if this whole thing is 60. so this radius is just going to be H tan 30. well so then we just need to substitute that in there well so Tangent 30 as it turns out is is root 3 over 3 or 1 over root 3. and then we just need to square that um 8 root 3 over 3 and just shove it in for r um for r squared and so When the Smoke Clears you get this expression for the volume of the cone purely in terms of how how high the water level is so it would be well one third Pi you get an H Cube because you have um this r squared would kind of give you an an H squared but then you have another H floating around here and then root 3 over 3 if you square it it'd be 3 9 so otherwise known as one-third so there's our volume of a cone purely in terms of the height um if it's a 60 degree base base angle and if you simplify that a bit you'll just get um Pi H cubed over nine so that relates the volume and the height well we're supposed to relate the rate of change of volume to the rate of change of height well so we can go for it and again it's going to be chain rule we want the rate of change of volume by Chain rule um that is going to be dvdh times DH DT so to gather that product will give you DV DT well so let's apply that to this expression so DV DT is going to be well we need to take the derivative of this expression with respect to H first so you bring the 3 down the exponent down in front reduce it by one so you get like 3 Pi H squared over 9 otherwise known as Pi H squared over three and then times DH DT well so we have this expression relating the rate of change of volume to the rate of change of the height well we're given the rate of change of the volume it's five centimeter cubed per second so we're just going to put a 5 in for DV DT and then the only unknown is going to be DH DT so we've basically solved the problem so let's do it for these two cases well so if in part A we have DV DT equals five um equals well pi times well the height is just um is 1 because the water is supposed to be one centimeter deep at the beginning 1 squared over three dhdt and so if we solve for the h DT what I multiply the 3 over you get 15 what over pi 15 over Pi happens to be about 4.77 so that's the rate at which the the level is instantaneously growing when this thing happens to be one centimeter High it can we can do Part B similarly um so again we have DV DT equals five because you're just adding water at a rate of five centimeter cubed per second equals pi times well now the depth is three centimeters so that's our new height 3 cubed um divided by three and then times DH DT so this is our only unknown the HDT and so it looks like we are going to get what 15 15 over 9 pi um and so if you work that out you'll get about 0.53 centimeters per second for dhdt and it makes sense that this thing would grow um slower um once you're at three centimeters high as opposed to one centimeter High because each additional chunk of volume would cause a smaller change in height so conceptually what I'm supposed to be drawing here is this magenta volume is supposed to be about the same or the same as this blue volume and so that's how you would find the rate of change of the height of a cone that's filling up with water hopefully you found this helpful and thanks for checking it out |
9581 | https://www.sciencedirect.com/science/article/abs/pii/S0891524596900711 | Protein-energy malnutrition (Kwashiorkor and Marasmus) - ScienceDirect
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Journal of Pediatric Health Care
Volume 10, Issue 1, January–February 1996, Pages 28-30
Clinical report
Protein-energy malnutrition (Kwashiorkor and Marasmus)
Author links open overlay panel Patricia T Castiglia
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G Parent et al.
In vitro lymphocyte-differentiating effects of thymulin (ZN-FTS) on lymphocyte subpopulations of severely malnourished children
American Journal of Clinical Nutrition (1994)
S.E Albers et al.
A case of Kwashiorkor
Cutis (1993)
R.E Behrman et al.
Nelson Textbook of pediatrics
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Adjunct use of honey in diabetes mellitus: A consensus or conundrum?
2020, Trends in Food Science and Technology Citation Excerpt :
Low HDL is known to be a risk factor for cardiovascular function and the subcutaneous fat area is found negatively correlated with HDL in type-1 diabetes (Abdulrhman et al., 2013; Jacob, Adams-Huet, & Raskin, 2006). SSFT has to do with the condition of subcutaneous fat (Castiglia, 1996) and reducing SSFT by using honey can be seen as a good impact and not a negative result. Various wound dressings containing Russian bee (Apis mellifera) honey, Malaysian tualang honey, and Manuka honey are found highly cost-effective and efficacious for diabetic ulcer/wound healing and pain reduction without adverse events (Alam, Islam, Gan, & Khalil, 2014; Gouda, 2016; Mohamed, El Lenjawi, Salma, & Abdi, 2014; Mohamed, El Lenjawi, et al., 2014; Al-Lenjawi et al., 2016; Mohamed et al., 2015; Simon et al., 2009; Tan et al., 2009). Show abstract Honey is being used in Complementary and Alternative Medicine, especially in Indian Ayurvedic Medicine, as an adjuvant and supplement in diabetes mellitus treatment since immemorial times. In recent times, the use of honey has experienced a renewed interest in the context of diabetes treatment because of the rise in the accessibility of evidence-based pharmacological and clinical findings, signifying its health benefits. There are differential opinions regarding the traditional use of honey in diabetes mellitus. The present review highlights various research propositions, hoisted issues, and misconceptions regarding the effects of honey in diabetes management and presents current challenges and future perspectives. A comprehensive critical review was performed by probing the traditional antidiabetic claims of honey, considering published reports in online databases. A total of 20 pre-clinical and 25 clinical studies investigated the antidiabetic effect of honey. Though in vivo studies are still limited, the findings reinforce the multi-targeted antidiabetic effect of honey, exerting antioxidant, nutritional, antihyperglycemic, immunomodulatory, anti-inflammatory, wound-healing, antihypertensive, hypolipidemic, and hypoglycaemic activities. Preclinical and clinical evidence suggests that honey may possess multi-faceted and adjunct effects to accomplish a better glycaemic control, ameliorate several metabolic derangements, and mitigate oxidative stress-evoked diabetic problems. Nevertheless, the findings remain inconclusive due to poor study designs and other limitations (e.g. short duration, few participants, the difference in type of study participants, varied honey sources, and administered doses). Overall, there is a significant gap in knowledge, and hence, carefully planned, detailed in vitro, in vivo, and clinical studies are warranted to reach better conclusions.
### Severe nutritional deficiencies in young infants with inappropriate plant milk consumption
2014, Archives De Pediatrie Show abstract Il existe une augmentation de l’utilisation de «jus» végétaux en remplacement des préparations infantiles chez les nourrissons. L’administration de ces boissons végétales, dont la composition ne respecte pas la réglementation européenne, est souvent motivée de façon injustifiée par un effet supposé néfaste du lait de vache. Le but de ce travail était de rapporter plusieurs cas de complications nutritionnelles sévères dues à cette substitution inadaptée. Dans un travail rétrospectif mené entre 2008 et 2011, nous rapportons 9 cas de nourrissons âgés de 4 à 14 mois ayant consommé des boissons végétales avec des complications imputables à leur administration. Quatre d’entre eux étaient âgés de moins de 6 mois et recevaient une boisson végétale de façon quasi exclusive au moment du diagnostic. L’abandon des préparations pour nourrissons avait été motivé par une allergie supposée aux protéines de lait de vache ou par des troubles digestifs divers. Les boissons incriminées étaient des jus de châtaignes, amande, riz et soja. Les symptômes d’appel avaient été dans 3 cas une dénutrition protéino-calorique, dans un cas un état de mal convulsif secondaire à une hypocalcémie et dans 5 cas un arrêt de la croissance pondérale. Biologiquement, 5 enfants avaient une anémie sévère, 3 des troubles du bilan phosphocalcique et 2 une hyponatrémie importante. L’administration de boissons végétales inadaptées en remplacement des formules adaptées aux nourrissons expose à un risque de carences nutritionnelles parfois responsables de complications pouvant mettre en jeu le pronostic vital. Des mesures réglementaires interdisant leur utilisation chez le nourrisson devraient être mises en place. Over the past few years, we have observed increasing consumption of inappropriate plant milks as an alternative to infant milk formula. Some families believe that foods labeled as natural are the most healthy and an appropriate nutritional choice. However, their composition does not respect European recommendations. They are always hypocaloric and protein, vitamin, and mineral concentrations are inadequate. The aim of this study was to report severe nutritional complications after inappropriate plant milk consumption. Between 2008 and 2011, we studied severe nutritional deficiencies caused by consumption of plant milks bought in health food stores or online shops. Infants were identified in our centers and examined through medical history, physical examination, and laboratory testing. Nine cases of infants aged from 4 to 14 months were observed. In all cases, these milks were used as an alternative to milk formulas for supposed cow's milk allergy. At diagnosis, four patients were aged 6 months or less. They had received plant milk exclusively for 1–3 months. The beverages consumed were rice, soya, almond and sweet chestnut milks. In three cases, infants presented severe protein-calorie malnutrition with substantial hypoalbuminemia (<20 g/L) and diffuse edema. In the other cases, the nutritional disorders were revealed by a refractory status epilepticus related to severe hypocalcemia (one case), growth arrest of both height and weight secondary to insufficient caloric intake (five cases), and severe cutaneous involvement (one case). Five children had severe iron deficiency anemia (<70 g/L), three children had a very low 25-hydroxy vitamin D level (nutritional rickets), and two had severe hyponatremia (<130 mmoL/L). Milk alternative beverages expose infants to severe nutritional deficiencies. Serious complications can occur. Early, exclusive, and extended use is riskier. These diseases are preventable, and parental education should be provided. Statutory measures forbidding their use in young infants should be organized to slow down the progress of this social trend.
### Global Hunger: A Challenge to Agricultural, Food, and Nutritional Sciences
2014, Critical Reviews in Food Science and Nutrition
### Protein energy malnutrition decreases immunity and increases susceptibility to influenza infection in mice
2013, Journal of Infectious Diseases
### Metabolic effects of honey in type 1 diabetes mellitus: A randomized crossover pilot study
2013, Journal of Medicinal Food
### Hospital hyponutrition
2011, Nutricion Hospitalaria
View all citing articles on Scopus
View full text
Copyright © 1996 Published by Mosby, Inc.
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9582 | https://www.youtube.com/watch?v=LBhRiTw_uXw | Solving Trigonometric Equations - How to Write General Solution
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Learn how to find the general solution when solving trigonometric equations. We go through 3 examples in this video to illustrate some different ways of writing the general solution.
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Transcript:
Intro in this video you're going to learn how to solve trigonometric equations and specifically we're going to talk about how to find the general solution so we're going to go through three examples First Example see if you can do these on your own and we'll go through them together the first one we've got 2 cosine X minus 1 equals 0 so when you're solving these trig equations is very similar to solving algebraic equations but you want to get the trig function by itself on one side of the equation so what we're gonna do here is we're gonna isolate this cosine of X by adding 1 to both sides of the equation so now we have 2 cosine x equals 1 we're going to divide both sides by 2 and so now you can see that we have cosine of x equals 1/2 so now we want to do is we want to go to the unit circle and when I ask ourselves where is cosine equal to 1/2 well remember on the unit circle cosine is the x-coordinate and so let's see where is the x-coordinate and 1/2 well we can see it's here as well as here so if I just draw a little sketch over here just to illustrate here's our unit circle and you can see cosine is 1/2 right here at PI over 3 or 60 degrees and over here at 5 PI over 3 or 300 degrees now notice these points here if we want to write a general solution what we have to do is we have to say X could equal PI over 3 plus if we add 2 pi or multiples of 2 pi we're going to be going around that unit circle and we're going to be ending up in the same spot on the unit circle and the value of the cosine is still going to be 1/2 so we're going to say PI over 3 plus 2 pi and and then you can write where n is an integer meaning like you can can't be 1/2 or 1/3 you have to be multiples of 2 pi you know can be positive or negative you can go either direction around the unit circle so it can also be 5 PI over 3 plus 2 pi n again where n is an integer so this is our general solution for the answer to this equation now we're gonna go into a little bit more challenging one in the next example so let's dive into that one next okay number two see if you can do Second Example this one we'd have 4 cosine squared of X minus 3 equals 0 so again we're trying to I isolate that trig function by itself on one side of the equation so we're going to add 3 to both sides so we get 4 cosine squared X is equal to 3 we're going to divide both sides by 4 and now we have cosine squared x equals 3/4 but we're going to want to do is take the square root of both sides to get the cosine by itself and when we do that when you take the square root of both sides you get two answers positive or negative square root of 3 we're going to leave a square root of 3 square root of 4 is equal to 2 so now what we're doing is we're saying cosine of what angle equals positive root 3 over 2 or negative root 3 over 2 and remember the cosine is the x-coordinate on the unit circle so you can see it's going to be positive 3 over 2 here and here negative root 3 over 2 here and here so if I just draw a sketch just to kind of illustrate here you can see we're at these 30 degree angles these PI over 6 degree reference angles so you can see it's actually going to be PI over 6 in each of these quadrants first second third and fourth but what I want you to notice is that see how over here at PI over six and seven PI over 6 see how they're like diametrically opposed it's like a diameter like across from each other and same thing over here with 5 PI over 6 and 11 PI over 6 these ones are diagonally across from each other diametrically opposed so when we write our general solution we can group those together to make it a little bit more compact of an answer so we have x equals PI over 6 plus because we're going halfway around the circle here that represents a 180 degrees or PI radians so we're going to say plus PI n where n is an integer and also we're going to take this one here 5 PI over 6 5 PI over 6 plus PI and we're and as an integer it can't be a half a third etc so we were able to group these together to make it a little bit more compact this would be our general solution if it just said you know find all the solutions between 0 to 2 pi ok like this then you would just say PI over 6 5 PI over 6 7 PI over 6 and 11 PI over 6 but what we're focusing on here is writing a general solution ok number three we have a little bit more Third Example challenging one we have cosine X minus secant of x equals zero how we solve that one and by the way if you like the way that I'm explaining these concepts and you want to go deeper into algebra to definitely check out my algebra 2 video course for sale that goes into all these different concepts plus more covering algebra 2 topics but let's jump into this one here what we're going to do is we're going to change the secant of X into 1 over cosine of X because remember secant is a reciprocal of cosine right and so now what we're going to do is we're gonna clear this denominator here by multiplying through by the cosine of X so I'm going to distribute that cosine of X to the left and right sides of the equation keeping it balanced right so this is going to give us cosine squared of X minus 1 okay because here these are cancelling numerator and denominator and 0 times anything of course is 0 so now I'm going to add 1 to both sides of the equation so this gives us 1 + cosine squared X and if we take the square root now of both sides to get the cosine by itself we get plus or minus 1 because remember when we take this credible sides we get those two answers positive or negative so now we go to the unit circle remember cosine is the x-coordinate on the unit circle and we say well where is the x-coordinate 1 or negative 1 so it's gonna be 1 here and it's going to be negative 1 here and those are the only two locations so just to draw a quick sketch it's going to be here and here and again notice that they're across from each other like a diet diag and all diameter has what I'm trying to say there so what we have is we actually have 0 plus we keep adding PI to get to this point and another PI to get to this point etc so if we keep going halfway around that circle right on up at one of these points where cosine is positive or negative 1 so you can see that x equals 0 plus pi + or since 0 is really nothing we could just say x equals PI n where n is an integer and this would be our general solution if you want to see more examples follow me over to that video right there where I show you how to solve even more trigonometric functions |
9583 | https://www.quora.com/Centripetal-force-F-mv-2%C3%B7r-What-does-it-mean-and-how-is-it-derived | Centripetal force F=mv^2÷r. What does it mean, and how is it derived? - Quora
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Centripetal force F=mv^2÷r. What does it mean, and how is it derived?
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John James
Lives in The United Kingdom · Author has 205 answers and 202.6K answer views
·6y
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Can anyone prove centripetal force= mv^2/r?
Consider a body with mass m m moving about a circular path of radius r r at velocity v v.
If the body commences its rotation at the top of the circle in the illustration above, its velocity vector will be pointing, say, to the left. Once the body reaches the bottom of the circle, the vector will have reversed its direction completely - therefore, the acceleration of the body during this period will be v−−v t=2 v t v−−v t=2 v t. The velocity is constant, and so t=x v t=x v, where x x is the displacement of the body. This will be equal to 2 r 2 r (the diameter of the motion).
Simplifying this, we ha
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Consider a body with mass m m moving about a circular path of radius r r at velocity v v.
If the body commences its rotation at the top of the circle in the illustration above, its velocity vector will be pointing, say, to the left. Once the body reaches the bottom of the circle, the vector will have reversed its direction completely - therefore, the acceleration of the body during this period will be v−−v t=2 v t v−−v t=2 v t. The velocity is constant, and so t=x v t=x v, where x x is the displacement of the body. This will be equal to 2 r 2 r (the diameter of the motion).
Simplifying this, we have that the centripetal acceleration is equal to 2 v⋅v 2 r=2 v 2 2 r=v 2 r 2 v⋅v 2 r=2 v 2 2 r=v 2 r
Assuming that the mass of the body is invariant, we can apply Newton’s second law of motion to find that F=m a=m v 2 r F=m a=m v 2 r. This force is causing the circular motion, and acts towards the centre of rotation.
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I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend.
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Can anyone prove centripetal force= mv^2/r?
What is this force that is also the subject of the equation F= (mv^2) /r?
What is the difference between mv^2/r and rm(w) ^2? They both denote the centripetal force but sometimes they don't give the same answer. Check comments for an example.
Is the radius directly proportional to the centripetal force or inversely proportional? The formulas are f=m v 2/f=m v 2/r and f=m r w 2 f=m r w 2.
How was the formula of centripetal force derived?
Anonymous
Updated 7y
Suppose we have uniform circular motion. (Any other kind of curved motion can be broken down into infinitesimal bits of uniform circular motion). Uniform circular motion about (0,0) can be described by the vector equation:
x=⟨r sin(ω t+ϕ),r cos(ω t+ϕ)⟩x=⟨r sin(ω t+ϕ),r cos(ω t+ϕ)⟩
The velocity is the derivative of this:
v=⟨r ω cos(ω t+ϕ),−r ω sin(ω t+ϕ)⟩v=⟨r ω cos(ω t+ϕ),−r ω sin(ω t+ϕ)⟩
And acceleration is the second derivative:
a=⟨−r ω 2 sin(ω t+ϕ),−r ω 2 cos(ω t+ϕ)⟩a=⟨−r ω 2 sin(ω t+ϕ),−r ω 2 cos(ω t+ϕ)⟩
Now if you calculate the magnitude of t
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Suppose we have uniform circular motion. (Any other kind of curved motion can be broken down into infinitesimal bits of uniform circular motion). Uniform circular motion about (0,0) can be described by the vector equation:
x=⟨r sin(ω t+ϕ),r cos(ω t+ϕ)⟩x=⟨r sin(ω t+ϕ),r cos(ω t+ϕ)⟩
The velocity is the derivative of this:
v=⟨r ω cos(ω t+ϕ),−r ω sin(ω t+ϕ)⟩v=⟨r ω cos(ω t+ϕ),−r ω sin(ω t+ϕ)⟩
And acceleration is the second derivative:
a=⟨−r ω 2 sin(ω t+ϕ),−r ω 2 cos(ω t+ϕ)⟩a=⟨−r ω 2 sin(ω t+ϕ),−r ω 2 cos(ω t+ϕ)⟩
Now if you calculate the magnitude of these vectors, you get:
|x|=r|x|=r
|v|=r ω|v|=r ω
|a|=r ω 2|a|=r ω 2
From which you can immediately see that
|a|=|v|2 r|a|=|v|2 r
Since F=m a F=m a, this means that:
|F|=m|a||F|=m|a|
|F|=m|v|2 r|F|=m|v|2 r
Bonus:
The direction of these vectors is:
^x=⟨sin(ω t+ϕ),cos(ω t+ϕ)⟩x^=⟨sin(ω t+ϕ),cos(ω t+ϕ)⟩
^v=⟨cos(ω t+ϕ),−sin(ω t+ϕ)⟩v^=⟨cos(ω t+ϕ),−sin(ω t+ϕ)⟩
^a=⟨−sin(ω t+ϕ),−cos(ω t+ϕ)⟩a^=⟨−sin(ω t+ϕ),−cos(ω t+ϕ)⟩
from which it can be seen that:
^r=−^x r^=−x^
This shows that the acceleration is always opposite the position; that is, towards the center. The same is true of the force.
^F=−^x F^=−x^
Also, notice that
F⋅v=0 F⋅v=0,
which is to say,
F⋅x d t=0 F⋅x d t=0
meaning that no work is done by centripetal force.
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Vir Narayan Singh
Former Professor (Retired) at Indian Institute of Technology, Roorkee (1972–2011) · Upvoted by
David Cousens
, PhD Physics, Griffith University (1982) · Author has 1K answers and 1.3M answer views
·6y
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How do we derive the formula for centripetal force?
Consider the following diagram:
An object is moving in a circular path of radius ‘r’. In a small amount of time Δt, it moves from A to B. At time t1, it is at A and its velocity is vi. At time t2, it is at B with velocity vf and Δt = t2-t1. The velocities vi and vf change only in direction. Their magnitudes remain the same. The average acceleration is
a = (vf-vi)/Δt = Δ v/Δt
Consider the two triangles with sides Δr, r, r and Δv, vi, vf. These are similar triangles. So, the following relation can be written:
Δv/v = Δr/r
So, Δv = vΔr/r
and a = (v/r)(Δr/Δt)
As point B approaches A, Δt tends to zero and Δ
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Consider the following diagram:
An object is moving in a circular path of radius ‘r’. In a small amount of time Δt, it moves from A to B. At time t1, it is at A and its velocity is vi. At time t2, it is at B with velocity vf and Δt = t2-t1. The velocities vi and vf change only in direction. Their magnitudes remain the same. The average acceleration is
a = (vf-vi)/Δt = Δ v/Δt
Consider the two triangles with sides Δr, r, r and Δv, vi, vf. These are similar triangles. So, the following relation can be written:
Δv/v = Δr/r
So, Δv = vΔr/r
and a = (v/r)(Δr/Δt)
As point B approaches A, Δt tends to zero and Δr/Δt = v. So,
a = v^2/r
In the two figures above, vi and vf are parallel and Δ v points to the centre of the circular path and so the acceleration also points to the centre of the circular path. The acceleration ‘a’ is then directed towards the centre of the circular path as the object moves along this path. This acceleration, a = v^2/r, is centripetal acceleration.
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Smitaj
Bsc in Physics&Mathematics, Monash University (Graduated 1973) · Author has 25.6K answers and 18.6M answer views
·1y
The MEANING of it is that it REQUIRES a force to keep something moving in a circle. Remove the force and the object goes straight.
Try turning a corner in your car on sheet ice and see it for yourself.
The AMOUNT of force required is given by the formula. And by rewriting we can get v= sqrt(F r / m) which states the highest speed you can get around a corner with a given force.
Or r = m v^2/ F which is the sharpest corner that can be used for a given force and speed.
The derivation is from the definitions. 1) F = ma = m dv / dt
You take two points that are close together and find the two velocities
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The MEANING of it is that it REQUIRES a force to keep something moving in a circle. Remove the force and the object goes straight.
Try turning a corner in your car on sheet ice and see it for yourself.
The AMOUNT of force required is given by the formula. And by rewriting we can get v= sqrt(F r / m) which states the highest speed you can get around a corner with a given force.
Or r = m v^2/ F which is the sharpest corner that can be used for a given force and speed.
The derivation is from the definitions. 1) F = ma = m dv / dt
You take two points that are close together and find the two velocities as vectors. Subtract them to get dv.
Now change in v = v theta ( from the triangle formed.) so dv / dt = v d theta / dt but d theta on dt is called the “angular velocity” w
Hence by substitution we get F = m v w and we can replace w with v/r to get F = mv^2/r or we can replace v with wr to get F = m w^2 r
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If we take formula FS= K.E, then we replace force with centripetal force and try to derive it will end up on 2S=r. What does it mean?
What is the reaction of Centripetal force?
What is an example of centripetal force, and what are its units?
What is the expression of centripetal force?
Why does centripetal force push outward?
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·
1y
Centripetal force is the force that keeps an object moving in a circular path. It acts towards the center of the circle around which the object is moving. The formula for centripetal force is given by:
F=m v 2 r F=m v 2 r
Meaning of the Formula
F: Centripetal force (in newtons, N)
m: Mass of the object (in kilograms, kg)
v: Tangential speed of the object (in meters per second, m/s)
r: Radius of the circular path (in meters, m)
This equation indicates that the centripetal force required to keep an object moving in a circle increases with:
The mass of the object (more massive objects require more for
Continue Reading
Centripetal force is the force that keeps an object moving in a circular path. It acts towards the center of the circle around which the object is moving. The formula for centripetal force is given by:
F=m v 2 r F=m v 2 r
Meaning of the Formula
F: Centripetal force (in newtons, N)
m: Mass of the object (in kilograms, kg)
v: Tangential speed of the object (in meters per second, m/s)
r: Radius of the circular path (in meters, m)
This equation indicates that the centripetal force required to keep an object moving in a circle increases with:
The mass of the object (more massive objects require more force).
The square of the speed (if the speed doubles, the required force increases by a factor of four).
The inverse of the radius (a smaller radius requires more force).
Derivation of the Formula
Understanding Circular Motion: When an object moves in a circle, it is constantly changing direction, which means it is accelerating. This acceleration is called centripetal acceleration (a c a c).
Centripetal Acceleration: The formula for centripetal acceleration can be derived from the definition of acceleration in circular motion. The acceleration is given by:
a c=v 2 r a c=v 2 r
where v v is the tangential velocity and r r is the radius of the circular path.
3. Newton's Second Law: According to Newton's second law, the net force acting on an object is equal to its mass times its acceleration:
F=m a F=m a
For centripetal motion, we can substitute the centripetal acceleration into this equation:
F=m⋅a c=m⋅v 2 r F=m⋅a c=m⋅v 2 r
4. Final Formula: This leads us directly to the centripetal force formula:
F=m v 2 r F=m v 2 r
Summary
The centripetal force is essential for maintaining circular motion, and it is directly related to the mass of the object, the speed of the object, and the radius of the path. Understanding this concept is crucial in physics, especially in mechanics and dynamics involving circular motion.
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KRay
Associate Professor of Physics (2002–present) · Author has 2.1K answers and 476.3K answer views
·1y
Related
How do you derive centripetal force?
I never use the term “centripetal force” in my classes because it tends to lead students to believe that it somehow a different force than any of the forces they know. It’s not.
Imagine a car going around in a circle. What’s the centripetal force? It’s the static friction force.
Imagine a ball attached to a string going around a vertical circle. What’s the centripetal force? It’s a combination of the tension force and gravity. At the top, it is gravity force minus the tension force. At the bottom it is tension force minus the gravity force. At the bottom, this is why the tension force must b lar
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I never use the term “centripetal force” in my classes because it tends to lead students to believe that it somehow a different force than any of the forces they know. It’s not.
Imagine a car going around in a circle. What’s the centripetal force? It’s the static friction force.
Imagine a ball attached to a string going around a vertical circle. What’s the centripetal force? It’s a combination of the tension force and gravity. At the top, it is gravity force minus the tension force. At the bottom it is tension force minus the gravity force. At the bottom, this is why the tension force must b large.
In other words, the “centripetal force” is ALWAYS a force or combination of forces that you are already familiar with.
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9 2
Anonymous
4y
Originally Answered: What is this force that is also the subject of the equation F= (mv^2) /r? ·
Centripetal force
The equation F = (mv^2)/r shows that force (centripetal) is equal to the product of the mass of an object and its velocity squared all divided by the radius of the circle (orbital path). When working with calculations related to circular motion, one must work in radians rather than in degrees. There is even an equation that shows the conversion of degrees into radians and that is 360 degrees = 2pi radians. As you can see, there are 2pi radians in a full circle and you will see 2pi popping up quite frequently in physics. One equation where it is noticeable is the relationshi
Continue Reading
Centripetal force
The equation F = (mv^2)/r shows that force (centripetal) is equal to the product of the mass of an object and its velocity squared all divided by the radius of the circle (orbital path). When working with calculations related to circular motion, one must work in radians rather than in degrees. There is even an equation that shows the conversion of degrees into radians and that is 360 degrees = 2pi radians. As you can see, there are 2pi radians in a full circle and you will see 2pi popping up quite frequently in physics. One equation where it is noticeable is the relationship between angular velocity (omega) and time period (T). The equation is T = (2pi)/(omega) and as you can see the important 2pi is present.
Angular velocity is the angle turned about a point per second and it has the units radians per second (rad s^-1).
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Ron Brown
Decades of teaching physics to undergrads · Author has 13.6K answers and 84.3M answer views
·4y
Related
How do you derive centripetal force?
The centripetal force on an object is just whatever force is acting on it perpendicular to it’s motion to make it change its direction. That is, it is the force that makes the object momentarily travel in a curved path. It is “centripetal” in that it points toward the center of curvature of the motion at that moment.
But, being a force, it is caused by some interaction with another object. For example, a ball on a string traveling in a circle is being pulled by the string - so the tension is the centripetal force. A car traveling in a circular path on a flat turn does so because the friction of
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The centripetal force on an object is just whatever force is acting on it perpendicular to it’s motion to make it change its direction. That is, it is the force that makes the object momentarily travel in a curved path. It is “centripetal” in that it points toward the center of curvature of the motion at that moment.
But, being a force, it is caused by some interaction with another object. For example, a ball on a string traveling in a circle is being pulled by the string - so the tension is the centripetal force. A car traveling in a circular path on a flat turn does so because the friction of the road on the tires turns it into a curved path. The Moon orbiting the Earth is being pulled into its orbit by the gravitational force, and so on.
So I think the question is asking about the centripetal force equation, commonly written
where m is the mass of the object, v is its speed, and r is the radius of curvature of the path at that particular moment.
But that equation is simply saying, whatever force is causing it to turn into a curved path, it must be equal to the mass of the object times the component of the acceleration perpendicular to its path. That is, F=mac, where ac here means the centripetal acceleration at that moment.
So your question is actually, “How do you derive the centripetal acceleration?”
The problem is, it is not a trivial question if you are not familiar with describing circular motion using vectors in polar coordinates and then using differential calculus to obtain the corresponding acceleration. I will set it up, but won’t solve it here:
To describe the location of a object in uniform circular motion in polar coordinates, one needs to describe the r vector as a function of time. That is,
where the i and j (with little ‘hats’) are called unit vectors in the directions of the x- and y-axes. But the angle is time dependent if the object is traveling around the circular path. That is,
where the omega symbol represents the rate at which the angle is increasing (called the angular velocity).
Given that location vector r(t), one needs to differentiate once with respect to time to get the velocity vector as a function of time and then again to get the acceleration vector. The resulting acceleration vector is
The minus sign means the acceleration is in the opposite direction as the radius vector, i.e., it points back toward the center of the circle, hence is centripetal. It’s magnitude is just
since the angular velocity is just the speed divided by the radius of curvature.
[I haven’t done all the steps, of course, and this is set up assuming uniform circular motion for simplicity, but the result is valid for any curved path at a particular moment.]
So pulling all of those pieces together says the “centripetal force” is the mass times the centripetal acceleration, hence can be written
which has a magnitude
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9 1
Ganesh
Illustrator, cartoonist, sculptor knowing a bit of Physics · Author has 2.2K answers and 2.4M answer views
·Updated 2y
Newton's first law tell us that an object continues to move in a straight line at an unchanging velocity unless an external force gets applied to the object forcing it to change its direction or magnitude. If the external force is applied in a direction perpendicular to the line of motion then the object will respond changing its direction and curving along an arc towards the direction of the force.
The radius of the arc is equal to m v 2 F m v 2 F meters, where m m is the object’s mass, v v is its velocity and F F is the Force applied.
Viewed in another way, if an object of mass m, moving in a circle
Continue Reading
Newton's first law tell us that an object continues to move in a straight line at an unchanging velocity unless an external force gets applied to the object forcing it to change its direction or magnitude. If the external force is applied in a direction perpendicular to the line of motion then the object will respond changing its direction and curving along an arc towards the direction of the force.
The radius of the arc is equal to m v 2 F m v 2 F meters, where m m is the object’s mass, v v is its velocity and F F is the Force applied.
Viewed in another way, if an object of mass m, moving in a circle of radius r, at a velocity of v meters (measured along the tangent to the circle), then it is certain that a Force equal to m v 2 r m v 2 r must be acting on the object at a direction perpendicular to its motion, which is causing to continuously change its direction. Such a Force is called a Centripetal Force.
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Richard Mentock
Studied geodesy and geophysics · Author has 1.7K answers and 707.4K answer views
·5y
Force is mass m times acceleration. With constant speed, the acceleration is just the change in direction of the velocity v, so accleration is just v times the rotational speed, which is v/r
F = m v v/r
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9 1
Yebraw Mada
Studied Mathematics and Physics at University of Bristol (Graduated 2021) · Author has 164 answers and 427K answer views
·8y
Related
Can anyone prove centripetal force= mv^2/r?
From Newton’s second Law, it can be shown that F=m a F=m a, where F is the net force acting on a body, m is the mass of the body and a is the acceleration. So to find the centripetal force we must first find the centripetal acceleration for a body undergoing circular motion.
The following is from a previous answer of mine where I derived the centripetal acceleration:
A particle in circular motion follows the graph of x 2+y 2=r 2 x 2+y 2=r 2, where r is the radius of the circular motion. This can be expressed in parametric equations with the time variable, allowing us to see where the particle would be at a specifi
Continue Reading
From Newton’s second Law, it can be shown that F=m a F=m a, where F is the net force acting on a body, m is the mass of the body and a is the acceleration. So to find the centripetal force we must first find the centripetal acceleration for a body undergoing circular motion.
The following is from a previous answer of mine where I derived the centripetal acceleration:
A particle in circular motion follows the graph of x 2+y 2=r 2 x 2+y 2=r 2, where r is the radius of the circular motion. This can be expressed in parametric equations with the time variable, allowing us to see where the particle would be at a specific point in time.
These are
x=r cos(2 π t T)x=r cos(2 π t T)
y=r sin(2 π t T)y=r sin(2 π t T)
Where T is the time period of the motion.
T=1 f T=1 f
Where f is the frequency and
ω=2 π f ω=2 π f
Where ω ω is angular velocity.
Therefore the parametric equations can be re-written as
x=r cos(ω t)x=r cos(ω t)
y=r sin(ω t)y=r sin(ω t)
Now we need to find the acceleration for the x and y components of position. We find these by double differentiating each position variable with respect to time.
∂2 x∂t 2=−r ω 2 cos(ω t)∂2 x∂t 2=−r ω 2 cos(ω t)
∂2 y∂t 2=−r ω 2 sin(ω t)∂2 y∂t 2=−r ω 2 sin(ω t)
Now we just need to find the resultant acceleration.
a 2=(∂2 x∂t 2)2+(∂2 y∂t 2)2 a 2=(∂2 x∂t 2)2+(∂2 y∂t 2)2
a 2=r 2 ω 4(c o s 2 ω t+s i n 2 ω t)a 2=r 2 ω 4(c o s 2 ω t+s i n 2 ω t)
a 2=r 2 ω 4 a 2=r 2 ω 4
a=r ω 2 a=r ω 2
Let’s now write this in a different form.
ω=v r ω=v r
Where v v is linear velocity. Therefore centripetal acceleration can be written as
a=v 2 r a=v 2 r
F=m a F=m a and so the centripetal force must therefore be
F=m v 2 r F=m v 2 r
Q.E.D.
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99 28
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Abhijeet Kumar
BS-MS Student at IISER · Author has 378 answers and 228.6K answer views
·3y
Originally Answered: What is this force that is also the subject of the equation F= (mv^2) /r? ·
F=(mv^2)/r
It is an expression for centripetal force.This gives us magnitude of force a body moving in circular motion experiences
The magnitude of centrifugal force is also equal to (mv^2)/r but direction is opposite to that of centripetal force
Centripetal force always act towards the centre while centrifugal always act away from the centre.
But the magnitude of both centripetal and centrifugal force is equal to (mv^2)/r
Thanks for A2A:)
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Francis Koh
Author has 86 answers and 54.7K answer views
·4y
Originally Answered: Why is centrifugal force F = m v² / r? ·
Newton’s second law.
F = ma
a is defined as velocity over time.
distance is defined as velocity times time, so
angular a = (angular v^2)r (in angular terms)
angular v = vr
a = (v^2)/r
so F = ma
F = m(v^2)/r
F = m(v^2)/r
Hope this helps.
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9 1
Related questions
Can anyone prove centripetal force= mv^2/r?
What is this force that is also the subject of the equation F= (mv^2) /r?
What is the difference between mv^2/r and rm(w) ^2? They both denote the centripetal force but sometimes they don't give the same answer. Check comments for an example.
Is the radius directly proportional to the centripetal force or inversely proportional? The formulas are f=m v 2/f=m v 2/r and f=m r w 2 f=m r w 2.
How was the formula of centripetal force derived?
If we take formula FS= K.E, then we replace force with centripetal force and try to derive it will end up on 2S=r. What does it mean?
What is the reaction of Centripetal force?
What is an example of centripetal force, and what are its units?
What is the expression of centripetal force?
Why does centripetal force push outward?
What is the definition of centripetal force? What is the direction of centripetal force? What is the magnitude of centripetal force?
If Centripetal force are high, what effect may be?
Why are the formulas for Centrifugal Force and Centripetal Force the same: F = mv²/r?
What is centripetal force? What is its function?
Why is the centripetal force directed inwards?
Related questions
Can anyone prove centripetal force= mv^2/r?
What is this force that is also the subject of the equation F= (mv^2) /r?
What is the difference between mv^2/r and rm(w) ^2? They both denote the centripetal force but sometimes they don't give the same answer. Check comments for an example.
Is the radius directly proportional to the centripetal force or inversely proportional? The formulas are f=m v 2/f=m v 2/r and f=m r w 2 f=m r w 2.
How was the formula of centripetal force derived?
If we take formula FS= K.E, then we replace force with centripetal force and try to derive it will end up on 2S=r. What does it mean?
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The word "allure" refers to the quality of being powerfully and mysteriously attractive or fascinating, often highlighting the concepts of charm, appeal, and enticement. It embodies the notions of drawing someone in through irresistible attraction, playing a significant role in various aspects of language, perception, and descriptive language.
This word, "allure," functions as both a noun and a verb. As a noun, it describes the quality of being powerfully and mysteriously attractive or fascinating. As a verb, it means to attract or charm.
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Contents
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2 Origin of the word
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4 Related Words or Synonyms
Definition of the word[]
The word "allure" is defined as a noun meaning the quality of being powerfully and mysteriously attractive or fascinating, such as in the sentence "The allure of the ancient ruins captivated the tourists." As a verb, it means to attract or charm, as in the sentence "The singer's voice allured the audience." This definition highlights its role in expressing the concepts of charm, appeal, and enticement in various contexts.
Origin of the word[]
The word "allure" originates from the Old French word "aleurer," meaning to attract, which is derived from the Latin word "ad-" (to) and "lura" (a lure). The term evolved through Middle English, maintaining its meaning related to attraction and charm. The historical evolution of "allure" reflects its consistent use in describing qualities that draw people in through fascination and appeal. Its etymology underscores the importance of expressing the ideas of charm and enticement in human language and interactions.
Usage of the word[]
This word "allure" is widely used to describe the quality of being powerfully and mysteriously attractive or fascinating, and related concepts in various contexts. In everyday conversation, it appears in sentences like "The new product's allure lies in its innovative design," indicating a common reference to charm and appeal. In literary and artistic contexts, "allure" often denotes discussions about elements that captivate and enchant, emphasizing the role of attraction and mystery in shaping human experiences and artistic expressions, such as "The allure of the unknown drew explorers to distant lands," highlighting its role in conveying thematic richness and practical significance. In broader metaphorical and descriptive contexts, it can refer to any situation involving the quality of being irresistibly attractive, as in "The allure of the city lights was impossible to resist," underscoring the idea of charm and enticement. Its usage spans different fields, underscoring its versatility in expressing the concepts of charm, appeal, and enticement.
Related Words or Synonyms[]
The word "allure" has several related words and synonyms that can be used depending on the context. Synonyms like "charm," "appeal," "attraction," and "enticement" convey similar meanings of the quality of being powerfully and mysteriously attractive or fascinating. Terms such as "fascination," "glamor," "magnetism," and "seduction" can also serve as alternatives when referring to allure, each bringing slight nuances to the expression. Additionally, phrases like "alluring beauty" and "alluring charm" capture the essence of the word and its various applications. These synonyms enrich the language by providing varied ways to articulate the concept of "allure," maintaining the core ideas of charm, appeal, and enticement.
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小学奥数专题---解析
周5s11adsqd980 2017-04-10 | 801阅读 | 32转藏
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一、填空选择题
1、_。
解析：该题型是竞赛中送分题型，不过也要仔细认真，不能出任何小差错。
根据分析可知：
，
，
...
提取公因式，即可得
原式==
2、_。
解析：此题同上。
根据分析：362+186=548，由此我们可以将式子进行转化，
原式= ==1
3、如果2×3^8能表示成K个连续正整数的和，则K的最大值为_____。（选自2016华杯赛）
解析：
此题在于考察数字求和与因式分解能力。难度不大，需要细心。
首先，可以将K个连续正整数之和写成一个式子，可设第一个数为t,那么就可以写成
(t+t+K-1)K÷2=2×3︺8
(K+2t-1)K=22×3︺8
观察左边的乘积式子，t作为第一个正整数，即t≥1,所以这个乘积式子的左边恒大于右边，并且两边之差是奇数(2K-1是奇数)，即左右两边有奇有偶。此时，我们再看等式的右边，要使得右边分解后的两个因式差值是奇数，那么22只能作为其中一个因式所用(因为两个因式都含2的话就为两个偶数，差值也为偶数)，要使得K为最大，即要使两个因式的差值最小化，这样分解得
22×33×3︺5
22×33与3︺5之间差值是最小的
此时，K值为22×33=108，t值为(3︺5-108+1)÷2=68，则他们就是：68.69.70.71......175。有时间的话可以可以验算一遍。
则K的最大值为108。
4、某校一二年级共有少先队员300人，二年级少先队员人数的2/5比一年级少先队员的1/4多55人，一年级少先队员_人，二年级少先队员___人。(小学六年级通用题)
解析：此题属单位标准量1的问题。
我们把两个比较的量列举如下：
二年级×2/5 = 一年级×1/4+55 （1）
（1）两边同时乘以4，即得，
二年级×8/5 = 一年级+220 （2）
由已知条件：一年级+二年级=300 （3）
（2）（3）两个式子综合，即
二年级×8/5 - 220 + 二年级 = 300，
即：二年级×13/5=520，得
二年级少先队员数量：520÷13/5=200，则一年级少先队员数量：300-200=100
5、一项工程，如果甲队单独做5天后，乙队再单独做7天，可以完成工程的1/5，如果甲队单独做7天，乙队再单独做5天，可以完成工程的1/4。如果甲队单独做完全部工程，需要_____天。(小学六年级通用题---工程问题)
解析：此题属于工程问题，实质是与上题单位标准量1的问题是同属一类问题。同样列举已知条件如下：
甲×5天 + 乙×7天 = 1/5 （1）
甲×7天 + 乙×5天 = 1/4 （2）
（2）式-（1）式，得
甲×2天-乙×2天=1/20，则
甲×1天-乙×1天=1/40 （3）
即甲队1天比乙队1天多做工程的1/40，
将（3）式代入（1）式，得
甲×5天 +（甲-1/40）×7天 = 1/5
即：甲×12天 = 1/5+7/40=3/8
—>甲×1天 = 1/32
—>甲队完成整个工程需要：单位1 ÷ 1/32 = 32（天）
6、甲数是乙数、丙数、丁数之和的1/2，乙数是甲数、丙数、丁数之和的1/3，丙数是甲数、乙数、丁数之和的1/4。已知丁数是260，那么甲数是_，乙数是，丙数是__。(小学六年级通用题---工程问题)
解析：同样列举已知条件来分析，获得更多关联的信息
甲 =（乙 + 丙 + 丁）×1/2 （1）
乙 =（甲 + 丙 + 丁）×1/3 （2）
丙 =（甲 + 乙 + 丁）×1/4 （3）
为了将以上3个式子综合在一起思考，观察式子特点，每个式子的右边都是3个数的和，并且缺的那个数就是式子左边的，这样的话，我们分别将3个数转化成4个数之和，由（1）（2）（3）分别对应得出结论：
甲 =（甲 + 乙 + 丙 + 丁）×1/3 （4）
乙 =（甲 + 乙 + 丙 + 丁）×1/4 （5）
丙 =（甲 + 乙 + 丙 + 丁）×1/5 （6）
由（4）（5）（6）可知，
丁 =（甲 + 乙 + 丙 + 丁）×（1 - 1/3 - 1/4 - 1/5）
=（ 甲 + 乙 + 丙 + 丁）×13/60
又 ∵ 丁 = 260
∴ 甲+乙+丙+丁=1200
∴ 甲 = 12001/4 = 400
乙 = 12001/4 = 300
丙 = 12001/5 = 240
7、一只野兔逃出80步后猎狗才发现，开始追它。已知野兔跑8步的路程，猎狗只需跑3步；猎狗跑4步的时间，野兔要跑9步。那么猎狗至少要跑_____步才能追上野兔。(小学六年级通用题---追及问题)
解析：这是一道追及问题，也是常用题型。
列举两个已知条件，尽可能得出更多的结论
野兔 × 8步 = 猎狗 × 3步 (距离相同)
T(猎狗4步) = T(野兔9步） (T代表时间)
此题的关键就是找出相同时间内猎狗的速度比野兔快多少步，而后知晓追上80步（距离差）所需要的时间。
由已知条件 T（猎狗4步） = T（野兔9步 ）
——> T（猎狗12步） = T（野兔27步) （1）
由已知条件 3步（猎狗） = 8步（野兔）
——> 12步（猎狗） = 32步（野兔) （2）
由（1）（2）两个式子分析
（1）式即是：在猎狗跑12步的时候，野兔跑27步
（2）式即是：猎狗跑12步的路程相当于野兔跑32步的路程
则在相同时间T（猎狗12步）内，猎狗比野兔多跑：
32步(野兔） - 27步(野兔） = 5步(野兔）
而野兔事先逃出80步，则可知猎狗要追上这80步需要跑：
80÷5×12=192（步）
8、如下图，正方形ABCD的边长为5，E、F为正方形外两点，满足AE=CF=4,BE=DF=3,那么EF'=_。（选自2016华杯赛）
解析：此题有两种方法：
第一种：计算比较繁琐，分过E，F做BC的垂线，垂足分别为M，N，计算出两个高EM，FN和MN的长度，即可计算EF2。
第二种：分别延长EB与FC交于点P，则根据分析可知∠P=90°，并且PB=4,PC=3，即可计算出EF2。
EF2=72+72=98
9、小明每个周末要去老师家，如图是小明从家到老师的路线，小明去老师家沿最短路线走，共有____条不同的路线。(小学奥数杂题)
解析：此题属于路线问题。一般这种题型掌握要点即可解决。
因为该题方格不多，很多同学可以用手画几下就可得出答案，但是这种只能解决这种简单的，当遇到方格数比较多的时候，这种用手画的方法已经不实用了，下面介绍一种用数字累加法则。
此题的前提是求出最短线路，首先要要明确的是每到一个交叉点不能走回头路也不能向与目标相反的方向走。
我们以小明家为起点，标记数字1为起点，有两个方向选择，即向右横向或者向下竖向出发，我们之所以给每个点标记数字，是因为所标记的数字都是代表起点到该点的路线数量(最短路线)。可以任意选取一个点做比方。从每个交叉点的角度上看，它所标记的数字是其上一个横向过来的数字与上一个竖向过来的数字之和，依次类推，依次给每个点标上数字，其结果如下：
即结果为13条不同线路且为最短路线。
拓展练习：
10、一张数学试卷，共有25道选择题，做对一题得4分，做错一题扣1分，如不做，不得分也不扣分。若某同学得78分，那么做对题,做错题,没做___题。(小学奥数通用题)
解析：该同学得分78，说明做对的题目所得分数至少不能少于78，204=80>78,说明至少答对20题，当做对20题时，204-78=2，即做错2题，25-20-2=3，即没做3题，当答对21题时，214-78=6，即做错6题，25-21-6=-2，与实际矛盾，即只有唯一答案：做对20题，做错2题，没做3题。
11、由于天气渐冷，牧场上的草每天以均匀的速度减少，经计算，现有牧场上的草可供20头牛吃5天，也可供16头牛吃6天，那么可供11头牛吃___天。(小学奥数通用题)
解析：此题为牛吃草问题，是小学奥数竞赛中的典型习题，也属于必考类型题，一定要掌握。
此种类型，我们可以考虑：1头牛1天共吃草为k，把k作为一个标准单位，就像我们常用的KG（质量单位）与KM（长度单位）一样。这样我们可以知道，20头牛吃5天即205=100k，100k是20头牛5天吃草的量，16头牛吃6天即166=96k，96k是16头牛6天吃草的量，之所以会有100k和96k两个不同的量，是因为他们的时间不一样，5天和6天，牧场的草量是以均匀的速度减少，那么6天减少的量比5天减少的量多一天，则可得1天牧场的草减少量：100k-96k=4k，这样就可以计算出原始牧场的总草量：100k+54k=120k或者96k+64k=120k。
第二步骤：利用已经所得结论：1天牧场的草减少量为4k，可看作是每天都有4K/1天=4头牛在吃草，即是120k/（20+4）头=5天，120k/（16+4）头=6天，当然答案也迎刃而解，即是120k/（11+4）头=8天
12、2^2016+2016^2除以7的余数是_____。（小学整除与余数）
解析：此题我把2的幂数改成今年的数字，没想到2016是7的倍数，等于2016^2已经不用考虑，但对这题而言，没有影响，解题思路还是一样。
因2^3=8,而8/7，余数为1，2^2016=2^（3672），则2^2016除以7的余数也为1，即所求2^2016+2016^2012除以7的余数是1。
13、某自然数可表示成9个连续自然数之和，也可表示成10个连续自然数之和，也可表示成11个连续自然数之和，那么符合条件的最小自然数是___。（公约数与公倍数）
解析：此题属于公约数、公倍数的基本题型，根据已知条件（连续自然数和的属性）可知，该自然数是9的倍数，5的是倍数，11的倍数，『可能有些同学会问10个连续自然数的和为什么不是10的倍数，关键问题在于10个数属于偶数个数，首尾相加刚好5对（奇数个数就不一样了，可以试写一下），5对数首尾相加的和都是一样，并且这个和都是奇数（5对首尾均是一个奇数与一个偶数相加），所以5对首尾相加的和即是一个奇数与5的乘积，即10个连续自然数的和是5的倍数，而不可能是10的倍数』，由于9、5、11两两互质，则该自然数是9511的倍数，即是495的倍数。495即为最小自然数。
14、1995的数字和是1+9+9+5=24，那么小于2000的四位数中数字和等于24的数有______个。（选自小学五年级奥数教程---数字和）
解析：小于2000的四位数，即说明千位上是1，则百位+十位+个位=24-1=23，可以由大到小排列一下：
（1）百位为9时， 十位9，个位5，即1995，十位8，个位6，即1986；十位7，个位7，即1977；十位6，个位8，即1968，十位5，个位9，即1959；
（2）百位为8时，十位9，个位6，即1896；十位8，个位7，即1887；十位7，个位8，即1878；十位6，个位9，即1869；
（3）百位为7时，十位9，个位7，即1797；十位8，个位9，即1789；
（4）百位为6时，十位为9，个位8，即1698，十位8，个位9，即1689；
（5）百位为5时，十位9，个位9.即1599
其他均不符合
则符合的四位数有：1995、1986、1977、1968、1959、1896、1887、1878、1869、1797、1788、1779、1698、1689、1599.
共计15个
15、一个三位数被13整除，其商恰好等于这个三位数各位数字之和，那么满足此条件的所有三位数之和是______。（选自小学五年级奥数教程）
解析：此题为常规题，是考验大家的应变能力。可设三位数为abc，则按条件可得，
100a+10b+c=13×（a+b+c）
即 87a = 3b+12c
29a = b+4c
因为 b≤9，C≤9，所以b+4c≤50，
即 29a≤50
则 a=1（a=0为两位数）
即 b+4c=29，
当c=7时，b=1；
当c=6时，b=5；
当c=5时，b=9；
则满足条件的三位数有117、156、195
117+156+195=468
16、小于1000且各位数字之和等于6的自然数共有______个。（小学数字和问题）
解析：根据题意可以得出两个结论:
(1)这个数可能是3位数或2位数或个位数;
(2)这个数个位数数字相加等于6;
可分情况讨论,如下:
1、当这个数为3位数时，按数字和等于6可以列举如下：
6 = 6+0+0
=5+1+0
=4+2+0=4+1+1
=3+3+0=3+2+1
=2+2+2
有且只有以上6种组合情况，那么，我们把所有符合条件的三位数列举出来：600；510，501，150，105；420，402，240，204；411，141，114；330，303，321，312，231，213，132，123；222。共计21个。
2、当这个数为2位数时，按数字和等于6可以列举如下：
6=6+0
=5+1
=4+2
=3+3
有且只有以上这4种组合情况，那么我们把所有符合条件的两位数列举出来：60；51，15；42，24；33.共计6个。
3、当这个数为个位数时，只有6。
则符合题意的个数为：21+6+1=28
17、从1～30这30个自然数中，任取3个不同的数，使得这3个数的和正好被3除尽，问共有____种不同的选法。(整除)
解析：1～30这个30个自然数，按照被3整除数的特点可以将他们归类分为3n+0，3n+1，3n+2这三种情况，例如：
7=32+1，10=33+1，即7、10同属3n+1这种情况
并且这三种情况的个数均为10，那么要使其中3个数的和能被3整除，共有下列几种组合：
（1）3n1+0、3n2+0、3n3+0；（2）3n1+0、3n2+1、3n3+2；
（3）3n1+1、3n2+1、3n3+1；（4）3n1+2、3n2+2、3n3+2；
有且只有这4种情况，那么计算出每种情况的数量，4种情况相加即得结果。
（1）3n1+0、3n2+0、3n3+0；
即从10个都被3整除的数中选出3个数的选法：C(3、10)=120
（2）3n1+0、3n2+1、3n3+2；
即从这三种数中各选出1个数的选法：C(1、10)×C(1、10)×C(1、10)=1000
（3）3n1+1、3n2+1、3n3+1；（与1同理）
即从10个都被3整除余1的数中选出3个数的选法：C(3、10)=120
（4）3n1+2、3n2+2、3n3+2；（与1同理）
即从10个都被3整除余2的数中选出3个数的选法：C(3、10)=120
综上所计，从1-30这30个自然数中，任选3个能被3整除的选法共有：120+1000+120+120=1360
18、海峡两岸总决赛于2016年8月21日在台湾大学举行，2016由数字0、1、2、6组成，由数字0、1、2、6组成(可以全用也可以不全用)的非0自然数中，从小到大排列，第2016个数字是_____。
解析：此题是第13、15题的升级版，同样是分类归纳总结：(从小到大分析)
(1)、当此数是个位数时：(非0自然数)
1、2、6
共3个数；
(2)、当此数为两位数时：
十位数有3种选择(即1、2、6)，个位数有4种选择(即0、1、2、6)
所以共有3×4=12个数
(3)、当为三位数时：
与上同理，共有：3?4?4=48个数
(4)当为四位数时：
与上同理，共有：3?4?4?4=192个数
(5)、当为五位数时：
与上同理，共有3?4?4?4?4=768
即此时是1-66666,共有768+192+48+12+3=1023个数，此时离所求的第2016个数还有一定的距离。
由前面所推理，那么从100000-166666共有4?4?4?4?4=1024个数，那么到此时共有1023+1024=2047,已经超出2016相差31个数，则考虑如下:
从166600-166666共有4?4=16个数,从166200-166266共有4?4=16个数，此时共计16+16=32,则减掉最小的一个数即是31个数，则第2016个数是166200。
19、甲、乙、丙、丁4个人参加数学竞赛，赛后猜测他们之间的成绩情况是：
甲说：“我考得最差。”
乙说：“我不是考得最差的。”
丙说：“我考得肯定是最好的。”
丁说：“我肯定没有丙考得好，但也不是最差的。”
成绩公布后，只有一人猜错了，则这4个人的实际成绩从高到低依次是______。
解析：此题可以用假设法，从已知结论反证四个人所说的话。则可知猜错的人是丙。他们的成绩从高到低依次是是乙-丙-丁-甲。
20、幼儿园里给小朋友分苹果，420个苹果正好均分，但今天刚好又新入一位小朋友，这样每个小朋友就要少分两个苹果，原来有______个小朋友。
解析：因此题是小学的的题型，尽量不要用方程解答。
分析：新入一位小朋友，所分的苹果少两个，即原来的小朋友每个人均少分两个苹果，可定义原来小朋友数量为a，那因新入一位小朋友，原来的小朋友一共少分a2个苹果，因总苹果不变，则新入的小朋友分得的苹果数量为a2,即是平均每个小朋友分得的苹果数量。则此时小朋友的数量是a+1,那么则等式成立:(a+1)(a2)=420，即(a+1)a=210,将210分解质因数:
210=2357，显然容易看出，210=1514,它们之间相差1,即a=14，即原来小朋友的数量是14。
21、A、B两地之间有条公路，小王步行从A地去B地，小张骑摩托车从B地出发不停地往返于A、B两地之间，若他们同时出发，前后速度保持不变，60分钟后两人第一次相遇，70分钟后小张第一次超过小王。当小王到达B地时，小张和小王迎面相遇过______次。
解析：此题可以借助画线段分析会更加清楚
(手机上不好勾画)
我就直接分析：1、确定前提：小王步行从A到B，小张骑摩托车从B到A且不停地往返AB之间(摩托车速度比人快)，他们前后的速度保持不变；
2、他们第一次相遇的时间是60分钟，即60分钟他们的速度和=AB的距离
3、相遇后，摩托车先到达A地返回B方向，从相遇到追上仅用70-60=10(分钟)，即摩托车用10分钟所走的路程等于小王步行60+70=130(分钟)的路程，即摩托车的速度是步行速度的130/10=13倍
此时可知小王步行全程所需时间:60(13+1)=840分钟，而此时摩托车共走的路程是步行路程的13倍，即往返BA两地13趟，奇次趟数为相遇，偶次趟数为追及，所以相遇的趟次为1.3.5.7.9.11.13。追及的趟次为2.4.6.8.10.12。
那么相遇的次数是7次，追及的次数是6次，当小王到达B地时，小张刚好到达A地。
22、设n是正整数，若从任意n个非负整数中一定能找到四个不同的数a、b、c、d使得a+b-c-d能被20整除，则n的最小值是_____。（选自2016华杯赛）
解析：
此题的关键在于以何种方法证明。
首先分析的是a+b-c-d=(a+b)-(c+d)，即这个差值能被20整除，也就是说a+b被20除的余数和c+d被20除的余数相同即可。
这样就引起我们对被20除后余数问题的思考。
因为被20除的余数有0.1.2...19,一共有20种，试想如果我们有超过21个数被20除，那么必定会有两个余数是相同的，则相同余数对应的两个数的差值能被20整除。
回到题目开头，(a+b)-(c+d),只要类似它们两数之和的个数足够多，就能够使得它们被20除的余数相同。
我们先看7个数，这7个数的两两组合个数是6+5+4+3+2+1=21，虽然总数是21，但是它们每一个数用到了(7-1)=6次，只要把其中两个数拿出来，剩下的和的总数只有4+3+2+1=10,就不一定有相同的余数，7个数两两和值虽然有21个，而且也至少有两个和值被20除的余数，但是这两个余数相同的和值只有在4个不同数的情况下才符合题意，所以不是必然情况。于是我们增加个数，当8个数的时候，拿出两个，剩余的和的个数5+4+3+2+1=15，加上自身也只有16个，还是不够，当为9个数的时候，拿出两个数，剩余7个数的和的个数是6+5+4+3+2+1=21,有21个和，再加上拿出的两个数之和，这时我们分两种情况讨论:
(1)当21个和值中有一个和与拿出的两个数之和被20除的余数相同，(a+b)-(c+d)被20整除即结果成立。
(2)当21个和值无一与拿出的两数和值被20除的余数相同，那么，这21个和值被20除的余数最多只有19种,即至少有两组余数相同或者一组里面有3个余数相同。那么继续再分两种情况考虑:
Ⅰ、当有两组余数相同；若两组余数中有其中一组是用4个不同的数，即结论已成立。
Ⅱ、主要考虑两组余数(对应的4个数)均有一个数用到两次，例如:a+b与a+c，他们被20除的余数相同，那么，b-c一定是20的倍数，此时，我们将b、c拿出，将原先的两个数和另外5个数组成7个数继续同理讨论，若新组成的7个数的21个和值被20除有一个余数与b、c 之和被20除后余数相同，即结论成立。若无一余数与b，c和值被20除相同，那么同样道理，必定会有a-d的差值被20值整除，这样就有(a+b)-(c+d)能被20整除，即结论成立。
23、如右图，三角形ABC中，AB=180厘米，AC=204厘米，D、F是AB上的点，E、G是AC上的点，连接CD，DE，EF，FG，将三角形ABC分成面积相等的五个小三角形。则AF+AG_____厘米。（选自2016华杯赛）
解析：
此题也比较容易，从外围的?BCD算起较容易些，由?AGF与?EGF的关系算起也可以，最终结果算出
AF=96,AG=76.5
则AF+AG=172.5(厘米)
24、在9×9的格子上，1×1的小方格的顶点叫做格点。如下图，三角形ABC的三个顶点都是格点，若一个格点P使得三角形PAB与三角形PAC的面积相等，就称P点为“好点”，那么在这张格子纸上共有______个“好点”。（选自2016华杯赛）
解析：
此题考察的是同学们的应变能力，在考场是，应变能力是非常重要的，应变能力同时也是基础功底的体现。
此题有两种解题思路。在这里我介绍一种。
首先必须要做的就是计算出两个线段的长度，即AB和AC，以每一个小格的对角线为基础，令每一个最小对角线为t，那么
AB=4t，AC=2t，AB=2AC
要使得?PAB和?PAC面积相等，那么，?PAB的高只能是?PAC的一半，这只是初步的计算。
下面分情况考虑P点的位置
(1)当P点在AB左上侧的时候；
不妨任意画一个点，连接PA，PB，PC，连接过后发现，要使S?PAB=S?PAC
其实就是S?PBC=S?ABC
那么只要过点A作BC的平行线就可以了，观察平行线与格子的交点
这样就同时将AC右侧的好点找出，一共就找出3个；
(2)当P点在AB右下侧的时候，这时，我们连接P与A，B，C
得知，PA必须平分线段BC，即经过BC中点时，才能使S?PAB=S?PAC
那么，连接BC的中点与A的联系并延长，此时，同样得到好点是3个(这时把最上方的好点也同时找出)
所有情况已经考虑完毕，即好点的总数是：3+3=6
25、如下图，边长为2cm的正方形ABCD中，E、F分别为AB、BC中点，连接C、E与A、F交于G点，则四边形ADCG的面积是______平方厘米。
解析：从问题的角度出发，此题的重点就是求出三角形AEG的面积。
第一步骤:因三角形ABF与三角形BCE的面积相等，即都等于21/2=1
并且这两个三角形有一个公共部分是四边形EBFG，那么除去公共部分就是两个三角AEG与CFG，所以这两个小三角形的面积相等。
第二步骤:可连接B、G
根据三角形AEG与BEG是等底同高的三角形，即他们的面积相等。
同理，三角形CFG=三角形BFG
所以，三角形AEG=三角形BEG=三角形BFG=三角形FCG(以上是指面积相等，手机上不好标注)
由此，三角形AEG的面积就迎刃而解，即
三角形ABF/3=1/3,则
四边形AGCD=22-1-1/3=8/3(平方厘米)
26、某商店第一天卖出一些笔，第二天每支笔降价1元后多卖出100支，第三天每支笔比以前一天涨价3元后比前一天少卖出200支。如果这三天每天卖得的钱相同，那么第一天每支笔售价是______。
解析：
此题可用假设法
可先设定第一天卖出a支，那么第二天卖出a+100支，第三天卖出a-100支，也就是说第二天与第三天卖出的平均数等于第一天卖出的数量，由3天所卖出的钱相同，则第二天第三天两天的平均卖价和第一天相同，此时用假设法思考
如果第二天的单价用第一天的单价计算，第二天会多得的钱为：a+100，同理如果第三天的单价以第一天的单价计算会少出的钱为：2(a-100)
则由题可知多出的钱和少出的钱相同，即
a+100=2(a-100)
→a=300
即第一天卖出300支笔
第二天卖出：300+100=400(支)
则第一天的单价：400×1÷100=4(元)
二、证明解答题
题1： 一百个和尚一百个馒头，大和尚一个人吃三个，小和尚三个人吃一个，问有几个大和尚，几个小和尚？（小学奥数基本题型---同属鸡兔同笼）
解析：此题与鸡兔同笼同属一类题型。首先从两个重要条件入手。
（1）一个大和尚吃三个；
（2）小和尚三个人吃一个（小和尚数量是3的倍数）
由这两个条件可知：
（1）1个小和尚吃的馒头数量：1/3；
1个大和尚吃的馒头数量：3；
（2）吃1个馒头需要3个小和尚；
吃1个馒头需要1/3个大和尚；
（3）1个大和尚相当于9个小和尚吃的馒头数量；
1个小和尚相当于1/9个大和尚吃的馒头数量；
由结论（1）（2）（3）可以推算（做多种假设），
1、假设一百个都是大和尚，那么至少需要馒头的数量为：1003=300，而实际是100个，即少300-100=200个，为什么会少200个是因为每一个小和尚比大和尚少吃3-1/3=8/3个，即得出小和尚的数量200÷（8/3）=75，那么大和尚的数量：100-75=25。
2、假设一百个都是小和尚，那么能吃馒头的数量为：100÷3=100/3，而实际是100个，即还有馒头的数量300-100/3=200/3，为什么会多200/3个是因为每一个大和尚比小和尚多吃3-1/3=8/3个，即得出大和尚的数量200/3÷（8/3）=25，那么小和尚的数量：100-25=75。（其实这种带分数的假设很别扭，纯粹是为了让读者更加明白如何通过各种渠道解题）
前面两种假设均是以吃馒头个数来分析的，下面我们用和尚数量再一次做具体分析，以确保该题型从多个角度均能彻底分析出结论。
3、假设100个馒头全部被小和尚吃完，那么需要小和尚数量：100×3=300，比实际和尚数量多300-100=200，这是因为9个小和尚相当于1个大和尚，这句话的精髓就是每存在一个大和尚，就会有9-1=8个小和尚多出，那么多200个小和尚，即可得大和尚数量：200÷8=25，那么小和尚数量：100-25=75。
4、假设100个馒头全部被大和尚吃完，那么需要大和尚数量：100÷3=100/3，比实际和尚数量少100-100/3=200/3，这同样是因为9个小和尚相当于1个大和尚（即1个小和尚相当于1/9个大和尚），只是精髓不同，就是每存在一个小和尚，就会有1-1/9=8/9个大和尚少出，那么少200/3个大和尚，即可得小和尚数量：200/3÷8/9=75，那么大和尚数量：100-25=75。
以上4种假设，预测喜欢用第3种假设的人最多，也最容易理解
此题解法当然也可以二元一次方程（小学没有学过），可以用X、Y分别代表大和尚和小和尚的数量，即得
(1)X+Y=100
(2)3X+Y/3=100
解得X=25,Y=75
拓展思考：
前面题目后面括号所写，此题属鸡兔同笼问题，那么我们将此题怎样转化成类似鸡兔同笼问题上来。我们可以假设小和尚为“特殊鸡'，大和尚为'特殊兔'，所吃的馒头为“特殊脚”，和尚的人数为“特殊头”。那么，此题就变成为：（“特殊”用字母“T”替代）
笼中“T鸡”和'T兔'共一百只，'T脚“共一百只，已知3只“T鸡”只有1只'T脚'，1只“T兔”有3只“T脚”，那么“T鸡”和“T兔”各多少只？
这样似乎还不能充分引起我们的思考，再进行变化，我们将T鸡和T兔的T脚数量同时扩大3倍，也就是原来100只T脚变成300只，那么题目就成：
笼中“T鸡”和'T兔'共一百只，'T脚“共三百只，已知每只“T鸡”只有1只'T脚'，1只“T兔”有9只“T脚”，那么“T鸡”和“T兔”各多少只？
这样就变得简单多了。
题2： 三个连续的自然数,均小于2016,已知最小的自然数能被13整除,中间的自然数能被15整除,最大的自然数能被17整除,那么,这三个自然数最小的是多少?(小学奥数基本题型---整除与余数)
解析：已知条件（1）三个连续自然数，（2）分别被13、15、17整除，此题的关键是怎么整合现有的已知条件。
利用条件（1），三个自然数只需用其中一个，可用最大的自然数，即可整合，即一个自然数，能被17整除，被15除余1，被13除余2，求这个自然数最小是多少？这就是整除与余数的问题。
这里例举两种解法：
一、一般教材的解法
步骤一、1、先求17与15的最小公倍数，即1715=255，
2、再计算出这个最小公倍数的N倍(N是具体要求出的数值)符合被13除的条件，即被13除余2，可算出25510=2550满足条件。
步骤二、1、先求17与13的最小公倍数，即1713=221，
2、再计算出这个最小公倍数的N倍(N是具体要求出的数值)符合被15除的条件，即被15除余1，可算出22111=2431满足条件。
步骤三、1、先求15与13的最小公倍数，即1513=195，
2、再计算出这个最小公倍数的N倍(N是具体要求出的数值)符合被17除的条件，即被17整除，可算出19517=3315满足条件。(因13、15、17两两互质，这个结果实际是不需要在计算的)
则得出符合条件的自然数：2550+2431+3315=8296，
因17、15、13的最小公倍数是171513=3315，所以最小的自然数是8296-33152=1666（1666<>
二、简便的解法
1、先求出17的倍数满足被15除的条件；
即178/15=9....1，即178=136满足被15除余1。
2、再计算出136加上[17,15]的倍数的和满足被13除的条件；（[ ]表示最小公倍数）
136/13，余数6，[17,15]=255，255/13，余数8，可计算出136+2556=1666，1666/13，余数2，满足条件。
因为1666<>
题3：现有三块草地，面积分别是5公顷、6公顷、8公顷，每块草地上的草一样厚，而且长得一样快，第一块草地可供11头牛吃10天，第二块草地可供12头牛吃14天，第三块草地可供19头牛吃多少天？（小学奥数升级版题型---牛吃草）
解析:此题为牛吃草问题。与普通牛吃草问题不同的是，这题有3块地，分别是5公顷，6公顷，8公顷。已知条件是每块草地长得一样厚，而且长得一样快。
可假设5公顷的草地总量为T，那么6公顷和8公顷的草地总量分别为6/5=1.2T和8/5=1.6T。
根据已知条件：第一块草地可供11头牛吃10天，第二块草地可供12头牛吃14天。
可定义1头牛1天吃的草量为M，1公顷草地1天长出的草量为N，那么就可得出以下等式:
1110M=1T+105N
1214M=1.2T+146N
两式同时扩大1.2倍后相减，可得
N=1.5M，
即1公顷草地1天长出的草量可供1.5头牛吃1天。
将N=1.5M任意代入其中一个式子得，
T=35M，
即5公顷草地原有总量是35M，那么8公顷草地原有总量为:35M8/5=56M
所求问题：8公顷草地可供19头牛吃几天？因每公顷草地每天新长草量是1.5M，则8公顷每天新长草量是1.5M8=12M，即8公顷草地每天新长的草量就可供12头牛吃。则可得求解：
56M/(19-12)=8天
(即19头牛中12头牛吃新长的草，另7头吃原有的草)。
题4：六年级有100名学生，他们都订阅了甲、乙、丙三种杂志中的一种、二种或三种。问至少有多少名学生订阅的杂志种类相同？(小学奥数基本题型---抽屉原理)
解析:学生订阅甲，乙，丙三类杂志的一种，二种或三种，可知共有几种类型:
即
C(1、3)+C(2、3)+C(3、3)=7种，列举如下:
1.甲；2.乙；3.丙；4.甲、乙；5.甲、丙；6.乙、丙；7.甲、乙、丙。
即可看成7种杂志。那么共100名学生分别订阅这7种杂志，即100/7=14…2,那么订相同杂志的学生至少有14+1=15 (名)
题5：有若干千克4%的盐水，蒸发了一些水分后变成了10%的盐水，再加300克4%的盐水，混合后变成6.4%的盐水。问最初的盐水是多少克？(小学奥数基本题型---浓度)
解析：我们先对后半句进行分析，
300克4%的盐水与若干克10%的盐水混合后变成6.4%的盐水，
我们可以从另外一个角度考虑这道题，
即是从若干克10%拿出一定的盐(不是盐水)放进300克4%的盐水，使其浓度从4%增加至6.4%，并且将自己的浓度从10%稀释到6.4%。
从这个角度出发，即有第一个问题：
需要多少克盐能将300克4%的盐水的浓度增加至6.4%？（盐变水不变）
300(1-4%)/(1-6.4%)=4000/13(克)
则增加的盐：4000/13-300=100/13（克）
即是从10%的盐水中拿出100/13克盐，使得它们均变成6.4%，那么第二个问题就是
从若干克10%的盐水中拿出有100/13克盐，使其浓度变成6.4%，那么10%的盐水有多少克？
此时我们把10%盐水中的水看成单位1（因为盐变水不变），则水的重量为：
100/13÷[(1-6.4%)÷(1-10%)]=180(克)
那么10%的盐水总重量：180÷(1-10%)=200(克),其盐的重量为:200-180=20或者20010%=20(克),
根据第一句话:最初是4%的盐水蒸发了一些水分后得到10%的盐水（水变盐不变），则最初的盐水为：
20÷4%=500(克)
题6：一辆汽车从A城开往B城，如果把车速提高20%，则可比原定时间提前1小时到达B城，如果按原速度行驶100千米后，将车速提高30%，也恰好比原定时间提前1小时到达B城。求A、B两城的距离。（小学奥数）
解析：通过第一句话提速至120%，时间减少1小时，可知，速度是原来的120%,行驶同样的路程，时间则是原来的100/120,即5/6,则原来所用的时间为:1÷(1-5/6)=6(小时),那么提速至120%所用时间为：6-1=5(小时)
题中第二句话可转化成：以原始速度行驶一段路程(100千米)后提速至130%行驶完全程，此时全程的速度是原始速度的120%，则令原始速度行驶的时间是T1,提速至130%行驶的时间是T2,则
T1(120%-1)=T2(130%-120%)
则T2=2T1,而
T1+T2=5小时
所以
T1=5/3小时，T2=10/3小时
即汽车以原始速度行驶5/3小时的路程是100千米，则其速度是
100÷5/3=60千米/小时
则A、B两地的距离为：60×6=360（千米）
题7：证明：对于任意自然数n，乘以一个适当的自然数，其结果均可由7与0组成。（小学五年级奥数）
解析：此题可由抽屉原理解答。
我们从结论出发，假设7、77、777、7777、…、7…7(n个7)被n除后均不能整除，那么余数可能是:
(n-1)、(n-2)......3、2、1，
不能整除的余数共有(n-1)种可能，那么n个数被n除后必定有n个余数，利用抽屉原理，至少有两个余数是一样的。
再次假设第q个数和第p个数被n除的余数相同(p>q)，则
7…7(p个7)-7…7(q个7)=7…70…0(p-q个7,q个0)
这就是说在前面n个数中如果没有被n整除的,必定会有其中两个数之差能被n整除。即n乘以一个适当的数，其乘积必定可以由7和0组成。
题8：已知S=，比较S与0.003的大小。（小学奥数）
解析：仔细观察这个式子，可知分母全为奇数且为等差为2的等差数列，分子全为偶数且为等差为2的等差数列，分母=分子+1。
我们试着再列两个式子：
S2=（1）
S3= （2）
这两个式子和S=有几个共同的特点是：
（1）它们公因数的个数一样多；
（2）分子与分母均为等差数列；
（3）分母=分子+1.
∵ > , ...,>;
<>,...,
∴ S>S3,S<>2
∵ SS2= ; SS3=
∴ SS2= ,>SS3=>
∴ S<0.002 ;="" s="">0.0017
即可知S的取值区间： 0.0017<><>, 则，
S<>
建议：此题的原出题者将S与0.003做比较，建议将0.003改成0.002或者0.0017，因为S更接近0.002或0.0017.
题9：如下图，长方形的面积为35平方厘米，左边直角三角形为5平方厘米，右上角直角三角形面积为7平方厘米，那么阴影部分面积是多少平方厘米？(小升初习题)
解析：先分析一下先决条件，挖掘出潜在数据关系。长方形的面积=长×宽，三角形面积=底×高÷2，两个都是直角三角形，并且其中一条边就是长方形的长或宽，至此结合三者面积的关系与已知分析，如拨云见日，迎刃而解。如下：
令长方形的AB=CD=X,AC=BD=Y,直角ABE三角形的AE=M,直角BDF三角形的DF=N,则
S□ABCD=X×Y=35 (1)
S△ABE=X×M÷2=5 (2)
S△ABE=Y×N÷2=7 (3)
由（1）（2）可得，M=2Y/7,则EC=AC-AE=5Y//7
由（1）（3）可得，N=2X/5,则CF=CD-FD=3X/5由前面两个结果可知
S△ECF=EC×FC÷2=(5Y/7)×(3X/5)÷2=X×Y×(3/14)=7.5
至此，只剩下最后一个阴影部分的面积，即是长方形面积与三个三角形面积和的差值，则
S阴影△BEF=35-5-7-7.5=15.5
题10：数一数，下图包含“”的长方形有多少个？（不含正方形）（小学奥数）
解析：根据题意，可分情况讨论，需要注意的是不含正方形
按照长方形的情况讨论 （1）2个小正方形组成：4个 （上下左右各1个）
（2）3个小正方形组成：5个 （横3，竖2）
（3）4个小正方形组成：4个 （横3，竖1）
（4）5个小正方形组成：2个 （横2）
（5）6个小正方形组成：11个 （横7，竖4）
（6）8个小正方形组成：8个 （横6，竖2）
（7）10个小正方形组成：4个 （横4）
（8）12个小正方形组成：11个 （横6+2(4×3，6×2)，竖3(3×4)）
（9）15个小正方形组成：4个 （横4）
（10）18个小正方形组成：2个 （横2）
（11）20个小正方形组成：2个 （横2）
（12）24个小正方形组成：1个 （横1）
即总数为4+5+4+2+11+8+4+11+4+2+2+1=58种
题11：如图正方形ABCD的面积是30平方厘米，M是AC的中点，求阴影部分的面积。（小升初题型）
解析：此题与填空题第20题如出一辙,可作AB的中点N,连接G、N，如图
∵四边形ABCD是正方形
∴△ADB与△ADC是关于AD所在直线对称的，AC与AB对称
又∵M、N分别为AC、AB的中点
∴M、N是关于直线AD对称的两个点
即△AGN与△AGM是关于直线AD对称的，则面积相等
S△AGN=S△AGM
而S△AGN=S△BGN(等底同高)
即S△AGN=S△AGM=S△BGN
∵S△AMB=S△AGN+S△AGM+S△BGN=30÷4=7.5
∴S△AGN=S△AGM=S△BGN=2.5
∴S阴影=S△ABM+S△ADM-2S△AGN
=7.5+7.5-2×2.5
=10
题12：如下图，已知四边形ABCD和AEGF均为正方形，
□ABCD的边长为4，连接B、F及D、F，求阴影部分面积。
分析：该题的核心条件只有两个，一是两正方形，二是已知其一正方形边长为4（4只能作为最终计算结果来用，因为可以是4，当然可以取更多的数值）。也就是说另一正方形的边长取值与结果无关，我们可作多条虚线，力求运用正方形四边相等的关系来寻求结果。利用众多虚线之后，得其中一条虚线是关联两个正方形的纽带。如下图
解析：连接两点A、F，假设□AEFG的边长为X,
∵ S△ABF=ABFG/2=4X/2=2X(平方单位)；
S△ADF=ADEF/2=4X/2=2X(平方单位)
∴ S△ABF=S△ADF
又∵ S△ABF=S△ABH+S△AHF S△ADF=S△AHF+S△DHF
这两个三角形有一个共同的区域，即△AHF，
等式两边同时减去共同区域，
∴ S△ABH=S△DHF
∵阴影S△BDF=S△DHF+S△BDH
按上述结果
∴阴影S△BDF=S△ABH+S△BDH=S△ABD=44/2=8(平方单位)
发散思维练习：若此题为填空题或选择题，为节省时间，我们可以取特殊值法，即两正方形的边长一样，再求解，就变得简单了，如下图
阴影S△BDE=44/2=8(平方单位)
拓展思维练习：为了加强思维训练，我们将此题延伸训练，根据此题与另一正方形的边长无关，我们画出如下图行
虽然看上去与题目的出入有点大，但经过推理，结果也是一样的。我简单的列举一下步骤程序(设□AEFG的边长为X）
S△ABF=4X/2 = S△ADF
即 S△ABH+S△AHF = S△DHF+S△AHF
---> S△ABH = S△DHF
阴影S△BDF = S△BDH+S△DHF
= S△BDH+S△ABH
= S△ABD
= 44/2
= 8(平方单位)
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Rheumatoid arthritis
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Glucocorticoids in rheumatoid arthritis: current status and future studies
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Abstract
Since their first use for treating rheumatoid arthritis (RA) in the late 1940s, glucocorticoids (GCs) have been representing a substantial part of the therapeutic arsenal for RA. However, even if GCs are still widely prescribed drugs, their toxicity is discussed controversially, so obtaining consensus on their use in RA is difficult. Hence, the most recent European League Against Rheumatism and American College of Rheumatology recommendations on early arthritis and RA management advocate the use of GCs as adjunct treatment to conventional synthetic disease-modifying antirheumatic drugs, at the lowest dose possible and for the shortest time possible. However, the recommendations remain relatively vague on dose regimens and routes of administration. Here, we describe literature data on which the current recommendations are based as well as data from recent trials published since the drafting of the guidelines. Moreover, we make proposals for daily practice and provide suggestions for studies that could help clarifying the place of GCs in RA management. Indeed, numerous items, including the benefit/risk ratio of low-dose and very low-dose GCs and optimal duration of GCs as bridging therapy, remain on the research agenda, and future studies are needed to guide the next recommendations for RA.
Key messages
What is already known about this subject?
Glucocorticoids are effective drugs in rheumatoid arthritis (RA), but because of their toxicity, their use requires care and vigilance.
What does this study add?
This review presents and discusses recent literature data on glucocorticoids use in RA and provides suggestions for future research in the field of glucocorticoids in RA.
How might this impact on clinical practice?
Glucocorticoids in RA must be used in combination with disease-modifying antirheumatic drugs (DMARDs), notably as bridging therapy with conventional synthetic DMARDs.
Evaluation of the benefit/risk ratio must be systematic, even for low-dose glucocorticoids.
Introduction
In this time of targeted therapies, therapeutic strategies and management of comorbidities in patients with rheumatoid arthritis (RA), the question of the future position glucocorticoids (GCs) may have in RA is worth asking. Indeed, although therapy with GCs was a major therapeutic advance in the 1950s for RA, the current emphasis is more on the disadvantages than the benefits of this treatment. However, in light of recent literature data, low-dose GCs still seem to have an important role in RA.1–3
The goal of treatments for patients with RA in the short term is to decrease disease activity and achieve clinical remission, and in the medium term to limit or prevent structural progression, disability and systemic manifestations. Available disease-modifying antirheumatic drugs (DMARDs) have shown their efficacy. Conventional synthetic (cs) DMARDs have a relatively long onset of action while most of biological (b) and targeted synthetic (ts) DMARDs (bDMARDs and tsDMARDs) act faster. GCs in RA provide the advantage of a rapid onset of action, which allows waiting for the onset of csDMARDs efficacy. Moreover, even if GCs are mainly widely used whenever clinicians need rapid symptomatic relief for their patients with RA, their structural effect must not be forgotten. However, the benefit/risk ratio of GCs remains precarious and their modalities of use in RA remain controversial.
In this review, we first detail the latest European and US recommendations on GCs use in RA and discuss the use of GCs in current practice. Then, we consider recent literature data on GCs efficacy (both clinical and structural) in RA and the various ways of using GCs. Finally, we address the adverse effects potentially associated with GCs. Our work being a narrative review and not a systematic review, we have not included all the existing studies but only the ones seeming the most relevant to us. Moreover, during this review, we propose topics that could be of interest to be evaluated in future studies of GCs use in RA.
International recommendations
European recommendations
Recent updates of the European recommendations for management of early arthritis and RA4–6 placed a greater focus on the benefit of GCs therapy than the previous versions. Short-term GCs therapy should now be considered as part of the initial treatment strategy and subsequently if an initial strategy has failed, as bridging therapy if a change in a csDMARD is considered.6 GCs should be tapered as rapidly as is clinically feasible: long-term use of GCs should be avoided, GCs should be gradually reduced and stopped, usually within 3 months and only exceptionally by 6 months.6 The term ‘low-dose’ GCs was replaced by ‘short-term’ GCs to take into account several current ways of using GCs, such as parenteral bolus.6
US recommendations
The 2015 American College of Rheumatology (ACR) guidelines for early and established RA recommend adding GCs to DMARDs during disease flares, at the lowest dose and for the shortest period possible.7 In contrast to the European recommendations, adding GCs when a csDMARD is initiated depends on disease activity.
Taken together, these international recommendations agree on the use of GCs for disease flare and possibly at the start of a new csDMARD. Specific advice concerning dosage, duration, route of administration and strategies is limited and less consensual, because reliable and detailed evidence is scarce. For US recommendations, a dose <10 mg/day is considered a low dose, and GCs should be tapered in less than 3 months, whereas for European recommendations, the threshold is 7.5 mg/day, and GCs could be prescribed in combination with csDMARDs for up to 6 months maximum, knowing that this duration is mainly expert-driven.8 Despite these differences, international guidelines underline the importance of GCs but also advocate the use of GCs at the lowest cumulative dose possible because of the high awareness of potentially associated adverse effects.
In these sets of recommendations as well as in this review, GCs dosages are expressed in prednisone equivalent.
Current practice
A recent study of an Australian cohort of patients with RA showed that the probability of GCs use throughout follow-up has decreased over time, from 55% in 2001 to 39% in 2012 (p<0.001).9 In this cohort, current csDMARD use but not bDMARD use was associated with increased current GCs use. In a recent analysis considering the years from 1980 up to 2004, the reduction of mean initial low-dose, for long-term GCs therapy in RA, was found from 10.3 mg/day up to 3.6 mg/day.10 In contrast, another observational cohort study showed that the proportion of patients initiating GCs was higher in the group from 1995 to 2007 compared with the earlier group from 1980 to 1994 (68% vs 36%) but the cumulative dose did not differ over the first year.11
Anyway, GCs are still widely used in RA. GCs appear to be used in approximately 50% of patients with RA,12 with varied duration and dosage among the studies. In the German CAPEA cohort of patients with early arthritis, 82% received methotrexate (MTX) within the first months, 77% received GCs and 20% of these received <7.5 mg/day prednisone but one third received >20 mg/day.13 After 2 years of follow-up, 12% of the patients received biologics, 52% were free of GCs and 41% were receiving <5 mg/day. In the French ESPOIR cohort of patients with early arthritis, 45% started GCs during the first 6 months and more than 50% received GCs at least once over 5 years after inclusion.14 Overall, the dose of GCs received during follow-up was very low, the mean was 3.1±2.9 mg/day.15 In the Canadian CATCH cohort of patients with RA, 42% were considered GCs users and the median oral daily dose was 5 mg (IQR 2.5–10).16
Glucocorticoids efficacy
Clinical efficacy
Current knowledge
For reasons of brevity, in this part, we will focus on only the most relevant and recent data on clinical efficacy of GCs in RA published during the last 6 years (table 1). In the following parts of our work, we will discuss other relevant studies published earlier.
Table 1
•
Characteristics of studies of the clinical efficacy of glucocorticoids published in the last 6 years
Overall, these results agreed with previous findings and with the conclusion of the most recent systematic literature reviews published on this topic and showed a beneficial effect of GCs when added to csDMARDs.5 17 The available data primarily relate to GCs in addition to csDMARDs and not, or not specifically, to bDMARDs or tsDMARDs. Moreover, the current literature concerns mainly patients with early arthritis, and studies reporting on GCs efficacy in patients with established RA are clearly less frequent.
In the CAPRA-2 trial, a study of patients with established RA (disease duration approximately 8 years) and active disease, low-dose (5 mg/day) prednisone with a modified release formulation (chronotherapy) added to existing DMARDs significantly ameliorated disease activity at 12 weeks as compared with placebo (PBO) added to existing DMARDs.18 19 DMARDs were almost exclusively csDMARDs because only one patient in each group received a bDMARD.
In the CareRA trial, patients with early RA without poor prognosis markers were randomised to one of two treatment arms.20 In one arm, GCs were initially associated with MTX (30 mg/day prednisone decreased to 5 mg/day in 6 weeks) while in the other arm, MTX was initiated without GCs. Disease Activity Score in 28 joints (DAS28) remission at 16 weeks was more frequently achieved in patients with than without GCs, although not significantly (65% vs 47%, p=0.08).20 According to data at 1 and 2 years, the rates of remission were still higher in the MTX+GCs than MTX-only arm but again not significantly.21 22
Ten-year data from the BeSt study were published in 2016.23 In this trial, 508 patients with early active RA were randomised to four arms: a pre-established maintenance treatment regimen prescribed as monotherapy beginning with MTX; a step-up group with sulfasalazine (SSZ) added to MTX in case of failure; a group following the COBRA scheme (SSZ, MTX and GCs initially at 60 mg/day, then progressively tapered to 7.5 mg/day in 6 weeks) and a group receiving MTX and infliximab from the beginning.24 In the initial study, patients following the COBRA strategy showed better clinical efficacy at 3 months than patients without GCs. At 10 years, approximately 50% of patients were in remission whatever their initial group of randomisation. In the meantime, strategy-driven changes have occurred and it is thus difficult to conclude on the long-term effect of initial GCs therapy.23
Results of the post-trial follow-up of the CAMERA-II study were published in 2017.25 In this study of 236 patients with early RA, the addition during 2 years of 10 mg/day prednisone to MTX was compared with a MTX and PBO group.26 Disease activity after 2 years of treatment improved more on average in the MTX-GCs arm than MTX-PBO arm, but the differences observed in the first months tended to decrease over time. In the follow-up study (median follow-up of approximately 6.6 years), significantly fewer patients of the MTX-GCs group had initiated a bDMARD than those from the MTX-PBO group (31% vs 50%, p<0.05).25
Data at 10 years from the BARFOT trial have been reported.27 In this open randomised trial of 250 patients with early RA, the addition during 2 years of 7.5 mg/day prednisolone to csDMARDs was compared with csDMARDs alone. Clinical outcomes were better in the GCs group at all time points (3, 6, 12, 18 and 24 months).28 A follow-up study at 4 years no longer found differences in the proportion of patients in remission between the groups.29 At 10 years, bDMARD use did not differ with and without GCs.27 Of note, the proportion of patients who used a bDMARD was very low (15% in each group), perhaps because patients were included in the BARFOT cohort between 1995 and 1999, before the era of biologics.
In light of these and other literature data, there is little doubt that GCs are effective for reducing disease activity in patients with RA, at least in the short term. However, confirming whether the clinical benefit of GCs persists in the medium and long term is difficult. Because of their toxicity and moderate structural effect (see below), GCs cannot be considered monotherapy and should always be used together with DMARDs.
Studies to perform
We notably lack studies evaluating the clinical effect in the medium and long term with short-term (<6 months) GCs therapy. Such studies should be RCTs including patients with RA for which a specific DMARD has to be initiated. These patients should be randomised into an arm receiving the DMARD with GCs discontinued over 3 months and an arm receiving the DMARD without GCs. Clinical efficacy measures should occur at medium and long term: after several months and even years of follow-up. As stated above, it would be of particular interest to conduct such studies in patients initiating a targeted DMARD and not only in patients initiating a csDMARD. Trials addressing the advantages provided by new formulations of GCs would also be of interest. Indeed, the pharmaceutical field of GCs continues to evolve, notably with the development of modified-release or delayed-release prednisone (chronotherapy) or GC receptor agonists to improve the efficacy and/or reduce toxicity.30–32 Modified night-release formulation of low-dose prednisone, although administered in the evening (acting like a replacement therapy), has been developed to contrast the circadian rise in proinflammatory cytokine levels (night), that contributes to RA disease activity and might represent the way to further optimise the DMARD activity exerted by low dose GCs in RA.33
Structural effect
Current knowledge
In this part, we have included all the studies on structural effects of GCs seeming relevant to us, without any time limit. In most trials, GCs were used in combination with csDMARDs, so concluding on the specific effect of GCs is difficult.
In 2005, the BARFOT randomised trial showed the slowing at 2 years of the structural progression more important with than without GCs (0.2% vs 0.4%, p=0.02).28 The follow-up at 4 years suggested a lower increase in total Sharp score during the 4 years with than without GCs, although not significantly (p=0.079).29 In 2007, a meta-analysis of 15 studies was in favour of a significant reduction of erosion progression by GCs given in addition to csDMARDs.34 In 2010, the structural impact of GCs added to csDMARDs was evaluated in another meta-analysis, showing approximately 70% reduced radiographic progression (p=0.0008).35 In this meta-analysis, the combination of GCs to csDMARDs had a similar effect as bDMARDs added to MTX (difference of 7% in radiographic progression between the two groups, p=0.44). However, the conclusions of this meta-analysis were limited by the high heterogeneity of included trials. In the 11-year follow-up study of the COBRA trial, structural damage progression was lower in the group initially treated according to the COBRA scheme (MTX+SSZ+GCs (60 mg/day to 7.5 mg/day in 6 weeks)) than in the group initially treated with SSZ monotherapy.36
More recently, in the CAMERA-II study, radiographic progression with GCs was significantly reduced, with 78% of patients in the MTX-GCs group remaining erosion-free after 2 years vs only 67% in the MTX-PBO group.26 Similarly, at the 2-year post-trial follow-up, the median erosion score was significantly lower in the former MTX-GCs group than in the former MTX-PBO group (p=0.002).25
In light of the literature, the structural effect of GCs seems clear, at least when GCs are added to csDMARDs. Moreover, this positive effect on radiographic progression seems to last even after discontinuation of GCs. Obviously, when combined with bDMARDs or tsDMARDs, showing any additional effect of GCs is difficult because these drugs already strongly inhibit structural progression.
Studies to perform
Studies comparing different durations of GCs treatment as bridging therapy in terms of structural protection are needed to increase the strength of the recommendations concerning the optimal period of GCs prescription. The approach of these studies would be to evaluate if a short-term (<6 months) GCs therapy added to MTX is as effective as a more prolonged GCs therapy (during 2 years, eg, as in CAMERA-II or BARFOT trials) on structural progression at medium and long term.
Glucocorticoids use strategies
For this part, selection of the studies was based only on their relevance and not on their publication date.
Bridging therapy
Current knowledge
When and how to introduce and taper GCs are important questions for rheumatologists and their patients. In view of the risks and benefits, the usual practice is to use an induction strategy followed by a progressive decrease. This is notably done when GCs are prescribed as bridging therapy along with the introduction of a csDMARD, to wait for its efficacy. This scheme was used in the COBRA study, with results published in 1997.37 In this study, 155 patients with early RA were randomised into two arms: combination treatment with SSZ, MTX and prednisolone (initially at 60 mg/day, then progressively tapered to 7.5 mg/day in 6 weeks) compared with SSZ only. The results favoured the GCs arm, both on the clinical and on the structural effects. Since this study, various strategies for the use of GCs have been evaluated, but few RCTs have been designed to specifically compare dosage, duration and tapering of GCs.
Recently, in two articles reporting on the same RCT, two different GCs strategies were compared: the COBRA-light strategy (prednisolone at 30 mg/day, tapered to 7.5 mg/day within 9 weeks) combined with MTX, and the original COBRA strategy.38 39 The outcomes were similar between the two arms in terms of reducing clinical disease activity, improving functional ability and preventing radiographic progression, showing that treatment regimens including medium-dose GCs are not inferior to regimens with initially high-dose GCs.40
A posthoc analysis of six phase III studies evaluating tofacitinib in RA, in monotherapy (ORAL start, ORAL Solo) or combined with csDMARDs (ORAL Scan, ORAL Standard, ORAL Sync, ORAL Step), examined the impact of the presence of GCs in the treatment groups.41 Mean dose of GCs was approximately of 6 mg/day. Across all studies, the concomitant use of GCs did not affect the clinical efficacy of tofacitinib. In contrast, response was frequently better for patients receiving MTX with GCs than without GCs, but the differences were often not statistically significant.
Moreover, in an analysis presented at the 2018 ACR meeting but not yet published of pooled data from four RCTs (AMBITION, ACT-RAY, ADACTA and FUNCTION) on intravenous tocilizumab (TCZ), concomitant GC therapy had no impact on the clinical efficacy of TCZ at 24 weeks.42 The authors assessed patients in the comparator arms: the clinical response of patients receiving adalimumab as well as those initiating MTX was not affected by the use of GCs. Data on the mean GCs dose were lacking in this abstract.
According to a posthoc analysis of the TOZURA trial, a phase IV study evaluating subcutaneous TCZ, the proportion of patients achieving DAS28 remission at 24 weeks was broadly similar between the groups receiving TCZ, with or without GCs (median dose 5 mg/day).43
However, concerning the quoted studies on tofacitinib and TCZ, it is necessary to specify that GCs use was not submitted to randomisation and could be prior to the beginning of the study making it difficult to compare the groups with and without GCs.
Studies to perform
The current literature mainly concerns the use of GCs as bridging therapy when a csDMARD is initiated, but we have few studies on the use of GCs as bridging therapy when a bDMARD or tsDMARD is initiated. Indeed, the onset of action might be faster for bDMARDs, and even more so for tsDMARDs, than csDMARDs, which could explain why the recent posthoc analyses of the tofacitinib and TCZ studies did not find a benefit of adding GCs. It would be useful to have RCTs designed to evaluate clinical and structural efficacy of a short-term GCs therapy prescribed when a targeted DMARD is initiated. Finally, studies on the best initiation dosage of GCs are lacking, as are those on the duration of GCs as bridging therapy and trials comparing several tapering strategies.
Flare treatment
Current knowledge
Data on this topic are scarce, among other reasons due to lack of an agreed-upon definition of RA flare.44 In an older study of 18 patients with RA, flares were treated with GCs (prednisone 25 mg/day tapered in 5 days) or PBO.45 Clinical outcomes were better at 6 months when GCs were used. In the BELIRA trial, patients with RA with active disease received a total of 250 mg/day prednisolone over 1 week added to their existing csDMARDs.46 This strategy resulted in a highly significant improvement in clinical, functional and serological measures at 1 week. A RCT comparing the efficacy of three intravenous pulses of 120 mg/day dexamethasone or 1 g/day methylprednisolone suggested that the two strategies were safe and effective for RA flares at 1 month.47
Studies to perform
Daily management of RA would be helped by studies investigating the modalities of treating disease flares with GCs, for example, by comparing pulse therapy to low-dose oral GCs.
Maintenance therapy
Current knowledge
Despite recommendations, many patients are already being treated for months and years with low doses GCs, without apparent excessive toxicity, and there is no consensus on whether the GCs therapy should be stopped or not. In the international, multicentre SEMIRA trial, 259 patients with RA in remission and treated for more than 6 months by TCZ±csDMARDs+GCs were randomly assigned to continued GCs or GCs taper.48 Even if continuing GCs provided better disease control than tapering GCs, almost two-thirds of patients still achieved treatment success at 24 weeks and could stop GCs entirely.
Studies to perform
Results of this trial are suggesting that tapering and discontinuation of GCs should be tested in patients in remission under TCZ and receiving long-term low-dose GCs. In case of a flare, reintroduction of GCs should be considered. Similar studies on patients with RA receiving other treatments than TCZ would be of interest in order to extrapolate the results.
Other administration routes
Parenteral route
Current knowledge
In a study published in 1993, 41 patients with RA were randomised to receive methylprednisolone orally (500 mg) or intramuscularly (120 mg) at baseline and 4 and 8 weeks.49 At 16 weeks, the intramuscular route seemed better than the oral administration on pain and Health Assessment Questionnaire (HAQ) score. In the tREACH trial, patients with early RA received csDMARDs (MTX monotherapy or MTX+SSZ+hydroxychloroquine (HCQ)) and oral GCs started at 15 mg/day and tapered during 10 weeks or MTX+SSZ+HCQ plus an initial intramuscular pulse of GCs.50 The two groups did not differ at 1 year in clinical response, structural progression or safety.
Studies to perform
From the literature data, concluding on the superiority of one GCs route of administration versus another is difficult and that is why current recommendations state that different dose regimens and route of administration can be used.6 Strategy trials, with designs similar to that of the tREACH study, comparing the efficacy of one or several parenteral (intravenous or intramuscular) GCs injections versus initiation and tapering of oral GCs as bridging therapy could answer this question. Besides efficacy issues, the parenteral form could be beneficial for preventing self-medication by patients and avoiding the risk with GCs tapering and mid-term to long-term GCs use.51
Intra-articular injections
Current knowledge
Subanalyses of the BeSt study showed that 3 months after intra-articular GCs injection, 50% of joints were no longer swollen. After 1 year, swelling had recurred in 14% of joints with initially good clinical response.52 Initially, disease activity scores significantly improved, but over time, DAS and HAQ scores became similar in injected and non-injected patients. After 8 years, the two groups did not differ in joint damage.40 In the CIMESTRA trial, patients with early RA received intra-articular GCs in any swollen joint (maximum four joints per visit) combined with step-up csDMARDs treatment over 2 years.53 Oral GCs were not allowed. At 2 years, the median cumulative number of injections for each patient was 13, and the cumulative dose of GCs was <1 mg/day. The injections had a rapid-onset anti-inflammatory action, and 2 weeks after inclusion, 39% of patients were in DAS28 remission. At the end of follow-up, 55.5% of joints injected at baseline did not show disease relapse, and intra-articular injections were well tolerated.
Recently, one small and open-label RCT showed greater clinical efficacy (ACR20/50/70 response) of csDMARDs combined with initial intra-articular GCs injections than csDMARDs alone in patients with early RA without oral GCs.54
Studies to perform
Larger RCTs with a design similar to the trial by Menon et al
54 and with compliance to blinding are needed, as are RCTs comparing intra-articular GCs injections with other routes of administration.
Glucocorticoids safety
In this part also, the publication date was not taken into account for the selection of the studies. Due to the considerable amount of data published on this topic, we have decided to notably quote meta-analysis and literature reviews.
Current knowledge
GCs toxicity is a major concern when higher dosages are given for a longer time. However, even low-dose to medium-dose GCs are associated with adverse effects.
In 2007, a EULAR taskforce published recommendations for the management of systemic GCs therapy in rheumatic diseases.55 According to a literature review, the taskforce identified the following main adverse events of GCs: cardiovascular diseases, infections, gastrointestinal diseases, psychological disorders, endocrine pathologies, dermatological issues, musculoskeletal disorders (including osteoporosis) and ophthalmological diseases.55
In a meta-analysis of trials and follow-up studies published in 2009, the rate of adverse effects linked to GCs (prescribed for RA) at <30 mg/day was 43/100 patient-years (95% CI 30 to 55).56 Another meta-analysis, also published in 2009, pooled the results of six RCTs comparing low-dose GCs (mean dose 5–10 mg/day) to PBO.57 According to this meta-analysis, the groups did not differ in any type of event or serious adverse events. According to the 11-year follow-up study of the COBRA trial, rates of hypertension and diabetes were higher in the group initially treated according to the COBRA scheme than in the group initially treated with SSZ monotherapy but rates of cardiovascular disease were similar between the two groups, knowing that patients from this group had also received GCs during follow-up.36 After 23 years of follow-up, the mortality rate of patients from the COBRA trial was similar to that of the general population, whatever the initial randomisation group.58 From the 10-year data of the BARFOT trial, incident cardiovascular events were evenly distributed with and without GCs (15% and 14%), but the risk of the first cerebrovascular event was almost four times higher with than without GCs (HR 3.7 (95% CI 1.2 to 11.4), with higher mortality with GCs but not significantly.27 Of note, 53% of the patients with GCs had stopped them after the first 2 years, and the mean dose of prednisolone for those taking GCs was 7.2±1.1 mg/day at the 2-year follow-up, 6.5±3.6 mg/day at 4 years and 4.9±3.3 mg/day at 8 years.27 Long-term (median follow-up almost 7 years) post-trial data from the CAMERA-II study are in agreement with these findings, showing no statistical difference in incident GCs-related comorbidities between the MTX-GCs and MTX-PBO-arms, but numerical differences in cardiovascular events (13 vs 8) and death (10 vs 6).25 Similar to the BARFOT post-trial study, the controlled design was stopped after the first 2 years, and only half of the patients from the MTX-GCs arm had stopped GCs at 3 years. In summary, data from RCTs are quite reassuring in terms of low-dose GCs safety, but most RCTs had only a short follow-up. However, in the two long-term post-trial studies of BARFOT and CAMERA-II, in which patients had initially taken GCs during 2 years, GCs were associated with more cardiovascular events, but not significantly.
Observational studies usually showed significantly more adverse events in patients with than without GCs, whether for cardiovascular events or for other adverse effects linked to GCs such as infections59 or osteoporosis, and even for low-dose GCs.60 However, data from observational studies should be considered with caution, notably because of population heterogeneity and confounding by indication. Nevertheless, cardiovascular toxicity with low-dose GCs was confirmed in a recent meta-analysis of RCTs and observational studies, noting 47% increased cardiovascular events in patients with RA receiving GCs.61 In 2014, a North American study reported a threshold of 8–15 mg/day for increased mortality linked to GCs in RA (adjusted HR 1.78 (95% CI 1.22 to 3.60)), after adjusting for potential confounders and the propensity to receive GCs.62 The minimum total cumulative dose associated with all-cause mortality was 40 g (adjusted HR 1.89 (95% CI 1.25 to 2.44)).62 Reaching this cumulative dose threshold would take approximately 21 years with a daily dose of 5 mg/day. Patients with RA can commonly reach such a duration of exposure.63
It is clear that adverse effects are related to the dose and duration of GCs but the data on the toxic effects of a low cumulative dose of GCs are scarce. In the CareRA trial, the proportion of patients with adverse events at 1 year was similar between patients who had received GCs according to the COBRA Slim scheme and patients without GCs.21 A recent study based on data from the French cohort of early arthritis patients (ESPOIR cohort) found no significant over-risk at 7 years of follow-up, despite numerical differences in cardiovascular events, infectious diseases or osteoporotic fracture between patients who received very low-dose GCs (mean dose 3.1±2.9 mg/day) and patients who never received GCs.15
Studies to perform
Studies comparing various strategies of GCs use in terms of the frequency of adverse effects could help practitioners to prescribe GCs in the most suitable way. The toxic effects of GCs are clearly dose-dependent but are also influenced by individual factors, and studies evaluating the toxicity of GCs should take into account the comorbidities of the patients with RA to whom GCs are prescribed so as to identify the patients for whom the risk/benefit ratio is favourable or not.
Conclusion
GCs clearly still have an important role to play in RA management. Their clinical (and structural) efficacy is widely acknowledged. Most studies have evaluated GCs efficacy as bridging therapy, combined with csDMARDs. Consequently, available studies have mainly focused on patients with early RA, and data for established RA are scarce. Nevertheless, GCs seem useful in this population to control flares. By contrast, GCs monotherapy does not represent an acceptable therapeutic option. The benefit of adding GCs to bDMARDs and tsDMARDs is most likely low or non-existent because of the fast onset of action of most of these DMARDs. GCs are also cheap, and their combination with csDMARDs could reduce or delay the use of bDMARDs or tsDMARDs.
However, there is also evidence of GCs toxic effects, notably with moderate-dose to high-dose GCs used for a longer time, but even low-dose GCs might have adverse effects. Besides, the long-term GCs safety might be linked to the cumulative dose, and, apart from the daily dose, the duration is also crucial. GCs must be used at the lowest possible dose and for the shortest possible time. With long-term (≥6 months) GCs therapy, EULAR has defined 5 mg/day or less (if needed for controlling the disease) as the acceptable daily intake in terms of cardiovascular risk and risk of hyperglycaemia/diabetes, osteoporosis and infection for the vast majority of patients, but the individual risk of harm must also be evaluated by taking into account patient-specific characteristics.64 The ongoing trial GLORIA, a 2-year pragmatic RCT aiming to assess the safety and efficacy of GCs at 5 mg/day vs PBO added to DMARDs in elderly patients with RA (≥65 years), could provide evidence to support further this threshold of 5 mg/day.65 In the same prospects of improving the risk/benefit ratio of GCs, studies of chronotherapy and GCs agonists seem promising.31 32
Current recommendations do not advocate a specific administration route versus another. Indeed, according to the available literature data, oral and parenteral use of GCs seem to have similar efficacy, but data are scarce and do not concern structural efficacy. However, the parenteral form might facilitate GCs withdrawal by preventing self-medication. Each GCs prescription must be preceded by an evaluation of the benefit/risk balance and by information provided to the patient in the context of a shared decision.
New studies with modern designs evaluating GCs in RA are still needed. Considering the existing literature, we have proposed in this review several leads to guide future research. Of note, future trials will also need to include cost-utility analyses and data on patient participation.66 In continuum with these future studies, further guideline updates will need to address specific conditions and circumstances for which GCs should be prescribed in order to improve the balance between efficacy and long-term safety.
Contributors: Conception or design of the work: CH, BC. Drafting the article: CH, BC. Critical revision of the article: BC, FB. Final approval of the version to be published: CH, FB, BC.
Funding: The authors have not declared a specific grant for this research from any funding agency in the public, commercial or not-for-profit sectors.
Competing interests: None declared.
Patient consent for publication: Not required.
Provenance and peer review: Commissioned; externally peer reviewed.
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Received: 27 June 2019
Accepted: 13 November 2019
First published: 7 January 2020
4 February 2020
The effect of glucocorticoids on bone health in rheumatoid arthritis
To the Editor,
I read the article by Hua et al.1 that was published in this journal with great interest. The authors provided an excellent review of the literature regarding the clinical efficacy and toxicity of glucocorticoids (GCs) in rheumatoid arthritis (RA). The review included comprehensive discussion about the efficacy of GCs as a bridging therapy in addition to conventional synthetic disease-modifying antirheumatic drugs (csDMARDs) based on the rapid onset of action of these drugs.1 The authors advocate that because even low-doses of GCs might have adverse effects, administration of these drugs should be restricted to the lowest dose for the shortest time.1 The first part about the effectiveness of GCs was well documented and convincing; however, the second part about the safety of these drugs seemed a little less convincing. This might be attributed to the fact that a small number of studies on the safety of GCs have been published. In particular, little evidence regarding bone-related adverse effects has been presented. We have obtained very preliminary data in our hospital about the effects of GCs on bone health, including fractures and osteoporosis, and would like to contribute these as a comment.
We retrospectively reviewed the medical records of 883 patients with RA who visited our hospital in 2018. Of these, 364 patients (41.2%, Figure 1A) were prescribed GCs. At the last visits in 2018, approximately 80% of the patients who were receiving GCs used 5 mg/day or less of GCs; the mean dosage was 4.1 mg/day (Figure 1B). Vertebral fractures were more frequently observed in patients who were receiving GCs than in those not receiving them (p = 0.028, Figure 1C). The incidence of non-vertebral fractures was the same regardless of GC use (Figure 1D). Osteoporosis, defined as <80% of the young-adult mean bone mineral density of the lumber spine or left femoral neck, was diagnosed in 195 patients (22.1%). The incidence of osteoporosis was twice as high among patients who were receiving GCs compared with those who were not receiving them (Figure 1E). The effect of GCs on osteoporosis was observed in both male and female patients (Figure 1F).
Figure is available upon request.
Patients with RA are at a high risk of fracture, and the use of GCs for more than 3 months is associated with an increased risk of vertebral fracture regardless of the dosage.2,3 Our data were very preliminary, and thus, it had several limitations. For example, we did not consider differences in patient backgrounds. However, this simplified retrospective study supports, at least in part, the concept that GCs should be used at minimal dosage and for the shortest duration possible.1
References
Hua C, Buttgereit F, Cobe B. Glucocorticoids in rheumatoid arthritis: current status and future studies. RMD Open 2020;6(1):e000536.
Xue AL, Wu SY, Jiang L, Feng AM, Guo HF, Zhao P. Bone fracture risk in patients with rheumatoid arthritis: A meta-analysis. Medicine (Baltimore) 2017;96:e6983.
Ozen G, Pedro S, Wolfe F, Michaud K. Medications associated with fracture risk in patients with rheumatoid arthritis. Ann Rheum Dis 2019;78:1041–1047.
Acknowledgments
The study protocol was approved by the Ethics Committee of the Osaki Citizen Hospital (No. 20190822-25) and performed in accordance with the Declaration of Helsinki.
Conflict of Interest
None declared.
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9588 | https://www.davidyding.com/navPages/rectangles | David Ding: Think More, Work Less
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David Ding
October 10, 2020
When he was only seven years old, there was a well-known urban legend about the German mathematician Carl Friedrich Gauss performing a tour de force right in front of his classmates and teacher. The story goes that presumably during a lesson in arithmetic, Gauss's teacher challenged his class to an exercise in addition: 1+2+3+⋯+100=? While Gauss's classmates immediately began scribbling on their notepads trying to solve the 99 addition problems, Gauss was not moving his pen. Instead, he stared at the expression and realized an interesting pattern: 1 and 100 make up 101, 2 and 99 make up 101, 3 and 98 add to 101, etc., all the way up to 50 and 51. It seemed that if one examined the terms from both ends, the elements involved would sum to 101. Then, all he had to do was to count how many pairs of "101"s appeared in the expression, which was 50, and 99 additions became one simple multiplication: 1+2+3+⋯+100=101×50=5050 Gauss arrived at the correct answer before his peers even made it past 20.
Of course, nowadays the result from the previous anecdote is simply known as the "Gauss Formula", which also works if the number of terms in the addition expression is odd instead of even. Namely, the sum of the first n natural numbers can be expressed as: ∑k=1 n k=n(n+1)2 He would end up contributing immensely in the field of mathematics throughout his life, having many more things named after him, and even be crowned as the "Prince of Mathematics" (Latin: Princeps mathematicorum). Furthermore, he would later show how his original "Gauss Formula" can also be proven using induction. However, induction is not the point of this blog. What I want everyone to learn is that sometimes it pays to take a step back, think a bit, and see your work cut in half (or more!). Thinking often is also a crucial component in developing habits to think critically, which as I echo here from my blog introduction, will be the tool that humanity will have in fighting for the advancement of our species. In math, as it is in life, there will be situations and problems where one way of arriving at the answer will be to use brute force. When impelled to do so, one would often discredit taking a step back to think about the problem first and instead immediately set out on the tedious task of taking the longest route to the answer. Doing so is not only physically and mentally draining and prone to errors, but one also misses out on the opportunity to discover something new and marvelous in the process. Sometimes knowledge is unearthed from solving a completely different set of problems, if only those problems are tackled using elegant solutions that would not have otherwise arose without a little thinking.
Let me illustrate this point by solving together the following problem. No advanced math, let alone calculus, is required.
"How many rectangles are there in an n by n grid of squares? (Note: squares are rectangles too)."
In the grid above for example, we have an 8 by 8 grid of squares, which for simplicity reasons, I will just call the grid a "chessboard". In order to make a rectangle from the chessboard, we need to trace out four grid lines to form the four sides of a rectangle, like so:
As you can see, we made a 2 by 3 rectangle in our chessboard by making the rectangle bounded vertically by columns 4 and 7 and horizontally by rows 2 and 4. One rectangle in the chessboard is considered distinct from another if those two rectangles are bounded by different coluns and/or rows (even if their areas are the same). So, back to our original question: how many distinct rectangles can we form from an n by n chessboard? Yes, brute force counting would certainly work in this case because this is a counting problem. So one without the patience to think first and rush into the work would take a very long time counting things up. This would probably not end well for the eager solver because for one thing, you are working with a generic integer n, and for another, double-counting and missed counts are a real issue, and for the third matter, we can do a lot better if we just have thought more.
Before we get to how we can do well by thinking about the problem more, I would just like to share the fact that YouTube content creator "blackpenredpen" has a very intuitive solution to this problem:
I will quickly recap his solution. The making of a rectangle out of the n by n chessboard is a two-step process: first you select the column boundaries and then the row boundaries. For column boundaries, there are n columns to choose from and you need to pick out two distinct columns (because a rectangle has two vertical sides). Therefore, the number of ways to do it is (n 2). By symmetry, the same process works for picking out two distinct rows out of n rows is also (n 2). Therefore, the total number of distinct rectangles that can be made out of the chessboard is: [(n 2)]2=(n(n+1)2)2+n 2(n+1)2 4
However, usually in math there are more than one way of solving a problem, and while blackpenredpen's solution is quite intuitive, for one thing it does involve some advanced knowledge in combinatorics. Furthermore, if we explore the problem deeper, perhaps we can start to see things in a different light, which then would help us extend what we learned solving this problem to other knowledge domains. In such a case, I would be happy to show how thinking more goes a long way.
I evoked the Hungarian mathematician George Pólya in my previous blog posts about problem solving, and I find relevant to mention him again here. In his How to solve it series, Pólya mentioned 14 different strategies when it comes to solving math problems. Three comes to my mind for this problem:
Solve a simpler problem
Make a table
Break problem in parts
The above three strategies are usually how mathematicians go about solving complex problems--play around with it, find a pattern (like how the seven-year old Gauss did), and generalize the pattern by seeing how solutions to smaller problems lead to solutions for bigger ones. Therefore, let us try a few values for n, starting with the smallest possible values.
| n | r(n) |
--- |
| 1 | 1 |
| 2 | 9 |
| 3 | 36 |
Here I let r(n) denote the number of distinct rectangles formed from an n by n chessboard, a.k.a. our answer. I encourage all readers to try out the values for 1, 2, and especially 3 to see for yourself the corresponding values for r(n) given. One observation is that r(n) seems to be a perfect square, which came to be of no surprise given blackpenredpen's solution. However, while I am finding a pattern and trying to turn the problem into smaller parts, looking at r(n) does not seem to help me, because the perfect squares in r(n) does not seem to tie with n that well:
| n | r(n) |
--- |
| 1 | 1 2=1 |
| 2 | 3 2=9 |
| 3 | 6 2=36 |
Aside from n=1, the fact that r(2)=3 2 and r(3)=6 2 just seemed to be lacking a direct connection between n and r(n). Therefore, as part of playing around, the next logical step, for me anyways, was to look at the difference between the r(n)'s, and more specifically, between consecutive r(n)'s. For completeness here, we'll let r(0)=0 since you cannot make any rectangles from a non-existent chessboard. r(1)−r(0)=1−0=1=1 3 r(2)−r(1)=9−1=8=2 3 r(3)−r(2)=36−9=27=3 3
Aha! It seems that we have found a much more connected pattern between r(n) and n by looking at the consecutive differences than looking at things directly. It seems that: r(n+1)−r(n)=(n+1)3 Ultimately, that's how mathematical discoveries are created. A bunch of child-hearted numberphiles playing around, finding patterns, getting an intuition, and finishing off with a rigorous proof. We seemed to have found a pattern--let us intepret the pattern first to see if it makes sense, then let us set about proving it.
So, what does our hypothesized equation mean? Going from an n by n chessboard to that of n+1 by n+1, our equation means that the number of new rectangles created is equal to (n+1)3. Now let us dive deeper by the aide of a diagram (which is also one of the strategies mentioned by Pólya!).
And there. We. Go. The above diagram illustrates perfectly the difference equation pertaining to our pattern. We notice that if a rectangle is contained entirely within the n by n chessboard, then it is also contained in our new n+1 by n+1 chessboard. Therefore, if we know the value of r(n), we don't have to consider that rectangle again for our new chessboard! The diagram also illustrates what we mean by new rectangles. The definition of a new rectangle created is one that contains at least one of the new squares introduced by upping the dimensions of our chessboard--i.e. containing at least one of red, green, or black squares.
Before I continue, let me denote the red squares as "row squares", because they are appended to each row of the old chessboard and considered as such because they are part of the (n+1)st column. Similarly, I am denoting the green squares as "column squares" because they are part of the (n+1)st row. For the black square that is not adjacent to the old chessboard, it is, by our definition, both a row and a column square.
So, our problem now becomes, instead of counting the total number of rectangles in our chessboard, we are counting the number of new rectangles only. Assuming the solution arrived at an earlier stage and only addressing the new ones are also the inspiration behind dynamic programming in computer science, so already with critical thinking and looking at the problem in a new light, we can see how that can be extended to other fields of math and science.
Let us first count the new rectangles involving the row squares, which are the red and black squares. There are three search dimensions:
The "width" of our new rectangle
The "height" of our new rectangle
The "location" of our new rectangle
It makes sense, since a rectangle has a width and height, and we can draw our rectangle anywhere in the chessboard. When we have multiple search dimensions, it often helps to fix all but one and look at the sole changing dimension for patterns or clues. In this case, let us fix the height to be 1 unit, and let us only look at rectangle formed in the first row, and look at how many x by 1 new rectangles we can form at the top of the chessboard.
Well, the answer is n+1 new rectangles. To see this, just let the red square in the first row be at the end of the new rectangle. We can prepend between 0 and n squares (inclusive) from the old chessboard at the top row in front of the red square to form our new rectangle, so the answer is (n+1).
By symmetry, there are nothing special about the rows, so fixing "height" to be of 1 unit, the total number of new rectangles we can form anywhere in our n+1 by n+1 chessboard is simply: (n+1)×(n+1) Here I don't combine the two (n+1) parts of the expression into a square term because those two (n+1)'s are different in meaning. One (n+1) is the number of height-1 rectangles we can form in a given row, and the other (n+1) is how many distinct rows in which we can form the new rectangles. The latter will change as the height changes. To see this, now we let height to be 2 units. How many rows can we form height-2 rectangles in our n+1 by n+1 chessboard? The answer is n rows because we cannot start creating our height-2 rectangles on the last row. Similar reasoning leads to the fact that there are n−1 rows in which we can form our height-3 rectangles, n−2 rows for height-4 rectangles, and so on, until there is only 1 row to form our height-(n+1) rectangles, which would span the entire height of our chessboard. Therefore, the total number of new rectangles that we can form entirely out of our new row squares (the red and black squares in our diagram) is: (n+1)(n+1)+(n+1)n+(n+1)(n−1)+⋯+(n+1), which we can write as: Answer=(n+1)∑k=1 n k=(n+1)[(n+1)(n+2)2]=(n+1)2(n+2)2
Notice how our drive to think and find patterns gets rewarded when we obtain an expression that can be simplified further via the Gauss Formula, which the Prince derived through pattern finding as well!
By symmetry, the number of new column rectangles that we can create, which are the ones containing the green and black squares, is also (n+1)2(n+2)2. Combining our results, the total number of new rectangles, which are defined as the ones containing the red, green, and black squares, becomes: Answer=(n+1)2(n+2)2+(n+1)2(n+2)2=2(n+1)2(n+2)2=(n+1)2(n+2)
However, let us put on our critical thinking hats and try to see whether we are done here. Is our answer for number of new rectangles created really (n+1)2(n+2)? If we think carefully, we can see that we double-counted here. Remember that our sole black square is both a row and a column square, so when we derived the number of new rectangles involving row and column squares, the ones that involve the black square got accounted for twice. We can summarize our findings with a Venn diagram.
So our answer is actually |A∪B|, or the number of elements in the union of the new rectangles created via row and column squares. This is same as adding up the total number of elements in both sets and subtracting the intersected members, i.e. the double-counted rectangles. In other words: |A∪B|=|A|+|B|−|A∩B| But what is A∩B? It is the set of a new rectangles created involving both row and column squares. Since row squares and column squares can only be joined by the black square, this set is equivalent to the set of all new rectangles that can be created if it contains the black square. So now we only need to worry about how many new squares we can create that contains the black square. This counting is now very straightforward, since the rectangle's location is fixed. Only dimensions of this new rectangle can vary and even then, there is only one possible rectangle per dimension, since the lower right square of the rectangle must be the black square (please try out yourself to see!). Therefore, since there are (n+1)2 possible dimensions in our n+1 by n+1 chessboard, |A∩B|=(n+1)2.
Putting everything together: Number of new rectangles created=|A∪B|=(|A|+|B|)−|A∩B|=(n+1)2(n+2)−(n+1)2=(n+1)2(n+1)=(n+1)3 As desired. Therefore, indeed, we have: r(n+1)−r(n)=(n+1)3
Since r(1)=1, by induction, we see that: r(n)=∑k=1 n k 3
So we arrived at two answers for the number of rectangles in an n by n chessboard: one by looking at number of new rectangles created, and the other by using combinatorics from blackpenredpen. Which one is correct? Well, the answer is both! As mentioned from the beginning, there are often multiple ways in solving math problems, and each solution brings about new light to the problem that can also be extended to other fields. Here, by combining our two solutions, we see a very interesting fact: 1 3+2 3+3 3+⋯+n 3=(1+2+3+⋯+n)2 This is the famous sum of cubes formula, in which the right-hand side is arrived by reverse engineering blackpenredpen's solution via the Gauss Formula (again!). The above equation can also be proven via induction (please try this!), but that is not the point of this blog. Through this very fascinating counting problem, I hope I can shed light into the fact that when we think critically when confronted with a problem instead of mindless delving into brute force work, not only do we make our work lighter, but we also uncover patterns that expand our knowledge and marvels at math and nature.
Epilogue
In the spirit of critical thinking and finding patterns, I challenge all my readers to the following problem, again involving a grid of squares:
"How many ways (distinct routes) can we take to traverse an M×N grid of squares, if we can only travel down and right, and we start from the top left corner and finish at the bottom right? We can only turn (change directions) at an intersecting point in the grid."
I will reveal the answer next time!
Bonus Question
If you solved the challenge problem, here is an extension--inside our grid network, there will be certain intersections in which you must visit during your journey from start to finish. In addition, there will be certain intersections in which you must avoid along your journey. Please describe a general solution for solving this extension problem, given the solution to the original challenge question.
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9589 | https://www.mathspanda.com/A2FM/Lessons/Changing_between_polar_and_Cartesian_coordinates_LESSON.pdf | Changing between polar and Cartesian coordinates Starter 1. By possibly considering complex number equivalents, convert these points from polar form, to Cartesian form : (a) (b) (c) 2. By possibly considering complex number equivalents, convert these points from Cartesian form to polar form, : (a) (b) (c) Notes Converting between polar and Cartesian coordinates is like converting between the and forms of complex numbers. Converting between Cartesian and polar coordinates Polar to Cartesian coordinates When converting from polar to Cartesian coordinates: i.e. Cartesian to polar form This is the same as converting from the , component form to the magnitude, direction form of a vector.
To find the distance to the pole: .
To find the angle: 1.
Calculate N.B. Notice that we ignore the signs of the components of the complex number when finding the initial angle.
2.
Sketch a quick diagram to decide which quadrant the point lies in and then decide what you need to do to the angle found in step 1. For example, if the point is in the 4th quadrant, subtract the angle found from (since ).
Converting between Cartesian and polar equations of curves Cartesian to polar form E.g. 1 Convert into polar form and state the family of curves from which it comes. Working: : To convert from Cartesian to polar form: replace by and by and simplify, possibly using trigonometric identities.
(r, θ) (x, y) (8, π 6 ) (6, −3π 4 ) (5, π 2 ) (x, y) (r, θ) (1, 3) (−4, 4) (0, −6) r(cos θ + sin θ) a + bi (r, θ) ≡(r cos θ, r sin θ) x = r cos θ y = r sin θ i j r = x2 + y2 tan−1 y x 2π 2π ≡360∘ (x2 + y2)2 = 4xy (x2 + y2)2 = 4xy (x2 + y2)2 = 4xy (r2)2 = 4r cos θ × r sin θ r4 = 2r2 sin 2θ r2 = 2 sin 2θ x r cos θ y r sin θ x y 1st Quadrant 2nd Quadrant 4th Quadrant 3rd Quadrant E.g. 2 Express these Cartesian equations in polar form: (a) (b) , where Polar to Cartesian form E.g. 3 Convert the polar equation to Cartesian form. Working: : To convert from polar to Cartesian form: replace by , by and by E.g. 4 Express these polar equations in Cartesian form: (a) (b) for Video: Converting Cartesian to polar coordinates Video: Converting polar to Cartesian coordinates Video: Converting polar equations to Cartesian equations Video: Converting Cartesian equations to polar equations Solutions to Starter and E.g.s Exercise p209 9C Qu 1i, 2i, 3i, 4i, 5-9 Summary When converting between Cartesian and polar coordinates and equations use: where N.B. When finding the angle, decide which quadrant the point and take this into account before finding the angle.
y = x2 x cos α + y sin α = p p > 0 r = 2a cos θ r = 2a cos θ x2 + y2 = 2a × x r x2 + y2 = 2ax r x2 + y2 cos θ x r sin θ y r r2 = a2 sin 2θ r = 2 sin θ 0 ≤θ < π x = r cos θ y = r sin θ r = x2 + y2 tan θ = y x 0 ≤θ < 2π |
9590 | https://www.youtube.com/watch?v=_NYsYuzMLs0 | Lecture 2: Contradiction and Induction
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Posted: 22 Jul 2025
MIT 6.1200J Mathematics for Computer Science, Spring 2024
Instructor: Zachary Abel
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Today we will explore some basic (but crucial!) proof techniques, and then two powerful techniques: proof by contradiction and proof by induction.
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Transcript:
Hello, good morning, good afternoon. Hope everyone's doing well. Welcome back to 61200. Um, today we are going to be talking more about proofs. Um, we spent all of last lecture going through twothirds of our definition of a mathematical proof. Um and the last term we haven't talked about yet uh is logical deductions. So we said that mathematical proofs are built from are verifications of propositions. We talked a lot about what propositions are um using a sequence of logical deductions starting from a base set of axioms and we talked about what axioms are. So finally we get to talk about logical deductions. This is going to be the heart of how we build and construct and check our proofs. Um, so this is where most of the proofing happens inside of our proofs. Um, so what is a logical deduction? Let's see. So an inference rule, let's see, is a rule for combining true propositions to form other true propositions. For example, one of the classic ones and one we've um we've talked about uh briefly is called modus ponins. And I'm writing this just to say let's not ever use those words again. Um the the concept is um is intuitive enough. If we know P and we know P implies Q. If we know these then we know Q. Yeah. If we know P and we know P implies Q then we know Q. If I think and thinking implies amming then I am. Um very intuitive. Let's again never mention this by name. That's way too uh precise for our uh for our purposes. Um there are some other famous ones like if we know P implies Q and we know not Q sorry not Q then all of this implies all of that implies not P. This is sort of the opposite direction. If we know P implies Q but Q is false then P better be false as well. Um here's another fun one. Uh if we know that not P implies false then that implies P. Basically saying if P isn't false then it's true. Hopefully intuitive but we have the tools to not just rely on intuition here. We can verify these statements are actual inference rules. Uh for example by drawing a truth table. Um let's do that last one just because it's smaller. Uh let's see. So uh we have our we have P and it has two possible values. Um and then we have our big old formula. So not P implies false. PNS imply P. I want to make sure that this formula on the right is always true no matter what P is. Um okay. Well, a useful way to fill out these uh these truth tables um is to go sort of one step at a time. And I like to do that by drawing directly under the symbol I'm working with. So like not P. When P is true, not P is false. And when P is false, not P is true. And I put that right under the the P bar. So I know I'm talking about the P bar. Now let's talk about this implies. Does false imply false? Yeah. Yeah, that's true. Does true imply false? No, that's the case we need to avoid for implication. Okay, so finally um this this implies is really this whole parenthetical um clause there. Does true imply true? Because remember this is true and false here. Does true imply true? Yeah. So this implies in this case is true. Does false imply false? also yes. Okay, so the column that corresponds to the whole formula is always true no matter what. And that is how we verify that this statement right here is always true. This is one possible way to verify that. Um these other um inference rules and lots of other ones that we're going to use in this class. You can also verify this way if you wish. Please don't like do this in the middle of a proof. Let's just use these rules. um as rules that we're familiar with. You don't have to reprove them every time. I'm just showing you that we have the tools to do it if we wanted to. All right. Um in general, when writing proofs in this class, um we're going to want each step to be clear and logical. We're going to want to use a valid um uh inference rule. Um but we just want to use it. Um, it should be clear from your writing what rule is being used, but you don't have to call it out every time. Like modus pollins, modus tolins, I didn't even tell you what that second one means. Um, just go ahead and use it. We're going to have a familiar set of tools and deductions uh when writing these proofs. So just um if you use those or you're doing something else that is clear enough um use that. And if it's less clear, explain it more. Um, of course, you should be explaining your steps. Um, usually you're relying on facts that we proved before or axioms that maybe we haven't explicitly written down yet. And you should be identifying those. When you're relying on something, tell us what you're relying on. Um, let's see. Please don't use what's called proof by intimidation. Clearly, this is true. This is obvious. This is intuitive. Uh, in fact, I would strongly recommend striking those words from your mathematical vocabulary entirely. Um, anything that is clear and obvious to you, well, first of all, it might not be clear and obvious to your readers, um, who might be your graders, who might be your peers. Um, and that can lose your points, u, because you didn't explain yourself fully. That can be demotivating for whoever is reading it, like, wait, this isn't obvious to me. Am I not as smart as I thought I was? Um, and of course, it runs the risk that the thing you said is obvious is wrong. Um, in fact, this is a common source of mistakes because instead of proving it, you just glossed over it. Um, the proof is there to help you check that things are correct and help you catch mistakes. And if you're skipping the proof, then you're skipping that redundancy. Um, so things are never obvious, things are never clear. Please explain it. Uh um I I mentioned last time that we're going to take basically all of high school math as our common set of axioms. Uh the the one exception to that is if we ask you to prove a theorem and you say, "I know this from high school math. It's an axiom. I'm done." That's maybe not um adhering to the spirit of the assignment. If we ask you to prove a statement that you already know is true, please please go and prove it. Um, we're asking you to think more deeply about why it's true. Does that make sense? Cool. All right. So, using these logical deductions and things like them, I want to talk about a couple very common proof outlines. Proof outlines. Oh, that's not a Q. That's a U. about lines for proving some of the most common shapes that our theorems might take. For example, if we have a theorem, there exists a natural number n such that n is at least 10 and uh n is prime. Here's our theorem. There exists a number that's at least 10 and is prime. There exists a prime um bigger than n. Can someone prove this for me? Yes. 17. That's actually the same example in my notes. Well done. Um proof n equals 17. Is that a full proof? I mean kind of. Yes. if we one more time. Okay. Um so the suggestion was we should be going through all of the possible factors of 17 and show that they're not factors to to verify that 17 is prime. Absolutely. Um that is something we haven't done. Um also we haven't checked that explicitly. We probably want to do that. Um, but let me back up and let me write down what the general proof usually looks like. Um, let's see. Proof. Oops. We'll show that n equals fill in the blank works. This n is in the set we asked for. It's in the natural numbers. In this case, we're going to secretly be using 17. Um, so it's pretty clear, but in general, it might be less clear. So, for reasons and n is prime because reasons. So this is in general how we recommend writing a proof of a of an exists. If you want to show something exists in a set with a property, you tell us what it is. You show us why it's in the set, you show us why it has the property. Pretty intuitive, right? Hopefully. I know I just used that word that I said we weren't we aren't supposed to use. Um but this is sort of inextricably tied with what this symbol means. In general, there exists a natural number. To to prove that there exists a number with this property, you have to show me a number with this property. That's the most straightforward way to do it. Um, and that's what this proof outline is letting us do. In this particular problem, we're going to say we'll show n= 17 works. This number 17 is a natural number because, well, yeah, that I I don't know how to convince you of that if you're not sure of that. Um and n equ= 17 is prime because and this is where we should if we want to be super careful check all the possible factors um it's not divisible by two by three by four etc. Uh Eric I am missing n greater equal to 10 um yeah we need to show that n satisfies this whole property um and n is prime and greater equal to 10 because reasons. All right. So more abstractly, if our theorem is um there exists X in some set such that P of X holds, then our proof can often look like um choose X equals some specific value that you have to put in. Um then X is in S because big to-do there. These are these are our to-dos and p of x is true because to-do. All right. Similarly, what if we're asked to prove a for all theorem for all numbers x in the reals. So for all real numbers x x^2 - 6x is greater than -10. So this theorem is about a for all not an exists. Um when we're asking about exists we need just a single example. But when we're asking about for all examples aren't enough. Remember last time when we talked about n^2 + n + 41 being prime lots of times? Well it's not prime every time. If we're trying to prove a for all, we need to show it that the property is true for every possible choice of X. And in general, let's see what this outline might look like. So if our theorem is for all X in some set, P of X is true. Our proof usually looks like this. Proof. Suppose X is a generic element of S then P of X is true because reasons, right? I'm think it makes sense to put little question marks in there so we know that we we have to figure out what's going on. There we go. this first line. Suppose X is some element of S. Which element? We're not allowed to know. We're not allowed to choose a particular one. We need to prove that this is true for every possible choice of X. Which means in our proof, the only thing we assume about X is that it's in S. So you start by taking something in S and assuming nothing else about it and then going and and concluding the property that we need to show. Again, this kind of like These might look very similar to you and they are. Um this says everything in S has this thing. This says a generic thing in S has this property. Um the difference is that suppose X is a generic element of S or however you want to phrase that makes it concrete. It gives us a particular X to talk about. So now we have something to use and talk about and reason about inside of our proof. Whereas up here, every how do we approach every that that's a little daunting. I I only know how to deal with one. So this gives us one. It's called x. We know nothing about it except that it's an s. But at least we have the one and we can work with it. So whenever you see a theorem that starts with a for all, your first line is almost always going to be this in order to bring some into scope that we can talk about. Sound good? Awesome. So how do we prove this theorem proof? Let's see. Let's see. Um suppose x is an element of the real numbers. Then let's see x - 3^ squared is greater equal 0. Um because all reals have non- negative squares. This can be a fact we're citing from high school. We know how the real numbers multiply to each other. All right. Right. And once we know this um let's see equivalently this thing says that x^2 - 6x uh is greater or equal to -9 which in turn is greater or equal to -10 as needed. So we started with pick some element of r without specifying which one. We ended with the property we need is true. QED. Sound good? QED by the way um is a common way to mark the end of a proof. It's Latin for quote erat demonstrandom which was to be shown. So it's a synonym for like as needed or um as desired. QED all the same thing. All right. Question. Sure. So the question was could we do this a different way? Could we look at like all the negative numbers and they behave one way and then look at all the numbers between zero and three or zero and six or something and they behave a different way and then look at all the rest. Yes, absolutely we can do something like that. That's called proof by cases. We're going to talk about it next lecture. Um but for now let's move on to the next common outline which is proving uh implications. If we have a theorem that says P implies Q. I mentioned we're going to be using implies all the time. And often theorems say if this then that. If P then Q. How do we prove this? Our proof often looks like assume P then Q is true because reasons again looks pretty similar to the kind of thing we're talking about. But this first sentence assume P assume P is true makes things concrete. It instantiates um the thing we're supposed to be assuming. Um it gives us something concrete to work with. We can now live in this hypothetical where we know for certain that P is true and then work with it as if it's true because inside of this proof for the rest of this proof it is. Assume P. Use that to show Q. Whenever we're trying to prove an implication, this is a common way to do it. Um this is called a direct proof. Um we're going to see indirect proofs a bit later. Um but there is a another common way uh actually sorry before that um let's do an example of proving something um in this fashion theorem. Um, if n is a multiple of 10, then it is even. If n is a multiple of 10, then it is even. We're proving an implication. N multiple of 10 implies even. P implies Q. So even without having to think very hard proof assume n is a multiple of 10. We know we're going to start with that. Assume the first thing we know we're we're going to end with the last thing. Therefore n is even. So question mark question mark question mark. Therefore, n is even. And if we can connect the dots from the first line to the last line, we'll have our proof. Notice, by the way, that was kind of an automatic process. If the theorem is an implication, then kind of this is what our proof is commonly going to look like. So, we can just write it down. And now we have a clearer, more concrete sense of the task we're supposed to do. Starting from the fact that n is even, we're supposed to derive the fact that it's sorry, starting from the fact that it's a multiple of 10, derive the fact that it's even. So, let's go ahead and do that. Assume n is a multiple of 10. Um, what does it mean to be a multiple of 10? So, n is 10 k for some integer k, right? Um, but then n equ= 2 5 k and so it's 2 an integer. Therefore, it's even. There's our proof. We're done. Just by identifying where we're supposed to start and where we're supposed to end, we were able to bridge the gap. Sound good? Nice. So, there's another common way of proving implications. Um this is called proof by contraositive. Recall last uh last lecture we said that P oops P implies Q is equivalent to its contraositive uh let's see not Q implies not P. This right here is the contraositive contraositive. And so if we're supposed to be proving P implies Q, sometimes it's more convenient to instead prove not Q implies not P. And so our proof outline um would say proof by contraositive. So assume Q is false. Assume not Q. See assume Q is false. Then P is false because question mark, right? Same as we did for for direct implications, but we're proving this implication instead. So assume not Q, use it to prove not P. And let's see an example of this theorem. Uh let's see. So for all integers n let's see if n^ squ is even then n is even for all natural numbers n. If n squ is even then n is even. Now automatically oh we're proving a for all. I know how to write that proof proof. Suppose n is an integer. Now we have an n to work with right we need to prove this implication. Um let's see. So proof by contraositive always nice to say what proof technique you're using if it's maybe not as self-evident as it might be. So now we're indicating we're proving this implication by contraositive. And once we've said that both you the writer and your audience the readers know what's supposed to come next. Proof of contraositive. So assume this thing on the right is false. Assume n is odd because that's the opposite of n being even. Want to show n squ is is not even. N squ is odd. This by the way wts um want to show this. This I find is really useful for me when planning out proofs, thinking about proofs, sometimes even when writing proofs. Um, this is an indicator that this is where I'm trying to go. This is the end of our proof. This is my goal. Um, and it signals both to you that this is work that's left to do. Um, it signals to your reader that okay, great. We we agree that we're working towards the right thing. So now let's go and do it. Um, and also as you're writing scratchwork and solving these proofs, um, and maybe you get distracted and have to come back later and you see a bunch of statements on your page. Wait, which of these statements did we prove and which are we trying to work towards? So, um, as I'm working on these proofs, I really like to be organized and say, we know this, we know that, we want to show this other thing. So, I know that we haven't proven it yet and we want to. All right. So proof a contraositive. So assume n is odd. We want to show that n^ squ is odd, right? Because that's that's what the proof method tells us to do. And now we can go do it. Okay. So assume n is odd. Um so n is 2k + 1. That's what it means to be odd for some integer k. Then n^2 is 4k ^2 + 4k + 1 which is 2 an integer + 1 which is odd q e d. And just to be precise about it 2k oop sorry 2k^ 2 + 2k but two times an integer plus one is what it means to be odd. So we're done. All right. So, we have proven implication. We've shown two different proof outlines for how to show P implies Q. One direct, you assume P, use it to conclude Q. Um, one sort of flipped around. Um, assume not Q, use it to derive not P. Uh, there's a related proof technique called proof by contradiction. Now, in growing up, you might have been told, "Don't use double negatives. They're not as clear. It's always much stronger to more directly say what you mean." But sometimes in math, a double negative isn't what we don't want to avoid. I think there were four negatives in there. Got it right. Yeah. Uh double negatives can sometimes work to our favor. uh and proof by contradiction is exactly that technique as a proof outline. So let's talk about what this does. Um proof by contradiction. So theorem P whatever P is proof by contradiction can be used to prove anything you want. You know sometimes sometimes it it works sometimes it doesn't sometimes it's nicer but it's not assuming a specific form of theorem. Our proof looks like this. Um let's see um for the sake of contradiction. Assume P is false. All right. Assume P is false. Then the proof continues. Then some other statement R. Any statement R we want we get to choose. Then r is both true and false, which is bad. That breaks math. Propositions are supposed to be true or false, not both. Um, that's a sign that our set of axioms is inconsistent. And remember, that's the the one thing we don't want them to be. Um, if something's true and false, then everything's true and everything's false. And and that's bad. let's not break math. Um, which is a contradiction. Um, and then to to explain it further, so our assumption is wrong. So, P is true. So we started by assuming P is false. We broke math. So the assumption that P is false must be wrong. So if P is false, it isn't true. Um now I put this last sentence in parenthesis because it's clear, it's clearer than if you leave it out. Um, but if you're writing between people that all agree on how proof by contradiction works and understand what's going on, then just including this word contradiction at the beginning and saying uh what the contradiction that we've reached is, that's usually enough. You don't have to have that that summary statement at the end if you don't want it. Um, that's a matter of personal style, but you should be saying right at the beginning that you're doing a proof by contradiction, and you should say at the end what your contradiction is. Does that make sense? Cool. Um, yes. Yeah. Um, that's exactly right. This uh inference rule that we analyzed at the beginning is exactly what's justifying proof by contradiction as a proof outline. Um, if not P implies that false is true, then P has to be true. That's exactly what this is saying and that's what the proof technique is doing. Very well spotted. Thank you. Um, I meant to say that and might have forgotten. So, thank you for the reminder. Um, a couple other things about this. Um, for the sake of contradiction, um, there are lots of other ways to say that. You can just say proof by contradiction and that's enough to signal to the reader that you're doing a proof by contradiction. Um, I've also seen block by way of contradiction assume P is false. Uh, I have seen that. I like it. I don't use it as much because I don't think it's as standard. Um, but it's silly and I like it. Um, at the end when you find your contradiction and you want to say this is a contradiction, we've broken math. We're done. However you want to say that part. Um, I've also seen this symbol equals x equals because it looks like two arrows pointing at each other contradicting each other. That's another fun embellishment if you want. But now that we've seen the proof outline, hey, that looks like a smiley face. Um, now that we've seen the proof outline, let's use it um on a real theorem. and that is that the square root of two is not a rational number. So remember this blackboard bold q is the set of rational numbers of um integer divided by non-zero integer the set of all such things. Uh and the theorem is that square root of two is not rational. So let's prove it. Often when your theorem has a knot on the outside, it is not true that property. Contradiction is a common way to handle that because if um because our proof by contradiction is going to say well proof by contradiction assume and again I want to emphasize the moment we say this is a proof by contradiction what we assume is fixed. There's only one thing we're supposed to assume. The outline tells us what we have to assume, which is that the entire theorem is false. That's the only thing we're supposed to do. We should still say it because we want you and the reader to agree that the correct procedure is being followed. You're assuming the correct thing. So assume that this is false. In other words, assume that square of two is a rational number. uh is not isn't do what I say not what I write. Um okay and now let's let's keep going. What does it mean for square root of two to be a rational number? Okay. Um so square of two equals a over b um for some integers a and b where b does not equal zero. And as we might recall from high school math, we can always assume that a fraction is in lowest terms. Um, in lowest terms, so a and b have no common divisor, right? We can always reduce a fraction to lowest terms. So let's take that fraction already written in lowest terms and let's keep going from there. Okay. So a over b= 2 means a = 2 b means a^2 = 2 b ^2 which means a^2 is even. And wait, we already proved something about if the square of a number is even then the number itself is even. It's very convenient that we already have that there. So since a^ squ is even this theorem shows us that a is even by the theorem over there on the previous board. Yeah. Okay. What does it mean for a to be even? So a equals 2 times some integer c for some c oop sorry for some integer I did it again c in the integers right that's what it means for a to be even it's two times an integer so let's keep going let's put this back into our formula um so now this says that 2 c^ 2 = 2 b ^2. Right? So this says that 4 c ^2 = 2 b ^2. We can divide by 2. This says 2 c ^2 = b ^2. But now b ^2 is even. And as we know by our by our lema over there again b is now even by the same theorem. Yeah. Okay. So, A and B are now both even. So, they have a common factor of two. I to be more precise, no common divisor is greater than one. um a and if a and b are in lowest terms then there's no integer bigger than one that divides both of them. But we just found that two divides both of them. So this is our contradiction. This statement here contradicts this statement here. Right? We have a single statement that we showed is both true and false. So we've reached a contradiction and our proof is done. Does that make sense? Awesome. Very nice. Very nice. Um, let me take an one more minute to talk about these proof outline techniques in general. Um, now these aren't meant to restrict the way you're allowed to prove things or restrict the way you're allowed to express things. They're there as scaffolds to make sure you're proving the right things um and to direct your thinking. The more you can break down a proof into smaller and smaller to-dos. Um first of all, the easier it is to make sure that your proof is correct. Um and secondly, the the easier it's going to be to think about those individual tasks. Smaller tasks are easier to handle than larger tasks. Um, also a common mistake when writing proofs in general is when you're presented with a theorem to dive right in. Why is this true? How am I going to figure this out without taking the time to realize that actually what's going on in your head, what you're trying to prove is not actually what the theorem is asking you to prove. And so that's just a bunch of wasted effort and lost points on your homework. Um if you had taken a few minutes um or with practice a few seconds to do these um standard outlining procedures to figure out what the structure of the proof is supposed to be, then now you know what the structure of the proof is supposed to be and you can spend your um your precious thinking cycles thinking about the things that you're actually supposed to prove. uh and let me demonstrate this um that with proof you can make this automatic. You can make this fast. You can really use it to help inform your uh your proof writing endeavors. So let's say we have a theorem. Let's see. So an integer n is foolish precisely when n+1 is barome. What do these underlying words mean? Absolutely nothing. But let's imagine they actually mean something. Let's imagine this theorem is meaningful and we've been asked to prove it. I claim there's a lot of progress we can make in writing and structuring this proof even before we go and look up what these two weird terms mean. So let's do that. Okay. So this theorem we can we can phrase it as for all n in z n is foolish if and only if n plus one is barome. Excuse me. Um, now this sorry I would I prefer to write it like this. FN if and only if BN plus one. I don't remember if we talked about if and only if on Tuesday. Great. Let's talk about it now. Um, so P if and only if Q also written as P if and only if Q um means that P and Q are both true or both false. um P and Q both true or both false. It means they're expressing the same condition. They're they're true in exactly the same times. They're false in exactly the same times. So they're logically equivalent. Um one way you can define this like we we could just draw the truth table like we did last time, but also commonly Oops. Um this is this can be defined as P implies Q and Q implies P. If P is true then Q is true and if Q is true then P is true. All right. Um and this if and only if um implies in both directions this concept is what we mean when we say precisely when these are true or false together. Sound good? Okay. So, we have our theorem. Let's prove it. Proof. Okay. The theorem starts with a for all. I know how to handle a for all. Um suppose n is an integer. We want to show fn if and only if bn plus one. Right? That was that was one of the first outlines we talked about. If you have a for all, you introduce a generic one and you want to show the property for that n. Right? Okay. Well, now we want to show an if and only if which we know is defined like this. So really instead of all of that, we can be a little more precise with it. We want to show that f(n) implies bn plus one and that bn plus one implies fn. And once we've done both of those tasks, we will have proved the ifolith. Okay. Well, now we're trying to prove implications. We know how to prove implications. Let let's use a direct method for each of these. So assume um fn is true. Now we want to show bn plus one. So these are sort of separate sections in our proof. In this part we are temporarily assuming that fn is true and using that to prove that bn plus one is true. And now down here uh assume instead that bn plus one is true. And now we want to show fn is true. And now our tasks are very clear to us. We need to show that if fn then bn plus one. We need to show um in addition to that we need to show that if bn plus1 then fn. And now at this point, there's not really much else we can do without going to look up what these terms mean, use their definitions, figure out why this thing is true. But this is where the interesting part of the proof happens, where the unique part of the proof happens, the part that's specific to this problem. Um, yeah, and that's a lot of progress that we made even before the theorem even makes sense. So once you get better at this task um the easier it will be to set things up correctly and set yourself up for success. Questions on that? Awesome. All right. So the other main topic for today, we talked a lot about proof outlines, proof by contradiction. The other main topic is proof by induction. [Applause] And interesting I didn't I didn't hear the groan that I usually get. Um proof by induction is one of those t one of those topics that um some students come in with the impression that it's difficult um that it's hard to grasp. Um and there are reasons for that. Um even before telling you what induction is, I can tell you that it's kind of a shame that we have to teach you induction so early. um because we're in we're now in the process of telling you that proofs are supposed to be clear and self-evident and um and easy to follow. And then we come to proof by induction which has a weird outline that has lots of things you have to say in weird orders for some arcane region like we're summoning an ancient demon. Why do we have to say this in that particular way in this place? I don't know. It's confusing and it is confusing and there's a reason it's confusing that sort of the reason induction works and the reason that um that the proof is structured the way it's structured isn't explained in every proof byinduction because everyone who's writing a proof by induction assumes that everyone who's reading their proof by induction knows what proof by induction is and why it works. So every proof by induction you see is completely opaque and makes absolutely no sense until you know what induction is. And I think that's why it can be difficult to approach. Um what we're going to talk about right now is first of all before I tell you what proof by induction is let's solve a problem together that's sort of going to lead us to the the why and the how and the what of induction. It's going to tell us what we need to do and what tool we need to do it and then induction will be there right at the end to say I'm the tool for the job. So here is a theorem theorem for all n and z for uh sorry for all natural numbers n 1 + 2 + 3 + up to n equals n n + 1 / 2. This might be familiar to some of you. There are lots of ways to prove it. You can prove it by induction. You can prove it by picture. You can prove it by counting. There are lots and lots of ways. Um, we're going to talk through one particular way of making sense of this theorem. And let's just go by intuition. I'm not we're not doing induction right now. We're just we're just figuring out why this thing is true. Uh, so first of all, let's make sure we understand what this thing is even saying. Um for example um so eg when n equals 4 this is claiming that 1 + 2 + 3 + 4 equals 4 5 over 2 which looks pretty good. This is 10 this is 20 over2. Yeah they're equal. And we're supposed to prove this um this kind of thing for every natural number n. Yeah. Um even all the way down to just n equals 1. eg n equals 1. What's the sum from one up to one? It's just the one itself. One is the sum of one term. Um, and that's supposed to equal 1 2 over 2, which looks like it does even down to n equals 0. What is the sum from one up to zero? That's weird. Um, that is a little weird. Turns out we can make sense of it. um in this in this context. So another way to write this is with what's called sigma notation k goes from one up to n of k. And you can think of this like a for loop. You start at k equals 1 and you write down the value of k in that case one. Then you go up to k equals 2 and write down the value. k equals 3 write down the value. All the way up to this part you go up to k equals n write down the value. And the sigma says you add all these together. Sigma is the the Greek s for sum. Um so this is notation for exactly this sum from 1 to n. Um and by convention and there are good reasons for this convention. But by convention the sum from k= 1 to 0 the sum of zero terms of whatever you want in here. But now we're talking about the sum k is just zero. The empty sum is zero. If you're adding nothing, you have a total of zero. And in that case, the sum the empty sum is 0, which is supposed to equal 0 1 / 2. And it does. Hooray. This theorem is true when n equals 0 and one. And four. We verified three cases, which is a pretty good 0% of all the cases we're trying to show. But it's a good start. So, let's see how to continue. [Applause] So, we are trying to show that 0 = 0 1 / 2. 1 = 1 2 / 2. 1 + 2 uh is 2 3 / 2. 1 + 2 + 3 uh is 3 4 / 2. I'll stop in a minute, I promise. 1 + 2 + 3 + 4 uh = 4 5 / 2. Let's stop there for now. But we're supposed to prove all of these statements. That's what the theorem is asking from us. And we can we can think about this a little bit. Um, if we stare at this, we can notice that on the left side, to get from this to this, we're adding one. To get from this to this, we're adding two. It's right there in how this sum is created. To get from here to here, we're adding three. Again, we're just adding one more term to the sum. Then we're adding four. Then we're adding five. Then six, then seven, then eight, and so on. There's a very simple pattern that explains how these numbers are changing from one row to the next. So, wouldn't it be convenient if the numbers on the right side followed that same pattern? And if they did, it would it would make a lot of sense, at least to me, that since they start at the same number and we're making the same changes to both sides every time we move down a row, that all the rows are going to be equal, that the left side is going to be the right side in every row. Yeah. So, we're hoping that that's a plus one and that's a plus two and that's a plus three and that's a plus4 and so on. These are question marks because we don't really know whether that's true yet. Uh, so let's write down what we're hoping is true. 1 + 2 + up to n is supposed to equal n n + 1 / 2. And then the next row, 1 plus 2 + n plus the next term, n + 1 is supposed to equal n + 1 n + 2 divided by two. These are two consecutive rows in our big list. And what we were saying is to get from here to there, we're definitely adding n plus one. We're just adding this one more term over here. So is it true that to get from here to here, we're adding n plus one? And we can check that. Let's see. Um, n + 1 n + 2 / 2 minus n n + 1 / 2. Well, we can factor in n+1 from both sides. That's n +1 n + 2 / 2 - n /2. That's looking pretty good. That's going to be n +1 2 over2 1. That's just n plus one. So our hunch about that pattern is true. To go from the nth row to the n plus first row, we're adding n plus one. Yeah. So what we said earlier is true. This pattern of add one then two then three then four then five that we saw on the left is recreated on the right. Add one, add two, add three, add four, add five and so on. So, at least to me, this feels like an explanation that should make sense and an explanation that we should be able to write down somehow as a mathematical proof. Does anyone have questions about this idea? I haven't said anything about how we're going to write this, but is the idea clear? Awesome. Um, let's see. So let's dig in a little more and see what exactly we accomplished over there. So this argument here on the right um that we did over here, let me write down a little more carefully what we accomplished. Um and it's going to look like this. Um, if 1 + 2 + n = n n + 1 over two. If that's true, then 1 + 2 + n + 1 equals n + 1 n + 2 over 2. because we just added n + one to both sides. You would also want the algebra work to show that going from this to this really is just adding n plus one. And that's exactly what we did over there. I don't want to write it again, so I just wrote the word algebra. Does it make sense that this is what we accomplished? If this row of our list is true, if left side equals right side, then the next row is also true. Yeah, cool. So, let's put some notation to this. I'm going to say that P of N is the statement. By the way, this colon equals is a way that we'll commonly say P of N is defined to be this thing. So not only is it equal to this, but this line right here is the definition of P of N. So let's define P of N to be the statement that 1 + 2 + up to N equals N N + 1 / 2. So PN is a predicate. It's not a theorem on its own. Uh what is a theorem is that for all n greater equal zero, p of n is true. That's what we're being asked to prove. That's that's just a restatement of our theorem from the top board over there. And now what we accomplished over on that board. Actually, I want a fresh board for this. Um, that's going to be this board up here. What we accomplished is first of all we proved P of 0. That's the the first line up here. And then um these these statements up here, what that's saying is that PN implies PN plus one. So we proved P 0 implies P1. We proved P1 implies P2. P2 implies P3. P3, I'll stop, I promise, implies P4, and all the way down. This is what we really accomplished on these last couple boards. But this isn't the theorem we're aiming for, right? The theorem we're aiming for asks for P of zero, oops, and P of one and P of two and P of three and P of four and all the way down. We want to know that all of these are true. These implications are are not what we're asking for. However, it kind of feels like this should be enough, right? Because once we know P of 0 and we know P of 0 implies P of one then modus ponins we know P of one and once we know P of one this implication tells us P of two once we know P of two pair with this we get P of three then we get P of four then all the way down we get all of them I don't know how to write that yet but that's the intuition that I have guiding this um and so it really feels like if we accomplish the left side we should be able to conclude the right side. Do we agree? All right. Um and now we have the principle of induction which just says what we wanted to say. The left hand side of this table implies the right hand side of this table. If you know all of these, then you conclude all of these. So when we're structuring a proof by induction, we must prove everything here on the left. That is our responsibility. And then everything on the right we get for free. by the principle of induction. To be more precise, principle of induction says if you know P of 0 and for all n greater equal z, p of n implies p of n plus one. That implies for all n greater equal z p of n. So that's exactly what's written in this table just in formula form. P of 0 is this first one over here. And then all of the PN's imply PN plus1's are all of the remaining stuff. And it says if we accomplish all of these then we can conclude these. Yes. Good question. So is the principle of induction a new axiom or is it just a name for this this intuitive fact that we have? Um in truth what it's really an axiom about is how the how the natural numbers are constructed. Um really what it's saying is that if you start from zero and then add one add one add one add one add one um to infinity you will hit all the natural numbers. That's one way of thinking about it. But in practice when we use the principle of induction we're using it like this. Um so yeah you can take it as an axiom you can take it as following from the definition of the natural numbers. We didn't really go that far down in um in defining our fundamentals. So we'll just leave it like this. Good question. All right. So we have the principle of induction. We know what it's accomplishing and why. So how do we actually write using the principle of induction? Let's actually write this proof now. And the thing to always keep in mind when writing a proof by induction is this table. We are responsible for proving everything on the left side and then we're able to conclude everything on the right side. So the theorem that we're trying to prove should be the right side and then we're going to do the left side to get it. So proof by induction um we need to say what p of n um let's see p of n is defined as the sum n + one over two same way we defined pn um before and we'll show that PN is true for every n greater equal 0 by induction. So the first thing we're responsible for is P 0. That's usually called the base case. Base case want to show P of zero. Well, P of 0 says that 0 = 0 1 / 2, which is true. So, we verified P of 0. Usually, the base cases are pretty straightforward. Um, it's the inductive step, inductive step that we need to watch out for. Um, we we now need to prove this whole thing. And we usually write it like this. um assume n is greater or equal zero and assume pn is true. We want to show pn plus one is true. And notice what we just wrote is exactly the unpacked outline for this theorem. It starts with for all. So assume n. It then has an implication. So assume the left side. Let's prove the right side. Right? The things we write when we're doing a proof by induction are not magic. They're just unpacking this. And now we can do it. The next thing I always recommend um when proving something by induction. What is P of N? What is P of N plus one? I know we defined it, but I can't remember what it is. I always find it much more useful to write these things down in context. So in other words, assume we're supposed to assume p of n. So assume 1 + 2 + up to n = n n + 1 / 2. And we want to show PN plus one which says that the next row uh n plus1 n plus2 over two. All right. And just by following the proof outline and by identifying the things we're supposed to identify now we have a very clear task in front of us. Assume this row, use it to prove this next row. And we've already done that. We know how to do that. Um, we know that this thing equals that thing and then this plus that equals that because of algebra. This is all the work we already did. And now we're seeing how the induction outline leads us to figuring out that that is the right work to do. Did that all make sense? Yes. Great question. Does this have to start at zero or could it start somewhere else? Um, yeah, it can start wherever we want. Um, if if we didn't know that P of zero really made sense, uh, and we wanted to start at P of one, we absolutely could have started at P of one as our base case and then all the implications above that. Just keep in mind that by not mentioning P of zero, we didn't prove P of zero and so we can't conclude it either. But wherever we started and all the implications after it um those fit together um to give us the thing on the right side. Um yes that's a slightly more general form of the proof of the induction principle. Um please go ahead and use that version as well. All right. So let's do one more proof by induction. Uh this is a really pretty problem. This seems to be a mandatory problem for any textbook on discrete math. Um, it's one of my favorites. And here is the problem. Which board did I leave on? This one. [Applause] So, the story is like this. We have a big old courtyard. um subdivided into lots and lots of tiny little cells. It's going to be 2 to the n tall and 2 to the n wide. So imagine like 64 by 64. It's a big old checkerboard. Um I'm just going to draw 8 by 8 because I don't have that kind of time. And now we're going to take a single cell and get rid of it. Which cell? I don't know. Doesn't matter. Um the question is, can we then fill the remaining cells with these LORs? You might know the word domino. It's two squares next to each other. It's a it's a fun game. Um this is like a domino, but it's three squares instead of two. and it's in the shape of an L. So it's called the L traumo. And the question is can we fill all of this board without that one missing cell um using L trainos and nothing else. So that is our theorem theorem um a 2 to the n by 2 to the n board uh without one cell without any one cell can be tiled with l prominos. So that's what we're supposed to prove. And let's see if I can set this up quickly. Um, yeah, that's all right. See if I can get the glare a little different. Okay, I do want to plug in. Thank you. I always forget this in Zoom meetings, too. Please share your screen. Yes. All right. So, I I I laser cut some toys to help us play with this puzzle. Um, so we're supposed to prove that a two to the n by two to the n board can always be tiled with elmos. Here, let me put something on there for you to enjoy. Well, let's phrase it like it were um like we were going to prove it with induction because we are. So let's say that P of N is going to be um a two to the N by two to the N board without one cell without any one cell. Doesn't matter which one you pick can be tiled with L's. Going to call them L's not L trromos. And notice I kind of just wrote the same thing twice. That often happens. P of N is not our theorem. It's the predicate um which is inside of the for all that is part of the theorem. So this is really every 2 to the n by two to the n board. Um so if p of n is this then our theorem is that for all n greater equal 1 p of n is true. All right. And now we need to prove this. Okay. Um let's start with n equals 1. You can think about what n equals 0 means. It turns out it's true. Um but let's start with n equals 1. So a 2 to the 1x 2 to the one board. That's a 2 x two board. Um and can you see the the little um crosshairs in there? It's a 2 x two grid. And we're supposed to say no matter which single piece no matter which single cell we remove, we can always tile what remains with these L-shapes. Anyone see how to do it? Yeah. All right. I'm gonna put that there. Solved. What if the missing cell was in a different place? Like there. Yep. Okay. Can still solve that. We're allowed to rotate. So the 2 by 2 case isn't very hard. Um, so let's keep track of our progress. P of one. Actually, I'm going to write this as the 2x two case. Not very hard. Let's let's step it up a little. What about the 4x4 case? And say we're trying to avoid that one cell. Well, we also need to solve it for avoiding that one cell and that one cell. We need to solve all 16 cases because the theorem says it doesn't matter which one you exclude. You can always tile the rest. Um, well, you can play around with this. I have lots and lots of of pieces to play with. You can you can tell I had way too much fun designing this thing. Um, and turns out you can you can make it work. Um, like this one. This is one of the 16 examples. Turns out if I put that one there and then that one there and there and there and there, then I've solved it. I've solved this one out of 16 cases. Um, but also it turns out it's not too difficult to generalize from there. Um, like if the missing cell is in one of these four positions, then I know how to do it. If it's in one of these four positions, then I also know how to do it. If it's there, I can just put that there. If it's there, I can put that there. If it's in one of these four spots, can still do it. if it's one of these four spots can still do it. So by looking at all of those 16 cases combined, we now have a way that can put that yellow cell anywhere we want and finish tiling the rest. Not a very nice or clean argument, but at least we did all 16 cases. So that's good enough for now. Let's call the 4x4 case solved for now and move on. Um, and I want to emphasize this is actually a key moment in this proof. We just solved the 4x4 case by looking at all 16 cases and showing that they're all possible. I want you to hold that deep in your heart that we have solved 4x4. Any 4x4 you ever see with a single missing cell can be tiled. Yeah. I want you to hold that deeply in your heart and then not worry too much about how we did all the 16 cases. Just know that it's true. Are we comfortable with that for now? 4x4 case is fully solved. Let's move on to 8 by8. We have an 8 by8. There's a single cell we're trying to avoid. Maybe it's there. Um maybe it's over here. Let's leave it there. I don't care. Um what we're really looking for here is sort of a generalizable method. I don't want to do 64 cases and in the 16 x 16 case I don't want to do what is that 256 cases. Please don't make me. Um we want a generalizable process that we can that we can use to solve every board size. Um and here is a cool idea that works nicely. Um so first thing we're going to do, we're going to we're going to shift our perspective. This isn't an 8 by8 board. This is for 4x4 boards. All right. And now um let me just make this one purple for now to remember that that's the special one. We cannot move that one. However, remember we we have solved the 4x4 case. So like if we if we're missing a cell somewhere in this quadrant and missing a cell somewhere in this quadrant and missing a cell somewhere in this quadrant, but now we just have four instances of the 4x4 problem. We have four 4x4s each missing a single cell. And we know how to solve all of those, right? So, we know how to solve all of the red that remains. Unfortunately, we're not actually allowed to use these individual cells. We're supposed to cover everything with our L-shaped tiles. Um, but remember, doesn't matter where we put these yellow ones. Anywhere we put this in here, we know how to solve that quadrant. Anywhere we put this in here, we know how to solve that quadrant. The purple one was the one that's given to us. We're not allowed to move that. Um, but the next clever idea is what if we put the yellow ones right there and then think of them as not three individual cells, but as an L in their own right. No, there we go. So now this is going to be our key idea. Um, if we're trying to tile everything except the single special cell, step one, think about it as four separate quadrants. Step two, put an L right in the middle, directly in the middle, to make sure that each of the four quadrants is now missing one cell. Yeah. If this initial cell that we're supposed to avoid is somewhere else, well, then we just put the L to touch the other three quadrants. So, we just rotate it this way instead. And once again, every single quadrant, all four quadrants are missing a single cell to really drive that home. Let's see. Uh, we have this shape here. That's a 4x4 missing a corner. This shape here is a 4x4 missing a corner. Here's a 4x4 missing a corner. Here's a 4x4 missing that initial cell. So we have four separate problems where if we can solve the four orange cases, we'll be able to piece them back together to form a solution. And this is why we stored that deep in our heart. We know we can solve 4x4s with a cell missing. So we know all four of these orange quadrants can be filled. So we already know that this bigger shape can also be filled. So we're done. No matter where this one started, um, we can do this thing. We, we put down the cross, we put the L in the middle. Now we have four quadrants, each missing a cell. We invoke the 4x4 case. And now we've solved any 8 by8 case that's given to us. Does that make sense? So if we believe that we can solve every 4x4, then we also believe that we can solve every 8 by 8. Yeah. And this keeps going. If you instead had an 8 by8, sorry, if you instead had a 16 x6, you can do the same thing. Draw the crosshairs, put the L in the middle. Now all four 8 by8 quadrants are missing a single cell. Invoke the 8 by8 case. Um, we can even see that here. Go away webcam. Um, so here here is a 16x6 with one cell we're supposed to avoid. So let's cut it into four quadrants and put an L right there in the middle. Now we should think of this as four separate 8 by8s each missing a single cell and we trust in our heart that we have solved the 8 by8 case. Right? Don't worry too much about how we solved it. Just be confident that we solved the 8 by8 case. Um that there is a way to tile each of the 8 by8 corners. Um so now we can piece them together to form our our 16x6 solution. And now it's all it's all L tile tiles and we win. So this is the idea behind this proof uh that we're that we're able to show that PN always implies P of N plus one. If you can tile a 2 to the N by two to the N missing one cell then you can tile Oops. Then you can tile a 2 to the n plus1 by 2 to the n plus1 missing a tile using our strategy of taking the 2 to the n plus1 by 2 to the n plus one and cutting it into four quadrants of 2 to the n by 2 to the n. Put an l in the middle. Make sure each quadrant is missing a cell. Invoke the the inductive hypothesis. That was a little sketchy. Um I didn't write this down carefully. uh you can check the lecture notes later if you want to see this written down a little more carefully but that is the core idea um and the problem I want you to see there's one more thing I want to mention before we go um so I I kept saying store it deep in your heart that we've solved the 4x4 and then use that to believe that we've solved 8 by8 um believe that we solved 8 by8 in order to believe that we've solved 16 x6 but sometimes belief isn't quite enough and you really want to actually run this algorithm. Um well, for then all all you have to do is um when you h when we have the 16 x 16 and you break it down into 8 by8s. Well, let's actually remember how we solved the 8 by8s. We did that by splitting them up into 4x4s. And then turns out you can do the same thing and split the 4x4s into 2 x2s. So now each 2x two is missing a single cell, but a 2 x two missing a single cell is just an L. So it's all L's. And then we just recombine uh reverse what we did before. And now we've tiled everything and we're done. And that's the algorithm. Thank you so much. [Applause] |
9591 | https://www.mimmocorrado.it/mat/esa/2008/2008ordq10.pdf | Esame di Stato 2008 - PNI www.mimmocorrado.it 1 ESAME DI STATO DI LICEO SCIENTIFICO Sessione Ordinaria 2008 CORSO DI ORDINAMENTO Quesito 10 Secondo il codice della strada il segnale di “salita ripida” (fig. a lato) preavverte di un tratto di strada con pendenza tale da costituire pericolo. La pendenza vi è espressa in percentuale e nell’esempio è 10%. Se si sta realizzando una strada rettilinea che, con un percorso di 1,2 km, supera un dislivello di 85 m, qual è la sua inclinazione (in gradi sessagesimali)? Quale la percentuale da riportare sul segnale? Soluzione La pendenza di una salita è definita come rapporto tra il dislivello BC e l’avanzamento orizzontale AB della strada. L’avanzamento orizzontale AB della strada è: 2 2 BC AC AB − = = 2 2 85 1200 − = 2 2 85 1200 − = 7225 1440000 − = 1432775 m 1197 ≅ La pendenza è: AB BC tg = α = 1197 85 ≅ 071 , 0 ≅ % 1 , 7 Pertanto la percentuale da riportare sul cartello è % 7 . 1200 m 85 m α A B C |
9592 | https://mechfamily-ju.com/storage/images/files/file_17314308026pQTy.pdf | Modern Control Engineering Fifth Edition Katsuhiko Ogata Prentice Hall Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City Sao Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo VP/Editorial Director, Engineering/Computer Science: Marcia J. Horton Assistant/Supervisor: Dolores Mars Senior Editor:Andrew Gilfillan Associate Editor:Alice Dworkin Editorial Assistant:William Opaluch Director of Marketing: Margaret Waples Senior Marketing Manager:Tim Galligan Marketing Assistant: Mack Patterson Senior Managing Editor: Scott Disanno Art Editor: Greg Dulles Senior Operations Supervisor:Alan Fischer Operations Specialist: Lisa McDowell Art Director: Kenny Beck Cover Designer: Carole Anson Media Editor: Daniel Sandin Credits and acknowledgments borrowed from other sources and reproduced, with permission, in this textbook appear on appropriate page within text.
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Library of Congress Cataloging-in-Publication Data on File 10 9 8 7 6 5 4 3 2 1 ISBN 10: 0-13-615673-8 ISBN 13: 978-0-13-615673-4 C iii Contents Preface ix Chapter 1 Introduction to Control Systems 1 1–1 Introduction 1 1–2 Examples of Control Systems 4 1–3 Closed-Loop Control Versus Open-Loop Control 7 1–4 Design and Compensation of Control Systems 9 1–5 Outline of the Book 10 Chapter 2 Mathematical Modeling of Control Systems 13 2–1 Introduction 13 2–2 Transfer Function and Impulse-Response Function 15 2–3 Automatic Control Systems 17 2–4 Modeling in State Space 29 2–5 State-Space Representation of Scalar Differential Equation Systems 35 2–6 Transformation of Mathematical Models with MATLAB 39 2–7 Linearization of Nonlinear Mathematical Models 43 Example Problems and Solutions 46 Problems 60 Chapter 3 Mathematical Modeling of Mechanical Systems and Electrical Systems 63 3–1 Introduction 63 3–2 Mathematical Modeling of Mechanical Systems 63 3–3 Mathematical Modeling of Electrical Systems 72 Example Problems and Solutions 86 Problems 97 Chapter 4 Mathematical Modeling of Fluid Systems and Thermal Systems 100 4–1 Introduction 100 4–2 Liquid-Level Systems 101 4–3 Pneumatic Systems 106 4–4 Hydraulic Systems 123 4–5 Thermal Systems 136 Example Problems and Solutions 140 Problems 152 Chapter 5 Transient and Steady-State Response Analyses 159 5–1 Introduction 159 5–2 First-Order Systems 161 5–3 Second-Order Systems 164 5–4 Higher-Order Systems 179 5–5 Transient-Response Analysis with MATLAB 183 5–6 Routh’s Stability Criterion 212 5–7 Effects of Integral and Derivative Control Actions on System Performance 218 5–8 Steady-State Errors in Unity-Feedback Control Systems 225 Example Problems and Solutions 231 Problems 263 iv Contents Chapter 6 Control Systems Analysis and Design by the Root-Locus Method 269 6–1 Introduction 269 6–2 Root-Locus Plots 270 6–3 Plotting Root Loci with MATLAB 290 6–4 Root-Locus Plots of Positive Feedback Systems 303 6–5 Root-Locus Approach to Control-Systems Design 308 6–6 Lead Compensation 311 6–7 Lag Compensation 321 6–8 Lag–Lead Compensation 330 6–9 Parallel Compensation 342 Example Problems and Solutions 347 Problems 394 Chapter 7 Control Systems Analysis and Design by the Frequency-Response Method 398 7–1 Introduction 398 7–2 Bode Diagrams 403 7–3 Polar Plots 427 7–4 Log-Magnitude-versus-Phase Plots 443 7–5 Nyquist Stability Criterion 445 7–6 Stability Analysis 454 7–7 Relative Stability Analysis 462 7–8 Closed-Loop Frequency Response of Unity-Feedback Systems 477 7–9 Experimental Determination of Transfer Functions 486 7–10 Control Systems Design by Frequency-Response Approach 491 7–11 Lead Compensation 493 7–12 Lag Compensation 502 7–13 Lag–Lead Compensation 511 Example Problems and Solutions 521 Problems 561 Chapter 8 PID Controllers and Modified PID Controllers 567 8–1 Introduction 567 8–2 Ziegler–Nichols Rules for Tuning PID Controllers 568 Contents v 8–3 Design of PID Controllers with Frequency-Response Approach 577 8–4 Design of PID Controllers with Computational Optimization Approach 583 8–5 Modifications of PID Control Schemes 590 8–6 Two-Degrees-of-Freedom Control 592 8–7 Zero-Placement Approach to Improve Response Characteristics 595 Example Problems and Solutions 614 Problems 641 Chapter 9 Control Systems Analysis in State Space 648 9–1 Introduction 648 9–2 State-Space Representations of Transfer-Function Systems 649 9–3 Transformation of System Models with MATLAB 656 9–4 Solving the Time-Invariant State Equation 660 9–5 Some Useful Results in Vector-Matrix Analysis 668 9–6 Controllability 675 9–7 Observability 682 Example Problems and Solutions 688 Problems 720 Chapter 10 Control Systems Design in State Space 722 10–1 Introduction 722 10–2 Pole Placement 723 10–3 Solving Pole-Placement Problems with MATLAB 735 10–4 Design of Servo Systems 739 10–5 State Observers 751 10–6 Design of Regulator Systems with Observers 778 10–7 Design of Control Systems with Observers 786 10–8 Quadratic Optimal Regulator Systems 793 10–9 Robust Control Systems 806 Example Problems and Solutions 817 Problems 855 vi Contents Appendix A Laplace Transform Tables 859 Appendix B Partial-Fraction Expansion 867 Appendix C Vector-Matrix Algebra 874 References 882 Index 886 Contents vii This page intentionally left blank P ix Preface This book introduces important concepts in the analysis and design of control systems.
Readers will find it to be a clear and understandable textbook for control system courses at colleges and universities. It is written for senior electrical, mechanical, aerospace, or chemical engineering students. The reader is expected to have fulfilled the following prerequisites: introductory courses on differential equations, Laplace transforms, vector-matrix analysis, circuit analysis, mechanics, and introductory thermodynamics.
The main revisions made in this edition are as follows: • The use of MATLAB for obtaining responses of control systems to various inputs has been increased.
• The usefulness of the computational optimization approach with MATLAB has been demonstrated.
• New example problems have been added throughout the book.
• Materials in the previous edition that are of secondary importance have been deleted in order to provide space for more important subjects. Signal flow graphs were dropped from the book. A chapter on Laplace transform was deleted. Instead, Laplace transform tables, and partial-fraction expansion with MATLAB are pre-sented in Appendix A and Appendix B, respectively.
• A short summary of vector-matrix analysis is presented in Appendix C; this will help the reader to find the inverses of n x n matrices that may be involved in the analy-sis and design of control systems.
This edition of Modern Control Engineering is organized into ten chapters.The outline of this book is as follows: Chapter 1 presents an introduction to control systems. Chapter 2 deals with mathematical modeling of control systems.A linearization technique for non-linear mathematical models is presented in this chapter. Chapter 3 derives mathematical models of mechanical systems and electrical systems. Chapter 4 discusses mathematical modeling of fluid systems (such as liquid-level systems, pneumatic systems, and hydraulic systems) and thermal systems.
Chapter 5 treats transient response and steady-state analyses of control systems.
MATLAB is used extensively for obtaining transient response curves. Routh’s stability criterion is presented for stability analysis of control systems. Hurwitz stability criterion is also presented.
Chapter 6 discusses the root-locus analysis and design of control systems, including positive feedback systems and conditionally stable systems Plotting root loci with MAT-LAB is discussed in detail. Design of lead, lag, and lag-lead compensators with the root-locus method is included.
Chapter 7 treats the frequency-response analysis and design of control systems.The Nyquist stability criterion is presented in an easily understandable manner.The Bode di-agram approach to the design of lead, lag, and lag-lead compensators is discussed.
Chapter 8 deals with basic and modified PID controllers. Computational approaches for obtaining optimal parameter values for PID controllers are discussed in detail, par-ticularly with respect to satisfying requirements for step-response characteristics.
Chapter 9 treats basic analyses of control systems in state space. Concepts of con-trollability and observability are discussed in detail.
Chapter 10 deals with control systems design in state space.The discussions include pole placement, state observers, and quadratic optimal control. An introductory dis-cussion of robust control systems is presented at the end of Chapter 10.
The book has been arranged toward facilitating the student’s gradual understanding of control theory. Highly mathematical arguments are carefully avoided in the presen-tation of the materials. Statement proofs are provided whenever they contribute to the understanding of the subject matter presented.
Special effort has been made to provide example problems at strategic points so that the reader will have a clear understanding of the subject matter discussed. In addition, a number of solved problems (A-problems) are provided at the end of each chapter, except Chapter 1.The reader is encouraged to study all such solved problems carefully; this will allow the reader to obtain a deeper understanding of the topics discussed. In addition, many problems (without solutions) are provided at the end of each chapter, except Chapter 1. The unsolved problems (B-problems) may be used as homework or quiz problems.
If this book is used as a text for a semester course (with 56 or so lecture hours),a good portion of the material may be covered by skipping certain subjects. Because of the abundance of example problems and solved problems (A-problems) that might answer many possible questions that the reader might have, this book can also serve as a self-study book for practicing engineers who wish to study basic control theories.
I would like to thank the following reviewers for this edition of the book:Mark Camp-bell, Cornell University; Henry Sodano, Arizona State University; and Atul G. Kelkar, Iowa State University.Finally,I wish to offer my deep appreciation to Ms.Alice Dworkin, Associate Editor, Mr. Scott Disanno, Senior Managing Editor, and all the people in-volved in this publishing project, for the speedy yet superb production of this book.
Katsuhiko Ogata x Preface 1 Introduction to Control Systems 1–1 INTRODUCTION Control theories commonly used today are classical control theory (also called con-ventional control theory), modern control theory, and robust control theory.This book presents comprehensive treatments of the analysis and design of control systems based on the classical control theory and modern control theory.A brief introduction of robust control theory is included in Chapter 10.
Automatic control is essential in any field of engineering and science. Automatic control is an important and integral part of space-vehicle systems, robotic systems, mod-ern manufacturing systems, and any industrial operations involving control of temper-ature, pressure, humidity, flow, etc. It is desirable that most engineers and scientists are familiar with theory and practice of automatic control.
This book is intended to be a text book on control systems at the senior level at a col-lege or university.All necessary background materials are included in the book. Math-ematical background materials related to Laplace transforms and vector-matrix analysis are presented separately in appendixes.
Brief Review of Historical Developments of Control Theories and Practices.
The first significant work in automatic control was James Watt’s centrifugal gover-nor for the speed control of a steam engine in the eighteenth century. Other significant works in the early stages of development of control theory were due to 1 2 Chapter 1 / Introduction to Control Systems Minorsky, Hazen, and Nyquist, among many others. In 1922, Minorsky worked on automatic controllers for steering ships and showed how stability could be deter-mined from the differential equations describing the system. In 1932, Nyquist developed a relatively simple procedure for determining the stability of closed-loop systems on the basis of open-loop response to steady-state sinusoidal inputs. In 1934, Hazen, who introduced the term servomechanisms for position control systems, discussed the design of relay servomechanisms capable of closely following a chang-ing input.
During the decade of the 1940s, frequency-response methods (especially the Bode diagram methods due to Bode) made it possible for engineers to design linear closed-loop control systems that satisfied performance requirements. Many industrial control systems in 1940s and 1950s used PID controllers to control pressure, temperature, etc.
In the early 1940s Ziegler and Nichols suggested rules for tuning PID controllers, called Ziegler–Nichols tuning rules. From the end of the 1940s to the 1950s, the root-locus method due to Evans was fully developed.
The frequency-response and root-locus methods, which are the core of classical con-trol theory, lead to systems that are stable and satisfy a set of more or less arbitrary per-formance requirements. Such systems are, in general, acceptable but not optimal in any meaningful sense. Since the late 1950s, the emphasis in control design problems has been shifted from the design of one of many systems that work to the design of one optimal system in some meaningful sense.
As modern plants with many inputs and outputs become more and more complex, the description of a modern control system requires a large number of equations. Clas-sical control theory, which deals only with single-input, single-output systems, becomes powerless for multiple-input, multiple-output systems. Since about 1960, because the availability of digital computers made possible time-domain analysis of complex sys-tems, modern control theory, based on time-domain analysis and synthesis using state variables, has been developed to cope with the increased complexity of modern plants and the stringent requirements on accuracy, weight, and cost in military, space, and in-dustrial applications.
During the years from 1960 to 1980, optimal control of both deterministic and sto-chastic systems, as well as adaptive and learning control of complex systems, were fully investigated. From 1980s to 1990s, developments in modern control theory were cen-tered around robust control and associated topics.
Modern control theory is based on time-domain analysis of differential equation systems. Modern control theory made the design of control systems simpler because the theory is based on a model of an actual control system. However, the system’s stability is sensitive to the error between the actual system and its model. This means that when the designed controller based on a model is applied to the actual system, the system may not be stable. To avoid this situation, we design the control system by first setting up the range of possible errors and then designing the con-troller in such a way that, if the error of the system stays within the assumed range, the designed control system will stay stable. The design method based on this principle is called robust control theory.This theory incorporates both the frequency-response approach and the time-domain approach.The theory is mathematically very complex.
Section 1–1 / Introduction 3 Because this theory requires mathematical background at the graduate level, inclu-sion of robust control theory in this book is limited to introductory aspects only. The reader interested in details of robust control theory should take a graduate-level control course at an established college or university.
Definitions.
Before we can discuss control systems, some basic terminologies must be defined.
Controlled Variable and Control Signal or Manipulated Variable.
The controlled variable is the quantity or condition that is measured and controlled.The control signal or manipulated variable is the quantity or condition that is varied by the controller so as to affect the value of the controlled variable. Normally, the controlled variable is the output of the system. Control means measuring the value of the controlled variable of the system and applying the control signal to the system to correct or limit deviation of the measured value from a desired value.
In studying control engineering, we need to define additional terms that are neces-sary to describe control systems.
Plants.
A plant may be a piece of equipment, perhaps just a set of machine parts functioning together, the purpose of which is to perform a particular operation. In this book, we shall call any physical object to be controlled (such as a mechanical device, a heating furnace, a chemical reactor, or a spacecraft) a plant.
Processes.
The Merriam–Webster Dictionary defines a process to be a natural, pro-gressively continuing operation or development marked by a series of gradual changes that succeed one another in a relatively fixed way and lead toward a particular result or end; or an artificial or voluntary, progressively continuing operation that consists of a se-ries of controlled actions or movements systematically directed toward a particular re-sult or end. In this book we shall call any operation to be controlled a process. Examples are chemical, economic, and biological processes.
Systems.
A system is a combination of components that act together and perform a certain objective. A system need not be physical. The concept of the system can be applied to abstract, dynamic phenomena such as those encountered in economics. The word system should,therefore,be interpreted to imply physical,biological,economic,and the like, systems.
Disturbances.
A disturbance is a signal that tends to adversely affect the value of the output of a system. If a disturbance is generated within the system, it is called internal, while an external disturbance is generated outside the system and is an input.
Feedback Control.
Feedback control refers to an operation that, in the presence of disturbances, tends to reduce the difference between the output of a system and some reference input and does so on the basis of this difference. Here only unpredictable dis-turbances are so specified, since predictable or known disturbances can always be com-pensated for within the system.
4 Chapter 1 / Introduction to Control Systems 1–2 EXAMPLES OF CONTROL SYSTEMS In this section we shall present a few examples of control systems.
Speed Control System.
The basic principle of a Watt’s speed governor for an en-gine is illustrated in the schematic diagram of Figure 1–1. The amount of fuel admitted to the engine is adjusted according to the difference between the desired and the actual engine speeds.
The sequence of actions may be stated as follows: The speed governor is ad-justed such that, at the desired speed, no pressured oil will flow into either side of the power cylinder. If the actual speed drops below the desired value due to disturbance, then the decrease in the centrifugal force of the speed governor causes the control valve to move downward, supplying more fuel, and the speed of the engine increases until the desired value is reached. On the other hand, if the speed of the engine increases above the desired value, then the increase in the centrifu-gal force of the governor causes the control valve to move upward. This decreases the supply of fuel, and the speed of the engine decreases until the desired value is reached.
In this speed control system, the plant (controlled system) is the engine and the controlled variable is the speed of the engine. The difference between the desired speed and the actual speed is the error signal.The control signal (the amount of fuel) to be applied to the plant (engine) is the actuating signal. The external input to dis-turb the controlled variable is the disturbance. An unexpected change in the load is a disturbance.
Temperature Control System.
Figure 1–2 shows a schematic diagram of tem-perature control of an electric furnace.The temperature in the electric furnace is meas-ured by a thermometer, which is an analog device.The analog temperature is converted Oil under pressure Power cylinder Close Open Pilot valve Control valve Fuel Engine Load Figure 1–1 Speed control system.
Section 1–2 / Examples of Control Systems 5 Thermometer Heater Interface Controller Interface Amplifier A/D converter Programmed input Electric furnace Relay Figure 1–2 Temperature control system.
to a digital temperature by an A/D converter. The digital temperature is fed to a con-troller through an interface.This digital temperature is compared with the programmed input temperature, and if there is any discrepancy (error), the controller sends out a sig-nal to the heater, through an interface, amplifier, and relay, to bring the furnace tem-perature to a desired value.
Business Systems.
A business system may consist of many groups. Each task assigned to a group will represent a dynamic element of the system. Feedback methods of reporting the accomplishments of each group must be established in such a system for proper operation.The cross-coupling between functional groups must be made a mini-mum in order to reduce undesirable delay times in the system. The smaller this cross-coupling, the smoother the flow of work signals and materials will be.
A business system is a closed-loop system.A good design will reduce the manageri-al control required.Note that disturbances in this system are the lack of personnel or ma-terials, interruption of communication, human errors, and the like.
The establishment of a well-founded estimating system based on statistics is manda-tory to proper management.It is a well-known fact that the performance of such a system can be improved by the use of lead time, or anticipation.
To apply control theory to improve the performance of such a system, we must rep-resent the dynamic characteristic of the component groups of the system by a relative-ly simple set of equations.
Although it is certainly a difficult problem to derive mathematical representations of the component groups, the application of optimization techniques to business sys-tems significantly improves the performance of the business system.
Consider, as an example, an engineering organizational system that is composed of major groups such as management, research and development, preliminary design, ex-periments, product design and drafting, fabrication and assembling, and tesing. These groups are interconnected to make up the whole operation.
Such a system may be analyzed by reducing it to the most elementary set of com-ponents necessary that can provide the analytical detail required and by representing the dynamic characteristics of each component by a set of simple equations. (The dynamic performance of such a system may be determined from the relation between progres-sive accomplishment and time.) 6 Chapter 1 / Introduction to Control Systems Required product Management Research and development Preliminary design Experiments Product design and drafting Fabrication and assembling Testing Product Figure 1–3 Block diagram of an engineering organizational system.
A functional block diagram may be drawn by using blocks to represent the func-tional activities and interconnecting signal lines to represent the information or product output of the system operation. Figure 1–3 is a possible block diagram for this system.
Robust Control System.
The first step in the design of a control system is to obtain a mathematical model of the plant or control object. In reality, any model of a plant we want to control will include an error in the modeling process.That is, the actual plant differs from the model to be used in the design of the control system.
To ensure the controller designed based on a model will work satisfactorily when this controller is used with the actual plant, one reasonable approach is to assume from the start that there is an uncertainty or error between the actual plant and its mathematical model and include such uncertainty or error in the design process of the control system. The control system designed based on this approach is called a robust control system.
Suppose that the actual plant we want to control is (s) and the mathematical model of the actual plant is G(s), that is, (s)=actual plant model that has uncertainty ¢(s) G(s)=nominal plant model to be used for designing the control system (s) and G(s) may be related by a multiplicative factor such as or an additive factor or in other forms.
Since the exact description of the uncertainty or error ¢(s) is unknown, we use an estimate of ¢(s) and use this estimate, W(s), in the design of the controller. W(s) is a scalar transfer function such that where is the maximum value of for and is called the H infinity norm of W(s).
0 v q W(jv) W(s)q ¢(s)q 6 W(s)q = max 0vq W(jv) G (s) = G(s) + ¢(s) G (s) = G(s)[1 + ¢(s)] G G G Section 1–3 / Closed-Loop Control versus Open-Loop Control 7 Using the small gain theorem, the design procedure here boils down to the deter-mination of the controller K(s) such that the inequality is satisfied, where G(s) is the transfer function of the model used in the design process, K(s) is the transfer function of the controller, and W(s) is the chosen transfer function to approximate ¢(s). In most practical cases, we must satisfy more than one such inequality that involves G(s), K(s), and W(s)’s. For example, to guarantee robust sta-bility and robust performance we may require two inequalities, such as for robust stability for robust performance be satisfied. (These inequalities are derived in Section 10–9.) There are many different such inequalities that need to be satisfied in many different robust control systems.
(Robust stability means that the controller K(s) guarantees internal stability of all systems that belong to a group of systems that include the system with the actual plant.
Robust performance means the specified performance is satisfied in all systems that be-long to the group.) In this book all the plants of control systems we discuss are assumed to be known precisely, except the plants we discuss in Section 10–9 where an introduc-tory aspect of robust control theory is presented.
1–3 CLOSED-LOOP CONTROL VERSUS OPEN-LOOP CONTROL Feedback Control Systems.
A system that maintains a prescribed relationship between the output and the reference input by comparing them and using the difference as a means of control is called a feedback control system. An example would be a room-temperature control system. By measuring the actual room temperature and comparing it with the reference temperature (desired temperature), the thermostat turns the heat-ing or cooling equipment on or off in such a way as to ensure that the room tempera-ture remains at a comfortable level regardless of outside conditions.
Feedback control systems are not limited to engineering but can be found in various nonengineering fields as well.The human body, for instance, is a highly advanced feed-back control system. Both body temperature and blood pressure are kept constant by means of physiological feedback. In fact, feedback performs a vital function: It makes the human body relatively insensitive to external disturbances, thus enabling it to func-tion properly in a changing environment.
ß W s(s) 1 + K(s)G(s) ß q 6 1 ß W m(s)K(s)G(s) 1 + K(s)G(s) ß q 6 1 ß W(s) 1 + K(s)G(s) ß q 6 1 8 Chapter 1 / Introduction to Control Systems Closed-Loop Control Systems.
Feedback control systems are often referred to as closed-loop control systems. In practice, the terms feedback control and closed-loop control are used interchangeably. In a closed-loop control system the actuating error signal, which is the difference between the input signal and the feedback signal (which may be the output signal itself or a function of the output signal and its derivatives and/or integrals), is fed to the controller so as to reduce the error and bring the output of the system to a desired value.The term closed-loop control always implies the use of feedback control action in order to reduce system error.
Open-Loop Control Systems.
Those systems in which the output has no effect on the control action are called open-loop control systems. In other words, in an open-loop control system the output is neither measured nor fed back for comparison with the input. One practical example is a washing machine. Soaking, washing, and rinsing in the washer operate on a time basis. The machine does not measure the output signal, that is, the cleanliness of the clothes.
In any open-loop control system the output is not compared with the reference input.
Thus, to each reference input there corresponds a fixed operating condition; as a result, the accuracy of the system depends on calibration. In the presence of disturbances, an open-loop control system will not perform the desired task. Open-loop control can be used, in practice, only if the relationship between the input and output is known and if there are neither internal nor external disturbances. Clearly, such systems are not feed-back control systems. Note that any control system that operates on a time basis is open loop. For instance, traffic control by means of signals operated on a time basis is another example of open-loop control.
Closed-Loop versus Open-Loop Control Systems.
An advantage of the closed-loop control system is the fact that the use of feedback makes the system response rela-tively insensitive to external disturbances and internal variations in system parameters.
It is thus possible to use relatively inaccurate and inexpensive components to obtain the accurate control of a given plant, whereas doing so is impossible in the open-loop case.
From the point of view of stability, the open-loop control system is easier to build be-cause system stability is not a major problem. On the other hand, stability is a major problem in the closed-loop control system, which may tend to overcorrect errors and thereby can cause oscillations of constant or changing amplitude.
It should be emphasized that for systems in which the inputs are known ahead of time and in which there are no disturbances it is advisable to use open-loop control.
Closed-loop control systems have advantages only when unpredictable disturbances and/or unpredictable variations in system components are present. Note that the output power rating partially determines the cost, weight, and size of a control system.
The number of components used in a closed-loop control system is more than that for a corresponding open-loop control system. Thus, the closed-loop control system is generally higher in cost and power.To decrease the required power of a system, open-loop control may be used where applicable. A proper combination of open-loop and closed-loop controls is usually less expensive and will give satisfactory overall system performance.
Most analyses and designs of control systems presented in this book are concerned with closed-loop control systems. Under certain circumstances (such as where no disturbances exist or the output is hard to measure) open-loop control systems may be desired.Therefore, it is worthwhile to summarize the advantages and disadvantages of using open-loop control systems.
The major advantages of open-loop control systems are as follows: 1. Simple construction and ease of maintenance.
2. Less expensive than a corresponding closed-loop system.
3. There is no stability problem.
4. Convenient when output is hard to measure or measuring the output precisely is economically not feasible.(For example,in the washer system,it would be quite ex-pensive to provide a device to measure the quality of the washer’s output, clean-liness of the clothes.) The major disadvantages of open-loop control systems are as follows: 1. Disturbances and changes in calibration cause errors, and the output may be different from what is desired.
2. To maintain the required quality in the output, recalibration is necessary from time to time.
1–4 DESIGN AND COMPENSATION OF CONTROL SYSTEMS This book discusses basic aspects of the design and compensation of control systems.
Compensation is the modification of the system dynamics to satisfy the given specifi-cations.The approaches to control system design and compensation used in this book are the root-locus approach, frequency-response approach, and the state-space ap-proach. Such control systems design and compensation will be presented in Chapters 6, 7, 9 and 10. The PID-based compensational approach to control systems design is given in Chapter 8.
In the actual design of a control system, whether to use an electronic, pneumatic, or hydraulic compensator is a matter that must be decided partially based on the nature of the controlled plant. For example, if the controlled plant involves flammable fluid, then we have to choose pneumatic components (both a compensator and an actuator) to avoid the possibility of sparks. If, however, no fire hazard exists, then electronic com-pensators are most commonly used.(In fact,we often transform nonelectrical signals into electrical signals because of the simplicity of transmission, increased accuracy, increased reliability, ease of compensation, and the like.) Performance Specifications.
Control systems are designed to perform specific tasks. The requirements imposed on the control system are usually spelled out as per-formance specifications.The specifications may be given in terms of transient response requirements (such as the maximum overshoot and settling time in step response) and of steady-state requirements (such as steady-state error in following ramp input) or may be given in frequency-response terms. The specifications of a control system must be given before the design process begins.
For routine design problems, the performance specifications (which relate to accura-cy, relative stability, and speed of response) may be given in terms of precise numerical values.In other cases they may be given partially in terms of precise numerical values and Section 1–4 / Design and Compensation of Control Systems 9 partially in terms of qualitative statements. In the latter case the specifications may have to be modified during the course of design, since the given specifications may never be satisfied (because of conflicting requirements) or may lead to a very expensive system.
Generally, the performance specifications should not be more stringent than neces-sary to perform the given task. If the accuracy at steady-state operation is of prime im-portance in a given control system, then we should not require unnecessarily rigid performance specifications on the transient response, since such specifications will require expensive components. Remember that the most important part of control system design is to state the performance specifications precisely so that they will yield an optimal control system for the given purpose.
System Compensation.
Setting the gain is the first step in adjusting the system for satisfactory performance. In many practical cases, however, the adjustment of the gain alone may not provide sufficient alteration of the system behavior to meet the given specifications. As is frequently the case, increasing the gain value will improve the steady-state behavior but will result in poor stability or even instability. It is then nec-essary to redesign the system (by modifying the structure or by incorporating addi-tional devices or components) to alter the overall behavior so that the system will behave as desired. Such a redesign or addition of a suitable device is called compensa-tion. A device inserted into the system for the purpose of satisfying the specifications is called a compensator. The compensator compensates for deficient performance of the original system.
Design Procedures.
In the process of designing a control system, we set up a mathematical model of the control system and adjust the parameters of a compensator.
The most time-consuming part of the work is the checking of the system performance by analysis with each adjustment of the parameters.The designer should use MATLAB or other available computer package to avoid much of the numerical drudgery neces-sary for this checking.
Once a satisfactory mathematical model has been obtained, the designer must con-struct a prototype and test the open-loop system. If absolute stability of the closed loop is assured, the designer closes the loop and tests the performance of the resulting closed-loop system. Because of the neglected loading effects among the components, nonlin-earities, distributed parameters, and so on, which were not taken into consideration in the original design work, the actual performance of the prototype system will probably differ from the theoretical predictions. Thus the first design may not satisfy all the re-quirements on performance. The designer must adjust system parameters and make changes in the prototype until the system meets the specificications. In doing this, he or she must analyze each trial, and the results of the analysis must be incorporated into the next trial.The designer must see that the final system meets the performance apec-ifications and, at the same time, is reliable and economical.
1–5 OUTLINE OF THE BOOK This text is organized into 10 chapters.The outline of each chapter may be summarized as follows: Chapter 1 presents an introduction to this book.
10 Chapter 1 / Introduction to Control Systems Chapter 2 deals with mathematical modeling of control systems that are described by linear differential equations. Specifically, transfer function expressions of differential equation systems are derived.Also, state-space expressions of differential equation sys-tems are derived. MATLAB is used to transform mathematical models from transfer functions to state-space equations and vice versa.This book treats linear systems in de-tail. If the mathematical model of any system is nonlinear, it needs to be linearized be-fore applying theories presented in this book. A technique to linearize nonlinear mathematical models is presented in this chapter.
Chapter 3 derives mathematical models of various mechanical and electrical sys-tems that appear frequently in control systems.
Chapter 4 discusses various fluid systems and thermal systems, that appear in control systems.Fluid systems here include liquid-level systems,pneumatic systems,and hydraulic systems. Thermal systems such as temperature control systems are also discussed here.
Control engineers must be familiar with all of these systems discussed in this chapter.
Chapter 5 presents transient and steady-state response analyses of control systems defined in terms of transfer functions. MATLAB approach to obtain transient and steady-state response analyses is presented in detail. MATLAB approach to obtain three-dimensional plots is also presented. Stability analysis based on Routh’s stability criterion is included in this chapter and the Hurwitz stability criterion is briefly discussed.
Chapter 6 treats the root-locus method of analysis and design of control systems. It is a graphical method for determining the locations of all closed-loop poles from the knowledge of the locations of the open-loop poles and zeros of a closed-loop system as a parameter (usually the gain) is varied from zero to infinity. This method was de-veloped by W. R. Evans around 1950. These days MATLAB can produce root-locus plots easily and quickly.This chapter presents both a manual approach and a MATLAB approach to generate root-locus plots. Details of the design of control systems using lead compensators, lag compensators, are lag–lead compensators are presented in this chapter.
Chapter 7 presents the frequency-response method of analysis and design of control systems. This is the oldest method of control systems analysis and design and was de-veloped during 1940–1950 by Nyquist, Bode, Nichols, Hazen, among others.This chap-ter presents details of the frequency-response approach to control systems design using lead compensation technique, lag compensation technique, and lag–lead compensation technique. The frequency-response method was the most frequently used analysis and design method until the state-space method became popular. However, since H-infini-ty control for designing robust control systems has become popular, frequency response is gaining popularity again.
Chapter 8 discusses PID controllers and modified ones such as multidegrees-of-freedom PID controllers.The PID controller has three parameters; proportional gain, integral gain, and derivative gain. In industrial control systems more than half of the con-trollers used have been PID controllers.The performance of PID controllers depends on the relative magnitudes of those three parameters. Determination of the relative magnitudes of the three parameters is called tuning of PID controllers.
Ziegler and Nichols proposed so-called “Ziegler–Nichols tuning rules” as early as 1942. Since then numerous tuning rules have been proposed.These days manufacturers of PID controllers have their own tuning rules. In this chapter we present a computer optimization approach using MATLAB to determine the three parameters to satisfy Section 1–5 / Outline of the Book 11 given transient response characteristics.The approach can be expanded to determine the three parameters to satisfy any specific given characteristics.
Chapter 9 presents basic analysis of state-space equations. Concepts of controllabil-ity and observability, most important concepts in modern control theory, due to Kalman are discussed in full. In this chapter, solutions of state-space equations are derived in detail.
Chapter 10 discusses state-space designs of control systems.This chapter first deals with pole placement problems and state observers.In control engineering,it is frequently desirable to set up a meaningful performance index and try to minimize it (or maximize it, as the case may be). If the performance index selected has a clear physical meaning, then this approach is quite useful to determine the optimal control variable.This chap-ter discusses the quadratic optimal regulator problem where we use a performance index which is an integral of a quadratic function of the state variables and the control vari-able.The integral is performed from t=0 to t= .This chapter concludes with a brief discussion of robust control systems.
q 12 Chapter 1 / Introduction to Control Systems 2 13 Mathematical Modeling of Control Systems 2–1 INTRODUCTION In studying control systems the reader must be able to model dynamic systems in math-ematical terms and analyze their dynamic characteristics.A mathematical model of a dy-namic system is defined as a set of equations that represents the dynamics of the system accurately, or at least fairly well. Note that a mathematical model is not unique to a given system.A system may be represented in many different ways and, therefore, may have many mathematical models, depending on one’s perspective.
The dynamics of many systems, whether they are mechanical, electrical, thermal, economic, biological, and so on, may be described in terms of differential equations.
Such differential equations may be obtained by using physical laws governing a partic-ular system—for example, Newton’s laws for mechanical systems and Kirchhoff’s laws for electrical systems. We must always keep in mind that deriving reasonable mathe-matical models is the most important part of the entire analysis of control systems.
Throughout this book we assume that the principle of causality applies to the systems considered.This means that the current output of the system (the output at time t=0) depends on the past input (the input for t<0) but does not depend on the future input (the input for t>0).
Mathematical Models.
Mathematical models may assume many different forms.
Depending on the particular system and the particular circumstances, one mathemati-cal model may be better suited than other models. For example, in optimal control prob-lems, it is advantageous to use state-space representations. On the other hand, for the 14 Chapter 2 / Mathematical Modeling of Control Systems transient-response or frequency-response analysis of single-input, single-output, linear, time-invariant systems, the transfer-function representation may be more convenient than any other. Once a mathematical model of a system is obtained, various analytical and computer tools can be used for analysis and synthesis purposes.
Simplicity Versus Accuracy.
In obtaining a mathematical model, we must make a compromise between the simplicity of the model and the accuracy of the results of the analysis. In deriving a reasonably simplified mathematical model, we frequently find it necessary to ignore certain inherent physical properties of the system. In particular, if a linear lumped-parameter mathematical model (that is, one employing ordinary dif-ferential equations) is desired, it is always necessary to ignore certain nonlinearities and distributed parameters that may be present in the physical system. If the effects that these ignored properties have on the response are small,good agreement will be obtained between the results of the analysis of a mathematical model and the results of the experimental study of the physical system.
In general,in solving a new problem,it is desirable to build a simplified model so that we can get a general feeling for the solution.A more complete mathematical model may then be built and used for a more accurate analysis.
We must be well aware that a linear lumped-parameter model,which may be valid in low-frequency operations,may not be valid at sufficiently high frequencies,since the neg-lected property of distributed parameters may become an important factor in the dynamic behavior of the system. For example, the mass of a spring may be neglected in low-frequency operations, but it becomes an important property of the system at high fre-quencies.(For the case where a mathematical model involves considerable errors,robust control theory may be applied. Robust control theory is presented in Chapter 10.) Linear Systems.
A system is called linear if the principle of superposition applies. The principle of superposition states that the response produced by the simultaneous application of two different forcing functions is the sum of the two individual responses. Hence, for the linear system, the response to several inputs can be calculated by treating one input at a time and adding the results. It is this principle that allows one to build up complicated solutions to the linear differential equation from simple solutions.
In an experimental investigation of a dynamic system, if cause and effect are pro-portional, thus implying that the principle of superposition holds, then the system can be considered linear.
Linear Time-Invariant Systems and Linear Time-Varying Systems.
A differ-ential equation is linear if the coefficients are constants or functions only of the in-dependent variable. Dynamic systems that are composed of linear time-invariant lumped-parameter components may be described by linear time-invariant differen-tial equations—that is, constant-coefficient differential equations. Such systems are called linear time-invariant (or linear constant-coefficient) systems. Systems that are represented by differential equations whose coefficients are functions of time are called linear time-varying systems. An example of a time-varying control sys-tem is a spacecraft control system. (The mass of a spacecraft changes due to fuel consumption.) Section 2–2 / Transfer Function and Impulse-Response Function 15 Outline of the Chapter.
Section 2–1 has presented an introduction to the math-ematical modeling of dynamic systems. Section 2–2 presents the transfer function and impulse-response function. Section 2–3 introduces automatic control systems and Sec-tion 2–4 discusses concepts of modeling in state space. Section 2–5 presents state-space representation of dynamic systems. Section 2–6 discusses transformation of mathemat-ical models with MATLAB. Finally, Section 2–7 discusses linearization of nonlinear mathematical models.
2–2 TRANSFER FUNCTION AND IMPULSE-RESPONSE FUNCTION In control theory, functions called transfer functions are commonly used to character-ize the input-output relationships of components or systems that can be described by lin-ear, time-invariant, differential equations. We begin by defining the transfer function and follow with a derivation of the transfer function of a differential equation system.
Then we discuss the impulse-response function.
Transfer Function.
The transfer function of a linear, time-invariant, differential equation system is defined as the ratio of the Laplace transform of the output (response function) to the Laplace transform of the input (driving function) under the assumption that all initial conditions are zero.
Consider the linear time-invariant system defined by the following differential equation: where y is the output of the system and x is the input.The transfer function of this sys-tem is the ratio of the Laplace transformed output to the Laplace transformed input when all initial conditions are zero, or By using the concept of transfer function, it is possible to represent system dynam-ics by algebraic equations in s. If the highest power of s in the denominator of the trans-fer function is equal to n, the system is called an nth-order system.
Comments on Transfer Function.
The applicability of the concept of the trans-fer function is limited to linear, time-invariant, differential equation systems.The trans-fer function approach, however, is extensively used in the analysis and design of such systems. In what follows, we shall list important comments concerning the transfer func-tion.(Note that a system referred to in the list is one described by a linear,time-invariant, differential equation.) = Y(s) X(s) = b0 sm + b1 sm-1 + p + bm-1 s + bm a0 sn + a1 sn-1 + p + an-1 s + an Transfer function = G(s) = l[output] l[input] 2 zero initial conditions = b0 x (m) + b1x (m-1) + p + bm-1 x # + bm x (n m) a0 y (n) + a1y (n-1) + p + an-1 y # + an y 16 Chapter 2 / Mathematical Modeling of Control Systems 1. The transfer function of a system is a mathematical model in that it is an opera-tional method of expressing the differential equation that relates the output vari-able to the input variable.
2. The transfer function is a property of a system itself, independent of the magnitude and nature of the input or driving function.
3. The transfer function includes the units necessary to relate the input to the output; however, it does not provide any information concerning the physical structure of the system. (The transfer functions of many physically different systems can be identical.) 4. If the transfer function of a system is known, the output or response can be stud-ied for various forms of inputs with a view toward understanding the nature of the system.
5. If the transfer function of a system is unknown, it may be established experimen-tally by introducing known inputs and studying the output of the system. Once established, a transfer function gives a full description of the dynamic character-istics of the system, as distinct from its physical description.
Convolution Integral.
For a linear, time-invariant system the transfer function G(s) is where X(s) is the Laplace transform of the input to the system and Y(s) is the Laplace transform of the output of the system, where we assume that all initial conditions in-volved are zero.It follows that the output Y(s) can be written as the product of G(s) and X(s), or (2–1) Note that multiplication in the complex domain is equivalent to convolution in the time domain (see Appendix A), so the inverse Laplace transform of Equation (2–1) is given by the following convolution integral: where both g(t) and x(t) are 0 for t<0.
Impulse-Response Function.
Consider the output (response) of a linear time-invariant system to a unit-impulse input when the initial conditions are zero. Since the Laplace transform of the unit-impulse function is unity, the Laplace transform of the output of the system is (2–2) Y(s) = G(s) = 3 t 0 g(t)x(t - t)dt y(t) = 3 t 0 x(t)g(t - t)dt Y(s) = G(s)X(s) G(s) = Y(s) X(s) Section 2–3 / Automatic Control Systems 17 The inverse Laplace transform of the output given by Equation (2–2) gives the impulse response of the system.The inverse Laplace transform of G(s), or is called the impulse-response function. This function g(t) is also called the weighting function of the system.
The impulse-response function g(t) is thus the response of a linear time-invariant system to a unit-impulse input when the initial conditions are zero.The Laplace trans-form of this function gives the transfer function. Therefore, the transfer function and impulse-response function of a linear, time-invariant system contain the same infor-mation about the system dynamics. It is hence possible to obtain complete informa-tion about the dynamic characteristics of the system by exciting it with an impulse input and measuring the response. (In practice, a pulse input with a very short dura-tion compared with the significant time constants of the system can be considered an impulse.) 2–3 AUTOMATIC CONTROL SYSTEMS A control system may consist of a number of components. To show the functions performed by each component, in control engineering, we commonly use a diagram called the block diagram. This section first explains what a block diagram is. Next, it discusses introductory aspects of automatic control systems, including various control actions.Then,it presents a method for obtaining block diagrams for physical systems,and, finally, discusses techniques to simplify such diagrams.
Block Diagrams.
A block diagram of a system is a pictorial representation of the functions performed by each component and of the flow of signals. Such a diagram de-picts the interrelationships that exist among the various components. Differing from a purely abstract mathematical representation, a block diagram has the advantage of indicating more realistically the signal flows of the actual system.
In a block diagram all system variables are linked to each other through functional blocks.The functional block or simply block is a symbol for the mathematical operation on the input signal to the block that produces the output.The transfer functions of the components are usually entered in the corresponding blocks, which are connected by ar-rows to indicate the direction of the flow of signals. Note that the signal can pass only in the direction of the arrows.Thus a block diagram of a control system explicitly shows a unilateral property.
Figure 2–1 shows an element of the block diagram.The arrowhead pointing toward the block indicates the input, and the arrowhead leading away from the block repre-sents the output. Such arrows are referred to as signals.
l-1CG(s)D = g(t) Transfer function G(s) Figure 2–1 Element of a block diagram.
18 Chapter 2 / Mathematical Modeling of Control Systems +– R(s) E(s) G(s) C(s) Summing point Branch point Figure 2–3 Block diagram of a closed-loop system.
Note that the dimension of the output signal from the block is the dimension of the input signal multiplied by the dimension of the transfer function in the block.
The advantages of the block diagram representation of a system are that it is easy to form the overall block diagram for the entire system by merely connecting the blocks of the components according to the signal flow and that it is possible to evaluate the contribution of each component to the overall performance of the system.
In general, the functional operation of the system can be visualized more readily by examining the block diagram than by examining the physical system itself. A block di-agram contains information concerning dynamic behavior, but it does not include any information on the physical construction of the system. Consequently, many dissimilar and unrelated systems can be represented by the same block diagram.
It should be noted that in a block diagram the main source of energy is not explicitly shown and that the block diagram of a given system is not unique.A number of different block diagrams can be drawn for a system, depending on the point of view of the analysis.
Summing Point.
Referring to Figure 2–2, a circle with a cross is the symbol that indicates a summing operation. The plus or minus sign at each arrowhead indicates whether that signal is to be added or subtracted. It is important that the quantities being added or subtracted have the same dimensions and the same units.
Branch Point.
A branch point is a point from which the signal from a block goes concurrently to other blocks or summing points.
Block Diagram of a Closed-Loop System.
Figure 2–3 shows an example of a block diagram of a closed-loop system. The output C(s) is fed back to the summing point, where it is compared with the reference input R(s). The closed-loop nature of the system is clearly indicated by the figure. The output of the block, C(s) in this case, is obtained by multiplying the transfer function G(s) by the input to the block,E(s).Any linear control system may be represented by a block diagram consisting of blocks, sum-ming points, and branch points.
When the output is fed back to the summing point for comparison with the input, it is necessary to convert the form of the output signal to that of the input signal. For example, in a temperature control system, the output signal is usually the controlled temperature.The output signal, which has the dimension of temperature, must be con-verted to a force or position or voltage before it can be compared with the input signal.
This conversion is accomplished by the feedback element whose transfer function is H(s), as shown in Figure 2–4.The role of the feedback element is to modify the output before it is compared with the input.(In most cases the feedback element is a sensor that measures + – a a – b b Figure 2–2 Summing point.
the output of the plant.The output of the sensor is compared with the system input, and the actuating error signal is generated.) In the present example, the feedback signal that is fed back to the summing point for comparison with the input is B(s) = H(s)C(s).
Open-Loop Transfer Function and Feedforward Transfer Function.
Refer-ring to Figure 2–4, the ratio of the feedback signal B(s) to the actuating error signal E(s) is called the open-loop transfer function.That is, The ratio of the output C(s) to the actuating error signal E(s) is called the feed-forward transfer function, so that If the feedback transfer function H(s) is unity, then the open-loop transfer function and the feedforward transfer function are the same.
Closed-Loop Transfer Function.
For the system shown in Figure 2–4, the output C(s) and input R(s) are related as follows: since eliminating E(s) from these equations gives or (2–3) The transfer function relating C(s) to R(s) is called the closed-loop transfer function. It relates the closed-loop system dynamics to the dynamics of the feedforward elements and feedback elements.
From Equation (2–3), C(s) is given by C(s) = G(s) 1 + G(s)H(s) R(s) C(s) R(s) = G(s) 1 + G(s)H(s) C(s) = G(s)CR(s) - H(s)C(s)D = R(s) - H(s)C(s) E(s) = R(s) - B(s) C(s) = G(s)E(s) Feedforward transfer function = C(s) E(s) = G(s) Open-loop transfer function = B(s) E(s) = G(s)H(s) Section 2–3 / Automatic Control Systems 19 R(s) B(s) E(s) G(s) H(s) C(s) + – Figure 2–4 Closed-loop system.
20 Chapter 2 / Mathematical Modeling of Control Systems G1(s) G1(s) G2(s) G2(s) C(s) R(s) C(s) C(s) R(s) R(s) + + G1(s) G2(s) +– (a) (b) (c) Figure 2–5 (a) Cascaded system; (b) parallel system; (c) feedback (closed-loop) system.
Thus the output of the closed-loop system clearly depends on both the closed-loop trans-fer function and the nature of the input.
Obtaining Cascaded, Parallel, and Feedback (Closed-Loop) Transfer Functions with MATLAB.
In control-systems analysis, we frequently need to calculate the cas-caded transfer functions, parallel-connected transfer functions, and feedback-connected (closed-loop) transfer functions. MATLAB has convenient commands to obtain the cas-caded, parallel, and feedback (closed-loop) transfer functions.
Suppose that there are two components G1(s) and G2(s) connected differently as shown in Figure 2–5 (a), (b), and (c), where To obtain the transfer functions of the cascaded system, parallel system, or feedback (closed-loop) system, the following commands may be used: [num, den] = series(num1,den1,num2,den2) [num, den] = parallel(num1,den1,num2,den2) [num, den] = feedback(num1,den1,num2,den2) As an example, consider the case where MATLAB Program 2–1 gives C(s)/R(s)=numden for each arrangement of G1(s) and G2(s). Note that the command printsys(num,den) displays the numden Cthat is,the transfer function C(s)/R(s)D of the system considered.
G1(s) = 10 s2 + 2s + 10 = num1 den1 , G2(s) = 5 s + 5 = num2 den2 G1(s) = num1 den1 , G2(s) = num2 den2 Section 2–3 / Automatic Control Systems 21 Automatic Controllers.
An automatic controller compares the actual value of the plant output with the reference input (desired value), determines the deviation, and produces a control signal that will reduce the deviation to zero or to a small value.
The manner in which the automatic controller produces the control signal is called the control action. Figure 2–6 is a block diagram of an industrial control system, which MATLAB Program 2–1 num1 = ; den1 = [1 2 10]; num2 = ; den2 = [1 5]; [num, den] = series(num1,den1,num2,den2); printsys(num,den) num/den = [num, den] = parallel(num1,den1,num2,den2); printsys(num,den) num/den = [num, den] = feedback(num1,den1,num2,den2); printsys(num,den) num/den = 10s + 50 s^3 + 7s^2 + 20s + 100 5s^2 + 20s + 100 s^3 + 7s^2 + 20s + 50 50 s^3 + 7s^2 + 20s + 50 Automatic controller Error detector Amplifier Actuator Plant Output Sensor Reference input Actuating error signal Set point + – Figure 2–6 Block diagram of an industrial control system, which consists of an automatic controller, an actuator, a plant, and a sensor (measuring element).
22 Chapter 2 / Mathematical Modeling of Control Systems consists of an automatic controller, an actuator, a plant, and a sensor (measuring ele-ment). The controller detects the actuating error signal, which is usually at a very low power level, and amplifies it to a sufficiently high level. The output of an automatic controller is fed to an actuator, such as an electric motor, a hydraulic motor, or a pneumatic motor or valve. (The actuator is a power device that produces the input to the plant according to the control signal so that the output signal will approach the reference input signal.) The sensor or measuring element is a device that converts the output variable into an-other suitable variable,such as a displacement,pressure,voltage,etc.,that can be used to compare the output to the reference input signal.This element is in the feedback path of the closed-loop system.The set point of the controller must be converted to a reference input with the same units as the feedback signal from the sensor or measuring element.
Classifications of Industrial Controllers.
Most industrial controllers may be classified according to their control actions as: 1. Two-position or on–off controllers 2. Proportional controllers 3. Integral controllers 4. Proportional-plus-integral controllers 5. Proportional-plus-derivative controllers 6. Proportional-plus-integral-plus-derivative controllers Most industrial controllers use electricity or pressurized fluid such as oil or air as power sources. Consequently, controllers may also be classified according to the kind of power employed in the operation, such as pneumatic controllers, hydraulic controllers, or electronic controllers. What kind of controller to use must be decided based on the nature of the plant and the operating conditions, including such considerations as safety, cost, availability, reliability, accuracy, weight, and size.
Two-Position or On–Off Control Action.
In a two-position control system, the actuating element has only two fixed positions, which are, in many cases, simply on and off.Two-position or on–off control is relatively simple and inexpensive and, for this rea-son, is very widely used in both industrial and domestic control systems.
Let the output signal from the controller be u(t) and the actuating error signal be e(t).
In two-position control, the signal u(t) remains at either a maximum or minimum value, depending on whether the actuating error signal is positive or negative, so that where U1 and U2 are constants. The minimum value U2 is usually either zero or –U1.
Two-position controllers are generally electrical devices, and an electric solenoid-oper-ated valve is widely used in such controllers.Pneumatic proportional controllers with very high gains act as two-position controllers and are sometimes called pneumatic two-position controllers.
Figures 2–7(a) and (b) show the block diagrams for two-position or on–off controllers.
The range through which the actuating error signal must move before the switching occurs = U 2 , for e(t) 6 0 u(t) = U 1 , for e(t) 7 0 Section 2–3 / Automatic Control Systems 23 is called the differential gap. A differential gap is indicated in Figure 2–7(b). Such a dif-ferential gap causes the controller output u(t) to maintain its present value until the ac-tuating error signal has moved slightly beyond the zero value.In some cases,the differential gap is a result of unintentional friction and lost motion; however, quite often it is inten-tionally provided in order to prevent too-frequent operation of the on–off mechanism.
Consider the liquid-level control system shown in Figure 2–8(a),where the electromag-netic valve shown in Figure 2–8(b) is used for controlling the inflow rate.This valve is either open or closed.With this two-position control,the water inflow rate is either a positive con-stant or zero. As shown in Figure 2–9, the output signal continuously moves between the two limits required to cause the actuating element to move from one fixed position to the other. Notice that the output curve follows one of two exponential curves, one correspon-ding to the filling curve and the other to the emptying curve. Such output oscillation be-tween two limits is a typical response characteristic of a system under two-position control.
(a) (b) U1 U2 u e U1 U2 u e Differential gap + – + – Figure 2–7 (a) Block diagram of an on–off controller; (b) block diagram of an on–off controller with differential gap.
115 V Float R C h (a) (b) qi Movable iron core Magnetic coil Figure 2–8 (a) Liquid-level control system; (b) electromagnetic valve.
h(t) t 0 Differential gap Figure 2–9 Level h(t)-versus-t curve for the system shown in Figure 2–8(a).
24 Chapter 2 / Mathematical Modeling of Control Systems From Figure 2–9, we notice that the amplitude of the output oscillation can be reduced by decreasing the differential gap. The decrease in the differential gap, however, increases the number of on–off switchings per minute and reduces the useful life of the component. The magnitude of the differential gap must be determined from such considerations as the accuracy required and the life of the component.
Proportional Control Action.
For a controller with proportional control action, the relationship between the output of the controller u(t) and the actuating error signal e(t) is or, in Laplace-transformed quantities, where Kp is termed the proportional gain.
Whatever the actual mechanism may be and whatever the form of the operating power, the proportional controller is essentially an amplifier with an adjustable gain.
Integral Control Action.
In a controller with integral control action, the value of the controller output u(t) is changed at a rate proportional to the actuating error signal e(t).That is, or where Ki is an adjustable constant.The transfer function of the integral controller is Proportional-Plus-Integral Control Action.
The control action of a proportional-plus-integral controller is defined by u(t) = K p e(t) + K p T i 3 t 0 e(t)dt U(s) E(s) = K i s u(t) = K i3 t 0 e(t)dt du(t) dt = K i e(t) U(s) E(s) = K p u(t) = K p e(t) Section 2–3 / Automatic Control Systems 25 or the transfer function of the controller is where is called the integral time.
Proportional-Plus-Derivative Control Action.
The control action of a proportional-plus-derivative controller is defined by and the transfer function is where is called the derivative time.
Proportional-Plus-Integral-Plus-Derivative Control Action.
The combination of proportional control action, integral control action, and derivative control action is termed proportional-plus-integral-plus-derivative control action. It has the advantages of each of the three individual control actions. The equation of a controller with this combined action is given by or the transfer function is where Kp is the proportional gain, is the integral time, and is the derivative time.
The block diagram of a proportional-plus-integral-plus-derivative controller is shown in Figure 2–10.
T d T i U(s) E(s) = K p a 1 + 1 T i s + T d s b u(t) = K p e(t) + K p T i 3 t 0 e(t)dt + K p T d de(t) dt T d U(s) E(s) = K pA1 + T d sB u(t) = K p e(t) + K p T d de(t) dt T i U(s) E(s) = K p a1 + 1 T i s b + – E(s) U(s) Kp(1 + Tis + Ti Tds2) Tis Figure 2–10 Block diagram of a proportional-plus-integral-plus-derivative controller.
26 Chapter 2 / Mathematical Modeling of Control Systems R(s) G1(s) G2(s) H(s) Disturbance D(s) C(s) + – ++ Figure 2–11 Closed-loop system subjected to a disturbance.
Closed-Loop System Subjected to a Disturbance.
Figure 2–11 shows a closed-loop system subjected to a disturbance. When two inputs (the reference input and dis-turbance) are present in a linear time-invariant system, each input can be treated independently of the other; and the outputs corresponding to each input alone can be added to give the complete output.The way each input is introduced into the system is shown at the summing point by either a plus or minus sign.
Consider the system shown in Figure 2–11. In examining the effect of the distur-bance D(s), we may assume that the reference input is zero; we may then calculate the response CD(s) to the disturbance only.This response can be found from On the other hand, in considering the response to the reference input R(s), we may assume that the disturbance is zero.Then the response CR(s) to the reference input R(s) can be obtained from The response to the simultaneous application of the reference input and disturbance can be obtained by adding the two individual responses. In other words, the response C(s) due to the simultaneous application of the reference input R(s) and disturbance D(s) is given by Consider now the case where |G1(s)H(s)| 1 and |G1(s)G2(s)H(s)| 1. In this case, the closed-loop transfer function CD(s)/D(s) becomes almost zero, and the effect of the disturbance is suppressed.This is an advantage of the closed-loop system.
On the other hand, the closed-loop transfer function CR(s)/R(s) approaches 1/H(s) as the gain of G1(s)G2(s)H(s) increases.This means that if |G1(s)G2(s)H(s)| 1, then the closed-loop transfer function CR(s)/R(s) becomes independent of G1(s) and G2(s) and inversely proportional to H(s), so that the variations of G1(s) and G2(s) do not affect the closed-loop transfer function CR(s)/R(s). This is another advantage of the closed-loop system.It can easily be seen that any closed-loop system with unity feedback, H(s)=1, tends to equalize the input and output. = G2(s) 1 + G1(s)G2(s)H(s) CG1(s)R(s) + D(s)D C(s) = CR(s) + CD(s) CR(s) R(s) = G1(s)G2(s) 1 + G1(s)G2(s)H(s) CD(s) D(s) = G2(s) 1 + G1(s)G2(s)H(s) Section 2–3 / Automatic Control Systems 27 Procedures for Drawing a Block Diagram.
To draw a block diagram for a sys-tem, first write the equations that describe the dynamic behavior of each component.
Then take the Laplace transforms of these equations, assuming zero initial conditions, and represent each Laplace-transformed equation individually in block form. Finally, as-semble the elements into a complete block diagram.
As an example, consider the RC circuit shown in Figure 2–12(a).The equations for this circuit are (2–4) (2–5) The Laplace transforms of Equations (2–4) and (2–5),with zero initial condition,become (2–6) (2–7) Equation (2–6) represents a summing operation, and the corresponding diagram is shown in Figure 2–12(b).Equation (2–7) represents the block as shown in Figure 2–12(c).
Assembling these two elements, we obtain the overall block diagram for the system as shown in Figure 2–12(d).
Block Diagram Reduction.
It is important to note that blocks can be connected in series only if the output of one block is not affected by the next following block. If there are any loading effects between the components, it is necessary to combine these components into a single block.
Any number of cascaded blocks representing nonloading components can be replaced by a single block, the transfer function of which is simply the product of the individual transfer functions. Eo(s) = I(s) Cs I(s) = Ei(s) - Eo(s) R eo = 1idt C i = ei - eo R (d) Ei(s) I(s) Eo(s) 1 R 1 Cs Eo(s) (b) Ei(s) I(s) 1 R (c) I(s) Eo(s) 1 Cs (a) R C eo ei i + – + – Figure 2–12 (a) RC circuit; (b) block diagram representing Equation (2–6); (c) block diagram representing Equation (2–7); (d) block diagram of the RC circuit.
28 Chapter 2 / Mathematical Modeling of Control Systems R G1 H1 H2 G3 G2 C R G1 H1 G3 G2 C R G3 C R C R C (a) (b) (c) (d) (e) H2 G1 H2 G1 G1G2 1 – G1G2H1 G1G2G3 1 – G1G2H1 + G2G3H2 G1G2G3 1 – G1G2H1 + G2G3H2 + G1G2G3 + – + – ++ + – + – ++ + – + – + – Figure 2–13 (a) Multiple-loop system; (b)–(e) successive reductions of the block diagram shown in (a).
A complicated block diagram involving many feedback loops can be simplified by a step-by-step rearrangement. Simplification of the block diagram by rearrangements considerably reduces the labor needed for subsequent mathematical analysis. It should be noted, however, that as the block diagram is simplified, the transfer functions in new blocks become more complex because new poles and new zeros are generated.
EXAMPLE 2–1 Consider the system shown in Figure 2–13(a). Simplify this diagram.
By moving the summing point of the negative feedback loop containing H2 outside the posi-tive feedback loop containing H1,we obtain Figure 2–13(b).Eliminating the positive feedback loop, we have Figure 2–13(c).The elimination of the loop containing H2/G1 gives Figure 2–13(d).Finally, eliminating the feedback loop results in Figure 2–13(e).
Section 2–4 / Modeling in State Space 29 Notice that the numerator of the closed-loop transfer function C(s)/R(s) is the product of the transfer functions of the feedforward path.The denominator of C(s)/R(s) is equal to (The positive feedback loop yields a negative term in the denominator.) 2–4 MODELING IN STATE SPACE In this section we shall present introductory material on state-space analysis of control systems.
Modern Control Theory.
The modern trend in engineering systems is toward greater complexity, due mainly to the requirements of complex tasks and good accu-racy. Complex systems may have multiple inputs and multiple outputs and may be time varying. Because of the necessity of meeting increasingly stringent requirements on the performance of control systems, the increase in system complexity, and easy access to large scale computers, modern control theory, which is a new approach to the analy-sis and design of complex control systems, has been developed since around 1960.This new approach is based on the concept of state. The concept of state by itself is not new, since it has been in existence for a long time in the field of classical dynamics and other fields.
Modern Control Theory Versus Conventional Control Theory.
Modern con-trol theory is contrasted with conventional control theory in that the former is appli-cable to multiple-input, multiple-output systems, which may be linear or nonlinear, time invariant or time varying, while the latter is applicable only to linear time-invariant single-input, single-output systems. Also, modern control theory is essen-tially time-domain approach and frequency domain approach (in certain cases such as H-infinity control), while conventional control theory is a complex frequency-domain approach. Before we proceed further, we must define state, state variables, state vector, and state space.
State.
The state of a dynamic system is the smallest set of variables (called state variables) such that knowledge of these variables at t=t0, together with knowledge of the input for t t0, completely determines the behavior of the system for any time t t0.
Note that the concept of state is by no means limited to physical systems. It is appli-cable to biological systems, economic systems, social systems, and others.
State Variables.
The state variables of a dynamic system are the variables mak-ing up the smallest set of variables that determine the state of the dynamic system. If at = 1 - G1 G2 H 1 + G2 G3 H 2 + G1 G2 G3 = 1 + A-G1 G2 H1 + G2 G3 H2 + G1 G2 G3B 1 + a (product of the transfer functions around each loop) 30 Chapter 2 / Mathematical Modeling of Control Systems least n variables x1, x2, p , xn are needed to completely describe the behavior of a dy-namic system (so that once the input is given for t t0 and the initial state at t=t0 is specified, the future state of the system is completely determined), then such n variables are a set of state variables.
Note that state variables need not be physically measurable or observable quantities.
Variables that do not represent physical quantities and those that are neither measura-ble nor observable can be chosen as state variables. Such freedom in choosing state vari-ables is an advantage of the state-space methods. Practically, however, it is convenient to choose easily measurable quantities for the state variables, if this is possible at all, be-cause optimal control laws will require the feedback of all state variables with suitable weighting.
State Vector.
If n state variables are needed to completely describe the behavior of a given system, then these n state variables can be considered the n components of a vector x. Such a vector is called a state vector.A state vector is thus a vector that deter-mines uniquely the system state x(t) for any time t t0, once the state at t=t0 is given and the input u(t) for t t0 is specified.
State Space.
The n-dimensional space whose coordinate axes consist of the x1 axis, x2 axis, p , xn axis, where x1, x2, p , xn are state variables, is called a state space.Any state can be represented by a point in the state space.
State-Space Equations.
In state-space analysis we are concerned with three types of variables that are involved in the modeling of dynamic systems: input variables, out-put variables, and state variables. As we shall see in Section 2–5, the state-space repre-sentation for a given system is not unique, except that the number of state variables is the same for any of the different state-space representations of the same system.
The dynamic system must involve elements that memorize the values of the input for t t1. Since integrators in a continuous-time control system serve as memory devices, the outputs of such integrators can be considered as the variables that define the inter-nal state of the dynamic system.Thus the outputs of integrators serve as state variables.
The number of state variables to completely define the dynamics of the system is equal to the number of integrators involved in the system.
Assume that a multiple-input,multiple-output system involves n integrators.Assume also that there are r inputs u1(t), u2(t), p , ur(t) and m outputs y1(t), y2(t), p , ym(t).
Define n outputs of the integrators as state variables: x1(t), x2(t), p , xn(t) Then the system may be described by (2–8) x # n(t) = f nAx1 , x2 , p , xn ; u1 , u2 , p , ur ; tB x # 2(t) = f 2Ax1 , x2 , p , xn ; u1 , u2 , p , ur ; tB x # 1(t) = f 1Ax1 , x2 , p , xn ; u1 , u2 , p , ur ; tB Section 2–4 / Modeling in State Space 31 The outputs y1(t), y2(t), p , ym(t) of the system may be given by (2–9) If we define ym(t) = gmAx1 , x2 , p , xn ; u1 , u2 , p , ur ; tB y2(t) = g2Ax1 , x2 , p , xn ; u1 , u2 , p , ur ; tB y1(t) = g1Ax1 , x2 , p , xn ; u1 , u2 , p , ur ; tB u(t) = F u1(t) u2(t) ur(t) V g(x, u, t) = F g1Ax1 , x2 , p , xn ; u1 , u2 , p , ur ; tB g2Ax1 , x2 , p , xn ; u1 , u2 , p , ur ; tB gmAx1 , x2 , p , xn ; u1 , u2 , p , ur ; tB V , y(t) = F y1(t) y2(t) ym(t) V , f(x, u, t) = F f 1Ax1 , x2 , p , xn ; u1 , u2 , p , ur ; tB f 2Ax1 , x2 , p , xn ; u1 , u2 , p , ur ; tB f nAx1 , x2 , p , xn ; u1 , u2 , p , ur ; tB V , x(t) = F x1(t) x2(t) xn(t) V , then Equations (2–8) and (2–9) become (2–10) (2–11) where Equation (2–10) is the state equation and Equation (2–11) is the output equation.
If vector functions f and/or g involve time t explicitly, then the system is called a time-varying system.
If Equations (2–10) and (2–11) are linearized about the operating state, then we have the following linearized state equation and output equation: (2–12) (2–13) where A(t) is called the state matrix, B(t) the input matrix, C(t) the output matrix, and D(t) the direct transmission matrix. (Details of linearization of nonlinear systems about y(t) = C(t)x(t) + D(t)u(t) x # (t) = A(t)x(t) + B(t)u(t) y(t) = g(x, u, t) x # (t) = f(x, u, t) 32 Chapter 2 / Mathematical Modeling of Control Systems m k b u(t) y(t) Figure 2–15 Mechanical system.
u(t) D(t) B(t) A(t) C(t) y(t) x(t) • dt Ú x(t) ++ ++ Figure 2–14 Block diagram of the linear, continuous-time control system represented in state space.
the operating state are discussed in Section 2–7.) A block diagram representation of Equations (2–12) and (2–13) is shown in Figure 2–14.
If vector functions f and g do not involve time t explicitly then the system is called a time-invariant system. In this case, Equations (2–12) and (2–13) can be simplified to (2–14) (2–15) Equation (2–14) is the state equation of the linear, time-invariant system and Equation (2–15) is the output equation for the same system. In this book we shall be concerned mostly with systems described by Equations (2–14) and (2–15).
In what follows we shall present an example for deriving a state equation and output equation.
EXAMPLE 2–2 Consider the mechanical system shown in Figure 2–15. We assume that the system is linear. The external force u(t) is the input to the system, and the displacement y(t) of the mass is the output.
The displacement y(t) is measured from the equilibrium position in the absence of the external force.This system is a single-input, single-output system.
From the diagram, the system equation is (2–16) This system is of second order.This means that the system involves two integrators. Let us define state variables x1(t) and x2(t) as Then we obtain or (2–17) (2–18) The output equation is (2–19) y = x1 x # 2 = - k m x1 - b m x2 + 1 m u x # 1 = x2 x # 2 = 1 m A-ky - by # B + 1 m u x # 1 = x2 x2(t) = y # (t) x1(t) = y(t) my $ + by # + ky = u y # (t) = Cx(t) + Du(t) x # (t) = Ax(t) + Bu(t) In a vector-matrix form, Equations (2–17) and (2–18) can be written as (2–20) The output equation, Equation (2–19), can be written as (2–21) Equation (2–20) is a state equation and Equation (2–21) is an output equation for the system.
They are in the standard form: where Figure 2–16 is a block diagram for the system. Notice that the outputs of the integrators are state variables.
Correlation Between Transfer Functions and State-Space Equations.
In what follows we shall show how to derive the transfer function of a single-input, single-output system from the state-space equations.
Let us consider the system whose transfer function is given by (2–22) This system may be represented in state space by the following equations: (2–23) (2–24) y = Cx + Du x # = Ax + Bu Y(s) U(s) = G(s) A = C 0 - k m 1 - b m S , B = C 0 1 m S , C = [1 0], D = 0 y = Cx + Du x # = Ax + Bu y = [1 0]Bx1 x2 R B x # 1 x # 2 R = C 0 - k m 1 - b m S Bx1 x2 R + C 0 1 m S u Section 2–4 / Modeling in State Space 33 u 1 m b m k m x2 x2 • x1 = y + – + + Figure 2–16 Block diagram of the mechanical system shown in Figure 2–15.
34 Chapter 2 / Mathematical Modeling of Control Systems where x is the state vector, u is the input, and y is the output.The Laplace transforms of Equations (2–23) and (2–24) are given by (2–25) (2–26) Since the transfer function was previously defined as the ratio of the Laplace transform of the output to the Laplace transform of the input when the initial conditions were zero, we set x(0) in Equation (2–25) to be zero.Then we have or By premultiplying to both sides of this last equation, we obtain (2–27) By substituting Equation (2–27) into Equation (2–26), we get (2–28) Upon comparing Equation (2–28) with Equation (2–22), we see that (2–29) This is the transfer-function expression of the system in terms of A, B, C, and D.
Note that the right-hand side of Equation (2–29) involves Hence G(s) can be written as where Q(s) is a polynomial in s. Notice that is equal to the characteristic poly-nomial of G(s). In other words, the eigenvalues of A are identical to the poles of G(s).
EXAMPLE 2–3 Consider again the mechanical system shown in Figure 2–15. State-space equations for the system are given by Equations (2–20) and (2–21).We shall obtain the transfer function for the system from the state-space equations.
By substituting A, B, C, and D into Equation (2–29), we obtain = [1 0]C s k m -1 s + b m S -1 C 0 1 m S = [1 0] c B s 0 0 s R - C 0 - k m 1 - b m S s -1 C 0 1 m S + 0 G(s) = C(s I - A)-1 B + D ∑s I - A∑ G(s) = Q(s) ∑s I - A∑ (s I - A)-1.
G(s) = C(s I - A)-1 B + D Y(s) = CC(s I - A)-1 B + DDU(s) X(s) = (s I - A)-1 BU(s) (s I - A)-1 (s I - A)X(s) = BU(s) s X(s) - AX(s) = BU(s) Y(s) = CX(s) + DU(s) sX(s) - x(0) = AX(s) + BU(s) Section 2–5 / State-Space Representation of Scalar Differential Equation Systems 35 Note that (Refer to Appendix C for the inverse of the 2 2 matrix.) Thus, we have which is the transfer function of the system. The same transfer function can be obtained from Equation (2–16).
Transfer Matrix.
Next, consider a multiple-input, multiple-output system.Assume that there are r inputs and m outputs Define The transfer matrix G(s) relates the output Y(s) to the input U(s), or where G(s) is given by [The derivation for this equation is the same as that for Equation (2–29).] Since the input vector u is r dimensional and the output vector y is m dimensional,the transfer ma-trix G(s) is an mr matrix.
2–5 STATE-SPACE REPRESENTATION OF SCALAR DIFFERENTIAL EQUATION SYSTEMS A dynamic system consisting of a finite number of lumped elements may be described by ordinary differential equations in which time is the independent variable. By use of vector-matrix notation, an nth-order differential equation may be expressed by a first-order vector-matrix differential equation. If n elements of the vector are a set of state variables, then the vector-matrix differential equation is a state equation. In this section we shall present methods for obtaining state-space representations of continuous-time systems.
G(s) = C(s I - A)-1 B + D Y(s) = G(s )U(s ) y = F y1 y2 ym V , u = F u1 u2 ur V y1 , y2 , p , ym .
u1 , u2 , p , ur , = 1 ms2 + bs + k G(s) = [1 0] 1 s2 + b m s + k m D s + b m - k m 1 s T C 0 1 m S C s k m -1 s + b m S -1 = 1 s2 + b m s + k m D s + b m - k m 1 s T 36 Chapter 2 / Mathematical Modeling of Control Systems State-Space Representation of nth-Order Systems of Linear Differential Equa-tions in which the Forcing Function Does Not Involve Derivative Terms.
Con-sider the following nth-order system: (2–30) Noting that the knowledge of together with the input u(t) for t 0, determines completely the future behavior of the system, we may take as a set of n state variables. (Mathematically, such a choice of state variables is quite convenient.Practically,however,because higher-order derivative terms are inaccurate, due to the noise effects inherent in any practical situations, such a choice of the state variables may not be desirable.) Let us define Then Equation (2–30) can be written as or (2–31) where B = G 0 0 0 1 W A = G 0 0 0 -an 1 0 0 -an-1 0 1 0 -an-2 p p p p 0 0 1 -a1 W , x = F x1 x2 xn V , x # = Ax + Bu x # n = -anx1 - p - a1xn + u x # n-1 = xn x # 2 = x3 x # 1 = x2 xn = y (n-1) x2 = y # x1 = y y(t), y # (t), p , y (n-1) (t) y(0), y # (0), p , y (n-1) (0), y (n) + a1y (n-1) + p + an-1 y # + an y = u Section 2–5 / State-Space Representation of Scalar Differential Equation Systems 37 The output can be given by or (2–32) where [Note that D in Equation (2–24) is zero.] The first-order differential equation, Equa-tion (2–31), is the state equation, and the algebraic equation, Equation (2–32), is the output equation.
Note that the state-space representation for the transfer function system is given also by Equations (2–31) and (2–32).
State-Space Representation of nth-Order Systems of Linear Differential Equa-tions in which the Forcing Function Involves Derivative Terms.
Consider the dif-ferential equation system that involves derivatives of the forcing function, such as (2–33) The main problem in defining the state variables for this case lies in the derivative terms of the input u. The state variables must be such that they will eliminate the de-rivatives of u in the state equation.
One way to obtain a state equation and output equation for this case is to define the following n variables as a set of n state variables: (2–34) xn = y (n-1) - b0u (n-1) - b1u (n-2) - p - bn-2 u # - bn-1 u = x # n-1 - bn-1 u x3 = y $ - b0 u $ - b1u # - b2 u = x # 2 - b2 u x2 = y # - b0 u # - b1 u = x # 1 - b1 u x1 = y - b0 u y (n) + a1 y (n-1) + p + an-1 y # + an y = b0 u (n) + b1 u (n-1) + p + bn-1 u # + bn u Y(s) U(s) = 1 sn + a1 sn-1 + p + an-1 s + an C = [1 0 p 0] y = Cx y = [1 0 p 0]F x1 x2 xn V 38 Chapter 2 / Mathematical Modeling of Control Systems where are determined from (2–35) With this choice of state variables the existence and uniqueness of the solution of the state equation is guaranteed. (Note that this is not the only choice of a set of state vari-ables.) With the present choice of state variables, we obtain (2–36) where is given by [To derive Equation (2–36), see Problem A–2–6.] In terms of vector-matrix equations, Equation (2–36) and the output equation can be written as y = [1 0 p 0]F x1 x2 xn V + b0 u + G b1 b2 bn-1 bn W u G x1 x2 xn-1 xn W G x # 1 x # 2 x # n-1 x # n W = G 0 0 0 -an 1 0 0 -an-1 0 1 0 -an-2 p p p p 0 0 1 -a1 W bn = bn - a1 bn-1 - p - an-1 b1 - an -1b0 bn x # n = -an x1 - an-1 x2 - p - a1 xn + bn u x # n-1 = xn + bn-1 u x # 2 = x3 + b2 u x # 1 = x2 + b1 u bn-1 = bn-1 - a1 bn-2 - p - an-2 b1 - an -1b0 b3 = b3 - a1 b2 - a2 b1 - a3 b0 b2 = b2 - a1 b1 - a2 b0 b1 = b1 - a1 b0 b0 = b0 b0 , b1 , b2 , p , bn-1 Section 2–6 / Transformation of Mathematical Models with MATLAB 39 or (2–37) (2–38) where In this state-space representation, matrices A and C are exactly the same as those for the system of Equation (2–30).The derivatives on the right-hand side of Equation (2–33) affect only the elements of the B matrix.
Note that the state-space representation for the transfer function is given also by Equations (2–37) and (2–38).
There are many ways to obtain state-space representations of systems. Methods for obtaining canonical representations of systems in state space (such as controllable canon-ical form, observable canonical form, diagonal canonical form, and Jordan canonical form) are presented in Chapter 9.
MATLAB can also be used to obtain state-space representations of systems from transfer-function representations, and vice versa.This subject is presented in Section 2–6.
2–6 TRANSFORMATION OF MATHEMATICAL MODELS WITH MATLAB MATLAB is quite useful to transform the system model from transfer function to state space, and vice versa. We shall begin our discussion with transformation from transfer function to state space.
Y(s) U(s) = b0 sn + b1 sn-1 + p + bn-1 s + bn sn + a1 sn-1 + p + an-1 s + an B = G b1 b2 bn-1 bn W , C = [1 0 p 0], D = b0 = b0 x = G x1 x2 xn-1 xn W , A = G 0 0 0 -an 1 0 0 -an-1 0 1 0 -an-2 p p p p 0 0 1 -a1 W y = Cx + Du x # = Ax + Bu 40 Chapter 2 / Mathematical Modeling of Control Systems Let us write the closed-loop transfer function as Once we have this transfer-function expression, the MATLAB command [A,B,C,D] = tf2ss(num,den) will give a state-space representation. It is important to note that the state-space repre-sentation for any system is not unique. There are many (infinitely many) state-space representations for the same system.The MATLAB command gives one possible such state-space representation.
Transformation from Transfer Function to State Space Representation.
Consider the transfer-function system (2–39) There are many (infinitely many) possible state-space representations for this system.
One possible state-space representation is Another possible state-space representation (among infinitely many alternatives) is (2–40) C x # 1 x # 2 x # 3 S = C -14 1 0 -56 0 1 -160 0 0 S C x1 x2 x3 S + C 1 0 0 S u y = [1 0 0]C x1 x2 x3 S + u C x # 1 x # 2 x # 3 S = C 0 0 -160 1 0 -56 0 1 -14 S C x1 x2 x3 S + C 0 1 -14 S u = s s3 + 14s2 + 56s + 160 Y(s) U(s) = s (s + 10)As2 + 4s + 16B Y(s) U(s) = numerator polynomial in s denominator polynomial in s = num den Section 2–6 / Transformation of Mathematical Models with MATLAB 41 (2–41) MATLAB transforms the transfer function given by Equation (2–39) into the state-space representation given by Equations (2–40) and (2–41). For the example system considered here, MATLAB Program 2–2 will produce matrices A, B, C, and D. y = [0 1 0]C x1 x2 x3 S + u MATLAB Program 2–2 num = [1 0]; den = [1 14 56 160]; [A,B,C,D] = tf2ss(num,den) A = -14 -56 -160 1 0 0 0 1 0 B = 1 0 0 C = 0 1 0 D = 0 Transformation from State Space Representation to Transfer Function.
To obtain the transfer function from state-space equations, use the following command: [num,den] = ss2tf(A,B,C,D,iu) iu must be specified for systems with more than one input. For example, if the system has three inputs (u1, u2, u3), then iu must be either 1, 2, or 3, where 1 implies u1, 2 implies u2, and 3 implies u3.
If the system has only one input, then either [num,den] = ss2tf(A,B,C,D) 42 Chapter 2 / Mathematical Modeling of Control Systems EXAMPLE 2–4 Obtain the transfer function of the system defined by the following state-space equations: MATLAB Program 2-3 will produce the transfer function for the given system.The transfer func-tion obtained is given by Y(s) U(s) = 25s + 5 s3 + 5s2 + 25s + 5 y = [1 0 0]C x1 x2 x3 S C x # 1 x # 2 x # 3 S = C 0 0 -5 1 0 -25 0 1 -5 S C x1 x2 x3 S + C 0 25 -120 S u MATLAB Program 2–3 A = [0 1 0; 0 0 1; -5 -25 -5]; B = [0; 25; -120]; C = [1 0 0]; D = ; [num,den] = ss2tf(A,B,C,D) num = 0 0.0000 25.0000 5.0000 den 1.0000 5.0000 25.0000 5.0000 % The same result can be obtained by entering the following command: [num,den] = ss2tf(A,B,C,D,1) num = 0 0.0000 25.0000 5.0000 den = 1.0000 5.0000 25.0000 5.0000 or [num,den] = ss2tf(A,B,C,D,1) may be used. For the case where the system has multiple inputs and multiple outputs, see Problem A–2–12.
Section 2–7 / Linearization of Nonlinear Mathematical Models 43 2–7 LINEARIZATION OF NONLINEAR MATHEMATICAL MODELS Nonlinear Systems.
A system is nonlinear if the principle of superposition does not apply.Thus, for a nonlinear system the response to two inputs cannot be calculated by treating one input at a time and adding the results.
Although many physical relationships are often represented by linear equations, in most cases actual relationships are not quite linear. In fact, a careful study of phys-ical systems reveals that even so-called “linear systems” are really linear only in lim-ited operating ranges. In practice, many electromechanical systems, hydraulic systems, pneumatic systems, and so on, involve nonlinear relationships among the variables.
For example, the output of a component may saturate for large input signals.There may be a dead space that affects small signals. (The dead space of a component is a small range of input variations to which the component is insensitive.) Square-law nonlin-earity may occur in some components. For instance, dampers used in physical systems may be linear for low-velocity operations but may become nonlinear at high veloci-ties, and the damping force may become proportional to the square of the operating velocity.
Linearization of Nonlinear Systems.
In control engineering a normal operation of the system may be around an equilibrium point, and the signals may be considered small signals around the equilibrium. (It should be pointed out that there are many ex-ceptions to such a case.) However, if the system operates around an equilibrium point and if the signals involved are small signals, then it is possible to approximate the non-linear system by a linear system. Such a linear system is equivalent to the nonlinear sys-tem considered within a limited operating range. Such a linearized model (linear, time-invariant model) is very important in control engineering.
The linearization procedure to be presented in the following is based on the ex-pansion of nonlinear function into a Taylor series about the operating point and the retention of only the linear term. Because we neglect higher-order terms of the Taylor series expansion, these neglected terms must be small enough; that is, the variables deviate only slightly from the operating condition. (Otherwise, the result will be inaccurate.) Linear Approximation of Nonlinear Mathematical Models.
To obtain a linear mathematical model for a nonlinear system, we assume that the variables deviate only slightly from some operating condition. Consider a system whose input is x(t) and out-put is y(t). The relationship between y(t) and x(t) is given by (2–42) If the normal operating condition corresponds to then Equation (2–42) may be expanded into a Taylor series about this point as follows: (2–43) = f(x –) + df dx (x - x –) + 1 2!
d2f dx2 (x - x –)2 + p y = f(x) x –, y –, y = f(x) 44 Chapter 2 / Mathematical Modeling of Control Systems where the derivatives are evaluated at If the variation is small, we may neglect the higher-order terms in Then Equation (2–43) may be written as (2–44) where Equation (2–44) may be rewritten as (2–45) which indicates that is proportional to Equation (2–45) gives a linear math-ematical model for the nonlinear system given by Equation (2–42) near the operating point Next, consider a nonlinear system whose output y is a function of two inputs x1 and x2, so that (2–46) To obtain a linear approximation to this nonlinear system,we may expand Equation (2–46) into a Taylor series about the normal operating point Then Equation (2–46) becomes where the partial derivatives are evaluated at Near the normal oper-ating point, the higher-order terms may be neglected.The linear mathematical model of this nonlinear system in the neighborhood of the normal operating condition is then given by y - y – = K 1Ax1 - x – 1B + K 2Ax2 - x – 2B x2 = x – 2 .
x1 = x – 1 , + 02f 0x2 2 Ax2 - x – 2B 2d + p + 1 2! c 02f 0x2 1 Ax1 - x – 1B 2 + 2 02f 0x1 0x2 Ax1 - x – 1BAx2 - x – 2B y = fAx – 1 , x – 2B + c 0f 0x1 Ax1 - x – 1B + 0f 0x2 Ax2 - x – 2B d x – 1 , x – 2 .
y = fAx1 , x2B y = y –.
x = x –, x - x –.
y - y – y - y – = K(x - x –) K = df dx 2 x=x – y – = f(x –) y = y – + K(x - x –) x - x –.
x - x – x = x –.
d2fdx2, p dfdx, Section 2–7 / Linearization of Nonlinear Mathematical Models 45 where The linearization technique presented here is valid in the vicinity of the operating condition.If the operating conditions vary widely,however,such linearized equations are not adequate, and nonlinear equations must be dealt with. It is important to remember that a particular mathematical model used in analysis and design may accurately rep-resent the dynamics of an actual system for certain operating conditions, but may not be accurate for other operating conditions.
EXAMPLE 2–5 Linearize the nonlinear equation z=xy in the region 5 x 7, 10 y 12. Find the error if the linearized equation is used to calcu-late the value of z when x=5, y=10.
Since the region considered is given by 5 x 7, 10 y 12, choose Then Let us obtain a linearized equation for the nonlinear equation near a point Expanding the nonlinear equation into a Taylor series about point and neglecting the higher-order terms, we have where Hence the linearized equation is z-66=11(x-6)+6(y-11) or z=11x+6y-66 When x=5, y=10, the value of z given by the linearized equation is z=11x+6y-66=55+60-66=49 The exact value of z is z=xy=50. The error is thus 50-49=1. In terms of percentage, the error is 2%.
b = 0(xy) 0y 2 x=x –, y=y – = x – = 6 a = 0(xy) 0x 2 x=x –, y=y – = y – = 11 z - z – = aAx - x –B + bAy - y –B y = y – x = x –, y – = 11.
x – = 6, z – = x –y – = 66.
y – = 11.
x – = 6, K 2 = 0f 0x2 2 x1=x – 1 , x2=x – 2 K 1 = 0f 0x1 2 x1=x – 1 , x2=x – 2 y – = fAx – 1 , x – 2B 46 Chapter 2 / Mathematical Modeling of Control Systems EXAMPLE PROBLEMS AND SOLUTIONS A–2–1.
Simplify the block diagram shown in Figure 2–17.
Solution. First, move the branch point of the path involving H1 outside the loop involving H2, as shown in Figure 2–18(a). Then eliminating two loops results in Figure 2–18(b). Combining two blocks into one gives Figure 2–18(c).
A–2–2.
Simplify the block diagram shown in Figure 2–19. Obtain the transfer function relating C(s) and R(s).
R(s) C(s) G H1 H2 + – ++ Figure 2–17 Block diagram of a system.
R(s) C(s) R(s) C(s) C(s) G H2 (a) (b) (c) H1 G G 1 + GH2 R(s) 1 + H1 G G + H1 1 + GH2 + – ++ Figure 2–18 Simplified block diagrams for the system shown in Figure 2–17.
G1 G2 R(s) C(s) X(s) + + ++ Figure 2–19 Block diagram of a system.
Example Problems and Solutions 47 G1 G2 R(s) C(s) G2 R(s) C(s) G1 + 1 R(s) C(s) G1G2 + G2 + 1 (a) (b) (c) ++ ++ ++ Figure 2–20 Reduction of the block diagram shown in Figure 2–19.
Solution. The block diagram of Figure 2–19 can be modified to that shown in Figure 2–20(a).
Eliminating the minor feedforward path, we obtain Figure 2–20(b), which can be simplified to Figure 2–20(c).The transfer function C(s)/R(s) is thus given by The same result can also be obtained by proceeding as follows: Since signal X(s) is the sum of two signals G1R(s) and R(s), we have The output signal C(s) is the sum of G2X(s) and R(s). Hence And so we have the same result as before: A–2–3.
Simplify the block diagram shown in Figure 2–21. Then obtain the closed-loop transfer function C(s)/R(s).
C(s) R(s) = G1 G2 + G2 + 1 C(s) = G2 X(s) + R(s) = G2CG1 R(s) + R(s)D + R(s) X(s) = G1 R(s) + R(s) C(s) R(s) = G1 G2 + G2 + 1 G1 G2 H3 G3 G4 H2 H1 +– + + +– R(s) C(s) Figure 2–21 Block diagram of a system.
48 Chapter 2 / Mathematical Modeling of Control Systems G1 G1 G2 H3 G4 G3 G4 H2 H1 ++ ++ +– +– R(s) R(s) C(s) C(s) H3 G1G4 G1 G2 1 + G1 G2 H1 R(s) C(s) G1 G2 G3 G4 1+ G1 G2 H1 + G3 G4 H2 – G2 G3 H3 + G1 G2 G3 G4 H1 H2 G3 G4 1 + G3 G4 H2 1 (a) (b) (c) Figure 2–22 Successive reductions of the block diagram shown in Figure 2–21.
G1 Gp ++ +– + + Gf C(s) D(s) R(s) E(s) U(s) H Gc Figure 2–23 Control system with reference input and disturbance input.
Solution. First move the branch point between G3 and G4 to the right-hand side of the loop con-taining G3, G4, and H2. Then move the summing point between G1 and G2 to the left-hand side of the first summing point. See Figure 2–22(a). By simplifying each loop, the block diagram can be modified as shown in Figure 2–22(b). Further simplification results in Figure 2–22(c), from which the closed-loop transfer function C(s)/R(s) is obtained as A–2–4.
Obtain transfer functions C(s)/R(s) and C(s)/D(s) of the system shown in Figure 2–23.
Solution. From Figure 2–23 we have (2–47) (2–48) (2–49) E(s) = R(s) - HC(s) C(s) = GpCD(s) + G1 U(s)D U(s) = Gf R(s) + Gc E(s) C(s) R(s) = G1 G2 G3 G4 1 + G1 G2 H 1 + G3 G4 H 2 - G2 G3 H 3 + G1 G2 G3 G4 H 1 H 2 Example Problems and Solutions 49 By substituting Equation (2–47) into Equation (2–48), we get (2–50) By substituting Equation (2–49) into Equation (2–50), we obtain Solving this last equation for C(s), we get Hence (2–51) Note that Equation (2–51) gives the response C(s) when both reference input R(s) and distur-bance input D(s) are present.
To find transfer function C(s)/R(s), we let D(s)=0 in Equation (2–51).Then we obtain Similarly, to obtain transfer function C(s)/D(s), we let R(s)=0 in Equation (2–51). Then C(s)/D(s) can be given by A–2–5.
Figure 2–24 shows a system with two inputs and two outputs. Derive C1(s)/R1(s), C1(s)/R2(s), C2(s)/R1(s), and C2(s)/R2(s). (In deriving outputs for R1(s), assume that R2(s) is zero, and vice versa.) C(s) D(s) = Gp 1 + G1 Gp Gc H C(s) R(s) = G1 GpAGf + GcB 1 + G1 Gp Gc H C(s) = Gp D(s) + G1 GpAGf + GcBR(s) 1 + G1 Gp Gc H C(s) + G1 Gp Gc HC(s) = Gp D(s) + G1 GpAGf + GcBR(s) C(s) = Gp D(s) + G1 GpEGf R(s) + GcCR(s) - HC(s)D F C(s) = Gp D(s) + G1 GpCGf R(s) + Gc E(s)D G1 C1 C2 R1 R2 G3 G4 + − + − G2 Figure 2–24 System with two inputs and two outputs.
50 Chapter 2 / Mathematical Modeling of Control Systems Solution. From the figure, we obtain (2–52) (2–53) By substituting Equation (2–53) into Equation (2–52), we obtain (2–54) By substituting Equation (2–52) into Equation (2–53), we get (2–55) Solving Equation (2–54) for C1, we obtain (2–56) Solving Equation (2–55) for C2 gives (2–57) Equations (2–56) and (2–57) can be combined in the form of the transfer matrix as follows: Then the transfer functions C1(s)/R1(s), C1(s)/R2(s), C2(s)/R1(s) and C2(s)/R2(s) can be obtained as follows: Note that Equations (2–56) and (2–57) give responses C1 and C2, respectively, when both inputs R1 and R2 are present.
Notice that when R2(s)=0, the original block diagram can be simplified to those shown in Figures 2–25(a) and (b). Similarly, when R1(s)=0, the original block diagram can be simplified to those shown in Figures 2–25(c) and (d). From these simplified block diagrams we can also ob-tain C1(s)/R1(s), C2(s)/R1(s), C1(s)/R2(s), and C2(s)/R2(s), as shown to the right of each corre-sponding block diagram. C2(s) R1(s) = - G1 G2 G4 1 - G1 G2 G3 G4 , C2(s) R2(s) = G4 1 - G1 G2 G3 G4 C1(s) R1(s) = G1 1 - G1 G2 G3 G4 , C1(s) R2(s) = - G1 G3 G4 1 - G1 G2 G3 G4 B C1 C2 R = D G1 1 - G1 G2 G3 G4 -G1 G2 G4 1 - G1 G2 G3 G4 -G1 G3 G4 1 - G1 G2 G3 G4 G4 1 - G1 G2 G3 G4 T B R1 R2 R C2 = -G1 G2 G4 R1 + G4 R2 1 - G1 G2 G3 G4 C1 = G1 R1 - G1 G3 G4 R2 1 - G1 G2 G3 G4 C2 = G4CR2 - G2 G1AR1 - G3 C2B D C1 = G1CR1 - G3 G4AR2 - G2 C1B D C2 = G4AR2 - G2 C1B C1 = G1AR1 - G3 C2B Example Problems and Solutions 51 +– R1 C1 R1 C1 1 – G1 G2 G3 G4 G1 G1 G3 G4 –G2 +– R1 C2 G3 G1 –G2 G4 = +– R2 C2 R2 C2 1 – G1 G2G3 G4 G4 G4 G2 G1 –G3 = R1 C2 1 – G1 G2 G3 G4 – G1 G2 G4 = +– R2 C1 G2 G4 –G3 G1 R2 C1 1 – G1 G2 G3 G4 – G1 G3 G4 = (a) (b) (c) (d) Figure 2–25 Simplified block diagrams and corresponding closed-loop transfer functions.
A–2–6.
Show that for the differential equation system (2–58) state and output equations can be given, respectively, by (2–59) and (2–60) where state variables are defined by x3 = y $ - b0 u $ - b1 u # - b2 u = x # 2 - b2 u x2 = y # - b0 u # - b1 u = x # 1 - b1 u x1 = y - b0 u y = [1 0 0]C x1 x2 x3 S + b0 u C x # 1 x # 2 x # 3 S = C 0 0 -a3 1 0 -a2 0 1 -a1 S C x1 x2 x3 S + C b1 b2 b3 S u y % + a1 y $ + a2 y # + a3 y = b0 u % + b1 u $ + b2 u # + b3 u 52 Chapter 2 / Mathematical Modeling of Control Systems and Solution. From the definition of state variables x2 and x3, we have (2–61) (2–62) To derive the equation for we first note from Equation (2–58) that Since we have Hence, we get (2–63) Combining Equations (2–61), (2–62), and (2–63) into a vector-matrix equation, we obtain Equa-tion (2–59). Also, from the definition of state variable x1, we get the output equation given by Equation (2–60).
A–2–7.
Obtain a state-space equation and output equation for the system defined by Solution. From the given transfer function, the differential equation for the system is Comparing this equation with the standard equation given by Equation (2–33), rewritten y % + a1 y $ + a2 y # + a3 y = b0 u % + b1 u $ + b2u # + b3 u y % + 4y $ + 5y # + 2y = 2u % + u $ + u # + 2u Y(s) U(s) = 2s3 + s2 + s + 2 s3 + 4s2 + 5s + 2 x # 3 = -a3 x1 - a2 x2 - a1 x3 + b3 u = -a1 x3 - a2 x2 - a3 x1 + b3 u = -a1 x3 - a2 x2 - a3 x1 + Ab3 - a1 b2 - a2 b1 - a3 b0Bu + Ab2 - b2 - a1 b1 - a2 b0Bu # + Ab3 - a1 b2 - a2 b1 - a3 b0Bu = -a1 x3 - a2 x2 - a3 x1 + Ab0 - b0Bu % + Ab1 - b1 - a1 b0Bu $ + b0 u % + b1 u $ + b2 u # + b3 u - b0 u % - b1 u $ - b2 u # -a2Ay # - b0 u # - b1 uB - a2 b0 u # - a2 b1 u - a3Ay - b0 uB - a3 b0 u = -a1Ay $ - b0 u $ - b1 u # - b2 uB - a1 b0 u $ - a1 b1 u # - a1 b2 u = A-a1 y $ - a2 y # - a3 yB + b0 u % + b1 u $ + b2 u # + b3 u - b0 u % - b1 u $ - b2 u # x # 3 = y % - b0 u % - b1 u $ - b2 u # x3 = y $ - b0 u $ - b1 u # - b2 u y % = -a1 y $ - a2 y # - a3 y + b0 u % + b1 u $ + b2 u # + b3 u x # 3 , x # 2 = x3 + b2 u x # 1 = x2 + b1 u b3 = b3 - a1 b2 - a2 b1 - a3 b0 b2 = b2 - a1 b1 - a2 b0 b1 = b1 - a1 b0 b0 = b0 Example Problems and Solutions 53 we find Referring to Equation (2–35), we have Referring to Equation (2–34), we define Then referring to Equation (2–36), Hence, the state-space representation of the system is This is one possible state-space representation of the system. There are many (infinitely many) others. If we use MATLAB, it produces the following state-space representation: See MATLAB Program 2-4. (Note that all state-space representations for the same system are equivalent.) C x1 x2 x3 S + 2u y = [-7 -9 -2] C x # 1 x # 2 x # 3 S = C -4 1 0 -5 0 1 -2 0 0 S C x1 x2 x3 S + C 1 0 0 S u y = [1 0 0]C x1 x2 x3 S + 2u C x # 1 x # 2 x # 3 S = C 0 0 -2 1 0 -5 0 1 -4 S C x1 x2 x3 S + C -7 19 -43 S u = -2x1 - 5x2 - 4x3 - 43u x # 3 = -a3 x1 - a2 x2 - a1 x3 + b3 u x # 2 = x3 + 19u x # 1 = x2 - 7u x3 = x # 2 - b2 u = x # 2 - 19u x2 = x # 1 - b1 u = x # 1 + 7u x1 = y - b0 u = y - 2u = 2 - 4 19 - 5 (-7) - 2 2 = -43 b3 = b3 - a1 b2 - a2 b1 - a3 b0 b2 = b2 - a1 b1 - a2 b0 = 1 - 4 (-7) - 5 2 = 19 b1 = b1 - a1 b0 = 1 - 4 2 = -7 b0 = b0 = 2 b0 = 2, b1 = 1, b2 = 1, b3 = 2 a1 = 4, a2 = 5, a3 = 2 54 Chapter 2 / Mathematical Modeling of Control Systems A–2–8.
Obtain a state-space model of the system shown in Figure 2–26.
Solution. The system involves one integrator and two delayed integrators. The output of each integrator or delayed integrator can be a state variable. Let us define the output of the plant as x1, the output of the controller as x2, and the output of the sensor as x3. Then we obtain Y(s) = X 1(s) X 3(s) X 1(s) = 1 s + 1 X 2(s) U(s) - X 3(s) = 1 s X 1(s) X 2(s) = 10 s + 5 U(s) Y(s) 1 s Controller Plant Sensor 10 s + 5 1 s + 1 + – Figure 2–26 Control system.
MATLAB Program 2–4 num = [2 1 1 2]; den = [1 4 5 2]; [A,B,C,D] = tf2ss(num,den) A = -4 -5 -2 1 0 0 0 1 0 B = 1 0 0 C = -7 -9 -2 D = 2 Example Problems and Solutions 55 which can be rewritten as By taking the inverse Laplace transforms of the preceding four equations, we obtain Thus, a state-space model of the system in the standard form is given by It is important to note that this is not the only state-space representation of the system. Infinite-ly many other state-space representations are possible. However, the number of state variables is the same in any state-space representation of the same system. In the present system, the num-ber of state variables is three, regardless of what variables are chosen as state variables.
A–2–9.
Obtain a state-space model for the system shown in Figure 2–27(a).
Solution. First, notice that (as+b)/s2 involves a derivative term. Such a derivative term may be avoided if we modify (as+b)/s2 as Using this modification, the block diagram of Figure 2–27(a) can be modified to that shown in Figure 2–27(b).
Define the outputs of the integrators as state variables, as shown in Figure 2–27(b).Then from Figure 2–27(b) we obtain which may be modified to Y(s) = X 1(s) sX 2(s) = -bX 1(s) + bU(s) sX 1(s) = X 2(s) + aCU(s) - X 1(s)D Y(s) = X 1(s) X 2(s) U(s) - X 1(s) = b s X 1(s) X 2(s) + aCU(s) - X 1(s)D = 1 s as + b s2 = aa + b s b 1 s C x1 x2 x3 S y = [1 0 0] C x # 1 x # 2 x # 3 S = C -5 0 1 10 0 0 0 -1 -1 S C x1 x2 x3 S + C 0 1 0 S u y = x1 x # 3 = x1 - x3 x # 2 = -x3 + u x # 1 = -5x1 + 10x2 Y(s) = X 1(s) sX 3(s) = X 1(s) - X 3(s) sX 2(s) = -X 3(s) + U(s) sX 1(s) = -5X 1(s) + 10X 2(s) 56 Chapter 2 / Mathematical Modeling of Control Systems Taking the inverse Laplace transforms of the preceding three equations, we obtain Rewriting the state and output equations in the standard vector-matrix form, we obtain A–2–10.
Obtain a state-space representation of the system shown in Figure 2–28(a).
Solution. In this problem, first expand (s+z)/(s+p) into partial fractions.
Next,convert K/Cs(s+a)D into the product of K/s and 1/(s+a).Then redraw the block diagram, as shown in Figure 2–28(b). Defining a set of state variables, as shown in Figure 2–28(b), we ob-tain the following equations: y = x1 x # 3 = -(z - p)x1 - px3 + (z - p)u x # 2 = -Kx1 + Kx3 + Ku x # 1 = -ax1 + x2 s + z s + p = 1 + z - p s + p y = [1 0]B x1 x2 R B x # 1 x # 2 R = B -a -b 1 0R B x1 x2 R + B a bR u y = x1 x # 2 = -bx1 + bu x # 1 = -ax1 + x2 + au U(s) Y(s) as + b 1 s2 (a) (b) a U(s) Y(s) b s 1 s X1(s) X2(s) + – + – ++ Figure 2–27 (a) Control system; (b) modified block diagram.
Example Problems and Solutions 57 Rewriting gives Notice that the output of the integrator and the outputs of the first-order delayed integrators C1/(s+a) and (z-p)/(s+p)D are chosen as state variables. It is important to remember that the output of the block (s+z)/(s+p) in Figure 2–28(a) cannot be a state variable, because this block involves a derivative term, s+z.
A–2–11.
Obtain the transfer function of the system defined by Solution. Referring to Equation (2–29), the transfer function G(s) is given by In this problem, matrices A, B, C, and D are A = C -1 0 0 1 -1 0 0 1 -2 S , B = C 0 0 1 S , C = [1 0 0], D = 0 G(s) = C(sI - A)-1B + D y = [1 0 0]C x1 x2 x3 S C x # 1 x # 2 x # 3 S = C -1 0 0 1 -1 0 0 1 -2 S C x1 x2 x3 S + C 0 0 1 S u y = [1 0 0]C x1 x2 x3 S C x # 1 x # 2 x # 3 S = C -a -K -(z - p) 1 0 0 0 K -p S C x1 x2 x3 S + C 0 K z - p S u u y u y (a) (b) s + z s + p K s(s + a) z – p s + p K s 1 s + a x1 x2 x3 + – + – ++ Figure 2–28 (a) Control system; (b) block diagram defining state variables for the system.
58 Chapter 2 / Mathematical Modeling of Control Systems Hence A–2–12.
Consider a system with multiple inputs and multiple outputs.When the system has more than one output, the MATLAB command [NUM,den] = ss2tf(A,B,C,D,iu) produces transfer functions for all outputs to each input. (The numerator coefficients are returned to matrix NUM with as many rows as there are outputs.) Consider the system defined by This system involves two inputs and two outputs.Four transfer functions are involved: and (When considering input u1, we assume that input u2 is zero and vice versa.) Solution. MATLAB Program 2-5 produces four transfer functions.
This is the MATLAB representation of the following four transfer functions: Y 2(s) U 2(s) = s - 25 s2 + 4s + 25 Y 1(s) U 2(s) = s + 5 s2 + 4s + 25 , Y 2(s) U 1(s) = -25 s2 + 4s + 25 Y 1(s) U 1(s) = s + 4 s2 + 4s + 25 , Y 2(s)U 2(s).
Y 1(s)U 2(s), Y 2(s)U 1(s), Y 1(s)U 1(s), B y1 y2 R = B 1 0 0 1R B x1 x2 R + B 0 0 0 0R B u1 u2 R Bx # 1 x # 2 R = B 0 -25 1 -4R B x1 x2 R + B 1 0 1 1R Bu1 u2 R = 1 (s + 1)2(s + 2) = 1 s3 + 4s2 + 5s + 2 = [1 0 0]F 1 s + 1 0 0 1 (s + 1)2 1 s + 1 0 1 (s + 1)2(s + 2) 1 (s + 1)(s + 2) 1 s + 2 V C 0 0 1 S G(s) = [1 0 0]C s + 1 0 0 -1 s + 1 0 0 -1 s + 2 S -1 C 0 0 1 S Example Problems and Solutions 59 A–2–13.
Linearize the nonlinear equation in the region defined by 8 x 10, 2 y 4.
Solution. Define Then where we choose Since the higher-order terms in the expanded equation are small, neglecting these higher-order terms, we obtain where z – = x –2 + 4x –y – + 6y –2 = 92 + 4 9 3 + 6 9 = 243 K 2 = 0f 0y 2 x=x –, y=y – = 4x – + 12y – = 4 9 + 12 3 = 72 K 1 = 0f 0x 2 x=x –, y=y – = 2x – + 4y – = 2 9 + 4 3 = 30 z - z – = K 1(x - x –) + K 2(y - y –) x – = 9, y – = 3.
z = f(x, y) = f(x –, y –) + c 0f 0x (x - x –) + 0f 0y (y - y –)d x=x –, y=y – + p f(x, y) = z = x2 + 4xy + 6y2 z = x2 + 4xy + 6y2 MATLAB Program 2–5 A = [0 1;-25 -4]; B = [1 1;0 1]; C = [1 0;0 1]; D = [0 0;0 0]; [NUM,den] = ss2tf(A,B,C,D,1) NUM = 0 1 4 0 0 –25 den = 1 4 25 [NUM,den] = ss2tf(A,B,C,D,2) NUM = 0 1.0000 5.0000 0 1.0000 -25.0000 den = 1 4 25 60 Chapter 2 / Mathematical Modeling of Control Systems Thus Hence a linear approximation of the given nonlinear equation near the operating point is z - 30x - 72y + 243 = 0 z - 243 = 30(x - 9) + 72(y - 3) R(s) C(s) G1 G2 G3 H1 H2 H3 + – + – + – ++ Figure 2–31 Block diagram of a system.
B–2–1. Simplify the block diagram shown in Figure 2–29 and obtain the closed-loop transfer function C(s)/R(s).
B–2–2. Simplify the block diagram shown in Figure 2–30 and obtain the closed-loop transfer function C(s)/R(s).
B–2–3. Simplify the block diagram shown in Figure 2–31 and obtain the closed-loop transfer function C(s)/R(s).
PROBLEMS R(s) C(s) G1 G2 G3 G4 + – + – + + Figure 2–29 Block diagram of a system.
R(s) C(s) G1 G2 H1 H2 + – ++ + – Figure 2–30 Block diagram of a system.
Problems 61 C(s) D(s) R(s) Gc(s) Gp(s) +– ++ Controller Plant Figure 2–32 Closed-loop system.
B–2–4. Consider industrial automatic controllers whose control actions are proportional,integral,proportional-plus-integral,proportional-plus-derivative,and proportional-plus-integral-plus-derivative. The transfer functions of these controllers can be given, respectively, by where U(s) is the Laplace transform of u(t), the controller output, and E(s) the Laplace transform of e(t), the actuat- U(s) E(s) = K p a1 + 1 T i s + T d s b U(s) E(s) = K pA1 + T d sB U(s) E(s) = K p a1 + 1 T i s b U(s) E(s) = K i s U(s) E(s) = K p ing error signal. Sketch u(t)-versus-t curves for each of the five types of controllers when the actuating error signal is (a) e(t)=unit-step function (b) e(t)=unit-ramp function In sketching curves, assume that the numerical values of Kp, Ki, and are given as proportional gain=4 integral gain=2 integral time=2 sec derivative time=0.8 sec B–2–5. Figure 2–32 shows a closed-loop system with a ref-erence input and disturbance input. Obtain the expression for the output C(s) when both the reference input and dis-turbance input are present.
B–2–6. Consider the system shown in Figure 2–33. Derive the expression for the steady-state error when both the ref-erence input R(s) and disturbance input D(s) are present.
B–2–7. Obtain the transfer functions C(s)/R(s) and C(s)/D(s) of the system shown in Figure 2–34. T d = T i = K i = K p = T d T i , C(s) R(s) E(s) D(s) +– ++ G2(s) G1(s) G2 H1 G3 G1 Gc R(s) C(s) D(s) +– +– ++ H2 Figure 2–33 Control system.
Figure 2–34 Control system.
62 Chapter 2 / Mathematical Modeling of Control Systems B–2–8. Obtain a state-space representation of the system shown in Figure 2–35.
B–2–9. Consider the system described by Derive a state-space representation of the system.
B–2–10. Consider the system described by Obtain the transfer function of the system. y = [1 0]B x1 x2 R Bx # 1 x # 2 R = B-4 3 -1 -1R Bx1 x2 R + B 1 1R u y % + 3y $ + 2y # = u u y s + z s + p 1 s2 + – Figure 2–35 Control system.
B–2–11. Consider a system defined by the following state-space equations: Obtain the transfer function G(s) of the system.
B–2–12. Obtain the transfer matrix of the system defined by B–2–13. Linearize the nonlinear equation z=x2+8xy+3y2 in the region defined by 2 x 4, 10 y 12.
B–2–14. Find a linearized equation for y=0.2x3 about a point x=2. B y1 y2 R = B1 0 0 1 0 0R C x1 x2 x3 S C x # 1 x # 2 x # 3 S = C 0 0 -2 1 0 -4 0 1 -6 S C x1 x2 x3 S + C 0 0 1 0 1 0 S Bu1 u2 R y = [1 2]B x1 x2 R B x # 1 x # 2 R = B -5 3 -1 -1R Bx1 x2 R + B 2 5R u 3 63 Mathematical Modeling of Mechanical Systems and Electrical Systems 3–1 INTRODUCTION This chapter presents mathematical modeling of mechanical systems and electrical systems. In Chapter 2 we obtained mathematical models of a simple electrical circuit and a simple mechanical system. In this chapter we consider mathematical modeling of a variety of mechanical systems and electrical systems that may appear in control systems.
The fundamental law govering mechanical systems is Newton’s second law. In Section 3–2 we apply this law to various mechanical systems and derive transfer-function models and state-space models.
The basic laws governing electrical circuits are Kirchhoff’s laws. In Section 3–3 we obtain transfer-function models and state-space models of various electrical circuits and operational amplifier systems that may appear in many control systems.
3–2 MATHEMATICAL MODELING OF MECHANICAL SYSTEMS This section first discusses simple spring systems and simple damper systems. Then we derive transfer-function models and state-space models of various mechanical systems.
64 Chapter 3 / Mathematical Modeling of Mechanical Systems and Electrical Systems EXAMPLE 3–1 Let us obtain the equivalent spring constants for the systems shown in Figures 3–1(a) and (b), respectively.
For the springs in parallel [Figure 3–1(a)] the equivalent spring constant keq is obtained from or For the springs in series [Figure–3–1(b)], the force in each spring is the same.Thus Elimination of y from these two equations results in or The equivalent spring constant keq for this case is then found as EXAMPLE 3–2 Let us obtain the equivalent viscous-friction coefficient for each of the damper systems shown in Figures 3–2(a) and (b).An oil-filled damper is often called a dashpot.A dashpot is a device that provides viscous friction,or damping.It consists of a piston and oil-filled cylinder.Any relative mo-tion between the piston rod and the cylinder is resisted by the oil because the oil must flow around the piston (or through orifices provided in the piston) from one side of the piston to the other.The dashpot essentially absorbs energy.This absorbed energy is dissipated as heat,and the dashpot does not store any kinetic or potential energy.
beq keq = F x = k1 k2 k1 + k2 = 1 1 k1 + 1 k2 k2 x = F + k2 k1 F = k1 + k2 k1 F k2 ax - F k1 b = F k1 y = F, k2(x - y) = F keq = k1 + k2 k1 x + k2 x = F = keq x k1 k2 y x F (a) (b) x F k1 k2 Figure 3–1 (a) System consisting of two springs in parallel; (b) system consisting of two springs in series.
Section 3–2 / Mathematical Modeling of Mechanical Systems 65 (a) The force f due to the dampers is In terms of the equivalent viscous-friction coefficient beq, force f is given by Hence (b) The force f due to the dampers is (3–1) where z is the displacement of a point between damper b1 and damper b2. (Note that the same force is transmitted through the shaft.) From Equation (3–1), we have or (3–2) In terms of the equivalent viscous-friction coefficient beq, force f is given by By substituting Equation (3–2) into Equation (3–1), we have Thus, Hence, beq = b1 b2 b1 + b2 = 1 1 b1 + 1 b2 f = beq(y # - x # ) = b1 b2 b1 + b2 (y # - x # ) = b1 b2 b1 + b2 (y # - x # ) f = b2(y # - z # ) = b2cy # - 1 b1 + b2 Ab2y # + b1x # B d f = beqAy # - x # B z # = 1 b1 + b2 Ab2 y # + b1 x # B Ab1 + b2Bz # = b2 y # + b1 x # f = b1(z # - x # ) = b2 (y # - z # ) beq = b1 + b2 f = beq(y # - x # ) f = b1 (y # - x # ) + b2(y # - x # ) = Ab1 + b2B(y # - x # ) x y (a) b2 x y z (b) b1 b1 b2 Figure 3–2 (a) Two dampers connected in parallel; (b) two dampers connected in series.
66 Chapter 3 / Mathematical Modeling of Mechanical Systems and Electrical Systems EXAMPLE 3–3 Consider the spring-mass-dashpot system mounted on a massless cart as shown in Figure 3–3. Let us obtain mathematical models of this system by assuming that the cart is standing still for t<0 and the spring-mass-dashpot system on the cart is also standing still for t<0. In this system, u(t) is the displacement of the cart and is the input to the system.At t=0, the cart is moved at a constant speed, or constant.The displacement y(t) of the mass is the output. (The displacement is relative to the ground.) In this system,m denotes the mass,b denotes the viscous-friction coefficient,and k de-notes the spring constant.We assume that the friction force of the dashpot is proportional to and that the spring is a linear spring; that is, the spring force is proportional to y-u.
For translational systems, Newton’s second law states that where m is a mass, a is the acceleration of the mass, and is the sum of the forces acting on the mass in the direction of the acceleration a. Applying Newton’s second law to the present system and noting that the cart is massless, we obtain or This equation represents a mathematical model of the system considered. Taking the Laplace transform of this last equation, assuming zero initial condition, gives Taking the ratio of Y(s) to U(s), we find the transfer function of the system to be Such a transfer-function representation of a mathematical model is used very frequently in control engineering.
Transfer function = G(s) = Y(s) U(s) = bs + k ms2 + bs + k Ams2 + bs + kBY(s) = (bs + k)U(s) m d2y dt2 + b dy dt + ky = b du dt + ku m d2y dt2 = -b a dy dt - du dt b - k(y - u) gF ma = a F y # - u # u # = m u y k b Massless cart Figure 3–3 Spring-mass-dashpot system mounted on a cart.
Section 3–2 / Mathematical Modeling of Mechanical Systems 67 Next we shall obtain a state-space model of this system. We shall first compare the differen-tial equation for this system with the standard form and identify a1, a2, b0, b1, and b2 as follows: Referring to Equation (3–35), we have Then, referring to Equation (2–34), define From Equation (2–36) we have and the output equation becomes or (3–3) and (3–4) Equations (3–3) and (3–4) give a state-space representation of the system. (Note that this is not the only state-space representation.There are infinitely many state-space representations for the system.) y = [1 0]B x1 x2 R Bx # 1 x # 2 R = C 0 - k m 1 - b m S B x1 x2 R + D b m k m - a b m b 2T u y = x1 x # 2 = -a2 x1 - a1 x2 + b2 u = - k m x1 - b m x2 + c k m - a b m b 2 du x # 1 = x2 + b1 u = x2 + b m u x2 = x # 1 - b1 u = x # 1 - b m u x1 = y - b0 u = y b2 = b2 - a1 b1 - a2 b0 = k m - a b m b 2 b1 = b1 - a1 b0 = b m b0 = b0 = 0 a1 = b m , a2 = k m , b0 = 0, b1 = b m , b2 = k m y $ + a1 y # + a2 y = b0 u $ + b1 u # + b2 u y $ + b m y # + k m y = b m u # + k m u 68 Chapter 3 / Mathematical Modeling of Mechanical Systems and Electrical Systems EXAMPLE 3–4 Obtain the transfer functions and of the mechanical system shown in Figure 3–4.
The equations of motion for the system shown in Figure 3–4 are Simplifying, we obtain Taking the Laplace transforms of these two equations, assuming zero initial conditions, we obtain (3–5) (3–6) Solving Equation (3–6) for and substituting it into Equation (3–5) and simplifying, we get from which we obtain (3–7) From Equations (3–6) and (3–7) we have (3–8) Equations (3–7) and (3–8) are the transfer functions and respectively.
EXAMPLE 3–5 An inverted pendulum mounted on a motor-driven cart is shown in Figure 3–5(a).This is a model of the attitude control of a space booster on takeoff. (The objective of the attitude control prob-lem is to keep the space booster in a vertical position.) The inverted pendulum is unstable in that it may fall over any time in any direction unless a suitable control force is applied.Here we consider X 2(s)U(s), X 1(s)U(s) X 2(s) U(s) = bs + k2 Am1 s2 + bs + k1 + k2BAm2 s2 + bs + k2 + k3B - Abs + k2B 2 X 1(s) U(s) = m2 s2 + bs + k2 + k3 Am1 s2 + bs + k1 + k2BAm2 s2 + bs + k2 + k3B - Abs + k2B 2 = Am2 s2 + bs + k2 + k3BU(s) CAm1 s2 + bs + k1 + k2BAm2 s2 + bs + k2 + k3B - Abs + k2B 2DX 1(s) X 2(s) Cm2 s2 + bs + Ak2 + k3B DX 2(s) = Abs + k2BX 1(s) Cm1 s2 + bs + Ak1 + k2B DX 1(s) = Abs + k2BX 2(s) + U(s) m2 x $ 2 + bx # 2 + Ak2 + k3Bx2 = bx # 1 + k2 x1 m1 x $ 1 + bx # 1 + Ak1 + k2Bx1 = bx # 2 + k2 x2 + u m2 x $ 2 = -k3 x2 - k2Ax2 - x1B - bAx # 2 - x # 1B m1 x $ 1 = -k1 x1 - k2Ax1 - x2B - bAx # 1 - x # 2B + u X 2(s)U(s) X 1(s)U(s) m1 m2 k2 x1 k1 k3 b u x2 Figure 3–4 Mechanical system.
only a two-dimensional problem in which the pendulum moves only in the plane of the page.The control force u is applied to the cart.Assume that the center of gravity of the pendulum rod is at its geometric center. Obtain a mathematical model for the system.
Define the angle of the rod from the vertical line as u. Define also the (x, y) coordinates of the center of gravity of the pendulum rod as AxG, yGB.Then yG = l cosu xG = x + l sinu Section 3–2 / Mathematical Modeling of Mechanical Systems 69 M P y x u O x (a) mg cos u u (b) u V V H H M y x u O x mg Figure 3–5 (a) Inverted pendulum system; (b) free-body diagram.
70 Chapter 3 / Mathematical Modeling of Mechanical Systems and Electrical Systems To derive the equations of motion for the system, consider the free-body diagram shown in Figure 3–5(b). The rotational motion of the pendulum rod about its center of gravity can be described by (3–9) where I is the moment of inertia of the rod about its center of gravity.
The horizontal motion of center of gravity of pendulum rod is given by (3–10) The vertical motion of center of gravity of pendulum rod is (3–11) The horizontal motion of cart is described by (3–12) Since we must keep the inverted pendulum vertical, we can assume that u(t) and are small quantities such that sinu u, cosu=1, and Then, Equations (3–9) through (3–11) can be linearized.The linearized equations are (3–13) (3–14) (3–15) From Equations (3–12) and (3–14), we obtain (3–16) From Equations (3–13), (3–14), and (3–15), we have or (3–17) Equations (3–16) and (3–17) describe the motion of the inverted-pendulum-on-the-cart system.
They constitute a mathematical model of the system.
EXAMPLE 3–6 Consider the inverted-pendulum system shown in Figure 3–6. Since in this system the mass is con-centrated at the top of the rod, the center of gravity is the center of the pendulum ball. For this case, the moment of inertia of the pendulum about its center of gravity is small, and we assume I=0 in Equation (3–17).Then the mathematical model for this system becomes as follows: (3–18) (3–19) Equations (3–18) and (3–19) can be modified to (3–20) (3–21) Mx $ = u - mgu Mlu $ = (M + m)gu - u ml2u $ + mlx $ = mglu (M + m)x $ + mlu $ = u AI + ml2Bu $ + mlx $ = mglu = mglu - l(mx $ + mlu $ ) Iu $ = mglu - Hl (M + m)x $ + mlu $ = u 0 = V - mg m(x $ + lu $ ) = H Iu $ = Vlu - Hl uu # 2 = 0.
u # (t) M d2x dt2 = u - H m d2 dt2 (lcosu) = V - mg m d2 dt2 (x + l sin u) = H Iu $ = Vl sinu - Hl cosu Section 3–2 / Mathematical Modeling of Mechanical Systems 71 Equation (3–20) was obtained by eliminating from Equations (3–18) and (3–19). Equation (3–21) was obtained by eliminating from Equations (3–18) and (3–19). From Equation (3–20) we obtain the plant transfer function to be The inverted-pendulum plant has one pole on the negative real axis and another on the positive real axis Hence,the plant is open-loop unstable.
Define state variables x1, x2, x3, and x4 by Note that angle u indicates the rotation of the pendulum rod about point P, and x is the location of the cart. If we consider u and x as the outputs of the system, then (Notice that both u and x are easily measurable quantities.) Then, from the definition of the state variables and Equations (3–20) and (3–21), we obtain x # 4 = - m M gx1 + 1 M u x # 3 = x4 x # 2 = M + m Ml gx1 -1 Ml u x # 1 = x2 y = B y1 y2 R = B u xR = B x1 x3 R x4 = x # x3 = x x2 = u # x1 = u Cs = A1M + m1MlB 1gD.
Cs = -A1M + m1MlB 1gD = 1 Ml as + A M + m Ml g b as - A M + m Ml g b Q (s) -U(s) = 1 Mls2 - (M + m)g u $ x $ 0 M P z u mg m sin u x x cos u u Figure 3–6 Inverted-pendulum system.
72 Chapter 3 / Mathematical Modeling of Mechanical Systems and Electrical Systems In terms of vector-matrix equations, we have (3–22) (3–23) Equations (3–22) and (3–23) give a state-space representation of the inverted-pendulum system.
(Note that state-space representation of the system is not unique.There are infinitely many such representations for this system.) B y1 y2 R = B 1 0 0 0 0 0 1 0R D x1 x2 x3 x4 T D x # 1 x # 2 x # 3 x # 4 T = F 0 M + m Ml g 0 - m M g 1 0 0 0 0 0 0 0 0 0 1 0 V D x1 x2 x3 x4 T + F 0 - 1 Ml 0 1 M V u 3–3 MATHEMATICAL MODELING OF ELECTRICAL SYSTEMS Basic laws governing electrical circuits are Kirchhoff’s current law and voltage law.
Kirchhoff’s current law (node law) states that the algebraic sum of all currents entering and leaving a node is zero. (This law can also be stated as follows:The sum of currents enter-ing a node is equal to the sum of currents leaving the same node.) Kirchhoff’s voltage law (loop law) states that at any given instant the algebraic sum of the voltages around any loop in an electrical circuit is zero. (This law can also be stated as follows:The sum of the volt-age drops is equal to the sum of the voltage rises around a loop.) A mathematical model of an electrical circuit can be obtained by applying one or both of Kirchhoff’s laws to it.
This section first deals with simple electrical circuits and then treats mathematical modeling of operational amplifier systems.
LRC Circuit.
Consider the electrical circuit shown in Figure 3–7. The circuit con-sists of an inductance L (henry), a resistance R (ohm), and a capacitance C (farad).
Applying Kirchhoff’s voltage law to the system, we obtain the following equations: (3–24) (3–25) 1 C 3 i dt = eo L di dt + Ri + 1 C 3 i dt = ei L eo R C ei i Figure 3–7 Electrical circuit.
Section 3–3 / Mathematical Modeling of Electrical Systems 73 Equations (3–24) and (3–25) give a mathematical model of the circuit.
A transfer-function model of the circuit can also be obtained as follows:Taking the Laplace transforms of Equations (3–24) and (3–25), assuming zero initial conditions, we obtain If ei is assumed to be the input and eo the output,then the transfer function of this system is found to be (3–26) A state-space model of the system shown in Figure 3–7 may be obtained as follows: First, note that the differential equation for the system can be obtained from Equation (3–26) as Then by defining state variables by and the input and output variables by we obtain and These two equations give a mathematical model of the system in state space.
Transfer Functions of Cascaded Elements.
Many feedback systems have com-ponents that load each other. Consider the system shown in Figure 3–8.Assume that ei is the input and eo is the output. The capacitances C1 and C2 are not charged initially.
y = [1 0]B x1 x2 R Bx # 1 x # 2 R = C 0 -1 LC 1 - R L S B x1 x2 R + C 0 1 LC S u y = eo = x1 u = ei x2 = e # o x1 = eo e $ o + R L e # o + 1 LC eo = 1 LC ei Eo(s) Ei(s) = 1 LCs2 + RCs + 1 1 C 1 s I(s) = Eo(s) LsI(s) + RI(s) + 1 C 1 s I(s) = Ei(s) R1 C1 eo R2 C2 ei i1 i2 Figure 3–8 Electrical system.
74 Chapter 3 / Mathematical Modeling of Mechanical Systems and Electrical Systems It will be shown that the second stage of the circuit (R2C2 portion) produces a loading effect on the first stage (R1C1 portion).The equations for this system are (3–27) and (3–28) (3–29) Taking the Laplace transforms of Equations (3–27) through (3–29), respectively, using zero initial conditions, we obtain (3–30) (3–31) (3–32) Eliminating I1(s) from Equations (3–30) and (3–31) and writing Ei(s) in terms of I2(s), we find the transfer function between Eo(s) and Ei(s) to be (3–33) The term R1C2s in the denominator of the transfer function represents the interaction of two simple RC circuits. Since the two roots of the denominator of Equation (3–33) are real.
The present analysis shows that, if two RC circuits are connected in cascade so that the output from the first circuit is the input to the second, the overall transfer function is not the product of and The reason for this is that, when we derive the transfer function for an isolated circuit, we implicitly as-sume that the output is unloaded. In other words, the load impedance is assumed to be infinite, which means that no power is being withdrawn at the output.When the sec-ond circuit is connected to the output of the first, however, a certain amount of power is withdrawn, and thus the assumption of no loading is violated.Therefore, if the trans-fer function of this system is obtained under the assumption of no loading, then it is not valid. The degree of the loading effect determines the amount of modification of the transfer function.
1AR2 C2 s + 1B.
1AR1 C1 s + 1B AR1 C1 + R2 C2 + R1 C2B 2 7 4R1 C1 R2 C2 , = 1 R1 C1 R2 C2 s2 + AR1 C1 + R2 C2 + R1 C2Bs + 1 Eo(s) Ei(s) = 1 AR1 C1 s + 1BAR2 C2 s + 1B + R1 C2 s 1 C2 s I 2(s) = Eo(s) 1 C1 sCI 2(s) - I 1(s)D + R2 I 2(s) + 1 C2 s I 2(s) = 0 1 C1 sCI 1(s) - I 2(s)D + R1 I 1(s) = Ei(s) 1 C2 3 i2 dt = eo 1 C1 3 Ai2 - i1B dt + R2 i2 + 1 C2 3 i2 dt = 0 1 C1 3 Ai1 - i2B dt + R1 i1 = ei Section 3–3 / Mathematical Modeling of Electrical Systems 75 Complex Impedances.
In deriving transfer functions for electrical circuits, we frequently find it convenient to write the Laplace-transformed equations directly, without writing the differential equations. Consider the system shown in Figure 3–9(a).
In this system, Z1 and Z2 represent complex impedances. The complex impedance Z(s) of a two-terminal circuit is the ratio of E(s), the Laplace transform of the voltage across the terminals, to I(s), the Laplace transform of the current through the element, under the assumption that the initial conditions are zero, so that Z(s)=E(s)/I(s). If the two-terminal element is a resistance R, capacitance C, or inductance L, then the complex impedance is given by R, 1/Cs, or Ls, respectively. If complex impedances are connected in series, the total impedance is the sum of the individual complex impedances.
Remember that the impedance approach is valid only if the initial conditions involved are all zeros. Since the transfer function requires zero initial conditions, the impedance approach can be applied to obtain the transfer function of the electrical circuit. This approach greatly simplifies the derivation of transfer functions of elec-trical circuits.
Consider the circuit shown in Figure 3–9(b).Assume that the voltages ei and eo are the input and output of the circuit, respectively. Then the transfer function of this circuit is For the system shown in Figure 3–7, Hence the transfer function Eo(s)/Ei(s) can be found as follows: which is, of course, identical to Equation (3–26).
Eo(s) Ei(s) = 1 Cs Ls + R + 1 Cs = 1 LCs2 + RCs + 1 Z1 = Ls + R, Z2 = 1 Cs Eo(s) Ei(s) = Z2(s) Z1(s) + Z2(s) i i i e2 e e1 eo ei Z1 Z1 Z2 Z2 (a) (b) Figure 3–9 Electrical circuits.
76 Chapter 3 / Mathematical Modeling of Mechanical Systems and Electrical Systems EXAMPLE 3–7 Consider again the system shown in Figure 3–8. Obtain the transfer function Eo(s)/Ei(s) by use of the complex impedance approach. (Capacitors C1 and C2 are not charged initially.) The circuit shown in Figure 3–8 can be redrawn as that shown in Figure 3–10(a), which can be further modified to Figure 3–10(b).
In the system shown in Figure 3–10(b) the current I is divided into two currents I1 and I2.
Noting that we obtain Noting that we obtain Substituting Z1=R1, Z2=1/AC1sB, Z3=R2, and Z4=1/AC2sB into this last equation, we get which is the same as that given by Equation (3–33). = 1 R1 C1 R2 C2 s2 + AR1 C1 + R2 C2 + R1 C2Bs + 1 Eo(s) Ei(s) = 1 C1 s 1 C2 s R1 a 1 C1 s + R2 + 1 C2 s b + 1 C1 s aR2 + 1 C2 s b Eo(s) Ei(s) = Z2 Z4 Z1AZ2 + Z3 + Z4B + Z2AZ3 + Z4B Eo(s) = Z4 I 2 = Z2 Z4 Z2 + Z3 + Z4 I Ei(s) = Z1 I + Z2 I 1 = cZ1 + Z2AZ3 + Z4B Z2 + Z3 + Z4 dI I 1 = Z3 + Z4 Z2 + Z3 + Z4 I, I 2 = Z2 Z2 + Z3 + Z4 I Z2 I 1 = AZ3 + Z4BI 2 , I 1 + I 2 = I Z1 Z3 Z2 Z4 Z1 I2 I1 Z2 Z3 Z4 I Ei(s) Eo(s) Eo(s) Ei(s) (a) (b) Figure 3–10 (a) The circuit of Figure 3–8 shown in terms of impedances; (b) equivalent circuit diagram.
Section 3–3 / Mathematical Modeling of Electrical Systems 77 Transfer Functions of Nonloading Cascaded Elements.
The transfer function of a system consisting of two nonloading cascaded elements can be obtained by elimi-nating the intermediate input and output. For example, consider the system shown in Figure 3–11(a).The transfer functions of the elements are and If the input impedance of the second element is infinite, the output of the first element is not affected by connecting it to the second element.Then the transfer function of the whole system becomes The transfer function of the whole system is thus the product of the transfer functions of the individual elements.This is shown in Figure 3–11(b).
As an example,consider the system shown in Figure 3–12.The insertion of an isolating amplifier between the circuits to obtain nonloading characteristics is frequently used in combining circuits. Since amplifiers have very high input impedances, an isolation amplifier inserted between the two circuits justifies the nonloading assumption.
The two simple RC circuits, isolated by an amplifier as shown in Figure 3–12, have negligible loading effects, and the transfer function of the entire circuit equals the prod-uct of the individual transfer functions.Thus, in this case, Electronic Controllers.
In what follows we shall discuss electronic controllers using operational amplifiers.We begin by deriving the transfer functions of simple operational-amplifier circuits.Then we derive the transfer functions of some of the operational-amplifier controllers.Finally,we give operational-amplifier controllers and their transfer functions in the form of a table.
= K AR1 C1 s + 1BAR2 C2 s + 1B Eo(s) Ei(s) = a 1 R1 C1 s + 1 b(K) a 1 R2 C2 s + 1 b G(s) = X 3(s) X 1(s) = X 2(s)X 3(s) X 1(s)X 2(s) = G1(s)G2(s) G2(s) = X 3(s) X 2(s) G1(s) = X 2(s) X 1(s) X1(s) G1(s) X2(s) X3(s) G2(s) (a) (b) X3(s) X1(s) G1(s) G2(s) Figure 3–11 (a) System consisting of two nonloading cascaded elements; (b) an equivalent system.
R1 C1 eo R2 C2 ei Isolating amplifier (gain K) Figure 3–12 Electrical system.
78 Chapter 3 / Mathematical Modeling of Mechanical Systems and Electrical Systems Operational Amplifiers.
Operational amplifiers, often called op amps, are frequently used to amplify signals in sensor circuits. Op amps are also frequently used in filters used for compensation purposes. Figure 3–13 shows an op amp. It is a common practice to choose the ground as 0 volt and measure the input voltages e1 and e2 relative to the ground. The input e1 to the minus terminal of the amplifier is inverted, and the input e2 to the plus terminal is not inverted.The total input to the amplifier thus becomes e2-e1. Hence, for the circuit shown in Figure 3–13, we have where the inputs e1 and e2 may be dc or ac signals and K is the differential gain (volt-age gain).The magnitude of K is approximately 105 ~ 106 for dc signals and ac signals with frequencies less than approximately 10 Hz. (The differential gain K decreases with the signal frequency and becomes about unity for frequencies of 1 MHz ~ 50 MHz.) Note that the op amp amplifies the difference in voltages e1 and e2. Such an amplifier is commonly called a differential amplifier. Since the gain of the op amp is very high, it is necessary to have a negative feedback from the output to the input to make the ampli-fier stable. (The feedback is made from the output to the inverted input so that the feed-back is a negative feedback.) In the ideal op amp, no current flows into the input terminals, and the output volt-age is not affected by the load connected to the output terminal. In other words, the input impedance is infinity and the output impedance is zero. In an actual op amp, a very small (almost negligible) current flows into an input terminal and the output can-not be loaded too much. In our analysis here, we make the assumption that the op amps are ideal.
Inverting Amplifier.
Consider the operational-amplifier circuit shown in Figure 3–14.
Let us obtain the output voltage eo.
eo = KAe2 - e1B = -KAe1 - e2B ei eo R2 i2 R1 i1 + – e9 Figure 3–14 Inverting amplifier.
e2 e1 eo + – Figure 3–13 Operational amplifier.
Section 3–3 / Mathematical Modeling of Electrical Systems 79 The equation for this circuit can be obtained as follows: Define Since only a negligible current flows into the amplifier, the current i1 must be equal to current i2. Thus Since and e¿ must be almost zero, or Hence we have or Thus the circuit shown is an inverting amplifier. If R1=R2, then the op-amp circuit shown acts as a sign inverter.
Noninverting Amplifier.
Figure 3–15(a) shows a noninverting amplifier.A circuit equivalent to this one is shown in Figure 3–15(b). For the circuit of Figure 3–15(b), we have where K is the differential gain of the amplifier. From this last equation, we get Since if then This equation gives the output voltage eo. Since eo and ei have the same signs,the op-amp circuit shown in Figure 3–15(a) is noninverting.
eo = a1 + R2 R1 bei R1AR1 + R2B 1K, K 1, ei = a R1 R1 + R2 + 1 K beo eo = K aei -R1 R1 + R2 eo b eo = - R2 R1 ei ei R1 = -eo R2 e¿ 0.
K 1, K(0 - e¿) = e0 ei - e¿ R1 = e¿ - eo R2 i1 = ei - e¿ R1 , i2 = e¿ - eo R2 eo ei R2 R1 + – eo ei R2 R1 – + (b) (a) Figure 3–15 (a) Noninverting operational amplifier; (b) equivalent circuit.
80 Chapter 3 / Mathematical Modeling of Mechanical Systems and Electrical Systems EXAMPLE 3–8 Figure 3–16 shows an electrical circuit involving an operational amplifier. Obtain the output eo.
Let us define Noting that the current flowing into the amplifier is negligible, we have Hence Since we have Taking the Laplace transform of this last equation, assuming the zero initial condition, we have which can be written as The op-amp circuit shown in Figure 3–16 is a first-order lag circuit.(Several other circuits involving op amps are shown in Table 3–1 together with their transfer functions. Table 3–1 is given on page 85.) Eo(s) Ei(s) = - R2 R1 1 R2 Cs + 1 Ei(s) R1 = - R2 Cs + 1 R2 Eo(s) ei R1 = -C deo dt - eo R2 e¿ 0, ei - e¿ R1 = C dAe¿ - eoB dt + e¿ - eo R2 i1 = i2 + i3 i1 = ei - e¿ R1 , i2 = C dAe¿ - eoB dt , i3 = e¿ - eo R2 ei eo R2 R1 C i1 i3 i2 + – e9 Figure 3–16 First-order lag circuit using operational amplifier.
Section 3–3 / Mathematical Modeling of Electrical Systems 81 Impedance Approach to Obtaining Transfer Functions.
Consider the op-amp circuit shown in Figure 3–17. Similar to the case of electrical circuits we discussed ear-lier, the impedance approach can be applied to op-amp circuits to obtain their transfer functions. For the circuit shown in Figure 3–17, we have Since we have (3–34) Eo(s) Ei(s) = - Z2(s) Z1(s) E¿(s) 0, Ei(s) - E¿(s) Z1 = E¿(s) - Eo(s) Z2 + – Eo(s) I(s) I(s) Ei(s) E9(s) Z1(s) Z2(s) Figure 3–17 Operational-amplifier circuit.
EXAMPLE 3–9 Referring to the op-amp circuit shown in Figure 3–16, obtain the transfer function Eo(s)/Ei(s) by use of the impedance approach.
The complex impedances Z1(s) and Z2(s) for this circuit are and The transfer function Eo(s)/Ei(s) is, therefore, obtained as which is, of course, the same as that obtained in Example 3-8.
Eo(s) Ei(s) = -Z2(s) Z1(s) = - R2 R1 1 R2 Cs + 1 Z2(s) = 1 Cs + 1 R2 = R2 R2 Cs + 1 Z1(s) = R1 82 Chapter 3 / Mathematical Modeling of Mechanical Systems and Electrical Systems Lead or Lag Networks Using Operational Amplifiers.
Figure 3–18(a) shows an electronic circuit using an operational amplifier. The transfer function for this circuit can be obtained as follows: Define the input impedance and feedback impedance as Z1 and Z2, respectively.Then Hence, referring to Equation (3–34), we have (3–35) Notice that the transfer function in Equation (3–35) contains a minus sign.Thus,this circuit is sign inverting. If such a sign inversion is not convenient in the actual application, a sign inverter may be connected to either the input or the output of the circuit of Figure 3–18(a).
An example is shown in Figure 3–18(b).The sign inverter has the transfer function of The sign inverter has the gain of Hence the network shown in Figure 3–18(b) has the following transfer function: (3–36) = K c a Ts + 1 aTs + 1 = K c s + 1 T s + 1 aT Eo(s) Ei(s) = R2 R4 R1 R3 R1 C1s + 1 R2 C2s + 1 = R4 C1 R3 C2 s + 1 R1 C1 s + 1 R2 C2 -R4R3 .
Eo(s) E(s) = - R4 R3 E(s) Ei(s) = - Z2 Z1 = - R2 R1 R1 C1s + 1 R2 C2s + 1 = - C1 C2 s + 1 R1 C1 s + 1 R2 C2 Z1 = R1 R1 C1 s + 1 , Z2 = R2 R2 C2 s + 1 + – + – + – (a) (b) Z1 C1 Z2 C2 R2 i2 i1 R1 Ei(s) E9(s) E(s) C1 C2 Ei(s) Eo(s) E(s) R1 R2 R3 R4 Lead or lag network Sign inverter Figure 3–18 (a) Operational-amplifier circuit; (b) operational-amplifier circuit used as a lead or lag compensator.
where Notice that This network has a dc gain of Note that this network, whose transfer function is given by Equation (3–36), is a lead network if or a<1. It is a lag network if PID Controller Using Operational Amplifiers.
Figure 3–19 shows an electronic proportional-plus-integral-plus-derivative controller (a PID controller) using opera-tional amplifiers.The transfer function is given by where Thus Noting that Eo(s) E(s) = - R4 R3 E(s) Ei(s) = - a R2 C2 s + 1 C2 s b a R1 C1 s + 1 R1 b Z1 = R1 R1 C1s + 1 , Z2 = R2 C2s + 1 C2s E(s) Ei(s) = - Z2 Z1 E(s)Ei(s) R1 C1 6 R2 C2 .
R1 C1 7 R2 C2 , K c a = R2 R4AR1 R3B.
K c a = R4 C1 R3 C2 R2 C2 R1 C1 = R2 R4 R1 R3 , a = R2 C2 R1 C1 T = R1 C1 , aT = R2 C2 , K c = R4 C1 R3 C2 Section 3–3 / Mathematical Modeling of Electrical Systems 83 + + – – Z1 C1 Z2 C2 R2 R1 Ei(s) Eo(s) E(s) R3 R4 Figure 3–19 Electronic PID controller.
84 Chapter 3 / Mathematical Modeling of Mechanical Systems and Electrical Systems we have (3–37) Notice that the second operational-amplifier circuit acts as a sign inverter as well as a gain adjuster.
When a PID controller is expressed as Kp is called the proportional gain, is called the integral time, and is called the derivative time. From Equation (3–37) we obtain the proportional gain Kp, integral time and derivative time to be When a PID controller is expressed as Kp is called the proportional gain, Ki is called the integral gain, and Kd is called the derivative gain. For this controller Table 3–1 shows a list of operational-amplifier circuits that may be used as con-trollers or compensators.
K d = R4 R2 C1 R3 K i = R4 R3 R1 C2 K p = R4AR1 C1 + R2 C2B R3 R1 C2 Eo(s) Ei(s) = K p + K i s + K d s T d = R1 C1 R2 C2 R1 C1 + R2 C2 T i = 1 R1 C1 + R2 C2 K p = R4AR1 C1 + R2 C2B R3 R1 C2 T d T i , T d T i Eo(s) Ei(s) = K p a1 + T i s + T d s b = R4AR1 C1 + R2 C2B R3 R1 C2 c1 + 1 AR1 C1 + R2 C2Bs + R1 C1 R2 C2 R1 C1 + R2 C2 sd = R4 R2 R3 R1 a R1 C1 + R2 C2 R2 C2 + 1 R2 C2 s + R1 C1 s b Eo(s) Ei(s) = Eo(s) E(s) E(s) Ei(s) = R4 R2 R3 R1 AR1 C1s + 1BAR2 C2s + 1B R2 C2s Section 3–3 / Mathematical Modeling of Electrical Systems 85 1 2 3 4 5 6 7 P I PD PI PID Lead or lag Lag–lead Control Action Operational-Amplifier Circuits G(s) = Eo(s) Ei(s) R4 R3 R2 R1 1 R1C2s R4 R3 R4 R3 R2 R1 (R1C1s + 1) R4 R3 R2 R1 R2C2s + 1 R2C2s R4 R3 R2 R1 (R1C1s + 1) (R2C2s + 1) R2C2s R4 R3 R2 R1 R1C1s + 1 R2C2s + 1 R6 R5 R4 R3 [(R1 + R3) C1s + 1] (R2C2s + 1) (R1C1s + 1) [(R2 + R4) C2s + 1] eo eo ei ei + – + – + – + – R1 R2 R2 R3 R4 R1 R3 R4 C2 eo ei + – + – R3 R4 C1 R2 R1 R1 eo ei + – + – R3 R4 C2 R2 R1 eo ei + – + – R3 R4 C2 C1 R2 R1 eo ei + – + – R3 R4 C2 C1 R4 R2 R1 R3 eo ei + – + – R5 R6 C2 C1 Table 3–1 Operational-Amplifier Circuits That May Be Used as Compensators 86 Chapter 3 / Mathematical Modeling of Mechanical Systems and Electrical Systems EXAMPLE PROBLEMS AND SOLUTIONS A–3–1.
Figure 3–20(a) shows a schematic diagram of an automobile suspension system.As the car moves along the road, the vertical displacements at the tires act as the motion excitation to the auto-mobile suspension system.The motion of this system consists of a translational motion of the cen-ter of mass and a rotational motion about the center of mass. Mathematical modeling of the complete system is quite complicated.
A very simplified version of the suspension system is shown in Figure 3–20(b).Assuming that the motion xi at point P is the input to the system and the vertical motion xo of the body is the output,obtain the transfer function (Consider the motion of the body only in the ver-tical direction.) Displacement xo is measured from the equilibrium position in the absence of input xi.
Solution. The equation of motion for the system shown in Figure 3–20(b) is or Taking the Laplace transform of this last equation, assuming zero initial conditions, we obtain Hence the transfer function Xo(s)/Xi(s) is given by X o(s) X i(s) = bs + k ms2 + bs + k Ams2 + bs + kBX o(s) = (bs + k)X i(s) mx $ o + bx # o + kxo = bxi # + kxi mx $ o + bAx # o - x # iB + kAxo - xiB = 0 X o(s)X i(s).
(a) k (b) xi Center of mass Auto body b P xo m Figure 3–20 (a) Automobile suspension system; (b) simplified suspension system.
Example Problems and Solutions 87 A–3–2.
Obtain the transfer function Y(s)/U(s) of the system shown in Figure 3–21. The input u is a displacement input. (Like the system of Problem A–3–1, this is also a simplified version of an automobile or motorcycle suspension system.) Solution. Assume that displacements x and y are measured from respective steady-state positions in the absence of the input u. Applying the Newton’s second law to this system, we obtain Hence, we have Taking Laplace transforms of these two equations, assuming zero initial conditions, we obtain Eliminating X(s) from the last two equations, we have which yields Y(s) U(s) = k1Abs + k2B m1 m2 s4 + Am1 + m2Bbs3 + Ck1 m2 + Am1 + m2Bk2Ds2 + k1 bs + k1 k2 Am1 s2 + bs + k1 + k2B m2 s2 + bs + k2 bs + k2 Y(s) = Abs + k2BY(s) + k1 U(s) Cm2 s2 + bs + k2DY(s) = Abs + k2BX(s) Cm1 s2 + bs + Ak1 + k2B DX(s) = Abs + k2BY(s) + k1 U(s) m2 y $ + by # + k2 y = bx # + k2 x m1 x $ + bx # + Ak1 + k2Bx = by # + k2 y + k1 u m2 y $ = -k2(y - x) - b(y # - x # ) m1 x $ = k2(y - x) + b(y # - x # ) + k1(u - x) y b x u m2 m1 k2 k1 Figure 3–21 Suspension system.
88 Chapter 3 / Mathematical Modeling of Mechanical Systems and Electrical Systems A–3–3.
Obtain a state-space representation of the system shown in Figure 3–22.
Solution. The system equations are The output variables for this system are y1 and y2. Define state variables as Then we obtain the following equations: Hence, the state equation is and the output equation is A–3–4.
Obtain the transfer function Xo(s)/Xi(s) of the mechanical system shown in Figure 3–23(a). Also obtain the transfer function Eo(s)/Ei(s) of the electrical system shown in Figure 3–23(b). Show that these transfer functions of the two systems are of identical form and thus they are analogous systems.
B y1 y2 R = B 1 0 0 0 0 1 0 0R D x1 x2 x3 x4 T D x # 1 x # 2 x # 3 x # 4 T = F 0 - k m1 0 k m2 1 - b m1 0 0 0 k m1 0 - k m2 0 0 1 0 V D x1 x2 x3 x4 T + E 0 0 0 1 m2 U u x # 4 = 1 m2 C-kAy2 - y1B + uD = k m2 x1 -k m2 x3 + 1 m2 u x # 3 = x4 x # 2 = 1 m1 C-by # 1 - kAy1 - y2B D = - k m1 x1 -b m1 x2 + k m1 x3 x # 1 = x2 x4 = y # 2 x3 = y2 x2 = y # 1 x1 = y1 m2 y $ 2 + kAy2 - y1B = u m1 y $ 1 + by # 1 + kAy1 - y2B = 0 m1 m2 k y1 b u y2 Figure 3–22 Mechanical system.
Example Problems and Solutions 89 Solution. In Figure 3–23(a) we assume that displacements xi, xo, and y are measured from their respective steady-state positions.Then the equations of motion for the mechanical system shown in Figure 3–23(a) are By taking the Laplace transforms of these two equations,assuming zero initial conditions,we have If we eliminate Y(s) from the last two equations, then we obtain or Hence the transfer function Xo(s)/Xi(s) can be obtained as For the electrical system shown in Figure 3–23(b), the transfer function Eo(s)/Ei(s) is found to be = AR1 C1 s + 1BAR2 C2 s + 1B AR1 C1 s + 1BAR2 C2 s + 1B + R2 C1 s Eo(s) Ei(s) = R1 + 1 C1 s 1 A1R2B + C2 s + R1 + 1 C1 s X o(s) X i(s) = a b1 k1 s + 1 b a b2 k2 s + 1 b a b1 k1 s + 1 b a b2 k2 s + 1 b + b2 k1 s Ab1 s + k1BX i(s) = ab1 s + k1 + b2 s - b2 s b2 s b2 s + k2 bX o(s) b1CsX i(s) - sX o(s)D + k1CX i(s) - X o(s)D = b2 sX o(s) - b2 s b2 sX o(s) b2 s + k2 b2CsX o(s) - sY(s)D = k2 Y(s) b1CsX i(s) - sX o(s)D + k1CX i(s) - X o(s)D = b2CsX o(s) - sY(s)D b2Ax # o - y # B = k2 y b1Ax # i - x # oB + k1Axi - xoB = b2Ax # o - y # B (a) (b) xi xo y k2 k1 b2 b1 R2 R1 eo ei C2 C1 Figure 3–23 (a) Mechanical system; (b) analogous electrical system.
90 Chapter 3 / Mathematical Modeling of Mechanical Systems and Electrical Systems A comparison of the transfer functions shows that the systems shown in Figures 3–23(a) and (b) are analogous.
A–3–5.
Obtain the transfer functions Eo(s)/Ei(s) of the bridged T networks shown in Figures 3–24(a) and (b).
Solution. The bridged T networks shown can both be represented by the network of Figure 3–25(a), where we used complex impedances.This network may be modified to that shown in Figure 3–25(b).
In Figure 3–25(b), note that I 1 = I 2 + I 3, I 2 Z1 = AZ3 + Z4BI 3 R R C1 C C C2 ei eo (a) R1 R2 ei eo (b) Figure 3–24 Bridged T networks.
Z1 Z4 Z3 Z2 Z1 Ei(s) Z4 Z3 Z2 I3 I2 I1 I1 I3 I3 I2 I1 I1 ei eo Eo(s) (a) (b) Figure 3–25 (a) Bridged T network in terms of complex impedances; (b) equivalent network.
Example Problems and Solutions 91 Hence Then the voltages Ei(s) and Eo(s) can be obtained as Hence, the transfer function Eo(s)/Ei(s) of the network shown in Figure 3–25(a) is obtained as (3–38) For the bridged T network shown in Figure 3–24(a), substitute into Equation (3–38).Then we obtain the transfer function Eo(s)/Ei(s) to be Similarly, for the bridged T network shown in Figure 3–24(b), we substitute into Equation (3–38).Then the transfer function Eo(s)/Ei(s) can be obtained as follows: = R1 CR2 Cs2 + 2R1 Cs + 1 R1 CR2 Cs2 + A2R1 C + R2 CBs + 1 Eo(s) Ei(s) = 1 Cs 1 Cs + R1 a 1 Cs + 1 Cs + R2 b R1 a 1 Cs + 1 Cs + R2 b + 1 Cs 1 Cs + R2 1 Cs Z1 = 1 Cs , Z2 = R1 , Z3 = 1 Cs , Z4 = R2 = RC1 RC2 s2 + 2RC2 s + 1 RC1 RC2 s2 + A2RC2 + RC1Bs + 1 Eo(s) Ei(s) = R2 + 1 C1 s aR + R + 1 C2 s b 1 C1 s aR + R + 1 C2 s b + R2 + R 1 C2 s Z1 = R, Z2 = 1 C1 s , Z3 = R, Z4 = 1 C2 s Eo(s) Ei(s) = Z3 Z1 + Z2 AZ1 + Z3 + Z4B Z2AZ1 + Z3 + Z4B + Z1 Z3 + Z1 Z4 = Z3 Z1 + Z2AZ1 + Z3 + Z4B Z1 + Z3 + Z4 I 1 = Z3 Z1 Z1 + Z3 + Z4 I 1 + Z2 I 1 Eo(s) = Z3 I 3 + Z2 I 1 = Z2AZ1 + Z3 + Z4B + Z1 AZ3 + Z4B Z1 + Z3 + Z4 I 1 = cZ2 + Z1AZ3 + Z4B Z1 + Z3 + Z4 dI 1 Ei(s) = Z1 I 2 + Z2 I 1 I 2 = Z3 + Z4 Z1 + Z3 + Z4 I 1 , I 3 = Z1 Z1 + Z3 + Z4 I 1 92 Chapter 3 / Mathematical Modeling of Mechanical Systems and Electrical Systems A–3–6.
Obtain the transfer function of the op-amp circuit shown in Figure 3–26.
Solution. The voltage at point A is The Laplace-transformed version of this last equation is The voltage at point B is Since and we must have Thus Hence A–3–7.
Obtain the transfer function Eo(s)/Ei(s) of the op-amp system shown in Figure 3–27 in terms of complex impedances Z1, Z2, Z3, and Z4. Using the equation derived, obtain the transfer function Eo(s)/Ei(s) of the op-amp system shown in Figure 3–26.
Solution. From Figure 3–27, we find Ei(s) - EA(s) Z3 = EA(s) - Eo(s) Z4 Eo(s) Ei(s) = - R2 Cs - 1 R2 Cs + 1 = -s -1 R2 C s + 1 R2 C 1 2 CEi(s) + Eo(s)D = 1 R2 Cs + 1 Ei(s) EA(s) = EB(s).
K 1, CEB(s) - EA(s)DK = Eo(s) EB(s) = 1 Cs R2 + 1 Cs Ei(s) = 1 R2 Cs + 1 Ei(s) EA(s) = 1 2 CEi(s) + Eo(s)D eA = 1 2 Aei - eoB + eo Eo(s)Ei(s) – + C A B R1 R1 R2 ei eo Figure 3–26 Operational-amplifier circuit.
Example Problems and Solutions 93 or (3–39) Since (3–40) by substituting Equation (3–40) into Equation (3–39), we obtain from which we get the transfer function Eo(s)/Ei(s) to be (3–41) To find the transfer function Eo(s)/Ei(s) of the circuit shown in Figure 3–26, we substitute into Equation (3–41).The result is which is, as a matter of course, the same as that obtained in Problem A–3–6.
Eo(s) Ei(s) = -R1 R2 - R1 1 Cs R1 a 1 Cs + R2 b = - R2 Cs - 1 R2 Cs + 1 Z1 = 1 Cs , Z2 = R2 , Z3 = R1 , Z4 = R1 Eo(s) Ei(s) = - Z4 Z2 - Z3 Z1 Z3AZ1 + Z2B c Z4 Z1 + Z4 Z2 - Z4 Z1 - Z3 Z1 Z4AZ1 + Z2B dEi(s) = - Z3 Z4 Eo(s) EA(s) = EB(s) = Z1 Z1 + Z2 Ei(s) Ei(s) - a1 + Z3 Z4 bEA(s) = - Z3 Z4 Eo(s) A B eo ei Z3 Z1 Z2 Z4 – + Figure 3–27 Operational-amplifier circuit.
94 Chapter 3 / Mathematical Modeling of Mechanical Systems and Electrical Systems A–3–8.
Obtain the transfer function of the operational-amplifier circuit shown in Figure 3–28.
Solution. We will first obtain currents i1, i2, i3, i4, and i5.Then we will use node equations at nodes A and B.
At node A, we have i1=i2+i3+i4, or (3–42) At node B, we get i4=i5, or (3–43) By rewriting Equation (3–42), we have (3–44) From Equation (3–43), we get (3–45) By substituting Equation (3–45) into Equation (3–44), we obtain Taking the Laplace transform of this last equation, assuming zero initial conditions, we obtain from which we get the transfer function as follows: Eo(s) Ei(s) = -1 R1 C1 R2C2 s2 + CR2 C2 + R1 C2 + AR1R3BR2 C2Ds + AR1R3B Eo(s)Ei(s) -C1 C2 R2 s2Eo(s) + a 1 R1 + 1 R2 + 1 R3 b A-R2 C2BsEo(s) - 1 R3 Eo(s) = Ei(s) R1 C1 a-R2 C2 d2eo dt2 b + a 1 R1 + 1 R2 + 1 R3 b A-R2 C2B deo dt = ei R1 + eo R3 eA = -R2 C2 deo dt C1 deA dt + a 1 R1 + 1 R2 + 1 R3 beA = ei R1 + eo R3 eA R2 = C2 -deo dt ei - eA R1 = eA - eo R3 + C1 deA dt + eA R2 i4 = eA R2 , i5 = C2 -deo dt i1 = ei - eA R1 ; i2 = eA - eo R3 , i3 = C1 deA dt Eo(s)Ei(s) i1 R1 i2 i4 i3 A C1 ei eo R3 i5 C2 B R2 – + Figure 3–28 Operational-amplifier circuit.
Example Problems and Solutions 95 A–3–9.
Consider the servo system shown in Figure 3–29(a).The motor shown is a servomotor,a dc motor de-signed specifically to be used in a control system.The operation of this system is as follows:A pair of potentiometers acts as an error-measuring device.They convert the input and output positions into proportional electric signals. The command input signal determines the angular position r of the wiper arm of the input potentiometer.The angular position r is the reference input to the system,and the electric potential of the arm is proportional to the angular position of the arm.The output shaft position determines the angular position c of the wiper arm of the output potentiometer.The differ-ence between the input angular position r and the output angular position c is the error signal e, or The potential difference is the error voltage, where er is proportional to r and ec is pro-portional to c; that is, and where K0 is a proportionality constant.The error volt-age that appears at the potentiometer terminals is amplified by the amplifier whose gain constant is K1.
The output voltage of this amplifier is applied to the armature circuit of the dc motor.A fixed volt-age is applied to the field winding. If an error exists, the motor develops a torque to rotate the out-put load in such a way as to reduce the error to zero. For constant field current, the torque developed by the motor is where K2 is the motor torque constant and ia is the armature current.
When the armature is rotating, a voltage proportional to the product of the flux and angular velocity is induced in the armature. For a constant flux, the induced voltage eb is directly propor-tional to the angular velocity or where eb is the back emf, K3 is the back emf constant of the motor, and u is the angular displace-ment of the motor shaft.
eb = K 3 du dt dudt, T = K 2 ia ec = K 0 c , er = K 0 r er - ec = ev e = r - c (a) Reference input Input device Input potentiometer Output potentiometer Feedback signal er ec r c c K1 ia T Ra La Error measuring device Amplifier Motor Gear train Load u K1ev ev (b) (c) Ev(s) E(s) R(s) C(s) U(s) K1K2 s(Las + Ra) (Jos + bo) + K2K3s K0 n C(s) R(s) K s(Js + B) + – + – Figure 3–29 (a) Schematic diagram of servo system; (b) block diagram for the system; (c) simplified block diagram.
96 Chapter 3 / Mathematical Modeling of Mechanical Systems and Electrical Systems Obtain the transfer function between the motor shaft angular displacement u and the error voltage ev. Obtain also a block diagram for this system and a simplified block diagram when La is negligible.
Solution. The speed of an armature-controlled dc servomotor is controlled by the armature volt-age ea. (The armature voltage is the output of the amplifier.) The differential equation for the armature circuit is or (3–46) The equation for torque equilibrium is (3–47) where J0 is the inertia of the combination of the motor, load, and gear train referred to the motor shaft and b0 is the viscous-friction coefficient of the combination of the motor, load, and gear train referred to the motor shaft.
By eliminating ia from Equations (3–46) and (3–47), we obtain (3–48) We assume that the gear ratio of the gear train is such that the output shaft rotates n times for each revolution of the motor shaft.Thus, (3–49) The relationship among Ev(s), R(s), and C(s) is (3–50) The block diagram of this system can be constructed from Equations (3–48), (3–49), and (3–50), as shown in Figure 3–29(b).The transfer function in the feedforward path of this system is When La is small, it can be neglected, and the transfer function G(s) in the feedforward path becomes (3–51) The term indicates that the back emf of the motor effectively increases the viscous friction of the system. The inertia J0 and viscous friction coefficient are b0 + AK 2 K 3RaB Cb0 + AK 2 K 3RaB Ds = K 0 K 1 K 2 nRa J 0 s2 + ab0 + K 2 K 3 Ra bs G(s) = K 0 K 1 K 2 n sCRaAJ 0 s + b0B + K 2 K 3D G(s) = C(s) Q (s) Q (s) Ev(s) Ev(s) E(s) = K 0 K 1 K 2 n sC ALa s + RaBAJ 0 s + b0B + K 2 K 3D Ev(s) = K 0CR(s) - C(s)D = K 0 E(s) C(s) = nQ (s) Q (s) Ev(s) = K 1 K 2 sALa s + RaBAJ 0 s + b0B + K 2 K 3 s J 0 d2u dt2 + b0 du dt = T = K 2 ia La dia dt + Ra ia + K 3 du dt = K 1 ev La dia dt + Ra ia + eb = ea ea = K 1 ev Problems 97 referred to the motor shaft. When J0 and are multiplied by 1/n2, the inertia and viscous-friction coefficient are expressed in terms of the output shaft. Introducing new parameters defined by moment of inertia referred to the output shaft viscous-friction coefficient referred to the output shaft the transfer function G(s) given by Equation (3–51) can be simplified, yielding or where The block diagram of the system shown in Figure 3–29(b) can thus be simplified as shown in Figure 3–29(c).
K m = K B , T m = J B = Ra J 0 Ra b0 + K 2 K 3 G(s) = K m sAT m s + 1B G(s) = K Js2 + Bs K = K 0 K 1 K 2nRa B = Cb0 + AK 2 K 3RaB Dn2 = J = J 0n2 = b0 + AK 2 K 3RaB PROBLEMS B–3–1. Obtain the equivalent viscous-friction coefficient beq of the system shown in Figure 3–30.
B–3–2. Obtain mathematical models of the mechanical sys-tems shown in Figures 3–31(a) and (b).
x b3 y b2 b1 m k (a) No friction x (Output) u(t) (Input force) m (b) No friction x (Output) u(t) (Input force) k1 k2 Figure 3–31 Mechanical systems.
Figure 3–30 Damper system.
98 Chapter 3 / Mathematical Modeling of Mechanical Systems and Electrical Systems y1 y2 u1 m2 b1 u2 k2 k1 m1 Figure 3–32 Mechanical system.
B–3–3. Obtain a state-space representation of the mechan-ical system shown in Figure 3–32, where u1 and u2 are the inputs and y1 and y2 are the outputs.
B–3–4. Consider the spring-loaded pendulum system shown in Figure 3–33. Assume that the spring force acting on the pendulum is zero when the pendulum is vertical, or u=0.
Assume also that the friction involved is negligible and the angle of oscillation u is small. Obtain a mathematical model of the system.
k k a mg u Figure 3–33 Spring-loaded pendulum system.
B–3–5. Referring to Examples 3–5 and 3–6, consider the inverted-pendulum system shown in Figure 3–34. Assume that the mass of the inverted pendulum is m and is evenly distributed along the length of the rod. (The center of gravity of the pendulum is located at the center of the rod.) Assuming that u is small, derive mathematical models for the system in the forms of differential equations, transfer functions, and state-space equations.
M y x u G O x y x u Figure 3–34 Inverted-pendulum system.
B–3–6. Obtain the transfer functions X1(s)/U(s) and X2(s)/U(s) of the mechanical system shown in Figure 3–35.
m1 m2 k3 k1 x1 x2 u b1 k2 b2 Figure 3–35 Mechanical system.
B–3–7. Obtain the transfer function Eo(s)/Ei(s) of the elec-trical circuit shown in Figure 3–36.
R1 eo R2 C L ei i1 i2 Figure 3–36 Electrical circuit.
B–3–8. Consider the electrical circuit shown in Figure 3–37.
Obtain the transfer function Eo(s)/Ei(s) by use of the block diagram approach.
R1 C1 eo R2 C2 ei i1 i2 Figure 3–37 Electrical circuit.
Problems 99 B–3–9. Derive the transfer function of the electrical circuit shown in Figure 3–38. Draw a schematic diagram of an analogous mechanical system.
R1 C1 R2 C2 eo ei Figure 3–38 Electrical circuit.
+ – C A R1 R2 ei eo Figure 3–39 Operational-amplifier circuit.
B–3–10. Obtain the transfer function of the op-amp circuit shown in Figure 3–39.
Eo(s)Ei(s) + – C A B R1 R2 R3 ei eo Figure 3–40 Operational-amplifier circuit.
B–3–11. Obtain the transfer function of the op-amp circuit shown in Figure 3–40.
Eo(s)Ei(s) B–3–12. Using the impedance approach, obtain the trans-fer function of the op-amp circuit shown in Figure 3–41.
Eo(s)Ei(s) + – C A B R1 R1 R2 ei eo Figure 3–41 Operational-amplifier circuit.
B–3–13. Consider the system shown in Figure 3–42. An armature-controlled dc servomotor drives a load consisting of the moment of inertia JL. The torque developed by the motor is T.The moment of inertia of the motor rotor is Jm.
The angular displacements of the motor rotor and the load element are um and u, respectively. The gear ratio is Obtain the transfer function Q (s)Ei(s).
n = uum .
L R T n ei Jm JL um u Figure 3–42 Armature-controlled dc servomotor system.
4 100 Mathematical Modeling of Fluid Systems and Thermal Systems 4–1 INTRODUCTION This chapter treats mathematical modeling of fluid systems and thermal systems.As the most versatile medium for transmitting signals and power, fluids—liquids and gases— have wide usage in industry.Liquids and gases can be distinguished basically by their rel-ative incompressibilities and the fact that a liquid may have a free surface, whereas a gas expands to fill its vessel. In the engineering field the term pneumatic describes fluid systems that use air or gases and hydraulic applies to those using oil.
We first discuss liquid-level systems that are frequently used in process control. Here we introduce the concepts of resistance and capacitance to describe the dynamics of such systems. Then we treat pneumatic systems. Such systems are extensively used in the au-tomation of production machinery and in the field of automatic controllers. For instance, pneumatic circuits that convert the energy of compressed air into mechanical energy enjoy wide usage.Also,various types of pneumatic controllers are widely used in industry.Next, we present hydraulic servo systems.These are widely used in machine tool systems,aircraft control systems, etc. We discuss basic aspects of hydraulic servo systems and hydraulic controllers.Both pneumatic systems and hydraulic systems can be modeled easily by using the concepts of resistance and capacitance. Finally, we treat simple thermal systems. Such systems involve heat transfer from one substance to another. Mathematical models of such systems can be obtained by using thermal resistance and thermal capacitance.
Outline of the Chapter.
Section 4–1 has presented introductory material for the chapter. Section 4–2 discusses liquid-level systems. Section 4–3 treats pneumatic systems—in particular, the basic principles of pneumatic controllers. Section 4–4 first discusses hydraulic servo systems and then presents hydraulic controllers. Finally, Section 4–5 analyzes thermal systems and obtains mathematical models of such systems.
Section 4–2 / Liquid-Level Systems 101 4–2 LIQUID-LEVEL SYSTEMS In analyzing systems involving fluid flow, we find it necessary to divide flow regimes into laminar flow and turbulent flow, according to the magnitude of the Reynolds num-ber. If the Reynolds number is greater than about 3000 to 4000, then the flow is turbu-lent.The flow is laminar if the Reynolds number is less than about 2000. In the laminar case, fluid flow occurs in streamlines with no turbulence. Systems involving laminar flow may be represented by linear differential equations.
Industrial processes often involve flow of liquids through connecting pipes and tanks.
The flow in such processes is often turbulent and not laminar. Systems involving turbu-lent flow often have to be represented by nonlinear differential equations. If the region of operation is limited, however, such nonlinear differential equations can be linearized.
We shall discuss such linearized mathematical models of liquid-level systems in this sec-tion.Note that the introduction of concepts of resistance and capacitance for such liquid-level systems enables us to describe their dynamic characteristics in simple forms.
Resistance and Capacitance of Liquid-Level Systems.
Consider the flow through a short pipe connecting two tanks. The resistance R for liquid flow in such a pipe or restriction is defined as the change in the level difference (the difference of the liquid levels of the two tanks) necessary to cause a unit change in flow rate; that is, Since the relationship between the flow rate and level difference differs for the laminar flow and turbulent flow, we shall consider both cases in the following.
Consider the liquid-level system shown in Figure 4–1(a). In this system the liquid spouts through the load valve in the side of the tank. If the flow through this restriction is laminar, the relationship between the steady-state flow rate and steady-state head at the level of the restriction is given by Q=KH R = change in level difference, m change in flow rate, m3sec Control valve Q + qo Q + qi H + h Load valve Capacitance C Resistance R (b) (a) Head H –H 0 h P q Q Flow rate tan–1Rt Slope = = 2H Q h q Figure 4–1 (a) Liquid-level system; (b) head-versus-flow-rate curve.
102 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems where steady-state liquid flow rate, m3sec coefficient, m2sec steady-state head, m For laminar flow, the resistance Rl is obtained as The laminar-flow resistance is constant and is analogous to the electrical resistance.
If the flow through the restriction is turbulent, the steady-state flow rate is given by (4–1) where steady-state liquid flow rate, m3sec coefficient, m2.5sec steady-state head, m The resistance Rt for turbulent flow is obtained from Since from Equation (4–1) we obtain we have Thus, The value of the turbulent-flow resistance Rt depends on the flow rate and the head.The value of Rt, however, may be considered constant if the changes in head and flow rate are small.
By use of the turbulent-flow resistance, the relationship between Q and H can be given by Such linearization is valid, provided that changes in the head and flow rate from their respective steady-state values are small.
In many practical cases,the value of the coefficient K in Equation (4–1),which depends on the flow coefficient and the area of restriction, is not known.Then the resistance may be determined by plotting the head-versus-flow-rate curve based on experimental data and measuring the slope of the curve at the operating condition.An example of such a plot is shown in Figure 4–1(b).In the figure,point P is the steady-state operating point.The tan-gent line to the curve at point P intersects the ordinate at point Thus, the slope of this tangent line is Since the resistance Rt at the operating point P is given by the resistance Rt is the slope of the curve at the operating point.
2H – Q –, 2H – Q –.
A0, -H – B.
Q = 2H Rt Rt = 2H Q dH dQ = 21H K = 21H 1H Q = 2H Q dQ = K 21H dH Rt = dH dQ H = K = Q = Q = K1H Rl = dH dQ = H Q H = K = Q = Section 4–2 / Liquid-Level Systems 103 Consider the operating condition in the neighborhood of point P. Define a small deviation of the head from the steady-state value as h and the corresponding small change of the flow rate as q.Then the slope of the curve at point P can be given by The linear approximation is based on the fact that the actual curve does not differ much from its tangent line if the operating condition does not vary too much.
The capacitance C of a tank is defined to be the change in quantity of stored liquid necessary to cause a unit change in the potential (head). (The potential is the quantity that indicates the energy level of the system.) It should be noted that the capacity (m3) and the capacitance (m2) are different. The capacitance of the tank is equal to its cross-sectional area. If this is constant, the capac-itance is constant for any head.
Liquid-Level Systems.
Consider the system shown in Figure 4–1(a). The vari-ables are defined as follows: steady-state flow rate (before any change has occurred), m3sec qi= small deviation of inflow rate from its steady-state value, m3sec qo= small deviation of outflow rate from its steady-state value, m3sec steady-state head (before any change has occurred), m h= small deviation of head from its steady-state value, m As stated previously, a system can be considered linear if the flow is laminar. Even if the flow is turbulent, the system can be linearized if changes in the variables are kept small.Based on the assumption that the system is either linear or linearized,the differential equation of this system can be obtained as follows:Since the inflow minus outflow during the small time interval dt is equal to the additional amount stored in the tank, we see that From the definition of resistance, the relationship between qo and h is given by The differential equation for this system for a constant value of R becomes (4–2) Note that RC is the time constant of the system.Taking the Laplace transforms of both sides of Equation (4–2), assuming the zero initial condition, we obtain where and Qi(s) = lCqiD H(s) = l[h] (RCs + 1)H(s) = RQi(s) RC dh dt + h = Rqi qo = h R C dh = Aqi - qoB dt H – = Q – = C = change in liquid stored, m3 change in head, m Slope of curve at point P = h q = 2H – Q – = Rt 104 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems Q + q Tank 1 Tank 2 H1 + h1 R1 H2 + h2 R2 Q + q2 C1 C2 Q + q1 Q : H1: H2: Steady-state flow rate Steady-state liquid level of tank 1 Steady-state liquid level of tank 2 Figure 4–2 Liquid-level system with interaction.
If qi is considered the input and h the output, the transfer function of the system is If, however, qo is taken as the output, the input being the same, then the transfer function is where we have used the relationship Liquid-Level Systems with Interaction.
Consider the system shown in Figure 4–2. In this system, the two tanks interact.Thus the transfer function of the system is not the product of two first-order transfer functions.
In the following, we shall assume only small variations of the variables from the steady-state values. Using the symbols as defined in Figure 4–2, we can obtain the following equations for this system: (4–3) (4–4) (4–5) (4–6) If q is considered the input and q2 the output, the transfer function of the system is (4–7) Q2(s) Q(s) = 1 R1 C1 R2 C2 s2 + AR1 C1 + R2 C2 + R2 C1Bs + 1 C2 dh2 dt = q1 - q2 h2 R2 = q2 C1 dh1 dt = q - q1 h1 - h2 R1 = q1 Qo(s) = 1 R H(s) Qo(s) Qi(s) = 1 RCs + 1 H(s) Qi(s) = R RCs + 1 Section 4–2 / Liquid-Level Systems 105 It is instructive to obtain Equation (4–7), the transfer function of the interacted system, by block diagram reduction. From Equations (4–3) through (4–6), we obtain the elements of the block diagram, as shown in Figure 4–3(a). By connecting signals prop-erly, we can construct a block diagram, as shown in Figure 4–3(b). This block diagram can be simplified, as shown in Figure 4–3(c). Further simplifications result in Figures 4–3(d) and (e). Figure 4–3(e) is equivalent to Equation (4–7).
(c) (d) (e) G3 (b) (a) Q(s) H1(s) H2(s) Q1(s) Q2(s) Q(s) Q1(s) Q2(s) Q(s) Q2(s) Q(s) Q2(s) 1 R1 1 R1 1 R1 1 R2 1 R2 1 R2 1 C1s 1 C1s 1 C2s G3 1 C2s G3 1 C2s R2C1s R2C1s 1 R1C1 s + 1 1 R2C2 s + 1 1 R1C1R2C2s2 + (R1C1 + R2C2 + R2C1)s + 1 H1(s) Q1(s) H2(s) 1 C1s H2(s) Q2(s) Q(s) H1(s) Q1(s) Q1(s) H2(s) Q2(s) + – + – + – + – + – + – + – + – + – + – Figure 4–3 (a) Elements of the block diagram of the system shown in Figure 4–2; (b) block diagram of the system; (c)–(e) successive reductions of the block diagram.
106 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems Notice the similarity and difference between the transfer function given by Equation (4–7) and that given by Equation (3–33).The term R2C1s that appears in the denominator of Equation (4–7) exemplifies the interaction between the two tanks.
Similarly, the term R1C2s in the denominator of Equation (3–33) represents the inter-action between the two RC circuits shown in Figure 3–8.
4–3 PNEUMATIC SYSTEMS In industrial applications pneumatic systems and hydraulic systems are frequently compared.Therefore, before we discuss pneumatic systems in detail, we shall give a brief comparison of these two kinds of systems.
Comparison Between Pneumatic Systems and Hydraulic Systems.
The fluid generally found in pneumatic systems is air; in hydraulic systems it is oil. And it is pri-marily the different properties of the fluids involved that characterize the differences between the two systems.These differences can be listed as follows: 1. Air and gases are compressible,whereas oil is incompressible (except at high pressure).
2. Air lacks lubricating property and always contains water vapor. Oil functions as a hydraulic fluid as well as a lubricator.
3. The normal operating pressure of pneumatic systems is very much lower than that of hydraulic systems.
4. Output powers of pneumatic systems are considerably less than those of hydraulic systems.
5. Accuracy of pneumatic actuators is poor at low velocities, whereas accuracy of hydraulic actuators may be made satisfactory at all velocities.
6. In pneumatic systems, external leakage is permissible to a certain extent, but in-ternal leakage must be avoided because the effective pressure difference is rather small. In hydraulic systems internal leakage is permissible to a certain extent, but external leakage must be avoided.
7. No return pipes are required in pneumatic systems when air is used, whereas they are always needed in hydraulic systems.
8. Normal operating temperature for pneumatic systems is 5° to 60°C (41° to 140°F).
The pneumatic system, however, can be operated in the 0° to 200°C (32° to 392°F) range. Pneumatic systems are insensitive to temperature changes, in contrast to hydraulic systems, in which fluid friction due to viscosity depends greatly on tem-perature. Normal operating temperature for hydraulic systems is 20° to 70°C (68° to 158°F).
9. Pneumatic systems are fire- and explosion-proof, whereas hydraulic systems are not, unless nonflammable liquid is used.
In what follows we begin with a mathematical modeling of pneumatic systems.Then we shall present pneumatic proportional controllers.
We shall first give detailed discussions of the principle by which proportional controllers operate.Then we shall treat methods for obtaining derivative and integral control actions. Throughout the discussions, we shall place emphasis on the Section 4–3 / Pneumatic Systems 107 Resistance R Capacitance C (a) (b) P + pi P + po 0 q q dq DP Slope = R d (DP) Figure 4–4 (a) Schematic diagram of a pressure system; (b) pressure-difference-versus-flow-rate curve.
fundamental principles, rather than on the details of the operation of the actual mechanisms.
Pneumatic Systems.
The past decades have seen a great development in low-pressure pneumatic controllers for industrial control systems, and today they are used extensively in industrial processes. Reasons for their broad appeal include an explosion-proof character, simplicity, and ease of maintenance.
Resistance and Capacitance of Pressure Systems.
Many industrial processes and pneumatic controllers involve the flow of a gas or air through connected pipelines and pressure vessels.
Consider the pressure system shown in Figure 4–4(a). The gas flow through the restriction is a function of the gas pressure difference pi-po. Such a pressure system may be characterized in terms of a resistance and a capacitance.
The gas flow resistance R may be defined as follows: or (4–8) where is a small change in the gas pressure difference and dq is a small change in the gas flow rate. Computation of the value of the gas flow resistance R may be quite time consuming. Experimentally, however, it can be easily determined from a plot of the pressure difference versus flow rate by calculating the slope of the curve at a given operating condition, as shown in Figure 4–4(b).
The capacitance of the pressure vessel may be defined by or (4–9) C = dm dp = V dr dp C = change in gas stored, lb change in gas pressure, lbfft2 d(¢P) R = d(¢P) dq R = change in gas pressure difference, lbfft2 change in gas flow rate, lbsec 108 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems where capacitance, lb-ft2lbf mass of gas in vessel, lb gas pressure, lbfft2 volume of vessel, ft3 density, lbft3 The capacitance of the pressure system depends on the type of expansion process involved. The capacitance can be calculated by use of the ideal gas law. If the gas ex-pansion process is polytropic and the change of state of the gas is between isothermal and adiabatic, then (4–10) where n=polytropic exponent.
For ideal gases, or where absolute pressure, lbfft2 volume occupied by 1 mole of a gas, ft3lb-mole universal gas constant, ft-lbflb-mole °R absolute temperature, °R specific volume of gas, ft3lb molecular weight of gas per mole, lblb-mole Thus (4–11) where Rgas=gas constant, ft-lbflb °R.
The polytropic exponent n is unity for isothermal expansion.For adiabatic expansion, n is equal to the ratio of specific heats cpcv,where cp is the specific heat at constant pres-sure and cv is the specific heat at constant volume. In many practical cases, the value of n is approximately constant, and thus the capacitance may be considered constant.
The value of drdp is obtained from Equations (4–10) and (4–11). From Equation (4–10) we have or Substituting Equation (4–11) into this last equation, we get dr dp = 1 nRgas T dr dp = 1 Knrn-1 = rn pnrn-1 = r pn dp = Knrn-1 dr pv = p r = R – M T = Rgas T M = v = T = R – = v – = p = pv = R – M T pv – = R –T p a V m b n = p rn = constant = K r = V = p = m = C = Section 4–3 / Pneumatic Systems 109 The capacitance C is then obtained as (4–12) The capacitance of a given vessel is constant if the temperature stays constant. (In many practical cases, the polytropic exponent n is approximately 1.0 ~ 1.2 for gases in unin-sulated metal vessels.) Pressure Systems.
Consider the system shown in Figure 4–4(a). If we assume only small deviations in the variables from their respective steady-state values, then this system may be considered linear.
Let us define gas pressure in the vessel at steady state (before changes in pressure have occurred), lbfft2 pi= small change in inflow gas pressure, lbfft2 po= small change in gas pressure in the vessel, lbfft2 V= volume of the vessel, ft3 m= mass of gas in the vessel, lb q= gas flow rate, lbsec r= density of gas, lb/ft3 For small values of pi and po,the resistance R given by Equation (4–8) becomes constant and may be written as The capacitance C is given by Equation (4–9), or Since the pressure change dpo times the capacitance C is equal to the gas added to the vessel during dt seconds, we obtain or which can be written as If pi and po are considered the input and output, respectively, then the transfer function of the system is where RC has the dimension of time and is the time constant of the system.
P o(s) P i(s) = 1 RCs + 1 RC dpo dt + po = pi C dpo dt = pi - po R C dpo = q dt C = dm dp R = pi - po q P – = C = V nRgas T 110 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems Air supply Orifice Input 0 Nozzle (a) (b) Flapper To control valve Ps Pb Ps Pb Pa X X Figure 4–5 (a) Schematic diagram of a pneumatic nozzle– flapper amplifier; (b) characteristic curve relating nozzle back pressure and nozzle–flapper distance.
Pneumatic Nozzle–Flapper Amplifiers.
A schematic diagram of a pneumatic nozzle–flapper amplifier is shown in Figure 4–5(a).The power source for this amplifier is a supply of air at constant pressure. The nozzle–flapper amplifier converts small changes in the position of the flapper into large changes in the back pressure in the noz-zle.Thus a large power output can be controlled by the very little power that is needed to position the flapper.
In Figure 4–5(a), pressurized air is fed through the orifice, and the air is ejected from the nozzle toward the flapper. Generally, the supply pressure for such a controller is 20 psig (1.4 kgfcm2 gage). The diameter of the orifice is on the order of 0.01 in.
(0.25 mm) and that of the nozzle is on the order of 0.016 in. (0.4 mm).To ensure prop-er functioning of the amplifier, the nozzle diameter must be larger than the orifice diameter.
In operating this system, the flapper is positioned against the nozzle opening. The nozzle back pressure is controlled by the nozzle–flapper distance X. As the flapper approaches the nozzle,the opposition to the flow of air through the nozzle increases,with the result that the nozzle back pressure increases. If the nozzle is completely closed by the flapper, the nozzle back pressure becomes equal to the supply pressure If the flapper is moved away from the nozzle, so that the nozzle–flapper distance is wide (on the order of 0.01 in.), then there is practically no restriction to flow, and the nozzle back pressure takes on a minimum value that depends on the nozzle–flapper device.
(The lowest possible pressure will be the ambient pressure ) Note that, because the air jet puts a force against the flapper, it is necessary to make the nozzle diameter as small as possible.
A typical curve relating the nozzle back pressure to the nozzle–flapper distance X is shown in Figure 4–5(b).The steep and almost linear part of the curve is utilized in the actual operation of the nozzle–flapper amplifier. Because the range of flapper dis-placements is restricted to a small value, the change in output pressure is also small, unless the curve is very steep.
The nozzle–flapper amplifier converts displacement into a pressure signal. Since industrial process control systems require large output power to operate large pneu-matic actuating valves, the power amplification of the nozzle–flapper amplifier is usually insufficient. Consequently, a pneumatic relay is often needed as a power amplifier in connection with the nozzle–flapper amplifier.
P b P a .
P b P s .
P b P b P b P s Section 4–3 / Pneumatic Systems 111 To atmosphere Pa Nozzle back pressure Pb Air supply Ps Pc To pneumatic valve (a) (b) To atmosphere Nozzle back pressure Pb Air supply Ps Pc To pneumatic valve Pneumatic Relays.
In practice, in a pneumatic controller, a nozzle–flapper amplifier acts as the first-stage amplifier and a pneumatic relay as the second-stage amplifier. The pneumatic relay is capable of handling a large quantity of airflow.
A schematic diagram of a pneumatic relay is shown in Figure 4–6(a).As the nozzle back pressure increases, the diaphragm valve moves downward. The opening to the atmosphere decreases and the opening to the pneumatic valve increases, thereby increasing the control pressure When the diaphragm valve closes the opening to the atmosphere, the control pressure becomes equal to the supply pressure When the nozzle back pressure decreases and the diaphragm valve moves upward and shuts off the air supply, the control pressure drops to the ambient pressure The control pressure can thus be made to vary from 0 psig to full supply pressure, usually 20 psig.
The total movement of the diaphragm valve is very small. In all positions of the valve, except at the position to shut off the air supply, air continues to bleed into the at-mosphere, even after the equilibrium condition is attained between the nozzle back pressure and the control pressure. Thus the relay shown in Figure 4–6(a) is called a bleed-type relay.
There is another type of relay, the nonbleed type. In this one the air bleed stops when the equilibrium condition is obtained and, therefore, there is no loss of pres-surized air at steady-state operation. Note, however, that the nonbleed-type relay must have an atmospheric relief to release the control pressure from the pneu-matic actuating valve.A schematic diagram of a nonbleed-type relay is shown in Fig-ure 4–6(b).
In either type of relay, the air supply is controlled by a valve, which is in turn controlled by the nozzle back pressure.Thus, the nozzle back pressure is converted into the control pressure with power amplification.
Since the control pressure changes almost instantaneously with changes in the nozzle back pressure the time constant of the pneumatic relay is negligible compared with the other larger time constants of the pneumatic controller and the plant.
P b , P c P c P c P a .
P c P b P s .
P c P c .
P b Figure 4–6 (a) Schematic diagram of a bleed-type relay; (b) schematic diagram of a nonbleed-type relay.
112 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems To atmosphere Nozzle back pressure Pb Air supply Ps Pc To pneumatic valve Figure 4–7 Reverse-acting relay.
It is noted that some pneumatic relays are reverse acting. For example, the relay shown in Figure 4–7 is a reverse-acting relay. Here, as the nozzle back pressure increases, the ball valve is forced toward the lower seat, thereby decreasing the control pressure Thus, this relay is a reverse-acting relay.
Pneumatic Proportional Controllers (Force-Distance Type).
Two types of pneu-matic controllers,one called the force-distance type and the other the force-balance type, are used extensively in industry.Regardless of how differently industrial pneumatic con-trollers may appear, careful study will show the close similarity in the functions of the pneumatic circuit.Here we shall consider the force-distance type of pneumatic controllers.
Figure 4–8(a) shows a schematic diagram of such a proportional controller.The nozzle– flapper amplifier constitutes the first-stage amplifier, and the nozzle back pressure is controlled by the nozzle–flapper distance.The relay-type amplifier constitutes the second-stage amplifier.The nozzle back pressure determines the position of the diaphragm valve for the second-stage amplifier, which is capable of handling a large quantity of airflow.
In most pneumatic controllers, some type of pneumatic feedback is employed. Feed-back of the pneumatic output reduces the amount of actual movement of the flapper.
Instead of mounting the flapper on a fixed point, as shown in Figure 4–8(b), it is often pivoted on the feedback bellows, as shown in Figure 4–8(c).The amount of feedback can be regulated by introducing a variable linkage between the feedback bellows and the flapper connecting point.The flapper then becomes a floating link. It can be moved by both the error signal and the feedback signal.
The operation of the controller shown in Figure 4–8(a) is as follows. The input sig-nal to the two-stage pneumatic amplifier is the actuating error signal. Increasing the actuating error signal moves the flapper to the left.This move will, in turn, increase the nozzle back pressure, and the diaphragm valve moves downward. This results in an in-crease of the control pressure. This increase will cause bellows F to expand and move the flapper to the right, thus opening the nozzle. Because of this feedback, the nozzle– flapper displacement is very small, but the change in the control pressure can be large.
It should be noted that proper operation of the controller requires that the feed-back bellows move the flapper less than that movement caused by the error signal alone.
(If these two movements were equal, no control action would result.) Equations for this controller can be derived as follows. When the actuating error is zero,or e=0,an equilibrium state exists with the nozzle–flapper distance equal to the X – , P c .
P b Section 4–3 / Pneumatic Systems 113 + – Orifice Actuating error signal Flapper Nozzle Pneumatic relay (a) (b) (c) Ps e a b F Pb + pb X + x Z + z Y + y Pc + pc Error signal Error signal Feedback signal E(s) X(s) Pc(s) Y(s) (e) b a + b a a + b A ks E(s) (f) Pc(s) Kp K b a + b e e e y y x a a b b a a + b y – = (d) displacement of bellows equal to the displacement of the diaphragm equal to the nozzle back pressure equal to and the control pressure equal to When an actuating error exists,the nozzle–flapper distance,the displacement of the bellows,the displacement of the diaphragm,the nozzle back pressure,and the control pressure deviate from their re-spective equilibrium values.Let these deviations be x, y, z, pb,and pc,respectively.(The pos-itive direction for each displacement variable is indicated by an arrowhead in the diagram.) P – c .
P – b , Z –, Y –, Figure 4–8 (a) Schematic diagram of a force-distance type of pneumatic proportional controller; (b) flapper mounted on a fixed point; (c) flapper mounted on a feedback bellows; (d) displacement x as a result of addition of two small displacements; (e) block diagram for the controller; (f) simplified block diagram for the controller.
114 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems Assuming that the relationship between the variation in the nozzle back pressure and the variation in the nozzle–flapper distance is linear, we have (4–13) where is a positive constant. For the diaphragm valve, (4–14) where K2 is a positive constant. The position of the diaphragm valve determines the control pressure. If the diaphragm valve is such that the relationship between and z is linear, then (4–15) where K3 is a positive constant. From Equations (4–13), (4–14), and (4–15), we obtain (4–16) where K=K1K3/K2 is a positive constant. For the flapper, since there are two small movements (e and y) in opposite directions, we can consider such movements separately and add up the results of two movements into one displacement x. See Figure 4–8(d).
Thus, for the flapper movement, we have (4–17) The bellows acts like a spring, and the following equation holds true: (4–18) where A is the effective area of the bellows and ks is the equivalent spring constant— that is, the stiffness due to the action of the corrugated side of the bellows.
Assuming that all variations in the variables are within a linear range, we can obtain a block diagram for this system from Equations (4–16), (4–17), and (4–18) as shown in Figure 4–8(e). From Figure 4–8(e), it can be clearly seen that the pneumatic controller shown in Figure 4–8(a) itself is a feedback system.The transfer function between and e is given by (4–19) A simplified block diagram is shown in Figure 4–8(f). Since and e are proportional, the pneumatic controller shown in Figure 4–8(a) is a pneumatic proportional controller.
As seen from Equation (4–19), the gain of the pneumatic proportional controller can be widely varied by adjusting the flapper connecting linkage. [The flapper connecting link-age is not shown in Figure 4–8(a).] In most commercial proportional controllers an ad-justing knob or other mechanism is provided for varying the gain by adjusting this linkage.
As noted earlier, the actuating error signal moved the flapper in one direction, and the feedback bellows moved the flapper in the opposite direction,but to a smaller degree.
pc P c(s) E(s) = b a + b K 1 + K a a + b A ks = K p pc Apc = ks y x = b a + b e -a a + b y pc = K 3 K 2 pb = K 1 K 3 K 2 x = Kx pc = K 3 z pc pb = K 2 z K 1 pb = K 1 x Section 4–3 / Pneumatic Systems 115 (a) (b) 0 0 X X Ps Ps Pb X Ps Pb Pa Pc Pa Pc The effect of the feedback bellows is thus to reduce the sensitivity of the controller.The principle of feedback is commonly used to obtain wide proportional-band controllers.
Pneumatic controllers that do not have feedback mechanisms [which means that one end of the flapper is fixed, as shown in Figure 4–9(a)] have high sensitivity and are called pneumatic two-position controllers or pneumatic on–off controllers. In such a con-troller, only a small motion between the nozzle and the flapper is required to give a complete change from the maximum to the minimum control pressure. The curves re-lating to X and to X are shown in Figure 4–9(b). Notice that a small change in X can cause a large change in which causes the diaphragm valve to be completely open or completely closed.
Pneumatic Proportional Controllers (Force-Balance Type).
Figure 4–10 shows a schematic diagram of a force-balance type pneumatic proportional controller. Force-balance type controllers are in extensive use in industry.Such controllers are called stack controllers.The basic principle of operation does not differ from that of the force-distance type controller.The main advantage of the force-balance type controller is that it elimi-nates many mechanical linkages and pivot joints,thereby reducing the effects of friction.
In what follows, we shall consider the principle of the force-balance type controller.
In the controller shown in Figure 4–10, the reference input pressure and the output pressure are fed to large diaphragm chambers. Note that a force-balance type pneu-matic controller operates only on pressure signals.Therefore, it is necessary to convert the reference input and system output to corresponding pressure signals.
P o P r P b , P c P b Output pressure Pr Po A1 A1 A2 Reference input pressure X + x Pc + pc Atmosphere Air supply Control pressure P1 = k (Pc + pc) Figure 4–10 Schematic diagram of a force-balance type pneumatic proportional controller.
Figure 4–9 (a) Pneumatic controller without a feedback mechanism; (b) curves versus X and versus X.
P c P b 116 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems As in the case of the force-distance type controller, this controller employs a flapper, nozzle, and orifices. In Figure 4–10, the drilled opening in the bottom chamber is the nozzle.The diaphragm just above the nozzle acts as a flapper.
The operation of the force-balance type controller shown in Figure 4–10 may be summarized as follows: 20-psig air from an air supply flows through an orifice, causing a reduced pressure in the bottom chamber. Air in this chamber escapes to the atmos-phere through the nozzle. The flow through the nozzle depends on the gap and the pressure drop across it. An increase in the reference input pressure while the out-put pressure remains the same, causes the valve stem to move down, decreasing the gap between the nozzle and the flapper diaphragm.This causes the control pressure to increase. Let (4–20) If pe=0, there is an equilibrium state with the nozzle–flapper distance equal to and the control pressure equal to At this equilibrium state, and (4–21) where a is a constant.
Let us assume that pe Z 0 and define small variations in the nozzle–flapper distance and control pressure as x and pc, respectively.Then we obtain the following equation: (4–22) From Equations (4–21) and (4–22), we obtain (4–23) At this point, we must examine the quantity x. In the design of pneumatic controllers, the nozzle–flapper distance is made quite small. In view of the fact that x/a is very much smaller than pc(1-k)A1 or peAA2-A1B—that is, for pe Z 0 we may neglect the term x in our analysis. Equation (4–23) can then be rewritten to reflect this assumption as follows: and the transfer function between pc and pe becomes where pe is defined by Equation (4–20). The controller shown in Figure 4–10 is a proportional controller.The value of gain Kp increases as k approaches unity. Note that the value of k depends on the diameters of the orifices in the inlet and outlet pipes of the feedback chamber. (The value of k approaches unity as the resistance to flow in the orifice of the inlet pipe is made smaller.) P c(s) P e(s) = A2 - A1 A1 1 1 - k = K p pc(1 - k)A1 = peAA2 - A1B x a peAA2 - A1B x a pc(1 - k)A1 x = aCpc(1 - k)A1 - peAA2 - A1BD X – + x = aC AP – c + pcBA1 - AP – c + pcBkA1 - peAA2 - A1B D X – = aAP – c A1 - P – c kA1B P 1 = P – c k (where k 6 1) P – c .
X – pe = P r - P o P c P o P r , Section 4–3 / Pneumatic Systems 117 C Pc + pc A k X + x Q + qi Figure 4–11 Schematic diagram of a pneumatic actuating valve.
Pneumatic Actuating Valves.
One characteristic of pneumatic controls is that they almost exclusively employ pneumatic actuating valves.A pneumatic actuating valve can provide a large power output. (Since a pneumatic actuator requires a large power input to produce a large power output, it is necessary that a sufficient quantity of pres-surized air be available.) In practical pneumatic actuating valves, the valve characteris-tics may not be linear; that is, the flow may not be directly proportional to the valve stem position, and also there may be other nonlinear effects, such as hysteresis.
Consider the schematic diagram of a pneumatic actuating valve shown in Figure 4–11.
Assume that the area of the diaphragm is A.Assume also that when the actuating error is zero, the control pressure is equal to and the valve displacement is equal to In the following analysis, we shall consider small variations in the variables and lin-earize the pneumatic actuating valve. Let us define the small variation in the control pressure and the corresponding valve displacement to be and x, respectively. Since a small change in the pneumatic pressure force applied to the diaphragm repositions the load, consisting of the spring, viscous friction, and mass, the force-balance equa-tion becomes where m=mass of the valve and valve stem b=viscous-friction coefficient k=spring constant If the force due to the mass and viscous friction are negligibly small, then this last equa-tion can be simplified to The transfer function between x and thus becomes X(s) P c(s) = A k = K c pc Apc = kx Apc = mx $ + bx # + kx pc X – .
P – c 118 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems where X(s)=l[x] and If qi, the change in flow through the pneumatic actuating valve, is proportional to x, the change in the valve-stem displacement, then where Qi(s)=lCqiD and Kq is a constant. The transfer function between qi and becomes where Kv is a constant.
The standard control pressure for this kind of a pneumatic actuating valve is between 3 and 15 psig. The valve-stem displacement is limited by the allowable stroke of the diaphragm and is only a few inches. If a longer stroke is needed, a piston–spring combination may be employed.
In pneumatic actuating valves, the static-friction force must be limited to a low value so that excessive hysteresis does not result. Because of the compressibility of air, the control action may not be positive; that is, an error may exist in the valve-stem position.
The use of a valve positioner results in improvements in the performance of a pneu-matic actuating valve.
Basic Principle for Obtaining Derivative Control Action.
We shall now present methods for obtaining derivative control action. We shall again place the emphasis on the principle and not on the details of the actual mechanisms.
The basic principle for generating a desired control action is to insert the inverse of the desired transfer function in the feedback path. For the system shown in Figure 4–12, the closed-loop transfer function is If @G(s)H(s)@ 1, then C(s)/R(s) can be modified to Thus, if proportional-plus-derivative control action is desired, we insert an element having the transfer function 1/(Ts+1) in the feedback path.
C(s) R(s) = 1 H(s) C(s) R(s) = G(s) 1 + G(s)H(s) Qi(s) P c(s) = K c K q = K v pc Qi(s) X(s) = K q P c(s) = lCpcD.
R(s) C(s) G(s) H(s) + – Figure 4–12 Control system.
Section 4–3 / Pneumatic Systems 119 + – e a b (a) (b) Pc(s) E(s) X(s) b a + b K a a + b A ks X + x Pc + pc Ps Figure 4–13 (a) Pneumatic proportional controller; (b) block diagram of the controller.
(a) (b) (c) e e a b Ps pc X + x Pc + pc R C Pc(s) E(s) X(s) K x t t t a a + b A ks b a + b 1 RCs + 1 + – Figure 4–14 (a) Pneumatic proportional-plus-derivative controller; (b) step change in e and the corre-sponding changes in x and pc plotted versus t; (c) block diagram of the controller.
Consider the pneumatic controller shown in Figure 4–13(a).Considering small changes in the variables,we can draw a block diagram of this controller as shown in Figure 4–13(b).
From the block diagram we see that the controller is of proportional type.
We shall now show that the addition of a restriction in the negative feedback path will modify the proportional controller to a proportional-plus-derivative controller, or a PD controller.
Consider the pneumatic controller shown in Figure 4–14(a).Assuming again small changes in the actuating error, nozzle–flapper distance, and control pressure, we can summarize the operation of this controller as follows: Let us first assume a small step change in e.
120 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems Then the change in the control pressure will be instantaneous.The restriction R will mo-mentarily prevent the feedback bellows from sensing the pressure change pc.Thus the feed-back bellows will not respond momentarily, and the pneumatic actuating valve will feel the full effect of the movement of the flapper.As time goes on,the feedback bellows will expand.
The change in the nozzle–flapper distance x and the change in the control pressure can be plotted against time t, as shown in Figure 4–14(b).At steady state, the feedback bellows acts like an ordinary feedback mechanism.The curve versus t clearly shows that this con-troller is of the proportional-plus-derivative type.
A block diagram corresponding to this pneumatic controller is shown in Figure 4–14(c). In the block diagram, K is a constant, A is the area of the bellows, and ks is the equivalent spring constant of the bellows.The transfer function between and e can be obtained from the block diagram as follows: In such a controller the loop gain is made much greater than unity.Thus the transfer function Pc(s)/E(s) can be simplified to give where Thus, delayed negative feedback, or the transfer function 1/(RCs+1) in the feedback path, modifies the proportional controller to a proportional-plus-derivative controller.
Note that if the feedback valve is fully opened, the control action becomes propor-tional. If the feedback valve is fully closed, the control action becomes narrow-band proportional (on–off).
Obtaining Pneumatic Proportional-Plus-Integral Control Action.
Consider the proportional controller shown in Figure 4–13(a). Considering small changes in the variables, we can show that the addition of delayed positive feedback will modify this proportional controller to a proportional-plus-integral controller, or a PI controller.
Consider the pneumatic controller shown in Figure 4–15(a).The operation of this con-troller is as follows:The bellows denoted by I is connected to the control pressure source without any restriction. The bellows denoted by II is connected to the control pressure source through a restriction.Let us assume a small step change in the actuating error.This will cause the back pressure in the nozzle to change instantaneously.Thus a change in the control pressure also occurs instantaneously. Due to the restriction of the valve in the path to bellows II, there will be a pressure drop across the valve.As time goes on, air will flow across the valve in such a way that the change in pressure in bellows II attains the value pc.Thus bellows II will expand or contract as time elapses in such a way as to move the flapper an additional amount in the direction of the original displacement e.This will cause the back pressure in the nozzle to change continuously, as shown in Figure 4–15(b).
pc pc K p = bks aA , T d = RC P c(s) E(s) = K pA1 + T d sB @KaAC(a + b)ks(RCs + 1)D @ P c(s) E(s) = b a + b K 1 + Ka a + b A ks 1 RCs + 1 pc pc pc pc Section 4–3 / Pneumatic Systems 121 (a) (b) (c) (d) e a b X + x Pc + pc Ps R C pc K K x e t t t E(s) X(s) P c(s) E(s) X(s) P c(s) a a + b b a + b b a + b A ks a a + b A ks a a + b A ks 1 RCs + 1 1 RCs + 1 I II + – ++ + – + – Figure 4–15 (a) Pneumatic proportional-plus-integral controller; (b) step change in e and the corre-sponding changes in x and pc plotted versus t; (c) block diagram of the controller; (d) simplified block diagram.
Note that the integral control action in the controller takes the form of slowly canceling the feedback that the proportional control originally provided.
A block diagram of this controller under the assumption of small variations in the variables is shown in Figure 4–15(c). A simplification of this block diagram yields Figure 4–15(d).The transfer function of this controller is P c(s) E(s) = b a + b K 1 + Ka a + b A ks a1 -1 RCs + 1 b 122 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems + – (a) (b) e a b X + x Ps Ri Rd C C Pc + pc (Ri Rd) Pc(s) E(s) X(s) K b a + b a a + b A ks 1 RdCs + 1 1 RiCs + 1 – + Figure 4–16 (a) Pneumatic proportional-plus-integral-plus-derivative controller; (b) block diagram of the controller.
where K is a constant,A is the area of the bellows,and ks is the equivalent spring constant of the combined bellows.If which is usually the case, the transfer function can be simplified to where Obtaining Pneumatic Proportional-Plus-Integral-Plus-Derivative Control Action.
A combination of the pneumatic controllers shown in Figures 4–14(a) and 4–15(a) yields a proportional-plus-integral-plus-derivative controller, or a PID con-troller. Figure 4–16(a) shows a schematic diagram of such a controller. Figure 4–16(b) shows a block diagram of this controller under the assumption of small variations in the variables.
K p = bks aA , T i = RC P c(s) E(s) = K p a1 + 1 T i s b @KaARCsC(a + b)ks(RCs + 1)D @ 1, Section 4–4 / Hydraulic Systems 123 The transfer function of this controller is By defining and noting that under normal operation and we obtain (4–24) where Equation (4–24) indicates that the controller shown in Figure 4–16(a) is a proportional-plus-integral-plus-derivative controller or a PID controller.
4–4 HYDRAULIC SYSTEMS Except for low-pressure pneumatic controllers,compressed air has seldom been used for the continuous control of the motion of devices having significant mass under external load forces. For such a case, hydraulic controllers are generally preferred.
Hydraulic Systems.
The widespread use of hydraulic circuitry in machine tool applications, aircraft control systems, and similar operations occurs because of such fac-tors as positiveness, accuracy, flexibility, high horsepower-to-weight ratio, fast starting, stopping, and reversal with smoothness and precision, and simplicity of operations.
The operating pressure in hydraulic systems is somewhere between 145 and 5000 lbfin.2 (between 1 and 35 MPa). In some special applications, the operating pressure may go up to 10,000 lbfin.2 (70 MPa). For the same power requirement, the weight and size of the hydraulic unit can be made smaller by increasing the supply pressure. With high-pressure hydraulic systems, very large force can be obtained. Rapid-acting, accurate positioning of heavy loads is possible with hydraulic systems. A combination of elec-tronic and hydraulic systems is widely used because it combines the advantages of both electronic control and hydraulic power.
K p = bks aA = K p a1 + 1 T i s + T d s b bks aA T d T i s2 + T i s + 1 T i s P c(s) E(s) bks aA AT d s + 1BAT i s + 1B AT i - T dBs T i T d , 1 @KaAAT i - T dBsC(a + b)ksAT d s + 1BAT i s + 1B D @ T i = Ri C, T d = Rd C P c(s) E(s) = bK a + b 1 + Ka a + b A ks ARi C - Rd CBs ARd Cs + 1BARi Cs + 1B 124 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems Advantages and Disadvantages of Hydraulic Systems.
There are certain advantages and disadvantages in using hydraulic systems rather than other systems.
Some of the advantages are the following: 1. Hydraulic fluid acts as a lubricant, in addition to carrying away heat generated in the system to a convenient heat exchanger.
2. Comparatively small-sized hydraulic actuators can develop large forces or torques.
3. Hydraulic actuators have a higher speed of response with fast starts, stops, and speed reversals.
4. Hydraulic actuators can be operated under continuous, intermittent, reversing, and stalled conditions without damage.
5. Availability of both linear and rotary actuators gives flexibility in design.
6. Because of low leakages in hydraulic actuators, speed drop when loads are applied is small.
On the other hand, several disadvantages tend to limit their use.
1. Hydraulic power is not readily available compared to electric power.
2. Cost of a hydraulic system may be higher than that of a comparable electrical system performing a similar function.
3. Fire and explosion hazards exist unless fire-resistant fluids are used.
4. Because it is difficult to maintain a hydraulic system that is free from leaks, the system tends to be messy.
5. Contaminated oil may cause failure in the proper functioning of a hydraulic system.
6. As a result of the nonlinear and other complex characteristics involved, the design of sophisticated hydraulic systems is quite involved.
7. Hydraulic circuits have generally poor damping characteristics.If a hydraulic circuit is not designed properly, some unstable phenomena may occur or disappear, de-pending on the operating condition.
Comments.
Particular attention is necessary to ensure that the hydraulic system is stable and satisfactory under all operating conditions. Since the viscosity of hydraulic fluid can greatly affect damping and friction effects of the hydraulic circuits, stability tests must be carried out at the highest possible operating temperature.
Note that most hydraulic systems are nonlinear. Sometimes, however, it is possible to linearize nonlinear systems so as to reduce their complexity and permit solutions that are sufficiently accurate for most purposes.A useful linearization technique for dealing with nonlinear systems was presented in Section 2–7.
Hydraulic Servo System.
Figure 4–17(a) shows a hydraulic servomotor. It is essentially a pilot-valve-controlled hydraulic power amplifier and actuator. The pilot valve is a balanced valve,in the sense that the pressure forces acting on it are all balanced.
A very large power output can be controlled by a pilot valve, which can be positioned with very little power.
In practice, the ports shown in Figure 4–17(a) are often made wider than the corre-sponding valves. In such a case, there is always leakage through the valves. Such leak-Section 4–4 / Hydraulic Systems 125 x y q q p0 p0 ps p1 p2 2 3 4 1 (a) x (b) 2 1 ps x0 2 + x x0 2 – x Load m b Figure 4–17 (a) Hydraulic servo system; (b) enlarged diagram of the valve orifice area.
age improves both the sensitivity and the linearity of the hydraulic servomotor. In the following analysis we shall make the assumption that the ports are made wider than the valves—that is, the valves are underlapped. [Note that sometimes a dither signal, a high-frequency signal of very small amplitude (with respect to the maximum displacement of the valve), is superimposed on the motion of the pilot valve. This also improves the sensitivity and linearity.In this case also there is leakage through the valve.] We shall apply the linearization technique presented in Section 2–7 to obtain a lin-earized mathematical model of the hydraulic servomotor. We assume that the valve is underlapped and symmetrical and admits hydraulic fluid under high pressure into a power cylinder that contains a large piston, so that a large hydraulic force is established to move a load.
In Figure 4–17(b) we have an enlarged diagram of the valve orifice area. Let us define the valve orifice areas of ports 1, 2, 3, 4 as A1, A2, A3, A4, respectively.Also, define the flow rates through ports 1, 2, 3, 4 as q1, q2, q3, q4, respectively. Note that, since the 126 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems valve is symmetrical, A1=A3 and A2=A4.Assuming the displacement x to be small, we obtain where k is a constant.
Furthermore, we shall assume that the return pressure po in the return line is small and thus can be neglected. Then, referring to Figure 4–17(a), flow rates through valve orifices are where and and g is the specific weight and is given by g=rg, where r is mass density and g is the acceleration of gravity. The flow rate q to the left-hand side of the power piston is (4–25) The flow rate from the right-hand side of the power piston to the drain is the same as this q and is given by In the present analysis we assume that the fluid is incompressible. Since the valve is symmetrical, we have q1=q3 and q2=q4. By equating q1 and q3, we obtain or If we define the pressure difference across the power piston as or ¢p = p1 - p2 ¢p ps = p1 + p2 ps - p1 = p2 q = q3 - q2 = C11p2 a x0 2 + x b - C21ps - p2 a x0 2 - x b q = q1 - q4 = C11ps - p1 a x0 2 + x b - C21p1 a x0 2 - x b C2 = c2 k12gg, C1 = c1 k12gg q4 = c2 A4 B 2g g Ap1 - p0B = C21p1 - p0 a x0 2 - x b = C21p1 a x0 2 - x b q3 = c1 A3 B 2g g Ap2 - p0B = C11p2 - p0 a x0 2 + x b = C11p2 a x0 2 + x b q2 = c2 A2 B 2g g Aps - p2B = C21ps - p2 a x0 2 - x b q1 = c1 A1 B 2g g Aps - p1B = C11ps - p1 a x0 2 + x b A2 = A4 = k a x0 2 - x b A1 = A3 = k a x0 2 + x b Section 4–4 / Hydraulic Systems 127 then For the symmetrical valve shown in Figure 4–17(a), the pressure in each side of the power piston is (1/2)ps when no load is applied, or As the spool valve is dis-placed, the pressure in one line increases as the pressure in the other line decreases by the same amount.
In terms of ps and we can rewrite the flow rate q given by Equation (4–25) as Noting that the supply pressure ps is constant. the flow rate q can be written as a func-tion of the valve displacement x and pressure difference or By applying the linearization technique presented in Section 3–10 to this case,the lin-earized equation about point is (4–26) where Coefficients a and b here are called valve coefficients. Equation (4–26) is a linearized mathematical model of the spool valve near an operating point The values of valve coefficients a and b vary with the operating point. Note that is negative and so b is negative.
Since the normal operating point is the point where near the normal operating point Equation (4–26) becomes (4–27) where K 2 = AC1 + C2B x0 412 1ps 7 0 K 1 = AC1 + C2B A ps 2 7 0 q = K1 x - K2 ¢p x – = 0, ¢p – = 0, q – = 0, 0f0¢p x = x –, ¢p = ¢p –, q = q –.
+ C2 212 1ps + ¢p – a x0 2 - x – b R 6 0 b = 0f 0¢p 2 x=x –, ¢p= ¢p – = - c C1 212 1ps - ¢p – a x0 2 + x – b a = 0f 0x 2 x=x –, ¢p= ¢p – = C1 A ps - ¢p – 2 + C2 A ps + ¢p – 2 q – = f(x –, ¢p –) q - q – = a(x - x –) + b(¢p - ¢p –) x = x –, ¢p = ¢p –, q = q – q = C1 A ps - ¢p 2 a x0 2 + x b - C2 A ps + ¢p 2 a x0 2 - x b = f(x, ¢p) ¢p, q = q1 - q4 = C1 A ps - ¢p 2 a x0 2 + x b - C2 A ps + ¢p 2 a x0 2 - x b ¢p, ¢p = 0.
p1 = ps + ¢p 2 , p2 = ps - ¢p 2 128 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems x = 2x1 x = x1 x = 0 x = –x1 0 q P x = –2x1 Figure 4–18 Characteristic curves of the linearized hydraulic servomotor.
Equation (4–27) is a linearized mathematical model of the spool valve near the origin Note that the region near the origin is most important in this kind of system, because the system operation usually occurs near this point.
Figure 4–18 shows this linearized relationship among q, x, and The straight lines shown are the characteristic curves of the linearized hydraulic servomotor.This family of curves consists of equidistant parallel straight lines, parametrized by x.
In the present analysis we assume that the load reactive forces are small, so that the leakage flow rate and oil compressibility can be ignored.
Referring to Figure 4–17(a), we see that the rate of flow of oil q times dt is equal to the power-piston displacement dy times the piston area A times the density of oil r.
Thus, we obtain Notice that for a given flow rate q the larger the piston area A is, the lower will be the velocity dydt. Hence, if the piston area A is made smaller, the other variables re-maining constant, the velocity dydt will become higher.Also, an increased flow rate q will cause an increased velocity of the power piston and will make the response time shorter.
Equation (4–27) can now be written as The force developed by the power piston is equal to the pressure difference times the piston area A or = A K 2 a K 1 x - Ar dy dt b Force developed by the power piston = A ¢P ¢P ¢P = 1 K 2 a K 1 x - Ar dy dt b Ar dy = q dt ¢P.
(x – = 0, ¢p – = 0, q – = 0.) Section 4–4 / Hydraulic Systems 129 For a given maximum force, if the pressure difference is sufficiently high, the piston area, or the volume of oil in the cylinder, can be made small. Consequently, to minimize the weight of the controller, we must make the supply pressure sufficiently high.
Assume that the power piston moves a load consisting of a mass and viscous friction.
Then the force developed by the power piston is applied to the load mass and friction, and we obtain or (4–28) where m is the mass of the load and b is the viscous-friction coefficient.
Assuming that the pilot-valve displacement x is the input and the power-piston displacement y is the output, we find that the transfer function for the hydraulic servo-motor is, from Equation (4–28), (4–29) where and From Equation (4–29) we see that this transfer function is of the second order.If the ratio is negligibly small or the time constant T is negligible, the transfer function Y(s)/X(s) can be simplified to give It is noted that a more detailed analysis shows that if oil leakage, compressibility (including the effects of dissolved air), expansion of pipelines, and the like are taken into consideration, the transfer function becomes where and are time constants.As a matter of fact, these time constants depend on the volume of oil in the operating circuit.The smaller the volume, the smaller the time constants.
T 2 T 1 Y(s) X(s) = K sAT 1 s + 1BAT 2 s + 1B Y(s) X(s) = K s mK 2AbK 2 + A 2rB T = mK 2 bK 2 + A 2r K = 1 bK 2 AK 1 + Ar K 1 = K s(Ts + 1) Y(s) X(s) = 1 sc a mK 2 AK 1 b s + bK 2 AK 1 + Ar K 1 d my $ + ab + A 2r K 2 by # = AK 1 K 2 x my $ + by # = A K 2 AK 1 x - Ary # B 130 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems x Port I Port II Power cylinder y Pilot valve Oil under pressure Figure 4–19 Hydraulic servomotor.
Hydraulic Integral Controller.
The hydraulic servomotor shown in Figure 4–19 is a pilot-valve-controlled hydraulic power amplifier and actuator. Similar to the hydraulic servo system shown in Figure 4–17, for negligibly small load mass the servomotor shown in Figure 4–19 acts as an integrator or an integral controller. Such a servomotor consti-tutes the basis of the hydraulic control circuit.
In the hydraulic servomotor shown in Figure 4–19, the pilot valve (a four-way valve) has two lands on the spool. If the width of the land is smaller than the port in the valve sleeve,the valve is said to be underlapped.Overlapped valves have a land width greater than the port width.A zero-lapped valve has a land width that is identical to the port width. (If the pilot valve is a zero-lapped valve,analyses of hydraulic servomotors become simpler.) In the present analysis, we assume that hydraulic fluid is incompressible and that the inertia force of the power piston and load is negligible compared to the hydraulic force at the power piston.We also assume that the pilot valve is a zero-lapped valve, and the oil flow rate is proportional to the pilot valve displacement.
Operation of this hydraulic servomotor is as follows. If input x moves the pilot valve to the right, port II is uncovered, and so high-pressure oil enters the right-hand side of the power piston. Since port I is connected to the drain port, the oil in the left-hand side of the power piston is returned to the drain. The oil flowing into the power cylinder is at high pressure; the oil flowing out from the power cylinder into the drain is at low pressure. The resulting difference in pressure on both sides of the power piston will cause it to move to the left.
Note that the rate of flow of oil q (kgsec) times dt (sec) is equal to the power-piston displacement dy (m) times the piston area A (m2) times the density of oil r (kgm3).
Therefore, (4–30) Because of the assumption that the oil flow rate q is proportional to the pilot-valve displacement x, we have (4–31) where K1 is a positive constant. From Equations (4–30) and (4–31) we obtain Ar dy dt = K 1 x q = K 1 x Ar dy = q dt Section 4–4 / Hydraulic Systems 131 (a) (b) e b a x y II I A B C Oil under pressure E(s) X(s) Y(s) a a + b b a + b K s + – Figure 4–20 (a) Servomotor that acts as a proportional controller; (b) block diagram of the servomotor.
The Laplace transform of this last equation, assuming a zero initial condition, gives or where K=K1/(Ar). Thus the hydraulic servomotor shown in Figure 4–19 acts as an integral controller.
Hydraulic Proportional Controller.
It has been shown that the servomotor in Figure 4–19 acts as an integral controller. This servomotor can be modified to a pro-portional controller by means of a feedback link. Consider the hydraulic controller shown in Figure 4–20(a).The left-hand side of the pilot valve is joined to the left-hand side of the power piston by a link ABC.This link is a floating link rather than one mov-ing about a fixed pivot.
The controller here operates in the following way. If input e moves the pilot valve to the right, port II will be uncovered and high-pressure oil will flow through port II into the right-hand side of the power piston and force this piston to the left.The power pis-ton, in moving to the left, will carry the feedback link ABC with it, thereby moving the pilot valve to the left.This action continues until the pilot piston again covers ports I and II.A block diagram of the system can be drawn as in Figure 4–20(b).The transfer func-tion between Y(s) and E(s) is given by Noting that under the normal operating conditions we have this last equation can be simplified to Y(s) E(s) = b a = K p @KaCs(a + b)D @ 1, Y(s) E(s) = b a + b K s 1 + K s a a + b Y(s) X(s) = K 1 Ars = K s ArsY(s) = K 1 X(s) 132 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems + – (a) (b) (c) R q P2 P1 A k y y z z t t Y(s) Z(s) 1 Ts T = RA2r k The transfer function between y and e becomes a constant.Thus,the hydraulic controller shown in Figure 4–20(a) acts as a proportional controller,the gain of which is Kp.This gain can be adjusted by effectively changing the lever ratio b/a. (The adjusting mechanism is not shown in the diagram.) We have thus seen that the addition of a feedback link will cause the hydraulic servomotor to act as a proportional controller.
Dashpots.
The dashpot (also called a damper) shown in Figure 4–21(a) acts as a differentiating element.Suppose that we introduce a step displacement to the piston po-sition y.Then the displacement z becomes equal to y momentarily.Because of the spring force,however,the oil will flow through the resistance R and the cylinder will come back to the original position.The curves y versus t and z versus t are shown in Figure 4–21(b).
Let us derive the transfer function between the displacement z and displacement y.
Define the pressures existing on the right and left sides of the piston as and respectively. Suppose that the inertia force involved is negligible.Then the force acting on the piston must balance the spring force.Thus where A=piston area, in.2 k=spring constant, lbfin.
The flow rate q is given by where q=flow rate through the restriction, lbsec R=resistance to flow at the restriction, lbf-secin.2-lb Since the flow through the restriction during dt seconds must equal the change in the mass of oil to the left of the piston during the same dt seconds, we obtain where r=density, lbin.3. (We assume that the fluid is incompressible or r=constant.) This last equation can be rewritten as dy dt - dz dt = q Ar = P 1 - P 2 RAr = kz RA 2r q dt = Ar(dy - dz) q = P 1 - P 2 R AAP 1 - P 2B = kz P 2(lbfin.2), P 1(lbfin.2) Figure 4–21 (a) Dashpot; (b) step change in y and the corresponding change in z plotted versus t; (c) block diagram of the dashpot.
Section 4–4 / Hydraulic Systems 133 + – (a) (b) Area = A Spring constant = k Density of oil = r Oil under pressure Resistance = R e x a b y z E(s) X(s) Y(s) b a + b a a + b K s Ts Ts + 1 Z(s) or Taking the Laplace transforms of both sides of this last equation, assuming zero initial conditions, we obtain The transfer function of this system thus becomes Let us define RA2rk=T. (Note that RA2rk has the dimension of time.) Then Clearly, the dashpot is a differentiating element. Figure 4–21(c) shows a block diagram representation for this system.
Obtaining Hydraulic Proportional-Plus-Integral Control Action.
Figure 4–22(a) shows a schematic diagram of a hydraulic proportional-plus-integral controller.A block diagram of this controller is shown in Figure 4–22(b). The transfer function Y(s)/E(s) is given by Y(s) E(s) = b a + b K s 1 + Ka a + b T Ts + 1 Z(s) Y(s) = Ts Ts + 1 = 1 1 + 1 Ts Z(s) Y(s) = s s + k RA 2r sY(s) = sZ(s) + k RA 2r Z(s) dy dt = dz dt + kz RA 2r Figure 4–22 (a) Schematic diagram of a hydraulic proportional-plus-integral controller; (b) block diagram of the controller.
134 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems + – (a) (b) e a b x y z R k q P2 P1 Area = A Density of oil = r X(s) Y(s) E(s) Z(s) b a + b a a + b K s 1 Ts + 1 In such a controller, under normal operation with the result that where Thus the controller shown in Figure 4–22(a) is a proportional-plus-integral controller (PI controller).
Obtaining Hydraulic Proportional-Plus-Derivative Control Action.
Figure 4–23(a) shows a schematic diagram of a hydraulic proportional-plus-derivative controller. The cylinders are fixed in space and the pistons can move. For this system, notice that Hence or Z(s) Y(s) = 1 Ts + 1 y = z + A k qR = z + RA 2r k dz dt q dt = rA dz q = P 2 - P 1 R k(y - z) = AAP 2 - P 1B K p = b a , T i = T = RA 2r k Y(s) E(s) = K p a1 + 1 T i s b @KaTC(a + b)(Ts + 1)D @ 1, Figure 4–23 (a) Schematic diagram of a hydraulic proportional-plus-derivative controller; (b) block diagram of the controller.
Section 4–4 / Hydraulic Systems 135 e a b x y R R k2 k1 Area = A z Figure 4–24 Schematic diagram of a hydraulic proportional-plus-integral-plus-derivative controller.
where A block diagram for this system is shown in Figure 4–23(b). From the block diagram the transfer function Y(s)/E(s) can be obtained as Under normal operation we have Hence where Thus the controller shown in Figure 4–23(a) is a proportional-plus-derivative controller (PD controller).
Obtaining Hydraulic Proportional-Plus-Integral-Plus-Derivative Control Action.
Figure 4–24 shows a schematic diagram of a hydraulic proportional-plus-integral-plus-derivative controller. It is a combination of the proportional-plus-integral controller and proportional-plus derivative controller.
If the two dashpots are identical except the piston shafts, the transfer function Z(s)/Y(s) can be obtained as follows: (For the derivation of this transfer function, refer to Problem A–4–9.) Z(s) Y(s) = T 1 s T 1 T 2 s2 + AT 1 + 2T 2Bs + 1 K p = b a , T = RA 2r k Y(s) E(s) = K p(1 + Ts) @aKC(a + b)s(Ts + 1)D @ 1.
Y(s) E(s) = b a + b K s 1 + a a + b K s 1 Ts + 1 T = RA 2r k 136 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems + – b a + b K s Y(s) E(s) X(s) Z(s) T1 s T1 T2 s2 + (T1 + 2T2)s + 1 a a + b Figure 4–25 Block diagram for the system shown in Figure 4–24.
A block diagram for this system is shown in Figure 4–25. The transfer function Y(s)/E(s) can be obtained as Under normal circumstances we design the system such that then where Thus, the controller shown in Figure 4–24 is a proportional-plus-integral-plus-derivative controller (PID controller).
4–5 THERMAL SYSTEMS Thermal systems are those that involve the transfer of heat from one substance to another. Thermal systems may be analyzed in terms of resistance and capacitance, although the thermal capacitance and thermal resistance may not be represented accurately as lumped parameters, since they are usually distributed throughout the sub-stance. For precise analysis, distributed-parameter models must be used. Here, however, to simplify the analysis we shall assume that a thermal system can be represented by a lumped-parameter model, that substances that are characterized by resistance to heat flow have negligible heat capacitance, and that substances that are characterized by heat capacitance have negligible resistance to heat flow.
K p = b a T 1 + 2T 2 T 1 , K i = b a 1 T 1 , K d = b a T 2 = K p + K i s + K d s Y(s) E(s) = b a T 1 T 2 s2 + AT 1 + 2T 2Bs + 1 T 1 s a a + b K s T 1 s T 1 T 2 s2 + AT 1 + 2T 2Bs + 1 1 Y(s) E(s) = b a + b K s 1 + a a + b K s T 1 s T 1 T 2 s2 + AT 1 + 2T 2Bs + 1 Section 4–5 / Thermal Systems 137 There are three different ways heat can flow from one substance to another: con-duction, convection, and radiation. Here we consider only conduction and convection.
(Radiation heat transfer is appreciable only if the temperature of the emitter is very high compared to that of the receiver.Most thermal processes in process control systems do not involve radiation heat transfer.) For conduction or convection heat transfer, where q=heat flow rate, kcalsec u=temperature difference, °C K=coefficient, kcalsec °C The coefficient K is given by where k=thermal conductivity, kcalm sec °C A=area normal to heat flow, m2 X=thickness of conductor, m H=convection coefficient, kcalm2 sec °C Thermal Resistance and Thermal Capacitance.
The thermal resistance R for heat transfer between two substances may be defined as follows: The thermal resistance for conduction or convection heat transfer is given by Since the thermal conductivity and convection coefficients are almost constant, the thermal resistance for either conduction or convection is constant.
The thermal capacitance C is defined by or where m=mass of substance considered, kg c=specific heat of substance, kcalkg °C C = mc C = change in heat stored, kcal change in temperature, °C R = d(¢u) dq = 1 K R = change in temperature difference, °C change in heat flow rate, kcalsec = HA, for convection K = kA ¢X , for conduction q = K ¢u 138 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems Heater Cold liquid Mixer Hot liquid (a) (b) Hi(s) R – 1 RCs Qi(s) Q(s) ++ – Thermal System.
Consider the system shown in Figure 4–26(a). It is assumed that the tank is insulated to eliminate heat loss to the surrounding air. It is also assumed that there is no heat storage in the insulation and that the liquid in the tank is perfectly mixed so that it is at a uniform temperature.Thus,a single temperature is used to describe the temperature of the liquid in the tank and of the outflowing liquid.
Let us define steady-state temperature of inflowing liquid, °C steady-state temperature of outflowing liquid, °C steady-state liquid flow rate, kgsec mass of liquid in tank, kg specific heat of liquid, kcalkg °C thermal resistance, °C seckcal thermal capacitance, kcal°C steady-state heat input rate, kcalsec Assume that the temperature of the inflowing liquid is kept constant and that the heat input rate to the system (heat supplied by the heater) is suddenly changed from to where hi represents a small change in the heat input rate.The heat outflow rate will then change gradually from to The temperature of the outflowing liq-uid will also be changed from to For this case, ho, C, and R are obtained, respectively, as The heat-balance equation for this system is C du = Ahi - hoB dt R = u ho = 1 Gc C = Mc ho = Gcu Q – o + u.
Q – o H – + ho .
H – H – + hi , H – H – = C = R = c = M = G = Q – o = Q – i = Figure 4–26 (a) Thermal system: (b) block diagram of the system.
Section 4–5 / Thermal Systems 139 or which may be rewritten as Note that the time constant of the system is equal to RC or M/G seconds.The transfer function relating u and hi is given by where and In practice, the temperature of the inflowing liquid may fluctuate and may act as a load disturbance. (If a constant outflow temperature is desired, an automatic controller may be installed to adjust the heat inflow rate to compensate for the fluctuations in the temperature of the inflowing liquid.) If the temperature of the inflowing liquid is sud-denly changed from to while the heat input rate H and the liquid flow rate G are kept constant, then the heat outflow rate will be changed from to and the temperature of the outflowing liquid will be changed from to The heat-balance equation for this case is or which may be rewritten The transfer function relating u and ui is given by where and If the present thermal system is subjected to changes in both the temperature of the inflowing liquid and the heat input rate, while the liquid flow rate is kept constant, the change u in the temperature of the outflowing liquid can be given by the following equation: A block diagram corresponding to this case is shown in Figure 4–26(b). Notice that the system involves two inputs.
RC du dt + u = ui + Rhi Qi(s) = lCui(t)D.
Q (s) = lCu(t)D Q (s) Q i(s) = 1 RCs + 1 RC du dt + u = ui C du dt = Gcui - ho C du = AGcui - hoB dt Q – o + u.
Q – o H – + ho , H – Q – i + ui Q – i H i(s) = lChi(t)D.
Q (s) = lCu(t)D Q (s) H i(s) = R RCs + 1 RC du dt + u = Rhi C du dt = hi - ho 140 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems Q Qi H Capacitance C Figure 4–27 Liquid-level system.
EXAMPLE PROBLEMS AND SOLUTIONS A–4–1.
In the liquid-level system of Figure 4–27 assume that the outflow rate Q m3sec through the out-flow valve is related to the head H m by Assume also that when the inflow rate Qi is 0.015 m3sec the head stays constant. For t<0 the system is at steady state AQi=0.015 m3secB. At t=0 the inflow valve is closed and so there is no inflow for t 0. Find the time necessary to empty the tank to half the original head. The capacitance C of the tank is 2 m2.
Solution. When the head is stationary, the inflow rate equals the outflow rate. Thus head Ho at t=0 is obtained from or The equation for the system for t>0 is or Hence Assume that, at t=t1, H=1.125 m. Integrating both sides of this last equation, we obtain It follows that or Thus, the head becomes half the original value (2.25 m) in 175.7 sec.
t1 = 175.7 21H 2 1.125 2.25 = 211.125 - 212.25 = -0.005t1 3 1.125 2.25 dH 1H = 3 t1 0 (-0.005)dt = -0.005t1 dH 1H = -0.005 dt dH dt = - Q C = -0.011H 2 -C dH = Q dt H o = 2.25 m 0.015 = 0.011H o Q = K1H = 0.011H Example Problems and Solutions 141 A–4–2.
Consider the liquid-level system shown in Figure 4–28. In the system, and are steady-state inflow rates and and are steady-state heads.The quantities qi1, qi2, h1, h2, q1, and qo are con-sidered small. Obtain a state-space representation for the system when h1 and h2 are the outputs and qi1 and qi2 are the inputs.
Solution. The equations for the system are (4–32) (4–33) (4–34) (4–35) Elimination of q1 from Equation (4–32) using Equation (4–33) results in (4–36) Eliminating q1 and qo from Equation (4–34) by using Equations (4–33) and (4–35) gives (4–37) Define state variables x1 and x2 by x1=h1 x2=h2 the input variables u1 and u2 by u1=qi1 u2=qi2 and the output variables y1 and y2 by y1=h1=x1 y2=h2=x2 Then Equations (4–36) and (4–37) can be written as x # 2 = 1 R1 C2 x1 - a 1 R1 C2 + 1 R2 C2 bx2 + 1 C2 u2 x # 1 = -1 R1 C1 x1 + 1 R1 C1 x2 + 1 C1 u1 dh2 dt = 1 C2 a h1 - h2 R1 + qi2 - h2 R2 b dh1 dt = 1 C1 aqi1 - h1 - h2 R1 b h2 R2 = qo C2 dh2 = Aq1 + qi2 - qoB dt h1 - h2 R1 = q1 C1 dh1 = Aqi1 - q1B dt H – 2 H – 1 Q – 2 Q – 1 C1 C2 R1 R2 Q1 + q1 Q2 + qi2 Q1 + qi1 Q1 + Q2 + qo H1 + h1 H2 + h2 Figure 4–28 Liquid-level system.
142 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems In the form of the standard vector-matrix representation, we have which is the state equation, and which is the output equation.
A–4–3.
The value of the gas constant for any gas may be determined from accurate experimental obser-vations of simultaneous values of p, v, and T.
Obtain the gas constant Rair for air. Note that at 32°F and 14.7 psia the specific volume of air is 12.39 ft3lb.Then obtain the capacitance of a 20-ft3 pressure vessel that contains air at 160°F.As-sume that the expansion process is isothermal.
Solution.
Referring to Equation (4–12), the capacitance of a 20-ft3 pressure vessel is Note that in terms of SI units, Rair is given by Rair=287 N-mkg K A–4–4.
In the pneumatic pressure system of Figure 4–29(a), assume that, for t<0, the system is at steady state and that the pressure of the entire system is Also, assume that the two bellows are identi-cal.At t=0, the input pressure is changed from to Then the pressures in bellows 1 and 2 will change from to and from to respectively.The capacity (volume) of each bellows is 510–4 m3, and the operating-pressure difference (difference between pi and p1 or difference between pi and p2) is between –0.5105 Nm2 and 0.5105 Nm2.The corresponding mass flow rates (kgsec) through the valves are shown in Figure 4–29(b).Assume that the bellows expand or contract linearly with the air pressures applied to them, that the equivalent spring con-stant of the bellows system is k=1105 Nm,and that each bellows has area A=1510–4 m2.
¢p P – + p2 , P – P – + p1 P – P – + pi .
P – P –.
C = V nRair T = 20 1 53.3 620 = 6.05 10-4 lb lbfft2 Rair = pv T = 14.7 144 12.39 460 + 32 = 53.3 ft-lbflb°R B y1 y2 R = B 1 0 0 1R Bx1 x2 R B x # 1 x # 2 R = D -1 R1 C1 1 R1 C2 1 R1 C1 - a 1 R1 C2 + 1 R2 C2 b T B x1 x2 R + D 1 C1 0 0 1 C2 T B u1 u2 R Bellows 1 Bellows 2 Valve 1 Valve 2 (a) (b) x Area A C C q1 q2 R1 R2 P+ p1 P+ p2 P + pi Valve 2 Valve 1 0.5 105 –3 10–5 1.5 10–5 – 0.5 105 Dp(N/m2) q(kg/sec) Figure 4–29 (a) Pneumatic pressure system; (b) pressure-difference-versus-mass-flow-rate curves.
Example Problems and Solutions 143 Defining the displacement of the midpoint of the rod that connects two bellows as x, find the transfer function Assume that the expansion process is isothermal and that the temperature of the entire system stays at 30°C.Assume also that the polytropic exponent n is 1.
Solution. Referring to Section 4–3, transfer function can be obtained as (4–38) Similarly, transfer function is (4–39) The force acting on bellows 1 in the x direction is and the force acting on bellows 2 in the negative x direction is The resultant force balances with kx, the equivalent spring force of the corrugated sides of the bellows.
or (4–40) Referring to Equations (4–38) and (4–39), we see that By substituting this last equation into Equation (4–40) and rewriting, the transfer function is obtained as (4–41) The numerical values of average resistances R1 and R2 are The numerical value of capacitance C of each bellows is where Rair=287 N-mkg K. (See Problem A–4–3.) Consequently, R1C=0.16710105.7510–9=9.60 sec R2C=0.33310105.7510–9=19.2 sec By substituting the numerical values for A, k, R1C, and R2C into Equation (4–41), we obtain X(s) P i(s) = 1.44 10-7s (9.6s + 1)(19.2s + 1) C = V nRair T = 5 10-4 1 287 (273 + 30) = 5.75 10-9 kg Nm2 R2 = d ¢p dq2 = 0.5 105 1.5 10-5 = 0.333 1010 Nm2 kgsec R1 = d ¢p dq1 = 0.5 105 3 10-5 = 0.167 1010 Nm2 kgsec X(s) P i(s) = A k AR2 C - R1 CBs AR1 Cs + 1BAR2 Cs + 1B X(s)P i(s) = R2 Cs - R1 Cs AR1 Cs + 1BAR2 Cs + 1B P i(s) P 1(s) - P 2(s) = a 1 R1 Cs + 1 -1 R2 Cs + 1 bP i(s) ACP 1(s) - P 2(s)D = kX(s) AAp1 - p2B = kx AAP – + p2B.
AAP – + p1B, P 2(s) P i(s) = 1 R2 Cs + 1 P 2(s)P i(s) P 1(s) P i(s) = 1 R1 Cs + 1 P 1(s)P i(s) X(s)P i(s).
144 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems A–4–5.
Draw a block diagram of the pneumatic controller shown in Figure 4–30.Then derive the transfer function of this controller.Assume that Assume also that the two bellows are identical.
If the resistance Rd is removed (replaced by the line-sized tubing),what control action do we get?
If the resistance Ri is removed (replaced by the line-sized tubing), what control action do we get?
Solution. Let us assume that when e=0 the nozzle–flapper distance is equal to and the con-trol pressure is equal to In the present analysis, we shall assume small deviations from the respective reference values as follows: small error signal small change in the nozzle–flapper distance small change in the control pressure small pressure change in bellows I due to small change in the control pressure small pressure change in bellows II due to small change in the control pressure small displacement at the lower end of the flapper In this controller, is transmitted to bellows I through the resistance Rd. Similarly, is trans-mitted to bellows II through the series of resistances Rd and Ri.The relationship between and is where derivative time. Similarly, pII and pI are related by the transfer function where integral time.The force-balance equation for the two bellows is where ks is the stiffness of the two connected bellows and A is the cross-sectional area of the bellows.The relationship among the variables e, x, and y is The relationship between and x is pc=Kx (K>0) pc x = b a + b e -a a + b y ApI - pIIBA = ks y T i = RiC = P II(s) P I(s) = 1 Ri Cs + 1 = 1 T i s + 1 T d = RdC = P I(s) P c(s) = 1 Rd Cs + 1 = 1 T d s + 1 pc pI pc pc y = pII = pI = pc = x = e = P – c .
X – Rd Ri .
e a b C C X + x Pc + pI Pc + pII Ps I II Ri Rd Pc + pc y Figure 4–30 Schematic diagram of a pneumatic controller.
Example Problems and Solutions 145 From the equations just derived, a block diagram of the controller can be drawn, as shown in Figure 4–31(a). Simplification of this block diagram results in Figure 4–31(b).
The transfer function between Pc(s) and E(s) is For a practical controller, under normal operation is very much greater than unity and Therefore, the transfer function can be simplified as follows: where Thus the controller shown in Figure 4–30 is a proportional-plus-integral-plus-derivative one.
If the resistance Rd is removed, or Rd=0, the action becomes that of a proportional-plus-integral controller.
K p = bks aA K p a1 + 1 T i s + T d s b = bks aA a T i + T d T i + 1 T i s + T d s b P c(s) E(s) bksAT i s + 1BAT d s + 1B aAT i s T i T d .
@KaAT i sC(a + b)ksAT i s + 1BAT d s + 1BD @ P c(s) E(s) = b a + b K 1 + K a a + b A ks a T i s T i s + 1 b a 1 T d s + 1 b + – + – + – E(s) X(s) Pc(s) K a a + b b a + b A ks PI(s) PII(s) 1 Td s + 1 1 Ti s + 1 (a) (b) K b a + b Pc(s) E(s) X(s) aATi s (a + b) ks(Ti s + 1) (Td s + 1) Figure 4–31 (a) Block diagram of the pneumatic controller shown in Figure 4–30; (b) simplified block diagram.
146 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems If the resistance Ri is removed, or Ri=0, the action becomes that of a narrow-band propor-tional, or two-position, controller. (Note that the actions of two feedback bellows cancel each other, and there is no feedback.) A–4–6.
Actual spool valves are either overlapped or underlapped because of manufacturing tolerances.
Consider the overlapped and underlapped spool valves shown in Figures 4–32(a) and (b). Sketch curves relating the uncovered port area A versus displacement x.
Solution. For the overlapped valve,a dead zone exists between and or The curve for uncovered port area A versus displacement x is shown in Figure 4–33(a). Such an overlapped valve is unfit as a control valve.
For the underlapped valve, the curve for port area A versus displacement x is shown in Figure 4–33(b). The effective curve for the underlapped region has a higher slope, meaning a higher sensitivity. Valves used for controls are usually underlapped.
A–4–7.
Figure 4–34 shows a hydraulic jet-pipe controller. Hydraulic fluid is ejected from the jet pipe. If the jet pipe is shifted to the right from the neutral position, the power piston moves to the left, and vice versa. The jet-pipe valve is not used as much as the flapper valve because of large null flow, slower response, and rather unpredictable characteristics. Its main advantage lies in its insensitivity to dirty fluids.
Suppose that the power piston is connected to a light load so that the inertia force of the load element is negligible compared to the hydraulic force developed by the power piston.What type of control action does this controller produce?
Solution. Define the displacement of the jet nozzle from the neutral position as x and the displacement of the power piston as y. If the jet nozzle is moved to the right by a small displace-- 1 2 x0 6 x 6 1 2 x0 .
1 2 x0 , - 1 2 x0 x0 2 x0 2 x0 2 x0 2 x x (a) (b) High pressure Low pressure High pressure Low pressure Figure 4–32 (a) Overlapped spool valve; (b) underlapped spool valve.
(a) (b) Effective area Area exposed to high pressure Area exposed to low pressure A x x0 2 A x x0 2 Figure 4–33 (a) Uncovered-port-area-A-versus displacement-x curve for the overlapped valve; (b) uncovered-port-area-A-versus-displacement-x curve for the underlapped valve.
Example Problems and Solutions 147 Oil under pressure A y x Figure 4–34 Hydraulic jet-pipe controller.
ment x, the oil flows to the right side of the power piston, and the oil in the left side of the power piston is returned to the drain.The oil flowing into the power cylinder is at high pressure; the oil flowing out from the power cylinder into the drain is at low pressure. The resulting pressure difference causes the power piston to move to the left.
For a small jet-nozzle displacement x, the flow rate q to the power cylinder is proportional to x; that is, For the power cylinder, where A is the power-piston area and r is the density of oil. Hence where constant.The transfer function Y(s)/X(s) is thus The controller produces the integral control action.
Y(s) X(s) = K s K = K 1(Ar) = dy dt = q Ar = K 1 Ar x = Kx Ar dy = q dt q = K 1 x 148 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems Engine Oil under pressure k b z e y a1 a2 v Figure 4–35 Speed control system.
+ – E(s) Y(s) Z(s) a2 a1 + a2 K s a1 a1 + a2 bs bs + k Figure 4–36 Block diagram for the speed control system shown in Figure 4–35.
A–4–8.
Explain the operation of the speed control system shown in Figure 4–35.
Solution. If the engine speed increases, the sleeve of the fly-ball governor moves upward. This movement acts as the input to the hydraulic controller.A positive error signal (upward motion of the sleeve) causes the power piston to move downward, reduces the fuel-valve opening, and decreases the engine speed.A block diagram for the system is shown in Figure 4–36.
From the block diagram the transfer function Y(s)/E(s) can be obtained as If the following condition applies, the transfer function Y(s)/E(s) becomes The speed controller has proportional-plus-integral control action.
Y(s) E(s) a2 a1 + a2 a1 + a2 a1 bs + k bs = a2 a1 a1 + k bs b 2 a1 a1 + a2 bs bs + k K s 2 1 Y(s) E(s) = a2 a1 + a2 K s 1 + a1 a1 + a2 bs bs + k K s Example Problems and Solutions 149 A–4–9.
Derive the transfer function Z(s)/Y(s) of the hydraulic system shown in Figure 4–37.Assume that the two dashpots in the system are identical ones except the piston shafts.
Solution. In deriving the equations for the system, we assume that force F is applied at the right end of the shaft causing displacement y. (All displacements y, w, and z are measured from re-spective equilibrium positions when no force is applied at the right end of the shaft.) When force F is applied, pressure becomes higher than pressure Similarly, For the force balance, we have the following equation: (4–42) Since (4–43) and we have Also, since q1 dt=A(dw-dz)r we have or Define A2Rr=B. (B is the viscous-friction coefficient.) Then (4–44) Also, for the right-hand-side dashpot we have Since or (4–45) Substituting Equations (4–43) and (4–45) into Equation (4–42), we have Taking the Laplace transform of this last equation, assuming zero initial condition, we obtain (4–46) k2 Y(s) = Ak2 + BsBW(s) + k1 Z(s) k2 y - k2 w = k1 z + Bw # AAP 2 - Pœ 2B = Bw # w # = q2 Ar = AAP 2 - Pœ 2B A 2Rr q2 = AP2 - Pœ 2BR, we obtain q2 dt = Ar dw w # - z # = k1 B z w # - z # = k1 z A 2Rr q1 = A(w # - z # )r k1 z = ARq1 q1 = P 1 - Pœ 1 R k1 z = AAP 1 - Pœ 1B k2(y - w) = AAP 1 - Pœ 1B + AAP 2 - Pœ 2B P 2 7 Pœ 2 .
Pœ 1 , or P 1 7 Pœ 1 .
P 1 R F R k2 k1 P1 q1 Area = A z q2 w w y P2 P2 9 P1 9 Figure 4–37 Hydraulic system.
150 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems Taking the Laplace transform of Equation (4–44), assuming zero initial condition, we obtain (4–47) By using Equation (4–47) to eliminate W(s) from Equation (4–46), we obtain from which we obtain the transfer function Z(s)/Y(s) to be Multiplying B/Ak1k2B to both the numerator and denominator of this last equation, we get Define Then the transfer function Z(s)/Y(s) becomes as follows: A–4–10.
Considering small deviations from steady-state operation, draw a block diagram of the air heat-ing system shown in Figure 4–38. Assume that the heat loss to the surroundings and the heat capacitance of the metal parts of the heater are negligible.
Solution. Let us define steady-state temperature of inlet air, °C steady-state temperature of outlet air, °C G=mass flow rate of air through the heating chamber, kgsec M=mass of air contained in the heating chamber, kg c=specific heat of air, kcalkg °C R=thermal resistance, °C seckcal C=thermal capacitance of air contained in the heating chamber=Mc, kcal°C steady-state heat input, kcalsec Let us assume that the heat input is suddenly changed from to and the inlet air temperature is suddenly changed from to Then the outlet air temperature will be changed from to The equation describing the system behavior is C duo = Ch + GcAui - uoB D dt Q – o + uo .
Q – o Q – i + ui .
Q – i H – + h H – H – = Q – o = Q – i = Z(s) Y(s) = T 1 s T 1 T 2 s2 + AT 1 + 2T 2Bs + 1 Bk1 = T 1 , Bk2 = T 2 .
Z(s) Y(s) = B k1 s B2 k1 k2 s2 + a 2B k2 + B k1 bs + 1 Z(s) Y(s) = k2 s Bs2 + A2k1 + k2Bs + k1 k2 B k2 Y(s) = Ak2 + BsB k1 + Bs Bs Z(s) + k1 Z(s) W(s) = k1 + Bs Bs Z(s) H + h Heater Qi + ui Qo + uo Figure 4–38 Air heating system.
Example Problems and Solutions 151 or Noting that we obtain or Taking the Laplace transforms of both sides of this last equation and substituting the initial condition that u0(0)=0, we obtain The block diagram of the system corresponding to this equation is shown in Figure 4–39.
A–4–11.
Consider the thin, glass-wall, mercury thermometer system shown in Figure 4–40.Assume that the thermometer is at a uniform temperature (ambient temperature) and that at t=0 it is immersed in a bath of temperature where ub is the bath temperature (which may be con-stant or changing) measured from the ambient temperature Define the instantaneous ther-mometer temperature by so that u is the change in the thermometer temperature satisfying the condition that u(0)=0. Obtain a mathematical model for the system.Also obtain an electri-cal analog of the thermometer system.
Solution. A mathematical model for the system can be derived by considering heat balance as fol-lows:The heat entering the thermometer during dt sec is q dt, where q is the heat flow rate to the thermometer.This heat is stored in the thermal capacitance C of the thermometer, thereby rais-ing its temperature by du.Thus the heat-balance equation is (4–48) C du = q dt Q – + u, Q –.
Q – + ub , Q – Qo(s) = R RCs + 1 H(s) + 1 RCs + 1 Qi(s) RC duo dt + uo = Rh + ui C duo dt = h + 1 R Aui - uoB Gc = 1 R C duo dt = h + GcAui - uoB H(s) 1 RCs + 1 R RCs + 1 Qi(s) Qo(s) ++ Figure 4–39 Block diagram of the air heating system shown in Figure 4–38.
Thermometer Bath Q + u Q + ub Figure 4–40 Thin, glass-wall, mercury thermo-meter system.
152 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems Since thermal resistance R may be written as heat flow rate q may be given, in terms of thermal resistance R, as where is the bath temperature and is the thermometer temperature. Hence, we can rewrite Equation (4–48) as or (4–49) Equation (4–49) is a mathematical model of the thermometer system.
Referring to Equation (4–49), an electrical analog for the thermometer system can be writ-ten as An electrical circuit represented by this last equation is shown in Figure 4–41.
RC deo dt + eo = ei RC du dt + u = ub C du dt = ub - u R Q – + u Q – + ub q = AQ – + ubB - AQ – + uB R = ub - u R R = d(¢u) dq = ¢u q R C eo ei Figure 4–41 Electrical analog of the thermometer system shown in Figure 4–40.
PROBLEMS B–4–1. Consider the conical water-tank system shown in Figure 4–42. The flow through the valve is turbulent and is related to the head H by where Q is the flow rate measured in m3sec and H is in meters.
Suppose that the head is 2 m at t=0. What will be the head at t=60 sec?
Q = 0.0051H 2m 3m 2m H r Figure 4–42 Conical water-tank system.
Problems 153 B–4–2. Consider the liquid-level control system shown in Figure 4–43.The controller is of the proportional type.The set point of the controller is fixed.
Draw a block diagram of the system, assuming that changes in the variables are small. Obtain the transfer func-tion between the level of the second tank and the distur-bance input qd. Obtain the steady-state error when the disturbance qd is a unit-step function.
C2 R1 C1 h2 R2 Q + qi qd Q + q0 H Proportional controller Figure 4–43 Liquid-level control system.
R C A X + x P + po P + pi k Figure 4–44 Pneumatic system.
B–4–3. For the pneumatic system shown in Figure 4–44, assume that steady-state values of the air pressure and the displacement of the bellows are and respectively.
Assume also that the input pressure is changed from to where pi is a small change in the input pressure.This change will cause the displacement of the bellows to change a small amount x.Assuming that the capacitance of the bel-lows is C and the resistance of the valve is R, obtain the transfer function relating x and pi.
P – + pi, P – X – , P – 154 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems B–4–4. Figure 4–45 shows a pneumatic controller.The pneu-matic relay has the characteristic that pc=Kpb, where K>0. What kind of control action does this controller produce? Derive the transfer function P c(s)E(s).
B–4–5. Consider the pneumatic controller shown in Figure 4–46.Assuming that the pneumatic relay has the char-acteristics that (where K>0),determine the con-trol action of this controller.The input to the controller is e and the output is pc.
pc = Kpb Actuating error signal Flapper Nozzle e a b X + x R I k Orifice Ps Pb + pb Pc + pc Figure 4–46 Pneumatic controller.
k Orifice Actuating error signal Flapper Nozzle Ps e a b Pb + pb X + x Y + y Pc + pc Figure 4–45 Pneumatic controller.
Problems 155 Actuating error signal Flapper Nozzle e a b k X + x R I II Orifice Ps Pb + pb Pc + pc Figure 4–47 Pneumatic controller.
Actuating error signal Flapper Nozzle e a b k X + x R2 I II R1 Orifice Ps Pb + pb Pc + pc Figure 4–48 Pneumatic controller.
B–4–6. Figure 4–47 shows a pneumatic controller.The sig-nal e is the input and the change in the control pressure is the output. Obtain the transfer function .
Assume that the pneumatic relay has the characteristics that where K>0.
pc = Kpb , P c(s)E(s) pc B–4–7. Consider the pneumatic controller shown in Figure 4–48. What control action does this controller pro-duce? Assume that the pneumatic relay has the character-istics that where K>0.
pc = Kpb , 156 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems B–4–8. Figure 4–49 shows a flapper valve. It is placed between two opposing nozzles.If the flapper is moved slight-ly to the right, the pressure unbalance occurs in the nozzles and the power piston moves to the left, and vice versa. Such a device is frequently used in hydraulic servos as the first-stage valve in two-stage servovalves. This usage occurs because considerable force may be needed to stroke larger spool valves that result from the steady-state flow force.To reduce or compensate this force, two-stage valve configura-tion is often employed; a flapper valve or jet pipe is used as the first-stage valve to provide a necessary force to stroke the second-stage spool valve.
diagram of the system of Figure 4–50 and then find the trans-fer function between y and x, where x is the air pressure and y is the displacement of the power piston.
x y Oil under pressure Oil under pressure y Flapper x Figure 4–49 Flapper valve.
u f l a b Oil under pressure Figure 4–51 Aircraft elevator control system.
B–4–9. Figure 4–51 is a schematic diagram of an aircraft elevator control system. The input to the system is the de-flection angle u of the control lever, and the output is the el-evator angle f. Assume that angles u and f are relatively small. Show that for each angle u of the control lever there is a corresponding (steady-state) elevator angle f.
Figure 4–50 Schematic diagram of a hydraulic servomotor.
Figure 4–50 shows a schematic diagram of a hydraulic servomotor in which the error signal is amplified in two stages using a jet pipe and a pilot valve. Draw a block Problems 157 B–4–10. Consider the liquid-level control system shown in Figure 4–52. The inlet valve is controlled by a hydraulic integral controller.Assume that the steady-state inflow rate is and steady-state outflow rate is also the steady-state head is steady-state pilot valve displacement is and steady-state valve position is We assume that the set point corresponds to the steady-state head The set point is fixed.Assume also that the disturbance inflow rate qd, which is a small quantity, is applied to the water tank at t=0.This disturbance causes the head to change from to This change results in a change in the outflow rate by qo.Through the hydraulic controller, the change in head causes a change in the inflow rate from to (The integral controller tends to keep the head constant as much as possible in the presence of disturbances.) We assume that all changes are of small quantities.
Q – + qi .
Q – H – + h.
H – H – .
R – Y –.
X – = 0, H – , Q –, Q – We assume that the velocity of the power piston (valve) is proportional to pilot-valve displacement x, or where K1 is a positive constant. We also assume that the change in the inflow rate qi is negatively proportional to the change in the valve opening y, or where Kv is a positive constant.
Assuming the following numerical values for the system, C=2 m2, R=0.5 secm2, Kv=1 m2sec a=0.25 m, b=0.75 m, K1=4 sec–1 obtain the transfer function H(s)/Qd(s).
qi = -K v y dy dt = K 1 x C (Capacitance) R (Resistance) a b h Y + y qd Q + qi H + h Q + qo x Figure 4–52 Liquid-level control system.
158 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems B–4–11. Consider the controller shown in Figure 4–53.The input is the air pressure pi measured from some steady-state reference pressure and the output is the displacement y of the power piston. Obtain the transfer function Y(s)P i(s).
P – B–4–12. A thermocouple has a time constant of 2 sec. A thermal well has a time constant of 30 sec.When the ther-mocouple is inserted into the well, this temperature-measuring device can be considered a two-capacitance system.
Determine the time constants of the combined thermo-couple–thermal-well system.Assume that the weight of the thermocouple is 8 g and the weight of the thermal well is 40 g.Assume also that the specific heats of the thermocouple and thermal well are the same.
a a b b Air pi (Input) y (Output) x k Bellows Figure 4–53 Controller.
5 159 Transient and Steady-State Response Analyses 5–1 INTRODUCTION In early chapters it was stated that the first step in analyzing a control system was to de-rive a mathematical model of the system. Once such a model is obtained, various meth-ods are available for the analysis of system performance.
In practice, the input signal to a control system is not known ahead of time but is random in nature, and the instantaneous input cannot be expressed analytically. Only in some special cases is the input signal known in advance and expressible analytically or by curves, such as in the case of the automatic control of cutting tools.
In analyzing and designing control systems, we must have a basis of comparison of performance of various control systems.This basis may be set up by specifying particular test input signals and by comparing the responses of various systems to these input signals.
Many design criteria are based on the response to such test signals or on the re-sponse of systems to changes in initial conditions (without any test signals).The use of test signals can be justified because of a correlation existing between the response char-acteristics of a system to a typical test input signal and the capability of the system to cope with actual input signals.
Typical Test Signals.
The commonly used test input signals are step functions, ramp functions, acceleration functions, impulse functions, sinusoidal functions, and white noise. In this chapter we use test signals such as step, ramp, acceleration and impulse signals.With these test signals, mathematical and experimental analyses of control sys-tems can be carried out easily, since the signals are very simple functions of time.
aa aa 160 Chapter 5 / Transient and Steady-State Response Analyses Which of these typical input signals to use for analyzing system characteristics may be determined by the form of the input that the system will be subjected to most frequently under normal operation. If the inputs to a control system are gradually changing functions of time, then a ramp function of time may be a good test signal. Sim-ilarly, if a system is subjected to sudden disturbances, a step function of time may be a good test signal; and for a system subjected to shock inputs, an impulse function may be best. Once a control system is designed on the basis of test signals, the performance of the system in response to actual inputs is generally satisfactory. The use of such test signals enables one to compare the performance of many systems on the same basis.
Transient Response and Steady-State Response.
The time response of a control system consists of two parts:the transient response and the steady-state response.
By transient response, we mean that which goes from the initial state to the final state.
By steady-state response, we mean the manner in which the system output behaves as t approaches infinity.Thus the system response c(t) may be written as where the first term on the right-hand side of the equation is the transient response and the second term is the steady-state response.
Absolute Stability, Relative Stability, and Steady-State Error.
In designing a control system, we must be able to predict the dynamic behavior of the system from a knowledge of the components. The most important characteristic of the dynamic behavior of a control system is absolute stability—that is,whether the system is stable or unstable.A control system is in equilibrium if,in the absence of any disturbance or input, the output stays in the same state.A linear time-invariant control system is stable if the output eventually comes back to its equilibrium state when the system is subjected to an initial condition. A linear time-invariant control system is critically stable if oscilla-tions of the output continue forever. It is unstable if the output diverges without bound from its equilibrium state when the system is subjected to an initial condition.Actually, the output of a physical system may increase to a certain extent but may be limited by mechanical “stops,” or the system may break down or become nonlinear after the out-put exceeds a certain magnitude so that the linear differential equations no longer apply.
Important system behavior (other than absolute stability) to which we must give careful consideration includes relative stability and steady-state error. Since a physical control system involves energy storage, the output of the system, when subjected to an input, cannot follow the input immediately but exhibits a transient response before a steady state can be reached. The transient response of a practical control system often exhibits damped oscillations before reaching a steady state. If the output of a system at steady state does not exactly agree with the input, the system is said to have steady-state error.This error is indicative of the accuracy of the system. In analyzing a control system, we must examine transient-response behavior and steady-state behavior.
Outline of the Chapter.
This chapter is concerned with system responses to aperiodic signals (such as step, ramp, acceleration, and impulse functions of time). The outline of the chapter is as follows: Section 5–1 has presented introductory material for the chapter. Section 5–2 treats the response of first-order systems to aperiodic inputs.
Section 5–3 deals with the transient response of the second-order systems. Detailed c(t) = ctr(t) + css(t) aa Section 5–2 / First-Order Systems 161 R(s) E(s) C(s) R(s) C(s) (a) (b) 1 Ts 1 Ts + 1 + – Figure 5–1 (a) Block diagram of a first-order system; (b) simplified block diagram.
analyses of the step response, ramp response, and impulse response of the second-order systems are presented. Section 5–4 discusses the transient-response analysis of higher-order systems.Section 5–5 gives an introduction to the MATLAB approach to the solution of transient-response problems. Section 5–6 gives an example of a transient-response problem solved with MATLAB. Section 5–7 presents Routh’s stability criterion. Section 5–8 discusses effects of integral and derivative control actions on system performance.
Finally, Section 5–9 treats steady-state errors in unity-feedback control systems.
5–2 FIRST-ORDER SYSTEMS Consider the first-order system shown in Figure 5–1(a). Physically, this system may represent an RC circuit, thermal system, or the like.A simplified block diagram is shown in Figure 5–1(b).The input-output relationship is given by (5–1) In the following, we shall analyze the system responses to such inputs as the unit-step, unit-ramp, and unit-impulse functions.The initial conditions are assumed to be zero.
Note that all systems having the same transfer function will exhibit the same output in response to the same input. For any given physical system, the mathematical response can be given a physical interpretation.
Unit-Step Response of First-Order Systems.
Since the Laplace transform of the unit-step function is 1/s, substituting R(s)=1/s into Equation (5–1), we obtain Expanding C(s) into partial fractions gives (5–2) Taking the inverse Laplace transform of Equation (5–2), we obtain for t 0 (5–3) Equation (5–3) states that initially the output c(t) is zero and finally it becomes unity.
One important characteristic of such an exponential response curve c(t) is that at t=T the value of c(t) is 0.632, or the response c(t) has reached 63.2% of its total change.This may be easily seen by substituting t=T in c(t).That is, c(T) = 1 - e-1 = 0.632 c(t) = 1 - e-tT, C(s) = 1 s -T Ts + 1 = 1 s -1 s + (1T) C(s) = 1 Ts + 1 1 s C(s) R(s) = 1 Ts + 1 aa 162 Chapter 5 / Transient and Steady-State Response Analyses c(t) 1 0 0.632 A B T 2T 3T 4T 5T t Slope = 1 T c(t) = 1 – e– (t/T) 63.2% 86.5% 95% 98.2% 99.3% Figure 5–2 Exponential response curve.
Note that the smaller the time constant T, the faster the system response. Another important characteristic of the exponential response curve is that the slope of the tangent line at t=0 is 1/T, since (5–4) The output would reach the final value at t=T if it maintained its initial speed of response.From Equation (5–4) we see that the slope of the response curve c(t) decreases monotonically from 1/T at t=0 to zero at t=q.
The exponential response curve c(t) given by Equation (5–3) is shown in Figure 5–2.
In one time constant,the exponential response curve has gone from 0 to 63.2% of the final value.In two time constants,the response reaches 86.5% of the final value.At t=3T,4T, and 5T, the response reaches 95%, 98.2%, and 99.3%, respectively, of the final value.Thus, for t 4T, the response remains within 2% of the final value. As seen from Equation (5–3), the steady state is reached mathematically only after an infinite time. In practice, however, a reasonable estimate of the response time is the length of time the response curve needs to reach and stay within the 2% line of the final value, or four time constants.
Unit-Ramp Response of First-Order Systems.
Since the Laplace transform of the unit-ramp function is 1/s2, we obtain the output of the system of Figure 5–1(a) as Expanding C(s) into partial fractions gives (5–5) Taking the inverse Laplace transform of Equation (5–5), we obtain for t 0 (5–6) The error signal e(t) is then = TA1 - e-tTB e(t) = r(t) - c(t) c(t) = t - T + Te-tT, C(s) = 1 s2 - T s + T2 Ts + 1 C(s) = 1 Ts + 1 1 s2 dc dt 2 t=0 = 1 T e-tT 2 t=0 = 1 T aa Section 5–2 / First-Order Systems 163 r(t) c(t) 6T 4T 2T 0 2T 4T 6T t T T r(t) = t c(t) Steady-state error Figure 5–3 Unit-ramp response of the system shown in Figure 5–1(a).
c(t) 0 2T T 4T 3T t 1 T c(t) = e– (t/T) 1 T Figure 5–4 Unit-impulse response of the system shown in Figure 5–1(a).
As t approaches infinity, e–t/T approaches zero, and thus the error signal e(t) approaches T or The unit-ramp input and the system output are shown in Figure 5–3. The error in following the unit-ramp input is equal to T for sufficiently large t.The smaller the time constant T, the smaller the steady-state error in following the ramp input.
Unit-Impulse Response of First-Order Systems.
For the unit-impulse input, R(s)=1 and the output of the system of Figure 5–1(a) can be obtained as (5–7) The inverse Laplace transform of Equation (5–7) gives for t 0 (5–8) The response curve given by Equation (5–8) is shown in Figure 5–4.
c(t) = 1 T e-tT, C(s) = 1 Ts + 1 e(q) = T aa 164 Chapter 5 / Transient and Steady-State Response Analyses An Important Property of Linear Time-Invariant Systems.
In the analysis above, it has been shown that for the unit-ramp input the output c(t) is for t 0 [See Equation (5–6).] For the unit-step input, which is the derivative of unit-ramp input, the output c(t) is for t 0 [See Equation (5–3).] Finally, for the unit-impulse input, which is the derivative of unit-step input, the output c(t) is for t 0 [See Equation (5–8).] Comparing the system responses to these three inputs clearly indicates that the response to the derivative of an input signal can be obtained by differentiating the response of the system to the original signal. It can also be seen that the response to the integral of the original signal can be obtained by integrating the response of the system to the original signal and by determining the integration constant from the zero-output initial condi-tion.This is a property of linear time-invariant systems. Linear time-varying systems and nonlinear systems do not possess this property.
5–3 SECOND-ORDER SYSTEMS In this section, we shall obtain the response of a typical second-order control system to a step input, ramp input, and impulse input. Here we consider a servo system as an example of a second-order system.
Servo System.
The servo system shown in Figure 5–5(a) consists of a proportional controller and load elements (inertia and viscous-friction elements). Suppose that we wish to control the output position c in accordance with the input position r.
The equation for the load elements is where T is the torque produced by the proportional controller whose gain is K. By taking Laplace transforms of both sides of this last equation, assuming the zero initial conditions, we obtain So the transfer function between C(s) and T(s) is By using this transfer function, Figure 5–5(a) can be redrawn as in Figure 5–5(b), which can be modified to that shown in Figure 5–5(c).The closed-loop transfer function is then obtained as Such a system where the closed-loop transfer function possesses two poles is called a second-order system. (Some second-order systems may involve one or two zeros.) C(s) R(s) = K Js2 + Bs + K = KJ s2 + (BJ)s + (KJ) C(s) T(s) = 1 s(Js + B) Js2C(s) + BsC(s) = T(s) Jc $ + Bc # = T c(t) = 1 T e-tT, c(t) = 1 - e-tT, c(t) = t - T + Te-tT, aa Section 5–3 / Second-Order Systems 165 + – r K 1 s(Js + B) e c T J B (a) + – R(s) R(s) C(s) C(s) T(s) (b) K K s(Js + B) + – (c) Figure 5–5 (a) Servo system; (b) block diagram; (c) simplified block diagram.
Step Response of Second-Order System.
The closed-loop transfer function of the system shown in Figure 5–5(c) is (5–9) which can be rewritten as The closed-loop poles are complex conjugates if B2-4JK<0 and they are real if B2-4JK 0. In the transient-response analysis, it is convenient to write where s is called the attenuation; vn, the undamped natural frequency; and z, the damp-ing ratio of the system. The damping ratio z is the ratio of the actual damping B to the critical damping or z = B Bc = B 21JK Bc = 21JK K J = v2 n , B J = 2zvn = 2s C(s) R(s) = K J cs + B 2J + B a B 2J b 2 - K J d cs + B 2J - B a B 2J b 2 - K J d C(s) R(s) = K Js2 + Bs + K aa 166 Chapter 5 / Transient and Steady-State Response Analyses R(s) E(s) C(s) vn s(s + 2zvn) 2 + – Figure 5–6 Second-order system.
In terms of z and vn, the system shown in Figure 5–5(c) can be modified to that shown in Figure 5–6, and the closed-loop transfer function C(s)/R(s) given by Equation (5–9) can be written (5–10) This form is called the standard form of the second-order system.
The dynamic behavior of the second-order system can then be described in terms of two parameters z and vn. If 0<z<1, the closed-loop poles are complex conjugates and lie in the left-half s plane. The system is then called underdamped, and the tran-sient response is oscillatory. If z=0, the transient response does not die out. If z=1, the system is called critically damped. Overdamped systems correspond to z>1.
We shall now solve for the response of the system shown in Figure 5–6 to a unit-step input.We shall consider three different cases: the underdamped (0<z<1), critically damped (z=1), and overdamped (z>1) cases.
(1) Underdamped case (0<z<1): In this case, C(s)/R(s) can be written where The frequency vd is called the damped natural frequency. For a unit-step input, C(s) can be written (5–11) The inverse Laplace transform of Equation (5–11) can be obtained easily if C(s) is writ-ten in the following form: Referring to the Laplace transform table in Appendix A, it can be shown that l-1c vd As + zvnB 2 + v2 d d = e-zvn t sin vd t l-1c s + zvn As + zvnB 2 + v2 d d = e-zvn t cos vd t = 1 s -s + zvn As + zvnB 2 + v2 d -zvn As + zvnB 2 + v2 d C(s) = 1 s -s + 2zvn s2 + 2zvn s + v2 n C(s) = v2 n As2 + 2zvn s + v2 nBs vd = vn21 - z2 .
C(s) R(s) = v2 n As + zvn + jvdBAs + zvn - jvdB C(s) R(s) = v2 n s2 + 2zvn s + v2 n aa Section 5–3 / Second-Order Systems 167 Hence the inverse Laplace transform of Equation (5–11) is obtained as for t 0 (5–12) From Equation (5–12), it can be seen that the frequency of transient oscillation is the damped natural frequency vd and thus varies with the damping ratio z.The error signal for this system is the difference between the input and output and is for t 0 This error signal exhibits a damped sinusoidal oscillation. At steady state, or at t=q, no error exists between the input and output.
If the damping ratio z is equal to zero, the response becomes undamped and oscillations continue indefinitely. The response c(t) for the zero damping case may be obtained by substituting z=0 in Equation (5–12), yielding for t 0 (5–13) Thus, from Equation (5–13), we see that vn represents the undamped natural frequen-cy of the system.That is, vn is that frequency at which the system output would oscillate if the damping were decreased to zero. If the linear system has any amount of damping, the undamped natural frequency cannot be observed experimentally. The frequency that may be observed is the damped natural frequency vd,which is equal to This frequency is always lower than the undamped natural frequency.An increase in z would reduce the damped natural frequency vd. If z is increased beyond unity, the response becomes overdamped and will not oscillate.
(2) Critically damped case (z=1): If the two poles of C(s)/R(s) are equal,the system is said to be a critically damped one.
For a unit-step input, R(s)=1/s and C(s) can be written (5–14) The inverse Laplace transform of Equation (5–14) may be found as for t 0 (5–15) This result can also be obtained by letting z approach unity in Equation (5–12) and by using the following limit: lim zS1 sinvd t 21 - z2 = lim zS1 sin vn21 - z2 t 21 - z2 = vn t c(t) = 1 - e-vn tA1 + vn tB, C(s) = v2 n As + vnB 2s vn21 - z2 .
c(t) = 1 - cosvn t, = e-zvn t acosvd t + z 21 - z2 sinvd t b, e(t) = r(t) - c(t) = 1 -e-zvn t 21 - z2 sin avd t + tan-1 21 - z2 z b , = 1 - e-zvn t acosvd t + z 21 - z2 sinvd t b l-1CC(s)D = c(t) aa 168 Chapter 5 / Transient and Steady-State Response Analyses (3) Overdamped case (z>1): In this case, the two poles of C(s)/R(s) are negative real and unequal. For a unit-step input, R(s)=1/s and C(s) can be written (5–16) The inverse Laplace transform of Equation (5–16) is for t 0 (5–17) where and Thus, the response c(t) includes two decaying exponential terms.
When z is appreciably greater than unity, one of the two decaying exponentials decreases much faster than the other, so the faster-decaying exponential term (which corresponds to a smaller time constant) may be neglected.That is, if –s2 is located very much closer to the jv axis than –s1 Awhich means @s2@ @s1@ B, then for an approximate solution we may neglect –s1.This is permissible because the effect of –s1 on the response is much smaller than that of –s2, since the term involving s1 in Equation (5–17) decays much faster than the term involving s2. Once the faster-decaying exponential term has disappeared, the response is similar to that of a first-order system, and C(s)/R(s) may be approximated by This approximate form is a direct consequence of the fact that the initial values and final values of both the original C(s)/R(s) and the approximate one agree with each other.
With the approximate transfer function C(s)/R(s), the unit-step response can be obtained as The time response c(t) is then for t 0 This gives an approximate unit-step response when one of the poles of C(s)/R(s) can be neglected.
c(t) = 1 - e-Az- 2z2-1Bvn t, C(s) = zvn - vn2z2 - 1 As + zvn - vn2z2 - 1Bs C(s) R(s) = zvn - vn2z2 - 1 s + zvn - vn2z2 - 1 = s2 s + s2 s2 = Az - 2z2 - 1Bvn .
s1 = Az + 2z2 - 1Bvn = 1 + vn 22z2 - 1 a e-s1 t s1 - e-s2 t s2 b , -1 22z2 - 1 Az - 2z2 - 1B e-Az- 2z2-1Bvnt c(t) = 1 + 1 22z2 - 1 Az + 2z2 - 1B e-Az+ 2z2-1Bvnt C(s) = v2 n As + zvn + vn2z2 - 1BAs + zvn - vn2z2 - 1Bs aa Section 5–3 / Second-Order Systems 169 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0 1 2 3 4 5 6 7 8 9 10 11 12 0.8 vnt c(t) z = 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 1.0 2.0 Figure 5–7 Unit-step response curves of the system shown in Figure 5–6.
A family of unit-step response curves c(t) with various values of z is shown in Fig-ure 5–7, where the abscissa is the dimensionless variable vnt. The curves are functions only of z. These curves are obtained from Equations (5–12), (5–15), and (5–17). The system described by these equations was initially at rest.
Note that two second-order systems having the same z but different vn will exhibit the same overshoot and the same oscillatory pattern. Such systems are said to have the same relative stability.
From Figure 5–7, we see that an underdamped system with z between 0.5 and 0.8 gets close to the final value more rapidly than a critically damped or overdamped system.
Among the systems responding without oscillation, a critically damped system exhibits the fastest response.An overdamped system is always sluggish in responding to any inputs.
It is important to note that, for second-order systems whose closed-loop transfer functions are different from that given by Equation (5–10), the step-response curves may look quite different from those shown in Figure 5–7.
Definitions of Transient-Response Specifications.
Frequently, the perform-ance characteristics of a control system are specified in terms of the transient response to a unit-step input, since it is easy to generate and is sufficiently drastic. (If the response to a step input is known,it is mathematically possible to compute the response to any input.) The transient response of a system to a unit-step input depends on the initial condi-tions. For convenience in comparing transient responses of various systems, it is a com-mon practice to use the standard initial condition that the system is at rest initially with the output and all time derivatives thereof zero. Then the response characteristics of many systems can be easily compared.
The transient response of a practical control system often exhibits damped oscilla-tions before reaching steady state. In specifying the transient-response characteristics of a control system to a unit-step input, it is common to specify the following: 1. Delay time, td 2. Rise time, tr aa 170 Chapter 5 / Transient and Steady-State Response Analyses c(t) 0.5 1 0 Allowable tolerance Mp td t 0.05 or 0.02 tr tp ts Figure 5–8 Unit-step response curve showing td, tr, tp, Mp, and ts.
3. Peak time, tp 4. Maximum overshoot, Mp 5. Settling time, ts These specifications are defined in what follows and are shown graphically in Figure 5–8.
1. Delay time, td: The delay time is the time required for the response to reach half the final value the very first time.
2. Rise time, tr:The rise time is the time required for the response to rise from 10% to 90%, 5% to 95%, or 0% to 100% of its final value. For underdamped second-order systems, the 0% to 100% rise time is normally used. For overdamped systems, the 10% to 90% rise time is commonly used.
3. Peak time, tp:The peak time is the time required for the response to reach the first peak of the overshoot.
4. Maximum (percent) overshoot, Mp: The maximum overshoot is the maximum peak value of the response curve measured from unity. If the final steady-state value of the response differs from unity, then it is common to use the maximum percent overshoot. It is defined by The amount of the maximum (percent) overshoot directly indicates the relative stability of the system.
5. Settling time, ts: The settling time is the time required for the response curve to reach and stay within a range about the final value of size specified by absolute per-centage of the final value (usually 2% or 5%). The settling time is related to the largest time constant of the control system.Which percentage error criterion to use may be determined from the objectives of the system design in question.
The time-domain specifications just given are quite important, since most control systems are time-domain systems; that is, they must exhibit acceptable time responses.
(This means that, the control system must be modified until the transient response is satisfactory.) Maximum percent overshoot = cAtpB - c(q) c(q) 100% aa Section 5–3 / Second-Order Systems 171 jv jvd vn s b zvn –s vn 1 – z2 0 Figure 5–9 Definition of the angle b.
Note that not all these specifications necessarily apply to any given case. For exam-ple, for an overdamped system, the terms peak time and maximum overshoot do not apply. (For systems that yield steady-state errors for step inputs, this error must be kept within a specified percentage level. Detailed discussions of steady-state errors are post-poned until Section 5–8.) A Few Comments on Transient-Response Specifications.
Except for certain applications where oscillations cannot be tolerated, it is desirable that the transient re-sponse be sufficiently fast and be sufficiently damped.Thus, for a desirable transient re-sponse of a second-order system, the damping ratio must be between 0.4 and 0.8. Small values of z(that is, z<0.4) yield excessive overshoot in the transient response, and a system with a large value of z(that is, z>0.8) responds sluggishly.
We shall see later that the maximum overshoot and the rise time conflict with each other.
In other words, both the maximum overshoot and the rise time cannot be made smaller simultaneously. If one of them is made smaller, the other necessarily becomes larger.
Second-Order Systems and Transient-Response Specifications.
In the fol-lowing, we shall obtain the rise time, peak time, maximum overshoot, and settling time of the second-order system given by Equation (5–10).These values will be obtained in terms of z and vn.The system is assumed to be underdamped.
Rise time tr: Referring to Equation (5–12),we obtain the rise time tr by letting cAtrB=1.
(5–18) Since we obtain from Equation (5–18) the following equation: Since and , we have Thus, the rise time tr is (5–19) where angle b is defined in Figure 5–9. Clearly, for a small value of tr, vd must be large.
tr = 1 vd tan-1 a vd -s b = p - b vd tan vd tr = - 21 - z2 z = - vd s zvn = s vn 21 - z2 = vd cosvd tr + z 21 - z2 sinvd tr = 0 e-zvn tr Z 0, cAtrB = 1 = 1 - e-zvn tr acosvd tr + z 21 - z2 sin vd tr b aa 172 Chapter 5 / Transient and Steady-State Response Analyses Peak time tp: Referring to Equation (5–12), we may obtain the peak time by differen-tiating c(t) with respect to time and letting this derivative equal zero. Since and the cosine terms in this last equation cancel each other, dcdt, evaluated at t=tp, can be simplified to This last equation yields the following equation: or Since the peak time corresponds to the first peak overshoot, Hence (5–20) The peak time tp corresponds to one-half cycle of the frequency of damped oscillation.
Maximum overshoot Mp: The maximum overshoot occurs at the peak time or at t=tp=pvd.Assuming that the final value of the output is unity, Mp is obtained from Equation (5–12) as (5–21) The maximum percent overshoot is If the final value c(q) of the output is not unity, then we need to use the following equation: Settling time ts: For an underdamped second-order system, the transient response is obtained from Equation (5–12) as for t 0 c(t) = 1 -e-zvn t 21 - z2 sin avd t + tan-1 21 - z2 z b , Mp = cAtpB - c(q) c(q) e-AsvdBp 100%. = e-AsvdBp = e-Az21-z2Bp = -e-zvnApvdB acosp + z 21 - z2 sin p b Mp = cAtpB - 1 tp = p vd vd tp = p.
vd tp = 0, p, 2p, 3p, p sin vd tp = 0 dc dt 2 t=tp = Asinvd tpB vn 21 - z2 e-zvn tp = 0 + e-zvn t a vd sin vd t -zvd 21 - z2 cosvd t b dc dt = zvn e-zvn t acosvd t + z 21 - z2 sinvd t b aa Section 5–3 / Second-Order Systems 173 1 c(t) 1 + 1 1 – z2 1 + e–zvnt 1 – z2 T = 1 zvn 1 – e–zvnt 1 – z2 0 1 – 1 1 – z2 3T 4T t T 2T Figure 5–10 Pair of envelope curves for the unit-step response curve of the system shown in Figure 5–6.
The curves are the envelope curves of the transient response to a unit-step input.The response curve c(t) always remains within a pair of the envelope curves, as shown in Figure 5–10.The time constant of these envelope curves is 1zvn.
The speed of decay of the transient response depends on the value of the time constant 1zvn. For a given vn, the settling time ts is a function of the damping ratio z. From Figure 5–7,we see that for the same vn and for a range of z between 0 and 1 the settling time ts for a very lightly damped system is larger than that for a properly damped system.For an overdamped system, the settling time ts becomes large because of the sluggish response.
The settling time corresponding to a ; 2% or ;5% tolerance band may be measured in terms of the time constant T=1zvn from the curves of Figure 5–7 for different values of z.The results are shown in Figure 5–11. For 0<z<0.9, if the 2% criterion is used, ts is approximately four times the time constant of the system. If the 5% criterion is used, then ts is approximately three times the time constant. Note that the settling time reaches a minimum value around z=0.76 (for the 2% criterion) or z=0.68 (for the 5% criterion) and then increases almost linearly for large values of z.
The discontinuities in the curves of Figure 5–11 arise because an infinitesimal change in the value of z can cause a finite change in the settling time.
For convenience in comparing the responses of systems, we commonly define the settling time ts to be (2% criterion) (5–22) or (5% criterion) (5–23) Note that the settling time is inversely proportional to the product of the damping ratio and the undamped natural frequency of the system. Since the value of z is usually determined from the requirement of permissible maximum overshoot, the settling time ts = 3T = 3 s = 3 zvn ts = 4T = 4 s = 4 zvn 1 ; Ae-zvn t21 - z2B aa 174 Chapter 5 / Transient and Steady-State Response Analyses z % 100 90 80 70 60 50 40 30 20 10 0 0.5 1.0 1.5 Mp Mp : Maximum overshoot C(s) R(s) = vn s2 + 2zvns + vn 2 2 Figure 5–12 Mp versus z curve.
2T 3T 4T T 5T 6T 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 z Settling time, ts 5% Tolerance band 2% Tolerance band Figure 5–11 Settling time ts versus z curves.
is determined primarily by the undamped natural frequency vn. This means that the duration of the transient period may be varied, without changing the maximum over-shoot, by adjusting the undamped natural frequency vn.
From the preceding analysis,it is evident that for rapid response vn must be large.To limit the maximum overshoot Mp and to make the settling time small,the damping ratio z should not be too small. The relationship between the maximum percent overshoot Mp and the damping ratio z is presented in Figure 5–12. Note that if the damping ratio is between 0.4 and 0.7, then the maximum percent overshoot for step response is between 25% and 4%.
aa Section 5–3 / Second-Order Systems 175 It is important to note that the equations for obtaining the rise time, peak time, max-imum overshoot, and settling time are valid only for the standard second-order system defined by Equation (5–10). If the second-order system involves a zero or two zeros, the shape of the unit-step response curve will be quite different from those shown in Figure 5–7.
EXAMPLE 5–1 Consider the system shown in Figure 5–6,where z=0.6 and vn=5 radsec.Let us obtain the rise time tr, peak time tp, maximum overshoot Mp, and settling time ts when the system is subjected to a unit-step input.
From the given values of z and vn, we obtain and s=zvn=3.
Rise time tr: The rise time is where b is given by The rise time tr is thus Peak time tp: The peak time is Maximum overshoot Mp: The maximum overshoot is The maximum percent overshoot is thus 9.5%.
Settling time ts: For the 2% criterion, the settling time is For the 5% criterion, Servo System with Velocity Feedback.
The derivative of the output signal can be used to improve system performance. In obtaining the derivative of the output position signal,it is desirable to use a tachometer instead of physically differentiating the output signal. (Note that the differentiation amplifies noise effects. In fact, if discontinuous noises are present,differentiation amplifies the discontinuous noises more than the useful signal. For example, the output of a potentiometer is a discontinuous voltage signal because, as the potentiometer brush is moving on the windings, voltages are induced in the switchover turns and thus generate transients.The output of the po-tentiometer therefore should not be followed by a differentiating element.) ts = 3 s = 3 3 = 1 sec ts = 4 s = 4 3 = 1.33 sec Mp = e-AsvdBp = e-(34)3.14 = 0.095 tp = p vd = 3.14 4 = 0.785 sec tr = 3.14 - 0.93 4 = 0.55 sec b = tan-1 vd s = tan-1 4 3 = 0.93 rad tr = p - b vd = 3.14 - b 4 vd = vn21 - z2 = 4 aa 176 Chapter 5 / Transient and Steady-State Response Analyses R(s) C(s) (a) 1 s K Js + B Kh R(s) C(s) (b) K s(Js + B + KKh) + – + – + – Figure 5–13 (a) Block diagram of a servo system; (b) simplified block diagram.
The tachometer, a special dc generator, is frequently used to measure velocity with-out differentiation process. The output of a tachometer is proportional to the angular velocity of the motor.
Consider the servo system shown in Figure 5–13(a). In this device, the velocity signal, together with the positional signal, is fed back to the input to produce the actuating error signal. In any servo system, such a velocity signal can be easily generated by a tachometer.The block diagram shown in Figure 5–13(a) can be simplified, as shown in Figure 5–13(b), giving (5–24) Comparing Equation (5–24) with Equation (5–9), notice that the velocity feedback has the effect of increasing damping.The damping ratio z becomes (5–25) The undamped natural frequency is not affected by velocity feedback.Not-ing that the maximum overshoot for a unit-step input can be controlled by controlling the value of the damping ratio z, we can reduce the maximum overshoot by adjusting the velocity-feedback constant Kh so that z is between 0.4 and 0.7.
It is important to remember that velocity feedback has the effect of increasing the damping ratio without affecting the undamped natural frequency of the system.
EXAMPLE 5–2 For the system shown in Figure 5–13(a), determine the values of gain K and velocity-feedback constant Kh so that the maximum overshoot in the unit-step response is 0.2 and the peak time is 1 sec.
With these values of K and Kh,obtain the rise time and settling time.Assume that J=1 kg-m2 and B=1 N-mradsec.
Determination of the values of K and Kh: The maximum overshoot Mp is given by Equation (5–21) as Mp = e-Az21-z2Bp vn = 1KJ z = B + KK h 21KJ C(s) R(s) = K Js2 + AB + KK hBs + K aa Section 5–3 / Second-Order Systems 177 This value must be 0.2.Thus, or which yields The peak time tp is specified as 1 sec; therefore, from Equation (5–20), or Since z is 0.456, vn is Since the natural frequency vn is equal to Then Kh is, from Equation (5–25), Rise time tr: From Equation (5–19), the rise time tr is where Thus, tr is Settling time ts: For the 2% criterion, For the 5% criterion, ts = 3 s = 1.86 sec ts = 4 s = 2.48 sec tr = 0.65 sec b = tan-1 vd s = tan-11.95 = 1.10 tr = p - b vd K h = 21KJz - B K = 21Kz - 1 K = 0.178 sec K = Jv2 n = v2 n = 12.5 N-m 1KJ, vn = vd 21 - z2 = 3.53 vd = 3.14 tp = p vd = 1 z = 0.456 zp 21 - z2 = 1.61 e-Az21-z2Bp = 0.2 aa 178 Chapter 5 / Transient and Steady-State Response Analyses 1.0 0.8 0.6 0.4 0.2 0 –0.2 –0.4 –0.6 –0.8 –1.0 0 2 4 6 8 10 12 vnt c(t) vn z = 0.1 z = 0.3 z = 0.5 z = 0.7 z = 1.0 Figure 5–14 Unit-impulse response curves of the system shown in Figure 5–6.
Impulse Response of Second-Order Systems.
For a unit-impulse input r(t),the corresponding Laplace transform is unity, or R(s)=1.The unit-impulse response C(s) of the second-order system shown in Figure 5-6 is The inverse Laplace transform of this equation yields the time solution for the response c(t) as follows: For 0 z<1, for t 0 (5–26) For z=1, for t 0 (5–27) For z>1, for t 0 (5–28) Note that without taking the inverse Laplace transform of C(s) we can also obtain the time response c(t) by differentiating the corresponding unit-step response, since the unit-impulse function is the time derivative of the unit-step function. A family of unit-impulse response curves given by Equations (5–26) and (5–27) with various val-ues of z is shown in Figure 5–14. The curves c(t)/vn are plotted against the dimen-sionless variable vnt, and thus they are functions only of z. For the critically damped and overdamped cases, the unit-impulse response is always positive or zero; that is, c(t) 0. This can be seen from Equations (5–27) and (5–28). For the underdamped case, the unit-impulse response c(t) oscillates about zero and takes both positive and negative values.
c(t) = vn 22z2 - 1 e-Az- 2z2-1Bvn t -vn 22z2 - 1 e-Az+ 2z2-1Bvn t, c(t) = v2 n te-vn t, c(t) = vn 21 - z2 e-zvn t sin vn21 - z2 t, C(s) = v2 n s2 + 2zvn s + v2 n aa Section 5–4 / Higher-Order Systems 179 c(t) 0 Unit-impulse response 1 + Mp tp t Figure 5–15 Unit-impulse response curve of the system shown in Figure 5–6.
From the foregoing analysis, we may conclude that if the impulse response c(t) does not change sign, the system is either critically damped or overdamped, in which case the corresponding step response does not overshoot but increases or decreases monot-onically and approaches a constant value.
The maximum overshoot for the unit-impulse response of the underdamped system occurs at where 0<z<1 (5–29) [Equation (5–29) can be obtained by equating dcdt to zero and solving for t.] The max-imum overshoot is where 0<z<1 (5–30) [Equation (5–30) can be obtained by substituting Equation (5–29) into Equation (5–26).] Since the unit-impulse response function is the time derivative of the unit-step response function, the maximum overshoot Mp for the unit-step response can be found from the corresponding unit-impulse response.That is, the area under the unit-impulse response curve from t=0 to the time of the first zero, as shown in Figure 5–15, is 1+Mp, where Mp is the maximum overshoot (for the unit-step response) given by Equation (5–21). The peak time tp (for the unit-step response) given by Equation (5–20) corresponds to the time that the unit-impulse response first crosses the time axis.
5–4 HIGHER-ORDER SYSTEMS In this section we shall present a transient-response analysis of higher-order systems in general terms. It will be seen that the response of a higher-order system is the sum of the responses of first-order and second-order systems.
c(t)max = vn exp a -z 21 - z2 tan-1 21 - z2 z b, t = tan-1 21 - z2 z vn21 - z2 , aa 180 Chapter 5 / Transient and Steady-State Response Analyses + – R(s) C(s) G(s) H(s) Figure 5–16 Control system.
Transient Response of Higher-Order Systems.
Consider the system shown in Figure 5–16.The closed-loop transfer function is (5–31) In general, G(s) and H(s) are given as ratios of polynomials in s, or and where p(s), q(s), n(s), and d(s) are polynomials in s.The closed-loop transfer function given by Equation (5–31) may then be written The transient response of this system to any given input can be obtained by a computer simulation. (See Section 5–5.) If an analytical expression for the transient response is de-sired, then it is necessary to factor the denominator polynomial. [MATLAB may be used for finding the roots of the denominator polynomial. Use the command roots(den).] Once the numerator and the denominator have been factored, C(s)/R(s) can be writ-ten in the form (5–32) Let us examine the response behavior of this system to a unit-step input. Consider first the case where the closed-loop poles are all real and distinct. For a unit-step input, Equation (5–32) can be written (5–33) where ai is the residue of the pole at s=–pi. (If the system involves multiple poles, then C(s) will have multiple-pole terms.) [The partial-fraction expansion of C(s), as given by Equation (5–33), can be obtained easily with MATLAB. Use the residue command. (See Appendix B.)] If all closed-loop poles lie in the left-half s plane, the relative magnitudes of the residues determine the relative importance of the components in the expanded form of C(s) = a s + a n i=1 ai s + pi C(s) R(s) = KAs + z1BAs + z2B p As + zmB As + p1BAs + p2B p As + pnB = b0 sm + b1 sm-1 + p + bm-1 s + bm a0 sn + a1 sn-1 + p + an-1 s + an (m n) C(s) R(s) = p(s)d(s) q(s)d(s) + p(s)n(s) H(s) = n(s) d(s) G(s) = p(s) q(s) C(s) R(s) = G(s) 1 + G(s)H(s) aa Section 5–4 / Higher-Order Systems 181 C(s). If there is a closed-loop zero close to a closed-loop pole, then the residue at this pole is small and the coefficient of the transient-response term corresponding to this pole becomes small. A pair of closely located poles and zeros will effectively cancel each other. If a pole is located very far from the origin, the residue at this pole may be small.
The transients corresponding to such a remote pole are small and last a short time.Terms in the expanded form of C(s) having very small residues contribute little to the transient response, and these terms may be neglected. If this is done, the higher-order system may be approximated by a lower-order one. (Such an approximation often enables us to es-timate the response characteristics of a higher-order system from those of a simplified one.) Next, consider the case where the poles of C(s) consist of real poles and pairs of complex-conjugate poles.A pair of complex-conjugate poles yields a second-order term in s. Since the factored form of the higher-order characteristic equation consists of first-and second-order terms, Equation (5–33) can be rewritten where we assumed all closed-loop poles are distinct. [If the closed-loop poles involve multiple poles, C(s) must have multiple-pole terms.] From this last equation, we see that the response of a higher-order system is composed of a number of terms involving the simple functions found in the responses of first- and second-order systems. The unit-step response c(t), the inverse Laplace transform of C(s), is then for t 0 (5–34) Thus the response curve of a stable higher-order system is the sum of a number of exponential curves and damped sinusoidal curves.
If all closed-loop poles lie in the left-half s plane, then the exponential terms and the damped exponential terms in Equation (5–34) will approach zero as time t increases.
The steady-state output is then c(q)=a.
Let us assume that the system considered is a stable one.Then the closed-loop poles that are located far from the jv axis have large negative real parts. The exponential terms that correspond to these poles decay very rapidly to zero. (Note that the hori-zontal distance from a closed-loop pole to the jv axis determines the settling time of tran-sients due to that pole.The smaller the distance is, the longer the settling time.) Remember that the type of transient response is determined by the closed-loop poles, while the shape of the transient response is primarily determined by the closed-loop zeros. As we have seen earlier, the poles of the input R(s) yield the steady-state response terms in the solution, while the poles of C(s)/R(s) enter into the exponential transient-response terms and/or damped sinusoidal transient-response terms.The zeros of C(s)/R(s) do not affect the exponents in the exponential terms, but they do affect the magnitudes and signs of the residues.
+ a r k=1 ck e-zk vk t sinvk21 - z2 k t, c(t) = a + a q j=1 aj e-pj t + a r k=1 bk e-zk vk t cosvk21 - z2 k t C(s) = a s + a q j=1 aj s + pj + a r k=1 bkAs + zk vkB + ck vk21 - z2 k s2 + 2zk vk s + v2 k (q + 2r = n) aa 182 Chapter 5 / Transient and Steady-State Response Analyses Dominant Closed-Loop Poles.
The relative dominance of closed-loop poles is determined by the ratio of the real parts of the closed-loop poles, as well as by the rel-ative magnitudes of the residues evaluated at the closed-loop poles.The magnitudes of the residues depend on both the closed-loop poles and zeros.
If the ratios of the real parts of the closed-loop poles exceed 5 and there are no zeros nearby, then the closed-loop poles nearest the jv axis will dominate in the transient-response behavior because these poles correspond to transient-response terms that decay slowly. Those closed-loop poles that have dominant effects on the transient-response behavior are called dominant closed-loop poles. Quite often the dominant closed-loop poles occur in the form of a complex-conjugate pair.The dominant closed-loop poles are most important among all closed-loop poles.
Note that the gain of a higher-order system is often adjusted so that there will exist a pair of dominant complex-conjugate closed-loop poles.The presence of such poles in a stable system reduces the effects of such nonlinearities as dead zone, backlash, and coulomb-friction.
Stability Analysis in the Complex Plane.
The stability of a linear closed-loop system can be determined from the location of the closed-loop poles in the s plane. If any of these poles lie in the right-half s plane, then with increasing time they give rise to the dominant mode, and the transient response increases monotonically or oscillates with increasing amplitude. This represents an unstable system. For such a system, as soon as the power is turned on, the output may increase with time. If no saturation takes place in the system and no mechanical stop is provided, then the system may eventually be subjected to damage and fail, since the response of a real physical sys-tem cannot increase indefinitely.Therefore, closed-loop poles in the right-half s plane are not permissible in the usual linear control system. If all closed-loop poles lie to the left of the jv axis, any transient response eventually reaches equilibrium. This repre-sents a stable system.
Whether a linear system is stable or unstable is a property of the system itself and does not depend on the input or driving function of the system.The poles of the input, or driving function, do not affect the property of stability of the system, but they con-tribute only to steady-state response terms in the solution.Thus, the problem of absolute stability can be solved readily by choosing no closed-loop poles in the right-half s plane, including the jv axis. (Mathematically, closed-loop poles on the jv axis will yield oscil-lations, the amplitude of which is neither decaying nor growing with time. In practical cases, where noise is present, however, the amplitude of oscillations may increase at a rate determined by the noise power level. Therefore, a control system should not have closed-loop poles on the jv axis.) Note that the mere fact that all closed-loop poles lie in the left-half s plane does not guarantee satisfactory transient-response characteristics.If dominant complex-conjugate closed-loop poles lie close to the jv axis, the transient response may exhibit excessive oscillations or may be very slow.Therefore,to guarantee fast,yet well-damped,transient-response characteristics, it is necessary that the closed-loop poles of the system lie in a particular region in the complex plane, such as the region bounded by the shaded area in Figure 5–17.
Since the relative stability and transient-response performance of a closed-loop con-trol system are directly related to the closed-loop pole-zero configuration in the s plane, aa Section 5–5 / Transient-Response Analysis with MATLAB 183 0 s s jv In this region z 0.4 s 4 ts Figure 5–17 Region in the complex plane satisfying the conditions z>0.4 and ts<4/s.
it is frequently necessary to adjust one or more system parameters in order to obtain suit-able configurations.The effects of varying system parameters on the closed-loop poles will be discussed in detail in Chapter 6.
5–5 TRANSIENT-RESPONSE ANALYSIS WITH MATLAB Introduction.
The practical procedure for plotting time response curves of systems higher than second order is through computer simulation. In this section we present the computational approach to the transient-response analysis with MATLAB.In particular, we discuss step response,impulse response,ramp response,and responses to other simple inputs.
MATLAB Representation of Linear Systems.
The transfer function of a system is represented by two arrays of numbers. Consider the system (5–35) This system can be represented as two arrays, each containing the coefficients of the polynomials in decreasing powers of s as follows: num = [2 25] den = [1 4 25] An alternative representation is num = [0 2 25] den = [1 4 25] C(s) R(s) = 2s + 25 s2 + 4s + 25 aa 184 Chapter 5 / Transient and Steady-State Response Analyses In this expression a zero is padded. Note that if zeros are padded, the dimensions of “num” vector and “den” vector become the same.An advantage of padding zeros is that the “num” vector and “den” vector can be directly added. For example, num + den = [0 2 25] + [1 4 25] = [1 6 50] If num and den (the numerator and denominator of the closed-loop transfer function) are known, commands such as step(num,den), step(num,den,t) will generate plots of unit-step responses (t in the step command is the user-specified time.) For a control system defined in a state-space form, where state matrix A, control matrix B, output matrix C, and direct transmission matrix D of state-space equations are known, the command step(A,B,C,D), step(A,B,C,D,t) will generate plots of unit-step responses. When t is not explicitly included in the step commands, the time vector is automatically determined.
Note that the command step(sys) may be used to obtain the unit-step response of a system. First, define the system by sys = tf(num,den) or sys = ss(A,B,C,D) Then, to obtain, for example, the unit-step response, enter step(sys) into the computer.
When step commands have left-hand arguments such as [y,x,t] = step(num,den,t) [y,x,t] = step(A,B,C,D,iu) [y,x,t] = step(A,B,C,D,iu,t) (5–36) no plot is shown on the screen. Hence it is necessary to use a plot command to see the response curves.The matrices y and x contain the output and state response of the sys-tem, respectively, evaluated at the computation time points t. (y has as many columns as outputs and one row for each element in t. x has as many columns as states and one row for each element in t.) Note in Equation (5–36) that the scalar iu is an index into the inputs of the system and specifies which input is to be used for the response, and t is the user-specified time.
If the system involves multiple inputs and multiple outputs, the step command, such as given by Equation (5–36), produces a series of step-response plots, one for each input and output combination of (For details, see Example 5–3.) y = Cx + Du x # = Ax + Bu aa Section 5–5 / Transient-Response Analysis with MATLAB 185 EXAMPLE 5–3 Consider the following system: Obtain the unit-step response curves.
Although it is not necessary to obtain the transfer-matrix expression for the system to obtain the unit-step response curves with MATLAB, we shall derive such an expression for reference.
For the system defined by the transfer matrix G(s) is a matrix that relates Y(s) and U(s) as follows: Taking Laplace transforms of the state-space equations, we obtain (5–37) (5–38) In deriving the transfer matrix, we assume that Then, from Equation (5–37), we get (5–39) Substituting Equation (5–39) into Equation (5–38), we obtain Thus the transfer matrix G(s) is given by The transfer matrix G(s) for the given system becomes Hence Since this system involves two inputs and two outputs, four transfer functions may be defined, depending on which signals are considered as input and output. Note that, when considering the BY 1(s) Y 2(s)R = ≥ s - 1 s2 + s + 6.5 s + 7.5 s2 + s + 6.5 s s2 + s + 6.5 6.5 s2 + s + 6.5 ¥ B U 1(s) U 2(s)R = 1 s2 + s + 6.5 B s - 1 s + 7.5 s 6.5R = 1 s2 + s + 6.5 B s 6.5 - 1 s + 1R B 1 1 1 0R = B 1 0 0 1R Bs + 1 - 6.5 1 sR -1 B 1 1 1 0R G(s) = C(s I - A)-1 B G(s) = C(s I - A)-1 B + D Y(s) = CC(s I - A)-1 B + DD U(s) X(s) = (s I - A)-1 BU(s) x(0) = 0. Y(s) = CX(s) + DU(s) s X(s) - x(0) = AX(s) + BU(s) Y(s) = G(s) U(s) y = Cx + Du x # = Ax + Bu B y1 y2 R = B 1 0 0 1R B x1 x2 R + B 0 0 0 0R B u1 u2 R Bx # 1 x # 2 R = B - 1 6.5 - 1 0R B x1 x2 R + B 1 1 1 0R Bu1 u2 R aa 186 Chapter 5 / Transient and Steady-State Response Analyses To: Y2 1.5 2 1 0.5 0 0 4 8 12 Time (sec) 1.5 2 1 0.5 0 0 4 8 12 To: Y1 0.4 0.6 0.2 −0.2 0 0 −0.4 0.4 0.6 0.2 −0.2 −0.4 0 4 8 12 0 4 8 12 From: U1 From: U2 Step Response Amplitude Figure 5–18 Unit-step response curves.
MATLAB Program 5–1 A = [–1 –1;6.5 0]; B = [1 1;1 0]; C = [1 0;0 1]; D = [0 0;0 0]; step(A,B,C,D) signal u1 as the input, we assume that signal u2 is zero, and vice versa.The four transfer functions are Assume that u1 and u2 are unit-step functions.The four individual step-response curves can then be plotted by use of the command step(A,B,C,D) MATLAB Program 5–1 produces four such step-response curves.The curves are shown in Figure 5–18.
(Note that the time vector t is automatically determined, since the command does not include t.) Y 2(s) U 2(s) = 6.5 s2 + s + 6.5 Y 2(s) U 1(s) = s + 7.5 s2 + s + 6.5 , Y 1(s) U 2(s) = s s2 + s + 6.5 Y 1(s) U 1(s) = s - 1 s2 + s + 6.5 , aa Section 5–5 / Transient-Response Analysis with MATLAB 187 MATLAB Program 5–2 % In this program we plot step-response curves of a system % having two inputs (u1 and u2) and two outputs (y1 and y2) % We shall first plot step-response curves when the input is % u1. Then we shall plot step-response curves when the input is % u2 % Enter matrices A, B, C, and D A = [-1 -1;6.5 0]; B = [1 1;1 0]; C = [1 0;0 1]; D = [0 0;0 0]; % To plot step-response curves when the input is u1, enter % the command 'step(A,B,C,D,1)' step(A,B,C,D,1) grid title ('Step-Response Plots: Input = u1 (u2 = 0)') text(3.4, -0.06,'Y1') text(3.4, 1.4,'Y2') % Next, we shall plot step-response curves when the input % is u2. Enter the command 'step(A,B,C,D,2)' step(A,B,C,D,2) grid title ('Step-Response Plots: Input = u2 (u1 = 0)') text(3,0.14,'Y1') text(2.8,1.1,'Y2') To plot two step-response curves for the input u1 in one diagram and two step-response curves for the input u2 in another diagram, we may use the commands step(A,B,C,D,1) and step(A,B,C,D,2) respectively. MATLAB Program 5–2 is a program to plot two step-response curves for the input u1 in one diagram and two step-response curves for the input u2 in another diagram.
Figure 5–19 shows the two diagrams, each consisting of two step-response curves. (This MATLAB program uses text commands. For such commands, refer to the paragraph following this example.) aa 188 Chapter 5 / Transient and Steady-State Response Analyses Step-Response Plots: Input = u2 (u1 = 0) Time (sec) Amplitude 0 1 2 3 4 5 6 7 8 9 10 1.6 1.2 0.8 0.4 0 –0.2 1.4 1 0.6 0.2 Y2 Y1 (b) Step-Response Plots: Input = u1 (u2 = 0) 2 1.5 1 0.5 0 –0.5 0 1 2 3 4 5 6 7 8 9 10 Time (sec) Amplitude Y2 Y1 (a) Figure 5–19 Unit-step response curves. (a) u1 is the input Au2=0B; (b) u2 is the input Au1=0B.
Writing Text on the Graphics Screen.
To write text on the graphics screen, enter, for example, the following statements: text(3.4, -0.06,'Y1') and text(3.4,1.4,'Y2') The first statement tells the computer to write ‘Y1’ beginning at the coordinates x=3.4, y=–0.06. Similarly, the second statement tells the computer to write ‘Y2’ beginning at the coordinates x=3.4, y=1.4. [See MATLAB Program 5–2 and Figure 5–19(a).] aa Section 5–5 / Transient-Response Analysis with MATLAB 189 MATLAB Program 5–3 wn = 5; damping_ratio = 0.4; [num0,den] = ord2(wn,damping_ratio); num = 5^2num0; printsys(num,den,'s') num/den = 25 S^2 + 4s + 25 Another way to write a text or texts in the plot is to use the gtext command. The syntax is gtext('text') When gtext is executed, the computer waits until the cursor is positioned (using a mouse) at the desired position in the screen. When the left mouse button is pressed, the text enclosed in simple quotes is written on the plot at the cursor’s position. Any number of gtext commands can be used in a plot. (See, for example, MATLAB Program 5–15.) MATLAB Description of Standard Second-Order System.
As noted earlier,the second-order system (5–40) is called the standard second-order system. Given vn and z, the command printsys(num,den) or printsys(num,den,s) prints num/den as a ratio of polynomials in s.
Consider,for example,the case where vn=5 radsec and z=0.4.MATLAB Program 5–3 generates the standard second-order system, where vn=5 radsec and z=0.4.
Note that in MATLAB Program 5–3,“num 0” is 1.
G(s) = v2 n s2 + 2zvn s + v2 n Obtaining the Unit-Step Response of the Transfer-Function System.
Let us consider the unit-step response of the system given by G(s) = 25 s2 + 4s + 25 aa 190 Chapter 5 / Transient and Steady-State Response Analyses MATLAB Program 5–4 % ------------- Unit-step response -------------% Enter the numerator and denominator of the transfer % function num = ; den = [1 4 25]; % Enter the following step-response command step(num,den) % Enter grid and title of the plot grid title (' Unit-Step Response of G(s) = 25/(s^2+4s+25)') 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 2 2.5 3 Time (sec) Unit-Step Response of G(s) = 25/(s2+4s+25) Amplitude Figure 5–20 Unit-step response curve.
Notice in Figure 5–20 (and many others) that the x-axis and y-axis labels are auto-matically determined. If it is desired to label the x axis and y axis differently, we need to modify the step command. For example, if it is desired to label the x axis as 't Sec' and the y axis as ‘Output,’ then use step-response commands with left-hand arguments, such as c = step(num,den,t) or, more generally, [y,x,t] = step(num,den,t) and use plot(t,y) command. See, for example, MATLAB Program 5–5 and Figure 5–21.
MATLAB Program 5–4 will yield a plot of the unit-step response of this system.A plot of the unit-step response curve is shown in Figure 5–20.
aa Section 5–5 / Transient-Response Analysis with MATLAB 191 MATLAB Program 5–5 % ------------- Unit-step response -------------num = ; den = [1 4 25]; t = 0:0.01:3; [y,x,t] = step(num,den,t); plot(t,y) grid title('Unit-Step Response of G(s)=25/(sˆ2+4s+25)') xlabel('t Sec') ylabel('Output') 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 2 2.5 3 t Sec Unit-Step Response of G(s) = 25/(s2+4s+25) Output Figure 5–21 Unit-step response curve.
Obtaining Three-Dimensional Plot of Unit-Step Response Curves with MATLAB.
MATLAB enables us to plot three-dimensional plots easily.The commands to obtain three-dimensional plots are “mesh” and “surf.” The difference between the “mesh” plot and “surf” plot is that in the former only the lines are drawn and in the lat-ter the spaces between the lines are filled in by colors. In this book we use only the “mesh” command.
EXAMPLE 5–4 Consider the closed-loop system defined by (The undamped natural frequency vn is normalized to 1.) Plot unit-step response curves c(t) when z assumes the following values: z=0, 0.2, 0.4, 0.6. 0.8, 1.0 Also plot a three-dimensional plot.
C(s) R(s) = 1 s2 + 2zs + 1 aa 192 Chapter 5 / Transient and Steady-State Response Analyses MATLAB Program 5–6 % ------- Two-dimensional plot and three-dimensional plot of unit-step % response curves for the standard second-order system with wn = 1 % and zeta = 0, 0.2, 0.4, 0.6, 0.8, and 1. -------t = 0:0.2:10; zeta = [0 0.2 0.4 0.6 0.8 1]; for n = 1:6; num = ; den = [1 2zeta(n) 1]; [y(1:51,n),x,t] = step(num,den,t); end % To plot a two-dimensional diagram, enter the command plot(t,y).
plot(t,y) grid title('Plot of Unit-Step Response Curves with \omega_n = 1 and \zeta = 0, 0.2, 0.4, 0.6, 0.8, 1') xlabel('t (sec)') ylabel('Response') text(4.1,1.86,'\zeta = 0') text(3.5,1.5,'0.2') text(3 .5,1.24,'0.4') text(3.5,1.08,'0.6') text(3.5,0.95,'0.8') text(3.5,0.86,'1.0') % To plot a three-dimensional diagram, enter the command mesh(t,zeta,y').
mesh(t,zeta,y') title('Three-Dimensional Plot of Unit-Step Response Curves') xlabel('t Sec') ylabel('\zeta') zlabel('Response') An illustrative MATLAB Program for plotting a two-dimensional diagram and a three-dimensional diagram of unit-step response curves of this second-order system is given in MATLAB Program 5–6. The resulting plots are shown in Figures 5–22(a) and (b), respectively. Notice that we used the command mesh(t,zeta,y') to plot the three-dimensional plot.We may use a command mesh(y') to get the same result. [Note that command mesh(t,zeta,y) or mesh(y) will produce a three-dimensional plot the same as Figure 5–22(b), except that x axis and y axis are interchanged.
See Problem A–5–15.] When we want to solve a problem using MATLAB and if the solution process involves many repetitive computations, various approaches may be conceived to simplify the MATLAB pro-gram.A frequently used approach to simplify the computation is to use “for loops.”MATLAB Pro-gram 5–6 uses such a “for loop.” In this book many MATLAB programs using “for loops” are presented for solving a variety of problems. Readers are advised to study all those problems care-fully to familiarize themselves with the approach.
aa Section 5–5 / Transient-Response Analysis with MATLAB 193 Plot of Unit-Step Response Curves with n = 1 and = 0, 0.2, 0.4, 0.6, 0.8, 1 Response 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 t (sec) 0 1 2 3 4 5 6 7 8 9 10 = 0 = 0 0.2 0.2 0.4 0.4 0.6 0.6 0.8 0.8 1.0 1.0 (a) 0 1 0.8 0.6 0.4 0.2 0 0 2 4 6 8 10 0.5 1 1.5 2 Response t Sec (b) Three-Dimensional Plot of Unit-Step Response Curves Figure 5–22 (a) Two-dimensional plot of unit-step response curves for z=0, 0.2, 0.4, 0.6, 0.8, and 1.0; (b) three-dimensional plot of unit-step response curves.
Obtaining Rise Time, Peak Time, Maximum Overshoot, and Settling Time with MATLAB.
MATLAB can conveniently be used to obtain the rise time,peak time, maximum overshoot, and settling time. Consider the system defined by MATLAB Program 5–7 yields the rise time,peak time,maximum overshoot,and settling time. A unit-step response curve for this system is given in Figure 5–23 to verify the C(s) R(s) = 25 s2 + 6s + 25 aa 194 Chapter 5 / Transient and Steady-State Response Analyses MATLAB Program 5–7 % ------- This is a MATLAB program to find the rise time, peak time, % maximum overshoot, and settling time of the second-order system % and higher-order system -------% ------- In this example, we assume zeta = 0.6 and wn = 5 -------num = ; den = [1 6 25]; t = 0:0.005:5; [y,x,t] = step(num,den,t); r = 1; while y(r) < 1.0001; r = r + 1; end; rise_time = (r - 1)0.005 rise_time = 0.5550 [ymax,tp] = max(y); peak_time = (tp - 1)0.005 peak_time = 0.7850 max_overshoot = ymax-1 max_overshoot = 0.0948 s = 1001; while y(s) > 0.98 & y(s) < 1.02; s = s - 1; end; settling_time = (s - 1)0.005 settling_time = 1.1850 Amplitude Time (sec) Step Response 0.6 0.4 0.2 0.8 1 1.2 1.4 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Figure 5–23 Unit-step response curve.
results obtained with MATLAB Program 5–7. (Note that this program can also be applied to higher-order systems. See Problem A–5–10.) aa Section 5–5 / Transient-Response Analysis with MATLAB 195 Impulse Response.
The unit-impulse response of a control system may be obtained by using any of the impulse commands such as impulse(num,den) impulse(A,B,C,D) [y,x,t] = impulse(num,den) [y,x,t] = impulse(num,den,t) (5–41) [y,x,t] = impulse(A,B,C,D) [y,x,t] = impulse(A,B,C,D,iu) (5–42) [y,x,t] = impulse(A,B,C,D,iu,t) (5–43) The command impulse(num,den) plots the unit-impulse response on the screen. The command impulse(A,B,C,D) produces a series of unit-impulse-response plots, one for each input and output combination of the system Note that in Equations (5–42) and (5–43) the scalar iu is an index into the inputs of the system and specifies which input to be used for the impulse response.
Note also that if the command used does not include “t” explicitly, the time vector is automatically determined. If the command includes the user-supplied time vector “t”, as do the commands given by Equations (5–41) and (5–43)], this vector specifies the times at which the impulse response is to be computed.
If MATLAB is invoked with the left-hand argument [y,x,t], such as in the case of [y,x,t] = impulse(A,B,C,D), the command returns the output and state responses of the system and the time vector t. No plot is drawn on the screen.The matrices y and x con-tain the output and state responses of the system evaluated at the time points t. (y has as many columns as outputs and one row for each element in t. x has as many columns as state variables and one row for each element in t.) To plot the response curve, we must include a plot command, such as plot(t,y).
y = Cx + Du x # = Ax + Bu EXAMPLE 5–5 Obtain the unit-impulse response of the following system: C(s) R(s) = G(s) = 1 s2 + 0.2s + 1 196 Chapter 5 / Transient and Steady-State Response Analyses aa Unit-Impulse Response of G(s) = 1/(s2+0.2s+1) Time (sec) Amplitude 0 5 10 15 20 25 30 35 40 45 50 1 0.8 0.2 –0.6 –0.8 0.6 0.4 0 –0.2 –0.4 Figure 5–24 Unit-impulse-response curve.
MATLAB Program 5–8 will produce the unit-impulse response. The resulting plot is shown in Figure 5–24.
MATLAB Program 5–8 num = ; den = [1 0.2 1]; impulse(num,den); grid title(‘Unit-Impulse Response of G(s) = 1/(s^2 + 0.2s + 1)‘) Alternative Approach to Obtain Impulse Response.
Note that when the initial conditions are zero, the unit-impulse response of G(s) is the same as the unit-step response of sG(s).
Consider the unit-impulse response of the system considered in Example 5–5. Since R(s)=1 for the unit-impulse input, we have We can thus convert the unit-impulse response of G(s) to the unit-step response of sG(s).
If we enter the following num and den into MATLAB, num = [0 1 0] den = [1 0.2 1] = s s2 + 0.2s + 1 1 s C(s) R(s) = C(s) = G(s) = 1 s2 + 0.2s + 1 Section 5–5 / Transient-Response Analysis with MATLAB 197 aa Unit-Step Response of sG(s) = s/(s2+0.2s+1) Time (sec) Amplitude 0 5 10 15 20 25 30 35 40 45 50 1 0.8 0.2 –0.6 –0.8 0.6 0.4 0 –0.2 –0.4 Figure 5–25 Unit-impulse-response curve obtained as the unit-step response of sG(s)= s/As2+0.2s+1B.
MATLAB Program 5–9 num = [1 0]; den = [1 0.2 1]; step(num,den); grid title(‘Unit-Step Response of sG(s) = s/(s^2 + 0.2s + 1)‘) and use the step-response command; as given in MATLAB Program 5–9, we obtain a plot of the unit-impulse response of the system as shown in Figure 5–25.
Ramp Response.
There is no ramp command in MATLAB. Therefore, we need to use the step command or the lsim command (presented later) to obtain the ramp re-sponse. Specifically, to obtain the ramp response of the transfer-function system G(s), divide G(s) by s and use the step-response command. For example, consider the closed-loop system For a unit-ramp input, R(s)=1/s2 . Hence To obtain the unit-ramp response of this system, enter the following numerator and de-nominator into the MATLAB program: num = [2 1]; den = [1 1 1 0]; C(s) = 2s + 1 s2 + s + 1 1 s2 = 2s + 1 (s2 + s + 1)s 1 s C(s) R(s) = 2s + 1 s2 + s + 1 aa 198 Chapter 5 / Transient and Steady-State Response Analyses MATLAB Program 5–10 % --------------- Unit-ramp response ---------------% The unit-ramp response is obtained as the unit-step % response of G(s)/s % Enter the numerator and denominator of G(s)/s num = [2 1]; den = [1 1 1 0]; % Specify the computing time points (such as t = 0:0.1:10) % and then enter step-response command: c = step(num,den,t) t = 0:0.1:10; c = step(num,den,t); % In plotting the ramp-response curve, add the reference % input to the plot. The reference input is t. Add to the % argument of the plot command with the following: t,t,'-'. Thus % the plot command becomes as follows: plot(t,c,'o',t,t,'-') plot(t,c,'o',t,t,'-') % Add grid, title, xlabel, and ylabel grid title('Unit-Ramp Response Curve for System G(s) = (2s + 1)/(s^2 + s + 1)') xlabel('t Sec') ylabel('Input and Output') and use the step-response command. See MATLAB Program 5–10. The plot obtained by using this program is shown in Figure 5–26.
Unit-Ramp Response Curve for System G(s) = (2s + 1)/(s2 + s +1) t Sec 0 1 2 3 4 5 6 7 8 9 10 Input and Output 12 0 4 2 6 8 10 Figure 5–26 Unit-ramp response curve.
aa Section 5–5 / Transient-Response Analysis with MATLAB 199 Unit-Ramp Response of a System Defined in State Space.
Next, we shall treat the unit-ramp response of the system in state-space form.Consider the system described by where u is the unit-ramp function. In what follows, we shall consider a simple example to explain the method. Consider the case where When the initial conditions are zeros, the unit-ramp response is the integral of the unit-step response. Hence the unit-ramp response can be given by (5–44) From Equation (5–44), we obtain (5–45) Let us define Then Equation (5–45) becomes (5–46) Combining Equation (5–46) with the original state-space equation, we obtain (5–47) (5–48) where u appearing in Equation (5–47) is the unit-step function.These equations can be written as where Note that x3 is the third element of x. A plot of the unit-ramp response curve z(t) can be obtained by entering MATLAB Program 5–11 into the computer.A plot of the unit-ramp response curve obtained from this MATLAB program is shown in Figure 5–27.
BB = C 0 1 0 S = B B 0 R , CC = [0 0 1], DD = AA = C 0 - 1 1 1 - 1 0 0 0 0 S = C A C 0 0 0 S z = CCx + DDu x # = AAx + BBu z = [0 0 1]C x1 x2 x3 S C x # 1 x # 2 x # 3 S = C 0 - 1 1 1 - 1 0 0 0 0 S C x1 x2 x3 S + C 0 1 0 S u x # 3 = x1 z = x3 z # = y = x1 z = 3 t 0 y dt D = C = [1 0], x(0) = 0 B = B 0 1R , A = B 0 - 1 1 - 1R , y = Cx + Du x # = Ax + Bu aa 200 Chapter 5 / Transient and Steady-State Response Analyses MATLAB Program 5–11 % --------------- Unit-ramp response ---------------% The unit-ramp response is obtained by adding a new % state variable x3. The dimension of the state equation % is enlarged by one % Enter matrices A, B, C, and D of the original state % equation and output equation A = [0 1;-1 -1]; B = [0; 1]; C = [1 0]; D = ; % Enter matrices AA, BB, CC, and DD of the new, % enlarged state equation and output equation AA = [A zeros(2,1);C 0]; BB = [B;0]; CC = [0 0 1]; DD = ; % Enter step-response command: [z,x,t] = step(AA,BB,CC,DD) [z,x,t] = step(AA,BB,CC,DD); % In plotting x3 add the unit-ramp input t in the plot % by entering the following command: plot(t,x3,'o',t,t,'-') x3 = [0 0 1]x'; plot(t,x3,'o',t,t,'-') grid title('Unit-Ramp Response') xlabel('t Sec') ylabel('Input and Output') Unit-Ramp Response t Sec Input and Output 0 1 2 3 4 5 6 7 8 9 10 9 5 1 0 8 6 3 2 4 7 10 Figure 5–27 Unit-ramp response curve.
aa Section 5–5 / Transient-Response Analysis with MATLAB 201 Obtaining Response to Arbitrary Input.
To obtain the response to an arbitrary input, the command lsim may be used.The commands like lsim(num,den,r,t) lsim(A,B,C,D,u,t) y = lsim(num,den,r,t) y = lsim(A,B,C,D,u,t) will generate the response to input time function r or u. See the following two examples.
(Also, see Problems A–5–14 through A–5–16.) EXAMPLE 5–6 Using the lsim command, obtain the unit-ramp response of the following system: We may enter MATLAB Program 5–12 into the computer to obtain the unit-ramp response.The resulting plot is shown in Figure 5–28.
C(s) R(s) = 2s + 1 s2 + s + 1 MATLAB Program 5–12 % ------- Ramp Response -------num = [2 1]; den = [1 1 1]; t = 0:0.1:10; r = t; y = lsim(num,den,r,t); plot(t,r,'-',t,y,'o') grid title('Unit-Ramp Response Obtained by Use of Command "lsim"') xlabel('t Sec') ylabel('Unit-Ramp Input and System Output') text(6.3,4.6,'Unit-Ramp Input') text(4.75,9.0,'Output') Unit-Ramp Response Obtained by use of Command “Isim” t Sec 0 1 2 3 4 5 6 7 8 9 10 Unit-Ramp Input and System Output 12 0 4 2 6 8 10 Output Unit-Ramp Input Figure 5–28 Unit-ramp response.
aa 202 Chapter 5 / Transient and Steady-State Response Analyses EXAMPLE 5–7 Consider the system Using MATLAB, obtain the response curves y(t) when the input u is given by 1. u=unit-step input 2. u=e–t Assume that the initial state is x(0)=0.
A possible MATLAB program to produce the responses of this system to the unit-step input Cu=1(t)D and the exponential input Cu=e–tD is shown in MATLAB Program 5–13.The result-ing response curves are shown in Figures 5–29(a) and (b), respectively. y = [1 0]B x1 x2 R B x # 1 x # 2 R = B - 1 - 1 0.5 0 R B x1 x2 R + B 0 1R u MATLAB Program 5–13 t = 0:0.1:12; A = [-1 0.5;-1 0]; B = [0;1]; C = [1 0]; D = ; % For the unit-step input u = 1(t), use the command "y = step(A,B,C,D,1,t)".
y = step(A,B,C,D,1,t); plot(t,y) grid title('Unit-Step Response') xlabel('t Sec') ylabel('Output') % For the response to exponential input u = exp(-t), use the command % "z = lsim(A,B,C,D,u,t)".
u = exp(-t); z = lsim(A,B,C,D,u,t); plot(t,u,'-',t,z,'o') grid title('Response to Exponential Input u = exp(-t)') xlabel('t Sec') ylabel('Exponential Input and System Output') text(2.3,0.49,'Exponential input') text(6.4,0.28,'Output') aa Section 5–5 / Transient-Response Analysis with MATLAB 203 Unit-Step Response t Sec 0 2 4 6 8 10 12 Output 1 0.2 0 1.2 0.6 0.4 0.8 1.4 (a) Response to Exponential Input u = e−t t Sec 0 2 4 6 8 10 12 −0.2 (b) Exponential Input and System Output 0.8 0 1 0.4 0.2 0.6 1.2 Exponential Input Ouput Figure 5–29 (a) Unit-step response; (b) response to input u=e–t.
Response to Initial Condition.
In what follows we shall present a few methods for obtaining the response to an initial condition. Commands that we may use are “step” or “initial”. We shall first present a method to obtain the response to the initial condi-tion using a simple example.Then we shall discuss the response to the initial condition when the system is given in state-space form. Finally, we shall present a command initial to obtain the response of a system given in a state-space form.
aa 204 Chapter 5 / Transient and Steady-State Response Analyses EXAMPLE 5–8 Consider the mechanical system shown in Figure 5–30, where m=1 kg, b=3 N-secm, and k=2 Nm. Assume that at t=0 the mass m is pulled downward such that x(0)=0.1 m and (0)=0.05 msec.The displacement x(t) is measured from the equilibrium position before the mass is pulled down. Obtain the motion of the mass subjected to the initial condition. (Assume no external forcing function.) The system equation is with the initial conditions x(0)=0.1 m and (x is measured from the equilib-rium position.) The Laplace transform of the system equation gives or Solving this last equation for X(s) and substituting the given numerical values, we obtain This equation can be written as Hence the motion of the mass m may be obtained as the unit-step response of the following system: MATLAB Program 5–14 will give a plot of the motion of the mass.The plot is shown in Figure 5–31.
G(s) = 0.1s2 + 0.35s s2 + 3s + 2 X(s) = 0.1s2 + 0.35s s2 + 3s + 2 1 s = 0.1s + 0.35 s2 + 3s + 2 X(s) = mx(0)s + mx # (0) + bx(0) ms2 + bs + k Ams2 + bs + kBX(s) = mx(0)s + mx # (0) + bx(0) mCs2X(s) - sx(0) - x # (0)D + bCsX(s) - x(0)D + kX(s) = 0 x # (0) = 0.05 msec.
mx $ + bx # + kx = 0 x # MATLAB Program 5–14 % --------------- Response to initial condition ---------------% System response to initial condition is converted to % a unit-step response by modifying the numerator polynomial % Enter the numerator and denominator of the transfer % function G(s) num = [0.1 0.35 0]; den = [1 3 2]; % Enter the following step-response command step(num,den) % Enter grid and title of the plot grid title('Response of Spring-Mass-Damper System to Initial Condition') m k b x Figure 5–30 Mechanical system.
aa Section 5–5 / Transient-Response Analysis with MATLAB 205 Response of Spring-Mass-Damper System to Initial Condition Amplitude 0.12 0.02 0 0.08 0.04 0.06 0.1 Time (sec) 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Figure 5–31 Response of the mechanical system considered in Example 5–8.
Response to Initial Condition (State-Space Approach, Case 1).
Consider the system defined by (5–49) Let us obtain the response x(t) when the initial condition x(0) is specified.Assume that there is no external input function acting on this system.Assume also that x is an n-vector.
First, take Laplace transforms of both sides of Equation (5–49).
This equation can be rewritten as (5–50) Taking the inverse Laplace transform of Equation (5–50), we obtain (5–51) (Notice that by taking the Laplace transform of a differential equation and then by taking the inverse Laplace transform of the Laplace-transformed equation we generate a differential equation that involves the initial condition.) Now define (5–52) Then Equation (5–51) can be written as (5–53) By integrating Equation (5–53) with respect to t, we obtain (5–54) where B = x(0), u = 1(t) z # = Az + x(0)1(t) = Az + Bu z $ = Az # + x(0) d(t) z # = x x # = Ax + x(0) d(t) s X(s) = AX(s) + x(0) s X(s) - x(0) = AX(s) x # = Ax, x(0) = x0 aa 206 Chapter 5 / Transient and Steady-State Response Analyses Referring to Equation (5–52), the state x(t) is given by Thus, (5–55) The solution of Equations (5–54) and (5–55) gives the response to the initial condition.
Summarizing,the response of Equation (5–49) to the initial condition x(0) is obtained by solving the following state-space equations: where MATLAB commands to obtain the response curves, where we do not specify the time vector t (that is, we let the time vector be determined automatically by MATLAB), are given next.
% Specify matrices A and B [x,z,t] = step(A,B,A,B); x1 = [1 0 0 ... 0]x'; x2 = [0 1 0 ... 0]x'; xn = [0 0 0 ... 1]x'; plot(t,x1,t,x2, ... ,t,xn) If we choose the time vector t (for example, let the computation time duration be from t = 0 to t = tp with the computing time increment of ), then we use the following MATLAB commands: t = 0: Δt: tp; % Specify matrices A and B [x,z,t] = step(A,B,A,B,1,t); x1 = [1 0 0 ... 0]x'; x2 = [0 1 0 ... 0]x'; xn = [0 0 0 ... 1]x'; plot(t,x1,t,x2, ... ,t,xn) (See, for example, Example 5–9.) ¢t B = x(0), u = 1(t) x = Az + Bu z # = Az + Bu x = z # = Az + Bu z # (t).
aa Section 5–5 / Transient-Response Analysis with MATLAB 207 Response to Initial Condition (State-Space Approach, Case 2).
Consider the system defined by (5–56) (5–57) (Assume that x is an n-vector and y is an m-vector.) Similar to case 1, by defining we can obtain the following equation: (5–58) where Noting that Equation (5–57) can be written as (5–59) By substituting Equation (5–58) into Equation (5–59), we obtain (5–60) The solution of Equations (5–58) and (5–60), rewritten here where gives the response of the system to a given initial condi-tion. MATLAB commands to obtain the response curves (output curves y1 versus t, y2 versus t, ... , ym versus t) are shown next for two cases: Case A.
When the time vector t is not specified (that is, the time vector t is to be de-termined automatically by MATLAB): % Specify matrices A, B, and C [y,z,t] = step(A,B,CA,CB); y1 = [1 0 0 ... 0]y'; y2 = [0 1 0 ... 0]y'; ym = [0 0 0 ... 1]y'; plot(t,y1,t,y2, ... ,t,ym) B = x(0) and u = 1(t), y = CAz + CBu z # = Az + Bu y = C(Az + Bu) = CAz + CBu y = Cz # x = z # , B = x(0), u = 1(t) z # = Az + x(0)1(t) = Az + Bu z # = x y = Cx x # = Ax, x(0) = x0 aa 208 Chapter 5 / Transient and Steady-State Response Analyses MATLAB Program 5–15 t = 0:0.01:3; A = [0 1;-10 -5]; B = [2;1]; [x,z,t] = step(A,B,A,B,1,t); x1 = [1 0]x'; x2 = [0 1]x'; plot(t,x1,'x',t,x2,'-') grid title('Response to Initial Condition') xlabel('t Sec') ylabel('State Variables x1 and x2') gtext('x1') gtext('x2') Case B.
When the time vector t is specified: t = 0: Δt: tp; % Specify matrices A, B, and C [y,z,t] = step(A,B,CA,CB,1,t) y1 = [1 0 0 ... 0]y'; y2 = [0 1 0 ... 0]y'; ym = [0 0 0 ... 1]y'; plot(t,y1,t,y2, ... ,t,ym) EXAMPLE 5–9 Obtain the response of the system subjected to the given initial condition.
or Obtaining the response of the system to the given initial condition resolves to solving the unit-step response of the following system: where Hence a possible MATLAB program for obtaining the response may be given as shown in MATLAB Program 5–15.The resulting response curves are shown in Figure 5–32.
B = x(0), u = 1(t) x = Az + Bu z # = Az + Bu x # = Ax, x(0) = x0 B x # 1 x # 2 R = B 0 - 10 1 - 5R B x1 x2 R , B x1(0) x2(0)R = B2 1R aa Section 5–5 / Transient-Response Analysis with MATLAB 209 Response to Initial Condition t Sec 0 0.5 1 1.5 2 2.5 3 State Variables x1 and x2 3 −2 −3 1 −1 0 2 x1 x2 Figure 5–32 Response of system in Example 5–9 to initial condition.
For an illustrative example of how to use Equations (5–58) and (5–60) to find the re-sponse to the initial condition, see Problem A–5–16.
Obtaining Response to Initial Condition by Use of Command Initial.
If the system is given in the state-space form, then the following command initial(A,B,C,D,[initial condition],t) will produce the response to the initial condition.
Suppose that we have the system defined by where x0 = B 2 1R A = B 0 - 10 1 - 5R , B = B 0 0R , C = [0 0], D = 0 y = Cx + Du x # = Ax + Bu, x(0) = x0 aa 210 Chapter 5 / Transient and Steady-State Response Analyses MATLAB Program 5–16 t = 0:0.05:3; A = [0 1;-10 -5]; B = [0;0]; C = [0 0]; D = ; [y,x] = initial(A,B,C,D,[2;1],t); x1 = [1 0]x'; x2 = [0 1]x'; plot(t,x1,'o',t,x1,t,x2,'x',t,x2) grid title('Response to Initial Condition') xlabel('t Sec') ylabel('State Variables x1 and x2') gtext('x1') gtext('x2') Response to Initial Condition t Sec 0 0.5 1 1.5 2 2.5 3 State Variables x1 and x2 3 −3 −2 −1 0 1 2 x1 x2 Figure 5–33 Response curves to initial condition.
EXAMPLE 5–10 Consider the following system that is subjected to the initial condition. (No external forcing function is present.) Obtain the response y(t) to the given initial condition. y(0) = 2, y # (0) = 1, y $(0) = 0.5 y % + 8y $ + 17y # + 10y = 0 Then the command “initial” can be used as shown in MATLAB Program 5–16 to obtain the response to the initial condition. The response curves x1(t) and x2(t) are shown in Figure 5–33.They are the same as those shown in Figure 5–32.
aa Section 5–5 / Transient-Response Analysis with MATLAB 211 MATLAB Program 5–17 t = 0:0.05:10; A = [0 1 0;0 0 1;-10 -17 -8]; B = [0;0;0]; C = [1 0 0]; D = ; y = initial(A,B,C,D,[2;1;0.5],t); plot(t,y) grid title('Response to Initial Condition') xlabel('t (sec)') ylabel('Output y') Output y t (sec) Response to Initial Condition 0.5 1 1.5 2 2.5 0 0 1 2 3 4 5 6 7 8 9 10 Figure 5–34 Response y(t) to initial condition.
By defining the state variables as we obtain the following state-space representation for the system: A possible MATLAB program to obtain the response y(t) is given in MATLAB Program 5–17.
The resulting response curve is shown in Figure 5–34.
y = [1 0 0]C x1 x2 x3 S C x # 1 x # 2 x # 3 S = C 0 0 - 10 1 0 - 17 0 1 - 8 S C x1 x2 x3 S, C x1(0) x2(0) x3(0) S = C 2 1 0.5 S x3 = y $ x2 = y # x1 = y aa 212 Chapter 5 / Transient and Steady-State Response Analyses 5–6 ROUTH’S STABILITY CRITERION The most important problem in linear control systems concerns stability.That is, under what conditions will a system become unstable? If it is unstable, how should we stabi-lize the system? In Section 5–4 it was stated that a control system is stable if and only if all closed-loop poles lie in the left-half s plane. Most linear closed-loop systems have closed-loop transfer functions of the form where the a’s and b’s are constants and m n. A simple criterion, known as Routh’s stability criterion, enables us to determine the number of closed-loop poles that lie in the right-half s plane without having to factor the denominator polynomial. (The polynomial may include parameters that MATLAB cannot handle.) Routh’s Stability Criterion.
Routh’s stability criterion tells us whether or not there are unstable roots in a polynomial equation without actually solving for them.
This stability criterion applies to polynomials with only a finite number of terms.When the criterion is applied to a control system, information about absolute stability can be obtained directly from the coefficients of the characteristic equation.
The procedure in Routh’s stability criterion is as follows: 1. Write the polynomial in s in the following form: (5–61) where the coefficients are real quantities.We assume that an Z 0; that is, any zero root has been removed.
2. If any of the coefficients are zero or negative in the presence of at least one posi-tive coefficient, a root or roots exist that are imaginary or that have positive real parts.Therefore, in such a case, the system is not stable. If we are interested in only the absolute stability, there is no need to follow the procedure further. Note that all the coefficients must be positive.This is a necessary condition, as may be seen from the following argument: A polynomial in s having real coefficients can al-ways be factored into linear and quadratic factors, such as (s+a) and As2+bs+cB, where a, b, and c are real. The linear factors yield real roots and the quadratic factors yield complex-conjugate roots of the polynomial.The factor As2+bs+cB yields roots having negative real parts only if b and c are both pos-itive. For all roots to have negative real parts, the constants a, b, c, and so on, in all factors must be positive.The product of any number of linear and quadratic factors containing only positive coefficients always yields a polynomial with positive coefficients. It is important to note that the condition that all the coefficients be positive is not sufficient to assure stability. The necessary but not sufficient condition for stability is that the coefficients of Equation (5–61) all be present and all have a positive sign. (If all a’s are negative, they can be made positive by multiplying both sides of the equation by –1.) a0 sn + a1 sn-1 + p + an-1 s + an = 0 C(s) R(s) = b0 sm + b1 sm-1 + p + bm-1 s + bm a0 sn + a1 sn-1 + p + an-1 s + an = B(s) A(s) aa Section 5–6 / Routh’s Stability Criterion 213 3. If all coefficients are positive, arrange the coefficients of the polynomial in rows and columns according to the following pattern: The process of forming rows continues until we run out of elements. (The total number of rows is n+1.) The coefficients b1, b2, b3, and so on, are evaluated as follows: The evaluation of the b’s is continued until the remaining ones are all zero. The same pattern of cross-multiplying the coefficients of the two previous rows is followed in evaluating the c’s, d’s, e’s, and so on.That is, c3 = b1 a7 - a1 b4 b1 c2 = b1 a5 - a1 b3 b1 c1 = b1 a3 - a1 b2 b1 b3 = a1 a6 - a0 a7 a1 b2 = a1 a4 - a0 a5 a1 b1 = a1 a2 - a0 a3 a1 sn sn-1 sn-2 sn-3 sn-4 s2 s1 s0 a0 a1 b1 c1 d1 e1 f 1 g1 a2 a3 b2 c2 d2 e2 a4 a5 b3 c3 d3 a6 a7 b4 c4 d4 p p p p p aa 214 Chapter 5 / Transient and Steady-State Response Analyses and This process is continued until the nth row has been completed.The complete array of coefficients is triangular. Note that in developing the array an entire row may be divid-ed or multiplied by a positive number in order to simplify the subsequent numerical calculation without altering the stability conclusion.
Routh’s stability criterion states that the number of roots of Equation (5–61) with positive real parts is equal to the number of changes in sign of the coefficients of the first column of the array. It should be noted that the exact values of the terms in the first col-umn need not be known; instead, only the signs are needed. The necessary and suffi-cient condition that all roots of Equation (5–61) lie in the left-half s plane is that all the coefficients of Equation (5–61) be positive and all terms in the first column of the array have positive signs.
EXAMPLE 5–11 Let us apply Routh’s stability criterion to the following third-order polynomial: where all the coefficients are positive numbers.The array of coefficients becomes The condition that all roots have negative real parts is given by EXAMPLE 5–12 Consider the following polynomial: Let us follow the procedure just presented and construct the array of coefficients. (The first two rows can be obtained directly from the given polynomial. The remaining terms are s4 + 2s3 + 3s2 + 4s + 5 = 0 a1 a2 7 a0 a3 s3 s2 s1 s0 a0 a1 a1 a2 - a0 a3 a1 a3 a2 a3 a0 s3 + a1 s2 + a2 s + a3 = 0 d2 = c1 b3 - b1 c3 c1 d1 = c1 b2 - b1 c2 c1 aa Section 5–6 / Routh’s Stability Criterion 215 obtained from these. If any coefficients are missing, they may be replaced by zeros in the array.) In this example, the number of changes in sign of the coefficients in the first column is 2. This means that there are two roots with positive real parts. Note that the result is unchanged when the coefficients of any row are multiplied or divided by a positive number in order to simplify the computation.
Special Cases.
If a first-column term in any row is zero, but the remaining terms are not zero or there is no remaining term, then the zero term is replaced by a very small positive number and the rest of the array is evaluated. For example, consider the following equation: (5–62) The array of coefficients is If the sign of the coefficient above the zero () is the same as that below it, it indicates that there are a pair of imaginary roots. Actually, Equation (5–62) has two roots at s=; j.
If, however, the sign of the coefficient above the zero () is opposite that below it, it indicates that there is one sign change. For example, for the equation the array of coefficients is One sign change: One sign change: There are two sign changes of the coefficients in the first column. So there are two roots in the right-half s plane. This agrees with the correct result indicated by the factored form of the polynomial equation.
s1 s0 - 3 - 2 2 s3 s2 1 0 L - 3 2 s3 - 3s + 2 = (s - 1)2(s + 2) = 0 s3 s2 s1 s0 1 2 0 L 2 1 2 s3 + 2s2 + s + 2 = 0 The second row is divided by 2.
6 s4 s3 s2 s1 s0 1 2 1 1 - 3 5 3 4 2 5 5 0 0 s4 s3 s2 s1 s0 1 2 1 - 6 5 3 4 5 5 0 6 ⁄ ⁄ ⁄ ⁄ aa 216 Chapter 5 / Transient and Steady-State Response Analyses If all the coefficients in any derived row are zero, it indicates that there are roots of equal magnitude lying radially opposite in the s plane—that is, two real roots with equal magnitudes and opposite signs and/or two conjugate imaginary roots. In such a case, the evaluation of the rest of the array can be continued by forming an auxiliary polynomi-al with the coefficients of the last row and by using the coefficients of the derivative of this polynomial in the next row. Such roots with equal magnitudes and lying radially op-posite in the s plane can be found by solving the auxiliary polynomial, which is always even.For a 2n-degree auxiliary polynomial,there are n pairs of equal and opposite roots.
For example, consider the following equation: The array of coefficients is The terms in the s3 row are all zero. (Note that such a case occurs only in an odd-numbered row.) The auxiliary polynomial is then formed from the coefficients of the s4 row.The auxiliary polynomial P(s) is which indicates that there are two pairs of roots of equal magnitude and opposite sign (that is, two real roots with the same magnitude but opposite signs or two complex-conjugate roots on the imaginary axis).These pairs are obtained by solving the auxiliary polynomial equation P(s)=0.The derivative of P(s) with respect to s is The terms in the s3 row are replaced by the coefficients of the last equation—that is, 8 and 96.The array of coefficients then becomes We see that there is one change in sign in the first column of the new array.Thus,the orig-inal equation has one root with a positive real part. By solving for roots of the auxiliary polynomial equation, we obtain or s = ;j5 s = ;1, s2 = -25 s2 = 1, 2s4 + 48s2 - 50 = 0 s5 s4 s3 s2 s1 s0 1 2 8 24 112.7 - 50 24 48 96 - 50 0 - 25 - 50 d Coefficients of dP (s)ds dP (s) ds = 8s3 + 96s P(s) = 2s4 + 48s2 - 50 s5 s4 s3 1 2 0 24 48 0 - 25 - 50 d Auxiliary polynomial P(s) s5 + 2s4 + 24s3 + 48s2 - 25s - 50 = 0 aa Section 5–6 / Routh’s Stability Criterion 217 These two pairs of roots of P(s) are a part of the roots of the original equation. As a matter of fact, the original equation can be written in factored form as follows: Clearly, the original equation has one root with a positive real part.
Relative Stability Analysis.
Routh’s stability criterion provides the answer to the question of absolute stability.This, in many practical cases, is not sufficient.We usu-ally require information about the relative stability of the system. A useful approach for examining relative stability is to shift the s-plane axis and apply Routh’s stability criterion.That is, we substitute into the characteristic equation of the system, write the polynomial in terms of and apply Routh’s stability criterion to the new polynomial in The number of changes of sign in the first column of the array developed for the polynomial in is equal to the num-ber of roots that are located to the right of the vertical line s=–s.Thus, this test reveals the number of roots that lie to the right of the vertical line s=–s.
Application of Routh’s Stability Criterion to Control-System Analysis.
Routh’s stability criterion is of limited usefulness in linear control-system analysis,mainly because it does not suggest how to improve relative stability or how to stabilize an unstable system. It is possible, however, to determine the effects of changing one or two parameters of a system by examining the values that cause instability. In the following, we shall consider the problem of determining the stability range of a parameter value.
Consider the system shown in Figure 5–35. Let us determine the range of K for stability.The closed-loop transfer function is The characteristic equation is The array of coefficients becomes s4 s3 s2 s1 s0 1 3 7 3 2 - 9 7 K K 3 2 K K 0 s4 + 3s3 + 3s2 + 2s + K = 0 C(s) R(s) = K sAs2 + s + 1B(s + 2) + K s ˆ s ˆ.
s ˆ; s = s ˆ - s (s = constant) (s + 1)(s - 1)(s + j5)(s - j5)(s + 2) = 0 + – R(s) C(s) K s(s2 + s + 1) (s + 2) Figure 5–35 Control system.
aa 218 Chapter 5 / Transient and Steady-State Response Analyses For stability, K must be positive, and all coefficients in the first column must be positive.
Therefore, When the system becomes oscillatory and, mathematically, the oscillation is sustained at constant amplitude.
Note that the ranges of design parameters that lead to stability may be determined by use of Routh’s stability criterion.
5–7 EFFECTS OF INTEGRAL AND DERIVATIVE CONTROL ACTIONS ON SYSTEM PERFORMANCE In this section, we shall investigate the effects of integral and derivative control actions on the system performance. Here we shall consider only simple systems, so that the effects of integral and derivative control actions on system performance can be clearly seen.
Integral Control Action.
In the proportional control of a plant whose transfer function does not possess an integrator 1s, there is a steady-state error, or offset, in the response to a step input. Such an offset can be eliminated if the integral control action is included in the controller.
In the integral control of a plant, the control signal—the output signal from the controller—at any instant is the area under the actuating-error-signal curve up to that instant.The control signal u(t) can have a nonzero value when the actuating error signal e(t) is zero, as shown in Figure 5–36(a).This is impossible in the case of the proportional controller, since a nonzero control signal requires a nonzero actuating error signal.
(A nonzero actuating error signal at steady state means that there is an offset.) Figure 5–36(b) shows the curve e(t) versus t and the corresponding curve u(t) versus t when the controller is of the proportional type.
Note that integral control action,while removing offset or steady-state error,may lead to oscillatory response of slowly decreasing amplitude or even increasing amplitude, both of which are usually undesirable.
K = 14 9 , 14 9 7 K 7 0 e(t) u(t) 0 0 t t e(t) u(t) 0 0 t t (a) (b) Figure 5–36 (a) Plots of e(t) and u(t) curves showing nonzero control signal when the actuating error signal is zero (integral control); (b) plots of e(t) and u(t) curves showing zero control signal when the actuating error signal is zero (proportional control).
aa Section 5–7 / Effects of Integral and Derivative Control Actions on System Performance 219 Proportional Control of Systems.
We shall show that the proportional control of a system without an integrator will result in a steady-state error with a step input.We shall then show that such an error can be eliminated if integral control action is included in the controller.
Consider the system shown in Figure 5–37. Let us obtain the steady-state error in the unit-step response of the system. Define Since the error E(s) is given by For the unit-step input R(s)=1/s, we have The steady-state error is Such a system without an integrator in the feedforward path always has a steady-state error in the step response. Such a steady-state error is called an offset. Figure 5–38 shows the unit-step response and the offset.
ess = lim tS q e(t) = lim sS0sE(s) = lim sS0 Ts + 1 Ts + 1 + K = 1 K + 1 E(s) = Ts + 1 Ts + 1 + K 1 s E(s) = 1 1 + G(s) R(s) = 1 1 + K Ts + 1 R(s) E(s) R(s) = R(s) - C(s) R(s) = 1 - C(s) R(s) = 1 1 + G(s) G(s) = K Ts + 1 1 Ts + 1 + – R(s) E(s) C(s) K Proportional controller Plant Figure 5–37 Proportional control system.
c(t) 1 0 t Offset Figure 5–38 Unit-step response and offset.
aa 220 Chapter 5 / Transient and Steady-State Response Analyses Integral Control of Systems.
Consider the system shown in Figure 5–39. The controller is an integral controller.The closed-loop transfer function of the system is Hence Since the system is stable, the steady-state error for the unit-step response can be obtained by applying the final-value theorem, as follows: Integral control of the system thus eliminates the steady-state error in the response to the step input. This is an important improvement over the proportional control alone, which gives offset.
Response to Torque Disturbances (Proportional Control).
Let us investigate the effect of a torque disturbance occurring at the load element. Consider the system shown in Figure 5–40.The proportional controller delivers torque T to position the load element, which consists of moment of inertia and viscous friction.Torque disturbance is denoted by D.
Assuming that the reference input is zero or R(s)=0, the transfer function between C(s) and D(s) is given by C(s) D(s) = 1 Js2 + bs + K p = 0 = lim sS0 s2(Ts + 1) Ts2 + s + K 1 s ess = lim sS0sE(s) E(s) R(s) = R(s) - C(s) R(s) = s(Ts + 1) s(Ts + 1) + K C(s) R(s) = K s(Ts + 1) + K 1 Ts + 1 + – R(s) C(s) E(s) K s Figure 5–39 Integral control system.
+ – ++ R D C E T Kp 1 s(Js + b) Figure 5–40 Control system with a torque disturbance.
aa Section 5–7 / Effects of Integral and Derivative Control Actions on System Performance 221 Hence The steady-state error due to a step disturbance torque of magnitude is given by At steady state, the proportional controller provides the torque which is equal in magnitude but opposite in sign to the disturbance torque The steady-state output due to the step disturbance torque is The steady-state error can be reduced by increasing the value of the gain Kp. Increasing this value, however, will cause the system response to be more oscillatory.
Response to Torque Disturbances (Proportional-Plus-Integral Control).
To eliminate offset due to torque disturbance, the proportional controller may be replaced by a proportional-plus-integral controller.
If integral control action is added to the controller, then, as long as there is an error signal, a torque is developed by the controller to reduce this error, provided the control system is a stable one.
Figure 5–41 shows the proportional-plus-integral control of the load element, consisting of moment of inertia and viscous friction.
The closed-loop transfer function between C(s) and D(s) is In the absence of the reference input, or r(t)=0, the error signal is obtained from E(s) = -s Js3 + bs2 + K p s + K p T i D(s) C(s) D(s) = s Js3 + bs2 + K p s + K p T i css = -ess = T d K p T d .
-T d , = - T d K p = lim sS0 -s Js2 + bs + K p T d s ess = lim sS0sE(s) T d E(s) D(s) = - C(s) D(s) = -1 Js2 + bs + K p ++ C E D R = 0 T Kp(1 + 1 Tis ) 1 s(Js + b) + – Figure 5–41 Proportional-plus-integral control of a load element consisting of moment of inertia and viscous friction.
aa 222 Chapter 5 / Transient and Steady-State Response Analyses If this control system is stable—that is, if the roots of the characteristic equation have negative real parts—then the steady-state error in the response to a unit-step disturbance torque can be obtained by applying the final-value theorem as follows: Thus steady-state error to the step disturbance torque can be eliminated if the controller is of the proportional-plus-integral type.
Note that the integral control action added to the proportional controller has converted the originally second-order system to a third-order one. Hence the control system may become unstable for a large value of Kp, since the roots of the characteristic equation may have positive real parts. (The second-order system is always stable if the coefficients in the system differential equation are all positive.) It is important to point out that if the controller were an integral controller, as in Figure 5–42, then the system always becomes unstable, because the characteristic equation will have roots with positive real parts. Such an unstable system cannot be used in practice.
Note that in the system of Figure 5–41 the proportional control action tends to stabilize the system,while the integral control action tends to eliminate or reduce steady-state error in response to various inputs.
Derivative Control Action.
Derivative control action, when added to a proportional controller, provides a means of obtaining a controller with high sensitivity. An advantage of using derivative control action is that it responds to the rate of change of the actuating error and can produce a significant correction before the magnitude of the actuating error becomes too large. Derivative control thus anticipates the actuating error, initiates an early corrective action, and tends to increase the stability of the system.
Js3 + bs2 + K = 0 = 0 = lim sS0 -s2 Js3 + bs2 + K p s + K p T i 1 s ess = lim sS0sE(s) Js3 + bs2 + K p s + K p T i = 0 + – ++ C E D R = 0 T K s 1 s(Js + b) Figure 5–42 Integral control of a load element consisting of moment of inertia and viscous friction.
aa Section 5–7 / Effects of Integral and Derivative Control Actions on System Performance 223 + – R(s) C(s) (a) (b) Kp 1 Js2 c(t) 1 0 t Figure 5–43 (a) Proportional control of a system with inertia load; (b) response to a unit-step input.
Although derivative control does not affect the steady-state error directly, it adds damping to the system and thus permits the use of a larger value of the gain K, which will result in an improvement in the steady-state accuracy.
Because derivative control operates on the rate of change of the actuating error and not the actuating error itself, this mode is never used alone. It is always used in combi-nation with proportional or proportional-plus-integral control action.
Proportional Control of Systems with Inertia Load.
Before we discuss further the effect of derivative control action on system performance, we shall consider the proportional control of an inertia load.
Consider the system shown in Figure 5–43(a). The closed-loop transfer function is obtained as Since the roots of the characteristic equation are imaginary, the response to a unit-step input continues to oscillate indefinitely, as shown in Figure 5–43(b).
Control systems exhibiting such response characteristics are not desirable.We shall see that the addition of derivative control will stabilize the system.
Proportional-Plus-Derivative Control of a System with Inertia Load.
Let us modify the proportional controller to a proportional-plus-derivative controller whose transfer function is The torque developed by the controller is proportional to Derivative control is essentially anticipatory,measures the instantaneous error velocity, and predicts the large overshoot ahead of time and produces an appropriate counteraction before too large an overshoot occurs.
K pAe + T d e # B.
K pA1 + T d sB.
Js2 + K p = 0 C(s) R(s) = K p Js2 + K p aa 224 Chapter 5 / Transient and Steady-State Response Analyses Consider the system shown in Figure 5–44(a). The closed-loop transfer function is given by The characteristic equation now has two roots with negative real parts for positive values of J, Kp, and Thus derivative control introduces a damping effect.A typical response curve c(t) to a unit-step input is shown in Figure 5–44(b). Clearly, the response curve shows a marked improvement over the original response curve shown in Figure 5–46(b).
Proportional-Plus-Derivative Control of Second-Order Systems.
A compromise between acceptable transient-response behavior and acceptable steady-state behavior may be achieved by use of proportional-plus-derivative control action.
Consider the system shown in Figure 5–45.The closed-loop transfer function is The steady-state error for a unit-ramp input is The characteristic equation is Js2 + AB + K dBs + K p = 0 ess = B K p C(s) R(s) = K p + K d s Js2 + AB + K dBs + K p T d .
Js2 + K p T d s + K p = 0 C(s) R(s) = K pA1 + T d sB Js2 + K p T d s + K p + – R(s) C(s) Kp + Kds 1 s(Js + B) Figure 5–44 (a) Proportional-plus-derivative control of a system with inertia load; (b) response to a unit-step input.
R(s) C(s) (a) (b) Kp (1 + Tds) c(t) 1 0 t 1 Js2 + – Figure 5–45 Control system.
aa Section 5–8 / Steady-State Errors in Unity-Feedback Control Systems 225 The effective damping coefficient of this system is thus B+Kd rather than B. Since the damping ratio z of this system is it is possible to make both the steady-state error ess for a ramp input and the maximum overshoot for a step input small by making B small,Kp large,and Kd large enough so that z is between 0.4 and 0.7.
5–8 STEADY-STATE ERRORS IN UNITY-FEEDBACK CONTROL SYSTEMS Errors in a control system can be attributed to many factors. Changes in the reference input will cause unavoidable errors during transient periods and may also cause steady-state errors.Imperfections in the system components,such as static friction,backlash,and amplifier drift, as well as aging or deterioration, will cause errors at steady state. In this section, however, we shall not discuss errors due to imperfections in the system com-ponents. Rather, we shall investigate a type of steady-state error that is caused by the incapability of a system to follow particular types of inputs.
Any physical control system inherently suffers steady-state error in response to certain types of inputs.A system may have no steady-state error to a step input, but the same system may exhibit nonzero steady-state error to a ramp input. (The only way we may be able to eliminate this error is to modify the system structure.) Whether a given system will exhibit steady-state error for a given type of input depends on the type of open-loop transfer function of the system, to be discussed in what follows.
Classification of Control Systems.
Control systems may be classified according to their ability to follow step inputs, ramp inputs, parabolic inputs, and so on. This is a reasonable classification scheme, because actual inputs may frequently be considered combinations of such inputs. The magnitudes of the steady-state errors due to these individual inputs are indicative of the goodness of the system.
Consider the unity-feedback control system with the following open-loop transfer function G(s): It involves the term sN in the denominator, representing a pole of multiplicity N at the origin.The present classification scheme is based on the number of integrations indicated by the open-loop transfer function.A system is called type 0, type 1, type 2, p , if N=0, N=1, N=2, p , respectively. Note that this classification is different from that of the order of a system. As the type number is increased, accuracy is improved; however, increasing the type number aggravates the stability problem. A compromise between steady-state accuracy and relative stability is always necessary.
We shall see later that, if G(s) is written so that each term in the numerator and denominator, except the term sN, approaches unity as s approaches zero, then the open-loop gain K is directly related to the steady-state error.
G(s) = KAT a s + 1BAT b s + 1B p AT m s + 1B sNAT 1 s + 1BAT 2 s + 1B p AT p s + 1B z = B + K d 22K p J aa 226 Chapter 5 / Transient and Steady-State Response Analyses Steady-State Errors.
Consider the system shown in Figure 5–46.The closed-loop transfer function is The transfer function between the error signal e(t) and the input signal r(t) is where the error e(t) is the difference between the input signal and the output signal.
The final-value theorem provides a convenient way to find the steady-state performance of a stable system. Since E(s) is the steady-state error is The static error constants defined in the following are figures of merit of control systems.
The higher the constants, the smaller the steady-state error. In a given system, the out-put may be the position, velocity, pressure, temperature, or the like. The physical form of the output, however, is immaterial to the present analysis.Therefore, in what follows, we shall call the output “position,” the rate of change of the output “velocity,” and so on.
This means that in a temperature control system “position” represents the output tem-perature,“velocity” represents the rate of change of the output temperature, and so on.
Static Position Error Constant Kp.
The steady-state error of the system for a unit-step input is The static position error constant Kp is defined by Thus, the steady-state error in terms of the static position error constant Kp is given by ess = 1 1 + K p K p = lim sS0G(s) = G(0) = 1 1 + G(0) ess = lim sS0 s 1 + G(s) 1 s ess = lim tS q e(t) = lim sS0sE(s) = lim sS0 sR(s) 1 + G(s) E(s) = 1 1 + G(s) R(s) E(s) R(s) = 1 - C(s) R(s) = 1 1 + G(s) C(s) R(s) = G(s) 1 + G(s) + – R(s) C(s) E(s) G(s) Figure 5–46 Control system.
aa Section 5–8 / Steady-State Errors in Unity-Feedback Control Systems 227 For a type 0 system, For a type 1 or higher system, for N 1 Hence, for a type 0 system, the static position error constant Kp is finite, while for a type 1 or higher system, Kp is infinite.
For a unit-step input, the steady-state error ess may be summarized as follows: for type 0 systems for type 1 or higher systems From the foregoing analysis, it is seen that the response of a feedback control system to a step input involves a steady-state error if there is no integration in the feedforward path. (If small errors for step inputs can be tolerated, then a type 0 system may be permissible, provided that the gain K is sufficiently large. If the gain K is too large, how-ever, it is difficult to obtain reasonable relative stability.) If zero steady-state error for a step input is desired, the type of the system must be one or higher.
Static Velocity Error Constant Kv.
The steady-state error of the system with a unit-ramp input is given by The static velocity error constant Kv is defined by Thus, the steady-state error in terms of the static velocity error constant Kv is given by The term velocity error is used here to express the steady-state error for a ramp input.The dimension of the velocity error is the same as the system error.That is,velocity error is not an error in velocity, but it is an error in position due to a ramp input.
For a type 0 system, K v = lim sS0 sKAT a s + 1BAT b s + 1B p AT 1 s + 1BAT 2 s + 1B p = 0 ess = 1 K v K v = lim sS0sG(s) = lim sS0 1 sG(s) ess = lim sS0 s 1 + G(s) 1 s2 ess = 0, ess = 1 1 + K , K p = lim sS0 KAT a s + 1BAT b s + 1B p sNAT 1 s + 1BAT 2 s + 1B p = q, K p = lim sS0 KAT a s + 1BAT b s + 1B p AT 1 s + 1BAT 2 s + 1B p = K aa 228 Chapter 5 / Transient and Steady-State Response Analyses For a type 1 system, For a type 2 or higher system, for N 2 The steady-state error ess for the unit-ramp input can be summarized as follows: for type 0 systems for type 1 systems for type 2 or higher systems The foregoing analysis indicates that a type 0 system is incapable of following a ramp input in the steady state.The type 1 system with unity feedback can follow the ramp input with a finite error.In steady-state operation,the output velocity is exactly the same as the input velocity, but there is a positional error.This error is proportional to the velocity of the input and is inversely proportional to the gain K.Figure 5–47 shows an example of the response of a type 1 system with unity feedback to a ramp input. The type 2 or higher system can follow a ramp input with zero error at steady state.
Static Acceleration Error Constant Ka.
The steady-state error of the system with a unit-parabolic input (acceleration input), which is defined by for t 0 for t<0 = 0, r(t) = t2 2 , ess = 1 K v = 0, ess = 1 K v = 1 K , ess = 1 K v = q, K v = lim sS0 sKAT a s + 1BAT b s + 1B p sNAT 1 s + 1BAT 2 s + 1B p = q, K v = lim sS0 sKAT a s + 1BAT b s + 1B p sAT 1 s + 1BAT 2 s + 1B p = K r(t) c(t) 0 t r(t) c(t) Figure 5–47 Response of a type 1 unity-feedback system to a ramp input.
aa Section 5–8 / Steady-State Errors in Unity-Feedback Control Systems 229 is given by The static acceleration error constant Ka is defined by the equation The steady-state error is then Note that the acceleration error, the steady-state error due to a parabolic input, is an error in position.
The values of Ka are obtained as follows: For a type 0 system, For a type 1 system, For a type 2 system, For a type 3 or higher system, for N 3 Thus, the steady-state error for the unit parabolic input is for type 0 and type 1 systems for type 2 systems for type 3 or higher systems ess = 0, ess = 1 K , ess = q, K a = lim sS0 s2KAT a s + 1BAT b s + 1B p sNAT 1 s + 1BAT 2 s + 1B p = q, K a = lim sS0 s2KAT a s + 1BAT b s + 1B p s2AT 1 s + 1BAT 2 s + 1B p = K K a = lim sS0 s2KAT a s + 1BAT b s + 1B p sAT 1 s + 1BAT 2 s + 1B p = 0 K a = lim sS0 s2KAT a s + 1BAT b s + 1B p AT 1 s + 1BAT 2 s + 1B p = 0 ess = 1 K a K a = lim sS0s2G(s) = 1 lim sS0s2G(s) ess = lim sS0 s 1 + G(s) 1 s3 aa 230 Chapter 5 / Transient and Steady-State Response Analyses Note that both type 0 and type 1 systems are incapable of following a parabolic input in the steady state. The type 2 system with unity feedback can follow a parabolic input with a finite error signal. Figure 5–48 shows an example of the response of a type 2 sys-tem with unity feedback to a parabolic input. The type 3 or higher system with unity feedback follows a parabolic input with zero error at steady state.
Summary.
Table 5–1 summarizes the steady-state errors for type 0, type 1, and type 2 systems when they are subjected to various inputs. The finite values for steady-state errors appear on the diagonal line.Above the diagonal, the steady-state errors are infinity; below the diagonal, they are zero.
r(t) c(t) 0 t r(t) c(t) Figure 5–48 Response of a type 2 unity-feedback system to a parabolic input.
Step Input Ramp Input Acceleration Input r(t)=1 r(t)=t Type 0 system q q Type 1 system 0 q Type 2 system 0 0 1 K 1 K 1 1 + K r(t) = 1 2 t2 Table 5–1 Steady-State Error in Terms of Gain K Remember that the terms position error, velocity error, and acceleration error mean steady-state deviations in the output position. A finite velocity error implies that after transients have died out, the input and output move at the same velocity but have a finite position difference.
The error constants Kp, Kv, and Ka describe the ability of a unity-feedback system to reduce or eliminate steady-state error.Therefore,they are indicative of the steady-state performance. It is generally desirable to increase the error constants, while maintaining the transient response within an acceptable range. It is noted that to improve the steady-state performance we can increase the type of the system by adding an integrator or integrators to the feedforward path. This, however, introduces an additional stability problem.The design of a satisfactory system with more than two integrators in series in the feedforward path is generally not easy.
aa Example Problems and Solutions 231 EXAMPLE PROBLEMS AND SOLUTIONS A–5–1.
In the system of Figure 5–49, x(t) is the input displacement and u(t) is the output angular displacement. Assume that the masses involved are negligibly small and that all motions are restricted to be small; therefore, the system can be considered linear.The initial conditions for x and u are zeros, or x(0–)=0 and u(0–)=0. Show that this system is a differentiating element.
Then obtain the response u(t) when x(t) is a unit-step input.
Solution. The equation for the system is or The Laplace transform of this last equation, using zero initial conditions, gives And so Thus the system is a differentiating system.
For the unit-step input X(s)=1s, the output becomes The inverse Laplace transform of gives u(t) = 1 L e-(kb)t Q(s) Q(s) = 1 L 1 s + (kb) Q(s) Q(s) X(s) = 1 L s s + (kb) aLs + k b L bQ(s) = sX(s) Lu # + k b Lu = x # bAx # - Lu # B = kLu No friction x b k u L Figure 5–49 Mechanical system.
aa 232 Chapter 5 / Transient and Steady-State Response Analyses Note that if the value of kb is large, the response u(t) approaches a pulse signal, as shown in Figure 5–50.
A–5–2.
Gear trains are often used in servo systems to reduce speed, to magnify torque, or to obtain the most efficient power transfer by matching the driving member to the given load.
Consider the gear-train system shown in Figure 5–51. In this system, a load is driven by a motor through the gear train.Assuming that the stiffness of the shafts of the gear train is infinite (there is neither backlash nor elastic deformation) and that the number of teeth on each gear is proportional to the radius of the gear, obtain the equivalent moment of inertia and equivalent viscous-friction coefficient referred to the motor shaft and referred to the load shaft.
In Figure 5–51 the numbers of teeth on gears 1, 2, 3, and 4 are N1, N2, N3, and N4, respectively.
The angular displacements of shafts, 1, 2, and 3 are u1, u2, and u3, respectively.Thus, and The moment of inertia and viscous-friction coefficient of each gear-train component are denoted by J1, b1; J2, b2; and J3, b3; respectively. (J3 and b3 include the moment of inertia and friction of the load.) u3 u2 = N 3 N 4 .
u2 u1 = N 1 N 2 x(t) t t 1 0 0 u(t) 1 L Figure 5–50 Unit-step input and the response of the mechanical system shown in Figure 5–49.
Shaft 1 Gear 2 Gear 1 Gear 3 Gear 4 Shaft 2 Shaft 3 J1, b1 N1 Input torque from motor Tm (t) u1 N2 N3 N4 u2 u3 Load torque TL (t) J2, b2 J3, b3 Figure 5–51 Gear-train system.
aa Example Problems and Solutions 233 Solution. For this gear-train system, we can obtain the following equations: For shaft 1, (5–63) where is the torque developed by the motor and is the load torque on gear 1 due to the rest of the gear train. For shaft 2, (5–64) where is the torque transmitted to gear 2 and is the load torque on gear 3 due to the rest of the gear train. Since the work done by gear 1 is equal to that of gear 2, or If the gear ratio reduces the speed as well as magnifies the torque. For shaft 3, (5–65) where is the load torque and is the torque transmitted to gear 4.
and are related by and u3 and u1 are related by Eliminating and from Equations (5–63), (5–64), and (5–65) yields Eliminating u2 and u3 from this last equation and writing the resulting equation in terms of u1 and its time derivatives, we obtain (5–66) Thus, the equivalent moment of inertia and viscous-friction coefficient of the gear train referred to shaft 1 are given, respectively, by Similarly,the equivalent moment of inertia and viscous-friction coefficient of the gear train referred to the load shaft (shaft 3) are given, respectively, by b3eq = b3 + a N 4 N 3 b 2 b2 + a N 2 N 1 b 2 a N 4 N 3 b 2 b1 J 3eq = J 3 + a N 4 N 3 b 2 J 2 + a N 2 N 1 b 2 a N 4 N 3 b 2 J 1 b1eq = b1 + a N 1 N 2 b 2 b2 + a N 1 N 2 b 2 a N 3 N 4 b 2 b3 J 1eq = J 1 + a N 1 N 2 b 2 J 2 + a N 1 N 2 b 2 a N 3 N 4 b 2 J 3 + cb1 + a N 1 N 2 b 2 b2 + a N 1 N 2 b 2 a N 3 N 4 b 2 b3du # 1 + a N 1 N 2 b a N 3 N 4 bT L = T m cJ 1 + a N 1 N 2 b 2 J 2 + a N 1 N 2 b 2 a N 3 N 4 b 2 J 3d u $ 1 J 1 u $ 1 + b1 u # 1 + N 1 N 2 AJ 2 u $ 2 + b2 u # 2B + N 1 N 3 N 2 N 4 AJ 3 u $ 3 + b3 u # 3 + T LB = T m T 4 T 1 , T 2 , T 3 , u3 = u2 N 3 N 4 = u1 N 1 N 2 N 3 N 4 T 4 = T 3 N 4 N 3 T 4 T 3 T 4 T L J 3 u $ 3 + b3 u # 3 + T L = T 4 N 1 N 2 6 1, T 2 = T 1 N 2 N 1 T 1 u1 = T 2 u2 T 3 T 2 J 2 u $ 2 + b2 u # 2 + T 3 = T 2 T 1 T m J 1 u $ 1 + b1 u # 1 + T 1 = T m aa 234 Chapter 5 / Transient and Steady-State Response Analyses The relationship between J1eq and J3eq is thus and that between b1eq and b3eq is The effect of J2 and J3 on an equivalent moment of inertia is determined by the gear ratios and For speed-reducing gear trains, the ratios, and are usually less than unity.
If and then the effect of J2 and J3 on the equivalent moment of inertia J1eq is negligible. Similar comments apply to the equivalent viscous-friction coefficient b1eq of the gear train. In terms of the equivalent moment of inertia J1eq and equivalent viscous-friction coefficient b1eq, Equation (5–66) can be simplified to give where A–5–3.
When the system shown in Figure 5–52(a) is subjected to a unit-step input, the system output responds as shown in Figure 5–52(b). Determine the values of K and T from the response curve.
Solution. The maximum overshoot of 25.4% corresponds to z=0.4. From the response curve we have Consequently, tp = p vd = p vn21 - z2 = p vn21 - 0.42 = 3 tp = 3 n = N 1 N 2 N 3 N 4 J 1eq u $ 1 + b1eq u # 1 + nT L = T m N 3 N 4 1, N 1 N 2 1 N 3 N 4 N 1 N 2 N 3 N 4 .
N 1 N 2 b1eq = a N 1 N 2 b 2 a N 3 N 4 b 2 b3eq J 1eq = a N 1 N 2 b 2 a N 3 N 4 b 2 J 3eq + – R(s) C(s) (a) (b) c(t) 1 0 3 t 0.254 K s(Ts + 1) Figure 5–52 (a) Closed-loop system; (b) unit-step response curve.
aa Example Problems and Solutions 235 It follows that From the block diagram we have from which Therefore, the values of T and K are determined as A–5–4.
Determine the values of K and k of the closed-loop system shown in Figure 5–53 so that the maximum overshoot in unit-step response is 25% and the peak time is 2 sec.Assume that J=1 kg-m2.
Solution. The closed-loop transfer function is By substituting J=1 kg-m2 into this last equation, we have Note that in this problem The maximum overshoot Mp is which is specified as 25%. Hence from which zp 21 - z2 = 1.386 e-zp21-z2 = 0.25 Mp = e-zp21-z2 vn = 1K, 2zvn = Kk C(s) R(s) = K s2 + Kks + K C(s) R(s) = K Js2 + Kks + K K = v2 n T = 1.142 1.09 = 1.42 T = 1 2zvn = 1 2 0.4 1.14 = 1.09 vn = A K T , 2zvn = 1 T C(s) R(s) = K Ts2 + s + K vn = 1.14 + – + – R(s) C(s) k 1 s K Js Figure 5–53 Closed-loop system.
aa 236 Chapter 5 / Transient and Steady-State Response Analyses or The peak time tp is specified as 2 sec.And so or Then the undamped natural frequency vn is Therefore, we obtain A–5–5.
Figure 5–54(a) shows a mechanical vibratory system.When 2 lb of force (step input) is applied to the system, the mass oscillates, as shown in Figure 5–54(b). Determine m, b, and k of the system from this response curve.The displacement x is measured from the equilibrium position.
Solution. The transfer function of this system is Since we obtain It follows that the steady-state value of x is x(q) = lim sS0sX(s) = 2 k = 0.1 ft X(s) = 2 sAms2 + bs + kB P(s) = 2 s X(s) P(s) = 1 ms2 + bs + k k = 2zvn K = 2 0.404 1.72 2.95 = 0.471 sec K = v2 n = 1.722 = 2.95 N-m vn = vd 21 - z2 = 1.57 21 - 0.4042 = 1.72 vd = 1.57 tp = p vd = 2 z = 0.404 k b x (a) (b) P(2-lb force) x(t) ft 0.1 0 1 2 3 4 5 t 0.0095 ft m Figure 5–54 (a) Mechanical vibratory system; (b) step-response curve.
aa Example Problems and Solutions 237 Hence Note that Mp=9.5% corresponds to z=0.6.The peak time tp is given by The experimental curve shows that tp=2 sec.Therefore, Since v2 n=km=20m, we obtain (Note that 1 slug=1 lbf-sec2ft.) Then b is determined from or A–5–6.
Consider the unit-step response of the second-order system The amplitude of the exponentially damped sinusoid changes as a geometric series. At time t=tp=pvd, the amplitude is equal to After one oscillation, or at t=tp+2p d=3pvd, the amplitude is equal to after another cycle of oscillation, the amplitude is The logarithm of the ratio of successive amplitudes is called the logarithmic decrement.Determine the logarithmic decrement for this second-order system.Describe a method for experimental determination of the damping ratio from the rate of decay of the oscillation.
Solution. Let us define the amplitude of the output oscillation at t=ti to be xi, where ti=tp+(i-1)T(T=period of oscillation). The amplitude ratio per one period of damped oscillation is Thus, the logarithmic decrement d is It is a function only of the damping ratio z. Thus, the damping ratio z can be determined by use of the logarithmic. decrement.
In the experimental determination of the damping ratio z from the rate of decay of the oscil-lation, we measure the amplitude x1 at t=tp and amplitude xn at t=tp+(n-1)T. Note that it is necessary to choose n large enough so that the ratio x1/xn is not near unity.Then x1 xn = e(n-1)2zp21-z2 d = ln x1 x2 = 2zp 21 - z2 x1 x2 = e-AsvdBp e-AsvdB3p = e2AsvdBp = e2zp21-z2 e-AsvdB5p.
e-AsvdB3p; e-AsvdBp.
C(s) R(s) = v2 n s2 + 2zvn s + v2 n b = 2zvn m = 2 0.6 1.96 5.2 = 12.2 lbfftsec 2zvn = b m m = 20 v2 n = 20 1.962 = 5.2 slugs = 167 lb vn = 3.14 2 0.8 = 1.96 radsec tp = p vd = p vn21 - z2 = p 0.8vn k = 20 lbfft aa 238 Chapter 5 / Transient and Steady-State Response Analyses or Hence A–5–7.
In the system shown in Figure 5–55, the numerical values of m, b, and k are given as m=1 kg, b=2 N-secm, and k=100 Nm. The mass is displaced 0.05 m and released without initial ve-locity.Find the frequency observed in the vibration.In addition,find the amplitude four cycles later.
The displacement x is measured from the equilibrium position.
Solution. The equation of motion for the system is Substituting the numerical values for m, b, and k into this equation gives where the initial conditions are x(0)=0.05 and From this last equation the undamped natural frequency vn and the damping ratio z are found to be The frequency actually observed in the vibration is the damped natural frequency vd.
In the present analysis, is given as zero.Thus, solution x(t) can be written as It follows that at t=nT, where T=2pvd, Consequently, the amplitude four cycles later becomes A–5–8.
Obtain both analytically and computationally the unit-step response of tbe following higher-order system: [Obtain the partial-fraction expansion of C(s) with MATLAB when R(s) is a unit-step function.] C(s) R(s) = 3s3 + 25s2 + 72s + 80 s4 + 8s3 + 40s2 + 96s + 80 = 0.05e-2.526 = 0.05 0.07998 = 0.004 m x(4T) = x(0)e-zvn 4T = x(0)e-(0.1)(10)(4)(0.6315) x(nT) = x(0)e-zvn nT x(t) = x(0)e-zvn t acos vd t + z 21 - z2 sinvd t b x # (0) vd = vn21 - z2 = 1011 - 0.01 = 9.95 radsec vn = 10, z = 0.1 x # (0) = 0.
x $ + 2x # + 100x = 0 mx $ + bx # + kx = 0 z = 1 n - 1 aln x1 xn b B4p2 + c 1 n - 1 aln x1 xn b d 2 ln x1 xn = (n - 1) 2zp 21 - z2 k m b x Figure 5–55 Spring-mass-damper system.
aa Example Problems and Solutions 239 Solution. MATLAB Program 5–18 yields the unit-step response curve shown in Figure 5–56. It also yields the partial-fraction expansion of C(s) as follows: - 0.4375 s + 2 -0.375 (s + 2)2 + 1 s = -0.5626(s + 2) (s + 2)2 + 42 + (0.3438) 4 (s + 2)2 + 42 + -0.4375 s + 2 + -0.375 (s + 2)2 + 1 s = -0.2813 - j0.1719 s + 2 - j4 + -0.2813 + j0.1719 s + 2 + j4 C(s) = 3s3 + 25s2 + 72s + 80 s4 + 8s3 + 40s2 + 96s + 80 1 s MATLAB Program 5–18 % ------- Unit-Step Response of C(s)/R(s) and Partial-Fraction Expansion of C(s) -------num = [3 25 72 80]; den = [1 8 40 96 80]; step(num,den); v = [0 3 0 1.2]; axis(v), grid % To obtain the partial-fraction expansion of C(s), enter commands % num1 = [3 25 72 80]; % den1 = [1 8 40 96 80 0]; % [r,p,k] = residue(num1,den1) num1 = [25 72 80]; den1 = [1 8 40 96 80 0]; [r,p,k] = residue(num1,den1) r = -0.2813- 0.1719i -0.2813+ 0.1719i -0.4375 -0.3750 -1.0000 p = -2.0000+ 4.0000i -2.0000- 4.0000i -2.0000 -2.0000 -0 k = [] aa 240 Chapter 5 / Transient and Steady-State Response Analyses Hence, the time response c(t) can be given by The fact that the response curve is an exponential curve superimposed by damped sinusoidal curves can be seen from Figure 5–56.
A–5–9.
When the closed-loop system involves a numerator dynamics, the unit-step response curve may exhibit a large overshoot. Obtain the unit-step response of the following system with MATLAB: Obtain also the unit-ramp response with MATLAB.
Solution. MATLAB Program 5–19 produces the unit-step response as well as the unit-ramp response of the system.The unit-step response curve and unit-ramp response curve, together with the unit-ramp input, are shown in Figures 5–57(a) and (b), respectively.
Notice that the unit-step response curve exhibits over 215% of overshoot. The unit-ramp response curve leads the input curve.These phenomena occurred because of the presence of a large derivative term in the numerator.
C(s) R(s) = 10s + 4 s2 + 4s + 4 - 0.4375e-2t - 0.375te-2t + 1 c(t) = -0.5626e-2t cos4t + 0.3438e-2t sin4t Amplitude Time (sec) Step Response 0.6 0.4 0.2 0.8 1 1.2 0 0 0.5 1 1.5 2 2.5 3 Figure 5–56 Unit-step response curve.
aa Example Problems and Solutions 241 Output t (sec) Unit-Step Response (a) 0.5 1 1.5 2 2.5 0 0 1 2 3 4 5 6 7 8 9 10 Unit-Ramp Input and Output t (sec) Unit-Ramp Response (b) 1 2 3 4 5 6 7 8 9 10 0 0 1 2 3 4 5 6 7 8 9 10 Unit-Ramp Input Output Figure 5–57 (a) Unit-step response curve; (b) unit-ramp response curve plotted with unit-ramp input.
MATLAB Program 5–19 num = [10 4]; den = [1 4 4]; t = 0:0.02:10; y = step(num,den,t); plot(t,y) grid title('Unit-Step Response') xlabel('t (sec)') ylabel('Output') num1 = [10 4]; den1 = [1 4 4 0]; y1 = step(num1,den1,t); plot(t,t,'--',t,y1) v = [0 10 0 10]; axis(v); grid title('Unit-Ramp Response') xlabel('t (sec)') ylabel('Unit-Ramp Input and Output') text(6.1,5.0,'Unit-Ramp Input') text(3.5,7.1,'Output') aa A–5–10.
Consider a higher-order system defined by Using MATLAB,plot the unit-step response curve of this system.Using MATLAB,obtain the rise time, peak time, maximum overshoot, and settling time.
Solution. MATLAB Program 5–20 plots the unit-step response curve as well as giving the rise time, peak time, maximum overshoot, and settling time.The unit-step response curve is shown in Figure 5–58.
C(s) R(s) = 6.3223s2 + 18s + 12.811 s4 + 6s3 + 11.3223s2 + 18s + 12.811 242 Chapter 5 / Transient and Steady-State Response Analyses MATLAB Program 5–20 % ------- This program is to plot the unit-step response curve, as well as to % find the rise time, peak time, maximum overshoot, and settling time.
% In this program the rise time is calculated as the time required for the % response to rise from 10% to 90% of its final value. -------num = [6.3223 18 12.811]; den = [1 6 11.3223 18 12.811]; t = 0:0.02:20; [y,x,t] = step(num,den,t); plot(t,y) grid title('Unit-Step Response') xlabel('t (sec)') ylabel('Output y(t)') r1 = 1; while y(r1) < 0.1, r1 = r1+1; end; r2 = 1; while y(r2) < 0.9, r2 = r2+1; end; rise_time = (r2-r1)0. 02 rise_time = 0.5800 [ymax,tp] = max(y); peak_time = (tp-1)0.02 peak_time = 1.6600 max_overshoot = ymax-1 max_overshoot = 0.6182 s = 1001; while y(s) > 0.98 & y(s) < 1.02; s = s-1; end; settling_time = (s-1)0.02 settling_time = 10.0200 aa Example Problems and Solutions 243 A–5–11.
Consider the closed-loop system defined by Using a “for loop,” write a MATLAB program to obtain unit-step response of this system for the following four cases: Solution. Define v2 n=a and 2zvn=b.Then, a and b each have four elements as follows: a=[1 4 16 36] b=[0.6 2 5.6 9.6] Case 4: z = 0.8, vn = 6 Case 3: z = 0.7, vn = 4 Case 2: z = 0.5, vn = 2 Case 1: z = 0.3, vn = 1 C(s) R(s) = v2 n s2 + 2zvn s + v2 n Output y(t) t (sec) Unit-Step Response 0.6 0.4 0.2 0.8 1 1.2 1.4 1.6 1.8 0 0 2 4 6 8 10 12 14 16 18 20 Figure 5–58 Unit-step response curve.
aa 244 Chapter 5 / Transient and Steady-State Response Analyses Using vectors a and b, MATLAB Program 5–21 will produce the unit-step response curves as shown in Figure 5–59.
Unit-Step Response Curves for Four Cases t Sec 0 1 2 3 4 5 6 7 8 Outputs 1.4 0 0.4 0.2 0.6 0.8 1 1.2 1 2 3 4 Figure 5–59 Unit-step response curves for four cases.
MATLAB Program 5–21 a = [1 4 16 36]; b = [0.6 2 5.6 9.6]; t = 0:0.1:8; y = zeros(81,4); for i = 1:4; num = [a(i)]; den = [1 b(i) a(i)]; y(:,i) = step(num,den,t); end plot(t,y(:,1),'o',t,y(:,2),'x',t,y(:,3),'-',t,y(:,4),'-.') grid title('Unit-Step Response Curves for Four Cases') xlabel('t Sec') ylabel('Outputs') gtext('1') gtext('2') gtext('3') gtext('4') aa Example Problems and Solutions 245 A–5–12.
Using MATLAB, obtain the unit-ramp response of the closed-loop control system whose closed-loop transfer function is Also, obtain the response of this system when the input is given by Solution. MATLAB Program 5–22 produces the unit-ramp response and the response to the exponential input r=e–0.5t.The resulting response curves are shown in Figures 5–60(a) and (b), respectively.
r = e-0.5t C(s) R(s) = s + 10 s3 + 6s2 + 9s + 10 MATLAB Program 5–22 % --------- Unit-Ramp Response ---------num = [1 10]; den = [1 6 9 10]; t = 0:0.1:10; r = t; y = lsim(num,den,r,t); plot(t,r,'-',t,y,'o') grid title('Unit-Ramp Response by Use of Command "lsim"') xlabel('t Sec') ylabel('Output') text(3.2,6.5,'Unit-Ramp Input') text(6.0,3.1,'Output') % --------- Response to Input r1 = exp(-0.5t). ---------num = [0 0 1 10]; den = [1 6 9 10]; t = 0:0.1:12; r1 = exp(-0.5t); y1 = lsim(num,den,r1,t); plot(t,r1,'-',t,y1,'o') grid title('Response to Input r1 = exp(-0.5t)') xlabel('t Sec') ylabel('Input and Output') text(1.4,0.75,'Input r1 = exp(-0.5t)') text(6.2,0.34,'Output') aa 246 Chapter 5 / Transient and Steady-State Response Analyses Unit-Ramp Response by Use of Command “lsim” t Sec Output 9 5 1 8 6 3 2 4 7 10 Unit-Ramp Input (a) 00 1 2 3 4 5 6 7 8 9 10 Output Response to Input r1 = e−0.5t Input r1 = e−0.5t Output t Sec 0 2 4 6 8 10 12 (b) Input and Output 1 0.1 0 0.4 0.2 0.3 0.5 0.6 0.7 0.8 0.9 Figure 5–60 (a) Unit-ramp response curve; (b) response to exponential input r1=e–0.5t.
A–5–13.
Obtain the response of the closed-loop system defined by when the input r(t) is given by r(t)=2+t [The input r(t) is a step input of magnitude 2 plus unit-ramp input.] C(s) R(s) = 5 s2 + s + 5 aa Example Problems and Solutions 247 MATLAB Program 5–23 num = ; den = [1 1 5]; t = 0:0.05:10; r = 2+t; c = lsim(num,den,r,t); plot(t,r,'-',t,c,'o') grid title('Response to Input r(t) = 2 + t') xlabel('t Sec') ylabel('Output c(t) and Input r(t) = 2 + t') Response to Input r(t) = 2 + t t Sec 0 1 2 3 4 5 6 7 8 9 10 Output c(t) and Input r(t) = 2 + t 12 0 4 2 6 8 10 Figure 5–61 Response to input r(t)=2+t.
2 s(s + 1) R(s) C(s) +− Figure 5–62 Control system.
Solution. A possible MATLAB program is shown in MATLAB Program 5–23. The resulting response curve, together with a plot of the input function, is shown in Figure 5–61.
A–5–14.
Obtain the response of the system shown in Figure 5–62 when the input r(t) is given by [The input r(t) is the unit-acceleration input.] r(t) = 1 2 t2 aa 248 Chapter 5 / Transient and Steady-State Response Analyses Solution. The closed-loop transfer function is MATLAB Program 5–24 produces the unit-acceleration response.The resulting response,together with the unit-acceleration input, is shown in Figure 5–63.
C(s) R(s) = 2 s2 + s + 2 MATLAB Program 5–24 num = ; den = [1 1 2]; t = 0:0.2:10; r = 0.5t.^2; y = lsim(num,den,r,t); plot(t,r,'-',t,y,'o',t,y,'-') grid title('Unit-Acceleration Response') xlabel('t Sec') ylabel('Input and Output') text(2.1,27.5,'Unit-Acceleration Input') text(7.2,7.5,'Output') Unit-Acceleration Response t Sec 0 1 2 3 4 5 6 7 8 9 10 Input and Output 50 0 10 5 15 20 25 30 35 40 45 Unit-Acceleration Input Ouput Figure 5–63 Response to unit-acceleration input.
A–5–15.
Consider the system defined by C(s) R(s) = 1 s2 + 2zs + 1 aa Example Problems and Solutions 249 MATLAB Program 5–25 t = 0:0.2:12; for n = 1:6; num = ; den = [1 2(n-1)0.2 1]; [y(1:61,n),x,t] = step(num,den,t); end plot(t,y) grid title('Unit-Step Response Curves') xlabel('t Sec') ylabel('Outputs') gtext('\zeta = 0'), gtext('0.2') gtext('0.4') gtext('0.6') gtext('0.8') gtext('1.0') % To draw a three-dimensional plot, enter the following command: mesh(y) or mesh(y').
% We shall show two three-dimensional plots, one using “mesh(y)” and the other using % "mesh(y')". These two plots are the same, except that the x axis and y axis are % interchanged.
mesh(y) title('Three-Dimensional Plot of Unit-Step Response Curves using Command "mesh(y)"') xlabel('n, where n = 1,2,3,4,5,6') ylabel('Computation Time Points') zlabel('Outputs') mesh(y') title('Three-Dimensional Plot of Unit-Step Response Curves using Command "mesh(y transpose)"') xlabel('Computation Time Points') ylabel('n, where n = 1,2,3,4,5,6') zlabel('Outputs') where z=0, 0.2, 0.4, 0.6, 0.8, and 1.0. Write a MATLAB program using a “for loop” to obtain the two-dimensional and three-dimensional plots of the system output. The input is the unit-step function.
Solution. MATLAB Program 5–25 is a possible program to obtain two-dimensional and three-dimensional plots. Figure 5–64(a) is the two-dimensional plot of the unit-step response curves for various values of z. Figure 5–64(b) is the three-dimensional plot obtained by use of the command “mesh(y)” and Figure 5–64(c) is obtained by use of the command “mesh(y¿)”. (These two three-dimensional plots are basically the same.The only difference is that x axis and y axis are in-terchanged.) aa 250 Chapter 5 / Transient and Steady-State Response Analyses A–5–16.
Consider the system subjected to the initial condition as given below.
(There is no input or forcing function in this system.) Obtain the response y(t) versus t to the given initial condition by use of Equations (5–58) and (5–60).
y = [1 0 0]C x1 x2 x3 S C x # 1 x # 2 x # 3 S = C 0 1 0 0 0 1 -10 -17 -8 S C x1 x2 x3 S , C x1(0) x2(0) x3(0) S = C 2 1 0.5 S (a) 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2 4 6 8 10 12 Outputs t Sec = 0 = 0 0.2 0.2 0.4 0.4 0.6 0.6 0.8 0.8 1.0 1.0 Unit-Step Response Curves Three-Dimensional Plot of Unit-Step Response Curves using Command “mesh(y)” 0 80 60 40 20 0 1 2 3 4 5 6 0.5 1 1.5 2 Outputs Computation Time Points n, where n = 1, 2, 3, 4, 5, 6 (b) 0 6 5 4 3 2 1 0 10 20 30 40 50 60 70 0.5 1 1.5 2 Outputs Three-Dimensional Plot of Unit-Step Response Curves using Command “mesh(y transpose)” n, where n = 1, 2, 3, 4, 5, 6 Computation Time Points (c) Figure 5–64 (a) Two-dimensional plot of unit-step response curves; (b) three-dimensional plot of unit-step response curves using command “mesh(y)”; (c) three-dimensional plot of unit-step response curves using command “mesh(y¿)”.
aa Example Problems and Solutions 251 Solution. A possible MATLAB program based on Equations (5–58) and (5–60) is given by MAT-LAB program 5–26.The response curve obtained here is shown in Figure 5–65. (Notice that this problem was solved by use of the command “initial”in Example 5–16.The response curve obtained here is exactly the same as that shown in Figure 5–34.) MATLAB Program 5–26 t = 0:0.05:10; A = [0 1 0;0 0 1;-10 -17 -8]; B = [2;1;0.5]; C=[1 0 0]; [y,x,t] = step(A,B,CA,CB,1,t); plot(t,y) grid; title('Response to Initial Condition') xlabel('t (sec)') ylabel('Output y') A–5–17.
Consider the following characteristic equation: Determine the range of K for stability.
Solution. The Routh array of coefficients is s4 s3 s2 s1 s0 1 K K - 1 K 1 -K2 K - 1 1 1 1 1 1 0 s4 + Ks3 + s2 + s + 1 = 0 Figure 5–65 Response y(t) to the given initial condition.
Output y t (sec) Response to Initial Condition 0.5 1 1.5 2 2.5 0 0 1 2 3 4 5 6 7 8 9 10 aa 252 Chapter 5 / Transient and Steady-State Response Analyses For stability, we require that From the first and second conditions, K must be greater than 1. For K>1, notice that the term 1-CK2/(K-1)D is always negative, since Thus, the three conditions cannot be fulfilled simultaneously.Therefore, there is no value of K that allows stability of the system.
A–5–18.
Consider the characteristic equation given by (5–67) The Hurwitz stability criterion, given next, gives conditions for all the roots to have negative real parts in terms of the coefficients of the polynomial.As stated in the discussions of Routh’s stability criterion in Section 5–6, for all the roots to have negative real parts, all the coefficients a’s must be positive.This is a necessary condition but not a sufficient condition. If this condition is not sat-isfied, it indicates that some of the roots have positive real parts or are imaginary or zero.A suf-ficient condition for all the roots to have negative real parts is given in the following Hurwitz stability criterion: If all the coefficients of the polynomial are positive, arrange these coefficients in the following determinant: where we substituted zero for as if s>n. For all the roots to have negative real parts, it is neces-sary and sufficient that successive principal minors of be positive. The successive principal minors are the following determinants: where as=0 if s>n. (It is noted that some of the conditions for the lower-order determinants are included in the conditions for the higher-order determinants.) If all these determinants are positive, and a0>0 as already assumed, the equilibrium state of the system whose characteristic ¢i = 5 a1 a0 0 0 a3 a2 a1 0 p p p p a2i-1 a2i-2 a2i-3 ai 5 (i = 1, 2, p , n - 1) ¢n ¢n = 7 a1 a0 0 0 0 a3 a2 a1 a0 0 a5 a4 a3 a2 0 p p p p p 0 an an-1 an-2 an-3 an-4 0 0 0 an an-1 an-2 0 0 0 0 0 an 7 a0 sn + a1 sn-1 + a2 sn-2 + p + an-1 s + an = 0 K - 1 - K2 K - 1 = -1 + K(1 - K) K - 1 6 0 K 7 0 K - 1 K 7 0 1 -K2 K - 1 7 0 aa Example Problems and Solutions 253 equation is given by Equation (5–67) is asymptotically stable. Note that exact values of determi-nants are not needed;instead,only signs of these determinants are needed for the stability criterion.
Now consider the following characteristic equation: Obtain the conditions for stability using the Hurwitz stability criterion.
Solution. The conditions for stability are that all the a’s be positive and that It is clear that, if all the a’s are positive and if the condition is satisfied, the condition is also satisfied.Therefore,for all the roots of the given characteristic equation to have neg-ative real parts, it is necessary and sufficient that all the coefficients a’s are positive and A–5–19.
Show that the first column of the Routh array of is given by where Solution. The Routh array of coefficients has the form 1 a1 b1 c1 a2 a3 b2 c2 a4 a5 b3 a6 p p p an ak = 0 if k 7 n ¢r = 7 a1 a3 a5 a2r-1 1 a2 a4 0 a1 a3 0 1 a2 0 0 0 ar 7 , (n r 1) 1, ¢1 , ¢2 ¢1 , ¢3 ¢2 , p , ¢n ¢n-1 sn + a1 sn-1 + a2 sn-2 + p + an-1 s + an = 0 ¢3 7 0.
¢2 7 0 ¢3 7 0 = a3Aa1 a2 - a0 a3B - a2 1 a4 7 0 = a1Aa2 a3 - a1 a4B - a0 a2 3 ¢3 = 3 a1 a0 0 a3 a2 a1 0 a4 a3 3 ¢2 = 2a1 a0 a3 a2 2 = a1 a2 - a0 a3 7 0 a0 s4 + a1 s3 + a2 s2 + a3 s + a4 = 0 aa 254 Chapter 5 / Transient and Steady-State Response Analyses The first term in the first column of the Routh array is 1.The next term in the first column is a1, which is equal to The next term is b1, which is equal to The next term in the first column is c1, which is equal to In a similar manner the remaining terms in the first column of the Routh array can be found.
The Routh array has the property that the last nonzero terms of any columns are the same; that is, if the array is given by then and if the array is given by then In any case, the last term of the first column is equal to an, or an = ¢n-1 an ¢n-1 = ¢n ¢n-1 a6 = b3 = d2 = f 1 a0 a1 b1 c1 d1 e1 f 1 a2 a3 b2 c2 d2 0 a4 a5 b3 0 a6 0 a7 = c3 = e2 = g1 a0 a1 b1 c1 d1 e1 f 1 g1 a2 a3 b2 c2 d2 e2 a4 a5 b3 c3 a6 a7 = ¢3 ¢2 = a1 a2 a3 - a2 3 - a2 1 a4 + a1 a5 a1 a2 - a3 b1 a3 - a1 b2 b1 = c a1 a2 - a3 a1 da3 - a1c a1 a4 - a5 a1 d c a1a2 - a3 a1 d a1 a2 - a3 a1 = ¢2 ¢1 ¢1 .
aa Example Problems and Solutions 255 For example, if n=4, then Thus it has been shown that the first column of the Routh array is given by A–5–20.
Show that the Routh’s stability criterion and Hurwitz stability criterion are equivalent.
Solution. If we write Hurwitz determinants in the triangular form where the elements below the diagonal line are all zeros and the elements above the diagonal line any numbers, then the Hurwitz conditions for asymptotic stability become which are equivalent to the conditions We shall show that these conditions are equivalent to where a1, b1, c1, p , are the elements of the first column in the Routh array.
Consider, for example, the following Hurwitz determinant, which corresponds to i=4: The determinant is unchanged if we subtract from the ith row k times the jth row. By subtracting from the second row a0/a1 times the first row, we obtain ¢4 = 4 a11 0 0 0 a3 a22 a1 a0 a5 a23 a3 a2 a7 a24 a5 a4 4 ¢4 = 4 a1 a0 0 0 a3 a2 a1 a0 a5 a4 a3 a2 a7 a6 a5 a4 4 a1 7 0, b1 7 0, c1 7 0, p a11 7 0, a22 7 0, p , ann 7 0 ¢i = a11 a22 p aii 7 0, (i = 1, 2, p , n) ¢i = 6 a11 a22 0 aii 6, (i = 1, 2, p , n) 1, ¢1 , ¢2 ¢1 , ¢3 ¢2 , p, ¢n ¢n-1 ¢4 = 4 a1 a3 a5 a7 1 a2 a4 a6 0 a1 a3 a5 0 1 a2 a4 4 = 4 a1 a3 0 0 1 a2 a4 0 0 a1 a3 0 0 1 a2 a4 4 = ¢3 a4 aa 256 Chapter 5 / Transient and Steady-State Response Analyses where Similarly, subtracting from the fourth row a0/a1 times the third row yields where Next, subtracting from the third row a1/a22 times the second row yields where Finally, subtracting from the last row times the third row yields where a44 = a ˆ44 - a ˆ43 a33 a34 ¢4 = 4 a11 0 0 0 a3 a22 0 0 a5 a23 a33 0 a7 a24 a34 a44 4 a ˆ43 a33 a34 = a5 - a1 a22 a24 a33 = a3 - a1 a22 a23 ¢4 = 4 a11 0 0 0 a3 a22 0 0 a5 a23 a33 a ˆ43 a7 a24 a34 a ˆ44 4 a ˆ44 = a4 - a0 a1 a5 a ˆ43 = a2 - a0 a1 a3 ¢4 = 4 a11 0 0 0 a3 a22 a1 0 a5 a23 a3 a ˆ43 a7 a24 a5 a ˆ44 4 a24 = a6 - a0 a1 a7 a23 = a4 - a0 a1 a5 a22 = a2 - a0 a1 a3 a11 = a1 aa Example Problems and Solutions 257 From this analysis, we see that The Hurwitz conditions for asymptotic stability reduce to the conditions The Routh array for the polynomial where a0>0 and n=4, is given by From this Routh array, we see that (The last equation is obtained using the fact that ) Hence the Hurwitz conditions for asymptotic stability become Thus we have demonstrated that Hurwitz conditions for asymptotic stability can be reduced to Routh’s conditions for asymptotic stability. The same argument can be extended to Hurwitz determinants of any order, and the equivalence of Routh’s stability criterion and Hurwitz stabil-ity criterion can be established.
A–5–21.
Consider the characteristic equation Using the Hurwitz stability criterion, determine the range of K for stability.
Solution. Comparing the given characteristic equation s4 + 2s3 + (4 + K)s2 + 9s + 25 = 0 s4 + 2s3 + (4 + K)s2 + 9s + 25 = 0 a1 7 0, b1 7 0, c1 7 0, d1 7 0 a34 = 0, a ˆ44 = a4 , and a4 = b2 = d1 .
a44 = a ˆ44 - a ˆ43 a33 a34 = a4 = d1 a33 = a3 - a1 a22 a23 = a3 b1 - a1 b2 b1 = c1 a22 = a2 - a0 a1 a3 = b1 a11 = a1 a0 a1 b1 c1 d1 a2 a3 b2 a4 a0 s4 + a1 s3 + a2 s2 + a3 s + a4 = 0 a11 7 0, a22 7 0, a33 7 0, a44 7 0, p ¢1 7 0, ¢2 7 0, ¢3 7 0, ¢4 7 0, p ¢1 = a11 ¢2 = a11 a22 ¢3 = a11 a22 a33 ¢4 = a11 a22 a33 a44 aa 258 Chapter 5 / Transient and Steady-State Response Analyses with the following standard fourth-order characteristic equation: we find The Hurwitz stability criterion states that is given by For all the roots to have negative real parts, it is necessary and sufficient that succesive principal minors of be positive.The successive principal minors are For all principal minors to be positive, we require that be positive.Thus, we require from which we obtain the region of K for stability to be A–5–22.
Explain why the proportional control of a plant that does not possess an integrating property (which means that the plant transfer function does not include the factor 1/s) suffers offset in response to step inputs.
Solution. Consider, for example, the system shown in Figure 5–66.At steady state, if c were equal to a nonzero constant r, then e=0 and u=Ke=0, resulting in c=0, which contradicts the assumption that c=r=nonzero constant.
A nonzero offset must exist for proper operation of such a control system. In other words, at steady state, if e were equal to r/(1+K), then u=Kr/(1+K) and c=Kr/(1+K), which results in the assumed error signal e=r/(1+K).Thus the offset of r/(1+K) must exist in such a system.
K 7 109 18 18K - 109 7 0 2K - 1 7 0 ¢i(i = 1, 2, 3) ¢3 = 3 a1 a0 0 a3 a2 a1 0 a4 a3 3 = 3 2 1 0 9 4 + K 2 0 25 9 3 = 18K - 109 ¢2 = 2a1 a0 a3 a2 2 = 22 1 9 4 + K2 = 2K - 1 ¢1 = @a1@ = 2 ¢4 ¢4 = 4 a1 a0 0 0 a3 a2 a1 a0 0 a4 a3 a2 0 0 0 a4 4 ¢4 a0 = 1, a1 = 2, a2 = 4 + K, a3 = 9, a4 = 25 a0 s4 + a1 s3 + a2s2 + a3 s + a4 = 0 + – r c e u K 1 Ts + 1 Figure 5–66 Control system.
aa Example Problems and Solutions 259 A–5–23.
The block diagram of Figure 5–67 shows a speed control system in which the output member of the system is subject to a torque disturbance. In the diagram, and D(s) are the Laplace transforms of the reference speed, output speed, driving torque, and disturbance torque, respectively. In the absence of a disturbance torque, the output speed is equal to the reference speed.
Vr(s), V(s), T(s), + – ++ D(s) E(s) T(s) V (s) Vr(s) K 1 Js Figure 5–67 Block diagram of a speed control system.
+ – K 1 Js VD(s) D(s) Figure 5–68 Block diagram of the speed control system of Figure 5–67 when Vr(s) = 0.
Investigate the response of this system to a unit-step disturbance torque. Assume that the reference input is zero, or Solution. Figure 5–68 is a modified block diagram convenient for the present analysis.The closed-loop transfer function is where is the Laplace transform of the output speed due to the disturbance torque.For a unit-step disturbance torque, the steady-state output velocity is From this analysis, we conclude that, if a step disturbance torque is applied to the output member of the system, an error speed will result so that the ensuing motor torque will exactly can-cel the disturbance torque.To develop this motor torque, it is necessary that there be an error in speed so that nonzero torque will result. (Discussions continue to Problem A–5–24.) = 1 K = lim sS0 s Js + K 1 s vD(q) = lim sS0sVD(s) VD(s) VD(s) D(s) = 1 Js + K Vr(s) = 0.
aa 260 Chapter 5 / Transient and Steady-State Response Analyses A–5–24.
In the system considered in Problem A–5–23, it is desired to eliminate as much as possible the speed errors due to torque disturbances.
Is it possible to cancel the effect of a disturbance torque at steady state so that a constant disturbance torque applied to the output member will cause no speed change at steady state?
Solution. Suppose that we choose a suitable controller whose transfer function is Gc(s), as shown in Figure 5–69. Then in the absence of the reference input the closed-loop transfer function between the output velocity and the disturbance torque D(s) is The steady-state output speed due to a unit-step disturbance torque is To satisfy the requirement that we must choose Gc(0)=q.This can be realized if we choose Integral control action will continue to correct until the error is zero. This controller, however, presents a stability problem, because the characteristic equation will have two imaginary roots.
One method of stabilizing such a system is to add a proportional mode to the controller or choose Gc(s) = K p + K s Gc(s) = K s vD(q) = 0 = 1 Gc(0) = lim sS0 s Js + Gc(s) 1 s vD(q) = lim sS0sVD(s) = 1 Js + Gc(s) VD(s) D(s) = 1 Js 1 + 1 Js Gc(s) VD(s) + – ++ D(s) E(s) T(s) V(s) Vr(s) Gc(s) 1 Js Figure 5–69 Block diagram of a speed control system.
aa Example Problems and Solutions 261 With this controller, the block diagram of Figure 5–69 in the absence of the reference input can be modified to that of Figure 5–70.The closed-loop transfer function becomes For a unit-step disturbance torque, the steady-state output speed is Thus, we see that the proportional-plus-integral controller eliminates speed error at steady state.
The use of integral control action has increased the order of the system by 1. (This tends to produce an oscillatory response.) In the present system, a step disturbance torque will cause a transient error in the output speed, but the error will become zero at steady state. The integrator provides a nonzero output with zero error. (The nonzero output of the integrator produces a motor torque that exactly cancels the disturbance torque.) Note that even if the system may have an integrator in the plant (such as an integrator in the transfer function of the plant), this does not eliminate the steady-state error due to a step distur-bance torque.To eliminate this,we must have an integrator before the point where the disturbance torque enters.
A–5–25.
Consider the system shown in Figure 5–71(a). The steady-state error to a unit-ramp input is ess=2zvn. Show that the steady-state error for following a ramp input may be eliminated if the input is introduced to the system through a proportional-plus-derivative filter, as shown in Figure 5–71(b), and the value of k is properly set. Note that the error e(t) is given by r(t)-c(t).
Solution. The closed-loop transfer function of the system shown in Figure 5–71(b) is Then R(s) - C(s) = a s2 + 2zvn s - v2 n ks s2 + 2zvn s + v2 n bR(s) C(s) R(s) = (1 + ks)v2 n s2 + 2zvn s + v2 n vD(q) = lim sS0sVD(s) = lim sS0 s2 Js2 + K p s + K 1 s = 0 VD(s) D(s) = s Js2 + K p s + K VD(s)D(s) 1 Js Kps + K s VD(s) D(s) + – Figure 5–70 Block diagram of the speed control system of Figure 5–69 when Gc(s)=Kp+(K/s) and Vr(s) = 0.
+ – + – R(s) C(s) (a) (b) 1 + ks vn s(s + 2zvn) 2 vn s(s + 2zvn) 2 Figure 5–71 (a) Control system; (b) control system with input filter.
aa 262 Chapter 5 / Transient and Steady-State Response Analyses If the input is a unit ramp, then the steady-state error is Therefore, if k is chosen as then the steady-state error for following a ramp input can be made equal to zero.Note that,if there are any variations in the values of z and/or vn due to environmental changes or aging, then a nonzero steady-state error for a ramp response may result.
A–5–26.
Consider the stable unity-feedback control system with feedforward transfer function G(s).
Suppose that the closed-loop transfer function can be written Show that where e(t) = r(t)-c(t) is the error in the unit-step response. Show also that Solution. Let us define and Then and For a unit-step input, R(s)=1/s and E(s) = Q(s) - P(s) sQ(s) E(s) = Q(s) - P(s) Q(s) R(s) C(s) R(s) = P(s) Q(s) AT 1 s + 1BAT 2 s + 1B p AT n s + 1B = Q(s) AT a s + 1BAT b s + 1B p AT m s + 1B = P(s) 1 K v = 1 lim sS0sG(s) = AT 1 + T 2 + p + T nB - AT a + T b + p + T mB 3 q 0 e(t) dt = AT 1 + T 2 + p + T nB - AT a + T b + p + T mB C(s) R(s) = G(s) 1 + G(s) = AT a s + 1BAT b s + 1B p AT m s + 1B AT 1 s + 1BAT 2 s + 1B p AT n s + 1B (m n) k = 2z vn = 2zvn - v2 n k v2 n = lim sS0s a s2 + 2zvn s - v2 n ks s2 + 2zvn s + v2 n b 1 s2 e(q) = r(q) - c(q) aa Problems 263 Since the system is stable, converges to a constant value. Noting that we have Since we have For a unit-step input r(t), since we have Note that zeros in the left half-plane (that is, positive ) will improve Kv. Poles close to the origin cause low velocity-error constants unless there are zeros nearby.
PROBLEMS T a , T b , p , T m 1 K v = 1 lim sS0sG(s) = AT 1 + T 2 + p + T nB - AT a + T b + p + T mB = lim sS0 1 1 + G(s) 1 s = 1 lim sS0sG(s) = 1 K v 3 q 0 e(t) dt = lim sS0E(s) = lim sS0 1 1 + G(s) R(s) 3 q 0 e(t) dt = AT 1 + T 2 + p + T nB - AT a + T b + p + T mB lim sS0Q¿(s) = T 1 + T 2 + p + T n lim sS0P¿(s) = T a + T b + p + T m = lim sS0 CQ¿(s) - P¿(s)D = lim sS0 Q¿(s) - P¿(s) Q(s) + sQ¿(s) 3 q 0 e(t)dt = lim sS0 Q(s) - P(s) sQ(s) 3 q 0 e(t)dt = lim sS0s E(s) s = lim sS0E(s) 1 q 0 e(t)dt B–5–1. A thermometer requires 1 min to indicate 98% of the response to a step input.Assuming the thermometer to be a first-order system, find the time constant.
If the thermometer is placed in a bath, the temperature of which is changing linearly at a rate of 10°min, how much error does the thermometer show?
B–5–2. Consider the unit-step response of a unity-feedback control system whose open-loop transfer function is G(s) = 1 s(s + 1) Obtain the rise time, peak time, maximum overshoot, and settling time.
B–5–3. Consider the closed-loop system given by Determine the values of z and vn so that the system responds to a step input with approximately 5% overshoot and with a settling time of 2 sec. (Use the 2% criterion.) C(s) R(s) = v2 n s2 + 2zvn s + v2 n aa 264 Chapter 5 / Transient and Steady-State Response Analyses x m k Impulsive force d(t) Figure 5–72 Mechanical system.
x1 T t t1 xn tn Figure 5–73 Decaying oscillation.
+ – + – R(s) C(s) R(s) C(s) (a) (b) 10 s(s + 1) Kh 10 s + 1 1 s + – Figure 5–74 (a) Control system; (b) control system with tachometer feedback.
B–5–4. Consider the system shown in Figure 5–72.The sys-tem is initially at rest. Suppose that the cart is set into mo-tion by an impulsive force whose strength is unity. Can it be stopped by another such impulsive force?
B–5–5. Obtain the unit-impulse response and the unit-step response of a unity-feedback system whose open-loop transfer function is B–5–6. An oscillatory system is known to have a transfer function of the following form: G(s) = v2 n s2 + 2zvn s + v2 n G(s) = 2s + 1 s2 B–5–7. Consider the system shown in Figure 5–74(a). The damping ratio of this system is 0.158 and the undamped nat-ural frequency is 3.16 radsec. To improve the relative sta-bility,we employ tachometer feedback.Figure 5–74(b) shows such a tachometer-feedback system.
Determine the value of Kh so that the damping ratio of the system is 0.5.Draw unit-step response curves of both the original and tachometer-feedback systems. Also draw the error-versus-time curves for the unit-ramp response of both systems.
Assume that a record of a damped oscillation is available as shown in Figure 5–73. Determine the damping ratio z of the system from the graph.
aa Problems 265 + – + – R(s) C(s) 1 s 16 s + 0.8 k R(s) C(s) 1 s K s + 2 k + – + – Figure 5–75 Closed-loop system.
Figure 5–76 Block diagram of a system.
B–5–8. Referring to the system shown in Figure 5–75, de-termine the values of K and k such that the system has a damping ratio z of 0.7 and an undamped natural frequency vn of 4 radsec.
B–5–9. Consider the system shown in Figure 5–76. Deter-mine the value of k such that the damping ratio z is 0.5.Then obtain the rise time tr, peak time tp, maximum overshoot Mp, and settling time ts in the unit-step response.
B–5–10. Using MATLAB, obtain the unit-step response, unit-ramp response, and unit-impulse response of the fol-lowing system: where R(s) and C(s) are Laplace transforms of the input r(t) and output c(t), respectively.
C(s) R(s) = 10 s2 + 2s + 10 B–5–11. Using MATLAB, obtain the unit-step response, unit-ramp response, and unit-impulse response of the fol-lowing system: where u is the input and y is the output.
B–5–12. Obtain both analytically and computationally the rise time, peak time, maximum overshoot, and settling time in the unit-step response of a closed-loop system given by C(s) R(s) = 36 s2 + 2s + 36 y = [1 0]B x1 x2 R B x # 1 x # 2 R = B -1 -0.5 1 0 R Bx1 x2 R + B 0.5 0 R u aa 266 Chapter 5 / Transient and Steady-State Response Analyses B–5–13. Figure 5–77 shows three systems. System I is a po-sitional servo system. System II is a positional servo system with PD control action. System III is a positional servo sys-tem with velocity feedback. Compare the unit-step, unit-impulse, and unit-ramp responses of the three systems.
Which system is best with respect to the speed of response and maximum overshoot in the step response?
B–5–14. Consider the position control system shown in Fig-ure 5–78. Write a MATLAB program to obtain a unit-step response and a unit-ramp response of the system.Plot curves x1(t) versus t, x2(t) versus t, x3(t) versus t, and e(t) versus t Cwhere e(t)=r(t)-x1(t)D for both the unit-step response and the unit-ramp response.
0.8 5 5 CIII(s) R(s) System III 1 5s + 1 1 s CII(s) R(s) System II 5(1 + 0.8s) 1 s(5s + 1) C(s) R(s) System I 1 s(5s + 1) + – + – + – + – Figure 5–77 Positional servo system (System I), positional servo system with PD control action (System II), and positional servo system with velocity feedback (System III).
5 4 x1 x2 x3 r e 1 s 1 s 2 0.1s + 1 + – + – Figure 5–78 Position control system.
Problems 267 aa B–5–15. Using MATLAB, obtain the unit-step response curve for the unity-feedback control system whose open-loop transfer function is Using MATLAB, obtain also the rise time, peak time, max-imum overshoot, and settling time in the unit-step response curve.
B–5–16. Consider the closed-loop system defined by where z=0.2, 0.4, 0.6, 0.8, and 1.0. Using MATLAB, plot a two-dimensional diagram of unit-impulse response curves.
Also plot a three-dimensional plot of the response curves.
B–5–17. Consider the second-order system defined by where z=0.2, 0.4, 0.6, 0.8, 1.0. Plot a three-dimensional diagram of the unit-step response curves.
B–5–18. Obtain the unit-ramp response of the system defined by where u is the unit-ramp input. Use the lsim command to obtain the response. y = [1 0]B x1 x2 R Bx # 1 x # 2 R = B 0 - 1 1 - 1R Bx1 x2 R + B 0 1R u C(s) R(s) = s + 1 s2 + 2zs + 1 C(s) R(s) = 2zs + 1 s2 + 2zs + 1 G(s) = 10 s(s + 2)(s + 4) B–5–19. Consider the differential equation system given by Using MATLAB, obtain the response y(t), subject to the given initial condition.
B–5–20. Determine the range of K for stability of a unity-feedback control system whose open-loop transfer function is B–5–21. Consider the following characteristic equation: Using the Routh stability criterion, determine the range of K for stability.
B–5–22. Consider the closed-loop system shown in Figure 5–79.
Determine the range of K for stability.Assume that K>0.
s4 + 2s3 + (4 + K)s2 + 9s + 25 = 0 G(s) = K s(s + 1)(s + 2) y $ + 3y # + 2y = 0, y(0) = 0.1, y # (0) = 0.05 + – R(s) C(s) K s – 2 (s + 1)(s2 + 6s + 25) Figure 5–79 Closed-loop system.
B–5–23. Consider the satellite attitude control system shown in Figure 5–80(a).The output of this system exhibits continued oscillations and is not desirable. This system can be stabilized by use of tachometer feedback, as shown in Figure 5–80(b). If K/J=4, what value of Kh will yield the damping ratio to be 0.6?
+ – + – R(s) C(s) (b) Kh K Js 1 s + – R(s) C(s) (a) K 1 Js2 Figure 5–80 (a) Unstable satellite attitude control system; (b) stabilized system.
aa 268 Chapter 5 / Transient and Steady-State Response Analyses Kh K C(s) R(s) 20 (s + 1) (s + 4) 1 s + – + – Figure 5–81 Servo system with tachometer feedback.
B–5–24. Consider the servo system with tachometer feedback shown in Figure 5–81. Determine the ranges of stability for K and Kh. (Note that Kh must be positive.) B–5–25. Consider the system where matrix A is given by (A is called Schwarz matrix.) Show that the first column of the Routh’s array of the characteristic equation |sI-A|=0 consists of 1, b1, b2, and b1b3.
B–5–26. Consider a unity-feedback control system with the closed-loop transfer function Determine the open-loop transfer function G(s).
Show that the steady-state error in the unit-ramp response is given by ess = 1 K v = a - K b C(s) R(s) = Ks + b s2 + as + b A = C 0 - b3 0 1 0 - b2 0 1 - b1 S x # = Ax B–5–27. Consider a unity-feedback control system whose open-loop transfer function is Discuss the effects that varying the values of K and B has on the steady-state error in unit-ramp response. Sketch typical unit-ramp response curves for a small value, medium value, and large value of K, assuming that B is constant.
B–5–28. If the feedforward path of a control system contains at least one integrating element, then the output continues to change as long as an error is present.The out-put stops when the error is precisely zero. If an external dis-turbance enters the system, it is desirable to have an integrating element between the error-measuring element and the point where the disturbance enters, so that the ef-fect of the external disturbance may be made zero at steady state.
Show that, if the disturbance is a ramp function, then the steady-state error due to this ramp disturbance may be eliminated only if two integrators precede the point where the disturbance enters.
G(s) = K s(Js + B) 6 269 Control Systems Analysis and Design by the Root-Locus Method 6–1 INTRODUCTION The basic characteristic of the transient response of a closed-loop system is closely related to the location of the closed-loop poles. If the system has a variable loop gain, then the location of the closed-loop poles depends on the value of the loop gain chosen.
It is important, therefore, that the designer know how the closed-loop poles move in the s plane as the loop gain is varied.
From the design viewpoint, in some systems simple gain adjustment may move the closed-loop poles to desired locations.Then the design problem may become the selec-tion of an appropriate gain value. If the gain adjustment alone does not yield a desired result, addition of a compensator to the system will become necessary. (This subject is discussed in detail in Sections 6–6 through 6–9.) The closed-loop poles are the roots of the characteristic equation. Finding the roots of the characteristic equation of degree higher than 3 is laborious and will need computer solution. (MATLAB provides a simple solution to this problem.) However, just finding the roots of the characteristic equation may be of limited value, because as the gain of the open-loop transfer function varies, the characteristic equation changes and the computations must be repeated.
A simple method for finding the roots of the characteristic equation has been developed by W. R. Evans and used extensively in control engineering. This method, called the root-locus method, is one in which the roots of the characteristic equation 270 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method H(s) G(s) C(s) R(s) + – Figure 6–1 Control system.
are plotted for all values of a system parameter. The roots corresponding to a par-ticular value of this parameter can then be located on the resulting graph. Note that the parameter is usually the gain, but any other variable of the open-loop transfer function may be used. Unless otherwise stated, we shall assume that the gain of the open-loop transfer function is the parameter to be varied through all values, from zero to infinity.
By using the root-locus method the designer can predict the effects on the location of the closed-loop poles of varying the gain value or adding open-loop poles and/or open-loop zeros.Therefore, it is desired that the designer have a good understanding of the method for generating the root loci of the closed-loop system, both by hand and by use of a computer software program like MATLAB.
In designing a linear control system, we find that the root-locus method proves to be quite useful, since it indicates the manner in which the open-loop poles and zeros should be modified so that the response meets system performance specifications.This method is particularly suited to obtaining approximate results very quickly.
Because generating the root loci by use of MATLAB is very simple, one may think sketching the root loci by hand is a waste of time and effort. However, experience in sketching the root loci by hand is invaluable for interpreting computer-generated root loci, as well as for getting a rough idea of the root loci very quickly.
Outline of the Chapter.
The outline of the chapter is as follows: Section 6–1 has presented an introduction to the root-locus method. Section 6–2 details the concepts underlying the root-locus method and presents the general procedure for sketching root loci using illustrative examples. Section 6–3 discusses generating root-locus plots with MATLAB. Section 6–4 treats a special case when the closed-loop system has positive feedback. Section 6–5 presents general aspects of the root-locus approach to the design of closed-loop systems. Section 6–6 discusses the control systems design by lead com-pensation. Section 6–7 treats the lag compensation technique. Section 6–8 deals with the lag–lead compensation technique. Finally, Section 6–9 discusses the parallel com-pensation technique.
6–2 ROOT-LOCUS PLOTS Angle and Magnitude Conditions.
Consider the negative feedback system shown in Figure 6–1.The closed-loop transfer function is (6–1) C(s) R(s) = G(s) 1 + G(s)H(s) Section 6–2 / Root-Locus Plots 271 The characteristic equation for this closed-loop system is obtained by setting the denominator of the right-hand side of Equation (6–1) equal to zero.That is, or (6–2) Here we assume that G(s)H(s) is a ratio of polynomials in s. [It is noted that we can extend the analysis to the case when G(s)H(s) involves the transport lag e–Ts.] Since G(s)H(s) is a complex quantity, Equation (6–2) can be split into two equations by equating the angles and magnitudes of both sides, respectively, to obtain the following: Angle condition: (6–3) Magnitude condition: (6–4) The values of s that fulfill both the angle and magnitude conditions are the roots of the characteristic equation, or the closed-loop poles. A locus of the points in the complex plane satisfying the angle condition alone is the root locus. The roots of the characteristic equation (the closed-loop poles) corresponding to a given value of the gain can be determined from the magnitude condition. The details of applying the angle and magnitude conditions to obtain the closed-loop poles are presented later in this section.
In many cases, G(s)H(s) involves a gain parameter K, and the characteristic equa-tion may be written as Then the root loci for the system are the loci of the closed-loop poles as the gain K is varied from zero to infinity.
Note that to begin sketching the root loci of a system by the root-locus method we must know the location of the poles and zeros of G(s)H(s). Remember that the angles of the complex quantities originating from the open-loop poles and open-loop zeros to the test point s are measured in the counterclockwise direction.For example,if G(s)H(s) is given by G(s)H(s) = KAs + z1B As + p1BAs + p2BAs + p3BAs + p4B 1 + KAs + z1BAs + z2B p As + zmB As + p1BAs + p2B p As + pnB = 0 ∑G(s)H(s)∑= 1 /G(s)H(s) = ;180°(2k + 1) (k = 0, 1, 2, p ) G(s)H(s) = -1 1 + G(s)H(s) = 0 272 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method Test point Test point –p4 –p3 –p2 –p1 s s –z1 f1 f1 jv u4 u2 u3 u4 u1 u3 u1 u2 0 –p4 –p2 A4 B1 A3 A2 A1 –p1 –p3 –z1 jv 0 (b) (a) where –p2 and –p3 are complex-conjugate poles, then the angle of G(s)H(s) is where f1, u1, u2, u3, and u4 are measured counterclockwise as shown in Figures 6–2(a) and (b).The magnitude of G(s)H(s) for this system is where A1, A2, A3, A4, and B1 are the magnitudes of the complex quantities s+p1, s+p2, s+p3, s+p4, and s+z1, respectively, as shown in Figure 6–2(a).
Note that, because the open-loop complex-conjugate poles and complex-conjugate zeros,if any,are always located symmetrically about the real axis,the root loci are always symmetrical with respect to this axis.Therefore,we only need to construct the upper half of the root loci and draw the mirror image of the upper half in the lower-half s plane.
Illustrative Examples.
In what follows, two illustrative examples for constructing root-locus plots will be presented. Although computer approaches to the construction of the root loci are easily available, here we shall use graphical computation, combined with inspection, to determine the root loci upon which the roots of the characteristic equation of the closed-loop system must lie. Such a graphical approach will enhance understanding of how the closed-loop poles move in the complex plane as the open-loop poles and zeros are moved.Although we employ only simple systems for illustrative purposes, the procedure for finding the root loci is no more complicated for higher-order systems.
Because graphical measurements of angles and magnitudes are involved in the analy-sis, we find it necessary to use the same divisions on the abscissa as on the ordinate axis when sketching the root locus on graph paper.
∑G(s)H(s)∑= KB1 A1 A2 A3 A4 /G(s)H(s) = f1 - u1 - u2 - u3 - u4 Figure 6–2 (a) and (b) Diagrams showing angle measurements from open-loop poles and open-loop zero to test point s.
Section 6–2 / Root-Locus Plots 273 R(s) C(s) K s(s + 1) (s + 2) + – Figure 6–3 Control system.
EXAMPLE 6–1 Consider the negative feedback system shown in Figure 6–3. (We assume that the value of gain K is nonnegative.) For this system, Let us sketch the root-locus plot and then determine the value of K such that the damping ratio z of a pair of dominant complex-conjugate closed-loop poles is 0.5.
For the given system, the angle condition becomes The magnitude condition is A typical procedure for sketching the root-locus plot is as follows: 1. Determine the root loci on the real axis. The first step in constructing a root-locus plot is to locate the open-loop poles, s=0, s=–1, and s=–2, in the complex plane. (There are no open-loop zeros in this system.) The locations of the open-loop poles are indicated by crosses. (The lo-cations of the open-loop zeros in this book will be indicated by small circles.) Note that the starting points of the root loci (the points corresponding to K=0) are open-loop poles. The number of individual root loci for this system is three, which is the same as the number of open-loop poles.
To determine the root loci on the real axis, we select a test point, s. If the test point is on the positive real axis, then This shows that the angle condition cannot be satisfied.Hence,there is no root locus on the positive real axis. Next, select a test point on the negative real axis between 0 and –1. Then Thus and the angle condition is satisfied.Therefore, the portion of the negative real axis between 0 and –1 forms a portion of the root locus. If a test point is selected between –1 and –2, then and - /s - /s + 1 - /s + 2 = -360° /s = /s + 1 = 180°, /s + 2 = 0° - /s - /s + 1 - /s + 2 = -180° /s = 180°, /s + 1 = /s + 2 = 0° /s = /s + 1 = /s + 2 = 0° ∑G(s)∑= 2 K s(s + 1)(s + 2) 2 = 1 = ;180°(2k + 1) (k = 0, 1, 2, p ) = - /s - /s + 1 - /s + 2 /G(s) = n K s(s + 1)(s + 2) G(s) = K s(s + 1)(s + 2) , H(s) = 1 274 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method It can be seen that the angle condition is not satisfied. Therefore, the negative real axis from –1 to –2 is not a part of the root locus. Similarly, if a test point is located on the negative real axis from –2 to –q, the angle condition is satisfied. Thus, root loci exist on the negative real axis between 0 and –1 and between –2 and –q.
2. Determine the asymptotes of the root loci. The asymptotes of the root loci as s approaches infinity can be determined as follows: If a test point s is selected very far from the origin, then and the angle condition becomes or Since the angle repeats itself as k is varied, the distinct angles for the asymptotes are determined as 60°, –60°, and 180°. Thus, there are three asymptotes. The one having the angle of 180° is the negative real axis.
Before we can draw these asymptotes in the complex plane, we must find the point where they intersect the real axis. Since if a test point is located very far from the origin, then G(s) may be written as For large values of s, this last equation may be approximated by (6–5) A root-locus diagram of G(s) given by Equation (6–5) consists of three straight lines.This can be seen as follows:The equation of the root locus is or which can be written as /s + 1 = ;60°(2k + 1) -3/s + 1 = ;180°(2k + 1) n K (s + 1)3 = ;180°(2k + 1) G(s) K (s + 1)3 G(s) = K s3 + 3s2 + p G(s) = K s(s + 1)(s + 2) Angles of asymptotes = ;180°(2k + 1) 3 (k = 0, 1, 2, p ) -3/s = ;180°(2k + 1) (k = 0, 1, 2, p ) lim sS q G(s) = lim sS q K s(s + 1)(s + 2) = lim sS q K s3 Section 6–2 / Root-Locus Plots 275 jv s v = 0 –1 – j 3 j 3 s + 1 – 3 v = 0 s + 1 + 3 v = 0 0 Figure 6–4 Three asymptotes.
By substituting s=s+jv into this last equation, we obtain or Taking the tangent of both sides of this last equation, which can be written as These three equations represent three straight lines,as shown in Figure 6–4.The three straight lines shown are the asymptotes. They meet at point s=–1. Thus, the abscissa of the intersection of the asymptotes and the real axis is obtained by setting the denominator of the right-hand side of Equation (6–5) equal to zero and solving for s. The asymptotes are almost parts of the root loci in regions very far from the origin.
3. Determine the breakaway point. To plot root loci accurately, we must find the breakaway point, where the root-locus branches originating from the poles at 0 and –1 break away (as K is increased) from the real axis and move into the complex plane.The breakaway point corresponds to a point in the s plane where multiple roots of the characteristic equation occur.
A simple method for finding the breakaway point is available. We shall present this method in the following: Let us write the characteristic equation as (6–6) f(s) = B(s) + KA(s) = 0 s + 1 -v 13 = 0, s + 1 + v 13 = 0, v = 0 v s + 1 = 13, -13 , 0 tan-1 v s + 1 = 60°, -60°, 0° /s + jv + 1 = ;60°(2k + 1) 276 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method where A(s) and B(s) do not contain K. Note that f(s)=0 has multiple roots at points where This can be seen as follows:Suppose that f(s) has multiple roots of order r,where .Then f(s) may be written as Now we differentiate this equation with respect to s and evaluate df(s)/ds at s=s1. Then we get (6–7) This means that multiple roots of f(s) will satisfy Equation (6–7). From Equation (6–6), we obtain (6–8) where The particular value of K that will yield multiple roots of the characteristic equation is obtained from Equation (6–8) as If we substitute this value of K into Equation (6–6), we get or (6–9) If Equation (6–9) is solved for s, the points where multiple roots occur can be obtained. On the other hand, from Equation (6–6) we obtain and If dK/ds is set equal to zero, we get the same equation as Equation (6–9). Therefore, the break-away points can be simply determined from the roots of It should be noted that not all the solutions of Equation (6–9) or of dK/ds=0 correspond to actual breakaway points. If a point at which dK/ds=0 is on a root locus, it is an actual breakaway or break-in point. Stated differently, if at a point at which dK/ds=0 the value of K takes a real positive value, then that point is an actual breakaway or break-in point.
dK ds = 0 dK ds = -B¿(s)A(s) - B(s)A¿(s) A 2(s) K = -B(s) A(s) B(s)A¿(s) - B¿(s)A(s) = 0 f(s) = B(s) -B¿(s) A¿(s) A(s) = 0 K = -B¿(s) A¿(s) A¿(s) = dA(s) ds , B¿(s) = dB(s) ds df(s) ds = B¿(s) + KA¿(s) = 0 df(s) ds 2 s=s1 = 0 f(s) = As - s1B rAs - s2B p As - snB r 2 df(s) ds = 0 Section 6–2 / Root-Locus Plots 277 For the present example, the characteristic equation G(s)+1=0 is given by or By setting dK/ds=0, we obtain or Since the breakaway point must lie on a root locus between 0 and –1, it is clear that s=–0.4226 corresponds to the actual breakaway point. Point s=–1.5774 is not on the root locus. Hence, this point is not an actual breakaway or break-in point. In fact, evaluation of the values of K corre-sponding to s=–0.4226 and s=–1.5774 yields 4. Determine the points where the root loci cross the imaginary axis. These points can be found by use of Routh’s stability criterion as follows: Since the characteristic equation for the present system is the Routh array becomes The value of K that makes the s1 term in the first column equal zero is K=6. The crossing points on the imaginary axis can then be found by solving the auxiliary equation obtained from the s2 row; that is, which yields The frequencies at the crossing points on the imaginary axis are thus The gain value corresponding to the crossing points is K=6.
An alternative approach is to let s=jv in the characteristic equation, equate both the real part and the imaginary part to zero, and then solve for v and K. For the present system, the char-acteristic equation, with s=jv, is or Equating both the real and imaginary parts of this last equation to zero, respectively, we obtain K - 3v2 = 0, 2v - v3 = 0 AK - 3v2B + jA2v - v3B = 0 (jv)3 + 3(jv)2 + 2(jv) + K = 0 v = ;12 .
s = ;j12 3s2 + K = 3s2 + 6 = 0 s3 s2 s1 s0 1 3 6 - K 3 K 2 K s3 + 3s2 + 2s + K = 0 K = -0.3849, for s = -1.5774 K = 0.3849, for s = -0.4226 s = -0.4226, s = -1.5774 dK ds = -A3s2 + 6s + 2B = 0 K = -As3 + 3s2 + 2sB K s(s + 1)(s + 2) + 1 = 0 278 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method jv j1 – j1 –1 –2 s + 1 s + 2 u2 u1 u3 0 s s Figure 6–5 Construction of root locus.
jv j1 – j1 1 –2 –3 0 s K = 6 K = 6 K = 1.0383 K = 1.0383 K K j2 – j2 60° –1 Figure 6–6 Root-locus plot.
from which Thus, root loci cross the imaginary axis at and the value of K at the crossing points is 6.
Also, a root-locus branch on the real axis touches the imaginary axis at v=0.The value of K is zero at this point.
5. Choose a test point in the broad neighborhood of the jv axis and the origin, as shown in Figure 6–5, and apply the angle condition. If a test point is on the root loci, then the sum of the three angles, u1+u2+u3, must be 180°. If the test point does not satisfy the angle condition, select another test point until it satisfies the condition. (The sum of the angles at the test point will indicate the direction in which the test point should be moved.) Continue this process and locate a sufficient number of points satisfying the angle condition.
6. Draw the root loci, based on the information obtained in the foregoing steps, as shown in Figure 6–6.
v = ;12 , v = ;12, K = 6 or v = 0, K = 0 Section 6–2 / Root-Locus Plots 279 7. Determine a pair of dominant complex-conjugate closed-loop poles such that the damping ratio z is 0.5. Closed-loop poles with z=0.5 lie on lines passing through the origin and making the angles with the negative real axis.From Figure 6–6,such closed-loop poles having z=0.5 are obtained as follows: The value of K that yields such poles is found from the magnitude condition as follows: Using this value of K, the third pole is found at s=–2.3326.
Note that, from step 4, it can be seen that for K=6 the dominant closed-loop poles lie on the imaginary axis at With this value of K, the system will exhibit sustained oscillations.
For K>6, the dominant closed-loop poles lie in the right-half s plane, resulting in an unstable system.
Finally, note that, if necessary, the root loci can be easily graduated in terms of K by use of the magnitude condition.We simply pick out a point on a root locus, measure the magnitudes of the three complex quantities s, s+1, and s+2, and multiply these magnitudes; the product is equal to the gain value K at that point, or Graduation of the root loci can be done easily by use of MATLAB. (See Section 6–3.) EXAMPLE 6–2 In this example, we shall sketch the root-locus plot of a system with complex-conjugate open-loop poles. Consider the negative feedback system shown in Figure 6–7. For this system, where K 0. It is seen that G(s) has a pair of complex-conjugate poles at A typical procedure for sketching the root-locus plot is as follows: 1. Determine the root loci on the real axis. For any test point s on the real axis, the sum of the angular contributions of the complex-conjugate poles is 360°, as shown in Figure 6–8.Thus the net effect of the complex-conjugate poles is zero on the real axis.The location of the root locus on the real axis is determined from the open-loop zero on the negative real axis.A simple test reveals that a section of the negative real axis, that between –2 and –q, is a part of the root locus. It is noted that, since this locus lies between two zeros (at s=–2 and s=–q), it is actually a part of two root loci, each of which starts from one of the two complex-conjugate poles. In other words, two root loci break in the part of the negative real axis between –2 and –q.
s = -1 + j12, s = -1 - j12 G(s) = K(s + 2) s2 + 2s + 3 , H(s) = 1 ∑s∑ ∑s + 1∑ ∑s + 2∑= K s = ;j12 . = 1.0383 K = ∑s(s + 1)(s + 2)∑s=-0.3337+j0.5780 s1 = -0.3337 + j0.5780, s2 = -0.3337 - j0.5780 ;cos-1z = ;cos-10.5 = ;60° R(s) C(s) K(s + 2) s2 + 2s + 3 + – Figure 6–7 Control system.
280 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method jv 0 s u9 2 s –p2 f1 f91 –z1 –p1 u2 u1 Figure 6–9 Determination of the angle of departure.
Since there are two open-loop poles and one zero,there is one asymptote,which coincides with the negative real axis.
2. Determine the angle of departure from the complex-conjugate open-loop poles. The pres-ence of a pair of complex-conjugate open-loop poles requires the determination of the angle of departure from these poles. Knowledge of this angle is important, since the root locus near a com-plex pole yields information as to whether the locus originating from the complex pole migrates toward the real axis or extends toward the asymptote.
Referring to Figure 6–9, if we choose a test point and move it in the very vicinity of the com-plex open-loop pole at s=–p1, we find that the sum of the angular contributions from the pole at s=p2 and zero at s=–z1 to the test point can be considered remaining the same. If the test point is to be on the root locus, then the sum of –u1, and must be where k=0, 1, 2, p .Thus, in the example, or The angle of departure is then u1 = 180° - u2 + f1 = 180° - 90° + 55° = 145° u1 = 180° - uœ 2 + fœ 1 = 180° - u2 + f1 fœ 1 - Au1 + uœ 2B = ;180°(2k + 1) ;180°(2k + 1), -uœ 2 fœ 1 , jv –1 0 –2 s j 2 – j 2 Test point u2 u1 Figure 6–8 Determination of the root locus on the real axis.
Section 6–2 / Root-Locus Plots 281 jv j1 – j1 1 –2 –3 –4 0 s z = 0.7 line j2 – j2 145° –1 Figure 6–10 Root-locus plot.
Since the root locus is symmetric about the real axis, the angle of departure from the pole at s=–p2 is –145°.
3. Determine the break-in point. A break-in point exists where a pair of root-locus branches coalesces as K is increased. For this problem, the break-in point can be found as follows: Since we have which gives or Notice that point s=–3.7320 is on the root locus. Hence this point is an actual break-in point.
(Note that at point s=–3.7320 the corresponding gain value is K=5.4641.) Since point s=–0.2680 is not on the root locus, it cannot be a break-in point. (For point s=–0.2680, the cor-responding gain value is K=–1.4641.) 4. Sketch a root-locus plot, based on the information obtained in the foregoing steps. To determine accurate root loci, several points must be found by trial and error between the break-in point and the complex open-loop poles. (To facilitate sketching the root-locus plot, we should find the direction in which the test point should be moved by mentally summing up the changes on the angles of the poles and zeros.) Figure 6–10 shows a complete root-locus plot for the system considered.
s = -3.7320 or s = -0.2680 s2 + 4s + 1 = 0 dK ds = -(2s + 2)(s + 2) - As2 + 2s + 3B (s + 2)2 = 0 K = - s2 + 2s + 3 s + 2 282 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method The value of the gain K at any point on root locus can be found by applying the magnitude condition or by use of MATLAB (see Section 6–3). For example, the value of K at which the complex-conjugate closed-loop poles have the damping ratio z=0.7 can be found by locating the roots, as shown in Figure 6–10, and computing the value of K as follows: Or use MATLAB to find the value of K. (See Section 6–4.) It is noted that in this system the root locus in the complex plane is a part of a circle. Such a circular root locus will not occur in most systems. Circular root loci may occur in systems that in-volve two poles and one zero, two poles and two zeros, or one pole and two zeros. Even in such systems, whether circular root loci occur depends on the locations of poles and zeros involved.
To show the occurrence of a circular root locus in the present system, we need to derive the equation for the root locus. For the present system, the angle condition is If s=s+jv is substituted into this last equation, we obtain which can be written as or Taking tangents of both sides of this last equation using the relationship (6–10) we obtain or which can be simplified to or This last equation is equivalent to or (s + 2)2 + v2 = A 13B 2 v = 0 vC(s + 2)2 + v2 - 3D = 0 2v(s + 1) (s + 1)2 - Av2 - 2B = v s + 2 v - 12 s + 1 + v + 12 s + 1 1 - a v - 12 s + 1 b a v + 12 s + 1 b = v s + 2 ; 0 1 < v s + 2 0 tan ctan-1 a v - 12 s + 1 b + tan-1 a v + 12 s + 1 b d = tan ctan-1 a v s + 2 b ; 180°(2k + 1)d tan (x ; y) = tanx ; tan y 1 < tanx tany tan-1 a v - 12 s + 1 b + tan-1 a v + 12 s + 1 b = tan-1 a v s + 2 b ; 180°(2k + 1) tan-1 a v s + 2 b - tan-1 a v - 12 s + 1 b - tan-1 a v + 12 s + 1 b = ;180°(2k + 1) /s + 2 + jv - /s + 1 + jv - j12 - /s + 1 + jv + j12 = ;180°(2k + 1) /s + 2 - /s + 1 - j12 - /s + 1 + j12 = ;180°(2k + 1) K = 2 As + 1 - j12BAs + 1 + j12B s + 2 2 s=-1.67+j1.70 = 1.34 Section 6–2 / Root-Locus Plots 283 These two equations are the equations for the root loci for the present system. Notice that the first equation, v=0, is the equation for the real axis. The real axis from s=–2 to s=–q corre-sponds to a root locus for K 0.The remaining part of the real axis corresponds to a root locus when K is negative. (In the present system, K is nonnegative.) (Note that K < 0 corresponds to the positive-feedback case.) The second equation for the root locus is an equation of a circle with center at s=–2, v=0 and the radius equal to That part of the circle to the left of the complex-conjugate poles corresponds to a root locus for K 0. The remaining part of the circle corresponds to a root locus when K is negative.
It is important to note that easily interpretable equations for the root locus can be derived for simple systems only. For complicated systems having many poles and zeros, any attempt to derive equations for the root loci is discouraged. Such derived equations are very complicated and their configuration in the complex plane is difficult to visualize.
General Rules for Constructing Root Loci.
For a complicated system with many open-loop poles and zeros, constructing a root-locus plot may seem complicated, but actually it is not difficult if the rules for constructing the root loci are applied. By locat-ing particular points and asymptotes and by computing angles of departure from com-plex poles and angles of arrival at complex zeros, we can construct the general form of the root loci without difficulty.
We shall now summarize the general rules and procedure for constructing the root loci of the negative feedback control system shown in Figure 6–11.
First, obtain the characteristic equation Then rearrange this equation so that the parameter of interest appears as the multiply-ing factor in the form (6–11) In the present discussions, we assume that the parameter of interest is the gain K, where K>0. (If K<0, which corresponds to the positive-feedback case, the angle condi-tion must be modified. See Section 6–4.) Note, however, that the method is still appli-cable to systems with parameters of interest other than gain. (See Section 6–6.) 1. Locate the poles and zeros of G(s)H(s) on the s plane.The root-locus branches start from open-loop poles and terminate at zeros (finite zeros or zeros at infinity). From the factored form of the open-loop transfer function, locate the open-loop poles and zeros in the s plane. CNote that the open-loop zeros are the zeros of G(s)H(s), while the closed-loop zeros consist of the zeros of G(s) and the poles of H(s).D 1 + KAs + z1BAs + z2B p As + zmB As + p1BAs + p2B p As + pnB = 0 1 + G(s)H(s) = 0 13.
H(s) G(s) C(s) R(s) + – Figure 6–11 Control system.
284 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method Note that the root loci are symmetrical about the real axis of the s plane, because the complex poles and complex zeros occur only in conjugate pairs.
A root-locus plot will have just as many branches as there are roots of the character-istic equation. Since the number of open-loop poles generally exceeds that of zeros, the number of branches equals that of poles. If the number of closed-loop poles is the same as the number of open-loop poles, then the number of individual root-locus branches terminating at finite open-loop zeros is equal to the number m of the open-loop zeros.
The remaining n-m branches terminate at infinity (n-m implicit zeros at infinity) along asymptotes.
If we include poles and zeros at infinity, the number of open-loop poles is equal to that of open-loop zeros. Hence we can always state that the root loci start at the poles of G(s)H(s) and end at the zeros of G(s)H(s), as K increases from zero to in-finity, where the poles and zeros include both those in the finite s plane and those at infinity.
2. Determine the root loci on the real axis. Root loci on the real axis are determined by open-loop poles and zeros lying on it. The complex-conjugate poles and complex-conjugate zeros of the open-loop transfer function have no effect on the location of the root loci on the real axis because the angle contribution of a pair of complex-conjugate poles or complex-conjugate zeros is 360° on the real axis. Each portion of the root locus on the real axis extends over a range from a pole or zero to another pole or zero.
In constructing the root loci on the real axis, choose a test point on it. If the total num-ber of real poles and real zeros to the right of this test point is odd, then this point lies on a root locus. If the open-loop poles and open-loop zeros are simple poles and sim-ple zeros, then the root locus and its complement form alternate segments along the real axis.
3. Determine the asymptotes of root loci. If the test point s is located far from the ori-gin,then the angle of each complex quantity may be considered the same.One open-loop zero and one open-loop pole then cancel the effects of the other. Therefore, the root loci for very large values of s must be asymptotic to straight lines whose angles (slopes) are given by where number of finite poles of G(s)H(s) number of finite zeros of G(s)H(s) Here, k=0 corresponds to the asymptotes with the smallest angle with the real axis.Al-though k assumes an infinite number of values, as k is increased the angle repeats itself, and the number of distinct asymptotes is n-m.
All the asymptotes intersect at a point on the real axis. The point at which they do so is obtained as follows:If both the numerator and denominator of the open-loop trans-fer function are expanded, the result is G(s)H(s) = KCsm + Az1 + z2 + p + zmBsm-1 + p + z1 z2 p zmD sn + Ap1 + p2 + p + pnBsn-1 + p + p1 p2 p pn m = n = Angles of asymptotes = ;180°(2k + 1) n - m (k = 0, 1, 2, p ) Section 6–2 / Root-Locus Plots 285 If a test point is located very far from the origin, then by dividing the denominator by the numerator, it is possible to write G(s)H(s) as or (6–12) The abscissa of the intersection of the asymptotes and the real axis is then obtained by setting the denominator of the right-hand side of Equation (6–12) equal to zero and solving for s, or (6–13) [Example 6–1 shows why Equation (6–13) gives the intersection.] Once this intersection is determined, the asymptotes can be readily drawn in the complex plane.
It is important to note that the asymptotes show the behavior of the root loci for A root-locus branch may lie on one side of the corresponding asymptote or may cross the corresponding asymptote from one side to the other side.
4. Find the breakaway and break-in points. Because of the conjugate symmetry of the root loci, the breakaway points and break-in points either lie on the real axis or occur in complex-conjugate pairs.
If a root locus lies between two adjacent open-loop poles on the real axis, then there exists at least one breakaway point between the two poles. Similarly, if the root locus lies between two adjacent zeros (one zero may be located at –q) on the real axis, then there always exists at least one break-in point between the two zeros. If the root locus lies be-tween an open-loop pole and a zero (finite or infinite) on the real axis, then there may exist no breakaway or break-in points or there may exist both breakaway and break-in points.
Suppose that the characteristic equation is given by The breakaway points and break-in points correspond to multiple roots of the charac-teristic equation.Hence,as discussed in Example 6–1,the breakaway and break-in points can be determined from the roots of (6–14) where the prime indicates differentiation with respect to s. It is important to note that the breakaway points and break-in points must be the roots of Equation (6–14), but not all roots of Equation (6–14) are breakaway or break-in points. If a real root of Equation (6–14) lies on the root-locus portion of the real axis, then it is an actual breakaway or break-in point. If a real root of Equation (6–14) is not on the root-locus portion of the real axis, then this root corresponds to neither a breakaway point nor a break-in point.
dK ds = - B¿(s)A(s) - B(s)A¿(s) A 2(s) = 0 B(s) + KA(s) = 0 ∑s∑ 1.
s = -Ap1 + p2 + p + pnB - Az1 + z2 + p + zmB n - m G(s)H(s) = K cs + Ap1 + p2 + p + pnB - Az1 + z2 + p + zmB n - m d n-m G(s)H(s) = K sn-m + C Ap1 + p2 + p + pnB - Az1 + z2 + p + zmB Dsn-m-1 + p 286 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method jv s Angle of departure u2 u1 f 0 Figure 6–12 Construction of the root locus. [Angle of departure =180°-(u1+u2)+f.] If two roots s=s1 and s=–s1 of Equation (6–14) are a complex-conjugate pair and if it is not certain whether they are on root loci, then it is necessary to check the corre-sponding K value. If the value of K corresponding to a root s=s1 of is pos-itive, point s=s1 is an actual breakaway or break-in point. (Since K is assumed to be nonnegative, if the value of K thus obtained is negative, or a complex quantity, then point s=s1 is neither a breakaway nor a break-in point.) 5. Determine the angle of departure (angle of arrival) of the root locus from a com-plex pole (at a complex zero). To sketch the root loci with reasonable accuracy, we must find the directions of the root loci near the complex poles and zeros.If a test point is cho-sen and moved in the very vicinity of a complex pole (or complex zero), the sum of the angular contributions from all other poles and zeros can be considered to remain the same. Therefore, the angle of departure (or angle of arrival) of the root locus from a complex pole (or at a complex zero) can be found by subtracting from 180° the sum of all the angles of vectors from all other poles and zeros to the complex pole (or complex zero) in question, with appropriate signs included.
Angle of departure from a complex pole=180° – (sum of the angles of vectors to a complex pole in question from other poles) ± (sum of the angles of vectors to a complex pole in question from zeros) Angle of arrival at a complex zero=180° – (sum of the angles of vectors to a complex zero in question from other zeros) ± (sum of the angles of vectors to a complex zero in question from poles) The angle of departure is shown in Figure 6–12.
6. Find the points where the root loci may cross the imaginary axis. The points where the root loci intersect the jv axis can be found easily by (a) use of Routh’s stability cri-terion or (b) letting s=jv in the characteristic equation,equating both the real part and the imaginary part to zero, and solving for v and K.The values of v thus found give the frequencies at which root loci cross the imaginary axis. The K value corresponding to each crossing frequency gives the gain at the crossing point.
7. Taking a series of test points in the broad neighborhood of the origin of the s plane, sketch the root loci. Determine the root loci in the broad neighborhood of the jv axis and the origin. The most important part of the root loci is on neither the real axis nor the asymptotes but is in the broad neighborhood of the jv axis and the origin.The shape dKds = 0 Section 6–2 / Root-Locus Plots 287 of the root loci in this important region in the s plane must be obtained with reasonable accuracy.(If accurate shape of the root loci is needed,MATLAB may be used rather than hand calculations of the exact shape of the root loci.) 8. Determine closed-loop poles. A particular point on each root-locus branch will be a closed-loop pole if the value of K at that point satisfies the magnitude condition. Con-versely, the magnitude condition enables us to determine the value of the gain K at any specific root location on the locus. (If necessary, the root loci may be graduated in terms of K.The root loci are continuous with K.) The value of K corresponding to any point s on a root locus can be obtained using the magnitude condition, or This value can be evaluated either graphically or analytically. (MATLAB can be used for graduating the root loci with K. See Section 6–3.) If the gain K of the open-loop transfer function is given in the problem, then by ap-plying the magnitude condition, we can find the correct locations of the closed-loop poles for a given K on each branch of the root loci by a trial-and-error approach or by use of MATLAB, which will be presented in Section 6–3.
Comments on the Root-Locus Plots.
It is noted that the characteristic equa-tion of the negative feedback control system whose open-loop transfer function is is an nth-degree algebraic equation in s. If the order of the numerator of G(s)H(s) is lower than that of the denominator by two or more (which means that there are two or more zeros at infinity), then the coefficient a1 is the negative sum of the roots of the equation and is independent of K. In such a case, if some of the roots move on the locus toward the left as K is increased, then the other roots must move toward the right as K is increased.This information is helpful in finding the general shape of the root loci.
It is also noted that a slight change in the pole–zero configuration may cause signif-icant changes in the root-locus configurations. Figure 6–13 demonstrates the fact that a slight change in the location of a zero or pole will make the root-locus configuration look quite different.
G(s)H(s) = KAsm + b1 sm-1 + p + bmB sn + a1 sn-1 + p + an (n m) K = product of lengths between point s to poles product of lengths between point s to zeros jv s jv s Figure 6–13 Root-locus plots.
288 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method C(s) R(s) (a) 1 s K (s + 1) (s + 2) C(s) R(s) (c) 1 s + 1 K s(s + 1) (s + 2) K s(s + 2) s + 1 H(s) C(s) R(s) (b) + – + – + – G(s) + – Figure 6–14 (a) Control system with velocity feedback; (b) and (c) modified block diagrams.
Cancellation of Poles of G(s) with Zeros of H(s).
It is important to note that if the denominator of G(s) and the numerator of H(s) involve common factors, then the corresponding open-loop poles and zeros will cancel each other, reducing the degree of the characteristic equation by one or more. For example, consider the system shown in Figure 6–14(a). (This system has velocity feedback.) By modifying the block diagram of Figure 6–14(a) to that shown in Figure 6–14(b), it is clearly seen that G(s) and H(s) have a common factor s+1. The closed-loop transfer function C(s)/R(s) is The characteristic equation is Because of the cancellation of the terms (s+1) appearing in G(s) and H(s), however, we have The reduced characteristic equation is The root-locus plot of G(s)H(s) does not show all the roots of the characteristic equa-tion, only the roots of the reduced equation.
To obtain the complete set of closed-loop poles, we must add the canceled pole of G(s)H(s) to those closed-loop poles obtained from the root-locus plot of G(s)H(s).
The important thing to remember is that the canceled pole of G(s)H(s) is a closed-loop pole of the system, as seen from Figure 6–14(c).
s(s + 2) + K = 0 = s(s + 2) + K s(s + 2) 1 + G(s)H(s) = 1 + K(s + 1) s(s + 1)(s + 2) Cs(s + 2) + KD(s + 1) = 0 C(s) R(s) = K s(s + 1)(s + 2) + K(s + 1) Section 6–2 / Root-Locus Plots 289 Typical Pole–Zero Configurations and Corresponding Root Loci.
In summa-rizing, we show several open-loop pole–zero configurations and their corresponding root loci in Table 6–1.The pattern of the root loci depends only on the relative separa-tion of the open-loop poles and zeros. If the number of open-loop poles exceeds the number of finite zeros by three or more, there is a value of the gain K beyond which root loci enter the right-half s plane, and thus the system can become unstable.A stable sys-tem must have all its closed-loop poles in the left-half s plane.
Table 6–1 Open-Loop Pole–Zero Configurations and the Corresponding Root Loci jv s jv s jv s jv s jv s jv s jv s jv s jv s jv s jv s jv s 290 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method Note that once we have some experience with the method,we can easily evaluate the changes in the root loci due to the changes in the number and location of the open-loop poles and zeros by visualizing the root-locus plots resulting from various pole–zero configurations.
Summary.
From the preceding discussions, it should be clear that it is possible to sketch a reasonably accurate root-locus diagram for a given system by following simple rules. (The reader should study the various root-locus diagrams shown in the solved problems at the end of the chapter.) At preliminary design stages, we may not need the precise locations of the closed-loop poles. Often their approximate locations are all that is needed to make an estimate of system performance. Thus, it is important that the designer have the capability of quickly sketching the root loci for a given system.
6–3 PLOTTING ROOT LOCI WITH MATLAB In this section we present the MATLAB approach to the generation of root-locus plots and finding relevant information from the root-locus plots.
Plotting Root Loci with MATLAB.
In plotting root loci with MATLAB we deal with the system equation given in the form of Equation (6–11), which may be written as where num is the numerator polynomial and den is the denominator polynomial.
That is, Note that both vectors num and den must be written in descending powers of s.
A MATLAB command commonly used for plotting root loci is rlocus(num,den) Using this command, the root-locus plot is drawn on the screen.The gain vector K is au-tomatically determined. (The vector K contains all the gain values for which the closed-loop poles are to be computed.) For the systems defined in state space, rlocus(A,B,C,D) plots the root locus of the system with the gain vector automatically determined.
Note that commands rlocus(num,den,K) and rlocus(A,B,C,D,K) use the user-supplied gain vector K. = sn + Ap1 + p2 + p + pnBsn-1 + p + p1 p2 p pn den = As + p1BAs + p2B p As + pnB = sm + Az1 + z2 + p + zmBsm-1 + p + z1 z2 p zm num = As + z1BAs + z2B p As + zmB 1 + K num den = 0 Section 6–3 / Plotting Root Loci with MATLAB 291 K(s + 3) s(s + 1)(s2 + 4s + 16) + – Figure 6–15 Control system.
If it is desired to plot the root loci with marks 'o' or 'x', it is necessary to use the fol-lowing command: r = rlocus(num,den) plot(r,'o') or plot(r,'x') Plotting root loci using marks o or x is instructive, since each calculated closed-loop pole is graphically shown; in some portion of the root loci those marks are densely placed and in another portion of the root loci they are sparsely placed.MATLAB supplies its own set of gain values used to calculate a root-locus plot. It does so by an internal adaptive step-size routine.Also,MATLAB uses the automatic axis-scaling feature of the plot command.
EXAMPLE 6–3 Consider the system shown in Figure 6–15. Plot root loci with a square aspect ratio so that a line with slope 1 is a true 45° line. Choose the region of root-locus plot to be where x and y are the real-axis coordinate and imaginary-axis coordinate, respectively.
To set the given plot region on the screen to be square, enter the command v = [-6 6 -6 6]; axis (v); axis('square') With this command, the region of the plot is as specified and a line with slope 1 is at a true 45°, not skewed by the irregular shape of the screen.
For this problem, the denominator is given as a product of first- and second-order terms. So we must multiply these terms to get a polynomial in s. The multiplication of these terms can be done easily by use of the convolution command, as shown next.
Define a = s (s + 1): a = [1 1 0] b = s2 + 4s + 16: b = [1 4 16] Then we use the following command: c = conv(a, b) Note that conv(a, b) gives the product of two polynomials a and b.See the following computer output: -6 x 6, -6 y 6 a = [1 1 0]; b = [1 4 16]; c = conv (a,b) c = 1 5 20 16 0 The denominator polynomial is thus found to be den = [1 5 20 16 0] 292 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method 0 −2 Real Axis Imag Axis 6 4 2 −4 −6 0 −2 6 4 2 −4 −6 Root-Locus Plot of G(s) = K(s + 3)/[s(s + 1)(s2 + 4s + 16)] Figure 6–16 Root-locus plot.
MATLAB Program 6–1 % --------- Root-locus plot ---------num = [1 3]; den = [1 5 20 16 0]; rlocus(num,den) v = [-6 6 -6 6]; axis(v); axis('square') grid; title ('Root-Locus Plot of G(s) = K(s + 3)/[s(s + 1)(s^2 + 4s + 16)]') To find the complex-conjugate open-loop poles (the roots of s2+4s+16=0), we may enter the roots command as follows: Note that in MATLAB Program 6–1, instead of den = [1 5 20 16 0] we may enter den = conv ([1 1 0], [1 4 16]) The results are the same.
r = roots(b) r = –2.0000 + 3.464li –2.0000 - 3.464li Thus, the system has the following open-loop zero and open-loop poles: Open-loop zero: s=–3 Open-loop poles: s=0, s=–1, s=–2 ; j3.4641 MATLAB Program 6–1 will plot the root-locus diagram for this system. The plot is shown in Figure 6–16.
Section 6–3 / Plotting Root Loci with MATLAB 293 EXAMPLE 6–4 Consider the negative feedback system whose open-loop transfer function G(s)H(s) is There are no open-loop zeros. Open-loop poles are located at s=–0.3+j3.1480, s=–0.3-j3.1480, s=–0.5, and s=0.
Entering MATLAB Program 6–2 into the computer, we obtain the root-locus plot shown in Figure 6–17. = K s4 + 1.1s3 + 10.3s2 + 5s G(s)H(s) = K s(s + 0.5)As2 + 0.6s + 10B MATLAB Program 6–2 % --------- Root-locus plot ---------num = ; den = [1 1.1 10.3 5 0]; r = rlocus(num,den); plot(r,'o') v = [-6 6 -6 6]; axis(v) grid title('Root-Locus Plot of G(s) = K/[s(s + 0.5)(s^2 + 0.6s + 10)]') xlabel('Real Axis') ylabel('Imag Axis') Real Axis –6 –4 6 4 2 –2 0 Imag Axis 6 –2 4 –6 2 0 –4 Root-Locus Plot of G(s) = K/[s(s+0.5)(s2+0.6s+10)] Figure 6–17 Root-locus plot.
Notice that in the regions near x=–0.3, y=2.3 and x=–0.3, y=–2.3 two loci approach each other.We may wonder if these two branches should touch or not.To explore this situation, we may plot the root loci using smaller increments of K in the critical region.
294 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method Real Axis –4 –2 –3 4 2 1 3 –1 0 Imag Axis 4 –1 3 –3 –4 2 0 1 –2 Root-Locus Plot of G(s) = K/[s(s+0.5)(s2+0.6s+10)] Figure 6–18 Root-locus plot.
MATLAB Program 6–3 % --------- Root-locus plot ---------num = ; den = [1 1.1 10.3 5 0]; K1 = 0:0.2:20; K2 = 20:0.1:30; K3 = 30:5:1000; K = [K1 K2 K3]; r = rlocus(num,den,K); plot(r, 'o') v = [-4 4 -4 4]; axis(v) grid title('Root-Locus Plot of G(s) = K/[s(s + 0.5)(s^2 + 0.6s + 10)]') xlabel('Real Axis') ylabel('Imag Axis') By a conventional trial-and-error approach or using the command rlocfind to be presented later in this section, we find the particular region of interest to be 20 K 30. By entering MATLAB Program 6–3, we obtain the root-locus plot shown in Figure 6–18. From this plot, it is clear that the two branches that approach in the upper half-plane (or in the lower half-plane) do not touch.
EXAMPLE 6–5 Consider the system shown in Figure 6–19.The system equations are u = r - y y = Cx + Du x # = Ax + Bu Section 6–3 / Plotting Root Loci with MATLAB 295 r u B y A C D x x • + – ++ ++ Figure 6–19 Closed-loop control system.
In this example problem we shall obtain the root-locus diagram of the system defined in state space.As an example let us consider the case where matrices A, B, C, and D are (6–15) The root-locus plot for this system can be obtained with MATLAB by use of the following command: rlocus(A,B,C,D) This command will produce the same root-locus plot as can be obtained by use of the rlocus (num,den) command, where num and den are obtained from [num,den] = ss2tf(A,B,C,D) as follows: num = [0 0 1 0] den = [1 14 56 160] MATLAB Program 6–4 is a program that will generate the root-locus plot as shown in Figure 6–20. C = [1 0 0], D = A = C 0 0 -160 1 0 -56 0 1 -14 S , B = C 0 1 -14 S MATLAB Program 6–4 % --------- Root-locus plot ---------A = [0 1 0;0 0 1;-160 -56 -14]; B = [0;1;-14]; C = [1 0 0]; D = ; K = 0:0.1:400; rlocus(A,B,C,D,K); v = [-20 20 -20 20]; axis(v) grid title('Root-Locus Plot of System Defined in State Space') 296 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method 0 jv jv vn vd f s 0 s 0.2 0.2 0.5 0.5 0.7 0.7 0.8 0.8 z = 0.9 z = 0.9 z = 0 z 0 z 0 z 1 (a) (b) Figure 6–21 (a) Complex poles; (b) lines of constant damping ratio z.
Real Axis –20 –15 20 0 15 –10 –5 5 10 Imag Axis 20 –5 –20 –10 15 0 –15 10 5 Root-Locus Plot of System Defined in State Space Figure 6–20 Root-locus plot of system defined in state space, where A, B, C, and D are as given by Equation (6–15).
Constant Z Loci and Constant Vn Loci.
Recall that in the complex plane the damping ratio z of a pair of complex-conjugate poles can be expressed in terms of the angle f, which is measured from the negative real axis, as shown in Figure 6–21(a) with In other words, lines of constant damping ratio z are radial lines passing through the origin as shown in Figure 6–21(b). For example, a damping ratio of 0.5 requires that the complex-conjugate poles lie on the lines drawn through the origin making angles of ;60° with the negative real axis. (If the real part of a pair of complex-conjugate poles is positive, which means that the system is unstable, the corresponding z is negative.) The damping ratio determines the angular location of the poles, while the z = cosf Section 6–3 / Plotting Root Loci with MATLAB 297 distance of the pole from the origin is determined by the undamped natural frequen-cy vn. The constant vn loci are circles.
To draw constant z lines and constant vn circles on the root-locus diagram with MATLAB, use the command sgrid.
Plotting Polar Grids in the Root-Locus Diagram.
The command sgrid overlays lines of constant damping ratio (z=0 ~ 1 with 0.1 increment) and circles of constant vn on the root-locus plot.See MATLAB Program 6–5 and the resulting diagram shown in Figure 6–22.
MATLAB Program 6–5 sgrid v = [-3 3 -3 3]; axis(v); axis('square') title('Constant \zeta Lines and Constant \omega_n Circles') xlabel('Real Axis') ylabel('Imag Axis') If only particular constant z lines (such as the z=0.5 line and z=0.707 line) and particular constant vn circles (such as the vn=0.5 circle, vn=1 circle, and vn=2 cir-cle) are desired, use the following command: sgrid([0.5, 0.707], [0.5, 1, 2]) If we wish to overlay lines of constant z and circles of constant vn as given above to a root-locus plot of a negative feedback system with num = [0 0 0 1] den = [1 4 5 0] 3 2 1 0 −3 −2 −1 0 1 3 2 −1 −3 −2 Real Axis Constant z Lines and Constant vn Circles Imag Axis 2 1 2 1 0.16 0.34 0.5 0.64 0.16 0.34 0.5 0.64 0.76 0.86 0.94 0.985 0.76 0.86 0.94 0.985 Figure 6–22 Constant z lines and constant vn circles.
298 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method 0.5 0.707 0.5 0.707 ωn = 1 ωn = 0.5 ωn = 2 −1 −0.5 0 0.5 1 −1.5 −2 −2.5 −3 0 0.5 1 1.5 2 −0.5 −1 −1.5 −2 Real Axis Root-Locus Plot with ζ = 0.5 and 0.707 Lines and ωn = 0.5, 1, and 2 Circles Imag Axis 0.5 1 2 Figure 6–23 Constant z lines and constant vn circles superimposed on a root-locus plot.
MATLAB Program 6–6 num = ; den = [1 4 5 0]; K = 0:0.01:1000; r = rlocus(num,den,K); plot(r,'-'); v = [-3 1 -2 2]; axis(v); axis('square') sgrid([0.5,0.707], [0.5,1,2]) grid title('Root-Locus Plot with \zeta = 0.5 and 0.707 Lines and \omega_n = 0.5,1, and 2 Circles') xlabel('Real Axis'); ylabel('Imag Axis') gtext('\omega_n = 2') gtext('\omega_n = 1') gtext('\omega_n = 0.5') % Place 'x' mark at each of 3 open-loop poles.
gtext('x') gtext('x') gtext('x') then enter MATLAB Program 6–6 into the computer. The resulting root-locus plot is shown in Figure 6–23.
If we want to omit either the entire constant z lines or entire constant vn circles, we may use empty brackets [] in the arguments of the sgrid command.For example,if we want to overlay only the constant damping ratio line corresponding to z=0.5 and no constant vn circles on the root-locus plot, then we may use the command sgrid(0.5, []) Section 6–3 / Plotting Root Loci with MATLAB 299 Conditionally Stable Systems.
Consider the negative feedback system shown in Figure 6–24.We can plot the root loci for this system by applying the general rules and procedure for constructing root loci, or use MATLAB to get root-locus plots. MAT-LAB Program 6–7 will plot the root-locus diagram for the system.The plot is shown in Figure 6–25.
R(s) C(s) K(s2 + 2s +4) s(s + 4) (s + 6)(s2 + 1.4s + 1) + – Figure 6–24 Control system.
MATLAB Program 6–7 num = [1 2 4]; den = conv(conv([1 4 0],[1 6]), [1 1.4 1]); rlocus(num, den) v = [-7 3 -5 5]; axis(v); axis('square') grid title('Root-Locus Plot of G(s) = K(s^2 + 2s + 4)/[s(s + 4)(s + 6)(s^2 + 1.4s + 1)]') text(1.0, 0.55,'K = 12') text(1.0,3.0,'K = 73') text(1.0,4.15,'K = 154') Real Axis −5 −4 −3 −2 −1 −6 −7 3 0 2 1 Imag Axis −5 5 4 3 2 −3 −2 −1 −4 0 1 Root-Locus Plot of G(s) = K(s2 + 2s + 4)/[s(s + 4)(s + 6)(s2 + 1.4s + 1)] K = 12 K = 73 K = 154 Figure 6–25 Root-locus plot of conditionally stable system.
It can be seen from the root-locus plot of Figure 6–25 that this system is stable only for limited ranges of the value of K—that is, 0<K<12 and 73<K<154. The sys-tem becomes unstable for 12<K<73 and 154K 1 Ta K = 1 Ta K = 1 Ta 1 T – s K(1 – Tas) s(Ts + 1) + – Figure 6–26 (a) Nonminimum-phase system; (b) root-locus plot.
Section 6–3 / Plotting Root Loci with MATLAB 301 To obtain a root-locus plot with MATLAB, enter the numerator and denominator as usual. For example, if T=1 sec and enter the following num and den in the program: num = [-0.5 1] den = [1 1 0] MATLAB Program 6–8 gives the plot of the root loci shown in Figure 6–27.
T a = 0.5 sec, MATLAB Program 6–8 num = [-0.5 1]; den = [1 1 0]; k1 = 0:0.01:30; k2 = 30:1:100; K3 = 100:5:500; K = [k1 k2 k3]; rlocus(num,den,K) v = [-2 6 -4 4]; axis(v); axis('square') grid title('Root-Locus Plot of G(s) = K(1 - 0.5s)/[s(s + 1)]') % Place 'x' mark at each of 2 open-loop poles.
% Place 'o' mark at open-loop zero.
gtext('x') gtext('x') gtext('o') Root-Locus Plot of G(s) = K(1 − 0.5s)/[s(s + 1)] Real Axis Imag Axis 1 −3 −4−2 −1 0 1 2 3 4 5 6 2 −1 −2 0 3 4 Figure 6–27 Root-locus plot of G(s) = K(1 - 0.5s) s(s + 1) .
Orthogonality of Root Loci and Constant-Gain Loci.
Consider the negative feedback system whose open-loop transfer function is G(s)H(s). In the G(s)H(s) plane, the loci of constant are circles centered at the origin, and the loci corre-sponding to (k=0, 1, 2, p) lie on the negative real axis /G(s)H(s) = ;180°(2k + 1) ∑G(s)H(s)∑= 302 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method of the G(s)H(s) plane, as shown in Figure 6–28. [Note that the complex plane employed here is not the s plane, but the G(s)H(s) plane.] The root loci and constant-gain loci in the s plane are conformal mappings of the loci of and of constant in the G(s)H(s) plane.
Since the constant-phase and constant-gain loci in the G(s)H(s) plane are orthog-onal, the root loci and constant-gain loci in the s plane are orthogonal. Figure 6–29(a) shows the root loci and constant-gain loci for the following system: G(s) = K(s + 2) s2 + 2s + 3 , H(s) = 1 ∑G(s)H(s)∑= /G(s)H(s) = ;180°(2k + 1) Re Im 0 G(s) H(s) Plane Re Im 0 G(s) H(s) Plane |G(s) H(s)| = constant G(s) H(s) = 180° (2k + 1) Figure 6–28 Plots of constant-gain and constant-phase loci in the G(s)H(s) plane.
(a) (b) s jv 0 K = 6 K = 6 j4 j6 –j4 K = 1 K = 2 K = 1 –6 –4 2 4 6 K = 10 K = 0.3 j2 –j2 –j6 K = 0.3 K = 0.3 –2 s jv 0 j2 j3 –j2 –3 –2 1 2 j1 –j1 –j3 –1 B C A Figure 6–29 Plots of root loci and constant-gain loci. (a) System with G(s)=K(s+2)/As2+2s+3B, H(s)=1; (b) system with G(s)=K/Cs(s+1)(s+2)D, H(s)=1.
Section 6–4 / Root-Locus Plots of Positive Feedback Systems 303 Notice that since the pole–zero configuration is symmetrical about the real axis, the constant-gain loci are also symmetrical about the real axis.
Figure 6–29(b) shows the root loci and constant-gain loci for the system: Notice that since the configuration of the poles in the s plane is symmetrical about the real axis and the line parallel to the imaginary axis passing through point (s=–1, v=0), the constant-gain loci are symmetrical about the v=0 line (real axis) and the s=–1 line.
From Figures 6–29(a) and (b), notice that every point in the s plane has the corre-sponding K value. If we use a command rlocfind (presented next), MATLAB will give the K value of the specified point as well as the nearest closed-loop poles corresponding to this K value.
Finding the Gain Value K at an Arbitrary Point on the Root Loci.
In MAT-LAB analysis of closed-loop systems, it is frequently desired to find the gain value K at an arbitrary point on the root locus. This can be accomplished by using the following rlocfind command: [K, r] = rlocfind(num, den) The rlocfind command, which must follow an rlocus command, overlays movable x-y co-ordinates on the screen. Using the mouse, we position the origin of the x-y coordinates over the desired point on the root locus and press the mouse button. Then MATLAB displays on the screen the coordinates of that point, the gain value at that point, and the closed-loop poles corresponding to this gain value.
If the selected point is not on the root locus, such as point A in Figure 6–29(a), the rlocfind command gives the coordinates of this selected point, the gain value of this point, such as K = 2, and the locations of the closed-loop poles, such as points B and C corresponding to this K value.[Note that every point on the s plane has a gain value.See, for example, Figures 6–29 (a) and (b).] 6–4 ROOT-LOCUS PLOTS OF POSITIVE FEEDBACK SYSTEMS Root Loci for Positive-Feedback Systems. In a complex control system, there may be a positive-feedback inner loop as shown in Figure 6–30. Such a loop is usually stabilized by the outer loop.In what follows,we shall be concerned only with the positive-feedback inner loop.The closed-loop transfer function of the inner loop is The characteristic equation is (6–17) 1 - G(s)H(s) = 0 C(s) R(s) = G(s) 1 - G(s)H(s) G(s) = K s(s + 1)(s + 2) , H(s) = 1 Reference W-4 304 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method C(s) G1(s) H1(s) R(s) H(s) G(s) + – + + Figure 6–30 Control system.
This equation can be solved in a manner similar to the development of the root-locus method for negative-feedback systems presented in Section 6–2. The angle condition, however, must be altered.
Equation (6–17) can be rewritten as which is equivalent to the following two equations: For the positive-feedback case, the total sum of all angles from the open-loop poles and zeros must be equal to Thus the root locus follows a 0° locus in contrast to the 180° locus considered previously.The magnitude condition remains unaltered.
To illustrate the root-locus plot for the positive-feedback system, we shall use the fol-lowing transfer functions G(s) and H(s) as an example.
The gain K is assumed to be positive.
The general rules for constructing root loci for negative-feedback systems given in Section 6–2 must be modified in the following way: Rule 2 is Modified as Follows: If the total number of real poles and real zeros to the right of a test point on the real axis is even, then this test point lies on the root locus.
Rule 3 is Modified as Follows: where number of finite poles of G(s)H(s) number of finite zeros of G(s)H(s) Rule 5 is Modified as Follows: When calculating the angle of departure (or angle of ar-rival) from a complex open-loop pole (or at a complex zero), subtract from 0° the sum of all angles of the vectors from all the other poles and zeros to the complex pole (or com-plex zero) in question, with appropriate signs included. m = n = Angles of asymptotes = ;k360° n - m (k = 0, 1, 2, p ) G(s) = K(s + 2) (s + 3)As2 + 2s + 2B , H(s) = 1 0° ; k360°. ∑G(s)H(s)∑= 1 /G(s)H(s) = 0° ; k360° (k = 0, 1, 2, p ) G(s)H(s) = 1 Section 6–4 / Root-Locus Plots of Positive Feedback Systems 305 Other rules for constructing the root-locus plot remain the same.We shall now apply the modified rules to construct the root-locus plot.
1. Plot the open-loop poles (s=–1+j, s=–1-j, s=–3) and zero (s=–2) in the complex plane.As K is increased from 0 to q, the closed-loop poles start at the open-loop poles and terminate at the open-loop zeros (finite or infinite), just as in the case of negative-feedback systems.
2. Determine the root loci on the real axis. Root loci exist on the real axis between –2 and ±q and between –3 and –q.
3. Determine the asymptotes of the root loci. For the present system, This simply means that asymptotes are on the real axis.
4. Determine the breakaway and break-in points. Since the characteristic equation is we obtain By differentiating K with respect to s, we obtain Note that Point s=–0.8 is on the root locus. Since this point lies between two zeros (a finite zero and an infinite zero), it is an actual break-in point. Points do not satisfy the angle condition and, therefore, they are neither breakaway nor break-in points.
5. Find the angle of departure of the root locus from a complex pole. For the com-plex pole at s=–1+j, the angle of departure u is or (The angle of departure from the complex pole at s=–1-j is 72°.) 6. Choose a test point in the broad neighborhood of the jv axis and the origin and apply the angle condition. Locate a sufficient number of points that satisfy the angle condition.
Figure 6–31 shows the root loci for the given positive-feedback system.The root loci are shown with dashed lines and a curve.
Note that if K 7 (s + 3)As2 + 2s + 2B s + 2 2 s=0 = 3 u = -72° u = 0° - 27° - 90° + 45° s = -2.35 ; j0.77 = 2(s + 0.8)(s + 2.35 + j0.77)(s + 2.35 - j0.77) 2s3 + 11s2 + 20s + 10 = 2(s + 0.8)As2 + 4.7s + 6.24B dK ds = 2s3 + 11s2 + 20s + 10 (s + 2)2 K = (s + 3)As2 + 2s + 2B s + 2 (s + 3)As2 + 2s + 2B - K(s + 2) = 0 Angles of asymptote = ;k360° 3 - 1 = ;180° 306 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method –5 –4 –3 –2 –1 1 2 s jv 0 j1 j2 –j1 –j2 Figure 6–31 Root-locus plot for the positive-feedback system with G(s)=K(s+2)/ C(s+3)As2+2s+2B D, H(s)=1.
one real root enters the right-half s plane. Hence, for values of K greater than 3, the sys-tem becomes unstable. (For K>3, the system must be stabilized with an outer loop.) Note that the closed-loop transfer function for the positive-feedback system is given by To compare this root-locus plot with that of the corresponding negative-feedback sys-tem,we show in Figure 6–32 the root loci for the negative-feedback system whose closed-loop transfer function is Table 6–2 shows various root-locus plots of negative-feedback and positive-feedback systems.The closed-loop transfer functions are given by for negative-feedback systems for positive-feedback systems C R = G 1 - GH , C R = G 1 + GH , C(s) R(s) = K(s + 2) (s + 3)As2 + 2s + 2B + K(s + 2) = K(s + 2) (s + 3)As2 + 2s + 2B - K(s + 2) C(s) R(s) = G(s) 1 - G(s)H(s) –5 –4 –3 –2 –1 1 2 s jv 0 j1 j2 j3 –j1 –j3 –j2 Figure 6–32 Root-locus plot for the negative-feedback system with G(s)=K(s+2)/ C(s+3)As2+2s+2B D, H(s)=1.
Section 6–4 / Root-Locus Plots of Positive Feedback Systems 307 where GH is the open-loop transfer function. In Table 6–2, the root loci for negative-feedback systems are drawn with heavy lines and curves,and those for positive-feedback systems are drawn with dashed lines and curves.
jv s jv s jv s jv s jv s jv s jv s jv s jv s jv s Table 6–2 Root-Locus Plots of Negative-Feedback and Positive-Feedback Systems Heavy lines and curves correspond to negative-feedback systems; dashed lines and curves correspond to positive-feedback systems.
308 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method 6–5 ROOT-LOCUS APPROACH TO CONTROL-SYSTEMS DESIGN Preliminary Design Consideration.
In building a control system, we know that proper modification of the plant dynamics may be a simple way to meet the performance specifications.This,however,may not be possible in many practical situations because the plant may be fixed and not modifiable.Then we must adjust parameters other than those in the fixed plant. In this book, we assume that the plant is given and unalterable.
In practice,the root-locus plot of a system may indicate that the desired performance cannot be achieved just by the adjustment of gain (or some other adjustable parameter).
In fact, in some cases, the system may not be stable for all values of gain (or other ad-justable parameter). Then it is necessary to reshape the root loci to meet the perform-ance specifications.
The design problems, therefore, become those of improving system performance by insertion of a compensator. Compensation of a control system is reduced to the design of a filter whose characteristics tend to compensate for the undesirable and unalterable characteristics of the plant.
Design by Root-Locus Method.
The design by the root-locus method is based on re-shaping the root locus of the system by adding poles and zeros to the system’s open-loop transfer function and forcing the root loci to pass through desired closed-loop poles in the s plane.The characteristic of the root-locus design is its being based on the assumption that the closed-loop system has a pair of dominant closed-loop poles.This means that the effects of zeros and additional poles do not affect the response characteristics very much.
In designing a control system, if other than a gain adjustment (or other parameter adjustment) is required,we must modify the original root loci by inserting a suitable com-pensator.Once the effects on the root locus of the addition of poles and/or zeros are fully understood,we can readily determine the locations of the pole(s) and zero(s) of the com-pensator that will reshape the root locus as desired. In essence, in the design by the root-locus method, the root loci of the system are reshaped through the use of a compensator so that a pair of dominant closed-loop poles can be placed at the desired location.
Series Compensation and Parallel (or Feedback) Compensation.
Figures 6–33(a) and (b) show compensation schemes commonly used for feedback control sys-tems. Figure 6–33(a) shows the configuration where the compensator Gc(s) is placed in series with the plant.This scheme is called series compensation.
An alternative to series compensation is to feed back the signal(s) from some ele-ment(s) and place a compensator in the resulting inner feedback path,as shown in Figure 6–33(b). Such compensation is called parallel compensation or feedback compensation.
In compensating control systems, we see that the problem usually boils down to a suitable design of a series or parallel compensator.The choice between series compen-sation and parallel compensation depends on the nature of the signals in the system, the power levels at various points, available components, the designer’s experience, eco-nomic considerations, and so on.
In general,series compensation may be simpler than parallel compensation;however, series compensation frequently requires additional amplifiers to increase the gain and/or to provide isolation.(To avoid power dissipation,the series compensator is inserted at the lowest energy point in the feedforward path.) Note that, in general, the number of com-ponents required in parallel compensation will be less than the number of components Section 6–5 / Root-Locus Approach to Control Systems Design 309 G1(s) G2(s) H(s) Gc(s) (b) + – + – Gc(s) G(s) H(s) (a) + – Figure 6–33 (a) Series compensation; (b) parallel or feed-back compensation.
in series compensation, provided a suitable signal is available, because the energy trans-fer is from a higher power level to a lower level. (This means that additional amplifiers may not be necessary.) In Sections 6–6 through 6–9 we first discuss series compensation techniques and then present a parallel compensation technique using a design of a velocity-feedback control system.
Commonly Used Compensators.
If a compensator is needed to meet the per-formance specifications, the designer must realize a physical device that has the pre-scribed transfer function of the compensator.
Numerous physical devices have been used for such purposes.In fact,many noble and useful ideas for physically constructing compensators may be found in the literature.
If a sinusoidal input is applied to the input of a network, and the steady-state output (which is also sinusoidal) has a phase lead, then the network is called a lead network.
(The amount of phase lead angle is a function of the input frequency.) If the steady-state output has a phase lag, then the network is called a lag network. In a lag–lead network, both phase lag and phase lead occur in the output but in different frequency regions; phase lag occurs in the low-frequency region and phase lead occurs in the high-frequency region.A compensator having a characteristic of a lead network,lag network,or lag–lead network is called a lead compensator, lag compensator, or lag–lead compensator.
Among the many kinds of compensators, widely employed compensators are the lead compensators, lag compensators, lag–lead compensators, and velocity-feedback (tachometer) compensators. In this chapter we shall limit our discussions mostly to these types. Lead, lag, and lag–lead compensators may be electronic devices (such as circuits using operational amplifiers) or RC networks (electrical, mechanical, pneumatic, hydraulic, or combinations thereof) and amplifiers.
Frequently used series compensators in control systems are lead, lag, and lag–lead compensators. PID controllers which are frequently used in industrial control systems are discussed in Chapter 8.
310 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method It is noted that in designing control systems by the root-locus or frequency-response methods the final result is not unique,because the best or optimal solution may not be pre-cisely defined if the time-domain specifications or frequency-domain specifications are given.
Effects of the Addition of Poles.
The addition of a pole to the open-loop transfer function has the effect of pulling the root locus to the right, tending to lower the system’s relative stability and to slow down the settling of the response. (Remember that the ad-dition of integral control adds a pole at the origin, thus making the system less stable.) Figure 6–34 shows examples of root loci illustrating the effects of the addition of a pole to a single-pole system and the addition of two poles to a single-pole system.
Effects of the Addition of Zeros.
The addition of a zero to the open-loop trans-fer function has the effect of pulling the root locus to the left,tending to make the system more stable and to speed up the settling of the response. (Physically, the addition of a zero in the feedforward transfer function means the addition of derivative control to the system.The effect of such control is to introduce a degree of anticipation into the sys-tem and speed up the transient response.) Figure 6–35(a) shows the root loci for a system (a) jv s (b) jv s (c) jv s Figure 6–34 (a) Root-locus plot of a single-pole system; (b) root-locus plot of a two-pole system; (c) root-locus plot of a three-pole system.
(a) jv s (b) jv s (c) jv s (d) jv s Figure 6–35 (a) Root-locus plot of a three-pole system; (b), (c), and (d) root-locus plots showing effects of addition of a zero to the three-pole system.
Section 6–6 / Lead Compensation 311 that is stable for small gain but unstable for large gain. Figures 6–35(b), (c), and (d) show root-locus plots for the system when a zero is added to the open-loop transfer function.
Notice that when a zero is added to the system of Figure 6–35(a), it becomes stable for all values of gain.
6–6 LEAD COMPENSATION In Section 6–5 we presented an introduction to compensation of control systems and dis-cussed preliminary materials for the root-locus approach to control-systems design and compensation. In this section we shall present control-systems design by use of the lead compensation technique. In carrying out a control-system design, we place a compen-sator in series with the unalterable transfer function G(s) to obtain desirable behavior.
The main problem then involves the judicious choice of the pole(s) and zero(s) of the compensator Gc(s) to have the dominant closed-loop poles at the desired location in the s plane so that the performance specifications will be met.
Lead Compensators and Lag Compensators.
There are many ways to realize lead compensators and lag compensators, such as electronic networks using operational amplifiers, electrical RC networks, and mechanical spring-dashpot systems.
Figure 6–36 shows an electronic circuit using operational amplifiers. The transfer function for this circuit was obtained in Chapter 3 as follows [see Equation (3–36)]: (6–18) where T = R1 C1 , aT = R2 C2 , K c = R4 C1 R3 C2 = K c a Ts + 1 aTs + 1 = K c s + 1 T s + 1 aT Eo(s) Ei(s) = R2 R4 R1 R3 R1 C1 s + 1 R2 C2 s + 1 = R4 C1 R3 C2 s + 1 R1 C1 s + 1 R2 C2 – + – + C1 C2 R1 R2 R3 R4 Ei(s) Eo(s) E(s) Figure 6–36 Electronic circuit that is a lead network if and a lag network if R1 C1 6 R2 C2 .
R1 C1 7 R2 C2 312 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method Notice that This network has a dc gain of From Equation (6–18), we see that this network is a lead network if or It is a lag network if The pole-zero configurations of this net-work when and are shown in Figure 6–37(a) and (b), respectively.
Lead Compensation Techniques Based on the Root-Locus Approach.
The root-locus approach to design is very powerful when the specifications are given in terms of time-domain quantities, such as the damping ratio and undamped natural frequency of the desired dominant closed-loop poles, maximum overshoot, rise time, and settling time.
Consider a design problem in which the original system either is unstable for all val-ues of gain or is stable but has undesirable transient-response characteristics. In such a case, the reshaping of the root locus is necessary in the broad neighborhood of the jv axis and the origin in order that the dominant closed-loop poles be at desired locations in the complex plane.This problem may be solved by inserting an appropriate lead com-pensator in cascade with the feedforward transfer function.
The procedures for designing a lead compensator for the system shown in Figure 6–38 by the root-locus method may be stated as follows: 1. From the performance specifications, determine the desired location for the dom-inant closed-loop poles.
R1 C1 6 R2 C2 R1 C1 7 R2 C2 R1 C1 6 R2 C2 .
a 6 1.
R1 C1 7 R2 C2 , K c a = R2 R4AR1 R3B.
K c a = R4 C1 R3 C2 R2 C2 R1 C1 = R2 R4 R1 R3 , a = R2 C2 R1 C1 jv s (a) 1 R2C2 – 1 R1C1 – jv s (b) 1 R2C2 – 1 R1C1 – 0 0 Figure 6–37 Pole-zero configurations: (a) lead network; (b) lag network.
Gc(s) G(s) + – Figure 6–38 Control system.
Section 6–6 / Lead Compensation 313 2. By drawing the root-locus plot of the uncompensated system (original system), ascertain whether or not the gain adjustment alone can yield the desired closed-loop poles. If not, calculate the angle deficiency f.This angle must be contributed by the lead compensator if the new root locus is to pass through the desired loca-tions for the dominant closed-loop poles.
3. Assume the lead compensator Gc(s) to be where a and T are determined from the angle deficiency. Kc is determined from the requirement of the open-loop gain.
4. If static error constants are not specified, determine the location of the pole and zero of the lead compensator so that the lead compensator will contribute the nec-essary angle f. If no other requirements are imposed on the system, try to make the value of a as large as possible.A larger value of a generally results in a larger value of Kv, which is desirable. Note that 5. Determine the value of Kc of the lead compensator from the magnitude condition.
Once a compensator has been designed, check to see whether all performance spec-ifications have been met. If the compensated system does not meet the performance specifications, then repeat the design procedure by adjusting the compensator pole and zero until all such specifications are met. If a large static error constant is required, cas-cade a lag network or alter the lead compensator to a lag–lead compensator.
Note that if the selected dominant closed-loop poles are not really dominant, or if the selected dominant closed-loop poles do not yield the desired result, it will be nec-essary to modify the location of the pair of such selected dominant closed-loop poles.
(The closed-loop poles other than dominant ones modify the response obtained from the dominant closed-loop poles alone.The amount of modification depends on the location of these remaining closed-loop poles.) Also, the closed-loop zeros affect the response if they are located near the origin.
EXAMPLE 6–6 Consider the position control system shown in Figure 6–39(a). The feedforward transfer function is The root-locus plot for this system is shown in Figure 6–39(b).The closed-loop transfer function for the system is = 10 (s + 0.5 + j3.1225)(s + 0.5 - j3.1225) C(s) R(s) = 10 s2 + s + 10 G(s) = 10 s(s + 1) Kv = lim sS0sGc(s)G(s) = Kca lim sS0sGc(s) Gc(s) = K c a Ts + 1 aTs + 1 = K c s + 1 T s + 1 aT , (0 6 a 6 1) 314 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method The closed-loop poles are located at The damping ratio of the closed-loop poles is .The undamped natural fre-quency of the closed-loop poles is 3.1623 radsec.Because the damping ratio is small, this system will have a large overshoot in the step response and is not desirable.
It is desired to design a lead compensator Gc(s) as shown in Figure 6–40(a) so that the dom-inant closed-loop poles have the damping ratio and the undamped natural frequency The desired location of the dominant closed-loop poles can be determined from as follows: s = -1.5 ; j2.5981 = (s + 1.5 + j2.5981)(s + 1.5 - j2.5981) s2 + 2zvns + vn 2 = s2 + 3s + 9 vn = 3radsec.
z = 0.5 vn = 210 = z = (12)210 = 0.1581 s = -0.5 ; j3.1225 R(s) C(s) (a) (b) 10 s(s + 1) G(s) Closed-loop pole jv 1 −3 −2 −1 j3 j2 j1 −j3 −j2 −j1 + – s Figure 6–39 (a) Control system; (b) root-locus plot.
(a) 10 s(s + 1) G(s) R(s) C(s) Gc(s) (b) Desired closed-loop pole jv 1 –3 –1.5 j2.5981 j2 j1 –j3 –j2 –j1 s 60° vn = 3 + – Figure 6–40 (a) Compensated system; (b) desired closed-loop pole location.
Section 6–6 / Lead Compensation 315 [See Figure 6–40 (b).] In some cases, after the root loci of the original system have been obtained, the dominant closed-loop poles may be moved to the desired location by simple gain adjustment.
This is, however, not the case for the present system.Therefore, we shall insert a lead compensator in the feedforward path.
A general procedure for determining the lead compensator is as follows: First, find the sum of the angles at the desired location of one of the dominant closed-loop poles with the open-loop poles and zeros of the original system, and determine the necessary angle f to be added so that the total sum of the angles is equal to The lead compensator must contribute this angle f. (If the angle f is quite large, then two or more lead networks may be needed rather than a single one.) Assume that the lead compensator Gc(s) has the transfer function as follows: The angle from the pole at the origin to the desired dominant closed-loop pole at s = –1.5+ j2.5981 is 120°.The angle from the pole at s=–1 to the desired closed-loop pole is 100.894°. Hence, the angle deficiency is Angle deficiency=180° – 120° – 100.894°=–40.894° Deficit angle 40.894° must be contributed by a lead compensator.
Note that the solution to such a problem is not unique. There are infinitely many solutions.
We shall present two solutions to the problem in what follows.
Method 1.
There are many ways to determine the locations of the zero and pole of the lead compensator. In what follows we shall introduce a procedure to obtain a largest possible value for a. (Note that a larger value of a will produce a larger value of Kv. In most cases, the larger the Kv is, the better the system performance.) First,draw a horizontal line passing through point P,the desired location for one of the dominant closed-loop poles.This is shown as line PA in Figure 6–41. Draw also a line connecting point P and the origin. Bisect the angle between the lines PA and PO, as shown in Figure 6–41.Draw two lines PC and PD that make angles with the bisector PB.The intersections of PC and PD with the negative real axis give the necessary locations for the pole and zero of the lead network.The compensator thus designed will make point P a point on the root locus of the compensated system.The open-loop gain is determined by use of the magnitude condition.
In the present system, the angle of G(s) at the desired closed-loop pole is n 10 s(s + 1) 2 s=-1.5+j2.5981 = -220.894° ;f2 Gc(s) = K c a Ts + 1 aTs + 1 = K c s + 1 T s + 1 aT , (0 6 a 6 1) ;180°(2k + 1).
jv s O A P C B D 1 aT – 1 T – f 2 f 2 Figure 6–41 Determination of the pole and zero of a lead network.
316 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method Thus, if we need to force the root locus to go through the desired closed-loop pole, the lead com-pensator must contribute f=40.894° at this point. By following the foregoing design procedure, we can determine the zero and pole of the lead compensator.
Referring to Figure 6–42, if we bisect angle APO and take 40.894°/2 each side, then the loca-tions of the zero and pole are found as follows: zero at s=–1.9432 pole at s=–4.6458 Thus, Gc(s) can be given as (For this compensator the value of a is a = 1.9432/4.6458 = 0.418.) The value of Kc can be determined by use of the magnitude condition.
or Hence, the lead compensator Gc(s) just designed is given by Then, the open-loop transfer function of the designed system becomes and the closed-loop transfer function becomes = 12.287s + 23.876 s3 + 5.646s2 + 16.933s + 23.876 C(s) R(s) = 12.287(s + 1.9432) s(s + 1)(s + 4.6458) + 12.287(s + 1.9432) Gc(s)G(s) = 1.2287 a s + 1.9432 s + 4.6458 b 10 s(s + 1) Gc(s) = 1.2287 s + 1.9432 s + 4.6458 Kc = 2 (s + 4.6458)s(s + 1) 10(s + 1.9432) 2 s=-1.5+j2.5981 = 1.2287 2Kc s + 1.9432 s + 4.6458 10 s(s + 1) 2 s=-1.5+j2.5981 = 1 Gc(s) = Kc s + 1 T s + 1 T = Kc s + 1.9432 s + 4.6458 jv 1 0 2 −1.9432 −4.6458 A P j3 j2 j1 −j2 −j1 s 20.447° 20.447° −3 Figure 6–42 Determination of the pole and zero of the lead compensator.
Section 6–6 / Lead Compensation 317 Figure 6–43 shows the root-locus plot for the designed system.
It is worthwhile to check the static velocity error constant Kv for the system just designed.
Note that the third closed-loop pole of the designed system is found by dividing the charac-teristic equation by the known factors as follows: The foregoing compensation method enables us to place the dominant closed-loop poles at the desired points in the complex plane. The third pole at s = 2.65 is fairly close to the added zero at 1.9432. Therefore, the effect of this pole on the transient response is relatively small.
Since no restriction has been imposed on the nondominant pole and no specification has been given concerning the value of the static velocity error coefficient, we conclude that the present de-sign is satisfactory.
Method 2.
If we choose the zero of the lead compensator at s = 1 so that it will cancel the plant pole at s = 1, then the compensator pole must be located at s = 3. (See Figure 6–44.) Hence the lead compensator becomes The value of Kc can be determined by use of the magnitude condition.
2Kc s + 1 s + 3 10 s(s + 1) 2 s=-1.5+j2.5981 = 1 Gc(s) = Kc s + 1 s + 3 s3 + 5.646s2 + 16.933s + 23.875 = (s + 1.5 + j2.5981)(s + 1.5 - j2.5981)(s + 2.65) = 5.139 = lim sS0 sc1.2287 s + 1.9432 s + 4.6458 10 s(s + 1) d Kv = lim sS0sGc(s)G(s) jv 1 –3 –1 –2 –4 –5 j2 j1 j3 –j3 –j2 –j1 s Figure 6–43 Root-locus plot of the designed system.
318 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method or Hence The open-loop transfer function of the designed system then becomes The closed-loop transfer function of the compensated system becomes Note that in the present case the zero of the lead compensator will cancel a pole of the plant, re-sulting in the second-order system,rather than the third-order system as we designed using Method 1.
The static velocity error constant for the present case is obtained as follows: Notice that the system designed by Method 1 gives a larger value of the static velocity error con-stant.This means that the system designed by Method 1 will give smaller steady-state errors in fol-lowing ramp inputs than the system designed by Method 2.
For different combinations of a zero and pole of the compensator that contributes 40.894°, the value of Kv will be different.Although a certain change in the value of Kv can be made by alter-ing the pole-zero location of the lead compensator, if a large increase in the value of Kv is desired, then we must alter the lead compensator to a lag–lead compensator.
Comparison of step and ramp responses of the compensated and uncompensated systems.
In what follows we shall compare the unit-step and unit-ramp responses of the three systems: the original uncompensated system, the system designed by Method 1, and the system designed by Method 2. The MATLAB program used to obtain unit-step response curves is given in = lim sS0 sc 9 s(s + 3) d = 3 Kv = lim sS0sGc(s)G(s) C(s) R(s) = 9 s2 + 3s + 9 Gc(s)G(s) = 0.9 s + 1 s + 3 10 s(s + 1) = 9 s(s + 3) Gc(s) = 0.9 s + 1 s + 3 Kc = 2 s(s + 3) 10 2 s=-1.5+j2.5981 = 0.9 jv 1 –3 –1 –2 –4 j2 j1 j3 –j2 –j1 s Desired closed-loop pole Compensator pole Compensator zero 60° 120° Figure 6–44 Compensator pole and zero.
Section 6–6 / Lead Compensation 319 MATLAB Program 6–9, where num1 and den1 denote the numerator and denominator of the system designed by Method 1 and num2 and den2 denote that designed by Method 2.Also, num and den are used for the original uncompensated system.The resulting unit-step response curves are shown in Figure 6–45.The MATLAB program to obtain the unit-ramp response curves of the MATLAB Program 6–9 % Unit-Step Response of Compensated and Uncompensated Systems num1 = [12.287 23.876]; den1 = [1 5.646 16.933 23.876]; num2 = ; den2 = [1 3 9]; num = ; den = [1 1 10]; t = 0:0.05:5; c1 = step(num1,den1,t); c2 = step(num2,den2,t); c = step(num,den,t); plot(t,c1,'-',t,c2,'.',t,c,'x') grid title('Unit-Step Responses of Compensated Systems and Uncompensated System') xlabel('t Sec') ylabel('Outputs c1, c2, and c') text(1.51,1.48,'Compensated System (Method 1)') text(0.9,0.48,'Compensated System (Method 2)') text(2.51,0.67,'Uncompensated System') Outputs c1, c2, and c 0.4 0.8 1.8 0 1 0.5 1.5 0 2 2.5 t Sec 3 3.5 4 4.5 5 1.2 0.6 1 0.2 1.4 1.6 Unit-Step Responses of Compensated Systems and Uncompensated System Compensated System (Method 1) Compensated System (Method 2) Uncompensated System Figure 6–45 Unit-step response curves of designed systems and original uncompensated system.
320 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method designed systems is given in MATLAB Program 6–10, where we used the step command to ob-tain unit-ramp responses by using the numerators and denominators for the systems designed by Method 1 and Method 2 as follows: num1 = [12.287 23.876] den1 = [1 5.646 16.933 23.876 0] num2 = den2 = [1 3 9 0] The resulting unit-ramp response curves are shown in Figure 6–46.
MATLAB Program 6–10 % Unit-Ramp Responses of Compensated Systems num1 = [12.287 23.876]; den1 = [1 5.646 16.933 23.876 0]; num2 = ; den2 = [1 3 9 0]; t = 0:0.05:5; c1 = step(num1,den1,t); c2 = step(num2,den2,t); plot(t,c1,'-',t,c2,'.',t,t,'-') grid title('Unit-Ramp Responses of Compensated Systems') xlabel('t Sec') ylabel('Unit-Ramp Input and Outputs c1 and c2') text(2.55,3.8,'Input') text(0.55,2.8,'Compensated System (Method 1)') text(2.35,1.75,'Compensated System (Method 2)') Unit-Ramp Input and Outputs c1 and c2 Unit-Ramp Responses of Compensated Systems Compensated System (Method 1) Input Compensated System (Method 2) 1 0.5 1.5 0 2 2.5 3 3.5 4 4.5 5 t Sec 5 2 0 3 4.5 1 0.5 4 2.5 3.5 1.5 Figure 6–46 Unit-ramp response curves of designed systems.
Section 6–7 / Lag Compensation 321 In examining these response curves notice that the compensated system designed by Method 1 exhibits a little bit larger overshoot in the step response than the compensated system designed by Method 2. However, the former has better response characteristics for the ramp input than the latter. So it is difficult to say which one is better.The decision on which one to choose should be made by the response requirements (such as smaller overshoots for step type inputs or smaller steady-state errors in following ramp or changing inputs) expected in the designed system. If both smaller overshoots in step inputs and smaller steady-state errors in following changing inputs are required, then we might use a lag–lead compensator. (See Section 6–8 for the lag–lead compen-sation techniques.) 6–7 LAG COMPENSATION Electronic Lag Compensator Using Operational Amplifiers.
The configuration of the electronic lag compensator using operational amplifiers is the same as that for the lead compensator shown in Figure 6–36. If we choose in the circuit shown in Figure 6–36, it becomes a lag compensator. Referring to Figure 6–36, the transfer function of the lag compensator is given by where Note that we use b instead of a in the above expressions. [In the lead compensator we used a to indicate the ratio which was less than 1, or 0<a<1.] In this book we always assume that 0<a<1 and b>1.
Lag Compensation Techniques Based on the Root-Locus Approach.
Consider the problem of finding a suitable compensation network for the case where the system exhibits satisfactory transient-response characteristics but unsatisfactory steady-state characteristics. Compensation in this case essentially consists of increasing the open-loop gain without appreciably changing the transient-response characteristics.This means that the root locus in the neighborhood of the dominant closed-loop poles should not be changed appreciably, but the open-loop gain should be increased as much as needed.
This can be accomplished if a lag compensator is put in cascade with the given feedforward transfer function.
To avoid an appreciable change in the root loci, the angle contribution of the lag network should be limited to a small amount, say less than 5°. To assure this, we place the pole and zero of the lag network relatively close together and near the origin of the s plane.Then the closed-loop poles of the compensated system will be shifted only slight-ly from their original locations. Hence, the transient-response characteristics will be changed only slightly.
R2 C2AR1 C1B, T = R1 C1 , bT = R2 C2 , b = R2 C2 R1 C1 7 1, K ˆ c = R4 C1 R3 C2 Eo(s) Ei(s) = K ˆ c b Ts + 1 bTs + 1 = K ˆ c s + 1 T s + 1 bT R2 C2 7 R1 C1 322 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method Consider a lag compensator where (6–19) If we place the zero and pole of the lag compensator very close to each other, then at where s1 is one of the dominant closed-loop poles,the magnitudes and are almost equal, or To make the angle contribution of the lag portion of the compensator small, we require This implies that if gain of the lag compensator is set equal to 1, the alteration in the transient-response characteristics will be very small,despite the fact that the overall gain of the open-loop transfer function is increased by a factor of b, where b>1. If the pole and zero are placed very close to the origin, then the value of b can be made large. (A large value of b may be used,provided physical realization of the lag compensator is possible.) It is noted that the value of T must be large, but its exact value is not critical. However, it should not be too large in order to avoid difficulties in realizing the phase-lag com-pensator by physical components.
An increase in the gain means an increase in the static error constants. If the open-loop transfer function of the uncompensated system is G(s), then the static velocity error constant Kv of the uncompensated system is If the compensator is chosen as given by Equation (6–19), then for the compensated system with the open-loop transfer function the static velocity error constant where Kv is the static velocity error constant of the uncompensated system.
Thus if the compensator is given by Equation (6–19), then the static velocity error constant is increased by a factor of where is approximately unity.
K ˆ c K ˆ c b, = K ˆ c bKv = lim sS0Gc(s)Kv K ˆ v = lim sS0sGc(s)G(s) K ˆ v becomes Gc(s)G(s) K v = lim sS0sG(s) K ˆ c -5° 6 n s1 + 1 T s1 + 1 bT 6 0° ∑GcAs1B∑= 4K ˆ c s1 + 1 T s1 + 1 bT 4 K ˆ c s1 + C1(bT)D s1 + (1T) s = s1 , Gc(s) = K ˆ c b Ts + 1 bTs + 1 = K ˆ c s + 1 T s + 1 bT Gc(s), Section 6–7 / Lag Compensation 323 The main negative effect of the lag compensation is that the compensator zero that will be generated near the origin creates a closed-loop pole near the origin.This closed-loop pole and compensator zero will generate a long tail of small amplitude in the step response, thus increasing the settling time.
Design Procedures for Lag Compensation by the Root-Locus Method.
The procedure for designing lag compensators for the system shown in Figure 6–47 by the root-locus method may be stated as follows (we assume that the uncompensated system meets the transient-response specifications by simple gain adjustment; if this is not the case, refer to Section 6–8): 1. Draw the root-locus plot for the uncompensated system whose open-loop trans-fer function is G(s). Based on the transient-response specifications, locate the dominant closed-loop poles on the root locus.
2. Assume the transfer function of the lag compensator to be given by Equation (6–19): Then the open-loop transfer function of the compensated system becomes 3. Evaluate the particular static error constant specified in the problem.
4. Determine the amount of increase in the static error constant necessary to satis-fy the specifications.
5. Determine the pole and zero of the lag compensator that produce the necessary increase in the particular static error constant without appreciably altering the original root loci. (Note that the ratio of the value of gain required in the spec-ifications and the gain found in the uncompensated system is the required ratio between the distance of the zero from the origin and that of the pole from the origin.) 6. Draw a new root-locus plot for the compensated system. Locate the desired dom-inant closed-loop poles on the root locus. (If the angle contribution of the lag net-work is very small—that is, a few degrees—then the original and new root loci are almost identical. Otherwise, there will be a slight discrepancy between them.Then locate, on the new root locus, the desired dominant closed-loop poles based on the transient-response specifications.) 7. Adjust gain of the compensator from the magnitude condition so that the dom-inant closed-loop poles lie at the desired location. A will be approximately 1.B K ˆ c K ˆ c Gc(s)G(s).
Gc(s) = K ˆ c b Ts + 1 bTs + 1 = K ˆ c s + 1 T s + 1 bT Gc(s) G(s) + – Figure 6–47 Control system.
324 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method EXAMPLE 6–7 Consider the system shown in Figure 6–48(a).The feedforward transfer function is The root-locus plot for the system is shown in Figure 6–48(b). The closed-loop transfer function becomes The dominant closed-loop poles are The damping ratio of the dominant closed-loop poles is The undamped natural frequency of the dominant closed-loop poles is 0.673 radsec.The static velocity error constant is 0.53 sec–1.
It is desired to increase the static velocity error constant Kv to about 5 sec–1 without appreciably changing the location of the dominant closed-loop poles.
To meet this specification, let us insert a lag compensator as given by Equation (6–19) in cascade with the given feedforward transfer function.To increase the static velocity error con-stant by a factor of about 10, let us choose b=10 and place the zero and pole of the lag com-pensator at s=–0.05 and s=–0.005, respectively.The transfer function of the lag compensator becomes Gc(s) = K ˆ c s + 0.05 s + 0.005 z = 0.491.
s = -0.3307 ; j0.5864 = 1.06 (s + 0.3307 - j0.5864)(s + 0.3307 + j0.5864)(s + 2.3386) C(s) R(s) = 1.06 s(s + 1)(s + 2) + 1.06 G(s) = 1.06 s(s + 1)(s + 2) 1.06 s(s + 1) (s + 2) Closed-loop poles j1 –j2 –j1 0 –1 –2 –3 1 jv s (a) (b) j2 + – Figure 6–48 (a) Control system; (b) root-locus plot.
Section 6–7 / Lag Compensation 325 The angle contribution of this lag network near a dominant closed-loop pole is about 4°. Because this angle contribution is not very small, there is a small change in the new root locus near the desired dominant closed-loop poles.
The open-loop transfer function of the compensated system then becomes where The block diagram of the compensated system is shown in Figure 6–49.The root-locus plot for the compensated system near the dominant closed-loop poles is shown in Figure 6–50(a),together with the original root-locus plot. Figure 6–50(b) shows the root-locus plot of the compensated system K = 1.06K ˆ c = K(s + 0.05) s(s + 0.005)(s + 1)(s + 2) Gc(s)G(s) = K ˆ c s + 0.05 s + 0.005 1.06 s(s + 1)(s + 2) Kc s + 0.05 s + 0.005 Kc = 0.966 1.06 s(s + 1) (s + 2) + – ^ ^ Figure 6–49 Compensated system.
Figure 6–50 (a) Root-locus plots of the compensated system and uncompensated system; (b) root-locus plot of compensated system near the origin.
Real Axis −2.5 −3 0 1 −0.5 0.5 −1.5 −2 −1 (a) Imag Axis 2 −2 1.5 −1 −1.5 1 0 0.5 −0.5 Root-Locus Plots of Compensated and Uncompensated Systems Uncompensated system Original closed-loop pole Compensated system New closed-loop pole 0.4 0.6 0.2 −0.2 −0.4 0 Root-Locus Plot of Compensated System near the Origin Real Axis Imag Axis −0.1 0.1 0.5 −0.3 −0.4 0.3 0 0.2 −0.2 0.4 −0.5 (b) 326 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method near the origin.The MATLAB program to generate the root-locus plots shown in Figures 6–50(a) and (b) is given in MATLAB Program 6–11.
If the damping ratio of the new dominant closed-loop poles is kept the same, then these poles are obtained from the new root-locus plot as follows: The open-loop gain K is determined from the magnitude condition as follows: = 1.0235 K = 2 s(s + 0.005)(s + 1)(s + 2) s + 0.05 2 s=-0.31+j0.55 s1 = -0.31 + j0.55, s2 = -0.31 - j0.55 MATLAB Program 6–11 % Root-locus plots of the compensated system and % uncompensated system % Enter the numerators and denominators of the % compensated and uncompensated systems numc = [1 0.05]; denc = [1 3.005 2.015 0.01 0]; num = [1.06]; den = [1 3 2 0]; % Enter rlocus command. Plot the root loci of both % systems rlocus(numc,denc) hold Current plot held rlocus(num,den) v = [-3 1 -2 2]; axis(v); axis('square') grid text(-2.8,0.2,'Compensated system') text(-2.8,1.2,'Uncompensated system') text(-2.8,0.58,'Original closed-loop pole') text(-0.1,0.85,'New closed-') text(-0.1,0.62,'loop pole') title('Root-Locus Plots of Compensated and Uncompensated Systems') hold Current plot released % Plot root loci of the compensated system near the origin rlocus(numc,denc) v = [-0.6 0.6 -0.6 0.6]; axis(v); axis('square') grid title('Root-Locus Plot of Compensated System near the Origin') Section 6–7 / Lag Compensation 327 Then the lag compensator gain is determined as Thus the transfer function of the lag compensator designed is (6–20) Then the compensated system has the following open-loop transfer function: The static velocity error constant Kv is In the compensated system, the static velocity error constant has increased to 5.12 sec–1, or 5.12/0.53=9.66 times the original value. (The steady-state error with ramp inputs has decreased to about 10% of that of the original system.) We have essentially accomplished the design objective of increasing the static velocity error constant to 5 sec–1.
Note that, since the pole and zero of the lag compensator are placed close together and are lo-cated very near the origin, their effect on the shape of the original root loci has been small. Except for the presence of a small closed root locus near the origin,the root loci of the compensated and the uncompensated systems are very similar to each other.However,the value of the static velocity error constant of the compensated system is 9.66 times greater than that of the uncompensated system.
The two other closed-loop poles for the compensated system are found as follows: The addition of the lag compensator increases the order of the system from 3 to 4, adding one ad-ditional closed-loop pole close to the zero of the lag compensator. (The added closed-loop pole at s=–0.0549 is close to the zero at s=–0.05.) Such a pair of a zero and pole creates a long tail of small amplitude in the transient response, as we will see later in the unit-step response. Since the pole at s=–2.326 is very far from the jv axis compared with the dominant closed-loop poles, the effect of this pole on the transient response is also small. Therefore, we may consider the closed-loop poles at to be the dominant closed-loop poles.
The undamped natural frequency of the dominant closed-loop poles of the compensated sys-tem is 0.631 radsec.This value is about 6% less than the original value, 0.673 radsec.This implies that the transient response of the compensated system is slower than that of the original system.
The response will take a longer time to settle down.The maximum overshoot in the step response will increase in the compensated system. If such adverse effects can be tolerated, the lag com-pensation as discussed here presents a satisfactory solution to the given design problem.
Next, we shall compare the unit-ramp responses of the compensated system against the uncompensated system and verify that the steady-state performance is much better in the compensated system than the uncompensated system.
To obtain the unit-ramp response with MATLAB, we use the step command for the system Since for the compensated system is = 1.0235s + 0.0512 s5 + 3.005s4 + 2.015s3 + 1.0335s2 + 0.0512s C(s) sR(s) = 1.0235(s + 0.05) sCs(s + 0.005)(s + 1)(s + 2) + 1.0235(s + 0.05)D C(s)CsR(s)D C(s)CsR(s)D.
s = -0.31 ; j0.55 s3 = -2.326, s4 = -0.0549 K v = lim sS0sG1(s) = 5.12 sec-1 = 5.12(20s + 1) s(200s + 1)(s + 1)(0.5s + 1) G1(s) = 1.0235(s + 0.05) s(s + 0.005)(s + 1)(s + 2) Gc(s) = 0.9656 s + 0.05 s + 0.005 = 9.656 20s + 1 200s + 1 K ˆ c = K 1.06 = 1.0235 1.06 = 0.9656 K ˆ c 328 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method we have numc = [1.0235 0.0512] denc = [1 3.005 2.015 1.0335 0.0512 0] Also, for the uncompensated system is Hence, num = [1.06] den = [1 3 2 1.06 0] MATLAB Program 6–12 produces the plot of the unit-ramp response curves. Figure 6–51 shows the result. Clearly, the compensated system shows much smaller steady-state error (one-tenth of the original steady-state error) in following the unit-ramp input. = 1.06 s4 + 3s3 + 2s2 + 1.06s C(s) sR(s) = 1.06 sCs(s + 1)(s + 2) + 1.06D C(s)CsR(s)D MATLAB Program 6–12 % Unit-ramp responses of compensated system and % uncompensated system % Unit-ramp response will be obtained as the unit-step % response of C(s)/[sR(s)] % Enter the numerators and denominators of C1(s)/[sR(s)] % and C2(s)/[sR(s)], where C1(s) and C2(s) are Laplace % transforms of the outputs of the compensated and un-% compensated systems, respectively. numc = [1.0235 0.0512]; denc = [1 3.005 2.015 1.0335 0.0512 0]; num = [1.06]; den = [1 3 2 1.06 0]; % Specify the time range (such as t= 0:0.1:50) and enter % step command and plot command. t = 0:0.1:50; c1 = step(numc,denc,t); c2 = step(num,den,t); plot(t,c1,'-',t,c2,'.',t,t,'--') grid text(2.2,27,'Compensated system'); text(26,21.3,'Uncompensated system'); title('Unit-Ramp Responses of Compensated and Uncompensated Systems') xlabel('t Sec'); ylabel('Outputs c1 and c2') Section 6–7 / Lag Compensation 329 t Sec 10 0 5 35 50 30 40 45 20 15 25 Outputs c1 and c2 50 0 15 5 35 25 30 20 45 40 10 Unit-Ramp Responses of Compensated and Uncompensated Systems Uncompensated system Compensated system Figure 6–51 Unit-ramp responses of compensated and uncompensated systems. [The compensator is given by Equation (6–20).] MATLAB Program 6–13 gives the unit-step response curves of the compensated and un-compensated systems. The unit-step response curves are shown in Figure 6–52. Notice that the lag-compensated system exhibits a larger maximum overshoot and slower response than the original uncompensated system. Notice that a pair of the pole at s=–0.0549 and zero at MATLAB Program 6–13 % Unit-step responses of compensated system and % uncompensated system % Enter the numerators and denominators of the % compensated and uncompensated systems numc = [1.0235 0.0512]; denc = [1 3.005 2.015 1.0335 0.0512]; num = [1.06]; den = [1 3 2 1.06]; % Specify the time range (such as t = 0:0.1:40) and enter % step command and plot command. t = 0:0.1:40; c1 = step(numc,denc,t); c2 = step(num,den,t); plot(t,c1,'-',t,c2,'.') grid text(13,1.12,'Compensated system') text(13.6,0.88,'Uncompensated system') title('Unit-Step Responses of Compensated and Uncompensated Systems') xlabel('t Sec') ylabel('Outputs c1 and c2') 330 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method Unit-Step Responses of Compensated and Uncompensated Systems Outputs c1 and c2 Uncompensated system Compensated system t Sec 5 0 30 40 25 35 15 10 20 1.4 0.4 0 1.2 0.8 1 0.6 0.2 Figure 6–52 Unit-step responses of compensated and uncompensated systems. [The compensator is given by Equation (6–20).] s=–0.05 generates a long tail of small amplitude in the transient response. If a larger maximum overshoot and a slower response are not desired, we need to use a lag–lead compensator as presented in Section 6–8.
Comments.
It is noted that under certain circumstances, however, both lead com-pensator and lag compensator may satisfy the given specifications (both transient-response specifications and steady-state specifications.) Then either compensation may be used.
6–8 LAG–LEAD COMPENSATION Lead compensation basically speeds up the response and increases the stability of the system. Lag compensation improves the steady-state accuracy of the system, but reduces the speed of the response.
If improvements in both transient response and steady-state response are desired, then both a lead compensator and a lag compensator may be used simultaneously.Rather than introducing both a lead compensator and a lag compensator as separate units, how-ever, it is economical to use a single lag–lead compensator.
Lag–lead compensation combines the advantages of lag and lead compensations.
Since the lag–lead compensator possesses two poles and two zeros, such a compensation increases the order of the system by 2, unless cancellation of pole(s) and zero(s) occurs in the compensated system.
Electronic Lag–Lead Compensator Using Operational Amplifiers.
Figure 6–53 shows an electronic lag–lead compensator using operational amplifiers. The transfer Section 6–8 / Lag-Lead Compensation 331 function for this compensator may be obtained as follows: The complex impedance Z1 is given by or Similarly, complex impedance Z2 is given by Hence, we have The sign inverter has the transfer function Thus the transfer function of the compensator shown in Figure 6–53 is (6–21) Let us define T 1 = AR1 + R3BC1 , T 1 g = R1 C1 , T 2 = R2 C2 , bT 2 = AR2 + R4BC2 Eo(s) Ei(s) = Eo(s) E(s) E(s) Ei(s) = R4 R6 R3 R5 c AR1 + R3BC1 s + 1 R1 C1 s + 1 d c R2 C2 s + 1 AR2 + R4BC2 s + 1 d Eo(s) E(s) = - R6 R5 E(s) Ei(s) = - Z2 Z1 = - R4 R3 AR1 + R3BC1 s + 1 R1 C1 s + 1 R2 C2 s + 1 AR2 + R4BC2 s + 1 Z2 = AR2 C2 s + 1BR4 AR2 + R4BC2 s + 1 Z1 = AR1 C1 s + 1BR3 AR1 + R3BC1 s + 1 1 Z1 = 1 R1 + 1 C1 s + 1 R3 – + – + C1 C2 R1 R5 Ei(s) Eo(s) E(s) Lag–lead network Sign inverter Z1 Z2 R2 R3 R4 R6 Figure 6–53 Lag–lead compensator.
332 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method Gc(s) G(s) + – Figure 6–54 Control system.
Then Equation (6–21) becomes (6–22) where Note that g is often chosen to be equal to b.
Lag–lead Compensation Techniques Based on the Root-Locus Approach.
Consider the system shown in Figure 6–54.Assume that we use the lag–lead compensator: (6–23) where and (Consider Kc to belong to the lead portion of the lag–lead compensator.) In designing lag–lead compensators, we consider two cases where and Case 1.
In this case, the design process is a combination of the design of the lead compensator and that of the lag compensator.The design procedure for the lag–lead compensator follows: 1. From the given performance specifications, determine the desired location for the dominant closed-loop poles.
2. Using the uncompensated open-loop transfer function G(s), determine the angle deficiency f if the dominant closed-loop poles are to be at the desired location.The phase-lead portion of the lag–lead compensator must contribute this angle f.
3. Assuming that we later choose sufficiently large so that the magnitude of the lag portion 4 s1 + 1 T 2 s1 + 1 bT 2 4 T 2 g Z b.
g = b.
g Z b g 7 1.
b 7 1 Gc(s) = K c b g AT 1 s + 1BAT 2 s + 1B a T 1 g s + 1 b AbT 2 s + 1B = K c ± s + 1 T 1 s + g T 1 ≤± s + 1 T 2 s + 1 bT 2 ≤ g = R1 + R3 R1 7 1, b = R2 + R4 R2 7 1, K c = R2 R4 R6 R1 R3 R5 R1 + R3 R2 + R4 Eo(s) Ei(s) = K c b g £ T 1 s + 1 T 1 g s + 1 ≥a T 2 s + 1 bT 2 s + 1 b = K c as + 1 T 1 b as + 1 T 2 b a s + g T 1 b as + 1 bT 2 b Section 6–8 / Lag-Lead Compensation 333 is approximately unity, where is one of the dominant closed-loop poles, choose the values of and g from the requirement that The choice of and g is not unique. (Infinitely many sets of and g are possible.) Then determine the value of Kc from the magnitude condition: 4. If the static velocity error constant Kv is specified, determine the value of b to satisfy the requirement for Kv.The static velocity error constant Kv is given by where Kc and g are already determined in step 3. Hence, given the value of Kv, the value of b can be determined from this last equation. Then, using the value of b thus deter-mined, choose the value of such that (The preceding design procedure is illustrated in Example 6–8.) Case 2.
If g=b is required in Equation (6–23), then the preceeding design procedure for the lag–lead compensator may be modified as follows: 1. From the given performance specifications, determine the desired location for the dominant closed-loop poles.
g = b.
-5° 6 n s1 + 1 T 2 s1 + 1 bT 2 6 0° 4 s1 + 1 T 2 s1 + 1 bT 2 4 1 T 2 = lim sS0sK c b g G(s) = lim sS0sK c ± s + 1 T 1 s + g T 1 ≤± s + 1 T 2 s + 1 bT 2 ≤G(s) K v = lim sS0sGc(s)G(s) 4K c s1 + 1 T 1 s1 + g T 1 GAs1B 4 = 1 T 1 T 1 n s1 + 1 T 1 s1 + g T 1 = f T 1 s = s1 334 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method 2. The lag–lead compensator given by Equation (6–23) is modified to (6–24) where b>1. The open-loop transfer function of the compensated system is Gc(s)G(s). If the static velocity error constant Kv is specified, determine the value of constant Kc from the following equation: 3. To have the dominant closed-loop poles at the desired location, calculate the angle contribution f needed from the phase-lead portion of the lag–lead compensator.
4. For the lag–lead compensator, we later choose sufficiently large so that is approximately unity, where is one of the dominant closed-loop poles. De-termine the values of and b from the magnitude and angle conditions: 5. Using the value of b just determined, choose so that The value of the largest time constant of the lag–lead compensator, should not be too large to be physically realized. (An example of the design of the lag–lead compen-sator when is given in Example 6–9.) g = b bT 2 , -5° 6 n s1 + 1 T 2 s1 + 1 bT 2 6 0° 4 s1 + 1 T 2 s1 + 1 bT 2 4 1 T 2 n s1 + 1 T 1 s1 + b T 1 = f 4K c ± s1 + 1 T 1 s1 + b T 1 ≤GAs1B 4 = 1 T 1 s = s1 4 s1 + 1 T 2 s1 + 1 bT 2 4 T 2 = lim sS0sK cG(s) K v = lim sS0sGc(s)G(s) Gc(s) = K c AT 1 s + 1BAT 2 s + 1B a T 1 b s + 1 b AbT 2 s + 1B = K c as + 1 T 1 b as + 1 T 2 b as + b T 1 b as + 1 bT 2 b Section 6–8 / Lag-Lead Compensation 335 EXAMPLE 6–8 Consider the control system shown in Figure 6–55.The feedforward transfer function is This system has closed-loop poles at The damping ratio is 0.125, the undamped natural frequency is 2 rad/sec, and the static velocity error constant is 8 sec–1.
It is desired to make the damping ratio of the dominant closed-loop poles equal to 0.5 and to increase the undamped natural frequency to 5 radsec and the static velocity error constant to 80 sec–1. Design an appropriate compensator to meet all the performance specifications.
Let us assume that we use a lag–lead compensator having the transfer function where g is not equal to b.Then the compensated system will have the open-loop transfer function From the performance specifications, the dominant closed-loop poles must be at Since the phase-lead portion of the lag–lead compensator must contribute 55° so that the root locus passes through the desired location of the dominant closed-loop poles.
To design the phase-lead portion of the compensator, we first determine the location of the zero and pole that will give 55° contribution. There are many possible choices, but we shall here choose the zero at s=–0.5 so that this zero will cancel the pole at s=–0.5 of the plant. Once the zero is chosen, the pole can be located such that the angle contribution is 55°. By simple calculation or graphical analysis, the pole must be located at s=–5.02. Thus, the phase-lead portion of the lag–lead compensator becomes Kc s + 1 T1 s + g T1 = Kc s + 0.5 s + 5.02 n 4 s(s + 0.5) 2 s=-2.50+j4.33 = -235° s = -2.50 ; j4.33 Gc(s)G(s) = K c ± s + 1 T 1 s + g T 1 ≤± s + 1 T 2 s + 1 bT 2 ≤G(s) Gc(s) = K c ± s + 1 T 1 s + g T 1 ≤± s + 1 T 2 s + 1 bT 2 ≤ (g 7 1, b 7 1) s = -0.2500 ; j1.9843 G(s) = 4 s(s + 0.5) 4 s(s + 0.5) + – Figure 6–55 Control system.
336 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method Thus Next we determine the value of Kc from the magnitude condition: Hence, The phase-lag portion of the compensator can be designed as follows: First the value of b is determined to satisfy the requirement on the static velocity error constant: Hence, b is determined as Finally, we choose the value such that the following two conditions are satisfied: We may choose several values for T2 and check if the magnitude and angle conditions are satis-fied.After simple calculations we find for T2 = 5 Since T2 = 5 satisfies the two conditions, we may choose Now the transfer function of the designed lag–lead compensator is given by = 10(2s + 1)(5s + 1) (0.1992s + 1)(80.19s + 1) = 6.26 a s + 0.5 s + 5.02 b a s + 0.2 s + 0.01247 b Gc(s) = (6.26) ± s + 1 2 s + 10.04 2 ≤± s + 1 5 s + 1 16.04 5 ≤ T 2 = 5 1 7 magnitude 7 0.98, -2.10° 6 angle 6 0° -5° 6 n s + 1 T 2 s + 1 16.04T 2 4 s=-2.5+j4.33 6 0° 4 s + 1 T 2 s + 1 16.04T 2 4 s=-2.5+j4.33 1, T 2 b = 16.04 = lim sS0s(6.26) b 10.04 4 s(s + 0.5) = 4.988b = 80 K v = lim sS0sGc(s)G(s) = lim sS0sK c b g G(s) Kc = 2 (s + 5.02)s 4 2 s=-2.5+j4.33 = 6.26 2Kc s + 0.5 s + 5.02 4 s(s + 0.5) 2 s=-2.5+j4.33 = 1 T1 = 2, g = 5.02 0.5 = 10.04 Section 6–8 / Lag-Lead Compensation 337 The compensated system will have the open-loop transfer function Because of the cancellation of the (s+0.5) terms,the compensated system is a third-order system.
(Mathematically, this cancellation is exact, but practically such cancellation will not be exact be-cause some approximations are usually involved in deriving the mathematical model of the sys-tem and, as a result, the time constants are not precise.) The root-locus plot of the compensated system is shown in Figure 6–56(a).An enlarged view of the root-locus plot near the origin is shown in Figure 6–56(b). Because the angle contribution of the phase lag portion of the lag–lead compensator is quite small, there is only a small change in the location of the dominant closed-loop poles from the desired location, The characteristic equation for the com-pensated system is or Hence the new closed-loop poles are located at The new damping ratio is z=0.491.Therefore the compensated system meets all the required per-formance specifications. The third closed-loop pole of the compensated system is located at Since this closed-loop pole is very close to the zero at the effect of this pole on the response is small. (Note that, in general, if a pole and a zero lie close to each other on the negative real axis near the origin, then such a pole-zero combination will yield a long tail of small amplitude in the transient response.) s = -0.2, s = -0.2078.
s = -2.4123 ; j4.2756 = (s + 2.4123 + j4.2756)(s + 2.4123 - j4.2756)(s + 0.2078) = 0 s3 + 5.0325s2 + 25.1026s + 5.008 s(s + 5.02)(s + 0.01247) + 25.04(s + 0.2) = 0 s = -2.5 ; j4.33.
Gc(s)G(s) = 25.04(s + 0.2) s(s + 5.02)(s + 0.01247) Root-Locus Plot of Compensated System Imag Axis 10 5 –5 –10 0 Real Axis (a) –2 2 8 6 0 4 10 –10 –4 –6 –8 Root-Locus Plot of Compensated System near the Origin RealAxis Imag Axis –0.5 0 –0.1 –0.3 –0.4 –0.2 –0.05 0.05 0.2 –0.15 –0.25 0.15 0 0.1 –0.1 0.25 –0.2 (b) Figure 6–56 (a) Root-locus plot of the compensated system; (b) root-locus plot near the origin.
338 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method t Sec 1 0 6 8 5 7 3 2 4 (a) Outputs 0.4 0.8 1.8 0 1.2 0.6 1 0.2 1.4 1.6 Unit-Step Responses of Compensated and Uncompensated Systems Uncompensated system Compensated system Steady-state error of compensated system = 0.0125 Steady-state error of uncompensated system = 0.125 Compensated system Uncompensated system t Sec 2 1 7 9 6 8 10 0 4 3 5 (b) Outputs 10 4 0 6 9 2 1 8 5 7 3 Unit-Ramp Responses of Compensated and Uncompensated Systems Figure 6–57 Transient-response curves for the compensated system and uncompensated system. (a) Unit-step response curves; (b) unit-ramp response curves.
The unit-step response curves and unit-ramp response curves before and after compensation are shown in Figure 6–57. (Notice a long tail of a small amplitude in the unit-step response of the compensated system.) EXAMPLE 6–9 Consider the control system of Example 6–8 again. Suppose that we use a lag–lead compensator of the form given by Equation (6–24), or Gc(s) = K c as + 1 T 1 b as + 1 T 2 b as + b T 1 b as + 1 bT 2 b (b 7 1) Section 6–8 / Lag-Lead Compensation 339 Assuming the specifications are the same as those given in Example 6–8, design a compensator Gc(s).
The desired locations for the dominant closed-loop poles are at The open-loop transfer function of the compensated system is Since the requirement on the static velocity error constant Kv is 80 sec–1, we have Thus The time constant and the value of b are determined from (The angle deficiency of 55° was obtained in Example 6–8.) Referring to Figure 6–58, we can easily locate points A and B such that (Use a graphical approach or a trigonometric approach.) The result is or The phase-lead portion of the lag–lead network thus becomes For the phase-lag portion, we choose such that it satisfies the conditions -5° 6 n s + 1 T2 s + 1 3.503T2 4 s=-2.50+j4.33 6 0° 4 s + 1 T2 s + 1 3.503T2 4 s=-2.50+j4.33 1, T 2 10 a s + 2.38 s + 8.34 b T 1 = 1 2.38 = 0.420, b = 8.34T 1 = 3.503 AO = 2.38, BO = 8.34 /APB = 55°, PA PB = 4.77 8 n s + 1 T 1 s + b T 1 4 s=-2.5+j4.33 = 55° 4 s + 1 T 1 s + b T 1 4 2 40 s(s + 0.5) 2 s=-2.5+j4.33 = 4 s + 1 T 1 s + b T 1 4 8 4.77 = 1 T 1 K c = 10 K v = lim sS0sGc(s)G(s) = lim sS0K c 4 0.5 = 8K c = 80 Gc(s)G(s) = K c as + 1 T 1 b as + 1 T 2 b as + b T 1 b as + 1 bT 2 b 4 s(s + 0.5) s = -2.50 ; j4.33 340 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method 0 jv s A B P 55° j5 j4 j3 j2 j1 –j4 –j3 –j2 –j1 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 1 2 Figure 6–58 Determination of the desired pole-zero location.
By simple calculations, we find that if we choose , then and if we choose , then Since is one of the time constants of the lag–lead compensator, it should not be too large. If can be acceptable from practical viewpoint, then we may choose . Then Thus, the lag–lead compensator becomes The compensated system will have the open-loop transfer function No cancellation occurs in this case, and the compensated system is of fourth order. Because the angle contribution of the phase lag portion of the lag–lead network is quite small, the dominant closed-loop poles are located very near the desired location. In fact, the location of the dominant closed-loop poles can be found from the characteristic equation as follows: The characteristic equation of the compensated system is which can be simplified to = (s + 2.4539 + j4.3099)(s + 2.4539 - j4.3099)(s + 0.1003)(s + 3.8604) = 0 s4 + 8.8685s3 + 44.4219s2 + 99.3188s + 9.52 (s + 8.34)(s + 0.0285)s(s + 0.5) + 40(s + 2.38)(s + 0.1) = 0 Gc(s)G(s) = 40(s + 2.38)(s + 0.1) (s + 8.34)(s + 0.0285)s(s + 0.5) Gc(s) = (10) a s + 2.38 s + 8.34 b a s + 0.1 s + 0.0285 b 1 bT 2 = 1 3.503 10 = 0.0285 T 2 = 10 T 2 = 10 T 2 1 7 magnitude 7 0.99, -1° 6 angle 6 0° T 2 = 10 1 7 magnitude 7 0.98, -1.5° 6 angle 6 0° T 2 = 5 Section 6–8 / Lag-Lead Compensation 341 The dominant closed-loop poles are located at The other closed-loop poles are located at Since the closed-loop pole at is very close to a zero at they almost cancel each other.Thus, the effect of this closed-loop pole is very small.The remaining closed-loop pole does not quite cancel the zero at The effect of this zero is to cause a larger overshoot in the step response than a similar system without such a zero. The unit-step response curves of the compensated and uncompensated systems are shown in Figure 6–59(a).The unit-ramp response curves for both systems are depicted in Figure 6–59(b).
s = -2.4.
(s = -3.8604) s = -0.1, s = -0.1003 s = -0.1003; s = -3.8604 s = -2.4539 ; j4.3099 t Sec 1 0.5 0 3.5 4.5 3 4 5 2 1.5 2.5 (a) Outputs 0.4 0.8 1.8 0 1.2 0.6 1 0.2 1.4 1.6 Unit-Step Responses of Compensated and Uncompensated Systems Uncompensated system Compensated system t Sec 0.5 0 3 4 2.5 3.5 1.5 1 2 (b) Outputs 1.5 2.5 4 0.5 0 3.5 2 3 1 Unit-Ramp Responses of Compensated and Uncompensated Systems Uncompensated system Compensated system Figure 6–59 (a) Unit-step response curves for the compensated and uncompensated systems; (b) unit-ramp response curves for both systems.
342 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method The maximum overshoot in the step response of the compensated system is approximately 38%. (This is much larger than the maximum overshoot of 21% in the design presented in Exam-ple 6–8.) It is possible to decrease the maximum overshoot by a small amount from 38%, but not to 20% if g=b is required, as in this example. Note that by not requiring g=b, we have an ad-ditional parameter to play with and thus can reduce the maximum overshoot.
6–9 PARALLEL COMPENSATION Thus far we have presented series compensation techniques using lead, lag, or lag–lead compensators.In this section we discuss parallel compensation technique.Because in the parallel compensation design the controller (or compensator) is in a minor loop, the de-sign may seem to be more complicated than in the series compensation case. It is, how-ever, not complicated if we rewrite the characteristic equation to be of the same form as the characteristic equation for the series compensated system. In this section we pres-ent a simple design problem involving parallel compensation.
Basic Principle for Designing Parallel Compensated System.
Referring to Figure 6–60(a), the closed-loop transfer function for the system with series compensa-tion is The characteristic equation is Given G and H, the design problem becomes that of determining the compensator Gc that satisfies the given specification.
1 + Gc GH = 0 C R = Gc G 1 + Gc GH G1(s) G2(s) H(s) Gc(s) Gc(s) G(s) H(s) (a) (b) + – + – + – C R R C Figure 6–60 (a) Series compensation; (b) parallel or feedback compensation.
Section 6–9 / Parallel Compensation 343 The closed-loop transfer function for the system with parallel compensation [Figure 6–60(b)] is The characteristic equation is By dividing this characteristic equation by the sum of the terms that do not involve Gc, we obtain (6–25) If we define then Equation (6–25) becomes Since Gf is a fixed transfer function, the design of Gc becomes the same as the case of series compensation. Hence the same design approach applies to the parallel compen-sated system.
Velocity Feedback Systems.
A velocity feedback system (tachometer feedback system) is an example of parallel compensated systems.The controller (or compensator) in such a system is a gain element.The gain of the feedback element in a minor loop must be determined properly so that the entire system satisfies the given design specifica-tions.The characteristic of such a velocity feedback system is that the variable parame-ter does not appear as a multiplying factor in the open-loop transfer function, so that direct application of the root-locus design technique is not possible. However, by rewrit-ing the characteristic equation such that the variable parameter appears as a multiply-ing factor, then the root-locus approach to the design is possible.
An example of control system design using parallel compensation technique is pre-sented in Example 6–10.
EXAMPLE 6–10 Consider the system shown in Figure 6–61. Draw a root-locus diagram.Then determine the value of k such that the damping ratio of the dominant closed-loop poles is 0.4.
Here the system involves velocity feedback.The open-loop transfer function is Open-loop transfer function = 20 s(s + 1)(s + 4) + 20ks 1 + Gc Gf = 0 Gf = G2 1 + G1 G2 H 1 + Gc G2 1 + G1 G2 H = 0 1 + G1 G2 H + G2 Gc = 0 C R = G1 G2 1 + G2 Gc + G1 G2 H C(s) R(s) 20 (s + 1) (s + 4) 1 s + – + – k Figure 6–61 Control system.
344 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method Notice that the adjustable variable k does not appear as a multiplying factor. The characteristic equation for the system is (6–26) Define Then Equation (6–26) becomes (6–27) Dividing both sides of Equation (6–27) by the sum of the terms that do not contain K, we get or (6–28) Equation (6–28) is of the form of Equation (6–11).
We shall now sketch the root loci of the system given by Equation (6–28). Notice that the open-loop poles are located at s=j2, s=–j2, s=–5, and the open-loop zero is located at s=0.
The root locus exists on the real axis between 0 and –5. Since we have The intersection of the asymptotes with the real axis can be found from as The angle of departure (angle u) from the pole at s=j2 is obtained as follows: Thus, the angle of departure from the pole s=j2 is 158.2°. Figure 6–62 shows a root-locus plot for the system. Notice that two branches of the root locus originate from the poles at and terminate on the zeros at infinity. The remaining one branch originates from the pole at s=–5 and terminates on the zero at s=0.
Note that the closed-loop poles with z=0.4 must lie on straight lines passing through the origin and making the angles with the negative real axis. In the present case, there are two intersections of the root-locus branch in the upper half s plane and the straight line of angle 66.42°.
Thus,two values of K will give the damping ratio z of the closed-loop poles equal to 0.4.At point P, the value of K is Hence k = K 20 = 0.4490 at point P K = 2 (s + j2)(s - j2)(s + 5) s 2 s=-1.0490+j2.4065 = 8.9801 ;66.42° s = ;j2 u = 180° - 90° - 21.8° + 90° = 158.2° s = -2.5 lim sS q Ks s3 + 5s2 + 4s + 20 = lim sS q K s2 + 5s + p = lim sS q K (s + 2.5)2 Angles of asymptote = ;180°(2k + 1) 2 = ; 90° lim sS q Ks (s + j2)(s - j2)(s + 5) = lim sS q K s2 1 + Ks (s + j2)(s - j2)(s + 5) = 0 1 + Ks s3 + 5s2 + 4s + 20 = 0 s3 + 5s2 + 4s + Ks + 20 = 0 20k = K s3 + 5s2 + 4s + 20ks + 20 = 0 Section 6–9 / Parallel Compensation 345 At point Q, the value of K is Hence Thus, we have two solutions for this problem. For k=0.4490, the three closed-loop poles are located at For k=1.4130, the three closed-loop poles are located at It is important to point out that the zero at the origin is the open-loop zero, but not the closed-loop zero. This is evident, because the original system shown in Figure 6–61 does not have a closed-loop zero, since G(s) R(s) = 20 s(s + 1)(s + 4) + 20(1 + ks) s = -2.1589 + j4.9652, s = -2.1589 - j4.9652, s = -0.6823 s = -1.0490 + j2.4065, s = -1.0490 - j2.4065, s = -2.9021 k = K 20 = 1.4130 at point Q K = 2 (s + j2)(s - j2)(s + 5) s 2 s=-2.1589+j4.9652 = 28.260 jv j6 j5 j4 j3 j2 j1 –j6 –j5 –j4 –j3 –j2 –j1 s –1 1 0 –2 –3 –4 –5 –6 –7 s = –2.1589 + j4.9652 Q P s = –1.0490 + j2.4065 s = –2.9021 66.42° Figure 6–62 Root-locus plot for the system shown in Figure 6–61.
346 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method The open-loop zero at s=0 was introduced in the process of modifying the characteristic equa-tion such that the adjustable variable K=20k was to appear as a multiplying factor.
We have obtained two different values of k to satisfy the requirement that the damping ratio of the dominant closed-loop poles be equal to 0.4. The closed-loop transfer function with k=0.4490 is given by The closed-loop transfer function with k=1.4130 is given by Notice that the system with k=0.4490 has a pair of dominant complex-conjugate closed-loop poles, while in the system with k=1.4130 the real closed-loop pole at s=–0.6823 is dominant, and the complex-conjugate closed-loop poles are not dominant. In this case, the response char-acteristic is primarily determined by the real closed-loop pole.
Let us compare the unit-step responses of both systems. MATLAB Program 6–14 may be used for plotting the unit-step response curves in one diagram. The resulting unit-step response curves for k=0.4490 and c2(t) for k=1.4130D are shown in Figure 6–63.
Cc1(t) = 20 (s + 2.1589 + j4.9652)(s + 2.1589 - j4.9652)(s + 0.6823) C(s) R(s) = 20 s3 + 5s2 + 32.26s + 20 = 20 (s + 1.0490 + j2.4065)(s + 1.0490 - j2.4065)(s + 2.9021) C(s) R(s) = 20 s3 + 5s2 + 12.98s + 20 MATLAB Program 6–14 % ---------- Unit-step response ----------% Enter numerators and denominators of systems with % k = 0.4490 and k = 1.4130, respectively. num1 = ; den1 = [1 5 12.98 20]; num2 = ; den2 = [1 5 32.26 20]; t = 0:0.1:10; c1 = step(num1,den1,t); c2 = step(num2,den2,t); plot(t,c1,t,c2) text(2.5,1.12,'k = 0.4490') text(3.7,0.85,'k = 1.4130') grid title('Unit-step Responses of Two Systems') xlabel('t Sec') ylabel('Outputs c1 and c2') Example Problems and Solutions 347 From Figure 6–63 we notice that the response of the system with k=0.4490 is oscillatory.
(The effect of the closed-loop pole at s=–2.9021 on the unit-step response is small.) For the system with k=1.4130, the oscillations due to the closed-loop poles at damp out much faster than purely exponential response due to the closed-loop pole at s=–0.6823.
The system with k=0.4490 (which exhibits a faster response with relatively small overshoot) has a much better response characteristic than the system with k=1.4130 (which exhibits a slow overdamped response).Therefore, we should choose k=0.4490 for the present system.
EXAMPLE PROBLEMS AND SOLUTIONS A–6–1.
Sketch the root loci for the system shown in Figure 6–64(a). (The gain K is assumed to be posi-tive.) Observe that for small or large values of K the system is overdamped and for medium val-ues of K it is underdamped.
Solution. The procedure for plotting the root loci is as follows: 1. Locate the open-loop poles and zeros on the complex plane. Root loci exist on the negative real axis between 0 and –1 and between –2 and –3.
2. The number of open-loop poles and that of finite zeros are the same.This means that there are no asymptotes in the complex region of the s plane.
3. Determine the breakaway and break-in points.The characteristic equation for the system is or K = -s(s + 1) (s + 2)(s + 3) 1 + K(s + 2)(s + 3) s(s + 1) = 0 s = -2.1589 ; j4.9652 t Sec 0 1 10 5 2 3 4 6 7 8 9 Outputs c1 and c2 1.2 0.4 0 0.6 0.2 1 0.8 Unit-Step Responses of Two Systems k = 1.4130 k = 0.4490 Figure 6–63 Unit-step response curves for the system shown in Figure 6–61 when the damping ratio z of the dominant closed-loop poles is set equal to 0.4. (Two possible values of k give the damping ratio z equal to 0.4.) 348 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method The breakaway and break-in points are determined from as follows: Notice that both points are on root loci. Therefore, they are actual breakaway or break-in points.At point s=–0.634, the value of K is Similarly, at s=–2.366, (Because point s=–0.634 lies between two poles, it is a breakaway point, and because point s=–2.366 lies between two zeros, it is a break-in point.) K = -(–2.366)(–1.366) (–0.366)(0.634) = 14 K = -(-0.634)(0.366) (1.366)(2.366) = 0.0718 s = -0.634, s = -2.366 = 0 = - 4(s + 0.634)(s + 2.366) C(s + 2)(s + 3)D 2 dK ds = - (2s + 1)(s + 2)(s + 3) - s(s + 1)(2s + 5) C(s + 2)(s + 3)D 2 (a) (b) R(s) C(s) jv s K = 0.0718 K = 14 –3 –2 –1 0 1 j1 j2 –j1 –j2 K(s + 2) s + 3 s(s + 1) + – Figure 6–64 (a) Control system; (b) root-locus plot.
Example Problems and Solutions 349 4. Determine a sufficient number of points that satisfy the angle condition. (It can be found that the root loci involve a circle with center at –1.5 that passes through the breakaway and break-in points.) The root-locus plot for this system is shown in Figure 6–64(b).
Note that this system is stable for any positive value of K since all the root loci lie in the left-half s plane.
Small values of K (0<K<0.0718) correspond to an overdamped system. Medium values of K (0.0718<K<14) correspond to an underdamped system. Finally, large values of K (14<K) correspond to an overdamped system.With a large value of K, the steady state can be reached in much shorter time than with a small value of K.
The value of K should be adjusted so that system performance is optimum according to a given performance index.
A–6–2.
Sketch the root loci for the system shown in Figure 6–65(a).
Solution. A root locus exists on the real axis between points s=–1 and s=–3.6. The asymp-totes can be determined as follows: The intersection of the asymptotes and the real axis is found from s = - 0 + 0 + 3.6 - 1 3 - 1 = -1.3 Angles of asymptotes = ;180°(2k + 1) 3 - 1 = 90°, -90° (a) (b) jv s –4 –3 –2 0 1 j3 j1 –j1 –j3 –1 –j2 j2 K(s + 1) s2(s + 3.6) + – Figure 6–65 (a) Control system; (b) root-locus plot.
350 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method Since the characteristic equation is we have The breakaway and break-in points are found from or from which we get Point s=0 corresponds to the actual breakaway point. But points are neither breakaway nor break-in points, because the corresponding gain values K become complex quantities.
To check the points where root-locus branches may cross the imaginary axis, substitute into the characteristic equation, yielding.
or Notice that this equation can be satisfied only if v=0, K=0. Because of the presence of a dou-ble pole at the origin, the root locus is tangent to the jv axis at v=0. The root-locus branches do not cross the jv axis. Figure 6–65(b) is a sketch of the root loci for this system.
A–6–3.
Sketch the root loci for the system shown in Figure 6–66(a).
Solution. A root locus exists on the real axis between point s=–0.4 and s=–3.6. The angles of asymptotes can be found as follows: The intersection of the asymptotes and the real axis is obtained from Next we shall find the breakaway points. Since the characteristic equation is we have K = - s3 + 3.6s2 s + 0.4 s3 + 3.6s2 + Ks + 0.4K = 0 s = - 0 + 0 + 3.6 - 0.4 3 - 1 = -1.6 Angles of asymptotes = ;180°(2k + 1) 3 - 1 = 90°, -90° AK - 3.6v2B + jvAK - v2B = 0 (jv)3 + 3.6(jv)2 + Kjv + K = 0 s = jv s = 1.65 ; j0.9367 s = 0, s = -1.65 + j0.9367, s = -1.65 - j0.9367 s3 + 3.3s2 + 3.6s = 0 dK ds = -A3s2 + 7.2sB(s + 1) - As3 + 3.6s2B (s + 1)2 = 0 K = - s3 + 3.6s2 s + 1 s3 + 3.6s2 + K(s + 1) = 0 Example Problems and Solutions 351 The breakaway and break-in points are found from from which we get or Thus, the breakaway or break-in points are at s=0 and s=–1.2. Note that s=–1.2 is a double root.When a double root occurs in at point s=–1.2, at this point.The value of gain K at point s=–1.2 is This means that with K=4.32 the characteristic equation has a triple root at point s=–1.2. This can be easily verified as follows: s3 + 3.6s2 + 4.32s + 1.728 = (s + 1.2)3 = 0 K = - s3 + 3.6s2 s + 4 2 s=-1.2 = 4.32 d2KAds2B = 0 dKds = 0 s(s + 1.2)2 = 0 s3 + 2.4s2 + 1.44s = 0 dK ds = -A3s2 + 7.2sB(s + 0.4) - As3 + 3.6s2B (s + 0.4)2 = 0 (a) (b) jv s –4 –3 –2 0 1 j3 j1 –j1 –j3 –j2 j2 K(s + 0.4) s2(s + 3.6) –1 –60° 60° + – Figure 6–66 (a) Control system; (b) root-locus plot.
352 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method Hence, three root-locus branches meet at point s=–1.2. The angles of departures at point s=–1.2 of the root locus branches that approach the asymptotes are that is, 60° and –60°. (See Problem A–6–4.) Finally,we shall examine if root-locus branches cross the imaginary axis.By substituting s=jv into the characteristic equation, we have or This equation can be satisfied only if v=0, K=0. At point v=0, the root locus is tangent to the jv axis because of the presence of a double pole at the origin.There are no points where root-locus branches cross the imaginary axis.
A sketch of the root loci for this system is shown in Figure 6–66(b).
A–6–4.
Referring to Problem A–6–3, obtain the equations for the root-locus branches for the system shown in Figure 6–66(a). Show that the root-locus branches cross the real axis at the breakaway point at angles ; 60°.
Solution. The equations for the root-locus branches can be obtained from the angle condition which can be rewritten as By substituting we obtain or By rearranging, we have Taking tangents of both sides of this last equation, and noting that we obtain which can be simplified to vs - v(s + 0.4) (s + 0.4)s + v2 = v(s + 3.6) + vs s(s + 3.6) - v2 v s + 0.4 - v s 1 + v s + 0.4 v s = v s + v s + 3.6 1 - v s v s + 3.6 tan ctan-1 a v s + 3.6 b ; 180°(2k + 1)d = v s + 3.6 tan-1 a v s + 0.4 b - tan-1 a v s b = tan-1 a v s b + tan-1 a v s + 3.6 b ;180°(2k + 1) tan-1 a v s + 0.4 b - 2 tan-1 a v s b - tan-1 a v s + 3.6 b = ;180°(2k + 1) /s + jv + 0.4 - 2/s + jv - /s + jv + 3.6 = ;180°(2k + 1) s = s + jv, /s + 0.4 - 2/s - /s + 3.6 = ;180°(2k + 1) n K(s + 0.4) s2(s + 3.6) = ;180°(2k + 1) A0.4K - 3.6v2B + jvAK - v2B = 0 (jv)3 + 3.6(jv)2 + K(jv) + 0.4K = 0 ;180°3, Example Problems and Solutions 353 or which can be further simplified to For s Z –1.6, we may write this last equation as which gives the equations for the root locus as follows: The equation v=0 represents the real axis. The root locus for 0 K q is between points s=–0.4 and s=–3.6. (The real axis other than this line segment and the origin s=0 corre-sponds to the root locus for –q K<0.) The equations (6–29) represent the complex branches for 0 K q.These two branches lie between s=–1.6 and s=0. [See Figure 6–66(b).] The slopes of the complex root-locus branches at the breakaway point (s=–1.2) can be found by evaluating of Equation (6–29) at point s=–1.2.
Since the root-locus branches intersect the real axis with angles A–6–5.
Consider the system shown in Figure 6–67(a). Sketch the root loci for the system. Observe that for small or large values of K the system is underdamped and for medium values of K it is overdamped.
Solution. A root locus exists on the real axis between the origin and –q. The angles of asymp-totes of the root-locus branches are obtained as The intersection of the asymptotes and the real axis is located on the real axis at The breakaway and break-in points are found from Since the characteristic equation is s3 + 4s2 + 5s + K = 0 dKds = 0.
s = - 0 + 2 + 2 3 = -1.3333 Angles of asymptotes = ;180°(2k + 1) 3 = 60°, -60°, -180° ;60°.
tan-1 13 = 60°, dv ds 2 s=-1.2 = ;A -s s + 1.6 2 s=-1.2 = ;A 1.2 0.4 = ;13 dvds v = ;(s + 1.2)A -s s + 1.6 v = -(s + 1.2)A -s s + 1.6 v = (s + 1.2)A -s s + 1.6 v = 0 vcv - (s + 1.2)A -s s + 1.6 d cv + (s + 1.2)A -s s + 1.6 d = 0 vCs(s + 1.2)2 + (s + 1.6)v2D = 0 vAs3 + 2.4s2 + 1.44s + 1.6v2 + sv2B = 0 354 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method we have Now we set which yields Since these points are on root loci,they are actual breakaway or break-in points. (At point s=–1, the value of K is 2, and at point s=–1.6667, the value of K is 1.852.) The angle of departure from a complex pole in the upper-half s plane is obtained from or The root-locus branch from the complex pole in the upper-half s plane breaks into the real axis at s=–1.6667.
Next we determine the points where root-locus branches cross the imaginary axis. By substi-tuting s=jv into the characteristic equation, we have or from which we obtain v = ; 15 , K = 20 or v = 0, K = 0 AK - 4v2B + jvA5 - v2B = 0 (jv)3 + 4(jv)2 + 5(jv) + K = 0 u = -63.43° u = 180° - 153.43° - 90° s = -1, s = -1.6667 dK ds = -A3s2 + 8s + 5B = 0 K = -As3 + 4s2 + 5sB (a) (b) K s(s2 + 4s + 5) jv s –4 –3 –2 0 1 j3 j2 j1 –j2 –j1 –j3 –1 K = 2 K = 1.852 + – Figure 6–67 (a) Control system; (b) root-locus plot.
Example Problems and Solutions 355 Root-locus branches cross the imaginary axis at and The root-locus branch on the real axis touches the jv axis at v=0. A sketch of the root loci for the system is shown in Figure 6–67(b).
Note that since this system is of third order, there are three closed-loop poles. The nature of the system response to a given input depends on the locations of the closed-loop poles.
For 0<K<1.852, there are a set of complex-conjugate closed-loop poles and a real closed-loop pole. For 1.852 K 2, there are three real closed-loop poles. For example, the closed-loop poles are located at For 2<K, there are a set of complex-conjugate closed-loop poles and a real closed-loop pole.
Thus, small values of K (0<K<1.852) correspond to an underdamped system. (Since the real closed-loop pole dominates, only a small ripple may show up in the transient response.) Medium values of K (1.852 K 2) correspond to an overdamped system. Large values of K (20 consist of portions of the straight lines as shown in Figure 6–69(b). (Note that each straight line starts from an open-loop pole and extends to infinity in the direction of 180°, 60°, or –60° measured from the real axis.) The remaining portion of each straight line corresponds to K<0.
A–6–8.
Consider a unity-feedback control system with the following feedforward transfer function: Using MATLAB, plot the root loci and their asymptotes.
Solution. We shall plot the root loci and asymptotes on one diagram. Since the feedforward trans-fer function is given by the equation for the asymptotes may be obtained as follows: Noting that the equation for the asymptotes may be given by Ga(s) = K (s + 1)3 lim sS q K s3 + 3s2 + 2s lim sS q K s3 + 3s2 + 3s + 1 = K (s + 1)3 = K s3 + 3s2 + 2s G(s) = K s(s + 1)(s + 2) G(s) = K s(s + 1)(s + 2) v = 0, s + 1 + 1 13 v = 0, s + 1 -1 13 v = 0 v as + 1 + 1 13 v b as + 1 -1 13 v b = 0 vA3s2 + 6s + 3 - v2B = 0 2v(s + 2)(s - 1) = -vAs2 + 4s + 7 - v2B 2v(s + 2) s2 + 4s + 4 - v2 + 3 = -v s - 1 v + 13 s + 2 + v - 13 s + 2 1 - a v + 13 s + 2 b a v - 13 s + 2 b = -v s - 1 360 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method Hence, for the system we have num = den = [1 3 2 0] and for the asymptotes, numa = dena = [1 3 3 1] In using the following root-locus and plot commands r = rlocus(num,den) a = rlocus(numa,dena) plot([r a]) the number of rows of r and that of a must be the same. To ensure this, we include the gain con-stant K in the commands. For example, K1 = 0:0.1:0.3; K2 = 0.3:0.005:0.5: K3 = 0.5:0.5:10; K4 = 10:5:100; K = [K1 K2 K3 K4] r = rlocus(num,den,K) a = rlocus(numa,dena,K) y = [r a] plot(y, '-') MATLAB Program 6–15 will generate a plot of root loci and their asymptotes as shown in Figure 6–70.
MATLAB Program 6–15 % ---------- Root-Locus Plots ----------num = ; den = [1 3 2 0]; numa = ; dena = [1 3 3 1]; K1 = 0:0.1:0.3; K2 = 0.3:0.005:0.5; K3 = 0.5:0.5:10; K4 = 10:5:100; K = [K1 K2 K3 K4]; r = rlocus(num,den,K); a = rlocus(numa,dena,K); y = [r a]; plot(y,'-') v = [-4 4 -4 4]; axis(v) grid title('Root-Locus Plot of G(s) = K/[s(s + 1)(s + 2)] and Asymptotes') xlabel('Real Axis') ylabel('Imag Axis') % Manually draw open-loop poles in the hard copy Example Problems and Solutions 361 Drawing two or more plots in one diagram can also be accomplished by using the hold com-mand. MATLAB Program 6–16 uses the hold command. The resulting root-locus plot is shown in Figure 6–71.
Root-Locus Plot of G(s) = K/[(s(s+1)(s+2)] and Asymptotes Imag Axis 4 –4 0 3 2 1 –1 –2 –3 Real Axis –4 1 –3 –2 –1 0 4 2 3 Figure 6–70 Root-locus plot.
MATLAB Program 6–16 % ------------ Root-Locus Plots ------------num = ; den = [1 3 2 0]; numa = ; dena = [1 3 3 1]; K1 = 0:0.1:0.3; K2 = 0.3:0.005:0.5; K3 = 0.5:0.5:10; K4 = 10:5:100; K = [K1 K2 K3 K4]; r = rlocus(num,den,K); a = rlocus(numa,dena,K); plot(r,'o') hold Current plot held plot(a,'-') v = [-4 4 -4 4]; axis(v) grid title('Root-Locus Plot of G(s) = K/[s(s+1)(s+2)] and Asymptotes') xlabel('Real Axis') ylabel('Imag Axis') 362 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method A–6–9.
Plot the root loci and asymptotes for a unity-feedback system with the following feedforward transfer function: Determine the exact points where the root loci cross the jv axis Solution. The feedforward transfer function G(s) can be written as Note that as s approaches infinity, can be written as where we used the following formula: The expression gives the equation for the asymptotes.
lim sS qG(s) = lim sS q K (s + 1)4 (s + a)4 = s4 + 4as3 + 6a2s2 + 4a3s + a4 = lim sS q K (s + 1)4 lim sS q K s4 + 4s3 + 6s2 + 4s + 1 lim sS qG(s) = lim sS q K s4 + 4s3 + 11s2 + 14s + 10 lim sS q G(s) G(s) = K s4 + 4s3 + 11s2 + 14s + 10 G(s) = K (s2 + 2s + 2)(s2 + 2s + 5) Root-Locus Plot of G(s) = K/[s(s+1)(s+2)] and Aysmptotes Imag Axis 4 –4 0 3 2 1 –1 –2 –3 Real Axis –4 1 –3 –2 –1 0 4 2 3 Figure 6–71 Root-locus plot.
Example Problems and Solutions 363 The MATLAB program to plot the root loci of G(s) and the asymptotes is given in MATLAB Program 6–17. Note that the numerator and denominator for G(s) are num = den = [1 4 11 14 10] For the numerator and denominator of the asymptotes we used numa = dena = [1 4 6 4 1] Figure 6–72 shows the plot of the root loci and asymptotes.
Since the characteristic equation for the system is (s2 + 2s + 2)(s2 + 2s + 5) + K = 0 lim sS q G(s) MATLAB Program 6–17 % Root-locus plot num = ; den = [1 4 11 14 10]; numa = ; dena = [1 4 6 4 1]; r = rlocus(num,den); plot(r,'-') hold Current plot held plot(r,'o') rlocus(numa,dena); v = [-6 4 -5 5]; axis(v); axis('square') grid title('Plot of Root Loci and Asymptotes') 4 2 0 −6 −4 −2 0 1 5 3 2 4 −2 −1 −5 −4 −3 Real Axis Imag Axis Plot of Root Loci and Asymptotes Figure 6–72 Plot of root loci and asymptotes.
364 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method the points where the root loci cross the imaginary axis can be found by substituting s=jv with the characteristic equation as follows: and equating the imaginary part to zero.The result is Thus the exact points where the root loci cross the jv axis are By equating the real part to zero, we get the gain value K at the crossing points to be 16.25.
A–6–10.
Consider a unity-feedback control system with the feed-forward transfer function G(s) given by Plot a root-locus diagram with MATLAB.
Solution. The feedforward transfer function G(s) can be written as A possible MATLAB program to plot a root-locus diagram is shown in MATLAB Program 6–18.
The resulting root-locus plot is shown in Figure 6–73.
G(s) = K(s + 1) s4 + 4s3 + 11s2 + 14s + 10 G(s) = K(s + 1) (s2 + 2s + 2)(s2 + 2s + 5) v = ;1.8708.
v = ;1.8708 = (v4 - 11v2 + 10 + K) + j(-4v3 + 14v) = 0 [(jv)2 + 2jv + 2][(jv)2 + 2jv + 5] + K MATLAB Program 6–18 num = [1 1]; den = [1 4 11 14 10]; K1 = 0:0.1:2; K2 = 2:0.0.2:2.5; K3 = 2.5:0.5:10; K4 = 10:1:50; K = [K1 K2 K3 K4] r = rlocus(num,den,K); plot(r, 'o') v = [-8 2 -5 5]; axis(v); axis('square') grid title('Root-Locus Plot of G(s) = K(s+1)/[(s^2+2s+2)(s^2+2s+5)]') xlabel('Real Axis') ylabel('Imag Axis') Example Problems and Solutions 365 A–6–11.
Obtain the transfer function of the mechanical system shown in Figure 6–74. Assume that the displacement xi is the input and displacement xo is the output of the system.
Solution. From the diagram we obtain the following equations of motion: Taking the Laplace transforms of these two equations, assuming zero initial conditions, and then eliminating Y(s), we obtain This is the transfer function between and By defining we obtain This mechanical system is a mechanical lead network.
X o(s) X i(s) = a Ts + 1 aTs + 1 = s + 1 T s + 1 aT b1 k = T, b2 b1 + b2 = a 6 1 X i(s).
X o(s) X o(s) X i(s) = b2 b1 + b2 b1 k s + 1 b2 b1 + b2 b1 k s + 1 b1Ax # o - y # B = ky b2Ax # i - x # oB = b1Ax # o - y # B −3 −2 −1 0 1 −4 −5 −8 2 −7 −6 0 1 3 2 4 5 −2 −1 −5 −4 −3 Real Axis Imag Axis Root-Locus Plot of G(s) = K(s + 1)/[(s2 + 2s + 2)(s2 + 2s + 5)] Figure 6–73 Plot of root loci.
b2 b1 k y xi xo Figure 6–74 Mechanical system.
366 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method A–6–12.
Obtain the transfer function of the mechanical system shown in Figure 6–75.Assume that the dis-placement xi is the input and displacement xo is the output.
Solution. The equations of motion for this system are By taking the Laplace transforms of these two equations, assuming zero initial conditions, we obtain If we eliminate Y(s) from the last two equations, the transfer function can be obtained as Define If k1, k2, b1, and b2 are chosen such that there exists a ı that satisfies the following equation: (6–30) then can be obtained as [Note that depending on the choice of k1, k2, b1, and b2, there does not exist a ı that satisfies Equation (6–30).] If such a ı exists and if for a given s1 (where s = s1 is one of the dominant closed-loop poles of the control system to which we wish to use this mechanical device) the following conditions are satisfied: then the mechanical system shown in Figure 6–75 acts as a lag–lead compensator.
-5° 6 n s1 + 1 T 2 s1 + 1 bT 2 6 0° 4 s1 + 1 T 2 s1 + 1 bT 2 4 1, X o(s) X i(s) = AT 1 s + 1BAT 2 s + 1B a T 1 b s + 1 b AbT 2 s + 1B = as + 1 T 1 b as + 1 T 2 b as + b T 1 b as + 1 bT 2 b X o(s)X i(s) b1 k1 + b2 k2 + b1 k2 = T1 b + bT2 (b 7 1) T1 = b1 k1 , T2 = b2 k2 , X o(s) X i(s) = a b1 k1 s + 1 b a b2 k2 s + 1 b a b1 k1 s + 1 b a b2 k2 s + 1 b + b1 k2 s X o(s)X i(s) b1CsX o(s) - sY(s)D = k1Y(s) b2CsX i(s) - sX o(s)D + k2CX i(s) - X o(s)D = b1CsX o(s) - sY(s)D b1Ax # o - y # B = k1y b2Ax # i - x # oB + k2Axi - xoB = b1Ax # o - y # B b1 b2 y xi xo k2 k1 Figure 6–75 Mechanical system.
Example Problems and Solutions 367 A–6–13.
Consider the model for a space-vehicle control system shown in Figure 6–76. Design a lead compensator Gc(s) such that the damping ratio z and the undamped natural frequency vn of the dominant closed-loop poles are 0.5 and 2 radsec, respectively.
Solution.
First Attempt: Assume the lead compensator Gc(s) to be From the given specifications, z=0.5 and vn=2 radsec, the dominant closed-loop poles must be located at We first calculate the angle deficiency at this closed-loop pole.
This angle deficiency must be compensated by the lead compensator. There are many ways to determine the locations of the pole and zero of the lead network. Let us choose the zero of the compensator at s=–1. Then, referring to Figure 6–77, we have the following equation: 1.73205 x - 1 = tan (90° - 70.8934°) = 0.34641 = -70.8934° Angle deficiency = -120° - 120° - 10.8934° + 180° s = -1 ; j13 Gc(s) = K c ± s + 1 T s + 1 aT ≤ (0 6 a 6 1) R(s) C(s) Lead compensator Gc(s) Space vehicle Sensor 1 s2 1 0.1s + 1 + – Figure 6–76 Space-vehicle control system.
jv s –1 0 19.1066° 70.8934° j1.73205 x Figure 6–77 Determination of the pole of the lead network.
368 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method or Hence, The value of can be determined from the magnitude condition as follows: Thus Since the open-loop transfer function becomes a root-locus plot of the compensated system can be obtained easily with MATLAB by entering num and den and using rlocus command.The result is shown in Figure 6–78. = 11.2(s + 1) 0.1s4 + 1.6s3 + 6s2 Gc(s)G(s)H(s) = 11.2 s + 1 (s + 6)s2(0.1s + 1) Gc(s) = 11.2 s + 1 s + 6 K c = 2 (s + 6)s2(0.1s + 1) s + 1 2 s=-1+j13 = 11.2000 K c 2 s + 1 s + 6 1 s2 1 0.1s + 1 2 s=-1+j13 = 1 K c Gc(s) = K c s + 1 s + 6 x = 1 + 1.73205 0.34641 = 6 Real Axis –10 5 10 –5 0 Imag Axis 0 10 –5 5 –10 Root-Locus Plot of Compensated System Figure 6–78 Root-locus plot of the compensated system.
Example Problems and Solutions 369 The closed-loop transfer function for the compensated system becomes Figure 6–79 shows the unit-step response curve. Even though the damping ratio of the dominant closed-loop poles is 0.5, the amount of overshoot is very much higher than expected.A closer look at the root-locus plot reveals that the presence of the zero at s=–1 is increasing the amount of the maximum overshoot. [In general, if a closed-loop zero or zeros (compensator zero or zeros) lie to the right of the dominant pair of the complex poles, then the dominant poles are no longer dominant.] If large maximum overshoot cannot be tolerated, the compensator zero(s) should be shifted sufficiently to the left.
In the current design, it is desirable to modify the compensator and make the maximum overshoot smaller. This can be done by modifying the lead compensator, as presented in the following second attempt.
Second Attempt: To modify the shape of the root loci, we may use two lead networks, each contributing half the necessary lead angle, which is Let us choose the location of the zeros at s=–3. (This is an arbitrary choice. Other choices such as s=–2.5 and s=–4 may be made.) Once we choose two zeros at s=–3, the necessary location of the poles can be determined as shown in Figure 6–80, or which yields y = 1 + 1.73205 0.09535 = 19.1652 = tan5.4466° = 0.09535 1.73205 y - 1 = tan (40.89334° - 35.4467°) 70.8934°2 = 35.4467°.
C(s) R(s) = 11.2(s + 1)(0.1s + 1) (s + 6)s2(0.1s + 1) + 11.2(s + 1) t Sec 2 1 0 7 9 6 8 10 4 3 5 Output 0.5 1.5 0 1 Unit-Step Response of Compensated System Figure 6–79 Unit-step response of the compensated system.
370 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method Hence, the lead compensator will have the following transfer function: The value of can be determined from the magnitude condition as follows: or Then the lead compensator just designed is Then the open-loop transfer function becomes A root-locus plot for the compensated system is shown in Figure 6–81(a). Notice that there is no closed-loop zero near the origin.An expanded view of the root-locus plot near the origin is shown in Figure 6–81(b).
The closed-loop transfer function becomes The closed-loop poles are found as follows: s = -27.9606 s = -9.1847 ; j7.4814 s = -1 ; j1.73205 C(s) R(s) = 174.3864(s + 3)2(0.1s + 1) (s + 19.1652)2s2(0.1s + 1) + 174.3864(s + 3)2 Gc(s)G(s)H(s) = 174.3864 a s + 3 s + 19.1652 b 2 1 s2 1 0.1s + 1 Gc(s) = 174.3864 a s + 3 s + 19.1652 b 2 K c = 174.3864 2K c a s + 3 s + 19.1652 b 2 1 s2 1 0.1s + 1 2 s=-1+j13 = 1 K c Gc(s) = K c a s + 3 s + 19.1652 b 2 –20 –16 –12 –8 –4 –1 0 jv s 35.4467° 40.89334° j1.73205 y Figure 6–80 Determination of the pole of the lead network.
Example Problems and Solutions 371 Figures 6–82(a) and (b) show the unit-step response and unit-ramp response of the compensated system.The unit-step response curve is reasonable and the unit-ramp response looks acceptable.
Notice that in the unit-ramp response the output leads the input by a small amount.This is because the system has a feedback transfer function 1/(0.1s+1). If the feedback signal versus t is plotted, together with the unit-ramp input, the former will not lead the input ramp at steady state. See Figure 6–82(c).
Real Axis −4 1 0 2 −2 −3 −1 (b) Imag Axis −2 1 3 −1 2 −3 0 Root-Locus Plot of Compensated System near Origin : Closed-loop poles Real Axis –25 –30 0 10 −5 5 –15 –20 –10 (a) Imag Axis –10 5 15 0 10 –15 −5 –20 20 Root-Locus Plot of Compensated System : Closed-loop poles Figure 6–81 (a) Root-locus plot of compensated system; (b) root-locus plot near the origin.
372 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method A–6–14.
Consider a system with an unstable plant as shown in Figure 6–83(a). Using the root-locus approach, design a proportional-plus-derivative controller that is, determine the values of Kp and such that the damping ratio z of the closed-loop system is 0.7 and the undamped natural frequency vn is 0.5 radsec.
Solution. Note that the open-loop transfer function involves two poles at s=1.085 and s=–1.085 and one zero at which is unknown at this point.
Since the desired closed-loop poles must have vn=0.5 radsec and z=0.7, they must be located at s = 0.5/180° ; 45.573° s = -1T d , T dB A t Sec 2 1 0 7 9 6 8 10 4 3 5 (a) Output 0.4 0.8 1.4 0 1.2 0.6 1 0.2 Unit-Step Response of Compensated System t Sec 1 0 5 4 2 3 (b) Unit-Ramp Input and Output 1.5 2.5 3.5 2 3 0 0.5 1 4 4.5 5 Unit-Ramp Response of Compensated System Output t Sec 1 0 5 4 2 3 (c) Unit-Ramp Input and Feedback Signal 1.5 2.5 3.5 2 3 0 0.5 1 4 4.5 5 Feedback Signal in Unit-Ramp Response Feedback Signal Figure 6–82 (a) unit-step response of the compensated system; (b) unit-ramp response of the compensated system; (c) a plot of feedback signal versus t in the unit-ramp response.
Example Problems and Solutions 373 (z=0.7 corresponds to a line having an angle of 45.573° with the negative real axis.) Hence, the desired closed-loop poles are at The open-loop poles and the desired closed-loop pole in the upper half-plane are located in the diagram shown in Figure 6–83(b).The angle deficiency at point s=–0.35+j0.357 is This means that the zero at must contribute 11.939°, which, in turn, determines the location of the zero as follows: s = - 1 T d = -2.039 s = -1T d -166.026° - 25.913° + 180° = -11.939° s = -0.35 ; j0.357 + – (a) (b) Kp(1 + Tds) 1 10000 (s2 – 1.1772) 0 jv s 45.573° j3 j2 j1 –j1 –j3 –j2 25.913° 166.026° Closed-loop pole 1.085 2 –1.085 – 4 –3 –2.039 Figure 6–83 (a) PD control of an unstable plant; (b) root-locus diagram for the system.
374 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method Hence, we have (6–31) The value of is The value of gain Kp can be determined from the magnitude condition as follows: or Hence, By substituting the numerical values of and into Equation (6–31), we obtain which gives the transfer function of the desired proportional-plus-derivative controller.
A–6–15.
Consider the control system shown in Figure 6–84. Design a lag compensator Gc(s) such that the static velocity error constant Kv is 50 sec–1 without appreciably changing the location of the orig-inal closed-loop poles, which are at Solution. Assume that the transfer function of the lag compensator is Gc(s) = K ˆ c s + 1 T s + 1 bT (b 7 1) s = -2 ; j16.
K pA1 + T d sB = 14,273(1 + 0.4904s) = 6999.5(s + 2.039) K p T d K p = 6999.5 0.4904 = 14,273 K p T d = 6999.5 2K p T d s + 2.039 10000As2 - 1.1772B 2 s=-0.35+j0.357 = 1 T d = 1 2.039 = 0.4904 T d K pA1 + T d sB = K p T d a 1 T d + s b = K p T d(s + 2.039) Gc(s) R(s) C(s) 10 s(s + 4) + – Figure 6–84 Control system.
Example Problems and Solutions 375 Since Kv is specified as 50 sec–1, we have Thus Now choose Then Choose T=10. Then the lag compensator can be given by The angle contribution of the lag compensator at the closed-loop pole is which is small.The magnitude of Gc(s) at is 0.981. Hence the change in the location of the dominant closed-loop poles is very small.
The open-loop transfer function of the system becomes The closed-loop transfer function is To compare the transient-response characteristics before and after the compensation,the unit-step and unit-ramp responses of the compensated and uncompensated systems are shown in Figures 6–85(a) and (b), respectively.The steady-state error in the unit-ramp response is shown in Figure 6–85(c).The designed lag compensator is acceptable.
C(s) R(s) = 10s + 1 s3 + 4.005s2 + 10.02s + 1 Gc(s)G(s) = s + 0.1 s + 0.005 10 s(s + 4) s = -2 + j6 = -1.3616° /Gc(s) 2 s=-2+j16 = tan-1 16 -1.9 - tan-1 16 -1.995 s = -2 + j16 Gc(s) = s + 0.1 s + 0.005 b = 20 K ˆ c = 1.
K ˆ c b = 20 K v = lim sS0sGc(s) 10 s(s + 4) = K ˆ c b2.5 = 50 376 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method t Sec 2 1 0 7 9 6 8 10 4 3 5 (a) Outputs 0.4 0.8 0 1.2 0.6 1 0.2 Unit-Step Responses of Compensated and Uncompensated Systems Uncompensated system Compensated system t Sec 2 1 0 7 9 6 8 10 4 3 5 (b) Input Ramp and Outputs 2 4 0 6 3 5 1 7 8 9 10 Unit-Ramp Responses of Compensated and Uncompensated Systems Uncompensated system has steady-state error of 0.4 Compensated system has steady-state error of 0.02 t Sec 36 35.5 35 38.5 39.5 38 39 40 37 36.5 37.5 (c) Input Ramp and Outputs 37.5 38.5 40 35 39.5 38 39 35.5 37 36.5 36 Unit-Ramp Response (35 t 40) Uncompensated system Compensated system Figure 6–85 (a) Unit-step responses of the compensated and uncompensated systems; (b) unit-ramp responses of both systems; (c) unit-ramp responses showing steady-state errors.
Example Problems and Solutions 377 A–6–16.
Consider a unity-feedback control system whose feedforward transfer function is given by Design a compensator such that the dominant closed-loop poles are located at and the static velocity error constant Kv is equal to 80 sec–1.
Solution. The static velocity error constant of the uncompensated system is Since Kv=80 is required, we need to increase the open-loop gain by 128. (This implies that we need a lag compensator.) The root-locus plot of the uncompensated system reveals that it is not possible to bring the dominant closed-loop poles to by just a gain adjustment alone.
See Figure 6–86. (This means that we also need a lead compensator.) Therefore, we shall employ a lag–lead compensator.
Let us assume the transfer function of the lag–lead compensator to be where Kc=128. This is because and we obtain Kc=128. The angle deficiency at the desired closed-loop pole is The lead portion of the lag–lead compensator must contribute . To choose we may use the graphical method presented in Section 6–8.
T 1 60° Angle deficiency = - 120° - 90° - 30° + 180° = - 60° s = -2 + j213 K v = lim sS0sGc(s)G(s) = lim sS0sK cG(s) = K c 10 16 = 80 Gc(s) = K c ± s + 1 T 1 s + b T 1 ≤± s + 1 T 2 s + 1 bT 2 ≤ -2 ; j213 K v = 10 16 = 0.625.
s = -2 ; j213 G(s) = 10 s(s + 2)(s + 8) Real Axis −10 5 10 −5 0 Imag Axis 10 −10 6 −6 −8 4 0 2 −2 8 −4 Root-Locus Plot of G(s) = 10/[s(s+2)(s+8)] Desired closed-loop pole Complex conjugate closed-loop pole Figure 6–86 Root-locus plot of Cs(s + 2)(s + 8)D.
G(s) = 10 The lead portion must satisfy the following conditions: and The first condition can be simplified as By using the same approach as used in Section 6–8, the zero and pole can be determined as follows: See Figure 6–87.The value of b is thus determined as For the lag portion of the compensator, we may choose 1 bT 2 = 0.01 b = 14.419 1 T 1 = 3.70, b T 1 = 53.35 As = bT 1B As = 1T 1B 4 s1 + 1 T 1 s1 + b T 1 4 s1=-2+j213 = 1 13.3333 n s1 + 1 T 1 s1 + b T 1 5 s1=-2+j213 = 60° 4128 ± s1 + 1 T 1 s1 + b T 1 ≤GAs1B 4 s1=-2+j213 = 1 378 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method –53.35 13.3333x –3.70 60° x jv s s1 0 Figure 6–87 Graphical determination of the zero and pole of the lead portion of the compensator.
Example Problems and Solutions 379 Then Noting that the angle contribution of the lag portion is –1.697° and the magnitude contribution is 0.9837.This means that the dominant closed-loop poles lie close to the desired location Thus the compensator designed, is acceptable.The feedforward transfer function of the compensated system becomes A root-locus plot of the compensated system is shown in Figure 6–88(a).An enlarged root-locus plot near the origin is shown in Figure 6–88(b).
Gc(s)G(s) = 1280(s + 3.7)(s + 0.1442) s(s + 53.35)(s + 0.01)(s + 2)(s + 8) Gc(s) = 128 a s + 3.70 s + 53.35 b a s + 0.1442 s + 0.01 b s = -2 ; j213.
n s1 + 0.1442 s1 + 0.01 2 s1=-2+j213 = -1.697° 2 s1 + 0.1442 s1 + 0.01 2 s1=-2+j213 = 0.9837 1 T 2 = 0.1442 Real Axis −40 −60 20 60 40 −20 0 (a) Imag Axis 60 −60 −40 40 0 20 −20 Root-Locus Plot of Compensated System Figure 6–88 (a) Root-locus plot of compensated system; (b) root-locus plot near the origin.
Real Axis −10 5 10 −5 0 (b) Imag Axis 10 −10 8 −6 −8 6 0 4 −2 2 −4 Root-Locus Plot of Compensated System near the Origin Desired closed-loop poles To verify the improved system performance of the compensated system, see the unit-step responses and unit-ramp responses of the compensated and uncompensated systems shown in Figures 6–89 (a) and (b), respectively.
A–6–17.
Consider the system shown in Figure 6–90. Design a lag–lead compensator such that the static velocity error constant Kv is 50 sec–1 and the damping ratio z of the dominant closed-loop poles is 0.5. (Choose the zero of the lead portion of the lag–lead compensator to cancel the pole at s=–1 of the plant.) Determine all closed-loop poles of the compensated system.
380 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method t Sec 2 1 0 7 9 6 8 10 4 3 5 (a) Outputs 0.4 0.8 1.4 0 1.2 0.6 1 0.2 Unit-Step Responses of Compensated and Uncompensated Systems Uncompensated system Compensated system t Sec 2 1 0 7 9 6 8 10 4 3 5 (b) Outputs 3 5 0 9 4 7 1 8 6 2 10 Unit-Ramp Responses of Compensated and Uncompensated Systems Uncompensated system Compensated system Figure 6–89 (a) Unit-step responses of compensated and uncompensated systems; (b) unit-ramp responses of both systems.
Example Problems and Solutions 381 1 s(s + 1) (s + 5) Gc(s) + – Figure 6–90 Control system.
Solution. Let us employ the lag–lead compensator given by where b>1. Then The specification that determines the value of Kc, or We now choose so that will cancel the (s+1) term of the plant. The lead portion then becomes For the lag portion of the lag–lead compensator we require where s=s1 is one of the dominant closed-loop poles. Noting these requirements for the lag por-tion of the compensator, at s=s1, the open-loop transfer function becomes GcAs1BGAs1B K c a s1 + 1 s1 + b b 1 s1As1 + 1BAs1 + 5B = K c 1 s1As1 + bBAs1 + 5B 4 s1 + 1 T 2 s1 + 1 bT 2 4 1, -5° 6 n s1 + 1 T 2 s1 + 1 bT 2 6 0° s + 1 s + b s + A1T 1B T 1 = 1 K c = 250 K v = 50 sec-1 = K c 5 = lim sS0s K cAT 1 s + 1BAT 2 s + 1B a T 1 b s + 1 b AbT 2 s + 1B 1 s(s + 1)(s + 5) K v = lim sS0sGc(s)G(s) Gc(s) = K c ± s + 1 T 1 s + b T 1 ≤± s + 1 T 2 s + 1 bT 2 ≤= K c AT 1 s + 1BAT 2 s + 1B a T 1 b s + 1 b AbT 2 s + 1B 382 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method Then at , the following magnitude and angle conditions must be satisfied: (6–32) (6–33) where k=0, 1, 2, p . In Equations (6–32) and (6–33), b and s1 are unknowns. Since the damping ratio z of the dominant closed-loop poles is specified as 0.5,the closed-loop pole s=s1 can be writ-ten as where x is as yet undetermined.
Notice that the magnitude condition, Equation (6–32), can be rewritten as Noting that we have (6–34) The angle condition, Equation (6–33), can be rewritten as or (6–35) We need to solve Equations (6–34) and (6–35) for b and x. By several trial-and-error calculations, it can be found that Thus The lag portion of the lag–lead compensator can be determined as follows:Noting that the pole and zero of the lag portion of the compensator must be located near the origin, we may choose That is, 1 T 2 = 0.16025 or T 2 = 6.25 1 bT 2 = 0.01 s1 = -1.9054 + j13(1.9054) = -1.9054 + j3.3002 b = 16.025, x = 1.9054 tan-1 a 13x -x + b b + tan-1 a 13x -x + 5 b = 60° = -120° - tan-1 a 13x -x + b b - tan-1 a 13x - x + 5 b = -180° nK c 1 A-x + j13xBA-x + b + j13xBA-x + 5 + j13xB x2(b - x)2 + 3x2 2(5 - x)2 + 3x2 = 125 K c = 250, 2 K c A-x + j13xBA-x + b + j13xBA-x + 5 + j13xB 2 = 1 s1 = -x + j13x nK c 1 s1As1 + bBAs1 + 5B = ;180°(2k + 1) 2K c 1 s1As1 + bBAs1 + 5B 2 = 1 s = s1 Example Problems and Solutions 383 With the choice of we find (6–36) and (6–37) Since our choice of is acceptable. Then the lag–lead compensator just designed can be writ-ten as Therefore, the compensated system has the following open-loop transfer function: A root-locus plot of the compensated system is shown in Figure 6–91(a).An enlarged root-locus plot near the origin is shown in Figure 6–91(b).
The closed loop transfer function becomes The closed-loop poles are located at Notice that the dominant closed-loop poles differ from the dominant closed-loop poles assumed in the computation of b and Small deviations of the dom-inant closed-loop poles from are due to the approximations involved in determining the lag portion of the compensator.[See Equations (6–36) and (6–37).] s = ;s1 = -1.9054 ; j3.3002 s = -1.8308 ; j3.2359 T 2 .
s = ;s1 s = -1.8308 ; j3.2359 s = -17.205 s = -0.1684 s = -1.8308 ; j3.2359 C(s) R(s) = 250(s + 0.16025) s(s + 0.01)(s + 5)(s + 16.025) + 250(s + 0.16025) Gc(s)G(s) = 250(s + 0.16025) s(s + 0.01)(s + 5)(s + 16.025) Gc(s) = 250 a s + 1 s + 16.025 b a s + 0.16025 s + 0.01 b T 2 = 6.25 -5° 6 -1.937° 6 0° = tan-1 a 3.3002 -1.74515 b - tan-1 a 3.3002 -1.89054 b = -1.937° n s1 + 1 T 2 s1 + 1 bT 2 = n -1.9054 + j3.3002 + 0.16025 -1.9054 + j3.3002 + 0.01 = 2 -1.74515 + j3.3002 -1.89054 + j3.3002 2 = 0.98 1 4 s1 + 1 T 2 s1 + 1 bT 2 4 = 2 -1.9054 + j3.3002 + 0.16025 -1.9054 + j3.3002 + 0.01 2 T 2 = 6.25, 384 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method Real Axis –20 5 0 10 –10 –15 –5 (a) Imag Axis –5 5 15 0 10 –10 –15 Root-Locus Plot of Compensated System Real Axis 1 0.5 –0.5 –1 0 (b) Imag Axis –0.6 0 0.6 –0.2 0.2 –1 1 0.8 0.4 –0.8 –0.4 Root-Locus Plot of Compensated System near the Origin Figure 6–91 (a) Root-locus plot of compensated system; (b) root-locus plot near the origin.
Figures 6–92(a) and (b) show the unit-step response and unit-ramp response of the designed system, respectively. Note that the closed-loop pole at s=–0.1684 almost cancels the zero at s=–0.16025. However, this pair of closed-loop pole and zero located near the origin pro-duces a long tail of small amplitude. Since the closed-loop pole at s=–17.205 is located very much farther to the left compared to the closed-loop poles at the effect of this real pole on the system response is very small. Therefore, the closed-loop poles at are indeed dominant closed-loop poles that determine the response characteristics of the closed-loop system. In the unit-ramp response, the steady-state error in following the unit-ramp input eventually becomes 1K v = 1 50 = 0.02.
s = -1.8308 ; j3.2359 s = -1.8308 ; j3.2359, Example Problems and Solutions 385 A–6–18.
Figure 6–93(a) is a block diagram of a model for an attitude-rate control system.The closed-loop transfer function for this system is The unit-step response of this system is shown in Figure 6–93(b). The response shows high-frequency oscillations at the beginning of the response due to the poles at The response is dominated by the pole at The settling time is approximately 240 sec.
s = -0.0167.
s = -0.0417 ; j2.4489. = 2(s + 0.05) (s + 0.0417 + j2.4489)(s + 0.0417 - j2.4489)(s + 0.0167) C(s) R(s) = 2s + 0.1 s3 + 0.1s2 + 6s + 0.1 t Sec 4 2 0 14 12 8 6 10 (a) Output 0.4 0.8 1.4 0 1.2 0.6 1 0.2 Unit-Step Response of Compensated System t Sec 2 1 0 7 9 6 8 10 4 3 5 (b) Output 3 7 0 9 4 8 2 6 5 1 10 Unit-Ramp Response of Compensated System Figure 6–92 (a) Unit-step response of the compensated system; (b) unit-ramp response of the compensated system.
386 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method Time (sec) 50 0 200 300 250 100 150 (b) Amplitude 0.3 0.6 0 1 0.4 0.7 0.1 0.9 0.8 0.5 0.2 Unit-Step Response of Uncompensated System R(s) C(s) 1 Rate gyro Hydraulic servo Aircraft 1 s 2s + 0.1 s2 + 0.1s + 4 (a) + – Figure 6–93 (a) Attitude-rate control system; (b) unit-step response.
It is desired to speed up the response and also eliminate the oscillatory mode at the beginning of the response. Design a suitable compensator such that the dominant closed-loop poles are at Solution. Figure 6–94 shows a block diagram for the compensated system.Note that the open-loop zero at and the open-loop pole at s=0 generate a closed-loop pole between s=0 and s=–0.05. Such a closed-loop pole becomes a dominant closed-loop pole and makes the re-sponse quite slow. Hence, it is necessary to replace this zero by a zero that is located far away from the jv axis—for example, a zero at s = -4.
s = -0.05 s = -2 ; j213.
Gc(s) R(s) C(s) 1 Rate gyro Hydraulic servo Aircraft 1 s 2s + 0.1 s2 + 0.1s + 4 + – Figure 6–94 Compensated attitude-rate control system.
Example Problems and Solutions 387 We now choose the compensator in the following form: Then the open-loop transfer function of the compensated system becomes To determine by the root-locus method, we need to find the angle deficiency at the desired closed-loop pole The angle deficiency can be found as follows: Hence, the lead compensator must provide 132.73°. Since the angle deficiency is –132.73°, we need two lead compensators, each providing 66.365°.Thus will have the following form: Suppose that we choose two zeros at s=–2. Then the two poles of the lead compensators can be obtained from or (See Figure 6–95.) Hence, G ˆ c(s) = K c a s + 2 s + 9.9158 b 2 = 9.9158 sp = 2 + 3.4641 0.4376169 3.4641 sp - 2 = tan(90° - 66.365°) = 0.4376169 = Kc a s + sz s + sp b 2 G ˆ c(s) G ˆ c(s) G ˆ c(s) = -132.73° Angle deficiency = -143.088° - 120° - 109.642° + 60° + 180° s = -2 + j213.
G ˆ c(s) = G ˆ c(s) s + 4 sAs2 + 0.1s + 4B Gc(s)G(s) = G ˆ c(s) s + 4 2s + 0.1 1 s 2s + 0.1 s2 + 0.1s + 4 Gc(s) = G ˆ c(s) s + 4 2s + 0.1 2 4 −6 −2 0 −4 −8 −10 −12 j4 j2 –j4 –j2 66.365° sp s = −2 + j2 3 jv s Figure 6–95 Pole and zero of .
G ˆ c(s) 388 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method The entire compensator for the system becomes The value of can be determined from the magnitude condition. Since the open-loop transfer function is the magnitude condition becomes Hence, Thus the compensator becomes The open-loop transfer function is given by A root-locus plot for the compensated system is shown in Figure 6–96.The closed-loop poles for the compensated system are indicated in the plot.The closed-loop poles, the roots of the charac-teristic equation (s + 9.9158)2sAs2 + 0.1s + 4B + 88.0227(s + 2)2(s + 4) = 0 Gc(s)G(s) = 88.0227(s + 2)2(s + 4) (s + 9.9158)2sAs2 + 0.1s + 4B Gc(s) = 88.0227 (s + 2)2(s + 4) (s + 9.9158)2(2s + 0.1) Gc(s) = 88.0227 K c = 2 (s + 9.9158)2sAs2 + 0.1s + 4B (s + 2)2(s + 4) 2 s=-2+j213 2K c (s + 2)2(s + 4) (s + 9.9158)2sAs2 + 0.1s + 4B 2 s=-2+j213 = 1 Gc(s)G(s) = K c (s + 2)2(s + 4) (s + 9.9158)2sAs2 + 0.1s + 4B K c Gc(s) = G ˆ c(s) s + 4 2s + 0.1 = K c (s + 2)2 (s + 9.9158)2 s + 4 2s + 0.1 Gc(s) Real Axis –15 10 5 15 –5 –10 0 Imag Axis –10 0 10 –5 5 15 –15 Root-Locus Plot of Compensated System Closed-loop poles Figure 6–96 Root-locus plot of the compensated system.
Example Problems and Solutions 389 are as follows: Now that the compensator has been designed, we shall examine the transient-response charac-teristics with MATLAB.The closed-loop transfer function is given by Figures 6–97(a) and (b) show the plots of the unit-step response and unit-ramp response of the compensated system.These response curves show that the designed system is acceptable.
C(s) R(s) = 88.0227(s + 2)2(s + 4) (s + 9.9158)2sAs2 + 0.1s + 4B + 88.0227(s + 2)2(s + 4) s = -0.8868 s = -7.5224 ; j6.5326 s = -2.0000 ; j3.4641 t Sec 1 0.5 0 3.5 4.5 3 4 5 2 1.5 2.5 (a) Output 0.4 0.8 1.4 0 1.2 0.6 1 0.2 Unit-Step Response of Compensated System t Sec 0 5 4 6 2 1 3 (b) Input and Output 2 4 6 0 3 5 1 Unit-Ramp Response of Compensated System Figure 6–97 (a) Unit-step response of the compensated system; (b) unit-ramp response of the compensated system.
390 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method A–6–19.
Consider the system shown in Figure 6–98(a).Determine the value of a such that the damping ratio z of the dominant closed poles is 0.5.
Solution. The characteristic equation is The variable a is not a multiplying factor. Hence, we need to modify the characteristic equation.
Since the characteristic equation can be written as we rewrite this equation such that a appears as a multiplying factor as follows: Define Then the characteristic equation becomes Notice that the characteristic equation is in a suitable form for the construction of the root loci.
1 + K sAs2 + 9s + 18B = 0 10a = K 1 + 10a sAs2 + 9s + 18B = 0 s3 + 9s2 + 18s + 10a = 0 1 + 10(s + a) s(s + 1)(s + 8) = 0 (a) (b) s + a s + 8 10 s(s + 1) j4 j3 j2 –j4 –j3 –j2 –j1 0 1 –1 –3 –5 –7 –2 2 –4 –6 jv s 60° K = 28 K = 28 j1 j6 –j6 j5 –j5 + – Figure 6–98 (a) Control system; (b) root-locus plot, where K=10a.
Example Problems and Solutions 391 This system involves three poles and no zero.The three poles are at s=0, s=–3, and s=–6.
A root-locus branch exists on the real axis between points s=0 and s=–3.Also,another branch exists between points s=–6 and s=–q.
The asymptotes for the root loci are found as follows: The intersection of the asymptotes and the real axis is obtained from The breakaway and break-in points can be determined from where Now we set which yields or Point s=–1.268 is on a root-locus branch.Hence,point s=–1.268 is an actual breakaway point.But point s=–4.732 is not on the root locus and therefore is neither a breakaway nor break-in point.
Next we shall find points where root-locus branches cross the imaginary axis. We substitute s=jv in the characteristic equation, which is as follows: or from which we get The crossing points are at and the corresponding value of gain K is 162.Also, a root-locus branch touches the imaginary axis at v=0. Figure 6–98(b) shows a sketch of the root loci for the system.
Since the damping ratio of the dominant closed-loop poles is specified as 0.5, the desired closed-loop pole in the upper-half s plane is located at the intersection of the root-locus branch in the upper-half s plane and a straight line having an angle of 60° with the negative real axis.The desired dominant closed-loop poles are located at At these points, the value of gain K is 28. Hence, a = K 10 = 2.8 s = -1 + j1.732, s = -1 - j1.732 v = ;312 v = ;312, K = 9v2 = 162 or v = 0, K = 0 AK - 9v2B + jvA18 - v2B = 0 (jv)3 + 9(jv)2 + 18(jv) + K = 0 s3 + 9s2 + 18s + K = 0 s = -1.268, s = -4.732 s2 + 6s + 6 = 0 dK ds = -A3s2 + 18s + 18B = 0 K = -As3 + 9s2 + 18sB dKds = 0, s = - 0 + 3 + 6 3 = -3 Angles of asymptotes = ;180°(2k + 1) 3 = 60°, -60°, 180° 392 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method Since the system involves two or more poles than zeros (in fact, three poles and no zero), the third pole can be located on the negative real axis from the fact that the sum of the three closed-loop poles is –9. Hence, the third pole is found to be at or A–6–20.
Consider the system shown in Figure 6–99(a). Sketch the root loci of the system as the velocity feedback gain k varies from zero to infinity. Determine the value of k such that the closed-loop poles have the damping ratio z of 0.7.
Solution. The open-loop transfer function is Since k is not a multilying factor, we modify the equation such that k appears as a multiplying factor. Since the characteristic equation is we rewrite this equation as follows: (6–38) Define Then Equation (6–38) becomes 1 + Ks s2 + s + 10 = 0 10k = K 1 + 10ks s2 + s + 10 = 0 s2 + s + 10ks + 10 = 0 Open-loop transfer function = 10 (s + 1 + 10k)s s = -7 s = -9 - (-1 + j1.732) - (-1 - j1.732) R(s) C(s) 1 s k jv s (a) (b) 10 s + 1 j4 j3 j2 −j1 j1 −j3 −j2 −j4 0 1 −1 −4 −6 −2 −3 −5 −7 2 K = 3.427 45.578 + – + – Figure 6–99 (a) Control system; (b) root-locus plot, where K=10k.
Example Problems and Solutions 393 Notice that the system has a zero at s=0 and two poles at Since this system involves two poles and one zero, there is a possibility that a circular root locus exists. In fact, this system has a circular root locus, as will be shown. Since the angle condition is we have By substituting into this last equation and rearranging, we obtain which can be rewritten as Taking tangents of both sides of this last equation, we obtain which can be simplified to or which yields Notice that v=0 corresponds to the real axis.The negative real axis (between s=0 and s=–q) corresponds to K 0, and the positive real axis corresponds to K<0. The equation is an equation of a circle with center at s=0, v=0 with the radius equal to A portion of this circle that lies to the left of the complex poles corresponds to the root locus for K>0. (The portion of the circle which lies to the right of the complex poles corresponds to the root locus for K<0.) Figure 6–99(b) shows a sketch of the root loci for K>0.
Since we require z=0.7 for the closed-loop poles, we find the intersection of the circular root locus and a line having an angle of 45.57° (note that cos45.57°=0.7) with the negative real axis. The intersection is at s=–2.214+j2.258. The gain K corresponding to this point is 3.427.
Hence, the desired value of the velocity feedback gain k is k = K 10 = 0.3427 110.
s2 + v2 = 10 v = 0 or s2 + v2 = 10 vAs2 - 10 + v2B = 0 2v(s + 0.5) (s + 0.5)2 - Av2 - 3.12252B = v s v + 3.1225 s + 0.5 + v - 3.1225 s + 0.5 1 - a v + 3.1225 s + 0.5 b a v - 3.1225 s + 0.5 b = v s tan-1 a v + 3.1225 s + 0.5 b + tan-1 a v - 3.1225 s + 0.5 b = tan-1 a v s b ; 180°(2k + 1) /s + 0.5 + j(v + 3.1225) + /s + 0.5 + j(v - 3.1225) = /s + jv ; 180°(2k + 1) s = s + jv /s - /s + 0.5 + j3.1225 - /s + 0.5 - j3.1225 = ;180°(2k + 1) n Ks s2 + s + 10 = ;180°(2k + 1) s = -0.5 ; j3.1225.
PROBLEMS 394 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method B–6–1. Plot the root loci for the closed-loop control sys-tem with B–6–2. Plot the root loci for the closed-loop control system with B–6–3. Plot the root loci for the system with B–6–4. Show that the root loci for a control system with are arcs of the circle centered at the origin with radius equal to B–6–5. Plot the root loci for a closed-loop control system with B–6–6. Plot the root loci for a closed-loop control system with Locate the closed-loop poles on the root loci such that the dominant closed-loop poles have a damping ratio equal to 0.5. Determine the corresponding value of gain K.
B–6–7. Plot the root loci for the system shown in Figure 6–100. Determine the range of gain K for stability.
G(s) = K(s + 9) sAs2 + 4s + 11B , H(s) = 1 G(s) = K(s + 0.2) s2(s + 3.6) , H(s) = 1 110 .
G(s) = KAs2 + 6s + 10B s2 + 2s + 10 , H(s) = 1 G(s) = K s(s + 0.5)As2 + 0.6s + 10B , H(s) = 1 G(s) = K s(s + 1)(s2 + 4s + 5) , H(s) = 1 G(s) = K(s + 1) s2 , H(s) = 1 B–6–8. Consider a unity-feedback control system with the following feedforward transfer function: Plot the root loci for the system. If the value of gain K is set equal to 2, where are the closed-loop poles located?
B–6–9. Consider the system whose open-loop transfer func-tion is given by Show that the equation for the asymptotes is given by Using MATLAB, plot the root loci and asymptotes for the system.
B–6–10. Consider the unity-feedback system whose feed-forward transfer function is The constant-gain locus for the system for a given value of K is defined by the following equation: Show that the constant-gain loci for 0 K q may be given by Sketch the constant-gain loci for K=1, 2, 5, 10, and 20 on the s plane.
B–6–11. Consider the system shown in Figure 6–101. Plot the root loci with MATLAB. Locate the closed-loop poles when the gain K is set equal to 2.
Cs(s + 1) + v2D 2 + v2 = K2 2 K s(s + 1) 2 = 1 G(s) = K s(s + 1) Ga(s)H a(s) = K s3 + 4.0068s2 + 5.3515s + 2.3825 G(s)H(s) = K(s - 0.6667) s4 + 3.3401s3 + 7.0325s2 G(s) = K sAs2 + 4s + 8B 2 s2 (s + 2) s + 1 s + 5 K R(s) C(s) + – Figure 6–100 Control system.
1 s + 1 K(s + 1) s(s2 + 2s + 6) + – Figure 6–101 Control system.
Problems 395 K(s – 1) (s + 2) (s + 4) + – G1(s) (a) (b) K(1 – s) (s + 2) (s + 4) + – G2(s) B–6–12. Plot root-locus diagrams for the nonminimum-phase systems shown in Figures 6–102(a) and (b), respectively.
B–6–14. Consider the system shown in Figure 6–104. Plot the root loci for the system. Determine the value of K such that the damping ratio z of the dominant closed-loop poles is 0.5. Then determine all closed-loop poles. Plot the unit-step response curve with MATLAB.
Figure 6–102 (a) and (b) Nonminimum-phase systems.
B–6–13. Consider the mechanical system shown in Figure 6–103. It consists of a spring and two dashpots. Obtain the transfer function of the system. The displacement xi is the input and displacement xo is the output. Is this system a mechanical lead network or lag network?
b2 b1 k xi xo Figure 6–103 Mechanical system.
K s(s2 + 4s + 5) + – Figure 6–104 Control system.
B–6–15. Determine the values of K, and of the system shown in Figure 6–105 so that the dominant closed-loop poles have the damping ratio z=0.5 and the undamped natural frequency vn=3 radsec.
T 2 T 1 , C R T1s + 1 T2s + 1 K 10 s(s + 1) + – Figure 6–105 Control system.
B–6–16. Consider the control system shown in Figure 6–106.
Determine the gain K and time constant T of the controller Gc(s) such that the closed-loop poles are located at s = -2 ; j2.
1 s(s + 2) K(Ts + 1) + – Gc(s) G(s) Figure 6–106 Control system.
Gc(s) 5 s(0.5s + 1) + – Figure 6–107 Control system.
B–6–17. Consider the system shown in Figure 6–107. De-sign a lead compensator such that the dominant closed-loop poles are located at Plot the unit-step re-sponse curve of the designed system with MATLAB.
s = -2 ; j213.
B–6–18. Consider the system shown in Figure 6–108. De-sign a compensator such that the dominant closed-loop poles are located at s = -1 ; j1.
Gc(s) 1 s2 Lead compensator Space vehicle + – Figure 6–108 Control system.
396 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method B–6–19. Referring to the system shown in Figure 6–109, de-sign a compensator such that the static velocity error con-stant is 20 sec–1 without appreciably changing the original location of a pair of the complex-conjugate closed-loop poles.
As = -2 ; j213B K v B–6–20. Consider the angular-positional system shown in Figure 6–110. The dominant closed-loop poles are located at The damping ratio z of the dominant closed-loop poles is 0.6.The static velocity error constant is 4.1 sec–1, which means that for a ramp input of 360°sec the steady-state error in following the ramp input is It is desired to decrease ev to one-tenth of the present value, or to increase the value of the static velocity error con-stant to 41 sec–1.It is also desired to keep the damping ratio z of the dominant closed-loop poles at 0.6.A small change in the undamped natural frequency vn of the dominant closed-loop poles is permissible.Design a suitable lag compensator to increase the static velocity error constant as desired.
K v ev = ui K v = 360°sec 4.1 sec-1 = 87.8° K v s = -3.60 ; j4.80.
B–6–21. Consider the control system shown in Figure 6–111.
Design a compensator such that the dominant closed-loop poles are located at and the static velocity error constant is 50 sec–1.
K v s = -2 ; j213 Gc(s) 16 s(s + 4) + – Figure 6–109 Control system.
Gc(s) 820 s(s + 10) (s + 20) + – Figure 6–110 Angular-positional system.
Gc(s) 10 s(s + 2) (s + 5) + – Figure 6–111 Control system.
B–6–24. Consider the system shown in Figure 6–114, which involves velocity feedback. Determine the values of the am-plifier gain K and the velocity feedback gain so that the following specifications are satisfied: 1. Damping ratio of the closed-loop poles is 0.5 2. Settling time 2 sec 3. Static velocity error constant 4. 0<Kh<1 K v 50 sec-1 K h R(s) C(s) 1 s Kh K 2s + 1 + – + – Figure 6–114 Control system.
B–6–23. Consider the control system shown in Figure 6–113.
Design a compensator such that the unit-step response curve will exhibit maximum overshoot of 25% or less and settling time of 5 sec or less.
Gc(s) 2s + 1 s(s + 1) (s + 2) + – Figure 6–112 Control system.
Gc(s) 1 s2 (s + 4) + – Figure 6–113 Control system.
B–6–22. Consider the control system shown in Figure 6–112.
Design a compensator such that the unit-step response curve will exhibit maximum overshoot of 30% or less and settling time of 3 sec or less.
Problems 397 B–6–25. Consider the system shown in Figure 6–115. The system involves velocity feedback. Determine the value of gain K such that the dominant closed-loop poles have a damping ratio of 0.5. Using the gain K thus determined, ob-tain the unit-step response of the system.
R(s) C(s) 1 s 0.2 K (s + 1) (s + 2) + – + – Figure 6–115 Control system.
B–6–26. Consider the system shown in Figure 6–116. Plot the root loci as a varies from 0 to q. Determine the value of a such that the damping ratio of the dominant closed-loop poles is 0.5.
s + a 2 s2 (s + 2) + – Figure 6–116 Control system.
B–6–28. Consider the system shown in Figure 6–118. As-suming that the value of gain K varies from 0 to q, plot the root loci when and 0.5.
Compare unit-step responses of the system for the following three cases: (1) K=10, Kh=0.1 (2) K=10, Kh=0.3 (3) K=10, Kh=0.5 K h = 0.1, 0.3, B–6–27. Consider the system shown in Figure 6–117. Plot the root loci as the value of k varies from 0 to q.What value of k will give a damping ratio of the dominant closed-loop poles equal to 0.5? Find the static velocity error constant of the system with this value of k.
s + 1.4 s + 5 10 s(s + 1) ks s + 10 + – + – Figure 6–117 Control system.
R(s) C(s) 1 s Kh K s + 1 + – + – Figure 6–118 Control system.
7 398 Control Systems Analysis and Design by the Frequency-Response Method 7–1 INTRODUCTION By the term frequency response, we mean the steady-state response of a system to a sinusoidal input. In frequency-response methods, we vary the frequency of the input signal over a certain range and study the resulting response.
In this chapter we present frequency-response approaches to the analysis and design of control systems.The information we get from such analysis is different from what we get from root-locus analysis. In fact, the frequency response and root-locus approaches complement each other. One advantage of the frequency-response approach is that we can use the data obtained from measurements on the physical system without deriving its mathematical model. In many practical designs of control systems both approaches are employed. Control engineers must be familiar with both.
Frequency-response methods were developed in 1930s and 1940s by Nyquist, Bode, Nichols, and many others. The frequency-response methods are most powerful in con-ventional control theory.They are also indispensable to robust control theory.
The Nyquist stability criterion enables us to investigate both the absolute and relative stabilities of linear closed-loop systems from a knowledge of their open-loop frequency-response characteristics. An advantage of the frequency-response approach is that frequency-response tests are, in general, simple and can be made accurately by use of readily available sinusoidal signal generators and precise measurement equipment.Often the transfer functions of complicated components can be determined experimentally by frequency-response tests. In addition, the frequency-response approach has the advan-tages that a system may be designed so that the effects of undesirable noise are negligible and that such analysis and design can be extended to certain nonlinear control systems.
Section 7–1 / Introduction 399 Although the frequency response of a control system presents a qualitative picture of the transient response,the correlation between frequency and transient responses is indirect,ex-cept for the case of second-order systems. In designing a closed-loop system, we adjust the frequency-response characteristic of the open-loop transfer function by using several de-sign criteria in order to obtain acceptable transient-response characteristics for the system.
Obtaining Steady-State Outputs to Sinusoidal Inputs.
We shall show that the steady-state output of a transfer function system can be obtained directly from the si-nusoidal transfer function—that is, the transfer function in which s is replaced by jv, where v is frequency.
Consider the stable,linear,time-invariant system shown in Figure 7–1.The input and out-put of the system,whose transfer function is G(s),are denoted by x(t) and y(t),respectively.
If the input x(t) is a sinusoidal signal, the steady-state output will also be a sinusoidal sig-nal of the same frequency, but with possibly different magnitude and phase angle.
Let us assume that the input signal to the system is given by [In this book “ ” is always measured in rad/sec. When the frequency is measured in cycle/sec, we use notation “f”.That is, .] Suppose that the transfer function G(s) of the system can be written as a ratio of two polynomials in s; that is, The Laplace-transformed output Y(s) of the system is then (7–1) where X(s) is the Laplace transform of the input x(t).
It will be shown that, after waiting until steady-state conditions are reached, the fre-quency response can be calculated by replacing s in the transfer function by jv. It will also be shown that the steady-state response can be given by where M is the amplitude ratio of the output and input sinusoids and f is the phase shift between the input sinusoid and the output sinusoid. In the frequency-response test, the input frequency v is varied until the entire frequency range of interest is covered.
The steady-state response of a stable, linear, time-invariant system to a sinusoidal input does not depend on the initial conditions. (Thus, we can assume the zero initial condition.) If Y(s) has only distinct poles, then the partial fraction expansion of Equa-tion (7–1) when x(t) = X yields (7–2) = a s + jv + a – s - jv + b1 s + s1 + b2 s + s2 + p + bn s + sn Y(s) = G(s)X(s) = G(s) vX s2 + v2 sin vt G(jv) = Mejf = M/f Y(s) = G(s)X(s) = p(s) q(s) X(s) G(s) = p(s) q(s) = p(s) As + s1BAs + s2B p As + snB v = 2pf v x(t) = X sinvt G(s) X(s) x(t) Y(s) y(t) Figure 7–1 Stable, linear, time-invariant system.
400 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method where a and the bi (where i=1, 2, p , n) are constants and is the complex conjugate of a.The inverse Laplace transform of Equation (7–2) gives (7–3) For a stable system, –s1, –s2, p , –sn have negative real parts.Therefore, as t approaches infinity, the terms and approach zero.Thus, all the terms on the right-hand side of Equation (7–3), except the first two, drop out at steady state.
If Y(s) involves multiple poles sj of multiplicity mj, then y(t) will involve terms such as For a stable system, the terms approach zero as t approaches infinity.
Thus, regardless of whether the system is of the distinct-pole type or multiple-pole type, the steady-state response becomes (7–4) where the constant a can be evaluated from Equation (7–2) as follows: Note that Since G(jv) is a complex quantity, it can be written in the following form: where @G(jv)@ represents the magnitude and f represents the angle of G(jv); that is, The angle f may be negative, positive, or zero. Similarly, we obtain the following expression for G(–jv): Then, noting that Equation (7–4) can be written (7–5) = Y sin (vt + f) = X@G(jv)@ sin (vt + f) yss(t) = X@G(jv)@ ej(vt+f) - e-j(vt+f) 2j a = -X@G(jv)@e-jf 2j , a – = X@G(jv)@ejf 2j G(-jv) = @G(-jv)@e-jf = @G(jv)@e-jf f = /G(jv) = tan-1 c imaginary part of G(jv) real part of G(jv) d G(jv) = @G(jv)@ejf a – = G(s) vX s2 + v2 (s - jv)2 s=jv = XG(jv) 2j a = G(s) vX s2 + v2 (s + jv)2 s=-jv = - XG(-jv) 2j yss(t) = ae-jvt + a –ejvt thje-sj t thje-sj t Ahj = 0, 1, 2, p , mj - 1B.
e-sn t e-s1 t, e-s2 t, p , y(t) = ae-jvt + a –ejvt + b1 e-s1 t + b2 e-s2 t + p + bn e-sn t (t 0) a – Section 7–1 / Introduction 401 X Y t Input x(t) = X sin vt Output y(t) = Y sin (vt + f) Figure 7–2 Input and output sinusoidal signals.
y x G(s) K Ts + 1 Figure 7–3 First-order system.
where Y=X@G(jv)@.We see that a stable, linear, time-invariant system subjected to a sinusoidal input will, at steady state, have a sinusoidal output of the same frequency as the input. But the amplitude and phase of the output will, in general, be different from those of the input. In fact, the amplitude of the output is given by the product of that of the input and @G(jv)@, while the phase angle differs from that of the input by the amount An example of input and output sinusoidal signals is shown in Figure 7–2.
On the basis of this, we obtain this important result: For sinusoidal inputs, Hence, the steady-state response characteristics of a system to a sinusoidal input can be obtained directly from The function G(jv) is called the sinusoidal transfer function. It is the ratio of Y(jv) to X(jv), is a complex quantity, and can be represented by the magnitude and phase angle with frequency as a parameter.The sinusoidal transfer function of any linear system is obtained by substituting jv for s in the transfer function of the system.
As already mentioned in Chapter 6,a positive phase angle is called phase lead,and a neg-ative phase angle is called phase lag.A network that has phase-lead characteristics is called a lead network, while a network that has phase-lag characteristics is called a lag network.
EXAMPLE 7–1 Consider the system shown in Figure 7–3.The transfer function G(s) is For the sinusoidal input x(t)=X sin vt, the steady-state output yss(t) can be found as follows: Substituting jv for s in G(s) yields G(jv) = K jTv + 1 G(s) = K Ts + 1 Y(jv) X(jv) = G(jv) /G(jv) = n Y(jv) X(jv) = phase shift of the output sinusoid with respect to the input sinusoid @G(jv)@ = 2 Y(jv) X(jv) 2 = amplitude ratio of the output sinuisoid to the input sinusoid f = /G(jv).
402 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method The amplitude ratio of the output to the input is while the phase angle f is Thus, for the input x(t)=X sin vt, the steady-state output yss(t) can be obtained from Equation (7–5) as follows: (7–6) From Equation (7–6), it can be seen that for small v, the amplitude of the steady-state output yss(t) is almost equal to K times the amplitude of the input.The phase shift of the output is small for small v. For large v, the amplitude of the output is small and almost inversely proportional to v.The phase shift approaches –90° as v approaches infinity.This is a phase-lag network.
EXAMPLE 7–2 Consider the network given by Determine whether this network is a lead network or lag network.
For the sinusoidal input x(t)=X sinvt, the steady-state output yss(t) can be found as follows: Since we have and Thus the steady-state output is From this expression, we find that if then Thus, if then the network is a lead network. If then the network is a lag network.
Presenting Frequency-Response Characteristics in Graphical Forms.
The sinusoidal transfer function, a complex function of the frequency v, is characterized by its magnitude and phase angle, with frequency as the parameter. There are three commonly used representations of sinusoidal transfer functions: T 1 6 T 2 , T 1 7 T 2 , tan-1T 1 v - tan-1 T 2 v 7 0.
T 1 7 T 2 , yss(t) = XT 221 + T2 1 v2 T 121 + T2 2 v2 sin Avt + tan-1 T 1 v - tan-1 T 2 vB f = /G(jv) = tan-1T 1 v - tan-1T 2 v @G(jv)@ = T 221 + T2 1 v2 T 121 + T2 2 v2 G(jv) = jv + 1 T 1 jv + 1 T 2 = T 2A1 + T 1 jvB T 1A1 + T 2 jvB G(s) = s + 1 T 1 s + 1 T 2 yss(t) = XK 21 + T2v2 sin Avt - tan-1 TvB f = /G(jv) = -tan-1Tv @G(jv)@ = K 21 + T2v2 Section 7–2 / Bode Diagrams 403 1. Bode diagram or logarithmic plot 2. Nyquist plot or polar plot 3. Log-magnitude-versus-phase plot (Nichols plots) We shall discuss these representations in detail in this chapter. We shall include the MATLAB approach to obtain Bode diagrams, Nyquist plots, and Nichols plots.
Outline of the Chapter.
Section 7–1 has presented introductory material on the frequency response. Section 7–2 presents Bode diagrams of various transfer-function systems. Section 7–3 treats polar plots of transfer functions. Section 7–4 discusses log-magnitude-versus-phase plots. Section 7–5 gives a detailed account of the Nyquist stability criterion. Section 7–6 discusses the stability analysis based on the Nyquist sta-bility criterion. Section 7–7 introduces measures of relative stability analysis. Sec-tion 7–8 presents a method for obtaining the closed-loop frequency response from the open-loop frequency response by use of the M and N circles. The Nichols chart is introduced here. Section 7–9 treats experimental determination of transfer func-tions. Section 7–10 presents introductory aspects of control systems design by the frequency-response approach. Sections 7–11, 7–12, and 7–13 give detailed accounts of lead compensation, lag compensation, and lag–lead compensation techniques, respectively.
7–2 BODE DIAGRAMS Bode Diagrams or Logarithmic Plots.
A Bode diagram consists of two graphs: One is a plot of the logarithm of the magnitude of a sinusoidal transfer function; the other is a plot of the phase angle; both are plotted against the frequency on a logarithmic scale.
The standard representation of the logarithmic magnitude of G(jv) is 20 log @G(jv)@, where the base of the logarithm is 10.The unit used in this representation of the magnitude is the decibel, usually abbreviated dB. In the logarithmic representation, the curves are drawn on semilog paper, using the log scale for frequency and the linear scale for either magnitude (but in decibels) or phase angle (in degrees). (The frequency range of inter-est determines the number of logarithmic cycles required on the abscissa.) The main advantage of using the Bode diagram is that multiplication of magni-tudes can be converted into addition. Furthermore, a simple method for sketching an approximate log-magnitude curve is available. It is based on asymptotic approxima-tions. Such approximation by straight-line asymptotes is sufficient if only rough in-formation on the frequency-response characteristics is needed. Should the exact curve be desired, corrections can be made easily to these basic asymptotic plots. Expanding the low-frequency range by use of a logarithmic scale for the frequency is highly advantageous, since characteristics at low frequencies are most important in practical systems. Although it is not possible to plot the curves right down to zero frequency because of the logarithmic frequency (log 0=–q), this does not create a serious problem.
Note that the experimental determination of a transfer function can be made simple if frequency-response data are presented in the form of a Bode diagram.
404 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Decibels (dB) Numbers 0.01 0.02 0.04 0.1 0.2 0.4 0.6 1 2 3 4 5 6 8 10 20 10 0 –10 –20 –30 –40 Figure 7–4 Number–decibel conversion line.
Basic Factors of G(jV)H(jV).
As stated earlier, the main advantage in using the logarithmic plot is the relative ease of plotting frequency-response curves. The basic factors that very frequently occur in an arbitrary transfer function G(jv)H(jv) are 1. Gain K 2. Integral and derivative factors (jv)<1 3. First-order factors (1+jvT)<1 4. Quadratic factors Once we become familiar with the logarithmic plots of these basic factors, it is possible to utilize them in constructing a composite logarithmic plot for any general form of G(jv)H(jv) by sketching the curves for each factor and adding individual curves graphically, because adding the logarithms of the gains corresponds to multiplying them together.
The Gain K.
A number greater than unity has a positive value in decibels, while a number smaller than unity has a negative value.The log-magnitude curve for a constant gain K is a horizontal straight line at the magnitude of 20 logK decibels.The phase angle of the gain K is zero. The effect of varying the gain K in the transfer function is that it raises or lowers the log-magnitude curve of the transfer function by the corresponding constant amount, but it has no effect on the phase curve.
A number–decibel conversion line is given in Figure 7–4. The decibel value of any number can be obtained from this line. As a number increases by a factor of 10, the corresponding decibel value increases by a factor of 20. This may be seen from the following: Similarly, 20 log AK 10nB = 20 log K + 20n 20 log (K 10) = 20 log K + 20 C1 + 2zAjvvnB + AjvvnB 2D <1 Section 7–2 / Bode Diagrams 405 Note that, when expressed in decibels, the reciprocal of a number differs from its value only in sign; that is, for the number K, Integral and Derivative Factors (jV)1.
The logarithmic magnitude of 1/jv in decibels is The phase angle of 1/jv is constant and equal to –90°.
In Bode diagrams, frequency ratios are expressed in terms of octaves or decades.An octave is a frequency band from v1 to 2v1, where v1 is any frequency value.A decade is a frequency band from v1 to 10v1, where again v1 is any frequency. (On the logarithmic scale of semilog paper, any given frequency ratio can be represented by the same hori-zontal distance. For example, the horizontal distance from v=1 to v=10 is equal to that from v=3 to v=30.) If the log magnitude –20 logv dB is plotted against v on a logarithmic scale, it is a straight line.To draw this straight line,we need to locate one point (0 dB,v=1) on it.Since the slope of the line is –20 dBdecade (or –6 dBoctave).
Similarly, the log magnitude of jv in decibels is The phase angle of jv is constant and equal to 90°.The log-magnitude curve is a straight line with a slope of 20 dBdecade. Figures 7–5(a) and (b) show frequency-response curves for 1/jv and jv, respectively. We can clearly see that the differences in the frequency responses of the factors 1/jv and jv lie in the signs of the slopes of the log-magnitude curves and in the signs of the phase angles. Both log magnitudes become equal to 0 dB at v=1.
If the transfer function contains the factor (1/jv)n or (jv)n, the log magnitude becomes, respectively, or The slopes of the log-magnitude curves for the factors (1/jv)n and (jv)n are thus –20n dBdecade and 20n dBdecade, respectively.The phase angle of (1/jv)n is equal to –90°n over the entire frequency range, while that of (jv)n is equal to 90°n over the entire frequency range. The magnitude curves will pass through the point (0 dB, v=1).
20 log @(jv)n@ = n 20 log ∑jv∑= 20n log v dB 20 log 2 1 (jv)n 2 = -n 20 log ∑jv∑= -20n log v dB 20 log ∑jv∑= 20 log v dB (-20 log 10v) dB = (-20 log v - 20) dB 20 log 2 1 jv 2 = -20 log v dB 20 log K = -20 log 1 K 406 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method First-Order Factors (1 jVT)1.
The log magnitude of the first-order factor 1/(1+jvT) is For low frequencies, such that v 1/T, the log magnitude may be approximated by Thus, the log-magnitude curve at low frequencies is the constant 0-dB line. For high frequencies, such that v 1/T, This is an approximate expression for the high-frequency range. At v=1/T, the log magnitude equals 0 dB; at v=10/T, the log magnitude is –20 dB. Thus, the value of –20 log vT dB decreases by 20 dB for every decade of v.For v 1/T, the log-magnitude curve is thus a straight line with a slope of –20 dB/decade (or –6 dB/octave).
Our analysis shows that the logarithmic representation of the frequency-response curve of the factor 1/(1+jvT) can be approximated by two straight-line asymptotes, one a straight line at 0 dB for the frequency range 0<v<1/T and the other a straight line with slope –20 dB/decade (or –6 dBoctave) for the frequency range 1/T<v<q.
The exact log-magnitude curve, the asymptotes, and the exact phase-angle curve are shown in Figure 7–6.
The frequency at which the two asymptotes meet is called the corner frequency or break frequency. For the factor 1/(1+jvT), the frequency v=1/T is the corner fre-quency, since at v=1/T the two asymptotes have the same value. (The low-frequency asymptotic expression at v=1/T is 20 log 1 dB=0 dB, and the high-frequency -20 log 21 + v2 T2 -20 log vT dB -20 log 21 + v2 T2 -20 log 1 = 0 dB 20 log 2 1 1 + jvT 2 = -20 log 21 + v2 T2 dB dB 40 20 0 –40 –20 0.1 10 1 100 v Slope = –20 dB/decade Bode diagram of G(jv) = 1/jv (a) f 0° –180° –90° 0.1 10 1 100 v dB 40 20 0 –40 –20 0.1 10 1 100 v Slope = 20 dB/decade Bode diagram of G(jv) = jv (b) f 180° 0° 90° 0.1 10 1 100 v Figure 7–5 (a) Bode diagram of G(jv)=1/jv; (b) Bode diagram of G(jv)=jv.
Section 7–2 / Bode Diagrams 407 10 0 –10 –20 0° –45° –90° f dB v 1 20T 1 10T 1 5T 1 T 1 2T 2 T 5 T 10 T 20 T Exact curve Asymptote Asymptote Corner frequency Figure 7–6 Log-magnitude curve, together with the asymptotes, and phase-angle curve of 1/(1+jvT).
asymptotic expression at v=1/T is also 20 log 1 dB=0 dB.) The corner frequency divides the frequency-response curve into two regions: a curve for the low-frequency re-gion and a curve for the high-frequency region.The corner frequency is very important in sketching logarithmic frequency-response curves.
The exact phase angle f of the factor 1/(1+jvT) is At zero frequency, the phase angle is 0°.At the corner frequency, the phase angle is At infinity, the phase angle becomes –90°. Since the phase angle is given by an inverse-tangent function, the phase angle is skew symmetric about the inflection point at f=–45°.
The error in the magnitude curve caused by the use of asymptotes can be calculated.
The maximum error occurs at the corner frequency and is approximately equal to –3 dB, since The error at the frequency one octave below the corner frequency—that is, at v=1/(2T)—is The error at the frequency one octave above the corner frequency—that is,at v=2/T— is -20 log 222 + 1 + 20 log 2 = -20 log 15 2 = -0.97 dB -20 log A 1 4 + 1 + 20 log 1 = -20 log 15 2 = -0.97 dB -20 log 11 + 1 + 20 log 1 = -10 log 2 = -3.03 dB f = -tan-1 T T = -tan-1 1 = -45° f = -tan-1 vT 408 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method 0 –1 –2 –3 –4 dB Corner frequency 1 10T 1 5T 1 2T 1 T 2 T 3 T 5 T 10 T v Figure 7–7 Log-magnitude error in the asymptotic expression of the frequency-response curve of 1/(1+jvT).
Thus, the error at one octave below or above the corner frequency is approximately equal to –1 dB. Similarly, the error at one decade below or above the corner frequency is approximately –0.04 dB. The error in decibels involved in using the asymptotic ex-pression for the frequency-response curve of 1/(1+jvT) is shown in Figure 7–7. The error is symmetric with respect to the corner frequency.
Since the asymptotes are quite easy to draw and are sufficiently close to the exact curve, the use of such approximations in drawing Bode diagrams is convenient in es-tablishing the general nature of the frequency-response characteristics quickly with a minimum amount of calculation and may be used for most preliminary design work. If accurate frequency-response curves are desired, corrections may easily be made by re-ferring to the curve given in Figure 7–7. In practice, an accurate frequency-response curve can be drawn by introducing a correction of 3 dB at the corner frequency and a correction of 1 dB at points one octave below and above the corner frequency and then connecting these points by a smooth curve.
Note that varying the time constant T shifts the corner frequency to the left or to the right, but the shapes of the log-magnitude and the phase-angle curves remain the same.
The transfer function 1/(1+jvT) has the characteristics of a low-pass filter. For frequencies above v=1/T, the log magnitude falls off rapidly toward –q.This is es-sentially due to the presence of the time constant. In the low-pass filter, the output can follow a sinusoidal input faithfully at low frequencies. But as the input frequen-cy is increased, the output cannot follow the input because a certain amount of time is required for the system to build up in magnitude. Thus, at high frequencies, the amplitude of the output approaches zero and the phase angle of the output approaches –90°.Therefore, if the input function contains many harmonics, then the low-frequency components are reproduced faithfully at the output, while the high-frequency components are attenuated in amplitude and shifted in phase.Thus, a first-order element yields exact, or almost exact, duplication only for constant or slowly varying phenomena.
An advantage of the Bode diagram is that for reciprocal factors—for example, the factor 1+jvT—the log-magnitude and the phase-angle curves need only be changed in sign, since 20 log ∑1 + jvT∑= -20 log 2 1 1 + jvT 2 Section 7–2 / Bode Diagrams 409 dB 40 20 0 –40 –20 v f 90° 0° 45° v Exact curve Asymptote 0.01 T 0.1 T 1 T 10 T 0.01 T 0.1 T 1 T 10 T Asymptote Figure 7–8 Log-magnitude curve, together with the asymptotes, and phase-angle curve for 1+jvT.
and The corner frequency is the same for both cases. The slope of the high-frequency as-ymptote of 1+jvT is 20 dBdecade,and the phase angle varies from 0° to 90° as the fre-quency v is increased from zero to infinity. The log-magnitude curve, together with the asymptotes, and the phase-angle curve for the factor 1+jvT are shown in Figure 7–8.
To draw a phase curve accurately, we have to locate several points on the curve.The phase angles of (1+jvT)<1 are For the case where a given transfer function involves terms like (1+jvT)<n,a similar asymptotic construction may be made.The corner frequency is still at v=1/T, and the asymptotes are straight lines.The low-frequency asymptote is a horizontal straight line <84.3° at v = 10 T <63.4° at v = 2 T <5.7° at v = 1 10T <26.6° at v = 1 2T <45° at v = 1 T /1 + jvT = tan-1vT = - n 1 1 + jvT 410 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method at 0 dB, while the high-frequency asymptote has the slope of –20n dBdecade or 20n dBdecade. The error involved in the asymptotic expressions is n times that for (1+jvT)<1.The phase angle is n times that of (1+jvT)<1 at each frequency point.
Quadratic Factors C1 2ZAjV/VnB AjV/VnB 2D 1.
Control systems often possess quadratic factors of the form G(j◊)= (7–7) If z>1, this quadratic factor can be expressed as a product of two first-order factors with real poles. If 0<z<1, this quadratic factor is the product of two complex-conjugate factors.Asymptotic approximations to the frequency-response curves are not accurate for a factor with low values of z. This is because the magnitude and phase of the quadratic factor depend on both the corner frequency and the damping ratio z.
The asymptotic frequency-response curve may be obtained as follows: Since for low frequencies such that v vn, the log magnitude becomes The low-frequency asymptote is thus a horizontal line at 0 dB. For high frequencies such that v vn, the log magnitude becomes The equation for the high-frequency asymptote is a straight line having the slope –40 dBdecade, since The high-frequency asymptote intersects the low-frequency one at v=vn, since at this frequency This frequency, vn, is the corner frequency for the quadratic factor considered.
The two asymptotes just derived are independent of the value of z. Near the frequency v=vn, a resonant peak occurs, as may be expected from Equation (7–7).
The damping ratio z determines the magnitude of this resonant peak. Errors obvi-ously exist in the approximation by straight-line asymptotes. The magnitude of the error depends on the value of z. It is large for small values of z. Figure 7–9 shows the exact log-magnitude curves, together with the straight-line asymptotes and the exact -40 log vn vn = -40 log 1 = 0 dB -40 log 10v vn = -40 - 40 log v vn -20 log v2 v2 n = -40 log v vn dB -20 log 1 = 0 dB 20 log 3 1 1 + 2z aj v vn b + a j v vn b 2 3 = -20 log B a1 - v2 v2 n b 2 + a2z v vn b 2 1 1 + 2z a j v vn b + aj v vn b 2 Section 7–2 / Bode Diagrams 411 20 10 –10 0 f dB 0° –90° –180° 0.4 0.6 0.8 1 2 4 6 8 10 0.1 0.2 v vn z = 0.1 z = 0.2 z = 0.3 z = 0.5 z = 0.7 z = 1.0 z = 0.1 z = 0.2 z = 0.3 z = 0.5 z = 0.7 z = 1.0 Asymptotes Figure 7–9 Log-magnitude curves, together with the asymptotes, and phase-angle curves of the quadratic transfer function given by Equation (7–7).
phase-angle curves for the quadratic factor given by Equation (7–7) with several values of z. If corrections are desired in the asymptotic curves, the necessary amounts of cor-rection at a sufficient number of frequency points may be obtained from Figure 7–9.
The phase angle of the quadratic factor C1+2zAjv/vnB+Ajv/vnB 2D –1 is (7–8) The phase angle is a function of both v and z. At v=0, the phase angle equals 0°. At the corner frequency v=vn, the phase angle is –90° regardless of z, since At v=q, the phase angle becomes –180°. The phase-angle curve is skew symmetric about the inflection point—the point where f=–90°.There are no simple ways to sketch such phase curves.We need to refer to the phase-angle curves shown in Figure 7–9.
f = -tan-1 a 2z 0 b = -tan-1 q = -90° f = n 1 1 + 2z aj v vn b + a j v vn b 2 = -tan-1 ≥ 2z v vn 1 - a v vn b 2 ¥ 412 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method The frequency-response curves for the factor can be obtained by merely reversing the sign of the log magnitude and that of the phase angle of the factor To obtain the frequency-response curves of a given quadratic transfer function, we must first determine the value of the corner frequency vn and that of the damping ratio z.
Then, by using the family of curves given in Figure 7–9, the frequency-response curves can be plotted.
The Resonant Frequency Vr and the Resonant Peak Value Mr. The magnitude of is (7–9) If has a peak value at some frequency, this frequency is called the resonant frequency. Since the numerator of is constant, a peak value of will occur when (7–10) is a minimum. Since Equation (7–10) can be written (7–11) the minimum value of g(v) occurs at Thus the resonant frequency vr is (7–12) As the damping ratio z approaches zero, the resonant frequency approaches vn. For 00.707, there is no resonant peak.The magnitude de-creases monotonically with increasing frequency v. (The magnitude is less than 0 dB for all values of v>0. Recall that, for 0.7<z<1, the step response is oscillatory, but the oscillations are well damped and are hardly perceptible.) @G(jv)@ vd = vn21 - z2 , vr = vn21 - 2z2 , for 0 z 0.707 v = vn21 - 2z2 .
g(v) = c v2 - v2 nA1 - 2z2B v2 n d 2 + 4z2A1 - z2B g(v) = a1 - v2 v2 n b 2 + a 2z v vn b 2 @G(jv)@ @G(jv)@ @G(jv)@ @G(jv)@ = 1 B a1 - v2 v2 n b 2 + a2z v vn b 2 G(jv) = 1 1 + 2z a j v vn b + a j v vn b 2 1 1 + 2z a j v vn b + aj v vn b 2 1 + 2z a j v vn b + aj v vn b 2 Section 7–2 / Bode Diagrams 413 14 12 10 8 6 4 2 0 z 0.2 1.0 0.8 0.6 0.4 Mr in dB Figure 7–10 Mr-versus-z curve for the second-order system 1/C1+2zAjvvnB+ AjvvnB 2D.
For 0 z 0.707, the magnitude of the resonant peak, Mr=|G(jvr)|, can be found from Equations (7–12) and (7–9). For 0 z 0.707, (7–13) For z>0.707, (7–14) As z approaches zero, Mr approaches infinity.This means that if the undamped system is excited at its natural frequency, the magnitude of G(jv) becomes infinity. The rela-tionship between Mr and z is shown in Figure 7–10.
The phase angle of G(jv) at the frequency where the resonant peak occurs can be obtained by substituting Equation (7–12) into Equation (7–8). Thus, at the resonant frequency vr, General Procedure for Plotting Bode Diagrams.
MATLAB provides an easy way to plot Bode diagrams. (The MATLAB approach is presented later in this section.) Here, however, we consider the case where we want to draw Bode diagrams manually without using MATLAB.
First rewrite the sinusoidal transfer function G(jv)H(jv) as a product of basic factors discussed above.Then identify the corner frequencies associated with these basic factors.
Finally,draw the asymptotic log-magnitude curves with proper slopes between the corner frequencies. The exact curve, which lies close to the asymptotic curve, can be obtained by adding proper corrections.
The phase-angle curve of G(jv)H(jv) can be drawn by adding the phase-angle curves of individual factors.
The use of Bode diagrams employing asymptotic approximations requires much less time than other methods that may be used for computing the frequency response of a transfer function. The ease of plotting the frequency-response curves for a given trans-fer function and the ease of modification of the frequency-response curve as compensation is added are the main reasons why Bode diagrams are very frequently used in practice.
/GAjvrB = -tan-1 21 - 2z2 z = -90° + sin-1 z 21 - z2 Mr = 1 Mr = @G(jv)@ max = @GAjvrB @ = 1 2z21 - z2 414 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method EXAMPLE 7–3 Draw the Bode diagram for the following transfer function: Make corrections so that the log-magnitude curve is accurate.
To avoid any possible mistakes in drawing the log-magnitude curve,it is desirable to put G(jv) in the following normalized form, where the low-frequency asymptotes for the first-order factors and the second-order factor are the 0-dB line: This function is composed of the following factors: The corner frequencies of the third, fourth, and fifth terms are v=3, v=2, and respectively. Note that the last term has the damping ratio of 0.3536.
To plot the Bode diagram, the separate asymptotic curves for each of the factors are shown in Figure 7–11.The composite curve is then obtained by algebraically adding the individual curves, also shown in Figure 7–11. Note that when the individual asymptotic curves are added at each fre-quency, the slope of the composite curve is cumulative. Below the plot has the slope of –20 dBdecade.At the first corner frequency the slope changes to –60 dBdecade and continues to the next corner frequency v=2, where the slope becomes –80 dBdecade. At the last corner frequency v=3, the slope changes to –60 dBdecade.
Once such an approximate log-magnitude curve has been drawn, the actual curve can be obtained by adding corrections at each corner frequency and at frequencies one octave below and above the corner frequencies. For first-order factors (1+jvT)<1, the corrections are ;3 dB at the corner frequency and ;1 dB at the frequencies one octave below and above the corner frequency. Corrections necessary for the quadratic factor are obtained from Figure 7–9.The exact log-magnitude curve for G(jv) is shown by a dashed curve in Figure 7–11.
Note that any change in the slope of the magnitude curve is made only at the corner frequencies of the transfer function G(jv). Therefore, instead of drawing individual magnitude curves and adding them up, as shown, we may sketch the magnitude curve without sketching individual curves. We may start drawing the lowest-frequency portion of the straight line (that is, the straight line with the slope –20 dBdecade for ). As the frequency is increased, we get the effect of the complex-conjugate poles (quadratic term) at the corner frequency The complex-conjugate poles cause the slopes of the magnitude curve to change from –20 to –60 dBdecade. At the next corner frequency, v=2, the effect of the pole is to change the slope to –80 dBdecade. Finally, at the corner frequency v=3, the effect of the zero is to change the slope from –80 to –60 dBdecade.
For plotting the complete phase-angle curve, the phase-angle curves for all factors have to be sketched. The algebraic sum of all phase-angle curves provides the complete phase-angle curve, as shown in Figure 7–11.
v = 12.
v 6 12 v = 12, v = 12, v = 12, 7.5, (jv)-1, 1 + j v 3 , a1 + j v 2 b -1 , c1 + j v 2 + (jv)2 2 d -1 G(jv) = 7.5 a jv 3 + 1 b (jv) a jv 2 + 1 b c (jv)2 2 + jv 2 + 1d G(jv) = 10(jv + 3) (jv)(jv + 2)C(jv)2 + jv + 2D Section 7–2 / Bode Diagrams 415 40 20 0 –20 dB –40 Exact curve 0.2 0.4 0.6 0.8 1 2 4 6 8 10 v –270° –180° –90° 0° 90° 0.2 0.4 0.6 0.8 1 2 4 6 8 10 v f G(jv) 2 2 5 5 4 4 3 1 G(jv) 3 1 Figure 7–11 Bode diagram of the system considered in Example 7–3.
Minimum-Phase Systems and Nonminimum-Phase Systems.
Transfer func-tions having neither poles nor zeros in the right-half s plane are minimum-phase trans-fer functions, whereas those having poles and/or zeros in the right-half s plane are nonminimum-phase transfer functions. Systems with minimum-phase transfer functions are called minimum-phase systems, whereas those with nonminimum-phase transfer functions are called nonminimum-phase systems.
For systems with the same magnitude characteristic, the range in phase angle of the minimum-phase transfer function is minimum among all such systems, while the range in phase angle of any nonminimum-phase transfer function is greater than this minimum.
It is noted that for a minimum-phase system, the transfer function can be uniquely determined from the magnitude curve alone. For a nonminimum-phase system, this is not the case. Multiplying any transfer function by all-pass filters does not alter the magnitude curve, but the phase curve is changed.
Consider as an example the two systems whose sinusoidal transfer functions are, respectively, G1(jv) = 1 + jvT 1 + jvT 1 , G2(jv) = 1 - jvT 1 + jvT 1 , 0 6 T 6 T 1 416 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method f 0° –90° –180° G1(jv) G2(jv) v Figure 7–13 Phase-angle characteristics of the systems G1(s) and G2(s) shown in Figure 7–12.
The pole–zero configurations of these systems are shown in Figure 7–12.The two sinu-soidal transfer functions have the same magnitude characteristics, but they have differ-ent phase-angle characteristics, as shown in Figure 7–13.These two systems differ from each other by the factor The magnitude of the factor (1-jvT)/(1+jvT) is always unity. But the phase angle equals –2 tan–1vT and varies from 0° to –180° as v is increased from zero to infinity.
As stated earlier, for a minimum-phase system, the magnitude and phase-angle char-acteristics are uniquely related. This means that if the magnitude curve of a system is specified over the entire frequency range from zero to infinity, then the phase-angle curve is uniquely determined, and vice versa. This, however, does not hold for a non-minimum-phase system.
Nonminimum-phase situations may arise in two different ways. One is simply when a system includes a nonminimum-phase element or elements. The other situation may arise in the case where a minor loop is unstable.
For a minimum-phase system, the phase angle at v=q becomes –90°(q-p), where p and q are the degrees of the numerator and denominator polynomials of the transfer function, respectively. For a nonminimum-phase system, the phase angle at v=q differs from –90°(q-p). In either system, the slope of the log-magnitude curve at v=q is equal to –20(q-p) dBdecade. It is therefore possible to detect whether the system is minimum phase by examining both the slope of the high-frequency asymptote of the log-magnitude curve and the phase angle at v=q. If the slope of the log-magnitude curve as v approaches infinity is –20(q-p) dBdecade and the phase angle at v=q is equal to –90°(q-p), then the system is minimum phase.
G(jv) = 1 - jvT 1 + jvT jv 1 T – 1 T1 – 1 T1 – s G1(s) = 1 + Ts 1 + T1s jv 1 T s G2(s) = 1 – Ts 1 + T1s 0 0 Figure 7–12 Pole–zero configurations of a minimum-phase system G1(s) and nonminimum-phase system G2(s).
Section 7–2 / Bode Diagrams 417 0° –100° –200° –300° – 400° –500° – 600° 0.1 0.2 0.4 0.6 0.8 1 10 2 4 6 8 vT e–jvT G G(jv) = e–jvT |G(jv)| = 0 dB Figure 7–14 Phase-angle characteristic of transport lag.
Nonminimum-phase systems are slow in responding because of their faulty behavior at the start of a response. In most practical control systems, excessive phase lag should be carefully avoided.In designing a system,if fast speed of response is of primary importance, we should not use nonminimum-phase components. (A common example of nonmini-mum-phase elements that may be present in control systems is transport lag or dead time.) It is noted that the techniques of frequency-response analysis and design to be presented in this and the next chapter are valid for both minimum-phase and nonminimum-phase systems.
Transport Lag.
Transport lag, which is also called dead time, is of nonminimum-phase behavior and has an excessive phase lag with no attenuation at high frequencies.
Such transport lags normally exist in thermal, hydraulic, and pneumatic systems.
Consider the transport lag given by The magnitude is always equal to unity, since Therefore, the log magnitude of the transport lag e–jvT is equal to 0 dB. The phase angle of the transport lag is The phase angle varies linearly with the frequency v.The phase-angle characteristic of transport lag is shown in Figure 7–14. = -57.3 vT (degrees) /G(jv) = -vT (radians) @G(jv)@ = ∑cos vT - j sinvT∑= 1 G(jv) = e-jvT 418 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method 0° –20 –10 0 10 20 –100° –200° –300° 0° –90° –180° –270° 0.1 0.2 0.4 0.6 0.8 1 10 2 4 6 8 v dB e–0.5 jv 1 + jv e–0.5 jv 1 1 + jv e–0.5 jv 1 + jv Figure 7–15 Bode diagram for the system e–jvL/(1+jvT) with L=0.5 and T=1.
EXAMPLE 7–4 Draw the Bode diagram of the following transfer function: The log magnitude is The phase angle of G(jv) is The log-magnitude and phase-angle curves for this transfer function with L=0.5 and T=1 are shown in Figure 7–15. = -vL - tan-1 vT /G(jv) = /e-jvL + n 1 1 + jvT = 0 + 20 log 2 1 1 + jvT 2 20 log @G(jv)@ = 20 log @e-jvL@ + 20 log 2 1 1 + jvT 2 G(jv) = e-jvL 1 + jvT Section 7–2 / Bode Diagrams 419 R(s) C(s) E(s) G(s) + – Figure 7–16 Unity-feedback control system.
dB 20 log Kp 0 –20 dB/decade –40 dB/decade v in log scale Figure 7–17 Log-magnitude curve of a type 0 system.
Relationship between System Type and Log-Magnitude Curve.
Consider the unity-feedback control system.The static position, velocity, and acceleration error con-stants describe the low-frequency behavior of type 0, type 1, and type 2 systems, respectively. For a given system, only one of the static error constants is finite and significant. (The larger the value of the finite static error constant, the higher the loop gain is as v approaches zero.) The type of the system determines the slope of the log-magnitude curve at low frequencies. Thus, information concerning the existence and magnitude of the steady-state error of a control system to a given input can be determined from the observation of the low-frequency region of the log-magnitude curve.
Determination of Static Position Error Constants.
Consider the unity-feedback control system shown in Figure 7–16. Assume that the open-loop transfer function is given by or Figure 7–17 shows an example of the log-magnitude plot of a type 0 system. In such a system, the magnitude of G(jv) equals Kp at low frequencies, or It follows that the low-frequency asymptote is a horizontal line at 20 log Kp dB.
lim vS0G(jv) = K = K p G(jv) = KAT a jv + 1BAT b jv + 1B p AT m jv + 1B (jv)NAT 1 jv + 1BAT 2 jv + 1B p AT p jv + 1B G(s) = KAT a s + 1BAT b s + 1B p AT m s + 1B sNAT 1 s + 1BAT 2 s + 1B p AT p s + 1B 420 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Determination of Static Velocity Error Constants.
Consider the unity-feedback control system shown in Figure 7–16.Figure 7–18 shows an example of the log-magnitude plot of a type 1 system. The intersection of the initial –20-dBdecade segment (or its extension) with the line v=1 has the magnitude 20 log Kv.This may be seen as follows: In a type 1 system Thus, The intersection of the initial –20-dBdecade segment (or its extension) with the 0-dB line has a frequency numerically equal to Kv. To see this, define the frequency at this intersection to be v1; then or As an example, consider the type 1 system with unity feedback whose open-loop transfer function is If we define the corner frequency to be v2 and the frequency at the intersection of the –40-dBdecade segment (or its extension) with 0-dB line to be v3, then v2 = F J , v2 3 = K J G(s) = K s(Js + F) K v = v1 2 K v jv1 2 = 1 20 log 2 K v jv 2 v=1 = 20 log K v G(jv) = K v jv , for v 1 dB 0 –20 dB/decade –40 dB/decade v in log scale 20 log Kv v1 v2 v3 v = 1 Figure 7–18 Log-magnitude curve of a type 1 system.
Section 7–2 / Bode Diagrams 421 dB 0 –20 dB/decade –40 dB/decade –60 dB/decade v in log scale 20 log Ka v = 1 va = Ka Figure 7–19 Log-magnitude curve of a type 2 system.
Since it follows that or On the Bode diagram, Thus, the v3 point is just midway between the v2 and v1 points.The damping ratio z of the system is then Determination of Static Acceleration Error Constants.
Consider the unity-feedback control system shown in Figure 7–16. Figure 7–19 shows an example of the log-magnitude plot of a type 2 system. The intersection of the initial –40-dBdecade segment (or its extension) with the v=1 line has the magnitude of 20 log Ka. Since at low frequencies it follows that 20 log 2 K a (jv)2 2 v=1 = 20 log K a G(jv) = K a (jv)2 , for v 1 z = F 21KJ = v2 2v3 log v1 - log v3 = log v3 - log v2 v1 v3 = v3 v2 v1 v2 = v2 3 v1 = K v = K F 422 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method The frequency va at the intersection of the initial –40-dBdecade segment (or its extension) with the 0-dB line gives the square root of Ka numerically.This can be seen from the following: which yields Plotting Bode Diagrams with MATLAB.
The command bode computes magni-tudes and phase angles of the frequency response of continuous-time, linear, time-invariant systems.
When the command bode (without left-hand arguments) is entered in the computer, MATLAB produces a Bode plot on the screen. Most commonly used bode commands are bode(num,den) bode(num,den,w) bode(A,B,C,D) bode(A,B,C,D,w) bode(A,B,C,D,iu,w) bode(sys) When invoked with left-hand arguments, such as [mag,phase,w] = bode(num,den,w) bode returns the frequency response of the system in matrices mag, phase, and w. No plot is drawn on the screen.The matrices mag and phase contain magnitudes and phase angles of the frequency response of the system, evaluated at user-specified frequency points.The phase angle is returned in degrees.The magnitude can be converted to deci-bels with the statement magdB = 20log10(mag) Other Bode commands with left-hand arguments are [mag,phase,w] = bode(num,den) [mag,phase,w] = bode(num,den,w) [mag,phase,w] = bode(A,B,C,D) [mag,phase,w] = bode(A,B.C,D,w) [mag,phase,w] = bode(A,B,C,D,iu,w) [mag,phase,w] = bode(sys) To specify the frequency range, use the command logspace(d1,d2) or logspace (d1,d2,n). logspace(d1,d2) generates a vector of 50 points logarithmically equally spaced between decades 10d1 and 10d2. (50 points include both endpoints. There are 48 points between the endpoints.) To generate 50 points between 0.1 radsec and 100 radsec, enter the command w = logspace(-1,2) va = 1K a 20 log 2 K a AjvaB 2 2 = 20 log 1 = 0 Section 7–2 / Bode Diagrams 423 MATLAB Program 7–1 num = ; den = [1 4 25]; bode(num,den) title('Bode Diagram of G(s) = 25/(s^2 + 4s + 25)') Frequency (rad/sec) Bode Diagram of G(s) = 25/(s2 + 4s + 25) −200 −50 −100 −150 0 −60 −40 −20 Phase (deg); Magnitude (dB) 20 0 100 101 102 Figure 7–20 Bode diagram of G(s) = 25 s2 + 4s + 25 .
logspace(dl,d2,n) generates n points logarithmically equally spaced between decades 10d1 and 10d2. (n points include both endpoints.) For example, to generate 100 points in-cluding both endpoints between 1 radsec and 1000 radsec,enter the following command: w = logspace(0,3,100) To incorporate the user-specified frequency points when plotting Bode diagrams, the bode command must include the frequency vector w, such as bode(num,den,w) and [mag,phase,w] = bode(A,B,C,D,w).
EXAMPLE 7–5 Consider the following transfer function: Plot a Bode diagram for this transfer function.
When the system is defined in the form use the command bode(num,den) to draw the Bode diagram. [When the numerator and denom-inator contain the polynomial coefficients in descending powers of s, bode(num,den) draws the Bode diagram.] MATLAB Program 7–1 shows a program to plot the Bode diagram for this sys-tem.The resulting Bode diagram is shown in Figure 7–20.
G(s) = num(s) den(s) G(s) = 25 s2 + 4s + 25 424 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Frequency (rad/sec) Bode Diagram of G(s) = 9(s2 + 0.2s + 1)/[s(s2 + 1.2s + 9)] −100 0 −50 100 50 −20 Phase (deg); Magnitude (dB) −10 40 0 10 20 30 10−2 10−1 100 101 Figure 7–22 Bode diagram of G(s) = 9As2 + 0.2s + 1B sAs2 + 1.2s + 9B .
EXAMPLE 7–6 Consider the system shown in Figure 7–21.The open-loop transfer function is Plot a bode diagram.
MATLAB Program 7–2 plots a Bode diagram for the system.The resulting plot is shown in Figure 7–22. The frequency range in this case is automatically determined to be from 0.01 to 10 radsec.
G(s) = 9As2 + 0.2s + 1B sAs2 + 1.2s + 9B 9(s2 + 0.2s + 1) s(s2 + 1.2s + 9) + – Figure 7–21 Control system.
MATLAB Program 7–2 num = [9 1.8 9]; den = [1 1.2 9 0]; bode(num,den) title('Bode Diagram of G(s) = 9(s^2 + 0.2s + 1)/[s(s^2 + 1.2s + 9)]') Section 7–2 / Bode Diagrams 425 Frequency (rad/sec) Bode Diagram of G(s) = 9(s2 + 0.2s + 1)/[s(s2 + 1.2s + 9)] −100 −50 0 50 100 −50 Phase (deg); Magnitude (dB) 0 50 10−2 10−1 100 101 102 103 Figure 7–23 Bode diagram of G(s) = 9As2 + 0.2s + 1B sAs2 + 1.2s + 9B .
MATLAB Program 7–3 num = [9 1.8 9]; den = [1 1.2 9 0]; w = logspace(-2,3,100); bode(num,den,w) title('Bode Diagram of G(s) = 9(s^2 + 0.2s + 1)/[s(s^2 + 1.2s + 9)]') If it is desired to plot the Bode diagram from 0.01 to 1000 radsec, enter the following command: w = logspace(-2,3,100) This command generates 100 points logarithmically equally spaced between 0.01 and 100 radsec.
(Note that such a vector w specifies the frequencies in radians per second at which the frequency response will be calculated.) If we use the command bode(num,den,w) then the frequency range is as the user specified, but the magnitude range and phase-angle range will be automatically determined. See MATLAB Program 7–3 and the resulting plot in Figure 7–23.
426 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Obtaining Bode Diagrams of Systems Defined in State Space.
Consider the system defined by where state vector (n-vector) output vector (m-vector) control vector (r-vector) state matrix (nn matrix) control matrix (nr matrix) output matrix (mn matrix) direct transmission matrix (mr matrix) A Bode diagram for this system may be obtained by entering the command bode(A,B,C,D) or others listed earlier in this section.
The command bode(A,B,C,D) produces a series of Bode plots, one for each input of the system, with the frequency range automatically determined. (More points are used when the response is changing rapidly.) The command bode(A,B,C,D,iu), where iu is the ith input of the system, produces the Bode diagrams from the input iu to all the outputs Ay1,y2, p , ym B of the system, with a frequency range automatically determined. (The scalar iu is an index into the in-puts of the system and specifies which input is to be used for plotting Bode diagrams).
If the control vector u has three inputs such that then iu must be set to either 1, 2, or 3.
If the system has only one input u, then either of the following commands may be used: bode(A,B,C,D) or bode(A,B,C,D,1) EXAMPLE 7–7 Consider the following system: This system has one input u and one output y. By using the command bode(A,B,C,D) y = [1 0]B x1 x2 R B x # 1 x # 2 R = B 0 -25 1 -4R Bx1 x2 R + B 0 25R u u = C u1 u2 u3 S D = C = B = A = u = y = x = y = Cx + Du x # = Ax + Bu Section 7–3 / Polar Plots 427 Frequency (rad/sec) Bode Diagram −200 −50 −100 −150 0 −60 −40 −20 Phase (deg); Magnitude (dB) 20 0 100 101 102 Figure 7–24 Bode diagram of the system considered in Example 7–7.
MATLAB Program 7–4 A = [0 1;-25 -4]; B = [0;25]; C = [1 0]; D = ; bode(A,B,C,D) title('Bode Diagram') and entering MATLAB Program 7–4 into the computer, we obtain the Bode diagram shown in Figure 7–24.
7–3 POLAR PLOTS The polar plot of a sinusoidal transfer function G(jv) is a plot of the magnitude of G(jv) versus the phase angle of G(jv) on polar coordinates as v is varied from zero to infin-ity.Thus, the polar plot is the locus of vectors as v is varied from zero to infinity. Note that in polar plots a positive (negative) phase angle is measured counter-clockwise (clockwise) from the positive real axis.The polar plot is often called the Nyquist plot.An example of such a plot is shown in Figure 7–25. Each point on the polar plot of G(jv) represents the terminal point of a vector at a particular value of v. In the polar plot, it is important to show the frequency graduation of the locus. The projections of G(jv) on the real and imaginary axes are its real and imaginary components.
@G(jv)@ /G(jv) If we replace the command bode(A,B,C,D) in MATLAB Program 7–4 with bode(A,B,C,D,1) then MATLAB will produce the Bode diagram identical to that shown in Figure 7–24.
428 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method MATLAB may be used to obtain a polar plot G(jv) or to obtain and accurately for various values of v in the frequency range of interest.
An advantage in using a polar plot is that it depicts the frequency-response charac-teristics of a system over the entire frequency range in a single plot. One disadvantage is that the plot does not clearly indicate the contributions of each individual factor of the open-loop transfer function.
Integral and Derivative Factors (jV)1.
The polar plot of G(jv)=1/jv is the negative imaginary axis, since The polar plot of G(jv)=jv is the positive imaginary axis.
First-Order Factors (1 jVT)1.
For the sinusoidal transfer function the values of G(jv) at v=0 and v=1/T are, respectively, and If v approaches infinity, the magnitude of G(jv) approaches zero and the phase angle approaches –90°.The polar plot of this transfer function is a semicircle as the frequen-cy v is varied from zero to infinity, as shown in Figure 7–26(a).The center is located at 0.5 on the real axis, and the radius is equal to 0.5.
To prove that the polar plot of the first-order factor is a semi-circle, define G(jv) = X + jY G(jv) = 1(1 + jvT) G aj 1 T b = 1 12 /-45° G(j0) = 1/0° G(jv) = 1 1 + jvT = 1 21 + v2 T2 /-tan-1 vT G(jv) = 1 jv = -j 1 v = 1 v /-90° /G(jv) @G(jv)@ Im Re G(jv) v = 0 v = v1 v2 v3 G( jv) Im [G( jv)] Re [G( jv)] Figure 7–25 Polar plot.
Section 7–3 / Polar Plots 429 where Then we obtain Thus, in the X-Y plane G(jv) is a circle with center at and with radius as shown in Figure 7–26(b). The lower semicircle corresponds to 0 v q, and the upper semicircle corresponds to –q v 0.
The polar plot of the transfer function 1+jvT is simply the upper half of the straight line passing through point (1,0) in the complex plane and parallel to the imaginary axis, as shown in Figure 7–27. The polar plot of 1+jvT has an appearance completely different from that of 1/(1+jvT).
Quadratic Factors C1 2ZAjV/VnB AjV/VnB 2D 1.
The low- and high-fre-quency portions of the polar plot of the following sinusoidal transfer function are given, respectively, by and The polar plot of this sinusoidal transfer function starts at and ends at as v increases from zero to infinity. Thus, the high-frequency portion of G(jv) is tangent to the negative real axis.
0/-180° 1/0° lim vS q G(jv) = 0/-180° lim vS0 G(jv) = 1/0° G(jv) = 1 1 + 2z a j v vn b + aj v vn b 2 , for z 7 0 1 2 , X = 1 2 , Y = 0 aX - 1 2 b 2 + Y2 = a 1 2 1 - v2 T2 1 + v2 T2 b 2 + a -vT 1 + v2 T2 b 2 = a 1 2 b 2 Y = -vT 1 + v2 T2 = imaginary part of G(jv) X = 1 1 + v2 T2 = real part of G(jv) Im Re v = 0 0 0.5 0.5 v = (a) (b) v v v 1 1 + v2T 2 1 1 vT = 1 G j1 T G j1 T 0 Y X v = – v = v = 0 vT 1 + v2T 2 Figure 7–26 (a) Polar plot of 1/(1+jvT); (b) plot of G(jv) in X-Y plane.
Im Re v = 0 v 1 0 Figure 7–27 Polar plot of 1+jvT.
430 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Im Re v = v = 0 Resonant peak vn vr 0 Figure 7–29 Polar plot showing the resonant peak and resonant frequency vr.
Examples of polar plots of the transfer function just considered are shown in Figure 7–28. The exact shape of a polar plot depends on the value of the damping ratio z, but the general shape of the plot is the same for both the underdamped case (1>z>0) and overdamped case (z>1).
For the underdamped case at v=vn, we have G(jvn)=1/(j2z), and the phase angle at v=vn is –90°. Therefore, it can be seen that the frequency at which the G(jv) locus intersects the imaginary axis is the undamped natural frequency vn. In the polar plot, the frequency point whose distance from the origin is maximum cor-responds to the resonant frequency vr. The peak value of G(jv) is obtained as the ratio of the magnitude of the vector at the resonant frequency vr to the magnitude of the vector at v=0.The resonant frequency vr is indicated in the polar plot shown in Figure 7–29.
For the overdamped case, as z increases well beyond unity, the G(jv) locus approaches a semicircle. This may be seen from the fact that, for a heavily damped system, the characteristic roots are real, and one is much smaller than the other. Since, for sufficiently large z, the effect of the larger root (larger in the absolute value) on the response becomes very small, the system behaves like a first-order one.
v = 0 Im Re 0 1 v = (z: Large) (z: Small) vn vn vn vn Figure 7–28 Polar plots of for z>0.
1 1 + 2z aj v vn b + aj v vn b 2 Section 7–3 / Polar Plots 431 Im Re v = 0 0 1 v Figure 7–30 Polar plot of for z>0.
1 + 2z aj v vn b + aj v vn b 2 Next, consider the following sinusoidal transfer function: The low-frequency portion of the curve is and the high-frequency portion is Since the imaginary part of G(jv) is positive for v>0 and is monotonically increasing, and the real part of G(jv) is monotonically decreasing from unity, the general shape of the polar plot of G(jv) is as shown in Figure 7–30. The phase angle is between 0° and 180°.
EXAMPLE 7–8 Consider the following second-order transfer function: Sketch a polar plot of this transfer function.
Since the sinusoidal transfer function can be written the low-frequency portion of the polar plot becomes and the high-frequency portion becomes lim vS q G(jv) = 0 - j0 lim vS0G(jv) = -T - jq G(jv) = 1 jv(1 + jvT) = - T 1 + v2 T2 - j 1 vA1 + v2 T2B G(s) = 1 s(Ts + 1) lim vS q G(jv) = q /180° lim vS0 G(jv) = 1/0° = a 1 - v2 v2 n b + j a 2zv vn b G(jv) = 1 + 2z a j v vn b + a j v vn b 2 432 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method The general shape of the polar plot of G(jv) is shown in Figure 7–31. The G(jv) plot is asymp-totic to the vertical line passing through the point (–T, 0). Since this transfer function involves an integrator (1/s),the general shape of the polar plot differs substantially from those of second-order transfer functions that do not have an integrator.
EXAMPLE 7–9 Obtain the polar plot of the following transfer function: Since G(jv) can be written the magnitude and phase angle are, respectively, and Since the magnitude decreases from unity monotonically and the phase angle also decreases monotonically and indefinitely, the polar plot of the given transfer function is a spiral, as shown in Figure 7–32.
/G(jv) = /e-jvL + n 1 1 + jvT = -vL - tan-1vT @G(jv)@ = @e-jvL@ 2 1 1 + jvT 2 = 1 21 + v2 T2 G(jv) = Ae-jvLB a 1 1 + jvT b G(jv) = e-jvL 1 + jvT Im Re 0 0 v v –T Figure 7–31 Polar plot of 1/Cjv(1+jvT)D.
Im Re 1 Figure 7–32 Polar plot of e-jvL(1 + jvT).
Section 7–3 / Polar Plots 433 Im Re 0 0 0 v = 0 v v v v v Type 2 system Type 1 system Type 0 system Figure 7–33 Polar plots of type 0, type 1, and type 2 systems.
General Shapes of Polar Plots.
The polar plots of a transfer function of the form where n>m or the degree of the denominator polynomial is greater than that of the numerator, will have the following general shapes: 1. For l=0 or type 0 systems: The starting point of the polar plot (which corre-sponds to v=0) is finite and is on the positive real axis. The tangent to the polar plot at v=0 is perpendicular to the real axis. The terminal point, which corresponds to v=q, is at the origin, and the curve is tangent to one of the axes.
2. For l=1 or type 1 systems: the jv term in the denominator contributes –90° to the total phase angle of G(jv) for 0 v q.At v=0, the magnitude of G(jv) is infinity, and the phase angle becomes –90°.At low frequencies, the polar plot is asymptotic to a line parallel to the negative imaginary axis.At v=q, the magni-tude becomes zero, and the curve converges to the origin and is tangent to one of the axes.
3. For l=2 or type 2 systems: The (jv)2 term in the denominator contributes –180° to the total phase angle of G(jv) for 0 v q. At v=0, the magni-tude of G(jv) is infinity, and the phase angle is equal to –180°. At low frequencies, the polar plot may be asymptotic to the negative real axis. At v=q, the magnitude becomes zero, and the curve is tangent to one of the axes.
The general shapes of the low-frequency portions of the polar plots of type 0, type 1, and type 2 systems are shown in Figure 7–33. It can be seen that, if the degree of the = b0(jv)m + b1(jv)m-1 + p a0(jv)n + a1(jv)n-1 + p G(jv) = KA1 + jvT aBA1 + jvT bB p (jv)lA1 + jvT 1BA1 + jvT 2B p 434 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method denominator polynomial of G(jv) is greater than that of the numerator, then the G(jv) loci converge to the origin clockwise.At v=q, the loci are tangent to one or the other axes, as shown in Figure 7–34.
Note that any complicated shapes in the polar plot curves are caused by the nu-merator dynamics—that is, by the time constants in the numerator of the transfer func-tion. Figure 7–35 shows examples of polar plots of transfer functions with numerator dynamics. In analyzing control systems, the polar plot of G(jv) in the frequency range of interest must be accurately determined.
Table 7–1 shows sketches of polar plots of several transfer functions.
Im Re 0 v = n – m = 1 n – m = 2 n – m = 3 G(jv) = bo(jv)m + … ao(jv)n + … Figure 7–34 Polar plots in the high-frequency range.
Im Re 0 0 v = v Im Re 0 0 v = v Figure 7–35 Polar plots of transfer functions with numerator dynamics.
Section 7–3 / Polar Plots 435 v Im Re Im Re 0 0 0 Im Re Im Re 0 0 Im Re Im Re 0 Im Im Im Im Re Re Re Re 0 0 0 0 1 0 v v v v 0 0 v v 0 0 v = v = v = v = v = v = v = 1 jv 1 1 1 1 + jvT jvT 1 + jvT jv jvT 1 + jvT 1 a v = v = 0 v = 0 v = 0 v = 0 v = 0 1 (jv)2 1 + jvT 1 + jvaT (a 1) 1 (1 + jvT1) (1 + jvT2) (1 + jvT3) vn2 jv[(jv)2 + 2zvn( jv) + vn2] 1 + jvT1 jv (1 + jvT2) (1 + jvT3) 1 Table 7–1 Polar Plots of Simple Transfer Functions Drawing Nyquist Plots with MATLAB.
Nyquist plots, just like Bode diagrams, are commonly used in the frequency-response representation of linear, time-invariant, feedback control systems. Nyquist plots are polar plots, while Bode diagrams are rectangular plots. One plot or the other may be more convenient for a particular opera-tion, but a given operation can always be carried out in either plot.
The MATLAB command nyquist computes the frequency response for continuous-time, linear, time-invariant systems.When invoked without left-hand arguments, nyquist produces a Nyquist plot on the screen.
436 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method The command nyquist(num,den) draws the Nyquist plot of the transfer function where num and den contain the polynomial coefficients in descending powers of s.Other commonly used nyquist commands are nyquist(num,den,w) nyquist(A,B,C,D) nyquist(A,B,C,D,w) nyquist(A,B,C,D,iu,w) nyquist(sys) The command involving the user-specified frequency vector w, such as nyquist(num,den,w) calculates the frequency response at the specified frequency points in radians per second.
When invoked with left-hand arguments such as [re,im,w] = nyquist(num,den) [re,im,w] = nyquist(num,den,w) [re,im,w] = nyquist(A,B,C,D) [re,im,w] = nyquist(A,B,C,D,w) [re,im,w] = nyquist(A,B,C,D,iu,w) [re,im,w] = nyquist(sys) MATLAB returns the frequency response of the system in the matrices re, im, and w.
No plot is drawn on the screen. The matrices re and im contain the real and imaginary parts of the frequency response of the system, evaluated at the frequency points speci-fied in the vector w. Note that re and im have as many columns as outputs and one row for each element in w.
EXAMPLE 7–10 Consider the following open-loop transfer function: Draw a Nyquist plot with MATLAB.
Since the system is given in the form of the transfer function, the command nyquist(num,den) may be used to draw a Nyquist plot. MATLAB Program 7–5 produces the Nyquist plot shown in Figure 7–36. In this plot, the ranges for the real axis and imaginary axis are automatically determined.
G(s) = 1 s2 + 0.8s + 1 G(s) = num(s) den(s) Section 7–3 / Polar Plots 437 MATLAB Program 7–5 num = ; den = [1 0.8 1]; nyquist(num,den) grid title('Nyquist Plot of G(s) = 1/(s^2 + 0.8s + 1)') MATLAB Program 7–6 % ---------- Nyquist plot ----------num = ; den = [1 0.8 1]; nyquist(num,den) v = [-2 2 -2 2]; axis(v) grid title('Nyquist Plot of G(s) = 1/(s^2 + 0.8s + 1)') Real Axis −0.5 −1 1.5 0.5 1 0 Imaginary Axis −1.5 1.5 −0.5 −1 0 0.5 1 Nyquist Plot of G(s) = 1/(s2 + 0.8s + 1) Figure 7–36 Nyquist plot of G(s) = 1 s2 + 0.8s + 1 .
If we wish to draw the Nyquist plot using manually determined ranges—for example, from –2 to 2 on the real axis and from –2 to 2 on the imaginary axis—enter the following command into the computer: v = [-2 2 -2 2]; axis(v); or, combining these two lines into one, axis([-2 2 -2 2]); See MATLAB Program 7–6 and the resulting Nyquist plot shown in Figure 7–37.
438 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Caution.
In drawing a Nyquist plot,where a MATLAB operation involves “Divide by zero,” the resulting Nyquist plot may have an erroneous or undesirable appearance.
For example, if the transfer function G(s) is given by then the MATLAB command num = ; den = [1 1 0]; nyquist(num,den) produces an undesirable Nyquist plot. An example of an undesirable Nyquist plot is shown in Figure 7–38. If such an undesirable Nyquist plot appears on the computer, G(s) = 1 s(s + 1) Real Axis −2 2 −1.5 1.5 1 0.5 0 −0.5 −1 Imaginary Axis −1 0.5 −2 2 −0.5 −1.5 0 1 1.5 Nyquist Plot of G(s) = 1/(s2 + 0.8s + 1) Figure 7–37 Nyquist plot of G(s) = 1 s2 + 0.8s + 1 .
Real Axis −1.2 −1.4 0 −0.4 −0.2 −0.8 −1 −0.6 Imaginary Axis −150 150 −50 −100 0 50 100 Nyquist Diagram Figure 7–38 Undesirable Nyquist plot.
Section 7–3 / Polar Plots 439 Real Axis −1.5 −2 1 2 0.5 1.5 −0.5 −1 0 Imaginary Axis −2 1 −5 5 −1 2 −3 −4 0 3 4 Nyquist Plot of G(s) = 1/[s(s+1)] Figure 7–39 Nyquist plot of G(s) = 1 s(s + 1) .
then it can be corrected if we specify the axis(v). For example, if we enter the axis command v = [-2 2 -5 5]; axis(v) in the computer,then a desirable form of Nyquist plot can be obtained.See Example 7–11.
EXAMPLE 7–11 Draw a Nyquist plot for the following G(s): MATLAB Program 7–7 will produce a desirable form of Nyquist plot on the computer, even though a warning message “Divide by zero” may appear on the screen.The resulting Nyquist plot is shown in Figure 7–39.
G(s) = 1 s(s + 1) MATLAB Program 7–7 % ---------- Nyquist plot----------num = ; den = [1 1 0]; nyquist(num,den) v = [-2 2 -5 5]; axis(v) grid title('Nyquist Plot of G(s) = 1/[s(s + 1)]') Notice that the Nyquist plot shown in Figure 7–39 includes the loci for both v>0 and v<0.
If we wish to draw the Nyquist plot for only the positive frequency region (v>0), then we need to use the command [re,im,w]=nyquist(num,den,w) 440 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Drawing Nyquist Plots of a System Defined in State Space.
Consider the system defined by where state vector (n-vector) output vector (m-vector) control vector (r-vector) state matrix (nn matrix) control matrix (nr matrix) output matrix (mn matrix) direct transmission matrix (mr matrix) D = C = B = A = u = y = x = y = Cx + Du x # = Ax + Bu MATLAB Program 7–8 % ---------- Nyquist plot----------num = ; den = [1 1 0]; w = 0.1:0.1:100; [re,im,w] = nyquist(num,den,w); plot(re,im) v = [-2 2 -5 5]; axis(v) grid title('Nyquist Plot of G(s) = 1/[s(s + 1)]') xlabel('Real Axis') ylabel('Imag Axis') Real Axis –1.5 –2 1 2 0.5 1.5 –0.5 –1 0 Imag Axis –2 1 –5 5 –1 2 –3 –4 0 3 4 Nyquist Plot of G(s) = 1/[s(s+1)] Figure 7–40 Nyquist plot of for v 7 0.
G(s) = 1 s(s + 1) A MATLAB program using this nyquist command is shown in MATLAB Program 7–8. The resulting Nyquist plot is presented in Figure 7–40.
Section 7–3 / Polar Plots 441 MATLAB Program 7–9 A = [0 1;-25 -4]; B = [0;25]; C = [1 0]; D = ; nyquist(A,B,C,D) grid title('Nyquist Plot') Nyquist plots for this system may be obtained by the use of the command nyquist(A,B,C,D) This command produces a series of Nyquist plots, one for each input and output com-bination of the system.The frequency range is automatically determined.
The command nyquist(A,B,C,D,iu) produces Nyquist plots from the single input iu to all the outputs of the system, with the frequency range determined automatically.The scalar iu is an index into the inputs of the system and specifies which input to use for the frequency response.
The command nyquist(A,B,C,D,iu,w) uses the user-supplied frequency vector w. The vector w specifies the frequencies in radians per second at which the frequency response should be calculated.
EXAMPLE 7–12 Consider the system defined by Draw a Nyquist plot.
This system has a single input u and a single output y. A Nyquist plot may be obtained by entering the command nyquist(A,B,C,D) or nyquist(A,B,C,D,1) MATLAB Program 7–9 will provide the Nyquist plot. (Note that we obtain the identical result by using either of these two commands.) Figure 7–41 shows the Nyquist plot produced by MATLAB Program 7–9. y = [1 0]B x1 x2 R + u B x # 1 x # 2 R = B 0 -25 1 -4R Bx1 x2 R + B 0 25R u 442 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method EXAMPLE 7–13 Consider the system defined by This system involves two inputs and two outputs. There are four sinusoidal output–input re-lationships: and Draw Nyquist plots for the system. (When considering input u1, we assume that input u2 is zero, and vice versa.) The four individual Nyquist plots can be obtained by the use of the command nyquist(A,B,C,D) MATLAB Program 7–10 produces the four Nyquist plots.They are shown in Figure 7–42.
Y 2(jv)U 2(jv).
Y 1(jv)U 2(jv), Y 2(jv)U 1(jv), Y 1(jv)U 1(jv), B y1 y2 R = B 1 0 0 1R Bx1 x2 R + B 0 0 0 0R B u1 u2 R Bx # 1 x # 2 R = B -1 6.5 -1 0R Bx1 x2 R + B 1 1 1 0R B u1 u2 R Real Axis –0.4 –0.6 0.6 1 0.4 0.8 0 –0.2 0.2 1.2 Imag Axis –1 0 1.5 –0.5 0.5 –1.5 1 Nyquist Plot Figure 7–41 Nyquist plot of system considered in Example 7–12.
MATLAB Program 7–10 A = [-1 -1;6.5 0]; B = [1 1;1 0]; C = [1 0;0 1]; D = [0 0;0 0]; nyquist(A,B,C,D) Section 7–4 / Log-Magnitude-versus-Phase Plots 443 7–4 LOG-MAGNITUDE-VERSUS-PHASE PLOTS Another approach to graphically portraying the frequency-response characteristics is to use the log-magnitude-versus-phase plot, which is a plot of the logarithmic magnitude in decibels versus the phase angle or phase margin for a frequency range of interest. [The phase margin is the difference between the actual phase angle f and –180°; that is, f-(–180°)=180°+f.] The curve is graduated in terms of the frequency v. Such log-magnitude-versus-phase plots are commonly called Nichols plots.
In the Bode diagram, the frequency-response characteristics of G(jv) are shown on semilog paper by two separate curves, the log-magnitude curve and the phase-angle curve, while in the log-magnitude-versus-phase plot, the two curves in the Bode dia-gram are combined into one. In the manual approach the log-magnitude-versus-phase plot can easily be constructed by reading values of the log magnitude and phase angle from the Bode diagram. Notice that in the log-magnitude-versus-phase plot a change in the gain constant of G(jv) merely shifts the curve up (for increasing gain) or down (for decreasing gain), but the shape of the curve remains the same.
Advantages of the log-magnitude-versus-phase plot are that the relative stability of the closed-loop system can be determined quickly and that compensation can be worked out easily.
The log-magnitude-versus-phase plot for the sinusoidal transfer function G(jv) and that for 1/G(jv) are skew symmetrical about the origin, since 2 1 G(jv) 2 in dB = - @G(jv)@ in dB 4 2 0 −2 −4 1 0.5 0 −0.5 −1 4 2 0 −2 −4 4 2 0 −2 −4 1 2 0 −1 Real Axis 3 1 2 0 −1 3 0 1 −1 −2 2 0 1 −1 −2 2 From: U1 From: U2 From: U1 From: U2 Real Axis Real Axis Real Axis To: Y2 Imaginary Axis To: Y1 To: Y2 To: Y1 Nyquist Diagrams Figure 7–42 Nyquist plot of system considered in Example 7–13.
444 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method and Figure 7–43 compares frequency-response curves of in three different representations. In the log-magnitude-versus-phase plot, the vertical distance between the points v=0 and v=vr,where vr is the resonant frequency,is the peak value of G(jv) in decibels.
Since log-magnitude and phase-angle characteristics of basic transfer functions have been discussed in detail in Sections 7–2 and 7–3, it will be sufficient here to give exam-ples of some log-magnitude-versus-phase plots.Table 7–2 shows such examples. (How-ever, more on Nichols charts will be discussed in Section 7–6.) G(jv) = 1 1 + 2z a j v vn b + a j v vn b 2 n 1 G(jv) = - /G(jv) 0 5 –5 –10 0° –90° –180° |G| in dB G Mr Mr 0.2vn 0.5vn vn 2vn vr (a) v = 0 v = 0 v = ∞ v v Im Re vr vn vr vn Mr (b) (c) –12 –15 6 3 0 1 –6 –3 –9 –90° –180° 0° |G| in dB G Figure 7–43 Three representations of the frequency response of for z>0.
(a) Bode diagram; (b) polar plot; (c) log-magnitude-versus-phase plot.
1 1 + 2z aj v vn b + aj v vn b 2 , Section 7–5 / Nyquist Stability Criterion 445 |G| in dB 20 10 0 –10 –20 –180° 0° 180° G |G| in dB 20 10 0 –10 –20 –180° 0° 180° G |G| in dB 20 10 0 –10 –20 –180° 0° 180° G |G| in dB 20 10 0 –10 –20 –180° 0° 180° G |G| in dB 20 10 0 –10 –20 –180° 0° 180° G |G| in dB 20 10 0 –10 –20 –180° 0° 180° G v v 0 v = 1 G = 1 jv G = 1 1 + jvT v v = 0 G = (jv)2 + 2zvn(jv) + vn2 vn2 G = 1 + jvT G = e–jvL G = 1 jv(1 + jvT) v v = 0 v v = 0 v v = 0 v v 0 Table 7–2 Log-Magnitude-versus-Phase Plots of Simple Transfer Functions 7–5 NYQUIST STABILITY CRITERION The Nyquist stability criterion determines the stability of a closed-loop system from its open-loop frequency response and open-loop poles.
This section presents mathematical background for understanding the Nyquist sta-bility criterion. Consider the closed-loop system shown in Figure 7–44.The closed-loop transfer function is C(s) R(s) = G(s) 1 + G(s)H(s) 446 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method For stability, all roots of the characteristic equation must lie in the left-half s plane.[It is noted that,although poles and zeros of the open-loop transfer function G(s)H(s) may be in the right-half s plane, the system is stable if all the poles of the closed-loop transfer function (that is, the roots of the characteristic equation) are in the left-half s plane.] The Nyquist stability criterion relates the open-loop frequency response G(jv)H(jv) to the number of zeros and poles of 1+G(s)H(s) that lie in the right-half s plane.This criterion,derived by H.Nyquist,is useful in control engineering be-cause the absolute stability of the closed-loop system can be determined graphically from open-loop frequency-response curves, and there is no need for actually determining the closed-loop poles. Analytically obtained open-loop frequency-response curves, as well as those experimentally obtained,can be used for the stability analysis.This is convenient be-cause, in designing a control system, it often happens that mathematical expressions for some of the components are not known; only their frequency-response data are available.
The Nyquist stability criterion is based on a theorem from the theory of complex variables.To understand the criterion, we shall first discuss mappings of contours in the complex plane.
We shall assume that the open-loop transfer function G(s)H(s) is representable as a ratio of polynomials in s. For a physically realizable system, the degree of the denom-inator polynomial of the closed-loop transfer function must be greater than or equal to that of the numerator polynomial.This means that the limit of G(s)H(s) as s approaches infinity is zero or a constant for any physically realizable system.
Preliminary Study.
The characteristic equation of the system shown in Figure 7–44 is We shall show that, for a given continuous closed path in the s plane that does not go through any singular points, there corresponds a closed curve in the F(s) plane. The number and direction of encirclements of the origin of the F(s) plane by the closed curve play a particularly important role in what follows, for later we shall correlate the number and direction of encirclements with the stability of the system.
Consider, for example, the following open-loop transfer function: The characteristic equation is (7–15) = 1 + 2 s - 1 = s + 1 s - 1 = 0 F(s) = 1 + G(s)H(s) G(s)H(s) = 2 s - 1 F(s) = 1 + G(s)H(s) = 0 1 + G(s)H(s) = 0 R(s) C(s) G(s) H(s) + – Figure 7–44 Closed-loop system.
Section 7–5 / Nyquist Stability Criterion 447 s Plane F(s) Plane 3 2 0 –2 2 3 4 –2 –3 Re Im jv v = –2 s = –1 s = –2 s = 1 s v = 0 v = 2 v = –1 v = 1 s = 2 s = 0 –2 –1 0 1 2 j2 j1 –j1 –j2 1 –1 3 –1 (a) (b) Figure 7–45 Conformal mapping of the s-plane grids into the F(s) plane, where F(s)=(s+1)/(s-1).
The function F(s) is analytic# everywhere in the s plane except at its singular points.
For each point of analyticity in the s plane, there corresponds a point in the F(s) plane.
For example, if s=2+j1, then F(s) becomes Thus, point s=2+j1 in the s plane maps into point 2-j1 in the F(s) plane.
Thus,as stated previously,for a given continuous closed path in the s plane,which does not go through any singular points, there corresponds a closed curve in the F(s) plane.
For the characteristic equation F(s) given by Equation (7–15), the conformal map-ping of the lines and the lines [see Figure 7–45(a)] yield cir-cles in the F(s) plane, as shown in Figure 7–45(b). Suppose that representative point s traces out a contour in the s plane in the clockwise direction. If the contour in the s plane encloses the pole of F(s), there is one encirclement of the origin of the F(s) plane by the locus of F(s) in the counterclockwise direction. [See Figure 7–46(a).] If the con-tour in the s plane encloses the zero of F(s), there is one encirclement of the origin of the F(s) plane by the locus of F(s) in the clockwise direction. [See Figure 7–46(b).] If the contour in the s plane encloses both the zero and the pole or if the contour enclos-es neither the zero nor the pole, then there is no encirclement of the origin of the F(s) plane by the locus of F(s). [See Figures 7–46(c) and (d).] From the foregoing analysis, we can say that the direction of encirclement of the ori-gin of the F(s) plane by the locus of F(s) depends on whether the contour in the s plane encloses a pole or a zero. Note that the location of a pole or zero in the s plane, whether in the right-half or left-half s plane, does not make any difference, but the enclosure of a pole or zero does. If the contour in the s plane encloses equal numbers of poles and zeros, then the corresponding closed curve in the F(s) plane does not encircle the ori-gin of the F(s) plane.The foregoing discussion is a graphical explanation of the mapping theorem, which is the basis for the Nyquist stability criterion.
s = 0, ;1, ;2 v = 0, ;1, ;2 F(2 + j1) = 2 + j1 + 1 2 + j1 - 1 = 2 - j1 #A complex function F(s) is said to be analytic in a region if F(s) and all its derivatives exist in that region.
448 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method jv s Plane j2 j1 0 –j1 –j2 3 1 –1 2 –2 A B C D s jv j2 j1 0 –j1 –j2 3 1 –1 2 A B C D s jv j2 0 –j2 3 1 –1 A B C D s jv 0 3 1 –1 2 –2 G H F E C D B A s Im F(s) Plane 2 1 0 –1 –2 1 2 A A D D A C B B C D E F G B C D C B Re Im 2 1 0 –1 –2 1 2 Re Im 2 1 0 –1 –2 1 –1 2 Re Im 2 1 0 –1 –2 3 1 –1 2 Re 3 –1 3 –1 –2 2 –2 j1 –j1 j2 j1 –j1 –j2 3 H A (a) (b) (c) (d) Figure 7–46 Closed contours in the s plane and their corresponding closed curves in the F(s) plane, where F(s)=(s+1)/(s-1).
Mapping Theorem.
Let F(s) be a ratio of two polynomials in s. Let P be the num-ber of poles and Z be the number of zeros of F(s) that lie inside some closed contour in the s plane, with multiplicity of poles and zeros accounted for. Let the contour be such that it does not pass through any poles or zeros of F(s).This closed contour in the s plane is then mapped into the F(s) plane as a closed curve. The total number N of clockwise encirclements of the origin of the F(s) plane, as a representative point s traces out the entire contour in the clockwise direction, is equal to Z-P. (Note that by this mapping theorem, the numbers of zeros and of poles cannot be found—only their difference.) Section 7–5 / Nyquist Stability Criterion 449 We shall not present a formal proof of this theorem here, but leave the proof to Problem A–7–6. Note that a positive number N indicates an excess of zeros over poles of the function F(s) and a negative N indicates an excess of poles over zeros. In control system applications, the number P can be readily determined for F(s)=1+G(s)H(s) from the function G(s)H(s). Therefore, if N is determined from the plot of F(s), the number of zeros in the closed contour in the s plane can be determined readily.Note that the exact shapes of the s-plane contour and F(s) locus are immaterial so far as encir-clements of the origin are concerned, since encirclements depend only on the enclosure of poles and/or zeros of F(s) by the s-plane contour.
Application of the Mapping Theorem to the Stability Analysis of Closed-Loop Systems.
For analyzing the stability of linear control systems, we let the closed con-tour in the s plane enclose the entire right-half s plane.The contour consists of the en-tire jv axis from v=–q to ±q and a semicircular path of infinite radius in the right-half s plane. Such a contour is called the Nyquist path. (The direction of the path is clockwise.) The Nyquist path encloses the entire right-half s plane and encloses all the zeros and poles of 1+G(s)H(s) that have positive real parts. [If there are no zeros of 1+G(s)H(s) in the right-half s plane, then there are no closed-loop poles there, and the system is stable.] It is necessary that the closed contour, or the Nyquist path, not pass through any zeros and poles of 1+G(s)H(s). If G(s)H(s) has a pole or poles at the origin of the s plane, mapping of the point s=0 becomes indeterminate. In such cases, the origin is avoided by taking a detour around it. (A detailed discussion of this special case is given later.) If the mapping theorem is applied to the special case in which F(s) is equal to 1+G(s)H(s), then we can make the following statement: If the closed contour in the s plane encloses the entire right-half s plane, as shown in Figure 7–47, then the num-ber of right-half plane zeros of the function F(s)=1+G(s)H(s) is equal to the num-ber of poles of the function F(s)=1+G(s)H(s) in the right-half s plane plus the number of clockwise encirclements of the origin of the 1+G(s)H(s) plane by the corresponding closed curve in this latter plane.
Because of the assumed condition that the function of 1+G(s)H(s) remains constant as s traverses the semicircle of infinite radius. Because of this, whether the locus of 1+G(s)H(s) encircles the origin of the 1+G(s)H(s) plane can be determined by considering only a part of the closed contour in the s plane—that is,the jv axis.Encirclements of the origin,if there are any,occur only lim sS q C1 + G(s)H(s)D = constant jv s 0 s Plane Figure 7–47 Closed contour in the s plane.
450 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Im Re 0 1 1 + G(jv) H(jv) Im Re 0 –1 1 + G(jv) H(jv) G(jv) H(jv) GH Plane 1 + GH Plane Figure 7–48 Plots of in the 1+GH plane and GH plane.
1 + G(jv)H(jv) while a representative point moves from –jq to ±jq along the jv axis, provided that no zeros or poles lie on the jv axis.
Note that the portion of the 1+G(s)H(s) contour from v=–q to v=q is sim-ply 1+G(jv)H(jv). Since 1+G(jv)H(jv) is the vector sum of the unit vector and the vector G(jv)H(jv), 1+G(jv)H(jv) is identical to the vector drawn from the –1+j0 point to the terminal point of the vector G(jv)H(jv), as shown in Figure 7–48.
Encirclement of the origin by the graph of 1+G(jv)H(jv) is equivalent to encir-clement of the –1+j0 point by just the G(jv)H(jv) locus.Thus,the stability of a closed-loop system can be investigated by examining encirclements of the –1+j0 point by the locus of G(jv)H(jv).The number of clockwise encirclements of the –1+j0 point can be found by drawing a vector from the –1+j0 point to the G(jv)H(jv) locus, starting from v=–q, going through v=0, and ending at v=±q, and by counting the number of clockwise rotations of the vector.
Plotting G(jv)H(jv) for the Nyquist path is straightforward.The map of the nega-tive jv axis is the mirror image about the real axis of the map of the positive jv axis.That is, the plot of G(jv)H(jv) and the plot of G(–jv)H(–jv) are symmetrical with each other about the real axis.The semicircle with infinite radius maps into either the origin of the GH plane or a point on the real axis of the GH plane.
In the preceding discussion, G(s)H(s) has been assumed to be the ratio of two poly-nomials in s. Thus, the transport lag e–Ts has been excluded from the discussion. Note, however, that a similar discussion applies to systems with transport lag, although a proof of this is not given here.The stability of a system with transport lag can be determined from the open-loop frequency-response curves by examining the number of encir-clements of the –1+j0 point, just as in the case of a system whose open-loop transfer function is a ratio of two polynomials in s.
Nyquist Stability Criterion.
The foregoing analysis, utilizing the encirclement of the –1+j0 point by the G(jv)H(jv) locus, is summarized in the following Nyquist stability criterion: Nyquist stability criterion [for a special case when G(s)H(s) has neither poles nor zeros on the jv axis]:In the system shown in Figure 7–44,if the open-loop transfer func-tion G(s)H(s) has k poles in the right-half s plane and then for stability,the G(jv)H(jv) locus,as v varies from –q to q,must encircle the –1+j0 point k times in the counterclockwise direction.
lim sS q G(s)H(s) = constant, Section 7–5 / Nyquist Stability Criterion 451 Im Re 0 –1 GH Plane Figure 7–49 Region enclosed by a Nyquist plot.
Remarks on the Nyquist Stability Criterion 1. This criterion can be expressed as where number of zeros of 1+G(s)H(s) in the right-half s plane number of clockwise encirclements of the –1+j0 point number of poles of G(s)H(s) in the right-half s plane If P is not zero,for a stable control system,we must have Z=0,or N=–P,which means that we must have P counterclockwise encirclements of the –1+j0 point.
If G(s)H(s) does not have any poles in the right-half s plane, then Z=N.
Thus, for stability there must be no encirclement of the –1+j0 point by the G(jv)H(jv) locus. In this case it is not necessary to consider the locus for the en-tire jv axis, only for the positive-frequency portion.The stability of such a system can be determined by seeing if the –1+j0 point is enclosed by the Nyquist plot of G(jv)H(jv).The region enclosed by the Nyquist plot is shown in Figure 7–49.
For stability, the –1+j0 point must lie outside the shaded region.
2. We must be careful when testing the stability of multiple-loop systems since they may include poles in the right-half s plane. (Note that although an inner loop may be unstable, the entire closed-loop system can be made stable by proper design.) Simple inspection of encirclements of the –1+j0 point by the G(jv)H(jv) locus is not sufficient to detect instability in multiple-loop systems. In such cases, how-ever, whether any pole of 1+G(s)H(s) is in the right-half s plane can be deter-mined easily by applying the Routh stability criterion to the denominator of G(s)H(s).
If transcendental functions,such as transport lag e–Ts,are included in G(s)H(s), they must be approximated by a series expansion before the Routh stability criterion can be applied.
3. If the locus of G(jv)H(jv) passes through the –1+j0 point, then zeros of the characteristic equation, or closed-loop poles, are located on the jv axis.This is not desirable for practical control systems. For a well-designed closed-loop system, none of the roots of the characteristic equation should lie on the jv axis. P = N = Z = Z = N + P 452 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method jv jv s s 0 s Plane s Plane j 0+ j 0– s = e e ju e Figure 7–50 Contour near the origin of the s plane and closed contour in the s plane avoiding poles and zeros at the origin.
Special Case when G(s)H(s) Involves Poles and/or Zeros on the jV Axis.
In the previous discussion, we assumed that the open-loop transfer function G(s)H(s) has neither poles nor zeros at the origin.We now consider the case where G(s)H(s) involves poles and/or zeros on the jv axis.
Since the Nyquist path must not pass through poles or zeros of G(s)H(s), if the func-tion G(s)H(s) has poles or zeros at the origin (or on the jv axis at points other than the origin), the contour in the s plane must be modified. The usual way of modifying the contour near the origin is to use a semicircle with the infinitesimal radius e, as shown in Figure 7–50. [Note that this semicircle may lie in the right-half s plane or in the left-half s plane. Here we take the semicircle in the right-half s plane.] A representative point s moves along the negative jv axis from –jq to j0–. From s=j0– to s=j0±, the point moves along the semicircle of radius e (where e 1) and then moves along the posi-tive jv axis from j0± to jq. From s=jq, the contour follows a semicircle with infinite radius, and the representative point moves back to the starting point, s=–jq.The area that the modified closed contour avoids is very small and approaches zero as the radius e approaches zero.Therefore, all the poles and zeros, if any, in the right-half s plane are enclosed by this contour.
Consider, for example, a closed-loop system whose open-loop transfer function is given by The points corresponding to s=j0± and s=j0– on the locus of G(s)H(s) in the G(s)H(s) plane are –jq and jq, respectively. On the semicircular path with radius e (where e 1), the complex variable s can be written where u varies from –90° to ±90°.Then G(s)H(s) becomes GAeejuBHAeejuB = K eeju = K e e-ju s = eeju G(s)H(s) = K s(Ts + 1) Section 7–5 / Nyquist Stability Criterion 453 jv s s Plane D C A B E F j 0+ j 0– +j – j (e 1) v = 0 + –1 D, E, F v = – v = GH Plane Re A B C Im v = 0– Figure 7–51 s-Plane contour and the G(s)H(s) locus in the GH plane, where G(s)H(s) = KCs(Ts + 1)D.
The value Ke approaches infinity as e approaches zero, and –u varies from 90° to –90° as a representative point s moves along the semicircle in the s plane. Thus, the points G(j0–)H(j0–)=jq and G(j0±)H(j0±)=–jq are joined by a semicircle of infinite radius in the right-half GH plane.The infinitesimal semicircular detour around the ori-gin in the s plane maps into the GH plane as a semicircle of infinite radius. Figure 7–51 shows the s-plane contour and the G(s)H(s) locus in the GH plane. Points A, B, and C on the s-plane contour map into the respective points A¿, B¿, and C¿ on the G(s)H(s) locus.As seen from Figure 7–51, points D, E, and F on the semicircle of infinite radius in the s plane map into the origin of the GH plane. Since there is no pole in the right-half s plane and the G(s)H(s) locus does not encircle the –1+j0 point, there are no zeros of the function 1+G(s)H(s) in the right-half s plane. Therefore, the system is stable.
For an open-loop transfer function G(s)H(s) involving a 1/sn factor (where n=2, 3, p ), the plot of G(s)H(s) has n clockwise semicircles of infinite radius about the origin as a representative point s moves along the semicircle of radius e (where e 1). For example, consider the following open-loop transfer function: Then As u varies from –90° to 90° in the s plane, the angle of G(s)H(s) varies from 180° to –180°, as shown in Figure 7–52. Since there is no pole in the right-half s plane and the locus encircles the –1+j0 point twice clockwise for any positive value of K, there are two zeros of 1+G(s)H(s) in the right-half s plane. Therefore, this system is always unstable.
lim sSeejuG(s)H(s) = K e2e2ju = K e2 e-2ju G(s)H(s) = K s2(Ts + 1) 454 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Note that a similar analysis can be made if G(s)H(s) involves poles and/or zeros on the jv axis.The Nyquist stability criterion can now be generalized as follows: Nyquist stability criterion [for a general case when G(s)H(s) has poles and/or zeros on the jv axis]:In the system shown in Figure 7–44,if the open-loop transfer function G(s)H(s) has k poles in the right-half s plane, then for stability the G(s)H(s) locus, as a representative point s traces on the modified Nyquist path in the clockwise di-rection, must encircle the –1+j0 point k times in the counterclockwise direction.
7–6 STABILITY ANALYSIS In this section, we shall present several illustrative examples of the stability analysis of control systems using the Nyquist stability criterion.
If the Nyquist path in the s plane encircles Z zeros and P poles of 1+G(s)H(s) and does not pass through any poles or zeros of 1+G(s)H(s) as a representative point s moves in the clockwise direction along the Nyquist path, then the corresponding con-tour in the G(s)H(s) plane encircles the –1+j0 point N=Z-P times in the clock-wise direction. (Negative values of N imply counterclockwise encirclements.) In examining the stability of linear control systems using the Nyquist stability crite-rion, we see that three possibilities can occur: 1. There is no encirclement of the –1+j0 point.This implies that the system is sta-ble if there are no poles of G(s)H(s) in the right-half s plane; otherwise, the sys-tem is unstable.
2. There are one or more counterclockwise encirclements of the –1+j0 point.In this case the system is stable if the number of counterclockwise encirclements is the same as the number of poles of G(s)H(s) in the right-half s plane; otherwise, the system is unstable.
3. There are one or more clockwise encirclements of the –1+j0 point. In this case the system is unstable.
In the following examples, we assume that the values of the gain K and the time con-stants (such as T, and ) are all positive.
T 2 T 1 , jv s Plane s GH Plane Re j 0+ j 0– +j –j e 1 v = 0+ v = 0– –1 Im v = – v = Figure 7–52 s-Plane contour and the G(s)H(s) locus in the GH plane, where G(s)H(s) = KCs2(Ts + 1)D.
Section 7–6 / Stability Analysis 455 Im Re –1 GH Plane G(jv) H(jv) v = 0 v = – v = Figure 7–53 Polar plot of G(jv)H(jv) considered in Example 7–14.
Im Re Re –1 –1 GH Plane v = 0– v = 0+ Im GH Plane v = 0– v = 0+ Small K Large K (Stable) (Unstable) P = 0 P = 0 N = 0 Z = 0 v = – v = v = – v = N = 2 Z = 2 Figure 7–54 Polar plots of the system considered in Example 7–15.
EXAMPLE 7–14 Consider a closed-loop system whose open-loop transfer function is given by Examine the stability of the system.
A plot of G(jv)H(jv) is shown in Figure 7–53. Since G(s)H(s) does not have any poles in the right-half s plane and the –1+j0 point is not encircled by the G(jv)H(jv) locus, this system is stable for any positive values of K, and T 2 .
T 1 , G(s)H(s) = K AT 1 s + 1BAT 2 s + 1B EXAMPLE 7–15 Consider the system with the following open-loop transfer function: Determine the stability of the system for two cases: (1) the gain K is small and (2) K is large.
The Nyquist plots of the open-loop transfer function with a small value of K and a large value of K are shown in Figure 7–54.The number of poles of G(s)H(s) in the right-half s plane is zero.
G(s)H(s) = K sAT 1 s + 1BAT 2 s + 1B 456 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Im Re GH Plane v = 0– v = 0+ v = – v = Im Re GH Plane v = 0– v = 0+ v = – v = Im Re GH Plane v = 0+ v = 0– v = v = – T1 T2 (Stable) T1 = T2 G(jv) H(jv) locus passes through the –1 + j0 point T1 T2 (Unstable) Figure 7–55 Polar plots of the system considered in Example 7–16.
Therefore, for this system to be stable, it is necessary that N=Z=0 or that the G(s)H(s) locus not encircle the –1+j0 point.
For small values of K, there is no encirclement of the –1+j0 point. Hence, the system is sta-ble for small values of K. For large values of K, the locus of G(s)H(s) encircles the –1+j0 point twice in the clockwise direction, indicating two closed-loop poles in the right-half s plane, and the system is unstable. (For good accuracy, K should be large. From the stability viewpoint, however, a large value of K causes poor stability or even instability.To compromise between accuracy and stability, it is necessary to insert a compensation network into the system. Compensating tech-niques in the frequency domain are discussed in Sections 7–11 through 7–13.) EXAMPLE 7–16 The stability of a closed-loop system with the following open-loop transfer function depends on the relative magnitudes of and Draw Nyquist plots and determine the stability of the system.
Plots of the locus G(s)H(s) for three cases, and are shown in Figure 7–55. For the locus of G(s)H(s) does not encircle the –1+j0 point, and the closed-loop system is stable. For , the G(s)H(s) locus passes through the –1+j0 point, which indicates that the closed-loop poles are located on the jv axis. For the locus of G(s)H(s) encircles the –1+j0 point twice in the clockwise direction.
Thus, the closed-loop system has two closed-loop poles in the right-half s plane, and the system is unstable.
T 1 7 T 2 , T 1 = T 2 T 1 6 T 2 , T 1 7 T 2 , T 1 6 T 2 , T 1 = T 2 , T 2 .
T 1 G(s)H(s) = KAT 2 s + 1B s2AT 1 s + 1B EXAMPLE 7–17 Consider the closed-loop system having the following open-loop transfer function: Determine the stability of the system.
G(s)H(s) = K s(Ts - 1) Section 7–6 / Stability Analysis 457 Im Re GH Plane v = 0– v = 0+ –1 v = v = –Figure 7–56 Polar plot of the system considered in Example 7–17.
Im Re GH Plane v = 0– v = 0+ –1 v = – v = Figure 7–57 Polar plot of the system considered in Example 7–18.
The function G(s)H(s) has one pole (s=1/T) in the right-half s plane.Therefore,P=1.The Nyquist plot shown in Figure 7–56 indicates that the G(s)H(s) plot encircles the –1+j0 point once clockwise.Thus, N=1. Since Z=N+P, we find that Z=2.This means that the closed-loop system has two closed-loop poles in the right-half s plane and is unstable.
EXAMPLE 7–18 Investigate the stability of a closed-loop system with the following open-loop transfer function: The open-loop transfer function has one pole (s=1) in the right-half s plane, or P=1.The open-loop system is unstable. The Nyquist plot shown in Figure 7–57 indicates that the –1+j0 point is encircled by the G(s)H(s) locus once in the counterclockwise direction. Therefore, N=–1. Thus, Z is found from Z=N+P to be zero, which indicates that there is no zero of 1+G(s)H(s) in the right-half s plane, and the closed-loop system is stable. This is one of the examples for which an unstable open-loop system becomes stable when the loop is closed.
G(s)H(s) = K(s + 3) s(s - 1) (K 7 1) 458 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method R(s) C(s) G(s) G1(s) G2(s) H1(s) H2(s) + – + – Figure 7–59 Multiple-loop system.
Conditionally Stable Systems.
Figure 7–58 shows an example of a G(jv)H(jv) locus for which the closed-loop system can be made unstable by varying the open-loop gain. If the open-loop gain is increased sufficiently, the G(jv)H(jv) locus encloses the –1+j0 point twice,and the system becomes unstable.If the open-loop gain is decreased sufficiently, again the G(jv)H(jv) locus encloses the –1+j0 point twice. For stable operation of the system considered here, the critical point –1+j0 must not be located in the regions between OA and BC shown in Figure 7–58. Such a system that is stable only for limited ranges of values of the open-loop gain for which the –1+j0 point is completely outside the G(jv)H(jv) locus is a conditionally stable system.
A conditionally stable system is stable for the value of the open-loop gain lying be-tween critical values, but it is unstable if the open-loop gain is either increased or de-creased sufficiently.Such a system becomes unstable when large input signals are applied, since a large signal may cause saturation, which in turn reduces the open-loop gain of the system. It is advisable to avoid such a situation.
Multiple-Loop System.
Consider the system shown in Figure 7–59.This is a mul-tiple-loop system.The inner loop has the transfer function G(s) = G2(s) 1 + G2(s)H 2(s) Im Re GH Plane 0 0 A B C v = v Figure 7–58 Polar plot of a conditionally stable system.
Section 7–6 / Stability Analysis 459 Figure 7–60 Control system.
If G(s) is unstable,the effects of instability are to produce a pole or poles in the right-half s plane.Then the characteristic equation of the inner loop,1+G2(s)H2(s)=0,has a zero or zeros in the right-half s plane. If G2(s) and H2(s) have poles here, then the number Z1 of right-half plane zeros of 1+G2(s)H2(s) can be found from where N1 is the number of clockwise encirclements of the –1+j0 point by the G2(s)H2(s) locus. Since the open-loop transfer function of the entire system is given by G1(s)G(s)H1(s), the stability of this closed-loop system can be found from the Nyquist plot of G1(s)G(s)H1(s) and knowledge of the right-half plane poles of G1(s)G(s)H1(s).
Notice that if a feedback loop is eliminated by means of block diagram reductions, there is a possibility that unstable poles are introduced; if the feedforward branch is eliminated by means of block diagram reductions, there is a possibility that right-half plane zeros are introduced.Therefore, we must note all right-half plane poles and zeros as they appear from subsidiary loop reductions. This knowledge is necessary in deter-mining the stability of multiple-loop systems.
EXAMPLE 7–19 Consider the control system shown in Figure 7–60.The system involves two loops. Determine the range of gain K for stability of the system by the use of the Nyquist stability criterion. (The gain K is positive.) To examine the stability of the control system,we need to sketch the Nyquist locus of G(s),where However, the poles of G(s) are not known at this point.Therefore, we need to examine the minor loop if there are right-half s-plane poles. This can be done easily by use of the Routh stability criterion. Since the Routh array becomes as follows: Notice that there are two sign changes in the first column. Hence, there are two poles of G2(s) in the right-half s plane.
Once we find the number of right-half s plane poles of G2(s),we proceed to sketch the Nyquist locus of G(s), where G(s) = G1(s)G2(s) = K(s + 0.5) s3 + s2 + 1 s3 s2 s1 s0 1 1 -1 1 0 1 0 G2(s) = 1 s3 + s2 + 1 G(s) = G1(s)G2(s) Z1 = N 1 + P 1 , P 1 R(s) C(s) K(s + 0.5) G1(s) G2(s) 1 s2(s + 1) + – + – 460 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Im v = 0.8 v = 0.7 j1.5 G K Plane v = 0.6 v = 0.9 j1 G(jv) K v = 0.4 v = 1 j0.5 v = 1.5 v = 1.4 v = 3 v = 0 v = v = 0.2 v = 0.1 –1 –0.5 0 0.5 1 Re v = 2 v = – –j0.5 –j1 –j1.5 Figure 7–61 Polar plot of G(jv)/K.
Our problem is to determine the range of the gain K for stability. Hence, instead of plotting Nyquist loci of G(jv) for various values of K, we plot the Nyquist locus of G(jv)/K. Figure 7–61 shows the Nyquist plot or polar plot of G(jv)/K.
Since G(s) has two poles in the right-half s plane, we have Noting that Z=N+P for stability, we require Z=0 or N=–2. That is, the Nyquist locus of G(jv) must encircle the –1+j0 point twice counterclockwise. From Figure 7–61, we see that, if the critical point lies between 0 and –0.5, then the G(jv)/K locus encircles the critical point twice counterclockwise.
Therefore, we require –0.5K<–1 The range of the gain K for stability is 20 and negative for g<0. For a minimum-phase system to be stable, the phase margin must be positive. In the logarithmic plots, the critical point in the complex plane corresponds to the 0-dB and –180° lines.
Gain margin: The gain margin is the reciprocal of the magnitude @G(jv)@ at the frequency at which the phase angle is –180°. Defining the phase crossover fre-quency v1 to be the frequency at which the phase angle of the open-loop transfer function equals –180° gives the gain margin Kg: In terms of decibels, The gain margin expressed in decibels is positive if Kg is greater than unity and nega-tive if Kg is smaller than unity.Thus, a positive gain margin (in decibels) means that the system is stable, and a negative gain margin (in decibels) means that the system is unstable.The gain margin is shown in Figures 7–67(a), (b), and (c).
For a stable minimum-phase system,the gain margin indicates how much the gain can be increased before the system becomes unstable. For an unstable system, the gain mar-gin is indicative of how much the gain must be decreased to make the system stable.
The gain margin of a first- or second-order system is infinite since the polar plots for such systems do not cross the negative real axis. Thus, theoretically, first- or second-order systems cannot be unstable. (Note, however, that so-called first- or second-order systems are only approximations in the sense that small time lags are neglected in de-riving the system equations and are thus not truly first- or second-order systems. If these small lags are accounted for, the so-called first- or second-order systems may become unstable.) It is noted that for a nonminimum-phase system with unstable open loop the stability condition will not be satisfied unless the G(jv) plot encircles the –1+j0 point. Hence, such a stable nonminimum-phase system will have negative phase and gain margins.
It is also important to point out that conditionally stable systems will have two or more phase crossover frequencies, and some higher-order systems with complicated numerator dynamics may also have two or more gain crossover frequencies, as shown in Figure 7–68. For stable systems having two or more gain crossover frequencies, the phase margin is measured at the highest gain crossover frequency.
A Few Comments on Phase and Gain Margins.
The phase and gain margins of a control system are a measure of the closeness of the polar plot to the –1+j0 point.
Therefore, these margins may be used as design criteria.
It should be noted that either the gain margin alone or the phase margin alone does not give a sufficient indication of the relative stability. Both should be given in the determination of relative stability.
For a minimum-phase system, both the phase and gain margins must be positive for the system to be stable. Negative margins indicate instability.
Proper phase and gain margins ensure us against variations in the system components and are specified for definite positive values.The two values bound the behavior of the K g dB = 20 log K g = -20 log @GAjv1B @ K g = 1 @GAjv1B @ Section 7–7 / Relative Stability Analysis 467 v = v = v Im Im Re Re 0 v 0 v1 v1 v2 v2 v3 v3 Phase crossover frequencies (v1, v 2, v3) Gain crossover frequencies (v1, v2, v3) Figure 7–68 Polar plots showing more than two phase or gain crossover frequencies.
Figure 7–69 Control system.
closed-loop system near the resonant frequency. For satisfactory performance, the phase margin should be between 30° and 60°, and the gain margin should be greater than 6 dB.
With these values, a minimum-phase system has guaranteed stability, even if the open-loop gain and time constants of the components vary to a certain extent.Although the phase and gain margins give only rough estimates of the effective damping ratio of the closed-loop system, they do offer a convenient means for designing control systems or adjusting the gain constants of systems.
For minimum-phase systems, the magnitude and phase characteristics of the open-loop transfer function are definitely related.The requirement that the phase margin be between 30° and 60° means that in a Bode diagram the slope of the log-magnitude curve at the gain crossover frequency should be more gradual than –40 dBdecade. In most practical cases, a slope of –20 dBdecade is desirable at the gain crossover frequency for stability. If it is –40 dBdecade, the system could be either stable or unstable. (Even if the system is stable,however,the phase margin is small.) If the slope at the gain crossover frequency is –60 dBdecade or steeper, the system is most likely unstable.
For nonminimum-phase systems, the correct interpretation of stability margins re-quires careful study.The best way to determine the stability of nonminimum-phase sys-tems is to use the Nyquist diagram approach rather than Bode diagram approach.
EXAMPLE 7–20 Obtain the phase and gain margins of the system shown in Figure 7–69 for the two cases where K=10 and K=100.
R(s) C(s) K s(s + 1) (s + 5) + – 468 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method The phase and gain margins can easily be obtained from the Bode diagram.A Bode diagram of the given open-loop transfer function with K=10 is shown in Figure 7–70(a).The phase and gain margins for K=10 are Phase margin=21°, Gain margin=8 dB Therefore, the system gain may be increased by 8 dB before the instability occurs.
Increasing the gain from K=10 to K=100 shifts the 0-dB axis down by 20 dB, as shown in Figure 7–70(b).The phase and gain margins are Phase margin=–30°, Gain margin=–12 dB Thus, the system is stable for K=10, but unstable for K=100.
Notice that one of the very convenient aspects of the Bode diagram approach is the ease with which the effects of gain changes can be evaluated.Note that to obtain satisfactory performance,we must increase the phase margin to 30° ~ 60°.This can be done by decreasing the gain K. Decreas-ing K is not desirable, however, since a small value of K will yield a large error for the ramp input.
This suggests that reshaping of the open-loop frequency-response curve by adding compensation may be necessary. Compensation techniques are discussed in detail in Sections 7–11 through 7–13.
Obtaining Gain Margin, Phase Margin, Phase-Crossover Frequency, and Gain-Crossover Frequency with MATLAB.
The gain margin,phase margin,phase-crossover frequency,and gain-crossover frequency can be obtained easily with MATLAB.The com-mand to be used is [Gm,pm,wcp,wcg] = margin(sys) 30 20 10 0 –30 –20 –10 0° –90° –180° –270° |G| in dB |G| in dB G 0.2 0.4 0.6 0.8 1 2 4 6 8 10 0° –90° –30° –180° –270° G 0.2 0.4 0.6 0.8 1 2 4 6 8 10 K = 10 K = 100 + 8 dB (Gain margin) (Phase margin) +21° v v 30 20 10 0 –10 50 40 (Gain margin) –12 dB (Phase margin) (a) (b) Figure 7–70 Bode diagrams of the system shown in Figure 7–69; (a) with K=10 and (b) with K=100.
Section 7–7 / Relative Stability Analysis 469 G(s) 20(s + 1) s(s + 5)(s2 + 2s + 10) + – Figure 7–71 Closed-loop system.
Frequency (rad/sec) Bode Diagram −300 −100 −150 −200 −250 0 −50 −100 Phase (deg); Magnitude (dB) 50 −50 0 10−1 100 101 4.0131 0.4426 102 9.9293 dB 103.6573 Figure 7–72 Bode diagram of G(s) shown in Figure 7–71.
MATLAB Program 7–11 num = [20 20]; den = conv([1 5 0],[1 2 10]); sys = tf(num,den); w = logspace(-1,2,100); bode(sys,w) [Gm,pm,wcp,wcg] = margin(sys); GmdB = 20log10(Gm); [GmdB pm wcp wcg] ans = 9.9293 103.6573 4.0131 0.4426 where Gm is the gain margin, pm is the phase margin, wcp is the phase-crossover fre-quency, and wcg is the gain-crossover frequency. For details of how to use this com-mand, see Example 7–21.
EXAMPLE 7–21 Draw a Bode diagram of the open-loop transfer function G(s) of the closed-loop system shown in Figure 7–71. Determine the gain margin, phase margin, phase-crossover frequency, and gain-crossover frequency with MATLAB.
A MATLAB program to plot a Bode diagram and to obtain the gain margin, phase margin, phase-crossover frequency, and gain-crossover frequency is shown in MATLAB Program 7–11.
The Bode diagram of G(s) is shown in Figure 7–72.
470 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Resonant Peak Magnitude Mr and Resonant Frequency Vr.
Consider the standard second-order system shown in Figure 7–73.The closed-loop transfer function is (7–16) where z and vn are the damping ratio and the undamped natural frequency, respectively.
The closed-loop frequency response is where As given by Equation (7–12), for 0 z 0.707, the maximum value of M occurs at the frequency vr, where (7–17) The frequency vr is the resonant frequency. At the resonant frequency, the value of M is maximum and is given by Equation (7–13), rewritten (7–18) where Mr is defined as the resonant peak magnitude. The resonant peak magnitude is related to the damping of the system.
The magnitude of the resonant peak gives an indication of the relative stability of the system. A large resonant peak magnitude indicates the presence of a pair of dominant closed-loop poles with small damping ratio, which will yield an undesirable transient response.A smaller resonant peak magnitude, on the other hand, indicates the absence of a pair of dominant closed-loop poles with small damping ratio, meaning that the system is well damped.
Remember that vr is real only if z<0.707.Thus, there is no closed-loop resonance if z>0.707. [The value of Mr is unity only if z>0.707. See Equation (7–14).] Since the values of Mr and vr can be easily measured in a physical system, they are quite useful for checking agreement between theoretical and experimental analyses.
Mr = 1 2z21 - z2 vr = vn21 - 2z2 M = 1 B a1 - v2 v2 n b 2 + a2z v vn b 2 , a = -tan-1 2z v vn 1 - v2 v2 n C(jv) R(jv) = 1 a 1 - v2 v2 n b + j2z v vn = Meja C(s) R(s) = v2 n s2 + 2zvn s + v2 n R(s) C(s) vn s(s + 2z vn) 2 + – Figure 7–73 Standard second-order system.
Section 7–7 / Relative Stability Analysis 471 It is noted, however, that in practical design problems the phase margin and gain margin are more frequently specified than the resonant peak magnitude to indicate the degree of damping in a system.
Correlation between Step Transient Response and Frequency Response in the Standard Second-Order System.
The maximum overshoot in the unit-step re-sponse of the standard second-order system, as shown in Figure 7–73, can be exactly correlated with the resonant peak magnitude in the frequency response. Hence, essen-tially the same information about the system dynamics is contained in the frequency re-sponse as is in the transient response.
For a unit-step input,the output of the system shown in Figure 7–73 is given by Equa-tion (5–12), or where (7–19) On the other hand, the maximum overshoot Mp for the unit-step response is given by Equation (5–21), or (7–20) This maximum overshoot occurs in the transient response that has the damped natural frequency The maximum overshoot becomes excessive for values of z<0.4.
Since the second-order system shown in Figure 7–73 has the open-loop transfer function for sinusoidal operation, the magnitude of G(jv) becomes unity when which can be obtained by equating @G(jv)@ to unity and solving for v.At this frequency, the phase angle of G(jv) is Thus, the phase margin g is (7–21) Equation (7–21) gives the relationship between the damping ratio z and the phase margin g. (Notice that the phase margin g is a function only of the damping ratio z.) = tan-1 2z 321 + 4z4 - 2z2 = 90° - tan-1 321 + 4z4 - 2z2 2z g = 180° + /G(jv) /G(jv) = - /jv - /jv + 2zvn = -90° - tan-1 321 + 4z4 - 2z2 2z v = vn321 + 4z4 - 2z2 G(s) = v2 n sAs + 2zvnB vd = vn21 - z2 .
Mp = e-Az21-z2Bp vd = vn21 - z2 c(t) = 1 - e-zvn t acosvd t + z 21 - z2 sin vd t b , for t 0 472 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method In the following,we shall summarize the correlation between the step transient response and frequency response of the standard second-order system given by Equation (7–16): 1. The phase margin and the damping ratio are directly related.Figure 7–74 shows a plot of the phase margin g as a function of the damping ratio z.It is noted that for the stan-dard second-order system shown in Figure 7–73,the phase margin g and the damping ratio z are related approximately by a straight line for 0 z 0.6, as follows: Thus a phase margin of 60° corresponds to a damping ratio of 0.6. For higher-order systems having a dominant pair of closed-loop poles, this relationship may be used as a rule of thumb in estimating the relative stability in the transient response (that is, the damping ratio) from the frequency response.
2. Referring to Equations (7–17) and (7–19), we see that the values of vr and vd are almost the same for small values of z.Thus, for small values of z, the value of vr is indicative of the speed of the transient response of the system.
3. From Equations (7–18) and (7–20), we note that the smaller the value of z is, the larger the values of Mr and Mp are.The correlation between Mr and Mp as a func-tion of z is shown in Figure 7–75.A close relationship between Mr and Mp can be seen for z>0.4. For very small values of z, Mr becomes very large AMr 1B, while the value of Mp does not exceed 1.
Correlation between Step Transient Response and Frequency Response in General Systems.
The design of control systems is very often carried out on the basis of the frequency response.The main reason for this is the relative simplicity of this ap-proach compared with others. Since in many applications it is the transient response of the system to aperiodic inputs rather than the steady-state response to sinusoidal in-puts that is of primary concern, the question of correlation between transient response and frequency response arises.
z = g 100° 90° 60° 30° 0°0 0.4 0.8 1.2 1.6 2.0 z g Straight-line approximation Figure 7–74 Curve g (phase margin) versus z for the system shown in Figure 7–73.
For the standard second-order system shown in Figure 7–73, mathematical rela-tionships correlating the step transient response and frequency response can be obtained easily.The time response of the standard second-order system can be predicted exactly from a knowledge of the Mr and vr of its closed-loop frequency response.
For nonstandard second-order systems and higher-order systems, the correlation is more complex, and the transient response may not be predicted easily from the fre-quency response because additional zeros and/or poles may change the correlation be-tween the step transient response and the frequency response existing for the standard second-order system. Mathematical techniques for obtaining the exact correlation are available, but they are very laborious and of little practical value.
The applicability of the transient-response–frequency-response correlation existing for the standard second-order system shown in Figure 7–73 to higher-order systems depends on the presence of a dominant pair of complex-conjugate closed-loop poles in the latter systems.
Clearly, if the frequency response of a higher-order system is dominated by a pair of com-plex-conjugate closed-loop poles, the transient-response– frequency-response correlation existing for the standard second-order system can be extended to the higher-order system.
For linear, time-invariant, higher-order systems having a dominant pair of complex-conjugate closed-loop poles, the following relationships generally exist between the step transient response and frequency response: 1. The value of Mr is indicative of the relative stability. Satisfactory transient per-formance is usually obtained if the value of Mr is in the range 1.0<Mr<1.4 A0 dB<Mr<3 dBB, which corresponds to an effective damping ratio of 0.4<z<0.7. For values of Mr greater than 1.5, the step transient response may exhibit several overshoots. (Note that, in general, a large value of Mr corresponds to a large overshoot in the step transient response. If the system is subjected to noise signals whose frequencies are near the resonant frequency vr, the noise will be amplified in the output and will present serious problems.) 2. The magnitude of the resonant frequency vr is indicative of the speed of the tran-sient response.The larger the value of vr, the faster the time response is. In other words, the rise time varies inversely with vr. In terms of the open-loop frequency Section 7–7 / Relative Stability Analysis 473 3 Mr 2 1 Mp 0 0.2 0.4 0.6 0.8 1.0 z Mr = 1 2z 1 – z 2 Mp = c(tp) –1 [Equation (5-21)] Figure 7–75 Curves Mr versus z and Mp versus z for the system shown in Figure 7–73.
474 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method response, the damped natural frequency in the transient response is somewhere between the gain crossover frequency and phase crossover frequency.
3. The resonant peak frequency vr and the damped natural frequency vd for the step transient response are very close to each other for lightly damped systems.
The three relationships just listed are useful for correlating the step transient re-sponse with the frequency response of higher-order systems, provided that they can be approximated by the standard second-order system or a pair of complex-conjugate closed-loop poles. If a higher-order system satisfies this condition, a set of time-domain specifications may be translated into frequency-domain specifications. This simplifies greatly the design work or compensation work of higher-order systems.
In addition to the phase margin, gain margin, resonant peak Mr, and resonant fre-quency vr, there are other frequency-domain quantities commonly used in performance specifications.They are the cutoff frequency, bandwidth, and the cutoff rate.These will be defined in what follows.
Cutoff Frequency and Bandwidth.
Referring to Figure 7–76, the frequency vb at which the magnitude of the closed-loop frequency response is 3 dB below its zero-fre-quency value is called the cutoff frequency.Thus For systems in which The closed-loop system filters out the signal components whose frequencies are greater than the cutoff frequency and transmits those signal components with frequencies lower than the cutoff frequency.
The frequency range 0 v vb in which the magnitude of is greater than –3 dB is called the bandwidth of the system. The bandwidth indicates the frequency where the gain starts to fall off from its low-frequency value.Thus, the bandwidth indicates how well the system will track an input sinusoid. Note that for a given vn, the rise time in-creases with increasing damping ratio z. On the other hand, the bandwidth decreases with the increase in z.Therefore, the rise time and the bandwidth are inversely proportional to each other.
C(jv)R(jv) 2 C(jv) R(jv) 2 6 -3 dB, for v 7 vb @C(j0)R(j0)@ = 0 dB, 2 C(jv) R(jv) 2 6 2 C(j0) R(j0) 2 - 3 dB, for v 7 vb dB 0 –3 Bandwidth vb v in log scale Figure 7–76 Plot of a closed-loop frequency response curve showing cutoff frequency vb and bandwidth.
Section 7–7 / Relative Stability Analysis 475 dB 0 –20 0.33 I I I II II II 1 1 1 1 v (in log scale) (a) (b) (c) 0 0 c(t) r(t) c(t) r(t) r(t) t t Figure 7–77 Comparison of dynamic characteristics of the two systems considered in Example 7–22.
(a) Closed-loop frequency-response curves; (b) unit-step response curves; (c) unit-ramp response curves.
The specification of the bandwidth may be determined by the following factors: 1. The ability to reproduce the input signal.A large bandwidth corresponds to a small rise time,or fast response.Roughly speaking,we can say that the bandwidth is proportional to the speed of response. (For example, to decrease the rise time in the step response by a factor of 2, the bandwidth must be increased by approximately a factor of 2.) 2. The necessary filtering characteristics for high-frequency noise.
For the system to follow arbitrary inputs accurately, it must have a large bandwidth.
From the viewpoint of noise,however,the bandwidth should not be too large.Thus,there are conflicting requirements on the bandwidth,and a compromise is usually necessary for good design. Note that a system with large bandwidth requires high-performance components, so the cost of components usually increases with the bandwidth.
Cutoff Rate.
The cutoff rate is the slope of the log-magnitude curve near the cutoff fre-quency.The cutoff rate indicates the ability of a system to distinguish the signal from noise.
It is noted that a closed-loop frequency response curve with a steep cutoff charac-teristic may have a large resonant peak magnitude, which implies that the system has a relatively small stability margin.
EXAMPLE 7–22 Consider the following two systems: Compare the bandwidths of these two systems.Show that the system with the larger bandwidth has a faster speed of response and can follow the input much better than the one with the smaller bandwidth.
Figure 7–77(a) shows the closed-loop frequency-response curves for the two systems. (Asymptot-ic curves are shown by dashed lines.) We find that the bandwidth of system I is 0 v 1 radsec and that of system II is 0 v 0.33 radsec. Figures 7–77(b) and (c) show, respectively, the unit-step re-sponse and unit-ramp response curves for the two systems.Clearly,system I,whose bandwidth is three times wider than that of system II,has a faster speed of response and can follow the input much better.
System I: C(s) R(s) = 1 s + 1 , System II: C(s) R(s) = 1 3s + 1 476 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method MATLAB Approach to Get Resonant Peak, Resonant Frequency, and Band-width.
The resonant peak is the value of the maximum magnitude (in decibels) of the closed-loop frequency response.The resonant frequency is the frequency that yields the maximum magnitude. MATLAB commands to be used for obtaining the resonant peak and resonant frequency are as follows: [mag,phase,w] = bode(num,den,w); or [mag,phase,w] = bode(sys,w); [Mp,k] = max(mag); resonant_peak = 20log10(Mp); resonant_frequency = w(k) The bandwidth can be obtained by entering the following lines in the program: n = 1; while 20log10(mag(n)) > = -3; n = n + 1; end bandwidth = w(n) For a detailed MATLAB program, see Example 7–23.
EXAMPLE 7–23 Consider the system shown in Figure 7–78.Using MATLAB,obtain a Bode diagram for the closed-loop transfer function. Obtain also the resonant peak, resonant frequency, and bandwidth.
MATLAB Program 7–12 produces a Bode diagram for the closed-loop system as well as the resonant peak, resonant frequency, and bandwidth. The resulting Bode diagram is shown in MATLAB Program 7–12 nump = ; denp = [0.5 1.5 1 0]; sysp = tf(nump,denp); sys = feedback(sysp,1); w = logspace(-1,1); bode(sys,w) [mag,phase,w] = bode(sys,w); [Mp,k] = max(mag); resonant_peak = 20log10(Mp) resonant_peak = 5.2388 resonant_frequency = w(k) resonant_frequency = 0.7906 n = 1; while 20log(mag(n))> = -3; n = n + 1; end bandwidth = w(n) bandwidth = 1.2649 Section 7–8 / Closed-Loop Frequency Response of Unity-Feedback Systems 477 Figure 7–79.The resonant peak is obtained as 5.2388 dB.The resonant frequency is 0.7906 radsec.
The bandwidth is 1.2649 radsec. These values can be verified from Figure 7–78.
7–8 CLOSED-LOOP FREQUENCY RESPONSE OF UNITY-FEEDBACK SYSTEMS Closed-Loop Frequency Response.
For a stable, unity-feedback closed-loop sys-tem, the closed-loop frequency response can be obtained easily from that of the open-loop frequency response. Consider the unity-feedback system shown in Figure 7–80(a).
The closed-loop transfer function is In the Nyquist or polar plot shown in Figure 7–80(b), the vector represents GAjv1B, where v1 is the frequency at point A. The length of the vector is @GAjv1B @ and the angle of the vector is The vector the vector from the –1+j0 point to the Nyquist locus, represents 1+GAjv1B. Therefore, the ratio of to repre-sents the closed-loop frequency response, or OA !
PA ! = GAjv1B 1 + GAjv1B = CAjv1B RAjv1B PA !
OA !
, PA !
, /GAjv1B.
OA !
OA !
OA !
C(s) R(s) = G(s) 1 + G(s) 1 s(0.5s + 1) (s + 1) + – R(s) C(s) Figure 7–78 Closed-loop system.
Frequency (rad/sec) Bode Diagram −300 −50 −100 −150 −200 −250 0 −60 −40 −20 Phase (deg); Magnitude (dB) 20 0 10−1 100 101 Figure 7–79 Bode diagram of the closed-loop transfer function of the system shown in Figure 7–78.
478 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method The magnitude of the closed-loop transfer function at v=v1 is the ratio of the magni-tudes of to . The phase angle of the closed-loop transfer function at v=v1 is the angle formed by the vectors to —that is f-u, shown in Figure 7–80(b). By measuring the magnitude and phase angle at different frequency points, the closed-loop frequency-response curve can be obtained.
Let us define the magnitude of the closed-loop frequency response as M and the phase angle as a, or In the following, we shall find the constant-magnitude loci and constant-phase-angle loci. Such loci are convenient in determining the closed-loop frequency response from the polar plot or Nyquist plot.
Constant-Magnitude Loci (M circles).
To obtain the constant-magnitude loci, let us first note that G(jv) is a complex quantity and can be written as follows: where X and Y are real quantities.Then M is given by and M2 is Hence (7–22) If M=1, then from Equation (7–22), we obtain This is the equation of a straight line parallel to the Y axis and passing through the point A- 1 2 , 0B.
X = - 1 2 .
X2A1 - M2B - 2M2X - M2 + A1 - M2BY2 = 0 M2 = X2 + Y2 (1 + X)2 + Y2 M = ∑X + jY∑ ∑1 + X + jY∑ G(jv) = X + jY C(jv) R(jv) = Meja PA !
OA !
PA !
OA !
(a) (b) G(s) Im Re O P –1 + j0 A G ( jv) u f v1 f – u + – Figure 7–80 (a) Unity-feedback system; (b) determination of closed-loop frequency response from open-loop frequency response.
Section 7–8 / Closed-Loop Frequency Response of Unity-Feedback Systems 479 – 4 –3 –2 –1 0 1 2 X Y M = 1.2 M = 1.3 M = 1 M = 1.4 M = 1.6 M = 2.0 M = 3.0 M = 5.0 –1 –2 1 2 M = 0.8 M = 0.4 M = 0.6 Figure 7–81 A family of constant M circles.
If Equation (7–22) can be written If the term M2/AM2-1B 2 is added to both sides of this last equation, we obtain (7–23) Equation (7–23) is the equation of a circle with center at X=–M2/AM2-1B, Y=0 and with radius @M/AM2-1B @.
The constant M loci on the G(s) plane are thus a family of circles.The center and ra-dius of the circle for a given value of M can be easily calculated. For example, for M=1.3, the center is at (–2.45, 0) and the radius is 1.88. A family of constant M cir-cles is shown in Figure 7–81. It is seen that as M becomes larger compared with 1, the M circles become smaller and converge to the –1+j0 point. For M>1, the centers of the M circles lie to the left of the –1+j0 point. Similarly, as M becomes smaller com-pared with 1, the M circle becomes smaller and converges to the origin. For 0<M<1, the centers of the M circles lie to the right of the origin. M=1 corresponds to the locus of points equidistant from the origin and from the –1+j0 point.As stated earlier, it is a straight line passing through the point and parallel to the imaginary axis. (The constant M circles corresponding to M>1 lie to the left of the M=1 line, and those corresponding to 0<M<1 lie to the right of the M=1 line.) The M circles are sym-metrical with respect to the straight line corresponding to M=1 and with respect to the real axis.
A- 1 2 , 0B aX + M2 M2 - 1 b 2 + Y2 = M2 AM2 - 1B2 X2 + 2M2 M2 - 1 X + M2 M2 - 1 + Y2 = 0 M Z 1, 480 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Constant-Phase-Angle Loci (N Circles).
We shall obtain the phase angle a in terms of X and Y. Since the phase angle a is If we define then Since we obtain or The addition of to both sides of this last equation yields (7–24) This is an equation of a circle with center at Y=1(2N) and with radius For example, if a=30°, then N=tan a=0.577, and the center and the radius of the circle corresponding to a=30° are found to be (–0.5, 0.866) and unity, respectively. Since Equation (7–24) is satisfied when X=Y=0 and X=–1, Y=0 regardless of the value of N, each circle passes through the origin and the –1+j0 point.
The constant a loci can be drawn easily,once the value of N is given.A family of constant N circles is shown in Figure 7–82 with a as a parameter.
It should be noted that the constant N locus for a given value of a is actually not the entire circle, but only an arc. In other words, the a=30° and a=–150° arcs are parts of the same circle.This is so because the tangent of an angle remains the same if ;180° (or multiples thereof) is added to the angle.
The use of the M and N circles enables us to find the entire closed-loop frequency response from the open-loop frequency response G(jv) without calculating the magni-tude and phase of the closed-loop transfer function at each frequency.The intersections 3 1 4 + 1(2N)2 .
X = -1 2 , aX + 1 2 b 2 + a Y -1 2N b 2 = 1 4 + a 1 2N b 2 A1 4B + 1(2N)2 X2 + X + Y2 - 1 N Y = 0 N = Y X -Y 1 + X 1 + Y X a Y 1 + X b = Y X2 + X + Y2 tan (A - B) = tan A - tanB 1 + tanA tanB N = tan ctan-1 a Y X b - tan-1 a Y 1 + X b d tan a = N a = tan-1 a Y X b - tan-1 a Y 1 + X b /eja = n X + jY 1 + X + jY Section 7–8 / Closed-Loop Frequency Response of Unity-Feedback Systems 481 –3 3 –3 –2 –2 –1 1 1 2 2 X Y a = 20° a = 30° a = 40° a = 100° 60° 120° 80° –60° –120° –80° a = –100° a = –40° a = –30° a = –20° Figure 7–82 A family of constant N circles.
of the G(jv) locus and the M circles and N circles give the values of M and N at fre-quency points on the G(jv) locus.
The N circles are multivalued in the sense that the circle for a=a1 and that for a=a1 ; 180°n (n=1, 2, p) are the same. In using the N circles for the determination of the phase angle of closed-loop systems, we must interpret the proper value of a. To avoid any error, start at zero frequency, which corresponds to a=0°, and proceed to higher frequencies.The phase-angle curve must be continuous.
Graphically, the intersections of the G(jv) locus and M circles give the values of M at the frequencies denoted on the G(jv) locus. Thus, the constant M circle with the smallest radius that is tangent to the G(jv) locus gives the value of the resonant peak magnitude Mr. If it is desired to keep the resonant peak value less than a certain value, then the system should not enclose the critical point (–1+j0 point) and, at the same time, there should be no intersections with the particular M circle and the G(jv) locus.
Figure 7–83(a) shows the G(jv) locus superimposed on a family of M circles. Figure 7–83(b) shows the G(jv) locus superimposed on a family of N circles. From these plots, it is possible to obtain the closed-loop frequency response by inspection. It is seen that the M=1.1 circle intersects the G(jv) locus at frequency point v=v1. This means that at this frequency the magnitude of the closed-loop transfer function is 1.1. In Fig-ure 7–83(a), the M=2 circle is just tangent to the G(jv) locus.Thus, there is only one point on the G(jv) locus for which @C(jv)/R(jv)@ is equal to 2. Figure 7–83(c) shows the closed-loop frequency-response curve for the system.The upper curve is the M-versus-frequency v curve, and the lower curve is the phase angle a-versus-frequency v curve.
The resonant peak value is the value of M corresponding to the M circle of small-est radius that is tangent to the G(jv) locus.Thus, in the Nyquist diagram, the resonant 482 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method peak value Mr and the resonant frequency vr can be found from the M-circle tangency to the G(jv) locus. (In the present example, Mr=2 and vr=v4.) Nichols Chart.
In dealing with design problems, we find it convenient to construct the M and N loci in the log-magnitude-versus-phase plane. The chart consisting of the M and N loci in the log-magnitude-versus-phase diagram is called the Nichols chart.
The G(jv) locus drawn on the Nichols chart gives both the gain characteristics and a = G 1 + G M = G 1 + G Im Re Im Re 2 0 0 –2 –2 –4 –4 –2 –4 2 –2 –4 M = 1.2 M = 1.4 M = 1.1 M = 1.1 M = 2 M = 0.6 M = 1.2 G(jv) G(jv) v1 (a) (b) (c) 20° 60° –20° –40° –10° v1 v2 v2 v3 v3 v4 v4 v5 v5 2 1.5 1 0.5 0 0° –90° –180° –270° v1 v2 v3 v4 v5 v v Figure 7–83 (a) G(jv) locus superimposed on a family of M circles; (b) G(jv) locus superimposed on a family of N circles; (c) closed-loop frequency-response curves.
Section 7–8 / Closed-Loop Frequency Response of Unity-Feedback Systems 483 + – R C G 0.25 dB 0.5 dB 1 dB 2 dB 3 dB 4 dB 5 dB 6 dB 9 dB –18 dB –12 dB –6 dB –5 dB –4 dB –3 dB –2 dB –1 dB –0.5 dB –0.25 dB –0.1 dB 0.1 dB 0 dB 12 dB 120° 150° –180° –150° –120° –90° –60° –30° –20° –10° –5° –2° 90° 60° 30° 20° 10° 5° 2° 0° –2° –5° –10° –20° –30° –60° 36 32 28 24 20 16 12 8 4 0 –16 –12 –8 –4 –240° –210° –180° –150° –120° –90° –60° –30° 0° GH |GH| in dB Figure 7–84 Nichols chart.
phase characteristics of the closed-loop transfer function at the same time.The Nichols chart is shown in Figure 7–84, for phase angles between 0° and –240°.
Note that the critical point (–1+j0 point) is mapped to the Nichols chart as the point (0 dB, –180°). The Nichols chart contains curves of constant closed-loop magni-tude and phase angle. The designer can graphically determine the phase margin, gain margin, resonant peak magnitude, resonant frequency, and bandwidth of the closed-loop system from the plot of the open-loop locus, G(jv).
The Nichols chart is symmetric about the –180° axis. The M and N loci repeat for every 360°, and there is symmetry at every 180° interval.The M loci are centered about the critical point (0 dB,–180°).The Nichols chart is useful for determining the frequency response of the closed loop from that of the open loop. If the open-loop frequency-re-sponse curve is superimposed on the Nichols chart, the intersections of the open-loop frequency-response curve G(jv) and the M and N loci give the values of the magni-tude M and phase angle a of the closed-loop frequency response at each frequency point. If the G(jv) locus does not intersect the M=Mr locus, but is tangent to it, then the resonant peak value of M of the closed-loop frequency response is given by Mr.The resonant frequency is given by the frequency at the point of tangency.
As an example, consider the unity-feedback system with the following open-loop transfer function: To find the closed-loop frequency response by use of the Nichols chart, the G(jv) locus is constructed in the log-magnitude-versus-phase plane by use of MATLAB or from G(jv) = K s(s + 1)(0.5s + 1) , K = 1 484 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method the Bode diagram.Figure 7–85(a) shows the G(jv) locus together with the M and N loci.
The closed-loop frequency-response curves may be constructed by reading the magni-tudes and phase angles at various frequency points on the G(jv) locus from the M and N loci, as shown in Figure 7–85(b). Since the largest magnitude contour touched by the G(jv) locus is 5 dB, the resonant peak magnitude Mr is 5 dB. The corresponding reso-nant peak frequency is 0.8 radsec.
Notice that the phase crossover point is the point where the G(jv) locus intersects the –180° axis (for the present system, v=1.4 radsec), and the gain crossover point is the point where the locus intersects the 0-dB axis (for the present system, v=0.76 radsec). The phase margin is the horizontal distance (measured in degrees) between the gain crossover point and the critical point (0 dB, –180°). The gain margin is the distance (in decibels) between the phase crossover point and the critical point.
The bandwidth of the closed-loop system can easily be found from the G(jv) locus in the Nichols diagram. The frequency at the intersection of the G(jv) locus and the M=–3 dB locus gives the bandwidth.
If the open-loop gain K is varied, the shape of the G(jv) locus in the log-magnitude-versus-phase diagram remains the same, but it is shifted up (for increasing K) or down (for decreasing K) along the vertical axis. Therefore, the G(jv) locus intersects the M 20 16 12 8 4 0 –16 –12 –8 –4 –240° –210° –180° –150° –120° –90° (a) (b) G 1 dB 3 dB 0.25 dB 5 dB 12 dB |G| in dB –1 dB –5 dB –12 dB 1.8 1.4 1.2 1 0.8 0.6 0.4 0.2 –30° –20° –10° –60° –120° –150° –90° v in rad/sec G 1 + G G 1 + G in dB –270° –180° –90° –15 –10 –5 0 5 10 0° 0.1 0.2 0.4 0.6 0.8 1 2 Figure 7–85 (a) Plot of G(jv) superimposed on Nichols chart; (b) closed-loop frequency-response curves.
Section 7–8 / Closed-Loop Frequency Response of Unity-Feedback Systems 485 |G| in dB G 15 10 5 0 –5 –10 –15 –90° –120° –150° –180° Mr = 1.4 20 log K = 4 G( jv) G( jv) K Figure 7–86 Determination of the gain K using the Nichols chart.
and N loci differently, resulting in a different closed-loop frequency-response curve. For a small value of the gain K,the G(jv) locus will not be tangent to any of the M loci,which means that there is no resonance in the closed-loop frequency response.
EXAMPLE 7–24 Consider the unity-feedback control system whose open-loop transfer function is Determine the value of the gain K so that Mr=1.4.
The first step in the determination of the gain K is to sketch the polar plot of Figure 7–86 shows the Mr=1.4 locus and the G(jv)/K locus. Changing the gain has no effect on the phase angle, but merely moves the curve vertically up for K>1 and down for K<1.
In Figure 7–86, the G(jv)/K locus must be raised by 4 dB in order that it be tangent to the desired Mr locus and that the entire G(jv)/K locus be outside the Mr=1.4 locus.The amount of vertical shift of the G(jv)/K locus determines the gain necessary to yield the desired value of Mr.Thus, by solving we obtain K = 1.59 20 logK = 4 G(jv) K = 1 jv(1 + jv) G(jv) = K jv(1 + jv) 486 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method 7–9 EXPERIMENTAL DETERMINATION OF TRANSFER FUNCTIONS The first step in the analysis and design of a control system is to derive a mathematical model of the plant under consideration.Obtaining a model analytically may be quite dif-ficult.We may have to obtain it by means of experimental analysis.The importance of the frequency-response methods is that the transfer function of the plant, or any other com-ponent of a system, may be determined by simple frequency-response measurements.
If the amplitude ratio and phase shift have been measured at a sufficient number of frequencies within the frequency range of interest, they may be plotted on the Bode di-agram.Then the transfer function can be determined by asymptotic approximations.We build up asymptotic log-magnitude curves consisting of several segments. With some trial-and-error juggling of the corner frequencies, it is usually possible to find a very close fit to the curve. (Note that if the frequency is plotted in cycles per second rather than radians per second, the corner frequencies must be converted to radians per sec-ond before computing the time constants.) Sinusoidal-Signal Generators.
In performing a frequency-response test, suitable sinusoidal-signal generators must be available. The signal may have to be in mechani-cal, electrical, or pneumatic form. The frequency ranges needed for the test are ap-proximately 0.001 to 10 Hz for large-time-constant systems and 0.1 to 1000 Hz for small-time-constant systems. The sinusoidal signal must be reasonably free from har-monics or distortion.
For very low frequency ranges (below 0.01 Hz), a mechanical signal generator (together with a suitable pneumatic or electrical transducer if necessary) may be used.
For the frequency range from 0.01 to 1000 Hz, a suitable electrical-signal generator (together with a suitable transducer if necessary) may be used.
Determination of Minimum-Phase Transfer Functions from Bode Diagrams.
As stated previously, whether a system is minimum phase can be determined from the frequency-response curves by examining the high-frequency characteristics.
To determine the transfer function, we first draw asymptotes to the experimental-ly obtained log-magnitude curve. The asymptotes must have slopes of multiples of ;20 dBdecade. If the slope of the experimentally obtained log-magnitude curve changes from –20 to –40 dBdecade at v=v1, it is clear that a factor 1/C1+jAv/v1BD exists in the transfer function. If the slope changes by –40 dBdecade at v=v2, there must be a quadratic factor of the form in the transfer function. The undamped natural frequency of this quadratic factor is equal to the corner frequency v2. The damping ratio z can be determined from the experimentally obtained log-magnitude curve by measuring the amount of resonant peak near the corner frequency v2 and comparing this with the curves shown in Figure 7–9.
Once the factors of the transfer function G(jv) have been determined, the gain can be determined from the low-frequency portion of the log-magnitude curve. Since such 1 1 + 2z aj v v2 b + aj v v2 b 2 terms as 1+jAv/v1B and 1+2zAjv/v2B+Ajv/v2B 2 become unity as v approaches zero, at very low frequencies the sinusoidal transfer function G(jv) can be written In many practical systems, l equals 0, 1, or 2.
1. For l=0, or type 0 systems, or The low-frequency asymptote is a horizontal line at 20 log K dB. The value of K can thus be found from this horizontal asymptote.
2. For l=1, or type 1 systems, or which indicates that the low-frequency asymptote has the slope –20 dBdecade.
The frequency at which the low-frequency asymptote (or its extension) intersects the 0-dB line is numerically equal to K.
3. For l=2, or type 2 systems, or The slope of the low-frequency asymptote is –40 dBdecade. The frequency at which this asymptote (or its extension) intersects the 0-dB line is numerically equal to Examples of log-magnitude curves for type 0, type 1, and type 2 systems are shown in Figure 7–87, together with the frequency to which the gain K is related.
The experimentally obtained phase-angle curve provides a means of checking the transfer function obtained from the log-magnitude curve. For a minimum-phase system, the experimental phase-angle curve should agree reasonably well with the theoretical phase-angle curve obtained from the transfer function just determined.These two phase-angle curves should agree exactly in both the very low and very high frequency ranges.
If the experimentally obtained phase angle at very high frequencies (compared with the corner frequencies) is not equal to –90°(q-p),where p and q are the degrees of the nu-merator and denominator polynomials of the transfer function, respectively, then the transfer function must be a nonminimum-phase transfer function.
1K .
20 log @G(jv)@ = 20 log K - 40 log v, for v 1 G(jv) = K (jv)2 , for v 1 20 log @G(jv)@ = 20 log K - 20 log v, for v 1 G(jv) = K jv , for v 1 20 log @G(jv)@ = 20 log K, for v 1 G(jv) = K, for v 1 lim vS0G(jv) = K (jv)l Section 7–9 / Experimental Determination of Transfer Functions 487 488 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Nonminimum-Phase Transfer Functions.
If,at the high-frequency end,the com-puted phase lag is 180° less than the experimentally obtained phase lag, then one of the zeros of the transfer function should have been in the right-half s plane instead of the left-half s plane.
If the computed phase lag differed from the experimentally obtained phase lag by a constant rate of change of phase,then transport lag,or dead time,is present.If we assume the transfer function to be of the form where G(s) is a ratio of two polynomials in s, then where we used the fact that constant. Thus, from this last equation, we can evaluate the magnitude of the transport lag T.
lim vS q /G(jv) = = 0 - T = -T = lim vS q d dv C /G(jv) - vTD lim vS q d dv /G(jv)e-jvT = lim vS q d dv C /G(jv) + /e-jvTD G(s)e-Ts (a) 0 –20 20 log K –40 –40 dB v = K v = K v in log scale (b) (c) 0 –20 –20 –20 –20 –40 –40 –40 –40 dB v in log scale 0 dB v in log scale 0 dB v in log scale 0 dB v in log scale v = K v = K Figure 7–87 (a) Log-magnitude curve of a type 0 system; (b) log-magnitude curves of type 1 systems; (c) log-magnitude curves of type 2 systems. (The slopes shown are in dBdecade.) A Few Remarks on the Experimental Determination of Transfer Functions 1. It is usually easier to make accurate amplitude measurements than accurate phase-shift measurements. Phase-shift measurements may involve errors that may be caused by instrumentation or by misinterpretation of the experimental records.
2. The frequency response of measuring equipment used to measure the system out-put must have a nearly flat magnitude-versus-frequency curve. In addition, the phase angle must be nearly proportional to the frequency.
3. Physical systems may have several kinds of nonlinearities. Therefore, it is nec-essary to consider carefully the amplitude of input sinusoidal signals. If the am-plitude of the input signal is too large, the system will saturate, and the frequency-response test will yield inaccurate results. On the other hand, a small signal will cause errors due to dead zone. Hence, a careful choice of the ampli-tude of the input sinusoidal signal must be made. It is necessary to sample the waveform of the system output to make sure that the waveform is sinusoidal and that the system is operating in the linear region during the test period. (The waveform of the system output is not sinusoidal when the system is operating in its nonlinear region.) 4. If the system under consideration is operating continuously for days and weeks, then normal operation need not be stopped for frequency-response tests. The si-nusoidal test signal may be superimposed on the normal inputs.Then,for linear sys-tems, the output due to the test signal is superimposed on the normal output. For the determination of the transfer function while the system is in normal opera-tion, stochastic signals (white noise signals) also are often used. By use of corre-lation functions, the transfer function of the system can be determined without interrupting normal operation.
EXAMPLE 7–25 Determine the transfer function of the system whose experimental frequency-response curves are as shown in Figure 7–88.
The first step in determining the transfer function is to approximate the log-magnitude curve by asymptotes with slopes ;20 dBdecade and multiples thereof, as shown in Figure 7–88. We then estimate the corner frequencies. For the system shown in Figure 7–88, the following form of the transfer function is estimated: The value of the damping ratio z is estimated by examining the peak resonance near v=6 radsec.
Referring to Figure 7–9, z is determined to be 0.5.The gain K is numerically equal to the frequency at the intersection of the extension of the low-frequency asymptote that has 20 dB/decade slope and the 0-dB line.The value of K is thus found to be 10.Therefore, G(jv) is tentatively determined as or G(s) = 320(s + 2) s(s + 1)As2 + 8s + 64B G(jv) = 10(1 + 0.5jv) jv(1 + jv)c1 + aj v 8 b + aj v 8 b 2 d G(jv) = K(1 + 0.5jv) jv(1 + jv)c1 + 2z aj v 8 b + aj v 8 b 2 d Section 7–9 / Experimental Determination of Transfer Functions 489 490 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method This transfer function is tentative because we have not examined the phase-angle curve yet.
Once the corner frequencies are noted on the log-magnitude curve, the corresponding phase-angle curve for each component factor of the transfer function can easily be drawn.The sum of these component phase-angle curves is that of the assumed transfer function. The phase-angle curve for G(jv) is denoted by in Figure 7–88. From Figure 7–88, we clearly notice a dis-crepancy between the computed phase-angle curve and the experimentally obtained phase-angle curve. The difference between the two curves at very high frequencies appears to be a constant rate of change. Thus, the discrepancy in the phase-angle curves must be caused by transport lag.
Hence, we assume the complete transfer function to be G(s)e–Ts. Since the discrepancy be-tween the computed and experimental phase angles is –0.2v rad for very high frequencies, we can determine the value of T as follows: or The presence of transport lag can thus be determined, and the complete transfer function deter-mined from the experimental curves is G(s)e-Ts = 320(s + 2)e-0.2s s(s + 1)As2 + 8s + 64B T = 0.2 sec.
lim vS q d dv /G(jv)e-jvT = -T = -0.2 /G 40 20 0 –20 –40 –60 dB –80 –100 0.1 0.2 0.4 0.6 1 4 2 6 10 20 40 –500° –400° –300° –200° –100° 0° v in rad/sec G Magnitude (asymptotic) (K = 10) Magnitude (experimental) Phase angle (experimental) Figure 7–88 Bode diagram of a system. (Solid curves are experimentally obtained curves.) Section 7–10 / Control Systems Design by Frequency-Response Approach 491 7–10 CONTROL SYSTEMS DESIGN BY FREQUENCY-RESPONSE APPROACH In Chapter 6 we presented root-locus analysis and design. The root-locus method was shown to be very useful to reshape the transient-response characteristics of closed-loop control systems. The root-locus approach gives us direct information on the tran-sient response of the closed-loop system.The frequency-response approach,on the other hand, gives us this information only indirectly. However, as we shall see in the remain-ing three sections of this chapter, the frequency-response approach is very useful in de-signing control systems.
For any design problem,the designer will do well to use both approaches to the design and choose the compensator that most closely produces the desired closed-loop response.
In most control systems design, transient-response performance is usually very im-portant. In the frequency-response approach, we specify the transient-response per-formance in an indirect manner.That is, the transient-response performance is specified in terms of the phase margin, gain margin, resonant peak magnitude (they give a rough estimate of the system damping);the gain crossover frequency,resonant frequency,band-width (they give a rough estimate of the speed of transient response); and static error constants (they give the steady-state accuracy). Although the correlation between the transient response and frequency response is indirect, the frequency-domain specifica-tions can be easily met in the Bode diagram approach.
After the open loop has been designed, the closed-loop poles and zeros can be de-termined.Then, the transient-response characteristics must be checked to see whether the designed system satisfies the requirements in the time domain. If it does not, then the compensator must be modified and the analysis repeated until a satisfactory result is obtained.
Design in the frequency domain is simple and straightforward. The frequency-response plot indicates clearly the manner in which the system should be modified, al-though the exact quantitative prediction of the transient-response characteristics cannot be made. The frequency-response approach can be applied to systems or components whose dynamic characteristics are given in the form of frequency-response data. Note that because of difficulty in deriving the equations governing certain components, such as pneumatic and hydraulic components, the dynamic characteristics of such compo-nents are usually determined experimentally through frequency-response tests.The ex-perimentally obtained frequency-response plots can be combined easily with other such plots when the Bode diagram approach is used. Note also that in dealing with high-frequency noises we find that the frequency-response approach is more convenient than other approaches.
There are basically two approaches in the frequency-domain design. One is the polar plot approach and the other is the Bode diagram approach. When a compensator is added, the polar plot does not retain the original shape, and, therefore, we need to draw a new polar plot,which will take time and is thus inconvenient.On the other hand,a Bode diagram of the compensator can be simply added to the original Bode diagram, and thus plotting the complete Bode diagram is a simple matter.Also, if the open-loop gain is varied, the magnitude curve is shifted up or down without changing the slope of the curve, and the phase curve remains the same. For design purposes, therefore, it is best to work with the Bode diagram.
A common approach to the design based on the Bode diagram is that we first adjust the open-loop gain so that the requirement on the steady-state accuracy is met.Then the magnitude and phase curves of the uncompensated open loop (with the open-loop gain just adjusted) are plotted. If the specifications on the phase margin and gain margin are not satisfied, then a suitable compensator that will reshape the open-loop transfer func-tion is determined. Finally, if there are any other requirements to be met, we try to sat-isfy them, unless some of them are mutually contradictory.
Information Obtainable from Open-Loop Frequency Response.
The low-frequency region (the region far below the gain crossover frequency) of the locus indi-cates the steady-state behavior of the closed-loop system.The medium-frequency region (the region near the gain crossover frequency) of the locus indicates relative stability.
The high-frequency region (the region far above the gain crossover frequency) indi-cates the complexity of the system.
Requirements on Open-Loop Frequency Response.
We might say that,in many practical cases, compensation is essentially a compromise between steady-state accura-cy and relative stability.
To have a high value of the velocity error constant and yet satisfactory relative sta-bility, we find it necessary to reshape the open-loop frequency-response curve.
The gain in the low-frequency region should be large enough, and near the gain crossover frequency, the slope of the log-magnitude curve in the Bode diagram should be –20 dBdecade.This slope should extend over a sufficiently wide frequency band to assure a proper phase margin. For the high-frequency region, the gain should be atten-uated as rapidly as possible to minimize the effects of noise.
Examples of generally desirable and undesirable open-loop and closed-loop frequency-response curves are shown in Figure 7–89.
Referring to Figure 7–90, we see that the reshaping of the open-loop frequency-response curve may be done if the high-frequency portion of the locus follows the G1(jv) locus,while the low-frequency portion of the locus follows the G2(jv) locus.The reshaped locus Gc(jv)G(jv) should have reasonable phase and gain margins or should be tangent to a proper M circle, as shown.
492 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Im Re –1 0 Desirable Undesirable Im dB Re Log v –1 0 Desirable Undesirable (a) (b) Desirable Undesirable Figure 7–89 (a) Examples of desirable and undesirable open-loop frequency-response curves; (b) examples of desirable and undesirable closed-loop frequency-response curves.
Basic Characteristics of Lead, Lag, and Lag–Lead Compensation.
Lead com-pensation essentially yields an appreciable improvement in transient response and a small change in steady-state accuracy.It may accentuate high-frequency noise effects.Lag compensation, on the other hand, yields an appreciable improvement in steady-state accuracy at the expense of increasing the transient-response time. Lag compensation will suppress the effects of high-frequency noise signals. Lag–lead compensation com-bines the characteristics of both lead compensation and lag compensation.The use of a lead or lag compensator raises the order of the system by 1 (unless cancellation occurs between the zero of the compensator and a pole of the uncompensated open-loop trans-fer function).The use of a lag–lead compensator raises the order of the system by 2 [un-less cancellation occurs between zero(s) of the lag–lead compensator and pole(s) of the uncompensated open-loop transfer function], which means that the system becomes more complex and it is more difficult to control the transient-response behavior.The par-ticular situation determines the type of compensation to be used.
7–11 LEAD COMPENSATION We shall first examine the frequency characteristics of the lead compensator. Then we present a design technique for the lead compensator by use of the Bode diagram.
Characteristics of Lead Compensators.
Consider a lead compensator having the following transfer function: where a is the attenuation factor of the lead compensator. It has a zero at s=–1/T and a pole at s=–1/(aT). Since 0<a<1, we see that the zero is always located to the right of the pole in the complex plane. Note that for a small value of a the pole is lo-cated far to the left. The minimum value of a is limited by the physical construction of K c a Ts + 1 aTs + 1 = K c s + 1 T s + 1 aT (0 6 a 6 1) Section 7–11 / Lead Compensation 493 Im Re –1 0 M Circle G2( jv) G1( jv) Gc( jv)G( jv) Figure 7–90 Reshaping of the open-loop frequency-response curve.
the lead compensator.The minimum value of a is usually taken to be about 0.05. (This means that the maximum phase lead that may be produced by a lead compensator is about 65°.) [See Equation (7–25).] Figure 7–91 shows the polar plot of with Kc=1. For a given value of a, the angle between the positive real axis and the tan-gent line drawn from the origin to the semicircle gives the maximum phase-lead angle, fm. We shall call the frequency at the tangent point vm. From Figure 7–91 the phase angle at v=vm is fm, where (7–25) Equation (7–25) relates the maximum phase-lead angle and the value of a.
Figure 7–92 shows the Bode diagram of a lead compensator when Kc=1 and a=0.1.
The corner frequencies for the lead compensator are v=1/T and v=1/(aT)=10/T.
By examining Figure 7–92, we see that vm is the geometric mean of the two corner fre-quencies, or log vm = 1 2 alog 1 T + log 1 aT b sinfm = 1 - a 2 1 + a 2 = 1 - a 1 + a K c a jvT + 1 jvaT + 1 (0 6 a 6 1) 494 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Im Re vm 0 1 v = v = 0 fm 1 2 (1 + a) a 1 2 (1 – a) Figure 7–91 Polar plot of a lead compensator a(jvT+1)/(jvaT+1), where 0<a<1.
10 0 v in rad/sec –10 –20 90° 0° dB 0.1 T 1 T 10 T 100 T 10 T fm Figure 7–92 Bode diagram of a lead compensator a(jvT+1)/(jvaT+1), where a=0.1.
Hence, (7–26) As seen from Figure 7–92, the lead compensator is basically a high-pass filter. (The high frequencies are passed, but low frequencies are attenuated.) Lead Compensation Techniques Based on the Frequency-Response Approach.
The primary function of the lead compensator is to reshape the frequency-response curve to provide sufficient phase-lead angle to offset the excessive phase lag associated with the components of the fixed system.
Consider the system shown in Figure 7–93.Assume that the performance specifica-tions are given in terms of phase margin, gain margin, static velocity error constants, and so on.The procedure for designing a lead compensator by the frequency-response approach may be stated as follows: 1. Assume the following lead compensator: Define Then The open-loop transfer function of the compensated system is where Determine gain K to satisfy the requirement on the given static error constant.
2. Using the gain K thus determined, draw a Bode diagram of G1(jv), the gain-adjusted but uncompensated system. Evaluate the phase margin.
3. Determine the necessary phase-lead angle to be added to the system. Add an additional 5° to 12° to the phase-lead angle required, because the addition of the G1(s) = KG(s) Gc(s)G(s) = K Ts + 1 aTs + 1 G(s) = Ts + 1 aTs + 1 KG(s) = Ts + 1 aTs + 1 G1(s) Gc(s) = K Ts + 1 aTs + 1 K c a = K Gc(s) = K c a Ts + 1 aTs + 1 = K c s + 1 T s + 1 aT (0 6 a 6 1) vm = 1 1aT Section 7–11 / Lead Compensation 495 Gc(s) G(s) + – Figure 7–93 Control system.
lead compensator shifts the gain crossover frequency to the right and decreases the phase margin.
4. Determine the attenuation factor a by use of Equation (7–25). Determine the frequency where the magnitude of the uncompensated system G1(jv) is equal to Select this frequency as the new gain crossover frequency. This frequency corresponds to and the maximum phase shift fm occurs at this frequency.
5. Determine the corner frequencies of the lead compensator as follows: Zero of lead compensator: Pole of lead compensator: 6. Using the value of K determined in step 1 and that of a determined in step 4, calculate constant Kc from 7. Check the gain margin to be sure it is satisfactory. If not, repeat the design process by modifying the pole–zero location of the compensator until a satisfactory result is obtained.
EXAMPLE 7–26 Consider the system shown in Figure 7–94.The open-loop transfer function is It is desired to design a compensator for the system so that the static velocity error constant Kv is 20 sec–1, the phase margin is at least 50°, and the gain margin is at least 10 dB.
We shall use a lead compensator of the form The compensated system will have the open-loop transfer function Gc(s)G(s).
Define where K=Kca.
G1(s) = KG(s) = 4K s(s + 2) Gc(s) = K c a Ts + 1 aTs + 1 = K c s + 1 T s + 1 aT G(s) = 4 s(s + 2) K c = K a v = 1 aT v = 1 T vm = 1A1aTB, -20 log A11aB.
496 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method 4 s(s + 2) + – Figure 7–94 Control system.
The first step in the design is to adjust the gain K to meet the steady-state performance spec-ification or to provide the required static velocity error constant. Since this constant is given as 20 sec–1, we obtain or With K=10, the compensated system will satisfy the steady-state requirement.
We shall next plot the Bode diagram of Figure 7–95 shows the magnitude and phase-angle curves of G1(jv). From this plot, the phase and gain margins of the system are found to be 17° and ±q dB, respectively. (A phase margin of 17° implies that the system is quite oscillatory.Thus, satisfying the specification on the steady state yields a poor transient-response performance.) The specification calls for a phase margin of at least 50°. We thus find the additional phase lead necessary to satisfy the relative stability re-quirement is 33°.To achieve a phase margin of 50° without decreasing the value of K,the lead com-pensator must contribute the required phase angle.
Noting that the addition of a lead compensator modifies the magnitude curve in the Bode di-agram, we realize that the gain crossover frequency will be shifted to the right.We must offset the increased phase lag of G1(jv) due to this increase in the gain crossover frequency. Considering the shift of the gain crossover frequency, we may assume that fm, the maximum phase lead re-quired, is approximately 38°. (This means that 5° has been added to compensate for the shift in the gain crossover frequency.) Since sin fm = 1 - a 1 + a G1(jv) = 40 jv(jv + 2) = 20 jv(0.5jv + 1) K = 10 K v = lim sS0sGc(s)G(s) = lim sS0s Ts + 1 aTs + 1 G1(s) = lim sS0 s4K s(s + 2) = 2K = 20 Section 7–11 / Lead Compensation 497 1 2 4 8 v in rad/sec 40 20 0 –20 –40 0° –90° –180° 10 20 40 60 100 17° dB Figure 7–95 Bode diagram for G1(jv)=10G(jv) = 40/Cjv(jv+2)D fm=38° corresponds to a=0.24. Once the attenuation factor a has been determined on the basis of the required phase-lead angle,the next step is to determine the corner frequencies v=1/T and v=1/(aT) of the lead compensator. To do so, we first note that the maximum phase-lead angle fm occurs at the geometric mean of the two corner frequencies,or [See Equa-tion (7–26).] The amount of the modification in the magnitude curve at due to the inclusion of the term (Ts+1)/(aTs+1) is Note that and @G1(jv)@=–6.2 dB corresponds to v=9 radsec.We shall select this frequency to be the new gain crossover frequency vc. Noting that this frequency corresponds to or we obtain and The lead compensator thus determined is where the value of Kc is determined as Thus, the transfer function of the compensator becomes Note that The magnitude curve and phase-angle curve for Gc(jv)/10 are shown in Figure 7–96. The compensated system has the following open-loop transfer function: Gc(s)G(s) = 41.7 s + 4.41 s + 18.4 4 s(s + 2) Gc(s) K G1(s) = Gc(s) 10 10G(s) = Gc(s)G(s) Gc(s) = 41.7 s + 4.41 s + 18.4 = 10 0.227s + 1 0.054s + 1 K c = K a = 10 0.24 = 41.7 Gc(s) = K c s + 4.41 s + 18.4 = K c a 0.227s + 1 0.054s + 1 1 aT = vc 1a = 18.4 1 T = 1avc = 4.41 vc = 1A 1aTB, 1A 1aTB, 1 1a = 1 10.24 = 1 0.49 = 6.2 dB 2 1 + jvT 1 + jvaT 2 v=1A1aTB = 4 1 + j 1 1a 1 + ja 1 1a 4 = 1 1a v = 1A1aTB v = 1A1aTB.
498 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method The solid curves in Figure 7–96 show the magnitude curve and phase-angle curve for the compen-sated system.Note that the bandwidth is approximately equal to the gain crossover frequency. The lead compensator causes the gain crossover frequency to increase from 6.3 to 9 radsec. The in-crease in this frequency means an increase in bandwidth.This implies an increase in the speed of response.The phase and gain margins are seen to be approximately 50° and ±q dB, respectively.
The compensated system shown in Figure 7–97 therefore meets both the steady-state and the relative-stability requirements.
Note that for type 1 systems, such as the system just considered, the value of the static veloc-ity error constant Kv is merely the value of the frequency corresponding to the intersection of the extension of the initial –20-dBdecade slope line and the 0-dB line, as shown in Figure 7–96.
Note also that we have changed the slope of the magnitude curve near the gain crossover frequency from –40 dBdecade to –20 dBdecade.
Section 7–11 / Lead Compensation 499 40 20 0 –20 –40 0° –90° –180° Gc 10 Kv 50° 1 2 4 6 v in rad/sec 10 20 40 60 100 Gc 10 GcG GcG –6 dB G1 = 10G G1 = 10G dB Figure 7–96 Bode diagram for the compensated system.
4 s(s + 2) 41.7(s + 4.41) s + 18.4 + – Figure 7–97 Compensated system.
Figure 7–98 shows the polar plots of the gain-adjusted but uncompensated open-loop trans-fer function G1(jv)=10 G(jv) and the compensated open-loop transfer function Gc(jv)G(jv).
From Figure 7–98, we see that the resonant frequency of the uncompensated system is about 6 radsec and that of the compensated system is about 7 radsec. (This also indicates that the bandwidth has been increased.) From Figure 7–98, we find that the value of the resonant peak Mr for the uncompensated sys-tem with K=10 is 3.The value of Mr for the compensated system is found to be 1.29.This clear-ly shows that the compensated system has improved relative stability.
Note that, if the phase angle of G1(jv) decreases rapidly near the gain crossover frequency, lead compensation becomes ineffective because the shift in the gain crossover frequency to the right makes it difficult to provide enough phase lead at the new gain crossover frequency. This means that, to provide the desired phase margin, we must use a very small value for a.The value of a, however, should not be too small (smaller than 0.05) nor should the maximum phase lead fm be too large (larger than 65°), because such values will require an additional gain of excessive value. [If more than 65° is needed, two (or more) lead networks may be used in series with an iso-lating amplifier.] Finally, we shall examine the transient-response characteristics of the designed system. We shall obtain the unit-step response and unit-ramp response curves of the compensated and uncompensated systems with MATLAB. Note that the closed-loop transfer functions of the uncompensated and compensated systems are given, respectively, by and C(s) R(s) = 166.8s + 735.588 s3 + 20.4s2 + 203.6s + 735.588 C(s) R(s) = 4 s2 + 2s + 4 500 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Mr = 1.29 –4 –3 –2 –1 –1 –2 –3 – 4 0 1 Mr = 3 Im Re 1 4 4 6 6 10 10 v =3 v =3 G1(jv) Gc( jv)G( jv) Figure 7–98 Polar plots of the gain-adjusted but uncompensated open-loop transfer function G1and compensated open-loop transfer function GcG.
MATLAB programs for obtaining the unit-step response and unit-ramp response curves are given in MATLAB Program 7–13. Figure 7–99 shows the unit-step response curves of the system before and after compensation. Also, Figure 7–100 depicts the unit-ramp response curves before and after compensation.These response curves indicate that the designed system is satisfactory.
Section 7–11 / Lead Compensation 501 MATLAB Program 7–13 %Unit-step responses num = ; den = [1 2 4]; numc = [166.8 735.588]; denc = [1 20.4 203.6 735.588]; t = 0:0.02:6; [c1,x1,t] = step(num,den,t); [c2,x2,t] = step(numc,denc,t); plot (t,c1,'.',t,c2,'-') grid title('Unit-Step Responses of Compensated and Uncompensated Systems') xlabel('t Sec') ylabel('Outputs') text(0.4,1.31,'Compensated system') text(1.55,0.88,'Uncompensated system') %Unit-ramp responses num1 = ; den1 = [1 2 4 0]; num1c = [166.8 735.588]; den1c = [1 20.4 203.6 735.588 0]; t = 0:0.02:5; [y1,z1,t] = step(num1,den1,t); [y2,z2,t] = step(num1c,den1c,t); plot(t,y1,'.',t,y2,'-',t,t,'--') grid title('Unit-Ramp Responses of Compensated and Uncompensated Systems') xlabel('t Sec') ylabel('Outputs') text(0.89,3.7,'Compensated system') text(2.25,1.1,'Uncompensated system') It is noted that the closed-loop poles for the compensated system are located as follows: Because the dominant closed-loop poles are located far from the jv axis, the response damps out quickly.
s = -6.4918 s = -6.9541 ; j8.0592 502 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method 7–12 LAG COMPENSATION In this section we first discuss the Nyquist plot and Bode diagram of the lag compensator.
Then we present lag compensation techniques based on the frequency-response approach.
Characteristics of Lag Compensators.
Consider a lag compensator having the following transfer function: Gc(s) = K c b Ts + 1 bTs + 1 = K c s + 1 T s + 1 bT (b 7 1) Outputs 5 2 0 3.5 4.5 1.5 0.5 3 4 2.5 1 t Sec 0 1 0.5 5 3.5 4.5 3 4 2 1.5 2.5 Unit-Ramp Responses of Compensated and Uncompensated Systems Compensated system Uncompensated system Figure 7–100 Unit-ramp response curves of the compensated and uncompensated systems.
Outputs 1.4 0.6 0 1 1.2 0.4 0.2 0.8 t Sec 0 1 6 4 5 2 3 Unit-Step Responses of Compensated and Uncompensated Systems Compensated system Uncompensated system Figure 7–99 Unit-step response curves of the compensated and uncompensated systems.
Section 7–12 / Lag Compensation 503 In the complex plane, a lag compensator has a zero at s=–1/T and a pole at s=–1/(bT). The pole is located to the right of the zero.
Figure 7–101 shows a polar plot of the lag compensator. Figure 7–102 shows a Bode diagram of the compensator, where Kc=1 and b=10. The corner frequencies of the lag compensator are at v=1/T and v=1/(bT). As seen from Figure 7–102, where the values of Kc and b are set equal to 1 and 10, respectively, the magnitude of the lag compensator becomes 10 (or 20 dB) at low frequencies and unity (or 0 dB) at high fre-quencies.Thus, the lag compensator is essentially a low-pass filter.
Lag Compensation Techniques Based on the Frequency-Response Approach.
The primary function of a lag compensator is to provide attenuation in the high-frequency range to give a system sufficient phase margin. The phase-lag characteristic is of no consequence in lag compensation.
The procedure for designing lag compensators for the system shown in Figure 7–93 by the frequency-response approach may be stated as follows: 1. Assume the following lag compensator: Gc(s) = K c b Ts + 1 bTs + 1 = K c s + 1 T s + 1 bT (b 7 1) Im 0 Re v = 0 v = Kc Kcb Figure 7–101 Polar plot of a lag compensator K cb(jvT + 1)(jvbT + 1).
30 20 v in rad/sec 10 0 0° –90° dB 0.01 T 0.1 T 1 T 10 T Figure 7–102 Bode diagram of a lag compensator b(jvT+1)/(jvbT+1), with b=10.
Define Then The open-loop transfer function of the compensated system is where Determine gain K to satisfy the requirement on the given static velocity error constant.
2. If the gain-adjusted but uncompensated system G1(jv)=KG(jv) does not sat-isfy the specifications on the phase and gain margins, then find the frequency point where the phase angle of the open-loop transfer function is equal to –180° plus the required phase margin. The required phase margin is the specified phase margin plus 5° to 12°. (The addition of 5° to 12° compensates for the phase lag of the lag compensator.) Choose this frequency as the new gain crossover frequency.
3. To prevent detrimental effects of phase lag due to the lag compensator, the pole and zero of the lag compensator must be located substantially lower than the new gain crossover frequency. Therefore, choose the corner frequency v=1/T (cor-responding to the zero of the lag compensator) 1 octave to 1 decade below the new gain crossover frequency. (If the time constants of the lag compensator do not become too large, the corner frequency v=1/T may be chosen 1 decade below the new gain crossover frequency.) Notice that we choose the compensator pole and zero sufficiently small.Thus the phase lag occurs at the low-frequency region so that it will not affect the phase margin.
4. Determine the attenuation necessary to bring the magnitude curve down to 0 dB at the new gain crossover frequency. Noting that this attenuation is de-termine the value of b. Then the other corner frequency (corresponding to the pole of the lag compensator) is determined from v=1/(bT).
5. Using the value of K determined in step 1 and that of b determined in step 4, cal-culate constant Kc from K c = K b -20 log b, G1(s) = KG(s) Gc(s)G(s) = K Ts + 1 bTs + 1 G(s) = Ts + 1 bTs + 1 KG(s) = Ts + 1 bTs + 1 G1(s) Gc(s) = K Ts + 1 bTs + 1 K cb = K 504 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Section 7–12 / Lag Compensation 505 EXAMPLE 7–27 Consider the system shown in Figure 7–103.The open-loop transfer function is given by It is desired to compensate the system so that the static velocity error constant Kv is 5 sec–1, the phase margin is at least 40°, and the gain margin is at least 10 dB.
We shall use a lag compensator of the form Define Define also The first step in the design is to adjust the gain K to meet the required static velocity error con-stant.Thus, or With K=5, the compensated system satisfies the steady-state performance requirement.
We shall next plot the Bode diagram of G1(jv) = 5 jv(jv + 1)(0.5jv + 1) K = 5 = lim sS0 sK s(s + 1)(0.5s + 1) = K = 5 K v = lim sS0sGc(s)G(s) = lim sS0s Ts + 1 bTs + 1 G1(s) = lim sS0sG1(s) G1(s) = KG(s) = K s(s + 1)(0.5s + 1) K c b = K Gc(s) = K c b Ts + 1 bTs + 1 = K c s + 1 T s + 1 bT (b 7 1) G(s) = 1 s(s + 1)(0.5s + 1) 1 s(s +1) (0.5s + 1) + – Figure 7–103 Control system.
The magnitude curve and phase-angle curve of G1(jv) are shown in Figure 7–104. From this plot, the phase margin is found to be –20°, which means that the gain-adjusted but uncompensated system is unstable.
Noting that the addition of a lag compensator modifies the phase curve of the Bode diagram, we must allow 5° to 12° to the specified phase margin to compensate for the modification of the phase curve.Since the frequency corresponding to a phase margin of 40° is 0.7 radsec,the new gain crossover frequency (of the compensated system) must be chosen near this value. To avoid overly large time constants for the lag compensator,we shall choose the corner frequency v=1/T (which corresponds to the zero of the lag compensator) to be 0.1 radsec.Since this corner frequency is not too far below the new gain crossover frequency,the modification in the phase curve may not be small.Hence,we add about 12° to the given phase margin as an allowance to account for the lag angle introduced by the lag compensator.The required phase margin is now 52°.The phase angle of the uncompensated open-loop transfer function is –128° at about v=0.5 radsec. So we choose the new gain crossover frequency to be 0.5 radsec.To bring the magnitude curve down to 0 dB at this new gain crossover frequency,the lag compensator must give the necessary attenuation, which in this case is –20 dB. Hence, or The other corner frequency v=1(bT), which corresponds to the pole of the lag compen-sator, is then determined as 1 bT = 0.01 radsec b = 10 20 log 1 b = -20 11 dB 0 dB 0° –90° –180° –270° v in rad/sec 0.02 0.004 G1 G1 GcG Gc K Gc 40 20 –20 –40 40° 0.01 0.04 0.1 0.6 0.2 0.4 1 2 4 GcG Figure 7–104 Bode diagrams for G1 (gain-adjusted but uncompensated open-loop transfer function), Gc (compensator), and GcG (compensated open-loop transfer function).
506 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Section 7–12 / Lag Compensation 507 Thus, the transfer function of the lag compensator is Since the gain K was determined to be 5 and b was determined to be 10, we have The open-loop transfer function of the compensated system is The magnitude and phase-angle curves of Gc(jv)G(jv) are also shown in Figure 7–104.
The phase margin of the compensated system is about 40°, which is the required value. The gain margin is about 11 dB, which is quite acceptable.The static velocity error constant is 5 sec–1, as required. The compensated system, therefore, satisfies the requirements on both the steady state and the relative stability.
Note that the new gain crossover frequency is decreased from approximately 1 to 0.5 radsec.
This means that the bandwidth of the system is reduced.
To further show the effects of lag compensation,the log-magnitude-versus-phase plots of the gain-adjusted but uncompensated system G1(jv) and of the compensated system Gc(jv)G(jv) are shown in Figure 7–105.The plot of G1(jv) clearly shows that the gain-adjusted but uncompensated system is unstable.The addition of the lag compensator stabilizes the system.The plot of Gc(jv)G(jv) is tan-gent to the M=3 dB locus. Thus, the resonant peak value is 3 dB, or 1.4, and this peak occurs at v=0.5 radsec.
Compensators designed by different methods or by different designers (even using the same ap-proach) may look sufficiently different.Any of the well-designed systems, however, will give similar transient and steady-state performance.The best among many alternatives may be chosen from the economic consideration that the time constants of the lag compensator should not be too large.
Gc(s)G(s) = 5(10s + 1) s(100s + 1)(s + 1)(0.5s + 1) K c = K b = 5 10 = 0.5 Gc(s) = K c(10) 10s + 1 100s + 1 = K c s + 1 10 s + 1 100 8 4 0 –4 –90 24 16 20 12 –8 –12 –16 –20 –240 –210 –180 –150 –120 G1 G1 in dB 0.6 0.4 0.8 1 0.1 0.2 2 0.6 0.4 0.8 G1 GcG 3 dB 1 2 4 Figure 7–105 Log-magnitude-versus-phase plots of G1 (gain-adjusted but uncompensated open-loop transfer function) and GcG (compensated open-loop transfer function).
508 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Finally, we shall examine the unit-step response and unit-ramp response of the compensated system and the original uncompensated system without gain adjustment. The closed-loop trans-fer functions of the compensated and uncompensated systems are and respectively. MATLAB Program 7–14 will produce the unit-step and unit-ramp responses of the compensated and uncompensated systems.The resulting unit-step response curves and unit-ramp response curves are shown in Figures 7–106 and 7–107, respectively. From the response curves we find that the designed system satisfies the given specifications and is satisfactory.
C(s) R(s) = 1 0.5s3 + 1.5s2 + s + 1 C(s) R(s) = 50s + 5 50s4 + 150.5s3 + 101.5s2 + 51s + 5 MATLAB Program 7–14 %Unit-step response num = ; den = [0.5 1.5 1 1]; numc = [50 5]; denc = [50 150.5 101.5 51 5]; t = 0:0.1:40; [c1,x1,t] = step(num,den,t); [c2,x2,t] = step(numc,denc,t); plot(t,c1,'.',t,c2,'-') grid title('Unit-Step Responses of Compensated and Uncompensated Systems') xlabel('t Sec') ylabel('Outputs') text(12.7,1.27,'Compensated system') text(12.2,0.7,'Uncompensated system') %Unit-ramp response num1 = ; den1 = [0.5 1.5 1 1 0]; num1c = [50 5]; den1c = [50 150.5 101.5 51 5 0]; t = 0:0.1:20; [y1,z1,t] = step(num1,den1,t); [y2,z2,t] = step(num1c,den1c,t); plot(t,y1,'.',t,y2,'-',t,t,'--'); grid title('Unit-Ramp Responses of Compensated and Uncompensated Systems') xlabel('t Sec') ylabel('Outputs') text(8.3,3,'Compensated system') text(8.3,5,'Uncompensated system') Section 7–12 / Lag Compensation 509 Note that the zero and poles of the designed closed-loop system are as follows: The dominant closed-loop poles are very close to the jv axis with the result that the response is slow. Also, a pair of the closed-loop pole at s=–0.1228 and the zero at s=–0.1 produces a slowly decreasing tail of small amplitude. Poles at s = -0.2859 ; j0.5196, s = -0.1228, s = -2.3155 Zero at s = -0.1 Outputs 1.4 0.6 0 1 1.2 0.4 0.2 0.8 t Sec 0 10 5 40 30 35 25 15 20 Unit-Step Responses of Compensated and Uncompensated Systems Compensated system Uncompensated system Figure 7–106 Unit-step response curves for the compensated and uncompensated systems (Example 7–27).
Outputs 20 8 0 12 18 4 2 16 10 14 6 t Sec 0 4 2 20 14 18 12 16 8 6 10 Unit-Ramp Responses of Compensated and Uncompensated Systems Compensated system Uncompensated system Figure 7–107 Unit-ramp response curves for the compensated and uncompensated systems (Example 7–27).
A Few Comments on Lag Compensation.
1. Lag compensators are essentially low-pass filters. Therefore, lag compensation permits a high gain at low frequencies (which improves the steady-state per-formance) and reduces gain in the higher critical range of frequencies so as to im-prove the phase margin. Note that in lag compensation we utilize the attenuation characteristic of the lag compensator at high frequencies rather than the phase-lag characteristic. (The phase-lag characteristic is of no use for compensation purposes.) 2. Suppose that the zero and pole of a lag compensator are located at s=–z and s=–p, respectively.Then the exact locations of the zero and pole are not critical provided that they are close to the origin and the ratio z/p is equal to the required multiplication factor of the static velocity error constant.
It should be noted, however, that the zero and pole of the lag compensator should not be located unnecessarily close to the origin, because the lag compen-sator will create an additional closed-loop pole in the same region as the zero and pole of the lag compensator.
The closed-loop pole located near the origin gives a very slowly decaying tran-sient response, although its magnitude will become very small because the zero of the lag compensator will almost cancel the effect of this pole. However, the tran-sient response (decay) due to this pole is so slow that the settling time will be ad-versely affected.
It is also noted that in the system compensated by a lag compensator the trans-fer function between the plant disturbance and the system error may not involve a zero that is near this pole. Therefore, the transient response to the disturbance input may last very long.
3. The attenuation due to the lag compensator will shift the gain crossover frequency to a lower frequency point where the phase margin is accept-able. Thus, the lag compensator will reduce the bandwidth of the system and will result in slower transient response. [The phase angle curve of Gc(jv)G(jv) is relatively unchanged near and above the new gain crossover frequency.] 4. Since the lag compensator tends to integrate the input signal, it acts approximately as a proportional-plus-integral controller. Because of this, a lag-compensated sys-tem tends to become less stable. To avoid this undesirable feature, the time con-stant T should be made sufficiently larger than the largest time constant of the system.
5. Conditional stability may occur when a system having saturation or limiting is compensated by use of a lag compensator.When the saturation or limiting takes place in the system, it reduces the effective loop gain.Then the system becomes less stable and unstable operation may even result, as shown in Figure 7–108.
To avoid this, the system must be designed so that the effect of lag compensa-tion becomes significant only when the amplitude of the input to the saturat-ing element is small. (This can be done by means of minor feedback-loop compensation.) 510 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Section 7–13 / Lag–Lead Compensation 511 7–13 LAG–LEAD COMPENSATION We shall first examine the frequency-response characteristics of the lag–lead compen-sator. Then we present the lag–lead compensation technique based on the frequency-response approach.
Characteristic of Lag–Lead Compensator.
Consider the lag–lead compensator given by (7–27) where g>1 and b>1. The term produces the effect of the lead network, and the term produces the effect of the lag network.
s + 1 T 2 s + 1 bT 2 = b a T 2 s + 1 bT 2 s + 1 b (b 7 1) s + 1 T 1 s + g T 1 = 1 g ° T 1 s + 1 T 1 g s + 1 ¢ (g 7 1) Gc(s) = K c ± s + 1 T 1 s + g T 1 ≤± s + 1 T 2 s + 1 bT 2 ≤ dB 40 30 20 10 0 –10 –20 –90° –180° –270° 0.7 1 2 4 6 8 10 20 Large gain Small gain v in rad/sec f 0 f 0 Figure 7–108 Bode diagram of a conditionally stable system.
In designing a lag–lead compensator, we frequently chose g=b. (This is not necessary. We can, of course, choose g Z b.) In what follows, we shall consider the case where g=b.The polar plot of the lag–lead compensator with Kc=1 and g=b becomes as shown in Figure 7–109. It can be seen that, for 0<v<v1, the compensator acts as a lag compensator, while for v1<vFigure 7–109 Polar plot of a lag–lead compensator given by Equation (7–27), with Kc=1 and g=b.
10 0 –10 –20 –30 90° 0° –90° v in rad/sec dB 0.01 T1 1 T1 10 T1 0.1 T1 0.001 T1 100 T1 Figure 7–110 Bode diagram of a lag–lead compensator given by Equation (7–27) with Kc=1, g=b=10, and T 2 = 10T 1 .
Section 7–13 / Lag–Lead Compensation 513 Lag–Lead Compensation Based on the Frequency-Response Approach.
The design of a lag–lead compensator by the frequency-response approach is based on the combination of the design techniques discussed under lead compensation and lag compensation.
Let us assume that the lag–lead compensator is of the following form: (7–28) where b>1. The phase-lead portion of the lag–lead compensator (the portion involv-ing ) alters the frequency-response curve by adding phase-lead angle and increasing the phase margin at the gain crossover frequency.The phase-lag portion (the portion in-volving ) provides attenuation near and above the gain crossover frequency and there-by allows an increase of gain at the low-frequency range to improve the steady-state performance.
We shall illustrate the details of the procedures for designing a lag–lead compen-sator by an example.
EXAMPLE 7–28 Consider the unity-feedback system whose open-loop transfer function is It is desired that the static velocity error constant be 10 sec–1, the phase margin be 50°, and the gain margin be 10 dB or more.
Assume that we use the lag–lead compensator given by Equation (7–28).[Note that the phase-lead portion increases both the phase margin and the system bandwidth (which implies increas-ing the speed of response).The phase-lag portion maintains the low-frequency gain.] The open-loop transfer function of the compensated system is Gc(s)G(s). Since the gain K of the plant is adjustable, let us assume that Kc=1. Then, From the requirement on the static velocity error constant, we obtain Hence, We shall next draw the Bode diagram of the uncompensated system with K=20, as shown in Figure 7–111. The phase margin of the gain-adjusted but uncompensated system is found to be –32°, which indicates that the gain-adjusted but uncompensated system is unstable.
The next step in the design of a lag–lead compensator is to choose a new gain crossover fre-quency.From the phase-angle curve for G(jv), we notice that at v=1.5 radsec.
It is convenient to choose the new gain crossover frequency to be 1.5 radsec so that the phase-lead angle required at v=1.5 radsec is about 50°, which is quite possible by use of a single lag–lead network.
Once we choose the gain crossover frequency to be 1.5 radsec, we can determine the corner frequency of the phase-lag portion of the lag–lead compensator. Let us choose the corner fre-quency (which corresponds to the zero of the phase-lag portion of the compensator) to be 1 decade below the new gain crossover frequency, or at v=0.15 radsec.
v = 1T 2 /G(jv) = -180° K = 20 K v = lim sS0sGc(s)G(s) = lim sS0sGc(s) K s(s + 1)(s + 2) = K 2 = 10 lim sS0Gc(s) = 1.
G(s) = K s(s + 1)(s + 2) T 2 T 1 Gc(s) = K c (T 1 s + 1)(T 2 s + 1) a T 1 b s + 1 b(bT 2 s + 1) = K c a s + 1 T 1 b as + 1 T 2 b as + b T 1 b as + 1 bT 2 b Recall that for the lead compensator the maximum phase-lead angle fm is given by Equation (7–25), where a is 1/b in the present case. By substituting a=1/b in Equation (7–25), we have Notice that b=10 corresponds to fm=54.9°. Since we need a 50° phase margin, we may choose b=10. (Note that we will be using several degrees less than the maximum angle, 54.9°.) Thus, b=10 Then the corner frequency (which corresponds to the pole of the phase-lag portion of the compensator) becomes v=0.015 radsec. The transfer function of the phase-lag portion of the lag–lead compensator then becomes The phase-lead portion can be determined as follows: Since the new gain crossover frequen-cy is v=1.5 radsec, from Figure 7–111, G(j1.5) is found to be 13 dB. Hence, if the lag–lead com-pensator contributes –13 dB at v=1.5 radsec, then the new gain crossover frequency is as desired. From this requirement, it is possible to draw a straight line of slope 20 dBdecade, pass-ing through the point (1.5 radsec, –13 dB). The intersections of this line and the 0-dB line and –20-dB line determine the corner frequencies. Thus, the corner frequencies for the lead portion s + 0.15 s + 0.015 = 10 a 6.67s + 1 66.7s + 1 b v = 1bT 2 sin fm = 1 - 1 b 1 + 1 b = b - 1 b + 1 514 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method v in rad/sec dB 60 40 20 0 –40 –20 90° 0 –90° –180° –270° 0.02 0.01 0.04 0.1 0.2 0.4 0.6 G GcG G GcG Gc Gc 16 dB 1 2 4 6 10 –32° 50° Figure 7–111 Bode diagrams for G (gain-adjusted but uncompensated open-loop transfer function), Gc (compensator), and GcG (compensated open-loop transfer function).
Section 7–13 / Lag–Lead Compensation 515 are v=0.7 radsec and v=7 radsec. Thus, the transfer function of the lead portion of the lag–lead compensator becomes Combining the transfer functions of the lag and lead portions of the compensator, we obtain the transfer function of the lag–lead compensator. Since we chose Kc=1, we have The magnitude and phase-angle curves of the lag–lead compensator just designed are shown in Figure 7–111.The open-loop transfer function of the compensated system is (7–29) The magnitude and phase-angle curves of the system of Equation (7–29) are also shown in Fig-ure 7–111.The phase margin of the compensated system is 50°, the gain margin is 16 dB, and the static velocity error constant is 10 sec–1. All the requirements are therefore met, and the design has been completed.
Figure 7–112 shows the polar plots of G(jv) (gain-adjusted but uncompensated open-loop transfer function) and Gc(jv)G(jv) (compensated open-loop transfer function).The Gc(jv)G(jv) locus is tangent to the M=1.2 circle at about v=2 radsec. Clearly, this indicates that the com-pensated system has satisfactory relative stability. The bandwidth of the compensated system is slightly larger than 2 radsec. = 10(1.43s + 1)(6.67s + 1) s(0.143s + 1)(66.7s + 1)(s + 1)(0.5s + 1) Gc(s)G(s) = (s + 0.7)(s + 0.15)20 (s + 7)(s + 0.015)s(s + 1)(s + 2) Gc(s) = a s + 0.7 s + 7 b a s + 0.15 s + 0.015 b = a 1.43s + 1 0.143s + 1 b a 6.67s + 1 66.7s + 1 b s + 0.7 s + 7 = 1 10 a 1.43s + 1 0.143s + 1 b M = 1.2 v = 0.15 v = 1 Im Re G GcG 0.2 0.4 –8 –7 –6 –5 –4 –3 –2 –1 0 2 1 1 2 –2 2 –1 –4 –3 –5 –8 –7 1 2 –6 Figure 7–112 Polar plots of G (gain adjusted) and GcG.
In the following we shall examine the transient-response characteristics of the compensated system. (The gain-adjusted but uncompensated system is unstable.) The closed-loop transfer func-tion of the compensated system is The unit-step and unit-ramp response curves obtained with MATLAB are shown in Figures 7–113 and 7–114, respectively.
C(s) R(s) = 95.381s2 + 81s + 10 4.7691s5 + 47.7287s4 + 110.3026s3 + 163.724s2 + 82s + 10 516 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Output 1.6 0.6 0 1 1.4 0.4 0.2 1.2 0.8 t Sec 0 4 2 20 14 18 12 16 8 6 10 Unit-Step Response of Compensated System Figure 7–113 Unit-step response of the compensated system (Example 7–28).
Output 20 8 0 12 18 4 2 16 10 14 6 t Sec 0 4 2 20 14 18 12 16 8 6 10 Unit-Ramp Response of Compensated System Figure 7–114 Unit-ramp response of the compensated system (Example 7–28).
Section 7–13 / Lag–Lead Compensation 517 Note that the designed closed-loop control system has the following closed-loop zeros and poles: The pole at s=–0.1785 and zero at s=–0.1499 are located very close to each other. Such a pair of pole and zero produces a long tail of small amplitude in the step response,as seen in Figure 7–113.
Also,the pole at s=–0.5425 and zero at s=–0.6993 are located fairly close to each other.This pair adds amplitude to the long tail.
Summary of Control Systems Design by Frequency-Response Approach.
The last three sections presented detailed procedures for designing lead, lag, and lag–lead compensators by the use of simple examples. We have shown that the design of a compensator to satisfy the given specifications (in terms of the phase margin and gain margin) can be carried out in the Bode diagram in a simple and straightforward manner. It is noted that not every system can be compensated with a lead, lag, or lag–lead compensator. In some cases compensators with complex poles and zeros may be used. For systems that cannot be designed by use of the root-locus or frequency-response methods, the pole-placement method may be used. (See Chapter 10.) In a given design problem if both conventional design methods and the pole-placement method can be used, conventional methods (root-locus or frequency-response methods) usually result in a lower-order stable compensator. Note that a satisfactory design of a compensator for a complex system may require a creative application of all available design methods.
Comparison of Lead, Lag, and Lag–Lead Compensation 1. Lead compensation is commonly used for improving stability margins. Lag com-pensation is used to improve the steady-state performance. Lead compensation achieves the desired result through the merits of its phase-lead contribution,where-as lag compensation accomplishes the result through the merits of its attenuation property at high frequencies.
2. In some design problems both lead compensation and lag compensation may sat-isfy the specifications. Lead compensation yields a higher gain crossover frequen-cy than is possible with lag compensation. The higher gain crossover frequency means a larger bandwidth.A large bandwidth means reduction in the settling time.
The bandwidth of a system with lead compensation is always greater than that with lag compensation.Therefore, if a large bandwidth or fast response is desired, lead compensation should be employed. If, however, noise signals are present, then a large bandwidth may not be desirable, since it makes the system more suscepti-ble to noise signals because of an increase in the high-frequency gain. Hence, lag compensation should be used for such a case.
3. Lead compensation requires an additional increase in gain to offset the attenua-tion inherent in the lead network.This means that lead compensation will require a larger gain than that required by lag compensation.A larger gain, in most cases, implies larger space, greater weight, and higher cost. s = -0.1785, s = -0.5425, s = -7.4923 Poles at s = -0.8973 ; j1.4439 Zeros at s = -0.1499, s = -0.6993 4. Lead compensation may generate large signals in the system. Such large signals are not desirable because they will cause saturation in the system.
5. Lag compensation reduces the system gain at higher frequencies without reduc-ing the system gain at lower frequencies. Since the system bandwidth is reduced, the system has a slower speed to respond. Because of the reduced high-frequen-cy gain, the total system gain can be increased, and thereby low-frequency gain can be increased and the steady-state accuracy can be improved. Also, any high-frequency noises involved in the system can be attenuated.
6. Lag compensation will introduce a pole-zero combination near the origin that will generate a long tail with small amplitude in the transient response.
7. If both fast responses and good static accuracy are desired,a lag–lead compensator may be employed.By use of the lag–lead compensator,the low-frequency gain can be increased (which means an improvement in steady-state accuracy), while at the same time the system bandwidth and stability margins can be increased.
8. Although a large number of practical compensation tasks can be accomplished with lead, lag, or lag–lead compensators, for complicated systems, simple compensation by use of these compensators may not yield satisfactory results.
Then, different compensators having different pole–zero configurations must be employed.
Graphical Comparison.
Figure 7–115(a) shows a unit-step response curve and unit-ramp response curve of an uncompensated system.Typical unit-step response and unit-ramp response curves for the compensated system using a lead, lag, and lag–lead compensator, respectively, are shown in Figures 7–115(b), (c), and (d).The system with a lead compensator exhibits the fastest response, while that with a lag compensator ex-hibits the slowest response, but with marked improvements in the unit-ramp response.
The system with a lag–lead compensator will give a compromise; reasonable improve-518 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method c(t) 1 0 t c(t) 1 0 t c(t) 1 0 t c(t) 1 0 t c(t) 0 t c(t) 0 t c(t) 0 t c(t) 0 t ess ess ess ess (a) (b) (c) (d) Figure 7–115 Unit-step response curves and unit-ramp response curves.
(a) Uncompensated system; (b) lead compensated system; (c) lag compensated system; (d) lag–lead compensated system.
Section 7–13 / Lag–Lead Compensation 519 ments in both the transient response and steady-state response can be expected. The response curves shown depict the nature of improvements that may be expected from using different types of compensators.
Feedback Compensation.
A tachometer is one of the rate feedback devices.
Another common rate feedback device is the rate gyro. Rate gyros are commonly used in aircraft autopilot systems.
Velocity feedback using a tachometer is very commonly used in positional servo systems. It is noted that, if the system is subjected to noise signals, velocity feedback may generate some difficulty if a particular velocity feedback scheme performs differentiation of the output signal. (The result is the accentuation of the noise effects.) Cancellation of Undesirable Poles.
Since the transfer function of elements in cascade is the product of their individual transfer functions, it is possible to cancel some undesirable poles or zeros by placing a compensating element in cascade, with its poles and zeros being adjusted to cancel the undesirable poles or zeros of the original system.
For example, a large time constant may be canceled by use of the lead network as follows: If is much smaller than we can effectively eliminate the large time constant Figure 7–116 shows the effect of canceling a large time constant in step transient response.
If an undesirable pole in the original system lies in the right-half s plane, this com-pensation scheme should not be used since, although mathematically it is possible to cancel the undesirable pole with an added zero, exact cancellation is physically impos-sible because of inaccuracies involved in the location of the poles and zeros. A pole in the right-half s plane not exactly canceled by the compensator zero will eventually lead to unstable operation, because the response will involve an exponential term that in-creases with time.
It is noted that if a left-half plane pole is almost canceled but not exactly can-celed, as is almost always the case, the uncanceled pole-zero combination will cause the response to have a small amplitude but long-lasting transient-response compo-nent. If the cancellation is not exact but is reasonably good, then this component will be small.
It should be noted that the ideal control system is not the one that has a transfer function of unity. Physically, such a control system cannot be built since it cannot T 1 .
T 1 , T 2 a 1 T 1 s + 1 b a T 1 s + 1 T 2 s + 1 b = 1 T 2 s + 1 AT 1 s + 1BAT 2 s + 1B T 1 x x y z y z t t t 1 T1s + 1 T1s + 1 T2s + 1 Figure 7–116 Step-response curves showing the effect of canceling a large time constant.
instantaneously transfer energy from the input to the output. In addition, since noise is almost always present in one form or another, a system with a unity transfer function is not desirable. A desired control system, in many practical cases, may have one set of dominant complex-conjugate closed-loop poles with a reasonable damping ratio and undamped natural frequency.The determination of the significant part of the closed-loop pole-zero configuration, such as the location of the domi-nant closed-loop poles, is based on the specifications that give the required system performance.
Cancellation of Undesirable Complex-Conjugate Poles.
If the transfer func-tion of a plant contains one or more pairs of complex-conjugate poles, then a lead, lag, or lag–lead compensator may not give satisfactory results. In such a case, a network that has two zeros and two poles may prove to be useful. If the zeros are chosen so as to cancel the undesirable complex-conjugate poles of the plant, then we can essentially replace the undesirable poles by acceptable poles. That is, if the undesirable complex-conjugate poles are in the left-half s plane and are in the form then the insertion of a compensating network having the transfer function will result in an effective change of the undesirable complex-conjugate poles to ac-ceptable poles. Note that even though the cancellation may not be exact, the com-pensated system will exhibit better response characteristics. (As stated earlier, this approach cannot be used if the undesirable complex-conjugate poles are in the right-half s plane.) Familiar networks consisting only of RC components whose transfer functions pos-sess two zeros and two poles are the bridged-T networks. Examples of bridged-T net-works and their transfer functions are shown in Figure 7–117. (The derivations of the transfer functions of the bridged-T networks were given in Problem A–3–5.) s2 + 2z1 v1 s + v2 1 s2 + 2z2 v2 s + v2 2 1 s2 + 2z1 v1 s + v2 1 520 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method C2 R R C1 R2 C C R1 ei eo ei eo (a) (b) Eo(s) Ei(s) RC1RC2s2 + 2RC2s + 1 RC1RC2s2 + (RC1 + 2RC2)s + 1 = Eo(s) Ei(s) R1CR2Cs2 + 2R1Cs + 1 R1CR2Cs2 + (R2C + 2R1C)s + 1 = Figure 7–117 Bridged-T networks.
Example Problems and Solutions 521 Concluding Comments.
In the design examples presented in this chapter, we have been primarily concerned only with the transfer functions of compensators. In ac-tual design problems, we must choose the hardware. Thus, we must satisfy additional design constraints such as cost, size, weight, and reliability.
The system designed may meet the specifications under normal operating condi-tions but may deviate considerably from the specifications when environmental changes are considerable. Since the changes in the environment affect the gain and time con-stants of the system, it is necessary to provide automatic or manual means to adjust the gain to compensate for such environmental changes, for nonlinear effects that were not taken into account in the design, and also to compensate for manufacturing tolerances from unit to unit in the production of system components. (The effects of manufactur-ing tolerances are suppressed in a closed-loop system; therefore, the effects may not be critical in closed-loop operation but critical in open-loop operation.) In addition to this, the designer must remember that any system is subject to small variations due mainly to the normal deterioration of the system.
15 10 5 0 –5 –10 –15 90° 45° 0° –45° –90° 0.2 0.4 0.6 1 2 4 6 10 20 40 v in rad/sec C(jv) R(jv) C(jv) R(jv) in dB Asymptote Figure 7–118 Bode diagram for 10(1 + jv)C(2 + jv)(5 + jv)D.
EXAMPLE PROBLEMS AND SOLUTIONS A–7–1.
Consider a system whose closed-loop transfer function is Clearly, the closed-loop poles are located at s=–2 and s=–5, and the system is not oscillatory.
Show that the closed-loop frequency response of this system will exhibit a resonant peak, al-though the damping ratio of the closed-loop poles is greater than unity.
Solution. Figure 7–118 shows the Bode diagram for the system. The resonant peak value is ap-proximately 3.5 dB. (Note that, in the absence of a zero, the second-order system with z>0.7 will not exhibit a resonant peak; however, the presence of a closed-loop zero will cause such a peak.) C(s) R(s) = 10(s + 1) (s + 2)(s + 5) A–7–2.
Consider the system defined by Obtain the sinusoidal transfer functions and In deriving and we assume that Simi-larly, in obtaining and we assume that Solution. The transfer matrix expression for the system defined by is given by where G(s) is the transfer matrix and is given by For the system considered here, the transfer matrix becomes Hence Assuming that U2(jv)=0, we find and as follows: Similarly, assuming that U1(jv)=0, we find and as follows: Notice that is a nonminimum-phase transfer function.
Y 2(jv)U 2(jv) Y 2(jv) U 2(jv) = jv - 25 (jv)2 + 4jv + 25 Y 1(jv) U 2(jv) = jv + 5 (jv)2 + 4jv + 25 Y 2(jv)U 2(jv) Y 1(jv)U 2(jv) Y 2(jv) U 1(jv) = -25 (jv)2 + 4jv + 25 Y 1(jv) U 1(jv) = jv + 4 (jv)2 + 4jv + 25 Y 2(jv)U 1(jv) Y 1(jv)U 1(jv) B Y 1(s) Y 2(s)R = D s + 4 s2 + 4s + 25 -25 s2 + 4s + 25 s + 5 s2 + 4s + 25 s - 25 s2 + 4s + 25 T B U 1(s) U 2(s)R = D s + 4 s2 + 4s + 25 -25 s2 + 4s + 25 s + 5 s2 + 4s + 25 s - 25 s2 + 4s + 25 T = 1 s2 + 4s + 25 B s + 4 -25 1 sR B1 0 1 1R C(s I - A)-1 B + D = B 1 0 0 1R B s 25 -1 s + 4R -1 B 1 0 1 1R G(s) = C(s I - A)-1 B + D Y(s) = G(s)U(s) y # = Cx + Du x # = Ax + Bu U 1(jv) = 0.
Y 2(jv)U 2(jv), Y 1(jv)U 2(jv) U 2(jv) = 0.
Y 2(jv)U 1(jv), Y 1(jv)U 1(jv) Y 2(jv)U 2(jv).
Y 1(jv)U 2(jv), Y 2(jv)U 1(jv), Y 1(jv)U 1(jv), B y1 y2 R = B 1 0 0 1R B x1 x2 R B x # 1 x # 2 R = B 0 -25 1 -4R B x1 x2 R + B 1 0 1 1R Bu1 u2 R 522 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Example Problems and Solutions 523 Frequency (rad/sec) Phase (deg); Magnitude (dB) Bode Diagrams 100 101 102 100 101 102 From: U1 From: U2 −40 −20 0 −100 0 100 −100 0 100 −200 0 200 To: Y1 To: Y2 Figure 7–119 Bode diagrams.
A–7–3.
Referring to Problem A–7–2, plot Bode diagrams for the system, using MATLAB.
Solution. MATLAB Program 7–15 produces Bode diagrams for the system. There are four sets of Bode diagrams: two for input 1 and two for input 2. These Bode diagrams are shown in Figure 7–119.
MATLAB Program 7–15 A = [0 1;-25 -4]; B = [1 1;0 1]; C = [1 0;0 1]; D = [0 0;0 0]; bode(A,B,C,D) 524 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method K s(s + 1) (s + 5) + – R(s) C(s) Figure 7–120 Closed-loop system.
A–7–4.
Using MATLAB, plot Bode diagrams for the closed-loop system shown in Figure 7–120 for K=1, K=10, and K=20. Plot three magnitude curves in one diagram and three phase-angle curves in another diagram.
Solution. The closed-loop transfer function of the system is given by Hence the numerator and denominator of C(s)R(s) are num = [K] den = [1 6 5 K] A possible MATLAB program is shown in MATLAB Program 7–16.The resulting Bode diagrams are shown in Figures 7–121(a) and (b). = K s3 + 6s2 + 5s + K C(s) R(s) = K s(s + 1)(s + 5) + K MATLAB Program 7–16 w = logspace(-1,2,200); for i = 1:3; if i = 1; K = 1;[mag,phase,w] = bode([K],[1 6 5 K],w); mag1dB = 20log10(mag); phase1 = phase; end; if i = 2; K = 10;[mag,phase,w] = bode([K],[1 6 5 K],w); mag2dB = 20log10(mag); phase2 = phase; end; if i = 3; K = 20;[mag,phase,w] = bode([K],[1 6 5 K],w); mag3dB = 20log10(mag); phase3 = phase; end; end semilogx(w,mag1dB,'-',w,mag2dB,'-',w,mag3dB,'-') grid title('Bode Diagrams of G(s) = K/[s(s + 1)(s + 5)], where K = 1, K = 10, and K = 20') xlabel('Frequency (rad/sec)') ylabel('Gain (dB)') text(1.2,-31,'K = 1') text(1.1,-8,'K = 10') text(11,-31,'K = 20') semilogx(w,phase1,'-',w,phase2,'-',w,phase3,'-') grid xlabel('Frequency (rad/sec)') ylabel('Phase (deg)') text(0.2,-90,'K = 1') text(0.2,-20,'K =10') text(1.6,-20,'K = 20') Example Problems and Solutions 525 Frequency (rad/sec) Bode Diagrams of G(s) = K/[s(s + 1)(s + 5)], where K = 1, K = 10, and K = 20 −140 Gain (dB) −120 −100 −80 −60 −40 −20 20 0 10−1 100 101 102 K = 10 K = 20 K = 1 (a) Frequency (rad/sec) −300 −200 −150 −100 −50 −250 0 Phase (deg) 10−1 100 101 102 K = 10 K = 20 K = 1 (b) Figure 7–121 Bode diagrams: (a) Magnitude-versus-frequency curves; (b) phase-angle-versus-frequency curves.
A–7–5.
Prove that the polar plot of the sinusoidal transfer function is a semicircle. Find the center and radius of the circle.
G(jv) = jvT 1 + jvT , for 0 v q 526 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Solution. The given sinusoidal transfer function G(jv) can be written as follows: where Then Hence, we see that the plot of G(jv) is a circle centered at (0.5,0) with radius equal to 0.5. The upper semicircle corresponds to 0 v q, and the lower semicircle corresponds to –q v 0.
A–7–6.
Prove the following mapping theorem: Let F(s) be a ratio of polynomials in s. Let P be the num-ber of poles and Z be the number of zeros of F(s) that lie inside a closed contour in the s plane, with multiplicity accounted for. Let the closed contour be such that it does not pass through any poles or zeros of F(s).The closed contour in the s plane then maps into the F(s) plane as a closed curve.The number N of clockwise encirclements of the origin of the F(s) plane, as a representa-tive point s traces out the entire contour in the s plane in the clockwise direction, is equal to Z-P.
Solution. To prove this theorem, we use Cauchy’s theorem and the residue theorem. Cauchy’s theorem states that the integral of F(s) around a closed contour in the s plane is zero if F(s) is analytic# within and on the closed contour, or Suppose that F(s) is given by where X(s) is analytic in the closed contour in the s plane and all the poles and zeros are located in the contour.Then the ratio F¿(s)/F(s) can be written (7–30) This may be seen from the following consideration: If is given by then has a zero of kth order at s=–z1. Differentiating F(s) with respect to s yields Hence, (7–31) We see that by taking the ratio , the kth-order zero of becomes a simple pole of .
F ˆ ¿(s)F ˆ (s) F ˆ (s) F ˆ ¿(s)F ˆ (s) F ˆ ¿(s) F ˆ (s) = k s + z1 + X¿(s) X(s) F ˆ ¿(s) = kAs + z1B k-1X(s) + As + z1B kX¿(s) F ˆ (s) F ˆ (s) = As + z1B kX(s) F ˆ (s) F¿(s) F(s) = a k1 s + z1 + k2 s + z2 + p b - a m1 s + p1 + m2 s + p2 + p b + X¿(s) X(s) F(s) = As + z1B k1As + z2B k2 p As + p1B m1As + p2B m2 p X(s) I F(s)ds = 0 aX - 1 2 b 2 + Y2 = Av2 T2 - 1B 2 4A1 + v2 T2B 2 + v2 T2 A1 + v2 T2B 2 = 1 4 X = v2 T2 1 + v2 T2 , Y = vT 1 + v2 T2 G(jv) = X + jY #For the definition of an analytic function, see the footnote on page 447.
Example Problems and Solutions 527 If the last term on the right-hand side of Equation (7–31) does not contain any poles or zeros in the closed contour in the s plane, F¿(s)/F(s) is analytic in this contour except at point s=–z1.
Then, referring to Equation (7–30) and using the residue theorem, which states that the integral of F¿(s)/F(s) taken in the clockwise direction around a closed contour in the s plane is equal to –2pj times the residues at the simple poles of F¿(s)/F(s), or we have where total number of zeros of F(s) enclosed in the closed contour in the s plane total number of poles of F(s) enclosed in the closed contour in the s plane [The k multiple zeros (or poles) are considered k zeros (or poles) located at the same point.] Since F(s) is a complex quantity, F(s) can be written and Noting that F¿(s)/F(s) can be written we obtain If the closed contour in the s plane is mapped into the closed contour in the F(s) plane, then The integral is zero since the magnitude is the same at the initial point and the final point of the contour Thus we obtain The angular difference between the final and initial values of u is equal to the total change in the phase angle of F¿(s)/F(s) as a representative point in the s plane moves along the closed contour. Noting that N is the number of clockwise encirclements of the origin of the F(s) plane and u2-u1 is zero or a multiple of 2p rad, we obtain u2 - u1 2p = -N u2 - u1 2p = P - Z G.
ln ∑F∑ D G d ln ∑F∑ I F¿(s) F(s) ds = I G d ln ∑F∑+ j I G du = j 3 du = 2pj(P - Z) G F¿(s) F(s) = d ln ∑F∑ ds + j du ds F¿(s) F(s) = d lnF(s) ds lnF(s) = ln ∑F∑+ ju F(s) = ∑F∑eju P = m1 + m2 + p = Z = k1 + k2 + p = I F¿(s) F(s) ds = -2pjCAk1 + k2 + pB - Am1 + m2 + pB D = -2pj(Z - P) I F¿(s) F(s) ds = -2pj a a residues b 528 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method 0 Re (a) Im v = –1 v = 0 Re Im –1 G Plane v = v = 0+ v = 0– (b) Figure 7–123 (a) Nyquist plot; (b) complete Nyquist plot in the G plane.
Thus, we have the relationship This proves the theorem.
Note that by this mapping theorem,the exact numbers of zeros and of poles cannot be found— only their difference. Note also that, from Figures 7–122(a) and (b), we see that if u does not change through 2p rad, then the origin of the F(s) plane cannot be encircled.
A–7–7.
The Nyquist plot (polar plot) of the open-loop frequency response of a unity-feedback control system is shown in Figure 7–123(a). Assuming that the Nyquist path in the s plane encloses the entire right-half s plane, draw a complete Nyquist plot in the G plane.Then answer the following questions: (a) If the open-loop transfer function has no poles in the right-half s plane, is the closed-loop system stable?
(b) If the open-loop transfer function has one pole and no zeros in right-half s plane,is the closed-loop system stable?
(c) If the open-loop transfer function has one zero and no poles in the right-half s plane, is the closed-loop system stable?
N = Z - P Re Im u1 u2 u2 Origin encircled u2 – u1 = 2p Origin not encircled u2 – u1 = 0 F(s) Plane F(s) Plane 0 (a) (b) Re Im 0 u1 Figure 7–122 Determination of encirclement of the origin of F(s) plane.
Example Problems and Solutions 529 Solution. Figure 7–123(b) shows a complete Nyquist plot in the G plane.The answers to the three questions are as follows: (a) The closed-loop system is stable, because the critical point (–1+j0) is not encircled by the Nyquist plot.That is, since P=0 and N=0, we have Z=N+P=0.
(b) The open-loop transfer function has one pole in the right-half s plane. Hence, P=1. (The open-loop system is unstable.) For the closed-loop system to be stable, the Nyquist plot must encircle the critical point (–1+j0) once counterclockwise. However, the Nyquist plot does not encircle the critical point.Hence,N=0.Therefore,Z=N+P=1.The closed-loop sys-tem is unstable.
(c) Since the open-loop transfer function has one zero, but no poles, in the right-half s plane, we have Z=N+P=0. Thus, the closed-loop system is stable. (Note that the zeros of the open-loop transfer function do not affect the stability of the closed-loop system.) A–7–8.
Is a closed-loop system with the following open-loop transfer function and with K=2 stable?
Find the critical value of the gain K for stability.
Solution. The open-loop transfer function is This open-loop transfer function has no poles in the right-half s plane. Thus, for stability, the –1+j0 point should not be encircled by the Nyquist plot. Let us find the point where the Nyquist plot crosses the negative real axis. Let the imaginary part of G(jv)H(jv) be zero, or from which Substituting into G(jv)H(jv), we obtain The critical value of the gain K is obtained by equating –2K/3 to –1, or Hence, The system is stable if Hence, the system with K=2 is unstable.
0 6 K 6 3 2 .
K = 3 2 - 2 3 K = -1 G aj 1 12 bH aj 1 12 b = - 2K 3 v = 112 v = ; 1 12 1 - 2v2 = 0 = K -3v2 + jvA1 - 2v2B G(jv)H(jv) = K jv(jv + 1)(2jv + 1) G(s)H(s) = K s(s + 1)(2s + 1) 530 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Re Im Re Re Im Im –1 –1 G Plane G Plane G Plane (Stable) (Unstable) (a) (b) v = v = v = v = – v = – v = – v = 0 v = 0 v = 0 P = 1 N = –1 Z = 0 P = 1 N = 0 K 1 K 1 Z = 1 K 2 K 2 – Figure 7–125 (a) Polar plot of K/(jv-1); (b) polar plots of K/(jv-1) for stable and unstable cases.
A–7–9.
Consider the closed-loop system shown in Figure 7–124. Determine the critical value of K for stability by the use of the Nyquist stability criterion.
Solution. The polar plot of is a circle with center at –K/2 on the negative real axis and radius K/2, as shown in Figure 7–125(a). As v is increased from –q to q, the G(jv) locus makes a counterclockwise rotation.
In this system, P=1 because there is one pole of G(s) in the right-half s plane. For the closed-loop system to be stable, Z must be equal to zero.Therefore, N=Z-P must be equal to –1, or there must be one counterclockwise encirclement of the –1+j0 point for stability. (If there is no encirclement of the –1+j0 point, the system is unstable.) Thus, for stability, K must be greater than unity,and K=1 gives the stability limit.Figure 7–125(b) shows both stable and unstable cases of G(jv) plots.
G(jv) = K jv - 1 R(s) C(s) K s – 1 + – Figure 7–124 Closed-loop system.
Example Problems and Solutions 531 Re Im 4 3 v = 2.45 v = 2 v = 1.5 –1 –2 –1 1 1 2 3 v = 1 v = 0 6 8 9 10 v = 0.5 2.65 1 + jv 2.65 e–0.8jv 1 + jv Figure 7–126 Polar plots of and 2.65/(1+jv).
2.65e-0.8jv(1 + jv) A–7–10.
Consider a unity-feedback system whose open-loop transfer function is Using the Nyquist plot, determine the critical value of K for stability.
Solution. For this system, The imaginary part of G(jv) is equal to zero if Hence, Solving this equation for the smallest positive value of v, we obtain Substituting v=2.4482 into G(jv), we obtain The critical value of K for stability is obtained by letting G(j2.4482) equal –1. Hence, or Figure 7–126 shows the Nyquist or polar plots of 2.65e–0.8jv/(1+jv) and 2.65/(1+jv).The first-order system without transport lag is stable for all values of K, but the one with a transport lag of 0.8 sec becomes unstable for K>2.65.
K = 2.65 0.378K = 1 G(j2.4482) = K 1 + 2.44822 (cos1.9586 - 2.4482 sin1.9586) = -0.378K v = 2.4482 v = -tan0.8v sin0.8v + v cos 0.8v = 0 = K 1 + v2C(cos 0.8v - v sin0.8v) - j(sin0.8v + v cos0.8v)D = K(cos0.8v - j sin 0.8v)(1 - jv) 1 + v2 G(jv) = Ke-0.8jv jv + 1 G(s) = Ke-0.8s s + 1 532 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method A–7–11.
Consider a unity-feedback system with the following open-loop transfer function: Draw a Nyquist plot with MATLAB and examine the stability of the closed-loop system.
Solution. MATLAB Program 7–17 produces the Nyquist diagram shown in Figure 7–127. From this figure, we see that the Nyquist plot does not encircle the –1+j0 point. Hence, N=0 in the Nyquist stability criterion. Since no open-loop poles lie in the right-half s plane, P=0. Therefore, Z=N+P=0. The closed-loop system is stable.
G(s) = 20As2 + s + 0.5B s(s + 1)(s + 10) A–7–12.
Consider the same system as discussed in Problem A–7–11. Draw the Nyquist plot for only the positive-frequency region.
Solution. Drawing a Nyquist plot for only the positive-frequency region can be done by the use of the following command: [re,im,w] = nyquist(num,den,w) The frequency region may be divided into several subregions by using different increments. For example, the frequency region of interest may be divided into three subregions as follows: w1 = 0.1:0.1:10; w2 = 10:2:100; w3 = 100:10:500; w = [w1 w2 w3] MATLAB Program 7–17 num = [20 20 10]; den = [1 11 10 0]; nyquist(num,den) v = [-2 3 -3 3]; axis(v) grid Real Axis −1 −0.5 −1.5 −2 3 1 1.5 0.5 2 2.5 0 Imaginary Axis −3 3 2 1 −2 −1 0 Nyquist Diagram Figure 7–127 Nyquist plot of G(s) = 20As2 + s + 0.5B s(s + 1)(s + 10) .
Example Problems and Solutions 533 Real Axis –3 3 2 1 0 –1 –2 Imag Axis –3 –5 1 –1 –4 –2 0 Nyquist Plot of G(s) = 20(s2+s+0.5)/[s(s+1)(s+10)] Figure 7–128 Nyquist plot for the positive-frequency region.
MATLAB Program 7–18 num = [20 20 10]; den = [1 11 10 0]; w1 = 0.1:0.1:10; w2 = 10:2:100; w3 = 100:10:500; w = [w1 w2 w3]; [re,im,w] = nyquist(num,den,w); plot(re,im) v = [-3 3 -5 1]; axis(v); grid title('Nyquist Plot of G(s) = 20(s^2 + s + 0.5)/[s(s + 1)(s + 10)]') xlabel('Real Axis') ylabel('Imag Axis') A–7–13.
Referring to Problem A–7–12, plot the polar locus of G(s) where Locate on the polar locus frequency points where v= 0.2, 0.3, 0.5, 1, 2, 6, 10, and 20 radsec.
Also, find the magnitudes and phase angles of G(jv) at the specified frequency points.
Solution. In MATLAB Program 7–19 we used the frequency vector w, which consists of three frequency subvectors: w1, w2, and w3. Instead of such a w, we may simply use the G(s) = 20As2 + s + 0.5B s(s + 1)(s + 10) MATLAB Program 7–18 uses this frequency region. Using this program, we obtain the Nyquist plot shown in Figure 7–128.
534 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method MATLAB Program 7–19 num = [20 20 10]; den = [1 11 10 0]; ww = logspace(-1,2,100); nyquist(num,den,ww) v = [-2 3 -5 0]; axis(v); grid hold Current plot held w = [0.2 0.3 0.5 1 2 6 10 20]; [re,im,w] = nyquist(num,den,w); plot(re,im,'o') text(1.1,-4.8,'w = 0.2') text(1.1,-3.1,'0.3') text(1.25,-1.7,'0.5') text(1.37,-0.4,'1') text(1.8,-0.3,'2') text(1.4,-1.1,'6') text(0.77,-0.8,'10') text(0.037,-0.8,'20') % ----- To get the values of magnitude and phase (in degrees) of G(jw) % at the specified w values, enter the command [mag,phase,w] % = bode(num,den,w) ------[mag,phase,w] = bode(num,den,w); % ----- The following table shows the specified frequency values w and % the corresponding values of magnitude and phase (in degrees) -----[w mag phase] ans = 0.2000 4.9176 -78.9571 0.3000 3.2426 -72.2244 0.5000 1.9975 -55.9925 1.0000 1.5733 -24.1455 2.0000 1.7678 -14.4898 6.0000 1.6918 -31.0946 10.0000 1.4072 -45.0285 20.0000 0.8933 -63.4385 frequency vector w = logscale(d1, d2, n). MATLAB Program 7–19 uses the following fre-quency vector: w = logscale(-1,2,100) This MATLAB program plots the polar locus and locates the specified frequency points on the polar locus, as shown in Figure 7–129.
Example Problems and Solutions 535 MATLAB Program 7–20 num = [-1 -4 -6]; den = [1 5 4]; nyquist(num,den); grid title('Nyquist Plot of G(s) = -(s^2 + 4s + 6)/(s^2 + 5s + 4)') A–7–14.
Consider a unity-feedback, positive-feedback system with the following open-loop transfer function: Draw a Nyquist plot.
Solution. The Nyquist plot of the positive-feedback system can be obtained by defining num and den as num = [-1 -4 -6] den = [1 5 4] and using the command nyquist(num,den). MATLAB Program 7–20 produces the Nyquist plot, as shown in Figure 7–130.
This system is unstable, because the –1+j0 point is encircled once clockwise. Note that this is a special case where the Nyquist plot passes through –1+j0 point and also encircles this point once clockwise.This means that the closed-loop system is degenerate; the system behaves as if it were an unstable first-order system.See the following closed-loop transfer function of the positive-feedback system: = s2 + 4s + 6 s - 2 C(s) R(s) = s2 + 4s + 6 s2 + 5s + 4 - As2 + 4s + 6B G(s) = s2 + 4s + 6 s2 + 5s + 4 Real Axis −1 −0.5 −1.5 −2 3 1 1.5 0.5 2 2.5 0 Imaginary Axis −5 0 −0.5 −1 −2.5 −3 −3.5 −4 −4.5 −2 −1.5 Nyquist Diagram w = 0.2 0.3 0.5 6 2 1 10 20 Figure 7–129 Polar plot of G(jv) given in Problem A–7–13.
536 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Note that the Nyquist plot for the positive-feedback case is a mirror image about the imaginary axis of the Nyquist plot for the negative-feedback case.This may be seen from Figure 7–131, which was obtained by use of MATLAB Program 7–21. (Note that the positive-feedback case is unsta-ble, but the negative-feedback case is stable.) MATLAB Program 7–21 num1 = [1 4 6]; den1 = [1 5 4]; num2 = [-1 -4 -6]; den2 = [1 5 4]; nyquist(num1,den1); hold on nyquist(num2,den2); v = [-2 2 -1 1]; axis(v); grid title('Nyquist Plots of G(s) and -G(s)') text(1.0,0.5,'G(s)') text(0.57,-0.48,'Use this Nyquist') text(0.57,-0.61,'plot for negative') text(0.57,-0.73,'feedback system') text(-1.3,0.5,'-G(s)') text(-1.7,-0.48,'Use this Nyquist') text(-1.7,-0.61,'plot for positive') text(-1.7,-0.73,'feedback system') Real Axis –1.4 –1.5 –0.9 –0.7 –1 –0.8 –1.2 –1.3 –1.1 Imag Axis –0.2 0.1 –0.5 0.5 –0.1 0.2 –0.3 –0.4 0 0.3 0.4 Nyquist Plot of G(s) = –(s2+4s+6)/(s2+5s+4) Figure 7–130 Nyquist plot for positive-feedback system.
Example Problems and Solutions 537 Real Axis –1.5 –2 1 2 0.5 1.5 –0.5 –1 0 Imag Axis –0.4 0.2 –1 1 –0.2 0.4 –0.6 –0.8 0 0.6 0.8 Nyquist Plots of G(s) and –G(s) –G(s) G(s) Use this Nyquist plot for positive feedback system Use this Nyquist plot for negative feedback system Figure 7–131 Nyquist plots for positive-feedback system and negative-feedback system.
A–7–15.
Consider the control system shown in Figure 7–60. (Refer to Example 7–19.) Using the inverse polar plot, determine the range of gain K for stability.
Solution.
Since we have Hence, the inverse of the feedforward transfer function is Notice that 1/G(s) has a pole at s=–0.5. It does not have any pole in the right-half s plane.
Therefore, the Nyquist stability equation reduces to Z=N since P=0. The reduced equation states that the number Z of the zeros of 1+C1/G(s)D in the right-half s plane is equal to N, the number of clockwise encirclements of the –1+j0 point. For stability, N must be equal to zero, or there should be no encirclement. Fig-ure 7–132 shows the Nyquist plot or polar plot of K/G(jv).
Notice that since = 0.5 - 0.5v2 - v4 + jvA-1 + 0.5v2B 0.25 + v2 K G(jv) = c (jv)3 + (jv)2 + 1 jv + 0.5 d a 0.5 - jv 0.5 - jv b Z = N + P 1 G(s) = s3 + s2 + 1 K(s + 0.5) G(s) = G1(s)G2(s) = K(s + 0.5) s3 + s2 + 1 G2(s) = 1 s3 + s2 + 1 538 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Figure 7–132 Polar plot of K/G(jv).
the K/G(jv) locus crosses the negative real axis at and the crossing point at the nega-tive real axis is –2.
From Figure 7–132, we see that if the critical point lies in the region between –2 and –q, then the critical point is not encircled. Hence, for stability, we require Thus, the range of gain K for stability is 2<K which is the same result as we obtained in Example 7–19.
A–7–16.
Figure 7–133 shows a block diagram of a space-vehicle control system. Determine the gain K such that the phase margin is 50°.What is the gain margin in this case?
Solution. Since we have The requirement that the phase margin be 50° means that must be equal to –130°, where vc is the gain crossover frequency, or /GAjvcB = -130° /GAjvcB /G(jv) = /jv + 2 - 2/jv = tan-1 v 2 - 180° G(jv) = K(jv + 2) (jv)2 -1 6 -2 K v = 12 , Im Re K G Plane K G Locus – –2 0 2 v = 0 v = 2 v v Example Problems and Solutions 539 Hence, we set from which we obtain Since the phase curve never crosses the –180° line, the gain margin is ±q dB. Noting that the magnitude of G(jv) must be equal to 0 dB at v=2.3835, we have from which we get This K value will give the phase margin of 50°.
A–7–17.
For the standard second-order system show that the bandwidth vb is given by Note that vb/vn is a function only of z. Plot a curve of vb/vn versus z.
Solution. The bandwidth vb is determined from @CAjvbB/RAjvbB @=–3 dB. Quite often, instead of –3 dB, we use –3.01 dB, which is equal to 0.707.Thus, Then from which we get v4 n = 0.5CAv2 n - v2 bB 2 + 4z2v2 n v2 bD v2 n 3Av2 n - v2 bB 2 + A2zvn vbB 2 = 0.707 2 CAjvbB RAjvbB 2 = 2 v2 n AjvbB 2 + 2zvnAjvbB + v2 n 2 = 0.707 vb = vnA1 - 2z2 + 24z4 - 4z2 + 2B 12 C(s) R(s) = v2 n s2 + 2zvn s + v2 n K = 2.38352 222 + 2.38352 = 1.8259 2 K(jv + 2) (jv)2 2 v=2.3835 = 1 vc = 2.3835 radsec tan-1 vc 2 = 50° G(s) K(s + 2) 1 s2 + – Figure 7–133 Space-vehicle control system.
540 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1.0 z vb vn Figure 7–134 Curve of vb/vn versus z, where vb is the bandwidth.
By dividing both sides of this last equation by v4 n, we obtain Solving this last equation for Avb/vnB 2 yields Since Avb/vnB 2>0, we take the plus sign in this last equation.Then or Figure 7–134 shows a curve relating vb/vn versus z.
A–7–18.
A Bode diagram of the open-loop transfer function G(s) of a unity-feedback control system is shown in Figure 7–135. It is known that the open-loop transfer function is minimum phase. From the diagram, it can be seen that there is a pair of complex-conjugate poles at v=2 radsec.
Determine the damping ratio of the quadratic term involving these complex-conjugate poles.
Also, determine the transfer function G(s).
Solution. Referring to Figure 7–9 and examining the Bode diagram of Figure 7–135, we find the damping ratio z and undamped natural frequency vn of the quadratic term to be z = 0.1, vn = 2 radsec vb = vnA1 - 2z2 + 24z4 - 4z2 + 2B 12 v2 b = v2 nA1 - 2z2 + 24z4 - 4z2 + 2B a vb vn b 2 = -2z2 + 1 ; 24z4 - 4z2 + 2 1 = 0.5 e c1 - a vb vn b 2 d 2 + 4z2 a vb vn b 2 f Example Problems and Solutions 541 40 20 –20 dB 0 –40 –60 –80 0.1 0.2 0.4 0.6 1 4 2 6 10 20 60 40 100 –270° –180° –90° 0° v in rad/sec Figure 7–135 Bode diagram of the open-loop transfer function of a unity-feedback control system.
Noting that there is another corner frequency at v=0.5 radsec and the slope of the magnitude curve in the low-frequency region is –40 dBdecade, G(jv) can be tentatively determined as follows: Since, from Figure 7–135 we find @G(j0.1)@=40 dB, the gain value K can be determined to be unity.Also, the calculated phase curve, versus v, agrees with the given phase curve. Hence, the transfer function G(s) can be determined to be A–7–19.
A closed-loop control system may include an unstable element within the loop.When the Nyquist stability criterion is to be applied to such a system, the frequency-response curves for the unsta-ble element must be obtained.
How can we obtain experimentally the frequency-response curves for such an unstable ele-ment? Suggest a possible approach to the experimental determination of the frequency response of an unstable linear element.
Solution. One possible approach is to measure the frequency-response characteristics of the un-stable element by using it as a part of a stable system.
G(s) = 4(2s + 1) s2As2 + 0.4s + 4B /G(jv) G(jv) = K a jv 0.5 + 1 b (jv)2c a jv 2 b 2 + 0.1(jv) + 1d 542 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Consider the system shown in Figure 7–136. Suppose that the element G1(s) is unstable.The complete system may be made stable by choosing a suitable linear element G2(s).We apply a si-nusoidal signal at the input.At steady state, all signals in the loop will be sinusoidal.We measure the signals e(t), the input to the unstable element, and x(t), the output of the unstable element.
By changing the frequency [and possibly the amplitude for the convenience of measuring e(t) and x(t)] of the input sinusoid and repeating this process, it is possible to obtain the frequency re-sponse of the unstable linear element.
A–7–20.
Show that the lead network and lag network inserted in cascade in an open loop act as proportional-plus-derivative control (in the region of small v) and proportional-plus-integral control (in the region of large v), respectively.
Solution. In the region of small v, the polar plot of the lead network is approximately the same as that of the proportional-plus-derivative controller.This is shown in Figure 7–137(a).
Similarly, in the region of large v, the polar plot of the lag network approximates the proportional-plus-integral controller, as shown in Figure 7–137(b).
A–7–21.
Consider a lag–lead compensator Gc(s) defined by Show that at frequency v1, where the phase angle of Gc(jv) becomes zero. (This compensator acts as a lag compensator for 0<v<v1 and acts as a lead compensator for v1<v<q.) (Refer to Figure 7–109.) v1 = 1 1T 1 T 2 Gc(s) = K c as + 1 T 1 b as + 1 T 2 b as + b T 1 b as + 1 bT 2 b G1(s) G2(s) r e x c + – Figure 7–136 Control system.
Im Im Re 0 PD controller Lead network a v = 0 v = v = v = 0 (a) (b) PI controller 1 Re 0 1 Lag network 1 b Figure 7–137 (a) Polar plots of a lead network and a proportional-plus-derivative controller; (b) polar plots of a lag network and a proportional-plus-integral controller.
Example Problems and Solutions 543 Solution. The angle of Gc(jv) is given by At we have Since or and also we have Thus, the angle of GcAjv1B becomes 0° at A–7–22.
Consider the control system shown in Figure 7–138. Determine the value of gain K such that the phase margin is 60°.What is the gain margin with this value of gain K?
Solution. The open-loop transfer function is = K(10s + 1) s3 + 1.5s2 + 0.5s G(s) = K s + 0.1 s + 0.5 10 s(s + 1) v = v1 = 11T 1 T 2 .
/GcAjv1B = 0° tan-1 1 b B T 1 T 2 + tan-1bB T 2 T 1 = 90° tan-1 B T 1 T 2 + tan-1 B T 2 T 1 = 90° tan atan-1 B T 1 T 2 + tan-1 B T 2 T 1 b = B T 1 T 2 + B T 2 T 1 1 - B T 1 T 2 B T 2 T 1 = q /GcAjv1B = tan-1 B T 1 T 2 + tan-1 B T 2 T 1 - tan-1 1 b B T 1 T 2 - tan-1bB T 2 T 1 v = v1 = 11T 1 T 2 , = tan-1vT 1 + tan-1vT 2 - tan-1vT 1b - tan-1vT 2 b /Gc(jv) = njv + 1 T 1 + njv + 1 T 2 - njv + b T 1 - njv + 1 bT 2 K s + 0.1 s + 0.5 10 s(s + 1) + – Figure 7–138 Control system.
544 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method MATLAB Program 7–22 num = [10 1]; den = [1 1.5 0.5 0]; bode(num,den) title('Bode Diagram of G(s) = (10s + 1)/[s(s + 0.5)(s + 1)]') Let us plot the Bode diagram of G(s) when K=1. MATLAB Program 7–22 may be used for this purpose. Figure 7–139 shows the Bode diagram produced by this program. From this diagram the required phase margin of 60° occurs at the frequency v=1.15 radsec.The magnitude of G(jv) at this frequency is found to be 14.5 dB.Then gain K must satisfy the following equation: or K = 0.188 20 logK = -14.5 dB Frequency (rad/sec) Bode Diagram of G(s) = (10s + 1)/[s(s + 0.5)(s + 1)] −200 −150 −50 −100 −50 Phase (deg); Magnitude (dB) 0 100 50 10−3 10−2 10−1 100 101 Figure 7–139 Bode diagram of G(s) = 10s + 1 s(s + 0.5)(s + 1) .
Thus, we have determined the value of gain K. Since the angle curve does not cross the –180° line, the gain margin is ±q dB.
To verify the results, let us draw a Nyquist plot of G for the frequency range w = 0.5:0.01:1.15 The end point of the locus (v=1.15 radsec) will be on a unit circle in the Nyquist plane.To check the phase margin, it is convenient to draw the Nyquist plot on a polar diagram, using polar grids.
To draw the Nyquist plot on a polar diagram, first define a complex vector z by z = re + iim = reiu where r and u (theta) are given by r = abs(z) theta = angle(z) The abs means the square root of the sum of the real part squared and imaginary part squared; angle means tan–1 (imaginary part/real part).
Example Problems and Solutions 545 If we use the command polar(theta,r) MATLAB will produce a plot in the polar coordinates. Subsequent use of the grid command draws polar grid lines and grid circles.
MATLAB Program 7–23 produces the Nyquist plot of G(jv), where v is between 0.5 and 1.15 radsec.The resulting plot is shown in Figure 7–140.Notice that point G(j1.15) lies on the unit MATLAB Program 7–23 %Nyquist plot in rectangular coordinates num = [1.88 0.188]; den = [1 1.5 0.5 0]; w = 0.5:0.01:1.15; [re,im,w] = nyquist(num,den,w); %Convert rectangular coordinates into polar coordinates % by defining z, r, theta as follows z = re + iim; r = abs(z); theta = angle(z); %To draw polar plot, enter command 'polar(theta,r)' polar(theta,r) text(-1,3,'Check of Phase Margin') text(0.3,-1.7,'Nyquist plot') text(-2.2,-0.75,'Phase margin') text(-2.2,-1.1,'is 60 degrees') text(1.45,-0.7,'Unit circle') Nyquist plot Phase margin is 60 degrees Unit circle 270 240 210 180 150 120 90 60 30 0 300 330 2 1 0.5 Check of Phase Margin 2.5 1.5 Figure 7–140 Nyquist plot of G(jv) showing that the phase margin is 60°.
circle, and the phase angle of this point is –120°. Hence, the phase margin is 60°. The fact that point G(j1.15) is on the unit circle verifies that at v=1.15 radsec the magnitude is equal to 1 or 0 dB. (Thus, v=1.15 is the gain crossover frequency.) Thus, K=0.188 gives the desired phase margin of 60°.
Note that in writing ‘text’ in the polar diagram we enter the text command as follows: text(x,y,' ') For example, to write ‘Nyquist plot’ starting at point (0.3, –1.7), enter the command text(0.3, –1.7,'Nyquist plot') The text is written horizontally on the screen.
A–7–23.
If the open-loop transfer function G(s) involves lightly damped complex-conjugant poles, then more than one M locus may be tangent to the G(jv) locus.
Consider the unity-feedback system whose open-loop transfer function is (7–32) Draw the Bode diagram for this open-loop transfer function.Draw also the log-magnitude-versus-phase plot,and show that two M loci are tangent to the G(jv) locus.Finally,plot the Bode diagram for the closed-loop transfer function.
Solution. Figure 7–141 shows the Bode diagram of G(jv). Figure 7–142 shows the log-magni-tude-versus-phase plot of G(jv). It is seen that the G(jv) locus is tangent to the M=8-dB locus at v=0.97 radsec, and it is tangent to the M=–4-dB locus at v=2.8 radsec.
G(s) = 9 s(s + 0.5)As2 + 0.6s + 10B 546 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method 40 20 0 dB –20 –40 0 –90 –180 –270 –360 0.1 0.2 0.4 1 2 4 10 v in rad/sec Figure 7–141 Bode diagram of G(s) given by Equation (7–32).
Example Problems and Solutions 547 Figure 7–143 shows the Bode diagram of the closed-loop transfer function. The magnitude curve of the closed-loop frequency response shows two resonant peaks. Note that such a case occurs when the closed-loop transfer function involves the product of two lightly damped second-order terms and the two corresponding resonant frequencies are sufficiently separated from each other.As a matter of fact, the closed-loop transfer function of this system can be written = 9 As2 + 0.487s + 1BAs2 + 0.613s + 9B C(s) R(s) = G(s) 1 + G(s) 30 24 18 12 6 0 –18 –12 –6 –360 –270 –180 –90 M = 0.5 dB M = –2 dB M = 8 dB M = 2 dB M = –4 dB 0.1 0.3 0.5 1 1.5 2 2.5 3 3.5 G G in dB Figure 7–142 Log-magnitude-versus-phase plot of G(s) given by Equation (7–32).
20 0 –20 –40 0° –90° –180° –270° –360° 0.1 0.2 0.4 0.6 1 2 4 6 10 v in rad/sec dB Figure 7–143 Bode diagram of where G(s) is given by Equation (7–32).
G(s)C1 + G(s)D, Clearly, the denominator of the closed-loop transfer function is a product of two lightly damped second-order terms (the damping ratios are 0.243 and 0.102), and the two resonant frequencies are sufficiently separated.
A–7–24.
Consider the system shown in Figure 7–144(a). Design a compensator such that the closed-loop system will satisfy the requirements that the static velocity error constant=20 sec–1, phase margin=50°, and gain margin G 10 dB.
Solution. To satisfy the requirements, we shall try a lead compensator Gc(s) of the form (If the lead compensator does not work, then we need to employ a compensator of different form.) The compensated system is shown in Figure 7–144(b).
Define where K=Kca. The first step in the design is to adjust the gain K to meet the steady-state per-formance specification or to provide the required static velocity error constant. Since the static ve-locity error constant Kv is given as 20 sec–1, we have or K=2 With K=2, the compensated system will satisfy the steady-state requirement.
We shall next plot the Bode diagram of G1(s) = 20 s(s + 1) = 10K = 20 = lim sS0 s10K s(s + 1) = lim sS0s Ts + 1 aTs + 1 G1(s) K v = lim sS0sGc(s)G(s) G1(s) = KG(s) = 10K s(s + 1) = K c s + 1 T s + 1 aT Gc(s) = K c a Ts + 1 aTs + 1 548 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Gc(s) G(s) G(s) 10 s(s + 1) (b) 10 s(s + 1) (a) + – + – Figure 7–144 (a) Control system; (b) compensated system.
Example Problems and Solutions 549 MATLAB Program 7–24 produces the Bode diagram shown in Figure 7–145. From this plot, the phase margin is found to be 14°.The gain margin is ±q dB.
Frequency (rad/sec) Bode Diagram of G1(s) = 20/[s(s + 1)] −200 −100 −150 −50 −100 Phase (deg); Magnitude (dB) 50 −50 0 10−1 100 101 102 Figure 7–145 Bode diagram of G1(s).
MATLAB Program 7–24 num = ; den = [1 1 0]; w = logspace(-1,2,100); bode(num,den,w) title('Bode Diagram of G1(s) = 20/[s(s + 1)]') Since the specification calls for a phase margin of 50°, the additional phase lead necessary to satisfy the phase-margin requirement is 36°.A lead compensator can contribute this amount.
Noting that the addition of a lead compensator modifies the magnitude curve in the Bode di-agram, we realize that the gain crossover frequency will be shifted to the right.We must offset the increased phase lag of G1(jv) due to this increase in the gain crossover frequency.Taking the shift of the gain crossover frequency into consideration, we may assume that fm, the maximum phase lead required, is approximately 41°. (This means that approximately 5° has been added to com-pensate for the shift in the gain crossover frequency.) Since fm=41° corresponds to a=0.2077. Note that a=0.21 corresponds to fm=40.76°. Whether we choose fm=41° or fm=40.76° does not make much difference in the final solution. Hence, let us choose a=0.21.
sin fm = 1 - a 1 + a Once the attenuation factor a has been determined on the basis of the required phase-lead angle, the next step is to determine the corner frequencies v=1/T and v=1/(aT) of the lead compensator. Notice that the maximum phase-lead angle fm occurs at the geometric mean of the two corner frequencies, or The amount of the modification in the magnitude curve at due to the inclusion of the term (Ts+1)/(aTs+1) is Note that We need to find the frequency point where, when the lead compensator is added, the total mag-nitude becomes 0 dB.
From Figure 7–145 we see that the frequency point where the magnitude of G1(jv) is –6.7778 dB occurs between v=1 and 10 radsec. Hence, we plot a new Bode diagram of G1(jv) in the frequency range between v=1 and 10 to locate the exact point where G1(jv)=–6.7778 dB. MATLAB Program 7–25 produces the Bode diagram in this frequency range, which is shown in Figure 7–146. From this diagram, we find the frequency point where occurs at v=6.5686 radsec. Let us select this frequency to be the new gain crossover frequency, or vc=6.5686 radsec. Noting that this frequency corresponds to or we obtain and 1 aT = vc 1a = 6.5686 10.21 = 14.3339 1 T = vc1a = 6.568610.21 = 3.0101 vc = 1 1aT 1A 1aTB, @G1(jv)@ = -6.7778 dB 1 1a = 1 10.21 = 6.7778 dB 2 1 + jvT 1 + jvaT 2 v= 1 1aT = 4 1 + j 1 1a 1 + ja 1 1a 4 = 1 1a v = 1A 1aTB v = 1A 1aTB.
550 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method MATLAB Program 7–25 num = ; den = [1 1 0]; w = logspace(0,1,100); bode(num,den,w) title('Bode Diagram of G1(s) = 20/[s(s + 1)]') Example Problems and Solutions 551 The lead compensator thus determined is where Kc is determined as Thus, the transfer function of the compensator becomes MATLAB Program 7–26 produces the Bode diagram of this lead compensator, which is shown in Figure 7–147.
Gc(s) = 9.5238 s + 3.0101 s + 14.3339 = 2 0.3322s + 1 0.06976s + 1 K c = K a = 2 0.21 = 9.5238 Gc(s) = Kc s + 3.0101 s + 14.3339 = Kc a 0.3322s + 1 0.06976s + 1 Frequency (rad/sec) Bode Diagram of G1(s) = 20/[s(s + 1)] −180 −140 −130 −150 −160 −170 −120 −20 −10 0 Phase (deg); Magnitude (dB) 40 30 20 10 100 101 Figure 7–146 Bode diagram of G1(s).
MATLAB Program 7–26 numc = [9.5238 28.6676]; denc = [1 14.3339]; w = logspace(-1,3,100); bode(numc,denc,w) title('Bode Diagram of Gc(s) = 9.5238(s + 3.0101)/(s + 14.3339') 552 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Frequency (rad/sec) Bode Diagram of Gc(s) = 9.5238(s + 3.0101)/(s + 14.3339) 0 40 30 20 10 60 50 5 Phase (deg); Magnitude (dB) 10 20 15 10−1 100 101 102 103 Figure 7–147 Bode diagram of Gc(s).
The open-loop transfer function of the designed system is MATLAB Program 7–27 will produce the Bode diagram of Gc(s)G(s), which is shown in Figure 7–148.
= 95.238s + 286.6759 s3 + 15.3339s2 + 14.3339s Gc(s)G(s) = 9.5238 s + 3.0101 s + 14.3339 10 s(s + 1) MATLAB Program 7–27 num = [95.238 286.6759]; den = [1 15.3339 14.3339 0]; sys = tf(num,den); w = logspace(–1,3,100); bode(sys,w); grid; title('Bode Diagram of Gc(s)G(s)') [Gm,pm,wcp,wcg] = margin(sys); GmdB = 20log10(Gm); [Gmdb,pm,wcp,wcg] ans = Inf 49.4164 Inf 6.5686 Example Problems and Solutions 553 From MATLAB Program 7–27 and Figure 7–148 it is clearly seen that the phase margin is ap-proximately 50° and the gain margin is ± q dB. Since the static velocity error constant Kv is 20 sec–1, all the specifications are met. Before we conclude this problem, we need to check the transient-response characteristics.
Unit-Step Response: We shall compare the unit-step response of the compensated system with that of the original uncompensated system.
The closed-loop transfer function of the original uncompensated system is The closed-loop transfer function of the compensated system is MATLAB Program 7–28 produces the unit-step responses of the uncompensated and compen-sated systems. The resulting response curves are shown in Figure 7–149. Clearly, the compensat-ed system exhibits a satisfactory response. Note that the closed-loop zero and poles are located as follows: Zero at s=–3.0101 Poles at s=–5.2880 ; j5.6824, s=–4.7579 Unit-Ramp Response: It is worthwhile to check the unit-ramp response of the compensated system. Since Kv=20 sec–1, the steady-state error following the unit-ramp input will be C(s) R(s) = 95.238s + 286.6759 s3 + 15.3339s2 + 110.5719s + 286.6759 C(s) R(s) = 10 s2 + s + 10 Frequency (rad/sec) Bode Diagram of Gc(s)G(s) −200 −100 −150 −50 −100 Phase (deg); Magnitude (dB) −50 50 0 10−1 100 101 102 103 Figure 7–148 Bode diagram of Gc(s)G(s).
MATLAB Program 7–29 produces the unit-ramp response curves. [Note that the unit-ramp response is obtained as the unit-step response of C(s)/sR(s).] The resulting curves are shown in Figure 7–150.The compensated system has a steady-state error equal to one-half that of the orig-inal uncompensated system.
554 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method Outputs 1.8 0.8 0 1.2 1.6 0.6 0.2 1 1.4 0.4 t Sec 0 1 6 4 5 2 3 Unit-Step Responses of Uncompensated System and Compensated System Uncompensated system Compensated system Figure 7–149 Unit-step responses of the uncompensated and compensated systems.
MATLAB Program 7–28 %Unit-step responses num1 = ; den1 = [1 1 10]; num2 = [95.238 286.6759]; den2 = [1 15.3339 110.5719 286.6759]; t = 0:0.01:6; [c1,x1,t] = step(num1,den1,t); [c2,x2,t] = step(num2,den2,t); plot(t,c1,'.',t,c2,'-') grid; title('Unit-Step Responses of Uncompensated System and Compensated System') xlabel('t Sec'); ylabel('Outputs') text(1.70,1.45,'Uncompensated System') text(1.1,0.5,'Compensated System') 1/Kv=0.05. The static velocity error constant of the uncompensated system is 10 sec–1. Hence, the original uncompensated system will have twice as large a steady-state error in following the unit-ramp input.
Example Problems and Solutions 555 A–7–25.
Consider a unity-feedback system whose open-loop transfer function is Design a lag–lead compensator Gc(s) such that the static velocity error constant is 10 sec–1, the phase margin is 50°, and the gain margin is 10 dB or more.
G(s) = K s(s + 1)(s + 4) MATLAB Program 7–29 %Unit-ramp responses num1 = ; den1 = [1 1 10 0]; num2 = [95.238 286.6759]; den2 = [1 15.3339 110.5719 286.6759 0]; t = 0:0.01:3; [c1,x1,t] = step(num1,den1,t); [c2,x2,t] = step(num2,den2,t); plot(t,c1,'.',t,c2,'-',t,t,'--'); grid; title('Unit-Ramp Responses of Uncompensated System and Compensated System'); xlabel('t Sec'); ylabel('Outputs') text(1.2,0.65,'Uncompensated System') text(0.1,1.3,'Compensated System') Outputs 3 0 2 2.5 1 0.5 1.5 t Sec 0 0.5 3 2 2.5 1 1.5 Unit-Ramp Responses of Uncompensated System and Compensated System Compensated System Uncompensated System Figure 7–150 Unit-ramp responses of the uncompensated and compensated systems.
Solution. We shall design a lag–lead compensator of the form Then the open-loop transfer function of the compensated system is Gc(s)G(s). Since the gain K of the plant is adjustable, let us assume that Kc=1. Then From the requirement on the static velocity error constant, we obtain Hence, K=40 We shall first plot a Bode diagram of the uncompensated system with K=40. MATLAB Pro-gram 7–30 may be used to plot this Bode diagram.The diagram obtained is shown in Figure 7–151. = K 4 = 10 K v = lim sS0sGc(s)G(s) = lim sS0sGc(s) K s(s + 1)(s + 4) lim sS0Gc(s) = 1.
Gc(s) = K c as + 1 T 1 b as + 1 T 2 b as + b T 1 b as + 1 bT 2 b 556 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method MATLAB Program 7–30 num = ; den = [1 5 4 0]; w = logspace(-1,1,100); bode(num,den,w) title('Bode Diagram of G(s) = 40/[s(s + 1)(s + 4)]') Frequency (rad/sec) Bode Diagram of G(s) = 40/[s(s + 1)(s + 4)] −250 −100 −150 −200 −50 −40 −20 0 Phase (deg); Magnitude (dB) 40 20 10−1 100 101 Figure 7–151 Bode diagram of G(s) = 40Cs(s + 1)(s + 4)D.
Example Problems and Solutions 557 From Figure 7–151, the phase margin of the gain-adjusted but uncompensated system is found to be –16°, which indicates that this system is unstable. The next step in the design of a lag–lead compensator is to choose a new gain crossover frequency. From the phase-angle curve for G(jv), we notice that the phase crossover frequency is v=2 radsec. We may choose the new gain crossover frequency to be 2 radsec so that the phase-lead angle required at v=2 radsec is about 50°. A single lag–lead compensator can provide this amount of phase-lead angle quite easily.
Once we choose the gain crossover frequency to be 2 radsec, we can determine the corner frequencies of the phase-lag portion of the lag–lead compensator. Let us choose the corner frequency (which corresponds to the zero of the phase-lag portion of the compensator) to be 1 decade below the new gain crossover frequency, or at v=0.2 radsec. For another corner frequency we need the value of b. The value of b can be determined from the consideration of the lead portion of the compensator, as shown next.
For the lead compensator, the maximum phase-lead angle fm is given by Notice that b=10 corresponds to fm=54.9°. Since we need a 50° phase margin, we may choose b=10. (Note that we will be using several degrees less than the maximum angle, 54.9°.) Thus, b=10 Then the corner frequency (which corresponds to the pole of the phase-lag portion of the compensator) becomes v=0.02 The transfer function of the phase-lag portion of the lag–lead compensator becomes The phase-lead portion can be determined as follows: Since the new gain crossover frequency is v=2 radsec, from Figure 7–151, @G(j2)@ is found to be 6 dB. Hence, if the lag–lead compen-sator contributes –6 dB at v=2 radsec,then the new gain crossover frequency is as desired.From this requirement, it is possible to draw a straight line of slope 20 dB/decade passing through the point (2 radsec, –6 dB). (Such a line has been manually drawn on Figure 7–151.) The intersec-tions of this line and the 0-dB line and –20-dB line determine the corner frequencies. From this consideration, the corner frequencies for the lead portion can be determined as v=0.4 radsec and v=4 radsec. Thus, the transfer function of the lead portion of the lag–lead compensator becomes Combining the transfer functions of the lag and lead portions of the compensator, we can obtain the transfer function Gc(s) of the lag–lead compensator. Since we chose Kc=1, we have Gc(s) = s + 0.4 s + 4 s + 0.2 s + 0.02 = (2.5s + 1)(5s + 1) (0.25s + 1)(50s + 1) s + 0.4 s + 4 = 1 10 a 2.5s + 1 0.25s + 1 b s + 0.2 s + 0.02 = 10 a 5s + 1 50s + 1 b v = 1AbT 2B sinfm = b - 1 b + 1 v = 1AbT 2B, v = 1T 2 The open-loop transfer function of the compensated system is Using MATLAB Program 7–32 the magnitude and phase-angle curves of the designed open-loop transfer function Gc(s)G(s) can be obtained as shown in Figure 7–153. Note that the denominator polynomial den1 was obtained using the conv command, as follows: = 40s2 + 24s + 3.2 s5 + 9.02s4 + 24.18s3 + 16.48s2 + 0.32s Gc(s)G(s) = (s + 0.4)(s + 0.2) (s + 4)(s + 0.02) 40 s(s + 1)(s + 4) 558 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method MATLAB Program 7–31 numc = [1 0.6 0.08]; denc = [1 4.02 0.08]; bode(numc,denc) title('Bode Diagram of Lag–Lead Compensator') Frequency (rad/sec) Bode Diagram of Lag-Lead Compensator −50 0 50 −20 −15 Phase (deg); Magnitude (dB) −10 0 −5 10−3 10−2 10−1 100 101 102 Figure 7–152 Bode diagram of the designed lag–lead compensator.
a = [1 4.02 0.08]; b = [1 5 4 0]; conv(a,b) ans = 1.0000 9.0200 24.1800 16.4800 0.320000 0 The Bode diagram of the lag–lead compensator Gc(s) can be obtained by entering MATLAB Program 7–31 into the computer.The resulting plot is shown in Figure 7–152.
Example Problems and Solutions 559 Since the phase margin of the compensated system is 50°, the gain margin is 12 dB, and the static velocity error constant is 10 sec–1, all the requirements are met.
We shall next investigate the transient-response characteristics of the designed system.
Unit-Step Response: Noting that we have To determine the denominator polynomial with MATLAB, we may proceed as follows: Define c(s) = 40(s + 0.4)(s + 0.2) = 40s2 + 24s + 3.2 b(s) = s(s + 1)(s + 4) = s3 + 5s2 + 4s a(s) = (s + 4)(s + 0.02) = s2 + 4.02s + 0.08 = 40(s + 0.4)(s + 0.2) (s + 4)(s + 0.02)s(s + 1)(s + 4) + 40(s + 0.4)(s + 0.2) C(s) R(s) = Gc(s)G(s) 1 + Gc(s)G(s) Gc(s)G(s) = 40(s + 0.4)(s + 0.2) (s + 4)(s + 0.02)s(s + 1)(s + 4) MATLAB Program 7–32 num1 = [40 24 3.2]; den1 = [1 9.02 24.18 16.48 0.32 0]; bode(num1,den1) title('Bode Diagram of Gc(s)G(s)') Frequency (rad/sec) Bode Diagram of Gc(s)G(s) −300 −250 −200 −150 −100 −50 0 −100 Phase (deg); Magnitude (dB) −50 0 50 100 10−4 10−3 10−2 10−1 100 101 102 Figure 7–153 Bode diagram of the open-loop transfer function Gc(s)G(s) of the compensated system.
Then we have a = [1 4.02 0.08] b = [1 5 4 0] c = [40 24 3.2] Using the following MATLAB program, we obtain the denominator polynomial.
560 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method a = [1 4.02 0.08]; b = [1 5 4 0]; c = [40 24 3.2]; p = [conv(a,b)] + [0 0 0 c] p = 1.0000 9.0200 24.1800 56.4800 24.3200 3.2000 MATLAB Program 7–33 %Unit-step response num = [40 24 3.2]; den = [1 9.02 24.18 56.48 24.32 3.2]; t = 0:0.2:40; step(num,den,t) grid title('Unit-Step Response of Compensated System') Amplitude 1.2 0.4 0 1 0.2 0.6 0.8 Unit-Step Response of Compensated System Time (sec) 0 10 5 40 30 35 25 15 20 Figure 7–154 Unit-step response curve of the compensated system.
MATLAB Program 7–33 is used to obtain the unit-step response of the compensated system.
The resulting unit-step response curve is shown in Figure 7–154. (Note that the gain-adjusted but uncompensated system is unstable.) Problems 561 Unit-Ramp Response: The unit-ramp response of the compensated system may be obtained by entering MATLAB Program 7–34 into the computer. Here we converted the unit-ramp response of GcG/A1+GcGB into the unit-step response of GcG/CsA1+GcGB D. The unit-ramp response curve obtained using this program is shown in Figure 7–155.
MATLAB Program 7–34 %Unit-ramp response num = [40 24 3.2]; den = [1 9.02 24.18 56.48 24.32 3.2 0]; t = 0:0.05:20; c = step(num,den,t); plot(t,c,'-',t,t,'.') grid title('Unit-Ramp Response of Compensated System') xlabel('Time (sec)') ylabel('Unit-Ramp Input and Output c(t)') Unit-Ramp Input and Output c(t) 20 8 0 12 18 4 2 16 10 14 6 Time (sec) 0 4 2 20 14 18 12 16 8 6 10 Unit-Ramp Response of Compensated System Figure 7–155 Unit-ramp response of the compensated system.
PROBLEMS B–7–1. Consider the unity-feedback system with the open-loop transfer function: G(s) = 10 s + 1 Obtain the steady-state output of the system when it is sub-jected to each of the following inputs: (a) r(t)=sin(t+30°) (b) r(t)=2 cos(2t-45°) (c) r(t)=sin(t+30°)-2 cos(2t-45°) 562 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method B–7–2. Consider the system whose closed-loop transfer function is Obtain the steady-state output of the system when it is sub-jected to the input r(t)=R sin vt.
B–7–3. Using MATLAB, plot Bode diagrams of G1(s) and G2(s) given below.
G1(s) is a minimum-phase system and G2(s) is a nonmini-mum-phase system.
B–7–4. Plot the Bode diagram of B–7–5. Given show that B–7–6. Consider a unity-feedback control system with the following open-loop transfer function: This is a nonminimum-phase system. Two of the three open-loop poles are located in the right-half s plane as follows: Plot the Bode diagram of G(s) with MATLAB.Explain why the phase-angle curve starts from 0° and approaches ±180°. s = 0.2328 - j0.7926 s = 0.2328 + j0.7926 Open-loop poles at s = -1.4656 G(s) = s + 0.5 s3 + s2 + 1 @GAjvnB @ = 1 2z G(s) = v2 n s2 + 2zvn s + v2 n G(s) = 10As2 + 0.4s + 1B sAs2 + 0.8s + 9B G2(s) = 1 - s 1 + 2s G1(s) = 1 + s 1 + 2s C(s) R(s) = KAT 2 s + 1B T 1 s + 1 B–7–7. Sketch the polar plots of the open-loop transfer function for the following two cases: (a) (b) B–7–8. Draw a Nyquist locus for the unity-feedback control system with the open-loop transfer function Using the Nyquist stability criterion, determine the stabili-ty of the closed-loop system.
B–7–9. A system with the open-loop transfer function is inherently unstable.This system can be stabilized by adding derivative control. Sketch the polar plots for the open-loop transfer function with and without derivative control.
B–7–10. Consider the closed-loop system with the following open-loop transfer function: Plot both the direct and inverse polar plots of G(s)H(s) with K=1 and K=10. Apply the Nyquist stability crite-rion to the plots, and determine the stability of the system with these values of K.
B–7–11. Consider the closed-loop system whose open-loop transfer function is Find the maximum value of K for which the system is stable.
B–7–12. Draw a Nyquist plot of the following G(s): B–7–13. Consider a unity-feedback control system with the following open-loop transfer function: Draw a Nyquist plot of G(s) and examine the stability of the system.
G(s) = 1 s3 + 0.2s2 + s + 1 G(s) = 1 sAs2 + 0.8s + 1B G(s)H(s) = Ke-2s s G(s)H(s) = 10K(s + 0.5) s2(s + 2)(s + 10) G(s)H(s) = K s2AT 1 s + 1B G(s) = K(1 - s) s + 1 T 7 T a 7 0, T 7 T b 7 0 T a 7 T 7 0, T b 7 T 7 0 G(s)H(s) = KAT a s + 1BAT b s + 1B s2(Ts + 1) Problems 563 B–7–14. Consider a unity-feedback control system with the following open-loop transfer function: Draw a Nyquist plot of G(s) and examine the stability of the closed-loop system.
B–7–15. Consider the unity-feedback system with the fol-lowing G(s): Suppose that we choose the Nyquist path as shown in Fig-ure 7–156. Draw the corresponding G(jv) locus in the G(s) plane. Using the Nyquist stability criterion, determine the stability of the system.
G(s) = 1 s(s - 1) G(s) = s2 + 2s + 1 s3 + 0.2s2 + s + 1 B–7–16. Consider the closed-loop system shown in Figure 7–157. G(s) has no poles in the right-half s plane.
If the Nyquist plot of G(s) is as shown in Figure 7–158(a), is this system stable?
If the Nyquist plot is as shown in Figure 7–158(b), is this system stable?
jv s e Figure 7–156 Nyquist path.
G(s) + – Figure 7–157 Closed-loop system.
B–7–17. A Nyquist plot of a unity-feedback system with the feedforward transfer function G(s) is shown in Figure 7–159.
If G(s) has one pole in the right-half s plane, is the sys-tem stable?
If G(s) has no pole in the right-half s plane, but has one zero in the right-half s plane, is the system stable?
Figure 7–158 Nyquist plots.
0 –1 Re Im G(jv) Figure 7–159 Nyquist plot.
0 Re Im –1 (a) Re Im 0 –1 (b) B–7–18. Consider the unity-feedback control system with the following open-loop transfer function G(s): Plot Nyquist diagrams of G(s) for K=1, 10, and 100.
B–7–19. Consider a negative-feedback system with the fol-lowing open-loop transfer function: Plot the Nyquist diagram of G(s). If the system were a pos-itive-feedback one with the same open-loop transfer func-tion G(s), what would the Nyquist diagram look like?
B–7–20. Consider the control system shown in Figure 7–160.
Plot Nyquist diagrams of G(s), where for k=0.3, 0.5, and 0.7. = 10 s3 + 6s2 + (5 + 10k)s G(s) = 10 sC(s + 1)(s + 5) + 10kD G(s) = 2 s(s + 1)(s + 2) G(s) = K(s + 2) s(s + 1)(s + 10) 564 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method k 1 s 10 (s + 1) (s + 5) + – + – Figure 7–160 Control system.
B–7–22. Referring to Problem B–7–21, it is desired to plot only for v>0. Write a MATLAB program to produce such a plot.
If it is desired to plot for –q<v<q, what changes must be made in the MATLAB program?
B–7–23. Consider the unity-feedback control system whose open-loop transfer function is Determine the value of a so that the phase margin is 45°.
B–7–24. Consider the system shown in Figure 7–161. Draw a Bode diagram of the open-loop transfer function G(s).
Determine the phase margin and gain margin.
G(s) = as + 1 s2 Y 1(jv)U 1(jv) Y 1(jv)U 1(jv) G(s) 25 s(s + 1) (s + 10) + – Figure 7–161 Control system.
G(s) 20(s + 1) s(s2 + 2s + 10) (s + 5) + – Figure 7–162 Control system.
B–7–21. Consider the system defined by There are four individual Nyquist plots involved in this sys-tem. Draw two Nyquist plots for the input u1 in one dia-gram and two Nyquist plots for the input u2 in another diagram. Write a MATLAB program to obtain these two diagrams. By1 y2 R = B1 0 0 1R Bx1 x2 R + B 0 0 0 0R Bu1 u2 R Bx # 1 x # 2 R = B -1 6.5 -1 0R B x1 x2 R + B 1 1 1 0R B u1 u2 R B–7–25. Consider the system shown in Figure 7–162.
Draw a Bode diagram of the open-loop transfer function G(s). Determine the phase margin and gain margin with MATLAB.
Problems 565 B–7–26. Consider a unity-feedback control system with the open-loop transfer function Determine the value of the gain K such that the phase margin is 50°.What is the gain margin with this gain K?
B–7–27. Consider the system shown in Figure 7–163. Draw a Bode diagram of the open-loop transfer function, and determine the value of the gain K such that the phase margin is 50°. What is the gain margin of this system with this gain K?
G(s) = K sAs2 + s + 4B 10 s(s + 1) K s + 0.1 s + 0.5 + – Figure 7–163 Control system.
B–7–28. Consider a unity-feedback control system whose open-loop transfer function is Determine the value of the gain K such that the resonant peak magnitude in the frequency response is 2 dB, or Mr=2 dB.
B–7–29. A Bode diagram of the open-loop transfer function G(s) of a unity-feedback control system is shown in Figure 7–164. It is known that the open-loop transfer function is minimum phase. From the diagram, it can be seen that there is a pair of complex-conjugate poles at v=2 radsec.
Determine the damping ratio of the quadratic term involv-ing these complex-conjugate poles. Also, determine the transfer function G(s).
G(s) = K sAs2 + s + 0.5B 40 20 –20 dB 0 –40 –60 –80 0.1 0.2 0.4 0.6 1 4 2 6 10 20 60 40 100 –270° –180° –90° 0° v in rad/sec Figure 7–164 Bode diagram of the open-loop transfer function of a unity-feedback control system.
566 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method B–7–30. Draw Bode diagrams of the PI controller given by and the PD controller given by B–7–31. Figure 7–165 shows a block diagram of a space-vehicle attitude-control system. Determine the proportional gain constant Kp and derivative time such that the band-width of the closed-loop system is 0.4 to 0.5 radsec. (Note that the closed-loop bandwidth is close to the gain crossover frequency.) The system must have an adequate phase mar-gin. Plot both the open-loop and closed-loop frequency re-sponse curves on Bode diagrams.
T d Gc(s) = 5(1 + 0.5s) Gc(s) = 5 a1 + 1 2s b Kp(1 + Tds) 1 s2 + – Gc(s) K s(0.1s + 1)(s + 1) + – Figure 7–166 Closed-loop system.
B–7–32. Referring to the closed-loop system shown in Fig-ure 7–166, design a lead compensator Gc(s) such that the phase margin is 45°,gain margin is not less than 8 dB,and the static velocity error constant Kv is 4.0 sec–1. Plot unit-step and unit-ramp response curves of the compensated system with MATLAB.
B–7–33. Consider the system shown in Figure 7–167. It is desired to design a compensator such that the static velocity error constant is 4 sec–1, phase margin is 50°, and gain margin is 8 dB or more. Plot the unit-step and unit-ramp response curves of the compensated system with MATLAB.
Gc(s) 1 Hydraulic servo 1 s Aircraft 2s + 0.1 s2 + 0.1s + 4 Rate gyro C R + – Figure 7–167 Control system.
Gc(s) 1 s(s + 1)(s + 5) + – Figure 7–168 Control system.
B–7–34. Consider the system shown in Figure 7–168. De-sign a lag–lead compensator such that the static velocity error constant Kv is 20 sec–1, phase margin is 60°, and gain margin is not less than 8 dB. Plot the unit-step and unit-ramp response curves of the compensated system with MATLAB.
Figure 7–165 Block diagram of space-vehicle attitude-control system.
8 567 PID Controllers and Modified PID Controllers 8–1 INTRODUCTION In previous chapters, we occasionally discussed the basic PID controllers. For example, we presented electronic, hydraulic, and pneumatic PID controllers. We also designed control systems where PID controllers were involved.
It is interesting to note that more than half of the industrial controllers in use today are PID controllers or modified PID controllers.
Because most PID controllers are adjusted on-site, many different types of tuning rules have been proposed in the literature.Using these tuning rules,delicate and fine tun-ing of PID controllers can be made on-site.Also, automatic tuning methods have been developed and some of the PID controllers may possess on-line automatic tuning capabilities. Modified forms of PID control, such as I-PD control and multi-degrees-of-freedom PID control, are currently in use in industry. Many practical methods for bump-less switching (from manual operation to automatic operation) and gain scheduling are commercially available.
The usefulness of PID controls lies in their general applicability to most control sys-tems. In particular, when the mathematical model of the plant is not known and there-fore analytical design methods cannot be used, PID controls prove to be most useful. In the field of process control systems,it is well known that the basic and modified PID con-trol schemes have proved their usefulness in providing satisfactory control, although in many given situations they may not provide optimal control.
In this chapter we first present the design of a PID controlled system using Ziegler and Nichols tuning rules.We next discuss a design of PID controller with the conventional 568 Chapter 8 / PID Controllers and Modified PID Controllers Plant Kp(1 + + Tds) 1 Tis + – Figure 8–1 PID control of a plant.
frequency-response approach, followed by the computational optimization approach to design PID controllers. Then we introduce modified PID controls such as PI-D control and I-PD control.Then we introduce multi-degrees-of-freedom control systems,which can satisfy conflicting requirements that single-degree-of-freedom control systems cannot.
(For the definition of multi-degrees-of-freedom control systems, see Section 8–6.) In practical cases, there may be one requirement on the response to disturbance input and another requirement on the response to reference input. Often these two re-quirements conflict with each other and cannot be satisfied in the single-degree-of-freedom case. By increasing the degrees of freedom, we are able to satisfy both. In this chapter we present two-degrees-of-freedom control systems in detail.
The computational optimization approach presented in this chapter to design con-trol systems (such as to search optimal sets of parameter values to satisfy given transient response specifications) can be used to design both single-degree-of-freedom control sys-tems and multi-degrees-of-freedom control systems, provided a fairly precice mathe-matical model of the plant is known.
Outline of the Chapter.
Section 8–1 has presented introductory material for the chapter. Section 8–2 deals with a design of a PID controller with Ziegler–Nichols Rules.
Section 8–3 treats a design of a PID controller with the frequency-response approach.
Section 8–4 presents a computational optimization approach to obtain optimal param-eter values of PID controllers. Section 8–5 discusses multi-degrees-of-freedom control systems including modified PID control systems.
8–2 ZIEGLER–NICHOLS RULES FOR TUNING PID CONTROLLERS PID Control of Plants.
Figure 8–1 shows a PID control of a plant. If a mathe-matical model of the plant can be derived, then it is possible to apply various design techniques for determining parameters of the controller that will meet the transient and steady-state specifications of the closed-loop system. However, if the plant is so com-plicated that its mathematical model cannot be easily obtained, then an analytical or computational approach to the design of a PID controller is not possible.Then we must resort to experimental approaches to the tuning of PID controllers.
The process of selecting the controller parameters to meet given performance spec-ifications is known as controller tuning. Ziegler and Nichols suggested rules for tuning PID controllers (meaning to set values and ) based on experimental step responses or based on the value of that results in marginal stability when only pro-portional control action is used. Ziegler–Nichols rules, which are briefly presented in the following, are useful when mathematical models of plants are not known. (These rules can, of course, be applied to the design of systems with known mathematical K p T d T i , K p , Section 8–2 / Ziegler–Nichols Rules for Tuning PID Controllers 569 Plant u(t) c(t) 1 Figure 8–2 Unit-step response of a plant.
Tangent line at inflection point K 0 c(t) t L T Figure 8–3 S-shaped response curve.
models.) Such rules suggest a set of values of and that will give a stable oper-ation of the system. However, the resulting system may exhibit a large maximum over-shoot in the step response, which is unacceptable. In such a case we need series of fine tunings until an acceptable result is obtained. In fact, the Ziegler–Nichols tuning rules give an educated guess for the parameter values and provide a starting point for fine tun-ing, rather than giving the final settings for and in a single shot.
Ziegler–Nichols Rules for Tuning PID Controllers.
Ziegler and Nichols pro-posed rules for determining values of the proportional gain integral time and de-rivative time based on the transient response characteristics of a given plant. Such determination of the parameters of PID controllers or tuning of PID controllers can be made by engineers on-site by experiments on the plant. (Numerous tuning rules for PID controllers have been proposed since the Ziegler–Nichols proposal.They are available in the literature and from the manufacturers of such controllers.) There are two methods called Ziegler–Nichols tuning rules: the first method and the second method.We shall give a brief presentation of these two methods.
First Method.
In the first method, we obtain experimentally the response of the plant to a unit-step input, as shown in Figure 8–2. If the plant involves neither integra-tor(s) nor dominant complex-conjugate poles, then such a unit-step response curve may look S-shaped, as shown in Figure 8–3. This method applies if the response to a step input exhibits an S-shaped curve. Such step-response curves may be generated experi-mentally or from a dynamic simulation of the plant.
The S-shaped curve may be characterized by two constants, delay time L and time constant T. The delay time and time constant are determined by drawing a tangent line at the inflection point of the S-shaped curve and determining the intersections of the tangent line with the time axis and line c(t)=K, as shown in Figure 8–3. The transfer T d T i , K p , T d K p , T i , T d K p , T i , 570 Chapter 8 / PID Controllers and Modified PID Controllers Kp Plant r(t) c(t) u(t) + – Figure 8–4 Closed-loop system with a proportional controller.
function C(s)/U(s) may then be approximated by a first-order system with a transport lag as follows: Ziegler and Nichols suggested to set the values of and according to the formula shown in Table 8–1.
Notice that the PID controller tuned by the first method of Ziegler–Nichols rules gives Thus, the PID controller has a pole at the origin and double zeros at s=–1/L.
Second Method.
In the second method, we first set and Using the proportional control action only (see Figure 8–4), increase Kp from 0 to a critical value Kcr at which the output first exhibits sustained oscillations. (If the output does not ex-hibit sustained oscillations for whatever value Kp may take, then this method does not apply.) Thus, the critical gain Kcr and the corresponding period are experimentally P cr T d = 0.
T i = q = 0.6T as + 1 L b 2 s = 1.2 T L a 1 + 1 2Ls + 0.5Ls b Gc(s) = K p a1 + 1 T i s + T d s b T d T i , K p , C(s) U(s) = Ke-Ls Ts + 1 Type of Controller P q 0 PI 0 PID 2L 0.5L 1.2 T L L 0.3 0.9 T L T L T d T i K p Table 8–1 Ziegler–Nichols Tuning Rule Based on Step Response of Plant (First Method) Section 8–2 / Ziegler–Nichols Rules for Tuning PID Controllers 571 Pcr 0 t c(t) Figure 8–5 Sustained oscillation with period ( is measured in sec.) P cr P cr .
Type of Controller P 0.5Kcr q 0 PI 0.45Kcr 0 PID 0.6Kcr 0.125P cr 0.5P cr 1 1.2 P cr T d T i K p Table 8–2 Ziegler–Nichols Tuning Rule Based on Critical Gain Kcr and Critical Period (Second Method) P cr determined (see Figure 8–5). Ziegler and Nichols suggested that we set the values of the parameters and according to the formula shown in Table 8–2.
T d K p , T i , Notice that the PID controller tuned by the second method of Ziegler–Nichols rules gives Thus, the PID controller has a pole at the origin and double zeros at Note that if the system has a known mathematical model (such as the transfer func-tion), then we can use the root-locus method to find the critical gain Kcr and the fre-quency of the sustained oscillations vcr, where These values can be found from the crossing points of the root-locus branches with the jv axis. (Obviously, if the root-locus branches do not cross the jv axis, this method does not apply.) 2pvcr = P cr .
s = -4P cr . = 0.075K cr P cr as + 4 P cr b 2 s = 0.6K cr a1 + 1 0.5P cr s + 0.125P cr s b Gc(s) = K p a 1 + 1 T i s + T d s b 572 Chapter 8 / PID Controllers and Modified PID Controllers Gc(s) PID controller 1 s(s + 1)(s + 5) C(s) R(s) + – Figure 8–6 PID-controlled system.
Comments.
Ziegler–Nichols tuning rules (and other tuning rules presented in the literature) have been widely used to tune PID controllers in process control systems where the plant dynamics are not precisely known. Over many years, such tuning rules proved to be very useful.Ziegler–Nichols tuning rules can,of course,be applied to plants whose dynamics are known. (If the plant dynamics are known, many analytical and graphical approaches to the design of PID controllers are available, in addition to Ziegler–Nichols tuning rules.) EXAMPLE 8–1 Consider the control system shown in Figure 8–6 in which a PID controller is used to control the system.The PID controller has the transfer function Although many analytical methods are available for the design of a PID controller for the pres-ent system, let us apply a Ziegler–Nichols tuning rule for the determination of the values of pa-rameters and Then obtain a unit-step response curve and check to see if the designed system exhibits approximately 25% maximum overshoot. If the maximum overshoot is excessive (40% or more), make a fine tuning and reduce the amount of the maximum overshoot to ap-proximately 25% or less.
Since the plant has an integrator, we use the second method of Ziegler–Nichols tuning rules.
By setting and we obtain the closed-loop transfer function as follows: The value of Kp that makes the system marginally stable so that sustained oscillation occurs can be obtained by use of Routh’s stability criterion. Since the characteristic equation for the closed-loop system is s3+6s2+5s+Kp=0 the Routh array becomes as follows: s3 s2 s1 s0 1 6 30 - Kp 6 Kp 5 Kp C(s) R(s) = K p s(s + 1)(s + 5) + K p T d = 0, T i = q T d .
T i , K p , Gc(s) = K p a1 + 1 T i s + T d s b Section 8–2 / Ziegler–Nichols Rules for Tuning PID Controllers 573 PID controller 1 s(s + 1)(s + 5) 6.3223 (s + 1.4235)2 s C(s) R(s) + – Figure 8–7 Block diagram of the system with PID controller designed by use of the Ziegler–Nichols tuning rule (second method).
Examining the coefficients of the first column of the Routh table, we find that sustained oscilla-tion will occur if Thus, the critical gain Kcr is Kcr=30 With gain Kp set equal to the characteristic equation becomes s3+6s2+5s+30=0 To find the frequency of the sustained oscillation, we substitute s=jv into this characteristic equation as follows: (jv)3+6(jv)2+5(jv)+30=0 or 6A5-v2B+jvA5-v2B=0 from which we find the frequency of the sustained oscillation to be or Hence, the period of sustained oscillation is Referring to Table 8–2, we determine and as follows: The transfer function of the PID controller is thus The PID controller has a pole at the origin and double zero at s=–1.4235. A block diagram of the control system with the designed PID controller is shown in Figure 8–7. = 6.3223(s + 1.4235)2 s = 18 a1 + 1 1.405s + 0.35124s b Gc(s) = K p a1 + 1 T i s + T d s b T d = 0.125P cr = 0.35124 T i = 0.5P cr = 1.405 K p = 0.6K cr = 18 T d T i , K p , P cr = 2p v = 2p 15 = 2.8099 v = 15.
v2 = 5 K cr(= 30), K p = 30.
574 Chapter 8 / PID Controllers and Modified PID Controllers Next, let us examine the unit-step response of the system. The closed-loop transfer function C(s)/R(s) is given by The unit-step response of this system can be obtained easily with MATLAB. See MATLAB Program 8–1. The resulting unit-step response curve is shown in Figure 8–8. The maximum overshoot in the unit-step response is approximately 62%.The amount of maximum overshoot is excessive. It can be reduced by fine tuning the controller parameters. Such fine tuning can be made on the computer. We find that by keeping and by moving the double zero of the PID controller to s=–0.65—that is, using the PID controller (8–1) the maximum overshoot in the unit-step response can be reduced to approximately 18% (see Figure 8–9). If the proportional gain Kp is increased to 39.42, without changing the location of the double zero (s=–0.65), that is, using the PID controller (8–2) Gc(s) = 39.42 a1 + 1 3.077s + 0.7692s b = 30.322 (s + 0.65)2 s Gc(s) = 18 a1 + 1 3.077s + 0.7692s b = 13.846 (s + 0.65)2 s K p = 18 C(s) R(s) = 6.3223s2 + 18s + 12.811 s4 + 6s3 + 11.3223s2 + 18s + 12.811 MATLAB Program 8–1 % ---------- Unit-step response ----------num = [6.3223 18 12.811]; den = [1 6 11.3223 18 12.811]; step(num,den) grid title('Unit-Step Response') Unit-Step Response Time (sec) 0 2 14 12 8 10 4 6 Amplitude 0 0.8 1.8 1.2 0.6 0.2 1.4 1.6 1 0.4 Figure 8–8 Unit-step response curve of PID-controlled system designed by use of the Ziegler–Nichols tuning rule (second method).
Section 8–2 / Ziegler–Nichols Rules for Tuning PID Controllers 575 Unit-Step Response Amplitude 0 0.6 1.2 0.8 0.4 0.2 1 Time (sec) 0 1 7 6 4 5 2 3 Figure 8–9 Unit-step response of the system shown in Figure 8–6 with PID controller having parameters and T d = 0.7692.
T i = 3.077, K p = 18, Amplitude 1.4 0.8 0.4 0 1 1.2 0.6 0.2 Unit-Step Response Time (sec) 0 0.5 5 4.5 3 3.5 4 1 1.5 2 2.5 Figure 8–10 Unit-step response of the system shown in Figure 8–6 with PID controller having parameters and T d = 0.7692.
T i = 3.077, K p = 39.42, then the speed of response is increased, but the maximum overshoot is also increased to approxi-mately 28%,as shown in Figure 8–10.Since the maximum overshoot in this case is fairly close to 25% and the response is faster than the system with given by Equation (8–1),we may consider as given by Equation (8–2) as acceptable.Then the tuned values of and become It is interesting to observe that these values respectively are approximately twice the values sug-gested by the second method of the Ziegler–Nichols tuning rule.The important thing to note here is that the Ziegler–Nichols tuning rule has provided a starting point for fine tuning.
It is instructive to note that, for the case where the double zero is located at s=–1.4235, in-creasing the value of Kp increases the speed of response, but as far as the percentage maximum overshoot is concerned, varying gain Kp has very little effect.The reason for this may be seen from K p = 39.42, T i = 3.077, T d = 0.7692 T d T i , K p , Gc(s) Gc(s) 576 Chapter 8 / PID Controllers and Modified PID Controllers the root-locus analysis. Figure 8–11 shows the root-locus diagram for the system designed by use of the second method of Ziegler–Nichols tuning rules. Since the dominant branches of root loci are along the lines for a considerable range of K, varying the value of K (from 6 to 30) will not change the damping ratio of the dominant closed-loop poles very much. However, varying the lo-cation of the double zero has a significant effect on the maximum overshoot, because the damping ratio of the dominant closed-loop poles can be changed significantly.This can also be seen from the root-locus analysis.Figure 8–12 shows the root-locus diagram for the system where the PID controller has the double zero at s=–0.65. Notice the change of the root-locus configuration.This change in the configuration makes it possible to change the damping ratio of the dominant closed-loop poles.
In Figure 8–12, notice that, in the case where the system has gain K=30.322, the closed-loop poles at s=–2.35_j4.82 act as dominant poles.Two additional closed-loop poles are very near the double zero at s=–0.65, with the result that these closed-loop poles and the double zero almost can-cel each other.The dominant pair of closed-loop poles indeed determines the nature of the response.
On the other hand, when the system has K=13.846, the closed-loop poles at s=–2.35_j2.62 are not quite dominant because the two other closed-loop poles near the double zero at s=–0.65 have considerable effect on the response.The maximum overshoot in the step response in this case (18%) is much larger than the case where the system is of second order and having only dominant closed-loop poles. (In the latter case the maximum overshoot in the step response would be approximately 6%.) It is possible to make a third, a fourth, and still further trials to obtain a better response. But this will take a lot of computations and time. If more trials are desired, it is desirable to use the computational approach presented in Section 10–3. Problem A–8–12 solves this problem with the computational approach with MATLAB. It finds sets of parameter values that will yield the maximum overshoot of 10% or less and the settling time of 3 sec or less.A solution to the present problem obtained in Problem A–8–12 is that for the PID controller defined by Gc(s) = K (s + a)2 s z = 0.3 1 s(s + 1)(s + 5) jv j3 j2 j1 –j3 –j2 –j1 –3 –2 –1 –4 –5 1 0 s K = 6.32 K = 6.32 K = 6.32 K = 6.32 z = 0.3 z = 0.3 K (s + 1.4235)2 s + – Figure 8–11 Root-locus diagram of system when PID controller has double zero at s=–1.4235.
Section 8–3 / Design of PID Controllers with Frequency-Response Approach 577 1 s(s + 1)(s +5) K (s + 0.65)2 s jv j8 j6 j4 j2 –j6 –j8 –j4 –j2 –6 –4 –2 –8 –10 2 0 s K = 60 K = 30.322 K = 30.322 K = 13.846 K = 13.846 K = 13.846 K = 60 z = 0.358 z = 0.67 + – Figure 8–12 Root-locus diagram of system when PID controller has double zero at s=–0.65.
K=13.846 corresponds to given by Equation (8–1) and K=30.322 corresponds to given by Equation (8–2).
Gc(s) Gc(s) the values of K and a are K=29, a=0.25 with the maximum overshoot equal to 9.52% and settling time equal to 1.78 sec.Another possible solution obtained there is that K=27, a=0.2 with the 5.5% maximum overshoot and 2.89 sec of settling time. See Problem A–8–12 for details.
8–3 DESIGN OF PID CONTROLLERS WITH FREQUENCY-RESPONSE APPROACH In this section we present a design of a PID controller based on the frequency-response approach.
Consider the system shown in Figure 8–13.Using a frequency-response approach,de-sign a PID controller such that the static velocity error constant is 4 sec−1, phase margin is 50° or more, and gain margin is 10 dB or more. Obtain the unit-step and unit-ramp response curves of the PID controlled system with MATLAB.
Let us choose the PID controller to be Gc(s) = K(as + 1)(bs + 1) s Since the static velocity error constant Kv is specified as 4 sec–1, we have Thus Next, we plot a Bode diagram of MATLAB Program 8–2 produces a Bode diagram of G(s).The resulting Bode diagram is shown in Figure 8–14.
G(s) = 4 sAs2 + 1B Gc(s) = 4(as + 1)(bs + 1) s = K = 4 Kv = lim sS0sGc(s) 1 s2 + 1 = lim sS0s K(as + 1)(bs + 1) s 1 s2 + 1 578 Chapter 8 / PID Controllers and Modified PID Controllers Figure 8–13 Control system.
Gc(s) 1 s2 + 1 + – MATLAB Program 8–2 num = ; den = [1 0.00000000001 1 0]; w = logspace(-1,1,200); bode(num,den,w) title('Bode Diagram of 4/[s(s^2+1)]') Frequency (rad/sec) Bode Diagram of 4/[s(s2 + 1)] −300 −100 −50 −150 −200 −250 0 −50 0 Phase (deg); Magnitude (dB) 50 10−1 100 101 Figure 8–14 Bode diagram of 4/CsAs2+1B D.
Section 8–3 / Design of PID Controllers with Frequency-Response Approach 579 MATLAB Program 8–3 num = [20 4]; den = [1 0.00000000001 1 0]; w = logspace(-2,1,101); bode(num,den,w) title('Bode Diagram of G(s) = 4(5s+1)/[s(s^2+1)]') Frequency (rad/sec) Bode Diagram of G(s) = 4(5s + 1)/[s(s2 + 1)] −200 −50 −100 −150 0 −20 0 Phase (deg); Magnitude (dB) 60 20 40 10−2 10−1 100 101 Figure 8–15 Bode diagram of G(s)=4(5s+1)/ CsAs2+1B D.
We need the phase margin of at least 50° and gain margin of 10 dB or more.
From the Bode diagram of Figure 8–14, we notice that the gain crossover frequency is approximately v=1.8 radsec. Let us assume the gain crossover frequency of the compensated system to be somewhere between v=1 and v=10 radsec.
Noting that we choose a=5. Then, (as+1) will contribute up to 90° phase lead in the high-frequency region. MATLAB Program 8–3 produces the Bode diagram of The resulting Bode diagram is shown in Figure 8–15.
4(5s + 1) sAs2 + 1B Gc(s) = 4(as + 1)(bs + 1) s Based on the Bode diagram of Figure 8–15, we choose the value of b. The term (bs+1) needs to give the phase margin of at least 50°. By simple MATLAB trials, we find b=0.25 to give the phase margin of at least 50° and gain margin of ±q dB.There-fore, by choosing b=0.25, we have and the open-loop transfer function of the designed system becomes MATLAB Program 8–4 produces the Bode diagram of the open-loop transfer function.
The resulting Bode diagram is shown in Figure 8–16. From it we see that the static ve-locity error constant is 4 sec–1, the phase margin is 55°, and the gain margin is ±q dB.
= 5s2 + 21s + 4 s3 + s Open-loop transfer function = 4(5s + 1)(0.25s + 1) s 1 s2 + 1 Gc(s) = 4(5s + 1)(0.25s + 1) s 580 Chapter 8 / PID Controllers and Modified PID Controllers MATLAB Program 8–4 num = [5 21 4]; den = [1 0 1 0]; w = logspace(-2,2,100); bode(num,den,w) title('Bode Diagram of 4(5s+1)(0.25s+1)/[s(s^2+1)]') Frequency (rad/sec) Bode Diagram of 4(5s + 1)(0.25s + 1)/[s(s2 + 1)] −200 −100 −50 0 50 −150 100 −50 Phase (deg); Magnitude (dB) 0 100 50 10–2 10–1 100 101 102 Figure 8–16 Bode diagram of 4(5s+1)(0.25s+1)/ CsAs2+1B D.
Section 8–3 / Design of PID Controllers with Frequency-Response Approach 581 Therefore, the designed system satisfies all the requirements.Thus, the designed system is acceptable. (Note that there exist infinitely many systems that satisfy all the require-ments.The present system is just one of them.) Next, we shall obtain the unit-step response and the unit-ramp response of the de-signed system.The closed-loop transfer function is Note that the closed-loop zeros are located at The closed-loop poles are located at Notice that the complex-conjugate closed-loop poles have the damping ratio of 0.5237.
MATLAB Program 8–5 produces the unit-step response and the unit-ramp response.
s = -0.1897 s = -2.4052 - j3.9119 s = -2.4052 + j3.9119 s = -4, s = -0.2 C(s) R(s) = 5s2 + 21s + 4 s3 + 5s2 + 22s + 4 MATLAB Program 8–5 % Unit-step response num = [5 21 4]; den = [1 5 22 4]; t = 0:0.01:14; c = step(num,den,t); plot(t,c) grid title('Unit-Step Response of Compensated System') xlabel('t (sec)') ylabel('Output c(t)') % Unit-ramp response num1 = [5 21 4]; den1 = [1 5 22 4 0]; t = 0:0.02:20; c = step(num1,den1,t); plot(t,c,'-',t,t,'--') title('Unit-Ramp Response of Compensated System') xlabel('t (sec)') ylabel('Unit-Ramp Input and Output c(t)') text(10.8,8,'Compensated System') 582 Chapter 8 / PID Controllers and Modified PID Controllers Output c(t) t (sec) Unit-Step Response of Compensated System 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 2 4 6 8 10 12 14 Figure 8–17 Unit-step response curve.
Unit-Ramp Input and Output c(t) t (sec) Unit-Ramp Response of Compensated System 20 18 16 14 12 10 8 6 4 2 0 0 2 4 6 8 10 12 14 16 18 20 Compensated System Figure 8–18 Unit-ramp input and the output curve.
The resulting unit-step response curve is shown in Figure 8–17 and the unit-ramp response curve in Figure 8–18. Notice that the closed-loop pole at s = 0.1897 and the zero at s = 0.2 produce a long tail of small amplitude in the unit-step response.
For an additional example of design of a PID controller based on the frequency-response approach, see Problem A–8–7.
Section 8–4 / Design of PID Controllers with Computational Optimization Approach 583 8–4 DESIGN OF PID CONTROLLERS WITH COMPUTATIONAL OPTIMIZATION APPROACH In this section we shall explore how to obtain an optimal set (or optimal sets) of parameter values of PID controllers to satisfy the transient response specifications by use of MATLAB.We shall present two examples to illustrate the approach in this section.
EXAMPLE 8–2 Consider the PID-controlled system shown in Figure 8–19.The PID controller is given by It is desired to find a combination of K and a such that the closed-loop system will have 10% (or less) maximum overshoot in the unit-step response. (We will not include any other condition in this problem.But other conditions can easily be included,such as that the settling time be less than a specified value. See, for example, Example 8–3.) There may be more than one set of parameters that satisfy the specifications. In this example, we shall obtain all sets of parameters that satisfy the given specifications.
To solve this problem with MATLAB, we first specify the region to search for appropriate K and a.We then write a MATLAB program that, in the unit-step response, will find a combination of K and a which will satisfy the criterion that the maximum overshoot is 10% or less.
Note that the gain K should not be too large, so as to avoid the possibility that the system re-quire an unnecessarily large power unit.
Assume that the region to search for K and a is 2 K 3 and 0.5 a 1.5 If a solution does not exist in this region, then we need to expand it. In some problems, however, there is no solution, no matter what the search region might be.
In the computational approach, we need to determine the step size for each of K and a. In the actual design process,we need to choose step sizes small enough.However,in this example,to avoid an overly large number of computations, we choose the step sizes to be reasonable—say, 0.2 for both K and a.
To solve this problem it is possible to write many different MATLAB programs.We present here one such program,MATLAB Program 8–6.In this program,notice that we use two “for”loops.We start the program with the outer loop to vary the “K” values. Then we vary the “a” values in the inner loop. We proceed by writing the MATLAB program such that the nested loops in the pro-gram begin with the lowest values of “K” and “a” and step toward the highest. Note that, depend-ing on the system and the ranges of search for “K” and “a” and the step sizes chosen, it may take from several seconds to a few minutes for MATLAB to compute the desired sets of the values.
In this program the statement solution(k,:) = [K(i) a(j) m] will produce a table of K, a, m values. (In the present system there are 15 sets of K and a that will exhibit m<1.10—that is, the maximum overshoot is less than 10%.) Gc(s) = K (s + a)2 s R(s) K C(s) PID controller 1.2 0.36s3 + 1.86s2 + 2.5s + 1 (s + a)2 s + – Figure 8–19 PID-controlled system.
584 Chapter 8 / PID Controllers and Modified PID Controllers To sort out the solution sets in the order of the magnitude of the maximum overshoot (starting from the smallest value of m and ending at the largest value of m in the table),we use the command sortsolution = sortrows(solution,3) MATLAB Program 8–6 %'K' and 'a' values to test K = [2.0 2.2 2.4 2.6 2.8 3.0]; a = [0.5 0.7 0.9 1.1 1.3 1.5]; % Evaluate closed-loop unit-step response at each 'K' and 'a' combination % that will yield the maximum overshoot less than 10% t = 0:0.01:5; g = tf([1.2],[0.36 1.86 2.5 1]); k = 0; for i = 1:6; for j = 1:6; gc = tf(K(i)[1 2a(j) a(j)^2], [1 0]); % controller G = gcg/(1 + gcg); % closed-loop transfer function y = step(G,t); m = max(y); if m < 1.10 k = k+1; solution(k,:) = [K(i) a(j) m]; end end end solution % Print solution table solution = 2.0000 0.5000 0.9002 2.0000 0.7000 0.9807 2.0000 0.9000 1.0614 2.2000 0.5000 0.9114 2.2000 0.7000 0.9837 2.2000 0.9000 1.0772 2.4000 0.5000 0.9207 2.4000 0.7000 0.9859 2.4000 0.9000 1.0923 2.6000 0.5000 0.9283 2.6000 0.7000 0.9877 2.8000 0.5000 0.9348 2.8000 0.7000 1.0024 3.0000 0.5000 0.9402 3.0000 0.7000 1.0177 sortsolution = sortrows(solution,3) % Print solution table sorted by % column 3 (continues on next page) Section 8–4 / Design of PID Controllers with Computational Optimization Approach 585 sortsolution = 2.0000 0.5000 0.9002 2.2000 0.5000 0.9114 2.4000 0.5000 0.9207 2.6000 0.5000 0.9283 2.8000 0.5000 0.9348 3.0000 0.5000 0.9402 2.0000 0.7000 0.9807 2.2000 0.7000 0.9837 2.4000 0.7000 0.9859 2.6000 0.7000 0.9877 2.8000 0.7000 1.0024 3.0000 0.7000 1.0177 2.0000 0.9000 1.0614 2.2000 0.9000 1.0772 2.4000 0.9000 1.0923 % Plot the response with the largest overshoot that is less than 10% K = sortsolution(k,1) K = 2.4000 a = sortsolution(k,2) a = 0.9000 gc = tf(K[1 2a a^2], [1 0]); G = gcg/(1 + gcg); step(G,t) grid % See Figure 8–20 % If you wish to plot the response with the smallest overshoot that is % greater than 0%, then enter the following values of 'K' and 'a' K = sortsolution(11,1) K = 2.8000 a = sortsolution(11,2) a = 0.7000 gc = tf(K[1 2a a^2], [1 0]); G = gcg/(1 + gcg); step(G,t) grid % See Figure 8–21 586 Chapter 8 / PID Controllers and Modified PID Controllers Amplitude Time (sec) Step Response 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Figure 8–20 Unit-step response of the system with K=2.4 and a=0.9.
(The maximum overshoot is 9.23%.) To plot the unit-step response curve of the last set of the K and a values in the sorted table, we enter the commands K = sortsolution (k,1) a = sortsolution (k,2) and use the step command. (The resulting unit-step response curve is shown in Figure 8–20.) To plot the unit-step response curve with the smallest overshoot that is greater than 0% found in the sorted table, enter the commands K = sortsolution (11,1) a = sortsolution (11,2) and use the step command. (The resulting unit-step response curve is shown in Figure 8–21.) Amplitude Time (sec) Step Response 1.2 1.4 1 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Figure 8–21 Unit-step response of the system with K=2.8 and a=0.7.
(The maximum overshoot is 0.24%.) Section 8–4 / Design of PID Controllers with Computational Optimization Approach 587 To plot the unit-step response curve of the system with any set shown in the sorted table, we spec-ify the K and a values by entering an appropriate sortsolution command.
Note that for a specification that the maximum overshoot be between 10% and 5%, there would be three sets of solutions: K=2.0000, a=0.9000, m=1.0614 K=2.2000, a=0.9000, m=1.0772 K=2.4000, a=0.9000, m=1.0923 Unit-step response curves for these three cases are shown in Figure 8–22. Notice that the sys-tem with a larger gain K has a smaller rise time and larger maximum overshoot.Which one of these three systems is best depends on the system’s objective.
EXAMPLE 8–3 Consider the system shown in Figure 8–23. We want to find all combinations of K and a values such that the closed-loop system has a maximum overshoot of less than 15%, but more than 10%, in the unit-step response. In addition, the settling time should be less than 3 sec. In this problem, assume that the search region is 3 K 5 and 0.1 a 3 Determine the best choice of the parameters K and a.
Amplitude Time (sec) Unit-Step Response Curves 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 K = 2.4, a = 0.9 K = 2.2, a = 0.9 K = 2, a = 0.9 Figure 8–22 Unit-step response curves of system with K=2, a=0.9; K=2.2, a=0.9; and K=2.4, a=0.9.
R(s) C(s) PID controller 4 s3 + 6s2 + 8s + 4 + – Plant (s + a)2 s K Figure 8–23 PID-controlled system with a simplified PID controller.
588 Chapter 8 / PID Controllers and Modified PID Controllers In this problem, we choose the step sizes to be reasonable, — say 0.2 for K and 0.1 for a. MATLAB Program 8–7 gives the solution to this problem. From the sortsolution table, it looks like the first row is a good choice.Figure 8–24 shows the unit step response curve for K = 3.2 and a = 0.9.Since this choice requires a smaller K value than most other choices,we may decide that the first row is the best choice.
MATLAB Program 8–7 t = 0:0.01:8; k = 0; for K = 3:0.2:5; for a = 0.1:0.1:3; num = [4K 8Ka 4Ka^2]; den = [1 6 8+4K 4+8Ka 4Ka^2]; y = step(num,den,t); s = 801;while y(s)>0.98 & y(s)<1.02; s = s – 1;end; ts = (s–1)0.01; % ts = settling time; m = max(y); if m<1.15 & m>1.10; if ts<3.00; k = k+1; solution(k,:) = [K a m ts]; end end end end solution solution = 3.0000 1.0000 1.1469 2.7700 3.2000 0.9000 1.1065 2.8300 3.4000 0.9000 1.1181 2.7000 3.6000 0.9000 1.1291 2.5800 3.8000 0.9000 1.1396 2.4700 4.0000 0.9000 1.1497 2.3800 4.2000 0.8000 1.1107 2.8300 4.4000 0.8000 1.1208 2.5900 4.6000 0.8000 1.1304 2.4300 4.8000 0.8000 1.1396 2.3100 5.0000 0.8000 1.1485 2.2100 sortsolution = sortrows(solution,3) sortsolution = 3.2000 0.9000 1.1065 2.8300 4.2000 0.8000 1.1107 2.8300 3.4000 0.9000 1.1181 2.7000 4.4000 0.8000 1.1208 2.5900 3.6000 0.9000 1.1291 2.5800 4.6000 0.8000 1.1304 2.4300 4.8000 0.8000 1.1396 2.3100 3.8000 0.9000 1.1396 2.4700 (continues on next page) Section 8–4 / Design of PID Controllers with Computational Optimization Approach 589 3.0000 1.0000 1.1469 2.7700 5.0000 0.8000 1.1485 2.2100 4.0000 0.9000 1.1497 2.3800 % Plot the response curve with the smallest overshoot shown in sortsolution table.
K = sortsolution(1,1), a = sortsolution(1,2) K = 3.2000 a = 0.9000 num = [4K 8Ka 4Ka^2]; den = [1 6 8+4K 4+8Ka 4Ka^2]; num num = 12.8000 23.0400 10.3680 den den = 1.0000 6.0000 20.8000 27.0400 10.3680 y = step(num,den,t); plot(t,y) % See Figure 8–24.
grid title('Unit-Step Response') xlabel('t sec') ylabel('Output y(t)') 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 2 4 6 8 t sec Output y(t) Unit-Step Response Figure 8–24 Unit-step response curve of the system with K=3.2 and a=0.9.
590 Chapter 8 / PID Controllers and Modified PID Controllers 8–5 MODIFICATIONS OF PID CONTROL SCHEMES Consider the basic PID control system shown in Figure 8–25(a), where the system is sub-jected to disturbances and noises.Figure 8–25(b) is a modified block diagram of the same system.In the basic PID control system such as the one shown in Figure 8–25(b),if the ref-erence input is a step function, then, because of the presence of the derivative term in the control action,the manipulated variable u(t) will involve an impulse function (delta func-tion). In an actual PID controller, instead of the pure derivative term , we employ where the value of g is somewhere around 0.1.Therefore, when the reference input is a step function,the manipulated variable u(t) will not involve an impulse function,but will involve a sharp pulse function. Such a phenomenon is called set-point kick.
PI-D Control.
To avoid the set-point kick phenomenon, we may wish to operate the derivative action only in the feedback path so that differentiation occurs only on the feedback signal and not on the reference signal.The control scheme arranged in this way is called the PI-D control. Figure 8–26 shows a PI-D-controlled system.
From Figure 8–26, it can be seen that the manipulated signal U(s) is given by U(s) = K p a 1 + 1 T i s bR(s) - K p a1 + 1 T i s + T d s b B(s) T d s 1 + gT d s T d s PID controller Plant Gp(s) 1 Tis 1 Tds Output Y(s) Noise N(s) Reference input R(s) (a) (b) Disturbance D(s) Gp(s) Y(s) N(s) R(s) E(s) B(s) Observed signal B(s) U(s) D(s) Kp + – ++ ++ ++ + + + ++ + – Figure 8–25 (a) PID-controlled system; (b) equivalent block diagram.
Section 8–5 / Modifications of PID Control Schemes 591 Notice that in the absence of the disturbances and noises, the closed-loop transfer function of the basic PID control system [shown in Figure 8–25(b)] and the PI-D control system (shown in Figure 8–26) are given, respectively, by and It is important to point out that in the absence of the reference input and noises, the closed-loop transfer function between the disturbance D(s) and the output Y(s) in either case is the same and is given by I-PD Control.
Consider the case where the reference input is a step function. Both PID control and PI-D control involve a step function in the manipulated signal. Such a step change in the manipulated signal may not be desirable in many occasions. There-fore, it may be advantageous to move the proportional action and derivative action to the feedback path so that these actions affect the feedback signal only.Figure 8–27 shows such a control scheme. It is called the I-PD control.The manipulated signal is given by Notice that the reference input R(s) appears only in the integral control part. Thus, in I-PD control, it is imperative to have the integral control action for proper operation of the control system.
U(s) = K p 1 T i s R(s) - K p a1 + 1 T i s + T d s b B(s) Y(s) D(s) = Gp(s) 1 + K p Gp(s) a 1 + 1 T i s + T d s b Y(s) R(s) = a 1 + 1 T i s b K p Gp(s) 1 + a1 + 1 T i s + T d s b K p Gp(s) Y(s) R(s) = a1 + 1 T i s + T d s b K p Gp(s) 1 + a 1 + 1 T i s + T d s bK p Gp(s) 1 Tis 1 Gp(s) Y(s) N(s) R(s) E(s) B(s) U(s) D(s) Kp Tds B(s) + – + – + ++ ++ Figure 8–26 PI-D-controlled system.
592 Chapter 8 / PID Controllers and Modified PID Controllers 1 Tis Gp(s) Y(s) N(s) R(s) B(s) B(s) U(s) D(s) Kp Tds 1 + – +– ++ ++ Figure 8–27 I-PD-controlled system.
The closed-loop transfer function Y(s)/R(s) in the absence of the disturbance input and noise input is given by It is noted that in the absence of the reference input and noise signals,the closed-loop transfer function between the disturbance input and the output is given by This expression is the same as that for PID control or PI-D control.
Two-Degrees-of-Freedom PID Control.
We have shown that PI-D control is ob-tained by moving the derivative control action to the feedback path, and I-PD control is obtained by moving the proportional control and derivative control actions to the feedback path. Instead of moving the entire derivative control action or proportional control action to the feedback path, it is possible to move only portions of these control actions to the feedback path, retaining the remaining portions in the feedforward path.
In the literature, PI-PD control has been proposed. The characteristics of this control scheme lie between PID control and I-PD control. Similarly, PID-PD control can be considered. In these control schemes, we have a controller in the feedforward path and another controller in the feedback path. Such control schemes lead us to a more gener-al two-degrees-of-freedom control scheme.We shall discuss details of such a two-degrees-of-freedom control scheme in subsequent sections of this chapter.
8–6 TWO-DEGREES-OF-FREEDOM CONTROL Consider the system shown in Figure 8–28, where the system is subjected to the disturbance input D(s) and noise input N(s), in addition to the reference input R(s).
is the transfer function of the controller and is the transfer function of the plant.We assume that is fixed and unalterable.
Gp(s) Gp(s) Gc(s) Y(s) D(s) = Gp(s) 1 + K p Gp(s) a 1 + 1 T i s + T d s b Y(s) R(s) = a 1 T i s b K p Gp(s) 1 + K p Gp(s) a1 + 1 T i s + T d s b Section 8–6 / Two-Degrees-of-Freedom Control 593 For this system, three closed-loop transfer functions Y(s)/R(s)=Gyr, Y(s)/D(s)=Gyd, and Y(s)/N(s)=Gyn may be derived.They are [In deriving Y(s)/R(s), we assumed D(s)=0 and N(s)=0. Similar comments apply to the derivations of Y(s)/D(s) and Y(s)/N(s).] The degrees of freedom of the control system refers to how many of these closed-loop transfer functions are independent. In the present case, we have Among the three closed-loop transfer functions Gyr, Gyn, and Gyd, if one of them is given, the remaining two are fixed.This means that the system shown in Figure 8–28 is a one-degree-of-freedom control system.
Next consider the system shown in Figure 8–29, where is the transfer function of the plant. For this system, closed-loop transfer functions Gyr, Gyn, and Gyd are given, respectively, by Gyn = Y(s) N(s) = - AGc1 + Gc2BGp 1 + AGc1 + Gc2BGp Gyd = Y(s) D(s) = Gp 1 + AGc1 + Gc2BGp Gyr = Y(s) R(s) = Gc1 Gp 1 + AGc1 + Gc2BGp Gp(s) Gyn = Gyd - Gp Gp Gyr = Gp - Gyd Gp Gyn = Y(s) N(s) = - Gc Gp 1 + Gc Gp Gyd = Y(s) D(s) = Gp 1 + Gc Gp Gyr = Y(s) R(s) = Gc Gp 1 + Gc Gp Gp(s) Y(s) N(s) R(s) B(s) U(s) D(s) Gc(s) + – ++ ++ Figure 8–28 One-degree-of-freedom control system.
594 Chapter 8 / PID Controllers and Modified PID Controllers Hence, we have In this case, if Gyd is given, then Gyn is fixed, but Gyr is not fixed, because Gc1 is independent of Gyd. Thus, two closed-loop transfer functions among three closed-loop transfer functions Gyr, Gyd, and Gyn are independent.Hence,this system is a two-degrees-of-freedom control system.
Similarly, the system shown in Figure 8–30 is also a two-degrees-of-freedom control system, because for this system Gyn = Y(s) N(s) = - Gc1 Gp 1 + Gc1 Gp Gyd = Y(s) D(s) = Gp 1 + Gc1 Gp Gyr = Y(s) R(s) = Gc1 Gp 1 + Gc1 Gp + Gc2 Gp 1 + Gc1 Gp Gyn = Gyd - Gp Gp Gyr = Gc1 Gyd Gp(s) Gc1(s) Y(s) N(s) R(s) B(s) U(s) D(s) Gc2(s) B(s) ++ ++ + – + – Figure 8–29 Two-degrees-of-freedom control system.
Gp(s) Gc1(s) Gc2(s) Y(s) N(s) U(s) D(s) R(s) B(s) + – ++ ++ ++ Figure 8–30 Two-degrees-of-freedom control system.
Section 8–7 / Zero-Placement Approach to Improve Response Characteristics 595 Hence, Clearly, if Gyd is given, then Gyn is fixed, but Gyr is not fixed, because Gc2 is independ-ent of Gyd.
It will be seen in Section 8–7 that, in such a two-degrees-of-freedom control system, both the closed-loop characteristics and the feedback characteristics can be adjusted independently to improve the system response performance.
8–7 ZERO-PLACEMENT APPROACH TO IMPROVE RESPONSE CHARACTERISTICS We shall show here that by use of the zero-placement approach presented later in this section, we can achieve the following: The responses to the ramp reference input and acceleration reference input exhibit no steady-state errors.
In high-performance control systems it is always desired that the system output follow the changing input with minimum error. For step, ramp, and acceleration inputs, it is desired that the system output exhibit no steady-state error.
In what follows, we shall demonstrate how to design control systems that will exhibit no steady-state errors in following ramp and acceleration inputs and at the same time force the response to the step disturbance input to approach zero quickly.
Consider the two-degrees-of-freedom control system shown in Figure 8–31. Assume that the plant transfer function is a minimum-phase transfer function and is given by Gp(s) = K A(s) B(s) Gp(s) Gyn = Gyd - Gp Gp Gyr = Gc2 Gyd + Gp - Gyd Gp Gp(s) Gc1(s) Y(s) R(s) D(s) Gc2(s) ++ + – + – Figure 8–31 Two-degrees-of-freedom control system.
596 Chapter 8 / PID Controllers and Modified PID Controllers where A(s)=As+z1B As+z2B p As+zmB B(s)=sNAs+pN±1B As+pN±2B p As+pnB where N may be 0, 1, 2 and n m. Assume also that Gc1 is a PID controller followed by a filter 1/A(s), or and Gc2 is a PID, PI, PD, I, D, or P controller followed by a filter 1/A(s). That is where some of a2, b2, and g2 may be zero.Then it is possible to write as (8–3) where a, b, and g are constants.Then Because of the presence of s in the numerator, the response y(t) to a step disturbance input approaches zero as t approaches infinity, as shown below. Since if the disturbance input is a step function of magnitude d, or and assuming the system is stable, then = 0 = lim sS0 sKA(0)d sB(0) + bK y(q) = lim sS0sc sKA(s) sB(s) + Aas + b + gs2BK d d s D(s) = d s Y(s) = sKA(s) sB(s) + Aas + b + gs2BK D(s) = sKA(s) sB(s) + Aas + b + gs2BK Y(s) D(s) = Gp 1 + AGc1 + Gc2BGp = K A(s) B(s) 1 + as + b + gs2 s K B(s) Gc1 + Gc2 = as + b + gs2 s 1 A(s) Gc1 + Gc2 Gc2(s) = a2 s + b2 + g2 s2 s 1 A(s) Gc1(s) = a1 s + b1 + g1 s2 s 1 A(s) The response y(t) to a step disturbance input will have the general form shown in Figure 8–32.
Note that Y(s)/R(s) and Y(s)/D(s) are given by Notice that the denominators of Y(s)/R(s) and Y(s)/D(s) are the same. Before we choose the poles of Y(s)/R(s), we need to place the zeros of Y(s)/R(s).
Zero Placement.
Consider the system If we choose p(s) as p(s)=a2s2+a1s+a0=a2As+s1B As+s2B that is, choose the zeros s=–s1 and s=–s2 such that, together with a2, the numerator polynomial p(s) is equal to the sum of the last three terms of the denominator polynomial—then the system will exhibit no steady-state errors in response to the step input, ramp input, and acceleration input.
Requirement Placed on System Response Characteristics.
Suppose that it is desired that the maximum overshoot in the response to the unit-step reference input be between arbitrarily selected upper and lower limits—for example, 2%<maximum overshoot<10% where we choose the lower limit to be slightly above zero to avoid having overdamped systems.The smaller the upper limit, the harder it is to determine the coefficient a’s. In some cases, no combination of the a’s may exist to satisfy the specification, so we must allow a higher upper limit for the maximum overshoot.We use MATLAB to search at least one set of the a’s to satisfy the specification.As a practical computational matter, instead of searching for the a’s, we try to obtain acceptable closed-loop poles by search-ing a reasonable region in the left-half s plane for each closed-loop pole. Once we determine all closed-loop poles,then all coefficients an, an–1, p , a1, a0 will be determined.
Y(s) R(s) = p(s) sn+1 + an sn + an-1 sn-1 + p + a2 s2 + a1 s + a0 Y(s) R(s) = Gc1 Gp 1 + AGc1 + Gc2BGp , Y(s) D(s) = Gp 1 + AGc1 + Gc2BGp Section 8–7 / Zero-Placement Approach to Improve Response Characteristics 597 0 t y Figure 8–32 Typical response curve to a step disturbance input.
598 Chapter 8 / PID Controllers and Modified PID Controllers Determination of Gc2.
Now that the coefficients of the transfer function Y(s)/R(s) are all known and Y(s)/R(s) is given by (8–4) we have Since Gc1 is a PID controller and is given by Y(s)/R(s) can be written as Therefore, we choose so that (8–5) The response of this system to the unit-step reference input can be made to exhibit the maximum overshoot between the chosen upper and lower limits, such as 2%<maximum overshoot<10% The response of the system to the ramp reference input or acceleration reference input can be made to exhibit no steady-state error.The characteristic of the system of Equa-tion (8–4) is that it generally exhibits a short settling time. If we wish to further shorten the settling time, then we need to allow a larger maximum overshoot—for example, 2%<maximum overshoot<20% The controller Gc2 can now be determined from Equations (8–3) and (8–5). Since Gc1 + Gc2 = as + b + gs2 s 1 A(s) Gc1 = a1 s + a0 + a2 s2 Ks 1 A(s) Kg1 = a2 , Ka1 = a1 , Kb1 = a0 Y(s) R(s) = KAa1 s + b1 + g1 s2B sn+1 + an sn + an-1 sn-1 + p + a2 s2 + a1 s + a0 Gc1 = a1 s + b1 + g1 s2 s 1 A(s) = Gc1 sKA(s) sn+1 + an sn + an-1 sn-1 + p + a2 s2 + a1 s + a0 = Gc1 sKA(s) sB(s) + Aas + b + gs2BK Y(s) R(s) = Gc1 Y(s) D(s) Y(s) R(s) = a2 s2 + a1 s + a0 sn+1 + an sn + an-1 sn-1 + p + a2 s2 + a1 s + a0 Section 8–7 / Zero-Placement Approach to Improve Response Characteristics 599 we have (8–6) The two controllers Gc1 and Gc2 can be determined from Equations (8–5) and (8–6).
EXAMPLE 8–4 Consider the two-degrees-of-freedom control system shown in Figure 8–33. The plant transfer function is given by Design controllers and such that the maximum overshoot in the response to the unit-step reference input be less than 19%, but more than 2%, and the settling time be less than 1 sec. It is desired that the steady-state errors in following the ramp reference input and acceler-ation reference input be zero.The response to the unit-step disturbance input should have a small amplitude and settle to zero quickly.
To design suitable controllers and first note that To simplify the notation, let us define Gc=Gc1+Gc2 Then = 10 s(s + 1) + 10Gc Y(s) D(s) = Gp 1 + Gp Gc = 10 s(s + 1) 1 + 10 s(s + 1) Gc Y(s) D(s) = Gp 1 + GpAGc1 + Gc2B Gc2(s), Gc1(s) Gc2(s) Gc1(s) Gp(s) = 10 s(s + 1) Gp(s) = AKa - a1Bs + AKb - a0B + AKg - a2Bs2 Ks 1 A(s) Gc2 = c as + b + gs2 s - a1 s + a0 + a2 s2 Ks d 1 A(s) Gp(s) Gc1(s) Y(s) R(s) U(s) D(s) Gc2(s) ++ + – + – Figure 8–33 Two-degrees-of-freedom control system.
600 Chapter 8 / PID Controllers and Modified PID Controllers Second, note that Notice that the characteristic equation for Y(s)/D(s) and the one for Y(s)/R(s) are identical.
We may be tempted to choose a zero of at s=–1 to cancel a pole at s=–1 of the plant However, the canceled pole s=–1 becomes a closed-loop pole of the entire system, as seen below. If we define as a PID controller such that (8–7) then The closed-loop pole at s=–1 is a slow-response pole, and if this closed-loop pole is included in the system, the settling time will not be less than 1 sec.Therefore, we should not choose as given by Equation (8–7).
The design of controllers and consists of two steps.
Design Step 1: We design to satisfy the requirements on the response to the step-disturbance input D(s). In this design stage, we assume that the reference input is zero.
Suppose that we assume that is a PID controller of the form Then the closed-loop transfer function Y(s)/D(s) becomes Note that the presence of “s”in the numerator of Y(s)/D(s) assures that the steady-state response to the step disturbance input is zero.
Let us assume that the desired dominant closed-loop poles are complex conjugates and are given by s=–a_jb = 10s s2(s + 1) + 10K(s + a)(s + b) = 10 s(s + 1) + 10K(s + a)(s + b) s Y(s) D(s) = 10 s(s + 1) + 10Gc Gc(s) = K(s + a)(s + b) s Gc(s) Gc(s) Gc2(s) Gc1(s) Gc(s) = 10s (s + 1)Cs2 + 10K(s + b)D Y(s) D(s) = 10 s(s + 1) + 10K(s + 1)(s + b) s Gc(s) = K(s + 1)(s + b) s Gc(s) Gp(s).
Gc(s) Y(s) R(s) = Gp Gc1 1 + Gp Gc = 10Gc1 s(s + 1) + 10Gc Section 8–7 / Zero-Placement Approach to Improve Response Characteristics 601 and the remaining closed-loop pole is real and is located at s=–c Note that in this problem there are three requirements. The first requirement is that the response to the step disturbance input damp out quickly.The second requirement is that the max-imum overshoot in the response to the unit-step reference input be between 19% and 2% and the settling time be less than 1 sec. The third requirement is that the steady-state errors in the re-sponses to both the ramp and acceleration reference inputs be zero.
A set (or sets) of reasonable values of a, b, and c must be searched using a computational approach.To satisfy the first requirement, we choose the search region for a, b, and c to be 2 a 6, 2 b 6, 6 c 12 This region is shown in Figure 8–34. If the dominant closed-loop poles s=–a_jb are located anywhere in the shaded region,the response to a step disturbance input will damp out quickly.(The first requirement will be met.) Notice that the denominator of Y(s)/D(s) can be written as = s3 + (2a + c)s2 + Aa2 + b2 + 2acBs + Aa2 + b2Bc = (s + a + jb)(s + a - jb)(s + c) = s3 + (1 + 10K)s2 + 10K(a + b)s + 10Kab s2(s + 1) + 10K(s + a)(s + b) 0 j6 j4 j2 –j6 –j4 –j2 –6 –4 –2 –8 –10 –12 2 s Region for a and b Region for c jv Figure 8–34 Search regions for a, b, and c.
602 Chapter 8 / PID Controllers and Modified PID Controllers Since the denominators of Y(s)/D(s) and Y(s)/R(s) are the same, the denominator of Y(s)/D(s) determines also the response characteristics for the reference input.To satisfy the third require-ment, we refer to the zero-placement method and choose the closed-loop transfer function Y(s)/R(s) to be of the following form: in which case the third requirement is automatically satisfied.
Our problem then becomes a search of a set or sets of desired closed-loop poles in terms of a, b, and c in the specified region, such that the system will satisfy the requirement on the re-sponse to the unit-step reference input that the maximum overshoot be between 19% and 2% and the settling time be less than 1 sec. (If an acceptable set cannot be found in the search region, we need to widen the region.) In the computational search, we need to assume a reasonable step size. In this problem, we assume it to be 0.2.
MATLAB Program 8–8 produces a table of sets of acceptable values of a, b, and c. Using this program, we find that the requirement on the response to the unit-step reference input is met by any of the 23 sets shown in the table in MATLAB Program 8–8. Note that the last row in the table corresponds to the last search point.This point does not satisfy the requirement and thus it should simply be ignored. (In the program written, the last search point produces the last row in the table whether or not it satisfies the requirement.) Y(s) R(s) = (2a + c)s2 + Aa2 + b2 + 2acBs + Aa2 + b2Bc s3 + (2a + c)s2 + Aa2 + b2 + 2acBs + Aa2 + b2Bc MATLAB Program 8–8 t = 0:0.01:4; k = 0; for i = 1:21; a(i) = 6.2-i0.2; for j = 1:21; b(j) = 6.2-j0.2; for h = 1:31; c(h) = 12.2-h0.2; num = [0 2a(i)+c(h) a(i)^2+b(j)^2+2a(i)c(h) (a(i)^2+b(j)^2)c(h)]; den = [1 2a(i)+c(h) a(i)^2+b(j)^2+2a(i)c(h) (a(i)^2+b(j)^2)c(h)]; y = step(num,den,t); m = max(y); s = 401; while y(s) > 0.98 & y(s) < 1.02; s = s-1; end; ts = (s-1)0.01; if m < 1.19 & m > 1.02 & ts < 1.0; k = k+1; table(k,:) = [a(i) b(j) c(h) m ts]; end end end end (continues on next page) Section 8–7 / Zero-Placement Approach to Improve Response Characteristics 603 As noted above, 23 sets of variables a, b, and c satisfy the requirement. Unit-step response curves of the system with any of the 23 sets are about the same.The unit-step response curve with a=4.2, b=2, c=12 is shown in Figure 8–35(a). The maximum overshoot is 18.96% and the settling time is 0.85 sec.
Using these values of a, b, and c, the desired closed-loop poles are located at s=–4.2_j2, s=–12 Using these closed-loop poles, the denominator of Y(s)/D(s) becomes or = s3 + 20.4s2 + 122.44s + 259.68 s3 + (1 + 10K)s2 + 10K(a + b)s + 10Kab = (s + 4.2 + j2)(s + 4.2 - j2)(s + 12) s2(s + 1) + 10K(s + a)(s + b) table(k,:) = [a(i) b(j) c(h) m ts] table = 4.2000 2.0000 12.0000 1.1896 0.8500 4.0000 2.0000 12.0000 1.1881 0.8700 4.0000 2.0000 11.8000 1.1890 0.8900 4.0000 2.0000 11.6000 1.1899 0.9000 3.8000 2.2000 12.0000 1.1883 0.9300 3.8000 2.2000 11.8000 1.1894 0.9400 3.8000 2.0000 12.0000 1.1861 0.8900 3.8000 2.0000 11.8000 1.1872 0.9100 3.8000 2.0000 11.6000 1.1882 0.9300 3.8000 2.0000 11.4000 1.1892 0.9400 3.6000 2.4000 12.0000 1.1893 0.9900 3.6000 2.2000 12.0000 1.1867 0.9600 3.6000 2.2000 11.8000 1.1876 0.9800 3.6000 2.2000 11.6000 1.1886 0.9900 3.6000 2.0000 12.0000 1.1842 0.9200 3.6000 2.0000 11.8000 1.1852 0.9400 3.6000 2.0000 11.6000 1.1861 0.9500 3.6000 2.0000 11.4000 1.1872 0.9700 3.6000 2.0000 11.2000 1.1883 0.9800 3.4000 2.0000 12.0000 1.1820 0.9400 3.4000 2.0000 11.8000 1.1831 0.9600 3.4000 2.0000 11.6000 1.1842 0.9800 3.2000 2.0000 12.0000 1.1797 0.9600 2.0000 2.0000 6.0000 1.2163 1.8900 604 Chapter 8 / PID Controllers and Modified PID Controllers By equating the coefficients of equal powers of s on both sides of this last equation, we obtain 1+10K=20.4 10K(a+b)=122.44 10Kab=259.68 Output t (sec) (a) Unit-Step Response (a = 4.2, b = 2, c = 12) 0.6 0.8 1 1.2 1.4 0.4 0.2 0 0 0.5 1 1.5 2 2.5 3 3.5 4 Figure 8–35 (a) Response to unit-step reference input (a=4.2, b=2, c=12); (b) response to unit-step disturbance input (a=4.2, b=2, c=12).
(b) Output t (sec) Response to Unit-Step Disturbance Input 0.02 0.03 0.04 0.05 0.06 0.07 0.01 0 −0.01 0 0.5 1 1.5 2 2.5 3 3.5 4 Section 8–7 / Zero-Placement Approach to Improve Response Characteristics 605 Hence Then can be written as The closed-loop transfer function Y(s)/D(s) becomes Using this expression, the response y(t) to a unit-step disturbance input can be obtained as shown in Figure 8–35(b).
Figure 8–36(a) shows the response of the system to the unit-step reference input when a, b, and c are chosen as a=3.2, b=2, c=12 Figure 8–36(b) shows the response of this system when it is subjected to a unit-step disturbance input. Comparing Figures 8–35(a) and Figure 8–36(a), we find that they are about the same. How-ever, comparing Figures 8–35(b) and 8–36(b), we find the former to be a little bit better than the latter. Comparing the responses of systems with each set in the table, we conclude the first set of values (a=4.2, b=2, c=12) to be one of the best.Therefore, as the solution to this problem, we choose a=4.2, b=2, c=12 Design Step 2: Next, we determine Gc1. Since Y(s)/R(s) can be given by = 10sGc1 s3 + 20.4s2 + 122.44s + 259.68 = 10 s(s + 1) Gc1 1 + 10 s(s + 1) 1.94s2 + 12.244s + 25.968 s Y(s) R(s) = Gp Gc1 1 + Gp Gc = 10s s3 + 20.4s2 + 122.44s + 259.68 = 10 s(s + 1) + 10 1.94s2 + 12.244s + 25.968 s Y(s) D(s) = 10 s(s + 1) + 10Gc = 1.94s2 + 12.244s + 25.968 s = KCs2 + (a + b)s + abD s Gc(s) = K (s + a)(s + b) s Gc(s) K = 1.94, a + b = 122.44 19.4 , ab = 259.68 19.4 606 Chapter 8 / PID Controllers and Modified PID Controllers our problem becomes that of designing to satisfy the requirements on the responses to the step, ramp, and acceleration inputs.
Since the numerator involves “s”, must include an integrator to cancel this “s”.
[Although we want “s” in the numerator of the closed-loop transfer function Y(s)/D(s) to obtain zero steady-state error to the step disturbance input, we do not want to have “s” in the numera-Gc1(s) Gc1(s) (a) Output t (sec) Unit-Step Response (a = 3.2, b = 2, c = 12) 0.6 0.8 1 1.2 1.4 0.4 0.2 0 0 0.5 1 1.5 2 2.5 3 3.5 4 Figure 8–36 (a) Response to unit-step reference input (a=3.2, b=2, c=12); (b) response to unit-step disturbance input (a=3.2, b=2, c=12).
(b) Output t (sec) Response to Unit-Step Disturbance Input 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.01 0 −0.01 0 0.5 1 1.5 2 2.5 3 3.5 4 Section 8–7 / Zero-Placement Approach to Improve Response Characteristics 607 tor of the closed-loop transfer function Y(s)/R(s).] To eliminate the offset in the response to the step reference input and eliminate the steady-state errors in following the ramp reference input and acceleration reference input, the numerator of Y(s)/R(s) must be equal to the last three terms of the denominator, as mentioned earlier.That is, or Thus, is a PID controller. Since is given as we obtain Thus, is a derivative controller. A block diagram of the designed system is shown in Figure 8–37.
The closed-loop transfer function Y(s)/R(s) now becomes Y(s) R(s) = 20.4s2 + 122.44s + 259.68 s3 + 20.4s2 + 122.44s + 259.68 Gc2(s) = -0.1s = a1.94s + 12.244 + 25.968 s b - a2.04s + 12.244 + 25.968 s b Gc2(s) = Gc(s) - Gc1(s) Gc(s) = Gc1(s) + Gc2(s) = 1.94s2 + 12.244s + 25.968 s Gc(s) Gc1(s) Gc1(s) = 2.04s + 12.244 + 25.968 s 10sGc1(s) = 20.4s2 + 122.44s + 259.68 Y(s) R(s) D(s) Gc2(s) Gc1(s) 0.1s + – ++ ++ 10 s(s + 1) 25.968 s 2.04s + 12.244 + Figure 8–37 Block diagram of the designed system.
608 Chapter 8 / PID Controllers and Modified PID Controllers The response to the unit-ramp reference input and that to the unit-acceleration reference input are shown in Figures 8–38(a) and (b), respectively.The steady-state errors in following the ramp input and acceleration input are zero. Thus, all the requirements of the problem are satisfied.
Hence, the designed controllers and are acceptable.
EXAMPLE 8–5 Consider the control system shown in Figure 8–39. This is a two-degrees-of-freedom system.In the design problem considered here, we assume that the noise input N(s) is zero. Assume that the plant transfer function is given by Gp(s) = 5 (s + 1)(s + 5) Gp(s) Gc2(s) Gc1(s) Unit-Ramp Response t (sec) 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Unit-Ramp Input and Output 2 0 0.4 0.2 0.6 0.8 1 1.2 1.4 1.6 1.8 (a) Output Unit-Ramp Input Figure 8–38 (a) Response to unit-ramp reference input; (b) response to unit-acceleration reference input.
Unit-Acceleration Response t (sec) 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Unit-Acceleration Input and Output 2.5 0 0.5 1 1.5 2 (b) Unit-Acceleration Input Output Section 8–7 / Zero-Placement Approach to Improve Response Characteristics 609 Assume also that the controller is of PID type.That is, The controller is of P or PD type. [If involves integral control action, then this will introduce a ramp component in the input signal, which is not desirable.Therefore, should not include the integral control action.] Thus, we assume that where may be zero.
Let us design controllers and such that the responses to the step-disturbance input and the step-reference input are of “desirable characteristics” in the sense that 1. The response to the step-disturbance input will have a small peak and eventually approach zero. (That is, there will be no steady-state error.) 2. The response to the step reference input will exhibit less than 25% overshoot with a settling time less than 2 sec.The steady-state errors to the ramp reference input and acceleration ref-erence input should be zero.
The design of this two-degrees-of-freedom control system may be carried out by following the steps 1 and 2 below.
1. Determine so that the response to the step-disturbance input is of desirable characteristics.
2. Design so that the responses to the reference inputs are of desirable characteristics without changing the response to the step disturbance considered in step 1.
Design of First, note that we assumed the noise input N(s) to be zero.To obtain the re-sponse to the step-disturbance input, we assume that the reference input is zero. Then the block diagram which relates Y(s) and D(s) can be drawn as shown in Figure 8–40. The transfer func-tion Y(s)/D(s) is given by Y(s) D(s) = Gp 1 + Gc1 Gp Gc1(s): Gc2(s) Gc1(s) Gc2(s) Gc1(s) T ˆ d Gc2(s) = K ˆ pA1 + T ˆ d sB Gc2(s) Gc2(s) Gc2(s) Gc1(s) = K p a1 + 1 T i s + T d s b Gc1(s) Gp(s) Gc1(s) Gc2(s) Y(s) N(s) U(s) D(s) R(s) B(s) + – ++ ++ ++ Figure 8–39 Two-degrees-of-freedom control system.
D(s) Y(s) + – Gp(s) Gc1(s) Figure 8–40 Control system.
610 Chapter 8 / PID Controllers and Modified PID Controllers where This controller involves one pole at the origin and two zeros. If we assume that the two zeros are located at the same place (a double zero), then can be written as Then the characteristic equation for the system becomes or s(s+1)(s+5)+5K(s+a)2=0 which can be rewritten as s3+(6+5K)s2+(5+10Ka)s+5Ka2=0 (8–8) If we place the double zero between s=–3 and s=–6, then the root-locus plot of may look like the one shown in Figure 8–41. The speed of response should be fast, but not faster than necessary, because faster response generally implies larger or more expensive components.
Therefore, we may choose the dominant closed-loop poles at s=–3_j2 (Note that this choice is not unique.There are infinitely many possible closed-loop poles that we may choose from.) Since the system is of third order, there are three closed-loop poles. The third one is located on the negative real axis to the left of point s=–5.
Let us substitute s=–3+j2 into Equation (8–8).
(–3+j2)3+(6+5K)(–3+j2)2+(5+10Ka)(–3+j2)+5Ka2=0 Gc1(s)Gp(s) 1 + Gc1(s)Gp(s) = 1 + K(s + a)2 s 5 (s + 1)(s + 5) = 0 Gc1(s) = K (s + a)2 s Gc1(s) Gc1(s) = K p a1 + 1 T i s + T d s b Root-Locus Plots of (s + a)2/(s3 + 6s2 + 5s) with a = 3, a = 4, a = 4.5, and a = 6 Real Axis Imag Axis 2 −4 −6 −8 −14 −12 −10 −8 −6 −4 −2 0 2 4 −2 0 6 8 a = 6 a = 4.5 a = 4 a = 3 Figure 8–41 Root-locus plots of 5K(s+a)2/Cs(s+1) (s+5)D when a=3, a=4, a=4.5, and a=6.
Section 8–7 / Zero-Placement Approach to Improve Response Characteristics 611 which can be simplified to 24+25K-30Ka+5Ka2+j(–16-60K+20Ka)=0 By equating the real part and imaginary part to zero, respectively, we obtain 24+25K-30Ka+5Ka2=0 (8–9) –16-60K+20Ka=0 (8–10) From Equation (8–10), we have (8–11) Substituting Equation (8–11) into Equation (8–9), we get a2=13 or a=3.6056 or –3.6056. Notice that the values of K become K=1.3210 for a=3.6056 K=–0.1211 for a=–3.6056 Since is in the feedforward path, the gain K should be positive. Hence, we choose K=1.3210, a=3.6056 Then can be given by To determine and we proceed as follows: (8–12) Thus, To check the response to a unit-step disturbance input, we obtain the closed-loop transfer function Y(s)/D(s). = 5s s3 + 12.605s2 + 52.63s + 85.8673 = 5s s(s + 1)(s + 5) + 5K(s + a)2 Y(s) D(s) = Gp 1 + Gc1 Gp K p = 9.5260, T i = 0.5547, T d = 0.1387 = 9.5260 a1 + 1 0.5547s + 0.1387s b Gc1(s) = 1.3210As2 + 7.2112s + 13B s T d , K p , T i , = 1.3210s2 + 9.5260s + 17.1735 s = 1.3210 (s + 3.6056)2 s Gc1(s) = K (s + a)2 s Gc1(s) Gc1(s) K = 4 5a - 15 612 Chapter 8 / PID Controllers and Modified PID Controllers The response to the unit-step disturbance input is shown in Figure 8–42. The response curve seems good and acceptable. Note that the closed-loop poles are located at s=–3_j2 and s=–6.6051.
The complex-conjugate closed-loop poles act as dominant closed-loop poles.
Design of We now design to obtain the desired responses to the reference inputs.
The closed-loop transfer function Y(s)/R(s) can be given by Zero placement.
We place two zeros together with the dc gain constant such that the numera-tor is the same as the sum of the last three terms of the denominator.That is, By equating the coefficients of s2 terms and s terms on both sides of this last equation, from which we get Therefore, (8–13) Gc2(s) = 1 + 1.2s K ˆ p = 1, T ˆ d = 1.2 47.63 + 5K ˆ p = 52.63 6.6051 + 5K ˆ p T ˆ d = 12.6051 = 12.6051s2 + 52.63s + 85.8673 A6.6051 + 5K ˆ p T ˆ dBs2 + A47.63 + 5K ˆ pBs + 85.8673 = A6.6051 + 5K ˆ p T ˆ dBs2 + A47.63 + 5K ˆ pBs + 85.8673 s3 + 12.6051s2 + 52.63s + 85.8673 = c 1.321s2 + 9.526s + 17.1735 s + K ˆ pA1 + T ˆ d sB d 5 (s + 1)(s + 5) 1 + 1.321s2 + 9.526s + 17.1735 s 5 (s + 1)(s + 5) Y(s) R(s) = AGc1 + Gc2BGp 1 + Gc1 Gp Gc2(s) Gc2(s): y d(t) t (sec) Unit-Step Response of Y(s)/D(s) 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.02 0.01 0 0 0.5 1 1.5 2 2.5 3 Figure 8–42 Response to unit-step disturbance input.
Section 8–7 / Zero-Placement Approach to Improve Response Characteristics 613 With this controller the closed-loop transfer function Y(s)/R(s) becomes The response to the unit-step reference input becomes as shown in Figure 8–43(a).
Y(s) R(s) = 12.6051s2 + 52.63s + 85.8673 s3 + 12.6051s2 + 52.63s + 85.8673 Gc2(s), (a) yr(t) t (sec) Response to Unit-Step Reference Input 0.6 0.8 1 1.2 1.4 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Figure 8–43 (a) Response to unit-step reference input; (b) response to unit-ramp reference input; (c) response to unit-acceleration reference input.
(b) yr(t) t (sec) Response to Unit-Ramp Reference Input 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Input Output 614 Chapter 8 / PID Controllers and Modified PID Controllers Figure 8–43 (continued) (c) yr(t) t (sec) Response to Unit-Acceleration Reference Input 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Input Output The response exhibits the maximum overshoot of 21% and the settling time is approximately 1.6 sec. Figures 8–43(b) and (c) show the ramp response and acceleration response. The steady-state errors in both responses are zero.The response to the step disturbance was satisfactory.Thus, the designed controllers and given by Equations (8–12) and (8–13), respectively, are satisfactory.
If the response characteristics to the unit-step reference input are not satisfactory, we need to change the location of the dominant closed-loop poles and repeat the design process.The domi-nant closed-loop poles should lie in a certain region in the left-half s plane (such as 2 a 6, 2 b 6, 6 c 12). If the computational search is desired, write a computer program (sim-ilar to MATLAB Program 8–8) and execute the search process.Then a desired set or sets of val-ues of a, b, and c may be found such that the system response to the unit-step reference input satisfies all requirements on maximum overshoot and settling time.
EXAMPLE PROBLEMS AND SOLUTIONS A–8–1.
Describe briefly the dynamic characteristics of the PI controller, PD controller, and PID controller.
Solution. The PI controller is characterized by the transfer function The PI controller is a lag compensator. It possesses a zero at and a pole at s=0. Thus, the characteristic of the PI controller is infinite gain at zero frequency. This improves the steady-state characteristics. However, inclusion of the PI control action in the system increases the s = -1T i Gc(s) = K p a1 + 1 T i s b Gc2(s) Gc1(s) Example Problems and Solutions 615 K E(s) U(s) 1 K0 T1s 1 + T1s 1 1 + T2s + – Figure 8–44 PID controller.
type number of the compensated system by 1, and this causes the compensated system to be less stable or even makes the system unstable.Therefore, the values of must be chosen care-fully to ensure a proper transient response. By properly designing the PI controller, it is possible to make the transient response to a step input exhibit relatively small or no overshoot.The speed of response, however, becomes much slower. This is because the PI controller, being a low-pass filter, attenuates the high-frequency components of the signal.
The PD controller is a simplified version of the lead compensator.The PD controller has the transfer function where The value of Kp is usually determined to satisfy the steady-state requirement. The corner frequency is chosen such that the phase lead occurs in the neighborhood of the gain crossover frequency. Although the phase margin can be increased, the magnitude of the com-pensator continues to increase for the frequency region (Thus, the PD controller is a high-pass filter.) Such a continued increase of the magnitude is undesirable, since it amplifies high-frequency noises that may be present in the system. Lead compensation can provide a sufficient phase lead, while the increase of the magnitude for the high-frequency region is very much smaller than that for PD control. Therefore, lead compensation is preferred over PD control.
Because the transfer function of the PD controller involves one zero, but no pole, it is not possible to electrically realize it by passive RLC elements only. Realization of the PD controller using op amps, resistors, and capacitors is possible, but because the PD controller is a high-pass filter, as mentioned earlier, the differentiation process involved may cause serious noise problems in some cases.There is,however,no problem if the PD controller is realized by use of the hydraulic or pneumatic elements.
The PD control, as in the case of the lead compensator, improves the transient-response characteristics, improves system stability, and increases the system bandwidth, which implies fast rise time.
The PID controller is a combination of the PI and PD controllers.It is a lag–lead compensator.
Note that the PI control action and PD control action occur in different frequency regions. The PI control action occurs at the low-frequency region and PD control action occurs at the high-frequency region.The PID control may be used when the system requires improvements in both transient and steady-state performances.
A–8–2.
Show that the transfer function U(s)/E(s) of the PID controller shown in Figure 8–44 is Assume that the gain K is very large compared with unity, or K 1.
U(s) E(s) = K 0 T 1 + T 2 T 1 c1 + 1 AT 1 + T 2Bs + T 1 T 2 s T 1 + T 2 d 1T d 6 v.
1T d Gc(s) = K pA1 + T d sB Gc(s), K p and T i 616 Chapter 8 / PID Controllers and Modified PID Controllers + – + – Ei(s) E(s) Eo(s) Z1 Z2 C1 C2 R1 R2 R4 R3 R5 Figure 8–45 Modified PID controller.
Solution A–8–3.
Consider the electronic circuit involving two operational amplifiers shown in Figure 8–45. This is a modified PID controller in that the transfer function involves an integrator and a first-order lag term. Obtain the transfer function of this PID controller.
Solution. Since and we have Also, Eo(s) E(s) = - R5 R4 E(s) Ei(s) = - Z2 Z1 = -AR2 C2s + 1BAR1 C1 s + 1B C2 sAR1 + R3 + R1 R3 C1 sB Z2 = R2 + 1 C2 s Z1 = 1 1 R1 + C1 s + R3 = R1 + R3 + R1 R3 C1 s 1 + R1 C1 s = K0 T 1 + T 2 T 1 c1 + 1 AT 1 + T 2Bs + T 1 T 2 s T 1 + T 2 d = K0 a1 + 1 T 1 s + T 2 s + T 2 T 1 b = K0 a1 + 1 T 1 s b A1 + T 2 sB = K0A1 + T 1 sBA1 + T 2 sB T 1 s K K a 1 K0 T 1 s 1 + T 1 s 1 1 + T 2 s b U(s) E(s) = K 1 + K a 1 K0 T 1 s 1 + T 1 s 1 1 + T 2 s b Example Problems and Solutions 617 R(s) C(s) 1 g 1 Tds + – Figure 8–46 Approximate differentiator.
PID controller Plant G(s) C(s) R(s) D(s) K(as + 1) (bs + 1) s 1 s2 + 3.6s + 9 + – ++ Figure 8–47 PID-controlled system.
Consequently, Notice that R1C1 and R2C2 determine the locations of the zeros of the controller, while R1, R3, and C1 affect the location of the pole on the negative real axis. R5/R4 adjusts the gain of the controller.
A–8–4.
In practice, it is impossible to realize the true differentiator. Hence, we always have to approxi-mate the true differentiator by something like One way to realize such an approximate differentiator is to utilize an integrator in the feedback path.
Show that the closed-loop transfer function of the system shown in Figure 8–46 is given by the pre-ceding expression. (In the commercially available differentiator, the value of g may be set as 0.1.) Solution. The closed-loop transfer function of the system shown in Figure 8–46 is Note that such a differentiator with first-order delay reduces the bandwidth of the closed-loop control system and reduces the detrimental effect of noise signals.
A–8–5.
Consider the system shown in Figure 8–47. This is a PID control of a second-order plant G(s). As-sume that disturbances D(s) enter the system as shown in the diagram. It is assumed that the ref-erence input R(s) is normally held constant, and the response characteristics to disturbances are a very important consideration in this system.
C(s) R(s) = 1 g 1 + 1 gT d s = T d s 1 + gT d s T d s 1 + gT d s T d s = R5 R2 R4 R3 as + 1 R1 C1 b as + 1 R2 C2 b s as + R1 + R3 R1 R3 C1 b Eo(s) Ei(s) = Eo(s) E(s) E(s) Ei(s) = R5 R4AR1 + R3BC2 AR1 C1 s + 1BAR2 C2 s + 1B s a R1 R3 R1 + R3 C1 s + 1 b 618 Chapter 8 / PID Controllers and Modified PID Controllers Design a control system such that the response to any step disturbance will be damped out quickly (in 2 to 3 sec in terms of the 2% settling time). Choose the configuration of the closed-loop poles such that there is a pair of dominant closed-loop poles. Then obtain the response to the unit-step disturbance input.Also, obtain the response to the unit-step reference input.
Solution. The PID controller has the transfer function For the disturbance input in the absence of the reference input, the closed-loop transfer function becomes (8–14) The specification requires that the response to the unit-step disturbance be such that the settling time be 2 to 3 sec and the system have a reasonable damping.We may interpret the specification as and vn=4 radsec for the dominant closed-loop poles.We may choose the third pole at s=–10 so that the effect of this real pole on the response is small. Then the desired charac-teristic equation can be written as (s+10)As2+20.54s+42B=(s+10)As2+4s+16B=s3+14s2+56s+160 The characteristic equation for the system given by Equation (8–14) is s3+(3.6+Kab)s2+(9+Ka+Kb)s+K=0 Hence, we require 3.6+Kab=14 9+Ka+Kb=56 K=160 which yields ab=0.065, a+b=0.29375 The PID controller now becomes With this PID controller, the response to the disturbance is given by = s (s + 10)As2 + 4s + 16B D(s) Cd(s) = s s3 + 14s2 + 56s + 160 D(s) = 10.4As2 + 4.5192s + 15.385B s = 160A0.065s2 + 0.29375s + 1B s Gc(s) = KCabs2 + (a + b)s + 1D s z = 0.5 = s s3 + (3.6 + Kab)s2 + (9 + Ka + Kb)s + K Cd(s) D(s) = s sAs2 + 3.6s + 9B + K(as + 1)(bs + 1) Gc(s) = K(as + 1)(bs + 1) s Example Problems and Solutions 619 Clearly, for a unit-step disturbance input, the steady-state output is zero, since The response to a unit-step disturbance input can be obtained easily with MATLAB. MATLAB Program 8–9 produces a response curve as shown in Figure 8–48(a). From the response curve, we see that the settling time is approximately 2.7 sec.The response damps out quickly.Therefore, the system designed here is acceptable.
lim tS q cd(t) = lim sS0sCd(s) = lim sS0 s2 (s + 10)As2 + 4s + 16B 1 s = 0 MATLAB Program 8–9 % Response to unit-step disturbance input numd = [1 0]; dend = [1 14 56 160]; t = 0:0.01:5; [c1,x1,t] = step(numd,dend,t); plot(t,c1) grid title('Response to Unit-Step Disturbance Input') xlabel('t Sec') ylabel('Output to Disturbance Input') % Response to unit-step reference input numr = [10.4 47 160]; denr = [1 14 56 160]; [c2,x2,t] = step(numr,denr,t); plot(t,c2) grid title('Response to Unit-Step Reference Input') xlabel('t Sec') ylabel('Output to Reference Input') For the reference input r(t), the closed-loop transfer function is The response to a unit-step reference input can also be obtained by use of MATLAB Program 8–9.
The resulting response curve is shown in Figure 8–48(b).The response curve shows that the max-imum overshoot is 7.3% and the settling time is 1.2 sec.The system has quite acceptable response characteristics. = 10.4s2 + 47s + 160 s3 + 14s2 + 56s + 160 Cr(s) R(s) = 10.4As2 + 4.5192s + 15.385B s3 + 14s2 + 56s + 160 620 Chapter 8 / PID Controllers and Modified PID Controllers A–8–6.
Consider the system shown in Figure 8–49. It is desired to design a PID controller such that the dominant closed-loop poles are located at For the PID controller, choose a=1 and then determine the values of K and b. Sketch the root-locus diagram for the designed system.
Solution. Since Gc(s)G(s) = K (s + 1)(s + b) s 1 s2 + 1 s = -1 ; j13.
Gc(s) Output to Disturbance Input 14 6 2 –4 8 12 4 10 0 –2 10–3 Response to Unit-Step Disturbance Input t Sec 0 0.5 5 4.5 3 3.5 4 1 1.5 2 2.5 (a) Output to Reference Input 1.2 0.6 0 0.8 1.0 0.4 0.2 Response to Unit-Step Reference Input t Sec 0 0.5 5 4.5 3 3.5 4 1 1.5 2 2.5 (b) Figure 8–48 (a) Response to unit-step disturbance input; (b) response to unit-step reference input.
Example Problems and Solutions 621 Real Axis –5 0 –1 1 –3 –4 –2 Imag Axis –2 0 3 –3 2 –1 1 Root-Locus Plot of Gc(s)G(s) Figure 8–50 Root-locus plot of the compensated system.
R(s) C(s) PID controller Plant Gc(s) G(s) (s + a ) (s + b) s K 1 s2 + 1 + – Figure 8–49 PID-controlled system.
the sum of the angles at one of the desired closed-loop poles, from the zero at s=–1 and poles at s=0, s=j, and s=–j is 90°-143.794°-120°-110.104°=–283.898° Hence the zero at s=–b must contribute 103.898°.This requires that the zero be located at b=0.5714 The gain constant K can be determined from the magnitude condition.
or K=2.3333 Then the compensator can be written as follows: The open-loop transfer function becomes From this equation a root-locus plot for the compensated system can be drawn. Figure 8–50 is a root-locus plot.
Gc(s)G(s) = 2.3333(s + 1)(s + 0.5714) s 1 s2 + 1 Gc(s) = 2.3333 (s + 1)(s + 0.5714) s 2K (s + 1)(s + 0.5714) s 1 s2 + 1 2 s=-1+j13 = 1 s = -1 + j13, 622 Chapter 8 / PID Controllers and Modified PID Controllers The closed-loop transfer function is given by The closed-loop poles are located at and s=–0.3333. A unit-step response curve is shown in Figure 8–51. The closed-loop pole at s=–0.3333 and a zero at s=–0.5714 produce a long tail of small amplitude.
A–8–7.
Consider the system shown in Figure 8–52. Design a compensator such that the static velocity error constant is 4 sec−1, phase margin is 50°, and gain margin is 10 dB or more. Plot unit-step and unit-ramp response curves of the compensated system with MATLAB.Also, draw a Nyquist plot of the compensated system with MATLAB. Using the Nyquist stability criterion, verify that the designed system is stable.
Solution. Since the plant does not have an integrator, it is necessary to have an integrator in the compensator. Let us choose the compensator to be where is to be determined later. Since the static velocity error constant is specified as 4 sec−1, we have Kv = lim sS0 sGc(s)s + 0.1 s2 + 1 = lim sS0 s K s G ˆc(s)s + 0.1 s2 + 1 = 0.1K = 4 G ˆc(s) Gc(s) = K s G ˆc(s), lim sS0 G ˆc(s) = 1 s = -1 ; j13 C(s) R(s) = 2.3333(s + 1)(s + 0.5714) s3 + s + 2.3333(s + 1)(s + 0.5714) Time (sec) 0 8 12 10 4 2 6 Amplitude 0.4 0.8 1.2 0.6 1 0.2 0 Unit-Step Response of Compensated System Figure 8–51 Unit-step response of the compensated system.
Gc(s) s + 0.1 s2 + 1 +− Figure 8–52 Control system.
Example Problems and Solutions 623 Thus, K = 40. Hence Next, we plot a Bode diagram of MATLAB Program 8–10 produces a Bode diagram of G(s) as shown in Figure 8–53.
G(s) = 40(s + 0.1) s(s2 + 1) Gc(s) = 40 s G ˆc(s) We need the phase margin of 50° and gain margin of 10 dB or more. Let us choose to be Then Gc(s) will contribute up to 90° phase lead in the high-frequency region.By simple MATLAB trials, we find that a = 0.1526 gives the phase margin of 50° and gain margin of dB.
+q G ˆc(s) = as + 1 (a 7 0) G ˆc(s) MATLAB Program 8–10 % Bode Diagram num = [40 4]; den = [1 0.000000001 1 0]; bode(num,den) title('Bode Diagram of G(s) = 40(s+0.1)/[s(s^2+1)]') Figure 8–53 Bode diagram of G(s)= 40(s+0.1)/[s(s2+1)].
Frequency (rad/sec) Bode Diagram of G(s) = 40(s + 0.1)/[s(s2 + 1)] −200 −50 −100 −150 0 −100 Phase (deg); Magnitude (dB) 0 300 200 100 10−3 10−2 10−1 100 101 624 Chapter 8 / PID Controllers and Modified PID Controllers MATLAB Program 8–11 % Bode Diagram num = conv([40 4],[0.1526 1]); den = [1 0.000000001 1 0]; sys = tf(num,den); w = logspace(-2,2,100); bode(sys,w) [Gm,pm,wcp,wcg] = margin(sys); GmdB = 20log10(Gm); [GmdB,pm,wcp,wcg] ans = Inf 50.0026 NaN 8.0114 title('Bode Diagram of G(s) = 40(s+0.1)(0.1526s+1)/[s(s^2+1)]') Figure 8–54 Bode diagram of G(s)=40(s+0.1) (0.1526s+1)/ [s(s2+1)].
Frequency (rad/sec) Bode Diagram of G(s) = 40(s + 0.1)(0.1526s + 1)/[s(s2 + 1)] −200 50 −50 0 −100 −150 100 −50 Phase (deg); Magnitude (dB) 0 100 50 10−2 10−1 100 101 102 The designed compensator has the following transfer function: Gc(s) = 40 s G ˆc(s) = 40(0.1526s + 1) s See MATLAB Program 8–11 and the resulting Bode diagram shown in Figure 8–54. From this Bode diagram we see that the static velocity error constant is 4 sec−1, phase margin is 50° and gain margin is dB.Therefore, the designed system satisfies all the requirements.
+q Example Problems and Solutions 625 The open-loop transfer function of the designed system is Open-loop transfer function We shall next check the unit-step response and the unit-ramp response of the designed system.
The closed-loop transfer function is The closed-loop poles are located at s=3.0032+j5.6573 s=3.0032-j5.6573 s=0.0975 MATLAB Program 8–12 will produce the unit-step response curve of the designed system.The re-sulting unit-step response curve is shown in Figure 8–55. Notice that the closed-loop pole at s = −0.0975 and the plant zero at s = −0.1 produce a long tail of small amplitude.
C(s) R(s) = 6.104s2 + 40.6104s + 4 s3 + 6.104s2 + 41.6104s + 4 = 6.104s2 + 40.6104s + 4 s(s2 + 1) = 40(0.1526s + 1) s s + 0.1 s2 + 1 MATLAB Program 8–12 % Unit-Step Response num = [6.104 40.6104 4]; den = [1 6.104 41.6104 4]; t = 0:0.01:10; step(num,den,t) grid Figure 8–55 Unit-step response of C(s)/R(s)=(6.104s2+ 40.6104s+4)/(s3+ 6.104s2+41.6104s+4).
Time (sec) 2 1 7 9 6 8 10 0 4 3 5 Amplitude 1.4 0.8 0 1.2 0.4 0.2 1 0.6 Step Response MATLAB Program 8–13 produces the unit-ramp response curve of the designed system. The resulting response curve is shown in Figure 8–56.
626 Chapter 8 / PID Controllers and Modified PID Controllers MATLAB Program 8–13 % Unit-Ramp Response num = [0 0 6.104 40.6104 4]; den = [1 6.104 41.6104 4 0]; t = 0:0.01:20; c = step(num,den,t); plot(t,c,'-.',t,t,'-') title('Unit-Ramp Response') xlabel('t(sec)') ylabel('Input Ramp Function and Output') text(3,11.5,'Input Ramp Function') text(13.8,11.2,'Output') Figure 8–56 Unit-ramp response of C(s)/R(s)= (6.104s2+40.6104s+ 4)/(s3+6.104s2+ 41.6104s+4).
Output Input Ramp Function t (sec) 4 2 14 18 12 16 20 0 8 6 10 Input Ramp Function and Output 20 8 0 12 18 4 2 16 10 14 6 Unit-Ramp Response Nyquist Plot.
Earlier we found that the three closed-loop poles of the designed system are all in the left-half s plane. Hence the designed system is stable. The purpose of plotting Nyquist diagram here is not to test the stability of the system, but to enhance our understanding of Nyquist stability analysis. For a complicated system, Nyquist plot will look complicated enough that it is not easy to count the number of encirclements of the −1+j0 point.
Example Problems and Solutions 627 Because the designed system involves three open-loop poles on the jw axis, the Nyquist dia-gram will look quite complicated as we will see in what follows: Define the open-loop transfer function of the designed system as G(s).Then Let us choose a modified Nyquist path in the s plane as shown in Figure 8–57(a). The modified path encloses three open-loop poles (s=0, s=j1, s=j1). Now define s1=s+ . Then, the Nyquist path in the s1 plane becomes as shown in Figure 8–57(b). In the s1 plane, the open-loop transfer function has three poles in the right-half s1 plane.
Let us choose Since we have Open-loop transfer function in the s1 plane A MATLAB program to obtain the Nyquist plot is shown in MATLAB Program 8–14. The re-sulting Nyquist plot is shown in Figure 8–58.
= 6.104s1 2 + 40.48832s1 + 3.5945064 s1 3 - 0.03s1 2 + 1.0003s1 - 0.010001 = 6.104(s1 2 - 0.02s1 + 0.0001) + 40.6104(s1 - 0.01) + 4 (s1 - 0.01)(s1 2 - 0.02s1 + 1.0001) G(s) = G(s1 - 0.01) s = s1 - s0, s0 = 0.01.
s0 G(s) = Gc(s)s + 0.1 s2 + 1 = 6.104s2 + 40.6104s + 4 s(s2 + 1) Figure 8–57 (a) Modified Nyquist path in the s plane; (b) Nyquist path in the s1 plane.
s plane s1 plane jv (a) (b) 0 s s0 jv 0 s MATLAB Program 8–14 % Nyquist Plot num = [6.104 40.48832 3.5945064]; den = [1 -0.03 1.0003 -0.010001]; nyquist(num,den) v = [-1500 1500 -2500 2500]; axis(v) 628 Chapter 8 / PID Controllers and Modified PID Controllers Figure 8–58 Nyquist plot.
−1500 −1000 −500 0 500 1000 1500 Real Axis Nyquest Diagram Imaginary Axis 2500 2000 1500 1000 500 0 −500 −1000 −1500 −2000 −2500 Figure 8–59 Redrawn Nyquist plot.
Im Re v = 0+ v = 0− v = −v = + Using the Nyquist plot obtained here, it is not easy to determine the encirclements of the −1+j0 point by the Nyquist locus. Therefore, we need to redraw this Nyquist plot qualitatively to show the details near the −1+j0 point. Such a redrawn Nyquist diagram is shown in Figure 8–59.
From this diagram we find that the −1+j0 point is encircled counterclockwise three times.
Hence, N −3. Since the open-loop transfer function has three poles in the right-half s1 plane, we have P 3.Then, we have Z N+P 0.This means that there are no closed-loop poles in the right-half s1 plane.The system is therefore stable.
A–8–8.
Show that the I-PD-controlled system shown in Figure 8–60(a) is equivalent to the PID-controlled system with input filter shown in Figure 8–60(b).
Example Problems and Solutions 629 Solution. The closed-loop transfer function C(s)/R(s) of the I-PD-controlled system is The closed-loop transfer function C(s)/R(s) of the PID-controlled system with input filter shown in Figure 8–60(b) is The closed-loop transfer functions of both systems are the same.Thus,the two systems are equivalent.
A–8–9.
The basic idea of the I-PD control is to avoid large control signals (which will cause a saturation phenomenon) within the system. By bringing the proportional and derivative control actions to the feedback path, it is possible to choose larger values for than those possible by the PID control scheme.
Compare,qualitatively,the responses of the PID-controlled system and I-PD-controlled system to the disturbance input and to the reference input.
Solution. Consider first the response of the I-PD-controlled system to the disturbance input.
Since, in the I-PD control of a plant, it is possible to select larger values for than those of the PID-controlled case, the I-PD-controlled system will attenuate the effect of disturbance faster than the PID-controlled case.
Next, consider the response of the I-PD-controlled system to a reference input. Since the I-PD-controlled system is equivalent to the PID-controlled system with input filter (refer to Prob-lem A–8–8),the PID-controlled system will have faster responses than the corresponding I-PD-con-trolled system, provided a saturation phenomenon does not occur in the PID-controlled system.
K p and T d K p and T d = K p T i s Gp(s) 1 + K p a1 + 1 T i s + T d s bGp(s) C(s) R(s) = 1 1 + T i s + T i T d s2 K p a1 + 1 T i s + T d s bGp(s) 1 + K p a1 + 1 T i s + T d s bGp(s) C(s) R(s) = K p T i s Gp(s) 1 + K p a1 + 1 T i s + T d s bGp(s) (b) Gp(s) C(s) R(s) Kp(1 + + Tds) 1 Tis 1 1 + Tis + TiTds2 + – (a) Kp Tis Gp(s) C(s) R(s) Kp(1 + Tds) + – + – Figure 8–60 (a) I-PD-controlled system; (b) PID-controlled system with input filter.
630 Chapter 8 / PID Controllers and Modified PID Controllers A–8–10.
In some cases it is desirable to provide an input filter as shown in Figure 8–61(a). Notice that the input filter is outside the loop. Therefore, it does not affect the stability of the closed-loop portion of the system.An advantage of having the input filter is that the zeros of the closed-loop transfer function can be modified (canceled or replaced by other zeros) so that the closed-loop response is acceptable.
Show that the configuration in Figure 8–61(a) can be modified to that shown in Figure 8–61(b), where The compensation structure shown in Figure 8–61(b) is some-times called command compensation.
Solution. For the system of Figure 8–61(a), we have (8–15) For the system of Figure 8–61(b), we have Thus or (8–16) By substituting into Equation (8–16), we obtain = Gf(s) Gc(s)Gp(s) 1 + Gc(s)Gp(s) C(s) R(s) = CGf(s)Gc(s) - Gc(s) + Gc(s)DGp(s) 1 + Gc(s)Gp(s) CGf(s) - 1DGc(s) Gd(s) = C(s) R(s) = CGd(s) + Gc(s)DGp(s) 1 + Gc(s)Gp(s) C(s) = Gp(s)EGd(s)R(s) + Gc(s)CR(s) - C(s)D F C(s) = Gp(s)U(s) E(s) = R(s) - C(s) U(s) = Gd(s)R(s) + Gc(s)E(s) C(s) R(s) = Gf(s) Gc(s)Gp(s) 1 + Gc(s)Gp(s) CGf(s) - 1DGc(s).
Gd(s) = Gf(s) (a) (b) Gc(s) C(s) R(s) Gp(s) Gf(s) Gc(s) C(s) R(s) E(s) Gd(s) Gp(s) U(s) + – + – ++ Figure 8–61 (a) Block diagram of control system with input filter; (b) modified block diagram.
which is the same as Equation (8–15). Hence, we have shown that the systems shown in Figures 8–61(a) and (b) are equivalent.
It is noted that the system shown in Figure 8–61(b) has a feedforward controller In such a case, does not affect the stability of the closed-loop portion of the system.
A–8–11.
A closed-loop system has the characteristic that the closed-loop transfer function is nearly equal to the inverse of the feedback transfer function whenever the open-loop gain is much greater than unity.
The open-loop characteristic may be modified by adding an internal feedback loop with a characteristic equal to the inverse of the desired open-loop characteristic. Suppose that a unity-feedback system has the open-loop transfer function Determine the transfer function H(s) of the element in the internal feedback loop so that the inner loop becomes ineffective at both low and high frequencies.
Solution. Figure 8–62(a) shows the original system. Figure 8–62(b) shows the addition of the in-ternal feedback loop around G(s). Since if the gain around the inner loop is large compared with unity, then is approximately equal to unity,and the transfer function C(s)/E(s) is approximately equal to 1/H(s).
On the other hand, if the gain is much less than unity, the inner loop becomes ineffective and C(s)/E(s) becomes approximately equal to G(s).
To make the inner loop ineffective at both the low- and high-frequency ranges, we require that Since, in this problem, G(jv) = K A1 + jvT 1BA1 + jvT 2B |G(jv)H(jv)| 1, for v 1 and v 1 @G(s)H(s)@ G(s)H(s)C1 + G(s)H(s)D C(s) E(s) = G(s) 1 + G(s)H(s) = 1 H(s) G(s)H(s) 1 + G(s)H(s) G(s) = K AT 1 s + 1BAT 2 s + 1B Gd(s) Gd(s).
Example Problems and Solutions 631 (a) (b) G(s) C R G(s) H(s) C E R GH(s) 1 H(s) C E R = + – + – + – + – + – Figure 8–62 (a) Control system; (b) addition of the internal feedback loop to modify the closed-loop characteristic.
632 Chapter 8 / PID Controllers and Modified PID Controllers the requirement can be satisfied if H(s) is chosen to be H(s)=ks because Thus, with H(s)=ks (velocity feedback), the inner loop becomes ineffective at both the low-and high-frequency regions. It becomes effective only in the intermediate-frequency region.
A–8–12.
Consider the control system shown in Figure 8–63. This is the same system as that considered in Example 8–1.In that example we designed a PID controller ,starting with the second method of the Ziegler–Nichols tuning rule. Here we design a PID controller using the computational approach with MATLAB.We shall determine the values of K and a of the PID controller such that the unit-step response will exhibit the maximum overshoot between 10% and 2% (1.02 maximum output 1.10) and the settling time will be less than 3 sec.The search region is 2 K 50, 0.05 a 2 Let us choose the step size for K to be 1 and that for a to be 0.05.
Write a MATLAB program to find the first set of variables K and a that will satisfy the given specifications.Also, write a MATLAB program to find all possible sets of variables K and a that will satisfy the given specifications. Plot the unit-step response curves of the designed system with the chosen sets of variables K and a.
Solution. The transfer function of the plant is The closed-loop transfer function C(s)/R(s) is given by A possible MATLAB program that will produce the first set of variables K and a that will satisfy the given specifications is given in MATLAB Program 8–15. In this program we C(s) R(s) = Ks2 + 2Kas + Ka2 s4 + 6s3 + (5 + K)s2 + 2Kas + Ka2 Gp(s) = 1 s3 + 6s2 + 5s Gc(s) = K (s + a)2 s Gc(s) lim vS q G(jv)H(jv) = lim vS q Kkjv A1 + jvT 1BA1 + jvT 2B = 0 lim vS0G(jv)H(jv) = lim vS0 Kkjv A1 + jvT 1BA1 + jvT 2B = 0 R(s) C(s) PID controller 1 s(s + 1) (s + 5) + – Gc(s) Figure 8–63 Control system.
Example Problems and Solutions 633 MATLAB Program 8–15 t = 0:0.01:5; for K = 50:-1:2; for a = 2:-0.05:0.05; num = [K 2Ka Ka^2]; den = [1 6 5+K 2Ka Ka^2]; y = step(num,den,t); m = max(y); s = 501; while y(s) > 0.98 & y(s) < 1.02; s = s-1; end; ts = (s-1)0.01; if m < 1.10 & m > 1.02 & ts < 3.0 break; end end if m < 1.10 & m > 1.02 & ts < 3.0 break end end plot(t,y) grid title('Unit-Step Response') xlabel('t sec') ylabel('Output') solution = [K;a;m;ts] solution = 32.0000 0.2000 1.0969 2.6400 use two ‘for’ loops. The specification for the settling time is interpreted by the following four lines: .
Note that for t=0:0.01:5, we have 501 computing time points. s=501 corresponds to the last computing time point.
The solution obtained by this program is K=32, a=0.2 with the maximum overshoot equal to 9.69% and the settling time equal to 2.64 sec.The resulting unit-step response curve is shown in Figure 8–64.
ts 6 3.0 ts = (s - 1) 0.01 s = s - 1; end; s = 501; while y(s) 7 0.98 and y(s) 6 1.02; 634 Chapter 8 / PID Controllers and Modified PID Controllers Output t (sec) Unit-Step Response 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Figure 8–64 Unit-step response curve.
MATLAB Program 8–16 t = 0:0.01:5; k = 0; for i = 1:49; K(i) = 51-i1; for j = 1:40; a(j) = 2.05-j0.05; num = [K(i) 2K(i)a(j) K(i)a(j)a(j)]; den = [1 6 5+K(i) 2K(i)a(j) K(i)a(j)a(j)]; y = step(num,den,t); m = max(y); s = 501; while y(s) > 0.98 & y(s) < 1.02; s = s-1; end; ts = (s-1)0.01; if m < 1.10 & m > 1.02 & ts < 3.0 k = k+1; table(k,:) = [K(i) a(j) m ts]; end end end table(k,:) = [K(i) a(j) m ts] table = (continues on next page) Next, we shall consider the case where we want to find all sets of variables that will satisfy the given specifications. A possible MATLAB program for this purpose is given in MATLAB Pro-gram 8–16. Note that in the table shown in the program, the last row of the table (k, :) or the first row of the sorttable should be ignored. (These are the last K and a values for searching purposes.) Example Problems and Solutions 635 32.0000 0.2000 1.0969 2.6400 31.0000 0.2000 1.0890 2.6900 30.0000 0.2000 1.0809 2.7300 29.0000 0.2500 1.0952 1.7800 29.0000 0.2000 1.0726 2.7800 28.0000 0.2000 1.0639 2.8300 27.0000 0.2000 1.0550 2.8900 2.0000 0.0500 0.3781 5.0000 sorttable = sortrows(table,3) sorttable = 2.0000 0.0500 0.3781 5.0000 27.0000 0.2000 1.0550 2.8900 28.0000 0.2000 1.0639 2.8300 29.0000 0.2000 1.0726 2.7800 30.0000 0.2000 1.0809 2.7300 31.0000 0.2000 1.0890 2.6900 29.0000 0.2500 1.0952 1.7800 32.0000 0.2000 1.0969 2.6400 K = sorttable(7,1) K = 29 a = sorttable(7,2) a= 0.2500 num = [K 2Ka Ka^2]; den = [1 6 5+K 2Ka Ka^2]; y = step(num,den,t); plot(t,y) grid hold Current plot held K = sorttable(2,1) K= 27 a = sorttable(2,2) a= 0.2000 (continues on next page) 636 Chapter 8 / PID Controllers and Modified PID Controllers From the sorttable, it seems that K=29, a=0.25 (max overshoot=9.52%, settling time=1.78 sec) and K=27, a=0.2 (max overshoot=5.5%, settling time=2.89 sec) are two of the best choices.The unit-step response curves for these two cases are shown in Figure 8–65.
From these curves,we might conclude that the best choice depends on the system objective.If a small maximum overshoot is desired,K=27, a=0.2 will be the best choice.If the shorter settling time is more important than a small maximum overshoot, then K=29, a=0.25 will be the best choice.
A–8–13.
Consider the two-degrees-of-freedom control system shown in Figure 8–66. The plant is given by Assuming that the noise input N(s) is zero, design controllers and such that the designed system satisfies the following: 1. The response to the step disturbance input has a small amplitude and settles to zero quickly (on the order of 1 sec to 2 sec).
Gc2(s) Gc1(s) Gp(s) = 100 s(s + 1) Gp(s) num = [K 2Ka Ka^2]; den = [1 6 5+K 2Ka Ka^2]; y = step(num,den,t); plot(t,y) title('Unit-Step Response Curves') xlabel('t (sec)') ylabel('Output') text(1.22,1.22,'K = 29, a = 0.25') text(1.22,0.72,'K = 27, a = 0.2') Output t (sec) Unit-Step Response Curves 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 K = 29, a = 0.25 K = 27, a = 0.2 Figure 8–65 Unit-step response curves.
Example Problems and Solutions 637 Gp(s) Gc1(s) Gc2(s) Y(s) N(s) U(s) D(s) R(s) B(s) + – ++ ++ ++ Figure 8–66 Two-degrees-of-freedom control system.
2. The response to the unit-step reference input has a maximum overshoot of 25% or less, and the settling time is 1 sec or less.
3. The steady-state errors in following ramp reference input and acceleration reference input are zero.
Solution. The closed-loop transfer functions for the disturbance input and reference input are given, respectively, by Let us assume that is a PID controller and has the following form: The characteristic equation for the system is Notice that the open-loop poles are located at s=0 (a double pole) and s=–1. The zeros are located at s=–a (a double zero).
In what follows, we shall use the root-locus approach to determine the values of a and K. Let us choose the dominant closed-loop poles at s=–5_j5.Then,the angle deficiency at the desired closed-loop pole at s=–5+j5 is –135°-135°-128.66°+180°=–218.66° The double zero at s=–a must contribute 218.66°. (Each zero must contribute 109.33°.) By a simple calculation, we find a=–3.2460 The controller is then determined as The constant K must be determined by use of the magnitude condition.This condition is @Gc1(s)Gp(s)@ s=-5+j5 = 1 Gc1(s) = K(s + 3.2460)2 s Gc1(s) 1 + Gc1(s)Gp(s) = 1 + K(s + a)2 s 100 s(s + 1) Gc1(s) = K(s + a)2 s Gc1(s) Y(s) R(s) = CGc1(s) + Gc2(s)DGp(s) 1 + Gc1(s)Gp(s) Y(s) D(s) = Gp(s) 1 + Gc1(s)Gp(s) 638 Chapter 8 / PID Controllers and Modified PID Controllers Since we obtain The controller thus becomes (8–17) Then, the closed-loop transfer function Y(s)/D(s) is obtained as follows: The response curve when D(s) is a unit-step disturbance is shown in Figure 8–67.
= 100s s3 + 12.403s2 + 74.028s + 120.148 = 100 s(s + 1) 1 + 0.11403(s + 3.2460)2 s 100 s(s + 1) Y(s) D(s) = Gp(s) 1 + Gc1(s)Gp(s) = 0.74028 + 1.20148 s + 0.11403s = 0.11403s2 + 0.74028s + 1.20148 s Gc1(s) = 0.11403(s + 3.2460)2 s Gc1(s) = 0.11403 K = 2 s2(s + 1) 100(s + 3.2460)2 2 s=-5+j5 Gc1(s)Gp(s) = K(s + 3.2460)2 s 100 s(s + 1) yd(t) t (sec) Response to Unit-Step Disturbance Input 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0.4 0.2 0 0 0.5 1 1.5 2 2.5 3 3.5 4 Figure 8–67 Response to unit-step disturbance input.
Example Problems and Solutions 639 (a) t (sec) Response to Unit-Step Reference Input 0.6 0.8 1 1.2 1.4 0.4 0.2 0 0 0.5 1 1.5 2 2.5 3 yr(t) Figure 8–68 (a) Response to unit-step reference input; (b) response to unit-ramp reference input; (c) response to unit-acceleration reference input.
Next, we consider the responses to reference inputs. The closed-loop transfer function Y(s)/R(s) is Let us define Then To satisfy the requirements on the responses to the ramp reference input and acceleration reference input, we use the zero-placement approach. That is, we choose the numerator of Y(s)/R(s) to be the sum of the last three terms of the denominator, or from which we get (8–18) Hence, the closed-loop transfer function Y(s)/R(s) becomes as The response curves to the unit-step reference input, unit-ramp reference input, and unit-acceleration reference input are shown in Figures 8–68(a), (b), and (c), respectively.The maximum Y(s) R(s) = 12.403s2 + 74.028s + 120.148 s3 + 12.403s2 + 74.028s + 120.148 = 0.74028 + 1.20148 s + 0.12403s Gc(s) = 0.12403s2 + 0.74028s + 1.20148 s 100sGc(s) = 12.403s2 + 74.028s + 120.148 = 100sGc(s) s3 + 12.403s2 + 74.028s + 120.148 Y(s) R(s) = Gc(s)Gp(s) 1 + Gc1(s)Gp(s) Gc1(s) + Gc2(s) = Gc(s) Y(s) R(s) = CGc1(s) + Gc2(s)DGp(s) 1 + Gc1(s)Gp(s) overshoot in the unit-step response is approximately 25% and the settling time is approximately 1.2 sec. The steady-state errors in the ramp response and acceleration response are zero. There-fore, the designed controller given by Equation (8–18) is satisfactory.
Finally, we determine Noting that Gc2(s) = Gc(s) - Gc1(s) Gc2(s).
Gc(s) 640 Chapter 8 / PID Controllers and Modified PID Controllers (b) t (sec) Response to Unit-Ramp Reference Input 1.5 2 2.5 3 1 0.5 0 0 0.5 1 1.5 2 2.5 3 Input Output yr(t) (c) yr(t) t (sec) Response to Unit-Acceleration Reference Input 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Input Output Figure 8–68 (continued) and from Equation (8–17) we obtain (8–19) Equations (8–17) and (8–19) give the transfer functions of the controllers re-spectively.The block diagram of the designed system is shown in Figure 8–69.
Note that if the maximum overshoot were much higher than 25% and/or the settling time were much larger than 1.2 sec, then we might assume a search region (such as 3 a 6, 3 b 6, and 6 c 12) and use the computational method presented in Example 8–4 to find a set or sets of variables that would give the desired response to the unit-step reference input.
PROBLEMS Gc1(s) and Gc2(s), = 0.01s - a0.7403 + 1.20148 s + 0.11403s b Gc2(s) = a0.7403 + 1.20148 s + 0.12403s b Gc1(s) = 0.7403 + 1.20148 s + 0.11403s Problems 641 100 s(s + 1) 0.01s Y(s) D(s) R(s) + – ++ ++ 1.20148 s 0.7403 + + 0.11403s Figure 8–69 Block diagram of the designed system.
B–8–1. Consider the electronic PID controller shown in Figure 8–70. Determine the values of R1, R2, R3, R4, C1, and C2 of the controller such that the transfer function is Gc(s) = Eo(s)Ei(s) = 30.3215 (s + 0.65)2 s Gc(s) = 39.42 a1 + 1 3.077s + 0.7692s b + – + – Ei(s) E(s) Eo(s) C1 C2 R1 R2 R3 R4 Figure 8–70 Electronic PID controller.
642 Chapter 8 / PID Controllers and Modified PID Controllers B–8–2. Consider the system shown in Figure 8–71.
Assume that disturbances D(s) enter the system as shown in the diagram. Determine parameters K, a, and b such that the response to the unit-step disturbance input and the response to the unit-step reference input satisfy the following specifications: The response to the step distur-bance input should attenuate rapidly with no steady-state error, and the response to the step reference input exhibits a maximum overshoot of 20% or less and a settling time of 2 sec.
B–8–3. Show that the PID-controlled system shown in Figure 8–72(a) is equivalent to the I-PD-controlled system with feedforward control shown in Figure 8–72(b).
B–8–4. Consider the systems shown in Figures 8–73(a) and (b). The system shown in Figure 8–73(a) is the system designed in Example 8–1. The response to the unit-step reference input in the absence of the disturbance input is shown in Figure 8–10. The system shown in Figure 8–73(b) is the I-PD-controlled system using the same Kp, as the system shown in Figure 8–73(a).
T i , and T d C(s) R(s) D(s) K(as + 1)(bs + 1) s 2(s + 2) (s + 1) (s + 10) + – ++ Figure 8–71 Control system.
(a) (b) Kp Tis Gp(s) C(s) R(s) Kp(1 + Tds) Kp(1 + Tds) Gp(s) C(s) R(s) Kp(1 + + Tds) 1 Tis + – + – + – + Figure 8–72 (a) PID-controlled system; (b) I-PD-controlled system with feedforward control.
Problems 643 Obtain the response of the I-PD-controlled system to the unit-step reference input with MATLAB. Compare the unit-step response curves of the two systems.
B–8–5. Referring to Problem B–8–4, obtain the response of the PID-controlled system shown in Figure 8–73(a) to the unit-step disturbance input.
Show that for the disturbance input, the responses of the PID-controlled system shown in Figure 8–73(a) and of the I-PD-controlled system shown in Figure 8–73(b) are exactly the same.[When considering D(s) to be the input,as-sume that the reference input R(s) is zero, and vice versa.] Also, compare the closed-loop transfer function C(s)/R(s) of both systems.
B–8–6. Consider the system shown in Figure 8–74.This sys-tem is subjected to three input signals: the reference input, disturbance input, and noise input. Show that the charac-teristic equation of this system is the same regardless of which input signal is chosen as input.
(a) (b) C(s) R(s) D(s) PID contoller 39.42 (1 + 1 3.077s + 0.7692s) 1 s(s + 1) (s + 5) D(s) C(s) R(s) 1 + 0.7692s 39.42 1 s(s + 1) (s + 5) 1 3.077s + – ++ + – + – ++ Figure 8–73 (a) PID-controlled system; (b) I-PD-controlled system.
G2(s) H(s) C(s) Noise N(s) R(s) Disturbance D(s) G1(s) ++ ++ + – Figure 8–74 Control system.
644 Chapter 8 / PID Controllers and Modified PID Controllers B–8–7. Consider the system shown in Figure 8–75. Obtain the closed-loop transfer function C(s)/R(s) for the refer-ence input and the closed-loop transfer function C(s)/D(s) for the disturbance input. When considering R(s) as the input, assume that D(s) is zero, and vice versa.
B–8–8. Consider the system shown in Figure 8–76(a), where K is an adjustable gain and G(s) and H(s) are fixed components. The closed-loop transfer function for the disturbance is To minimize the effect of disturbances, the adjustable gain K should be chosen as large as possible.
Is this true for the system in Figure 8–76(b), too?
C(s) D(s) = 1 1 + KG(s)H(s) G1(s) G2(s) R(s) C(s) D(s) G3(s) H1(s) H2(s) + – + – ++ Figure 8–75 Control system.
G(s) R(s) C(s) D(s) D(s) K H(s) G(s) R(s) C(s) K H(s) (a) (b) ++ + – + + + – Figure 8–76 (a) Control system with disturbance entering in the feedforward path; (b) control system with disturbance entering in the feedback path.
D(s) R(s) Y(s) N(s) Gc2 Gc1 Gc3 G1 G2 + – – ++ – – ++ ++ Problems 645 B–8–9. Show that the control systems shown in Fig-ures 8–77(a), (b), and (c) are two-degrees-of-freedom systems. In the diagrams, Gc1 and Gc2 are controllers and Gp is the plant.
B–8–10. Show that the control system shown in Figure 8–78 is a three-degrees-of freedom system. The transfer func-tions Gc1, Gc2, and Gc3 are controllers.The plant consists of transfer functions G1 and G2.
(c) D(s) Y(s) N(s) Gp Gc1 Gc2 R(s) ++ + – ++ (b) D(s) Y(s) N(s) Gp Gc1 Gc2 R(s) + – + + ++ D(s) R(s) Y(s) Gp Gc1 Gc2 (a) + – ++ ++ N(s) Figure 8–77 (a), (b), (c) Two degrees-of-freedom systems.
Figure 8–78 Three-degrees-of-freedom system.
646 Chapter 8 / PID Controllers and Modified PID Controllers B–8–11. Consider the control system shown in Figure 8–79.
Assume that the PID controller is given by It is desired that the unit-step response of the system exhibit the maximum overshoot of less than 10%, but more than 2% (to avoid an almost overdamped system), and the settling time be less than 2 sec.
Using the computational approach presented in Section 8–4, write a MATLAB program to determine the values of K and a that will satisfy the given specifications. Choose the search region to be 1 K 4, 0.4 a 4 Choose the step size for K and a to be 0.05. Write the program such that the nested loops start with the highest values of K and a and step toward the lowest.
Using the first-found solution, plot the unit-step response curve.
B–8–12. Consider the same control system as treated in Problem B–8–11 (Figure 8–79).The PID controller is given by It is desired to determine the values of K and a such that the unit-step response of the system exhibits the maximum Gc(s) = K (s + a)2 s Gc(s) = K (s + a)2 s overshoot of less than 8%,but more than 3%,and the settling time is less than 2 sec. Choose the search region to be 2 K 4, 0.5 a 3 Choose the step size for K and a to be 0.05.
First, write a MATLAB program such that the nested loops in the program start with the highest values of K and a and step toward the lowest and the computation stops when a successful set of K and a is found for the first time.
Next, write a MATLAB program that will find all pos-sible sets of K and a that will satisfy the given specifications.
Among multiple sets of K and a that satisfy the given specifications,determine the best choice.Then,plot the unit-step response curves of the system with the best choice of K and a.
B–8–13. Consider the two-degrees-of-freedom control system shown in Figure 8–80.The plant is given by Design controllers and such that the response to the unit-step disturbance input should have small amplitude and settle to zero quickly (in approximately 2 sec).The response to the unit-step reference input should be such that the maximum overshoot is 25% (or less) and the settling time is 2 sec.Also, the steady-state errors in the response to the ramp and acceleration reference inputs should be zero.
Gc2(s) Gc1(s) Gp(s) = 3(s + 5) s(s + 1)As2 + 4s + 13B Gp(s) R(s) C(s) PID controller 1.2 (0.3s + 1) (s + 1) (1.2s + 1) + – Gc(s) Figure 8–79 Control system.
Gp(s) Gc1(s) Gc2(s) Y(s) U(s) D(s) R(s) B(s) + – ++ ++ Figure 8–80 Two-degrees-of-freedom control system.
Problems 647 B–8–14. Consider the system shown in Figure 8–81. The plant is given by Determine the controllers and such that, for the step disturbance input, the response shows a small am-plitude and approaches zero quickly (in a matter of 1 to 2 sec). For the response to the unit-step reference input, it is desired that the maximum overshoot be 20% or less and the settling time 1 sec or less. For the ramp reference input and acceleration reference input, the steady-state errors should be zero.
Gc2(s) Gc1(s) Gp(s) = 2(s + 1) s(s + 3)(s + 5) Gp(s) B–8–15. Consider the two-degrees-of-freedom control system shown in Figure 8–82. Design controllers and such that the response to the step disturbance input shows a small amplitude and settles to zero quickly (in 1 to 2 sec) and the response to the step reference input ex-hibits 25% or less maximum overshoot and the settling time is less than 1 sec.The steady-state error in following the ramp reference input or acceleration reference input should be zero.
Gc2(s) Gc1(s) Gp(s) Gc1(s) Y(s) R(s) U(s) D(s) Gc2(s) ++ + – + – Figure 8–81 Two-degrees-of-freedom control system.
1 s2 C1(s) Y(s) R(s) D(s) C2(s) ++ + – + – Figure 8–82 Two-degrees-of-freedom control system.
9 648 Control Systems Analysis in State Space 9–1 INTRODUCTION A modern complex system may have many inputs and many outputs, and these may be interrelated in a complicated manner.To analyze such a system, it is essential to reduce the complexity of the mathematical expressions,as well as to resort to computers for most of the tedious computations necessary in the analysis.The state-space approach to system analysis is best suited from this viewpoint.
While conventional control theory is based on the input–output relationship,or trans-fer function, modern control theory is based on the description of system equations in terms of n first-order differential equations, which may be combined into a first-order vector-matrix differential equation.The use of vector-matrix notation greatly simplifies the mathematical representation of systems of equations.The increase in the number of state variables, the number of inputs, or the number of outputs does not increase the complexity of the equations.In fact,the analysis of complicated multiple-input,multiple-output systems can be carried out by procedures that are only slightly more compli-cated than those required for the analysis of systems of first-order scalar differential equations.
This chapter and the next deal with the state-space analysis and design of control sys-tems. Basic materials of state-space analysis, including the state-space representation of It is noted that in this book an asterisk used as a superscript of a matrix, such as A, implies that it is a con-jugate transpose of matrix A.The conjugate transpose is the conjugate of the transpose of a matrix. For a real matrix (a matrix whose elements are all real), the conjugate transpose A is the same as the transpose AT.
Section 9–2 / State-Space Representations of Transfer-Function Systems 649 systems, controllability, and observability are presented in this chapter. Useful design methods based on state-feedback control are given in Chapter 10.
Outline of the Chapter.
Section 9–1 has presented an introduction to state-space analysis of control systems. Section 9–2 deals with the state-space representation of transfer-function systems. Here we present various canonical forms of state-space equa-tions. Section 9–3 discusses the transformation of system models (such as from transfer-function to state-space models, and vice versa) with MATLAB. Section 9–4 presents the solution of time-invariant state equations. Section 9–5 gives some useful results in vector-matrix analysis that are necessary in studying the state-space analysis of control systems. Section 9–6 discusses the controllability of control systems and Section 9–7 treats the observability of control systems.
9–2 STATE-SPACE REPRESENTATIONS OF TRANSFER-FUNCTION SYSTEMS Many techniques are available for obtaining state-space representations of transfer-function systems. In Chapter 2 we presented a few such methods. This section presents state-space representations in the controllable, observable, diagonal, or Jordan canonical forms. (Methods for obtaining such state-space representations from transfer functions are discussed in detail in Problems A–9–1 through A–9–4.) State-Space Representations in Canonical Forms.
Consider a system defined by (9–1) where u is the input and y is the output.This equation can also be written as (9–2) In what follows we shall present state-space representations of the system defined by Equation (9–1) or (9–2) in controllable canonical form, observable canonical form, and diagonal (or Jordan) canonical form.
Controllable Canonical Form.
The following state-space representation is called a controllable canonical form: (9–3) G x # 1 x # 2 x # n-1 x # n W = G 0 0 0 -an 1 0 0 -an-1 0 1 0 -an-2 p p p p 0 0 1 -a1 W G x1 x2 xn-1 xn W + G 0 0 0 1 W u Y(s) U(s) = b0 sn + b1 sn-1 + p + bn-1 s + bn sn + a1 sn-1 + p + an-1 s + an y (n) + a1 y (n-1) + p + an-1 y # + an y = b0 u (n) + b1 u (n-1) + p + bn-1 u # + bn u 650 Chapter 9 / Control Systems Analysis in State Space (9–4) The controllable canonical form is important in discussing the pole-placement approach to control systems design.
Observable Canonical Form.
The following state-space representation is called an observable canonical form: (9–5) (9–6) Note that the nn state matrix of the state equation given by Equation (9–5) is the transpose of that of the state equation defined by Equation (9–3).
Diagonal Canonical Form.
Consider the transfer-function system defined by Equa-tion (9–2). Here we consider the case where the denominator polynomial involves only distinct roots. For the distinct-roots case, Equation (9–2) can be written as (9–7) The diagonal canonical form of the state-space representation of this system is given by = b0 + c1 s + p1 + c2 s + p2 + p + cn s + pn Y(s) U(s) = b0 sn + b1 sn-1 + p + bn-1 s + bn As + p1BAs + p2B p As + pnB y = [0 0 p 0 1]G x1 x2 xn-1 xn W + b0 u F x # 1 x # 2 x # n V = F 0 1 0 0 0 0 p p p 0 0 1 -an -an-1 -a1 V F x1 x2 xn V + F bn - an b0 bn-1 - an-1 b0 b1 - a1 b0 V u y = Cbn - an b0 bn-1 - an-1 b0 p b1 - a1 b0D F x1 x2 xn V + b0 u (9–8) (9–9) Jordan Canonical Form.
Next we shall consider the case where the denominator polynomial of Equation (9–2) involves multiple roots. For this case, the preceding diagonal canonical form must be modified into the Jordan canonical form. Suppose, for example, that the pi’s are different from one another, except that the first three pi’s are equal, or p1=p2=p3. Then the factored form of Y(s)/U(s) becomes The partial-fraction expansion of this last equation becomes A state-space representation of this system in the Jordan canonical form is given by (9–10) (9–11) y = Cc1 c2 p cnD F x1 x2 xn V + b0 u H x # 1 x # 2 x # 3 x # 4 x # n X = H -p1 0 0 0 0 1 -p1 0 p p 0 1 -p1 0 0 0 : 0 -p4 0 p p 0 : 0 0 -pn X H x1 x2 x3 x4 xn X + H 0 0 1 1 1 X u Y(s) U(s) = b0 + c1 As + p1B 3 + c2 As + p1B 2 + c3 s + p1 + c4 s + p4 + p + cn s + pn Y(s) U(s) = b0 sn + b1 sn-1 + p + bn-1 s + bn As + p1B 3As + p4BAs + p5B p As + pnB y = Cc1 c2 p cnD F x1 x2 xn V + b0 u F x # 1 x # 2 x # n V = F -p1 0 -p2 0 -pn V F x1 x2 xn V + F 1 1 1 V u Section 9–2 / State-Space Representations of Transfer-Function Systems 651 652 Chapter 9 / Control Systems Analysis in State Space EXAMPLE 9–1 Consider the system given by Obtain state-space representations in the controllable canonical form, observable canonical form, and diagonal canonical form.
Controllable Canonical Form: Observable Canonical Form: Diagonal Canonical Form: Eigenvalues of an n n Matrix A.
The eigenvalues of an nn matrix A are the roots of the characteristic equation |lI-A|=0 The eigenvalues are also called the characteristic roots.
Consider, for example, the following matrix A: The characteristic equation is |lI-A|= =l3+6l2+11l+6 =(l+1)(l+2)(l+3)=0 The eigenvalues of A are the roots of the characteristic equation, or –1, –2, and –3.
Diagonalization of n n Matrix.
Note that if an nn matrix A with distinct eigenvalues is given by 3 l 0 6 -1 l 11 0 -1 l + 6 3 A = C 0 0 -6 1 0 -11 0 1 -6 S y(t) = [2 -1]B x1(t) x2(t)R B x # 1(t) x # 2(t)R = B -1 0 0 -2R B x1(t) x2(t)R + B1 1R u(t) y(t) = [0 1]B x1(t) x2(t)R B x # 1(t) x # 2(t)R = B 0 1 -2 -3R B x1(t) x2(t)R + B 3 1R u(t) y(t) = [3 1]B x1(t) x2(t)R B x # 1(t) x # 2(t)R = B 0 -2 1 -3R B x1(t) x2(t)R + B0 1R u(t) Y(s) U(s) = s + 3 s2 + 3s + 2 Section 9–2 / State-Space Representations of Transfer-Function Systems 653 (9–12) the transformation x=Pz, where P= l1, l2, p , ln=n distinct eigenvalues of A will transform P–1AP into the diagonal matrix, or If the matrix A defined by Equation (9–12) involves multiple eigenvalues, then diagonalization is impossible. For example, if the 33 matrix A, where has the eigenvalues l1, l1, l3, then the transformation x=Sz, where will yield This is in the Jordan canonical form.
S-1 AS = C l1 0 0 1 l1 0 0 0 l3 S S = C 1 l1 l1 2 0 1 2l1 1 l3 l3 2 S A = C 0 0 -a3 1 0 -a2 0 1 -a1 S P-1 AP = F l1 0 l2 0 ln V G 1 l1 l1 2 l1 n-1 1 l2 l2 2 l2 n-1 p p p p 1 ln ln 2 ln n-1 W A = G 0 0 0 -an 1 0 0 -an-1 0 1 0 -an-2 p p p p 0 0 1 -a1 W 654 Chapter 9 / Control Systems Analysis in State Space EXAMPLE 9–2 Consider the following state-space representation of a system.
(9–13) (9–14) Equations (9–13) and (9–14) can be put in a standard form as (9–15) (9–16) where The eigenvalues of matrix A are l1=–1, l2=–2, l3=–3 Thus, three eigenvalues are distinct. If we define a set of new state variables z1, z2, and z3 by the transformation or x=Pz (9–17) where (9–18) then, by substituting Equation (9–17) into Equation (9–15), we obtain By premultiplying both sides of this last equation by P–1, we get (9–19) or + C 3 -3 1 2.5 -4 1.5 0.5 -1 0.5 S C 0 0 6 S u C z # 1 z # 2 z # 3 S = C 3 -3 1 2.5 -4 1.5 0.5 -1 0.5 S C 0 0 -6 1 0 -11 0 1 -6 S C 1 -1 1 1 -2 4 1 -3 9 S C z1 z2 z3 S z # = P-1 APz + P-1 Bu Pz # = APz + Bu P = C 1 l1 l1 2 1 l2 l2 2 1 l3 l3 2 S = C 1 -1 1 1 -2 4 1 -3 9 S C x1 x2 x3 S = C 1 -1 1 1 -2 4 1 -3 9 S C z1 z2 z3 S A = C 0 0 -6 1 0 -11 0 1 -6 S , B = C 0 0 6 S , C = [1 0 0] y = Cx x # = Ax + Bu y = [1 0 0]C x1 x2 x3 S C x # 1 x # 2 x # 3 S = C 0 0 -6 1 0 -11 0 1 -6 S C x1 x2 x3 S + C 0 0 6 S u Section 9–2 / State-Space Representations of Transfer-Function Systems 655 Simplifying gives (9–20) Equation (9–20) is also a state equation that describes the same system as defined by Equation (9–13).
The output equation, Equation (9–16), is modified to y=CPz or (9–21) Notice that the transformation matrix P, defined by Equation (9–18), modifies the coefficient matrix of z into the diagonal matrix.As is clearly seen from Equation (9–20), the three scalar state equations are uncoupled. Notice also that the diagonal elements of the matrix P–1AP in Equation (9–19) are identical with the three eigenvalues of A. It is very important to note that the eigen-values of A and those of P–1AP are identical.We shall prove this for a general case in what follows.
Invariance of Eigenvalues.
To prove the invariance of the eigenvalues under a linear transformation, we must show that the characteristic polynomials ∑lI-A∑and @lI-P–1AP@ are identical.
Since the determinant of a product is the product of the determinants, we obtain Noting that the product of the determinants @P–1@ and ∑P∑is the determinant of the prod-uct @P–1P@, we obtain Thus, we have proved that the eigenvalues of A are invariant under a linear transformation.
Nonuniqueness of a Set of State Variables.
It has been stated that a set of state vari-ables is not unique for a given system.Suppose that x1, x2, p , xn are a set of state variables. = ∑l I - A∑ @l I - P-1 AP@ = @P-1 P@ @l I - A@ = @P-1@ @P@ @l I - A@ = @P-1@ @l I - A@ @P@ = @P-1(l I - A) P@ @l I - P-1 AP@ = @l P-1 P - P-1 AP@ = [1 1 1]C z1 z2 z3 S y = [1 0 0]C 1 -1 1 1 -2 4 1 -3 9 S C z1 z2 z3 S C z # 1 z # 2 z # 3 S = C -1 0 0 0 -2 0 0 0 -3 S C z1 z2 z3 S + C 3 -6 3 S u 656 Chapter 9 / Control Systems Analysis in State Space Then we may take as another set of state variables any set of functions provided that, for every set of values there corresponds a unique set of values x1, x2, p , xn, and vice versa.Thus, if x is a state vector, then where is also a state vector,provided the matrix P is nonsingular.Different state vectors convey the same information about the system behavior.
9–3 TRANSFORMATION OF SYSTEM MODELS WITH MATLAB In this section we shall consider the transformation of the system model from transfer function to state space, and vice versa. We shall begin our discussion with the transformation from transfer function to state space.
Let us write the closed-loop transfer function as Once we have this transfer-function expression, the MATLAB command [A, B, C, D] = tf2ss(num,den) will give a state-space representation. It is important to note that the state-space repre-sentation for any system is not unique.There are many (indeed, infinitely many) state-space representations for the same system.The MATLAB command gives one possible such state-space representation.
State-Space Formulation of Transfer-Function Systems.
Consider the transfer-function system (9–22) There are many (again, infinitely many) possible state-space representations for this system. One possible state-space representation is y = [1 0 0]C x1 x2 x3 S + u C x # 1 x # 2 x # 3 S = C 0 0 -10 1 0 -5 0 1 -6 S C x1 x2 x3 S + C 0 10 -50 S u Y(s) U(s) = 10s + 10 s3 + 6s2 + 5s + 10 Y(s) U(s) = numerator polynomial in s denominator polynomial in s = num den x ˆ = Px x ˆ , x ˆ 1 , x ˆ 2 , p , x ˆ n , x ˆ n = X nAx1 , x2 , p , xnB x ˆ 2 = X 2Ax1 , x2 , p , xnB x ˆ 1 = X 1Ax1 , x2 , p , xnB Section 9–3 / Transformation of System Models with MATLAB 657 Another possible state-space representation (among infinitely many alternatives) is (9–23) (9–24) MATLAB transforms the transfer function given by Equation (9–22) into the state-space representation given by Equations (9–23) and (9–24).For the example system considered here, MATLAB Program 9–1 will produce matrices A, B, C, and D. y = [0 10 10]C x1 x2 x3 S + u C x # 1 x # 2 x # 3 S = C -6 1 0 -5 0 1 -10 0 0 S C x1 x2 x3 S + C 1 0 0 S u MATLAB Program 9–1 num = [10 10]; den = [1 6 5 10]; [A,B,C,D] = tf2ss(num,den) A = -6 -5 -10 1 -0 - 0 0 -1 - 0 B = 1 0 0 C = 0 10 10 D = 0 Transformation from State Space to Transfer Function.
To obtain the transfer function from state-space equations, use the following command: [num,den] = ss2tf(A,B,C,D,iu) iu must be specified for systems with more than one input. For example, if the system has three inputs (u1, u2, u3), then iu must be either 1, 2, or 3, where 1 implies u1, 2 implies u2, and 3 implies u3.
If the system has only one input, then either [num,den] = ss2tf(A,B,C,D) or [num,den] = ss2tf(A,B,C,D,1) may be used. (See Example 9–3 and MATLAB Program 9–2.) 658 Chapter 9 / Control Systems Analysis in State Space For the case where the system has multiple inputs and multiple outputs, see Example 9–4.
EXAMPLE 9–3 Obtain the transfer function of the system defined by the following state-space equations: MATLAB Program 9–2 will produce the transfer function for the given system. The transfer function obtained is given by Y(s) U(s) = 25.04s + 5.008 s3 + 5.0325s2 + 25.1026s + 5.008 y = [1 0 0]C x1 x2 x3 S C x # 1 x # 2 x # 3 S = C 0 0 -5.008 1 0 -25.1026 0 1 -5.03247 S C x1 x2 x3 S + C 0 25.04 -121.005 S u MATLAB Program 9–2 A = [0 1 0;0 0 1;-5.008 -25.1026 -5.03247]; B = [0;25.04; -121.005]; C = [1 0 0]; D = ; [num,den] = ss2tf(A,B,C,D) num = 0 -0.0000 25.0400 5.0080 den = 1.0000 5.0325 25.1026 5.0080 % The same result can be obtained by entering the following command [num,den] = ss2tf(A,B,C,D,1) num = 0 -0.0000 25.0400 5.0080 den = 1.0000 5.0325 25.1026 5.0080 Section 9–3 / Transformation of System Models with MATLAB 659 EXAMPLE 9–4 Consider a system with multiple inputs and multiple outputs.When the system has more than one output, the command [NUM,den] = ss2tf(A,B,C,D,iu) produces transfer functions for all outputs to each input. (The numerator coefficients are returned to matrix NUM with as many rows as there are outputs.) Consider the system defined by This system involves two inputs and two outputs.Four transfer functions are involved: and (When considering input u1, we assume that input u2 is zero and vice versa.) See the output of MATLAB Program 9–3.
Y 2(s)U 2(s).
Y 1(s)U 2(s), Y 2(s)U 1(s), Y 1(s)U 1(s), B y1 y2 R = B 1 0 0 1R B x1 x2 R + B 0 0 0 0R Bu1 u2 R B x # 1 x # 2 R = B 0 -25 1 -4R B x1 x2 R + B 1 0 1 1R Bu1 u2 R MATLAB Program 9–3 A = [0 1;-25 -4]; B = [1 1;0 1]; C = [1 0;0 1]; D = [0 0;0 0]; [NUM,den] = ss2tf(A,B,C,D,1) NUM = 0 1 4 0 0 -25 den = 1 4 25 [NUM,den] = ss2tf(A,B,C,D,2) NUM = 0 1.0000 5.0000 0 1.0000 -25.0000 den = 1 4 25 This is the MATLAB representation of the following four transfer functions: Y 1(s) U 2(s) = s + 5 s2 + 4s + 25 , Y 2(s) U 2(s) = s - 25 s2 + 4s + 25 Y 1(s) U 1(s) = s + 4 s2 + 4s + 25 , Y 2(s) U 1(s) = -25 s2 + 4s + 25 660 Chapter 9 / Control Systems Analysis in State Space 9–4 SOLVING THE TIME-INVARIANT STATE EQUATION In this section,we shall obtain the general solution of the linear time-invariant state equa-tion.We shall first consider the homogeneous case and then the nonhomogeneous case.
Solution of Homogeneous State Equations.
Before we solve vector-matrix differential equations, let us review the solution of the scalar differential equation (9–25) In solving this equation, we may assume a solution x(t) of the form x(t)=b0+b1t+b2t2+p+bktk+p (9–26) By substituting this assumed solution into Equation (9–25), we obtain (9–27) If the assumed solution is to be the true solution, Equation (9–27) must hold for any t.
Hence, equating the coefficients of the equal powers of t, we obtain The value of b0 is determined by substituting t=0 into Equation (9–26), or x(0)=b0 Hence, the solution x(t) can be written as We shall now solve the vector-matrix differential equation (9–28) where By analogy with the scalar case, we assume that the solution is in the form of a vector power series in t, or x(t)=b0+b1t+b2t2+p+bktk+p (9–29) A = n n constant matrix x = n-vector x # = Ax = eatx(0) x(t) = a1 + at + 1 2! a2t2 + p + 1 k! aktk + p bx(0) bk = 1 k! akb0 b3 = 1 3 ab2 = 1 3 2 a3b0 b2 = 1 2 ab1 = 1 2 a2b0 b1 = ab0 = aAb0 + b1 t + b2 t2 + p + bk tk + pB b1 + 2b2 t + 3b3 t2 + p + kbk tk-1 + p x # = ax Section 9–4 / Solving the Time-Invariant State Equation 661 By substituting this assumed solution into Equation (9–28), we obtain (9–30) If the assumed solution is to be the true solution, Equation (9–30) must hold for all t. Thus, by equating the coefficients of like powers of t on both sides of Equation (9–30),we obtain By substituting t=0 into Equation (9–29), we obtain x(0)=b0 Thus, the solution x(t) can be written as The expression in the parentheses on the right-hand side of this last equation is an nn matrix. Because of its similarity to the infinite power series for a scalar exponential, we call it the matrix exponential and write In terms of the matrix exponential, the solution of Equation (9–28) can be written as (9–31) Since the matrix exponential is very important in the state-space analysis of linear systems, we shall next examine its properties.
Matrix Exponential.
It can be proved that the matrix exponential of an nn matrix A, converges absolutely for all finite t. (Hence, computer calculations for evaluating the elements of eAt by using the series expansion can be easily carried out.) eAt = a q k=0 A ktk k!
x(t) = eAt x(0) I + At + 1 2! A 2t2 + p + 1 k! A ktk + p = eAt x(t) = a I + At + 1 2! A 2t2 + p + 1 k! A ktk + p b x(0) bk = 1 k! A k b0 b3 = 1 3 Ab2 = 1 3 2 A 3 b0 b2 = 1 2 Ab1 = 1 2 A 2 b0 b1 = Ab0 = AAb0 + b1 t + b2 t2 + p + bk tk + p B b1 + 2 b2 t + 3 b3 t2 + p + k bk tk-1 + p 662 Chapter 9 / Control Systems Analysis in State Space Because of the convergence of the infinite series the series can be differentiated term by term to give The matrix exponential has the property that This can be proved as follows: In particular, if s=–t, then Thus, the inverse of is Since the inverse of always exists, is nonsingular.
It is very important to remember that To prove this, note that + A 2 Bt3 2!
+ AB2t3 2!
+ B3t3 3!
+ p = I + (A + B)t + A 2t2 2!
+ ABt2 + B2t2 2!
+ A 3t3 3! eAteBt = a I + At + A 2t2 2!
+ A 3t3 3!
+ p b aI + Bt + B2t2 2!
+ B3t3 3!
+ p b e(A+B)t = I + (A + B)t + (A + B)2 2! t2 + (A + B)3 3! t3 + p e(A+B)t Z eAteBt, if AB Z BA e(A+B)t = eAteBt, if AB = BA eAt eAt e- At.
eAt eAte- At = e- AteAt = eA(t-t) = I = eA(t+s) = a q k=0 A k (t + s)k k! = a q k=0 A k a a q i=0 tisk-i i!(k - i)! b eAteAs = a a q k=0 A ktk k! b a a q k=0 A ksk k! b eA(t+s) = eAteAs = cI + At + A 2t2 2!
+ p + A k-1tk-1 (k - 1)! + pd A = eAt A = AcI + At + A 2t2 2!
+ p + A k-1tk-1 (k - 1)! + pd = AeAt d dt eAt = A + A 2t + A 3t2 2!
+ p + A ktk-1 (k - 1)! + p g q k=0 A ktkk!, Section 9–4 / Solving the Time-Invariant State Equation 663 Hence, The difference between and vanishes if A and B commute.
Laplace Transform Approach to the Solution of Homogeneous State Equations.
Let us first consider the scalar case: (9–32) Taking the Laplace transform of Equation (9–32), we obtain sX(s)-x(0)=aX(s) (9–33) where Solving Equation (9–33) for X(s) gives The inverse Laplace transform of this last equation gives the solution x(t)=eatx(0) The foregoing approach to the solution of the homogeneous scalar differential equation can be extended to the homogeneous state equation: (9–34) Taking the Laplace transform of both sides of Equation (9–34), we obtain sX(s)-x(0)=AX(s) where Hence, (sI-A)X(s)=x(0) Premultiplying both sides of this last equation by (sI-A)–1, we obtain X(s)=(sI-A)–1x(0) The inverse Laplace transform of gives the solution Thus, x(t)=l–1C(sI-A)–1D x(0) (9–35) Note that Hence, the inverse Laplace transform of (sI-A)–1 gives (9–36) l-1C(s I - A)-1D = I + At + A 2t2 2!
+ A 3t3 3!
+ p = eAt (s I - A)-1 = I s + A s2 + A 2 s3 + p x(t).
X(s) X(s) = l[x].
x # (t) = Ax(t) X(s) = x(0) s - a = (s - a)-1x(0) X(s) = l[x].
x # = ax eAteBt e(A+B)t + BA 2 + ABA + B2 A + BAB - 2 A 2 B - 2 AB2 3!
t3 + p e(A+B)t - eAteBt = BA - AB 2! t2 664 Chapter 9 / Control Systems Analysis in State Space (The inverse Laplace transform of a matrix is the matrix consisting of the inverse Laplace transforms of all elements.) From Equations (9–35) and (9–36), the solution of Equation (9–34) is obtained as The importance of Equation (9–36) lies in the fact that it provides a convenient means for finding the closed solution for the matrix exponential.
State-Transition Matrix.
We can write the solution of the homogeneous state equation (9–37) as (9–38) where is an nn matrix and is the unique solution of To verify this, note that and We thus confirm that Equation (9–38) is the solution of Equation (9–37).
From Equations (9–31), (9–35), and (9–38), we obtain Note that From Equation (9–38), we see that the solution of Equation (9–37) is simply a transformation of the initial condition. Hence, the unique matrix is called the state-transition matrix.The state-transition matrix contains all the information about the free motions of the system defined by Equation (9–37).
If the eigenvalues l1, l2, p , ln of the matrix A are distinct, than will contain the n exponentials In particular, if the matrix A is diagonal, then (t) = eAt = F el1 t 0 el2 t 0 eln t V (A: diagonal) el1 t, el2 t, p ,eln t (t) (t) -1(t) = e- At = (-t) (t) = eAt = l-1C(s I - A)-1D x # (t) = # (t) x(0) = A(t) x(0) = Ax(t) x(0) = (0) x(0) = x(0) # (t) = A(t), (0) = I (t) x(t) = (t) x(0) x # = Ax x(t) = eAt x(0) Section 9–4 / Solving the Time-Invariant State Equation 665 If there is a multiplicity in the eigenvalues—for example, if the eigenvalues of A are l1, l1, l1, l4, l5, p , ln, then will contain, in addition to the exponentials terms like and Properties of State-Transition Matrices.
We shall now summarize the important properties of the state-transition matrix For the time-invariant system for which we have the following: 1.
2.
or 3.
4.
5.
EXAMPLE 9–5 Obtain the state-transition matrix of the following system: Obtain also the inverse of the state-transition matrix, For this system, The state-transition matrix is given by Since the inverse of (sI-A) is given by = D s + 3 (s + 1)(s + 2) -2 (s + 1)(s + 2) 1 (s + 1)(s + 2) s (s + 1)(s + 2) T (s I - A)-1 = 1 (s + 1)(s + 2) B s + 3 -2 1 s R s I - A = B s 0 0 s R - B 0 -2 1 -3R = B s 2 -1 s + 3R (t) = eAt = l-1C(s I - A)-1D (t) A = B 0 -2 1 -3R -1(t).
B x # 1 x # 2 R = B 0 -2 1 -3R B x1 x2 R (t) At2 - t1B At1 - t0B = At2 - t0B = At1 - t0B At2 - t1B C(t)D n = (nt) At1 + t2B = eAAt1+t2B = eAt1eAt2 = At1B At2B = At2B At1B -1(t) = (-t) (t) = eAt = Ae- AtB -1 = C(-t)D -1 (0) = eA0 = I (t) = eAt x # = Ax (t).
t2el1 t.
tel1 t el1 t, el4 t, el5 t, p ,eln t, (t) 666 Chapter 9 / Control Systems Analysis in State Space Hence, Noting that we obtain the inverse of the state-transition matrix as follows: Solution of Nonhomogeneous State Equations.
We shall begin by considering the scalar case (9–39) Let us rewrite Equation (9–39) as Multiplying both sides of this equation by e–at, we obtain Integrating this equation between 0 and t gives or The first term on the right-hand side is the response to the initial condition and the second term is the response to the input u(t).
Let us now consider the nonhomogeneous state equation described by (9–40) where By writing Equation (9–40) as and premultiplying both sides of this equation by e–At, we obtain e- AtCx # (t) - Ax(t)D = d dt Ce- At x(t)D = e- At Bu(t) x # (t) - Ax(t) = Bu(t) B = n r constant matrix A = n n constant matrix u = r-vector x = n-vector x # = Ax + Bu x(t) = eatx(0) + eat 3 t 0 e-atbu(t)dt e-atx(t) - x(0) = 3 t 0 e-atbu(t)dt e-atCx # (t) - ax(t)D = d dt Ce-atx(t)D = e-atbu(t) x # - ax = bu x # = ax + bu -1(t) = e- At = B 2et - e2t -2et + 2e2t et - e2t -et + 2e2tR -1(t) = (-t), = B 2e-t - e-2t -2e-t + 2e-2t e-t - e-2t -e-t + 2e-2tR (t) = eAt = l-1C(s I - A)-1D Section 9–4 / Solving the Time-Invariant State Equation 667 Integrating the preceding equation between 0 and t gives or (9–41) Equation (9–41) can also be written as (9–42) where Equation (9–41) or (9–42) is the solution of Equation (9–40). The solution x(t) is clearly the sum of a term consisting of the transition of the initial state and a term arising from the input vector.
Laplace Transform Approach to the Solution of Nonhomogeneous State Equations.
The solution of the nonhomogeneous state equation can also be obtained by the Laplace transform approach.The Laplace transform of this last equation yields sX(s)-x(0)=AX(s)+BU(s) or (sI-A)X(s)=x(0)+BU(s) Premultiplying both sides of this last equation by (sI-A)–1, we obtain X(s)=(sI-A)–1x(0)+(sI-A)–1BU(s) Using the relationship given by Equation (9–36) gives X(s)=lCeAtD x(0)+lCeAtD BU(s) The inverse Laplace transform of this last equation can be obtained by use of the convolution integral as follows: Solution in Terms of Thus far we have assumed the initial time to be zero.
If, however, the initial time is given by t0 instead of 0, then the solution to Equation (9–40) must be modified to (9–43) x(t) = eAAt-t0B xAt0B + 3 t t0 eA(t-t) Bu(t)dt xAt0B.
x(t) = eAt x(0) + 3 t 0 eA(t-t) Bu(t)dt x # = Ax + Bu (t) = eAt.
x(t) = (t) x(0) + 3 t 0 (t - t) Bu(t)dt x(t) = eAt x(0) + 3 t 0 eA(t-t) Bu(t)dt e- At x(t) - x(0) = 3 t 0 e- At Bu(t)dt 668 Chapter 9 / Control Systems Analysis in State Space EXAMPLE 9–6 Obtain the time response of the following system: where u(t) is the unit-step function occurring at t=0, or u(t)=1(t) For this system, The state-transition matrix was obtained in Example 9–5 as The response to the unit-step input is then obtained as or If the initial state is zero, or x(0)=0, then x(t) can be simplified to 9–5 SOME USEFUL RESULTS IN VECTOR-MATRIX ANALYSIS In this section we present some useful results in vector-matrix analysis that we use in Section 9–6. Specifically, we present the Cayley–Hamilton theorem, the minimal poly-nomial,Sylvester’s interpolation method for calculating and the linear independence of vectors.
Cayley–Hamilton Theorem.
The Cayley–Hamilton theorem is very useful in proving theorems involving matrix equations or solving problems involving matrix equations.
Consider an nn matrix A and its characteristic equation: |lI-A|=ln+a1ln–1+p+an–1l+an=0 The Cayley–Hamilton theorem states that the matrix A satisfies its own characteristic equation, or that An+a1 An–1+p+an–1 A+an I=0 (9–44) To prove this theorem, note that is a polynomial in l of degree n-1.
That is, adj(l I - A) = B1 ln-1 + B2 ln-2 + p + Bn-1 l + Bn adj(l I - A) eAt, B x1(t) x2(t)R = C 1 2 - e-t + 1 2 e-2t e-t - e-2t S Bx1(t) x2(t)R = B 2e-t - e-2t -2e-t + 2e-2t e-t - e-2t -e-t + 2e-2tR B x1(0) x2(0)R + B 1 2 - e-t + 1 2 e-2t e-t - e-2t R x(t) = eAt x(0) + 3 t 0 B 2e-(t-t) - e-2(t-t) -2e-(t-t) + 2e-2(t-t) e-(t-t) - e-2(t-t) -e-(t-t) + 2e-2(t-t)R B 0 1R dt (t) = eAt = B 2e-t - e-2t -2e-t + 2e-2t e-t - e-2t -e-t + 2e-2tR (t) = eAt A = B 0 -2 1 -3R , B = B 0 1R B x # 1 x # 2 R = B 0 -2 1 -3R Bx1 x2 R + B 0 1R u Section 9–5 / Some Useful Results in Vector-Matrix Analysis 669 where Since (lI-A) adj(lI-A)=Cadj(lI-A)D(lI-A)=|lI-A|I we obtain From this equation, we see that A and (i=1, 2, p , n) commute. Hence, the product of (lI-A) and becomes zero if either of these is zero. If A is substitut-ed for l in this last equation, then clearly lI-A becomes zero. Hence, we obtain An+a1 An–1+p+an–1 A+an I=0 This proves the Cayley–Hamilton theorem, or Equation (9–44).
Minimal Polynomial.
Referring to the Cayley–Hamilton theorem, every nn matrix A satisfies its own characteristic equation. The characteristic equation is not, however, necessarily the scalar equation of least degree that A satisfies.The least-degree polynomial having A as a root is called the minimal polynomial. That is, the minimal polynomial of an nn matrix A is defined as the polynomial of least degree, f(l)=lm+a1lm–1+p+am–1l+am, m n such that or f(A)=Am+a1 Am–1+p+am–1 A+am I=0 The minimal polynomial plays an important role in the computation of polynomials in an nn matrix.
Let us suppose that a polynomial in l, is the greatest common divisor of all the elements of We can show that if the coefficient of the highest-degree term in l of is chosen as 1, then the minimal polynomial is given by (9–45) [See Problem A–9–8 for the derivation of Equation (9–45).] It is noted that the minimal polynomial of an nn matrix A can be determined by the following procedure: 1. Form and write the elements of as factored polynomials in l.
2. Determine as the greatest common divisor of all the elements of Choose the coefficient of the highest-degree term in l of to be 1.If there is no common divisor, 3. The minimal polynomial is then given as divided by Matrix Exponential In solving control engineering problems, it often becomes necessary to compute If matrix A is given with all elements in numerical values, MATLAB provides a simple way to compute , where T is a constant.
eAT eAt.
eAt.
d(l).
∑l I - A∑ f(l) d(l) = 1.
d(l) adj(l I - A).
d(l) adj(l I - A) adj(l I - A) f(l) f(l) = ∑l I - A∑ d(l) f(l) d(l) adj(l I - A).
d(l), f(A) = 0, f(l) adj(l I - A) Bi = AB1 ln-1 + B2 ln-2 + p + Bn-1 l + BnB(l I - A) = (l I - A)AB1 ln-1 + B2 ln-2 + p + Bn-1 l + BnB ∑l I - A∑ I = Iln + a1 Iln-1 + p + an-1 Il + an I B1 = I.
670 Chapter 9 / Control Systems Analysis in State Space Aside from computational methods, several analytical methods are available for the computation of We shall present three methods here.
Computation of eAt: Method 1.
If matrix A can be transformed into a diagonal form, then can be given by (9–46) where P is a diagonalizing matrix for A. [For the derivation of Equation (9–46), see Problem A–9–11.] If matrix A can be transformed into a Jordan canonical form,then can be given by where S is a transformation matrix that transforms matrix A into a Jordan canonical form J.
As an example, consider the following matrix A: The characteristic equation is |lI-A|=l3-3l2+3l-1=(l-1)3=0 Thus,matrix A has a multiple eigenvalue of order 3 at It can be shown that matrix A has a multiple eigenvector of order 3.The transformation matrix that will transform matrix A into a Jordan canonical form can be given by The inverse of matrix S is Then it can be seen that = C 1 0 0 1 1 0 0 1 1 S = J S-1 AS = C 1 -1 1 0 1 -2 0 0 1 S C 0 0 1 1 0 -3 0 1 3 S C 1 1 1 0 1 2 0 0 1 S S-1 = C 1 -1 1 0 1 -2 0 0 1 S S = C 1 1 1 0 1 2 0 0 1 S l = 1.
A = C 0 0 1 1 0 -3 0 1 3 S eAt = SeJt S-1 eAt eAt = PeDt P-1 = P F el1 t 0 el2 t 0 eln t V P-1 eAt eAt.
Section 9–5 / Some Useful Results in Vector-Matrix Analysis 671 Noting that we find Computation of eAt: Method 2.
The second method of computing uses the Laplace transform approach. Referring to Equation (9–36), can be given as follows: Thus, to obtain first invert the matrix This results in a matrix whose elements are rational functions of s. Then take the inverse Laplace transform of each element of the matrix.
EXAMPLE 9–7 Consider the following matrix A: Compute by use of the two analytical methods presented previously.
Method 1.
The eigenvalues of A are 0 and –2 A necessary transformation matrix P may be obtained as Then, from Equation (9–46), is obtained as follows: Method 2.
Since we obtain (s I - A)-1 = D 1 s 0 1 s(s + 2) 1 s + 2 T s I - A = B s 0 0 s R - B 0 0 1 -2R = B s 0 -1 s + 2R eAt = B 1 0 1 -2R B e0 0 0 e-2tR B1 0 1 2 - 1 2 R = B 1 0 1 2 A1 - e-2tB e-2t R eAt P = B 1 0 1 -2R Al1 = 0, l2 = -2B.
eAt A = B 0 0 1 -2R (s I - A).
eAt, eAt = l-1C(s I - A)-1D eAt eAt = C et - tet + 1 2 t2et 1 2 t2et tet + 1 2 t2et tet - t2et et - tet - t2et -3tet - t2et 1 2 t2et tet + 1 2 t2et et + 2tet + 1 2 t2et S = C 1 1 1 0 1 2 0 0 1 S C et 0 0 tet et 0 1 2 t2et tet et S C 1 -1 1 0 1 -2 0 0 1 S eAt = SeJt S-1 eJt = C et 0 0 tet et 0 1 2 t2et tet et S 672 Chapter 9 / Control Systems Analysis in State Space Hence, Computation of eAt: Method 3.
The third method is based on Sylvester’s interpo-lation method.(For Sylvester’s interpolation formula,see Problem A–9–12.) We shall first consider the case where the roots of the minimal polynomial of A are distinct.
Then we shall deal with the case of multiple roots.
Case 1: Minimal Polynomial of A Involves Only Distinct Roots.
We shall assume that the degree of the minimal polynomial of A is m. By using Sylvester’s interpolation formula, it can be shown that can be obtained by solving the following determinant equation: (9–47) By solving Equation (9–47) for can be obtained in terms of the Ak (k=0, 1, 2, p , m-1) and the (i=1, 2, 3, p , m). [Equation (9–47) may be expanded, for ex-ample, about the last column.] Notice that solving Equation (9–47) for is the same as writing (9–48) and determining the (k=0, 1, 2, p , m-1) by solving the following set of m equations for the If A is an nn matrix and has distinct eigenvalues, then the number of to be determined is m=n. If A involves multiple eigenvalues, but its minimal polynomial has only simple roots, however, then the number m of to be determined is less than n.
Case 2: Minimal Polynomial of A Involves Multiple Roots.
As an example, consider the case where the minimal polynomial of A involves three equal roots and has other roots that are all distinct. By applying Sylvester’s interpolation formula, it can be shown that can be obtained from the following determinant equation: eAt Al4 , l5 , p , lmB Al1 = l2 = l3B ak(t)’s ak(t)’s a0(t) + a1(t)lm + a2(t)lm 2 + p + am-1(t)lm m-1 = elm t a0(t) + a1(t)l2 + a2(t)l2 2 + p + am-1(t)l2 m-1 = el2 t a0(t) + a1(t)l1 + a2(t)l1 2 + p + am-1(t)l1 m-1 = el1 t ak(t): ak(t) eAt = a0(t) I + a1(t) A + a2(t) A 2 + p + am-1(t) A m-1 eAt eli t eAt eAt, 7 1 1 1 I l1 l2 lm A l1 2 l2 2 lm 2 A 2 p p p p l1 m-1 l2 m-1 lm m-1 A m-1 el1 t el2 t elm t eAt 7 = 0 eAt f(l) eAt = l-1C(s I - A)-1D = B 1 0 1 2 A1 - e-2tB e-2t R Section 9–5 / Some Useful Results in Vector-Matrix Analysis 673 =0 (9–49) Equation (9–49) can be solved for by expanding it about the last column.
It is noted that, just as in case 1, solving Equation (9–49) for is the same as writing (9–50) and determining the ak(t)’s (k=0, 1, 2, p , m-1) from The extension to other cases where, for example, there are two or more sets of multiple roots will be apparent.Note that if the minimal polynomial of A is not found,it is possible to substitute the characteristic polynomial for the minimal polynomial.The number of computations may, of course, be increased.
EXAMPLE 9–8 Consider the matrix Compute using Sylvester’s interpolation formula.
From Equation (9–47), we get 3 1 1 I l1 l2 A el1 t el2 t eAt 3 = 0 eAt A = B 0 0 1 -2R a0(t) + a1(t)lm + a2(t)lm 2 + p + am-1(t)lm m-1 = elm t a0(t) + a1(t)l4 + a2(t)l4 2 + p + am-1(t)l4 m-1 = el4 t a0(t) + a1(t)l1 + a2(t)l1 2 + p + am-1(t)l1 m-1 = el1 t a1(t) + 2a2(t)l1 + 3a3(t)l1 2 + p + (m - 1)am-1(t)l1 m-2 = tel1 t a2(t) + 3a3(t)l1 + p + (m - 1)(m - 2) 2 am-1(t)l1 m-3 = t2 2 el1 t eAt = a0(t) I + a1(t) A + a2(t) A 2 + p + am-1(t) A m-1 eAt eAt (m - 1)(m - 2) 2 l1 m-3 (m - 1)l1 m-2 l1 m-1 l4 m-1 lm m-1 A m-1 t2 2 el1 t tel1 t el1 t el4 t elmt eAt 0 0 1 1 1 I 0 1 l1 l4 lm A 1 2l1 l1 2 l4 2 lm 2 A 2 3l1 3l1 2 l1 3 l4 3 lm 3 A 3 p p p p p p p p p 674 Chapter 9 / Control Systems Analysis in State Space Substituting 0 for l1 and –2 for l2 in this last equation, we obtain Expanding the determinant, we obtain or An alternative approach is to use Equation (9–48).We first determine a0(t) and a1(t) from Since l1=0 and l2=–2, the last two equations become Solving for a0(t) and a1(t) gives Then can be written as Linear Independence of Vectors.
The vectors x1, x2, p , xn are said to be linearly independent if where c1, c2, p , cn are constants, implies that Conversely, the vectors x1, x2, p , xn are said to be linearly dependent if and only if xi can be expressed as a linear combination of xj (j=1, 2, p , n; j Z i), or xi = a n j=1 jZi cj xj c1 = c2 = p = cn = 0 c1 x1 + c2 x2 + p + cn xn = 0 eAt = a0(t) I + a1(t) A = I + 1 2 A1 - e-2tB A = B1 0 1 2 A1 - e-2tB e-2t R eAt a0(t) = 1, a1(t) = 1 2 A1 - e-2tB a0(t) - 2a1(t) = e-2t a0(t) = 1 a0(t) + a1(t)l2 = el2 t a0(t) + a1(t)l1 = el1 t = B 1 0 1 2 A1 - e-2tB e-2t R = 1 2 b B0 0 1 -2R + B 2 0 0 2R - B 0 0 1 -2R e-2t r eAt = 1 2 AA + 2 I - Ae-2tB -2eAt + A + 2 I - Ae-2t = 0 3 1 1 I 0 -2 A 1 e-2t eAt 3 = 0 Section 9–6 / Controllability 675 for some set of constants cj.This means that if xi can be expressed as a linear combination of the other vectors in the set,it is linearly dependent on them or it is not an independent member of the set.
EXAMPLE 9–9 The vectors are linearly dependent since The vectors are linearly independent since implies that Note that if an nn matrix is nonsingular (that is, the matrix is of rank n or the determinant is nonzero) then n column (or row) vectors are linearly independent.If the nn matrix is singular (that is, the rank of the matrix is less than n or the determinant is zero), then n column (or row) vectors are linearly dependent.To demonstrate this, notice that 9–6 CONTROLLABILITY Controllability and Observability.
A system is said to be controllable at time t0 if it is possible by means of an unconstrained control vector to transfer the system from any initial state x(t0) to any other state in a finite interval of time.
A system is said to be observable at time t0 if,with the system in state x(t0), it is possible to determine this state from the observation of the output over a finite time interval.
The concepts of controllability and observability were introduced by Kalman.They play an important role in the design of control systems in state space. In fact, the conditions of controllability and observability may govern the existence of a complete solution to the control system design problem. The solution to this problem may not Cy1 y2 y3D = C 1 2 3 1 0 1 2 2 2 S = nonsingular Cx1 x2 x3D = C 1 2 3 1 0 1 2 2 4 S = singular c1 = c2 = c3 = 0 c1 y1 + c2 y2 + c3 y3 = 0 y1 = C 1 2 3 S , y2 = C 1 0 1 S , y3 = C 2 2 2 S x1 + x2 - x3 = 0 x1 = C 1 2 3 S , x2 = C 1 0 1 S , x3 = C 2 2 4 S 676 Chapter 9 / Control Systems Analysis in State Space exist if the system considered is not controllable. Although most physical systems are controllable and observable, corresponding mathematical models may not possess the property of controllability and observability.Then it is necessary to know the conditions under which a system is controllable and observable.This section deals with controlla-bility and the next section discusses observability.
In what follows, we shall first derive the condition for complete state controllability.
Then we derive alternative forms of the condition for complete state controllability followed by discussions of complete output controllability.Finally,we present the concept of stabilizability.
Complete State Controllability of Continuous-Time Systems.
Consider the continuous-time system.
(9–51) where The system described by Equation (9–51) is said to be state controllable at t=t0 if it is possible to construct an unconstrained control signal that will transfer an initial state to any final state in a finite time interval If every state is controllable, then the system is said to be completely state controllable.
We shall now derive the condition for complete state controllability.Without loss of generality, we can assume that the final state is the origin of the state space and that the initial time is zero, or t0=0.
The solution of Equation (9–51) is Applying the definition of complete state controllability just given, we have or (9–52) Referring to Equation (9–48) or (9–50), can be written (9–53) Substituting Equation (9–53) into Equation (9–52) gives (9–54) x(0) = - a n-1 k=0 A k B 3 t1 0 ak(t)u(t)dt e-At = a n-1 k=0 ak(t) A k e-At x(0) = -3 t1 0 e-At Bu(t)dt xAt1B = 0 = eAt1 x(0) + 3 t1 0 eA(t1-t) Bu(t)dt x(t) = eAt x(0) + 3 t 0 eA(t-t) Bu(t)dt t0 t t1 . B = n 1 matrix A = n n matrix u = control signal (scalar) x = state vector (n-vector) x # = Ax + Bu Section 9–6 / Controllability 677 Let us put Then Equation (9–54) becomes (9–55) If the system is completely state controllable, then, given any initial state x(0), Equation (9–55) must be satisfied.This requires that the rank of the nn matrix be n.
From this analysis,we can state the condition for complete state controllability as fol-lows: The system given by Equation (9–51) is completely state controllable if and only if the vectors are linearly independent, or the nn matrix is of rank n.
The result just obtained can be extended to the case where the control vector u is r-dimensional. If the system is described by where u is an r-vector, then it can be proved that the condition for complete state controllability is that the nnr matrix be of rank n, or contain n linearly independent column vectors.The matrix is commonly called the controllability matrix.
EXAMPLE 9–10 Consider the system given by Since the system is not completely state controllable.
CB ABD = B 1 0 1 0R = singular B x # 1 x # 2 R = B 1 0 1 -1R B x1 x2 R + B 1 0R u CB AB p A n-1 BD CB AB p A n-1 BD x # = Ax + Bu CB AB p A n-1 BD B, AB, p , A n-1 B CB AB p A n-1 BD = - CB AB p A n-1 BD F b0 b1 bn-1 V x(0) = - a n-1 k=0 A k Bbk 3 t1 0 ak(t)u(t)dt = bk 678 Chapter 9 / Control Systems Analysis in State Space EXAMPLE 9–11 Consider the system given by For this case, The system is therefore completely state controllable.
Alternative Form of the Condition for Complete State Controllability.
Consider the system defined by (9–56) where If the eigenvectors of A are distinct, then it is possible to find a transformation matrix P such that Note that if the eigenvalues of A are distinct,then the eigenvectors of A are distinct;how-ever, the converse is not true. For example, an nn real symmetric matrix having multiple eigenvalues has n distinct eigenvectors. Note also that each column of the P matrix is an eigenvector of A associated with Let us define (9–57) Substituting Equation (9–57) into Equation (9–56), we obtain (9–58) By defining P-1 B = F = Af ijB z # = P-1 APz + P-1 Bu x = Pz li (i = 1, 2, p , n).
P-1 AP = D = F l1 0 l2 0 ln V B = n r matrix A = n n matrix u = control vector (r-vector) x = state vector (n-vector) x # = Ax + Bu CB ABD = B 0 1 1 -1R = nonsingular B x # 1 x # 2 R = B 1 2 1 -1R B x1 x2 R + B 0 1R[u] Section 9–6 / Controllability 679 we can rewrite Equation (9–58) as If the elements of any one row of the nr matrix F are all zero, then the corresponding state variable cannot be controlled by any of the ui. Hence, the condition of complete state controllability is that if the eigenvectors of A are distinct, then the system is com-pletely state controllable if and only if no row of has all zero elements. It is im-portant to note that, to apply this condition for complete state controllability, we must put the matrix in Equation (9–58) in diagonal form.
If the A matrix in Equation (9–56) does not possess distinct eigenvectors, then diagonalization is impossible.In such a case,we may transform A into a Jordan canonical form. If, for example, A has eigenvalues and has n-3 distinct eigenvectors, then the Jordan canonical form of A is The square submatrices on the main diagonal are called Jordan blocks.
Suppose that we can find a transformation matrix S such that If we define a new state vector z by (9–59) then substitution of Equation (9–59) into Equation (9–56) yields (9–60) The condition for complete state controllability of the system of Equation (9–56) may then be stated as follows: The system is completely state controllable if and only if (1) = Jz + S-1 Bu z # = S-1 ASz + S-1 Bu x = Sz S-1 AS = J J = I l1 0 0 0 1 l1 0 0 1 l1 l4 0 1 l4 l6 0 ln Y l1 ,l1 ,l1 ,l4 ,l4 ,l6 , p , ln P-1 AP P-1 B z # n = ln zn + f n1 u1 + f n2 u2 + p + f nr ur z # 2 = l2 z2 + f 21 u1 + f 22 u2 + p + f 2r ur z # 1 = l1 z1 + f 11 u1 + f 12 u2 + p + f 1r ur 680 Chapter 9 / Control Systems Analysis in State Space no two Jordan blocks in J of Equation (9–60) are associated with the same eigenvalues, (2) the elements of any row of that correspond to the last row of each Jordan block are not all zero, and (3) the elements of each row of that correspond to distinct eigenvalues are not all zero.
EXAMPLE 9–12 The following systems are completely state controllable: The following systems are not completely state controllable: Condition for Complete State Controllability in the s Plane.
The condition for complete state controllability can be stated in terms of transfer functions or transfer matrices.
It can be proved that a necessary and sufficient condition for complete state con-trollability is that no cancellation occur in the transfer function or transfer matrix. If cancellation occurs, the system cannot be controlled in the direction of the canceled mode.
EXAMPLE 9–13 Consider the following transfer function: Clearly, cancellation of the factor (s+2.5) occurs in the numerator and denominator of this transfer function. (Thus one degree of freedom is lost.) Because of this cancellation, this system is not completely state controllable.
X(s) U(s) = s + 2.5 (s + 2.5)(s - 1) E x # 1 x # 2 x # 3 x # 4 x # 5 U = E -2 0 0 0 1 -2 0 0 1 -2 -5 0 0 1 -5 U E x1 x2 x3 x4 x5 U + E 4 2 1 3 0 U u C x # 1 x # 2 x # 3 S = C -1 0 0 1 -1 0 0 0 -2 S C x1 x2 x3 S + C 4 0 3 2 0 0 S B u1 u2 R B x # 1 x # 2 R = B -1 0 0 -2R B x1 x2 R + B 2 0R u E x # 1 x # 2 x # 3 x # 4 x # 5 U = E -2 0 0 0 1 -2 0 0 1 -2 -5 0 0 1 -5 U E x1 x2 x3 x4 x5 U + E 0 0 3 0 2 1 0 0 0 1 U B u1 u2 R C x # 1 x # 2 x # 3 S = C -1 0 0 1 -1 0 0 0 -2 S C x1 x2 x3 S + C 0 4 3 Su Bx # 1 x # 2 R = B -1 0 0 -2R Bx1 x2 R + B 2 5R u S-1 B S-1 B Section 9–6 / Controllability 681 The same conclusion can be obtained by writing this transfer function in the form of a state equation.A state-space representation is Since the rank of the matrix is 1.Therefore, we arrive at the same conclusion:The system is not completely state controllable.
Output Controllability.
In the practical design of a control system, we may want to control the output rather than the state of the system. Complete state controllability is neither necessary nor sufficient for controlling the output of the system. For this reason, it is desirable to define separately complete output controllability.
Consider the system described by (9–61) (9–62) where The system described by Equations (9–61) and (9–62) is said to be completely output controllable if it is possible to construct an unconstrained control vector u(t) that will transfer any given initial output yAt0B to any final output yAt1B in a finite time interval t0 t t1.
It can be proved that the condition for complete output controllability is as follows: The system described by Equations (9–61) and (9–62) is completely output controllable if and only if the m(n+1)r matrix is of rank m. (For a proof, see Problem A–9–16.) Note that the presence of the Du term in Equation (9–62) always helps to establish output controllability.
Uncontrollable System.
An uncontrollable system has a subsystem that is physically disconnected from the input.
CCB CAB CA 2 B p CA n-1 B DD D = m r matrix C = m n matrix B = n r matrix A = n n matrix y = output vector (m-vector) u = control vector (r-vector) x = state vector (n-vector) y = Cx + Du x # = Ax + Bu CB ABD CB ABD = B 1 1 1 1R Bx # 1 x # 2 R = B 0 2.5 1 -1.5R B x1 x2 R + B 1 1R u 682 Chapter 9 / Control Systems Analysis in State Space Stabilizability.
For a partially controllable system, if the uncontrollable modes are stable and the unstable modes are controllable, the system is said to be stabilizable. For example, the system defined by is not state controllable.The stable mode that corresponds to the eigenvalue of –1 is not controllable.The unstable mode that corresponds to the eigenvalue of 1 is controllable.
Such a system can be made stable by the use of a suitable feedback.Thus this system is stabilizable.
9–7 OBSERVABILITY In this section we discuss the observability of linear systems. Consider the unforced system described by the following equations: (9–63) (9–64) where The system is said to be completely observable if every state xAt0B can be determined from the observation of y(t) over a finite time interval, The system is, there-fore, completely observable if every transition of the state eventually affects every ele-ment of the output vector.The concept of observability is useful in solving the problem of reconstructing unmeasurable state variables from measurable variables in the mini-mum possible length of time. In this section we treat only linear, time-invariant systems.
Therefore, without loss of generality, we can assume that t0=0.
The concept of observability is very important because, in practice, the difficulty encountered with state feedback control is that some of the state variables are not accessible for direct measurement, with the result that it becomes necessary to estimate the unmeasurable state variables in order to construct the control signals. It will be shown in Section 10–5 that such estimates of state variables are possible if and only if the system is completely observable.
In discussing observability conditions, we consider the unforced system as given by Equations (9–63) and (9–64).The reason for this is as follows: If the system is described by then x(t) = eAt x(0) + 3 t 0 eA(t-t) Bu(t)dt y = Cx + Du x # = Ax + Bu t0 t t1 . C = m n matrix A = n n matrix y = output vector (m-vector) x = state vector (n-vector) y = Cx x # = Ax B x # 1 x # 2 R = B 1 0 0 -1R B x1 x2 R + B1 0R u Section 9–7 / Observability 683 and y(t) is Since the matrices A, B, C, and D are known and u(t) is also known, the last two terms on the right-hand side of this last equation are known quantities. Therefore, they may be subtracted from the observed value of y(t). Hence, for investigating a necessary and sufficient condition for complete observability,it suffices to consider the system described by Equations (9–63) and (9–64).
Complete Observability of Continuous-Time Systems.
Consider the system described by Equations (9–63) and (9–64).The output vector y(t) is Referring to Equation (9–48) or (9–50), we have where n is the degree of the characteristic polynomial. [Note that Equations (9–48) and (9–50) with m replaced by n can be derived using the characteristic polynomial.] Hence, we obtain or (9–65) If the system is completely observable, then, given the output y(t) over a time interval x(0) is uniquely determined from Equation (9–65). It can be shown that this requires the rank of the nmn matrix to be n. (See Problem A–9–19 for the derivation of this condition.) From this analysis, we can state the condition for complete observability as follows: The system described by Equations (9–63) and (9–64) is completely observable if and only if the nnm matrix is of rank n or has n linearly independent column vectors. This matrix is called the observability matrix.
CC AC p (A)n-1 CD F C CA CA n-1 V 0 t t1 , y(t) = a0(t) Cx(0) + a1(t) CAx(0) + p + an-1(t) CA n-1 x(0) y(t) = a n-1 k=0 ak(t) CA k x(0) eAt = a n-1 k=0 ak(t) A k y(t) = CeAt x(0) y(t) = CeAt x(0) + C 3 t 0 eA(t-t) Bu(t)dt + Du 684 Chapter 9 / Control Systems Analysis in State Space EXAMPLE 9–14 Consider the system described by Is this system controllable and observable?
Since the rank of the matrix is 2, the system is completely state controllable.
For output controllability, let us find the rank of the matrix Since the rank of this matrix is 1. Hence, the system is completely output controllable.
To test the observability condition, examine the rank of Since the rank of is 2. Hence, the system is completely observable.
Conditions for Complete Observability in the s Plane.
The conditions for com-plete observability can also be stated in terms of transfer functions or transfer matrices.
The necessary and sufficient conditions for complete observability is that no cancella-tion occur in the transfer function or transfer matrix. If cancellation occurs, the canceled mode cannot be observed in the output.
EXAMPLE 9–15 Show that the following system is not completely observable: where Note that the control function u does not affect the complete observability of the system.To examine complete observability, we may simply set u=0. For this system, we have CC AC (A)2 CD = C 4 5 1 -6 -7 -1 6 5 -1 S x = C x1 x2 x3 S , A = C 0 0 -6 1 0 -11 0 1 -6 S , B = C 0 0 1 S , C = [4 5 1] y = Cx x # = Ax + Bu CC ACD CC ACD = B 1 0 1 1R [C AC].
CCB CABD = [0 1] CCB CABD.
CB ABD = B0 1 1 -1R y = [1 0]B x1 x2 R B x # 1 x # 2 R = B 1 -2 1 -1R Bx1 x2 R + B 0 1R u Section 9–7 / Observability 685 Note that Hence, the rank of the matrix is less than 3.Therefore, the system is not completely observable.
In fact, in this system, cancellation occurs in the transfer function of the system.The transfer function between X1(s) and U(s) is and the transfer function between Y(s) and X1(s) is Therefore, the transfer function between the output Y(s) and the input U(s) is Clearly, the two factors (s+1) cancel each other.This means that there are nonzero initial states x(0), which cannot be determined from the measurement of y(t).
Comments.
The transfer function has no cancellation if and only if the system is com-pletely state controllable and completely observable.This means that the canceled transfer function does not carry along all the information characterizing the dynamic system.
Alternative Form of the Condition for Complete Observability.
Consider the system described by Equations (9–63) and (9–64), rewritten (9–66) (9–67) Suppose that the transformation matrix P transforms A into a diagonal matrix, or where D is a diagonal matrix. Let us define Then Equations (9–66) and (9–67) can be written Hence, y(t) = CPeDt z(0) y = CPz z # = P-1 APz = Dz x = Pz P-1 AP = D y = Cx x # = Ax Y(s) U(s) = (s + 1)(s + 4) (s + 1)(s + 2)(s + 3) Y(s) X 1(s) = (s + 1)(s + 4) X 1(s) U(s) = 1 (s + 1)(s + 2)(s + 3) CC AC (A)2 CD 3 4 5 1 -6 -7 -1 6 5 -1 3 = 0 or The system is completely observable if none of the columns of the mn matrix CP consists of all zero elements.This is because, if the ith column of CP consists of all zero elements, then the state variable zi(0) will not appear in the output equation and there-fore cannot be determined from observation of y(t). Thus, x(0), which is related to z(0) by the nonsingular matrix P,cannot be determined.(Remember that this test applies only if the matrix is in diagonal form.) If the matrix A cannot be transformed into a diagonal matrix,then by use of a suitable transformation matrix S, we can transform A into a Jordan canonical form, or where J is in the Jordan canonical form.
Let us define Then Equations (9–66) and (9–67) can be written Hence, The system is completely observable if (1) no two Jordan blocks in J are associated with the same eigenvalues, (2) no columns of CS that correspond to the first row of each Jordan block consist of zero elements, and (3) no columns of CS that correspond to distinct eigenvalues consist of zero elements.
To clarify condition (2), in Example 9–16 we have encircled by dashed lines the columns of CS that correspond to the first row of each Jordan block.
EXAMPLE 9–16 The following systems are completely observable. E x # 1 x # 2 x # 3 x # 4 x # 5 U = E 2 0 0 0 1 2 0 0 1 2 -3 0 0 1 -3 U E x1 x2 x3 x4 x5 U , c y1 y2d = B 1 0 1 1 1 1 0 1 0 0R E x1 x2 x3 x4 x5 U C x # 1 x # 2 x # 3 S = C 2 0 0 1 2 0 0 1 2 S C x1 x2 x3 S , c y1 y2d = B 3 4 0 0 0 0R C x1 x2 x3 S B x # 1 x # 2 R = B -1 0 0 -2R Bx1 x2 R , y = [1 3]Bx1 x2 R y(t) = CSeJt z(0) y = CSz z # = S-1 ASz = Jz x = Sz S-1 AS = J P-1 AP y(t) = CPF el1 t 0 el2 t 0 eln t V z(0) = CPF el1 tz1(0) el2 tz2(0) eln tzn(0) V 686 Chapter 9 / Control Systems Analysis in State Space The following systems are not completely observable.
Principle of Duality.
We shall now discuss the relationship between controllability and observability.We shall introduce the principle of duality, due to Kalman, to clarify apparent analogies between controllability and observability.
Consider the system S1 described by where and the dual system S2 defined by where The principle of duality states that the system S1 is completely state controllable (observable) if and only if system S2 is completely observable (state controllable).
To verify this principle, let us write down the necessary and sufficient conditions for complete state controllability and complete observability for systems S1 and S2. C = conjugate transpose of C B = conjugate transpose of B A = conjugate transpose of A n = output vector (r-vector) v = control vector (m-vector) z = state vector (n-vector) n = Bz z # = Az + Cv C = m n matrix B = n r matrix A = n n matrix y = output vector (m-vector) u = control vector (r-vector) x = state vector (n-vector) y = Cx x # = Ax + Bu E x # 1 x # 2 x # 3 x # 4 x # 5 U = E 2 0 0 0 1 2 0 0 1 2 -3 0 0 1 -3 U E x1 x2 x3 x4 x5 U , c y1 y2d = B 1 0 1 1 1 1 0 0 0 0R E x1 x2 x3 x4 x5 U C x # 1 x # 2 x # 3 S = C 2 0 0 1 2 0 0 1 2 S C x1 x2 x3 S , c y1 y2d = B 0 0 1 2 3 4R C x1 x2 x3 S B x # 1 x # 2 R = B -1 0 0 -2R Bx1 x2 R , y = [0 1]Bx1 x2 R Section 9–7 / Observability 687 688 Chapter 9 / Control Systems Analysis in State Space For system S1: 1. A necessary and sufficient condition for complete state controllability is that the rank of the nnr matrix be n.
2. A necessary and sufficient condition for complete observability is that the rank of the nnm matrix be n.
For system S2: 1. A necessary and sufficient condition for complete state controllability is that the rank of the nnm matrix be n.
2. A necessary and sufficient condition for complete observability is that the rank of the nnr matrix be n.
By comparing these conditions, the truth of this principle is apparent. By use of this principle, the observability of a given system can be checked by testing the state con-trollability of its dual.
Detectability.
For a partially observable system, if the unobservable modes are stable and the observable modes are unstable, the system is said to be detectable. Note that the concept of detectability is dual to the concept of stabilizability.
EXAMPLE PROBLEMS AND SOLUTIONS A–9–1.
Consider the transfer function system defined by Equation (9–2), rewritten (9–68) Derive the following controllable canonical form of the state-space representation for this transfer-function system: (9–69) G x # 1 x # 2 x # n-1 x # n W = G 0 0 0 -an 1 0 0 -an-1 0 1 0 -an-2 p p p p 0 0 1 -a1 W G x1 x2 xn-1 xn W + G 0 0 0 1 W u Y(s) U(s) = b0 sn + b1 sn-1 + p + bn-1 s + bn sn + a1 sn-1 + p + an-1 s + an CB AB p A n-1 BD CC AC p (A)n-1 CD CC AC p (A)n-1 CD CB AB p A n-1 BD Example Problems and Solutions 689 (9–70) Solution. Equation (9–68) can be written as which can be modified to (9–71) where Let us rewrite this last equation in the following form: From this last equation, the following two equations may be obtained: (9–72) (9–73) Now define state variables as follows: Then, clearly, sX n-1(s) = X n(s) sX 2(s) = X 3(s) sX 1(s) = X 2(s) X n(s) = sn-1Q(s) X n-1(s) = sn-2Q(s) X 2(s) = sQ(s) X 1(s) = Q(s) + Abn - an b0BQ(s) Y ˆ (s) = Ab1 - a1 b0Bsn-1Q(s) + p + Abn-1 - an-1 b0BsQ(s) snQ(s) = -a1 sn-1Q(s) - p - an-1 sQ(s) - an Q(s) + U(s) = U(s) sn + a1 sn-1 + p + an-1 s + an = Q(s) Y ˆ (s) Ab1 - a1 b0Bsn-1 + p + Abn-1 - an-1 b0Bs + Abn - an b0B Y ˆ (s) = Ab1 - a1 b0Bsn-1 + p + Abn-1 - an-1 b0Bs + Abn - an b0B sn + a1 sn-1 + p + an-1 s + an U(s) Y(s) = b0 U(s) + Y ˆ (s) Y(s) U(s) = b0 + Ab1 - a1 b0Bsn-1 + p + Abn-1 - an-1 b0Bs + Abn - an b0B sn + a1 sn-1 + p + an-1 s + an y = Cbn - an b0 bn-1 - an-1 b0 p b1 - a1 b0D F x1 x2 xn V + b0 u 690 Chapter 9 / Control Systems Analysis in State Space which may be rewritten as (9–74) Noting that we can rewrite Equation (9–72) as or (9–75) Also, from Equations (9–71) and (9–73), we obtain The inverse Laplace transform of this output equation becomes (9–76) Combining Equations (9–74) and (9–75) into one vector–matrix differential equation, we obtain Equation (9–69). Equation (9–76) can be rewritten as given by Equation (9–70). Equations (9–69) and (9–70) are said to be in the controllable canonical form. Figure 9–1 shows the block diagram representation of the system defined by Equations (9–69) and (9–70).
y = Abn - an b0Bx1 + Abn-1 - an-1 b0Bx2 + p + Ab1 - a1 b0Bxn + b0 u + Abn - an b0BX 1(s) = b0 U(s) + Ab1 - a1 b0BX n(s) + p + Abn-1 - an-1 b0BX 2(s) + Abn - an b0BQ(s) Y(s) = b0 U(s) + Ab1 - a1 b0Bsn-1Q(s) + p + Abn-1 - an-1 b0BsQ(s) x # n = -an x1 - an-1 x2 - p - a1 xn + u sX n(s) = -a1 X n(s) - p - an-1 X 2(s) - an X 1(s) + U(s) snQ(s) = sX n(s), x # n-1 = xn x # 2 = x3 x # 1 = x2 b0 y u a1 a2 an–1 an xn–1 xn x1 x2 b1 – a1b0 b2 – a2b0 bn–1 – an–1b0 bn – anb0 ++ ++ ++ ++ ++ ++ ++ + – Figure 9–1 Block diagram representation of the system defined by Equations (9–69) and (9–70) (controllable canonical form).
Example Problems and Solutions 691 A–9–2.
Consider the following transfer-function system: (9–77) Derive the following observable canonical form of the state-space representation for this transfer-function system: (9–78) (9–79) Solution. Equation (9–77) can be modified into the following form: By dividing the entire equation by sn and rearranging, we obtain (9–80) Now define state variables as follows: (9–81) X 1 (s) = 1 s Cbn U(s) - an Y(s)D X 2 (s) = 1 s Cbn-1 U(s) - an-1 Y(s) + X 1 (s)D X n-1 (s) = 1 s Cb2 U(s) - a2 Y(s) + X n-2 (s)D X n (s) = 1 s Cb1 U(s) - a1 Y(s) + X n-1 (s)D + 1 sn-1 Cbn-1 U(s) - an-1 Y(s)D + 1 sn Cbn U(s) - an Y(s)D Y(s) = b0 U(s) + 1 s Cb1 U(s) - a1 Y(s)D + p + sCan-1 Y(s) - bn-1 U(s)D + an Y(s) - bn U(s) = 0 snCY(s) - b0 U(s)D + sn-1Ca1 Y(s) - b1 U(s)D + p y = [0 0 p 0 1]G x1 x2 xn-1 xn W + b0 u F x # 1 x # 2 x # n V = F 0 1 0 0 0 0 p p p 0 0 1 -an -an-1 -a1 V F x1 x2 xn V + F bn - an b0 bn-1 - an-1 b0 b1 - a1 b0 V u Y(s) U(s) = b0 sn + b1 sn-1 + p + bn-1 s + bn sn + a1 sn-1 + p + an-1 s + an 692 Chapter 9 / Control Systems Analysis in State Space Then Equation (9–80) can be written as (9–82) By substituting Equation (9–82) into Equation (9–81) and multiplying both sides of the equations by s, we obtain Taking the inverse Laplace transforms of the preceding n equations and writing them in the reverse order, we get Also, the inverse Laplace transform of Equation (9–82) gives Rewriting the state and output equations in the standard vector-matrix forms gives Equations (9–78) and (9–79). Figure 9–2 shows a block diagram representation of the system defined by Equations (9–78) and (9–79).
y = xn + b0u x # n = xn-1 - a1 xn + Ab1 - a1 b0Bu x # n-1 = xn-2 - a2 xn + Ab2 - a2 b0Bu x # 2 = x1 - an-1 xn + Abn-1 - an-1 b0Bu x # 1 = -an xn + Abn - an b0Bu sX 1(s) = -an X n(s) + Abn - an b0BU(s) sX 2(s) = X 1(s) - an-1 X n(s) + Abn-1 - an-1 b0BU(s) sX n-1(s) = X n-2(s) - a2 X n(s) + Ab2 - a2 b0BU(s) sX n(s) = X n-1(s) - a1 X n(s) + Ab1 - a1 b0BU(s) Y(s) = b0 U(s) + X n(s) y u an–1 a1 an xn–1 x1 x2 xn b0 bn – anb0 bn–1 – an–1b0 b1 – a1b0 + – ++ – ++ – ++ Figure 9–2 Block diagram representation of the system defined by Equations (9–78) and (9–79) (observable canonical form).
Example Problems and Solutions 693 A–9–3.
Consider the transfer-function system defined by (9–83) where Derive the state-space representation of this system in the following diagonal canonical form: (9–84) (9–85) Solution. Equation (9–83) may be written as (9–86) Define the state variables as follows: which may be rewritten as sX n(s) = -pn X n(s) + U(s) sX 2(s) = -p2 X 2(s) + U(s) sX 1(s) = -p1 X 1(s) + U(s) X n(s) = 1 s + pn U(s) X 2(s) = 1 s + p2 U(s) X 1(s) = 1 s + p1 U(s) Y(s) = b0 U(s) + c1 s + p1 U(s) + c2 s + p2 U(s) + p + cn s + pn U(s) y = Cc1 c2 p cnD F x1 x2 xn V + b0 u F x # 1 x # 2 x # n V = F -p1 0 -p2 0 -pn V F x1 x2 xn V + F 1 1 1 V u pi Z pj . = b0 + c1 s + p1 + c2 s + p2 + p + cn s + pn Y(s) U(s) = b0 sn + b1 sn-1 + p + bn-1 s + bn As + p1BAs + p2B p As + pnB 694 Chapter 9 / Control Systems Analysis in State Space The inverse Laplace transforms of these equations give (9–87) These n equations make up a state equation.
In terms of the state variables X1(s), X2(s), p , Xn(s), Equation (9–86) can be written as The inverse Laplace transform of this last equation is (9–88) which is the output equation.
Equation (9–87) can be put in the vector-matrix equation as given by Equation (9–84). Equa-tion (9–88) can be put in the form of Equation (9–85).
Figure 9–3 shows a block diagram representation of the system defined by Equations (9–84) and (9–85).
It is noted that if we choose the state variables as X ˆ n(s) = cn s + pn U(s) X ˆ 2(s) = c2 s + p2 U(s) X ˆ 1(s) = c1 s + p1 U(s) y = c1 x1 + c2 x2 + p + cn xn + b0 u Y(s) = b0 U(s) + c1 X 1(s) + c2 X 2(s) + p + cn X n(s) x # n = -pn xn + u x # 2 = -p2 x2 + u x # 1 = -p1 x1 + u u y xn x2 x1 c2 1 s + p2 c1 b0 cn 1 s + p1 … … 1 s + pn ++ + ++ Figure 9–3 Block diagram representation of the system defined by Equations (9–84) and (9–85) (diagonal canonical form).
Example Problems and Solutions 695 then we get a slightly different state-space representation.This choice of state variables gives from which we obtain (9–89) Referring to Equation (9–86), the output equation becomes from which we get (9–90) Equations (9–89) and (9–90) give the following state-space representation for the system: A–9–4.
Consider the system defined by (9–91) where the system involves a triple pole at s=–p1. (We assume that, except for the first three pi’s being equal, the pi’s are different from one another.) Obtain the Jordan canonical form of the state-space representation for this system.
Y(s) U(s) = b0 sn + b1 sn-1 + p + bn-1 s + bn As + p1B 3As + p4BAs + p5B p As + pnB y = [1 1 p 1]F x ˆ 1 x ˆ 2 x ˆ n V + b0 u F x ˆ # 1 x ˆ # 2 x ˆ # n V = F -p1 0 -p2 0 -pn V F x ˆ 1 x ˆ 2 x ˆ n V + F c1 c2 cn V u y = x ˆ 1 + x ˆ 2 + p + x ˆ n + b0 u Y(s) = b0 U(s) + X ˆ 1(s) + X ˆ 2(s) + p + X ˆ n(s) x ˆ # n = -pn x ˆ n + cn u x ˆ # 2 = -p2 x ˆ 2 + c2 u x ˆ # 1 = -p1 x ˆ 1 + c1 u sX ˆ n(s) = -pn X ˆ n(s) + cn U(s) sX ˆ 2(s) = -p2 X ˆ 2(s) + c2 U(s) sX ˆ 1(s) = -p1 X ˆ 1(s) + c1 U(s) 696 Chapter 9 / Control Systems Analysis in State Space Solution. The partial-fraction expansion of Equation (9–91) becomes which may be written as (9–92) Define Notice that the following relationships exist among X1(s), X2(s), and X3(s): Then, from the preceding definition of the state variables and the preceding relationships, we obtain sX n(s) = -pn X n(s) + U(s) sX 4(s) = -p4 X 4(s) + U(s) sX 3(s) = -p1 X 3(s) + U(s) sX 2(s) = -p1 X 2(s) + X 3(s) sX 1(s) = -p1 X 1(s) + X 2(s) X 2(s) X 3(s) = 1 s + p1 X 1(s) X 2(s) = 1 s + p1 X n(s) = 1 s + pn U(s) X 4(s) = 1 s + p4 U(s) X 3(s) = 1 s + p1 U(s) X 2(s) = 1 As + p1B 2 U(s) X 1(s) = 1 As + p1B 3 U(s) + c3 s + p1 U(s) + c4 s + p4 U(s) + p + cn s + pn U(s) Y(s) = b0 U(s) + c1 As + p1B 3 U(s) + c2 As + p1B 2 U(s) Y(s) U(s) = b0 + c1 As + p1B 3 + c2 As + p1B 2 + c3 s + p1 + c4 s + p4 + p + cn s + pn Example Problems and Solutions 697 The inverse Laplace transforms of the preceding n equations give The output equation, Equation (9–92), can be rewritten as The inverse Laplace transform of this output equation is Thus, the state-space representation of the system for the case when the denominator polynomial involves a triple root –p1 can be given as follows: (9–93) (9–94) The state-space representation in the form given by Equations (9–93) and (9–94) is said to be in the Jordan canonical form. Figure 9–4 shows a block diagram representation of the system given by Equations (9–93) and (9–94).
A–9–5.
Consider the transfer-function system Obtain a state-space representation of this system with MATLAB.
Y(s) U(s) = 25.04s + 5.008 s3 + 5.03247s2 + 25.1026s + 5.008 y = Cc1 c2 p cnD F x1 x2 xn V + b0 u H x # 1 x # 2 x # 3 x # 4 x # n X = H -p1 0 0 0 0 1 -p1 0 p p 0 1 -p1 0 0 0 0 -p4 0 p p 0 0 -pn X H x1 x2 x3 x4 xn X + H 0 0 1 1 1 X u y = c1 x1 + c2 x2 + c3 x3 + c4 x4 + p + cn xn + b0 u Y(s) = b0 U(s) + c1 X 1(s) + c2 X 2(s) + c3 X 3(s) + c4 X 4(s) + p + cn X n(s) x # n = -pn xn + u x # 4 = -p4 x4 + u x # 3 = -p1 x3 + u x # 2 = -p1 x2 + x3 x # 1 = -p1 x1 + x2 698 Chapter 9 / Control Systems Analysis in State Space y u c1 1 s + p1 1 s + p1 1 s + p1 c4 1 s + p4 c2 c3 x3 x4 xn x2 x1 b0 cn … … 1 s + pn ++ ++ + + + ++ Figure 9–4 Block diagram representation of the system defined by Equations (9–93) and (9–94) (Jordan canonical form).
Solution. MATLAB command [A,B,C,D] = tf2ss(num,den) will produce a state-space representation for the system. See MATLAB Program 9–4.
MATLAB Program 9–4 num = [25.04 5.008]; den = [1 5.03247 25.1026 5.008]; [A,B,C,D] = tf2ss(num,den) A = -5.0325 -25.1026 -5.0080 1.0000 0 0 0 1.0000 0 B = 1 0 0 C = 0 25.0400 5.0080 D = 0 Example Problems and Solutions 699 This is the MATLAB representation of the following state-space equations: A–9–6.
Consider the system defined by where Obtain the response of the system to each of the following inputs: (a) The r components of u are impulse functions of various magnitudes.
(b) The r components of u are step functions of various magnitudes.
(c) The r components of u are ramp functions of various magnitudes.
Solution.
(a) Impulse response: Referring to Equation (9–43), the solution to the given state equation is Substituting t0=0– into this solution, we obtain Let us write the impulse input u(t) as where w is a vector whose components are the magnitudes of r impulse functions applied at t=0.The solution of the state equation when the impulse input d(t)w is given at t=0 is (9–95) (b) Step response: Let us write the step input u(t) as u(t) = k = eAt x(0-) + eAt Bw x(t) = eAt x(0-) + 3 t 0-eA(t-t) Bd(t) w dt u(t) = d(t) w x(t) = eAt x(0 -) + 3 t 0-eA(t-t) Bu(t)dt x(t) = eAAt-t0B xAt0B + 3 t t0 eA(t-t) Bu(t)dt B = n r constant matrix A = n n constant matrix u = control vector (r-vector) x = state vector (n-vector) x # = Ax + Bu y = [0 25.04 5.008]C x1 x2 x3 S + u C x # 1 x # 2 x # 3 S = C -5.0325 1 0 -25.1026 0 1 -5.008 0 0 S C x1 x2 x3 S + C 1 0 0 S u 700 Chapter 9 / Control Systems Analysis in State Space where k is a vector whose components are the magnitudes of r step functions applied at t=0.The solution to the step input at t=0 is given by If A is nonsingular, then this last equation can be simplified to give (9–96) (c) Ramp response: Let us write the ramp input u(t) as where v is a vector whose components are magnitudes of ramp functions applied at t=0.The solution to the ramp input tv given at t=0 is If A is nonsingular, then this last equation can be simplified to give (9–97) A–9–7.
Obtain the response y(t) of the following system: where u(t) is the unit-step input occurring at t=0, or u(t)=1(t) Solution. For this system The state transition matrix can be obtained as follows: (t) = eAt = l-1C(s I - A)-1D (t) = eAt A = B -1 1 -0.5 0 R , B = B 0.5 0 R y = [1 0]C x1 x2 S B x # 1 x # 1 R = B -1 1 -0.5 0 R B x1 x2 R + B 0.5 0 R u, Bx1(0) x2(0)R = B 0 0R = eAt x(0) + CA -2AeAt - IB - A -1tD Bv x(t) = eAt x(0) + AA -2BAeAt - I - AtB Bv = eAt x(0) + eAt a I 2 t2 - 2 A 3! t3 + 3 A 2 4! t4 - 4 A 3 5! t5 + p b Bv = eAt x(0) + eAt 3 t 0 e-Att dt Bv x(t) = eAt x(0) + 3 t 0 eA(t-t) Bt v dt u(t) = t v = eAt x(0) + A -1AeAt - IB Bk x(t) = eAt x(0) + eAtC-AA -1BAe-At - IB D Bk = eAt x(0) + eAt aIt - At2 2!
+ A 2t3 3!
- p b Bk = eAt x(0) + eAtc 3 t 0 aI - At + A 2t2 2!
- p b dtd Bk x(t) = eAt x(0) + 3 t 0 eA(t-t) Bk dt Example Problems and Solutions 701 Since we have Since x(0)=0 and k=1, referring to Equation (9–96), we have Hence, the output y(t) can be given by A–9–8.
The Cayley–Hamilton theorem states that every nn matrix A satisfies its own characteristic equation. The characteristic equation is not, however, necessarily the scalar equation of least degree that A satisfies. The least-degree polynomial having A as a root is called the minimal polynomial. That is, the minimal polynomial of an nn matrix A is defined as the polynomial f(l) of least degree, such that or The minimal polynomial plays an important role in the computation of polynomials in an nn matrix.
Let us suppose that d(l), a polynomial in l, is the greatest common divisor of all the elements of adj(lI-A). Show that, if the coefficient of the highest-degree term in l of d(l) is chosen as 1, then the minimal polynomial f(l) is given by Solution. By assumption,the greatest common divisor of the matrix adj(lI-A) is d(l).Therefore, adj(lI-A)=d(l)B(l) f(l) = 2 l I - A d(l) 2 f(A) = A m + a1 A m-1 + p + am-1 A + am I = 0 f(A) = 0, f(l) = lm + a1 lm-1 + p + am-1 l + am , m n y(t) = [1 0]B x1 x2 R = x1 = e-0.5t sin0.5t = B e-0.5t sin 0.5t -e-0.5t(cos0.5t + sin0.5t) + 1R = B 0 -2 1 -2R B 0.5e-0.5t(cos0.5t - sin0.5t) - 0.5 e-0.5t sin0.5t R = A -1AeAt - IB B x(t) = eAt x(0) + A -1AeAt - IB Bk = B e-0.5t(cos0.5t - sin0.5t) 2e-0.5t sin0.5t -e-0.5t sin0.5t e-0.5t(cos0.5t + sin0.5t)R (t) = eAt = l-1C(s I - A)-1D = D s + 0.5 - 0.5 (s + 0.5)2 + 0.52 1 (s + 0.5)2 + 0.52 -0.5 (s + 0.5)2 + 0.52 s + 0.5 + 0.5 (s + 0.5)2 + 0.52 T (s I - A)-1 = B s + 1 -1 0.5 s R -1 = 1 s2 + s + 0.5 B s 1 -0.5 s + 1R 702 Chapter 9 / Control Systems Analysis in State Space where the greatest common divisor of the n2 elements (which are functions of l) of B(l) is unity.
Since (lI-A) adj(lI-A)=|lI-A|I we obtain (9–98) from which we find that is divisible by d(l). Let us put (9–99) Because the coefficient of the highest-degree term in l of d(l) has been chosen to be 1, the coefficient of the highest-degree term in l of c(l) is also 1. From Equations (9–98) and (9–99), we have Hence, Note that c(l) can be written as where a(l) is of lower degree than f(l). Since c(A)=0 and f(A)=0, we must have a(A)=0.
Also, since f(l) is the minimal polynomial, a(l) must be identically zero, or Note that because f(A)=0, we can write Hence, Noting that , we obtain Since the greatest common divisor of the n2 elements of B(l) is unity, we have Therefore, Then, from this last equation and Equation (9–99), we obtain f(l) = ∑l I - A∑ d(l) c(l) = f(l) g(l) = 1 B(l) = g(l) C(l) (l I - A) B(l) = c(l) I c(l) I = g(l)f(l) I = g(l)(l I - A) C(l) f(l) I = (l I - A) C(l) c(l) = g(l)f(l) c(l) = g(l)f(l) + a(l) c(A) = 0 (l I - A) B(l) = c(l) I ∑l I - A∑= d(l)c(l) ∑l I - A∑ d(l)(l I - A) B(l) = ∑l I - A∑ I Example Problems and Solutions 703 A–9–9.
If an nn matrix A has n distinct eigenvalues, then the minimal polynomial of A is identical to the characteristic polynomial.Also, if the multiple eigenvalues of A are linked in a Jordan chain, the minimal polynomial and the characteristic polynomial are identical. If, however, the multiple eigenvalues of A are not linked in a Jordan chain, the minimal polynomial is of lower degree than the characteristic polynomial.
Using the following matrices A and B as examples, verify the foregoing statements about the minimal polynomial when multiple eigenvalues are involved: Solution. First, consider the matrix A.The characteristic polynomial is given by Thus, the eigenvalues of A are 2, 2, and 1. It can be shown that the Jordan canonical form of A is and the multiple eigenvalues are linked in the Jordan chain as shown.
To determine the minimal polynomial, let us first obtain adj(lI-A). It is given by Notice that there is no common divisor of all the elements of Hence, d(l)=1.
Thus, the minimal polynomial f(l) is identical to the characteristic polynomial, or A simple calculation proves that = C 0 0 0 0 0 0 0 0 0 S = 0 = C 8 0 0 72 8 21 28 0 1 S - 5C 4 0 0 16 4 9 12 0 1 S + 8C 2 0 0 1 2 3 4 0 1 S - 4C 1 0 0 0 1 0 0 0 1 S A3 - 5 A2 + 8 A - 4 I = l3 - 5l2 + 8l - 4 f(l) = ∑l I - A∑= (l - 2)2(l - 1) adj(l I - A).
adj(l I - A) = C (l - 2)(l - 1) 0 0 (l + 11) (l - 2)(l - 1) 3(l - 2) 4(l - 2) 0 (l - 2)2 S C 2 0 0 1 2 0 0 0 1 S ∑l I - A∑= 3 l - 2 0 0 -1 l - 2 -3 -4 0 l - 1 3 = (l - 2)2(l - 1) A = C 2 0 0 1 2 3 4 0 1 S , B = C 2 0 0 0 2 3 0 0 1 S 704 Chapter 9 / Control Systems Analysis in State Space but Thus, we have shown that the minimal polynomial and the characteristic polynomial of this matrix A are the same.
Next, consider the matrix B.The characteristic polynomial is given by A simple computation reveals that matrix B has three eigenvectors, and the Jordan canonical form of B is given by Thus, the multiple eigenvalues are not linked.To obtain the minimal polynomial, we first compute : from which it is evident that Hence, As a check, let us compute : For the given matrix B, the degree of the minimal polynomial is lower by 1 than that of the char-acteristic polynomial.As shown here, if the multiple eigenvalues of an nn matrix are not linked in a Jordan chain, the minimal polynomial is of lower degree than the characteristic polynomial.
f(B) = B2 - 3 B + 2 I = C 4 0 0 0 4 9 0 0 1 S - 3C 2 0 0 0 2 3 0 0 1 S + 2C 1 0 0 0 1 0 0 0 1 S = C 0 0 0 0 0 0 0 0 0 S = 0 f(B) f(l) = ∑l I - B∑ d(l) = (l - 2)2(l - 1) l - 2 = l2 - 3l + 2 d(l) = l - 2 adj(l I - B) = C (l - 2)(l - 1) 0 0 0 (l - 2)(l - 1) 3(l - 2) 0 0 (l - 2)2 S adj(l I - B) C 2 0 0 0 2 0 0 0 1 S ∑l I - B∑= 3 l - 2 0 0 0 l - 2 -3 0 0 l - 1 3 = (l - 2)2(l - 1) = C 0 0 0 13 0 0 0 0 0 S Z 0 = C 4 0 0 16 4 9 12 0 1 S - 3C 2 0 0 1 2 3 4 0 1 S + 2C 1 0 0 0 1 0 0 0 1 S A2 - 3 A + 2 I Example Problems and Solutions 705 A–9–10.
Show that by use of the minimal polynomial, the inverse of a nonsingular matrix A can be ex-pressed as a polynomial in A with scalar coefficients as follows: (9–100) where a1, a2, p , am are coefficients of the minimal polynomial Then obtain the inverse of the following matrix A: Solution. For a nonsingular matrix A, its minimal polynomial f(A) can be written as where am Z 0. Hence, Premultiplying by A–1, we obtain which is Equation (9–100).
For the given matrix A, adj(lI-A) can be given as Clearly, there is no common divisor d(l) of all elements of adj(lI-A). Hence, d(l)=1.
Consequently, the minimal polynomial f(l) is given by Thus, the minimal polynomial f(l) is the same as the characteristic polynomial.
Since the characteristic polynomial is we obtain f(l) = l3 + 3l2 - 7l - 17 ∑l I - A∑= l3 + 3l2 - 7l - 17 f(l) = ∑l I - A∑ d(l) = ∑l I - A∑ adj(I - A) = C l2 + 4l + 3 3l + 7 l + 1 2l + 6 l2 + 2l - 3 2 -4 -2l + 2 l2 - 7 S A -1 = - 1 am AA m-1 + a1 A m-2 + p + am-2 A + am-1 IB I = - 1 am AA m + a1 A m-1 + p + am-2 A 2 + am-1 AB f(A) = A m + a1 A m-1 + p + am-1 A + am I = 0 A = C 1 3 1 2 -1 0 0 -2 -3 S f(l) = lm + a1 lm-1 + p + am-1 l + am A -1 = - 1 am AA m-1 + a1 A m-2 + p + am-2 A + am-1 IB 706 Chapter 9 / Control Systems Analysis in State Space By identifying the coefficients ai of the minimal polynomial (which is the same as the characteristic polynomial in this case), we have The inverse of A can then be obtained from Equation (9–100) as follows: A–9–11.
Show that if matrix A can be diagonalized, then where P is a diagonalizing transformation matrix that transforms A into a diagonal matrix, or P–1AP=D, where D is a diagonal matrix.
Show also that if matrix A can be transformed into a Jordan canonical form, then where S is a transformation matrix that transforms A into a Jordan canonical form J,or S–1AS=J.
Solution. Consider the state equation If a square matrix can be diagonalized, then a diagonalizing matrix (transformation matrix) exists and it can be obtained by a standard method. Let P be a diagonalizing matrix for A. Let us define Then where D is a diagonal matrix.The solution of this last equation is Hence, x(t) = Px ˆ(t) = PeDt P-1 x(0) x ˆ(t) = eDt x ˆ(0) x ˆ # = P-1 APx ˆ = Dx ˆ x = Px ˆ x # = Ax eAt = SeJt S-1 eAt = PeDt P-1 = C 3 17 7 17 1 17 6 17 - 3 17 2 17 - 4 17 2 17 - 7 17 S = 1 17 C 3 7 1 6 -3 2 -4 2 -7 S = 1 17 c C 7 -2 -2 0 7 2 -4 8 9 S + 3C 1 3 1 2 -1 0 0 -2 -3 S - 7C 1 0 0 0 1 0 0 0 1 S s A -1 = - 1 a3 AA 2 + a1 A + a2 IB = 1 17 AA 2 + 3 A - 7 IB a1 = 3, a2 = -7, a3 = -17 Example Problems and Solutions 707 Noting that x(t) can also be given by the equation we obtain or (9–101) Next, we shall consider the case where matrix A may be transformed into a Jordan canonical form. Consider again the state equation First obtain a transformation matrix S that will transform matrix A into a Jordan canonical form so that where J is a matrix in a Jordan canonical form. Now define Then The solution of this last equation is Hence, Since the solution x(t) can also be given by the equation we obtain Note that eJt is a triangular matrix [which means that the elements below (or above, as the case may be) the principal diagonal line are zeros] whose elements are elt, telt, , and so forth. For example, if matrix J has the following Jordan canonical form: then eJt = C el1 t 0 0 tel1 t el1 t 0 1 2 t2el1 t tel1 t el1 t S J = C l1 0 0 1 l1 0 0 1 l1 S 1 2 t2elt eAt = SeJt S-1 x(t) = eAt x(0) x(t) = Sx ˆ(t) = SeJt S-1 x(0) x ˆ(t) = eJt x ˆ(0) x ˆ # = S-1 AS x ˆ = Jx ˆ x = Sx ˆ S-1 AS = J x # = Ax eAt = PeDt P-1 = PF el1 t 0 el2 t 0 eln t V P-1 eAt = PeDt P-1, x(t) = eAt x(0) 708 Chapter 9 / Control Systems Analysis in State Space Similarly, if then A–9–12.
Consider the following polynomial in l of degree m-1, where we assume l1, l2, p , lm to be distinct: where k=1, 2, p , m. Notice that Then the polynomial f(l) of degree m-1, takes on the values fAlkB at the points lk. This last equation is commonly called Lagrange’s interpolation formula.The polynomial f(l) of degree m-1 is determined from m independent data fAl1B, fAl2B, p , fAlmB. That is, the polynomial f(l) passes through m points fAl1B, fAl2B, p , fAlmB. Since f(l) is a polynomial of degree m-1, it is uniquely determined.
Any other representations of the polynomial of degree m-1 can be reduced to the Lagrange polynomial f(l). = a m k=1 fAlkB Al - l1B p Al - lk-1BAl - lk+1B p Al - lmB Alk - l1B p Alk - lk-1BAlk - lk+1B p Alk - lmB f(l) = a m k=1 fAlkBpk(l) pkAliB = b1, 0, if i = k if i Z k pk(l) = Al - l1B p Al - lk-1BAl - lk+1B p Al - lmB Alk - l1B p Alk - lk-1BAlk - lk+1B p Alk - lmB eJt = G el1 t 0 0 0 tel1 t el1 t 0 1 2 t2el1 t tel1 t el1 t el4 t 0 tel4 t el4 t el6 t 0 0 0 el7 t W J = G l1 0 0 0 1 l1 0 0 1 l1 l4 0 1 l4 l6 0 l7 W Example Problems and Solutions 709 Assuming that the eigenvalues of an nn matrix A are distinct, substitute A for l in the polynomial pk(l).Then we get Notice that pk(A) is a polynomial in A of degree m-1. Notice also that Now define (9–102) Equation (9–102) is known as Sylvester’s interpolation formula. Equation (9–102) is equivalent to the following equation: (9–103) Equations (9–102) and (9–103) are frequently used for evaluating functions f(A) of matrix A— for example, (lI-A)–1, eAt, and so forth. Note that Equation (9–103) can also be written as (9–104) Show that Equations (9–102) and (9–103) are equivalent.To simplify the arguments, assume that m=4.
7 1 1 1 I l1 l2 lm A l2 1 l2 2 l2 m A 2 p p p p p lm-1 1 lm-1 2 lm-1 m A m-1 fAl1B fAl2B fAlmB f(A) 7 = 0 8 1 l1 l2 1 lm-1 1 fAl1B 1 l2 l2 2 lm-1 2 fAl2B p p p p p 1 lm l2 m lm-1 m fAlmB I A A 2 A m-1 f(A) 8 = 0 = a m k=1 fAlkB AA - l1 IB p AA - lk-1 IBAA - lk+1 IB p AA - lm IB Alk - l1B p Alk - lk-1BAlk - lk+1B p Alk - lmB f(A) = a m k=1 fAlkBpk(A) pkAli IB = bI, 0, if i = k if i Z k pk(A) = AA - l1 IB p AA - lk-1 IBAA - lk+1 IB p AA - lm IB Alk - l1B p Alk - lk-1BAlk - lk+1B p Alk - lmB 710 Chapter 9 / Control Systems Analysis in State Space Solution. Equation (9–103), where m=4, can be expanded as follows: Since and we obtain = 0 + fAl1B CAA - l4 IBAA - l3 IBAA - l2 IBAl4 - l3BAl4 - l2BAl3 - l2B D - fAl2B CAA - l4 IBAA - l3 IBAA - l1 IBAl4 - l3BAl4 - l1BAl3 - l1B D + fAl3B CAA - l4 IBAA - l2 IBAA - l1 IBAl4 - l2BAl4 - l1BAl2 - l1B D - fAl4B CAA - l3 IBAA - l2 IBAA - l1 IBAl3 - l2BAl3 - l1BAl2 - l1B D ¢ = f(A)CAl4 - l3BAl4 - l2BAl4 - l1BAl3 - l2BAl3 - l1BAl2 - l1B D 4 1 li l2 i l3 i 1 lj l2 j l3 j 1 lk l2 k l3 k I A A 2 A 3 4 = AA - lk IBAA - lj IBAA - li IBAlk - ljBAlk - liBAlj - liB 4 1 l1 l2 1 l3 1 1 l2 l2 2 l3 2 1 l3 l2 3 l3 3 1 l4 l2 4 l3 4 4 = Al4 - l3BAl4 - l2BAl4 - l1BAl3 - l2BAl3 - l1BAl2 - l1B + fAl1B 4 1 l2 l2 2 l3 2 1 l3 l2 3 l3 3 1 l4 l2 4 l3 4 I A A 2 A 3 4 + fAl3B 4 1 l1 l2 1 l3 1 1 l2 l2 2 l3 2 1 l4 l2 4 l3 4 I A A 2 A 3 4 - fAl2B 4 1 l1 l2 1 l3 1 1 l3 l2 3 l3 3 1 l4 l2 4 l3 4 I A A 2 A 3 4 = f(A)4 1 l1 l2 1 l3 1 1 l2 l2 2 l3 2 1 l3 l2 3 l3 3 1 l4 l2 4 l3 4 4 - fAl4B 4 1 l1 l2 1 l3 1 1 l2 l2 2 l3 2 1 l3 l2 3 l3 3 I A A 2 A 3 4 ¢ = 5 1 l1 l2 1 l3 1 fAl1B 1 l2 l2 2 l3 2 fAl2B 1 l3 l2 3 l3 3 fAl3B 1 l4 l2 4 l3 4 fAl4B I A A 2 A 3 f(A) 5 Example Problems and Solutions 711 Solving this last equation for f(A), we obtain where m=4.Thus, we have shown the equivalence of Equations (9–102) and (9–103).Although we assumed m=4, the entire argument can be extended to an arbitrary positive integer m. (For the case when the matrix A involves multiple eigenvalues, refer to Problem A–9–13.) A–9–13.
Consider Sylvester’s interpolation formula in the form given by Equation (9–104): This formula for the determination of f(A) applies to the case where the minimal polynomial of A involves only distinct roots.
Suppose that the minimal polynomial of A involves multiple roots. Then the rows in the determinant that correspond to the multiple roots become identical, and therefore modification of the determinant in Equation (9–104) becomes necessary.
Modify the form of Sylvester’s interpolation formula given by Equation (9–104) when the minimal polynomial of A involves multiple roots. In deriving a modified determinant equation, assume that there are three equal roots in the minimal polynomial of A and that there are other roots that are distinct.
Solution. Since the minimal polynomial of A involves three equal roots, the minimal polynomial f(l) can be written as An arbitrary function f(A) of an nn matrix A can be written as where the minimal polynomial f(A) is of degree m and a(A) is a polynomial in A of degree m-1 or less. Hence we have where a(l) is a polynomial in l of degree m-1 or less, which can thus be written as (9–105) a(l) = a0 + a1 l + a2 l2 + p + am-1 lm-1 f(l) = g(l)f(l) + a(l) f(A) = g(A)f(A) + a(A) = Al - l1B 3Al - l4BAl - l5B p Al - lmB f(l) = lm + a1 lm-1 + p + am-1 l + am Al4, l5, p , lmB Al1 = l2 = l3B 7 1 1 1 I l1 l2 lm A l2 1 l2 2 l2 m A 2 p p p p p lm-1 1 lm-1 2 lm-1 m A m-1 fAl1B fAl2B fAlmB f(A) 7 = 0 = a m k=1 fAlkB AA - l1 IB p AA - lk-1 IBAA - lk+1 IB p AA - lm IB Alk - l1B p Alk - lk-1BAlk - lk+1B p Alk - lmB + fAl3B AA - l1 IBAA - l2 IBAA - l4 IB Al3 - l1BAl3 - l2BAl3 - l4B + fAl4B AA - l1 IBAA - l2 IBAA - l3 IB Al4 - l1BAl4 - l2BAl4 - l3B f(A) = fAl1B AA - l2 IBAA - l3 IBAA - l4 IB Al1 - l2BAl1 - l3BAl1 - l4B + fAl2B AA - l1 IBAA - l3 IBAA - l4 IB Al2 - l1BAl2 - l3BAl2 - l4B 712 Chapter 9 / Control Systems Analysis in State Space In the present case we have (9–106) By substituting l1, l4, p , lm for l in Equation (9–106), we obtain the following m-2 equations: (9–107) By differentiating Equation (9–106) with respect to l, we obtain (9–108) where Substitution of l1 for l in Equation (9–108) gives Referring to Equation (9–105), this last equation becomes (9–109) Similarly, differentiating Equation (9–106) twice with respect to l and substituting l1 for l, we obtain This last equation can be written as (9–110) Rewriting Equations (9–110), (9–109), and (9–107), we get (9–111) a0 + a1 lm + a2 l2 m + p + am-1 lm-1 m = fAlmB a0 + a1 l4 + a2 l2 4 + p + am-1 lm-1 4 = fAl4B a0 + a1 l1 + a2 l2 1 + p + am-1 lm-1 1 = fAl1B a1 + 2a2 l1 + p + (m - 1)am-1 lm-2 1 = f¿Al1B a2 + 3a3 l1 + p + (m - 1)(m - 2) 2 am-1 lm-3 1 = f–Al1B 2 f–Al1B = 2a2 + 6a3 l1 + p + (m - 1)(m - 2)am-1 lm-3 1 d2 d2l f(l)2 l=l1 = f–Al1B = d2 dl2 a(l)2 l=l1 f¿Al1B = a1 + 2a2 l1 + p + (m - 1)am-1 lm-2 1 d dl f(l)2 l=l1 = f¿Al1B = d dl a(l)2 l=l1 Al - l1B 2h(l) = d dl Cg(l)Al - l1B 3Al - l4B p Al - lmB D d dl f(l) = Al - l1B 2h(l) + d dl a(l) fAlmB = aAlmB fAl4B = aAl4B fAl1B = aAl1B = g(l)CAl - l1B3Al - l4B p Al - lmB D + a(l) f(l) = g(l)f(l) + a(l) Example Problems and Solutions 713 These m simultaneous equations determine the ak values (where k=0, 1, 2, p , m-1). Noting that f(A)=0 because it is a minimal polynomial, we have f(A) as follows: Hence, referring to Equation (9–105), we have (9–112) where the ak values are given in terms of fAl1B, f¿Al1B, f–Al1B, fAl4B, fAl5B, p , fAlmB. In terms of the determinant equation, f(A) can be obtained by solving the following equation: (9–113) Equation (9–113) shows the desired modification in the form of the determinant. This equation gives the form of Sylvester’s interpolation formula when the minimal polynomial of A involves three equal roots. (The necessary modification of the form of the determinant for other cases will be apparent.) A–9–14.
Using Sylvester’s interpolation formula, compute eAt, where Solution. Referring to Problem A–9–9, the characteristic polynomial and the minimal polynomial are the same for this A.The minimal polynomial (characteristic polynomial) is given by Note that l1=l2=2 and l3=1. Referring to Equation (9–112) and noting that f(A) in this problem is eAt, we have where a0(t), a1(t), and a2(t) are determined from the equations a0(t) + a1(t)l3 + a2(t)l2 3 = el3 t a0(t) + a1(t)l1 + a2(t)l2 1 = el1 t a1(t) + 2a2(t)l1 = tel1 t eAt = a0(t) I + a1(t) A + a2(t) A 2 f(l) = (l - 2)2(l - 1) A = C 2 0 0 1 2 3 4 0 1 S f–Al1B 2 f¿Al1B fAl1B fAl4B fAlmB f(A) = 0 0 0 1 1 1 I 0 1 l1 l4 lm A 1 2l1 l2 1 l2 4 l2 m A 2 3l1 3l2 1 l3 1 l3 4 l3 m A 3 p p p p p p (m - 1)(m - 2) 2 lm-3 1 (m - 1)lm-2 1 lm-1 1 lm-1 4 lm-1 m A m-1 f(A) = a(A) = a0 I + a1 A + a2 A 2 + p + am-1 A m-1 f(A) = g(A)f(A) + a(A) = a(A) 714 Chapter 9 / Control Systems Analysis in State Space Substituting l1=2, and l3=1 into these three equations gives Solving for a0(t), a1(t), and a2(t), we obtain Hence, A–9–15.
Show that the system described by (9–114) (9–115) where is completely output controllable if and only if the composite mnr matrix P, where is of rank m. (Notice that complete state controllability is neither necessary nor sufficient for complete output controllability.) Solution. Suppose that the system is output controllable and the output y(t) starting from any y(0), the initial output, can be transferred to the origin of the output space in a finite time interval 0 t T.That is, (9–116) y(T) = Cx(T) = 0 P = CCB CAB CA 2 B p CA n-1 BD C = m n matrix B = n r matrix A = n n matrix y = output vector (m-vector) (m n) u = control vector (r-vector) x = state vector (n-vector) y = Cx x # = Ax + Bu = C e2 t 0 0 12et - 12e2 t + 13te2 t e2t -3et + 3e2t -4et + 4e2t 0 et S + Aet - e2t + te2tB C 4 0 0 16 4 9 12 0 1 S eAt = A4et - 3e2t + 2te2tB C 1 0 0 0 1 0 0 0 1 S + A-4et + 4e2 t - 3te2 tB C 2 0 0 1 2 3 4 0 1 S a2(t) = et - e2 t + te2 t a1(t) = -4et + 4e2t - 3te2t a0(t) = 4et - 3e2t + 2te2t a0(t) + a1(t) + a2(t) = et a0(t) + 2a1(t) + 4a2(t) = e2t a1(t) + 4a2(t) = te2t Example Problems and Solutions 715 Since the solution of Equation (9–114) is at t=T, we have (9–117) Substituting Equation (9–117) into Equation (9–116), we obtain (9–118) On the other hand, y(0)=Cx(0). Notice that the complete output controllability means that the vector Cx(0) spans the m-dimensional output space. Since eAT is nonsingular, if Cx(0) spans the m-dimensional output space, so does CeATx(0), and vice versa. From Equation (9–118) we obtain Note that can be expressed as the sum of AiBj; that is, where and ai(t) satisfies (p: degree of the minimal polynomial of A) and Bj is the jth column of B.Therefore, we can write CeATx(0) as From this last equation, we see that is a linear combination of CAiBj (i=0, 1, 2, p , p-1; j=1, 2, p , r). Note that if the rank of Q, where is m, then so is the rank of P, and vice versa. [This is obvious if p=n. If p0, the system dynamics can be described by Equations (10–19) and (10–21), or (10–22) We shall design the type 1 servo system such that the closed-loop poles are located at desired positions. The designed system will be an asymptotically stable system, y(q) will approach the constant value r, and u(q) will approach zero. (r is a step input.) Notice that at steady state we have (10–23) Noting that r(t) is a step input, we have r(q)=r(t)=r(constant) for t>0. By subtracting Equation (10–23) from Equation (10–22), we obtain (10–24) Define Then Equation (10–24) becomes (10–25) Equation (10–25) describes the error dynamics.
The design of the type 1 servo system here is converted to the design of an asymp-totically stable regulator system such that e(t) approaches zero, given any initial condi-tion e(0). If the system defined by Equation (10–19) is completely state controllable, then, by specifying the desired eigenvalues m1, m2, p , mn for the matrix A-BK, matrix K can be determined by the pole-placement technique presented in Section 10–2.
The steady-state values of x(t) and u(t) can be found as follows: At steady state (t=q), we have, from Equation (10–22), x # (q) = 0 = (A - BK) x(q) + Bk1 r e # = (A - BK) e x(t) - x(q) = e(t) x # (t) - x # (q) = (A - BK)Cx(t) - x(q)D x # (q) = (A - BK) x(q) + Bk1 r(q) x # = Ax + Bu = (A - BK) x + Bk1 r K = Ck1 k2 p knD = -Kx + k1 r u = - C0 k2 k3 p knD F x1 x2 xn V + k1Ar - x1B Section 10–4 / Design of Servo Systems 741 Since the desired eigenvalues of A-BK are all in the left-half s plane, the inverse of matrix A-BK exists. Consequently, x(q) can be determined as Also, u(q) can be obtained as (See Example 10–4 to verify this last equation.) EXAMPLE 10–4 Design a type 1 servo system when the plant transfer function has an integrator.Assume that the plant transfer function is given by The desired closed-loop poles are and s=–10. Assume that the system configuration is the same as that shown in Figure 10–4 and the reference input r is a step function.
Obtain the unit-step response of the designed system.
Define state variables x1, x2, and x3 as follows: Then the state-space representation of the system becomes (10–26) (10–27) where Referring to Figure 10–4 and noting that n=3, the control signal u is given by (10–28) where The state-feedback gain matrix K can be obtained easily with MATLAB. See MATLAB Program 10–4.
K = Ck1 k2 k3D u = -Ak2 x2 + k3 x3B + k1Ar - x1B = -Kx + k1 r A = C 0 0 0 1 0 -2 0 1 -3 S , B = C 0 0 1 S , C = [1 0 0] y = Cx x # = Ax + Bu x3 = x # 2 x2 = x # 1 x1 = y s = -2 ; j213 Y(s) U(s) = 1 s(s + 1)(s + 2) u(q) = -Kx(q) + k1 r = 0 x(q) = -(A - BK)-1 Bk1 r MATLAB Program 10–4 A = [0 1 0;0 0 1;0 -2 -3]; B = [0;0;1]; J = [-2+j2sqrt(3) -2-j2sqrt(3) -10]; K = acker(A,B,J) K = 160.0000 54.0000 11.0000 742 Chapter 10 / Control Systems Design in State Space The state feedback gain matrix K is thus K = [160 54 11] Unit-Step Response of the Designed System: The unit-step response of the designed system can be obtained as follows: Since from Equation (10–22) the state equation for the designed system is (10–29) and the output equation is (10–30) Solving Equations (10–29) and (10–30) for y(t) when r is a unit-step function gives the unit-step response curve y(t) versus t. MATLAB Program 10–5 yields the unit-step response. The result-ing unit-step response curve is shown in Figure 10–5.
y = [1 0 0]C x1 x2 x3 S C x # 1 x # 2 x # 3 S = C 0 0 -160 1 0 -56 0 1 -14 S C x1 x2 x3 S + C 0 0 160 S r A - BK = C 0 0 0 1 0 -2 0 1 -3 S - C 0 0 1 S [160 54 11] = C 0 0 -160 1 0 -56 0 1 -14 S MATLAB Program 10–5 % ---------- Unit-step response ----------% Enter the state matrix, control matrix, output matrix, % and direct transmission matrix of the designed system AA = [0 1 0;0 0 1;-160 -56 -14]; BB = [0;0;160]; CC = [1 0 0]; DD = ; % Enter step command and plot command t = 0:0.01:5; y = step(AA,BB,CC,DD,1,t); plot(t,y) grid title('Unit-Step Response') xlabel('t Sec') ylabel('Output y') Section 10–4 / Design of Servo Systems 743 Unit-Step Response Output y 0 0.6 1.2 0.8 0.4 0.2 1 t Sec 0 3.5 1 0.5 2.5 5 4 4.5 1.5 2 3 Figure 10–5 Unit-step response curve y(t) versus t for the system designed in Example 10–4.
Note that since we have At steady state the control signal u becomes zero.
Design of Type 1 Servo System when the Plant Has No Integrator.
If the plant has no integrator (type 0 plant), the basic principle of the design of a type 1 servo sys-tem is to insert an integrator in the feedforward path between the error comparator and the plant, as shown in Figure 10–6. (The block diagram of Figure 10–6 is a basic form of the type 1 servo system where the plant has no integrator.) From the diagram, we obtain (10–31) (10–32) (10–33) (10–34) where x = state vector of the plant (n-vector) j # = r - y = r - Cx u = -Kx + kI j y = Cx x # = Ax + Bu = -[160 54 11]C r 0 0 S + 160r = 0 u(q) = -[160 54 11]C x1(q) x2(q) x3(q) S + 160r u(q) = -Kx(q) + k1 r(q) = -Kx(q) + k1 r 744 Chapter 10 / Control Systems Design in State Space We assume that the plant given by Equation (10–31) is completely state controllable.The transfer function of the plant can be given by To avoid the possibility of the inserted integrator being canceled by the zero at the origin of the plant, we assume that Gp(s) has no zero at the origin.
Assume that the reference input (step function) is applied at t=0. Then, for t>0, the system dynamics can be described by an equation that is a combination of Equations (10–31) and (10–34): (10–35) We shall design an asymptotically stable system such that x(q),j(q),and u(q) approach constant values, respectively.Then, at steady state, and we get y(q)=r.
Notice that at steady state we have (10–36) Noting that r(t) is a step input, we have r(q)=r(t)=r (constant) for t>0. By subtracting Equation (10–36) from Equation (10–35), we obtain (10–37) B x # (t) - x # (q) j # (t) - j # (q)R = B A -C 0 0R B x(t) - x(q) j(t) - j(q)R + B B 0 R Cu(t) - u(q)D Bx # (q) j # (q)R = B A -C 0 0R B x(q) j(q)R + B B 0 R u(q) + B 0 1R r(q) j # (t) = 0, B x # (t) j # (t)R = B A -C 0 0R B x(t) j(t)R + B B 0 R u(t) + B 0 1R r(t) Gp(s) = C(s I - A)-1 B C = 1 n constant matrix B = n 1 constant matrix A = n n constant matrix r = reference input signal (step function, scalar) j = output of the integrator (state variable of the system, scalar) y = output signal (scalar) u = control signal (scalar) y K A B kI C x r j .
j u + – + – ++ Figure 10–6 Type 1 servo system.
Section 10–4 / Design of Servo Systems 745 Define Then Equation (10–37) can be written as (10–38) where (10–39) Define a new (n+1)th-order error vector e(t) by -vector Then Equation (10–38) becomes (10–40) where and Equation (10–39) becomes (10–41) where The state error equation can be obtained by substituting Equation (10–41) into Equation (10–40): (10–42) If the desired eigenvalues of matrix (that is, the desired closed-loop poles) are specified as m1, m2, p , mn+1, then the state-feedback gain matrix K and the integral gain constant kI can be determined by the pole-placement technique presented in Section 10–2, provided that the system defined by Equation (10–40) is completely state controllable. Note that if the matrix has rank n+1, then the system defined by Equation (10–40) is completely state controllable. (See Problem A–10–12.) B A -C B 0 R A ˆ - B ˆ K ˆ e # = AA ˆ - B ˆ K ˆ Be K ˆ = CK -kID ue = -K ˆ e A ˆ = B A -C 0 0R , B ˆ = B B 0 R e # = A ˆ e + B ˆ ue e(t) = B xe(t) je(t) R = (n + 1) ue(t) = -Kxe(t) + kI je(t) B x # e(t) j # e(t)R = B A -C 0 0R B xe(t) je(t) R + B B 0 R ue(t) u(t) - u(q) = ue(t) j(t) - j(q) = je(t) x(t) - x(q) = xe(t) 746 Chapter 10 / Control Systems Design in State Space 0 M P z u mg m sin u x x cos u u Figure 10–8 Inverted-pendulum control system.
As is usually the case, not all state variables can be directly measurable. If this is the case, we need to use a state observer. Figure 10–7 shows a block diagram of a type 1 servo system with a state observer. [In the figure, each block with an integral symbol represents an integrator (1/s).] Detailed discussions of state observers are given in Section 10–5.
EXAMPLE 10–5 Consider the inverted-pendulum control system shown in Figure 10–8. In this example, we are concerned only with the motion of the pendulum and motion of the cart in the plane of the page.
It is desired to keep the inverted pendulum upright as much as possible and yet control the position of the cart—for instance, move the cart in a step fashion. To control the position of the cart, we need to build a type 1 servo system. The inverted-pendulum system mounted on a cart does not have an integrator.Therefore, we feed the position signal y (which indicates the po-sition of the cart) back to the input and insert an integrator in the feedforward path, as shown y K A B kI C x r j .
j u Observer + – + – ++ Figure 10–7 Type 1 servo system with state observer.
Section 10–4 / Design of Servo Systems 747 x = Ax + Bu .
y = Cx k1 kI k2 k3 k4 r u x y j j .
+ – + – Figure 10–9 Inverted-pendulum control system. (Type 1 servo system when the plant has no integrator.) in Figure 10–9.We assume that the pendulum angle u and the angular velocity are small, so that and We also assume that the numerical values for M, m, and l are given as Earlier in Example 3–6 we derived the equations for the inverted-pendulum system shown in Figure 3–6, which is the same as that in Figure 10–8. Referring to Figure 3–6, we started with the force-balance and torque-balance equations and ended up with Equations (3–20) and (3–21) to model the inverted-pendulum system. Referring to Equations (3–20) and (3–21), the equations for the inverted-pendulum control system shown in Figure 10–8 are (10–43) (10–44) When the given numerical values are substituted, Equations (10–43) and (10–44) become (10–45) (10–46) Let us define the state variables x1, x2, x3, and x4 as Then,referring to Equations (10–45) and (10–46) and Figure 10–9 and considering the cart position x as the output of the system, we obtain the equations for the system as follows: (10–47) (10–48) (10–49) (10–50) j # = r - y = r - Cx u = -Kx + kI j y = Cx x # = Ax + Bu x4 = x # x3 = x x2 = u # x1 = u x $ = 0.5u - 0.4905u u $ = 20.601u - u Mx $ = u - mgu Mlu $ = (M + m)gu - u M = 2 kg, m = 0.1 kg, l = 0.5 m uu # 2 0.
cosu 1, sinu u, u # 748 Chapter 10 / Control Systems Design in State Space where For the type 1 servo system, we have the state error equation as given by Equation (10–40): (10–51) where and the control signal is given by Equation (10–41): where To obtain a reasonable speed and damping in the response of the designed system (for example, the settling time of approximately 4 ~ 5 sec and the maximum overshoot of 15% ~ 16% in the step response of the cart), let us choose the desired closed-loop poles at s=mi (i=1, 2, 3, 4, 5), where We shall determine the necessary state-feedback gain matrix by the use of MATLAB.
Before we proceed further, we must examine the rank of matrix P, where Matrix P is given by (10–52) The rank of this matrix can be found to be 5.Therefore, the system defined by Equation (10–51) is completely state controllable, and arbitrary pole placement is possible. MATLAB Program 10–6 produces the state feedback gain matrix K ˆ .
P = B A -C B 0R = E 0 20.601 0 -0.4905 0 1 0 0 0 0 0 0 0 0 -1 0 0 1 0 0 0 -1 0 0.5 0 U P = B A -C B 0R m1 = -1 + j13, m2 = -1 - j13, m3 = -5, m4 = -5, m5 = -5 K ˆ = CK -kID = Ck1 k2 k3 k4 -kID ue = -K ˆ e A ˆ = B A -C 0 0R = E 0 20.601 0 -0.4905 0 1 0 0 0 0 0 0 0 0 -1 0 0 1 0 0 0 0 0 0 0 U , B ˆ = B B 0 R = E 0 -1 0 0.5 0 U e # = A ˆ e + B ˆ ue A = D 0 20.601 0 -0.4905 1 0 0 0 0 0 0 0 0 0 1 0 T , B = D 0 -1 0 0.5 T , C = [0 0 1 0] Section 10–4 / Design of Servo Systems 749 MATLAB Program 10–6 A = [0 1 0 0; 20.601 0 0 0; 0 0 0 1; -0.4905 0 0 0]; B = [0;-1;0;0.5]; C = [0 0 1 0]; Ahat = [A zeros(4,1); -C 0]; Bhat = [B;0]; J = [-1+jsqrt(3) -1-jsqrt(3) -5 -5 -5]; Khat = acker(Ahat,Bhat,J) Khat = -157.6336 -35.3733 -56.0652 -36.7466 50.9684 Thus, we get and Unit Step-Response Characteristics of the Designed System.
Once we determine the feed-back gain matrix K and the integral gain constant kI, the step response in the cart position can be obtained by solving the following equation, which is obtained by substituting Equation (10–49) into Equation (10–35): (10–53) The output y(t) of the system is x3(t), or (10–54) Define the state matrix, control matrix, output matrix, and direct transmission matrix of the system given by Equations (10–53) and (10–54) as AA, BB, CC, and DD, respectively. MATLAB Program 10–7 may be used to obtain the step-response curves of the designed system. Notice that, to obtain the unit-step response, we entered the command [y,x,t] = step(AA,BB,CC,DD,1,t) Figure 10–10 shows curves x1 versus t, x2 versus t, x3 (= output y) versus t, x4 versus t, and x5 (= j) versus t. Notice that y(t) C= x3(t)D has approximately 15% overshoot and the settling time is approximately 4.5 sec. j(t) C= x5(t)D approaches 1.1.This result can be derived as follows: Since or D 0 0 0 0 T = D 0 20.601 0 -0.4905 1 0 0 0 0 0 0 0 0 0 1 0 T D 0 0 r 0 T + D 0 -1 0 0.5 Tu(q) x # (q) = 0 = Ax(q) + Bu(q) y = [0 0 1 0 0]B x jR + r B x # j # R = B A - BK - C BkI 0 R B x jR + B 0 1R r kI = -50.9684 K = Ck1 k2 k3 k4D = [-157.6336 -35.3733 -56.0652 -36.7466] 750 Chapter 10 / Control Systems Design in State Space MATLAB Program 10–7 % The following program is to obtain step response % of the inverted-pendulum system just designed A = [0 1 0 0;20.601 0 0 0;0 0 0 1;-0.4905 0 0 0]; B = [0;-1;0;0.5]; C = [0 0 1 0] D = ; K = [-157.6336 -35.3733 -56.0652 -36.7466]; KI = -50.9684; AA = [A - BK BKI;-C 0]; BB = [0;0;0;0;1]; CC = [C 0]; DD = ; % To obtain response curves x1 versus t, x2 versus t, % x3 versus t, x4 versus t, and x5 versus t, separately, enter % the following command t = 0:0.02:6; [y,x,t] = step(AA,BB,CC,DD,1,t); x1 = [1 0 0 0 0]x'; x2 = [0 1 0 0 0]x'; x3 = [0 0 1 0 0]x'; x4 = [0 0 0 1 0]x'; x5 = [0 0 0 0 1]x'; subplot(3,2,1); plot(t,x1); grid title('x1 versus t') xlabel('t Sec'); ylabel('x1') subplot(3,2,2); plot(t,x2); grid title('x2 versus t') xlabel('t Sec'); ylabel('x2') subplot(3,2,3); plot(t,x3); grid title('x3 versus t') xlabel('t Sec'); ylabel('x3') subplot(3,2,4); plot(t,x4); grid title('x4 versus t') xlabel('t Sec'); ylabel('x4') subplot(3,2,5); plot(t,x5); grid title('x5 versus t') xlabel('t Sec'); ylabel('x5') Section 10–5 / State Observers 751 we get Since u(q)=0, we have, from Equation (10–33), and so Hence, for r=1, we have It is noted that, as in any design problem, if the speed and damping are not quite satisfactory, then we must modify the desired characteristic equation and determine a new matrix Computer simulations must be repeated until a satisfactory result is obtained.
10–5 STATE OBSERVERS In the pole-placement approach to the design of control systems, we assumed that all state variables are available for feedback. In practice, however, not all state variables are available for feedback. Then we need to estimate unavailable state variables.
K ˆ .
j(q) = 1.1 j(q) = 1 kI CKx(q)D = 1 kI k3 x3(q) = -56.0652 -50.9684 r = 1.1r u(q) = 0 = -Kx(q) + kI j(q) u(q) = 0 0 0.2 x1 versus t 0 6 4 2 t Sec 0 2 –1 1 x3 versus t 0 6 4 2 0.5 1.5 x5 versus t 0 6 4 2 0 1 t Sec t Sec x1 x3 x5 0 0.5 –0.5 x2 versus t 0 6 4 2 t Sec 0 2 –1 1 x4 versus t 0 6 4 2 t Sec x2 x4 Figure 10–10 Curves x1 versus t, x2 versus t, x3 ( =output y) versus t, x4 versus t, and x5 ( =j) versus t.
752 Chapter 10 / Control Systems Design in State Space Estimation of unmeasurable state variables is commonly called observation.A device (or a computer program) that estimates or observes the state variables is called a state observer, or simply an observer. If the state observer observes all state variables of the system, regardless of whether some state variables are available for direct measurement, it is called a full-order state observer.There are times when this will not be necessary,when we will need observation of only the unmeasurable state variables, but not of those that are directly measurable as well. For example, since the output variables are observable and they are linearly related to the state variables,we need not observe all state variables, but observe only n-m state variables, where n is the dimension of the state vector and m is the dimension of the output vector.
An observer that estimates fewer than n state variables, where n is the dimension of the state vector, is called a reduced-order state observer or, simply, a reduced-order observer. If the order of the reduced-order state observer is the minimum possible, the observer is called a minimum-order state observer or minimum-order observer. In this section, we shall discuss both the full-order state observer and the minimum-order state observer.
State Observer.
A state observer estimates the state variables based on the measurements of the output and control variables. Here the concept of observability discussed in Section 9–7 plays an important role. As we shall see later, state observers can be designed if and only if the observability condition is satisfied.
In the following discussions of state observers, we shall use the notation to designate the observed state vector. In many practical cases, the observed state vector is used in the state feedback to generate the desired control vector.
Consider the plant defined by (10–55) (10–56) The observer is a subsystem to reconstruct the state vector of the plant. The mathe-matical model of the observer is basically the same as that of the plant, except that we include an additional term that includes the estimation error to compensate for inaccuracies in matrices A and B and the lack of the initial error.The estimation error or observation error is the difference between the measured output and the estimated output.The initial error is the difference between the initial state and the initial estimated state.Thus, we define the mathematical model of the observer to be (10–57) where is the estimated state and is the estimated output.The inputs to the observer are the output y and the control input u. Matrix Ke, which is called the observer gain matrix, is a weighting matrix to the correction term involving the difference between the measured output y and the estimated output This term continuously corrects the model output and improves the performance of the observer.Figure 10–11 shows the block diagram of the system and the full-order state observer.
C x .
C x x = AA - Ke CB x + Bu + Ke y x = A x + Bu + Ke(y - C x ) y = Cx x # = Ax + Bu x x Section 10–5 / State Observers 753 u y y ~ Full-order state observer A B C Ke A B C x x ~ ++ ++ ++ + – Figure 10–11 Block diagram of system and full-order state observer, when input u and output y are scalars.
Full-Order State Observer.
The order of the state observer that will be discussed here is the same as that of the plant. Assume that the plant is defined by Equations (10–55) and (10–56) and the observer model is defined by Equation (10–57).
To obtain the observer error equation, let us subtract Equation (10–57) from Equation (10–55): (10–58) Define the difference between x and as the error vector e, or Then Equation (10–58) becomes (10–59) From Equation (10–59), we see that the dynamic behavior of the error vector is deter-mined by the eigenvalues of matrix A-Ke C. If matrix A-Ke C is a stable matrix, the error vector will converge to zero for any initial error vector e(0). That is, will converge to x(t) regardless of the values of x(0) and If the eigenvalues of matrix A-Ke C are chosen in such a way that the dynamic behavior of the error vector is asymptotically stable and is adequately fast, then any error vector will tend to zero (the origin) with an adequate speed.
If the plant is completely observable, then it can be proved that it is possible to choose matrix Ke such that A-Ke C has arbitrarily desired eigenvalues. That is, the observer gain matrix Ke can be determined to yield the desired matrix A-Ke C. We shall discuss this matter in what follows.
x (0).
x (t) e # = AA - Ke CBe e = x - x x = AA - Ke CB(x - x ) x # - x = Ax - A x - Ke(Cx - C x ) 754 Chapter 10 / Control Systems Design in State Space Dual Problem.
The problem of designing a full-order observer becomes that of de-termining the observer gain matrix Ke such that the error dynamics defined by Equation (10–59) are asymptotically stable with sufficient speed of response. (The asymptotic stability and the speed of response of the error dynamics are determined by the eigenvalues of matrix A-Ke C.) Hence, the design of the full-order observer becomes that of determining an appropriate Ke such that A-Ke C has desired eigenvalues.Thus, the problem here becomes the same as the pole-placement problem we discussed in Section 10–2. In fact, the two problems are mathematically the same. This property is called duality.
Consider the system defined by In designing the full-order state observer, we may solve the dual problem, that is, solve the pole-placement problem for the dual system assuming the control signal v to be If the dual system is completely state controllable, then the state feedback gain matrix K can be determined such that matrix A-CK will yield a set of the desired eigenvalues.
If m1, m2, p , mn are the desired eigenvalues of the state observer matrix, then by taking the same mi’s as the desired eigenvalues of the state-feedback gain matrix of the dual system, we obtain Noting that the eigenvalues of A-CK and those of A-KC are the same, we have Comparing the characteristic polynomial and the characteristic poly-nomial for the observer system [refer to Equation (10–57)], we find that Ke and K are related by Thus,using the matrix K determined by the pole-placement approach in the dual system, the observer gain matrix Ke for the original system can be determined by using the relationship Ke=K. (See Problem A–10–10 for the details.) Necessary and Sufficient Condition for State Observation.
As discussed, a necessary and sufficient condition for the determination of the observer gain matrix Ke for the desired eigenvalues of A-Ke C is that the dual of the original system z # = A z + Cv Ke = K @s I - AA - Ke CB @ @s I - (A - K C)@ @s I - (A - C K)@ = @s I - (A - K C)@ @s I - (A - C K)@ = As - m1BAs - m2B p As - mnB v = -Kz n = B z z # = A z + Cv y = Cx x # = Ax + Bu Section 10–5 / State Observers 755 be completely state controllable. The complete state controllability condition for this dual system is that the rank of be n. This is the condition for complete observability of the original system defined by Equations (10–55) and (10–56).This means that a necessary and sufficient condition for the observation of the state of the system defined by Equations (10–55) and (10–56) is that the system be completely observable.
Once we select the desired eigenvalues (or desired characteristic equation), the full-order state observer can be designed, provided the plant is completely observable.The desired eigenvalues of the characteristic equation should be chosen so that the state observer responds at least two to five times faster than the closed-loop system considered.As stated earlier, the equation for the full-order state observer is (10–60) It is noted that thus far we have assumed the matrices A, B, and C in the observer to be exactly the same as those of the physical plant. If there are discrepancies in A, B, and C in the observer and in the physical plant, the dynamics of the observer error are no longer governed by Equation (10–59). This means that the error may not approach zero as expected.Therefore, we need to choose Ke so that the observer is stable and the error remains acceptably small in the presence of small modeling errors.
Transformation Approach to Obtain State Observer Gain Matrix Ke.
By following the same approach as we used in deriving the equation for the state feedback gain matrix K, we can obtain the following equation: (10–61) where Ke is an n1 matrix, and [Refer to Problem A–10–10 for the derivation of Equation (10–61).] W = G an-1 an-2 a1 1 an-2 an-3 1 0 p p p p a1 1 0 0 1 0 0 0 W N = CC A C p (A)n-1CD Q = (WN)-1 Ke = QF an - an an-1 - an-1 a1 - a1 V = (WN)-1F an - an an-1 - an-1 a1 - a1 V x = AA - Ke CB x + Bu + Ke y CC A C p (A)n-1 CD 756 Chapter 10 / Control Systems Design in State Space Direct-Substitution Approach to Obtain State Observer Gain Matrix Ke.
Similar to the case of pole placement,if the system is of low order,then direct substitution of matrix Ke into the desired characteristic polynomial may be simpler. For example, if x is a 3-vector, then write the observer gain matrix Ke as Substitute this Ke matrix into the desired characteristic polynomial: By equating the coefficients of the like powers of s on both sides of this last equation, we can determine the values of ke1, ke2, and ke3. This approach is convenient if n=1, 2, or 3, where n is the dimension of the state vector x. (Although this approach can be used when n=4, 5, 6, p , the computations involved may become very tedious.) Another approach to the determination of the state observer gain matrix Ke is to use Ackermann’s formula.This approach is presented in the following.
Ackermann’s Formula.
Consider the system defined by (10–62) (10–63) In Section 10–2 we derived Ackermann’s formula for pole placement for the system defined by Equation (10–62).The result was given by Equation (10–18), rewritten thus: For the dual of the system defined by Equations (10–62) and (10–63), the preceding Ackermann’s formula for pole placement is modified to (10–64) As stated earlier, the state observer gain matrix Ke is given by K, where K is given by Equation (10–64).Thus, (10–65) Ke = K = f(A)G C CA CA n-2 CA n-1 W -1 G 0 0 0 1 W = f(A)G C CA CA n-2 CA n-1 W -1 G 0 0 0 1 W K = [0 0 p 0 1]CC A C p (A)n-1 CD -1f(A) n = B z z # = A z + Cv K = [0 0 p 0 1]CB AB p A n-1 BD -1f(A) y = Cx x # = Ax + Bu @s I - AA - Ke CB @ = As - m1BAs - m2BAs - m3B Ke = C ke1 ke2 ke3 S Section 10–5 / State Observers 757 where f(s) is the desired characteristic polynomial for the state observer, or where m1, m2, p , mn are the desired eigenvalues. Equation (10–65) is called Ackermann’s formula for the determination of the observer gain matrix Ke.
Comments on Selecting the Best Ke.
Referring to Figure 10–11, notice that the feedback signal through the observer gain matrix Ke serves as a correction signal to the plant model to account for the unknowns in the plant. If significant unknowns are involved, the feedback signal through the matrix Ke should be relatively large. Howev-er, if the output signal is contaminated significantly by disturbances and measurement noises, then the output y is not reliable and the feedback signal through the matrix Ke should be relatively small. In determining the matrix Ke, we should carefully examine the effects of disturbances and noises involved in the output y.
Remember that the observer gain matrix Ke depends on the desired characteristic equation The choice of a set of is, in many instances, not unique.As a general rule, however, the observer poles must be two to five times faster than the controller poles to make sure the observation error (estimation error) converges to zero quickly. This means that the observer estimation error decays two to five times faster than does the state vector x. Such faster decay of the observer error compared with the desired dynamics makes the controller poles dominate the system response.
It is important to note that if sensor noise is considerable,we may choose the observer poles to be slower than two times the controller poles, so that the bandwidth of the sys-tem will become lower and smooth the noise. In this case the system response will be strongly influenced by the observer poles. If the observer poles are located to the right of the controller poles in the left-half s plane, the system response will be dominated by the observer poles rather than by the control poles.
In the design of the state observer, it is desirable to determine several observer gain matrices Ke based on several different desired characteristic equations. For each of the several different matrices Ke, simulation tests must be run to evaluate the resulting system performance. Then we select the best Ke from the viewpoint of overall system performance. In many practical cases, the selection of the best matrix Ke boils down to a compromise between speedy response and sensitivity to disturbances and noises.
EXAMPLE 10–6 Consider the system where We use the observed state feedback such that u = -K x A = B 0 1 20.6 0 R , B = B 0 1R , C = [0 1] y = Cx x # = Ax + Bu m1 , m2 , p , mn As - m1BAs - m2B p As - mnB = 0 f(s) = As - m1BAs - m2B p As - mnB 758 Chapter 10 / Control Systems Design in State Space Design a full-order state observer, assuming that the system configuration is identical to that shown in Figure 10–11.Assume that the desired eigenvalues of the observer matrix are The design of the state observer reduces to the determination of an appropriate observer gain matrix Ke.
Let us examine the observability matrix.The rank of is 2.Hence,the system is completely observable and the determination of the desired observer gain matrix is possible.We shall solve this problem by three methods.
Method 1: We shall determine the observer gain matrix by use of Equation (10–61). The given system is already in the observable canonical form. Hence, the transformation matrix Q=(WN)–1 is I. Since the characteristic equation of the given system is we have The desired characteristic equation is Hence, Then the observer gain matrix Ke can be obtained from Equation (10–61) as follows: Method 2: Referring to Equation (10–59): the characteristic equation for the observer becomes Define Then the characteristic equation becomes (10–66) = s2 + ke2 s - 20.6 + ke1 = 0 2 B s 0 0 s R - B 0 1 20.6 0 R + B ke1 ke2 R [0 1]2 = 2 s -1 -20.6 + ke1 s + ke2 2 Ke = Bke1 ke2 R @s I - A + Ke C@ = 0 e # = AA - Ke CBe Ke = (WN)-1Ba2 - a2 a1 - a1 R = B 1 0 0 1R B 100 + 20.6 20 -0 R = B 120.6 20 R a1 = 20, a2 = 100 (s + 10)2 = s2 + 20s + 100 = s2 + a1 s + a2 = 0 a1 = 0, a2 = -20.6 ∑s I - A∑= 2 s -1 -20.6 s 2 = s2 - 20.6 = s2 + a1 s + a2 = 0 [C A C] = B 0 1 1 0R m1 = -10, m2 = -10 Section 10–5 / State Observers 759 Since the desired characteristic equation is by comparing Equation (10–66) with this last equation, we obtain or Method 3: We shall use Ackermann’s formula given by Equation (10–65): where Thus, and As a matter of course, we get the same Ke regardless of the method employed.
The equation for the full-order state observer is given by Equation (10–57), or Finally, it is noted that, similar to the case of pole placement, if the system order n is 4 or higher, methods 1 and 3 are preferred, because all matrix computations can be carried out by a computer, while method 2 always requires hand computation of the characteristic equation involving unknown parameters ke1, ke2, p , ken.
Effects of the Addition of the Observer on a Closed-Loop System.
In the pole-placement design process, we assumed that the actual state x(t) was available for feedback. In practice, however, the actual state x(t) may not be measurable, so we will need to design an observer and use the observed state for feedback as shown in Fig-ure 10–12. The design process, therefore, becomes a two-stage process, the first stage being the determination of the feedback gain matrix K to yield the desired characteristic equation and the second stage being the determination of the observer gain matrix Ke to yield the desired observer characteristic equation.
Let us now investigate the effects of the use of the observed state rather than the actual state x(t), on the characteristic equation of a closed-loop control system.
x (t), x (t) B x 1 x 2 R = B 0 1 -100 -20 R B x 1 x 2 R + B 0 1R u + B 120.6 20 R y x = AA - Ke CB x + Bu + Ke y = B 120.6 20 412 120.6R B 0 1 1 0R B 0 1R = B120.6 20 R Ke = AA 2 + 20A + 100IB B0 1 1 0R -1 B0 1R f(A) = A 2 + 20A + 100I f(s) = As - m1BAs - m2B = s2 + 20s + 100 Ke = f(A)B C CAR -1 B 0 1R Ke = B 120.6 20 R ke1 = 120.6, ke2 = 20 s2 + 20s + 100 = 0 760 Chapter 10 / Control Systems Design in State Space Consider the completely state controllable and completely observable system defined by the equations For the state-feedback control based on the observed state With this control, the state equation becomes (10–67) The difference between the actual state x(t) and the observed state has been defined as the error e(t): Substitution of the error vector e(t) into Equation (10–67) gives (10–68) Note that the observer error equation was given by Equation (10–59), repeated here: (10–69) Combining Equations (10–68) and (10–69), we obtain (10–70) B x # e # R = B A - BK 0 BK A - Ke CR B x eR e # = AA - Ke CBe x # = (A - BK) x + BKe e(t) = x(t) - x (t) x (t) x # = Ax - BK x = (A - BK) x + BK(x - x ) u = -K x x , y = Cx x # = Ax + Bu u y y ~ A B C Ke –K A B C x x ~ ++ ++ ++ + – Figure 10–12 Observed-state feedback control system.
Section 10–5 / State Observers 761 Equation (10–70) describes the dynamics of the observed-state feedback control system.
The characteristic equation for the system is or Notice that the closed-loop poles of the observed-state feedback control system consist of the poles due to the pole-placement design alone and the poles due to the observer design alone. This means that the pole-placement design and the observer design are independent of each other.They can be designed separately and combined to form the observed-state feedback control system. Note that, if the order of the plant is n, then the observer is also of nth order (if the full-order state observer is used), and the resulting characteristic equation for the entire closed-loop system becomes of order 2n.
Transfer Function of the Observer-Based Controller.
Consider the plant defined by Assume that the plant is completely observable. Assume that we use observed-state feedback control Then, the equations for the observer are given by (10–71) (10–72) where Equation (10–71) is obtained by substituting into Equation (10–57).
By taking the Laplace transform of Equation (10–71), assuming a zero initial condition, and solving for we obtain By substituting this into the Laplace transform of Equation (10–72), we obtain (10–73) Then the transfer function U(s)/Y(s) can be obtained as Figure 10–13 shows the block diagram representation for the system. Notice that the transfer function acts as a controller for the system. Hence, we call the transfer function (10–74) U(s) -Y(s) = num den = KAs I - A + Ke C + BKB -1 Ke KAs I - A + Ke C + BKB -1 Ke U(s) Y(s) = -KAs I - A + Ke C + BKB -1 Ke U(s) = -KAs I - A + Ke C + BKB-1Ke Y(s) X (s) X (s) = As I - A + Ke C + BKB -1 Ke Y(s) X (s), u = -K x u = -K x x = AA - Ke C - BKB x + Ke y u = -K x . y = Cx x # = Ax + Bu @s I - A + BK@ @s I - A + Ke C@ = 0 2s I - A + BK 0 -BK s I - A + Ke C2 = 0 762 Chapter 10 / Control Systems Design in State Space the observer-based controller transfer function or,simply,the observer-controller transfer function.
Note that the observer-controller matrix may or may not be stable, although A-BK and A-Ke C are chosen to be stable. In fact, in some cases the matrix A-Ke C-BK may be poorly stable or even unstable.
EXAMPLE 10–7 Consider the design of a regulator system for the following plant: (10–75) (10–76) where Suppose that we use the pole-placement approach to the design of the system and that the desired closed-loop poles for this system are at s=mi (i=1, 2), where m1=–1.8+j2.4 and m2=–1.8-j2.4. The state-feedback gain matrix K for this case can be obtained as follows: Using this state-feedback gain matrix K, the control signal u is given by Suppose that we use the observed-state feedback control instead of the actual-state feedback control, or where we choose the observer poles to be at s=–8, s=–8 Obtain the observer gain matrix Ke and draw a block diagram for the observed-state feedback control system.Then obtain the transfer function for the observer controller, and draw another block diagram with the observer controller as a series controller in the feedforward path. Finally, obtain the response of the system to the following initial condition: x(0) = B 1 0R , e(0) = x(0) - x (0) = B 0.5 0 R U(s)[-Y(s)] u = -K x = -[29.6 3.6]B x 1 x 2 R u = -Kx = -[29.6 3.6]B x1 x2 R K = [29.6 3.6] A = B 0 20.6 1 0R , B = B 0 1R , C = [1 0] y = Cx x # = Ax + Bu A - Ke C - BK R(s) = 0 Y(s) U(s) Plant –Y(s) K(sI – A + KeC + BK)–1Ke + – Figure 10–13 Block diagram representation of system with a controller-observer.
Section 10–5 / State Observers 763 For the system defined by Equation (10–75), the characteristic polynomial is Thus, The desired characteristic polynomial for the observer is Hence, For the determination of the observer gain matrix, we use Equation (10–61), or where Hence, (10–77) Equation (10–77) gives the observer gain matrix Ke. The observer equation is given by Equation (10–60): (10–78) Since Equation (10–78) becomes or The block diagram of the system with observed-state feedback is shown in Figure 10–14(a). = B -16 -93.6 1 -3.6R B x 1 x 2 R + B 16 84.6R y B x 1 x 2 R = b B 0 20.6 1 0R - B 16 84.6R [1 0] - B 0 1R [29.6 3.6] r B x 1 x 2 R + B 16 84.6R y x = AA - Ke C - BKB x + Ke y u = -K x x = AA - Ke CB x + Bu + Ke y = B 0 1 1 0R B84.6 16 R = B 16 84.6R Ke = b B 0 1 1 0R B 1 0 0 1R r -1 B 64 + 20.6 16 - 0 R W = Ba1 1 1 0R = B 0 1 1 0R N = [C AC] = B 1 0 0 1R Ke = (WN)-1B a2 - a2 a1 - a1 R a1 = 16, a2 = 64 = s2 + a1 s + a2 As - m1BAs - m2B = (s + 8)(s + 8) = s2 + 16s + 64 a1 = 0, a2 = -20.6 ∑s I - A∑= 2 s -20.6 -1 s 2 = s2 - 20.6 = s2 + a1 s + a2 764 Chapter 10 / Control Systems Design in State Space Referring to Equation (10–74), the transfer function of the observer-controller is As a matter of course, the same transfer function can be obtained with MATLAB. For example, MATLAB Program 10–8 produces the transfer function of the observer controller.Figure 10–14(b) shows a block diagram of the system.
= 778.2s + 3690.7 s2 + 19.6s + 151.2 = [29.6 3.6]B s + 16 93.6 -1 s + 3.6R -1 B 16 84.6R U(s) -Y(s) = KAs I - A + Ke C + BKB-1 Ke ++ –+ R(s) = 0 Y(s) U(s) –Y(s) 1 s2 – 20.6 + – (b) u y x x ~ C A C –K B B 0 1 0 1 1 0 1 0 0 20.6 1 0 0 20.6 1 0 16 84.6 –29.6 –3.6 ++ ++ (a) 778.2s + 3690.7 s2 + 19.6s + 151.2 Figure 10–14 (a) Block diagram of system with observed-state feedback; (b) block diagram of transfer-function system.
Section 10–5 / State Observers 765 MATLAB Program 10–8 % Obtaining transfer function of observer controller --- full-order observer A = [0 1;20.6 0]; B = [0;1]; C = [1 0]; K = [29.6 3.6]; Ke = [16;84.6]; AA = A-KeC-BK; BB = Ke; CC = K; DD = 0; [num,den] = ss2tf(AA,BB,CC,DD) num = 1.0e+003 0 0.7782 3.6907 den = 1.0000 19.6000 151.2000 The dynamics of the observed-state feedback control system just designed can be described by the following equations: For the plant, For the observer, The system, as a whole, is of fourth order.The characteristic equation for the system is The characteristic equation can also be obtained from the block diagram for the system shown in Figure 10–14(b). Since the closed-loop transfer function is Y(s) R(s) = 778.2s + 3690.7 As2 + 19.6s + 151.2BAs2 - 20.6B + 778.2s + 3690.7 = s4 + 19.6s3 + 130.6s2 + 374.4s + 576 = 0 @s I - A + BK@ @s I - A + Ke C@ = As2 + 3.6s + 9BAs2 + 16s + 64B u = -[29.6 3.6]B x 1 x 2 R B x 1 x 2 R = B -16 -93.6 1 -3.6R B x 1 x 2 R + B 16 84.6R y y = [1 0]B x1 x2 R B x # 1 x # 2 R = B 0 20.6 1 0R B x1 x2 R + B0 1R u 766 Chapter 10 / Control Systems Design in State Space MATLAB Program 10–9 A = [0 1; 20.6 0]; B = [0;1]; C = [1 0]; K = [29.6 3.6]; Ke = [16; 84.6]; sys = ss([A-BK BK; zeros(2,2) A-KeC],eye(4),eye(4),eye(4)); t = 0:0.01:4; z = initial(sys,[1;0;0.5;0],t); x1 = [1 0 0 0]z'; x2 = [0 1 0 0]z'; e1 = [0 0 1 0]z'; e2 = [0 0 0 1]z'; subplot(2,2,1); plot(t,x1 ),grid title('Response to Initial Condition') ylabel('state variable x1') subplot(2,2,2); plot(t,x2),grid title('Response to Initial Condition') ylabel('state variable x2') subplot(2,2,3); plot(t,e1),grid xlabel('t (sec)'), ylabel('error state variable e1') subplot(2,2,4); plot(t,e2),grid xlabel('t (sec)'), ylabel('error state variable e2') the characteristic equation is As a matter of course, the characteristic equation is the same for the system in state-space representation and in transfer-function representation.
Finally, we shall obtain the response of the system to the following initial condition: Referring to Equation (10–70), the response to the initial condition can be determined from A MATLAB Program to obtain the response is shown in MATLAB Program 10–9.The resulting response curves are shown in Figure 10–15.
B x # e # R = B A - BK 0 BK A - Ke CR B x eR , B x(0) e(0)R = D 1 0 0.5 0 T x(0) = B1 0R , e(0) = B0.5 0 R = s4 + 19.6s3 + 130.6s2 + 374.4s + 576 = 0 As2 + 19.6s + 151.2BAs2 - 20.6B + 778.2s + 3690.7 Section 10–5 / State Observers 767 u y x y Plant C A B –K x ~ Minimum-order observer Transformation ++ Figure 10–16 Observed-state feedback control system with a minimum-order observer.
Minimum-Order Observer.
The observers discussed thus far are designed to reconstruct all the state variables. In practice, some of the state variables may be accu-rately measured. Such accurately measurable state variables need not be estimated.
Suppose that the state vector x is an n-vector and the output vector y is an m-vector that can be measured. Since m output variables are linear combinations of the state variables, m state variables need not be estimated. We need to estimate only n-m state variables.Then the reduced-order observer becomes an (n-m)th-order observ-er. Such an (n-m)th-order observer is the minimum-order observer. Figure 10–16 shows the block diagram of a system with a minimum-order observer.
Response to Initial Condition Response to Initial Condition state variable x1 state variable x2 error state variable e1 error state variable e2 1.5 1 0.5 0 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 t (sec) t (sec) 3 4 −0.5 −0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 −1.5 −1 −0.5 0 0.5 0 −0.5 −1.5 −1 −2 −2.5 Figure 10–15 Response curves to initial condition.
768 Chapter 10 / Control Systems Design in State Space It is important to note, however, that if the measurement of output variables involves significant noises and is relatively inaccurate, then the use of the full-order observer may result in a better system performance.
To present the basic idea of the minimum-order observer, without undue mathe-matical complications, we shall present the case where the output is a scalar (that is, m=1) and derive the state equation for the minimum-order observer. Consider the system (10–79) (10–80) where the state vector x can be partitioned into two parts xa (a scalar) and xb Can (n-1)-vectorD. Here the state variable xa is equal to the output y and thus can be directly measured, and xb is the unmeasurable portion of the state vector. Then the partitioned state and output equations become (10–81) (10–82) where From Equation (10–81), the equation for the measured portion of the state becomes or (10–83) The terms on the left-hand side of Equation (10–83) can be measured. Equation (10–83) acts as the output equation. In designing the minimum-order observer, we consider the left-hand side of Equation (10–83) to be known quantities.Thus,Equation (10–83) relates the measurable quantities and unmeasurable quantities of the state.
From Equation (10–81), the equation for the unmeasured portion of the state becomes (10–84) Noting that terms Abaxa and Bbu are known quantities, Equation (10–84) describes the dynamics of the unmeasured portion of the state.
x # b = Aba xa + Abb xb + Bb u x # a - Aaa xa - Ba u = Aab xb x # a = Aaa xa + Aab xb + Ba u Bb = (n - 1) 1 matrix Ba = scalar Abb = (n - 1) (n - 1) matrix Aba = (n - 1) 1 matrix Aab = 1 (n - 1) matrix Aaa = scalar y = [1 0]cxa xb d cx # a x # b d = c Aaa Aba Aab Abb d cxa xb d + c Ba Bb du y = Cx x # = Ax + Bu Section 10–5 / State Observers 769 Full-Order State Observer Minimum-Order State Observer A Abb Bu y C Aab Ke (n1 matrix) Ke [(n-1)1 matrix] x # a - Aaa xa - Ba u Aba xa + Bb u x b x Table 10–1 List of Necessary Substitutions for Writing the Observer Equation for the Minimum-Order State Observer In what follows we shall present a method for designing a minimum-order observer.
The design procedure can be simplified if we utilize the design technique developed for the full-order state observer.
Let us compare the state equation for the full-order observer with that for the minimum-order observer.The state equation for the full-order observer is and the “state equation” for the minimum-order observer is The output equation for the full-order observer is and the “output equation” for the minimum-order observer is The design of the minimum-order observer can be carried out as follows: First, note that the observer equation for the full-order observer was given by Equation (10–57), which we repeat here: (10–85) Then, making the substitutions of Table 10–1 into Equation (10–85), we obtain (10–86) where the state observer gain matrix Ke is an (n-1)1 matrix. In Equation (10–86), notice that in order to estimate ,we need the derivative of xa.This presents a difficulty, because differentiation amplifies noise. If xa (= y) is noisy, the use of is unacceptable.
x # a x b x b = AAbb - Ke AabB x b + Aba xa + Bb u + KeAx # a - Aaa xa - Ba uB x = AA - Ke CB x + Bu + Ke y x # a - Aaa xa - Ba u = Aab xb y = Cx x # b = Abb xb + Aba xa + Bb u x # = Ax + Bu 770 Chapter 10 / Control Systems Design in State Space To avoid this difficulty, we eliminate in the following way. First rewrite Equation (10–86) as (10–87) Define and (10–88) Then Equation (10–87) becomes (10–89) Define Then Equation (10–89) becomes (10–90) Equation (10–90) and Equation (10–88) together define the minimum-order observer.
Since where 0 is a row vector consisting of (n-1) zeros, if we define then we can write in terms of and y as follows: (10–91) This equation gives the transformation from Figure 10–17 shows the block diagram of the observed-state feedback control system with the minimum-order observer,based on Equations (10–79),(10–80),(10–90),(10–91) and Next we shall derive the observer error equation. Using Equation (10–83), Equation (10–86) can be modified to (10–92) x b = AAbb - Ke AabB x b + Aba xa + Bb u + Ke Aab xb u = -K x .
H to x .
x = C ˆ H + D ˆ y H x C ˆ = c 0 In-1 d, D ˆ = c 1 Ke d x = c xa x b d = c y x b d = c 0 In-1 d C x b - Ke yD + c 1 Ke dy y = C1 0D cxa xb d H = A ˆ H + B ˆ y + F ˆ u F ˆ = Bb - Ke Ba B ˆ = A ˆ Ke + Aba - Ke Aaa A ˆ = Abb - Ke Aab + Aba - Ke AaaDy + ABb - Ke BaBu H = AAbb - Ke AabBH + C AAbb - Ke AabBKe x b - Ke y = x b - Ke xa = H xb - Ke y = xb - Ke xa = H + ABb - Ke BaBu + CAAbb - Ke AabBKe + Aba - Ke AaaDy = AAbb - Ke AabBA x b - Ke yB x b - Ke x # a = AAbb - Ke AabB x b + AAba - Ke AaaBy + ABb - Ke BaBu x # a Section 10–5 / State Observers 771 u y x x C B –K Minimum-order observer Transformation ~ ~ h ~ h .
++ + ++ + + D C F A B x .
^ ^ ^ ^ A ^ By subtracting Equation (10–92) from Equation (10–84), we obtain (10–93) Define Then Equation (10–93) becomes (10–94) This is the error equation for the minimum-order observer. Note that e is an (n-1)-vector.
The error dynamics can be chosen as desired by following the technique developed for the full-order observer, provided that the rank of matrix is n-1. (This is the complete observability condition applicable to the minimum-order observer.) F Aab Aab Abb Aab An-2 bb V e # = AAbb - Ke AabBe e = xb - x b = H - H x # b - x b = AAbb - Ke AabBAxb - x bB Figure 10–17 System with observed-state feedback, where the observer is the minimum-order observer.
772 Chapter 10 / Control Systems Design in State Space The characteristic equation for the minimum-order observer is obtained from Equation (10–94) as follows: (10–95) where are desired eigenvalues for the minimum-order observer. The observer gain matrix Ke can be determined by first choosing the desired eigenvalues for the minimum-order observer [that is, by placing the roots of the characteristic equation, Equation (10–95), at the desired locations] and then using the procedure developed for the full-order observer with appropriate modifications. For example, if the formula for determining matrix Ke given by Equation (10–61) is to be used, it should be modified to (10–96) where Ke is an matrix and Note that are coefficients in the characteristic equation for the state equation Also, if Ackermann’s formula given by Equation (10–65) is to be used, then it should be modified to (10–97) Ke = fAAbbB G Aab Aab Abb Aab An-3 bb Aab An-2 bb W -1 G 0 0 0 1 W @s I - Abb@ = sn-1 + a ˆ1 sn-2 + p + a ˆn-2 s + a ˆn-1 = 0 a ˆ1 , a ˆ2 , p , a ˆn-2 W ˆ = G a ˆn-2 a ˆn-3 a ˆ1 1 a ˆn-3 a ˆn-4 1 0 p p p p a ˆ1 1 0 0 1 0 0 0 W = (n - 1) (n - 1) matrix N ˆ = CAab Abb Aab p AAbbB n-2 AabD = (n - 1) (n - 1) matrix (n - 1) 1 Ke = Q ˆ F a ˆ n-1 - a ˆn-1 a ˆ n-2 - a ˆn-2 a ˆ 1 - a ˆ1 V = AW ˆ N ˆ B -1F a ˆ n-1 - a ˆn-1 a ˆ n-2 - a ˆn-2 a ˆ 1 - a ˆ1 V m1 , m2 , p , mn-1 = sn-1 + a ˆ 1 sn-2 + p + a ˆ n-2 s + a ˆ n-1 = 0 @s I - Abb + Ke Aab@ = As - m1BAs - m2B p As - mn-1B Section 10–5 / State Observers 773 where Observed-State Feedback Control System with Minimum-Order Observer.
For the case of the observed-state feedback control system with full-order state observer, we have shown that the closed-loop poles of the observed-state feedback control system consist of the poles due to the pole-placement design alone, plus the poles due to the observer design alone. Hence, the pole-placement design and the full-order observer design are independent of each other.
For the observed-state feedback control system with minimum-order observer, the same conclusion applies.The system characteristic equation can be derived as (10–98) (See Problem A–10–11 for the details.) The closed-loop poles of the observed-state feed-back control system with a minimum-order observer comprise the closed-loop poles due to pole placement Cthe eigenvalues of matrix (A-BK)D and the closed-loop poles due to the minimum-order observer Cthe eigenvalues of matrix (Abb-Ke Aab)D.There-fore, the pole-placement design and the design of the minimum-order observer are independent of each other.
Determining Observer Gain Matrix Ke with MATLAB.
Because of the duality of pole-placement and observer design, the same algorithm can be applied to both the pole-placement problem and the observer-design problem. Thus, the commands acker and place can be used to determine the observer gain matrix Ke.
The closed-loop poles of the observer are the eigenvalues of matrix A-Ke C.The closed-loop poles of the pole-placement are the eigenvalues of matrix A-BK.
Referring to the duality problem between the pole-placement problem and observer-design problem,we can determine Ke by considering the pole-placement problem for the dual system. That is, we determine Ke by placing the eigenvalues of A-CKe at the desired place. Since Ke=K, for the full-order observer we use the command Ke = acker(A',C',L)' where L is the vector of the desired eigenvalues for the observer. Similarly, for the full-order observer, we may use Ke = place(A',C',L)' provided L does not include multiple poles. [In the above commands, prime (') indicates the transpose.] For the minimum-order (or reduced-order) observers, use the following commands: Ke = acker(Abb',Aab',L)' or Ke = place(Abb',Aab',L)' @s I - A + BK@ @s I - Abb + Ke Aab@ = 0 fAAbbB = An-1 bb + a ˆ 1 An-2 bb + p + a ˆ n-2 Abb + a ˆ n-1 I 774 Chapter 10 / Control Systems Design in State Space EXAMPLE 10–8 Consider the system where Let us assume that we want to place the closed-loop poles at Then the necessary state-feedback gain matrix K can be obtained as follows: K=[90 29 4] (See MATLAB Program 10–10 for a MATLAB computation of this matrix K.) Next, let us assume that the output y can be measured accurately so that state variable x1 (which is equal to y) need not be estimated. Let us design a minimum-order observer. (The minimum-order observer is of second order.) Assume that we choose the desired observer poles to be at s=–10, s=–10 Referring to Equation (10–95), the characteristic equation for the minimum-order observer is In what follows, we shall use Ackermann’s formula given by Equation (10–97).
(10–99) where Since we have Abb = B 0 -11 1 -6R , Ba = 0, Bb = B 0 1R Aaa = 0, Aab = [1 0], Aba = B 0 -6R x = c xa x b d = D x1 x 2 x 3 T , A = D 0 0 -6 1 0 -11 0 1 -6 T , B = D 0 0 1 T fAAbbB = A2 bb + a ˆ 1 Abb + a ˆ 2 I = A2 bb + 20Abb + 100I Ke = fAAbbB C Aab Aab Abb S -1 B 0 1R = s2 + 20s + 100 = 0 = (s + 10)(s + 10) @s I - Abb + Ke Aab@ = As - m1BAs - m2B s1 = -2 + j213, s2 = -2 - j213, s3 = -6 A = C 0 0 -6 1 0 -11 0 1 -6 S , B = C 0 0 1 S , C = [1 0 0] y = Cx x # = Ax + Bu Section 10–5 / State Observers 775 MATLAB Program 10–10 A = [0 1 0;0 0 1;-6 -11 -6]; B = [0;0;1]; J = [-2+j2sqrt(3) -2-j2sqrt(3) -6]; K = acker(A,B,J) K = 90.0000 29.0000 4.0000 Abb = [0 1;-11 -6]; Aab = [1 0]; L = [-10 -10]; Ke = acker(Abb',Aab',L)' Ke = 14 5 Equation (10–99) now becomes (A MATLAB computation of this Ke is given in MATLAB Program 10–10.) = B 89 -154 14 5R B 0 1R = B 14 5R Ke = b B 0 -11 1 -6R 2 + 20B 0 -11 1 -6R + 100B 1 0 0 1R r B 1 0 0 1R -1 B 0 1R Referring to Equations (10–88) and (10–89),the equation for the minimum-order observer can be given by (10–100) where Noting that the equation for the minimum-order observer, Equation (10–100), becomes or B h 2 h 3 R = B -14 -16 1 -6R B h 2 h 3 R + B-191 -260R y + B 0 1R u + B 0 -6R - B 14 5R 0 ry + b B 0 1R - B 14 5R 0 ru B h 2 h 3 R = B -14 -16 1 -6R B h 2 h 3 R + b B -14 -16 1 -6R B 14 5R Abb - Ke Aab = B 0 -11 1 -6R - B 14 5R [1 0] = B -14 -16 1 -6R H = x b - Ke y = x b - Ke x1 H = AAbb - Ke AabBH + CAAbb - Ke AabBKe + Aba - Ke AaaDy + ABb - Ke BaBu 776 Chapter 10 / Control Systems Design in State Space where or If the observed-state feedback is used, then the control signal u becomes where K is the state feedback gain matrix. Figure 10–18 is a block diagram showing the configu-ration of the system with observed-state feedback, where the observer is the minimum-order observer.
u = -K x = -KC x1 x 2 x 3 S B x 2 x 3 R = B h 2 h 3 R + Ke x1 B h 2 h 3 R = B x 2 x 3 R - Ke y u y x x Plant C A B Minimum-order observer Transformation 0 1 0 h ~ ~ ~ 0 1 0 0 0 1 x1 Kex1 1 Ke 1 14 5 0 –6 14 5 –14 –16 1 –6 Bb – KeBa Abb – KeAab Aba – KeAaa h ~ h .
Ke ++ + ++ + + + + [ –90 –29 –4 ] –K Figure 10–18 System with observed state feedback, where the observer is the minimum-order observer designed in Example 10–8.
Section 10–5 / State Observers 777 Transfer Function of Minimum-Order Observer-Based Controller.
In the minimum-order observer equation given by Equation (10–89): define, similar to the case of the derivation of Equation (10–90), Then, the following three equations define the minimum-order oberver: (10–101) (10–102) (10–103) Since Equation (10–103) can be rewritten as (10–104) by substituting Equation (10–104) into Equation (10–101), we obtain (10–105) Define Then Equations (10–105) and (10–104) can be written as (10–106) (10–107) Equations (10–106) and (10–107) define the minimum-order observer-based controller.
By considering u as the output and –y as the input, U(s) can be written as Since the input to the observer controller is –Y(s), rather than Y(s), the transfer function of the observer controller is (10–108) This transfer function can be easily obtained by using the following MATLAB statement: [num,den] = ss2tf(Atilde, Btilde, -Ctilde, -Dtilde) (10–109) U(s) -Y(s) = num den = - C C As I - A B -1B + D D = - C C As I - A B -1B + D D[-Y(s)] U(s) = C C As I - A B -1B + D DY(s) u = C H + D y H = A H + B y D = -AK a + Kb KeB C = -Kb B = B ˆ - F ˆ AK a + Kb KeB A = A ˆ - F ˆ Kb = AA ˆ - F ˆ KbBH + CB ˆ - F ˆ AK a + Kb KeB Dy H = A ˆ H + B ˆ y + F ˆ C-KbH - AK a + Kb KeByD = -KbH - AK a + Kb KeBy u = -K x = - CK a KbD B y x b R = -K a y - Kb x b u = -K x H = x b - Ke y H = A ˆ H + B ˆ y + F ˆ u F ˆ = Bb - Ke Ba B ˆ = A ˆ Ke + Aba - Ke Aaa A ˆ = Abb - Ke Aab H = AAbb - Ke AabBH + CAAbb - Ke AabBKe + Aba - Ke AaaDy + ABb - Ke BaBu 778 Chapter 10 / Control Systems Design in State Space 10–6 DESIGN OF REGULATOR SYSTEMS WITH OBSERVERS In this section we shall consider a problem of designing regulator systems by using the pole-placement-with-observer approach.
Consider the regulator system shown in Figure 10–19. (The reference input is zero.) The plant transfer function is Using the pole-placement approach, design a controller such that when the system is subjected to the following initial condition: where x is the state vector for the plant and e is the observer error vector, the maximum undershoot of y(t) is 25 to 35% and the settling time is about 4 sec.Assume that we use the minimum-order observer. (We assume that only the output y is measurable.) We shall use the following design procedure: 1. Derive a state-space model of the plant.
2. Choose the desired closed-loop poles for pole placement. Choose the desired observer poles.
3. Determine the state feedback gain matrix K and the observer gain matrix Ke.
4. Using the gain matrices K and Ke obtained in step 3, derive the transfer function of the observer controller.If it is a stable controller,check the response to the given ini-tial condition. If the response is not acceptable, adjust the closed-loop pole location and/or observer pole location until an acceptable response is obtained.
Design step 1: We shall derive the state-space representation of the plant. Since the plant transfer function is the corresponding differential equation is Referring to Section 2–5, let us define the state variables x1, x2, and x3 as follows: x3 = x # 2 - b2 u x2 = x # 1 - b1 u x1 = y - b0 u y % + 10y $ + 24y # = 10u # + 20u Y(s) U(s) = 10(s + 2) s(s + 4)(s + 6) x(0) = C 1 0 0 S , e(0) = B 1 0R G(s) = 10(s + 2) s(s + 4)(s + 6) r = 0 y u Plant –y Controller + – Figure 10–19 Regulator system.
Section 10–6 / Design of Regulator Systems with Observers 779 Also, is defined by where and [See Equation (2–35) for the calculation of b’s.] Then the state-space equation and out-put equation can be obtained as Design step 2: As the first trial, let us choose the desired closed-loop poles at s=–1+j2, s=–1-j2, s=–5 and choose the desired observer poles at s=–10, s=–10 Design step 3: We shall use MATLAB to compute the state feedback gain matrix K and the observer gain matrix Ke. MATLAB Program 10–11 produces matrices K and Ke. y = [1 0 0]C x1 x2 x3 S + u C x # 1 x # 2 x # 3 S = C 0 0 0 1 0 -24 0 1 -10 S C x1 x2 x3 S + C 0 10 -80 S u b3 = -80.
b2 = 10, b1 = 0, b0 = 0, = -24x2 - 10x3 + b3 u x # 3 = -a3x1 - a2x2 - a1x3 + b3u x # 3 MATLAB Program 10–11 % Obtaining the state feedback gain matrix K A = [0 1 0;0 0 1;0 -24 -10]; B = [0;10;-80]; C = [1 0 0]; J = [-1+j2 -1-j2 -5]; K = acker(A,B,J) K = 1.2500 1.2500 0.19375 % Obtaining the observer gain matrix Ke Aaa = 0; Aab = [1 0]; Aba = [0;0]; Abb = [0 1;-24 -10];Ba = 0; Bb = [10;-80]; L = [-10 -10]; Ke = acker(Abb',Aab',L)' Ke = 10 -24 780 Chapter 10 / Control Systems Design in State Space MATLAB Program 10–12 % Determination of transfer function of observer controller A = [0 1 0;0 0 1;0 -24 -10]; B = [0;10;-80]; Aaa = 0; Aab = [1 0]; Aba = [0;0]; Abb = [0 1;-24 -10]; Ba = 0; Bb = [10;-80]; Ka = 1.25; Kb = [1.25 0.19375]; Ke = [10;-24]; Ahat = Abb - KeAab; Bhat = AhatKe + Aba - KeAaa; Fhat = Bb - KeBa; Atilde = Ahat - FhatKb; Btilde = Bhat - Fhat(Ka + KbKe); Ctilde = -Kb; Dtilde = -(Ka + KbKe); [num,den] = ss2tf(Atilde, Btilde, -Ctilde, -Dtilde) num = 9.1000 73.5000 125.0000 den = 1.0000 17.0000 -30.0000 In the program, matrices J and L represent the desired closed-loop poles for pole place-ment and the desired poles for the observer, respectively. The matrices K and Ke are obtained as Design step 4: We shall determine the transfer function of the observer controller.
Referring to Equation (10–108), the transfer function of the observer controller can be given by We shall use MATLAB to calculate the transfer function of the observer controller.
MATLAB Program 10–12 produces this transfer function.The result is Define the system with this observer controller as System 1. Figure 10–20 shows the block diagram of System 1. = 9.1(s + 5.6425)(s + 2.4344) (s + 18.6119)(s - 1.6119) Gc(s) = 9.1s2 + 73.5s + 125 s2 + 17s - 30 Gc(s) = U(s) -Y(s) = num den = - C C As I - A B -1B + D D Ke = B 10 -24R K = [1.25 1.25 0.19375] Section 10–6 / Design of Regulator Systems with Observers 781 MATLAB Program 10–13 % Obtaining the characteristic equation [num1,den1] = ss2tf(A-BK,eye(3),eye(3),eye(3),1); [num2,den2] = ss2tf(Abb-KeAab,eye(2),eye(2),eye(2),1); charact_eq = conv(den1,den2) charact_eq = 1.0e+003 0.0010 0.0270 0.2550 1.0250 2.0000 2.5000 r = 0 y u + – 9.1s2 + 73.5s + 125 s2 + 17s – 30 10(s + 2) s(s + 4) (s + 6) Observer controller Plant The observer controller has a pole in the right-half s plane (s=1.6119). The exis-tence of an open-loop right-half s plane pole in the observer controller means that the system is open-loop unstable, although the closed-loop system is stable. The latter can be seen from the characteristic equation for the system: (See MATLAB Program 10–13 for the calculation of the characteristic equation.) A disadvantage of using an unstable controller is that the system becomes unstable if the dc gain of the system becomes small. Such a control system is neither desirable nor acceptable. Hence, to get a satisfactory system, we need to modify the closed-loop pole location and/or observer pole location. = (s + 1 + j2)(s + 1 - j2)(s + 5)(s + 10)(s + 10) = 0 = s5 + 27s4 + 255s3 + 1025s2 + 2000s + 2500 ∑s I - A + BK∑ @s I - Abb + Ke Aab@ Figure 10–20 Block diagram of System 1.
Second trial: Let us keep the desired closed-loop poles for pole placement as before, but modify the observer pole locations as follows: s=–4.5, s=–4.5 Thus, L=[–4.5 –4.5] Using MATLAB, we find the new Ke to be Ke = B -1 6.25R 782 Chapter 10 / Control Systems Design in State Space MATLAB Program 10–14 % Determination of transfer function of observer controller.
A = [0 1 0;0 0 1;0 -24 -10]; B = [0;10;-80]; Aaa = 0; Aab = [1 0]; Aba = [0;0]; Abb = [0 1;-24 -10]; Ba = 0; Bb = [10;-80]; Ka = 1.25; Kb = [1.25 0.19375]; Ke = [-1;6.25]; Ahat = Abb - KeAab; Bhat = AhatKe + Aba - KeAaa; Fhat = Bb - KeBa; Atilde = Ahat - FhatKb; Btilde = Bhat - Fhat(Ka + KbKe); Ctilde = -Kb; Dtilde = -(Ka + KbKe); [num,den] = ss2tf(Atilde,Btilde,-Ctilde,-Dtilde) num = 1.2109 11.2125 25.3125 den = 1.0000 6.0000 2.1406 Next, we shall obtain the transfer function of the observer controller. MATLAB Program 10–14 produces this transfer function as follows: = 1.2109(s + 5.3582)(s + 3.9012) (s + 5.619)(s + 0.381) Gc(s) = 1.2109s2 + 11.2125s + 25.3125 s2 + 6s + 2.1406 Notice that this is a stable controller. Define the system with this observer controller as System 2. We shall proceed to obtain the response of System 2 to the given initial condition: By substituting into the state-space equation for the plant, we obtain (10–110) = Ax - BK bx - B 0 eR r = Ax - BKx + BCKa KbD B 0 eR x # = Ax - BK x = Ax - BKB xa x b R = Ax - BKB xa xb - eR u = -K x x(0) = C 1 0 0 S , e(0) = B 1 0R Section 10–6 / Design of Regulator Systems with Observers 783 The error equation for the minimum-order observer is (10–111) By combining Equations (10–110) and (10–111), we get with the initial condition MATLAB Program 10–15 produces the response to the given initial condition. The response curves are shown in Figure 10–21.They seem to be acceptable.
B x(0) e(0)R = E 1 0 0 1 0 U Bx # e # R = B A - BK 0 BKb Abb - Ke Aab R B x eR e # = AAbb - Ke AabBe MATLAB Program 10–15 % Response to initial condition.
A = [0 1 0;0 0 1;0 -24 -10]; B = [0;10;-80]; K = [1.25 1.25 0.19375]; Kb = [1.25 0.19375]; Ke = [-1;6.25]; Aab = [1 0]; Abb = [0 1;-24 -10]; AA = [A-BK BKb; zeros(2,3) Abb-KeAab]; sys = ss(AA,eye(5),eye(5),eye(5)); t = 0:0.01:8; x = initial(sys,[1;0;0;1;0],t); x1 = [1 0 0 0 0]x'; x2 = [0 1 0 0 0]x'; x3 = [0 0 1 0 0]x'; e1 = [0 0 0 1 0]x'; e2 = [0 0 0 0 1]x'; subplot(3,2,1); plot(t,x1); grid xlabel ('t (sec)'); ylabel('x1') subplot(3,2,2); plot(t,x2); grid xlabel ('t (sec)'); ylabel('x2') subplot(3,2,3); plot(t,x3); grid xlabel ('t (sec)'); ylabel('x3') subplot(3,2,4); plot(t,e1); grid xlabel('t (sec)'); ylabel('e1') subplot(3,2,5); plot(t,e2); grid xlabel('t (sec)'); ylabel('e2') 784 Chapter 10 / Control Systems Design in State Space Next,we shall check the frequency-response characteristics.The Bode diagram of the open-loop system just designed is shown in Figure 10–22.The phase margin is about 40° and the gain margin is ±q dB.The Bode diagram of the closed-loop system is shown in Figure 10–23.The bandwidth of the system is approximately 3.8 rad/sec.
x1 0 2 4 6 8 t (sec) 0 2 4 6 8 t (sec) 0 2 4 6 8 t (sec) 0 2 4 6 8 t (sec) 0 2 4 6 8 t (sec) −0.5 0 0.5 1 e2 −3 −2 −1 0 x2 −1.5 −0.5 −1 0 0.5 x3 −5 5 0 10 15 e1 0 1 0.5 1.5 Figure 10–21 Response to the given initial condition; x1(0)=1, x2(0)=0, x3(0)=0, e1(0)=1, e2(0)=0.
Frequency (rad/sec) Bode Diagram of System 2 — Open Loop −200 −150 −100 −50 −100 Phase (deg); Magnitude (dB) −50 0 50 100 10−3 10−2 10−1 100 101 102 Figure 10–22 Bode diagram for the open-loop transfer function of System 2.
Section 10–6 / Design of Regulator Systems with Observers 785 Finally,we shall compare the root-locus plots of the first system with L=[–10 –10] and the second system with L=[–4.5 –4.5]. The plot for the first system given in Figure 10–24(a) shows that the system is unstable for small dc gain and becomes stable for large dc gain.The plot for the second system given in Figure 10–24(b), on the other hand, shows that the system is stable for any positive dc gain.
Frequency (rad/sec) Bode Diagram of System 2 — Closed Loop −200 −50 −100 −150 0 −60 −40 Phase (deg); Magnitude (dB) 20 −20 0 10−1 100 101 102 Figure 10–23 Bode diagram for the closed-loop transfer function of System 2.
Root-Locus Plot of (91s3 + 917s2 + 2720s + 2500)/ (s5 + 27s4 + 164s3 + 108s2 − 720s) Real Axis Imag Axis 2 −4 −6 −8 −14 −12 −10 −8 −6 −4 −2 0 2 4 −2 0 6 8 (a) Root-Locus Plot of (12.109s3 + 136.343s2 + 477.375s + 506.25)/ (s5 + 16s4 + 86.1406s3 + 165.406s2 + 51.3744s) Real Axis Imag Axis 2 −3 −4 −5−8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 −2 −1 0 1 4 5 (b) Figure 10–24 (a) Root-locus plot of the system with observer poles at s=–10 and s=–10; (b) root-locus plot of the system with observer poles at s=–4.5 and s=–4.5.
786 Chapter 10 / Control Systems Design in State Space Comments 1. In designing regulator systems,note that if the dominant controller poles are placed far to the left of the jv axis, the elements of the state feedback gain matrix K will become large. Large gain values will make the actuator output become large, so that saturation may take place. Then the designed system will not behave as designed.
2. Also, by placing the observer poles far to the left of the jv axis, the observer controller becomes unstable,although the closed-loop system is stable.An unstable observer controller is not acceptable.
3. If the observer controller becomes unstable, move the observer poles to the right in the left-half s plane until the observer controller becomes stable.Also,the desired closed-loop pole locations may need to be modified.
4. Note that if the observer poles are placed far to the left of the jv axis, the band-width of the observer will increase and will cause noise problems. If there is a serious noise problem, the observer poles should not be placed too far to the left of the jv axis.The general requirement is that the bandwidth should be sufficiently low so that the sensor noise will not become a problem.
5. The bandwidth of the system with the minimum-order observer is higher than that of the system with the full-order observer, provided that the multiple observer poles are placed at the same place for both observers. If the sensor noise is a seri-ous problem, use of a full-order observer is recomnended.
10–7 DESIGN OF CONTROL SYSTEMS WITH OBSERVERS In Section 10–6 we discussed the design of regulator systems with observers.(The systems did not have reference or command inputs.) In this section we consider the design of control systems with observers when the systems have reference inputs or command inputs. The output of the control system must follow the input that is time varying. In following the command input, the system must exhibit satisfactory performance (a reasonable rise time, overshoot, settling time, and so on).
In this section we consider control systems that are designed by use of the pole-placement-with-observer approach. Specifically, we consider control systems using observer controllers. In Section 10–6 we discussed regulator systems, whose block diagram is shown in Figure 10–25. This system has no reference input, or r=0. When the system has a reference input, several different block diagram configurations are conceivable, each having an observer controller.Two of these configurations are shown in Figures 10–26 (a) and (b); we shall consider them in this section.
r = 0 y u Plant –y Observer controller + – Figure 10–25 Regulator system.
Section 10–7 / Design of Control Systems with Observers 787 r y Plant r + u Observer controller + – –u r y u Plant r – y Observer controller + – (a) (b) r y u Observer controller + – 1 s(s2 + 1) Plant Figure 10–27 Control system with observer controller in the feedforward path.
Configuration 1: Consider the system shown in Figure 10–27. In this system the refer-ence input is simply added at the summing point.We would like to design the observer controller such that in the unit-step response the maximum overshoot is less than 30% and the settling time is about 5 sec.
In what follows we first design a regulator system.Then,using the observer controller designed, we simply add the reference input r at the summing point.
Before we design the observer controller, we need to obtain a state-space represen-tation of the plant. Since we obtain By choosing the state variables as we get y = Cx x # = Ax + Bu x3 = y $ x2 = y # x1 = y y % + y # = u Y(s) U(s) = 1 sAs2 + 1B Figure 10–26 (a) Control system with observer controller in the feedforward path; (b) Control system with observer controller in the feedback path.
788 Chapter 10 / Control Systems Design in State Space where Next, we choose the desired closed-loop poles for pole placement at s=–1+j, s=–1-j, s=–8 and the desired observer poles at s=–4, s=–4 The state feedback gain matrix K and the observer gain matrix Ke can be obtained as follows: See MATLAB Program 10–16. Ke = B 8 15R K = [16 17 10] A = C 0 0 0 1 0 -1 0 1 0 S , B = C 0 0 1 S , C = [1 0 0] MATLAB Program 10–16 A = [0 1 0;0 0 1;0 -1 0]; B = [0;0;1]; J = [-1+j -1-j -8]; K = acker(A,B,J) K = 16 17 10 Aab = [1 0]; Abb = [0 1;-1 0]; L = [-4 -4]; Ke = acker(Abb',Aab',L)' Ke = 8 15 The transfer function of the observer controller is obtained by use of MATLAB Program 10–17.The result is = 302(s + 0.5017 + j0.772)(s + 0.5017 - j0.772) (s + 9 + j5.6569)(s + 9 - j5.6569) Gc(s) = 302s2 + 303s + 256 s2 + 18s + 113 Section 10–7 / Design of Control Systems with Observers 789 Figure 10–28 shows the block diagram of the regulator system just designed. Figure 10–29 shows the block diagram of a possible configuration of the control system based on the regulator system shown in Figure 10–28. The unit-step response curve for this control system is shown in Figure 10–30.The maximum overshoot is about 28% and the settling time is about 4.5 sec.Thus, the designed system satisfies the design requirements.
MATLAB Program 10–17 % Determination of transfer function of observer controller A = [0 1 0;0 0 1;0 -1 0]; B = [0;0;1]; Aaa = 0; Aab = [1 0]; Aba = [0;0]; Abb = [0 1;-1 0]; Ba = 0; Bb = [0;1]; Ka = 16; Kb=[17 10]; Ke = [8;15]; Ahat = Abb - KeAab; Bhat = AhatKe + Aba - KeAaa; Fhat = Bb - KeBa; Atilde = Ahat - FhatKb; Btilde = Bhat - Fhat(Ka + KbKe); Ctilde = -Kb; Dtilde = -(Ka + KbKe); [num,den] = ss2tf(Atilde,Btilde,-Ctilde,-Dtilde) num = 302.0000 303.0000 256.0000 den = 1 18 113 y –y u – 302s2 + 303s + 256 s2 + 18s + 113 Observer controller 1 s(s2 + 1) Plant y r – y r u – + 302s2 + 303s + 256 s2 + 18s + 113 1 s(s2 + 1) Observer controller Plant Figure 10–29 Control system with observer controller in the feedforward path.
Figure 10–28 Regulator system with observer controller.
790 Chapter 10 / Control Systems Design in State Space Output y t (sec) Unit-Step Response of (302s2 + 303s + 256)/(s5 +18s4 + 114s3 + 320s2 + 416s + 256) 0.6 0.8 1 1.2 1.4 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 Figure 10–30 Unit-step response of the control system shown in Figure 10–29.
Configuration 2: A different configuration of the control system is shown in Figure 10–31.The observer controller is placed in the feedback path.The input r is introduced into the closed-loop system through the box with gain N. From this block diagram, the closed-loop transfer function is obtained as We determine the value of constant N such that for a unit-step input r, the output y is unity as t approaches infinity.Thus we choose The unit-step response of the system is shown in Figure 10–32. Notice that the maxi-mum overshoot is very small, approximately 4%.The settling time is about 5 sec.
N = 256 113 = 2.2655 Y(s) R(s) = NAs2 + 18s + 113B sAs2 + 1BAs2 + 18s + 113B + 302s2 + 303s + 256 y Nr + u –u r – + 302s2 + 303s + 256 s2 + 18s + 113 1 s(s2 + 1) N Figure 10–31 Control system with observer controller in the feedback path.
Section 10–7 / Design of Control Systems with Observers 791 Output y t (sec) Unit-Step Response of (2.2655s2 + 40.779s + 256)/(s5 + 18s4 + 114s3 + 320s2 + 416s + 256) 0.6 0.8 1 1.2 1.4 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 Figure 10–32 The unit-step response of the system shown in Figure 10–31. (The closed-loop poles for pole placement are at s=–1 ; j, s=–8. The observer poles are at s=–4, s=–4.) Comments.
We considered two possible configurations for the closed-loop control systems using observer controllers.As stated earlier, other configurations are possible.
The first configuration,which places the observer controller in the feedforward path,gen-erally gives a fairly large overshoot.The second configuration,which places the observer con-troller in the feedback path,gives a smaller overshoot.This response curve is quite similar to that of the system designed by the pole-placement approach without using the observer con-troller. See the unit-step response curve of the system, shown in Figure 10–33, designed by the pole-placement approach without observer.Here the desired closed-loop poles used are s = -1 + j, s = -1 - j, s = -8 Output y t (sec) Unit-Step Response of System without Observer 0.6 0.8 1 1.2 1.4 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 Figure 10–33 The unit-step response of the control system designed by the pole placement approach without observer.
(The closed-loop poles are at s=–1 ; j, s=–8.) 792 Chapter 10 / Control Systems Design in State Space Frequency (rad/sec) Bode Diagrams of Closed-Loop Systems −300 0 −100 −200 100 −150 −100 Phase (deg); Magnitude (dB) 50 −50 0 10−1 100 101 102 System 1 System 2 System 1 System 2 Figure 10–34 Bode diagrams of closed-loop system 1 (shown in Figure 10–29) and closed-loop system 2 (shown in Figure 10–31).
Note that, in these two systems, the rise time and settling time are determined primari-ly by the desired closed-loop poles for pole placement. (See Figures 10–32 and 10–33.) The Bode diagrams of closed-loop system 1 (shown in Figure 10–29) and closed-loop system 2 (shown in Figure 10–31) are shown in Figure 10–34. From this figure, we find that the bandwidth of system 1 is 5 radsec and that of system 2 is 1.3 radsec.
Summary of State-Space Design Method 1. The state-space design method based on the pole-placement-combined-with-observer approach is very powerful.It is a time-domain method.The desired closed-loop poles can be arbitrarily placed, provided the plant is completely state controllable.
2. If not all state variables can be measured, an observer must be incorporated to estimate the unmeasurable state variables.
3. In designing a system using the pole-placement approach, several different sets of desired closed-loop poles need be considered, the response characteristics compared, and the best one chosen.
4. The bandwidth of the observer controller is generally large, because we choose observer poles far to the left in the s plane. A large bandwidth passes high-frequency noises and causes the noise problem.
5. Adding an observer to the system generally reduces the stability margin. In some cases, an observer controller may have zero(s) in the right-half s plane, which means that the controller may be stable but of nonminimum phase. In other cases, the controller may have pole(s) in the right-half s plane—that is, the controller is unstable.Then the designed system may become conditionally stable.
6. When the system is designed by the pole-placement-with-observer approach, it is advisable to check the stability margins (phase margin and gain margin), using a Section 10–8 / Quadratic Optimal Regulator Systems 793 frequency-response method. If the system designed has poor stability margins, it is possible that the designed system may become unstable if the mathematical model involves uncertainties.
7. Note that for nth-order systems, classical design methods (root-locus and frequency-response methods) yield low-order compensators (first or second order).
Since the observer-based controllers are nth-order Cor (N-m)th-order if the minimum-order observer is usedD for an nth-order system, the designed system will become 2nth order Cor (2n-m)th orderD.Since lower-order compensators are cheaper than higher-order ones, the designer should first apply classical methods and, if no suitable compensators can be determined, then try the pole-placement-with-observer design approach presented in this chapter.
10–8 QUADRATIC OPTIMAL REGULATOR SYSTEMS An advantage of the quadratic optimal control method over the pole-placement method is that the former provides a systematic way of computing the state feedback control gain matrix.
Quadratic Optimal Regulator Problems.
We shall now consider the optimal regulator problem that, given the system equation (10–112) determines the matrix K of the optimal control vector (10–113) so as to minimize the performance index (10–114) where Q is a positive-definite (or positive-semidefinite) Hermitian or real symmetric matrix and R is a positive-definite Hermitian or real symmetric matrix. Note that the second term on the right-hand side of Equation (10–114) accounts for the expenditure of the energy of the control signals. The matrices Q and R determine the relative importance of the error and the expenditure of this energy. In this problem, we assume that the control vector u(t) is unconstrained.
As will be seen later,the linear control law given by Equation (10–113) is the optimal control law. Therefore, if the unknown elements of the matrix K are determined so as to minimize the performance index, then u(t)=–Kx(t) is optimal for any initial state x(0).The block diagram showing the optimal configuration is shown in Figure 10–35.
J = 3 q 0 (xQx + u Ru)dt u(t) = -Kx(t) x # = Ax + Bu x = Ax + Bu .
x u –K Figure 10–35 Optimal regulator system.
794 Chapter 10 / Control Systems Design in State Space Now let us solve the optimization problem. Substituting Equation (10–113) into Equation (10–112), we obtain In the following derivations, we assume that the matrix A-BK is stable, or that the eigenvalues of A-BK have negative real parts.
Substituting Equation (10–113) into Equation (10–114) yields Let us set where P is a positive-definite Hermitian or real symmetric matrix.Then we obtain Comparing both sides of this last equation and noting that this equation must hold true for any x, we require that (10–115) It can be proved that if A-BK is a stable matrix, there exists a positive-definite ma-trix P that satisfies Equation (10–115). (See Problem A–10–15.) Hence our procedure is to determine the elements of P from Equation (10–115) and see if it is positive definite. (Note that more than one matrix P may satisfy this equation.
If the system is stable, there always exists one positive-definite matrix P to satisfy this equation.This means that, if we solve this equation and find one positive-definite matrix P,the system is stable.Other P matrices that satisfy this equation are not positive definite and must be discarded.) The performance index J can be evaluated as Since all eigenvalues of A-BK are assumed to have negative real parts, we have Therefore, we obtain (10–116) Thus, the performance index J can be obtained in terms of the initial condition x(0) and P.
To obtain the solution to the quadratic optimal control problem, we proceed as follows: Since R has been assumed to be a positive-definite Hermitian or real symmetric matrix, we can write R = T T J = x(0) Px(0) x(q) S 0.
J = 3 q 0 x(Q + K RK)x dt = -x Px2 q 0 = -x(q) Px(q) + x(0) Px(0) (A - BK) P + P(A - BK) = -(Q + K RK) x(Q + K RK) x = -x # Px - x Px # = -xC(A - BK) P + P(A - BK)D x x(Q + KRK) x = - d dt (xPx) = 3 q 0 x(Q + KRK) x dt J = 3 q 0 (xQx + xKRKx)dt x # = Ax - BKx = (A - BK) x Section 10–8 / Quadratic Optimal Regulator Systems 795 where T is a nonsingular matrix.Then Equation (10–115) can be written as which can be rewritten as The minimization of J with respect to K requires the minimization of with respect to K. (See Problem A–10–16.) Since this last expression is nonnegative, the minimum occurs when it is zero, or when Hence, (10–117) Equation (10–117) gives the optimal matrix K.Thus,the optimal control law to the quad-ratic optimal control problem when the performance index is given by Equation (10–114) is linear and is given by The matrix P in Equation (10–117) must satisfy Equation (10–115) or the following reduced equation: (10–118) Equation (10–118) is called the reduced-matrix Riccati equation.The design steps may be stated as follows: 1. Solve Equation (10–118), the reduced-matrix Riccati equation, for the matrix P.
[If a positive-definite matrix P exists (certain systems may not have a positive-definite matrix P), the system is stable, or matrix A-BK is stable.] 2. Substitute this matrix P into Equation (10–117). The resulting matrix K is the optimal matrix.
A design example based on this approach is given in Example 10–9. Note that if the matrix A-BK is stable, the present method always gives the correct result.
Finally, note that if the performance index is given in terms of the output vector rather than the state vector, that is, then the index can be modified by using the output equation to (10–119) and the design steps presented in this section can be applied to obtain the optimal matrix K.
J = 3 q 0 (x C QCx + u Ru)dt y = Cx J = 3 q 0 (y Qy + u Ru)dt A P + PA - PBR-1 B P + Q = 0 u(t) = -Kx(t) = -R-1 B Px(t) K = T-1(T)-1 B P = R-1 B P TK = (T)-1 B P xCTK - (T)-1 B PDCTK - (T)-1 B PD x AP + PA + CTK - (T)-1B PDCTK - (T)-1B PD - PBR-1BP + Q = 0 (A - K B) P + P(A - BK) + Q + K T TK = 0 796 Chapter 10 / Control Systems Design in State Space EXAMPLE 10–9 Consider the system shown in Figure 10–36.Assuming the control signal to be determine the optimal feedback gain matrix K such that the following performance index is minimized: where (m 0) From Figure 10–36, we find that the state equation for the plant is where We shall demonstrate the use of the reduced-matrix Riccati equation in the design of the optimal control system. Let us solve Equation (10–118), rewritten as Noting that matrix A is real and matrix Q is real symmetric, we see that matrix P is a real sym-metric matrix. Hence, this last equation can be written as This equation can be simplified to B 0 p11 0 p12 R + B 0 0 p11 p12 R - B p2 12 p12 p22 p12 p22 p2 22 R + B 1 0 0 mR = B 0 0 0 0R - B p11 p12 p12 p22 R B 0 1R [0 1]B p11 p12 p12 p22 R + B 1 0 0 mR = B 0 0 0 0R B 0 1 0 0R B p11 p12 p12 p22 R + B p11 p12 p12 p22 R B 0 0 1 0R A P + PA - PBR-1 B P + Q = 0 A = B 0 0 1 0R , B = B0 1R x # = Ax + Bu Q = B 1 0 0 mR J = 3 q 0 AxT Qx + u2B dt u(t) = -Kx(t) u x1 Plant x2 –K Figure 10–36 Control system.
Section 10–8 / Quadratic Optimal Regulator Systems 797 from which we obtain the following three equations: Solving these three simultaneous equations for p11, p12, and p22, requiring P to be positive definite, we obtain Referring to Equation (10–117), the optimal feedback gain matrix K is obtained as Thus, the optimal control signal is (10–120) Note that the control law given by Equation (10–120) yields an optimal result for any initial state under the given performance index. Figure 10–37 is the block diagram for this system.
Since the characteristic equation is if m=1, the two closed-loop poles are located at These correspond to the desired closed-loop poles when m=1.
Solving Quadratic Optimal Regulator Problems with MATLAB.
In MATLAB, the command lqr(A,B,Q,R) s = -0.866 + j 0.5, s = -0.866 - j 0.5 ∑s I - A + BK∑= s2 + 1m + 2 s + 1 = 0 u = -Kx = -x1 - 1m + 2 x2 = C1 1m + 2D = Cp12 p22D = [0 1]B p11 p12 p12 p22 R K = R-1 B P P = B p11 p12 p12 p22 R = B 1m + 2 1 1 1m + 2R m + 2p12 - p2 22 = 0 p11 - p12 p22 = 0 1 - p2 12 = 0 – – u x1 Plant x2 m + 2 Figure 10–37 Optimal control of the plant shown in Figure 10–36.
798 Chapter 10 / Control Systems Design in State Space solves the continuous-time,linear,quadratic regulator problem and the associated Riccati equation. This command calculates the optimal feedback gain matrix K such that the feedback control law minimizes the performance index subject to the constraint equation Another command [K,P,E] = lqr(A,B,Q,R) returns the gain matrix K,eigenvalue vector E,and matrix P,the unique positive-definite solution to the associated matrix Riccati equation: If matrix A-BK is a stable matrix,such a positive-definite solution P always exists.The eigenvalue vector E gives the closed-loop poles of A-BK.
It is important to note that for certain systems matrix A-BK cannot be made a sta-ble matrix, whatever K is chosen. In such a case, there does not exist a positive-definite matrix P for the matrix Riccati equation. For such a case, the commands K = lqr(A,B,Q,R) [K,P,E] = lqr(A,B,Q,R) do not give the solution. See MATLAB Program 10–18.
EXAMPLE 10–10 Consider the system defined by Show that the system cannot be stabilized by the state-feedback control scheme whatever matrix K is chosen. (Notice that this system is not state controllable.) Define Then = B -1 - k1 0 1 - k2 2 R A - BK = B -1 0 1 2R - B 1 0R Ck1 k2D K = Ck1 k2D u = -Kx B x # 1 x # 2 R = B -1 0 1 2R B x1 x2 R + B 1 0R u PA + A P - PBR-1 B P + Q = 0 x # = Ax + Bu J = 3 q 0 (x Qx + u Ru)dt u = -Kx Section 10–8 / Quadratic Optimal Regulator Systems 799 Hence, the characteristic equation becomes The closed-loop poles are located at Since the pole at s=2 is in the right-half s plane, the system is unstable whatever K matrix is chosen. Hence, quadratic optimal control techniques cannot be applied to this system.
Let us assume that matrices Q and R in the quadratic performance index are given by and that we write MATLAB Program 10–18.The resulting MATLAB solution is K = [NaN NaN] (NaN means ‘not a number.’) Whenever the solution to a quadratic optimal control problem does not exist, MATLAB tells us that matrix K consists of NaN.
Q = B 1 0 0 1R , R = s = -1 - k1 , s = 2 = As + 1 + k1B(s - 2) = 0 ∑s I - A + BK∑= 2 s + 1 + k1 0 -1 + k2 s - 2 2 MATLAB Program 10–18 % ---------- Design of quadratic optimal regulator system ----------A = [-1 1;0 2]; B = [1;0]; Q = [1 0;0 1]; R = ; K = lqr(A,B,Q,R) Warning: Matrix is singular to working precision.
K = NaN NaN % If we enter the command [K,P,E] = lqr(A,B,Q,R), then [K,P,E] = lqr(A,B,Q,R) Warning: Matrix is singular to working precision.
K = NaN NaN P = -Inf -Inf -Inf -Inf E = -2.0000 -1.4142 800 Chapter 10 / Control Systems Design in State Space EXAMPLE 10–11 Consider the system described by where The performance index J is given by where Assume that the following control u is used.
Determine the optimal feedback gain matrix K.
The optimal feedback gain matrix K can be obtained by solving the following Riccati equation for a positive-definite matrix P: The result is Substituting this P matrix into the following equation gives the optimal K matrix: Thus, the optimal control signal is given by MATLAB 10–19 also yields the solution to this problem.
u = -Kx = -x1 - x2 = [0 1]B 2 1 1 1R = [1 1] K = R-1 B¿ P P = B 2 1 1 1R A¿ P + PA - PBR-1 B¿ P + Q = 0 u = -Kx Q = B 1 0 0 1R , R = J = 3 q 0 (x¿ Qx + u¿ Ru) dt A = B 0 0 1 -1R , B = B 0 1R x # = Ax + Bu MATLAB Program 10–19 % ---------- Design of quadratic optimal regulator system ----------A = [0 1;0 -1]; B = [0;1]; Q = [1 0; 0 1]; R = ; K = lqr(A,B,Q,R) K = 1.0000 1.0000 Section 10–8 / Quadratic Optimal Regulator Systems 801 EXAMPLE 10–12 Consider the system given by where The performance index J is given by where Obtain the positive-definite solution matrix P of the Riccati equation, the optimal feedback gain matrix K, and the eigenvalues of matrix A-BK.
MATLAB Program 10–20 will solve this problem.
Q = C 1 0 0 0 1 0 0 0 1 S , R = J = 3 q 0 (x¿ Qx + u¿ Ru) dt A = C 0 0 -35 1 0 -27 0 1 -9 S , B = C 0 0 1 S x # = Ax + Bu MATLAB Program 10–20 % ---------- Design of quadratic optimal regulator system ----------A = [0 1 0;0 0 1;-35 -27 -9]; B = [0;0;1]; Q = [1 0 0;0 1 0;0 0 1]; R = ; [K,P,E] = lqr(A,B,Q,R) K = 0.0143 0.1107 0.0676 P = 4.2625 2.4957 0.0143 2.4957 2.8150 0.1107 0.0143 0.1107 0.0676 E = -5.0958 -1.9859 + 1.7110i -1.9859 - 1.7110i 802 Chapter 10 / Control Systems Design in State Space Next, let us obtain the response x of the regulator system to the initial condition x(0), where With state feedback u=–Kx, the state equation for the system becomes Then the system, or sys, can be given by sys = ss(A-BK, eye(3), eye(3), eye(3)) MATLAB Program 10–21 produces the response to the given initial condition. The response curves are shown in Figure 10–38.
x # = Ax + Bu = (A - BK) x x(0) = C 1 0 0 S EXAMPLE 10–13 Consider the system shown in Figure 10–39. The plant is defined by the following state-space equations: where The control signal u is given by u = k1Ar - x1B - Ak2 x2 + k3 x3B = k1 r - Ak1 x1 + k2 x2 + k3 x3B A = C 0 0 0 1 0 -2 0 1 -3 S , B = C 0 0 1 S , C = [1 0 0], D = y = Cx + Du x # = Ax + Bu MATLAB Program 10–21 % Response to initial condition.
A = [0 1 0;0 0 1;-35 -27 -9]; B = [0;0;1]; K = [0.0143 0.1107 0.0676]; sys = ss(A-BK, eye(3),eye(3),eye(3)); t = 0:0.01:8; x = initial(sys,[1;0;0],t); x1 = [1 0 0]x'; x2 = [0 1 0]x'; X3 = [0 0 1]x'; subplot(2,2,1); plot(t,x1); grid xlabel('t (sec)'); ylabel('x1') subplot(2,2,2); plot(t,x2); grid xlabel('t (sec)'); ylabel('x2) subplot(2,2,3); plot(t,x3); grid xlabel('t (sec)'); ylabel('x3') Section 10–8 / Quadratic Optimal Regulator Systems 803 x1 x2 x3 0.6 0.8 1 1.2 0.4 0.2 0 0 2 4 6 8 0 2 4 6 8 0 2 4 6 8 −0.2 −3 −2 −1 0 1 2 0 0.2 −0.2 −0.4 −0.8 −0.6 −1 −1.2 t (sec) t (sec) t (sec) Figure 10–38 Response curves to initial condition.
In determining an optimal control law, we assume that the input is zero, or r=0.
Let us determine the state-feedback gain matrix K, where such that the following performance index is minimized: where Q = C q11 0 0 0 q22 0 0 0 q33 S , R = 1, x = C x1 x2 x3 S = C y y # y $ S J = 3 q 0 (x¿ Qx + u¿ Ru) dt K = Ck1 k2 k3D + – + – x = Ax + Bu .
k2 k3 y = Cx k1 r u x x2 x3 y = x1 Figure 10–39 Control system.
804 Chapter 10 / Control Systems Design in State Space To get a fast response, q11 must be sufficiently large compared with q22, q33, and R. In this problem, we choose To solve this problem with MATLAB, we use the command K = lqr(A,B,Q,R) MATLAB Program 10–22 yields the solution to this problem.
q11 = 100, q22 = q33 = 1, R = 0.01 MATLAB Program 10–22 % ---------- Design of quadratic optimal control system ----------A = [0 1 0;0 0 1;0 -2 -3]; B = [0;0;1]; Q = [100 0 0;0 1 0;0 0 1]; R = [0.01]; K = Iqr(A,B,Q,R) K = 100.0000 53.1200 11.6711 Next we shall investigate the step-response characteristics of the designed system using the matrix K thus determined.The state equation for the designed system is and the output equation is To obtain the unit-step response, use the following command: [y,x,t] = step(AA,BB,CC,DD) where AA=A-BK, BB=Bk1, CC=C, DD=D MATLAB Program 10–23 produces the unit-step response of the designed system. Figure 10–40 shows the response curves x1, x2, and x3 versus t on one diagram.
y = Cx = [1 0 0]C x1 x2 x3 S = (A - BK) x + Bk1 r = Ax + BA-Kx + k1 rB x # = Ax + Bu Section 10–8 / Quadratic Optimal Regulator Systems 805 MATLAB Program 10–23 % ---------- Unit-step response of designed system ----------A = [0 1 0;0 0 1;0 -2 -3]; B = [0;0;1] C = [1 0 0]; D = ; K = [100.0000 53.1200 11.6711]; k1 = K(1); k2 = K(2); k3 = K(3); % Define the state matrix, control matrix, output matrix, % and direct transmission matrix of the designed systems as AA, % BB, CC, and DD AA = A - BK; BB = Bk1; CC = C; DD = D; t = 0:0.01:8; [y,x,t] = step (AA,BB,CC,DD,1,t); plot(t,x) grid title('Response Curves x1, x2, x3, versus t') xlabel('t Sec') ylabel('x1,x2,x3') text(2.6,1.35,'x1') text(1.2,1.5,'x2') text(0.6,3.5,'x3') Response Curves x1, x2, x3 versus t x1, x2, x3 5 –2 1 4 2 0 –1 3 t Sec 0 5 2 3 1 4 8 6 7 x3 x2 x1 Figure 10–40 Response curves x1 versus t, x2 versus t, and x3 versus t.
806 Chapter 10 / Control Systems Design in State Space Concluding Comments on Optimal Regulator Systems 1. Given any initial state x(t0), the optimal regulator problem is to find an allowable control vector u(t) that transfers the state to the desired region of the state space and for which the performance index is minimized. For the existence of an optimal control vector u(t), the system must be completely state controllable.
2. The system that minimizes (or maximizes, as the case may be) the selected performance index is, by definition, optimal. Although the controller may have nothing to do with “optimality” in many practical applications, the important point is that the design based on the quadratic performance index yields a stable control system.
3. The characteristic of an optimal control law based on a quadratic performance index is that it is a linear function of the state variables, which implies that we need to feed back all state variables.This requires that all such variables be available for feedback. If not all state variables are available for feedback, then we need to employ a state observer to estimate unmeasurable state variables and use the es-timated values to generate optimal control signals.
Note that the closed-loop poles of the system designed by the use of the quadratic optimal regulator approach can be found from Since these closed-loop poles correspond to the desired closed-loop poles in the pole-placement approach, the transfer functions of the observer controllers can be obtained from either Equation (10–74) if the observer is of full-order type or Equation (10–108) if the observer is of minimum-order type.
4. When the optimal control system is designed in the time domain, it is desirable to investigate the frequency-response characteristics to compensate for noise effects.
The system frequency-response characteristics must be such that the system at-tenuates highly in the frequency range where noise and resonance of components are expected.(To compensate for noise effects,we must in some cases either modify the optimal configuration and accept suboptimal performance or modify the performance index.) 5. If the upper limit of integration in the performance index J given by Equation (10–114) is finite, then it can be shown that the optimal control vector is still a linear function of the state variables,but with time-varying coefficients.(Therefore, the determination of the optimal control vector involves that of optimal time-varying matrices.) 10–9 ROBUST CONTROL SYSTEMS Suppose that given a control object (i.e., a system with a flexible arm) we wish to de-sign a control system. The first step in the design of a control system is to obtain a mathematical model of the control object based on the physical law. Quite often the model may be nonlinear and possibly with distributed parameters. Such a model may be difficult to analyze. It is desirable to approximate it by a linear constant-coefficient system that will approximate the actual object fairly well. Note that even though the ∑s I - A + BK∑= 0 model to be used for design purposes may be a simplified one, it is necessary that such a model must include any intrinsic character of the actual object.Assuming that we can get a model that approximates the actual system quite well, we must get a simplified model for the purpose of designing the control system that will require a compensator of lowest order possible. Thus, a model of a control object (whatever it may be) will probably include an error in the modeling process. Note that in the frequency-response approach to control systems design, we use phase and gain margins to take care of the modeling errors. However, in the state-space approach, which is based on the dif-ferential equations of the plant dynamics, no such “margins” are involved in the design process.
Since the actual plant differs from the model used in the design, a question arises whether the controller designed using a model will work satisfactorily with the actu-al plant. To ensure that it will do so, robust control theory has been developed since around 1980.
Robust control theory uses the assumption that the models we use in designing con-trol systems have modeling errors.We shall present an introduction to this theory in this section. Basically, the theory assumes that there is an uncertainty or error between the actual plant and its mathematical model and includes such uncertainty or error in the design process of the control system.
Systems designed based on the robust control theory will possess the following properties: (1) Robust stability.
The control system designed is stable in the presence of perturbation.
(2) Robust performance.
The control system exhibits predetermined response characteristics in the presence of perturbation.
This theory requires considerations based on frequency-response analysis and time-domain analysis. Because of the mathematical complications associated with robust con-trol theory, detailed discussion of robust control theory is beyond the scope of the senior engineering student. In this section, only introductory discussion of robust control the-ory is presented.
Uncertain Elements in Plant Dynamics.
The term uncertainty refers to the dif-ferences or errors between the model of the plant and the actual plant.
Uncertain elements that may appear in practical systems may be classified as struc-tured uncertainty and unstructured uncertainty.An example of structured uncertainty is any parametric variation in the plant dynamics, such as variations in poles and zeros of the plant transfer function. Examples of unstructured uncertainty include frequency-dependent uncertainty, such as high-frequency modes that we normally neglect in mod-eling plant dynamics. For example, in the modeling of a flexible-arm system, the model may include a finite number of modes of oscillation.The modes of oscillation that are not included in the modeling behave as uncertainty of the system. Another example of un-certainty occurs in the linearization of a nonlinear plant. If the actual plant is nonlinear and its model is linear, then the difference acts as unstructured uncertainty.
In this section we consider the case where the uncertainty is unstructured. In addi-tion we assume that the plant involves only one uncertainty. (Some plants may involve multiple uncertain elements.) Section 10–9 / Robust Control Systems 807 In the robust control theory, we define unstructured uncertainty as . Since the exact description of is unknown, we use an estimate of (as to the magnitude and phase characteristics) and use this estimate in the design of the controller that sta-bilizes the control system. Stability of a system with unstructured uncertainty can then be examined by use of the small gain theorem to be given following the definition of the norm.
Norm.
The norm of a stable single-input–single-output system is the largest possible amplification factor of the steady-state response to sinusoidal excitation.
For a scalar (s), gives the maximum value of .It is called the norm.
See Figure 10–41.
In robust control theory we measure the magnitude of the transfer function by the norm. Assume that the transfer function is proper and stable. [Note that a transfer function is called proper if is limited and definite. If = 0, it is called strictly proper.] The norm of is defined by means the maximum singular value of . ( means .) Note that the singular value of a transfer function is defined by where is the ith largest eigenvalue of and it is always a non-negative real value. By making smaller, we make the effect of input w on the output z smaller.
It is frequently the case that instead of using the maximum singular value , we use the inequality and limit the magnitude of (s) by .To make the magnitude of small, we choose a small and require that .
7 £ 7 q 6 g g 7 £ 7 q g £ 7 £ 7 q 6 g 7 £ 7 q 7 £ 7 q ££ li(££) si(£) = 2li(££) £ smax s [£(jv)] s [£(jv)] 7 £ 7 q = s [£(jv)] £(s) Hq £(q) £(q) £(s) £(s) Hq Hq £(jv) 7 £ 7 q £ Hq H Hq ¢(s) ¢(s) ¢(s) 808 Chapter 10 / Control Systems Design in State Space F(s) ||F|| |F(jv)| in db v z v Figure 10–41 Bode diagram and the norm .
7 £7 q Hq Small-Gain Theorem.
Consider the closed-loop system shown in Figure 10–42. In the figure and M(s) are stable and proper transfer functions.
The small-gain theorem states that if then this closed-loop system is stable. That is, if the norm of M(s) is smaller than 1, this closed-loop system is stable. This theorem is an extension of the Nyquist stability criterion.
It is important to note that the small-gain theorem gives a sufficient condition for sta-bility.That is, a system may be stable even if it does not satisfy this theorem. However, if a system satisfies the small-gain theorem, it is always stable.
System with Unstructured Uncertainty.
In some cases an unstructured uncer-tainty error may be considered multiplicative such that where is the true plant dynamics and G is the model plant dynamics. In other cases an unstructured uncertainty error may be considered additive such that In either case we assume that the norm of or is bounded such that where and are positive constants.
EXAMPLE 10–14 Consider a control system with unstructured multiplicative uncertainty.We shall consider robust stability and robust performance of the system. (A system with unstructured additive uncertain-ty will be discussed in Problem A–10–18.) Robust Stability.
Let us define true plant dynamics G=model of plant dynamics unstructured multiplicative uncertainty We assume that is stable and its upper bound is known.We also assume that and G are related by = G(I + ¢m) G G ¢m ¢m = G = ga gm 7 ¢m7 6 gm , 7 ¢a7 6 ga ¢a ¢m G = G + ¢a G G = G(1 + ¢m) ¢(s) Hq 7 ¢(s)M(s)7 q 6 1 ¢(s) Section 10–9 / Robust Control Systems 809 (s) M(s) Figure 10–42 Closed-loop system.
Consider the system shown in Figure 10–43(a). Let us examine the transfer function between point A and point B. Notice that Figure 10–43(a) can be redrawn as shown in Figure 10-43(b).The transfer function between point A and point B can be given by Define (10–121) Using Equation (10–121) we can redraw Figure 10–43(b) as Figure 10–43(c).Applying the small-gain theorem to the system consisting of and T as shown in Figure 10–43(c), we obtain the condition for stability to be (10–122) In general, it is impossible to precisely model Therefore, let us use a scalar transfer function such that where is the largest singular value of .
Consider, instead of Inequality (10–122), the following inequality: (10–123) If Inequality (10–123) holds true, Inequality (10–122) will always be satisfied. By making the norm of to be less than 1, we obtain the controller K that will make the system stable.
Suppose that we cut the line at point A in Figure 10–43(a). Then we obtain Figure 10–43(d).
Replacing by , we obtain Figure 10–43(e). Redrawing Figure 10–43(e), we obtain Figure 10–43(f). Figure 10–43(f) is called a generalized plant diagram.
Referring to Equation (10–121), T is given by (10–124) Then Inequality (10–123) can be rewritten as (10–125) Clearly, for a stable plant model G(s), K(s)=0 will satisfy Inequality (10–125). However, K(s)=0 is not the desirable transfer function for the controller. To find an acceptable trans-fer function for K(s), we may add another condition—for example, that the resulting system will have robust performance such that the system output follows the input with minimum error, or another reasonable condition. In what follows we shall obtain the condition for robust performance.
ß W mK(s)G(s) 1 + K(s)G(s) ß q 6 1 T = KG 1 + KG W mI ¢m W mT Hq 7W mT7 q 6 1 ¢m(jv) s{¢m(jv)} s{¢m(jv)} 6 W m(jv) W m(jv) ¢m.
7 ¢mT7 q 6 1 ¢m (1 + KG)-1 KG = T KG 1 + KG = (1 + KG)-1 KG 810 Chapter 10 / Control Systems Design in State Space Section 10–9 / Robust Control Systems 811 K (f) w u z y P G w u y K WmI (e) y z ++ – m T B A (c) G K m (d) y u B z w A ++ – m K G y u +− A B G K m (a) (b) y u A ++ – B Figure 10–43 (a) Block diagram of a system with unstructured multiplicative uncertainty; (b)–(d) successive modifications of the block diagram of (a); (e) block diagram showing a generalized plant with unstructured multiplicative uncertainty; (f) generalized plant diagram.
Robust Performance.
Consider the system shown in Figure 10–44. Suppose that we want the output y(t) to follow the input r(t) as closely as possible, or we wish to have Since the transfer function Y(s)/R(s) is we have Define where S is commonly called the sensitivity function and T defined by Equation (10–124) is called the complementary sensitivity function. In this robust performance problem we want to make the norm of S smaller than the desired transfer function or which can be written as (10–126) Combining Inequalities (10–123) and (10–126), we get where T+S=1, or (10–127) Our problem then becomes to find K(s) that will satisfy Inequality (10–127). Note that depend-ing on the chosen Wm(s) and Ws(s) there may be many K(s) that satisfy Inequality (10–127), or may be no K(s) that satisfies Inequality (10–127). Such a robust control problem using Inequality (10–127) is called a mixed-sensitivity problem.
Figure 10–45(a) is a generalized plant diagram, where two conditions (robust stability and ro-bust performance) are specified.A simplified version of this diagram is shown in Figure 10–45(b).
∑ W m(s) K(s)G(s) 1 + K(s)G(s) W s(s) 1 1 + K(s)G(s) ∑ q 6 1 gW mT W sS g q 6 1 7W s S7 q 6 1 7 S7 q 6 Ws -1 Ws -1 Hq 1 1 + KG = S E(s) R(s) = R(s) - Y(s) R(s) = 1 -Y(s) R(s) = 1 1 + KG Y(s) R(s) = KG 1 + KG lim tS q [r(t) - y(t)] = lim tS q e(t) S 0 812 Chapter 10 / Control Systems Design in State Space r e y K(s) G(s) + – Figure 10–44 Closed-loop system.
Section 10–9 / Robust Control Systems 813 K (b) w u z y P G w u y K WmI WsI (a) y z2 z1 z ++ – Figure 10–45 (a) Generalized plant diagram; (b) simplfied version of the generalized plant diagram shown in (a).
Finding Transfer Function z(s)/w(s) from a Generalized Plant Diagram.
Consider the generalized plant diagram shown in Figure 10–46.
In this diagram w(s) is the exogenous disturbance and u(s) is the manipulated vari-able. z(s) is the controlled variable and y(s) is the observed variable.
Consider this control system consisting of the generalized plant P(s) and the con-troller K(s).The equation that relates the outputs z(s) and y(s) and the inputs w(s) and u(s) of the generalized plant P(s) is The equation that relates u(s) and y(s) is given by u(s)=K(s)y(s) Define the transfer function that relates the controlled variable z(s) to the exogenous disturbance w(s) as (s).Then z(s) = £(s)w(s) £ B z(s) y(s)R = B P11 P21 P12 P22 R Bw(s) u(s) R Note that can be determined as follows: Since z(s)=P11w(s)+P12u(s) y(s)=P21w(s)+P22u(s) u(s)=K(s)y(s) we obtain y(s)=P21w(s)+P22K(s)y(s) Hence or Therefore, Hence, (10–128) EXAMPLE 10–15 Let us determine the P matrix in the generalized plant diagram of the control system considered in Example 10–14. We derived Inequality (10–125) for the control system to be robust stable.
Rewriting Inequality (10–125), we have (10–129) g W mKG 1 + KG g q 6 1 £(s) = P 11 + P 12K(s)[I - P 22K(s)]-1P 21 = {P 11 + P 12K(s)[I - P 22K(s)]-1P 21}w(s) z(s) = P 11w(s) + P 12K(s)[I - P 22K(s)]-1P 21w(s) y(s) = [I - P 22K(s)]-1P 21w(s) [I - P 22K(s)]y(s) = P 21w(s) £(s) 814 Chapter 10 / Control Systems Design in State Space K(s) P(s) w u z y P11 P21 P12 P22 Figure 10–46 A generalized plant diagram.
If we define (10–130) then Inequality (10–129) can be written as Referring to Equation (10–128), rewritten as notice that if we choose the generalized plant P matrix as (10–131) Then we obtain which is exactly the same as in Equation (10–130).
We derived in Example 10–14 that if we wished to have the output y follow the input r as close as possible, we needed to make the norm of (s), where (10–132) less than 1. [See Inequality (10–126).] Note that the controlled variable z is related to the exogenous disturbance w by and referring to Equation (10–128) Notice that if we choose the P matrix as (10–133) then we obtain which is the same as in Equation (10–132).
£2 = W s c 1 1 + KG d = W s c1 -KG 1 + KG d = W s - W sKG(I + KG)-1 £ = P 11 + P 12K(I - P 22K)-1P 21 P = cW s -W sG I -G d £(s) = P 11 + P 12K(I - P 22K)-1P 21 z = £(s)w £2 = W s I + KG £2 Hq £1 = W mKG(I + KG)-1 £ = P 11 + P 12K(I - P 22K)-1P 21 P = c 0 W mG I -G d £ = P 11 + P 12K(I - P 22K)-1P 21 7 £17 q 6 1 £1 = W mKG 1 + KG Section 10–9 / Robust Control Systems 815 If both the robust stability and robust performance conditions are required, the control sys-tem must satisfy the condition given by Inequality (10–127), rewritten as (10–134) For the P matrix, we combine Equations (10–133) and (10–131) and get (10–135) If we construct P(s) as given by Equation (10–135), then the problem of designing a control system to satisfy both robust stability and robust performance conditions can be formulated by using the generalized plant represented by Equation (10–135). As mentioned earlier, such a problem is called a mixed-sensitivity problem. By using the generalized plant given by Equation (10–135) we are able to determine the controller K(s) that satisfies Inequality (10–134). The generalized plant diagram for the system considered in Example 10–14 becomes as shown in Figure 10–47.
H Infinity Control Problem.
To design a controller K of a control system to sat-isfy various stability and performance specifications, we utilize the concept of the gen-eralized plant.
As mentioned earlier a generalized plant is a linear model consisting of a model of the plant and weighting functions corresponding to the specifications for the required performance. Referring to the generalized plant shown in Figure 10–48, the H infinity control problem is a problem to design a controller K that will make the norm of the transfer function from the exogenous disturbance w to the controlled variable z less than a specified value.
Hq P = C Ws 0 I -WsG WmG -G S ∑ W m KG 1 + KG W s 1 1 + KG ∑6 1 816 Chapter 10 / Control Systems Design in State Space K w u z2 y O I WmG −G z1 Ws −WsG Figure 10–47 Generalized plant of the system discussed in Example 10–15.
The reason to use generalized plants, rather than individual block diagrams of con-trol systems, is that a number of control systems with uncertain elements have been designed using generalized plants and, consequently, established design approaches using such plants are available.
Note that any weighting function, such as W(s), is an important parameter to in-fluence the resulting controller K(s). In fact, the goodness of the resulting designed system depends on the choice of the weighting function or functions used in the de-sign process.
Note that the controller that is the solution to the H infinity control problem is com-monly called the H infinity controller.
Solving Robust Control Problems.
There are three established approaches to solve robust control problems.They are 1. Solve robust control problems by deriving the Riccati equations and solving them.
2. Solve robust control problems by using the linear matrix inequality approach.
3. Solve robust control problems that involve structural uncertainties by using the analysis and synthesis approach.
Solving robust control problems by use of any of the above methods requires a broad mathematical background.
In this section we have presented only an introduction to the robust control theory.
Solving any robust control problem requires mathematical background beyond the scope of the senior engineering student. Therefore, an interested reader may take a graduate-level control course at an established college or university and study this sub-ject in detail.
EXAMPLE PROBLEMS AND SOLUTIONS A–10–1.
Consider the system defined by Suppose that this system is not completely state controllable.Then the rank of the controllability matrix is less than n, or (10–136) rank CB AB p A n-1 BD = q 6 n x # = Ax + Bu m m Example Problems and Solutions 817 K w u z y Generalized plant Figure 10–48 A generalized plant diagram.
This means that there are q linearly independent column vectors in the controllability matrix. Let us define such q linearly independent column vectors as f1, f2, p , fq. Also, let us choose n-q additional n-vectors vq+1, vq+2, p , vn such that is of rank n. By using matrix P as the transformation matrix, define Show that can be given by where A11 is a qq matrix,A12 is a q(n-q) matrix,A22 is an (n-q)(n-q) matrix, and 0 is an (n-q)q matrix. Show also that matrix can be given by where B11 is a q1 matrix and 0 is an (n-q)1 matrix.
Solution. Notice that or (10–137) Also, (10–138) Since we have q linearly independent column vectors f1,f2,p ,fq, we can use the Cayley–Hamilton theorem to express vectors Af1, Af2, p , Afq in terms of these q vectors.That is, Afq = a1q f1 + a2q f2 + p + aqq fq Af2 = a12 f1 + a22 f2 + p + aq2 fq Af1 = a11 f1 + a21 f2 + p + aq1 fq B = PB ˆ = Cf1 f2 p fq vq+1 p vnD A ˆ CAf1 Af2 p Afq Avq+1 p AvnD AP = PA ˆ B ˆ = cB11 0 d B ˆ A ˆ = cA11 0 A12 A22 d A ˆ P-1 AP = A ˆ , P-1 B = B ˆ P = Cf1 f2 p fq vq+1 vq+2 p vnD 818 Chapter 10 / Control Systems Design in State Space Example Problems and Solutions 819 Hence, Equation (10–137) may be written as follows: Define Then Equation (10–137) can be written as = Cf1 f2 p fq vq+1 p vnD cA11 0 A12 A22 d CAf1 Af2 p Afq Avq+1 p AvnD E aq+1q+1 anq+1 p p aq+1n ann U = A22 E 0 0 p p 0 0 U = A21 = (n - q) q zero matrix F a1q+1 a2q+1 aqq+1 p p p a1n a2n aqn V = A12 F a11 a21 aq1 p p p a1q a2q aqq V = A11 = Cf1 f2 p fq vq+1 p vnD a11 a21 aq1 0 0 p p p p p a1q a2q aqq 0 0 a1q+1 a2q+1 aqq+1 aq+1q+1 anq+1 p p p p p a1n a2n aqn aq+1n ann CAf1 Af2 p Afq Avq+1 p AvnD Thus, Hence, Next, referring to Equation (10–138), we have (10–139) Referring to Equation (10–136), notice that vector B can be written in terms of q linearly independent column vectors f1, f2, p , fq. Thus, we have Consequently, Equation (10–139) may be written as follows: Thus, where A–10–2.
Consider a completely state controllable system Define the controllability matrix as M: M = CB AB p A n-1 BD x # = Ax + Bu B11 = F b11 b21 bq1 V B ˆ = cB11 0 d b11 f1 + b21 f2 + p + bq1 fq = Cf1 f2 p fq vq+1 p vnD b11 b21 bq1 0 0 B = b11 f1 + b21 f2 + p + bq1 fq B = Cf1 f2 p fq vq+1 p vnD B ˆ P-1 AP = A ˆ = cA11 0 A12 A22 d AP = PcA11 0 A12 A22 d 820 Chapter 10 / Control Systems Design in State Space Show that where a1, a2, p , an are the coefficients of the characteristic polynomial Solution. Let us consider the case where n=3. We shall show that (10–140) The left-hand side of Equation (10–140) is The right-hand side of Equation (10–140) is (10–141) The Cayley–Hamilton theorem states that matrix A satisfies its own characteristic equation or, in the case of n=3, (10–142) Using Equation (10–142), the third column of the right-hand side of Equation (10–141) becomes Thus, Equation (10–141) becomes Hence, the left-hand side and the right-hand side of Equation (10–140) are the same. We have thus shown that Equation (10–140) is true. Consequently, The preceding derivation can be easily extended to the general case of any positive integer n.
A–10–3.
Consider a completely state controllable system Define M = CB AB p A n-1 BD x # = Ax + Bu M-1 AM = C 0 1 0 0 0 1 -a3 -a2 -a1 S CB AB A 2 BD C 0 1 0 0 0 1 -a3 -a2 -a1 S = CAB A 2 B A 3 BD -a3 B - a2 AB - a1 A 2 B = A-a3 I - a2 A - a1 A 2BB = A 3 B A 3 + a1 A 2 + a2 A + a3 I = 0 CB AB A 2 BD C 0 1 0 0 0 1 -a3 -a2 -a1 S = CAB A 2 B -a3 B - a2 AB - a1 A 2 BD AM = ACB AB A 2 BD = CAB A 2 B A 3 BD AM = MC 0 1 0 0 0 1 -a3 -a2 -a1 S ∑s I - A∑= sn + a1 sn-1 + p + an-1 s + an M-1 AM = G 0 1 0 0 0 0 1 0 p p p p 0 0 0 1 -an -an-1 -an-2 -a1 W Example Problems and Solutions 821 822 Chapter 10 / Control Systems Design in State Space and where the ai’s are coefficients of the characteristic polynomial Define also Show that Solution. Let us consider the case where n=3. We shall show that (10–143) Referring to Problem A–10–2, we have Hence, Equation (10–143) can be rewritten as Therefore, we need to show that (10–144) The left-hand side of Equation (10–144) is C 0 1 0 0 0 1 -a3 -a2 -a1 S C a2 a1 1 a1 1 0 1 0 0 S = C -a3 0 0 0 a1 1 0 1 0 S C 0 1 0 0 0 1 -a3 -a2 -a1 S W = WC 0 0 -a3 1 0 -a2 0 1 -a1 S W-1C 0 1 0 0 0 1 -a3 -a2 -a1 S W = C 0 0 -a3 1 0 -a2 0 1 -a1 S M-1 AM = C 0 1 0 0 0 1 -a3 -a2 -a1 S T-1 AT = (MW)-1 A(MW) = W-1(M-1 AM) W = C 0 0 -a3 1 0 -a2 0 1 -a1 S T-1 AT = G 0 0 0 -an 1 0 0 -an-1 0 1 0 -an-2 p p p p 0 0 1 -a1 W , T-1 B = G 0 0 0 1 W T = MW ∑s I - A∑= sn + a1 sn-1 + p + an-1 s + an W = G an-1 an-2 a1 1 an-2 an-3 1 0 p p p p a1 1 0 0 1 0 0 0 W Example Problems and Solutions 823 The right-hand side of Equation (10–144) is Clearly, Equation (10–144) holds true.Thus, we have shown that Next, we shall show that (10–145) Note that Equation (10–145) can be written as Noting that we have The derivation shown here can be easily extended to the general case of any positive integer n.
A–10–4.
Consider the state equation where The rank of the controllability matrix M, is 2.Thus, the system is completely state controllable.Transform the given state equation into the controllable canonical form.
Solution. Since = s2 + 2s + 1 = s2 + a1 s + a2 ∑s I - A∑= 2s - 1 4 -1 s + 32 = (s - 1)(s + 3) + 4 M = CB ABD = B 0 2 2 -6R A = B 1 -4 1 -3R , B = B 0 2R x # = Ax + Bu T-1 B = C 0 0 1 S TC 0 0 1 S = CB AB A 2 BD C a2 a1 1 a1 1 0 1 0 0 S C 0 0 1 S = CB AB A 2 BD C 1 0 0 S = B B = TC 0 0 1 S = MWC 0 0 1 S T-1 B = C 0 0 1 S T-1 AT = C 0 0 -a3 1 0 -a2 0 1 -a1 S C a2 a1 1 a1 1 0 1 0 0 S C 0 0 -a3 1 0 -a2 0 1 -a1 S = C -a3 0 0 0 a1 1 0 1 0 S 824 Chapter 10 / Control Systems Design in State Space we have Define where Then and Define Then the state equation becomes Since and we have which is in the controllable canonical form.
A–10–5.
Consider a system defined by where A = B 0 -2 1 -3R , B = B 0 2R , C = [1 0] y = Cx x # = Ax + Bu B x ˆ # 1 x ˆ # 2 R = B 0 -1 1 -2R B x ˆ 1 x ˆ 2 R + B 0 1R u T-1 B = B 0.5 0.5 0 0.5R B 0 2R = B 0 1R T-1AT = B 0.5 0.5 0 0.5R B 1 -4 1 -3R B 2 -2 0 2R = B 0 -1 1 -2R x ˆ # = T-1 ATx ˆ + T-1Bu x = Tx ˆ T-1 = B 0.5 0.5 0 0.5R T = B 0 2 2 -6R B 2 1 1 0R = B 2 -2 0 2R M = B0 2 2 -6R , W = B 2 1 1 0R T = MW a1 = 2, a2 = 1 Example Problems and Solutions 825 The characteristic equation of the system is The eigenvalues of matrix A are –1 and –2.
It is desired to have eigenvalues at –3 and –5 by using a state-feedback control u=–Kx.
Determine the necessary feedback gain matrix K and the control signal u.
Solution. The given system is completely state controllable, since the rank of is 2. Hence, arbitrary pole placement is possible.
Since the characteristic equation of the original system is we have The desired characteristic equation is Hence, It is important to point out that the original state equation is not in the controllable canonical form, because matrix B is not Hence, the transformation matrix T must be determined.
Hence, Referring to Equation (10–13), the necessary feedback gain matrix is given by Thus, the control signal u becomes u = -Kx = -[6.5 2.5]B x1 x2 R = C15 - 2 8 - 3D B 0.5 0 0 0.5R = [6.5 2.5] K = Ca2 - a2 a1 - a1D T-1 T-1 = B 0.5 0 0 0.5R T = MW = CB ABD B a1 1 1 0R = B 0 2 2 -6R B 3 1 1 0R = B 2 0 0 2R B 0 1R a1 = 8, a2 = 15 (s + 3)(s + 5) = s2 + 8s + 15 = s2 + a1 s + a2 = 0 a1 = 3, a2 = 2 s2 + 3s + 2 = s2 + a1 s + a2 = 0 M = CB ABD = B 0 2 2 -6R ∑s I - A∑= 2 s 2 -1 s + 32 = s2 + 3s + 2 = (s + 1)(s + 2) = 0 826 Chapter 10 / Control Systems Design in State Space A–10–6.
A regulator system has a plant Define state variables as By use of the state-feedback control u=–Kx, it is desired to place the closed-loop poles at Obtain the necessary state-feedback gain matrix K with MATLAB.
Solution. The state-space equations for the system become Hence, (Note that, for the pole placement, matrices C and D do not affect the state-feedback gain matrix K.) Two MATLAB programs for obtaining state-feedback gain matrix K are given in MATLAB Programs 10–24 and 10–25. C = [1 0 0], D = A = C 0 0 -6 1 0 -11 0 1 -6 S , B = C 0 0 10 S y = [1 0 0]C x1 x2 x3 S + 0u C x # 1 x # 2 x # 3 S = C 0 0 -6 1 0 -11 0 1 -6 S C x1 x2 x3 S + C 0 0 10 S u s = -2 + j213, s = -2 - j213, s = -10 x3 = x # 2 x2 = x # 1 x1 = y Y(s) U(s) = 10 (s + 1)(s + 2)(s + 3) MATLAB Program 10–24 A = [0 1 0;0 0 1;-6 -11 -6]; B = [0;0;10]; J = [-2+j2sqrt(3) -2-j2sqrt(3) -10]; K = acker(A,B,J) K = 15.4000 4.5000 0.8000 Example Problems and Solutions 827 A–10–7.
Consider a completely observable system Define the observability matrix as N: Show that (10–146) where a1, a2, p , an are the coefficients of the characteristic polynomial Solution. Let us consider the case where n=3. Then Equation (10–146) can be written as (10–147) Equation (10–147) may be rewritten as (10–148) We shall show that Equation (10–148) holds true.The left-hand side of Equation (10–148) is (10–149) NA = C C CA CA 2 S A = C CA CA 2 CA 3 S N A = C 0 0 -a3 1 0 -a2 0 1 -a1 S N NA(N)-1 = C 0 0 -a3 1 0 -a2 0 1 -a1 S ∑s I - A∑= sn + a1 sn-1 + p + an-1 s + an N A(N)-1 = G 0 0 0 -an 1 0 0 an-1 0 1 0 -an-2 p p p p 0 0 1 -a1 W N = CC AC p (A)n-1 CD y = Cx x # = Ax MATLAB Program 10–25 A = [0 1 0;0 0 1; -6 -11 -6]; B = [0;0;10]; J = [-2+j2sqrt(3) -2-J2Sqrt(3) -10]; K = place(A,B,J) place: ndigits= 15 K = 15.4000 4.5000 0.8000 828 Chapter 10 / Control Systems Design in State Space The right-hand side of Equation (10–148) is (10–150) The Cayley–Hamilton theorem states that matrix A satisfies its own characteristic equation, or Hence, Thus, the right-hand side of Equation (10–150) becomes the same as the right-hand side of Equation (10–149). Consequently, which is Equation (10–148).This last equation can be modified to The derivation presented here can be extended to the general case of any positive integer n.
A–10–8.
Consider a completely observable system defined by (10–151) (10–152) Define and where the a’s are coefficients of the characteristic polynomial Define also Q = (WN)-1 ∑s I - A∑= sn + a1 sn-1 + p + an-1 s + an W = G an-1 an-2 a1 1 an-2 an-3 1 0 p p p p a1 1 0 0 1 0 0 0 W N = CC A C p (A)n-1 CD y = Cx + Du x # = Ax + Bu NA(N)-1 = C 0 0 -a3 1 0 -a2 0 1 -a1 S NA = C 0 0 -a3 1 0 -a2 0 1 -a1 S N -a1 CA 2 - a2 CA - a3 C = CA 3 A 3 + a1 A 2 + a2 A + a3 I = 0 = C CA CA 2 -a3 C - a2 CA - a1 CA 2 S C 0 0 -a3 1 0 -a2 0 1 -a1 S N = C 0 0 -a3 1 0 -a2 0 1 -a1 S C C CA CA 2 S Example Problems and Solutions 829 Show that where the bk’s (k=0, 1, 2, p , n) are those coefficients appearing in the numerator of the transfer function when C(sI-A)–1B+D is written as follows: where D=b0.
Solution. Let us consider the case where n=3. We shall show that (10–153) Note that, by referring to Problem A–10–7, we have Hence, we need to show that or (10–154) WC 0 0 -a3 1 0 -a2 0 1 -a1 S = C 0 1 0 0 0 1 -a3 -a2 -a1 S W WC 0 0 -a3 1 0 -a2 0 1 -a1 S W-1 = C 0 1 0 0 0 1 -a3 -a2 -a1 S (WN) A(WN)-1 = WCN A(N)-1D W-1 = WC 0 0 -a3 1 0 -a2 0 1 -a1 S W-1 Q-1 AQ = (WN) A(WN)-1 = C 0 1 0 0 0 1 -a3 -a2 -a1 S C(s I - A)-1 B + D = b0 sn + b1 sn-1 + p + bn-1 s + bn sn + a1 sn-1 + p + an-1 s + an Q-1 B = F bn - an b0 bn-1 - an-1 b0 b1 - a1 b0 V CQ = [0 0 p 0 1] Q-1 AQ = G 0 1 0 0 0 0 1 0 p p p p 0 0 0 1 -an -an-1 -an-2 -a1 W 830 Chapter 10 / Control Systems Design in State Space The left-hand side of Equation (10–154) is The right-hand side of Equation (10–154) is Thus, we see that Equation (10–154) holds true. Hence, we have proved Equation (10–153).
Next we shall show that or Notice that Hence, we have shown that Next define Then Equation (10–151) becomes (10–155) and Equation (10–152) becomes (10–156) Referring to Equation (10–153), Equation (10–155) becomes C x ˆ # 1 x ˆ # 2 x ˆ # 3 S = C 0 1 0 0 0 1 -a3 -a2 -a1 S C x ˆ 1 x ˆ 2 x ˆ 3 S + C g3 g2 g1 S u y = CQx ˆ + Du x ˆ # = Q-1 AQx ˆ + Q-1 Bu x = Qx ˆ [0 0 1] = C(WN )-1 = CQ = [1 0 0]C C CA CA 2 S = C 0 0 1 = [0 0 1]C a2 a1 1 a1 1 0 1 0 0 S C C CA CA 2 S C(WN)-1 = [0 0 1] CQ = [0 0 1] = C -a3 0 0 0 a1 1 0 1 0 S C 0 1 0 0 0 1 -a3 -a2 -a1 S W = C 0 1 0 0 0 1 -a3 -a2 -a1 S C a2 a1 1 a1 1 0 1 0 0 S = C -a3 0 0 0 a1 1 0 1 0 S W C 0 0 -a3 1 0 -a2 0 1 -a1 S = C a2 a1 1 a1 1 0 1 0 0 S C 0 0 -a3 1 0 -a2 0 1 -a1 S Example Problems and Solutions 831 where The transfer function G(s) for the system defined by Equations (10–155) and (10–156) is Noting that we have Note that D=b0. Since we have Hence, Thus, we have shown that Note that what we have derived here can be easily extended to the case when n is any positive integer.
A–10–9.
Consider a system defined by y = Cx x # = Ax + Bu Q-1 B = C g3 g2 g1 S = C b3 - a3 b0 b2 - a2 b0 b1 - a1 b0 S g1 = b1 - a1 b0 , g2 = b2 - a2 b0 , g3 = b3 - a3 b0 = b0 s3 + b1 s2 + b2 s + b3 s3 + a1 s2 + a2 s + a3 = b0 s3 + Ag1 + a1 b0Bs2 + Ag2 + a2 b0Bs + g3 + a3 b0 s3 + a1 s2 + a2 s + a3 = g1 s2 + g2 s + g3 s3 + a1 s2 + a2 s + a3 + b0 G(s) = 1 s3 + a1 s2 + a2 s + a3 C1 s s2D C g3 g2 g1 S + D C s -1 0 0 s -1 a3 a2 s + a1 S -1 = 1 s3 + a1 s2 + a2 s + a3 C s2 + a1 s + a2 s + a1 1 -a3 s2 + a1 s s -a3 s -a2 s - a3 s2 S G(s) = [0 0 1]C s -1 0 0 s -1 a3 a2 s + a1 S -1 C g3 g2 g1 S + D CQ = [0 0 1] G(s) = CQAs I - Q-1 AQB -1 Q-1 B + D C g3 g2 g1 S = Q-1 B 832 Chapter 10 / Control Systems Design in State Space where The rank of the observability matrix N, is 2. Hence, the system is completely observable. Transform the system equations into the ob-servable canonical form.
Solution. Since we have Define where Then and Define Then the state equation becomes or (10–157) The output equation becomes y = CQx ˆ = B0 1 -1 -2R B x ˆ 1 x ˆ 2 R + B 0 2R u B x ˆ # 1 x ˆ # 2 R = B-1 1 0 1R B 1 -4 1 -3R B -1 1 0 1R Bx ˆ 1 x ˆ 2 R + B -1 1 0 1R B 0 2R u x ˆ # = Q-1 AQx ˆ + Q-1 Bu x = Qx ˆ Q-1 = B -1 1 0 1R Q = b B 2 1 1 0R B 1 -3 1 -2R r -1 = B -1 1 0 1R -1 = B -1 1 0 1R N = B 1 1 -3 -2R , W = B a1 1 1 0R = B 2 1 1 0R Q = (WN)-1 a1 = 2, a2 = 1 ∑s I - A∑= s2 + 2s + 1 = s2 + a1 s + a2 N = CC A CD = B 1 1 -3 -2R A = B 1 -4 1 -3R , B = B 0 2R , C = [1 1] Example Problems and Solutions 833 or (10–158) Equations (10–157) and (10–158) are in the observable canonical form.
A–10–10.
For the system defined by consider the problem of designing a state observer such that the desired eigenvalues for the observer gain matrix are m1, m2, p , mn.
Show that the observer gain matrix given by Equation (10–61), rewritten as (10–159) can be obtained from Equation (10–13) by considering the dual problem. That is, the matrix Ke can be determined by considering the pole-placement problem for the dual system, obtaining the state-feedback gain matrix K, and taking its conjugate transpose, or Ke=K.
Solution. The dual of the given system is (10–160) Using the state-feedback control Equation (10–160) becomes Equation (10–13), which is rewritten here, is (10–161) where For the original system, the observability matrix is Hence, matrix T can also be written as Since we have and (T)-1 = (WN)-1 T = W N = WN W = W, T = NW CC A C p (A)n-1 CD = N T = MW = CC A C p (A)n-1 CD W K = Can - an an-1 - an-1 p a2 - a2 a1 - a1D T-1 z # = (A - C K) z v = -Kz n = B z z # = A z + C v Ke = (WN)-1F an - an an-1 - an-1 a1 - a1 V y = Cx x # = Ax + Bu y = [1 1]B -1 1 0 1R B x ˆ 1 x ˆ 2 R = [0 1]B x ˆ 1 x ˆ 2 R 834 Chapter 10 / Control Systems Design in State Space Taking the conjugate transpose of both sides of Equation (10–146), we have Since Ke=K, this last equation is the same as Equation (10–159).Thus, we obtained Equation (10–159) by considering the dual problem.
A–10–11.
Consider an observed-state feedback control system with a minimum-order observer described by the following equations: (10–162) (10–163) where Axa is the state variable that can be directly measured, and corresponds to the observed state variables.B Show that the closed-loop poles of the system comprise the closed-loop poles due to pole placement Cthe eigenvalues of matrix (A-BK)] and the closed-loop poles due to the minimum-order observer [the eigenvalues of matrix Solution. The error equation for the minimum-order observer may be derived as given by Equation (10–94), rewritten thus: (10–164) where From Equations (10–162) and (10–163), we obtain (10–165) Combining Equations (10–164) and (10–165) and writing we obtain (10–166) Equation (10–166) describes the dynamics of the observed-state feedback control system with a minimum-order observer.The characteristic equation for this system is or @s I - A + BK@ @s I - Abb + Ke Aab@ = 0 2s I - A + BK 0 -BKb s I - Abb + Ke Aab 2 = 0 B x # e # R = B A - BK 0 BKb Abb - Ke Aab R B x eR K = CK a KbD = Ax - BK ex - c0 ed f = (A - BK) x + BKc0 ed x # = Ax - BK x = Ax - BKc xa x b d = Ax - BKc xa xb - ed e = xb - x b e # = AAbb - Ke AabB e AAbb - Ke AabB D x b x = cxa xb d, x = c xa x b d u = -Kx y = Cx x # = Ax + Bu K = AT-1BF an - an an-1 - an-1 a1 - a1 V = (T)-1F an - an an-1 - an-1 a1 - a1 V = (WN)-1F an - an an-1 - an-1 a1 - a1 V Example Problems and Solutions 835 The closed-loop poles of the observed-state feedback control system with a minimum-order observer consist of the closed-loop poles due to pole placement and the closed-loop poles due to the minimum-order observer. (Therefore, the pole-placement design and the design of the minimum-order observer are independent of each other.) A–10–12.
Consider a completely state controllable system defined by (10–167) where Suppose that the rank of the following matrix is n+1. Show that the system defined by (10–168) where is completely state controllable.
Solution. Define Because the system given by Equation (10–167) is completely state controllable,the rank of matrix M is n.Then the rank of is n+1. Consider the following equation: (10–169) Since matrix is of rank n+1, the left-hand side of Equation (10–169) is of rank n+1.Therefore,the right-hand side of Equation (10–169) is also of rank n+1. Since = CA ˆ B ˆ A ˆ 2 B ˆ p A ˆ n B ˆ B ˆ D = B AB -CB A 2 B -CAB p p A n B -CA n-1 B B 0 R B AM -CM B 0R = B ACB AB p A n-1 BD -CCB AB p A n-1 BD B 0 R B A -C B 0R B A -C B 0R B M 0 0 1R = B AM -CM B 0R BM 0 0 1R M = CB AB p A n-1 BD A ˆ = B A -C 0 0R , B ˆ = B B 0 R , ue = u(t) - u(q) e # = A ˆ e + B ˆ ue B A -C B 0R (n + 1) (n + 1) C = 1 n constant matrix B = n 1 constant matrix A = n n constant matrix y = output signal (scalar) u = control signal (scalar) x = state vector (n-vector) y = Cx x # = Ax + Bu 836 Chapter 10 / Control Systems Design in State Space we find that the rank of is n+1. Thus, the system defined by Equation (10–168) is completely state controllable.
A–10–13.
Consider the system shown in Figure 10–49. Using the pole-placement-with-observer approach, design a regulator system such that the system will maintain the zero position Ay1=0 and y2=0B in the presence of disturbances. Choose the desired closed-loop poles for the pole-placement part to be and the desired poles for the minimum-order observer to be First, determine the state feedback gain matrix K and observer gain matrix Ke. Then, obtain the response of the system to an arbitrary initial condition—for example, where e1 and e2 are defined by Assume that m1=1 kg, m2=2 kg, k=36 Nm, and b=0.6 N-sm.
Solution. The equations for the system are By substituting the given numerical values for m1, m2, k, and b and simplifying, we obtain Let us choose the state variables as follows: x4 = y # 2 x3 = y # 1 x2 = y2 x1 = y1 y $ 2 = 18y1 - 18y2 + 0.3y # 1 - 0.3y # 2 y $ 1 = -36y1 + 36y2 - 0.6y # 1 + 0.6y # 2 + u m2 y $ 2 = kAy1 - y2B + bAy # 1 - y # 2B m1 y $ 1 = kAy2 - y1B + bAy # 2 - y # 1B + u e2 = y2 - y 2 e1 = y1 - y 1 e1(0) = 0.1, e2(0) = 0.05 y1(0) = 0.1, y2(0) = 0, y # 1(0) = 0, y # 2(0) = 0 s = -15, s = -16 s = -2 + j213, s = -2 - j213, s = -10, s = -10 CB ˆ A ˆ B ˆ A ˆ 2 B ˆ p A ˆ n B ˆ D m1 m2 y1 y2 u k b Regulator Figure 10–49 Mechanical system.
Example Problems and Solutions 837 Then, the state-space equations become Define The state feedback gain matrix K and observer gain matrix Ke can be obtained easily by use of MATLAB as follows: (See MATLAB Program 10–26.) Ke = B 14.4 0.3 0.6 15.7R K = [130.4444 -41.5556 23.1000 15.4185] A = E 0 0 -36 18 0 0 36 -18 1 0 -0.6 0.3 0 1 0.6 -0.3 U = C Aaa Aba Aab Abb S , B = E 0 0 1 0 U = C Ba Bb S B y1 y2 R = B 1 0 0 1 0 0 0 0R D x1 x2 x3 x4 T D x # 1 x # 2 x # 3 x # 4 T = D 0 0 -36 18 0 0 36 -18 1 0 -0.6 0.3 0 1 0.6 -0.3 T D x1 x2 x3 x4 T + D 0 0 1 0 T u MATLAB Program 10–26 A = [0 0 1 0;0 0 0 1;-36 36 -0.6 0.6;18 -18 0.3 -0.3]; B = [0;0;1;0]; J = [-2+j2sqrt(3) -2-j2sqrt(3) -10 -10]; K = acker(A,B,J) K = 130.4444 -41.5556 23.1000 15.4185 Aab = [1 0;0 1]; Abb = [-0.6 0.6;0.3 -0.3]; L = [-15 -16]; Ke = place(Abb',Aab',L)' place: ndigits= 15 Ke = 14.4000 0.6000 0.3000 15.7000 Response to Initial Condition: Next, we obtain the response of the designed system to the given initial condition. Since x = B xa x b R = B y x b R u = -K x x # = Ax + Bu we have (10–170) Note that where Then, Equation (10–170) can be written as (10–171) Since, from Equation (10–94), we have (10–172) by combining Equations (10–171) and (10–172) into one equation, we have The state matrix here is a 66 matrix.The response of the system to the given initial condition can be obtained easily with MATLAB. (See MATLAB Program 10–27.) The resulting response curves are shown in Figure 10–50.The response curves seem to be acceptable.
cx # e # d = cA - BK 0 BKF Abb - Ke Aab d cx ed e # = AAbb - Ke AabB e x # = (A - BK) x + BKFe F = c0 I d x - x = c xa xb d - c xa x b d = c 0 xb - x b d = c0 ed = c0 I de = Fe x # = Ax - BK x = (A - BK) x + BKAx - x B 838 Chapter 10 / Control Systems Design in State Space x1 0 1 2 3 4 t (sec) 0 1 2 3 4 t (sec) 0 1 2 3 4 t (sec) 0 1 2 3 4 t (sec) 0 1 2 3 4 t (sec) −0.05 0 0.1 0.05 0.15 e1 0 0.05 0.1 0 1 2 3 4 t (sec) e2 0 0.02 0.04 0.06 x2 −0.02 0.02 0 0.04 0.06 x3 −0.6 −0.2 −0.4 0 0.2 x4 −0.2 0.1 0 −0.1 0.2 Response to initial condition Response to initial condition Figure 10–50 Response curves to initial condition.
Example Problems and Solutions 839 MATLAB Program 10–27 % Response to initial condition A = [0 0 1 0;0 0 0 1;-36 36 -0.6 0.6;18 -18 0.3 -0.3]; B = [0;0;1;0]; K = [130.4444 -41.5556 23.1000 15.4185]; Ke = [14.4 0.6;0.3 15.7]; F = [0 0;0 0;1 0;0 1]; Aab = [1 0;0 1]; Abb = [-0.6 0.6;0.3 -0.3]; AA = [A-BK BKF; zeros(2,4) Abb-KeAab]; sys = ss(AA,eye(6),eye(6),eye(6)); t = 0:0.01:4; y = initial(sys,[0.1;0;0;0;0.1;0.05],t); x1 = [1 0 0 0 0 0]y'; x2 = [0 1 0 0 0 0]y'; x3 = [0 0 1 0 0 0]y'; x4 = [0 0 0 1 0 0]y'; e1 = [0 0 0 0 1 0]y'; e2 = [0 0 0 0 0 1]y'; subplot(3,2,1); plot(t,x1); grid; title('Response to initial condition'), xlabel('t (sec)'); ylabel('x1') subplot(3,2,2); plot(t,x2); grid; title('Response to initial condition'), xlabel('t (sec)'); ylabel('x2') subplot(3,2,3); plot(t,x3); grid; xlabel('t (sec)'); ylabel('x3') subplot(3,2,4); plot(t,x4); grid; xlabel('t (sec)'); ylabel('x4') subplot(3,2,5); plot(t,e1); grid; xlabel('t (sec)');ylabel('e1') subplot(3,2,6); plot(t,e2); grid; xlabel('t (sec)'); ylabel('e2') r = 0 y u –y Observer controller + – 4 s(s + 2) Plant Figure 10–51 Regulator system.
A–10–14.
Consider the system shown in Figure 10–51.Design both the full-order and minimum-order observers for the plant.Assume that the desired closed-loop poles for the pole-placement part are located at Assume also that the desired observer poles are located at (a) s=–8, s=–8 for the full-order observer (b) s=–8 for the minimum-order observer Compare the responses to the initial conditions specified below: (a) for the full-order observer: x1(0) = 1, x2(0) = 0, e1(0) = 1, e2(0) = 0 s = -2 + j213, s = -2 - j2 13 840 Chapter 10 / Control Systems Design in State Space (b) for the minimum-order observer: Also, compare the bandwidths of both systems.
Solution. We first determine the state-space representation of the system. By defining state variables x1 and x2 as we obtain For the pole-placement part,we determine the state feedback gain matrix K.Using MATLAB, we find K to be K=[4 0.5] (See MATLAB Program 10–28.) Next, we determine the observer gain matrix Ke for the full-order observer. Using MATLAB, we find Ke to be (See MATLAB Program 10–28.) Ke = B 14 36R y = [1 0]B x1 x2 R B x # 1 x # 2 R = B 0 0 1 -2R B x1 x2 R + B 0 4R u x2 = y # x1 = y x1(0) = 1, x2(0) = 0, e1(0) = 1 MATLAB Program 10–28 % Obtaining matrices K and Ke.
A = [0 1;0 -2]; B = [0;4]; C = [1 0]; J = [-2+j2sqrt(3) -2-j2sqrt(3)]; L = [-8 -8]; K = acker(A,B,J) K = 4.0000 0.5000 Ke = acker(A',C',L)' Ke = 14 36 Now we find the response of this system to the given initial condition. Referring to Equation (10–70), we have This equation defines the dynamics of the designed system using the full-order observer.MATLAB Program 10–29 produces the response to the given initial condition.The resulting response curves are shown in Figure 10–52.
B x # e # R = B A - BK 0 BK A - Ke CR Bx eR Example Problems and Solutions 841 MATLAB Program 10–29 % Response to initial condition ---- full-order observer A = [0 1;0 -2]; B = [0;4]; C = [1 0]; K = [4 0.5]; Ke = [14;36]; AA = [A-BK BK; zeros(2,2) A-KeC]; sys = ss(AA, eye(4), eye(4), eye(4)); t = 0:0.01:8; x = initial(sys, [1;0;1;0],t); x1 = [1 0 0 0]x'; x2 = [0 1 0 0]x'; e1 = [0 0 1 0]x'; e2 = [0 0 0 1]x'; subplot(2,2,1); plot(t,x1); grid xlabel('t (sec)'); ylabel('x1') subplot(2,2,2); plot(t,x2); grid xlabel('t (sec)'); ylabel('x2') subplot(2,2,3); plot(t,e1); grid xlabel('t (sec)'); ylabel('e1') subplot(2,2,4); plot(t,e2); grid xlabel('t (sec)'); ylabel('e2') x1 x2 e1 e2 0.4 0.6 0.8 1 0.2 0 −0.2 0 2 4 6 8 0 2 4 6 8 0 2 4 6 8 −0.4 0.6 0.8 1 1.2 0.4 0.2 0 −0.2 0 1 −1 −2 −0.5 0 −1 −1.5 −3 −2 t (sec) t (sec) 0 2 4 6 8 t (sec) t (sec) Figure 10–52 Response curves to initial condition.
842 Chapter 10 / Control Systems Design in State Space MATLAB Program 10–31 % Obtaining Ke ---- minimum-order observer Aab = ; Abb = [-2]; LL = [-8]; Ke = acker(Abb',Aab',LL)' Ke = 6 MATLAB Program 10–30 % Determination of transfer function of observer controller ---- full-order observer A = [0 1;0 -2]; B = [0;4]; C = [1 0]; K = [4 0.5]; Ke = [14;36]; [num,den] = ss2tf(A-KeC-BK, Ke,K,0) num = 0 74.0000 256.0000 den = 1 18 108 To obtain the transfer function of the observer controller, we use MATLAB. MATLAB Program 10–30 produces this transfer function.The result is num den = 74s + 256 s2 + 18s + 108 = 74(s + 3.4595) (s + 9 + j5.1962)(s + 9 - j5.1962) Next, we obtain the observer gain matrix Ke for the minimum-order observer. MATLAB Program 10–31 produces Ke. The result is K e = 6 The response of the system with minimum-order observer to the initial condition can be ob-tained as follows: By substituting into the plant equation given by Equation (10–79) u = -K x Example Problems and Solutions 843 MATLAB Program 10–32 % Response to intial condition ---- minimum-order observer A = [0 1;0 -2]; B = [0;4]; K = [4 0.5]; Kb = 0.5; Ke = 6; Aab = 1; Abb = -2; AA = [A-BK BKb; zeros(1,2) Abb-KeAab]; sys = ss(AA,eye(3),eye(3),eye(3)); t = 0:0.01:8; x = initial(sys,[1;0;1],t); x1 = [1 0 0]x'; x2 = [0 1 0]x'; e = [0 0 1]x'; subplot(2,2,1); plot(t,x1); grid xlabel('t (sec)'); ylabel('x1') subplot(2,2,2); plot(t,x2); grid xlabel('t (sec)'); ylabel('x2') subplot(2,2,3); plot(t,e); grid xlabel('t (sec)'); ylabel('e') we find or The error equation is Hence the system dynamics are defined by Based on this last equation, MATLAB Program 10–32 produces the response to the given initial condition.The resulting response curves are shown in Figure 10–53.
B x # e # R = B A - BK 0 BK b Abb - K e Aab R B x e R e # = AAbb - K e AabB e x # = (A - BK) x + BK b e = (A - BK) x + BCKa KbD B 0 e R x # = Ax - BK x = Ax - BKx + BK(x - x ) 844 Chapter 10 / Control Systems Design in State Space e 0 2 4 6 8 0 0.2 0.4 0.6 0.8 1 t (sec) x1 x2 0.6 0.8 1 1.2 0.4 0.2 0 0 2 4 6 8 0 2 4 6 8 −0.2 0 0.5 −0.5 −1 −2 −1.5 −2.5 t (sec) t (sec) Figure 10–53 Response curves to initial condition.
The transfer function of the observer controller, when the system uses the minimum-order observer, can be obtained by use of MATLAB Program 10–33.The result is num den = 7s + 32 s + 10 = 7(s + 4.5714) s + 10 MATLAB Program 10–33 % Determination of transfer function of observer controller ---- minimum-order observer A = [0 1;0 -2]; B = [0;4]; Aaa = 0; Aab = 1; Aba = 0; Abb = -2; Ba = 0; Bb = 4; Ka = 4; Kb = 0.5; Ke = 6; Ahat = Abb - KeAab; Bhat = AhatKe + Aba - KeAaa; Fhat = Bb - KeBa; Atilde = Ahat - FhatKb; Btilde = Bhat - Fhat(Ka + KbKe); Ctilde = -Kb; Dtilde = -(Ka + KbKe); [num,den] = ss2tf(Atilde, Btilde, -Ctilde, -Dtilde) num = 7 32 den = 1 10 Example Problems and Solutions 845 The observer controller is clearly a lead compensator.
The Bode diagrams of System 1 (closed-loop system with full-order observer) and of Sys-tem 2 (closed-loop system with minimum-order observer) are shown in Figure 10–54. Clearly, the bandwidth of System 2 is wider than that of System 1. System 1 has a better high-frequency noise-rejection characteristic than System 2.
A–10–15.
Consider the system where x is a state vector (n-vector) and A is an nn constant matrix.We assume that A is non-singular. Prove that if the equilibrium state x=0 of the system is asymptotically stable (that is, if A is a stable matrix), then there exists a positive-definite Hermitian matrix P such that where Q is a positive-definite Hermitian matrix.
Solution. The matrix differential equation.
has the solution Integrating both sides of this matrix differential equation from t=0 to t=q, we obtain X(q) - X(0) = A a 3 q 0 X dt b + a 3 q 0 X dt b A X = eA t QeAt X # = A X + XA, X(0) = Q A P + PA = -Q x # = Ax Frequency (rad/sec) Bode Diagrams of Systems −300 −100 −200 −250 −50 −150 0 −100 −50 Phase (deg); Magnitude (dB) 50 0 10−1 100 101 102 System 1 System 2 System 1 System 2 Figure 10–54 Bode diagrams of System 1 (system with full-order observer) and System 2 (system with minimum-order observer).
System 1= (296s+1024) (s4+20s3+144s2 +512s+1024); System 2= (28s+128) (s3+12s2+48s+128).
846 Chapter 10 / Control Systems Design in State Space Noting that A is a stable matrix and, therefore, we obtain Let us put Note that the elements of are finite sums of terms like where the li are the eigenvalues of A and mi is the multiplicity of li. Since the li possess negative real parts, exists. Note that Thus P is Hermitian (or symmetric if P is a real matrix).We have thus shown that for a stable A and for a positive-definite Hermitian matrix Q, there exists a Hermitian matrix P such that We now need to prove that P is positive definite. Consider the following Her-mitian form: Hence, P is positive definite.This completes the proof.
A–10–16.
Consider the control system described by (10–173) where Assuming the linear control law (10–174) determine the constants k1 and k2 so that the following performance index is minimized: J = 3 q 0 xT x dt u = -Kx = -k1 x1 - k2 x2 A = B 0 0 1 0R , B = B0 1R x # = Ax + Bu = 0, for x = 0 = 3 q 0 AeAt xB QAeAt xB dt 7 0, for x Z 0 x Px = x 3 q 0 eA t QeAt dt x A P + PA = -Q.
P = 3 q 0 eA t QeAt dt = P 3 q 0 eA t QeAt dt teli t p , tmi-1 eli t, eli t, eAt P = 3 q 0 X dt = 3 q 0 eA t QeAt dt -X(0) = -Q = A a 3 q 0 X dt b + a 3 q 0 X dt b A X(q) = 0, Example Problems and Solutions 847 Consider only the case where the initial condition is Choose the undamped natural frequency to be 2 radsec.
Solution. Substituting Equation (10–174) into Equation (10–173), we obtain or (10–175) Thus, Elimination of x2 from Equation (10–175) yields Since the undamped natural frequency is specified as 2 radsec, we obtain Therefore, is a stable matrix if k2>0. Our problem now is to determine the value of k2 so that the performance index is minimized, where the matrix P is determined from Equation (10–115), rewritten Since in this system Q=I and R=0, this last equation can be simplified to (10–176) Since the system involves only real vectors and real matrices, P becomes a real symmetric matrix.
Then Equation (10–176) can be written as B 0 1 -4 -k2 R B p11 p12 p12 p22 R + B p11 p12 p12 p22 R B 0 -4 1 -k2 R = B-1 0 0 -1R (A - BK) P + P(A - BK) = -I (A - BK) P + P(A - BK) = -(Q + K RK) J = 3 q 0 xT x dt = xT(0) P(0) x(0) A - BK A - BK = B 0 -4 1 -k2 R k1 = 4 x $ 1 + k2 x # 1 + k1 x1 = 0 A - BK = B 0 -k1 1 -k2 R = B 0 -k1 1 -k2 R B x1 x2 R B x # 1 x # 2 R = B 0 0 1 0R B x1 x2 R + B0 1R C-k1 x1 - k2 x2D x # = Ax - BKx x(0) = B c 0R 848 Chapter 10 / Control Systems Design in State Space Solving for the matrix P, we obtain The performance index is then (10–177) To minimize J, we differentiate J with respect to k2 and set equal to zero as follows: Hence, With this value of k2, we have Thus, the minimum value of J is obtained by substi-tuting into Equation (10–177), or The designed system has the control law The designed system is optimal in that it results in a minimum value for the performance index J under the assumed initial condition.
A–10–17.
Consider the same inverted-pendulum system as discussed in Example 10–5.The system is shown in Figure 10–8, where M=2 kg, m=0.1 kg, and l=0.5 m.The block diagram for the system is shown in Figure 10–9.The system equations are given by j # = r - y = r - Cx u = -Kx + kI j y = Cx x # = Ax + Bu u = -4x1 - 120x2 J min = 15 2 c2 k2 = 120 02J0k2 2 7 0.
k2 = 120 0J 0k2 = a -5 2k2 2 + 1 8 b c2 = 0 0J0k2 = a 5 2k2 + k2 8 b c2 = [c 0]B p11 p12 p12 p22 R B c 0R = p11 c2 J = xT(0) Px(0) P = B p11 p12 p12 p22 R = D 5 2k2 + k2 8 1 8 1 8 5 8k2 T Example Problems and Solutions 849 where Referring to Equation (10–51), the error equation for the system is given by where and the control signal is given by Equation (10–41): where Using MATLAB, determine the state feedback gain matrix such that the following performance index J is minimized: J = 3 q 0 (e Qe + u Ru) dt K ˆ x = D x1 x2 x3 x4 T = D u u # x x # T e = B xe je R = B x(t) - x(q) j(t) - j(q) R K ˆ = CK -kID = Ck1 k2 k3 k4 -kID ue = -K ˆ e A ˆ = B A -C 0 0R = E 0 20.601 0 -0.4905 0 1 0 0 0 0 0 0 0 0 -1 0 0 1 0 0 0 0 0 0 0 U , B ˆ = B B 0 R = E 0 -1 0 0.5 0 U e # = A ˆ e + B ˆ ue A = D 0 20.601 0 -0.4905 1 0 0 0 0 0 0 0 0 0 1 0 T , B = D 0 -1 0 0.5 T , C = [0 0 1 0] 850 Chapter 10 / Control Systems Design in State Space where Obtain the unit-step response of the system designed.
Solution. A MATLAB program to determine is given in MATLAB Program 10–34.The result is k1 = -188.0799, k2 = -37.0738, k3 = -26.6767, k4 = -30.5824, kI = -10.0000 K ˆ Q = E 100 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 U , R = 0.01 MATLAB Program 10–34 % Design of quadratic optimal control system A = [0 1 0 0;20.601 0 0 0;0 0 0 1;-0.4905 0 0 0]; B = [0;-1;0;0.5]; C = [0 0 1 0]; D = ; Ahat = [A zeros(4,1);-C 0]; Bhat = [B;0]; Q = [100 0 0 0 0;0 1 0 0 0;0 0 1 0 0;0 0 0 1 0;0 0 0 0 1]; R = [0.01]; Khat = lqr(Ahat,Bhat,Q,R) Khat = -188.0799 -37.0738 -26.6767 -30.5824 10.0000 Unit-Step Response.
Once we have determined the feedback gain matrix K and the integral gain constant kI, we can determine the unit-step response of the designed system.The system equation is (10–178) [Refer to Equation (10–35).] Since Equation (10–178) can be written as follows: (10–179) The output equation is y = [C 0]B x jR + r B x # j # R = B A - BK -C BkI 0 R B x jR + B 0 1R r u = -Kx + kI j Bx # j # R = B A -C 0 0R B x jR + B B 0 R u + B 0 1R r MATLAB Program 10–35 gives the unit-step response of the system given by Equation (10–179).
The resulting response curves are presented in Figure 10–55. It shows response curves versus t, versus t, versus t, versus t, and versus t, where the input r(t) to the cart is a unit-step function All initial conditions are set equal to zero. Figure 10–56 is an enlarged version of the cart position versus t. The cart moves backward a very small amount for the first 0.6 sec or so. (Notice that the cart velocity is negative for the first 0.4 sec.) This is due to the fact that the inverted-pendulum-on-the-cart system is a nonminimum-phase system.
y C= x3(t)D Cr(t) = 1 mD.
j C= x5(t)D y # C= x4(t)D y C= x3(t)D u # C= x2(t)D u C= x1(t)D Example Problems and Solutions 851 MATLAB Program 10–35 % Unit-step response A = [0 1 0 0;20.601 0 0 0;0 0 0 1;-0.4905 0 0 0]; B = [0;-1;0;0.5]; C = [0 0 1 0]; D = ; K = [-188.0799 -37.0738 -26.6767 -30.5824]; kI = -10.0000; AA = [A-BK BkI; -C 0]; BB = [0;0;0;0;1]; CC= [C 0]; DD = D; t = 0:0.01:10; [y,x,t] = step(AA,BB,CC,DD,1,t); x1 = [1 0 0 0 0]x'; x2 = [0 1 0 0 0]x'; x3 = [0 0 1 0 0]x'; x4 = [0 0 0 1 0]x'; x5 = [0 0 0 0 1]x'; subplot(3,2,1); plot(t,x1); grid; xlabel('t (sec)'); ylabel('x1') subplot(3,2,2); plot(t,x2); grid; xlabel('t (sec)'); ylabel('x2') subplot(3,2,3); plot(t,x3); grid; xlabel('t (sec)'); ylabel('x3') subplot(3,2,4); plot(t,x4); grid; xlabel('t (sec)'); ylabel('x4') subplot(3,2,5); plot(t,x5); grid; xlabel('t (sec)'); ylabel('x5') Comparing the step-response characteristics of this system with those of Example 10–5, we notice that the response of the present system is less oscillatory and exhibits less maximum overshoot in the position response Ax3 versus tB. The system designed by use of the quadratic optimal regulator approach generally gives such characteristics—less oscillatory and well damped.
852 Chapter 10 / Control Systems Design in State Space x1 0 2 4 6 8 10 t (sec) 0 2 4 6 8 10 t (sec) 0 2 4 6 8 10 t (sec) 0 2 4 6 8 10 t (sec) 0 2 4 6 8 10 t (sec) −0.02 0 0.02 0.04 x5 0 1 2 3 x2 −0.05 0.05 0 0.1 0.15 x3 −0.5 0.5 0 1 1.5 x4 −0.2 0.2 0 0.4 0.6 Figure 10–55 Response curves to a unit-step input.
Cart Position x3 versus t Cart Position x3 t (sec) 0 1 2 3 4 5 6 7 8 9 10 1.2 1 0.8 0.6 0.4 0.2 0 −0.2 Figure 10–56 Cart position versus t curve.
A–10–18.
Consider the stability of a system with unstructured additive uncertainty as shown in Figure 10–57(a). Define true plant dynamics G=model of plant dynamics unstructured additive uncertainty ¢a = G = Example Problems and Solutions 853 G w u K WaI y z −+ K (f) (e) w u z y P G K a (d) u z w −+ a Ta B A (c) G K a (a) y u A ++ – B a K G y u +− A B (b) Figure 10–57 (a) Block diagram of a system with unstructured additive uncertainty; (b)–(d) successive modifications of the block diagram of (a); (e) block diagram showing a generalized plant with unstructured additive uncertainty; (f) generalized plant diagram.
854 Chapter 10 / Control Systems Design in State Space Assume that is stable and its upper bound is known.Assume also that and G are related by =G+ a Obtain the condition that the controller K must satisfy for robust stability.Also, obtain a gener-alized plant diagram for this system.
Solution. Let us obtain the transfer function between point A and point B in Figure 10–57(a).
Redrawing Figure 10–57(a), we obtain Figure 10–57(b).Then the transfer function between points A and B can be obtained as Define Then Figure 10–57(b) can be redrawn as Figure 10–57(c).By using the small-gain theorem,the con-dition for the robust stability of the closed-loop system can be obtained as (10–180) Since it is impossible to model precisely, we need to find a scalar transfer function such that for all v and use this instead of a. Then, the condition for the robust stability of the closed-loop system can be given by (10–181) If Inequality (10–181) holds true, then it is evident that Inequality (10–180) also holds true. So this is the condition to guarantee the robust stability of the designed system. In Figure 10–57(e), a in Figure 10–57(d) was replaced by .
To summarize, if we make the norm of the transfer function from w to z to be less than 1, the controller K that satisfies Inequality (10–181) can be determined.
Figure 10–57(e) can be redrawn as that shown in Figure 10–57(f), which is the generalized plant diagram for the system considered.
Note that for this problem the matrix that relates the controlled variable z and the exoge-nous disturbance w is given by Noting that u(s)=K(s)y(s) and referring to Equation (10–128), is given by the elements of the P matrix as follows: To make this equal to we may choose P11=0, P12=Wa, P21=I, and P22=G.Then, the P matrix for this problem can be obtained as P = B 0 I Wa -GR W aK(I + GK)-1, £(s) £(s) = P 11 + P 12K(I - P 22K)-1P21 £(s) z = £(s)w = (W aT a)w = [W aK(I + GK)-1]w £ Hq WaI ¢ 7W aT a7 q 6 1 ¢ Wa(jv) s{¢a(jv)} 6 W a(jv) Wa(jv) ¢a 7 ¢aTa7 q 6 1 K(1 + GK)-1 = Ta K 1 + GK = K(1 + GK)-1 ¢ G G ¢a Problems 855 x = Ax + Bu .
y = Cx x2 x3 k2 k1 k3 r u x y = x1 + – + – Figure 10–58 Type 1 servo system.
PROBLEMS B–10–1. Consider the system defined by where Transform the system equations into (a) controllable canon-ical form and (b) observable canonical form.
B–10–2. Consider the system defined by where Transform the system equations into the observable canon-ical form.
B–10–3. Consider the system defined by where By using the state-feedback control it is desired to have the closed-loop poles at Deter-mine the state-feedback gain matrix K.
B–10–4. Solve Problem B–10–3 with MATLAB.
s = -10.
s = -2 ; j4, u = -Kx, A = C 0 0 -1 1 0 -5 0 1 -6 S , B = C 0 1 1 S x # = Ax + Bu A = C -1 1 0 0 -2 0 1 0 -3 S , B = C 0 1 1 S , C = [1 1 1] y = Cx x # = Ax + Bu C = [1 1 0] A = C -1 1 0 0 -2 0 1 0 -3 S , B = C 0 0 1 S , y = Cx x # = Ax + Bu B–10–5. Consider the system defined by Show that this system cannot be stabilized by the state-feedback control whatever matrix K is chosen.
B–10–6. A regulator system has a plant Define state variables as By use of the state-feedback control it is desired to place the closed-loop poles at Determine the necessary state-feedback gain matrix K.
B–10–7. Solve Problem B–10–6 with MATLAB.
B–10–8. Consider the type 1 servo system shown in Figure 10–58. Matrices A, B, and C in Figure 10–58 are given by Determine the feedback gain constants k1, k2, and k3 such that the closed-loop poles are located at Obtain the unit-step response and plot the output y(t)-versus-t curve.
s = -2 + j4, s = -2 - j4, s = -10 A = C 0 0 0 1 0 -5 0 1 -6 S , B = C 0 0 1 S , C = [1 0 0] s = -2 + j213, s = -2 - j213, s = -10 u = -Kx, x3 = x # 2 x2 = x # 1 x1 = y Y(s) U(s) = 10 (s + 1)(s + 2)(s + 3) u = -Kx, B x # 1 x # 2 R = B 0 0 1 2R B x1 x2 R + B1 0R u 856 Chapter 10 / Control Systems Design in State Space B–10–9. Consider the inverted-pendulum system shown in Figure 10–59.Assume that M=2 kg, m=0.5 kg, l=1 m Define state variables as and output variables as Derive the state-space equations for this system.
It is desired to have closed-loop poles at Determine the state-feedback gain matrix K.
Using the state-feedback gain matrix K thus determined, examine the performance of the system by computer simu-lation.Write a MATLAB program to obtain the response of the system to an arbitrary initial condition. Obtain the response curves x1(t) versus t, x2(t) versus t, x3(t) versus t, and x4(t) versus t for the following set of initial condition: x1(0) = 0, x2(0) = 0, x3(0) = 0, x4(0) = 1 ms s = -4 + j4, s = -4 - j4, s = -20, s = -20 y1 = u = x1 , y2 = x = x3 x1 = u, x2 = u # , x3 = x, x4 = x # where Design a full-order state observer. The desired observer poles are s=–5 and s=–5.
B–10–11. Consider the system defined by where Design a full-order state observer, assuming that the desired poles for the observer are located at s=–10, s=–10, s=–15 B–10–12. Consider the system defined by Given the set of desired poles for the observer to be design a full-order observer.
B–10–13. Consider the double integrator system defined by If we choose the state variables as then the state-space representation for the system becomes as follows: y = [1 0]B x1 x2 R B x # 1 x # 2 R = B 0 0 1 0R B x1 x2 R + B0 1R u x2 = y # x1 = y y $ = u s = -5 + j513, s = -5 - j513, s = -10 y = [1 0 0]C x1 x2 x3 S + C 0 0 1.244 S u C x # 1 x # 2 x # 3 S = C 0 0 1.244 1 0 0.3956 0 1 -3.145 S C x1 x2 x3 S A = C 0 0 -5 1 0 -6 0 1 0 S , B = C 0 0 1 S , C = [1 0 0] y = Cx x # = Ax + Bu A = B -1 1 1 -2R , C = [1 0] 0 M P z u mg m sin u x x cos u u Figure 10–59 Inverted-pendulum system.
B–10–10. Consider the system defined by y = Cx x # = Ax Problems 857 Y(s) R(s) U(s) Observer controller + – s2 + 2s + 50 s(s + 4) (s + 6) Figure 10–60 Control system with observer controller in the feedforward path.
It is desired to design a regulator for this system. Using the pole-placement-with-observer approach,design an observer controller.
Choose the desired closed-loop poles for the pole-placement part to be s=–0.7071+j0.7071, s=–0.7071-j0.7071 and assuming that we use a minimum-order observer,choose the desired observer pole at s=–5 B–10–14. Consider the system where Design a regulator system by the pole-placement-with-observer approach. Assume that the desired closed-loop poles for pole placement are located at s=–1+j, s=–1-j, s=–5 The desired observer poles are located at s=–6, s=–6, s=–6 Also, obtain the transfer function of the observer controller.
B–10–15. Using the pole-placement-with-observer approach, design observer controllers (one with a full-order observer and the other with a minimum-order observer) for the system shown in Figure 10–60.The desired closed-loop poles for the pole-placement part are s=–1+j2, s=–1-j2, s=–5 A = C 0 0 -6 1 0 -11 0 1 -6 S , B = C 0 0 1 S , C = [1 0 0] y = Cx x # = Ax + Bu The desired observer poles are s=–10, s=–10, s=–10 for the full-order observer s=–10, s=–10 for the minimum-order observer.
Compare the unit-step responses of the designed systems.
Compare also the bandwidths of both systems.
B–10–16. Using the pole-placement-with-observer approach, design the control systems shown in Figures 10–61(a) and (b).
Assume that the desired closed-loop poles for the pole place-ment are located at s=–2+j2, s=–2-j2 and the desired observer poles are located at s=–8, s=–8 Obtain the transfer function of the observer controller.
Compare the unit-step responses of both systems.[In System (b), determine the constant N so that the steady-state out-put y(q) is unity when the input is a unit-step input.] Y(s) R(s) Observer controller + – 1 s(s + 1) 1 s(s + 1) Y(s) Observer controller + – (b) R(s) N (a) Plant Figure 10–61 Control systems with observer controller: (a) observer controller in the feedforward path; (b) observer controller in the feedback path.
B–10–17. Consider the system defined by where a = adjustable parameter 7 0 A = C 0 0 -1 1 0 -2 0 1 -a S x # = Ax 858 Chapter 10 / Control Systems Design in State Space Determine the value of the parameter a so as to minimize the following performance index: Assume that the initial state x(0) is given by B–10–18. Consider the system shown in Figure 10–62.
Determine the value of the gain K so that the damping ratio z of the closed-loop system is equal to 0.5. Then determine also the undamped natural frequency vn of the closed-loop system.Assuming that e(0)=1 and evaluate 3 q 0 e2(t) dt e #(0) = 0, x(0) = C c1 0 0 S J = 3 q 0 xT x dt B–10–21. Consider the inverted-pendulum system shown in Figure 10–59. It is desired to design a regulator system that will maintain the inverted pendulum in a vertical po-sition in the presence of disturbances in terms of angle u and/or angular velocity The regulator system is required to return the cart to its reference position at the end of each control process. (There is no reference input to the cart.) The state-space equation for the system is given by where We shall use the state-feedback control scheme Using MATLAB, determine the state-feedback gain matrix such that the following performance index J is minimized: where Then obtain the system response to the following initial condition: Plot response curves u versus t, versus t, x versus t, and versus t.
x # u # D x1(0) x2(0) x3(0) x4(0) T = D 0.1 0 0 0 T Q = D 100 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 T , R = 1 J = 3 q 0 Ax Qx + u RuB dt K = Ck1 k2 k3 k4D u = -Kx B = D 0 -1 0 0.5 T , x = D u u # x x # T A = D 0 20.601 0 -0.4905 1 0 0 0 0 0 0 0 0 0 1 0 T x # = Ax + Bu u # .
+ – r = 0 c u e K 5 (s + 1) (2s + 1) Figure 10–62 Control system.
B–10–19. Determine the optimal control signal u for the system defined by where such that the following performance index is minimized: B–10–20. Consider the system It is desired to find the optimal control signal u such that the performance index is minimized. Determine the optimal signal u(t).
J = 3 q 0 AxT Qx + u2B dt, Q = B1 0 0 mR Bx # 1 x # 2 R = B0 0 1 0R B x1 x2 R + B0 1R u J = 3 q 0 AxT x + u2B dt A = B0 0 1 -1R , B = B 0 1R x # = Ax + Bu A 859 Appendix Appendix A first presents the complex variable and complex function.Then it presents tables of Laplace transform pairs and properties of Laplace transforms.Finally,it presents frequently used Laplace transform theorems and Laplace transforms of pulse function and impulse function.
Complex Variable.
A complex number has a real part and an imaginary part, both of which are constant. If the real part and/or imaginary part are variables, a complex quantity is called a complex variable. In the Laplace transformation we use the notation s as a complex variable; that is, where s is the real part and v is the imaginary part.
Complex Function.
A complex function G(s), a function of s, has a real part and an imaginary part or where Gx and Gy are real quantities. The magnitude of G(s) is and the angle u of G(s) is The angle is measured counterclockwise from the pos-itive real axis.The complex conjugate of G(s) is Complex functions commonly encountered in linear control systems analysis are single-valued functions of s and are uniquely determined for a given value of s.
G –(s) = Gx - jGy .
tan-1 AGyGxB.
2Gx 2 + Gy 2 , G(s) = Gx + jGy s = s + jv Laplace Transform Tables A complex function G(s) is said to be analytic in a region if G(s) and all its deriva-tives exist in that region.The derivative of an analytic function G(s) is given by Since can approach zero along an infinite number of different paths. It can be shown, but is stated without a proof here, that if the derivatives taken along two particular paths, that is, and are equal, then the deriva-tive is unique for any other path and so the derivative exists.
For a particular path (which means that the path is parallel to the real axis), For another particular path (which means that the path is parallel to the imaginary axis), If these two values of the derivative are equal, or if the following two conditions are satisfied,then the derivative dG (s)/ ds is uniquely determined.These two conditions are known as the Cauchy–Riemann conditions.If these conditions are satisfied,the func-tion G(s) is analytic.
As an example, consider the following G(s): Then G(s + jv) = 1 s + jv + 1 = Gx + jGy G(s) = 1 s + 1 0Gx 0s = 0Gy 0v and 0Gy 0s = - 0Gx 0v 0Gx 0s + j 0Gy 0s = 0Gy 0v - j 0Gx 0v d ds G(s) = lim j¢vS0 a ¢Gx j¢v + j ¢Gy j¢v b = -j 0Gx 0v + ¢Gy 0v ¢s = j¢v d ds G(s) = lim ¢sS0 a ¢Gx ¢s + j ¢Gy ¢s b = 0Gx 0s + j 0Gy 0s ¢s = ¢s ¢s = ¢s + j¢v ¢s = j¢v, ¢s = ¢s ¢s = ¢s + j¢v, ¢s d ds G(s) = lim ¢sS0 G(s + ¢s) - G(s) ¢s = lim ¢sS0 ¢G ¢s 860 Appendix A / Laplace Transform Tables where It can be seen that, except at s=–1 (that is, s=–1, v=0), G(s) satisfies the Cauchy–Riemann conditions: Hence G(s)=1/(s+1) is analytic in the entire s plane except at s=–1.The deriva-tive dG (s)/ ds, except at s=1, is found to be Note that the derivative of an analytic function can be obtained simply by differentiat-ing G(s) with respect to s. In this example, Points in the s plane at which the function G(s) is analytic are called ordinary points, while points in the s plane at which the function G(s) is not analytic are called singular points. Singular points at which the function G(s) or its derivatives approach infinity are called poles. Singular points at which the function G(s) equals zero are called zeros.
If G(s) approaches infinity as s approaches –p and if the function for n=1, 2, 3, p has a finite, nonzero value at s=–p, then s=–p is called a pole of order n. If n=1, the pole is called a simple pole. If n=2, 3, p , the pole is called a second-order pole, a third-order pole, and so on.
To illustrate, consider the complex function G(s) = K(s + 2)(s + 10) s(s + 1)(s + 5)(s + 15)2 G(s)(s + p)n, d ds a 1 s + 1 b = -1 (s + 1)2 = -1 (s + jv + 1)2 = -1 (s + 1)2 d ds G(s) = 0Gx 0s + j 0Gy 0s = 0Gy 0v - j 0Gx dv 0Gy 0s = - 0Gx 0v = 2v(s + 1) C(s + 1)2 + v2D 2 0Gx 0s = 0Gy 0v = v2 - (s + 1)2 C(s + 1)2 + v2D 2 Gx = s + 1 (s + 1)2 + v2 and Gy = -v (s + 1)2 + v2 Appendix A / Laplace Transform Tables 861 G(s) has zeros at s=–2, s=–10, simple poles at s=0, s=–1, s=–5, and a double pole (multiple pole of order 2) at s=–15. Note that G(s) becomes zero at s=q. Since for large values of s G(s) possesses a triple zero (multiple zero of order 3) at s=q. If points at infinity are included, G(s) has the same number of poles as zeros.To summarize, G(s) has five zeros (s=–2, s=–10, s=q, s=q, s=q) and five poles (s=0, s=–1, s=–5, s=–15, s=–15).
Laplace Transformation.
Let us define f(t)=a function of time t such that f(t)=0 for t<0 s=a complex variable l=an operational symbol indicating that the quantity that it prefixes is to be transformed by the Laplace integral F(s)=Laplace transform of f(t) Then the Laplace transform of f(t) is given by The reverse process of finding the time function f(t) from the Laplace transform F(s) is called the inverse Laplace transformation.The notation for the inverse Laplace trans-formation is l–1, and the inverse Laplace transform can be found from F(s) by the fol-lowing inversion integral: where c, the abscissa of convergence, is a real constant and is chosen larger than the real parts of all singular points of F(s).Thus, the path of integration is parallel to the jv axis and is displaced by the amount c from it.This path of integration is to the right of all sin-gular points.
Evaluating the inversion integral appears complicated.In practice,we seldom use this integral for finding f(t).We frequently use the partial-fraction expansion method given in Appendix B.
In what follows we give Table A–1, which presents Laplace transform pairs of com-monly encountered functions, and Table A–2, which presents properties of Laplace transforms.
l-1CF(s)D = f(t) = 1 2pj 3 c+jq c-jq F(s)est ds, for t 7 0 lCf(t)D = F(s) = 3 q 0 e-st dt Cf(t)D = 3 q 0 f(t)e-st dt 1 q 0 e-st dt G(s) K s3 862 Appendix A / Laplace Transform Tables Appendix A / Laplace Transform Tables 863 f(t) F(s) 1 Unit impulse d(t) 1 2 Unit step 1(t) 3 t 4 5 tn (n=1, 2, 3, p) 6 e–at 7 te–at 8 9 tne–at (n=1, 2, 3, p) 10 sinvt 11 cosvt 12 sinhvt 13 coshvt 14 15 16 17 1 s(s + a)(s + b) 1 ab c1 + 1 a - b Abe-at - ae-btB d s (s + a)(s + b) 1 b - a Abe-bt - ae-atB 1 (s + a)(s + b) 1 b - a Ae-at - e-btB 1 s(s + a) 1 a A1 - e-atB s s2 - v2 v s2 - v2 s s2 + v2 v s2 + v2 n!
(s + a)n+1 1 (s + a)n 1 (n - 1)! tn-1e-at (n = 1, 2, 3, p ) 1 (s + a)2 1 s + a n!
sn+1 1 sn tn-1 (n - 1)! (n = 1, 2, 3, p ) 1 s2 1 s Table A–1 Laplace Transform Pairs (continues on next page) 864 Appendix A / Laplace Transform Tables 18 19 20 e–at sin vt 21 e–at cos vt 22 23 24 25 1-cos vt 26 vt-sin vt 27 sinvt-vt cos vt 28 29 t cos vt 30 31 s2 As2 + v2B 2 1 2v (sinvt + vt cosvt) s As2 + v2 1BAs2 + v2 2B 1 v2 2 - v2 1 Acosv1 t - cosv2 tB Av2 1 Z v2 2B s2 - v2 As2 + v2B 2 s As2 + v2B 2 1 2v t sinvt 2v3 As2 + v2B 2 v3 s2As2 + v2B v2 sAs2 + v2B (0 6 z 6 1, 0 6 f 6 p2) v2 n sAs2 + 2zvn s + v2 nB f = tan-1 21 - z2 z 1 -1 21 - z2 e-zvn t sin Avn21 - z2 t + fB (0 6 z 6 1, 0 6 f 6 p2) s s2 + 2zvn s + v2 n f = tan-1 21 - z2 z -1 21 - z2 e-zvn t sin Avn21 - z2 t - fB v2 n s2 + 2zvn s + v2 n vn 21 - z2 e-zvn t sinvn21 - z2 t (0 6 z 6 1) s + a (s + a)2 + v2 v (s + a)2 + v2 1 s2(s + a) 1 a2 Aat - 1 + e-atB 1 s(s + a)2 1 a2 A1 - e-at - ate-atB Table A–1 (continued) Appendix A / Laplace Transform Tables 865 1 2 3 4 5 where 6 7 8 9 10 11 12 13 14 15 16 17 18 lCf(t)g(t)D = 1 2pj 3 c+jq c-jq F(p)G(s - p)dp lc 3 t 0 f 1(t - t)f 2(t)dtd = F 1(s)F 2(s) lcf a 1 a b d = aF(as) lc 1 t f(t)d = 3 q s F(s) ds if lim tS0 1 t f(t) exists lCtnf(t)D = (-1)n dn dsn F(s) (n = 1, 2, 3, p ) lCt2f(t)D = d2 ds2 F(s) lCtf(t)D = -dF(s) ds lCf(t - a)1(t - a)D = e-asF(s) a 0 lCe-atf(t)D = F(s + a) 3 q 0 f(t)dt = lim sS0F(s) if 3 q 0 f(t) dt exists lc 3 t 0 f(t)dtd = F(s) s l; c 3 p 3 f(t)(dt)nd = F(s) sn + a n k=1 1 sn-k+1 c 3 p 3 f(t)(dt)kd t=0 ; l; c 3 f(t)dtd = F(s) s + 1 s c 3 f(t) dtd t=0 ; f(t) (k-1) = dk-1 dtk-1 f(t) l; c dn dtn f(t)d = snF(s) - a n k=1 sn-kf(0 ;) (k-1) l; c d2 dt2 f(t)d = s2F(s) - sf(0 ;) - f # (0 ;) l; c d dt f(t)d = sF(s) - f(0 ;) lCf 1(t) ; f 2(t)D = F 1(s) ; F 2(s) lCAf(t)D = AF(s) Table A–2 Properties of Laplace Transforms Finally, we present two frequently used theorems, together with Laplace transforms of the pulse function and impulse function.
866 Appendix A / Laplace Transform Tables Initial value theorem Final value theorem Pulse function Impulse function = As s = A = lim t0S0 d dt0 [A(1 - e-st0)] d dt0 (t0s) = 0, for t 6 0, t0 6 t lCg(t)D = lim t0S0 c A t0s (1 - e-st0)d g(t) = lim t0S0 A t0 , for 0 6 t 6 t0 lCf(t)D = A t0s - A t0s e-st0 f(t) = A t0 1(t) - A t0 1(t - t0) f(q) = lim tS q f(t) = lim sS0 sF(s) f(0+) = lim tS0+f(t) = lim sS qsF(s) B 867 Appendix Before we present MATLAB approach to the partial-fraction expansions of transfer functions, we discuss the manual approach to the partial-fraction expansions of transfer functions.
Partial-Fraction Expansion when F(s) Involves Distinct Poles Only.
Consider F(s) written in the factored form for m<n where p1, p2,p ,pn and z1, z2,p ,zm are either real or complex quantities,but for each com-plex pi or zj there will occur the complex conjugate of pi or zj,respectively.If F(s) involves distinct poles only,then it can be expanded into a sum of simple partial fractions as follows: (B–1) where ak (k=1, 2,p ,n) are constants.The coefficient ak is called the residue at the pole at s=–pk. The value of ak can be found by multiplying both sides of Equation (B–1) by As+pkB and letting s=–pk, which gives = ak + p + ak s + pk As + pkB + p + an s + pn As + pkBR s=-pk c As + pkB B(s) A(s) d s=-pk = c a1 s + p1 As + pkB + a2 s + p2 As + pkB F(s) = B(s) A(s) = a1 s + p1 + a2 s + p2 + p + an s + pn F(s) = B(s) A(s) = KAs + z1BAs + z2B p As + zmB As + p1BAs + p2B p As + pnB , Partial-Fraction Expansion We see that all the expanded terms drop out with the exception of ak.Thus the residue ak is found from Note that, since f(t) is a real function of time, if p1 and p2 are complex conjugates, then the residues a1 and a2 are also complex conjugates. Only one of the conjugates, a1 or a2, needs to be evaluated, because the other is known automatically.
Since f(t) is obtained as for t 0 EXAMPLE B–1 Find the inverse Laplace transform of The partial-fraction expansion of F(s) is where a1 and a2 are found as Thus for t 0 EXAMPLE B–2 Obtain the inverse Laplace transform of Here, since the degree of the numerator polynomial is higher than that of the denominator poly-nomial, we must divide the numerator by the denominator.
G(s) = s + 2 + s + 3 (s + 1)(s + 2) G(s) = s3 + 5s2 + 9s + 7 (s + 1)(s + 2) = 2e-t - e-2t, = l-1c 2 s + 1 d + l-1c -1 s + 2 d f(t) = l-1CF(s)D a2 = c(s + 2) s + 3 (s + 1)(s + 2) d s=-2 = c s + 3 s + 1 d s=-2 = -1 a1 = c(s + 1) s + 3 (s + 1)(s + 2) d s=-1 = c s + 3 s + 2 d s=-1 = 2 F(s) = s + 3 (s + 1)(s + 2) = a1 s + 1 + a2 s + 2 F(s) = s + 3 (s + 1)(s + 2) f(t) = l-1CF(s)D = a1 e-p1 t + a2 e-p2 t + p + an e-pn t, l-1c ak s + pk d = ak e-pk t ak = c As + pkB B(s) A(s) d s=-pk 868 Appendix B / Partial-Fraction Expansion Note that the Laplace transform of the unit-impulse function d(t) is 1 and that the Laplace transform of dd(t)/ dt is s.The third term on the right-hand side of this last equation is F(s) in Example B–1. So the inverse Laplace transform of G(s) is given as for t 0– EXAMPLE B–3 Find the inverse Laplace transform of Notice that the denominator polynomial can be factored as If the function F(s) involves a pair of complex-conjugate poles, it is convenient not to expand F(s) into the usual partial fractions but to expand it into the sum of a damped sine and a damped cosine function.
Noting that s2+2s+5=(s+1)2+22 and referring to the Laplace transforms of e–at sinvt and e–at cos vt, rewritten thus, the given F(s) can be written as a sum of a damped sine and a damped cosine function: It follows that for t 0 Partial-Fraction Expansion when F(s) Involves Multiple Poles.
Instead of dis-cussing the general case, we shall use an example to show how to obtain the partial-fraction expansion of F(s).
Consider the following F(s): The partial-fraction expansion of this F(s) involves three terms, F(s) = B(s) A(s) = b1 s + 1 + b2 (s + 1)2 + b3 (s + 1)3 F(s) = s2 + 2s + 3 (s + 1)3 = 5e-t sin2t + 2e-t cos2t, = 5l-1c 2 (s + 1)2 + 22 d + 2l-1c s + 1 (s + 1)2 + 22 d f(t) = l-1CF(s)D = 5 2 (s + 1)2 + 22 + 2 s + 1 (s + 1)2 + 22 F(s) = 2s + 12 s2 + 2s + 5 = 10 + 2(s + 1) (s + 1)2 + 22 lCe-at cosvtD = s + a (s + a)2 + v2 lCe-at sinvtD = v (s + a)2 + v2 s2 + 2s + 5 = (s + 1 + j2)(s + 1 - j2) F(s) = 2s + 12 s2 + 2s + 5 g(t) = d dt d(t) + 2d(t) + 2e-t - e-2t, Appendix B / Partial-Fraction Expansion 869 where b3, b2, and b1 are determined as follows. By multiplying both sides of this last equation by (s+1)3, we have (B–2) Then letting s=–1, Equation (B–2) gives Also, differentiation of both sides of Equation (B–2) with respect to s yields (B–3) If we let s=–1 in Equation (B–3), then By differentiating both sides of Equation (B–3) with respect to s, the result is From the preceding analysis it can be seen that the values of b3, b2, and b1 are found systematically as follows: = 1 2 (2) = 1 = 1 2! c d2 ds2 As2 + 2s + 3B d s=-1 b1 = 1 2! e d2 ds2 c(s + 1)3B(s) A(s) d f s=-1 = 0 = (2s + 2)s=-1 = c d ds As2 + 2s + 3B d s=-1 b2 = e d ds c(s + 1)3 B(s) A(s) d f s=-1 = 2 = As2 + 2s + 3Bs=-1 b3 = c(s + 1)3 B(s) A(s) d s=-1 d2 ds2 c(s + 1)3 B(s) A(s) d = 2b1 d ds c(s + 1)3 B(s) A(s) d s=-1 = b2 d ds c(s + 1)3 B(s) A(s) d = b2 + 2b1(s + 1) c(s + 1)3 B(s) A(s) d s=-1 = b3 (s + 1)3 B(s) A(s) = b1(s + 1)2 + b2(s + 1) + b3 870 Appendix B / Partial-Fraction Expansion We thus obtain for t 0 Comments.
For complicated functions with denominators involving higher-order polynomials, partial-fraction expansion may be quite time consuming. In such a case, use of MATLAB is recommended.
Partial-Fraction Expansion with MATLAB.
MATLAB has a command to obtain the partial-fraction expansion of B(s)/A(s). Consider the following function B(s)/A(s): where some of ai and bj may be zero. In MATLAB row vectors num and den specify the coefficients of the numerator and denominator of the transfer function.That is, num = [b0 b1 ... bn] den = [1 a1 ... an] The command [r,p,k] = residue(num,den) finds the residues (r), poles (p), and direct terms (k) of a partial-fraction expansion of the ratio of two polynomials B(s) and A(s).
The partial-fraction expansion of B(s)/A(s) is given by (B–4) Comparing Equations (B–1) and (B–4), we note that p(1)=–p1, p(2)=–p2, p , p(n)=–pn; r(1)=a1, r(2)=a2, p , r(n)=an. [k(s) is a direct term.] EXAMPLE B–4 Consider the following transfer function, B(s) A(s) = 2s3 + 5s2 + 3s + 6 s3 + 6s2 + 11s + 6 B(s) A(s) = r(1) s - p(1) + r(2) s - p(2) + p + r(n) s - p(n) + k(s) B(s) A(s) = num den = b0 sn + b1 sn-1 + p + bn sn + a1 sn-1 + p + an = A1 + t2Be-t, = e-t + 0 + t2e-t = l-1c 1 s + 1 d + l-1c 0 (s + 1)2 d + l-1c 2 (s + 1)3 d f(t) = l-1CF(s)D Appendix B / Partial-Fraction Expansion 871 For this function, num= [2 5 3 6] den = [1 6 11 6] The command [r,p,k] = residue(num,den) gives the following result: 872 Appendix B / Partial-Fraction Expansion [r,p,k] = residue(num,den) r = -6.0000 -4.0000 3.0000 p = -3.0000 -2.0000 -1.0000 k = 2 (Note that the residues are returned in column vector r, the pole locations in column vector p, and the direct term in row vector k.) This is the MATLAB representation of the following partial-fraction expansion of B(s)/A(s): Note that if p(j)=p(j+1)=p=p(j+m-1) Cthat is, pj=pj+1=p=pj+m-1D, the pole p(j) is a pole of multiplicity m. In such a case, the expansion includes terms of the form For details, see Example B–5.
r(j) s - p(j) + r(j + 1) Cs - p(j)D 2 + p + r(j + m - 1) Cs - p(j)D m = -6 s + 3 + -4 s + 2 + 3 s + 1 + 2 B(s) A(s) = 2s3 + 5s2 + 3s + 6 s3 + 6s2 + 11s + 6 EXAMPLE B–5 Expand the following B(s)/A(s) into partial fractions with MATLAB.
For this function, we have num = [1 2 3] den = [1 3 3 1] The command [r,p,k] = residue(num,den) gives the result shown next: B(s) A(s) = s2 + 2s + 3 (s + 1)3 = s2 + 2s + 3 s3 + 3s2 + 3s + 1 Appendix B / Partial-Fraction Expansion 873 It is the MATLAB representation of the following partial-fraction expansion of B(s)/A(s): Note that the direct term k is zero.
B(s) A(s) = 1 s + 1 + 0 (s + 1)2 + 2 (s + 1)3 num = [1 2 3]; den = [1 3 3 1]; [r,p,k] = residue(num,den) r = 1.0000 0.0000 2.0000 p = -1.0000 -1.0000 -1.0000 k = [] C 874 Appendix In this appendix we first review the determinant of a matrix, then we define the adjoint matrix, the inverse of a matrix, and the derivative and integral of a matrix.
Determinant of a Matrix.
For each square matrix, there exists a determinant.The determinant of a square matrix A is usually written as or det A. The determinant has the following properties: 1. If any two consecutive rows or columns are interchanged,the determinant changes its sign.
2. If any row or any column consists only of zeros, then the value of the dererminant is zero.
3. If the elements of any row (or any column) are exactly k times those of another row (or another column), then the value of the determinant is zero.
4. If, to any row (or any column), any constant times another row (or column) is added, the value of the determinant remains unchanged.
5. If a determinant is multiplied by a constant, then only one row (or one column) is multiplied by that constant. Note, however, that the determinant of k times an nn matrix A is kn times the determinant of A, or @kA@ = kn@A@ @A@ Vector-Matrix Algebra This is because 6. The determinant of the product of two square matrices A and B is the product of determinants, or If B=nm matrix and C=mn matrix, then det(In+BC)=det(Im+CB) If and D=mm matrix, then where S=D-CA1 B.
If , then where T=A-BD1 C.
If or then Rank of Matrix.
A matrix A is said to have rank m if there exists an mm sub-matrix M of A such that the determinant of M is nonzero and the determinant of every rr submatrix (where ) of A is zero.
As an example, consider the following matrix: A = D 1 2 3 4 0 1 -1 0 1 0 1 2 1 1 0 2 T r m + 1 detcA B 0 Dd = det Adet D detcA 0 C Dd = det Adet D C = 0, B = 0 detcA B C Dd = det Ddet T D Z 0 detcA B C Dd = det Adet S A Z 0 @AB@ = @A@ @B@ kA = D ka11 ka12 p ka1m ka21 ka22 p ka2m o o o kan1 kan2 p kanm T Appendix C / Vector-Matrix Algebra 875 Note that =0. One of a number of largest submatrices whose determinant is not equal to zero is Hence, the rank of the matrix A is 3.
Minor Mij.
If the ith row and jth column are deleted from an nn matrix A, the resulting matrix is an (n-1)(n-1) matrix. The determinant of this (n-1) (n-1) matrix is called the minor Mij of the matrix A.
Cofactor Aij.
The cofactor Aij of the element aij of the nn matrix A is defined by the equation Aij=(1)ijMij That is, the cofactor Aij of the element aij is (1)ij times the determinant of the matrix formed by deleting the ith row and the jth column from A. Note that the cofactor Aij of the element aij is the coefficient of the term aij in the expansion of the determinant , since it can be shown that If are replaced by then because the determinant of A in this case possesses two identical rows. Hence, we obtain Similarly, Adjoint Matrix.
The matrix B whose element in the ith row and jth column equals Aji is called the adjoint of A and is denoted by adj A, or B=(bij)=(Aji)=adj A That is, the adjoint of A is the transpose of the matrix whose elements are the cofactors of A, or adj A = D A11 A21 p An1 A12 A22 p An2 o o o A1n A2n p Anm T a n k=1 akiAkj = dij@A@ a n k=1 ajkAik = dji@A@ aj1Ai1 + aj2Ai2 + p + ajnAin = 0 i Z j aj1, aj2, p , ajn, ai1, ai2, p , ain ai1Ai1 + ai2Ai2 + p + ainAin = @A@ @A@ C 1 2 3 0 1 -1 1 0 1 S @A@ 876 Appendix C / Vector-Matrix Algebra Note that the element of the jth row and ith column of the product A(adj A) is Hence, A(adj A) is a diagonal matrix with diagonal elements equal to , or A(adj A)= I Similarly, the element in the jth row and ith column of the product (adj A)A is Hence, we have the relationship A(adj A)=(adj A)A= I (C–1) Thus where Aij is the cofactor of aij of the matrix A.Thus, the terms in the ith column of A1 are l/ times the cofactors of the ith row of the original matrix A. For example, if then the adjoint of A and the determinant are respectively found to be @A@ A = C 1 2 0 3 -1 -2 1 0 -3 S @A@ G A11 @A@ A21 @A@ p An1 @A@ A12 @A@ A22 @A@ p An2 @A@ o o o A1n @A@ A2n @A@ p Ann @A@ W A -1 = adj A @A@ = @A@ a n k=1 bjkaki = a n k=1 Akjaki = dij @A@ @A@ @A@ a n k=1 ajkbki = a n k=1 ajkAik = dji @A@ Appendix C / Vector-Matrix Algebra 877 G W adj A= = C 3 6 -4 7 -3 2 1 2 -7 S 3 -1 1 0 - 1 2 1 0 1 2 3 -1 - 3 -2 1 -3 1 0 1 -3 - 1 0 3 -2 -1 -2 0 -3 - 2 0 0 -3 2 0 -1 -2 and =17 Hence, the inverse of A is In what follows, we give formulas for finding inverse matrices for the 22 matrix and the 33 matrix. For the 22 matrix the inverse matrix is given by For the 33 matrix the inverse matrix is given by A = C a b c d e f g h i S where @A@ Z 0 A-1 = 1 ad - bc c d -b -c a d A = ca b c dd where ad - bc Z 0 A-1 = adj A @A@ = C 3 17 6 17 - 4 17 7 17 - 3 17 2 17 1 17 2 17 - 7 17 S @A@ 878 Appendix C / Vector-Matrix Algebra G W d e g h - a b g h a b d e - d f g i a c g i - a c d f A-1 = 1 @A@ e f h i - b c h i b c e f Note that There are several more useful formulas available. Assume that A=nn matrix, B=nm matrix, C=mn matrix, and D=mm matrix.Then [A + BC]-1 = A-1 - A-1 B[Im + CA-1 B]-1 CA-1 (A-1) = (A)-1 (A-1)¿ = (A¿)-1 (A-1)-1 = A If and then If then If then Finally, we present the MATLAB approach to obtain the inverse of a square matrix.
If all elements of the matrix are given as numerical values, this approach is best.
MATLAB Approach to Obtain the Inverse of a Square Matrix.
The inverse of a square matrix A can be obtained with the command inv(A) For example, if matrix A is given by then the inverse of matrix A is obtained as follows: A = C 1 1 2 3 4 0 1 2 5 S cA B C Dd -1 = c T-1 -T-1 BD-1 -D-1 CT-1 D-1 + D-1 CT-1 BD-1d @D@ Z 0, T = A - BD -1 C, and @T@ Z 0, cA B C Dd -1 = cA-1 + A-1 BS-1 CA-1 -A-1 BS-1 -S-1 CA-1 S-1 d @A@ Z 0, S = D - CA-1 B, @S@ Z 0, cA 0 C Dd -1 = c A-1 0 -D-1 CA-1 D-1d cA B 0 Dd -1 = cA-1 -A-1 BD-1 0 D-1 d @D@ Z 0, @A@ Z 0 Appendix C / Vector-Matrix Algebra 879 A = [1 1 2;3 4 0;1 2 5]; inv(A) ans = 2.2222 0.1111 0.8889 1.6667 0.3333 0.6667 0.2222 0.1111 0.1111 That is MATLAB Is Case Sensitive.
It is important to note that MATLAB is case sen-sitive.That is, MATLAB distinguishes between upper- and lowercase letters.Thus, x and X are not the same variable. All function names must be in lowercase, such as inv(A), eig(A), and poly(A).
Differentiation and Integration of Matrices.
The derivative of an nm matrix A(t) is defined to be the nm matrix, each element of which is the derivative of the corresponding element of the original matrix, provided that all the elements aij(t) have derivatives with respect to t.That is, Similarly, the integral of an nm matrix A(t) is defined to be Differentiation of the Product of Two Matrices.
If the matrices A(t) and B(t) can be differentiated with respect to t, then Here again the multiplication of A(t) and dB(t)/dt [or dA(t)/dt and B(t)] is, in gener-al, not commutative.
d dt [A(t)B(t)] = dA(t) dt B(t) + A(t) dB(t) dt G 3 a11(t) dt 3 a12(t) dt p 3 a1m(t) dt 3 a21(t) dt 3 a22(t) dt p 3 a2m(t) dt o o o 3 an1(t) dt 3 a2n(t) dt p 3 anm(t) dt W 3 A(t) dt = a 3 aij(t) dt b = G d dt a11(t) d dt a12(t) p d dt a1m(t) d dt a21(t) d dt a22(t) p d dt a2m(t) o o o d dt an1(t) d dt an2(t) p d dt anm(t) W d dt A(t) = a d dt aij(t) b = A-1 = C 2.2222 -0.1111 -0.8889 -1.6667 0.3333 0.6667 0.2222 -0.1111 0.1111 S 880 Appendix C / Vector-Matrix Algebra Differentiation of A21(t).
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References 885 I 886 Index A Absolute stability, 160 Ackermann’s formula: for observer gain matrix, 756–57 for pole placement, 730–31 Actuating error, 8 Actuator, 21–22 Adjoint matrix, 876 Air heating system, 150 Aircraft elevator control system, 156 Analytic function, 860 Angle: of arrival, 286 of departure, 280, 286 Angle condition, 271 Asymptotes: Bode diagram, 406–07 root loci, 274–75, 284–85 Attenuation, 165 Attitude-rate control system, 386 Automatic controller, 21 Automobile suspension system, 86 Auxiliary polynomial, 216 B Back emf, 95 constant, 95 Bandwidth, 474, 539 Basic control actions: integral, 24 on-off, 22 proportional, 24 proportional-plus-derivative, 25 proportional-plus-integral, 24 proportional-plus-integral-plus-derivative, 35 two-position, 22–23 Bleed-type relay, 111 Block, 17 Block diagram, 17–18 reduction, 27–28, 48 Bode diagram, 403 error in asymptotic expression of, 403 of first-order factors, 406–07, 409 general procedure for plotting, 413 plotting with MATLAB, 422–25 of quadratic factors, 410–12 of system defined in state space, 426–27 Branch point, 18 Break frequency, 406 Breakaway point, 275–76, 285–86, 351 Break-in point, 276, 281, 285–86, 351 Bridged-T networks, 90, 520 Business system, 5 Index 887 C Canonical forms: controllable, 649 diagonal, 650 Jordan, 651, 653 observable, 650 Capacitance: of pressure system, 107–09 of thermal system, 137 of water tank, 103 Cancellation of poles and zeros, 288 Cascaded system, 20 Cascaded transfer function, 20 Cauchy–Riemann conditions, 860–61 Cauchy’s theorem, 526 Cayley–Hamilton theorem, 668, 701 Characteristic equation, 652 Characteristic polynomial, 34 Characteristic roots, 652 Circular root locus, 282 Classical control theory, 2 Classification of control systems, 225 Closed-loop control system, 8 Closed-loop system, 20 Closed-loop frequency response, 477 Closed-loop frequency response curves: desirable shapes of, 492 undesirable shapes of, 492 Closed-loop transfer function, 19–20 Cofactor, 876 Command compensation, 630 Compensation: feedback, 308 parallel, 308 series, 308 Compensator: lag, 323, 503–04 lag–lead, 332–34, 511–13 lead, 312–13, 495–96 Complete observability, 683–84 conditions for, 684–85 in the s plane, 684 Complete output controllablility, 714 Complete state controllability, 676–81 in the s plane, 680–81 Complex-conjugate poles: cancellation of undesirable, 520 Complex function, 859 Complex impedence, 75 Complex variable, 859 Computational optimization approach to design PID controller, 583–89 Conditional stability, 299–300, 510–11 Conditionally stable system, 299–300, 458, 510–11 Conduction heat transfer, 137 Conformal mapping, 447, 462–64 Conical water tank system, 152 Constant-gain loci, 302–03 Constant-magnitude loci (M circles), 478–79 Constant phase-angle loci (N circles), 480–81 Constant vn loci, 296 Constant z lines, 298 Constant z loci, 296 Control actions, 21 Control signal, 3 Controllability, 675–81 matrix, 677 output, 681 Controllable canonical form, 649, 688 Controlled variable, 3 Controller, 22 Convection heat transfer, 137 Conventional control theory, 29 Convolution, integral, 16 Corner frequency, 406 Critically damped system, 167 Cutoff frequency, 474 Cutoff rate, 475 D Damped natural frequency, 167 Damper, 64, 132 Damping ratio, 165 lines of constant, 296 Dashpot, 64, 132–33 Dead space, 43 Decade, 405 Decibel, 403 Delay time, 169–70 Derivative control action, 118–20, 222 Derivative gain, 84 Derivative time, 25, 61 Detectability, 688 Determinant, 874 Diagonal canonical form, 694 Diagonalization of nn matrix, 652 Differential amplifier, 78 Differential gap, 23, 24 Differentiating system, 231 Differentiation: of inverse matrix, 881 of matrix, 880 of product of two matrices, 880 Differentiator: approximate, 617 Direct transmission matrix, 31 Disturbance, 3, 26 Dominant closed-loop poles, 182 Duality, 754 E eAt: computation of, 670–71 Eigenvalue, 652 invariance of, 655 Electromagnetic valve, 23 Electronic controller, 77, 83 Engineering organizational system, 5–6 Equivalent moment of inertia, 234 Equivalent spring constant, 64 Equivalent viscous-friction coefficient, 65, 234 Evans,W. R., 2, 11, 269 Exponential response curve, 162 F Feedback compensation, 308–09, 342, 519 Feedback control, 3 Feedback control system, 7 Feedback system, 20 Feedforward transfer function, 19 Final value theorem, 866 First-order lag circuit, 80 First-order system, 161–64 unit-impulse response of, 163 unit-ramp response of, 162–63 unit-step response of, 161–62 Flapper, 110 valve, 156 Fluid systems: mathematical modeling of, 100 Free-body diagram, 69–70 Frequency response, 398 correlation between step response and, 471–74 lag compensation based on, 502–11 lag–lead compensation based on, 511–17 lead compensation based on, 493–502 Full-order state observer, 752–53 Functional block, 17 G Gain crossover frequency, 467–69 Gain margin, 464–67 Gas constant, 108 for air, 142 universal, 108 Gear train, 232 system, 232–34 Generalized plant, 813, 815–17 diagram, 810–16, 853–54 H H infinity control problem, 816 H infinity norm, 6, 808 888 Index Hazen, 2, 11 High-pass filter, 495 Higher-order systems, 179 transient response of, 180–81 Hurwitz determinants, 252–58 Hurwitz stability criterion, 252–53, 255–58 equivalence of Routh’s stability criterion and, 255–57 Hydraulic controller: integral, 130 jet-pipe, 147 proportional, 131 proportional-plus-derivative, 134–35 proportional-plus-integral, 133–34 proportional-plus-integral-plus-derivative, 135–36 Hydraulic servo system, 124–25 Hydraulic servomotor, 128, 130, 156 Hydraulic system, 106, 123–39, 149 advantages and disadvantages of, 124 compared with pneumatic system, 106 I Ideal gas law, 108 Impedance: approach to obtain transfer function, 75–76 Impulse function, 866 Impulse response, 163, 178–79, 195–97 function, 16–17 Industrial controllers, 22 Initial condition: response to, 203–11 Initial value theorem, 866 Input filter, 261, 630 Input matrix, 31 Integral control, 220 Integral control action, 24–25, 218 Integral controller, 22 Integral gain, 61 Integral time, 25, 61 Integration of matrix, 880 Inverse Laplace transform: partial-fraction expansion method for obtaining, 867–73 Inverse Laplace transformation, 862 Inverse of a matrix: MATLAB approach to obtain, 879 Inverse polar plot, 461–62, 537–38 Inverted-pendulum system, 68–72, 98 Inverted-pendulum control system, 746–51 Inverting amplifier, 78 I-PD control, 591–92 I-PD-controlled system, 592, 628–29, 643 with feedforward control, 642 Index 889 J Jet-pipe controller, 146–47 Jordan blocks, 679 Jordan canonical form, 651, 695, 706–07 K Kalman, R. E., 12, 675 Kirchhoff’s current law, 72 Kirchhoff’s loop law, 72 Kirchhoff’s node law, 72 Kirchhoff’s voltage law, 72 L Lag compensation, 321 Lag compensator, 311, 321, 502 Bode diagram of, 503 design by frequency-response method, 502–11 design by root-locus method, 321, 323 polar plot of, 503 Lag network, 82, 542 Lag–lead compensation, 330, 335, 338, 377, 511–18 Lag–lead compensator: Bode diagram of, 558 design by frequency-response method, 513–17 design by root-locus method, 331–32, 380–82 electronic, 330–32 polar plot of, 512 Lag–lead network: electronic, 330–32 mechanical, 366 Lagrange polynomial, 708 Lagrange’s interpolation formula, 708 Laminar-flow resistance, 102 Laplace transform, 862 properties of, 865 table of, 863–64 Lead compensator, 311, 493 Bode diagram of, 494 design by frequency-response method, 493–502 design by root-locus method, 311–18 polar plot of, 494 Lead, lag, and lag–lead compensators: comparison of, 517–18 Lead network, 542 electronic, 82 mechanical, 365 Lead time, 5 Linear approximation: of nonlinear mathematical models, 43 Linear system, 14 constant coefficient, 14 Linear time-invariant system, 14, 164 Linear time-varying system, 14 Linearization: of nonlinear systems, 43 Liquid-level control system, 157 Liquid-level systems, 101, 103–04, 140–41 Log-magnitude curves of quadratic transfer function, 411 Logarithmic decrement, 237 Logarithmic plot, 403 Log-magnitude versus phase plot, 403, 443–44 LRC circuit, 72–73 M M circles, 478–79 a family of constant, 479 Magnitude condition, 271 Manipulated variable, 3 Mapping theorem, 448–49 Mathematical model, 13 MATLAB commands: MATLAB: obtaining maximum overshoot with, 194 obtaining peak time with, 194 obtaining response to initial condition with, 266 partial-fraction expansion with, 871–73 plotting Bode diagram with, 422–23 plotting root loci with, 290–91 writing text in diagrams with, 188–89 [A,B,C,D] = tf2ss(num,den), 40, 656, 698 bode(A,B,C,D), 422, 426 bode(A,B,C,D,iu), 426–27 bode(A,B,C,D,iu,w), 422 bode(A,B,C,D,w), 422 bode(num,den), 422 bode(num,den,w), 422, 425, 551 bode(sys), 422 bode(sys,w), 552 c = step(num,den,t), 190 for loop, 243, 249, 584 [Gm,pm,wcp,wcg,] = margin(sys), 468–69 gtext ('text'), 189 impulse(A,B,C,D), 195 impulse(num, den), 195 initial(A,B,C,D,[initial condition],t), 209 inv(A), 879 K = acker(A,B,J), 736 K = lqr(A,B,Q,R), 798 K = place(A,B,J), 736 MATLAB commands (Cont.) Ke = acker(A',C',L)', 773 Ke = acker(Abb,Aab,L)', 773 Ke = place(A',C',L)', 773 Ke = place(Abb',Aab',L)', 773 [K,P,E] = lqr(A,B,Q,R), 798 [K,r] = rlocfind(num,den), 303 logspace(d1,d2), 422 logspace(d1,d2,n), 422–23 lqr(A,B,Q,R), 797 lsim(A,B,C,D,u,t), 201 lsim(num,den,r,t), 201 magdB = 20log10(mag), 422 [mag,phase,w] = bode(A,B,C,D), 422 [mag,phase,w] = bode(A,B,C,D,iu,w), 422 [mag,phase,w] = bode(A,B,C,D,w), 422 [mag,phase,w] = bode(num,den), 422 [mag,phase,w] = bode(num,den,w), 422, 476 [mag,phase,w] = bode(sys), 422 [mag,phase,w] = bode(sys,w), 476 mesh, 192 mesh(y), 192, 249 mesh(y'), 192, 249 [Mp,k] = max(mag), 476 NaN, 799 [num,den] = feedback(num1,den1, num2,den2), 20–21 [num,den] = parallel(num1,den1, num2,den2), 20–21 [num,den] = series(num1,den1, num2,den2), 20–21 [num,den] = ss2tf(A,B,C,D), 41, 657 [num,den] = ss2tf(A,B,C,D,iu), 41–42, 58, 657 [NUM,den] = ss2tf(A,B,C,D,iu), 59, 659 nyquist(A,B,C,D), 436, 441–42 nyquist(A,B,C,D,iu), 441 nyquist(A,B,C,D,iu,w), 436, 441 nyquist(A,B,C,D,w), 436 nyquist(num,den), 436 nyquist(num, den,w), 436 nyquist(sys), 436 polar(theta,r), 545 printsys(num,den), 20–21, 189 printsys(num,den,'s'), 189 r = abs(z), 544 [r,p,k] = residue(num,den), 239, 871–72 [re,im,w] = nyquist(A,B,C,D), 436 [re,im,w] = nyquist(A,B,C,D,iu,w), 436 [re,im,w] = nyquist(A,B,C,D,w), 436 [re,im,w] = nyquist(num,den), 436 [re,im,w] = nyquist(num,den,w), 436 890 Index [re,im,w] = nyquist(sys), 436 residue, 867 resonant_frequency = w(k), 476 resonant_peak = 20log10(Mp), 476 rlocfind, 303 rlocus(A,B,C,D), 295 rlocus(A,B,C,D,K), 290, 295 rlocus(num,den), 290–91 rlocus(num,den,K), 290 sgrid, 297 sortsolution, 584 step(A,B,C,D), 184, 186 step(A,B,C,D,iu), 184 step(num,den), 184 step(num,den,t), 184 step(sys), 184 sys = ss(A,B,C,D), 184 sys = tf(num,den), 184 text, 188 theta = angle(z), 544 w = logspace(d2,d3,100), 425 y = lsim(A,B,C,D,u,t), 201 y = lsim(num,den,r,t), 201 [y, x, t] = impulse(A,B,C,D), 195 [y, x, t] = impulse(A,B,C,D,iu), 195 [y, x, t] = impulse(A,B,C,D,iu,t), 195 [y, x, t] = impulse(num,den), 195 [y, x, t] = impulse(num,den,t), 195 [y, x, t] = step(A,B,C,D,iu), 184 [y, x, t] = step(A,B,C,D,iu,t), 184 [y, x, t] = step(num,den,t), 184, 190 z = re+jim, 544 End of MATLAB commands Matrix exponential, 661, 669–674 closed solution for, 663 Matrix Riccati equation, 798, 800 Maximum overshoot: in unit-impulse response, 179 in unit-step response, 170, 172 versus z curve, 174 Maximum percent overshoot, 170 Maximum phase lead angle, 494, 498 Measuring element, 21 Mechanical lag–lead system, 366 Mechanical lead system, 365 Mechanical vibratory system, 236 Mercury thermometer system, 151 Minimal polynomial, 669, 704–06 Minimum-order observer, 767–77 based controller, 777 Minimum-order state observer, 752 Minimum-phase system, 415–16 Minimum-phase transfer function, 415 Minor, 876 Modern control theory, 7, 29 versus conventional control theory, 29 Index 891 Motor torque constant, 95 Motorcycle suspension system, 87 Multiple-loop system, 458–59 N N circles, 480–81 a family of constant, 481 Newton’s second law, 66 Nichols, 2, 11, 398 Nichols chart, 482–85 Nichols plots, 403 Nonbleed-type relay, 111 Nonhomogeneous state equation: solution of, 666–67 Noninverting amplifier, 79 Nonlinear mathematical models: linear approximation of, 43–45 Nonlinear system, 43 Nonminimum-phase systems, 300–01, 415, 417 Nonminimum-phase transfer function, 415, 488 Nonuniqueness: of a set of state variables, 655 Nozzle-flapper amplifier, 110 Number-decibel conversion line, 404 Nyquist, H., 2, 11, 398 Nyquist path, 545 Nyquist plot, 403, 439–40, 443 of positive-feedback system, 535–37 of system defined in state space, 440–43 Nyquist stability analysis, 454–62 Nyquist stability criterion, 445–54 applied to inverse polar plots, 461–62 O Observability, 675, 682–88 complete, 683–85 matrix, 653 Observable canonical form, 650, 692 Observation, 752 Observed-state feedback control system, 761 Observer, 753 design of control system with, 786–93 full-order, 753 mathematical model of, 752 minimum-order, 767–73 Observer-based controller: transfer function of, 761 Observer controller: in the feedback path of control system, 787, 790–93 in the feedforward path of control system, 787–90 Observer-controller matrix, 762 Observer-controller transfer function, 761–62 Observer error equation, 753 Observer gain matrix, 755 MATLAB determination of, 773 Octave, 405 Offset, 258 On-off control action, 22–23 On-off controller, 22 One-degree-of-freedom control system, 593 op amps, 78 Open-loop control system, 8 advantages of, 9 disadvantages of, 9 Open-loop frequency response curves: reshaping of, 493 Open-loop transfer function, 19 Operational amplifier, 78 Operational amplifier circuits, 93–94 for lead or lag compensator: table of, 85 Optimal regulator problem, 806 Ordinary point, 861 Orthogonality: of root loci and constant gain loci, 301–02 Output controllability, 681 Output equation, 31 Output matrix, 31 Overdamped system, 168–69 Overlapped spool valve, 146 Overlapped valve, 130 P Parallel compensation, 308–09, 342–43 Partial-fraction expansion, 867–73 with MATLAB, 871–73 PD control, 373 PD controller, 614–15 Peak time, 170, 172, 193 Performance index, 793 Performance specifications, 9 Phase crossover frequency, 467–69 Phase margin, 464–67 versus z curve, 472 PI controller, 2, 614–15 PI-D control, 590–92 PID control system, 572–77, 583, 587, 617–21, 628–29, 642–43 basic, 590 with input filter, 629 two-degrees-of-freedom, 592–95 PID controller, 567, 577, 614–16, 620, 632 modified, 616 using operational amplifiers, 83–84 Pilot valve, 124, 130 PI-PD control, 592 PID-PD control, 592 Plant, 3 Pneumatic actuating valve, 117–18 Pneumatic controllers, 144–45, 154–55 Pneumatic nozzle-flapper amplifier, 110 Pneumatic on-off controller, 115 Pneumatic pressure system, 142 Pneumatic proportional controller, 112–16 force-balance type, 115–16 force-distance type, 112–15 Pneumatic proportional-plus-derivative controller, 119–20 Pneumatic proportional-plus-integral control action, 120–22 Pneumatic proportional-plus-integral-plus-derivative control action, 122–23 Pneumatic relay, 111 bleed type, 111 nonbleed type, 111 reverse acting, 112 Pneumatic systems, 106–23, 153 compared with hydraulic system, 106 Pneumatic two-position controller, 115 Polar grids, 297 Polar plot, 403, 427–28, 430, 432 Pole: 861 of order n, 861 simple, 861 Pole assignment technique, 723 Pole-placement: necessary and sufficient conditions for arbitrary, 725 Pole placement problem, 723–35 solving with MATLAB, 735–36 Positive-feedback system: Nyquist plot for, 536–37 root loci for, 303–07 Positional servo system, 95–97 Pressure system, 107, 109 Principle of duality, 687 Principle of superposition, 43 Process, 3 Proportional control, 219 Proportional control action, 24 Proportional controller, 22 Proportional gain, 25, 61 Proportional-plus-derivative control: of second-order system, 224 of system with inertia load, 223 Proportional-plus-derivative control action, 25 Proportional-plus-derivative controller, 22, 542 892 Index Proportional-plus-integral control action, 24 Proportional-plus-integral controller, 22, 121, 542 Proportional-plus-integral-plus-derivative control action, 25 Proportional-plus-integral-plus-derivative controller, 22 Pulse function, 866 Q Quadratic factor, 410 log-magnitude curves of, 411 phase-angle curves of, 411 Quadratic optimal control problem: MATLAB solution of, 804 Quadratic optimal regulator system, 793–95 MATLAB design of, 797 R Ramp response, 197 Rank of matrix, 875 Reduced-matrix Riccati equation, 795–97 Reduced-order observer, 752 Reduced-order state observer, 752 Reference input, 21 Regulator system with observer controller, 778–86, 789 Relative stability, 160, 217, 462 Residue, 867 Residue theorem, 527 Resistance: gas-flow, 107 laminar-flow, 101–02 of pressure system, 107, 109 of thermal system, 137 turbulent-flow, 102 Resonant frequency, 430, 470 Resonant peak, 413, 430, 470 versus z curve, 413 Resonant peak magnitude, 413, 470 Response: to arbitrary input, 201 to initial condition, 203–11 to torque disturbance, 221 Reverse-acting relay, 112 Riccati equation, 795 Rise time, 169–171 obtaining with MATLAB, 193–94 Robust control: system, 16, 806–17 theory, 2, 7 Robust performance, 7, 807, 812 Robust pole placement, 735 Robust stability, 7, 807, 809 Index 893 Root loci: general rules for constructing, 283–87 for positive-feedback system, 303–07 Root locus, 271 method, 269–70 Routh’s stability criterion, 212–18 S Schwarz matrix, 268 Second-order system, 164 impulse response of, 178–79 standard form of, 166 step response of, 165–75 transient-response specification of, 171 unit-step response curves of, 169 Sensor, 21 Series compensation, 308–09, 342 Servo system, 95, 164–65 design of, 739–51 with tachometer feedback, 268 with velocity feedback, 175–77 Servomechanism, 2 Set point, 21 Set-point kick, 590 Settling time, 170, 172–73 obtaining with MATLAB, 194 versus z curve, 174 Sign inverter, 79 Simple pole, 861 Singular points, 861 Sinusoidal signal generator, 486 Sinusoidal transfer function, 401 Small gain theorem, 809 Space vehicle control system, 367, 538–39 Speed control system, 4, 148 Spool valve: linealized mathematical model of, 127 Spring-loaded pendulum system, 98 Spring-mass-dashpot system, 66 Square-law nonlinearity, 43 S-shaped curve, 569 Stability analysis, 454–62 in the complex plane, 182 Stabilizability, 688 Stack controller, 115 Standard second-order system, 189 State, 29 State controllability: complete, 676, 678, 680 State equation, 31 solution of homogeneous, 660 solution of nonhomogeneous, 666–67 Laplace transform solution of, 663 State-feedback gain matrix, 724 MATLAB approach to determine, 735–36 State matrix, 31 State observation: necessary and sufficient conditions for, 754–55 State observer, 751–77 design with MATLAB, 773 type 1 servo system with, 746 State observer gain matrix: 755 Ackermann’s formula to obtain, 756–57 direct substitution approach to obtain, 756 transformation approach to obtain, 755 State space, 30 State-space equation, 30 correlation between transfer function and, 649, 656 solution of, 660 State-space representation: in canonical forms, 649 of nth order system, 36–39 State-transition matrix, 664 properties of, 665 State variable, 29 State vector, 30 Static acceleration error constant, 228, 421 determination of, 421–22 Static position error constant, 226, 419 Static velocity error constant, 227, 420 Steady-state error, 160, 226 for unit parabolic input, 229 for unit ramp input, 228 in terms of gain K, 230 Steady-state response, 160 Step response, 699–700 of second-order system, 165–69 Summing point, 18 Suspension system: automobile, 86–87 motorcycle, 87 Sylvester’s interpolation formula, 673, 709–713 System, 3 Sytem types, 419 type 0, 225, 230, 419, 433, 487–88 type 1, 225, 230, 420, 433, 487–88 type 2, 225, 230, 421, 433, 487–88 System response to initial condition: MATLAB approach to obtain, 203–11 T Tachometer, 176 feedback, 343 Taylor series expansion, 43–45 Temperature control systems, 4–5 Test signals, 159 Text: writing on the graphic screen, 188 Thermal capacitance, 137 Thermal resistance, 137 Thermal systems, 100,136–39 Thermometer system, 151–52 Three-degrees-of-freedom system, 645 Three-dimensional plot, 192 of unit-step response curves with MATLAB, 191–93 Traffic control system, 8 Transfer function, 15 of cascaded elements, 73–74 of cascaded systems, 20 closed-loop, 20 of closed-loop system, 20 experimental determination of, 489–90 expression in terms of A, B, C, and D, 34 of feedback system, 19 feedforward, 19 of minimum-order observer-based controller, 777 of nonloading cascaded elements, 77 observer-controller, 762, 780–82 open-loop, 19 of parallel systems, 20 sinusoidal, 401 Transfer matrix, 35 Transformation: from state space to transfer function, 41–42, 657 from transfer function to state space, 40–41, 656 Transient response, 160 analysis with MATLAB, 183–211 of higher-order system, 180 specifications, 169, 171 Transport lag, 417 phase angle characteristics of, 417 Turbulent-flow resistance, 102 Two-degrees-of-freedom control system, 593–95, 599–614, 636–41, 646–47 Two-position control action, 22–23 Two-position controller, 22 Type 0 system, 225, 230, 488 log-magnitude curve for, 419, 488 polar plot of, 433 Type 1 servo system: design of, 743–51 pole-placement design of, 739–46 Type 1 system, 420 log-magnitude curve for, 420, 488 polar plot of, 433 894 Index Type 2 system, 421 log-magnitude curve for, 421, 488 polar plot of, 433 U Uncontrollable system, 681 Undamped natural frequency, 165 Underdamped system, 166–67 Underlapped spool valve, 146 Unit acceleration input, 247 Unit-impulse response: of first-order system, 163 of second-order system, 178 Unit-impulse response curves: a family of, 178 obtained by use of MATLAB, 196–97 Unit-ramp response: of first-order system, 162–63 of second-order system, 197–200 of system defined in state space, 199–200 Unit-step response: of first-order system, 161 of second-order system, 163, 167, 169 Universal gas constant, 108 Unstructured uncertainty: additive, 852–53 multiplicative, 809 system with, 809 V Valve: overlapped, 130 underlapped, 130 zero-lapped, 130 Valve coefficient, 127 Vectors: linear dependence of, 674 linear independence of, 674 Velocity error, 227 Velocity feedback, 176, 343, 519 W Watt’s speed governor, 4 Weighting function, 17 Z Zero, 861 of order m, 862 Zero-lapped valve, 130 Zero placement, 595, 597, 612 approach to improve response charac-teristics, 595–97 Ziegler–Nichols tuning rules, 11, 568–77 first method, 569–70 second method, 570–71 |
9593 | https://www.desmos.com/calculator/ajq7w5xeru | Polar Equations with Sliders | Desmos
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9594 | https://medium.com/@rajat01221/transfer-learning-and-the-mathematics-behind-it-0988153178aa | Transfer Learning and the Mathematics Behind It | by Rajat Sharma | Medium
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Transfer Learning and the Mathematics Behind It
Rajat Sharma
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Jul 21, 2024
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In the ever-evolving field of machine learning and artificial intelligence, transfer learning has emerged as a powerful paradigm. Unlike traditional machine learning approaches, where models are trained from scratch on a specific task, transfer learning leverages pre-trained models on related tasks to improve performance and reduce training time on new tasks. This article delves into the concept of transfer learning, its mathematical foundations, and its applications across various domains.
Understanding Transfer Learning
Transfer learning is a machine learning technique where a model developed for one task is reused as the starting point for a model on a second task. It is particularly useful when the second task has limited labeled data, making it challenging to train a model from scratch. The primary goal of transfer learning is to transfer knowledge gained from solving one problem to a different but related problem.
Types of Transfer Learning
Inductive Transfer Learning: Here, the source and target tasks are different, but the domains may be the same or different. The objective is to induce the correct hypothesis for the target task.
Transductive Transfer Learning: In this type, the source and target tasks are the same, but the domains are different. The goal is to adapt the model learned in the source domain to the target domain.
Unsupervised Transfer Learning: Both the source and target tasks are different and the domains may also be different. The focus is on transferring knowledge when no labeled data is available in both tasks.
Mathematical Formulation of Transfer Learning
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Key Concepts in Transfer Learning
Feature Representation Transfer: This approach focuses on finding a common feature representation for both the source and target domains. The goal is to learn a new feature space where the distributions of the source and target domains are similar.
Parameter Transfer: This technique transfers parameters or parts of models (e.g., layers in a neural network) from the source task to the target task. Fine-tuning is often used to adapt these parameters to the target domain.
Relational Knowledge Transfer: This method transfers the relationship between data points from the source to the target domain, which is particularly useful in tasks involving graphs and networks.
Mathematical Logic Behind Feature Representation Transfer
Feature representation transfer aims to find a feature space where the source and target domains have similar distributions. Mathematically, this can be achieved through methods like Maximum Mean Discrepancy (MMD) and adversarial training.
Maximum Mean Discrepancy (MMD)
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Adversarial Training
Adversarial training involves training a discriminator to distinguish between the source and target domain data while simultaneously training the feature extractor to confuse the discriminator. This approach is inspired by Generative Adversarial Networks (GANs).
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Mathematically, let D(⋅) be the discriminator and F(⋅) be the feature extractor. The objective functions for the discriminator and feature extractor are:
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Mathematical Logic Behind Parameter Transfer
Parameter transfer focuses on reusing and fine-tuning the parameters of a pre-trained model from the source task for the target task. This approach is particularly useful in deep learning, where large neural networks are trained on massive datasets.
Fine-Tuning
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Regularization
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Applications of Transfer Learning
Transfer learning has found applications in various domains, showcasing its versatility and effectiveness. Here are some notable applications:
Computer Vision
In computer vision, transfer learning is widely used due to the availability of large pre-trained models like VGG, ResNet, and Inception. These models, trained on large datasets like ImageNet, serve as feature extractors for various vision tasks, including object detection, segmentation, and image classification.
Natural Language Processing (NLP)
Transfer learning has revolutionized NLP with models like BERT, GPT, and RoBERTa. These models are pre-trained on massive text corpora and fine-tuned for specific tasks such as sentiment analysis, text classification, and machine translation.
Medical Imaging
In medical imaging, labeled data is often scarce and expensive to obtain. Transfer learning enables the use of pre-trained models on natural images to improve diagnostic tasks such as disease detection and segmentation in medical scans.
Speech Recognition
Transfer learning is employed in speech recognition to adapt pre-trained models on large speech datasets to new languages or dialects, improving the accuracy and robustness of speech-to-text systems.
Challenges and Future Directions
While transfer learning offers significant benefits, it also presents challenges that need to be addressed:
Domain Adaptation: Ensuring that the model generalizes well to the target domain despite differences in data distribution.
Negative Transfer: Preventing the transfer of irrelevant or harmful knowledge from the source to the target task.
Interpretability: Understanding the transferred knowledge and its impact on the target task.
Future research directions in transfer learning include developing more robust domain adaptation techniques, exploring transfer learning for unsupervised and semi-supervised learning, and improving the interpretability of transferred models.
Conclusion
Transfer learning represents a significant advancement in machine learning, allowing models to leverage existing knowledge to solve new and related tasks more efficiently. The mathematical foundations of transfer learning, including feature representation transfer and parameter transfer, provide a rigorous framework for understanding and applying this technique. With its wide range of applications and ongoing research, transfer learning continues to be a pivotal area in the quest for more intelligent and adaptable AI systems.
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9595 | https://sired.udenar.edu.co/278/1/85369.pdf | EL PRINCIPIO DEL M´ AXIMO APLICADO A ECUACIONES DIFERENCIALES PARCIALES SEMI-LINEALES DE SEGUNDO ORDEN DE TIPO PARAB´ OLICO Carlos Andr´ es Cer´ on Erazo UNIVERSIDAD DE NARI˜ NO FACULTAD DE CIENCIAS EXACTAS Y NATURALES DEPARTAMENTO DE MATEM´ ATICAS Y ESTAD´ ISTICA SAN JUAN DE PASTO 2012 EL PRINCIPIO DEL M´ AXIMO APLICADO A ECUACIONES DIFERENCIALES PARCIALES SEMI-LINEALES DE SEGUNDO ORDEN DE TIPO PARAB´ OLICO Carlos Andr´ es Cer´ on Erazo Trabajo de grado presentado como requisito parcial para optar al t´ ıtulo de Licenciado en Matem´ aticas Director: Miller Orlando Cer´ on Gom´ ez Magister en Ciencias Matem´ aticas UNIVERSIDAD DE NARI˜ NO FACULTAD DE CIENCIAS EXACTAS Y NATURALES DEPARTAMENTO DE MATEM´ ATICAS Y ESTAD´ ISTICA SAN JUAN DE PASTO 2012 Nota de Responsabilidad “Las ideas y conclusiones aportadas en el trabajo son responsabilidad exclusiva de sus autores” Art´ ıculo 1o de acuerdo 324 de octubre 11 de 1966 emanado por el Honorable Consejo Directivo de la Universidad de Nari˜ no Nota de Aceptaci´ on: Miller Orlando Cer´ on G´ omez Presidente de Tesis Saulo Mosquera L´ opez Jurado 1 Eduardo Ibarg¨ uen Mondrag´ on Jurado 2 San Juan de Pasto, agosto de 2012 Agradecimientos Quiero agradecer en primera instancia a Dios por haberme dado la oportunidad de seguir mi carrera, poder ingresar a la mejor universidad del departamento de Nari˜ no y formarme profesionalmente. Gracias a ´ el por tenerme con vida para poder llegar hasta este punto y por tener a mi lado a mis padres y hermanos a quienes dedico este trabajo por todo el cari˜ no que me han entregado, los valores que de ellos recib´ ı y el infinito apoyo que en gran parte mi madre me brind´ o; fue con ella con quien habl´ e muchas veces cuando estaba tomando la decisi´ on de entrar a Licenciatura en Matem´ aticas, sin duda que esas discusiones fueron muy importantes para tomar la decisi´ on correcta y estoy seguro que as´ ı fue, ahora pues, gracia a ello soy quien soy.
Agradezco con todo el coraz´ on a mi familia quien me poyo y siempre quiso lo mejor para m´ ı, a Mar´ ıa quien me ha brindado la compa˜ n´ ıa perfecta, el apoyo y el amor incondicional que necesit´ e durante todo este tiempo. Desearnos suerte en lo que vendr´ a para nuestras vidas que, estoy seguro seguir´ an unidas m´ as fuertemente que nunca.
Quisiera agradecer a mi asesor y amigo Msc. Miller Cer´ on G´ omez por trabajar conmigo y guiar cada una de las etapas del presente trabajo con su conocimiento y trayectoria. Sus valiosos consejos sirvieron de mucho y en verdad que fue un honor el haber trabajado a su lago. Gratificar tambi´ en a los profesores del Departamento de Matem´ aticas y Estad´ ıstica por sus valiosas ense˜ nanzas y todo el apoyo que me brindaron durante mis a˜ nos en la carrera y por supuesto durante el desarrollo de mi memoria de matem´ atico.
Finalmente es el momento de agradecer a quienes me hicieron pasar unos a˜ nos espectaculares mientras estudi´ abamos, mis compa˜ neros, en general a todo el grupo, porque ha sido muy enriquecedor nutrirme de sus diferentes especialidades porque no todos ten´ ıamos las mismas habilidades.
Resumen El principio del m´ aximo es uno de los instrumentos m´ as conocidos y ´ utiles en el campo de las ecuaciones diferenciales; aporta elementos matem´ aticos valiosos para estudiar ciertas caracter´ ısticas de las soluciones de ecuaciones diferenciales en derivadas ordinarias y par-ciales de los tipos el´ ıpticas, parab´ olicas e hiperb´ olicas y nos ayuda a obtener informaci´ on del comportamiento de su soluci´ on. En este trabajo se pretende, mediante el estudio y an´ alisis de la aplicabilidad del principio a las ecuaciones diferenciales lineales ordinarias-parciales de segundo orden de tipo parab´ olico, obtener condiciones iniciales o propiedades de las funciones que intervienen en una ecuaci´ on diferencial semi-lineal parcial de segundo orden de tipo parab´ olico de la forma: ut + a(u, x, t)ux + g(u, x, t) = uxx, (1) para aplicar dicho principio en ella y obtener informaci´ on de su soluci´ on.
Abstract The maximum principle is one of the best known and most useful instruments in the field of differential equations, it provides valuable mathematical elements to study certain fea-tures for the solutions of ordinary differential equations and partial elliptic, parabolic and hyperbolic type and it helps in the information of the solution. This paper seeks, through the study and analysis of the applicability principle to ordinary-partial linear differen-tial equations of second order parabolic type, to obtain initial conditions or properties of the functions involved in a semi-linear differential partial equation of the second order of parabolic type of the form: ut + a(u, x, t)ux + g(u, x, t) = uxx, to apply that principle and obtain information for its solution.
Contenido Introducci´ on 11 1. Preliminares 13 1.1. Ecuaci´ on Diferencial Ordinaria . . . . . . . . . . . . . . . . . . . . . . . . .
14 1.2. Ecuaci´ on Diferencial Parcial . . . . . . . . . . . . . . . . . . . . . . . . . .
15 1.2.1.
Orden de una Ecuaci´ on Diferencial . . . . . . . . . . . . . . . . . .
15 1.2.2.
Ecuaci´ on Diferencial Parcial Lineal de Segundo Orden . . . . . . .
15 1.2.3.
Ecuaci´ on Diferencial Parcial Semi-lineal de Segundo Orden . . . . .
16 1.2.4.
Ecuaci´ on Diferencial Parcial Lineal de Segundo Orden de Tipo Parab´ oli-co . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17 1.2.5.
La Ecuaci´ on del Calor General . . . . . . . . . . . . . . . . . . . . .
17 2. El Principio del M´ aximo en Ecuaciones Diferenciales Ordinarias Lineales de Segundo Orden 18 2.1. El Principio del M´ aximo en una Dimensi´ on . . . . . . . . . . . . . . . . . .
18 2.1.1.
Teorema del Principio del M´ aximo en una dimensi´ on . . . . . . . .
19 3. EL Principio del M´ aximo en Ecuaciones Diferenciales Parciales Lineales de Segundo Orden de Tipo Parab´ olico 25 3.1. Existencia Local . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25 3.1.1.
Localidad Lipschitz . . . . . . . . . . . . . . . . . . . . . . . . . . .
25 3.1.2.
Teorema de Existencia Local . . . . . . . . . . . . . . . . . . . . . .
26 3.2. El Operador Parab´ olico Unidimensional . . . . . . . . . . . . . . . . . . . .
26 3.2.1.
El Principio del M´ aximo Aplicado a la Ecuaci´ on del Calor . . . . .
26 3.3. El Operador Parab´ olico General . . . . . . . . . . . . . . . . . . . . . . . .
29 3.3.1.
Lemas Auxiliares Generales . . . . . . . . . . . . . . . . . . . . . .
29 8 3.3.2.
Teoremas Generales . . . . . . . . . . . . . . . . . . . . . . . . . . .
36 3.4. Lemas Auxiliares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38 3.5. Teoremas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38 4. EL Principio del M´ aximo Aplicado a Ecuaciones Diferenciales Parciales Semi-lineales de Segundo Orden de Tipo Parab´ olico 43 Conclusiones 49 Bibliograf´ ıa 50 Lista de Figuras 2.1. Funci´ on z(x) ≡eα(x−c) −1 . . . . . . . . . . . . . . . . . . . . . . . . . . .
20 2.2. Aspecto de una funci´ on con m´ aximo en el extremo izquierdo . . . . . . . .
21 2.3. Aspecto de una funci´ on con m´ aximo en el extremo derecho . . . . . . . . .
22 2.4. Soluci´ on de la igualdad u ′′ −cot xu′ = 0, en el intervalo (−1, 1) . . . . . . .
24 2.5. Soluci´ on de la igualdad u ′′ −cot xu′ = 0, en el intervalo [0, 1] . . . . . . . .
24 3.1. Regi´ on E ∪∂E del calor . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27 3.2. Ilustraci´ on Lema General 1 . . . . . . . . . . . . . . . . . . . . . . . . . . .
30 3.3. Ilustraci´ on Lema General 2 . . . . . . . . . . . . . . . . . . . . . . . . . . .
33 3.4. Ilustraci´ on Lema General 3 . . . . . . . . . . . . . . . . . . . . . . . . . . .
34 3.5. Ilustraci´ on de la demostraci´ on, teorema 4 . . . . . . . . . . . . . . . . . . .
37 3.6. Ilustraci´ on de la demostraci´ on, teorema 7 . . . . . . . . . . . . . . . . . . .
40 3.7. Derivada direccional en una direcci´ on exterior desde Et0 en P . . . . . . .
42 4.1. Regi´ on de estudio E = R × [0, T]; 0 < T < ∞de la ecuaci´ on (1) . . . . . .
44 4.2. Franja (−L, L) × (0, T) . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47 Introducci´ on El campo de las ecuaciones diferenciales resulta ser muy amplio y complejo, de hecho, exis-ten ecuaciones diferenciales que no se pueden resolver expl´ ıcitamente por lo que se requiere una descripci´ on matem´ atica del problema y un intento por obtener informaci´ on de la soluci´ on a partir de la ecuaci´ on diferencial, es por eso que surge la necesidad de desarrollar y manejar herramientas ´ utiles que nos permitan abordar este tipo de problemas, una de estas herramientas es el principio del m´ aximo, ´ util en la aprox-imaci´ on de soluciones; puede ser aplicado tanto a ecuaciones diferenciales ordinarias como a ecuaciones diferenciales parciales; dentro de ´ esta ´ ultima clase se encuentran ecuaciones muy particulares que son especiales debido a su aplicabilidad y estudio en la f´ ısica como lo son la ecuaci´ on del calor y la ecuaci´ on de onda. En t´ erminos generales, una funci´ on f satisface el principio del m´ aximo, si esta cumple una desigualdad diferencial f ′′ ≥0 en un intervalo (a, b) que alcanza su m´ aximo valor en uno de los extremos ya sea a o b, en otras palabras a´ un m´ as generales las funciones que satisfacen una desigualdad diferencial en un dominio Ωalcanzan su m´ aximo en la frontera de ese dominio. Concretamente, existen dos tipos de principios: el principio del m´ aximo fuerte dice que si una funci´ on alcanza su m´ aximo en el interior de su dominio, la funci´ on es una constante de manera uniforme, es decir, que el m´ aximo no se alcanza en ning´ un punto del interior a menos que la funci´ on sea constante; an´ alogamente el principio del m´ aximo d´ ebil, dice que el m´ aximo de la funci´ on se encuentra en la frontera del dominio, pero puede volver a ocurrir en el interior.
Suele haber una interpretaci´ on f´ ısica natural del principio del m´ aximo en los problemas de ecuaciones diferenciales que se plantean en la f´ ısica, por ejemplo, las condiciones de contorno se pueden cambiar de una manera que sugiera intuitivamente que la temperatura resultante deba ser m´ as peque˜ na, esto puede ser demostrado utilizando el principio del m´ aximo, tambi´ en se ha utilizado este principio en la teor´ ıa de control ´ optimo, conocida como optimizaci´ on din´ amica, para obtener condiciones necesarias (condiciones iniciales, 11 condiciones de frontera o condiciones de funciones que intervienen)y lograr la optimizaci´ on de sistemas din´ amicos o sistemas que evolucionan con el tiempo, igualmente en soluciones viscosas y otras aplicaciones. Adem´ as este principio es una propiedad importante de las ecuaciones parab´ olicas que se utiliza para deducir la singularidad de sus soluciones. De esta manera nuestro trabajo busc´ o condiciones iniciales y propiedades de las funciones en la ecuaci´ on (1) para obtener informaci´ on de su soluci´ on a trav´ es del principio del m´ aximo.
Inicialmente hizo un recorrido sobre la teor´ ıa de ecuaciones diferenciales tanto ordinarias como parciales para conceptualizar y manejar t´ erminos importantes que se utilizaron a lo largo del desarrollo de este trabajo y que fueron de gran ayuda para su comprensi´ on. Luego estudi´ o y analiz´ o la aplicabilidad del principio del m´ aximo en las ecuaciones diferenciales lineales ordinarias de segundo orden mostrando un caso simple de dicho principio y la metodolog´ ıa que se aplico sobre el mismo; esta parte fue importante ya que una buena forma de estudiar esta tem´ atica es empezando por ecuaciones sencillas que den una luz al problema. Posteriormente estudi´ o y analiz´ o la aplicabilidad del principio sobre ecuaciones diferenciales lineales parciales de segundo orden de tipo parab´ olico y enfatiz´ o de manera particular en la aplicabilidad y metodolog´ ıa de dicho principio sobre una ecuaci´ on llamada la ecuaci´ on del calor; por lo que se hab´ ıa propuesto el estudio de una ecuaci´ on diferencial no lineal parcial de tipo parab´ olico de segundo orden la cual se someti´ o a un tratamiento y una transformaci´ on que hizo de ella una ecuaci´ on diferencial lineal parcial de tipo parab´ olico, prototipo de la ecuaci´ on del calor. Finalmente se logr´ o obtener condiciones necesarias, se aplic´ o el principio y se obtuvo informaci´ on de la soluci´ on de la ecuaci´ on (1) 12 Cap´ ıtulo 1 Preliminares Este cap´ ıtulo se fundamenta en los cap´ ıtulos 1, 3 del libro y establece los conceptos generales que se utilizan a lo largo del desarrollo de este trabajo. Al mismo tiempo, se mencionan algunos ejemplos de cada concepto, aplicaciones de la ecuaci´ on del calor como una de las Ecuaciones Diferenciales Parciales importantes y de nuestro inter´ es. Se obvia la mayor´ ıa de la notaci´ on debido a que son est´ andares manejados en muchos de los libros de Ecuaciones Diferenciales.
Abreviaturas y Notaci´ on ux: Primera derivada parcial de u respecto de x.
utt: Segunda derivada parcial de u respecto de t.
∂E: Frontera del conjunto E.
L: Operador lineal.
≡: Definirse como.
E.D.O: Ecuaci´ on Diferencial Ordinaria.
E.D.P: Ecuaci´ on Diferencial Parcial.
E.D.P.L: Ecuaci´ on Diferencial Parcial Lineal.
Dominio: conjunto abierto y conexo (en el espacio euclidiano).
13 Conceptos Generales Ecuaci´ on Diferencial Una ecuaci´ on diferencial es una igualdad en la que intervienen derivadas de una o m´ as funciones desconocidas. Dependiendo del n´ umero de variables independientes respecto de las que se deriva, las ecuaciones diferenciales se dividen en Ecuaciones Diferenciales Ordi-narias, que contienen derivadas respecto a una sola variable independiente y Ecuaciones Diferenciales Parciales, que contienen derivadas respecto a dos o m´ as variables indepen-diente; estas ultimas ser´ an las que trataremos en este trabajo. La resoluci´ on de ecuaciones diferenciales es un tipo de problema matem´ atico que consiste en buscar una funci´ on que cumpla una determinada ecuaci´ on diferencial, se puede llevar a cabo mediante un m´ etodo espec´ ıfico si es que existe para la ecuaci´ on diferencial en cuesti´ on, o se busca informaci´ on de la soluci´ on a partir de herramientas matem´ aticas dise˜ nadas o adaptadas para este fin, como lo es el principio del m´ aximo.
1.1.
Ecuaci´ on Diferencial Ordinaria Definici´ on 1. Una E.D.O de variable independiente x y variable dependiente u es una igualdad que puede ser expresada de la forma: F(x, u, u′, u′′, u′′′ . . . , un) = 0, donde n es un entero positivo y F es una funci´ on de las (n + 2) variables indicadas.
Un ejemplo de este tipo de ecuaci´ on es: u′ + x2u = ex, 8u′′′ + x2u′′ + u′ + 3u = 0.
Se ha dedicado mucho estudio a la resoluci´ on de este tipo de ecuaciones, estando casi completamente desarrollada la teor´ ıa para ecuaciones diferenciales ordinarias lineales.
Son importantes en diversas ´ areas de estudio como la geometr´ ıa, mec´ anica y astronom´ ıa, adem´ as de muchas otras aplicaciones por ejemplo, el movimiento de un sistema masa-resorte donde el resorte tiene masa despreciable, el cual solo depende de la variable tiempo.
14 1.2.
Ecuaci´ on Diferencial Parcial Definici´ on 2. Una E.D.P de variables independientes x,t y variable dependiente u es una igualdad que puede ser expresada de la forma: F(x, t, u, ux, ut, uxx, uxut, uxxx, uxxut, ...) = 0, donde F es una funci´ on de las variables indicadas y al menos una derivada parcial ocurre.
Un ejemplo de este tipo de ecuaci´ on es: uxx −utt = 0, 2xuxxtt + 4uxxx + uxx + utt = 0.
No todas las E.D.P resultan tener el mismo inter´ es; las hay que tienen tan s´ olo un inter´ es puramente acad´ emico, mientras que hay otras cuyo inter´ es reside en tener su origen en problemas de la F´ ısica y otras ciencias. Problemas t´ ıpicos son la propagaci´ on del sonido o del calor, la electrost´ atica, la electrodin´ amica, la din´ amica de fluidos, la elasticidad, la mec´ anica cu´ antica y muchos m´ as.
1.2.1.
Orden de una Ecuaci´ on Diferencial Definici´ on 3. El orden de una ecuaci´ on diferencial se define como la mayor derivada que aparece en la ecuaci´ on.
Por ejemplo: 8u′′′ + x2u′ + 3u = 0, ux + x2u = ex, la primera es una ecuaci´ on diferencial ordinaria de orden tres y la segunda es una ecuaci´ on diferencial parcial de primer orden.
1.2.2.
Ecuaci´ on Diferencial Parcial Lineal de Segundo Orden Definici´ on 4. Una E.D.P.L de segundo orden con variable dependiente u y variables independientes x,t sobre un conjunto Ωen el plano es una igualdad que puede ser expresada 15 en la forma:: Auxx + Buxt + Cutt + Dux + Eut + Fu + G = 0, donde A, B, C, D, F, G son funciones lineales de x, t y A2 + B2 + C2 ̸= 0 sobre Ω.
Algunos ejemplos de este tipo de ecuaci´ on son: uxx −utt + u = 0, uxx −ut = 0, ´ esta ultima se conoce como la Ecuaci´ on del Calor. Ambos son ejemplos de ecuaciones de segundo orden lineales sobre Ω= R2. Observemos que lo que define que una ecuaci´ on diferencial sea lineal es que no aparecen productos de la funci´ on inc´ ognita consigo misma ni ninguna de sus derivadas, tampoco aparecen en forma de funciones compuestas (por ejemplo, sin(u)).
1.2.3.
Ecuaci´ on Diferencial Parcial Semi-lineal de Segundo Or-den Definici´ on 5. Una E.D.P de segundo orden con variable dependiente u y variables inde-pendientes x, t se dice semi-lineal sobre un conjunto Ωen el plano si y solo si puede ser expresada de la forma: Auxx + Buxt + Cutt + Φ(x, t, u, ux, ut) = 0, donde A, B, C son funciones de x, t, A2 + B2 + C2 ̸= 0 sobre Ωy Φ es una funci´ on de algunas de las variables indicadas.
Un ejemplo de este tipo de ecuaci´ on es: uxx + xutt −ux −ext sin u = 0, y la ecuaci´ on objeto de estudio ut + a(u, x, t)ux + g(u, x, t) = uxx, las cuales son E.D.P semi-lineal de segundo orden sobre Ω= R2. Observemos que en la caracterizaci´ on de una ecuaci´ on diferencial semi-lineal aparecen productos de la funci´ on inc´ ognita consigo misma y con sus derivadas, asimismo aparecen funciones compuestas (por ejemplo, sin(u)).
16 1.2.4.
Ecuaci´ on Diferencial Parcial Lineal de Segundo Orden de Tipo Parab´ olico Definici´ on 6. Una E.D.P.L de segundo orden sobre un punto (x0, t0) de un conjunto Ω de la forma Auxx + Buxt + Cutt + Dux + Eut + Fu + G = 0, se dice de tipo parab´ olico en dicho punto, si cumple que B2(x0, t0) −4A(x0, t0)C(x0, t0) = 0, y es parab´ olico en Ωsi para todo punto del dominio se cumple esta condici´ on.
Nosotros nos vamos a interesar por una de las m´ as cl´ asicas e importantes de este tipo de E.D.P de segundo orden, la ecuaci´ on del calor que mas adelante se define. Adem´ as a las Ecuaciones Diferenciales Parciales Semi-lineales de Segundo Orden de Tipo Parab´ olico les rige la misma definici´ on.
1.2.5.
La Ecuaci´ on del Calor General Definici´ on 7. Es una E.D.P de tipo parab´ olico de la forma: ut(x, t) −∆u(x, t) = f(x, t), (1.1) donde ∆u es el laplaciano de u en las variables x = (x1, . . . , xn), y viene dado por ∆u = n X i=1 uxixi siendo n ≥1.
La ecuaci´ on (1.1) para n ≤3, describe bajo hip´ otesis f´ ısicas adecuadas la evoluci´ on de la temperatura u(x, t) en cada instante t y en cada punto x ∈Rn, de un cuerpo sometido a una fuente de calor determinada por f(x, t).
Como se mencion´ o anteriormente la ecuaci´ on sobre la cual nos interesamos es la ecuaci´ on de segundo orden homog´ enea uxx −ut = 0, en donde la fuente f(x, t) = 0, y el laplaciano ∆u en la variable x ∈R es uxx.
Para otros ejemplos de EDP que aparecen en las aplicaciones, se pueden consultar los libros y .
17 Cap´ ıtulo 2 El Principio del M´ aximo en Ecuaciones Diferenciales Ordinarias Lineales de Segundo Orden Este cap´ ıtulo se fundamenta en el primer cap´ ıtulo del libro y presenta una introduc-ci´ on a un caso simple del principio del m´ aximo en ecuaciones diferenciales ordinarias con funciones que satisfacen una desigualdad diferencial no estricta en un dominio, muestra este principio en su forma mas conocida utilizando una funci´ on auxiliar diferente y al mismo tiempo exhibe condiciones necesarias para establecer dicho principio. Complemen-tariamente se desarrolla en cada teorema su respectiva prueba, similarmente a como lo presenta el libro, aunque de manera mas detallada para su comprensi´ on.
2.1.
El Principio del M´ aximo en una Dimensi´ on Una funci´ on u(x) que es continua en el intervalo [a, b] toma su valor m´ aximo en un punto de este intervalo. Si u(x) tiene segundas derivadas continuas y un m´ aximo relativo en un punto c entre a y b, entonces del calculo elemental conocemos que: u ′(c) = 0 y u ′′(c) ≤0.
(2.1) Ahora bien, suponga que u(x) satisface la desigualdad diferencial estricta L[u] ≡u ′′ + g(x)u ′ > 0, (2.2) 18 en el intervalo abierto (a, b), donde g(x) es cualquier funci´ on acotada. Es claro que la relaci´ on (2.1) no puede establecerse en un punto c ∈(a, b) ya que en ´ el, L[u] ≤0 lo cual contradice la desigualdad (2.2). Una caracter´ ıstica esencial de este argumento es que la desigualdad (2.2) sea estricta, es decir, se asuma que u ′′ +g(x)u ′ nunca es cero, o en pocas palabras que u ̸= constante.
En conclusi´ on, cada vez que (2.2) se mantenga, u(x) no puede alcanzar su valor m´ aximo en cualquier punto interior c ∈(a, b) excepto en los puntos extremos a o b.
Observemos sin embargo, que si u(x) satisface la desigualdad diferencial no estricta L[u] ≡u ′′ + g(x)u ′ ≥0, (2.3) en un intervalo (a, b) con g(x) acotado, se admite la soluci´ on u = constante, ya que para todo punto, en particular para c se tiene u ′ = 0 y u ′′ = 0, para lo cual L[u] = 0 que satisface la desigualdad (2.3).
En s´ ıntesis, cada vez que (2.3) se mantenga, el m´ aximo de u(x) puede ser alcanzado en los punto extremos a, b o en un punto interior c.
Estos son casos simples de un principio del m´ aximo. El siguiente teorema muestra la conclusi´ on para cuando se admite el m´ aximo en un punto interior 2.1.1.
Teorema del Principio del M´ aximo en una dimensi´ on Teorema 1. Sea u una funci´ on que depende de x y satisface la desigualdad diferencial L[u] ≡u ′′ + g(x)u ′ ≥0, (2.4) sobre un intervalo abierto (a, b), donde g(x) es cualquier funci´ on acotada.
Si u(x) ≤M en (a, b) y si el m´ aximo M de u se alcanza en un punto interior c ∈(a, b), entonces u ≡M, es decir, u es constante.
Demostraci´ on. Supongamos que u alcanza su m´ aximo valor M en un punto interior c de (a, b), luego u(c) = M, 19 escojamos un punto d ∈(a, b) tal que d > c, es decir que u(d) < M, y definamos la funci´ on auxiliar z(x) ≡eα(x−c) −1, con α una constante positiva a ser elegida, Figura 2.1: Funci´ on z(x) ≡eα(x−c) −1 claramente z(x) < 0 en (a, c), z(c) = 0 y z(x) > 0 en (c, b). Aplicando el operador L a la funci´ on z(x) se tiene que: L[z] ≡z′′ + g(x)z′, L[z] ≡α2eα(x−c) + g(x)αeα(x−c), L[z] ≡αeα(x−c)[α + g(x)], con α > −g(x) de modo que L[z] > 0, para a < x < d, esto lo podemos hacer puesto que g(x) es acotada.
Adem´ as, sea w(x) = u(x) + εz(x), (2.5) donde ε es una constante positiva que cumple con ε < M −u(d) z(d) , (2.6) esto se puede encontrar debido a que z(d) > 0 y u(d) < M, de este modo • En (a, c), como z(x) < 0 y u(x) < M, se tiene en (2.5) que w(x) < M.
20 • En c, como z(c) = 0 y u(c) = M, se tiene en (2.5) que w(c) = M.
• En d, como u(d) < M y εz(d) < M −u(d) seg´ un (2.6), se tiene en (2.5) que w(d) < u(d) + M −u(d), es decir w(d) < M.
De lo anterior se tiene, que w alcanza su m´ aximo en un punto interior c de (a, d).
Sin embargo, por la desigualdad (2.4) y al ser L[z], ε > 0 tenemos que L[w] = L[u] + εL[z] > 0, lo cual contradice el hecho de tener un valor m´ aximo M en un punto interior c de (a, d) de acuerdo a (2.2).
Por lo tanto u ≡M, es decir u = constante.
Si d < c usamos la funci´ on auxiliar z(x) ≡e−α(x−c) −1, con α > g(x) y tenemos la misma conclusi´ on.
NOTA. Observe que si el m´ aximo de la funci´ on u(x) ocurre en un extremo, entonces Si es el extremo izquierdo, la funci´ on debe tener el siguiente aspecto Figura 2.2: Aspecto de una funci´ on con m´ aximo en el extremo izquierdo y su pendiente en el punto debe ser negativa.
Si es el extremo derecho, la funci´ on debe tener el siguiente aspecto 21 Figura 2.3: Aspecto de una funci´ on con m´ aximo en el extremo derecho y su pendiente en el punto debe ser positiva.
El siguiente teorema establece dicho resultado.
Teorema 2. Supongamos que u(x) es una funci´ on no constante que satisface la desigualdad diferencial L[u] ≡u ′′ + g(x)u ′ ≥0, sobre un intervalo abierto (a, b) y tiene derivadas laterales en a y en b, adem´ as g(x) es una funci´ on acotada en cada sub-intervalo cerrado de (a, b), entonces: a.) Si el m´ aximo de u ocurre en x = a y g(x) esta acotada a la izquierda de x = a, entonces u ′(a) < 0.
b.) Si el m´ aximo de u ocurre en x = b y g(x) esta acotada a la derecha de x = b, entonces u ′(b) > 0.
Demostraci´ on. Supongamos que u ≤M para a ≤x ≤b y que alcanza su m´ aximo valor M en un punto x = a, es decir que u(a) = M, escojamos un punto d ∈(a, b) tal que u(d) < M, y definamos la funci´ on auxiliar z(x) ≡eα(x−a) −1, con α > 0.
Aplicando el operador L a la funci´ on z(x) se tiene que: L[z] ≡z′′ + g(x)z′, L[z] ≡α2ex−a + g(x)αex−a, L[z] ≡αex−a[α + g(x)], 22 con α > −g(x) para a ≤x ≤d; de modo que L[z] > 0.
Ahora, sea w(x) = u(x) + εz(x), donde ε es una constante positiva como en (2.6) ε < M −u(d) z(d) .
Debido a que L[w] ≡L[u] + L[z] > 0, el m´ aximo de w en el intervalo [a, d] debe ocurrir en uno de lo extremos a o d, en efecto: • En a, como z(a) = 0 y u(a) = M, se tiene que w(a) = M.
• En d, como u(d) < M y εz(d) < M −u(d) se tiene que w(d) < u(d) −M + u(d), es decir w(d) < M.
De modo que el m´ aximo de w se alcanza en a. Por consiguiente, la derivada por la derecha de a de la funci´ on w no puede ser positiva, es decir que w′(a) ≤0, (2.7) pero como u ′(a) = 0, z′(a) = α > 0 y ε > 0, entonces: w′(a) = u′(a) + εz′(a) > 0, (2.8) lo cual contradice la desigualdad (2.7). Por lo tanto u′(a) < 0.
Si el m´ aximo ocurre en b el argumento es similar.
NOTA. El requerimiento de que g sea acotado es necesario en los teoremas anteriores, por ejemplo si consideramos el problema: u ′′ −cot xu′ = 0, 23 en el intervalo (−1, 1). Claramente su soluci´ on esta dada por u = cos x y su m´ aximo se obtiene en x = 0 (0 punto interior de (−1, 1)), lo cual es una clara violaci´ on al teorema (1) como se muestra en la figura (2.4).
Figura 2.4: Soluci´ on de la igualdad u ′′ −cot xu′ = 0, en el intervalo (−1, 1) El mismo problema ocurre para el teorema (2) item a si consideremos u ′′ −cot xu′ = 0, sobre el intervalo [0, 1]. Claramente su soluci´ on est´ a dada por u = cos x y su m´ aximo se obtiene en x = 0 (0 punto extremo de [0, 1]), para la cual u′(0) = −sin(0) = 0.
Figura 2.5: Soluci´ on de la igualdad u ′′ −cot xu′ = 0, en el intervalo [0, 1] 24 Cap´ ıtulo 3 EL Principio del M´ aximo en Ecuaciones Diferenciales Parciales Lineales de Segundo Orden de Tipo Parab´ olico Este cap´ ıtulo se fundamenta en el cap´ ıtulo 3 del libro y en el articulo . Presenta definiciones y teoremas importantes que son de gran ayuda para la comprensi´ on del principio del m´ aximo en ecuaciones diferenciales lineales parab´ olicas, tales como: el Teorema de Existencia Local; que argumenta la existencia de la soluci´ on u, el operador parab´ olico unidimensional; que es utilizado para dar la aplicaci´ on del principio del m´ aximo a una ecuaci´ on particular de tipo parab´ olico llamada la ecuaci´ on del calor y mas adelante nos ayudar´ a en el tratamiento de la ecuaci´ on problema. Adicional a esto, se construye la prueba general de algunos Lemas Auxiliares y Teoremas que posteriormente se mencionan en este cap´ ıtulo.
3.1.
Existencia Local 3.1.1.
Localidad Lipschitz Sea E un subconjunto abierto de Rn y F : E →Rn, se dice que f es localmente lipschitz en E si para cada punto x0 ∈E, existe una vecindad de x0, Nε(x0) ⊂E y una constante L0 > 0 tal que para todo x, y ∈Nε(x0) |f(x) −f(y)| ≤L0|x −y|.
25 3.1.2.
Teorema de Existencia Local Supongamos que g es una funci´ on localmente lipschitz para u, acotada para (x, t) ∈R2 y f ∈C1(R), entonces ut + f(u)x + g(u, x, t) = uxx con dato inicial |u(x, 0)| ≤M tiene ´ unica soluci´ on uε(x, t) ∈C∞(R × (0, τ0)) para τ0 peque˜ no el cual depende ´ unicamente de la norma L∞del dato inicial y, |uε(x, t)| ≤2M ∀(x, t) ∈R × [0, τ0).
3.2.
El Operador Parab´ olico Unidimensional Definici´ on 8. El operador diferencial L[u] ≡a(x, t)uxx + b(x, t)ux −ut, se dice parab´ olico en un punto (x, t) si a(x, t) > 0 y se dice uniformemente parab´ olico en un dominio Ωdel plano xt si existe una constante positiva η tal que a(x, t) ≥η para todo (x, t) ∈Ω.
3.2.1.
El Principio del M´ aximo Aplicado a la Ecuaci´ on del Calor Teorema 3. Sea u(x, t) una funci´ on que satisface la desigualdad diferencial L[u] ≡uxx −ut ≥0, (3.1) en la regi´ on rectangular E = {0 < x < l; 0 < t ≤T} del plano xt y sean S1,S2,S3 y S4 los lados de dicha regi´ on, definidos por: S1 : {x = 0, 0 ≤t ≤T} S2 : {0 ≤x ≤l, t = 0}, S3 : {x = l, 0 ≤t ≤T} S4 : {0 < x < l, t = T}.
26 Figura 3.1: Regi´ on E ∪∂E del calor Entonces, el m´ aximo de u sobre E ∪∂E debe ocurrir en uno de los tres lados S1, S2 o S3.
Demostraci´ on. Supongamos que M es el m´ aximo de los valores de u que ocurren sobre S1, S2, S3 y adem´ as que existe un punto P(x0, y0) de E donde u tiene un valor M1 > M.
Definamos la funci´ on auxiliar w(x) ≡M1 −M 2l2 (x −x0)2, (3.2) si tomamos a x −x0 como la longitud l se tendr´ ıa: w(l) = M1 −M 2l2 (l)2, w(l) = M1 −M 2 , entonces, ya que u ≤M para todos los puntos de S1, S2 y S3 tenemos que v(x, t) ≡u(x, t) + w(x) ≤M + M1 −M 2 < M + M1 2 y por la suposici´ on M < M1 M + M1 < M1 + M1, M + M1 < 2M1, M + M1 2 < M1, por lo cual v(x, t) < M1 (3.3) para todos los puntos de S1, S2 y S3. Adem´ as v(x0, y0) ≡u(x0, t0) + w(x0) = u(x0, t0) = M1.
(3.4) 27 Aplicando el operador L a la funci´ on w(x) se tiene que: L[w] ≡wxx −wt = M1 −M l2 > 0.
(3.5) Por otra parte, seg´ un (3.5) y la desigualdad (3.1) L[v] ≡L[u] + L[w] = L[u] + M1 −M l2 > 0, de all´ ı que L[v] = vxx −vt > 0, (3.6) a trav´ es de E. Las condiciones (3.3) y (3.4) muestran que v debe asumir su m´ aximo ya sea en un punto interior de E o a lo largo del intervalo abierto S4 : {0 < x < l, t = T}, Sin embargo, debido a la desigualdad (3.6) la funci´ on v no puede tener un m´ aximo interior.
Entonces si el m´ aximo ocurriera sobre S4 se tendr´ ıa que vxx ≤0, (3.7) adem´ as, como wt = 0 seg´ un (3.2), se tiene que vt = ut + wt = ut, (3.8) as´ ı, de la desigualdad (3.6) vxx > vt, (3.9) luego, debido a (3.7) y (3.9) vt < 0 entonces, podemos decir seg´ un (3.8), que ut < 0 y el m´ aximo no puede ocurrir sobre S4.
En s´ ıntesis, la asunci´ on de que existe un punto P(x0, t0) de E donde u tiene un valor M1 > M nos lleva a una contradicci´ on. Por lo tanto M es el valor m´ aximo de u y ocurre sobre ∂E : S1, S2 o S3.
NOTA. El operador del calor unidimensional (3.1) es uniformemente parab´ olico en todo el plano xt ya que en este caso se ha establecido que a(x, t) ≡1, b(x, t) ≡0, y existe una constante positiva η donde a(x, t) ≡1 ≥η para todo (x, t) del plano.
28 3.3.
El Operador Parab´ olico General Definici´ on 9. El operador diferencial L[u] ≡ n X i,j=1 aij(x, t)uxixj + n X i=1 bi(x, t)uxi −ut se dice parab´ olico a (x, t) ≡(x1, x2, . . . , xn, t) si existe un n´ umero positivo η tal que n X i,j=1 aij(x, t)εiεj ≥η n X i=1 εi2 para todas las n-uplas de numeros reales (ε1, ε2, . . . , εn). Adem´ as el operador L es uniformemente parab´ olico en una regi´ on ET si n X i,j=1 aij(x, t)εiεj ≥η n X i=1 εi2 se cumple con el mismo n´ umero positivo η para todo (x, t) en ET .
3.3.1.
Lemas Auxiliares Generales Lema General 1. Supongamos que u(x, t) satisface la desigualdad diferencial L[u] ≡ n X i,j=1 aij(x, t)uxixj + n X i=1 bi(x, t)uxi −ut ≥0 en un dominio E del plano xt, donde (x, t) = (x1, x2, . . . , xn, t), aij(x, t), bi(x, t) son acotados y L[u] es uniformemente parab´ olico. Si el m´ aximo M de u(x, t) en E ocurre en un punto P(xo, t0) de la frontera de una bola (n + 1)-dimensional K cualquiera, contenido en E y u < M en el interior de K, entonces la tangente a K en P es paralela al eje x. (es decir, P est´ a arriba o abajo de K).
Demostraci´ on. Construyamos una bola (n+1)-dimensional K contenida en E, centrada en (x, t) y de radio R KR(x, t) = {(x, t) ∈Rn+1; d[(x, t), (x, t)] < R} donde (x, t) = (x1, x2, . . . , xn, t), luego d[(x, t), (x, t)] = v u u t n X i=1 (xi −xi)2 + (t −t)2] < R.
(3.10) Supongamos que P(xo, t0) = P(xo1, xo2, . . . , xon, t0) no est´ a en la parte superior o inferior de K, es decir xo ̸= x, ahora construyamos otra bola (n+1)-dimensional K1 centrada en P y radio R1 < d[x, xo], tal que K1 se encuentre completamente en E. Luego la frontera de K1 consiste de dos regiones: 29 C′, es la intersecci´ on de ∂K1 con la bola cerrada K ∪∂K (incluye extremos).
C′′, es el complemento de C′ con respecto a ∂K1.
Como podemos apreciar en el caso unidimensional en la figura (3.2) Figura 3.2: Ilustraci´ on Lema General 1 Por hip´ otesis u < M sobre C′, luego existe una constante positiva η peque˜ na tal que u + η ≤Msobre C′, (3.11) y dado que u ≤M a trav´ es de E se tiene u ≤Msobre C′′.
(3.12) En principio, definamos la funci´ on auxiliar v(x, t) ≡e−α Pn i=1(xi−xi)2+(t−t)2 −e−αR2, (3.13) entonces para valores positivos de α v > 0 en K, v = 0 en ∂K, v < 0 en el exterior de K, (3.14) luego L[v] ≡ n X i,j=1 aij(x, t)vxixj + n X i=1 bi(x, t)vxi −vt, 30 de donde vxi = e−α Pn i=1(xi−xi)2+(t−t)2 [−2α(xi −xi)], vxjxi = e−α Pn i=1(xi−xi)2+(t−t)2 [−2α(xj −xj)][−2α(xi −xi)], vxjxi = e−α Pn i=1(xi−xi)2+(t−t)2 [4α2(xi −xi)(xj −xj)], vt = e−α Pn i=1(xi−xi)2+(t−t)2 [−2α(t −t)], entonces L[v] ≡ n X ij=1 aije−α Pn i=1(xi−xi)2+(t−t)2 [4α2(xi −xi)(xj −xj)] − n X i=1 bie−α Pn i=1(xi−xi)2+(t−t)2 [2α(xi −xi)] + e−α Pn i=1(xi−xi)2+(t−t)2 [2α(t −t)].
La bola K1 y su frontera cumple con que d[x, x] ≥d[xo, x] −R1 > 0, por lo que es posible escoger α tan grande que L[v] > 0 para (x, t) en K1 ∪∂K1.
(3.15) NOTA. Como se supone que u es soluci´ on, entonces impl´ ıcitamente u es de clase C2 con respecto a la variable x, por tanto sus derivadas mixtas son iguales.
Ahora, sea w(x, t) = u(x, t) + εv(x, t), donde ε es una constante positiva.
Observe que En C′: Por la desigualdad (3.11) podemos seleccionar ε tan peque˜ no que w = u + εv < Msobre C′.
En C′′: Debido a (3.12)y (3.14), w = u + εv < Msobre C′′.
De esto w < M sobre ∂K1. Por otra parte 31 En P: w(xo, t0) = u(xo, t0) + εv(xo, t0), y como P ∈∂K se tiene que w(xo, t0) = u(xo, t0), luego, por hip´ otesis w(xo, t0) = M.
As´ ı que el m´ aximo de w ocurre en el punto interior P de K1, sin embargo En K1: L[w] = L[u] + εL[v], y por hip´ otesis y (3.15) L[w] > 0.
En s´ ıntesis, la asunci´ on de que P(xo, t0) = P(xo1, xo2, . . . , xon, t0) no est´ a en la parte superior o inferior de K nos lleva a una contradicci´ on. Por lo tanto, P(xo, t0) est´ a en la parte superior o inferior de K. (es decir, P est´ a arriba o abajo de K).
Lema General 2.
Supongamos que u(x, t) satisface la desigualdad diferencial L[u] ≥0 como en el lema anterior en un dominio E del plano xt . Si u(x, t) ≤M a trav´ es de E y en alg´ un punto interior P(xo, t0) de E se tiene que u(P) < M, entonces para cualquier x sobre el segmento horizontal l del interior de E que contiene a P se cumple que u(x, t0) < M.
De la misma manera, si u(x, t) ≤M a trav´ es de E y en alg´ un punto interior P(xo, t0) de E se tiene que u(P) = M, entonces para cualquier x sobre el segmento horizontal l del interior de E que contiene a P se cumple que u(x, t0) = M.
32 Figura 3.3: Ilustraci´ on Lema General 2 Lema General 3.
Supongamos que u(x, t) satisface la desigualdad diferencial L[u] ≥0 como en el lema 1 en la mitad inferior Kt = {(x, t); d[(x, t), (x, t)] < R2, con t ≤t}, de una bola K centrada en P(x, t) y radio R, (KR(x, t)). Si u(x, t) < M en la parte de K donde t < t, entonces u(P) < M.
Demostraci´ on. Definamos la funci´ on v(x, t) ≡e− Pn i=1(xi−xi)2+α(t−t) −1, (3.16) luego L[v] ≡ n X i,j=1 aij(x, t)vxixj + n X i=1 bi(x, t)vxi −vt, de donde vxi = e− Pn i=1(xi−xi)2+α(t−t) [−2(xi −xi)], vxjxi = e− Pn i=1(xi−xi)2+α(t−t) [−2(xj −xj)][−2(xi −xi)], vxjxi = e− Pn i=1(xi−xi)2+α(t−t) [4(xi −xi)(xj −xj)], vt = e− Pn i=1(xi−xi)2+α(t−t) [−α], entonces L[v] ≡ n X ij=1 aije− Pn i=1(xi−xi)2+α(t−t) [4(xi −xi)(xj −xj)] 33 − n X i=1 bie− Pn i=1(xi−xi)2+α(t−t) [2(xi −xi)] + e− Pn i=1(xi−xi)2+α(t−t) [α].
NOTA. De la misma manera al suponer que u es soluci´ on, entonces impl´ ıcitamente u es de clase C2 con respecto a la variable x, por tanto sus derivadas mixtas son iguales.
Elegimos α positivo y tan grande que L[v] > 0 para (x, t) en K para t ≤t.
(3.17) Por otra parte, el hiperparaboloide n X i=1 (xi −xi)2 + α(t −t) = 0, (3.18) es tangente a la linea t = t en el punto P(x, t).
Denotemos por C′ a la parte de ∂K que esta debajo del hiperparaboloide (incluyendo los extremos).
Denotemos por C′′ a la parte del hiperparaboloide contenida en la bola (n+1) dimensional K.
Y D ser´ a la regi´ on encerrada por C′ y C′′.
Como se muestra en la figura (3.4) para el caso unidimensional Figura 3.4: Ilustraci´ on Lema General 3 Por hip´ otesis u < M sobre C′, luego existe una constante positiva η peque˜ na tal que u + η ≤Msobre C′.
(3.19) Ahora, sea w(x, t) = u(x, t) + εv(x, t), donde ε es una constante positiva. Debido a (3.18) v(x, t) = 0 sobre C′′ y observe que si tomamos ε peque˜ no: 34 En D L[w] = L[u] + εL[v], por hip´ otesis y (3.17) L[w] > 0.
De esto, w no puede alcanzar su m´ aximo en D, adem´ as En C′ Debido a que v es acotado w = u + εv < Msobre C′ En C′′ por la igualdad (3.18) w = u + εv = u, o sea que w ≤Msobre C′′.
Por lo que w tampoco puede alcanzar su m´ aximo en C′, pero s´ ı puede alcanzarlo en el punto P de C′′, as´ ı que wt ≥0 en P.
De este modo, como wt(x, t) = ut(x, t) + εvt(x, t), se tiene ut(x, t) + εvt(x, t) ≥0, en P luego, ut(x, t) ≥−εvt(x, t).
(3.20) Por otro lado vt(x, t) = e− Pn i=1(xi−xi)2+α(t−t) [−α] = −α, es decir que vt(x, t) < 0, y reemplazando en (3.20), ut(x, t) > 0.
35 Adem´ as, las desigualdades uxi(x, t) = 0, uxixj(x, t) ≤0.
llevan a que L[u] ≤0 lo cual contradice la hip´ otesis.
Con la tenencia de estos lemas, podemos establecer la demostraci´ on de los siguientes resultados: 3.3.2.
Teoremas Generales Teorema 4. Suponga que u(x, t) satisface la desigualdad diferencial L[u] ≡ n X i,j=1 aij(x, t)uxixj + n X i=1 bi(x, t)uxi −ut ≥0 en Et1 = {(x, t) ∈E; t ≤t1} del dominio E donde (x, t) = (x1, x2, . . . , xn, t), aij(x, t), bi(x, t) son acotados y L es uniforme-mente parab´ olico en Et1. Si el m´ aximo de u ocurre en un punto P(x1, t1) de Et1 y si Q(x1, t0) es un punto de E que se puede conectar mediante segmentos horizontales y verticales, entonces u(x1, t0) = M.
Demostraci´ on. Supongamos que u(x1, t0) < M y que el segmento l = {(x, t) : x = x1; t0 ≤t ≤t1}, se encuentra en E.
Sea r la cota superior de los valores de u sobre l tal que u(x1, t) < M; t0 ≤t < r.
As´ ı por el lema General 2 tendr´ ıamos que t0 ≤t < r y u < M, en cada punto de los segmentos horizontales del interior de E que contienen a (x1, t), como se muestra para el caso unidimen-sional en la figura (10). Luego, existe un radio R > 0 tal que u < M para d[x, x1] < R, t0 ≤t < r.
Por continuidad u(x1, r) = M, 36 de lo contrario, el ser discontinua en el punto (x1, r) implicar´ ıa que u no es diferenciable y de esta manera, no cumplir´ ıa con la desigualdad diferencial en este punto. Ahora construyamos la bola (n + 1)-dimensional K centrada en (x1, r) y radio R, (KR(x1, r)), Figura 3.5: Ilustraci´ on de la demostraci´ on, teorema 4 debido a que u < M en la porci´ on de K donde t < r se concluye por el lema General 3 que u(x1, r) < M, lo cual lleva a una contradicci´ on.
Teorema 5. Suponga que u(x, t) satisface la desigualdad diferencial L[u] ≡ n X i,j=1 aij(x, t)uxixj + n X i=1 bi(x, t)uxi −ut ≥0 en un dominio E, donde a(x, t), b(x, t) son acotados y L es uniformemente parab´ olico y sea Et0 = {(x, t) ∈E; t ≤t0}.
Adem´ as suponga que u es continuamente diferenciable en el punto frontera P(xo, t0), que u(P) = M, que u(x, t) < M para (x, t) ∈ Et0, que K es una bola (n + 1)-dimensional centrado en (x1, t1) con x1 ̸= xo y es tangente a ∂E en el punto P, adem´ as que la porci´ on de K por debajo de t = t0 se encuentra en Et0. Si ∂ ∂ν se refiere a cualquier derivada direccional exterior desde Et0 a P, entonces uν > 0 en P.
A continuaci´ on se presentar´ a el caso unidimensional para los lemas y teoremas.
37 3.4.
Lemas Auxiliares Lema 1. Supongamos que u(x, t) satisface la desigualdad diferencial L[u] ≡a(x, t)uxx + b(x, t)ux −ut ≥0 en un dominio E del plano xt, donde a(x, t), b(x, t) son acotados y L[u] es uniformemente parab´ olico. Si el m´ aximo M de u(x, t) en E ocurre en un punto P(x0, t0) de la frontera de un disco K cualquiera, contenido en E y u < M en el interior de K, entonces la tangente a K en P es paralela al eje x. (es decir, P est´ a arriba o abajo de K).
Lema 2. Supongamos que u(x, t) satisface la desigualdad diferencial L[u] ≥0 como en el lema anterior en un dominio E del plano xt. Si u(x, t) ≤M a trav´ es de E y en alg´ un punto interior P(x0, t0) de E se tiene que u(P) < M, entonces para cualquier x sobre el segmento horizontal l del interior de E que contiene a P se cumple que u(x, t0) < M.
De la misma manera, si u(x, t) ≤M a trav´ es de E y en alg´ un punto interior P(x0, t0) de E se tiene que u(P) = M, entonces para cualquier x sobre el segmento horizontal l del interior de E que contiene a P se cumple que u(x, t0) = M.
Lema 3. Supongamos que u(x, t) satisface la desigualdad diferencial L[u] ≥0 como en el lema 1 en la mitad inferior Kt = {(x, t); d[(x, t), (x, t)] < R2, con t ≤t } de un disco K centrado en P(x, t) y radio R, (KR(x, t)). Si u(x, t) < M en la parte de K donde t < t, entonces u(P) < M.
Con la tenencia de estos lemas, podemos establecer los siguientes resultados algunos de los cuales se mostraron anteriormente en forma general.
3.5.
Teoremas Teorema 6. Suponga que u(x, t) satisface la desigualdad diferencial L[u] ≡a(x, t)uxx + b(x, t)ux −ut ≥0, 38 en Et1 = {(x, t) ∈E; t ≤t1}, del dominio E donde a(x, t), b(x, t) son acotados y L es uniformemente parab´ olico en Et1. Si u ≤M en Et1 y u(x1, t1) = M, entonces u = M en todo punto de Et1 que se puede conectar con (x1, t1) mediante segmentos horizontales y verticales que se encuentran en Et1.
Teorema 7. Suponga que u(x, t) satisface la desigualdad diferencial L[u] ≡a(x, t)uxx + b(x, t)ux −ut ≥0, en una regi´ on E, donde a(x, t), b(x, t) son acotados y L es uniformemente parab´ olico y sea Et0 = {(x, t) ∈E; t ≤t0}.
Adem´ as suponga que u es continuamente diferenciable en el punto frontera P(x0, t0), que u(P) = M, que u(x, t) < M para (x, t) ∈Et0, que K es un disco centrado en (x1, t1) con x1 ̸= x0 y es tangente a ∂E en el punto P, adem´ as que la porci´ on de K por debajo de t = t0 se encuentra en Et0. Si ∂ ∂ν se refiere a cualquier derivada direccional en una direcci´ on exterior desde Et0 en P, entonces uν > 0 en P.
Demostraci´ on. Construyamos un disco K1 centrado en P y radio R1 < d[x0, x1].
Llamemos C′ a la porci´ on de ∂K1 contenida en Et0 junto con los extremos.
Llamemos C′′ al arco de ∂K que est´ a en K1 ∩Et0.
y llamemos D a la regi´ on formada por C′, C′′ y el segmento t = t0.
Como se muestra en la figura (3.6).
Al elegir K1 mas peque˜ no que K podemos hacer que u < M sobre C′′ excepto en P, adem´ as u < M sobre C′ y podemos afirmar los siguientes tres hechos 1.) u < M sobre C′′ excepto en P.
(3.21) 2.) u = M en P.
(3.22) 39 3.) Existe una constante positiva η suficientemente peque˜ na tal que u + η ≤M sobre C′.
(3.23) Figura 3.6: Ilustraci´ on de la demostraci´ on, teorema 7 Ahora definamos la funci´ on v(x, t) ≡e−α[(x−x1)2+(t−t1)2] −e−αR2, (3.24) luego, aplicando el operador L a la funci´ on v(x, t) se tiene que: L[v] ≡a(x, t)vxx + b(x, t)vx −vt, de donde vx = e−α[(x−x1)2+(t−t1)2][−2α(x −x1)], vxx = e−α[(x−x1)2+(t−t1)2][−2α(x −x1)][−2α(x −x1)] + e−α[(x−x1)2+(t−t1)2][−2α], vxx = e−α[(x−x1)2+(t−t1)2][4α2(x −x1)2] −e−α[(x−x1)2+(t−t1)2][2α], vt = e−α[(x−x1)2+(t−t1)2][−2α(t −t1)], entonces L[v] ≡a e−α[(x−x1)2+(t−t1)2][4α2(x −x1)2] −e−α[(x−x1)2+(t−t1)2][2α] +be−α[(x−x1)2+(t−t1)2][−2α(x −x1)] −e−α[(x−x1)2+(t−t1)2][−2α(t −t1)], 40 L[v] ≡ae−α[(x−x1)2+(t−t1)2][4α2(x −x1)2 −2α] −be−α[(x−x1)2+(t−t1)2][2α(x −x1)] + e−α[(x−x1)2+(t−t1)2][2α(t −t1)], de aqu´ ı L[u] ≡2αe−α[(x−x1)2+(t−t1)2][2αa(x −x1)2 −a −b(x −x1) + (t −t1)], y si escogemos α lo suficientemente grande tenemos que L[v] > 0 para (x, t) ∈D ∪∂D.
(3.25) Adem´ as, sea w(x, t) = u(x, t) + εv(x, t), donde ε es una constante positiva tan peque˜ na que u < M sobre C′, (3.26) luego En D L[w] = L[u] + εL[v], por hip´ otesis y (3.25) L[w] > 0.
De esto, w no puede alcanzar su m´ aximo en D, ahora En ∂K v = e−αR2 −e−αR2, v = 0, por lo tanto w = u + εv = u, y de acuerdo a (3.21) w < Msobre C′′ excepto en P.
En P w(x0, t0) = u(x0, t0) + εv(x0, t0), como P ∈∂K, entonces w(x0, t0) = u(x0, t0), 41 y por (3.22) w(x0, t0) = M en P.
Si centramos nuestra atenci´ on en la regi´ on D, por hip´ otesis adem´ as de (3.21) y (3.26), podemos concluir que el m´ aximo de w sobre D ∪∂D ocurre en el ´ unico punto P Por lo tanto en P: wν = uν + εvν ≥0, (3.27) Figura 3.7: Derivada direccional en una direcci´ on exterior desde Et0 en P ahora, vν = ν · grad(v), vν = ν · ⟨−2α(x −x1)e−α[(x−x1)2+(t−t1)2] , −2α(t −t1)e−α[(x−x1)2+(t−t1)2]⟩, o sea que vν(x0, t0) = ν · ⟨−2α(x0 −x1)e−αR2 , −2α(t0 −t1)e−αR2⟩, vν(x0, t0) = ν · −2αe−αR2⟨(x0 −x1) , (t0 −t1)⟩, vν(x0, t0) = −2ν · nαe−αR2, de donde vν(x0, t0) < 0, Es decir, de acuerdo a (3.27) se concluye que uν > 0 en P.
42 Cap´ ıtulo 4 EL Principio del M´ aximo Aplicado a Ecuaciones Diferenciales Parciales Semi-lineales de Segundo Orden de Tipo Parab´ olico Este cap´ ıtulo se fundamenta en los libros , y en el articulo citados en la bibliograf´ ıa.
Muestra un principio del m´ aximo mas general aplicado a la ecuaci´ on (1) ut + a(u, x, t)ux + g(u, x, t) = uxx, en una regi´ on E = R × [0, T]. Inicialmente determina condiciones necesarias mediante la teor´ ıa desarrollada en los cap´ ıtulos 2 y 3 y crea una transformaci´ on que linealiza el problema, para en adelante aprovechar la teor´ ıa antes mencionada (lineal), aplicar dicho principio y establecer su comportamiento.
El siguiente teorema constituye el resultado Teorema 8. Sea u(x,t) una soluci´ on de la ecuaci´ on diferencial parcial semi-lineal de tipo parab´ olico ut + a(u, x, t)ux + g(u, x, t) = uxx, (4.1) con dato inicial |u(x, 0)| ≤M definida en una regi´ on E = R × [0, T]; 0 < T < ∞, adem´ as |g(u, x, t)| ≤c|u| + e c; con c, e c > 0 y a(u, x, t) localmente acotada. Entonces para todo T > 0, existe M(T) > 0 tal que |u(x, t)| ≤M(T) sobre R × [0, T].
43 Figura 4.1: Regi´ on de estudio E = R × [0, T]; 0 < T < ∞de la ecuaci´ on (1) Demostraci´ on. Multiplicando la ecuaci´ on (4.1) por 2u, tenemos que 2uut + 2ua(u, x, t)ux + 2ug(u, x, t) = 2uuxx, 2uut + a(u, x, t)2uux = 2uuxx −2ug(u, x, t), y sumando y restando el termino 2(ux)2, 2uut + a(u, x, t)2uux = 2(ux)2 + 2uuxx −2(ux)2 −2ug(u, x, t), (u2)t + a(u, x, t)(u2)x = (2uux)x −2(ux)2 −2ug(u, x, t), (u2)t + a(u, x, t)(u2)x ≤(2uux)x −2(ux)2 + 2|u|(c|u| + e c), o sea que (u2)t + a(u, x, t)(u2)x ≤(u2)xx + (2c + 1)u2 + e c2.
(4.2) Sea v = u2e−(2c+1)t, (4.3) entonces u2 = v e−(2c+1)t = ve(2c+1)t, de donde (u2)t = vte(2c+1)t + ve(2c+1)t(2c + 1), (u2)x = vxe(2c+1)t, (u2)xx = vxxe(2c+1)t, as´ ı, reemplazando estos t´ erminos en (4.2) se tiene que vte(2c+1)t + ve(2c+1)t(2c + 1) + a(u, x, t)vxe(2c+1)t ≤vxxe(2c+1)t + (2c + 1)e(2c+1)t + e c2, e(2c+1)t[vt + (2c + 1)v + a(u, x, t)vx] ≤e(2c+1)t[vxx + (2c + 1)v + e c2 e(2c+1)t ], 44 vt + (2c + 1)v + a(u, x, t)vx ≤vxx + (2c + 1)v + e c2 e(2c+1)t , vt + a(u, x, t)vx ≤vxx + e c2e−(2c+1)t.
(4.4) Ahora haciendo w = v + e c2 2c + 1e−(2c+1)t, (4.5) tenemos v = w − e c2 2c+1e−(2c+1)t, de donde vt = wt + e c2 2c + 1e−(2c+1)t(2c + 1), vx = wx, vxx = wxx, por lo que (4.4) se transforma en wt + e c2 2c + 1e−(2c+1)t(2c + 1) + a(u, x, t)wx ≤wxx + e c2e−(2c+1)t, wt + a(u, x, t)wx ≤wxx + e c2e−(2c+1)t − e c2 2c + 1e−(2c+1)t(2c + 1), wt + a(u, x, t)wx ≤wxx + e c2e−(2c+1)t −e c2e−(2c+1)t, wt + a(u, x, t)wx ≤wxx.
(4.6) Adem´ as, w(x, 0) = v(x, 0) + e c2 2c + 1e−(2c+1)0, que de acuerdo a (4.3), w(x, 0) = u2(x, 0)e−(2c+1)0 + e c2 2c + 1e−(2c+1)0, w(x, 0) = u2(x, 0) + e c2 2c + 1, luego, por hip´ otesis w(x, 0) ≤M2 + e c2 2c + 1.
(4.7) Aplicamos a (4.6) la siguiente transformaci´ on w = r + M2 + e c2 2c + 1 + N(x2 + cLet) L2 , 45 donde L, c, N son constantes positivas y adem´ as N es la cota superior de w sobre R × [0, T]; (N puede ser obtenido por la existencia local). Entonces reemplazando w en (4.6), tenemos rt + NcLet L2 + a(u, x, t)[rx + 2Nx L2 ] = rxx + 2N L2 , rt + NcLet L2 + a(u, x, t)rx + a(u, x, t)2Nx L2 = rxx + 2N L2 , rt + NcLet L2 + a(u, x, t)rx + a(u, x, t)2Nx L2 −2N L2 = rxx, rt + a(u, x, t)rx + NcLet L2 + a(u, x, t)2Nx L2 −2N L2 = rxx, rt + a(u, x, t)rx + [cLet + a(u, x, t)2x −2] N L2 = rxx.
(4.8) Ahora, w(x, 0) = r(x, 0) + M2 + e c2 2c + 1 + N(x2 + cLe0) L2 , r(x, 0) = w(x, 0) −M2 − e c2 2c + 1 −N(x2 + cL) L2 , que de acuerdo a la desigualdad (4.7) r(x, 0) ≤M2 + e c2 2c + 1 −M2 − e c2 2c + 1 −N(x2 + cL) L2 , r(x, 0) ≤−N(x2 + cL) L2 , y debido a que N, c, L > 0 r(x, 0) < 0.
(4.9) De all´ ı, si tomamos una franja (−L, L) × (0, T) 46 Figura 4.2: Franja (−L, L) × (0, T) entonces w(L, t) = r(L, t) + M2 + e c2 2c + 1 + N(L2 + cLet) L2 , r(L, t) = w(L, t) −M2 − e c2 2c + 1 −N(L2 + cLet) L2 , y por ser N la cota superior de w, w(L, t) ≤Nsobre (−L, L) × (0, T), es decir que r(L, t) ≤N −M2 − e c2 2c + 1 −N(L2 + cLet) L2 , r(L, t) ≤N −M2 − e c2 2c + 1 −NL2 L2 −NcLet L2 , r(L, t) ≤N −M2 − e c2 2c + 1 −N −Ncet L , r(L, t) ≤−M2 − e c2 2c + 1 −Ncet L , y como N, c, L, e c > 0 r(L, t) < 0.
(4.10) De la misma manera se obtiene que r(−L, t) < 0.
(4.11) As´ ı desde (4.8)-(4.11) y aplicando el principio del m´ aximo podemos concluir que r(x, t) < 0 sobre (−L, L) × (0, T).
(4.12) 47 En efecto, supongamos que (4.12) no se cumple, es decir que r(x, t) ≥0, luego, sea t la cota superior de los valores de t donde r < 0, entonces por continuidad rt ≥0 rx = 0 rxx ≤0 en (x, t).
(4.13) Por otro lado, si escogemos c lo suficientemente grande tal que cLet + a(u, x, t)2x −2 > 0 en (−L, L) × (0, T) tendr´ ıamos que la ecuaci´ on (4.8) se contradice, as´ ı que r(x, t) < 0 sobre (−L, L) × (0, T), de esta manera se prueba (4.12). No obstante, para un punto (x0, t0) ∈(−L, L) × (0, T) se tiene que w(x0, t0) = r(x0, t0) + M2 + e c2 2c + 1 + N(x2 0 + cLet0) L2 , o sea que w(x0, t0) ≤M2 + e c2 2c + 1 + N(x2 0 + cLet0) L2 , para lo cual w ≤M2 + e c2 2c+1 si hacemos que L vaya a infinito.
Finalmente, por como se hab´ ıa definido v en (4.3) y w en (4.5) se tiene que u2e−(2c+1)t + e c2 2c + 1e−(2c+1)t ≤M2 + e c2 2c + 1, e−(2c+1)t u2 + e c2 2c + 1 ≤M2 + e c2 2c + 1, u2 + e c2 2c + 1 ≤ M2 + e c2 2c + 1 e(2c+1)t, u2 ≤ M2 + e c2 2c + 1 e(2c+1)t − e c2 2c + 1, y como c, e c > 0 u2 ≤ M2 + e c2 2c + 1 e(2c+1)t, de all´ ı que, |u| ≤ h M2 + e c2 2c + 1 e(2c+1)ti 1 2 , por tanto |u| ≤M(T).
48 Conclusiones El desarrollo de la teor´ ıa lineal, el estudio y an´ alisis de la aplicabilidad del principio del m´ aximo en ecuaciones lineales tanto ordinarias como parciales de tipo parab´ olico ha sido determinante y ha permitido abordar un nuevo problema de estudio sobre teor´ ıa no lineal, es decir, sobre la ecuaci´ on (1). Hemos podido utilizar el resultado y la metodolog´ ıa que tiene la aplicabilidad del principio del m´ aximo sobre ecuaciones parciales de tipo parab´ olico como la ecuaci´ on del calor, en una ecuaci´ on m´ as general (resultado de una transformaci´ on, de la cual la ecuaci´ on del calor es el prototipo) y en un dominio m´ as general para lograr el objetivo propuesto que es obtener informaci´ on del comportamiento de su soluci´ on.
Una buena forma de estudiar este tipo de problem´ atica, es empezando por abordar teor´ ıas y ecuaciones mas sencillas que den luz al problema y que nos ayuden a determinar condi-ciones o propiedades para la misma La dificultad del problema expuesto aqu´ ı radica en que la teor´ ıa no lineal es muy escasa y en este tipo de problema se ha recurrido a linealizarlo mediante una transformaci´ on, para en adelante abordar y utilizar la aplicabilidad del principio del m´ aximo sobre la teor´ ıa lineal que ya ha sido expuesta.
Aunque el trabajo expuesto aqu´ ı es de tipo monogr´ afico, permite al lector comprender de una mejor manera la metodolog´ ıa y aplicabilidad del principio del m´ aximo sobre ecuaciones lineales ya que se ha profundizado en ella, se han dado los detalles de las pruebas de una manera minuciosa para finalmente entender el tratamiento que se ha hecho sobre la ecuaci´ on de estudio.
Se sugiere estudiar o indagar, si el principio del m´ aximo podr´ ıa ser aplicado a otra ecuaci´ on o modificaci´ on de la ecuaci´ on (1), asimismo podr´ ıa pensarse en estudiarla en mas dimen-siones, deducir la singularidad de su soluci´ on o pensar en un sistema de ecuaciones.
49 Bibliograf´ ıa Greenspan, Donald. Introduction to Partial Differential Equations. Dover Edition. Mineola, New york: Dover Publications, Inc. (2000), 204.p Lawrence C, Evans. Partial Differential Equations. American Mathematical Society. Rhode Island: (1998), 662.p Friedman, Avner. Partial Differential Equations of Parabolic Type. Prentice-Hall. Engle-wood Cliffs, N.J: (1964), 347.p Bers L., F. John, y M. Schechter. Partial Differential Equations. New York: Interscience Publishers. (1964), 343.p Garabedian. P. Partial Differential Equations. New York: Interscience Publishers. (1964), 672.p O.A, Ladyzenskaja; V.A. Solonnikov; N.N. Ural´ceva.
Linear and Quasilinear Equations of parabolic Type. American Math. Society, (1968), 648.p Protter, Murray y Weinberger, Hans. Maximun Principles in Differential Equations. New York: Springer-Verlag. (1984), 261.p Cer´ on, Miller. O. El Principio del M´ aximo. Revita Sigma, Departamento de Matem´ aticas.
Univeridad de Nari˜ no. Vol VII (2008), p. 28-34 Klingenberg, Christian, Lu, Yunguang y Rend´ on Leonardo. On Global Lipschitz-continuous Solutions of Isentropic Gas Dynamics. Taylor & Francis Group. Vol 82 (2003), p. 35-43 Cer´ on, Miller. O. Soluciones Viscosas para un Sistema de Leyes de Conservaci´ on. Tesis de Maestr´ ıa para la obtenci´ on del titulo de Acad´ emico de Magister en Matem´ aticas, Facultad de Ciencias, Universidad Nacional, Bogot´ a, Colombia.(2007), p. 15-25 Aronson D. G y Serrin James. A Maximum Principle for Nonlinear Parabolic Equations, University of Minnesota. Tomo 21, No2 (1967), p. 291-305 50 R. Dautray, J.L. Lions. Analyse Mathematique et Calcul Numerique pour les Sciences et les Techniques, Masson, (1985), 1062.p I. Peral Alonso Primer Curso de Ecuaciones en Derivadas Parciales, Addison Wesley. Uni-versidad Aut´ onoma de Madrid, (1993), 326.p 51 |
9596 | https://math.stackexchange.com/questions/3813208/definition-or-proof-of-a-frac-mn-sqrtnam | Stack Exchange Network
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Definition or proof of $a^{\frac mn}=\sqrt[n]{a^m}$? [duplicate]
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Just a curiosity in this book of an high school with the title The powers with exponent rational
the $$\large \color{red}{a^{\frac mn}=\sqrt[n]{a^m}}, \quad a\geq 0$$ is given like a definition. I don't remember if there was a proof. But is there one or should we give it as a definition?
algebra-precalculus
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asked Sep 3, 2020 at 16:47
SebastianoSebastiano
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$\begingroup$ Questa è una definizione, ma poi si deve dimostrare che è ben definita, e le varie proprietà delle potenze anche per questi esponenti razionali. $\endgroup$
Crostul
– Crostul
2020-09-03 16:53:38 +00:00
Commented Sep 3, 2020 at 16:53
$\begingroup$ @Crostul Hi, from Sicily. I am happy to receive also your answer. $\endgroup$
Sebastiano
– Sebastiano
2020-09-03 19:04:22 +00:00
Commented Sep 3, 2020 at 19:04
6
$\begingroup$ Does this answer your question? Can you raise a number to an irrational exponent? $\endgroup$
Joe
– Joe
2020-09-03 20:49:06 +00:00
Commented Sep 3, 2020 at 20:49
$\begingroup$ @Sebastiano There is a post from Andre Nicolas that you might find to be of interest. He states that this can indeed be a definition. However, we can also use the natural logarithm to define exponentiation, which tends to be what is done in practice. $\endgroup$
Joe
– Joe
2020-09-03 20:50:54 +00:00
Commented Sep 3, 2020 at 20:50
1
$\begingroup$ @Sebastiano Ah, I seethis approach would not be appropriate. Best of luck with your teaching! $\endgroup$
Joe
– Joe
2020-09-03 21:16:40 +00:00
Commented Sep 3, 2020 at 21:16
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Let $r$ be a positive rational number, represented as the quotient of two positive integers, $r=m/n$. We will show that two definitions of $a^r$, where $a\ge0$ are equivalent: $a^r=\sqrt[n]{a^m}$, and $a^r=(\sqrt[n]{a})^m$. Furthermore, we shall show that either definition is independent of the representation of $r$ as the quotient of positive integers.
PART $(1)$:
Let $y\ge 0$ be defined
$$y=\sqrt[n]{a^m}=(a^m)^{1/n}\tag1$$
By definition of the $n$'th root, we have from $(1)$
$$y^n=a^m\tag2$$
Now, let $y=z^m$ for some number $z\ge 0$. Then, substituting into $(2)$ reveals
$$(z^m)^n=a^m\tag3$$
Using $(z^m)^n=(z^n)^m$ in $(3)$, we find that $z^n=a$, which implies that
$$y=(a^{1/n})^m\tag4$$
Equating $(1)$ and $(4)$ yields
$$(a^m)^{1/n}=(a^{1/n})^m$$
So, the two definitions of $a^r$, $(a^m)^{1/n}$ and $(a^{1/n})^m$ that are equivalent. That is to say, that
$$\bbox[5px,border:2px solid #C0A000]{\sqrt[n]{a^m}=(\sqrt[n]{a})^m}\tag5$$
PART $(2)$:
Now, we shall show that the definitions of $a^r=a^{m/n}$ in $(5)$ are independent of the representation of $r$.
Suppose $\ell$ is an arbitrary integer. Then, $a^r=a^{m/n}=a^{(\ell m)/(\ell n)}$. We wish to show that
$$a^r=\sqrt[\ell n]{a^{\ell m}}=\sqrt[ n]{a^{ m}}$$
We let $w$ be defined as
$$w=(a^{\ell m})^{1/(\ell n)}\tag6$$
so that $a^{\ell m}=w^{\ell n}$, which implies $a^m=w^n$, from which we have
$$w=(a^m)^{1/n}$$
And similarly, we wish to show that
$$(\sqrt[\ell n]{a})^{\ell m}=(\sqrt[n]{a})^{ m}$$
Let $v= (\sqrt[\ell n]{a})^{\ell m}$. But from $(5)$ we have
$$v=\sqrt[\ell n]{a^{\ell m}}$$
which we just showed was equal to $\sqrt[n]{a^m}$. Therefore, we have proven the independence of the representation of $r$ as the quotient of positive integers.
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edited Sep 3, 2020 at 21:04
answered Sep 3, 2020 at 17:03
Mark ViolaMark Viola
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$\begingroup$ Do you have a cristal ball? ahaha. I was logging now. I prefer adopt another character instead of $z$ that I give only to complex numbers. Ok and it is perfect your proof. But I not understand because not exist a proof. $\endgroup$
Sebastiano
– Sebastiano
2020-09-03 19:06:32 +00:00
Commented Sep 3, 2020 at 19:06
$\begingroup$ What is the definition that you are using for $a^{m/n}$ in the last line of your answer? $\endgroup$
posilon
– posilon
2020-09-03 19:15:39 +00:00
Commented Sep 3, 2020 at 19:15
$\begingroup$ @posilon Since we have shown that $(a^m)^{1/n}$ is equal to $(a^{1/n})^m$, then evidently $a^{m/n}$ is unambiguous. $\endgroup$
Mark Viola
– Mark Viola
2020-09-03 19:47:58 +00:00
Commented Sep 3, 2020 at 19:47
$\begingroup$ @MarkViola Hi, excuse me very much. Why I find anothe downvoted that I not understood? $\endgroup$
Sebastiano
– Sebastiano
2020-09-12 15:44:39 +00:00
Commented Sep 12, 2020 at 15:44
$\begingroup$ Hi Sebasiano. I don't know why there are so many down voters. $\endgroup$
Mark Viola
– Mark Viola
2020-09-12 15:47:41 +00:00
Commented Sep 12, 2020 at 15:47
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$a^{\frac mn}$ can be defined as $\sqrt[n]{a^m}$, in which case there is no point of talking about a proof of $a^{\frac mn} = \sqrt[n]{a^m}$. What you should prove is that this is well defined, i.e. if $\frac mn = \frac{m'}{n'}$ then $\sqrt[n]{a^m} = \sqrt[n']{a^{m'}}$. Indeed \begin{align} &&mn' &= m'n\ \Rightarrow &&a^{mn'} &= a^{m'n}\ \Rightarrow &&(a^m)^{n'} &= (a^{m'})^n\ \Rightarrow &&((a^m)^{n'})^{1/nn'} &= ((a^{m'})^n)^{1/nn'}\ \stackrel{()}{\Rightarrow} &&(((a^m)^{n'})^{1/n'})^{1/n} &= (((a^{m'})^n)^{1/n})^{1/n'}\ \Rightarrow &&(a^m)^{1/n} &= (a^{m'})^{1/n'}. \end{align} $()$ Here we make use of the property $b^{1/\kappa\lambda}=(b^{1/\kappa})^{1/\lambda}$ for $b \geq 0$ and $\kappa,\lambda \in \mathbb{Z}_{>0}$. This follows from the fact that both LHS and RHS are non-negative solutions of the equation $x^{\kappa\lambda}=b$.
You could also talk about the motivation behind the definition: We would like $(a^x)^y = a^{x\cdot y}$ to hold for rational exponents, as it holds when $x,y \in \mathbb{N}$. So we would like to define $a^{\frac mn}$ in a way so that $(a^{\frac mn})^n = a^m$.
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edited Sep 4, 2020 at 4:01
answered Sep 3, 2020 at 16:57
posilonposilon
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$\begingroup$ That's not quite true. One needs to show that $(a^m)^{1/n}=(a^{1/n})^m$. I've posted a solution accordingly. $\endgroup$
Mark Viola
– Mark Viola
2020-09-03 17:05:03 +00:00
Commented Sep 3, 2020 at 17:05
$\begingroup$ What is not true? And what's wrong with defining $a^{m/n} := \sqrt[n]{a^m}$ without first showing the equality you suggest? $\endgroup$
posilon
– posilon
2020-09-03 19:06:25 +00:00
Commented Sep 3, 2020 at 19:06
$\begingroup$ You need to show that $\sqrt[n]{a^m}=(\sqrt[n]{a})^m$. Have you done that? If not, then you cannot define $a^{m/n}$ uniquely. Is it $(a^m)^{1/n}$ or is it $(a^{1/n})^m$? $\endgroup$
Mark Viola
– Mark Viola
2020-09-03 19:44:17 +00:00
Commented Sep 3, 2020 at 19:44
$\begingroup$ What's wrong with making an arbitrary choice between the two (for example $(a^m)^{1/n}$ as I suggested)? $\endgroup$
posilon
– posilon
2020-09-03 19:48:43 +00:00
Commented Sep 3, 2020 at 19:48
$\begingroup$ Suppose that $r$ is a rational number which can be represented by $r=m/n$ for integers $m$ and $n$. Then, $r$ is also given by $r=(qm)/(qn)$ for any integer $q$. So, what is $a^r$? Is it $(a^m)^{1/n}$? Is it $(a^{2m})^{1/(2n)}$? How can you be sure these are equal? $\endgroup$
Mark Viola
– Mark Viola
2020-09-03 20:02:42 +00:00
Commented Sep 3, 2020 at 20:02
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Deutsch 101 Handout: The
Perfekt
Tense
We‟re finally going to learn to talk about p ast events in German! You‟re about to learn what‟s called the „present perfect‟ tense in English, or
Das Perfekt
in German. (Don‟t be confused by t he word „present‟
-- this is a past tense that talks about past events --
it‟s the English terminology that‟s confusing.) Although there are two
different past tenses in German (as in English), the
Perfekt
that you are learning is used most often in conversational (spoken) German. You will learn the other past tense (the narrative or simple past tense) in second-semester German.The first thing we need to talk about when learning to form the
Perfekt
is the difference between strong and weak verbs in German. We have the same difference in English, too. Consider the following examples:play - played spielen - gespielt speak - spoken sprechen - gesprochen learn - learned lernen - gelernt give - given geben - gegeben
Like English, German has a group of „regular‟ (termed „weak‟) verbs that always add a
-t
ending for the past
participle, but another very different group of „irregular‟ (termed „strong‟) verbs that add an
-en
. We will be
learning both kinds today, but for just now, we‟re going to focus on the (easier) weak verbs.
To form the past participle of the German
Perfekt
tense for
weak verbs
, you need to take the stem of the verb(the infinitive minus the
-en
ending) and add a
ge-
prefix and a
-t
suffix.spielen - gespielt machen - gemacht fragen - gefragt wohnen - gewohnt arbeiten - gearbeitet reden - geredet There are two exceptions to this nice regularity. The first ist that verbs that end in
-ieren
,like
fotografieren
,
diskutieren
,
studieren
, etc., do NOT get a
ge-
prefix. (These verbs can be easily recognized as English-French cognates, and they all end in
-ieren
, so it‟s a pretty easy group to remember.)
studieren - studiert manipulieren - manipuliert reparieren - repariert The other exception is for inseparable prefixes, like
ver-
,
be-
and
miss-
. When a verb has an inseparable prefix, it does NOT get a
ge-
prefix for the participle form.besuchen - besucht erleben - erlebt verkaufen - verkauft What happens with separable-
prefix verbs? As you might expect, the prefix gets „stuck‟ back on to the participle at
the end, but the
ge-
prefix ends up in the middle of the word:Wir haben die Tür zugemacht.
We closed the door.
Hast du dein Zimmer aufgeräumt?
Did you clean up your room?
Now that you can form th e participle form of the verb, we need to learn the complete syntax for a past tense sentence. To form a complete German past tense sentence, you need to add a helping verb,either
‘haben’
or
‘sein’
. Let‟s look
at the verbs that take
‘haben’
:Ich habe Fußball gespielt.
I played football.
Maria hat zwei Semester Deutsch gelernt.
Maria learned German for two semesters.
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Hast du deine Hausaufgaben gemacht?
Did you do your homework?
As you can see, the helping verb (
haben
in these sentences) is conjugated to match the subject, while the participle (
ge
-stem-
t
) remains constant. Also note that the participle occurs at the very end of the sentence, while the conjugated helping verb is in the normal verb position (second element for statements, first for questions).
Let‟s try a few simple sentences for practice. These are all weak („regular‟) verbs.
1.
to make = _______ __
participle: _______ __
Did you make the b ed? ______ ___ _____
2.
to clean up = ________ _
participle: _______ __
I cleaned up my room yesterday. _____ ___
3.
to have = ___ ______
participle: _______ __
Tom had a party on Monday. ___ _____ _______
4.
to cost= ___ ______
participle: _______ __
How much did your bike cost? ____ ___ _______
5.
to wait = ___
participle: _______ __
I waited for 20 minutes! ______ ____ ___
Now we need to learn about the other type of verb, the
strong verb
. These verbs are harder, because they often change their stem vowel in unpredictable ways, so they need t o be memorized. Still, they do build their participle form in a regular fashion: a
ge-
prefix is added, and an
-en
(NOT
-t
) suffix. The stem vowel will often change, but not always. See the separate chart (Verb Classes) for details on stem vowel changes.sing - sung fly - flown give - given singen - gesungen fliegen - geflogen geben - gegeben There are a
handful of strong verbs that don‟t just change their vowel, but the whole stem. These irregular verbs just need to be memorized, but fortunately they‟re the most common verbs (g o, come, be, do), so you‟ll see t hem
a lot and get familiar with them very quickly.gehen - gegangen stehen - gestanden sein - gewesen The same rules for inseparable prefixes (no
ge-
added to participle) and separable prefixes (added back on before the
ge-
) hold true for all verbs, strong, mixed and weak.weggehen - weggegangen mitbringen - mitgebracht verstehen - verstanden
In addition to the strong verbs, t here is a very small handful (about six) of verbs that are called „mixed‟ verbs,
because they act like a mix between strong and weak verbs. They take a
ge
-+-
t
form like weak verbs, but their
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stem vowels change. Again, these verbs just need to be memorized.denken - gedacht bringen - gebracht kennen - gekannt
Let‟s try a few sentences again. These are all strong or mixed verbs, so you‟ll need to look at your c
hart/list.1.
to write = __ _
participle: _______ __
Yesterday I wrote a letter. _______ ________ __
2.
to speak = _______ __
participle: _______ __
My grandfather spoke German. ___ ___ _______
3.
to drink = ___
participle: _______ __
Did you guys drink a lot of beer last night? _____ ____
4.
to see = ___
participle: _______ __
Whom
(accusative)
did you see? ___ _____ _____
5.
to know a fact = ___
participle: _______ __
I didn’t know the answer. _____ _____ _
Almost done! The last thing to learn about the
Perfekt
tense is that sometimes
‘haben’
is not the right helping verb to use. Rather, you need to use the helping verb
‘sein’
(
er ist
, etc) for verbs that meet both of these criteria:a) the verb indicates a
change of position or condition
, or a crossing of a „boundary‟
e.g.
gehen, kommen, wandern, sterben
(=to die),
einschlafen
(=to fall asleep)b) the verb is
intransitive
(= does NOT have a direct object)e.g.
fahren
(
ich bin nach Milwaukee gefahren
, BUT
ich habe mein Auto gefahren
)In addition, the three verbs
sein
(to be) and
bleiben
(to stay) and
passieren
(to happen) both take
‘sein’
as a
helping verb, although they don‟t match the criteria above. Consider these examples:
Anna ist nach Deutschland geflogen.
Anna flew to Germany.
Ich bin um 7 Uhr nach Hause gekommen.
I came home at 7 o’
clock.
Bist du schon eingeschlafen?
Have you fallen asleep already?
Paul ist ein fleißiger Student gewesen.
Paul was a hard-working student.
Once again, practice by making complete sentences. All of these verbs take
‘sein’
as a helping verb, but some are strong verbs (
ge
-stem-
en
) while others are weak (
ge
-stem-
t
).1.
to come = ______ ___
participle: _______ __
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Sandra didn’t come to cl ass yesterday. _____ _______
2.
to travel = ______ ___
participle: _______ __
We travelled to Europe last year. _____ ___ ___
3.
to fly = ______ ___
participle: _______ __
Have you ever
(=jemals)
flown to Australia? _____ ____
4.
to be = ______ ___
participle: _______ __
I have never
(=niemals)
been in China. ____ ____ ____
5.
to go = ___ ______
participle: _______ __
They went to the movies on Saturday. ___ ____ ______
Now we need to mix things up. Below are blanked out sentences: some verbs take
‘sein’
as a helping verb, others take
‘haben’
. Try to determine which helping verb to use.1. Wir __ nach Hause gegangen.2. Paul __ uns gesehen.3. Wir __ Pizza gegessen.4. __ ihr um zehn Uhr eingeschlafen?5. __ du gestern Fußball gespielt?6. Tante Uschi __ Pharmazie studiert.7. __ Sie zur Uni gelaufen?8. Meine Großmutter __ im Jahre 1978 gestorben.9. Ich ___ nach Madison gefahren.10. Sie __ _______ ihr F ahrrad gefahren.
Finally, let‟s mix everything together. Below are sentences with missing verbs. Fill in the helping verbs
(either
‘haben’
or
‘sein’
) and also the participles (either strong or weak). Take it slow! It‟s hard at first
, but it will get much better with a little bit of practice.1.
arbeiten / lernen:
Wir ___ den ganzen Tag __ , aber wir __ nichts _ __.2.
aufstehen / essen:
Ich ___ heute um 6 Uhr __ , und dann __ ich Frühstück _ __.3.
fragen / sagen:
Du __ mich _____, und ich __
“nein” ___ .
4.
mitkommen / bleiben:
__ Georg zur Party __ ___? -- Nein, er __ zu Hause _ ____.5.
passieren / fahren:
Was __ hier ____ _? -- Mein Auto __ gegen einen Baum _ ____.
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Using Cofunction Identities
Trigonometry Skills Practice
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00:04 Using cofunction…
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Nicole Koenig, Aaron Gaffney
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Example SolutionsPractice Questions
Using Cofunction Identities
Step 1: Determine what cofunction identities are needed, and apply them accordingly.
Step 2: If necessary, use other known formulas and trigonometric identities, such as reciprocal identities or Pythagorean identities, to simplify completely.
Using Cofunction Identities: Equations and Vocabulary
Cofunction Identities Cofunction identities are trigonometric identities that show the relationship of complementary angles and trigonometric functions/ratios between sine and cosine, tangent and cotangent, and secant and cosecant.
We know that the sum of the angles of a triangle are 180 degrees (π radians). In a right triangle, the right angle is 90 degrees (π 2 radians), so the remaining two angles are complementary angles and must also add up to 90 degrees (π 2 radians). Therefore, the measures of the two acute angles of a right triangle are x and π 2−x radians.
Using this right triangle and trigonometric ratios, we can see that the sine of an angle is the same as the cosine of the complement of that angle.
sin(π 2−x)=opposite hypotenuse=a c cos(x)=adjacent hypotenuse=a c
We see that sin(π 2−x)=cos(x). Similarly, this concept is true for all of the cofunction pairs:
Let's try putting these steps, definitions, and cofunction identities into practice by working through four examples of using cofunction identities to simplify trigonometric expressions.
Using Cofunction Identities: Example 1
Use cofunction identities to simplify the expression fully:
cos(π 2−x)cscx
Step 1: Determine what cofunction identities are needed, and apply them accordingly.
We will use the cofunction identity of sinx=cos(π 2−x) to rewrite the expression as follows:
cos(π 2−x)cscx=sinx cscx
Step 2: If necessary, use other known formulas and trigonometric identities, such as reciprocal identities or Pythagorean identities, to simplify completely.
We will use the reciprocal identity of cscx=1 sinx to rewrite the expression and simplify completely as follows:
sinx cscx=sinx(1 sinx)=1
Thus, if we simplify cos(π 2−x)cscx completely, we get 1.
Using Cofunction Identities: Example 2
Use cofunction identities to simplify the expression fully:
cosx cos(π 2−x)
Step 1: Determine what cofunction identities are needed, and apply them accordingly.
We will use the cofunction identity of sinx=cos(π 2−x) to rewrite the expression as follows:
cosx cos(π 2−x)=cosx sinx
Step 2: If necessary, use other known formulas and trigonometric identities, such as reciprocal identities or Pythagorean identities, to simplify completely.
We will use the trigonometric identity of cotx=cosx sinx to simplify completely.
cosx sinx=cotx
Thus, if we simpify cosx cos(π 2−x) completely, we get cotx.
Using Cofunction Identities: Example 3
Use cofunction identities to simplify the expression fully:
tan(π 2−x)secx
Step 1: Determine what cofunction identities are needed, and apply them accordingly.
We will use the cofunction identity cotx=tan(π 2−x) to rewrite the expression as follows:
tan(π 2−x)secx=cotx secx
Step 2: If necessary, use other known formulas and trigonometric identities, such as reciprocal identities or Pythagorean identities, to simplify completely.
We will use the identities of cotx=cosx sinx,secx=1 cosx,and cscx=1 sinx to rewrite and simplify the expression as follows:
cotx secx=(cosx sinx)(1 cosx)=(1 sinx)=cscx
Thus, if we simplify tan(π 2−x)secx completely, we get cscx.
Using Cofunction Identities: Example 4
Use cofunction identities to simplify the expression fully:
sin(π 2−x)−sin 2x cosx
Note: Use the identity cos 2x=1−sin 2x in step 2.
Step 1: Determine what cofunction identities are needed, and apply them accordingly.
We will use the cofunction identity cosx=sin(π 2−x) to rewrite the expression as follows:
sin(π 2−x)−sin 2x cosx=cosx−sin 2x cosx
Step 2: If necessary, use other known formulas and trigonometric identities, such as reciprocal identities or Pythagorean identities, to simplify completely.
We will use factoring and the trigonometric identity cos 2x=1−sin 2x to rewrite the expression and simplify as follows:
cosx−sin 2x cosx=cosx(1−sin 2x)=cosx(cos 2x)=cos 3x
Thus, if we simplify sin(π 2−x)−sin 2x cosx completely, we get cos 3x.
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9599 | https://www.jaad.org/article/S0190-9622(03)00896-X/abstract | Predisposing factors and clinical features of eczema herpeticum: a retrospective analysis of 100 cases - Journal of the American Academy of Dermatology
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ReportVolume 49, Issue 2p198-205 August 2003
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Predisposing factors and clinical features of eczema herpeticum: a retrospective analysis of 100 cases
Andreas Wollenberg, MD
Andreas Wollenberg, MD
Affiliations
Department of Dermatology and Allergy, Ludwig Maximilian University, MunichGermany
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∙ Claudia Zoch, MD
Claudia Zoch, MD
Affiliations
Department of Dermatology and Allergy, Ludwig Maximilian University, MunichGermany
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∙ Stefanie Wetzel, MD
Stefanie Wetzel, MD
Affiliations
Department of Dermatology and Allergy, Ludwig Maximilian University, MunichGermany
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∙ Gerd Plewig, MD
Gerd Plewig, MD
Affiliations
Department of Dermatology and Allergy, Ludwig Maximilian University, MunichGermany
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∙ Bernhard Przybilla, MD
Bernhard Przybilla, MD
Affiliations
Department of Dermatology and Allergy, Ludwig Maximilian University, MunichGermany
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Affiliations & Notes Article Info
Department of Dermatology and Allergy, Ludwig Maximilian University, MunichGermany
Publication History:
Accepted November 30, 2002
Footnotes:
☆
Funding sources: None.
Conflicts of interest: None identified.
DOI: 10.1067/S0190-9622(03)00896-X External LinkAlso available on ScienceDirect External Link
Copyright: © 2003 American Academy of Dermatology, Inc.
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Abstract
Background
Eczema herpeticum (EH) is a widespread herpes simplex virus infection of inflamed skin, most often occurring in patients with atopic dermatitis (AD). A monomorphic eruption of dome-shaped blisters and pustules in the eczematous lesions along with severe systemic illness lead to the clinical diagnosis, but atypical variants with disseminated slits may also occur. Topical use of corticosteroids is alleged to be a pathogenetic factor for EH, but predisposing factors for EH are largely unknown.
Objective and methods
We sought to characterize the clinical features and predisposing factors for EH. A retrospective analysis of 100 patients with EH seen from 1980 through 1996 and of 105 control patients with AD was performed.
Results
Fever and lymphopenia were associated with EH, whereas an increased erythrocyte sedimentation rate was frequently seen in patients with EH and control patients who were impetiginized. In 100 patients with EH, primary herpes simplex virus infection was likely in 20 patients, and a secondary herpes simplex virus infection was suggestive in 26 patients. In all, 13 patients had a second EH, whereas 3 patients had a third EH. Patients with EH had a significantly earlier onset of AD and a significantly higher total serum IgE level than the control patients. More than 75% of the patients with EH had not received corticosteroid treatment in the 4 weeks before onset of EH.
Conclusions
The characteristics of patients with EH are those associated with severe manifestations of AD. The majority of EH occurs in patients with untreated AD, arguing against a role for topical corticosteroids in the development of EH.
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Fig 1A, Early onset of atopic dermatitis (AD) is risk factor for eczema herpeticum (EH). Histogram plot of patients with EH (black, n = 100) and control patients with AD (striped, n = 105), classified for respective age of onset of AD in decades. Patients with EH have significantly earlier age of onset of AD (P< .05) as compared with control patients (Mann-Whitney U test). B, No seasonal dependent onset of EH. Histogram plot showing month of onset of EH (n = 100), demonstrating highest incidence in November but no apparent seasonal preference in incidence of EH.
Fig 2 Patients with eczema herpeticum (EH) and control patients show similar personal history of atopic disease. Concomitant atopic diseases in personal history of patients with EH (n = 75) and control patients with atopic dermatitis (n = 93). Patients with EH tend to have concomitant atopic diseases more frequently (not significant, chi-square test).
Fig 3 Primary and secondary herpes simplex virus (HSV) infection may cause eczema herpeticum (EH). Classification of patients with EH: primary or secondary HSV infection as a result of serologic tests, medical history, or both.
Fig 4A, Patients with eczema herpeticum (EH) have increased body temperature independent of additional heavy bacterial colonization. Mean body temperature (°C) on admission of patients with EH (black, n = 93) and control patients with atopic dermatitis (AD) (striped, n = 47), separately shown for patients with and without additional bacterial colonization. Patients with EH had significantly higher mean body temperature independent of additional bacterial colonization than control patients (P = .05, chi-square test). B, Patients with EH have increased erythrocyte sedimentation rate (ESR) independent of additional heavy bacterial colonization. ESR (mm) on admission of patients with EH (black, n = 88) and control patients with AD (striped, n = 72), separately shown for patients with and without additional bacterial colonization. Patients with EH without additional bacterial colonization had significantly higher ESR after 1 hour and after 2 hours than control patients without additional bacterial colonization (chi-square test). ESR in individuals with additional bacterial colonization tended to be higher compared with control group (not significant, chi-square test).
Fig 5A, High total serum IgE level is risk factor for eczema herpeticum (EH). Patients with EH (black, n = 45) and control patients with atopic dermatitis (AD) (striped, n = 50), classified for total serum IgE level <100 kU/L, 100 to 999 kU/L, 1000 to 9999 kU/L, or ≥ 10,000 kU/L. The patients with EH show significantly higher IgE level in Mann-Whitney U test (P< .05) as compared with control patients. B, Corticosteroid pretreatment is uncommon in patients with EH. Histogram display of evaluable number of patients with EH (black, n = 93) and control patients with AD (striped, n = 101) with no corticosteroids (none), topical or systemic corticosteroid treatment in 4 weeks before admission.
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