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(cid:2)(cid:4)(cid:3) (cid:2)(cid:6)(cid:3) (cid:21)(cid:20) (cid:23)(cid:22) (4.129) (cid:23)(cid:22) (cid:23)(cid:20) (cid:8) , and similarly for the other functions. This quadrature where (cid:23) (cid:8) . scheme is exact for linear variations of the kernel (cid:21) In the present case (cid:24) (cid:17)(cid:20)(cid...
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:17)(cid:20)(cid:19) (cid:4)(cid:2) (cid:17)(cid:20)(cid:19) (cid:4)(cid:2) (4.131) (cid:17)(cid:9)(cid:19) Here it is interesting to note that Eq. (4.130) is identical to Eq. (4.108) except for the simple change in integration weight from (cid:23) , basically applying a sinc-function squared to the field amplitude vs r...
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) " $ (cid:13) (cid:2) (cid:21) (cid:3) (cid:8) (cid:28) (cid:5) (cid:1) (cid:9) (cid:11) (cid:21) (cid:3) (cid:8) (cid:28) (cid:5) (cid:0) (cid:9) (cid:16) (cid:6) (cid:23) (cid:2) (cid:10) $ (cid:6) (cid:21) (cid:10) (cid:21) (cid:3) (cid:8) (cid:25) (cid:21) (cid:3) (cid:10) (cid:9) (cid:8) (cid:10) (cid:6) (cid:10)...
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ities are usually the controlling factor, and since the number of singularities increase with frequency, and since singularities are not necessarily bet- ter represented by a linear than a constant kernel, the improvement in computational efficiency is much less pronounced for underwater acoustic problems, and the Filon...
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be rather insignificant. Further, for computation of transmission loss, usually performed on a dense spatial grid, the fact that the adaptive sampling has to be performed individually for each receiver makes it rather inten- sive computationally. However, for time domain computations for a small number of receivers, it ...
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Lecture 1 Overview of some probability distributions. In this lecture we will review several common distributions that will be used often throughtout the class. Each distribution is usually described by its probability function (p.f.) in the case of discrete distributions or probability density function (p.d.f.) i...
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is defined by ∞ R we have P(X = a) = 0. Given a function � : R, the X � E�(X) = � �(x)p(x)dx. � −� Notation. The fact that a random variable X has distribution P will be denoted by P. Normal (Gaussian) Distribution N(�, π2). Normal distribution is a continuous dis­ X � tribution on R with p.d.f. p(x) = 1 ≤...
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y 1 ≤2ϕ 2 y 2 dy = 0 e− since the integrand is an odd function. To compute the second moment EY 2 , let us first note 1 is a probability density function, it integrates to 1, i.e. that since � 2 y 2 2� e− If we integrate this by parts, we get, � −� � 1 = 1 ≤2ϕ 2 y dy. 2 e− 1 = � 2 y 2 dy = e− 1 ≤2ϕ...
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�i, π2) then their sum will also have a normal distribution Xi, 1 i n, such that Xi � ≈ ≈ X1 + . . . + Xn � N(�1 + . . . + �n, π1 2 + . . . + π2 ).n Normal distribution appears in one of the most important results that one learns in probabil­ ity class, namely, a Central Limit Theorem (CLT), which states the foll...
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= X p(1) = P(X = 1) = p, p(0) = P(X = 0) = 1 − p for some p [0, 1]. ∞ It is easy to check that EX = p, Var(X) = p(1 − p). Binomial Distribution B(n, p). This distribution describes a random variable X that is a number of successes in n trials with probability of success p. In other words, X is a sum of n in...
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that X will exceed level t + s given that it will exceed level t can be computed as follows: P(X t + s X | ∼ ∼ t) = P(X t + s, X ∼ P(X t) ∼ �t = e− �(t+s)/e− ∼ = e− t) = P(X t + s) t) ∼ P(X �s = P(X ∼ s), ∼ i.e. P(X X t + s | t) = P(X s). ∼ ∼ If X represent a lifetime of some object in s...
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phone number; distribution of bacteria on some surface or weed in the field. All these examples share some common properties that give rise to a Poisson distribution. Suppose that we count a number of random objects in a certain region T and this counting process has the following properties: 4 1. Average nu...
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0 + P(Xi = 1) + αn, 0 k � � k 2 kP(Xi = k), and by the last property above we assume that αn becomes where αn = small with n, since the probability to observe more that two objects on the interval of size T /n becomes small as n becomes large. Combining two equations above gives, P(Xi = 1) � T . n 2 is small, ...
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associated with this distribution. For example, ’normrnd’ generates random numbers from distribution ’norm’, ’normpdf’ gives p.d.f., ’norm­ cdf’ gives c.d.f., ’normfit’ fits the normal distribution for a given dataset (we will look at this last type of functions when we discuss Maximum Likelihood Estimators). Please, l...
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MIT 6.972 Algebraic techniques and semidefinite optimization February 28, 2006 Lecturer: Pablo A. Parrilo Scribe: ??? Lecture 6 Last week we learned about explicit conditions to determine the number of real roots of a univariate polynomial. Today we will expand on these themes, and study two mathematical objects o...
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. ⎢ ⎢ ⎢ p(x0)x0 ⎢ ⎢ p(x0) ⎢ ⎢ q(x0)x0 ⎢ ⎢ q(x0)x0 ⎢ ⎢ . . . ⎢ ⎢ ⎢ ⎣ q(x0)x0 q(x0) n−2 n−1 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = 0. This implies that the matrix on the left­hand side, called the Sylvester matrix Sylx(p, q) associated to p and q, is singular and thus its determinant must vanish. It is not too...
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q(αj ) = p det q(Cp) n j=1 k=1 j=1 = (−1)nm q n m m � p(βk ) = (−1)nm q det p(Cq ) n m k=1 (1) • Kronecker products: Using a well­known connection to Kronecker products, we can also write (1) as • B´ezout matrix To be completed m n pn q det(Cp ⊗ Im − I m n ⊗ Cq ). ToDo If can be shown that all these const...
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easily checkable condition for the simultaneous vanishing of two univariate polynomials. Can we use the resultant to produce a condition for a polynomial to have a double root? Recall that if a polynomial p(x) as a double root at x0 (which can be real or complex), then its derivative p�(x) also vanishes at x0. Thus,...
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3.1 Polynomial equations One of the most natural applications of resultants is in the solution of polynomial equations in two variables. For this, consider a polynomial system p(x, y) = 0, q(x, y) = 0, (2) with only a finite number of solutions (which is generically the case). Consider a fixed value of y0, and the ...
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the corresponding value of x� ≈ −1.3853. 3.2 Implicitization of rational curves To be completed 3.3 Random matrices To be completed ToDo ToDo 4 The set of nonnegative polynomials One of the main reasons why nonnegativity conditions about polynomials are difficult is because these sets can have a quite complicate...
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but it also vanishes at points inside the set. Why is this? Example 5. Consider the univariate polynomial p(x) = x4 + 2ax2 + b. For what values of a, b does it hold that p(x) ≥ 0 ∀x ∈ R? Since the leading term x4 has even degree and is strictly positive, p(x) is strictly positive if and only if it has no real roots...
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511.52b0042-2-1.5-1-0.50.511.52a-1-0.50.511.52b-3-2-1123a-1123b0220444-3-2-1123a-1123b Figure 3: A three­dimensional convex set, described by one quadratic and one linear inequality, whose projection on the (a, b) plane is equal to the set in Figure 1. One has to do with its algebraic structure, and the other one wit...
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5 and Figure 1. The presence of “extraneous” components of the discriminant inside the feasible set is an important roadblock for the availability of “easily computable” barrier functions. Indeed, every polynomial that vanishes on the boundary of the set Pn must necessarily have the discriminant as a factor. This is...
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F ⊆ S, with the property that x, y ∈ S, 1 F is exposed if it can be written as F = S ∩ H, where H is a supporting hyperplane of S. 2 (x + y) ∈ F ⇒ x, y ∈ F . A face 6­5 −20246024012345abt Figure 4: The discriminant of the polynomial x4 + 4ax3 + 6bx2 + 4cx + 1. The convex set inside the “bowl” corresponds to the re...
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In Proceedings of the American Control Conference, 2002. [RG95] M. Ramana and A. J. Goldman. Some geometric results in semidefinite programming. J. Global Optim., 7(1):33–50, 1995. [Stu98] B. Sturmfels. Introduction to resultants. In Applications of computational algebraic geometry (San Diego, CA, 1997), volume 53 ...
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MIT OpenCourseWare http://ocw.mit.edu 18.917 Topics in Algebraic Topology: The Sullivan Conjecture Fall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Steenrod Operations (Lecture 2) The objective of today’s lecture is to introduce the Steenrod operations a...
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-module spectrum is the cochain complex C ∗(X; F2) of a topological space X. The cohomology groups of this F2-module spectrum are simply the cohomology groups of X. The cohomology H∗(X; F2) has the structure of a graded commutative ring. The multiplication on H∗(X; F2) arises from a multiplication which exists on t...
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we will denote by Dn(V ). Remark 3. In concrete terms, Dn(V ) may be computed in the following way. Let M denote the vector space F2, with the trivial action of Σn. Choose a resolution . . . → P −1 → → P 0 M by free F2[Σn]-modules. We let EΣn denote the complex P •. (We can think of EΣn as a contractible comple...
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the consequences of the existence of a symmetric multiplication on a complex V . Notation 8. Let n be an integer. We let F2[−n] denote the complex which consists of a 1-dimensional vector space in cohomological degree n, and zero elsewhere. Let en denote a generator for the F2-vector space Hn F2[−n], so we have iso...
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homology H (RP ∞; F2): this is just a one-dimensional vector space in each degree m, with a unique generator which we will denote by xm. ∗ Definition 9. Let V be a complex, and let v ∈ Hn V , so that v determines a homotopy class of maps For i ≤ n, we let denote the image of under the induced map η : F2[−n] → V. ...
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9 yields operations Sqi : Hn(X; F2) → Hn+i(X; F2). These are the usual Steenrod operations. i v completely account for the cohomology groups of any extended Remark 12. The operations v �→ Sq square D2(V ). More precisely, let us suppose that V is an F2-module spectrum, and that {vi}i∈I is an ordered basis for π∗V...
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to prove. If k = n, then k Sq k (v + v�) = (v + v�)2 = Sq k (v) + Sq (v�) + (vv� + v�v). Since the multiplication map V ⊗ V → D2(V ) is commutative on the level of homotopy, we have vv� + v�v = 2vv� = 0. Now suppose that k < n. By functoriality, it will suffice to treat the universal case where V � F[−n] ⊕ F[−n]. Us...
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8.022 (E&M) – Lecture 9 Topics: „ RC circuits Thevenin’s theorem „ Last time „ Electromotive force: „ How does a battery work and its internal resistance „ How to solve simple circuits: „ „ Kirchhoff’s first rule: at any node, sum of the currents in = sum of the currents out (conservation of charge at node...
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at any moment in time tage drop on the resistor: -IR „ Vol Æ Q(t) Æ V(t) Q C − IR 0 = „ Not useful in this form since I=I(Q) „ I=-dQ/dt (- sign because C is losing charge) Q dQ + dt C 0R = „ Easy integral yields to exponential decay of the charge: − ( ) Q t Q e = 0 t RC G. Sciolla – MIT 8.022 –...
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t RC d dt − e ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Q 0 RC t RC − e „ Same exponential decay as for Q(t) G. Sciolla – MIT 8.022 – Lecture 9 7 Charging capacitors „ Now 3 elements in circuit: EMF, capacitor and resistor „ Capacitor starts uncharged C C - - - - - - - + + + + + + + s s I V + - R R „ What happens wh...
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( − RC ) ' dQ Q ' Integrating between t=0 and t: ⇒ = − dt RC Q Q t = ( ) ∫ Q = 0 t dQ ' ∫ = − Q ' t t = = 0 dt RC ⇒ ln Q t CV ( ) - CV - = − t RC ⇒ Q CV ( ) - t V C − t RC = e − Q t ( ) = C t RC − e V ⎛ 1 −⎜ ⎝ ⎞ ⎟ ⎠ G. Sciolla – MIT 8.022 – Lecture 9 10 5 ...
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I ( ) t = t R C − e V R „ Are Kirchhoff’s laws valid at any moment in time? V − Q C − IR V = − V t R C − e ⎛ 1 ⎜ ⎝ − ⎞ ⎟ ⎠ − R t R C − e V R = 0 O K ! „ Asymptotic behavior of the capacitor: „ At t=0: I=V/R as if C were a short circuit „ At t=infinity , I=0 as if C were an open circu it...
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9 13 Verify time constant (E8) „ „ RC circuit with „ VEMF = squared 5 V pulses Variable C initially = 0.3 µF Variable R initially = 400 Ω „ „ 2 „ R1 = 100 Ω Display on scope V C and I(R1) Verify that time constant is RC „ VC(t) 5V 1-e-t/RC IAG(t) 10mA -t/RC e R2 V EMF C G R1 A G. Sciolla – ...
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8.022 – Lecture 9 16 8 Thevenin equivalence Thevenin’s theorem: Any combination of resistors and EMFs with 2 terminals can be replaced with a series of a battery V and a resistor R OC is the open circuit voltage T where „ VOC „ RT=VOC/I or RT=Req wi th all the EMF shorted short where I short is the curren...
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„ „ There is on y one current go ng through the reduced c rcuit At t=0, C behaves like a short At t=0 I short=VOC/RT Æ i l with I i Æ RT=VOC/I short G. Sciolla – MIT 8.022 – Lecture 9 18 9 Solve the actual problem Calculate VOC and RT=VOC/I short for our problem: R1 V + - R2 + - C VOC ≡ +...
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: VOC≡ + - RT Any unknown combination of Rs and EMFs Careful: „ Thevenin works only when the elements in the box follow Ohm’s law, i.e. on between V and I linear relati G. Sciolla – MIT 8.022 – Lecture 9 20 10 Oscillating circuit (E13) „ RC circuit with: „ VEMF = 1 kV „ C = 0.1 µF „ R = 2.5 M...
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022 – Lecture 9 22 11 Norton’s theorem Any combination of resistors and EMFs with 2 terminals can be replaced with a parallel of a current generator IN „ RT and a resistor R is the equivalent resistance of the circuit w th all the EMF shorted and all the T where i current sources open (same as Thevenin!) IN ...
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Lecture 8 Primitive Roots (Prime Powers), Index Calculus Recap - if prime p, then there’s a primitive root g mod p and it’s order mod p is p − 1 = qe1 e2 1 q2 . . . qr . We showed that there are integers gi mod p with order i − 1 ≡ 0 mod p). Set g = (cid:81) gi - qei (counting number of solutions to xq i exactly i (cid...
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Then there are infinitely many primes p for which a is a primite root mod p. This is an open question. Hooley proved this conditional on GRH, and Heath- Brown showed that if a is a prime, then there are at most 2 values of a which fail the conjecture (Definition) Discrete Log: Say p is a prime, and g is a primitive root ...
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0 mod p), we can write a ≡ gl mod p so if x ≡ gk as before then gkd ≡ gl mod p. This means that gkd−l ≡ 1 mod p ↔ p − 1|kd − l ↔ kd ≡ l mod p − 1 (k is variable), which has a solution iff (d, p − 1) divides l, in which case it has exactly (d, p − 1) solutions. Note: (d, p − 1) divides l ←→ p − 1 divides l(p − 1) (d, p ...
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. 2 Since there are p possible values of t(0 ≤ t ≤ p − 1), any of these remaining ones give a g + tp which is a primitive root mod p2. Consider f (x) = xp−1 1: mod − and f (cid:48)(g) = (p − 1)gp−2 (cid:54)≡ 0 mod p, p it has the root g. Since f (cid:48)(x) = (p − 1)xp 2 by Hensel’s Lemma there is a unique lift g + tp ...
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−1 ≡ 1 mod pe−1 assuming that ) = 1 + bpe−1 with p (cid:45) b. Need to We know that gpe−2(p−1) = 1 + bpe−1. Raising to power p we get gpe−1(p−1) = (1 + bpe−1)p = 1 + pbpe−1 + (bpe−1 )2 + (cid:18) (cid:19) p 3 (bpe−1)3 + . . . (cid:18) (cid:19) p 2 pe ≡ 1 + bpe mod +1 (cid:0) (cid:1) (cid:0) (cid:1) 2 so p b2p2e (becaus...
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1 mod m and aφ(m(cid:48)) ≡ 1 mod m(cid:48). So aφ(m)φ(m(cid:48))/2 ≡ (aφ(m))φ(m(cid:48))/2 ≡ 1 mod m aφ(m)φ(m(cid:48))/2 ≡ 1 mod m(cid:48) Similarly so, aφ(m)φ(m(cid:48))/2 ≡ 1 mod n but φ(n) = φ(m)φ(m(cid:48)) so ordn(a) < φ(n). So a can’t be a primitive root mod n. Only remaining candidate is n = 2k for k ≥ 3. No pr...
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MIT OpenCourseWare http://ocw.mit.edu 6.189 Multicore Programming Primer, January (IAP) 2007 Please use the following citation format: Saman Amarasinghe, 6.189 Multicore Programming Primer, January (IAP) 2007. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). ...
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ism ● Interleaved Concurrency „ Logically simultaneous processing B „ Interleaved execution on a single C processor A ● Parallelism „ Physically simultaneous processing „ Requires a multiprocessors or a multicore system A B C Time Time Prof. Saman Amarasinghe, MIT. 4 6.189 IAP 2007 MIT Account and Ban...
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String pass = in.readLine(); if (!acc.is_password(pass)) throw new Exception(); out.print(“yourbalance is“ + acc.getbal()); out.print("Depositor withdraw amount > “ ); int val = in.read(); if (acc.getbal() + val > 0) acc.post(val); else throw new Exception(); out.print(“yourbalance is“ + acc.getbal()); } cat...
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{ out.println("Invalidinput, restart“ ); I need to run multiple ATM machines from my program, how do I do that? 8 Prof. Saman Amarasinghe, MIT. 6.189 IAP 2007 MIT } } } } Concurrency in Java ● Java has a predefined class java.lang.Thread which provides the mechanism by which threads are created public class ...
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(); out.print("Password>“ ); String pass = in.readLine(); if (!acc.is_password(pass)) throw new Exception(); out.print(“yourbalance is“ + acc.getbal()); out.print("Depositor withdraw amount > “ ); int val = in.read(); if (acc.getbal() + val > 0) acc.post(val); else throw new Exception(); out.print(“yourbala...
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} catch(Exception e) { out.println("Invalidinput, restart“ ); } } I need to run multiple ATM machines from my program, how do I do that? 12 Prof. Saman Amarasinghe, MIT. 6.189 IAP 2007 MIT Activity trace ATM 1 Account ID > allyssa Password > MITROCKS ATM 2 Account ID > ben Password > i T m e Your account balanc...
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acc.getbal()); Your account balance is 10 out.print(“yourbalance is“ + acc.getbal()); Your account balance is 10 Prof. Saman Amarasinghe, MIT. 15 6.189 IAP 2007 MIT Activity trace II balanc ATM 1 e 100 void post(int v) { 100 100 10 10 v } ATM 2 void post(int v) { balance = balance + v -90 balance = + v ...
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6.189 IAP 2007 MIT Why You Need Locks thread A thread B if (no milk && no note) if (no milk && no note) leave note buy milk remove note leave note buy milk remove note Milk Vitamin D Milk Vitamin D Image by MIT OpenCourseWare. Image by MIT OpenCourseWare. ● Does this work?too much milk Prof. Saman Amarasinghe,...
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code can be labeled as synchronized ● The synchronized keyword takes as a parameter an object whose lock the system needs to obtain before it can continue ● Example: synchronized (acc) { if (acc.getbal() + val > 0) acc.post(val); else throw new Exception(); out.print(“yourbalance is “ + acc.getbal()); } Prof...
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IAP 2007 MIT Activity trace II balance ATM 1 100 100 100 100 10 10 10 out.print(“yourbalance is“ + acc.getbal()); Your account balance is 100 out.print("Depositor withdraw amount >“); Deposit or Withdraw amount > -90 int val = in.read(); synchronized(acc) if (acc.getbal() + val > 0) acc.post(val); o...
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bnk.get(id); if (acc == null) throw new Exception(); out.print("Password>“ ); String pass = in.readLine(); if (!acc.is_password(pass)) throw new Exception(); synchronized (acc) { out.print(“yourbalance is “ + acc.getbal()); out.print("Depositor withdraw amount >“ ); int val = in.read(); if (acc.getbal() + val > ...
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() > val) from.post(-val); synchronized(to) Waiting for Allyssa’s account to be released to perform DEADLOCKED! Prof. Saman Amarasinghe, MIT. 29 6.189 IAP 2007 MIT Avoiding Deadlock ● Cycle in locking graph = deadlock ● Standard solution: canonical order for locks „ Acquire in increasing order „ Release in...
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2007 MIT Races Race conditions – insidious bugs „ Non-deterministic, timing dependent „ Cause data corruption, crashes „ Difficult to detect, reproduce, eliminate ● Many programs contain races „ Inadvertent programming errors „ Failure to observe locking discipline Prof. Saman Amarasinghe, MIT. 33 6.189 IAP...
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philosopher thinks for a while. „ When the philosopher becomes hungry, she stops thinking and… – Picks up left and right chopstick – He cannot eat until he has both chopsticks, has to wait until both chopsticks are available – When the philosopher gets the two chopsticks she eats „ When the philosopher is done ...
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; public void run() { try { while(true) { 1 2 3 4 synchronized(table) { synchronized(left) { synchronized(right) { System.out.println(times + ": Philosopher " + position + " is done eating"); } } } } } catch (Exception e) { System.out.println("Philosopher " + position + "'s meal got disturbed"); } } ...
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. 44 6.189 IAP 2007 MIT Conclusion ● Concurrency and Parallelism are important concepts in Computer Science ● Concurrency can simplify programming „ However it can be very hard to understand and debug concurrent programs ● Parallelism is critical for high performance „ From Supercomputers in national labs to ...
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3.044 MATERIALS PROCESSING LECTURE 9 Example 1: Casting into low conductivity molds Greatly simplified if: 1. Mold is thick → can neglect air → semi-infinite geometry → erf 2. Mold controls heat loss → Tmold increases → kmold decreases → 3. Superheating is negligible → liquid poured at Tm gradients are in mold D...
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) k mρmcp,m t (cid:3) 1 2 s = 2(T0 − Tm) ρsHf Boundary Condition: @t = 0, s = 0 Boundary Condition: @s = L, t = tf 3.044 MATERIALS PROCESSING 3 tf ∝ L2 V A (cid:2) L ≈ tf ∝ V A (cid:3) 2 ⇒ Chvorinov’s Rule Example 2: Thin castings / Cold molds / Highly conductive molds out (cid:9) (cid:6) (cid:7)(cid:8) −h(Tm − Tmold ...
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Machine Learning for Healthcare HST.956, 6.S897 Lecture 5: Risk stratification (continued) David Sontag 1 Outline for today’s class 1. Risk stratification (continued) – Deriving labels – Evaluation – Subtleties with ML-based risk stratification 2. Survival modeling 2 Where ...
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AUC = Probability that algorithm ranks a positive patient over a negative patient Invariant to amount of class imbalance Diabetes 1-year gap False positive rate 8 3-.-4?-1H9F-17/91).N717./-145/4.).01?- !"#$% #&'(&")"*(&"+, 0507LLD)89.05-5) 9:)c05/...
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-fair-use/ 12 Non-stationarity: Top 100 lab measurements over time s b a L Time (in months, from 1/2005 up to 1/2014) → Significance of features may change over time © Narges Razavian. All rights reserved. This content is excluded from our Creative Commons lic...
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Caruana et al., Intelligible Models for Healthcare: Predicting Pneumonia Risk and Hospital 30- day Readmission. KDD 2015.] 16 Intervention-tainted outcomes • Formally, this is what’s happening: ED triage Tr...
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No big wins from deep models on structured data/text Supplemental Table 1: Prediction accuracy of each task of deep learning model compared to baselines Inpatient Mortality, AUROC1 (95% CI) Deep learning 24 hours after admission Full feature enhanced baseline at 24 hours after admission Full feature simple baseline ...
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.86) 0.83 (0.83-0.84) 0.81 (0.80-0.82) 0.74 (0.73-0.75) Courtesy of Rajkomar et al. Used under CC BY. 21 No big wins from deep models on structured data/text Comparison to Razavian et al. ‘15 Supplemental Table 1: Prediction accuracy of each task of deep learning model compared to baselines In...
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7 days AUROC (95% CI) Deep learning 24 hours after admission Full feature enhanced baseline at 24 hours after admission Full feature simple baseline at 24 hours after admission Baseline (mLiu4) at 24 hours after admission 0.86((0.86 -0.87 0.85 (0.84-0.85) 0.83 (0.82-0.84 ( ((0.75 -0.77 0.76 -0.86 (0.85 0.85( 0.83 ( (0....
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reserved. This content is excluded from our Creative Commons license. For more information, see https://ocw.mit.edu/help/faq-fair-use/ 25 Survival modeling • Why not use classification, as before? – Less data for training (due to exclusions) – Pessimistic estimates due to choice of...
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12 IEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS, VOL. 3, NO. 1, JANUARY 2004 A Laguerre Polynomial-Based Bound on the Symbol Error Probability for Adaptive Antennas With Optimum Combining Marco Chiani, Senior Member, IEEE, Moe Z. Win, Fellow, IEEE, Alberto Zanella, Member, IEEE, and Jack H. Winters, Fellow, IEEE ...
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ing (PSK) have been derived for the case of a single nonfading interferer with Rayleigh fading of the desired signal in [1] and [2] and with Rayleigh fading of the desired signal and a single interferer in [3]. In [4], two dif­ ferent methods (direct and moment generation function-based ap­ proaches), requiring a si...
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Digital Object Identifier 10.1109/TWC.2003.821165 multiple interferers of arbitrary powers, closed-form expressions of the BEP for PSK with OC are not available in the literature: Thus, Monte Carlo simulation has been used to determine the BEP in [2]. Unfortunately, such simulations are computation intensive and n...
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ibel), and therefore, in general, significantly better than previous bounds for the case of multiple equal-power interferers. In Section II, we describe the system model. Performance and upper bounds are derived in Section III, and in Section IV we compare our analytical bounds with Monte Carlo simulations. II. S...
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can be written as where denoted by values with as those of the are the eigenvalues of a complex Wishart matrix,2 eigen­ can be thought of [9]. Hence, the first of matrix (1) (6) and and and are the mean (over fading) energies of the where desired and interfering signals, respectively; are the desired...
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ers and thermal noise in a fading environment is now ob­ tained by averaging the conditional SEP over the (desired and , interfering signals) channel ensemble as is the SEP conditioned on the random variable where . This can be accomplished by using the chain rule of condi­ tional expectation as (7) where we fir...
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where together with the fact that conditional SEP antennas and , conditioned on interferers, becomes . Using (8) and is Gaussian with i.i.d. elements, the , in the general case of Then, by remembering that , and the fact that ’s are real and nonnegative, the following inequality and holds: Therefore, by ...
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, eq. (39)], we see that (14) is the -ary exact expression of the SEP for coherent detection of -branch MRC in the absence of interference. Note PSK using that in (13) is independent of interference, and de­ pends only upon the SNR and the number of antenna elements. Other factors in (13) are independent of SNR,...
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20 22 24 26 28 30 SNR [dB] Fig. 2. Comparison of the bounds on SEP for coherent detection of binary PSK using OC with four antennas and SIR � �� dB in the presence of one and four interferers. Also shown in the figure are the results obtained by Monte Carlo simulations. is a monic polynomial of degree with nonn...
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known bound, e.g., the difference is more than 4 dB at SEP . Fig. 3 shows the SEP as a function of SIR for several values of SNR, for , and binary phase-shift keying (BPSK) modulation. The comparison with simulation results shows that, for a reasonable range of SIR values, the proposed upper bound is tight for ...
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to three, an 8-PSK modulation scheme is considered with dB. Performance with MRC is evaluated by using SIR is given , where by (14). Also shown in the figure are Monte Carlo simulation results for OC. The results show that, as expected, for small SNR, the thermal noise is dominant and, therefore, MRC and OC per...
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SIR)-plane has been obtained for . Note that the 8 PSK. The curves are for two asymptotes (vertical and horizontal) give the values of SIR and SNR without thermal noise and interference, respectively. The region below each curve represents the outage domain re­ gion in which all points produce an SEP higher than t...
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,” Telecommun. Radio Eng., vol. 34/35, pp. 83–85, Oct. 1980. [2] J. H. Winters, “Optimum combining in digital mobile radio with cochannel interference,” IEEE J. Select. Areas Commun., vol. SAC-2, pp. 528–539, July 1984. [3] V. A. Aalo and J. Zhang, “Performance of antenna array systems with optimum combining in a...
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multiple antennas,” IEEE Wireless Pers. Commun., vol. 6, pp. 311–335, Mar. 1998. [9] M. Chiani, M. Z. Win, A. Zanella, and J. H. Winters, “Exact symbol error probability for optimum combining in the presence of multiple co-channel interferers and thermal noise,” in Proc. IEEE Global Telecommunications Conf., vol. ...
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M. Z. Win and J. H. Winters, “Virtual branch analysis of symbol error probability for hybrid selection/maximal-ratio combining in Rayleigh fading,” IEEE Trans. Commun., vol. 49, pp. 1926–1934, Nov. 2001. [16] M. K. Simon, S. M. Hinedi, and W. C. Lindsey, Digital Communication Techniques: Signal Design and Detection...
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6.720J/3.43J Integrated Microelectronic Devices - Spring 2007 Lecture 9-1 Lecture 9 - Carrier Flow (cont.) February 23, 2007 Contents: 1. Shockley’s Equations 2. Simplifications of Shockley equations to 1D quasi-neutral situations 3. Majority-carrier type situations Reading assignment: del Alamo, Ch. 5, §§5.3-5...
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2)ve drift + qDe∇n (cid:2) −) + − NA Hole current equation: (cid:2) Jh = qp (cid:2)vh drift − qDh∇p (cid:2) Electron continuity equation: ∇ · J(cid:2) (cid:2) ∂t = Gext − U (n, p) + 1 ∂n e q Hole continuity equation: ∇ · J(cid:2) (cid:2) ∂p ∂t = Gext − U (n, p) − 1 h q Total current equation: t = J(...
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of Technology. Downloaded on [DD Month YYYY]. 6.720J/3.43J Integrated Microelectronic Devices - Spring 2007 Lecture 9-5 Shockley’s equations in 1D: Gauss’ law: E = q ∂ ∂x (cid:2)(p − n + ND − NA) Electron current equation: Je = −qnvdrift (E) + qDe e ∂n ∂x Hole current equation: drift (E) − qDh Jh = ...
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situation (field independent of n, p) Majority-carrier type situation (V=0, n'=p'=0) Minority-carrier type situation (V=0, n'=p'=0, LLI) Cite as: Jesús del Alamo, course materials for 6.720J Integrated Microelectronic Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of T...
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N D −−N NA + D −− NA | (cid:6) 1 which implies n o − (cid:4) p o + −N D N − A • Additionally, quasi-neutrality outside equilibrium: | p(cid:5) − n(cid:5) n (cid:5) | (cid:4) | p(cid:5) − n(cid:5) p (cid:5) | (cid:6) 1 which implies: p (cid:5) (cid:4) n (cid:5) • QN approximation good if n, p high ⇒ carriers ...
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://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.720J/3.43J Integrated Microelectronic Devices - Spring 2007 Lecture 9-10 2. Subtract one continuity equation from the other: ∂Jt ∂x = q ∂(n − p) ∂t = − ∂ρ ∂t continuity equation for net volume charge: if Jt changes with...
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Jt (cid:4) 0 ∂x Jt = Je + Jh Cite as: Jesús del Alamo, course materials for 6.720J Integrated Microelectronic Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.720J/3.43J Integrated Microelectronic Devices - Spring 2007 Lecture 9...
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carrier current (n-type): Must distinguish between internal field in TE (Eo) and total field outside equilibrium (E). For simplicity, do in low-field limit (exact case done in notes). In equilibrium: Jeo = qμenoEo + qDe dno dx = 0 Out of equilibrium: Hence: dno Je (cid:4) qμenoE + qDe dx Je = qμeno(E − Eo) = qμenoE...
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� Example 1: Integrated Resistor with uniform doping (n-type) z y L W n p n+ n+ top view x x n+ t n n+ cross-sectional view p Uniform doping ⇒ Eo = 0, then: Jt = −qNDve drift (E) • If E not too high, I-V characteristics: Jt (cid:4) qNDμeE V I = W tqNDμe L Cite as: Jesús del Alamo, course materials ...
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8)= Jt(x) • Majority carrier-type situations characterized by application of external voltage without perturbing carrier concentrations. • Majority-carrier type situations dominated by drift of majority carriers. • Integrated resistor: – for low voltages, current proportional to voltage across – for high voltage...
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Introduction to Algorithms: 6.006 Massachusetts Institute of Technology Instructors: Erik Demaine, Jason Ku, and Justin Solomon Lecture 1: Introduction Lecture 1: Introduction The goal of this class is to teach you to solve computation problems, and to communicate that your solutions are correct and efficient. Pr...
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