ECE 438 Exam 2 Solutions, 11/08/2006.
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1 NAME: ECE 438 Exam Solutions, /08/006. This is a closed-book exam, but you are allowed one standard (8.5-by-) sheet of notes. No calculators are allowed. Total number of points: 50. This exam counts for 0% of your final grade. You have 75 minutes to complete SEVEN problems. Be sure to fully and clearly explain all your answers. There will not be any discussion of grades. All re-grade requests must be submitted in writing, as stated in the course information handout. Score Grader Total score:
2 Formula Sheet Z-Transform. Definition.. Some transform pairs. x(n)z n n= x(n) = δ(n), ROC is the whole z-plane a n u(n),a 0 az,roc z > a a n u( n ),a 0 az,roc z < a 3. Some transform properties. a x (n) + a x (n) x(n n 0 ) x y(n) a X (z) + a X (z) z n 0 X(z) X(z)Y (z) If x(n) is a right-sided sequence, the ROC of X(z) is z > p where p is the outermost pole of X(z). A system is BIBO stable if and only if the unit circle is contained in the ROC of its transfer function. Random Variables and Random Processes If X is a continuous random variable, uniformly distributed between a and b, then f X (x) = E[X] = var(x) = { b a, a x b 0, otherwise b + a (b a) A random process X(n) is wide-sense stationary if E[X(n)] = E[X(0)] for any n and E[X(n)X(m)] = E[X(0)X(m n)] for any n and m.
3 Problem (5 points). Find the Z-transform of the sequence n u(n 00) where u(n) is the discrete-time unit step. Specify the region of convergence. Solution. The z-transform of n u(n) is z, with ROC z >. Shifting the signal by 00 results in the time domain signal n 00 u(n 00) whose Z-transform, because of the time-shift property, is z 00 z. Since n u(n 00) = 00 n 00 u(n 00), its Z-transform, by the linearity property, is 00 z 00 z. The region of convergence is the same as the original transform, z >. Problem (0 points). (a) Suppose X(n) is a discrete-time random process such that E[X(n)] = 3 for all integer n and E[X(n)X(m)] = e n m for all integer n and m. Is this process wide-sense stationary? (b) Suppose W(n) is a discrete-time random process such that W(0) is a continuous random variable, uniformly distributed between -0.5 and 0.5, and W(5) is a continuous random variable, uniformly distributed between - and. Could the random process W(n) be wide-sense stationary? (c) Suppose Y (n) is a random process such that Y (0) is a uniform random variable, uniformly distributed between -0.5 and 0.5, and Y (5) is a Gaussian random variable with mean 0 and variance /. Could the random process Y (n) be wide-sense stationary? Solution. (a) Yes, this process is WSS because both parts of the definition of the WSS are satisfied: first, the mean sequence is constant, and second, since E[X(n)X(m)] = e n m and E[X(0)X(m n)] = e n m, we have E[X(n)X(m)] = E[X(0)X(m n)] for any m and n. (b) If W is a WSS process, then we must have E[W(n)W(m)] = E[W(0)W(m n)] for all m and n. Taking m = n = 5, this yields E[W(5)W(5)] = E[W(0)W(0)] which is not true. Indeed, W(5) is a random variable which is uniform between - and, and therefore its variance E[W(5)W(5)] is equal to /3 (as given on the formula sheet), whereas W(0) is a random variable which is uniform between -0.5 and 0.5, and therefore its variance E[W(0)W(0)] is equal to /. Therefore, this process cannot be WSS. (c) Y (0) has mean 0 and variance /, and so does Y (5). This information is consistent with Y being a WSS sequence. We have no further information about the mean function or the autocorrelation function of Y. We therefore conclude that the process Y could be WSS. (But based on the provided information alone, we cannot say that it is definitely WSS or definitely not WSS.) 3
4 Problem 3 (0 points). Let W(n) be a sequence of independent zero-mean random variables with var(w(n)) = for all n. Let Y (n) = W(n)W(n ). Determine the mean sequence E[Y (n)] and the autocorrelation function E[Y (m)y (n)] of the process Y (n). Is this process wide-sense stationary? Solution. Since W(n) and W(n ) are independent, they are uncorrelated, and the expected value of their product is equal to the product of their expected values. Therefore, E[Y (n)] = E[W(n)W(n )] = E[W(n)]E[W(n )] = 0. The autocorrelation function is found using the same property: r Y Y (m,n) = E[Y (m)y (n)] = E[W(m )W(m)W(n )W(n)] When m < n and when m > n, we have: When m = n, we have: r Y Y (m,n) = E[W(m )]E[W(m)]E[W(n )]E[W(n)] = 0. r Y Y (m,n) = E[W(n )W (n )W(n)] = E[W(n )]E[W (n )]E[W(n)] = 0 When m = n +, we have: r Y Y (m,n) = E[W(n )W (n)w(n + )] = E[W(n )]E[W (n)]e[w(n + )] = 0 When m = n, we have: r Y Y (m,n) = E[W (n )W (n)] = E[W (n )]E[W (n)] =. Thus, r Y Y (m,n) = δ(n m). Since the mean sequence is constant and since E[Y (m)y (n)] = E[Y (0)Y (n m)], Y (n) is a wide-sense stationary process. 4
5 Problem 4 (0 points). Consider the z-transform 5z 5z + 6z Sketch the three different regions of convergence (ROC s) that are possible for this z-transform. For each ROC, find the corresponding inverse transform x(n). Solution. The poles of X(z) are the roots of the denominator: z = and z = 3, therefore where A z + B 3z A = B = X(z)( z ) z= = 5 3 = X(z)( 3z ) z=3 = = Therefore, z + 3z Since the poles are z = and z = 3, the three possible regions of convergence are: Region : z > 3 which is the outside of the circle of radius 3 centered at the origin of the z-plane. In this case, the ROC is outside of both poles which means both terms in X(z) correspond to right-sided sequences: x(n) = n u(n) + 3 n u(n). Region : < z < 3 which is the ring between the circles of radii and 3 centered at the origin of the z-plane. In this case, the ROC is outside of the pole at but inside the pole at 3 which means that the first term in X(z) corresponds to a right-sided sequence and the second term corresponds to a left-sided sequence: x(n) = n u(n) 3 n u( n ). Region 3: z < which is the inside the circle of radius centered at the origin of the z-plane. In this case, the ROC is inside of both poles which means both terms in X(z) correspond to left-sided sequences: x(n) = n u( n ) 3 n u( n ). 5
6 Problem 5 (5 points). When Al throws a dart, the probability density function of the distance between the center of the target and the point where the dart lands is uniformly distributed between 0 and foot. When Bob throws a dart, the probability density function of the distance between the center of the target and the point where the dart lands is uniformly distributed between 0 and feet. They each throw a dart once and the one whose dart is closer to the center of the target wins. Al s throw and Bob s throw are independent. What is the probability that Al wins? Fully justify your answer. Solution. Let the two random variables representing Al s and Bob s distance from the origin be A and B, respectively. Since they are independent random variables which are uniform(0,) and uniform(0,) respectively, their joint pdf is: { f A,B (a,b) =, 0 a and 0 b 0, otherwise The event that Al wins is the event that A < B, therefore, its probability is the integral of f A,B (a,b) over the region a > b, which is simply (/) times the area of the [0,] [0,] rectangle that intersects the half-plane a > b. This is 3/4. Problem 6 (5 points). A causal LTI system is described by the following input-output equation: 3 4 y(n) + Ay(n ) A y(n ) = x(n), where y(n) is the response to the input signal x(n), and where the coefficient A is a continuous realvalued random variable, uniformly distributed between 0 and. What is the probability that this system is BIBO stable? Fully justify your answer. Solution. Taking the Z-transform of both sides yields the following transfer function for this system: The poles are the roots of the denominator: H(z) = Y (z) (3/4) + Az A z z, = A ± A + 4(3/4)A A = A ± A (3/4), where we use the fact that A is nonnegative, and so A = A. Therefore, z = (/3)A and z = A. Since the system is causal, its impulse response is a right-sided sequence, which means that the ROC of the transfer function is outside of the outermost pole, i.e., the ROC is z > A. In order for the system to be BIBO stable, the ROC must include the unit circle: in other words, we must have A <, or A < /. Since A is uniform between 0 and, the probability of this event is /4. 6
7 Problem 7 (5 points). Determine the optimal quantization intervals [x 0,x ) and [x,x ] and corresponding quantization levels q and q for the -bit Max quantizer to quantize a random variable which has the following probability density function: { x, 0 x f(x) = 0, otherwise In other words, select the parameters x 0,x,x,q,q to minimize the mean-square error E[(X Y ) ] where X is the random variable to be quantized and Y is the result of quantization. Solution. Since the probability of all observations outside of the interval [0,] is zero, we can set x 0 = and x =. As shown in class by differentiating E[(X Y ) ] with respect to the parameters and setting the partial derivatives to zero, the following identities hold: q = q = x 0 x 0 xf(x)dx f(x)dx x xf(x)dx x f(x)dx x = q + q Using the expression for f(x) in the integrals above and simplifying, we get: q = 3 x q = 3 + x + x + x x = q + q Substituting q and q from the first two equations into the third one and simplifying, leads to the following quadratic equation for x : x + x = 0. This equation has two real roots only one of which is positive, x = ( + 5)/. Substituting this back into the other two equations leads to the following values for the quantization levels: q = ( + 5)/3 and q = ( + 5)/3. 7
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