ECE302 Spring 2006 HW12 Solutions April 27, 2006 1 Solutions to HW12 Note: These solutions are D. J. Goodman, the authors of our textbook. I have annotated and corrected them as necessary. Text in italics is mine. Problem 10.10.2 Let A be a nonnegative• random variable that is independent of any collection of samples X(t1),...,X(tk) of a wide sense stationary random process X(t). Is Y (t) = A + X(t) a wide sense stationary process? Problem 10.10.2 Solution To show that Y (t) is wide-sense stationary we must show that it meets the two requirements of Definition 10.15, namely that its expected value and autocorelation function must be independent of t. Since Y (t)= A + X(t), the mean of Y (t) is E [Y (t)] = E [A]+ E [X(t)] = E [A]+ µX (1) The autocorrelation of Y (t) is RY (t,τ)= E [(A + X(t)) (A + X(t + τ))] (2) = E A2 + E [A] E [X(t)]+ AE [X(t + τ)] + E [X(t)X(t + τ)] (3) 2 = E A + 2E [A] µX + RX (τ), (4) where the last equality is justified by the fact that we are given that X(t) is wide sense stationary. We see that neither E[Y (t)] nor RY (t,τ) depend on t. Thus Y (t) is a wide sense stationary process. Problem 10.11.1 X(t) and Y (t) are independent• wide sense stationary processes with expected values µX and µY and autocorrelation functions RX (τ) and RY (τ) respectively. Let W (t)= X(t)Y (t). (a) Find µW and RW (t,τ) and show that W (t) is wide sense stationary. (b) Are W (t) and X(t) jointly wide sense stationary? Problem 10.11.1 Solution (a) Since X(t) and Y (t) are independent processes, E [W (t)] = E [X(t)Y (t)] = E [X(t)] E [Y (t)] = µX µY . (1) In addition, RW (t,τ)= E [W (t)W (t + τ)] (2) = E [X(t)Y (t)X(t + τ)Y (t + τ)] (3) = E [X(t)X(t + τ)] E [Y (t)Y (t + τ)] (4) = RX (τ)RY (τ) (5) We can conclude that W (t) is wide sense stationary. ECE302 Spring 2006 HW12 Solutions April 27, 2006 2 (b) To examine whether X(t) and W (t) are jointly wide sense stationary, we calculate RW X (t,τ)= E [W (t)X(t + τ)] = E [X(t)Y (t)X(t + τ)] . (6) By independence of X(t) and Y (t), RW X(t,τ)= E [X(t)X(t + τ)] E [Y (t)] = µY RX (τ). (7) Since W (t) and X(t) are both wide sense stationary and since RW X (t,τ) depends only on the time difference τ, we can conclude from Definition 10.18 that W (t) and X(t) are jointly wide sense stationary. Problem 10.11.2 X(t) is a wide sense stationary random process. For each process Xi(t) defined below, determine whether Xi(t) and X(t) are jointly wide sense stationary. (a) X1(t)= X(t + a) (b) X2(t)= X(at) Problem 10.11.2 Solution To show that X(t) and Xi(t) are jointly wide sense stationary, we must first show that Xi(t) is wide sense stationary and then we must show that the cross correlation RXXi (t,τ) is only a function of the time difference τ. For each Xi(t), we have to check whether these facts are implied by the fact that X(t) is wide sense stationary. (a) Since E[X1(t)] = E[X(t + a)] = µX and RX1 (t,τ)= E [X1(t)X1(t + τ)] (1) = E [X(t + a)X(t + τ + a)] (2) = RX (τ), (3) we have verified that X1(t) is wide sense stationary. Now we calculate the cross correlation RXX1 (t,τ)= E [X(t)X1(t + τ)] (4) = E [X(t)X(t + τ + a)] (5) = RX (τ + a). (6) Since RXX1 (t,τ) depends on the time difference τ but not on the absolute time t, we conclude that X(t) and X1(t) are jointly wide sense stationary. (b) Since E[X2(t)] = E[X(at)] = µX and RX2 (t,τ)= E [X2(t)X2(t + τ)] (7) = E [X(at)X(a(t + τ))] (8) = E [X(at)X(at + aτ)] = RX (aτ), (9) ECE302 Spring 2006 HW12 Solutions April 27, 2006 3 we have verified that X2(t) is wide sense stationary. Now we calculate the cross correlation RXX2 (t,τ)= E [X(t)X2(t + τ)] (10) = E [X(t)X(a(t + τ))] (11) = R ((a 1)t + τ). (12) X − Except for the trivial case when a = 1 and X2(t)= X(t), RXX2 (t,τ) depends on both the absolute time t and the time difference τ, we conclude that X(t) and X2(t) are not jointly wide sense stationary. Problem 10.11.3 X(t) is a wide sense stationary stochastic process with autocorrelation function RX (τ) = 10 sin(2π1000τ)/(2π1000τ). The process Y (t) is a version of X(t) delayed by 50 microsec- −5 onds: Y (t)= X(t t0) where t0 = 5 10 s. − × (a) Derive the autocorrelation function of Y (t). (b) Derive the cross-correlation function of X(t) and Y (t). (c) Is Y (t) wide sense stationary? (d) Are X(t) and Y (t) jointly wide sense stationary? Problem 10.11.3 Solution (a) Y (t) has autocorrelation function RY (t,τ)= E [Y (t)Y (t + τ)] (1) = E [X(t t0)X(t + τ t0)] (2) − − = RX (τ). (3) (b) The cross correlation of X(t) and Y (t) is RXY (t,τ)= E [X(t)Y (t + τ)] (4) = E [X(t)X(t + τ t0)] (5) − = R (τ t0). (6) X − (c) We have already verified that RY (t,τ) depends only on the time difference τ. Since E[Y (t)] = E[X(t t0)] = µ , we have verified that Y (t) is wide sense stationary. − X (d) Since X(t) and Y (t) are wide sense stationary and since we have shown that RXY (t,τ) depends only on τ, we know that X(t) and Y (t) are jointly wide sense stationary. Comment: This problem is badly designed since the conclusions don’t depend on the specific RX (τ) given in the problem text. (Sorry about that!) ECE302 Spring 2006 HW12 Solutions April 27, 2006 4 Problem 11.2.1 The random sequence• Xn is the input to a discrete-time filter. The output is X +1 + X + X 1 Y = n n n− . n 3 (a) What is the impulse response hn? (b) Find the autocorrelation of the output Yn when Xn is a wide sense stationary random sequence with µX = 0 and autocorrelation 1 n = 0, R [n]= X 0 otherwise. Problem 11.2.1 Solution (a) Note that ∞ 1 1 1 Y = h X = X +1 + X + X 1 (1) i n i−n 3 i 3 i 3 i− n=X−∞ By matching coefficients, we see that 1/3 n = 1, 0, 1 h = − (2) n 0 otherwise (b) By Theorem 11.5, the output autocorrelation is ∞ ∞ R [n]= h h R [n + i j] (3) Y i j X − i=X−∞ j=X−∞ 1 1 1 = R [n + i j] (4) 9 X − iX=−1 jX=−1 1 = (R [n + 2] + 2R [n + 1] + 3R [n] + 2R [n 1] + R [n 2]) (5) 9 X X X X − X − We see that the filter is linear and time invariant. Substituting in RX [n] yields 1/3 n = 0 2/9 n = 1 RY [n]= | | (6) 1/9 n = 2 | | 0 otherwise ECE302 Spring 2006 HW12 Solutions April 27, 2006 5 Problem 11.2.2 X(t) is a wide sense• stationary process with autocorrelation function sin(2000πt) + sin(1000πt) R (τ) = 10 . X 2000πt The process X(t) is sampled at rate 1/Ts = 4,000 Hz, yielding the discrete-time process Xn. What is the autocorrelation function RX [k] of Xn? Problem 11.2.2 Solution Applying Theorem 11.4 with sampling period Ts = 1/4000 s yields sin(2000πkTs) + sin(1000πkTs) RX [k]= RX (kTs) = 10 (1) 2000πkTs sin(0.5πk) + sin(0.25πk) = 20 (2) πk = 10 sinc(0.5k) + 5 sinc(0.25k) (3) Problem 11.3.1 Xn is a stationary• Gaussian sequence with expected value E[Xn] = 0 and autocorrelation −|k| ′ function RX [k] = 2 . Find the PDF of X = X1 X2 X3 . Problem 11.3.1 Solution −|k| Since the process Xn has expected value E[Xn] = 0, we know that CX (k)= RX (k) = 2 . ′ Thus X = X1 X2 X3 has covariance matrix 20 2−1 2−2 1 1/2 1/4 1 0 1 CX = 2− 2 2− = 1/2 1 1/2 . (1) 2−2 2−1 20 1/4 1/2 1 From Definition 5.17, the PDF of X is 1 1 ′ −1 fX (x)= exp x CX x . (2) (2π)n/2[det (CX)]1/2 −2 Equivalently, we can write out the PDF in terms of the variables x1, x2 and x3. To do so we find that the inverse covariance matrix is 4/3 2/3 0 −1 − CX = 2/3 5/3 2/3 (3) − − 0 2/3 4/3 − A little bit of algebra will show that det(CX) = 9/16 and that 2 2 2 1 ′ −1 2x1 5x2 2x3 2x1x2 2x2x3 x CX x = + + .