ECE302 Spring 2006 HW12 Solutions April 27,

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1 ECE02 Spring 2006 HW2 Solutions April 27, 2006 Solutions to HW2 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 Let A be a nonnegative random variable that is independent of any collection of samples Xt ),...,Xt k ) of a wide sense stationary random process Xt). Is Y t) = A + Xt) a wide sense stationary process? Problem Solution To show that Y t) is wide-sense stationary we must show that it meets the two requirements of Definition 0.5, namely that its expected value and autocorelation function must be independent of t. Since Y t) = A + Xt), the mean of Y t) is The autocorrelation of Y t) is E [Y t)] = E [A] + E [Xt)] = E [A] + µ X ) R Y t,τ) = E [A + Xt))A + Xt + τ))] 2) = E [ A 2] + E [A] E [Xt)] + AE [Xt + τ)] + E [Xt)Xt + τ)] ) = E [ A 2] + 2E [A]µ X + R X τ), 4) where the last equality is justified by the fact that we are given that Xt) is wide sense stationary. We see that neither E[Y t)] nor R Y t,τ) depend on t. Thus Y t) is a wide sense stationary process. Problem 0.. Xt) and Y t) are independent wide sense stationary processes with expected values µ X and µ Y and autocorrelation functions R X τ) and R Y τ) respectively. Let Wt) = Xt)Y t). a) Find µ W and R W t,τ) and show that Wt) is wide sense stationary. b) Are Wt) and Xt) jointly wide sense stationary? Problem 0.. Solution a) Since Xt) and Y t) are independent processes, E [Wt)] = E [Xt)Y t)] = E [Xt)]E [Y t)] = µ X µ Y. ) In addition, R W t,τ) = E [Wt)Wt + τ)] 2) = E [Xt)Y t)xt + τ)y t + τ)] ) = E [Xt)Xt + τ)]e [Y t)y t + τ)] 4) = R X τ)r Y τ) 5) We can conclude that Wt) is wide sense stationary.

2 ECE02 Spring 2006 HW2 Solutions April 27, b) To examine whether Xt) and Wt) are jointly wide sense stationary, we calculate R WX t,τ) = E [Wt)Xt + τ)] = E [Xt)Y t)xt + τ)]. 6) By independence of Xt) and Y t), R WX t,τ) = E [Xt)Xt + τ)]e [Y t)] = µ Y R X τ). 7) Since Wt) and Xt) are both wide sense stationary and since R WX t,τ) depends only on the time difference τ, we can conclude from Definition 0.8 that Wt) and Xt) are jointly wide sense stationary. Problem 0..2 Xt) is a wide sense stationary random process. For each process X i t) defined below, determine whether X i t) and Xt) are jointly wide sense stationary. a) X t) = Xt + a) b) X 2 t) = Xat) Problem 0..2 Solution To show that Xt) and X i t) are jointly wide sense stationary, we must first show that X i t) is wide sense stationary and then we must show that the cross correlation R XXi t,τ) is only a function of the time difference τ. For each X i t), we have to check whether these facts are implied by the fact that Xt) is wide sense stationary. a) Since E[X t)] = E[Xt + a)] = µ X and R X t,τ) = E [X t)x t + τ)] ) = E [Xt + a)xt + τ + a)] 2) = R X τ), ) we have verified that X t) is wide sense stationary. Now we calculate the cross correlation R XX t,τ) = E [Xt)X t + τ)] 4) = E [Xt)Xt + τ + a)] 5) = R X τ + a). 6) Since R XX t,τ) depends on the time difference τ but not on the absolute time t, we conclude that Xt) and X t) are jointly wide sense stationary. b) Since E[X 2 t)] = E[Xat)] = µ X and R X2 t,τ) = E [X 2 t)x 2 t + τ)] 7) = E [Xat)Xat + τ))] 8) = E [Xat)Xat + aτ)] = R X aτ), 9)

3 ECE02 Spring 2006 HW2 Solutions April 27, 2006 we have verified that X 2 t) is wide sense stationary. Now we calculate the cross correlation R XX2 t,τ) = E [Xt)X 2 t + τ)] 0) = E [Xt)Xat + τ))] ) = R X a )t + τ). 2) Except for the trivial case when a = and X 2 t) = Xt), R XX2 t,τ) depends on both the absolute time t and the time difference τ, we conclude that Xt) and X 2 t) are not jointly wide sense stationary. Problem 0.. Xt) is a wide sense stationary stochastic process with autocorrelation function R X τ) = 0 sin2π000τ)/2π000τ). The process Y t) is a version of Xt) delayed by 50 microseconds: Y t) = Xt t 0 ) where t 0 = s. a) Derive the autocorrelation function of Y t). b) Derive the cross-correlation function of Xt) and Y t). c) Is Y t) wide sense stationary? d) Are Xt) and Y t) jointly wide sense stationary? Problem 0.. Solution a) Y t) has autocorrelation function b) The cross correlation of Xt) and Y t) is R Y t,τ) = E [Y t)y t + τ)] ) = E [Xt t 0 )Xt + τ t 0 )] 2) = R X τ). ) R XY t,τ) = E [Xt)Y t + τ)] 4) = E [Xt)Xt + τ t 0 )] 5) = R X τ t 0 ). 6) c) We have already verified that R Y t,τ) depends only on the time difference τ. Since E[Y t)] = E[Xt t 0 )] = µ X, we have verified that Y t) is wide sense stationary. d) Since Xt) and Y t) are wide sense stationary and since we have shown that R XY t,τ) depends only on τ, we know that Xt) and Y t) are jointly wide sense stationary. Comment: This problem is badly designed since the conclusions don t depend on the specific R X τ) given in the problem text. Sorry about that!)

4 ECE02 Spring 2006 HW2 Solutions April 27, Problem.2. The random sequence X n is the input to a discrete-time filter. The output is a) What is the impulse response h n? Y n = X n+ + X n + X n. b) Find the autocorrelation of the output Y n when X n is a wide sense stationary random sequence with µ X = 0 and autocorrelation { n = 0, R X [n] = 0 otherwise. Problem.2. Solution a) Note that Y i = h n X i n = X i+ + X i + X i ) n= By matching coefficients, we see that h n = { / n =,0, 0 otherwise 2) b) By Theorem.5, the output autocorrelation is R Y [n] = h i h j R X [n + i j] ) = 9 i= j= i= j= R X [n + i j] 4) = 9 R X [n + 2] + 2R X [n + ] + R X [n] + 2R X [n ] + R X [n 2]) 5) We see that the filter is linear and time invariant. Substituting in R X [n] yields / n = 0 2/9 n = R Y [n] = /9 n = 2 0 otherwise 6)

5 ECE02 Spring 2006 HW2 Solutions April 27, Problem.2.2 Xt) is a wide sense stationary process with autocorrelation function sin2000πt) + sin000πt) R X τ) = πt The process Xt) is sampled at rate /T s = 4,000 Hz, yielding the discrete-time process X n. What is the autocorrelation function R X [k] of X n? Problem.2.2 Solution Applying Theorem.4 with sampling period T s = /4000 s yields R X [k] = R X kt s ) = 0 sin2000πkt s) + sin000πkt s ) 2000πkT s ) sin0.5πk) + sin0.25πk) = 20 πk 2) = 0 sinc0.5k) + 5 sinc0.25k) ) Problem.. X n is a stationary Gaussian sequence with expected value E[X n ] = 0 and autocorrelation function R X [k] = 2 k. Find the PDF of X = [ X X 2 X ]. Problem.. Solution Since the process X n has expected value E[X n ] = 0, we know that C X k) = R X k) = 2 k. Thus X = [ ] X X 2 X has covariance matrix /2 /4 C X = = /2 /2. ) /4 /2 From Definition 5.7, the PDF of X is f X x) = 2π) n/2 exp ) [det C X )] /2 2 x C X x. 2) Equivalently, we can write out the PDF in terms of the variables x, x 2 and x. To do so we find that the inverse covariance matrix is 4/ 2/ 0 C X = 2/ 5/ 2/ ) 0 2/ 4/ A little bit of algebra will show that detc X ) = 9/6 and that It follows that f X x) = 2 x C X x = 2x2 + 5x x2 2x x 2 2x 2x. 4) 4 exp 2x2 2π) /2 5x x2 + 2x x 2 + 2x ) 2x. 5)

6 ECE02 Spring 2006 HW2 Solutions April 27, Problem..2 X n is a sequence of independent random variables such that X n = 0 for n < 0 while for n 0, each X n is a Gaussian 0,) random variable. Passing X n through the filter h = [ ] yields the output Yn. Find the PDFs of: a) Y = [ Y Y 2 Y ], b) Y 2 = [ Y Y 2 ]. Problem..2 Solution The sequence X n is passed through the filter The output sequence is Y n. h = [ h 0 h h 2 ] = [ ] ) a) Following the approach of Equation.58), we can write the output Y = [ ] Y Y 2 Y as X Y h h X Y = Y 2 = h 2 h h 0 0 X = 0 X X Y 0 h 2 h h ) X 2 X } {{ } H X }{{} X We note that the components of X are iid Gaussian 0, ) random variables. Hence X has covariance matrix C X = I, the identity matrix. Since Y = HX, 2 2 C Y = HC X H = HH = 2 2. ) 2 Some calculation by hand or by Matlab) will show that detc Y ) = and that C Y = ) 2 2 Some algebra will show that This implies Y has PDF y C Y y = 5y2 + 5y y2 + 8y y 2 + 2y y + 4y 2 y. 5) f Y y) = 2π) /2 [detc Y )] = 2π) /2 exp exp /2 2 y C Y y 5y2 + 5y y2 + 8y y 2 + 2y y + 4y 2 y 6 ) 6) ). 7)

7 ECE02 Spring 2006 HW2 Solutions April 27, b) To find the PDF of Y 2 = [ ], Y Y 2 we start by observing that the covariance matrix of Y 2 is just the upper left 2 2 submatrix of C Y. That is, [ ] [ ] 2 2 C Y2 = and C /2 2 Y 2 =. 8) Since detc Y2 ) = 2, it follows that f Y2 y) = 2π) /2 [detc Y2 )] = 2π) /2 2 exp exp /2 2 y C 2 y2 2y y 2 y2 2 ) Y 2 y 9) ). 0) Problem.5. Xt) is a wide sense stationary process with autocorrelation function R X τ) = 0 What is the power spectral density of Xt)? sin2000πτ) + sin000πτ). 2000πτ Problem.5. Solution To use Table., we write R X τ) in terms of the autocorrelation In terms of the sinc ) function, we obtain sincx) = sinπx) πx. ) R X τ) = 0sinc2000τ) + 5sinc000τ). 2) From Table., S X f) = 0 ) f 2,000 rect + 5 ) f 2000,000 rect,000 ) Here is a graph of the PSD S X f) f

8 ECE02 Spring 2006 HW2 Solutions April 27, Problem.6. X n is a wide sense stationary discrete-time random sequence with autocorrelation function { δ[k] + 0.) k k = 0, ±, ±2,..., R X [k] = 0 otherwise. Find the power spectral density S X f). Problem.6. Solution Since the random sequence X n has autocorrelation function R X [k] = δ k + 0.) k, ) We can find the PSD directly from Table.2 with 0. k corresponding to a k. The table yields 0.) cos 2πφ S X φ) = ) 2 = 20.)cos 2πφ.0 0.2cos 2πφ. 2)