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 = + + . (4) 2 3 6 3 − 3 − 3 It follows that 2 2 2 4 2x1 5x2 2x3 2x1x2 2x2x3 fX (x)= exp + + . (5) 3(2π)3/2 − 3 − 6 − 3 3 3 ECE302 Spring 2006 HW12 Solutions April 27, 2006 6

Problem 11.3.2

Xn is a sequence• of independent random variables such that Xn = 0 for n < 0 while for n 0, each Xn is a Gaussian (0, 1) random variable. Passing Xn through the ﬁlter ≥ ′ h = 1 1 1 yields the output Y . Find the PDFs of: − n ′ (a) Y3 = Y1 Y2 Y3 ,

′ (b) Y2 = Y1 Y2 . Problem 11.3.2 Solution

The sequence Xn is passed through the ﬁlter ′ ′ h = h0 h1 h2 = 1 1 1 (1) − The output sequence is Yn. ′ (a) Following the approach of Equation (11.58), we can write the output Y3 = Y1 Y2 Y3 as X0 X0 Y1 h1 h0 0 0 11 00 X1 − X1 Y3 = Y2 = h2 h1 h0 0  =  1 1 1 0 . (2) X2 − X2 Y3 0 h2 h1 h0   0 1 1 1       X3  −  X3   H   X | {z } We note that the components of X are iid Gaussian (0, 1) random variables.| {z } Hence X has covariance matrix CX = I, the identity matrix. Since Y3 = HX, 2 2 1 ′ ′ − CY3 = HCXH = HH =  2 3 2 . (3) − − 1 2 3  − 

Some calculation (by hand or by Matlab) will show that det(CY3 ) = 3 and that

5 4 1 −1 1 CY = 4 5 2 . (4) 3 3 1 2 2   Some algebra will show that 2 2 2 5y + 5y + 2y + 8y1y2 + 2y1y3 + 4y2y3 y′C−1 y = 1 2 3 . (5) Y3 3

This implies Y3 has PDF

1 1 ′ −1 fY (y)= exp y C y (6) 3 3/2 1/2 Y3 (2π) [det (CY3 )] −2 2 2 2 1 5y + 5y + 2y + 8y1y2 + 2y1y3 + 4y2y3 = exp 1 2 3 . (7) (2π)3/2√3 − 6 ECE302 Spring 2006 HW12 Solutions April 27, 2006 7

′ (b) To ﬁnd the PDF of Y2 = Y1 Y2 , we start by observing that the covariance matrix

of Y2 is just the upper left 2 2 submatrix of CY3 . That is, ×

2 2 −1 3/2 1 CY2 = − and CY = . (8) 2 3 2 1 1 −

Since det(CY2 ) = 2, it follows that

1 1 ′ −1 Y y y C y f 2 ( )= 3/2 1/2 exp Y2 (9) (2π) [det (CY2 )] −2

1 3 2 2 = exp y1 2y1y2 y2 . (10) (2π)3/2√2 −2 − −

Problem 11.5.1 X(t) is a wide sense• stationary process with autocorrelation function

sin(2000πτ) + sin(1000πτ) R (τ) = 10 . X 2000πτ What is the power spectral density of X(t)? Problem 11.5.1 Solution

To use Table 11.1, we write RX (τ) in terms of the autocorrelation sin(πx) sinc(x)= . (1) πx In terms of the sinc( ) function, we obtain ·

RX (τ) = 10 sinc(2000τ) + 5 sinc(1000τ). (2)

From Table 11.1, 10 f 5 f S (f)= rect + rect (3) X 2,000 2000 1,000 1,000 Here is a graph of the PSD.

0.012 0.01 0.008 (f)

X 0.006 S 0.004 0.002 0 −1500 −1000 −500 0 500 1000 1500 f ECE302 Spring 2006 HW12 Solutions April 27, 2006 8

Problem 11.6.1

Xn is a wide sense stationary• discrete-time random sequence with autocorrelation function

δ[k] + (0.1)|k| k = 0, 1, 2,..., RX [k]= ± ± 0 otherwise.

Find the power spectral density SX (f). Problem 11.6.1 Solution

Since the random sequence Xn has autocorrelation function

|k| RX [k]= δk + (0.1) , (1)

We can ﬁnd the PSD directly from Table 11.2 with 0.1|k| corresponding to a|k|. The table yields 1 (0.1)2 2 0.2 cos 2πφ S (φ)=1+ − = − . (2) X 1 + (0.1)2 2(0.1) cos 2πφ 1.01 0.2 cos 2πφ − −