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 annotatedand 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) awide 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)

= E[A2

]+ 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 widesense stationary process.

Problem 10.11.1 •X(t) and Y (t) are independent wide sense stationary processes with expected values µX andµY 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)] = µXµY . (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

RWX(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),

RWX(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 RWX(t, τ) dependsonly on the time difference τ , we can conclude from Definition 10.18 that W (t) andX(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 thatXi(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 thesefacts 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 crosscorrelation

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 crosscorrelation

RXX2(t, τ) = E [X(t)X2(t + τ)] (10)

= E [X(t)X(a(t + τ))] (11)

= RX((a − 1)t + τ). (12)

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) arenot 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-onds: Y (t) = X(t − t0) where t0 = 5 × 10−5s.

(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)

= RX(τ − t0). (6)

(c) We have already verified that RY (t, τ) depends only on the time difference τ . SinceE[Y (t)] = E[X(t − t0)] = µX , we have verified that Y (t) is wide sense stationary.

(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 thespecific 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

Yn =Xn+1 + Xn + Xn−1

3.

(a) What is the impulse response hn?

(b) Find the autocorrelation of the output Yn when Xn is a wide sense stationary randomsequence with µX = 0 and autocorrelation

RX [n] =

{1 n = 0,0 otherwise.

Problem 11.2.1 Solution

(a) Note that

Yi =∞∑

n=−∞

hnXi−n =1

3Xi+1 +

1

3Xi +

1

3Xi−1 (1)

By matching coefficients, we see that

hn =

{1/3 n = −1, 0, 10 otherwise

(2)

(b) By Theorem 11.5, the output autocorrelation is

RY [n] =

∞∑

i=−∞

∞∑

j=−∞

hihjRX [n + i − j] (3)

=1

9

1∑

i=−1

1∑

j=−1

RX [n + i − j] (4)

=1

9(RX [n + 2] + 2RX [n + 1] + 3RX [n] + 2RX [n − 1] + RX [n − 2]) (5)

We see that the filter is linear and time invariant. Substituting in RX [n] yields

RY [n] =

1/3 n = 02/9 |n| = 11/9 |n| = 20 otherwise

(6)

ECE302 Spring 2006 HW12 Solutions April 27, 2006 5

Problem 11.2.2 •X(t) is a wide sense stationary process with autocorrelation function

RX(τ) = 10sin(2000πt) + sin(1000πt)

2000πt.

The process X(t) is sampled at rate 1/Ts = 4,000 Hz, yielding the discrete-time processXn. 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

RX [k] = RX(kTs) = 10sin(2000πkTs) + sin(1000πkTs)

2000πkTs(1)

= 20sin(0.5πk) + sin(0.25πk)

πk(2)

= 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

function RX [k] = 2−|k|. Find the PDF of X =[X1 X2 X3

]′.

Problem 11.3.1 Solution

Since the process Xn has expected value E[Xn] = 0, we know that CX(k) = RX(k) = 2−|k|.Thus X =

[X1 X2 X3

]′has covariance matrix

CX =

20 2−1 2−2

2−1 20 2−1

2−2 2−1 20

=

1 1/2 1/41/2 1 1/21/4 1/2 1

. (1)

From Definition 5.17, the PDF of X is

fX (x) =1

(2π)n/2[det (CX)]1/2exp

(

−1

2x′C−1

Xx

)

. (2)

Equivalently, we can write out the PDF in terms of the variables x1, x2 and x3. To do sowe find that the inverse covariance matrix is

C−1

X=

4/3 −2/3 0−2/3 5/3 −2/3

0 −2/3 4/3

(3)

A little bit of algebra will show that det(CX) = 9/16 and that

1

2x′C−1

Xx =

2x21

3+

5x22

6+

2x23

3− 2x1x2

3− 2x2x3

3. (4)

It follows that

fX (x) =4

3(2π)3/2exp

(

−2x21

3− 5x2

2

6− 2x2

3

3+

2x1x2

3+

2x2x3

3

)

. (5)

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 filterh =

[1 −1 1

]′yields the output Yn. Find the PDFs of:

(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

Y3 =

Y1

Y2

Y3

=

h1 h0 0 0h2 h1 h0 00 h2 h1 h0

X0

X1

X2

X3

=

−1 1 0 01 −1 1 00 1 −1 1

︸ ︷︷ ︸

H

X0

X1

X2

X3

︸ ︷︷ ︸

X

. (2)

We note that the components of X are iid Gaussian (0, 1) random variables. HenceX has covariance matrix CX = I, the identity matrix. Since Y3 = HX,

CY3= HCXH′ = HH′ =

2 −2 1−2 3 −21 −2 3

. (3)

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

C−1

Y3=

1

3

5 4 14 5 21 2 2

. (4)

Some algebra will show that

y′C−1

Y3y =

5y21 + 5y2

2 + 2y23 + 8y1y2 + 2y1y3 + 4y2y3

3. (5)

This implies Y3 has PDF

fY3(y) =

1

(2π)3/2[det (CY3)]1/2

exp

(

−1

2y′C−1

Y3y

)

(6)

=1

(2π)3/2√

3exp

(

−5y21 + 5y2

2 + 2y23 + 8y1y2 + 2y1y3 + 4y2y3

6

)

. (7)

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,

CY2=

[2 −2−2 3

]

and C−1

Y2=

[3/2 11 1

]

. (8)

Since det(CY2) = 2, it follows that

fY2(y) =

1

(2π)3/2[det (CY2)]1/2

exp

(

−1

2y′C−1

Y2y

)

(9)

=1

(2π)3/2√

2exp

(

−3

2y21 − 2y1y2 − y2

2

)

. (10)

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

RX(τ) = 10sin(2000πτ) + sin(1000πτ)

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

sinc(x) =sin(πx)

πx. (1)

In terms of the sinc(·) function, we obtain

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

From Table 11.1,

SX (f) =10

2,000rect

(f

2000

)

+5

1,000rect

(f

1,000

)

(3)

Here is a graph of the PSD.

−1500 −1000 −500 0 500 1000 15000

0.002

0.004

0.006

0.008

0.01

0.012

f

SX(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

RX [k] =

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

Find the power spectral density SX(f).

Problem 11.6.1 Solution

Since the random sequence Xn has autocorrelation function

RX [k] = δk + (0.1)|k|, (1)

We can find the PSD directly from Table 11.2 with 0.1|k| corresponding to a|k|. The tableyields

SX (φ) = 1 +1 − (0.1)2

1 + (0.1)2 − 2(0.1) cos 2πφ=

2 − 0.2 cos 2πφ

1.01 − 0.2 cos 2πφ. (2)