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 ))]
= E A2 + E [A] E [X(t)] + AE [X(t + τ )] + E [X(t)X(t + τ )]
= E A2 + 2E [A] µX + RX (τ ),
(2)
(3)
(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
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April 27, 2006
(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
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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)
= 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) 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 microseconds: Y (t) = X(t − t0 ) where t0 = 5 × 10−5 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 )]
= E [X(t − t0 )X(t + τ − t0 )]
= RX (τ ).
(1)
(2)
(3)
(b) The cross correlation of X(t) and Y (t) is
RXY (t, τ ) = E [X(t)Y (t + τ )]
= E [X(t)X(t + τ − t0 )]
= RX (τ − t0 ).
(4)
(5)
(6)
(c) We have already verified that RY (t, τ ) depends only on the time difference τ . Since
E[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 the
specific RX (τ ) given in the problem text. (Sorry about that!)
ECE302 Spring 2006
HW12 Solutions
April 27, 2006
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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 random
sequence with µX = 0 and autocorrelation
1 n = 0,
RX [n] =
0 otherwise.
Problem 11.2.1 Solution
(a) Note that
Yi =
∞
X
hn Xi−n =
n=−∞
1
1
1
Xi+1 + Xi + Xi−1
3
3
3
By matching coefficients, we see that
1/3 n = −1, 0, 1
hn =
0
otherwise
(1)
(2)
(b) By Theorem 11.5, the output autocorrelation is
RY [n] =
∞
X
∞
X
i=−∞ j=−∞
=
hi hj RX [n + i − j]
(3)
1
1
1 X X
RX [n + i − j]
9
(4)
1
(RX [n + 2] + 2RX [n + 1] + 3RX [n] + 2RX [n − 1] + RX [n − 2])
9
(5)
i=−1 j=−1
=
We see that the filter is linear and time




RY [n] =



invariant. Substituting in RX [n] yields
1/3
2/9
1/9
0
n=0
|n| = 1
|n| = 2
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
sin(2000πt) + sin(1000πt)
.
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 ?
RX (τ ) = 10
Problem 11.2.2 Solution
Applying Theorem 11.4 with sampling period Ts = 1/4000 s yields
sin(2000πkTs ) + sin(1000πkTs )
2000πkTs
sin(0.5πk) + sin(0.25πk)
= 20
πk
= 10 sinc(0.5k) + 5 sinc(0.25k)
RX [k] = RX (kTs ) = 10
(1)
(2)
(3)
Problem 11.3.1 •
Xn is a stationary Gaussian sequence with expected
value E[X
′ n ] = 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 hasexpected value E[Xn ] = 0, we know that CX (k) = RX (k) = 2−|k| .
′
Thus X = X1 X2 X3 has covariance matrix
 0
 

2
2−1 2−2
1 1/2 1/4
CX = 2−1 20 2−1  = 1/2 1 1/2 .
(1)
−2
−1
0
2
2
2
1/4 1/2 1
From Definition 5.17, the PDF of X is
1 ′ −1
1
exp − x CX x .
fX (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

C−1
5/3 −2/3
(3)
X = −2/3
0
−2/3 4/3
A little bit of algebra will show that det(CX ) = 9/16 and that
1 ′ −1
2x2 5x2 2x2 2x1 x2 2x2 x3
x CX x = 1 + 2 + 3 −
−
.
2
3
6
3
3
3
(4)
4
2x21 5x22 2x23 2x1 x2 2x2 x3
−
−
+
+
.
fX (x) =
exp −
3
6
3
3
3
3(2π)3/2
(5)
It follows that
ECE302 Spring 2006
HW12 Solutions
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April 27, 2006
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 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
 
 
  
 X0

 X0
Y1
h1 h0 0 0  
−1 1
0 0  
X1  
X1 
Y3 = Y2  = h2 h1 h0 0  
(2)
=
1
−1
1
0 
X2 
X2  .
Y3
0 h2 h1 h0
0
1 −1 1
X3
|
{z
} X3
| {z }
H
X
We note that the components of X are iid Gaussian (0, 1) random variables. Hence
X has covariance matrix CX = I, the identity matrix. Since Y3 = HX,


2 −2 1
CY3 = HCX H′ = HH′ = −2 3 −2 .
(3)
1 −2 3
Some calculation (by hand or by Matlab) will

5 4
1
−1
CY3 =
4 5
3
1 2
show that det(CY3 ) = 3 and that

1
2 .
(4)
2
Some algebra will show that
y′ C−1
Y3 y =
5y12 + 5y22 + 2y32 + 8y1 y2 + 2y1 y3 + 4y2 y3
.
3
(5)
This implies Y3 has PDF
1 ′ −1
1
exp − y CY3 y
fY3 (y) =
2
(2π)3/2 [det (CY3 )]1/2
2
2
1
5y1 + 5y2 + 2y32 + 8y1 y2 + 2y1 y3 + 4y2 y3
√ exp −
=
.
6
(2π)3/2 3
(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,
2 −2
3/2 1
−1
CY2 =
and CY2 =
.
(8)
−2 3
1 1
Since det(CY2 ) = 2, it follows that
1
1 ′ −1
fY2 (y) =
exp − y CY2 y
2
(2π)3/2 [det (CY2 )]1/2
3 2
1
2
√
exp − y1 − 2y1 y2 − y2 .
=
2
(2π)3/2 2
(9)
(10)
Problem 11.5.1 •
X(t) is a wide sense stationary process with autocorrelation function
RX (τ ) = 10
sin(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,
10
SX (f ) =
rect
2,000
f
2000
5
rect
+
1,000
f
1,000
(3)
Here is a graph of the PSD.
0.012
0.01
SX(f)
0.008
0.006
0.004
0.002
0
−1500
−1000
−500
0
f
500
1000
1500
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
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 table
yields
2 − 0.2 cos 2πφ
1 − (0.1)2
=
.
(2)
SX (φ) = 1 +
2
1 + (0.1) − 2(0.1) cos 2πφ
1.01 − 0.2 cos 2πφ