MAE 3020 Information Processing for Automation
Assignment 7 and solution
1. (3.25 in the textbook) X(t) is a complex-valued WSS random process defined as
X(t) = A exp(2jYt + j)
where, A, Y and are independent random variables with the following pdfs:
a exp(  a 2 2)
a0
f A ( )  
0
elsewhere

 1

10,000  y  11,000
f Y ( y )  1000
 0
elsewhere
1

   
f ( )   2
 0
elsewhere
Find the psd of X(t).
S: First, let us calculate the autocorrelation function:
RXX() = E{X*(t)X(t + )}
= E{(A exp(  2jYt  j))(A exp(2jY(t + ) + j))}
= E{A2}E{exp(2jY}
Note that
E{A2} = 2
In addition, RXX() is independent on . Hence:
2 11000
R XX ( ) 
exp( j 2y )dy
1000 10000
2
1
11000

exp( j 2y ) 10000
1000 j 2
1 j
exp( j 22000 )  exp( j 20000 )

1000 
From the Fourier transform table, it is seen
 j
F   sgn( f )
t 
F{exp(j2f0t)} = (f – f0)
1
 j

F
exp  j 2200t  
sgn( f  1100)
 2000
 1000
1
 j

F
exp  j 2000t  
sgn( f  1000)
 2000
 1000
It then follows that:
1
U ( f  10,00)  U ( f  11,000)
S

xx( f ) 500
2. (3.35 in the textbook) Let Z(t) = x(t) + Y(t) where x(t) is a deterministic, periodic
power signal with a period T and Y(t) is a zero mean ergodic random process. Find
the autocorrelation function and also the psd function of Z(t) using time averages.
S: Z(t) = x(t) + Y(t)
1 T /2
RZZ ( ) T    x(t ) x(t   )  x(t ) y (t   )  x(t   ) y (t )  x(t   ) y (t   )dt 

T  T / 2
1 T /2
   x(t ) x(t   )dt   RYY ( )

T  T / 2
Note that the cross terms are dropped because x(t) is periodic (not because Y(t) has a
zero mean). Since x(t) is periodic, it can be written as follows:
2kt
2kt
x(t )   ak cos
 bk sin
T
T
k
Hence,
2kt
2kt  
2k (t   )
2k (t   ) 

x(t ) x(t   )   a j cos
 b j sin
a k cos
 bk sin



T
T 
T
T
l
k 
Because of the orthogonality, it can be shown that:
1 t
2k
x(t ) x(t   )dt   ak2  bk2 cos


T
/
2
T
T
k
Therefore:
2k
RZZ ( )   ak2  bk2 cos
 RYY ( )
T
k
Using the Fourier transform table:
ak2  bk2  
k
k 

S ZZ ( f )  
  f      f    RYY ( f )

2  
T
T 

k




3. (3.39 in the textbook) A stationary zero-mean Gaussian random process X(t) has an
autocorrelation function:
RXX() = 10 exp( )
Show that X(t) is ergodic in the mean and autocorrelation function.
S: (i) The mean
T /2
1
E   X T   E  X (t )dt   X  0
T / 2
T
1 T /2   
var x(t )   1  C XX ( )d
T T / 2  T 
2 T  
  1  10 exp(  )d
10 0  T 


2 2

 10  2 1  exp( T ) 
T T

since:
lim var[ x(t )]  0
T 
X(t) is ergodic in the mean
(ii) The autocorrelation function
1 T /2
E R XX ( ) T   E{ X (t ) X (t   )dt
T T / 2
1 T /2
  R XX ( )d
T T / 2
 R XX ( )






R XX ( )d   10 exp(   )d

0


0
  10e d   10e  d
 10  10  20  
Therefore, it is ergodic in the autocorrelation function.
4. (3.46 in the textbook) The probability density function of a random variable X is
shown in the figure below. (a) If X is quantized into four levels using a uniform
quantizing rule, find the MSE (Mean Square Error); (b) If X is quantized into four
levels using a minimum (MSE) non-uniform quantizer, find the quantizer end points
and output levels as well as the MSE.
fX(x)
1/2
-2
0
2
x
S: This problem deals with the quantilization. From the figure, it is seen that the
probability density function is:
1
f X ( x)  2  x ,
x 2
4
(a) Using a four-level uniform quantizer, we have:
-2 < x < -1: m = (-2 - 1) / 2 = -1.5
-1 < x < 0:
m = (-1 - 0) / 2 = -0.5
0 < x < 1:
m = (0 + 1) / 2 = 0.5
-2 < x < -1: m = (2 + 1) / 2 = 1.5
Hence:
N q  E{( X  m) 2 }
4
xi
i 1
xi 1
   ( x  mi ) 2 f X ( x)dx
0
1
2
1  1
( x  1.5) 2 (2  x)dx   ( x  0.5) 2 (2  x)dx   ( x  0.5) 2 (2  x)dx   ( x  1.5)( 2  x)dx 


1
0
1
2  2
 1 / 12

 
S q  E m 
2
4
  mi 
2
i 1

xi
xi 1
f X ( x)dx
 3/ 4
Therefore, the mean squares error (MSE) is:
N q 1 / 12

 1/ 9
Sq
3/ 4
(b) When nonuniform quantizer is used, the quantizer values, mi, cannot be
analytically determined. However, assuming that X is normally distributed (the
triangle is similar to normal after all),
Nq = (2.2)Q-1.96, when (Q >> 1)
Since Q = 4 (not much greater than 1), Nq = 0.1453. Thus,
N q 0.1453

 0.1938
Sq
3/ 4
5. (4.6 in the textbook) The input-output relationship of a discrete-time linear timeinvariant causal system is given by:
Y(n) = h(0)X(n) + h(1)X(n-1) + … + h(k)X(n-k)
The input sequence X(n) is stationary, zero mean, Gaussian with
1 j  0
E{ X (n) X (n  j )}  
0 j  0
(a) Find the pdf of Y(n)
(b) Find RYY(n) and SYY(f).
S: (a) Since X(n) are Gaussian, Y(n) will be Gaussian. Its mean is:
E{Y(n)} = E{h(0)X(n) + h(1)X(n-1) + … + h(k)X(n-k)}
= h(0)E{X(n)} + h(1)E{X(n-1)} + … + h(k)E{X(n-k)}
=0
and its variance is:
E{Y2(n) } = E{[h2(0)X2(n) + h2(1)X2(n-1) +… + h2(k)X2(n-k)]}
= h2(0) + h2(1) +… + h2(k)
(c) The autocorrelation is:
R yy (n)  R XX (n) * h(n) * h(n)
 h( n) * h(  n)

  h( n  j ) h(  j )
j 


 h( n  j ) h( j )
j  
Hence:
J
 h( n  j ) h( j ) n  J
RYY (n)  
j o

0
nJ
In addition:
SYY ( f )  F RYY ( ) 
K
K
  h(n   )h( ) exp(  j 2nf )
n   K  0