sm8_53
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sm8_53


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Problem 8.53 A baseband signal is disturbed by a noise process N(t) as shown by 
 
 ( ) )(3.0sin)( tNtAtX += \u3c0 
 
where N(t) is a stationary Gaussian process of zero mean and variance \u3c32. 
 
(a) What are the density functions of the random variables X1 and X2 where 
 
11
)( == ttXX 
 
22
)( == ttXX 
 
(b) The noise process N(t) has an autocorrelation function given by 
 ( )\u3c4\u3c3\u3c4 \u2212= exp)( 2NR 
 
What is the joint density function of X1 and X2, that is, ? ),( 21, 21 xxf XX
 
Solution 
(a) The random variable X1 has a mean 
 [ ] ( )[ ]
[ ]
( )\u3c0
\u3c0
\u3c0
3.0sin
)()3.0sin(
)(3.0sin)(
1
11
A
tNA
tNAtX
=
+=
+=
E
EE
 
 
Since X1 is equal to N(t1) plus a constant, the variance of X1 is the same as that of N(t1). In 
addition, since N(t1) is a Gaussian random variable, X1 is also Gaussian with a density 
given by 
{ }21 2/)(exp21)(1 \u3c3µ\u3c3\u3c0 \u2212\u2212= xxf X 
 
where [ )( 11 tXE= ]µ . By a similar argument, the density function of X2 is 
 
{ }22 2/)(exp21)(2 \u3c3µ\u3c3\u3c0 \u2212\u2212= xxf X 
 
where ( )\u3c0µ 6.0sin2 A= . 
 
 
Continued on next slide 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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page\u20268-66 
Problem 8-53 continued 
 
(b) First note that since the mean of X(t) is not constant, X(t) is not a stationary random 
process. However, X(t) is still a Gaussian random process, so the joint distribution of 
N Gaussian random variables may be written as Eq. (8.90). For the case of N = 2, this 
equation reduces to 
 
{ }2/)()(exp
2
1)( 12/1
Tf µxµxxX \u2212\u39b\u2212\u2212\u39b=
\u2212
\u3c0 
 
where \u39b is the 2x2 covariance matrix. Recall that cov(X1,X2) =E[(X1-µ1)(X2-µ2)], so that 
 
( ) (
( ) \u23a5\u23a6
\u23a4\u23a2\u23a3
\u23a1
\u2212
\u2212=
\u23a5\u23a6
\u23a4\u23a2\u23a3
\u23a1=
\u23a5\u23a6
\u23a4\u23a2\u23a3
\u23a1=\u39b
22
22
2212
2111
1exp
)1exp(
)0()1(
)1()0(
,cov,cov
),cov(),cov(
\u3c3\u3c3
\u3c3\u3c3
NN
NN
RR
RR
XXXX
XXXX
)
 
 
If we let \u3c1 = exp(-1) then 
 
)1( 24 \u3c1\u3c3 \u2212=\u39b 
 
and 
 
\u23a5\u23a6
\u23a4\u23a2\u23a3
\u23a1
1\u2212
\u2212
\u2212=\u39b
\u2212
\u3c1
\u3c1
\u3c1\u3c3
1
)1(
1
22
1 
 
Making these substitutions into the above expression, we obtain upon simplification 
 
\u23ad\u23ac
\u23ab
\u23a9\u23a8
\u23a7
\u2212
\u2212\u2212\u2212\u2212+\u2212\u2212\u2212= )1(2
))((2)()(exp
12
1),( 22
2211
2
22
2
11
2221, 21 \u3c1\u3c3
µµ\u3c1µµ
\u3c1\u3c0\u3c3
xxxxxxf XX 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
page\u20268-67