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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 permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 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