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part42

# part42 - EE 5375/7375 Random Processes Homework#7 Solutions...

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EE 5375/7375 Random Processes October 23, 2003 Homework #7 Solutions Problem 1. textbook problem 7.2 Let p ( x ) be the rectangular function shown in Fig. P7.2. Is R X ( τ ) = p ( τ/T ) a valid autocorrelation function? R X ( τ ) is a rectangle with height A in the interval ( - T, T ). According to Appendix B in the textbook, its Fourier transform is A 2 T sin(2 πfT ) 2 πfT . Since this is negative sometimes, but power spectral densities must always be non-negative, p ( τ/T ) cannot be a valid autocorrelation function. Problem 2. textbook problem 7.5 Let Z ( t ) = X ( t ) + Y ( t ). Under what conditions does S Z ( f ) = S X ( f ) + S Y ( f )? We know (from notes and Example 7.4 in the textbook) that R Z ( τ ) = R X ( τ ) + R Y,X ( τ ) + R X,Y ( τ ) + R Y ( τ ) and S Z ( f ) = S X ( f ) + S Y,X ( f ) + S X,Y ( f ) + S Y ( f ) It is clear that S Z ( f ) = S X ( f ) + S Y ( f ) only if R Y,X ( τ ) + R X,Y ( τ ) = 0. This would be true, for instance, if R Y,X ( τ ) = R X,Y ( τ ) = 0 for all τ ; then X and Y are said to be orthogonal to each other. Problem 3. textbook problem 7.6 Show that (a) R X,Y ( τ ) = R Y,X ( - τ ) (b) S X,Y ( f ) = S * Y,X ( f ). (a) R X,Y ( τ ) = E ( X ( t + τ ) Y ( t )) = E ( Y ( t ) X ( t + τ )) = R Y,X ( - τ ) (b) The spectral density is the Fourier transform of the autocorrrelation function: S X,Y ( f ) = -∞ R X,Y ( τ ) e - j 2 πfτ = -∞ R Y,X ( - τ ) e - j 2 πfτ = -∞ R Y,X ( τ ) e j 2 πfτ = S * Y,X ( f ) The last step follows from observing that the usual Fourier transform uses e - j 2 πfτ = cos(2 πfτ ) - j sin(2 πfτ ) in the integral, but the second to last step has e j 2 πfτ = cos(2 πfτ ) + j sin(2 πfτ ). Therefore the result will be the complex conjugate of the usual Fourier transform. Problem 4. textbook problem 7.7 Let Y ( t ) = X ( t ) - X ( t - d ). (a) Find R X,Y ( τ ) and S X,Y ( f ). (b) Find R Y ( τ ) and S Y ( f ). (a) We find R X,Y ( τ ) = E ( X ( t + τ )( X ( t ) - X ( t - d ))) = E ( X ( t + τ ) X ( t )) - E ( X ( t + τ ) X ( t - d )) = R X ( τ ) - R X ( τ + d ) Taking the Fourier transform, S X,Y ( f ) = S X ( f ) - S X ( f ) e j 2 πfd 1

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where we have used the property that a time shift of T in the time domain corresponds to an additional exponential factor in the frequency domain, for example, a Fourier transform pair is R ( τ - T ) S ( f ) e - j 2 πfT (b) Similarly, we find R Y ( τ ) = E (( X ( t + τ ) - X ( t + τ - d ))( X ( t ) - X ( t - d ))) = 2 R X ( τ ) - R X ( τ + d ) - R X ( τ - d ) Taking the Fourier transform, S Y ( f ) = 2 S X ( f ) - S X ( f ) e j 2 πfd - S X ( f ) e - j 2 πfd = 2 S X ( f )(1 - cos(2 πfd )) Problem 5. textbook problem 7.8 Let X ( t ) and Y ( t ) be independent wide-sense stationary random processes, and define Z ( t ) = X ( t ) Y ( t ). (a) Show that Z ( t ) is wide-sense stationary. (b) Find R Z ( τ
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part42 - EE 5375/7375 Random Processes Homework#7 Solutions...

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