This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 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 ( ) = p ( /T ) a valid autocorrelation function? R X ( ) 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 fT ) 2 fT . Since this is negative sometimes, but power spectral densities must always be non-negative, p ( /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 ( ) = R X ( ) + R Y,X ( ) + R X,Y ( ) + R Y ( ) 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 ( ) + R X,Y ( ) = 0. This would be true, for instance, if R Y,X ( ) = R X,Y ( ) = 0 for all ; then X and Y are said to be orthogonal to each other. Problem 3. textbook problem 7.6 Show that (a) R X,Y ( ) = R Y,X (- ) (b) S X,Y ( f ) = S * Y,X ( f ). (a) R X,Y ( ) = E ( X ( t + ) Y ( t )) = E ( Y ( t ) X ( t + )) = R Y,X (- ) (b) The spectral density is the Fourier transform of the autocorrrelation function: S X,Y ( f ) = Z - R X,Y ( ) e- j 2 f d = Z - R Y,X (- ) e- j 2 f d = Z - R Y,X ( ) e j 2 f d = S * Y,X ( f ) The last step follows from observing that the usual Fourier transform uses e- j 2 f = cos(2 f )- j sin(2 f ) in the integral, but the second to last step has e j 2 f = cos(2 f ) + j sin(2 f ). 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 ( ) and S X,Y ( f ). (b) Find R Y ( ) and S Y ( f ). (a) We find R X,Y ( ) = E ( X ( t + )( X ( t )- X ( t- d ))) = E ( X ( t + ) X ( t ))- E ( X ( t + ) X ( t- d )) = R X ( )- R X ( + d ) Taking the Fourier transform, S X,Y ( f ) = S X ( f )- S X ( f ) e j 2 fd 1 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 ( - T ) S ( f ) e- j 2 fT (b) Similarly, we find R Y ( ) = E (( X ( t + )- X ( t + - d ))( X ( t )- X ( t- d ))) = 2 R X ( )- R X ( + d )- R X ( - d ) Taking the Fourier transform,...
View Full Document
This note was uploaded on 11/29/2009 for the course EE 131A taught by Professor Lorenzelli during the Fall '08 term at UCLA.
- Fall '08