sol.HW2

# sol.HW2 - EE 7615 Solutions Problem Set 2 2010 Sept 1 The...

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EE 7615 Solutions Problem Set # 2 Sept. 2010 1. The following fact regarding Gaussian random variables is used in the solution. Fact: If X 1 ,X 2 ,X 3 and X 4 are four Gaussian random variables, then E [ X 1 X 2 X 3 X 4 ] = E [ X 1 X 2 ] E [ X 3 X 4 ]+ E [ X 1 X 3 ] E [ X 2 X 4 ]+ E [ X 1 X 4 ] E [ X 2 X 3 ] . (a) E [ Y ( t )] = E [ X 2 ( t )] = R X (0) = Z -∞ S X ( f ) df = 2 E [ Z ( t )] = m Y Z -∞ h ( τ ) where h ( τ ) is the impulse response of the ﬁlter. From Fourier transforms we know Z -∞ h ( τ ) = H (0) = 1 Thus E [ Z ( t )] = 2. (b) R Y ( t,s ) = E [ Y ( t ) Y ( s )] = E [ X 2 ( t ) X 2 ( s )] Using the fact above we get R Y ( t,s ) = E [ X 2 ( t )] E [ X 2 ( s )]+2 { E [ X ( t ) X ( s ) } 2 = R X (0) R X (0)+2 { R X ( t - s ) } 2 Thus R Y ( τ ) = 2 R 2 X ( τ ) + R 2 X (0) We see that the process { Y ( t ) } is a WSS process. S Y ( f ) = F { R Y ( τ ) } = 2 S X ( f ) * S X ( f ) + R 2 X (0) δ ( f ) where * is the convolution operation. Therefore S Y ( f ) = G ( f ) + R 2 X (0) δ ( f ) where G ( f ) is given by G ( f ) = ( 2 - | f | | f | ≤ 2 0 otherwise. 1

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(c) Since Z ( t ) is the response of a linear time-invariant system to a WSS input, it is WSS. We know that S Z ( f ) = S Y ( f ) | H ( f ) | 2 . Thus
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sol.HW2 - EE 7615 Solutions Problem Set 2 2010 Sept 1 The...

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