Review_Quiz_Slns_Spring_08

# Review_Quiz_Slns_Spring_08 - ECE161A Solutions To Review...

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ECE161A : Solutions To Review Quiz Problem 1 Note that the Fourier Transform of y * [ - n ] is Y * ( e ) . Proof: Y ( e ) = k = -∞ y [ k ] e - jωk , and hence Y * ( e ) = X k = -∞ y * [ k ] e jωk = X k = -∞ y * [ - n ] e - jωn Now to find an expression for v [ n ] , we use the above result and the inverse Fourier Transform formula. v [ n ] = 1 2 π π Z - π V ( e ) e jωn = 1 2 π π Z - π X ( e ) Y * ( e ) e jωn = 1 2 π π Z - π X ( e ) X k = -∞ y * [ k ] e jωk e jωn from above = X k = -∞ y * [ k ] 1 2 π Z -∞ X ( e ) e jωk e jωn = X k = -∞ y * [ k ] 1 2 π Z -∞ X ( e ) e ( n + k ) = X k = -∞ y * [ k ] x [ n + k ] 1

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Problem 2 a) H 1 ( z ) : y 1 [ n ] = 2 x [ n ] + 3 x [ n - 2] H 1 ( z ) = 2 + 3 z - 2 , ROC: | z | > 0 H 2 ( z ) : y [ n ] = y 1 [ n - 1] - 2 5 y [ n - 1] H 2 ( z ) : z - 1 1 + 2 5 z - 1 , | z | > 2 5 H ( z ) = H 1 ( z ) H 2 ( z ) = z - 1 (2 + 3 z - 2 ) 1 + 2 5 z - 1 , | z | > 2 5 b) H ( z )is causal. ROC > | r 1 | and includes infinity. Also, cascade of causal systems is causal. c) H ( z )is stable. ROC includes unit circle. d) h [ n ] = 2( - 2 5 ) n - 1 u [ n - 1] + 3( - 2 5 ) n - 3 u [ n - 3] follows from the fact that 1 1 + 2 5 z - 1 ( - 2 5 ) n u [ n ] and the numerator only scales and delays the sequence. e) x [ n ] = e j π 4 - e j π 5 + e - j π 4 e - j π 5 n + 3 2 j · e j π 6 n - 3 2 j e - j π 6 n y [ n ] = e j π 4 H ( e j π 5 ) e j π 5 n + e - j π 4 H ( e - j π 5 ) e - j π 5 n + 3 2 j e j π 6 n H ( e j π 6 ) - 3 2 j e - j π 6 n H ( e j π 6 ) = 2 | H
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