Review_Quiz_Slns_Spring_08

Review_Quiz_Slns_Spring_08 - ECE161A : Solutions To Review...

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Unformatted text preview: ECE161A : Solutions To Review Quiz Problem 1 Note that the Fourier Transform of y * [- n ] is Y * ( e jω ) . Proof: Y ( e jω ) = ∞ ∑ k =-∞ y [ k ] e- jωk , and hence Y * ( e jω ) = ∞ 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 jω ) e jωn dω = 1 2 π π Z- π X ( e jω ) Y * ( e jω ) e jωn dω = 1 2 π π Z- π X ( e jω ) ∞ X k =-∞ y * [ k ] e jωk e jωn dω from above = ∞ X k =-∞ y * [ k ] 1 2 π ∞ Z-∞ X ( e jω ) e jωk e jωn dω = ∞ X k =-∞ y * [ k ] 1 2 π ∞ Z-∞ X ( e jω ) e jω ( n + k ) dω = ∞ X k =-∞ y * [ k ] x [ n + k ] 1 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 | > 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 ) =...
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Review_Quiz_Slns_Spring_08 - ECE161A : Solutions To Review...

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