Review_Quiz_Slns_Fall_08

Review_Quiz_Slns_Fall_08 - Review Quiz Solution 1. (a) x[n]...

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Review Quiz Solution 1. (a) x [ n ] is a stable, causal, complex sequence and X ( z ) = n = -∞ x [ n ] z - n , with ROC { z : | z | > r 1 } , with r 1 < 1 . The sequence r [ n ] = k = -∞ x * [ k ] x [ n + k ] . Then R ( z ) = X n = -∞ r [ n ] z - n = X n = -∞ X k = -∞ x * [ k ] x [ n + k ] z - n = X k = -∞ x * [ k ] X n = -∞ x [ n + k ] z - n = X k = -∞ x * [ k ] X l = -∞ x [ l ] z - ( l - k ) = X k = -∞ x * [ k ] X l = -∞ x [ l ] z - l z k = X k = -∞ x * [ k ] z k X ( z ) = X k = -∞ x * [ k ]( z - 1 ) - k X ( z ) = ˆ X k = -∞ x [ k ]( z *- 1 ) - k ! * X ( z ) = X * ( z *- 1 ) X ( z ) = X * ( 1 z * ) X ( z ) Since the ROC associated with X * ( 1 z * ) is { z : | z | < 1 r 1 } , the Region of convergence is equal to ROC = { z : | z | < r 1 } ∩ { z : | z | < 1 r 1 } = { z : r 1 < | z | < 1 r 1 } . (b) r 1 < 1 1 r 1 > 1 . Hence the ROC includes the unit circle and the Fourier transform exists. 2. (a) Since the systems is stable, the region of convergence includes the unit circle.
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This document was uploaded on 05/26/2010.

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Review_Quiz_Slns_Fall_08 - Review Quiz Solution 1. (a) x[n]...

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