Exam 2 Solution Spring 209

Exam 2 Solution Spring 209

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Unformatted text preview: his is a highpass filter. The z-transform of the time-domain representation is Y.z/ D X.z/ z 1 X.z/ D X.z/.1 z 1 /, so the system function is H.z/ D Y.z/ D1 X.z/ z 1 : The frequency response is H.e j! / D H.z/jzDe j! D 1 j! e ; so the squared magnitude response is jH.e j! /j2 D .1 e j! /.1 2 j! .e C e 2 Thus we have a highpass filter characteristic: D2 e j! / D 1 j! /D2 e j! e j! C1 2 cos.!/: |H(ej ω)|2 4 3 2 1 0 −6 −4 −2 0 ω 2 4 6 or 2 |H(ej ω)| 1.5 1 0.5 0 −6 −4 −2 0 ω 2 4 6 Fo...
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This note was uploaded on 11/14/2013 for the course EEE 4001F taught by Professor Nicolls during the Summer '13 term at University of Cape Town.

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