Solution4

Solution4 - EE 200 Fall 2009 (Weber) Homework 4 Solutions...

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Unformatted text preview: EE 200 Fall 2009 (Weber) Homework 4 Solutions 1. a. The individual frequency responses are given by H 1 ( e j ) = 1- e- j H 2 ( e j ) = 1 + e- j 2 H 3 ( e j ) = e- j + e- j 2 The overall frequency response is then H ( e j ) = H 1 ( e j ) H 2 ( e j ) H 3 ( e j ) = (1- e- j )(1 + e- j 2 )( e- j + e- j 2 ) = (1 + e- j 2- e- j - e- j 3 )( e- j + e- j 2 ) = e- j + e- j 3- e- j 2- e- j 4 + e- j 2 + e- j 4- e- j 3- e- j 5 = e- j - e- j 5 b. The difference equation for the cascaded systems is y [ n ] = x [ n- 1]- x [ n- 5] 2. a. From the impulse response: H ( e j ) = 1- e- j 3 = e- j 3 / 2 parenleftBig e j 3 / 2- e- j 3 / 2 parenrightBig = e- j 3 / 2 (2 j sin( 3 / 2)) = e- j 3 / 2 (2 e j/ 2 sin( 3 / 2)) = 2 e- j ( 3 / 2- / 2) sin( 3 / 2) b. For an input frequency of 2...
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This note was uploaded on 02/26/2010 for the course EE 30446 at USC.

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Solution4 - EE 200 Fall 2009 (Weber) Homework 4 Solutions...

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