Solution5

Solution5 - EE 200 – Fall 2009(Weber Homework 5 Solutions...

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Unformatted text preview: EE 200 – Fall 2009 (Weber) Homework 5 Solutions 1. a. y [ n ] = 1 2 y [ n- 1]- 1 3 y [ n- 2]- x [ n ] + 3 x [ n- 1]- 2 x [ n- 2] Y ( z ) = 1 2 z- 1 Y ( z )- 1 3 z- 2 Y ( z )- X ( z ) + 3 z- 1 X ( z )- 2 z- 2 X ( z ) H ( z ) = Y ( z ) X ( z ) =- 1 + 3 z- 1- 2 z- 2 1- 1 2 z- 1 + 1 3 z- 2 =- z 2 + 3 z- 2 z 2- 1 2 z + 1 3 Zeros:- z 2 + 3 z- 2 = 0 → z = 1 , z = 2 Poles: z 2- 1 2 z + 1 3 = 0 → z = 1 4 ± j 1 4 radicalBig 13 3 = 0 . 25 ± . 52 j b. y [ n ] =- . 9 y [ n- 6] + x [ n ] Y ( z ) =- . 9 z- 6 Y ( z ) + X ( z ) H ( z ) = Y ( z ) X ( z ) = 1 1 + 0 . 9 z- 6 = z 6 z 6 + 0 . 9 Zeros: six at z = 0 Poles: z 6 =- . 9 → z = (0 . 9) 1 / 6 e j ( π 6 + n π 3 ) where n = 0 , . . ., 5 6 1 2. a. Y ( z ) = . 8 z- 1 Y ( z )- . 8 X ( z ) + z- 1 X ( z ) H ( z ) = Y ( z ) X ( z ) = z- 1- . 8 1- . 8 z- 1 = 1- . 8 z z- . 8 b. Zero at z = 1 . 25, pole at z = 0 . 8 c. H ( e j ˆ ω ) = e- j ˆ ω- . 8 1- . 8 e- j ˆ ω = 1- . 8 e j ˆ ω e j ˆ ω- . 8 d....
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This note was uploaded on 02/26/2010 for the course EE 30446 at USC.

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Solution5 - EE 200 – Fall 2009(Weber Homework 5 Solutions...

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