Solution5

# Solution5 - EE 200 Fall 2009(Weber Homework 5 Solutions 1 a...

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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 ± 0 . 52 j b. y [ n ] = - 0 . 9 y [ n - 6] + x [ n ] Y ( z ) = - 0 . 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 = - 0 . 9 z = (0 . 9) 1 / 6 e j ( π 6 + n π 3 ) where n = 0 , . . . , 5 6 1

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2. a. Y ( z ) = 0 . 8 z - 1 Y ( z ) - 0 . 8 X ( z ) + z - 1 X ( z ) H ( z ) = Y ( z ) X ( z ) = z - 1 - 0 . 8 1 - 0 . 8 z - 1 = 1 - 0 . 8 z z - 0 . 8 b. Zero at z = 1 . 25, pole at z = 0 . 8 c. H ( e j ˆ ω ) = e - j ˆ ω - 0 . 8 1 - 0 . 8 e - j ˆ ω = 1 - 0 . 8 e j ˆ ω e j ˆ ω - 0 . 8 d. vextendsingle vextendsingle H ( e j ˆ ω ) vextendsingle vextendsingle 2 = H ( e j ˆ ω ) H ( e j ˆ ω ) * = e - j ˆ ω - 0 . 8 1 - 0 . 8 e - j ˆ ω e j ˆ ω - 0 . 8 1 - 0 . 8 e j ˆ ω = 1 - 0 . 8 e - j ˆ ω - 0 . 8 e j ˆ ω + 0 . 64 1 - 0 . 8 e j ˆ ω - 0 . 8 e - j ˆ ω + 0 .. 64 = 1 e. The filter will have no effect on the magnitude of the input signal, but will change the phase of the frequencies.
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