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Solution4 - EE 200 Fall 2009(Weber Homework 4 Solutions 1 a...

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EE 200 – Fall 2009 (Weber) Homework 4 Solutions 1. a. The individual frequency responses are given by H 1 ( e j ˆ ω ) = 1 - e - j ˆ ω H 2 ( e j ˆ ω ) = 1 + e - j ˆ ω 2 H 3 ( e j ˆ ω ) = e - j ˆ ω + e - j ˆ ω 2 The overall frequency response is then H ( e j ˆ ω ) = H 1 ( e j ˆ ω ) H 2 ( e j ˆ ω ) H 3 ( e j ˆ ω ) = (1 - e - j ˆ ω )(1 + e - j ˆ ω 2 )( e - j ˆ ω + e - j ˆ ω 2 ) = (1 + e - j ˆ ω 2 - e - j ˆ ω - e - j ˆ ω 3 )( e - j ˆ ω + e - j ˆ ω 2 ) = e - j ˆ ω + e - j ˆ ω 3 - e - j ˆ ω 2 - e - j ˆ ω 4 + e - j ˆ ω 2 + e - j ˆ ω 4 - e - j ˆ ω 3 - e - j ˆ ω 5 = e - j ˆ ω - e - j ˆ ω 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 ˆ ω ) = 1 - e - j ˆ ω 3 = e - j ˆ ω 3 / 2 parenleftBig e j ˆ ω 3 / 2 - e - j ˆ ω 3 / 2 parenrightBig = e - j ˆ ω 3 / 2 (2 j sin(ˆ ω 3 / 2)) = e - j ˆ ω 3 / 2 (2 e jπ/ 2 sin(ˆ ω 3 / 2)) = 2 e - j ω 3 / 2 - π/ 2) sin(ˆ ω 3 / 2) b. For an input frequency of 2 π/ 3 the response is H ( e j 2 π/ 3 ) = 2 e - j ((2 π/ 3)3 / 2 - π/ 2) sin((2 π/ 3)3 / 2) = 2 e - j ( π/ - π/ 2) sin( π ) = 0 The output signal is then y [ n ] = 0.
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