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Unformatted text preview: ECE 2025 Summer 2010, Problem Set 6, Solutions 6.1 We are given a system function that simplifies according to H ( z ) = z 2 ( 1 z 1 )( 1 + j 5 z 1 )( 1 j 5 z 1 ) = z 2 ( 1 z 1 )( 1 + 25 z 2 ) = z 2 ( 1 z 1 + 25 z 2 25 z 3 ) = z 2 z 3 + 25 z 4 25 z 5 (a) The difference equation is obtained by taking the inverse ztransform of the inputoutput relation Y ( z ) = ( z 2 z 3 + 25 z 4 25 z 5 ) X ( z ) which yields y [ n ] = x [ n 2] x [ n 3] + 25 x [ n 4] 25 x [ n 5] (b) The impulse response is obtained by taking the inverse ztransform of the system function H ( z ) = z 2 z 3 + 25 z 4 25 z 5 which yields h [ n ] = δ [ n 2] δ [ n 3] + 25 δ [ n 4] 25 δ [ n 5] (c) The ztransform of the input signal x [ n ] = δ [ n ] + 2 δ [ n 1] + 3 δ [ n 2] + 2 δ [ n 3] + δ [ n 4] is given by X ( z ) = 1 + 2 z 1 + 3 z 2 + 2 z 3 + z 4 so the output signal is obtained by taking the inverse ztransform of Y ( z ) = ( z 2 z 3 + 25 z 4 25 z 5 )( 1 + 2 z 1 + 3 z 2 + 2 z 3 + z 4 ) = z 2 + z 3 + 26 z 4 + 24 z 5 + 24 z 6 26 z 7 25 z 8 25 z 9 which yields y [ n ] = δ [ n 2]+ δ [ n 3]+26 δ [ n 4]+24 δ [ n 5]+24 δ [ n 6] 26 δ [ n 7] 25 δ [ n 8] 25 δ [ n 9] (d) The system function is equal to H ( z ) = z 2 ( 1 z 1 )( 1 + j 5 z 1 )( 1 j 5 z 1 ) The zeros of the system are located at points in the zplane at which H ( z ) = 0. In this case, there are three zeros located at z = 1 , ± j 5, and only the zero located at z = 1 is on the unit circle (at angle ˆ ω = 0). Consequently, the only frequency which would be nulled by this system is ˆ ω = 0 (and the corresponding alias frequencies ˆ ω = ± 2 π, ± 4 π,... ). This can also be determined by evaluating the frequency response function H ( e j ˆ ω ) = e j 2ˆ ω ( 1 e j ˆ ω )( 1 + j 5 e j ˆ ω )( 1 j 5 e j ˆ ω ) where the second factor vanishes at ˆ ω = 0....
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 Summer '08
 JUANG
 Digital Signal Processing, Signal Processing, LTI system theory, Complex number, Impulse response, Maximum length sequence

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