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hw06su10_soln

# hw06su10_soln - ECE 2025 Summer 2010 Problem Set 6...

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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 z-transform of the input-output 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 z-transform 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 z-transform 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 z-transform 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 z-plane 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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hw06su10_soln - ECE 2025 Summer 2010 Problem Set 6...

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