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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 ] = [ n- 2]- [ n- 3] + 25 [ n- 4]- 25 [ n- 5] (c) The z-transform of the input signal x [ n ] = [ n ] + 2 [ n- 1] + 3 [ n- 2] + 2 [ n- 3] + [ 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 ] = [ n- 2]+ [ n- 3]+26 [ n- 4]+24 [ n- 5]+24 [ n- 6]- 26 [ n- 7]- 25 [ n- 8]- 25 [ 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 = 0). Consequently, the only frequency which would be nulled by this system is = 0 (and the corresponding alias frequencies = 2 , 4 ,... ). This can also be determined by evaluating the frequency response function H ( e j ) = e- j 2 ( 1- e- j )( 1 + j 5 e- j )( 1- j 5 e- j ) where the second factor vanishes at = 0....

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