soln6 - EECS 451 SOLUTIONS TO PROBLEM SET#6 1 Input x(n...

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EECS 451 SOLUTIONS TO PROBLEM SET #6 1. Input x ( n ) clearly has frequency components only at ω = 0; ω = ± π 2 ; and ω = π . Impulse resp. h ( n ) = 1 2 ( δ ( n ) - δ ( n - 2)) transfer function H ( e ) = 1 2 (1 - e - 2 ). ω = 0 : X ( e j 0 ) = 5 e j 00 o ; H ( e j 0 ) = 0 Y ( e j 0 ) = 0. Recall EECS 210? ω = π 2 : X ( e j π 2 ) = 3 e j 60 o ; H ( e j π 2 ) = 1 Y ( e j π 2 ) = 3 e j 60 o ; Y ( e - j π 2 ) = 3 e - j 60 o . ω = π : X ( e ) = 4 e - j 45 o ; H ( e ) = 0 Y ( e ) = 0 . y ( t ) = 3 cos( π 2 n + 60 o ). 2a. H ( z ) = C z - 1 z +0 . 9 for some constant C . No constant signals pass zero at z = 1. High-pass with 1 pole pole= | p | e = - 0 . 9. 2c. 1 = H ( e ) = C e - 1 e +0 . 9 = - 2 - . 1 C = 1 20 . 2d. H ( z ) = 1 20 1 - z - 1 1+0 . 9 z - 1 = Y ( z ) X ( z ) . Cross-multiply: Y ( z )(1 + 0 . 9 z - 1 ) = 1 20 X ( z )(1 - z - 1 ) y ( n ) + 0 . 9 y ( n - 1) = 1 20 x ( n ) - 1 20 x ( n - 1). 2e.

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soln6 - EECS 451 SOLUTIONS TO PROBLEM SET#6 1 Input x(n...

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