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**Unformatted text preview: **Problem 7.1 Our solutions will use the fundamental z-transform pair α n u [ n ] ←→ 1 1- αz- 1 . a) X 1 ( z ) = 1 1- 1 2 z- 1 + z- 1 1- 1 3 z- 1 (sum of rational functions) = 1 + 2 3 z- 1- 1 2 z- 2 1- 5 6 z- 1 + 1 6 z- 2 (ratio of polynomials) = (1- z 1 z- 1 )(1- z 2 z- 1 ) (1- 1 2 z- 1 )(1- 1 3 z- 1 ) (product of factors) , where z 1 =- 1 / 3 + p 11 / 18 ≈ . 4484 and z 2 =- 1 / 3- p 11 / 18 ≈ - 1 . 1151. b) X 2 ( z ) = 1 2 1 1- . 9 z- 1 + 1 1 + 0 . 9 z- 1 (sum of rational functions) = 1 1- . 81 z- 2 (ratio of polynomials) = 1 (1- . 9 z- 1 )(1 + 0 . 9 z- 1 ) (product of factors) c) X 3 ( z ) = 1 2 1 1- . 9 z- 1- 1 1 + 0 . 9 z- 1 (sum of rational functions) = . 9 z- 1 1- . 81 z- 2 (ratio of polynomials) = . 9 z- 1 (1- . 9 z- 1 )(1 + 0 . 9 z- 1 ) (product of factors) d) Notice that x 4 [ n ] = (0 . 9) n u [ n ], so X 4 ( z ) = 1 1- . 9 z- 1 which is the same in all three forms. e) Notice that x 5 [ n ] = (- 1 / 2) n ( e j π 3 n / 2 + e- j π 3 n / 2) u [ n ] = 1 2- 1 2 e j π 3 n +- 1 2 e- j π 3 n 1 ECE 2025 HW #4 Solutions, Summer 2010 – July 9, 2010 and so X 5 ( z ) = 1 2 1 1 + 1 2 e jπ/ 3 z- 1 + 1 1 + 1 2 e- jπ/ 3 z- 1 ! (sum of rational functions) = 1 2 2 + cos( π/ 3) z- 1 1 + cos( π/ 3) z- 1 + 1 4 z- 2 ! = 1 2 2 + 1 2 z- 1 1 + 1 2 z- 1 + 1 4 z- 2 ! (ratio of polynomials) = 1 + 1 4 z- 1 (1 + 1 2 e jπ/ 3 z- 1 )(1 + 1 2 e- jπ/ 3 z- 1 ) (product of factors) 2 ECE 2025 HW #4 Solutions, Summer 2010 – July 9, 2010 Problem 7.2 a) We have H ( z ) = Y ( z ) X ( z ) = 1- 2 . 7 z- 3 + 1 . 8 z- 4 6- . 3 z- 1- . 5 z- 5 and so Y ( z ) = 1 20 z- 1 Y ( z ) + 1 12 z- 5 Y ( z ) + 1 6 X ( z )- 9 20 z- 3 X ( z ) + 3 10 z- 4 X ( z ) ....

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