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**Unformatted text preview: **UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN Department of Electrical and Computer Engineering ECE 410 Digital Signal Processing Quiz 4 Solutions Thursday, October 16, 2008 Student Name: Section: D. Jones (1pm), Y. Bresler (3pm) Problem 1 (10 points) Determine the two-sided z -transform for each of the following sequences. Specify the poles, zeros, and the region of convergence of the z-transform in each case. (a) (5 points) x 1 [ n ] = ( 1 2 ) | n | (b) (5 points) x 2 [ n ] = ( 1 2 ) | 5( n- 2) | Solution: (a) x 1 [ n ] = 1 2 | n | = 1 2 u [ n ] + 2 n u [- n- 1] ⇒ X 1 ( z ) = z z- 1 2- z z- 2 = (- 3 2 ) z ( z- 2)( z- 1 2 ) , ROC: 1 2 < | z | < 2 X 1 ( z ) has a zero at z = 0 and poles at z = 2 and z = 0 . 5. (b) Define ˜ x 2 [ n ] = 1 2 | 5 n | = 1 32 u [ n ] + 32 n u [- n- 1] ⇒ ˜ X 1 ( z ) = z z- 1 32- z z- 32 = (- 32 + 1 32 ) z ( z- 32)( z- 1 32 ) , ROC: 1 32 < | z | < 32 Now x 2 [ n ] 4 = ˜ x 2 [ n- 2] ⇒ X 2 ( z ) = z- 2 ˜ X 2 ( z ) = (- 32 + 1 32 ) z ( z- 32)( z- 1 32 ) , ROC: 1 32 < | z | < 32 1 We can also derive the same result directly from the definition of z-transform: x 2 [ n ] = 1 2 | 5( n- 2) | ⇒ X 2 ( z ) = ∞ X n =-∞ 1 2 | 5( n- 2) | z- n = z- 2 ∞ X n =-∞ 1 2 5 | n- 2 | z- ( n- 2) = z- 2 ∞ X m =-∞ 1 32 | m | z- m = z- 2 " ∞ X m =0 1 32 m z- m +- 1 X m =-∞ 32 m z- m # = z- 2 " ∞ X m =0 1 32 z m + ∞ X m =1 z 32 m # = z- 2 " 1 1- 1 32 z + z 32 1- z 32 # , 1 32 | z | < 1; | z | 32 < 1 = z- 2 " z z- 1 32 + z 32- z # , ROC: 1 32 < | z | < 32 = (- 32 + 1 32 ) z ( z- 32)( z- 1 32 ) , ROC: 1 32 < | z | < 32 Hence, X 2 ( z ) has no zeros and poles at z = 0, z = 32 and z = 1 32 . 2 Problem 2 (10 points) Let the algebraic form of a system transfer function H ( z ) be: H ( z ) = z- 1 1- 3 2 z- 1- z- 2 (a) Determine all possible ROCs for H ( z ). (b) For each possible choice of ROC, specify if the system has a right-sided, left-sided, or two- sided unit pulse response, and if it is BIBO stable. ( Note: You are not required to compute the unit pulse response. ) Solution: H ( z ) = z ( z- 2)( z + 1 2 ) H ( z ) has poles at z = 2 and z =- 1 2 . Therefore H ( z ) has three possible ROCs ROC h [ n ] BIBO stable | z | < 1 2 Left-sided...

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