1 - 1/29/10 ECE 483 Homework #1 Solution Problem 1. Problem...

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Unformatted text preview: 1/29/10 ECE 483 Homework #1 Solution Problem 1. Problem 2-3. Note that the closed-form expressions of the z-transforms should be in terms of the constants a and T in addition to z . Solution: (a) E ( z ) = z z- e- aT . (b) E ( z ) = 1 z- e- T . (c) E ( z ) = z- 4 z- e- T . Problem 2. Given the z-transform of a discrete-time signal e ( k ), k = 0 , 1 , . . . , as follows: E ( z ) = . 5( z + 1) z ( z . 6) . (a) Use the inversion formula method to find e (0), e (1), and e (2). (b) Use the long-division method to find e (0), e (1), and e (2). (c) Use the partial fraction expansion method to find e ( k ) for k = 0 , 1 , . . . . (d) Use Matlab to find and plot e ( k ) for k = 0 , 1 , . . . . Solution: (a) We compute e (0) = summationdisplay bracketleftbigg Residue of . 5( z + 1) z 2 ( z . 6) bracketrightbigg = . 8 ( z . 6) 2 vextendsingle vextendsingle vextendsingle vextendsingle z =0 + . 5( z + 1) z 2 vextendsingle vextendsingle vextendsingle vextendsingle z =0 . 6 = 0 e (1) = summationdisplay bracketleftbigg Residue of . 5( z + 1) z ( z . 6) bracketrightbigg = . 5( z + 1) z . 6 vextendsingle vextendsingle vextendsingle vextendsingle z =0 + . 5( z + 1) z vextendsingle vextendsingle vextendsingle vextendsingle z =0 . 6 = 0 . 5 e (2) = summationdisplay bracketleftbigg Residue of . 5( z + 1) z . 6 bracketrightbigg = 0 . 5( z + 1) vextendsingle vextendsingle z =0 . 6 = 0 . 8 ....
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1 - 1/29/10 ECE 483 Homework #1 Solution Problem 1. Problem...

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