SampleMidterm #1

SampleMidterm #1 - ECE 483 Midterm 1 Solution Problem 1(20...

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Unformatted text preview: ECE 483 Midterm 1 Solution Problem 1 (20 points) Given the sample period T = 0 . 5 seconds. Find the z-transform E ( z ) of a continuous-time signal with the following Laplace transform: E ( s ) = 1- e- s- e- 1 . 1 s s 2 + 4 . Solution: E ( z ) = Z bracketleftbigg 1- e- 2 sT s 2 + 4- e- 2 . 2 sT s 2 + 4 bracketrightbigg = (1- z- 2 ) Z bracketleftbigg 1 s 2 + 4 bracketrightbigg- z- 2 Z m bracketleftbigg 1 s 2 + 4 bracketrightbigg m =0 . 8 = (1- z- 2 ) . 5 z sin(1) z 2- 2 z cos(1) + 1- z- 2 . 5 z sin(0 . 8) + 0 . 5sin(0 . 2) z 2- 2 z cos(1) + 1 = ( z 2- 1) sin(1)- sin(0 . 8)- z- 1 sin(0 . 2) 2 z ( z 2- 2 z cos(1) + 1) . Problem 2 (25 points) Consider the following discrete-time linear system: y ( k )- 1 . 3 y ( k- 1) + 0 . 3 y ( k- 2) = e ( k- 1)- . 5 e ( k- 2) , k = 0 , 1 , 2 ,... . (a) (5 pts) What is the transfer function from the input e ( k ) to the output y ( k ), assuming zero initial conditions: y (- 1) = y (- 2) = e (- 1) = e (- 2) = 0? Solution: Taking the z transform of both sides, we have Y ( z ) E ( z ) = z- 1- . 5 z- 2 1- 1 . 3 z- 1 + 0 . 3 z- 2 = z- . 5 ( z- 1)( z- . 3) ....
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This note was uploaded on 04/15/2011 for the course ECE 483 taught by Professor Evens during the Spring '08 term at Purdue.

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SampleMidterm #1 - ECE 483 Midterm 1 Solution Problem 1(20...

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