Quiz_3_sol

# Quiz_3_sol - ECE 483 Quiz 3 Solution Problem 1(6 points...

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Unformatted text preview: ECE 483 Quiz # 3 Solution Problem 1. (6 points) Suppose T = 0.1 s. Find E ∗ (s) and E (z ) for the following E (s): s+3 . E (s) = s(s + 2) (You do NOT need to simplify the answer.) Solution: Using the formula E ∗ (s) = 0,−2 Residue λ+3 1.5 0.5 1 = − . −T s −T (s−λ) λ(λ + 2) 1 − e 1−e 1 − e−T (s+2) Therefore, E (z ) = 0.5 1.5z 1.5 0.5z − = . − −1 −1 e−2T 1−z 1−z z − 1 z − e−0.2 Problem 2. (14 points) For both parts, assume T = 0.5 s for the samplers and the data holds are the 0-th order data hold. Suppose the input e(t) = e−t , t ≥ 0. (a) (7 pts) Find C1 (s) and C1 (z ). 1 Solution: Since E (s) = s+1 , we have E (z ) = Z [E (s)] = E ∗ (s) = E (z ) Therefore, z →esT z z − e −T , hence = esT . esT − e−T 2 1 − e−sT 2 esT − 1 · · E ∗ (s) = 2 · sT . s s s e − e−T Since the second factor is already a star transform, we can take the star transform and then convert to z -transform as: 2 z−1 2T z z−1 2T z z C1 (z ) = Z 2 · = · = = . −T 2 z − e−T −T ) s z−e (z − 1) (z − 1)(z − e (z − 1)(z − e−0.5 ) C1 (s) = (b) (7 pts) Find C2 (s) and C2 (z ). Solution: Note that F (s) = 2 E (s) = s F (z ) = Z 2 s(s+1) . Thus, 2 2z (1 − e−T ) = , s(s + 1) (z − 1)(z − e−T ) 1 − e−sT ∗ 1 − e−sT 2esT (1 − e−T ) 2 1 − e−T C2 (s) = F (s) = · sT = · sT . s s (e − 1)(esT − e−T ) s e − e−T Taking the z -transform to obtain: C2 (z ) = Z 2 1 − e−T 2z 1 − e−T 2(1 − e−0.5 )z · = · = . s z − e−T z − 1 z − e−T (z − 1)(z − e−0.5 ) 1 ...
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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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