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# SampleFinal - ECE 483 Final Exam Solution T T 1 0.5 u k y k...

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Unformatted text preview: 05/03/10 ECE 483 Final Exam Solution T T 1 0.5 u ( k ) y ( k ) + + + +- Problem 1 (10 points). Consider the above simulation diagram with two time delay units. (a) (5 pts) Find a state space representation of the system. Identify the matrices A,B,C,D . Solution: Choose the outputs of the top and bottom time delay units as x 1 ( k ) and x 2 ( k ): x ( k + 1) = bracketleftbigg- 1 0 . 5 1 bracketrightbigg x ( k ) + bracketleftbigg 1 . 5 bracketrightbigg u ( k ) y ( k ) = bracketleftbig 0 0 . 5 bracketrightbig x ( k ) + 0 . 5 u ( k ) . (b) (5 pts) Find the transfer function Y ( z ) U ( z ) . Solution: Y ( z ) U ( z ) = C ( zI- A ) − 1 B + D = . 75 z 2 + z- . 5 + 0 . 5 . D 1 (z) R 1 ( s ) G 1 ( s ) R 2 ( s ) G 2 ( s ) H 2 ( s ) T H 1 ( s ) T C 1 ( s ) C 2 ( s ) +-- +-- 1 Problem 2 (10 points). Assume R 2 ( s ) = 0 in the above system. Find the transfer function C 1 ( z ) R 1 ( z ) . If such a transfer function does not exist, find C 1 ( z ) instead. Solution: Denote the input to the top sampler as E 1 ( s ). Then C 1 = G 1 D ∗ 1 E ∗ 1 ⇒ C ∗ 1 = G ∗ 1 D ∗ 1 E ∗ 1 E 1 = R 1- C 2- H 1 C 1 = R 1- C 2- D ∗ 1 G 1 H 1 E ∗ 1 ⇒ E ∗ 1 = R ∗ 1- C ∗ 2- D ∗ 1 [ G 1 H 1 ] ∗ E ∗ 1 C 2 = G 2 ( R 2- H 2 C ∗ 2- C 1 ) = G 2 (- H 2 C ∗ 2- G 1 D ∗ 1 E ∗ 1 ) ⇒ C ∗ 2 =- [ G 2 H 2 ] ∗ C ∗ 2- [ G 1 G 2 ] ∗ D ∗ 1 E ∗ 1 Eliminating the intermediate variables E 1 and C 2 , we have C 1 ( z ) R 1 ( z ) = G 1 ( z ) D 1 ( z )[1 + G 2 H 2 ( z )] 1 + D 1 ( z ) G 1 H 1 ( z ) + G 2 H 2 ( z ) + D 1 ( z ) G 1 H 1 ( z ) G 2 H 2 ( z )- G 1 G 2 ( z ) D 1 ( z ) ....
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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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SampleFinal - ECE 483 Final Exam Solution T T 1 0.5 u k y k...

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