Final_Fa10

Final_Fa10 - ECE 483 Final Exam Solution Problem 1 (15...

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Unformatted text preview: ECE 483 Final Exam Solution Problem 1 (15 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 . (b) (5 pts) Find the transfer function Y ( z ) U ( z ) . (c) (5 pts) Suppose the input u ( k ) is a unit step signal. Determine y ( ∞ ), the final value of the output y ( k ), if it exists. If the final value does not exist, state your reason. Solution: (a) Identify the outputs of the left and right time-delay units as x 1 ( k ) and x 2 ( k ), respectively. x ( k + 1) = bracketleftbigg- . 2- 1 1- . 5 bracketrightbigg x ( k ) + bracketleftbigg 1 2 bracketrightbigg u ( k ) y ( k ) = bracketleftbig 0 1 bracketrightbig x ( k ) . (b) The transfer function is Y ( z ) U ( z ) = C ( zI- A ) − 1 B + D = bracketleftbig 0 1 bracketrightbig bracketleftbigg z + 0 . 2 1- 1 z + 0 . 5 bracketrightbigg − 1 bracketleftbigg 1 2 bracketrightbigg = bracketleftbig 0 1 bracketrightbig 1 ( z + 0 . 2)( z + 0 . 5) + 1 bracketleftbigg z + 0 . 5- 1 1 z + 0 . 2 bracketrightbiggbracketleftbigg 1 2 bracketrightbigg = 2 z + 1 . 4 z 2 + 0 . 7 z + 1 . 1 (c) y ( ∞ ) does not exist since Y ( z ) U ( z ) is not stable. Problem 2 (10 points). Suppose T = 0 . 5 s, the data hold is the zero-order data hold, and G p ( s ) = s + 1 s ( s + 2) . (a) (6 pts) Find G ( z ) = Z bracketleftBig 1 − e- sT s G p ( s ) bracketrightBig . (b) (2 pts) Find the closed-loop transfer function C ( z ) R ( z ) . You may write the answer in terms of G ( z ) in part (a) without further simplification. (c) (2 pts) Find the steady-state errors e ss of the closed-loop system when tracking the unit step input and the unit ramp input, respectively. Solution: (a) Since G p ( s ) s = . 25 s + . 5 s 2- . 25 s +2 , we have G ( z ) = (1- z − 1 ) Z bracketleftbigg G p ( s ) s bracketrightbigg = z- 1 z bracketleftbigg . 25 z z- 1 + . 25 z ( z- 1) 2- . 25 z z- e − 1 bracketrightbigg = 0 . 25 + . 25 z- 1- . 25( z- 1) z- e − 1 (b) The closed-loop transfer function is H ( z ) = C ( z ) R ( z ) = G ( z ) 1 + G ( z ) , where G ( z ) is as computed in part (a).) is as computed in part (a)....
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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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Final_Fa10 - ECE 483 Final Exam Solution Problem 1 (15...

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