This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 timedelay 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 zeroorder 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 closedloop 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 steadystate errors e ss of the closedloop 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 closedloop 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)....
View
Full
Document
This note was uploaded on 04/15/2011 for the course ECE 483 taught by Professor Evens during the Spring '08 term at Purdue.
 Spring '08
 evens

Click to edit the document details