157aq1 - UNIVERSITY OF CALIFORNIA, DAVIS Department of...

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UNIVERSITY OF CALIFORNIA, DAVIS Department of Electrical and Computer Engineering EEC 157A CONTROL SYSTEMS QUIZ 1 FALL 2005 80 Minutes; Closed Book; 2 pages of notes allowed NAME : ALL PROBLEMS ARE BASED ON THE UNITY-FEEDBACK CONTROL SYSTEM: - R ( s ) + - h - E ( s ) C ( s ) - P ( s ) - Y ( s ) 6 input: r ( t ) , output: y ( t ) error: e ( t ) = r ( t ) - y ( t ) input-output transfer-function: H yr ( s ) = Y ( s ) R ( s ) input-error transfer-function: H er ( s ) = E ( s ) R ( s ) Problem 1 (25 points): The closed-loop step-response is y ( t ) = 0 . 2 - 0 . 6 e - 5 t +0 . 4 e - 6 t , t 0. a) What is the transfer-function H yr ( s ) ? b) Is the system BIBO stable? c) What is the steady-state error e ss due to a unit-step input? a) step-response is given, therefore, R ( s ) = 1 s Y ( s ) = 0 . 2 s - 0 . 6 s + 5 + 0 . 4 s + 6 = 0 . 6 s + 6 s ( s + 5)( s + 6) = H yr ( s ) = Y ( s ) R ( s ) = sY ( s ) = 0 . 6 s + 6 ( s + 5)( s + 6) b) the transfer function H yr is proper and has poles at - 5 , - 6, both in the open left-half-plane . Therefore, BIBO stable. c) E ( s ) = R ( s ) - Y ( s ) = 1 s - 0 . 6 s + 6 s ( s + 5)( s + 6) = s 2 + 10 . 4 s + 24 s ( s + 5)( s + 6) can use Final Value Theorem since system is BIBO stable. e ss = lim t →∞ e ( t ) = lim s 0 sE ( s ) = lim s 0 s s 2 + 10 . 4 s + 24 s ( s + 5)( s + 6) = 4 / 5 alternately, you can find y ss y ss = lim t →∞ y ( t ) = lim s 0 sY ( s ) = lim s 0 s 0 . 6 s + 6 s ( s + 5)( s + 6) = 1 / 5 and e ss = 1 - y ss = 4 / 5 1
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Problem 2 (25 points): The input-output transfer-function is H yr ( s ) = 5 s + 10 s 3 + 4 s 2 + 3 s + 10 . a) What is the steady-state step-response y ss ? b) What is the input-error transfer function H er ( s ) ? c) What is the steady-state error due to a unit-ramp input? d) What is the type of this system? a) y ss = lim t →∞ y ( t ) = lim s 0 sY ( s ) = lim s 0 sH yr ( s ) R ( s ) = lim s 0 s 5 s + 10 s 3 + 4 s 2 + 3 s + 10 · 1 s = 1 b) E ( s ) = R ( s ) - Y ( s ) = H er ( s ) = 1 - H yr ( s ) = 1 - 5 s + 10 s 3 + 4 s 2 + 3 s
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This note was uploaded on 06/23/2008 for the course ECE 157A taught by Professor Chang during the Spring '08 term at Case Western.

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157aq1 - UNIVERSITY OF CALIFORNIA, DAVIS Department of...

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