HW5 solution

# HW5 solution - EML4312 Spring 2011 HOMEWORK 5 SOLUTION...

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EML4312 Spring 2011 HOMEWORK 5 SOLUTION Lead/Lag Control via MC/AC and Coe¢ cient Matching The due date for this assignment is Friday 3/14. 1. For the following plant model: G p ( s ) = 1 s ( s + 1) ; develop a controller to place closed-loop poles at s 1 ; 2 = 3 ± 5 j; with a steady state error of E ss = 0.01 (for a step input). (a) (10 points) MC/AC Check if closed-loop poles satisfy AC 6 s 6 s + 1 j s = 3+5 j = 6 ( 3 + 5 j ) 6 ( 2 + 5 j ) = 120 : 96 111 : 80 = 232 : 76 6 = 180 + 360 n Use root locus to see if we need lead or lag compensator. Since the closed-loop poles lie on the left side of the current roots, we need lead compensator. Place the zero of the compensator directly below the desired location at s = 3 . G c ( s ) = K ( s + 3) ( s + p ) : Using AC: 6 s + 3 6 s + p 6 s 6 s + 1 j s = 3+5 j = 6 5 j 6 p 3 + 5 j 6 ( 3 + 5 j ) 6 ( 2 + 5 j ) = 90 tan 1 ( 5 p 3 ) 120 : 96 111 : 80 = 180 + 360 n choose n = 1 ! tan 1 ( 5 p 3 ) = 37 : 24 ! p = 9 : 58 Using MC: K s + 3 s ( s + 1)( s + 9 : 58) s = 3+5 j = 1 ! K 5 j ( 3 + 5 j )( 2 + 5 j )(6 : 58 + 5 j ) = 1 ! K = p 5 2 + 3 2 p 5 2 + 2 2 p 5 2 + 6 : 58 2 5 = 51 : 9 Check steady-state error requirement: E ss = lim s ! 0 sE ( s ) = lim s ! 0 sR ( s ) 1 + G c ( s ) G p ( s ) (since R ( s ) = 1 s ) = lim s ! 0 1 1 + 51 : 9( s +3) s ( s +1)( s +9 : 58) = lim s ! 0 s ( s + 1)( s + 9 : 58) s ( s + 1)( s + 9 : 58) + 51 : 9( s + 3) = 0 < 0 : 01 : Hence, G c ( s ) = 51 : 9 s + 3 s + 9 : 58 : 1

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(b) Coe¢ cient Matching (10 points) G c ( s ) = K s + z s + p ! G c ( s ) G p ( s ) = K ( s + z ) s ( s + 1)( s + p ) : A = s ( s + 1)( s + p ) + K ( s + z ) = s 3 + ( p + 1) s 2 + ( p + K ) s + Kz D = ( s + 3 5 j )( s + 3 + 5 j )( s + a ) = ( s 2 + 6 s + 34)( s + a ) = s 3 + (6 + a ) s 2 + (34 + 6 a ) s + 34 a By coe¢ cient matching and assuming z = 3 ; p + 1 = 6 + a p + K = 34 + 6 a ! K = 5 a + 29 3 K = 34 a ! a = 4 : 58 ; K = 51 : 9 ; p = 9 : 58 Similarly check stead-state error requirement. Hence G c ( s ) = 51 : 9 s + 3 s + 9 : 58 : 2. For the following plant model: G p ( s ) = 1 ( s + 1)( s + 2) ; develop a controller to place closed-loop poles at s 1 ; 2 = 3 ± j; with a steady state error of E ss = 0.01 (for a step input). (a) (10 points) MC/AC Check if closed-loop poles satisfy AC 6 s + 1 6 s + 2 j s = 3+ j = 6 ( 2 + j ) 6 ( 1 + j ) = 153 : 43 135 = 288 : 43 6 = 180 + 360 n Use root locus to see if we need lead or lag compensator. Since the closed-loop poles lie on the left side of the current roots, we need lead compensator. Place the zero of the compensator at s = 2 : 5 . G c ( s ) = K ( s + 2 : 5) ( s + p ) : Using AC: 6 s + 2 : 5 6 s + p 6 s + 1 6 s + 2 j s = 3+ j = 6 & 0 : 5 + j 6 p 3 + j 6 ( 2 + j ) 6 ( 1 + j ) = 116 : 57 tan 1 ( 1 p 3 ) 153 : 43 135 = 180 + 360 n choose n = 1 ! tan 1 ( 1 p 3 ) = 8 : 14 ! p = 9 : 99 : 2
Using MC: K s + 2 :

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## This note was uploaded on 09/05/2011 for the course EML 4312 taught by Professor Dixon during the Spring '07 term at University of Florida.

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HW5 solution - EML4312 Spring 2011 HOMEWORK 5 SOLUTION...

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