上更大半 SHANGHAI JIAO TONG UNIVERSITY Chapter 5: Root Locus Nov.13-15,2012
Chapter 5: Root Locus Nov. 13-15, 2012
Two Conditions for Plotting root locus Given open-loop transfer function Gk(S) Characteristic equation K4(+ ∏(+p,) Magnitude Condition and argument Condition II(s+p) J=1 K ∠∑ (s+)-2∑(s+p1)=±(2k+1)z,k=012 i=1 S+Z 上潇人字 SHANGHAI ILAO TONG UNIVIRSITY
Two Conditions for Plotting Root Locus ( ) 1 1 ( ) 1 ( ) m g i i k n j j K s z G s s p = = + = = − + Given open-loop transfer function Gk (s) Characteristic equation = = + + = m i i n j j g s z s p K 1 1 | ( )| | ( )| ( ) ( ) (2 1) , 0,1,2.. 1 1 + − + = + = = = s z s p k k n j j m i i Magnitude Condition and Argument Condition
Rules for plotting root locus Content Rules 1 Continuity and Symmetry Symmetry rule Starting and end points n segments start from n open-loop Number of segments poles, and end at m open-loopzeros and(n-m) zeros at infinity Segments on real axis On the left of an odd number of poles or zeros Asymptote n-m segments N(2k+1) 丌,k=0,±1,+2, n-m 5 Asymptote ∑ (-p)-∑(-=) 上大字
Content Rules 1 Continuity and Symmetry Symmetry Rule 2 Starting and end points Number of segments n segments start from n open-loop poles, and end at m open-loop zeros and (n-m) zeros at infinity. 3 Segments on real axis On the left of an odd number of poles or zeros 4 Asymptote n-m segments： 5 Asymptote 3 Rules for Plotting Root Locus n m p z n j m i j i − − − − = =1 =1 ( ) ( ) , 0, 1, 2, (2 1) = − + k n m k ＝
Breakaway dIF(s)1-0 F(S)=P(s)+K,Z(s)=o and break -in as points P(s)2(s)-P(sz()=0 P1 Angle of emergence Angle of 千(2k+)+∑-∑ 7 emergence and Angle of entry J≠P entry ±(k+1)+∑-∑ 8 Cross on the Substitute s=ja to characteristic equation imaginary axis and solve Routh’ s formula 上潇人字 SHANGHAI ILAO TONG UNIVIRSITY
4 7 Angle of emergence and entry Angle of emergence Angle of entry 8 Cross on the imaginary axis Substitute s = j to characteristic equation and solve Routh’s formula = = = + + − n j m i z i z j i k 1 1 (2 1) = = = + + − m i n j p j p i j k 1 1 (2 1) 6 Breakaway and break-in points ( ) 0 [ ] = ds d F s F(s) = P(s)+ Kg Z(s) = 0 P(s)Z(s)− P(s)Z(s) = 0 = = − = − m i n i j pi 1 z 1 1 1
Rule 6: Brea kaway and break-in Points on the Realaxis Break-in point O Breakaway point Use the following necessary condition dg(sh(s-o of dsL(s)H(s) dk 0 or B=0 ds ds P61(6)P6)=02 S+Ei Stpi 上潇人字 SHANGHAI ILAO TONG UNIVIRSITY 2021/1/26
2021/1/26 5 j Break-in point j Breakaway point Rule 6: Breakaway and Break-in Points on the Real Axis Use the following necessary condition P(s)Z(s)− P(s)Z(s) = 0 ( ) ( ) ( ) ( ) 0 d d 0 or 1 d d 0 or d d = = = s K s s G s H s G s H s g i pj s z s + = + 1 1