上海交通大学：《自动控制理论》课程教学资源（PPT课件讲稿）Chapter 05 根轨迹分析法 Root Locus

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

Example 5.3.1: Given the open-loop transfer function please draw the root locus. G(SH(S k(s+3) (S+1)(S+2 ↑/o k=5.18°poek=02 ±0(2k+=士1801 2± Break (6如如马平2 breaka way 2)+3 s+pn-m a+1a+2 k at 上潇人字 SHANGHAI ILAO TONG UNIVIRSITY 2021/1/26

2021/1/26 6 3.5.Asymptote Symmetry 6.Breakaway and break-in points 1.Draw the open 4. Segments on real axis -loop poles and zeros -3 -2 -1 2. Two segments pole pole zero   180 2 1 180 (2 1) =  −  + = k  ( ) ( ) 0 2 1 1 1 ( 1 2) 3 = − − − + = − − − − =   = = n m p z n j m i j i  breaka way Break￾in a = −3 2 a a − a + − = − 2 1 1 1 3 1 Example 5.3.1: Given the open-loop transfer function, please draw the root locus. ( 1)( 2) ( 3) ( ) ( ) + + + = s s k s G s H s 3 1 2 1 1 + + + = + + =   = = a a a s z s p k m i i n j j k =5.818 k =0.172

Example 5.3. 2: Given the open-loop transfer function GK(S) K(S+ Ja S please prove that the root locus in the complex plane is a circle. Conclusion: For the open-loop transfer function with one zero and two poles, the root locus of characteristic equation is probably a circle in the complex plane. 上潇人字 SHANGHAI ILAO TONG UNIVIRSITY

2 * ( 1) ( ) s K s G s k + = Example 5.3.2: j  0 s1 s2 Conclusion: For the open-loop transfer function with one zero and two poles, the root locus of characteristic equation is probably a circle in the complex plane. please prove that the root locus in the complex plane is a circle. Given the open-loop transfer function

Example 5.3.3: K G(S)H(S) S(S+1)(S+2) O i1.414 y化尽飞+≡的 60°180° K=6 solution、S 0花=-158 0+1+2 ON K 632-R 3s2+6=0 K=6 K=6 K -j1.414 上潇人字 SHANGHAI ILAO TONG UNIVIRSITY

2021/1/26 8 Example 5.3.3: ( 1)( 2) ( ) ( ) + + = s s s K G s H s 1. Open-loop poles and zeros -2 -1 2. Segments on real axis 3. Asymptotes 1 3 0 0 1 2 60 ,180 3 0 180 (2 1) = − − + + = − =  −  + =      k 4. Breakaway and break-in points ( ) ( ) ( ) ( ) 0.42 1.58 [3 6 2] 0 0 1 2 2 = − = − + + = − = solution s s yields s s P s Z s P s Z s   -0.42 5. Points across the imaginary axis s K K s s K s s s s K 0 1 2 3 3 2 3 6 3 1 2 3 2 0 − + + + = K=6 3 6 0 2 s + = j1.414 K=6 K=6 -j1.414

Example 5.3.4 k G(SH(S=- (s+1)2(s+1+18s+1-j18 across tie miami points s4+4s3+24s2+40s+19+k=0 O 18 2419+k i3.16 40 Breakaway 3 14 19+k 484-4k 14 14s2+140=0 19+ks=±j3.16 3. × 18 4844k=0ge=121

2021/1/26 9 3. Symmetry 4. No segments on real axis 6. Breakaway and break 5. Asymptote -in points 7.The point where the locus across the imaginary axis Example 5.3.4: 1. Find poles and zeros -1 2. 4 segments 1 4 1 1 1 1 1 1 = − + + + = − − − = −   = = n m p z n j m i j i    =    + = 45 135 4 180 (2 1) 或 k   1 1 3 3 12 10 0 0 [( 1) ( 1 18)( 1 18)] 1 2,3 3 2 2 Solutions s s j yields s s s ds d s s j s j = − = −  + + + = = + + + + − Breakaway s k k s s k s s k s s s s k + − + + + + + + + = 19 0 14 484 4 14 19 4 40 0 1 24 19 4 24 40 19 0 0 1 2 3 4 4 3 2 ( 1) ( 1 18)( 1 18) ( ) ( ) 2 s s j s j k G s H s + + + + − = 484-4 =0 get k k =121 14 140 0 2 s + = =121 s =  j3.16 -j3.16 k j3.16 k =121

Example 5.3.5 R() 0.5 s(05+1 2 Please sketch the root locus with respect to K=/0, too) 上潇人字 SHANGHAI ILAO TONG UNIVIRSITY

Example 5.3.5 s K (0.5 1) 0.5 s s + R(s) C(s) 2 + − − + Please sketch the root locus with respect to K=[0,+∞)