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 Breakin 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