第五章《根轨迹法》 =1S+2;=S+P 因为 (s)Q(s)-P(S)Q(S)=0 P(s)S as 其中P(s)=(s+1Xs+z2)…(s+m) Q(s)=(s+p1)(S+p2)…(s+p S s+Z 1+2 S+2 IIn O(s) S+ pI S+ p2 st p 所以 S+z st p10 (2) 因为 即 其中 即 所以 - 第五章《根轨迹法》   = = + = + n j i m i 1 s zi 1 s p 1 1 Q(s) Q (s) P(s) P (s)  =  [ln ( )] [ln Q(s)] ds d P s ds d = ( ) ( )( ) ( ) ( ) ( )( ) ( ) 1 2 1 2 n m Q s s p s p s p P s s z s z s z = + + + = + + +   n m s p s p s p Q s ds d s z s z s z P s ds d + + + + + + = + + + + + + = 1 1 1 [ln ( )] 1 1 1 [ln ( )] 1 2 1 2     = = + = + n j i m i 1 s zi 1 s p 1 1 P(s)Q(s) − P(s)Q(s) = 0