例3.G(S)=C+)(T2>T1) 解: K1+T IGGo= O1+7,2o2 Ir ∠G(jo)=-90°+ arct7o- arctiI2O 0=0G(j0)=0∠Gjo)=-90 K(T1-T2)Re O=00Gj0)=0∠G(jo)=-90 G()= k(71-T2)K(1+7T2O 1+T22a(1+T2o2) mU()=K(T1-72) im(o)=-00
: 3. G(S) (T T ) ( 1) 2 1 K(TS 1 ) 2 1 解 例 = + + S T S = − = − + + − + − = = = = = = = = + − + + = → → lim ( ) lim ( ) ( ) (1 T ) (1 ) 1 T ( ) G(j ) | G(j ) | 0 G(j ) -90 0 | G(j ) | G(j ) -90 G(j ) -90 1 K 1 T | G(j ) | 0 1 2 0 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 2 1 V U K T T K T T j k T T arctgT arctgT T Re K(T1 -T2 ) Im =
二单反馈系统的频率咂 (S)G(S) R()1+G(s) G(S (1)向量作图法 CUo G(o) RGo 1+G(o) A(o)e(@) 在开环频率响应(j) Nyquist图中 ∠G(o1)=p(1) Q4=1+G(o1)Z1+G(m1)=v(o1) Im G(jO, OA A(@1-11+G(o, )Q4 v(a1) O Rm 0(01)=<G(o) =00)-(.8(x P(a 1+G(o) A
( ) ( ) 1 ( ) ( ) ( ) 1 ( ) ( ) ( ) 1 ( ) [1 ( )] ( ) ( ) ( ) ( ) ( ) 1 ( ) ( ) ( ) ( ) (1) 1 ( ) ( ) ( ) ( ) . 1 1 1 1 1 1 1 1 1 1 1 ( ) = − + = = + = = + + = = = + = + = G j G j QA OA G j G j A QA G j G j G j G j Nyquist A e G j G j R j C j G s G s R s C s j 在开环频率响应 图 中 向量作图法 二 单位反馈系统的频率响应 G(s) A Q O -1 ( ) 1 ( ) 1 ( ) 1 Im Rm