例3.GS)= K(TS+ S(12S+1) (T2>T1) 解 K1+T22 IGGo= O1+7,2o2 Im ∠G(jo)=-90°+ actg71o- arctiI2O O=0|G(j)=∞0∠G(j0)=-90 K(T1-T2)Re O=00Gj0)=0∠G()=-90 GGo k(71-T2)K(1+7T2O 1+T2o o(1+T2o2 mU(0)=K(T1=72) →>0 m()=-00 O→>0
: 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 =
二单位反馈系統的频率呃立 C(s) G(s) R(S)1+G(s) G(S) (1)向量作图法 Co) G(o) RGo 1+G(O A(@)e 在开环频率响应G(0) Nyquist图中 ∠G(o1)=p(o1) Q4=+G(o1)∠[1+G(m1)=v(1) Im G(O OA A(O1)1+G(j0 )Q4 y(a1) O Rm G(o e(1)=∠ 1+G() =(O1)-v(1 p(o1) 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