《机械控制工程基础》课程教学资源（PPT课件）第四章 频率特性分析

(2)积分环节 G(j@)= Im jo G[(jo)] GUo)-1 ∠G(j0)=-90° 0=0,G(j0)→0，∠G(j0)=-90° 81 Re 0→o,G(U0)=0,∠G(j0)=-90° 0=∞ -90° 0=0

（2）积分环节 →  =  = −  = →   = −   = −  = = , ( ) 0, ( ) 90 0, ( ) , ( ) 90 ( ) 90 1 ( ) 1 ( )            G j G j G j G j G j G j j G j 0 Re Im -90° ω=0 G[(jω)] ω=∞

(3)微分环节 G(jo)=j@ G(j0)=0 ∠G(j0)=90° 0=0,G(U0)=0,∠G(j0)=90° Im 0→0，G(j0)→0，∠G(j0)=90° +90° 0 0=0 Re

（3） 微分环节 →  →   =  = =  =   =  = = , ( ) , ( ) 90 0, ( ) 0, ( ) 90 ( ) 90 ( ) ( )            G j G j G j G j G j G j G j j 0 Re Im +90° ω=0

(4)惯性环节 K Im G[(jw)] G(j@)= 1+iTo K 0 0=0 K G(j01= 0=∞ -45° Re 1+T2o2 ∠G(jo)=-arctan To w=1/T0 0=0,G(jo=K,∠G(jo)=0° 0=1/T,G(Ujw)=K1V2,∠G(jo)=-45 w→0，G(j0)=0,∠G(jw)=-90°

（4） 惯性环节 →  =  = −  = =  = −  = =  =   = − + = + = , ( ) 0, ( ) 90 1/ , ( ) / 2, ( ) 45 0, ( ) , ( ) 0 ( ) arctan 1 ( ) 1 ( ) 2 2                G j G j T G j K G j G j K G j G j T T K G j j T K G j →  =  = −  = =  = −  = =  =   = − + = + = , ( ) 0, ( ) 90 1/ , ( ) / 2, ( ) 45 0, ( ) , ( ) 0 ( ) arctan 1 ( ) 1 ( ) 2 2                G j G j T G j K G j G j K G j G j T T K G j j T K G j 0 Re Im -45° ω=0 G[(jω)] ω=∞ ω K ω=1/T

(5)一阶微分环节（导前环节） G[(j@)] G(j@)=1+jT@ Im G(j0)=V1+T2o2 ∠G(jo)=arctan To w=0,G(j0=1,∠G(j0)=0° 0=0 0 (1,j0) 0=1/T,G(jo)=V2,∠G(Ujo)=45° Re ∠G[(jo)] 0=o,G(j0)-→0，∠G(j0)=90°

（5）一阶微分环节（导前环节） =  →   =  = =  =  = =  =   = = + = + , ( ) , ( ) 90 1/ , ( ) 2, ( ) 45 0, ( ) 0, ( ) 0 ( ) arctan ( ) 1 ( ) 1 2 2                G j G j T G j G j G j G j G j T G j T G j j T =  →   =  = =  =  = =  =   = = + = + , ( ) , ( ) 90 1/ , ( ) 2, ( ) 45 0, ( ) 1, ( ) 0 ( ) arctan ( ) 1 ( ) 1 2 2                G j G j T G j G j G j G j G j T G j T G j j T 0 Re Im G[(jω)] ω (1,j0) ∠G[(jω)] ω=0

(6)振荡环节 o 元=0/0n G(S)= 1 s2+250,s+o G(j@)=- -2)+j25入 2 0 1 G(j@)= -02+o2+j250.0 G(j@)= V1-2+4528 ∠G(jo)=-arctan 251 1-22 2=0,0=0,G(jo)=1,∠G(jo)=0° 入=1,0=⊙n,G(jo)=1/25,∠G(jo)=-90° 元=0,0=∞，G(jo)=0,∠G(j0)=-180°

（6）振荡环节 =  =  =  = −  = = =  = −  = = =  =  −  = − − + = − + = = − + + = + + = , , ( ) 0, ( ) 180 1, , ( ) 1/ 2 , ( ) 90 0, 0, ( ) 1, ( ) 0 1 2 ( ) arctan (1 ) 4 ( ) (1 ) 2 1 ( ) / 2 ( ) 2 ( ) 2 2 2 2 2 2 2 2 2 2 2 2                                         G j G j G j G j G j G j G j K G j j G j j G j s s G s n n n n n n n n =  =  =  = −  = = =  = −  = = =  =  −  = − − + = − + = = − + + = + + = , , ( ) 0, ( ) 180 1, , ( ) 1/ 2 , ( ) 90 0, 0, ( ) 1, ( ) 0 1 2 ( ) arctan (1 ) 4 ( ) (1 ) 2 1 ( ) / 2 ( ) 2 ( ) 2 2 2 2 2 2 2 2 2 2 2 2                                         G j G j G j G j G j G j G j K G j j G j j G j s s G s n n n n n n n n =  =  =  = −  = = =  = −  = = =  =  −  = − − + = − + = = − + + = + + = , , ( ) 0, ( ) 180 1, , ( ) 1/ 2 , ( ) 90 0, 0, ( ) 1, ( ) 0 1 2 ( ) arctan (1 ) 4 1 ( ) (1 ) 2 1 ( ) / 2 ( ) 2 ( ) 2 2 2 2 2 2 2 2 2 2 2 2                                         G j G j G j G j G j G j G j G j j G j j G j s s G s n n n n n n n n