西安科技大学：《自动控制原理》课程教学资源（PPT课件）第五章 线性系统的频域分析

6振荡环节 G(S)= 0<5<1 S+250nS+On GO)=_0+0“,2n+2m 2=o 2-j2 (1-x2)2+(2)2 U(O=n (1-2)2+(22) >52>3 V(o) (1-x2)2+(29)2 Ggo) ∠G(O)=- arct22 (1-x2)+45 =0=0G(o)=1∠G()=0 =1O=0n G(jo)=∠G(o)=90° G(O)0∠G()=-180 2的取值不同极坐标图型的形状不同

的取值不同极坐标图型的形状不同 振荡环节 , | G(j ) | 0 G(j ) 180 1 | G(j ) | G(j ) 90 0 0 | G(j ) | 1 G(j ) 0 | G(j ) | G(j ) -arctg V( ) U( ) G(j ) G(S) 0 1 6. 2 1 1- 2 (1- ) 4 1 (1 ) (2 ) 2 (1 ) (2 ) 1 (1 ) (2 ) 1 2 1 ( ) 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2                                                       =  =  =  = − = = =  = − = = =  = =  = = = = = = =   + − + − − + − − + − − = − + + + − + + + n j j j S S n n n n n n n n n  1  2 3   1   2   3  n  n  n

7二阶微分环节 G(S)=T S+2ETS+I T-7 GjO)=-T2o2+j2To+1=(1-2)+j252x= n (G0)=√(1-x2)2+42 ∠G(jO)= arct 25 1- =00=0G(j0)=1∠G)=0° 1=10=0n(m)25∠G(j0)=90° 元=0D=0G(m)=∠GjO)=180 j0)

| G(j ) | G(j ) 180 1 | G(j ) | 2 G(j ) 90 0 0 | G(j ) | 1 G(j ) 0 1- 2 G(j ) arctg | G(j ) | (1- ) 4 G(j ) - T 2 1 (1 ) 2 1 G(S) T 2 1 T 7. 2 2 2 2 2 n 2 2 2 n 2 2    =  =  =   = = = =  = = = =  =  = = + = + + = − + = = + + =                                   n j T j S TS 二阶微分环节 (1,j0)

8延时环节 G(s)=e GGo)= eto cos to-sn to u(@)=costo o)=-sinta G(jo)=1∠G(jo)=-O u(o)+v2(O) 极坐标图为一单位圆端点在单位圆上元限不 9不稳定环节 G(S)=TSI GGo= -I-jOT (-1j0) 0 jOT-1 1+02T Ggo) Re ∠Gio)=-180+ arctgoTu(o)=-1√(o)=-jbT O=0G(j)=1∠G(jO)=-180 O=|G(o)=K2∠G(o)=-135 O G(a)|=0∠G(j)=-90°

    | G(j ) | 0 G(j ) -90 | G(j ) | G(j ) -135 0 | G(j ) | 1 G(j ) -180 G(j ) -180 u( ) - 1 v( ) - j T | G(j ) | G(j ) G(S) 9. , u ( ) v ( ) 1 | G(j ) | 1 G(j ) - u( ) cos v( ) -sin G(j ) e cos - jsin G(S) e 8. 2 1 1 1 T 1 1 T -1-j T j T-1 1 TS-1 1 2 2 -j - s 2 2 2 2 =  =  = = =  = = =  =  = + = = = = = = + = =  = = = = = = + +                                    T arctg T 不稳定环节 极坐标图为一单位圆端点在单位圆上无限循环 延时环节 =0 Re Im (-1,j0) 0  = 0  = 

四,极坐标图举例 例解 Gs)=x+试绘制其Mwu图。 G(jo) K jo(+jTo) G() O√1+T2O ∠G(0)=-90°-arcg7O O=01(Gio)=∞0∠G()=-90 Im O=0G()=0∠G(i)=-180 G(O)=,KT2-j一K U(o)=Re[Ggo) KT (kT,jO) o Re 1+T-o 0三 V(o)=MiGDol o(1+T2o2) lm U(o)=-kT lim V(o)=0 O→>0 0→>0

: 1. G(s) s(Ts 1 ) K 解 例 = + 试绘制其Nyquist图 。 lim U( ) lim V( ) 0 V( ) Im[G(j )] U( ) Re[G(j )] - G(j ) - j | G(j ) | 0 G(j ) -180 0 | G(j ) | G(j ) -90 G(j ) -90 | G(j ) | G(j ) 0 0 (1 T ) -k 1 T K T (1 T ) K 1 T -K T 1 T K j (1 j T ) K 2 2 2 2 2 2 2 2 2 2 = − = = = = = = =  =  = = =   =  = − = = → → + + + + + +                              k T arctgT    -(kT,j0) Re Im 0  =    四．极坐标图举例

例2.G(S) K S(1+TS)(1+T2S) 解 K GGo) (jO)2(1+1)1+m2O) K IGgo) 2y1+To2√1+T2a2 ZGGo=-180-arctgToarctgT,0 O=0|G()|=∞∠G(a)=-180 O=∞ G(o)=0 ∠G(o)=-360 GGo)=RelGgo+Im[ o) 令Re[G(O)=0得= √T 这时ImG(o)= K(TT + T 由此得出 Nyquist图与虚轴的交点

: 2. G(S) S (1 T S)(1 T S) K 1 2 2 解 例 + + = 由此得出 图与虚轴的交点 这 时 令 得 Nyquist K(T ) Im[G(j )] T 1 Re[G(j )] 0 G(j ) Re[G(j )] Im[G(j )] | G(j ) | 0 G(j ) -360 0 | G(j ) | G(j ) -180 G(j ) -180 T 1 T 1 T | G(j ) | (j ) (1 T )(1 T ) G(j ) 1 2 1 2 1 2 1 2 2 2 2 2 2 1 2 1 2 2 2 3 T T T T arctgT arctg K j j K + = = = = + =  =  = = =   =  = − − + + = + + =                             = 