6振荡环节 G(S)=o2 20<5<1 S2+250nS+0n GGo) es2 72= 0-+O j25 oont )2+J20 A=2 2-j2 (1-x2)2+(2)2 U()=a-a1=42a 91>52>53 V(o)= (1-2)2+(2)2 I GGo) 1∠G(O)=- arct22 (1-2-)+45 =00=0G()=1∠G()=0 1=10=0n1G(a)=k∠G0)=-90° oO G(jo)|=0 ∠G()=-180 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 GS)=TS2+257S+1T= G()=-To2+j2o+1=(1-2)+j2xx= G(io)|=√(1-2)2+42x2 ∠G(jO)= arct 25 1- =00=0(0)=1∠Gjo)=0 10=0n|G(j)=25∠Gjo)=90 元=00=0(()=∞∠Go)=180 (1j0)
| 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=e=cos to-jsin TO u(@)=costo a)=-sinta G(j)=1∠GO)=-0 u2(O)+y2(o)=1 极坐标图为一单位圆端点在单位圆上无限不 9不稳定环节 G(S)=-1 T-1 G(O)= -I-IOT (-1j0) 0 jOT-1 1+02T2 =0 GGo Re ∠Gio)=-180+ arctgoTu(o)=-1√(o)=-jbT =0G(j)=1∠G(j)=-180 O=|G(o)=K2∠G(o)=-135 G(jO)=0∠G()=-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.G(s)=试绘制其 gust图。 解 G(jo) K o(l+To) GGo O√1+T2o ∠G(O)=-90- arcto O=0|Gj)=∞∠Gjo)=90 Im 0=00 G(O)|=0∠G(i0)=-180 GGo=kn2-j-K 1+T2O o(+T o U(o=Re[GGo)=-k. (kT,jO) o Re 0三 V(o=ImIGGo=4k O(+To 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+T1s)(1+T2S) 解 K G(a)= a)(1+/o)(1+2O) I GGo)l= 2y1+To2y√1+T2o2 ZGGo=-180-arctgTo-arctgT,0 )=0G()o∠G(0)=-180° O=∞ G(o)=0 ∠G(o)=-360° GGo)=RelGgo)+Im[GG o) 令Re[Go)=0得a 这时ImG(o)= K(TT2 T,+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 + = = = = + = = = = = = = − − + + = + + = =