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5-2目录1、典型环节2、典型环节的频率特性3、开环幅相曲线4、开环对数频率特性曲线5、延迟环节和延迟系统6、传递函数的频域实验确定(实验课讲)典型环节1Ts2sT)s(G221Ts2sT1)sG22G(s)=k比例环节G(s)=s微分环节积分环节一阶微分二阶微分惯性环节振荡环节G(s)=Ts+1欠阻尼二阶系统sdtds1dt一阶系统1Ts2sT1)s(221Ts1)s(G1Ts1)s(s1)s(G1Ts2sT)s(G221Ts2sT1)sG22G(s)=k比例环节G(s)=s微分环节积分环节一阶微分二阶微分惯性环节振荡环节G(s)=Ts+1sdtd1Ts1)s(Gs1)s(Gj0不稳定的…1Ts)s(G1Ts1)s(G1Ts2sT)s(G221Ts2sT1)s(G22典型环节零极点分布图(补充)微分环节的幅相曲线G(s)=sj)j(G这是一个正的纯虚矢量o90~0均为变化时,各矢量的角度从jIm[G(jω)]Re[G(jω)]01234矢量的模随着ω的增大而增大积分环节的幅相曲线(P192)j1)j(G这是一个负的纯虚矢量o90~0均为-变化时,各矢量的角度从+jIm[G(jω)]Re[G(jω)]0矢量的模随着ω的增大而减小G(s)=s11j=一阶微分的幅相曲线(P192)1Tj)j(G+这是一个实部恒为1变化矢量的角度从oo90~0矢量的模随着ω的增大从1变化到无穷G(s)=Ts+1jIm[G(jω)]Re[G(jω)]012341虚部随ω增大而增大的矢量20854210.50)()(A惯性环节G(jω)G(s)=0.5s+11j01Im[G(jω)]Re[G(jω)]0°1-14.5°0.97-26.6°0.89-45°0.710.450.370.240.115.0j1)j(G5.0tg)(1125.01)(A2-63.4°-68.2°-76°-84°二阶微分的幅相曲线T2j)T1()j(G22矢量的虚部始终为正Tω1时,实部为正,矢量在第一象限Tω=1时,实部为零,矢量在正虚轴上Tω1时,实部为负,矢量在第二象限jIm[G(jω)]Re[G(jω)]012j)T1j(Gooo180~90~0~0时,矢量的角度从从1Ts2sT)s(G22振荡环节G(jω)分析101Ts2sT1s2s)s(G222nn22n22222222T1T2arctgT4)T1(1)j(Go01)0j(Go1800)j(Gon9021)T1j(G)j(G得令,0d)(dA2nr212mr121A)(A)707.00(振荡环节G(jω)曲线(P194)(Nyquist曲线)0j121)(An2r121)(A2nr21典型环节相角小结(补充)1Ts2sT)s(G22G(s)=s微分环节积分环节一阶微分二阶微分惯性环节振荡环节G(s)=Ts+1G(s)=s1G(s)=Ts+110==恒定正90o恒定负90o0o~+90o0o~-90o0o~90o~180o0o~-90o~-180o1Ts2sT1)sG22非最小相位环节相角小结(幅频特性相同,相频特性相反)1Ts2sT)s(G22G(s)=k(k0)G(s)=-Ts+1G(s)=-Ts+11不稳定的不稳定的不稳定的不稳定的不稳定的比例环节一阶微分惯性环节二阶微分振荡环节名称G(s)0==恒定-180o0o~-90o0o~+90o0o~-180o0o~+180o1Ts2sT1)s(G22延迟环节Tse)s(GTje)j(G1)(AT)()]j(GIm[j)]j(GRe[10-1响角度不影响模与其它环节串联时只影开环幅相曲线的绘制例1)3s)(2s(60)s(G起点终点和交点10)s(G起点2s60)s(G终点1800交点:5j)6(60)j(G29.4j)6j(G分子分母保留最低次方起点:分子分母保留最高次方终点:则与虚轴相交有解若,0]GHRe[则与实轴相交有解若,0]GHIm[开环幅相曲线的绘制例2)5s(s10GHs2)s(G起点分子分母保留最低次方起点:分子分母保留最高次方终点:则与虚轴相交有解若,0]GHRe[则与实轴相交有解若,0]GHIm[902s10)s(G终点1800起点终点和交点交点:5j10)j(G2无交点)]j(GRe[0)]j(GIm[j开环幅相曲线的绘制例3起点终点和交点32s)4s5s(2GH3s8GH起点270s2GH终点90032j]5j)4[(2)j(H)j(G交点5.2)2j(H)2j(G]GHRe[]GHIm[j05.2开环幅相曲线的绘制例4)1s2(s)1s(10GH2180,s10GH2起点1800,s5GH2终点]j)12[()1(10)j(H)j(G222无交点]GHRe[]GHIm[j0)j3(2j320GH1时)12j()1j(10)j(H)j(G2时保留最低次方0js的幅相曲线绘制)1s(s)3s)(2s(5)s(G2解:o180)0j(Go900)j(G求交点:0)]j(GIm[令处。,与负实轴相交于25252105)1j(G曲线如图所示:-251无实数解,所以与虚轴无交点2s30)s(G,0)]j(GRe[令0642=时保留最高次方jss5)s(G)1()]64()1(j[5)j(G22221,即012MATLAB绘制的图)1s)(1s(s)1s)(3s)(2s(5)s(G20)]j(GRe[)]j(GIm[j①G(s)=1s[-20][-20][-20],1.0j1lg2010lg20dB20,1j1lg201lg20dB0②G(s)=10s1j10lg20dB2010lg201③G(s)=5s点过)0,2.0(100.2210.1L(ω)dBω0dB2040-40-20201002s1001[-40]积分环节L(ω)(图5-11)①G(s)=s100.2210.1L(ω)dBω0dB2040-40-2020100[+20][+20][+20]微分环节L(ω)(图5-11),1jlg201lg20dB0,10jlg2010lg20dB20②G(s)=2s5.025.0jlg20dB0③G(s)=0.1s对数曲线求斜率(补充)ωL(ω)dB0dBabLaLb斜率=对边邻边=La-Lb×babalglgab斜率例题(补充)求截止频率ωcωc=0.4L(ω)dBω0dB-7.96-21.94ωc15斜率=-7.96lg1sk)s(G∴∵ω=1时,cj4.0s4.0)s(G则有令=1得:–(-21.94)–lg5L(1)=-7.96=20lgk,∴k=0.4699.098.1320惯性环节对数幅频渐近曲线的分析(图5-11)1Ts1)s(G1T1lg20)(Alg20)(L22时,1TdB01lg20)(L时,1TT1lg20)(L时,T11Ts1)s(G1时,T11Ts1)s(GTs1水平线斜率为[-20]过(1/T,0)的斜线时,T1dB321lg20)(LdB321lg20)(L①G(s)=10.5s+1100②G(s)=s+5惯性环节L(ω)(图5-11)[-20][-20]26dB1s2.0204段直线方程怎么求得?100.2210.1ω0dB2040-40-2020100dB)(L0o-30o-45o-60o-90o)(一阶微分L(ω)(图5-11)[+20][+20])1s5.2(03.0)s(G100.2210.1ω0dB2040-40-2020100dB)(L0o+30o+45o+60o+90o°)(1s5.0)s(G1、)1.0s25.0(3.0)s(G2、振荡环节L(ω)渐近线分析(P195)1Ts2sT1)s(G22时,1T时,1T1)(A22T1)(AdB0)(L,Tlg40)(L,)(A222222T4)T1(1T1或n1)s(G,=T1或n22sT1)s(G,注意:要在ωn或ωr处修正!!!这项总是去掉的!振荡环节L(ω)(P195)100.2210.1L(ω)dBω0dB2040-40-2020100422.0242)(2222sssssGnnndB14.8121lg20Alg202m92.1212nr[-40]二阶微分(图5-11)2n2nn222s2s1Ts2sT)s(GT1no180)j(G,01)0j(Go,902)j(Gonj01幅相曲线o902对数幅频渐近曲线[+40]2nr212m12lg20L2lg20)(Lnn时有峰值707.000dBωdB)(L90)(nr峰值-渐近线值100.2210.1L(ω)dBω0dB2040-40-2020100[-20][-40])130/s)(1s2(s)1s5.0(40)s(H)s(G绘制的L(ω)曲线转折频率:0.5230斜率:-40-20-40[-20][-40]dB385.0dB521.0s40时为,时为低频段:开环的L曲线绘制(P202)的对数曲线。绘制)100s4s)(1s(s)15s(2000)s(G22解:db14.8L,59.9,10,2.0mrn对数相频:相频特性的画法为:起始角,终止角,转折频率处的角。o45o6.22o3.2o90o7.78o90o15o90o3.84o8.126o90o90例题(补充)-90o-114.7o-93.7o-137.5o-180o对数幅频:低频段:20/s[-20]转折频率:1510斜率:-400-40修正值:01510频率:1tg2.0tg21211004tgs/1s202s80低频段:20/s[-20]转折频率:1510斜率:-400-4001510-90o-114.7o-93.7o-137.5o-180o[-20][-40][-40]1510绘制曲线ω0dB20dB-20dB-90o-120o-150o-180odB)(L)(db14.8L,59.9mr)100s4s)(1s(s)1s2.0(2000)s(G22由L(ω)求G(s)例1(P205)ω0L(ω)dB-20203[+20][-20]0ωL(ω)[-20][-40]1002002)1s301()1s3.01(1.0)1s02.0(s200k)s(G)(1sT122)1sT(11.0k3.01T1301T220klg20:由斜率的几何意义读出30.30dB20,dB203slg2021,读出或者由s200)s(G1Ts102.0T1Ts200100js2解得由50s100s20022解出或
本文标题:自动控制原理5-2典型环节与开环系统的频率特性
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