实验二北京科技大学自控实验.docx

50-50-WO-150-90Syslem: sysFrequency (rad/sJrO.CilSSMagnitude (dB): 432Sysicrn: sysSy siem; sysFreq uency (rad's): 0.5& Magnitude (dB): 8 94Fr-equency (rad'sj.Magnitude (dB):-44.2Syslem: sys Frequency (rad's):4& Manftude (dB); -69.4(ll5E-e.d-270-135-ISO-2251010io-2io2【自我实践4-1】某单位负反馈系统的开环传递函数G(s)=-，求(1)当k=4时， s(s + 1)(s + 2)计算系统的增益裕度，相位裕度，在Bode图上标注低频段斜率，高频段斜率及低频段、高频段的渐近相位角。(2)如果希望增益裕度为16dB，求出响应的k值，并验证。当 K=4 时>> num=4;den=1,320；G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridtitle(' Bode Diagram of G(s)=4/s(s+1)(s+2)')G =sA3 + 3 sA2 + 2 sContinuous-time transfer function.Gm =1.5000,Pm =11.4304,Wcg =1.4142,Wcp =1.1431 title(' Bode Diagram of G(s)=4/s(s+1)(s+2)')100Bode DiagramFrequency低频段斜率为-20dB/dec，高频段斜率为-60dB/dec，低频段渐近相位角为-90度，高频段的渐 近相位角为-270度。增益裕度GM=1.5000dB/dec，相位裕度Pm=11.4304度(2)当增益裕度为16dB时，算得K=0.951,对应的伯德图为：>> num=0.951;den=1,32。;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridtitle(' Bode Diagram of G(s)=4/s(s+1)(s+2)')G =0.951 sA3 + 3 sA2 + 2 sContinuous-time transfer function.Gm =6.3091,Pm =54.7839,Wcg =1.4142,Wcp =0.4276 title(' Bode Diagram ')10-£io-11 硝1O1记Fnequenc/(留)招nl'E落臣 京以 心急£【自我实践4-2】系统开环传递函数G(s)=，试分析系统的稳定性。s(0.5s + 1)(0.1s +1)计算可得当K=12时系统的增益裕度，相位裕度为0.对应的程序为：>> num=12;den=0.05,0.6,1,0;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridG =120.05 sA3 + 0.6 sA2 + sContinuous-time transfer function.Gm =1Pm =9.5374e-06Wcg =4.4721Wcp =4.4721Bode- Diag ram5010_|1 砂io1It/Froquancy (rid/s)当K=10时即当k<12时的特例，对应的程序为：>> num=10;den=0.05,0.6,1,0;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridG =100.05 sA3 + 0.6 sA2 + sContinuous-time transfer function.Bode DiagramGm =1.2000Pm =3.9431Wcg =4.4721Wcp =4.07765-00-10010-'讨1011十1(?Frequ enc y (r ad/s)50系统产生衰减震荡，增益裕度和相角裕度都大于0，系统稳定。当K=20时即当k>12时的特例，对应的程序为：>> num=20;den=0.05,0.6,1,0;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridG =200.05 sA3 + 0.6 sA2 + sContinuous-time transfer function.Gm =0.6000, Pm =-10.5320, Wcg =4.4721, Wcp =5.7247Bode Diagram 配10-'佰io1讨FnequrlCy此时的增益裕度和相角裕度都小于0，系统不稳定。【自我实践4-3】某单位负反馈系统的开环传递函数G(s) =316，求绘制5(0.015 + 1)(0.k +1)Bode图，在幅频特性曲线上标出低频段斜率、高频段斜率、开环截止频率和中频段穿越频 率；在幅频特性曲线标出：低频段渐近相位角、高频段渐近相位角和-180。线的穿越频率。(2) 计算系统的相位裕度y和幅值裕度h，并确定系统的稳定性。程序为：>> num=31.6;den=0.001,0.11,1,0;G=tf(num,den) Gm,Pm,Wcg,Wcp=margin(G) bode(num,den) grid31.60.001 sA3 + 0.11 sA2 + sContinuous-time transfer function.Gm =3.4810, Pm =22.2599, Wcg =31.6228, Wcp =16.3053Bode diagram1QO Sysiein: s/s&0Magnitude (dB): -71.6Frequency (rad/s);0.123Magnitude (dB): 48.2Syslem: sysFrequency (rad's): 16Magnitude dBV 0.149Syslem: s/sFrequency (railE):4911 (y Syslem: sys/ Frequency (rati即：7 1 1e+03 1/S. Phase (deg): -2&&(Bap) #E-_cd Sysiem: sysFrequency (rad/s):0.fi52Magnitude (dBO： 31.4Syslem: sysSysiem: sysFrequency (racfs); 0.185Phase (deg); -91,2Syslem: sysFrequency Crad/s): 3-1低频段斜率为-20dB/dec,高频段斜率为-60dB/dec,低频段渐近相位角为-90度，高频段的渐 近相位角为-270度。增益裕度GM=3.4810dB/dec，相位裕度Pm=22.2599度，-180度线穿越 频率为Wcg =31.6228。由伯德图可知系统的相位裕度和幅值裕度都大于零，故系统稳定。【自我实践4-4】某单位负反馈系统的开环传递函数G(s)= 轮+ 1)，令k=1作bode图， s 2(0.1s +1)应用频域稳定判据确定系统的稳定性，并确定使系统获得最大相位裕度的增益k值。当K=1时的伯德图程序为：>> num=1,1;den=0.1,1,0,0;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridG =s + 10.1 sA3 + sA2Continuous-time transfer function.Gm =0，Pm =44.4594，Wcg =0，Wcp =1.2647最大相位裕度.【综合实践】试观察下列典型环节BODE图形状，分析参数变化时对BODE图的影响，填写 下表。(1)比例环节：K (K=10、K=30)K=10 num=10;den=0,1;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridtitle(Bode Diagram)K=30num=30;den=0,1;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)grid title(Bode Diagram)Frequency (rad/s)(胃)制 euclK(2)惯性环节：(K=1、K=10、T=0.1、1)Ts 1K=1,T=0.1cp=margin(G) bode(num,den) gridtitle- Bode Diagram ')0.1 s + 1Continuous-time transfer function10-'诵1O1讨1O3Frequency (rad/s)K=1,T=1>> num=1;den=1,1;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridtitle- Bode Diagram ')G =1s + 1Continuous-time transfer function.Gm =InfPm = -180Wcg =NaNFrequmc/ (rad/s)(BaputnELIQ.K=10,T=0.1>> num=10;den=0.1,1;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G) bode(num,den)gridtitle- Bode Diagram ')G =100.1 s + 1Continuous-time transfer function.Gm = InfPm =95.7406Wcg =NaNWcp =99.4731Bode Diagram(60,0留£Frequency (rad/s)K=10,T=1>> num=10;den=1,1;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridtitle- Bode Diagram ')G =10s + 1Continuous-time transfer function.Gm =InfPm =95.7406Wcg =NaNWcp =9.9473201 俨10-1061011(?Frequency (rad/s)(3)积分环节： (K=1、K=10)sK=1>> num=1;den=1,0;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridtitle- Bode Diagram ')G =1 _ sContinuous-time transfer function.Gm = InfPm =90Wcg = NaNWcp =1寸101Frequency (radfs)K=10>> num=10;den=1,0;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridtitle- Bode Diagram ')G =10-sContinuous-time transfer function.Gm =InfPm =90Wcg =NaNWcp =10.0000(4)微分环节：Ks (K=1、K=10)K=1>> num=1,0;den=0,1;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridtitle- Bode Diagram ')G =sContinuous-time transfer function.Gm =InfPm =-90Wcg =NaNWcp =1(8m epmEBEs10°101Frqufehcy 仲5 o9059.8K=10>> num=10,0;den=0,1;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridtitle- Bode Diagram ')G =10 sContinuous-time transfer function.Gm = InfPm =-90Wcg =NaNWcp =0.1000rB-ode Diagram1(?101Frequency (rad闾K® 2,(5)二阶惯性环节：s2 + 2孙 +.2(K=1、K=1°、& =0二、& =» & =5、=1)K=1, & =0.1,=1>> num=1;den=1,0.2,1;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridtitle- Bode Diagram ')G =1sA2 + 0.2 s + 1Continuous-time transfer function.Gm =InfPm =16.2591Wcg =InfWcp =1.4000(tapn 菖莹K=1, & =1, ® =1>> num=1;den=1,2,1;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridtitle- Bode Diagram ')G =1sA2 + 2 s + 1Continuous-time transfer function.Gm = InfPm =-180Wcg =InfWcp = 0Bode Diagram1 俨io-1*佰it?Frequency rad/s)K=1, & =5, ® =1>> num=1;den=1,10,1;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridtitle- Bode Diagram ')G =1sA2 + 10 s + 1Continuous-time transfer function.Gm = InfPm =-180Wcg = InfWcp = 0Diag ram010-31fl-£10-' 谄 1O11 护Frequency (rad/s)K=10, & =0.1, ® =1>> num=10;den=1,0.2,1;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridtitle- Bode Diagram ')G =10sA2 + 0.2 s + 1Continuous-time transfer function.Gm =IfPm = 3.7990Wcg = InfWcp =3.3137Bod Diag ram10-'iS101Frequency (radys)K=10, & =1, s =1>> num=10;den=1,2,1;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridtitle- Bode Diagram ')G = 10sA2 + 2 s + 1Continuous-time transfer function.Gm = InfPm = 36.8699Wcg =InfWcp =3.0000£ apn-ESEs(BagWWE-CCLFnequMCy (伯d/s)K=10, & =5, ® =1>> num=10;den=1,10,1;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridtitle- Bode Diagram ')G =10sA2 + 10 s + 1Continuous-time transfer function.Gm =InfPm =90.0000Wcg =InfWcp =1.000010-310-21 矿 io6 1O1io3Freque-ncy (rad/'s)实验结果分析:k变大，系统的幅频特性曲线向上平移,相频特性不变增加积分环节，系 统性能改善，但是会使系统稳定性变差，比例系数增大可以使幅频特性的幅值增益增加，高 频段的斜率也增大。时间常数较大，则先出现拐点。【综合实践】已知系统开环传递函数：试根据Bode图比较上述各幅频和相频特性曲线有 什么变化，并计算出幅穿频率厂相位余量Y、相穿频率3”幅值余量Kg。1)0型系统:K(Ts + 1)(T s +1)>> num=10; den=12,13,1;G=tf(num,den) Gm,Pm,Wcg,Wcp=margin(G) bode(num,den) gridtitle- Bode Diagram ')G =1012 sA2 + 13 s + 1Continuous-time transfer function.Gm = Inf, Pm =62.6208, Wcg = Inf, Wcp = 0.683010-31 广 10-' 讨 io1Frequency rad/s)2)1型系统:Ks (Ts +1)1Ks (Ts + 1)(Ts +1)12(K=10、W、T2 =15)K:当 K=10,T1=1 时：s (T s +1)1>> num=10;den=1,1,0;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridtitle- Bode Diagram ')G =10sA2 + sContinuous-time transfer function.Gm = InfPm = 17.9642Wcg = InfWcp =3.0842io-2io-110°10'廿Frequency (rad/s)K:当 K=10,T1=1,T2=15 时s(Ts + 1)(Ts +1)12>> num=10;den=15,16,1,0;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridtitle- Bode Diagram ')G =1015 sA3 + 16 sA2 + sContinuous-time transfer function.Gm = 0.1067Pm =-30.9978Wcg = 0.2582Wcp =0.7319K2& =0.4、® =1)时:7 (K=2、S ( S 2 + 2 W S + 2 )>> num=2; den=1,0.8,1,0; G=tf(num,den) Gm,Pm,Wcg,Wcp=margin(G) bode(num,den) gridtitle- Bode Diagram ')G = 2sA3 + 0.8 sA2 + sContinuous-time transfer function.Gm =0.4000Pm =-39.7436Wcg =1Wcp =1.386510210T*101讨Frequency (rad/s)3) 2 型系统：、(K=10、T =5)时：s 2(T s 1)11>> num=10;den=5,1,0,0;G=tf(num,den)Gm,Pm,Wcg,Wcp=margin(G)bode(num,den)gridtitle- Bode Diagram ')G = 105 sA3 + sA2Continuous-time transfer function.Gm =0Pm =-80.9429Wcg = 0Wcp =1.2547150Bode Diagram顽10"10°101Frequency （rad/s）结果说明：系统类型越高，相角裕度越小，增益裕度越小，稳定性越差。三、思考题1）典型环节有哪些？分析典型环节中的参数T、K、3、（参数变化对BODE图的影响。答：典型环节有比例环节，一阶微分环节，一阶积分环节，一阶惯性环节，二阶微分环节， 二阶积分环节，二阶震荡环节。T增大，使伯德图的拐点向左偏移，T减小拐点向右偏移；K 增大使幅值增益增加，不影响相频特性。（增加使幅频增益更平滑。K主要影响开环增益， 使幅频曲线上下平移。2）系统类型对系统BODE图有哪些影响？对系统的相位稳定余量有什么影响？答：系统类型增加使伯德图的相频特性曲线更加陡峭，稳定裕量减小。