鲁棒控制作业答案解析

Exercise1.DerivetheclassicalfeedbackcontroltransfermatrixanddrawLTIblockdiagram典型闭环系统方框图解：定义如下的双端口结构：定义：procsensornn，pyzu由闭环系统的框图可知控制输入为uu，系统测量输出为()psensoryyn由框图得0()*pprocyunP，0()*procsensoryunPn综上可得：00000001()1pprocsensorpsensoryPPnzunPyuynPPu其中000000011zzuyyuPPPPPPPPP根据公式1((1))zzuyuyHPPKPKP，可得：1((1))zzuyuyHPPKPKP通过上面传递函数，我们可以得到如下的双端口网络：0PKprocsensornnpyzuuy典型闭环系统二端口网络结构图2.Derivetheclassicalone-degreefeedbackcontroltransfermatrixanddrawLTIblockdiagram.K0Pupyprocnsensornr伺服系统结构方框图解：定义procsensornnr，pyzu由闭环系统的框图可知控制器的输入()psensoryryn，其输出uu其中0()*pprocyunP，0()*procsensoryrunPn综上可得：0000000001()11procpsensorpsensornyPPnzuPryuynPPu其中000000000111zzuyyuPPPPPPPPP根据公式1((1))zzuyuyHPPKPKP，可得：1110000001110000(1)(1)(1)(1)(1)(1)PKPKPKPKPKPHKPKPKKPKKP1110000001110000(1)(1)(1)(1)(1)(1)procpsensornyPKPKPKPKPKPznHuKPKPKKPKKPr通过以上分析，现绘制二端口网络结构图，如下图所示。K0Pprocsensornnrpyzuuy伺服系统二端口网络结构图3.Derivetheclassicaltwo-degreefeedbackcontroltransfermatrixanddrawLTIblockdiagramK0Pupyprocnsensornr0双自由度控制系统结构方框图解：定义procsensornnr，pyzu由闭环系统的框图可知控制器的输入()psensoroutrryyny，其输出uu其中0()*pprocyunP，0()*procsensorryunPn()rypsensorruKyKKyn综上可得：00000000010010()10procpsensorpsensornyPPnuzPrryuuynPP其中0000000001001010zzuyyuPPPPPPPPP根据公式zH，1((1))zzuyuyHPPKPKP，可得：1110000001110000(1)(1)(1)(1)(1)(1)yyyryyyyyryPKPKPKPKPKPHKPKPKKPKKP1110000001110000(1)(1)(1)(1)(1)(1)procyyyrypsensoryyyyrynPKPKPKPKPKPyznHKPKPKKPKKPur通过以上分析，现绘制二端口网络结构图，如下图所示。K0Pprocsensornnrpyzuuoutryy双自由度控制系统二端口网络结构图4.ProgrammingtosynthesizeanH∞controllerinμ-toolboxandLMI-toolbox.Consideradisturbanceprocessdescribedby:22001100(),()101(0.051)101dGsGssssInthisproblem,thedisturbancerejectionisanimportantobjectiveinadditiontocommandtracking.Solution:（1）ProgrammingtosynthesizeanH∞controllerinμ-toolbox：%********************************************************************%*file：μ_toolbox.m%*function：ProgrammingtosynthesizeanH∞controllerinμ_toolbox%********************************************************************clearall;clc;G=nd2sys(1,conv([101],conv([0.051],[0.051])),200);m=1.5;wb=10;A=1.e-4;wu=1;wp=nd2sys([1/mwb],[1wb*A]);systemnames='Gwpwu';inputvar='[r(1);u(1)]';outputvar='[wp;wu;r-G]';input_to_G='[u]';input_to_wp='[r-G]';input_to_wu='[u]';sysoutname='p';cleanupsysic='yes';sysic;[khinf,ghinf,gopt]=hinfsyn(p,1,1,0.5,20,0.001);运行结果为：ResettingvalueofGammaminbasedonD_11,D_12,D_21termsTestbounds:0.6667gamma=20.0000gammahamx_eigxinf_eighamy_eigyinf_eignrho_xyp/f20.0009.6e+0001.3e-0051.0e-003-1.8e-0120.0000p10.3339.6e+0001.3e-0051.0e-0030.0e+0000.0000p5.5009.5e+0001.3e-0051.0e-0030.0e+0000.0000p3.0839.5e+0001.4e-0051.0e-003-1.8e-0120.0000p1.8759.4e+0001.5e-0051.0e-0030.0e+0000.0000p1.2719.1e+000-2.1e+001#1.0e-003-1.8e-0120.0000f1.5739.3e+0001.5e-0051.0e-0030.0e+0000.0000p1.4229.2e+0001.6e-0051.0e-0030.0e+0000.0000p1.3469.2e+000-1.1e+002#1.0e-0030.0e+0000.0000f1.3849.2e+0001.6e-0051.0e-0030.0e+0000.0000p1.3659.2e+000-3.3e+003#1.0e-003-1.8e-0120.0000f1.3759.2e+0001.6e-0051.0e-0030.0e+0000.0000p1.3709.2e+0001.6e-0051.0e-0030.0e+0000.0000p1.3689.2e+0001.6e-0051.0e-0030.0e+0000.0000p1.3669.2e+0001.6e-0051.0e-0030.0e+0000.0000p1.3669.2e+000-2.2e+004#1.0e-0030.0e+0000.0000fGammavalueachieved:1.3664（2）ProgrammingtosynthesizeanH∞controllerinLMI-toolbox：%***********************************************************************%*file：LMI_toolbox.m%*function：ProgrammingtosynthesizeanH∞controllerinLMI_toolbox%***********************************************************************clearall;clc;s=tf('s');Gd=200*(s+2)/(s*(0.005*s+1)*(10*s+1)*(0.05*s+1));%desiredmodelG1=200/((10*s+1)*(0.05*s+1)^2);[K,cl,g,in]=loopsyn(G1,Gd);%Hinf_loop-shapingcontrollersynthesissm=ss(cl.a,cl.b,cl.c,cl.d);figure(1);bode(sm,'r');holdon;bode(G1,'b');figure(2);step(sm,'r');holdon;d=1/s;Y1=G1/(1+G1);step(Y1,'b');Y1=(G1/(1+G1))*(1/s);Y2=100*d*(s^2+40*s+400)/(80400+10*s^3+401*s^2+4040*s);[n,r]=tfdata(((Y1+Y2)*s),'v');[A,B,C,D]=tf2ss(n,r);Y=ss(A,B,C,D);figure(3);t=0:0.01:10;step(Y,t);holdon;k=0.5*(s+2)*(0.05*s+1)/(s*(0.005*s+1));Y3=100*d*s*(400/(s+20)^2)*(s*(400/(s+20)^2)+200)/(1001*s*(400/(s+20)^2)^2+22200*s*(400/(s+20)^2)+40000+10*s*(400/(s+20)^2)^2*s+2000*s*(400/(s+20)^2)*s);t=0:0.01:10;[Q1,t1]=step(Y3,t);Y4=Gd/(1+Gd);[Q2,t1]=step(Y4,t);plot(t,(Q1+Q2),'r');运行结果为：-300-200-1000100Magnitude(dB)10-2100102104-360-1800180360Phase(deg)BodeDiagramFrequency(rad/sec)图1Shaping前后的Bode图比较00.511.522.500.20.40.60.811.21.41.6StepResponseTime(sec)Amplitude图2无扰动时Shaping前后的阶跃响应图的比较01234567891000.511.522.5Ste