大學(xué)交通運輸工程學(xué)院《機械工程控制基礎(chǔ)實驗報告》上機題及答案解析

0第3題............................................11s2kk-forj=1:6gtext('k=0.1','linewidth',1.5,'fontsize',10)%用鼠標(biāo)放置文本gtext('k=0.2','linewidth',1.5gtext('k=0.5','linewidth',1.5gtext('k=0.8','linewidth',1.5gtext('k=1.0','linewidth',1.5gtext('k=2.4','linewidth',1.523forj=1:6gtext('k=0.1','linewidth',1.5gtext('k=0.2','linewidth',1.5gtext('k=0.5','linewidth',1.5gtext('k=0.8','linewidth',1.5gtext('k=1.0','linewidth',1.5gtext('k=2.4','linewidth',1.5331sys=zpk([],[-1+3*i-finalvalue=polyval(n,0)/polyvalwhile(y(l)>0.98*finalvalue)&(y(l)<1.02*finalval45EQ\*jc3\*hps34\o\al(\s\up13(0),5)2s2650.2s2s(s50.2s2s(s2ess1=polyval(n1,0)/ess2=polyval(n2,0)/7桿鉸接處無摩擦。假定系統(tǒng)中所有變量變化范圍很小，則可以將該系統(tǒng)視ssssX(s)s10(s1s1s10(s10(s5ttitle('thestepresponseofthesysxlabel('time(second8xkc54m707my2x(t)2.452t92222sX(s)x2=0.707*ones(size(ttitle('ResponseofHangingxlabel('Time(second)');ylabel('inputx(t)andoutputy(t)EQ\*jc3\*hps34\o\al(\s\up17(0),5)EQ\*jc3\*hps34\o\al(\s\up17(K),s)s2s222n2nw=[0,logspace(-2,2,20forj=1:5sys=tf([wn*wn],[1,2*e(j)*K,并判斷閉環(huán)系統(tǒng)的穩(wěn)定性。K,并判斷閉環(huán)系統(tǒng)的穩(wěn)定性。EQ\*jc3\*hps34\o\al(\s\up17(16),s)Wc=interp1(mag1,W,0,'splineW_3db=interp1(mag1,W,-3,'splineW_90=interp1(pha1,W,-90,'spline第5題已知單位反饋系統(tǒng)開環(huán)傳遞函數(shù)5(s2)5(s2)title('NyquistDiagraKK221xlabel('Frequency/(rad/title('feedbackbodediatitle('feedbackbodediafinalvalue=polyval(num,0)/polyval(dewhile(y(l1)>0.98*finalvalue)&(y(l1)<1.02*finalvalwhile(y(l2)>0.95*finalvalue)&(y(l2)<1.05*finalval1+ccoo_ylabel('x_o(t)')title('StepResponseofGK(s)=k/(s+1)(s+2)(s+5)')1H(s)為單位反饋，對系統(tǒng)采用比例微分控制，比例系title('StepResponseofGK(s)=k/(s+1)(s+2)(s+5)')1title('StepResponseofGK(s)=k/(s+1)(s+2)(s+5)')K,試用MATLAB,試用MATLAB設(shè)計PID校正裝置，使得4s－1。若要校正更大的相位角，則值要更大，故選β=10。于是相位滯后校正裝置另一轉(zhuǎn)角頻deng=conv([1,0],conv([0title(['校正后系統(tǒng)的相位裕度=',num2str(PmEQ\*jc3\*hps34\o\al(\s\up18(K),5s)解：首先設(shè)計超前校正環(huán)節(jié)，并繪制校正前后的開環(huán)伯德[mag,phase,w]=bode(sys_oT=1/(wcg*sqrt(alpha));Tz=title(['校正后系統(tǒng)的相位裕度=',num2str(Pmstep(sys_old,'r',sys解：設(shè)計滯后校正環(huán)節(jié)，并繪制校正前后的開環(huán)伯德den=conv([1,0],conv([1wcg=spline(phase(1,:),w',dPm-18step(Gy_close,'--b',Gx_ctitle(['校正后系統(tǒng)的相位裕度=',num2str(Pmfori=find((mag1<=-19.8)&(mag1>=-w_1=w1(ii)fori=find((mag1<=0.2)&(mag1>=jj=input('enterindexfordesiredmians=0.67810.68000.68180.68370.68560.6875在上述數(shù)據(jù)組中選擇數(shù)值308，得到轉(zhuǎn)角頻率之一ians=在上述數(shù)據(jù)組中選擇數(shù)值808，得到轉(zhuǎn)角頻率之一sysc=tf(numc1,denc1)*tf(nu,設(shè)計要求：相位裕度Wc=13.5;%rad/seconddisp('Modeloforigdisp('Result:modeloflag-leadisp('Result:modelofnewopendisp('Thesystem