1、控制系統仿真與CAD實驗課程報告一、實驗教學目標與基本要求上機實驗是本課程重要的實踐教學環節。實驗的目的不僅僅是驗證理論知識，更重要的是通過上機加強學生的實驗手段與實踐技能，掌握應用MATLAB/Simulink求解控制問題的方法，培養學生分析問題、解決問題、應用知識的能力和創新精神，全面提高學生的綜合素質。通過對MATLAB/Simulink進行求解，基本掌握常見控制問題的求解方法與命令調用，更深入地認識和了解MATLAB語言的強大的計算功能與其在控制領域的應用優勢。上機實驗最終以書面報告的形式提交，作為期末成績的考核內容。二、題目及解答第一部分：MATLAB必備基礎知識、控制系統模型與轉換

2、、線性控制系統的計算機輔助分析1.考慮茗名的Mssol化學反應方程組&=*InyE=61（XC）3選定口=6=02c=5.且工1（3=工式0）=工式0）=0,繪制仿真結果的三維相軌跡，并得出其在K-y平面上的投影.f=inline(-x(2)-x(3);x(1)+a*x(2);b+(x(1)-c)*x(3),t,x,flag,a,b,c);t,x=ode45(f,0,100,0;0;0,0.2,0.2,5.7);plot3(x(:,1),x(:,2),x(:,3),grid,figure,plot(x(:,1),x(:,2),grid2.4解二之,最優化:口:用1H1Dy=(x)x(1)A2-

3、2*x(1)+x(2);ff=optimset;ff.LargeScale=off;ff.TolFun=1e-30;ff.TolX=1e-15;ff.TolCon=1e-20;x0=1;1;1;xm=0;0;0;xM=;A=;B=;Aeq=;Beq=;x,f,c,d=fmincon(y,x0,A,B,Aeq,Beq,xm,xM,wzhfc1,ff)Warning:OptionsLargeScale=offandAlgorithm=.學習參trust-region-reflectiveconflict.IgnoringAlgorithmandrunningactive-setalgorithm.

4、Toruntrust-region-reflective,setLargeScale=on.Torunactive-setwithoutthiswarning,useAlgorithm=active-set.Infminconat456Localminimumpossible.Constraintssatisfied.fminconstoppedbecausethesizeofthecurrentsearchdirectionislessthantwicetheselectedvalueofthestepsizetoleranceandconstraintsaresatisfiedtowith

5、intheselectedvalueoftheconstrainttolerance.Activeinequalities(towithinoptions.TolCon=1e-20):lowerupperineqlinineqnonlin2x=1.000001.0000f=-1.0000iterations:5funcCount:20Issteplength:1stepsize:3.9638e-26algorithm:medium-scale:SQP,Quasi-Newton,line-searchfirstorderopt:7.4506e-09constrviolation:0message

6、:1x766char3.請珞r面的傳遞由數模型輸入乳MATLAB環境.3）的133十2言答：十53H十0四，丁1秒(a) s=tf(s);G=(sA3+4*s+2)/(sA3*(sA2+2)*(sA2+1)A3+2*s+5)sA3+4s+2sA11+5sA9+9sA7+2sA6+12sA5+4sA4+12sA3Continuous-timetransferfunction.(b)z=tf(z,0.1);H=(zA2+0.568)/(z-1)*(zA2-0.2*z+0.99)H=zA2+0.568zA3-1.2zA2+1.19z-0.99Sampletime:0.1secondsDiscrete

7、-timetransferfunction.4.假設描述系統的常微分方程為y十1時十的陰十的閔=2班小請選擇一綱狀態變量，并將此方程在MATLAE工作空間中表示出來,如果想得到系統的傳遞函數和零極點模型，我門將如何求取“得出的結果又是怎樣的？由微分方程模型能否直接寫出系統的傳遞函數模型？A=010;001;-15-4-13;B=002,;C=100;D=0;G=ss(A,B,C,D),Gs=tf(G),Gz=zpk(G)G=a=x1x2x3x1010x2001x3-15-4-13b=u1x10x20x32c=x1x2x3y1100d=u1y10Continuous-timestate-spac

8、emodel.Gs=2sA3+13sA2+4s+15Continuous-timetransferfunction.Gz=(s+12.78)(sA2+0.2212s+1.174)Continuous-timezero/pole/gainmodel.5.已知某系統的差分方程模型為十？)十式/十1)十01辦)=十1)十2試即).試將其輸入到MATLAB工乍空間.設采樣周期為0.01sz=tf(z,0.01);H=(z+2)/(zA2+z+0.16)H=z+2zA2+z+0.16Sampletime:0.01secondsDiscrete-timetransferfunction.6.假設某單位負反

9、饋系統中，G=/十十區十5),&=+試用MATLAB推導出閉環系統的傳遞函數模型.symsJKpKis;G=(s+1)/(J*sA2+2*s+5);Gc=(Kp*s+Ki)/s;GG=feedback(G*Gc,1)GG=(Ki+Kp*s)*(s+1)/(J*sA3+(Kp+2)*sA2+(Ki+Kp+5)*s+Ki)7.從：面給出的典型反饋控制系統結構子模型中一求出總系統的狀態方程與傳遞函數模型；并得出各個模型的零極點模型表示.(Ai仃.1694十III211.875+317.64+211)(+94.34)(s+II.1684)35786.7Z-1十108444z-十4j(二T十20)(-1

10、十74.04)(a)s=tf(s);G=(211.87*s+317.64)/(s+20)*(s+94.34)*(s+0.1684);Gc=(169.6*s+400)/(s*(s+4);H=1/(0.01*s+1);GG=feedback(G*Gc,H),Gd=ss(GG),Gz=zpk(GG)GG=359.3sA3+3.732e04sA2+1.399e05s+1270560.01sA6+2.185sA5+142.1sA4+2444sA3+4.389e04sA2+1.399e05s+127056Continuous-timetransferfunction.Gd=x1x2x3x4x5x6x1-2

11、18.5-111.1-29.83-16.74-6.671-3.029x212800000x30640000x40032000x5000800x6000020b=u1x14x20x30x40x50x60yix1x2x3x4x5x6001.0973.5591.6680.7573u1y10Continuous-timestate-spacemodel.Gz=35933.152(s+100)(s+2.358)(s+1.499)(sA2+3.667s+3.501)(sA2+11.73s+339.1)(sA2+203.1s+1.07e04)Continuous-timezero/pole/gainmode

12、l.(b)設采樣周期為0.1sz=tf(z,0.1);G=(35786.7*zA2+108444*zA3)/(1+4*z)*(1+20*z)*(1+74.04*z);Gc=z/(1-z);H=z/(0.5-z);GG=feedback(G*Gc,H),Gd=ss(GG),Gz=zpk(GG)GG=-108444zA5+1.844e04zA4+1.789e04*31.144e05zA5+2.876e04zA4+274.2zA3+782.4zA2+47.52z+0.5Sampletime:0.1secondsDiscrete-timetransferfunction.Gd=a=x1x2x3x4x5

13、x1-0.2515-0.00959-0.1095-0.05318-0.01791x20.250000x300.25000x4000.12500x50000.031250b=u1x11x20x30x40x50x1x2x3x4x5y10.39960.63490.10380.050430.01698d=u1y1-0.9482Sampletime:0.1secondsDiscrete-timestate-spacemodel.Gz=-0.94821zA3(z-0.5)(z+0.33)(z+0.3035)(z+0.04438)(z+0.01355)(zA2-0.11z+0.02396)Sampletim

14、e:0.1secondsDiscrete-timezero/pole/gainmodel.8.巳知系統的方樞圖如圖所示，試推導出從輸入信號r(i)到輸出信號y(t)的總系統模型.s=tf(s);g1=1/(s+1);g2=s/(sA2+2);g3=1/sA2;g4=(4*s+2)/(s+1)A2;g5=50;g6=(sA2+2)/(sA3+14);G1=feedback(g1*g2,g4);G2=feedback(g3,g5);GG=3*feedback(G1*G2,g6)GG=3sA6+6sA5+3sA4+42sA3+84sA2+42ssA10+3sA9+55sA8+175sA7+300sA

15、6+1323sA5+2656sA4+3715sA3+7732sA2+5602s+1400Continuous-timetransferfunction.9.已知傳遞函數模型G(s)=(s+l)2(s2+2fi+400)(s+5)氣岸-3s+100)*(s2十3后十2500)對不同采樣周期了=和T=1秒對之進行離散化，比我原系統的階躍喃應與各離散系統的階躍嘀應曲線.提示：后面將介紹，如果已知系統模型為G,則用smp(G)即可繪制出其階躍響應曲援.s=tf(s);T0=0.01;T1=0.1;T2=1;G=(s+1)A2*(sA2+2*s+400)/(s+5)A2*(sA2+.學習參3*s+100

16、)*(sA2+3*s+2500);Gd1=c2d(G,T0),Gd2=c2d(G,T1),Gd3=c2d(G,T2),step(G),figure,step(Gd1),figure,step(Gd2),figure,step(Gd3)Gd1=4.716e-05zA5-0.0001396zA4+9.596e-05zA3+8.18e-05zA2-0.0001289z+4.355e-05zA6-5.592zA5+13.26zA4-17.06z3+12.58zA2-5.032z+0.8521Sampletime:0.01secondsDiscrete-timetransferfunction.Gd2=

17、0.0003982zA5-0.0003919zA4-0.000336*3+0.0007842zA2-0.000766z+0.0003214zA6-2.644zA5+4.044zA4-3.94*3+2.549zA2-1.056z+0.2019Sampletime:0.1secondsDiscrete-timetransferfunction.Gd3=8.625e-05zA5-4.48e-05zA4+6.545e-06zA3+1.211e-05zA2-3.299e-06z+1.011e-07zA6-0.0419zA5-0.07092zA4-0.0004549*3+0.002495zA2-3.347

18、e-05z+1.125e-07Sampletime:1secondsDiscrete-timetransferfunction.SIjbpRea-DDn-xerunelS4I：Qrrd9110.判定下列差繳傳遞函數模型的穩定性.11,1“十以2一內十2也4十及：十2十。十JU）國4+匹一3M營十2(a)G=tf(1,1212);eig(G),pzmap(G)ans=-2.0000-0.0000+1.0000i-0.0000-1.0000i系統為臨界穩止o(b) G=tf(1,63211);eig(G),pzmap(G)ans=-0.4949+0.4356i-0.4949-0.4356i0.24

19、49+0.5688i0.2449-0.5688isplkjubim)04rUAruj-Rea*Axe(MwneiB:1有一對共腕復根在右半平面，所以系統不穩定(c) G=tf(1,11-3-12);eig(G),pzmap(G)ans=-2.0000-1.00001.00001.0000PnhB-ZflroiMur有兩根在右半平面，故系統不穩定。11.判定下面采樣系統的穩定柢3-3-0.39ff-0.09(b)及=J一1,711.04;2+0-268;十0,024-3二十2（力=,二口,2二0.2二|0.田(1) H=tf(-32,1-0.2-0.250.05);pzmap(H),abs(ei

20、g(H)ans=0.50000.50000.2000-MMSUS.-sxy.ca_a_nE-FlealAxnIseconits1)系統穩定。(2) H=tf(3-0.39-0.09,1-1.71.040.2680.024);pzmap(H),abs(eig(H)ans=1.19391.19390.12980.1298系統不穩定12.治出連續系統的狀態方程模型，請判定系統的穩定性-0.215000010-0.51.600.T(i)=00-11.3S5.80x(t)十0四）000-33.31000.0000-10-:打(1)A=-0.20.5000;0-0.51.600;00-14.385.80;

21、000-33.3100;0000-10;B=000030;C=zeros(1,5);D=0;G=ss(A,B,C,D),eig(G).學習參x1x2x3x4x5x1-0.20.5000x20-0.51.600x300-14.385.80x4000-33.3100x50000-10b=u1x10ans=-0.2000-0.5000-14.3000-33.3000-10.0000x20x40x530c=x1x2x3x4x5y100000d=u1y10Continuous-timestate-spacemodel.系統穩定。13.請求出下面自治系統狀態方程的解析解-4-1;-320-4;A=sym(

22、A);syms-5出=_3.-3并和數值解得出的曲線比較一A=-5200;0-400;-32t;x=expm(A*t)*1;2;0;1x=4*exp(-4*t)-3*exp(-5*t)2*exp(-4*t)12*exp(-4*t)-18*exp(-5*t)+3*t*exp(-4*t)-4*tA2*(exp(-4*t)/(4*t)+exp(-4*t)/(2*tA2)+8*tA2*(exp(-4*t)/2-exp(-4*t)/(2*t)-16*t*(exp(-4*t)-exp(-4*t)/(2*t)6*exp(-4*t)-9*exp(-5*t)-8*t*(exp(-4*t)-exp(-4*t)/(

23、2*t)G=ss(-5200;0-400;-32-4-1;-320-4,1;2;0;1,eye(4),zeros(4,1);tt=0:0,01:2;xx=;fori=1:length(tt)t=tt(i);xx=xxeval(x);endy=impulse(G,tt);plot(tt,xx,tt,y,:)解析解和數值解的脈沖響應曲線如圖所示，可以看出他們完全一致14.試繪制不列開環系統的根軌跡曲線，并大致確定使單位負反饋系統穩定的K值范圍K(s十6)s(s+3)(A-F4-Lj)(s+4-4j)(a)s=tf(s);G=(s+6)*(s-6)/(s*(s+3)*(s+4-4j)*(s+4+4j

24、);rlocus(G),grid不存在K使得系統穩定(b)G=tf(1,2,2,111480);rlocus(G),gridRealAxafaeccnds1)放大根軌跡圖像，可以看到，根軌跡與虛軸交點處，K伯:為5.53,因此，,-3E一0Ktau=2;n,d=paderm(tau,1,3);s=tf(s);G=tf(n,d)*(s-1)/(s+1)A5,rlocus(G)-1.5sA2+4.5s-3sA8+8sA7+29.5sA6+65.5sA5+95sA4+91sA3+55.5sA2+19.5s+3Continuous-timetransferfunction.Rx刈Locus7歪Reo:

25、Ldd箏由圖得0Ks=tf(s);G=8*(s+1)/(sA2*(s+15)*(sA2+6*s+10);bode(G),figure,nyquist(G),figure,nichols(G),Gm,y,wcg,wcp=margin(G),figure,step(feedback(G,1)Gm=30.4686y=4.2340wcg=1.5811wcp=0.2336BodeOogr-am5*0sflfwMTO翁I1工21-MlSE省nEnw5o5)-2-2-3用JB雪左Frquflz=tf(z);G=0.45*(z+1.31)*(z+0.054)*(z-0.957)/(z*(z-1)*(z-0.3

26、68)*(z-0.99);bode(G),figure,nyquist(G),figure,nichols(G),Gm,y,wcg,wcp=margin(G),figure,step(feedback(G,1)Warning:Theclosed-loopsystemisunstable.Inwarningat26InDynamicSystem.marginat63Gm=0.9578y=-1.7660wcg=1.0464wcp=1.0734FTtqMsney什Mrt)270215135Opti-LaoeFhaie(deojNgu&iDdigramReelAxd10001如0200025W3期3如

27、040M延帆I5000Un*IBflGDnil系統不穩定17.假設受控對象模型為G=并假設由某種方法設計出串聯控制器模s(I十s/(L5)(I十s/50)型為G式身)=絆坐共空加，試用頻域響應的方法判定閉環系統的性能，并用時娥晌(s十U.5)(十50)應檢驗得出的結論.s=tf(s);G=100*(1+s/2.5)/(s*(1+s/0.5)*(1+s/50);Gc=1000*(s+1)*(s+2.5)/(s+0.5)*(s+50);GG=G*Gc;nyquist(GG),grid,figure,bode(GG),figure,nichols(GG),grid,figure,step(feedb

28、ack(GG,1)由奈氏圖可得，曲線不包圍（-1,j0）點，而開環系統不含有不穩定極點，所以根據奈氏穩定判據閉環系統是穩定的。L8SbepResparrie0.40.2Tm(sflcandaji用階躍響應來驗證，可得系統是穩定的第二部分：Simulink在系統仿真中的應用、控制系統計算機輔助設計、控制工程中的仿真技術應用2.2考慮簡單的踐性微分手程峭如+5y+的十每十2y-b3t十唱4st疝in+開;，3)且子程的初值為y(0)=1,諷=式0)=l/2,y(0)=02試用Simulink搭建起系統的仿真模型，井繪制出仿真結果曲線.由第2章介絹的知識，該方程可以用微分方程數值解的形式進行分析，試

29、比較二者的分析結果，symsyt;y=dsolve(D4y+5*D3y+6*D2y+4*Dy+2*y=exp(-3*t)+exp(-5*t)*sin(4*t+pi/3),y(0)=1,Dy(0)=1/2,D2y(0)=1/2,D3y(0)=1/5);tt=0:.05:10;yy=;fork=1:length(tt)yy=yysubs(y,t,ti)；endplot(tout,yout,tt,yy,:)3.建立起如圖印7所示非線性系統網的Sinmlink框圖，并觀察在單位除跋信號輸入下系統的輸出曲線和誤差曲線.圖齊了習題5的系統方框圖輸出曲線及誤差曲線4.已知某系統的Simulink仿真框密如圖

30、5-15所示，試由謨框圖寫出系統的數學橫型公式.*1/51/SloregrataeImegratcrl【口舊察皿口1加記富耳噂四p5in(u(1)*exp(2.3*(-i(2)J)4FenPit-AoductClock圖5-15習題7的SimMink仿直框圖X1=tsin(x2e2.3(4)X2=XiX3=sin(x2e2.3(4)5.假設已知直流電機拖動模型方框圖如圖二17所示，試利用Simulink提供的工具提取該系統的總模型，并利用該工具繪制系統的階躍響鹿、頑域響魔曲線”圖5-17習題10的臺流出機拖劫系統方椎.圖A,B,C,D=linmod(part2_5);G=ss(A,B,C,D)

31、Warning:Usingadefaultvalueof0.2formaximumstepsize.Thesimulationstepsizewillbeequaltoorlessthanthisvalue.YoucandisablethisdiagnosticbysettingAutomaticsolverparameterselectiondiagnostictononeintheDiagnosticspageoftheconfigurationparametersdialog.Indlinmodat172Inlinmodat60a=x1x2x3x4x5.學習參考x6x1000000x20

32、-1000000x31300-100000x40200-0.88-10000x50000-1000x6000294.1-29.41-149.3x70100-0.44000x8-27.5600001.045e+004x9000100-100x1000000x7x8x9x10x101.400x20000x30000x411.76000x501.400x60019.610x70000x80-6.66700x90000x100000b=u1x10x21x30x40x50x60x70x80x9x100c=x1x2x3x4x5x6x7x8x9x10y1130000000000d=u1y10Continuo

33、us-timemodel.subplot(221),step(G),grid,subplot(222),bode(G),grid,subplot(223),nyquist(G),grid,subplot(224),nichols(G),grid階躍響應和頻率響應曲線6.HyguifilDifigrflm口1口。2DQ3004GQReaLAxitBodeuagrm一岳B強一dMiad口rirluudo假設系統的對象模型為G(s)=-一并已知我們可以設!t-個控制器為&=請觀察在該控制器式系統的動態特性-比較原系統和校正后系統的幅值和相匝雷專堂給出進一步改進系統性能的建議.s=tf(s);G=21

34、0*(s+1.5)/(s+1.75)*(s+16)*(sA2+3*s+11.25);Gc=52.5*(s+1.5)/(s+14.86);GG=feedback(G*Gc,1);step(feedback(G,1),figure,step(GG),xlim(8595)口tU4niM2月口與Gm,garma,wcg,wcp=margin(G)Gm=4.8921garma=60.0634wcg=7.9490wcp=3.9199Gm,garma,wcg,wcp=margin(G*Gc)Warning:Theclosed-loopsystemisunstable.Inwarningat26InDynam

35、icSystem.marginat63Gm=0.8090garma=-6.0615wcg=17.1659wcp=7.若系統的狀態方程模型為()3.選擇加權矩陣Q=由電(1,2,3,4)及冗=心，則設計出這一線性二次型指標的最優控制搟及在最優控制下的閉環系統極點位置，并繪制出閉環系統各個狀態的曲線.A=0100;0010;-3123;2100;B=10;21;32;43;Q=diag(1234);R=eye(2);K,P=lqr(A,B,Q,R),eig(A-B*K)-0.09781.21181.87670.7871-3.8819-0.46682.67131.0320P=5.44000.6152-2.31630.04520.61521.8354-0.0138-0.7582-2.3163-0.01