控制系統(tǒng)仿真與CAD_實驗報告

1、控制系統(tǒng)仿真與CAD實驗課程報告一、實驗教學(xué)目標(biāo)與基本要求上機(jī)實驗是本課程重要的實踐教學(xué)環(huán)節(jié)。實驗的目的不僅僅是驗證理論知識，更重要的是通過上機(jī)加強學(xué)生的實驗手段與實踐技能，掌握應(yīng)用MATLAB/Simulink求解控制問題的方法，培養(yǎng)學(xué)生分析問題、解決問題、應(yīng)用知識的能力和創(chuàng)新精神，全面提高學(xué)生的綜合素質(zhì)。通過對MATLAB/Simulink進(jìn)行求解，基本掌握常見控制問題的求解方法與命令調(diào)用，更深入地認(rèn)識和了解MATLAB語言的強大的計算功能與其在控制領(lǐng)域的應(yīng)用優(yōu)勢。上機(jī)實驗最終以書面報告的形式提交，作為期末成績的考核內(nèi)容。二、題目及解答第一部分：MATLAB必備基礎(chǔ)知識、控制系統(tǒng)模型與轉(zhuǎn)換

2、、線性控制系統(tǒng)的計算機(jī)輔助分析1.考慮茗名的Mssol化學(xué)反應(yīng)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(jié)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(yōu)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(xué)習(xí)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(shù)模型輸入乳MATLAB環(huán)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(shè)描述系統(tǒng)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(tài)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(tǒng)的傳遞函數(shù)和零極點模型，我門將如何求取“得出的結(jié)果又是怎樣的？由微分方程模型能否直接寫出系統(tǒng)的傳遞函數(shù)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(tǒng)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(shè)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(shè)某單位負(fù)反

9、饋系統(tǒng)中，G=/十十區(qū)十5),&=+試用MATLAB推導(dǎo)出閉環(huán)系統(tǒng)的傳遞函數(shù)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(tǒng)結(jié)構(gòu)子模型中一求出總系統(tǒng)的狀態(tài)方程與傳遞函數(shù)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(shè)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(tǒng)的方樞圖如圖所示，試推導(dǎo)出從輸入信號r(i)到輸出信號y(t)的總系統(tǒng)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(shù)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(jìn)行離散化，比我原系統(tǒng)的階躍喃應(yīng)與各離散系統(tǒng)的階躍嘀應(yīng)曲線.提示：后面將介紹，如果已知系統(tǒng)模型為G,則用smp(G)即可繪制出其階躍響應(yīng)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(xué)習(xí)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(shù)模型的穩(wěn)定性.11,1“十以2一內(nèi)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(tǒng)為臨界穩(wěn)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(fù)根在右半平面，所以系統(tǒng)不穩(wěn)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(tǒng)不穩(wěn)定。11.判定下面采樣系統(tǒng)的穩(wěn)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(tǒng)穩(wěn)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(tǒng)不穩(wěn)定12.治出連續(xù)系統(tǒng)的狀態(tài)方程模型，請判定系統(tǒng)的穩(wěn)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(xué)習(xí)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(tǒng)穩(wěn)定。13.請求出下面自治系統(tǒng)狀態(tài)方程的解析解-4-1;-320-4;A=sym(

22、A);syms-5出=_3.-3并和數(shù)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(shù)值解的脈沖響應(yīng)曲線如圖所示，可以看出他們完全一致14.試?yán)L制不列開環(huán)系統(tǒng)的根軌跡曲線，并大致確定使單位負(fù)反饋系統(tǒng)穩(wěn)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(tǒng)穩(wěn)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I(xiàn)5000Un*IBflGDnil系統(tǒng)不穩(wěn)定17.假設(shè)受控對象模型為G=并假設(shè)由某種方法設(shè)計出串聯(lián)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(yīng)的方法判定閉環(huán)系統(tǒng)的性能，并用時娥晌(s十U.5)(十50)應(yīng)檢驗得出的結(jié)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(huán)系統(tǒng)不含有不穩(wěn)定極點，所以根據(jù)奈氏穩(wěn)定判據(jù)閉環(huán)系統(tǒng)是穩(wěn)定的。L8SbepResparrie0.40.2Tm(sflcandaji用階躍響應(yīng)來驗證，可得系統(tǒng)是穩(wěn)定的第二部分：Simulink在系統(tǒng)仿真中的應(yīng)用、控制系統(tǒng)計算機(jī)輔助設(shè)計、控制工程中的仿真技術(shù)應(yīng)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(tǒng)的仿真模型，井繪制出仿真結(jié)果曲線.由第2章介絹的知識，該方程可以用微分方程數(shù)值解的形式進(jìn)行分析，試

29、比較二者的分析結(jié)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(tǒng)網(wǎng)的Sinmlink框圖，并觀察在單位除跋信號輸入下系統(tǒng)的輸出曲線和誤差曲線.圖齊了習(xí)題5的系統(tǒng)方框圖輸出曲線及誤差曲線4.已知某系統(tǒng)的Simulink仿真框密如圖

30、5-15所示，試由謨框圖寫出系統(tǒng)的數(shù)學(xué)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(xí)題7的SimMink仿直框圖X1=tsin(x2e2.3(4)X2=XiX3=sin(x2e2.3(4)5.假設(shè)已知直流電機(jī)拖動模型方框圖如圖二17所示，試?yán)肧imulink提供的工具提取該系統(tǒng)的總模型，并利用該工具繪制系統(tǒng)的階躍響鹿、頑域響魔曲線”圖5-17習(xí)題10的臺流出機(jī)拖劫系統(tǒng)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(xué)習(xí)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(yīng)和頻率響應(yīng)曲線6.HyguifilDifigrflm口1口。2DQ3004GQReaLAxitBodeuagrm一岳B強一dMiad口rirluudo假設(shè)系統(tǒng)的對象模型為G(s)=-一并已知我們可以設(shè)!t-個控制器為&=請觀察在該控制器式系統(tǒng)的動態(tài)特性-比較原系統(tǒng)和校正后系統(tǒng)的幅值和相匝雷專堂給出進(jìn)一步改進(jìn)系統(tǒng)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(tǒng)的狀態(tài)方程模型為()3.選擇加權(quán)矩陣Q=由電(1,2,3,4)及冗=心，則設(shè)計出這一線性二次型指標(biāo)的最優(yōu)控制搟及在最優(yōu)控制下的閉環(huán)系統(tǒng)極點位置，并繪制出閉環(huán)系統(tǒng)各個狀態(tài)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