控制系统仿真与CAD实验报告.docx

1、控制系统仿真与CAD实验报告控制系统仿真与CAD实验课程报告、实验教学目标与基本要求上机实验是本课程重要的实践教学环节。实验的目的不仅仅是验证理论知 识，更重要的是通过上机加强学生的实验手段与实践技能，掌握应用 MATLAB/Simulink求解控制问题的方法，培养学生分析问题、解决问题、应用 知识的能力和创新精神，全面提高学生的综合素质。通过对MATLAB/Simulink进行求解，基本掌握常见控制问题的求解方法与命 令调用，更深入地认识和了解MATLAB语言的强大的计算功能与其在控制领域的 应用优势。上机实验最终以书面报告的形式提交，作为期末成绩的考核内容。二、题目及解答第一部分：MATL

2、AB必备基础知识、控制系统模型与转换、线性控制系统的计 算机辅助分析1.考虑碧名的Rossol化学反应方程组选定d = c = 且T1(O)= r2(0) = x3(0) =0.绘制仿育勢果的三维相轨:卜 并傅 出其在脣丫平面上的投彭.f=i nlin e(-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.求解下囲的最优化问题”(

3、a) min - 2jti 十 ar空)(4谥+鬭* X firtn 0y=(x)x(1F2-2*x(1)+x(2);ff=optimset;ff.LargeScale=off;ff.TolFu n=1e-30;ff.TolX =1e-15;ff.TolCo n=1e-20;x0=1;1;1;xm=0;0;0;xM=;A=;B=;Aeq=;Beq=;x,f,c,d =fmi nco n(y,x0,A,B,Aeq,Beq,xm,xM,wzhfc1,ff)Warning: Opti ons LargeScale = off and Algorithm = trust-region-reflecti

4、ve conflict.Ignoring Algorithm and running active-set algorithm. To run trust-regio n-reflective, set LargeScale = on. To run active-set without this warning, use Algorithm = active-set. In fmincon at 456Local minimum possible. Constraints satisfied.fmincon stopped because the size of the current se

5、arch direction is less than twice the selected value of the step size tolerance and constraints are satisfied to within the selected value of the constraint tolerance.Active inequalities (to within options.TolCon = 1e-20):lower upper ineqlin ineqnonlin2x =1.000001.0000f =-1.0000c =iterations: 5 fun

6、cCou nt: 20Issteple ngth: 1stepsize: 3.9638e-26algorithm: medium-scale: SQP, Quasi-Newt on, li ne-search firstorderopt: 7.4506e-09con strviolati on: 0 message: 1x766 char3.请持下面的传递函數欖理输入到MATLAB环境闾*)=科晶舞FT恥)=时备瞎而护秒(a) s=tf(s);G=(sA3+4*s+2)/(sA3*(sA2+2)*(sA2+1)A3+2*s+5)sA3 + 4 s + 2sA11 + 5 sA9 + 9 sA7

7、 + 2 sA6 + 12 sA5 + 4 sA4 + 12 sA3 Con ti nu ous-time tran sfer function.(b) z=tf(z,0.1);H=(zA2+0.568)/(z-1)*(zA2-0.2*z+0.99)H =zA2 + 0.568zA3 - 1.2 zA2 + 1.19 z - 0.99Sample time: 0.1 sec ondsDiscrete-time tran sfer function.4.假设描述系统的常微分方程为期+ 13y(t)十4讥站+ = 请选择一组状态变量，并将此方程在MATLAB工作空闻中表示出来.如果想得到系统的传递

8、菌数和零极点模型： 我V将如何求取？得出的结果又是怎样西？由徴分方程模型能否直接写岀系统的传递函数模 型？ A=0 1 0;0 0 1;-15 -4 -13;B=0 0 2;C=1 00;D=0;G=ss(A,B,C,D),Gs=tf(G),Gz=zpk(G)x1x2x3x1010x2001x3-15-4-13b =u1x10x20x32c =x1 x2 x3y1 1 0 0d =u1y1 0Con ti nu ous-time state-space model.Gs =sA3 + 13 sA2 + 4 s + 15Con ti nu ous-time tran sfer function.

9、Gz =(s+12.78) (sA2 + 0.2212s + 1.174)Con ti nu ous-time zero/pole/ga in model.5.已知某系统的差分方程摸型为+ 2) + y(k + 1.)十0】切的=卷依+ 1.) v2u(k).试将其输入 到MATLAB X作空间设采样周期为0.01s z=tf(z,0.01);H=(z+2)/(zA2+z+0.16)H =zA2 + z + 0.16Sample time: 0.01 sec ondsDiscrete-time tran sfer function.6.假设某单位负反馈系统申，=(3十+劄+ 5) Gc(3)

10、= (KFS七瓦订2、试用 MAT1 - A 13推导岀闭环系统的传递函数模型。 syms J Kp Ki s;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.从二面给出西典型反馈控制系统结构子模型中一求出总系统的狀态方程与传递函数模型并 得出各个模型的零极点模型表示”(b) G(z169.6.-十IIII“(g 十 4；i211.87s + 317.64357S6.7-1 十 10844

11、4（1十住一1十勿）心一1十74.04+ 211)+ 94.34 )(s + II.16S4)(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 =359.3 sA3 + 3.732e04 sA2 + 1.399e05 s + 1270560.01 sA6 + 2.185 sA5 + 142.1 sA4 + 2444 sA3 + 4.389e04 sA2 + 1

12、.399e05 s + 127056Continuous-time transfer function.Gd =a =x1x2x3x4x5x6x1-218.5-111.1 -29.83-16.74-6.671-3.029x212800000x30640000x40032000x5000800x6000020b =u1x14x20x30x40x50x60x1x2x3x4x5 x6y1001.0973.5591.668 0.7573y1u10Continuous-time state-space model.Gz =35933.152 (s+100) (s+2.358) (s+1.499)(sA2

13、 + 3.667s + 3.501) (sA2 + 11.73s + 339.1) (sA2 + 203.1s + 1.07e04) Continuous-time zero/pole/gain model.(b)设采样周期为0.1s z=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 =-108444 zA5 + 1.844e04 zA4 + 1.789e04 zA31.144

14、e05 zA5 + 2.876e04 zA4 + 274.2 zA3 + 782.4 zA2 + 47.52 z + 0.5Sample time: 0.1 secondsDiscrete-time transfer function.Gd =x1x2x3x4x5x1x2x3-0.2515 -0.00959 -0.1095 -0.053180.250.250.125x4-0.017910.03125x5b =u1 x1 1 x2 0 x3 0 x4 0 x5 0x1x2x3x4x5y1 0.3996 0.6349 0.1038 0.05043 0.01698u1y1 -0.9482Sample

15、 time: 0.1 sec ondsDiscrete-time state-space model.Gz =-0.94821 zA3 (z-0.5) (z+0.33)(z+0.3035) (z+0.04438) (z+0.01355)宫2 - 0.11z + 0.02396)Sample time: 0.1 sec ondsDiscrete-time zero/pole/ga in model.8.lL知系统的方框图如图匸3 :斤几 试推导岀从输入信号到输出倍号以的总杀统模型。s=tf(s);g 仁 1/(s+1);g2=s/(sA2+2);g3=1/sA2;g4=(4*s+2)/(s+1)

16、A2;g5=50;g6=(sA2+2)/(sW+14);G 仁feedback(g1*g2,g4);G2=feedback(g3,g5);GG=3*feedback(G1*G2,g6)GG =3 sA6 + 6 sA5 + 3 sA4 + 42 sA3 + 84 sA2 + 42 ssA10 + 3 sA9 + 55 sA8 + 175 sA7 + 300 sA6 + 1323 sA5 + 2656 sA4 + 3715 sA3 + 7732sA2 + 5602 s + 1400Con ti nu ous-time tran sfer fun cti on.9.is- l)a(sa 十 2s

17、十 400)砾传递函数模型&7仆亠汕二亠十皿对不同采擇周取“0.01,0.1和秒对之进行高散化，比较原系统的廉跃响应与各离散系统的阶跃啊应曲域.提示：后面将介绍，如果已知系统撲型为G,则Step(G即可绘制出其阶跃响应曲线.s=tf(s);T0=0.01;T1=0.1;T2=1;G=(s+1)A2*(sA2+2*s+400)/(s+5)A2*(sA2+3*s+100 )*(sA2+3*s+2500);Gd仁c2d(G,T0),Gd2=c2d(G,T1),Gd3=c2d(G,T2),step(G),figure,ste p(Gd1),figure,step(Gd2),figure,step(Gd

18、3)Gd1 =4.716e-05zA5 - 0.0001396乙八4 + 9.596e-05zA3 + 8.18e-05zA2 - 0.0001289z + 4.355e-05zA6 - 5.592 zA5 + 13.26 乙八4 - 17.06 乙八3 + 12.58 乙八2 - 5.032 z + 0.8521Sample time: 0.01 secondsDiscrete-time transfer function.Gd2 =0.0003982zA5 - 0.0003919zM - 0.000336zW + 0.0007842zA2 - 0.000766z +0.0003214zA6

19、 - 2.644 zA5 + 4.044 zA4 - 3.94 zA3 + 2.549 zA2 - 1.056 z + 0.2019Sample time: 0.1 secondsDiscrete-time transfer function.Gd3 =8.625e-05 zA5 - 4.48e-05 zA4 + 6.545e-06 zA3 + 1.211e -05 zA2 - 3.299e-06 z +1.011e-07zA6 - 0.0419 zA5 - 0.07092 zA4 - 0.0004549zA3 + 0.002495zA2 - 3.347e-05z +1.125e-07Samp

20、le time: 1 secondsDiscrete-time transfer function.辛.=O!.? T-Jr.T*lfiF10.判定下列逹续传递函数模型的稳定性。(町嗣+ 2辭十应十2 庆4十艮3 + 2辭十書十 J)曲十昇3屛一总十2(a) G=tf(1,1 2 1 2);eig(G),pzmap(G) ans =-2.0000-0.0000 + 1.0000i-0.0000 - 1.0000i系统为临界稳定(b) G=tf(1,6 3 2 1 1);eig(G),pzmap(G)ans =-0.4949 + 0.4356i-0.4949 - 0.4356i0.2449 +

21、0.5688i0.2449 - 0.5688iDle-ZeTi Map期 4.1有一对共轭复根在右半平面，所以系统不稳定arJL4L -CQUS - JeE-JffE-(c) G=tf(1,1 1 -3 -1 2);eig(G),pzmap(G) ans =-2.0000-1.00001.0000 1.0000有两根在右半平面，故系统不稳定。11.判足F雹采样系统的伦足世(a)坦二)=迟十2d 二 0.2日二。25匸 I 0.05()卅(=)=3尹0.3恥-0.09-1.7z3 lMz2 + 0.268c +0,02J H=tf(-3 2,1 -0.2 -0.25 0.05);pzmap(H)

22、,abs(eig(H) ans =0.50000.50000.2000系统稳定 H=tf(3 -0.39 -0.09,1 -1.7 1.04 0.268 0.024);pzm ap(H),abs(eig(H) ans =1.19391.19390.12980.1298系统不稳疋12.給出连续系统的状态方程模型，请判定系统的稳定性.-0.2000 1 0 10-0.51.6000(a). =001-1.3S5.80x(t)十0ti(f)00033.31000.0000-10-.30 J(1) A=-0.2 0.5 0 0 0;0 -0.5 1.6 0 0;0 0 -14.3 85.8 0;0 0 0 -33.3 100;0 0 0 0 -10;B=0 0 0 0 30;C=zeros(1,5);D=0;G=ss(A,B,C,D),eig(G)x1x2x3x4x5x1-0.20.5000x20-0.51.600x300-14.385.80x4000-33.3100x50000-10b =u1x1 0 ans =-0.2000-0.5000-14.3000