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1、控制系统仿真与CAD实验课程报告 一、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：MATLAB 必备基础知识、控制系统模型
2、与转换、线性控制系统的计算机辅助分析1. 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.y=(x)x(1)2-2*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
3、;0;xM=;A=;B=;Aeq=;Beq=;x,f,c,d=fmincon(y,x0,A,B,Aeq,Beq,xm,xM,wzhfc1,ff)Warning: Options LargeScale = off and Algorithm =trust-region-reflective conflict.Ignoring Algorithm and running active-set algorithm. To runtrust-region-reflective, setLargeScale = on. To run active-set without this warning, us
4、eAlgorithm = active-set. In fmincon at 456 Local minimum possible. Constraints satisfied.fmincon stopped because the size of the current search direction is less thantwice the selected value of the step size tolerance and constraints are satisfied to within the selected value of the constraint toler
5、ance.Active inequalities (to within options.TolCon = 1e-20): lower upper ineqlin ineqnonlin 2 x = 1.0000 0 1.0000f = -1.0000c = 4d = iterations: 5funcCount: 20lssteplength: 1stepsize: 3.9638e-26algorithm: medium-scale: SQP, Quasi-Newton, line-searchfirstorderopt: 7.4506e-09constrviolation: 0message:
6、 1x766 char3.(a) s=tf(s);G=(s3+4*s+2)/(s3*(s2+2)*(s2+1)3+2*s+5)G = s3 + 4 s + 2 - s11 + 5 s9 + 9 s7 + 2 s6 + 12 s5 + 4 s4 + 12 s3 Continuous-time transfer function.(b) z=tf(z,0.1);H=(z2+0.568)/(z-1)*(z2-0.2*z+0.99)H = z2 + 0.568 - z3 - 1.2 z2 + 1.19 z - 0.99Sample time: 0.1 secondsDiscrete-time tran
7、sfer function.4. A=0 1 0;0 0 1;-15 -4 -13;B=0 0 2;C=1 0 0;D=0;G=ss(A,B,C,D),Gs=tf(G),Gz=zpk(G)G = a = x1 x2 x3 x1 0 1 0 x2 0 0 1 x3 -15 -4 -13 b = u1 x1 0 x2 0 x3 2 c = x1 x2 x3 y1 1 0 0 d = u1 y1 0Continuous-time state-space model.Gs = 2 - s3 + 13 s2 + 4 s + 15 Continuous-time transfer function.Gz
8、= 2 - (s+12.78) (s2 + 0.2212s + 1.174) Continuous-time zero/pole/gain model.5.设采样周期为0.01s z=tf(z,0.01);H=(z+2)/(z2+z+0.16)H = z + 2 - z2 + z + 0.16 Sample time: 0.01 secondsDiscrete-time transfer function.6. syms J Kp Ki s;G=(s+1)/(J*s2+2*s+5);Gc=(Kp*s+Ki)/s;GG=feedback(G*Gc,1) GG = (Ki + Kp*s)*(s +
9、 1)/(J*s3 + (Kp + 2)*s2 + (Ki + Kp + 5)*s + Ki)7.(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 s3 + 3.732e04 s2 + 1.399e05 s + 127056 - 0.01 s6 + 2.185 s5 + 142.1 s4 + 2444 s3 + 4.389e04 s2 + 1.3
10、99e05 s + 127056Continuous-time transfer function.Gd = a = x1 x2 x3 x4 x5 x6 x1 -218.5 -111.1 -29.83 -16.74 -6.671 -3.029 x2 128 0 0 0 0 0 x3 0 64 0 0 0 0 x4 0 0 32 0 0 0 x5 0 0 0 8 0 0 x6 0 0 0 0 2 0 b = u1 x1 4 x2 0 x3 0 x4 0 x5 0 x6 0 c = x1 x2 x3 x4 x5 x6 y1 0 0 1.097 3.559 1.668 0.7573 d = u1 y1 0Continuous-time state-space model.Gz = 35933.152 (s+100) (s+2.358) (s+1.499) - (s2 + 3.667s + 3.501) (s2 + 11.73s + 339.1) (s2 + 203.1s + 1.07e04)Continuous-time zero/pole/gain model.(b)设采样周期为0.1sz=tf(z,0.1);G=(35786.7*z2+108444*z3)/(1+4*z)*(1+20*z)*(1+74.04*z);Gc=z/(1-z);H=z/(0.5-z);
