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1、-控制系统仿真与CAD实验课程报告一、实验教学目标与根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进展求解，根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容。二、题目及解答第一局部：MATLAB 必备根底知识、控制系统模型与
2、转换、线性控制系统的计算机辅助分析1.f=inline(-*(2)-*(3);*(1)+a*(2);b+(*(1)-c)*(3),t,*,flag,a,b,c);t,*=ode45(f,0,100,0;0;0,0.2,0.2,5.7);plot3(*(:,1),*(:,2),*(:,3),grid,figure,plot(*(:,1),*(:,2),grid2.y=(*)*(1)2-2*(1)+*(2);ff=optimset;ff.LargeScale=off;ff.TolFun=1e-30;ff.Tol*=1e-15;ff.TolCon=1e-20;*0=1;1;1;*m=0;0;0;*M
3、=;A=;B=;Aeq=;Beq=;*,f,c,d=fmincon(y,*0,A,B,Aeq,Beq,*m,*M,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, useAlgo
4、rithm = 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 tolerance.
5、Active inequalities (to within options.TolCon = 1e-20): lower upper ineqlin ineqnonlin 2 * = 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: 1*76
6、6 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 transfer
7、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 = *1 *2 *3 *1 0 1 0 *2 0 0 1 *3 -15 -4 -13b = u1 *1 0 *2 0 *3 2c = *1 *2 *3 y1 1 0 0d = u1 y1 0Continuous-time state-space model.Gs = 2 - s3 + 13 s2 + 4 s + 15 Continuous-time transfer function.Gz = 2 - (s+
8、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 + 1)/(J*s3 +
9、 (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 + 1.399e05 s + 12
10、7056Continuous-time transfer function.Gd = a = *1 *2 *3 *4 *5 *6 *1 -218.5 -111.1 -29.83 -16.74 -6.671 -3.029 *2 128 0 0 0 0 0 *3 0 64 0 0 0 0 *4 0 0 32 0 0 0 *5 0 0 0 8 0 0 *6 0 0 0 0 2 0 b = u1 *1 4 *2 0 *3 0 *4 0 *5 0 *6 0 c = *1 *2 *3 *4 *5 *6 y1 0 0 1.097 3.559 1.668 0.7573 d = u1 y1 0Continuou
11、s-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);GG=feedback(G*Gc,H),Gd=ss(GG),Gz=zpk(GG)GG = -108444 z5 + 1.844e04 z4 + 1.789e04 z3 - 1.144e05 z5 + 2.876e04 z4 + 274.2 z3 + 782.4 z2 + 47.52 z + 0.5 Sample time: 0.1 secondsDiscrete-time transfer function.Gd = a = *1 *2 *3 *4 *5 *1 -0.2515 -0.00959 -0.1095 -0.05318 -0.0179
