实验二 控制系统建模与分析.doc

实验二 控制系统建模与分析一、 实验目的熟悉MATLAB对控制系统的建模以及时域和频域的分析函数。二、 实验题目1. 表示下列传递函数模型,并转化成其他的数学模型 num1=conv(4,8,conv(1,6,6,1,6,6);den1=conv(1,0,conv(1,3,3,1,1,3,2,5);sys1=tf(num1,den1)sys=zpk(sys1)sys=ss(sys1) num2=4,-2;den2=1,0,2,5;sys2=tf(num2,den2)sys=zpk(sys2)sys=ss(sys2) num3=1;den3=conv(2,1,1,2);sys3=tf(num3,den3)sys=zpk(sys3)sys=ss(sys3) a=0,1;-1,-2,b=0;1,c=0,1,d=0;sys4=ss(a,b,c,d)sys=tf(sys4)sys=zpk(sys4)（1）Transfer function: 4 s5 + 56 s4 + 288 s3 + 672 s2 + 720 s + 288-s7 + 6 s6 + 14 s5 + 21 s4 + 24 s3 + 17 s2 + 5 s Zero/pole/gain: 4 (s+4.732)2 (s+2) (s+1.268)2-s (s+2.904) (s+1)3 (s2 + 0.09584s + 1.722) a = x1 x2 x3 x4 x5 x6 x7 x1 -6 -3.5 -2.625 -1.5 -1.063 -0.3125 0 x2 4 0 0 0 0 0 0 x3 0 2 0 0 0 0 0 x4 0 0 2 0 0 0 0 x5 0 0 0 1 0 0 0 x6 0 0 0 0 1 0 0 x7 0 0 0 0 0 0.5 0 b = u1 x1 16 x2 0 x3 0 x4 0 x5 0 x6 0 x7 0 c = x1 x2 x3 x4 x5 x6 x7 y1 0 0.0625 0.4375 1.125 2.625 2.813 2.25 d = u1 y1 0 Continuous-time model. （2）Transfer function: 4 s - 2-s3 + 2 s + 5 Zero/pole/gain: 4 (s-0.5)-(s+1.328) (s2 - 1.328s + 3.764) a = x1 x2 x3 x1 0 -1 -1.25 x2 2 0 0 x3 0 2 0 b = u1 x1 2 x2 0 x3 0 c = x1 x2 x3 y1 0 1 -0.25 d = u1 y1 0 Continuous-time model. （3）Transfer function: 1-2 s2 + 5 s + 2 Zero/pole/gain: 0.5-(s+2) (s+0.5) a = x1 x2 x1 -2.5 -1 x2 1 0 b = u1 x1 0.5 x2 0 c = x1 x2 y1 0 1 d = u1 y1 0 （4）Continuous-time model.a = 0 1 -1 -2b = 0 1c = 0 1 a = x1 x2 x1 0 1 x2 -1 -2 b = u1 x1 0 x2 1 c = x1 x2 y1 0 1 d = u1 y1 0 Continuous-time model. Transfer function:s - 2.22e-016-s2 + 2 s + 1 Zero/pole/gain: s-(s+1)22. 一个系统的结构框图如下所示。（1）试绘出系统闭环的根轨迹图；并在根轨迹图上任选一点，试计算该点的增益k及其所有极点的位置；g1=tf(1,0.5,1,0.1);g2=tf(20,conv(1,0,conv(1,2,1,10);g3=series(g1,g2);g4=feedback(g3,1);g5=tf(1.5,1);g6=series(g5,g4)rlocus(g6)k,poles=rlocfind(g6)Transfer function: 30 s + 15-s4 + 12.1 s3 + 21.2 s2 + 22 s + 10 Select a point in the graphics windowselected_point = -18.4597 + 0.0920ik = 86.9298poles = -18.4601 3.4306 +11.4139i 3.4306 -11.4139i -0.5011 （2）试绘出系统的bode图，并计算系统的幅值裕量和相角裕量，并给出几种方法判断系统的稳定性。bode(g6)gm,pm,wcg,wcp=margin(g6)gm = 5.4038pm = 42.5806wcg = 3.9008wcp =1.78123. 已知系统传递函数为：(1) 绘出系统在阶跃信号为0.11(t)时的响应曲线，并求系统性能指标：稳态值、上升时间、调节时间、超调量。g0=tf(20,1,8,36,40,20),step(g0*0.1)pos,ta,ts,tp=stepchar(g0*0.1,0.005)Transfer function: 20-s4 + 8 s3 + 36 s2 + 40 s + 20pos = 2.5945ta = 4.5988ts = 8.5632tp = 5.7881 (2) 绘出离散化系统在阶跃信号为0.11(t)时的响应曲线，采样周期为T=0.3秒。figuresysd=c2d(g0,0.3)step(sysd)Transfer function:0.004074 z3 + 0.02648 z2 + 0.01603 z + 0.0009526 z4 - 1.902 z3 + 1.268 z2 - 0.4099 z + 0.09072Sampling time: 0.3 4. 对于教材例9.4.3中引用了构造Routh判据的列表程序，在运行时发现当劳斯表出现全零行和第一列元素为零时，程序有错误。改正并完善此程序。原routh函数见教材p374.function rtab,info=routh0(den) if any(den=0 F(g)=n-k-f-1; f=f+2; g=g+1; end a3=vec2.*F; %求导后的系数即为原系数乘以相应阶数所得 info=info,All elements in row . int2str(k+2) are zeros; %输出提示信息 elseif abs(a3(1)eps %若第一行元素为0，其余不为0 a3(1)=1e-6; info=info,Replaced first element; end if k=length(den)-2 a3=den(end); end vec2=vec2,zeros(1,nrT-length(vec2); %补零 a3=a3,zeros(1,nrT-length(a3); rt