用MATLAB 实现 控制系统的模型转换.doc

实验 控制系统的模型转换一实验目的：掌握控制系统的微分方程、状态方程、传递函数、零极点增益、部分分式描述及转换；掌握常用数据拟合与插值方法。二实验方法及预习内容：1利用Matlab工具箱中常用的五种模型转换命令进行模型描述和转换；2利用Matlab工具箱中的多项式拟合命令对实验数据进行拟合。三实验内容：1用Matlab语言求下列系统的状态方程、传递函数、零极点增益、和部分分式形式的模型参数，并分别写出其相应的数学模型表达式：（1） G（s）= 程序：num=1 14 48 48;den=1 20 70 100 48;A,B,C,D=tf2ss(num,den)Z,P,K=tf2zp(num,den)R,P,H=residue(num,den)sys=ZPK(Z,P,K)sys=ss(A,B,C,D)运行结果：A = -20 -70 -100 -48 1 0 0 0 0 1 0 0 0 0 1 0B = 1 0 0 0C = 1 14 48 48D = 0Z = -9.4641 -2.5359 -2.0000P = -16.0051 -1.5269 + 0.9247i -1.5269 - 0.9247i -0.9412 K = 1R = 0.3892 -0.0932 - 0.3754i -0.0932 + 0.3754i 0.7973 P = -16.0051 -1.5269 + 0.9247i -1.5269 - 0.9247i -0.9412 H = Zero/pole/gain: (s+9.464) (s+2.536) (s+2)-(s+16.01) (s+0.9412) (s2 + 3.054s + 3.187) a = x1 x2 x3 x4 x1 -20 -70 -100 -48 x2 1 0 0 0 x3 0 1 0 0 x4 0 0 1 0 b = u1 x1 1 x2 0 x3 0 x4 0 c = x1 x2 x3 x4 y1 1 14 48 48 d = u1 y1 0 Continuous-time model.（2） =y=0 2 0 2 X程序：A=2.25 -5 -1.25 -0.5;2.25 -4.25 -1.25 -0.25;0.25 -0.5 -1.25 -1;1.25 -1.75 -0.25 -0.75;B=4;2;2;0;C=0 2 0 2;D=0;num,den=ss2tf(A,B,C,D)Z,P,K=ss2zp(A,B,C,D)R,P,H=residue(num,den)sys=tf(num,den)sys=ZPK(Z,P,K)运行结果：num = 0 4.0000 14.0000 22.0000 15.0000den = 1.0000 4.0000 6.2500 5.2500 2.2500Z = -1.0000 + 1.2247i -1.0000 - 1.2247i -1.5000 P = -0.5000 + 0.8660i -0.5000 - 0.8660i -1.5000 -1.5000 K = 4.0000R = 4.0000 -0.0000 0.0000 - 2.3094i 0.0000 + 2.3094iP = -1.5000 -1.5000 -0.5000 + 0.8660i -0.5000 - 0.8660iH = Transfer function: 4 s3 + 14 s2 + 22 s + 15-s4 + 4 s3 + 6.25 s2 + 5.25 s + 2.25 Zero/pole/gain:4 (s+1.5) (s2 + 2s + 2.5)- (s+1.5)2 (s2 + s + 1) 2已知元件的实验数据如下，拟合这一数据，并尝试给出其特性方程。 X 0.0100 1.0100 2.0100 3.0100 4.0100 Y 2.5437 7.8884 9.6242 11.6071 11.9727 X 5.0100 6.0100 7.0100 8.0100 9.0100 y 13.2189 14.2679 14.6134 15.4045 15.0805 采用一元线性回归：X=0.0100 1.0100 2.0100 3.0100 4.0100 5.0100 6.0100 7.0100 8.0100 9.0100;Y=2.5437 7.8884 9.6242 11.6071 11.9727 13.2189 14.2679 14.6134 15.4045 15.0805;N=2;x=ones(10,1) Xb,bint,r,rint,stats=regress(Y,x)for i=1:1:10 h(i)=1.2098*i+6.1659;endplot(X,Y,*,X,h)实验结果：b =6.1659 1.2098拟合优度：0.8314采用非线性回归：X=0.0100 1.0100 2.0100 3.0100 4.0100 5.0100 6.0100 7.0100 8.0100 9.0100;Y=2.5437 7.8884 9.6242 11.6071 11.9727 13.2189 14.2679 14.6134 15.4045 15.0805;N=2;polyfit(X,Y,N)for i=1:1:10 h(i)=-0.1905*i*i+2.9283*i+3.8625;endplot(X,Y,*,X,h)得到的结果：ans = -0.1905 2.9283 3.8625