由实验数据求传递函数范例

阶跃响应采样数据如下所示，求出传递的估计值。 0 0 0.20 0.0338 0.90 0.4409 0.02 0.0001 0.22 0.0395 1.00 0.4924 0.04 0.0005 0.24 0.0480 1.50 0.6904 0.06 0.0014 0.26 0.0571 2.00 0.8121 0.08 0.0031 0.28 0.0668 2.50 0.8860 0.10 0.0057 0.30 0.0771 3.00 0.9309 0.12 0.0091 0.50 0.1979 3.50 0.9581 0.14 0.0135 0.60 0.2624 4.00 0.9746 0.16 0.0187 0.70 0.3253 5.00 0.9907 0.18 0.0248 0.80 0.3851                 解: MATLAB程序如下： t=[0:0.02:0.30 0.50:0.10:1.00 1.50:0.50:4.00 5.00]; yt=[0 0.0001 0.0005 0.0014 0.0031 0.0057 0.0091 0.0135 0.0187 0.0248 0.0338 ... 0.0395 0.0480 0.0571 0.0668 0.0771 0.1979 0.2624 0.3253 0.3851 0.4409 ... 0.4924 0.6904 0.8121 0.8860 0.9309 0.9581 0.9746 0.9907]; yinf=1; %firstly fitting euqation y1=log(yinf-yt); p=polyfit(t,y1,1); alpha=-p(1); A=-exp(p(2)); yt1=yinf+A*exp(-alpha*t); figure(1) plot(t,yt,'*',t,yt1) grid on; figure(2) plot(t,p(1)*t+p(2),t,y1,'*'); grid on; e1=yt1-yt; figure(3) plot(t,e1) grid on; shg a1=find(abs(e1)<=0.05); a11=find(t(a1)>1.0); y11=log(yinf-yt(a1(a11(1)):29)); p=polyfit(t(a1(a11(1)):29),y11,1); alpha=-p(1); A=-exp(p(2)); figure(4) plot(t,p(1)*t+p(2),t,y1,'*'); grid on; yt11=yinf+A*exp(-alpha*t); figure(5) plot(t,yt,'*',t,yt11) grid on; %secondly fitting euqation ys2=yt-yt11; a2=find(ys2>=0.0001); y2=log(ys2(a2)); t2=t(a2); p2=polyfit(t2,y2,1); beta=-p2(1); B=exp(p2(2)); yt2=yt11+B*exp(-beta*t); figure(6) plot(t,yt,'*',t,yt2) grid on; e2=yt2-yt; figure(7) plot(t,e2) grid on; shg b=find(abs(e2)<=0.001); b2=find(t(b)>0.5); y22=log(ys2(b(b2(1)):29)); p=polyfit(t(b(b2(1)):29),y22,1); beta=-p2(1); B=exp(p2(2)); yt22=yt11+B*exp(-beta*t); figure(8) plot(t,e2) grid on; shg figure(9) plot(t,yt,'*',t,yt22) grid on; hold on; e2=yt22-yt; syms xf f=(x+alpha)*(x+beta)+A*(x+beta)*x+B*(x+alpha)*x; f=collect(f); num=sym2poly(f); den=conv(conv([1,0],[1,alpha]),[1,beta]); tf(num,den(1:3)) [y,t3]=step(tf(num,den(1:3))); figure(10) plot(t3,y,'k',t,yt,'*') 结果如下：（只给出主要的结果）