由实验数据求传递函数范例

阶跃响应采样数据如下表所示，求出传递函数的估计值。

t y(t)t y(t)t y(t)

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 0.240.0480 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）