阶跃响应采样数据如下
表
关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf
所示，求出传递函数的估计值。
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）
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