Why the order of the system is reduced while using PEM in System Identification?

1 次查看(过去 30 天)
Hi All,
I used the following code for System Identification using PEM and iddata. But the initial system used to generate the step response was a 4th order system and the system identified using the following commands gave a 3rd order system. Why does this disparity happen?
Regards, Raghu
clc
clear all
close all
A=[-0.6129 0 .0645 0 ; 0 .7978 0 -.4494; -6.4516 0 -.7419 0;0 0.4494 0 .8876];
B=[.0323;.8989;.1290;.2247];
C=[.96774 .1124 1.6129 .4719];
D=0;
% statespace model
sys1=ss(A,B,C,D);
sys5=c2d(sys1,.001,'zoh')
step(sys1);
%figure;
resp=step(sys5);
[m,n]=size(resp);
k=10000/2;
% dividing data into two datasets
for j=1:k
sys2(j,1)=resp(j,1);
sys3(j,1)=resp(k,1);
k=k+1;
end
% modelling system
u=ones((m-1)/2,1);
sys4=iddata(sys2,u,.001);
sysid=pem(sys4);
% system parameters from the model
figure,step(sysid)
[ad bd cd dd]=ssdata(sysid)
ssfinal=ss(ad,bd,cd,dd,.001)
figure,plot(sysid)
figure,plot(sys3)

回答(1 个)

Arkadiy Turevskiy
Arkadiy Turevskiy 2014-3-27
If you follow the process described here

类别

Help CenterFile Exchange 中查找有关 Linear Model Identification 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by