# about why in pole placement gives me error?

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azam ghamari on 26 Dec 2018
Commented: azam ghamari on 29 Dec 2018
Hi guys
Does every body knows why in the system :
s=tf('s');
G=(s-2.44)/((s+5)*(s+6)*(s+7));
[A,B,C,D]=tf2ss([1 2.44],[1 18 107 210]);
k_max=1001;
for j=1:k_max
p(j,:)=[-j-1 -3*j -j-2];
K(j,:)=place(A,B,p(j,:));
end
with differeing the p value it gives me this error?
Error using place (line 78)
The "place" command cannot place poles with multiplicity greater than rank(B).
Error in test2 (line 33)
K(j,:)=place(A,B,p(j,:));

Aquatris on 27 Dec 2018
It is just the error says. Place command cannot place poles with multiplicity greater than rank(B). In your case, rank(B) is 1. However when j=1, you want to place your poles at [-2 -3 -3]. You see the multiplicity of pole at -3 is 2. So place command cannot be used. If you start j from 2 instead of 1, there won't be an issue.
Aquatris on 28 Dec 2018
I dont know why you think it does not work. You do realize place is for state feedback not output feedback so you need an observer right?

azam ghamari on 28 Dec 2018
no, i used it for state feedbackany way when we have somethink like this:
P=p(j,:)=[-j-1 -3*j -j-2];
and B is for example 4x2;
it doesn't work and give me the result.
It gives me this error:
Subscripted assignment dimension mismatch.
Error in test2 (line 31)
K(j,:)=place(A,B,p(j,:));
A=[0.5 0 0.025 0;0 -0.01 0 0.017;0 0 -0.025 0;0 0 0 -0.0178];
B=[0.48 0;0 0.35;0 0.077;0.055 0];
C=[0.5 0 1 0];
D=0;
sys=ss(A,B,C,D);
G=tf(sys)
k_max=1001;
for j=1:k_max
p(j,:)=[-10*rand(1) -20*rand(1) -15*rand(1) -rand(1)];
K(j,:)=place(A,B,p(j,:));
end
##### 2 CommentsShowHide 1 older comment
azam ghamari on 29 Dec 2018
it works. thanks

R2017b

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