Place function for observer gain L
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I tried to construct a simple observer and here is my code.
clc
clear
%%Initializing
A =[-1 0; 1 0];
B = [0.9; 0];
C =[0 0.5];
D = 0;
L=place(A',C',[-10 -15])';
eig(A-L*C)
eig(A)
x=[-1;1]; % initial state
xhat=[0;0]; % initial estimate
XX=x;
XXhat=xhat;
T=40;
UU=ones(1,T); % input signal
for k=1:T,
u=UU(k);
y(k)=C*x+D*u;
yhat(k)=C*xhat+D*u;
x=A*x+B*u;
error(k+1)=(y(k)-yhat(k));
xhat=A*xhat+B*u+L*(y(k)-yhat(k));
XX=[XX,x];
XXhat=[XXhat,xhat];
end
figure
plot(1:T,yhat);
hold on
plot(1:T,y);
hold off
figure
plot(error);
And I got really confused. In theory i should assign poles on the far left plane in s domain but when i assign poles at [-10 -15] the observer didn't work. However, if assign poles less than 1 the observer works. Does anyone know whats going on?
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Jeroen
2016년 5월 9일
If your model is discrete the poles should lie in the unit circle (between -1 and 1) to make the system stable. If the model is continuous the poles must be less than zero, for the system to be stable.
답변 (1개)
Kiran Khunte
2018년 7월 2일
The value of poles should be inside the unit circle to let your esitmate to reach zero as k increases. As you see the poles are the eigen values of matrix 'A-L*C' which is the multiplying factor for err(k). i.e. err(k+1) = (([A-L*C])^k)*err(k). So to make err(k+1) tends to zero you will need eigen values of matrix [A-L*C] as small as possible.
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