How to plot values from a for loop?

Maybe my question isn't so specific, but I'll try to explain what's giving me a hard time.

I have to calculate

(C/(s*I-A))*B

where A,B and C are matrices and s is a eigenvalue of some given matrix. I have to calculate it for every eigenvalue and plot it. This is what I did

svA=eig(A);
for P=1:50
fr=(CN/(i*svA(P)*eye(n)-A))*B;
fr_org{P}=fr;
end
fr_original=fr_org{1:50};
plot(fr_original)

but I didn't get anything. Any advice?

Answer 1

[Walter Roberson]

Try

fr_original = cell2mat(fr_org);

[Ljix]

Using the above I got something that looks like a graph of a function (some connected line) and I want just to point out values in coordinate system, nothing connected. Any further advice?

[Walter Roberson]

plot(fr_original, '*')

[Ljix]

Now it's working. Thank you.

Answer 2

[Suresh Garimella]

Try this, Hope it works

svA=eig(A); 
for P=1:50 fr=(CN/(i*svA(P)*eye(n)-A))*B; 
fr_org{P}=fr; 
end 
%fr_original=fr_org{1:50}; comment this line 
fr_original=fr_org; % directly assign 
%plot(fr_original) comment this line 
plot(cell2mat(fr_original)) % convert class using cell2mat to plot

[Ljix]

Almost what I want. Using

plot(cell2mat(fr_original),'+') gives what I want.

Answer 3

[Star Strider]

Actually, this is incorrect:

Y = (C/(s*I-A))*B

The correct expression is:

Y = C*((s*I-A)\B)

that you then invert to create:

y = C*expm(A*t)*B

You cannot do what you want to do in the Laplace domain, especially since s = σ + j*ω. You have to invert it to the time domain to do anything with it.

If you want to do a pole-zero plot, there are functions for that such as pzplot in the Control Systems Toolbox.

[Ljix]

Thank you. Isn't frequency response given by Y = C*((s*I-A)\B), where s=i*ω and ω is eigenvalue of A? How to plot y = C*expm(A*t)*B? I tried defining t=linspace(0,100,25) and then calculate y as function (as defined) and I got message about dimension error.

[Star Strider]

My pleasure.

In a pole-zero plot, σ is the real part, and jω is the imaginary part. This becomes equivalent to the Fourier transform if you set σ=0.

If you want the Bode plot (frequency response only, along the σ=0 line), use the transfer function representation of your system created from the core MATLAB ss2tf function, then plot it using the Signal Processing Toolbox freqs function.

EDIT — I Answered your question about y = C*expm(A*t)*B*u in How to plot y = C*expm(A*t)*B?