Daniel_cleveland asked . 2021-10-04

Lyapunov Plotting: using 'mesh' and 'meshgrid' with matricies

I'm trying to plot a 3d graph of a Lyapunov function of a control system I've created. The function is:

V(x)=xT*P*x

where x is a 2x1 matrix of the errors, e and de/dt:

    x = [e;ed];

Therefore x transpose is a 1x2 matrix:

    xT = [e, ed];

and the P matrix is 2x2 of constants (ie p1=const,p2=const,p3=const,p4=const):

    P = [ p1 p2 ; p3 p4];

"e" and "ed" are a 1xn set of data from my simulation thus V ends up being a 1xn matrix

This is my code right now, I don't know what to do from here:

e = ErrorData.signals.values(:,1);
ed = ErrorData.signals.values(:,2);

 for i=1:3420
     V(i) = [e(i) , ed(i)] * P * [e(i) ; ed(i)] ;
 end

I'm trying to plot (x,y,z) = (e,ed,V) in 3 dimensions but I can't seem to be able make my final vector V suitable for plotting in 3d. for e and ed you can use meshgrid, but I can't get V in the proper form.

I was using this code as an example from ( Lyapunov Example ):

x=[-4:.04:4];
y=x;
[X,Y]=meshgrid(x,y);
z=X.^2 + Y.^2;
mesh(X,Y,z)

mesh , meshgrid , lyapunov , plot , matrix , matrix manipulation , matrix mulitplication

Expert Answer

John Williams answered . 2024-11-26 04:34:49

Found the solution. I had to replicate what mesh does.

e = ErrorData.signals.values(:,1);
ed = ErrorData.signals.values(:,2);
[E,ED] = meshgrid(e,ed);

for i=1:3420
  for j=1:3420
      V(i,j) = [E(i,j) , ED(i,j)] * P * [E(i,j) ; ED(i,j)] ;
  end
end

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