Strange answer form feasp (LMI Solvers)
2 views (last 30 days)
Show older comments
Hey everyone
I was testing the feasp command of matlab in order to confirm the stability of a matrix. But then when I run the code I wrote, it turns out that such command returns a P matrix which has one negative eigenvalue. Could someone please point out whether there is any problem on my code?
m1 = 290;
m2 = 59;
k1 = 16812;
k2 = 190000;
b1 = 1000;
alfa = 4.515*(10^13);
beta = 1;
gama = 1.545*(10^9);
tau = (1/30);
Ps = 10342500;
A = 3.35*(10^(-4));
Aa = [0 1 0 0 0 0
-k1/m1 -(b1)/m1 (k1/m1) (b1/m1) A/m1 0
0 0 0 1 0 0
k1/m2 b1/m2 -(k1+k2)/m2 -b1/m2 -A/m2 0
0 -alfa*A 0 alfa*A -beta gama*sqrt(Ps)
0 0 0 0 0 -1/tau ];
Ba =[0
0
0
0
0
1/tau];
setlmis([])
P = lmivar(1,[6 1]);
lmiterm([1 1 1 P],1,Aa,'s');
lmiterm([-2 1 1 P],1,1);
lmiterm([2 1 1 0],0);
lmisys = getlmis;
[tmin,xfeas] = feasp(lmisys);
Pf = dec2mat(lmisys,xfeas,P);
eig(Pf)
I believe that one source of such problem could be that a few of the eigenvalues of Aa are rather small, and perhaps due to numerical approximation, the program is giving a wrong answer... can it be a source of the mentioned error?
1 Comment
Star Strider
on 25 Jun 2016
I don’t have the Robust Control Toolbox, so I can’t run your code.
When I evaluate it up to the ‘Aa’ assignment and do:
Aa_cond = cond(Aa)
Aa_cond =
829.7659e+021
With such a poorly-conditioned matrix, I would not trust any results.
Answers (0)
See Also
Categories
Find more on LMI Solvers in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!