Strange answer form feasp (LMI Solvers)
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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 件のコメント
Star Strider
2016 年 6 月 25 日
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.
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