Strange answer form feasp (LMI Solvers)

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Allan Andre do Nascimento 2016-6-25

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评论： Star Strider 2016-6-25

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?

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Star Strider 2016-6-25

在 MATLAB Online 中打开

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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Control Systems Robust Control Toolbox Linear Matrix Inequalities LMI Solvers

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