[MATLAB数学相关] 不确定系统状态反馈控制器求解中的可行性算例，以下是我做的一个算例，但不可行，哪位高人能帮我看看啊，谢谢

[junzh]

>> A=[-3 0 4;2 0 5;2 3 4];
>> B=[-0.1 0.3 0.2;0.7 0.5 0.4;0.2 0.3 0.2];
>> D1=[-0.1;0;-0.2];
>> D2=[0;-0.1;-0.4];
>> D3=[-0.3;-0.2;-0.1];
>> E1=[0 0.1 0.2];
>> E2=[0.1 0.2 0.1];
>> E3=[0.3 0 0.1];
>> setlmis([]);
>> X=lmivar(1,[3,1]);
>> Y=lmivar(2,[3,3]);
>> lmiterm([1,1,1,X],A,1,'s');
>> lmiterm([1,1,1,Y],B,1,'s');
>> lmiterm([1,1,2,0],0.1*D1);
>> lmiterm([1,1,3,0],0.1*B*D2);
>> lmiterm([1,1,4,-X],1,E1');
>> lmiterm([1,1,4,-Y],1,E2');
>> lmiterm([1,1,5,X],1,E3');
>> lmiterm([1,1,6,0],0.02*D1);
>> lmiterm([1,1,8,X],1,E3');
>> lmiterm([1,2,2,0],-0.1);
>> lmiterm([1,2,3,0],-0.1);
>> lmiterm([1,3,4,0],-0.1);
>> lmiterm([1,3,5,0],-0.1);
>> lmiterm([1,4,6,0],-0.02);
>> lmiterm([1,5,7,0],-0.02);
>> lmiterm([1,6,8,0],-0.03);
>> lmiterm([1,7,9,0],-0.03);
>> lmiterm([-2,1,1,X],1,1);
>> lmis=getlmis;
>> [tmin,feas]=feasp(lmis);

Solver for LMI feasibility problems L(x) < R(x)
This solver minimizes t subject to L(x) < R(x) + t*I
The best value of t should be negative for feasibility

Iteration : Best value of t so far

1 0.189678
2 0.174185
* switching to QR
3 0.174185
4 0.165606
5 0.165606
6 0.165606
7 0.164023
8 0.164023
9 0.162923
10 0.162923
11 0.162923
12 0.162882
13 0.162843
14 0.162827
15 0.162827
16 0.162827
17 0.162827
18 0.162827
19 0.162827
20 0.162827
21 0.162827
22 0.162827
23 0.162827
24 0.162827

Result: could not establish feasibility nor infeasibility
f-radius saturation: 71.233% of R = 1.00e+009
Termination due to SLOW PROGRESS:
t was decreased by less than 10.000% during
the last 10 iterations.


These LMI constraints were found infeasible