本文主要是介绍LMI实例3,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
` %描述常数项
A=[0.2 2.1 0.5;0 0.1 0.4;0 0 1];
W=[1 0 0;0.0196 -1.3514 -2.2837;-0.0218 0 0.4376];
C=[0.05 0 1.29;0.01 0.87 0;0.13 0.12 0.5];
D1=[4 0.6 0.4;1.3 1.5 0.3;0.5 0.6 0.7];
D2=[0.02 0.07 0;0.02 0 0.03;0.14 0.2 0.3];
a=0.8;
%初始化LMI
setlmis([]);
%定义变量
P=lmivar(1,[3,1]);
Q=lmivar(2,[3,3]);
%定义每个矩阵不等式中各项的内容
lmiterm([1 1 1 P],A’*W’,1,‘S’);
lmiterm([1 1 1 Q],-C’,1,‘S’);
lmiterm([1 1 1 P],D1’,1,‘S’);
lmiterm([1 1 1 Q],D2’,-W’,‘S’);
lmiterm([1 1 1 0],a^2);
lmiterm([1 1 2 P],1,W);
lmiterm([1 2 2 0],-1);
%描数第二个想
lmiterm([2 1 1 P],-1,1);
%获得LMI的内部系统描述
lmiis=getlmis;
[tmin,xx]=feasp(lmiis)
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
- switching to QR
1 -0.017774
Result: best value of t: -0.017774
f-radius saturation: 0.000% of R = 1.00e+09
tmin =
-0.0178
xx =
0.3379
0.0007
0.0831
0.0116
0.1201
0.9061
-5.0916
2.1495
1.3202
1.5630
10.8244
-8.4637
9.4894
53.2440
0.0431
P=dec2mat(lmiis,xx,P)
P =
0.3379 0.0007 0.0116
0.0007 0.0831 0.1201
0.0116 0.1201 0.9061
Q=dec2mat(lmiis,xx,Q)
Q =
-5.0916 2.1495 1.3202
1.5630 10.8244 -8.4637
9.4894 53.2440 0.0431
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