求解matlab中的lmi应用

>> A0=[0,0.2;0.1,-0.3]

>> A1=[-0.3,0.2;0.1,0]

>> A2=[0,0.2;0.3,-0.1]

>> H0=[-0.3,0;0.2,0.1]

>> H1=[0.3,0.1;0.2,-0.1]

>> H2=[0,0.1;-0.3,0.2]

>> Q=[0.7,0.5;0.5,1]

>> R=[1,0.2;0.2,0.6]

>> h=0.15

>> setlmis([]);

1

>> P=lmivar(1,[2,1]);

>> lmiterm([1 1 1 P],1,A0,'s');

>> lmiterm([1 1 1 0],Q+h*h*R);

>> lmiterm([1 1 2 P],1,A1);

>> lmiterm([1 1 2 P],-A0',C);

>> lmiterm([1 1 3 P],1 A2);

>> lmiterm([1 1 3 P],1,A2);

>> lmiterm([1 1 4 P],H0',1);

>> lmiterm([1 2 2 P],-C',A1,'s');

>> lmiterm([1 2 2 0],-Q);

>> lmiterm([1 2 3 P],-C',A2);

>> lmiterm([1 2 4 P],H1',1);

>> lmiterm([1 3 3 0],-R);

2

>> lmiterm([1 3 4 P],H2',1);

>> lmiterm([1 4 4 P],-1,1);

>> lmisys=getlmis;

>> [tmin,xfeas]=feasp(lmisys)

>> pmat=dec2mat(lmisys,xfeas,P)

运行结果

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 1.420862

2 1.228024

3 1.228024

3

*** new lower bound: 0.070826

Result: best value of t: 1.228024

f-radius saturation: 0.000% of R = 1.00e+009

These LMI constraints were found infeasible

tmin =

1.2280

xfeas =

-0.1338

-0.0107

0.8971

pmat =

-0.1338 -0.0107

-0.0107 0.8971

4

第二个程序

>> C=[0.2,0;1,0.2]

>> A0=[0.5,0;0,0.3]

>> A1=[-1,0;-1,-1]

>> A2=[0,0.2;0.3,-0.1]

>> H2=[0,0.1;-0.3,0.2]

>> H0=[0.2,0;0,0.2]

>> H1=[0.3,0;0,0.3]

>> Q=[0.5,0.3;0.3,1]

>> R=[0.2,0.1;0.1,1]

>> h=0.35

>> setlmis([]);

>> P=lmivar(1,[2,1]);

5

>> lmiterm([1 1 1 P],1,A0,'s');

>> lmiterm([1 1 1 0],Q+h*h*R);

>> lmiterm([1 1 2 P],1,A1);

>> lmiterm([1 1 2 P],-A0',C);

>> lmiterm([1 1 3 P],1,A2);

>> lmiterm([1 1 4 P],H0',1);

>> lmiterm([1 2 2 P],-C',A1,'s');

>> lmiterm([1 2 2 0],-Q);

>> lmiterm([1 2 3 P],-C',A2);

>> lmiterm([1 2 4 P],H1',1);

>> lmiterm([1 3 3 0],-R);

>> lmiterm([1 3 4 P],H2',1);

>> lmiterm([1 4 4 P],-1,1);

6

>> lmisys=getlmis;

>> [tmin,xfeas]=feasp(lmisys)

>> pmat=dec2mat(lmisys,xfeas,P)

运行结果

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 1.258002

*** new lower bound: -0.115517

2 1.204465

*** new lower bound: 0.148583

Result: best value of t: 1.204465

7

f-radius saturation: 0.000% of R = 1.00e+009

These LMI constraints were found infeasible

tmin =

1.2045

xfeas =

-0.1942

-0.0264

-0.0850

pmat =

-0.1942 -0.0264 8

-0.0264

-0.0850