求解matlab中的lmi应用

1、Matlab程序尸4+4,产+0+为2区 P-DrPA0I 也卫4 -里 FDPA20尸4.4产00 -DrPA2 产。-R网ph2 C=0.2,0;0. L-O.ll A0=0A2;0.L-0.3 A1=|-O.3,O.2;O.LO1 A2=0A2;0.30.1 HO=|-O.3,O:O.2,O.l H l=0.3,0.1：0.2.-0.1 H2=0,0.l;-0.3,0,2 Q=0.7,0.5;0.5J R= 1,020.206 h=0.15 setlmis(J); P=Imivar(L2J ); lmiterm( 1 1 1 P,LA0,s); lmiterm( 1 1 1 0,Q+h*

2、h*R); lmiterm( 1 1 2 PJ,A1);lmiterm(l 1 2P,-A0,C);lmiterm( 1 1 3 PJ A2); lmitcrm( 1 1 3 PJ,A2);lmiterm(l 1 4P,H0,l);lmiterm( 1 2 2 P,CAl；s); lmiterm( 1 2 2 0,-Q);lmiterm( 1 2 3 PrCA2); lmiterm( 1 2 4 lmiterm( 1 3 3 0,-R); lmiterm(l 3 4 P,H2,1); lmiterm( 1 4 4 P,-1,1);lmisys=getlmis;tmin,xfeas=feasp(

3、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 feasibilityIteration : Best value of t so far1.4208621.22802431.228024*new lower bound:0.070826Result: best value of t:1.

4、228024f-radius saturation: 0.000% of R = 1.00e+009These LMI constraints were found infeasible tmin =1.2280 xfeas =-0.1338-0.01070.8971pmat =-0.1338-0.0107-0.01070.8971第二个程序 C=0,2.0; 1.0.2AO=(O.5,O;O,O.3Al=(-L0;-lrlA2=|0,0.2;0.3r0.1 H2=0.0.l;-0.3,0.2 HO=O.2,O;O0,2H l=0.3,0A0.3 Q=|O.5,O.3;O.3J R=0.2Al

5、：0.1J h=0.35 setlmis(); P=lmivar(l,2J); lmiterm( 1 1 1 PJ,A0；s*); lmiterm( 1 1 1 0,Q+h*h*R); Imiterm( 1 I 2 PJ,A1); lmiterm( 1 1 2 PA0C); lmiterm( 1 1 3 PJ,A2);lmiterm(l 1 4P,H0l); lmiterm( 1 2 2 P,-C,Al,s)； lmiterm( 1 2 2 0,-Q); lmiterm( 1 2 3 P,C,A2); lmiterm( 1 2 4 P,H7,1);lmiterm( 1 3 3 0,-R);lm

6、iterm(l 3 4P,H2,1); lmiterm( 1 4 4 P,-1,1);lmisys=getlmis;tmin,xfeas=feasp(lmisys) pmat=dec2mat(lniisys,xfcas,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 feasibilityIteration1*2*Best value of t so far1.258002new lower bound:-0.1155171.204465new lower bound:0.148583Result: best value of t:1.204465f-radius saturation: 0.000% of R = 1.00e+009These LMI constra