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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
>> 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 |
所有时间均为北京时间。现在的时间是 03:19。 |
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