錐補(bǔ)線性化問題和應(yīng)用線性矩陣不等式解決控制問題

1、3線性矩陣不等式應(yīng)用實(shí)例采用MATLAB中LM I工具街求解矩陣不等式的問題對(duì)了對(duì)稱矩陣不等式可以采用徘補(bǔ)線性化方法，ZuX *k解AnA22 * A i.】)A 2<jr-l) A 2n (rr- Dn< 0(1)*尸*-1*即在滿足A 12小J*A 224 2/f-D4 2 < 0(2)* 一MA ( n 1) n*KX、*、Np/0,Q/VO1.1N(3)的條件下使tra(r(PA/4- 0V)取得最小值錐補(bǔ)線性化的LMI算法為1)首先找到滿足和中3個(gè)知陣不等式的所有未知矩陣變最的一個(gè)可行射(丸.Qs必“ No),令迭代次數(shù)Jt = O,2)對(duì)印陣變量(A 0.MN)求

2、解如下最小化問題:M inimi?p| traref PM + MtP + 0N+ NQ)|Subject to 2 X 3)令求出的最優(yōu)解為；?上一，。葉1 A/h 1, Ni).一.3)3證所求出的最優(yōu)解是否滿足1)"；游足,則得解，若果不滿足，檢立k是否達(dá)到規(guī)定的迭代 次數(shù),如果達(dá)到則系統(tǒng)無(wú)解：否則.令k-k+ 1.轉(zhuǎn)到步驟2繼續(xù)執(zhí)行程序.以卜對(duì)一具體問題所做的仿肉所需要描述的LM I為：S-P 000A + BK、FC*-5000FC*共R-Q0BKFC- A< 0* *-R00* *px0* *，Lp>aa /?> a s> a 扭> a ao

3、 a f. k 是適維矩陣按照上面的誰(shuí)補(bǔ)線性化方法算法，所描述的LM1如下：A=l 0.25;0 1;B=0.0063;0.0500;C=l 0;D=0;U=0 0;0 0;setlmis()P=lmivar(l,2 1);Q=lmivar(l,2 1);R=lmivar(l,2 1);S=lmivar(l,2 1);M=lmivar(l,2 1);N=lmivar(l,2 1);K=lmivar(2,l 2);F=lmivar(2,2 1);lmiterm( 1 1 1 S,l,l);lmiterm( 1 1 1lmiterm( 1 5 1 0,A);lmiterm( 1 5 1 K,B)；

4、 lmiterm( 1 6 1 F,-1,C);lmiterm( 1 2 2lmiterm( 1 6 2 F,1,C);lmiterm( 1 3 3 R, 1,1);lmiterm( 1 3 3 lmiterm( 1 5 3 lmiterm( 1 6 3 F,1,C);lmiterm( 1 6 3 0,-A);lmiterm( 1 4 4 lmiterm( 1 5 5 lmiterm( 1 6 6 lmiterm( -2 1 1 P,l,l);lmiterm( -2 2 1 0,l);lmiterm( -2 2 2 M, 1,1);lmiterm( -3 1 1lmiterm( -3 2 1

5、0,l);lmiterm( -3 2 2 N, 1,1);lmiterm( -4 1 1lmiterm( -5 1 1 Q,l,l);lmiterm( -6 1 1 R,l,l);lmiterm( -7 1 1 S,l,l);lmiterm( -8 1 1lmiterm( -9 1 1 N,l,l); linisys =getlmis ;tmin,xfeasp =feasp(lmisys)PP =dec2mat(lmisys,xfeasp,P)QQ =dec2mat(lmisys,xfeasp,Q)RR =dec2mat(lmisys,xfeasp,R) SS =dec2mat(lmisys,

6、xfeasp,S) MM =dec2mat(lmisys,xfeasp,M) NN =dec2mat(lmisys,xfeasp,N) KK =dec2mat(lmisys,xfeasp,K)FF =dec2mat(lmisys,xfeasp,F) fori =1:100 n =decnbr(lmisys);c =zeros(n ,1);forj=l:nPj,Qj,Rj,Sj,Mj,Nj,Kj,Fj =defcx(lmisys,j,P,Q,R,S,M,N,K,F); c(j)=trace(PP*Mj+MM*Pj +QQ*Nj+NN*Qj);endoptions = le-4,0,0,0,0；c

7、opt, xopt=mincx(lmisys,c,options)PPP =dec2mat(lmisys,xopt,P);QQQ =dec2mat(lmisys,xopt,Q);RRR =dec2mat(lmisys,xopt,R);SSS =dec2mat(lmisys,xopt,S);MMM =dec2mat(lmisys,xopt,M);NNN =dec2mat(lmisys,xopt,N);KKK =dec2mat(lmisys,xopt,K);FFF =dec2mat(lmisys,xopt,F);Z = SSS-PPP,U,U,U,(A+B*KKK)',-(FFF*C)&#

8、39;U,-SSS,U,U,U,(FFF*C),;U,U,RRR-QQQ,U,-(B*KKK)：-(A-FFF*Cy； U,U,U,-RRR,U,U;A+B*KKK,U,-B*KKK,U,-inv(PPP),U;-FFF*C,FFF*C,-(A -FFF*C),U,U,-inv(QQQ);Y =eig(Z);i2 =0;for il =1 :length(Y),if(Y(il , l)<0),12 =i2 +1 ;endendif(i2 =length(Y), break ;endPP =PPPQQ=QQQRR =RRRSS =SSSMM=MMMNN =NNNKK=KKKFF =FFFe

9、ndif(i=10),disp(' There is no result'); endSolver for linear objective mmiimzation under LMI constrauitsIterations :Best objective value so far149.439545236.063874323.102877414.567845514.567845613.442476new lower bound:0.296420713.442476*new lower bound:1.679189812.669593*new lower bound:3.5

10、15934912.669593*new lower bound:4.2857311011.916238*new lower bound:4.3956051111.916238*new lower bound:5.063925129.631247*new lower bound:5.242941139.178411*new lower bound:5.879022148.769358*new lower bound:6390317158.502939*new lower bound:6.790746168.326950*new lower bound:7.100688178.238591*new

11、 lower bound:7338510188.176522*new lower bound:7.520225198.161872*new lower bound:7.658027208.161872*new lower bound:7.790606218.147783*new lower bound:8.066778228.114291238.106330*new lower bound:8.074037248.104751*new lower bound:8.080386258.104153*new lower bound:8.085205268.103217*new lower boun

12、d:8.089119278.102513*new lower bound:8.091864288.101985*new lower bound:8.093950298.101589*new lower bound:8.095533308.101291*new lower bound:8.096731318.101177*new lower bound:8.097637328.100996*new lower bound:8.098334338.100859*new lower bound:8.098850348.100806*new lower bound:8.099240358.100721

13、*new lower bound:8.099951Result: feasible solution of required accuracybest objective value:8.100721guaranteed relative accuracy: 9.50e-05f-radiiis saniration: 0.000% of R= 1.00e+09copt =8.1007xopt =0.16390.00301.35730.36631.027718.70110.0038-0.01730.08120.0001-0.00070.00616.1002-0.01350.73683.2584-0.17090.0646-0.5553-2.78230.0001-0.0000PP =0.16390.0