good books for learning linear matrix inequality with MATLAB examples

Question

Ayesha ayub 2018-11-14

0
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https://ww2.mathworks.cn/matlabcentral/answers/429889-good-books-for-learning-linear-matrix-inequality-with-matlab-examples

回答： Waseem Abbasi 2019-7-3

Hello,

i am in my reasurch phase and badly stuck in LMI coding using matlab. can anyone suggest me some good books on LMI with matlab example please. iam trying to solve the following problem (from "Monitoring of stealthy attack in networked control systems") :

so far i have coded this but it gives me infeasible sol whereas in paper thr range of lambda is [0,0.72]:

 
 
clc
close all
clear all
%%%%%%%%%%%%%%%%%%%%%%%%%%SYSTEM DYNAMICS%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
A=[0.6 0    0.34 0.35 ;
   0   0.8  0    0.37    ;
   0   0    0.5  0    ;
   0   0    0    0.9];
B=[1 0 0; 1 0 1; 0 0 2; 0 1 1];
C=[1 0 0 0; 0 1 0 0; 0 0 0 1];
f=[0 0 1 0]';
F=A*f;
 H=pinv(C*F);
Ab=A-A*F*H*C;
bta=[ 1  0  1;
      2  1  1 ];
sigma=bta*(eye(3)-C*F*H);
% rank(sigma);
cb=sigma*C;
Kb =[0.0057   -0.0028;
   -0.0275    0.0618;
    0.0041   -0.0178;
    0.0077    0.0239];
K =[0.0276    0.0230    0.0229;
    0.0252    0.0612    0.0338;
    0.0015   -0.0001   -0.0001;
    0.0162    0.0235    0.0313 ]; 
%%
lambda=0;
k=1;
l=zeros(1,91);
n=zeros(1,91);
      
%%
 %LMI PART
for i=1:91
i
setlmis([]);
Y=lmivar(1,[4,1]) ;  
    
%defining first lmi
%   |Y                                     sqrt(1-lamda)(YA+KC)   sqrt( lamda)(YA_bar+K_bar*C_bar)|
%   | sqrt(1-lamda)(YA+KC)'                Y                      0                               |   >0
%   | sqrt( lamda)(YA_bar+K_bar*C_bar))'   0                      Y                               |
%%
%row 1   (%   |Y    sqrt(1-lamda)(YA+KC)  sqrt( lamda)(YA_bar+K_bar*C_bar)|)
lmiterm([-1 1 1 Y], 1,1);      % Y>0      
lmiterm([-1 1 2 Y], 1,sqrt(1-lambda)*A);   %sqrt(1-lmd)*(YA)
lmiterm([-1 1 2 0], sqrt(1-lambda)*K*C);   %sqrt(1-lmd)*(KC)
lmiterm([-1 1 3 Y], 1,sqrt(lambda)*Ab);    %sqrt(lmd)*(YA_bar)
lmiterm([-1 1 3 0], sqrt(lambda)*Kb*cb);   %sqrt(lmd)*(K_bar C_bar)
%% 
%row 2    (|Y   0|)
  lmiterm([-1 2 2 Y], 1,1);  %Y>0
lmiterm([-1 2 3 0],zeros(4,4));     %4X4 block of zeros
%%
%row 3 (| Y |)
 lmiterm([-1 3 3 Y], 1,1);   %Y>0
 %%
% defining 2nd lmi  (Y<I )    (ignoring this LMI is feasible)
lmiterm([2 1 1 Y],1,1);    %Y <  ... 
lmiterm([-2 1 1 0],eye(4));    % ...  < I
%% 
  syslmis=getlmis ;
 [tmin,Yfes] = feasp(syslmis);
  Y=dec2mat(syslmis,Yfes,Y) ;
    Y1(:,:,k)=Y;
  K1(:,:,k)=K;
  Kb1(:,:,k)=Kb;
  L(:,k)=lambda;
%%
  
  a=Y1(:,:,i);
  b=Y1(:,:,i)*A+(K1(:,:,i)*C);
  c=Y1(:,:,i)*Ab+(Kb1(:,:,i)*cb);
  
  
  D=[a                    sqrt(1-L(:,i))*b     sqrt(L(:,i))*c;
     sqrt(1-L(:,i))*b'    a                    zeros(4,4);
     sqrt(L(:,i))*c'      zeros(4,4)           a];
            
%%       
          [R1,p1] = chol(D);    %p1=0 implies D>0
          [R2,p2] = chol(Y);    %p2=0 implies Y>0
          [p1,p2]
        if(p1==0 && p2==0)
display('__________________________________positive definite____________________________________________')
       dummy(:,k)=k ;  %for picking lamda from L with index in dummy for wich D>0 & Y>0
       lmd(:,k)=lambda;
        end
   if(p1~=0 || p2~=0)
       dummy(:,k)=0;
%  lmd(:,k)=-1;s
display('__________________________________not positive definite________________________________________')
   end
      lambda=lambda+0.01;
 
k=k+1;
display('_______________________________________________________________________________________________')
end
 display(' all lambdas for which Y>0 and D(Y)>0')
lmd 