H Infinity contrl ---LMI方法（Yalmip 求解）和Riccati 方法

参考书籍：Yu, Hai-Hua_ Duan, Guangren - LMIs in control systems _ analysis, design and applications (2013, CRC Press) - libgen.lc(一书的附录有详细介绍LMI的编写方式-有很多例子)

参考网站：http://www.doc88.com/p-8718494012962.html

                     YALMIP

                    Download - YALMIP

有问题可以上google论坛--在matlab 问答模块可以搜到网址--可以提问相关问题，

 关于LMI 本身求解参考：

线性矩阵不等式(LMI)的_MATLAB求解-    http://www.doc88.com/p-7874379836838.html

Matlab中LMI(线性矩阵不等式)工具箱使用教程        https://www.doc88.com/p-304944065894.html

线性矩阵不等式的LMI工具箱求解-LMI三个函数的应用实例；

 https://www.doc88.com/p-3925042294217.html

https://www.doc88.com/p-660150804881.html

https://blog.csdn.net/qq_28093585/article/details/69358180

ceshi_Hinf_lmi

clc;clear; close all;
A  = [2 1 -2;1 -1 -3; 4 0 -1];
B = [1 0;0 3;3 1];
E1 = [1 0.2 -0.5]';
C = [2 1 -0.5];
D = [1 1];
E2 = [0.05]

P = 10e+7;
X0 = [8.7047 -6.5321 4.2500;
     -6.5321 8.8601 -4.2821;
     4.2500 -4.2821 5.0227];
W0 = [-5.2786 2.9364 -7.7991;
      -3.4736 -0.8733 6.0925];
W =P*W0; X = X0*P;
K1 =W*inv(X)

Ac =A-B*K1;
Bc = B;
Cc = C;
Dc = D;
sys_cl = ss(Ac,Bc,Cc,Dc);
step(sys_cl)
% figure(1)
% t = 0:0.01:15;
% r =[0.5*ones(size(t));
%     0.5*ones(size(t))];
% [y,t,x]=lsim(sys_cl,r,t);
% plot(t,y)
% 
% 
% K2 = [1.2850 0.5978 -2.0283;
%      -1.4101 -0.7807 1.4861]
% Ac =A-B*K2;
% Bc = B;
% Cc = C;
% Dc = D;
% sys_cl = ss(Ac,Bc,Cc,Dc);
% 
% figure(2)
% t = 0:0.01:15;
% r =[0.5*ones(size(t));
%     0.5*ones(size(t))];
% [y,t,x]=lsim(sys_cl,r,t);
% plot(t,y)

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