Solve LMIs control in delay system

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Hoang Vu Huy
Hoang Vu Huy 2023 年 11 月 6 日
I need to simulation this paper https://doi.org/10.1016/j.ifacol.2016.10.403
Find maximum to exist matrices P>0, Z>0 and Q symmetric such that
and
where
How to solve this problem? I try as follows, but it has some warning that
%% Clear
clear; clc; close;
%% Constant & Initial Condition
t = 0:0.01:20;
K = [-2 3 0; 1 1 0; -3 1 -3];
alp = [1 0.5 -1];
f = sin(t);
%% Solve LMIs
% Declare Variables
setlmis([])
A1 = diag(sign(alp))*K*diag(alp);
P = lmivar(2,[3 3]);
Z = lmivar(2,[3 3]);
Q = lmivar(1,[3 1]);
tau = 0.1;
% Definitions of the LMI
%LMI#1
lmiterm([-1 1 1 P], 1, 1); % LMI #1: P
lmiterm([-1 1 1 Z], 1, 1); % LMI #1: Z
lmiterm([-1 1 2 -Z], 1, 1); % LMI #1: -Z
lmiterm([-1 2 1 -Z], 1, 1); % LMI #1: -Z
lmiterm([-1 2 2 Q], tau,1); % LMI #1: -tau_dash*Q
lmiterm([-1 2 2 Z], 1, 1); % LMI #1: Z
%LMI#2
lmiterm([2 1 1 Q], 1, 1); % LMI #2: Q
lmiterm([2 1 1 Z], 1/tau, -1); % LMI #2: -Z/tau
lmiterm([2 2 1 P], A1', 1); %LMI #2: A1'*P
lmiterm([2 2 1 Z], 1/tau, 1); %LMI #2: Z/tau
lmiterm([2 3 1 0], 0); %LMI #2: 0
lmiterm([2 1 2 P], 1, A1 ); % LMI #2: P*A1
lmiterm([2 1 2 Z], 1/tau, 1); % LMI #2: Z/tau
lmiterm([2 2 2 Q], 1, -1); %LMI #2: -Q
lmiterm([2 2 2 Z], 1/tau, -1); %LMI #2: -Z/tau
lmiterm([2 3 2 Z], tau, A1); %LMI #2: tau*Z*A1
lmiterm([2 1 3 0], 0); % LMI #2: 0
lmiterm([2 2 3 Z], tau*A1', 1); % LMI #2: tau*A1'*Z
lmiterm([2 3 3 Z], -tau, 1); %LMI #2:
%LMI#3
lmiterm([-3 1 1 P],1,1); % LMI #3: P
%LMI#4
lmiterm([-4 1 1 Z],1,1); % LMI #4: Z
lmis = getlmis;
[tmin,xfeas] = feasp(lmis);

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