資源簡介
1、考慮了非線性項滿足Lipschitz條件的非線性系統觀測器設計問題,利用Lyapunov方法給出新的判斷觀測誤差穩定性的條件,并由所給的條件通過求解線性矩陣不等式來設計觀測器。
2、利用matlab lmi工具箱對原文例子進行了求解

代碼片段和文件信息
%參考文獻:Lipschitz非線性系統觀測器設計新方法,馬克茂,馬萍
%1、Lipschitz常數的求解方法??2、非線性觀測器增益的求解
clear
clc
A=[0?1;1?-1];
C=[0?1];
gamma=0.49;?%Lipschitz常數
i=0;
%-----------initial?a?LMI?system------------
setlmis([]);
%----------define?variables----------------
%?P?is?a?2?by?2?matrix
P=lmivar(1[2?1]);
%?W?is?a?2?by?1?matrix
W=lmivar(2[2?1]);
%?q?is?1?by?1?matrix
q=lmivar(1[1?0]);
while?1
????P0=P;
????W0=W;
????gamma0=gamma;
????%-----------initial?a?LMI?system------------
????setlmis([]);
????%----------define?variables----------------
????%?P?is?a?2?by?2?symmetric?matrix
????P=lmivar(1[2?1]);
????%?W?is?a?2?by?1?matrix
????W=lmivar(2[2?1]);
????%?q?is?1?by?1?matrix
????q=lmivar(1[1?0]);
????%P>0
????lmiterm([-2?1?1?P]11);
????lmiterm([2?1?1?0]0);
????%
????lmiterm([1?1?1?P]A‘1‘s‘);
????%lmiterm([1?1?1?P]1A);
????lmiterm([1?1?1?W]1C‘s‘);
????%lmiterm([1?1?1?-W]C‘1);
????lmiterm([1?1?1?P]11);
????lmiterm([1?1?1?q]1eye(2));
????%
????lmiterm([3?1?1?P]gamma^21);
????lmiterm([3?1?1?q]-1eye(2));
????lmis?=?getlmis;
????[tmin?feas]?=?feasp(lmis);
????P?=?dec2mat(lmis?feas?P);
????W?=?dec2mat(lmis?feas?W);
????q?=?dec2mat(lmis?feas?q);
????if?tmin>0
????????if?i==1
????????????gamma=gamma/2;
????????????continue;
????????else
????????????P=P0;
????????????W=W0;
????????????gamma=gamma0;
????????????break;
????????end
????else
????????Q=-(A‘*P+P*A+C‘*W‘+W*C+P);
????????gamma=sqrt(min(eig(Q))/max(eig(P)))
????end
????i=i+1;
end
disp(‘反饋增益:‘)
L=inv(P)*W
disp(‘Lipschitz常數:‘)
gamma
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件?????257724??2010-07-20?17:58??Lipschitz非線性系統觀測器設計新方法\Lipschitz非線性系統觀測器設計新方法.pdf
?????文件???????1651??2010-07-22?11:44??Lipschitz非線性系統觀測器設計新方法\lmi1.m
?????文件?????????38??2010-07-22?13:04??Lipschitz非線性系統觀測器設計新方法\說明.txt
?????目錄??????????0??2010-07-22?11:42??Lipschitz非線性系統觀測器設計新方法
-----------?---------??----------?-----??----
???????????????259413????????????????????4
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