How to solve affine Linear Matrix Inequlaity?

조회 수: 2 (최근 30일)
Vishal Agrawal
Vishal Agrawal 2014년 1월 16일
답변: Johan Löfberg 2014년 1월 16일
The problem is
I want to solve following LMIs which is affine in h(t) :
A(h(t))'*P-C'*R+P*A(h(t))-R'*C<0; and P>0
where A(h)=[0 1;h(t) -0.1]; C=[0.2 1];
h(t) is a time varying parameter whose maximum and minimum value is 0 and -45.86 P is a symmetric matrix variable of dimension 2x2 R is a matrix variable of dimension 1x2
I am able to solve LMI without varying parameter h(t) using LMI Toolbox, but not able to solve above one. I go through the document of functions pvec, psys, quadstab in MATLAB's Robust Control Toolbox , but they are not addressing the type of problem at which I am stuck.
I want to find the value of matrices P and R such that above LMI is feasible. I am solving the above problem while going through paper for designing observer using differential mean value theorem based approach, the above condition is sufficient to successfully design observer, paper link is : http://ieeexplore.ieee.org/xpl/articleDetails.jsp?tp=&arnumber=1583180&queryText%3Dobserver+design+using+diffrential+mean+value+theorem+approach
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Vishal Agrawal
Vishal Agrawal 2014년 1월 16일
h(t)=3∗z(t)^2 , whose maximum and minimum value I find through simulation of 3∗x(t)^2 , where x(t) is system trajectory. z(t) always lies between maximum and minimum value of x(t), this is explained in paper whose link I given above....

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답변 (1개)

Johan Löfberg
Johan Löfberg 2014년 1월 16일
Standard approach is to use a common Lyapunov function, i.e., define two LMIs, one for h = 0 and one for h = -45.86. From convexity, the Lyapunov inequality is satisfied for any h inbetween.

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