optimization_in_scilab.pdf

Note that lmi is also a list of lists containing the

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Note that LMI is also a list of lists containing the values of the LMI matrices. This is just a matter of convenience. Now, we can solve the problem in Scilab as follows (assuming lists A and B , and matrix PI have already been defined). First we should initialize Q and Y . --> N=size(A); [n,nu]=size(B(1)); Q_init=list(); Y_init=list(); --> for i=1:N, Q_init(i)=zeros(n,n);Y_init(i)=zeros(nu,n);end Then, we can use lmisolver as follows: --> XLIST0=list(Q_init,Y_init) --> XLISTF=lmisolver(XLIST0,jump_sf_eval) --> [Q,Y]=XLISTF(:); The above commands can be encapsulated in a solver function, say jump_sf , in which case we simply need to type: --> [Q,Y]=jump_sf(A,B,PI) to obtain the solution. Descriptor Lyapunov inequalities In the study of descriptor systems, it is sometimes necessary to find (or find out that it does not exist) an n × n matrix X satisfying E T X = X T E 0 A T X + X T A + I 0 where E and A are n × n matrices such that E, A is a regular pencil. In this problem, which clearly is a Σ problem, the LME functions play important role. The evaluation function can be written as follows 45
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function [LME,LMI,OBJ]=dscr_lyap_eval(XLIST) X=XLIST(:) LME=E’*X-X’*E LMI=list(-A’*X-X’*A-eye(),E’*X) OBJ=[] and the problem can be solved by (assuming E and A are already defined) --> XLIST0=list(zeros(A)) --> XLISTF=lmisolver(XLIST0,dscr_lyap_eval) --> X=XLISTF(:) Mixed H 2 /H Control Consider the linear system ˙ x = Ax + B 1 w + B 2 u z 1 = C 1 x + D 11 w + D 12 u z 2 = C 2 x + D 22 u The mixed H 2 /H control problem consists in finding a stabilizing feedback which yields k T z 1 w k < γ and minimizes k T z 2 w k 2 where k T z 1 w k and k T z 2 w k 2 denote respectively the closed-loop transfer functions from w to z 1 and z 2 . In [ 22 ], it is shown that the solution to this problem can be expressed as K = LX - 1 where X and L are obtained from the problem of minimizing Trace( Y ) subject to: X - X T = 0 , Y - Y T = 0 , and - AX + B 2 L + ( AX + B 2 L ) T + B 1 B T 1 XC T 1 + L T D T 12 + B 1 D T 11 C 1 X + D 12 L + D 11 B T 1 - γ 2 I + D 11 D T 11 > 0 Y C 2 X + D 22 L ( C 2 X + D 22 L ) T X > 0 To solve this problem with lmisolver , we define the evaluation function: function [LME,LMI,OBJ]=h2hinf_eval(XLIST) [X,Y,L]=XLIST(:) LME=list(X-X’,Y-Y’); LMI=list(-[A*X+B2*L+(A*X+B2*L)’+B1*B1’,X*C1’+L’*D12’+B1*D11’;... (X*C1’+L’*D12’+B1*D11’)’,-gamma^2*eye()+D11*D11’],... [Y,C2*X+D22*L;(C2*X+D22*L)’,X]) OBJ=trace(Y); and use it as follows: --> X_init=zeros(A); Y_init=zeros(C2*C2’); L_init=zeros(B2’) --> XLIST0=list(X_init,Y_init,L_init); --> XLISTF=lmisolver(XLIST0,h2hinf_eval); --> [X,Y,L]=XLISTF(:) 46
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Descriptor Riccati equations In Kalman filtering for descriptor system Ex ( k + 1) = Ax ( k ) + u ( k ) y ( k + 1) = Cx ( k + 1) + r ( k ) where u and r are zero-mean, white Gaussian noise sequences with covariance Q and R respec- tively, one needs to obtain the positive solution to the descriptor Riccati equation (see [ 33 ]) P = - ( 0 0 I ) APA T + Q 0 E 0 R C E T C T 0 - 1 0 0 I .
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