coefficients.m

国外专家做的求解LMI鲁棒控制的工具箱,可以相对高效的解决LMI问题

function [base,v] = coefficients(p,x)
%COEFFICIENTS Extract coefficients and monomials from polynomials
%
%   [c,v] = COEFFICIENTS(p,x) extracts the coefficents
%   of a polynomial p(x) = c'*v(x)
%
%   INPUT
%    p : SDPVAR object
%    x : SDPVAR object
%
%   OUTPUT
%    c : SDPVAR object
%    v : SDPVAR object
%
%   EXAMPLE
%    sdpvar x y s t                
%    p = x^2+x*y*(s+t)+s^2+t^2;     % define p(x,y), parameterized with s and t
%    [c,v] = coefficients(p,[x y]); 
%    sdisplay([c v]) 
%
%   See also SDPVAR

% Author Johan L鰂berg
% $Id: coefficients.m,v 1.7 2005/05/10 15:04:51 joloef Exp $  

allvar = depends(p);
if nargin==1
    xvar = allvar;
    x = recover(xvar);
else
    xvar = depends(x);
end

% Try to debug this!
t = setdiff(allvar,xvar);
[exponent_p,p_base] = getexponentbase(p,recover(depends(p)));
ParametricIndicies = find(ismember(allvar,t));
% FIX : don't define it here, wait until sparser below. Speed!!
tempbase = parameterizedbase(p,[],recover(t),ParametricIndicies,exponent_p,p_base);
[i,j,k] = unique(full(exponent_p(:,find(~ismember(allvar,t)))),'rows');
%V = sparse(max(k),length(tempbase));
%for i = 1:max(k)    
%    V(i,find(k==i)) = 1;
%end
V = sparse(1:length(k),k,1,length(tempbase),max(k))';
base = V*tempbase;
if nargout == 2
    keepthese = j(1:max(k));
    v = recovermonoms(exponent_p(keepthese,find(~ismember(allvar,t))),recover(xvar));
end


function p_base_parametric = parameterizedbase(p,z, params,ParametricIndicies,exponent_p,p_base)

% Check for linear parameterization
parametric_basis = exponent_p(:,ParametricIndicies);
if all(sum(parametric_basis,2)==0)
    p_base_parametric = full(p_base(:));
    return
end
if all(sum(parametric_basis,2)<=1)
    p_base_parametric = full(p_base(:));
    n = length(p_base_parametric);
    ii = [];
    vars = [];
    js = sum(parametric_basis,1);
    for i = 1:size(parametric_basis,2)
        if js(i)
            j = find(parametric_basis(:,i));
            ii = [ii j(:)'];
            vars = [vars repmat(i,1,js(i))];
        end
    end
    k = setdiff1D(1:n,ii);
    if isempty(k)
        p_base_parametric = p_base_parametric.*sparse(ii,repmat(1,1,n),params(vars));
    else
        pp = params(vars); % Must do this, bug in ML 6.1 (x=sparse(1);x([1 1]) gives different result in 6.1 and 7.0!)
        p_base_parametric = p_base_parametric.*sparse([ii k(:)'],repmat(1,1,n),[pp(:)' ones(1,1,length(k))]);
    end
else
    % Bummer, nonlinear parameterization sucks...
    for i = 1:length(p_base)
        j = find(exponent_p(i,ParametricIndicies));
        if ~isempty(j)
            temp = p_base(i);
            for k = 1:length(j)
                if exponent_p(i,ParametricIndicies(j(k)))==1
                    temp = temp*params(j(k));
                else
                    temp = temp*params(j(k))^exponent_p(i,ParametricIndicies(j(k)));
                end
            end
            xx{i} = temp;
        else
            xx{i} = p_base(i);
        end
    end
    p_base_parametric = stackcell(sdpvar(1,1),xx)';
end