--- /dev/null
+%% Copyright (C) 2005 Julius O. Smith III <jos@ccrma.stanford.edu>
+%%
+%% This program is free software; you can redistribute it and/or modify it under
+%% the terms of the GNU General Public License as published by the Free Software
+%% Foundation; either version 3 of the License, or (at your option) any later
+%% version.
+%%
+%% This program is distributed in the hope that it will be useful, but WITHOUT
+%% ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
+%% FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more
+%% details.
+%%
+%% You should have received a copy of the GNU General Public License along with
+%% this program; if not, see <http://www.gnu.org/licenses/>.
+
+%% -*- texinfo -*-
+%% @deftypefn {Function File} {[@var{r}, @var{p}, @var{f}, @var{m}] =} residuez (@var{B}, @var{A})
+%% Compute the partial fraction expansion of filter @math{H(z) = B(z)/A(z)}.
+%%
+%% INPUTS:
+%% @var{B} and @var{A} are vectors specifying the digital filter @math{H(z) = B(z)/A(z)}.
+%% Say @code{help filter} for documentation of the @var{B} and @var{A}
+%% filter coefficients.
+%%
+%% RETURNED:
+%% @itemize
+%% @item @var{r} = column vector containing the filter-pole residues@*
+%% @item @var{p} = column vector containing the filter poles@*
+%% @item @var{f} = row vector containing the FIR part, if any@*
+%% @item @var{m} = column vector of pole multiplicities
+%% @end itemize
+%%
+%% EXAMPLES:
+%% @example
+%% Say @code{test residuez verbose} to see a number of examples.
+%% @end example
+%%
+%% For the theory of operation, see
+%% @indicateurl{http://ccrma.stanford.edu/~jos/filters/residuez.html}
+%%
+%% @seealso{residue residued}
+%% @end deftypefn
+
+function [r, p, f, m] = residuez(B, A, tol)
+ % RESIDUEZ - return residues, poles, and FIR part of B(z)/A(z)
+ %
+ % Let nb = length(b), na = length(a), and N=na-1 = no. of poles.
+ % If nb<na, then f will be empty, and the returned filter is
+ %
+ % r(1) r(N)
+ % H(z) = ---------------- + ... + ----------------- = R(z)
+ % [ 1-p(1)/z ]^m(1) [ 1-p(N)/z ]^m(N)
+ %
+ % If, on the other hand, nb >= na, the FIR part f will not be empty.
+ % Let M = nb-na+1 = order of f = length(f)-1). Then the returned filter is
+ %
+ % H(z) = f(1) + f(2)/z + f(3)/z^2 + ... + f(M+1)/z^M + R(z)
+ %
+ % where R(z) is the parallel one-pole filter bank defined above.
+ % Note, in particular, that the impulse-response of the one-pole
+ % filter bank is in parallel with that of the the FIR part. This can
+ % be wasteful when matching the initial impulse response is important,
+ % since F(z) can already match the first N terms of the impulse
+ % response. To obtain a decomposition in which the impulse response of
+ % the IIR part R(z) starts after that of the FIR part F(z), use RESIDUED.
+ %
+ % J.O. Smith, 9/19/05
+
+ if nargin==3
+ warning("tolerance ignored");
+ end
+ NUM = B(:)'; DEN = A(:)';
+ % Matlab's residue does not return m (since it is implied by p):
+ [r,p,f,m]=residue(conj(fliplr(NUM)),conj(fliplr(DEN)));
+ p = 1 ./ p;
+ r = r .* ((-p) .^m);
+ if f, f = conj(fliplr(f)); end
+end
+
+%!test
+%! B=[1 -2 1]; A=[1 -1];
+%! [r,p,f,m] = residuez(B,A);
+%! assert(r,0,100*eps);
+%! assert(p,1,100*eps);
+%! assert(f,[1 -1],100*eps);
+%! assert(m,1,100*eps);
+
+%!test
+%! B=1; A=[1 -1j];
+%! [r,p,f,m] = residuez(B,A);
+%! assert(r,1,100*eps);
+%! assert(p,1j,100*eps);
+%! assert(f,[],100*eps);
+%! assert(m,1,100*eps);
+
+%!test
+%! B=1; A=[1 -1 .25];
+%! [r,p,f,m] = residuez(B,A);
+%! [rs,is] = sort(r);
+%! assert(rs,[0;1],1e-7);
+%! assert(p(is),[0.5;0.5],1e-8);
+%! assert(f,[],100*eps);
+%! assert(m(is),[1;2],100*eps);
+
+%!test
+%! B=1; A=[1 -0.75 .125];
+%! [r,p,f,m] = residuez(B,A);
+%! [rs,is] = sort(r);
+%! assert(rs,[-1;2],100*eps);
+%! assert(p(is),[0.25;0.5],100*eps);
+%! assert(f,[],100*eps);
+%! assert(m(is),[1;1],100*eps);
+
+%!test
+%! B=[1,6,2]; A=[1,-2,1];
+%! [r,p,f,m] = residuez(B,A);
+%! [rs,is] = sort(r);
+%! assert(rs,[-10;9],1e-7);
+%! assert(p(is),[1;1],1e-8);
+%! assert(f,[2],100*eps);
+%! assert(m(is),[1;2],100*eps);
+
+%!test
+%! B=[6,2]; A=[1,-2,1];
+%! [r,p,f,m] = residuez(B,A);
+%! [rs,is] = sort(r);
+%! assert(rs,[-2;8],1e-7);
+%! assert(p(is),[1;1],1e-8);
+%! assert(f,[],100*eps);
+%! assert(m(is),[1;2],100*eps);
+
+%!test
+%! B=[1,6,6,2]; A=[1,-2,1];
+%! [r,p,f,m] = residuez(B,A);
+%! [rs,is] = sort(r);
+%! assert(rs,[-24;15],2e-7);
+%! assert(p(is),[1;1],1e-8);
+%! assert(f,[10,2],100*eps);
+%! assert(m(is),[1;2],100*eps);
+
+%!test
+%! B=[1,6,6,2]; A=[1,-(2+j),(1+2j),-j];
+%! [r,p,f,m] = residuez(B,A);
+%! [rs,is] = sort(r);
+%! assert(rs,[-2+2.5j;7.5+7.5j;-4.5-12j],1E-6);
+%! assert(p(is),[1j;1;1],1E-6);
+%! assert(f,-2j,1E-6);
+%! assert(m(is),[1;2;1],1E-6);
+
+%!test
+%! B=[1,0,1]; A=[1,0,0,0,0,-1];
+%! [r,p,f,m] = residuez(B,A);
+%! [as,is] = sort(angle(p));
+%! rise = [ ...
+%! 0.26180339887499 - 0.19021130325903i; ...
+%! 0.03819660112501 + 0.11755705045849i; ...
+%! 0.4; ...
+%! 0.03819660112501 - 0.11755705045849i; ...
+%! 0.26180339887499 + 0.19021130325903i;];
+%! pise = [ ...
+%! -0.80901699437495 - 0.58778525229247i; ...
+%! 0.30901699437495 - 0.95105651629515i; ...
+%! 1; ...
+%! 0.30901699437495 + 0.95105651629515i; ...
+%! -0.80901699437495 + 0.58778525229247i];
+%! assert(r(is),rise,100*eps);
+%! assert(p(is),pise,100*eps);
+%! assert(f,[],100*eps);
+%! assert(m,[1;1;1;1;1],100*eps);
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