function [r, p, k, e] = residue(b,a,toler)
#[r p k e] = residue(b,a)
#If b and a are vectors of polynomial coefficients, then residue
#calculates the partial fraction expansion corresponding to the
#ratio of the two polynomials. The vector r contains the residue
#terms, p contains the pole values, k contains the coefficients of
#a direct polynomial term (if it exists) and e is a vector containing
#the powers of the denominators in the partial fraction terms.
#Assuming b and a represent polynomials P(s) and Q(s) we have:
#
# P(s)    M       r(m)         N
# ---- =  #  -------------  +  # k(n)*s^(N-n)
# Q(s)   m=1 (s-p(m))^e(m)    n=1
#
#(# represents summation) where M is the number of poles (the length of
#the r, p, and e vectors) and N is the length of the k vector.
#
#[r p k e] = residue(b,a,tol)
#This form of the function call may be used to set a tolerance value.
#The default value is 0.001. The tolerance value is used to determine
#whether poles with small imaginary components are declared real. It is
#also used to determine if two poles are distinct. If the ratio of the
#imaginary part of a pole to the real part is less than tol, the imaginary
#part is discarded. If two poles are farther apart than tol they are
#distinct.
#
#Example:
# b = [1 1 1];
# a = [1  -5   8  -4];
#
# [r p k e] = residue(b,a)
#
# returns
#
# r = [-2 7 3]; p = [2 2 1]; k = []; e = [1 2 1];
#
# which implies the following partial fraction expansion
#
#       s^2 + s + 1         -2       7        3
#   ------------------- = ----- + ------- + -----
#   s^3 - 5s^2 + 8s - 4   (s-2)   (s-2)^2   (s-1)
#
#SEE ALSO: poly, roots, conv, deconv, polyval, polyderiv, polyinteg

# Author:
#  Tony Richardson
#  amr@mpl.ucsd.edu
#  June 1994

# Here's the method used to find the residues.
# The partial fraction expansion can be written as:
#
#          
#   P(s)    D   M(k)      A(k,m)
#   ---- =  #    #    -------------
#   Q(s)   k=1  m=1   (s - pr(k))^m
#
# (# is used to represent a summation) where D is the number of
# distinct roots, pr(k) is the kth distinct root, M(k) is the
# multiplicity of the root, and A(k,m) is the residue cooresponding
# to the kth distinct root with multiplicity m.  For example,
#
#        s^2            A(1,1)   A(2,1)    A(2,2)
# ------------------- = ------ + ------ + -------
# s^3 + 4s^2 + 5s + 2    (s+2)    (s+1)   (s+1)^2
#
# In this case there are two distinct roots (D=2 and pr = [-2 -1]),
# the first root has multiplicity one and the second multiplicity
# two (M = [1 2]) The residues are actually stored in vector format as
# r = [ A(1,1) A(2,1) A(2,2) ].
#
# We then multiply both sides by Q(s).  Continuing the example:
#
# s^2 = r(1)*(s+1)^2 + r(2)*(s+1)*(s+2) + r(3)*(s+2)
#
# or
#
# s^2 = r(1)*(s^2+2s+1) + r(2)*(s^2+3s+2) +r(3)*(s+2)
#
# The coefficients of the polynomials on the right are stored
# in a row vector called rhs, while the coefficients of the
# polynomial on the left is stored in a row vector called lhs.
# If the multiplicity of any root is greater than one we'll
# also need derivatives of this equation of order up to the
# maximum multiplicity minus one.  The derivative coefficients
# are stored in successive rows of lhs and rhs.
#
# For our example lhs and rhs would be:
#
#       | 1 0 0 |
# lhs = |       |
#       | 0 2 0 |
#
#       | 1 2 1 1 3 2 0 1 2 |
# rhs = |                   |
#       | 0 2 2 0 2 3 0 0 1 |
#
# We then form a vector B and a matrix A obtained by evaluating the
# polynomials in lhs and rhs at the pole values. If a pole has a
# multiplicity greater than one we also evaluate the derivative
# polynomials (successive rows) at the pole value.
#
# For our example we would have
#
# | 4|   | 1 0 0 |   | r(1) |
# | 1| = | 0 0 1 | * | r(2) |
# |-2|   | 0 1 1 |   | r(3) |
#
# We then solve for the residues using matrix division.

  if(nargin < 2 || nargin > 3)
    error("usage: residue(b,a[,toler])");
  endif

  if (nargin == 2)
    # Set the default tolerance level
    toler = .001;
  endif

  # Make sure both polynomials are in reduced form.
  a = polyreduce(a);
  b = polyreduce(b);

  b = b/a(1);
  a = a/a(1);

  la = length(a);
  lb = length(b);

  # Handle special cases here.
  if(la == 0 || lb == 0)
    k = r = p = e = [];
    return;
  elseif (la == 1)
    k = b/a;
    r = p = e = [];
    return;
  endif

  # Find the poles.
  p = roots(a);
  lp = length(p);

  # Determine if the poles are (effectively) real.
  index = find(abs(imag(p) ./ real(p)) < toler);
  if (length(index) != 0)
    p(index) = real(p(index));
  endif

  # Find the direct term if there is one.
  if(lb>=la)
    # Also returns the reduced numerator.
    [k, b] = deconv(b,a);
    lb = length(b);
  else
    k = [];
  endif

  if(lp == 1)
    r = polyval(b,p);
    e = 1;
    return;
  endif


  # We need to determine the number and multiplicity of the roots.
  # D is the number of distinct roots.
  # M is a vector of length D containing the multiplicity of each root.
  # pr is a vector of length D containing only the distinct roots.
  # e is a vector of length lp which indicates the power in the partial
  # fraction expansion of each term in p.

  # Set initial values.  We'll remove elements from pr as we find
  # multiplicities.  We'll shorten M afterwards.
  e = ones(lp,1);
  M = zeros(lp,1);
  pr = p;
  D = 1; M(1) = 1;

  old_p_index = 1; new_p_index = 2;
  M_index = 1; pr_index = 2;
  while(new_p_index<=lp)
    if(abs(p(new_p_index) - p(old_p_index)) < toler)
      # We've found a multiple pole.
      M(M_index) = M(M_index) + 1;
      e(new_p_index) = e(new_p_index-1) + 1;
      # Remove the pole from pr.
      pr(pr_index) = [];
    else
      # It's a different pole.
      D++; M_index++; M(M_index) = 1;
      old_p_index = new_p_index; pr_index++;
    endif
    new_p_index++;
  endwhile

  # Shorten M to it's proper length
  M = M(1:D);

  # Now set up the polynomial matrices.

  MM = max(M);
  # Left hand side polynomial
  lhs = zeros(MM,lb);
  rhs = zeros(MM,lp*lp);
  lhs(1,:) = b;
  rhi = 1;
  dpi = 1;
  mpi = 1;
  while(dpi<=D)
    for ind = 1:M(dpi)
      if(mpi>1 && (mpi+ind)<=lp)
        cp = [p(1:mpi-1); p(mpi+ind:lp)];
      elseif (mpi==1)
        cp = p(mpi+ind:lp);
      else
        cp = p(1:mpi-1);
      endif
      rhs(1,rhi:rhi+lp-1) = prepad(poly(cp),lp);
      rhi = rhi + lp;
    endfor
    mpi = mpi + M(dpi);
    dpi++;
  endwhile
  if(MM > 1)
    for index = 2:MM
      lhs(index,:) = prepad(polyderiv(lhs(index-1,:)),lb);
      ind = 1;
      for rhi = 1:lp
        cp = rhs(index-1,ind:ind+lp-1);
        rhs(index,ind:ind+lp-1) = prepad(polyderiv(cp),lp);
        ind = ind + lp;
      endfor
    endfor
  endif

  # Now lhs contains the numerator polynomial and as many derivatives as are
  # required.  rhs is a matrix of polynomials, the first row contains the
  # corresponding polynomial for each residue and successive rows are
  # derivatives.

  # Now we need to evaluate the first row of lhs and rhs at each distinct
  # pole value.  If there are multiple poles we will also need to evaluate
  # the derivatives at the pole value also.

  B = zeros(lp,1);
  A = zeros(lp,lp);

  dpi = 1;
  row = 1;
  while(dpi<=D)
    for mi = 1:M(dpi)
      B(row) = polyval(lhs(mi,:),pr(dpi));
      ci = 1;
      for col = 1:lp
        cp = rhs(mi,ci:ci+lp-1);
        A(row,col) = polyval(cp,pr(dpi));
        ci = ci + lp;
      endfor
      row++;
    endfor
    dpi++;
  endwhile

  # Solve for the residues.
  r = A\B;

endfunction