% Diary of MATLAB commands for partial fraction expansion.
% ELEC 320, R. Kozick

>> help roots

 ROOTS  Find polynomial roots.
    ROOTS(C) computes the roots of the polynomial whose coefficients
    are the elements of the vector C. If C has N+1 components,
    the polynomial is C(1)*X^N + ... + C(N)*X + C(N+1).
 
    See also POLY, RESIDUE, FZERO.

>> roots([1 5 6])

ans =

    -3
    -2

>> help residue

 RESIDUE Partial-fraction expansion (residues).
    [R,P,K] = RESIDUE(B,A) finds the residues, poles and direct term of
    a partial fraction expansion of the ratio of two polynomials B(s)/A(s).
    If there are no multiple roots,
       B(s)       R(1)       R(2)             R(n)
       ----  =  -------- + -------- + ... + -------- + K(s)
       A(s)     s - P(1)   s - P(2)         s - P(n)
    Vectors B and A specify the coefficients of the numerator and
    denominator polynomials in descending powers of s.  The residues
    are returned in the column vector R, the pole locations in column
    vector P, and the direct terms in row vector K.  The number of
    poles is n = length(A)-1 = length(R) = length(P). The direct term
    coefficient vector is empty if length(B) < length(A), otherwise
    length(K) = length(B)-length(A)+1.
 
    If P(j) = ... = P(j+m-1) is a pole of multplicity m, then the
    expansion includes terms of the form
                 R(j)        R(j+1)                R(j+m-1)
               -------- + ------------   + ... + ------------
               s - P(j)   (s - P(j))^2           (s - P(j))^m
 
    [B,A] = RESIDUE(R,P,K), with 3 input arguments and 2 output arguments,
    converts the partial fraction expansion back to the polynomials with
    coefficients in B and A.
 
    Warning: Numerically, the partial fraction expansion of a ratio of
    polynomials represents an ill-posed problem.  If the denominator
    polynomial, A(s), is near a polynomial with multiple roots, then
    small changes in the data, including roundoff errors, can make
    arbitrarily large changes in the resulting poles and residues.
    Problem formulations making use of state-space or zero-pole
    representations are preferable.
 
    See also POLY, ROOTS, DECONV.

>> disp('Example 1')
Example 1
>> num = [1 17];
>> den = [1 4 -5];
>> roots(den)

ans =

    -5
     1

>> [r, p] = residue(num, den)

r =

    -2
     3


p =

    -5
     1

>> disp('Example 2')
Example 2
>> num = [3 -5];
>> den = [1 3 7 5];
>> [r, p] = residue(num, den)

r =

   1.0000 - 0.7500i
   1.0000 + 0.7500i
  -2.0000          


p =

  -1.0000 + 2.0000i
  -1.0000 - 2.0000i
  -1.0000          

>> abs(r)

ans =

    1.2500
    1.2500
    2.0000

>> angle(r)/pi*180

ans =

  -36.8699
   36.8699
  180.0000

>> disp('Example 3')
Example 3
>> num = [16 43];
>> den = [1 4 -3 -18];
>> [r, p] = residue(num, den)

r =

   -3.0000
    1.0000
    3.0000


p =

   -3.0000
   -3.0000
    2.0000

>> disp('Example 4')
Example 4
>> num = [1 0 2 -4];
>> den = [1 4 -2];
>> [r, p] = residue(num, den)

r =

   20.6145
   -0.6145


p =

   -4.4495
    0.4495