Partial fraction expansion (partial fraction decomposition)
The inputs to
residue are vectors of coefficients
of the polynomials
b = [bm ... b1 b0] and
= [an ... a1 a0]. The outputs are the residues
= [rn ... r2 r1], the poles
p = [pn ... p2 p1],
and the polynomial
k. For most textbook problems,
Find the partial fraction expansion of the following ratio of polynomials F(s) using
b = [-4 8]; a = [1 6 8]; [r,p,k] = residue(b,a)
r = 2×1 -12 8
p = 2×1 -4 -2
k = 
This represents the partial fraction expansion
Convert the partial fraction expansion back to polynomial coefficients using
[b,a] = residue(r,p,k)
b = 1×2 -4 8
a = 1×3 1 6 8
This result represents the original fraction F(s).
If the degree of the numerator is equal to the degree of the denominator, the output
k can be nonzero.
Find the partial fraction expansion of a ratio of two polynomials F(s) with complex roots and equal degree of numerator and denominator, where F(s) is
b = [2 1 0 0]; a = [1 0 1 1]; [r,p,k] = residue(b,a)
r = 3×1 complex 0.5354 + 1.0390i 0.5354 - 1.0390i -0.0708 + 0.0000i
p = 3×1 complex 0.3412 + 1.1615i 0.3412 - 1.1615i -0.6823 + 0.0000i
k = 2
residue returns the complex roots and poles, and a constant term in
k, representing the partial fraction expansion
When the degree of the numerator is greater than the degree of the denominator, the output
k is a vector that represents the coefficients of a polynomial in s.
Perform the following partial fraction expansion of F(s) using
b = [2 0 0 1 0]; a = [1 0 1]; [r,p,k] = residue(b,a)
r = 2×1 complex 0.5000 - 1.0000i 0.5000 + 1.0000i
p = 2×1 complex 0.0000 + 1.0000i 0.0000 - 1.0000i
k = 1×3 2 0 -2
k represents the polynomial .
b— Coefficients of numerator polynomial
Coefficients of the polynomial in the numerator, specified as a vector of numbers representing the coefficients of the polynomial in descending powers of s.
Complex Number Support: Yes
a— Coefficients of denominator polynomial
Coefficients of the polynomial in the denominator, specified as a vector of numbers representing the coefficients of the polynomial in descending powers of s.
Complex Number Support: Yes
r— Residues of partial fraction expansion
Residues of partial fraction expansion, returned as a column vector of numbers.
p— Poles of partial fraction expansion
Poles of partial fraction expansion, returned as a column vector of numbers.
k— Direct term
Direct term, returned as a row vector of numbers that specify the coefficients of the polynomial in descending powers of s.
Consider the fraction F(s) of two polynomials b and a of degree n and m, respectively
The fraction F(s) can be represented as a sum of simple fractions
This sum is called the partial fraction expansion of F. The values rm,...,r1 are the residues, the values pm,...,p1 are the poles, and k(s) is a polynomial in s. For most textbook problems, k(s) is 0 or a constant.
The number of poles
n = length(a)-1 = length(r) = length(p)
The direct term vector is empty if
length(k) = length(b)-length(a)+1
p(j) = ... = p(j+m-1) is a pole of multiplicity
then the expansion includes terms of the form
residue first obtains the poles using
Next, if the fraction is nonproper, the direct term
deconv, which performs polynomial long
residue determines the residues
by evaluating the polynomial with individual roots removed. For repeated
resi2 computes the residues at the repeated
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 result in arbitrarily large changes in the resulting poles and residues. Problem formulations making use of state-space or zero-pole representations are preferable.
 Oppenheim, A.V. and R.W. Schafer. Digital Signal Processing. Prentice-Hall, 1975, p. 56.