# Documentation

### This is machine translation

Translated by
Mouseover text to see original. Click the button below to return to the English verison of the page.

To view all translated materals including this page, select Japan from the country navigator on the bottom of this page.

# residuez

Z-transform partial-fraction expansion

## Syntax

[r,p,k] = residuez(b,a)
[b,a] = residuez(r,p,k)

## Description

residuez converts a discrete time system, expressed as the ratio of two polynomials, to partial fraction expansion, or residue, form. It also converts the partial fraction expansion back to the original polynomial coefficients.

### Note

Numerically, the partial fraction expansion of a ratio of polynomials is an ill-posed problem. If the denominator polynomial is near a polynomial with multiple roots, then small changes in the data, including round-off errors, can cause arbitrarily large changes in the resulting poles and residues. You should use state-space or pole-zero representations instead.

[r,p,k] = residuez(b,a) finds the residues, poles, and direct terms of a partial fraction expansion of the ratio of two polynomials, b(z) and a(z). Vectors b and a specify the coefficients of the polynomials of the discrete-time system b(z)/a(z) in descending powers of z.

$\begin{array}{c}B\left(z\right)={b}_{0}+{b}_{1}{z}^{-1}+{b}_{2}{z}^{-2}+\cdots +{b}_{m}{z}^{-m}\\ A\left(z\right)={a}_{0}+{a}_{1}{z}^{-1}+{a}_{2}{z}^{-2}+\cdots +{a}_{n}{z}^{-n}\end{array}$

If there are no multiple roots and a > n-1,

$\frac{B\left(z\right)}{A\left(z\right)}=\frac{r\left(1\right)}{1-p\left(1\right){z}^{-1}}+\cdots +\frac{r\left(n\right)}{1-p\left(n\right){z}^{-1}}+k\left(1\right)+k\left(2\right){z}^{-1}+\cdots +k\left(m-n+1\right){z}^{-\left(m-n\right)}$

The returned column vector r contains the residues, column vector p contains the pole locations, and row vector k contains the direct terms. The number of poles is

n = length(a)-1 = length(r) = length(p)

The direct term coefficient vector k is empty if length(b) is less than length(a); otherwise:

length(k) = length(b) - length(a) + 1

If p(j) = ... = p(j+s-1) is a pole of multiplicity s, then the expansion includes terms of the form

$\frac{r\left(j\right)}{1-p\left(j\right){z}^{-1}}+\frac{r\left(j+1\right)}{{\left(1-p\left(j\right){z}^{-1}\right)}^{2}}+\cdots +\frac{r\left(j+{s}_{r}-1\right)}{{\left(1-p\left(j\right){z}^{-1}\right)}^{s}}$

[b,a] = residuez(r,p,k) with three input arguments and two output arguments, converts the partial fraction expansion back to polynomials with coefficients in row vectors b and a.

The residue function in the standard MATLAB® language is very similar to residuez. It computes the partial fraction expansion of continuous-time systems in the Laplace domain (see reference [1]), rather than discrete-time systems in the z-domain as does residuez.

## Examples

collapse all

Compute the partial-fraction expansion corresponding to the third-order IIR lowpass filter described by the transfer function

Express the numerator and denominator as polynomial convolutions.

b0 = 0.05634;
b1 = [1  1];
b2 = [1 -1.0166 1];
a1 = [1 -0.683];
a2 = [1 -1.4461 0.7957];

b = b0*conv(b1,b2);
a = conv(a1,a2);

Compute the residues, poles, and direct terms of the partial-fraction expansion.

[r,p,k] = residuez(b,a)
r =
-0.1153 - 0.0182i
-0.1153 + 0.0182i
0.3905 + 0.0000i

p =
0.7230 + 0.5224i
0.7230 - 0.5224i
0.6830 + 0.0000i

k = -0.1037

Plot the poles and zeros of the transfer function and overlay the poles you just found.

zplane(b,a)
hold on
plot(p,'^r')
hold off

Use residuez again to reconstruct the transfer function.

[bn,an] = residuez(r,p,k)
bn =

0.0563   -0.0009   -0.0009    0.0563

an =

1.0000   -2.1291    1.7834   -0.5435

## Algorithms

residuez applies standard MATLAB functions and partial fraction techniques to find r, p, and k from b and a. It finds

• The direct terms a using deconv (polynomial long division) when length(b) > length(a)-1.

• The poles using p = roots(a).

• Any repeated poles, reordering the poles according to their multiplicities.

• The residue for each nonrepeating pole pj by multiplying b(z)/a(z) by 1/(1 - pjz−1) and evaluating the resulting rational function at z = pj.

• The residues for the repeated poles by solving

S2*r2 = h - S1*r1

for r2 using \. h is the impulse response of the reduced b(z)/a(z), S1 is a matrix whose columns are impulse responses of the first-order systems made up of the nonrepeating roots, and r1 is a column containing the residues for the nonrepeating roots. Each column of matrix S2 is an impulse response. For each root pj of multiplicity sj, S2 contains sj columns representing the impulse responses of each of the following systems.

$\frac{1}{1-{p}_{j}{z}^{-1}},\frac{1}{{\left(1-{p}_{j}{z}^{-1}\right)}^{2}},\cdots ,\frac{1}{{\left(1-{p}_{j}{z}^{-1}\right)}^{{s}_{j}}}$

The vector h and matrices S1 and S2 have n + xtra rows, where n is the total number of roots and the internal parameter xtra, set to 1 by default, determines the degree of over-determination of the system of equations.

## References

[1] Oppenheim, Alan V., Ronald W. Schafer, and John R. Buck. Discrete-Time Signal Processing. 2nd Ed. Upper Saddle River, NJ: Prentice Hall, 1999.