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Z-transform partial-fraction expansion

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

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

`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.

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*.

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

If there are no multiple roots and `a`

> `n-1`

,

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

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

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

`[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`

.

`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

*p*by multiplying_{j}*b*(*z*)/*a*(*z*) by 1/(1 -*p*_{j}*z*^{−1}) and evaluating the resulting rational function at*z*=*p*._{j}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*p*of multiplicity_{j}*s*,_{j}`S2`

contains*s*columns representing the impulse responses of each of the following systems._{j}$$\frac{1}{1-{p}_{j}{z}^{-1}},\frac{1}{{(1-{p}_{j}{z}^{-1})}^{2}},\cdots ,\frac{1}{{(1-{p}_{j}{z}^{-1})}^{{s}_{j}}}$$

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.

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

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