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# residuez

Z-transform partial-fraction expansion

## Syntax

``[ro,po,ko] = residuez(bi,ai)``
``[bo,ao] = residuez(ri,pi,ki) ``

## Description

example

````[ro,po,ko] = residuez(bi,ai)` finds the residues, poles, and direct terms of a partial fraction expansion of the ratio of numerator and denominator polynomials, `b` and `a`.```
````[bo,ao] = residuez(ri,pi,ki) ` with three input arguments and two output arguments, converts the partial fraction expansion back to polynomials with coefficients in row vectors `b` and `a`.```

## Examples

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

`$H\left(z\right)=\frac{0.05634\left(1+{z}^{-1}\right)\left(1-1.0166{z}^{-1}+{z}^{-2}\right)}{\left(1-0.683{z}^{-1}\right)\left(1-1.4461{z}^{-1}+0.7957{z}^{-2}\right)}.$`

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 = 3×1 complex -0.1153 - 0.0182i -0.1153 + 0.0182i 0.3905 + 0.0000i ```
```p = 3×1 complex 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 = 1×4 0.0563 -0.0009 -0.0009 0.0563 ```
```an = 1×4 1.0000 -2.1291 1.7834 -0.5435 ```

## Input Arguments

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Polynomial coefficients, specified as vectors. 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 multiple roots and `a` > `n-1`,

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

Residues of the partial fraction, specified as a vector.

Poles of the partial fraction, specified as a vector.

Direct terms, specified as a row vector.

## Output Arguments

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Residues of the partial fraction, returned as a vector.

Pole of the partial fraction, returned as a vector. The number of poles is

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

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

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

Direct terms, returned as a row vector. The direct term coefficient vector `k` is empty if `length(b)` is less than `length(a)`; otherwise:

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

Polynomial coefficients, returned as vectors.

## Algorithms

`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.

`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.

Note

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`.

## 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.

## See Also

Introduced before R2006a

## Support

#### Deep Learning for Signal Processing with MATLAB

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