# 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.

residualz将离散时间系统（表示为两个多项式的比率）转换为部分分数扩展或留数形式。 它还将部分分数展开转换回原始多项式系数。

`[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.

`[r,p,k] = residuez(b,a)` 求出两个多项式b（z）和a（z）的比率的部分分数展开的留数，极点和直接项。 向量b和a以z的下降幂指定离散时间系统b（z）/ a（z）的多项式的系数。

• n.残渣；残余物；【数】残数；【化】残基

• 网络残留物；留数；剩余物

• 变形复数：residues；

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

`[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`.

原文作者：Z字形编排问题
原文地址: https://blog.csdn.net/Reborn_Lee/article/details/83446437
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