remarkRemark hypothesisHypothesis claimClaim Inverse $\mathcal{Z}$-Transform
of Rational FunctionsM. Vaez, A. Hosseini, and K. Jamshidi
[][nocite]ex_supplement

# An Alternative Approach to
Inverse $\mathcal{Z}$-Transform of Rational Functions

MohammadJavad Vaez  A master student in the School of Mathematics, Statistics,
and Computer Science, College of Science, University of Tehran, Tehran, Iran
<EMAIL_ADDRESS>Alireza Hosseini School of Mathematics,
Statistics, and Computer Science, College of Science, University of Tehran,
Tehran, Iran<EMAIL_ADDRESS>Kamal Jamshidi Faculty of Computer
Engineering, University of Isfahan, Isfahan, Iran<EMAIL_ADDRESS>
###### Abstract

Our paper introduces a novel method for calculating the inverse
$\mathcal{Z}$-transform of rational functions. Unlike some existing approaches
that rely on partial fraction expansion and involve dividing by $z$, our
method allows for the direct computation of the inverse
$\mathcal{Z}$-transform without such division. Furthermore, our method expands
the rational functions over real numbers instead of complex numbers. Hence, it
doesn’t need algebraic manipulations to obtain a real-valued answer.
Furthermore, it aligns our method more closely with established techniques
used in integral, Laplace, and Fourier transforms. In addition, it can lead to
fewer calculations in some cases.

###### keywords:

inverse $\mathcal{Z}$-transform, partial fraction expansion, discrete-time
signals, discrete-time systems, Cauchy residue calculus, teaching methodology

93-08, 93C55, 93C62, 94A12

##  1 Introduction

The inverse $\mathcal{Z}$-transform plays a fundamental role in the field of
digital signal processing and system analysis. It allows us to recover the
original time-domain representation of a discrete-time signal from its
corresponding frequency-domain representation in the $z$-domain. This
transformation is invaluable in various areas, including digital filter
design, control systems, and communication systems, as well as in the analysis
and solution of difference equations (recurrence relations).

Traditionally, the computation of the inverse $\mathcal{Z}$-transform of
rational functions has been tackled using established techniques, such as long
division, contour integration, and partial fraction expansion [8, 6].

One commonly adopted approach, proposed by [8], employs a comprehensive method
based on partial fraction expansion. However, unlike conventional methods for
integration, Laplace transform, and Fourier transform, this approach requires
an initial division of the assumed function by $z$. Interestingly, a similar
division was previously proposed in an earlier paper by [5], leading to
discussions on the seeming differences in results. Recently, an alternative
method was proposed in [6], which eliminates the need for partial fraction
expansion but still requires division by $z$. Additionally, if the assumed
function has complex poles, further algebraic manipulations are required to
obtain a real-valued result.

On the other hand, in a prior work by [3], the authors mentioned that dividing
by $z$ increases the burden of calculation for some cases. Moreover, they
attempted to align partial fraction expansion with the Laplace transform by
avoiding such division and finding some new formulas for inverse
$\mathcal{Z}$-transform of rational functions. Nonetheless, they did not
address cases involving complex poles of degree higher than one.

In this paper, we aim to generalize the overlooked method proposed by [3] to
accommodate arbitrary degrees of complex poles. By doing so, we introduce a
novel and versatile method for computing the inverse $\mathcal{Z}$-transform
of rational functions.

In the forthcoming sections of this paper, we will briefly review previous
methods, followed by a detailed presentation of our proposed method. We will
highlight its underlying theorems and demonstrate its applicability through
illustrative examples. Additionally, we will compare the performance of our
method with existing approaches.

##  2 computation of inverse $\mathcal{Z}$-transform

In this paper, we primarily focus on the unilateral $\mathcal{Z}$-transform,
which is particularly important in digital signal processing, as it deals with
signals in the positive time domain. As mentioned in prior research [8], there
are at least three well-known methods for computing the inverse
$\mathcal{Z}$-transform of rational functions. Recently, a new method has been
proposed by[6]. In the sequel, we introduce the main methods proposed
previously in this literature.

###  2.1 Contour Integration

The more basic method, applicable to any function, involves contour
integration. If we denote the $\mathcal{Z}$-transform of the discrete-time
signal $x[n]$ as $X(z)$, the inverse $\mathcal{Z}$-transform can be expressed
as:

(1) $x[n]=\frac{1}{2\pi i}\oint_{C}\,X(z)z^{n-1}\,dz$

where $C$ is a counterclockwise contour that encircles the origin and lies
entirely in the region of convergence (ROC). For unilateral
$\mathcal{Z}$-transform, the interior of $C$ must contain all poles of $X(z)$.
Assuming that $X(z)$ has $K$ distinct poles $z_{1},z_{2},\ldots,z_{K}$, we can
evaluate (1) by applying Cauchy’s residue theorem

(2)
$x[n]=\sum_{k=1}^{K}\displaystyle\underset{z=z_{k}}{\mathrm{Res}}\left(X(z)z^{n-1}\right)$

if $z_{i}$ is a pole of multiplicity $m$, then

(3)
$\displaystyle\underset{z=z_{k}}{\mathrm{Res}}\left(X(z)z^{n-1}\right)=\frac{1}{(m-1)!}\lim_{z\to
z_{i}}\frac{d^{m-1}}{{dz}^{m-1}}\left[(z-z_{i})^{m}X(z)z^{n-1}\right]$

The other two methods mentioned in [8] specifically trigger the rational
functions 111Of course, partial fraction expansion can also be applied to non-
rational functions involving infinite series and the Mittag-Leffler theorem
[2]. Here, however, we focus solely on rational functions.. A rational
function is nothing but the quotient of two polynomials, say

(4)
$X(z)=\frac{b_{0}z^{p}+b_{1}z^{p-1}+b_{2}z^{p-2}+\ldots+b_{p}}{z^{q}+a_{1}z^{q-1}+a_{2}z^{q-2}+\ldots+a_{q}}$

###  2.2 long division

The next approach is the long division method. This method generally does not
yield a closed-form expression for $x[n]$. Thus, for the sake of brevity,
interested readers can refer to Example 2.12 in [9]. Nonetheless, it is
important to mention that, according to [8], using this method allows us to
easily deduce the following relationships:

(5) $\text{If }q>p,\text{ then }x[0]=x[1]=\ldots=x[q-p-1]=0.$

(6) $\text{If }q\geq p,\text{ then }x[q-p]=b_{0}.$

###  2.3 Partial Fraction Expansion

The primary method of focus in both [8] and our paper is partial fraction
expansion, also known as partial fraction decomposition. However, there are
two notable distinctions between our approach and the one proposed in [8]:

  * •

our approach decomposes polynomials over the field of real numbers while their
approach decompose them over complex numbers.

  * •

our approach expands $X(z)$, whereas their approaches expand $\frac{X(z)}{z}$.

In fact, these differences align our method with the conventional methods used
for integral, Laplace transform, and Fourier transform.

Table 1, originally presented in [8], is utilized as a reference for comparing
the inverse $\mathcal{Z}$-transform of a rational function with our proposed
method.

Table 1: Inverse $\mathcal{Z}$-Transforms for the General Terms in Partial-Fraction Expansion, Taken from [8] Type of poles | Term | Inverse $\mathcal{Z}$-transform
---|---|---
Single/multiple poles at $z=0$ | $\frac{A}{z^{n_{0}}}$ | $A\delta(n-n_{0})$
Single real pole | $\frac{Az}{z-r}$ | $Ar^{n},\text{ }n\geq 0$
Multiple real pole | $\frac{Az}{(z-r)^{q}}$ | $\frac{Ar^{n-q+1}}{(q-1)!}\prod_{i=0}^{q-2}(n-i),\text{ }n\geq 0$
Single complex poles | $\begin{aligned} &\begin{tabular}[]{@{}l@{}}$\frac{Ae^{i\phi}z}{z-re^{i\theta}}+\frac{Ae^{-i\phi}z}{z-re^{-i\theta}},$\\\ $A,r\in\mathbb{R}_{+}$\end{tabular}\end{aligned}$ | $\begin{aligned} &\begin{tabular}[]{@{}l@{}}$2Ar^{n}\cos(n\theta+\phi),\text{ }n\geq 0$\end{tabular}\end{aligned}$
Multiple complex poles | $\begin{aligned} &\begin{tabular}[]{@{}l@{}}$\frac{Ae^{i\phi}z}{(z-re^{i\theta})^{q}}+\frac{Ae^{-i\phi}z}{(z-re^{-i\theta})^{q}},$\\\ $A,r\in\mathbb{R}_{+}$\end{tabular}\end{aligned}$ | $\begin{aligned} &\begin{tabular}[]{@{}l@{}}$\frac{2Ar^{n-q+1}}{(q-1)!}\cos\left((n-q+1)\theta+\phi\right)$\\\ $\times\prod_{i=0}^{q-2}(n-i),\text{ }n\geq 0$\end{tabular}\end{aligned}$

###  2.4 The Method Proposed by [6]

The primary method proposed by [6] for the unilateral $\mathcal{Z}$-transform
is outlined as follows:

Let $\frac{X(z)}{z}=\frac{N(z)}{D(z)}$ and $D(z)$ has distinct roots
$z_{1},z_{2},\ldots,z_{K}$ with multiplicity $m_{1},m_{2},\ldots,m_{K}$
respectively. Then, the expression for $x[n]$ is given by:

(7)
$x[n]=\sum_{k=1}^{K}\sum_{j=0}^{m_{k}-1}c_{k,m_{k}-1-j}\binom{n}{j}z_{k}^{n-j}$

If $z_{k}=0$, $\binom{n}{j}z_{k}^{n-j}$ in Equation (7) should be replaced
with $\delta[n-j]$.The coefficients $c_{k,j}$, where $k=1,2,\ldots,K$ and
$j=0,1,\ldots,m_{k}-1$, are calculated using the following formula:

(8)
$c_{k,j}=\frac{1}{j!D_{k}(z_{k})}\left(N^{(j)}(z_{k})-\sum_{l=0}^{j-1}c_{k,l}(j)_{l}D_{k}^{(j-l)}(z_{k})\right)$

Here, $(j)_{l}$ represents the falling factorial:

(9) $\left(j\right)_{l}:={\overbrace{j(j-1)\dots(j-l+1)}^{\text{$l$ terms}}}$

Moreover, $N^{(j)}(z),D_{k}^{(j)}(z)$ denote the $j$-th derivatives of
$N(z),D_{k}(z)$ respectively, and $D_{k}(z)=\lim_{z\to
z_{k}}(z-z_{k})^{m_{k}}D(z)$. The equation (8) is recursive and complicated.
For this reason, the first few terms of that has been calculated in [6]:

(10) $c_{k,0}=\frac{N(z_{k})}{D_{k}(z_{k})}$ (11)
$c_{k,1}=\frac{N^{\prime}(z_{k})-c_{k,0}D_{k}^{\prime}(z_{k})}{D_{k}(z_{k})}$
(12)
$c_{k,2}=\frac{N^{\prime\prime}(z_{k})-c_{k,0}D_{k}^{\prime\prime}(z_{k})-2c_{k,1}D_{k}^{\prime}(z_{k})}{2D_{k}(z_{k})}$

In the subsequent part, we will explain our proposed method.

##  3 Our Proposed Method

In our approach, we utilize the conventional partial fraction expansion of a
rational function over the field of real numbers, as described in [7]. Let
$X(z)=\frac{N(z)}{D(z)}$. If $deg(N)\geq deg(D)$, divide the numerator by the
denominator to obtain
$\frac{N(z)}{D(z)}=\text{polynomial}+\frac{N_{1}(z)}{D(z)}$ (where $N_{1}(z)$
represents the remainder from the division of $N(z)$ by $D(z)$). Next,
factorize the denominator into factors of the following forms: $(z-r)^{u}$ and
$(z^{2}-2az+(a^{2}+b^{2}))^{k}$, where $r,a,b\in\mathbb{R}$ (Note that the
second form of factors is irreducible over the real numbers).

For each factor of the form $(z-r)^{u}$, the partial fraction expansion should
include a sum of $u$ fractions as follows:

$\frac{A_{1}}{z-r}+\frac{A_{2}}{(z-r)^{2}}+\ldots+\frac{A_{u}}{(z-r)^{u}}$

For each factor of the form $(z^{2}-2az+(a^{2}+b^{2}))^{k}$, the partial
fraction expansion should include a sum of $k$ fractions:

$\frac{B_{1}z+C_{1}}{z^{2}-2az+(a^{2}+b^{2})}+\frac{B_{2}z+C_{2}}{(z^{2}-2az+(a^{2}+b^{2}))^{2}}+\ldots+\frac{B_{k}z+C_{k}}{(z^{2}-2az+(a^{2}+b^{2}))^{k}}$

That is, the partial fraction expansion of $X(z)$ is the following:

(13)
$X(z)=\frac{N(z)}{D(z)}=P(z)+\sum_{h=1}^{v}\sum_{j=1}^{u_{h}}\frac{A_{hj}}{(z-r_{h})^{j}}+\sum_{h=1}^{w}\sum_{j=1}^{k_{h}}\frac{B_{hj}z+C_{hj}}{(z^{2}-2a_{h}z+(a_{h}^{2}+b_{h}^{2}))^{j}}$

There are several approaches to determine the coefficients
$A_{hj},B_{hj},C_{hj}$. The most direct method involves multiplying both sides
of the equation by the common denominator, $D(z)$. This yields a polynomial
equation where the left-hand side simplifies to $N(z)$, while the right-hand
side consists of coefficients expressed as linear combinations of the
constants $A_{hj},B_{hj},C_{hj}$. By equating the coefficients of
corresponding terms, we obtain a system of linear equations. This system
always possesses a unique solution. Standard methods of linear algebra can be
employed to find this solution. Alternatively, limits can be utilized, as
demonstrated in [1].

Therefore, by leveraging the linearity property of the inverse
$\mathcal{Z}$-transform, it is sufficient to calculate the inverse
$\mathcal{Z}$-transform of each term. In the following sections, we will
determine these inverse transforms.

###  3.1 Powers of $z$

For the polynomial terms and terms containing poles at the origin, it is
sufficient to know the inverse $\mathcal{Z}$-transform of the powers of $z$,
which can be easily found.

(14) $\mathcal{Z}^{-1}(z^{k})=\delta[n+k]$

where $k$ can be an arbitrary integer, and $\delta[\cdot]$ represents the
discrete-time unit impulse function. This equation was also addressed in [9,
8].

###  3.2 Terms Containing Real Non-zero Poles

######  Theorem 3.1.

(15)
$\mathcal{Z}^{-1}\left(\frac{1}{\left(z-a\right)^{k}}\right)=\binom{n-1}{k-1}a^{n-k}u\left[n-k\right]$
(16) $=\binom{n-1}{n-k}a^{n-k}u\left[n-k\right]$

where $u[\cdot]$ is the discrete-time unit step function.222Note that as
mentioned in [6], $\lim_{\epsilon\to
0}\binom{n-1}{k-1}\epsilon^{n-k}=\delta[n-k]$, This implies that the terms
containing poles at the origin are the limiting case of the real non-zero
poles.

######  Proof 3.2.

Since $\binom{n-1}{k-1}=\binom{n-1}{n-k}$, it suffices to prove Equation (15).
According to the first theorem of [3], we know that

(17)
$\mathcal{Z}^{-1}\left(\frac{1}{z-a}\right)=a^{n-1}u\left[n-1\right]=\binom{n-1}{1-1}a^{n-1}u\left[n-1\right]$

Hence, the statement is held for $k=1$. Moreover, based on the second theorem
of [3], for $k\geq 2$

$\displaystyle\mathcal{Z}^{-1}\left(\frac{1}{\left(z-a\right)^{k}}\right)$
$\displaystyle=\frac{1}{(k-1)!}\mathcal{Z}^{-1}\left(\frac{(k-1)!}{\left(z-a\right)^{k}}\right)$
(18)
$\displaystyle=\frac{(n-1)(n-2)\ldots(n-(k-1))}{(k-1)!}a^{n-k}u\left[n-1\right]$

Note that for each integer $n$, we have
$(n-1)(n-2)\ldots(n-(k-1))u\left[n-1\right]=(n-1)(n-2)\ldots(n-(k-1))u\left[n-k\right]$
Therefore,

$\displaystyle\mathcal{Z}^{-1}\left(\frac{1}{\left(z-a\right)^{k}}\right)$
$\displaystyle=\frac{(n-1)(n-2)\ldots(n-(k-1))}{(k-1)!}a^{n-k}u\left[n-k\right]$
(19) $\displaystyle=\binom{n-1}{k-1}a^{n-k}u\left[n-k\right]$

###  3.3 Terms Containing Complex Poles

######  Theorem 3.3.

Let $a\pm ib=re^{i\pm\theta}$. Then

(20)
$\mathcal{Z}^{-1}\left(\frac{1}{\left(z^{2}-2az+\left(a^{2}+b^{2}\right)\right)^{k}}\right)=\left(\frac{-1}{4}\right)^{k-1}\frac{r^{n-2k}u\left[n-2k\right]}{\left(\sin{\theta}\right)^{2k-1}}\scalebox{1.25}{S}$

where

(21)
$\scalebox{1.25}{S}=\sum_{j=0}^{k-1}{\left(-1\right)^{j}\left(\begin{matrix}k-1\\\
j\\\
\end{matrix}\right)\left(n-1\right)_{j}\left(n-\left(k+1+j\right)\right)_{k-1-j}\sin{\left(\left(n-2j-1\right)\theta\right)}}$

where the notation $\left(x\right)_{n}$ is the falling factorial introduced in
(9).

Since the proof of this theorem is too long and technical, we have provided it
in appendix A.

######  Corollary 3.4.

Let $a\pm ib=re^{i\pm\theta}$. Then

(22)
$\mathcal{Z}^{-1}\left(\frac{z}{\left(z^{2}-2az+\left(a^{2}+b^{2}\right)\right)^{k}}\right)=\left(\frac{-1}{4}\right)^{k-1}\frac{r^{n+1-2k}u\left[n+1-2k\right]}{\left(\sin{\theta}\right)^{2k-1}}\scalebox{1.25}{S}$

where

(23)
$\scalebox{1.25}{S}=\sum_{j=0}^{k-1}{\left(-1\right)^{j}\left(\begin{matrix}k-1\\\
j\\\
\end{matrix}\right)\left(n\right)_{j}\left(n-\left(k+j\right)\right)_{k-1-j}\sin{\left(\left(n-2j\right)\theta\right)}}$

######  Proof 3.5.

We know that $\mathcal{Z}(f[n+1])=zF(z)-zf[0]$. Furthermore, if

(24)
$f[n]=\mathcal{Z}^{-1}\left(\frac{1}{\left(z^{2}-2az+\left(a^{2}+b^{2}\right)\right)^{k}}\right)$

then based on (20), for each $k\geq 1,f[0]=0$. As a result,

(25)
$\mathcal{Z}(f[n+1])=zF(z)=\frac{z}{\left(z^{2}-2az+\left(a^{2}+b^{2}\right)\right)^{k}}$

so, if we replace $n$ with $n+1$ in (20), we will get the desired result.

Now that we have obtained the formulas Eqs. 20 and 22, we can find the inverse
$\mathcal{Z}$-transform of our building blocks of partial fraction expansion.

######  Corollary 3.6.

Let
$x[n]=\mathcal{Z}^{-1}\left(\frac{Az+B}{\left(z^{2}-2az+\left(a^{2}+b^{2}\right)\right)^{k}}\right)$.
Then

(26)
$x[n]=\left(\frac{-1}{4}\right)^{k-1}\frac{Ar^{n+1-2k}u\left[n+1-2k\right]}{\left(\sin{\theta}\right)^{2k-1}}S_{\mathrm{\scalebox{1.5}{$\scriptstyle
1$}}}+\left(\frac{-1}{4}\right)^{k-1}\frac{Br^{n-2k}u\left[n-2k\right]}{\left(\sin{\theta}\right)^{2k-1}}S_{\mathrm{\scalebox{1.5}{$\scriptstyle
2$}}}$

where

(27) $S_{\mathrm{\scalebox{1.5}{$\scriptstyle
1$}}}=\sum_{j=0}^{k-1}{\left(-1\right)^{j}\left(\begin{matrix}k-1\\\ j\\\
\end{matrix}\right)\left(n\right)_{j}\left(n-\left(k+j\right)\right)_{k-1-j}\sin{\left(\left(n-2j\right)\theta\right)}}$

and

(28) $S_{\mathrm{\scalebox{1.5}{$\scriptstyle
2$}}}=\sum_{j=0}^{k-1}{\left(-1\right)^{j}\left(\begin{matrix}k-1\\\ j\\\
\end{matrix}\right)\left(n-1\right)_{j}\left(n-\left(k+1+j\right)\right)_{k-1-j}\sin{\left(\left(n-2j-1\right)\theta\right)}}$

Therefore,

(29)
$x[n]=x[2k-1]\delta[n+1-2k]+\left(\frac{-1}{4}\right)^{k-1}\frac{r^{n-2k}u[n-2k]\text{
}\scalebox{1.25}{S}}{(\sin\theta)^{2k-1}}$

where

(30)
$\displaystyle\scalebox{1.2}{$S$}=\sum_{j=0}^{k-1}(-1)^{j}\binom{k-1}{j}\Bigg{[}$
$\displaystyle Ar(n)_{j}\left(n-(k+j)\right)_{k-1-j}\sin((n-2j)\theta)$
$\displaystyle+B(n-1)_{j}\left(n-(k+1+j)\right)_{k-1-j}\sin((n-2j-1)\theta)\Bigg{]}$

The term $x[2k-1]\delta[n+1-2k]$ is used to handle the difference between the
arguments of unit step functions. But since $x[2k-1]$ is unknown, we must
calculate it. Let us denote the the degree of the numerator by $p$, and the
degree of the denominator by $q$. Obviously,

$q=2k\text{ and }p=\left\\{\begin{array}[]{ll}1&A\neq 0\\\ 0&A=0\\\
\end{array}\right.$

Then, according to (6), if $A\neq 0$, $x[2k-1]=x[q-p]=A$. And if $A=0$, based
on (5), $x[2k-1]=x[q-p-1]=0=A$. Therefore, $x[2k-1]=A$. To summarize,

(31)
$\mathcal{Z}^{-1}\left(\frac{Az+B}{\left(z^{2}-2az+\left(a^{2}+b^{2}\right)\right)^{k}}\right)=A\delta[n+1-2k]+\left(\frac{-1}{4}\right)^{k-1}\frac{r^{n-2k}u[n-2k]\text{
}\scalebox{1.25}{S}}{(\sin\theta)^{2k-1}}$

where  S is obtained through (30). Although the equation (31) may come across
as intricate, it covers a broad class of functions.

##  4 Examples

Now, we give some examples to illustrate how to use Eqs. 20 and 22.

######  Example 4.0.

In this example, we will compute the inverse $\mathcal{Z}$-transform of
$\frac{1}{z^{2}+1}$. Hence, $k=1$, and $a\pm ib=e^{i\pm\frac{\pi}{2}}$; i.e.
$r=1,\theta=\frac{\pi}{2}$. Therefore, the summation used in (20) will have
just one summand for $u=0$. Then, both falling factorial terms will be equal
to 1. To sum up,

(32)
$\mathcal{Z}^{-1}\left(\frac{1}{z^{2}+1}\right)=\frac{u[n-1]}{\sin\frac{\pi}{2}}\sin\left((n-1)\frac{\pi}{2}\right)=-u[n-1]\cos\left(\frac{n\pi}{2}\right)$

Now, let’s find the answer with the method used in [8]. According to that
method, first, we must decompose $Y(z)=\frac{X(z)}{z}$ over the field of
complex numbers; that is

(33) $\frac{1}{z(z^{2}+1)}=\frac{A}{z}+\frac{B}{z-i}+\frac{B^{*}}{z+i}\text{
where }B=B_{1}e^{i\phi}$

and the coefficients are obtained via these limits

$A=\lim_{z\to 0}zY(z)=\lim_{z\to 0}\frac{1}{z^{2}+1}=1$

and

$B=\lim_{z\to i}(z-i)Y(z)=\lim_{z\to i}\frac{1}{z(z+i)}=\frac{-1}{2}$

Therefore, by multiplying both sides of (33) by $z$, we will have

$\frac{1}{z^{2}+1}=1+\frac{-z/2}{z-i}+\frac{-z/2}{z+i}$

so, using the Table 1, we obtain

(34)
$f[n]=\delta[n]+2\left(\frac{-1}{2}\right)\cos\left(\frac{\pi}{2}n\right)=\delta[n]-\cos\left(\frac{n\pi}{2}\right)$

It’s easy to see that the results of Eqs. 32 and 34 are identical.

Also, let’s solve it by the method proposed by [6] mentioned in Eqs. 7 and 8.
Since $\frac{X(z)}{z}=\frac{1}{z(z^{2}+1)}$, we have
$z_{1}=0,z_{2}=i,z_{3}=-i,m_{1}=m_{2}=m_{3}=1,P(z)=1,Q(z)=z(z^{2}+1)$.
Therefore,

(35) $x[n]=c_{1,0}\delta[n]+\sum_{k=2}^{3}c_{k,0}z_{k}^{n}\text{ where
}c_{k,0}=\frac{P(z_{k})}{Q_{k}(z_{k})}$

That is,

$\displaystyle x[n]$
$\displaystyle=c_{1,0}\delta[n]+c_{2,0}z_{2}^{n}+c_{3,0}z_{3}^{n}$
$\displaystyle=\left[\frac{1}{z^{2}+1}\right]_{z=0}\delta[n]+\left[\frac{1}{z(z+i)}\right]_{z=i}i^{n}+\left[\frac{1}{z(z-i)}\right]_{z=i}(-i)^{n}$
$\displaystyle=\delta[n]-\frac{i^{n}}{2}-\frac{(-i)^{n}}{2}=\delta[n]-\frac{(i^{n}+(-i)^{n})}{2}=\delta[n]-\mathfrak{Re}(i^{n})$
(36) $\displaystyle=\delta[n]-\cos(\frac{n\pi}{2})$

######  Example 4.0.

Let’s consider a more complex example with two parameters $A,B$. We want to
find the inverse $\mathcal{Z}$-transform of
$\frac{Az+B}{(z^{2}-2az+(a^{2}+b^{2}))^{3}}$. If we want to solve it using our
proposed method, we can use Eqs. 30 and 31. Based on them, we’ll have

$\mathcal{Z}^{-1}\left(\frac{Az+B}{\left(z^{2}-2az+\left(a^{2}+b^{2}\right)\right)^{3}}\right)=A\delta[n-5]+\left(\frac{-1}{4}\right)^{2}\frac{r^{n-6}u[n-6]\text{
}\scalebox{1.25}{S}}{(\sin\theta)^{5}}$

where

$\displaystyle\scalebox{1.2}{$S$}=\sum_{j=0}^{2}(-1)^{j}\binom{2}{j}\Bigg{[}$
$\displaystyle Ar(n)_{j}\left(n-(3+j)\right)_{2-j}\sin((n-2j)\theta)$
$\displaystyle+B(n-1)_{j}\left(n-(4+j)\right)_{2-j}\sin((n-2j-1)\theta)\Bigg{]}$

As a result,

$x[n]=A\delta[n-5]+\frac{r^{n-6}u\left[n-6\right]}{16\left(\sin{\theta}\right)^{5}}\left(S_{0}+S_{1}+S_{2}\right)$

where

$S_{0}=Ar(n-3)(n-2)\sin(n\theta)+B(n-4)(n-5)\sin((n-1)\theta)$
$S_{1}=-2\left(Arn(n-4)\sin((n-2)\theta)+B(n-1)(n-5)\sin((n-3)\theta)\right)$
$S_{2}=Arn(n-1)\sin((n-4)\theta)+B(n-1)(n-2)\sin((n-5)\theta)$

If we want to obtain it via the method proposed by [8]. First, we must get the
partial fraction decomposition of $Y(z)=\frac{X(x)}{z}$.

$\displaystyle\frac{Az+B}{z(z^{2}-2az+(a^{2}+b^{2}))^{3}}=\frac{A_{1,1}}{z}$
$\displaystyle+\frac{A_{2,1}}{z-(a+ib)}+\frac{A_{2,1}^{*}}{z-(a-ib)}$
$\displaystyle+\frac{A_{2,2}}{(z-(a+ib))^{2}}+\frac{A_{2,2}^{*}}{(z-(a-ib))^{2}}$
$\displaystyle+\frac{A_{2,3}}{(z-(a+ib))^{3}}+\frac{A_{2,3}^{*}}{(z-(a-ib))^{3}}$

where

$A_{1,1}=\lim_{z\to 0}zY(z)=\lim_{z\to
0}\frac{Az+B}{(z^{2}-2az+(a^{2}+b^{2}))^{3}}=\frac{B}{(a^{2}+b^{2})^{3}}=\frac{B}{r^{6}}\text{
,}$ $A_{2,p}=\frac{1}{(3-p)!}\lim_{z\to
a+ib}\frac{d^{3-p}}{dz^{3-p}}\left((z-(a+ib))^{3}Y(z)\right)\text{ for
}p=1,2,3$

It’s easy to see that using this method, the rest of calculations will be too
long and cumbersome.

Furthermore, if we want to get the result using the method proposed by [6]
(Eqs. 7 and 8), it will be as follows: First, let
$z_{1}=0,z_{2}=a+ib,z_{3}=a-ib$. Then

$x[n]=c_{1,0}\delta[n]+\sum_{k=2}^{3}\sum_{j=0}^{2}c_{k,2-j}\binom{n}{j}z_{k}^{n-j}$

Since $z_{3}=z_{2}^{*}$ and $(z_{k}^{n-j})^{*}=(z_{k}^{*})^{n-j}$, we can
write

$x[n]=c_{1,0}\delta[n]+2\mathfrak{Re}\left(\sum_{j=0}^{2}c_{2,2-j}\binom{n}{j}z_{2}^{n-j}\right)$

For the sake of brevity, we leave the rest of the computations to interested
readers! Note that since this example was parametric, and two of its roots
were complex numbers, computation of the coefficients and simplifying the
results to get a neat real-valued function as the result will be cumbersome
and time-consuming.

##  5 Conclusions

In this paper, we presented a novel technique for computing the inverse
$\mathcal{Z}$-transform of rational functions. Through a couple of examples,
we demonstrated that compared to existing methods, our approach can result in
fewer calculations in some cases. Unlike the method proposed by [8], our
approach eliminates the need for dividing by $z$ and instead utilizes the
conventional partial fraction expansion over real numbers. These two
characteristics make our method resemble the approaches used for computing
well-known integral transforms. Furthermore, compared to the work done by [6],
our method offers the advantage of not requiring additional algebraic
manipulations to obtain a real-valued solution.

##  Appendix A Proof of the Equation (20)

In this section, we prove equation (20). According to the following lemma, it
can be easily seen that the first few terms of the time-domain sequence are
zero.

######  Lemma A.1.

Let $f[n]$ be the inverse $\mathcal{Z}$-transform of
$F(z)=\frac{1}{\left(z^{2}-2az+\left(a^{2}+b^{2}\right)\right)^{k}}$. Then for
every $n$ such that $n<2k$, $f[n]=0$.

######  Proof A.2.

It’s an immediate result of (5). Nevertheless, we have provide another proof
for that in appendix B.

Now assume $n\geq 2k$. For these $n$’s, we are going to find the inverse
$\mathcal{Z}$-transform of
$F(z)=\frac{1}{\left(z^{2}-2az+\left(a^{2}+b^{2}\right)\right)^{k}}$ using
another method, that is the residue method. Note that

$f[n]=\mathcal{Z}^{-1}\left(\frac{1}{\left(z-\left(a+bi\right)\right)^{k}}\times\frac{1}{\left(z-\left(a-bi\right)\right)^{k}}\right)$

Since its both poles are of degree $k$, based on (3), we have:

(37)
$\displaystyle\underset{z=a+ib}{\mathrm{Res}}f(z)=\frac{1}{\left(k-1\right)!}\left.\left[\frac{d^{k-1}}{dz^{k-1}}\left(\frac{z^{n-1}}{\left(z-\left(a-ib\right)\right)^{k}}\right)\right]\right\rvert_{z=a+ib}$

and

(38)
$\displaystyle\underset{z=a-ib}{\mathrm{Res}}f(z)=\frac{1}{\left(k-1\right)!}\left.\left[\frac{d^{k-1}}{dz^{k-1}}\left(\frac{z^{n-1}}{\left(z-\left(a+ib\right)\right)^{k}}\right)\right]\right\rvert_{z=a-ib}$

Therefore, according to (2), $f[n]$ is some of the results of Eqs. 37 and 38.
Now we calculate a general case for the terms appeared in Eqs. 37 and 38.
Suppose $\xi$ and $\eta$ are complex conjugates. Furthermore, define

(39)
$D=\left.\left[\frac{d^{k-1}}{dz^{k-1}}\left(\frac{z^{n-1}}{\left(z-\eta\right)^{k}}\right)\right]\right|_{z=\xi}$

then, according to general Leibniz rule, we have:

(40) $D=\sum_{t=0}^{k-1}{\left(\begin{matrix}k-1\\\ t\\\
\end{matrix}\right)\left.\left[\frac{d^{t}}{dz^{t}}\left(z^{n-1}\right)\right]\right|_{z=\xi}}\left.\left[\frac{d^{k-1-t}}{dz^{k-1-t}}\left(\left(z-\eta\right)^{-k}\right)\right]\right|_{z=\xi}$

Note that

(41)
$\left.\left[\frac{d^{t}}{dz^{t}}\left(z^{n-1}\right)\right]\right|_{z=\xi}=u\left[n-1-t\right]\left(n-1\right)\ldots\left(n-t\right)\xi^{n-t-1}$

Since we have supposed that $n\geq 2k$, and according to the summation upper
bound, $t\leq k-1$, we conclude that $n-t-1\geq k\geq 1$. Therefore,
$u[n-t-1]=1$. Hence

(42) $\displaystyle D$
$\displaystyle=\sum_{t=0}^{k-1}\binom{k-1}{t}(n-1)\ldots(n-t)\xi^{n-t-1}$
$\displaystyle\quad\times(-1)^{k-1-t}k(k+1)\ldots(2k-2-t)(\xi-\eta)^{-(2k-1-t)}$

Also note that:

$\binom{k-1}{t}(n-1)\ldots(n-t)\times k(k+1)\ldots(2k-2-t)$
$\displaystyle=\frac{(k-1)!}{t!(k-1-t)!}\times\frac{(n-1)!}{(n-t-1)!}\times\frac{(2k-2-t)!}{(k-1)!}$
$\displaystyle=\frac{(n-1)!}{t!(n-t-1)!}\times\frac{(2k-2-t)!}{(k-1-t)!}$ (43)
$\displaystyle=\binom{n-1}{t}\times\frac{(2k-2-t)!}{(k-1-t)!}$

Substituting equation A in 42 we’ll have:

$D=\xi^{n-1}\left(-1\right)^{k-1}\left(\xi-\eta\right)^{-\left(2k-1\right)}\sum_{t=0}^{k-1}{\left(\begin{matrix}n-1\\\
t\\\
\end{matrix}\right)\frac{\left(2k-2-t\right)!}{\left(k-1-t\right)!}\xi^{-t}\times\left(-1\right)^{-t}\left(\xi-\eta\right)^{t}}$
(44)
$=\xi^{n-1}\left(-1\right)^{k-1}\left(\xi-\eta\right)^{1-2k}\sum_{t=0}^{k-1}{\left(\begin{matrix}n-1\\\
t\\\
\end{matrix}\right)\frac{\left(2k-2-t\right)!}{\left(k-1-t\right)!}}\left(\frac{\eta-\xi}{\xi}\right)^{t}$

Now, since we supposed $\xi$ and $\eta$ are complex conjugates, we can write
$\xi=re^{i\theta}$ and $\eta=re^{-i\theta}$. Substituting these values, we
will have:

$D=\left.\left[\frac{d^{k-1}}{dz^{k-1}}\left(\frac{z^{n-1}}{\left(z-re^{-i\theta}\right)^{k}}\right)\right]\right|_{z=re^{i\theta}}$
$=r^{n-1}e^{i\left(n-1\right)\theta}\left(-1\right)^{k-1}\left(2ri\
\sin{\theta}\right)^{1-2k}\sum_{t=0}^{k-1}{\left(\begin{matrix}n-1\\\ t\\\
\end{matrix}\right)\frac{\left(2k-2-t\right)!}{\left(k-1-t\right)!}\left(e^{-2i\theta}-1\right)^{t}}$
(45) $=r^{n-2k}e^{i\left(n-1\right)\theta}\left(-1\right)^{k-1}\left(2i\
\sin{\theta}\right)^{1-2k}\sum_{t=0}^{k-1}{\left(\begin{matrix}n-1\\\ t\\\
\end{matrix}\right)\frac{\left(2k-2-t\right)!}{\left(k-1-t\right)!}\left(e^{-2i\theta}-1\right)^{t}}$

Similarly, 38 is as follows, where $D^{*}$ refers to the complex conjugate of
$D$:

$D^{*}=\left.\left[\frac{d^{k-1}}{dz^{k-1}}\left(\frac{z^{n-1}}{\left(z-re^{i\theta}\right)^{k}}\right)\right]\right|_{z=re^{-i\theta}}$
(46) $=r^{n-2k}e^{i\left(1-n\right)\theta}\left(-1\right)^{k-1}\left(-2i\
\sin{\theta}\right)^{1-2k}\sum_{t=0}^{k-1}{\left(\begin{matrix}n-1\\\ t\\\
\end{matrix}\right)\frac{\left(2k-2-t\right)!}{\left(k-1-t\right)!}\left(e^{2i\theta}-1\right)^{t}}$

Therefore, using the Eqs. 37, 38, 39, 45, and 46, we conclude that

(47) $f\left[n\right]=r^{n-2k}\left(-1\right)^{k-1}\left(2i\
\sin{\theta}\right)^{1-2k}\left(A-A^{*}\right)$

where

(48) $A=e^{i\left(n-1\right)\theta}\sum_{t=0}^{k-1}{\left(\begin{matrix}n-1\\\
t\\\
\end{matrix}\right)\frac{\left(2k-2-t\right)!}{\left(k-1-t\right)!}\left(e^{-2i\theta}-1\right)^{t}}$

and

(49)
$A^{*}=e^{i\left(1-n\right)\theta}\sum_{t=0}^{k-1}{\left(\begin{matrix}n-1\\\
t\\\
\end{matrix}\right)\frac{\left(2k-2-t\right)!}{\left(k-1-t\right)!}\left(e^{2i\theta}-1\right)^{t}}$

Since $\left(A-A^{*}\right)=2i\mathfrak{Im}(A)$, we can write:

(50) $f\left[n\right]=r^{n-2k}\left(-1\right)^{k-1}\left(2i\
\sin{\theta}\right)^{1-2k}\left(2i\mathfrak{Im}(A)\right)$

Now, based on binomial expansion of $\left(e^{2i\theta}-1\right)^{t}$, we
have:

$A=e^{i\left(n-1\right)\theta}\sum_{t=0}^{k-1}{\left(\begin{matrix}n-1\\\ t\\\
\end{matrix}\right)\frac{\left(2k-2-t\right)!}{\left(k-1-t\right)!}\sum_{j=0}^{t}{\left(\begin{matrix}t\\\
j\\\ \end{matrix}\right)\left(e^{-2i\theta}\right)^{j}\left(-1\right)^{t-j}}}$
(51)
$=e^{i\left(n-1\right)\theta}\sum_{t=0}^{k-1}\sum_{j=0}^{t}{\left(\begin{matrix}n-1\\\
t\\\
\end{matrix}\right)\frac{\left(2k-2-t\right)!}{\left(k-1-t\right)!}\left(\begin{matrix}t\\\
j\\\ \end{matrix}\right)\left(e^{-2i\theta}\right)^{j}\left(-1\right)^{t-j}}$

By interchanging the order of summations:

(52)
$A=e^{i\left(n-1\right)\theta}\sum_{j=0}^{k-1}\sum_{t=j}^{k-1}{\left(\begin{matrix}n-1\\\
t\\\
\end{matrix}\right)\frac{\left(2k-2-t\right)!}{\left(k-1-t\right)!}\left(\begin{matrix}t\\\
j\\\ \end{matrix}\right)\left(e^{-2i\theta}\right)^{j}\left(-1\right)^{t-j}}$

Therefore

$A=e^{i\left(n-1\right)\theta}\sum_{j=0}^{k-1}{\left(e^{-2i\theta}\right)^{j}\sum_{t=j}^{k-1}{\left(\begin{matrix}n-1\\\
t\\\
\end{matrix}\right)\frac{\left(2k-2-t\right)!}{\left(k-1-t\right)!}\left(\begin{matrix}t\\\
j\\\ \end{matrix}\right)\left(-1\right)^{t-j}}}$ (53)
$=\sum_{j=0}^{k-1}{e^{\left(n-2j-1\right)i\theta}\sum_{t=j}^{k-1}{\left(\begin{matrix}n-1\\\
t\\\
\end{matrix}\right)\frac{\left(2k-2-t\right)!}{\left(k-1-t\right)!}\left(\begin{matrix}t\\\
j\\\ \end{matrix}\right)\left(-1\right)^{t-j}}}$ (54)
$\Longrightarrow\mathfrak{Im}(A)=\sum_{j=0}^{k-1}{\sin{(\left(n-2j-1\right)\theta)}\sum_{t=j}^{k-1}{\left(\begin{matrix}n-1\\\
t\\\
\end{matrix}\right)\frac{\left(2k-2-t\right)!}{\left(k-1-t\right)!}\left(\begin{matrix}t\\\
j\\\ \end{matrix}\right)\left(-1\right)^{t-j}}}$

Using equation (50), we conclude that:

$\displaystyle f[n]$ $\displaystyle=r^{n-2k}(-1)^{k-1}(2i\sin{\theta})^{1-2k}$
$\displaystyle\quad\times\left(2i\sum_{j=0}^{k-1}\sin{((n-2j-1)\theta)}\sum_{t=j}^{k-1}\left(\binom{n-1}{t}\frac{(2k-2-t)!}{(k-1-t)!}\binom{t}{j}(-1)^{t-j}\right)\right)$
$\displaystyle=\frac{r^{n-2k}}{(\sin{\theta})^{2k-1}2^{2k-2}}$ (55)
$\displaystyle\quad\times\sum_{j=0}^{k-1}(-1)^{j}\sin{((n-2j-1)\theta)}\sum_{t=j}^{k-1}\left(\binom{n-1}{t}\frac{(2k-2-t)!}{(k-1-t)!}\binom{t}{j}(-1)^{t}\right)$

Note that we had assumed that $n\geq 2k$. To combine this with the fact that
$f[n]=0$ for $n<2k$ (obtained from lemma A.1), we can write

$\displaystyle f[n]$
$\displaystyle=\frac{r^{n-2k}u[n-2k]}{(\sin{\theta})^{2k-1}2^{2k-2}}$ (56)
$\displaystyle\quad\times\sum_{j=0}^{k-1}(-1)^{j}\sin{((n-2j-1)\theta)}\sum_{t=j}^{k-1}\left(\binom{n-1}{t}\frac{(2k-2-t)!}{(k-1-t)!}\binom{t}{j}(-1)^{t}\right)$

######  Lemma A.3.

$\sum_{t=j}^{k-1}\binom{n-1}{t}\frac{(2k-2-t)!}{(k-1-t)!}\binom{t}{j}(-1)^{t}$
(57) $\displaystyle=c_{j,k}(n-1)\ldots(n-j)\times(n-(k+1+j))\ldots(n-(2k-1))$
$\displaystyle=c_{j,k}(n-1)_{j}(n-(k+1+j))_{k-1-j}$

where $c_{j,k}$ is a fixed number (with respect to $n$) and we will find it in
the next theorem. Furthermore, the notation $\left(x\right)_{n}$ is the
falling factorial defined in equation (9).

######  Proof A.4.

First of all, we know that the result of the series is a $(k-1)$ degree
polynomial in terms of $n$. So, it suffices to find $k-1$ roots for it. Now,
note that

(58) $\left(\begin{matrix}n-1\\\ t\\\
\end{matrix}\right)=\frac{\left(n-1\right)\ldots(n-t)}{t!}$

Therefore, for each $t\geq j$, we have:

(59) $\left(\begin{matrix}n-1\\\ t\\\
\end{matrix}\right)=\left(n-1\right)\ldots\left(n-j\right)f\left(n\right)$

where $f(n)$ is a polynomial in terms of $n$. Therefore

(60)
$\displaystyle\sum_{t=j}^{k-1}\binom{n-1}{t}\frac{(2k-2-t)!}{(k-1-t)!}\binom{t}{j}(-1)^{t}$
$\displaystyle=(n-1)\ldots(n-j)\sum_{t=j}^{k-1}f_{t}(n)$
$\displaystyle=(n-1)\ldots(n-j)g(n)$

where $f_{t}(n),g$ are also polynomials of $n$.

Now, we prove that $k+1+j,k+2+j,\ldots,2k-1$ are some roots of

(61) $\Phi\left(n\right)=\sum_{t=j}^{k-1}{\left(\begin{matrix}n-1\\\ t\\\
\end{matrix}\right)\frac{\left(2k-2-t\right)!}{\left(k-1-t\right)!}\left(\begin{matrix}t\\\
j\\\ \end{matrix}\right)\left(-1\right)^{t}}$

i.e. if $n=k+\rho$ where $1+j\leq\rho\leq k-1$, then $\Phi(n)=0$.

(62) $\displaystyle\Phi(k+\rho)$
$\displaystyle=\sum_{t=j}^{k-1}\binom{k+\rho-1}{t}\frac{(2k-2-t)!}{(k-1-t)!}\binom{t}{j}(-1)^{t}$
$\displaystyle=\sum_{t=j}^{k-1}\frac{(k+\rho-1)!}{t!(k+\rho-1-t)!}\frac{(2k-2-t)!}{(k-1-t)!}\frac{t!}{j!(t-j)!}(-1)^{t}$
$\displaystyle=\frac{(k+\rho-1)!}{j!}\sum_{t=j}^{k-1}\frac{1}{(k+\rho-1-t)!}\frac{(2k-2-t)!}{(k-1-t)!}\frac{1}{(t-j)!}(-1)^{t}$
$\displaystyle=\frac{(k+\rho-1)!}{j!(k-1-j)!}\sum_{t=j}^{k-1}\frac{(2k-2-t)!}{(k+\rho-1-t)!}\frac{(k-1-j)!}{(k-1-t)!(t-j)!}(-1)^{t}$
$\displaystyle=\frac{(k+\rho-1)!}{j!(k-1-j)!}\sum_{t=j}^{k-1}\frac{(2k-2-t)!}{(k+\rho-1-t)!}\binom{k-1-j}{t-j}(-1)^{t}$

By considering $w=t-j$, we’ll have:

(63)
$\Phi\left(k+\rho\right)=\frac{\left(-1\right)^{j}\left(k+\rho-1\right)!}{j!\left(k-1-j\right)!}\sum_{w=0}^{k-1-j}{\frac{\left(2k-2-j-w\right)!}{\left(k+\rho-1-j-w\right)!}\left(\begin{matrix}k-1-j\\\
w\\\ \end{matrix}\right)\left(-1\right)^{w}}$

Again, $\sigma=k-1-j$ implies that:

(64)
$\Phi\left(k+\rho\right)=\frac{\left(-1\right)^{j}\left(k+\rho-1\right)!}{j!\left(k-1-j\right)!}\sum_{w=0}^{\sigma}{\frac{\left(2\sigma+j-w\right)!}{\left(\sigma+\rho-w\right)!}\left(\begin{matrix}\sigma\\\
w\\\ \end{matrix}\right)\left(-1\right)^{w}}$

So, it suffices to prove that

(65)
$\sum_{w=0}^{\sigma}{\frac{\left(2\sigma+j-w\right)!}{\left(\sigma+\rho-w\right)!}\left(\begin{matrix}\sigma\\\
w\\\ \end{matrix}\right)\left(-1\right)^{w}}=0$

where $1+j\leq\rho\leq\sigma+j$. Note that

(66)
$\frac{\left(2\sigma+j-w\right)!}{\left(\sigma+\rho-w\right)!}=\left(2\sigma+j-w\right)\left(2\sigma+j-w-1\right)\ldots\left(2\sigma+j-w-\nu\right)$

where $\nu=\sigma+j-\rho-1$. Therefore, $-1\leq\nu\leq\sigma-2$. We can
rewrite this equation like this:

(67)
$\frac{\left(2\sigma+j-w\right)!}{\left(\sigma+\rho-w\right)!}=\left(\sigma-w+\left(\sigma+j\right)\right)\left(\sigma-w+\left(\sigma+j-1\right)\right)\ldots\left(\sigma-w+\left(\sigma+j-\nu\right)\right)$

which is equal to

$\sum_{p=0}^{\nu+1}{a_{p}{(\sigma-w)}^{p}}$

where $a_{p}$’s are some coefficients independent of $w$ and they can be
calculated by Vieta’s formulas. Therefore, it suffices to show that for each
$p$ such that $0\leq p\leq\sigma-1$:

(68)
$\sum_{w=0}^{\sigma}{\sum_{p=0}^{\nu+1}{a_{p}{(\sigma-w)}^{p}}\left(\begin{matrix}\sigma\\\
w\\\ \end{matrix}\right)\left(-1\right)^{w}}=0$

By changing the order of summation, and the fact that $a_{p}$’s are
independent of $w$, it’s equivalent to prove:

(69)
$\sum_{p=0}^{\nu+1}{a_{p}\sum_{w=0}^{\sigma}{(\sigma-w)}^{p}\left(\begin{matrix}\sigma\\\
w\\\ \end{matrix}\right)\left(-1\right)^{w}}=0$

which is true, since

$\sum_{w=0}^{\sigma}{{(\sigma-w)}^{p}\left(\begin{matrix}\sigma\\\ w\\\
\end{matrix}\right)\left(-1\right)^{w}}$

represents the number of ways of putting $p$ distinct balls into $\sigma$
distinct boxes such that none of the boxes is empty [4]. Since $0\leq
p\leq\sigma-1$:

(70) $\sum_{w=0}^{\sigma}{{(\sigma-w)}^{p}\left(\begin{matrix}\sigma\\\ w\\\
\end{matrix}\right)\left(-1\right)^{w}}=0$

######  Corollary A.5.

(71) $c_{j,k}=\left(-1\right)^{k-1}\left(\begin{matrix}k-1\\\ j\\\
\end{matrix}\right)$

######  Proof A.6.

Since both sides of the equation (57) are polynomials of $n$, their leading
coefficient must agree. The leading coefficient of the LHS of (57), i.e. the
coefficient of $n^{k-1}$, is:

(72)
$\frac{1}{\left(k-1\right)!}\frac{\left(2k-2-\left(k-1\right)\right)!}{\left(k-1-\left(k-1\right)\right)!}\left(\begin{matrix}k-1\\\
j\\\
\end{matrix}\right)\left(-1\right)^{k-1}=\left(-1\right)^{k-1}\left(\begin{matrix}k-1\\\
j\\\ \end{matrix}\right)$

Hence, The leading coefficient of the RHS of (57), which is $c_{j,k}$, is
equal to

$\left(-1\right)^{k-1}\left(\begin{matrix}k-1\\\ j\\\ \end{matrix}\right)$

######  Corollary A.7.

using Appendices A, 57, and 71, we conclude that

$\mathcal{Z}^{-1}\left(\frac{1}{\left(z^{2}-2az+\left(a^{2}+b^{2}\right)\right)^{k}}\right)$
$\displaystyle=\frac{r^{n-2k}u[n-2k]}{\left(\sin{\theta}\right)^{2k-1}2^{2k-2}}$
$\displaystyle\quad\times\sum_{j=0}^{k-1}{\left(-1\right)^{j}\sin{\left(\left(n-2j-1\right)\theta\right)}\sum_{t=j}^{k-1}{\left(\begin{matrix}n-1\\\
t\\\
\end{matrix}\right)\frac{\left(2k-2-t\right)!}{\left(k-1-t\right)!}\left(\begin{matrix}t\\\
j\\\ \end{matrix}\right)\left(-1\right)^{t}}}$
$\displaystyle=\left(\frac{-1}{4}\right)^{k-1}\frac{r^{n-2k}j\left[n-2k\right]}{\left(\sin{\theta}\right)^{2k-1}}$
(73)
$\displaystyle\quad\times\sum_{j=0}^{k-1}{\left(-1\right)^{j}\left(\begin{matrix}k-1\\\
j\\\
\end{matrix}\right)\left(n-1\right)_{j}\left(n-\left(k+1+j\right)\right)_{k-1-j}\sin{\left(\left(n-2j-1\right)\theta\right)}}$

##  Appendix B Another Proof for the Lemma A.1

######  Proof B.1.

We use the fact that $\mathcal{Z}(x[n]*y[n])=X(z)Y(z)$, where $*$ denotes the
convolution operator.

(74) $\displaystyle f[n]$
$\displaystyle=\mathcal{Z}^{-1}\left(\frac{1}{\left(z^{2}-2az+\left(a^{2}+b^{2}\right)\right)^{k}}\right)$
$\displaystyle=\mathcal{Z}^{-1}\left(\frac{1}{\left(z-(a+bi)\right)^{k}}\times\frac{1}{\left(z-(a-bi)\right)^{k}}\right)$
$\displaystyle=\binom{n-1}{n-k}(a+ib)^{n-k}u[n-k]\ast\binom{n-1}{n-k}(a-ib)^{n-k}u[n-k]$
$\displaystyle=\sum_{m=0}^{\infty}\binom{m-1}{m-k}(a+ib)^{m-k}u[m-k]\binom{n-m-1}{n-m-k}(a-ib)^{n-m-k}u[n-m-k]$
$\displaystyle=\sum_{m=k}^{n-k}\binom{m-1}{m-k}(a+ib)^{m-k}\binom{n-m-1}{n-m-k}(a-ib)^{n-m-k}$

Therefore, if $n<2k$, then $n-k<k$, so $f[n]$ will be $0$.

## References

  * [1] G. W. Bluman, Problem book for first year calculus, 1 1984, https://doi.org/10.1007/978-1-4612-1132-7, https://doi.org/10.1007/978-1-4612-1132-7.
  * [2] M. P. Chaudhary, M. Qureshi, and I. H. Khan, Hypergeometric forms of well known partial fraction expansions of some meromorphic functions, Global Journal of Science Frontier Research, 11 (2011), http://globaljournals.org/GJSFR_Volume11/6-Hypergeometric-Forms-of-Well-Known-Partial-Fraction.pdf.
  * [3] C.-C. Chen, T.-L. Chien, and C.-J. Wu, New results of rational inverseztransformation via partial fraction expansion and some application, Journal Of The Chinese Institute Of Engineers, (2004), https://doi.org/10.1080/02533839.2004.9670873, https://doi.org/10.1080/02533839.2004.9670873.
  * [4] B. S. Duran and P. L. Odell, Cluster Analysis ; a Survey [By] Benjamin S. Duran [And] Patrick L. Odell, Springer Berlin Heidelberg, 1 ed., 1 1974.
  * [5] M. Hassul and B. Shahian, Simplified inverse z‐transform, International Journal of Mathematical Education in Science and Technology, 22 (1991), pp. 915–918, https://doi.org/10.1080/0020739910220608.
  * [6] Z. Juric, A simple and elegant method for computing the inverse z-transform of rational functions, in 2023 XXIX International Conference on Information, Communication and Automation Technologies (ICAT), 2023, pp. 1–6, https://doi.org/10.1109/ICAT57854.2023.10171336.
  * [7] R. Larson, Algebra and trigonometry, Cengage Learning, 1 2010.
  * [8] M. V. Moreira and J. C. Basilio, Fair and Square Computation of Inverse ${\cal Z}$-Transforms of Rational Functions, IEEE Transactions on Education, 55 (2012), pp. 285–290, https://doi.org/10.1109/te.2011.2171185, https://doi.org/10.1109/te.2011.2171185.
  * [9] C. L. Phillips, H. T. Nagle, and A. Chakrabortty, Digital Control System Analysis & Design, Pearson, 4th ed., 2014.