There are quite a few methods two find inverse Z-transforms, a few of which we will discuss here.

I) Identify familiar forms. 


We have used particular forms involving Z-transforms over and over, so we can use pattern recognition:

\begin{example}

\begin{align*}
\hat{X}(z) &= \frac{1}{1 - \alpha Z^{-1}} \quad  \abs{\alpha} \lt \abs{z} \\
x(n) &= \alpha^n u(n)
\end{align*}


\end{example}

\begin{example}

\begin{align*}
\hat{X}(z) &= z^{-5} \\
x(n) &= \delta(n-5)
\end{align*}


II) Power Series Expansions


Recall the formula
$$
\sum \limits_{n=0}^{\infty} \beta^n = \frac{1}{1 - \beta} \quad  \abs{\beta} \lt 1
$$

\end{example}

\begin{example}

\begin{align*}
\hat{X}(z) &= \frac{1}{1 - \alpha z^{-1}} \quad  \abs{\alpha} \lt \abs{z} \\
\hat{X}(z) &= \sum \limits_{n=0}^{\infty} \lr{\alpha z^{-1}}^n \\
\hat{X}(z) &= \sum \limits_{n=0}^{\infty} \alpha^n z^{-n}\\
\hat{X}(z) &= \sum \limits_{n=-\infty}^{\infty} \alpha^n u(n) z^{-n}\\
x(n) &= \alpha^n u(n) \\
\end{align*}

Recall the expansion for $\log(1 + \beta)$:

$$
\log(1 + \beta) = \sum \limits_{n=1}^{\infty} \frac{ (-1)^{n+1} \beta^n } { n} \quad \abs{\beta} \lt 1 
$$

\end{example}

\begin{example}

\begin{align*}
\hat{X}(z) &= \log( 1 + \alpha z^{-1}) \quad \abs{\alpha} \lt \abs{z} \\
\hat{X}(z) &= \sum \limits_{n=1}^{\infty} \frac{ (-1)^{n+1} \lr{\alpha z^{-1}}^n } { n} \\
\hat{X}(z) &= \sum \limits_{n=1}^{\infty} \lr{\frac{ (-1)^{n+1} \alpha^n } { n}} z^{-n} \\
\end{align*}

This tells us that
$$x(n) = \begin{cases} 0 & n\leq 0 \\ \frac{(-1)^{n+1}\alpha^n}{n} & n \geq 1 \\ \end{cases}$$

Hence, $x(n) = \frac{(-1)^{n+1}\alpha^n}{n} u(n-1)$. Note that $\log( 1 + \alpha z^{-1})$ is not rational in $Z$. $x(n) = \frac{(-1)^{n+1}\alpha^n}{n} u(n-1)$ is not a linear combination of shifted one-sided exponentials multiplied by polynomials.


III) Differentiation


From the above example, let $y(n) = nx(n) = (-1)^{n+1}\alpha^n u(n-1)$.

\begin{align*}
y(n) &= (-1)^{n+1} \alpha^n u(n-1) \\
y(n) &= -(-1)^n \alpha^n u(n-1) \\
y(n) &= -(-\alpha)^n u(n-1) \\
y(n) &= \alpha (-\alpha)^{n-1} u(n-1) \\
\end{align*}

We have $y(n)$ is a shifted and scaled version of $\hat{G}(z) = \frac{1}{1 + \alpha z^{-1}}$.

\begin{align*}
\hat{Y}(z) = \frac{\alpha z^{-1}}{1 + \alpha z^{-1}} \\
\end{align*}

Or, we could use the differentiation property:

\begin{nicebox}
\begin{align*}
x(n) &\ztp \hat{X}(z) \\
n x(n) &\ztp -z \frac{d}{dz} \hat{X}(z) \\
\end{align*}
\end{nicebox}

\begin{align*}
\hat{X} &= \log(1 + \alpha z^{-1}) \\
\frac{d}{dz}\hat{X} &= \frac{1}{1 + \alpha z^{-1}} \lr{-\alpha z^{-2}}  \\
-z \frac{d}{dz}\hat{X} &= \frac{\alpha z^{-1}}{1 + \alpha z^{-1}} \\
\end{align*}

We arrive at the same answer.


IV) Complex Integration (outside the scope of this course)


$$
x(n) = \frac{1}{2\pi i} \oint_c \hat{X}(z)z^{n-1}dz \ztp \hat{X}(z) = \sum \limits_{n\in\Z} x(n) z^{-n}
$$

Where $c$ is the closed contour within the $RoC$.

\end{example}


Inverse Transforms