We can summarize the Z-Transform as follows:

\begin{nicebox}
\begin{align*}
x(n) &\ztp \hat{X}(z) = \sum \limits_{n=-\infty}^{\infty} x(n) z^{-n} \\
RoC(x) &= \{ z \in \C \st \sum \limits_{n=-\infty}^{\infty} \abs{x(n) z^{-n}} \lt \infty \}
\end{align*}
\end{nicebox}

Since the convention used is $z = Re^{i\omega}$, we can also write the region of convergence as

\begin{nicebox}
\begin{align*}
RoC(x) &= \{ z \in \C \st \sum \limits_{n=-\infty}^{\infty} \abs{x(n) R^{-n}} \lt \infty \}
\end{align*}
\end{nicebox}

This is because only $R \bydef \abs{z}$ is important in defining the region of convergence. The math is also very clear:

\begin{align*}
\abs{x(n)z^{-n}} = \abs{x(n)R^{-n}e^{-i\omega n}} = \abs{x(n)R^{-n}}
\end{align*}

\begin{example}

$x(n) = \alpha^n u(n)$ is causal. Determine the Z-Transform.

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

We find the the Z-Transform is $\frac{1}{1 - \alpha z^{-1} }$ with a radius of convergence $\abs{\alpha} \lt \abs{z}$.
%\img{images/roc/roc/causal.ps}

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\end{example}

\begin{example}

Given $y(n) = -\alpha^n u(-n-1)$, determine the Z-Transform.

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

We find the the Z-Transform is $\frac{1}{1 - \alpha z^{-1} }$ with a radius of convergence $\abs{\alpha} \gt \abs{z}$.

%\img{images/roc/roc/anticausal.ps}

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\end{example}


\begin{claim}

If $x$ is causal ($x(n) = 0 \mbox{ } \forall n  \lt0$), and $R \in RoC(x)$, then $\tilde{R} \gt R \in Roc(x)$. In other words, causal signals have $RoC$'s that extend from outside a circle all of the way to infinity. 

\end{claim}

\begin{proof}

First determine a relationship between $\tilde{R}^{-n}$ and $R^{-n}$:
\begin{align*}
\tilde{R} &\gt R \\
\tilde{R}^{-1} &\lt R^{-1} \\
\tilde{R}^{-n} &\lt R^{-n} \\
\end{align*}

We need to consider the absolute summability of the tranfer function. $\sum \limits_{n=-\infty}^{\infty} \abs{x(n) z^{-n}} = \sum \limits_{n=0}^{\infty} \abs{x(n) R^{-n}}$

If sum converges absolutely for $R$, then certainly when $R^{-n}$ is replaced with a smaller term $\tilde{R}^{-n}$, the sum will still converge. Therefore, if $R \in RoC(x)$, then $\tilde{R} \in Roc(x)$.

\end{proof}

\begin{example}

Given our signal from a few examples ago, $x(n) = \alpha^n u(n)$, find the Z-transform of $g(n) = x(n+1) = \alpha^{n+1} u(n+1)$. This example brings out one of the properties of the Z-transform which you will use frequently, which is the shifting property. Lets solve this for an arbitrary shift.

\begin{align*}
\hat{G}(z) &= \sum \limits_{n=-\infty}^{\infty} g(n) z^{-n} \\
\hat{G}(z) &= \sum \limits_{n=-\infty}^{\infty} x(n-N) z^{-n} \\
\hat{G}(z) &= \sum \limits_{n=-\infty}^{\infty} x(m) z^{-\lr{m+N}} \quad \mbox{ let } m = n-N \\
\hat{G}(z) &= z^{-N}\sum \limits_{n=-\infty}^{\infty} x(m) z^-m = z^{-N}\hat{X}(z)\\
\hat{G}(z) &= z^{-N}\hat{X}(z)\\
\end{align*}

This defines the shifting property for the Z-transform:

\begin{nicebox}
\begin{align*}
x(n) &\ztp \hat{X}(z) \\
x(n-N) &\ztp z^{-N}\hat{X}(z) \\
\end{align*}
\end{nicebox}

In our case we have $N = -1$, which means that $\hat{G}(z) = z\hat{X}(z) = \frac{z}{1-\alpha z^{-1}}$, where $\abs{\alpha z^{-1}} \lt 1$. What happens at infinity? Lets write out $\hat{G}(z)$ term by term:

$$
\hat{G}(z) = z + \alpha + \alpha^2 z^{-1} + \alpha^3 z^{-2} + \cdots 
$$

Notice that as $z \to \infty$, it blows up. This demonstrates how right-sided signals that are not causal have $RoC$s of the form $R \lt \abs{z} \lt \infty$. An analogous statement can be made about left-sided signals that are not anti-causal, they are of the form $0 \lt \abs{z} \lt R$. The first term that blows up these types of signals is $x(1)z^{-1}$.

\end{example}

Region of Convergence