Finite and Infinite Signals
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$\newcommand{\op}[1]{\mathsf #1}$
$\newcommand{\ztarrow}{\stackrel{\quad\op Z\quad}{\longrightarrow}}$

Consider a finite signal (different from zero in a finite number of samples):

.. math::
   x[n] = \begin{matrix}2& \underline{3}& 5& 3& 1 \end{matrix}

where we use the convention that the origin ($n=0$) is denoted with
the underlining and signal values that are not given are equal to
zero. The Z-transform equals:

.. math::
   X(z) = 2 z^1 + 3 + 5 z^{-1} + 3 z^{-2} + z^{-3}

with ROC the entire complex plane without the origin. Now consider an
infinite signal

.. math::
   x[n] = \begin{matrix}\underline{1}& 0.8& 0.8^2& 0.8^3& 0.8^4& \cdots\end{matrix}

with Z-transform:

.. math::
   X(z) = \sum_{n=0}^{\infty} 0.8^n z^{-n} = \sum_{n=0}^{\infty} \left(0.8 z^{-1}\right)^n

Remember your geometric sequences (meetkundige reeks) from math class?
Without proof we state:

.. math::
   X(z) = \frac{1}{1-0.8 z^{-1}},\quad |z|>0.8

the ROC is $|z|>0.8$. The ROC follows from the observation that the
geometric sequence only converges in case $|0.8 z^{-1}|<1$. Note that
not all infinite signals result in a finite Z-transform (in fact, most
infinite signals do not have bounded Z-transform).

The second example can be summarized as:

.. math::
   0.8^n u[n] \ZTright \frac{1}{1-0.8 z^{-1}},\quad |z|>0.8.