Pulse and Shift
===============
$\newcommand{\op}[1]{\mathsf #1}$
$\newcommand{\ztarrow}{\stackrel{\op Z}{\longrightarrow}}$

Let's start with a simple signal $x[n]=\delta[n]$. Then we have:

.. math::
   X(z) &= \sum_{n=-\infty}^{\infty} x[n] z^{-n} \\
        &= \sum_{n=-\infty}^{\infty} \delta[n] z^{-n} \\
	&= z^0 = 1

The last step uses the sifting property of the delta pulse. A pulse in
the time domain corresponds with a constant value 1 in the Z-domain:

.. math::
   \delta[n]\ztarrow 1


In case we take a shifted pulse $\delta[n-n_0]$ we get:

.. math::
   X(z) = \sum_{n=\infty}^{\infty} x[n] z^{-n}
        = \sum_{n=\infty}^{\infty} \delta[n-n_0] z^{-n}
	= z^{-n_0}

i.e.:

.. math::
   \delta[n-n_o]\ztarrow z^{-n_0}
   
Shifting over $n_0$ in the time domain thus corresponds with
multiplication with $z^{-n_0}$ in the Z-domain. We have seen that in
the above equation for the pulse but we can prove it for any signal
$x[n]$.

Consider the shifted signal $x[n-n_0]$, then by definition the
Z-transform of the shifted signal is:

.. math::
   \sum_{n=\infty}^{\infty} x[n-n_0] z^{-n}

changing from variable $m=n-n_0$ we get

.. math::
   \sum_{m=\infty}^{\infty} x[m] z^{-(m+n_0)}
   = z^{-n_0} \sum_{m=\infty}^{\infty} x[m] z^{-m)}
   = z^{-n_0} X(z)

where $X(z)$ is the Z-transform of the original signal $x[n]$.

Summarizing, given the \Zt{} pair:

.. math::
   x[n]\stackrel{\op Z}{\longrightarrow}X(z)

we have that:

.. math::
   x[n-n_0]\stackrel{\op Z}{\longrightarrow}z^{-n_0}X(z)