The Z-Operator
==============
$\newcommand{\op}[1]{\mathsf #1}$
$\newcommand{\ztarrow}{\stackrel{\op Z}{\longrightarrow}}$
$\newcommand{\z}{\op z}$


The Z-transform takes a sequence $x[n]$ and returns a function
$X(z)$. In the z-domain we have seen that shifting a signal $x[n]$
over $n_0$, resulting in $x[n-n_0]$, leads to the Z-transform
$z^{-n_0}X(z)$ where $X(z)$ is the transform of the original signal.

This observation has lead to the introduction of the $\z$-operator
that takes a discrete signal $x$ and shifts it:

.. math::
   (\z x)[n] = x[n-1]

With this algebraic notation of a shift, together with multiplications
and summations this is all we need to characterize /all/ linear
translation invariant filters.

Note that shifting over 3 samples corresponds with the operator $\z \z
\z = \z^3$ or in general:

.. math::
   (\z^{m} x)[n] = x[n-m]

Evidently the operator $\z^{-1}$ corresponds with a shift in the
opposite direction and thus the above equation may be interpreted for
both positive and negative (integer) values of $m$. 

A simple non-causal running average filter over 5 samples can be
defined as the operator:

.. math::
   \op f = \tfrac{1}{5}\left( \z^2 + \z^1 + 1 + \z^{-1} + \z^{-2}\right)

Using this filter to process a signal $x$ is denoted as $\op f x$ or
$\op f(x)$ where we assume $x$ is a discrete signal.

A recursive band stop filter is defined with:

.. math::
   y[n] = 1.8 y[n-1] - 0.9 y[n-2] + x[n] - 1.9 x[n-1] + x[n-2]


Using the $\z$ operator we may write:

.. math::
   y[n] = 1.8 (\z^{-1} y)[n] - 0.9 (\z^{-2}y)[n] + x[n] - 1.9 (\z^{-1}x)[n] +
   (\z^{-2}x)[n]

because now all signals are indexed at $n$ we may leave the index
out to obtain the expression that is valid for all $n$:

.. math::
   y = 1.8 \z^{-1} y - 0.9 \z^{-2}y + x - 1.9 \z^{-1}x + \z^{-2}x
   
or equivalently:

.. math::
   y = \frac{1 - 1.9 \z^{-1} + \z^{-2}}{1 - 1.8 \z^{-1} + 0.9 z^{-2}} x
   = \frac{\z^2-1.9\z+1}{\z^2-1.8\z+0.9} x

This band stop filter thus can be formally defined as:

.. math::
   \op f = \frac{\z^2-1.9\z+1}{\z^2-1.8\z+0.9}

and of course in the definition of the filter $\op f$ you recognize
the Z-transform of the impulse response of the filter characterized
with the Z-transform:

.. math::
   H(z) = \frac{z^2-1.9z+1}{z^2-1.8z+0.9}

Please note however that $H(z)$ is *not* the filter $\op f$. In the
expression for $H(z)$, $z$ is a complex number whereas in the
expression for $\op f$, $\z$ is an operator. The operator notation for
$\op f$ is a symbolic notation. It can be implemented in practice
using the difference equation we started with. 

Using the $\z$ operator to define discrete LTI filters is nowadays
often used. In fact even software is written in which the $\z$
operator notation plays a central role. Below we give some examples
using the 'audiolazy' Python package.

.. exec_python:: discreteharmonics  DTP
   :linenumbers:
   :code: show
   :Code_label: Show code for figures
   :results: hide

   import numpy as np
   import matplotlib.pyplot as plt
   from audiolazy import z

   H = (z**2-1.9*z+1)/(z**2-1.8*z+0.9);
   H.zplot().savefig('source/figures/hzplot.pdf');
   H.plot().savefig('source/figures/hfplot.pdf');


.. image:: /figures/hzplot.png
   :width: 40%

.. image:: /figures/hfplot.png
   :width: 58%


Note that in the code given here the function `savefig()` is used to
save the figure in a file. In an interactive session you can use
`show()` instead.

Although conceptually nice, audiolazy still lacks soms of the standard
filters. A lot of these standard filters are available in
scipy.signal. In the lecture notes on filter design some examples of
the use of scipy will be given.

.. todo:: look at the status of audiolazy