IIR Filters
-----------

**Infinite impulse response filters** are characterized with a
difference equation like:

.. math::
   y[n] = \sum_{k=0}^M b_k x[n-k] - \sum_{k=1}^N a_k y[n-k]

as discussed in the second chapter on LTI systems. There we have shown
that for instance a simple system characterized with

.. math::
   y[n] = x[n] + \alpha y[n-1]

corresponds with an LTI system with an infinite impulse response:

.. math::
   h[n] = (1\;\alpha\;\alpha^2\;\alpha^3\;\cdots)

and we have established that it is a stable filter in case
$|\alpha|<1$. In this section we will analyze these IIR filters in the
Z-domain.

Setting $a_0=1$ the difference equation at the top can be rewritten
as:

.. math::
   \sum_{k=0}^N a_k y[n-k] = \sum_{k=0}^M b_k x[n-k]

Applying the Z-transform to both sides (and using the shift property)
we get

.. math::
   \left(\sum_{k=0}^N a_k z^{-k} \right) Y(z) =
   \left(\sum_{k=0}^M b_k z^{-k} \right) X(z)

or

.. math::
   H(z) = \frac{Y(z)}{X(z)} = \frac{\sum_{k=0}^M b_k z^{-k}}{\sum_{k=0}^N a_k z^{-k}}


In these notes we are going to restrict ourselves to the case $M\leq
2$ and $N\leq2$. In audio processing such a digital filter is called a
**biquad filter**.

.. math::
   H(z) &= \frac{b_0 + b_1 z^{-1} + b_2 z^{-2}}{a_0 + a_1 z^{-1} + a_2 z^{-2}}\\
   &= \frac{b_0 z^2 + b_1 z + b_2}{a_0 z^2 + a_1 z + a_2}

The first expression if the starting point to represent the filter as
a block diagram. The second form is starting point for poles-zeros
analysis of the filter.

The analysis in the Z-domain is done through some examples of digital
filters. Consider the filter defined with difference equation

.. math::
   y[n] = \alpha y[n-1] + b_0 x[n]

Leading to the transfer function $H(z)$:

.. math::
   H(z) = \frac{Y(z)}{X(z)} &= \frac{b_0}{1-\alpha z^{-1}}\\
   &=\frac{b_0 z}{z-\alpha}

The pole of this filter is at $z=\alpha$ and thus the filter is stable
for $\alpha<1$.

The frequency response of the filter can be found by substituting
$e^{j\W}$ for $z$ in the transfer function:

.. math::
   H(\W) = H(z)\bigg\rvert_{z=e^{j\W}} = \frac{b_0}{1-\alpha e^{-j\W}}


.. exec_python:: lowpass  IIRfilters
   :linenumbers:
   :code: shutter
   :Code_label: Show code for figure
   :results: hide

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

   Hlow = 1/(1-0.8*z**(-1))
   Hlow.plot(freq_scale='log', rate=44100)
   plt.savefig('source/figures/lowpassresppnse.png')


.. figure:: /figures/lowpassresppnse.png

   **Lowpass Response of Filter** with transfer function $H(z) =
   \frac{1}{1-0.8 z^{-1}}$