Poles and Zeros
===============

The generic form of the transfer function of a constant coefficient
LTI system is:

.. math::
   H(z) = \frac{N(z)}{D(z)} 

where $N(z)$ and $D(z)$ are polynomials in $z$. Assuming $M$-th order
polynomials we have:

.. math::
   H(z) = \frac{b_0 + b_1 z + b_2 z^2 + \cdots + b_M z^M}{
                a_0 + a_1 z + a_2 z^2 + \cdots + a_M z^M}

In the complex domain a $M$-th order polynomial has exactly $M$ zeros
and we thus may write:

.. math::
   H(z) = K \frac{(z-z_1)(z-z_2)\cdots(z-z_M)}{
   (z-p_1)(z-p_2)\cdots(z-p_M)}

where the **zeros** $z_i$'s are the zeros of $N(z)$ and the **poles** of the system, 
$p_i$'s, are the zeros of $D(z)$. 

Because a LTI system is completely characterized by its transfer
function $H(z)$, the system is also completely characterized by its
set of zeros and poles (together with a gain factor $K$). Plotting the
zeros and poles in the complex plane gives the *Argand* diagram of the
LTI system. In the Argand diagram we can also indicate the ROC of
$H(z)$.

Consider the LTI system with transfer function $H(z)$:

.. math::
   H(z) &= \frac{z^2 - 1.9 z + 1}{z^2 - 1.8 z + 0.9} \\
        &= \frac{(z - 0.95 - 0.31j)(z - 0.95 + 0.31j)}{
               (z - 0.9 - 0.3j)(z - 0.9 + 0.3 j)}

The Argand diagram (plot of poles and zeros in the complex plane) and
the frequency response $H(e^{j\Omega})$ are sketched below (see later
section on the $\mathsf z$-operator for the Python code).

.. exec_python:: discreteharmonics  DTP
   :linenumbers:
   :code: shutter
   :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.png')
   H.plot().savefig('source/figures/hfplot.png')


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

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