Editing Infinite impulse response (section)

=== Bilinear Transform === 
The bilinear transform is a special case of a conformal mapping, often used to convert a transfer function <math>H_a(s)</math> of a linear, time-invariant (LTI) filter in the continuous-time domain (often called an analog filter) to a transfer function <math>H_d(z)</math> of a linear, shift-invariant filter in the discrete-time domain.
The bilinear transform is a first-order approximation of the natural logarithm function that is an exact mapping of the ''z''-plane to the ''s''-plane.  When the Laplace transform is performed on a discrete-time signal (with each element of the discrete-time sequence attached to a correspondingly delayed unit impulse), the result is precisely the Z transform of the discrete-time sequence with the substitution of

:<math>
\begin{align}
z &= e^{sT}   \\
  &= \frac{e^{sT/2}}{e^{-sT/2}} \\
  &\approx \frac{1 + s T / 2}{1 - s T / 2}
\end{align}
</math>

where <math> T </math> is the numerical integration step size of the trapezoidal rule used in the bilinear transform derivation; or, in other words, the sampling period. The above bilinear approximation can be solved for <math> s </math> or a similar approximation for <math> s = (1/T) \ln(z) </math> can be performed.

The inverse of this mapping (and its first-order bilinear approximation) is

:<math>
\begin{align}
s &= \frac{1}{T} \ln(z)  \\
  &= \frac{2}{T} \left[\frac{z-1}{z+1} + \frac{1}{3} \left( \frac{z-1}{z+1} \right)^3  + \frac{1}{5} \left( \frac{z-1}{z+1} \right)^5  + \frac{1}{7} \left( \frac{z-1}{z+1} \right)^7 + \cdots \right] \\
  &\approx  \frac{2}{T} \frac{z - 1}{z + 1} \\
  &=  \frac{2}{T} \frac{1 - z^{-1}}{1 + z^{-1}}
\end{align}
</math>

This relationship is used in the Laplace transfer function of any analog filter or the digital infinite impulse response (IIR) filter T(z) of the analog filter.<br />
The bilinear transform essentially uses this first order approximation and substitutes into the continuous-time transfer function, <math> H_a(s) </math>

:<math>s \leftarrow \frac{2}{T} \frac{z - 1}{z + 1}.</math>

That is

:<math>H_d(z) = H_a(s) \bigg|_{s = \frac{2}{T} \frac{z - 1}{z + 1}}= H_a \left( \frac{2}{T} \frac{z-1}{z+1} \right). \ </math>
which is used to calculate the IIR digital filter, starting from the Laplace transfer function of the analog filter. <br />