Editing Bilinear transform (section)

== Transformation for a general first-order continuous-time filter ==
It is possible to relate the coefficients of a continuous-time, analog filter with those of a similar discrete-time digital filter created through the bilinear transform process. Transforming a general, first-order continuous-time filter with the given transfer function

:<math>H_a(s) = \frac{b_0 s + b_1}{a_0 s + a_1} = \frac{b_0 + b_1 s^{-1}}{a_0 + a_1 s^{-1}}</math>

using the bilinear transform (without prewarping any frequency specification) requires the substitution of

:<math>s \leftarrow K \frac{1 - z^{-1}}{1 + z^{-1}}</math>

where

:<math>K \triangleq \frac{2}{T} </math>.

However, if the frequency warping compensation as described below is used in the bilinear transform, so that both analog and digital filter gain and phase agree at frequency <math>\omega_0</math>, then

:<math>K \triangleq \frac{\omega_0}{\tan\left(\frac{\omega_0 T}{2}\right)} </math>.

This results in a discrete-time digital filter with coefficients expressed in terms of the coefficients of the original continuous time filter:
:<math>H_d(z)=\frac{(b_0 K + b_1) + (-b_0 K + b_1)z^{-1}}{(a_0 K + a_1) + (-a_0 K + a_1)z^{-1}}</math>

Normally the constant term in the denominator must be normalized to 1 before deriving the corresponding [[difference equation]].  This results in

:<math>H_d(z)=\frac{\frac{b_0 K + b_1}{a_0 K + a_1} + \frac{-b_0 K + b_1}{a_0 K + a_1}z^{-1}}{1 + \frac{-a_0 K + a_1}{a_0 K + a_1}z^{-1}}. </math>

The difference equation (using the [[Digital filter#Direct form I|Direct form I]]) is

:<math>
y[n] = \frac{b_0 K + b_1}{a_0 K + a_1} \cdot x[n] + \frac{-b_0 K + b_1}{a_0 K + a_1} \cdot x[n-1] - \frac{-a_0 K + a_1}{a_0 K + a_1} \cdot y[n-1] \ .
</math>