Bilinear Transform

The bilinear transform is a method for designing digital filters from analog models. More precisely, it brings filters from the Laplace domain to the z-domain:

$$ H(s) \xrightarrow[\text{Transform}]{\text{Bilinear}} H(z) $$

In contrast to the impulse invariance method (that operates in the time domain), the bilinear transform is performed in the frequency domain. To do this, the continuous-time s-plane is morphed to the disrete-time z-plane by bending the frequency axis to become the unit circle:

To get the bilinear transform for a given $H(s)$, we need to substitute $s$ with:

$$ \boxed{ s = \frac{2}{T} \frac{1-z^{-1}}{1+z^{-1}} } $$

with the sampling period $T$:

$$ T=\frac{1}{f_s} $$

Passive RC Lowpass Example¶

A simple example for the bilinear transform can be shown for the first order RC lowpass:

Transfer Function¶

The resulting filter has a low pass characteristic with the following frequency response:

In the Laplace domain its transfer function is defined as (see Laplace Section):

$$ H(s) = \frac{\omega_c}{\omega_c + s} = \frac{1}{1 + \frac{s}{\omega_c}} $$

$s$ represents the Laplace operator:

$$ s = \sigma + j \omega $$

The cutoff frequency $\omega_c$ of the passive filter is:

$$ \omega_c = \frac{1}{RC} $$

Substitution¶

For a transform to the z-domain, the Laplace operator $s$ is subsstituted by the $z$ operator:

$$ s = \frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}} $$

The following steps rearrange the resulting transfer function to be comprised of two polynomials with $z^{-n}$:

$$ \begin{align} % % H(z) = & \frac{\omega_c}{\omega_c + \frac{2}{T} \frac{1-z^{-1}}{1+z^{-1}}} \\ % = & \frac{\frac{T}{2} \omega_c}{ \frac{T}{2} \omega_c + \frac{1-z^{-1}}{1+z^{-1}}} \\ % = & \frac{\frac{T}{2} \omega_c (1+ z^{-1})}{ \frac{T}{2} \omega_c (1+ z^{-1}) + 1-z^{-1}} \\ % = & \frac{\frac{T}{2} \omega_c + \frac{T}{2} \omega_c z^{-1}}{ \frac{T}{2} \omega_c (1+ z^{-1}) + 1-z^{-1}} \\ % = & \frac{\frac{T}{2} \omega_c + \frac{T}{2} \omega_c z^{-1}}{ \frac{T}{2} \omega_c + \frac{T}{2} \omega_c z^{-1} + 1-z^{-1}} \\ % = & \frac{\frac{T}{2} \omega_c + \frac{T}{2} \omega_c z^{-1}}{1 + \frac{T}{2} \omega_c + (\frac{T}{2} \omega_c -1 )z^{-1}} \\ % \end{align} $$

For getting the coefficients of our digital filter, the transfer function needs to meet the following structure with $a_0=1$:

$$ H(z) = \frac{b_0 + b_1 z^{-1}}{1+ a_1 z^{-1}} $$

This is achieved by expanding and simplifying:

$$ \begin{align} % % H(z) = & \frac{\frac{T}{2} \omega_c + \frac{T}{2} \omega_c z^{-1}}{ \underbrace{ \mathbf{\color{gray} 1 + \frac{T}{2} \omega_c }}_{\text{must be 1}} + \left(\frac{T}{2} \omega_c -1 \right)z^{-1}} \\ % = & \frac{\frac{\frac{T}{2} \omega_c}{ 1 + \frac{T}{2} \omega_c} + \frac{ \frac{T}{2} \omega_c}{ 1 + \frac{T}{2} \omega_c} z^{-1}}{1+ \left(\frac{\frac{T}{2} \omega_c}{1 + \frac{T}{2} \omega_c} - \frac{1}{1 + \frac{T}{2} \omega_c} \right)z^{-1}} \\ % = & \frac{\frac{\frac{T}{2} \omega_c}{ 1 + \frac{T}{2} \omega_c} + \frac{ \frac{T}{2} \omega_c}{ 1 + \frac{T}{2} \omega_c} z^{-1}}{1+ \left(\frac{\frac{T}{2} \omega_c -1}{ \frac{T}{2} \omega_c + 1} \right)z^{-1}} \\ \end{align} $$

Coefficients¶

Following the above rearrangements, the coefficients can be directly extracted from the transfer funcion:

$$ H(z) = \frac{\frac{\frac{T}{2} \omega_c}{ 1 + \frac{T}{2} \omega_c} + \frac{ \frac{T}{2} \omega_c}{ 1 + \frac{T}{2} \omega_c} z^{-1}}{1+ \left(\frac{\frac{T}{2} \omega_c -1}{ \frac{T}{2} \omega_c + 1} \right)z^{-1}} = \frac{b_0 + b_1 z^{-1}}{1+ a_1 z^{-1}} $$

$$ \begin{align} \mathbf b_0 = & \frac{\frac{T}{2} \omega_c}{ 1 + \frac{T}{2} \omega_c} \\ \mathbf b_1 = & \frac{\frac{T}{2} \omega_c}{ 1 + \frac{T}{2} \omega_c} \\ \mathbf a_1 = & \frac{\frac{T}{2} \omega_c -1}{ \frac{T}{2} \omega_c + 1} \\ \end{align} $$

Filter Topology¶

The original dependence of the electrical components R and C is not important at this point. However, with $\omega_c$ = $2 \pi f_0$ we can get the exact coefficients for any cutoff frequency which can be used for the difference equation or applied in the following topology:

Frequency Distortion¶

With the continuous frequency $\omega_C$ and the discrete frequency $\omega_D$, analog and digital frequencies have the following relation:

$$ \omega_C = \frac{2}{T} \tan \left( \frac{\omega_D T}{2} \right) $$

For small values, this can be considered linear. The larger the values, the more distortion.

Pre-Warping¶

The following pre-warping is required, considering the cutoff frequency of the analog filter $\hat{\omega}_c$ and the cutoff frequency $\omega_0$ of the digital filter:

$$\begin{eqnarray} jω_a &=& \frac{2}{T} \frac{1−e^{−jω_dT}}{1+e^{−jω_dT}} \\ &=& \frac{2}{T} \frac{1−e^{−jω_dT}}{1+e^{−jω_dT}} \\ &=& \frac{2}{T} \frac{e^{-jωd T/2}(e^{jωd T/2} − e^{−jωd T/2})}{e^{−jωd T/2}(e^{2jωd T/2} + e^{−jωd T/2})}\\ &=& \frac{2}{T} \cos\left( ω_d\frac{T}{2} \right) j \sin\left( ω_d\frac{T}{2} \right)\\ &=& j \frac{2}{T} \tan \left( ω_d\frac{T}{2} \right) \end{eqnarray}$$

With pre-warping, any frequency - for example $\omega_c$ from the RC low-pass example - can be corrected with the following equation:

$$ \hat{\omega}_c = \frac{2}{T} \tan \left( \frac{\omega_c T}{2} \right) $$

The bilinear transform results in:

$$ s = \frac{\omega_c}{\tan \left( \frac{\omega_c T}{2} \right)} \frac{1-z^{-1}}{1+z^{-1}} $$

Exercise¶

By swapping R and C we can turn our first order lowpass into a highpass filter: