# Bilinear Transform: RC Lowpass

### Passive RC Lowpass¶

A simple example for the bilinear transform is the 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:

$$ H(s) = \frac{\omega_c}{\omega_c + s} = \frac{1}{1 + \frac{s}{\omega_c}} $$$s$ represents the Laplace operator

$$ s = \sigma + j \omega $$and the cutoff frequency 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:

$$ s = \frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}} $$$$ z = e^{sT} $$The following steps rearrange the resulting transfer function to be comprised of summands with the factor $z^{-n}$:

$$ \begin{align} % % H(z) = & \frac{\omega_c}{\omega_c + \frac{2}{T} \frac{1-z^{-1}}{1+z^{-1}}} \\ % = & \frac{\frac{T}{2} \omega_c}{ \frac{T}{2} \omega_c + \frac{1-z^{-1}}{1+z^{-1}}} \\ % = & \frac{\frac{T}{2} \omega_c (1+ z^{-1})}{ \frac{T}{2} \omega_c (1+ z^{-1}) + 1-z^{-1}} \\ % = & \frac{\frac{T}{2} \omega_c + \frac{T}{2} \omega_c z^{-1}}{ \frac{T}{2} \omega_c (1+ z^{-1}) + 1-z^{-1}} \\ % = & \frac{\frac{T}{2} \omega_c + \frac{T}{2} \omega_c z^{-1}}{ \frac{T}{2} \omega_c + \frac{T}{2} \omega_c z^{-1} + 1-z^{-1}} \\ % = & \frac{\frac{T}{2} \omega_c + \frac{T}{2} \omega_c z^{-1}}{1 + \frac{T}{2} \omega_c + (\frac{T}{2} \omega_c -1 )z^{-1}} \\ % \end{align} $$For getting the coefficients of our digital filter, the transfer function needs to meet the following strucutre where $a_0=1$:

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