Butterworth IIR Filter

Whew! That's the groundwork in place, we're now able to move onto something a bit more interesting and hopefully get back into code at some point too.

In the previous example we looked at the general normalised second order low-pass filter function:

$$H(s) = \frac {1}{s^2 + \frac {s}{Q} + 1}$$

When $Q = \frac {1}{\sqrt 2}$  this is known as the Butterworth function and gives a magnitude of $\frac {1}{\sqrt 2}$ at the passband frequency (in the normalised function where $\omega = 1$ ). i.e.

$$|H(\omega)|^2 = \frac {1}{(\omega^4+(\frac {1}{Q^2}-2)\omega^2+1)}$$
$$|H(j\omega)|_{\omega=1}^2 = \frac {1}{1^4 + (\sqrt 2^2 - 2)1^2 + 1}= \frac {1}{\sqrt 2}$$

The magnitude response for an analogue low-pass filter can be generalised into function of $\frac{n}{2}$ of terms of $\omega$ in the denominator where the Q term in the 2nd order case has been brought into a general purpose coefficient $D_n$.

$$|H(j\omega)|^2 = \frac {1}{D_{2n}\omega^n+...+D_2\omega^2+1}$$

Note that as this is a squared function it only contains even terms, which can be written:

$$|H(j\omega)|^2 = \frac {1}{1+D_{2n}\omega^{2n}}$$

Graphing this function shows the Butterworth response giving the maximally flat magnitude response in the passband:



This clearly shows for the prototype filter, when the passband frequency is normalised $\frac {\omega}{\omega_p}=1$ that the magnitude response is $|H(j\omega)|^2 = \frac {1}{2}$.



Translating this back from $j\omega \rightarrow s$ domains gives the following transfer function:

$$\frac {1}{1+D_{2n}\omega^{2n}}|_{\omega \rightarrow s} \rightarrow H(s)H(-s) = \frac {1}{1+(-1)^ns^{2n}}$$

Giving 2n roots (poles) of the denominator:

$$1+(-1)^ns^{2n}=0$$

$$s^{2n} = \begin{cases} 1=e^{j2k\pi} & \text{n odd} \\ -1=e^{j(2k+1)\pi} & \text{n even} \\ \end{cases}$$

Which gives us 2n poles:

$$s_k = e^{j[\frac {(2k+n-1)}{2n}]\pi} \qquad \text{   k=1,2,....,(2n)}$$

Each of the poles have unitary magnitude and are spread evenly around the left side of the s-plane. In all cases (except n=1) there are a set of matched conjugate pairs and in odd orders an additional pole at $s=-j\omega$.

The poles have the following polar angle and are positioned as follows on the s-plane where the circle has unitary magnitude:

The denominator polynomial for the transfer function can then be described as:

$$H(s) = \frac {1}{D_{B}(s)} \qquad  \text{where} \qquad D_{B}(s)=\prod^n_{k=1}(s-s_k) = 1+d_1s+d_2s^2+...+d_ns^n$$

Where $D_B(s)$ is the Butterworth polynomial which can be expressed as a set of roots ($s_k$) or expanded form which can be found in many places and has the following values expressed in factored or non-factored form.



The factored forms show sets of quadratics each representing the complex-conjugate pairs. The polynomial terms can be calculated by:

$$d_k = \frac { cos \left( (k-1) \frac {\pi}{2} \right)}{sin \left( \frac {k\pi}{2n} \right)}d_{k-1} \qquad \text {k=1,2,3...,n}$$

So, with these numbers we just need to apply the s-plane to z-plane mapping, using the Bilinear Transform as described before to calculate the appropriate filter. Well, we didn't quite get to code tonight. Let's see how tomorrow evening goes.