Butterworth Filters

Figure 1: The Butterworth low-pass filter with $N=5$.

Butterworth filters are a class of all-pole filters, where the poles of the normalized transfer function are equally spaced along the unit circle ($\omega _{0}=1\,{\text{rad/s}}$). This results in a maximally flat pass-band magnitude response, or equivalently:

$\left.{\frac {\partial ^{N}\left|H\left(j\omega \right)\right|}{\partial \omega ^{N}}}\right|_{\omega =0}=0$

(1)

This means that the $N^{\text{th}}$ derivative of the magnitude at DC is zero.

The Low-Pass Butterworth Filter

The low-pass Butterworth filter has the following magnitude response:

$\left|H\left(j\omega \right)\right|^{2}={\frac {1}{1+\left({\frac {\omega }{\omega _{0}}}\right)^{2N}}}$

(2)

Where $N$ is the filter order and $\omega _{0}$ is the $-3\,{\text{dB}}$ frequency. Note that $\left|H\left(j\omega \right)\right|^{2}=\left|H\left(s\right)\right|^{2}=H\left(s\right)\cdot H^{*}\!\left(s\right)=H\left(s\right)\cdot H\left(-s\right)$ at $s=j\omega$. Thus:

$H\left(s\right)\cdot H\left(-s\right)={\frac {1}{1+\left({\frac {-s^{2}}{\omega _{0}^{2}}}\right)^{N}}}$

(3)

Thus, the poles are the roots of:

$1+\left({\frac {-s^{2}}{\omega _{0}^{2}}}\right)^{N}=0$

(4)

Or equivalently:

${\frac {-s^{2}}{\omega _{0}^{2}}}=\left(-1\right)^{\frac {1}{N}}$

(5)

Since we can write $-1=e^{j\pi }$, the $N$ roots of $\left(-1\right)^{\frac {1}{N}}$ can be written as:

$\left(-1\right)^{\frac {1}{N}}=e^{j\pi {\frac {2k-1}{N}}}$

(6)

For $k=1\ldots N$. Thus, we get:

$-{\frac {s^{2}}{\omega _{0}^{2}}}=e^{j\pi {\frac {2k-1}{N}}}$

(7)

Solving for $s$, we get the poles of the low-pass Butterworth filter:

${\begin{aligned}s&=\omega _{0}\left(-1\cdot e^{j\pi {\frac {2k-1}{N}}}\right)^{\frac {1}{2}}\\&=\omega _{0}\left(e^{j\pi }\cdot e^{j\pi {\frac {2k-1}{N}}}\right)^{\frac {1}{2}}\\&=\omega _{0}\cdot e^{j\pi {\frac {2k+N-1}{2N}}}\\&=p_{k}\end{aligned}}$

(8)

We can then write:

$H\left(s\right)={\frac {1}{\displaystyle \prod _{k=1}^{N}{\frac {s-p_{k}}{\omega _{0}}}}}$

(9)