This page will cover the derivation of the transfer functions of low-pass and high-pass Butterworth filters. Butterworth filters are designed to have a very flat frequency response in the passband.

### Definition

Normalized Butterworth filters are defined in the frequency domain as follows: $\begin{array}{}\text{(1)}& |{H}_{n}\left(j\omega \right)|\triangleq \frac{1}{\sqrt{1+{\omega }^{2n}}}\end{array}$ In order to determine the transfer function, we'll start from the frequency response squared. We'll assume that the transfer function ${H}_{n}\left(s\right)$ is a rational function with real coefficients. Therefore, $\stackrel{―}{{H}_{n}\left(s\right)}={H}_{n}\left(\stackrel{―}{s}\right)$. $\begin{array}{rl}{|{H}_{n}\left(j\omega \right)|}^{2}\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& {H}_{n}\left(j\omega \right)\stackrel{―}{{H}_{n}\left(j\omega \right)}\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& {H}_{n}\left(j\omega \right){H}_{n}\left(\stackrel{―}{j\omega }\right)\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& {H}_{n}\left(j\omega \right){H}_{n}\left(-j\omega \right)\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& \frac{1}{1+{\omega }^{2n}}\end{array}$ We're looking for the transfer function ${H}_{n}\left(s\right)$, so we'll use the identity $s=j\omega ⇔\omega =\frac{s}{j}$. ${H}_{n}\left(s\right){H}_{n}\left(-s\right)=\frac{1}{1+{\left(\frac{s}{j}\right)}^{2n}}$

### Poles of ${H}_{n}\left(s\right){H}_{n}\left(-s\right)$

The poles of this transfer function are given by: $\begin{array}{rl}& {\left(\frac{s}{j}\right)}^{2n}=-1\\ ⇔\phantom{\rule{1em}{0ex}}& {s}^{2n}=-1{\left(j\right)}^{2n}\\ ⇔\phantom{\rule{1em}{0ex}}& {s}^{2n}=-1{\left(-1\right)}^{n}\\ ⇔\phantom{\rule{1em}{0ex}}& {s}^{2n}={\left(-1\right)}^{n+1}\\ ⇔\phantom{\rule{1em}{0ex}}& {s}^{2n}={e}^{j\pi \left(n+1\right)}\end{array}$ Keep in mind that this is a polynomial of order $2n$, so it has $2n$ complex roots. ${s}_{k}={e}^{j2\pi \frac{2k+n+1}{4n}}\phantom{\rule{1em}{0ex}}k\in \left\{0,1,\dots 2n-1\right\}$ For example, for $n=3$, the poles are: $\begin{array}{rl}{s}_{0}\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& {e}^{j2\pi \frac{0+3+1}{12}}\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}{e}^{j2\pi \frac{2}{6}}\\ {s}_{1}\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& {e}^{j2\pi \frac{2+3+1}{12}}\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}{e}^{j2\pi \frac{3}{6}}\\ {s}_{2}\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& {e}^{j2\pi \frac{4+3+1}{12}}\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}{e}^{j2\pi \frac{4}{6}}\\ {s}_{3}\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& {e}^{j2\pi \frac{6+3+1}{12}}\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}{e}^{j2\pi \frac{5}{6}}\\ {s}_{4}\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& {e}^{j2\pi \frac{8+3+1}{12}}\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}{e}^{j2\pi \frac{6}{6}}\\ {s}_{5}\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& {e}^{j2\pi \frac{10+3+1}{12}}\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}{e}^{j2\pi \frac{1}{6}}\end{array}$ These are all points on the unit circle, $\pi /3=60°$ apart. The poles are stable if they are in the left half plane, if their complex argument is between 90° and 270°: $\begin{array}{rl}& 2\pi \frac{2k+n+1}{4n}\in \left(\frac{\pi }{2},\frac{3\pi }{2}\right)\\ ⇔\phantom{\rule{1em}{0ex}}& 2k+n+1\in \left(n,3n\right)\\ ⇔\phantom{\rule{1em}{0ex}}& k\in \left(-\frac{1}{2},n-\frac{1}{2}\right)\\ ⇒\phantom{\rule{1em}{0ex}}& k\in \left(-\frac{1}{2},n-\frac{1}{2}\right)\cup \left\{0,1,\dots 2n-1\right\}\\ ⇔\phantom{\rule{1em}{0ex}}& k\in \left\{0,1,\dots n-1\right\}\end{array}$ $\begin{array}{}\text{(2)}& {s}_{k,stable}={e}^{j2\pi \frac{2k+n+1}{4n}}\phantom{\rule{1em}{0ex}}k\in \left\{0,1,\dots n-1\right\}\end{array}$

### Poles of ${H}_{n}\left(s\right)$

We want our filter ${H}_{n}\left(s\right)$ to be stable, so we pick the poles in the left half plane to be the poles of ${H}_{n}\left(s\right)$. The unstable poles, for $k\in \left\{n,n+1,\dots 2n-1\right\}$ are the poles of ${H}_{n}\left(-s\right)$. They are the opposites of the poles of ${H}_{n}\left(s\right)$: $\begin{array}{rlr}{s}_{k,unstable}\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& {e}^{j2\pi \frac{2k+n+1}{4n}}\phantom{\rule{1em}{0ex}}& k\in \left\{n,n+1,\dots 2n-1\right\}\\ & l\triangleq k-n\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& {e}^{j2\pi \frac{2\left(l+n\right)+n+1}{4n}}\phantom{\rule{1em}{0ex}}& l\in \left\{0,1,\dots n-1\right\}\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& {e}^{j\left(2\pi \frac{2l+n+1}{4n}+\pi \right)}\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& {e}^{j\pi }\cdot {e}^{j2\pi \frac{2l+n+1}{4n}}\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& -1\cdot {e}^{j2\pi \frac{2l+n+1}{4n}}\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& -{s}_{l,stable}\end{array}$

### Butterworth Polynomials

We'll define the normalized Butterworth polynomial as follows: $\begin{array}{}\text{(3)}& {B}_{n}\left(s\right)\triangleq \prod _{k=0}^{n-1}\left(s-{e}^{j2\pi \frac{2k+n+1}{4n}}\right)\end{array}$ We'll rearrange the product to group each pole with its complex conjugate. Then, using the identity ${e}^{j\theta }+{e}^{-j\theta }=2\mathrm{cos}\theta$, we can further simplify this expression:
Even order $n$: $\begin{array}{rl}{B}_{n}\left(s\right)\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& \prod _{k=0}^{n-1}\left(s-{e}^{j2\pi \frac{2k+n+1}{4n}}\right)\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& \prod _{k=0}^{\frac{n}{2}-1}\left(s-{e}^{j2\pi \frac{2k+n+1}{4n}}\right)\prod _{l=\frac{n}{2}}^{n-1}\left(s-{e}^{j2\pi \frac{2l+n+1}{4n}}\right)\\ & l=n-k-1\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& \prod _{k=0}^{\frac{n}{2}-1}\left(s-{e}^{j2\pi \frac{2k+n+1}{4n}}\right)\left(s-{e}^{j2\pi \frac{2\left(n-k-1\right)+n+1}{4n}}\right)\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& \prod _{k=0}^{\frac{n}{2}-1}\left(s-{e}^{j2\pi \frac{2k+n+1}{4n}}\right)\left(s-\frac{{e}^{j2\pi \frac{-2k+3n-1}{4n}}}{1}\right)\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& \prod _{k=0}^{\frac{n}{2}-1}\left(s-{e}^{j2\pi \frac{2k+n+1}{4n}}\right)\left(s-\frac{{e}^{j2\pi \frac{-2k+3n-1}{4n}}}{{e}^{j2\pi }}\right)\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& \prod _{k=0}^{\frac{n}{2}-1}\left(s-{e}^{j2\pi \frac{2k+n+1}{4n}}\right)\left(s-{e}^{j2\pi \left(\frac{-2k+3n-1}{4n}-1\right)}\right)\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& \prod _{k=0}^{\frac{n}{2}-1}\left(s-{e}^{j2\pi \frac{2k+n+1}{4n}}\right)\left(s-{e}^{j2\pi \frac{-2k+3n-4n-1}{4n}}\right)\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& \prod _{k=0}^{\frac{n}{2}-1}\left(s-{e}^{j2\pi \frac{2k+n+1}{4n}}\right)\left(s-{e}^{j2\pi \frac{-2k-n-1}{4n}}\right)\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& \prod _{k=0}^{\frac{n}{2}-1}\left(s-{e}^{j2\pi \frac{2k+n+1}{4n}}\right)\left(s-{e}^{-j2\pi \frac{2k+n+1}{4n}}\right)\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& \prod _{k=0}^{\frac{n}{2}-1}\left({s}^{2}-s{e}^{j2\pi \frac{2k+n+1}{4n}}-s{e}^{-j2\pi \frac{2k+n+1}{4n}}+1\right)\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& \prod _{k=0}^{\frac{n}{2}-1}\left({s}^{2}-2\mathrm{cos}\left(2\pi \frac{2k+n+1}{4n}\right)s+1\right)\end{array}$

Odd order $n$:
In this case, $n-1$ is even, and you get a special pole for $k=\frac{n-1}{2}$: $\begin{array}{rl}{s}_{\frac{n-1}{2}}\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& {e}^{j2\pi \frac{2\frac{n-1}{2}+n+1}{4n}}\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& {e}^{j2\pi \frac{2n}{4n}}\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& {e}^{j\pi }\\ \phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}& -1\end{array}$ After isolating this pole, we're left with an even number of complex conjugate poles, just like in the case where $n$ was even.

In conclusion, the normalized Butterworth polynomial of degree $n$ is given by:

### Butterworth Transfer Function ${H}_{n}\left(s\right)$

The transfer function ${H}_{n}\left(s\right)$ has no zeros, so the numerator is a constant. The poles of ${H}_{n}\left(s\right)$ are given by Equation $\text{(2)}$, so the denominator is given by Equation $\text{(3)}$. ${H}_{n}\left(s\right)=\frac{c}{{B}_{n}\left(s\right)}$

We wanted a DC gain of $1$ $\left(=0dB\right)$ for $\omega =0$: $\begin{array}{rl}& |{H}_{n}\left(0j\right)|\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}1\\ ⇔\phantom{\rule{1em}{0ex}}& |\frac{c}{{B}_{n}\left(0\right)}|\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}1\\ ⇔\phantom{\rule{1em}{0ex}}& |\frac{c}{\prod _{k=0}^{n-1}\left(0-{e}^{j2\pi \frac{2k+n+1}{4n}}\right)}|\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}1\\ ⇔\phantom{\rule{1em}{0ex}}& \frac{|c|}{\prod _{k=0}^{n-1}|-{e}^{j2\pi \frac{2k+n+1}{4n}}|}\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}1\\ ⇔\phantom{\rule{1em}{0ex}}& \frac{|c|}{1}\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}1\end{array}$ If we want no phase offset for low frequencies, we can postulate that $\mathrm{\angle }{H}_{n}\left(0j\right)=0$: $\begin{array}{rl}& \mathrm{\angle }{H}_{n}\left(0j\right)\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}0\\ ⇔\phantom{\rule{1em}{0ex}}& \mathrm{\angle }\left(\frac{c}{{B}_{n}\left(0\right)}\right)\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}0\\ ⇔\phantom{\rule{1em}{0ex}}& \mathrm{\angle }c-\mathrm{\angle }\left(\prod _{k=0}^{\frac{n}{2}-1}\left({0}^{2}-2\mathrm{cos}\left(2\pi \frac{2k+n+1}{4n}\right)\cdot 0+1\right)\right)\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}0\\ ⇔\phantom{\rule{1em}{0ex}}& \mathrm{\angle }c-\mathrm{\angle }1\phantom{\rule{0.278em}{0ex}}=\phantom{\rule{0.278em}{0ex}}0\end{array}$ The derivation is analogous for odd $n$.
Therefore, $c=1$, and we've eliminated all unknown parameters from the transfer function: $\begin{array}{}\text{(5)}& {H}_{n}\left(s\right)=\frac{1}{{B}_{n}\left(s\right)}\end{array}$

### High-Pass Butterworth filters

Up until now, we only looked at the low-pass Butterworth filter. There's also a high-pass version: $\begin{array}{}\text{(6)}& |{H}_{n,hp}\left(j\omega \right)|\triangleq \frac{1}{\sqrt{1+{\omega }^{-2n}}}\end{array}$

We can just multiply the numerator and the denominator by ${\omega }^{n}$ to get a more familiar form: $|{H}_{n,hp}\left(j\omega \right)|=\frac{{\omega }^{n}}{\sqrt{1+{\omega }^{2n}}}$ As you can see, the poles will be the same as for the low-pass version. On top of that, there now are $n$ zeros for $s=0$.
So the transfer function becomes: $\begin{array}{}\text{(7)}& {H}_{n,hp}\left(s\right)=\frac{{s}^{n}}{{B}_{n}\left(s\right)}\end{array}$

### Non-normalized Butterworth Filters

Up until now, we only looked at normalized Butterworth filters, that have a corner frequency of . To get a specific corner frequency ${\omega }_{c}$, we can just scale $\omega$, so the definitions become: $\begin{array}{}\text{(8)}& |{H}_{n,lp}\left(j\omega \right)|\triangleq \frac{1}{\sqrt{1+{\left(\frac{\omega }{{\omega }_{c}}\right)}^{2n}}}\end{array}$ $\begin{array}{}\text{(9)}& |{H}_{n,hp}\left(j\omega \right)|\triangleq \frac{1}{\sqrt{1+{\left(\frac{{\omega }_{c}}{\omega }\right)}^{2n}}}\end{array}$ If you start recalculating the transfer functions, you'll quickly realize that this just scales everything by a factor of ${\omega }_{c}$. The poles no longer lie on the unit circle, but on a circle with radius $|{s}_{k}|={\omega }_{c}$.
This results in the following transfer functions: $\begin{array}{}\text{(10)}& {H}_{n,lp}\left(s\right)=\frac{1}{{B}_{n}\left(\frac{s}{{\omega }_{c}}\right)}\end{array}$ $\begin{array}{}\text{(11)}& {H}_{n,hp}\left(s\right)=\frac{{s}^{n}}{{\omega }_{c}^{n}\phantom{\rule{0.278em}{0ex}}{B}_{n}\left(\frac{s}{{\omega }_{c}}\right)}\end{array}$

The gain at the corner frequency can easily be determined from the definitions: This is often called the -point or the half-power point, because a sinusoidal input signal at that frequency will result in an output signal that has only half of the power of the input signal: ${|{H}_{n}\left(j{\omega }_{c}\right)|}^{2}=\frac{1}{2}$.