Butterworth Filters

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: $$\begin{array}{}\text{(1)}& |H_n(j\omega )|\triangleq \frac{1}{\sqrt{1+{\omega }^{2n}}}\end{array}$$ In order to determine the transfer function, we'll start from the frequency response squared. We'll assume that the transfer function $H_n(s)$ is a rational function with real coefficients. Therefore, $\overline{H_n(s)}=H_n\left(\bar{s}\right)$. $$\begin{array}{rl}{|H_n(j\omega )|}^2=& H_n(j\omega )\overline{H_n(j\omega )}\\ =& H_n(j\omega )H_n(\overline{j\omega })\\ =& H_n(j\omega )H_n(-j\omega )\\ =& \frac{1}{1+{\omega }^{2n}}\end{array}$$ We're looking for the transfer function $H_n(s)$, so we'll use the identity $s=j\omega \iff \omega =\frac{s}{j}$. $$H_n(s)H_n(-s)=\frac{1}{1+{\left(\frac{s}{j}\right)}^{2n}}$$

Poles of $H_n(s)H_n(-s)$

The poles of this transfer function are given by: $$\begin{array}{rl}& {\left(\frac{s}{j}\right)}^{2n}=-1\\ \iff & s^{2n}=-1{\left(j\right)}^{2n}\\ \iff & s^{2n}=-1{(-1)}^n\\ \iff & s^{2n}={(-1)}^{n+1}\\ \iff & s^{2n}=e^{j\pi (n+1)}\end{array}$$ Keep in mind that this is a polynomial of order $2n$, so it has $2n$ complex roots. $$s_k=e^{j2\pi \frac{2k+n+1}{4n}}k\in \{0,1,\dots 2n-1\}$$ For example, for $n=3$, the poles are: $$\begin{array}{rl}{s}_0=& e^{j2\pi \frac{0+3+1}{12}}=e^{j2\pi \frac{2}{6}}\\ s_1=& e^{j2\pi \frac{2+3+1}{12}}=e^{j2\pi \frac{3}{6}}\\ s_2=& e^{j2\pi \frac{4+3+1}{12}}=e^{j2\pi \frac{4}{6}}\\ s_3=& e^{j2\pi \frac{6+3+1}{12}}=e^{j2\pi \frac{5}{6}}\\ s_4=& e^{j2\pi \frac{8+3+1}{12}}=e^{j2\pi \frac{6}{6}}\\ s_5=& e^{j2\pi \frac{10+3+1}{12}}=e^{j2\pi \frac{1}{6}}\end{array}$$ 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°: $$\begin{array}{rl}& 2\pi \frac{2k+n+1}{4n}\in (\frac{\pi }{2},\frac{3\pi }{2})\\ \iff & 2k+n+1\in (n,3n)\\ \iff & k\in (-\frac{1}{2},n-\frac{1}{2})\\ \Rightarrow & k\in (-\frac{1}{2},n-\frac{1}{2})\cup \{0,1,\dots 2n-1\}\\ \iff & k\in \{0,1,\dots n-1\}\end{array}$$ $$\begin{array}{}\text{(2)}& s_{k,stable}=e^{j2\pi \frac{2k+n+1}{4n}}k\in \{0,1,\dots n-1\}\end{array}$$

Poles of $H_n(s)$

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

Butterworth Polynomials

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

Odd order $n$:
In this case, $n-1$ is even, and you get a special pole for $k=\frac{n-1}{2}$: $$\begin{array}{rl}{s}_{\frac{n-1}{2}}=& e^{j2\pi \frac{2\frac{n-1}{2}+n+1}{4n}}\\ =& e^{j2\pi \frac{2n}{4n}}\\ =& e^{j\pi }\\ =& -1\end{array}$$ 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: $$\begin{array}{}\text{(4)}& B_n(s)=\{\begin{array}{ll}\prod _{k=0}^{\frac{n}{2}-1}(s^2-2\cos \left(2\pi \frac{2k+n+1}{4n}\right)s+1)& \text{even}n\\ (s+1)\prod _{k=0}^{\frac{n-1}{2}-1}(s^2-2\cos \left(2\pi \frac{2k+n+1}{4n}\right)s+1)& \text{odd}n\end{array}\end{array}$$

Butterworth Transfer Function $H_n(s)$

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

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

High-Pass Butterworth filters

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

We can just multiply the numerator and the denominator by ${\omega }^n$ to get a more familiar form: $$|H_{n,hp}(j\omega )|=\frac{{\omega }^n}{\sqrt{1+{\omega }^{2n}}}$$ 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: $$\begin{array}{}\text{(7)}& H_{n,hp}(s)=\frac{s^n}{B_n(s)}\end{array}$$

Non-normalized Butterworth Filters

Up until now, we only looked at normalized Butterworth filters, that have a corner frequency of $1\mathrm{rad}/s$. To get a specific corner frequency ${\omega }_c$, we can just scale $\omega$, so the definitions become: $$\begin{array}{}\text{(8)}& |H_{n,lp}(j\omega )|\triangleq \frac{1}{\sqrt{1+{\left(\frac{\omega }{{\omega }_c}\right)}^{2n}}}\end{array}$$ $$\begin{array}{}\text{(9)}& |H_{n,hp}(j\omega )|\triangleq \frac{1}{\sqrt{1+{\left(\frac{{\omega }_c}{\omega }\right)}^{2n}}}\end{array}$$ 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 $\left|s_k\right|={\omega }_c$.
This results in the following transfer functions: $$\begin{array}{}\text{(10)}& H_{n,lp}(s)=\frac{1}{B_n\left(\frac{s}{{\omega }_c}\right)}\end{array}$$ $$\begin{array}{}\text{(11)}& H_{n,hp}(s)=\frac{s^n}{{\omega }_c^n{B}_n\left(\frac{s}{{\omega }_c}\right)}\end{array}$$

The gain at the corner frequency can easily be determined from the definitions: $$\begin{array}{rl}|H_{n,lp}(j{\omega }_c)|=|H_{n,hp}(j{\omega }_c)|=& \frac{1}{\sqrt{1+{\left(\frac{{\omega }_c}{{\omega }_c}\right)}^{2n}}}\\ =& \frac{1}{\sqrt{2}}\\ =& \frac{\sqrt{2}}{2}\\ \approx & 0.707\\ 20{\log }_{10}|H_n(j{\omega }_c)|=& 20{\log }_{10}\left(\frac{\sqrt{2}}{2}\right)\\ =& 10{\log }_{10}\left(\frac{1}{2}\right)\\ \approx & -3.01dB\end{array}$$ This is often called the $-3dB$-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: ${|H_n(j{\omega }_c)|}^2=\frac{1}{2}$.