Butterworth Lowpass Poles and Zeros Next  |  Prev  |  Up  |  Top  |  Index  |  JOS Index  |  JOS Pubs  |  JOS Home  |  Search

Butterworth Lowpass Poles and Zeros

When the maximally flat optimality criterion is applied to the general (analog) squared amplitude response $ G_a^2(\omega_a)\isdef \left\vert H_a(j\omega_a)\right\vert^2$, a surprisingly simple result is obtained [63]:

$\displaystyle G_a^2(\omega_a) = \frac{1}{1+\omega_a^{2N}} \protect$ (G.1)

where $ N$ is the desired order (number of poles). This simple result is obtained when the response is taken to be maximally flat at $ \omega_a=\infty$ as well as dc (i.e., when both $ G_a^2(\omega_a)$ and $ G_a^2(1/\omega_a)$ are maximally flat at dc).G.1Also, an arbitrary scale factor for $ \omega_a$ has been set such that the cut-off frequency (-3dB frequency) is $ \omega_c = 1$ rad/sec.

The analytic continuationB.2) of $ G_a^2(\omega_a)$ to the whole $ s$-plane may be obtained by substituting $ \omega_a = s/j$ to obtain

$\displaystyle H_a(s)H_a(-s) = \frac{1}{1+\left(\frac{s}{j}\right)^{2N}} =

The $ 2N$ poles of this expression are simply the roots of unity when $ N$ is odd, and the roots of $ -1$ when $ N$ is even. Half of these poles $ s_k$ are in the left-half $ s$-plane ( re$ \left\{s_k\right\}<0$) and thus belong to $ H_a(s)$ (which must be stable). The other half belong to $ H_a(-s)$. In summary, the poles of an $ N$th-order Butterworth lowpass prototype are located in the $ s$-plane at $ s_k = \sigma_k +
j\omega_k = e^{j\theta_k}$, where [63, p. 168]

\begin{displaymath}\begin{array}{rcrl} \sigma_k &=&-\!&\sin(\theta_k)\\ \omega_k &=&&\cos(\theta_k) \end{array} \protect\end{displaymath} (G.2)


$\displaystyle \theta_k \isdef \frac{(2k+1)\pi}{2N}

for $ k=0,1,2,\dots,N-1$. These poles may be quickly found graphically by placing $ 2N$ poles uniformly distributed around the unit circle (in the $ s$ plane, not the $ z$ plane--this is not a frequency axis) in such a way that each complex poles has a complex-conjugate counterpart.

A Butterworth lowpass filter additionally has $ N$ zeros at $ s=\infty$. Under the bilinear transform $ s = c(z-1)/(z+1)$, these all map to the point $ z = -1$, which determines the numerator of the digital filter as $ (1+z^{-1})^N$.

Given the poles and zeros of the analog prototype, it is straightforward to convert to digital form by means of the bilinear transformation.

Next  |  Prev  |  Up  |  Top  |  Index  |  JOS Index  |  JOS Pubs  |  JOS Home  |  Search

[How to cite this work] [Order a printed hardcopy]

``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (August 2006 Edition).
Copyright © 2007-02-02 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
CCRMA  [Automatic-links disclaimer]