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Introduction to Digital Filters
    Recursive Digital Filter Design
      
             Butterworth Lowpass Poles and Zeros

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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 [64]:

$\displaystyle G_a^2(\omega_a) = \frac{1}{1+\omega_a^{2N}} \protect$ (I.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).I.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 continuation (§D.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}} = \frac{1}{1+(-1)^Ns^{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 [64, p. 168]

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

with

$\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.

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Next: Example: Second-Order Butterworth Lowpass
written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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