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]:

$$G_a^2(\omega_a) = \frac{1}{1+\omega_a^{2N}} \protect$$ (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 continuation (§B.2) of $G_a^2(\omega_a)$ to the whole $s$-plane may be obtained by substituting $\omega_a = s/j$ to obtain

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

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 [63, p. 168]

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

with

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

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.