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Butterworth Lowpass Design

Almost all methods for filter design are optimal in some sense, and the choice of optimality determines nature of the design. Butterworth filters are optimal in the sense of having a maximally flat amplitude response, as measured using a Taylor series expansion about dc [64, p. 162]. Of course, the trivial filter $ H(z)=1$ has a perfectly flat amplitude response, but that's an allpass, not a lowpass filter. Therefore, to constrain the optimization to the space of lowpass filters, we need constraints on the design, such as $ H(1)=1$ and $ H(-1)=0$. That is, we may require the dc gain to be 1, and the gain at half the sampling rate to be 0.

It turns out Butterworth filters (as well as Chebyshev and Elliptic Function filter types) are much easier to design as analog filters which are then converted to digital filters. This means carrying out the design over the $ s$ plane instead of the $ z$ plane, where the $ s$ plane is the complex plane over which analog filter transfer functions are defined. The analog transfer function $ H_a(s)$ is very much like the digital transfer function $ H(z)$, except that it is interpreted relative to the analog frequency axis $ s=j\omega_a$ (the ``$ j\omega$ axis'') instead of the digital frequency axis $ z=e^{j\omega_d T}$ (the ``unit circle''). In particular, analog filter poles are stable if and only if they are all in the left-half of the $ s$ plane, i.e., their real parts are negative. An introduction to Laplace transforms is given in Appendix D, and an introduction to converting analog transfer functions to digital transfer functions using the bilinear transform appears in §I.3.

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 continuationD.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 [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)


$\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 pole 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.

Example: Second-Order Butterworth Lowpass

In the second-order case, we have, for the analog prototype,

$\displaystyle H_a(s) = \frac{1}{(s + a)(s + \overline{a})}

where, from Eq.$ \,$(I.2), $ a = e^{j\pi/4}$, so that

$\displaystyle H_a(s) = \frac{1}{(s + e^{j\pi/4})(s + e^{-j\pi/4})} = \frac{1}{s^2 + \sqrt{2}s + 1} \protect$ (I.3)

To convert this to digital form, we apply the bilinear transform

$\displaystyle s = c\frac{1-z^{-1}}{1+z^{-1}}

(from Eq.$ \,$(I.9)), where, as discussed in §I.3, we set

$\displaystyle c = \cot(\omega_cT/2) \isdef \frac{\cos(\omega_cT/2)}{\sin(\omega_cT/2)}

to obtain a digital cut-off frequency at $ \omega_c$ radians per second. For example, choosing $ \omega_c T = \pi/2$ (a cut off at one-fourth the sampling rate), we get

$\displaystyle c = \frac{\cos(\pi/4)}{\sin(\pi/4)} = 1

and the digital filter transfer function is
$\displaystyle H_d(z)$ $\displaystyle =$ $\displaystyle H_a\left(\frac{1-z^{-1}}{1+z^{-1}}\right) =
\frac{1}{\left(\frac{1-z^{-1}}{1+z^{-1}}\right)^2 + \sqrt{2}\left(\frac{1-z^{-1}}{1+z^{-1}}\right) + 1}$ (I.4)
  $\displaystyle =$ $\displaystyle \frac{(1+z^{-1})^2}{(1-2z^{-1}+z^{-2}) + (\sqrt{2} - \sqrt{2}z^{-2}) + (1+2z^{-1}+z^{-2})}$ (I.5)
  $\displaystyle =$ $\displaystyle \frac{(1+z^{-1})^2}{(2+\sqrt{2}) + (2-\sqrt{2})z^{-2}}$ (I.6)
  $\displaystyle =$ $\displaystyle \frac{1}{2+\sqrt{2}}\frac{(1+z^{-1})^2}{1 + \frac{2-\sqrt{2}}{2+\sqrt{2}}z^{-2}}$ (I.7)

Note that the numerator is $ (1+z^{-1})^2$, as predicted earlier. As a check, we can verify that the dc gain is 1:

$\displaystyle H_d(1) = \frac{2^2}{2+\sqrt{2} + 2-\sqrt{2}} = 1

It is also immediately verified that $ H_d(-1) = 0$, i.e., that there is a (double) notch at half the sampling rate.

In the analog prototype, the cut-off frequency is $ \omega_a=1$ rad/sec, where, from Eq.$ \,$(I.1), the amplitude response is $ G_a(j)=1/\sqrt{2}$. Since we mapped the cut-off frequency precisely under the bilinear transform, we expect the digital filter to have precisely this gain. The digital frequency response at one-fourth the sampling rate is

$\displaystyle H_d(j) = \frac{(1-j)^2}{2+\sqrt{2} - (2-\sqrt{2})} = -\frac{j}{\sqrt{2}}, \protect$ (I.8)

and $ 20\log_{10}(\left\vert H_d(j)\right\vert)=-3$ dB as expected.

Note from Eq.$ \,$(I.8) that the phase at cut-off is exactly -90 degrees in the digital filter. This can be verified against the pole-zero diagram in the $ z$ plane, which has two zeros at $ z = -1$, each contributing +45 degrees, and two poles at $ z=\pm
j\sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}}$, each contributing -90 degrees. Thus, the calculated phase-response at the cut-off frequency agrees with what we expect from the digital pole-zero diagram.

In the $ s$ plane, it is not as easy to use the pole-zero diagram to calculate the phase at $ \omega_a=1$, but using Eq.$ \,$(I.3), we quickly obtain

$\displaystyle H_a(j\cdot 1) = \frac{1}{j^2 + \sqrt{2}j + 1} = -\frac{j}{\sqrt{2}},

and exact agreement with $ H_d(e^{j\pi/2})$ [Eq.$ \,$(I.8)] is verified.

A related example appears in §9.2.4.

Next Section:
Digitizing Analog Filters with the Bilinear Transformation
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Lowpass Filter Design