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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.


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``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (September 2007 Edition).
Copyright © 2014-03-23 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
CCRMA