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Introduction to Digital Filters
    Recursive Digital Filter Design

             Example: Second-Order Butterworth Lowpass

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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

, 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 plane, which has two zeros at , 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 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.

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Previous: Butterworth Lowpass Poles and Zeros
Next: Digitizing Analog Filters with the Bilinear Transformation

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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