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RE: What is the difference equation of a Butterworth filter?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg27568] RE: What is the difference equation of a Butterworth filter?
  • From: bruce.detterich at ieee.org (Bruce Detterich)
  • Date: Wed, 7 Mar 2001 04:08:09 -0500 (EST)
  • Organization: The Math Forum
  • References: <972ge7$4kr@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

You'll have to do a little leg work, but here's the basic approach:

1. Find the desired order of Butterworth filter and frequency scale it
in the S-domain.  As a hint to make your life easy, look for
"Butterworth Standard Forms".

2. Once you have the S-domain expression, make the "bilateral"
Z-domain substitution for S.  I believe the substitution is something
like s=(z-1)/(z+1).

3. Simplify the result to a Z-domain fraction (polynomials in Z in the
numerator and denominator), from which you can go directly to the
difference equation.

Mathematica should make short work of this.  The whole thing should be
covered in any good introductory text on sampled data signal
processing.

BEWARE: Some tables of Butterworth Standard Forms have incorrect
values for orders above the 3rd.  This problem persisted in many texts
for many years (case of authors not checking their sources), so you're
best off if you find a text that describes the Butterworth derivation
procedure (it's easy), and capture it in Mathematica.


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