[music-dsp] Bessel poles?

Wen X xue.wen at elec.qmul.ac.uk
Thu Jun 11 07:12:17 EDT 2009


If the poles are different then so are the filters. If the poles are related
by constant scaling then the filters are related by linear time- (or
frequency-) stretch, i.e. they have different bandwidths yet are otherwise
the same, in particular if one is Bessel then so is the other.

-----Original Message-----
From: music-dsp-bounces at music.columbia.edu
[mailto:music-dsp-bounces at music.columbia.edu] On Behalf Of Vinnie
Sent: 10 June 2009 13:17
To: music-dsp at music.columbia.edu
Subject: [music-dsp] Bessel poles?


I've developed a Bessel low-pass analog prototype. For filter order 3 my
reverse Bessel polynomial is

  B(s) = 15 + 15*s + 6*s^2 + s^3

I'm using a root finder to find all s where B(s)=0. The root finder
definitely works, I plug the roots back into the equation and the result is
sufficiently close to zero (within 1e-15).

The resulting filter is similar to a Bessel response but my poles are off by
a constant scale factor, according to
http://www.rfcafe.com/references/electrical/bessel-poles.htm

For example, my poles for order=3 are:
  (-2.3221)
  (-1.8389,1.7544)
  (-1.8389,-1.7544)

But according to that page the poles should be
  (-1.3270)
  (-1.0509,1.0025)
  (-1.0509,-1.0025)

I know that I am close, my poles are off by a scale factor that varies with
order. In this case the factor is 1.75.

Maybe this has something to do with the numerator? Isn't the transfer
function:

 H(s) = 15 / ( 15 + 15*s + 6*s^2 + s^3 ) 

But what does that matter for finding the poles (i.e. roots of B(s))? I
don't think it matters since a root is a root regardless of what is in the
numerator.

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