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MathGroup Archive 2004

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Re: Bilinear Transforms

  • To: mathgroup at smc.vnet.net
  • Subject: [mg50784] Re: Bilinear Transforms
  • From: Chris Williams <chrisw at tartarus.uwa.edu.au>
  • Date: Tue, 21 Sep 2004 03:48:58 -0400 (EDT)
  • References: <cijen5$hmb$1@smc.vnet.net> <cild86$r3d$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Thanks Steve,

The main problem is the Together[] part (it doesn't seem to want to 
finish any time this year) - I have had some success breaking the filter 
into smaller parts though - hopefully this will work out well.

Steve Luttrell wrote:

> Try this (not sure whether you are using z or z^(-1) but this doesn't affect 
> the basic idea below):
> 
> Build a little FIR filter
> q = Sum[a[i]/z^i,{i,10}]
> 
> Warp it.
> 
> r = q /. z -> (1 - \[Lambda]/z)/(z^(-1) - \[Lambda])
> 
> Gather together as numerator over denominator.
> 
> s = Together[r]
> 
> Get the coefficients in the numerator.
> 
> CoefficientList[Numerator[s],z]
> 
> Steve Luttrell
> 
> "Chris Williams" <chrisw at tartarus.uwa.edu.au> wrote in message 
> news:cijen5$hmb$1 at smc.vnet.net...
> 
>>Hi everyone,
>>
>>I have an warped FIR filter with order around 1400 - however to unwarp
>>this filter configuration for use on my data I need to apply the
>>bilinear transform:
>>
>>u' ->  (z^-1 - lambda)/(1-lambda*z^-1)
>>
>>So essentially each unit delay becomes an all-pass element.
>>I've substituted in the transform into a transfer function expression
>>for my filter, and now I want to get the transfer function back into the
>>normal form of a fraction of polynomials of z.
>>
>>I've tried Together[] and Simplify[] without much success. Has anyone
>>accomplished something similar - any help would be greatly appreciated ;)
>>
>>Cheers and thanks,
>>
>>Chris.
>>
> 
> 
> 


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