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MathGroup Archive 2004

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Re: Bilinear Transforms-> Möbius transforms

  • To: mathgroup at smc.vnet.net
  • Subject: [mg50807] Re: Bilinear Transforms-> Möbius transforms
  • From: Roger Bagula <tftn at earthlink.net>
  • Date: Wed, 22 Sep 2004 00:11:10 -0400 (EDT)
  • References: <cijen5$hmb$1@smc.vnet.net>
  • Reply-to: tftn at earthlink.net
  • Sender: owner-wri-mathgroup at wolfram.com

It appear to be Fuchsian with determinant in lamda squared.
I usually similfy these by hand:
u'=(z-1/Lamda)/(z/Lamda+1)=z/(z/Lamda+1)-(1/Lamda)/(z/Lamda+1)
I think that's what you want.
These are called Möbius transforms and
are associated with the Poincare
nonEuclidean plane.
Fuchsian groups are a big well studied area.
A tip is to multiply by (z/z) and (-1/Lamda)/(-1/Lamda)
to get a form where it is:
u'=(a*z+b)/(c*z+b)
By bilinear they mean that the 2by2:
M={{a,b},{c,d}}
multiplies when the the functions are taken like f(g(z)).
These matrices are also associated as well with Klein groups and 
Teichmuller space
in Mathematics. 
Chris Williams wrote:

>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.
>
>  
>

-- 
Respectfully, Roger L. Bagula
tftn at earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
URL :  http://home.earthlink.net/~tftn
URL :  http://victorian.fortunecity.com/carmelita/435/ 



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