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. >> > > >