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MathGroup Archive 2007

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Re: numerical inversion of laplace transform

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74849] Re: [mg74827] numerical inversion of laplace transform
  • From: "dan siegal-gaskins" <dantimatter at gmail.com>
  • Date: Sun, 8 Apr 2007 05:04:47 -0400 (EDT)

i have that G(t) = F(t)*p(t) (where that asterisk is for convolution).  the
function p(t)=UnitStep[30-t]. for G(t), i've got a list of data points that
i fit to a sum of sines and cosines, and i'm calling that fit G(t).  it's
periodic if that helps.

i only decided to use the laplace transform because of the step function.
maybe there's a different transform that would be more appropriate to my
specific problem??

thanks!

dan


On 4/7/07, bsyehuda at gmail.com <bsyehuda at gmail.com> wrote:
> >
> > Hi Dan,
> > Inversion is not so straightforward as the direct transformation. Many
> > times tricks are involved.
> > Therefore, it would help if you provide the expressions for the
> > functions
> > regards
> > yehuda
> >
> > On 4/7/07, dantimatter <dantimatter at gmail.com> wrote:
> > >
> > >
> > > hello all,
> > >
> > > i have a function G(t) which is a convolution of F(t) and p(t):
> > > G=F*p .
> > > now p(t) is a step function (p(t)=UnitStep[30-t]) and i know what G(t)
> > > is, but i'd like to get at F.  what i've been trying to do is take the
> > >
> > > laplace transform of G and dividing by the laplace transform of p,
> > > then inverting that to get F. i'm having a lot of trouble.   the built
> > > in inverseLaplaceTransform function is unable to do it, and i haven't
> > > found any numerical routines that are reliable.  is there some reason
> > > why i shouldn't be able to do this?  perhaps there are better ways to
> > > get at F(t)?  your advice is much appreciated.
> > >
> > > thanks,
> > > dan
> > >
> > >
> > >
> >
>


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