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MathGroup Archive 2007

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Re: numerical inversion of laplace transform

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74861] Re: numerical inversion of laplace transform
  • From: "Roman" <rschmied at gmail.com>
  • Date: Mon, 9 Apr 2007 06:09:11 -0400 (EDT)
  • References: <ev7j3g$l2j$1@smc.vnet.net>

Dan,
Since your p[t] is a step function, your problem can be solved
exactly.
   G[t] = Integrate[p[x]*F[t-x], {x,-Infinity,Infinity}]
Given your choice of p this becomes
   G[t] = Integrate[F[t-x], {x,-Infinity,a}]
where a=30. Substitute z=t-x, you get
   G[t] = Integrate[F[z], {z,t-a,Infinity}]
The first derivative of this is
   G'[t] = -F[t-a]
So you can find F simply from
   F[t] = -G'[t+a] = -G'[t+30]

Roman.

On Apr 7, 10:00 am, "dantimatter" <dantimat... 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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