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MathGroup Archive 2007

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Re: numerical inversion of laplace transform

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74881] Re: numerical inversion of laplace transform
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Tue, 10 Apr 2007 05:11:43 -0400 (EDT)

Try to post it again; as you see your mathematica code appeared
in unreadable format.

Dimitris

PS

Select the cells of Mathematica code (both inputs/outputs) you are
interested in
posting and press Ctrl+Shift+I; this will convert everything to
InputForm
which is preferable for posting issues.

=CF/=C7 dantimatter =DD=E3=F1=E1=F8=E5:
> G(t) ==.06 - 4.17 Cos[\(\[Pi]\ x\)\/84] + 1.18 Cos[\(\[Pi]\ x\)\/42] -
> 2==2E95 Sin[\(\[Pi]\ x\)\/84] + 0.71 Sin[\(\[Pi]\ x\)\/42]
> p(t)==UnitStep[30-t]
>
> the function i'd like to invert is then:
>
> G(s)/p(s)==  6.06/s - 0.11/(==CF==80^2/7056 + s^2) - (4.17s)/(==CF==80^2/=
7056 ==
> + s^2)
> + 0.05/(==CF==80^2/1764 + s^2) + (1.18s)/(==CF==80^2/1764 + s^2)
>
> thanks,
> dan
>
> On Apr 8, 4:16 am, "dimitris" <dimmec... at yahoo.com> wrote:
> > No helpful assistance can be given if you don't show us your involved
> > functions to be inverted.
> >
> > Dimitris
> >
> > ==CF/==C7 dantimatter ==DD==E3==F1==E1==F8==E5:
> >
> > > 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 th=
e
> > > 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 buil=
t
> > > 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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