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Re: numerical inversion of laplace transform

• To: mathgroup at smc.vnet.net
• Subject: [mg74877] Re: numerical inversion of laplace transform
• From: "dantimatter" <dantimatter at gmail.com>
• Date: Mon, 9 Apr 2007 06:17:27 -0400 (EDT)
• References: <ev7j3g$l2j$1@smc.vnet.net><evabti$dg5$1@smc.vnet.net>

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 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