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Re: numerical inversion of laplace transform

Hi Roman,

I should have added that my function is only valid for positive times,
so that:    G[t] = Integrate[F[t-x], {x,0,a}]
Substituting z=t-x as you suggest, and  G[t] = Integrate[F[z], {z,t-
a,t}], so that

G'[t] = F[t] - F[t-a]

Now is there a way to get F[t] from this??   The only other
information I have is that F is periodic with a period of 150:  F[t

Thanks for all the help!


On Apr 9, 5:09 am, "Roman" <rschm... at> wrote:
> 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> 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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