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Re: numerical inversion of laplace transform
*To*: mathgroup at smc.vnet.net
*Subject*: [mg74899] Re: numerical inversion of laplace transform
*From*: "dantimatter" <dantimatter at gmail.com>
*Date*: Tue, 10 Apr 2007 05:21:02 -0400 (EDT)
*References*: <ev7j3g$l2j$1@smc.vnet.net><evd3d9$58r$1@smc.vnet.net>
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
+150]=F[t].
Thanks for all the help!
dan
On Apr 9, 5:09 am, "Roman" <rschm... at gmail.com> 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 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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