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