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