Re: numerical inversion of laplace transform

*To*: mathgroup at smc.vnet.net*Subject*: [mg74881] Re: numerical inversion of laplace transform*From*: "dimitris" <dimmechan at yahoo.com>*Date*: Tue, 10 Apr 2007 05:11:43 -0400 (EDT)

Try to post it again; as you see your mathematica code appeared in unreadable format. Dimitris PS Select the cells of Mathematica code (both inputs/outputs) you are interested in posting and press Ctrl+Shift+I; this will convert everything to InputForm which is preferable for posting issues. =CF/=C7 dantimatter =DD=E3=F1=E1=F8=E5: > 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 th= e > > > 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 buil= t > > > 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