Re: numerical inversion of laplace transform

*To*: mathgroup at smc.vnet.net*Subject*: [mg74849] Re: [mg74827] numerical inversion of laplace transform*From*: "dan siegal-gaskins" <dantimatter at gmail.com>*Date*: Sun, 8 Apr 2007 05:04:47 -0400 (EDT)

i have that G(t) = F(t)*p(t) (where that asterisk is for convolution). the function p(t)=UnitStep[30-t]. for G(t), i've got a list of data points that i fit to a sum of sines and cosines, and i'm calling that fit G(t). it's periodic if that helps. i only decided to use the laplace transform because of the step function. maybe there's a different transform that would be more appropriate to my specific problem?? thanks! dan On 4/7/07, bsyehuda at gmail.com <bsyehuda at gmail.com> wrote: > > > > Hi Dan, > > Inversion is not so straightforward as the direct transformation. Many > > times tricks are involved. > > Therefore, it would help if you provide the expressions for the > > functions > > regards > > yehuda > > > > On 4/7/07, dantimatter <dantimatter 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 > > > > > > > > > > > >