Re: numerical inversion of laplace transform

*To*: mathgroup at smc.vnet.net*Subject*: [mg75143] Re: numerical inversion of laplace transform*From*: dantimatter <dantimatter at gmail.com>*Date*: Wed, 18 Apr 2007 05:05:58 -0400 (EDT)*References*: <evfk8c$5bk$1@smc.vnet.net><evksdj$ppp$1@smc.vnet.net>

Hi Roman et al., I don't seem to be getting the same answer that you indicated I should get in the April 12th response. Is there something about Fourier transforms I'm not quite 'getting'?? The Fourier transform of the step function p[t]=UnitStep[33.6-t] is FTp = -((I*Cos[33.6*=CF=89])/(Sqrt[2*Pi]*=CF=89)) + Sqrt[Pi/2]*DiracDelta= [=CF=89] + Sin[33.6*=CF=89]/(Sqrt[2*Pi]*=CF=89) The function that I've fit to my data is g[t]: g[t] = 6.06 - 4.17*Cos[(Pi*t)/84] + 1.19*Cos[(Pi*t)/42] - 2=2E95*Sin[(Pi*t)/84] + 0.71*Sin[(Pi*t)/42] so that the Fourier transform of g is FTg = (-439.34 - 310.621*I)*DiracDelta[Pi - 84*=CF=89] + (62.45 + 37.59*I)*DiracDelta[Pi - 42*=CF=89] + 15.20*DiracDelta[=CF=89] + (62.45 - 37.59*I)*DiracDelta[Pi + 42*=CF=89] - (439.34 - 310.62*I)*DiracDelta[Pi + 84*=CF=89] I want to invert FTg/FTp, but when I use the InverseFourierTransform routine, I get -0.29*Cos[(Pi*t)/84] + 0.24*Cos[(Pi*t)/42] - 0.38*Sin[(Pi*t)/84] - 0=2E10*Sin[(Pi*t)/42] What is it that I'm doing wrong? I don't get an answer with a constant term. Overall, your answer looks a lot better than mine. I suspect the problem comes partly from me having a DiracDelta in my expression for FTp, or - and this is more likely - I'm not using the Fourier transform properly. Can you see what I'm doing wrong? As always, your help is greatly appreciated. Dan On Apr 12, 3:59 am, "Roman" <rschm... at gmail.com> wrote: > Dan, > > Given your G(t), what I wrote can be simplified drastically. First, > transform to exponential writing using > TrigToExp[G[t]] > Then, you can manually do the transformation I wrote at 3) on the > prefactors of the exponentials. For example, the term with Exp[-i*Pi*t/ > 84], which has w=Pi/84, acquires an extra factor of -i*w/(1- > Exp[i*w*a]) = 0.0257383 - 0.0187*i. The constant term gets a prefactor > of 1/a. > > Putting everything together, you get > > F[t] = 303/(50*a) > - Cos[(Pi*t)/84]*((59*Pi)/3360 + (139*Pi*Cot[(a*Pi)/168])/5600) > + Cos[(Pi*t)/42]*((71*Pi)/8400 + (17*Pi*Cot[(a*Pi)/84])/1200) > + ((139*Pi)/5600 - (59*Pi*Cot[(a*Pi)/168])/3360)*Sin[(Pi*t)/84] > - ((17*Pi)/1200 - (71*Pi*Cot[(a*Pi)/84])/8400)*Sin[(Pi*t)/42] > > With a = 33.6 as you specify, you get > > F[t] = 0.18035714285714283 > - 0.16249350579592095*Cos[(Pi*t)/84] > + 0.04101478009014459*Cos[(Pi*t)/42] > + 0.0020508865613427935*Sin[(Pi*t)/84] > - 0.035877998487864125*Sin[(Pi*t)/42] > > Roman. > > On Apr 11, 8:12 am, "dantimatter" <dantimat... at gmail.com> wrote: > > > hi dimitris, > > > sorry about that posting. here we go again: > > > G(t)=6.06 - 4.17*Cos[(Pi*t)/84] + 1.19*Cos[(Pi*t)/42] - > > 2.95*Sin[(Pi*t)/84] + 0.71*Sin[(Pi*t)/42] > > p(t) = UnitStep[33.6 - t] > > > so that the Laplace-transformed function that I'd like to invert, G(s)/ > > p(s) is > > > G(s)/p(s) = > > (3.83*^7*E^(33.6*s)*(0.0029 - 0.049*s + s^2)*(0.0054 + 0.031*s + s^2))/ > > ((-1. + E^(33.6*s))*(Pi^2 + 1764.*s^2)*(Pi^2 + 7056.*s^2)) > > > thanks! > > dan