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MathGroup Archive 2007

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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




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