2
$\begingroup$

When I calculate the Fourier transform of the function $f(t) = \mathrm e^{-|t|/\tau} \text{ with } \tau >0$ in Mathematica once via the function FourierTransform and once by hand, I get different results. And with "by hand" I mean letting Mathematica calculate the integral $ft_2(\omega) = \int_\mathbb{R} f(t) \mathrm e^{-2 \pi \mathrm i \omega t} \, \mathrm dt$

My input is:

f[t_] = Exp[-Abs[t]/\[Tau]];  ft1[\[Omega]_] = FourierTransform[f[t], t, \[Omega]]  ft2[\[Omega]_] =   Integrate[f[t]*Exp[-2*Pi*I*\[Omega]*t], {t, -Infinity, Infinity},   Assumptions -> {{\[Omega], \[Tau]} \[Element] Reals, \[Tau] > 0}]  \[Omega] = 0.123; \[Tau] = 0.456; ft1[\[Omega]] ft2[\[Omega]] 

And the generated output is:

$ft_1(\omega) = \sqrt{\frac{2}{\pi}} \cdot \frac{\tau}{1+\tau^2\omega^2}$ $ft_2(\omega) = \frac{2 \tau}{(-\mathrm i + 2 \pi \tau \omega)(\mathrm i + 2 \pi \tau \omega)} = \frac{2 \tau}{1 + 4 \pi^2 \omega^2 \tau^2}$ and

0.362694 0.811248+ 0. I 

As you can see, we get different values for $\omega = 0.123$ and $\tau = 0.456$. I am most certain that there is some error with the integral, as the result from FourierTransform can also be obtained from rule 207 in the List from Wikipedia.

  • 1
    You do know that `FourierTransform[]` has an unconventional normalization, don't you? Did you check the `FourierParameters` option?2011-12-03

1 Answers 1

4

Does FourierTransform[Exp[-Abs[t]/\[Tau]], t, \[Omega], FourierParameters -> {0, -2 Pi}] work for you? It does for me...

  • 0
    I did have a look through the documentation but wasn't sure.. will take another look through it!2013-04-02