Talk:Continuous Fourier transform

I think the table and the definition is not consistent -- there are some <math>2\pi<math> factors that don't tally. I won't fix it for fear of making it worse .... fiddly stuff. Lupin 15:16, 12 Jun 2004 (UTC)

It's a pain to keep things consistent through convention changes in the text, but nothing is jumping out at me right now. Which rules concern you? —Steven G. Johnson 15:48, Jun 12, 2004 (UTC)

There is at least an inconsistency with the Dirac <math>\delta<math>:

"Furthermore, the useful Dirac delta is a tempered distribution but not a function; its Fourier transform is the constant function 1."

Yet in the table the <math>\frac{1}{\sqrt{2\pi}}<math> convention is used. Eldacan 20:57, 6 Jul 2004 (UTC)

The text is inconsistent with the FT definition used here; I'll correct it. —Steven G. Johnson 04:04, Jul 7, 2004 (UTC)

Convolution Theorem

Looks like I forgot about this page for a while :) Isn't the Fourier transform of a convolution the product of the fourier transforms, without any <math>2\pi<math> factors? Unless you want your convolutions to have constant factors too... this appears to need fixage on the convolution page as well. Lupin 10:21, 7 Jul 2004 (UTC)

Whether you have constant factors in the convolution theorem depends upon your FT definition (and the convolution definition). With the FT and convolution definitions here, there are constant factors. Proof:

Let:

<math>h(t) = \int_{-\infty}^{\infty} f(\tau) g(t - \tau) d\tau<math>

<math>f(t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F(\omega) e^{i\omega t} d\omega<math>

<math>g(t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} G(\omega) e^{i\omega t} d\omega<math>

Then the Fourier transform of h is:

<math>H(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} h(t) e^{-i\omega t} dt<math>

<math>= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} dt \int_{-\infty}^{\infty} d\tau \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} d\omega' F(\omega') e^{i\omega' \tau} \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} d\omega'' G(\omega'') e^{i\omega'' (t-\tau)} e^{-i\omega t} <math>

<math>= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} d\omega' F(\omega') \int_{-\infty}^{\infty} d\omega'' G(\omega'') \left( \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{i(\omega'' - \omega)t} dt \right) \left( \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{i(\omega' - \omega'')t} d\tau \right) <math>

<math>= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} d\omega' F(\omega') \int_{-\infty}^{\infty} d\omega'' G(\omega'') \left( \sqrt{2\pi} \delta(\omega'' - \omega) \right) \left( \sqrt{2\pi} \delta(\omega' - \omega'') \right) <math>

<math>= \sqrt{2\pi} F(\omega) G(\omega)<math>

—Steven G. Johnson 17:11, Jul 7, 2004 (UTC)

Quite right, I dropped a constant in my rough calculation :-) By the way, there's a quicker proof where you don't need to use icky delta functions. The second line is obtained from the first with the change of variables <math>x'=x-t<math>.

<math>H(y)=\frac{1}{\sqrt{2\pi}}\int\int f(t)g(x-t)\,dt\, e^{-ixy}\,dx<math>

<math>=\left(\frac{1}{\sqrt{2\pi}}\int f(t)e^{-ity}\,dt\right)\left(\frac{1}{\sqrt{2\pi}}\int g(x')\ e^{-ix'y}\,dx'\right)\cdot\sqrt{2\pi}<math>

<math>=F(y)G(y)\cdot\sqrt{2\pi}.<math>

Change normalization to simplify formulas?

Hello. I wonder if there's any support for changing the normalization convention to whatever makes the formulas come out simplest. For example, so that the convolution thm becomes F(f*g) = F(f) F(g). I believe that such a convention is commonly followed. Comments? Wile E. Heresiarch 02:02, 26 Oct 2004 (UTC)

main definition

I'm wondering why there is a <math>\frac{1}{\sqrt{2\pi}}<math> in front of the transform and inverse integrals. Most definitions that I've seen are:

<math>X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-i\omega t}dt<math>

and

<math>x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} X(\omega) e^{i\omega t}d\omega<math>

Ok, thanks!

As noted in the article, the normalization is somewhat arbitrary and varies between authors; the only important thing is that the product of the two constants is <math>1/2\pi<math>. (The one here is common among physicists and mathematicians; see e.g. Mathematical Methods for Physicists by Arfken and Weber. It has the nice property that the transform is unitary without scaling, so that the forward and backwards transforms are conjugates.) —Steven G. Johnson 18:06, Nov 15, 2004 (UTC)

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I don't understand that: "The Fourier transform is close to a self-inverse mapping: if F(?) is defined as above, and f is sufficiently smooth, then..." if F(w) is defined as above - what does it refer to? ("as above") "f is sufficiently smooth" - what's the idea? f will be the result of the (inverse) transformation. How we can say before the transformation if f is sufficiently smooth? Jun 10, 2005 <btw, how to make a timestamp without registration?>

Typing four tildes (i.e, ~~~~) should do. Cburnett 05:17, Jun 11, 2005 (UTC)