2
$\begingroup$

I know that the idea of the Fourier Transform is to break a function into a sum of trigonometric functions. Consider the following function: $$ f_{\alpha}(t) = e^{-\alpha|t|}$$

The Fourier Transform of this is $$\tilde{f}_{\alpha}(\omega) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{-i \omega t} e^{-\alpha |t|} \ dt $$

$$ = \frac{\alpha}{\pi(\alpha^{2}+\omega^{2})}$$

What precisely does this mean? How does did relate to the question of breaking a function into a sum of trigonometric functions?

$\endgroup$
  • 1
    $\begingroup$ A Fourier transform on the circle (or some interval) breaks a function up into trigonometric pieces. It's a little harder to interpret the Fourier transform on the line, which you have above. $\endgroup$ – Potato Jun 24 '13 at 3:18
  • 2
    $\begingroup$ It's best to think of $\mathcal{F}f(\omega)$ as being analogous to the coefficients of the Fourier series expansion. In fact, when deriving the Fourier transform from the Fourier series (when making appropriate limits), that is exactly what role it plays. So you can (VERY) loosely think of $\frac{\alpha}{\pi(\alpha^2+\omega^2)}$ as being the amplitude of the function with Fourier component $e^{i\omega t}$. But @Potato is right. It's a little harder to interpret. Even with what I said above, your indexing set is no longer integers but the real numbers so it gets to be a little opaque. $\endgroup$ – Cameron Williams Jun 24 '13 at 3:19
  • 1
    $\begingroup$ Just to add the wikipedia entry's nice animation (basically says the same thing with Cameron Williams): upload.wikimedia.org/wikipedia/commons/7/72/… $\endgroup$ – Shuhao Cao Jun 24 '13 at 3:35
  • $\begingroup$ @ShuhaoCao that is a really neat animation, I have to say. I'll be using that in the future :) $\endgroup$ – Cameron Williams Jun 24 '13 at 3:47
  • $\begingroup$ I'd suggest asking engineers that deal with DSP rather than mathematicians. The latter will undoubtedly complicate things for you on this one. $\endgroup$ – AnonSubmitter85 Jun 24 '13 at 3:48
1
$\begingroup$

I had this same question when learning about the Fourier transform on the real line. The Fourier transform on the circle (or some interval) is very clearly motivated. It breaks a function up into periodic pieces so it's easier to handle. But it's hard to see what the Fourier transform on the real line is doing.

The only thing I could find that fully answered my questions was this article by Terrance Tao. It's excellent.

$\endgroup$
1
$\begingroup$

The value of the Fourier transform at a given frequency (i.e., $\omega$) is simply the contribution of that frequency to the signal. If you consider something basic, say a sine wave (which has just a single frequency), then $\tilde{f}(\omega)$ will be zero everywhere except where $\omega$ is equal to the frequency of the sine wave, at which point $\tilde{f}(\omega)$ will have a magnitude equal to the amplitude of the wave and a phase equal to its argument at $t=0$.

The signal in your example, however, is made up of a continuum of frequencies and $\tilde{f}(\omega)$ just tells you the contribution of each.

$\endgroup$
0
$\begingroup$

I brought an extensive answer to a very similar question over here and hope that it would halp you. The function there is targeting a discrete inverse Fourier but is general enough to also cover your question that is quite generic.

How to interpret Fourier Transform result?

$\endgroup$
0
$\begingroup$

The idea of the Fourier series is that every periodic function can be decomposed into an infinite series of sines and cosines.

Fourier transform is generalization of this result.Any function $F$ (satisfying some conditions) can be written in the form

$$F(x)= \int_{- \infty}^{\infty}\bar F(t)e^{2\pi f i x}df$$

You can interpret this above integral as "decomposing" the function $F$ in terms of $e^{2 \pi f i x}$. Here $\bar F$ is the fourier transform of $f$.So, the value of $\bar F$ at $f$ gives the "contribution" of $e^{2 \pi f i x}$ in this integral. So, Fourier transform is analogous to Fourier coefficients in a Fourier series.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.