0
$\begingroup$

We know that one of the important Fourier transform properties is that, the Fourier transform of a narrow function has a broad spectrum, and vice versa,

We can easily see this in this example, the Fourier transform $F(k)$ of the Delta function $\delta (x-x_0)$ which is a narrow function, has a broad spectrum (sinusoidal function), i.e. $$F(k) = \int_{\infty}^{\infty}\delta (x-x_0)e^{-2\pi i kx}dx = e^{-2\pi i kx_0}$$

Is there a more general way to prove that a Fourier transform of a narrow (or Broad) function has a Broad (or Narrow) spectrum?

I'ma physics student and I think the Uncertainty principle (position-momentum) in some ways proves the above by taking standard deviations, where position and momentum eigen function turn out to be Fourier pairs

$\endgroup$
  • $\begingroup$ en.wikipedia.org/wiki/Fourier_transform#Uncertainty_principle ? $\endgroup$ – Batman Apr 8 '16 at 11:02
  • $\begingroup$ Thanks @Batman that seems to be what I'm looking for, but I'll still keep the quick open if anyone else has a different approach $\endgroup$ – Oswald Apr 8 '16 at 11:28
  • $\begingroup$ I would like to know the specific definitions of "broad" and "narrow" here. $\endgroup$ – user247327 May 16 '17 at 14:06
0
$\begingroup$

You could argue that any narrow (finite) function is just the product of a periodic version of this function with a $\mathrm{rect}$ function. In the Fourier domain, this becomes a convolution of the transform of the original function (which is narrow) with an $\mathrm{si}$ function, thus broadening the original transform. The narrower the original function, the broader the $\mathrm{si}$ function becomes, the broader the transform of the orignal function.

$\endgroup$

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.