1
$\begingroup$

I am having issues trying to evaluate a particular Fourier transform. The signal of the Gaussian pulse is:

$$ u(t) = \frac{d^2 (e^{\frac{-t^2}{2\sigma^2}}e^{j2\pi f_0 t})}{dt^2} $$

Using the definition of the continuous F.T:

$$ \mathcal{F} [u(t)] = U(f) = \int_{-\infty}^{\infty} (e^{-i2\pi ft})u(t) dt $$

I am sure that it is just a manipulation of F.T. pairs and properties. My first attempt at it was to assume that the end result would just have some parameters being multiplied by U(f) much like the property of derivatives of the F.T.:

$$ \mathcal{F} [\frac{d^2u(t)}{dt^2}] = -(2\pi f)^2U(f) $$

I greatly appreciate the help or advice!

$\endgroup$
  • $\begingroup$ Your first attempt is largely correct. Is there something you think is wrong about it? $\endgroup$ – eyeballfrog Sep 27 '17 at 19:25
  • $\begingroup$ No I do not think that there was something wrong about it. I couldn't quite think of where to go from there after making that association. $\endgroup$ – Nanners Sep 27 '17 at 19:30
  • $\begingroup$ $e^{-\pi t^2 }$ is its own Fourier transform so $U(f) = \ldots$ $\endgroup$ – reuns Sep 27 '17 at 19:39
1
$\begingroup$

So, as you correctly stated, $$ \mathcal{F}\left[\frac{d^2u}{dt^2}\right] = -(2\pi f)^2 \mathcal{F}[u(t)] = -(2\pi f)^2 \mathcal{F}\left[e^{-t^2/(2\sigma^2)}e^{2\pi if_0t}\right]. $$ The next step is to use the translation identity: $\mathcal{F}[e^{2\pi i f_0 t}u(t)](f) = \mathcal{F}[u(t)](f-f_0)$ to get $$ \mathcal{F}\left[\frac{d^2u}{dt^2}\right] = -(2\pi f)^2\mathcal{F}\left[e^{-t^2/(2\sigma^2)}\right](f-f_0) $$ Lastly, we check our handy table of Fourier Transforms to see that $\mathcal{F}[e^{-t^2/(2\sigma^2)}] = \sqrt{2\pi}\sigma e^{-(2\pi f\sigma)^2/2}$. So the result is $$ \mathcal{F}\left[\frac{d^2u}{dt^2}\right] =-\sqrt{2\pi}\sigma(2\pi f)^2 e^{-(2\pi (f-f_0)\sigma)^2/2} $$ And here's Wolfram Alpha confirming it.

$\endgroup$
  • $\begingroup$ Thanks for the help! For the most part I can follow along with the process, but I was wondering if you could elaborate a little? Primarily on the translation identity, so when you take the F.T. of that one it is essentially a frequency shift property that you must apply for the F.T. of the Gaussian that you did last. Am I understanding that correctly? That is why the final result has a (f - f0) rather than just f in the exponential $\endgroup$ – Nanners Sep 27 '17 at 20:20
  • 1
    $\begingroup$ Yes. The reason for the identity is that the complex exponential shifts the kernel of the Fourier transform: $\mathcal{F}[u(t)e^{2\pi if_0t}](f) = \int_{-\infty}^\infty [u(t) e^{2\pi i f_0 t}]e^{-2\pi i f t} dt = \int_{-\infty}^\infty [u(t)] e^{-2\pi i(f-f_0) t} dt = \mathcal{F}[u(t)](f-f_0)$. $\endgroup$ – eyeballfrog Sep 27 '17 at 20:57

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.