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!
Best Answer
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.