# Finding the Fourier Transform of Second Derivative of a Gaussian Signal

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!

• Your first attempt is largely correct. Is there something you think is wrong about it? – eyeballfrog Sep 27 '17 at 19:25
• 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. – Nanners Sep 27 '17 at 19:30
• $e^{-\pi t^2 }$ is its own Fourier transform so $U(f) = \ldots$ – reuns Sep 27 '17 at 19:39

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.
• Yes. The reason for the identity is that the complex exponential shifts the kernel of the Fourier transform: $\mathcal{F}[u(t)e^{2\pi if_0t}](f) = \int_{-\infty}^\infty [u(t) e^{2\pi i f_0 t}]e^{-2\pi i f t} dt = \int_{-\infty}^\infty [u(t)] e^{-2\pi i(f-f_0) t} dt = \mathcal{F}[u(t)](f-f_0)$. – eyeballfrog Sep 27 '17 at 20:57