Fourier Transform of a Generic Gaussian

The following computation was done out of an exercise from ELEC 221 Signals and Systems. The method does require some trickery to compute, but once it's done, we have a stunning result: the Fourier Transform of a Gaussian is another Gaussian!

Result: The Fourier Transform of the Gaussian \(f(t) = ae^{-bt^2}\) defined on \(-\infty < t < \infty\) for some constants \(a, b \in \mathbb{R}\) is \[\mathcal{F}[ ae^{-bt^2}] = a \sqrt{\frac{\pi}{b}}e^{-\omega^2/4b}\]

where we have used the following Fourier Transform definition:

\[ \mathcal{F}[f(t)] = \int_{-\infty}^{\infty} f(t)e^{-j\omega t}\,dt\]

Proof: Let \(f(t)\) and \(\mathcal{F}\) be as stated. Then we compute:

\[ \begin{aligned} \mathcal{F}[ ae^{-bt^2}] &= \int_{-\infty}^{\infty} ae^{-bt^2}e^{-j\omega t}\,dt\\ &= a\int_{-\infty}^{\infty} e^{-bt^2}e^{-j\omega t}\,dt\\ &= a\int_{-\infty}^{\infty} \exp[-(bt^2 + j\omega t)]\,dt\\ &= a\int_{-\infty}^{\infty} \exp\left[-b\left(t^2 + \frac{j\omega t}{b}\right)\right]\,dt\\ &= a\int_{-\infty}^{\infty} \exp\left[-b\left(t^2 + \frac{j\omega t}{b} + \left(\frac{j\omega}{2b}\right)^2 - \left(\frac{j\omega}{2b}\right)^2 \right)\right]\,dt\\ &= a\int_{-\infty}^{\infty} \exp\left[-b\left(\left(t + \frac{j\omega}{2b}\right)^2 + \left(\frac{j\omega}{2b}\right)^2 \right)\right]\,dt\\ &= a\int_{-\infty}^{\infty} \exp\left[-b\left(t + \frac{j\omega}{2b}\right)^2 - \frac{\omega^2}{4b}\right]\,dt\\ &= a\exp\left[- \frac{\omega^2}{4b}\right] \int_{-\infty}^{\infty} \exp\left[-b\left(t + \frac{j\omega}{2b}\right)^2\right]\,dt \end{aligned} \]

Substitute \(x = t + \frac{j\omega}{2b} \Rightarrow dx = dt\).

\[ \begin{aligned} &= a\exp\left[- \frac{\omega^2}{4b}\right] \int_{-\infty}^{\infty} \exp\left[-bx^2\right]\,dx\\ &= a\exp\left[- \frac{\omega^2}{4b}\right] I \end{aligned} \]

To simplify, we will focus on the definite integral:

\[ \begin{aligned} I &= \int_{-\infty}^{\infty} e^{-bx^2}\,dx\\ I^2 &= \int_{-\infty}^{\infty} e^{-bx^2}\,dx \int_{-\infty}^{\infty} e^{-bx^2}\,dx\\ \end{aligned} \]

Since the two integrals are independent of each other, we will do a change of variables for one of them

\[ \begin{aligned} I^2 &= \int_{-\infty}^{\infty} e^{-bx^2}\,dx \int_{-\infty}^{\infty} e^{-by^2}\,dy\\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-b(x^2 + y^2)}\,dx\,dy\\ &= \int_{0}^{\infty}\int_{0}^{2\pi} e^{-br^2}r\,dr\,d\theta \end{aligned} \]

Substitute \(u = r^2 \Rightarrow du = 2r\,dr\). Bounds are the same. We also evaluate the integral with respect to \(\theta\).

\[ \begin{aligned} I^2 &= 2\pi\int_{0}^{\infty}\frac{e^{-bu}}{2}\,du = \pi\int_{0}^{\infty}e^{-bu}\,du\\ &= \pi \left[\frac{e^{-bu}}{-b}\right]_{0}^{\infty}\\ &= \frac{\pi}{b} \left(0 + 1\right) = \frac{\pi}{b} \end{aligned} \]

Therefore \(\displaystyle I = \sqrt{\frac{\pi}{b}}\). Collecting all our work, we arrive at the final result: \[ \mathcal{F}[ ae^{-bt^2}] = a \sqrt{\frac{\pi}{b}}e^{-\omega^2/4b} \qquad \blacksquare \]

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