GATE 2023 EC – Question 28

Networks, Signals and Systems · Continuous-time Signals · 1 mark · Multiple choice

The Fourier transform $X(\omega)$ of $x(t)=e^{-t^2}$ is

Note: $\int_{-\infty}^{\infty}e^{-y^2}dy=\sqrt\pi$

$\sqrt\pi\,e^{\frac{\omega^2}{2}}$
$\dfrac{e^{-\frac{\omega^2}{4}}}{2\sqrt\pi}$
$\sqrt\pi\,e^{-\frac{\omega^2}{4}}$
$\sqrt\pi\,e^{-\frac{\omega^2}{2}}$

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Correct answer: (C) $\sqrt\pi\,e^{-\frac{\omega^2}{4}}$

Explanation

$X(\omega)=\int e^{-t^2}e^{-j\omega t}dt$. Completing the square, $-t^2-j\omega t=-(t+j\omega/2)^2-\omega^2/4$, so $X(\omega)=e^{-\omega^2/4}\int e^{-y^2}dy=\sqrt\pi\,e^{-\omega^2/4}$.