Question

Fourier transform of the Gaussian pulse $x\left(t\right)=e^{-4t^2}$ is given to be $A{e}^{-(\frac{\pi f}{2}{)}^2}$ where A is _____

• A. 0.886

Answer 1

The correct option is A 0.886

$X\left(\omega \right)={\int }_{-\infty }^{\infty }x\left(t\right)e^{-j\omega t}dt$

Where, $x\left(t\right)=e^{-b^2}{t}^2(b=2)$

$X\left(\omega \right)={\int }_{-\infty }^{\infty }{e}^{-b^2{t}^2-j\omega t}dt$

Substituting $b^2{t}^2+j\omega t={(bt+\frac{j\omega }{2b})}^2+\frac{{\omega }^2}{4b^2}$

$\therefore X\left(\omega \right)={\int }_{-\infty }^{\infty }{e}^{{(bt+\frac{j\omega }{2b})}^2}{e}^{\left(\frac{-{\omega }^2}{4b^2}\right)}dt$

Substituting $u=bt+\frac{j\omega }{2b}$

$du=bdt$

$X\left(\omega \right)=e^{\frac{-{\omega }^2}{4b^2}}.\frac{2}{b}{\int }_0^{\infty }{e}^{-u^2}du$

$\because {\int }_0^{\infty }{e}^{-u^2}du=\frac{\sqrt{\pi }}{2}$

$\therefore X\left(\omega \right)=\frac{\sqrt{\pi }}{b}.e^{\frac{-{\omega }^2}{4b^2}}=\frac{\sqrt{\pi }}{b}.e^{{(-\frac{\pi f}{b})}^2}$

$\therefore A=\frac{\sqrt{\pi }}{b}=\frac{\sqrt{\pi }}{2}=0.8862=0.886$