Question

The Fourier transform of the below function is

• A. $\omega \sin \omega$
• B. $\frac{2\sin \omega }{\omega }$
• C. $\frac{\omega }{\sin \omega }$
• D. $\frac{\cos \omega }{2\omega }$

Answer 1

The correct option is B $\frac{2\sin \omega }{\omega }$

Given function is $f\left(t\right)=\{\begin{array}{c}1,1\le t\le 1\\ 0,otherwise\end{array}$
F{f(t)} = ${\int }_{-\infty }^{\infty }{e}^{-j\omega t}f\left(t\right)dt$
(Standard Result of Fourier Transform)
$={\int }_{-1}^1{e}^{-j\omega t}\left(1\right)dt$
$={\left[\frac{e^{-j\omega t}}{-j\omega }\right]}_{-1}^1$
$=\frac{1}{j\omega }[e^{j\omega }-e^{-j\omega }]$
$=\frac{2j\sin \omega }{j\omega }$
$=\frac{2\sin \omega }{\omega }$