Question

Consider a signal defined by

$x\left(t\right)=\{\begin{array}{c}{e}^{j10t}\text{for}|t|\le 1\\ 0\text{for}|t|>1\end{array}$

Its Fourier Transform is

Answer 1

Since, $x\left(t\right)=\{\begin{array}{cc}{e}^{j10t}& \text{for}|t|\le 1\\ 0& \text{for}|t|>1\end{array}$

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

So, $X\left(\omega \right)={\int }_{-1}^1{e}^{j10t}{e}^{-j\omega t}dt={\int }_{-1}^1{e}^{jt(10-\omega )}dt$

$=\frac{1}{j(01-\omega )}[e^{j(10-\omega )}-e^{-j(10-\omega )}]$

$X\left(\omega \right)=\left\{\frac{2\sin (\omega -10)}{(\omega -10)}\right\}$