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Consider a signal defined by

$x(t)= \begin{cases} e^{j10 t} & \text{for } \mid t \mid \leq 1 \\ 0& \text{for } \mid t \mid > 1\end{cases}$

Its Fourier Transform is

- $\dfrac{2 \sin (\omega -10)}{\omega - 10}\\$
- $2e^{j10}\dfrac{ \sin (\omega -10)}{\omega - 10} \\$
- $\dfrac{2 \sin \omega}{\omega - 10} \\$
- $e^{j10 \omega }\dfrac{2 \sin\omega }{\omega }$