Question

Let the signal x(t) have the Fourier transform $X\left(\omega \right)$.
Consider the signal, $y\left(t\right)=\frac{d}{dt}\left[x\right(t-t_d\left)\right]$

(where $t_d$ is a arbitrary delay.)
The magnitude of the Fourier transform of y(t) is given by the expression

• A. $|\omega ||X\left(\omega \right)|e^{-j\omega t_d}$
• B. ${\omega }^2\cdot |X\left(\omega \right)|$
• C. $X\left(\omega \right)|\cdot \omega$
• D. $|X\left(\omega \right)||\omega |$

Answer 1

The correct option is D

$|X\left(\omega \right)||\omega |$

$x\left(t\right)⟷X\left(\omega \right)$

$x(t-t_d)⟷e^{-j\omega t_d}X\left(\omega \right)$

$\frac{d\left[x\right(t-t_d\left)\right]}{dt}⟷j\omega e^{-j\omega t_d}X\left(\omega \right)$

$\frac{d\left[x\right(t-t_d\left)\right]}{dt}⟷|j\omega |X\left(\omega \right)|\left|e^{-j\omega t_d}\right|$

$=|\omega ||X\left(\omega \right)|=|X\left(\omega \right)||\omega |$

Hence correct option is (a).