Question

A signal x(t) having Fourier transform X$\left(\omega \right)$ is defined by $X\left(\omega \right)$ = $\frac{1}{(2+j\omega {)}^2}$, another signal y(t) having Fourier transform $Y\left(\omega \right)$ is given by $y\left(t\right)=\frac{16}{t}x\left(t\right).$ Then value of Y(2) is_____

• A. 5.65

Answer 1

The correct option is A 5.65
$x\left(t\right)⟷X\left(\omega \right)$

$x\left(t\right)=t{e}^{-2t}u\left(t\right)$

$y\left(t\right)=\frac{16}{t}\times t{e}^{-2t}u\left(t\right)=16e^{-2t}u\left(t\right)$

$Y\left(\omega \right)=\frac{16}{2+j\omega }$

$|Y\left(2\right)|=\left|\frac{16}{2+j2}\right|=\frac{16}{\sqrt{2^2+2^2}}=5.65$