Question

The Fourier Transform of a signal x(t), denoted by $X\left(j\omega \right)$ is shown in the figure.


Let $y\left(t\right)=x\left(t\right)+e^{jt}x\left(t\right).$ The value of Fourier transform of y(t) evaluated at the angular frequency $\omega$ = 0.5 rad/s is

Answer 1

$y\left(t\right)=x\left(t\right)+e^{jt}x\left(t\right)$

$Y\left(\omega \right)=X\left(\omega \right)+X(\omega -1)$

$Y\left(\omega \right){|}_{\omega =0.5}=1.5$