0 votes 0 votes The Fourier transform $\mathrm{F}\left\{e^{-t} u(t)\right\}$ is equal to $\frac{1}{1+j 2 \pi f}$. Therefore, $\mathrm{F}\left\{\frac{1}{1+j 2 \pi t}\right\}$ is $e^f u(f)$ $e^{-f} u(f)$ $e^{f} u(-f)$ $e^{-f} u(-f)$ Others gate2002-ec + – admin asked Sep 27, 2022 • edited Nov 30, 2023 by makhdoom ghaya admin 46.4k points 106 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.