0 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 $e^f u(f)$ $e^{-f} u(f)$ $e^{f} u(-f)$ $e^{-f} u(-f)$ Continuous-time Signals gate2002-ec continuous-time-signals fourier-transform signals-and-systems + – admin 318 views answer comment Share Follow Add Sync Questions Print 0 reply Please log in or register to add a comment.