0 votes 0 votes If $G(f)$ represents the Fourier transform of a signal $g$ (t) which is real and odd symmetric in time, then $\mathrm{G}(f)$ is complex $G(f)$ is imaginary $\mathrm{G}(f)$ is real $\mathrm{G}(f)$ is real and non-negative. Others gate1992-ec + – admin asked Sep 13, 2022 • edited Dec 1, 2023 by makhdoom ghaya admin 46.4k points 61 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.