0 0 votes The Laplace transform of a continuous-time signal $x(t)$ is $X(s)=\frac{5-s}{s^{2}-s-2}$. If the Fourier transform of this signal exists, then $x(t)$ is $e^{2 t} u(t)-2 e^{-t} u(t)$ $-e^{2t} u(-t)+2 e^{-t} u(t)$ $-e^{2 t} u(-t)-2 e^{-t} u(t)$ $e^{2 t} u(-t)-2 e^{-t} u(t)$ Continuous-time Signals gate2002-ec continuous-time-signals laplace-transform signals-and-systems + – admin 526 views answer comment Share Follow Add Sync Questions Print 0 reply Please log in or register to add a comment.