The Laplace transform of a continuous-time signal x(t) is $$X\left( s \right) = {{5 - s} \over {{s^2} - s - 2}}.$$ If the Fourier transform of this signal exists, then x(t) is
A. e2tu(t) - 2e-tu(t)
B. -e2tu(-t) + 2e-tu(t)
C. -e2tu(-t) - 2e-tu(t)
D. e2tu(-t) - 2e-tu(t)
Answer: Option C

Join The Discussion