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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


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