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The Laplace transform of a continuous-time signal x(t) is,
X(s) = 5 - s s² - s - 2
If the Fourier transform of this signal exists, then x(t) is
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- e2t u(t) – 2e– t u(t)
- – e2t u(– t) + 2e– t u(t)
- – e2t u(– t) – 2e– tu(t)
- none of these
Correct Option: A
X(s) = | = | + | |||
(s + 1)(s - 2) | (s + 1) | (s - 2) |
A = | ![]() | = - 2; and B | ![]() | = 1 | ||||
s - 1 | s=-1 | s + 1 | s=2 |
then, X(t) = L– 1 X(s) = L– 1 | ![]() | + | ![]() | |||
s + 1 | s - 2 |
= [e2t – 2e– t]u(t)