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Signals and systems electrical engineering miscellaneous

  1. 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
    1. e2t u(t) – 2e– t u(t)
    2. – e2t u(– t) + 2e– t u(t)
    3. – e2t u(– t) – 2e– tu(t)
    4. none of these
Correct Option: A

X(s) =
5 - s
=
A
+
B
(s + 1)(s - 2)(s + 1)(s - 2)

A =
5 - s
= - 2; and B
5 - s
= 1
s - 1s=-1s + 1s=2

then, X(t) = L– 1 X(s) = L– 1
- 2
+
1
s + 1s - 2

= [e2t – 2e– t]u(t)



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