For the continuous-time signal 0, x(t) = 5te(t-2) – 10e(t-2), for t<2 for t>2 its Laplace transform X(s) is: Hint: You may want to write the signal x(t) in terms of the unit step function u(t) before taking the Laplace transform O A. 5e2s (s+1)2 O B. 5e -2s (s-1)2 5e -2s (s+1)? 5e2s (s- 1)2 OD.

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Author:Robert L. Boylestad
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For the continuous-time signal
0,
x(t) =
5te(t-2) – 10et-2),
for t<2
for t>2
its Laplace transform X(s) is:
Hint: You may want to write the signal x(t) in terms of the unit step function u(t) before taking the Laplace transform
A.
5e2s
(s+1)2
O B.
5e -2s
(s-1)2
О.
5e -2s
(s+ 1)?
OD.
5e2s
(s- 1)2
CLEAR MY CHOICE
Transcribed Image Text:For the continuous-time signal 0, x(t) = 5te(t-2) – 10et-2), for t<2 for t>2 its Laplace transform X(s) is: Hint: You may want to write the signal x(t) in terms of the unit step function u(t) before taking the Laplace transform A. 5e2s (s+1)2 O B. 5e -2s (s-1)2 О. 5e -2s (s+ 1)? OD. 5e2s (s- 1)2 CLEAR MY CHOICE
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