Questions
ECEN-314:200,501 Final Exam- Requires Respondus LockDown Browser
Single choice
Which of the following is the correct inverse Laplace Transform of ๐ ( ๐ ) = 2 ๐ + 1 ๐ 2 + 2 ๐ + 5 ?
Options
A.๐
โ
2
๐ก
[
1
2
๐
๐
๐
(
๐ก
)
โ
2
๐
๐
๐
(
๐ก
)
]
๐ข
(
๐ก
)
B.๐
โ
๐ก
[
2
๐
๐
๐
(
2
๐ก
)
โ
1
2
๐
๐
๐
(
2
๐ก
)
]
๐ข
(
๐ก
)
C.๐
โ
2
๐ก
[
2
๐
๐
๐
(
๐ก
)
+
1
2
๐
๐
๐
(
๐ก
)
]
๐ข
(
๐ก
)
D.๐
โ
๐ก
[
2
๐
๐
๐
(
2
๐ก
)
+
1
2
๐
๐
๐
(
2
๐ก
)
]
๐ข
(
๐ก
)
E.๐
โ
2
๐ก
[
2
๐
๐
๐
(
๐ก
)
โ
1
2
๐
๐
๐
(
๐ก
)
]
๐ข
(
๐ก
)
F.๐
โ
๐ก
[
1
2
๐
๐
๐
(
2
๐ก
)
โ
2
๐
๐
๐
(
2
๐ก
)
]
๐ข
(
๐ก
)
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Step-by-Step Analysis
{
"reasoning": "To tackle this inverse Laplace transform, Iโll break X(s) into parts whose time-domain forms we know.\n\nFirst, consider the term 2/s. Its inverse Laplace transform is simply the constant 2 for t โฅ 0, i.e., L^{-1}{2/s} = 2. This contributes a constant component 2ยทu(t) in the time domain.\n\nNext, examine 1/(s^2 + 2s + 5). Completing the square gives s^2 + 2s + 5 = (s + 1)^2 + 4. This matches the standard form for e^{-at} sin(bt) and e^{-at} cos(bt) terms:\n- L{e^{-t} cos(2t)} = (s + 1)/[(s + 1)^2 + 4]\n- L{e^{-t} sin(2t)} = 2/[(s + 1)^2 + 4]\n\nWe can express 1/[(s + 1)^2 + 4] as (1/2) ยท [ 2/((s + 1)^2 + 4) ], which corresponds to (1/2)ยทL{e^{-t} sin(2t)}. Therefore, L^{-1}{1/[(s + 1)^2 + 4]} = (1/2) e^{-t} sin(2t).\n\nPutting t......Login to view full explanationLog in for full answers
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