题目
题目

ECEN-314:200,501 Final Exam- Requires Respondus LockDown Browser

单项选择题

Which of the following is the correct inverse Laplace Transform of 𝑋 ( 𝑠 ) = 2 𝑠 + 1 𝑠 2 + 2 𝑠 + 5 ?

选项
A.𝑒 − 2 𝑡 [ 1 2 𝑐 𝑜 𝑠 ( 𝑡 ) − 2 𝑠 𝑖 𝑛 ( 𝑡 ) ] 𝑢 ( 𝑡 )
B.𝑒 − 𝑡 [ 2 𝑐 𝑜 𝑠 ( 2 𝑡 ) − 1 2 𝑠 𝑖 𝑛 ( 2 𝑡 ) ] 𝑢 ( 𝑡 )
C.𝑒 − 2 𝑡 [ 2 𝑐 𝑜 𝑠 ( 𝑡 ) + 1 2 𝑠 𝑖 𝑛 ( 𝑡 ) ] 𝑢 ( 𝑡 )
D.𝑒 − 𝑡 [ 2 𝑐 𝑜 𝑠 ( 2 𝑡 ) + 1 2 𝑠 𝑖 𝑛 ( 2 𝑡 ) ] 𝑢 ( 𝑡 )
E.𝑒 − 2 𝑡 [ 2 𝑐 𝑜 𝑠 ( 𝑡 ) − 1 2 𝑠 𝑖 𝑛 ( 𝑡 ) ] 𝑢 ( 𝑡 )
F.𝑒 − 𝑡 [ 1 2 𝑐 𝑜 𝑠 ( 2 𝑡 ) − 2 𝑠 𝑖 𝑛 ( 2 𝑡 ) ] 𝑢 ( 𝑡 )
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思路分析
{ "reasoning": "To tackle this inverse Laplace transform, I’ll 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 ≥ 0, i.e., L^{-1}{2/s} = 2. This contributes a constant component 2·u(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) · [ 2/((s + 1)^2 + 4) ], which corresponds to (1/2)·L{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 explanation

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