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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‘’ โˆ’ 2 ๐‘ก [ 1 2 ๐‘ ๐‘œ ๐‘  ( ๐‘ก ) โˆ’ 2 ๐‘  ๐‘– ๐‘› ( ๐‘ก ) ] ๐‘ข ( ๐‘ก )
B.๐‘’ โˆ’ ๐‘ก [ 2 ๐‘ ๐‘œ ๐‘  ( 2 ๐‘ก ) โˆ’ 1 2 ๐‘  ๐‘– ๐‘› ( 2 ๐‘ก ) ] ๐‘ข ( ๐‘ก )
C.๐‘’ โˆ’ 2 ๐‘ก [ 2 ๐‘ ๐‘œ ๐‘  ( ๐‘ก ) + 1 2 ๐‘  ๐‘– ๐‘› ( ๐‘ก ) ] ๐‘ข ( ๐‘ก )
D.๐‘’ โˆ’ ๐‘ก [ 2 ๐‘ ๐‘œ ๐‘  ( 2 ๐‘ก ) + 1 2 ๐‘  ๐‘– ๐‘› ( 2 ๐‘ก ) ] ๐‘ข ( ๐‘ก )
E.๐‘’ โˆ’ 2 ๐‘ก [ 2 ๐‘ ๐‘œ ๐‘  ( ๐‘ก ) โˆ’ 1 2 ๐‘  ๐‘– ๐‘› ( ๐‘ก ) ] ๐‘ข ( ๐‘ก )
F.๐‘’ โˆ’ ๐‘ก [ 1 2 ๐‘ ๐‘œ ๐‘  ( 2 ๐‘ก ) โˆ’ 2 ๐‘  ๐‘– ๐‘› ( 2 ๐‘ก ) ] ๐‘ข ( ๐‘ก )
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{ "reasoning": "To tackle this inverse Laplace transform, Iโ€™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 โ‰ฅ 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 explanation

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