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Using the Convolution Theorem, We Find That L−1{1/((s+1)(s2+1))}=\mathcal { L } ^ { - 1 } \left\{ 1 / \left( ( s + 1 ) \left( s ^ { 2 } + 1 \right) \right) \right\} =

Question 9

Multiple Choice

Using the convolution theorem, we find that L−1{1/((s+1) (s2+1) ) }=\mathcal { L } ^ { - 1 } \left\{ 1 / \left( ( s + 1 ) \left( s ^ { 2 } + 1 \right) \right) \right\} =


A) (e−t+sin⁡t−cos⁡t) /2\left( e ^ { - t } + \sin t - \cos t \right) / 2
B) (et+sin⁡t−cos⁡t) /2\left( e ^ { t } + \sin t - \cos t \right) / 2
C) (e−t+sin⁡t+cos⁡t) /2\left( e ^ { - t } + \sin t + \cos t \right) / 2
D) (et−sin⁡t−cos⁡t) /2\left( e ^ { t } - \sin t - \cos t \right) / 2
E) (e−t−sin⁡t−cos⁡t) /2\left( e ^ { - t } - \sin t - \cos t \right) / 2

Correct Answer:

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