Laplace inverse of (e^(−s)(3s^2−s+2))/((s−1)(s^2+1))

Vrbljanovwu 2022-09-07 Answered
Laplace inverse of e s ( 3 s 2 s + 2 ) ( s 1 ) ( s 2 + 1 )
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Answers (1)

Colin Dougherty
Answered 2022-09-08 Author has 8 answers
I prefer the method of residues to partial fractions. If you recall, the residue of a simple pole s 0 of a function f(s) is
lim s s 0 [ ( s s 0 ) f ( s ) ]
Here
f ( s ) = 3 s 2 s + 2 ( s 1 ) ( s 2 + 1 ) e s t
Therefore are poles at s=1 and s = ± i. The sum of the residues at these poles are
2 e t + 1 i ( 1 + i ) ( 2 i ) e i t + 1 + i ( 1 i ) ( 2 i ) e i t = 2 e t + cos t
and that is the ILT.
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