Need guidance with inverse laplace transform (2s)/(s^2+6s+13)

Alisa Taylor

Alisa Taylor

Answered question

2022-10-22

Need guidance with inverse laplace transform
I have the following formula, which I need to get the inverse laplace transform of:
2 s s 2 + 6 s + 13
I've managed to get 2 e 3 t cos ( 2 t ), that's rather simple - but according to matlab there's one more term: 3 e 3 t sin ( 2 t )
I don't really understand why.

Answer & Explanation

silenthunter440

silenthunter440

Beginner2022-10-23Added 19 answers

We have (through the residue theorem or partial fraction decomposition):
L 1 ( 2 s ( s + 3 ) 2 + 4 ) = L 1 ( 1 3 2 i s ( 3 2 i ) + 1 + 3 2 i s ( 3 + 2 i ) )
and since:
L 1 ( 1 s α ) = e α t
it follows that:
L 1 ( 2 s ( s + 3 ) 2 + 4 ) = ( 1 3 2 i ) e ( 3 + 2 i ) t + ( 1 + 3 2 i ) e ( 3 2 i ) t = e 3 t ( 2 cos ( 2 t ) 3 sin ( 2 t ) ) .

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