Find inverse Laplace transformation L^(-1)=(s)/((s−3)(s−4)(s−12))

cimithe4c 2022-10-15 Answered
I have the following inverse laplace transformation:
L 1 = s ( s 3 ) ( s 4 ) ( s 12 )
After looking at the laplace transformations the closest I've found is:
a e a t b e b t a b = s ( s a ) ( s b )
I've been working out a solution for having three variables, but I cant seem to get the correct solution. Is there an identity for this type of transformation?
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Answers (2)

Dobricap
Answered 2022-10-16 Author has 14 answers
Hint: The Partial Fraction Expansion is:
s ( s 3 ) ( s 4 ) ( s 12 ) = 1 2 ( s 4 ) + 1 3 ( s 3 ) + 1 6 ( s 12 )
L 1 ( s ( s 3 ) ( s 4 ) ( s 12 ) ) = ( 1 / 2 ) e 4 t + ( 1 / 3 ) e 3 t + ( 1 / 6 ) e 12 t
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Emmy Swanson
Answered 2022-10-17 Author has 1 answers
Using partial-fraction expansion, we have
s ( s 3 ) ( s 4 ) ( s 12 ) = A ( s 3 ) + B ( s 4 ) + C ( s 12 )
where A,B and C are obtained as follows:
A = s ( s 4 ) ( s 12 ) | s = 3 = 1 3 , B = s ( s 3 ) ( s 12 ) | s = 4 = 1 2 , C = s ( s 3 ) ( s 4 ) | s = 12 = 1 6 .
As a result, we get
s ( s 3 ) ( s 4 ) ( s 12 ) = 1 3 ( s 3 ) 1 2 ( s 4 ) + 1 6 ( s 12 )
Now, it is straightforward to use the Laplace inverse table, specifically, L 1 [ 1 s + a ] = e a t u ( t ), so we get
L 1 [ s ( s 3 ) ( s 4 ) ( s 12 ) ] = L 1 [ 1 3 ( s 3 ) ] L 1 [ 1 2 ( s 4 ) ] + L 1 [ 1 6 ( s 12 ) ] = 1 3 L 1 [ 1 ( s 3 ) ] 1 2 L 1 [ 1 ( s 4 ) ] + 1 6 L 1 [ 1 ( s 12 ) ] = 1 3 e 3 t 1 2 e 4 t + 1 6 e 12 t
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