How can I find the Laplace transform of −e^(bt)sin(omega_0 t)u(−t) ?

Addyson Bright 2022-09-25 Answered
Laplace transform of e b t sin ( ω 0 t ) u ( t )
I know that Laplace transform of
e α t sin ( ω 0 t ) u ( t )
is ω 0 ( s + α ) 2 + ω 0 2 where R e ( s ) > R e ( α ) and u ( t ) = 1 t 0 . Knowing that, how can I find the Laplace transform of e b t sin ( ω 0 t ) u ( t )     ?
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Answers (1)

xjiaminhoxy4
Answered 2022-09-26 Author has 9 answers
Let
f ( t ) = e b t sin ( ω 0 t ) u ( t )
Notice that
f ( t ) = e b t sin ( ω 0 t ) u ( t )
Laplace transforms have a nice property that
L { f ( t ) } ( s ) = L { f ( t ) } ( s )
Using your known result of
L { f ( t ) } ( s ) = ω 0 ( s + b ) 2 + ω 0
we can use the property above to see that
L { f ( t ) } ( s ) = ω 0 ( s + b ) 2 + ω 0 = ω 0 ( s b ) 2 + ω 0
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