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Use a table of Laplace transforms to find the Inverse Laplace transform of

F(s)=
(3s+9)/(s^2+9)


f(t)=_____

1 Answer

5 votes

Answer:

The inverse Laplace transform of
F(s) = (3\cdot s + 9)/(s^(2)+9) is
f(t) = 3\cdot \cos \omega t + 3\cdot \sin \omega t.

Explanation:

In this case, we should use the following direct and inverse Laplace Transforms:


F(s) = G(s) + H(s) (1)


\mathcal{L}^(-1) \{(s)/(s^(2)+\omega^(2)) \} = \cos \omega t (2)


\mathcal{L}^(-1)\{(\omega)/(s^(2)+\omega^(2)) \} = \sin \omega t (3)


\mathcal\{L\}^(-1)\{c\cdot F(s)\} = c\cdot \mathcal\{L\}^(-1)\{F(s)\} (4)

Then, we apply all these trasforms:


F(s) = (3\cdot s + 9)/(s^(2)+9)


F(s) = 3\cdot \left((s)/(s^(2)+9) \right)+3\cdot \left((3)/(s^(2)+9) \right)


f(t) = 3\cdot \cos \omega t + 3\cdot \sin \omega t

The inverse Laplace transform of
F(s) = (3\cdot s + 9)/(s^(2)+9) is
f(t) = 3\cdot \cos \omega t + 3\cdot \sin \omega t.

User Nazmul Haque
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