What is the inverse Laplace transormation of this please?

Hi I'm trying to work out the inverse Laplace transform of 4/[(s-2)^2 + 16]

Could anyone help please? I have tried factorising the denominator part which can also be written as s^2 - 4s + 20 but aren't there only surd factors to satisfy this? I think I'm better leaving it in its original completed square form but I'm if I do I'm not sure how to go about doing the partial fractions stage to find a standard inverse transform...

Re: What is the inverse Laplace transormation of this please?

you have to split the denominator in (s-2-4i)(s-2+4i)

$\displaystyle -1/8 i e^{(2-4 i) t} (-1+e^{8 i t})$

Re: What is the inverse Laplace transormation of this please?

Quote:

Originally Posted by

**chunkylumber111** Hi I'm trying to work out the inverse Laplace transform of 4/[(s-2)^2 + 16]

Could anyone help please? I have tried factorising the denominator part which can also be written as s^2 - 4s + 20 but aren't there only surd factors to satisfy this? I think I'm better leaving it in its original completed square form but I'm if I do I'm not sure how to go about doing the partial fractions stage to find a standard inverse transform...

$\displaystyle \displaystyle \begin{align*} \mathbf{L}^{-1}\left\{\frac{4}{(s-2)^2 + 16}\right\} &= e^{2t}\,\mathbf{L}^{-1}\left\{ \frac{4}{s^2 + 4^2} \right\} \\ &= e^{2t}\sin{\left(4t\right)} \end{align*} $

Re: What is the inverse Laplace transormation of this please?

Thanks Prove It, that's a great help. Is there an intermediate step you did to get e^2t L^-1 (4/s^2 + 4^2)?

Re: What is the inverse Laplace transormation of this please?

Quote:

Originally Posted by

**chunkylumber111** Thanks Prove It, that's a great help. Is there an intermediate step you did to get e^2t L^-1 (4/s^2 + 4^2)?

It's the horizontal shifting theorem.

$\displaystyle \displaystyle \begin{align*} \mathbf{L}\left\{ e^{a\,t}f(t) \right\} = F(s - a) \end{align*} $.