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)
 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...
^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.
 \right\} = F(s - a) \end{align*} )
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