Inverse Laplace Transform

**kjchauhan** · February 29th 2012, 01:27 PM

Need help to solve the following:

Find the inverse Laplace transpose using convolution theorem:

$L^{-1} \left\{ \frac{s}{ $s^2 + 4$^3}\right\}$

Thanks.

**Kiwi_Dave** · March 13th 2012, 11:19 PM

Let

$L^{-1} \left\{ \frac{s}{ $s^2 + 4$^3}\right\}=L^{-1} \left\{ \frac{s}{ $s^2 + 4$}\frac{1}{ $s^2 + 4$}\frac{1}{ $s^2 + 4$}\right\}=f(t)*g(t)*h(t)$

Then by the convolution theorem we can choose

$L^{-1} \left\{ \frac{s}{ $s^2 + 4$}\right\}=f(t)$

$L^{-1} \left\{ \frac{1}{ $s^2 + 4$}\right\}=g(t)=h(t)$

And you will find these inverse Laplace transforms on Wikipedia.

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