Question

Find the inverse Laplace transform for X(s) = s/(2s^2-8)

Answer 1

Answer:

The inverse Laplace transform of X(s) is $\frac{1}{2}cosh2t$

Step-by-step explanation:

The given Laplace transform is

$X(s)=\frac{s}{(2s^{2}-8 )}$

The inverse Laplace transform formula is given by

$L^{-1}({\frac{s}{s^{2}-a^{2} } })= coshat$

To find the inverse Laplace transform of X(s) consider

$X(s)=\frac{s}{2(s^{2}-4 )}$

$X(s)=\frac{s}{2(s^{2}-2^{2} )}$

To find the inverse Laplace transform first we reduce the given equation to a certain form is given by

$L^{-1}(\frac{s}{2(s^{2}-2^{2} )}) = \frac{1}{2} L^{-1} (\frac{s}{(s^{2}-2^{2} )})$

                     $=\frac{1}{2}cosh2t$

Therefore the inverse Laplace transform of X(s) is given by $\frac{1}{2}cosh2t$