Question

Find the inverse Laplace transform for X(s) = \(\frac{s}{2s^2-8}\).

(a) cosh⁡2t

(b) \(\frac{1}{2}\) cosh⁡2t

(c) sinh⁡2t

(d) \(\frac{1}{2}\) sinh⁡2t

I have been asked this question in a national level competition.

The above asked question is from Inverse Laplace Transform topic in division Laplace Transform and System Design of Signals and Systems

Answer 1

The correct option is (b) \(\frac{1}{2}\) cosh⁡2t

Easiest explanation: Given X(s) = \(\frac{s}{2s^2-8} = \frac{1}{2}  \frac{s}{(s^2-2^2)}\)

We know that cosh⁡ωt =  \(\frac{s}{(s^2-w^2)}\)

∴x(t)=L^-1 [X(s)] = \(\frac{1}{2} L^{-1} \left(\frac{s}{(s^2-2^2)}\right) = \frac{1}{2} cosh⁡2t.\)