Question

Find the inverse laplace transform of \(\frac{s}{(s^2+ 4)^2}\).

(a) ^1⁄4 sin(2t)

(b) ^t^2⁄4 sin(2t)

(c) ^t⁄4 sin(2t)

(d) ^t⁄4 sin(2t^2)

I have been asked this question by my college professor while I was bunking the class.

Question is taken from Laplace Transform by Properties in section Laplace Transform of Engineering Mathematics

Answer 1

Right answer is (c) ^t⁄4 sin(2t)

Explanation: Given, \(Y(s)=\frac{s}{(s^2+ 4)^2}\)

Inverse Laplace transform of \(\frac{1}{s^2+4}\)=sin⁡(2t)

Now, \(\frac{d}{ds} (\frac{1}{s^2+4})\)=-tsin(2t)

Inverse lapalce of \(\frac{-2s}{(s^2+4)^2}=-\frac{t}{2} sin(2t)\)

Inverse lapalce of \(\frac{s}{(s^2+4)^2}=\frac{t}{4} sin(2t)\)