Question

Find the inverse laplace transform of \(Y(s)=\frac{2s}{1-s^2}e^{-s}\).

(a) -e^-t + 1 + e^t – 1

(b) -e^-t + 1 – e^t + 1

(c) -e^-t + 1 + e^t + 1

(d) -e^-t + 1 – e^t – 1

This question was addressed to me in unit test.

I'd like to ask this question from Laplace Transform by Properties topic in section Laplace Transform of Engineering Mathematics

Answer 1

The correct option is (d) -e^-t + 1 – e^t – 1

Explanation: Given,

Y(s)=\(\frac{2s}{1-s^2}e^{-s}\)

Let,G(s)=\(\frac{2s}{1-s^2}=-\frac{1}{s – 1}-\frac{1}{s + 1}\)

hence,g(t)=\(-e^{-t} – e^t\)

Since,Y(s)=\(e^{-s} G(s)=>y(t)=g(t-1)\)

hence,y(t)=\(-e^{-t+1}-e^{t-1}\)