Question

Find the inverse Laplace transform for the function X(s) = \(\frac{1+e^{-2s}}{3s^2+2s}\).

(a) e^-(2/3)t u(t) – u(t) + e^-(2/3)(t-2) u(t-2)-u(t-2)

(b) e^-(2/3)t u(t) + e^-(2/3)(t-2) u(t-2)

(c) e^-(2/3)(t-2) u(t-2) – u(t-2)

(d) e^-(2/3)t u(t) – u(t)

The question was asked during an online interview.

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

Answer 1

The correct answer is (a) e^-(2/3)t u(t) – u(t) + e^-(2/3)(t-2) u(t-2)-u(t-2)

To elaborate: Given function X(s) = \(\frac{1+e^{-2s}}{3s^2+2s}\)

x(t) = L^-1 [X(s)] = \(L^{-1} \Big[\frac{1+e^{-2s}}{3s^2+2s}\Big] = L^{-1} \Big[\frac{1}{3s^2+2s}\Big] + L^{-1} \Big[\frac{e^{-2s}}{3s^2+2s}\Big]\)

\(L^{-1} \Big[\frac{1}{3s^2+2s}\Big] = L^{-1} \Big\{\frac{1}{3s[s+(2/3)]}\Big\} = L^{-1} \Big\{\frac{-1}{s} + \frac{1}{[s+(2/3)]}\Big\}\) = e^-(2/3)t u(t) – u(t)

\(L^{-1} \Big[\frac{e^{-2s}}{3s^2+2s}\Big] = L^{-1} \left(\frac{1}{3s^2+2s}\right)_{t=t-2}\) = e^-(2/3)(t-2) u(t-2)-u(t-2)

∴x(t) = e^-(2/3)t u(t) – u(t) + e^-(2/3)(t-2) u(t-2)-u(t-2).