Question

Given x(t)=e^-t u(t). Find the inverse Laplace transform of e^-3s X(2s).

(a) \(\frac{1}{2}\) e^-(t-3)/2 u(t+3)

(b) \(\frac{1}{2}\) e^-(t-3)/2  u(t-3)

(c) \(\frac{1}{2}\) e^(t-3)/2 u(t-3)

(d) \(\frac{1}{2}\) e^(t-3)/2  u(t+3)

This question was addressed to me during an interview.

This intriguing question originated from Inverse Laplace Transform topic in section Laplace Transform and System Design of Signals and Systems

Answer 1

The correct option is (b) \(\frac{1}{2}\) e^-(t-3)/2  u(t-3)

To explain I would say: Given x(t) = e^-t u(t)

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

X(2s) = \(\frac{1}{2s+1} = \frac{1/2}{s+(1/2)}\)

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

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

∴L^-1 [e^-3s X(2s)] = \(\frac{1}{2}\) e^-(t-3)/2  u(t-3) if x(t) = e^-t u(t).