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

+1 vote
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

by (42.1k points)
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).

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