Question

Find the inverse Laplace transform for X(s) = \(ln ⁡(\frac{s+a}{s+b})\).

(a) \(\frac{e^{-at} – e^{-bt}}{t}\)

(b) \(\frac{e^{-bt} – e^{-at}}{t}\)

(c) \(\frac{e^{-at} + e^{-bt}}{t}\)

(d) \(\frac{e^{bt} + e^{-at}}{t}\)

I have been asked this question in an interview for internship.

My question is from Inverse Laplace Transform topic in division Laplace Transform and System Design of Signals and Systems

Answer 1

Right option is (b) \(\frac{e^{-bt} – e^{-at}}{t}\)

Explanation: Given X(s) = \(ln ⁡(\frac{s+a}{s+b})\)

x(t) = L^-1 [X(s)] = L^-1 \(\Big[ln (\frac{s+a}{s+b})\Big]\)

L[x(t)] = \(ln ⁡(\frac{s+a}{s+b})\) = ln⁡(s+a)-ln⁡(s+b)

L[tx(t)] = –\(\frac{d}{ds}\) [ln⁡(s+a)-ln⁡(s+b)] = \(\frac{-1}{s+a} + \frac{1}{s+b} = \frac{1}{s+b} – \frac{1}{s+a}\)

tx(t) = \(L^{-1}(\frac{1}{s+b} – \frac{1}{s+a})\) = e^-bt – e^-at

x(t) = \(\frac{e^{-bt} – e^{-at}}{t}\).