Question

Find the inverse Laplace transform of X(s) = \(\frac{s}{s^2 a^2+b^2}\).

(a) \(\frac{1}{a^2} \,cos⁡(\frac{a}{b})t\)

(b) \(\frac{1}{a^2} \,cos⁡(\frac{b}{a})t\)

(c) \(\frac{1}{a^2} \,sin⁡(\frac{b}{a})t\)

(d) \(\frac{1}{a^2} \,sin⁡(\frac{a}{b})t\)

I have been asked this question by my school principal while I was bunking the class.

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

Answer 1

The correct option is (b) \(\frac{1}{a^2} \,cos⁡(\frac{b}{a})t\)

The best explanation: Given X(s) = \(\frac{s}{s^2 a^2+b^2} = \frac{1}{a^2} \Big[\frac{s}{s^2+(b/a)^2}\Big]\)

We know that L^-1 \(\left(\frac{s}{s^2+ω^2}\right)\) = cos⁡ωt

∴x(t) = L^-1 [X(s)] = \(\frac{1}{a^2} L^{-1} \Big[\frac{s}{s^2+(b/a)^2}\Big] = \frac{1}{a^2} \,cos⁡(\frac{b}{a})t\).