Question

Find the inverse Laplace transform for \(\frac{s}{(s+2)^2+1}\).

(a) [2e^-2t  cos⁡t + e^-2t   sin⁡t]u(t)

(b) [e^-2t cos⁡t + 2e^-2t  sin⁡t]u(t)

(c) [2e^-2t  cos⁡t – e^-2t  sin⁡t]u(t)

(d) [e^-2t  cos⁡t – 2e^-2t  sin⁡t]u(t)

The question was asked in an interview for internship.

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

Answer 1

Correct answer is (d) [e^-2t  cos⁡t – 2e^-2t  sin⁡t]u(t)

Explanation: Given X(s) = \(\frac{s}{(s+2)^2+1}\)

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

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

= [e^-2t  cos⁡t – 2e^-2t  sin⁡t]u(t).