Question

If F1 (s) = \(\frac{1}{s+2}\) and F2 (s) = \(\frac{1}{s+3}\), find the inverse Laplace transform of F(s) = F1 (s) F2 (s).

(a) [e^-2t + e^-3t]u(t)

(b) [e^-2t – e^-3t]u(t)

(c) [e^2t + e^3t]u(t)

(d) [e^2t + e^-3t]u(t)

I got this question in an internship interview.

The above asked question is from Inverse Laplace Transform topic in chapter Laplace Transform and System Design of Signals and Systems

Answer 1

Correct choice is (b) [e^-2t – e^-3t]u(t)

Best explanation: Given F1 (s) = \(\frac{1}{s+2}\) and F2 (s) = \(\frac{1}{s+3}\).

F(s) = F1 (s) F2 (s) = \(\left(\frac{1}{s+2}\right)\left(\frac{1}{s+3}\right) = \frac{1}{s+2} – \frac{1}{s+3}\)

Applying inverse Laplace transform, we get

f(t) = [e^-2t – e^-3t]u(t).