Question

The inverse Laplace transform of F(s) = \(\frac{e^{-3s}}{s(s^2+3s+2)}\) is ______________

(a) {0.5 + 0.5e^-2(t+3) – e^-(t+3)} u (t+3)

(b) {0.5 + 0.5e^-2(t+3) – e^-(t-3)} u (t-3)

(c) {0.5 – 0.5e^-2(t-3) – e^-(t-3)} u (t-3)

(d) 0.5 + 0.5e^-2t – e^-t)

I had been asked this question in a national level competition.

I would like to ask this question from Problems on Initial and Final Value Theorem topic in portion Application of the Laplace Transform in Circuit Analysis of Network Theory

Answer 1

Correct option is (b) {0.5 + 0.5e^-2(t+3) – e^-(t-3)} u (t-3)

The explanation: Let G(s) = \(\frac{1}{s(s^2+3s+2)}\)

Or, F(s) = G(s) e^-3s

G (t) = L^-1{G(s)}

= L^-1\({\frac{A}{s} + \frac{B}{s+2} + \frac{C}{s+1}}\)

Solving we get, A = 0.5, B = 0.5, C = -1

So, G (t) = 0.5 + 0.5e^-2t-e^-t

The inverse Laplace transform is F (t) = {0.5 + 0.5e^-2(t+3) – e^-(t-3)} u (t-3).