Question

Find the inverse Laplace transform for the function X(s) = \(\frac{2s-1}{s^2+4s+8}\).

(a) e^-2t  cos⁡2t u(t) – e^-2t  sin⁡2t u(t)

(b) 2e^-2t  cos⁡2t u(t) – \(\frac{5}{2}\) e^-2t  sin⁡2t u(t)

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

(d) e^-2t cos⁡2t u(t) – \(\frac{5}{2}\) e^-2t  sin⁡2t u(t)

I have been asked this question in unit test.

This intriguing question comes from Inverse Laplace Transform in chapter Laplace Transform and System Design of Signals and Systems

Answer 1

Right choice is (b) 2e^-2t  cos⁡2t u(t) – \(\frac{5}{2}\) e^-2t  sin⁡2t u(t)

Explanation: Given function X(s) = \(\frac{2s-1}{s^2+4s+8} = \frac{2s-1}{(s+2)^2+2^2} = \frac{2(s+2)-5}{(s+2)^2+2^2}\)

= \(\frac{2(s+2)}{(s+2)^2+2^2} – \frac{5}{2} \frac{2}{(s+2)^2+2^2}\)

Applying inverse Laplace transform, we get

x(t) = 2e^-2t  cos⁡2t u(t) – \(\frac{5}{2}\) e^-2t  sin⁡2t u(t).