Question

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

(a) \(\frac{1}{a}\)  t sin⁡at

(b) \(\frac{1}{2a}\)  t sin⁡at

(c) \(\frac{1}{a}\)  t cos⁡at

(d) \(\frac{1}{2a}\)  t cos⁡at

The question was posed to me in a dream while sleeping

Asked question is from Inverse Laplace Transform topic in chapter Laplace Transform and System Design of Signals and Systems

Answer 1

Right option is (b) \(\frac{1}{2a}\)  t sin⁡at

Best explanation: Given X(s) = \(\frac{s}{(s^2+a^2)^2}\)

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

= \(\frac{1}{2a} tL^{-1} \Big[\frac{a}{s^2+a^2}\Big] = \frac{1}{2a}t sin⁡at\).