Find the inverse Laplace transform for the function X(s) = \(\frac{2s-1}{s^2+4s+8}\).
Posted: Thu Jul 14, 2022 11:09 am
a) e-2t cos2t u(t) – e-2t sin2t u(t)
b) 2e-2t cos2t u(t) – \(\frac{5}{2}\) e-2t sin2t u(t)
c) 2e-2t cos2t u(t) – e-2t sin2t u(t)
d) e-2t cos2t u(t) – \(\frac{5}{2}\) e-2t sin2t u(t)
b) 2e-2t cos2t u(t) – \(\frac{5}{2}\) e-2t sin2t u(t)
c) 2e-2t cos2t u(t) – e-2t sin2t u(t)
d) e-2t cos2t u(t) – \(\frac{5}{2}\) e-2t sin2t u(t)