We have:
We solve this and find
We can then use the inverse Laplace transform, and then we get:
h(t) = (1 – 2te -t – e -2t)u(t)
Initial and final value:
The limit of sH(s) when s approaches infinity is 0, and when we look at
h(t) we get h(0) = 0.
The limit of sH(s) when s approaches 0 is 1, and when we look at h(t)
we get the limit of h(t) when t approaches infinity is 1 also.
Thus the initial and final value theorems are verified here.