Question 9.22: Use the initial and final value properties to find the initi......
Use the initial and final value properties to find the initial and final values of the waveform whose transform is
F(s) = 2 \frac{(s + 3)}{s(s + 1)(s + 2)}
The given F(s) is a proper rational function, so the initial value property can be applied as
f(0) = \underset{s→∞}{\lim} sF(s) = \underset{s→∞}{\lim}\left[2 \frac{(s + 3)}{(s + 1)(s + 2)}\right] = 0
The poles of sF(s) are located in the left half plane at s = −1 and s = −2; hence, the final value property can be applied as
f(∞) = \underset{s→0}{\lim} sF(s) = \underset{s→0}{\lim} \left[2 \frac{(s + 3)}{(s + 1)(s + 2)}\right] = 3
In Example 9–11, the waveform corresponding to this transform was found to be
f(t) = [3 − 4e-t + e-2t]u(t)
from which we find
f(0) = 3 − 4e-0 + e-0 = 0
f(∞) = 3 − 4e-∞ + e-∞ = 3
which confirms the results found directly from f(s).