Question 9.11: Find the waveform corresponding to the transform F(s) = 2 (s......

F(s) is a proper rational function and has simple poles at s = 0, s = −1, s = −2. Its partial fraction expansion is

F(s) = \frac{k_{1}}{s} + \frac{k_{2}}{s  +  1} + \frac{k_{3}}{s  +  2}

The cover-up algorithm yields the residues as

k_{1} = sF(s) \mid_{s =  0}  = \frac{2(s  +  3)}{(s  +  1)(s  +  2)}\mid_{s =  0}  = 3

k_{2} = (s + 1) F(s)\mid_{s =  -1}  = \frac{2(s  +  3)}{s(s  +  2)}\mid_{s =  -1}  = -4

k_{3} = (s + 2)F(s) \mid_{s =  -2}  = \frac{2(s  +  3)}{s(s  +  1)}\mid_{s =  -2}  = 1

The inverse transform f(t) is

f(t) = [3  –  4 e^{-t} + e^{-2t}] u(t)