Question

The inverse Laplace transform of F(s) = \(\frac{2}{s^2+3s+2}\) is ______________

(a) -2e^-2t + 2e^-t

(b) 2e^-2t + 2e^-t

(c) -2e^-2t – 2e^-t

(d) 2e^-t + e^-2t

The question was asked in exam.

Query is from Common Laplace Transforms topic in section Laplace Transform and System Design of Signals and Systems

Answer 1

Correct choice is (a) -2e^-2t + 2e^-t

Explanation: s^2 + 3s + 2 = (s+2) (s+1)

Now, F(s) = \(\frac{A}{(s+2)} + \frac{B}{(s+1)}\)

Hence, A = (s+2) F(s) |s=-2

= \(\frac{2}{s+1}\)|s=-2 = -2

And, B = (s+1) F(s) |s=-1

= \(\frac{2}{s+2}\)| s=-1 = 2

∴ F(s) = \(\frac{-2}{(s+2)} + \frac{2}{(s+1)}\)

∴ F (t) = L^-1{F(s)}

 = -2e^-2t + 2e^-t  for t≥0