Question

What is the steady state value of F (t), if it is known that F(s) = \(\frac{2}{s(S+1)(s+2)(s+3)}\)?

(a) \(\frac{1}{2}\)

(b) \(\frac{1}{3}\)

(c) \(\frac{1}{4}\)

(d) Cannot be determined

The question was asked in a job interview.

The query is from Problems on Initial and Final Value Theorem in chapter Application of the Laplace Transform in Circuit Analysis of Network Theory

Answer 1

The correct answer is (b) \(\frac{1}{3}\)

Easy explanation: From the equation of F(s), we can infer that, a simple pole is at origin and all other poles are having negative real part.

∴ F(∞) = lims→0 s F(s)

= lims→0  \(\frac{2s}{s(S+1)(s+2)(s+3)}\)

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

= \(\frac{2}{6} = \frac{1}{3}\).