Question

What is the steady state value of F (t), if it is known that F(s) = \(\frac{1}{(s+2)^2 (s+4)}\)?

(a) \(\frac{1}{16}\)

(b) Cannot be determined

(c) 0

(d) \(\frac{1}{8}\)

I got this question by my school teacher while I was bunking the class.

This intriguing question originated from Problems on Initial and Final Value Theorem topic in section Application of the Laplace Transform in Circuit Analysis of Network Theory

Answer 1

Right option is (c) 0

To elaborate: The steady state value of F(s) exists since all poles of the given Laplace transform have negative real part.

∴F(∞) = lims→0 s F(s)

 = lims→0  \(\frac{s}{(s+2)^2 (s+4)}\)

= 0.