Question

Given a system function H(s) = \(\frac{V(s)}{I(s)} = \frac{(s+4)}{(s+3)^2}\). And i(t) is a unit step, then V(t) in the steady state is ___________

(a) \(\frac{4}{9}\)

(b) \(\frac{4}{3}\)

(c) 0

(d) ∞

I got this question in an interview for job.

This interesting question is from Laplace Transform of a Periodic Function topic in chapter Application of the Laplace Transform in Circuit Analysis of Network Theory

Answer 1

Correct answer is (a) \(\frac{4}{9}\)

The explanation is: V(s) = \(\frac{(s+4)}{(s+3)^2}, I(s) = \frac{(s+4)}{s(s+3)^2}\)

At steady state, sV(s) = \(\frac{0+4}{(0+3)^2}\)

= \(\frac{4}{9}\).