Question

What is the steady state value of The DT signal 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) $\frac{1}{8}$

The question was asked in quiz.

The query is from Discrete-Time Fourier Transform in division Fourier Transform of Signals and Systems

Answer 1

Right option is (c) 0

For explanation I would say: The steady state value of the DT signal 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.