Consider the following Laplace transforms of certain signals. For which of the following, final value theorem is not applicable?

Concept:

Final value theorem:

• A final value theorem allows the time domain behaviour to be directly calculated by taking a limit of a frequency domain expression
• Final value theorem states that the final value of a system can be calculated by

\(f\left( \infty \right)=\underset{s\to 0}{\mathop{\lim }}\,sF\left( s \right)\)

             Where F(s) is the Laplace transform of the function.

• For final value theorem to be applicable system should be stable in steady-state and for that real part of poles should lie in the left side of s plane.

Initial value theorem:

\(C\left( 0 \right)=\underset{t\to 0}{\mathop{\lim }}\,c\left( t \right)=\underset{s\to \infty }{\mathop{\lim }}\,sC\left( s \right)\)

It is applicable only when the number of poles of C(s) is more than the number of zeros of C(s).

Application:

1) \(F\left( s \right)=\frac{s-1}{s+2}\), the pole is lying in the left half of the s plane. So, the final value theorem is applicable.

2) \(F\left( s \right)=\frac{s+1}{s+2}\), the pole is lying in the left half of the s plane. So, the final value theorem is applicable.

3) \(F\left( s \right)=\frac{s+1}{\left( s+2 \right)\left( s+3 \right)}\), all the poles are lying in the left half of the s plane. So, the final value theorem is applicable.