Laplace Final Value Analysis
Your expression looks like the Laplace-domain transfer function:
\[ \frac{\omega}{s^2 + \omega^2} \]
If you are asking for the final value (steady-state value) using the Final Value Theorem, we compute it like this.
Step 1: Apply Final Value Theorem
For a Laplace function \(F(s)\):
\[ \lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s) \]
Step 2: Substitute \(F(s)\)
\[ sF(s) = s \cdot \frac{\omega}{s^2 + \omega^2} \]
Step 3: Take \(s \to 0\)
\[ \lim_{s \to 0} \frac{s\omega}{s^2 + \omega^2} = \frac{0}{\omega^2} = 0 \]
Final value = 0
Why?
The inverse Laplace of
\[ \frac{\omega}{s^2 + \omega^2} \]
is \(\sin(\omega t)\), which oscillates forever and its average steady value is 0.
Answer: 0
Second Expression
Consider the Laplace expression:
\[ F(s) = \frac{s}{s^2 + \omega^2} \]
1. Inverse Laplace
This is a standard Laplace pair:
\[ \mathcal{L}^{-1}\left(\frac{s}{s^2+\omega^2}\right) = \cos(\omega t) \]
2. Final Value
Using the Final Value Theorem:
\[ \lim_{t\to\infty} f(t) = \lim_{s\to0} sF(s) \]
Substitute \(F(s)\):
\[ sF(s) = s\left(\frac{s}{s^2+\omega^2}\right) = \frac{s^2}{s^2+\omega^2} \]
Now take \(s \to 0\):
\[ \lim_{s\to0} \frac{s^2}{s^2+\omega^2} = 0 \]
Final value = 0
Reason
Since the inverse Laplace is \(\cos(\omega t)\), the signal oscillates forever and does not settle to a constant value, so its steady-state value is 0.
Answer: 0
Quick Memory Trick for Exams
| Laplace Form | Time Function |
|---|---|
| \(\frac{s}{s^2+\omega^2}\) | \(\cos(\omega t)\) |
| \(\frac{\omega}{s^2+\omega^2}\) | \(\sin(\omega t)\) |
Both oscillate → Final value = 0