Final Value of \(F(s)=\frac{1}{s(s-1)}\)
We use the Final Value Theorem:
\[ \lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s) \]
Step 1: Given
\[ F(s) = \frac{1}{s(s-1)} \]
Step 2: Apply Final Value Theorem
\[ \lim_{t\to\infty} f(t) = \lim_{s\to0} sF(s) \]
Substitute \(F(s)\):
\[ sF(s) = s\left(\frac{1}{s(s-1)}\right) = \frac{1}{s-1} \]
Step 3: Take \(s \to 0\)
\[ \lim_{s\to0} \frac{1}{s-1} = \frac{1}{0-1} = -1 \]
Important Note: The Final Value Theorem is not valid here because \(F(s)\) has a pole at \(s=1\) (right half plane), meaning the system is unstable.
Verification via Inverse Laplace
Partial fraction expansion:
\[ F(s) = \frac{1}{s(s-1)} = \frac{-1}{s} + \frac{1}{s-1} \]
Inverse Laplace transform:
\[ f(t) = -1 + e^{t} \]
As \(t \to \infty\):
\[ e^{t} \to \infty \]
So the function diverges and does not have a finite final value.
Conclusion
- Final Value Theorem (applied blindly) gives −1 (invalid)
- Actual final value: ∞ (diverges) due to pole at \(s=1\)