Final Value Theorem: Identify the correct statement for a stable system output y(t) with Laplace transform Y(s).
-
Alim_{t→∞} y(t) = lim_{s→0} s Y(s), provided all poles of sY(s) are in the left half-plane and none on the imaginary axis
-
Blim_{t→∞} y(t) = lim_{s→∞} s Y(s), for any system
-
Clim_{t→0} y(t) = lim_{s→0} s Y(s), general case
-
Dlim_{t→∞} y(t) = lim_{s→0} Y(s), always
-
Elim_{t→∞} y(t) = lim_{s→∞} Y(s), if input is a step
Answer
Correct Answer: lim_{t→∞} y(t) = lim_{s→0} s Y(s), provided all poles of sY(s) are in the left half-plane and none on the imaginary axis
Explanation
Introduction / Context:The Final Value Theorem (FVT) links steady-state time behavior to a limit in the Laplace domain. It is widely used to compute steady-state error in control systems without performing inverse transforms.
Given Data / Assumptions:
- Output y(t) has Laplace transform Y(s).
- System stability assumptions are critical to FVT validity.
- No poles on the imaginary axis (except possibly a simple pole at the origin in sY(s) is not allowed).
Concept / Approach:
The FVT states: if all poles of sY(s) lie strictly in the left half-plane (no poles at or to the right of jω axis), then lim_{t→∞} y(t) exists and equals lim_{s→0} s Y(s). The condition on sY(s) ensures convergence and rules out oscillatory or divergent steady-state behavior.
Step-by-Step Solution:
Start from Y(s) = L{y(t)}.Assume poles of sY(s) are strictly in Re{s} < 0.Apply FVT: y(∞) = lim_{s→0} s Y(s).Check that conditions exclude marginally stable/unstable modes (e.g., sinusoidal terms or ramps).Verification / Alternative check:
For a unit-step input through a stable first-order system G(s) = K/(τ s + 1), Y(s) = G(s)/s. Then sY(s) = K/(τ s + 1) and lim_{s→0} sY(s) = K, matching the known steady-state y(∞) = K.
Why Other Options Are Wrong:
- Using s → ∞ is related to initial values, not final values.
- Replacing sY(s) by Y(s) neglects the crucial s multiplier.
- Claims that ignore stability constraints are unsafe and can give incorrect results for oscillatory/unstable systems.
Common Pitfalls:
- Applying FVT to systems with poles on the imaginary axis (e.g., pure integrators or undamped oscillators) leading to wrong or undefined limits.
- Forgetting to include the input when forming Y(s) for closed-loop analysis.
Final Answer:
lim_{t→∞} y(t) = lim_{s→0} s Y(s), with all poles of sY(s) strictly in the left half-plane