Control Theory #36 — Final Value Theorem (Worked Example 13)

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In this worked example, we apply the Final Value Theorem (FVT) to two transfer functions differing by a single sign. We learn why stability is a strict prerequisite and what happens when we apply FVT blindly to an unstable system.

Topics covered:
- FVT statement: lim y(t) as t→∞ equals lim s·Y(s) as s→0
- For step input: lim y(t) = G(0) when G(s) is stable
- Critical condition: all poles in the open left-half plane
- Quick stability check for second-order polynomials
- What goes wrong when FVT is applied to an unstable system

Problem setup:
- G₁(s) = 3(s+4) / (2s² + 4s + 1) (all coefficients positive → stable)
- G₂(s) = 3(s+4) / (2s² + 4s − 1) (negative constant → one pole in RHP, unstable)

Key results:
- G₁: stable, FVT applies → lim y(t) = G₁(0) = 12
- G₂: unstable, FVT does NOT apply. G₂(0) = −12 is a wrong number. Actual y(t) contains e^(+0.22t), diverges. Limit does not exist.

Every step shown clearly — no shortcuts.

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Видео Control Theory #36 — Final Value Theorem (Worked Example 13) канала EngineeringENG - AcEdumy