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

Instructor: Dr. Süleyman Burak ÇELİK

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. Playlist: Control Theory - AcEdumy GitHub: https://github.com/acedumy #ControlTheory #WorkedExample #FinalValueTheorem #Stability #StepResponse #Laplace #ControlEngineering #AcEdumy