Control Theory #38 — Step Response & Controller Stability (Worked Example 15)

Instructor: Dr. Süleyman Burak ÇELİK

In this worked example, we answer five questions about a single problem — all driven by reading pole-zero locations on the s-plane. Three transfer functions show three different step-response shapes: overshoot, undershoot, and oscillation. Then we switch to a feedback loop with three candidate controllers and decide which one keeps the closed loop stable and which one drives the steady-state error to zero. Topics covered: - Part (a): Overshoot from a left-half-plane zero close to the origin - Part (b): Undershoot from a right-half-plane (non-minimum-phase) zero - Part (c): Oscillation from complex-conjugate poles (discriminant test, damping ratio) - Part (d): Closed-loop stability — why you must never cancel an unstable plant pole - Part (e): Zero steady-state error from an integrator in the loop (system Type 1) Transfer functions: - G₁(s) = (s − 4) / [(s + 2)(s + 4)] - G₂(s) = (s + 1) / [(s + 2)(s + 4)] - G₃(s) = (s + 4) / (s² + 2s + 4) Plant + controllers: - G(s) = (s + 4) / (s − 3) (unstable plant) - C₁(s) = (s − 3) / (s + 4), C₂(s) = (1 + s) / s, C₃(s) = 1 / (1 + s) Key results: - Part (a): G₂ — zero at −1 sits between origin and slowest pole → overshoot - Part (b): G₁ — zero at +4 in RHP, initial slope and final value have opposite signs → undershoot - Part (c): G₃ — Δ = 4 − 16 = −12 ‹ 0 → poles −1 ± j√3, ζ = ½, underdamped → oscillation - Part (d): C₁ cancels the plant's RHP pole at +3 → standard rule: never cancel an unstable pole - Part (e): C₂ contains 1/s → integrator → Type 1 system → e_∞ = 0 Big picture: step-response shape is read directly from the s-plane (zero/pole locations); closed-loop behavior is decided by 1 + L(s) = 0 plus the integrator count in L(s). Every step shown clearly — no shortcuts. Playlist: Control Theory - AcEdumy GitHub: https://github.com/acedumy