Control Theory #43 — Root Locus + K Range for Stability and Oscillation (Worked Example 20)

Instructor: Dr. Süleyman Burak ÇELİK

A free gain K hidden in an open-loop transfer function with one right-half-plane pole and two purely imaginary zeros. We answer three questions about the closed loop — root locus sketch, K range for stability, K range for oscillation — using a single characteristic equation and four short calculations. Topics covered: - Stability of a unity-feedback closed loop ⟺ closed-loop poles in the open left half plane - Building the closed-loop characteristic equation from 1 + G_o(s) = 0 - Quadratic stability shortcut: all coefficients › 0 ⟺ stable (Routh-Hurwitz necessary + sufficient at order 2) - Discriminant analysis Δ = b² − 4ac ‹ 0 ⟺ complex conjugate poles ⟺ oscillation - Root locus rules: number of branches, real-axis segments, asymptotes, jω-axis crossing, departure/arrival angles, behavior at infinity - Combining stability and oscillation ranges to find when the system is both Problem (Sample Midterm A, last problem): G_o(s) = K (s² + 4) / [(s + 3)(s − 1)]. - (a) Sketch the root locus and write down the rules used. - (b) For which K is the closed loop stable? - (c) For which K is the closed loop oscillatory? Open-loop characteristics: - Poles: s = −3, +1 (one in right half plane — open-loop unstable) - Zeros: s = ±2j (on the jω axis) - n = m = 2 (no asymptotes) Step-by-step solution: - Closed-loop CE: (1 + K)·s² + 2s + (4K − 3) = 0 - Coefficients: a = 1+K, b = 2, c = 4K − 3 - (b) Stability — all coefficients › 0: - 1 + K › 0 ⇒ K › −1 - 2 › 0 always - 4K − 3 › 0 ⇒ K › 3/4 - Combined: K › 3/4 - (c) Oscillation — discriminant ‹ 0: - Δ = 4 − 4(1+K)(4K−3) = 16 − 16K² − 4K - Δ ‹ 0 ⇔ 4K² + K − 4 › 0 - Roots in K: K = (−1 ± √65)/8 ≈ 0.883 and −1.133 - Combined with stability: K › (√65 − 1)/8 ≈ 0.883 - (a) Root locus: - 2 branches (n = m = 2) - Real-axis segment: [−3, +1] - No asymptotes (n − m = 0) - jω crossing at origin when K = 3/4 (stability boundary) - Branches start at OL poles ±2j, slide along real-axis segment, break away at K ≈ 0.883, curve into the complex plane and end at OL zeros ±2j as K → ∞ Final answers: - (a) Two branches starting at −3 and +1, sliding along the real-axis segment between them, breaking away at K = (√65 − 1)/8, curving to ±2j as K → ∞. Crosses the jω axis at the origin when K = 3/4. - (b) Closed loop is stable for K › 3/4. - (c) Closed loop is oscillatory for K › (√65 − 1)/8 ≈ 0.883. Big picture: closed-loop poles tell every story — stability, oscillation, transient shape — and the root locus is the picture that binds them together as a function of the gain. Three questions, one characteristic equation, four short calculations. This concludes the midterm worked-example series — twenty problems covering plant modeling, transfer functions, state space, linearization, stability, time specifications, and root locus. Every step shown clearly — no shortcuts. Topics: 00:00 Cover 00:03 Problem — root locus + K range for stability and oscillation 01:24 Strategy — one CE, three different facts 02:42 Step 1 — build the closed-loop characteristic equation 04:18 Step 2 — stability via quadratic coefficients ⇒ K › 3/4 06:00 Step 3 — oscillation via discriminant ⇒ K › 0.883 08:48 Step 4 — root locus rules and sketch 12:14 Summary Playlist: Control Theory - AcEdumy GitHub: https://github.com/acedumy