Control Theory #42 — Routh-Hurwitz Parametric Stability (Worked Example 19)

Instructor: Dr. Süleyman Burak ÇELİK

A transfer function with a free parameter K hidden inside its denominator — and a clean inequality that pins down the stable range. We use Routh-Hurwitz in two stages — first the necessary condition, then the full Routh table — and never compute a single pole. Topics covered: - Stability of a transfer function ⟺ every pole sits in the open left half plane - Stage 1 — necessary condition: every coefficient must be present and share the same sign - Stage 2 — Routh table: build from coefficients, check first column for sign changes - Computing parametric Routh entries with a 2×2 determinant rule, divided by the leading element of the row above - Reading off the inequalities when entries depend on K - Spot-checking the boundary numerically with simulated step responses Problem (Sample Midterm A, Q10): G(s) = (s + 3) / (3s⁴ + 3K s³ + 4s² + K s + 1/3). For which values of K is the transfer function stable? Step-by-step solution: - The numerator does not affect stability — only the denominator polynomial matters. - Necessary condition (Stage 1): coefficients are 3, 3K, 4, K, 1/3. Three are fixed positive; the other two demand K › 0. Necessary condition collapses to K › 0. - Routh table (Stage 2): - s⁴ row: 3, 4, 1/3 - s³ row: 3K, K, 0 - s² row: b₁ = (3K·4 − 3·K)/(3K) = 9K/3K = 3, b₂ = (3K·1/3 − 0)/(3K) = 1/3 - s¹ row: c₁ = (3·K − 3K·1/3)/3 = 2K/3 - s⁰ row: d₁ = b₂ = 1/3 - First column: 3, 3K, 3, 2K/3, 1/3. Two entries depend on K; both demand K › 0. - Verification with simulated step responses: - K = 2 → stable (settles to a finite value) - K = 0.1 → stable but lightly damped (long ringing) - K = 0 → marginally unstable (one coefficient becomes zero) - K = −0.5 → strongly unstable (output explodes) Final answer: G(s) is stable if and only if K › 0. Big idea: when a transfer function carries a free parameter inside its denominator, the Routh table converts the stability question into a system of inequalities in that parameter. Solve the inequalities — get the stable range. No root finding needed. The same method scales to any parameter, any number of coefficients, any order of polynomial. Every step shown clearly — no shortcuts. Topics: 00:00 Cover 00:03 Problem — find every K for which G(s) is stable 01:22 Strategy — Routh-Hurwitz in two stages 02:37 Stage 1 — necessary condition reduces to K › 0 03:39 Stage 2 — set up the Routh table 04:45 s² row: compute b₁ = 3 and b₂ = 1/3 06:11 s¹ and s⁰ rows: compute c₁ = 2K/3 and d₁ = 1/3 07:41 First column inequalities → final answer K › 0 09:10 Verification — step responses for four K values 10:37 Summary Playlist: Control Theory - AcEdumy GitHub: https://github.com/acedumy