Control Theory #37 — State-Space + BIBO + Routh-Hurwitz (Worked Example 14)

Instructor: Dr. Süleyman Burak ÇELİK

In this worked example, we run three different stability checks on a single state-space system. We see how internal stability, BIBO stability, and Routh-Hurwitz can each reveal instability — sometimes even when the transfer function looks deceptively simple. Topics covered: - Part (a): Internal (state-space) stability from eigenvalues of A - Part (b): BIBO stability via transfer function G(s) = cᵀ(sI−A)⁻¹b - Part (c): Routh-Hurwitz when the s² coefficient is zero — the ε-trick Problem setup: - State-space: ẋ = [[4, 2], [0, −3]] x + [[1], [0]] u, y = [2, 3] x - For part (c): G(s) = (s+4) / (s³ + 2s + 1) Key results: - Part (a): A upper triangular → eigenvalues 4 and −3. λ = 4 in RHP → INSTABLE - Part (b): G(s) derives to 2/(s−4) after (s+3) cancellation. Pole at +4 → NOT BIBO stable. Cancellation reveals an uncontrollable mode at −3 - Part (c): ε-trick gives first column {1, ε, −∞, 1} → 2 sign changes → 2 RHP poles → UNSTABLE Key insight: BIBO pole-zero cancellation can hide unstable internal modes. A missing s² coefficient creates a zero in the first Routh column — the ε-trick reveals hidden right-half-plane poles. Every step shown clearly — no shortcuts. Playlist: Control Theory - AcEdumy GitHub: https://github.com/acedumy #ControlTheory #WorkedExample #StateSpace #BIBO #RouthHurwitz #Stability #TransferFunction #AcEdumy