Routh-Hurwitz Stability Worked Example
A step-by-step construction of the Routh array for the characteristic equation s³ + 6s² + 11s + 6, showing how to compute each table entry and read the first column for sign changes. The animation links zero sign changes to zero right-half-plane roots, then confirms stability by plotting all three poles in the left-half of the s-plane. Useful for control systems students learning to apply the Routh-Hurwitz criterion to polynomial stability analysis.
16:9 · every frame verified for overlaps, spacing and edges before rendering
The prompt that made it
Create a 90-second animated worked example titled: “Routh Criterion — Stable System Example” Characteristic equation: P(s) = s³ + 6s² + 11s + 6 Construct the Routh table step-by-step. First two rows: s³ | 1 11 s² | 6 6 Calculate the s¹ first entry: (6×11 − 1×6)/6 = (66 − 6)/6 = 10 Complete table: s³ | 1 11 s² | 6 6 s¹ | 10 0 s⁰ | 6 Highlight the first column: 1 6 10 6 Animate signs beside each value: • • • • Count sign changes: 0 Display: Number of RHP roots = 0 Therefore: SYSTEM IS STABLE Show an s-plane and animate three pole markers in the left-half plane. Narration: “All elements of the first column have the same sign. Therefore, there are no sign changes and no roots in the right-half plane.” Final screen: FIRST COLUMN: • o + SIGN CHANGES = 0 RHP ROOTS = 0 STABLE ✓
Make the next one in this series
- Show a Routh criterion example with an unstable system that has sign changes
- Animate the special case when a first-column entry is zero
- Compare Routh criterion results with actual pole locations from factoring the polynomial
