Routh-Hurwitz stability criterion

Why it matters

Before tuning a servo for speed or accuracy you must know whether it is stable at all — and up to what gain. Finding the roots of a fourth- or fifth-order characteristic equation by hand is impractical, but the Routh–Hurwitz test tells you how many roots lie in the right half-plane, and the range of a gain for stability, using only the polynomial coefficients. It is the first check in almost every gain-selection problem.

Key ideas

Stability. An LTI system is BIBO stable if and only if every root of its characteristic equation (every closed-loop pole) has a negative real part. A root with positive real part gives a growing exponential; simple roots on the jω-axis give sustained oscillation (marginal stability); repeated jω-axis roots give growing oscillation.

Characteristic equation. For a feedback loop it is 1 + G(s)H(s) = 0, cleared to polynomial form a₀sⁿ + a₁sⁿ⁻¹ + … + aₙ = 0 with a₀ > 0.

Necessary condition (quick screen). All coefficients must be present and of the same sign. A missing term or a sign change means the system is not stable — stop. But all-positive coefficients are not sufficient for third and higher order; you still need the array. (For first- and second-order polynomials, all-positive coefficients are sufficient.)

Building the Routh array.

1. Row sⁿ: a₀, a₂, a₄, … (starting with the highest power, every second coefficient).
2. Row sⁿ⁻¹: a₁, a₃, a₅, …
3. Each new element uses the two rows above: b₁ = (a₁a₂ − a₀a₃)/a₁, b₂ = (a₁a₄ − a₀a₅)/a₁, and so on — the first-column elements of the two rows above, cross-multiplied with the next column.
4. Continue down to row s⁰. Any row may be multiplied by a positive constant to simplify arithmetic.

The criterion. The number of sign changes in the first column equals the number of roots in the right half-plane. The system is stable if and only if every first-column element is positive (no sign changes, no zeros).

Special case 1 — a zero first element, rest of row not all zero. Replace the zero by a small positive ε, continue, and count sign changes as ε → 0⁺. (Alternatively replace s by 1/s and test the reversed polynomial.) This case always means at least one root is not in the left half-plane.

Special case 2 — an entire row of zeros. The polynomial has roots placed symmetrically about the origin: a real pair ±σ, an imaginary pair ±jω, or a complex quadruple. Form the auxiliary polynomial A(s) from the row just above the zero row (it uses only every second power), replace the zero row with the coefficients of dA/ds, and continue. The roots of A(s) are the symmetric roots themselves and also a factor of the original polynomial. Sign changes below the zero row count right-half-plane roots; jω-axis roots come from solving A(s) = 0.

Parameter range. With a gain K in the coefficients, require every first-column entry > 0 to get the range of K. The value that makes the s¹ row zero is the marginal (critical) gain K_mar; the s² row's auxiliary equation at that gain gives the frequency of sustained oscillation — the same point where the root locus crosses the jω-axis.

Limits. Routh tells you whether roots are in the left half-plane, not how far in. To check relative stability (all roots left of s = −σ), substitute s = z − σ and test the new polynomial. It applies to polynomials only — not to systems with pure time delay (use Nyquist or Bode).

Formulas

a₀sⁿ + a₁sⁿ⁻¹ + a₂sⁿ⁻² + … + aₙ = 0 (a₀ > 0)

b₁ = (a₁·a₂ − a₀·a₃)/a₁, b₂ = (a₁·a₄ − a₀·a₅)/a₁ c₁ = (b₁·a₃ − a₁·b₂)/b₁

• aᵢ: polynomial coefficients; b, c: entries of the third and fourth rows.

Third-order stability condition for a₀s³ + a₁s² + a₂s + a₃ = 0: all aᵢ > 0 and a₁·a₂ > a₀·a₃

Auxiliary polynomial: A(s) from the row above an all-zero row; replace the zero row with dA/ds.

Relative stability: substitute s = z − σ and apply Routh in z.

Worked examples

Example 1 (standard — gain range). A unity-feedback loop has G(s) = K/(s(s + 1)(s + 2)). Find the range of K for stability, the marginal gain and the frequency of oscillation.

1. Characteristic equation: s(s + 1)(s + 2) + K = 0, i.e. s³ + 3s² + 2s + K = 0.
2. Array: s³: 1, 2; s²: 3, K; s¹: (3 × 2 − 1 × K)/3 = (6 − K)/3; s⁰: K.
3. Need (6 − K)/3 > 0 and K > 0: 0 < K < 6.
4. Marginal gain K_mar = 6 (the s¹ row becomes zero).
5. Auxiliary equation from the s² row: 3s² + 6 = 0 → s = ±j√2, so the loop oscillates at ω = 1.414 rad/s.

Example 2 (GATE level — row of zeros). Determine the location of the roots of s⁵ + 2s⁴ + 24s³ + 48s² − 25s − 50 = 0.

1. A coefficient is negative, so the system is unstable; Routh tells us where the roots are.
2. s⁵: 1, 24, −25; s⁴: 2, 48, −50.
3. s³: (2 × 24 − 1 × 48)/2 = 0 and (2 × (−25) − 1 × (−50))/2 = 0 — a full row of zeros.
4. Auxiliary polynomial from s⁴: A(s) = 2s⁴ + 48s² − 50; dA/ds = 8s³ + 96s. New s³ row: 8, 96.
5. s²: (8 × 48 − 2 × 96)/8 = 24, and (8 × (−50) − 2 × 0)/8 = −50 → 24, −50.
6. s¹: (24 × 96 − 8 × (−50))/24 = 2704/24 = 112.7; s⁰: −50.
7. First column: 1, 2, 8, 24, 112.7, −50 → one sign change → one root in the right half-plane.
8. A(s) = 0 → s⁴ + 24s² − 25 = (s² + 25)(s² − 1) = 0 → s = ±1 and s = ±j5.
9. Result: one root at s = +1 (RHP), two on the jω-axis at ±j5, two in the LHP (the remaining factor is (s + 1)(s + 2)). The system is unstable.

Common mistakes

• Starting row 1 with "even powers": row 1 always starts with the coefficient of the highest power, whatever its parity.
• Concluding stability from all-positive coefficients for a third- or higher-order polynomial.
• Arithmetic sign slips in the cross-multiplication; write b₁ = (a₁a₂ − a₀a₃)/a₁ explicitly each time.
• Treating a single zero in the first column as a row of zeros (or vice versa) — they need different fixes.
• Forgetting to also require K > 0 from the s⁰ row when finding a gain range.
• Reading the oscillation frequency from the wrong row: use the row just above the zero row at K = K_mar.

For GATE ME

Expect: number of RHP roots for a given polynomial; range of K for stability of a unity-feedback loop; marginal gain and oscillation frequency; special cases with a zero element or a row of zeros; and conceptual MCQs on necessary versus sufficient conditions. Practise the third-order shortcut a₁a₂ > a₀a₃ — it saves time on many questions.

Quick check

1. Is s³ + 5s² + 8s + 10 = 0 stable?
2. For s⁴ + 3s³ + 5s² + 7s + 9 = 0, what is the first element of the s² row?
3. What does an entire row of zeros indicate?
4. For s³ + 3s² + 2s + K = 0, what is the marginal gain?
5. Can s⁴ + 2s² + 3 = 0 be stable?

Answers: 1. Yes — 5 × 8 = 40 > 10 and the first column 1, 5, 6, 10 is all positive. 2. (3 × 5 − 1 × 7)/3 = 2.67. 3. Roots symmetric about the origin (for example a pair on the jω-axis), found from the auxiliary polynomial. 4. K = 6. 5. No — the s³ and s terms are missing.