Stability: Routh-Hurwitz and Root Locus

BIBO stability, Routh-Hurwitz criterion and special cases, range of K for stability, root locus rules (asymptotes, centroid, breakaway points, jω crossing) and effect of adding poles and zeros.

A closed-loop system is stable when all its poles lie in the left half of the s-plane. Routh's test checks this without finding the roots; the root locus shows how the poles move as the gain changes.

Routh–Hurwitz criterion

For the characteristic equation a₀sⁿ + a₁sⁿ⁻¹ + … + aₙ = 0:

  1. Necessary: all coefficients present and of the same sign.
  2. Build the Routh array: first two rows from alternate coefficients; each next entry = (product cross-difference)/(first element of the row above).
  3. The number of sign changes in the first column = the number of right-half-plane roots.

Special cases:

  • A zero in the first column (rest non-zero): replace it by a small ε and continue.
  • A whole row of zeros: form the auxiliary polynomial from the row above, differentiate it, and use its coefficients. It signals symmetric roots (for example on the jω axis → sustained oscillation).

For s³ + a s² + b s + c: stable if a, b, c > 0 and ab > c.

Root locus

The root locus traces the closed-loop poles of 1 + K G(s)H(s) = 0 as K goes from 0 to ∞.

  • Starts at the open-loop poles (K = 0), ends at the open-loop zeros (or at infinity).
  • Number of branches = number of poles (n).
  • On the real axis: a point is on the locus if the number of real poles and zeros to its right is odd.
  • Asymptotes: n − m of them, at angles (2q + 1)180°/(n − m).
  • Centroid: σ = (Σ poles − Σ zeros)/(n − m).
  • Breakaway points: solve dK/ds = 0.
  • jω-axis crossing: from the Routh array (marginal K and the oscillation frequency).
  • Angle condition ∠GH = ±180°(2q + 1); magnitude condition |KGH| = 1.

Adding a pole pushes the locus to the right (less stable); adding a zero pulls it to the left (more stable).

Worked example

G(s)H(s) = K/(s(s + 2)(s + 4)), unity feedback.

  • Characteristic equation: s³ + 6s² + 8s + K = 0.
  • Stable if K > 0 and 6 × 8 > K: 0 < K < 48.
  • At K = 48 the auxiliary equation 6s² + 48 = 0 gives oscillation at ω = 2√2 = 2.83 rad/s.
  • Asymptote centroid = (0 − 2 − 4)/3 = −2, at ±60° and 180°.

Practice

GATE-styleNAT · numerical answer

For s³ + 4s² + 5s + K = 0, what is the largest K for stability?

Show answer and explanationAnswer: 20

Stable when 4 × 5 > K, so K < 20 (the marginal value is 20).

GATE-styleNAT · numerical answer

G(s)H(s) = K(s + 1)/(s(s + 3)(s + 5)). Where is the centroid of the asymptotes?

Show answer and explanationAnswer: -3.5

σ = (Σp − Σz)/(n − m) = (−8 + 1)/2 = −3.5.

GATE-styleNAT · numerical answer

How many roots of s³ − s² + 2s + 8 = 0 lie in the right half-plane? (Use Routh's array.)

Show answer and explanationAnswer: 2

First column: 1, −1, (−2 − 8)/(−1) = 10, 8. Two sign changes: two RHP roots.

GATE-styleMCQ · one correct answer

On the real axis, a point lies on the root locus if the total number of real poles and zeros to its right is:

Show answer and explanationAnswer: A

Only then does the angle condition (an odd multiple of 180°) hold.