Routh stability criterion and root locus

Why it matters

Feedback can destabilise a perfectly stable process if the controller gain is too high. Before tuning any loop you need the gain at which it starts to oscillate without decay: the ultimate gain. The Routh test finds it algebraically from the characteristic equation, and the root locus shows how every closed-loop pole moves as the gain rises.

Key ideas

Stability. A linear closed loop is stable if every root of the characteristic equation 1 + G_OL(s) = 0 has a negative real part. A root on the imaginary axis gives sustained oscillation (marginal stability); a root in the right half-plane gives growth.

Routh–Hurwitz test. Write the characteristic polynomial a₀sⁿ + a₁sⁿ⁻¹ + … + aₙ = 0 with a₀ > 0.

• Necessary condition: all coefficients present and positive. A missing or negative coefficient means instability at once.
• Routh array: row 1 holds a₀, a₂, a₄, …; row 2 holds a₁, a₃, a₅, …. Each further element is b = (a₁·a₂ − a₀·a₃)/a₁, and so on, using the two rows above it.
• Criterion: the system is stable if and only if every element of the first column is positive. The number of sign changes in the first column equals the number of roots in the right half-plane.
• Special cases: a zero first-column element with other non-zero elements in its row is replaced by a small ε > 0 and the test continued. A whole row of zeros means roots symmetric about the origin (often a pair on the imaginary axis); form the auxiliary polynomial from the row above and solve it to find them.
• Ultimate gain: with K_c as a parameter, the K_c that makes a first-column element zero is the ultimate gain K_cu; the auxiliary equation of the row above then gives the oscillation frequency ω_u and ultimate period P_u = 2π/ω_u. These feed directly into Ziegler–Nichols tuning.
• Limitation: Routh cannot handle e^(−θs) exactly; use a Padé approximation or switch to frequency response.

Root locus. The root locus is the path of the closed-loop poles as K (usually K_c) goes from 0 to ∞ for 1 + K·G(s) = 0, with G having n poles and m zeros. Rules for sketching:

1. Branches start at the open-loop poles (K = 0) and end at the open-loop zeros or at infinity (K → ∞); there are n branches.
2. A point on the real axis lies on the locus if the number of real poles plus zeros to its right is odd.
3. n − m branches go to infinity along asymptotes at angles (2q + 1)·180°/(n − m), q = 0, 1, …, meeting the real axis at the centroid σ = (Σ poles − Σ zeros)/(n − m).
4. Breakaway points from the real axis satisfy dK/ds = 0.
5. Imaginary-axis crossings come from the Routh test (K_cu and ω_u).
6. The locus is symmetric about the real axis. Reading the locus: poles moving toward the right mean less damping; lines of constant ζ are rays from the origin at angle cos⁻¹ζ to the negative real axis.

Process insight. First- and second-order processes under P control can never go unstable (their loci stay in the left half-plane). Three or more lags, or any dead time, produce a finite ultimate gain. The more lags, the lower K_cu.

Formulas

1 + G_c·G_v·G_p·G_m = 0

• Characteristic equation.

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

• Routh array elements.

a₀s³ + a₁s² + a₂s + a₃ = 0 stable if all aᵢ > 0 and a₁·a₂ > a₀·a₃

• Cubic shortcut.

ω_u = √(a₃/a₁) (from the auxiliary equation a₁s² + a₃ = 0 at K_cu), P_u = 2π/ω_u

• Cubic at the ultimate gain; ω_u in rad per unit time.

σ = (Σp − Σz)/(n − m), θ_q = (2q + 1)·180°/(n − m)

• Root-locus centroid and asymptote angles.

dK/ds = 0

• Breakaway points.

Worked examples

Example 1 (standard): root locus and ultimate gain. G_OL = K/(s(s + 2)(s + 3)). Find the real-axis segments, asymptotes, breakaway point, and the gain and frequency at which the locus crosses the imaginary axis.

1. Poles 0, −2, −3; no zeros; n − m = 3.
2. Real-axis locus: between 0 and −2, and left of −3 (odd number of poles to the right).
3. Centroid σ = (0 − 2 − 3)/3 = −1.667; asymptote angles 60°, 180°, 300°.
4. K = −s(s + 2)(s + 3) = −(s³ + 5s² + 6s); dK/ds = −(3s² + 10s + 6) = 0 gives s = −0.785 or −2.549. Only −0.785 lies on the locus (between 0 and −2); K there = 2.11.
5. Characteristic equation s³ + 5s² + 6s + K = 0. Routh: row s¹ = (5·6 − K)/5. It is zero at K = 30.
6. Auxiliary equation 5s² + 30 = 0 gives s = ±j√6, so ω = 2.449 rad/s.

Stable for 0 < K < 30; crossing at s = ±j2.449 rad/s; breakaway at s ≈ −0.785

Example 2 (GATE level): ultimate gain of three lags. G_p = 1/((s + 1)(2s + 1)(4s + 1)) (time in min) under P control, G_v = G_m = 1. Find K_cu and P_u.

1. Expand: (s + 1)(2s + 1)(4s + 1) = 8s³ + 14s² + 7s + 1.
2. Characteristic equation: 8s³ + 14s² + 7s + (1 + K_c) = 0.
3. Routh rows: [8, 7], [14, 1 + K_c], s¹ element = (14·7 − 8·(1 + K_c))/14, s⁰ element = 1 + K_c.
4. Stability: 98 − 8(1 + K_c) > 0, so 1 + K_c < 12.25 and K_c < 11.25 (and K_c > −1).
5. At K_cu = 11.25: auxiliary equation 14s² + 12.25 = 0 gives ω_u = √(12.25/14) = √0.875 = 0.935 rad/min.
6. P_u = 2π/0.935 = 6.72 min.

K_cu = 11.25; ω_u ≈ 0.935 rad/min; P_u ≈ 6.72 min

Example 3 (short): counting unstable roots. For s⁴ + 2s³ + 3s² + 4s + 5 = 0: rows [1, 3, 5], [2, 4], [(2·3 − 1·4)/2 = 1, (2·5 − 1·0)/2 = 5], [(1·4 − 2·5)/1 = −6], [5]. First column 1, 2, 1, −6, 5 has two sign changes: two roots in the right half-plane.

Common mistakes

• Concluding stability from positive coefficients alone; that is necessary but not sufficient beyond second order.
• Counting a sign change as a pair of unstable roots; each sign change is one root.
• Forgetting to add 1 to K_c in the constant term (1 + K_c) of the characteristic equation.
• Using ω_u in rad/min but reporting P_u as if ω were cycles per minute (P_u = 2π/ω_u).
• Applying Routh to e^(−θs) without approximation.
• Putting real-axis root-locus segments where an even number of poles and zeros lie to the right.

For GATE CH

Expect: the range of K_c for stability by Routh (cubic shortcut a₁a₂ > a₀a₃), the ultimate gain and period, the number of right-half-plane roots, and root-locus features (centroid, asymptotes, breakaway, crossing). Many tuning questions start from K_cu and P_u obtained here. Practise expanding products of lags quickly.

Quick check

1. Is s³ + s² + 2s + 8 = 0 stable?
2. A loop has characteristic equation s³ + 3s² + 3s + 1 + K_c = 0. Find K_cu.
3. How many asymptotes does a root locus with 4 poles and 1 zero have, and at what angles?
4. What does a whole row of zeros in a Routh array indicate?

Answers: 1. No, 1·2 < 1·8 (two right-half-plane roots). 2. 3·3 = 1 + K_c gives K_cu = 8. 3. Three, at 60°, 180° and 300°. 4. Roots symmetric about the origin, typically a pair on the imaginary axis, found from the auxiliary polynomial.