Stability: Routh, Root Locus and Margins

Poles in the left half-plane, the Routh test, root locus and how delay erodes margin.

What stability means

A system is stable if a bounded input gives a bounded output and any disturbance dies out; unstable if the output grows without limit (until something saturates or breaks). Stability is the first requirement of any control loop: a loop that is too aggressive turns a corrective action into a self-reinforcing oscillation, as in a shower that swings between scalding and freezing, a robot arm that shakes, or a heating loop that cycles.

Poles and the characteristic equation

The closed loop has the transfer function T = C G ÷ (1 + C G). Its denominator, the characteristic equation 1 + L(s) = 0, has roots that are the closed-loop poles. The loop is stable if and only if all poles have negative real parts (left half of the s-plane, see Laplace Transforms and Transfer Functions); one pole on the imaginary axis means sustained oscillation (the stability limit) and one in the right half-plane means exponential growth. Finding the roots of a high-order polynomial by hand is tedious, so tests were devised that tell the answer without them.

The Routh–Hurwitz test

For a characteristic polynomial a₃s³ + a₂s² + a₁s + a₀ the Routh array shows whether any root is in the right half-plane without solving it: all coefficients must be positive, and for a cubic the product a₂ a₁ must exceed a₃ a₀. The number of sign changes in the first column of the full array equals the number of unstable roots.

Worked example: a third-order plant under proportional control

Take a plant G(s) = 1 ÷ ((s + 1)(s + 2)(s + 3)) with a proportional controller of gain Kc. The characteristic equation is (s + 1)(s + 2)(s + 3) + Kc = s³ + 6 s² + 11 s + (6 + Kc) = 0.

  • Routh: a₂ a₁ > a₃ a₀ means 6 × 11 > 6 + Kc, so Kc < 60. Above that, two poles cross to the right half-plane and the loop is unstable.
  • At the limit Kc = 60 the polynomial factors as (s + 6)(s² + 11), so the poles are s = −6 and s = ±j 3.317: the loop oscillates steadily at 3.32 rad/s (period 1.89 s).
  • Check: the first Routh column of the array is 1, 6, (66 − 6 − Kc) ÷ 6 = 8.3 for Kc = 10 (stable), 0.0 for Kc = 60 (marginal), -3.3 for Kc = 80 (a sign change: two unstable roots).

Root locus

The root locus shows how the closed-loop poles move across the s-plane as a gain K is increased from 0 to infinity, starting at the open-loop poles and ending at the zeros. It lets you see at once how much gain gives a good damping, and when the loop goes unstable. The rules (branches start at poles, end at zeros or run to infinity along asymptotes, the locus lies on the real axis to the left of an odd number of poles and zeros, and so on) are in any control text, and software draws it instantly.

Worked example: L(s) = K ÷ (s (s + 4))

The characteristic equation is s² + 4 s + K = 0, with the poles s = −2 ± √(4 − K):

Gain KClosed-loop polesDamping ζ = 2 ÷ √KOvershoot
2-0.59, -3.411.000 %
4-2.00, -2.001.000 %
8−2 ± j2.000.714 %
20−2 ± j4.000.4521 %
40−2 ± j6.000.3235 %

The poles start at 0 and −4, meet at the breakaway point s = −2 when K = 4 (ζ = 1) and then move vertically as a complex pair at a constant real part of −2: raising the gain never makes this loop unstable, but the damping falls and the overshoot grows. With one more pole in the plant (the oven’s dead time adds many) the branches bend to the right and cross the imaginary axis.

Worked example: pushing the oven loop to instability

Worked example: the ultimate gain of the oven

  • With P-only control the loop turns unstable when ∠L = −180° and |L| = 1. The phase of the oven is −arctan(ω τ) − ω θ, which reaches −180° at ωu = 0.110 rad/s (period Pu = 57 s). The gain there is K ÷ √(1 + (ωu τ)²), so the ultimate gain is Ku = √(1 + (ωu τ)²) ÷ K = 6.61.
  • A simulation with a unit setpoint step (heater limits removed) confirms it:
KpGainRinging amplitude at 600 to 1 000 sAmplitude at 2 600 to 3 000 sVerdict
3.300.50 × Ku0.0000.000settles: stable
5.280.80 × Ku0.0050.000settles: stable
6.280.95 × Ku0.7450.033decays slowly: nearly marginal
6.941.05 × Ku41.12122 169.187grows: unstable

Well below Ku the ringing dies away quickly, just below it very slowly, and at about Ku and above it grows (period about 57 s) until the actuator saturates; the simulation’s half-second steps add a trifle of delay, so it tips over a hair below the exact value. This is the basis of the ultimate-gain tuning method of PID Control and Tuning. Note how close the gain limit is: with Kp = 6.6 the oven oscillates, a P-only gain of 2 is a gain margin of 3.3.

Delay and non-minimum-phase systems

Dead time is the commonest cause of loops going unstable: it costs phase without reducing the gain. For the lambda-tuned oven loop (crossover at 0.0333 rad/s) each extra second of delay takes 1.91° off the phase margin:

Total dead timePhase marginVerdict
15 s61°comfortable
25 s42°rings
35 s23°barely stable
55 s-15°unstable

The loop reaches the stability limit when the extra delay is about 32 s (the delay margin). Slow sensors, long pipes, filters and slow sampling all add to the delay (Digital Control and Sampling). Non-minimum-phase systems (a right-half-plane zero, as in a boiler drum level where the water first rises when steam is increased, or an aircraft’s elevator) show an inverse response: the output first goes the wrong way. They limit the bandwidth to about half the zero’s frequency and no controller can remove the limit.

Stability margins in practice

  • Design for a phase margin of 45° to 60° and a gain margin of at least 2: the process changes (valves wear, fluids change, loads vary), so a loop that is only just stable on paper will oscillate in service.
  • Test the margins on the real loop by increasing the gain by the factor you want as a margin, and check that the loop remains stable; or by adding a delay.
  • Nonlinearities (saturation, backlash, stiction) can cause limit cycles in loops that the linear theory says are stable (Tuning and Troubleshooting Control Loops).
  • Safety: an unstable controller can drive a machine to its limits. Always have independent limits and protection that do not rely on the control loop (see System Integration).

Related

FAQ

Can a stable open-loop system become unstable in closed loop?

Yes, if the loop gain is high enough at a frequency where the phase reaches −180°. Feedback can stabilize an unstable plant and destabilize a stable one.

What is a limit cycle?

A steady oscillation of fixed amplitude caused by a nonlinearity, such as a valve with stiction or an on-off controller. Linear theory predicts growth or decay; the nonlinearity limits the amplitude.

Why is the Routh test used?

It gives the range of a gain for which a loop is stable, directly, without finding the roots. Software now finds the roots easily, but the idea of the stability boundary is still the one to understand.

What does an unstable pole in the plant mean?

The plant itself diverges (an inverted pendulum, a magnetically levitated mass). Feedback is essential, and there is a minimum gain as well as a maximum.

References

  • Ogata, K., Modern Control Engineering, Pearson (stability, Routh, root locus); Nise, N. S., Control Systems Engineering, Wiley.
  • Franklin, G. F., et al., Feedback Control of Dynamic Systems, Pearson.
  • Dorf, R. C., Bishop, R. H., Modern Control Systems, Pearson.

Last reviewed: October 2026. The simulated oven and the textbook examples are illustrative.