Root Locus

The root locus plots how closed-loop poles move as a gain varies. Learn the sketching rules, asymptotes, breakaway points and jω crossings, and how to pick a gain for a desired damping ratio.

Changing the controller gain moves the closed-loop poles. The root locus is the path traced by every pole as K goes from 0 to ∞. From a single sketch you can see the gain range for stability, where the response becomes oscillatory, and which gain gives the damping you want.

Evans' rules let you sketch it by hand in minutes; software draws it exactly.

An analogy

The root locus is like a map of every route a car could take as you turn one steering knob — you pick the spot on the map that gives the ride you want, then read off the knob setting.

Start and end

Branches start (K = 0) at the open-loop poles and end (K → ∞) at the open-loop zeros or head to infinity along asymptotes. There are as many branches as poles, and the locus is symmetric about the real axis.

Real-axis segments and asymptotes

A real-axis point lies on the locus if the number of real poles and zeros to its right is odd. The (n − m) branches going to infinity follow straight asymptotes radiating from the centroid σ_a at angles (2q + 1)·180°/(n − m).

Breakaway and jω crossings

Where two branches meet on the real axis they break away into the complex plane; solve dK/ds = 0. Crossings of the imaginary axis mark the stability limit; find them from the Routh array (critical K and ω).

Design with the locus

Draw radial lines of constant ζ (angle arccos ζ from the negative real axis) and circles of constant ω_n. The intersection with the locus gives the gain for that damping. If the locus doesn't pass through the desired region, add a compensator: lead (zero–pole pair) pulls the locus left for speed and damping; lag improves steady-state error without moving the locus much.

Formulas

Worked example: Sketch rules for K/(s(s + 1)(s + 5))

Given: Poles 0, −1, −5; no zeros

Find: Asymptotes, real-axis segments, breakaway and jω crossing.

Answer: Breakaway −0.47, asymptotes from −2 at ±60°/180°, crosses jω at K = 30

Worked example: Gain for ζ = 0.5

Given: G(s) = K/(s(s + 4)), unity feedback

Find: K for ζ = 0.5 and the resulting ω_n.

Answer: K = 16

Common mistakes

Applications

Key takeaways

Going further