Root locus technique

Why it matters

When you turn up the gain of a controller, the closed-loop poles move — and with them the damping, speed and eventually the stability of the loop. The root locus is a map of that movement. With a quick sketch you can see the gain range for stability, the gain that gives a required damping ratio, and whether a compensator pole or zero will help, before running a single simulation.

Key ideas

What the root locus is: the set of all closed-loop pole locations as the gain K of G(s)H(s) = K·N(s)/D(s) varies from 0 to ∞. The closed-loop poles satisfy 1 + K·N(s)/D(s) = 0, i.e. D(s) + K·N(s) = 0.

Angle and magnitude conditions

• Angle condition (decides whether a point is on the locus): ∠G(s)H(s) = ±180°(2q + 1), i.e. Σ(angles from zeros) − Σ(angles from poles) = odd multiple of 180°.
• Magnitude condition (gives the gain at a point on the locus): |K·N(s)/D(s)| = 1, so K = (product of distances from the poles)/(product of distances from the zeros).

Construction rules (for K ≥ 0, with n open-loop poles and m zeros, n ≥ m)

1. Number of branches = n. The locus is symmetric about the real axis.
2. Start and end: branches start at the open-loop poles (K = 0) and end at the open-loop zeros (K → ∞); n − m branches go to infinity.
3. Real-axis segments: a point on the real axis is on the locus if the total number of real poles and zeros to its right is odd.
4. Asymptotes: n − m of them, at angles (2q + 1)·180°/(n − m), q = 0, 1, …, n − m − 1, meeting the real axis at the centroid σa.
5. Breakaway / break-in points: on real-axis segments where dK/ds = 0, with K = −D(s)/N(s). A root is valid only if it lies on the locus and gives K > 0.
6. Imaginary-axis crossing: from Routh (critical gain and auxiliary equation) or by putting s = jω into the characteristic equation and equating real and imaginary parts to zero.
7. Angle of departure from a complex pole: φd = 180° − Σ(angles from other poles) + Σ(angles from zeros). Angle of arrival at a complex zero: φa = 180° − Σ(angles from other zeros) + Σ(angles from poles).

Reading the locus

• Lines from the origin at angle θ = cos⁻¹ ζ to the negative real axis are lines of constant damping ratio. Where the locus crosses such a line, the magnitude condition gives the gain for that ζ.
• Adding a pole (e.g. an integrator or a lag) bends the locus to the right, toward instability, and reduces the critical gain.
• Adding a zero (e.g. derivative or lead action) bends the locus to the left, improving relative stability and speed.

Formulas

• D(s) + K·N(s) = 0 — closed-loop characteristic equation.
• Σ∠(s − zi) − Σ∠(s − pj) = ±180°(2q + 1) — angle condition.
• K = Π|s − pj| / Π|s − zi| — magnitude condition.
• θa = (2q + 1)·180°/(n − m) — asymptote angles.
• σa = (Σ poles − Σ zeros)/(n − m) — centroid (real parts).
• dK/ds = 0, with K = −D(s)/N(s) — breakaway and break-in points.
• φd = 180° − Σθp + Σθz — departure angle (angles measured at the complex pole).

Symbols: pj open-loop poles, zi open-loop zeros (in s⁻¹, i.e. rad/s), n, m their counts, K the root-locus gain (in the units that make KN/D dimensionless). The rules here are for negative feedback and K ≥ 0; for K < 0 (complementary locus) the angle condition becomes an even multiple of 180°.

Worked examples

Example 1 — sketching a type-1 third-order locus Given: G(s)H(s) = K/[s(s + 2)(s + 4)].

1. Poles at 0, −2, −4; no zeros; n − m = 3 branches go to infinity.
2. Real-axis segments: between 0 and −2 (one pole to the right) and left of −4 (three poles to the right).
3. Asymptotes at 60°, 180°, 300°; centroid σa = (0 − 2 − 4)/3 = −2.
4. Breakaway: K = −(s³ + 6s² + 8s), dK/ds = −(3s² + 12s + 8) = 0 → s = −0.845 or −3.155. Only −0.845 lies on the segment (0, −2). Gain there: K = 0.845 × 1.155 × 3.155 = 3.08.
5. jω crossing: characteristic equation s³ + 6s² + 8s + K = 0; Routh gives K = 48 with 6s² + 48 = 0 → ω = 2.83 rad/s.

Result: branches leave 0 and −2, break away at s = −0.845 (K = 3.08), follow the ±60° asymptotes and cross into the RHP at ±j2.83 rad/s when K = 48; the third branch runs from −4 to −∞.

Example 2 — departure angle and break-in (GATE level) Given: G(s)H(s) = K(s + 1)/(s² + 4s + 5), poles at −2 ± j1, zero at −1.

1. At the pole −2 + j1: angle from the other pole −2 − j1 is 90°; angle from the zero −1 (vector −1 + j1) is 135°.
2. φd = 180° − 90° + 135° = 225°, i.e. −135°: the branch leaves heading down and to the left.
3. Real-axis locus: left of −1 (one zero to the right).
4. Break-in: K = −(s² + 4s + 5)/(s + 1); dK/ds = 0 → (2s + 4)(s + 1) − (s² + 4s + 5) = s² + 2s − 1 = 0 → s = −1 − √2 = −2.414 (the other root, −1 + √2, is not on the locus).
5. Gain at break-in: K = −(5.828 − 9.657 + 5)/(−1.414) = 0.828.
6. Characteristic equation s² + (4 + K)s + (5 + K): all coefficients positive for any K > 0, so the loop is stable for all positive gain.

Result: φd = −135° (225°), break-in at s = −2.414 with K = 0.828, stable for all K > 0. (The complex part of this locus is a circle of radius √2 about the zero at −1.)

Common mistakes

• Using the sum of zeros in place of the sum of poles in the centroid, or dividing by n instead of n − m.
• Accepting a dK/ds = 0 root that is not on a real-axis locus segment, or that gives negative K.
• Writing the departure angle as 180° − (Σθp − Σθz) with the signs swapped, which gives +135° instead of −135° in Example 2.
• Counting complex poles when testing real-axis segments — complex pairs contribute zero net angle there.
• Believing that adding a pole improves stability; it pushes the locus toward the RHP.

For GATE IN

Expect questions on the number of asymptotes and their centroid, breakaway points, departure angles, the jω-axis crossing gain and frequency, the gain for a given damping ratio via the magnitude condition, and the effect of adding poles or zeros. Practise checking each candidate breakaway point against the real-axis rule, and combining Routh with the locus for crossing points.

Quick check

1. G(s)H(s) = K/[s(s + 1)(s + 3)]. Where is the centroid?
2. How many asymptotes does K(s + 2)/[s²(s + 5)] have, and at what angles?
3. For K/[s(s + 4)], find K for ζ = 0.5.
4. Does adding a zero in the left half-plane generally pull the locus left or right?

Answers: 1. σa = (0 − 1 − 3)/3 = −1.33; 2. Two, at ±90°; 3. Closed loop s² + 4s + K: 2ζωn = 4 → ωn = 4, K = 16; 4. Left.