Root locus technique

Why it matters

Turning up the gain of a position loop makes it faster and more accurate — until it overshoots, rings and finally goes unstable. The root locus shows exactly how the closed-loop poles travel as the gain changes, so you can pick the gain that gives a required damping, see where stability is lost, and judge whether a compensator pole or zero is needed. It links the open-loop transfer function you know to the closed-loop behaviour you want.

Key ideas

What the root locus is. For the characteristic equation 1 + K·G(s)H(s) = 0, with K varied from 0 to ∞, the root locus is the set of paths traced in the s-plane by the closed-loop poles. It is drawn from the open-loop poles and zeros, which are usually known in factored form.

Angle and magnitude conditions. A point s lies on the locus if and only if the angle condition holds; the gain that puts a closed-loop pole there comes from the magnitude condition:

• Angle: (sum of angles from the open-loop zeros to s) − (sum of angles from the open-loop poles to s) = ±180°(2q + 1).
• Magnitude: K = (product of distances from the poles to s)/(product of distances from the zeros to s).

Construction rules (n open-loop poles, m zeros, K ≥ 0):

1. Branches. There are n branches (if n ≥ m). The locus is symmetric about the real axis.
2. Start and end. Branches start (K = 0) at the open-loop poles. m of them end (K → ∞) at the finite zeros; the other n − m go to infinity along asymptotes.
3. Real axis. A real-axis point is on the locus if the total number of real poles and zeros to its right is odd.
4. Asymptotes. Angles θ = (2q + 1)·180°/(n − m), q = 0, 1, …, n − m − 1; they meet the real axis at the centroid σ_A.
5. Breakaway/break-in points. Write K = −1/(G(s)H(s)) and solve dK/ds = 0; keep only roots that lie on a real-axis segment of the locus (and give K > 0).
6. jω-axis crossing. Use Routh: the marginal gain makes the s¹ row zero, and the auxiliary equation gives the crossing frequency.
7. Angles of departure/arrival at complex poles/zeros from the angle condition.

Reading the locus. Constant-ζ lines are rays from the origin at angle cos⁻¹ζ to the negative real axis; where the locus cuts the ray, the magnitude condition gives the gain for that damping. Points further left mean faster decay.

Effect of added poles and zeros. Adding an open-loop pole pushes the locus to the right — less stable, slower; for example K/(s(s + 2)) is stable for all K, but K/(s(s + 2)(s + 4)) goes unstable at K = 48. Adding a left-half-plane zero pulls the locus to the left — more stable; this is the basis of PD and lead compensation.

Formulas

1 + K·G(s)H(s) = 0 — closed-loop characteristic equation; K ≥ 0.

∠G(s)H(s) = ±180°(2q + 1) — angle condition.

K = Π|s − pᵢ| / Π|s − zⱼ| — magnitude condition (pᵢ: open-loop poles; zⱼ: open-loop zeros).

θ_A = (2q + 1)·180°/(n − m) — asymptote angles.

σ_A = (Σ poles − Σ zeros)/(n − m) — centroid (real parts; s⁻¹).

dK/ds = 0, with K = −1/(G(s)H(s)) — breakaway and break-in points.

φ_dep = 180° − Σ(angles from other poles) + Σ(angles from zeros) — departure from a complex pole.

φ_arr = 180° + Σ(angles from poles) − Σ(angles from other zeros) — arrival at a complex zero.

Worked examples

Example 1 (standard). Sketch the root locus of G(s)H(s) = K/(s(s + 2)(s + 4)).

1. Poles 0, −2, −4; no zeros; n − m = 3 branches go to infinity.
2. Real-axis locus: between 0 and −2, and left of −4.
3. Asymptotes: 60°, 180°, 300°; centroid σ_A = (0 − 2 − 4)/3 = −2.
4. Breakaway: K = −(s³ + 6s² + 8s); dK/ds = 0 → 3s² + 12s + 8 = 0 → s = −0.845 or −3.155. Only −0.845 lies on the locus: breakaway at s = −0.845, with K = 0.845 × 1.155 × 3.155 = 3.08.
5. jω-crossing: s³ + 6s² + 8s + K = 0; s¹ row (48 − K)/6 = 0 → K_mar = 48; auxiliary 6s² + 48 = 0 → ω = 2.83 rad/s.
6. So for 0 < K < 3.08 all closed-loop poles are real (overdamped), for 3.08 < K < 48 the dominant pair is complex, and K > 48 is unstable.

Example 2 (GATE level, departure and break-in). For G(s)H(s) = K(s + 2)/(s² + 2s + 2), find the angle of departure from the pole at −1 + j1, the break-in point and the gain there.

1. Poles −1 ± j1; zero −2; n − m = 1, so one branch ends at −2 and the other goes to −∞ along 180°.
2. At −1 + j1: angle from zero −2 is tan⁻¹(1/1) = 45°; angle from the other pole −1 − j1 is 90°.
3. φ_dep = 180° − 90° + 45° = 135°.
4. Break-in: K = −(s² + 2s + 2)/(s + 2); dK/ds = 0 → (2s + 2)(s + 2) − (s² + 2s + 2) = 0 → s² + 4s + 2 = 0 → s = −2 ± √2.
5. The real-axis locus is left of −2, so break-in at s = −3.414.
6. K = (s² + 2s + 2)/|s + 2| at s = −3.414: (11.657 − 6.828 + 2)/1.414 = 6.828/1.414 = 4.83.
7. The complex branches lie on a circle centred at the zero (−2) with radius √2; the system is stable for all K > 0.

Common mistakes

• Counting only poles (or only zeros) to the right when testing a real-axis point — count both.
• Forgetting that n − m branches go to infinity; saying "all branches end at the zeros".
• Keeping a dK/ds = 0 root that is not on a real-axis segment of the locus.
• Using the magnitude condition without the gain K, or forgetting that K must be positive.
• Mixing up departure (subtract angles from other poles, add from zeros) and arrival (the opposite).
• Using the open-loop poles as closed-loop poles for K > 0.

For GATE ME

Expect: number of branches and asymptotes, centroid and asymptote angles; real-axis segments; breakaway point; jω-crossing gain and frequency (usually through Routh); the gain for a closed-loop pole at a given point; and conceptual MCQs on the effect of adding a pole or a zero. Practise the centroid and breakaway computations until they take under two minutes.

Quick check

1. G(s)H(s) = K/(s² + 4s + 5): what are the asymptote angles?
2. For K/(s(s + 2)(s + 4)), what K puts a closed-loop pole at s = −5?
3. How many branches does K(s + 1)/(s(s + 2)(s + 4)) have, and how many go to infinity?
4. Is the point s = −3 on the root locus of K/(s(s + 2)(s + 4))?
5. Does adding a left-half-plane zero tend to stabilise or destabilise a loop?

Answers: 1. 90° and 270°. 2. K = 5 × 3 × 1 = 15. 3. Three branches; two go to infinity. 4. No — two poles lie to its right (an even number). 5. Stabilise; it bends the locus to the left.