Short notes

Root Locus Short Notes

Vary the gain K in a unity-feedback system with G(s) = K/[s(s+2)(s+4)] from 0 to infinity and watch the three closed-loop poles move — at K=0 they start at the open-loop poles 0, −2, −4, and as K increases two of them move toward each other, break away from the real axis, and sweep into the right half-plane somewhere above K=48. That migration path of closed-loop poles plotted in the s-plane as K varies is the root locus, and reading it tells you the maximum stable gain before the poles cross into the right half-plane.

EEE, ECE

How it works

Root locus construction rules: locus starts at open-loop poles (K=0) and ends at open-loop zeros (K=∞). Number of branches equals number of open-loop poles n; (n−m) branches go to infinity along asymptotes, where m is the number of finite zeros. Asymptote angles = (2q+1)×180°/(n−m) for q = 0, 1, 2, ... Centroid of asymptotes σ_a = (Σpoles − Σzeros)/(n−m). Points on the real axis are on the locus if the total number of open-loop poles and zeros to the right is odd. Breakaway point: solve dK/ds = 0 where K = −1/G(s)H(s).

Key points to remember

For G(s) = K/[s(s+2)(s+4)]: n=3, m=0, three asymptotes at angles 60°, 180°, 300°, centroid = (0+−2+−4−0)/3 = −2. Gain at a specific point s₀ on the locus: K = 1/|G(s₀)|. The jω-axis crossing (marginally stable gain) is found by substituting s = jω into the characteristic equation and solving for ω and K — or directly from the Routh array's auxiliary polynomial. Adding a zero to G(s) pulls the locus toward the left half-plane (improving stability); adding a pole pushes it rightward. PD control effectively adds a zero, while PI control adds a pole at the origin.

Exam tip

The examiner always asks you to sketch the root locus of a given G(s)H(s) showing asymptotes, centroid, real-axis segments, and the jω crossing — compute the centroid numerically and find the breakaway point using dK/ds = 0.

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