Question 3 of 8: Root Locus Analysis and Proportional Gain Selection

Question 3: Root Locus Analysis and Proportional Gain Selection (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Unit-feedback loop with proportional gain $K_p$ acting on $G(s) = (s+4)(s+8)/[s^2(s+6)]$: open-loop poles at $s = 0$ (double) and $s = -6$, open-loop zeros at $s = -4$ and $s = -8$, so $n = 3$ and $m = 2$. Target overshoot 5%.

Find. The full root-locus geometry (asymptote angles, centroid, real-axis segments, break points, and any imaginary-axis crossing with its critical gain); the gain $K_{op}$ giving 5% overshoot together with the resulting settling time, rise time and steady-state error; and a comment on how the true third-order response departs from the dominant-pair estimate.

Approach. Apply the standard root-locus construction rules, then use the Routh array to settle the question of an imaginary-axis crossing, search the locus for the gain whose complex pair carries $\zeta = 0.69$, and finally weigh the third pole and the two finite zeros against that pair to judge dominance.

1. Count branches and asymptotes. With $n = 3$ poles and $m = 2$ finite zeros there are three branches; two terminate on the finite zeros at $-4$ and $-8$ and the remaining $n-m = 1$ branch escapes to infinity. A single asymptote means the asymptote angle is $$\theta = \frac{(2q+1)180^\circ}{n-m} = 180^\circ$$ i.e. the escaping branch runs out along the negative real axis. The centroid formula still evaluates, to $$\sigma_a = \frac{\sum p_i - \sum z_j}{n-m} = \frac{(0+0-6)-(-4-8)}{1} = +6$$ though with only one asymptote this value simply anchors a horizontal line rather than a fan.
2. Identify the real-axis segments. A point on the real axis belongs to the locus when the count of real poles and zeros strictly to its right is odd. Counting rightwards from each interval gives: to the right of the origin, zero items (even, not on the locus); between $-4$ and $0$, the double pole at the origin counts twice (even, not on the locus); between $-6$ and $-4$, two poles plus the zero at $-4$ is three (odd, on the locus); between $-8$ and $-6$, four items (even, not on the locus); left of $-8$, five items (odd, on the locus). So the real-axis locus is the segment $[-6,\,-4]$ together with the ray $(-\infty,\,-8]$.
3. Departure from the double pole at the origin. Two branches leave the repeated pole at $s = 0$. For a pole of multiplicity two the departure angles are separated by $360^\circ/2 = 180^\circ$ and, by the angle criterion applied just off the origin, they leave vertically at $\pm 90^\circ$. These two branches swing into the left half plane as a complex pair.
4. Locate the break point. Break points satisfy $dK/ds = 0$ where $K(s) = -1/G(s) = -s^2(s+6)/[(s+4)(s+8)]$. Differentiating and clearing denominators leaves a quartic whose only admissible real root is $$s_b = -13.90$$ which does lie on the locus ray $(-\infty,-8]$. The associated gain is $$K_b = -\frac{s_b^2(s_b+6)}{(s_b+4)(s_b+8)} = 26.13$$ This is a break-in point: the complex pair that left the origin returns to the real axis there, after which one branch heads left to infinity and the other moves right to terminate on the zero at $-8$.
5. Test for an imaginary-axis crossing. The closed-loop characteristic equation is $$s^2(s+6) + K(s+4)(s+8) = s^3 + (6+K)s^2 + 12K\,s + 32K = 0$$ Building the Routh array, the $s^1$ element is $$\frac{(6+K)(12K) - 32K}{6+K} = \frac{K\,(12K + 40)}{6+K}$$ Every coefficient and this element are strictly positive for all $K \gt 0$, so no sign change ever occurs. Therefore $$\boxed{\text{there is no imaginary-axis crossing: } \omega_{osc} \text{ and } K_{crit} \text{ do not exist, and the loop is stable for every } K \gt 0}$$ This is consistent with the geometry: with a single asymptote at 180° and both finite zeros in the left half plane, no branch can ever reach the $j\omega$ axis.
Figure 3.1 — Root locus of G(s) = (s+4)(s+8)/[s^2(s+6)]. Crosses mark open-loop poles, circles the zeros. The branches leave the double pole at the origin vertically, break in at -13.90, and no branch ever reaches the imaginary axis, so the loop is stable for every positive gain.
6. Find the gain giving 5% overshoot. Inverting the overshoot relation, $$\zeta = \frac{-\ln(0.05)}{\sqrt{\pi^2+\ln^2(0.05)}} = 0.690$$ Searching the locus for the gain whose complex pair sits on the $\zeta = 0.690$ ray gives $$\boxed{K_{op} = 10.14}$$ at which the three closed-loop roots are $$s_{1,2} = -5.646 \pm j5.921, \qquad s_3 = -4.847$$ so the dominant pair has $\omega_n = |s_{1,2}| = 8.182$ rad/s and $\zeta = 5.646/8.182 = 0.690$ as required.
7. Estimate the transient specifications. Using the dominant pair, $$T_{s(\pm5\%)} = \frac{3}{\zeta\omega_n} = \frac{3}{5.646} = \boxed{0.531\ \text{s}}$$ $$T_{r(0-100\%)} = \frac{\pi - \cos^{-1}\zeta}{\omega_d} = \frac{\pi - 0.808}{5.921} = \boxed{0.394\ \text{s}}$$ Note the question asks for the $\pm5\%$ settling band, hence the numerator 3 rather than 4.
8. Steady-state error. The open loop carries $s^2$ in its denominator, so the system is Type 2. The position constant is infinite and $$e_{ss(step)} = \frac{1}{1+K_{pos}} = \boxed{0\%}$$ independently of the gain — the double integrator tracks a step (and indeed a ramp) with no residual error.
9. Compare the dominant model with the true response (part 3). Three effects make the actual response differ from the clean second-order estimate. First, the third pole at $-4.847$ lies closer to the imaginary axis than the real part of the "dominant" pair at $-5.646$, so on a naive separation test the pair is not dominant at all and the estimates above should be suspect. What rescues them is the open-loop zero at $-4$: it sits only 0.85 units from that third pole and very nearly cancels it, so the residue attached to the slow real mode is small and its contribution to the step response is correspondingly weak. Second, the surviving zeros at $-4$ and $-8$ differentiate the response, which typically raises the overshoot above the nominal 5% and shortens the rise time relative to the pole-only formula. Third, because the near-cancellation is imperfect, a small slowly decaying tail persists that the second-order model does not predict, so the true settling time tends to be slightly longer than 0.531 s. The estimates remain useful engineering approximations, but the dominance claim here rests on the pole–zero near-cancellation, not on pole separation, and that should be stated explicitly.