Question 8 of 8: Root Locus Analysis, Gain Selection and Response Estimation

Question 8: Root Locus Analysis, Gain Selection and Response Estimation (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.

Find. The complete root-locus geometry (asymptote angles and centroid, break points, imaginary-axis crossing and critical gain); the gain meeting the 5% overshoot target with the associated settling time, rise time and steady-state error; and a critique of the dominant-pole prediction.

Approach. Apply the standard construction rules, then solve $dK/ds = 0$ for the break points and test the Routh array for imaginary-axis crossings; select the gain where the locus meets the required damping ray; and finally check the third pole and the zero before trusting the estimates.

1. Count branches and locate real-axis segments. With $n = 3$ poles and $m = 1$ zero there are three branches; one terminates on the finite zero at $-3$ and the other two run to infinity. A point on the real axis belongs to the locus when the number of real poles and zeros strictly to its right is odd. That gives the segments $$[-3,\,-2] \quad\text{and}\quad [-10,\,-5].$$ The interval $[-5,-3]$ has two singularities to its right and so is excluded.
2. Asymptote angles and centroid. The two branches heading to infinity follow asymptotes at $$\theta_a = \frac{(2q+1)180^\circ}{n - m} = \frac{180^\circ}{2},\ \frac{540^\circ}{2} = \pm 90^\circ,$$ radiating from the centroid $$\sigma_a = \frac{\sum p_i - \sum z_j}{n - m} = \frac{(-2 - 5 - 10) - (-3)}{2} = \frac{-14}{2} = -7.$$ So the pair leaves the real axis and climbs vertically near $\mathrm{Re}\,s = -7$.
3. Break points. On the locus $K_p = -1/G(s)$, i.e. $$K_p(s) = -\frac{(s+2)(s+5)(s+10)}{5(s+3)}.$$ Setting $dK_p/ds = 0$ and solving gives a single real root inside a locus segment, $$\boxed{s_b = -7.367 \text{ (break-away)}, \qquad K_p(s_b) = 1.53.}$$ It lies in $[-10,-5]$, so the two branches from the poles at $-5$ and $-10$ meet there at gain 1.53 and depart perpendicular to the real axis. The segment $[-3,-2]$ carries no break point: that branch simply travels from the pole at $-2$ to the zero at $-3$ as the gain rises.
4. Test for an imaginary-axis crossing. The closed-loop characteristic polynomial is $$1 + K_p G(s) = 0 \;\Rightarrow\; s^3 + 17 s^2 + (80 + 5K_p)s + (100 + 15K_p) = 0.$$ Its Routh array has first column $1$, $17$, $\dfrac{17(80+5K_p) - (100+15K_p)}{17} = \dfrac{1260 + 70K_p}{17}$, and $100 + 15K_p$. Every one of these is strictly positive for all $K_p \gt 0$; the $s^1$ entry never vanishes.
5. State the stability conclusion. Because no Routh entry changes sign, $$\boxed{\text{the locus never crosses the imaginary axis: the loop is stable for every } K_p \gt 0.}$$ There is therefore no finite $K_{crit}$ and no $\omega_{osc}$ — formally $K_{crit} \to \infty$. The geometry says the same thing: the two infinite branches follow vertical asymptotes anchored at $\sigma_a = -7$, deep in the left half-plane, so they can never reach the axis. This is a legitimate and complete answer to the part of the question that asks for the crossover coordinates.
6. Part 2 — convert the overshoot target into a damping ratio. With $PO = 5\%$ and $L = \ln 0.05 = -2.9957$, $$\zeta = \frac{2.9957}{\sqrt{9.8696 + 8.9744}} = \frac{2.9957}{4.3410} = 0.690.$$ The design point is where the locus crosses the ray at $\theta = \cos^{-1} 0.690 = 46.4^\circ$ from the negative real axis.
7. Find the gain at that point. Sweeping $K_p$ and factoring the cubic, the complex pair reaches $\zeta = 0.690$ at $$\boxed{K_{op} = 13.12,}$$ where the closed-loop roots are $s_{1,2} = -7.097 \pm j\,7.443$ and $s_3 = -2.806$. The dominant pair therefore has $\omega_n = |s_1| = 10.28$ rad/s and $\sigma = \zeta\omega_n = 7.097$ s$^{-1}$.
8. Settling time and rise time. Using the five-percent criterion requested, $$T_{s(\pm 5\%)} = \frac{3}{\zeta\omega_n} = \frac{3}{7.097} = 0.423\ \text{s},$$ and with $\omega_d = 7.443$ rad/s the rise time is $$T_{r(0-100\%)} = \frac{\pi - \cos^{-1}\zeta}{\omega_d} = \frac{3.1416 - 0.8097}{7.443} = 0.313\ \text{s}.$$
9. Steady-state error. The loop is Type 0, so $$K_{pos} = K_{op}\,G(0) = 13.12 \times \frac{5 \times 3}{2 \times 5 \times 10} = 13.12 \times 0.15 = 1.968,$$ and $$e_{ss(step)\%} = \frac{100}{1 + K_{pos}} = \frac{100}{2.968} = 33.7\%.$$ That is a very large error — the inevitable consequence of choosing the gain purely for damping on a Type-0 plant.
10. Part 3 — where the model and reality part company. Three effects will make the measured response differ from the second-order prediction. First, the third closed-loop pole at $-2.806$ is actually closer to the imaginary axis than the complex pair at $-7.097$, so the usual dominance argument is inverted; the only thing that rescues the model is the open-loop zero at $-3$, which sits just $0.194$ away from that pole and very nearly cancels it. Second, that near-cancellation is not exact, so a small slow exponential tail rides on the fast oscillation and stretches the effective settling time beyond the predicted 0.42 s. Third, the closed-loop zero itself adds derivative action, which tends to increase overshoot slightly above the nominal 5% while quickening the initial rise. In short, the predicted specifications are sound as engineering estimates but should be treated as approximate, and the large 33.7% steady-state error — which the model reports correctly — is the real limitation of proportional-only control here.