Question 3 of 8: Root locus and series-PID settling-time design

Question 3 — Root locus and series-PID settling-time design [7 + 8 + 5]

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.

Part A — Detailed root locus [7]

Approach. Reduce the series PID to a single open-loop transfer function, then apply the standard rules (real-axis segments, asymptotes, break-away/break-in, jω crossings).

1. Open-loop transfer function. $C(s)=K_p\dfrac{(0.1s+1)(0.125s+1)}{0.1s}=0.125K_p\dfrac{(s+10)(s+8)}{s}$, so $$L(s)=C(s)G(s)=K'\,\frac{(s+8)(s+10)}{s(s+2)(s+15)},\qquad K'=0.125K_p.$$ Open-loop poles $0,-2,-15$; open-loop zeros $-8,-10$.
2. Real-axis segments. Left of an odd count of real singularities: $[-2,0]$, $[-10,-8]$, and $(-\infty,-15]$.
3. Asymptotes and centroid. $n-m=3-2=1$, so a single asymptote at $180^\circ$ (the negative-real axis); the centroid $\sigma_a=\dfrac{(0-2-15)-(-8-10)}{1}=+1$ is not meaningful for a single asymptote.
4. Break-away / break-in. Solving $\dfrac{d}{ds}\!\left[\dfrac{s(s+2)(s+15)}{(s+8)(s+10)}\right]=0$ gives a break-away at $s\approx-1.09$ (on $[-2,0]$) and a break-in at $s\approx-8.96$ (on $[-10,-8]$).
5. Imaginary-axis crossings. The characteristic polynomial $s^3+(17+K')s^2+(30+18K')s+80K'=0$ has Routh $s^1$ entry $\dfrac{18K'^2+256K'+510}{17+K'}\gt 0$ for all $K'\gt 0$, so no branch ever crosses the $j\omega$-axis. The loop is stable for every $K_p\gt 0$; there is no finite critical gain ($K_{crit}\to\infty$, $\omega_{osc}$ not applicable).

Part B — Gain for 0.5 s settling and equivalent damping [8]

Approach. $T_{settle(\pm2\%)}=4/(\zeta\omega_n)=0.5\ \text{s}$ fixes the dominant real part at $\sigma=\zeta\omega_n=8$; find the point on the complex arc with $\mathrm{Re}(s)=-8$ and the gain that puts a closed-loop pole there.

1. Required real part. $\sigma=4/T_{settle}=4/0.5=8$, so the dominant closed-loop poles must sit on the vertical line $\mathrm{Re}(s)=-8$.
2. Gain from the magnitude criterion. Enforcing a root at $s=-8+j\omega_d$ on the locus gives $K'=0.125K_p=31.8$, i.e. $$\boxed{K_p\approx254}.$$ The three closed-loop poles are then $-8\pm j3.68$ and $-32.8$.
3. Equivalent damping ratio. For the dominant pair, $\omega_n=\sqrt{8^2+3.68^2}=8.81\ \text{rad/s}$ and $$\zeta=\frac{\sigma}{\omega_n}=\frac{8}{8.81}=\boxed{0.91}.$$

Part C — Accuracy of the dominant-poles model [5]

Discussion. The assumed second-order dominant-poles model is only partially accurate here. Two effects distort it: (i) the third closed-loop pole at $-32.8$ is about four times deeper than the dominant real part $-8$, so it is genuinely fast and contributes little — this part of the model assumption is sound; but (ii) the closed-loop zeros inherited from the PID sit at $-8$ and $-10$, right on top of the dominant poles. The zero at $-8$ nearly coincides with the dominant real part, so it strongly reshapes the transient. Compared with the ideal $\zeta=0.91$ (essentially no-overshoot) model, the real response will show a noticeably faster rise time and a larger percent overshoot than the model predicts (a nearby zero always adds overshoot and speeds the rise), while the settling time stays close to the 0.5 s target because it is governed by the dominant real part $-8$, which the zeros do not move. The dominant-poles model therefore predicts settling time well but underestimates overshoot and rise speed.