Question 7 of 8: Root-locus design, gain selection, step specs
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, 16-Elec-A2 Systems & Control, May 2017, 3 hours, closed book (approved Casio/Sharp calculator plus one signed, double-sided 8.5 × 11" formula sheet; a Laplace-transform table and the standard $\zeta$–overshoot and $\zeta$–resonant-peak design plots are supplied on pages 2–3). Questions 1 and 2 are compulsory; a complete paper is five questions, so a candidate chooses three of Q3–Q8. Each question is worth 20 marks. All eight are worked below so the set is a complete study resource.
Reference texts: N. S. Nise, Control Systems Engineering (7th ed., Wiley) — Routh–Hurwitz (Ch. 6), root locus (Ch. 8), steady-state error and static error constants (Ch. 7), frequency response, Nyquist, gain/phase margins (Ch. 10), PID and compensator design (Ch. 9–11), state space, controllability/observability and Mason’s rule (Ch. 3, 5, 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling, second-order correlations and pole-placement. All block diagrams, root loci, pole–zero maps, Nyquist plots, step responses and the state-realization diagrams below are redrawn as inline figures.
Reading the supplied design charts. The percent-overshoot chart uses $PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}$ and the second-order model is $G_m(s)=K_{dc}\,\dfrac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$. Every result here is derived analytically in the $s$-domain and cross-checked; the design charts are used only to corroborate $\zeta$ from $PO$.
Question 7 — Root-locus design, gain selection, step specs [20]
Given. open-loop poles $-2,-4,-7$; no finite zeros. Find. the locus geometry and marginal gain.
Asymptotes. $n-m=3$ branches to infinity at $\pm60^\circ,180^\circ$; centroid $\sigma_a=\dfrac{-2-4-7}{3}=\boxed{-4.33}$.
Break-away. With $K_p=-\tfrac1{10}(s+2)(s+4)(s+7)$, $\dfrac{dK_p}{ds}=0$ gives $3s^2+26s+50=0$, roots $-2.88$ and $-5.79$. Only $s=-2.88$ lies on the real-axis locus segment (between $-2$ and $-4$), so $\boxed{s_b=-2.88}$ with break-away gain $K_b=0.41$. ($-5.79$ sits between $-4$ and $-7$, which is not on the locus, so it is rejected — there is no break-in.)
jw-axis crossover. Characteristic $(s+2)(s+4)(s+7)+10K_p=s^3+13s^2+50s+56+10K_p=0$; the Routh $s^1$ row is $\dfrac{594-10K_p}{13}$. Marginal at $\boxed{K_{crit}=59.4}$; the $s^2$ auxiliary $13s^2+650=0$ gives $\omega_{osc}=\sqrt{50}=\boxed{7.07\ \text{rad/s}}$.
[Figure not reproduced: Q7.2 redrawn: real-axis segments $[-4,-2]$ and $(-\infty,-7]$; break-away at $-2.88$; three asymptotes from centroid $-4.33$; jw crossing $\pm j7.07$ at $K_{crit}=59.4$. See the official exam paper.]
Item (2) — gain for $PO\approx5\%$ and step specs
Damping from overshoot. $\zeta=\dfrac{-\ln(0.05)}{\sqrt{\pi^2+\ln^2 0.05}}=\boxed{0.690}$.
Gain on the $\zeta=0.69$ ray. Intersecting the locus with the $\cos^{-1}0.69=46.4^\circ$ ray (angle condition $\sum\angle(s-p_i)=180^\circ$) gives the dominant pair $s=-2.32\pm j2.43$ ($\omega_n=3.35$ rad/s); the magnitude condition $K_p=\tfrac1{10}\prod|s-p_i|$ there yields $\boxed{K_{op}=3.82}$, with the third (real) pole at $\approx-8.4$.
Closed-loop poles at $K_{op}=3.82$: dominant pair on the $\zeta=0.69$ ray ($-2.32\pm j2.43$) plus a real pole near $-8.4$ ($\sim2.6\times$ further left).
Settling time ($\pm5\%$). $\sigma=\zeta\omega_n=2.32$, $T_{settle(5\%)}=\dfrac{3}{\sigma}=\boxed{1.30\ \text{s}}$.
Rise time (0–100\%). $T_{rise}=\dfrac{\pi-\beta}{\omega_d}$ with $\beta=\cos^{-1}\zeta=0.808$ rad and $\omega_d=2.43$: $T_{rise}=\boxed{0.96\ \text{s}}$.
Steady-state error. Type 0 plant, $K_{pos}=K_{op}\,G(0)=3.82\cdot\dfrac{10}{56}=0.682$, so $e_{ss(step)}=\dfrac{1}{1+K_{pos}}=\boxed{59.5\%}$ — large, because there is no integrator and $K_{op}$ is modest.
Item (3) — dominant model vs actual
The third pole at $\approx-8.4$ is only $\sim2.6\times$ further from the jw-axis than the dominant pair (well short of the usual $5\times$ rule), so it is not negligible: the actual response overshoots slightly less than the $5\%$ target and rises a little more slowly than the pure second-order estimate. There are no closed-loop zeros to speed it up, so the dominant-pole model is a good but mildly optimistic predictor. The very large steady-state error would in practice motivate adding integral action.
