Question 5 of 8: Root locus, gain selection, step specs
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, 07-Elec-A2 Systems & Control, December 2016, 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 ζ–overshoot and ζ–resonant-peak design plots are supplied). 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, Bode, Nyquist, gain/phase margins (Ch. 10), lead–lag and PID design (Ch. 9–11), state space and Mason’s rule (Ch. 3, 5); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling and second-order correlations. All block diagrams, root loci, pole–zero maps, Bode sketches, Nyquist plots and the 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 $G_m(s)=K_{dc}\,\dfrac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$. Questions 4, 6 and 7 are read from printed Bode/response plots, so those values carry normal chart-reading tolerance; the s-domain results (Q1, Q2, Q3, Q5, Q8) are exact.
Question 5 — Root locus, gain selection, step specs [20]
Figure Q5.1 — proportional control of $G(s)=50/[s(s+5)(s+10)]$.
Item (1) — root locus
Given. open-loop poles $0,-5,-10$; no finite zeros. Find. the locus geometry and marginal gain.
Asymptotes. $n-m=3$ branches to infinity at $\pm60^\circ,180^\circ$; centroid $\sigma_a=\dfrac{0-5-10}{3}=\boxed{-5}$.
Break-away. With $K=-s(s+5)(s+10)/50$, $\dfrac{dK}{ds}=0$ gives $3s^2+30s+50=0$, roots $-2.11$ and $-7.89$. Only $s=-2.11$ lies on a real-axis locus segment (between $0$ and $-5$), so $\boxed{s_b=-2.11}$, with break-away gain $K_b=0.96$.
[Figure not reproduced: Q5.2 redrawn: real-axis segments $[-5,0]$ and $(-\infty,-10]$; break-away at $-2.11$; three asymptotes from centroid $-5$; jw crossing at $\pm j7.07$ ($K_{crit}=15$). See the official exam paper.]
Item (2) — gain for $PO\approx5\%$ and step specs
Damping from overshoot. $\zeta=\dfrac{-\ln(0.05)}{\sqrt{\pi^2+\ln^2 0.05}}=0.69$.
Gain on the $\zeta=0.69$ ray. Placing the dominant pair on the $\cos^{-1}0.69=46.4^\circ$ ray and solving the characteristic equation gives $\boxed{K_{op}=1.69}$, dominant pair $-1.89\pm j1.99$ ($\omega_n=2.74$) and a far third pole at $-11.2$.
Closed-loop poles at $K_{op}=1.69$: dominant pair on the $\zeta=0.69$ ray plus a well-separated real pole at $-11.2$ ($\sim5.9\times$ further left).
Settling time ($\pm5\%$). $\sigma=\zeta\omega_n=1.89$, $T_{settle(5\%)}=\dfrac{3}{\sigma}=\boxed{1.58\ \text{s}}$.
Rise time (0–100\%). $T_{rise}=\dfrac{\pi-\beta}{\omega_d}$ with $\beta=\cos^{-1}\zeta=0.810$ rad, $\omega_d=\omega_n\sqrt{1-\zeta^2}=1.99$: $T_{rise}=\boxed{1.17\ \text{s}}$.
Steady-state error. The plant has a free integrator (Type 1), so for a step input $e_{ss(step)}=\boxed{0\%}$.
Item (3) — dominant model vs actual
The third pole at $-11.2$ is only $\sim5.9\times$ further left than the dominant pair, so it is close enough to add a little extra lag: the actual response is slightly slower to rise and overshoots a touch less than the pure second-order estimate. There are no closed-loop zeros to speed the response up, so the dominant-pole model is a good but mildly optimistic predictor of speed.
