Question 7 of 8: Root-locus design, gain for $\zeta=0.5$, 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, December 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, $\zeta$–resonant-peak and phase-margin 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), lead/lag and PID design (Ch. 9–11), state space and controllability/observability (Ch. 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling, second-order correlations, pole placement and frequency-domain compensator design. All block diagrams, root loci, pole–zero maps, Bode and Nyquist plots, closed-loop magnitude curves and step responses below are redrawn as inline figures.
Reading the supplied design charts. The percent-overshoot chart uses $PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}$, the phase-margin chart uses $\Phi_m\approx100\,\zeta$, 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 from the exact transfer functions given in the paper; the supplied plots are used only to corroborate $\zeta$, $\Phi_m$ and the crossover frequencies. Where a printed figure’s hand-read tick differs from the exact value, the exact value governs and the gap is flagged.
Question 7 — Root-locus design, gain for $\zeta=0.5$, step specs [20]
Figure Q7.1 loop — proportional $K_p$ ahead of $G(s)=100/[(s+1)(s+2)(s+5)]$.
Given. open-loop poles $-1,-2,-5$; no finite zeros; loop gain $100K_p$. Find. the locus geometry, the $\zeta=0.5$ gain, the gain margin and step specs.
Item (1) — root locus
Asymptotes. $n-m=3$ branches to infinity at $\pm60^\circ,180^\circ$; centroid $\sigma_a=\dfrac{-1-2-5}{3}=\boxed{-2.67}$.
Break-away. $\dfrac{dK_p}{ds}=0$ on $(s+1)(s+2)(s+5)=s^3+8s^2+17s+10$ gives $3s^2+16s+17=0$, roots $-1.46$ and $-3.87$. Only $s=-1.46$ lies on the real-axis segment $[-2,-1]$, so $\boxed{s_b=-1.46}$ (no break-in; $-3.87$ is not on the locus).
jw crossover. Characteristic $s^3+8s^2+17s+10+100K_p=0$; the Routh $s^1$ row is $(126-100K_p)/8$, zero at $\boxed{K_{crit}=1.26}$, and the $s^2$ auxiliary $8s^2+136=0$ gives $\omega_{osc}=\sqrt{17}=\boxed{4.12\ \text{rad/s}}$.
[Figure not reproduced: Q7.1 redrawn: real-axis segments $[-2,-1]$ and $(-\infty,-5]$; break-away $-1.46$; three asymptotes from centroid $-2.67$; jw crossing $\pm j4.12$ at $K_{crit}=1.26$. See the official exam paper.]
Item (2) — gain for $\zeta=0.5$ and gain margin
Damping ray. $\zeta=0.5\Rightarrow$ the pole lies on the $\cos^{-1}0.5=60^\circ$ ray. Enforcing the angle condition $\sum\angle(s-p_i)=180^\circ$ on that ray gives the dominant pair $s=-1.06\pm j1.84$ ($\omega_n=2.12$ rad/s).
Operating gain. The magnitude condition $K_{op}=\tfrac{1}{100}\prod|s-p_i|$ there yields $\boxed{K_{op}=0.165}$, with the third (real) pole at $-5.87$.
Gain margin. $G_m=\dfrac{K_{crit}}{K_{op}}=\dfrac{1.26}{0.165}=\boxed{7.63}\;(17.7\ \text{dB}).$
Closed-loop poles at $K_{op}=0.165$: dominant pair on the $\zeta=0.5$ ray ($-1.06\pm j1.84$) plus a real pole near $-5.87$ ($\sim5.5\times$ further left).
Second-order model. Keep the dominant pair: $\omega_n=\boxed{2.12\ \text{rad/s}}$, $\zeta=0.5$; the model DC gain equals the loop DC gain $K_{dc}=\dfrac{K_{pos}}{1+K_{pos}}=\dfrac{1.65}{2.65}=\boxed{0.623}$ ($K_{pos}=K_{op}\cdot100/10=1.65$). Thus $G_m(s)=\dfrac{0.623(4.51)}{s^2+2.13s+4.51}=\dfrac{2.81}{s^2+2.13s+4.51}$.
Specs. $PO=100e^{-0.5\pi/\sqrt{0.75}}=\boxed{16.3\%}$; $T_{settle(\pm2\%)}=4/(\zeta\omega_n)=\boxed{3.77\ \text{s}}$; $T_{rise(0-100\%)}=(\pi-\cos^{-1}\zeta)/\omega_d=\boxed{1.14\ \text{s}}$ ($\omega_d=1.84$); $e_{ss(step)}=1/(1+1.65)=\boxed{37.7\%}$ — large, since the plant is Type 0 and $K_{op}$ is small.
Dominant-model unit-step response at $K_{op}=0.165$: $16.3\%$ overshoot, final value $0.623$ (steady-state error $37.7\%$).
Item (3), comment. The third pole at $-5.87$ is only $\sim5.5\times$ deeper than the dominant pair — right at the usual dominance threshold — so the true response overshoots slightly less and rises a little slower than the pure second-order estimate; the large steady-state error is exact.
