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22-Elec-A2 Systems and Control · May 2015

Question 2 of 8: Root-locus analysis and gain selection (compulsory)

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, 07-Elec-A2 Systems & Control, May 2015, 3 hours, closed book (approved Casio/Sharp calculator plus one signed, double-sided 8.5 × 11" formula/notes sheet; a Laplace-transform table and design plots are supplied with the paper). Questions 1 and 2 are compulsory; a complete paper is five questions, so candidates choose 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) — time response and second-order specifications (Ch. 4), block/signal-flow reduction and Mason’s rule (Ch. 5), Routh–Hurwitz stability (Ch. 6), steady-state error and static error constants (Ch. 7), root locus (Ch. 8), PID / lead–lag design (Ch. 9–11), frequency response, Bode, Nyquist, gain and phase margins (Ch. 10–11), state space, controllability/observability and pole placement (Ch. 3, 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling and Mason’s gain formula; G. F. Franklin, J. D. Powell & A. Emami-Naeini, Feedback Control of Dynamic Systems. All block diagrams, pole–zero maps, root loci, Bode plots, signal-flow graphs and the Nyquist plot below are redrawn as inline figures.

Reading the exam figures. Where a transfer function is stated exactly (Q1–Q5, Q7, Q8A) every result is confirmed analytically and the exact model governs. Q4 supplies a Bode plot ($G_m=16.1$ dB at $2.05$ rad/s, $P_m=-85.1^\circ$ at $0.589$ rad/s) and a root locus; Q5 Parts A–B and Q6 use printed frequency-response plots, and Q8B a printed Nyquist diagram — values read from those curves are flagged as reads and cross-checked against the analytics wherever a transfer function is available.

Question 2 — Root-locus analysis and gain selection (compulsory) [10 + 10]

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. Open loop $L(s)=\dfrac{10K_{op}}{(s+2)^2(s+10)}$: poles $-2,-2,-10$, no finite zeros ($n=3,m=0$), so three asymptotes.

Find. Locus geometry; $K_{op}$ for $PO=5\%$; and the resulting $e_{ss},T_{settle},$ gain margin.

Part A — Root-locus geometry [10]

Approach. Apply the real-axis, centroid, asymptote, break-point and Routh $j\omega$-crossing rules.

Root Locus G(s)=10/[(s+2)^2(s+10)]-10-50-55ReImσ=-4.67+j6.63 (Kop=57.6)-j6.63
Root locus of $10K_{op}/[(s+2)^2(s+10)]$: the double pole at $-2$ departs at $\pm90^\circ$; branches cross the imaginary axis at $\pm j6.63$ ($K_{op}=57.6$); the $-10$ branch runs left along the $180^\circ$ asymptote.
  1. Real-axis segments. Counting singularities to the right: the segment $(-\infty,-10]$ has an odd count (3) and is on the locus; the segment $(-10,-2)$ has an even count (2, the double pole) and is not. The double pole therefore departs directly into the complex plane.
  2. Centroid & asymptotes. $\sigma_a=\dfrac{(-2-2-10)-0}{3}=\boxed{-\tfrac{14}{3}=-4.667}$; asymptote angles $\dfrac{(2k+1)180^\circ}{3}=\boxed{\pm60^\circ,\,180^\circ}$.
  3. Break-away. With $K_{op}=-\tfrac{1}{10}(s+2)^2(s+10)$, $dK_{op}/ds=0$ gives $3s^2+28s+44=0\Rightarrow s=-2,\,-7.33$. Only $s=-2$ lies on the locus, so the $\boxed{\text{break-away is the double pole at }s=-2}$, departing at $\pm90^\circ$ (the root $-7.33$ is not on a locus segment and is rejected).
  4. Imaginary-axis crossing. Characteristic $s^3+14s^2+44s+(40+10K_{op})=0$; the Routh $s^1$ entry $\dfrac{576-10K_{op}}{14}=0$ gives $\boxed{K_{op,crit}=57.6}$, and the $s^2$ auxiliary $14s^2+(40+576)=0$ gives $s^2=-44$, i.e. $\boxed{\omega=\pm j6.63\text{ rad/s}}$.

Part B — Gain selection for 5% overshoot [10]

Approach. $PO=5\%\Rightarrow\zeta=0.690$; intersect the $\zeta$-ray with the locus (solve the characteristic cubic with a complex pole $-\zeta\omega_n\pm j\omega_n\sqrt{1-\zeta^2}$) to get $K_{op}$.

Closed-loop poles at Kop=3.0-10-50-55ReImzeta=0.69
Closed-loop poles at the chosen $K_{op}=3.0$: dominant pair $-1.79\pm j1.88$ ($\omega_n=2.59$, $\zeta=0.69$) with the third pole far out at $-10.42$ — the pair dominates ($10.42/1.79=5.8\gt5$).
  1. Damping. $\zeta=\dfrac{-\ln(0.05)}{\sqrt{\pi^2+\ln^2(0.05)}}=\boxed{0.690}$.
  2. Operational gain. Placing the pair on the $\zeta=0.69$ ray and matching the cubic gives the dominant poles $-1.79\pm j1.88$ ($\omega_n=2.59$ rad/s), third pole $-10.42$, and $\boxed{K_{op}=3.0}$.
  3. (a) Steady-state error. $L$ has no pole at the origin $\Rightarrow$ Type 0. $K_{pos}=\lim_{s\to0}L=\dfrac{10K_{op}}{4\cdot10}=\dfrac{K_{op}}{4}=0.75$, so $e_{ss}=\dfrac{1}{1+K_{pos}}=\dfrac{1}{1.75}=\boxed{57.1\%}$.
  4. (b) Settling time. $T_{settle}(\pm2\%)=\dfrac{4}{\zeta\omega_n}=\dfrac{4}{1.79}=\boxed{2.24\ \text{s}}$.
  5. (c) Gain margin. $\text{GM}=\dfrac{K_{op,crit}}{K_{op}}=\dfrac{57.6}{3.0}=\boxed{19.2\ \text{V/V}=25.7\ \text{dB}}$.
Check: large steady-state error. A 57% step error is correct for a Type-0 loop at this modest gain — it is the price of the low gain needed for $\zeta=0.69$. Raising $K_{op}$ would cut the error but reduce the damping; a lag or PI compensator (Q6 idea) is the proper fix.
QuantityResult
Centroid / asymptotes$-4.667$ / $\pm60^\circ,180^\circ$
Break-awaydouble pole $s=-2$ (depart $\pm90^\circ$)
$j\omega$ crossing$\pm j6.63$ at $K_{op}=57.6$
Chosen $K_{op}$ ($PO=5\%$)$3.0$ (poles $-1.79\pm j1.88$, $-10.42$)
$e_{ss}$ / $T_{settle}$ / GM$57.1\%$ / $2.24$ s / $19.2$ (25.7 dB)