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22-Elec-A2 Systems and Control · May 2016

Question 4 of 8: Root-locus analysis, gain selection and dominant model

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 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 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 specs (Ch. 4), block reduction (Ch. 5), Routh–Hurwitz (Ch. 6), steady-state error and static error constants (Ch. 7), root locus (Ch. 8), PID and lead–lag design (Ch. 9–11), frequency response, Bode, Nyquist, gain/phase margins (Ch. 10–11), state space, controllability/observability (Ch. 3, 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling. All step responses, block diagrams, pole–zero maps, root loci, the Bode sketches and the Nyquist plot 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 $G_m(s)=K_{dc}\,\dfrac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$; the figures on the exam paper (step responses, root loci, Bode plots) are read graphically, so the values below carry normal chart-reading tolerance.

Question 4 — Root-locus analysis, gain selection and dominant model [5 + 5 + 7 + 3]

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 poles at $s=0$ and $s=-5\pm j\sqrt{11}=-5\pm j3.32$, no finite zeros.

Find. locus geometry, $(\omega_{osc},K_{crit})$, $K_{op}$ and $T_{rise}$ for $5\%$ overshoot, and the trace identities.

Q4 — Root locus of G(s)=10/[s(s^2+10s+36)]-15-10-505-10-5510ReImσ=-3.33wosc=6, Kcrit=36
Q4.1: root locus with asymptote fan at centroid $\sigma=-3.33$ and the $j\omega$ crossing at $\omega=6$.

Part (1) — asymptotes, centroid, departure

  1. Asymptote angles. $n-m=3$ branches to infinity: $\theta_a=\dfrac{(2k+1)180^\circ}{3}=\boxed{\pm60^\circ,\,180^\circ}$.
  2. Centroid. $\sigma_a=\dfrac{\sum p_i-\sum z_i}{n-m}=\dfrac{0+(-5+j3.32)+(-5-j3.32)}{3}=\boxed{-3.33}$.
  3. Departure angle from $-5+j3.32$. $\theta_d=180^\circ-\big[\angle(\text{from }0)+\angle(\text{from }-5-j3.32)\big]=180^\circ-(146.4^\circ+90^\circ)=\boxed{-56.4^\circ}$ (and $+56.4^\circ$ from the conjugate pole).

Part (2) — imaginary-axis crossover

  1. Characteristic equation. $s(s^2+10s+36)+10K=s^3+10s^2+36s+10K=0$.
  2. Routh marginal condition. $s^1$ row $=36-K=0\Rightarrow \boxed{K_{crit}=36}$; auxiliary $10s^2+10K=0\Rightarrow s^2=-36\Rightarrow \boxed{\omega_{osc}=6\text{ rad/s}}$.

Part (3) — gain for 5% overshoot and rise time

  1. Target damping. $PO=5\%\Rightarrow\zeta=\dfrac{-\ln0.05}{\sqrt{\pi^2+\ln^20.05}}=\boxed{0.69}$.
  2. Gain on the $\zeta=0.69$ ray. Solving the characteristic equation for the dominant pair with $\zeta=0.69$ (third pole kept far, at $-5.4$) gives $\boxed{K_{op}\approx6.0}$, dominant poles $-2.30\pm j2.42$ ($\omega_n=3.34$ rad/s).
  3. Rise time (0–100%). $T_{rise}=\dfrac{\pi-\beta}{\omega_d}$ with $\beta=\cos^{-1}\zeta=0.809$ rad, $\omega_d=2.42$: $T_{rise}=\dfrac{\pi-0.809}{2.42}=\boxed{0.96\text{ s}}$.
Q4 — actual vs dominant-model step response0.00.20.40.60.81.01.201234Trace A: actual 3rd-order (K=6)Trace B: 2nd-order modeltime (s)
Q4.2: reproduced step responses — the pure 2nd-order model (dashed) is faster with slightly more overshoot; the actual 3rd-order response (solid) lags.

Part (4) — identify the traces

  1. Effect of the extra pole. The dominant-poles model drops the real pole at $-5.4$; adding it back inserts lag, so the actual system rises slower and overshoots slightly less. In Fig Q4.3 the faster solid curve is therefore the model and the slower dash-dot curve is the actual system: actual = Trace A, model = Trace B.
QuantityValue
Asymptotes / centroid$\pm60^\circ,180^\circ$ / $\sigma=-3.33$
Departure angles$\pm56.4^\circ$
$\omega_{osc}$ / $K_{crit}$$6$ rad/s / $36$
$K_{op}$ ($PO=5\%$) / $T_{rise}$$\approx6.0$ / $0.96$ s
Trace identityactual = A, model = B