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22-Elec-A2 Systems and Control · December 2017

Question 6 of 8: Second-order dominant-pole models from three sources

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 6 — Second-order dominant-pole models from three sources [20]

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.

−+R(s)K_p = 3proportionalG(s)10(s+2)/[(s+0.1)^2(s+20)^2]Y(s)Figure Q6 loop - proportional K_p=3
Figure Q6 loop — proportional $K_p=3$ ahead of $G(s)=10(s+2)/[(s+0.1)^2(s+20)^2]$; open loop equals $30(s+2)/[(s+0.1)^2(s+20)^2]$.

Given. $K_p=3$; closed-loop poles as listed; open loop $=$ Q2 plant. Find. three second-order models and the step specs. Approach. each model matches a lightly-damped pair; the DC gain is fixed by the closed-loop $G_{cl}(0)$.

Item (1) — model from the pole locations ($G_{m1}$)

  1. DC gain. $G_{cl}(0)=\dfrac{30\cdot2}{(0.1)^2(20)^2+30\cdot2}=\dfrac{60}{64}=\boxed{0.9375}$ (Type-0, $K_{pos}=15$, $K_{dc}=15/16$).
  2. Dominant pair. From $p_{1,2}=-0.1308\pm j0.3794$: $\omega_n=|p_{1,2}|=\boxed{0.401\ \text{rad/s}}$, $\zeta=-\operatorname{Re}(p)/\omega_n=\boxed{0.326}$. The far poles $-18.8,-21.1$ are $\gt45\times$ deeper, so they are negligible.
  3. Model. $$\boxed{G_{m1}(s)=\frac{K_{dc}\,\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}=\frac{0.151}{s^2+0.262\,s+0.161}.}$$

Item (2) — model from the open-loop Bode ($G_{m2}$)

  1. Read the margins. The open-loop Bode (Q2.1) gives gain crossover $\omega_{cp,u}=0.378$ rad/s and phase margin $\Phi_{m,u}=38.2^\circ$.
  2. Convert. $\zeta\approx\Phi_{m,u}/100=\boxed{0.38}$ and $\omega_n=\omega_{cp,u}/\sqrt{\sqrt{1+4\zeta^4}-2\zeta^2}=\boxed{0.44\ \text{rad/s}}$, with the same $K_{dc}=0.9375$: $G_{m2}(s)=\dfrac{0.179}{s^2+0.334\,s+0.191}$.

Item (3) — model from the closed-loop Bode ($G_{m3}$)

[Figure not reproduced: Figure Q6.1 redrawn (exact closed loop): low-frequency magnitude $0.94$ ($=K_{dc}$), resonant peak $M_r\approx1.55$ at $\omega_r\approx0.36$ rad/s, matching the printed plot. See the official exam paper.]

  1. Resonant-peak reads. On Fig Q6.1 the curve starts at $\approx0.94$ ($=K_{dc}=0.9375$) and peaks at $M_r\approx1.55$ at $\omega_r\approx0.36$ rad/s (between the $0.3$ and $0.4$ grid lines).
  2. Damping from $M_r$. $\dfrac{M_r}{K_{dc}}=\dfrac{1.55}{0.9375}=1.653=\dfrac{1}{2\zeta\sqrt{1-\zeta^2}}\Rightarrow\zeta=\boxed{0.319}$; then $\omega_r=\omega_n\sqrt{1-2\zeta^2}\Rightarrow\omega_n=\dfrac{0.36}{\sqrt{1-2(0.319)^2}}=\boxed{0.403\ \text{rad/s}}$: $G_{m3}(s)=\dfrac{0.152}{s^2+0.257\,s+0.163}$.
Q6 closed-loop poles: dominant pair -0.131+/-j0.379, far poles -18.8,-21.1; zero -2-20-15-10-50ReIm
Q6 closed-loop pole–zero map: dominant pair $-0.131\pm j0.379$, far real poles $-18.8,-21.1$, plant zero $-2$ — the pair clearly dominates.

Item (4) — comparison and step specs

  1. Compare. $(\zeta,\omega_n)=(0.326,0.401),\,(0.38,0.44),\,(0.319,0.403)$. $G_{m1}$ and $G_{m3}$ are practically identical, because the closed-loop magnitude is shaped by the same dominant pair. $G_{m2}$ comes out a little more damped and faster, since $\Phi_m\approx100\zeta$ is only an approximate correlation. $G_{m1}$ is the most reliable: it uses the exact pole coordinates rather than a chart read.
  2. Step specs (from $G_{m1}$, $\zeta=0.326,\omega_n=0.401$). $PO=100e^{-\zeta\pi/\sqrt{1-\zeta^2}}=\boxed{33.9\%}$; $T_{settle(\pm2\%)}=4/(\zeta\omega_n)=\boxed{30.6\ \text{s}}$; $T_{rise(0-100\%)}=(\pi-\cos^{-1}\zeta)/\omega_d=\boxed{5.0\ \text{s}}$ ($\omega_d=0.379$); $e_{ss(step)}=1-K_{dc}=\boxed{6.25\%}$.
QuantityValue
$K_{dc}$$0.9375$
$G_{m1}$ ($\zeta,\omega_n$)$0.326,\ 0.401$ rad/s (exact poles)
$G_{m2}$ ($\zeta,\omega_n$)$0.38,\ 0.44$ (open-loop Bode)
$G_{m3}$ ($\zeta,\omega_n$)$0.319,\ 0.403$ (closed-loop Bode)
$PO$ / $e_{ss}$$33.9\%$ / $6.25\%$
$T_{settle}$ / $T_{rise}$$30.6$ s / $5.0$ s