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22-Elec-A2 Systems and Control · December 2017

Question 2 of 8: Lag-controller design in the frequency domain (compulsory)

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 2 — Lag-controller design in the frequency domain (compulsory) [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)G_c(s)lag controllerG(s)30(s+2)/[(s+0.1)^2(s+20)^2]Y(s)Figure Q2 loop - lag controller G_c ahead of plant G
Figure Q2 loop — lag controller $G_c(s)=K_c(\tau\alpha s+1)/(\tau s+1)$ ahead of the double-lag-double-lead plant $G(s)$.

Given.

Process$G(s)=30(s+2)/[(s+0.1)^2(s+20)^2]$ (Type 0)
Controllerlag $G_c=K_c(\tau\alpha s+1)/(\tau s+1)$, $\alpha\lt1$
Error spec$e_{ss,c}=\tfrac12 e_{ss,u}$
Overshoot spec$PO\le10\%$

Find. $K_{pos,u},K_{pos,c}$; $\Phi_{m,u},\Phi_{m,c}$; $\omega_{cpc}$ and $G_c(s)$; the step specs. Approach. a lag network raises the low-frequency gain (fixing the error constant through $K_c$) while its attenuation $\alpha$ pulls the gain crossover down to a frequency where the plant phase still leaves the required margin.

Item (1) — position constants

  1. Uncompensated. Type 0, so $K_{pos,u}=G(0)=\dfrac{30\cdot2}{(0.1)^2(20)^2}=\dfrac{60}{4}=\boxed{15}$, giving $e_{ss,u}=\dfrac{1}{1+15}=0.0625$.
  2. Compensated. Halving the error, $e_{ss,c}=0.03125$, needs $1+K_{pos,c}=1/0.03125=32$, so $\boxed{K_{pos,c}=31}$. The lag DC gain supplies the boost: $K_c=K_{pos,c}/K_{pos,u}=31/15=\boxed{2.07}$.

Item (2) — phase margins

  1. Uncompensated margin. The uncompensated gain crossover ($|G|=1$) is at $\omega_{cp,u}=0.378$ rad/s where $\angle G=-141.8^\circ$, so $\Phi_{m,u}=180^\circ-141.8^\circ=\boxed{38.2^\circ}$.
  2. Target margin. $PO\le10\%\Rightarrow\zeta\ge0.591$; the phase-margin chart gives $\Phi_{m,c}\approx100\zeta=59^\circ$. Round up and design to $\boxed{\Phi_{m,c}\approx60^\circ}$; a further $\approx5^\circ$ allowance for the lag network's residual phase is added when locating the new crossover.

Item (3) — crossover, controller parameters and sketch

  1. New crossover. Choose $\omega_{cpc}$ where the plant phase equals $-180^\circ+\Phi_{m,c}+5^\circ=-115^\circ$; from $G(j\omega)$ this is $\boxed{\omega_{cpc}=0.17\ \text{rad/s}}$.
  2. Required attenuation. There $|K_cG(j\omega_{cpc})|=8.07$ ($+18.1$ dB), so the lag must attenuate by that factor: $\alpha=\dfrac{1}{|K_cG(j\omega_{cpc})|}=\boxed{0.124}$ ($-18.1$ dB).
  3. Place the corner. Put the upper (zero) corner one decade below crossover, $\dfrac{1}{\tau\alpha}=\dfrac{\omega_{cpc}}{10}=0.0169$ rad/s, so $\tau\alpha=59.2$ and $\tau=59.2/\alpha=478$. The lag pole sits at $\dfrac1\tau=0.00209$ rad/s.
  4. Controller. $$\boxed{G_c(s)=2.07\,\frac{59.2\,s+1}{478\,s+1}}\qquad(\text{zero }-0.0169,\ \text{pole }-0.00209).$$ A numerical check gives compensated crossover $0.17$ rad/s with phase margin $60^\circ$ and $K_{pos}=31$ — both specs met.
Q2 Bode: uncompensated G (blue) vs lag-compensated G_cG (red dashed)-100-80-60-40-2002040Magnitude (dB)-270-225-180-135-90-45010^-310^-210^-110^010^110^2Phase (deg)Frequency (rad/s)ω_pc=0.17
Q2.1 with the design overlaid: uncompensated $G$ (solid) vs lag-compensated $G_cG$ (dashed). The lag lifts the DC gain to $K_{pos}=31$ and drops the crossover to $0.17$ rad/s, where the phase margin is $60^\circ$.

Item (4) — step-response specifications

  1. Second-order model. $\zeta\approx\Phi_{m,c}/100=0.60$ and $\omega_n\approx\omega_{cpc}/\sqrt{\sqrt{1+4\zeta^4}-2\zeta^2}=0.237$ rad/s.
  2. Specs. $PO=100e^{-\zeta\pi/\sqrt{1-\zeta^2}}=\boxed{9.5\%}$ ($\le10\%$ ✓); $T_{settle(\pm2\%)}=\dfrac{4}{\zeta\omega_n}=\boxed{28.2\ \text{s}}$; $T_{rise(0-100\%)}=\dfrac{\pi-\beta}{\omega_d}=\boxed{11.7\ \text{s}}$ ($\beta=\cos^{-1}\zeta$); $e_{ss(step)}=1/(1+31)=\boxed{3.13\%}$.
  3. Comment. The lag meets both requirements but the loop is slow — the attenuation pulls crossover to $0.17$ rad/s, so settling is tens of seconds. That is the accepted price of a lag: better accuracy, lower bandwidth.
QuantityValue
$K_{pos,u}$ / $K_{pos,c}$$15$ / $31$ ($K_c=2.07$)
$\Phi_{m,u}$ / $\Phi_{m,c}$$38.2^\circ$ / $\approx60^\circ$
$\omega_{cpc}$ / $\alpha$$0.17$ rad/s / $0.124$
$G_c(s)$$2.07(59.2s+1)/(478s+1)$
$PO$ / $e_{ss}$$9.5\%$ / $3.13\%$
$T_{settle}$ / $T_{rise}$$28.2$ s / $11.7$ s
Check: design is not unique. The corner placement (one decade below crossover) and the $5^\circ$ lag allowance are standard engineering choices; other reasonable placements give slightly different $\tau,\alpha$ but the same $K_{pos}=31$, $\Phi_m\approx60^\circ$ and step specs. The printed Fig Q2.1 (landscape page) agrees with the exact $G(s)$: flat $+23.5$ dB ($=20\log_{10}15$), $0^\circ$ at low frequency, 0 dB crossing near $0.38$ rad/s with phase $\approx-142^\circ$. The step specifications are the dominant-pair estimates the question asks for. The lag zero–pole pair also leaves a slow closed-loop pole near the lag zero, so a full-order simulation overshoots less (about $4\%$) but takes much longer (over $100$ s) to enter the $\pm2\%$ band. That slow tail is typical of lag networks.