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22-Elec-A2 Systems and Control · December 2016

Question 7 of 8: Lag-controller design in the frequency domain

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

Notes on this paper

Paper format. National Exams, 07-Elec-A2 Systems & Control, December 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 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, Bode, Nyquist, gain/phase margins (Ch. 10), lead–lag and PID design (Ch. 9–11), state space and Mason’s rule (Ch. 3, 5); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling and second-order correlations. All block diagrams, root loci, pole–zero maps, Bode sketches, Nyquist plots and the realization diagrams 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 $G_m(s)=K_{dc}\,\dfrac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$. Questions 4, 6 and 7 are read from printed Bode/response plots, so those values carry normal chart-reading tolerance; the s-domain results (Q1, Q2, Q3, Q5, Q8) are exact.

Question 7 — Lag-controller design in the frequency domain [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)Gc(s)Lag controllerG(s)30(s+2)/[(s+0.1)^2(s+20)^2]Y(s)Figure Q7.1 - lag-compensated unity feedback
Figure Q7.1 — lag-compensated loop, $G(s)=30(s+2)/[(s+0.1)^2(s+20)^2]$.

Item (1) — uncompensated reads and specs

Given. $G(0)=\dfrac{30\cdot2}{0.1^2\cdot20^2}=15$ (Type 0). Find. $\Phi_m,\omega_{cp}$ and the uncompensated step specs.

[Figure not reproduced: Figure Q7.2 redrawn: flat at $+23.5$ dB to the double corner $0.1$, then $-40$ dB/dec; gain crossover $\omega_{cp}\approx0.38$ rad/s where the phase is $\approx-142^\circ$, i.e. $\Phi_m\approx38^\circ$. See the official exam paper.]

  1. Read the margins. Gain crossover $\omega_{cp}\approx0.38$ rad/s; phase there $\approx-142^\circ$, so $\Phi_m\approx\boxed{38^\circ}$.
  2. Overshoot. $\zeta\approx\Phi_m/100=0.38\Rightarrow PO\approx27\%$.
  3. Steady-state error. position constant $K_p=G(0)=15$, so $e_{ss(step)}=\dfrac{1}{1+K_p}=\dfrac{1}{16}=\boxed{6.25\%}$.
  4. Settling time. $T_{settle(2\%)}=\dfrac{4}{\zeta\omega_{cp}}\approx\dfrac{4}{0.38\cdot0.38}\approx28\ \text{s}$ — slow, because crossover is low.

Item (2) — required $\Phi_{mc}$ and controller DC gain

  1. DC gain for the error spec. Compensated error is half: $e_{ss,c}=\tfrac12\cdot\tfrac1{16}=\tfrac1{32}$, so $1+K_{p,c}=32\Rightarrow K_{p,c}=31$. Since $K_{p,c}=K_{dc}\,G(0)=15K_{dc}$, $\boxed{K_{dc}=K_c=31/15=2.07}$.
  2. Required phase margin. $PO=10\%\Rightarrow\zeta=0.591$, so $\Phi_{mc}\approx100\zeta=\boxed{59^\circ}$ (add $\approx5^\circ$ for the lag’s residual phase, target $\approx64^\circ$).

Item (3) — lag design

  1. New crossover. Choose $\omega_{cpc}$ where the plant phase equals $-180^\circ+\Phi_{mc}+5^\circ=-116^\circ$: from the phase curve, $\boxed{\omega_{cpc}\approx0.17}$ rad/s.
  2. Required attenuation. At $\omega_{cpc}$, $|K_{dc}G(j\omega_{cpc})|\approx7.8$ ($+17.9$ dB). The lag high-frequency gain must drop this to $0$ dB: $\alpha=\dfrac{1}{K_{dc}|G(j\omega_{cpc})|}=\boxed{0.128}$ ($-17.9$ dB).
  3. Place the corners. Put the lag zero a decade below the new crossover: $\dfrac{1}{\tau\alpha}=\dfrac{\omega_{cpc}}{10}\approx0.017$ rad/s ($\tau\alpha\approx58$ s); the pole is then $\dfrac{1}{\tau}=\alpha\cdot0.017\approx0.0022$ rad/s ($\tau\approx453$ s).
  4. Controller. $$\boxed{G_c(s)=2.07\,\frac{58\,s+1}{453\,s+1}}\quad(K_c=2.07,\ \alpha=0.128).$$
Q7 - lag Gc: pole -0.0022 nearer origin than zero -0.017 => LAG0ReIm
Lag controller pole/zero: pole $-0.0022$ closer to the origin than the zero $-0.017$ — the signature of a lag (phase-lag, low-frequency gain boost).

Item (4) — compensated specs

By construction $\Phi_{mc}\approx59^\circ\Rightarrow PO\approx10\%$; $e_{ss(step)}=3.125\%$ (halved); $T_{settle(2\%)}=\dfrac{4}{\zeta\omega_{cpc}}\approx\dfrac{4}{0.591\cdot0.17}\approx40\ \text{s}$. The lag trades speed for accuracy: overshoot and error improve while settling slows.

QuantityUncompensatedCompensated
$\Phi_m$$38^\circ$$59^\circ$
$PO$$\approx27\%$$\approx10\%$
$e_{ss(step)}$$6.25\%$$3.125\%$
$T_{settle(2\%)}$$\approx28$ s$\approx40$ s
Controller$G_c=2.07\dfrac{58s+1}{453s+1}$, $\alpha=0.128$
Check: $\Phi_m,\omega_{cp}$ and the new crossover $\omega_{cpc}$ are read from Figure Q7.2; the controller corner frequencies follow the standard “zero one decade below crossover” rule and carry chart-reading tolerance. The very long settling time is intrinsic to this low-bandwidth plant.