Question 7 of 8: Lead Controller Design in the Frequency Domain

Question 7: Lead Controller Design in the Frequency Domain (20 marks)

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. Uncompensated Bode data: $K_{dc,u}=2.5$ dB (low-frequency plateau); gain-crossover $\omega_{cpu}=3.44$ rad/s with $\Phi_{m,u}=26.9^{\circ}$; phase-crossover $\omega=6.2$ rad/s with $G_m=10.1$ dB. Design bounds: $e_{ss}\le25\%$, $PO\le20\%$, $T_{settle(\pm2\%)}\ge1$ s.

Find. $K_{pos\_u}$, $e_{ss(step)\%\_u}$, $K_{pos\_c}$, $a_0$; a design choice of $\Phi_{m\_c}$ and $\omega_{cp\_c}$; and $a_1,b_1,G_c(s)$.

Approach. Use the plotted low-frequency magnitude for $K_{pos\_u}$; invert the required $e_{ss}$ bound for $K_{pos\_c}$ and hence $a_0$; choose a design point $(\Phi_{m\_c},\omega_{cp\_c})$ satisfying $PO\le20\%$ and $T_{settle}\ge1$ s; then apply the standard maximum-phase Lead-network formulas.

1. Part 1) — position constants and $a_0$. $K_{pos\_u}=10^{2.5/20}=\boxed{1.334}$, so $e_{ss(step)\%\_u}=\dfrac{100}{1+K_{pos\_u}}=\boxed{42.9\%}$ — far above the $25\%$ ceiling, confirming compensation is needed. The requirement $e_{ss}\le25\%$ needs $K_{pos\_c}\ge\dfrac{100}{25}-1=\boxed{3.0}$. Since the compensator adds a DC gain factor $a_0$ in series with the uncompensated loop, $a_0=K_{pos\_c}/K_{pos\_u}=\boxed{2.25}$.
2. Part 2) — design choice of $\Phi_{m\_c}$ and $\omega_{cp\_c}$. $PO\le20\%$ needs $\zeta\ge\dfrac{-\ln0.20}{\sqrt{\pi^2+\ln^20.20}}=0.456$; using the standard design approximation $\Phi_m({}^{\circ})\approx100\zeta$, a comfortable choice is $\boxed{\Phi_{m\_c}=45^{\circ}}$ ($\zeta\approx0.45$, just above the minimum, for design margin). The settling-time floor $T_{settle(\pm2\%)}=4/(\zeta\omega_n)\ge1$ s, with $\omega_n\approx\omega_{cp\_c}$ for a well-damped loop, needs $\omega_{cp\_c}\lesssim4/\zeta\approx8.9$ rad/s; choosing $\boxed{\omega_{cp\_c}=5\ \text{rad/s}}$ leaves comfortable margin on both bounds.

   Check: interpolated Bode readings. The uncompensated magnitude/phase AT $\omega=5$ rad/s are not printed directly — they are estimated by log-linear interpolation between the two given chart points ($\omega=3.44$, $0$ dB, $-153.1^{\circ}$) and ($\omega=6.2$, $-10.1$ dB, $-180^{\circ}$), giving approximately $-6.4$ dB and $-170.2^{\circ}$ at $\omega=5$; a reader with the original chart in hand should re-read these two numbers directly off it.

3. Part 3) — remaining Lead parameters. The uncompensated system already supplies about $180-170.2=9.8^{\circ}$ of phase margin at $\omega_{cp\_c}=5$; the Lead network must supply the rest plus a safety margin: $\phi_{max}=\Phi_{m\_c}-9.8^{\circ}+5^{\circ}\approx\boxed{40^{\circ}}$. The standard Lead formulas give $$\alpha=\frac{1-\sin\phi_{max}}{1+\sin\phi_{max}}=\boxed{0.216},\qquad \omega_z=\omega_{cp\_c}\sqrt\alpha=\boxed{2.32\ \text{rad/s}},\qquad \omega_p=\omega_{cp\_c}/\sqrt\alpha=\boxed{10.77\ \text{rad/s}}.$$ With $a_0=2.25$ fixed by Part 1, $a_1=a_0/\omega_z=\boxed{0.969}$ and $b_1=1/\omega_p=\boxed{0.0929}$, giving $$\boxed{G_c(s)=\frac{0.969\,s+2.25}{0.0929\,s+1}}$$ — a genuine Lead network ($\omega_z=2.32<\omega_p=10.77$). Overlaid on Figure Q7.1, this compensator raises the low-frequency magnitude by $20\log_{10}a_0\approx7.0$ dB (fixing the steady-state error) and adds a phase “bump” peaking near $\omega=5$ rad/s, pushing the compensated crossover out toward the chosen design frequency while lifting the phase margin there from about $10^{\circ}$ to roughly the target $45^{\circ}$.