Question 6 of 8: Lag Controller Design in the Frequency Domain

Question 6: Lag 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.

Find. $K_{pos,u}$ and $K_{pos,c}$; the uncompensated margins and the compensated targets; the three lag-controller parameters and $G_c(s)$; and the resulting closed-loop step specifications.

Approach. Set the low-frequency gain from the error specification, translate the overshoot limit into a phase-margin target, find the frequency where the plant already supplies that phase, and let the lag section attenuate the magnitude down to 0 dB there.

1. Part 1 — position constants. The uncompensated loop is the plant alone, so $$K_{pos,u} = G(0) = \frac{20 \times 3}{(0.1)^2 (20)^2} = \frac{60}{4} = 15,$$ giving $e_{ss,u} = 1/(1 + 15) = 0.0625$, i.e. $6.25\%$. The requirement is $e_{ss,c} = 0.0625/5 = 0.0125$, so $$\boxed{K_{pos,c} = \frac{1}{0.0125} - 1 = 79, \qquad e_{ss,c} = 1.25\%.}$$ Since $K_{pos,c} = K_c \cdot G(0)$, the controller DC gain must be $K_c = 79/15 = 5.267$ — and note that $a_0 = K_c$ in the alternative parameterisation the question gives.
2. Part 2 — read the uncompensated margins. From Figure Q6.2 the magnitude of $G(j\omega)$ crosses 0 dB at $$\omega_{cp,u} = 0.376\ \text{rad/s},$$ where the phase curve reads $-145.2^\circ$, so $\Phi_{m,u} = 180^\circ - 145.2^\circ = 34.8^\circ$. That margin corresponds to $\zeta \approx 0.348$ and an overshoot near $31\%$ — well outside the 15% requirement, which is why compensation is needed even before the error specification is considered.
3. Choose the compensated phase margin. Inverting the overshoot formula at $PO = 15\%$ with $L = \ln 0.15 = -1.8971$, $$\zeta = \frac{1.8971}{\sqrt{9.8696 + 3.5990}} = 0.517,$$ and the page-3 correlation $\Phi_m \approx 100\zeta$ gives a required margin of $51.7^\circ$. Rounding to a convenient design target, $$\boxed{\Phi_{m,c} = 52^\circ.}$$
4. Choose the compensated crossover frequency. A lag network is placed so that its own phase lag has almost died away at the new crossover, but some residue always remains; the standard allowance is an extra $5^\circ$. So the new crossover must be put where the plant phase is $-180^\circ + 52^\circ + 5^\circ = -123^\circ$. Reading along the Q6.2 phase curve, that occurs at $$\boxed{\omega_{cp,c} = 0.195\ \text{rad/s},}$$ roughly half the uncompensated crossover — the price a lag design always pays for its extra low-frequency gain.
5. Part 3 — find the attenuation factor $\alpha$. At the new crossover the plant magnitude is $|G(j0.195)| = 3.141$, and the controller contributes $K_c\alpha$ there (its high-frequency asymptote, since the corner frequencies are placed well below). Requiring the loop magnitude to be exactly unity, $$K_c\,\alpha\,|G(j\omega_{cp,c})| = 1 \;\Longrightarrow\; \alpha = \frac{1}{5.267 \times 3.141} = 0.0605.$$ As required for a lag network, $\alpha \lt 1$; the attenuation is $20\log_{10}(1/0.0605) = 24.4$ dB.
6. Place the controller corner frequencies. The controller zero is put one decade below the new crossover so that its phase lag is spent before the crossover is reached: $$\frac{1}{\tau\alpha} = \frac{\omega_{cp,c}}{10} = 0.0195 \;\Longrightarrow\; \tau\alpha = 51.40\ \text{s},$$ and therefore $\tau = 51.40/0.0605 = 850.2$ s.
7. Write the controller. Substituting the three parameters, $$\boxed{G_c(s) = 5.267\,\frac{51.40\,s + 1}{850.2\,s + 1} = \frac{270.7\,s + 5.267}{850.2\,s + 1},}$$ so in the question’s alternative notation $a_1 = 270.7$, $a_0 = 5.267$ and $b_1 = 850.2$. The controller zero sits at $-0.0195$ and its pole at $-0.00118$, both far below the crossover, exactly as a lag design intends.
8. Confirm the achieved margin. Evaluating the compensated loop $G_c(j\omega)G(j\omega)$ numerically, the magnitude crosses unity at $\omega = 0.195$ rad/s with a phase of $-128.5^\circ$, giving an achieved phase margin of $51.5^\circ$ — within half a degree of the $52^\circ$ target, which confirms that the $5^\circ$ lag allowance was well chosen.
9. Part 4 — compensated step-response specifications. From the achieved margin, $\zeta \approx 0.01 \times 51.5 = 0.515$, and from the crossover relation $\omega_n = \omega_{cp,c}/\sqrt{\sqrt{1+4\zeta^4} - 2\zeta^2} = 0.195/0.7755 = 0.252$ rad/s. Then $PO = 100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}} = 15.1\%$ (meeting the specification), $e_{ss(step)} = 1/(1+79) = 1.25\%$ (meeting it as well), $T_{s(\pm 2\%)} = 4/(\zeta\omega_n) = 4/0.1297 = 30.9$ s and $T_{r(0-100\%)} = (\pi - \cos^{-1}0.515)/(0.252 \times 0.857) = 9.79$ s. $$\boxed{PO = 15.1\%,\ e_{ss} = 1.25\%,\ T_r = 9.79\ \text{s},\ T_s = 30.9\ \text{s}.}$$