Question 2 of 8: Lead Controller Design in the Frequency Domain

Question 2: Lead Controller Design in the Frequency Domain (20 marks, compulsory)

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. The closed-loop step-response estimates for the uncompensated loop; the uncompensated and required position constants, target phase margin and target crossover frequency; and the three lead-controller parameters $K_c$, $\tau$ and $\alpha$ written as a transfer function.

[Figure not reproduced: Figure 2.1 — Redrawn frequency response of the process G(jw). The low-frequency asymptote sits at +23.5 dB (= 20 log 15), the gain crossover at 2.99 rad/s with a 22 degree phase margin, and the phase crossover at 5.27 rad/s with a 9.64 dB gain margin. See the official exam paper.]

Approach. Convert each frequency-domain read into its second-order time-domain equivalent using the supplied $\Phi_m \approx 100\zeta$ correlation, size $K_c$ from the position-constant requirement, translate the transient specifications into a required damping ratio and natural frequency (hence a target crossover), and finally set the lead ratio $\alpha$ from the magnitude condition at the new crossover, checking the phase margin the design actually delivers.

1. Damping ratio and overshoot of the uncompensated loop. The formula sheet gives the second-order correlation $\zeta \approx 0.01\,\Phi_m$, so a 22° margin implies $$\zeta_u \approx 0.01 \times 22 = 0.22$$ and the overshoot follows from $$PO = 100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}} = 100\,e^{-0.22\pi/\sqrt{1-0.0484}} = \boxed{49.2\%}$$ A 22° margin is very lightly damped, and this large overshoot is the first specification the lead network has to fix.
2. Steady-state error of the uncompensated loop. The process has no free integrator, so the loop is Type 0 and the position constant is simply the DC gain: $$K_{pos\_u} = \lim_{s\to 0} G(s) = \frac{150 \times 0.6}{0.4 \times 1^2 \times 15} = \frac{90}{6} = 15$$ (equivalently the $+23.5$ dB low-frequency asymptote visible on Figure Q2.1), giving $$e_{ss(step)} = \frac{1}{1+K_{pos\_u}} = \frac{1}{16} = \boxed{6.25\%}$$
3. Natural frequency implied by the crossover. For a standard second-order loop the gain-crossover frequency and the natural frequency are related by $$\omega_{cp} = \omega_n\sqrt{\sqrt{1+4\zeta^4}-2\zeta^2}$$ With $\zeta_u = 0.22$ the bracket evaluates to $0.9528$, so $$\omega_{n\_u} = \frac{2.99}{0.9528} = 3.138\ \text{rad/s}, \qquad \omega_{d} = \omega_n\sqrt{1-\zeta^2} = 3.061\ \text{rad/s}$$
4. Rise time and settling time. Using the standard underdamped expressions, $$T_{r(0-100\%)} = \frac{\pi - \cos^{-1}\zeta}{\omega_d} = \frac{\pi - 1.349}{3.061} = \boxed{0.586\ \text{s}}, \qquad T_{s(\pm2\%)} = \frac{4}{\zeta\omega_n} = \frac{4}{0.690} = \boxed{5.79\ \text{s}}$$ The uncompensated loop therefore fails every requirement: 49.2% against 15%, 6.25% against 5%, 0.586 s against 0.25 s and a settling time more than ten times the 0.5 s allowance.
5. Required position constant and hence $K_c$. To hold the step error at 5%, $$e_{ss} = \frac{1}{1+K_{pos\_c}} \le 0.05 \;\Rightarrow\; K_{pos\_c} \ge 19$$ Because the lead network contributes unity gain at DC apart from its own $K_c$, the required scaling is $$K_c = \frac{K_{pos\_c}}{K_{pos\_u}} = \frac{19}{15} = \boxed{1.267}$$
6. Target phase margin. Inverting the overshoot relation for $PO = 15\%$, $$\zeta_c = \frac{-\ln(0.15)}{\sqrt{\pi^2 + \ln^2(0.15)}} = \frac{1.8971}{3.6699} = 0.517$$ so the correlation $\Phi_m \approx 100\zeta$ calls for $$\Phi_{m\_c} \approx 51.7^\circ \;\;(\text{use } 52^\circ)$$
7. Target crossover frequency. Both transient specifications must be satisfied, so take the more demanding. Settling time requires $$\frac{4}{\zeta_c\omega_n} \le 0.5 \;\Rightarrow\; \omega_n \ge \frac{4}{0.5 \times 0.517} = 15.48\ \text{rad/s}$$ while rise time requires $$\omega_n \ge \frac{\pi - \cos^{-1}(0.517)}{0.25\sqrt{1-0.517^2}} = 9.88\ \text{rad/s}$$ Settling time binds, so $\omega_n = 15.48$ rad/s. Converting back to a crossover frequency with the same bracket (now $0.7742$ at $\zeta_c = 0.517$), $$\boxed{\,\omega_{cp\_c} = 15.48 \times 0.7742 = 11.98\ \text{rad/s}\,}$$ The crossover must move out by a factor of four, from 2.99 to about 12 rad/s.
8. Size the lead ratio from the magnitude condition. The lead network has magnitude $1/\sqrt{\alpha}$ at its peak frequency $\omega_m = 1/(\tau\sqrt{\alpha})$. Placing that peak at the target crossover and demanding unit open-loop magnitude there gives $|K_cG(j\omega_{cp\_c})|\cdot(1/\sqrt{\alpha}) = 1$, i.e. $\sqrt{\alpha} = |K_cG(j\omega_{cp\_c})|$. Evaluating the process at 11.98 rad/s gives $|K_cG| = 0.06853$ ($-23.3$ dB), so $$\alpha = (0.06853)^2 = \boxed{0.00470}$$
9. Complete the network and check the delivered margin. The lead time constant follows from the peak-placement condition, $$\tau = \frac{1}{\omega_{cp\_c}\sqrt{\alpha}} = \frac{1}{11.98\sqrt{0.00470}} = 1.218\ \text{s}, \qquad \alpha\tau = 0.00572\ \text{s}$$ so that $$\boxed{\,G_c(s) = 1.267\,\frac{1.218\,s + 1}{0.00572\,s + 1}\,}$$ Recomputing the compensated loop $G_c(j\omega)G(j\omega)$ numerically confirms that the crossover does land at 11.98 rad/s and that the phase margin there is 52.1°, just above the 51.7° target — so all four specifications are met. The maximum phase the network can add is $\phi_{max} = \sin^{-1}\!\frac{1-\alpha}{1+\alpha} = 82.2^\circ$, of which the design consumes most.