Question 5 of 8: Controller Design in Frequency Domain — Lead Controller (20 marks)

Question 5: Controller Design in Frequency Domain — Lead Controller (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. A unity-feedback loop under lead control, $G_c(s)=K_c\dfrac{\tau s+1}{\alpha\tau s+1}=\dfrac{a_1s+a_0}{b_1s+1}$ with $\alpha\lt 1$.

Find. (1) uncompensated $\Phi_{m\_u}$, $\omega_{cp\_u}$ and the resulting step-response estimates; (2) $K_{pos\_u}$, $K_{pos\_c}$, $\Phi_{m\_c}$ and $\omega_{cp\_c}$; (3) the lead controller parameters.

[Figure not reproduced: Figure S5.2 - Redrawn Figure Q5.1: uncompensated process G(jw). Flat 20.6 dB at low frequency (Type 0) and phase 0 to -270 deg. Gain crossover 2.407 rad/s with phase margin 29.4 deg. See the official exam paper.]

Approach. Convert each transient specification into a target phase margin and crossover frequency, use the error specification to fix the DC gain $K_c$, then size $\alpha$ from the magnitude condition at the intended crossover and place the lead’s phase peak there.

1. Read the uncompensated margins (part 1). The magnitude curve crosses 0 dB at $\omega_{cp\_u}=2.41$ rad/s, where the phase is about $-150.6^\circ$, so$$\Phi_{m\_u}=180^\circ-150.6^\circ=29.4^\circ.$$
2. Estimate the uncompensated closed-loop specs. Using the chart approximation $\zeta\approx0.01\Phi_m=0.294$ and $\omega_n=\omega_{cp}/\sqrt{\sqrt{1+4\zeta^4}-2\zeta^2}=2.62$ rad/s:$$PO=100e^{-\zeta\pi/\sqrt{1-\zeta^2}}=38.1\%,\qquad T_{s(\pm2\%)}=\frac{4}{\zeta\omega_n}=5.20\ \text{s},$$$$T_{r(0-100\%)}=\frac{\pi-\arccos\zeta}{\omega_n\sqrt{1-\zeta^2}}=0.746\ \text{s}.$$The position constant is $K_{pos\_u}=G(0)=\dfrac{100(0.8)}{0.5(1)^2(15)}=10.67$, so$$e_{ss(step\%)}=\frac{100}{1+K_{pos\_u}}=8.57\%.$$Every specification is violated: the loop is too slow, too oscillatory and too inaccurate.
3. Fix the DC gain from the error specification (part 2). Requiring $e_{ss}\le4\%$ means $1/(1+K_{pos\_c})\le0.04$, i.e. $\boxed{K_{pos\_c}\ge24}$. Since $G_c(0)=K_c$,$$K_c=\frac{K_{pos\_c}}{K_{pos\_u}}=\frac{24}{10.667}=2.25.$$
4. Convert the transient specs to $\Phi_{m\_c}$ and $\omega_{cp\_c}$. From $PO\le15\%$, $\zeta\ge0.5169$, so by the chart $\Phi_{m\_c}\approx100\zeta=51.7^\circ$. The speed specs give$$T_s:\ \omega_n\ge\frac{4}{0.5169\times0.7}=11.05,\qquad T_r:\ \omega_n\ge\frac{\pi-\arccos(0.5169)}{0.3\sqrt{1-0.5169^2}}=8.23,$$so settling time binds at $\omega_n\ge11.05$ rad/s. Converting to crossover, $\omega_{cp}=\omega_n\sqrt{\sqrt{1+4\zeta^4}-2\zeta^2}=0.774\omega_n$, giving $\boxed{\omega_{cp\_c}\approx8.6\ \text{rad/s}}$.
5. Size $\alpha$ from the magnitude condition (part 3). At the intended crossover the compensated loop gain must be unity. A lead with its phase peak at $\omega_{cp\_c}$ contributes a magnitude boost of $1/\sqrt{\alpha}$ there, so $|K_cG(j\omega_{cp\_c})|\cdot\dfrac{1}{\sqrt{\alpha}}=1$, i.e. $\sqrt{\alpha}=|K_cG(j\omega_{cp\_c})|$. Evaluating the plant at $8.6$ rad/s gives $|K_cG|=0.17405$, hence$$\boxed{\alpha=0.17405^2=0.03030}$$
6. Place the phase peak at the crossover. The maximum phase lead occurs at $\omega=1/(\tau\sqrt{\alpha})$, so setting that equal to $\omega_{cp\_c}$,$$\tau=\frac{1}{\omega_{cp\_c}\sqrt{\alpha}}=\frac{1}{8.6\times0.17405}=0.6681\ \text{s}.$$The corner frequencies are $1/\tau=1.497$ rad/s and $1/(\alpha\tau)=49.41$ rad/s, and the peak lead available is $\phi_{max}=\arcsin\dfrac{1-\alpha}{1+\alpha}=70.2^\circ$.
7. Write the controller and check the achieved margin.$$\boxed{G_c(s)=2.25\,\frac{0.6681s+1}{0.02024s+1}=\frac{1.5031s+2.25}{0.02024s+1}}$$so $a_1=1.5031$, $a_0=2.25$, $b_1=0.02024$. Recomputing the margin on the exact product $G_c(j\omega)G(j\omega)$ gives a crossover at $8.60$ rad/s with $\Phi_m=51.7^\circ$ — the target is met exactly.