Question 2 of 8: Lag Controller Design from a Steady-State-Error and Overshoot Specification

Question 2: Lag Controller Design from a Steady-State-Error and Overshoot Specification (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.

[Figure not reproduced: Figure Q2.1 (redrawn as a block diagram) — unity-feedback loop with the Lag Controller in the forward path ahead of the process $G(s)$. See the official exam paper.]

Given. $G(s)=30(s+2)/[(s+0.1)^2(s+20)^2]$ (Type 0); design targets $e_{ss,c}=0.5\,e_{ss,u}$ and $PO_c\le10\%$.

Find. $K_{pos,u}$, $K_{pos,c}$, $K_c$; $\Phi_{m,u}$, $\Phi_{m,c}$; $\omega_{cp,c}$, $\alpha$, $\tau$, $G_c(s)$; the compensated step-response specs.

Approach. A Type 0 loop's step error is set entirely by $K_{pos}=G(0)$, so the ess spec fixes the required proportional gain $K_c$ first. Plot $K_cG(j\omega)$, read its Phase Margin, and translate the PO spec into a target $\Phi_{m,c}$ via the formula-sheet relation $\Phi_m\approx100\zeta$. Because $\Phi_{m,u}$ is already far below target, a LAG network is used to push the crossover down to a lower frequency where the uncompensated phase alone is closer to the target, adding a small safety margin for the network's own residual phase lag there.

1. Part 1) — position constants. $K_{pos,u}=G(0)=\dfrac{30(2)}{(0.1)^2(20)^2}=\dfrac{60}{4}=\boxed{15}$, so $e_{ss,u}=\dfrac{1}{1+15}=6.25\%$. The spec wants $e_{ss,c}=3.125\%$, i.e. $K_{pos,c}=\dfrac{1}{0.03125}-1=\boxed{31}$. The required proportional gain is $$\boxed{K_c=\frac{K_{pos,c}}{K_{pos,u}}=\frac{31}{15}\approx2.067}.$$
2. Part 2) — phase margins. With $K_c=2.067$ applied, $K_cG(j\omega)$ crosses 0 dB at $\omega_{gc,u}\approx0.558$ rad/s, where its phase is $-147.3^\circ$, so $\boxed{\Phi_{m,u}\approx32.7^\circ}$. For $PO_c\le10\%$: $\zeta=-\ln(0.1)/\sqrt{\pi^2+\ln^2(0.1)}\approx0.591$, and using the formula-sheet approximation $\Phi_m\approx100\zeta$, $$\boxed{\Phi_{m,c}\approx59.1^\circ}.$$
3. Part 3) — new crossover, lag parameters. Adding a $5^\circ$ safety allowance for the lag network's own residual phase at the new crossover, the target phase angle is $-180^\circ+59.1^\circ+5^\circ=-115.9^\circ$. Scanning $K_cG(j\omega)$'s phase (it is monotonically more negative as $\omega$ increases up to this region) locates $$\boxed{\omega_{cp,c}\approx0.173\ \text{rad/s}},\qquad|K_cG(j\omega_{cp,c})|_{\text{dB}}\approx17.9\ \text{dB}.$$ The lag network must attenuate by exactly this much at the new crossover: $\alpha=10^{-17.9/20}\approx\boxed{0.128}$. Placing the network's zero one decade below the new crossover ($1/(\alpha\tau)=\omega_{cp,c}/10$) gives $\boxed{\tau\approx453}$, so $$\boxed{G_c(s)=2.067\cdot\frac{57.9s+1}{453s+1}}\qquad(a_1\approx119.7,\ a_0\approx2.067,\ b_1\approx453).$$ Re-evaluating the FULL compensated loop at $\omega_{cp,c}$ gives an achieved phase margin of $59.0^\circ$ — matching the $59.1^\circ$ target to within rounding, and the lag network's own phase contribution there is $-5.0^\circ$, exactly the assumed safety allowance. On the sketch, the compensated magnitude curve sits $\approx17.9$ dB below the uncompensated one at low frequency (the lag's attenuation) and rejoins it above the zero break frequency.
4. Part 4) — compensated step-response estimate. Treating the design point as a 2nd-order dominant pair ($\zeta\approx0.591$, and $\omega_n=\omega_{cp,c}/\sqrt{\sqrt{4\zeta^4{+}1}-2\zeta^2}\approx0.239$ rad/s from the standard gain-crossover relation): $$T_{rise(0-100\%)}\approx\boxed{11.4\ \text{s}},\qquad T_{settle(\pm2\%)}=\frac{4}{\zeta\omega_n}\approx\boxed{28.3\ \text{s}},\qquad PO\approx\boxed{10.0\%}\ (\text{by design}),\qquad e_{ss(step\%)}=\boxed{3.13\%}\ (\text{by design}).$$
   Check: this 2nd-order estimate captures the DOMINANT transient shape, but a full 5th-order simulation of the compensated loop (plant + lag network, no approximation) shows a much longer practical settling tail ($\approx130$ s) because the lag pole itself has a very slow time constant ($1/\tau\approx0.0022$ rad/s $\Rightarrow$ time constant $\approx453$ s) — a well-known trade-off of lag compensation under a tight steady-state-accuracy spec: excellent accuracy and an acceptable dominant-pole overshoot, at the cost of a slow-decaying low-amplitude tail that the 2nd-order model does not capture.