Question 3 of 7: PI-controller design from a dominant-poles model

Question 3 — PI-controller design from a dominant-poles model [7 + 7 + 6]

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. Process $\dfrac{10}{s+10}$; PI controller $C(s)=K_p\dfrac{T_i s+1}{T_i s}$; specs PO $=15\%$, $T_{s(\pm2\%)}=0.1$ s, $e_{ss}=0$.

Find. $K_{dc},\zeta,\omega_n$ (Part A); $G_{cl}(s)$ and $K_p,T_i$ (Part B); the $G_{cl}$–$G_m$ comparison (Part C).

[Figure not reproduced: Figure Q3.1 (redrawn) — PI controller in cascade with the first-order plant. See the official exam paper.]

Part A — Dominant second-order model [7]

Approach. Overshoot fixes $\zeta$; the $2\%$ settling time then fixes $\omega_n$; zero steady-state error to a step means $K_{dc}=1$.

1. Damping ratio from overshoot. $\zeta=\dfrac{-\ln(0.15)}{\sqrt{\pi^2+\ln^2(0.15)}}=\dfrac{1.897}{\sqrt{9.87+3.60}}=\boxed{0.517}.$
2. Natural frequency from settling time. $T_{s(\pm2\%)}=\dfrac{4}{\zeta\omega_n}=0.1$ $\Rightarrow \zeta\omega_n=40\Rightarrow \omega_n=\dfrac{40}{0.517}=\boxed{77.4\ \text{rad/s}}.$
3. DC gain and model. The integral term forces $e_{ss}=0$, so $K_{dc}=1$ and $$G_m(s)=\frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}=\boxed{\frac{5988}{s^2+80s+5988}}.$$

Part B — Closed-loop transfer function and PI values [7]

Approach. Form the unity-feedback closed loop of $C(s)G(s)$ and match its characteristic polynomial to $s^2+2\zeta\omega_n s+\omega_n^2$.

1. Open loop. $C(s)G(s)=K_p\dfrac{T_i s+1}{T_i s}\cdot\dfrac{10}{s+10}=\dfrac{10K_p(T_i s+1)}{T_i s(s+10)}.$
2. Closed loop. $G_{cl}(s)=\dfrac{10K_p(T_i s+1)}{T_i s(s+10)+10K_p(T_i s+1)}=\dfrac{10K_p(T_i s+1)}{T_i s^2+10T_i(1+K_p)s+10K_p}.$ Dividing the denominator by $T_i$: $s^2+10(1+K_p)s+\dfrac{10K_p}{T_i}$.
3. Match the characteristic polynomial. $10(1+K_p)=2\zeta\omega_n=80\Rightarrow \boxed{K_p=7}$; and $\dfrac{10K_p}{T_i}=\omega_n^2=5988\Rightarrow \boxed{T_i=\dfrac{70}{5988}=0.0117\ \text{s}}.$

Part C — Substitution and comparison [6]

Approach. Insert $K_p,T_i$ and compare the resulting $G_{cl}$ with the target $G_m$; the only structural difference is a controller zero.

1. Compensated closed loop. With $10K_p=70$ and $10K_p/T_i=5988$, $$G_{cl}(s)=\frac{70s+5988}{s^2+80s+5988}\quad\text{versus}\quad G_m(s)=\frac{5988}{s^2+80s+5988}.$$
2. The difference. $G_{cl}$ carries an extra finite zero at $s=-1/T_i=-85.5$ that the pure model $G_m$ lacks; the denominators (hence the poles $-40\pm j66.2$) are identical.
3. Effect on the three specs. $e_{ss}$: unchanged — both have DC gain $1$, so $e_{ss}=0$ is preserved. PO: the added left-half-plane zero increases overshoot above the design $15\%$ (a zero near the poles adds a derivative "kick"). $T_s$: the zero speeds the rise and peak times, so settling is marginally faster but the $\pm2\%$ envelope (set by $\zeta\omega_n=40$) is essentially unchanged.