Question 4 of 8: PID design by pole placement with rate feedback

Question 4 — PID design by pole placement with rate feedback [5 + 5 + 10]

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. $T(s)$; the dominant-model $\zeta,\omega_n$; and the three gains.

[Figure not reproduced: Figure Q4.1 redrawn — P+I forward control with inner rate feedback $K_ds+1$. See the official exam paper.]

Part 1 — Closed-loop transfer function [5]

Approach. One loop with forward path $G_{fwd}=\dfrac{K_p(s+K_i)}{s(s^2+8s+4)}$ and feedback $H=K_ds+1$.

1. Assemble $T=\dfrac{G_{fwd}}{1+G_{fwd}H}$. Multiplying through by $s(s^2+8s+4)$: $$T(s)=\frac{K_p(s+K_i)}{s^3+(8+K_pK_d)s^2+\bigl(4+K_p+K_pK_dK_i\bigr)s+K_pK_i}.$$
2. Structure. The free integrator from the PI term makes the loop Type 1 and the feedback is unity at DC ($H(0)=1$), so $\boxed{e_{ss}(\text{step})=0}$. The closed-loop zero is at $s=-K_i$.

Part 2 — Dominant-pole model [5]

1. Damping from overshoot. $\zeta=\dfrac{-\ln(0.10)}{\sqrt{\pi^2+\ln^2(0.10)}}=\boxed{0.591}$.
2. Natural frequency from settling. $T_s=\dfrac{4}{\zeta\omega_n}=1$ s gives $\zeta\omega_n=4$, so $\omega_n=\dfrac{4}{0.591}=\boxed{6.77\ \text{rad/s}}$ ($\omega_n^2=45.8$). Dominant pair: $s=-4\pm j5.46$.

Part 3 — Gains by pole–zero cancellation [10]

Approach. Desired characteristic polynomial $(s^2+8s+45.8)(s+K_i)$; the closed-loop zero at $-K_i$ cancels the assigned third pole. Match coefficients term by term.

1. Constant term. $K_pK_i=45.8\,K_i\Rightarrow\boxed{K_p=45.8}$ ($=\omega_n^2$).
2. $s^2$ term. $8+K_pK_d=8+K_i\Rightarrow K_pK_d=K_i$.
3. $s^1$ term. $4+K_p+K_pK_dK_i=45.8+8K_i$. Using $K_pK_d=K_i$ gives $K_pK_dK_i=K_i^2$, so $49.8+K_i^2=45.8+8K_i$, i.e. $$K_i^2-8K_i+4=0\Rightarrow K_i=7.46\ \text{or}\ 0.536.$$
4. Selection. Take $\boxed{K_i=7.46}$ so the cancelled third pole ($-7.46$) sits well beyond the dominant pair — robust cancellation and genuine dominance. Then $K_d=\dfrac{K_i}{K_p}=\dfrac{7.46}{45.8}=\boxed{0.163}$. (The root $K_i=0.536$ would place the cancelled pole at $-0.536$, inside the dominant pair — fragile to any mismatch.)