PID design from a dominant-poles model (compulsory): 22-Elec-A2 Systems and Control, May 2013, Q2

Question 2 — PID design from a dominant-poles model (compulsory) [5 + 8 + 7]

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 Q2.1 — Unit-feedback loop with parallel PID $C(s)=K_p(1+\tfrac{1}{T_i s}+T_d s)$ and process $G(s)=20/(s^2+5s+3)$.

Part A — Second-order model from the step response [5]

Given (read from Figure Q2.2). Steady-state value $=1.0$ (so $K_{dc}=1$); peak $\approx1.10$ (percent overshoot $\text{PO}\approx10\%$) at peak time $t_p\approx4\ \text{s}$; the response enters the $\pm2\%$ band at $t\approx7\ \text{s}$.

Find. $K_{dc},\zeta,\omega_n$ and $G_m(s)$.

Damping ratio from overshoot. $\text{PO}=e^{-\zeta\pi/\sqrt{1-\zeta^2}}=0.10\Rightarrow \zeta=\dfrac{-\ln 0.10}{\sqrt{\pi^2+\ln^2 0.10}}=\boxed{0.591}.$
Natural frequency from peak time. $t_p=\dfrac{\pi}{\omega_n\sqrt{1-\zeta^2}}=4\ \text{s}\Rightarrow \omega_n=\dfrac{\pi}{4\sqrt{1-0.591^2}}\approx\boxed{1.0\ \text{rad/s}}$. (Check: $T_s=4/(\zeta\omega_n)=6.8\ \text{s}\approx7\ \text{s}$, and $K_{dc}=1$.)
Model transfer function. $$G_m(s)=\frac{K_{dc}\,\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}=\boxed{\frac{1}{s^2+1.18s+1}}.$$

Part B — Closed-loop transfer function and PID values [8]

Approach. Write the parallel PID over a common denominator, form the unity-feedback closed loop, and match its characteristic polynomial to the dominant pair from Part A plus one insignificant pole ten times deeper.

Closed-loop transfer function. With $C(s)=\dfrac{K_d s^2+K_p s+K_i}{s}$ (where $K_i=K_p/T_i,\ K_d=K_pT_d$) and $G(s)=\dfrac{20}{s^2+5s+3}$, $$G_{cl}(s)=\frac{20(K_d s^2+K_p s+K_i)}{s^3+(5+20K_d)s^2+(3+20K_p)s+20K_i}.$$
Desired characteristic polynomial. Dominant pair $s^2+1.18s+1$ from Part A; place the third pole at $10\zeta\omega_n=5.91$: $$(s^2+1.18s+1)(s+5.91)=s^3+7.09s^2+7.99s+5.91.$$
Match coefficients. $5+20K_d=7.09\Rightarrow K_d=0.105$; $\;3+20K_p=7.99\Rightarrow K_p=0.249$; $\;20K_i=5.91\Rightarrow K_i=0.296$.
Controller time constants. $$\boxed{K_p=0.249},\quad T_i=\frac{K_p}{K_i}=\frac{0.249}{0.296}=\boxed{0.844\ \text{s}},\quad T_d=\frac{K_d}{K_p}=\frac{0.105}{0.249}=\boxed{0.420\ \text{s}}.$$

Part C — Poles, zeros, DC gain and discussion [7]

Poles. Roots of the matched denominator: $s=-5.91$ and $s=-0.591\pm j0.807$ (the intended dominant pair, $\zeta=0.591$, $\omega_n=1.0$).
Zeros. Roots of $K_d s^2+K_p s+K_i=0$: $s=-1.19\pm j1.18$.
DC gain. $G_{cl}(0)=\dfrac{20K_i}{20K_i}=\boxed{1}$ — the integral term forces exact unit DC gain (zero steady-state error to a step).

Figure Q2.3 — Compensated closed-loop pole–zero map: dominant poles $-0.59\pm j0.81$, far pole $-5.91$, and a pair of complex zeros $-1.19\pm j1.18$.

Discussion. The design places the dominant poles exactly on the second-order model, but the actual $G_{cl}(s)$ carries two extra features the model ignores: a pair of complex zeros at $-1.19\pm j1.18$, sitting close to (just left of) the dominant poles, and a real pole at $-5.91$. The nearby zeros speed up the rise and add extra overshoot, so the true percent overshoot will exceed the model's 10%; the far pole at $-5.91$ (about ten times the dominant real part) adds only a small, fast lag and slightly slows the very first part of the rise. Net effect: rise time a little shorter, overshoot somewhat larger, settling time close to the model's ~7 s.

Quantity	Result
2nd-order model	$G_m(s)=\dfrac{1}{s^2+1.18s+1}$ ($K_{dc}=1,\ \omega_n=1.0,\ \zeta=0.591$)
PID parameters	$K_p=0.249,\ T_i=0.844\ \text{s},\ T_d=0.420\ \text{s}$
Closed-loop poles	$-5.91,\ -0.591\pm j0.807$
Closed-loop zeros	$-1.19\pm j1.18$
Closed-loop DC gain	$\boxed{K_{dc}=1}$

Concept & Theory

Topic. Dominant-poles PID synthesis by characteristic-polynomial matching.

Key relations. $\text{PO}=e^{-\zeta\pi/\sqrt{1-\zeta^2}}$, $t_p=\pi/(\omega_n\sqrt{1-\zeta^2})$, $T_s\approx4/(\zeta\omega_n)$; parallel PID $C(s)=(K_d s^2+K_p s+K_i)/s$; match $1+C(s)G(s)$ to $(s^2+2\zeta\omega_n s+\omega_n^2)(s+10\zeta\omega_n)$.

Why this works. A PID adds one pole (the integrator at the origin) and up to two zeros, raising the loop order by one. Choosing the extra closed-loop pole a decade deeper keeps it insignificant, so the two dominant poles govern the response and can be dictated by the desired $\zeta,\omega_n$. Coefficient matching then delivers unique $K_p,K_i,K_d$.

Common pitfall. Forgetting that the PID zeros appear in the closed-loop numerator and materially reshape the response — the pole locations alone do not tell the whole story (Part C).

Source. Nise §9 (PID / cascade compensation), §4 (second-order response, overshoot–damping relations).

Question 2 of 8: PID design from a dominant-poles model (compulsory)

Question 2 — PID design from a dominant-poles model (compulsory) [5 + 8 + 7]

Part A — Second-order model from the step response [5]

Part B — Closed-loop transfer function and PID values [8]

Part C — Poles, zeros, DC gain and discussion [7]