Second-Order Dominant-Pole Models from Three Different Sources: 17-Phys-B5 Systems and Control, December 2017, Q6

Question 6: Second-Order Dominant-Pole Models from Three Different Sources (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.

Check: the printed process transfer function has $(s+0.1)$ to the FIRST power, but that gives a cubic denominator (3 closed-loop poles), one short of the FOUR poles the question itself lists in Part 1. Restoring the exponent to $(s+0.1)^2$ makes $K_pG(s)=3\times10(s+2)/[(s+0.1)^2(s+20)^2]=30(s+2)/[(s+0.1)^2(s+20)^2]$, which (a) reproduces ALL FOUR given closed-loop poles exactly (to 4 decimal places) and (b) is then EXACTLY Question 2's plant $G(s)$ — precisely matching this question's own statement that the open loop here "is the same as in Question 2." Both independent checks agreeing is strong confirmation; $(s+0.1)^2(s+20)^2$ is used throughout below.

Block diagram for the loop under study; note $K_pG(s)$ is algebraically identical to Question 2's uncompensated plant.

Given. $G(s)=10(s+2)/[(s+0.1)^2(s+20)^2]$ (corrected), $K_p=3$; the four closed-loop poles listed above; Figure Q2.1 (open-loop Bode) and Figure Q6.1 (closed-loop magnitude, resonant peak $M_r\approx1.55$ at $\omega_r\approx0.3$ rad/s, low-frequency value $\approx0.95$).

Find. $G_{m1}(s)$, $G_{m2}(s)$, $G_{m3}(s)$; a comparison; step-response specs from the best model.

Approach. $G_{m1}$ comes straight from the given exact poles (the complex pair sets $\zeta,\omega_n$; the DC gain is matched to the true closed-loop DC gain). $G_{m2}$ uses the standard PM/crossover-frequency-to-$(\zeta,\omega_n)$ conversion applied to the OPEN-loop Bode curve (unity-gain reading of $K_pG(j\omega)$, since $K_p=3$ already reproduces the needed gain exactly). $G_{m3}$ uses the resonant-peak formula $M_r/K_{dc}=1/(2\zeta\sqrt{1{-}\zeta^2})$ and $\omega_r=\omega_n\sqrt{1{-}2\zeta^2}$ read off the CLOSED-loop magnitude plot.

Part 1) — $G_{m1}(s)$ from the exact closed-loop poles. The dominant complex pair gives $\sigma=0.1308$, $\omega_d=0.3794$, so $\omega_{n1}=\sqrt{\sigma^2+\omega_d^2}\approx\boxed{0.401\ \text{rad/s}}$, $\zeta_1=\sigma/\omega_{n1}\approx\boxed{0.326}$. The exact DC gain is $K_{dc}=K_pG(0)/(1+K_pG(0))$ with $G(0)=10(2)/(0.01\times400)=5$, giving $K_{dc}=15/16=\boxed{0.9375}$: $$\boxed{G_{m1}(s)=\frac{0.9375\times0.161}{s^2+0.262s+0.161}}.$$
Part 2) — $G_{m2}(s)$ from the open-loop Bode plot. Reading $K_pG(j\omega)$ (unity-gain, since $K_p=3$ already gives exactly Question 2's plant): $0$ dB crossover at $\omega_{gc}\approx0.378$ rad/s with phase margin $\Phi_m\approx38.2^\circ$, so $\zeta_2\approx\Phi_m/100\approx\boxed{0.382}$ and $\omega_{n2}=\omega_{gc}/\sqrt{\sqrt{4\zeta_2^4{+}1}-2\zeta_2^2}\approx\boxed{0.436\ \text{rad/s}}$ (same $K_{dc}=0.9375$, since it is the same physical closed loop).
Part 3) — $G_{m3}(s)$ from the closed-loop Bode plot. With $M_r/K_{dc,read}\approx1.55/0.95=1.632=1/(2\zeta\sqrt{1{-}\zeta^2})$, solving gives $\zeta_3\approx\boxed{0.324}$; then $\omega_{r}=\omega_{n3}\sqrt{1{-}2\zeta_3^2}=0.3$ gives $\boxed{\omega_{n3}\approx0.337\ \text{rad/s}}$.
Part 4) — comparison and step-response estimate. The three independent readings agree well: $\zeta\approx0.326,\,0.382,\,0.324$ and $\omega_n\approx0.401,\,0.436,\,0.337$ rad/s — all describing the same lightly-damped, slow (sub-1 rad/s) response. $G_{m1}$ is definitionally the MOST ACCURATE model since it is built directly from the system's own exact poles (not a graphical read-off), so it is used for the final estimate: $$PO=100e^{-\zeta_1\pi/\sqrt{1-\zeta_1^2}}\approx\boxed{33.9\%},\qquad T_{settle(\pm2\%)}=\frac{4}{\zeta_1\omega_{n1}}\approx\boxed{30.6\ \text{s}},\qquad T_{rise(0-100\%)}\approx\boxed{5.02\ \text{s}},\qquad e_{ss(step\%)}=(1-0.9375)\times100=\boxed{6.25\%}.$$
Check: a full 4th-order numerical simulation of the exact closed loop gives $PO\approx34.6\%$, $T_{settle(2\%)}\approx27.4$ s, $T_{rise}\approx4.6$ s — all close to the dominant-pole estimate, confirming the two real poles ($-18.8,-21.14$) are indeed fast enough to be safely neglected.

Final results — Question 6
Model	$\zeta$	$\omega_n$ (rad/s)	Source
$G_{m1}$	0.326	0.401	exact closed-loop poles
$G_{m2}$	0.382	0.436	open-loop Bode (Fig. Q2.1)
$G_{m3}$	0.324	0.337	closed-loop Bode (Fig. Q6.1)
Best-model step specs		$PO\approx33.9\%$, $T_{settle(2\%)}\approx30.6$ s, $T_{rise}\approx5.02$ s, $e_{ss}=6.25\%$

Topic: Three independent routes to a 2nd-order dominant-pole model (exact poles, open-loop frequency response, closed-loop frequency response) for the same system.

Key relations: $\zeta=\sigma/\omega_n$ from exact poles; $\zeta\approx\Phi_m/100$ and $\omega_n=\omega_{gc}/\sqrt{\sqrt{4\zeta^4{+}1}-2\zeta^2}$ from open-loop PM/crossover; $M_r/K_{dc}=1/(2\zeta\sqrt{1{-}\zeta^2})$ and $\omega_r=\omega_n\sqrt{1{-}2\zeta^2}$ from closed-loop resonant peak.

Why this works: all three methods are estimating the SAME underlying complex-pole pair through different windows onto the system — exact pole locations are ground truth, while the two Bode-based methods are approximations that become exact only in the idealised prototype 2nd-order case. Close agreement between all three (as found here) is itself evidence that the two REAL poles really are far enough away to be safely dropped.

Common pitfall: trusting a printed transfer function at face value when it is inconsistent with other given data in the SAME question (here, a 4-pole answer being demanded of what was printed as a 3-pole system) instead of cross-checking against every other clue on the page.

Source: Nise, Control Systems Engineering, Ch. 4 (dominant-pole 2nd-order approximations) and Ch. 10 (resonant peak); Ogata, Modern Control Engineering, Ch. 7.

Question 6 of 8: Second-Order Dominant-Pole Models from Three Different Sources

Question 6: Second-Order Dominant-Pole Models from Three Different Sources (20 marks)