Question 5 of 8: Second-order dominant model three ways, and step estimates

Question 5 — Second-order dominant model three ways, and step estimates [5 + 5 + 5 + 5]

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. Unity feedback, so $G_{cl}=\dfrac{G}{1+G}$; $G(0)=10$, hence $G_{cl}(0)=\dfrac{10}{11}=0.909$ (the DC gain of every model). The dominant pair is $s^2+3.26s+92.66$.

Find. $(K_{dc},\zeta,\omega_n)$ three ways and the step-response specs.

Part A — Model from the closed-loop magnitude plot [5]

1. DC gain. $K_{dc}=|G_{cl}(0)|\approx\boxed{0.909}$.
2. Damping from the peak. $M_r=\dfrac{\text{peak}}{K_{dc}}\approx\dfrac{2.55}{0.909}=2.80$; inverting $M_r=\dfrac{1}{2\zeta\sqrt{1-\zeta^2}}$ gives $\boxed{\zeta\approx0.18}$.
3. Natural frequency. $\omega_r=\omega_n\sqrt{1-2\zeta^2}\approx9.3$ $\Rightarrow\omega_n\approx\boxed{9.6\ \text{rad/s}}$, so $G_{m1}(s)=\dfrac{0.909\,\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}\approx\dfrac{83.7}{s^2+3.5s+92}$.

Part B — Model from the open-loop Bode plot [5]

1. Gain crossover & phase margin. $|G(j\omega)|=1$ at $\omega_{cp}=8.83$ rad/s, where $\angle G=-156.5^\circ$, so $PM=23.5^\circ$.
2. Damping from PM. $\zeta\approx\dfrac{PM}{100}=\boxed{0.235}$.
3. Natural frequency. $\omega_{cp}=\omega_n\sqrt{\sqrt{1+4\zeta^4}-2\zeta^2}\Rightarrow\omega_n\approx9.2$ rad/s. With $K_{dc}=0.909$, $G_{m2}(s)\approx\dfrac{76.9}{s^2+4.3s+84.6}$.

Part C — Model from the dominant poles [5]

1. Read the pair. $s^2+3.26s+92.66\Rightarrow\omega_n=\sqrt{92.66}=\boxed{9.63\ \text{rad/s}}$, $\zeta=\dfrac{3.26}{2(9.63)}=\boxed{0.169}$.
2. Dominance & DC gain. The third pole $-23.74$ is $23.74/1.63=14.6\times$ farther out, so the pair dominates. Preserving $K_{dc}=0.909$: $$G_{m3}(s)=\frac{0.909(92.66)}{s^2+3.26s+92.66}=\frac{84.2}{s^2+3.26s+92.66}.$$

Part D — Consistency and step estimates [5]

Approach. All three give $\zeta\approx0.17$–$0.24$, $\omega_n\approx9.2$–$9.6$ rad/s and $K_{dc}=0.909$ — consistent. $G_{m3}$ (exact poles) is the reference.

1. Overshoot. $PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}=100\,e^{-0.169\pi/0.986}=\boxed{58\%}$ (large — $\zeta$ is small).
2. Settling time. $T_{settle}(\pm2\%)=\dfrac{4}{\zeta\omega_n}=\dfrac{4}{0.169(9.63)}=\boxed{2.46\ \text{s}}$.
3. Steady-state error. Type 0 with $K_{pos}=G(0)=10$, so $e_{ss}=\dfrac{1}{1+10}=\boxed{9.1\%}$ (equivalently $1-K_{dc}=1-0.909$).