Question 5 of 8: Three Second-Order Dominant-Pole Models — s-Domain, Closed Loop and Open Loop Frequency Response

Question 5: Three Second-Order Dominant-Pole Models — s-Domain, Closed Loop and Open Loop Frequency Response (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.

Given.

Find. Three independent second-order models — from the pole locations, from the closed-loop resonance, and from the open-loop margins — then the step-response specifications from whichever is most trustworthy.

Approach. Establish dominance from the pole geometry and match DC gain for the first model; invert the resonant-peak relation for the second; use the phase-margin and crossover-frequency correlations for the third; then compare the three before quoting specifications.

1. Part 1 — establish dominance. The dominant pair has real part $0.19$, whereas $p_3 = -17.83$ and $p_4 = -22.0$ have real parts $94$ and $116$ times larger. Their transients decay by a factor $e^{-94} $ within one time constant of the dominant mode, so they are irrelevant to the visible response. The closed-loop zero at $-3$ is $15.8$ times further out than the dominant real part, which is comfortably beyond the usual factor-of-five threshold, so it adds no appreciable overshoot. A second-order model is therefore well justified — more so than in Question 3, where the separations were much tighter.
2. Extract the model parameters from the pole locations. The dominant pair gives directly $$\omega_n = |p_1| = \sqrt{0.19^2 + 0.86^2} = 0.881\ \text{rad/s}, \qquad \zeta = \frac{0.19}{0.881} = 0.216.$$ The DC gain follows from the Type-0 open loop: $G_{open}(0) = K_p G(0) = 5 \times \dfrac{20 \times 3}{(0.1)^2 (20)^2} = 5 \times 15 = 75$, so $K_{dc} = \dfrac{75}{1 + 75} = 0.9868$. Hence $$\boxed{G_{m1}(s) = 0.9868\,\frac{0.776}{s^2 + 0.380\,s + 0.776}, \quad \zeta = 0.216,\ \omega_n = 0.881\ \text{rad/s}.}$$
3. Part 2 — read the closed-loop frequency response. From Figure Q5.2 the curve is flat at $|G_{cl}| \approx 1$ at low frequency, confirming $K_{dc} \approx 1$, and peaks at $M_r = 2.43$ at $\omega_r = 0.85$ rad/s. The second-order resonant-peak relation supplied on page 3 is $$\frac{M_r}{K_{dc}} = \frac{1}{2\zeta\sqrt{1-\zeta^2}} = 2.43.$$ Writing $x = \zeta^2$ this becomes $4x(1-x) = 1/2.43^2 = 0.1694$, i.e. $x^2 - x + 0.04234 = 0$, whose physically meaningful (lightly damped) root is $x = 0.04432$.
4. Convert the resonance to model parameters. Taking the square root, $\zeta = 0.211$. The resonant frequency of a second-order system is $\omega_r = \omega_n\sqrt{1 - 2\zeta^2}$, so $$\omega_n = \frac{\omega_r}{\sqrt{1 - 2(0.211)^2}} = \frac{0.85}{0.9556} = 0.890\ \text{rad/s},$$ giving $$\boxed{G_{m2}(s) = \frac{0.792}{s^2 + 0.375\,s + 0.792}, \quad \zeta = 0.211,\ \omega_n = 0.890\ \text{rad/s}.}$$
5. Part 3 — read the open-loop frequency response. Figure Q6.2 shows $G(j\omega)$ alone, so the proportional gain must be added: multiplying by $K_p = 5$ raises the magnitude curve by $20\log_{10}5 = 13.98$ dB without touching the phase. The gain crossover of $|K_p G(j\omega)| = 1$ then occurs at $\omega_{cp} = 0.877$ rad/s, where the plot reads a phase of $-155.7^\circ$, so the phase margin is $$\Phi_m = 180^\circ - 155.7^\circ = 24.3^\circ.$$
6. Convert the margins into model parameters. The page-3 chart gives the approximate correlation $\zeta \approx 0.01\,\Phi_m$, so $\zeta = 0.243$. The gain crossover of a second-order loop satisfies $\omega_{cp} = \omega_n\sqrt{\sqrt{1 + 4\zeta^4} - 2\zeta^2}$, hence $$\omega_n = \frac{0.877}{\sqrt{\sqrt{1 + 4(0.243)^4} - 2(0.243)^2}} = \frac{0.877}{0.9424} = 0.931\ \text{rad/s},$$ and with the same $K_{dc} = 0.987$, $$\boxed{G_{m3}(s) = 0.9868\,\frac{0.867}{s^2 + 0.452\,s + 0.867}, \quad \zeta = 0.243,\ \omega_n = 0.931\ \text{rad/s}.}$$
7. Compare the three models. The damping ratios are $0.216$, $0.211$ and $0.243$ (a spread of about 15%) and the natural frequencies $0.881$, $0.890$ and $0.931$ rad/s (a spread under 6%). Such close agreement is itself evidence that the dominant-pole assumption is sound. $G_{m1}$ is the most accurate of the three because it is built from exact pole locations rather than from chart readings, and because the $\zeta \approx 0.01\Phi_m$ correlation behind $G_{m3}$ is only approximate at low damping.
8. Part 4 — step-response specifications from $G_{m1}$. With $\zeta = 0.2157$, $\omega_n = 0.8807$ rad/s, $\zeta\omega_n = 0.190$ and $\omega_d = 0.860$ rad/s: the steady-state error is $e_{ss(step)\%} = (1 - K_{dc}) \times 100 = 1.32\%$ (equivalently $1/(1+K_{pos})$ with $K_{pos} = 75$); the overshoot is $PO = 100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}} = 49.95\%$; the settling time is $T_{s(\pm2\%)} = 4/0.190 = 21.1$ s; and the rise time is $T_{r(0-100\%)} = (\pi - \cos^{-1}0.2157)/0.860 = 2.08$ s. $$\boxed{e_{ss} = 1.32\%,\ PO = 50.0\%,\ T_{s} = 21.1\ \text{s},\ T_{r} = 2.08\ \text{s}.}$$ The loop is fast to rise but rings for a very long time — the classic signature of a proportional controller pushed for accuracy on a plant with a near-double pole close to the origin.

[Figure not reproduced: Figure Q5.2 (redrawn) — closed-loop magnitude for K p = 5, with the resonant peak and the low-frequency level marked. These are the two readings that produce G m2 . See the official exam paper.]