Question 3 of 8: Second Order Dominant Poles Model and Step-Response Specifications

Question 3: Second Order Dominant Poles Model and Step-Response Specifications (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. The fourth-order closed loop of the Question 2 servo at its operating gain,

Find. The DC gain; the dominant second-order model parameters $K_{dc}$, $\zeta$, $\omega_n$ and $G_m(s)$; and five unit-step response specifications from that model.

Approach. Identify which factors actually shape the response, match the model DC gain to the true DC gain, extract $\zeta$ and $\omega_n$ from the dominant quadratic, and then evaluate the standard second-order response formulas.

1. Evaluate the DC gain. Setting $s = 0$ in the polynomial form, $$G_{cl}(0) = \frac{250 \times 0.5}{125} = \frac{125}{125} = 1.00.$$ $$\boxed{G_{cl}(0) = 1.00\ \text{V/V}.}$$ This is the integral action of the PI controller showing up in the closed loop: the equivalent open loop is Type 1, so the position error constant is infinite and the DC gain is exactly unity.
2. Decide which poles dominate. Three observations settle it. First, the slow real pole at $-0.54$ sits almost exactly on top of the closed-loop zero at $-0.50$: their separation is only 0.04, so they form a near-dipole whose residue is tiny and whose net contribution to the response nearly cancels. Second, the fast pole at $-12.57$ is $13.3$ times further from the imaginary axis than the complex pair’s real part ($0.945$), so its transient has died long before the oscillation does. Third, what remains is the lightly damped pair from $s^2 + 1.89s + 18.47$, which is therefore dominant.
3. Extract the model parameters. Comparing the dominant quadratic with the standard form $s^2 + 2\zeta\omega_n s + \omega_n^2$, $$\omega_n = \sqrt{18.47} = 4.298\ \text{rad/s}, \qquad \zeta = \frac{1.89}{2 \times 4.298} = 0.220.$$ The model DC gain must reproduce the true DC gain, so $K_{dc} = G_{cl}(0) = 1.00$. Hence $$\boxed{G_m(s) = \frac{18.47}{s^2 + 1.89 s + 18.47}, \quad K_{dc} = 1.00,\ \zeta = 0.220,\ \omega_n = 4.298\ \text{rad/s}.}$$
4. Percent overshoot. With $\zeta = 0.2199$, $$PO = 100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}} = 100\,e^{-0.6908/0.9755} = 100\,e^{-0.7082} = 49.3\%.$$ A light damping ratio of about 0.22 always produces this kind of large overshoot; the response will ring visibly.
5. Settling time and damped frequency. The envelope decays as $e^{-\zeta\omega_n t}$ with $\zeta\omega_n = 0.945\ \text{s}^{-1}$, so $$T_{s(\pm 2\%)} = \frac{4}{\zeta\omega_n} = \frac{4}{0.945} = 4.23\ \text{s}.$$ The damped natural frequency is $\omega_d = \omega_n\sqrt{1-\zeta^2} = 4.298 \times 0.9755 = 4.192$ rad/s.
6. Period of oscillation. One full cycle of the ringing takes $$T_{period} = \frac{2\pi}{\omega_d} = \frac{2\pi}{4.192} = 1.50\ \text{s},$$ so roughly three visible cycles fit inside the settling time — consistent with the low damping.
7. Rise time (0–100%). For an under-damped second-order system the output first reaches its final value at $$T_{r(0-100\%)} = \frac{\pi - \theta}{\omega_d}, \qquad \theta = \cos^{-1}\zeta = \cos^{-1} 0.2199 = 1.349\ \text{rad},$$ giving $T_r = (3.1416 - 1.349)/4.192 = 0.428$ s. The rise is fast precisely because the damping is poor.
8. Steady-state error. Since $K_{dc} = 1.00$ exactly, $$e_{ss(step)\%} = |1 - K_{dc}| \times 100 = 0\%.$$ $$\boxed{PO = 49.3\%,\ T_s = 4.23\ \text{s},\ T_{period} = 1.50\ \text{s},\ T_r = 0.428\ \text{s},\ e_{ss} = 0\%.}$$