22-Elec-A2 Systems and Control · May 2018

Question 8 of 8: Three Routes to a Second-Order Dominant-Poles Model

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, 16-Elec-A2 Systems and Control, May 2018 — 3 hours, closed book (approved Casio or Sharp calculator plus one double-sided handwritten 8.5 × 11″ formula sheet). A short Laplace-transform table is printed on page 2, and the $\zeta$–overshoot, $\zeta$–resonant-peak and phase-margin design charts plus the second-order model, controllability/observability and state-space transfer-function definitions are on page 3. Questions 1 and 2 are compulsory; five questions constitute a complete paper, so a candidate chooses three of Q3–Q8. Each question carries 20 marks, for 100 in total. All eight questions are worked below so the set is a complete study resource.

Reference texts. N. S. Nise, Control Systems Engineering (7th ed., Wiley) — block-diagram and signal-flow reduction with Mason’s rule (Ch. 5), steady-state error and the static error constants (Ch. 7), Routh–Hurwitz (Ch. 6), root locus (Ch. 8), cascade compensator design (Ch. 9), frequency response and gain/phase margins (Ch. 10), frequency-domain lead design (Ch. 11), state space and controllability/observability (Ch. 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — second-order correlations, dominant-poles modelling and pole placement by state feedback. Every block diagram, root locus, pole–zero map, Bode plot, closed-loop magnitude curve and step response below is redrawn as an inline figure.

Reading the supplied design charts. Page 3 gives $PO = 100\,e^{-\zeta\pi/\sqrt{1-\zeta^{2}}}$ for percent overshoot, $M_r/K_{dc} = 1/\left(2\zeta\sqrt{1-\zeta^{2}}\right)$ for the closed-loop resonant peak, and the phase-margin correlation $\Phi_m \approx 100\zeta$ (equivalently $\zeta \approx 0.01\,\Phi_m$). Those three relations, together with $T_{settle(\pm 2\%)} \approx 4/(\zeta\omega_n)$ and $T_{rise(0-100\%)} = (\pi - \cos^{-1}\zeta)/\omega_d$, carry most of the numerical work in Questions 2, 5, 7 and 8.

Question 8: Three Routes to a Second-Order Dominant-Poles Model (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. Unity-feedback loop with $K_pG_p(s) = 100/[s(s+5)^{2}]$ at $K_p = 1$; closed-loop poles $-0.7791 \pm j3.3524$ and $-8.4418$; open-loop Bode plots (Figure Q8.2) and the closed-loop magnitude plot (Figure Q8.3), from which the resonant peak reads $M_r \approx 2.15\ \text{V/V}$ at $\omega_r \approx 4\ \text{rad/s}$ with a low-frequency magnitude of 1.

Find. Three second-order models — from the pole coordinates, from the open-loop margins, and from the closed-loop resonance — then a comparison and the step-response specifications from the most accurate one.

[Figure not reproduced: Figure Q8.1 (redrawn) — unity-feedback loop with proportional controller $K_p$ and plant $100/[s(s+5)^{2}]$. See the official exam paper.]

Approach. Model 1 comes straight from the given poles once dominance is verified. Model 2 uses the open-loop phase margin through $\zeta \approx 0.01\Phi_m$ and the crossover-to-$\omega_n$ relation. Model 3 inverts the resonant-peak formula from page 3. All three share $K_{dc}=1$ because the loop is Type 1.

Verify dominance and build model 1. The complex pair has $$\omega_n = |p_1| = \sqrt{0.7791^{2}+3.3524^{2}} = 3.4417\ \text{rad/s}, \qquad \zeta = \frac{0.7791}{3.4417} = 0.22637$$ and the separation from the real pole is $$\frac{|p_3|}{|\mathrm{Re}\,p_1|} = \frac{8.4418}{0.7791} = 10.8 \;\gt\; 5$$ so the dominant-pair approximation is valid. The loop is Type 1, hence $K_{dc}=1$, and $$\boxed{\;G_{m1}(s) = \frac{11.845}{s^{2}+1.5582\,s+11.845}\;}$$ As a check, the three given poles multiply out to $s^{3}+10s^{2}+25s+100$, exactly the closed-loop denominator $s(s+5)^{2}+100$ — so the supplied factorisation is exact.
Model 2 from the open-loop margins. Solving $|G_p(j\omega)| = 100/[\omega(\omega^{2}+25)] = 1$ gives the gain crossover, and the phase there is $-90^{\circ} - 2\tan^{-1}(\omega/5)$: $$\omega_{cp} = 2.961\ \text{rad/s}, \qquad \Phi_m = 28.73^{\circ}$$ Applying the page-3 correlation $\zeta \approx 0.01\Phi_m = 0.2873$ and converting crossover to natural frequency, $$\omega_n = \frac{\omega_{cp}}{\sqrt{\sqrt{1+4\zeta^{4}}-2\zeta^{2}}} = \frac{2.961}{0.9211} = 3.215\ \text{rad/s}$$ $$\boxed{\;G_{m2}(s) = \frac{10.335}{s^{2}+1.8468\,s+10.335}\;}$$
Model 3 from the closed-loop resonance. The magnitude plot starts at 1 at low frequency, confirming $K_{dc}=1$, and peaks at $M_r \approx 2.15$. Inverting the page-3 resonant-peak relation, $$\frac{M_r}{K_{dc}} = \frac{1}{2\zeta\sqrt{1-\zeta^{2}}} = 2.15 \;\Longrightarrow\; 4\zeta^{2}(1-\zeta^{2}) = \frac{1}{2.15^{2}} \;\Longrightarrow\; \zeta = 0.2395$$ (taking the lightly damped root, as the peak requires $\zeta \lt 0.707$). From $\omega_r = \omega_n\sqrt{1-2\zeta^{2}}$ with $\omega_r \approx 4$: $$\omega_n = \frac{4}{\sqrt{1-2(0.2395)^{2}}} = 4.251\ \text{rad/s}$$ $$\boxed{\;G_{m3}(s) = \frac{18.074}{s^{2}+2.0365\,s+18.074}\;}$$
Compare the three models. All three agree that the loop is lightly damped with $\zeta$ between 0.23 and 0.29 and $\omega_n$ between 3.2 and 4.3 rad/s, which is reassuring, but they are not equally trustworthy:
- $G_{m1}$ is exact in its damping and natural frequency, because it is built from the true closed-loop poles rather than from a chart reading or an empirical correlation. Its only approximation is discarding the pole at $-8.44$, and the 10.8:1 separation makes that safe.
- $G_{m2}$ relies on $\Phi_m \approx 100\zeta$, which the page-3 chart itself shows to be the approximate (straight-line) relationship; it overestimates $\zeta$ by about 27 %.
- $G_{m3}$ depends on reading a peak off a printed curve. Computing the true closed-loop magnitude gives $M_r = 2.116$ at $\omega_r = 3.235\ \text{rad/s}$ — the peak height was read well, but the peak frequency read of 4 rad/s is about 24 % high, which is what inflates $\omega_n$ to 4.25.
Model 1 is therefore the one to quote.
Step-response specifications from $G_{m1}$. The loop is Type 1, so $$e_{ss}(\text{step}\%) = \boxed{\;0\%\;}$$ With $\zeta = 0.22637$ and $\omega_d = \omega_n\sqrt{1-\zeta^{2}} = 3.3524$ (exactly the imaginary part of the given pole, as it must be), $$T_{rise(0-100\%)} = \frac{\pi - \cos^{-1}\zeta}{\omega_d} = \frac{3.1416-1.3423}{3.3524} = \boxed{\;0.537\ \text{s}\;}$$ $$PO = 100\,e^{-\zeta\pi/\sqrt{1-\zeta^{2}}} = 100\,e^{-0.7301} = \boxed{\;48.2\%\;}$$ A 48 % overshoot on a joint under plain proportional control is the practical message: this loop needs derivative action or a lead network before it is usable.

Closed-loop pole map at $K_p=1$. The pair $-0.7791 \pm j3.3524$ lies 10.8 times closer to the imaginary axis than the real pole at $-8.4418$, which is what licenses the second-order reduction.

Exact closed-loop magnitude $|G_{cl}(j\omega)|$ for $G_{cl}=100/(s^{3}+10s^{2}+25s+100)$. The true resonant peak is 2.116 V/V at 3.235 rad/s: Figure Q8.3’s peak height of 2.15 is a good read, but its peak frequency of 4 rad/s is about 24 % high — the source of the discrepancy in model 3.

Quantity	Result
Dominance ratio	$10.8 \gt 5$ — model applies
Model 1 (from poles): $\zeta$, $\omega_n$	$0.2264$, $3.442\ \text{rad/s}$
$G_{m1}(s)$	$\dfrac{11.845}{s^{2}+1.5582s+11.845}$
Model 2 (open-loop Bode): $\omega_{cp}$, $\Phi_m$	$2.961\ \text{rad/s}$, $28.73^{\circ}$
$G_{m2}(s)$	$\dfrac{10.335}{s^{2}+1.8468s+10.335}$
Model 3 (closed-loop $M_r$): $\zeta$, $\omega_n$	$0.2395$, $4.251\ \text{rad/s}$
$G_{m3}(s)$	$\dfrac{18.074}{s^{2}+2.0365s+18.074}$
Most accurate model	$G_{m1}$ (built from exact poles)
$e_{ss}(\text{step}\%)$	$0\%$ (Type 1)
$T_{rise(0-100\%)}$	$0.537\ \text{s}$
$PO$	$48.2\%$

Check: Figure Q8.2 is inconsistent with the stated plant. The printed open-loop chart shows 20 dB at 0.1 rad/s and a 0 dB crossing at 1 rad/s, whereas $100/[s(s+5)^{2}]$ has 52 dB at 0.1 rad/s and crosses at 2.961 rad/s. As in Question 5, the printed chart appears to be idealised, so $\Phi_m$ and $\omega_{cp}$ for model 2 were computed analytically from the given transfer function. The closed-loop chart (Figure Q8.3) is by contrast broadly consistent — its unit low-frequency magnitude and 2.15 peak height both check out against the exact response — so only its peak-frequency read was replaced. A candidate working purely graphically would reach the same three-model comparison and the same conclusion that the pole-based model is the one to trust.

Topic. Three independent routes to the same second-order dominant-poles model — from s-plane pole coordinates, from open-loop frequency margins, and from the closed-loop resonant peak — and how to judge between them.

Key relations. From poles: $\omega_n = |p|$, $\zeta = -\mathrm{Re}(p)/|p|$. From open loop: $\zeta \approx 0.01\Phi_m$ and $\omega_{cp} = \omega_n\sqrt{\sqrt{1+4\zeta^{4}}-2\zeta^{2}}$. From closed loop: $M_r/K_{dc} = 1/(2\zeta\sqrt{1-\zeta^{2}})$ and $\omega_r = \omega_n\sqrt{1-2\zeta^{2}}$ (a peak exists only for $\zeta \lt 0.707$). Step specifications: $PO = 100e^{-\zeta\pi/\sqrt{1-\zeta^{2}}}$, $T_{rise(0-100\%)} = (\pi-\cos^{-1}\zeta)/\omega_d$.

Why this works. All three routes describe the same underlying resonance, so they must agree to within the accuracy of their inputs, and comparing them is really a comparison of evidence quality. The pole-based model uses exact algebraic data and needs only the dominance assumption. The margin-based model adds the empirical correlation $\Phi_m \approx 100\zeta$, which the page-3 chart presents explicitly as a straight-line approximation to a curved true relationship, so it carries a known bias — visible here as a 27 % overestimate of $\zeta$. The resonance-based model uses an exact formula but feeds it two numbers read off a printed curve, and errors in the graphical read propagate directly; the peak height was read well and the peak frequency was not, which is exactly why $\zeta$ from model 3 nearly matches model 1 while $\omega_n$ does not. Recognising that the exact-data model wins, and saying why, is the substance of part 4.

Common pitfall. Taking the wrong root when inverting the resonant-peak relation. Writing $x=\zeta^{2}$ turns it into $4x(1-x) = 1/M_r^{2}$, a quadratic with two positive roots; the physical one is the smaller, since a resonant peak requires light damping. Another frequent error is confusing $\omega_r$ with $\omega_n$ or with $\omega_d$ — the three differ by the factors $\sqrt{1-2\zeta^{2}}$ and $\sqrt{1-\zeta^{2}}$. Finally, do not average the three models to "split the difference": they have unequal reliability, and the correct engineering answer is to identify the best-founded one and use it.

Source. Nise, Control Systems Engineering 7th ed., §10.8–10.11 (relation between closed-loop frequency response, phase margin and the transient specifications) and §4.6–4.8; Ogata 5th ed., §7.2 and §7.10.

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