22-Elec-A2 Systems and Control · May 2018

Question 2 of 8: Error Constants and the 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 2: Error Constants and the Second-Order Dominant-Poles Model (20 marks, compulsory)

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. $K_{op} = 3.0$, no disturbance, and the two transfer functions supplied by the paper: $G_{open}(s) = 0.6(s+10)/[s(s^{2}+2s+3)]$ and $G_{cl}(s) = 0.6(s+10)/[(s+1.827)(s^{2}+0.173s+3.284)]$. Both agree with the reduction of Question 1 at $K_p = 3$.

Find. The three static error constants and their steady-state errors; the closed-loop DC gain; the dominant second-order model $(K_{dc},\zeta,\omega_n)$ and its transfer function; and the resulting unit-step percent overshoot, 2 % settling time and 0–100 % rise time.

Closed-loop pole–zero map at $K_{op}=3$. The lightly damped pair $-0.0865 \pm j1.8101$ sits far closer to the imaginary axis than the real pole at $-1.827$ (a separation of 21.1 to 1), so the pair dominates; the zero at $-10$ is remote and is absorbed into the DC gain.

Approach. Read the system type off $G_{open}$ to get the error constants by the standard limits, evaluate $G_{cl}(0)$ directly, then identify the dominant pole pair from the given factorisation and match it to the page-3 second-order template before applying the step-response correlations.

Classify the loop and evaluate the error constants. $G_{open}$ has exactly one pole at the origin, so the loop is Type 1. The three static constants follow from their defining limits: $$K_{pos} = \lim_{s\to 0} G_{open}(s) = \infty, \qquad K_v = \lim_{s\to 0} sG_{open}(s) = \frac{0.6\times 10}{3} = 2.0, \qquad K_a = \lim_{s\to 0} s^{2}G_{open}(s) = 0$$ The velocity constant is the only finite one, which is the signature of a Type 1 system.
Convert the constants into steady-state errors. For unity feedback, $e_{ss} = 1/(1+K_{pos})$ for a step, $1/K_v$ for a ramp and $1/K_a$ for a parabola: $$e_{ss(step)} = \frac{1}{1+\infty} = \boxed{\;0\%\;}, \qquad e_{ss(ramp)} = \frac{1}{K_v} = \frac{1}{2.0} = \boxed{\;0.5\ \text{V/V}\;}, \qquad e_{ss(parab)} = \frac{1}{0} = \boxed{\;\infty\;}$$ Physically: the arm reaches any commanded constant speed exactly, tracks a constantly accelerating speed command with a fixed 0.5 V/V lag, and cannot follow a parabolic command at all.
Closed-loop DC gain. Substituting $s = 0$ into the given factorisation, $$G_{cl}(0) = \frac{0.6 \times 10}{1.827 \times 3.284} = \frac{6}{6.000} = \boxed{\;1.000\;}$$ Unity DC gain is the direct consequence of the zero step error found above — the two answers are the same statement seen from either side of the loop.
Test dominance and extract the model parameters. The quadratic factor $s^{2}+0.173s+3.284$ gives, by comparison with $s^{2}+2\zeta\omega_n s + \omega_n^{2}$, $$\omega_n = \sqrt{3.284} = 1.8122\ \text{rad/s}, \qquad \zeta = \frac{0.173}{2\omega_n} = \frac{0.173}{3.6244} = 0.04773$$ so the pair sits at $-\zeta\omega_n \pm j\omega_d = -0.0865 \pm j1.8101$. The third pole is at $-1.827$, and the separation ratio is $1.827/0.0865 = 21.1$ — far beyond the usual factor-of-five rule, so the complex pair is strongly dominant and the model is justified.
Write the model transfer function. Matching the page-3 template $G_m(s) = K_{dc}\,\omega_n^{2}/(s^{2}+2\zeta\omega_n s + \omega_n^{2})$ with $K_{dc} = G_{cl}(0) = 1$: $$\boxed{\;G_m(s) = \frac{3.284}{s^{2} + 0.173\,s + 3.284}\;}$$ Note that $K_{dc}$ is taken from the exact $G_{cl}(0)$, not from the quadratic alone; this is what makes the remote zero at $-10$ and the third pole cancel out of the DC behaviour.
Step-response specifications. With $\zeta = 0.04773$ the page-3 overshoot correlation gives $$PO = 100\,e^{-\zeta\pi/\sqrt{1-\zeta^{2}}} = 100\,e^{-0.15011} = \boxed{\;86.06\%\;}$$ and the settling and rise times follow from $\zeta\omega_n = 0.0865$ and $\omega_d = \omega_n\sqrt{1-\zeta^{2}} = 1.8101$: $$T_{settle(\pm 2\%)} = \frac{4}{\zeta\omega_n} = \frac{4}{0.0865} = \boxed{\;46.24\ \text{s}\;}, \qquad T_{rise(0-100\%)} = \frac{\pi - \cos^{-1}\zeta}{\omega_d} = \frac{3.1416 - 1.5231}{1.8101} = \boxed{\;0.894\ \text{s}\;}$$ The contrast between them is the engineering message: the arm gets to the commanded speed in well under a second, but then rings for the better part of a minute.

Quantity	Result
System type	Type 1
$K_{pos}$ / $e_{ss(step)}$	$\infty$ / $0\%$
$K_v$ / $e_{ss(ramp)}$	$2.0$ / $0.5\ \text{V/V}$
$K_a$ / $e_{ss(parab)}$	$0$ / $\infty$
DC gain $G_{cl}(0) = K_{dc}$	$1.000$
Damping ratio $\zeta$	$0.04773$
Natural frequency $\omega_n$	$1.8122\ \text{rad/s}$
Model $G_m(s)$	$\dfrac{3.284}{s^{2}+0.173s+3.284}$
Percent overshoot $PO$	$86.06\%$
$T_{settle(\pm 2\%)}$	$46.24\ \text{s}$
$T_{rise(0-100\%)}$	$0.894\ \text{s}$

Check: the dominant-pair model is mathematically sound here (21:1 separation), but $\zeta = 0.0477$ means the loop is operating at 86 % overshoot and rings for roughly 46 s. For a robot joint that is not a usable tuning — $K_{op}=3$ sits at 80 % of the $K_{crit}=3.75$ found in Question 1, i.e. a gain margin of only 1.25. Question 3 shows why the paper nevertheless selects it (the ramp-error specification forces it), and the honest engineering conclusion is that this specification and a well-damped response cannot both be met by proportional gain alone.

Topic. Static error constants for a unity-feedback loop, and reduction of a high-order closed loop to an equivalent second-order dominant-poles model.

Key relations. $K_{pos}=\lim_{s\to0}G$, $K_v=\lim_{s\to0}sG$, $K_a=\lim_{s\to0}s^{2}G$, with $e_{ss}=1/(1+K_{pos})$, $1/K_v$, $1/K_a$ for step, ramp and parabolic inputs; $s^{2}+2\zeta\omega_n s+\omega_n^{2}$ as the standard second-order denominator; $PO=100e^{-\zeta\pi/\sqrt{1-\zeta^{2}}}$; $T_{settle(\pm2\%)}\approx 4/(\zeta\omega_n)$; $T_{rise(0-100\%)}=(\pi-\cos^{-1}\zeta)/\omega_d$.

Why this works. The static error constants are just the final-value theorem applied to $E(s)=R(s)/(1+G_{open}(s))$ for the three canonical inputs, so the system type — the number of free integrators — determines which constant is finite and therefore which input the loop can track with bounded error. The dominant-poles idea rests on the fact that each pole contributes a term decaying like $e^{-\sigma t}$: a pole ten or twenty times further left has already vanished while the slow pair is still ringing, so the transient is effectively second order. Taking $K_{dc}$ from the exact $G_{cl}(0)$ rather than from the quadratic keeps the model faithful at DC, so the remote zero and fast pole are accounted for in amplitude even though they are dropped from the dynamics.

Common pitfall. Reporting $e_{ss(step)}$ as a small non-zero number because $K_{pos}$ was computed from $G_{cl}$ rather than $G_{open}$ — the error constants are always defined on the open-loop function. A second common slip is forming $\zeta$ from the wrong coefficient: the linear coefficient is $2\zeta\omega_n$, not $\zeta\omega_n$, so forgetting the factor of two doubles the damping ratio and understates the overshoot dramatically. Finally, do not assume $K_{dc}=1$ by reflex; it is unity here only because the loop is Type 1.

Source. Nise, Control Systems Engineering 7th ed., Ch. 7 (system type and static error constants) and §4.6–4.8 (second-order response, overshoot and settling/rise-time correlations); Ogata 5th ed., §5.3 and §5.8.

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