22-Elec-A2 Systems and Control · May 2018

Question 3 of 8: Steady-State Error by Superposition and the Final Value Theorem

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 3: Steady-State Error by Superposition and the Final Value Theorem (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 Question 1 servo, with $G_{open}(s) = 0.2K_p(s+10)/[s(s^{2}+2s+3)]$ and $G_{dist}(s) = -0.1s(s+1)/\Delta(s)$, where $\Delta(s) = s^{3}+2s^{2}+(3+0.2K_p)s+2K_p$. Specifications: zero step error and $e_{ss(ramp)} = 0.5\ \text{V/V}$. Test signals for parts 3 and 4: $\omega_{ref} = 2t\cdot 1(t)$ (a ramp of slope 2) and $T_{dist} = 10t\cdot 1(t)$ (a ramp of slope 10). From Question 1, $K_{crit} = 3.75$.

Find. The operating gain that meets the ramp specification; whether the loop is still stable there and its gain margin in V/V; and the total steady-state error under simultaneous reference and disturbance ramps at $K_p = 1$ and at $K_p = K_{op}$.

Approach. The step specification is automatic for a Type 1 loop, so the ramp specification alone fixes $K_v$ and hence $K_{op}$; compare that gain with $K_{crit}$ for the margin. For parts 3 and 4, superposition lets each input be pushed through its own transfer function with the final value theorem, and the two error contributions are then added with the correct signs.

Express $K_v$ as a function of the gain. From the open-loop function of Question 1, $$K_v(K_p) = \lim_{s\to 0} s\,G_{open}(s) = \lim_{s\to 0} \frac{0.2K_p(s+10)}{s^{2}+2s+3} = \frac{0.2K_p \times 10}{3} = \frac{2K_p}{3}$$ The zero-step-error requirement needs no work: the loop is Type 1 for every $K_p \gt 0$, so $K_{pos}=\infty$ and the step error is identically zero. The ramp specification is therefore the only binding one.
Solve for the operating gain. Setting $e_{ss(ramp)} = 1/K_v = 0.5$ requires $K_v = 2$, hence $$\frac{2K_p}{3} = 2 \;\Longrightarrow\; \boxed{\;K_{op} = 3.0\;}$$ This is exactly the operating gain that Question 2 was handed as a given, which closes the loop between the two questions.
Stability and gain margin at the operating point. Question 1 established stability for $0 \lt K_p \lt 3.75$. Since $K_{op} = 3.0 \lt 3.75$, the system is still stable. The gain margin is the factor by which the gain may be raised before the loop reaches the stability boundary: $$G_m = \frac{K_{crit}}{K_{op}} = \frac{3.75}{3.0} = \boxed{\;1.25\ \text{V/V}\;} \quad (= 20\log_{10}1.25 = 1.94\ \text{dB})$$ A margin of 1.25 is very thin — a 25 % gain drift, or any unmodelled lag, would take this joint unstable. That is the price of meeting the ramp specification with proportional gain alone.
Set up the superposition at $K_p = 1$. With two inputs acting at once the total error is the sum of the two separate contributions. The reference ramp of slope 2 contributes, through the velocity constant, $$e_{ref} = \frac{2}{K_v(1)} = \frac{2}{2(1)/3} = \frac{2}{0.6667} = 3.000$$ The disturbance contributes an output deviation, obtained from the final value theorem applied to $G_{dist}$ with the ramp $10/s^{2}$: $$\omega_{load,dist}(\infty) = \lim_{s\to 0} s\,G_{dist}(s)\,\frac{10}{s^{2}} = 10\lim_{s\to 0}\frac{-0.1(s+1)}{\Delta(s)} = \frac{-10 \times 0.1}{2K_p} = \frac{-1}{2(1)} = -0.500$$
Combine the two contributions with the correct sign. Because the error is $e = \omega_{ref} - \omega_{load}$, an output pushed down by the load torque increases the error. The disturbance contribution to the error is therefore $-(-0.500) = +0.500$, and $$e_{ss,total}\big|_{K_p=1} = 3.000 + 0.500 = \boxed{\;3.500\ \text{V/V}\;}$$ The sign reasoning is the graded step here: a resisting torque and a tracking lag both hold the arm behind its command, so the two effects add rather than partially cancel.
Repeat at the operating gain $K_p = K_{op} = 3$. Both terms scale inversely with the gain: $$e_{ref} = \frac{2}{K_v(3)} = \frac{2}{2.0} = 1.000, \qquad e_{dist} = \frac{10 \times 0.1}{2(3)} = \frac{1}{6} = 0.1667$$ $$e_{ss,total}\big|_{K_p=3} = 1.000 + 0.1667 = \boxed{\;1.1667\ \text{V/V}\;}$$ Tripling the gain cuts the total error by a factor of three, from 3.500 to 1.167 — but Question 2 showed that the same change drives the overshoot to 86 % and leaves a gain margin of only 1.25. This is the classic accuracy-versus-stability trade of proportional control, and it is the reason the remaining questions on this paper turn to lead, PD and state-feedback compensation.

Quantity	Result
Required operating gain $K_{op}$	$3.0$
Stable at $K_{op}$?	Yes ($K_{op}=3.0 \lt K_{crit}=3.75$)
Gain margin $G_m$	$1.25\ \text{V/V}$ ($1.94\ \text{dB}$)
$K_p=1$: reference-ramp error	$3.000$
$K_p=1$: disturbance-ramp error	$+0.500$
$K_p=1$: total $e_{ss}$	$3.500\ \text{V/V}$
$K_p=K_{op}$: reference-ramp error	$1.000$
$K_p=K_{op}$: disturbance-ramp error	$+0.1667$
$K_p=K_{op}$: total $e_{ss}$	$1.1667\ \text{V/V}$

Topic. Superposition of reference-tracking and disturbance-rejection errors in a linear feedback loop, evaluated with the final value theorem.

Key relations. $K_v = \lim_{s\to0}sG_{open}$ and $e_{ss(ramp)} = R/K_v$ for a ramp of slope $R$; the final value theorem $x(\infty) = \lim_{s\to0}sX(s)$, valid whenever the closed loop is stable; gain margin $G_m = K_{crit}/K_{op}$; and, for two simultaneous inputs, $e_{total} = e_{ref} + e_{dist}$ with $e_{dist} = -\lim_{s\to0}sG_{dist}(s)T_d(s)$.

Why this works. Superposition holds because the system is linear: the response to reference and disturbance acting together is the sum of the responses to each acting alone, so two independent final-value calculations can simply be added. The final value theorem converts each of those into a single limit, which is why no time-domain solution is ever needed. The essential subtlety is bookkeeping on signs. The reference contribution is naturally an error, but the disturbance calculation returns an output deviation, and error is defined as command minus output — so that result must be negated before the two are combined. Here the load torque enters with gain $-1$ and drags the speed down, so its error contribution is positive and adds to the tracking lag. Note also that the theorem is only legitimate because Question 1 confirmed stability at these gains; applying it to an unstable loop returns a finite number that means nothing.

Common pitfall. Subtracting the disturbance term instead of adding it, on the intuition that "feedback opposes disturbances" — feedback reduces the disturbance error but does not reverse its sign. The other frequent error is forgetting the ramp slope: $e_{ss}=1/K_v$ applies to a unit ramp, so a slope-2 command gives $2/K_v$, and using the unit-ramp formula halves the answer here.

Source. Nise, Control Systems Engineering 7th ed., §7.5–7.6 (steady-state error for disturbances, and error in terms of the static constants) and §10.7 (gain and phase margins); Ogata 5th ed., §5.8.

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