22-Elec-B2 Advanced Control Systems · December 2019

Question 4 of 5: Servo motor with rate feedback

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

Notes on this paper

Paper format. 16-Elec-B2 Advanced Control Systems, December 2019 — a three-hour open-book examination; any non-communicating calculator is permitted. The cover page states “Any four questions constitute a complete paper. Only the first four questions as they appear in your answer book will be marked” and “All questions are of equal value (25%)”. The sitting prints five questions, so each carries 25 marks and a candidate answers four. Question 1 is a seventeen-item multiple-choice block whose per-item weights are printed in the margin as [1], [2] or [3] and total exactly 25. A table of inverse Laplace transforms and a table of Laplace/z transforms are appended. All five questions are worked below, because this set is a study resource rather than a timed sitting.

Reference texts. N. S. Nise, Control Systems Engineering, 8th ed. (Ch. 4 time response, percent overshoot, peak time and settling time; Ch. 6 Routh–Hurwitz stability; Ch. 7 steady-state error and system type; Ch. 8 root-locus sketching rules including break-away/break-in points and angles of departure; Ch. 10 frequency response and stability margins); R. C. Dorf and R. H. Bishop, Modern Control Systems, 13th ed. (Ch. 5, 6, 7, 9); K. Ogata, Modern Control Engineering, 5th ed. (Ch. 5 transient response, Ch. 6 root-locus analysis, Ch. 7 frequency-response analysis); G. F. Franklin, J. D. Powell and A. Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed. (Ch. 3, 5, 6). These are the references listed for exam code 16-Elec-B2 in the Engineers Canada syllabus, and this paper’s vocabulary (“break-in point”, “intercept of the asymptotes”, “angle of departure”) follows Nise closely.

Check — three source readings checked against the printed paper.

(1) Question 1, item 2. The printed sentence reads “where $\theta(s)$, $T(s)$, $J$ and $K$ represent torque, angular displacement, moment of inertia and spring constant, respectively”. The paper really does print the first two descriptions in the wrong order. The drawing beside it settles the physics — $T(t)$ is the applied torque acting on the inertia $J$, and $\theta$ is the resulting angular displacement, which is also the only reading under which $\theta(s)/T(s)$ has the units of a compliance. The answer does not depend on the naming at all, because only the denominator is used.

(2) Question 1, item 16. The margins are read off a printed Bode pair, so they are inherently approximate. The printed Bode pair gives a gain crossover near $\omega \approx 3$ rad/s where the phase curve sits near $-116^\circ$, and a phase crossover near $\omega \approx 10$ rad/s where the magnitude curve is near $-21$ dB. Both readings point at the same option. The graph is coarse enough that the honest statement is “approximately”, exactly as the question words it.

(3) Question 2. The pole-zero map prints its two crosses to the right of the imaginary axis, at $s = 1 \pm j1$, and its two circles on the negative real axis at $-2$ and $-3$. This is the whole point of the question, because an open-loop plant with two right-half-plane poles is unstable until enough gain is applied, which is the opposite of the usual root-locus habit.

Question 4: Servo motor with rate feedback (25 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. A position servo consisting of a torque path $K/(Js+B)$ and an integrator $1/s$, with a rate (velocity) feedback $K_f$ taken from between them and an outer unity position loop.

Given data
Quantity	Symbol	Value
Moment of inertia	$J$	$1\ \text{kg}\cdot\text{m}^{2}$
Viscous damping	$B$	$1\ \text{N}\cdot\text{m}\cdot\text{s/rad}$
Maximum overshoot to a unit step	$M_p$	$0.1$ (10 %)
Peak time	$T_p$	$0.2\ \text{s}$
Unknowns	$K,\ K_f$	forward gain and rate-feedback gain

The printed servo block diagram: the rate feedback $K_f$ is tapped between $K/(Js+B)$ and the integrator, so it closes around the velocity only, while the outer unity loop closes around position.

Find. (a) $\zeta$ and $\omega_n$, (b) the settling time, (c) the overall transfer function in standard second-order form, and (d) the two gains.

Approach. The two transient specifications determine $\zeta$ and $\omega_n$ uniquely; reducing the inner loop first gives a two-parameter second-order closed loop whose coefficients are then matched to $\omega_n^{2}$ and $2\zeta\omega_n$.

Part (a) — damping ratio from the overshoot. Overshoot depends on $\zeta$ alone: $M_p = \exp\!\left(-\pi\zeta/\sqrt{1-\zeta^{2}}\right)$. Taking logarithms and solving, $$\zeta = \frac{-\ln M_p}{\sqrt{\pi^{2}+\ln^{2}M_p}} = \frac{-\ln 0.1}{\sqrt{\pi^{2}+(\ln 0.1)^{2}}} = \frac{2.3026}{\sqrt{9.8696+5.3019}} = \frac{2.3026}{3.8951}$$ $$\boxed{\zeta = 0.5912}$$
Natural frequency from the peak time. The peak occurs after exactly half a cycle of the damped oscillation, $T_p = \pi/\omega_d$, so $$\omega_d = \frac{\pi}{T_p} = \frac{\pi}{0.2} = 15.708\ \text{rad/s}, \qquad \omega_n = \frac{\omega_d}{\sqrt{1-\zeta^{2}}} = \frac{15.708}{0.80656}$$ $$\boxed{\omega_n = 19.475\ \text{rad/s}}$$ Substituting back reproduces $M_p = 0.100$ and $T_p = 0.200$ s, so the pair is consistent.
Part (b) — settling time from the envelope. The transient decays inside an envelope $e^{-\zeta\omega_n t}$, with $\zeta\omega_n = 0.5912 \times 19.475 = 11.513\ \text{s}^{-1}$. Using the usual 2 % criterion, $$T_s = \frac{4}{\zeta\omega_n} = \frac{4}{11.513} = \boxed{0.3474\ \text{s}}$$ On the 5 % criterion the same envelope gives $T_s = 3/(\zeta\omega_n) = 0.2606$ s. Either is acceptable provided the criterion is stated; the 2 % figure is quoted here because it is the convention used throughout Nise.
Part (c) — reduce the inner rate loop first. The inner loop has forward path $K/(Js+B)$ and feedback $K_f$, so $$G_{\text{inner}}(s) = \frac{K/(Js+B)}{1 + K_f\,K/(Js+B)} = \frac{K}{Js+B+KK_f}$$ The rate feedback therefore does nothing but add $KK_f$ to the mechanical damping $B$ — that is the whole engineering point of tachometer feedback.
Cascade the integrator and close the outer loop. With the integrator in series, $$G(s) = \frac{K}{s\left(Js+B+KK_f\right)}, \qquad \frac{Y(s)}{X(s)} = \frac{G(s)}{1+G(s)} = \frac{K}{Js^{2}+\left(B+KK_f\right)s+K}$$ Dividing through by $J$ puts it in standard form: $$\boxed{\frac{Y(s)}{X(s)} = \frac{K/J}{s^{2} + \dfrac{B+KK_f}{J}\,s + \dfrac{K}{J}} = \frac{\omega_n^{2}}{s^{2}+2\zeta\omega_n s + \omega_n^{2}}}$$ Comparing coefficients gives the two design equations $\omega_n^{2} = K/J$ and $2\zeta\omega_n = (B+KK_f)/J$. With $J = B = 1$ these reduce to $T(s) = K/[s^{2}+(1+KK_f)s+K]$.
Part (d) — the forward gain. The first design equation involves only $K$: $$K = J\,\omega_n^{2} = (1)(19.475)^{2} = \boxed{379.29}$$ Its units are $\text{N}\cdot\text{m/rad}$, since $K/J$ must have units of $\text{s}^{-2}$.
The rate-feedback gain. Substituting $K$ into the damping equation, $$B + KK_f = 2\zeta\omega_n J = 2(0.5912)(19.475)(1) = 23.026$$ $$K_f = \frac{2\zeta\omega_n J - B}{K} = \frac{23.026-1}{379.29} = \boxed{K_f = 0.05807\ \text{s}}$$ Note how little of the required damping the motor supplies on its own: $B = 1$ out of the 23.03 needed, so the tachometer loop provides 96 % of it. Without the rate feedback ($K_f = 0$) the same $K$ would give $\zeta = 1/(2\sqrt{379.29}) = 0.0257$ — about 92 % overshoot — which is why the inner loop is there.
Confirm the design against the two specifications. With $K = 379.29$ and $K_f = 0.05807$ the closed-loop denominator is $s^{2}+23.026s+379.29$, whose roots are $-11.513 \pm j15.708$; their magnitude is $19.475 = \omega_n$ and $-\operatorname{Re}(s)/|s| = 0.5912 = \zeta$, as required. Simulating the unit-step response gives a peak of 1.100 at $t = 0.200$ s, reproducing both specifications exactly, and the dc gain $K/K = 1$ confirms zero steady-state position error.

Question 4: the designed closed-loop step response. The peak of 1.10 at $t = 0.2$ s is the pair of specifications, and the response is inside the 2 % band from $T_s = 0.347$ s onward.

Question 4 — results
Part	Quantity	Symbol	Result
(a)	Damping ratio	$\zeta$	$0.5912$
(a)	Natural frequency	$\omega_n$	$19.475\ \text{rad/s}$
(a)	Damped frequency	$\omega_d$	$15.708\ \text{rad/s}$
(b)	Settling time (2 %)	$T_s$	$0.3474\ \text{s}$
(b)	Settling time (5 %)	$T_s$	$0.2606\ \text{s}$
(c)	Overall transfer function	$Y/X$	$K/[Js^{2}+(B+KK_f)s+K]$
(d)	Forward gain	$K$	$379.29$
(d)	Rate-feedback gain	$K_f$	$0.05807\ \text{s}$

Topic. Specification-driven design of a position servo, and the role of tachometer (rate) feedback in an inner loop.

Key relations. $M_p = e^{-\pi\zeta/\sqrt{1-\zeta^{2}}}$; $T_p = \pi/(\omega_n\sqrt{1-\zeta^{2}})$; $T_s = 4/(\zeta\omega_n)$ (2 %) or $3/(\zeta\omega_n)$ (5 %); inner-loop reduction $G/(1+GH)$; coefficient matching $\omega_n^{2} = K/J$ and $2\zeta\omega_n = (B+KK_f)/J$.

Why this works. The two transient specifications carry exactly two pieces of information, and a second-order loop has exactly two free parameters, so the design is a square problem with a unique solution. Rate feedback is the mechanism that decouples them: $K$ alone fixes $\omega_n$, and $K_f$ then buys whatever damping is missing without touching $\omega_n$. Raising $B$ physically would do the same job but costs efficiency and cannot be tuned after the machine is built.

Common pitfall. Closing the outer loop first, or treating $K_f$ as if it fed back position rather than velocity — both give a first-order-looking denominator and a wrong $K_f$. The other frequent slip is forgetting $J$ in $2\zeta\omega_n = (B+KK_f)/J$, which is invisible here because $J = 1$ but is fatal on any variant with $J \ne 1$.

Source. Ogata, Modern Control Engineering, 5th ed., §5.4 (servo system with velocity feedback, worked in the same block-diagram form); Nise, Control Systems Engineering, 8th ed., §4.6 and §5.2 (block-diagram reduction).

← Question 3 Next: Question 5 →

Quantity	Symbol	Value
Moment of inertia	\(J\)	\(1\ \text{kg}\cdot\text{m}^{2}\)
Viscous damping	\(B\)	\(1\ \text{N}\cdot\text{m}\cdot\text{s/rad}\)
Maximum overshoot to a unit step	\(M_p\)	\(0.1\) (10 %)
Peak time	\(T_p\)	\(0.2\ \text{s}\)
Unknowns	\(K,\ K_f\)	forward gain and rate-feedback gain

Part	Quantity	Symbol	Result
(a)	Damping ratio	\(\zeta\)	\(0.5912\)
(a)	Natural frequency	\(\omega_n\)	\(19.475\ \text{rad/s}\)
(a)	Damped frequency	\(\omega_d\)	\(15.708\ \text{rad/s}\)
(b)	Settling time (2 %)	\(T_s\)	\(0.3474\ \text{s}\)
(b)	Settling time (5 %)	\(T_s\)	\(0.2606\ \text{s}\)
(c)	Overall transfer function	\(Y/X\)	\(K/[Js^{2}+(B+KK_f)s+K]\)
(d)	Forward gain	\(K\)	\(379.29\)
(d)	Rate-feedback gain	\(K_f\)	\(0.05807\ \text{s}\)