22-Elec-B2 Advanced Control Systems · December 2018

Question 5 of 5: Design on a printed circular locus — and the limits of a PD compensator

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

Notes on this paper

Paper format. 16-Elec-B2 Advanced Control Systems, December 2018 — a three-hour open-book examination. 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%)”. This 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] or [2] and total exactly 25. Tables of inverse Laplace transforms and of Laplace/z transforms are appended to the paper. 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 and the rise-time/peak-time/settling-time relations, Ch. 6 Routh–Hurwitz stability, Ch. 7 steady-state error and system type, Ch. 8 root-locus sketching rules, Ch. 9 root-locus design of cascade compensators, 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 design, Ch. 7 frequency-response design); 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 (percent overshoot, settling time, asymptote intercept, “summation of the angles”) follows Nise closely.

Check — two source readings that were checked against the printed paper.

(1) Question 1, item e. The Routh table is confirmed from the printed paper, because the $\varepsilon$-row entries are the whole content of the item. The paper prints the $s^2$ first-column entry as $(1-4\varepsilon)/\varepsilon$, which is exactly what the Routh–Hurwitz recursion gives, and the printed $s^1$ entry $(2\varepsilon^2+1-4\varepsilon)/(1-4\varepsilon)$ is derivable only from that value — so the table is correct and internally consistent, and the two rows cross-check each other. The count of sign changes, and hence the answer, is the same for $\varepsilon\to0^+$ and $\varepsilon\to0^-$.

(2) Question 4. The block diagram prints the plant as $10/[s(s+2)(s+10)]$ with no $K$ shown, while part (a) says “the open-loop transfer function when $K = 20$”. This solution takes the direct reading, $G(s)=K/[s(s+2)(s+10)]$ with $K = 20$, so the printed 10 is one sample value of the same numerator gain. The alternative reading — a fixed plant gain of 10 multiplied by $K$, i.e. $G(s)=10K/[s(s+2)(s+10)]$ — is carried through in a callout under part (b) so that both mark schemes are covered.

Question 5: Design on a printed circular locus — and the limits of a PD compensator (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.

Quantity	Value
Loop transfer function	$G(s) = K(s+2)(s+1)/[(s-2)(s-1)]$, unity feedback
Open-loop poles	$s = +1,\ +2$ (both unstable)
Open-loop zeros	$s = -1,\ -2$
Printed locus	circle centred at the origin, radius $\approx 1.41$
Damping demanded, (a)	$\zeta = 0.707$, i.e. the $135^\circ$ ray
Speed-up demanded, (d)	settling time reduced by a factor of 3, same $\zeta$
PI zero, (g)	$s = -0.1$

Find. (a)–(c) the design poles on the printed locus at $\zeta = 0.707$, the gain there, and the resulting settling time; (d) the relocated design poles for a threefold speed-up at the same damping; (e) the angle sum at that new point; (f) the angle a PD zero would have to contribute, and whether that is possible; (g) a PI controller with its zero at $-0.1$ that removes the steady-state error.

Approach. Exploit the structure the printed graph is hinting at: because the plant’s poles and zeros are mirror images through the origin, the product of the closed-loop roots is constant, so every complex closed-loop pair lies on a circle of fixed radius. That turns (a) into reading a lattice point off the graph, (b) into one coefficient comparison, and it is also what makes the part-(f) claim impossible.

Part (a) — why the locus is a circle, and where the $\zeta$ ray meets it. The characteristic polynomial is $$(s-2)(s-1) + K(s+2)(s+1) = (s^2 - 3s + 2) + K(s^2 + 3s + 2) = (1+K)s^2 + 3(K-1)s + 2(1+K).$$ The product of the two roots is the ratio of the constant to the leading coefficient, $$s_1 s_2 = \frac{2(1+K)}{1+K} = 2 \qquad\text{for every } K,$$ so whenever the roots are a complex conjugate pair, $|s|^2 = s_1s_2 = 2$ and $$\boxed{|s| = \sqrt{2} = 1.414 \text{ for all } K}$$ That is exactly the circle printed on the paper. A damping ratio of $\zeta = 0.707 = 1/\sqrt2$ puts the poles on the ray at $\cos^{-1}(0.707) = 45^\circ$ from the negative real axis, i.e. at $135^\circ$, so $$s = \sqrt{2}\,\bigl(\cos 135^\circ + j\sin 135^\circ\bigr) = \sqrt2\left(-\tfrac{1}{\sqrt2} + j\tfrac{1}{\sqrt2}\right),$$ $$\boxed{s = -1 \pm j1}$$ This is why the question says to use the scaled graph: the intersection of the $45^\circ$ ray with the circle of radius $\sqrt2$ is a lattice point, readable by eye without any arithmetic.
Part (b) — the gain at those poles. Compare the coefficient of $s$ in the characteristic polynomial with the required root sum $s_1+s_2 = 2\,\mathrm{Re}(s) = -2$: $$-\frac{3(K-1)}{1+K} = -2 \quad\Longrightarrow\quad 3(K-1) = 2(1+K) \quad\Longrightarrow\quad 3K - 3 = 2 + 2K,$$ $$\boxed{K = 5}$$ Check by the magnitude condition: $$\left|\frac{5(s+2)(s+1)}{(s-2)(s-1)}\right|_{s=-1+j} = 5\,\frac{|1+j|\,|j|}{|-3+j|\,|-2+j|} = 5\,\frac{1.4142 \times 1}{3.1623 \times 2.2361} = 1.000 ,$$ and by direct substitution: at $K=5$ the polynomial is $6s^2 + 12s + 12$, i.e. $s^2 + 2s + 2$, whose roots are $-1 \pm j1$ with $\zeta = 1/\sqrt2 = 0.7071$. Note in passing that stability requires $3(K-1) > 0$, i.e. $K > 1$: with both open-loop poles unstable, this loop needs a minimum gain merely to function.
Part (c) — settling time. The 2% settling time depends only on the real part of the dominant pair: $$\boxed{T_s = \frac{4}{|\mathrm{Re}(s)|} = \frac{4}{1} = 4\ \text{s}}$$ For completeness, $\omega_n = \sqrt2 = 1.414$ rad/s, $\omega_d = 1$ rad/s, peak time $T_p = \pi/1 = 3.14$ s and percent overshoot $100e^{-\pi(0.707)/\sqrt{1-0.5}} = 4.3\%$.
Part (d) — the relocated design poles. A threefold reduction in settling time means $$T_{s,\text{new}} = \frac{4}{3} = 1.333\ \text{s} \quad\Longrightarrow\quad \left|\mathrm{Re}(s_{\text{new}})\right| = \frac{4}{T_{s,\text{new}}} = 3 ,$$ and holding $\zeta = 0.707$ keeps the $135^\circ$ ray, so the imaginary part must equal the real part in magnitude: $$\boxed{s_{\text{new}} = -3 \pm j3}$$ Its radius is $|s_{\text{new}}| = 3\sqrt2 = 4.243$, three times the old radius.
Part (d) continued — note immediately that this point is off the locus. Every complex closed-loop pair of the uncompensated loop satisfies $|s| = \sqrt2$, and $4.243 \ne 1.414$. So no value of $K$ whatsoever reaches $-3 \pm j3$: the demand for a faster response at fixed damping is a demand for a larger radius, and the radius of this locus is gain-invariant. Compensation of some kind is unavoidable, which is precisely what part (f) puts to the test.

Part (e) — summation of the angles at the new design point. With $s_{\text{new}} = -3 + j3$, the four vectors and their arguments are

Singularity	Vector $s_{\text{new}} - (\cdot)$	Angle
zero at $-1$	$-2 + j3$	$180^\circ - \tan^{-1}(3/2) = 123.69^\circ$
zero at $-2$	$-1 + j3$	$180^\circ - \tan^{-1}(3/1) = 108.44^\circ$
pole at $+1$	$-4 + j3$	$180^\circ - \tan^{-1}(3/4) = 143.13^\circ$
pole at $+2$	$-5 + j3$	$180^\circ - \tan^{-1}(3/5) = 149.04^\circ$

Zeros count positive and poles negative, so $$\sum\angle = (123.69^\circ + 108.44^\circ) - (143.13^\circ + 149.04^\circ) = 232.13^\circ - 292.17^\circ ,$$ $$\boxed{\sum\angle = -60.05^\circ}$$ Since this is not an odd multiple of $180^\circ$, the point is confirmed off the locus — the same conclusion the radius argument gave, now in the currency part (f) needs.

Part (f) — the angle a PD zero would have to supply. Adding a compensator zero contributes $\theta_{z_c}$ to the sum, and the angle criterion demands an odd multiple of $180^\circ$: $$-60.05^\circ + \theta_{z_c} = -180^\circ \quad\Longrightarrow\quad \boxed{\theta_{z_c} = -119.96^\circ}$$ or, taking the $+180^\circ$ branch, $$-60.05^\circ + \theta_{z_c} = +180^\circ \quad\Longrightarrow\quad \theta_{z_c} = +240.05^\circ .$$ Both are impossible. A real zero placed anywhere on the real axis, seen from a point in the upper half-plane, subtends an angle strictly between $0^\circ$ and $180^\circ$: the vector from the zero to $-3+j3$ always has a positive imaginary part, so its argument can be made arbitrarily close to $0^\circ$ (zero far to the right) or to $180^\circ$ (zero far to the left), but it can never be negative and never exceed $180^\circ$.
Part (f) continued — the one-line verdict, and what would work. The claim is invalid: the PD zero would have to contribute $-119.96^\circ$ (equivalently $+240.05^\circ$), and a real zero can only ever contribute between $0^\circ$ and $180^\circ$ at an upper-half-plane point, so no PD compensator reaches $-3 \pm j3$. The structural reason is worth stating: the loop is already $60^\circ$ ahead of the angle criterion at this point, so it needs lag, not lead. A real pole enters the criterion with a minus sign and therefore can supply it. Requiring $-60.05^\circ - \theta_p = -180^\circ$ gives $\theta_p = 119.96^\circ$, and a pole at $s = -p$ subtends that angle when $$\tan\left(180^\circ - 119.96^\circ\right) = \frac{3}{3 - p} \;\Longrightarrow\; 3 - p = \frac{3}{\tan 60.04^\circ} = 1.729 \;\Longrightarrow\; p = 1.271 ,$$ so an added pole at $s = -1.271$ — a lag rather than a lead element — does place the design point on the compensated locus. Recomputing the sum with it in place: $-60.05^\circ - 119.96^\circ = -180.01^\circ$ ✓
Part (g) — PI controller with its zero at $-0.1$. The proportional design of part (b) leaves a steady-state step error, because the loop is Type 0: with $K = 5$, $G(0) = 5(2)(1)/[(-2)(-1)] = 5$, so $e_{ss} = 1/(1+5) = 0.167$, a 17% offset. A PI controller supplies the missing integrator: $$\boxed{C(s) = K_p\,\frac{s + 0.1}{s} = K_p + \frac{0.1K_p}{s}}$$ The loop becomes Type 1, $K_v$ is finite and non-zero, and the steady-state step error is exactly zero for any stabilising $K_p$ — which is what “make steady state error diminish” asks for. The compensated characteristic polynomial is $$s(s-2)(s-1) + K_p(s+0.1)(s+2)(s+1) = (1+K_p)s^3 + (3.1K_p - 3)s^2 + (2.3K_p + 2)s + 0.2K_p .$$
Part (g) continued — choose the gain and check stability. Applying the cubic Routh conditions to that polynomial, all coefficients are positive and $a_2a_1 > a_3a_0$ only for $K_p > 0.998$; so, as in part (b), a minimum gain is needed merely to stabilise the two right-half-plane poles. Choosing $K_p = 5$ — the same proportional gain as part (b), so the comparison is clean — gives $$6s^3 + 12.5s^2 + 13.5s + 1.0 = 0 \quad\Longrightarrow\quad s = -1.002 \pm j1.042,\quad s = -0.0797 .$$ The dominant pair has barely moved from the part-(b) design ($-1 \pm j1$), with $\zeta = 0.693$ against 0.707, so the transient shape is essentially preserved while the steady-state error is eliminated. Verify the DC gain: the closed-loop numerator and denominator constants are both $1.0$, so $T(0) = 1$ exactly.
Part (g) continued — state the price honestly. The integrator introduces a third closed-loop pole at $-0.0797$, close to (but not cancelled by) the compensator zero at $-0.1$. Its residue in the unit-step response is $-0.193$, which is $47\%$ of the dominant pair’s residue magnitude of $0.412$ — not negligible. Since that mode decays as $e^{-0.0797t}$, the 2% settling time becomes $$0.193\,e^{-0.0797\,T_s} = 0.02 \quad\Longrightarrow\quad T_s = \frac{\ln(0.193/0.02)}{0.0797} = \frac{2.268}{0.0797} = 28.5\ \text{s},$$ against 4 s for the proportional design. So the PI removes the 17% offset but stretches settling by a factor of seven — a genuine engineering trade, and the mark-earning observation. Moving the PI zero further out (say to $-0.5$) would shorten the tail at the cost of a larger transient excursion; the paper fixes the zero at $-0.1$, so the slow tail is part of the specified answer.

Question 5: the printed locus of $K(s+2)(s+1)/[(s-2)(s-1)]$ is the circle $|s| = \sqrt2$, because the root product is 2 for every $K$. The $\zeta = 0.707$ ray meets it at the lattice point $-1 \pm j1$ ($K = 5$). The part-(d) target $-3 + j3$ lies far outside the circle, so no gain — and no PD compensator — can reach it.

Part (g): unit-step responses. The proportional design ($K = 5$) settles at 0.833, a 17% offset; the PI controller with its zero at $-0.1$ reaches 1.000 exactly, but the closed-loop pole it adds at $-0.0797$ leaves a slow tail that stretches the 2% settling time from 4 s to about 28 s.

Part	Quantity	Result
—	Locus radius (all $K$)	$\|s\| = \sqrt2 = 1.414$
(a)	Design poles at $\zeta = 0.707$	$s = -1 \pm j1$
(b)	Gain at the design poles	$K = 5$ (stability needs $K > 1$)
(c)	Settling time	$T_s = 4/1 = 4$ s ($T_p = 3.14$ s, $\%OS = 4.3$)
(d)	New design poles	$s = -3 \pm j3$ ($T_s = 1.333$ s), radius $4.243$
(e)	Summation of angles at $-3+j3$	$-60.05^\circ$
(f)	Required PD zero angle	$-119.96^\circ$ (or $+240.05^\circ$)
(f)	Validity of the claim	Invalid — a real zero subtends only $0^\circ$–$180^\circ$ there
(f)	What would work instead	an added pole at $s = -1.271$
(g)	PI controller	$C(s) = 5(s+0.1)/s$; stable for $K_p > 0.998$
(g)	Closed-loop poles with PI	$-1.002 \pm j1.042$, $-0.0797$
(g)	Steady-state step error	$0.167 \to 0$ exactly; $T_s$ degrades $4 \to 28.5$ s

Topic. Root-locus design on a mirror-image (all-pass-like) plant: the gain-invariant circular locus, the angle criterion as a feasibility test for a compensator structure, and PI compensation of a Type-0 loop.

Key relations. For $K(s+a)(s+b)/[(s-a)(s-b)]$ the closed-loop polynomial is $(1+K)s^2 - (a+b)(1-K)s + ab(1+K)$, so the root product is $ab$ and every complex pair lies on $|s| = \sqrt{ab}$; stability requires $K > 1$. Design-point relations $\zeta = -\mathrm{Re}(s)/|s|$, $T_s = 4/|\mathrm{Re}(s)|$, $T_p = \pi/\omega_d$. Angle criterion $\sum\angle z - \sum\angle p = \pm180^\circ(2k+1)$; a real singularity seen from an upper-half-plane point subtends an angle in $(0^\circ, 180^\circ)$, with zeros entering positively and poles negatively. PI controller $C = K_p(s+z)/s$ raises the system type by one, forcing zero steady-state step error.

Why this works. The mirror symmetry of this plant makes the constant term and the leading coefficient of the characteristic polynomial move together, which pins the root product and hence the radius. That single fact answers four sub-parts at once: it explains the printed circle, it makes the $\zeta$-ray intersection a lattice point that can be read off the graph, it shows that a faster response at fixed damping is unreachable by gain, and — through the angle sum — it shows why lead compensation cannot help either. The angle criterion is the right tool for part (f) because it is a statement about geometry alone: one evaluates the required contribution and then asks whether the proposed structure can produce it, without ever computing a gain. And a PI controller works on the steady-state error for a reason independent of all this — the pole it puts at the origin makes the loop Type 1, and the final-value theorem then forces $e_{ss} = 0$ for a step regardless of the gain.

Common pitfall. The dominant trap in part (f) is to answer with arithmetic instead of with sign: candidates compute the required angle, find a value that is not between $0^\circ$ and $180^\circ$, and then place a zero anyway by taking an absolute value or a modulo — producing a confident location that does not satisfy the criterion. Compute the sum, look at whether the deficiency is positive or negative, and let that decide zero versus pole. A second trap is part (d): reading “three times faster” as tripling $\omega_n$ rather than tripling $|\mathrm{Re}(s)|$. At fixed $\zeta$ the two happen to coincide, but the specification given is on settling time, which is a statement about the real part, and on a problem where $\zeta$ changes the two readings diverge. Finally, in part (g), reporting only that the error goes to zero and stopping: the added slow pole is barely cancelled by the zero at $-0.1$, and quoting the resulting settling time is what distinguishes a design from an algebraic exercise.

Source. Nise, Control Systems Engineering, 8th ed., §8.5–8.7 (locus geometry, design by the angle criterion, gain from the magnitude condition), §9.2 (ideal integral compensation and steady-state error), §9.3 (ideal derivative compensation and its angle bookkeeping), §4.6 (settling time, peak time, percent overshoot); Ogata, Modern Control Engineering, 5th ed., §6.6–6.8; Dorf and Bishop, Modern Control Systems, 13th ed., Ch. 9.

← Question 4 Back to the paper →

Singularity	Vector \(s_{\text{new}} - (\cdot)\)	Angle
zero at \(-1\)	\(-2 + j3\)	\(180^\circ - \tan^{-1}(3/2) = 123.69^\circ\)
zero at \(-2\)	\(-1 + j3\)	\(180^\circ - \tan^{-1}(3/1) = 108.44^\circ\)
pole at \(+1\)	\(-4 + j3\)	\(180^\circ - \tan^{-1}(3/4) = 143.13^\circ\)
pole at \(+2\)	\(-5 + j3\)	\(180^\circ - \tan^{-1}(3/5) = 149.04^\circ\)

Quantity	Value
Loop transfer function	\(G(s) = K(s+2)(s+1)/[(s-2)(s-1)]\), unity feedback
Open-loop poles	\(s = +1,\ +2\) (both unstable)
Open-loop zeros	\(s = -1,\ -2\)
Printed locus	circle centred at the origin, radius \(\approx 1.41\)
Damping demanded, (a)	\(\zeta = 0.707\), i.e. the \(135^\circ\) ray
Speed-up demanded, (d)	settling time reduced by a factor of 3, same \(\zeta\)
PI zero, (g)	\(s = -0.1\)

Part	Quantity	Result
—	Locus radius (all \(K\))	\(\|s\| = \sqrt2 = 1.414\)
(a)	Design poles at \(\zeta = 0.707\)	\(s = -1 \pm j1\)
(b)	Gain at the design poles	\(K = 5\) (stability needs \(K > 1\))
(c)	Settling time	\(T_s = 4/1 = 4\) s (\(T_p = 3.14\) s, \(\%OS = 4.3\))
(d)	New design poles	\(s = -3 \pm j3\) (\(T_s = 1.333\) s), radius \(4.243\)
(e)	Summation of angles at \(-3+j3\)	\(-60.05^\circ\)
(f)	Required PD zero angle	\(-119.96^\circ\) (or \(+240.05^\circ\))
(f)	Validity of the claim	Invalid — a real zero subtends only \(0^\circ\)–\(180^\circ\) there
(f)	What would work instead	an added pole at \(s = -1.271\)
(g)	PI controller	\(C(s) = 5(s+0.1)/s\); stable for \(K_p > 0.998\)
(g)	Closed-loop poles with PI	\(-1.002 \pm j1.042\), \(-0.0797\)
(g)	Steady-state step error	\(0.167 \to 0\) exactly; \(T_s\) degrades \(4 \to 28.5\) s