22-Elec-B2 Advanced Control Systems · May 2017

Question 3 of 5: Placing the zero of a PD compensator by the angle criterion

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

Notes on this paper

Paper format. 16-Elec-B2 Advanced Control Systems, May 2017 — 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 5. The paper prints the $s^2$ first-column entry of the Routh table as $(1-4\varepsilon)/\varepsilon$, which is exactly what the Routh–Hurwitz recursion gives, and which is confirmed independently by the printed $s^1$ entry. The table on the paper is correct and internally consistent. The answer is unaffected by the alternative form $\varepsilon/(1-4\varepsilon)$, because both are positive for small $\varepsilon>0$.

(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 3: Placing the zero of a PD compensator by the angle criterion (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 gain	$G(s) = 10/[s(s+10)(s+20)]$
Open-loop poles	$s = 0,\ -10,\ -20$
Open-loop zeros	none
Design point	$s_d = -12.78 + j24.94$ (and its conjugate)
Controller	PD, $G_c(s) = K(s + z_c)$ — one real zero, no pole
Implied damping	$\zeta = 12.78/\|s_d\| = 12.78/28.02 = 0.4561$ (about 20% overshoot)

Find. (a) the net angle the three open-loop poles subtend at the design point; (b) the angle the compensator zero must supply so that the total obeys the locus angle criterion; (c) the real-axis location of that zero.

Approach. A point lies on the compensated root locus only if the angles of all open-loop singularities, taken as zeros minus poles, sum to an odd multiple of $180^\circ$. Sum the three pole angles at $s_d$, read off the deficiency, and place the single PD zero on the real axis at the position that subtends exactly that deficiency.

Part (a) — angle from each open-loop pole to the design point. The angle contributed by a pole at $-p$ is the argument of the vector $s_d - (-p) = s_d + p$, measured from the positive real direction. With $s_d = -12.78 + j24.94$: $$\theta_1 = \angle(s_d - 0) = \angle(-12.78 + j24.94) = 180^\circ - \tan^{-1}\!\frac{24.94}{12.78} = 180^\circ - 62.87^\circ = 117.13^\circ,$$ $$\theta_2 = \angle(s_d + 10) = \angle(-2.78 + j24.94) = 180^\circ - \tan^{-1}\!\frac{24.94}{2.78} = 180^\circ - 83.64^\circ = 96.36^\circ,$$ $$\theta_3 = \angle(s_d + 20) = \angle(7.22 + j24.94) = \tan^{-1}\!\frac{24.94}{7.22} = 73.85^\circ .$$ Note that the pole at $-20$ lies to the left of the design point’s real part, so its vector points up and to the right and its angle is acute, while the other two are obtuse. The sum is $$\boxed{\sum \theta_{\text{poles}} = 117.13^\circ + 96.36^\circ + 73.85^\circ = 287.34^\circ}$$ Poles enter the angle criterion with a minus sign, so the uncompensated plant contributes a net $-287.34^\circ$ at $s_d$.
Part (b) — angle the compensator zero must supply. The angle criterion requires $$\sum \angle\text{zeros} - \sum \angle\text{poles} = \pm 180^\circ(2k+1).$$ With one compensator zero contributing $\theta_z$ and nothing else, $$\theta_z - 287.34^\circ = -180^\circ \quad\Longrightarrow\quad \boxed{\theta_z = 287.34^\circ - 180^\circ = 107.34^\circ}$$ The plant is $287.34^\circ$ “too lagging” by $107.34^\circ$ relative to the $180^\circ$ it needs, and the zero must make up exactly that deficiency. The choice of the $-180^\circ$ branch rather than $+180^\circ$ matters: the $+180^\circ$ branch would demand $\theta_z = 467.34^\circ \equiv 107.34^\circ$ modulo $360^\circ$, the same ray, so the answer is unambiguous here.
Part (c) — convert that angle into a location. A real zero at $s = -z_c$ subtends, at the design point, $$\theta_z = \angle\bigl(s_d + z_c\bigr) = \angle\bigl[(z_c - 12.78) + j\,24.94\bigr].$$ Since $\theta_z = 107.34^\circ$ is obtuse, the real part must be negative, so $z_c < 12.78$ and $$\tan\left(180^\circ - 107.34^\circ\right) = \frac{24.94}{12.78 - z_c} \quad\Longrightarrow\quad \tan 72.66^\circ = 3.2064 = \frac{24.94}{12.78 - z_c},$$ $$12.78 - z_c = \frac{24.94}{3.2064} = 7.778 \quad\Longrightarrow\quad z_c = 12.78 - 7.778 = 5.002 .$$ So the PD compensator zero goes at $$\boxed{s = -5.00,\qquad G_c(s) = K\,(s + 5)}$$ The result is a round number, which is the intended reading: the design point printed on the paper was itself generated from a zero at $-5$.
Check the angle criterion end to end. With the zero at $-5$, the vector $s_d + 5 = -7.78 + j24.94$ has angle $180^\circ - \tan^{-1}(24.94/7.78) = 180^\circ - 72.66^\circ = 107.34^\circ$, and $$107.34^\circ - 287.34^\circ = -180.00^\circ ,$$ so $s_d$ lies exactly on the compensated locus. The gain is not required, but for completeness the magnitude condition would give $K = |s_d||s_d+10||s_d+20| / \bigl(10\,|s_d+5|\bigr)$.
Interpret the design. The zero at $-5$ sits between the origin pole and the pole at $-10$, which is what allows it to contribute a large positive angle without moving the Type-1 character of the loop: the free integrator survives, so the compensated system still tracks a step with zero error, while the added derivative action pulls the dominant pair from wherever the uncompensated locus put it out to $-12.78 \pm j24.94$ — a settling time of $4/12.78 = 0.313$ s and a damped frequency of 24.94 rad/s. A useful sanity check is that a PD zero can only ever supply an angle strictly between $0^\circ$ and $180^\circ$ as seen from an upper-half-plane point, so a demand of $107.34^\circ$ is comfortably achievable; the same test applied in Question 5 will show a demand that is not.

Question 3: the three open-loop pole vectors to the design point $-12.78 + j24.94$ subtend $117.13^\circ$, $96.36^\circ$ and $73.85^\circ$ (sum $287.34^\circ$). The PD zero at $-5.00$ supplies the missing $107.34^\circ$, closing the angle criterion at $-180^\circ$; the blue curve is the resulting compensated locus.

Quantity	Result
Angle from the pole at $0$	$117.13^\circ$
Angle from the pole at $-10$	$96.36^\circ$
Angle from the pole at $-20$	$73.85^\circ$
(a) Summation of pole angles	$287.34^\circ$ (net contribution $-287.34^\circ$)
(b) Required zero angle	$107.34^\circ$
(c) PD zero location	$s = -5.00$, i.e. $G_c(s) = K(s+5)$
Angle-criterion check	$107.34^\circ - 287.34^\circ = -180.00^\circ$ ✓
Resulting dominant metrics	$\zeta = 0.456$, $T_s = 0.313$ s, $T_p = 0.126$ s

Topic. Root-locus design of an ideal derivative (PD) compensator: using the angle criterion to place a single real zero so that a chosen point becomes a closed-loop pole.

Key relations. Angle criterion $\sum\angle(s_d - z_j) - \sum\angle(s_d - p_i) = \pm180^\circ(2k+1)$; magnitude criterion $K = \prod|s_d-p_i| / \prod|s_d-z_j|$. Angle of a real singularity at $-a$ seen from $s_d = -\sigma + j\omega_d$: $\tan^{-1}\!\bigl[\omega_d/(a-\sigma)\bigr]$ if $a > \sigma$, otherwise $180^\circ - \tan^{-1}\!\bigl[\omega_d/(\sigma-a)\bigr]$. Placement formula $z_c = \sigma - \omega_d/\tan(180^\circ - \theta_z)$.

Why this works. The closed-loop poles are the roots of $1 + G_c(s)G(s) = 0$, i.e. the points where $G_cG$ equals $-1$. Splitting that complex equation into modulus and argument decouples the design: the argument condition involves only the pole–zero geometry and so decides where the zero must go, while the modulus condition involves $K$ and so decides how hard to drive it. Because a PD compensator adds one zero and no pole, the argument condition is a single scalar equation in the single unknown $z_c$ — which is why the design is exact rather than iterative, and why the question can ask for the zero without asking for $K$.

Common pitfall. Two errors recur. The first is a quadrant slip: the pole at $-20$ lies to the left of the design point, so its vector has a positive real part and an acute angle, whereas the poles at $0$ and $-10$ give obtuse angles. Taking $\tan^{-1}(\omega_d/\text{real part})$ blindly, without checking the sign of the real part, turns $117.13^\circ$ into $-62.87^\circ$ and destroys the sum. The second is forgetting that poles enter the criterion with a minus sign, which flips the deficiency and puts the zero on the wrong side of the design point. A cheap guard is the one applied in step 4: recompute the full sum with the proposed zero in place and confirm it lands on $-180^\circ$.

Source. Nise, Control Systems Engineering, 8th ed., §9.3 (“improving transient response: ideal derivative compensation”, worked with exactly this angle bookkeeping) and §8.5; Ogata, Modern Control Engineering, 5th ed., §6.6 (lead compensation by the root-locus method); Franklin, Powell and Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed., §5.4.

← Question 2 Next: Question 4 →

Quantity	Result
Angle from the pole at \(0\)	\(117.13^\circ\)
Angle from the pole at \(-10\)	\(96.36^\circ\)
Angle from the pole at \(-20\)	\(73.85^\circ\)
(a) Summation of pole angles	\(287.34^\circ\) (net contribution \(-287.34^\circ\))
(b) Required zero angle	\(107.34^\circ\)
(c) PD zero location	\(s = -5.00\), i.e. \(G_c(s) = K(s+5)\)
Angle-criterion check	\(107.34^\circ - 287.34^\circ = -180.00^\circ\) ✓
Resulting dominant metrics	\(\zeta = 0.456\), \(T_s = 0.313\) s, \(T_p = 0.126\) s

Quantity	Value
Loop gain	\(G(s) = 10/[s(s+10)(s+20)]\)
Open-loop poles	\(s = 0,\ -10,\ -20\)
Open-loop zeros	none
Design point	\(s_d = -12.78 + j24.94\) (and its conjugate)
Controller	PD, \(G_c(s) = K(s + z_c)\) — one real zero, no pole
Implied damping	\(\zeta = 12.78/\|s_d\| = 12.78/28.02 = 0.4561\) (about 20% overshoot)