22-Elec-B2 Advanced Control Systems · May 2017

Question 2 of 5: Root-locus geometry — asymptotes, break points and a $j\omega$-axis crossing

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 2: Root-locus geometry — asymptotes, break points and a $j\omega$-axis crossing (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.

Part	Plant	Open-loop poles	Open-loop zeros	$n-m$
A	$K(s-0.5)/(s-1)^2$	$+1$ (double)	$+0.5$	1
B	$K(s-1)(s-2)/[s(s+1)]$	$0,\ -1$	$+1,\ +2$	0

Both loops are unity feedback with $K \ge 0$; Part B additionally supplies the break points $-0.37\ (K=0.07)$ and $1.37\ (K=13.93)$.

Find. For Part A the asymptote angle and the break-away/break-in coordinates; for Part B the closed-loop transfer function, the imaginary-axis crossing and its gain from the Routh array, and an approximate sketch of the locus.

Approach. Asymptote count and direction come from $n-m$; break points come from solving $dK/ds = 0$ along the real-axis locus, where $K(s)$ is the gain implied by the magnitude condition; and the imaginary-axis crossing comes from forcing a Routh row to vanish and solving the resulting auxiliary equation.

Part A(a) — count the branches and place the asymptotes. Part A has $n = 2$ finite poles (a double pole at $s=+1$) and $m = 1$ finite zero (at $s=+0.5$), so $n - m = 1$: exactly one branch escapes to infinity, along one asymptote at $$\theta_a = \frac{(2k+1)180^\circ}{n-m} = \frac{180^\circ}{1} = \boxed{180^\circ}$$ with the companion centroid $$\sigma_a = \frac{\sum \text{poles} - \sum \text{zeros}}{n-m} = \frac{(1+1) - 0.5}{1} = 1.5 .$$ For a single asymptote the centroid adds nothing geometrically — a ray at $180^\circ$ through any real point is the negative real direction — but it is worth quoting, because the same formula is what fixes the asymptote intersection when $n-m \ge 2$.
Part A(b) — establish which stretch of the real axis carries the locus. Apply the odd-count rule to the three real singularities $1, 1, 0.5$, reading from the right: to the right of $s>1$ there are none (even, not on the locus); between $0.5$ and $1$ there are two, the double pole (even, not on the locus); to the left of $0.5$ there are three (odd, on the locus). So the real-axis locus is the ray $s \le 0.5$, and the double pole at $s=+1$ must leave the axis immediately.
Part A(b) continued — solve $dK/ds = 0$. The magnitude condition on the real axis gives the gain as a function of position, $$K(s) = -\frac{(s-1)^2}{s-0.5},$$ and break points are the stationary points of this function. Differentiating and clearing the denominator, $$\frac{dK}{ds} = 0 \;\Longrightarrow\; 2(s-1)(s-0.5) - (s-1)^2 = 0 \;\Longrightarrow\; (s-1)\bigl[2(s-0.5) - (s-1)\bigr] = 0 \;\Longrightarrow\; (s-1)\,s = 0 .$$ The two roots are $$\boxed{s = +1 \quad\text{and}\quad s = 0}$$
Part A(b) continued — identify which is which, and at what gain. Evaluating $K(s)$ at each root separates them: $$K(1) = -\frac{0}{0.5} = 0 \qquad\text{(the breakaway, at the double pole itself)},$$ $$K(0) = -\frac{(-1)^2}{-0.5} = +2 \qquad\text{(the break-in)}.$$ So the two branches break away from the real axis at $s = 1$ with $K = 0$ and break back in at $s = 0$ with $K = 2$. The departure angles at the double pole follow from the angle condition: the zero at $0.5$ subtends $0^\circ$ from a point just above $s=1$, so $0 - 2\theta_d = \pm 180^\circ$ and $\theta_d = \mp 90^\circ$ — the branches leave vertically.
Part A(b) continued — confirm from the closed-loop polynomial. The characteristic polynomial is $$(s-1)^2 + K(s-0.5) = s^2 + (K-2)s + (1 - 0.5K),$$ whose discriminant is $$(K-2)^2 - 4(1-0.5K) = K^2 - 4K + 4 - 4 + 2K = K(K-2).$$ This is negative exactly for $0 < K < 2$, so the roots are complex on that interval and real outside it — independent confirmation that the excursion off the axis starts at $K = 0$ and ends at $K = 2$. Note also that the real part of the complex pair is $(2-K)/2$, which is positive for all $K<2$: this loop is unstable for every $K<2$, and for $K>2$ one real root heads for the right-half-plane zero at $+0.5$ while the other runs left. The loop is in fact unstable at every gain, which the sketch makes immediate.

Part A: root locus of $K(s-0.5)/(s-1)^2$. The branches leave the double pole at $s=+1$ vertically, arc left, and break in at $s = 0$ when $K = 2$; thereafter one branch terminates on the zero at $+0.5$ and the other runs left along the single $180^\circ$ asymptote.

Part B(a) — close the loop. With unity feedback, $$T(s) = \frac{G(s)}{1+G(s)} = \frac{K(s-1)(s-2)}{s(s+1) + K(s-1)(s-2)} .$$ Expanding the denominator, $s^2 + s + K(s^2 - 3s + 2)$, and collecting powers of $s$, $$\boxed{T(s) = \frac{K(s-1)(s-2)}{(1+K)s^2 + (1-3K)s + 2K}}$$ Both open-loop zeros are in the right half-plane, so they are also closed-loop zeros for every $K$ — the loop is irreducibly non-minimum-phase.
Part B(a) continued — build the Routh array. For the quadratic characteristic polynomial $(1+K)s^2 + (1-3K)s + 2K$:

Row First column Second column

$s^2$ $1+K$ $2K$

$s^1$ $1-3K$ $0$

$s^0$ $2K$ $0$

All three first-column entries must be positive for stability: $1+K>0$, $1-3K>0$ and $2K>0$, which together give the stability range $$0 < K < \tfrac13 .$$
Part B(a) continued — force the $s^1$ row to vanish. An imaginary-axis crossing occurs where the $s^1$ row is zero: $$1 - 3K = 0 \quad\Longrightarrow\quad \boxed{K = \tfrac13 = 0.3333}$$ The crossing frequency comes from the auxiliary equation formed from the row above, $$(1+K)s^2 + 2K = 0 \;\Longrightarrow\; \tfrac43 s^2 + \tfrac23 = 0 \;\Longrightarrow\; s^2 = -\tfrac12 ,$$ $$\boxed{s = \pm j\,0.7071}$$ So the locus crosses the imaginary axis at $\pm j0.707$ when $K = 1/3$, and the loop is stable only below that gain. Substituting $K=1/3$ back into the closed-loop denominator returns $1.3333s^2 + 0.6667$, whose roots are $\pm j0.7071$ exactly — the check closes.
Part B(b) — assemble the sketch from the four ingredients. (i) Real-axis segments. The real singularities, from the right, are the zeros at $+2$ and $+1$ and the poles at $0$ and $-1$. Odd counts occur on $[1,2]$ (one singularity to the right) and on $[-1,0]$ (three), so those two segments carry the locus; $[0,1]$ and $(-\infty,-1]$ do not. (ii) Branches at infinity. $n = m = 2$, so no branch escapes — both branches start on the poles and finish on the zeros, and the locus is bounded. (iii) Break points. The paper supplies breakaway at $-0.37\ (K = 0.07)$ on the $[-1,0]$ segment and break-in at $1.37\ (K = 13.93)$ on the $[1,2]$ segment; both are confirmed by $dK/ds = 0$, which reduces to $2s^2 - 2s - 1 = 0$ with roots $s = (1\pm\sqrt3)/2 = -0.3660,\ 1.3660$, and $K(-0.366) = 0.0718$, $K(1.366) = 13.928$. (iv) Imaginary-axis crossing. $\pm j0.707$ at $K = 1/3$ from part (a). The branches therefore leave $0$ and $-1$, meet at $-0.366$, swing out into the complex plane, cross the imaginary axis at $\pm j0.707$, return to the real axis at $+1.366$, and split into the zeros at $+1$ and $+2$.

Row	First column	Second column
\(s^2\)	\(1+K\)	\(2K\)
\(s^1\)	\(1-3K\)	\(0\)
\(s^0\)	\(2K\)	\(0\)

Part B: root locus of $K(s-1)(s-2)/[s(s+1)]$. Breakaway at $-0.37$ ($K=0.07$), imaginary-axis crossing at $\pm j0.707$ ($K=1/3$), break-in at $+1.37$ ($K=13.93$). With $n=m=2$ the whole locus is a bounded closed curve; the loop is stable only on the short interval $0<K<1/3$.

Quantity	Result
A(a) asymptote angle ($n-m=1$)	$180^\circ$ (one asymptote); centroid $\sigma_a = 1.5$
A(b) breakaway point	$s = +1.0$ at $K = 0$, departure angles $\pm 90^\circ$
A(b) break-in point	$s = 0$ at $K = 2$
A closed-loop pair complex for	$0 < K < 2$; unstable at every $K$
B(a) closed-loop transfer function	$T(s) = \dfrac{K(s-1)(s-2)}{(1+K)s^2+(1-3K)s+2K}$
B(a) $j\omega$-axis crossing	$s = \pm j0.7071$
B(a) gain at the crossing	$K = 1/3 = 0.3333$
B(a) stability range	$0 < K < 1/3$
B(b) break points (confirmed)	breakaway $-0.3660\ (K=0.0718)$; break-in $1.3660\ (K=13.928)$

Topic. Root-locus construction for non-minimum-phase and right-half-plane-pole plants: asymptote counting, break-point calculation by $dK/ds = 0$, and the Routh–Hurwitz route to an imaginary-axis crossing.

Key relations. Asymptote angles $\theta_a = (2k+1)180^\circ/(n-m)$ and centroid $\sigma_a = (\sum p_i - \sum z_j)/(n-m)$. Real-axis rule: a real point is on the locus if the number of real poles and zeros to its right is odd. Break points: $dK/ds = 0$ with $K(s) = -D(s)/N(s)$. Departure angle from a pole of multiplicity $q$: $q\theta_d = \sum\angle\text{zeros} - \sum\angle\text{other poles} - 180^\circ(2k+1)$. Imaginary-axis crossing: set the Routh $s^1$ row to zero and solve the auxiliary equation formed from the row above.

Why this works. Every point of the locus satisfies $1 + KG(s) = 0$, which splits into an angle condition (independent of $K$) and a magnitude condition (which then defines $K$ at that point). The real-axis rule is the angle condition evaluated on the axis, where each singularity to the right contributes exactly $180^\circ$ and each to the left exactly $0^\circ$. Break points are where two branches coincide, i.e. where the characteristic polynomial has a repeated root; since $K$ is a smooth function of position along the axis and turns around at such a point, $dK/ds = 0$ locates them without ever factoring a polynomial. And because a Routh first-column entry vanishing signals a root pair on the imaginary axis, the auxiliary equation extracts that pair exactly — which is why the paper asks for the crossing “using Routh-Hurwitz” rather than by trial.

Common pitfall. In Part A, treating $s = 0$ as the breakaway and $s = +1$ as something to discard: both roots of $dK/ds = 0$ are genuine break points, and the way to tell them apart is to evaluate $K$ at each — $K = 0$ marks the departure from the open-loop pole, $K = 2$ the return. The related error is to report a “breakaway” that lies off the real-axis locus; always apply the odd-count rule first, because a stationary point of $K(s)$ on a stretch that is not on the locus corresponds to a negative gain. In Part B, the trap is to assume that a locus with $n = m$ still has asymptotes; it has none, and a sketch showing branches running off to infinity contradicts the pole–zero count.

Source. Nise, Control Systems Engineering, 8th ed., §8.4 (the five sketching rules), §8.5 (refining the sketch: break points, departure angles, $j\omega$-axis crossings via Routh), §6.2–6.3 (Routh array and the auxiliary polynomial); Ogata, Modern Control Engineering, 5th ed., §6.2.

← Question 1 Next: Question 3 →

Part	Plant	Open-loop poles	Open-loop zeros	\(n-m\)
A	\(K(s-0.5)/(s-1)^2\)	\(+1\) (double)	\(+0.5\)	1
B	\(K(s-1)(s-2)/[s(s+1)]\)	\(0,\ -1\)	\(+1,\ +2\)	0

Quantity	Result
A(a) asymptote angle (\(n-m=1\))	\(180^\circ\) (one asymptote); centroid \(\sigma_a = 1.5\)
A(b) breakaway point	\(s = +1.0\) at \(K = 0\), departure angles \(\pm 90^\circ\)
A(b) break-in point	\(s = 0\) at \(K = 2\)
A closed-loop pair complex for	\(0 < K < 2\); unstable at every \(K\)
B(a) closed-loop transfer function	\(T(s) = \dfrac{K(s-1)(s-2)}{(1+K)s^2+(1-3K)s+2K}\)
B(a) \(j\omega\)-axis crossing	\(s = \pm j0.7071\)
B(a) gain at the crossing	\(K = 1/3 = 0.3333\)
B(a) stability range	\(0 < K < 1/3\)
B(b) break points (confirmed)	breakaway \(-0.3660\ (K=0.0718)\); break-in \(1.3660\ (K=13.928)\)

Question 2 of 5: Root-locus geometry — asymptotes, break points and a \(j\omega\)-axis crossing

Question 2: Root-locus geometry — asymptotes, break points and a \(j\omega\)-axis crossing (25 marks)