Question 2 of 5: Root locus from a pole-zero map

Question 2: Root locus from a pole-zero map (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 unity-feedback loop whose open-loop pole-zero map prints two crosses in the right half-plane and two circles on the negative real axis, together with the break-in coordinate for part (d).

Find. (a) the open-loop transfer function, (b) the angle of departure from the complex poles, (c) the closed-loop transfer function and, from a Routh table, the imaginary-axis crossing and the gain that produces it, and (d) the completed root-locus sketch.

Approach. Write \(G(s)\) directly from the map, apply the angle condition at a point infinitesimally close to a pole to get the departure angle, form the characteristic polynomial and use the Routh array to locate the gain at which its \(s^{1}\) row vanishes, then assemble the sketch from the real-axis rule, the departure angles, the crossing and the given break-in.

1. Part (a) — read the singularities off the map and assemble \(G(s)\). A cross marks a pole and a circle marks a zero, so the open-loop function has poles at \(s = 1 \pm j1\) and zeros at \(s = -2\) and \(s = -3\). Building the factored form and expanding, $$G(s) = \frac{K(s+2)(s+3)}{(s-1-j1)(s-1+j1)} = \frac{K(s+2)(s+3)}{(s-1)^{2}+1} = \boxed{\frac{K\left(s^{2}+5s+6\right)}{s^{2}-2s+2}}$$ Both open-loop poles lie in the right half-plane, so the plant itself is unstable; the loop will have to supply enough gain to pull them across.
2. Part (b) — angles from each singularity to the departing pole. The angle condition states that every point on the locus satisfies \(\sum \angle(\text{zeros}) - \sum \angle(\text{poles}) = (2k+1)180^\circ\). Applied at a test point a hair away from \(p_1 = 1+j1\), every vector is measured from the singularity to \(p_1\): $$\angle(p_1+2) = \arctan\frac{1}{3} = 18.435^\circ, \qquad \angle(p_1+3) = \arctan\frac{1}{4} = 14.036^\circ$$ $$\angle(p_1-p_2) = \angle(j2) = 90^\circ$$
3. Solve the angle condition for the unknown departure angle. Writing the condition with \(\theta_d\) as the contribution of the departing pole itself, $$\theta_d = 180^\circ + \sum \angle(\text{zeros}) - \sum \angle(\text{other poles}) = 180^\circ + 18.435^\circ + 14.036^\circ - 90^\circ$$ $$\boxed{\theta_{d} = +122.47^\circ \ \text{from } 1+j1, \qquad \theta_{d} = -122.47^\circ \ \text{from } 1-j1}$$ The conjugate pole departs at the mirror-image angle, as it must, because the locus is symmetric about the real axis. Taking \(k = \pm 1, \pm 2, \dots\) merely adds multiples of \(360^\circ\) and returns the same direction, which is what the note in the question is pointing at.
4. Part (c) — form the closed-loop transfer function. For unity negative feedback, $$T(s) = \frac{G(s)}{1+G(s)} = \frac{K\left(s^{2}+5s+6\right)}{\left(s^{2}-2s+2\right)+K\left(s^{2}+5s+6\right)}$$ Collecting powers of \(s\) in the denominator gives the characteristic polynomial $$\boxed{(1+K)s^{2} + (5K-2)s + (2+6K) = 0}$$
5. Build the Routh array and find where the \(s^{1}\) row vanishes. For a quadratic the array is short, and the first column is simply the three coefficients: $$\begin{array}{c|cc} s^{2} & 1+K & 2+6K \\ s^{1} & 5K-2 & 0 \\ s^{0} & 2+6K & \end{array}$$ An entire row of zeros — the signature of a pair of roots on the imaginary axis — occurs when \(5K-2 = 0\), i.e. $$\boxed{K = 0.4}$$ Positive \(K\) below 0.4 makes the \(s^{1}\) entry negative, so the loop is unstable there: it is the lower bound on gain, not an upper one, because the open-loop poles start in the right half-plane.
6. Extract the crossing frequency from the auxiliary polynomial. The row above the vanishing row forms the auxiliary polynomial, whose roots are the imaginary-axis pair. With \(K = 0.4\), $$(1+K)s^{2} + (2+6K) = 1.4\,s^{2} + 4.4 = 0 \;\Longrightarrow\; s^{2} = -\frac{4.4}{1.4} = -3.1429$$ $$\boxed{s = \pm j1.7728 \ \text{at } K = 0.4}$$ Substituting \(K = 0.4\) back into the quadratic and solving numerically returns \(s = \pm j1.7728\) exactly, with a real part of zero to machine precision — the cheapest possible check on the Routh work.
7. Part (d) — assemble the sketch. Four construction facts fix the whole picture. There are \(n = m = 2\), so no branch runs to infinity and there are no asymptotes; every branch starts on a pole and ends on a zero. On the real axis the locus exists only where an odd number of real singularities lie to the right, which is the segment \(-3 \le \sigma \le -2\) between the two zeros. The branches leave \(1 \pm j1\) at \(\pm 122.47^\circ\), sweep leftward across the imaginary axis at \(\pm j1.7728\) when \(K = 0.4\), and meet the real axis at the given break-in point \(\sigma_b = -2.43\) — recomputing it from \(dK/ds = 0\) with \(K = -(s^{2}-2s+2)/[(s+2)(s+3)]\) gives \(7s^{2}+8s-22 = 0\), whose admissible root is \(-2.4341\), confirming the printed value. The gain there is \(K = 52.08\); beyond it the two roots separate along the real axis and run into the zeros at \(-2\) and \(-3\).
8. State the stability conclusion the sketch implies. Because both branches start in the right half-plane and cross into the left at \(K = 0.4\), and no branch ever returns, $$\boxed{\text{the closed loop is stable for } K \gt 0.4}$$ This is the mark-earning observation: for this plant more gain is stabilising, which reverses the habit built on minimum-phase plants where increasing gain eventually destabilises the loop.