Question 5 of 6: Root Locus of a Third-Order Loop — Segments, Asymptotes and the Imaginary-Axis Crossing

Question 5: Root Locus of a Third-Order Loop — Segments, Asymptotes and the Imaginary-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. The characteristic equation $1+KL_{0}(s)=0$ with $L_{0}(s)=1/[s(s+1)(s+5)]$, so the locus has $n=3$ open-loop poles at $s=0,\;-1,\;-5$, no finite zeros ($m=0$), and $K$ is swept from $0$ to $+\infty$.

Find. (a) the portions of the real axis that belong to the locus, (b) the asymptotes followed by the three branches as $K\to\infty$, and (c) the gain at which a pair of branches crosses the imaginary axis.

Approach. Apply the standard root-locus construction rules — the real-axis test counts poles and zeros to the right, the asymptote centroid and angles follow from $n-m=3$ — and pin the imaginary-axis crossing with Routh’s criterion on the expanded characteristic polynomial.

Part (a) — real-axis segments

1. State the real-axis rule. A point on the real axis lies on the $180^{\circ}$ root locus if and only if the total number of real open-loop poles and zeros strictly to its right is odd. This follows from the angle condition: each real singularity to the right contributes $180^{\circ}$ to $\angle L_{0}(s)$, and each to the left contributes $0^{\circ}$.
2. Test each interval between the poles. The real poles divide the axis into four intervals. To the right of the origin there are no singularities (count $0$, even — not on the locus). Between $-1$ and $0$ the only singularity to the right is the pole at the origin (count $1$, odd — on the locus). Between $-5$ and $-1$ the poles at $0$ and $-1$ lie to the right (count $2$, even — not on the locus). To the left of $-5$ all three poles lie to the right (count $3$, odd — on the locus).
3. Collect the answer. The real-axis portions of the locus are$$\boxed{\;-1\le\sigma\le0\quad\text{and}\quad\sigma\le-5\;}$$drawn as the heavy purple segments in the figure. Two branches start at $s=0$ and $s=-1$, approach each other along the first segment, and break away into the complex plane; the third branch runs from $s=-5$ leftward to infinity along the negative real axis.
4. Locate the breakaway point (a useful refinement). Writing $K=-s(s+1)(s+5)=-(s^{3}+6s^{2}+5s)$ and setting $dK/ds=0$ gives $3s^{2}+12s+5=0$, whose roots are $s=-0.4725$ and $s=-3.528$. Only the first lies on a locus segment, so the two branches break away at $s=-0.472$, where $K=1.128$.

Part (b) — asymptotes as $K\to\infty$

1. Count the branches that escape to infinity. With $n=3$ poles and $m=0$ finite zeros, $n-m=3$ branches must run off to infinity, and they do so along $3$ straight-line asymptotes.
2. Compute the asymptote angles. The angles are equally spaced and given by$$\phi_{\ell}=\frac{(2\ell+1)180^{\circ}}{n-m},\qquad \ell=0,1,2$$Substituting $n-m=3$ gives$$\phi=60^{\circ},\;180^{\circ},\;300^{\circ}\;(\equiv-60^{\circ})$$
3. Compute the centroid. The asymptotes all radiate from the centre of gravity of the finite singularities,$$\sigma_{a}=\frac{\sum\text{poles}-\sum\text{zeros}}{n-m}=\frac{(0)+(-1)+(-5)-0}{3}=\frac{-6}{3}$$so that$$\boxed{\;\sigma_{a}=-2,\qquad \phi=\pm60^{\circ},\;180^{\circ}\;}$$
4. Describe the resulting picture. The dashed green rays in the figure show the three asymptotes meeting at $\sigma_{a}=-2$. The complex pair that breaks away at $s=-0.472$ curves upward and to the right, asymptotically approaching the $\pm60^{\circ}$ rays, while the third branch tracks the $180^{\circ}$ ray leftward. Because two asymptotes point into the right half-plane, the loop is guaranteed to go unstable at sufficiently high gain — part (c) finds exactly where.

Part (c) — the imaginary-axis crossing

1. Expand the characteristic polynomial. Clearing the fraction,$$s(s+1)(s+5)+K=0\quad\Longrightarrow\quad s^{3}+6s^{2}+5s+K=0$$
2. Apply Routh’s criterion. The array for the cubic is$$\begin{array}{c|cc} s^{3} & 1 & 5\\ s^{2} & 6 & K\\ s^{1} & \dfrac{30-K}{6} & 0\\ s^{0} & K & \end{array}$$so the loop is stable for $0<K<30$, and the $s^{1}$ entry vanishes exactly at$$\boxed{\;K=30\;}$$
3. Find the crossing frequency from the auxiliary polynomial. With the $s^{1}$ row zero, the row above supplies the factor containing the imaginary roots:$$6s^{2}+K=0\;\Longrightarrow\;s^{2}=-\frac{30}{6}=-5\;\Longrightarrow\;s=\pm j\sqrt{5}=\pm j2.236$$
4. Confirm by direct factorisation. At $K=30$ the cubic factors as $s^{3}+6s^{2}+5s+30=(s+6)(s^{2}+5)$, whose roots are $s=-6$ and $s=\pm j2.236$ — precisely the orange markers on the figure. The closed loop is therefore marginally stable at $K=30$, sustaining an undamped oscillation at $2.236\ \text{rad/s}$ (about $0.356\ \text{Hz}$), and unstable for any larger gain.