Root Locus and Critical Gain of a Three-Pole Loop: 22-Mec-A3 System Analysis and Control, December 2019, Q6

Question 6: Root Locus and Critical Gain of a Three-Pole Loop (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-negative-feedback loop with $G(s)=\dfrac{K}{s\left(1+0.1s\right)\left(1+s\right)}$, $K$ variable from 0 upward.

Find. (a) the root locus with all its defining features — poles, zeros, real-axis segments, asymptotes, centroid, breakaway point and imaginary-axis crossing; (b) the gain at which the closed loop first becomes unstable.

Approach. Convert to pole–zero (root-locus) form, apply the standard construction rules in order, and obtain the imaginary-axis crossing from the Routh array, which gives both the critical gain and the frequency of oscillation.

Convert to pole–zero form. Factor the coefficients out of the two lags: $$G(s)=\frac{K}{s\left(0.1\right)\left(s+10\right)\left(s+1\right)} = \frac{10K}{s\left(s+1\right)\left(s+10\right)}.$$ Write $K^{*}=10K$ for the root-locus gain. The open loop has three poles at $s=0,\;-1,\;-10$ and no finite zeros.
Number of branches and real-axis segments. With $n=3$ poles and $m=0$ zeros there are three branches, all ending at infinity. A point on the real axis belongs to the locus when the total number of real poles and zeros to its right is odd, which gives the segments $-1\le\sigma\le0$ and $\sigma\le-10$.
Asymptotes. The $n-m=3$ asymptotes leave at $$\theta_k=\frac{\left(2k+1\right)180^\circ}{n-m}=60^\circ,\;180^\circ,\;300^\circ, \qquad k=0,1,2,$$ from the centroid $$\sigma_0=\frac{\sum\text{poles}-\sum\text{zeros}}{n-m} = \frac{0+(-1)+(-10)}{3} = \boxed{\;-3.67\;}$$
Breakaway point. On the segment between $0$ and $-1$ the two branches meet and leave the axis where $dK^{*}/ds=0$, i.e. where $$\frac{d}{ds}\left[s\left(s+1\right)\left(s+10\right)\right] = 3s^{2}+22s+10 = 0 \;\Longrightarrow\; s=-0.487 \text{ or } s=-6.85.$$ Only $s=-0.487$ lies on a locus segment, so $\boxed{\;\sigma_b=-0.487\;}$, reached at $K^{*}=2.377$, i.e. $K=0.238$. Beyond that gain the pair is complex and moves toward the $\pm60^\circ$ asymptotes.
Imaginary-axis crossing and critical gain. The characteristic equation is $$s\left(s+1\right)\left(s+10\right)+10K = s^{3}+11s^{2}+10s+10K = 0,$$ with Routh array $$\begin{array}{c|cc} s^{3} & 1 & 10\\ s^{2} & 11 & 10K\\ s^{1} & \dfrac{11(10)-10K}{11} & 0\\ s^{0} & 10K & \end{array}$$ The $s^{1}$ entry vanishes when $110=10K$, so the loop is stable for $0<K<11$ and $\boxed{\;K_{\text{crit}} = 11\;}$
Frequency at the crossing. Substituting $K=11$ into the row above gives the auxiliary equation $$11s^{2}+10K = 11s^{2}+110 = 0 \;\Longrightarrow\; s = \pm j\sqrt{10} = \pm j3.16\ \text{rad}\,\text{s}^{-1},$$ so at the critical gain the system sustains an undamped oscillation at $3.16\ \text{rad}\,\text{s}^{-1}$ (period 1.99 s); the third pole is then at $s=-11$.

Root locus of K/[s(1 + 0.1s)(1 + s)] as K rises from 0: the two low-frequency branches break away at σ = −0.487 and cross the imaginary axis at ±j3.16 when K = 11, while the third branch runs left from the pole at −10. The dashed mark is the asymptote centroid σ₀ = −3.67.

The plotted locus assembles all of these features. Two branches leave the poles at 0 and −1, meet on the axis at −0.487, and turn upward and downward toward the ±60° asymptotes, crossing the imaginary axis at ±j3.16 exactly when K = 11; the third branch runs left from −10 along the 180° asymptote. Note that the fast pole at −10 barely participates at low gain, which is why the loop behaves like a second-order system until the gain approaches its critical value.

Question 6 — root-locus characteristics
Feature	Value
Root-locus form	$10K/\left[s\left(s+1\right)\left(s+10\right)\right]$
Open-loop poles / zeros	$s=0,\,-1,\,-10$ / none
Real-axis segments	$-1\le\sigma\le0$ and $\sigma\le-10$
Asymptote angles	$60^\circ,\;180^\circ,\;300^\circ$
Asymptote centroid	$\sigma_0=-3.67$
Breakaway point	$\sigma_b=-0.487$ (at $K=0.238$)
Imaginary-axis crossing	$s=\pm j3.16\ \text{rad}\,\text{s}^{-1}$
(b) Critical gain	$K=11$ (stable for $0<K<11$)

Topic: Root-locus construction and the determination of critical gain from the Routh array.

Key relations: $1+KG_0(s)=0$; branches $=n$; asymptote angles $(2k+1)180^\circ/(n-m)$; centroid $\sigma_0=\left(\sum p_i-\sum z_i\right)/(n-m)$; breakaway where $dK/ds=0$; imaginary-axis crossing from the vanishing $s^{1}$ Routh row and the auxiliary polynomial of the row above.

Why this works: The locus is the set of points where the open-loop phase equals $180^\circ$, so every construction rule is a phase or magnitude statement in disguise. The Routh array and the locus give the same critical gain by different routes: the array detects the moment a pole pair sits on the imaginary axis, while the auxiliary equation formed from the row above that zero row supplies the crossing frequency free of charge — a much faster route than reading coordinates off a sketch.

Common pitfall: Forgetting to normalise $\left(1+0.1s\right)\left(1+s\right)$ into $0.1(s+10)(s+1)$, which inflates the reported critical gain by a factor of ten. The other standard slip is accepting both roots of the breakaway equation; only roots lying on an actual locus segment are breakaway points, and here $s=-6.85$ falls in the gap between $-1$ and $-10$ where the locus does not exist.

Source: Ogata, Modern Control Engineering, 5th ed., Ch. 6.2 (root-locus plots) and Ch. 5.6; Nise, Control Systems Engineering, 8th ed., Ch. 8; Franklin, Powell & Emami-Naeini, Ch. 5.

Question 6 of 6: Root Locus and Critical Gain of a Three-Pole Loop

Question 6: Root Locus and Critical Gain of a Three-Pole Loop (25 marks)