22-Mec-A3 System Analysis and Control · December 2017

Question 4 of 6: Root Locus of a Four-Pole Plant and its Imaginary-Axis Crossings

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

Notes on this paper

Paper format. National Exams, December 2017 — 16-Mec-A3, System Analysis and Control. Three hours, closed book; a Casio or Sharp approved calculator and semi-log graph paper are the only permitted aids. Six questions are printed; any four constitute a complete paper and all questions are of equal value (25 marks each). A table of Laplace transforms is supplied as page 5. All six questions are solved here.

Reference texts. K. Ogata, Modern Control Engineering, 5th ed. (Prentice Hall) — the source of this paper's problem style; N. S. Nise, Control Systems Engineering, 7th ed. (Wiley); R. C. Dorf and R. H. Bishop, Modern Control Systems, 13th ed. (Pearson); G. F. Franklin, J. D. Powell and A. Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed. (Pearson).

Check: conventions used throughout. All transfer functions are written in the Laplace variable s with zero initial conditions unless a question states otherwise. “Unity feedback” means $H(s)=1$, so the closed-loop transfer function is $T(s)=G(s)/[1+G(s)]$ and the error signal is $E(s)=R(s)-C(s)$. Root-locus gains are the gain $K$ exactly as it appears in the printed $G(s)$.

Question 4: Root Locus of a Four-Pole Plant and its Imaginary-Axis Crossings (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 transfer function is $G(s)H(s)=K/\left[(s^2+2s+2)(s^2+2s+5)\right]$, with variable gain $K\ge 0$. There are four open-loop poles and no finite zeros.

Find. The complete root locus for $K>0$, and in particular the points at which the branches cross the imaginary axis together with the gain at which that happens.

Approach. Locate the open-loop poles, apply the standard construction rules (number of branches, real-axis segments, asymptotes, break points, departure angles), then obtain the imaginary-axis crossings exactly from the Routh array of the closed-loop characteristic polynomial rather than by reading them off a sketch.

Factor the open-loop poles. Each quadratic is solved by completing the square: $$s^2+2s+2=(s+1)^2+1\;\Rightarrow\;s=-1\pm j1,\qquad s^2+2s+5=(s+1)^2+4\;\Rightarrow\;s=-1\pm j2$$ All four poles therefore lie on the vertical line $\operatorname{Re}(s)=-1$, at $-1\pm j1$ and $-1\pm j2$. There are $n=4$ poles, $m=0$ zeros, hence four branches, all of which run off to infinity.
Real-axis segments and asymptotes. Since no open-loop pole or zero lies on the real axis, no part of the real axis belongs to the locus for $K>0$. With $n-m=4$, there are four asymptotes; their centroid and angles are $$\sigma_a=\frac{\sum \text{poles}-\sum \text{zeros}}{n-m}=\frac{(-1)+(-1)+(-1)+(-1)}{4}=-1,\qquad \theta_a=\frac{(2k+1)180^\circ}{4}=\pm45^\circ,\ \pm135^\circ$$
Departure angles from the complex poles. Using the angle criterion at $p=-1+j2$, the angles contributed by the other three poles are $90^\circ$ (from $-1+j1$), $90^\circ$ (from $-1-j1$) and $90^\circ$ (from $-1-j2$), summing to $270^\circ$. Hence $$\theta_d=180^\circ-270^\circ=-90^\circ$$ i.e. the branch leaves $-1+j2$ heading downwards. Repeating at $p=-1+j1$ gives $90^\circ-90^\circ+90^\circ=90^\circ$ and $\theta_d=180^\circ-90^\circ=+90^\circ$, heading upwards. The two branches therefore move toward one another along the line $\operatorname{Re}(s)=-1$ and must meet.
Find the break point where they meet. The characteristic equation is $$\left(s^2+2s+2\right)\left(s^2+2s+5\right)+K=0\;\Longrightarrow\;s^4+4s^3+11s^2+14s+10+K=0$$ so $K=-\left(s^4+4s^3+11s^2+14s+10\right)$, and break points satisfy $dK/ds=0$: $$4s^3+12s^2+22s+14=0\;\Longrightarrow\;2s^3+6s^2+11s+7=0$$ Since $s=-1$ is a root, factoring gives $(s+1)(2s^2+4s+7)=0$, whose quadratic roots are $s=-1\pm j\sqrt{2.5}=-1\pm j1.5811$. Evaluating the gain there (using $s^2+2s=(s+1)^2-1=-3.5$): $$K=-\left(-3.5+2\right)\left(-3.5+5\right)=-(-1.5)(1.5)=2.25$$ $$\boxed{\;\text{break point at } s=-1\pm j1.5811,\ \ K=2.25\;}$$ The real root $s=-1$ returns $K=-4<0$ and so belongs to the negative-gain locus — consistent with there being no real-axis locus for $K>0$.
Confirm the vertical-line segment analytically. Putting $s=-1+j\omega$ makes $s^2+2s=-1-\omega^2$, so $$K=-\left(1-\omega^2\right)\left(4-\omega^2\right)$$ This is positive exactly for $1<\omega<2$, which proves that the segment of the line $\operatorname{Re}(s)=-1$ between the poles $-1\pm j1$ and $-1\pm j2$ is part of the locus, and is maximised at $\omega^2=2.5$, returning $K_{\max,\text{seg}}=2.25$ — the break point already found.
Locate the imaginary-axis crossings by Routh. Applying the array to $s^4+4s^3+11s^2+14s+(10+K)$: $$\begin{array}{c|ccc} s^4 & 1 & 11 & 10+K\\ s^3 & 4 & 14 & \\ s^2 & \dfrac{4(11)-14}{4}=7.5 & 10+K & \\ s^1 & \dfrac{7.5(14)-4(10+K)}{7.5}=\dfrac{65-4K}{7.5} & & \\ s^0 & 10+K & & \end{array}$$ The $s^1$ entry vanishes when $65-4K=0$, i.e. at $$K=\frac{65}{4}=16.25$$ and the auxiliary polynomial from the $s^2$ row gives the crossing frequency: $$A(s)=7.5s^2+(10+16.25)=7.5s^2+26.25=0\;\Longrightarrow\;s^2=-3.5$$ $$\boxed{\;\text{the loci cross the } j\omega \text{ axis at } s=\pm j\sqrt{3.5}=\pm j1.8708,\ \text{at } K=16.25\;}$$
State the resulting stability range and sketch the locus. Since the first column also requires $10+K>0$, the closed loop is stable for $$\boxed{\;0<K<16.25\;}$$ The complete picture: four branches start at the poles, the inner and outer pairs approach one another along $\operatorname{Re}(s)=-1$ and break away at $-1\pm j1.5811$ when $K=2.25$; thereafter one pair curves left and one pair curves right, the right-moving pair crossing the imaginary axis at $\pm j1.8708$ when $K=16.25$, all four branches ultimately following the $\pm45^\circ$, $\pm135^\circ$ asymptotes centred on $\sigma_a=-1$.

Figure 4.1 — Root locus of $K/[(s^2+2s+2)(s^2+2s+5)]$ for $K>0$, drawn with equal scaling on both axes. Crosses mark the four open-loop poles on $\operatorname{Re}(s)=-1$; green dots mark the break points at $-1\pm j1.5811$ ($K=2.25$) and magenta dots the imaginary-axis crossings at $\pm j1.8708$ ($K=16.25$). Dashed grey lines are the $\pm45^\circ$/$\pm135^\circ$ asymptotes through $\sigma_a=-1$.

Quantity	Result
Open-loop poles	$-1\pm j1$ and $-1\pm j2$ (four; no finite zeros)
Characteristic equation	$s^4+4s^3+11s^2+14s+10+K=0$
Asymptote centroid / angles	$\sigma_a=-1$; $\pm45^\circ$, $\pm135^\circ$
Departure angles	$-90^\circ$ from $-1+j2$; $+90^\circ$ from $-1+j1$
Break point	$s=-1\pm j1.5811$ at $K=2.25$
Imaginary-axis crossings	$s=\pm j1.8708$ $(\sqrt{3.5})$ at $K=16.25$
Stability range	$0<K<16.25$

Topic. Root-locus construction for a plant with two lightly damped complex pole pairs, and exact determination of imaginary-axis crossings by the Routh auxiliary polynomial.

Key relations. Magnitude and angle criteria $|KG(s)H(s)|=1$, $\angle G(s)H(s)=\pm(2k+1)180^\circ$; asymptote centroid $\sigma_a=(\sum p_i-\sum z_i)/(n-m)$ and angles $(2k+1)180^\circ/(n-m)$; departure angle $\theta_d=180^\circ-\sum\angle(\text{other poles})+\sum\angle(\text{zeros})$; break points from $dK/ds=0$; imaginary-axis crossing from the vanishing Routh row and its auxiliary polynomial.

Why this works. The locus is the set of points where the open-loop phase equals an odd multiple of $180^\circ$; every construction rule is a consequence of that single condition. The symmetry of this particular plant — all four poles sharing $\operatorname{Re}(s)=-1$ — makes the substitution $s=-1+j\omega$ unusually powerful, because it reduces the quartic gain expression to a quadratic in $\omega^2$ and yields the break point in one line. The Routh route to the crossing is important precisely because a hand sketch cannot give a trustworthy number: the array converts the geometric question “where does the locus meet the axis?” into the algebraic question “at what gain does a row vanish?”, and the auxiliary polynomial then supplies $\omega$ exactly.

Common pitfall. The classic mistake is assuming there must be a real-axis break-away point and reporting $s=-1$ as the answer; here $dK/ds=0$ does have the real root $s=-1$, but it corresponds to $K=-4$ and so lies on the negative-gain locus, not the one asked for. Always evaluate the gain at each candidate and discard those returning $K\le0$ or complex $K$. A second trap is estimating the crossing from the asymptotes: the $\pm45^\circ$ asymptotes cross the imaginary axis at $\pm j1$, whereas the true crossing is at $\pm j1.871$ — the asymptotes describe only the far-field behaviour. Finally, sketch the locus with equal scaling on the two axes; unequal scaling distorts the $45^\circ$ asymptotes and makes the departure angles look wrong.

Source. Ogata, Modern Control Engineering, 5th ed., §6–2 and §6–3 (root-locus plots, construction rules and worked examples); Nise, Control Systems Engineering, 7th ed., Ch. 8.

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Quantity	Result
Open-loop poles	\(-1\pm j1\) and \(-1\pm j2\) (four; no finite zeros)
Characteristic equation	\(s^4+4s^3+11s^2+14s+10+K=0\)
Asymptote centroid / angles	\(\sigma_a=-1\); \(\pm45^\circ\), \(\pm135^\circ\)
Departure angles	\(-90^\circ\) from \(-1+j2\); \(+90^\circ\) from \(-1+j1\)
Break point	\(s=-1\pm j1.5811\) at \(K=2.25\)
Imaginary-axis crossings	\(s=\pm j1.8708\) \((\sqrt{3.5})\) at \(K=16.25\)
Stability range	\(0<K<16.25\)