Root locus — breakaway and break-in points: 22-Mec-A3 System Analysis and Control, May 2018, Q4

Question 4: Root locus — breakaway and break-in points (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. Open-loop poles at $s=0$ and $s=\pm j1$; open-loop zeros at $s=\pm j0.325$ (note $0.325^{2}=0.105625$); unity negative feedback; $K>0$.

Find. The root-locus sketch, and the exact real-axis breakaway and break-in points with their gains.

Question 4 — root locus near the origin. Crosses are open-loop poles, circles are open-loop zeros; the third branch continues along the negative real axis to $-\infty$.

Approach. Apply the standard construction rules (real-axis segments, asymptote count, departure and arrival angles), then locate the real-axis break points from $dK/ds=0$ and classify each as breakaway or break-in by whether $K$ is locally maximum or minimum there.

Count the branches and asymptotes. There are $n=3$ poles and $m=2$ finite zeros, so three branches start at the poles, two terminate on the zeros $\pm j0.325$, and $n-m=1$ branch runs to infinity. A single asymptote must have angle $\theta=180^{\circ}/(n-m)=180^{\circ}$: the escaping branch travels out along the negative real axis.
Find the real-axis segments. A point on the real axis belongs to the locus when the number of real poles and zeros to its right is odd. All the poles and zeros except $s=0$ are complex conjugates and contribute nothing to that count, so every point with $s<0$ has exactly one (the pole at the origin) to its right. Hence the entire negative real axis is on the locus and no part of the positive real axis is.
Check the departure and arrival angles. At the pole $p=+j1$ the angle criterion gives $\theta_{d}=180^{\circ}+\sum\angle(p-z_{i})-\sum\angle(p-p_{j})=180^{\circ}+(90^{\circ}+90^{\circ})-(90^{\circ}+90^{\circ})=180^{\circ}$: the branch leaves the pole horizontally, straight to the left. The same construction at the zero $+j0.325$ gives an arrival angle of $180^{\circ}$, so the branches also approach the zeros horizontally. Both agree with the sketch above.
Set up the break-point condition. On the locus $1+G(s)=0$, so$$K(s)=-\frac{s\left(s^{2}+1\right)}{s^{2}+0.105625}.$$Break points occur where $K$ is stationary along the real axis, $dK/ds=0$. Differentiating with the quotient rule and clearing the denominator gives the quartic$$s^{4}-\left(1-3\times 0.105625\right)s^{2}+0.105625=0\quad\Longleftrightarrow\quad s^{4}-0.683125\,s^{2}+0.105625=0,$$or, cleared of decimals by multiplying through by $-2{,}560{,}000$, $-2{,}560{,}000\,s^{4}+1{,}748{,}800\,s^{2}-270{,}400=0$ — a quadratic in $s^{2}$.
Solve for the real break points. Solving the quadratic in $s^{2}$ gives $s^{2}=0.236492$ and $s^{2}=0.446634$, hence four real roots $\pm0.48630$ and $\pm0.66831$. Only the negative pair lies on the locus (the positive real axis is not part of it), so the two break points are$$\boxed{\,s=-0.4863\ \ (K=1.7576),\qquad s=-0.6683\ \ (K=1.7506)\,}$$The gains follow by substituting each root back into $K(s)$, for example $K(-0.4863)=0.4863\left(1+0.2365\right)/\left(0.2365+0.1056\right)=0.6013/0.3421=1.7576$.
Classify each point. Along the negative real axis $K$ rises from 0 at the origin to a local maximum of 1.7576 at $s=-0.4863$, dips to a local minimum of 1.7506 at $s=-0.6683$, and then increases without bound as $s\to-\infty$. A local maximum of $K$ is a breakaway point (branches leave the axis) and a local minimum is a break-in point (branches arrive):
- $s=-0.6683$, $K=1.7506$ — break-in: the pair that departed from $\pm j1$ meets the real axis here and splits, one root heading left toward $-\infty$ and the other heading right.
- $s=-0.4863$, $K=1.7576$ — breakaway: that right-moving root meets the branch coming out of the pole at the origin, and the pair leaves the axis to curve back to the zeros $\pm j0.325$.
Between the two gains, $1.7506<K<1.7576$, all three closed-loop roots are real.
Comment on stability. No branch ever crosses into the right half-plane: the two branches that leave $\pm j1$ move immediately left, and the terminating zeros sit on the imaginary axis only in the limit $K\to\infty$. The closed loop is therefore stable for every finite $K>0$, although the damping is poor at large gain because the roots crowd toward $\pm j0.325$.

Quantity	Result
Open-loop poles / zeros	$0,\ \pm j1$ / $\pm j0.325$
Branches to infinity	1, asymptote at $180^{\circ}$
Real-axis locus	the entire negative real axis
Break-in point	$s=-0.6683$ at $K=1.7506$
Breakaway point	$s=-0.4863$ at $K=1.7576$
Departure angle at $+j1$ / arrival at $+j0.325$	$180^{\circ}$ / $180^{\circ}$
Stability	stable for all $K>0$

Topic: Root-locus construction for a plant whose poles and zeros both sit on the imaginary axis, and the breakaway/break-in classification test.

Key relations: Angle criterion $\sum\angle(s-z)-\sum\angle(s-p)=180^{\circ}$; magnitude criterion $K=\prod|s-p|/\prod|s-z|$; real-axis rule (odd count to the right); asymptote angles $(2q+1)180^{\circ}/(n-m)$; break points from $dK/ds=0$, with $K$ locally maximum at a breakaway and locally minimum at a break-in.

Why this works: Writing the locus condition as $K=-1/G_{0}(s)$ turns the geometric construction into a one-variable calculus problem on the real axis: the closed-loop roots move continuously with $K$, so a point where $K$ is stationary is exactly where two branches meet and must leave the axis (or where two arriving branches merge). Because the imaginary-axis zeros lie inside the imaginary-axis poles, the pair leaving $\pm j1$ is pulled around and back to the axis, which is why this plant has two real break points instead of the single one a beginner expects.

Common pitfall: Assuming a single break point and stopping at the first root of $dK/ds=0$, or failing to test whether each root is a maximum or a minimum of $K$ — the two are only 0.007 apart in gain here, so guessing which is which from the sketch alone is unreliable. Discarding the positive roots $+0.4863$ and $+0.6683$ is also essential: they solve the algebra but do not lie on the locus.

Source: Dorf & Bishop, Modern Control Systems, 14th ed. §7.2–7.3 (root-locus rules, breakaway points); Nise, Control Systems Engineering, 8th ed. §8.4–8.5 (sketching the root locus, break-in and breakaway).

Question 4 of 6: Root locus — breakaway and break-in points

Question 4: Root locus — breakaway and break-in points (25 marks)