Question 3 of 7: Root Locus of $k/[s(s^{2}+6s+12)]$

Question 3: Root Locus of $k/[s(s^{2}+6s+12)]$ (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. Unity-feedback loop with $kG(s)=\dfrac{k}{s(s^{2}+6s+12)}$, so the characteristic equation is $s^{3}+6s^{2}+12s+k=0$ and $k$ is swept over $0\le k<\infty$. There are three open-loop poles and no finite zeros.

Find. (a) the open-loop pole/zero locations and the real-axis segments of the locus; (b) the departure angles from the complex poles; (c) all breakaway and break-in points; (d) the complete locus; and (e) the gain $k$ at which the closed-loop complex pair has $\zeta=0.5$.

Approach. Factor the plant to place the poles, apply the real-axis rule and the angle-of-departure rule, differentiate $k(s)$ for the break points, then impose the $\zeta=0.5$ ray as an algebraic condition on the branch equation rather than reading it off a sketch.

1. (a) Locate the open-loop poles and the real-axis segments. The quadratic factor has roots

There are no finite zeros, so all three branches must run off to infinity. A point on the real axis belongs to the locus when the number of real poles and zeros strictly to its right is odd. Only $s=0$ is a real pole, so every point to its left qualifies and no point to its right does. The real-axis portion of the locus is therefore the entire half-line $-\infty<\sigma\le0$.

2. (b) Apply the angle-of-departure rule at $p_{2}=-3+j1.7321$. The angle condition $\sum\angle(\text{zeros})-\sum\angle(\text{poles})=180^{\circ}$ evaluated just off the pole gives $\theta_{d}=180^{\circ}-\sum\theta_{\text{other poles}}$. The two contributions are

The first follows because $\tan^{-1}(1.7321/3)=30^{\circ}$ in the second quadrant, and the second because the conjugate pole lies directly below, so the vector between them points straight up. Hence

and by conjugate symmetry the departure from $p_{3}=-3-j1.7321$ is $\boxed{\theta_{d}=\mp60^{\circ}}$, i.e. $-60^{\circ}$ from the upper pole and $+60^{\circ}$ from the lower one. Both branches head down and to the right, towards the real axis.

3. (c) Find the break points from $dk/ds=0$. Solving the characteristic equation for the gain gives $k=-(s^{3}+6s^{2}+12s)$, so

There is exactly one break point, at $s=-2$, and it is a breakaway rather than a break-in: it lies on the real-axis segment, and the gain there is

There are no break-in points, which is what one should expect: break-in points are where branches return to the real axis to terminate on real zeros, and this plant has none.

4. (d) Assemble the locus — and notice it is exact. The asymptote centroid and angles are

The centroid coincides with the breakaway point, which is the clue that something special is happening. Indeed $dk/ds$ has a double root at $s=-2$, meaning three branches meet there simultaneously, and

so the characteristic equation can be rewritten exactly as $(s+2)^{3}=8-k$. For $k>8$ its solutions are $s=-2+\sqrt[3]{k-8}\,e^{\,j(180^{\circ},\,\pm60^{\circ})}$: three perfectly straight rays leaving $s=-2$ at exactly $180^{\circ}$ and $\pm60^{\circ}$. The locus does not merely approach its asymptotes, it lies on them. For $0<k<8$ one root travels left along the real axis from $s=0$ while the complex pair descends from $s=-3\pm j1.7321$ (at $k=4$, for instance, the roots are $-0.4126$ and $-2.7937\pm j1.3747$); at $k=8$ all three coincide in a triple root at $s=-2$; beyond that one root runs left along the real axis while a complex pair opens outward at $\pm60^{\circ}$.

The imaginary axis is crossed when $s^{3}+6s^{2}+12s+k$ has purely imaginary roots; the Routh condition $6\times12>k$ gives $k_{\max}=72$ at $\omega=\sqrt{12}=3.46$ rad/s, so the whole region of interest here is comfortably stable.

5. (e) Impose $\zeta=0.5$ algebraically. A damping ratio of $0.5$ means the pole sits on a ray at $\cos^{-1}(0.5)=60^{\circ}$ from the negative real axis. Writing the complex branch as $s=-2+r\,e^{\,j60^{\circ}}=(-2+0.5r)+j0.8660r$ with $r=\sqrt[3]{k-8}$, the $\zeta=0.5$ condition $|\operatorname{Im}|=\tan60^{\circ}\,|\operatorname{Re}|$ becomes

and therefore

Substituting back, the closed-loop roots at $k=16$ are $s=-1\pm j1.7321$ and $s=-4$. Two checks confirm this: the pair has $\omega_n=\sqrt{1+3}=2$ and $\zeta=1/2$ as required, and expanding $(s+4)(s^{2}+2s+4)=s^{3}+6s^{2}+12s+16$ reproduces the characteristic polynomial with $k=16$ exactly.

One honest caveat is worth recording rather than glossing over. The real root at $s=-4$ is farther from the origin than the complex pair at $|s|=2$, but only by a factor of two, so calling the pair “dominant” is a judgement call rather than a clear-cut one. Measured the way that matters for the transient — by the real part, which sets the decay rate — the real mode is a factor of four faster ($\sigma=4$ against $\sigma=1$, i.e. $\tau=0.25$ s against $1$ s), and that head start is what makes the second-order approximation usable at all. It is still only an approximation: the extra pole slows the leading edge, so the step response of $T(s)=16/[(s+4)(s^{2}+2s+4)]$ peaks at $1.139$ at $t=2.12$ s — 13.9% overshoot, appreciably below the $16.3\%$ a bare $\zeta=0.5$ pair would give.