Root locus and the gain for $\zeta=0.5$: 22-Mec-A3 System Analysis and Control, May 2013, Q5

Question 5: Root locus and the gain for $\zeta=0.5$ (equal value)

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 system with

$$G(s)=\frac{K}{s\,(s+1)\,(s+2)}.$$

Open-loop poles at $0,\,-1,\,-2$; no finite zeros.

Find. the root-locus diagram, and the value of $K$ giving the dominant complex pole pair a damping ratio $\zeta=0.5$.

Root locus of $K/[s(s+1)(s+2)]$. Three branches leave the poles at $0,-1,-2$ (×); two break away at $\sigma=-0.423$ and follow the $60^\circ$ asymptotes (centroid $-1$), crossing the $j\omega$ axis at $\pm j\sqrt2$ when $K=6$. The dashed ray is the $\zeta=0.5$ line; the design points ($\circ$) at $-\tfrac13\pm j0.577$ correspond to $K=1.04$.

Approach. Sketch the locus from its rules (real-axis segments, asymptotes, breakaway, $j\omega$ crossing), then impose the $\zeta=0.5$ geometry on the characteristic cubic and solve for the pole locations and $K$.

Locus skeleton. With $n=3$ poles and $m=0$ zeros there are three branches to infinity. The real-axis locus lies where an odd number of real poles/zeros sit to the right: on $[-1,0]$ and $(-\infty,-2]$. The three asymptotes have $$\text{centroid }\sigma_a=\frac{0-1-2}{3}=-1,\qquad \text{angles }\pm60^\circ,\,180^\circ.$$
Breakaway point. On $[-1,0]$ the two branches meet and leave the axis where $dK/ds=0$ with $K=-s(s+1)(s+2)$: $$3s^{2}+6s+2=0\;\Rightarrow\;s=-0.423\ (\text{the root in }[-1,0]).$$
Imaginary-axis crossing. The characteristic equation is $s^{3}+3s^{2}+2s+K=0$; Routh marginal stability requires $3\cdot2=K$, so the locus crosses the $j\omega$ axis at $$K=6,\qquad s=\pm j\sqrt2.$$ Thus the loop is stable for $0<K<6$; the $\zeta=0.5$ design will sit well inside this.
Impose $\zeta=0.5$. Let the complex pair be $s=-a\pm jb$ and the third (real) pole $-c$. A damping ratio $\zeta=0.5$ means the poles lie on rays at $60^\circ$ from the negative real axis, so $b=\sqrt3\,a$. Matching the cubic $s^{3}+3s^{2}+2s+K$ by Vieta: $$\underbrace{2a+c=3}_{\text{sum}},\qquad \underbrace{a^{2}+b^{2}+2ac=2}_{\text{pair sum}}.$$ Substituting $b^{2}=3a^{2}$ and $c=3-2a$ into the second equation gives $3a=1$, so $$a=\tfrac13,\qquad c=\tfrac73,\qquad b=\tfrac{\sqrt3}{3}=0.577.$$ The dominant poles are $s=-0.333\pm j0.577$ (check $\omega_n=\sqrt{a^{2}+b^{2}}=\tfrac23$, $\zeta=a/\omega_n=0.5$ &checkmark).
Gain at the design point. $K$ equals the product of the pole magnitudes (constant term of the cubic): $$K=(a^{2}+b^{2})\,c=\left(\tfrac49\right)\!\left(\tfrac73\right)=\frac{28}{27}=\boxed{1.037}.$$ Because $1.04\ll6$, the design is comfortably stable and the complex pair (closest to the $j\omega$ axis) dominates the response.

Quantity	Value
Asymptote centroid / angles	$-1$ / $\pm60^\circ,180^\circ$
Breakaway point	$\sigma=-0.423$
$j\omega$-axis crossing	$\pm j\sqrt2$ at $K=6$
Dominant poles at $\zeta=0.5$	$-0.333\pm j0.577$ ($\omega_n=0.667$)
Gain for $\zeta=0.5$	$K=28/27\approx1.04$

Topic: Root-locus construction and design for a specified damping ratio.

Key relations: $1+KG_0=0$; asymptote centroid $\sigma_a=(\sum p-\sum z)/(n-m)$, angles $(2q+1)180^\circ/(n-m)$; breakaway $dK/ds=0$; $\zeta=\cos\theta$ where $\theta$ is the pole angle from the negative real axis; $K=\prod|\,s-p_i|$ on the locus.

Why this works: Every point on the locus satisfies the angle condition, so a constant-$\zeta$ ray intersects it at the design poles; Vieta's relations then tie the two complex poles and the third real pole to the cubic's coefficients, giving $a$, $c$ and $K$ in closed form without trial magnitudes.

Common pitfall: equating $\zeta=0.5$ with a $45^\circ$ ray — it is $60^\circ$ from the negative real axis ($\cos60^\circ=0.5$). Also, forgetting the third real pole when computing $K$: the magnitude condition must include the distance to every open-loop pole, not just the complex pair.

Question 5 of 6: Root locus and the gain for $\zeta=0.5$

Question 5: Root locus and the gain for $\zeta=0.5$ (equal value)