Exercises — Root locus — Evans' method, rules for sketching

Poles are the roots of the bottom: $s=0,-2,-5$. So $n=3$. Zeros are the roots of the top: $s=-3$. So $m=1$. Number of branches $=n=3$ (one branch per closed-loop root of the degree-3 polynomial — see the WHY box above). Branches ending on finite zeros $=m=1$. Branches escaping to infinity $=n-m=3-1===2==$.

Set the denominator to zero: $(s+1{)}^2+4=0\Rightarrow (s+1{)}^2=-4\Rightarrow s+1=\pm 2j$. So the poles are $s=-1+2j$ and $s=-1-2j$ — a complex-conjugate pair (mirror images across the real axis, exactly as Rule 2 symmetry promises). $n=2$, $m=0$.

The rule: a real point is on the locus if the count of real poles and zeros strictly to its right is odd.

WHY the odd-count criterion? (geometry — look at Figure s02.) A point is on the locus only if it satisfies the angle condition: the total angle of $G(s)H(s)$ must be an odd multiple of $180°$. The angle of $G(s)H(s)$ is (sum of angles from each zero to the test point) minus (sum of angles from each pole to the test point). Now picture a test point sitting on the real axis:

• Any complex-conjugate pair of poles/zeros contributes two angles that are exact mirror images (one $+\varphi$, one $-\varphi$). They cancel — so complex singularities never affect a real-axis test point.
• A real singularity to the left of the test point points along the positive real direction → contributes angle $0°$.
• A real singularity to the right of the test point points along the negative real direction → contributes exactly $180°$.

So the total angle is $180°\times (\text{number of real singularities to the right})$. For that total to be an odd multiple of $180°$ (the angle condition), the count to the right must be odd. That is the entire origin of the rule.

Apply it. Mark the axis from right to left: pole $0$, then zero $-2$, then pole $-4$.

• Point at $s=-1$ (between $0$ and $-2$): one singularity to the right ($s=0$) → count $=1$, odd → ON.
• Point at $s=-3$ (between $-2$ and $-4$): two to the right ($0,-2$) → count $=2$, even → OFF.
• Point at $s=-5$ (left of $-4$): three to the right → odd → ON. On the locus: the segment $[-2,0]$ and the ray $(-\infty ,-4]$ — the amber segments in Figure s02.

$n=3$ poles $\{0,-2,-4\}$, $m=0$ zeros, so $n-m=3$ branches escape.

WHY these angles are odd multiples of $180°/(n-m)$. Very far from all the poles and zeros, every one of them looks like it sits at roughly the same spot, so $G(s)H(s)\approx \frac{1}{s^{n-m}}$ (here $n-m$ poles minus zeros left over in the denominator). The angle condition then reads $\angle \frac{1}{s^{n-m}}=-(n-m)\cdot \angle s=\pm 180°(2k+1)$. Solving for the angle of $s$ gives $\angle s=\frac{\pm 180°(2k+1)}{n-m}$. The odd multiples $(2k+1)$ are inherited directly from the "$-1$ has angle $180°,540°,\dots$" fact — that is why the numerator is always an odd multiple of $180°$, and dividing by $(n-m)$ fans them out into $(n-m)$ equally-spaced escape directions.

WHY the centroid is the average. Divide the polynomials: $\frac{N}{D}$ for large $s$ behaves like $\frac{1}{(s-{\sigma }_a{)}^{n-m}}$ where matching the first two leading coefficients forces ${\sigma }_a=\frac{\sum \text{poles}-\sum \text{zeros}}{n-m}$. Physically: the far-field can't tell the poles apart, so it replaces them by a single equivalent pole of multiplicity $n-m$ sitting at their balance point (their average) — that balance point is where the $(n-m)$ ramps all intersect.

Compute: Centroid: ${\sigma }_a=\frac{(0-2-4)-0}{3}=\frac{-6}{3}===-2==$. Angles: ${\theta }_a=\frac{180°(2k+1)}{3}$ for $k=0,1,2\Rightarrow {\theta }_a=60°,180°,300°$. Three straight ramps leave the point $s=-2$ at those angles.

Here $D(s)=s(s+4)=s^2+4s$, $N(s)=1$. Why $dK/ds=0$? On the real segment two branches approach, collide into a double root, then peel off. A double root is where the gain function $K(s)$ turns around — a local extremum — so its derivative is zero. $K=-\frac{D(s)}{N(s)}=-(s^2+4s).$ $\frac{dK}{ds}=-(2s+4)=0\Rightarrow s===-2==.$ Gain there: $K=-((-2{)}^2+4(-2))=-(4-8)===4==.$ So at $K=4$ we get a repeated pole at $s=-2$; beyond that the poles go complex with constant real part $-2$.

Characteristic equation $D(s)+KN(s)=0$: $s(s+1)(s+3)+K=s^3+4s^2+3s+K=0.$ Build the Routh array:

WHY a zero row means a $j\omega$-crossing. The Routh test counts how many roots sit in the right half plane. A root exactly on the imaginary axis ($s=\pm j\omega$, real part $0$) is the knife-edge between stable and unstable — the moment a root is crossing over. Routh signals that exact instant by making a whole row vanish, because the array has lost the ability to distinguish "just stable" from "just unstable": a pair of purely imaginary roots has appeared. So the $K$ that zeros a row is precisely the $K$ at which a branch touches the axis. WHY the auxiliary polynomial gives the frequency. When a row zeros out, the row directly above it is the coefficient list of a factor of the characteristic polynomial — the auxiliary polynomial — and that factor is exactly the one carrying the imaginary-axis roots. Solving it hands you $\pm j\omega$ directly.

Stability needs the $s^1$ entry $>0$: $\frac{12-K}{4}>0\Rightarrow K<12$. Critical gain: $K===12==$ (the $s^1$ row becomes zero). Crossing frequency: the row above the zero row is the $s^2$ row, giving auxiliary $4s^2+K=0\Rightarrow 4s^2+12=0\Rightarrow s^2=-3\Rightarrow s=\pm j\sqrt{3}$. So $\omega ===\sqrt{3}\approx 1.732==$ at $K=12$.

WHERE the formula comes from (look at Figure s03). Put a test point $s$ an infinitesimal step away from the pole $p$, in the (unknown) departure direction ${\theta }_d$. Any point on the locus must obey the angle condition $\sum \angle (\text{zeros}\to s)-\sum \angle (\text{poles}\to s)=180°.$ The pole $p$ itself contributes the term $-\angle (p\to s)=-{\theta }_d$ (the vector from $p$ to the nearby test point points in the departure direction). Every other pole and zero is so far away compared to the tiny step that the angle it makes is just its angle to $p$. Substitute and isolate ${\theta }_d$: $-{\theta }_d+\sum \angle (\text{zeros}\to p)-\sum \angle (\text{other poles}\to p)=180°,$ $\boxed{{\theta }_d=-180°+\sum \angle (\text{zeros}\to p)-\sum \angle (\text{other poles}\to p)}$ and since $-180°$ and $+180°$ name the same direction, the standard form writes $+180°$. That $180°$ is nothing but the "$-1$ has angle $180°$" fact from the characteristic equation — the same odd-$180°$ that shaped every other rule.

Apply it. No zeros, so that sum is $0$. Two "other poles": the conjugate $-1-2j$ and $0$.

• Vector from $-1-2j$ to $p=-1+2j$: change $(0,+4)$ → points straight up → angle $=90°$.
• Vector from $0$ to $p=-1+2j$: change $(-1,+2)$ → quadrant II → $\arctan \left(\frac{2}{-1}\right)=180°-63.43°=116.57°$. ${\theta }_d=180°+0-(90°+116.57°)=180°-206.57°=-26.57°.$ Naming it positively: ${\theta }_d===333.43°==(\equiv -26.57°)$. The branch leaves the upper complex pole heading down-and-right — the cyan arrow in Figure s03.

A second-order pole $s=-\sigma \pm j{\omega }_d$ has natural frequency ${\omega }_n=\sqrt{{\sigma }^2+{\omega }_d^2}$ and $\zeta =\frac{\sigma }{{\omega }_n}$ (the cosine of the angle the pole makes with the negative real axis). Here $\sigma =1$ and ${\omega }_d=\sqrt{K-1}$. $\zeta =\frac{\sigma }{\sqrt{{\sigma }^2+{\omega }_d^2}}=\frac{1}{\sqrt{1+(K-1)}}=\frac{1}{\sqrt{K}}.$ Set $\frac{1}{\sqrt{K}}=0.5\Rightarrow \sqrt{K}=2\Rightarrow K===4==.$ Check: at $K=4$, poles are $-1\pm j\sqrt{3}$; the angle from negative real axis has $\cos \theta =1/2$, i.e. $\theta =60°$, giving $\zeta =0.5$. ✔

Now $n=3$ poles $\{0,-1,-2\}$, $m=1$ zero $\{-0.5\}$, so $n-m=2$ branches escape (down from 3). Centroid: ${\sigma }_a=\frac{(0-1-2)-(-0.5)}{2}=\frac{-3+0.5}{2}=\frac{-2.5}{2}===-1.25==$. Asymptote angles: $\frac{180°(2k+1)}{2}=90°,270°$ — the escaping branches now go straight up and down instead of bending into the right half plane at $60°$.

New Routh test (the edge case we must not skip). Characteristic equation $D(s)+KN(s)=0$: $s(s+1)(s+2)+K(s+0.5)=s^3+3s^2+(2+K)s+0.5K=0.$ Routh array:

The $s^1$ entry is $\frac{6+3K-0.5K}{3}=\frac{6+2.5K}{3}$. For $K>0$ this is always positive, and the $s^0$ entry $0.5K>0$ too. So no positive $K$ makes a row vanish — the locus never crosses the imaginary axis. Comment: the zero at $-0.5$ has turned an unstable-at-$K=6$ system into one that is stable for all $K>0$. This is precisely why a PD controller (which contributes a left-half-plane zero) is a stabilising tool — the zero drags a branch leftward and steers the escaping pair up the vertical asymptotes rather than into the right half plane.

Poles/zeros: $n=3$ poles $\{0,-2,-4\}$, $m=0$. Three branches; all $n-m=3$ escape to infinity.

Real-axis segments (Rule 3). From the right: pole $0$, pole $-2$, pole $-4$.

• $(-2,0)$: one to the right → odd → ON.
• $(-4,-2)$: two to the right → even → OFF.
• $(-\infty ,-4)$: three to the right → odd → ON.

Asymptotes (Rule 4). ${\sigma }_a=\frac{0-2-4}{3}=-2$; angles $\frac{180°(2k+1)}{3}=60°,180°,300°$.

Breakaway (Rule 5). $K=-D/N=-(s)(s+2)(s+4)=-(s^3+6s^2+8s)$. $\frac{dK}{ds}=-(3s^2+12s+8)=0\Rightarrow 3s^2+12s+8=0.$ $s=\frac{-12\pm \sqrt{144-96}}{6}=\frac{-12\pm \sqrt{48}}{6}=-2\pm \frac{4\sqrt{3}}{6}=-2\pm \mathrm{1.1547.}$ Roots: $s=-0.845$ and $s=-3.155$. Only $s=-0.845$ lies on a real segment that is ON the locus ($(-2,0)$), so breakaway $=-0.845$. (The $-3.155$ root sits in the OFF gap $(-4,-2)$, so it is rejected.) Gain at breakaway: $K=-((-0.845{)}^3+6(-0.845{)}^2+8(-0.845))===3.079==$ (to 3 d.p.).

$j\omega$-crossing (Rule 6). Char. eqn: $s^3+6s^2+8s+K=0$. Routh: the $s^1$ entry is $\frac{6\cdot 8-K}{6}=\frac{48-K}{6}$. Zero at $K===48==$. Auxiliary from the $s^2$ row $6s^2+K=0\Rightarrow 6s^2+48=0\Rightarrow s^2=-8\Rightarrow \omega =\sqrt{8}=2\sqrt{2}\approx ==2.828==$.

Full picture: two branches leave $0$ and $-2$, meet and break away at $-0.845$, then curve up/down to the $60°/300°$ asymptotes crossing the imaginary axis at $\pm j2\sqrt{2}$ when $K=48$; the third branch runs left from $-4$ to $-\infty$ along the $180°$ asymptote. Stable for $0<K<48$.

$n=2$ poles both at $s=0$ (a double pole), $m=0$. Characteristic equation: $s^2+K=0\Rightarrow s=\pm j\sqrt{K}$.

• $K\to 0$: both poles sit at the origin (they start there).
• As $K$ increases: poles move straight up and down the imaginary axis to $\pm j\sqrt{K}$.
• $K\to \infty$: they escape to $\pm j\infty$ along the axis. Centroid check: ${\sigma }_a=\frac{0+0}{2}=0$, angles $90°,270°$ — the asymptotes ARE the imaginary axis, consistent with the exact answer. Stability: the poles are on the imaginary axis for every $K>0$ (real part $=0$), so the system is marginally stable / oscillatory for all gains — never asymptotically stable. This is the degenerate case where the entire locus lives on the stability boundary. A pure double integrator cannot be stabilised by gain alone; you must add a zero (PD action).