Worked examples — Root locus — Evans' method, rules for sketching

Before we start, one reminder of the alphabet so nothing is used unearned:

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The scenario matrix

Every root-locus problem lives in one of these cells. The worked examples afterward each hit a labelled cell so together they cover the whole grid.

Cell	Case class	What is special	Example
A	Two real poles, no zero	Simplest breakaway, vertical asymptotes	Ex 1
B	Real pole at origin ($s=0$)	Integrator; starts on the danger axis	Ex 1, Ex 4
C	Three real poles → $j\omega$ crossing	Locus enters right-half-plane; finite max gain	Ex 2
D	Adding a finite zero	Zero pulls a branch back; re-stabilises	Ex 3
E	Complex open-loop poles	Need angle-of-departure (Rule 7)	Ex 5
F	Degenerate — zero cancels a pole	Pole–zero cancellation, branch count drops	Ex 6
G	Unstable open-loop pole (right-half-plane start)	Branch must be dragged left to stabilise	Ex 7
H	Real-world word problem	Translate hardware → $G(s)H(s)$	Ex 4 (GNC autopilot)
I	Exam twist — solve for a required damping	Combine locus + Damping ratio and settling time	Ex 8

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Ex 1 — Cells A + B: two real poles, one at the origin

Forecast: guess before reading — where do the two branches go, and at what value of $K$ do the poles first become oscillatory?

1. Count poles/zeros. $n=2$ (poles at $s=0,-2$), $m=0$. Why this step? Every rule keys off $n$ and $m$; get them first.
2. Real-axis segment (Rule 3). Pick a test point in $(-2,0)$. Only the pole at $s=0$ lies to its right → count $=1$ (odd) → segment $[-2,0]$ is on the locus. Why this step? The odd-count rule is the fast way to know which real stretch the branches ride before they split.
3. Asymptotes (Rule 4). $n-m=2$, so ${\theta }_a=\frac{\pm 180°(2k+1)}{2}=90°,270°$. Centroid ${\sigma }_a=\frac{(0)+(-2)-0}{2}=-1$. Why this step? Both branches escape to infinity; the asymptotes tell us the escaping direction is straight up/down from $\sigma =-1$.
4. Breakaway (Rule 5). $K=-\frac{D}{N}=-s(s+2)$. Then $\frac{dK}{ds}=-(2s+2)=0\Rightarrow s=-1$. Why this step? Two branches collide on the real axis; at a double root $K$ is at an extremum, so $dK/ds=0$ pinpoints the split.
5. Gain at breakaway. $K=-(-1)(-1+2)=1$. Why this step? It labels the split point — for $K>1$ the poles leave the axis and become complex (oscillatory).

See the branches leave $0$ and $-2$, meet at $-1$, then rise vertically:

Verify: at $K=1$ the characteristic equation is $s^2+2s+1=(s+1{)}^2=0$ → double pole at $-1$. ✔ For $K=2$: $s^2+2s+2=0\Rightarrow s=-1\pm j$, real part exactly $-1$ (on the vertical asymptote). ✔ Stable for all $K>0$.

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Ex 2 — Cell C: three real poles, $j\omega$ crossing

Forecast: with three branches all fleeing to infinity, at least one must curve into the right-half-plane. At what $K$?

1. Poles/zeros. $n=3$ ($0,-1,-2$), $m=0$.
2. Asymptotes. ${\theta }_a=\frac{\pm 180°(2k+1)}{3}=60°,180°,300°$; ${\sigma }_a=\frac{0-1-2}{3}=-1$. Why this step? The $60°$ and $300°$ asymptotes point rightward and up/down — a warning that branches head toward the unstable region.
3. Characteristic equation. $s(s+1)(s+2)+K=s^3+3s^2+2s+K=0$. Why this step? Routh–Hurwitz needs the polynomial coefficients.
4. Routh array. $\begin{array}{ccc}{s}^3& 1& 2\\ s^2& 3& K\\ s^1& \frac{3\cdot 2-K}{3}& 0\\ s^0& K& \end{array}$ First-column positivity needs $\frac{6-K}{3}>0\Rightarrow K<6$ and $K>0$. Why this step? A sign change in the first column = a right-half-plane pole = instability.
5. Crossing frequency. At $K=6$ the $s^1$ row vanishes; the auxiliary polynomial is the $s^2$ row: $3s^2+6=0\Rightarrow s^2=-2\Rightarrow s=\pm j\sqrt{2}$. Why this step? A zero row means poles sit exactly on the imaginary axis; the auxiliary equation reads off their frequency.

Verify: substitute $K=6,s=j\sqrt{2}$ into $s^3+3s^2+2s+6$: $(j\sqrt{2}{)}^3=-2\sqrt{2}j$, $3(j\sqrt{2}{)}^2=-6$, $2(j\sqrt{2})=2\sqrt{2}j$, $+6$. Sum $=(-6+6)+(-2\sqrt{2}+2\sqrt{2})j=0$. ✔ Critical gain $K=6$, $\omega =\sqrt{2}\approx 1.414\text{ rad/s}$.

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Ex 3 — Cell D: a zero re-stabilises the system

Forecast: does adding a left-half-plane zero make the escaping branches steeper or shallower? Does it help stability?

1. Poles/zeros. $n=3$ ($0,-1,-2$), $m=1$ (zero at $-3$).
2. Branches to infinity. $n-m=2$, so only two branches escape (down from three in Ex 2). One branch now terminates on the finite zero at $-3$. Why this step? Rule 1 — each finite zero "catches" one branch, so the zero is a target that removes an escaping branch.
3. New asymptotes. ${\theta }_a=\frac{\pm 180°(2k+1)}{2}=90°,270°$ (vertical, not $60°$). Centroid ${\sigma }_a=\frac{(0-1-2)-(-3)}{2}=\frac{-3+3}{2}=0$. Why this step? Compare with Ex 2's $60°/300°$ tilt: the zero straightened the asymptotes to vertical — the branches no longer lean into the right-half-plane as aggressively.
4. Stability check (Routh). Char. eqn: $s^3+3s^2+2s+K(s+3)=s^3+3s^2+(2+K)s+3K=0$. Routh $s^1$ term: $\frac{3(2+K)-3K}{3}=\frac{6}{3}=2>0$ for all $K>0$. Why this step? This is the payoff: the $s^1$ Routh entry is a positive constant, so no $K$ makes it change sign — the system is stable for every gain. The zero (a PD action) bought unconditional stability.

Verify: Routh first column entries: $1,3,2,3K$. All positive for $K>0$ — no sign changes → stable for all $K>0$. ✔ Centroid ${\sigma }_a=0$.

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Ex 4 — Cells B + H: real-world GNC autopilot (word problem)

Forecast: a double integrator means two poles glued to the origin. Intuition says "unstable" — check it.

1. Poles/zeros. $n=3$: a repeated pole at $s=0$ (multiplicity 2) and a pole at $-4$. $m=0$. Why this step? The repeated origin pole is the special feature — two branches start from the very same point on the danger axis.
2. Asymptotes. ${\sigma }_a=\frac{0+0-4}{3}=-\frac{4}{3}$, ${\theta }_a=\frac{\pm 180°(2k+1)}{3}=60°,180°,300°$. Why this step? All three branches escape ($n-m=3$). The $60°$ and $300°$ asymptotes point up-right and down-right — straight toward the unstable region — while the $180°$ asymptote sends the third branch along the negative real axis to the left. That single left-going branch is the only well-behaved one; the two others fly into the right-half-plane. WHAT the $180°$ branch looks like: the pole at $-4$ moves left along the real axis toward $-\infty$ as $K$ grows, staying safely stable — but it cannot rescue the two branches leaving the origin, which curve rightward across the danger axis almost immediately.
3. Routh test. Char. eqn: $s^3+4s^2+0\cdot s+K=0$. The $s^1$ coefficient is $0$. $\begin{array}{ccc}{s}^3& 1& 0\\ s^2& 4& K\\ s^1& \frac{4\cdot 0-K}{4}=-\frac{K}{4}& \\ s^0& K& \end{array}$ Why this step? The $s^1$ entry $-K/4$ is negative for every $K>0$ → a first-column sign change → at least one right-half-plane pole for all $K$.
4. Conclusion. Pure gain can never stabilise a double integrator. You must add a zero (rate feedback / PD) — exactly why real autopilots use derivative terms. Why this step? This is the engineering lesson: the matrix cell "pole on the axis + no zero" is a guaranteed-unstable trap.

Verify: for $K=1$, $s^3+4s^2+1=0$ has a root at $s\approx +0.24$ (positive real part) → unstable. ✔ (checked numerically below).

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Ex 5 — Cell E: complex open-loop poles, angle of departure

Forecast: the branch leaves a slanted pole — does it head up-left (stabilising) or up-right (toward danger)?

1. Locate poles. $s^2+2s+2=0\Rightarrow s=-1\pm j$. Plus a pole at $s=0$. So poles: $0,-1+j,-1-j$; $m=0$, $n=3$. Why this step? Rule 7 (angle of departure) applies only to complex poles; find them first.
2. Set up Rule 7 at $p=-1+j$. ${\theta }_d=180°+\underset{=0}{\underbrace{\sum \angle (\text{zeros}\to p)}}-\sum \angle (\text{other poles}\to p).$ Why this step? A test point a hair away from $p$ must still satisfy the angle condition; solving for the departure direction gives ${\theta }_d$.
3. Angle from pole at $0$ to $p$. Vector $p-0=-1+j$ → angle $=135°$.
4. Angle from pole at $-1-j$ to $p$. Vector $(-1+j)-(-1-j)=2j$ → angle $=90°$. Why these steps? Each "other pole" contributes the angle of the arrow drawn from that pole to $p$; we subtract their sum.
5. Combine. ${\theta }_d=180°-(135°+90°)=180°-225°=-45°$ (equivalently $315°$).

Verify: ${\theta }_d=-45°$ means the branch leaves $-1+j$ heading down-right toward the real axis — consistent with the symmetric partner leaving $-1-j$ at $+45°$; their conjugate symmetry (Rule 2) is respected since $-45°$ and $+45°$ are mirror images. ✔

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Ex 6 — Cell F: degenerate pole–zero cancellation

Forecast: the factor $(s+2)$ appears top and bottom. Does the pole at $-2$ still start a branch?

1. Cancel first. $\frac{s+2}{s(s+2)(s+5)}=\frac{1}{s(s+5)}$ for $s\ne -2$. Why this step? A common factor is a pole–zero cancellation: the pole at $-2$ is exactly annihilated by the zero at $-2$. It contributes no moving branch.
2. Effective count. After cancellation $n=2$ ($0,-5$), $m=0$. Only two branches, not three. Why this step? The degenerate cell shrinks the branch count; forgetting to cancel would give a fake extra branch.
3. Real-axis segment. Test point in $(-5,0)$: pole at $0$ to the right → count 1 (odd) → $[-5,0]$ on the locus.
4. Breakaway. $K=-s(s+5)$, $dK/ds=-(2s+5)=0\Rightarrow s=-2.5$; gain $K=-(-2.5)(2.5)=6.25$. Why this step? Same procedure as Ex 1 but on the reduced system — the breakaway sits at the midpoint $-2.5$.

Verify: breakaway of $\frac{1}{s(s+5)}$: midpoint $-2.5$, and $K=(2.5)(2.5)=6.25$. At $K=6.25$: $s^2+5s+6.25=(s+2.5{)}^2=0$ ✔ double pole at $-2.5$.

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Ex 7 — Cell G: unstable open-loop pole must be dragged left

Forecast: the system is born unstable. Can the zero at $-3$ pull the runaway pole into the left half-plane, and above what gain?

1. Poles/zeros. Poles $0,+1$; zero $-3$. $n=2$, $m=1$. Why this step? One escaping branch ($n-m=1$) and one branch that must land on the zero at $-3$.
2. Real-axis segments. Test $x$ just right of $0$ (in $(0,1)$): to its right sits the pole at $+1$ → count 1 (odd) → segment $(0,1)$ on the locus. Test $x<-3$: poles at $0,+1$ and zero at $-3$ all to the right → count 3 (odd) → $(-\infty ,-3)$ on the locus. Why this step? This shows a branch travels from $+1$ leftward and eventually reaches $-3$: the zero is the rescue target.
3. Characteristic equation. $s(s-1)+K(s+3)=s^2+(K-1)s+3K=0$. Why this step? We need conditions for both roots to have negative real part.
4. Stability conditions (2nd-order: all coefficients positive).
   • $s^1$ coefficient $K-1>0\Rightarrow K>1$.
   • $s^0$ coefficient $3K>0\Rightarrow K>0$ (auto-satisfied). Why this step? For a quadratic $s^2+bs+c$, both roots lie in the left half-plane iff $b>0$ and $c>0$. The binding one is $K>1$.
5. Conclusion. Stable for $K>1$. At $K=1$ a pole sits exactly at $s=0$ (marginal); below that, the right-half-plane pole is not yet captured.

Verify: at $K=1$: $s^2+0\cdot s+3=0\Rightarrow s=\pm j\sqrt{3}$ — marginal ($\omega =\sqrt{3}$). At $K=2$: $s^2+s+6=0\Rightarrow s=-0.5\pm j2.398$ — both left-half-plane, stable. ✔

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Ex 8 — Cell I: exam twist, hit a target damping ratio

Forecast: we already know for $K>1$ the poles are $-1\pm j\beta$. Which $\beta$ gives $\zeta =0.5$, and does it need a higher or lower $K$ than the breakaway?

1. Closed-loop poles. $s^2+2s+K=0\Rightarrow s=-1\pm \sqrt{1-K}$. For $K>1$: $s=-1\pm j\sqrt{K-1}$. Why this step? We need the pole in the standard 2nd-order form $s=-\zeta {\omega }_n\pm j{\omega }_n\sqrt{1-{\zeta }^2}$.
2. Match natural frequency. ${\omega }_n^2=K$ (constant term) so ${\omega }_n=\sqrt{K}$; real part $-\zeta {\omega }_n=-1\Rightarrow \zeta {\omega }_n=1$. Why this step? These two identities let us solve for $K$ directly from $\zeta$.
3. Solve for $K$. $\zeta =\frac{1}{{\omega }_n}=\frac{1}{\sqrt{K}}=0.5\Rightarrow \sqrt{K}=2\Rightarrow K=4$. Why this step? $\zeta =0.5$ fixes ${\omega }_n=2$, hence $K={\omega }_n^2=4$. (Note $K=4>1$: past the breakaway, as forecast.)
4. Damping-ray geometry. $\zeta =\cos \theta$ where $\theta$ is the angle from the negative real axis. $\zeta =0.5\Rightarrow \theta =60°$. Why this step? On the locus you'd draw a $60°$ ray from the origin and read off where it hits the vertical branch — a purely geometric way to set gain (see Damping ratio and settling time).
5. Settling time. $t_s\approx \frac{4}{\zeta {\omega }_n}=\frac{4}{1}=4\text{ s}$ (2% criterion). Why this step? $\zeta {\omega }_n=1$ is the pole's real part; settling time depends only on that.

Verify: $K=4\Rightarrow s^2+2s+4=0\Rightarrow s=-1\pm j\sqrt{3}$. Then ${\omega }_n=|s|=\sqrt{1+3}=2$ ✔, $\zeta =1/{\omega }_n=0.5$ ✔, $t_s=4/(0.5\cdot 2)=4$ s ✔.

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Recall

Which example cell teaches you that pure gain can never fix a double integrator? ::: Ex 4 (cells B+H) — the $s^1$ Routh entry is $-K/4<0$ for all $K>0$. In Ex 3, what did adding the zero at $-3$ do to the asymptote centroid? ::: Moved it from $-1$ (Ex 2) to $0$ and straightened the asymptotes to vertical, giving stability for all $K>0$. In Ex 8, what gain gives $\zeta =0.5$ and what is the settling time? ::: $K=4$, poles $-1\pm j\sqrt{3}$, $t_s\approx 4$ s.

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Connections

• Root locus — Evans' method, rules for sketching — parent rules used throughout
• Characteristic equation & closed-loop poles — the single equation behind every example
• Routh–Hurwitz stability criterion — Ex 2, 3, 4 crossings & stability ranges
• PID / PD controller design — Ex 3 & 4: adding a zero to re-stabilise
• Damping ratio and settling time — Ex 8 target-damping design
• GNC control loops — Ex 4 autopilot context
• Nyquist criterion — encirclement view of the same $L(s)=-1$
• Bode plot & frequency response — frequency-domain companion