Root locus — Evans' method, rules for sketching: Visual walkthrough

Before line one, four plain words:

Step 1 — What " $L (s) = - 1$ " actually asks

WHAT. The parent note showed closed-loop poles satisfy $1 + L (s) = 0$ , i.e.

loop formula at the point s L (s) = a specific complex number - 1 .

So we are not solving for a size — we are asking: at which points $s$ does the loop formula equal the exact number $- 1$ ?

WHY $- 1$ and not, say, $0$ ? Because feedback subtracts. The loop must return a signal that cancels the input ( $1 + L = 0$ ). The number that means "same size, flipped direction" is $- 1$ .

PICTURE. Every complex number is an arrow from the origin: a length and a direction (angle $∠$ ). The number $- 1$ is the arrow of length $1$ pointing straight left — direction $180°$ .

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

So " $L (s) = - 1$ " secretly carries two demands: the arrow's length must be $1$ , and its angle must be $180°$ . Split them and you get Evans' two conditions.

Step 2 — Splitting one demand into two (length and angle)

WHAT. Any complex number $z$ can be written as (length) $\times$ (unit arrow at its angle):

z = length = magnitude ∣ z ∣ \cdot unit arrow = e^{j a r g (z)} ∠ z .

Matching $L (s) = - 1$ :

magnitude condition ∣ L (s) ∣ = 1 AND angle condition ∠ L (s) = 180° .

WHY split? Because $L (s) = K G (s) H (s)$ and $K$ is a positive real number — a pure stretch, no spin. A positive number has angle $0°$ . So $K$ cannot change the direction of the arrow; it only changes its length.

The angle of $L (s)$ depends only on $G (s) H (s)$ , never on $K$ .
The length is where $K$ lives, to be dialled so the total length becomes $1$ .

PICTURE. Turning the knob $K$ slides the arrowhead along its own direction (a stretch), but never rotates it.

Here $k$ ranges over all integers $Z = {\dots, - 2, - 1, 0, 1, 2, \dots}$ . That is what covers both turning directions at once: $k = 0$ gives $+ 180°$ , $k = - 1$ gives $- 180°$ , $k = 1$ gives $+ 540°$ , and so on. The factor $(2 k + 1)$ (an odd number) is there because pointing "straight left" happens at every odd multiple of $180°$ ; letting $k$ run over all integers is why we need no separate $\pm$ sign.

Step 3 — What one pole/zero contributes: a single angle

WHAT. Write $G (s) H (s)$ using its factors. If there is a zero at $z_{1}$ and poles at $p_{1}, p_{2}$ :

G (s) H (s) = \frac{s - z _{1}}{( s - p _{1} ) ( s - p _{2} )} .

The term $s - p_{1}$ is the arrow from the pole $p_{1}$ to our test point $s$ . Its angle $∠ (s - p_{1})$ is how steeply you look up from $p_{1}$ to $s$ .

WHY arrows? Because " $s - p_{1}$ " literally means "start at $p_{1}$ , walk to $s$ " — that's a directed arrow, and directed arrows have an angle.

Angle bookkeeping — for a product/quotient, angles add and subtract:

∠ G (s) H (s) = zero adds its angle ∠ (s - z_{1}) - each pole subtracts its angle ∠ (s - p_{1}) - ∠ (s - p_{2}) .

PICTURE. Drop a test point $s$ anywhere. From each ✕ and ◯ draw an arrow to $s$ . Measure each arrow's angle off the horizontal. Zeros count plus, poles count minus.

Step 4 — The real-axis rule falls out for free

WHAT. Test a point that sits on the horizontal (real) axis. Which real poles/zeros put it on the locus?

WHY only count things to the RIGHT? Look at the arrows:

A real dot to the left of $s$ points rightward → angle $0°$ → contributes nothing (whether it is a ✕ or a ◯).
A real dot to the right of $s$ points leftward → angle $180°$ → contributes $\pm 180°$ .
A complex-conjugate pair (one above, one below) sends up two arrows whose angles are $+ α$ and $- α$ → they cancel to $0°$ .

WHY do poles ( $- 180°$ ) and zeros ( $+ 180°$ ) both pass the same odd test? From Step 3, a right-side zero adds $+ 180°$ and a right-side pole subtracts $180°$ , i.e. contributes $- 180°$ . But $+ 180°$ and $- 180°$ are the same direction — they differ by $360°$ , a full turn. For the angle condition we only care whether the total is an odd multiple of $180°$ , and both $+ 180°$ and $- 180°$ are odd multiples. So each right-side dot — pole or zero — flips the parity by one, and only the count (not the sign) matters.

So the total angle is an odd multiple of $180°$ exactly when the number of real dots (poles + zeros) to the right is odd. ✔

PICTURE. Below: a point between two poles sees exactly one dot on its right → total $180°$ → on the locus. A point to the right of both sees two → total $360°$ → not on the locus.

odd count to the right

point is on the locus (total angle is an odd multiple of

180°

)

even count to the right

point is off the locus (total angle is

0°

360°

)

Step 5 — Where beads leave the axis: the breakaway

WHAT. Between the two poles $s = 0$ and $s = - 2$ (parent's Example 1), both beads slide toward each other, collide, then peel off vertically. The collision point is the breakaway.

WHY does $d K / d s = 0$ mark it? Recall from the definitions box that $N (s)$ is the numerator polynomial (roots = zeros) and $D (s)$ the denominator polynomial (roots = poles); here $N (s) = 1$ and $D (s) = s (s + 2)$ . The characteristic equation $1 + K \frac{N}{D} = 0$ rearranges to $D + K N = 0$ , so we can solve for the gain:

K = - \frac{D ( s )}{N ( s )} = - \frac{s ( s + 2 )}{1} = - s (s + 2) .

Along the real segment, as $s$ moves from $0$ inward, $K$ rises; past the meeting point it would have to fall to keep two real poles — impossible for real gain, so instead the poles go complex. The switch from rising to falling is a peak: $\frac{d K}{d s} = 0$ .

\frac{d K}{d s} = - (2 s + 2) = 0 \Rightarrow s = - 1, K = - (- 1) (1) = 1.

PICTURE. Plot $K$ against $s$ on the segment $[- 2, 0]$ : a smooth hill peaking at $s = - 1$ , $K = 1$ . That peak is the breakaway; beyond it the beads climb straight up and down.

Step 6 — Where escaping branches aim: asymptotes from the angle condition

WHAT. The two beads that peel off at $s = - 1$ must go somewhere as $K \to \infty$ . Using the counting letters from the definitions box ( $n = 2$ poles, $m = 0$ zeros), the number of branches that escape to infinity is $n - m = 2$ . They approach straight rays. This step derives their angles and their launch point straight from the angle condition — no memorising.

WHY $n - m$ rays, and why these angles? Look at a test point $s$ enormously far away. From that distance all the poles and zeros huddle at nearly the same spot, so every arrow drawn to $s$ has almost the same angle $θ$ . With $m$ zeros (each $+$ ) and $n$ poles (each $-$ ), the angle bookkeeping from Step 3 becomes

∠ G (s) H (s) \approx zeros, + m θ - poles, - n θ = - (n - m) θ .

Setting this equal to the required $180° (2 k + 1)$ and solving for the ray angle:

θ_{a} = \frac{180° ( 2 k + 1 )}{n - m} .

For $n - m = 2$ , stepping $k$ gives $k = 0 \Rightarrow 90°$ , $k = 1 \Rightarrow 270°$ , and $k = 2$ returns to $90° + 360°$ — so exactly $n - m$ distinct directions appear before they repeat. That is why there are $n - m$ asymptotes.

WHY the launch point is the centroid? The rays don't leave from the origin; they leave from the average of the dots. Here is the "next-term matching" spelled out. The escaping branches are governed, for large $∣ s ∣$ , by the ratio $D (s) / N (s)$ :

\frac{D ( s )}{N ( s )} = \frac{s ^{n} + d _{1} s ^{n - 1} + \dots}{s ^{m} + e _{1} s ^{m - 1} + \dots} = s^{n - m} + (d_{1} - e_{1}) s^{n - m - 1} + \dots

where $d_{1} = - \sum (poles)$ and $e_{1} = - \sum (zeros)$ are the negatives of the root-sums (Vieta). We want a single pole of multiplicity $n - m$ at some point $σ_{a}$ that matches this far-field, i.e. $D / N \approx (s - σ_{a})^{n - m}$ . Expanding that binomial,

(s - σ_{a})^{n - m} = s^{n - m} - (n - m) σ_{a} s^{n - m - 1} + \dots

Matching the second coefficient of the two expressions:

- (n - m) σ_{a} = d_{1} - e_{1} = - \sum (poles) + \sum (zeros) \Rightarrow σ_{a} = \frac{\sum ( poles ) - \sum ( zeros )}{n - m} .

Each symbol: $\sum (poles)$ adds up every ✕-location, $\sum (zeros)$ subtracts every ◯-location, and dividing by $n - m$ takes their balance point along the real axis. For us $σ_{a} = \frac{0 + ( - 2 ) - 0}{2} = - 1$ .

PICTURE. The centroid at $- 1$ with two rays climbing at $90°$ and $270°$ — the beads' highway signs.

n - m

for

1/ [s (s + 2)]

2

→ two asymptotes at

90°

and

270°

centroid

σ_{a}

(0 + (- 2)) /2 = - 1

Step 7 — Degenerate & edge cases (never leave a gap)

WHAT. Four scenarios the beginner trips on:

A pole sits exactly on the test point ( $s = p_{1}$ ). Then $s - p_{1} = 0$ : the arrow has zero length, its angle is undefined. That's fine — it's the start of a branch ( $K = 0$ ), not an interior test point.
A zero sits exactly on the test point ( $s = z_{1}$ ). Then the numerator arrow $s - z_{1} = 0$ vanishes, so $∠ (s - z_{1})$ is likewise undefined and $∣ G (s) H (s) ∣ = 0$ . Via the magnitude condition $K = 1/∣ G (s) H (s) ∣ \to \infty$ : this is the end of a branch ( $K \to \infty$ ), the target the bead is aiming at — not an interior point to test either.
A test point exactly on a real singularity. The angle jumps discontinuously (arrow flips as you cross). The rule is stated for points between singularities, never on them.
All poles cancel above and below (pure conjugate stack, no real dots to the right). Count $= 0$ , which is even → that stretch of axis is not on the locus. Zero is even.

WHY track these? Because the sketch is only trustworthy if every square of the map is accounted for — including the empty ones.

PICTURE. A number line coloured green where the locus lives, gray where it doesn't, with the odd/even count printed under each stretch, plotting the ✕ start-points explicitly.

zero real dots to the right

even count → NOT on the locus

s = z_{1}

(zero on test point)

branch END,

K \to \infty

— not an interior test point

Step 8 — Why the danger line is never crossed (angle + magnitude, not assertion)

WHAT. The imaginary axis (up–down through the origin) is the border: a closed-loop pole on it means steady oscillation, and to its right means blow-up. For $1/ [s (s + 2)]$ we now prove the locus stays strictly to the left of it.

WHY no crossing? Two facts we already earned combine:

Real-axis piece: the only on-locus real stretch is $[- 2, 0]$ (Step 4) — entirely at real part $\leq 0$ .
Escaping pieces: both branches leave the breakaway at $s = - 1$ along the vertical asymptote $σ_{a} = - 1$ (Step 6). A vertical line at real part $- 1$ never reaches real part $0$ .

Put a would-be crossing point on the axis, $s = j ω$ , and test the magnitude condition: $K = ∣ s ∣ ∣ s + 2∣ = ω^{2} \cdot ω^{2} + 4$ . This is a positive real number for every $ω$ , but the branch's real part is pinned at $- 1$ by the asymptote — the pole is at $- 1 + j ω$ , not on the axis. There is simply no $(s, K \geq 0)$ pair that is both on the locus and on the axis.

Conclusion (stability): every closed-loop pole has real part $\leq 0$ for all $K > 0$ — the system is unconditionally stable. (Contrast Routh–Hurwitz stability criterion, which finds a finite critical gain whenever a real crossing does exist — as in the parent's Example 2.)

PICTURE. The red danger line drawn, with the locus visibly hugging real part $- 1$ and never touching it.

The one-picture summary

WHAT. Everything above, on one map for $G (s) H (s) = \frac{1}{s ( s + 2 )}$ : the two ✕ poles, the green on-locus segment $[- 2, 0]$ (Step 4), the breakaway at $s = - 1, K = 1$ (Step 5), the two branches peeling vertically along the centroid $σ_{a} = - 1$ with asymptote angles $90°/270°$ (Step 6), and the red danger line (imaginary axis) which the locus never crosses (Step 8) — so stable for all $K > 0$ .

Recall Feynman: the whole walk in plain words

A feedback loop needs its formula to equal the number negative one — an arrow of length one pointing straight left. Left-pointing is a direction (an angle), and the knob $K$ is just a positive stretch that can never rotate an arrow. So the direction part ( $180°$ ) is fixed by geometry alone, and $K$ only fine-tunes the length to $1$ . To test if you're standing on a path, look back at every ✕ (pole) and ◯ (zero), read the angle of the arrow pointing at you, add the zeros and subtract the poles — if the total is $180°$ , you're on it. On the flat real axis this simplifies beautifully: dots to your left point away (0°, ignore them), dots to your right point at you (180°), and mirror-image complex pairs cancel — so you just count dots to the right, and want an odd number (poles and zeros count the same because $+ 180°$ and $- 180°$ are the same direction). Two beads starting on two poles slide together, collide where the gain $K$ hits its little hilltop ( $d K / d s = 0$ ), then fly off along straight ramps. To find where those ramps aim, stand far away: all the dots blur into one spot, every arrow shares one angle $θ$ , and the angle condition forces $θ = 180° (2 k + 1) / (n - m)$ — here $90°$ and $270°$ , launched from the balance-point of the dots at $- 1$ . Because that ramp is a vertical wall at $- 1$ , and the only real path is left of zero too, no bead can ever touch the red line (the imaginary axis) — the machine stays calm at every knob setting.

Connections

Root locus — Evans' method, rules for sketching — the parent this page derives
Characteristic equation & closed-loop poles — where $1 + L (s) = 0$ comes from
Routh–Hurwitz stability criterion — finds the $j ω$ crossing when one exists
PID / PD controller design — adds zeros (◯) to bend the locus leftward
Damping ratio and settling time — the vertical branch's real part sets the damping
Nyquist criterion — the same $L (s) = - 1$ seen as an encirclement
Bode plot & frequency response — magnitude/angle split, viewed vs frequency
GNC control loops — the physical loops this tunes