Controls and Systems · Chapter 7 of 10 · Advanced
The Root-Locus Method
As you turn up the gain, the closed-loop poles slide across the s-plane. The root locus draws their path, turning a tuning knob into a picture of speed and stability.
Readiness check
This chapter tracks poles versus gain. Tick only what you can do closed-notes.
- Find the poles and zeros of an open-loop transfer function.
- Write the characteristic equation 1 + KG = 0.
- Recall that poles set speed and stability.
- Add angles and magnitudes of complex factors.
- Use a Routh array to find a gain limit.
The core idea
The root locus is the path the closed-loop poles trace as the gain runs from zero to infinity. A few rules sketch it from the open-loop poles and zeros alone.
1 + KG(s) = 0branches start at poles, end at zeros or ∞centroid σa = (Σpoles − Σzeros)/(n − m)Every closed-loop pole satisfies 1 + KG(s) = 0, so as K varies the roots move continuously, starting at the open-loop poles when K = 0 and ending at the open-loop zeros (or off to infinity) as K grows. The locus lies where the angle of KG equals 180 degrees. Simple rules, real-axis segments, asymptote angles, and a centroid, sketch the whole picture without solving the polynomial at every gain. Reading where the branches sit shows the designer what damping and speed a given gain delivers, and where the locus crosses into the right half-plane, the gain that destabilises the loop.
The skills, taught in order
Five skills define the locus, its sketching rules, the angle and magnitude conditions, the imaginary-axis crossing, and design use.
7.1 What the locus shows
The root locus plots the closed-loop poles as the gain K sweeps from 0 to infinity. There are as many branches as open-loop poles. Each branch starts at a pole (K = 0) and ends at a zero or runs to infinity (K → ∞).
7.2 The sketching rules
A point on the real axis is on the locus if the total number of real poles and zeros to its right is odd. Branches that escape to infinity follow asymptotes whose angles are ±180°(2k+1)/(n − m), all meeting the real axis at the centroid σa = (Σpoles − Σzeros)/(n − m).
| Feature | Rule |
|---|---|
| Number of branches | equal to the number of poles n |
| Start and end | poles at K = 0, zeros or infinity at K → ∞ |
| Real-axis segment | odd count of poles and zeros to the right |
| Asymptote angles | ±180°(2k+1)/(n − m) |
| Centroid | (Σpoles − Σzeros)/(n − m) |
7.3 Angle and magnitude conditions
A point is on the locus if the angles from the poles and zeros to it sum to 180 degrees (the angle condition). Once on the locus, the gain there is found from the magnitude condition K = 1/|G|, the product of distances from the poles over the product from the zeros.
7.4 Crossing the imaginary axis
Where the locus crosses into the right half-plane, the system goes unstable. That crossing gain and frequency come straight from a Routh array on the characteristic polynomial, tying this chapter to the last.
7.5 Designing with the locus
To meet a damping or settling target, mark the desired pole region and read the gain where the locus passes through it. If the locus does not reach the target, a compensator (a pole-zero pair) reshapes it, the subject of lead and lag design.
Engineering connection: the locus turns the response specs of Chapter 4 into a gain choice, and its imaginary-axis crossing matches the Routh limit of Chapter 6.
Worked example 1: sketching a root locus
For the open-loop transfer function G(s) = K/[s(s + 2)(s + 4)], find the number of branches, the real-axis segments, the asymptote angles, and the centroid.
- ProblemFind the branches, real-axis segments, asymptotes, and centroid for the locus in Figure 1.
- Given / findG = K/[s(s + 2)(s + 4)], poles at 0, −2, −4, no zeros. Find the locus features.
- AssumptionsStandard negative-feedback root locus for K ≥ 0.
- ModelApply the counting, real-axis, asymptote, and centroid rules.
- Equationsangles = ±180°(2k+1)/(n − m)σa = (Σpoles − Σzeros)/(n − m)
- Solven = 3 poles, m = 0 zeros, so 3 branches. Real-axis locus is [−2, 0] and (−∞, −4] (odd count to the right). Asymptotes: n − m = 3, angles ±60° and 180°, centroid σa = (0 − 2 − 4)/3 = −2.
- CheckTwo branches leave 0 and −2, break away between them, and follow the ±60° asymptotes into the right half-plane; the third runs left from −4 along 180°. The picture is consistent.
- ConclusionA handful of rules sketched the locus from the poles alone. The branches heading into the right half-plane warn that high gain will destabilise the loop, quantified next.
Worked example 2: the imaginary-axis crossing
For the same system, find the gain K at which the root locus crosses the imaginary axis and the frequency of oscillation there.
- ProblemFind the crossing gain and frequency for the locus in Figure 2.
- Given / findCharacteristic equation 1 + K/[s(s + 2)(s + 4)] = 0. Find K and ω at the imaginary-axis crossing.
- AssumptionsThe crossing is where a first-column Routh entry vanishes.
- ModelForm the characteristic polynomial, build a Routh array, and set the s¹ entry to zero; the auxiliary polynomial gives ω.
- Equationss³ + 6s² + 8s + K = 0s¹ entry = (6·8 − K)/6, auxiliary 6s² + K = 0
- SolveThe s¹ entry (48 − K)/6 = 0 gives K = 48. The auxiliary 6s² + 48 = 0 gives s² = −8, so ω = √8 = 2.83 rad/s.
- CheckSubstituting s = j2.83 with K = 48 into the polynomial gives zero, confirming the poles sit on the imaginary axis at that gain. This matches the same Routh boundary method from Chapter 6.
- ConclusionThe locus and the Routh array agree: K = 48 is the stability limit. A design would choose a gain well below it for adequate damping.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Branches to nonexistent zeros | Locus ends in the wrong place | "How many finite zeros are there?" | Branches with no zero run to infinity along the asymptotes. |
| Real-axis rule misapplied | Wrong segments on the locus | "Is the count to the right odd?" | A real-axis point is on the locus only with an odd count of poles and zeros to its right. |
| Wrong centroid | Asymptotes meet at the wrong point | "Did I subtract the zeros and divide by n − m?" | σa = (Σpoles − Σzeros)/(n − m). |
| Reading gain off the wrong branch | Design gain mismatched to damping | "Is the target point actually on the locus?" | Check the angle condition before using the magnitude condition for K. |
Practice ladder
For G = K/[s(s + 6)], find the centroid and asymptote angles.
Show answer
n − m = 2, so angles are ±90°. Centroid σa = (0 − 6)/2 = −3. The branches go straight up and down from −3.
For G = K/[s(s + 2)(s + 4)] (the worked system), what gain places a closed-loop pole on the real axis at the breakaway? The breakaway is near s = −0.85.
Show answer
K = |s||s + 2||s + 4| at s = −0.85 = 0.85 × 1.15 × 3.15 ≈ 3.08. This is the gain where two real poles meet before becoming complex.
For G = K/[(s + 1)(s + 3)], describe the locus and state whether it can ever go unstable.
Show answer
Two poles, no zeros, centroid at −2, asymptotes ±90°. The branches meet on the real axis and rise vertically through −2, staying in the left half-plane for all K, so it never goes unstable.
Take a real system with a gain knob and describe how its dominant poles would move as you turn the gain up, and what you would watch for.
What good work looks like
Poles moving along a locus toward less damping, a recognition of the gain where overshoot grows or the locus nears the imaginary axis, and a stopping point chosen for margin.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
For a real plant, sketch the root locus from its poles and zeros, choose a gain for a damping target, and find the gain at which it would go unstable.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. What does the root locus plot?
The closed-loop poles as the gain K runs from 0 to infinity.
2. Where do branches start and end?
At the open-loop poles (K = 0) and end at the zeros or at infinity (K → ∞).
3. Give the centroid formula.
σa = (Σpoles − Σzeros)/(n − m).
4. What is the angle condition?
The angles from poles and zeros to a locus point sum to 180 degrees.
5. How do you find the destabilising gain?
A Routh array on the characteristic polynomial gives the crossing gain and frequency.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Ogata, Modern Control Engineering, Chapter 6 (Root-Locus Method) |
| Cross-reference | Nise, Ch. 8 · Dorf and Bishop, Ch. 7 |
| Core topics | 7.1 What the locus shows · 7.2 Sketching rules · 7.3 Angle and magnitude · 7.4 Imaginary-axis crossing · 7.5 Design |
| Engineering connection | The locus turns response specs into a gain and matches the Routh stability limit. |
| Read next | Chapter 8: Frequency Response and Bode Plots. |