Lieu des racines
Introduit| English | Sens |
|---|---|
| root locus | lieu des racines |
| asymptote/ˈæsɪmptəʊt/ | asymptote |
| breakaway point | point de rupture |
Turning the gain knob
- Gain is the knob every amplifier and every controller turns. As it turns, the closed-loop poles move.
- The root locus 根轨迹 shows where they go, before anything is built.
The angle condition
- A point $s$ is on the locus when the angles from the zeros minus the angles from the poles add to $180°(2k+1)$.
- Each branch starts at an open-loop pole (at $K = 0$) and ends at a zero or at infinity.
At K = 0 every branch of the root locus sits on an open-loop pole.
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With no gain the loop is open, so the closed-loop poles are the open-loop poles. As K grows they move towards the zeros or to infinity.
Sketch rules
- Real axis: a segment is on the locus when an odd number of poles plus zeros lies to its right.
- Asymptote 渐近线 angles are $\dfrac{180°(2k+1)}{n - m}$, starting from the centroid of the poles minus the zeros.
Open-loop poles sit at 0, −2 and −4, with no zeros. Which real-axis segments are on the locus?
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Between −2 and 0 one pole lies to the right (odd); between −4 and −2, two (even); left of −4, three (odd).
Poles at 0, −3 and −6, no zeros. Where do the asymptotes start?
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The centroid is $(0 - 3 - 6)/3 = -3$. Three asymptotes leave it at ±60° and 180°.
Breakaway and crossing
- The breakaway point 分离点, where branches leave the real axis, solves $\dfrac{d}{ds}\Big[\dfrac{\prod(s - p_{i})}{\prod(s - z_{i})}\Big] = 0$. Keep only roots on the locus.
- The $j\omega$ crossing: put $s = j\omega$ into the characteristic equation and solve the real and imaginary parts.
For L = K/[s(s + 4)], where is the breakaway point?
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$\dfrac{d}{ds}(s^{2} + 4s) = 2s + 4 = 0$ gives $s = -2$, halfway between the poles. The gain there is $|{-2}| \times |2| = 4$.
Worked example: the running plant
- $L = \dfrac{K}{s(s+2)(s+4)}$: asymptotes at ±60° and 180° from $-2$. Breakaway at $s = -0.845$ ($K = 3.08$); the other root, −3.155, is off the locus.
- With $s = j\omega$: $\omega^{2} = 8$ and $K = 6\omega^{2} = 48$. The locus crosses at ±j2.83, the same 48 as Routh.
![Root locus of K/[s(s+2)(s+4)] with the ζ = 0.5 design point marked.](/handout-media/university_electrical_engineering/assets/10-root-locus.png?v=1793069544)
The running plant K/[s(s + 2)(s + 4)] crosses into the right half plane at what gain?
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Putting $s = j\omega$ gives $\omega^{2} = 8$ et $K = 48$, the same boundary as the Routh test.
Worked example: design for ζ = 0.5
- $\zeta = 0.5$ is the ray at 60°. Along it, the angle condition holds at $s^{*} = -0.667 + j1.155$.
- The gain is $K = |s^{*}|\,|s^{*} + 2|\,|s^{*} + 4| = 1.334 \times 1.764 \times 3.527 = 8.30$. The third pole, at −4.67, is far to the left, so the pair dominates.
A design point has |s*| = 1.334, |s* + 2| = 1.764 and |s* + 4| = 3.527. What gain puts the poles there? Give 2 decimal places.
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The magnitude condition: $K = 1.334 \times 1.764 \times 3.527 = 8.30$.
Three common slips
- Keeping an off-locus breakaway root. Only roots on a real-axis segment count.
- Damping as distance. Circles are constant $\omega_{n}$; rays are constant $\zeta$.
- Asymptote for crossing. Compute the $j\omega$ crossing; asymptotes only guide the sketch.
All poles with the same damping ratio lie on
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ζ is the cosine of the pole angle, so a ray is constant ζ. A circle is constant ω_n; a vertical line is constant settling time.
Recap
- Angle condition in; segments, asymptotes, breakaway and crossing out.
- Branches start at poles and end at zeros or infinity.
- Design: walk the locus to the damping ray and read off the gain.