Skip to main content

Root locus of s³+9s²+Ks+K = 0

Short answer

The root locus has a triple meeting point at s=-3 (K=27), with asymptotes at -4 (±90°). Branches from the double pole at the origin leave vertically, meet the real branch from -9 at -3, then one goes to the zero at -1 and two go to infinity along the asymptotes. The system is stable for all K>0.

The characteristic equation of the system is s³ + 9s² + sK + K = 0. Sketch the complete root locus and comment on stability.

Converting to root-locus form

s3+9s2+K(s+1)=0  ⇒  1+K(s+1)s2(s+9)=0s^3 + 9s^2 + K(s + 1) = 0 \;\Rightarrow\; 1 + \frac{K(s+1)}{s^2(s+9)} = 0

Equivalent open-loop transfer function: G(s)H(s)=K(s+1)s2(s+9)G(s)H(s) = \dfrac{K(s+1)}{s^2(s+9)}.

Root locus construction

  1. Poles: 0,0,−90, 0, -9. Zero: −1-1. With n=3n = 3 and m=1m = 1, two branches go to infinity.
  2. Real-axis locus: [−9,−1][-9, -1].
  3. Asymptotes:
σA=(0+0−9)−(−1)2=−4,θA=±90∘\sigma_A = \frac{(0 + 0 - 9) - (-1)}{2} = -4, \qquad \theta_A = \pm 90^\circ
  1. Departure from the double pole at 0: 2ϕ=180∘2\phi = 180^\circ, so the branches leave vertically at ±90∘\pm 90^\circ and curve into the LHP.
  2. Breakaway point:
K=−s3+9s2s+1,dKds=0⇒2s3+12s2+18s=0K = -\frac{s^3 + 9s^2}{s + 1}, \qquad \frac{dK}{ds} = 0 \Rightarrow 2s^3 + 12s^2 + 18s = 0 2s(s+3)2=0⇒s=0, −3, −32s(s + 3)^2 = 0 \Rightarrow s = 0,\ -3,\ -3

s=−3s = -3 is a triple root point:

K=−−27+81−2=27K = -\frac{-27 + 81}{-2} = 27

Check: (s+3)3=s3+9s2+27s+27(s + 3)^3 = s^3 + 9s^2 + 27s + 27, which matches the characteristic equation with K=27K = 27. ✓ 6. Branch angles at s=−3s = -3: three branches meet, so arriving and leaving branches are spaced 60∘60^\circ apart. The two complex branches from the origin and the real branch from −9-9 arrive (from directions ±60∘\pm 60^\circ and 180∘180^\circ). They leave at 0∘0^\circ (to the zero at −1-1) and ±120∘\pm 120^\circ (toward the asymptotes σ=−4\sigma = -4). 7. jωj\omega crossing (Routh):

RowCol 1Col 2
s3s^31KK
s2s^29KK
s1s^18K/98K/9
s0s^0KK

No sign change for K>0K > 0, so there is no jωj\omega crossing.

                           .                 |
                           .                 |
                           .                 |
                           ..                |
                            .                |
                            .                |
                            ..       .....   |
                             .. ......   ....|
----x..........................*........o---.x----
                             .. ......   ....|
                            ..       .....   |
                            .                |
                            .                |
                           ..                |
                           .                 |
                           .                 |
                           .                 |
   -9                     -4  -3             0
x pole (0, 0, -9)  o zero (-1)  * triple point (-3, K=27)

Comment on stability

  • For 0<K<270 < K < 27: one real pole lies between −9-9 and −3-3, plus a complex pair moving from the origin toward −3-3. The system is stable and underdamped.
  • At K=27K = 27: there is a triple pole at s=−3s = -3. The system is stable with a critically damped type response.
  • For K>27K > 27: one real pole lies between −3-3 and −1-1, and a complex pair moves toward σ=−4\sigma = -4. The system is stable and underdamped.
  • The first column of the Routh array is positive for all K>0K > 0.

Answer: The locus has a triple meeting point at s = −3 (K = 27), asymptotes at σ = −4 (±90°), and stays entirely in the LHP. The system is stable for all K > 0.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Control Systems chapter 5 questions →