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Root locus of K/[s(s+1)(s²+4s+5)] with ξ = 0.5

Short answer

For ξ = 0.5, the dominant closed-loop poles are s = −0.288 ± j0.500 (ωn = 0.577 rad/s) with gain K ≈ 2.04. The system is stable for 0 < K < 8.

Plot the root loci for closed loop system with G(s) = K/[s(s+1)(s²+4s+5)], H(s) = 1. Also determine the dominant closed loop pole with ξ = 0.5.

G(s)=Ks(s+1)(s2+4s+5),H(s)=1G(s) = \frac{K}{s(s+1)(s^2+4s+5)}, \qquad H(s) = 1

Characteristic equation: s4+5s3+9s2+5s+K=0s^4 + 5s^3 + 9s^2 + 5s + K = 0.

1. Poles, zeros, branches

Poles: 0, −1, −2±j10,\ -1,\ -2 \pm j1 (n=4n = 4). No zeros (m=0m = 0). All four branches go to infinity.

2. Real-axis locus

Only between 00 and −1-1.

3. Asymptotes

σA=0−1−2−24=−1.25,θA=±45∘, ±135∘\sigma_A = \frac{0 - 1 - 2 - 2}{4} = -1.25, \qquad \theta_A = \pm 45^\circ,\ \pm 135^\circ

4. Breakaway point

dKds=−(4s3+15s2+18s+5)=0\frac{dK}{ds} = -(4s^3 + 15s^2 + 18s + 5) = 0

Real root: s=−0.393s = -0.393. Gain there: K=0.393×0.607×3.583=0.855K = 0.393 \times 0.607 \times 3.583 = 0.855. The other roots, −1.68±j0.60-1.68 \pm j0.60, are not on the locus.

5. Angle of departure from −2+j1-2 + j1

From 00: 153.43∘153.43^\circ; from −1-1: 135∘135^\circ; from −2−j1-2-j1: 90∘90^\circ.

ϕd=180∘−378.43∘=−198.43∘≡+161.57∘\phi_d = 180^\circ - 378.43^\circ = -198.43^\circ \equiv +161.57^\circ

6. jωj\omega crossing (Routh)

RowCol 1Col 2Col 3
s4s^419KK
s3s^355
s2s^28KK
s1s^1(40−5K)/8(40-5K)/8
s0s^0KK

Kmar=8K_{mar} = 8. Auxiliary equation: 8s2+8=0⇒s=±j18s^2 + 8 = 0 \Rightarrow s = \pm j1.

Sketch

                                    |
                                    |
    ...                             |       ..
      .....                         |    ....
          ....                      | ....
             .....x                .X..
                                 ...|
                                .#  |
---------------------------x....*...x-------------
                                .#  |
                                 ...|
             .....x                .X..
          ....                      | ....
      .....                         |    ....
    ...                             |       ..
                                    |
                                    |
                 -2       -1        0
x pole   * breakaway (-0.393)   X jw crossing (+-j1)
# zeta = 0.5 poles (-0.288 +- j0.5, K = 2.04)
  • Branches from 00 and −1-1 break away at −0.393-0.393, curve toward the ±45° asymptotes, and cross the jωj\omega-axis at ±j1\pm j1.
  • Branches from −2±j1-2 \pm j1 leave at ±161.6∘\pm 161.6^\circ and follow the ±135° asymptotes.

7. Dominant closed-loop poles for ξ = 0.5

Constant-ξ line: θ=cos⁡−10.5=60∘\theta = \cos^{-1}0.5 = 60^\circ from the negative real axis, i.e. s=ωn(−0.5+j0.866)s = \omega_n(-0.5 + j0.866). Search along it for ∠G=−180∘\angle G = -180^\circ. The solution is at ωn=0.577\omega_n = 0.577, giving s=−0.288+j0.500s = -0.288 + j0.500.

FromAngle to ssDistance
pole 00120.00∘120.00^\circ0.577
pole −1-135.07∘35.07^\circ0.870
pole −2+j1-2+j1−16.30∘-16.30^\circ1.783
pole −2−j1-2-j141.22∘41.22^\circ2.276
∠G=−(120.00+35.07−16.30+41.22)∘=−180.0∘  ✓\angle G = -(120.00 + 35.07 - 16.30 + 41.22)^\circ = -180.0^\circ \;\checkmark

Gain from the magnitude criterion:

K=0.577×0.870×1.783×2.276=2.04K = 0.577 \times 0.870 \times 1.783 \times 2.276 = 2.04

With K=2.04K = 2.04, the other two closed-loop poles are at −2.21±j1.11-2.21 \pm j1.11. They are about 7.7 times farther from the jωj\omega-axis than the dominant pair, so the pair at −0.288±j0.5-0.288 \pm j0.5 is truly dominant.

Answer: The dominant closed-loop poles for ξ = 0.5 are s = −0.288 ± j0.500 (ωn=0.577\omega_n = 0.577 rad/s), at K ≈ 2.04. The system is stable for 0 < K < 8.

✓ Answer checked by Claude Opus 5.5 (high effort)

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