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Root locus of K(s+9)/[s(s²+4s+11)]

Short answer

Real-axis locus: [-9,0]. Asymptotes: σ=+2.5, ±90°. No breakaway point. Departure angle from -2+j2.646 is -16.38°. Stable range: 0<K<8.8. At the marginal limit K=8.8, the undamped natural frequency is 4.45 rad/s.

A unity feedback control system has an open loop transfer function G(s) = K(s+9)/[s(s²+4s+11)]. Sketch the root locus and determine: i) The range of 'K' for system to be stable ii) Undamped natural frequency of oscillation.

G(s)=K(s+9)s(s2+4s+11)G(s) = \frac{K(s+9)}{s(s^2+4s+11)}

Characteristic equation: s3+4s2+(11+K)s+9K=0s^3 + 4s^2 + (11+K)s + 9K = 0.

Root locus data

  1. Poles: 0, −2±j2.6460,\ -2 \pm j2.646. Zero: −9-9. With n=3n = 3 and m=1m = 1, two branches go to infinity.
  2. Real-axis locus: [−9,0][-9, 0].
  3. Asymptotes:
σA=(0−2−2)−(−9)2=+2.5,θA=±90∘\sigma_A = \frac{(0 - 2 - 2) - (-9)}{2} = +2.5, \qquad \theta_A = \pm 90^\circ

The asymptotes lie in the right half-plane, so the complex branches must cross the jωj\omega-axis. 4. Breakaway: dK/ds=0⇒2s3+31s2+72s+99=0dK/ds = 0 \Rightarrow 2s^3 + 31s^2 + 72s + 99 = 0. Roots: s=−13.03s = -13.03 (gives K<0K < 0; not on the locus) and −1.24±j1.51-1.24 \pm j1.51 (complex). So there is no breakaway point. The pole at 00 moves along the real axis to the zero at −9-9. 5. Angle of departure from −2+j2.646-2 + j2.646: from zero −9-9: 20.71∘20.71^\circ; from pole 00: 127.09∘127.09^\circ; from the conjugate pole: 90∘90^\circ.

ϕd=180∘+20.71∘−(127.09∘+90∘)=−16.38∘\phi_d = 180^\circ + 20.71^\circ - (127.09^\circ + 90^\circ) = -16.38^\circ
  1. jωj\omega crossing (Routh):
RowCol 1Col 2
s3s^3111+K11+K
s2s^249K9K
s1s^14(11+K)−9K4=44−5K4\dfrac{4(11+K) - 9K}{4} = \dfrac{44 - 5K}{4}
s0s^09K9K

Sketch

                                   | ..
                                   |..
                                   ..
                                  .X
                                ...|
                            x....  |
                                   |
                                   |
----o----..........................x--------------
                                   |
                                   |
                            x....  |
                                ...|
                                  .X
                                   ..
                                   |..
                                   | ..
   -9               -4             0       2.5
x pole   o zero (-9)
X jw crossing (+-j4.45, K = 8.8)

(i) Range of K for stability

All first-column entries must be positive:

  • 9K>0⇒K>09K > 0 \Rightarrow K > 0
  • 44−5K>0⇒K<8.844 - 5K > 0 \Rightarrow K < 8.8

Answer (i): 0 < K < 8.8

(ii) Undamped natural frequency of oscillation

At K=8.8K = 8.8 the auxiliary equation is

4s2+9(8.8)=0  ⇒  s2=−19.8  ⇒  s=±j4.454s^2 + 9(8.8) = 0 \;\Rightarrow\; s^2 = -19.8 \;\Rightarrow\; s = \pm j4.45

Answer (ii): ω = 4.45 rad/s (sustained oscillation when K = 8.8).

✓ Answer checked by Claude Opus 5.5 (high effort)

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