Skip to main content

Root locus with poles 0, −6, −2 ± j3

Short answer

Open-loop TF: K/[s(s+6)(s²+4s+13)]. Real-axis breakaway at s=-4.20 with K≈104.6. Departure angles from complex poles ∓70.6°. Stable for 0<K<227.8; at marginal K=227.8 it oscillates at 2.79 rad/s.

For a system having open loop pole-zero plot as in figure, sketch Root locus. [Figure: s-plane with open-loop poles at s = 0, s = −6 and s = −2 ± j3; no zeros]

Open-loop poles at 0,−6,−2±j30, -6, -2 \pm j3 and no zeros give

G(s)H(s)=Ks(s+6)(s2+4s+13)G(s)H(s) = \frac{K}{s(s+6)(s^2+4s+13)}

Characteristic equation: s4+10s3+37s2+78s+K=0s^4 + 10s^3 + 37s^2 + 78s + K = 0.

1. Branches

n=4n = 4, m=0m = 0: 4 branches, all ending at infinity.

2. Real-axis locus

Between 00 and −6-6 (one real pole to the right is an odd count). There is no locus left of −6.

3. Asymptotes

σA=0−6−2−24=−2.5θA=±45∘, ±135∘\begin{aligned} \sigma_A &= \frac{0 - 6 - 2 - 2}{4} = -2.5 \\ \theta_A &= \pm 45^\circ,\ \pm 135^\circ \end{aligned}

4. Breakaway point

K=−(s4+10s3+37s2+78s)K = -(s^4 + 10s^3 + 37s^2 + 78s)

dKds=−(4s3+30s2+74s+78)=0\frac{dK}{ds} = -(4s^3 + 30s^2 + 74s + 78) = 0

Roots: s=−4.20s = -4.20 and s=−1.65±j1.39s = -1.65 \pm j1.39. Only s=−4.20s = -4.20 is on the real-axis locus. Gain there:

K=4.20×1.80×13.84≈104.6K = 4.20 \times 1.80 \times 13.84 \approx 104.6

5. Angle of departure from −2+j3-2 + j3

Angles to −2+j3-2+j3: from 00: 123.69∘123.69^\circ; from −6-6: 36.87∘36.87^\circ; from −2−j3-2-j3: 90∘90^\circ.

ϕd=180∘−(123.69∘+36.87∘+90∘)=−70.56∘\phi_d = 180^\circ - (123.69^\circ + 36.87^\circ + 90^\circ) = -70.56^\circ

At −2−j3-2-j3, it is +70.56∘+70.56^\circ.

6. Imaginary-axis crossing

RowCol 1Col 2Col 3
s4s^4137KK
s3s^31078
s2s^229.2KK
s1s^1(2277.6−10K)/29.2(2277.6 - 10K)/29.2
s0s^0KK

Kmar=227.76K_{mar} = 227.76. Auxiliary equation: 29.2s2+227.76=0⇒ω=7.8=2.7929.2s^2 + 227.76 = 0 \Rightarrow \omega = \sqrt{7.8} = 2.79 rad/s.

Sketch

                                       |
    ...                                |      ....
      ....                             |   ....
         ....                x.        .....
            ...               .........X
              ...                      |
                ...                    |
                  ..                   |
----------x........*...................x----------
                  ..                   |
                ...                    |
              ...                      |
            ...               .........X
         ....                x.        .....
      ....                             |   ....
    ...                                |      ....
                                       |
         -6      -4.2       -2         0
x pole   * breakaway (-4.2)
X jw crossing (+-j2.79, K = 227.8)

The branches from 00 and −6-6 meet at −4.2-4.2 and break away at ±90°, then bend to the ±135° asymptotes. The complex poles leave at ∓70.6∘\mp 70.6^\circ, curve right, cross the jωj\omega-axis at ±j2.79\pm j2.79 and follow the ±45° asymptotes.

Answer: Breakaway at s = −4.20 (K ≈ 104.6). The departure angle is ∓70.6°. The system is stable for 0 < K < 227.8 and oscillates at 2.79 rad/s when K = 227.8.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Control Systems chapter 5 questions →