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Root locus of k(s+2)/[s(s+1)(s²+8s+64)]

Short answer

G(s)H(s)=k(s+2)/[s(s+1)(s²+8s+64)]. Real-axis breakaway at -0.572, break-in at -3.51. Stable for 0<k<442.5. Marginally stable at k=442.5 with sustained oscillation frequency 7.50 rad/s.

Sketch Root locus plot for the system open loop transfer function G(s)H(s) = k(s+2)/[s(s+1)(s²+8s+64)]. Discuss the region for stability, instability and marginal stability. What is frequency of oscillation at the point of marginal stability?

G(s)H(s)=k(s+2)s(s+1)(s2+8s+64)G(s)H(s) = \frac{k(s+2)}{s(s+1)(s^2+8s+64)}

Characteristic equation: s4+9s3+72s2+(64+k)s+2k=0s^4 + 9s^3 + 72s^2 + (64+k)s + 2k = 0.

1. Poles, zeros, branches

  • Poles: 0,−1,−4±j6.930, -1, -4 \pm j6.93 (n=4n = 4). Zero: −2-2 (m=1m = 1).
  • There are 4 branches: one ends at −2-2 and three go to infinity.

2. Real-axis locus

[−1,0][-1, 0] and (−∞,−2](-\infty, -2].

3. Asymptotes

σA=(0−1−4−4)−(−2)3=−2.33θA=60∘, 180∘, 300∘\begin{aligned} \sigma_A &= \frac{(0 - 1 - 4 - 4) - (-2)}{3} = -2.33 \\ \theta_A &= 60^\circ,\ 180^\circ,\ 300^\circ \end{aligned}

4. Breakaway / break-in points

dkds=0⇒3s4+26s3+126s2+288s+128=0\frac{dk}{ds} = 0 \Rightarrow 3s^4 + 26s^3 + 126s^2 + 288s + 128 = 0

Real roots: s=−0.572s = -0.572 (k=10.24k = 10.24), which is a breakaway between 0 and −1, and s=−3.51s = -3.51 (k=281.5k = 281.5), which is a break-in on (−∞,−2](-\infty, -2]. The other two roots are complex and are not on the locus.

5. Angle of departure from −4+j6.93-4 + j6.93

Angles: from zero −2-2: 106.1∘106.1^\circ; from pole 00: 120∘120^\circ; from −1-1: 113.4∘113.4^\circ; from −4−j6.93-4 - j6.93: 90∘90^\circ.

ϕd=180∘+106.1∘−(120∘+113.4∘+90∘)=−37.3∘\phi_d = 180^\circ + 106.1^\circ - (120^\circ + 113.4^\circ + 90^\circ) = -37.3^\circ

6. Imaginary-axis crossing (Routh)

RowCol 1Col 2Col 3
s4s^41722k2k
s3s^3964+k64+k
s2s^2(584−k)/9(584-k)/92k2k
s1s^1(584−k)(64+k)−162k584−k\dfrac{(584-k)(64+k) - 162k}{584-k}
s0s^02k2k

Setting the s1s^1 row to zero: k2−358k−37376=0⇒k=442.5k^2 - 358k - 37376 = 0 \Rightarrow k = 442.5. Auxiliary equation: 584−442.59s2+885=0⇒ω=7.50\dfrac{584 - 442.5}{9}s^2 + 885 = 0 \Rightarrow \omega = 7.50 rad/s.

Shape of the locus

  • The branches from 00 and −1-1 meet at −0.572-0.572 and break away into the complex plane. They loop to the left and re-enter the real axis at the break-in point −3.51-3.51. From there, one branch goes right to the zero at −2-2 and the other goes left to −∞-\infty.
  • The complex poles leave at ∓37.3∘\mp 37.3^\circ (moving right), cross the jωj\omega-axis at ±j7.50\pm j7.50, and follow the ±60° asymptotes.
                                      |     ....
                                      |  ....
                                     .X...
                       x..............|
                                      |
                                      |
                                      |
                                      |
                         ...........  |
........................*.....o---x...x-----------
                         ...........  |
                                      |
                                      |
                                      |
                                      |
                       x..............|
                                     .X...
                                      |  ....
                                      |     ....
       -8             -4     -2       0
x pole   o zero (-2)   * break-in (-3.51)
X jw crossing (+-j7.5, k = 442.5)

Stability regions

Range of kkNature
0<k<442.50 < k < 442.5Stable (all poles in LHP)
k=442.5k = 442.5Marginally stable, poles at ±j7.50\pm j7.50
k>442.5k > 442.5Unstable (two poles in RHP)

Answer: Marginal stability at k = 442.5. The frequency of oscillation there is ω = 7.50 rad/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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