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Root locus of K/[(s²+2s+2)(s²+2s+5)]

Short answer

No real-axis locus. Asymptotes intersect at σ=-1 with angles ±45°,±135°. Departure angles: +90° from -1+j1, -90° from -1+j2. Complex breakaway at -1±j1.581 (K=2.25). Imaginary-axis crossing occurs at ω=1.871 rad/s (K=16.25). System stable for 0<K<16.25.

Sketch the root locus for the unity feedback system having the forward path transfer function G(s) = K/[(s²+2s+2)(s²+2s+5)].

G(s)=K(s2+2s+2)(s2+2s+5)G(s) = \frac{K}{(s^2+2s+2)(s^2+2s+5)}

Characteristic equation: s4+4s3+11s2+14s+10+K=0s^4 + 4s^3 + 11s^2 + 14s + 10 + K = 0.

1. Poles and zeros

  • Poles: −1±j1-1 \pm j1 and −1±j2-1 \pm j2 (n=4n = 4). No zeros.
  • All 4 branches go to infinity.

2. Real-axis locus

There are no real poles or zeros, so no part of the real axis is on the locus.

3. Asymptotes

σA=−1−1−1−14=−1,θA=±45∘, ±135∘\sigma_A = \frac{-1-1-1-1}{4} = -1, \qquad \theta_A = \pm 45^\circ,\ \pm 135^\circ

4. Angles of departure

  • At −1+j1-1 + j1: angles from −1−j1-1-j1: 90∘90^\circ; from −1+j2-1+j2: −90∘-90^\circ; from −1−j2-1-j2: 90∘90^\circ.
ϕd=180∘−(90∘−90∘+90∘)=90∘\phi_d = 180^\circ - (90^\circ - 90^\circ + 90^\circ) = 90^\circ

The branch leaves straight up.

  • At −1+j2-1 + j2: the three angles are all 90∘90^\circ.
ϕd=180∘−270∘=−90∘\phi_d = 180^\circ - 270^\circ = -90^\circ

The branch leaves straight down.

So the two branches move toward each other along the line σ=−1\sigma = -1 and meet.

5. Breakaway points

K=−(s4+4s3+11s2+14s+10)K = -(s^4 + 4s^3 + 11s^2 + 14s + 10)

dKds=−(4s3+12s2+22s+14)=−2(s+1)(2s2+4s+7)=0\frac{dK}{ds} = -(4s^3 + 12s^2 + 22s + 14) = -2(s+1)(2s^2+4s+7) = 0
  • s=−1s = -1: K=−(1)(4)=−4<0K = -(1)(4) = -4 < 0, so it is not on the locus.
  • s=−1±j1.581s = -1 \pm j1.581: at s=−1+jbs = -1 + jb, (s2+2s+2)=1−b2=−1.5(s^2+2s+2) = 1 - b^2 = -1.5 and (s2+2s+5)=4−b2=1.5(s^2+2s+5) = 4 - b^2 = 1.5. Hence K=2.25>0K = 2.25 > 0, a valid complex breakaway point.

At K=2.25K = 2.25 the poles meet at −1±j1.581-1 \pm j1.581 and then split horizontally (±90° from the vertical) toward the 45° and 135° asymptotes.

6. Imaginary-axis crossing (Routh)

RowCol 1Col 2Col 3
s4s^411110+K10+K
s3s^3414
s2s^27.510+K10+K
s1s^1(65−4K)/7.5(65 - 4K)/7.5
s0s^010+K10+K

Kmar=16.25K_{mar} = 16.25. Auxiliary equation: 7.5s2+26.25=0⇒ω=3.5=1.8717.5s^2 + 26.25 = 0 \Rightarrow \omega = \sqrt{3.5} = 1.871 rad/s.

Sketch

                                 |
                                 |
     ....                        |       ....
        ......                   |  ......
             ......     .x     ..X...
                  ......*....... |
                        x.       |
                                 |
                                 |
---------------------------------+----------------
                                 |
                                 |
                        x.       |
                  ......*....... |
             ......     .x     ..X...
        ......                   |  ......
     ....                        |       ....
                                 |
                                 |
       -3              -1        0
x pole   * breakaway (-1 +- j1.58, K = 2.25)
X jw crossing (+-j1.87, K = 16.25)

Each pair moves along σ=−1\sigma = -1, meets at −1+j1.581-1 + j1.581, and turns. One branch heads to the 135° asymptote; the other heads right, crossing the jωj\omega-axis at j1.871j1.871 and following the 45° asymptote.

Answer: Complex breakaway at −1 ± j1.581 (K = 2.25). Departure angles are +90° from −1+j1 and −90° from −1+j2. The system is stable for 0 < K < 16.25 and oscillates at 1.871 rad/s when K = 16.25.

✓ Answer checked by Claude Opus 5.5 (high effort)

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