Root locus of K(s+1)/[s²(s+3.6)]

$$G(s) = \frac{K(s+1)}{s^2(s+3.6)}$$

Characteristic equation: $s^3 + 3.6s^2 + Ks + K = 0$.

Root locus construction

1. Poles: $0, 0$ (double) and $-3.6$. Zero: $-1$. With $n = 3$ and $m = 1$, two branches go to infinity.
2. Real-axis locus: $[-3.6, -1]$ only. Left of a double pole the count is even, so $(-1, 0)$ is not on the locus.
3. Asymptotes:

$$\sigma_A = \frac{(0 + 0 - 3.6) - (-1)}{2} = -1.3, \qquad \theta_A = \pm 90^\circ$$

4. Departure from the double pole at the origin: $2\phi = 180^\circ + \angle(\text{zero}) - \angle(\text{pole } {-3.6}) = 180^\circ + 0 - 0$, so $\phi = \pm 90^\circ$. The two branches leave the origin vertically.
5. Breakaway points:

$$K = -\frac{s^3 + 3.6s^2}{s + 1}, \qquad \frac{dK}{ds} = 0 \Rightarrow 2s^3 + 6.6s^2 + 7.2s = 0$$ $$s(2s^2 + 6.6s + 7.2) = 0 \Rightarrow s = 0 \text{ or } s = -1.65 \pm j0.94$$

Apart from the double pole at the origin, the roots are complex. So there is no breakaway or break-in point on the real axis. A real break-in point exists only if the pole lies at more than 9 times the zero (here $3.6 < 9 \times 1$). 6. Shape: the two branches from the origin go up and down into the LHP, bend left, and approach the vertical asymptotes $\sigma = -1.3$ from the right. The branch from $-3.6$ moves right to the zero at $-1$. 7. $j\omega$ crossing (Routh):

All entries are positive for every $K > 0$, so the locus never crosses into the RHP.

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  -3.6                      -1         0
x pole (0, 0, -3.6)   o zero (-1)
asymptotes at sigma = -1.3; no real breakaway

The branches from 0 bend into the LHP toward the asymptotes σ = −1.3; the branch from −3.6 ends at −1.

(i) Range of K for overdamped response

Overdamping needs all closed-loop poles real. Because there is no real breakaway point, the two poles that leave the origin stay complex for every $K > 0$. (Numerical check: for K from 0.001 to 10 000, the cubic always has a complex pair.) Hence there is no positive value of K for an overdamped response; the response is always underdamped.

(ii) Range of K for an unstable system

For $K > 0$ all poles are in the LHP, so the system is stable for every positive $K$. At $K = 0$ there is a double pole at the origin. For $K < 0$ the constant term is negative and a root enters the RHP.

Answer: Overdamped: no value of K > 0. Unstable: K ≤ 0 (K < 0 unstable, K = 0 has a double pole at the origin). Stable for all K > 0.