Root locus with poles 0, −1 and zero −3

Open-loop poles at $0$ and $-1$ and a zero at $-3$ give

$$G(s)H(s) = \frac{K(s+3)}{s(s+1)}$$

Characteristic equation: $s^2 + (1+K)s + 3K = 0$.

Root locus construction

1. Branches: $n = 2$, $m = 1$. One branch ends at $-3$; the other goes to $-\infty$ along $180^\circ$.
2. Real-axis locus: $[-1, 0]$ and $(-\infty, -3]$.
3. Breakaway and break-in points:

$$K = -\frac{s^2 + s}{s + 3}, \qquad \frac{dK}{ds} = 0 \Rightarrow s^2 + 6s + 3 = 0$$ $$s = -3 \pm \sqrt6 = -0.551,\ -5.449$$

• Breakaway at $s = -0.551$: $K = \dfrac{0.551 \times 0.449}{2.449} = 0.101$
• Break-in at $s = -5.449$: $K = \dfrac{5.449 \times 4.449}{2.449} = 9.90$

4. Complex part: a circle centred at the zero $-3$ with radius $\sqrt{3 \times (3 - 1)} = \sqrt6 = 2.449$ (distance from the zero to the breakaway point). Its highest point is $-3 + j2.449$.
5. $j\omega$ crossing: for $K > 0$ both coefficients $(1+K)$ and $3K$ are positive, so there is no crossing.

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..............*............o----------x..*..x-----
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            -5.45         -3           -.55
x pole (0, -1)  o zero (-3)  * breakaway / break-in

The poles at 0 and −1 move toward each other, break away at −0.551, travel around the circle, and break in at −5.449. Then one branch goes to −3 and the other to −∞.

Range of K for stability

By Routh–Hurwitz for $s^2 + (1+K)s + 3K$: we need $1 + K > 0$ and $3K > 0$, i.e. $K > 0$.

Answer: The system is stable for all K > 0. The breakaway point is at −0.551 (K = 0.101) and the break-in point at −5.449 (K = 9.90).