Root locus: poles −2±j3, zero at origin

Open-loop poles at $-2 \pm j3$ and a zero at the origin give

$$G(s)H(s) = \frac{Ks}{(s+2)^2 + 9} = \frac{Ks}{s^2 + 4s + 13}$$

Characteristic equation: $s^2 + (4+K)s + 13 = 0$.

Root locus construction

1. Branches: $n = 2$, $m = 1$. One branch ends at the zero ($s = 0$); the other goes to $-\infty$.
2. Real-axis locus: $(-\infty, 0]$, the whole negative real axis.
3. Asymptote: $n - m = 1$, angle $180^\circ$ (the negative real axis).
4. Angle of departure from $-2 + j3$: angle from zero $0$: $123.69^\circ$; from pole $-2-j3$: $90^\circ$.

$$\phi_d = 180^\circ + 123.69^\circ - 90^\circ = 213.69^\circ \;(\equiv -146.31^\circ)$$

The branch leaves moving left and downward. 5. Break-in point:

$$K = -\frac{s^2 + 4s + 13}{s}, \qquad \frac{dK}{ds} = 0 \;\Rightarrow\; s^2 - 13 = 0 \;\Rightarrow\; s = -\sqrt{13} = -3.606$$

($s = +3.606$ is not on the locus.) 6. Shape: a pole-pair with one zero gives a circle centred at the zero, with radius $\sqrt{13} = 3.606$ (the distance from the zero to the poles). The complex poles move along this circle and meet the real axis at $-3.606$. One branch then moves right to the zero at $0$ and the other goes to $-\infty$. 7. $j\omega$ crossing: none for $K > 0$, since all coefficients of $s^2 + (4+K)s + 13$ are positive. The system is stable for all $K > 0$.

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......................*................o----------
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                      ..               |
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                            .x         |
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         -6         -3.6               0
x pole (-2+-j3)  o zero (0)  * break-in (-3.606)

Gain at critical damping

Critical damping corresponds to the break-in point $s = -3.606$, where the two closed-loop poles are real and equal.

$$K = \frac{|s^2 + 4s + 13|}{|s|} = \frac{13 - 14.422 + 13}{3.606} = \frac{11.578}{3.606} = 3.21$$

Check: critical damping means $(4+K)^2 = 4 \times 13 = 52$, so $K = \sqrt{52} - 4 = 3.21$. ✓

Answer: K ≈ 3.21. Both closed-loop poles are then at s = −3.606.