Root locus of K(s+2)/(s²−4s+13)

$$G(s) = \frac{K(s+2)}{s^2 - 4s + 13}$$

Poles: $s = 2 \pm j3$ (in the right half-plane). Zero: $s = -2$. Characteristic equation: $s^2 + (K-4)s + (13 + 2K) = 0$.

(i) Sketch of the root locus

1. Branches: $n = 2$, $m = 1$. One branch ends at the zero $-2$; the other goes to $-\infty$.
2. Real-axis locus: $(-\infty, -2]$, i.e. to the left of the zero.
3. Asymptote: one, at $180^\circ$.
4. Angle of departure from $2 + j3$: angle from zero $-2$: $\tan^{-1}(3/4) = 36.87^\circ$; from pole $2-j3$: $90^\circ$.

$$\phi_d = 180^\circ + 36.87^\circ - 90^\circ = 126.87^\circ$$

5. Shape: the complex part is a circle centred at the zero $-2$ with radius $\sqrt{4^2 + 3^2} = 5$. The poles move left along this circle, cross the $j\omega$-axis, and meet on the real axis at $-7$. From there, one branch goes right to $-2$ and the other goes left to $-\infty$.

                                   |
                        .........  |
                  .......       ...X...
               ....                |  ....
             ...                   |     .x
            ..                     |
           ..                      |
           .                       |
..........*.................o------+--------------
           .                       |
           ..                      |
            ..                     |
             ...                   |     .x
               ....                |  ....
                  .......       ...X...
                        .........  |
                                   |
         -7                -2      0      2
x pole (2 +- j3)   o zero (-2)   * break-in (-7)
X jw crossing (+-j4.58, K = 4)

(ii) Imaginary-axis crossing

At $s = j\omega$, the real and imaginary parts of the characteristic equation must both vanish:

$$\begin{aligned} \text{Imag: } & (K-4)\omega = 0 \Rightarrow K = 4 \\ \text{Real: } & -\omega^2 + 13 + 2K = 0 \Rightarrow \omega^2 = 21 \end{aligned}$$

The locus crosses at $s = \pm j4.583$. (Check: the circle $|s+2| = 5$ at $\sigma = 0$ gives $\omega = \sqrt{25 - 4} = \sqrt{21}$.)

(iii) Gain at the crossing

K = 4. The system is unstable for $K < 4$ and stable for $K > 4$.

(iv) Break-in point

$$K = -\frac{s^2 - 4s + 13}{s + 2}, \qquad \frac{dK}{ds} = 0 \Rightarrow s^2 + 4s - 21 = 0 \Rightarrow s = -7 \text{ or } 3$$

$s = 3$ is not on the locus. Break-in at s = −7, where

$$K = -\frac{49 + 28 + 13}{-5} = 18$$

Answer: $j\omega$ crossing at ±j4.583 rad/s with K = 4. The break-in point is s = −7 (K = 18). The departure angle is ±126.87°.