Root locus of K(s+1)/[(s²+2s+2)(s²+2s+5)]

The root locus is drawn for $0 \le k < \infty$ with the standard construction rules.

$$G(s)H(s) = \frac{k(s+1)}{(s^2+2s+2)(s^2+2s+5)}$$

Step 1: Poles and zeros

• Poles: $s = -1\pm j1$ and $s = -1 \pm j2$, so $n = 4$.
• Zero: $s = -1$, so $m = 1$.
• Number of branches = 4. One branch ends at the zero and three go to infinity.

Step 2: Real-axis locus

There are no real poles. The only real-axis singularity is the zero at $-1$. Points to the left of $-1$ have one zero to their right (odd), so the real axis from $-1$ to $-\infty$ is on the locus.

Step 3: Asymptotes

$$\begin{aligned} \theta_q &= \frac{(2q+1)180^\circ}{n-m} = 60^\circ,\ 180^\circ,\ 300^\circ \\ \sigma_A &= \frac{(-1-1-1-1)-(-1)}{3} = \frac{-3}{3} = -1 \end{aligned}$$

Step 4: Angles of departure

At $p_1 = -1+j1$, angles from the other poles and the zero:

• from $-1-j1$: $90^\circ$; from $-1+j2$: $-90^\circ$; from $-1-j2$: $90^\circ$
• from zero $-1$: $90^\circ$

$$\phi_{d1} = 180^\circ - (90^\circ - 90^\circ + 90^\circ) + 90^\circ = 180^\circ$$

At $p_2 = -1+j2$:

• from $-1+j1$, $-1-j1$, $-1-j2$: $90^\circ$ each; from zero: $90^\circ$

$$\phi_{d2} = 180^\circ - 270^\circ + 90^\circ = 0^\circ$$

So the branch from $-1+j1$ leaves horizontally to the left, and the branch from $-1+j2$ leaves horizontally to the right. The lower half is the mirror image.

Step 5: Break-in point

$$k = -\frac{(s^2+2s+2)(s^2+2s+5)}{s+1}$$

Put $p = s+1$, so $k = -\frac{(p^2+1)(p^2+4)}{p}$:

$$\frac{dk}{dp} = 0 \Rightarrow 3p^4 + 5p^2 - 4 = 0 \Rightarrow p^2 = 0.591 \ \text{(other root negative)}$$

$p = -0.769$ gives $s = -1.77$ (on the locus) with $k = 9.50 > 0$. So $s = -1.77$ is a break-in point. The branches from $-1\pm j1$ meet here. After that, one goes right to the zero at $-1$ and the other goes left to $-\infty$. The roots $p=\pm j1.50$ give a complex $k$, so they are not on the locus.

Step 6: Imaginary-axis crossing

Characteristic equation: $s^4 + 4s^3 + 11s^2 + (14+k)s + (10+k) = 0$

$s^1$ row $=0$: $260 - k^2 = 0 \Rightarrow k = 16.12$.

Auxiliary equation: $\frac{30-16.12}{4}s^2 + 26.12 = 0 \Rightarrow s = \pm j2.74$.

Sketch

 Upper half shown; lower half is its mirror image.

                         jw     /
                          |    /  60 deg asymptote
                          *   /   (from -1)
                         /| j2.74 (k=16.1)
              x-------- ' |
            -1+j2  (0 deg)|
                          |
         .--x (180 deg)   |
        /  -1+j1          |
       /                  |
 <====o=====o-------------+-----> sigma
   -1.77   -1             0
  (k=9.5)  zero

Results