Root locus of k/[s(s+4)(s²+4s+20)]

The root locus is the path traced by the closed-loop poles (roots of $1+G(s)H(s)=0$) as $K$ varies from 0 to $\infty$. It is sketched using the standard construction rules.

$$G(s)H(s) = \frac{K}{s(s+4)(s^2+4s+20)}$$

Step 1: Poles and zeros

• Poles: $s=0$, $s=-4$, $s=-2\pm j4$ (from $s^2+4s+20=0$). So $n=4$.
• Zeros: none, $m=0$. All 4 branches go to infinity.

Step 2: Real-axis locus

A point on the real axis is on the locus if the number of poles + zeros to its right is odd. So the locus lies between $0$ and $-4$.

Step 3: Asymptotes

$$\begin{aligned} \theta_q &= \frac{(2q+1)180^\circ}{n-m} = 45^\circ,\ 135^\circ,\ 225^\circ,\ 315^\circ \\ \sigma_A &= \frac{\sum \text{poles} - \sum\text{zeros}}{n-m} = \frac{0-4-2-2}{4} = -2 \end{aligned}$$

Step 4: Breakaway points

$$\begin{aligned} K &= -s(s+4)(s^2+4s+20) = -(s^4+8s^3+36s^2+80s) \\ \frac{dK}{ds} &= -(4s^3+24s^2+72s+80) = 0 \\ &\Rightarrow s^3 + 6s^2 + 18s + 20 = 0 \\ &\Rightarrow (s+2)(s^2+4s+10) = 0 \\ s &= -2,\quad s = -2\pm j2.45 \end{aligned}$$

• At $s=-2$: $K = -(16-64+144-160) = 64 > 0$, a valid breakaway point (on the real-axis segment).
• At $s=-2\pm j2.45$: $K = 100$ (real and positive), so these are also valid break points.

So the branches from $0$ and $-4$ meet at $-2$ ($K=64$) and leave vertically along $s=-2$. They meet the branches coming down from $-2\pm j4$ at $-2\pm j2.45$ ($K=100$). Then they break away again towards the $45^\circ$ and $135^\circ$ asymptotes.

Step 5: Angle of departure from complex pole $-2+j4$

Angles from the other poles to $-2+j4$:

• From $s=0$: $180^\circ - \tan^{-1}(4/2) = 116.57^\circ$
• From $s=-4$: $\tan^{-1}(4/2) = 63.43^\circ$
• From $s=-2-j4$: $90^\circ$

$$\phi_d = 180^\circ - (116.57^\circ + 63.43^\circ + 90^\circ) = -90^\circ$$

So the locus leaves $-2+j4$ straight downwards ($-90^\circ$), and leaves $-2-j4$ at $+90^\circ$.

Step 6: Crossing of the imaginary axis (Routh)

Characteristic equation: $s^4+8s^3+36s^2+80s+K=0$

$s^1$ row $=0$: $2080 - 8K = 0 \Rightarrow K = 260$. Auxiliary equation: $26s^2 + 260 = 0 \Rightarrow s = \pm j3.16$.

Step 7: Sketch

 Upper half shown; lower half is its mirror image.

  \                          jw        /
   \  135 deg asymptote      |        /  45 deg
    \                        |       /   asymptote
     \__                     * j3.16 (K=260)
        \      x -2+j4     _/|
         \__   |         _/  |
            \  v       _/    |
             `-o------'      |
         -2+j2.45 (K=100)    |
               ^             |
               |             |
   ----x=======o=============x---- sigma
      -4      -2 (K=64)      0

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