Root Locus Technique

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

Results

Result: at $\zeta=0.7$, $K \approx 28.4$, with $\omega_n \approx 1.98$ rad/s and $\omega_d \approx 1.42$ rad/s.

The root locus is sketched with the standard rules. The $\zeta = 0.7$ line is then drawn from the origin, and the magnitude condition at its intersection with the locus gives $K$.

$$G(s)H(s) = \frac{K}{s(s^2+10s+24)} = \frac{K}{s(s+4)(s+6)}$$

Step 1: Poles and zeros

Poles: $0,\ -4,\ -6$ ($n=3$). No zeros ($m=0$), so 3 branches go to infinity.

Step 2: Real-axis segments

Between $0$ and $-4$, and from $-6$ to $-\infty$.

Step 3: Asymptotes

$$\theta = \frac{(2q+1)180^\circ}{3} = 60^\circ,\ 180^\circ,\ 300^\circ, \qquad \sigma_A = \frac{0-4-6}{3} = -3.33$$

Step 4: Breakaway point

$$\begin{aligned} K &= -(s^3+10s^2+24s) \\ \frac{dK}{ds} &= -(3s^2+20s+24) = 0 \\ s &= \frac{-20\pm\sqrt{400-288}}{6} = -1.57 \ \text{or}\ -5.10 \end{aligned}$$

$s=-5.10$ is not on the locus. So the breakaway point is $s=-1.57$, where $K = |s||s+4||s+6| = 16.9$.

Step 5: Imaginary-axis crossing

Characteristic equation: $s^3+10s^2+24s+K=0$

$K = 240$ makes the $s^1$ row zero, and $10s^2+240=0$ gives $s = \pm j4.90$. The system is stable for $0<K<240$.

Step 6: Point with $\zeta = 0.7$

The constant-$\zeta$ line makes angle $\theta = \cos^{-1}0.7 = 45.57^\circ$ with the negative real axis. Points on it are $s = -0.7\omega_n + j0.714\omega_n$. Using the angle condition $\angle s + \angle(s+4) + \angle(s+6) = 180^\circ$ (solved by trial):

$$\omega_n = 1.984\ \text{rad/s}, \qquad s = -1.389 \pm j1.417$$

Magnitude condition:

$$K = |s|\,|s+4|\,|s+6| = 1.984 \times 2.971 \times 4.824 = 28.4$$

The third closed-loop pole is then at $s \approx -7.22$, far to the left, so the complex pair is dominant.

Sketch

 Upper half shown; lower half is its mirror image.

                            jw      /
                             |     / 60 deg asymptote
                             *    /  (from -3.33)
                            /| j4.90 (K=240)
                           / |
                          /  |
     zeta=0.7 point -->  *   |  -1.39+j1.42 (K=28.4)
                         |.  |
                         | . |   . = zeta line
                         |  .|
  <====x------x==========o===x----> sigma
      -6     -4       -1.57  0
                     (K=16.9)

Answer: at $\zeta = 0.7$: $K \approx 28.4$, natural frequency $\omega_n \approx 1.98$ rad/s (damped frequency $\omega_d = \omega_n\sqrt{1-\zeta^2} \approx 1.42$ rad/s). Closed-loop dominant poles are at $-1.39 \pm j1.42$.

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

The root locus is the path of the closed-loop poles in the s-plane as $K$ varies from 0 to $\infty$. The characteristic equation is $1 + G(s) = 0$, i.e.

$$s(s+1)(s^2+4s+5) + K = 0 \;\Rightarrow\; s^4 + 5s^3 + 9s^2 + 5s + K = 0$$

Step 1: Poles, zeros and branches

• Open-loop poles: $s = 0,\ -1,\ -2 \pm j1$ ($n = 4$); zeros: none ($m = 0$).
• Number of branches = 4; all four go to infinity.

Step 2: Real-axis locus

A real-axis point is on the locus if the number of real poles + zeros to its right is odd. Only the segment between 0 and −1 qualifies.

Step 3: Asymptotes

$$\begin{aligned} \sigma_A &= \frac{\sum p - \sum z}{n-m} = \frac{0-1-2-2}{4} = -1.25 \\ \theta_A &= \frac{(2q+1)180^\circ}{4} = \pm 45^\circ,\ \pm 135^\circ \end{aligned}$$

Step 4: Breakaway point

$K = -(s^4+5s^3+9s^2+5s)$, so

$$\frac{dK}{ds} = -(4s^3+15s^2+18s+5) = 0$$

Roots: $s = -0.393$ and $s = -1.679 \pm j0.603$. Only $s = -0.393$ lies on the locus (between 0 and −1). Gain there:

$$K = 0.393 \times 0.607 \times |(-0.393)^2 + 4(-0.393) + 5| = 0.855$$

Step 5: Angle of departure from $-2+j1$

Angles from the other poles to $-2+j1$: from $0$: $153.43^\circ$; from $-1$: $135^\circ$; from $-2-j1$: $90^\circ$.

$$\phi_d = 180^\circ - (153.43^\circ + 135^\circ + 90^\circ) = -198.43^\circ \equiv 161.57^\circ$$

By symmetry, the angle from $-2-j1$ is $-161.57^\circ$.

Step 6: Imaginary-axis crossing (Routh array)

Marginal stability: $40 - 5K = 0 \Rightarrow K = 8$. Auxiliary equation: $8s^2 + 8 = 0 \Rightarrow s = \pm j1$. The locus crosses the $j\omega$ axis at $\pm j1$ rad/s when $K = 8$.

Sketch

                                    |
                                    |
    ...                             |       ..
      .....                         |    ....
          ....                      | ....
             .....x                .X..
                                 ...|
                                ..  |
---------------------------x....*...x-------------
                                ..  |
                                 ...|
             .....x                .X..
          ....                      | ....
      .....                         |    ....
    ...                             |       ..
                                    |
                                    |
                 -2       -1        0
x pole   * breakaway (-0.393)
X jw crossing (+-j1, K = 8)

Branches from 0 and −1 meet at −0.393, break away at ±90°, and bend toward the ±45° asymptotes, crossing at ±j1 (K = 8). Branches from $-2\pm j1$ leave at $\pm 161.6^\circ$ and approach the ±135° asymptotes.

Gain for critical damping

Critical damping occurs when the dominant closed-loop poles are real and equal, i.e. at the breakaway point $s = -0.393$. The other two poles are then at $-2.107 \pm j1.046$.

Answer: Critical damping at K ≈ 0.855 (double pole at s = −0.393). The system is stable for $0 < K < 8$.

Criteria of root locus

Every point $s$ on the root locus satisfies the characteristic equation $1 + G(s)H(s) = 0$, i.e. $G(s)H(s) = -1$. This gives two conditions:

1. Angle criterion: $\angle G(s)H(s) = \pm(2q+1)180^\circ$, $q = 0, 1, 2, \dots$ The sum of angles from the zeros minus the sum of angles from the poles must be an odd multiple of $180^\circ$. A point is on the locus only if it meets this condition.
2. Magnitude criterion: $|G(s)H(s)| = 1$. This gives the value of $K$ at a point already known to be on the locus:

$$K = \frac{\prod |s - p_i|}{\prod |s - z_j|}$$

Open-loop transfer function

Poles at $0$ and $-1 \pm j2$, no zeros:

$$G(s) = \frac{K}{s\left[(s+1)^2 + 4\right]} = \frac{K}{s(s^2+2s+5)}$$

Characteristic equation: $s^3 + 2s^2 + 5s + K = 0$.

Sketching steps

1. Branches: $n = 3$, $m = 0$, so there are 3 branches and all go to infinity.
2. Real-axis locus: the whole negative real axis, from $0$ to $-\infty$ (one pole to the right).
3. Asymptotes:

$$\sigma_A = \frac{0 - 1 - 1}{3} = -0.667, \qquad \theta_A = 60^\circ,\ 180^\circ,\ 300^\circ$$

4. Breakaway: $dK/ds = -(3s^2 + 4s + 5) = 0 \Rightarrow s = -0.667 \pm j1.106$. These are complex, so there is no breakaway point. The branch from the origin moves straight along the negative real axis to $-\infty$.
5. Angle of departure from $-1 + j2$: from pole $0$: $116.57^\circ$; from $-1-j2$: $90^\circ$.

$$\phi_d = 180^\circ - (116.57^\circ + 90^\circ) = -26.57^\circ$$

At $-1-j2$, it is $+26.57^\circ$. 6. $j\omega$-axis crossing (Routh):

$K_{mar} = 10$; auxiliary equation $2s^2 + 10 = 0 \Rightarrow s = \pm j\sqrt5 = \pm j2.236$.

Sketch

                                    |         ....
                                    |      ....
                                    |   ....
                                    .....
                           x........X
                                    |
                                    |
                                    |
....................................x-------------
                                    |
                                    |
                                    |
                           x........X
                                    .....
                                    |   ....
                                    |      ....
                                    |         ....
        -3                -1        0
x pole (0, -1 +- j2)
X jw crossing (+-j2.24, K = 10)

The complex poles leave at $\mp 26.6^\circ$, bend toward the ±60° asymptotes from $-0.667$, and cross the imaginary axis at $\pm j2.236$.

Answer: $G(s) = K/[s(s^2+2s+5)]$; stable for $0 < K < 10$; at $K = 10$ it oscillates at $\omega = 2.236$ rad/s.

Open-loop poles at $0, -6, -2 \pm j3$ and no zeros give

$$G(s)H(s) = \frac{K}{s(s+6)(s^2+4s+13)}$$

Characteristic equation: $s^4 + 10s^3 + 37s^2 + 78s + K = 0$.

1. Branches

$n = 4$, $m = 0$: 4 branches, all ending at infinity.

2. Real-axis locus

Between $0$ and $-6$ (one real pole to the right is an odd count). There is no locus left of −6.

3. Asymptotes

$$\begin{aligned} \sigma_A &= \frac{0 - 6 - 2 - 2}{4} = -2.5 \\ \theta_A &= \pm 45^\circ,\ \pm 135^\circ \end{aligned}$$

4. Breakaway point

$K = -(s^4 + 10s^3 + 37s^2 + 78s)$

$$\frac{dK}{ds} = -(4s^3 + 30s^2 + 74s + 78) = 0$$

Roots: $s = -4.20$ and $s = -1.65 \pm j1.39$. Only $s = -4.20$ is on the real-axis locus. Gain there:

$$K = 4.20 \times 1.80 \times 13.84 \approx 104.6$$

5. Angle of departure from $-2 + j3$

Angles to $-2+j3$: from $0$: $123.69^\circ$; from $-6$: $36.87^\circ$; from $-2-j3$: $90^\circ$.

$$\phi_d = 180^\circ - (123.69^\circ + 36.87^\circ + 90^\circ) = -70.56^\circ$$

At $-2-j3$, it is $+70.56^\circ$.

6. Imaginary-axis crossing

$K_{mar} = 227.76$. Auxiliary equation: $29.2s^2 + 227.76 = 0 \Rightarrow \omega = \sqrt{7.8} = 2.79$ rad/s.

Sketch

                                       |
    ...                                |      ....
      ....                             |   ....
         ....                x.        .....
            ...               .........X
              ...                      |
                ...                    |
                  ..                   |
----------x........*...................x----------
                  ..                   |
                ...                    |
              ...                      |
            ...               .........X
         ....                x.        .....
      ....                             |   ....
    ...                                |      ....
                                       |
         -6      -4.2       -2         0
x pole   * breakaway (-4.2)
X jw crossing (+-j2.79, K = 227.8)

The branches from $0$ and $-6$ meet at $-4.2$ and break away at ±90°, then bend to the ±135° asymptotes. The complex poles leave at $\mp 70.6^\circ$, curve right, cross the $j\omega$-axis at $\pm j2.79$ and follow the ±45° asymptotes.

Answer: Breakaway at s = −4.20 (K ≈ 104.6). The departure angle is ∓70.6°. The system is stable for 0 < K < 227.8 and oscillates at 2.79 rad/s when K = 227.8.

$$G(s) = \frac{K(s+1)}{(s+2)(s+4)(s+6)}$$

Characteristic equation: $s^3 + 12s^2 + 44s + 48 + K(s+1) = 0$.

Root locus data

• Poles: $-2, -4, -6$; zero: $-1$. So $n = 3$, $m = 1$, with 2 branches going to infinity.
• Real-axis locus: $[-2, -1]$ and $[-6, -4]$.
• Asymptotes:

$$\sigma_A = \frac{(-2-4-6) - (-1)}{2} = -5.5, \qquad \theta_A = \pm 90^\circ$$

• Breakaway: $\dfrac{dK}{ds} = 0 \Rightarrow 2s^3 + 15s^2 + 24s - 4 = 0$. The root on the locus is $s = -5.04$, where $K = 0.751$.
• The branch from $-2$ moves to the zero at $-1$. The branches from $-4$ and $-6$ meet at $-5.04$, break away, and go vertically along the asymptote $\sigma = -5.5$.

Locating the ζ = 0.5 point

Constant-ζ line: $\theta = \cos^{-1}(0.5) = 60^\circ$ from the negative real axis. Points on it are $s = \omega_n(-0.5 + j0.866)$. Search along this line for $\angle G(s) = -180^\circ$:

Angle check at $s = -5.43 + j9.41$:

$$115.22^\circ - (110.04^\circ + 98.65^\circ + 86.54^\circ) = -180.0^\circ \;\checkmark$$

Gain by magnitude criterion

$$K = \frac{10.012 \times 9.514 \times 9.423}{10.397} = 86.3$$

Check

With $K = 86.3$: $s^3 + 12s^2 + 130.3s + 134.3 = 0$. Its roots are $-5.43 \pm j9.41$ and $-1.14$. The third pole at $-1.14$ lies very close to the zero at $-1$, so the two nearly cancel. The complex pair therefore governs the response (second-order approximation valid).

                 .                          |
                 .                          |
                 *.                         |
                  .                         |
                  .                         |
                  .                         |
                  .                         |
                  ..                        |
                   .                        |
---------------x.........x--------x....o----+-----
                   .                        |
                  ..                        |
                  .                         |
                  .                         |
                  .                         |
                  .                         |
                 *.                         |
                 .                          |
                 .                          |
              -6       -4        -2         0
x pole   o zero (-1)
* zeta = 0.5 poles (-5.43 +- j9.41, K = 86.3)

Answer: K ≈ 86.3. Dominant poles: s = −5.43 ± j9.41 ($\omega_n = 10.86$ rad/s, ζ = 0.5).

A point $s_1$ in the s-plane lies on the root locus only if it is a root of the characteristic equation $1 + KG(s)H(s) = 0$ for some $K \ge 0$. Writing $KG(s_1)H(s_1) = -1 = 1\angle \pm 180^\circ$ gives two criteria.

1. Angle criterion (decides whether the point is on the locus)

$$\angle G(s_1)H(s_1) = \sum_{j} \angle(s_1 - z_j) - \sum_{i} \angle(s_1 - p_i) = \pm(2q+1)\,180^\circ$$

• Draw vectors from every open-loop pole and zero to $s_1$ and measure their angles from the positive real axis.
• If (sum of zero angles) − (sum of pole angles) is an odd multiple of $180^\circ$, the point is on the locus. Otherwise it is not.
• $K$ does not appear in this condition, so it alone fixes the shape of the locus.

2. Magnitude criterion (gives K at that point)

$$|K\,G(s_1)H(s_1)| = 1 \;\Rightarrow\; K = \frac{\prod_i |s_1 - p_i|}{\prod_j |s_1 - z_j|}$$

That is, K is the product of the distances from the poles divided by the product of the distances from the zeros (1 if there are no zeros). This is used only after the angle criterion is satisfied.

Example

$G(s) = \dfrac{K}{s(s+4)}$, test point $s_1 = -2 + j2$:

• $\angle s_1 = 135^\circ$ and $\angle(s_1 + 4) = 45^\circ$.
• Sum $= -(135^\circ + 45^\circ) = -180^\circ$, so the point is on the locus.
• $K = |s_1|\,|s_1 + 4| = 2.83 \times 2.83 = 8$.

Related real-axis rule

A real-axis point is on the locus if the number of real poles plus zeros to its right is odd. This rule follows directly from the angle criterion.

The root locus passes through a point $s_1$ only if $s_1$ satisfies the angle criterion:

$$\angle G(s_1)H(s_1) = \sum \angle(s_1 - z_j) - \sum \angle(s_1 - p_i) = \pm(2q+1)\,180^\circ$$

with $q = 0, 1, 2, \dots$ To apply it, draw vectors from all open-loop zeros and poles to $s_1$. Then take the sum of the zero angles minus the sum of the pole angles. If the result is an odd multiple of $180^\circ$, the point lies on the root locus for $K > 0$; otherwise it does not. Example: for $G = K/[s(s+2)]$, the point $-1 + j1$ gives $-(135^\circ + 45^\circ) = -180^\circ$, so it is on the locus.

$$G(s)H(s) = \frac{k(s+2)}{s(s+1)(s^2+8s+64)}$$

Characteristic equation: $s^4 + 9s^3 + 72s^2 + (64+k)s + 2k = 0$.

1. Poles, zeros, branches

• Poles: $0, -1, -4 \pm j6.93$ ($n = 4$). Zero: $-2$ ($m = 1$).
• There are 4 branches: one ends at $-2$ and three go to infinity.

2. Real-axis locus

$[-1, 0]$ and $(-\infty, -2]$.

3. Asymptotes

$$\begin{aligned} \sigma_A &= \frac{(0 - 1 - 4 - 4) - (-2)}{3} = -2.33 \\ \theta_A &= 60^\circ,\ 180^\circ,\ 300^\circ \end{aligned}$$

4. Breakaway / break-in points

$$\frac{dk}{ds} = 0 \Rightarrow 3s^4 + 26s^3 + 126s^2 + 288s + 128 = 0$$

Real roots: $s = -0.572$ ($k = 10.24$), which is a breakaway between 0 and −1, and $s = -3.51$ ($k = 281.5$), which is a break-in on $(-\infty, -2]$. The other two roots are complex and are not on the locus.

5. Angle of departure from $-4 + j6.93$

Angles: from zero $-2$: $106.1^\circ$; from pole $0$: $120^\circ$; from $-1$: $113.4^\circ$; from $-4 - j6.93$: $90^\circ$.

$$\phi_d = 180^\circ + 106.1^\circ - (120^\circ + 113.4^\circ + 90^\circ) = -37.3^\circ$$

6. Imaginary-axis crossing (Routh)

Setting the $s^1$ row to zero: $k^2 - 358k - 37376 = 0 \Rightarrow k = 442.5$. Auxiliary equation: $\dfrac{584 - 442.5}{9}s^2 + 885 = 0 \Rightarrow \omega = 7.50$ rad/s.

Shape of the locus

• The branches from $0$ and $-1$ meet at $-0.572$ and break away into the complex plane. They loop to the left and re-enter the real axis at the break-in point $-3.51$. From there, one branch goes right to the zero at $-2$ and the other goes left to $-\infty$.
• The complex poles leave at $\mp 37.3^\circ$ (moving right), cross the $j\omega$-axis at $\pm j7.50$, and follow the ±60° asymptotes.

                                      |     ....
                                      |  ....
                                     .X...
                       x..............|
                                      |
                                      |
                                      |
                                      |
                         ...........  |
........................*.....o---x...x-----------
                         ...........  |
                                      |
                                      |
                                      |
                                      |
                       x..............|
                                     .X...
                                      |  ....
                                      |     ....
       -8             -4     -2       0
x pole   o zero (-2)   * break-in (-3.51)
X jw crossing (+-j7.5, k = 442.5)

Stability regions

Answer: Marginal stability at k = 442.5. The frequency of oscillation there is ω = 7.50 rad/s.

$$G(s)H(s) = \frac{k(s-1)}{s(s-1)}$$

The factor $(s-1)$ appears in both numerator and denominator. It must not be cancelled before forming the characteristic equation, because the cancelled pole at $s = +1$ is still a real mode of the system.

Characteristic equation

$$\begin{aligned} 1 + G(s)H(s) &= 0 \\ s(s-1) + k(s-1) &= 0 \\ (s-1)(s+k) &= 0 \end{aligned}$$

Closed-loop poles: $s = +1$ (for every $k$) and $s = -k$.

Root locus construction

• Open-loop poles: $0$ and $+1$. Open-loop zero: $+1$.
• At $s = +1$ the pole and zero coincide, so a branch "starts and ends" at the same point. One closed-loop pole stays fixed at $s = +1$ for all $k$.
• The reduced part $k/s$ has a pole at $0$ and no finite zero. Its real-axis locus is the whole negative real axis, with asymptote angle $180^\circ$.
• That branch starts at $s = 0$ ($k = 0$) and moves left along the negative real axis, with the pole at $s = -k$ (e.g. $-5$ for $k = 5$). There is no breakaway point and no $j\omega$ crossing.

               jw
               |
 <=============x-----⊗------> σ
  s = -k       0    +1
 (moves left)       (fixed pole,
                    pole-zero overlap)

Comments

• If the pair is cancelled first, one gets $G = k/s$, which seems stable for all $k > 0$. That conclusion is wrong.
• The closed-loop pole at $s = +1$ lies in the right half-plane for every $k$. The system is therefore unstable for all values of $k$. Its output contains a growing mode $e^{t}$ that the gain cannot move.
• Pole-zero cancellation in the right half-plane is never used to stabilise a system. Exact cancellation is impossible in practice, and the hidden unstable mode still exists internally.

Answer: The root locus is a fixed point at $s = +1$ plus a branch from $0$ to $-\infty$ on the negative real axis. The system is unstable for all $k$.

The root locus passes through a point $P$ (the desired dominant closed-loop pole) only if $P$ satisfies the angle criterion. The gain needed there is then found from the magnitude criterion.

Test steps

1. Mark all open-loop poles ($p_i$) and zeros ($z_j$) of $G(s)H(s)$ and the point $P = -\sigma_d + j\omega_d$.
2. Draw vectors from every pole and zero to $P$ and measure their angles from the positive real axis.
3. Compute

$$\phi = \sum \angle(P - z_j) - \sum \angle(P - p_i)$$

4. If $\phi = \pm 180^\circ(2q+1)$, then P lies on the root locus. If not, P cannot be a closed-loop pole for any $K$. The difference $\phi_c = \pm 180^\circ - \phi$ is the angle a compensator must supply (the starting point of lead/lag design).
5. If P is on the locus, find the gain:

$$K = \frac{\prod |P - p_i|}{\prod |P - z_j|}$$

6. Check that the other closed-loop poles at this $K$ lie well to the left (about 5 times farther from the $j\omega$-axis) or are nearly cancelled by zeros. Only then is P truly dominant.

Example

$G(s) = \dfrac{K}{s(s+2)}$, $P = -1 + j\sqrt3$ (ζ = 0.5, $\omega_n = 2$):

• $\angle P = 120^\circ$ and $\angle(P + 2) = 60^\circ$.
• $\phi = -(120^\circ + 60^\circ) = -180^\circ$, so P is on the locus.
• $K = 2 \times 2 = 4$.