Root locus of K(s²−2s+5)/(s²+1.5s−1)

$$G(s)H(s) = \frac{K(s^2 - 2s + 5)}{s^2 + 1.5s - 1} = \frac{K(s - 1 - j2)(s - 1 + j2)}{(s + 2)(s - 0.5)}$$

Characteristic equation: $(1+K)s^2 + (1.5 - 2K)s + (5K - 1) = 0$.

1. Poles, zeros, branches

• Poles: $-2,\ +0.5$. Zeros: $1 \pm j2$.
• $n = m = 2$, so both branches start at the poles and end at the complex zeros. There are no asymptotes.

2. Real-axis locus

$[-2,\ 0.5]$: one real pole lies to the right of every point in this segment.

3. Breakaway point

$$K = -\frac{s^2 + 1.5s - 1}{s^2 - 2s + 5}, \qquad \frac{dK}{ds} = 0 \;\Rightarrow\; 3.5s^2 - 12s - 5.5 = 0$$

$s = -0.409$ or $s = 3.838$. Only $s = -0.409$ is on the locus. Gain there:

$$K = -\frac{0.1676 - 0.614 - 1}{0.1676 + 0.819 + 5} = 0.242$$

4. Angle at the complex zeros (angle of arrival)

There are no complex poles, so the angle asked for is the arrival angle at the complex zeros. At $1 + j2$: angle from the other zero $1 - j2$ is $90^\circ$; from pole $0.5$: $75.96^\circ$; from pole $-2$: $33.69^\circ$.

$$\phi_a = 180^\circ - 90^\circ + (75.96^\circ + 33.69^\circ) = 199.65^\circ \;(\equiv -160.35^\circ)$$

At $1 - j2$ it is $-199.65^\circ$ ($+160.35^\circ$).

5. Imaginary-axis crossing and stability

For a second-order polynomial, all coefficients must have the same sign:

• $1 + K > 0$: always true for $K > 0$.
• $1.5 - 2K > 0 \Rightarrow K < 0.75$
• $5K - 1 > 0 \Rightarrow K > 0.2$

At $K = 0.75$: $1.75s^2 + 2.75 = 0 \Rightarrow s = \pm j1.254$. At $K = 0.2$: one closed-loop pole is at $s = 0$.

Stable for 0.2 < K < 0.75.

Sketch

                           |
                           |
                           |   .....o
                           |....
                          .X.
                        ...|
                        .  |
                       ..  |
---------x.............*.......x------------------
                       ..  |
                        .  |
                        ...|
                          .X.
                           |....
                           |   .....o
                           |
                           |
        -2           -0.41          1
x pole (-2, 0.5)   o zero (1 +- j2)
* breakaway (-0.41)   X jw crossing (+-j1.25)

The poles at −2 and 0.5 meet at −0.41, break away into a loop, cross the $j\omega$-axis at $\pm j1.254$ ($K = 0.75$), and end at the zeros $1 \pm j2$.

6. K for poles at $-0.35 \pm j0.6$

Distances from $s_1 = -0.35 + j0.6$: to pole $0.5$: 1.040; to pole $-2$: 1.756; to zero $1+j2$: 1.945; to zero $1-j2$: 2.930. The angle sum is $\approx -181^\circ$, so the point is (very nearly) on the locus.

$$K = \frac{1.040 \times 1.756}{1.945 \times 2.930} = 0.32$$

Check: with $K = 0.32$ the roots are $-0.325 \pm j0.592 \approx -0.35 \pm j0.6$.

Answer: Breakaway at −0.409 (K = 0.242). Arrival angle is ±199.65° at $1 \pm j2$. The system is stable for 0.2 < K < 0.75. The required gain is K ≈ 0.32.