---
title: "Root locus of K(s²−2s+5)/(s²+1.5s−1)"
subject: "Control Systems"
chapter: 5
key: "semester_3/control_systems:rl-k-s2-2s5-s2-15s"
source: https://www.bctengineer.com/q/semester_3/control_systems/rl-k-s2-2s5-s2-15s
site: BCT Engineering
---

# Root locus of K(s²−2s+5)/(s²+1.5s−1)

Control Systems · Chapter 5: Root Locus Technique

- Asked in: 2071 Chaitra, 8 marks
- Video: [Root locus solved example 2](https://www.youtube.com/watch?v=ZjeAY8TGzH0) — S&T Dude
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/rl-k-s2-2s5-s2-15s
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-5

## Question

The open loop transfer function of a control system is given by G(s)H(s) = K(s² − 2s + 5)/(s² + 1.5s − 1). Sketch the root locus for 0 ≤ K ≤ ∞ and determine the breakaway point, the angle of departure from complex poles and the stability conditions. Also find value of K that gives poles at (−0.35 ± j0.6).

## Short answer

Real-axis locus segment: [-2, 0.5]. Breakaway at -0.409 (K=0.242). Arrival angles at complex zeros 1±j2 are ±199.65°. Stable for 0.2<K<0.75; jω crossing at ω=1.254 rad/s (K=0.75). For closed-loop poles -0.35±j0.6, the required gain is K≈0.32.

## Answer

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

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 $s = 0$.

| K | Location of poles |
|---|---|
| $0 < K < 0.2$ | one real pole in RHP, so unstable |
| $0.2 < K < 0.242$ | both real, in LHP |
| $0.242 < K < 0.75$ | complex, in LHP |
| $K = 0.75$ | $\pm j1.254$, marginal |
| $K > 0.75$ | complex, in RHP, so unstable |

**Stable for 0.2 < K < 0.75.**

#### Sketch
```text
                           |
                           |
                           |   .....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**.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With n = m = 2 here, where do the branches end?
   A. At infinity
   B. At the complex zeros 1 ± j2
   C. At the origin
   D. At the poles
   Answer: B. Both branches start at the poles and end at the zeros, so there are no asymptotes. (Key point: Branch ends)

2. Which root of 3.5s^2 - 12s - 5.5 = 0 is the breakaway?
   A. s = 3.838
   B. s = -0.409
   C. s = 0.5
   D. s = -2
   Answer: B. Only -0.409 lies on the real-axis locus [-2, 0.5]; K there is 0.242. (Key point: Valid breakaway)

3. Since there are no complex poles, the angle computed is:
   A. Angle of departure of 0°
   B. Angle of arrival at the complex zeros
   C. Asymptote angle
   D. Nothing is computed
   Answer: B. The arrival angle at 1 ± j2 is ±199.65°. (Key point: Angle asked)

4. For (1+K)s^2 + (1.5-2K)s + (5K-1), the system is stable for:
   A. 0 < K < 0.2
   B. K > 0.75
   C. 0.2 < K < 0.75
   D. 0 < K < 0.75
   Answer: C. All coefficients positive requires K < 0.75 and K > 0.2. (Key point: Stability range)

## Close variants set in other exams

- [Root locus of K/[s(s+6)(s²+4s+13)]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arl-k-s-s6-s2-4s) — Control Systems, chapter 5, asked 6×
- [Root locus of k/[s(s+4)(s²+4s+20)]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arl-k-s-s4-s2-4s) — Control Systems, chapter 5, asked 6×

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