---
title: "Root locus with poles 0, −1 and zero −3"
subject: "Control Systems"
chapter: 5
key: "semester_3/control_systems:rl-poles-0-1-zero-3"
source: https://www.bctengineer.com/q/semester_3/control_systems/rl-poles-0-1-zero-3
site: BCT Engineering
---

# Root locus with poles 0, −1 and zero −3

Control Systems · Chapter 5: Root Locus Technique

- Asked in: 2081 Baisakh, 8 marks
- Video: [root locus in control system](https://www.youtube.com/watch?v=MIVkzOiFvhI) — Smart Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/rl-poles-0-1-zero-3
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-5

## Question

Sketch the root locus of the system having open loop poles and zero plot as shown below, and the range of parameter K for stability. [Figure: open-loop poles at s = 0 and s = −1; zero at s = −3]

## Short answer

The root locus has a breakaway at -0.551 (K=0.101) and a break-in at -5.449 (K=9.90), with a circular complex segment centered at -3. The system is stable for all K>0. Overdamped response occurs for 0<K<0.101 and K>9.90, underdamped for 0.101<K<9.90.

## Answer

Open-loop poles at $0$ and $-1$ and a zero at $-3$ give

$$
G(s)H(s) = \frac{K(s+3)}{s(s+1)}
$$

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

#### Root locus construction
1. **Branches:** $n = 2$, $m = 1$. One branch ends at $-3$; the other goes to $-\infty$ along $180^\circ$.
2. **Real-axis locus:** $[-1, 0]$ and $(-\infty, -3]$.
3. **Breakaway and break-in points:**
$$
K = -\frac{s^2 + s}{s + 3}, \qquad \frac{dK}{ds} = 0 \Rightarrow s^2 + 6s + 3 = 0
$$
$$
s = -3 \pm \sqrt6 = -0.551,\ -5.449
$$
- Breakaway at $s = -0.551$: $K = \dfrac{0.551 \times 0.449}{2.449} = 0.101$
- Break-in at $s = -5.449$: $K = \dfrac{5.449 \times 4.449}{2.449} = 9.90$
4. **Complex part:** a circle centred at the zero $-3$ with radius $\sqrt{3 \times (3 - 1)} = \sqrt6 = 2.449$ (distance from the zero to the breakaway point). Its highest point is $-3 + j2.449$.
5. **$j\omega$ crossing:** for $K > 0$ both coefficients $(1+K)$ and $3K$ are positive, so there is no crossing.

```text
                                            |
                          ....              |
                    .......  ......         |
                  ...             ....      |
                ...                  ..     |
               ..                     ...   |
              ..                        .   |
              .                         ..  |
..............*............o----------x..*..x-----
              .                         ..  |
              ..                        .   |
               ..                     ...   |
                ...                  ..     |
                  ...             ....      |
                    .......  ......         |
                          ....              |
                                            |
            -5.45         -3           -.55
x pole (0, -1)  o zero (-3)  * breakaway / break-in
```
The poles at 0 and −1 move toward each other, break away at −0.551, travel around the circle, and break in at −5.449. Then one branch goes to −3 and the other to −∞.

#### Range of K for stability
By Routh–Hurwitz for $s^2 + (1+K)s + 3K$: we need $1 + K > 0$ and $3K > 0$, i.e. $K > 0$.

| K | Closed-loop poles |
|---|---|
| $0 < K < 0.101$ | real, between 0 and −1 (overdamped) |
| $0.101 < K < 9.90$ | complex, on the circle (underdamped) |
| $K > 9.90$ | real, one between −3 and −5.45, the other left of −5.45 |

**Answer:** The system is **stable for all K > 0**. The breakaway point is at −0.551 (K = 0.101) and the break-in point at −5.449 (K = 9.90).

## Check yourself

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

1. For poles at 0 and -1 and a zero at -3, which real-axis segments belong to the locus?
   A. [-1, 0] only
   B. [-3, -1] and (-infinity, -3]
   C. [-1, 0] and (-infinity, -3]
   D. (-infinity, -1] only
   Answer: C. Counting poles and zeros to the right: points between -1 and 0 have one (odd); points between -3 and -1 have two (even); points left of -3 have three (odd). (Key point: Real-axis locus)

2. Which equation gives the breakaway and break-in points, and what are they?
   A. s^2 + 6s + 3 = 0, giving -0.551 and -5.449
   B. s^2 + 3s + 1 = 0, giving -0.38 and -2.62
   C. s^2 + 6s + 3 = 0, giving 0.551 and 5.449
   D. s^2 + 4s + 3 = 0, giving -1 and -3
   Answer: A. From K = -(s^2 + s)/(s + 3), dK/ds = 0 gives s^2 + 6s + 3 = 0, so s = -3 +- sqrt6. (Key point: Breakaway and break-in points)

3. What shape does the complex part of the locus take?
   A. A straight vertical line at s = -0.5
   B. A parabola opening to the left
   C. A circle centred at the zero (-3) with radius sqrt6
   D. A circle centred at the origin with radius 3
   Answer: C. The complex part is a circle centred at the zero -3 whose radius equals the distance from the zero to the breakaway point, sqrt(6) = 2.449. (Key point: Complex locus shape)

4. From the characteristic equation s^2 + (1+K)s + 3K = 0, for which K is the system stable?
   A. 0 < K < 9.9
   B. K > 0 only
   C. K > 3
   D. 0 < K < 0.101
   Answer: B. For a second-order polynomial all coefficients must be positive: 1 + K > 0 and 3K > 0 give K > 0, so there is no jw crossing and it is stable for all K > 0. (Key point: Stability range)

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