---
title: "Root locus criteria; poles 0, −1±j2"
subject: "Control Systems"
chapter: 5
key: "semester_3/control_systems:rl-criteria-poles-0-1j2"
source: https://www.bctengineer.com/q/semester_3/control_systems/rl-criteria-poles-0-1j2
site: BCT Engineering
---

# Root locus criteria; poles 0, −1±j2

Control Systems · Chapter 5: Root Locus Technique

- Asked in: 2081 Baisakh, 2+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-criteria-poles-0-1j2
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-5

## Question

Mention the criteria of Root locus plot. Write open loop transfer function and hence sketch the root locus of unity feedback system whose open loop poles and zeros are as shown in s plane below. [Figure: s-plane with open-loop poles at s = 0 and s = −1 ± j2; no zeros shown]

## Short answer

Root locus uses angle criterion (sum of zero angles minus pole angles = odd multiple of 180°) and magnitude criterion (|GH|=1 to find K). For poles at 0, -1±j2, G(s)=K/[s(s²+2s+5)]. System stable for 0<K<10; at K=10 it oscillates at 2.236 rad/s.

## Answer

#### 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):**

| Row | Col 1 | Col 2 |
|---|---|---|
| $s^3$ | 1 | 5 |
| $s^2$ | 2 | $K$ |
| $s^1$ | $(10-K)/2$ | |
| $s^0$ | $K$ | |

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

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

## Check yourself

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

1. What does the angle criterion say?
   A. Angle is a multiple of 360°
   B. Angle of G(s)H(s) is an odd multiple of 180° at points on the locus
   C. Magnitude is 1
   D. Angle is 90°
   Answer: B. Points on the locus satisfy G(s)H(s) = -1, so its angle is ±(2q+1)180°. (Key point: Angle criterion)

2. What is the magnitude criterion used for?
   A. Finding K at a point already known to be on the locus
   B. Deciding whether a point is on the locus
   C. Finding asymptote angles
   D. Finding the number of branches
   Answer: A. |G H| = 1 gives K = product of pole distances over zero distances. (Key point: Magnitude criterion)

3. Which G(s) matches poles at 0 and -1±j2?
   A. K/[s(s^2+2s+3)]
   B. K/[s(s+1)(s+2)]
   C. K/[s(s^2+2s+5)]
   D. K(s+1)/[s(s^2+4)]
   Answer: C. (s+1)^2+4 = s^2+2s+5. (Key point: Transfer function)

4. Why is there no breakaway point?
   A. K is negative there
   B. The real axis is not on the locus
   C. The roots of dK/ds=0 are complex (-0.667±j1.106)
   D. There are no poles on the real axis
   Answer: C. Complex roots are not on the real axis, so no breakaway occurs. (Key point: Breakaway)

5. Where does the locus cross jω?
   A. ±j5 at K = 5
   B. ±j1 at K = 2
   C. ±j2 at K = 4
   D. ±j2.236 at K = 10
   Answer: D. Routh gives K=10; 2s^2+10=0 gives s=±j√5. (Key point: Stability limit)

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