---
title: "Angle and magnitude criteria for k/[(s+1)(s+3)]"
subject: "Control Systems"
chapter: 5
key: "semester_3/control_systems:rl-angle-mag-k-s1-s3"
source: https://www.bctengineer.com/q/semester_3/control_systems/rl-angle-mag-k-s1-s3
site: BCT Engineering
---

# Angle and magnitude criteria for k/[(s+1)(s+3)]

Control Systems · Chapter 5: Root Locus Technique

- Asked in: 2071 Chaitra, 4 marks
- Video: [Numerical : Angle Condition of Root Locus](https://www.youtube.com/watch?v=ysRHvpp30WQ) — Learning Electronics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/rl-angle-mag-k-s1-s3
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-5

## Question

For the unity feedback system with open loop transfer function (OLTF) G(s) = k/[(s+1)(s+3)], use angle criteria to check whether the root locus passes from point sd = −2 + j3.5. If yes, use magnitude criteria to select the appropriate value of gain parameter.

## Short answer

Yes, the point lies on the locus because the angle criterion gives -180° (sd is on the vertical line σ=-2). The required gain from the magnitude criterion is k = |sd+1|·|sd+3| = 13.25.

## Answer

$$
G(s) = \frac{k}{(s+1)(s+3)}, \qquad s_d = -2 + j3.5
$$

#### Angle criterion
The locus passes through $s_d$ only if $\angle G(s_d) = \pm 180^\circ(2q+1)$.

| Vector | Value | Angle |
|---|---|---|
| $s_d + 1$ | $-1 + j3.5$ | $180^\circ - \tan^{-1}(3.5) = 105.95^\circ$ |
| $s_d + 3$ | $1 + j3.5$ | $\tan^{-1}(3.5) = 74.05^\circ$ |

$$
\angle G(s_d) = -(105.95^\circ + 74.05^\circ) = -180^\circ
$$

The angle criterion is satisfied, so **the root locus passes through $-2 + j3.5$**. This is expected: for two real poles, the complex part of the locus is the vertical line through their midpoint, $\sigma = -2$.

#### Magnitude criterion
$$
\begin{aligned}
k &= |s_d + 1|\,|s_d + 3| \\
&= \sqrt{1 + 3.5^2} \times \sqrt{1 + 3.5^2} \\
&= 3.640 \times 3.640 = 13.25
\end{aligned}
$$

#### Check
With $k = 13.25$: $s^2 + 4s + 3 + 13.25 = s^2 + 4s + 16.25 = 0$, so $s = -2 \pm j3.5$. ✓

**Answer:** Yes, the point lies on the root locus. The required gain is **k = 13.25**.

## Check yourself

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

1. At s_d = -2 + j3.5, the angle of G(s_d) is:
   A. -90°
   B. -180°
   C. -135°
   D. 0°
   Answer: B. -(105.95° + 74.05°) = -180°, so s_d is on the locus. (Key point: Angle test)

2. For two real poles, the complex part of the locus is:
   A. A circle around the origin
   B. The real axis
   C. A vertical line through the midpoint of the poles
   D. A line at 45°
   Answer: C. The line sigma = -2 (midpoint of -1 and -3) is the complex part. (Key point: Why it is on the locus)

3. What gain k is selected by the magnitude criterion?
   A. 3.64
   B. 7.28
   C. 13.25
   D. 16.25
   Answer: C. k = 3.640 x 3.640 = 13.25. (Key point: Gain)

## Related in Control Systems

- [Checking whether root locus passes through point P](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arl-point-p-dominant) — Control Systems, chapter 5
- [Root locus criteria; poles 0, −1±j2](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arl-criteria-poles-0-1j2) — Control Systems, chapter 5
- [Root locus: poles −2±j3, zero at origin](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arl-poles-2j3-zero-0-critical) — Control Systems, chapter 5
- [Root locus with poles 0, −6, −2 ± j3](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arl-poles-0-6-2j3) — Control Systems, chapter 5
- [Transfer function with G3 feedforward, H1 and unity loop](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1g2g3-h1-unity-2070o) — Control Systems, chapter 3
- [Criteria for a point to lie on root locus](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arl-point-criteria) — Control Systems, chapter 5

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