---
title: "Proof that TE10 is the dominant mode"
subject: "RF and Microwave Engineering"
chapter: 4
key: "semester_7/rf_and_microwave_engineering:te10-dominant-mode-proof"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/te10-dominant-mode-proof
site: BCT Engineering
---

# Proof that TE10 is the dominant mode

RF and Microwave Engineering · Chapter 4: RF and Microwave Components and Devices

- Asked in: 2081 Bhadra, 3 marks
- Video: [Dominant Mode of Waveguide in Microwave Engineering by Engineering Funda](https://www.youtube.com/watch?v=w4LtkYdkNdY) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/te10-dominant-mode-proof
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-4

## Question

Show that the dominant mode in rectangular waveguides operating in TE mode is TE10, when b>a.

## Short answer

For a>b, cutoff is fc,mn=(c/2)sqrt((m/a)^2+(n/b)^2). TE10 gives c/2a, TE01 gives c/2b, and since a>b, TE10 has the lowest cutoff; higher modes and TM11 have larger fc. Hence TE10 is dominant, with λc=2a.

## Answer

The dominant mode is the mode with the **lowest cut-off frequency**. The condition "b > a" in the question is taken as a misprint for the usual convention **a > b**, where *a* is the wider (broad) wall and the index *m* counts half-wave variations along it.

For an air-filled rectangular guide, the cut-off frequency of the TEmn mode is
```text
 fc,mn = (c/2) √((m/a)² + (n/b)²)
```
For TE modes m and n can be 0, 1, 2 … but not both zero (TE00 has no field).

The candidates for the lowest cut-off are:
```text
 TE10: fc = c/(2a)
 TE01: fc = c/(2b)
 TE11: fc = (c/2)√(1/a² + 1/b²)
```
Since a > b, 1/a < 1/b, so c/(2a) < c/(2b) < (c/2)√(1/a² + 1/b²). Any higher m or n only increases fc, and TM modes start from TM11, which also has a higher fc. Hence **TE10 has the lowest cut-off frequency and is the dominant mode**, with λc = 2a.

(If the indices were assigned the other way, i.e. the wider wall were along *y*, the same argument would make TE01 the dominant mode; physically it is always the mode with one half-wave across the **wider** wall.)

## Check yourself

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

1. The dominant mode is the mode with
   A. the lowest cut-off frequency
   B. the highest cut-off frequency
   C. the highest attenuation
   D. the shortest guide wavelength
   Answer: A. It is the first mode to propagate as frequency rises. (Key point: Dominant mode definition)

2. For a > b, which ordering is correct?
   A. c/(2b) < c/(2a) < (c/2) sqrt(1/a^2 + 1/b^2)
   B. c/(2a) < c/(2b) < (c/2) sqrt(1/a^2 + 1/b^2)
   C. (c/2) sqrt(1/a^2 + 1/b^2) < c/(2a) < c/(2b)
   D. c/(2a) = c/(2b)
   Answer: B. 1/a < 1/b, and TE11 adds both terms, so TE10 is lowest. (Key point: Comparison)

3. Why can TM modes and TE00 not be the dominant mode?
   A. TM modes have lower cut-off but are attenuated
   B. TE00 has the highest cut-off
   C. TE00 has no field, and TM modes start at TM11, which has higher fc
   D. TM modes require m = 0
   Answer: C. TM11 is the lowest TM mode and its cut-off exceeds that of TE10. (Key point: Why not TE00 or TM)

## Related in RF and Microwave Engineering

- [Rectangular waveguide TE mode field equations](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Arectangular-waveguide-te-equations) — RF and Microwave Engineering, chapter 4, asked 11×
- [Rectangular TM equations; TE10 vs TE20](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Arectangular-tm-te10-te20) — RF and Microwave Engineering, chapter 4
- [Dominant mode and 2.254 cm waveguide at 6 GHz](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Awaveguide-dominant-mode-2-254cm) — RF and Microwave Engineering, chapter 4
- [Circular waveguide TM field strength and dominant mode](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Acircular-waveguide-tm-field) — RF and Microwave Engineering, chapter 4
- [Magic tee, degenerate modes, dominant mode for a = 3b](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Amagic-tee-modes-a-3b) — RF and Microwave Engineering, chapter 4
- [Short note: Dominant mode in waveguide](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Adominant-mode-waveguide-note) — RF and Microwave Engineering, chapter 4

## Related in other subjects

- [Define cutoff frequency, dominant mode](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Acutoff-frequency-dominant-mode-define) — Electromagnetism, chapter 6
- [Cut-off of TM11 and dominant mode](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Awaveguide-tm11-cutoff-dielectric-1-07) — Electromagnetism, chapter 6
- [Cut-off of dominant and TE11, 2.5×1.2 cm](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Awaveguide-2-5x1-2-cutoff-te11) — Electromagnetism, chapter 6
- [Short note: TE and TM modes](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Ate-tm-modes-rectangular-waveguide) — Electromagnetism, chapter 6, asked 2×