---
title: "Rectangular TM equations; TE10 vs TE20"
subject: "RF and Microwave Engineering"
chapter: 4
key: "semester_7/rf_and_microwave_engineering:rectangular-tm-te10-te20"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/rectangular-tm-te10-te20
site: BCT Engineering
---

# Rectangular TM equations; TE10 vs TE20

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

- Asked in: 2071 Bhadra, 8+2 marks
- Video: [Rectangular Waveguide (Basics, Modes, Cut-Off Frequency, Wave Impedance, Phase & Group Velocity)](https://www.youtube.com/watch?v=Y3szgmhuXeg) — 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/rectangular-tm-te10-te20
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-4

## Question

Describe field equations and other related parameters of a rectangular waveguide in TM mode. Compare TE10 and TE20 in terms of cut-off frequency and dominant mode.

## Short answer

TM modes in a rectangular guide have Ez = E0 sin(mπx/a) sin(nπy/b) e^{-jβz} with m,n ≥ 1; the lowest is TM11. TE10 has cutoff fc=c/(2a) and is the dominant mode; TE20 has fc=c/a, twice TE10's cutoff, making it a higher-order mode.

## Answer

#### Field equations of a rectangular waveguide in TM mode
For a guide of inner dimensions a × b (a > b), TM modes have Hz = 0. Solving ∇t²Ez + kc²Ez = 0 with Ez = 0 on all four walls gives kx = mπ/a, ky = nπ/b (m, n ≥ 1):
```text
 Ez = E0 sin(mπx/a) sin(nπy/b) e^(-jβz)
 Ex = -(jβ/kc²)(mπ/a) E0 cos(mπx/a) sin(nπy/b) e^(-jβz)
 Ey = -(jβ/kc²)(nπ/b) E0 sin(mπx/a) cos(nπy/b) e^(-jβz)
 Hx =  (jωε/kc²)(nπ/b) E0 sin(mπx/a) cos(nπy/b) e^(-jβz)
 Hy = -(jωε/kc²)(mπ/a) E0 cos(mπx/a) sin(nπy/b) e^(-jβz)
 Hz = 0
```
**Related parameters:**
- kc² = (mπ/a)² + (nπ/b)²; β = √(ω²με − kc²)
- Cut-off frequency: fc = (c/2)√((m/a)² + (n/b)²) (air)
- Cut-off wavelength: λc = 2/√((m/a)² + (n/b)²)
- Guide wavelength: λg = λ0/√(1 − (fc/f)²)
- Phase velocity vp = c/√(1 − (fc/f)²); group velocity vg = c√(1 − (fc/f)²)
- Wave impedance: Z_TM = η√(1 − (fc/f)²), always less than η
- Lowest TM mode: **TM11** (TM10 and TM01 do not exist because Ez would vanish everywhere).

#### TE10 vs TE20
```text
 TE10: fc = c/(2a),  λc = 2a
 TE20: fc = 2c/(2a) = c/a,  λc = a
```
| Point | TE10 | TE20 |
|---|---|---|
| Cut-off frequency | c/(2a) | c/a = 2 × fc(TE10) |
| Half-wave variations across a | One | Two |
| Dominant? | Yes (lowest fc for a > b) | No, a higher-order mode |
| X-band guide (a = 2.286 cm) | 6.56 GHz | 13.12 GHz |

TE10 is the **dominant mode**; the band between fc(TE10) and fc(TE20) (if a ≥ 2b) is the single-mode operating range of the guide.

## Check yourself

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

1. In rectangular TM modes, which field component is zero?
   A. Ez
   B. Hz
   C. Ex
   D. Ey
   Answer: B. TM modes have Hz = 0 and Ez ≠ 0. (Key point: TM longitudinal field)

2. The TM wave impedance Z_TM compared with η is:
   A. Always greater than η
   B. Equal to η
   C. η√(1 − (fc/f)²), always less than η
   D. Infinite at all frequencies
   Answer: C. Z_TM = η√(1 − (fc/f)²) which is below η above cut-off. (Key point: TM impedance)

3. How does fc of TE20 compare with that of TE10?
   A. Equal
   B. Half
   C. Four times
   D. Twice: c/a versus c/(2a)
   Answer: D. fc(TE20) = c/a = 2 × fc(TE10). (Key point: TE10 versus TE20 cut-off)

4. Which of TE10 and TE20 is dominant, and why?
   A. TE20, as it has more variations
   B. TE10, since it has the lowest cut-off for a > b
   C. Neither
   D. Both equally
   Answer: B. TE10 has the lowest cut-off frequency, so it propagates first. (Key point: Dominant mode)

## Related in RF and Microwave Engineering

- [Rectangular waveguide TM mode field equations](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Arectangular-waveguide-tm-equations) — RF and Microwave Engineering, chapter 4, asked 11×
- [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
- [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 waveguide cut-off and phase velocity](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Arectangular-waveguide-assumed-dimensions) — RF and Microwave Engineering, chapter 4
- [Proof that TE10 is the dominant mode](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Ate10-dominant-mode-proof) — RF and Microwave Engineering, chapter 4
- [Circular waveguide TE mode field equations](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Acircular-waveguide-te-equations) — RF and Microwave Engineering, chapter 4, asked 2×

## Related in other subjects

- [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
- [Dominant cut-off, dielectric-filled guide](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Awaveguide-dielectric-filled-dominant-cutoff) — Electromagnetism, chapter 6
- [Define TM mode; TM11 cut-off 5×3 cm](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Atm-mode-define-cutoff-5x3-guide) — 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