---
title: "Circular waveguide TE equations and degenerate modes"
subject: "RF and Microwave Engineering"
chapter: 4
key: "semester_7/rf_and_microwave_engineering:circular-te-rectangular-modes"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/circular-te-rectangular-modes
site: BCT Engineering
---

# Circular waveguide TE equations and degenerate modes

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

- Asked in: 2082 Baisakh, 7+3 marks
- Video: [Circular wave Guide | TE mode analysis | Microwave Engineering | Lec-34](https://www.youtube.com/watch?v=OBw-sXJ0fSM) — Education 4u
- 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/circular-te-rectangular-modes
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-4

## Question

Write the electromagnetic field equations of a circular waveguide operating at TE Mode. Explain degenerative and dominant modes of a rectangular waveguide with suitable examples.

## Short answer

For circular TE modes, Ez=0 and Hz=(A sin nφ + B cos nφ) Jn(kcρ)e^-jβz; cutoff is fc=p'nm c/(2πa), with TE11 dominant (p'11=1.841). In rectangular guides, TE10 is dominant with fc=c/(2a); degenerate modes share cutoff, for example TE11 and TM11, and TE10/TE01 in a square guide.

## Answer

#### Field equations of a circular waveguide in TE mode
For a circular guide of radius *a* (cylindrical coordinates ρ, φ, z) with Ez = 0, Hz satisfies Bessel's equation and the solution finite at the axis is
```text
 Hz = (A sin nφ + B cos nφ) Jn(kcρ) e^(-jβz)
```
The transverse fields from Maxwell's equations are:
```text
 Eρ = -(jωμn/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)
 Eφ =  (jωμ/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)
 Hρ = -(jβ/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)
 Hφ = -(jβn/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)
 Ez = 0
```
Boundary condition Eφ(ρ = a) = 0 gives J'n(kc a) = 0, so kc = p'nm/a.

**Characteristic equations:**
- fc = p'nm·c/(2πa), λc = 2πa/p'nm
- β = √(k² − kc²), λg = λ0/√(1 − (fc/f)²)
- Z_TE = η/√(1 − (fc/f)²)
- Dominant mode TE11: p'11 = 1.841, λc = 3.41a.

#### Dominant and degenerate modes of a rectangular waveguide
**Dominant mode:** the mode with the **lowest cut-off frequency** (longest cut-off wavelength). For a > b:
```text
 fc,mn = (c/2) √((m/a)² + (n/b)²)
 TE10: fc = c/(2a)  (lowest)
```
TE10 is the dominant mode. Example: a = 2.286 cm, b = 1.016 cm gives fc,TE10 = 6.56 GHz, next mode TE20 at 13.12 GHz. So between 6.56 and 13.12 GHz only TE10 propagates, giving single-mode operation (X-band, 8.2–12.4 GHz).

**Degenerate modes:** **different modes (different field patterns) that have the same cut-off frequency**. In a rectangular guide, **TEmn and TMmn** (m, n ≥ 1) always share the same fc, e.g. TE11 and TM11, TE21 and TM21. In a square guide (a = b), TE10 and TE01 are also degenerate.

Example: in the X-band guide above, TE11 and TM11 both have fc = (3×10⁸/2)√((1/0.02286)² + (1/0.01016)²) ≈ 16.16 GHz.

Degenerate modes are undesirable because energy can transfer between them at discontinuities, causing distortion; guides are therefore operated in the dominant-mode band.

## Check yourself

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

1. For TE modes in a circular guide, the cut-off condition is
   A. J'n(kc a) = 0, giving kc = p'nm/a
   B. Jn(kc a) = 0
   C. Hz = 0 at ρ = 0
   D. Eρ = 0 everywhere
   Answer: A. Tangential Eφ vanishes at the wall and Eφ is proportional to J'n. (Key point: Boundary condition)

2. The dominant mode in a rectangular waveguide (a > b) is
   A. TE01, with fc = c/(2b)
   B. TE10, with fc = c/(2a)
   C. TE11, with fc = (c/2) sqrt(1/a^2 + 1/b^2)
   D. TM11
   Answer: B. TE10 has the lowest cut-off frequency because a is the larger dimension. (Key point: Dominant mode rectangular)

3. Degenerate modes are
   A. modes with the lowest cut-off frequency
   B. modes that cannot propagate
   C. different modes with the same cut-off frequency, such as TE11 and TM11
   D. the same mode at different frequencies
   Answer: C. TEmn and TMmn (m, n ≥ 1) share fc in a rectangular guide. (Key point: Degenerate modes)

4. For a = 2.286 cm, b = 1.016 cm, only TE10 propagates between
   A. 6.56 GHz and 16.15 GHz
   B. 13.12 GHz and 16.15 GHz
   C. 0 and 6.56 GHz
   D. 6.56 GHz and 13.12 GHz
   Answer: D. TE10 cut-off is 6.56 GHz and the next mode, TE20, begins at 13.12 GHz. (Key point: Single-mode band)

## 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×
- [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×
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- [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×
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