---
title: "TE vs TM mode in circular waveguide"
subject: "RF and Microwave Engineering"
chapter: 4
key: "semester_7/rf_and_microwave_engineering:circular-waveguide-te-vs-tm"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/circular-waveguide-te-vs-tm
site: BCT Engineering
---

# TE vs TM mode in circular waveguide

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

- Asked in: 2071 Magh, 10 marks
- 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-waveguide-te-vs-tm
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-4

## Question

Describe how TE mode is different from TM mode in a circular waveguide.

## Short answer

TE modes have Hz≠0, Ez=0 with boundary condition Jn'(kc a)=0 giving kc = p'nm/a; TM modes have Ez≠0, Hz=0 with Jn(kc a)=0 giving kc = pnm/a. TE11 is the dominant mode (lowest cut-off, λc=3.41a), TM01 is next (λc=2.61a). TE wave impedance exceeds η, TM wave impedance is less than η.

## Answer

In a circular waveguide of radius *a*, both TE and TM modes are solutions of Bessel's equation, but they differ in which longitudinal field exists, which boundary condition applies and therefore which Bessel roots set the cut-off.

#### TE modes (Ez = 0)
```text
 Hz = (A sin nφ + B cos nφ) Jn(kcρ) e^(-jβz)
 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)
```
Boundary condition Eφ = 0 at ρ = a gives **J'n(kc·a) = 0**, kc = p'nm/a.

#### TM modes (Hz = 0)
```text
 Ez = (A sin nφ + B cos nφ) Jn(kcρ) e^(-jβz)
 Eρ = -(jβ/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)
 Eφ = -(jβn/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)
 Hρ =  (jωεn/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)
 Hφ = -(jωε/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)
```
Boundary condition Ez = 0 at ρ = a gives **Jn(kc·a) = 0**, kc = pnm/a.

#### Differences
| Point | TE mode | TM mode |
|---|---|---|
| Longitudinal field | Hz ≠ 0, Ez = 0 | Ez ≠ 0, Hz = 0 |
| Boundary condition | J'n(kc a) = 0 | Jn(kc a) = 0 |
| Cut-off wave number | kc = p'nm/a | kc = pnm/a |
| Cut-off frequency | fc = p'nm c/(2πa) | fc = pnm c/(2πa) |
| Lowest mode | TE11 (p'11 = 1.841), λc = 3.41a | TM01 (p01 = 2.405), λc = 2.61a |
| Dominant? | TE11 is dominant mode of the guide | TM01 is the second mode |
| Wave impedance | Z_TE = η/√(1 − (fc/f)²) > η | Z_TM = η√(1 − (fc/f)²) < η |
| Low-loss mode | TE01: wall loss falls with frequency | No such mode |
| Typical use | TE11 general transmission; TE01 long low-loss links | TM01 rotary joints (circular symmetry) |

#### Mode roots and degeneracy
| Mode | Root | Mode | Root |
|---|---|---|---|
| TE11 | 1.841 | TM01 | 2.405 |
| TE21 | 3.054 | TM11 | 3.832 |
| TE01 | 3.832 | TM21 | 5.135 |

TE01 and TM11 share the root 3.832, so they are **degenerate** (same cut-off). Also, each mode with n ≥ 1 has two polarizations (sin nφ and cos nφ), another form of degeneracy; this makes TE11 polarization sensitive to bends and ellipticity.

#### Example
For a = 1 cm (air): fc(TE11) = 1.841 × 3 × 10⁸/(2π × 0.01) = 8.79 GHz; fc(TM01) = 11.48 GHz. Between 8.79 and 11.48 GHz only TE11 propagates.

## Check yourself

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

1. Which condition sets the cut-off for TE modes in a circular guide?
   A. Jn(kc a) = 0
   B. J'n(kc a) = 0 (from Eφ = 0 at the wall)
   C. Hz = 0 at the wall
   D. Ez = 0 at the axis
   Answer: B. For TE modes Eφ = 0 at ρ = a gives J'n(kc a) = 0 and kc = p'nm/a. (Key point: Boundary condition difference)

2. Which pair is correct for the lowest TE and TM modes of a circular guide?
   A. TE01 and TM11
   B. TE11 (1.841) and TM01 (2.405)
   C. TM01 (1.841) and TE11 (2.405)
   D. TE21 and TM21
   Answer: B. TE11 has p'11 = 1.841 and TM01 has p01 = 2.405. (Key point: Lowest modes)

3. How do the wave impedances compare with η above cut-off?
   A. Z_TE < η and Z_TM > η
   B. Both equal η
   C. Z_TE > η and Z_TM < η
   D. Both are greater than η
   Answer: C. Z_TE = η/√(1 − (fc/f)²) > η and Z_TM = η√(1 − (fc/f)²) < η. (Key point: Wave impedance)

4. Why are TE01 and TM11 degenerate in a circular guide?
   A. They have the same polarization
   B. They share the same root 3.832, so same cut-off
   C. They both have Hz = 0
   D. They are both dominant
   Answer: B. J'0 and J1 share the zero 3.832, giving the same cut-off frequency. (Key point: Degeneracy)

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