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

# Circular waveguide TE mode field equations

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

- Asked in: 2079 Bhadra, 10 marks; 2073 Bhadra, 6 marks (asked 2 times)
- 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-waveguide-te-equations
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-4

## Question

Provide the fundamental field and characteristic equations of a circular waveguide for TE mode.

## Short answer

Circular TE modes: Hz ∝ Jn(kcρ)·cos(nφ)e^{-jβz}, with kc = p'_{nm}/a from J'n(kc a)=0. Cut-off frequency fc = p'_{nm}c/(2πa). Dominant TE11 has p'_{11}=1.841, λc≈3.41a. Phase constant β=√(k²−kc²), wave impedance Z_TE=ωμ/β.

## Answer

A circular waveguide is a hollow metal tube of inner radius *a*. Fields are written in cylindrical coordinates (ρ, φ, z) with propagation e^(−jβz). In a **TE mode**, Ez = 0 and Hz ≠ 0.

```text
          .-----.
        /    ρ    \
       |  o---->   |   radius a
        \         /    wave along z
          '-----'
```

#### Wave equation and solution
```text
 ∂²Hz/∂ρ² + (1/ρ)∂Hz/∂ρ + (1/ρ²)∂²Hz/∂φ² + kc² Hz = 0
 kc² = k² - β²
```
Separation of variables, Hz = R(ρ)·Φ(φ), gives Φ = A sin nφ + B cos nφ and Bessel's equation for R, whose finite solution at ρ = 0 is Jn(kc ρ):
```text
 Hz = (A sin nφ + B cos nφ) Jn(kc ρ) e^(-jβz)
```

**Boundary condition:** Eφ = 0 at ρ = a. Since Eφ ∝ J'n(kc ρ):
```text
 J'n(kc a) = 0  ->  kc = p'nm / a
```
where p'nm is the m-th root of J'n(x).

#### TEnm field equations
```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)
 Ez = 0
```

#### Characteristic equations
- **Cut-off frequency:** fc = p'nm·c/(2πa) (air-filled)
- **Cut-off wavelength:** λc = 2πa/p'nm
- **Phase constant:** β = √(k² − (p'nm/a)²)
- **Guide wavelength:** λg = λ0/√(1 − (fc/f)²)
- **Phase and group velocity:** vp = c/√(1 − (fc/f)²), vg = c√(1 − (fc/f)²)
- **Wave impedance:** Z_TE = ωμ/β = η/√(1 − (fc/f)²)

**Roots of J'n(x):**
| Mode | p'nm | λc |
|---|---|---|
| TE11 | 1.841 | 3.41a |
| TE21 | 3.054 | 2.06a |
| TE01 | 3.832 | 1.64a |
| TE31 | 4.201 | 1.50a |

#### Important modes
- **TE11** has the smallest root, so it is the **dominant mode** of the circular waveguide: fc = 1.841c/(2πa), λc = 3.41a. Example: a = 1 cm gives fc = 8.79 GHz.
- **TE01** has circularly symmetric fields and its wall loss falls as frequency rises, so it is used for low-loss long-distance waveguide links.

## Check yourself

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

1. In the TE mode of a circular waveguide,
   A. Ez = 0 and Hz is the generating field
   B. Hz = 0 and Ez is the generating field
   C. Both Ez and Hz are zero
   D. Both Ez and Hz are nonzero
   Answer: A. TE means no electric field along propagation, so all other fields derive from Hz. (Key point: TE mode definition)

2. What boundary condition gives the cut-off values for TE modes, and what results?
   A. Hz = 0 at ρ = a, so Jn(kc a) = 0 and kc = pnm/a
   B. Eφ = 0 at ρ = a, so J'n(kc a) = 0 and kc = p'nm/a
   C. Eρ = 0 at ρ = 0, so kc = 0
   D. Eφ = 0 at ρ = a, so Jn(kc a) = 1
   Answer: B. Eφ is proportional to J'n(kcρ), so its zero at the wall gives roots of the derivative of the Bessel function. (Key point: Boundary condition)

3. Why is only Jn(kcρ) kept in the solution for Hz?
   A. It is the only solution that is zero at the wall
   B. It is the only solution periodic in φ
   C. It is finite at ρ = 0, whereas the second solution is infinite on the axis
   D. It makes Ez nonzero
   Answer: C. The Neumann function diverges at the axis, so the finite Bessel function is chosen. (Key point: Why Bessel Jn)

4. The cut-off frequency of TEnm in an air-filled circular guide of radius a is
   A. fc = pnm·c/(2πa) using roots of Jn
   B. fc = c/(2a)
   C. fc = p'nm·c/(πa^2)
   D. fc = p'nm·c/(2πa)
   Answer: D. kc = p'nm/a and fc = kc c/(2π). (Key point: Cut-off)

## 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×
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