---
title: "Rectangular waveguide cut-off and phase velocity"
subject: "RF and Microwave Engineering"
chapter: 4
key: "semester_7/rf_and_microwave_engineering:rectangular-waveguide-assumed-dimensions"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/rectangular-waveguide-assumed-dimensions
site: BCT Engineering
---

# Rectangular waveguide cut-off and phase velocity

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

- Asked in: 2077 Chaitra, 6 marks
- Video: [Impedance Rectangular Wave Guide Numerical Example](https://www.youtube.com/watch?v=E2ZiKZMSZWM) — TutorialsPoint
- 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-waveguide-assumed-dimensions
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-4

## Question

For a rectangular waveguide, with suitably assumed breadth, width and frequency (for dominant mode), determine cut-off frequency, phase velocity, propagation wavelength in the waveguide and the characteristic impedance.

## Short answer

For WR-90 air-filled waveguide at 10 GHz in TE10 mode: cut-off frequency 6.562 GHz, phase velocity 3.976e8 m/s, guide wavelength 3.976 cm, characteristic impedance about 500 ohms.

## Answer

**Assumption:** a standard **WR-90 (X-band)** air-filled waveguide, a = 2.286 cm (breadth), b = 1.016 cm (height), operating at **f = 10 GHz** in the dominant **TE10** mode. Constants: c = 3 × 10⁸ m/s, η0 = 120π ≈ 377 Ω.

**Cut-off frequency (TE10):**
```text
 fc = c/(2a) = (3 × 10⁸)/(2 × 0.02286) = 6.562 GHz
```
Next mode TE20 (13.12 GHz) and TE01 (c/2b = 14.76 GHz) are above 10 GHz, so only TE10 propagates.

**Common factor:**
```text
 √(1 - (fc/f)²) = √(1 - (6.562/10)²)
               = √(1 - 0.4306) = 0.7546
 λ0 = c/f = 3 cm
```
**Phase velocity:**
```text
 vp = c/0.7546 = 3.976 × 10⁸ m/s
```
**Guide (propagation) wavelength:**
```text
 λg = λ0/0.7546 = 3/0.7546 = 3.976 cm
```
**Characteristic (wave) impedance:**
```text
 Z_TE = η0/0.7546 = 377/0.7546 = 499.6 Ω
```

| Quantity | Value |
|---|---|
| Cut-off frequency fc | 6.562 GHz |
| Phase velocity vp | 3.976 × 10⁸ m/s |
| Guide wavelength λg | 3.976 cm |
| Wave impedance Z_TE | ≈ 500 Ω |

**Answer: fc = 6.562 GHz, vp = 3.976 × 10⁸ m/s, λg = 3.976 cm, Z_TE ≈ 500 Ω** (for a = 2.286 cm, b = 1.016 cm, f = 10 GHz). Group velocity for reference: vg = c × 0.7546 = 2.264 × 10⁸ m/s.

## Check yourself

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

1. For a = 2.286 cm, the TE10 cut-off is
   A. 13.12 GHz
   B. 3.28 GHz
   C. 10 GHz
   D. 6.562 GHz
   Answer: D. fc = c/(2a) = 3×10^8/0.04572 = 6.562 GHz. (Key point: Cut-off)

2. At 10 GHz the factor sqrt(1-(fc/f)^2) is
   A. 0.7546
   B. 0.4306
   C. 1.325
   D. 0.6562
   Answer: A. 1-(0.6562)^2 = 0.5694, square root 0.7546. (Key point: Common factor)

3. The guide wavelength λg is
   A. 3 × 0.7546 = 2.26 cm
   B. λ0/0.7546 ≈ 3.976 cm, longer than λ0 = 3 cm
   C. 3 cm
   D. 6.8 cm
   Answer: B. λg = λ0/sqrt(1-(fc/f)^2) is always larger than λ0. (Key point: Guide wavelength)

4. The TE10 wave impedance is
   A. 377×0.7546 ≈ 285 Ω
   B. 377 Ω
   C. η0/0.7546 ≈ 500 Ω
   D. 50 Ω
   Answer: C. Z_TE = η0/sqrt(1-(fc/f)^2). (Key point: Wave impedance)

## Related in RF and Microwave Engineering

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## Related in other subjects

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