---
title: "Cut-off and single-mode range, 4×2 cm"
subject: "Electromagnetism"
chapter: 6
key: "semester_3/electromagnetism:waveguide-4x2cm-cutoff-single-mode-range"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/waveguide-4x2cm-cutoff-single-mode-range
site: BCT Engineering
---

# Cut-off and single-mode range, 4×2 cm

Electromagnetism · Chapter 6: Wave guides

- Asked in: 2076 Chaitra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/waveguide-4x2cm-cutoff-single-mode-range
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-6

## Question

A rectangular waveguide has dimension a = 4 cm and b = 2 cm. Determine the cut-off frequency and range of frequencies over which the guide will operate in single mode.

## Short answer

Cut-off of TE_10: f_c = c/(2a) = 3×10⁸/(2×0.04) = 3.75 GHz. Next modes TE_20 and TE_01 have f_c = 7.5 GHz. Single-mode operation range: 3.75 GHz < f < 7.5 GHz.

## Answer

Given: a = 4 cm, b = 2 cm (air-filled, c = 3×10⁸ m/s).

Cut-off frequency of mode (m, n):

f_c(mn) = (c/2) √[(m/a)² + (n/b)²]

#### Cut-off of the dominant mode (TE₁₀)

```text
f_c(10) = c/(2a) = 3×10⁸ / (2 × 0.04) = 3.75×10⁹ Hz
```

**Cut-off frequency = 3.75 GHz** (dominant mode TE₁₀).

#### Other low modes

```text
f_c(20) = c/a   = 3×10⁸/0.04 = 7.5 GHz
f_c(01) = c/2b  = 3×10⁸/(2×0.02) = 7.5 GHz
f_c(11) = (c/2)√[(1/0.04)² + (1/0.02)²] = 1.5×10⁸ × √(625 + 2500) = 8.385 GHz
```

The next higher mode after TE₁₀ is TE₂₀ (and TE₀₁, equal at 7.5 GHz because a = 2b).

#### Single-mode range

Only TE₁₀ propagates when its cut-off is below f and the next cut-off (7.5 GHz) is above f:

**3.75 GHz < f < 7.5 GHz**

(Bandwidth 3.75 GHz.)

## Check yourself

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

1. For a = 4 cm, b = 2 cm, what is the cut-off frequency of the dominant TE10 mode?
   A. 7.5 GHz
   B. 1.875 GHz
   C. 3.75 GHz
   D. 8.385 GHz
   Answer: C. f_c = c/(2a) = 3×10^8/(2×0.04) = 3.75 GHz. (Key point: Dominant mode cut-off)

2. Which modes have the next cut-off after TE10 in this guide, and why do they coincide?
   A. TE11 and TM11 at 8.385 GHz
   B. TE20 and TE01 at 7.5 GHz, since a = 2b
   C. TE02 at 15 GHz
   D. TM11 only at 7.5 GHz
   Answer: B. f_c(20) = c/a = 7.5 GHz and f_c(01) = c/(2b) = 7.5 GHz, equal because a = 2b. (Key point: Next higher mode)

3. Over what range does only the TE10 mode propagate?
   A. 3.75 GHz < f < 7.5 GHz
   B. 0 < f < 3.75 GHz
   C. 3.75 GHz < f < 8.385 GHz
   D. 7.5 GHz < f < 8.385 GHz
   Answer: A. TE10 must be above its own cut-off (3.75 GHz) and f must stay below the next cut-off (7.5 GHz). (Key point: Single-mode range)

4. Which expression gives f_c(11) correctly for this guide?
   A. (c/2)√[(1/0.04) + (1/0.02)]
   B. (c/2)(1/0.04 + 1/0.02)
   C. c/(2×0.04) + c/(2×0.02)
   D. (c/2)√[(1/0.04)^2 + (1/0.02)^2] = 8.385 GHz
   Answer: D. The squares of m/a and n/b are added under the root: 1.5×10^8 × √3125 = 8.385 GHz. (Key point: TE11 cut-off method)

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