---
title: "Air-filled 3 cm waveguide at 6 GHz; coupler parameters"
subject: "RF and Microwave Engineering"
chapter: 4
key: "semester_7/rf_and_microwave_engineering:waveguide-3cm-6ghz-directional-coupler"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/waveguide-3cm-6ghz-directional-coupler
site: BCT Engineering
---

# Air-filled 3 cm waveguide at 6 GHz; coupler parameters

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

- Asked in: 2080 Baisakh, 6+4 marks
- Video: [Important Problem solved in Rectangular Waveguide](https://www.youtube.com/watch?v=kVYDgThPYdM) — Padmasri Naban
- 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/waveguide-3cm-6ghz-directional-coupler
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-4

## Question

For an air-filled rectangular waveguide with a width of 3 cm and a desired frequency of operation of 6 GHz (for dominant mode), determine cut-off frequency, cut-off wavelength, group velocity, phase velocity, propagation wavelength in the waveguide and the characteristic impedance. Explain the four basic parameters used to describe the performance of a directional coupler.

## Short answer

For a=3 cm, TE10 gives λc=6 cm and fc=5 GHz; at 6 GHz, λg=9.045 cm, vp=5.427e8 m/s, vg=1.658e8 m/s, and Z_TE≈682 Ω. Directional coupler performance is described by coupling, directivity, isolation, and insertion loss.

## Answer

#### Part 1: Air-filled waveguide, a = 3 cm, f = 6 GHz, TE10 mode
Constants: c = 3 × 10⁸ m/s, η0 = 120π ≈ 377 Ω.

**Cut-off wavelength:**
```text
 λc = 2a = 2 × 3 = 6 cm
```
**Cut-off frequency:**
```text
 fc = c/λc = (3 × 10⁸)/(0.06) = 5 × 10⁹ Hz = 5 GHz
```
Since f = 6 GHz > 5 GHz, TE10 propagates.

**Common factor:**
```text
 λ0 = c/f = 3 × 10⁸ / 6 × 10⁹ = 5 cm
 √(1 - (fc/f)²) = √(1 - (5/6)²) = √0.3056 = 0.5528
```
**Guide (propagation) wavelength:**
```text
 λg = λ0/0.5528 = 5/0.5528 = 9.045 cm
```
**Phase velocity:**
```text
 vp = c/0.5528 = 5.427 × 10⁸ m/s
```
**Group velocity:**
```text
 vg = c × 0.5528 = 1.658 × 10⁸ m/s
 (check: vp·vg = 9 × 10¹⁶ = c²)
```
**Characteristic (wave) impedance:**
```text
 Z_TE = η0/0.5528 = 377/0.5528 = 682 Ω
```

| Quantity | Value |
|---|---|
| fc | 5 GHz |
| λc | 6 cm |
| λg | 9.045 cm |
| vp | 5.427 × 10⁸ m/s |
| vg | 1.658 × 10⁸ m/s |
| Z_TE | 682 Ω |

**Answer: fc = 5 GHz, λc = 6 cm, vg = 1.658 × 10⁸ m/s, vp = 5.427 × 10⁸ m/s, λg = 9.045 cm, Z_TE ≈ 682 Ω.**

#### Part 2: Four basic parameters of a directional coupler
Ports: 1 = input, 2 = through, 3 = coupled, 4 = isolated.
```text
 (1) Input =================== Through (2)
             coupling region
 (4) Isolated ---------------- Coupled (3)
```
1. **Coupling factor:** C = 10 log(P1/P3) dB = −20 log|S31|. The fraction of input power sent to the coupled port (e.g. 10 dB coupler sends 10%).
2. **Directivity:** D = 10 log(P3/P4) dB = 20 log(|S31|/|S41|). How well the coupler separates forward and reverse waves; ideally infinite, typically 30–40 dB.
3. **Isolation:** I = 10 log(P1/P4) dB = −20 log|S41| = C + D. Power leaking to the isolated port.
4. **Insertion loss:** IL = 10 log(P1/P2) dB = −20 log|S21|. Power lost in the main line, including the coupled power.

## Check yourself

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

1. For a = 3 cm TE10, the cut-off frequency is
   A. 10 GHz
   B. 5 GHz (λc = 2a = 6 cm)
   C. 2.5 GHz
   D. 6 GHz
   Answer: B. fc = c/(2a) = 3×10^8/0.06 = 5 GHz. (Key point: Cut-off)

2. At 6 GHz, with sqrt(1-(5/6)^2) = 0.5528, the phase velocity is
   A. c×0.5528 ≈ 1.66×10^8 m/s
   B. 3×10^8 m/s
   C. c/0.5528 ≈ 5.43×10^8 m/s, greater than c
   D. c/0.5528 ≈ 1.66×10^8 m/s
   Answer: C. Phase velocity exceeds c; group velocity is c×0.5528, and vp·vg = c^2. (Key point: Phase velocity)

3. The TE10 wave impedance here is
   A. η0×0.5528 ≈ 208 Ω
   B. 377 Ω
   C. 120 Ω
   D. η0/0.5528 ≈ 682 Ω
   Answer: D. Z_TE = η0/sqrt(1-(fc/f)^2), larger than 377 Ω. (Key point: Wave impedance)

4. Isolation of a directional coupler equals
   A. coupling + directivity (I = C + D)
   B. coupling - directivity
   C. insertion loss + coupling
   D. directivity only
   Answer: A. I = 10 log(P1/P4) = C + D. (Key point: Coupler parameters)

## Related in RF and Microwave Engineering

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