---
title: "Reuse distance and spectral efficiency"
subject: "Wireless Communication"
chapter: 2
key: "semester_7/wireless_communication:reuse-distance-spectral-efficiency"
source: https://www.bctengineer.com/q/semester_7/wireless_communication/reuse-distance-spectral-efficiency
site: BCT Engineering
---

# Reuse distance and spectral efficiency

Wireless Communication · Chapter 2: Cellular Mobile Communication Concepts

- Asked in: 2074 Bhadra, 4 marks
- Video: [L10-1: Frequency Reuse in Mobile Computing | Types of Frequency Reuse | Reuse Factor Formula](https://www.youtube.com/watch?v=k-CnNXuQ0pU) — Easy Engineering Classes
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/wireless_communication/reuse-distance-spectral-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-2

## Question

Why does minimizing reuse distance maximize spectral efficiency of a cellular system?

## Short answer

Spectral efficiency is proportional to 1/N and 1/D², so smaller reuse distance gives more channels per cell and more frequency reuse over an area. However, co-channel interference limits D. The minimum D that still meets the required S/I gives maximum spectral efficiency.

## Answer

**Spectral efficiency** of a cellular system is the traffic (or number of channels) carried per unit bandwidth per unit area, e.g. channels/MHz/km². It depends directly on how often the same frequency is reused in space.

**Relation between reuse distance and cluster size:**
```text
D = R √(3N)     →     N = (D/R)² / 3
```
A smaller reuse distance D (for a given R) means a smaller cluster size N.

**Effect on capacity:** with S total duplex channels,
```text
Channels per cell  k = S / N
Total capacity     C = M × S = (number of cells / N) × S
```
So C ∝ 1/N ∝ 1/D². If D is minimised:
1. Each cell gets more channels (S/N larger).
2. The same frequency is reused more times over the same area (more clusters M).
3. More users are served per MHz per km², so spectral efficiency is maximised.

**Example:** with S = 420 channels and 70 cells:
| N | D/R | Channels/cell | Total capacity |
|---|---|---|---|
| 12 | 6.00 | 35 | 2450 |
| 7 | 4.58 | 60 | 4200 |
| 4 | 3.46 | 105 | 7350 |

**Limit:** D cannot be reduced indefinitely, because co-channel interference rises:
```text
S/I = (√(3N))ⁿ / i₀ = (D/R)ⁿ / i₀
```
Therefore the **minimum reuse distance that still meets the required S/I** (e.g. 18 dB for AMPS, giving N = 7 for n = 4) gives the maximum spectral efficiency. Sectoring, power control and better modulation allow a smaller D and hence higher efficiency.

## Check yourself

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

1. From D = R sqrt(3N), reducing D (for fixed R) means:
   A. Larger cluster size N
   B. Smaller cluster size N
   C. Smaller cell radius
   D. No change in N
   Answer: B. N = (D/R)^2/3. (Key point: Relation of D and N)

2. Capacity C is proportional to:
   A. N
   B. D^2
   C. N^2
   D. 1/N (so 1/D^2)
   Answer: D. Smaller N gives more channels per cell and more clusters. (Key point: Capacity vs N)

3. With S = 420 and 70 cells, N = 4 gives total capacity of:
   A. 7350
   B. 4200
   C. 2450
   D. 1680
   Answer: A. 105 channels per cell x 70 cells = 7350. (Key point: Example)

4. Why can D not be reduced indefinitely?
   A. The cell radius becomes zero
   B. Channels run out
   C. Co-channel interference rises, so S/I falls below the required value
   D. Hexagons no longer tile
   Answer: C. The minimum D that still meets required S/I gives maximum efficiency. (Key point: Limit)

## Related in Wireless Communication

- [Numerical: co-channel distance for 7-cell reuse](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acochannel-distance-seven-cell) — Wireless Communication, chapter 2
- [CCI/ACI, reuse ratio proof and channels per cell](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acci-aci-reuse-duplex-channels) — Wireless Communication, chapter 2
- [Numerical: cellular vs non-cellular voice channels](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acellular-vs-noncellular-channels) — Wireless Communication, chapter 2
- [Numerical: 32-cell system area and channels](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acellular-32-cells-capacity) — Wireless Communication, chapter 2

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