---
title: "Derivation of S/I expression"
subject: "Wireless Communication"
chapter: 2
key: "semester_7/wireless_communication:signal-interference-ratio-expression"
source: https://www.bctengineer.com/q/semester_7/wireless_communication/signal-interference-ratio-expression
site: BCT Engineering
---

# Derivation of S/I expression

Wireless Communication · Chapter 2: Cellular Mobile Communication Concepts

- Asked in: 2074 Magh, 4 marks
- Video: [Frequency Reuse Factor | Co-Channel Interference & Signal To Interference Ratio](https://www.youtube.com/watch?v=JfIIiecbEg8) — ECIT Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/wireless_communication/signal-interference-ratio-expression
- Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-2

## Question

Obtain the expression S/I = (√(3N))ⁿ / i₀, where symbols have their usual meaning.

## Short answer

S/I = R⁻ⁿ / Σ Dᵢ⁻ⁿ. With all first-tier interferers at equal distance D, this becomes (D/R)ⁿ / i₀. Using the hexagonal co-channel reuse ratio D/R = √(3N) yields S/I = (√(3N))ⁿ / i₀.

## Answer

**Assumptions:** all base stations transmit equal power, path loss exponent n is the same everywhere, the mobile is at the cell edge (distance R from its own base station), and only the first tier of i₀ co-channel cells at distance D is considered.

**Signal-to-interference ratio:**
```text
S / I = S / Σ(i=1..i₀) Iᵢ
```

**Received power falls with distance as d⁻ⁿ** (log-distance path loss):
```text
Pr = P₀ (d/d₀)⁻ⁿ
S  ∝ R⁻ⁿ
Iᵢ ∝ Dᵢ⁻ⁿ
```

**Substituting:**
```text
S / I = R⁻ⁿ / Σ(i=1..i₀) Dᵢ⁻ⁿ
```

**All first-tier interferers at equal distance Dᵢ = D:**
```text
S / I = R⁻ⁿ / (i₀ D⁻ⁿ)
      = (D/R)ⁿ / i₀
```

**Using the co-channel reuse ratio D/R = Q = √(3N):**
```text
S / I = (√(3N))ⁿ / i₀
```

```text
      I       I
        \   /
   I ── [BS]·Mobile ── I     i₀ = 6 first-tier
        /   \              co-channel cells at D
      I       I
```

**Example:** N = 7, n = 4, i₀ = 6:
```text
S/I = (√21)⁴ / 6 = 441 / 6 = 73.5 = 18.66 dB
```
which meets the 18 dB requirement of AMPS. For sectoring, i₀ = 2 (120°) or 1 (60°).

## Check yourself

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

1. The derivation assumes the mobile is:
   A. At the cell centre
   B. At the cell edge, distance R from its base station
   C. At distance D from its base station
   D. Moving at high speed
   Answer: B. Worst case S. (Key point: Assumptions)

2. Received power varies with distance as:
   A. d^n
   B. d^2 always
   C. log d
   D. d^-n
   Answer: D. S ~ R^-n and I ~ D^-n. (Key point: Path loss)

3. With all i0 interferers at distance D:
   A. S/I = (D/R)^n / i0
   B. S/I = (R/D)^n / i0
   C. S/I = i0 (D/R)^n
   D. S/I = (D/R)^n x i0
   Answer: A. Sum of i0 equal terms gives i0 D^-n. (Key point: Equal distances)

4. For N = 7, n = 4, i0 = 6, S/I is:
   A. 441 (26.4 dB)
   B. 12 (10.8 dB)
   C. 73.5 (18.66 dB)
   D. 24 (13.8 dB)
   Answer: C. 441/6 = 73.5. (Key point: Example)

## Related in Wireless Communication

- [Derivation of co-channel reuse ratio](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aderive-cochannel-reuse-ratio) — Wireless Communication, chapter 2
- [Doppler spread and Doppler shift derivation](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Adoppler-spread-shift-derivation) — Wireless Communication, chapter 3

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