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Derivation of S/I expression

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.

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

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:

S / I = S / Σ(i=1..i₀) Iᵢ

Received power falls with distance as d⁻ⁿ (log-distance path loss):

Pr = P₀ (d/d₀)⁻ⁿ
S  ∝ R⁻ⁿ
Iᵢ ∝ Dᵢ⁻ⁿ

Substituting:

S / I = R⁻ⁿ / Σ(i=1..i₀) Dᵢ⁻ⁿ

All first-tier interferers at equal distance Dᵢ = D:

S / I = R⁻ⁿ / (i₀ D⁻ⁿ)
      = (D/R)ⁿ / i₀

Using the co-channel reuse ratio D/R = Q = √(3N):

S / I = (√(3N))ⁿ / i₀

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

Example: N = 7, n = 4, i₀ = 6:

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°).

✓ Answer checked by Claude Opus 5.5 (high effort)

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