Prove Q = √3N; optimal N for 15 dB S/I (n = 4)

Proof that Q = √(3N)

In a hexagonal layout, to reach the nearest co-channel cell you move i cells along one chain of hexagons, turn 60°, and move j cells more (this is how N = i² + ij + j² is defined).

• Distance between centres of two adjacent hexagons = √3·R (R = centre-to-vertex radius).
• Using axes 60° apart and the cosine rule on the triangle with sides i·√3R and j·√3R and the angle between them 120°:

D² = (i√3R)² + (j√3R)² − 2(i√3R)(j√3R)cos120°
   = 3R²(i² + j² + ij)
   = 3R²·N
D  = R√(3N)
Q  = D/R = √(3N)

start --(i cells)--> turn 60° --(j cells)--> co-ch cell
        \___________________ D ______________________/

Optimal N for S/I = 15 dB, n = 4

Required S/I = 15 dB = 10^1.5 = 31.62. Using S/I = Qⁿ / i₀ = (√(3N))⁴ / i₀ = (3N)²/i₀. Valid N: 1, 3, 4, 7, 9, 12, …

a) Omnidirectional antennas (i₀ = 6)

(3N)²/6 ≥ 31.62  →  3N ≥ 13.77  →  N ≥ 4.59
N = 4: Q = 3.464, S/I = 144/6 = 24
       = 13.80 dB < 15 (fails)
N = 7: Q = 4.583, S/I = 441/6 = 73.5
       = 18.66 dB > 15 (ok)

N = 7

b) 120° sectoring (i₀ = 2)

(3N)²/2 ≥ 31.62  →  3N ≥ 7.95  →  N ≥ 2.65
Try N = 3: Q = 3, S/I = 81/2 = 40.5 = 16.07 dB > 15 (ok)

(The exact worst-case geometry, S/I = 1/[Q⁻⁴ + (Q+0.7)⁻⁴], gives 17.52 dB for N = 3, also acceptable.) N = 3

c) 60° sectoring (i₀ = 1)

(3N)² ≥ 31.62  →  3N ≥ 5.62  →  N ≥ 1.87
Try N = 3: S/I = 81 = 19.08 dB > 15 (ok)
(N = 1 gives 9 = 9.54 dB, fails)

N = 3

Effect of trunking efficiency

Sectoring splits each cell's channels among the sectors, so each sector is a smaller trunk and trunking efficiency drops. 60° sectoring gives the same N = 3 as 120° sectoring, but splits channels into 6 groups instead of 3, losing more trunking efficiency for no gain in cluster size.

Answer: N = 7 (omni), N = 3 (120°), N = 3 (60°). Considering trunking efficiency, 120° sectoring with N = 3 is the best choice.