- 2080 Bhadra · 4 marks
Derivation of co-channel reuse ratio
Short answer
With hexagonal cells, moving i and j cells along axes and using cosine rule gives D² = 3R²(i²+ij+j²). Cluster size N = i²+ij+j² from area scaling, so D = R√(3N), and co‑channel reuse ratio Q = D/R = √(3N).
Derive the formula for co-channel reuse ratio: Q = √(3N).
The co-channel reuse ratio Q = D/R, where D is the distance between centres of the nearest co-channel cells and R is the cell radius.
Assumptions: hexagonal cells of radius R (centre to vertex); distance between centres of two adjacent cells = √3 R.
Locating a co-channel cell: in a hexagonal layout, the nearest co-channel cell is found by moving i cells along a chain of hexagons, turning 60° anticlockwise, and moving j cells.
* Co-channel cell
/
/ j cells
/ (after a 60° turn)
A *-------------*
i cells
D = straight-line distance from A to the co-channel cell
The two paths have lengths i√3R and j√3R, and the angle between them (inside the triangle) is 120°. By the cosine rule:
D² = (i√3R)² + (j√3R)² − 2(i√3R)(j√3R) cos 120°
= 3R²i² + 3R²j² + 3R²ij (cos 120° = −1/2)
= 3R² (i² + ij + j²)
Relating to cluster size: the area of a hexagon is proportional to the square of the distance between centres of neighbouring hexagons. A cluster of N cells repeats with centre spacing D, while single cells repeat with spacing √3R, so
N = (D / √3R)² = i² + ij + j²
Substituting:
D² = 3R² N
D = R √(3N)
Q = D / R = √(3N)
| N | i, j | Q = √(3N) |
|---|---|---|
| 3 | 1, 1 | 3.00 |
| 4 | 2, 0 | 3.46 |
| 7 | 2, 1 | 4.58 |
| 12 | 2, 2 | 6.00 |
A larger Q means more separation between co-channel cells, so less co-channel interference but fewer channels per cell (less capacity).
✓ Answer checked by Claude Opus 5.5 (high effort)
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