- 2079 Bhadra · 3+5 marks
GoS and optimal N for 15 dB S/I (n = 3)
Short answer
GoS is the blocking probability in a trunked system, measured by Erlang B. With n=3 and required S/I=15 dB: omnidirectional antennas need N=12 (15.56 dB), 120° sectoring N=7 (16.82 dB), and 60° sectoring N=4 (16.19 dB) because sectoring cuts co‑channel interferers, enabling smaller clusters.
Define Grade of Service (GoS) and explain how can it be measured in a 'blocked call cleared' type of trunking system. A cellular service provider decides to use a digital TDMA scheme that can tolerate a signal-to-interference ratio of 15 dB in the worst case. The mobile radio channel provided a propagation path loss exponent of n = 3. Find the optimal value of N for (a) Omni-directional antennas, (b) 120° sectoring, and (c) 60° sectoring. Comment on your results.
Grade of Service (GoS)
GoS is a measure of the ability of a user to access a trunked system during the busiest hour. It is usually expressed as the probability that a call is blocked (or delayed beyond a set time). Example: GoS = 2 % means 2 calls in 100 are blocked during the busy hour.
Measurement in a blocked-calls-cleared system: a blocked call is not queued; the user must try again later. Assuming Poisson call arrivals, exponential holding times and infinite users, the blocking probability is given by the Erlang B formula:
Aᶜ / C!
Pr[block] = ─────────────
Σ (k=0..C) Aᵏ / k!
where A = offered traffic (Erlangs) = U·λ·H and C = number of channels. In practice GoS is measured as the ratio of blocked call attempts to total call attempts in the busy hour, and the Erlang B chart/table gives A for a given C and GoS.
Optimal N for S/I ≥ 15 dB, n = 3
Using S/I = (√(3N))ⁿ / i₀, with i₀ = 6 (omni), 2 (120°), 1 (60°) first-tier interferers. Valid cluster sizes are N = i² + ij + j² = 3, 4, 7, 9, 12, …
(a) Omnidirectional (i₀ = 6)
N = 7 : (√21)³ / 6 = 96.23 / 6 = 16.04 → 12.05 dB < 15 ✗
N = 9 : (√27)³ / 6 = 140.30 / 6 = 23.38 → 13.69 dB < 15 ✗
N = 12: (√36)³ / 6 = 216 / 6 = 36.00 → 15.56 dB ≥ 15 ✓
Optimal N = 12.
(b) 120° sectoring (i₀ = 2)
N = 4 : (√12)³ / 2 = 41.57 / 2 = 20.78 → 13.18 dB < 15 ✗
N = 7 : 96.23 / 2 = 48.12 → 16.82 dB ≥ 15 ✓
Optimal N = 7.
(c) 60° sectoring (i₀ = 1)
N = 3 : (√9)³ = 27.00 → 14.31 dB < 15 ✗
N = 4 : (√12)³ = 41.57 → 16.19 dB ≥ 15 ✓
Optimal N = 4.
| Antenna | i₀ | Optimal N | S/I |
|---|---|---|---|
| Omni | 6 | 12 | 15.56 dB |
| 120° | 2 | 7 | 16.82 dB |
| 60° | 1 | 4 | 16.19 dB |
Comment
Sectoring reduces the number of first-tier co-channel interferers, so the required S/I is met with a smaller cluster. A smaller N gives more channels per cell (S/N) and hence more capacity: going from N = 12 to N = 4 gives three times as many channels per cell. The cost is more antennas, more handoffs between sectors and lower trunking efficiency. With n = 3 (less path loss than n = 4), interference falls off more slowly, so larger clusters are needed than in the usual n = 4 case.
✓ Answer checked by Claude Opus 5.5 (high effort)
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