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Proper vs improper handoff; cellular concept N = 7

Short answer

An improper handoff drops the call because the signal falls below the minimum usable level before transfer. A proper handoff completes between the handoff threshold and minimum level. For N=7, cluster uses i=2, j=1, reuse ratio √21=4.58, giving S/I about 18.7 dB, meeting AMPS requirements.

Compare and contrast between improper handoff situation and proper handoff situation. Explain cellular concept for N = 7.

Improper vs proper handoff

A handoff starts when the signal from the serving base station falls to the handoff threshold Pr,handoff, which is kept above the minimum usable level Pr,min by a margin Δ.

 Signal  BS1                          BS2
   │ \                                /
   │   \ A: Pr,handoff reached     /
   │     \                  ____/
   │.......\ B: Pr,min ....../...........
   │  (improper: call drops at B)
   │  (proper: transferred to BS2 at C, before B)
   └───────────────────────────────── distance
PointImproper handoffProper handoff
What happensSignal falls below Pr,min before transferTransfer completes between Pr,handoff and Pr,min
ResultCall is droppedCall continues smoothly
Typical causesΔ too small, slow MSC processing, no free channel in new cellAdequate Δ, fast processing, free channel available
User experienceDisconnectionNo noticeable break

Too large a Δ is also bad: it gives unnecessary handoffs and loads the MSC.

Cellular concept for N = 7

In the cellular concept, the service area is divided into small hexagonal cells, each served by a low-power base station. The total S channels are divided among N = 7 cells forming a cluster, so each cell gets S/7 channels. The cluster is repeated across the area (frequency reuse).

        ___
    ___/ 2 \___
   / 7 \___/ 3 \
   \___/ 1 \___/
   / 6 \___/ 4 \
   \___/ 5 \___/
       \___/
  • N = 7 is obtained with i = 2, j = 1 (N = i² + ij + j² = 4 + 2 + 1).
  • Reuse ratio Q = √(3×7) = 4.58, so co-channel cells are 4.58R apart.
  • With n = 4 and six interferers, S/I = 4.58⁴/6 ≈ 73.5 = 18.7 dB, which meets the 18 dB needed by AMPS. That is why N = 7 is the classic choice.
  • Frequency reuse factor = 1/7.

✓ Answer checked by Claude Opus 5.5 (high effort)

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