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CDMA merits/demerits; GSM frame efficiency

Short answer

Multiple access allows many users to share limited radio spectrum. CDMA merits include universal reuse, soft capacity, and soft handoff, but suffers from near-far problem and self-jamming. GSM frame efficiency: overhead per slot is 40.25 bits out of 156.25 bits, so useful data fraction is 116/156.25 = 74.24%.

Define multiple access. What are the merits and demerits of Code Division Multiple Access? If a normal GSM time slot consists of 6 trailing bits, 8.25 guard bits, 26 training bits, and 2 traffic bursts of 58 bits of data, find the frame efficiency.

Multiple access

Multiple access is the technique by which many users share a common, limited radio spectrum at the same time without causing serious interference to each other. The shared resource is divided in frequency (FDMA), time (TDMA), code (CDMA) or space (SDMA).

Merits of CDMA

  1. Universal frequency reuse (N = 1): all cells use the same carrier; no frequency planning.
  2. Soft capacity: more users can be added with a gradual drop in quality.
  3. Soft handoff: the mobile connects to two base stations at once; fewer dropped calls.
  4. Multipath advantage: RAKE receiver combines multipath components (diversity).
  5. Voice activity and sectorisation gains directly increase capacity.
  6. Privacy and interference rejection due to PN spreading and processing gain.
  7. Wide bandwidth reduces the effect of frequency-selective fading.

Demerits of CDMA

  1. Near-far problem – needs fast and accurate power control.
  2. Self-jamming – codes not perfectly orthogonal; interference from other users limits capacity.
  3. Complex receivers – code acquisition, tracking and RAKE processing.
  4. Strict timing – IS-95 base stations need GPS synchronisation.
  5. Cell breathing – coverage shrinks as load increases.

GSM frame efficiency

Given: tail (trailing) bits = 6, guard bits = 8.25, training bits = 26, data = 2 × 58 = 116 bits.

Bits per slot   = 6 + 8.25 + 26 + 2×58 = 156.25 bits
Bits per frame  b_T  = 8 slots × 156.25 = 1250 bits
Overhead/slot   = 6 + 8.25 + 26 = 40.25 bits
Overhead/frame  b_OH = 8 × 40.25 = 322 bits

η_f = (1 − b_OH / b_T) × 100%
    = (1 − 322 / 1250) × 100%
    = (1 − 0.2576) × 100%
    = 74.24 %

(Equivalently, useful data per slot = 116/156.25 = 0.7424.)

Answer: Frame efficiency η = 74.24 %

✓ Answer checked by Claude Opus 5.5 (high effort)

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