Wireless 101
M10 · L01
Module 10 · Lesson 1

One Medium, Many Users

Every budget in Module 9 described one link, alone in its band. That link does not exist. Three ways exist to cut one shared resource — and only three.

01 / 11
Wireless 101
M10 · L01
The rectangle

Frequency, Time, or Neither

Three cuts of one resource
\text{FDMA: split } B \quad \text{TDMA: split } T \quad \text{CDMA: share both}

Two transmissions that overlap in both axes collide. A multiple access scheme is a rule for cutting the rectangle. Space is the fourth axis — sectors (M7-L3), MIMO (M10-L3).

02 / 11
Wireless 101
M10 · L01
FDMA — worked: AMPS

A Private Strip, Forever

  • 25 MHz at 30 kHz spacing: 25e6/30e3 = 833 channels
  • Per 12.5 MHz operator, 10 kHz edge guards: (12.5e6 − 20e3)/30e3 = 416
  • Guard bands cost 40 kHz of 25 MHz = 0.16%, not the great waste
  • 30 kHz is Carson-full: 2(12 + 3) = 30 kHz exactly (M4-L2)
  • With N = 7 reuse (M9-L3), a cell gets 416/7 = 59 channels

The real cost is an idle channel is wasted, and one full RF chain per user.

03 / 11
Wireless 101
M10 · L01
TDMA — worked: GSM

Take Turns

  • 200 kHz carrier, 1625/6 = 270.833 kbit/s gross
  • 4.615 ms frame, 8 slots → slot = 4.615/8 = 577 µs
  • 1250 bits per frame → 1250/8 = 156.25 bits per slot
  • Per user 270.833/8 = 33.85 kbit/s for a 13 kbit/s codec
  • Guard 8.25 bits × 3.692 µs = 30.5 µs = 5.28% of the slot

30.5 µs is 9.1 km of propagation. Timing advance stretches it to 34.9 km.

04 / 11
Wireless 101
M10 · L01
CDMA — worked: IS-95

Below the Noise Floor

Processing gain
G_p = \frac{R_c}{R_b} = \frac{W}{R_b}

1.2288 Mcps / 9.6 kbit/s = 128, and 10log₁₀128 = 21.07 dB (2⁷ × 3.0103 ✓). So the in-band SNR is 7 − 21.07 = −14.07 dB — buried, and still recovered.

05 / 11
Wireless 101
M10 · L01
Try it — 1.25 MHz × 1250 kbit/s gross

Cut the Rectangle

Scheme FDMA
Users 6 users
usable 94.4% 196.7 kbit/s each guard 70 kHz user 7 refused
06 / 11
Wireless 101
M10 · L01
Hard versus soft

User 42 Is Admitted

Soft uplink capacity
N \approx 1 + \frac{G_p}{(E_b/N_0)\,\nu\,(1+f)}
  • 128/5.012 = 25.5 → 26 users, continuous talkers
  • Voice activity ν = 0.4 → 63.9, so ≈65
  • Other-cell f = 0.6 → 63.9/1.6 = ≈41 per sector
  • Same 1.25 MHz as FDMA with reuse 7: 41/7 = 5.9 per cell

123/5.9 = 21× — the optimistic end of a real advantage. Capacity is a target, not a count.

07 / 11
Wireless 101
M10 · L01
Why CDMA is hard

The Near-Far Problem

100 m against 1 km at n = 4 is 10 × 4 × log₁₀(10) = 40 dB. In CDMA the near handset is in your band by design — there is no filter for it (M9-L3).

20 dB hot
100 users
Design total
41
Loop rate
800 Hz

Two loops, ~1 dB steps. What costs capacity is the residual scatter, not the mean — all users 1 dB high is harmless.

08 / 11
Wireless 101
M10 · L01
The illuminating asymmetry

Orthogonal Only When Aligned

  • Walsh codes correlate to zero — but only in perfect step
  • Downlink: one transmitter, one clock → 64 Walsh codes
  • Uplink: many clocks, so alignment is abandoned — long PN codes of period 2⁴²−1, repeating every 41 days
  • A chip is 814 ns = 244 m. Aligning uplinks to a fraction of that beats TDMA’s guard time for difficulty
  • Echoes desynchronise the downlink against itself → the rake
09 / 11
Wireless 101
Multiple Access

Check what carries the users

Four questions on how FDMA, TDMA and CDMA carve up one shared resource.

Question 1 of 0
Score 0/0

10 / 11
Wireless 101
M10 · L01
Recap

What you learned

  • FDMA splits B, TDMA splits T, CDMA splits neither
  • AMPS: 833 channels, 416 per operator, 59 per cell at reuse 7
  • GSM: 577 µs slots, 156.25 bits, 30.5 µs of guard = 5.28%
  • IS-95: Gp = 128 = 21.07 dB, signal 14 dB under the floor
  • Hard capacity refuses; soft capacity admits and degrades
Up next in Module 10
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