Wireless 101
M10 · L03
Module 10 · Lesson 3

The Enemy Becomes the Resource

M8-L3 spent a whole lesson proving multipath wrecks a link. This lesson says: multipath is the resource. Rich scattering is what creates the independent channels MIMO sells.

01 / 11
Wireless 101
M10 · L03
One coefficient per antenna pair

The Channel Is a Matrix

The MIMO channel
\mathbf{y} = \mathbf{H}\mathbf{x} + \mathbf{n}, \quad \mathbf{H} \in \mathbb{C}^{N_r \times N_t}

Nr measurements of Nt signals. Solvable — but only if the equations are independent, and that is a property of the environment, not the hardware.

02 / 11
Wireless 101
M10 · L03
Same antennas, three purchases

Three Gains That Compete

  • Multiplexing — rate, ×min(Nt,Nr). Wants decorrelated
  • Diversity — reliability, order up to NtNr. Wants decorrelated
  • Array gain — SNR in dB (M7-L4). Wants correlated

A beam is a coherent sum; streams need incoherence. So eight antennas cannot be one sharp beam and eight streams at once — the multiplexing–diversity tradeoff.

03 / 11
Wireless 101
M10 · L03
B = 20 MHz, SNR = 20 dB

Four Times the Bits

Capacity, rich scattering, high SNR
C \approx \min(N_t, N_r)\cdot B \log_2(1 + \mathrm{SNR})
  • log₂(101) = 4.6151/0.6931 = 6.658 b/s/Hz
  • 1×1: 20e6 × 6.658 = 133.2 Mbps; 2×2 266.3; 4×4 532.7
  • 4×2 → min(4,2) = 2, so still 266.3 — the client binds
04 / 11
Wireless 101
M10 · L03
The two missing assumptions

Rank Is Not Free

  • Pure line of sight → every row a copy → rank 1
  • Rank 1 on a 4×4 array: 133.2 Mbps, not 532.7
  • Fixed total power, four streams: −10 log₁₀4 = −6.02 dB each
  • 4 × log₂(26) = 4 × 4.700 → 376.0 Mbps, 29% under

A datasheet promises antennas. It cannot promise scatterers.

05 / 11
Wireless 101
M10 · L03
Try it — B = 20 MHz, SNR = 20 dB

The MIMO Gain Explorer

Nt 4 TX
Nr 4 RX
Scatter 100% rich
rank 4 532.7 Mbps div order 16 P(fade) 4.6e-17 multiplexing dominates
06 / 11
Wireless 101
M10 · L03
The other gain

Diversity Is an Exponent

Diversity order and the outage tail
P(\text{all faded}) = p^{\,N_t N_r}
  • M8-L3: p = 1 − e−0.1 = 0.0952 for a 10 dB fade
  • Order 2: 0.0952² = 0.906% ✓ matches M8-L3
  • Order 4 (2×2): 0.0952⁴ = 0.0082%, one moment in 12,000
  • Alamouti STBC: order 2 from 2×1, no feedback needed
07 / 11
Wireless 101
M10 · L03
What actually ships

2×2, 4×4, and 7.5 cm

  • 802.11n up to 4 streams; ac and ax up to 8, with MU-MIMO
  • LTE: 2×2 baseline, 4×4 in later releases
  • Decorrelation wants ≈ λ/2: at 2 GHz, 0.15/2 = 7.5 cm
  • M7-L4’s λ/2 was an upper bound — aliasing. This is a lower one

Two antennas span a phone. A third and fourth have nowhere to go — so designers decorrelate by polarisation instead (M7-L2).

08 / 11
Wireless 101
M10 · L03
64 to 256 elements (M7-L4)

Massive MIMO: Many Users

Not 64 streams to one phone — min(64, 2) = 2 caps that. Many users on the same time–frequency resource, each with a beam that nulls the others. Space becomes the access dimension.

  • Needs accurate channel knowledge → reciprocity matters (M10-L4)
  • One full RF chain per antenna — power, cost, heat
  • Pilot overhead grows with antenna count
  • Pilot contamination does not average away as the array grows
09 / 11
Wireless 101
MIMO Basics

Check what the antennas buy

Four questions on spatial multiplexing, diversity, and what actually sets MIMO rate.

Question 1 of 0
Score 0/0

10 / 11
Wireless 101
M10 · L03
Recap

What you learned

  • Multipath becomes the resource: H, and its rank is the prize
  • C ≈ min(Nt,Nr) B log₂(1+SNR) → 133.2 / 266.3 / 532.7 Mbps
  • Rank 1 collapses 4×4 back to one stream
  • Diversity order NtNr: 0.0952⁴ = 0.0082%
  • λ/2 = 7.5 cm at 2 GHz; massive MIMO serves users, not streams
Up next in Module 10
Duplexing →
All lessons Read article
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