For every memoryless channel, there is a definite number C - - PowerPoint PPT Presentation

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For every memoryless channel, there is a definite number C - - PowerPoint PPT Presentation

Jun 06, 2023 221 likes •305 views

Channel coding theorem For every memoryless channel, there is a definite number C (computable) such that: If the data rate R < C , then there exists a coding scheme that can deliver data at rate R over the channel with vanishing error

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SLIDE 1

62

Channel coding theorem

  • For every memoryless channel, there is a definite number C (computable) such that:
  • If the data rate R < C, then there exists a coding scheme that can deliver data at rate R
  • ver the channel with vanishing error probability as the block length
  • Conversely, if the data rate R > C, then no matter what coding scheme is used, the error

probability will converge to 1 as n → ∞ n → ∞

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SLIDE 2

AWGN channel model (discrete-time, real-valued)

63

Codebook: Vm = um + Zm, m = 1, . . . , n, Zm

  • ∼ N(0, σ2)

Zm um C comprises 2nR length-n codewords u ∈ Rn

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SLIDE 3

Sphere packing interpretation

64

Rn V = u + Z

  • By LLN, as , the received V will lie

at the surface of the n-dimensional sphere centered at u with radius with probability 1 n → ∞ √ nσ2

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SLIDE 4

Necessary condition: capacity upper bound

  • maximum # of non-overlapping spheres =

maximum # of codewords that can be reliably delivered

  • A necessary condition is

65

Rn

p n(P + σ2)

V = u + Z

√ nσ2

2nR ≤ √

n(P +σ2)

n

√ nσ2n

⇐ ⇒ R ≤ 1

n log

√

n(P +σ2)

n

√ nσ2n

  • =

1 2 log

  • 1 + P

σ2

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SLIDE 5

Achieving capacity (1)

  • Prove the existence of good codebook

by random coding, as we did before for linear block codes:

  • Randomly generate 2nR length-n codewords

uniformly inside the “u-sphere” of radius

  • Goal: ensure the average-over-random-code

average probability of error vanishes as

  • Decoding:

66

C u-sphere

√ nP u1 u2

√ nP n → ∞

α ,

P P +σ2 : the MMSE coefficient

V − → MMSE − → αV − → Nearest Neighbor − → ˆ u

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SLIDE 6

Achieving capacity (3)

  • Pairwise probability of error
  • Ratio of the volume of the two spheres:
  • Union bound:
  • Sufficient condition for vanishing :

68

u-sphere

√ nP

  • n P σ2

P +σ2

u1 u2 αV

P {Eu1→u2}

P {Eu1→u2} = √

nP σ2/(P +σ2)

n

√ nP

n

=

  • σ2

P +σ2

n/2 P(n)

e

≤ (2nR − 1)P {Eu1→u2} ≤ 2nR

σ2 P +σ2

n/2

P(n)

e

= 2n(R− 1

2 log(1+ P σ2 ))

R < 1 2 log ✓ 1 + P σ2 ◆

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SLIDE 7

Resources in AWGN channel

71

30 5 Bandwidth W (MHz) Capacity Limit for W → ∞ Power limited region 0.2 1 Bandwidth limited region (Mbps) C(W ) 0.4 25 20 15 10 1.6 1.4 1.2 0.8 0.6 P N0log2 e

C(W) = W log ✓ 1 + P N0W ◆ ≈ W P N0W log2 e = P N0 log2 e

Fix P