for every memoryless channel there is a definite number c
play

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

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


  1. 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 probability as the block length n → ∞ ‣ Conversely, if the data rate R > C , then no matter what coding scheme is used, the error probability will converge to 1 as n → ∞ 62

  2. ������ AWGN channel model (discrete-time, real-valued) Z m ∼ N (0 , σ 2 ) V m = u m + Z m , m = 1 , . . . , n, Z m u m C comprises 2 nR length- n codewords u ∈ R n Codebook: 63

  3. Sphere packing interpretation • By LLN, as , the received V will lie V = u + Z n → ∞ at the surface of the n -dimensional sphere R n centered at u with radius with √ n σ 2 probability 1 64

  4. Necessary condition: capacity upper bound • maximum # of non-overlapping spheres = V = u + Z maximum # of codewords that can be R n reliably delivered • A necessary condition is p n ( P + σ 2 ) √ n n ( P + σ 2 ) 2 nR ≤ √ n σ 2 n � √ � n n ( P + σ 2 ) ⇒ R ≤ 1 1 � 1 + P � n log = 2 log √ ⇐ n σ 2 σ 2 √ n σ 2 n 65

  5. Achieving capacity (1) • Prove the existence of good codebook C u -sphere by random coding , as we did before for linear block codes: ‣ Randomly generate 2 nR length- n codewords uniformly inside the “ u -sphere” of radius √ √ nP nP ‣ Goal: ensure the average-over-random-code average probability of error vanishes as n → ∞ • u 1 Decoding: P + σ 2 : the MMSE coefficient α , P u 2 → MMSE − → Nearest Neighbor − V − → α V − u → ˆ 66

  6. Achieving capacity (3) • Pairwise probability of error P {E u 1 → u 2 } u -sphere ‣ Ratio of the volume of the two spheres: � n/ 2 n √ � nP σ 2 / ( P + σ 2 ) σ 2 P {E u 1 → u 2 } = √ = P + σ 2 n nP • √ Union bound: nP � n/ 2 ≤ (2 nR − 1) P {E u 1 → u 2 } ≤ 2 nR � P ( n ) σ 2 e P + σ 2 � n P σ 2 P + σ 2 = 2 n ( R − 1 2 log ( 1+ P σ 2 )) u 1 • Su ffi cient condition for vanishing : P ( n ) α V e u 2 ✓ ◆ R < 1 1 + P 2 log σ 2 68

  7. Resources in AWGN channel Fix P ✓ ◆ P N 0 W log 2 e = P P C ( W ) = W log 1 + log 2 e ≈ W N 0 W N 0 1.6 P N 0 log 2 e 1.4 Power limited region 1.2 1 0.8 Capacity C ( W ) 0.6 (Mbps) Limit for W → ∞ 0.4 Bandwidth limited region 0.2 0 0 5 10 15 20 25 30 Bandwidth W (MHz) 71

Download Presentation
Download Policy: The content available on the website is offered to you 'AS IS' for your personal information and use only. It cannot be commercialized, licensed, or distributed on other websites without prior consent from the author. To download a presentation, simply click this link. If you encounter any difficulties during the download process, it's possible that the publisher has removed the file from their server.

Recommend


More recommend