Channel coding

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Channel coding

  1. 1. Introduction to Informationtheorychannel capacity and models A.J. Han Vinck University of Essen May 2011
  2. 2. This lecture Some models Channel capacity  Shannon channel coding theorem  converse
  3. 3. some channel models Input X P(y|x) output Y transition probabilitiesmemoryless:- output at time i depends only on input at time i- input and output alphabet finite
  4. 4. Example: binary symmetric channel (BSC) 1-p Error Source 0 0 E p X Y = X ⊕E + 1 1 Input Output 1-pE is the binary error sequence s.t. P(1) = 1-P(0) = pX is the binary information sequenceY is the binary output sequence
  5. 5. from AWGNto BSC pHomework: calculate the capacity as a function of A and σ2
  6. 6. Other models 1-e0 0 (light on) 0 0 e X p Y 1-p 1 (light off)1 e 1 1 P(X=0) = P0 P(X=0) = P0Z-channel (optical) Erasure channel(MAC)
  7. 7. Erasure with errors 1-p-e 0 0 e p p e 1 1 1-p-e
  8. 8. burst error model (Gilbert-Elliot) Random error channel; outputs independent Error Source P(0) = 1- P(1); Burst error channel; outputs dependent P(0 | state = bad ) = P(1|state = bad ) = 1/2; Error Source P(0 | state = good ) = 1 - P(1|state = good ) = 0.999 State info: good or bad transition probability PgbPgg good bad Pbb Pbg
  9. 9. channel capacity: I(X;Y) = H(X) - H(X|Y) = H(Y) – H(Y|X) (Shannon 1948) X Y H(X) channel H(X|Y) max I(X; Y) = capacity P( x ) notes: capacity depends on input probabilities because the transition probabilites are fixed
  10. 10. Practical communication system design Code book Code receivemessage word in estimate 2k channel decoder Code book with errors n There are 2k code words of length n k is the number of information bits transmitted in n channel uses
  11. 11. Channel capacity Definition: The rate R of a code is the ratio k/n, where k is the number of information bits transmitted in n channel uses Shannon showed that: : for R ≤ C encoding methods exist with decoding error probability 0
  12. 12. Encoding and decoding according to ShannonCode: 2k binary codewords where p(0) = P(1) = ½Channel errors: P(0 →1) = P(1 → 0) = p i.e. # error sequences ≈ 2nh(p)Decoder: search around received sequence for codeword with ≈ np differences space of 2n binary sequences
  13. 13. decoding error probability1. For t errors: |t/n-p|> Є → 0 for n → ∞ (law of large numbers)2. > 1 code word in region (codewords random) 2 nh ( p) P(> 1) ≈ (2 k − 1) 2n → 2 − n (1− h ( p)− R ) = 2 − n (C BSC − R ) → 0 k for R = < 1 − h (p) n and n → ∞
  14. 14. channel capacity: the BSC 1-p I(X;Y) = H(Y) – H(Y|X) 0 0 the maximum of H(Y) = 1 X p Y since Y is binary 1 1 H(Y|X) = h(p) 1-p = P(X=0)h(p) + P(X=1)h(p) Conclusion: the capacity for the BSC CBSC = 1- h(p) Homework: draw CBSC , what happens for p > ½
  15. 15. channel capacity: the BSC Explain the behaviour!1.0 Channel capacity 0.5 1.0 Bit error p
  16. 16. channel capacity: the Z-channel Application in optical communications0 0 (light on) H(Y) = h(P0 +p(1- P0 ) ) X p Y H(Y|X) = (1 - P0 ) h(p) 1-p 1 (light off)1 For capacity, P(X=0) = P0 maximize I(X;Y) over P0
  17. 17. channel capacity: the erasure channel Application: cdma detection 1-e0 0 I(X;Y) = H(X) – H(X|Y) e H(X) = h(P0 )X Y H(X|Y) = e h(P0) e1 1 Thus Cerasure = 1 – e P(X=0) = P0 (check!, draw and compare with BSC and Z)
  18. 18. Erasure with errors: calculate the capacity! 1-p-e 0 0 e p p e 1 1 1-p-e
  19. 19. 0 0 1/3 example 1 1 1/3 2 2 Consider the following example For P(0) = P(2) = p, P(1) = 1-2p H(Y) = h(1/3 – 2p/3) + (2/3 + 2p/3); H(Y|X) = (1-2p)log23 Q: maximize H(Y) – H(Y|X) as a function of p Q: is this the capacity? hint use the following: log2x = lnx / ln 2; d lnx / dx = 1/x
  20. 20. channel models: general diagram P1|1 y1 x1 P2|1 Input alphabet X = {x1, x2, …, xn} P1|2 x2 y2 P2|2 Output alphabet Y = {y1, y2, …, ym} : Pj|i = PY|X(yj|xi) : : : : In general: : xn calculating capacity needs more Pm|n theory ymThe statistical behavior of the channel is completely defined bythe channel transition probabilities Pj|i = PY|X(yj|xi)
  21. 21. * clue: I(X;Y) is convex ∩ in the input probabilities i.e. finding a maximum is simple
  22. 22. Channel capacity: converse For R > C the decoding error probability > 0 Pe k/n C
  23. 23. Converse: For a discrete memory less channel channel Xi Yi n n n n I ( X ; Y ) = H (Y ) − ∑ H (Yi | X i ) ≤ ∑ H (Yi ) − ∑ H (Yi | X i ) = ∑ I ( X i ; Yi ) ≤ nC n n n i =1 i =1 i =1 i =1 Source generates one source encoder channel decoder out of 2k equiprobable m Xn Yn m‘ messages Let Pe = probability that m‘ ≠ m
  24. 24. converse R := k/nk = H(M) = I(M;Yn)+H(M|Yn) 1 – C n/k - 1/k ≤ Pe ≤ I(Xn;Yn) + 1 + k Pe ≤ nC + 1 + k Pe Pe ≥ 1 – C/R - 1/nR Hence: for large n, and R > C, the probability of error Pe > 0
  25. 25. We used the data processing theoremCascading of Channels I(X;Z) X Y Z I(X;Y) I(Y;Z)The overall transmission rate I(X;Z) for the cascade cannot be larger than I(Y;Z), that is: I(X; Z) ≤ I(Y; Z)
  26. 26. Appendix:Assume: binary sequence P(0) = 1 – P(1) = 1-p t is the # of 1‘s in the sequenceThen n → ∞ , ε > 0 Weak law of large numbers Probability ( |t/n –p| > ε ) → 0i.e. we expect with high probability pn 1‘s
  27. 27. Appendix: Consequence: 1. n(p- ε) < t < n(p + ε) with high probability n ( p + ε) n n 2. ∑   ≈ 2nε  ≈ 2nε2 nh ( p) t  pn  n ( p −ε)     1 log 2nε  n  lim n 2   → h ( p) h (p) = − p log 2 p − (1 − p) log 2 (1 − p) 3. n→ ∞  pn   Homework: prove the approximation using ln N! ~ N lnN for N large. N −NOr use the Stirling approximation: N ! → 2π N N e
  28. 28. Binary Entropy: h(p) = -plog2p – (1-p) log2 (1-p) 1h 0.9 Note: 0.8 h(p) = h(1-p) 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 p
  29. 29. Capacity for Additive White Gaussian Noise Noise Input X Output Y Cap := sup [H(Y) − H( Noise)] p( x ) x 2 ≤S / 2 W W is (single sided) bandwidth Input X is Gaussian with power spectral density (psd) ≤S/2W; Noise is Gaussian with psd = σ2noise Output Y is Gaussian with psd = σy2 = S/2W + σ2noise For Gaussian Channels: σy2 = σx2 +σnoise2
  30. 30. NoiseX Y X Y Cap = 1 log 2 (2πe(σ 2 + σ 2 )) − 1 log 2 (2πeσ 2 ) bits / trans. 2 x noise 2 noise σ2 + σ2 noise x = 1 log 2 ( 2 ) bits / trans. σ 2 noise σ2 + S / 2W noise Cap = W log 2 ( ) bits / sec . σ2 noise 1 −z2 / 2 σ2p(z) = e z ; H(Z) = 2 log2 (2πeσ2 ) bits 1 z 2πσ2 z
  31. 31. Middleton type of burst channel model 0 0 1 1 Transition probability P(0) channel 1 channel 2Select channel k …with probability channel k hasQ(k) transition probability p(k)
  32. 32. Fritzman model:multiple states G and only one state B Closer to an actual real-world channel 1-p G1 … Gn B Error probability 0 Error probability h
  33. 33. Interleaving: from bursty to random burstyMessage interleaver channel interleaver -1 messageencoder decoder „random error“ Note: interleaving brings encoding and decoding delay Homework: compare the block and convolutional interleaving w.r.t. delay
  34. 34. Interleaving: block Channel models are difficult to derive: - burst definition ? - random and burst errors ? for practical reasons: convert burst into random errorread in row wise 1 0 1 0 1 transmit 0 1 0 0 0 0 0 0 1 0 column wise 1 0 0 1 1 1 1 0 0 1
  35. 35. De-Interleaving: block read in column 1 0 1 e 1 read out wise 0 1 e e 0 0 0 e 1 0 this row contains 1 error 1 0 e 1 1 row wise 1 1 e 0 1
  36. 36. Interleaving: convolutional input sequence 0 input sequence 1 delay of b elements ••• input sequence m-1 delay of (m-1)b elements in Example: b = 5, m = 3 out

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