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Channel Capacity and Coding ECE 515 Information Theory Channel Capacity and Coding 1 Information Theory Problems • How to transmit or store information as efficiently as possible. • What is the maximum amount of information that can be transmitted or stored reliably? • How can information be kept secure? 2 Digital Communications System W X PX Data source PY|X Data sink Ŵ Y PY Ŵ is an estimate of W 3 Communication Channel • Cover and Thomas Chapter 7 4 Discrete Memoryless Channel X Y x1 y1 x2 y2 . xK yJ J≥K≥2 5 Discrete Memoryless Channel Channel Transition Matrix py(|)11 x py (|)1 xkK py (|)1 x P= py(|)j x py (|)jk x py (|)jK x 1 py(|)J x1 py (|)Jk x py (|)JK x 6 Binary Symmetric Channel x1 y1 x2 y2 channel matrix pp pp 7 Binary Errors with Erasure Channel 1-p-q x 0 0 y 1 q 1 p e y2 p q x2 1 1 y3 1-p-q 8 BPSK Modulation K=2 2E x t = + b cos 2π ft 0 bit 1 ( ) ( c ) Eb is the energy per bit Tb Tb is the bit duration 2E f is the carrier x t = − b cos 2π ft 1 bit c 2 ( ) ( c ) frequency Tb 9 BPSK Demodulation in AWGN − Eb 0 + Eb x2 x1 p(y|x2) p(y|x1) 0 x2 x1 2 1 (yx− k ) pYX (yx |= xk ) = exp− | 2 2 2πσ 2σ 10 BPSK Demodulation K=2 − + Eb 0 Eb J=3 y3 y2 y1 − + Eb 0 Eb J=4 y4 y3 y2 y1 − + Eb 0 Eb J=8 y8 y1 11 Mutual Information for a BSC crossover probability p X BSC Y pp=1 − channel matrix pp PY|X = pp X Y p w 0 0 p p(xw= 0) = p p(x==−= 1) 1 ww w 1 1 p 12 13 Convex Functions Concave Convex 14 Concave Function 15 Convex Function 16 Mutual Information 17 18 input probabilities 19 20 21 channel transition probabilities 22 BSC I(X;Y) w p 23 Channel Capacity X channel Y 24 Binary Symmetric Channel x1 y1 x2 y2 C=−== 1h()forp ww 1/2 25 Properties of the Channel Capacity • C ≥ 0 since I(X;Y) ≥ 0 • C ≤ log|X| = log(K) since C = max I(X;Y) ≤ max H(X) = log(K) • C ≤ log|Y| = log(J) for the same reason • I(X;Y) is a concave function of p(X), so a local maximum is a global maximum 26 Symmetric Channels A discrete memoryless channel is said to be symmetric if the set of output symbols {yj}, j = 1, 2, ..., J, can be partitioned into subsets such that for each subset of the matrix of transition probabilities, each column is a permutation of the other columns, and each row is also a permutation of the other rows. 27 Binary Channels Symmetric channel matrix Non-symmetric channel matrix 1 − pp 1 − pp12 P = P= pp12≠ pp1 − pp121 − 28 Binary Errors with Erasure Channel 1-p-q x 0 0 y 1 q 1 p e y2 p q x2 1 1 y3 1-p-q 29 Symmetric Channels • No partition required → strongly symmetric • Partition required → weakly symmetric 30 Capacity of a Strongly Symmetric Channel Theorem KJ I(X;Y)= ∑∑ p(xk ) p( yx jk | )logp( yx jk | )+ H(Y) kj=11 = J = ∑p(yxjk | )logp( yx jk | )+ H(Y) j=1 31 Example J = K = 3 x .7 y 1 .1 .2 1 .2 .7 x2 y2 .1 .2 .1 x3 .7 y3 32 Example .7 .2 .1 P = Y|X .1 .7 .2 .2 .1 .7 KJ ∑∑p(xk ) p( yx jk | )logp( yx jk | ) kj=11 = =p(x1 )[ .7log.7 ++ .1log.1 .2log.2] +p(x2 )[ .2log.2 ++ .7log.7 .1log.1] +p(x3 )[ .1log.1 ++ .2log.2 .7log.7] =.7log.7 ++ .2log.2 .1log.1 J = ∑p(yxjk | )logp( yx jk | ) j=1 33 Example J H(Y)= −∑ p(yyjj )logp( ) j=1 K p(y11 )= ∑ p( yx |kk )p( x ) k=1 K p(y22 )= ∑ p( yx |kk )p( x ) k=1 K p(yJ )= ∑ p( yx Jk | )p( x k ) k=1 p(y1 )=++ .7p( xxx 123 ) .2p( ) .1p( ) p(y2 )=++ .1p( xxx 123 ) .7p( ) .2p( ) p(y3 )=++ .2p( xxx 123 ) .1p( ) .7p( ) 34 r-ary Symmetric Channel pp p 1 − p rr−− r − 11 1 p pp 1 − p r−1 rr −− 11 Pp= p p 1 − p rr−−11 r − 1 ppp 1 − p −−−rrr111 35 r-ary Symmetric Channel pp C=− (1p )log(1 −+− pr ) ( 1) log( )+ logr rr−−11 p =logr +− (1 p )log(1 −+ pp ) log( ) r −1 =logr +− (1 p )log(1 −+ pp ) log( pp ) − log( r − 1) =logr −− hp ( ) p log( r − 1) • r = 2 C = 1 – h(p) • r = 3 C = log23 – h(p) – p • r = 4 C = 2 – h(p) – plog23 36 Binary Errors with Erasure Channel 1-p-q 0 0 q p e p q 1 1 1-p-q 37 Binary Errors with Erasure Channel .8 .05 = PY|X .15 .15 .05 .8 .425 .5 = = ×= PX PY P Y|X P X .15 .5 .425 38 Capacity of a Weakly Symmetric Channel L CC= ∑qii i=1 • qi – probability of channel i • Ci – capacity of channel i 39 Binary Errors with Erasure Channel 1 −−pq 1 − q q 0 0 0 p q 1 − q p e − 1 q q 1 1 −−pq 1 1 q 1 − q .9412 .0588 P1 = P2 = [ 1.0 1.0] .0588 .9412 40 Binary Errors with Erasure Channel L CC= ∑qii i=1 C1 = .9412log.9412 + .0588log.0588 += log2 .6773 C2 = 1log1 += log1 0.0 C= .85(.6773) += .15(0.0) .5757 41 Binary Erasure Channel 1-p 0 0 p e p 1 1 1-p C = 1 - p 42 Z Channel (Optical) 1 0 0 (light on) p 1 1 (light off) 1-p 43 Z Channel (Optical) 22 p(yxjk | ) I(X;Y)= ∑∑ p(xk )p( yx jk | )log kj=11 = p(y j ) 1 I(x1 ;Y)= log w+ pw p 1 I(xp2 ;Y)= log+− (1p )log w+ pw w I(X;Y) = w× I(xx12 ;Y) +× w I( ;Y) 44 Z Channel (Optical) I(X;Y) = w× I(xx12 ;Y) +× w I( ;Y) 1 w*1= − (1−+p )(1 2h(pp )/(1− ) ) pp/(1− ) C= log2 ( 1 +− (1pp ) ) 45 Mutual Information for the Z Channel • p = 0.15 I(X;Y) w 46 Channel Capacity for the Z, BSC and BEC p 47 Blahut-Arimoto Algorithm KJ p(yx | ) = jk I(X;Y)∑∑ p(xk ) p( yx jk | )log K kj=11 = px( )( py | x ) ∑ l jl l=1 • An analytic solution for the capacity can be very difficult to obtain • The alternative is a numerical solution – Arimoto Jan. 1972 – Blahut Jul. 1972 • Exploits the fact that I(X;Y) is a concave function of p(xk) 48 Blahut-Arimoto Algorithm 49 Blahut-Arimoto Algorithm • Update the probabilities 50 51 Symmetric Channel Example 52 53 Non-Symmetric Channel Example 0.7000 54 55 56 57 58 Binary Symmetric Channel x1 y1 x2 y2 channel matrix pp pp=1 − pp 59 Binary Symmetric Channel • Consider a block of N = 1000 bits – if p = 0, 1000 bits are received correctly – if p = 0.01, 990 bits are received correctly – if p = 0.5, 500 bits are received correctly • When p > 0, we do not know which bits are in error – if p = 0.01, C = .919 bit – if p = 0.5, C = 0 bit 60 Triple Repetition Code • N = 3 message w codeword c 0 000 1 111 61 Binary Symmetric Channel Errors • If N bits are transmitted, the probability of an m bit error pattern is Nm− ppm (1 − ) • The probability of exactly m errors is Nm− N ppm 1 − ( m) ( ) • The probability of m or more errors is N Ni− N ppi 1 − ∑( i ) ( ) im= 62 Triple Repetition Code • N = 3 • The probability of 0 errors is (1− p )3 • The probability of 1 error is 3p(1− p )2 • The probability of 2 errors is 3pp2 (1− ) • The probability of 3 errors is p3 63 Triple Repetition Code • For p = 0.01 – The probability of 0 errors is .970 – The probability of 1 error is 2.94×10-2 – The probability of 2 errors is 2.97×10-4 – The probability of 3 errors is 10-6 1 • If p 2 p(0 errors) p(1 error) p(2 errors) p(3 errors) 64 Triple Repetition Code – Decoding Received Word Codeword Error Pattern 000 000 000 001 000 001 010 000 010 100 000 100 111 111 000 110 111 001 101 111 010 011 111 100 65 Triple Repetition Code • Majority vote or nearest neighbor decoding will correct all single errors 000, 001, 010, 100 → 000 111, 110, 101, 011 → 111 • The probability of a decoding error is then 2 323 Pe = 3p (1 −+ pp ) = 3 p − 2 p < p • If p = 0.01, then Pe = 0.000298 and only one word in 3356 will be in error after decoding. • A reduction by a factor of 33. 66 Code Rate • After compression, the data is (almost) memoryless and uniformly distributed (equiprobable) • Thus the entropy of the messages (codewords) is H(WM ) = logb • The blocklength of a codeword is N 67 Code Rate • The code rate is given by log M R = b bits per channel use N • M is the number of codewords • N is the block length • For the triple repetition code log 2 1 R = 2 = 33 68 Repetition Code rate R 69 YN XN 70 Shannon’s Noisy Coding Theorem For any ε > 0 and for any rate R less than the channel capacity C, there is an encoding and decoding scheme that can be used to ensure that the probability of decoding error is less than ε for a sufficiently large block length N.
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