Lecture 3: Shannon s Theorem

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1 CSE 533: Error-Correctng Codes (Autumn 006 Lecture 3: Shannon s Theorem October 9, 006 Lecturer: Venkatesan Guruswam Scrbe: Wdad Machmouch 1 Communcaton Model The communcaton model we are usng conssts of a source that generates dgtal nformaton. Ths nformaton s sent to a destnaton through a channel. The communcaton can happen n the spatal doman (.e., we need to send nformaton over a physcal dstance on a channel or n the tme doman (.e., we want to retreve data that we stored at an earler pont of tme. The channel can be assocated wth nose. So we have two cases : Noseless case: The channel n ths case transmts symbols wthout causng any errors. One would need to explot the redundancy n the source to economze the length of the transmsson. Ths s done through data compresson, also called source codng. The nformaton s decompressed at destnaton. Nosy case: The channel n ths case ntroduces nose that causes errors n the receved symbols at the destnaton. To reduce the errors ncurred due to nose, one should add systematc redundancy to the nformaton to be sent. Ths s done through channel codng. It s known that relable communcaton s possble n the above model f the entropy of the source,.e., the amount of non-redundant nformaton t generates per unt of tme, s less than the capacty of the channel,.e., the maxmum number of nformaton bts that can be communcated relable per channel use. We wll focus on the channel codng problem n the presence of nose, assumng that the nformaton s generated at the source s source-coded or compressed nto a strng of non-redundant symbols over a fnte alphabet pror to communcaton on the channel. We can communcate such a two-stage scheme as n the followng fgure, and treat source and channel codng n solaton, due to what s known as the Source-Channel Codng theorem. The followng dagram shows the modules of the communcaton model: Source-Channel Codng Theorem: For a source wth entropy no greater than the capacty of the channel, dvdng the transmsson process to source codng followed by channel codng can acheve a probablty of error tendng to zero for a large block length. Then the part of the prevous scheme we ll be consderng s: 1

2 Shannon put forth a stochastc model of the channel. For us, t suffces to talk about dscrete memoryless channels. Such channels have an nput alphabet X, an output alphabet Y and a probablty transton matrx descrbng the output dstrbuton for every nput. Each symbol s sent over the channel ndependently of the prevous symbols sent. Thus the channel s prescrbed by a X Y stochastc matrx where each row sums to 1. Probablty transton matrx: Y { }} { X p(y/x Examples of Channels Channels are often descrbed by nput-output dagrams..1 Bnary Symmetrc Channel (BSC The BSC takes as nput one bt (0/1 and flps t wth probablty p, 0 p 1. p s called the crossover probablty; we wrte BSC p to denote the bnary symmetrc channel wth crossover probablty p.

3 . Bnary Erasure Channel (BEC Fgure 1: Dagram for BSC p The BEC takes as nput one bt (0/1and erases t to? wth probablty ε, 0 ε 1,. ε s called the erasure probablty; we wrte BEC ε to denote the bnary erasure channel wth erasure probablty ε. Fgure : Dagram for BEC ε.3 Contnuous Output Channel The contnuous output channel takes as nput a symbol from a fnte alphabet and maps t, accordng to a specfc nose dstrbuton, to a real number. One example s the Bnary Input Addtve Whte Gaussan Nose (BIAWGN channel, where the nose has a normal dstrbuton and acts addtvely. Σ = { 1, 1} Channel R x y = x + z; z N(0, σ Hence the condtonal probablty densty functon of the channel output y on nput x s gven by : Pr(y x = 1 (y x exp ( πσ σ.4 Nosy Typewrter Channel The nosy typewrter channel s gven by the followng dagram: 3

4 Zero-error communcaton s possble f we just send A, C, E,..., Y or B, D, F,..., Z. So we wll end up wth only 13 possbltes for the sent symbols. Thus the capacty of the channel s at least log 13 bts. In fact, one can prove that ths rate s the maxmum possble and the capacty of the channel s exactly log 13 bts. Zero-error communcaton at a postve rate s not possble wth BSC p, snce for every par of strngs x and y, there s a postve probablty that x gets dstorted nto y. Ths probablty of mscommuncaton can be reduced arbtrarly by hgh enough order repetton code. Consder mappng 0 to m zeroes and 1 to m ones. At the destnaton, decodng s done by majorty,.e. f the number of receved 0 s s greater than m, we decde on a 0, else we decde on a 1. Hence, the probablty of error s gven by: Pr(error = m ( m = m p (1 p m. Ths probablty tends to 0 as m tends to but ths causes the rate to tend to zero!! Can we acheve any desred probablty of error whle mantanng a postve rate? The answer s yes. In fact, the largest possble rate was precsely characterzed and descrbed n Shannon s work. 3 Shannon Capacty Theorem We wll start by defnng the bnary entropy functon. Defnton 3.1. For 0 x 1, the entropy bnary functon, denoted H(x s gven by Note that we have H(xn = x xn (1 x (1 xn. 1 H(x = x log x + (1 x log 1 1 x. Theorem 3. (Shannon Capacty Theorem for the BSC. For every p, 0 p < 1, and 0 < ε < 1/ p, there exsts δ > 0 such that for all large n, there exst an encodng functon E : {0, 1} k {0, 1} n and a decodng functon D : {0, 1} n {0, 1} k for k = (1 H(p + εn such that m {0, 1} k, Pr nose of the BSC p D(E(m + nose m δn. 4

5 The occurence of the entropy functon H(p n the statement of the capacty theorem arses snce we wll see that H(pn s an accurate asymptotc estmate of the volume of a Hammng ball of radus pn. Lemma 3.3. For 0 p 1, Vol (B(0, pn = pn H(pn. Proof. 1 = (p + (1 p n = = = p (1 p n (1 p n ( p 1 p (1 p n ( p 1 p p pn (1 p (1 pn H(pn. pn We wll frst prove the converse of Shannon theorem to gve an ntuton why 1 H(p s the best rate one can hope for. For ths purpose we wll use the lower bound Vol (B(0, pn pn H(pn o(n. Ths fact follows from applyng Strlng approxmaton of n! gven by: n! = ( ( n 1 πn 1 + Θ. e n We wll also need the Chernoff bound. Chernoff bound If X 1, X,... X n are..d 0/1 random varables wth Pr[X 0 < ε < 1 n Pr[ X (p + εn] ε n 3 Pr[ =1 n =1 X (p εn] ε n 3 = 1] = p, then Proof sketch of the converse of Shannon theorem Let E : {0, 1} k {0, 1} n be an encodng functon, and let D be the decodng functon D : 5

6 {0, 1} n {0, 1} k. For a message m {0, 1} k, let S m be the nverse mage of m under D,.e., S m = {y D(y = m}. When m s transmtted, by the Chernoff bound, wth hgh probablty y = E(m + nose les n a shell S of rad ((p εn, (p + εn around E(m. To acheve small decodng error probablty, most of the strngs n S must be decoded to m,.e., belong to S m. We know that S pn H(pn o(n. It follows that S m H(pn o(n for every m. Ths mples that n m {0,1} k S m k+h(pn o(n, whch yelds k (1 H(p + o(1n. We wll contnue the proof of Shannon theorem next lecture. 6

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