S Advanced Digital Communication (4 cr) Targets today

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1 S Advanced Dtal Communcaton (4 cr) Convolutonal Codes Tarets today Why to apply convolutonal codn? Defnn convolutonal codes Practcal encodn crcuts Defnn qualty of convolutonal codes Decodn prncples Vterb decodn 2

2 k bts Convolutonal encodn (n,k,l) (n,k,l) encoder encoder n bts nput bt n(l+) output bts messae bts encoded bts Convolutonal codes are appled n applcatons that requre ood performance wth low mplementaton complexty. They operate on code streams (not n blocks) Convoluton codes have memory that utlzes prevous bts to encode or decode follown bts (block codes are memoryless) Convolutonal codes are denoted by (n,k,l), where L s code (or encoder) Memory depth (number of rester staes) Constrant lenth C=n(L+) s defned as the number of encoded bts a messae bt can nfluence to Convolutonal codes acheve ood performance by expandn ther memory depth 3 Example: Convolutonal encoder, k =, n = 2 x' m m m x'' m m 2 2 memory depth L = number of states x x' x'' x' x'' x' x''... out (n,k,l) = (2,,2) encoder Convolutonal encoder s a fnte state machne (FSM) processn nformaton bts n a seral manner Thus the enerated code s a functon of nput and the state of the FSM In ths (n,k,l) = (2,,2) encoder each messae bt nfluences a span of C= n(l+)=6 successve output bts = constrant lenth C Thus, for eneraton of n-bt output, we requre n ths example n shft resters n k = convolutonal encoder 4 2

3 Example: (n,k,l)=(3,2,) Convolutonal encoder x m m m ' 3 2 x m m m '' 3 x m m ''' 2 After each new block of k nput bts follows a transton nto new state Hence, from each nput state transton, 2 k dfferent output states may follow Each messae bt nfluences a span of C = n(l+) = 3(+) = 6 successve output bts 5 Generator sequences k bts (n,k,l) (n,k,l) encoder encoder n bts (n,k,l) Convolutonal code can be descrbed by the enerator sequences ( 2) ( n ),,... that are the mpulse responses for each coder n output branches: 0 2 m (2,,2) encoder ( ) ( ) ( ) ( ) n [ n n n 0 m ] [ 0 ] Note that the enerator sequence lenth ( 2) [] exceeds rester depth always by Generator sequences specfy convolutonal code completely by the assocated enerator matrx Encoded convoluton code s produced by matrx multplcaton of the nput and the enerator matrx 6 3

4 Convoluton pont of vew n encodn and enerator matrx Encoder outputs are formed by modulo-2 dscrete convolutons: v u*, v u*... v u* where u s the nformaton sequence: u ( u, u, ) ( 2) ( 2) ( ) ( ) Therefore, the l:th bt of the :th output branch s* m v u u u... u Hence, for ths crcut the follown equatons result, (assume: ) 0 ( ) ( ) ( ) ( ) ( ) l 0 l l l 0 l lm m where m L, u 0, l L2 ul 2 2 l 3 ul 3 branches v u u u l l l2 l3 ( 2 ) v u u u u l l l l2 l3 x y( u) x( k) y( u k) v ka xy encoder output: nput bt n(l+) output bts ( 2 ) [ 0 ] [ ] ( 2 ) ( 2 ) ( 2 ) [ v v v v v v...] *note that u s reversed n tme as n the defnton of convoluton top rht 7 Example: Usn enerator matrx ( 2 ) [ 0 ] [ ] v ( ) ( ) l l 0 u u ( ) l... u ( ) lm m ul m ( 2) 0 0 ( 2) ( 2) m m Verfy that you can obtan the result shown! 8 4

5 S.Ln, D.J. Costello: Error Control Codn, II ed, p Representn convolutonal codes: Code tree Number of braches devatn from each node equals 2 k (n,k,l) = (2,,2) encoder x' m m m 2 x'' m m 2 x x' x'' x' x'' x' x''... out x' 0 0 x'' 0 x' 0 0 x'' 0 0 m 2m 0 Ths tells how one nput bt s transformed nto two output bts (ntally rester s all zero) x' 0 x'' 0 0 5

6 Representn convolutonal codes compactly: code trells and state daram Input state ndcated by dashed lne Code trells State daram Shft rester states Inspectn state daram: Structural propertes of convolutonal codes Each new block of k nput bts causes a transton nto new state Hence there are 2 k branches leavn each state Assumn encoder zero ntal state, encoded word for any nput of k bts can thus be obtaned. For nstance, below for u=( 0 ), encoded word v=(, 0, 0, 0,, 0,, ) s produced: Verfy that you obtan the same result! Input state - encoder state daram for (n,k,l)=(2,,2) code - note that the number of states s 8 = 2 L+ => L = 2 (two state bts) 2 6

7 Code weht, path an, and eneratn functon The state daram can be modfed to yeld nformaton on code dstance propertes (= tells how ood the code s to detect or correct errors) Rules (example on the next slde): Splt S 0 nto ntal and fnal state, remove self-loop (2) Label each branch by the branch an X. Here s the weht* of the n encoded bts on that branch (3) Each path connectn the ntal state and the fnal state represents a nonzero code word that dveres and re-emeres wth S 0 only once The path an s the product of the branch ans alon a path, and the weht of the assocated code word s the power of X n the path an Code weh dstrbuton s obtaned by usn a wehted an formula to compute ts eneratn functon (nput-output equaton) T( X ) A X where A s the number of encoded words of weht *In lnear codes, weht s the number of :s n the encoder output 3 branch an weht: 2 weht: Example: The path representn the state sequence S 0 S S 3 S 7 S 6 S 5 S 2 S 4 S 0 has the path an X 2 X X X X 2 X X 2 X 2 =X 2 and the correspondn code word has the weht of 2 Where does these terms come from? T( X ) X 3X 5X A X X 25 X

8 Dstance propertes of convolutonal codes Code strenth s measured by the mnmum free dstance: d free where v and v are the encoded words correspondn nformaton sequences u and u. Code can correct up to t d free / 2 errors. The mnmum free dstance d free denotes: The mnmum weht of all the paths n the state daram that dvere from and remere wth the all-zero state S 0 The lowest power of the code-eneratn functon T(X) T( X ) A X * for dervaton, see Carlson s, p. 583 mn d( v ', v '') : u' u'' X 3X 5X Code an*: Gc kd / 2 n R d c / 2 free free 9 0 X 25 X... d free 6 5 Codn an for some selected convolutonal codes Here s a table of some selected convolutonal codes and ther code ans R C d free /2 expressed for hard decodn also by 0lo ( R d 0 c free / 2) db 6 8

9 Decodn of convolutonal codes Maxmum lkelhood decodn of convolutonal codes means fndn the code branch n the code trells that was most lkely transmtted Therefore maxmum lkelhood decodn s based on calculatn code Hammn dstances for each branch potentally formn encoded word Assume that the nformaton symbols appled nto an AWGN channel are equally alke and ndependent Let s denote by x encoded symbols (no errors) and by y receved (potentally erroneous) symbols: x x x... x... y y y... y... Probablty to decode the symbols s then p( y, x) p( y x ) 0 x The most lkely path throuh the trells wll maxmze ths metrc. Often ln() s taken from both sdes, because probabltes are often small numbers, yeldn: ln p( y, x) ln p( y x ) (note ths corresponds equavalently also the smallest Hammn dstance) receved code: non - erroneous code: m y x Decoder (=dstance calculaton) bt decsons 7 Example of exhaustve maxmal lkelhood detecton Assume a three bt messae s transmtted and encoded by (2,,2) convolutonal encoder. To clear the decoder, two zero-bts are appended after messae. Thus 5 bts are encoded resultn 0 bts of code. Assume channel error probablty s p = 0.. After the channel 0,0,0,,00 s produced (ncludn some errors). What comes after the decoder, e.. what was most lkely the transmtted code and what were the respectve messae bts? a b states c d decoder outputs f ths path s selected 8 9

10 p( y, x) p( y x ) 0 ln p( y, x) ln p( y x ) 0 weht for prob. to receve bt n-error errors correct 9 correct: =8;8 ( 0.) 0.88 false: =2;2 ( 2.30) 4.6 total path metrc: 5.48 Note also the Hammn dstances! The larest metrc, verfy that you et the same result! 20 0

11 The Vterb alorthm Problem of optmum decodn s to fnd the mnmum dstance path from the ntal state back to the ntal state (below from S 0 to S 0 ). The mnmum dstance s one of the sums of all path metrcs from S 0 to S 0 Exhaustve maxmum lkelhood method must search all the paths n phase trells (2 k paths emern/ entern from 2 L+ states for an (n,k,l) code) The Vterb alorthm ets mprovement n computatonal effcency va concentratn nto survvor paths of the trells 2 The survvor path Assume for smplcty a convolutonal code wth k=, and thus up to 2 k = 2 branches can enter each state n trells daram Assume optmal path passes S. Metrc comparson s done by addn the metrc of S and S 2 to S. At the survvor path the accumulated metrc s naturally smaller (otherwse t could not be the optmum path) For ths reason the non-survved path can be dscarded -> all path alternatves need not to be further consdered Note that n prncple the whole transmtted sequence must be receved before decson. However, n practce storn of states for L nput lenth of 5L s qute adequate 2 nodes, determned by memory depth k 2 branches enter each node branch of larer metrc dscarded 22

12 Example of usn the Vterb alorthm Assume the receved sequence s y and the (n,k,l)=(2,,2) encoder shown below. Determne the Vterb decoded output sequence! states (Note that for ths encoder code rate s /2 and memory depth equals L = 2) 23 The maxmum lkelhood path After rester lenth L+=3 branch pattern bens to repeat Smaller accumulated metrc selected (2) (0) (Branch Hammn dstances n parenthess) Frst depth wth two entres to the node The decoded ML code sequence s whose Hammn dstance to the receved sequence s 4 and the respectve decoded sequence s (why?). Note that ths s the mnmum dstance path. (Black crcles denote the deleted branches, dashed lnes: '' was appled) 24 2

13 How to end-up decodn? In the prevous example t was assumed that the rester was fnally flled wth zeros thus fndn the mnmum dstance path In practce wth lon code words zeron requres feedn of lon sequence of zeros to the end of the messae bts: ths wastes channel capacty & ntroduces delay To avod ths path memory truncaton s appled: Trace all the survvn paths to the depth where they mere Fure rht shows a common pont at a memory depth J J s a random varable whose applcable mantude shown n the fure (5L) has been expermentally tested for nelble error rate ncrease Note that ths also ntroduces the delay of 5L! J 5L staes of the trells 25 Lessons learned You understand the dfferences between cyclc codes and convolutonal codes You can create state daram for a convolutonal encoder You know how to construct convolutonal encoder crcuts based on known the enerator sequences You can analyze code strenths based on known code eneraton crcuts / state darams or enerator sequences You understand how to realze maxmum lkelhood convolutonal decodn by usn exhaustve search You understand the prncple of Vterb decodn 26 3