Digital Communication I: Modulation and Coding Course-Lecture 10 potx

Digital Communications I: Modulation and Coding Course Period 3 - 2007 Catharina Logothetis Lecture 10 Lecture 10 2 Last time, we talked about:  Channel coding  Linear block codes  The error detection and correction capability  Encoding and decoding  Hamming codes  Cyclic codes Lecture 10 3 Today, we are going to talk about:  Another class of linear codes, known as Convolutional codes.  We study the structure of the encoder.  We study different ways for representing the encoder. Lecture 10 4 Convolutional codes  Convolutional codes offer an approach to error control coding substantially different from that of block codes.  A convolutional encoder:  encodes the entire data stream, into a single codeword.  does not need to segment the data stream into blocks of fixed size ( Convolutional codes are often forced to block structure by periodic truncation ).  is a machine with memory.  This fundamental difference in approach imparts a different nature to the design and evaluation of the code.  Block codes are based on algebraic/combinatorial techniques.  Convolutional codes are based on construction techniques. Lecture 10 5 Convolutional codes-cont’d  A Convolutional code is specified by three parameters or where  is the coding rate, determining the number of data bits per coded bit.  In practice, usually k=1 is chosen and we assume that from now on.  K is the constraint length of the encoder a where the encoder has K-1 memory elements.  There is different definitions in literatures for constraint length. ),,( Kkn ),/( Knk nkR c /= Lecture 10 6 Block diagram of the DCS Information source Rate 1/n Conv. encoder Modulator Information sink Rate 1/n Conv. decoder Demodulator    sequenceInput 21 , ), ,,( i mmm=m       bits) coded ( rdBranch wo 1 sequence Codeword 321 , ), ,,,( n nijiii i , ,u, ,uuU UUUU = = = G(m)U , ) ˆ , , ˆ , ˆ ( ˆ 21 i mmm=m        dBranch worper outputs 1 dBranch worfor outputsr Demodulato sequence received 321 , ), ,,,( n nijii i i i , ,z, ,zzZ ZZZZ = =Z C h a n n e l Lecture 10 7 A Rate ½ Convolutional encoder  Convolutional encoder (rate ½, K=3)  3 shift-registers where the first one takes the incoming data bit and the rest, form the memory of the encoder. Input data bits Output coded bits m 1 u 2 u First coded bit Second coded bit 21 ,uu (Branch word) Lecture 10 8 A Rate ½ Convolutional encoder 1 0 0 1 t 1 u 2 u 11 21 uu 0 1 0 2 t 1 u 2 u 01 21 uu 1 0 1 3 t 1 u 2 u 00 21 uu 0 1 0 4 t 1 u 2 u 01 21 uu )101(=m Time Output OutputTime Message sequence: (Branch word) (Branch word) Lecture 10 9 A Rate ½ Convolutional encoder Encoder)101(=m )1110001011(=U 0 0 1 5 t 1 u 2 u 11 21 uu 0 0 0 6 t 1 u 2 u 00 21 uu Time Output Time Output (Branch word) (Branch word) Lecture 10 10 Effective code rate  Initialize the memory before encoding the first bit (all- zero)  Clear out the memory after encoding the last bit (all- zero)  Hence, a tail of zero-bits is appended to data bits.  Effective code rate :  L is the number of data bits and k=1 is assumed: data Encoder codewordtail ceff R KLn L R < −+ = )1( [...]... 1 0 0 00 10 11 0/00 0/00 0/00 Output bits 11 10 0/00 0/00 1/11 1/11 0/11 1/00 1/01 0/11 1/00 0 /10 0/01 t1 1/11 1/01 0/11 1/00 0 /10 0/01 t2 1/11 1/01 0/11 1/00 0 /10 0/01 t3 1/11 1/01 0/11 1/00 0 /10 1/01 0/01 t4 Lecture 10 0 /10 0/01 t5 t6 19 Trellis – cont’d Tail bits Input bits 1 0 1 0 0 00 10 11 0/00 0/00 0/00 0/11 0/11 Output bits 11 10 0/00 0/00 1/11 1/11 1/11 0/11 1/00 0 /10 1/01 1/01 0 /10 0/01 t1... contents Input sequence : 1 0 0 Output sequence : 11 10 11 Input m 100 010 001 Branch word u1 u2 1 1 1 0 1 1 Output 1 11 10 11 0 00 00 00 1 Modulo-2 sum: 11 10 11 11 10 00 10 11 Lecture 10 12 Encoder representation – cont’d  Polynomial representation:  We define n generator polynomials, one for each modulo-2 adder Each polynomial is of degree K-1 or less and describes the connection of the shift registers... Current state S3 11 1 /10 S3 11 Lecture 10 input Next state output 0 1 0 1 0 1 0 1 S0 S2 S0 S2 00 11 11 00 10 01 01 10 S1 S3 S1 S3 17 Trellis – cont’d  Trellis diagram is an extension of the state diagram that shows the passage of time  Example of a section of trellis for the rate ½ code State 0/00 S 0 = 00 1/11 S 2 = 10 0/11 S1 = 01 1/01 S3 = 11 1/00 0 /10 0/01 1 /10 ti ti +1 Lecture 10 Time 18 Trellis... −1 Lecture 10 15 State diagram – cont’d  A state diagram is a way to represent the encoder  A state diagram contains all the states and all possible transitions between them  Only two transitions initiating from a state  Only two transitions ending up in a state Lecture 10 16 State diagram – cont’d 0/00 1/11 S2 S0 00 Input Output (Branch word) 0/11 S0 00 01 1/00 10 S1 01 0/01 S2 10 S1 0 /10 1/01 Current... representation:  We define n binary vector with K elements (one vector for each modulo-2 adder) The i:th element in each vector, is “1” if the i:th stage in the shift register is connected to the corresponding modulo2 adder, and “0” otherwise  Example: g1 = (111) g 2 = (101 ) u1 m u1 u2 u2 Lecture 10 11 Encoder representation – cont’d  Impulse response representaiton:  The response of encoder to... with m( X )g 2 ( X ) Lecture 10 13 Encoder representation –cont’d In more details: m( X )g1 ( X ) = (1 + X 2 )(1 + X + X 2 ) = 1 + X + X 3 + X 4 m( X )g 2 ( X ) = (1 + X 2 )(1 + X 2 ) = 1 + X 4 m( X )g1 ( X ) = 1 + X + 0 X 2 + X 3 + X 4 m( X )g 2 ( X ) = 1 + 0 X + 0 X 2 + 0 X 3 + X 4 U( X ) = (1,1) + (1,0) X + (0,0) X 2 + (1,0) X 3 + (1,1) X 4 U = 11 10 00 10 11 Lecture 10 14 State diagram     A... 10 0 /10 0/01 t5 t6 19 Trellis – cont’d Tail bits Input bits 1 0 1 0 0 00 10 11 0/00 0/00 0/00 0/11 0/11 Output bits 11 10 0/00 0/00 1/11 1/11 1/11 0/11 1/00 0 /10 1/01 1/01 0 /10 0/01 t1 t2 t3 0 /10 0/01 t4 Lecture 10 t5 t6 20 . Digital Communications I: Modulation and Coding Course Period 3 - 2007 Catharina Logothetis Lecture 10 Lecture 10 2 Last time, we talked about:  Channel coding  Linear block. two transitions ending up in a state Lecture 10 17 State diagram – cont’d 10 01 00 11 outputNext state inputCurrent state 101 010 11 011 100 10 001 110 01 111 000 00 0 S 1 S 2 S 3 S 0 S 2 S 0 S 2 S 1 S 3 S 3 S 1 S 0 S 1 S 2 S 3 S 1/11 1/00 1/01 1 /10 0/11 0/00 0/01 0 /10 Input Output (Branch. otherwise.  Example: m 1 u 2 u 21 uu )101 ( )111( 2 1 = = g g Lecture 10 12 Encoder representation – cont’d  Impulse response representaiton:  The response of encoder to a single “one” bit that goes through it.  Example: 1100 1 0101 0 1 1100 1 1101 1 :sequenceOutput