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EEE436

EEE436. DIGITAL COMMUNICATION Coding. Error-Correcting capability of the Convolutional Code.

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EEE436

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  1. EEE436 DIGITAL COMMUNICATION Coding EEE377 Lecture Notes

  2. Error-Correcting capability of the Convolutional Code The error-correcting capability of a convolutional code is determined by its constraint length K = L + 1 where L is the number of bits of message sequence in the shift registers and also the free distance, dfree The constraint length of a convolutional code, expressed in terms of message bits is equal to the number of shifts over which a single message bit can influence the encoder output. In an encoder with an L-stage shift register, the memory of the encoder equals L message bits and K=L+1 shifts are required for a message bit to enter the shift register and finally come out. Thus the constraint length = K. The constraint length determines the maximum free distance of a code Free distance is equal to the minimum Hamming distance between any two codewords in the code. A convolutional code can correct t errors if dfree > 2t EEE377 Lecture Notes

  3. Error-correction The free distance can be obtained from the state diagram by splitting node a into ao and a1 EEE377 Lecture Notes

  4. Rules A branch multiplies the signal at its input node by the transmittance characterising that branch A node with incoming branches sums the signals produced by all of those branches The signal at a node is applied equally to all the branches outgoing from that node The transfer function of the graph is the ratio of the output signal to the input signal EEE377 Lecture Notes

  5. D on a branch describes the Hamming weight of the encoder output corresponding to that branch The exponent of L is always equal to one since the length of each branch is one. Let T(D,L) denote the transfer function of the signal flow graph Using rules 1, 2 and 3 to obtain the following input-output relations b = D2Lao + Lc c= DLb + DLd d=DLb + DLd a1=D2Lc EEE377 Lecture Notes

  6. Solving the set of equations for the ratio a1/a0, the transfer function of the graph is given by T(D,L) = D5L3 / 1-DL(1+L) We can use the binomial expansion and power series to get the expression for the distance transfer function (with L=1) as T(D,1)=D5 + 2D6 + 4D7 + …….. Since the free distance is the minimum Hamming distance between any two codewords in the code and the distance transfer function T(D,1) enumerates the number of codewords that are a given distance apart, it follows that the exponent of the first term in the expansion T(D,1) defines the free distance=5. Therefore the (2,1,2) convolutional encoder with constraint length K = 3, can only correct up to 2 errors. EEE377 Lecture Notes

  7. Individual Assignment • Given that the (2,1,2) convolutional encoder is used to encode a message, m = 0000 • After encoding the codeword, c is sent to the receiver. • The receiver received codeword, r = 110001000000 • Using Viterbi algorithm, decode the received codeword and recover the original message. • Show your progression of the Trellis diagrams • Give some comments and explanation on your answers • Individual submissions by Friday 5:00pm. Submit to my room. EEE377 Lecture Notes

  8. Error-correction EEE377 Lecture Notes

  9. Turbo Codes • A relatively new class of convolutional codes first introduced in 1993 • A basic turbo encoder is a recursive systematic encoder that employs two convolutional encoders (recursive systematic convolutional or RSC in parallel, where the second encoder is preceded by a pseudorandom interleaver to permute the symbol sequence • Turbo code is also known as Parallel Concatenated Codes (PCC) Systematic bits, xk Message bits Parity bits, y1k RSC encoder 1 Puncture & MUX Totransmitter Interleaver RSC encoder 2 Parity bits, y2k EEE377 Lecture Notes

  10. Turbo Codes • The input data stream is applied directly to encoder 1 and the pseudorandomly reordered version of the same data stream is applied to encoder 2 • Both encoders produce the parity bits. • The parity bits and the original bit stream are multiplexed and then transmitted • The block size is determined by the size of the interleaver for example 65, 536 is common) • Puncture is applied to remove some parity bits to maintain the code rate at 1/2 . For example, by eliminating odd parity bits from the first RSC and the even parity bits from the second RSC Systematic bits, xk Message bits Parity bits, y1k RSC encoder 1 Puncture & MUX Totransmitter Interleaver RSC encoder 2 Parity bits, y2k EEE377 Lecture Notes

  11. RSC encoder for Turbo encoding EEE377 Lecture Notes

  12. RSC encoder for Turbo encoding 0011 0001 0001 0001 0001 1111 Recursive Non-recursive More 1’s for recursive gives better error performance EEE377 Lecture Notes

  13. Turbo decoding Turbo decoder consists of two maximum a posterior (MAP) decoders and a feedback path Decoding operates on the noisy version of the systematic bits and the two sets of parity bits in two decoding stages to produce estimate of the original message bits The first decoder takes the information from the received signal and calculates the A posterior probability (APP) value This value is then used as the a priori probability value for the second decoder The output is then fedback to the first decoder where the process repeats in an iterative fashion with each iteration producing more refined estimates. EEE377 Lecture Notes

  14. Turbo decoding uses BCJR algorithm • BCJR ( Bahl, Cocke, Jelinek and Raviv, 1972) algorithm is a maximum a posteriori probability (MAP) decoder which minimizes the bit errors by estimating the a posteriori probabilitities of the individual bits in a code word. • It takes into account the recursive character of the RSC codes and computes a log-likelihood ratio to estimate the APP for each bit. EEE377 Lecture Notes

  15. Low Density Parity Check (LDPC) codes • An LDPC code is specified in terms of its parity-check matrix, H that has the following structural properties • Each row consists of ρ 1’s • Each column consists of γ 1’s • The number of 1’s in common between any two columns is no greater than 1; ie λ= 0 or 1 • Both ρ and γ are small compared with the length of the code • LDPC codes are recognised in the form of (n, γ, ρ) • H is said to be a low density parity check matrix • H has constant row and column weights (ρ and γ ) • Density of H = total number of 1’s divided by total number of entries in H EEE377 Lecture Notes

  16. Low Density Parity Check (LDPC) codes • Example (15, 4, 4) LDPC code • Each row consists of ρ=4 1’s • Each column consists of γ=4 1’s • The number of 1’s in common between any two columns is no greater than 1; ie λ= 0 or 1 • Both ρ and γ are small compared with the length of the code • Density = 4/15 = 0.267 EEE377 Lecture Notes

  17. Low Density Parity Check (LDPC) codes – Constructing H • For a given choice of ρ and γ, form a kγby kρmatrix H (where k = a positive integer > 1) that consists of γ k-by-kρ submatrix, H1, H2,….Hγ • Each row of a submatrix has ρ 1’s and each column of a submatrix contains a single 1 • Therefore each submatrix has a total of kρ 1’s. • Based on this, construct H1 by appropriately placing the 1’s. • For ,the ith row of H1 contains all its ρ 1’s in columns (i-1) ρ+1 to iρ • The other submatrices are merely column permutations of H1 EEE377 Lecture Notes

  18. Low Density Parity Check (LDPC) codes – Example Constructing H • Choice of ρ=4 and γ=3 and k=5 • Form a kγby kρ (15-by-20) matrix H that consists of γ=3 k-by-kρ (5-by-20) submatrix, H1, H2,Hγ=3 • Each row of a submatix has ρ=4 1’s and each column of a submatrix contains a single 1 • Therefore each submatrix has a total of kρ=20 1’s. • Based on this, construct H1 by appropriately placing the 1’s. • For ,the ith row of H1 contains all its ρ=4 1’s in columns (i-1) ρ+1 to iρ • The other submatrices are merely column permutations of H1 EEE377 Lecture Notes

  19. Low Density Parity Check (LDPC) codes – • Example Constructing H for (20, 3, 4) LDPC code EEE377 Lecture Notes

  20. 0 • Construction of Low Density Parity Check (LDPC) codes • There are many techniques of constructing the LDPC codes • Constructing LDPC codes with shorter block is easier than the longer ones • For large block sizes, LDPC codes are commonly constructed by first studying the behaviour of decoders. • Among the techniques are Pseudo-random techniques, Combinatorial approaches and finite geometry. These are beyond the scope of this lecture. • For this lecture, we see how short LDPC codes are constructed from a given parity check matrix. • For example a (6,3) linear LDPC code given by the following H EEE377 Lecture Notes

  21. 0 • Construction of Low Density Parity Check (LDPC) codes • For example a (6,3) linear LDPC code given by the following H • The 8 codewords can be obtained by putting the parity-check matrix H into this form H=[PT I In-k ]; where P=coefficient matrix and In-k = identity matrix • The generator matrix is, G = [In-k I P] • At the receiver, H=[PT I In-k ] is used to check the error syndrome. Exercise: Generate the codeword for m=001 and show how the receiver performs the error checking EEE377 Lecture Notes

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