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Modulation and coding course- lecture 11

Another class of linear codes, known as Convolutional codes. -We studied the structure of the encoder and different ways for representing it. Modulation, Demodulation and Coding Course Period 3 - 2007 Catharina Logothetis Lecture 11 Last time, we talked about: Another class of linear codes, known as Convolutional codes. We studied the structure of the encoder and different ways for representing it. Lecture 11 2 Today, we are going to talk about: What are the state diagram and trellis representat

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  1. Modulation, Demodulation and Coding Course Period 3 - 2007 Catharina Logothetis Lecture 11
  2. Last time, we talked about: Another class of linear codes, known as Convolutional codes. We studied the structure of the encoder and different ways for representing it. Lecture 11 2
  3. Today, we are going to talk about: What are the state diagram and trellis representation of the code? How the decoding is performed for Convolutional codes? What is a Maximum likelihood decoder? What are the soft decisions and hard decisions? How does the Viterbi algorithm work? Lecture 11 3
  4. Block diagram of the DCS Information Rate 1/n Modulator source Conv. encoder m = (m1 , m2 ,..., mi ,...) U = G(m) 144 44 2 3 = (U1 , U 2 , U 3 ,..., U i ,...) Channel Input sequence 144 2444 4 3 Codeword sequence U i = u1i ,...,u ji ,...,u ni 14 2444 3 Branch word ( n coded bits) Information Rate 1/n Demodulator sink Conv. decoder m = (m1 , m2 ,..., mi ,...) ˆ ˆ ˆ ˆ Z = ( Z1 , Z 2 , Z 3 ,..., Z i ,...) 144 2444 4 3 received sequence Zi = z1i ,...,z ji ,...,z ni { 14 2444 3 Demodulator outputs n outputs per Branch word for Branch word i Lecture 11 4
  5. State diagram A finite-state machine only encounters a finite number of states. State of a machine: the smallest amount of information that, together with a current input to the machine, can predict the output of the machine. In a Convolutional encoder, the state is represented by the content of the memory. Hence, there are 2 states. K −1 Lecture 11 5
  6. 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 11 6
  7. State diagram – cont’d Current input Next output 0/00 Output state state (Branch word) S0 Input S0 0 S0 00 1/11 00 0/11 00 1 S2 11 1/00 S1 0 S0 11 S2 S1 10 01 01 1 S2 00 0/10 S2 0 S1 10 1/01 S3 0/01 10 1 S3 01 0 01 11 S3 S1 1/10 11 1 S3 10 Lecture 11 7
  8. 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 S 0 = 00 0/00 1/11 S 2 = 10 0/11 1/00 S1 = 01 1/01 0/10 0/01 S3 = 11 1/10 ti ti +1 Time Lecture 11 8
  9. Trellis –cont’d A trellis diagram for the example code Input bits Tail bits 1 0 1 0 0 Output bits 11 10 00 10 11 0/00 0/00 0/00 0/00 0/00 1/11 1/11 1/11 1/11 1/11 0/11 0/11 0/11 0/11 0/11 1/00 1/00 1/00 1/00 1/00 0/10 0/10 0/10 0/10 0/10 1/01 1/01 1/01 1/01 1/01 0/01 0/01 0/01 0/01 0/01 t1 t 2 t 3 t 4 t 5 t 6 Lecture 11 9
  10. Trellis – cont’d Input bits Tail bits 1 0 1 0 0 Output bits 11 10 00 10 11 0/00 0/00 0/00 0/00 0/00 1/11 1/11 1/11 0/11 0/11 0/11 0/10 1/00 0/10 0/10 1/01 1/01 0/01 0/01 t1 t 2 t 3 t 4 t 5 t 6 Lecture 11 10
  11. Optimum decoding If the input sequence messages are equally likely, the optimum decoder which minimizes the probability of error is the Maximum likelihood decoder. ML decoder, selects a codeword among all the possible codewords which maximizes the likelihood (m′ ) function p (Z | U ) where Z is the received sequence and U (m′) is one of the possible codewords: 2 L codewords to search!!! ML decoding rule: Choose U ( m′) if p(Z | U ( m′) ) = max(m) p(Z | U ( m ) ) over all U Lecture 11 11
  12. ML decoding for memory-less channels Due to the independent channel statistics for memoryless channels, the likelihood function becomes ∞ ∞ n p(Z | U (m) ) = p z1 , z2 ,..., zi ,... ( Z1 , Z 2 ,..., Z i ,... | U ( m) ) = ∏ p( Z i | U i (m) ) =∏∏ p ( z ji | u (jim ) ) i =1 i =1 j =1 and equivalently, the log-likelihood function becomes ∞ ∞ n γ U (m) = log p(Z | U (m) ) = ∑ log p ( Z i | U i (m) ) = ∑∑ log p ( z ji | u (jim ) ) i =1 i =1 j =1 Path metric Branch metric Bit metric The path metric up to time index "i", is called the partial path metric. ML decoding rule: Choose the path with maximum metric among all the paths in the trellis. This path is the “closest” path to the transmitted sequence. Lecture 11 12
  13. Binary symmetric channels (BSC) 1 1 p Modulator Demodulator p = p(1 | 0) = p(0 | 1) input output p 1 − p = p(1 | 1) = p (0 | 0) 0 1-p 0 If d m = d (Z, U ) is the Hamming distance between Z (m ) and U, then p (Z | U ( m ) ) = p d m (1 − p ) Ln − d m Size of coded sequence ⎛1− p ⎞ γ U (m) = − d m log⎜ ⎜ p ⎟ + Ln log(1 − p ) ⎟ ⎝ ⎠ ML decoding rule: Choose the path with minimum Hamming distance from the received sequence. Lecture 11 13
  14. AWGN channels For BPSK modulation the transmitted sequence corresponding to the codeword U (m ) is denoted by where S ( m ) = ( S1( m ) , S 2( m ) ,..., Si( m ) ,...) and Si ( m ) = ( s1(im ) ,..., s (jim ) ,..., snim ) ) ( and sij = ± Ec . The log-likelihood function becomes ∞ n γ U (m) = ∑∑ z ji s (jim ) =< Z, S ( m ) > Inner product or correlation between Z and S i =1 j =1 Maximizing the correlation is equivalent to minimizing the Euclidean distance. ML decoding rule: Choose the path which with minimum Euclidean distance to the received sequence. Lecture 11 14
  15. Soft and hard decisions In hard decision: The demodulator makes a firm or hard decision whether one or zero is transmitted and provides no other information for the decoder such that how reliable the decision is. Hence, its output is only zero or one (the output is quantized only to two level) which are called “hard- bits”. Decoding based on hard-bits is called the “hard-decision decoding”. Lecture 11 15
  16. Soft and hard decision-cont’d In Soft decision: The demodulator provides the decoder with some side information together with the decision. The side information provides the decoder with a measure of confidence for the decision. The demodulator outputs which are called soft- bits, are quantized to more than two levels. Decoding based on soft-bits, is called the “soft-decision decoding”. On AWGN channels, 2 dB and on fading channels 6 dB gain are obtained by using soft-decoding over hard-decoding. Lecture 11 16
  17. The Viterbi algorithm The Viterbi algorithm performs Maximum likelihood decoding. It find a path through trellis with the largest metric (maximum correlation or minimum distance). It processes the demodulator outputs in an iterative manner. At each step in the trellis, it compares the metric of all paths entering each state, and keeps only the path with the smallest metric, called the survivor, together with its metric. It proceeds in the trellis by eliminating the least likely paths. K −1 It reduces the decoding complexity to L 2 ! Lecture 11 17
  18. The Viterbi algorithm - cont’d Viterbi algorithm: A. Do the following set up: For a data block of L bits, form the trellis. The trellis has L+K-1 sections or levels and starts at time t1 and ends up at time t L + K . Label all the branches in the trellis with their corresponding branch metric. For each state in the trellis at the time ti which is denoted by S (ti ) ∈ {0,1,...,2 K −1} , define a parameter Γ(S (ti ), ti ) B. Then, do the following: Lecture 11 18
  19. The Viterbi algorithm - cont’d 1. Set Γ(0, t1 ) = 0 and i = 2. 2. At time ti , compute the partial path metrics for all the paths entering each state. 3. Set Γ(S (ti ), ti ) equal to the best partial path metric entering each state at time ti . Keep the survivor path and delete the dead paths from the trellis. 1. If i < L + K , increase i by 1 and return to step 2. A. Start at state zero at time t L + K . Follow the surviving branches backwards through the trellis. The path thus defined is unique and correspond to the ML codeword. Lecture 11 19
  20. Example of Hard decision Viterbi decoding m = (101) U = (11 10 00 10 11) Z = (11 10 11 10 01) 0/00 0/00 0/00 0/00 0/00 1/11 1/11 1/11 0/11 0/11 0/11 0/10 1/00 0/10 0/10 1/01 1/01 0/01 0/01 t1 t2 t3 t4 t5 t6 Lecture 11 20
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