Transcript EEE436 - Universiti Sains Malaysia

EEE436
DIGITAL COMMUNICATION
Coding
En. Mohd Nazri Mahmud
MPhil (Cambridge, UK)
BEng (Essex, UK)
[email protected]
Room 2.14
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Convolutional Codes
Unlike block codes that operate on block-by-block basis, the
convolutional codes are based on message bits that come in serially
rather than in blocks.
The structure of a generic convolutional encoder can be written in a
compact form of (n, k, L) ; n= no of streams, or no of modulo-2 adder;
k=number of group of shift register; L=number of state of the shift register
The encoder takes an L-bit message sequence to produce a coded
output sequence of length n(L+M) bits, where n=no of stream of the
encoded bits, M=no of bits of message sequence
The encoder is implemented by a tapped shift register with L+1
stages.
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Convolutional Codes
cj
The message bits in the register are combined by modulo-2 addition
to form the encoded bit, cj
Cj = mj-L.gL + …..+ mj-1.g1 + mj.g0
L
=
 m j  i .g i
i 0
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Convolutional Codes – Example
A (2,1,2) convolutional encoder with n=2, k=1 and L=2
c’j
c’’j
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Convolutional Codes
To provide the extra check bits for error control, the encoded bits
from multiple streams are interleaved.
For example, a convolutional encoder with n=2 (ie two streams of
encoded bits)
c’j
c’’j
Encoded bits from stream 1, c’j = mj-2 + mj-1 + mj
Encoded bits from stream 2, c’’j = mj-2 + mj
interleaved
C= c’1c’’1c’2c’’2c’3c’’3………..
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Convolutional Codes
Each stream may be characterised in terms of its impulse response
( the response of that stream to a symbol 1 with zero initial
condition).
Every stream is also characterised in terms of a generator
polynomial, which is the unit-delay transform of the impulse
response
Let the following generator sequence denotes the impulse
response of the i-th path
 g i  , g i  , g i  ,........,g i  
 0
M 
1
2

The generator polynomial of the i-th path is given by
 g i  D   g i   g i D  g i D 2  ........ g i  D M 


M
0
1
2


Where D denotes the unit-delay variable
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Convolutional Codes – Example
A convolutional encoder with two streams of encoded bits with
message sequence 10011 as an input
Stream 1
c’j
c’’j
Stream 2
First, we find the impulse response of both streams to a symbol 1.
Impulse response of stream 1 = (1 1 1)
Impulse response of stream 2 = (1 0 1)
Then, write the corresponding generator polynomial of both streams
 g i  D   g i   g i D  g i D 2  ........ g i  D M 


M
0
1
2


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Convolutional Codes – Example
First, we find the impulse response of both streams to a symbol 1.
Impulse response of stream 1 = (1 1 1)
Impulse response of stream 2 = (1 0 1)
Then, write the corresponding generator polynomials of both streams
 g 1D   1  D  D 2 




 g 2D   1  D 2 




Then, write the message polynomial for input message (10011)
 mD   1  D 3  D 4 


Then, find the output polynomial for both streams by multiplying the
generator polynomial and the message polynomial
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Convolutional Codes – Example
Then, find the output polynomial for both streams by multiplying the
generator polynomial and the message polynomial
 c 2D   g 2D .mD 




 c 1D   g 1D .mD 




= (1 + D + D2)(1 + D3 + D4)
= (1 + D2)(1 + D3 + D4)
= 1 + D + D2 + D3 + D6
= 1 + D2 + D3 + D4 + D5 + D6
So, the output sequence for stream 1
is 1111001
The output sequence for stream 2
is 1011111
Interleave
C = 11,10,11,11,01,01,11
Original message (10011)
Encoded sequence (11101111010111)
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Convolutional Codes – Exercise
A convolutional encoder with two streams of encoded bits with
message sequence 110111001 as an input
Stream 1
c’j
c’’j
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Stream 2
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Convolutional Codes – Exercise
First, we find the impulse response of both streams to a symbol 1.
Impulse response of stream 1 = (1 1 1)
Impulse response of stream 2 = (1 0 1)
Then, write the corresponding generator polynomials of both streams
 g 1D   1  D  D 2 




 g 2D   1  D 2 




Then, write the message polynomial for input message (110111001)
 mD   1  D  D 3  D 4  D 5  D8 


Then, find the output polynomial for both streams by multiplying the
generator polynomial and the message polynomial
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Convolutional Codes – Exercise
Then, find the output polynomial for both streams by multiplying the
generator polynomial and the message polynomial
 c 2D   g 2D .mD 




 c 1D   g 1D .mD 




= 1 + D5 + D7 + D8 + D9 + D10
= 1 + D +D2 + D4 + D6 + D7 + D8 + D10
So, the output sequence for stream 1
is 10000101111
The output sequence for stream 2
is 11101011101
Interleave
C = 11 01 01 00 01 10 01 11 11 10 11
Original message (110111001)
Encoded sequence (11 01 01 00 01 10 01 11 11 10 11 )
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Convolutional Codes – Code Tree, Trellis and State Diagram
Draw a code tree for the convolutional encoder below.
Stream 1
c’j
c’’j
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Stream 2
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Code Tree
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Trellis
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State Diagram
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Maximum Likelihood Decoding of a Convolutional Code
Let m denotes a message vector
c denotes the corresponding code vector
r the received code vector (which may differ from c due to channel noise)
Given the received vector r, the decoder is required to make an estimate m^
of the message vector
Since there is a one-to-one correspondence between the message vector m and
the code vector c, therefore
m^ = m only if c^ = c , otherwise a decoding error is committed
The decoding rule is said to be optimum when the probability of decoding error is
minimised
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Maximum Likelihood Decoding of a Convolutional Code
Let say both the transmitted code vector c and the received vector r represent
binary sequence of length N.
These two sequences may differ from each other in some locations because of
errors due to channel noise.
Let ci and ri denote the ith element of c and r respectively
We then have
pr c    pri ci 
N
i 1
The decoding rule is said to be optimum when the probability of decoding error is
minimised
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pr c    pri ci 
N
i 1
Correspondingly, the log-likelihood is
log pr c    log pri ci 
N
i 1
The probability of decoding error is minimised if the log-likelihood
function is maximized (the estimate c^ is chosen to maximise the function)
Let the transition probability
pri ci 
pri ci  

be defined as
p
1 p
If ri  ci
If ri  ci
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Suppose also that the received vector r differs from the transmitted code vector c in
exactly d positions (where d is the Hamming distance between vectors r and c)
We may rewrite the log-likelihood function
log p r c    log p ri ci 
N
i 1
as
log pr c   d log p  N  d  log1  p 
 p 
  N log1  p 
 d log
1

p


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log pr c   d log p  N  d  log1  p 
 p 
  N log1  p 
 d log
1

p


In general, the probability of error occurring is low enough for us to assume p < ½
And also that Nlog(1-p) is constant for all c.
The function is maximised when d is minimised. (ie the smallest Hamming distance)
“ Choose the estimate c^ that minimise the Hamming distance between
The received vector r and the transmitted vector c.”
The maximum likelihood decoder is reduced to a minimum distance decoder.
The received vector r is compared with each possible transmitted code vector c
and the particular one “closest” to r is chosen as the correct transmitted code vector.
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The Viterbi Algorithm
The equivalence between maximum likelihood decoding and minimum distance
decoding implies that we may decode a convolutional code by choosing a path
in the code tree whose coded sequence differs from the received sequence in the
fewest number of places
Since a code tree is equivalent to a trellis, we may limit our choice to the possible
Paths in the trellis representation of the code.
Viterbi algorithm operates by computing a metric or discrepancy for every possible
Path in the trellis
The metric for a particular path is defined as the Hamming distance between the
Coded sequence represented by that path and the received sequence.
For each state in the trellis, the algorithm compares the two paths entering the node
And the path with the lower metric is retained.
The retained paths are called survivors.
The sequence along the path with the smallest metric is the maximum likelihood
Choice and represent the transmitted sequence
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The Viterbi Algorithm -example
In the circuit beliow, suppose that the encoder generates an all-zero sequence and
that the received sequence is (0100010000…) in which the are two errors due to
channel noise: one in second bit and the other in the sixth bit.
We can show that this double-error pattern is correctable using the Viterbi algorithm.
Stream 1
c’j
c’’j
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Stream 2
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The Viterbi Algorithm -example
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The Viterbi Algorithm -exercise
Using the same circuit, suppose that the received sequence is 110111.
Using the Viterbi algorithm, what is the corresponding encoded sequence
transmitted by the receiver? What is the original message bit?
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