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8/8/2019 3320 Conv Coding
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S.723320 Advanced Digital Communication (4 cr)
Convolutional Codes
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Timo O. Korhonen, HUT Communication Laboratory
Targets today
x Why to apply convolutional coding?
x Defining convolutional codes
x Practical encoding circuits
x Defining quality of convolutional codes
x
Decoding principlesx Viterbi decoding
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Timo O. Korhonen, HUT Communication Laboratory
Convolutional encoding
x Convolutional codes are applied in applications that require good performance with
low implementation complexity. They operate on code streams (not in blocks)
x Convolution codes have memory that utilizes previous bits to encode or decode
following bits (block codes are memoryless)
x Convolutional codes are denoted by (n,k,L), where L is code (or encoder) Memory
depth (number of register stages)
x Constraint length C=n(L+1) is defined as the number of encoded bits a message
bit can influence tox Convolutional codes achieve good performance by expanding their memory depth
n(L+1) output bits
input bit
(n,k,L)
encoder
(n,k,L)
encoder
k bits n bits
encoded bits
message bits
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Timo O. Korhonen, HUT Communication Laboratory
Example: Convolutional encoder, k = 1, n = 2
x Convolutional encoder is a finite state machine (FSM) processing information bits
in a serial manner
x Thus the generated code is a function of input and the state of the FSM
x In this (n,k,L) = (2,1,2) encoder each message bit influences a span of C= n(L+1)=6 successive output bits = constraint length C
x Thus, for generation of nbit output, we require in this example n shift registers in k =
1 convolutional encoder
2 1
2
'
''
j j j j
j j j
x m m m
x m m
− −
−
= ⊕ ⊕
= ⊕
1 1 2 2 3 3' '' ' '' ' '' ...
out x x x x x x x=
(n,k,L) = (2,1,2) encoder
memory
depth L
= number
of states
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5Timo O. Korhonen, HUT Communication Laboratory
Example: (n,k,L)=(3,2,1) Convolutional encoder
3 2' j j j j
x m m m− −
= ⊕ ⊕
3 1''
j j j j x m m m
− −= ⊕ ⊕
2'''
j j j x m m
−= ⊕
•After each new block of k input bitsfollows a transition into new state•Hence, from each input state
transition, 2k different output states
may follow•Each message bit influences a span
of C = n(L+1) = 3(1+1) = 6
successive output bits
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6Timo O. Korhonen, HUT Communication Laboratory
Generator sequences
x (n,k,L) Convolutional code can be described by the generator sequences
that are the impulse responses for each coder n output branches:
x Generator sequences specify convolutional code completely by the
associated generator matrix
x Encoded convolution code is produced by matrix multiplication of the
input and the generator matrix
(1) (2) ( )
, ,...n
g g g
(1)
( 2)
[1 0 11]
[1111]
=
=
g
g
(2,1,2) encoder
Note that the generator sequence length
exceeds register depth always by 1
(n,k,L)
encoder
(n,k,L)
encoder
k bits n bits
(1)
0g (1)
2g (1)
mg
( ) ( ) ( ) ( )
0 1[ ]n n n n
m g g g =g L
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Convolution point of view in encoding and
generator matrixx Encoder outputs are formed by modulo2
discrete convolutions:
where u is the information sequence:
x Therefore, the l :th bit of the j:th output branch is*
x Hence, for this circuit the following equations result,
(assume: )
(1) (1) ( 2) (2) ( ) ( )* , * ... * j j= = =v u g v u g v u g
( ) ( ) ( ) ( ) ( )0 1 10
...
where 1, 0,
m j j j j jl l i l l l l m mi
l i
v u g u g u g u g
m L u l i
− − −=
−
= = + + +
= + <∑
@
0 1( , , )u u=u L
(1)
2 3
( 2)
1 2 3
l l l l
l l l l l
v u u uv u u u u
− −
− − −
= + += + + +
(1)
( 2)
[1 0 11]
[111 1]
= =
g
g
(1) ( 2) (1) ( 2) (1) ( 2)
0 0 1 1 2 2[ ...]v v v v v v=v
encoder output: j
branches
n(L+1) output bits
input bit
2 L =
*note that u is reversed in time as in the definition of convolution top right
( ) ( ) ( ) xyk A
x y u x k y u k ∈∑⊗ = −
3l u −
(1)
3 g (1)
2 g 2l u −
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Timo O. Korhonen, HUT Communication Laboratory
Example: Using generator matrix
(1)
( 2)
[1 0 11]
[111 1]
= =
g
g
Verify that you can obtain the result shown!
11 10
01
11 00 01 11 01⊕ ⊕ ⊕ =1 2 3 1 2 3
1 4 4 2 4 4 3(1) ( 2)
0 0g g (1) ( 2)
1 1g g
( ) ( )
0
( )
1 1
( )...
j j
l l
j
l
j
l m m
v u g
u g
u g
−
−
=
+
+ +
(1) ( 2)
m mg g
l m
u−
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Timo O. Korhonen, HUT Communication Laboratory S.Lin, D.J. Costello: Error Control Coding, II ed, p. 456
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Timo O. Korhonen, HUT Communication Laboratory
2 1
2
'
''
j j j j
j j j
x m m m
x m m
− −
−
= ⊕ ⊕
= ⊕
1 1 2 2 3 3' '' ' '' ' '' ...out x x x x x x x=
This tells how one input bit
is transformed into two output bits
(initially register is all zero)
Representing convolutional codes: Code tree
2 1 0 1 j jm m− −
=
' 1
''
0
1
1
0
j
j
x
x
= ⊕ ⊕
= ⊕
' 0
''
0
0
1
0
j
j
x
x
= ⊕ ⊕ = ⊕
(n,k,L) = (2,1,2) encoder
' 0
''
1
0
0
1
j
j
x
x
= ⊕ ⊕
= ⊕
Number of brachesdeviating from each node
equals 2k
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Timo O. Korhonen, HUT Communication Laboratory
Representing convolutional codes compactly:
code trellis and state diagram
Shift register states
Input state ‘1’
indicated by dashed line
Code trellisState diagram
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Timo O. Korhonen, HUT Communication Laboratory
Inspecting state diagram: Structural properties of
convolutional codesx Each new block of k input bits causes a transition into new state
x Hence there are 2k branches leaving each statex Assuming encoder zero initial state, encoded word for any input of k
bits can thus be obtained. For instance, below for u=(1 1 1 0 1),
encoded word v=(1 1, 1 0, 0 1, 0 1, 1 1, 1 0, 1 1, 1 1) is produced:
 encoder state diagram for (n,k,L)=(2,1,2) code
 note that the number of states is 8 = 2L+1 => L = 2 (two state bits)
Verify that you obtain the same result!
Input state
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13Timo O. Korhonen, HUT Communication Laboratory
Code weight, path gain, and generating function
x The state diagram can be modified to yield information on code distance
properties (= tells how good the code is to detect or correct errors)x Rules (example on the next slide):
– (1) Split S0into initial and final state, remove selfloop
– (2) Label each branch by the branch gain X i.Here i is the weight* of the n
encoded bits on that branch
– (3) Each path connecting the initial state and the final state represents anonzero code word that diverges and reemerges with S
0only once
x The path gain is the product of the branch gains along a path, and the weight of
the associated code word is the power of X in the path gain
x Code weigh distribution is obtained by using a weighted gain formula to
compute its generating function (inputoutput equation)
where Ai is the number of encoded words of weight i
( )i
i
AT X X ∑= i
*In linear codes, weight is the number of ‘1’:s in the encoder output
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14Timo O. Korhonen, HUT Communication Laboratory
Example: The path representing the state
sequence S0S1S3S7S6S5S2S4S0 has the
path gain X2X1X1X1X2X1X2X2=X12
and the corresponding code word
has the weight of 12
6 7 8
9 10
( )
3 5
11 25 ....
i
ii
T X A X
X X X
X X
∑=
= + +
+ + +
Where does these terms come from?
weight: 1weight: 2 branch
gain
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15Timo O. Korhonen, HUT Communication Laboratory
Distance properties of convolutional codes
x Code strength is measured by the minimum free distance:
where v’ and v’’ are the encoded words corresponding information
sequences u’ and u’’. Code can correct up to errors.
x The minimum free distance d free denotes:
w The minimum weight of all the paths in the state diagram thatdiverge from and remerge with the allzero state S
0
w The lowest power of the codegenerating function T(X)
{ }min ( ', '') : ' '' free
d d = ≠v v u u
6 7 8
9 10
( )
3 511 25 ....
i
ii
T X A X
X X X X X
∑=
= + ++ + +
6 free
d ⇒ =
( )/ 2 / 2 1c free c free
G kd n R d = = >
Code gain*:
/ 2 freet d <
* for derivation, see Carlson’s, p. 583
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16Timo O. Korhonen, HUT Communication Laboratory
Coding gain for some selected convolutional codes
x Here is a table of some selected convolutional codes and their
code gains RC d free /2 expressed for hard decoding also by
1010log ( / 2) dB
freec R d γ =
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Decoding of convolutional codes
x Maximum likelihood decoding of convolutional codes means finding the
code branch in the code trellis that was most likely transmittedx Therefore maximum likelihood decoding is based on calculating code
Hamming distances for each branch potentially forming encoded word
x Assume that the information symbols applied into an AWGN channel areequally alike and independent
x
Let’s denote by x encoded symbols (no errors) and by y received(potentially erroneous) symbols:
x Probability to decode the symbols is then
x
The most likely path through the trellis will maximize this metric.Often ln() is taken from both sides, because probabilities are oftensmall numbers, yielding:
(note this corresponds equavalently also the smallest Hamming distance)
0 1 2... ...
j x x x x=x
0 1... ...
j y y y=y
0
( , ) (  ) j j
j
p p y x∞
=∏=y x
Decoder
(=distance
calculation)x
yreceived code:
non 
erroneous code:
bit
decisions
{ } { }1
ln ( , ) ln ( ) j mj
j
p p y x∞
=∑=y x
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18Timo O. Korhonen, HUT Communication Laboratory
Example of exhaustive maximal likelihood detection
x Assume a three bit message is transmitted and encoded by (2,1,2)
convolutional encoder. To clear the decoder, two zerobits are appendedafter message. Thus 5 bits are encoded resulting 10 bits of code. Assume
channel error probability is p = 0.1. After the channel 10,01,10,11,00 is
produced (including some errors). What comes after the decoder, e.g. what
was most likely the transmitted code and what were the respective message
bits?
a
b
c
d
states
decoder outputs
if this path is selected
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19Timo O. Korhonen, HUT Communication Laboratory
0
0
( , ) (  )
ln ( , ) ln (  )
j j j
j j j
p p y x
p p y x
∞
=
∞=
∏
∑
=
=
y x
y x
errorscorrect
weight for prob. toreceive bit inerror
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20Timo O. Korhonen, HUT Communication Laboratory
Note also the Hamming distances!
correct:1+1+2+2+2=8;8 ( 0.11) 0.88
false:1+1+0+0+0=2;2 ( 2.30) 4.6
total path metric: 5.48
⋅ − = −⋅ − = −
−
The largest metric, verify
that you get the same result!
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21Timo O. Korhonen, HUT Communication Laboratory
The Viterbi algorithm
x Problem of optimum decoding is to find the minimum distance path
from the initial state back to the initial state (below from S 0 to S 0). Theminimum distance is one of the sums of all path metrics from S
0to S
0
x Exhaustive maximum likelihood
method must search all the paths
in phase trellis (2k paths emerging/
entering from 2 L+1
states for an (n,k,L) code)
x The Viterbi algorithm gets
improvement in computational
efficiency via concentrating into
survivor paths of the trellis
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22Timo O. Korhonen, HUT Communication Laboratory
The survivor pathx Assume for simplicity a convolutional code with k =1, and thus up to 2k = 2
branches can enter each state in trellis diagram
x Assume optimal path passes S . Metric comparison is done by adding the metric of
S 1 and S 2 to S . At the survivor path the accumulated metric is naturally smaller (otherwise it could not be the optimum path)
x For this reason the nonsurvived path can
be discarded > all path alternatives need not
to be further considered
x Note that in principle the whole transmitted
sequence must be received before decision.
However, in practice storing of states for
input length of 5 L is quite adequate
2 branches enter each nodek
2 nodes, determined L
by memory depth
branch of larger metric discarded
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23Timo O. Korhonen, HUT Communication Laboratory
Example of using the Viterbi algorithm
x Assume the received sequence is
and the (n,k,L)=(2,1,2) encoder shown below. Determine the Viterbi
decoded output sequence!
01101111010001 y =
(Note that for this encoder code rate is 1/2 and memory depth equals L = 2)
states6 44 7 4 48
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25Timo O. Korhonen, HUT Communication Laboratory
How to endup decoding?
x In the previous example it was assumed that the register was finally
filled with zeros thus finding the minimum distance pathx In practice with long code words zeroing requires feeding of long
sequence of zeros to the end of the message bits: this wastes channel
capacity & introduces delay
x To avoid this path memory truncation is applied:
– Trace all the surviving paths to thedepth where they merge
– Figure right shows a common point
at a memory depth J
– J is a random variable whose applicable
magnitude shown in the figure (5 L)has been experimentally tested for
negligible error rate increase
– Note that this also introduces the
delay of 5 L! 5 stages of the trellis J L>
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26Timo O. Korhonen, HUT Communication Laboratory
Lessons learned
x You understand the differences between cyclic codes and
convolutional codesx You can create state diagram for a convolutional encoder
x You know how to construct convolutional encoder circuits
based on knowing the generator sequences
x You can analyze code strengths based on known code
generation circuits / state diagrams or generator sequences
x You understand how to realize maximum likelihood
convolutional decoding by using exhaustive search
x You understand the principle of Viterbi decoding