Communication engineering -UNIT IV .pptx

UNIT IV - INFORMATION THEORY AND CODING
2/27/2024 MAHENDRA COLLEGE OF ENGINEERING 1
SYLLABUS
• Measure of information
• Entropy
• Source coding theorem
• Shannon–Fano coding, Huffman Coding, LZ Coding
• Channel capacity
• Shannon-Hartley law ,Shannon's limit
• Error control codes – Cyclic codes, Syndrome
calculation
• Convolution Coding, Sequential and Viterbi decoding

INFORMATION THEORY INTRODUCTION
• The purpose of a communication system is to
facilitate the transmission of signals generated by a
information source to the receiver end over a
communication channel.
• Information theory is a branch of probability
theory which may be applied to the study of
communication systems. Information theory allows
us to determine the information content in a
message signal leading to different source coding
techniques for efficient transmission of message.
• In the context of communication, information theory
deals with mathematical modeling and analysis of
communication rather than with physical sources

INFORMATION THEORY INTRODUCTION
• In particular, it provides answers to two
fundamental questions,
• What is the irreducible complexity below which a signal
cannot be compressed?
• What is the ultimate transmission rate for reliable
communication over a noisy channel?
• The answers to these questions lie on the entropy
of the source and capacity of a channel
respectively.
• Entropy is defined in terms of the probabilistic
behavior of a source of information.
• Channel Capacity is defined as the intrinsic ability
of a channel to convey information; it is naturally

INFORMATION
• Information is the source of a communication system,
whether it is analog or digital communication. Information is
related to the probability of occurrence of the event. The unit
of information is called the bit.
• Based on memory, information source can be classified as
follows:
• A source with memory is one for which a current
symbol depends on the previous symbols.
• A memory less source is one for which each symbol
produced is independent of the previous symbols.
• A Discrete Memory less Source (DMS) can be
characterized by list of symbols, the probability
assignment to these symbols, and the specification of

DISCRETE MEMORY LESS SOURCE (DMS)
• Consider a probabilistic experiment involves the observation of the
output emitted by a discrete source during every unit of time (Signaling
Interval). The source output is modeled as a discrete random variable
(S), which takes on symbols from a fixed finite alphabet.
S= { s0,s1,s2,… ,sK-1}
with probabilities,
P(S=sk)= pk , k=0,1,2,…….,K-1.
The set of probabilities that must satisfy the condition
• We assume that the symbols emitted by the source during successive
signaling intervals are statistically independent.
• A source having the alone described properties is called a Discrete
Memory less Source (DMS).

INFORMATION AND UNCERTAINTY
• Information is related to the probability of
occurrence of the event. More is the uncertainty,
more is the information associated or related with
it.
• Consider the event S=sk, describing the emission
of symbol sk by the source with probability pk
clearly, if the probability pk=1 and pi=0 for all i≠k,
then
• , there is no surprise and therefore no information
when symbol sk is emitted.
• If, on the other hand, the source symbols occur
with different probabilities and the probability pk is
low, then there is more surprise and, therefore,
information when symbol sk is emitted by the

INFORMATION AND UNCERTAINTY
• Example:
• Sun rises in East : Here
Uncertainty is zero there is no
surprise in the statement. The
probability of occurrence is 1
(pk=1).
• Sun does not rise in East :
Here uncertainty is high, because
there is maximum surprise,
maximum information is not
possible.
• The amount of information is
related to the inverse of the
probability of occurrence of
the event S = sk as shown in
Figure. The amount of
information gained after
observing the event S = sk,
which occurs with probability

PROPERTIES OF INFORMATION
• Even if we are absolutely certain of the outcome of an
event, even before it occurs, there is no information
gained.
• The occurrence of an event S = sk either provides some
or no information, but never brings about a loss of
information.
• That is, the less probable an event is, the more
information we gain when it occurs.

• This additive property follow from the logarithmic
definition of I(sk).
• If sk and sl are statistically independent.
• It is standard practice in information theory to use a
logarithm to base 2 with binary signalling in mind. The
resulting unit of information is called the bit, which is a
contraction of the words binary digit.

• Hence , one bit is the amount of information
that we gain when one of two possible and
equally likely (i.e..equiprobable) event
occurs.
• Note that the information I(sk) is positive,
because the logarithm of a number less than
one, such as a probability, is negative.

ENTROPHY (AVERAGE INFORMATION)
• The entropy of a discrete random variable, representing the
output of a source of information, is a measure of the
average information content per source symbol.
• The amount of information I(sk) produced by the source
during an arbitrary signalling interval depends on the symbol
sk emitted by the source at the time. The self-information
I(sk) is a discrete random variable that takes on the values
I(s0), I(s1), …, I(sK – 1) with probabilities p0, p1, ….., pK – 1
respectively
• The expectation of I(sk) over all the probable values taken by
the random variable S is given by

PROPERTIES OF ENTROPY
Consider a discrete memory less source whose
mathematical model is defined by
S= { s0,s1,s2,… ,sK-1}
with probabilities,
P(S=sk)= pk , k=0,1,2,…….,K-1.
The entropy H(S) of the discrete random variable S
is bounded as follows:
0 ≤ H (s ) ≤ log2 (K )
where K is the number of symbols in the alphabet

Property 1: H(S) = 0, if, and only if, the probability pk = 1 for
some k, and the remaining probabilities in the set are all
zero; this lower bound on entropy corresponds to no
uncertainty

Property 2: H(S) = log K, if, and only if, pk = 1/K for all k ; this
upper bound on entropy corresponds to maximum
uncertainty.

Property 3: The upper bound (Hmax) on
Entropy is given as

INFORMATION RATE (R)
Information rate (R) is represented in average
number of bits of information per second.
R = r H(S)
Where,
R - information rate ( Information bits / second )
H(S) - the Entropy or average information (bits / symbol)
and
r - the rate at which messages are generated (symbols /
second).

MUTUAL INFORMATION
• The mutual information I(X;Y) is a measure of the uncertainty
about the channel input, which is resolved by observing the
channel output.
• Mutual information I(xi,yj) of a channel is defined as the
amount of information transferred when xi transmitted and yj
received.

DISCRETE MEMORYLESS CHANNELS
Discrete Memoryless Channel is a statistical model
with an input X and output Y (i.e., noisy version of X).
Both X and Y are random variables.
The channel is said to be discrete ,when both X and Y
have finite sizes. It is said to be memoryless when the
current output symbol depends only on the current

DISCRETE MEMORYLESS CHANNELS

TYPES OF CHANNELS
• LOSSLESS CHANNELS
A channel described by a channel matrix with only one
non-zero element in each column is called a lossless
channel. The channel matrix of a lossless channel will be
like,

TYPES OF CHANNELS
DETERMINISTIC CHANNEL
• A channel described by a channel matrix with
one non-zero element in each row is called a
deterministic channel

TYPES OF CHANNELS
BINARY SYMMETRIC CHANNEL
• Binary Symmetric Channel has two input symbols x0=0
and x1=1 and two output symbols y0=0 and y1=1. The
channel is symmetric because the probability of receiving a 1
if a 0 is sent is the same as the probability of receiving a 0 if
a 1 is sent.
• In other words correct bit transmitted with probability 1-p,
wrong bit transmitted with probability p. It is also called
“cross-over probability”. The conditional probability of error is
denoted as p.

TYPES OF CHANNELS
BINARY ERASURE CHANNEL (BEC)
A Binary erasure channel (BEC) has two inputs (0,1)
and three outputs (0,y,1). The symbol y indicates that, due to
noise, no deterministic decision can be made as to whether
the received symbol is a 0 or 1. In other words the symbol y
represents that the output is erased. Hence the name Binary
Erasure Channel (BEC)

CHANNEL CAPACITY
Channel Capacity of a discrete memory less
channel as the maximum average mutual information
I(X:Y) in any single use of channel i.e., signaling interval,
where the maximization is over all possible input
probability distribution {p(xi)} on X. It is measured in bits
per channel use.
Channel capacity (C) =max I (X:Y)
The channel capacity C is a function only of the
transition probabilities p(yj|xi), which defines the channel.
The calculation of channel capacity (C) involves
maximization of average mutual information I(X:Y) over
K variables

CHANNEL CAPACITY

CHANNEL CODING THEOREM
• The design goal of channel coding is to increase the
resistance of a digital communication system to channel
noise.
• Specifically, channel coding consists of mapping the
incoming data sequence into a channel input sequence and
inverse mapping the channel output sequence into an output
data sequence in such a way that the overall effect of
channel noise on the system is minimized.

• Mapping operation is performed in the transmitter by a
channel encoder, whereas the inverse mapping operation is
performed in the receiver by a channel decoder.
THEOREM
The channel-coding theorem for a discrete memoryless
channel is stated as follows:
• Let a discrete memoryless source with an alphabet 𝒮 have
entropy H(S) for random variable S and produce symbols
once every Ts seconds. Let a discrete memoryless channel
have capacity C and be used once every Tc seconds. Then,
if

there exists a coding scheme for which the source
output can be transmitted over the channel and be
reconstructed with an arbitrarily small probability of error.
The parameter C/Tc is called the critical rate.
When the system is said to be signalling at the critical
rate then it need to satisfy the below condition
Conversely, if

it is not possible to transmit information over the channel and
reconstruct it with an arbitrarily small probability of error.
The ratio Tc/Ts equals the code rate of encoder denoted by ‘r’,
where
• In short, the Channel Coding Theorem states that if a discrete
memoryless channel has capacity C and a source generates
information at a rate less than C, then there exists a coding
technique such that the output of the source maybe
transmitted over the channel with an arbitrarily low probability
of symbol error.
• Conversely, it is not possible to find such a code if the code
rate ‘r’ is greater than the channel capacity C. If r ≤ C, there
exists a code capable of achieving an arbitrarily low

Limitations
It does not show us how to construct a good code.
The theorem does not have a precise result for the
probability of symbol error after decoding the channel
output.

MAXIMUM ENTROPY FOR GAUSSIAN CHANNEL

CHANNEL (INFORMATION ) CAPACITY THEOREM OR
SHANNON’S THEOREM
The information capacity (C) of a continuous
channel of bandwidth B hertz, affected by AWGN of total
noise power with in the channel bandwidth N, is given by
the formula,
where, C - Information Capacity.
B - Channel Bandwidth
S - Signal power (or) Average Transmitted power
N – Total noise power within the channel bandwidth (B)
It is easier to increase the information capacity of
a continuous communication channel by expanding its
bandwidth than by increasing the transmitted power for a
prescribed noise variance.

SHANNON’S THEOREM

SHANNON’S THEOREM
Trade off
Noiseless Channel has infinite bandwidth.
If there is no noise in the channel , then N=0
hence Then
Infinite bandwidth has limited capacity. If B
is infinite, channel capacity is limited,
because noise power (N) increases, (S/N)
ratio decreases. Hence, even B approaches
infinity, capacity does not approach infinity.

SOURCE CODING THEOREM
Efficient representation of data generated by a discrete
source is accomplished by some encoding process. The
device that performs the representation is called a source
encoder.
Efficient source encoder must satisfy two functional
requirements
• Code Nodes produced by the encoder are in binary form.
• The source code is uniquely decodable, so that the original
source sequence can be reconstructed perfectly from the
encoded binary sequence.

THEOREM STATEMENT
• Given a discrete memory less source of entropy H(S) or
H(X) then the average code word length L for any distortion
less source encoding is bounded as,
• where, the Entropy H(S) or H(X) is the fundamental limit on
the average number of bits/source symbol.

AVERAGE CODE WORD LENGTH

CODING EFFICIENCY ( Ƞ )
The value of coding efficiency (Ƞ) is always
lesser or equal to 1. The source encoder is said to be
efficient when Ƞ approaches unity.

VARIANCE
The variance is defined as,

CODING TECHNIQUES
The two popular coding technique used in Information theory
are
(i) Shannon-Fano Coding
(ii) Huffman Coding.
ALGORITHM OF SHANNON-FANO CODING:
(i) List the source symbols in order of decreasing probability.
(ii) Partition the set into two sets that are as close to
equiprobables as possible.
(iii) Now assign “0” to the upper set and assign “1” to the lower
set.
(iv) Continue this process, each time partitioning the sets with
as nearly equal probabilities as possible until further

SHANNON-FANO CODING TECHNIQUES
STEP 1
(i) A discrete memoryless source has symbols x1, x2, x3, x4,
x5 with probabilities of 0.4, 0.2, 0.1, 0.2, 0.1 respectively.
Construct a Shannon-Fano code for the source and
calculate code efficiency η

STEP 2

STEP 3

STEP 4

2 .For a discrete memoryless source ‘X” with 6 symbols x1,
x2,….., x6 the probabilities are p(x1) =0.3, p(x2) =0.25, p(x3)
= 0.2, p(x4) = 0.12, p(x5) = 0.08, p(x6) = 0.05 respectively.
Using Shannon-Fano coding, calculate entropy, average
length of code, efficiency and redundancy of the code.
SYMBOL PROBABILITY
X1 p(x1) =0.3
X2 p(x2) =0.25
X3 p(x3) = 0.2
X4 p(x4) = 0.12
X5 p(x5) = 0.08
X6 p(x6) = 0.05

• A discrete memoryless source has five symbols
x1,x2,x3,x4 and x5 with probabilities
0.4,0.19,0.16,0.15 and 0.15 respectively attached
to every symbol.Construct a Shannon-Fano code
for the source and calculate code efficiency.

HUFFMAN CODING
• Huffman coding is a source coding method used for
data compression applications
• It is used to generate variable length codes and gives
lowest possible values of average length of code.
• Huffman coding provides maximum efficiency and
minimum redundancy
• Huffman coding is also known as minimum
redundancy code or optimum code

ALGORITHM FOR HUFFMAN CODING
• Step 1 : Arrange given messages in decreasing order of
probabilities.
• Step 2: Group the least probable M messages and
assign them symbols for coding.
• Step 3 : Add the probabilities of grouped messages and
place them as high as possible and rearrange them in
decreasing order again. This process is known as
reduction.
• Step 4: Repeat the steps 2 & 3 until only M or less than
M probabilities remains.
• Step 5: obtain the code words for each message by
tracing the probabilities from left to right in direction of the

PROBLEMS OF HUFFMAN CODING
• 1)Using Huffman algorithm find the average code
word length and code efficiency. Consider the 5
source symbols of a DMS with probabilities
p(m1)=0.4, p(m2)=0.2, p(m3)=0.2, p(m4)=0.1 and
p(m5)=0.1.
Message (mk) probabilities(Pk)
m1 0.4
m2 0.2
m3 0.2
m4 0.1
m5 0.1

Message (mk) probabilities(Pk) Stage I
m1 0.4 0.4
m2 0.2 0.2
m3 0.2 0.2
m4 0.1 0.2
m5 0.1
0
1

Message (mk)
probabilities(P
k)
Stage I Stage II
m1 0.4 0.4 0.4
m2 0.2 0.2 0.4
m3 0.2 0.2 0.2
m4 0.1 0.2
m5 0.1
0
1
0
1

Message
(mk) probabilitie
s(Pk)
Stage I Stage II Stage III
m1 0.4 0.4 0.4 0.6
m2 0.2 0.2 0.4 0.4
m3 0.2 0.2 0.2
m4 0.1 0.2
m5 0.1
0
1
0
1
0
0
1
1

Message
(mk) probabilit
ies(Pk)
Stage I Stage II Stage III Code
Word
m1 0.4 0.4 0.4 0.6 00
m2 0.2 0.2 0.4 0.4
m3 0.2 0.2 0.2
m4 0.1 0.2
m5 0.1
0
1
0
1
0
0
1
1

Message
(mk) probabilit
ies(Pk)
Word
m1 0.4 0.4 0.4 0.6 00
m2 0.2 0.2 0.4 0.4 10
m3 0.2 0.2 0.2
m4 0.1 0.2
m5 0.1
0
1
0
1
0
0
1
1

Message
(mk) probabilit
ies(Pk)
Word
m1 0.4 0.4 0.4 0.6 00
m2 0.2 0.2 0.4 0.4 10
m3 0.2 0.2 0.2 11
m4 0.1 0.2
m5 0.1
0
1
0
1
0
0
1
1

Message
(mk) probabilit
ies(Pk)
Word
m1 0.4 0.4 0.4 0.6 00
m2 0.2 0.2 0.4 0.4 10
m3 0.2 0.2 0.2 11
m4 0.1 0.2 010
m5 0.1
0
1
0
1
0
0
1
1

Message
(mk) probabilit
ies(Pk)
Word
m1 0.4 0.4 0.4 0.6 00
m2 0.2 0.2 0.4 0.4 10
m3 0.2 0.2 0.2 11
m4 0.1 0.2 010
m5 0.1 011
0
1
0
1
0
0
1
1

Messag
e (mk) probabi
lities(Pk
)
Stage I Stage II Stage
III
Code
Word
No of bit
per
message
(Ik)
m1 0.4 0.4 0.4 0.6 00 2
m2 0.2 0.2 0.4 0.4 10 2
m3 0.2 0.2 0.2 11 2
m4 0.1 0.2 010 3
m5 0.1 011 3

• For a discrete memoryless source ‘X” with 6 symbols x1,
x2,….., x6 the probabilities are p(x1) =0.3, p(x2) =0.25, p(x3) =
0.2, p(x4) = 0.12, p(x5) = 0.08, p(x6) = 0.05 respectively.
Using Huffman coding, calculate entropy, average length of
code, efficiency and redundancy of the code.
Message (mk)
probabilities(Pk
)
X1 0.3
X2 0.25
X3 0.2
X4 0.12
X5 0.08
x6 0.05

Message (mk) probabilities(Pk) Stage I
X1 0.3 0.3
X2 0.25 0.25
X3 0.2 0.2
X4 0.12 0.13
X5 0.08 0.12
x6 0.05
0
1

Message (mk)
probabilities(P
k)
Stage I Stage II
X1 0.3 0.3 0.3
X2 0.25 0.25 0.25
X3 0.2 0.2 0.25
X4 0.12 0.13 0.2
X5 0.08 0.12
x6 0.05
0
1
0
1

Messag
e (mk) probabilities(P
k)
Stage I Stage II Stage III
X1 0.3 0.3 0.3 0.45
X2 0.25 0.25 0.25 0.3
X3 0.2 0.2 0.25 0.25
X4 0.12 0.13 0.2
X5 0.08 0.12
x6 0.05
0
1
0
1
0
1

Messa
ge
(mk)
probabilities
(Pk)
Stage I Stage II Stage III Stage IV
X1 0.3 0.3 0.3 0.45 0.55
X2 0.25 0.25 0.25 0.3 0.45
X3 0.2 0.2 0.25 0.25
X4 0.12 0.13 0.2
X5 0.08 0.12
x6 0.05
0
1
0
1
0
1
0
1
0
1

Mess
age
(mk)
probabilitie
s(Pk)
III
Stage
IV
Code
word
X1 0.3 0.3 0.3 0.45 0.55 00
X2 0.25 0.25 0.25 0.3 0.45
X3 0.2 0.2 0.25 0.25
X4 0.12 0.13 0.2
X5 0.08 0.12
x6 0.05
0
1
0
1
0
1
0
1
0
1

Mess
age
(mk)
probabilitie
s(Pk)
III
Stage
IV
Code
word
X1 0.3 0.3 0.3 0.45 0.55 00
X2 0.25 0.25 0.25 0.3 0.45 10
X3 0.2 0.2 0.25 0.25
X4 0.12 0.13 0.2
X5 0.08 0.12
x6 0.05
0
1
0
1
0
1
0
1
0
1

Mess
age
(mk)
probabilitie
s(Pk)
III
Stage
IV
Code
word
X1 0.3 0.3 0.3 0.45 0.55 00
X2 0.25 0.25 0.25 0.3 0.45 10
X3 0.2 0.2 0.25 0.25 11
X4 0.12 0.13 0.2
X5 0.08 0.12
x6 0.05
0
1
0
1
0
1
0
1
0
1

Mess
age
(mk)
probabilitie
s(Pk)
III
Stage
IV
Code
word
X1 0.3 0.3 0.3 0.45 0.55 00
X2 0.25 0.25 0.25 0.3 0.45 10
X3 0.2 0.2 0.25 0.25 11
X4 0.12 0.13 0.2 011
X5 0.08 0.12
x6 0.05
0
1
0
1
0
1
0
1
0
1

Mess
age
(mk)
probabilitie
s(Pk)
III
Stage
IV
Code
word
X1 0.3 0.3 0.3 0.45 0.55 00
X2 0.25 0.25 0.25 0.3 0.45 10
X3 0.2 0.2 0.25 0.25 11
X4 0.12 0.13 0.2 011
X5 0.08 0.12 0100
x6 0.05
0
1
0
1
0
1
0
1
0
1

Mess
age
(mk)
probabilitie
s(Pk)
III
Stage
IV
Code
word
X1 0.3 0.3 0.3 0.45 0.55 00
X2 0.25 0.25 0.25 0.3 0.45 10
X3 0.2 0.2 0.25 0.25 11
X4 0.12 0.13 0.2 011
X5 0.08 0.12 0100
x6 0.05 0101
0
1
0
1
0
1
0
1
0
1

Mes
sag
e
(mk)
probabili
ties(Pk)
Stage
I
Stage
II
Stage
III
Stage
IV
Cod
e
wor
d
No of bit
per
messag
e (Ik)
X1 0.3 0.3 0.3 0.45 0.55 00 2
X2 0.25 0.25 0.25 0.3 0.45 10 2
X3 0.2 0.2 0.25 0.25 11 2
X4 0.12 0.13 0.2 011 3
X5 0.08 0.12 010
0
4

LEMPEL ZIV CODING

ALGORITHM FOR LEMPEL ZIV CODING
• Principle: The source data streams is parsed
into segments that are the shortest redundancies
not encountered previously
Step 1: Divide the given input data stream into sub
sequences
Step 2: Assign numerical position for each sub
sequences
Step 3: Assign numerical representation for each

PROBLEMS FOR LEMPEL ZIV CODING
• Example 1: Given data stream :
0 0 0 1 0 1 1 1 0 0 1 0 1 0 0 1 0 1
Step 1: Divide the given input data stream into
sub sequences
0 1 00 01 011 10 010 100 101
Numerical
Position
1 2 3 4 5 6 7 8 9
Sub
sequences
0 1 00 01 011 10 010 100 101

Numerical
Position
1 2 3 4 5 6 7 8 9
Sub
sequences
0 1 0 0 0 1 01 1 1 0 01 0 10 0 10 1
Numerical
Position
1 2 3 4 5 6 7 8 9
Sub sequences 0 1 0 0 0 1 01 1 1 0 01 0 10 0 10 1
Numerical
Representatio
n
11 12 4 2 2 1 4 1 6 1 6 2

•
Numerical
Position
1 2 3 4 5 6 7 8 9
Sub sequences 0 1 0 0 0 1 01 1 1 0 01 0 10 0 10 1
Numerical
Representatio
n
- - 11 12 4 2 2 1 4 1 6 1 6 2
Binary
Encoded
Blocks
0 1 001 0 001 1 1001 010
0
100
0
110
0
110
1
Final Encoded Block sequence for the given data stream :
0010 0011 1001 0100 1000 1100 1101

• Example 2: Given data stream :
1 1 1 0 1 0 0 0 1 1 0 1 0 1 1 0 1 0
sub sequences
1 0 11 10 100 01 101 011 010
Numerical
Position
1 2 3 4 5 6 7 8 9
Sub
sequences
1 0 11 10 100 01 101 011 010

Numerical
Position
1 2 3 4 5 6 7 8 9
Sub
sequences
1 0 11 10 100 01 101 011 010
Numerical
Position
1 2 3 4 5 6 7 8 9
Sub sequences 1 0 11 10 100 01 101 011 010
Numerical
Representatio
n
11 12 4 2 2 1 4 1 6 1 6 2

•
Numerical
Position
1 2 3 4 5 6 7 8 9
Sub sequences 1 0 0 0 0 1 01 1 1 0 01 0 10 0 10 1
Numerical
Representatio
n
- - 11 12 4 2 2 1 4 1 6 1 6 2
Binary
Encoded
Blocks
1 0 001 0 001 1 1001 010
0
100
0
110
0
110
1
0010 0011 1001 0100 1000 1100 1101

• Example 3 : Given data stream :
A A B A B B B A B A A B A B B B A
B B A B B
sub sequences
A A B A B B B A B A A B A B B B A B
B A BB
Numerical
Position
1 2 3 4 5 6 7 8 9
Sub
sequences
A A B A B B B ABA ABAB BB ABBA BB

Numerical
Position
1 2 3 4 5 6 7 8 9
Sub
sequences
A A
B
A B
B
B ABA ABA
B
BB ABB
A
BB
Numerical
Position
1 2 3 4 5 6 7 8 9
Sub
sequences
A A
B
A B
B
B ABA ABA
B
BB ABB
A
BB
Numerical
representati
on
φA 1B 2B φB 2A 5B 4B 3A 7

Numerical
Position
1 2 3 4 5 6 7 8 9
Sub
sequences
A A
B
A B
B
B ABA ABA
B
BB ABB
A
BB
Numerical
representati
on
Binary
Encoded
Blocks
0 11 101 1 100 1011 1001 110 111
Max no of bits :
4
Apply Zero Padding to all other codes to get decoding
without redudancies

Numerical
Position
1 2 3 4 5 6 7 8 9
Sub
sequences
A A
B
A B
B
B ABA ABAB BB ABBA BB
Numerical
representatio
n
Binary
Encoded
Blocks
0 11 101 1 100 1011 1001 110 111
Binary
Encoded
Blocks
000
0
0011 0101 0001 0100 1011 1001 0110 0111
0000 0011 0101 0001 0100 1011 1001
0110 0111

• Example 4 : Given data stream :
B B A B A A A B A B B A B A A A B A
A B A A
sub sequences
B B A BAA A B A B B A B A A A B A
A B AA
Numerical
Position
1 2 3 4 5 6 7 8 9
Sub
sequences
B BA BAA A BAB BABA AA BAAB AA

Numerical
Position
1 2 3 4 5 6 7 8 9
Sub
sequences
B B
A
B A
A
A BAB BAB
A
AA BAA
B
AA
Numerical
Position
1 2 3 4 5 6 7 8 9
Sub
sequences
B B
A
B A
A
A BAB BAB
A
AA BAA
B
AA
Numerical
representati
on
φB 1A 2A φA 2B 5A 4A 3B 7

Numerical
Position
1 2 3 4 5 6 7 8 9
Sub
sequences
B B
A
B A
A
A BAB BAB
A
AA BAA
B
AA
Numerical
representati
on
Binary
Encoded
Blocks
0 11 101 1 100 1011 1001 110 111
Max no of bits :
4
Apply Zero Padding to all other codes to get decoding
without redudancies

Numerical
Position
1 2 3 4 5 6 7 8 9
Sub
sequences
B B
A
B A
A
A BAB BABA AA BAAB AA
Numerical
representatio
n
Binary
Encoded
Blocks
0 11 101 1 100 1011 1001 110 111
Binary
Encoded
Blocks
000
0
0011 0101 0001 0100 1011 1001 0110 0111
0000 0011 0101 0001 0100 1011 1001
0110 0111

ERROR CONTROL CODES
ERROR:
Error is a condition when the output information does not
match with the input information. During transmission, digital
signals suffer from noise that can introduce errors in the
binary bits travelling from one system to other. That means a
0 bit may change to 1 or a 1 bit may change to 0
Digital
Source
Source
Encod
er
Error
Control
Coding
Line
Coding
Modulat
or
Chann
el
Noise
Digital
Sink
Source
Decod
er
Error
Control
Decoding
Line
Decoding
Demod
ulator
+
Transmitter
Receiver
X(w)
N(w)
Y(w)

ERROR CONTROL CODES
CLASSIFICATION OF ERRORS:
Content Error:
These are errors in the content of
messageintroduced due to noise during transmission
Flow integrity Error
These errors are missing blocks of data, data lost in
network or data delivered to wrong destination
CLASSIFICATION OF
CODES:
 Error Detecting
 Error Correcting

IMPORTANT DEFINITIONS RELATED TO CODES
 Code Word:
The code word is the n bit encoded block
of bits. It contains message bits and parity or
redundant bits.
 Block Length:
The number of bits ‘n’ after coding is known as
block length.
 Code Rate (or) Code Efficiency:
The code rate is defined astheratio of
the number of message bits (k) to the total number
of bits (n) in a code word.
𝐶𝑜𝑑𝑒 𝑟𝑎𝑡𝑒 (𝑟 )= number of message bits (k)

 Code Vector:
An ‘n’ bit code word can be visualized in an n – dimensional
space as a vector whose elements or coordinates are bits in the code
word.

Hamming Distance:
The Hamming distance between two code words is equal
to the number of differences between them, e.g.,
1 0 0 1 1 0 1 1
1 1 0 1 0 0 1 0
Hamming distance = 3
Minimum Distance dmin:
It is defined as the smallest Hamming distance between
any pair of code vectors in the code.
Hamming Weight of a Code Word [w(x)]:
It is defined as the number of non – zero elements in the
code word. It is denoted by w(x)
for eg: X= 01110101, then w(x) = 5

ERROR CONTROL CODES
Error Control Coding
The main purpose of Error control coding is to enable the
receiver to detect and ever correct the errors by introducing
some redundancies into the data which is to be transmitted.
There are basically two mechanisms to for adding redundancy.
They are
 Block Coding
 Convolutional coding

CLASSIFICATION OF ERROR–CORRECTING
CODES:
Block Codes (No
memory is required):
(n,k) block code is
generated when the
channel encoder
accepts information in
successive k bit blocks.
At the end of each
such block, (n- k) parity
bit is added, which
contains no information
and termed as
Convolutional Codes
(Memory is required)
Here the code words
are generated by discrete –
time convolution of the
input sequence with impulse
response of the encoder.
Unlike block codes,
channel encoder accepts
messages as a continuous
sequence and generates a
continuous sequence of
encoded bits at the output

CLASSIFICATION OF ERROR–CORRECTING
CODES:
Linear Codes and non linear Codes
Linear codes have the unique property that when any
two code words of linear code are added in modulo – 2 adder, a
third code word is produced which is also a code, which is not
the case for non – linear codes.

CLASSIFICATION OF ERROR CONTROL CODING
TECHNIQUES

TYPES OF ERROR CONTROL TECHNIQUES
AUTOMATIC REPEAT REQUEST (ARQ)
 Here, when an error is detected, a request is made
for retransmission of signal.
 A feedback channel is necessary for this
retransmission.
 It differs from the FEC system in the following three
aspects.
1. Less number of check bits (parity bits) are required
increasing the (k/n) ratio for (n, k) block code.
2. Additional hardware to implement feedback path is
required.
3. But rate at forward transmission needs to make

AUTOMATIC REPEAT REQUEST (ARQ)
 For each message signal at the input, the encoder produces code words
which are stored temporarily at encoder output and transmitted over forward
transmission channel.
 At the destination, decoders decode the signals and give a positive
acknowledgement (ACK) and negative acknowledgement (NAK)
respectively in case no error and error is detected.
 On receipt of NAK, the controller retransmits the appropriate word
stored in the input buffer.

FORWARD ERROR CORRECTION (FEC)
TECHNIQUE
 It is a digital modulation system where discrete source generates
information in binary form.
 The channel encoder accepts these message bits and add redundant
bits to them leaving higher bit rate for transmission.
 Channel decoders uses the redundant bits to check for
actually and erroneous transmitted messages.

ERROR DETECTION TECHNIQUE
Whenever a message is transmitted, it may get
scrambled by noise or data may get corrupted. To avoid
this, we use error-detecting codes which are additional
data added to a given digital message to help us detect if
an error occurred during transmission of the message. A
simple example of error-detecting code is parity check.
i) Parity Checking
ii) Check sum Error Detection
iii) Cyclic Redundancy Check (CRC)

PARITY CHECKING OF ERROR DETECTION
• It is the simplest technique for detecting and correcting
errors. The MSB of an 8-bits word is used as the parity bit
and the remaining 7 bits are used as data or message bits.
The parity of 8-bits transmitted word can be either even parity
or odd parity.
• Even parity -- Even parity means the number of 1's in the
given word including the parity bit should be even (2,4,6,....).
• Odd parity -- Odd parity means the number of 1's in the
given word including the parity bit should be odd (1,3,5,....).

USE OF PARITY BIT
• The parity bit can be set to 0 and 1 depending on the type of
the parity required.
• For even parity, this bit is set to 1 or 0 such that the no. of "1
bits" in the entire word is even. Shown in fig. (a).
• For odd parity, this bit is set to 1 or 0 such that the no. of "1
bits" in the entire word is odd. Shown in fig (b)
fig. (a).
fig. (b).
0 0 0 1 1 1 0 1 1 1
0 0 1 1 1 1 0 0 1 1
Message or data bits
Parity
bit
Parity
bit
Parity
bit
Parity
bit

• ERROR DETECTION USING PARITY BIT
Parity checking at the receiver can detect the presence
of an error if the parity of the receiver signal is different from
the expected parity. That means, if it is known that the parity
of the transmitted signal is always going to be "even" and if
the received signal has an odd parity, then the receiver can
conclude that the received signal is not correct. If an error is
detected, then the receiver will ignore the received byte and
request for retransmission of the same byte to the transmitter.

CHECK SUM ERROR DETECTION
• This is a block code method where a checksum is created
based on the data values in the data blocks to be transmitted
using some algorithm and appended to the data.
• When the receiver gets this data, a new checksum is
calculated and compared with the existing checksum. A non-
match indicates an error.
Error Detection by Checksums
• For error detection by checksums, data is divided into fixed
sized frames or segments.
• Sender’s End − The sender adds the segments using 1’s
complement arithmetic to get the sum. It then complements
the sum to get the checksum and sends it along with the data
frames.
• Receiver’s End − The receiver adds the incoming segments
along with the checksum using 1’s complement arithmetic to

CHECK SUM ERROR DETECTION
Example
• Suppose that the sender wants to
send 4 frames each of 8 bits,
where the frames are 11001100,
10101010, 11110000 and
11000011.
• The sender adds the bits using 1s
complement arithmetic. While
adding two numbers using 1s
complement arithmetic, if there is
a carry over, it is added to the
sum.
• After adding all the 4 frames, the
sender complements the sum to
get the checksum, 11010011, and
sends it along with the data
frames.
• The receiver performs 1s
complement arithmetic sum of all
the frames including the

CYCLIC REDUNDANCY CHECK (CRC)
 The concept of parity checking can be extended from
detection to correction of single error by arranging the data
block in rectangular matrix.
 This will lead to two set of parity bits, viz.
•Longitudinal Redundancy Check
(LRC) and
• Vertical Redundancy Check (VRC).

LONGITUDINAL REDUNDANCY CHECK (LRC)
 In Longitudinal Redundancy Check, one row is taken
up at a time and counting the number of 1s, the
parity bit is adjusted to achieve even parity.
 Here for checking the message block, a complete
character known as Block Check Character (BCC)
is added at the end of the block of information,
which may be of even or odd parity

VERTICAL REDUNDANCY CHECK (LRC)
In VRC, the ASCII code for individual alphabets are
considered arranged vertically and then counting the
number of 1s, the parity bit is adjusted to achieve even
parity.

LONGITUDINAL REDUNDANCY CHECK (LRC)
 A single error in any bit will result in a non – correct LRC in
the row and a non – correct VRC in the column. The bit
which is common to both the row and column is the bit in
error. The limitation is though it can detect multiple errors
but is capable to correct only a single error as for multiple
errors it is not suitable to locate the position of the errors.
 1 in the square box in the next table, is the bit in error as it is
evident from the erroneous results both in the LRC and the
VRC columns
1

LINEAR BLOCK CODES
PRINCIPLE OF BLOCK CODING:
 For the block of k message bits, (n-k) parity bits or check bits
are added.
 Hence the total bits at the output of channel encoder are ‘n’.
 Such codes are called (n,k) block codes
Message
Block
input
CHANNEL
ENCODER
Code Block
output
MESSAGE
k bits
MESSAGE Check bits
k bits (n – k)
bits

LINEAR BLOCK CODES
SYSTEMATIC CODES:
In the systematic
block code, the message
bits appear at the
beginning of the code
word. Then the check bits
are transmitted in a block.
This type of code is called
NON-SYSTEMATIC
CODES:
In non-systematic
codes it is not possible
to identify message
bits and check bits.
They are mixed in the
block code.

LINEAR BLOCK CODES

LINEAR BLOCK CODES
i.e q are the number of redundant bits added by the encoder.
The above code vector can also written as,
X =(M│C)
Here,
M = k-bit message vector
C = q –bit check vector
The check bits play the role of error detection and
correction. The job of the linear block code is to generate
those “check bits”.

MATRIX DESCRIPTION OF LINEAR BLOCK CODES:
The code vector can be represented as,
X = MG
Here, X = Code vector of 1 × n size or n bits
M = Message vector of 1 × k size bits
G = Generator matrix of k × n size.

Note: All the additions are mod – 2 additions

STRUCTURE OF LINEAR BLOCK CODER

LINEAR BLOCK CODES:

LINEAR BLOCK CODES:
ii) To obtain the equation for check bits:
Here k= 3, q = 3, and n = 6
Here the block size of the message vector is 3 bits, Hence 2k =
23 = 8 message vectors are possible as shown below,

LINEAR BLOCK CODES:

LINEAR BLOCK CODES:
iii) To determine check bits and code vectors
for every message vectors:

LINEAR BLOCK CODES:

PROPERTIES OF LINEAR BLOCK CODES

above
eqn

ERROR DETECTION AND CORRECTION OF LINEAR
BLOCK CODES

HAMMING CODES

ERROR DETECTION AND CORRECTION CAPABILITIES OF
HAMMING CODES:
For Hamming Code , dmin = 3
So ,
S ≤ 3 -1
No of error detected by Hamming code S ≤ 2
No of error corrected by Hamming code t≤ 1

HAMMING CODES:

iv) To determine check bits and code vectors for
every message vectors:

HAMMING CODES

ERROR DETECTION AND CORRECTION CAPABILITIES OF
HAMMING CODES:
For Hamming Code , dmin = 3
So ,
S ≤ 3 -1
No of error detected by Hamming code S ≤ 2
No of error corrected by Hamming code t≤ 1

HAMMING CODES:

iv) To determine Error Detection and Error Correction

SYNDROME DECODING
To avoid such complexity, syndrome decoding is used in
linear block codes.

SYNDROME DECODING

SYNDROME DECODING
S = [1 0 1]

SYNDROME DECODING

ERROR CORRECTION USING SYNDROME VECTOR

• Here the block size of the
message vector is 3 bits,
Hence 2k = 23 = 8 message
vectors are possible as shown
below,

Similarly by calculating for other bit error vectors we
have the following decoding table

CYCLIC CODES
CYCLIC CODES:
 Cyclic codes are the sub class of linear block codes.
 Cyclic codes can be in systematic or non-systematic form.
 In systematic form, check bits are calculated separately and
the code vector is X= (M:C)form. Here ‘M’ represents
message bits and ‘C represents check bits.
Definition:
A linear code is called cyclic code if every cyclic shift of
the code vector produces some other code vector.
Properties of Cyclic Codes:
Cyclic codes exhibit two fundamental properties:
(i) Linearity
(ii) Cyclic property

CYCLIC CODES

CYCLIC CODES
Algebraic structures of cyclic codes:
The code words can be represented by a polynomial
For eg., let us consider the n-bit codeword,
X = {xn-1, xn-2….. x1,x0}
The above code word can be represented by a polynomial of degree
less than or equal to (n-1).i.e.,
X(p) = xn-1pn-1 + xn-2pn-2 + …..x1p + x0
Here X(p) is the polynomial of degree (n-1)
p is the arbitrary variable of the polynomial.
The power of ‘p’ represents the positions of the code words. i.e.,
pn-1 represents MSB
p0 represents LSB
p1 represents second bit from LSB side.

CYCLIC CODES
Generation of Code Vectors in Non-Systematic
form:
Let M = {mk-1, mk-2, …m1,m0} be ‘k’ bits of message
vector. Then it can be represented by the polynomial
as,
M(p) = mk-1pk-1 + mk-2pk-2 +…..m1p +m0
Let X(p) represents the codeword polynomial. It is
given as,
X(p) = M(p)G(p)
Here G(p) is the generating polynomial of degree
‘q’.
For an (n,k) cyclic code, q =n-k represents the
number of parity bits. The generating polynomial is
given as,

GENERATION OF CODE VECTORS IN NON-
SYSTEMATIC FORM:
If M1, M2, M3,…..etc are the other message
vectors, then the corresponding code vectors
can be calculated as,
X1(p) = M1(p)G(p)
X2(p) = M2(p)G(p)
X3(p) = M3(p)G(p) and so on.
All the above code vectors X1, X2, X3…..are in
non-systematic form and they satisfy cyclic
property.
Note: The generator polynomial G(p) remains

Example: The generator polynomial of (7,4) cyclic code
is G(p) = p3+p+1.
Find all the code vectors for the code in non-
systematic form.
Sol:
Here n = 7 and k = 4
, q= n-k =3
To find possible
message vectors,
2k = 24 = 16 message
vectors of four bits
are possible

SYSTEMATIC FORM:

SYSTEMATIC FORM:
To check whether cyclic property is satisfied:
Now Code vector X9 is considered form the
above table
X9 = (1 0 1 1 0 0 0)
By shifting the above code vector to left side by
1 bit position,
X’ = (0 1 1 0 0 0 1)
From the table , it is shown that,
X’ = X8 = ( 0 1 1 0 0 0 1)
Thus cyclic shift of X9 produces X8. This
cyclic propert can be verified for other code

GENERATION OF CODE VECTORS IN SYSTEMATIC
FORM:

FORM:
Above equation implies,
• (i) multiply message polynomial by pq.
• (ii) Divide pq M(p) by generator polynomial
• (iii) Remainder of the division is C(p)

Example: The generator polynomial of a (7,4) cyclic
code is G(p) = p3+p+1.
Find all the code vectors in systematic form.
Sol:
Here n = 7 and k = 4 ,
q= n-k =3
To find possible
message vectors,
2k = 24 = 16 message
vectors of four bits are
possible

FORM:
To find message polynomial,
Consider any message vector M = (m3 m2
m1 m0) = (0 1 0 1)
Then, M(p) = m3p3 + m2 p2 + m1p1 + m0
M(p) = p2 + 1
The given generator polynomial is,
G(p) = p3 + p + 1

FORM:
To obtain pq M(p):
Since q =3, pq M(p) will be,
pq M(p) = p3 M(p)
= p3 (p2 + 1)
pq M(p) = p5 + p3
pq M(p) = p5 + 0p4 + p3 + 0p2 + 0p +1
G(p) = p3 + p + 1
and G(p) = p3 + 0p2 + p + 1

FORM:

FORM:
This is the required code vector for the message vector ( 0 1 0
1) in systematic form. The code vectors for other message
bits can be obtained by following the same procedure and
listed in the table as below

FORM:

ENCODING USING AN (N - K) BIT SHIFT REGISTER:

OPERATION:
 The feedback switch is first closed. The output switch is
connected to message input.
 All the shift registers are initialized to all zero state. The k
message bits are shifted to the transmitter as well as shifted into
the registers.
 After the shift of k messages bits the registers contain q check
bits. The feedback switch is now opened and output switch is
connected to check bits position. With the every shift, the check
bits are then shifted to the transmitter.

Here q = 3 , so there are 3 flipflops in shift register to hold
check bits c1, c2,c0
Since g2 = 0 , its link is not connected. g1 = 1, hence its link is
connected.

Shift register bit position for input message 1100

Operation of (7,4) cyclic code encoder

CONVOLUTIONAL CODES
• A Convolutional coding is done by combining the fixed
number of input bits.
• The input bits are stored in the fixed length shift register
and they are combined with the help of mod-2 adders. This
operation is equivalent to binary covolution and hence its
called Covolutional coding.

CONVOLUTIONAL CODES

STATES OF THE ENCODER
• In the above convolutional encoder, the previous two
successive message bits m1 and m2 represents state.
• The input message bits m affects the state of the encoder as
well as outputs x1 and x2 during that state.
• Whenever new message bit is shifted to m, the contents of m1
and m2 define new state. And outputs x1 and x2 are also
changed according to new state m1,m2 and message bit m.
• Let the initial values of bits stored in m1 and m2 be zero. That
is m1m2 = 00 initially and the encoder is in state ‘a’

STATES OF THE ENCODER
m2 m1 States of encoder
0 0 a
0 1 b
1 0 c
1 1 d

DEVELOPMENT OF THE CODE TREE

• The development of code tree for the message sequence m
= 110 is given below. Initially let us assume that encoder is
in state ‘a’ . i.e., m1m2 = 00
(i) When message bit m = 1, (first bit)
The first message input is m = 1 With this input x1 and x2 is
calculated as follows,
For the given convolutional encoder, the outputs x1 and x2
are given as,

Input bits Present state Next state Output of the
encoder
0 a (00) a (00) 00
1 a (00) b (01) 11
1 b (01) d (11) 01
0 d(11) c (10) 01

Inp
ut
bits
Pres
ent
sta
te
Ne
xt
stat
e
Outp
ut of
the
enco
der
0 a
(00
)
a
(00
)
00
1 a
(00
)
b
(01
)
11
1 b
(01
d
(11
01

CODE TRELLIS
• Code trellis is the more compact representation of
the code tree.
• In code tree there are fore states or nodes. Every
states goes to some other state depending upon
the input code.
• Trellis represents the single, an unique diagram for
such transitions.

STATE TABLE
Input bits Present state Next state Output of the
encoder
0 a (00) a (00) 00
1 b (10) 11
0
b (10)
c (01) 10
1
d (11) 01
0 c (01) a (00) 11
1 b (10) 00
0 d(11) c (01) 01
1 d(11) 10

CODE TRELLIS
00
11
11 00
10
01
01
10

CODE TRELLIS
• The node on the left side denotes four
possible current states and those on the right
represents the next states.
• The solid transition line represents input m =
0 and broken line represents input m = 1.
• Along with each transition line the output x1x2
is represented during the transition.
• For eg., let the encoder be in current state of
‘a’. If input m = 0, the next state will be ‘a’,
with the output x1x2 = 11.
• Thus code trellis is compact representation

STATE DIAGRAM
• If the current and next states of the encoder
are combined, then it can be represented by
a state diagram.
• For eg., let the encoder is in state ’a’. If input
bit m = 0, then the next state is same i.e., ‘a’,
with the output x1x2 = 00. This is shown as
self loop at node ‘a’ in the state diagram.
• If input m =1 , then state diagram shows the
next state as ‘b’, with outputs x1x2 = 11.

STATE DIAGRAM
b
c
10
10
11
11
0
0
01
01

SOLVED EXAMPLE
Example: A code rate 1 / 3 covolutional
encoder has generating vectors as
g1 = (1 0 0), g2 = (1 1 1) and g3 = (1 0 1)
(i) Sketch the encoder configuration.
(ii) Draw the code tree, state diagram, and
trellis diagram.
(iii) If input message sequence is 10110,
determine the output sequence of the
encoder.

SOLVED EXAMPLE

SOLVED EXAMPLE
d

SOLVED EXAMPLE

CODE TRELLIS
000
11
1
011 10
0
010
101
001
110

STATE DIAGRAM
b
c
110
010
11
1
01
1
10
0
10
1
001
0 0
0

CODE TREE
000
111
m =
0
m =
1
000
000
000
111
111

TO FIND OUTPUT SEQUENCE

VITERBI DECODING ALGORITHM

ALGORITHM STEPS

Example :
Viterbi Decoding
Input data m: 1 1 0 1 1
Codeword U: 11 01 01 00 01
Received seq. Z: 11 01 01 10 01
a = 00
b = 10
t1 t2
2
0
2

a
0

b

Example :
Viterbi Decoding
Codeword U: 11 01 01 00 01
Received seq. Z: 11 01
a = 00
b = 10
t1 t2 t3
2 1
0 1
2
0
c = 01
d = 11
2

a
0

b
3

a
3

b
2

c
0

d

Example :
Viterbi Decoding
Codeword U: 11 01 01 00 01
Received seq. Z: 11 01 01
a = 00
b = 10
t1 t2 t3 t4
2 1 1
0 1 1
2
2
0
0
1
1
2
0
c = 01
d = 11
3

a
3

b
2

c
0

d

Example :
Viterbi Decoding
Codeword U: 11 01 01 00 01
a = 00
b = 10
t1 t2 t3 t4
2 1 1
0 1 1
2
2
0
0
1
1
2
0
c = 01
d = 11
Path metric
4
3

Example :
Viterbi Decoding
Codeword U: 11 01 01 00 01
a = 00
b = 10
t1 t2 t3 t4
2 1 1
0 1 1
2
2
0
0
1
1
2
0
c = 01
d = 11
Path metric
4
0

a = 00
b = 10
t1 t2 t3 t4
2 1 1
0 1 1
2
2
0
0
1
1
2
0
c = 01
d = 11
Path metric
3
2
Example :
Viterbi Decoding
Codeword U: 11 01 01 00 01

Example :
Viterbi Decoding
Codeword U: 11 01 01 00 01
a = 00
b = 10
t1 t2 t3 t4
0
2
0
0
1
1
2
c = 01
d = 11
3

a
3

b
0

c
t5
01
1
1 1
1
0
2 2
0

Example :
Viterbi Decoding
Codeword U: 11 01 01 00 01
a = 00
b = 10
t1 t2 t3 t4
0
2
0
0
1
1
2
c = 01
d = 11
t5
01
1
1 1
1
0
2 2
0
Path metric
4
1

Example :
Viterbi Decoding
Codeword U: 11 01 01 00 01
a = 00
b = 10
t1 t2 t3 t4
0
2
0
0
1
1
2
c = 01
d = 11
t5
01
1 1
1
0
2 2
0
Path metric
4
1

Example :
Viterbi Decoding
Codeword U: 11 01 01 00 01
a = 00
b = 10
t1 t2 t3 t4
0
2
0
0
1
2
c = 01
d = 11
t5
01
1
1
0
2 2
0
Path metric
3
4

Example :
Viterbi Decoding
a = 00
b = 10
t1 t2 t3 t4
0
2
0
0
1
2
c = 01
d = 11
t5
1
1
0
2
0
5
Path metric
2
Codeword U: 11 01 01 00 01
Received seq. Z: 11 01 01 01

Example :
Viterbi Decoding
Codeword U: 11 01 01 00 01
a = 00
b = 10
t1 t2 t3 t4
0
2
0
0
1
2
c = 01
d = 11
t5
10
1
1
0
0
1

a
1

b
3

c
2

d
01
t6
1
1 1
1
2
0
0
2

a = 00
b = 10
t1 t2 t3 t4
0
2
0
0
1
2
c = 01
d = 11
t5
1
1
0
0
t6
1
1 1
1
2
0
0
2
Path metric
2
4
Example :
Viterbi Decoding
Codeword U: 11 01 01 00 01
Received seq. Z: 11 01 01 10 01

a = 00
b = 10
t1 t2 t3 t4
0
0
0
2
c = 01
d = 11
t5
1
1
0
t6
1
1
2
0
0
2
Path metric
3
2
Example :
Viterbi Decoding
Codeword U: 11 01 01 00 01
Received seq. Z: 11 01 01 10 01

a = 00
b = 10
t1 t2 t3 t4
0
0
0
2
c = 01
d = 11
t5
1
1
0
t6
1
1
0
0
2
Path metric
2
4
Example :
Viterbi Decoding
Codeword U: 11 01 01 00 01
Received seq. Z: 11 01 01 10 01

a = 00
b = 10
t1 t2 t3 t4
0
0
0
2
c = 01
d = 11
t5
1
1
0
t6
1
1
0
0
2

a
2

b
2

c
1

d
Example :
Viterbi Decoding
Codeword U: 11 01 01 00 01
Received seq. Z: 11 01 01 10 01

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