LECTURE-10 (Data Communication) ~www.fida.com.bd

LECTURE-10

REFERENCE :
Chapter 10
Error Detection

and
Correction
10.1 www.fida.com.bd

Note

Data can be corrupted
during transmission.

Some applications require that
errors be detected and corrected.

10.2

10-1 INTRODUCTION

Let us first discuss some issues related, directly or
indirectly, to error detection and correction.

Topics discussed in this section:
Types of Errors
Redundancy
Detection Versus Correction
Forward Error Correction Versus Retransmission
Coding
Modular Arithmetic

10.3

Note

In a single-bit error, only 1 bit in the data
unit has changed.

10.4

Figure 10.1 Single-bit error

10.5

Note

A burst error means that 2 or more bits
in the data unit have changed.

10.6

Figure 10.2 Burst error of length 8

10.7

Note

To detect or correct errors, we need to
send extra (redundant) bits with data.

10.8

Figure 10.3 The structure of encoder and decoder

10.9

Note

In this book, we concentrate on block
codes; we leave convolution codes
to advanced texts.

10.10

Note

In modulo-N arithmetic, we use only the
integers in the range 0 to N −1,
inclusive.

10.11

Figure 10.4 XORing of two single bits or two words

10.12

10-2 BLOCK CODING

In block coding, we divide our message into blocks,
each of k bits, called datawords. We add r redundant
bits to each block to make the length n = k + r. The
resulting n-bit blocks are called codewords.

Error Detection
Error Correction
Hamming Distance
Minimum Hamming Distance

10.13

Figure 10.5 Datawords and codewords in block coding

10.14

Example 10.1

The 4B/5B block coding discussed in Chapter 4 is a good
example of this type of coding. In this coding scheme,
k = 4 and n = 5. As we saw, we have 2k = 16 datawords
and 2n = 32 codewords. We saw that 16 out of 32
codewords are used for message transfer and the rest are
either used for other purposes or unused.

10.15

Figure 10.6 Process of error detection in block coding

10.16

Example 10.2

Let us assume that k = 2 and n = 3. Table 10.1 shows the
list of datawords and codewords. Later, we will see
how to derive a codeword from a dataword.

Assume the sender encodes the dataword 01 as 011 and
sends it to the receiver. Consider the following cases:

1. The receiver receives 011. It is a valid codeword. The
receiver extracts the dataword 01 from it.

10.17

Example 10.2 (continued)

2. The codeword is corrupted during transmission, and
111 is received. This is not a valid codeword and is
discarded.

3. The codeword is corrupted during transmission, and
000 is received. This is a valid codeword. The receiver
incorrectly extracts the dataword 00. Two corrupted
bits have made the error undetectable.

10.18

Table 10.1 A code for error detection (Example 10.2)

10.19

Note

An error-detecting code can detect
only the types of errors for which it is
designed; other types of errors may
remain undetected.

10.20

Figure 10.7 Structure of encoder and decoder in error correction

10.21

Example 10.3

Let us add more redundant bits to Example 10.2 to see if
the receiver can correct an error without knowing what
was actually sent. We add 3 redundant bits to the 2-bit
dataword to make 5-bit codewords. Table 10.2 shows the
datawords and codewords. Assume the dataword is 01.
The sender creates the codeword 01011. The codeword is
corrupted during transmission, and 01001 is received.
First, the receiver finds that the received codeword is not
in the table. This means an error has occurred. The
receiver, assuming that there is only 1 bit corrupted, uses
the following strategy to guess the correct dataword.

10.22

1. Comparing the received codeword with the first
codeword in the table (01001 versus 00000), the
receiver decides that the first codeword is not the one
that was sent because there are two different bits.

2. By the same reasoning, the original codeword cannot
be the third or fourth one in the table.

3. The original codeword must be the second one in the
table because this is the only one that differs from the
received codeword by 1 bit. The receiver replaces
01001 with 01011 and consults the table to find the
dataword 01.
10.23

Table 10.2 A code for error correction (Example 10.3)

10.24

Note

The Hamming distance between two
words is the number of differences
between corresponding bits.

10.25

Example 10.4

Let us find the Hamming distance between two pairs of
words.

1. The Hamming distance d(000, 011) is 2 because

2. The Hamming distance d(10101, 11110) is 3 because

10.26

Note

The minimum Hamming distance is the
smallest Hamming distance between
all possible pairs in a set of words.

10.27

Example 10.5

Find the minimum Hamming distance of the coding
scheme in Table 10.1.
Solution
We first find all Hamming distances.

The dmin in this case is 2.

10.28

Example 10.6

Find the minimum Hamming distance of the coding
scheme in Table 10.2.

Solution
We first find all the Hamming distances.

The dmin in this case is 3.

10.29

Note

To guarantee the detection of up to s
errors in all cases, the minimum
Hamming distance in a block
code must be dmin = s + 1.

10.30

Example 10.7

The minimum Hamming distance for our first code
scheme (Table 10.1) is 2. This code guarantees detection
of only a single error. For example, if the third codeword
(101) is sent and one error occurs, the received codeword
does not match any valid codeword. If two errors occur,
however, the received codeword may match a valid
codeword and the errors are not detected.

10.31

Example 10.8

Our second block code scheme (Table 10.2) has dmin = 3.
This code can detect up to two errors. Again, we see that
when any of the valid codewords is sent, two errors create
a codeword which is not in the table of valid codewords.
The receiver cannot be fooled.

However, some combinations of three errors change a
valid codeword to another valid codeword. The receiver
accepts the received codeword and the errors are
undetected.

10.32

Figure 10.8 Geometric concept for finding dmin in error detection

10.33

Figure 10.9 Geometric concept for finding dmin in error correction

10.34

Note

To guarantee correction of up to t errors
in all cases, the minimum Hamming
distance in a block code
must be dmin = 2t + 1.

10.35

Example 10.9

A code scheme has a Hamming distance dmin = 4. What is
the error detection and correction capability of this
scheme?

Solution
This code guarantees the detection of up to three errors
(s = 3), but it can correct up to one error. In other words,
if this code is used for error correction, part of its capability
is wasted. Error correction codes need to have an odd
minimum distance (3, 5, 7, . . . ).

10.36

10-3 LINEAR BLOCK CODES

Almost all block codes used today belong to a subset
called linear block codes. A linear block code is a code
in which the exclusive OR (addition modulo-2) of two
valid codewords creates another valid codeword.

Minimum Distance for Linear Block Codes
Some Linear Block Codes

10.37

Note

In a linear block code, the exclusive OR
(XOR) of any two valid codewords
creates another valid codeword.

10.38

Example 10.10

Let us see if the two codes we defined in Table 10.1 and
Table 10.2 belong to the class of linear block codes.

1. The scheme in Table 10.1 is a linear block code
because the result of XORing any codeword with any
other codeword is a valid codeword. For example, the
XORing of the second and third codewords creates the
fourth one.

2. The scheme in Table 10.2 is also a linear block code.
We can create all four codewords by XORing two
other codewords.
10.39

Example 10.11

In our first code (Table 10.1), the numbers of 1s in the
nonzero codewords are 2, 2, and 2. So the minimum
Hamming distance is dmin = 2. In our second code (Table
10.2), the numbers of 1s in the nonzero codewords are 3,
3, and 4. So in this code we have dmin = 3.

10.40

Note

A simple parity-check code is a
single-bit error-detecting
code in which
n = k + 1 with dmin = 2.

10.41

Table 10.3 Simple parity-check code C(5, 4)

10.42

Figure 10.10 Encoder and decoder for simple parity-check code

10.43

Example 10.12

Let us look at some transmission scenarios. Assume the
sender sends the dataword 1011. The codeword created
from this dataword is 10111, which is sent to the receiver.
We examine five cases:

1. No error occurs; the received codeword is 10111. The
syndrome is 0. The dataword 1011 is created.
2. One single-bit error changes a1 . The received
codeword is 10011. The syndrome is 1. No dataword
is created.
3. One single-bit error changes r0 . The received codeword
is 10110. The syndrome is 1. No dataword is created.
10.44


4. An error changes r0 and a second error changes a3 .
The received codeword is 00110. The syndrome is 0.
The dataword 0011 is created at the receiver. Note that
here the dataword is wrongly created due to the
syndrome value.
5. Three bits—a3, a2, and a1—are changed by errors.
The received codeword is 01011. The syndrome is 1.
The dataword is not created. This shows that the simple
parity check, guaranteed to detect one single error, can
also find any odd number of errors.

10.45

Note

A simple parity-check code can detect
an odd number of errors.

10.46

Note

All Hamming codes discussed in this
book have dmin = 3.

The relationship between m and n in
these codes is n = 2m − 1.

10.47

Figure 10.11 Two-dimensional parity-check code

10.48


10.49


10.50

Table 10.4 Hamming code C(7, 4)

10.51

Figure 10.12 The structure of the encoder and decoder for a Hamming code

10.52

Table 10.5 Logical decision made by the correction logic analyzer

10.53

Example 10.13

Let us trace the path of three datawords from the sender
to the destination:
1. The dataword 0100 becomes the codeword 0100011.
The codeword 0100011 is received. The syndrome is
000, the final dataword is 0100.
The syndrome is 011. After flipping b2 (changing the 1
to 0), the final dataword is 0111.
The syndrome is 101. After flipping b0, we get 0000,
the wrong dataword. This shows that our code cannot
correct two errors.
10.54

Example 10.14

We need a dataword of at least 7 bits. Calculate values of
k and n that satisfy this requirement.
Solution
We need to make k = n − m greater than or equal to 7, or
2m − 1 − m ≥ 7.
1. If we set m = 3, the result is n = 23 − 1 and k = 7 − 3,
or 4, which is not acceptable.
2. If we set m = 4, then n = 24 − 1 = 15 and k = 15 − 4 =
11, which satisfies the condition. So the code is
C(15, 11)

10.55

Figure 10.13 Burst error correction using Hamming code

10.56

10-4 CYCLIC CODES

Cyclic codes are special linear block codes with one
extra property. In a cyclic code, if a codeword is
cyclically shifted (rotated), the result is another
codeword.

Cyclic Redundancy Check
Hardware Implementation
Polynomials
Cyclic Code Analysis
Advantages of Cyclic Codes
Other Cyclic Codes
10.57

Table 10.6 A CRC code with C(7, 4)

10.58

Figure 10.14 CRC encoder and decoder

10.59

Figure 10.15 Division in CRC encoder

10.60

Figure 10.16 Division in the CRC decoder for two cases

10.61

Figure 10.17 Hardwired design of the divisor in CRC

10.62

Figure 10.18 Simulation of division in CRC encoder

10.63

Figure 10.19 The CRC encoder design using shift registers

10.64

Figure 10.20 General design of encoder and decoder of a CRC code

10.65

Figure 10.21 A polynomial to represent a binary word

10.66

Figure 10.22 CRC division using polynomials

10.67

Note

The divisor in a cyclic code is normally
called the generator polynomial
or simply the generator.

10.68

Note

In a cyclic code,
If s(x) ≠ 0, one or more bits is corrupted.
If s(x) = 0, either

a. No bit is corrupted. or
b. Some bits are corrupted, but the
decoder failed to detect them.

10.69

Note

In a cyclic code, those e(x) errors that
are divisible by g(x) are not caught.

10.70

Note

If the generator has more than one term
and the coefficient of x0 is 1,
all single errors can be caught.

10.71

Example 10.15

Which of the following g(x) values guarantees that a
single-bit error is caught? For each case, what is the
error that cannot be caught?
a. x + 1 b. x3 c. 1
Solution
a. No xi can be divisible by x + 1. Any single-bit error can
be caught.
b. If i is equal to or greater than 3, xi is divisible by g(x).
All single-bit errors in positions 1 to 3 are caught.
c. All values of i make xi divisible by g(x). No single-bit
error can be caught. This g(x) is useless.

10.72

Figure 10.23 Representation of two isolated single-bit errors using polynomials

10.73

Note

If a generator cannot divide xt + 1
(t between 0 and n – 1),
then all isolated double errors
can be detected.

10.74

Example 10.16

Find the status of the following generators related to two
isolated, single-bit errors.
a. x + 1 b. x4 + 1 c. x7 + x6 + 1 d. x15 + x14 + 1
Solution
a. This is a very poor choice for a generator. Any two
errors next to each other cannot be detected.
b. This generator cannot detect two errors that are four
positions apart.
c. This is a good choice for this purpose.
d. This polynomial cannot divide xt + 1 if t is less than
32,768. A codeword with two isolated errors up to
32,768 bits apart can be detected by this generator.
10.75

Note

A generator that contains a factor of
x + 1 can detect all odd-numbered
errors.

10.76

Note

❏ All burst errors with L ≤ r will be
detected.
❏ All burst errors with L = r + 1 will be
detected with probability 1 – (1/2) r–1.
❏ All burst errors with L > r + 1 will be
detected with probability 1 – (1/2) r.

10.77

Example 10.17

Find the suitability of the following generators in relation
to burst errors of different lengths.
a. x6 + 1 b. x18 + x7 + x + 1 c. x32 + x23 + x7 + 1

Solution
a. This generator can detect all burst errors with a length
less than or equal to 6 bits; 3 out of 100 burst errors
with length 7 will slip by; 16 out of 1000 burst errors of
length 8 or more will slip by.

10.78


b. This generator can detect all burst errors with a length
less than or equal to 18 bits; 8 out of 1 million burst
errors with length 19 will slip by; 4 out of 1 million
burst errors of length 20 or more will slip by.

c. This generator can detect all burst errors with a length
less than or equal to 32 bits; 5 out of 10 billion burst
errors with length 33 will slip by; 3 out of 10 billion
burst errors of length 34 or more will slip by.

10.79

Note

A good polynomial generator needs to
have the following characteristics:
1. It should have at least two terms.
2. The coefficient of the term x0 should
be 1.
3. It should not divide xt + 1, for t
between 2 and n − 1.
4. It should have the factor x + 1.

10.80

Table 10.7 Standard polynomials

10.81

10-5 CHECKSUM

The last error detection method we discuss here is
called the checksum. The checksum is used in the
Internet by several protocols although not at the data
link layer. However, we briefly discuss it here to
complete our discussion on error checking

Idea
One’s Complement
Internet Checksum

10.82

Example 10.18

Suppose our data is a list of five 4-bit numbers that we
want to send to a destination. In addition to sending these
numbers, we send the sum of the numbers. For example,
if the set of numbers is (7, 11, 12, 0, 6), we send (7, 11, 12,
0, 6, 36), where 36 is the sum of the original numbers.
The receiver adds the five numbers and compares the
result with the sum. If the two are the same, the receiver
assumes no error, accepts the five numbers, and discards
the sum. Otherwise, there is an error somewhere and the
data are not accepted.

10.83

Example 10.19

We can make the job of the receiver easier if we send the
negative (complement) of the sum, called the checksum.
In this case, we send (7, 11, 12, 0, 6, −36). The receiver
can add all the numbers received (including the
checksum). If the result is 0, it assumes no error;
otherwise, there is an error.

10.84

Example 10.20

How can we represent the number 21 in one’s
complement arithmetic using only four bits?

Solution
The number 21 in binary is 10101 (it needs five bits). We
can wrap the leftmost bit and add it to the four rightmost
bits. We have (0101 + 1) = 0110 or 6.

10.85

Example 10.21

How can we represent the number −6 in one’s
complement arithmetic using only four bits?

Solution
In one’s complement arithmetic, the negative or
complement of a number is found by inverting all bits.
Positive 6 is 0110; negative 6 is 1001. If we consider only
unsigned numbers, this is 9. In other words, the
complement of 6 is 9. Another way to find the
complement of a number in one’s complement arithmetic
is to subtract the number from 2n − 1 (16 − 1 in this case).

10.86

Example 10.22

Let us redo Exercise 10.19 using one’s complement
arithmetic. Figure 10.24 shows the process at the sender
and at the receiver. The sender initializes the checksum
to 0 and adds all data items and the checksum (the
checksum is considered as one data item and is shown in
color). The result is 36. However, 36 cannot be expressed
in 4 bits. The extra two bits are wrapped and added with
the sum to create the wrapped sum value 6. In the figure,
we have shown the details in binary. The sum is then
complemented, resulting in the checksum value 9 (15 − 6
= 9). The sender now sends six data items to the receiver
including the checksum 9.
10.87


The receiver follows the same procedure as the sender. It
adds all data items (including the checksum); the result
is 45. The sum is wrapped and becomes 15. The wrapped
sum is complemented and becomes 0. Since the value of
the checksum is 0, this means that the data is not
corrupted. The receiver drops the checksum and keeps
the other data items. If the checksum is not zero, the
entire packet is dropped.

10.88

Figure 10.24 Example 10.22

10.89

Note

Sender site:
1. The message is divided into 16-bit words.
2. The value of the checksum word is set to 0.
3. All words including the checksum are
added using one’s complement addition.
4. The sum is complemented and becomes the
checksum.
5. The checksum is sent with the data.

10.90

Note

Receiver site:
1. The message (including checksum) is
divided into 16-bit words.
2. All words are added using one’s
complement addition.
3. The sum is complemented and becomes the
new checksum.
4. If the value of checksum is 0, the message
is accepted; otherwise, it is rejected.

10.91

Example 10.23

Let us calculate the checksum for a text of 8 characters
(“Forouzan”). The text needs to be divided into 2-byte
(16-bit) words. We use ASCII (see Appendix A) to change
each byte to a 2-digit hexadecimal number. For example,
F is represented as 0x46 and o is represented as 0x6F.
Figure 10.25 shows how the checksum is calculated at the
sender and receiver sites. In part a of the figure, the value
of partial sum for the first column is 0x36. We keep the
rightmost digit (6) and insert the leftmost digit (3) as the
carry in the second column. The process is repeated for
each column. Note that if there is any corruption, the
checksum recalculated by the receiver is not all 0s. We
leave this an exercise.
10.92

Figure 10.25 Example 10.23

10.93

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