MD05 Error Detection and Correction

8/12/2019 MD05 Error Detection and Correction

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CNT 3004

Module 5


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Chapter 10

Error Detectionand

Correction

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.


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Chapter 10

Error

Detection

And

Correction

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.


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10.4

INTRODUCTION

Data can be corrupted during transmission.

Some applications require that errors be detected andcorrected.

In a single-bit error, only one bit in the data unit haschanged.


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10.5

Burst of errors

A burst of error means that two or more bits in the

data unit have changed.


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10.6

Detection versus Correction

The correction of errors is more difficult than the

detection.

In error detection, we are only looking to see if any

error has occurred. The answer is a simple yes or no.

We are not even interested in the number of corrupted

bits. A single-bit error is the same for us as a burst of

errors.

In error correction, we need to know the exact numberof bits that are corrupted and, more importantly, their

location in the message.


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10.7

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

Redundancy is achieved through various codingschemes. The sender uses an encoderand the

receiver uses a decoder.

The encoder at the sender site adds redundant bits

whose values are based on the actual data bits.The checker at the receiver site examines the

redundant and data bits to detect or correct errors.

Redundancy


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10.8

The structure of encoder and decoder


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10.9

In modulo-N arithmetic, we use only the

integers in the range 0 to N 1, inclusive.

In modulo-2 (binary) arithmetic, we use only

two integer values 0 and 1.

Modular arithmetic


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10.10

XOR of two single bits XOR of two binary strings

Exclusive OR (XOR)

XOR is used for both addition and subtraction in modulo-2

arithmetic. The result of XOR is 0 if the two bits are the same and 1

if they are different.


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10.11

There are two methods of error correction: Forward error correction

If there are errors, the receiver can instantaneously recoverthe correct data using the redundant bits. We will study thistype of error correction in this module.

Correction by retransmission

If there are errors, the sender must retransmit the data to thereceiver. We will study this type of error correction in thenext module.

In CNT 3004, we concentrate on block codes for errordetection and correction; we will not coverconvolution codes.


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10.12

BLOCK CODING

In block coding, we divide our message into blocks, each of k

bits, calleddatawords. We add r redundant bits to each block tomake the length n = k + r. The resulting n-bit blocks are called

codewords.

With k bits, we can create 2^k different datawords and with nbits we can create 2^n different codewords.

The encoder assigns a unique codeword for each dataword. The

same dataword is always encoded as the same codeword. This

means that we have 2^n - 2^k codewords that are not used and

are called invalid codeword.


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10.13

Datawords and codewords in block coding


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10.14

Process of error detection in block coding

Errors can be detected if

The receiver knows the list of valid codewords

The original valid codeword has changed to an invalid one.


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10.15

Let us assume that k = 2 and n = 3. The table below shows the list of datawordsand codewords. Since valid codewords have even number of 1s, an odd number

of 1s indicates that the codeword is invalid.

Assume the sender encodes the dataword 01 as 011 and sends it to the receiver.

Consider the followingthree cases:

1. The receiver receives 011. It is a valid codeword. The receiver correctly

extracts the dataword 01from it.

2. The first bit of the codeword is corrupted during transmission, and 111 is

received. This is not a valid codeword and is discarded.

3. The 2nd

and 3rd

bits of the codeword are corrupted during transmission, and000 is received. This is a valid codeword. The receiver incorrectly extracts the

dataword 00. Two corrupted bits have made the error undetectable.

Example

Three cases1. 011011

2. 011111

3. 011000


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10.16

An error-detecting code can detect only the types of errors forwhich it is designed; other types of errors may remain

undetected.

Error correction is much more difficult than error detection. Inaddition to detecting that the received codeword is invalid, the

receiver needs to find (determine) the value of the original

codeword.

More redundant bits are needed for error correction than forerror detection.

In the case of error correction, the encoder at the sender site

adds error-correcting redundant bits whose values are basedon the actual data bits. The checker at the receiver site

examines the redundant bits and the data bits to detect then

correct the errors.

Error detection vs. error correction


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10.17

Structure of encoder and decoder in error correction


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10.18

Let us add more redundant bits to see if the receiver can correct anerror without knowing what was actually sent. We add 3 redundant

bits to the 2-bit dataword to make 5-bit codewords. The Table below

shows the datawords and codewords. Assume the dataword is 01.

The sender creates the codeword 01011. The codeword is corruptedduring transmission, and 01001 is received. First, the receiver finds

that the received codeword is not in the table. This means an error

has occurred. Assuming that there is at most one bit corrupted, the

receiver, finds the correct codeword as shown on the next slide.

Example


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Assuming that there is at most one bit corrupted, the receiver usesthe following strategy to guess the correct dataword.

1.After 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 twodifferent 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.

Example (continued)

The above strategy is

further illustrated on the

next slide.


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The receiver received the codeword 01001

The receiver compares the received codeword with every valid codeword

in the table and count the number of different bits.

2 bits

01001

1 bit

3 bits

4 bits

Example (continued)

Action:codeword 01001 is corrected to codeword 01011because it gives the minimum bit difference of one bit.

The correct dataword is 01

Minimum


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Another Example

The valid codeword 01011 differs from the received codeword 00011 by

only in one bit. The valid codeword 01011 is selected and the dataword 01

is extracted.

The receiver received the codeword 00011

Remark

The two invalid codewords 01001 and 00011 differ from the valid codeword

01011 by only one bit but differ from any other valid codeword by more than

one bit.


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10.22

We can use XOR to find the Hamming distance between two pairs

of words.

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

Example

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

Hamming distance

The Hamming distance between two words of the same size

is the number of differences between the corresponding bits.


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10.23

Find the minimum Hamming distance of the coding scheme

given in the above Table.Solution

We first find all Hamming distances.

Example

The dmin in this case is 2.

The minimum Hamming distance

is the smallest Hamming distance

between all possible pairs in a set

of codewords.

Minimum Hamming Distance


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10.24

Find the minimum

Hamming distance of the

coding scheme given in

the shown Table.

SolutionWe first find all the Hamming distances.

The dmin in this case is 3.

Example


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10.25

There are three main parameters for a coding scheme:the codeword sizen, the dataword sizek, and the minimum

Hamming distancedmin. The Hamming distance between the sent codeword and the

received codeword is the number of bits that are corrupted

during transmission.

To guarantee the detection of up to s errors in all

cases, the minimum Hamming distance in a blockcode must be dmin = s + 1.

Minimum Hamming distance for error detection

Notation: the variables is used to denote the number of bit errors that

can be detected and the variablet is used to denote the number of bit

errors that can be corrected.


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10.27

Example

The block code scheme given in the Table below has dmin = 3. This code can

detect up to two errors. Again, we see that when any of the valid codewords is

sent, two bit errors create a codeword which is not in the table of valid

codewords. The receiver cannot be fooled in this case and will be able to detect

the two bit errors.

However, some combinations of three bit errors change a valid codeword to

another valid codeword, for example, when 01011 is changed to 00000. The

receiver accepts the received codeword and the errors are undetected.


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10.28

Geometric concept for finding dmin in error detection

Let us assume the sent codeword is atthe center of a circle with radius s.

All other codewords created by s or

less bit errors are points inside thecircle or on its perimeter.

All other valid codewords (having atleast s+1 different bits) are outside

the circle.


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10.29

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 createsanother valid codeword.


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10.32

Note: In linear block codes, the minimum Hamming distance is the

number of 1s in the nonzero valid codeword with the smallest

number of 1s.

In our first code (Table A), 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 B), the numbers of 1s in the

nonzero codewords are 3, 3, and 4. So in this code we have dmin = 3.

Minimum Hamming Distance in Linear

Block Codes

Table A Table B


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10.33

In a simple parity-check code, a k-bit dataword is changed to k+1

bit codeword. In this case, n=k+1. The extra bit is called the parity

bit and is selected to make the total number of 1s even. The

minimum Hamming distance for this type of block codes is 2.

Simple parity-check codes

A simple parity-check code is a single-bit error-detecting code

in which n = k+ 1 with dmin = 2.

The Table shows a parity-check

code with k=2 and n=3.


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Simple parity-check code C(5, 4)

Parity-check code with k=4 and n=5, denoted C(5,4)

The encoder uses a generator that takes a copy of the 4-bit dataworda3a2a1a0and generates a parity bit r0 using XOR. The parity bit makes the

number of 1s in the 5-bit codeworda3a2a1a0r0 even. The transmitted

codeworda3a2a1a0r0arrives at the receiver possibly with bit errors in the

formb3b2b1b0q0. The receiver constructs a one-bit syndromesto detect any

single bit error. Details are shown on the next slide.


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10.35

Encoder and decoder for simple parity-check code

r0 = a3 a2 a1 a0r0 makes total number of 1s even

s0 = b3 b2 b1 b0 q0s0is called the syndrome

s0 must be 0 to accept the codeword

k = 4 , n = 5

Simple parity check code C(5 4)


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10.36

Let us look at some transmission scenarios. Assume the sender sends the dataword 1011.

The codeword created from this dataword is a3a2a1a0r0=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 errorchanges a1. The received codeword is 10011. The syndrome is 1.

No dataword is created. The error is detected and the codeword is rejected.

3. One single-bit errorchanges r0. The received codeword is 10110. The syndrome is 1.

No dataword is created. The error is detected and the codeword is rejected.

4. Two bit errors: two errors change r0and 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 because the syndrome value s0 =0.

Simple parity-check code C(5, 4)

5. Three bits errorsa3, a2, and a1are

changed by errors. The received

codeword is 01011. The syndrome is 1.

The dataword is not created.This shows that the simple parity

check code, guaranteed to detect one

single bit error, can also detect any

odd number of errors.


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10.38

Two-dimensional parity-check code

In the 4th Edition, this topic is covered in Section 10.3, pages 280-281.

In the 5th Edition, this topic is introduced in the Problem P10-11 on page 289 at the

end of Chapter 10.


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10.39


In the two-dimensional parity-check code, the data is organized as a table. In theexample, below four 7-bit datawordsare organized as rows of the table. A parity bit is

computed for each dataword and stored in the 8thcolumn. The augmented 8-bit row

represents a codeword.A fifth row is created by computing a parity bit for each column.

In the diagram below, the boxes mark the position of bit errors and the arrows point to

the locations of nonzero syndromes when errors occur in the marked positions.

The two-dimensional parity-check code can detect up to three errors that

occur anywhere in the table. Errors affecting four bits may not be always

detected.


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10.40


0 0 0

0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0

0

1

0

0

0

0

0

0

1

0

0

0

Action at the receiverThe receiver adds a parity bit (9th bit) for each received 8-bit codeword.

The receiver creates a 6th column by adding a parity bit for each column.

Parity bits with value 1 indicates the presence of an error in the

corresponding row or column.


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10.41


Two-dimensional parity-check codes can detect up to three

errors that occur anywhere in the group of codewords.Notice that two bit errors within a codeword can always bedetected by a two-dimensional parity-check code butcannot be detected by a simple parity-check code.


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10.42

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.

In CNT 3004, we only cover a category of cyclic codes called

Cyclic Redundancy Check (CRC) codes which are used in LANs

and WANs. In CRC, the generation of the redundant bits is done

using modulo-2 division.


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C C d


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10.45

CRC DecoderThe checker uses the same agreed upon divisord3d2d1d0 to generate the syndrome

bits. The received codeword b3b2b1b0q2q1q0 is divided by the divisor modulo 2. Theremainder of the modulo 2 division s2s1s0 is the syndrome. If there are no errors, thevalue of the syndrome is 000.


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Modulo 2 Division in the CRC decoder for two cases


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Modulo 2 Division in the CRC decoder for two cases

The checker uses the same agreed upon divisorto divide the received codeword

using modulo 2 arithmetic. The remainder of the modulo 2 division S2S1S0 is calledthe syndrome. The codeword is accepted if the syndrome is 0. A non-zerosyndrome indicates the presence of bit errors; in this case, the receiver discards

the dataword.

P l i l


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10.48

A better way to understand cyclic codes and how they can beanalyzed is to represent them as polynomials. A pattern (string) of 0s

and 1s can be represented as a polynomial with coefficients of 0 and

1. The power of each term indicates the position of the bit and the

coefficient shows the value of the bit.

Thedegreeof the polynomial is the highest power in the polynomial.

For example, the degree of the polynomial shown below is 6.

Polynomials

Addi d S bt ti P l i l


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Adding and Subtracting Polynomials

Operations are done in modulo-2 and coefficients areonly 0 and 1

Addition and subtraction give the same result

Adding (or subtracting) is done by combining terms

and deleting pairs of identical terms

Example: Add x^5 + x^4 + x^2 to x^6 + x^4 + x^2

Adding x^5 + x^4 + x^2

and x^6 + x^4 + x^2

gives x^6 + x^5

10.49

M lti l i d Di idi P l i l


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Multiplying and Dividing Polynomials

Multiplying a term by another term is done by adding

their powers.

For example x^5 * x^4 = x^9.

For the division operation, we subtract the power of

the second term from the power of the first term.For example, x^7 / x^3 = x^4.

10.50

Cyclic Code Encoder Using Polynomials


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Cyclic Code Encoder Using Polynomials

Example:The divisor 1011 is represented as x^3+x+1 and the dataword

1001 is represented as x^3+1. We first multiply the dataword by

x^3 to shift it to the left by three positions (i.e., store zeros in the

three redundant bit positions). This gives the dividend polynomial

x^6+x^3 that will be divided by the divisor x^3+x+1.

We divide the term x^6 by the first term of the divisor x^3. Thisgives the first term of the quotient = x^6/x^3 = x^3. Then we

multiply the term x^3 by the divisor and subtract the result from

the dividend. We continue the division operation until the degree

of the dividend is less than the degree of the divisor. Theremainder of the division is used as the three redundant bits r2r1r0.

10.51

The division using polynomials is shown on the next slide.

CRC di i i i l i l


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10.52

CRC division using polynomials

The divisor in a cycliccode is normally called

the generator polynomialor simply the generator.

divisor = 1011

codeword = 1001000

r2r1r0 = 110

Notations for Polynomials in Cyclic Codes


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Notations for Polynomials in Cyclic Codes

Dataword: d(x) degree < k

Codeword: c(x) degree < n

Generator: g(x) degree < n-k +1Syndrome: s(x) degree < n-k

Error: e(x) degree < n

Example: n=7

codeword= 0001100 c(x) = x^3+x^2

received as 1001000 two bits in error

error = 1000100 e(x) = x^6 + x^2

Example1:


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10.54

Example1:

Polynomials for some 5-bit codewords

0 1 1 0 1 x^3 + x^2 + 1

0 0 0 1 1 x + 1

1 0 0 0 0 x^40 0 0 0 1 1

1 1 1 1 0 x^4 + x^3 + x^2 + x

Example 2:The transmitted 5-bit codeword

1 1 1 1 0 x^4 + x^3 + x^2 + xwas received as

1 0 1 0 0 x^4 + x^2

The error polynomial is

The above error polynomial indicates two error bits.

The errorpolynomial is

obtained by

computing the

XOR of the

polynomial of

the transmittedcodeword and

the polynomial

of the received

codeword.

e(x) = x^3 + x

Cyclic Code Analysis


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10.55

In a cyclic code,

Ifs(x) 0, then at least one bit is corrupted.

Ifs(x) = 0, then no bit is corrupted assuming the number of

corrupted bits does not exceed the errordetection capability of the code.

s(x) = syndrome computed by the receiver

Cyclic Code Analysis


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10.56

Rule: if the generatorg(x) has more than one term and thecoefficient ofx0 is 1, then all single bit errors can be caught

because the remainder of e(x)/g(x) = x^i/g(x) is not zero.

For a single-bit error

e(x) = x^iwhere i is the position of the bit error.

In a cyclic code, errors represented by the error polynomial e(x) are notcaught if e(x) is divisible byg(x), i.e., if the remainder of e(x) / g(x) is zero.

To check if the generatorg(x) can detect errors of type e(x):

divide e(x) byg(x)If the remainder of e(x) / g(x) is zero, then the generatorg(x) cannot

guarantee the detection of errors of type e(x).

Example


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Which of the following divisor polynomials g(x) 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. The generator polynomial x+1 can detect any single-bit error in any positionbecause the polynomial has two terms and the coefficient of x0 is 1. For any i,

xi is not divisible by x + 1 because xi/ (x+1) always has a remainder.

b.If i=0, 1, or 2 then xi/ x3 has a non zero remainder equal to xi. Sosingle bit

errors in positions 0, 1 and 2 can be detected. If i is equal to 3 or greater than3, xi is divisible by g(x)= x3, i.e., xi/ x3 does not have a remainder when i 3.

Sosingle bit errors in positions 3 or higher cannot be caughtas shown below.

c. The polynomial consists of a single term and cannot detect any single-biterror. All values of i make xidivisible by g(x)=1. No single-bit error can be

caught.

n-bit codewordcn-1c5c4c3c2c1c0cannot be caught can be caught


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10.59

Standard polynomials

This slide is only for your info

Advantages of Cyclic Codes


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10.60

Cyclic codes have a very good performance indetecting single-bit errors, double errors, an odd

number of errors, and burst errors.

Cyclic codes can easily be implemented insoftware. They can also be implemented in

hardware by using a handful of electronic

devices. Cyclic codes are especially fast when

implemented in hardware. This has made cyclic

codes a good candidate for many networks.


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CHECKSUM


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10.62

CHECKSUM

The last error detection method we discuss here is called the

checksum. The checksum is used in the Internet by several

protocols.

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, thereceiver assumes no error, accepts the five numbers, and discards

the sum. Otherwise, there is an error somewhere and the data are

not accepted

Example

Example


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10.63

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.

The diagram on the next slide shows the logic of the

checksum method.

Checksum


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10.64

The checksum must be represented using the same

number of bits (say m bits) used to represent each

data number.


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Example(7, 11, 12, 0, 6, 36)

checksum


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10.66

The Figure on the next slide 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 red

color). The result is 36. However, 36 cannot be expressed in 4 bits. The extra

two bits are wrapped and added to the sum to create the wrapped sum value 6.Details of computing the wrapped sum are shown in binary. The wrapped sum

is then complemented, resulting in the checksum value 15 6=9. The sender

now sends six data items to the receiver including the 4-bitchecksum 9.

The receiver follows the same procedure as the sender. It adds all data items

(including the checksum); the result is the 6-bit value 45 which has two extra

bits over the length of a data item. The sum is wrapped with the two extra bits

and becomes 15. The wrapped sum is complemented and becomes 0. Since the

value of the checksum computed by the receiver 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.

( , , , , , )

Example


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10.67

1 1 1 1

0 0 0 0

0 1 1 0

1 0 0 1

Initialized checksum Received checksum

Example


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In the previous example, the decimal number 36 in binary is(100100)2. To change it to a 4-bit number we add the extra

two leftmost bits to the right four bits as shown below.

10.68

1 0 0 1 0 01 0

+

0 1 1 0 wrapped sum = 6

1 0 0 1 complemented wrapped sum =checksum = 9

Internet 16-bit Checksum


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10.69

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 ones complement addition.

4. The wrapped sum is complemented and becomes thechecksum.

5. The checksum is sent with the data.

Receiver site:1. The message (including the checksum) isdivided into 16-bit words.

2. All words are added using onescomplement addition.

3. The wrapped sum is complemented and becomes thenew checksum.

4. If the value of checksum is 0, the messageis accepted; otherwise, it is rejected.

Other Approaches


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10.70

This illustrates one major problem with the traditional

checksum calculation. If two items are transposed in

transmission, the checksum cannot catch this error. The

reason is that the traditional checksum is not weighted: it

treats each data item equally. In other words, the order of

data items is immaterial to the calculation. Several

approaches have been used to prevent this problem. Two ofthem are the Fletcher Checksum and Adler Checksum.

They will not be covered in CNT 3004.

(7, 11, 12, 0, 6) (7, 12, 11, 0, 6) (7, 10, 10, 3, 6)

The following sets of four numbers produce the samechecksum

Error Detection Fields in Various Protocols


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TCP Header Format

This Slide is for Information Only



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UDP Header Format




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IPv4 Packet Format

IP Checksum (16 bits) covers only the header (not the data) and is recalculated by each

router visited by the packet.



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ATM Cell Format

ATM Header Error Control or HEC (8 bits) covers only the 5-byte header (not the data)and is recalculated by each ATM switch visited by the cell.

ATM cell is a packet of fixed size (54 bytes) consisting of a header (5 bytes) and a

data payload (48 bytes).



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FCS: the frame check sum is a 4-byte cyclical redundancy check (CRC) value

used for error checking.



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802.11: is the Wireless WiFi MAC protocol.

FORWARD ERROR CORRECTION


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10.77

We discussed error detection and

retransmission in the previous sections.However, retransmission of corrupted and

lost packets is not useful for real-time

multimedia transmission. We need to correctthe error or reproduce the packet

immediately.

Using Hamming Distance


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10.78

We earlier discussed the Hamming distance for

error detection. For error correction, we definitely

need more distance. It can be shown that to correct

t errors, we need to have dmin = 2t + 1. Forexample, to correct 2 bit errors, we need dmin = 5.

Example: if we want to correct 10 bits in a packet,we need to make the minimum hamming distance

21 bits, which means a lot of redundant bits need

to be sent with the data. The geometricalrepresentation of this concept is shown on the next

slide.

Geometric concept for finding dmin in error correction


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10.79

Each valid codeword is at the center of a circular territory with radius t.

If a codeword is corrupted by t or less bit errors, the corrupted codeword is apoint inside the circle or on its perimeter.

If the receiver receives a codeword that belongs to a circular territory, it

decides that it is the codeword at the center of this circle under the assumption

that no more than t bit errors have occurred.

Concept of correcting t errors

Minimum Hamming distance for error correction


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To guarantee correction of up tot bit errors in allcases, the minimum Hamming distance in a block

code must be dmin = 2t + 1.

A code scheme has a minimum Hamming distance dmin = 4. What is the error

detection and correction capability of this scheme?

This code guarantees the detection of up to three bit errors

(s=3), but it can correct up to one bit error (t=1). In other words,

if this code is used for error correction, part of its capability is wasted because aminimum Hamming distance dmin = 3 is sufficient to correct any one bit error.

Efficient error correction codes need to have an odd minimum Hamming

distance (3, 5, 7, 9, 11, ).

Example

Solution dmin = 4=2t + 1 t = 1dmin = 4=s + 1 s = 3

Reminder:to guarantee the detection of up tosbit errors, we must have dmin = s + 1

Error-Correcting Hamming Codes

Source: 4thEdition Textbook only


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In CNT 3004, we only cover Hamming codes withminimum distance dmin = 3. These codes can

correct only a single bit error.

In these codes, we addrredundant (check) bits to

each dataword of length k bits to make the lengthof the codeword n=k+r bits.

The relationship between r and n in these codes isn = (2 ^ r ) 1.

Hamming code C(7, 4)



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The notation C(7, 4) means that each dataword in the code is of length k=4

bits and the number of redundant bits is r=3 bits, making the length of the

codeword n=7 bits. Notice that 7 = 2^3 - 1

The structure of the encoder and decoder for a Hamming code



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Generating the three parity check bits



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r0 = a2 a1 a0

r1 = a3 a2 a1

r2 = a1 a0 a3

Note: the above three equations are not unique

Example:

Consider the dataword 0011 a3=0, a2=0, a1=1, a0=1.

We compute the redundant bits as follows

r0 = a2 a1 a0 = 0 1 1 = 0

r1 = a3 a2 a1 = 0 0 1 = 1r2 = a1 a0 a3 = 1 1 0 = 0

The 4-bit dataword 0011 is transmitted as the 7-bit codeword 0011010.

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General Logic



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r0 = a2 a1 a0r1 = a3 a2 a1

r2 = a1 a0 a3

s0 = b2 b1 b0 q0s1 = b3 b2 b1 q1s2 = b1 b0 b3 q2

Example 1: error in bit a0, i.e., b0 is corrupted

This causes s0 and s2 to be 1. Syndrome value is 101

Syndrome is the 3-bit number S2 S1 S0. An error in bit aj causescertain syndrome bits to be 1 and results in a unique syndrome value.

Example 2: error in bit a1, i.e., b1 is corruptedThis causes all syndrome bits to be 1. Syndrome is 111

Example 3: error in bit a2, i.e., b2 is corruptedThis causes s0 and s1 to be 1. Syndrome value is 011

See next slide for all bit error positions


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Let us trace the path of the following dataword from the sender to the

Example Hamming code C(7, 4) Source: 4thEdition only


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destination:

Case 1: No Errors

The dataword 0100becomes the codeword 0100011.

No errors occurred and the received codeword is 0100011 = b3b2b1b0q2q1q0Using the equations, we find the syndrome bits are s0 = s1= s2 = 0.

The syndrome is 000 implying no errors. The final dataword is 0100.

syndrome=s2s1s0s0 =b2 b1 b0 q0s1 =b3 b2 b1 q1s2 =b1 b0 b3 q2


destination:



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destination:

Case 2: One bit Error

The dataword0111becomes the codeword 0111001.

One bit error occurred. The received codeword is 0011001 = b3b2b1b0q2q1q0

The computed syndrome is 011 which correctly indicates an error in bit b2.After flipping b2 (changing the 0 to 1), the final dataword is 0111.

syndrome=s2s1s0s0 =b2 b1 b0 q0s1 =b3 b2 b1 q1s2 =b1 b0 b3 q2


destination:



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destination:

Case 2: Two bit Errors

The dataword 1101 is coded as 1101000.

Two bit errors occurred. The received codeword is 0001000 = b3b2b1b0q2q1q0

The computed syndrome is 101 indicating an error in b0.After flipping b0, we get 0000, the wrong dataword.

This shows that our code cannot correct two bit errors.

syndrome=s2s1s0s0 =b2 b1 b0 q0s1 =b3 b2 b1 q1

s2 =b1

b0

b3

q2

A Hamming code uses r = 4 redundant bits. Calculate the

maximum value of k (i e maximum length of the dataword) that

Example


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maximum value of k (i.e., maximum length of the dataword) thatsatisfies the Hamming code requirement for single bit error

correction.

SolutionThe number of redundant bits is r =4 and we know that n = 2^r-1

Thus we have

n = 2 ^ 4 1 = 15 and k = n - r = 15 - 4 = 11

This means that the maximum length of the dataword = k = 11 bits

The Hamming code is

C(15, 11)

Forward Error Correction Using XOR


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For real-time multimedia applications, the retransmission of

lost packets is not practical. A simple forward error

correction (FEC) scheme can be used by transmitting one

additional redundant packet for each N packets of multimedia traffic. The redundant packet is computed as the

XOR of the N data packets.

If one packet is lost out of the N+1 transmitted packets, thelost packet can be immediately reproduced by computing the

XOR of the N received packets. This is shown on the next

slide.

Forward Error Correction Using XOR


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The simple FEC method uses the property of the

exclusive OR (XOR) operation as shown below.

If packet Pi is lost, the receiver can compute it as

follows:

Redundant packet R

MD05 Error Detection and Correction

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