error detection (1)(4)

8/14/2019 Error Detection (1)(4)

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MSRIT INFORMATION SCIENCE 1

ERROR DETECTIONERROR DETECTION&&

ERROR CORRECTIONERROR CORRECTION


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AMAKAMAK

A-> ANKITA (1MS07IS133)

M-> MAYANK (1MS07IS047)

A-> ANSHUJ (1MS07IS011)

K-> KRISH (1MS07IS038)


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INTRODUCTION TO ERROR

REDUNDANCY

CODING

LINEAR BLOCK CODING

HAMMING CODE

CYCLIC REDUNDANCY CHECK(CRC)

IMPLEMENTATION OF HAMMING CODE

TABLE OF CONTENTS:TABLE OF CONTENTS:


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What is an error???Unpredictable change of bits from 1->0 or 0->1.

TypesoSingle bit error

oBurst error(Multiple)

o

Redundancy: Correction or detection of errors.

INTRODUCTIONINTRODUCTION


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Forward error correction: Method of Guessing theactual message using the redundant bits.

Retransmission: Repeated sending of message untilerror free.

Modulo Arithmetic: Modulo 2- Remainder after division can be either 0or1.

Modulo 3- Remainder after division can be either0,1 or 2.

. . .

Modulo n- Remainder after division can be either0,1,2.n-1.


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Adding : 0+0=0 0+1=1 1+0=1 1+1=0 Subt: 0-0=0 0-1=1 1-0=1 1-1=0

XOR operation: 1 0 1 1 0 1 1 1 0 0 0 1 0 1 0


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CODING: Redundancy is achieved through

coding.

qBLOCK CODING: Message divided into blocks.

k bit datawords

r redundant bits n= k+r, n>k,n bit codeword.

2k total datawords(equal to number of valid

codewords.) 2n total codewords

2n -2k invalid codewords



ERROR DETECTION

oIf resulting codeword is invalid.

oIf multiple errors in the codeword result invalid codeword.

ERROR CORRECTIONoMore difficult.

oNeed more number of redundant bits than for

detection.oInvolves error detection as well as finding the

position(s) where error has occurred.



Generation of codewords for each dataword: C(7,4)

n ,k

Codeword is generated by the generator whichappends 3 redundant bits at the end of the

dataword.Ro =a2 + a1 + a0 Modulo-2R1= a3 + a2 + a1 Modulo-2

R2= a

1+ a

0+ a

3Modulo-2

HAMMING CODE GENERATIONHAMMING CODE GENERATION



DATAWORD

CODEWORD

DATAWORD

CODEWORD

0000 0000000 1000 1000110

0001 0001101 1001 1001011

0010 0010111 1010 10100010011 0011010 1011 1011100

0100 0100011 1100 1100101

0101 0101110 1101 1101000

0110 0110100 1110 11100100111 0111001 1111 1111111

A CRC CODE WITH C(7,4)A CRC CODE WITH C(7,4)



checker on the receiver side will generate a

3bit syndrome by the formulae given below:s0 = b2 + b1 + b0 + q0 modulo-2

s1 = b3 + b2 + b1 + q1 modulo-2

s2 = b1 + b0 + b3 + q2 modulo-2

If the syndrome i.e s2s1s0 is 000 then data is

error freeor undetected.Otherwise, received codeword has an error(s).



MAGIC TABLE

Depending upon the value of syndrome we can find theposition of occurrence of error and then the bitposition where error has occurred is flipped.

SYNDROME

000 001 010 011 100 101 110 111ERROR None q0 q1 b2 q2 b0 b3 b1



CODEWORD NOTATION ONCODEWORD NOTATION ON

SENDERS AND RECEIVERS SIDESENDERS AND RECEIVERS SIDE

a3 a2 a1 a0 R2 R1 R0

b3 b2 b1 b0 q2 q1 q0

RECEIVEDCODEWORD

0 1 0 1 0 0 0

CODEWORD SENT

1 1 0 1 0 0 0



Number of bit change occurring between twocodewords.

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

The total number of ones is equal to the number of bitchanges between two codewords.

HAMMING DISTANCEHAMMING DISTANCE



Smallest Hamming Distance between all sets ofcodewords.

Ex-

d(0000000,0001101) = 3d(0001100,0111001) = 4d(0110100,0111001) =3

d(11111111,0000000) =7. & so on..

Dmin = 3 for the above set of codewords.

MINIMUM HAMMING DISTANCEMINIMUM HAMMING DISTANCE



Suppose s errors are to be detected, then dmin

should be s+1.for the example taken, it can detect upto a

maximum of 2 errors.

Suppose t errors are to be corrected, then thedmin should be 2t+1.

in the above example, it can correct upto only 1

error.



Linear block code?

Linear block code with an extra property: code

word is cyclically rotated that generatesanother codeword.

1010110 is a codeword on rotating 0101101 which is another codeword.

CYCLIC CODECYCLIC CODE


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Type of linear block code which only detects

errors.

Its computation resembles a long division

operation in which the quotient is discarded

and the remainder becomes the result.

CYCLIC REDUNDANCY CHECKCYCLIC REDUNDANCY CHECK

(CRC)(CRC)


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CRC ENCODER AND DECODERCRC ENCODER AND DECODER


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Encoder on senders side generates codeword.

Dataword size is k bits.

Desired codeword is n bits.

Augment dataword by appending n-k 0s.

Divisor (predefined) of size n-k+1, dividesaugmented dataword in generator.

Obtained remainder is appended to dataword.


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The generated codeword is sent to receiver via

some transmission medium.

Decoder on receivers side checks for errors. The checker divides the codeword by the same

divisor.

This generates a remainder which is called asyndrome.

If the syndrome is 0 then there is no error or

the error is undetected. If syndrome is non zero, error has been

detected and data is discarded.


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The given dataword can be represented inpolynomial terms.

Multiply the dataword with xn-k to generateaugmented dataword.

The augmented dataword is divided by the

generator polynomial g(x) and the resultingremainder is added to the augmented dataword.

Note in division when we subtract we actually

perform XOR operation.

CODEWORD GENERATIONCODEWORD GENERATION


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The divisor on the receiving side divides thereceived code word and generates a

remainder.

Remainder is also called as a syndrome. If the syndrome generated is 0 then there is no

error in transmission or undetected error.

Non zero syndrome means that error has beendetected.

No error correction is possible using CRC.

ERROR DETECTIONERROR DETECTION


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Received codeword can be represented as Received codeword=c(x)+e(x) where c(x) is

original codeword e(x) is the error. The error is detected if

received codeword=c(x)+e(x) is not divisible.

g(x) If e(x) is divisible by g(x) then error goes

undetected.Single bit error:

e(x)=xi.

xi should not be divisible by g(x). x0 term should be 1 so that we can catch

the error.


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Two Isolated bit errors: e(x)=xi+xj.

e(x)=xi(1+xj-i ) where i


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Polynomial should contain more than oneterm.

Polynomial should have the x0 term equal to1.

Polynomial should contain x+1 as a factor. Polynomial should not divide 1+xt for

0


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QUESTIONSQUESTIONS

???

error detection (1)(4)

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