data link layer-megh

8/9/2019 Data Link Layer-Megh

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Data Link Layer


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Physical: Describes the transmission of raw bits in terms of

mechanical and electrical issues.

Data Link: Describes how a shared communication channel can

be accessed, and how a data frame can be reliably transmitted.

Network: Describes how routing is to be done. Mostly needed insubnets.

Transport: The hardest one: generally offers connection-oriented

as well as connectionless services, and varying degrees of

reliability. This layer provides the actual network interface to

applications.

Application: Contains the stuff that users see: e-mail, remote

logins, the Webs exchange protocol, etc.


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Fig 1: Simplified representation of the data link andassociated layers


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The overall task of the data link layer is to make

the physical line appear error free to the higherlayers (this is usually called a virtual circuit).

Thus the higher layer of the protocol hierarchy

can pass data down to the lower layers and beable to assume that if the message reaches thedestination it will error free.


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DLLpurpose?

The goal of the data link layeris to provide reliable, efficient communication

between adjacent machines connected by a single

communication channel

1.Group the physical layer bit stream into units calledframes

2. Sender checksums the frame and sends checksumtogether with data. The checksum allows the receiver todetermine when a frame has been damaged in transit.


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3. Receiver recomputes the checksum and comparesit with the received value. If they differ, an errorhas

occurred and the frame is discarded.

4.P

erhaps return a positive or negativeacknowledgment to the sender.

5. Flow control. Prevent a fast sender from

overwhelming a slower receiver.

6. Error control


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P

lacement of DLLPlacement of the data link protocol.


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Routing

The network layer consists of routing

they are connected through point=to-point links

The router would really its packet to be sent

correctly, guaranteed,and in the order it was

issued. It is up to the data link to make unreliable

connections look perfect, or at least, fairly good


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Data Link Layer Design Issues Services Provided to the

Network Layer

Framing

Error Control

Flow Control


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F

unctions of th

e Data Link Layer Provide service interface to the

network layer

Dealing with transmission errors

Regulating data flow

Slow receivers not swamped by fastsenders


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Functions of the Data Link Layer

(2)Relationship between packets and frames.


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Services Provided to Network

Layer(a) Virtual communication.

(b)Actual communication.


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Types of services provided to the

Network Layer

Unacknowledged Connectionless service

Acknowledged Connectionless service

Acknowledged Connection-Orientedservice


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Unacknowledged Connectionless

service

Losses are taken care of at higher layers

No recovering of lost or corrupted frame

Used on reliable medium like coax cables

or optical fiber, where the error rate is low.

Appropriate for voice, where delay is

worse than bad data.


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Acknowledged Connectionless

service

Useful on unreliable medium like wireless.

Acknowledgements add delays.

returns information a frame has safely arrived.

time-out, resend, frames received twice

Adding ack in the DLL rather than in the NL. Leaving it

for the NL is inefficient as a large message (packet) has

to be resent in that case in contrast to small frames here.


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Acknowledged Connection-

oriented service

Most reliable

Established connection before any data is sent.

Guaranteed service

Each frame sent is indeed received

Each frame is received exactly once

Frames are received in order

Special care has to be taken to ensure this in

connectionless services


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Placement of DLL

Placement of the data link protocol.


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Framing

Character Count

Flag bytes with byte stuffing

F

lag bytes with

bit stuffing

Physical layer coding violations


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Framing, character count

Uses a field in the header to specify the number of

characters in the frame when the Data Link Layer at

the destination sees the character count.

It knows how many character follow and hence

where the end of the frame is.


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Problem withFraming with CC

What if the count is garbled.

Even if with checksum, the receiver knows

that the frame is bad there is no way to tellwhere the next frame starts.

Asking for retransmission doesnt help

eith

er because th

e start of th

eretransmitted frame is not known

No longer used


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Byte stuffing or character stuffing.

The senders DLL insert a special escape byte(ESC) just before each accidental flag byte inthe data.

The Data Link Layer on the receiving endremoves the escape byte before the data aregiven to the network layer.

A framing flag byte can be distinguished fromone in the data by the absence or presence ofan escape byte before it.

When an escape byte occurs in the middle ofthe data, it too is stuffed with an escape byte .


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Framing with byte stuffing

(a) A frame delimited by flag bytes.

(b) Four examples of byte sequences

before and after stuffing.


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Framing with byte stuffing

Problem : fixed character size : assumes

character size to be 8 bits : cant handle

heterogeneous environment.


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Framing with bit stuffing

Bit stuffing(a) The original data.

(b) The data as they appear on the line.

(c) The data as they are stored in receivers memoryafter destuffing.


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Encoding Violations:

Send an signal that doesn't conform to any legalbit representation.

In Manchester encoding, for instance, 1-bits

are represented by a high-low sequence, and 0-bits by low-high sequences.

The start/end of a frame could be representedby the signal low-low orhigh-high.

Finally, some systems use a combination of these techniques. IEEE 802.3, for instance, hasboth a length field and special frame start andframe end patterns.


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Error Control

Positive and Negative feedback

Timers : what happens when a frame completelyvanishes : receiver neither sends a +ack nor

ack then timer comes to help.

It may result in a frame being sent more than once

and received more than once :

solution : assign sequence numbers to frames


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Error Control

Error control is concerned with insuring

that all frames are eventually delivered

(possibly in order) to a destination. How?Three items are required.

Acknowledgements:

Timers:

Sequence Numbers:


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Acknowledgements:

The receiver returns a special acknowledgment

(ACK) frame to the sender indicating the correct

receipt of a frame. In some systems, the receiver also returns a

negative acknowledgment (NACK) for

incorrectly-received frames.

It can retransmit a frame right away withoutwaiting for a timer to expire.


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Timer

Retransmission timers are used to resend

frames that don't produce an ACK.

When sending a frame, schedule a timer

to expire at some time after the ACK

should have been returned.

If the timer goes off, retransmit the frame.


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Sequence Numbers

Retransmissions introduce the possibility

of duplicate frames.

It suppress duplicates, add sequence

numbers to each frame.

so that a receiver can distinguish between

new frames and old copies.


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Flow Control

Flow control deals with throttling the speed

of the sender to match that of the receiver.

Usually, this is a dynamic process, as the

receiving speed depends on such

changing factors as the load, and

availability of buffer space.


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Two approaches to prevent the sender from

pumping the frames

Feedback based flow control: the receiver

sends back information to the sender giving it

permission to send more data or at least tellingthe senderhow the receiver is doing.

Rate based flow control: the protocol has a

built-in mechanism that limits the rate at which

senders may transmit data, without usingfeedback from the receiver.


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Types of errors

1. Single- - bit error

2. Burst error


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Redundancy Check

Shorter group of bits may be appended to

the end of each unit. This technique is

called redundancy .

The extra bits are redundant to the

information; they are discarded as soon as

the accuracy of the transmission has beendetermined.


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1.Parity check:

Add a bit to a bit string such that the total

number of 1-bits is even (or odd).

e.g :(even parity) 1011010 ) 10110100.

(odd parity) 1011010 ) 10110101


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Two-dimensional parity check

1 1 0 0 1 1 1 11 0 1 1 1 0 1 1

0 1 1 1 0 0 1 0

0 1 0 1 0 0 1 10 1 0 1 0 1 0 1

11001111 10111011 01110010 01010011

1100111 1011101 0111001 0101001


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Error detecting code

2.Polynomial code or CRC( Cyclic

Redundancy Check )


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CRC generator and checker

Zero, accept; nonzero, reject

Data CRC

Divisor

Remainder

Sender

Data.. 00...0

Divisor

CRC

Data CRC

Receiver


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Division

Modulo 2 division can be performed in amanner similar to arithmetic long division.

Subtract the denominator (the bottomnumber) from the leading parts of theenumerator (the top number).

Proceed along the enumerator until itsend is reached. Remember that we areusing modulo 2 subtraction.


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Binary division in a CRC generator

Data=100100=x^2+x^5

Divisor = 1101 =x^0+x^2+x^3

111101

Divisor 1101) 100100000 (data plus extra zeros)1101

1000

1101

1010

1101

11101101

0110

0000

1100

1101

0001 (remainder)


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A sequence of redundant bits, called the CRC

or the CRC remainder, is appended to the end

of a data unit So that the resulting data unit becomes exactly

divisible by a second, predetermined binary

number.

The redundancy bits used by CRC are derivedby dividing the data unit by a predetermined

divisor


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To be valid, a CRC must have two

qualities:

It must have exactly one less bit than thedivisor.

Append it to the end of the data string must

make the resulting bit sequence exactly

divisible by the divisor.


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Three steps:

A string of n 0s is appended to the data unit. Thenumber n is one less that the number of bits inthe predetermined divisor, which is n+1 bits.

The newly elongated data unit is divided by thedivisor, using a process called binary division.The remainder resulting from this division is theCRC.

The CRC of n bits derived in step 2 replaces t

heappended 0s at the end of the data unit. (CRC

may consist of all 0s).

CRC generator: uses modulo-2 division.


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CRC checker:

It functions exactly as the generator does.

After receiving the data appended with the

CRC, it does the same modulo-2 division. If the remainder is all 0s, the CRC is dropped

and the data are accepted

otherwise, the received stream of bits is

discarded and data are sent.


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Error-Detecting Codes


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Withn-bit code words, we have 2 n possible code

words consisting of 2 m data bits (where n = m + r).

This gives us the inequality:

(n + 1) v 2 m e 2 n

Because n = m + r, we can rewrite the inequality

as:

(m

+r

+ 1)v

2

m e

2

m + r

or (m

+r

+ 1)e

2

r

This inequality gives us a lower limit on the number

of check bits that we need in our code words.

2.8 ErrorDetection and Correction


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Suppose we have a set ofn-bit code words

consisting ofm data bits and r(redundant) parity

bits.

An error could occur in any of the n bits, so eac

hcode word can be associated withn erroneous

words at a Hamming distance of 1.

Therefore,we have n + 1 bit patterns for each

code word: one valid code word, and n erroneouswords.



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Suppose we have data words of lengthm = 8.

Then:

(8 + r+ 1) e 2 r

implies that r must be greater than or equal to 4. This means to build a code with 8-bit data words

that will correct single-bit errors, we must add 4

check bits, creating code words of length 12.

So how do we assign values to these checkbits?



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Checksum generator

Sender:

The unit is divided into k sections, each of n bits.All sections are added using 1s complement to get the

sum.

The sum is complemented and becomes the checksum.

The checksum is sent with the data


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Checksum checker:

Receiver:

The unit is divided into k sections, eachof n bits.

All sections are added using 1scomplement to get the sum.

The sum is complemented.If the result is zero, the data are

accepted; otherwise they are rejected.


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Checksum calculation-sender

Data=10101001 00111001

10101001+

00111001

11100010

Checksum=00011101

Data sent is 10101001 00111001 00011101


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Checksum calculation-Receiver

Data sent is 10101001 00111001 00011101

Data=10101001 00111001

10101001+

00111001

0001110111111111

Complement=00000000


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Error Correction

Error correction has two most common ways:

Error correction by retransmission: when theerror is discovered, the receiver can have the

sender retransmit the entire data unit.

Forward Error Correction (FEC): a receiver can

use an error-correcting code, which

automatically corrects certain errors.


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To calculate the number of redundancy

bits r required to correct a given number of

data bits m, we must find a relationshipbetween m & r. With m bits of data and r

bits of redundancy added to them, the

length

of th

e resulting code is m+r.


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If the total number of bits in a transmittable unit

is m+r, then r must be able to indicate at least

m+r+1 different states. Of these, one state means no error, & m+r

states indicate the location of an error in each of

the m+r positions.

So m+r+1 states must be discoverable by r bits.Therefore, 2r must be equal to or greater than

m+r+1: 2r>=m+r+1.


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Hamming Code

1

1

10 9 8 7 6 5 4 3 2 1

d d d r 8 d d d r 4 d r2 r1

Positions of redundancy bits in hamming code: 7bit ASCII

code with 4 redundancy bits

Redundancy bits are in the position 1, 2, 4 & 8.


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In the hamming code, each r bit is the

parity bit for one combination of data bits,as shown below:

r1 : bits 1,3,5,7,9,11

r2 : bits 2, 3, 6, 7, 10, 11

r4 : bits 4,5,6,7

r8 : bits 8, 9, 10, 11


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Redundancy bits calculation:

r1 will take care of these bits




11 9 7 5 3 1


r

1

r2dr4dddr8ddd

236710

11

7 6 5 4


11 1

0

9 8



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11 10 9 8 7 6 5 4 3 2 1

1 0 0 1 1 0 1Data: 1001101

1 0 0 1 1 0 1 1Adding r1

1 0 0 1 1 0 1 0 1 Adding r2

1 0 0 1 1 0 0 1 0 1Adding r4

1 0 0 1 1 1 0 0 1 0 1 Adding r8


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The use of error correcting code is often referred to asforward error correction.

Received Data: 1 0 0 1 0 1 0 0 1 0 1

Redundancy bits: 0 1 1 1

(the bit position 7 is in error)


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Using our code words of length 12, number each

bit position starting with 1 in the low-order bit.

Each bit position corresponding to an even

power of 2 will be occupied by a check bit.

These check bits contain the parity of each bit

position for which it participates in the sum.



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Since 2 (= 21) contributes to the digits, 2, 3, 6, 7, 10,

and 11. Position 2 will contain the parity for bits 3,

6, 7, 10, and 11.

When we use even parity, this is the modulo 2 sum

of the participating bit values.

For the bit values shown, we have a parity value of

0 in the second bit position.


What are the values for the other parity

bits?


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The completed code word is shown above. Bit 1checks the digits, 3, 5, 7, 9, and 11, so its

value is 1. Bit 2 checks digits 2, 3, 6, 7, 10, and 11. Bit 4

checks the digits, 5, 6, 7, and 12, so its value is 1. Bit 8 checks the digits, 9, 10, 11, and 12, so its

value is also 1.

Using the Hamming algorithm, we can not onlydetect single bit errors in this code word, but alsocorrect them!



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Suppose an error occurs in bit 5, as shown above.

Our parity bit values are: Bit 1 checks digits, 3, 5, 7, 9, and 11. Its value is 1,

butshould be zero.

Bit 2 checks digits 2, 3, 6, 7, 10, and 11. The zero is

correct.

Bit 4 checks digits, 5, 6, 7, and 12. Its value is 1, but

should be zero.

Bit 8 checks digits, 9, 10, 11, and 12. This bit is

correct.



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Hamming codes can correct burst errors:

Arrange k consecutive codewords in a

single matrix.

Transmit the data one column at a time(normally the data would be transmitted

one row at a time).


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CRC contd..

At the receiving end, receiver again divides the

polynomial corresponding to the received bits by

G(x) and accepts it iff the remainder is zero.

Now let E(x) denote the polynomial

corresponding to the errored bits. Then receiver

receives

T(x) = T(x) + E(x) G(x) divides T(x) iff it divides E(x)


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CRC

Detecting single bit errors

E(x) = x^i

Choose G(x) = any polynomial with at least

two terms


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Detecting 2 single bit errors

E(x) = x^i + x^j = x^i (x ^ (j-i) + 1)

Choose G(x) s.t it neither divides x nor

divides x^k + 1 for any k


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Detecting odd number of single bit errors

E(x) cant be of the form (x + 1) Q(x)

Choose G(x) of the type (x + 1) Q(x)


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G(x) = a general polyn of degree r

Will detect single burst of length


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Detecting single burst of length k


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Hamming Codes : for ED n EC

m data bits together with r error check bits forman n = (m + r) bit codeword

The number of bits two codewords differ in iscalled the hamming distance between the twocodewords

Significance : If two codewords are at HD d thenit requires d single bit errors to convert one intothe other


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HD of a coding scheme

For m bit data .. All the 2^m possiblecombinations are legal

But not all the 2^n codewords are used

-- in a coding scheme (algorithm to compute thecheck bits) some of these codewords are legaland others are illegal

For eq .. Consider parity : 1(r = 1) parity bit isappended with value so that the total number of1s in the codeword is even ..

Th

en 11011 is a legal codeword in th

is sch

eme but


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HD of a list of legal codewords

Minimum HD between any pair of legal

codewords in the list

Remember : Each algorithm to compute the

check bits create a different list of legalcodewords


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Use of HD for error detection

To detect d single bit errors , we need (an

algorithm that creates) a code list with HD

at least d + 1

For eg . For the parity scheme .. HD is 2

..hence it can be used to detect single biterrors (d=1)


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Continued

If the recvd codeword is legal .. We accept it ,

And if it is illegal we report (detect) an error

Q1 : Can ithappen t

hat we recv a legalcodeword when d single bit errors have ocurred

this is eqwt to saying can we get a legal codefrom another legal code by d single bit errors?

A1 : No, since the HD of the code is at least d +1. So a legal CW can be genearted from anotherLCW by inerting at least d + 1 bits and not byinverting d or less bits.


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Continued

Q2 : Can we get an illegal CW when no

errorhas occurred ?

A2 : Obviously not ..since the legal CWwas sent by the sender and if no errorhas

occurred then the recver must recv a legal

CW


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Use of HD for error correction

To correct d single bit errors , we need (analgorithm that creates) a code list with HD atleast 2d + 1.

For eg. Consider the following legal CWs:

0000000000,0000011111,1111100000,1111111111

HD is 5 .. It can be used to correct 2 single biterrors


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Continued..

Claim : Suppose we recv an illegal code C .. Then

there is a unique legal code which is at a

distance d or less from C

Proof : Suppose there are 2 codes C1 and C2 at

distance d (or less) from C .. Then C1 can be

obtained from C2 by 2d (or less) inversions .. Acontradiction to (code has HD at least 2d + 1)


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Hamming Code to correct one bit


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Hamming Code to correct one bit

errors

The bits of the CW are numbered left to

right , starting from 1 the bits that are

powers of 2 are check bits (1,2,4,8 ) andthe remaining are data bits.

Expand the position of each data bit in

powers of 2 ..for eg. 11 = 1 + 2 + 8 .. So11th bit contributes to the computation of

value of these check bits I.e. 1,2, 8


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Continued

We do this for each data bit ..

The value of a check bit is computed so

that the parity of the all the data bits thatcontribute to it together with the check bititself is even.

For eg .

data bits 1001000 will be sent as thecodeword 00110010000

Hamming Code to correct burst


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Hamming Code to correct burst

errors


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IEEE 802 LANs use

For eg.

X^32 + x^26 + x^2

3+ x^22 ..x^2 + x + 1

Detects single burst of length


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I Acknowledge

Help from the following site

http://www.cs.vu.nl/~ast/

In preparing this lecture.

data link layer-megh

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