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Chapter 13

MultipleAccess

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Figure 13.1 Multiple-access protocols

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13.1 Random Access13.1 Random Access

MA CSMA

CSMA/CD

CSMA/CA

Each Station has the right to the medium without being controlled by any other station

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Figure 13.3 ALOHA network

The earliest random-access method ,developed in 1970sDesigned to be used on wireless Local area Network at 9600bps

Based on Following rules

•Multiple Access

•Any station can send a frame when it has to send a frame

•Acknowledgment

•After sending a frame the station waits for an acknowledgment .If it does not receive the Ack within 2times maximum propagation delay ,It tries sending the frame again after a random amount of time

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Figure 13.4 Procedure for ALOHA protocol

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Figure 13.5 Collision in CSMA

Sense before transmitThe possibility of collision still exist because of the propagation delay

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Figure 13.6 Persistence strategies

It defines the procedure for a station that senses a busy medium

•Station senses the line ,If the line is Idle station sends the frame immediately

•If the line is not Idle station waits for a random period of time and then senses the line again

This method has two variationsI-PERSISTANTStation senses the line ,If the line is Idle station sends the frame immediately(with a probability of 1) ,this method increases the chance of collisionp-PERSISTANTIf the line is Idle station sends the frame with probability p ,and refrain from sending with a probability 1-pStation runs a Random number between 1-100

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13.7 CSMA/CD procedure

In the exponential back of method the station waits an amount of time between 0 and 2N x Maximum propagation time N is the number of attempted transmission

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Figure 13.8 CSMA/CA procedure

IFG Time =Inter frame Gap

Used in Wireless LAN

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13.2 Control Access13.2 Control Access

Reservation

Polling

Token Passing

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Figure 13.9 Reservation access method

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Figure 13.10 Select

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Figure 13.11 Poll

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Figure 13.12 Token-passing network

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Figure 13.13 Token-passing procedure

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13.3 Channelization13.3 Channelization

FDMA

TDMA

CDMA

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In FDMA, the bandwidth is divided into channels.

NoteNote::

In TDMA, the bandwidth is just one channel that is timeshared.

In CDMA, one channel carries all transmissions simultaneously.

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Figure 13.14 Chip sequences

•Each station is assigned a code ,which is a sequence of numbers called CHIPS

•Chip period is always less than a bit period

•Chip rate is always an integral multiple of bit rate

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Figure 13.15 Encoding rules

•If a station needs to send 0 bit ,it sends a –1

•If a station needs to send 1 bit ,it sends a +1

•If a station needs to send no signal ,it sends a 0

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Figure 13.16 CDMA multiplexer

In the following example Four stations sharing the link during the 1-bit interval is shown below

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Figure 13.17 CDMA demultiplexer

There is only one sequence flowing through the channel ,the sum of the sequences But each receiver can detect its data from the sum

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ORTHOGONAL SEQUENCES The sequences are not chosen randomly

They should be orthogonal to each other Orthogonal sequences satisfies following

properties If we multiply a sequence by –1,every

element in the sequence is complemented If we multiply two sequences element by

element and add the results , we get a number called inner product If two sequences are same we get N (N=Number of sequences) ,if they are different we get 0 (A.A=N A.B=0)

A.(-A) = -N For the generation of the

sequences ,WALSH TABLE method is used

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WALSH TABLE It is a two dimensional table with

equal number of rows and columns Each row is a sequence of chip For W1 ( one Row one Column Walsh

Table) ,we can choose –1 or +1 If we know the table for N sequences

WN ,we can create the table for 2N sequences W2N

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Figure 13.18 W1 and W2N

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Figure 13.19 Sequence generation

Now each row can be used for each station as its chip code

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Example 1Example 1

Check to see if the second property about orthogonal codes holds for our CDMA example.

SolutionSolution

The inner product of each code by itself is N. This is shown for code C; you can prove for yourself that it holds true for the other codes.

C . C =

If two sequences are different, the inner product is 0.

B . C =

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Example 2Example 2

Check to see if the third property about orthogonal codes holds for our CDMA example.

SolutionSolution

The inner product of each code by its complement is N. This is shown for code C; you can prove for yourself that it holds true for the other codes. C . (C ) =

The inner product of a code with the complement of another code is 0.

B . (C ) =

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Orthogonal Codes of CDMA In Our previous example At the Multiplexer the sequence S that comes out is

S=-A-B+D Station 1 is sending –A (-1 was multiplied by A) Station 2 is sending –B Station 3 is sending an empty sequence Station 4 is sending D

At the De multiplexer the sequence S is received Station 1 finds the inner product of S and A

S.A=(-A-B+D).A=-4+0+0=-4 The result is divided by 4 (–4/ 4 = -1) , And –1 represents bit 0

Same is true for stations 2 ,3 and 4