Chapter 12 Multiple Access

Figure 12.1 Data link layer divided into two functionality-oriented sublayers

Figure 12.2 Taxonomy of multiple-access protocols

12-1 RANDOM ACCESS12-1 RANDOM ACCESS

In In random accessrandom access or or contentioncontention methods, no station is superior to methods, no station is superior to another station and none is assigned the control over another. No another station and none is assigned the control over another. No station permits, or does not permit, another station to send. At each station permits, or does not permit, another station to send. At each instance, a station that has data to send uses a procedure defined by instance, a station that has data to send uses a procedure defined by the protocol to make a decision on whether or not to send. the protocol to make a decision on whether or not to send. Figure 12.3 Frames in a pure ALOHA network

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

Figure 12.4 Procedure for pure ALOHA protocol

CSMA (Carrier Sense Multiple Access)

CSMA: listen before transmit: If channel sensed idle: transmit entire frame If channel sensed busy, defer transmission Human analogy: don’t interrupt others!

CSMA collisions collisions can still occur:

propagation delay means two nodes may not hear each other’s transmission

collision: entire packet transmission time wasted

•Sense before transmit•The possibility of collision still exist because of the propagation delay

Figure 12.8 Space/time model of the collision in CSMA

Figure 12.9 Vulnerable time in CSMA

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

Figure 12.10 Behavior of three persistence methods

Figure 12.11 Flow diagram for three persistence methods

Figure 12.15 Energy level during transmission, idleness, or collision

Figure 12.12 Collision of the first bit in CSMA/CD

Figure 12.13 Collision and abortion in CSMA/CD

Figure 12.14 Flow diagram for the CSMA/CD

Figure 12.16 Timing in CSMA/CA

In CSMA/CA, the IFS can also be used to define the priority of a station or a frame.

In CSMA/CA, if the station finds the channel busy, it does not restart the timer of the contention window;

it stops the timer and restarts it when the channel becomes idle.

Figure 12.17 Flow diagram for CSMA/CA

12-2 CONTROLLED ACCESS12-2 CONTROLLED ACCESS

In In controlled accesscontrolled access, the stations consult one another to find , the stations consult one another to find which station has the right to send. A station cannot send which station has the right to send. A station cannot send unless it has been authorized by other stations. unless it has been authorized by other stations.

ReservationPollingToken Passing

Topics discussed in this section:Topics discussed in this section:

Figure 12.18 Reservation access method

Figure 12.19 Select and poll functions in polling access method

Token passing:control token passed from one node to next sequentially.concerns:•token overhead •single point of failure (token)

Polling: master node “invites” slave nodes to transmit in turnconcerns:

•polling overhead •single point of failure (master)

Figure 12.20 Logical ring and physical topology in token-passing access method

12-3 CHANNELIZATION12-3 CHANNELIZATION

ChannelizationChannelization is a multiple-access method in which the is a multiple-access method in which the available bandwidth of a link is shared in time, frequency, or available bandwidth of a link is shared in time, frequency, or

through code, between different stations.through code, between different stations.

Frequency-Division Multiple Access (FDMA)Time-Division Multiple Access (TDMA)Code-Division Multiple Access (CDMA)

Topics discussed in this section:Topics discussed in this section:

In FDMA, the available bandwidth of the common channel is divided into bands that are separated by

guard bands.

Figure 12.21 Frequency-division multiple access (FDMA)

Figure 12.22 Time-division multiple access (TDMA)

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

Figure 12.23 Simple idea of communication with code

In CDMA, one channel carries all transmissions simultaneously.

unique “code” assigned to each user; i.e., code set partitioning used mostly in wireless broadcast channels (cellular, satellite, etc) all users share same frequency, but each user has own “chipping”

sequence (i.e., code) to encode data encoded signal = (original data) X (chipping sequence) decoding: inner-product of encoded signal and chipping sequence allows multiple users to “coexist” and transmit simultaneously

with minimal interference (if codes are “orthogonal”)

•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

Figure 12.24 Chip sequences

Figure 12.25 Data representation in CDMA

Figure 12.26 Sharing channel in CDMA

There is only one sequence flowing through the

channel ,the sum of the sequences But each

receiver can detect its data from the sum

In the following example Four stations sharing the

link during the 1-bit interval is shown below

Figure 12.27 Digital signal created by four stations in CDMA

Figure 12.28 Decoding of the composite signal for one in CDMA

The number of sequences in a Walsh table needs to be N = 2m.

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

Find the chips for a network witha. Two stations b. Four stations

Example 12.6

SolutionWe can use the rows of W2 and W4 in Figure 12.29:a. For a two-station network, we have [+1 +1] and [+1 −1].

b. For a four-station network we have [+1 +1 +1 +1], [+1 −1 +1 −1], [+1 +1 −1 −1], and [+1 −1 −1 +1].

Example 12.7What is the number of sequences if we have 90 stations in our network?

SolutionThe number of sequences needs to be 2m. We need to choose m = 7 and N = 27 or 128.We can then use 90 of the sequences as the chips.

Prove that a receiving station can get the data sent by a specific sender if it multiplies the entire data on the channel by the sender’s chip code and then divides it by the number of stations.

Example 12.8

SolutionLet us prove this for the first station, using our previous four-station example. We can say that the data on the channel D = (d1 c⋅ 1 + d2 c⋅ 2 + d3 c⋅ 3 + d4 c⋅ 4). The receiver which wants to get the data sent by station 1 multiplies these data by c1.

When we divide the result by N, we get d1 .

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