It is a channelization protocol that allows the total usable bandwidth in a shared channel to be shared across multiple stations based on their time, distance and codes. It can access all the stations at the same time to send the data frames to the channel.Following are the various methods to access the channel based on their time, distance and codes:
FDMA (Frequency Division Multiple Access)
TDMA (Time Division Multiple Access)
CDMA (Code Division Multiple Access)
Frequency Division Multiple Access (FDMA): FDMA is a type of channelization protocol. This bandwidth is divided into various frequency bands. Each station is allocated a band to send data and that band is reserved for the particular station for all the time which is as follows.
The frequency bands of different stations are separated by small bands of unused frequency and unused frequency bands are called as guard bands that prevent the interference of stations.
It is like the access method in the data link layer in which the data link layer at each station tells its physical layer to make a bandpass signal from the data passed to it. The signal is created in the allocated band and there is no physical multiplexer at the physical layer.
Time Division Multiple Access (TDMA) : TDMA is the channelization protocol in which bandwidth of channel is divided into various stations on the time basis. There is a time slot given to each station, the station can transmit data during that time slot only which is as follows.
Each station must aware of its beginning of time slot and the location of the time slot.
TDMA requires synchronization between different stations.
It is type of access method in the data link layer. At each station data link layer tells the station to use the allocated time slot.
Code Division Multiple Access (CDMA) : In CDMA, all the stations can transmit data simultaneously. It allows each station to transmit data over the entire frequency all the time. Multiple simultaneous transmissions are separated by unique code sequence. Each user is assigned with a unique code sequence.
In the below figure, there are 4 stations marked as 1, 2, 3 and 4. Data assigned with respective stations as d1, d2, d3 and d4 and the code assigned with respective stations as c1, c2, c3 and c4.
Special property of the assigned code in CDMA is :
1.Multiply each code by another we get 0
2.Multiply each code by itself we get 4
For Example: Suppose station 2 wants to receive the data from station 1 then the data = (d1.c1+d2.c2 + d3.c3 +d4.c4) . c1
= d1.c1.c1+d2.c2.c1 + d3.c3.c1 +d4.c4.c1
= d1.4+d2.0 + d3.0 +d4.0
= d1.4
therefore the data = d1.4/total no of station
= d1.4/4
= d1
Chip Sequence: CDMA is based on coding theory, sequence of numbers is called as code and they are called as chips, so they are called chip sequence
c1 = [ +1+1+1 +1 ]
c2 = [ +1 -1 +1 -1 ]
c3 = [ +1 +1 -1 -1 ]
c4 = [ +1 -1 -1 +1 ]
Properties of Chip Sequence:
1. Each sequence is made up of N elements, where N is Number of station
2.Multiply a sequence by a number i.e 2 . c3
2 . [ +1 +1 -1 -1 ]
[ +2 +2 -2 -2 ]
3.Multiply two equal sequences i.e c3 .c3
[ +1 +1 -1 -1 ] . [ +1 +1 -1 -1 ]
[ +1 +1 +1 +1 ]
[ 4 ]
4.Multiply two different sequences i.e c3 .c1
[ +1 +1 -1 -1 ] . [ +1 +1 +1 +1 ]
[ +1 +1 -1 -1 ]
[ 0 ]
5.Adding two sequences i.e c3 + c1
[ +1 +1 -1 -1 ] + [ +1 +1 +1 +1 ]
[ +2 +2 0 0 ]
Data Representation in CDMA:
Data bit 0 -> -1
Data bit 1 -> +1
Data bit Silent -> 0
consider the data bit across station 1 is 0 = -1
i.e d1 . c1
-1 . [ +1+1+1+1]
-1 -1 -1 -1
consider the data bit across station 2 is 0 = -1
i.e d2 . c2
-1 . [ +1 -1 +1 -1 ]
-1 +1 -1 +1
consider the data bit across station 3 is Silent = 0
i.e d3 . c3
0 . [ +1 +1 -1 -1 ]
0 0 0 0
consider the data bit across station 4 is 1 = +1
i.e d4 . c4
+1 . [ +1 -1 -1 +1 ]
+1 -1 -1 +1
Sum of all four sequences is
[ -1 -1 -1 -1 ] + [ -1 +1 -1 +1 ] + [ 0 0 0 0 ] + [ +1 -1 -1 +1 ]
[ (-1-1+0+1) (-1+1+0-1) (-1-1+0-1) (-1+1+0+1) ]
[ (-1) (-1) (-3) (+1) ]
For Example: Suppose if station 3 wants to receive the data from station 2 then the data = (d1.c1+d2.c2 + d3.c3 +d4.c4) . c2
= [ (-1) (-1) (-3) (+1) ] . [ +1 -1 +1 -1 ]
= [ -1+1-3-1 ]
= [ -4 ]
therefore the data = -4/4
= -1
= 0
Hence the bit 0 is received from the station 2
For Example: Suppose if station 1 wants to receive the data from station 4 then the data = (d1.c1+d2.c2 + d3.c3 +d4.c4) . c4
= [ (-1) (-1) (-3) (+1) ] . [ +1 -1 -1 +1 ]
= [ -1+1+3+1 ]
= [ +4 ]
therefore the data = +4/4
= +1
= 1
Hence the bit 1 is received from the station 4
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