COMMUNICATION SYSTEM EECB353Chapter 7 Part IIIMULTIPLE ACCESS

Intan Shafinaz MustafaDept of Electrical Engineering

Universiti Tenaga Nasionalhttp://metalab.uniten.edu.my/~shafinaz

MULTIPLE ACCESS TECHNIQUE

For multiple users to be able to share a common resource in a managed and effective way requires some form of access protocol that defines how or when the sharing is to take place and the means for identifying individual messages.

Process is known as multiplexing in wired networks and multiple access in wireless digital communications.

2

MULTIPLE ACCESS TECHNIQUE Three classes of multiple access techniques will

be considered: techniques where individual users are identified by

assigning different frequency slots (FDMA) techniques where individual users are given different

time slots (TDMA) and techniques where individual users are given the same

time and frequency slots but are identified by a different code (CDMA).

3

FDMA “Some of the bandwidth all of the time”.

Simplest and oldest method Bandwidth is divided into N non-overlapping

frequency bands Guard bands minimize interference between channels Each station is assigned a different frequency No. of channel

Can be inefficient if more than N station want to transmit (resulting in delays) or traffic is bursty (resulting in unused bandwidth and delays)

Lower channel bit rate than TDMA means less susceptible to multipath ISI (Intersymbol Interference)

4

c

guardt

B

BBN

2

FDMA

5

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

6

Example 1 : If a US AMPS cellular operator is allocated 12.5 MHz for each simplex band, and if Bt is 12.5 MHz, Bguard is 10 kHz, and Bc is 30 kHz, find the number of channels available in an FDMA system. (Ans: 416)

Example 2 : Determine the number of channel available in an FDMA system if the total spectrum allocation is 15 MHz, guard band is 10 kHz and the channel bandwidth is 25 kHz. (Ans: 599)

TDMA “All of the bandwidth some of the time” Users share same frequency band in non-

overlapping time intervals. Guard time can be as small as the

synchronization of the network permits All users must be synchronized with base station

to within a fraction of guard time Guard time of 30-50 microsec common in TDMA

No. of Channel:

where m = max number of TDMA user

7

c

guardtot

B

BBmN

)2(

TDMA

8

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

CDMA CDMA is a mulplexing technique used with SS

Narrowband message signal is multiplied by very large bandwidth spreading signal (pseudo-noise)

All users can use same carrier frequency and may transmit simultaneously

Each user has own pseudorandom codeword which is approximately orthogonal to other codewords

Receiver performs time correlation operation to detect only specific codeword, other codewords appear as noise due to decorrelation

9

Advantages No timing coordination unlike

TDMA CDMA uses spread spectrum,

resistant to interference (multipath fading) CDMA can provide more users per cell

No hard limit on number of users

Disadvantages Implementation complexity

of spread spectrum Power control is essential

for practical operation

CDMA

10

In Code Division Multiple Access (CDMA), one channel carries all transmissions simultaneously.

CDMA - Analogy

CDMA simply means communication with different codes.

For example, in a large room with many people, two people can talk in English if nobody else understands English.

Another two people can talk in Chinese if they are the only ones who understand Chinese, and so on.

In other words, the common channel (space of the room), can easily allow communication between several couples, but in different languages (codes)

11

CDMA - Idea

Assume we have 4 different stations 1,2,3 and 4 connected to the same channel.

The data from station 1 are d1, from station 2 are d2 and so on.

The code assigned to station 1 is c1, to station 2 is c2 and so on. Assume that the assigned codes have two properties:

1. If we multiply each code by another, we get 0.2. If we multiply each code by itself, we get 4 (number of stations)

The data sent is:d1.c1 + d2.c2 + d3.c3 + d4.c4

12

CDMA

13

Suppose station 1 and 2 are talking to each other. Station 2 wants to hear what station 1 is saying. It multiplies the data on the channel by c1 (the code

from station 1) Since (c1.c1) is 4, but (c2.c1), (c3.c1) and (c4.c1) are all

0s, station 2 divides the result by 4 to get the data from station 1.

Data = (d1.c1 + d2.c2 + d3.c3 + d4.c4). c1

= d1.c1.c1 + d2.c2 .c1+ d3.c3 .c1+ d4.c4.c1 = 4.d1

CDMA – Chip Sequences

The sequence were not chosen randomly, they were carefully selected. They are called orthogonal sequence and have the following properties:1. C2 * C2 = [ +1 -1 +1 -1] * [ +1 -1 +1 -1] = 4 (inner product of two equal

sequence equal to number of element in each sequence)2. C2 * C4 = [+1 -1 +1 -1] * [+1 -1 -1 +1] = 0 (inner product of two different

sequence equal to 0)3. C2+C4 = [+1 -1 +1 -1] + [+1 -1 -1 +1] = [+2 -2 0 0] (adding two sequence

will result another sequence)4. A * C4 = [+A –A –A +A] (scalar multiplication)

14

CDMA – Data Representation

15

Silence == idle

Example – Four stations share the link during a 1-bit interval.

Station 1 and 2 are sending a 0 bit and station 4 is sending 1 bit while station 3 is silent.

The data at the sender site are translated to be [ -1, -1, 0, +1].

Note:1. Each station multiplies the correspondingg number

by its chip (its orthogonal sequence), which is unique for each station.

2. The sequence on the channel is the sum of all four sequences.

CDMA – Sharing Channel

16

CDMA – Digital signal created by four stations

17

CDMA - Decoding of the composite signal

18

CDMA - Summary

CDMA (Code Division Multiple Access) unique “code” assigned to each user; ie, code set partitioning used mostly in wireless broadcast channels (cellular, satellite,etc) all users share same frequency, but each user has own

“chipping” sequence (ie, 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”)

CDMA Encode/Decode

CDMA: two-sender interference

sender 1

sender 2

uses sender 1 codeto receive sender 1 data

CDMA - Example Given a binary chip sequence and a bipolar chip sequence

for 4 stations as below :A : 00011011 A : (-1 –1 –1 +1 +1 –1 +1 +1 )B : 00101110 B : ( -1 –1 +1 –1 +1 +1 +1 –1 )C : 01011100 C : ( -1 +1 –1 +1 +1 +1 –1 –1 )D : 01000010 D : ( -1 +1 –1 –1 –1 –1 +1 –1 )

i. What is the resulting chip sequence if :a. Only C transmit 1b. B and C transmit 1c. A, B and D transmit 1, C transmit 0

ii. Prove that these 4 stations sequences are Orthogonal Sequences

22

CDMA – FE sem 1 09/10

Station A and station B with bipolar chip sequence of 00011011 and 00101110 respectively, are transmitting signals simultaneously. a) What is the resulting bipolar chip sequence if station A transmits 1 but station B silence?b) What is the resulting bipolar chip sequence if both stations transmit 0?c) If station C with bipolar chip sequence 10110001 appears in the communication link, what would happen to station A and B. Determine whether there is any interference experienced for both stations. Justify your answer with calculations.d) If station C wants to communicate with station B, suggest a new sequence station C should use.

23

References:

Data Communications and Networking, 4th Edition, Behrouz A. Forouzan

Wireless Communications Principles and Practice, 2nd Edition, Thedore S Rappaport

24