COMMUNICATION SYSTEM EECB353 Chapter 7 Part III MULTIPLE ACCESS Intan Shafinaz Mustafa Dept of Electrical Engineering Universiti Tenaga Nasional

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

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 B t is 12.5 MHz, B guard is 10 kHz, and B c 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 microsec common in TDMA No. of Channel: where m = max number of TDMA user 7

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 d 1, from station 2 are d 2 and so on.  The code assigned to station 1 is c 1, to station 2 is c 2 and so on.  Assume that the assigned codes have two properties: 1. If we multiply each code by another, we get If we multiply each code by itself, we get 4 (number of stations)  The data sent is: d 1.c 1 + d 2.c 2 + d 3.c 3 + d 4.c 4 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 c 1 (the code from station 1)  Since (c 1.c 1 ) is 4, but (c 2.c 1 ), (c 3.c 1 ) and (c 4.c 1 ) are all 0s, station 2 divides the result by 4 to get the data from station 1. Data = (d 1.c 1 + d 2.c 2 + d 3.c 3 + d 4.c 4 ). c 1 = d 1.c 1.c 1 + d 2.c 2.c 1 + d 3.c 3.c 1 + d 4.c 4.c 1 = 4.d 1

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 = [ ] * [ ] = 4 (inner product of two equal sequence equal to number of element in each sequence) 2. C2 * C4 = [ ] * [ ] = 0 (inner product of two different sequence equal to 0) 3. C2+C4 = [ ] + [ ] = [ ] (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 code to receive sender 1 data

CDMA - Example Given a binary chip sequence and a bipolar chip sequence for 4 stations as below : A : A : (-1 –1 – – ) B : B : ( -1 –1 +1 – –1 ) C : C : ( – –1 –1 ) D : D : ( –1 –1 –1 –1 +1 –1 ) i. What is the resulting chip sequence if : a. Only C transmit 1 b. B and C transmit 1 c. 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 and 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 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, 4 th Edition, Behrouz A. Forouzan Wireless Communications Principles and Practice, 2 nd Edition, Thedore S Rappaport 24