Cmda At The Cross Road

Multiple Access Techniques Guest Lecture Muhammad Adnan M.Sc Computer Engineering

Why Mobile Communication? Question: Why we need a new technology when we have such a developed public telephone network?

Answer:

MOBILITY

Challenges of Mobility Challenges of using a radio channel:  The use of radio channels necessitates methods of sharing them – channel access. (FDMA, TDMA, CDMA)  The wireless channel–poses a more challenging problem than with wires.  Bandwidth: it is possible to add wires but not bandwidth. So it is important to develop technologies that provide for spectrum reuse.  Privacy and security-a more difficult issue than with wired phone. Others: low energy (battery), hand off, roaming, etc.

Multiple Access Overview-A party! 

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Consider a number of students at a party. The goal of the students is to have intelligible conversation. The house at which the party is being held is the resource available. FDMA: Each pair of students has a separate room to talk TDMA : Everyone is in the same room and each pair has a limited time slot to converse CDMA : Everyone is in the same room, talking at the same time, but each pair talks in a different language.

View of the CDMA Concept

FDMA (Frequency Division Multiple Access) 

In FDMA, the available bandwidth is divided into channels and each channel is allocated to a user. The allocated channel is then in the user’s permanent use. It is used in cellular telephone and satellite networks.

Problems in FDMA?  

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Channel is wasted if there is no user As channel division is fixed if new user come we have to divide the whole channel again. Problem of frequency guard (guard band)

TDMA 

In TDMA, the entire bandwidth is just one channel. The stations share the capacity of channel and time. Each station is allocated a time slot during which it can send data. TDMA is used in cellular telephone network.

Problems in TDMA?  

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Time slot is wasted if there is no user As channel division (in terms of time) is fixed if new user come we have to divide the whole channel again. It remove the Problem of frequency guard (guard band)

CMDA (Code Division Multiple Access) 

In CDMA one channel carries all transmissions simultaneously. CDMA employs analog-to-digital conversion (ADC) in combination with Spread Spectrum technology. It is used in ultra-high-frequency (UHF) cellular telephone systems in the 800-MHz and 1.9-GHz bands.

History 

CDMA is a military technology first used during World War II by the English allies to foil German attempts at jamming transmissions. The allies decided to transmit over several frequencies, instead of one, making it difficult for the Germans to pick up the complete signal.

HOW CDMA DIFFERS FROM FDMA AND TDMA 

CDMA differs from FDMA because only one channel occupies the entire bandwidth of the link. It differs from TDMA because all stations can send data simultaneously i.e. there is no time sharing.

MAC ALGORITHM Suppose we have four stations, each has a sequence of chips which we designate as A, B, C and D. Chip Sequences:A= +1,+1,+1,+1 B= +1,-1, +1,-1 C= +1,-1,-1,-1 D= +1,-1,-1,+1

ENCODING RULES • • •

If a station needs to send a 0 bit, it sends a -1. If a station needs to send a 1 bit, it sends a +1. When a station is idle it sends no signal which is represented by 0. Data bit 0 → -1 Data bit 1 → +1 Silence → 0

AN EXAMPLE Consider the four stations that share the link during 1 bit interval.   

stations 1 and 2 are sending a 0 bit channel 4 is sending a 1 bit Station 3 is silent

SENDER

STEPS FOLLOWED 

The multiplexer receives 1 encoded number from each station (-1, -1, 0 and +1).

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The encoded number sent by station 1 is multiplied by each chip in sequence A. A new sequence is the result (-1,-1,-1,-1).

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Likewise, the encoded number sent by station 2 is multiplied by each chip in sequence B. The same result is true for the remaining two encoded numbers. The result is four new sequences.

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All first chips are added, as are all second, third and fourth chips. The result is one new sequence.

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The sequence is transmitted through the link.

RECEIVER

STEPS FOLLOWED 

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The demultiplexer receives the sequence sent across the link. It multiplies the sequence by the code for each receiver. The multiplication is done chip by chip. The chips in each sequence are added. The result is always +4,-4 or 0. The result of step 3 is divided by 4 to get +1, -1 or 0. The number in step 4 is decoded to 0, 1 or silence by the receiver.

OBSERVATION 

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Each station receives what is sent by the sender. Third receiver does not receive data because the sender was idle. There is only one sequence flowing through the channel the sum of the sequences. each receiver detects its own data from the sum.

ORTHOGONAL SEQUENCES 

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The sequences in our example are called Orthogonal Sequences. The sequence were not chosen randomly; they were carefully selected.

SEQUENCE GENERATION 

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To generate sequences, Walsh Table is used, which is a two-dimensional table with an equal number of rows and columns Each row is a sequence of chips. The Walsh Table W1 for a one-chip sequence has one row and one column. We can choose -1 or +1 for the chip for this trivial table (here, we choose +1). According to Walsh, if we know the table for n sequences Wn, we can create a table for sequences W2n, as shown in the following figure. The Wn with the overhead bar stands for the complement of the Wn, where each +1 is changed to -1 and vice versa.

CREATING W2 &W4 FROM W1 

The following figure shows the process.

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After we select W1, W2 can be made from for W1’s with the last one the complement of W1. After W2 is generated, W4 can be made of four W2,s with the last one the complement of W2.

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Similarly we can create W8, composed of four w4’s.

PROPERTIES OF ORTHOGONAL SEQUENCES 

If we multiply a sequence by -1, every element in the sequence is complemented (+1 becomes -1 and -1 becomes +1).

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If we multiply two sequences, element by element, and add the results, we get a number called the Inner Product. If the two sequences are the same, we get N, where N is the number of sequences. If they are different, we get 0. The inner product uses a dot as the operator. So A.A is N but A.B is 0.

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The inner product of a sequence by its complement is – N, So A.(-A) is –N.

PROVING 2nd PROPERTY Consider the following: A = [+1,+1,-1,-1] A.A = [+1,+1,+1,+1] . [+1,+1,+1,+1] = 1+1+1+1 =4 So it is proved that A.A is always equal to N.

Now,

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A = [+1,+1,+1,+1] B = [+1,-1,+1,-1] A.B = [+1,+1,+1,+1] . [+1,-1,+1,-

= 1-1+1-1 =0 So it is proved that A.B is always equal to 0.

PROVING 3rd PROPERTY Consider the following A = [+1,+1,+1,+1] A.(-A) = [+1,+1,+1,+1] . [-1,-1,1,-1] = -1-1-1-1 = -4 So the third property is also proved.

Spread Spectrum Access  

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Two techniques Frequency Hopped Multiple Access (FHMA) Direct Sequence Multiple Access (DSMA)

Frequency Hopped Multiple Access (FHMA  

Digital multiple access technique

A wideband radio channel is used.

Same wideband spectrum is used by all users 

Carrier freq of users are varied in a pseudo-random fashion.

Each user uses a narrowband channel at a specific instance of time. The random change in frequency ensures small probability of using the same narrowband channel 

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The sender receiver change frequency (through hopping) using similar PN sequence, hence they are synchronized.

Rate of hopping versus Symbol rate

If hopping rate is greater: Called Fast Frequency Hopping One bit transmitted in multiple hops. If symbol rate is greater: Called Slow Frequency Hopping Multiple bits are transmitted in a hopping period

Concept of Frequency Hopping Spread Spectrum

An Example of Frequency Hopping Pattern

Case Study-Bluetooth

Case Study: Bluetooth –Pico net and FHSS

Code Division Multiple Access (CDMA)

Concept of Direct Sequence Spread Spectrum

Some Simulation Results

Simulation Results RAKE vs MMSE Equalizer Diff. Spreading Factors (No of users = 1) RAKE SF = 4 MMSE SF = 4 RAKE SF = 8 MMSE SF = 8 RAKE SF = 32 MMSE SF = 32

Teb

Eb/N0

RAKE vs MMSE Equalizer Different Number of users (Spreading Factor = 32) RAKE 16 Users MMSE 16 Users RAKE 8 Users MMSE 8 Users RAKE 4 Users MMSE 4 Users

Teb

Eb/N0

Conclusions • A CDMA receiver performs better when used with an MMSE equalizer in place of a RAKE receiver for different Spreading Factors • Similarly, the MMSE equalized CDMA receiver performs better when the network is occupied i.e. there are more number of users

References   

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Wireless Communication, By Rappaport Digital communication by B. Sklar The MAI performance of orthogonal codes for channel cover in Asynchronous CDMA system by, Miftadi Sudjai and Mosa N.A. Abu Rgheff, CDMA Research Group, Department of Electronics and Communication Engineering, University of Plymouth, UK Simple MMSE equilizer for CDMA downlink, to restore chip sequence: comparison to zero forcing and RAKE, by T.P Kranussl, Michael D.Z. and Gerrt Leus. SEE, Purdue University, Deptt of Electrical Engg, K.U.Leuven. And a lot of websites…..

Questions….?? ?