cmda at the cross road

8/14/2019 CMDA at the Cross Road

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Multiple Access Techniques

Guest Lecture

Muhammad Adnan

M.Sc Computer Engineering


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Why Mobile Communication?

Question:Why we need a new technology when we havesuch a developed public telephone network?

Answer:

MOBILITY


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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 channelposes a more challenging

problem than with wires. Bandwidth: it is possible to add wires but not

bandwidth. So it is important to develop technologiesthat provide for spectrum reuse.

Privacy and security-a more difficult issue than withwired phone. Others: low energy (battery), hand off,roaming, etc.


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Multiple Access Overview-A party!

Consider a number of students at a party. Thegoal of the students is to have intelligibleconversation. The house at which the party isbeing held is the resource available.

FDMA: Each pair of students has a separateroom to talk

TDMA : Everyone is in the same room and eachpair has a limited time slot to converse

CDMA : Everyone is in the same room, talkingat the same time, but each pair talks in adifferent language.


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View of the CDMA Concept


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FDMA (Frequency Division MultipleAccess)

In FDMA, the availablebandwidth is dividedinto channels and eachchannel is allocated to

a user. The allocatedchannel is then in theusers permanent use.It is used in cellulartelephone and satellitenetworks.


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Problems in FDMA?

Channel is wasted if there is no user

As channel division is fixed if new

user come we have to divide thewhole channel again.

Problem of frequency guard (guardband)


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TDMA

In TDMA, the entirebandwidth is just onechannel. The stationsshare the capacity of

channel and time.Each station isallocated a time slotduring which it cansend data. TDMA isused in cellulartelephone network.


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Problems in TDMA?

Time slot is wasted if there is no user

As channel division (in terms of time)

is fixed if new user come we have todivide the whole channel again.

It remove the Problem of frequencyguard (guard band)


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CMDA (Code Division MultipleAccess)

In CDMA one channel carries alltransmissions simultaneously. CDMAemploys analog-to-digital conversion(ADC) in combination with SpreadSpectrum technology. It is used inultra-high-frequency (UHF) cellular

telephone systems in the 800-MHzand 1.9-GHz bands.


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History

CDMA is a military technology firstused during World War II by theEnglish allies to foil German attemptsat jamming transmissions. The alliesdecided to transmit over severalfrequencies, instead of one, making it

difficult for the Germans to pick upthe complete signal.


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MAC ALGORITHM

Suppose we have four stations, eachhas a sequence of chips which wedesignate as A, B, C and D.

Chip Sequences:-A= +1,+1,+1,+1

B= +1,-1, +1,-1C= +1,-1,-1,-1D= +1,-1,-1,+1


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ENCODING RULES

If a station needs to send a 0 bit, itsends 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


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ANEXAMPLE

Consider the four stations that sharethe 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


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SENDER


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STEPS FOLLOWED The multiplexer receives 1 encoded number from each

station (-1, -1, 0 and +1).

The encoded number sent by station 1 is multiplied byeach chip in sequence A. A new sequence is the result

(-1,-1,-1,-1). Likewise, the encoded number sent by station 2 is

multiplied by each chip in sequence B. The same resultis true for the remaining two encoded numbers. Theresult is four new sequences.

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

The sequence is transmitted through the link.


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RECEIVER


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STEPS FOLLOWED The demultiplexer receives the sequence sent

across the link.

It multiplies the sequence by the code for eachreceiver. The multiplication is done chip by chip.

The chips in each sequence are added. Theresult is always +4,-4 or 0.

The result of step 3 is divided by 4 to get +1, -1or 0.

The number in step 4 is decoded to 0, 1 orsilence by the receiver.


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OBSERVATION 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 datafrom the sum.


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ORTHOGONAL SEQUENCES

The sequences in our example arecalled Orthogonal Sequences.

The sequence were not chosenrandomly; they were carefullyselected.


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SEQUENCE GENERATION

To generate sequences, Walsh Table is used, which isa two-dimensional table with an equal number of rowsand columns

Each row is a sequence of chips.

The Walsh Table W1 for a one-chip sequence has onerow 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 sequencesW2n, 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.


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CREATING W2 &W4 FROM W1 The following figure shows the process.

After we select W1, W2 can be made from for W1swith the last one the complement of W1.

After W2 is generated, W4 can be made of four W2,swith the last one the complement of W2.

Similarly we can create W8, composed of four w4s.


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PROPERTIESOFORTHOGONALSEQUENCES If we multiply a sequence by -1, every element in the

sequence is complemented (+1 becomes -1 and -1becomes +1).

If we multiply two sequences, element by element, and

add the results, we get a number called the InnerProduct. 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.

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


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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

= 4So it is proved that A.A is always

equal to N.


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Now,

A = [+1,+1,+1,+1]

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

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

= 1-1+1-1

= 0So it is proved that A.B is always

equal to 0.


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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

= -4So the third property is also

proved.


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Spread Spectrum Access

Two techniques

Frequency Hopped Multiple Access(FHMA)

Direct Sequence Multiple Access(DSMA)


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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 oftime.The random change in frequency ensures small probability of usingthe same narrowband channel

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

Rate of hopping versus Symbol rateIf hopping rate is greater: Called Fast Frequency HoppingOne bit transmitted in multiple hops.If symbol rate is greater: Called Slow Frequency HoppingMultiple bits are transmitted in a hopping period


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Concept of Frequency HoppingSpread Spectrum


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An Example of FrequencyHopping Pattern


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Case Study-Bluetooth


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Case Study: Bluetooth Pico netand FHSS


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Code Division Multiple Access(CDMA)


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Concept of Direct SequenceSpread Spectrum


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Some Simulation Results


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Simulation ResultsRAKE vs MMSE Equalizer

Teb

Eb/N0

RAKE SF = 4

MMSE SF = 4

RAKE SF = 8

MMSE SF = 8RAKE SF = 32

MMSE SF = 32

Diff. Spreading Factors

(No of users = 1)


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RAKE vs MMSE Equalizer

Teb

Eb/N0

RAKE 16 Users

MMSE 16 Users

RAKE 8 Users

MMSE 8 Users

RAKE 4 Users

MMSE 4 Users

Different Number of users

(Spreading Factor = 32)


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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


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References

Wireless Communication, By Rappaport

Digital communication by B. Sklar

The MAI performance of orthogonal codes for channel cover inAsynchronous CDMA system by, Miftadi Sudjai and Mosa N.A. AbuRgheff, 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 ofElectrical Engg, K.U.Leuven.

And a lot of websites..


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Questions.??

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