Transcript No Slide Title

Multiple Access Techniques
for wireless communication
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Multiple access schemes allow many mobile
users to share a finite amount of radio
spectrum
High quality of communications must be
maintained during the sharing process
Multiple Access Techniques
Multiple Access Techniques
PR
FDMA
TDMA CDMA SDMA
•Packet Radio
•Frequency Division Multiple Access
•Time Division Multiple Access
•Code Division Multiple Access
•Space Division Multiple Access
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Multiple Access (MA) Technologies
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Cellular System
MA Technique
AMPS ( Advanced Mobile Phone
system )
GSM ( Global System for Mobile )
FDMA / FDD
US DC ( U. S Digital Cellular )
TDMA / FDD
JDC ( Japanese Digital Cellular )
TDMA / FDD
IS – 95 ( U.S Narrowband Spread
Spectrum )
CDMA / FDD
TDMA / FDD
Frequency Division Multiple Access (FDMA)
code
C1 C2 CN
frequency
time
C1
4
C2
CN
frequency
Principle of FDMA Operation
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Each user is allocated a unique frequency band
or channel. These channels are assigned on
demand to users who request service
In FDD, the channel has two frequencies –
forward channel & reverse channel
Properties of FDMA
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Bandwidth of FDMA channels is narrow (30
KHz)
No equalization is required, since the symbol
time is large compared to average delay spread
FDMA systems have higher cost
o Costly band pass filters to eliminate
spurious radiation
o Duplexers in both T/R increase
subscriber costs
Number Of channels in FDMA System
Bg
Bg
N
Bt  2Bg
Bt
Bc
Bg  GuardBand
Bc  ChannelBandwidth
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Example
In the US, each cellular carrier is allocated 416
channels,
Bt  12.5MHz
Bg  10KHz
Bc  30KHz
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(12.5  106 )  2(10  103 ) 
N
 416
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30  10
Time Division Multiple Access (TDMA)
code
C1
CN
frequency
time
C1
9
C2
CN
time
TDMA Operating principle
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TDMA systems divide each FDMA channel into
time slots
Each user occupies a cyclically repeating time
slot.
TDMA can allow different number of time slots
for separate user
TDMA Frame Structure
Preamble
Slot 1
Trail Bit
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Information Trail Bits
message
Slot 2
Sync Bit
Slot N
Information Guard Bits
Bit
Components of TDMA Frame
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Preamble  Address and synchronization
information for base station and subscriber
identification
Guard times  Synchronization of receivers
between different slots and frames
TDMA properties
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Data Transmission for user of TDMA system
occurs in discrete bursts
o The result is low battery consumption.
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o Handoff process is simpler
Since different slots are used for T and R,
duplexers are not required.
Equalization is required, since transmission
rates are higher than FDMA channels
Efficiency of TDMA
Frame Efficiency
No.ofbits / frame containingtransmitted data
f 
Total Numberof bits / frame
 (1  bOH / bT )  100
(bT  bOH )

 100
bT
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Frame efficiency parameters
bT  Total Number of bits per frame
=Tf  R
Tf =Frame duration
R=Channel bit rate
bOH =Number of overhead bits /frame
=Nr  br  Nt  bp  Nt  b g  Nr  b g
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Frame efficiency parameter definition
Nr  Number of reference bits per frame
Nt  Number of traffic bits per frame
br  Number of overhead bits per reference burst
bp  Number of overhead bits per preamble in each slots
b g  Number of equivalent bits in each guard time interval
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Number of channels in TDMA System
N=
m(Btot -2Bguard )
Bc
m  Maximum number of TDMA users supported on each radio channel
Bguard  Guard band to present user at the edge of the band
from 'bleeding over' to an adjacent radio service
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Example
The GSM System uses a TDMA frame structure
where each frame consist of 8 time slots, and
each time slot contains 156.25 bits, and data is
transmitted at 270.833 kbps in the channel.
• Time duration of a bit
• Time duration of a slot
• Time duration of a frame
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Solution
•
Time duration of a bit
1
1
=Tb =

 3.692 s
3
bit-rate 270.833  10
•
Time duration of a slot
 Tslot  156.25  Tb  0.577 
s
ms
•
Time duration of a frame
 8  Tslot  4.615ms
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Example
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If a normal GSM timeslot consists of 6 trailing bits,
8.25 guard bits, 26 training bits, and 2 traffic bursts
of 58 bits of data, find the frame efficiency.
Solution
 Time slot has 6+ 8.25+ 26 + 2(58) = 156.25 bits.
 A frame has 8 * 156.25 = 1250 bits / frame.
• The number of overhead bits per frame is:
bOH = 8(6) + 8(8.25) + 8(26) = 322 bits
Frame efficiency = (1250 – 322 )/1250 = 74.24 %
Capacity of Cellular Systems
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Channel capacity of a wireless system is the
maximum number of users possible in the
system
Channel capacity depends on:
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Bandwidth available
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Signal to Noise ratio (SNR) in the channel
Calculation of cell capacity
For a Cellular System
 m = Capacity/cell =
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Bt
Bc * N
Bt = Total spectrum for the system
BC = Channel bandwidth
N = Number of cells / cluster
Co-channel cell interference
CELL A
CELL A
CELL A
CELL A
CELL A
CELL A
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Channel capacity calculation
m
Bt
2/n
 6 S  
Bc  n / 2   
 3  I  min 
where n is the path loss exponent
S 
 I  is the minimum required Signal to Interferen ce ratio
  min
S 
 I   12 dB for digital transmiss ion
  min
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S/I for digital cellular system
 S  Eb Rb Ec Rc
I  I  I
 
R b  Channel bit rate
Eb  Energy per bit
R c  Rate of channel code
Ec  Energy per code symbol
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Capacity of Digital Cellular CDMA
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Capacity of FDMA and TDMA system is
bandwidth limited.
Capacity of CDMA system is interference
limited.
The link performance of CDMA increases as
the number of users decreases.
Number of possible users in CDMA


 WR  
N  1 
 S
  Eb  
  No  
 
 
 R   Pr oces sing Gain
where W
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• is the background thermal noise
•S is the average user power
•W is the total RF bandwidth
•R is the information bit rate
Techniques to improve capacity
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Antenna Sectorization
A cell site with 3 antennas, each having a
beamwidth of 120 degrees , has one-third of
the interference received by omnidirectional antenna. This increases the
capacity by a factor of 3
Monitoring or Voice activity
Each transmitter is switched off during
period of no voice activity. Voice activity is
denoted by a factor a
SNR Improvement
Eb

No
W R
 S

Ns  1 a 
where Ns  Number of users per sector
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SNR Improvement …
W 
 R 
1
Ns  1 

,)0  a  1
a  Eb 
(S  a )
 N 
o 

 
If a = 3/8 and number of sector is equal to 3 ,
SNR increases by a factor of 8.
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Example
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If W = 1.25 MHz, R= 9600 bps, and a
minimum acceptable Eb/ No is 10 dB,
determine the maximum number of users
that can be supported in a single cell CDMA
system using
omni directional base station antennas and
no voice activity detection
3 sectors at base station and a = 3/8.
Assume the system is interference limited.
 = 0.
Solution
(a)
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

 WR  
N  1 
 S
  Eb  
  No  
 1.25  10

9600   0
 1 
10




 1  13.02  14
 
 
Solution …
(b) Users per sector
W 
 R 
1
Ns  1 

,)0  a  1


a Eb
(S  a
 N 
o

 
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 1   1.25  10

9600   0

 1 
 3 
10

 8 

 35.7
Solution …
Total users N in 3 sectors
 3Ns
 3  35.7
 107 users / cell
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