Multiple Access Techniques

Multiple access schemes are used to allow many

mobile users to share simultaneously a finite
amount of radio spectrum.
The sharing of spectrum is required to achieve
high capacity by simultaneously allocating the
available bandwidth (or the available amount of
channels) to multiple users.
For high quality communications, this must be
done without severe degradation in the
performance of the system.
1
 Difference between multiplexing and multiple access

2
 Multiple Access Techniques

Multiple Access Techniques

FDMA TDMA CDMA SDMA

3
 Multiple Access (MA) Technologies
used in Different Wireless Systems
Cellular Systems MA Technique
AMPS ( Advanced Mobile FDMA / FDD
Phone system )
GSM ( Global System for TDMA / FDD
Mobile )
US DC ( U. S Digital TDMA / FDD
Cellular )
JDC ( Japanese Digital TDMA / FDD
Cellular )

4
 …Multiple Access (MA) Technologies
used in Different Wireless Systems
Cellular Systems MA Technique
DECT ( Digital FDMA / FDD
European Cordless
Telephone )
IS – 95 ( U.S Narrowband CDMA / FDD
Spread Spectrum )

5
 Frequency Division
Multiple Access (FDMA)
code C1 C2 Cn

frequency

time frequency
C1 C2 Cn
6
 Principles Of Operation

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.

7
 …Principles Of Operation

During the period of the call, no other user

can share the same frequency band.
If the FDMA channel is not in use, then it
sits idle and cannot be used by other users
to increase or share capacity. This is a
wasted resource.

8
 Properties of FDMA
The bandwidth of FDMA channels is narrow
(30 KHz) since it supports only one call/
carrier.
FDMA systems have higher cost
Costly band pass filters to eliminate
spurious radiation
Duplexers in both T/R increase
subscriber costs

9
 Number Of Channel Supported
By FDMA System

Bg Bg

Bt − 2Bg Bt
N=
Bc
Bg → GuardBand
Bc → ChannelBandwidth
10
 Example
In the US, each cellular carrier is allocated 416
channels,

Bt = 12.5MHz
Bg = 10KHz
Bc = 30KHz
(12.5 × 106 ) − 2(10 × 103 ) 
N= = 416
30 × 10 3

11
 Time Division
Multiple Access (TDMA)
code
C1
Cn
frequency

time time
C1 C2 Cn
12
 Principles Of Operation
TDMA systems divide the radio spectrum
into time slots and each user is allowed to
either transmit or receive in each time slots.
Each user occupies a cyclically repeating
time slots. TDMA can allow different number
of time slots for separate user.

13
 TDMA Frame Structure
Preamble Information Trail Bits
message

Slot 1 Slot 2 Slot N

Trail Bit Sync Bit Information Guard Bits

Bit

14
 Components of 1 TDMA Frame

Preamble  Address and synchronization

information for base station and subscriber
identification
Guard times  Synchronization of
receivers between a different slots and
frames

15
 Principles Of Operation
TDMA shares the single carrier frequency
with several users, where each user
makes use of non-overlapping timeslots.
Data Transmission for user of TDMA
system is discrete bursts
• The result is low battery consumption.
• Handoff process is simpler, since it is
able to listen for other base stations
during idle time slots.
16
 …Principles Of Operation

Since different slots are used for T and R,

duplexers are not required.
Equalization is required, since transmission
rates are generally very high as compared to
FDMA channels.

17
 Efficiency of TDMA
Frame Efficiency :
No.ofbits / frame containingtransmitted data
ηf =
Total Numberof bits / frame

= (1 − bOH / bT ) × 100
(bT − bOH )
= × 100
bT

18
 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
19
 …Frame efficiency parameters

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

20
 Number of channels in TDMA System
m(Btot -2Bguard )
N=
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

21
 Example

GSM System uses a TDMA / FDD system.

The GSM System uses a 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.
Find: ……

22
 …Example

1. Time duration of a bit

2. Time duration of a slot
3. Time duration of a frame and
4. How long must a user occupying a single
slot must wait between two simultaneous
transmissions?

23
 Solution
• Time duration of a bit
1 1
=Tb = = = 3.692 µs
bit-rate 270.833 × 10 3

• Time duration of a slot

= Tslot = 156.25 × Tb = 0.577 µms

s

24
 …Solution
• Time duration of a frame

= 8 × Tslot = 4.615ms

• A user has to wait 4.615 ms before next

transmission

25
 Example
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 slots have 6 + 8.25 + 26 + 2(58) =
156.25 bits.
A frame has 8 * 156.25 = 1250 bits / frame.
26
 …Example

The number of overhead bits per frame is

given by

bOH = 8(6) + 8(8.25) + 8(26) = 322 bits

Frame efficiency = (1250 – 322 ) / 1250
= 74.24 %

27
 Spread Spectrum Multiple Access
Technologies (SSMA)
SSMA technologies uses techniques which
has a transmission bandwidth that is much
greater than maximum required RF
bandwidth.
This is achieved by pseudo noise (PN)
sequence that contents a narrowband signal
to a wideband noise-like signal before
transmission.

28
 …Spread Spectrum Multiple Access
Technologies (SSMA)

SSMA provides immunity to multiple

interference and has robust multiple access
capability.

29
 Types Of Spread
Spectrum Techniques

Frequency Hopped Multiple Access

( FHMA )

Direct Sequence Multiple Access ( CDMA )

30
 Direct Sequence Spread
Spectrum (DS-SS)
code

C1
C2
frequency
C3
Cn

time
31
 Direct Sequence Spread
Spectrum (DS-SS)
S1 (t)
τ1
m1 (t)
r(t)
PN1 (t)
cos(2πfc t + ϕ1 )
Σ

τk
m2 (t)
PNK (t) cos(2πfc t + ϕk )
32
 Principles of operation-transmitter
The narrowband message signal
mi(t) is multiplied by a pseudo noise
code sequence that has a chip rate >>
data rate of message.
All users use the same carrier
frequency and may transmit
simultaneously. The kth transmitted
signal is given by:
Sk (t) = (2E s / Ts )1/ 2mk (t)pk (t)cos(2πfc t + ϕk )

33
 CDMA Receiver
k
Z i (t)
>
∫ (.)dt
r(t) < m (t)
k

PNK (t) cos(2πfc t + ϕk )

34
 Principles of operation-receiver
At the receiver, the received signal is
correlated with the appropriate signature
sequence to produce desire variable.
iT +τ1

Z (t) = ∫ r(t)p1 (t − τ1 )cos[2πfc (t − τ1 ) + ϕ1 ]dt

1
i
(i −1)T +τ1

35
 Message Signal
m(t) is a time sequence of non-overlapping
pulses of duration T, each of which has an
amplitude (+/-) 1.
The PN waveform consists of N pulses or
chips for message symbol period T.
NTC = T
where TC is the chip period.

36
 Example: Assume N=4
1

-1
PN Wave for N =4
1

-1
37
 Correlator output for first user
iT +τ1

Z i1 (t) = ∫
(i −1)T +τ1
r(t)p1 (t − τ1 )cos[2πfc (t − τ1 ) + ϕ1 ]dt

•The multiplied signal will be p2(t) = 1 for the

correct signal and will yield the dispersed
signal and can be demodulated to yield the
message signal mi(t).
S1 (t) = (2Es / Ts ) m1 (t)p1(t)cos(2πfc t + ϕ1 )
1/ 2

38
 Probability of bit error
Probability of bit error
Pe = Q {1/ [(K –1)/3N + (N0/2Eb)]1/2}
K = Number of users
N = Number of chips/ symbol

Now when, Eb/No  ∝

Pe = Q{[3N/(K-1)]1/2 }

39
 Important Advantages of CDMA

Many users of CDMA use the same

frequency. Either TDD or FDD may be used.
Multipath fading may be substantially
reduced because of large signal bandwidth.
There is no absolute limit on the number of
users in CDMA. The system performance
gradually degrades for all users as the
number of users is increased.

40
 Drawbacks of CDMA

Self-jamming is a problem in a CDMA

system. Self-jamming arise because the PN
sequence are not exactly orthogonal, non-
zero contributions from other users in the
system arise
The near- far problem occurs at a CDMA
receiver if an undesired user has high
detected power as compared to the desired
user.
41
 Capacity of Cellular Systems

Channel capacity for a radio system is

defined as the maximum number of
channels or users that can be provided in a
fixed frequency band  spectrum
efficiency of wireless system.

42
 …Capacity of Cellular Systems

For a Cellular System

m = Radio Capacity Matrix = Bt / (BC * N)
Bt = Total allocated spectrum for the
system
BC = Channel bandwidth
N = Number of cells in frequency reuse
pattern

43
 Channel capacity design

CELL A

CELL A
CELL A

CELL A CELL A
CELL A

44
 …Channel capacity design
for given C/I ratio
Carrier to Interference ratio
C = Do − n
I 6 × D−n
Do =Distance from desired base station to mobile
For maximum interference D0 = R
( I) (6 R D )
−n
C = 1
min

( CI ) min
= 1
6 ( Q)
−n

45
 …Channel capacity design
for given C/I ratio
Q = Co- Channel reuse ratio

{ ( I) }
−µ
n
= 6× C
min

Also ,Q = (3 × N) 0.5

{= ( I ) }
2/n
2 6× C
Therefore , N = (Q) min
− eqn2
3 3

46
 …Channel capacity design
for given C/I ratio
Bt
substituting, m =
 
( ) 
2/n

Bc 6 × C I  ( 1/ 3 ) 
 min 
 

Bt
When n = 4, m = radio channels / cells
 
( ) 
(1/ 2)

Bc ( 2 / 3 ) × C I  
 min 
 

Typical Values of C ( I) min

= 18 dB for Analog FM and12 dB for Digital

47
 Equation of C/I for digital
cellular system

( I)=
C (Eb × R b )
I
(Ec × R c )
=
I
R b = Channel bit rate
Eb = Energy per bit
R c = Rate of channel code
Ec = Energy per code symbol
48
 Comparison of FDMA and
TDMA systems - FDMA
The total bandwidth Bt is divided into M
channels, each with Bandwidth Bc. The
radio capacity for FDMA is given by
M
m=
( )
0.5
2 C
3 I
C = Eb R b
I = IoBc
where Io = Interference power / Hz
49
 TDMA
Assume FDMA occupies the same
spectrum as a single channel TDMA.
C′ = EbR′b
I′ = IoB′c
where R′b = Transmission rate of TDMA system
R b = Transmission rate of FDMA system
Eb = Energy per bit

50
 Example

Consider a FDMA system with 3 channels,

each having a bandwidth of 10 KHz and
transmission rate of 10 kbps. A TDMA
system has 3 time slots, channel bandwidth
of 30 KHz, and a transmission rate of 30
kbps.

51
 Example …
For the TDMA scheme, the received carrier
to interference ratio for a single user is
measured for 1/3 of the time the channel is
in use.
Compare the radio capacity of the 2
systems.

52
 Solution
For FDMA system

C = EbR b = Eb × 10 4

I = IoBc = Io × 10 kHz
4
C =  Eb  × 10 =  Eb 
I  Io ÷  I ÷
 10  o 
4

53
 Solution …
For TDMA system

C′ = E R ′ =
( E b × 10 4
)
b b 0.333s
′I = IoB′c = Io × 30 × 103

C = (
 Eb × 104
÷
)
 
×
1   Eb 
= ÷
I  0.333s ÷ I × 30 × 10 ÷ I
  o 
3
  o

54
 Capacity of Digital Cellular CDMA

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.

55
 Single Cell System

The cellular network consists of a large

number of mobile users communicating
with a base station.
The cell site transmitter consist of linear
combiner which adds spread signal of
individual users for the forward channel.

56
 Single Cell System …

A pilot signal is also included in the cell-site

transmitter and is used by each mobile to
set its own power control for the reverse
link.

57
 Capacity of single cell system
Let the number of users be N and the
signal power from each of N users be S

Signal to noise ratio SNR= S

[( N-1) × S]
1
=
N −1

58
 Bit energy-to-noise ratio
of single cell system
The bit energy to noise ratio is an important
factor in communication systems
S
Eb R
=
No  S η
(N − 1) × W  + W
R= Baseband information bit rate
W= Total RF bandwidth, W
η= Background thermal Noise
59
 Number of possible of
single cell system

W
Eb R
=
No η
(N − 1) +
S

60
 Number of users that
can access the system

 
( )
 WR  η
N = 1+ 
  Eb ÷
( )
− S
  No  

( R ) = Pr oces sing Gain

where W

61
 Number of users that
can access the system …..

In order to increase the capacity, the

interference due to other users should be
reduced. There are mainly two techniques.

62
 Techniques to improve capacity

Antenna Sectorization:
A cell site with 3 antennas, each having a
beamwidth of 1200 , has interference No’,
which is 1/3 of the interference received by
omni-directional antenna. This increases the
capacity by a factor of 3.

63
 Techniques to improve capacity …

Monitoring or Voice activity:

Each transmitter is switched off during
period of no voice activity. Voice activity is
denoted by a factor α

64
 SNR Improvement

Eb
=
( W R)
N′o
( Ns − 1) α + ( S)
η

where Ns = Number of users per sector

65
 SNR Improvement …

W 
( )
Ns = 1 + 1  R η
α  Eb 
− ,)0 < α < 1
(S × α )
 N′ 
 o 

If α = 3/8 and number of sector is equal to 3 ,

SNR increases by a factor of 8.

66
 CDMA Power Control

In CDMA, the system capacity is controlled

if each mobile transmitter power level is
controlled so that its signal arrives at the
cell site with minimum required S/I.

67
 CDMA Power Control ...

If the signal power of all mobile transmitters

within the area of cell site are controlled,
then total signal power received at all site
from all mobile will be equal to average
received power times the number of mobiles
operating in region of coverage.
Optimal power is desired, never too weak or
too strong.

68
 Example
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
(a) omni directional base station antennas
and no voice activity detection
(b) 3 sectors at base station and α = 3/8.
Assume the system is interference limited.
η = 0.
69
 Solution
(a)  
( )
 WR  η
N = 1+ 
  Eb ÷
( )
− S
  No  
 1.25 × 10 
= 1+  9600  − 0
 10 
 
= 1 + 13.02 = 14

70
 Solution …
(b)
 
( )W

N = 1+  R
  Eb ÷
 η
( )
− S
  No  
 1   1.25 × 10 
= 1+  ÷ 9600  − 0
 3 ÷ 10 
 8  
= 35.7

71
 Solution …

Total amount of
users N = 3Ns
= 3 × 35.7
= 107 users / cell

72