Electronics And Communication

1. 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

2. Difference between multiplexing and multiple access
2

3. Multiple Access Techniques
Multiple Access Techniques
FDMA TDMA CDMA SDMA
3

4. 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

5. …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

6. Frequency Division
Multiple Access (FDMA)
code C1 C2 Cn
frequency
time frequency
C1 C2 Cn
6

7. 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

8. …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

9. 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

10. Number Of Channel Supported
By FDMA System
Bg Bg
Bt − 2Bg Bt
N=
Bc
Bg → GuardBand
Bc → ChannelBandwidth
10

11. Example
In the US, each cellular carrier is allocated 416
channels,
Bt = 12.5MHz
Bg = 10KHz
Bc = 30KHz
(12.5 × 106 ) − 2(10 × 103 ) 
  = 416
N=
30 × 10 3
11

12. Time Division
Multiple Access (TDMA)
code
C1
Cn
frequency
time time
C1 C2 Cn
12

13. 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

14. TDMA Frame Structure
Preamble Information Trail Bits
message
Slot 1 Slot 2 Slot N
Trail Bit Sync Bit Information Guard Bits
Bit
14

15. 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

16. 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

17. …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

18. Efficiency of TDMA
Frame Efficiency :
No.ofbits / frame containingtransmitted data
ηf =
Total Numberof bits / frame
= (1 − bOH / bT ) × 100
(bT − bOH )
= × 100
bT
18

19. 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

20. …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

21. 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

22. 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

23. …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

24. 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 µs
ms
24

25. …Solution
• Time duration of a frame
= 8 × Tslot = 4.615ms
• A user has to wait 4.615 ms before next
transmission
25

26. 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

27. …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

28. 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

29. …Spread Spectrum Multiple Access
Technologies (SSMA)
SSMA provides immunity to multiple
interference and has robust multiple access
capability.
29

30. Types Of Spread
Spectrum Techniques
Frequency Hopped Multiple Access
( FHMA )
Direct Sequence Multiple Access ( CDMA )
30

31. Direct Sequence Spread
Spectrum (DS-SS)
code
C1
C2
frequency
C3
Cn
time
31

32. 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

33. 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

34. CDMA Receiver
k
Z i (t)
>
∫ (.)dt
r(t) < m (t)
k
PNK (t) cos(2πfc t + ϕk )
34

35. 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

36. 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

37. Example: Assume N=4
1
-1
PN Wave for N =4
1
-1
37

38. 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

39. 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

40. 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

41. 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

42. 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

43. …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

44. Channel capacity design
CELL A
CELL A
CELL A
CELL A CELL A
CELL A
44

45. …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) ( R D)
−n
C = 1
min 6
( CI ) min
= 1
6 ( Q)
−n
45

46. …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

47. …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

48. 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

49. 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

50. TDMA
Assume FDMA occupies the same
spectrum as a single channel TDMA.
C′ = E b R ′
b
I′ = IoB′
c
where R′ = Transmission rate of TDMA system
b
R b = Transmission rate of FDMA system
Eb = Energy per bit
50

51. 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

52. 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

53. 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

54. Solution …
For TDMA system
C′ = E R ′ =
(E b × 10 4
)
b b 0.333s
I′ = IoB′ = Io × 30 × 103
c
C = (
 Eb × 104 )
 
÷× 
1   Eb 
=
I  0.333s ÷ I × 30 × 10 3 ÷  Io ÷
  o   
54

55. 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

56. 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

57. 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

58. 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

59. 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

60. Number of possible of
single cell system
W
Eb R
=
No η
(N − 1) +
S
60

61. Number of users that
can access the system
 
( )
 WR  η
N = 1+ 
  Eb  
 N ÷
( )
− S

 o 

( R ) = Pr oces sing Gain
where W
61

62. 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

63. 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

64. 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

65. SNR Improvement
Eb
=
( W R)
N′
o
( Ns − 1) α + ( S)
η
where Ns = Number of users per sector
65

66. 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

67. 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

68. 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

69. 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

70. Solution
(a)  
( )
 WR  η
N = 1+ 
  Eb  
 N ÷
( )
− S

 o 
 1.25 × 10 
= 1+  9600  − 0
 10 
 
= 1 + 13.02 = 14
70

71. Solution …
(b)
 
( )
W

N = 1+ 

R − η
 Eb  
 S( )

 No ÷

 1   1.25 × 10 
= 1+  ÷ 9600  − 0
 3 ÷ 10 
 8  
= 35.7
71

72. Solution …
Total amount of
users N = 3Ns
= 3 × 35.7
= 107 users / cell
72