1. Fundamentals of Cellular and Wireless Networks
Lecture ID: ET- IDA-113/114
, v10
Prof. W. Adi
Lecture-12
Multiple Access and Duplexing
for Wireless Communication
Text book Rappaport Ch. 8

2. Multiple Access Techniques
Channel Duplexing
User Multiple Access Techniques
Capacity of Cellular Systems

3. Channel Duplexing Duplexing: to send and receive at the same time
(Required for telephone systems to listen and talk at the same time
by using two simplex channels: forward and reverse channel respectively)
Frequency Division Duplexing FDD:
Uses two separate frequency bands forward band and reverse band
Reverse channel
Forward channel
Reverse
Band
Frequency
separation
Forward
Band
Frequency
Time Division Duplexing TDD:
Uses two separate time slots: forward time slot and reverse time slot
Reverse channel
Forward channel
Reverse
Time Slot
Time
separation
Forward
Time Slot
Time

4. User Multiple Access Techniques
1. Frequency Division Multiple Access: FDMA
2. Time Division Multiple Access: TDMA
3. Spread Spectrum Multiple Access (SSMA):
3.1 - Frequency Hopped Multiple Access: FHMA
3.2 - Code Division Multiple Access: CDMA
( also called Direct Sequence Multiple Access)
3.3- Hybrid SS techniques:
. Hybrid FDMA/CDMA (FCDMA)
. Frequency Hopped CDMA (FH-CDMA)
. Time Division CDMA (TCDMA)
. Time Division Frequency Hopping (TDFH) (used in GSM)
4. Space Division Multiple Access: SDMA
5. Packet Radio
- Protocols: ALOHA, Slotted ALOHA
- Carrier Sense Multiple Access (CSMA):

5. 1. Frequency Division Multiple Access: FDMA
Code Bt Bg 1 2
... N Bg
Frequency Bc
Guard channel
Bt – 2Bg
N =
Time Bc
Example: US AMPS allocates 12.5 MHz for each simplex band.
Each voice channel requires Bc=30 KHz. Guard channel bandwidth is Bg=10 kHz.
The number of channels available in this FDMA system is:
N = [12.5 x 106 – 2 (10x103)] / 30 x 103 = 416 channels for each cellular carrier

6. 2. Time Division Multiple Access: TDMA
Code Bc 1
Frequency 2
... m
Time
Principle: m Channels each assigned a certain time slot, but all are using the same frequency band Bc

7. Example: GSM Combines Time & Frequency Division Multiple Access
Code Bt Bc Bg 1 2
... .. Bg
Frequency 2 2
...
... m m
m ( Bt – 2Bg)
N = Bc
Time
Example: GSM uses TDD/FDD and allocates Bt=25 MHz for forward link, which is broken into channels of 200 kHz. On each channel 8 TDM speech channels are supported (m=8). Assuming no guard channels: The number of simultaneous user in GSM system is:
N = 8 x [25 x 106 / 200 x 103] = 1000 channels
FDM Channels
Each with m time slots

8. TDMA Frame Structure and its EfficiencybT
Preamble
Information Message
Trail Bits
Slot 1
Slot 2
Slot 3 …
Slot N
Trail Bits
Sync. Bits
Information Data
Guard Bits bu
Overhead bits
bT -  bOH
N bu
User data
Frame Efficiency =
 = = =
Total data bT bT

9. Example 1: GSM System

10. GSM TDMA Frame Structure and its Efficiency
25 MHz/200 kHz = 125
control=124
Bit rate= kbps
bits /slot
bits/slot
4.125 Bits
4.125 Bits
Example: in GSM total Frame length is bt= = Bits.
user data are bu=2x57+2=116 bits, GSM Frame Efficency  = 116/ = %
Source: Prof. Rohling Hamburg Harburg University-Germany

11. Example 2: European Cordless Phone System DECT
Source: Prof. Rohling Hamburg Harburg University-Germany

12. Source: Prof. Rohling Hamburg Harburg University-Germany

13. Source: Prof. Rohling Hamburg Harburg University-Germany

14. 3.1 Frequency Hopped Multiple Access: FHMA
Spread Spectrum Multiple Access
3.1 Frequency Hopped Multiple Access: FHMA
Principle: Many users are jumping between same narrow different frequency bands. Every user jumps according to his own unique sequence. (two or more users may occupy the same frequency band at the same time).
Users are recovered at the receiver by using their individual sequences.
Code 1
Frequency 1 N
User 1 2 3
User 2 2 5
Time
Example: FHMA is used in Bluetooth wireless network

15. 3.2 Code Division Multiple Access: CDMA
Spread Spectrum Multiple Access
3.2 Code Division Multiple Access: CDMA
Principle of CDMA: Every users signal is multiplied by a unique spreading signal and all users send at the same time using the same wide frequency band.
Code N
... 2 1
Frequency
Essential requirement to effectively recover the mixed channels is power control:
Power control (within a cell) assures that all mobiles within the base station coverage area provide the same signal level to the base station receiver.
Time

16. 3.3 Hybrid SS Multiple Access Techniques
Spread Spectrum Multiple Access
3.3 Hybrid SS Multiple Access Techniques
. Hybrid FDMA/CDMA (FCDMA)
. Frequency Hopped CDMA (FH-CDMA)
. Time Division CDMA (TCDMA)
. Time Division Frequency Hopping (TDFH) (used in GSM)
Example:Time Division - Frequency Hopping Multiple Access: TDFH (used in GSM)
Code 1 1
Frequency 1 2 2 2
...
...
... N N N
Time

17. 4. Space Division Multiple Access : SDMA
Principle of SDMA: Every user is served by using spot beam antenna at the same frequency. As the beam is directed, the interference between users having the same frequency is minimized.

18. Capacity of Cellular Systems
Capacity of a cellular system: Number of users in a cell
Capacity of Analog and Digital Cellular Systems
Capacity of Cellular CDMA Systems

19. Radio Capacity of Cellular Systems
Reminder: in a cellular system of cluster size N:
Frequency reuse factor is 1/N
Frequency reuse distance D = R  3N
Co-Channel reuse ratio Q = D/R = 3N
Co-channel interference ratio S/I:
S/I = R-n /  Di–n
Simplified S/I = R-n / 6 D–n = 1/6 (R/D)-n= Qn/6 = 1/6 ( (3N)-1/2 )-n = 1/6 (3N)n/2
=> N = 1/3 (6 S/I)2/n
The Radio Capacity m of a cell is interference dependant:
Where:
(C/I)min: minimum acceptable Carrier to Interference ratio of the system
C/I is considered as S/I (see Rappaport ch. 8.7)
Bt : total allocated spectrum
Bc: channel bandwidth
n: Path loss exponent Bt
m =
Bc (1/3) [6 (C/I)min ] 2/n Bt
m = N
For n=4 
Bc [2/3 (C/I)min ]1/2
* See Rappaport Ch. 8.7

20. Capacity of Cellular CDMA Systems
If we have a perfect power control, and the received signal power of every one of the N CDMA users is S. Then the SNR of the reverse link:
SNR = Signal power / Noise power = S / (N -1) S = 1 / (N-1)
If the bit rate is R and the allocated bandwidth is W then Eb/N0 is :
Eb/N0 = = (if the thermal noise  is neglected)
S/R
W/R
(N-1) (S/W) + 
(N-1)
W/R
N = 1 +
Eb/N0
If the senders are switched off during the periods of no voice activity. And voice activity factor is denoted by . And if the cell is sectored to have Ns users per sector, then the new average signal to noise ratio is :
and the number of users/sector
W/R
W/R
Eb/N0‘ =
Ns= 1 +
1/
 (Ns-1)
(Eb/N0‘)

21. Capacity of Cellular CDMA Systems
Example: In a CDMA system if the allocated bandwidth is W=1.25 MHz and
the data bit rate is R=9600 bps and the minimum acceptable Eb/N0 = 10 dB.
Determine the maximum number of users that can be supported in a single cell for:
Omnidirectional base station antenna and no voice activity detection
Using three-sectors at the base station and voice activity detection with =3/8. Assume the system is interference limited and ignore thermal noise.
Solution:
Substituting in :
Substituting in:
Every cell has 3 sectors, thus the number of users in the cell is 3 Ns= 3 x 35.7 = 107 users/cell
W/R
1.25 x 106 / 9600
N = 1 + 
N = 1 +
= 14 users
Eb/N0 10
W/R
1.25 x 106 / 9600
Ns= 1 +
(1/) 
Ns = 1 +
8/3
(Eb/N0‘) 10
Ns= 35.7 users/sector