Fundamentals of Cellular and Wireless Networks Lecture ID: ET- IDA-113/114 27.07.2010 , v07 Prof. W. Adi Lecture-12 Multiple Access and Duplexing for Wireless Communication Text book Rappaport Ch. 8

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

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

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):

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

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

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

TDMA Frame Structure and its Efficiency bT 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

Example 1: GSM System

GSM TDMA Frame Structure and its Efficiency 25 MHz/200 kHz = 125 125 -1 control=124 Bit rate= 270.833 kbps 156.25 bits /slot 156.25 bits/slot 4.125 Bits 4.125 Bits Example: in GSM total Frame length is bt= 4.125 + 3 + 57 + 1 + 26 + 1 + 57 + 3 + 4.125 = 156.25 Bits. user data are bu=2x57+2=116 bits, GSM Frame Efficency  = 116/156.25 = 74.24 % Source: Prof. Rohling Hamburg Harburg University-Germany

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

Source: Prof. Rohling Hamburg Harburg University-Germany

Source: Prof. Rohling Hamburg Harburg University-Germany

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

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

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

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.

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

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

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‘)

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