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4/11/20151 Mobile Computing COE 446 Wireless Multiple Access Tarek Sheltami KFUPM CCSE COE Principles.

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Presentation on theme: "4/11/20151 Mobile Computing COE 446 Wireless Multiple Access Tarek Sheltami KFUPM CCSE COE Principles."— Presentation transcript:

1 4/11/20151 Mobile Computing COE 446 Wireless Multiple Access Tarek Sheltami KFUPM CCSE COE Principles of Wireless Networks K. Pahlavan and P. Krishnamurth

2 Outline CDMA capacity calculations With/without sectorization gain, voice of active interference reduction factor and interference increase factor Comparison of the Capacity of different 2G Systems 4/11/20152

3 3 How many users can simultaneously use a CDMA System before the system collapses? CDMA systems are implemented based on the spread spectrum technology A spread spectrum transmitter spreads the signal power over a spectrum N times wider than the spectrum of the message An information BW of R occupies a transmission of BW W W= NR  (1) The spread spectrum receiver processes the received signal with a processing gain of N

4 4/11/20154 How many users can simultaneously use a CDMA System before the system collapses?.. During the processing at the receiver, the power of received signal having the code of that particular receiver will be increased N times beyond the value before processing Let us consider a single cell with M simultaneous users on the uplink channel Let us assume that we have an ideal power control enforced on the channel, so that the received power of signals from all MTs has the same value of P Then, the received power from the target user after processing at the receiver is NP, and the received power interference from M-1 other MTs is (M-1)P

5 4/11/20155 How many users can simultaneously use a CDMA System before the system collapses?.. Let us further assume that a cellular system is limited and the background noise is dominated by interference noise from other users, then, the receiver Signal-to- interference ratio for a target receiver:  (2)

6 4/11/20156 How many users can simultaneously use a CDMA System before the system collapses?.. All users always have a requirement for the acceptable error rate of the received data stream For a given modulation and coding specification of the system, the error rate requirement will be supported by minimum of S r requirements From EQs #1 & 2  (3)

7 4/11/20157 Using QPSK modulation and convolution coding, the IS- 95 digital cellular systems require 3 dB < S r < 9 dB. The bandwidth of the channel is 1.25 MHz, and the transmission rate is R = 9600 bps. Find the capacity of a single IS-95 cell. Solution: Using Equation (4.3) we can support: Problem 2: Capacity of One Carrier in a Single-Cell CDMA System

8 4/11/20158 Practical Consideration In practice, design of digital cellular systems, there are other parameters affect the number of users that can be supported by the system 1. Number of Sectors: The use of sectored antenna is an important factor in maximizing BW efficiency Cell sectorization using directional antennas reduces the overall interference and increasing M Sectorization gain = G A

9 4/11/20159 Practical Consideration.. With ideal sectorization the user in one sector of a BS antenna do not interfere with the users operating in other sectors G A = N sec, N sec is the number of sectors in the cell In practice antenna patterns can not be designed to have ideal characteristics, and due to multipath reflection, users in general communicate with more than one sector The sector BS gain assumed to be G A = 2.5 (4 dB) 2. The Voice of activity interference reduction factor (G v ):  The ratio of the total connection time to the active talkspurt  On the average, in two-way conversation, each user talks roughly 50% of the time

10 4/11/ Practical Consideration..  The short pause in the flow of natural speech reduce the activity factor further to about 40% of connection time in each direction As a result, the typical number used for G v = 2.5 (4 dB) 3. The Interference Increase Factor (H o ) Accounts for the users in other cells in CDMA system Because all neighboring cells in a CDMA cellular network operate at the same frequency, they will cause additional interference This interference is relatively small due to the processing gain and the distance involved H o = 1.6 (2 dB), is commonly used in industry

11 4/11/ Practical Consideration.. From EQs 1, 2 and 3  (4) The last part of EQ.# 4 is called Performance improving factor: K=(G A G V /H o ) = 4 (6 dB)

12 4/11/ Determine the multi-cell IS-95 CDMA capacity with correction for sectorization and voice activity. Use the numbers from Problem 2. Solution: If we continue the previous example with the new correction factor included, the range for the number of simultaneous users becomes 64 < M < 260. Problem 3: Capacity of One Carrier in a Multi-Cell CDMA System with Correction Factors

13 4/11/ Compare the capacity of IS-95 CDMA with AMPS FDMA and IS-136 TDMA systems. For the CDMA system, assume an acceptable signal to interference ratio of 6 dB, data rate of 9600 bps, voice duty cycle of 50 percent, effective antenna separation factor of 2.75 (close to ideal 3-sector antenna), and neighboring cell interference factor of Problem 4: Comparison of the Capacity of Different 2G Systems

14 4/11/ For the IS-95 CDMA using Equation (4.4) for each carrier with W = 1.25 MHz, R = 9600 bps, S r = 4 (6dB), G v = 2 (50 percent voice activity). G A = 2.75, and H o = 1.67 we have: For the IS-136 with a carrier bandwidth of W c = 30 kHz, the number of users per carrier of N u = 3, and frequency reuse factor of K = 4 (commonly used in these systems), each W = 1.25 MHz of bandwidth provides for For the AMPS analog system with carrier bandwidth of W c = 30 kHz, and frequency reuse factor of K = 7 (commonly used in these systems), each W = 1.25 MHz of bandwidth provides for Solution

15 4/11/ Determine the capacity of GSM for K = 3. Solution: For the GSM system with a carrier bandwidth of W c = 200 kHz, the number of users per carrier of N u = 8, and frequency reuse factor of K = 3 (commonly used in these systems), each W = 1.25MHz of bandwidth provides for Problem 4: Comparison of the Capacity of Different 2G Systems


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