Presentation on theme: "Multiple Access Techniques for wireless communication"— Presentation transcript:
1Multiple Access Techniques for wireless communication Multiple access schemes allow many mobile users to share a finite amount of radio spectrumHigh quality of communications must be maintained during the sharing process
3Multiple Access (MA) Technologies Cellular SystemMA TechniqueAMPS ( Advanced Mobile Phone system )FDMA / FDDGSM ( Global System for Mobile )TDMA / FDDUS DC ( U. S Digital Cellular )JDC ( Japanese Digital Cellular )IS – 95 ( U.S Narrowband Spread Spectrum )CDMA / FDD
5Principle of FDMA Operation Each user is allocated a unique frequency band or channel. These channels are assigned on demand to users who request serviceIn FDD, the channel has two frequencies – forward channel & reverse channel
6Properties of FDMA Bandwidth of FDMA channels is narrow (30 KHz) No equalization is required, since the symbol time is large compared to average delay spreadFDMA systems have higher costCostly band pass filters to eliminate spurious radiationDuplexers in both T/R increase subscriber costs
12Components of TDMA Frame Preamble Address and synchronization information for base station and subscriber identificationGuard times Synchronization of receivers between different slots and frames
13TDMA propertiesData Transmission for user of TDMA system occurs in discrete burstsThe result is low battery consumption.Handoff process is simplerSince different slots are used for T and R, duplexers are not required.Equalization is required, since transmission rates are higher than FDMA channels
18ExampleThe GSM System uses a TDMA frame structure where each frame consist of 8 time slots, and each time slot contains bits, and data is transmitted at kbps in the channel.Time duration of a bitTime duration of a slotTime duration of a frame
19Solution Time duration of a bit Time duration of a slot ms Time duration of a frame
20Frame efficiency = (1250 – 322 )/1250 = 74.24 % ExampleIf 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.SolutionTime slot has (58) = bits.A frame has 8 * = 1250 bits / frame.The number of overhead bits per frame is:bOH = 8(6) + 8(8.25) + 8(26) = 322 bitsFrame efficiency = (1250 – 322 )/1250 = %
21Capacity of Cellular Systems Channel capacity of a wireless system is the maximum number of users possible in the systemChannel capacity depends on:Bandwidth availableSignal to Noise ratio (SNR) in the channel
22Calculation of cell capacity For a Cellular Systemm = Capacity/cell =Bt = Total spectrum for the systemBC = Channel bandwidthN = Number of cells / cluster
26Capacity 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.
27Number of possible users in CDMA h is the background thermal noiseS is the average user powerW is the total RF bandwidthR is the information bit rate
28Techniques to improve capacity Antenna SectorizationA cell site with 3 antennas, each having a beamwidth of 120 degrees , has one-third of the interference received by omni-directional antenna. This increases the capacity by a factor of 3Monitoring or Voice activityEach transmitter is switched off during period of no voice activity. Voice activity is denoted by a factor a
31ExampleIf 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 usingomni directional base station antennas and no voice activity detection3 sectors at base station and a = 3/8. Assume the system is interference limited. h = 0.