Computer EngineeringKyungpook National University
Multiplexing Multiple channels on 1 physical link Common on long-haul, high capacity, links
Computer EngineeringKyungpook National University
Frequency Division Multiplexing 넓은주파수대역을여러개의좁은채널로분할하여독립적으로사용하는방식
모든사람이같은언어를사용하고, 큰회의실을작은대화방으로나눈다음, 각대화방에들어가서동시에대화를함
통신보호대역 (guard band)을두어채널간간섭을피함 수신측에서는대역통과필터를거친신호만을통과시키므로상호교란을방지할수있음
Computer EngineeringKyungpook National University
AM/FM 라디오방송 AM 방송할당대역 : 530 ~ 1,700KHz
모든방송국들이이대역을같이사용해야함
각 AM 방송국은 10KHz의대역폭이필요각방송국은서로다른반송파를사용방송국이신호를편이시키고다중화시킴공중으로발산되는신호는이모든신호를합한것수신장치가이모든신호를받아서원하는신호만필터링
FM 방송은 200KHz 더넓은대역이필요 88 ~ 108MHz 대역을사용
TV 방송 각 TV 채널은각각 6MHz의대역폭을가짐
Example
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각 100KHz 대역폭을갖는 5개의채널을 FDM으로다중화할경우필요한대역폭은? 단, 보호대역은 10KHz.
(5 X 100KHz) + (4 X 10KHz) = 540KHz
Example
Computer EngineeringKyungpook National University
1Mbps로전송하는 4개의데이터채널(디지털)이 1MHz의아날로그위성채널을사용하기위한 FDM을설계하라. 1 MHz의위성채널을 4개의채널로분할하면각채널은 250KHz. 따라서, 1 Mbps의각디지털채널에서 4비트가 1Hz로변조되어야함 16-QAM 사용
Example
4비트가하나의신호로인코딩
아날로그위성채널
Computer EngineeringKyungpook National University
FDM Voiceband Example Group
12 voice channels (4khz each) = 48khz 60khz ~ 108khz
Supergroup FDM of 5 group signals supports 60 channels 420khz ~ 612 khz
Mastergroup FDM of 10 supergroups supports 600 channels
Computer EngineeringKyungpook National University
Asymmetrical Digital Subscriber Line (ADSL) Currently installed twisted pair cable Uses frequency division multiplexing
Reserve lowest 25Khz for voice (POTS : Plain Old Telephone Service)
Is asymmetric - bigger downstream than upstream Has a range of up to 5.5km
Computer EngineeringKyungpook National University
Wavelength Division Multiplexing Architecture similar to FDM systems
Multiplexer consolidates laser sources for transmission over single fiber 다중빛소스를단일빛으로결합
프리즘이용 : 임계각과주파수기반
Demux separates channels at the destination 단일빛은다중빛소스로분리
Computer EngineeringKyungpook National University
Synchronous Time Division Multiplexing 시간축에서여러개의시간슬롯을나누어전송매체를공유하는방식
자신에게할당된시간슬롯동안데이터전송
회의실전체를사용하여자기에게할당된시간동안에만이야기를함
통상 FDM에의하여분할된하나의주파수대역을시간축에서다시분할
Computer EngineeringKyungpook National University
Example 네개의 800 bps 채널을 TDM으로다중화하는데각채널마다 1 byte씩교대(interleave)할경우프레임길이와비트시간은얼마인가? 프레임전송율과비트전송율은? 각채널의전송율이 800 bps이므로다중화링크의전송율은 4*800 =
3200 bps 비트시간은 1/3200 s 프레임의크기 = 4*8 = 32 비트 프레임길이는 32 * 1/3200 = 1/100 s 프레임전송율 = 100 frame/s
Computer EngineeringKyungpook National University
Data Link Control on TDMFlow and error control on individual channel
No data link control protocol is needed over multiplexed link
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Framing
One control bit added to each TDM frame Identifiable bit pattern used on control channel
Eg. Alternating 01010101…unlikely on a data channel
Compare incoming bit patterns on each channel with known sync pattern
01010101
Computer EngineeringKyungpook National University
Statistical TDM
Statistical TDM allocates time slots dynamically based on demand In synch TDM, many
slots are wasted Multiplexer scans
input lines and collects data until frame full
May have problems during peak periods Must buffer inputs
Computer EngineeringKyungpook National University
FDD & TDD
다중화를사용하여 full duplex를구현하는방식 FDD (Frequency Division Duplex)서로다른 주파수를이용하여양방향전송
TDD (Time Division Duplex)하나의주파수대역을 2개의시분할채널로분할하여양방향전송
송신과수신이시간을달리하기때문에 부가적인지연이발생
Computer EngineeringKyungpook National University
Cable Modems
Dedicate two cable TV channels to data transfer Each channel shared by number of subscribers,
using statistical TDM Downstream
Cable scheduler delivers data in small packets Active subscribers share downstream capacity Headend scheduler notifies subscriber of slots to use
Allocates upstream time slots to subscribers
Upstream User requests timeslots on shared upstream channel