Multiplexingnetworking.khu.ac.kr/.../Chapter6(Multiplexing).ppt · PPT file · Web view2015-06-12 · Multiplexing Prof. Choong Seon HONG 6 장 다중화(Multiplexing) Dividing a

11Kyung Hee Universit

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Prof. Choong Seon HONG

MultiplexingMultiplexing


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6 6 장 다중화장 다중화 (Multiplexing)(Multiplexing)

Dividing a link into channels

• Word link refers to the physical path• Channel refers to the portion of a link that carries a transmission between a given pair of lines.


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다중화다중화 (Multiplexing)(Multiplexing) 다중화 (Multiplexing)

is the set of techniques that allows the simultaneous transmission of multiple signals across a single data link.


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6.1 FDM6.1 FDM

다중화기 (Multiplexer) transmission streams combine into a single stream(man

y to one)

역다중화기 (Demultiplexer) stream separates into its component transmission(one t

o many) and directs them to their intended receiving devices


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8.2 8.2 Many to One/One to ManyMany to One/One to Many

Categories of Multiplexing


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FDMFDM

FDM(Frequency-Division Multiplexing) is an analog technique that can be applied when the

bandwidth of a link is greater than the combined bandwidths of the signals to be transmitted


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FDM (cont’d)FDM (cont’d)

FDM process each telephone generates a signal of a similar frequency

range these signals are modulated onto different carrier

frequencies(f1, f2, f3)


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FDM (cont’d)FDM (cont’d)

FDM multiplexing process, time-domain


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FDM(cont’d)FDM(cont’d)

FDM multiplexing process, frequency-domain


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Demultiplexing separates the individual signals from their carries and pa

sses them to the waiting receivers.


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FDM demultiplexing process, time-domain


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FDM demultiplexing, frequency-domain


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Example 1Example 1

Example 1Example 1

Assume that a voice channel occupies a bandwidth of 4 KHz. We need to combine three voice channels into a link with a bandwidth of 12 KHz, from 20 to 32 KHz. Show the configuration using the frequency domain without the use of guard bands.

SolutionSolution

Shift (modulate) each of the three voice channels to a different bandwidth, as shown in Figure 6.6.


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Example 1 (cont’d)Example 1 (cont’d)


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Example 2Example 2

Example 2Example 2

Five channels, each with a 100-KHz bandwidth, are to be multiplexed together. What is the minimum bandwidth of the link if there is a need for a guard band of 10 KHz between the channels to prevent interference?

SolutionSolution

For five channels, we need at least four guard bands. This means that the required bandwidth is at least 5 x 100 + 4 x 10 = 540 KHz, as shown in Figure 6.7.


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Example 3Example 3Example 3Example 3

Four data channels (digital), each transmitting at 1 Mbps, use a satellite channel of 1 MHz. Design an appropriate configuration using FDM

SolutionSolution

The satellite channel is analog. We divide it into four channels, each channel having a 250-KHz bandwidth. Each digital channel of 1 Mbps is modulated such that each 4 bits are modulated to 1 Hz. One solution is 16-QAM modulation. Figure 6.8 shows one possible configuration.


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Example : Cable Televisioncoaxial cable has a bandwidth of approximately 500Mhzindividual television channel require about 6Mhz of

bandwidth for transmissioncan carry 83 channels theoretically


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Analog HierarchyAnalog Hierarchy

To maximize the efficiency of their infrastructure, telephone companies have traditionally multiplexed signals from lower bandwidth lines onto higher bandwidth lines.


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Applications of FDMApplications of FDM

Example 4Example 4The Advanced Mobile Phone System (AMPS) uses two bands. The first band, 824 to 849 MHz, is used for sending; and 869 to 894 MHz is used for receiving. Each user has a bandwidth of 30 KHz in each direction. The 3-KHz voice is modulated using FM, creating 30 KHz of modulated signal. How many people can use their cellular phones simultaneously?

SolutionSolutionEach band is 25 MHz. If we divide 25 MHz into 30 KHz, we get 833.33. In reality, the band is divided into 832 channels.


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6.2 Wave Division Multiplexing (WDM)6.2 Wave Division Multiplexing (WDM)

WDM is conceptually same as FDM except that the multiplexing and demultiplexing involve li

ght signals transmitted through fiber-optic channels


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WDM (cont’d)WDM (cont’d)

WDM is an analog multiplexing technique to combine optical signals.

Very narrow bands of light from different sources are combined to make a wider band of light


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WDM (cont’d)WDM (cont’d)

Combining and splitting of light sources are easily handled by a prism Prism bends a beam of light based on the angle of

incidence and the frequency. One application is the SONET.


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6.3 TDM6.3 TDM

TDM(Time-Division Multiplexing) is a digital process that can be applied when the data

rate capacity of the transmission medium is greater than the data rate required by the sending and receiving device

TDM is a digital multiplexing technique to combine data.


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TDM(cont’d)TDM(cont’d)


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TDM (cont’d)TDM (cont’d)

In a TDM, the data rate of the link is n times faster, and the unit duration is n times shorter.In a TDM, the data rate of the link is n times faster, and the unit duration is n times shorter.


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TDM (cont’d)TDM (cont’d)

Example 5Example 5Four 1-Kbps connections are multiplexed together. A unit is 1 bit. Find (1) the duration of 1 bit before multiplexing, (2) the transmission rate of the link, (3) the duration of a time slot, and (4) the duration of a frame?

SolutionSolutionWe can answer the questions as follows:

1. The duration of 1 bit is 1/1 Kbps, or 0.001 s (1 ms).2. The rate of the link is 4 Kbps.3. The duration of each time slot 1/4 ms or 250 s. 4. The duration of a frame 1 ms.


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TDM can be implemented in two ways

Synchronous TDM

Asynchronous TDM


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Synchronous TDMthe multiplexer allocates exactly the same time slot to

each device at all times, whether or not a device has anything to transmit.


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FrameTime slots are grouped into framesA frame consists of one complete cycle of time slots,

including one or more slots dedicated to each sending device, plus framing bits.


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Synchronous TDM


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TDM(cont’d)TDM(cont’d)Interleaving

synchronous TDM can be compared to a very fast rotating switch

switch moves from device to device at a constant rate and in a fixed order

6 empty slots out of 24 are being wasted


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Demultiplexer decomposes each frame by discarding the framing bits and extracting each character in turn

Synchronous TDM, demultiplexing process


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Example 6Example 6Four channels are multiplexed using TDM. If each channel sends 100 bytes/s and we multiplex 1 byte per channel, show the frame traveling on the link, the size of the frame, the duration of a frame, the frame rate, and the bit rate for the link. SolutionSolution


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Example 7Example 7

A multiplexer combines four 100-Kbps channels using a time slot of 2 bits. Show the output with four arbitrary inputs. What is the frame rate? What is the frame duration? What is the bit rate? What is the bit duration?

SolutionSolution

Figure 6.16 shows the output for four arbitrary inputs.


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Framing bits

~ allows the demultiplexer to synchronize with the incoming stream so that it can separate the time slots accurately

(ex: 01010101 ….)


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Synchronous TDM Example4 characters + 1 framing bit


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Asynchronous TDM : statistical time-division multiplexing

Synchronous or Asynchronous : Not flexible or Flexible


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Examples of asynchronous TDM frames

a. Case 1: Only three lines sending data


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b. Case 2: Only four lines sending data


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c. Case 3: All five lines sending data


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Multiplexing application(cont’d)Multiplexing application(cont’d)

Analog HierarchyTo maximize the efficiency of their infrastructure,

telephone companies have traditionally multiplexed signals from lower bandwidth lines onto higher bandwidth lines.


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Digital Servicesadvantage

- less sensitive than analog service to noise- lower cost


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Multiplexing application(cont’d)Multiplexing application(cont’d)DS(Digital Signal) Service

~ is a hierarchy of digital signal


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DS Service

DS-0 : single digital channel of 64Kbps

DS-1 : 1,544Mbps, 24 개의 64Kbps + 8Kbps 의 overhead

DS-2 : 6,312Mbps, 96 개의 64Kbps+168Kbps 의 overhead

DS-3 : 44,376Mbps, 672 개의 64Kbps+1.368Mbps 의 overhead

DS-4 : 274,176Mbps,4032 개의 64Kbps+16.128Mbps 의 overhead


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T Lines

Service Line Rate(Mbps) Voice Channels

DS-1

DS-2

DS-3

DS-4

T-1

T-2

T-3

T-4

1,544

6,312

44,736

274,176

24

96

672

4032


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T Lines for Analog Transmission

T-1 line for multiplexing telephone lines


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T-1 FrameT-1 Frame

193 = 24 x 8 + 1(1 bit for synchronization)


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E LinesE Lines E line rates

E LineRate

(Mbps)Voice

Channels

E-1E-1 2.0482.048 3030

E-2E-2 8.4488.448 120120

E-3E-3 34.36834.368 480480

E-4E-4 139.264139.264 19201920


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Inverse Multiplexingtakes the data stream from one high-speed line and

breaks it into portion that can be sent across several lower speed lines simultaneously, with no loss in the collective data rate


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Multiplexing and inverse multiplexing


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Why do we need inverse multiplexing ?wants to send data, voice, and video each of which

requires a different data rate.

[example]voice - 64 Kbps linkdata - 128 Kbps linkvideo - 1,544 Mbps link

Multiplexingnetworking.khu.ac.kr/.../Chapter6(Multiplexing).ppt · PPT file · Web view2015-06-12 · Multiplexing Prof. Choong Seon HONG 6 장 다중화(Multiplexing) Dividing a

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