Data Communication and Networks. Multiplexing. Prof. Dr. Abdul Qadeer Khan Rajput

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Transcription:

Data Communication and Networks Multiplexing By Prof. Dr. Abdul Qadeer Khan Rajput

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Multiplexing Multiplexing is a set of techniques that allows the simultaneous transmission of multiple signals across a single data link. Multiplexer combines the incoming lines into one [single stream]. De-multiplexer (DEMUX) separates the stream back into its component transmissions (one to many) and directs them to their corresponding lines. 4

Basic Definitions Multiplexer(MUX): device to combine signals to one composite signal. Demultiplexer(DEMUX): device to separate the composite signal into signal components Path: refers to the physical link. Channel: refers to a portion of a path. 5

Types of Multiplexing There are two basic types of multiplexing in use: Frequency Division Multiplexing (FDM) & Time Division Multiplexing (TDM). Wavelength Division Multiplexing (WDM) is one special kinds of FDM used in fiber optical transmission. TDM can be further divided into synchronous TDM and asynchronous TDM. 6

Categories of multiplexing 7

FDM Definition Frequency Division Multiplexing (FDM) is the technique to combine signals from different bands. Each signal modulates a separate carrier frequency. 8

FDM Multiplexing time domain 9

FDM Demultiplexing time domain 10

FDM Multiplexing frequency domain 11

FDM Demultiplexing frequency domain 12

Concept of Guard Band To prevent channel overlapping, they must be separated by unused bandwidth called Guard Band. 13

Figure Dividing a link into channels 14

Frequency Division Multiplexing Figure FDM Frequency-division multiplexing is an analog technique that can be applied when the bandwidth of a link (in hertz) is greater than the combined bandwidths of the signals to be transmitted. Signals generated by each sending device modulate different carrier frequencies. FDM is an analog multiplexing technique that combines signals 15

Frequency Division Multiplexing FDM Useful bandwidth of medium exceeds required bandwidth of channel Each signal is modulated to a different carrier frequency Carrier frequencies separated so signals do not overlap (guard bands) e.g. broadcast radio Channel allocated even if no data 16

Figure FDM process 17

FDM System 18

FDM Example 19

Analog Carrier Systems 20

Example 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. Solution Shift (modulate) each of the three voice channels to a different bandwidth, as shown in Figure. 21

Figure Example 1 22

Example 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? Solution 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. 23

Figure Example 2 24

Example 3 Four data channels (digital), each transmitting at 1 Mbps, use a satellite channel of 1 MHz. Design an appropriate configuration using FDM Solution 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. 25

Figure Example 3 26

In analog hierarchy, 12 voice channels are multiplexed onto a higher-bandwidth line to create a group. The join them as group, supergroup, master group and jumbo group. 27

Analog hierarchy Figure Analog hierarchy 28

Example 4 The 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? Solution Each 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. 29

Example 5 Four 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? Solution We 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 ms. 4. The duration of a frame 1 ms. 30

TDM Definition Time Division Multiplex (TDM) is the process of combining signals together in the time domain. 31

TDM Basics 3T The basic principle of TDM is Sampling Theorem. The basic T unit in TDM is Time Slot. The data in TDM is grouped into Frames, which consists of one cycle of time slots. 32

TDM TDM is a digital multiplexing technique to combine data Instead of sharing a portion of the bandwidth as in FDM, time is shared. Each connection occupies a portion of time in the link. In a TDM, the data rate of the link is n times faster, and the unit duration is n times shorter 33

Interleaving Interleaving is the process of multiplexing. In TDM, synchronization between the sender and receiver is very important. 34

Interleaving TDM can be visualized as two fast rotating switches, one on the multiplexing side and the other on the demultiplexing side. The switches are synchronized and rotate at the same speed, but in opposite directions. On the multiplexing side, as the switch opens in front of a connection, that connection has the opportunity to send a unit onto the path. This process is called interleaving. 35

Example 6 Four 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. Solution The multiplexer is shown in Figure. 36

T-1 line for multiplexing telephone lines T lines are digital lines designed for the transmission of digital data, audio, or video. T lines also can be used for analog transmission (regular telephone connections), provided the analog signals are sampled first, then timedivision multiplexed. 37

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? Solution Example 7 Figure shows the output for four arbitrary inputs. 38

Framing bits For synchronization between multiplexer and demultiplexer, one or more synchronization bits are usually added to the beginning of each frame. These bits, called framing bits, follow a pattern, frame to frame, that allows the demultiplexer to synchronize with the incoming stream so that it can separate the time slots accurately. In bit padding, the multiplexer adds extra bits to a device s source stream to force the speed relationships among the various devices into integer multiples of each other. 39

Example 8 We have four sources, each creating 250 characters per second. If the interleaved unit is a character and 1 synchronizing bit is added to each frame, find (1) the data rate of each source, (2) the duration of each character in each source, (3) the frame rate, (4) the duration of each frame, (5) the number of bits in each frame, and (6) the data rate of the link. 40

Solution We can answer the questions as follows: 1. The data rate of each source is 2000 bps = 2 Kbps. 2. The duration of a character is 1/250 s, or 4 ms. 3. The link needs to send 250 frames per second. 4. The duration of each frame is 1/250 s, or 4 ms. 5. Each frame is 4 x 8 + 1 = 33 bits. 6. The data rate of the link is 250 x 33, or 8250 bps. 41

Example 9 Two channels, one with a bit rate of 100 Kbps and another with a bit rate of 200 Kbps, are to be multiplexed. How this can be achieved? What is the frame rate? What is the frame duration? What is the bit rate of the link? Solution We can allocate one slot to the first channel and two slots to the second channel. Each frame carries 3 bits. The frame rate is 100,000 frames per second because it carries 1 bit from the first channel. The frame duration is 1/100,000 s, or 10 ms. The bit rate is 100,000 frames/s x 3 bits/frame, or 300 Kbps. 42

Digital Signal (DS) hierarchy DS-0 service is a single digital channel of 64 Kbps. DS-1 is a 1.544-Mbps service [24 times 64 Kbps plus 8Kbps of overhead] And so on. DS-0, DS-1, and so on are the names of services. To implement those services, the telephone companies use T lines (T-1 to T-4). 43

Table DS and T lines rates Service Line Rate (Mbps) Voice Channels DS-1 T-1 1.544 24 DS-2 T-2 6.312 96 DS-3 T-3 44.736 672 DS-4 T-4 274.176 4032 44

Europeans use a version of T lines called E lines The two systems are conceptually identical, but their capacities differ. E Line Rate (Mbps) Voice Channels E-1 2.048 30 E-2 8.448 120 E-3 34.368 480 Table. E line rates E-4 139.264 1920 45

Synchronous TDM In synchronous TDM, each source is preassigned a fixed location of time slot. Each source can and only can send information at the time slot given to it. If a source has no data to send, its time slot remains empty. This can lead to inefficiency. If n sources are grouped together, the total data rate of the path is n times the original data rate of each source. 46

Synchronous TDM Example 47

Synchronous TDM - Multiplexing 48

Synchronous TDM - Demultiplexing 49

TDM Example We have four sources, each creating 250 characters per second. If the interleaved unit is a character and 1 synchronizing bit is added to each frame, find (1) the data rate of each source, (2) the duration of each character in each source, (3) the frame rate, (4) the duration of each frame, (5) the number of bits in each frame, and (6) the data rate of the link. 50

Solution We can answer the questions as follows: 1. The data rate of each source is (250 chars x 8 bits) = 2000 bps = 2 kbps. 2. The duration of a character is 1/250 s, or 4 ms. 3. The link needs to send 250 frames per second. 4. The duration of each frame is 1/250 s, or 4 ms. 5. Each frame is (4sources x 8bits) + 1sync = 33 bits. 6. The data rate of the link is 250 x 33, or 8250 bps. (strictly, bit rate of the link is 8250bps the actual data rate is only 8000 data bits/s) 51

Asynchronous TDM In asynchronous TDM or statistical TDM, only sources containing data will be sent with time slot. Therefore, asynchronous TDM can avoid bandwidth waste in synchronous TDM. But, in order to distinguish data from different sources, address should be added into the frame structure, increasing the overhead of the transmission. 52

Asynchronous TDM Example 53

Address and Overhead As shown below, address is added before the data from each source. It is practical only when the data size for each time slot is relatively larger than the address. In this example, the addressing information takes up as much space as the actual data - inefficient 54

TDM in Telephone System FDM was used in the original telephone system, but now the telephone lines (except the subscriber line) are all in digital form, so FDM is not in use now. There are two types of TDM used in telephone system. In US, T-1 line with basic rate of 1.544 Mbps is used, while in Europe and China, E-1 Line of 2.048 Mbps is in use. DS is a service, T-n is a line 55

Digital Signal (DS) Hierarchy in US 8k sample/sec x 8bit samples gives 64kbps basic line DS1 = 24 x DS0 = 24x64k=1536k 1536k + 8k(sync) =1544k 56

T-1 Frame Structure Data sync 57

Sync bit One frame = 24 * 8 + 1 sync bit Frame rate = 193 * 8000 = 1.544 Mbps 58