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Multiplexing

Multiplexing. 6 장 다중화 (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. 다중화 (Multiplexing). 다중화 ( Multiplexing )

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Multiplexing

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  1. Multiplexing

  2. 6 장 다중화(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.

  3. 다중화(Multiplexing) • 다중화(Multiplexing) • is the set of techniques that allows the simultaneous transmission of multiple signals across a single data link.

  4. 6.1 FDM • 다중화기(Multiplexer) • transmission streams combine into a single stream(many to one) • 역다중화기(Demultiplexer) • stream separates into its component transmission(one to many) and directs them to their intended receiving devices

  5. 8.2 Many to One/One to Many • Categories of Multiplexing

  6. FDM • 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

  7. 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)

  8. FDM (cont’d) • FDM multiplexing process, time-domain

  9. FDM(cont’d) • FDM multiplexing process, frequency-domain

  10. FDM(cont’d) • Demultiplexing • separates the individual signals from their carries and passes them to the waiting receivers.

  11. FDM(cont’d) • FDM demultiplexing process, time-domain

  12. FDM(cont’d) • FDM demultiplexing, frequency-domain

  13. Example 1 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 6.6.

  14. Example 1 (cont’d)

  15. Example 2 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 6.7.

  16. Example 2 (cont’d)

  17. Example 3 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.

  18. Example 3 (cont’d)

  19. FDM(cont’d) • Example : Cable Television • coaxial cable has a bandwidth of approximately 500Mhz • individual television channel require about 6Mhz of bandwidth for transmission • can carry 83 channels theoretically

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

  21. Applications of FDM 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.

  22. 6.2 Wave Division Multiplexing (WDM) • WDM is conceptually same as FDM • except that the multiplexing and demultiplexing involve light signals transmitted through fiber-optic channels

  23. 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

  24. 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.

  25. 6.3 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.

  26. TDM(cont’d)

  27. TDM (cont’d) In a TDM, the data rate of the link is n times faster, and the unit duration is n times shorter.

  28. TDM (cont’d) 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.

  29. TDM(cont’d) • TDM can be implemented in two ways • Synchronous TDM • Asynchronous TDM

  30. TDM(cont’d) • Synchronous TDM • the multiplexer allocates exactly the same time slot to each device at all times, whether or not a device has anything to transmit.

  31. TDM(cont’d) • Frame • Time slots are grouped into frames • A frame consists of one complete cycle of time slots, including one or more slots dedicated to each sending device, plus framing bits.

  32. TDM(cont’d) • Synchronous TDM

  33. 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

  34. TDM(cont’d) • Demultiplexer decomposes each frame by discarding the framing bits and extracting each character in turn • Synchronous TDM, demultiplexing process

  35. TDM(cont’d)

  36. TDM(cont’d) 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

  37. TDM(cont’d) Example 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? Solution Figure 6.16 shows the output for four arbitrary inputs.

  38. TDM(cont’d)

  39. TDM(cont’d) • Framing bits ~ allows the demultiplexer to synchronize with the incoming stream so that it can separate the time slots accurately (ex: 01010101 ….)

  40. TDM(cont’d) • Synchronous TDM Example 4 characters + 1 framing bit

  41. TDM(cont’d) • Asynchronous TDM : statistical time-division multiplexing Synchronous or Asynchronous : Not flexible or Flexible

  42. TDM(cont’d) • Examples of asynchronous TDM frames a. Case 1: Only three lines sending data

  43. TDM(cont’d) b. Case 2: Only four lines sending data

  44. TDM(cont’d) c. Case 3: All five lines sending data

  45. Multiplexing application(cont’d) • Analog Hierarchy • To maximize the efficiency of their infrastructure, telephone companies have traditionally multiplexed signals from lower bandwidth lines onto higher bandwidth lines.

  46. Multiplexing application(cont’d) • Digital Services • advantage - less sensitive than analog service to noise - lower cost

  47. Multiplexing application(cont’d) • DS(Digital Signal) Service ~ is a hierarchy of digital signal

  48. Multiplexing application(cont’d) • DS Service • DS-0 : single digital channel of 64Kbps • DS-1 : 1,544Mbps, 24개의 64Kbps + 8Kbps의 overhead • DS-2 : 6,312Mbps, 96개의 64Kbps+168Kbps의 overhead • DS-3 : 44,376Mbps, 672개의 64Kbps+1.368Mbps의 overhead • DS-4 : 274,176Mbps,4032개의 64Kbps+16.128Mbps의 overhead

  49. 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 Multiplexing application(cont’d) • T Lines

  50. T Lines for Analog Transmission • T-1 line for multiplexing telephone lines

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