Ch 06

Ch 06

• 1.
  Chapter 6 Multiplexing
• 2.
  Figure 6.1 Dividing a link into channels
• 3.
  Figure 6.2 Categories of multiplexing
• 4.
  6.1 FDMMultiplexing Process Demultiplexing Process The Analog Hierarchy Other Applications of FDM Implementation
• 5.
  Figure 6.3 FDM
• 6.
  FDM is ananalog multiplexing technique that combines signals. Note :
• 7.
  Figure 6.4 FDM process
• 8.
  Figure 6.5 FDM demultiplexing example
• 9.
  Example 1 Assumethat 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.
• 10.
  Figure 6.6 Example 1
• 11.
  Example 2 Fivechannels, 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.
• 12.
  Figure 6.7 Example 2
• 13.
  Example 3 Fourdata 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.
• 14.
  Figure 6.8 Example 3
• 15.
  Figure 6.9 Analog hierarchy
• 16.
  Example 4 TheAdvanced 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.
• 17.
  6.2 WDMWave Division Multiplexing
• 18.
  Figure 6.10 WDM
• 19.
  WDM is ananalog multiplexing technique to combine optical signals. Note:
• 20.
  Figure 6.11 Prisms in WDM multiplexing and demultiplexing
• 21.
  6.3 TDMTime Slots and Frames Interleaving Synchronizing Bit Padding Digital Signal (DS) Service T Lines Inverse TDM More TDM Applications
• 22.
  Figure 6.12 TDM
• 23.
  TDM is adigital multiplexing technique to combine data. Note:
• 24.
  Figure 6.13 TDM frames
• 25.
  Example 5 Four1-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  s. 4. The duration of a frame 1 ms.
• 26.
  In a TDM,the data rate of the link is n times faster, and the unit duration is n times shorter. Note:
• 27.
  Figure 6.14 Interleaving
• 28.
  Example 6 Fourchannels 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 6.15.
• 29.
  Figure 6.15 Example 6
• 30.
  Example 7 Amultiplexer 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.
• 31.
  Figure 6.16 Example 7
• 32.
  Figure 6.17 Framing bits
• 33.
  Example 8 Wehave 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. Solution See next slide.
• 34.
  Solution (continued) Wecan 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.
• 35.
  Example 9 Twochannels, 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.
• 36.
  Figure 6.18 DS hierarchy
• 37.
  Table 6.1 DS and T lines rates 4032 274.176 T-4 DS-4 T-3 T-2 T-1 Line 672 44.736 DS-3 96 6.312 DS-2 24 1.544 DS-1 Voice Channels Rate (Mbps) Service
• 38.
  Figure 6.19 T-1 line for multiplexing telephone lines
• 39.
  Figure 6.20 T-1 frame structure
• 40.
  Table 6.2 E line rates 1920 139.264 E-4 34.368 8.448 2.048 Rate (Mbps) 480 E-3 120 E-2 30 E-1 Voice Channels E Line
• 41.
  Figure 6.21 Multiplexing and inverse multiplexing