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Part II. Physical Layer and Media

Part II. Physical Layer and Media. Chapter 6. Bandwidth Utilization: Multiplexing. COMP 3270 Computer Networks Computing Science Thompson Rivers University. Learning Objectives. Use of Frequency-Division Multiplexing to transfer multiple channels.

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Part II. Physical Layer and Media

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  1. Part II. Physical Layer and Media Chapter 6. Bandwidth Utilization: Multiplexing COMP 3270 Computer Networks Computing Science Thompson Rivers University

  2. Learning Objectives • Use of Frequency-Division Multiplexing to transfer multiple channels. • Use of Synchronous Time-Division Multiplexing to transfer multiple channels, with or without synchronization bit. • Use of Synchronous Time-Division Multiplexing to transfer multiple channels, when the inputs have different transmission rates.

  3. Bandwidth utilization is the wise use of available bandwidth to achieve specific goals for multiple data channels. Efficiency can be achieved by multiplexing;

  4. 1. MULTIPLEXING Whenever the bandwidth of a medium linking two devices is greater than the bandwidth needs of the devices, the link can be shared. Multiplexing is the set of techniques that allows the simultaneous transmission of multiple signals across a single data link. As data and telecommunications use increases, so does traffic. Topics discussed in this section: • Frequency-Division Multiplexing (FDM) • Wavelength-Division Multiplexing (WDM) • Synchronous Time-Division Multiplexing (TDM)

  5. Dividing a link into channels ☺ Examples? - Protocol stack - T1 line - Wireless communications, including cell phones - Telephone line for voice and ADSL services,

  6. Categories of multiplexing

  7. Frequency Division Multiplexing (FDM)

  8. Different frequency bands

  9. FDM is an analog multiplexing technique that combines analog signals.

  10. FDM process ??? Different frequency bands

  11. FDM demultiplexing example ???

  12. Example: 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. Assume there are no guard bands. Solution:

  13. Not time difference

  14. Example: 5 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:

  15. Example: 4 data channels (digital), each transmitting at 1 Mbps, use a satellite channel of 1 MHz. Design an appropriate configuration, using FDM. Solution: How much bandwidth can each channel use? What is transmission rate of each channel? What kind of modulation can be used? n-QAM?

  16. Bandwidth, not carrier signals The minimum required bandwidth of QAM is equal to …

  17. Example: The Advanced Mobile Phone System (AMPS) uses two bands. The first band of 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. How many people can use their cellular phones simultaneously? • Solution: • Each band is 25 MHz. If we divide 25 MHz by 30 kHz, we get 833.33. In reality, • the band is divided into 832 channels. • Of these, 42 channels are used for control, • which means only 790 channels are available for cellular phone users.

  18. Wavelength-division multiplexing (WDM) Conceptually equal to FDM Not time difference Optical fiber

  19. Prisms in wavelength-division multiplexing and demultiplexing

  20. Up to now • Multiplexing • FDM • ☺ Advantage? • Simple • ☺ Disadvantage? • Inefficient • Possible idle channels • WDM • TDM

  21. Time Division Multiplexing (TDM) One logical channel frame One frequency band is used for all the data channels.

  22. TDM is a digital multiplexing technique for combining several low-rate channels into one high-rate one.

  23. Synchronous time-division multiplexing The data rate can be smaller than the sum of all the input rates. True or false? Is the frame rate equal to the data rate of an input channel?

  24. empty Each frame duration is 3T. How much is the data rate of the output link faster than the data rate of an input channel?

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

  26. Example: Four 1 Mbps inputs and one data stream for the output. The unit of data is 1 bit. Find (a) the output frame rate, and (b) the output bit rate. Solution:

  27. Example: 4 1-Kbps connections are multiplexed together. A unit is 1 bit. Find the transmission rate of the link. Solution:

  28. Example: 4 channels are multiplexed using TDM. If each channel sends 800 bps and we multiplex 1 byte per channel, show the frame traveling on the link, the size of the frame, the frame rate, and the bit rate for the link. Solution:

  29. Example: A multiplexer combines 4 100-Kbps channels using a time slot of 2 bits. Show the output with 4 arbitrary inputs. What is the frame rate? What is the bit rate? Solution:

  30. ☺ What if input data rates are different?

  31. Frame Synchronizing For framing, synchronization between the multiplexer and demultiplexer is a major issue.

  32. Framing bits

  33. Example: We have 4 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, (3) the frame rate, (5) the number of bits in each frame, and (6) the data rate of the link. A frame of length 4B + 1b Solution:

  34. Example: 2 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 bit rate of the link? A frame of length 3b Solution:

  35. T-1 line for multiplexing telephone lines ☺ Data unit: 8 bits, 1 synchronization bit => frame rate, bit rate, …?

  36. T-1 frame structure

  37. Multiplexing and inverse multiplexing ☺ Example? - data communication of telephone networks - ADSL - OFDM

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