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Net 222: Communications and networks fundamentals ( Practical Part)

Net 222: Communications and networks fundamentals ( Practical Part). Tutorial 5 : Chapter 8 Data & computer communications. Chapter 8 pages 271 ( Data & computer communications). Problems: 8.1 (a) 8.9 8.11(b) 8.13. Question.

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Net 222: Communications and networks fundamentals ( Practical Part)

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  1. Net 222: Communications and networks fundamentals (Practical Part) Tutorial 5: Chapter 8 Data &computer communications Networks and Communication Department

  2. Chapter 8 pages 271 (Data & computer communications) • Problems: • 8.1 (a) • 8.9 • 8.11(b) • 8.13 Networks and Communication Department

  3. Question • 8.1 The information in four analog signals is to be multiplexed and transmitted over a telephone channel that has a 400- to 3100-Hz bandpass. Each of the analog baseband signals is bandlimited to 500 Hz. Design a communication system (block diagram) that will allow the transmission of these four sources over the telephone channel using • a. Frequency division multiplexing Show the block diagrams of the complete system, including the transmission, channel, and reception portions. Include the bandwidths of the signals at the various points in the systems. Networks and Communication Department

  4. Answer Networks and Communication Department

  5. Question • 8.9 Twenty-four voice signals are to be multiplexed and transmitted over twisted pair. What is the bandwidth required for FDM? • NOTE: Assume that the voice signal frequency is 4 KHz. Networks and Communication Department

  6. Answer • The required bandwidth for FDM is 24 × 4 = 96 kHz. Networks and Communication Department

  7. Question • 8.11 A time division multiplexer is used to combine the data streams of a number of terminals, each terminal needs 110-bps for data transmission over a 2400-bps digital line. Assume that at least 3% of the line capacity is reserved for some uses. • b. Determine the number of terminals that can be accommodated by the multiplexer. Networks and Communication Department

  8. Answer • Available capacity = 2400 bps - 3% =2400 – (2400 - 3%) = 2328 bps OR 2400 – (2400 – 3/100)= 2328 bps • Number of terminals =2328/110=21.6=21 terminals Networks and Communication Department

  9. Question • 8.13Ten 9600-bps lines are to be multiplexed using TDM. What is the total capacity required for synchronous TDM? Networks and Communication Department

  10. Answer • Total capacity = 9600 bps × 10 =96000 bps =96 kbps Networks and Communication Department

  11. The End Any Questions ? Networks and Communication Department

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