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COMMUNICATION TECHNOLOGY

COMMUNICATION TECHNOLOGY. by Shashi Bhushan School of Computer and Information Sciences. Q. Discuss the relationship between data rate and bandwidth. Answer: Direct relationship Higher data rate of signal, the greater the bandwidth.

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COMMUNICATION TECHNOLOGY

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  1. COMMUNICATION TECHNOLOGY by Shashi Bhushan School of Computer and Information Sciences

  2. Q. Discuss the relationship between data rate and bandwidth.

  3. Answer: • Direct relationship • Higher data rate of signal, the greater the bandwidth. • The greater the bandwidth of a signal the higher the data rate that can be transmitted using the signal.

  4. Q. What is the difference between bandwidth, baudrate and channel capacity?

  5. Answer: Bandwidth Bandwidth is an electrical characteristic of transmission line or circuit. It indicates the range of frequencies measured in Hz. (cycles per second) which can be successfully transmitted over the line.

  6. Baudrate Baudrate is the number of signal elements. This defines the signalling rate on the transmission line. A signal element is a discrete voltage, phase or frequency value.

  7. Channel Capacity Channel Capacity is the rate at which data can be transmitted over a given Communication Path or channel under given condition.

  8. In case of noise free channel, the limitation of data rate is simply the bandwidth of the channel.

  9. Measuring the Channel Capacity • Nyquist Law • Shannon’s Law

  10. Nyquist Law Given a bandwidth of w, the highest signal rate that can be carried is 2w.

  11. Nyquist Law If the signal to be transmitted are binary (two voltage levels) then the data rate that can be supported by whz is 2 wbps

  12. Nyquist Law Q. Suppose a bandwidth is 3100 hz then what is the channel capacity if (a) Signal to be transmitted are two voltage levels? (b) If the signal to be transmitted are 8 voltage levels?

  13. Nyquist Law • Answer: • For two voltage levels • C =2 * 3100 = 6200 bps • (b) For 8 voltage levels with multilevel signalling the Nyquist formulation becomes • C = 2 W log2M • M = digital levels = 8 • W = bandwidth

  14. Nyquist Law For multivoltage level C = 2 * 3100 * log28 = 2 * 3100 * 3 = 18,600 bps

  15. For a given bandwidth the data rate can be increased by increasing the number of different level of signals.

  16. Shannon’s Law • Used for noisy channel • C = W log2(1 + S/N) • S/N = Signal Power/Noise Power

  17. Shannon’s Law Q. Calculate a channel capacity of W = 3100 Hz, S/N =1000:1 or 30db Answer: C = 3100 log2 (1 +1000) = 30894 bps

  18. Shannon’s Law Observation: Data rate could be increased by either increasing signal strength or bandwidth. However, as the signal strength increases, so do the nonlinearities in the system; leading to an increase intermodulation noise.

  19. Shannon’s Law • More amplifiers needed for increasing S/N ratio. • Since noise is constant throughout the line the location of the amplifiers must be closely inserted.

  20. Shannon’s Law • Frequent spacing of amplifiers increases costs. • Amplifiers must be carefully designed to minimise the amount of noise that is amplified along with the signal.

  21. Transmission Impairements • Attenuation and attenuation distortions • Delay distortion • Noise

  22. Network Devices Methods for connecting to a network or an independently connected computer to another computer are: • Through a modem connection • Through an ISDN connection • Through a gateway • Through a bridge or router

  23. Modem Modem converts digital data into an analog form that can be transmitted over a standard telephone line

  24. Modem The term modem is a composite word that refers to the two functional entities that make up the device; a modulator and a demodulator.

  25. Modem Normally, telephone type connections are unsuitable for digital data as they have a limited bandwidth of 300 to 400 hz. A modem must then be used to convert the digital information into an analog form that is transmittable over the telephone line.

  26. Modem A modulator treats a digital signal as a series of 1s and 0s and so can transform it into a completely analog signal by using the digital to analog mechanisms of ASK, FSK, PSX & QAM.

  27. Modem • ASK manipulates amplitude • FSK manipulates frequency • PSK manipulates phase • QAM manipulates both phase and amplitude

  28. Theoretical Bit Rates for Modems

  29. Modem Standards Today, many of the most popular modems available are based on standards published by ITU-T.

  30. Routers • Examine the network address field to determine the best route for the packets • Supports several different types of network layer protocols

  31. Routers • Routers communicate with other routers to exchange routing information. • Most network O.S. have associated routing protocols to support the transfer of routing information.

  32. Routing Protocols • BGP (Border Gateway Protocol) • EGP (Exterior Gateway Protocol) • OSPF (Open Shortest Path First) • RIP (Routing Information Protocol)

  33. Routing Protocols • RIP – Most popular protocols in the past. • Distributed with UNIX.

  34. Routing Protocols • RIP is based on distance vector algorithm which measures the number of hops to the destination router. • Disadvantage – The smallest hops may not be the best route.

  35. Routing Protocols • With distance vector routing each router maintains a table by communicating with neighbouring routers. • Updating the table takes time when changes occur.

  36. Routing Protocols Link state algorithm is based, not only on hops, but on other parameters such as delay capacity, reliability and throughput.

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