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4123702 Data Communications System

4123702 Data Communications System. By Ajarn Preecha Pangsuban. Chapter 5 – Analog Transmission. Analog Signals. Infinite number of values in a range More susceptible to noise and is less precise than a digital signal

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4123702 Data Communications System

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  1. 4123702Data Communications System By Ajarn Preecha Pangsuban

  2. Chapter 5 – Analog Transmission

  3. Analog Signals • Infinite number of values in a range • More susceptible to noise and is less precise than a digital signal • Since they are more variable than digital signals, they can convey greater subtleties 4123702 Data Communications System @YRU

  4. Bit Rate and Baud Rate • Bit rate – number of bits transmitted in 1 second • Computer efficiency • Baud rate – number of signal units per second required to represent those bits; equal to or less than bit rate • Network transmission efficiency • Determines bandwidth required 4123702 Data Communications System @YRU

  5. Example 1 An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, find the baud rate and the bit rate. Baud rate = 1000 bauds per second (baud/s) Bit rate = 1000 x 4 = 4000 bps Solution 4123702 Data Communications System @YRU

  6. Example 2 The bit rate of a signal is 3000. If each signal unit carries 6 bits, what is the baud rate? Baud rate = bit rate/bits per signal = 3000 / 6 = 500 baud/s Solution 4123702 Data Communications System @YRU

  7. Modulation of Digital Data • Carrier Signal - High-frequency signal used for digital-to-analog or analog-to-digital modulation • May change any aspect of an analog signal to represent digital data: amplitude, frequency, and phase • Modulation - process of changing one of the characteristics of an analog signals based on information in digital signal 4123702 Data Communications System @YRU

  8. Digital-to-Analog Conversion 4123702 Data Communications System @YRU

  9. Methods • Three mechanisms • Amplitude shift keying (ASK) • Frequency shift keying (FSK) • Phase shift keying (PSK) • Combined approach: quadrature amplitude modulation (QAM) 4123702 Data Communications System @YRU

  10. Amplitude Shift Keying (ASK) • Strength of carrier signal is varied to represent a 1 or 0 • Frequency and phase remain constant while amplitude is changed • Amplitude remains constant during bit duration • Highly susceptible to noise interference since noise usually affects amplitude • Minimum bandwidth required is equal to baud rate 4123702 Data Communications System @YRU

  11. Amplitude Shift Keying (ASK) 4123702 Data Communications System @YRU

  12. Relationship between baud rate and bandwidth in ASK BW = (1+d)Nbaud 4123702 Data Communications System @YRU

  13. Example 3 Find the minimum bandwidth for an ASK signal transmitting at 2000 bps. The transmission mode is half-duplex. Solution In ASK the baud rate and bit rate are the same. The baud rate is therefore 2000. An ASK signal requires a minimum bandwidth equal to its baud rate. Therefore, the minimum bandwidth is 2000 Hz. 4123702 Data Communications System @YRU

  14. Example 4 Given a bandwidth of 5000 Hz for an ASK signal, what are the baud rate and bit rate? Solution In ASK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But because the baud rate and the bit rate are also the same for ASK, the bit rate is 5000 bps. 4123702 Data Communications System @YRU

  15. Example 5 Given a bandwidth of 10,000 Hz (1000 to 11,000 Hz), draw the full-duplex ASK diagram of the system. Find the carriers and the bandwidths in each direction. Assume there is no gap between the bands in the two directions. Solution For full-duplex ASK, the bandwidth for each direction is BW = 10000 / 2 = 5000 Hz The carrier frequencies can be chosen at the middle of each band (see Fig. 5.5). fc (forward) = 1000 + 5000/2 = 3500 Hz fc (backward) = 11000 – 5000/2 = 8500 Hz 4123702 Data Communications System @YRU

  16. Solution to Example 5 4123702 Data Communications System @YRU

  17. Frequency Shift Keying (FSK) • Frequency is varied to represent a 1 or 0 • Frequency during bit duration is constant • Amplitude and phase remain constant • Avoids most of noise problems of ASK; can ignore voltage spikes • Limited by physical capabilities of medium • Bandwidth required is equal to baud rate plus frequency shift 4123702 Data Communications System @YRU

  18. Frequency Shift Keying (FSK) 4123702 Data Communications System @YRU

  19. Relationship between baud rate and bandwidth in FSK BW = fc1-fc0+Nbaud 4123702 Data Communications System @YRU

  20. Example 6 Find the minimum bandwidth for an FSK signal transmitting at 2000 bps. Transmission is in half-duplex mode, and the carriers are separated by 3000 Hz. Solution For FSK BW = baud rate + fc1- fc0 BW = bit rate + fc1 - fc0 = 2000 + 3000 = 5000 Hz 4123702 Data Communications System @YRU

  21. Example 7 Find the maximum bit rates for an FSK signal if the bandwidth of the medium is 12,000 Hz and the difference between the two carriers is 2000 Hz. Transmission is in full-duplex mode. Solution • Because the transmission is full duplex, only 6000 Hz is allocated for each direction. • BW = baud rate + fc1 - fc0 • Baud rate = BW - (fc1 - fc0 ) = 6000 - 2000 = 4000 • But because the baud rate is the same as the bit rate, the bit rate is 4000 bps. 4123702 Data Communications System @YRU

  22. Phase Shift Keying (PSK) • Phase of carrier is varied to represent 1 or 0 • Peak amplitude and frequency remain constant • Phase remains constant during each bit duration • Not as susceptible to noise degradation as ASK or to bandwidth limitations of FSK 4123702 Data Communications System @YRU

  23. Phase Shift Keying (PSK) 2-PSK or Binary PSK 4123702 Data Communications System @YRU

  24. PSK constellation (Phase state diagram) Constellationis a graphical representation of the phase and amplitude of different bit combinations in digital-to-analog modulation. 4123702 Data Communications System @YRU

  25. 4-PSK; 8-PSK • 4-PSK • May utilize four variations of phase shift by 90 degrees • Each phase shift represents 2 bits (dibit); technique is referred to as 4-PSK • Allows data transmission two times as fast as 2-PSK • 8-PSK • Vary signal by shifts of 45 degrees; each shift may then represent three bits (tribit) and send data three times as fast 4123702 Data Communications System @YRU

  26. The 4-PSK method (Q-PSK) 4123702 Data Communications System @YRU

  27. The 4-PSK characteristics 4123702 Data Communications System @YRU

  28. The 8-PSK characteristics 4123702 Data Communications System @YRU

  29. Relationship between baud rate and bandwidth in PSK BW = (1+d)Nbaud 4123702 Data Communications System @YRU

  30. Example 8 Find the bandwidth for a 4-PSK signal transmitting at 2000 bps. Transmission is in half-duplex mode. Solution For 4-PSK the baud rate is one –half of the bit rate (1 baud : 2 bit). The baud rate is therefore 1000. A PSK signal requires a bandwidth equal to its baud rate . Therefore, the bandwidth is 1000 Hz 4123702 Data Communications System @YRU

  31. Example 9 Given a bandwidth of 5000 Hz for an 8-PSK signal, what are the baud rate and bit rate? Solution For PSK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps. 4123702 Data Communications System @YRU

  32. Quadrature Amplitude Modulation (QAM) • Combination of ASK and PSK so that a maximum contrast between each signal unit (bit, dibit, tribit, and so on) is achieved • x number of variations in phase and y variations in amplitude • Number of phase shifts is always larger than number of amplitude shifts due to amplitude susceptibility to noise • QAM is therefore less susceptible to noise than ASK • Same bandwidth is required for ASK and PSK 4123702 Data Communications System @YRU

  33. The 4-QAM and 8-QAM constellations 4123702 Data Communications System @YRU

  34. Time domain for an 8-QAM signal 4123702 Data Communications System @YRU

  35. 16-QAM constellations 4123702 Data Communications System @YRU

  36. Bit/Baud Comparison 4123702 Data Communications System @YRU

  37. Bit and baud rate comparison 4123702 Data Communications System @YRU

  38. Example 10 A constellation diagram consists of eight equally spaced points on a circle. If the bit rate is 4800 bps, what is the baud rate? Solution The constellation indicates 8-PSK with the points 45 degrees apart. Since 23 = 8, 3 bits are transmitted with each signal unit. Therefore, the baud rate is 4800/3 = 1600 baud 4123702 Data Communications System @YRU

  39. Example 11 Compute the bit rate for a 1000-baud 16-QAM signal. Solution A 16-QAM signal has 4 bits per signal unit since log216 = 4. Thus, (1000)(4) = 4000 bps 4123702 Data Communications System @YRU

  40. Example 12 Compute the baud rate for a 72,000-bps 64-QAM signal. Solution A 64-QAM signal has 6 bits per signal unit since log2 64 = 6. Thus, 72000/6 = 12,000 baud 4123702 Data Communications System @YRU

  41. 5.2 Telephone Modems • Traditional phone lines carry frequencies between 300 and 3300 Hz; bandwidth of 3000 Hz • All this range is used for transmitting voice • Data signals require higher degree of accuracy to ensure integrity (the edges of this range are not used) • Effective bandwidth for data transmission is 2400 Hz (between 600 and 3000 Hz) • MODEM is a composite word refer to signal modulator and signal demodulator • modulator – creates a band-pass analog signal from binary data • demodulator – recovers binary data from modulated signal 4123702 Data Communications System @YRU

  42. Telephone line bandwidth 4123702 Data Communications System @YRU

  43. Modulation/demodulation Telco is Telephone Company or Your local telephone service provider (such as TOT, TT&T) 4123702 Data Communications System @YRU

  44. Modem Standards • The most popular MODEMs available are based on the V-series standards by the ITU-T • V-series standards that define data transmission over telephone lines. (ITU-T : International Telecommunications Union-Telecommunication Standardization Sector) 4123702 Data Communications System @YRU

  45. V.32 (32-QAM) • A combined modulation and encoding technique called Trellis Coded Modulation (TCM) • Trellis is essentially QAM plus a redundant bit • A pentabit is transmitted (4 bit data stream and 1 extra bit ) 4123702 Data Communications System @YRU

  46. The V.32 constellation and bandwidth 4123702 Data Communications System @YRU

  47. V.32 bis (128-QAM) • The first of the ITU-T standards to support 14,400 bps transmission • Uses 128-QAM transmission (7 bit/baud with 1 bit for error control) • Data rate of 2400 baud is 2400*6 = 14,400 bps • Automatic speed adjustment depending on quality of line or signal 4123702 Data Communications System @YRU

  48. The V.32bis constellation and bandwidth 4123702 Data Communications System @YRU

  49. V.34 bis • For bit rate of 28,800 with a 960 point constellation to a bit rate 33,600 with 1664 point constellation 4123702 Data Communications System @YRU

  50. V.90 • V.90 MODEM with a bit rate 56,00bps, called 56 k MODEMs • The downloading rate is a maximum of 56kbps, while the uploading rate can be maximum of 33.6 kbps (asymmetric) • These MODEM may be used only if one party is using digital signaling (such as through an ISP) 4123702 Data Communications System @YRU

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