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Analog Transmission and Digital Modulation

This chapter discusses the modulation of digital data onto analog signals, including amplitude shift keying (ASK), frequency shift keying (FSK), and phase shift keying (PSK).

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Analog Transmission and Digital Modulation

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  1. Chapter 5 Analog Transmission

  2. Analog Transmission • Digital data to Analog signal • Digital Modulation • Analog data to Analog signal • Analog Modulation

  3. 5.1 Modulation of Digital Data Digital-to-Analog Conversion Amplitude Shift Keying (ASK) Frequency Shift Keying (FSK) Phase Shift Keying (PSK) Quadrature Amplitude Modulation Bit/Baud Comparison

  4. Figure 5.1Digital-to-analog modulation

  5. Digital Modulation พาบิตข้อมูลไปบนชุดของสัญญาณอนาลอก (signal unit) สัญญาณอนาลอกที่ใช้พาข้อมูล สัญญาณพาหะ (Carrier signal) ส่วนใหญ่ที่ใช้เป็น sine wave ทำได้โดยเปลี่ยนแปลงคุณสมบัติของสัญญาณพาหะ Amplitude -> Amplitude Shift Keying (ASK) Frequency -> Frequency Shift Keying (FSK) Phase -> Phase Shift Keying (PSK) Digital Demodulation การถอดบิตข้อมูลจากชุดสัญญาณอนาลอกที่ได้รับ Digital-to-Analog Modulation(Digital Modulation)

  6. Bit rate (bps) the number of bits per second Baud rate (signal unit/sec: baud / sec: baud) the number of signal units per second less than or equal to the bit rate Bit Rate vs Baud Rate

  7. 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 Solution Baud rate = 1000 bauds per second (baud/s) Bit rate = 1000 x 4 = 4000 bps

  8. Example 2 The bit rate of a signal is 3000. If each signal unit carries 6 bits, what is the baud rate? Solution Baud rate = 3000 / 6 = 500 baud/s

  9. Bit representation Changing Amplitude of Career Signal One bit, One signal unit Ex ‘0’ -> A1 ‘1’ -> A2 Benefit Simple (normally used for fiber optic) Require Less Bandwidth Disadvantage Easily effected by noise Amplitude Shift Keying(ASK)

  10. Figure 5.3ASK Bit rate = baud rate

  11. Digital Data m X LPF X ASK Modulation 2 cosA cosB = cos (A + B) + cos (A - B) Decision making LPF X ASK Demodulation

  12. Figure 5.4Relationship between baud rate and bandwidth in ASK

  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.

  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.

  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

  16. Figure 5.5Solution to Example 5

  17. VCO • Bit representation • Changing Frequency of Career Signal • One bit, One signal unit • Ex ‘0’ -> f1 ‘1’ -> f2 • Benefit • Less effected by noise • Normally used in high frequency radio transmission or coaxial cable • Disadvantage • Require Large Bandwidth FSK Modulation

  18. Figure 5.6FSK Bit rate = baud rate

  19. Figure 5.7Relationship between baud rate and bandwidth in FSK

  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

  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.

  22. Bit representation Changing Phase of Career Signal One bit, One signal unit Ex ‘0’ -> Φ1 ‘1’ -> Φ2 Benefit Less effected by noise compared to ASK Normally used in MODEM Require Bandwidth less than FSK Disadvantage Difficult to detect phase shift in case of phase difference (Φ1- Φ2) is too small Phase Shift Keying(PSK)

  23. Figure 5.8PSK Bit rate = baud rate

  24. Figure 5.9PSK constellation

  25. Figure 5.10The 4-PSK method 4-PSK = 2n-PSK=22-PSK Bit rate = n x baud rate

  26. Figure 5.11The 4-PSK characteristics

  27. Figure 5.12The 8-PSK characteristics

  28. x X X BPSK Modulation

  29. LPF X 2 cosA cosB = cos (A + B) + cos (A - B) Decision making BPSK Demodulation

  30. Figure 5.13Relationship between baud rate and bandwidth in PSK

  31. Example 8 Find the bandwidth for a 4-PSK signal transmitting at 2000 bps. Transmission is in half-duplex mode. Solution For PSK the baud rate is the same as the bandwidth, which means the baud rate is 1000. But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 3,000 bps.

  32. 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.

  33. Bit representation Combination of ASK and PSK Changing Amplitude & Phase of Career Signal One bit, One signal unit Ex ‘0’ -> A1, Φ1 ‘1’ -> A2, Φ2 Benefit Less effected by noise compared to ASK Require less bandwidth Disadvantage Complex demodulation technique Quadrature Amplitude Modulation(QAM)

  34. Figure 5.14The 4-QAM and 8-QAM constellations

  35. Figure 5.15Time domain for an 8-QAM signal 8-QAM = 2n-QAM=23-PSK Bit rate = n x baud rate

  36. Figure 5.1616-QAM constellations ITU-T recommendation OSI recommendation

  37. QAM Modulation X Series-to-Parallel X QAM signal + Y X Digital Data “11 01 00 10 01”

  38. QAM Demodulation Parallel-to-Serial LPF X Digital Data “11 01 00 10 01” LPF X 2 sinA sinB = - cos (A + B) + cos (A - B) 2 cosA cosB = cos (A + B) + cos (A - B)

  39. Figure 5.17Bit and baud

  40. Table 5.1 Bit and baud rate comparison

  41. 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? 3 bits/baud 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/s

  42. 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

  43. 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

  44. 5.2 Telephone Modems Modem Standards

  45. Figure 5.18Telephone line bandwidth A telephone line has a bandwidth of almost 2400 Hz for data transmission.

  46. Note: Modem stands for modulator/demodulator.

  47. Figure 5.19Modulation/demodulation True True

  48. Baud Rate for Half-Duplex ASK BWASK = Baud rateASK = 2400 Baud/s Bit rateASK = Baud rateASK = 2400 bps

  49. Baud Rate for Full-Duplex ASK BWASK = Baud rateASK = 1200 Baud/s Bit rateASK = Baud rateASK = 1200 bps

  50. Baud Rate for Half-Duplex FSK BWFSK = Baud rateFSK + (fc1 – fc0) Bit rateFSK = Baud rateFSK = BWFSK – (fc1 – fc0)

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