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Chapter 5. Analog Transmission

Chapter 5. Analog Transmission. 5.1 Digital-to-Analog 부호화. ASK(Amplitude Shift Keying) FSK(Frequency Shift Keying) PSK(Phase Shift Keying) QAM(Quadrature Amplitude Modulation) : related to Amplitude and Phase. Shift Keying = modulation. Digital-to-Analog 부호화 (cont’d).

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Chapter 5. Analog Transmission

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  1. Chapter 5. Analog Transmission

  2. 5.1 Digital-to-Analog 부호화 • ASK(Amplitude Shift Keying) • FSK(Frequency Shift Keying) • PSK(Phase Shift Keying) • QAM(Quadrature Amplitude Modulation) : related to Amplitude and Phase Shift Keying = modulation

  3. Digital-to-Analog 부호화(cont’d) • Type of Digital-to-Analog encoding

  4. Digital-to-Analog 부호화(cont’d) • Bit rate : the number of bits per second. • Baud rate : the number of signal units per second. • Baud rate is less than or equal to the bit rate. • Bit rate equals the baud rate x the number of bits represented by each signal unit • 반송신호 또는 주파수 (Carrier Signal or Carrier Frequency) • base signal for the information signal

  5. Digital-to-Analog 부호화(cont’d) • Example 1 • An analog signal carries 4 bits in each signal element. If 1000 signal elements are sent per second, find the baud rate and the bit rate. • Solution • Baud rate = Number of signal elements = 1000 bauds per second • Bit rate = Baud rate x Number of bits per signal element = 1000 x 4 = 4000 bps

  6. Digital-to-Analog 부호화(cont’d) • Example 2 • The bit rate of a signal : 3000 • If each signal element carries six bits, what is the baud rate ? • Solution • Baud rate = Bit rate/ number of bits per signal element = 3000/6 = 500 baud per second

  7. Digital-to-Analog 부호화(cont’d) • ASK(Amplitude Shift Keying) • Both frequency and phase remain constant while the amplitude changes. • Highly susceptible to noise interference • Noise usually affects the amplitude.

  8. Digital-to-Analog 부호화(cont’d) • ASK encoding

  9. Digital-to-Analog 부호화(cont’d) • Relationship between baud rate and bandwidth in ASK • BW = (1 + d) x N baud N baud : Baud rate d : factor related to the modulation process (with a minimum value of 0)

  10. Digital-to-Analog 부호화(cont’d) • Example 3 • Find the minimum bandwidth for an ASK signal transmitting at 2000 bps. 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 2000Hz

  11. Digital-to-Analog 부호화(cont’d) • 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.

  12. Digital-to-Analog 부호화(cont’d) • 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 bandwidth in each direction. Assume there is no gap between the bands in two directions. • Solution • Bandwidth for each direction : 10000/2 = 5000 Hz • Carrier frequencies : fc (forward) = 1000 + 5000/2 = 3500 Hz fc (backward) = 11000 – 5000/2 = 8500 Hz

  13. Digital-to-Analog 부호화(cont’d) Solution to Example 5

  14. Digital-to-Analog 부호화(cont’d) • FSK(Frequency Shift Keying) • The frequency of the signal is varied to represent binary 1 or 0. • Avoiding most of the problems from noise - Can ignore voltage spikes

  15. Digital-to-Analog 부호화(cont’d) • FSK encoding • Peak amplitude and phase remain constant

  16. Digital-to-Analog 부호화(cont’d) • Bandwidth for FSK

  17. Digital-to-Analog 부호화(cont’d) • PSK

  18. Digital-to-Analog 부호화(cont’d) • PSK(Phase Shift Keying) • the phase is varied to represent binary 1 or 0. phase 0 bit 1 0 180 0 1 Constellation diagram

  19. The 4-PSK method

  20. The 4-PSK Characteristics

  21. The 8-PSK Characteristics • Find the bandwidth for a 4-PSK signal transmitting at 2000 bps. Transmission is in half-duplex mode.

  22. Relationship between baud rate and bandwidth in PSK

  23. Bandwidth for PSK Example 4: Find the bandwidth for a 4-PSK signal transmitting at 2000 bps. Transmission is in half-duplex mode. Answer) For 4-PSK baud rate is one-half of the bit rate. The baud rate is therefore 1000. A PSK signal requires a bandwidth equal to its baud rate. Therefore, the bandwidth is 1000 Hz. Example 5 : Given a bandwidth of 5000 Hz for an 8-PSK signal, what are the baud rate and bit rate? 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.

  24. Digital-to-Analog 부호화(cont’d) • QAM(Quadrature Amplitude Modulation) • Quadrature amplitude modulation is a combination of ASK and PSK so that a maximum contrast between each signal unit (bit, dibit, tribit, and so on) is achieved

  25. The 4-QAM and 8-QAM constellations

  26. Digital-to-Analog 부호화(cont’d) • Time domain for an 8-QAM signal

  27. 16-QAM constellations • The minimum bandwidth for QAM is the same as that required for ASK and PSK

  28. Bit/Baud Comparison

  29. Bit and Baud rate comparison

  30. Bit and Baud rate comparison • 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

  31. Bit and Baud rate comparison • 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 • 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

  32. 5.2 Telephone Modems • A telephone line has a bandwidth of almost 2400 Hz for data transmission

  33. Telephone Modems • Modem stands for modulator/demodulator. • Modulator – creates a band-pass analog signal from binary data • Demodulator – recovers the binary data from the modulated signal

  34. Modulation/demodulation

  35. Modem Standards • V.32 – 9600 bps : called Trellis-coded modulation • V.32bis – 14,400 bps • V.34bis – 28,800 bps & 33,600 bps • V.90 – 33,600 upload, 56,000 bps download • V.92 – 48 Kpbs up, 56 Kpbs down

  36. Modem Standards -The V.32 constellation and bandwidth • 32-QAM with a baud rate of 2400 • 4 data bits x 2400 = 9600 bps

  37. Modem Standards -The V.32bis constellation and bandwidth • 128-QAM • 6 data bits x 2400 baud = 14,400 bps • fall-back, fall-forward feature enabling modem to adjust speed depending on line and/or signal quality

  38. Modem Standards -The V.34bis and V.90 • V.34bis • bit rate of 28,800 with 960-point constellation • bit rate of 33,600 with 1664-point constellation • V.90 • download 56K, upload 33.6K

  39. Traditional Modems

  40. 56K Modems

  41. 56K Modems • telephone company samples 8,000 times per second • 8 bits per sample (7 data bits) • rate = 8,000 x 7 = 56,000 bps

  42. 5.3 Modulation of Analog Signals • Analog-to-Analog encoding is the representation of analog information by an analog signal. • Analog-to-Analog encoding

  43. Analog-to-Analog 부호화(cont’d) • Type of analog-to-analog encoding

  44. Analog-to-Analog 부호화(cont’d) • AM(Amplitude Modulation) ~ The frequency and phase of the carrier remain the same; only the amplitude changes to follow variations in the information.

  45. Analog-to-Analog 부호화(cont’d) • Amplitude modulation

  46. Analog-to-Analog 부호화(cont’d) • AM bandwidth • The total bandwidth required for AM can be determined from the bandwidth of the audio signal. The total bandwidth required for AM can be determined from the bandwidth of the audio signal: BWtotal = 2 x BWmod.

  47. Analog-to-Analog 부호화(cont’d) • AM bandwidth

  48. Analog-to-Analog 부호화(cont’d) • Audio signal(음성과 음악) bandwidth : 5 KHz • Minimum bandwidth : 10 KHz (bandwidth for AM radio station) • AM stations are allowed carrier frequencies anywhere between 530 and 1700 KHz(1.7 MHz) • each frequency must be separated by 10 KHz

  49. Analog-to-Analog 부호화(cont’d) • AM band allocation

  50. Analog-to-Analog 부호화(cont’d) • We have an audio signal with a bandwidth of 4 KHz. What is the bandwidth needed if we modulate the signal using AM? Ignore FCC regulations. • An AM signal requires twice the bandwidth of the original signal: BW = 2 x 4 KHz = 8 KHz

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