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 (QAM)
Bit/Baud Comparison

4. Figure 5.1 Digital-to-Analog Modulation
Frequency Shift Keying
Phase Shift Keying
Amplitude Shift Keying
Quadrature Amplitude Modulation

5. Digital-to-Analog Modulation (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

6. Bit Rate vs Baud Rate Bit rate (bps)
Number of bits per second
Baud rate (signal unit/sec: baud / sec: baud)
Number of signal units per second
less than or equal to the bit 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 If each signal unit carries 6 bits,
what is the baud rate?
Solution
Baud rate = 3000 / 6 = 500 baud/s

9. Amplitude Shift Keying (ASK)
Bit representation
Changing Amplitude of Carrier 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

10. Figure ASK
Bit rate = baud rate

11. ASK Modulation ASK Demodulation Carrier Signal กรองความถี่เฉพาะที่ผ่าน
Low Pass Filter
กรองความถี่เฉพาะที่ผ่าน
transmission ได้เท่านั้น X X
Digital Data
LPF m
ASK Demodulation
2 cosA cosB = cos (A + B) + cos (A - B) X
LPF
Decision making

12. Figure 5.4 Relationship between baud rate and bandwidth in ASK
ขึ้นกับ เทคนิคการ modulation & filtering process
= S (baud/s)

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 = / 2 = 5000 Hz
The carrier frequencies can be chosen at the middle of each band (see Fig. 5.5).
fc (forward) = /2 = 3500 Hz
fc (backward) = – 5000/2 = 8500 Hz

16. Figure 5.5 Solution to Example 5

17. FSK Modulation Bit representation Benefit Disadvantage
Voltage Controlled Oscillator
VCO
Bit representation
Changing Frequency of Carrier 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 FSK
Bit rate = Baud rate

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

20. Example 6 Solution For FSK BW = Baud rate + fc1 - fc0
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
Bit Rate = Baud Rate
For FSK
BW = Baud rate + fc1 - fc0
BW = Bit rate + fc1 - fc0 = = 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
Bit Rate = Baud Rate
Because the transmission is full duplex, only 6000 Hz is allocated for each direction.
BW = Baud rate + fc1 - fc0
Baud rate = BW - (fc1 - fc0 ) = = 4000
But because the baud rate is the same as the bit rate, the bit rate is 4000 bps.

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

23. Figure PSK
Bit rate = Baud rate
180o 0o

24. Figure 5.9 PSK constellation

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

26. Figure 5.11 The 4-PSK characteristics

27. 4-PSK = 2n-PSK=23-PSK Bit rate = n x Baud rate
Figure The 8-PSK characteristics
4-PSK = 2n-PSK=23-PSK
Bit rate = n x Baud rate
Bit rate = 3 x Baud rate

28. X x
Pulse 1 X
[0,1]
[-1,1] -1 1
Data
BPSK Modulation

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

30. Figure 5.13 Relationship 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.
4-PSK = 22-PSK
Bit rate = 2 x Baud rate
2000 = 2 x Baud rate
Baud rate = 1000 baud/s
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?
Bandwidth = Baud rate
= Baud rate
8-PSK = 23-PSK
Bit rate = 3 x Baud rate
= 3 x 5000
= 15,000 bps
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. Quadrature Amplitude Modulation (QAM)
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

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

35. 8-QAM = 2n-QAM=23-QAM Figure 5.15 Time domain for an 8-QAM signal
Bit rate = n x Baud rate
90o
180o 0o
270o

36. Figure 5.16 16-QAM constellations
ITU-T recommendation
OSI recommendation

37. QAM Modulation 10 00 Series-to-Parallel 11 01 Pulse Digital Data X X
y=1 00 X +
QAM signal
(Pulse -1,1)
(Pulse 1,1)
X=-1
X=1
(Pulse 1,1)
Series-to-Parallel X Y 11 01
(Pulse -1,-1)
y=-1
Digital Data
“ ”
Pulse
Digital Data
QAM Modulation

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

39. Figure Bit and baud

40. Table 5.1 Bit and baud rate comparison ASK, FSK, 2-PSK Bit 1 N
Modulation
Units
Bits/Baud
Baud rate
Bit Rate
ASK, FSK, 2-PSK
Bit 1 N
4-PSK, 4-QAM
Dibit 2 2N
8-PSK, 8-QAM
Tribit 3 3N
16-QAM
Quadbit 4 4N
32-QAM
Pentabit 5 5N
64-QAM
Hexabit 6 6N
128-QAM
Septabit 7 7N
256-QAM
Octabit 8 8N

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 = log224 = 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 = log226 = 6.
Thus,
72000 / 6 = 12,000 baud

44. 5.2 Telephone Modems
Modem Standards

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

46. Modem stands for MOdulator/DEModulator.
Note:
Modem stands for MOdulator/DEModulator.

47. Figure 5.19 Modulation/demodulation
True
True
Analog
Analog
ASK, FSK, PSK, QAM
Digital
Digital

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

51. Baud Rate for Full-Duplex FSK
Bandwidth = Baud Rate + (fc1-fc0)
Bit Rate = Baud Rate
Baud Rate for Full-Duplex FSK
Max baud rate = Max bit rate = (fc1-fc0)
1200 Hz
1200 Hz
BWFSK = Baud rateFSK + (fc1 – fc0)
Bit rateFSK = Baud rateFSK = BWFSK – (fc1 – fc0)

52. Bell Modems Guard Band or Guard Frequency Bit rate = 2 x Baud rate 500
200
455
200
600
600
600
Bit rate = 2 x Baud rate

53. Bell Modems Bit rate = 2 x Baud rate Bit rate = 3 x Baud rate

54. ITU Modems International Telecommunication Union
(Four-Phase Differential PSK)

55. ITU Modems

56. V.22bis 16-QAM Constellation

57. V.32 Constellation

58. V.33 Constellation

59. Figure 5.20 The V.32 constellation and bandwidth

60. Figure 5.21 The V.32bis constellation and bandwidth

61. Figure 5.22 Traditional modems
Analog
Analog
Digital
Digital
Analog
Analog
Digital
Digital

62. Figure K modems
Digital
Analog
Digital
Digital
Analog
Digital

63. The 56K Digital Modem A 56K modem is based upon one of two standards:
- V.90 - Upstream speed is maximum 33,600 bps
- V.92 - Newer standard which allows maximum upstream speed of 48 Kbps (under ideal conditions) and can place a data connection on hold if the telephone service accepts call waiting and a voice telephone call arrives

64. 5.3 Modulation of Analog Signals
Amplitude Modulation (AM)
Frequency Modulation (FM)
Phase Modulation (PM)

65. Figure 5.24 Analog-to-Analog Modulation

66. Note:
The total bandwidth required for AM can be determined from the bandwidth of the audio signal: BWt = 2 x BWm.
Audio bandwidth
(4 KHz)
AM bandwidth

67. Figure 5.26 Amplitude Modulation

68. Amplitude Modulation

69. Figure AM bandwidth

70. Peak-to-Peak
m=1 : 100% AM

71. Figure 5.28 AM band allocation
535
545
540
550

72. Example 13 Solution 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 (Federal Communications Commission) regulations.
Solution
BWt = 2 x BWm
An AM signal requires twice the bandwidth of the original signal:
BW = 2 x 4 KHz = 8 KHz

73. Stereo Audio bandwidth
Note:
The total bandwidth required for FM can be determined from the bandwidth of the audio signal: BWt = 10 x BWm.
Stereo Audio bandwidth
(15 KHz)
FM bandwidth

74. Figure 5.29 Frequency modulation

75. Figure FM bandwidth

76. The bandwidth of a stereo audio signal is usually 15 KHz.
Note:
The bandwidth of a stereo audio signal is usually 15 KHz.
Therefore, an FM station needs at least a bandwidth of 150 KHz.
The FCC requires the minimum bandwidth to be at least 200 KHz (0.2 MHz).

77. Figure 5.31 FM band allocation
88.1
88.5
88.2
88.4
88.6

78. Example 14 Solution We have an audio signal with a bandwidth of 4 MHz.
What is the bandwidth needed if we modulate the signal using FM?
Ignore FCC (Federal Communications Commission) regulations.
Solution
BWt = 10 x BWm
An FM signal requires 10 times the bandwidth of the original signal:
BW = 10 x 4 MHz = 40 MHz