1. Chapter 4 Digital Transmission Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. Chapter 4: Outline 4.1 DIGITAL-TO-DIGITAL CONVERSION 4.1 DIGITAL-TO-DIGITAL CONVERSION 4.2 ANALOG-TO-DIGITAL CONVERSION 4.2 ANALOG-TO-DIGITAL CONVERSION 4.1 TRANSMISSION MODES 4.1 TRANSMISSION MODES

3. 4.3 4-1 DIGITAL-TO-DIGITAL CONVERSION In this section, we explore how digital data is represented using digital signals.

4. 4.4 4-1 DIGITAL-TO-DIGITAL CONVERSION The conversion involves three techniques: line coding, line coding, block coding, and block coding, and scrambling. scrambling.

5. 4.5 4-1 DIGITAL-TO-DIGITAL CONVERSION Line coding is always needed; block coding and scrambling may or may not be needed.

6. 4.6 4.4.1 Line Coding Line coding converts a sequence of bits to a digital signal. At the sender, digital data are encoded into a digital signal At the receiver, the digital data are recreated by decoding the digital signal.

7. 4.7 Figure 4.1: Line coding and decoding

8. 4.8 4.4.1 Line Coding Define the ratio of data elements to signal elements with r = (data elements) / (signal elements).

9. 4.9 Figure 4.2: Signal elements versus data elements

10. 4.10 4.4.1 Line Coding Let S be the number of signals per second. This is baud rate. The relationship between bandwidth, N-bps, and baud rate, S-signals/sec, is: S = c * N * 1/r The value c is the case factor: 0 < c <=1

11. A signal is carrying data in which one data element is encoded as one signal element (r = 1). If the bit rate is 100 kbps, what is the average value of the baud rate if c is ½ (the average value between 0 and 1)? Example 4.1 4.11

12. A signal is carrying data in which one data element is encoded as one signal element (r = 1). If the bit rate is 100 kbps, what is the average value of the baud rate if c is ½ ? Example 4.1 Solution The baud rate is then 4.12

13. The maximum data rate of a channel (see Chapter 3) is N max = 2 × B × log2 L (defined by the Nyquist formula). Does this agree with the previous formula for N max ? Example 4.2 4.13

14. The maximum data rate of a channel (see Chapter 3) is N max = 2 × B × log2 L (defined by the Nyquist formula). Does this agree with the previous formula for N max ? S = ½ N / r solve for N, N = 2 S r Example 4.2 4.14

15. The maximum data rate of a channel (see Chapter 3) is N max = 2 × B × log2 L (defined by the Nyquist formula). Does this agree with the previous formula for N max ? S = ½ N / r solve for N, N = 2 S r, by substitution, 2 B log2 ( L ) = 2 S r r = log2( L ) Example 4.2 4.15

16. The maximum data rate of a channel (see Chapter 3) is N max = 2 × B × log2 L (defined by the Nyquist formula). Does this agree with the previous formula for N max ? 2 B log2 ( L ) = 2 S r r = log2( L ) B in cycles /sec is the same as S signals/sec, where a cycle is a signal for this case. Example 4.2 4.16

17. 4.17 Figure 4.3: Synchronization

18. In a digital transmission, the receiver clock is 0.1 percent faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is 1 kbps? Example 4.3 4.18

19. In a digital transmission, the receiver clock is 0.1 percent faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is 1 kbps? 0.1% is.001 Total bits received = (1 +.001)*1000 bits/sec Total bits received = 1001 (one extra bit). Example 4.3 4.19

20. In a digital transmission, the receiver clock is 0.1 percent faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is 1 kbps? Example 4.3 Solution At 1 kbps, the receiver receives 1001 bps instead of 1000 bps. 4.20

21. In a digital transmission, the receiver clock is 0.1 percent faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is 1 Mbps? Example 4.3 At 1 Mbps, the receiver receives 1,001,000 bps instead of 1,000,000 bps. 4.21

22. 4.22 Figure 4.3: Effect of lack of synchronization

23. 4.23 4.4.2 Line Coding Schemes We can roughly divide line coding schemes into five broad categories, as shown in Figure 4.4..

24. 4.24 Figure 4.4: Line coding scheme

25. 4.25 Figure 4.5: Unipolar NRZ scheme

26. 4.26 Figure 4.6: Polar schemes (NRZ-L and NRZ-I)

27. A system is using NRZ-I to transfer 10-Mbps data. What are the signal rate and bandwidth? Case factor c = ½ Example 4.4 4.27

28. A system is using NRZ-I to transfer 10-Mbps data. What are the signal rate and bandwidth? Case factor c = ½ NRZ-I has r = 1 (1 bit / signal) S = c N / r Example 4.4 4.28

29. A system is using NRZ-I to transfer 1-Mbps data. What are the signal rate and bandwidth? Case factor c = ½ NRZ-I has r = 1 (1 bit / signal) S = c N / r S = ½ (1Mbps) / 1 =.5 Mbaud = 500 Kbaud Example 4.4 4.29

30. A system is using NRZ-I to transfer 1-Mbps data. What are the signal rate and bandwidth? r = log2(L) N = 2 B log2( L) solve for B B = ½ N / r B = 1Mbps/2 =.5 MHz = 500 KHz Example 4.4 4.30

31. 4.31 Figure 4.7: Polar schemes (RZ)

32. 4.32 Figure 4.8: Polar biphase

33. Bipolar Bipolar – positive and negative amplitudes above and below the time axis. AMI Pseudoternary

34. AMI Alternate Mark Inversion Every other 1-bit has a 180 degree phase change The 0-bit has zero amplitude Used for some long distance communications.

35. Pseudoternary “three” states: 0-bits alternate phase 1-bits have zero amplitude

36. 4.36 Figure 4.9: Bipolar schemes: AMI and pseudoternary

37. Multilevel Schemes The goal is to send more bits per signal. We designate the different schemes by this notation: mBnL m = number of bits per signal element B = two possible data elements (0 or 1) binary L = number of levels

38. Multilevel Schemes We designate the different schemes by this notation: mBnL B^m <= L^n

39. mBnL Schemes Example: 2B1Q (used for DSL) 2=2 B=2 n=1 L=4 (Q is for quad)

40. 4.40 Figure 4.10: Multilevel: 2B1Q (the diagram is wrong!)

41. 8B6T 2^8 <= 3^6, 256 <= 729 this is the early line-code implementation of fast Ethernet. It does not use Manchester coding.

42. 4.42 Figure 4.11: Multilevel: 8B6T (early version of fast Ethernet)

43. 4D-PAM5 A different category of multilevel line coding: 4-dimensional 5-level pulse amplitude modulation.

44. 4D-PAM5 Four wires transmit in parallel using 8B4Q over each wire. This is used for Gigabit Ethernet.

45. 4.45 Figure 4.12: Multilevel: 4D-PAMS scheme

46. 4.46 Figure 4.13: Multi-transition MLT-3 scheme

47. MLT-3 see page 108 for the rules

48. Table 4.1 : Summary of line coding schemes 4.48

49. 4.49 4.4.3 Block Coding Block coding provides redundancy to ensure synchronization and error detecting.

50. 4.50 4.4.3 Block Coding In general, block coding changes a block of m bits into a block of n bits, where n is larger than m. Block coding is referred to as an mB/nB encoding technique.

51. 4.51 Figure 4.14: Block coding concept

52. 4.52 Figure 4.15: Using block coding 4B/5B with NRZ-I line coding scheme

53. Table 4.2 : 4B/5B mapping codes 4.53

54. Block Coding nB/mB Coding 2^n bits as 2^m bits, m > n.

55. 4.55 Figure 4.16: Substitution in 4B/5B block coding

56. We need to send data at a 1-Mbps rate. What is the minimum required bandwidth, using 1.a combination of 4B/5B and NRZ-I; or 2. Manchester coding? C = ½ for both Example 4.5 4.56

57. We need to send data at a 1-Mbps rate. What is the minimum required bandwidth, using 1.a combination of 4B/5B and NRZ-I; or 2. Manchester coding C = ½ for both cases There is a 25% increase in bits due to the overhead in case-1 The link must therefore support a 1.25 Mbps rate to move 1- Mbps data. Example 4.5 4.57

58. We need to send data at a 1-Mbps rate. What is the minimum required bandwidth, using 1.a combination of 4B/5B and NRZ-I; or, 2. Manchester coding NRZ-I encodes one bit per signal, r = 1. Manchester encoding encodes one bit per two signals: r = ½ Example 4.5 4.58

59. We need to send data at a 1-Mbps rate. What is the minimum required bandwidth, using 1.a combination of 4B/5B and NRZ-I; or, 2.Manchester coding? C = ½ for both cases. B = c N / r NRZ-I : B = ½ (1.25Mbps) / 1 = 626KHz Man : B = ½ ( 1 Mbps) /.5 = 1. MHz Example 4.5 4.59

60. 4.60 Figure 4.17: 8B/10B block encoding

61. 4.61 4.4.4 Scrambling Scrambling is used with long distance bipolar AMI line coding.

62. 4.62 4.4.4 Scrambling Scrambling helps synchronize when the data stream contains a sequence of 8 consecutive zeros.

63. 4.63 4.4.4 Scrambling B8ZS is the designation of AMI with scrambling. 8 consecutive zeros is coded as 000VB0VB.

64. 4.64 Figure 4.19: Two cases of B8ZS scrambling technique

65. 4.65 4-2 ANALOG-TO-DIGITAL CONVERSION The tendency today is to change an analog signal to digital data. Example: Voice is converted to a digital stream by a phone then converted to an analog signal representing digital data.

66. 4.66 4.2.1 Pulse Code Modulation (PCM) The most common technique to change an analog signal to digital data (digitization) is called pulse code modulation (PCM).

67. 4.67 4.2.1 Pulse Code Modulation (PCM) A PCM encoder has three steps: Sampling Encoding Quantizing

68. 4.68 Figure 4.21: Components of PCM encoder

69. Sampling Sampling – a snapshot of the analog signal is recorded at regular time intervals. If the time interval is short, the digital signal is a good estimate of the original analog signal.

70. 4.70 Figure 4.22: An example sampling method for PCM

71. Nyquist Theorem To reproduce the original analog signal, one necessary condition is that the sampling rate must be twice the highest frequency of the analog signal.

72. 4.72 Figure 4.23: Nyquist sampling rate for low-pass and bandpass signals

73. 4.73 Figure 4.24: Recovery of a sine wave with different sampling rates.

74. Telephone companies digitize voice by assuming a maximum frequency of 4000 Hz. What is the needed sample rate for voice transmission? Sample rate = 2*fmax Example 4.9 4.74

75. Telephone companies digitize voice by assuming a maximum frequency of 4000 Hz. What is the needed sample rate for voice transmission? Sample rate = 2*fmax Sample rate = 2 * 4KHz = 8 K samples/sec Example 4.9 4.75

76. A complex low-pass signal has a bandwidth of 200 kHz. What is the minimum sampling rate for this signal? Example 4.10 4.76

77. A complex low-pass signal has a bandwidth of 200 kHz. What is the minimum sampling rate for this signal? Example 4.10 Solution Sample rate = 2 * fmax = 400 Ksamples/sec 4.77

78. A complex bandpass signal has a bandwidth of 200 kHz. What is the minimum sampling rate for this signal? Example 4.11. 4.78

79. A complex bandpass signal has a bandwidth of 200 kHz. What is the minimum sampling rate for this signal? Example 4.11 Solution We cannot find the minimum sampling rate in this case because we do not know where the bandwidth starts or ends. We do not know the maximum frequency in the signal. 4.79

80. PCM: Quantization 1. Analog signal is between Vmin and Vmax

81. PCM: Quantization 1. Analog signal is between Vmin and Vmax 2. Divide the range into L equal zones delta = (Vmax-Vmin)/L

82. PCM: Quantization 1. Analog signal is between Vmin and Vmax 2. Divide the range into L equal zones 3. Normalize using the delta value.

83. PCM: Quantization 1. Analog signal is between Vmin and Vmax 2. Divide the range into L equal zones 3. Normalize using the delta value 4. Assign a quantized value from 0 to L-1 to the mid-point of each zone.

84. 4.84 Figure 4.26: Quantization and encoding of a sampled signal

85. SNR-db for PCM SNR db = 6.02 nb + 1.76 nb = number of bits per sample

86. What is the SNR dB in the example of Figure 4.26? Example 4.12 4.86

87. What is the SNR dB in the example of Figure 4.26? Example 4.12 Solution We can use the formula to find the quantization. We have eight levels and 3 bits per sample, so SNR dB = 6.02(3) + 1.76 = 19.82 dB. Increasing the number of levels increases the SNR. 4.87

88. A telephone subscriber line must have an SNR dB above 40. What is the minimum number of bits per sample? Example 4.13 4.88

89. A telephone subscriber line must have an SNR dB above 40. What is the minimum number of bits per sample? SNR-db = 6.02 nb + 1.76 db, solve for r Example 4.13 4.89

90. A telephone subscriber line must have an SNR dB above 40. What is the minimum number of bits per sample? SNRdb = 6.02 nb + 1.76 db, solve for r nb = (SNRdb – 1.76 db)/6.02 nb = (40 -1.76)/6.02 = 6.4 nb = 6.4, must be at least 7 bits Example 4.13 4.90

91. We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample? The human voice has frequencies from 0 to about 4KHz. Example 4.14 4.91

92. We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample? The human voice has frequencies from 0 to about 4KHz. Sample rate = 2 * fmax Sample rate = 2 * 4KHz = 8Ksamples/sec Bit rate = 8 bits/sample * 8Ksamples/sec = 64kbps Example 4.14 4.92

93. 4.93 4-3 TRANSMISSION MODES The transmission of binary data across a link can be accomplished in either –parallel mode –serial mode

94. 4.94 Figure 4.31: Data transmission modes

95. 4.95 4.3.1 Parallel Transmission Bit streams are sent in parallel over multiple wires.

96. 4.96 Figure 4.32: Parallel transmission

97. 4.97 4.3.2 Serial Transmission In serial transmission one bit follows another, so we need only one communication channel rather than n to transmit data between two communicating devices

98. 4.98 Figure 4.33: Serial transmission

99. Serial Transmission Types Asynchronous Synchronous Isochronous

100. Serial Transmission Types Asynchronous – signal timing is not important due to the start bit, stop bit, and gap for each 8-bit frame. inefficient due to overhead

101. 4.101 Figure 4.34: Asynchronous transmission

102. Serial Transmission Synchronous – continuous stream of bits without overhead or gaps. Timing at both ends must be correct.

103. 4.103 Figure 4.35: Synchronous transmission