1. Chapter 4
Digital Transmission

2. Digital Transmission Digital data to digital signal
Line coding
Analog data to digital signal
PCM (Pulse code modulation)

3. 4.1 Line Coding
Some Characteristics
Line Coding Schemes
Some Other Schemes

4. Figure 4.1 Digital Transmission: Line coding
Actual signal: digital pulse
Digital data: Abstract Data
วัตถุประสงค์ของการทำ Line coding
เพื่อแทน digital data ด้วย ระดับ voltage ที่กำหนด
เพื่อกำจัดหรือลดปัญหา
เพื่อเพิ่มความเร็วในการส่งข้อมูล

5. ‘1’: +5V ‘0’: 0V ‘1’: +5V or -5V ‘0’: 0V + + - two
Figure Signal level versus data level
‘1’: +5V
‘0’: 0V
‘1’: +5V or -5V
two
‘0’: 0V + + -

6. Problems Digital data to Digital signal conversion
DC component
สัญญาณดิจิตอลที่มีองค์ประกอบ DC ทำให้ระดับสัญญาณเพี้ยนไป
Lack of Synchronization
การตรวจจับบิตข้อมูลผิดตำแหน่ง ทำให้อ่านค่าของระดับสัญญาณผิดไป
ปัญหาเหล่านี้ ส่งผลให้
ตัวรับแปลงสัญญาณกลับเป็นบิตข้อมูลผิดเพี้ยนไป

7. Figure DC component + + + + + + + + - - - -

8. DC Component Problem
สัญญาณดิจิตอลที่มีองค์ประกอบ DC ทำให้ระดับสัญญาณเพี้ยนไป
Stray Capacitance
Coupling
Voltage level is constant, spectrum creates very low frequencies
These frequencies around zero called DC (direct current) component
DC component present problems for system that cannot
pass low frequencies or system that uses electrical coupling
(via transformer)

9. Figure 4.4 Lack of synchronization
Fast clock

10. Example 3
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?
How many if the data rate is 1 Mbps?
Solution
At 1 Kbps:
1000 bits sent 1001 bits received1 extra bps
At 1 Mbps:
1,000,000 bits sent 1,001,000 bits received1000 extra bps

11. Signal level (Element) versus Data level (Element)
r means number of data elements carried by each signal element

12. Example 1
A signal has two data levels with a pulse duration of 1 ms. We calculate the pulse rate and bit rate as follows:
Pulse duration = signal duration (sec) = 1 ms
Pulse Rate = 1/Pulse duration = 1/ 10-3= 1000 pulses/s
L = data level
L = 2; Bit Rate = Pulse Rate x log2 L
= 1000 x log2 2 = 1000 bps
L = 4; Bit Rate = 1000 x log2 4 = 2000 bps
จำนวน Bit (data) ต่อ 1 Signal Element
(Pulse)

13. Figure 4.5 Line coding schemes
Multi-level

14. Unipolar encoding uses only one voltage level (positive or negative).
Figure Unipolar encoding
Note:
Unipolar encoding uses only one voltage level
(positive or negative). +

15. Polar encoding uses two voltage levels (positive and negative).
Figure Types of polar encoding
Note:
Polar encoding uses two voltage levels (positive and negative).
Non-Return-to-Zero
Return-to-Zero

16. In NRZ-I (NRZ-M) the signal is inverted if a 1 is encountered.
Note:
In NRZ-L the level of the signal is dependent upon the state of the bit.
In NRZ-I (NRZ-M) the signal is inverted if a 1 is encountered.

17. NRZ-I: Non-Return-to-Zero-Inversion NRZ-M: Non-Return-to-Zero-Mark
Figure NRZ-L and NRZ-I encoding
NRZ-L: Non-Return-to-Zero-Level
‘0’: +5V
‘1’: -5V
NRZ-I: Non-Return-to-Zero-Inversion
NRZ-M: Non-Return-to-Zero-Mark
‘1’: Inversion + + + + +
‘0’: Noninversion - - - -

18. Figure 4.9 Return-to-Zero (RZ) encoding
Bipolar-Return-to-Zero (BIP)
A good encoded digital signal must contain a provision for synchronization.
‘0’: -V +V
‘1’: +V -V

19. Note:
In Manchester encoding,
the transition at the middle of the bit is used for both
Synchronization and
Bit representation.

20. Figure 4.10 Manchester encoding+V -V

21. Note: In differential Manchester encoding,
The transition at the middle of the bit
is used only for synchronization The bit representation is defined by
the inversion or noninversion at the
beginning of the bit.

22. For Synchronization ‘0’: Inversion ‘0’: Inversion
Figure Differential Manchester encoding
For Synchronization
‘0’: Inversion
‘1’: Noninversion
‘0’: Inversion

23. Applications of Line Coding
NRZ encoding:
RS232 based protocols
Manchester encoding:
Ethernet networks
Hard drive
Differential Manchester encoding:
token-ring networks
NRZ-Inverted encoding:
Fiber Distributed Data Interface (FDDI)

24. Figure 4.12 Bipolar AMI encoding
AMI stands for Alternate Mark Inversion,
Variation of AMI is Pseudoternary
In bipolar encoding, we use three levels: positive, zero, and negative.
‘1’: Inversion
‘0’: 0 V +V

25. (Bipolar with Eight-Zero Substitution)
Figure 5-11
B8ZS Encoding
(Bipolar with Eight-Zero Substitution)
V B 0 V B
V B 0 V B
V B 0 V B
V = Violation, B = Bipolar

26. Example: B8ZS Encoding 0 0 0 V B 0 V B V = Violation, B = Bipolar
Figure 5-13
Example: B8ZS Encoding
V B V B +V +V -V
V = Violation, B = Bipolar

27. (High Density Bipolar 3-zero Encoding)
Figure 5-12
Pattern 1 : V
HDB3 Encoding
(High Density Bipolar 3-zero Encoding)
Pattern 2 :
B V V V
B V
B V

28. Figure 5-14
Example: HDB3 Encoding V
B V

30. Line Coding Multi-level coding m B n L 2m ≤ Ln.
Length of binary pattern
Length of signal pattern
Binary data
Number of Level in the signaling
m B n L
B (binary) for L=2
T(trenary) for L=3
Q(quaternary) for L=4
2m ≤ Ln.
Line Coding
Data
Elements
Signal
Elements
Multi-level coding
1) 2B1Q (2 Binary 1 Quaternary)
2) 8B/6T (8 data bits as six ternary)
3) MLT-3 (Multi-Level Transition 3

31. Used in ISDN 64 kbps or 128 kbps via telephone line
Figure B1Q
(2 Binary 1 Quaternary) - -
Used in ISDN 64 kbps or 128 kbps via telephone line

32. 2B1Q (two binary, one quaternary)
Uses data patterns of size 2
one signal element belonging to four-level signal
2B1Q is used in DSL (Digital Subscriber Line) technology

33. 28=256 different data patterns
Figure Example of 8B/6T encoding
36=478 different signal patterns
ใช้ Signal 3 ระดับ
Data 8 Bits
ความยาวของ Signal = 6
=222
redundant signal elements
Used in 100BASET4 Ethernet (100 Mbps) via UTP in Star Topology

34. ‘0’: Non-Inversion
‘1’: Inversion ( -V, 0, +V)
First introduced by Cisco System for FDDI (Fiber Distributed Data Interface: Token Ring) Used in 100Base-TX (100 Mbit/s Ethernet)

35. ‘0’: Non-Inversion Figure 4.14 MLT-3 signal
‘1’: Inversion ( -V, 0, +V)

36. 4D-PAM5: Gigabit Ethernet: 1000Base-T
(Four-Dimensional Five-level Pulse Amplitude Modulation) 00 01 11 10

37. Polar

38. 4.2 Block Coding
Steps in Transformation
Some Common Block Codes

39. Figure Block coding
m < n
Combination

40. Figure 4.16 Substitution in block coding
25 = 32
24 = 16

41. No more than one leading zero (left bit) and
No more than two trailing zeros (right bits)
Table B/5B encoding
Block-coded stream does not have
more than three consecutive 0s
Data
Code
0000
11110
1000
10010
0001
01001
1001
10011
0010
10100
1010
10110
0011
10101
1011
10111
0100
01010
1100
11010
0101
01011
1101
11011
0110
01110
1110
11100
0111
01111
1111
11101

42. Table 4.1 4B/5B encoding (Continued)
Data
Code
Q (Quiet)
00000
I (Idle)
11111
H (Halt)
00100
J (start delimiter)
11000
K (start delimiter)
10001
T (end delimiter)
01101
S (Set)
11001
R (Reset)
00111

44. Pulse Code Modulation (PCM)
Analog to Digital Conversion (A/D)

45. 4.3 Sampling
Pulse Amplitude Modulation (PAM)
Pulse Code Modulation (PCM)
Sampling Rate: Nyquist Theorem
How Many Bits per Sample?
Bit Rate

46. Figure 4.22 From analog signal to PCM digital code
Continuous
Discrete

47. Note:
Pulse Amplitude Modulation has some applications, but it is not used by itself in data communication.
However, it is the first step in another very popular conversion method called Pulse Code Modulation.

48. Components of PCM encoder

49. Three different sampling methods for PCM

50. According to the Nyquist theorem,
Nyquist sampling rate for low-pass and bandpass signals
According to the Nyquist theorem,
the sampling rate must be at least 2 times the highest frequency contained in the signal.

51. Recovery of a sampled sine wave for different sampling rates
Sampling at the Nyquist rate can create a good approximation of the original sine wave.
Oversampling can also create the same approximation, but is redundant and unnecessary.
Sampling below the Nyquist rate does not produce a signal that looks like the original sine wave.

52. Sampling of a clock with only one hand
The second hand of a clock has a period of 60 s.
According to the Nyquist theorem, we need to sample hand every 30 s

54. Figure 4.23 Nyquist theorem
According to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequency.

55. Example 4
What sampling rate is needed for a signal with a bandwidth of 10,000 Hz (1000 to 11,000 Hz)?
Solution
The sampling rate must be twice the highest frequency in the signal:
Sampling rate = 2 x (11,000) = 22,000 samples/s

56. Examples
An example of under-sampling is the seemingly backward rotation of the wheels of a forward-moving car in a movie.
A movie is filmed at 24 frames per second.
If a wheel is rotating more than 12 times per second, the under- sampling creates the impression of a backward rotation.
Telephone companies digitize voice by assuming a maximum frequency of 4000 Hz.
The sampling rate therefore is 8000 samples per second.

57. Example A complex low-pass signal has a bandwidth of 200 kHz.
What is the minimum sampling rate for this signal?
Solution
The bandwidth of a low-pass signal is between 0 and f, where f is the maximum frequency in the signal.
Therefore, we can sample this signal at 2 times the highest frequency (200 kHz).
The sampling rate is therefore 400,000 samples per second.

58. Quantization and encoding of a sampled signal

59. Figure 4.19 Quantized PAM signal
** Signed Integer **

60. Figure 4.20 Quantizing by using sign and magnitude

61. Figure PCM

62. Components of a PCM decoder

63. Example 5
A signal is sampled. Each sample requires at least 12 levels of precision (+0 to +5 and -0 to -5).
How many bits should be sent for each sample?
Solution
L >= > L = 16
We need 4 bits; 1 bit for the sign and 3 bits for the value.
A 3-bit value can represent 23 = 8 levels (000 to 111), which is more than what we need.
A 2-bit value is not enough since 22 = 4.
A 4-bit value is too much because 24 = 16.

64. Example 6
We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample?
Solution
Human voice normally contains frequencies from 0 to 4000 Hz.
Sampling rate = 4000 x 2 = 8000 samples/s
Bit rate = sampling rate x number of bits per sample = 8000 x 8
= 64,000 bps = 64 Kbps

65. In this case, the sampling rate is twice the bandwidth.
Note:
Note that we can always change a band-pass signal to a low-pass signal before sampling.
In this case, the sampling rate is twice the bandwidth.

66. The process of delta modulation

67. Delta modulation components

68. 4.4 Transmission Mode
Parallel Transmission
Serial Transmission

69. Figure 4.24 Data transmission

70. Figure 4.25 Parallel transmission

71. Figure 4.26 Serial transmission
Universal Asynchronous Receiver/Transmitter (UART)

72. Note: In asynchronous transmission, we send
1 start bit (0) at the beginning and
1 or more stop bits (1s) at the end of each byte.
There may be a gap between each byte.
Asynchronous here means “asynchronous at the byte level,” but the bits are still synchronized; their durations are the same.

73. Figure 4.27 Asynchronous transmission

74. It is the responsibility of the receiver to group the bits.
Note:
In synchronous transmission, we send bits one after another without start/stop bits or gaps.
It is the responsibility of the receiver to group the bits.

75. Figure 4.28 Synchronous transmission