Chapter 4 Digital Transmission Http netwk hannam ac

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Chapter 4 디지털 전송 (Digital Transmission) Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 1

Chapter 4 디지털 전송 (Digital Transmission) Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 1

4 -1 DIGITAL-TO-DIGITAL CONVERSION In this section, we see how we can represent digital

4 -1 DIGITAL-TO-DIGITAL CONVERSION In this section, we see how we can represent digital data by using digital signals. The conversion involves three techniques: line coding, block coding, and scrambling. Line coding is always needed; block coding and scrambling may or may not be needed. Topics discussed in this section: Line Coding Schemes Block Coding Scrambling Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 3

Example 4. 1 A signal is carrying data in which one data element is

Example 4. 1 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 between 0 and 1? Solution We assume that the average value of c is 1/2. The baud rate is then Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 7

회선 코딩(Line Coding) ■ 대역폭 Although the actual bandwidth of a digital signal is

회선 코딩(Line Coding) ■ 대역폭 Although the actual bandwidth of a digital signal is infinite, the effective bandwidth is finite. n 최소 대역폭 n 채널의 대역폭이 주어지면, 최대 데이터 율 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 8

Example 4. 2 The maximum data rate of a channel (see Chapter 3) is

Example 4. 2 The maximum data rate of a channel (see Chapter 3) is Nmax = 2 × B × log 2 L (defined by the Nyquist formula). Does this agree with the previous formula for Nmax? Solution A signal with L levels actually can carry log 2 L bits per level. If each level corresponds to one signal element and we assume the average case (c = 1/2), then we have Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 9

회선 코딩(Line Coding) n 동기화 결핍 효과 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY

회선 코딩(Line Coding) n 동기화 결핍 효과 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 11

Example 4. 3 In a digital transmission, the receiver clock is 0. 1 percent

Example 4. 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, the receiver receives 1001 bps instead of 1000 bps. At 1 Mbps, the receiver receives 1, 000 bps instead of 1, 000 bps. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 12

회선 코딩 방식 단극형 ■ 비영복귀 ■ Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY

회선 코딩 방식 단극형 ■ 비영복귀 ■ Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 13

단극형(Unipolar) ▶ 양 전압 비트 1, 음 전압 비트 0 Http: //netwk. hannam. ac.

단극형(Unipolar) ▶ 양 전압 비트 1, 음 전압 비트 0 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 14

비영복귀 (NRZ; non-return-to-zero) Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 16

비영복귀 (NRZ; non-return-to-zero) Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 16

NRZ-L and NRZ-I both have an average signal rate of N/2 Bd. Http: //netwk.

NRZ-L and NRZ-I both have an average signal rate of N/2 Bd. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 18

NRZ-L and NRZ-I both have a DC component problem. Http: //netwk. hannam. ac. kr

NRZ-L and NRZ-I both have a DC component problem. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 19

Example 4. 4 A system is using NRZ-I to transfer 10 -Mbps data. What

Example 4. 4 A system is using NRZ-I to transfer 10 -Mbps data. What are the average signal rate and minimum bandwidth? Solution The average signal rate is S = N/2 = 500 kbaud. The minimum bandwidth for this average baud rate is Bmin = S = 500 k. Hz. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 20

영복귀(RZ; return to zero) Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 21

영복귀(RZ; return to zero) Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 21

In Manchester and differential Manchester encoding, the transition at the middle of the bit

In Manchester and differential Manchester encoding, the transition at the middle of the bit is used for synchronization. The minimum bandwidth of Manchester and differential Manchester is 2 times that of NRZ. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 24

양극형(Bipolar) In bipolar encoding, we use three levels: positive, zero, and negative. Http: //netwk.

양극형(Bipolar) In bipolar encoding, we use three levels: positive, zero, and negative. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 25

AMI와 가삼진수(Pseudoternary) q 양극형 교대표시반전(AMI: Alternate Mark Inversion) Http: //netwk. hannam. ac. kr HANNAM

AMI와 가삼진수(Pseudoternary) q 양극형 교대표시반전(AMI: Alternate Mark Inversion) Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 26

In m. Bn. L schemes, a pattern of m data elements is encoded as

In m. Bn. L schemes, a pattern of m data elements is encoded as a pattern of n signal elements in which 2 m ≤ Ln. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 28

다준위 방식(Multilevel Schemes) q 4 D-PAM 5(4차원 5준위 펄스 진폭 변조) ■ Four-dimensional five-level

다준위 방식(Multilevel Schemes) q 4 D-PAM 5(4차원 5준위 펄스 진폭 변조) ■ Four-dimensional five-level pulse amplitude modulation ■ 4 D : 데이터가 4개의 회선으로 동시에 전송 ■ 5개의 준위 : -2, -1, 0, 1, 2 ■ 1 G bps LAN에 사용 ■ 478 – 256 = 222개의 신호는 동기화나 오류 검색에 사용 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 31

다준위 방식(Multilevel Schemes) q 다중회선 전송 : MLT-3 ■ Multiline transmission three level ■

다준위 방식(Multilevel Schemes) q 다중회선 전송 : MLT-3 ■ Multiline transmission three level ■ 3개의 준위( +1, 0, -1) 사용 ■ 100 M bps 전송에 적합한 방식 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 32

회선 부호화 요약 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 33

회선 부호화 요약 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 33

블록 코딩(Block Coding) q m 비트를 n 비트 블록으로 바꾼다 q n 은 m

블록 코딩(Block Coding) q m 비트를 n 비트 블록으로 바꾼다 q n 은 m 보다 크다 q m. B/n. B 부호화 Block coding is normally referred to as m. B/n. B coding; it replaces each m-bit group with an n-bit group. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 34

블록 코딩(Block Coding) q 4 B/5 B(4 Binary / 5 Binary) ■ NRG-I 방식과

블록 코딩(Block Coding) q 4 B/5 B(4 Binary / 5 Binary) ■ NRG-I 방식과 혼합하여 사용 ■ 4비트 데이터를 5비트 코드로 바꾼다 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 36

블록 코딩(Block Coding) q 4 B/5 B Mapping Code Http: //netwk. hannam. ac. kr

블록 코딩(Block Coding) q 4 B/5 B Mapping Code Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 37

블록 코딩(Block Coding) q 4 B/5 B(4 Binary / 5 Binary) 블록 코딩 예

블록 코딩(Block Coding) q 4 B/5 B(4 Binary / 5 Binary) 블록 코딩 예 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 38

Example 4. 5 We need to send data at a 1 -Mbps rate. What

Example 4. 5 We need to send data at a 1 -Mbps rate. What is the minimum required bandwidth, using a combination of 4 B/5 B and NRZ-I or Manchester coding? Solution First 4 B/5 B block coding increases the bit rate to 1. 25 Mbps. The minimum bandwidth using NRZ-I is N/2 or 625 k. Hz. The Manchester scheme needs a minimum bandwidth of 1 MHz. The first choice needs a lower bandwidth, but has a DC component problem; the second choice needs a higher bandwidth, but does not have a DC component problem. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 39

블록 코딩(Block Coding) q 8 B/10 B ■ 5 B/6 B와 3 B/4 B를

블록 코딩(Block Coding) q 8 B/10 B ■ 5 B/6 B와 3 B/4 B를 합한 것 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 40

스크램블링(Scrambling, 뒤섞기) q B 8 ZS ■ 양극 8열 대치(Bipolar with 8 zero substitution)

스크램블링(Scrambling, 뒤섞기) q B 8 ZS ■ 양극 8열 대치(Bipolar with 8 zero substitution) ■ 8개의 연속된 0을 000 VB 신호로 대치 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 42

스크램블링(Scrambling, 뒤섞기) q HDB 3 ■ 고밀도 양극 3 영(High-density bipolar 3 -zero) ■

스크램블링(Scrambling, 뒤섞기) q HDB 3 ■ 고밀도 양극 3 영(High-density bipolar 3 -zero) ■ 4개의 연속된 0을 000 V 나 B 00 V로 대치 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 43

스크램블링(Scrambling, 뒤섞기) q HDB 3 ■ HDB 3은 연속된 4개의 0을 마지막 대체 이후의

스크램블링(Scrambling, 뒤섞기) q HDB 3 ■ HDB 3은 연속된 4개의 0을 마지막 대체 이후의 0이 아닌 펄스의 개 수에 따라 000 V 나 B 00 V 로 대체한다 HDB 3 substitutes four consecutive zeros with 000 V or B 00 V depending on the number of nonzero pulses after the last substitution. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 44

4 -2 ANALOG-TO-DIGITAL CONVERSION We have seen in Chapter 3 that a digital signal

4 -2 ANALOG-TO-DIGITAL CONVERSION We have seen in Chapter 3 that a digital signal is superior to an analog signal. The tendency today is to change an analog signal to digital data. In this section we describe two techniques, pulse code modulation and delta modulation. Topics discussed in this section: Pulse Code Modulation (PCM) Delta Modulation (DM) Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 45

나이퀴스트 정리에 의하면, 채집율은 신호에 포 함된 최대 주파수의 최소한 두 배가 되어야 한다.

나이퀴스트 정리에 의하면, 채집율은 신호에 포 함된 최대 주파수의 최소한 두 배가 되어야 한다. According to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequency contained in the signal. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 49

Low-pass 와 bandpass 신호의 나이퀴스트 채집율 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 50

Low-pass 와 bandpass 신호의 나이퀴스트 채집율 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 50

Example 4. 6 For an intuitive example of the Nyquist theorem, let us sample

Example 4. 6 For an intuitive example of the Nyquist theorem, let us sample a simple sine wave at three sampling rates: fs = 4 f (2 times the Nyquist rate), fs = 2 f (Nyquist rate), and fs = f (one-half the Nyquist rate). Figure 4. 24 shows the sampling and the subsequent recovery of the signal. It can be seen that sampling at the Nyquist rate can create a good approximation of the original sine wave (part a). Oversampling in part b can also create the same approximation, but it is redundant and unnecessary. Sampling below the Nyquist rate (part c) does not produce a signal that looks like the original sine wave. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 51

Figure 4. 24 Recovery of a sampled sine wave for different sampling rates Http:

Figure 4. 24 Recovery of a sampled sine wave for different sampling rates Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 52

Example 4. 7 Consider the revolution of a hand of a clock. The second

Example 4. 7 Consider the revolution of a hand of a clock. The second hand of a clock has a period of 60 s. According to the Nyquist theorem, we need to sample the hand every 30 s (Ts = T or fs = 2 f ). In Figure 4. 25 a, the sample points, in order, are 12, 6, 12, and 6. The receiver of the samples cannot tell if the clock is moving forward or backward. In part b, we sample at double the Nyquist rate (every 15 s). The sample points are 12, 3, 6, 9, and 12. The clock is moving forward. In part c, we sample below the Nyquist rate (Ts = T or fs = f ). The sample points are 12, 9, 6, 3, and 12. Although the clock is moving forward, the receiver thinks that the clock is moving backward. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 53

Figure 4. 25 Sampling of a clock with only one hand Http: //netwk. hannam.

Figure 4. 25 Sampling of a clock with only one hand Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 54

Example 4. 8 An example related to Example 4. 7 is the seemingly backward

Example 4. 8 An example related to Example 4. 7 is the seemingly backward rotation of the wheels of a forward-moving car in a movie. This can be explained by under-sampling. A movie is filmed at 24 frames per second. If a wheel is rotating more than 12 times per second, the undersampling creates the impression of a backward rotation. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 55

Example 4. 9 Telephone companies digitize voice by assuming a maximum frequency of 4000

Example 4. 9 Telephone companies digitize voice by assuming a maximum frequency of 4000 Hz. The sampling rate therefore is 8000 samples per second. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 56

Example 4. 10 A complex low-pass signal has a bandwidth of 200 k. Hz.

Example 4. 10 A complex low-pass signal has a bandwidth of 200 k. Hz. 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 k. Hz). The sampling rate is therefore 400, 000 samples per second. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 57

Example 4. 11 A complex bandpass signal has a bandwidth of 200 k. Hz.

Example 4. 11 A complex bandpass signal has a bandwidth of 200 k. Hz. What is the minimum sampling rate for this signal? 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. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 58

채집(Sampling) q 펄스 코드 변조(PCM, Pulse Code Modulation) ■ 부호화(Encoding) ▶각 표본이 nb 비트의

채집(Sampling) q 펄스 코드 변조(PCM, Pulse Code Modulation) ■ 부호화(Encoding) ▶각 표본이 nb 비트의 부호로 바뀌는 것 ▶Bit rate = sampling rate X number of bits per sample = fs X nb Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 61

Quantization and encoding of a sampled signal Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY

Quantization and encoding of a sampled signal Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 62

Example 4. 12 What is the SNRd. B in the example of Figure 4.

Example 4. 12 What is the SNRd. B in the example of Figure 4. 26? Solution We can use the formula to find the quantization. We have eight levels and 3 bits per sample, so SNRd. B = 6. 02(3) + 1. 76 = 19. 82 d. B Increasing the number of levels increases the SNR. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 63

Example 4. 13 A telephone subscriber line must have an SNRd. B above 40.

Example 4. 13 A telephone subscriber line must have an SNRd. B above 40. What is the minimum number of bits per sample? Solution We can calculate the number of bits as Telephone companies usually assign 7 or 8 bits per sample. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 64

Example 4. 14 We want to digitize the human voice. What is the bit

Example 4. 14 We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample? Solution The human voice normally contains frequencies from 0 to 4000 Hz. So the sampling rate and bit rate are calculated as follows: Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 65

PCM 복호기 구성 요소 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 66

PCM 복호기 구성 요소 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 66

Example 4. 15 We have a low-pass analog signal of 4 k. Hz. If

Example 4. 15 We have a low-pass analog signal of 4 k. Hz. If we send the analog signal, we need a channel with a minimum bandwidth of 4 k. Hz. If we digitize the signal and send 8 bits per sample, we need a channel with a minimum bandwidth of 8 × 4 k. Hz = 32 k. Hz. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 67

델타 변조(DM, Delta Modulation) Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 68

델타 변조(DM, Delta Modulation) Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 68

델타 변조 구성 요소 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 69

델타 변조 구성 요소 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 69

델타 복조 구성 요소 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 70

델타 복조 구성 요소 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 70

4 -3 전송 모드(TRANSMISSION MODE) The transmission of binary data across a link can

4 -3 전송 모드(TRANSMISSION MODE) The transmission of binary data across a link can be accomplished in either parallel or serial mode. In parallel mode, multiple bits are sent with each clock tick. In serial mode, 1 bit is sent with each clock tick. While there is only one way to send parallel data, there are three subclasses of serial transmission: asynchronous, and isochronous. Topics discussed in this section: Parallel Transmission Serial Transmission Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 71

데이터 전송과 모드 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 72

데이터 전송과 모드 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 72

병렬 전송 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 73

병렬 전송 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 73

직렬 전송 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 75

직렬 전송 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 75

비동기식 전송 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 77

비동기식 전송 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 77

In asynchronous transmission, we send 1 start bit (0) at the beginning and 1

In asynchronous transmission, we send 1 start bit (0) at the beginning and 1 or more stop bits (1 s) at the end of each byte. There may be a gap between each byte. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 79

Asynchronous here means “asynchronous at the byte level, ” but the bits are still

Asynchronous here means “asynchronous at the byte level, ” but the bits are still synchronized; their durations are the same. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 80

동기식 전송 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 81

동기식 전송 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 81

In synchronous transmission, we send bits one after another without start or stop bits

In synchronous transmission, we send bits one after another without start or stop bits or gaps. It is the responsibility of the receiver to group the bits. Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 82

요약 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 84

요약 Http: //netwk. hannam. ac. kr HANNAM UNIVERSITY 84