数据通信基础：第4章 数字传输

1、数据通信基础第4章 (数字传输)20110916第4章内容提要第4章 数字传输4.1 线路码型4.2 分块编码4.3 采样4.4 传输模式201109164.1 Line CodingTransmission can be digital or analog. In this chapter we will discuss digital form. Definition:Line Coding is the process of converting a series of binary data bits to a digital signal.WHY do we need Line co

2、ding?Answer: Digital signal should be converted to a special required form, which is suitable for the transmission in media.Figure 4.1 Line codingDigital dataDigital signalCan be sent to media2011091620110916Some Characteristics of Line CodingTo design a special kind of line coding, we need to defin

3、e some key termsSignal level 信号电平 data level 数据电平Pulse rate 脉冲速率 bit rate 比特速率Pulse rate = numbers of PULSEs per secondL numbers of signal levelsBitRate = PulseRate * log2LDC (Direct Current) componentsSelf-synchronizationFigure 4.2 Signal level versus data leveltwoNote error in bookData LevelSignal

4、 Level2011091620110916Example of multi-pulse multi-bitWe have seen this example in chapter 3.There are FOUR signal levels, which contain log24 = 2 bits per pulse.In above example, PulseRate = 4 pulses/s, while BitRate = 8 bits/s = 4*2 = PulseRate * log2L0V3V2V1V011110001 s20110916Example 1A signal h

5、as two data levels with a pulse duration of 1 ms. We calculate the pulse rate and bit rate as follows:Pulse Rate = 1/ 10-3= 1000 pulses/sBit Rate = Pulse Rate x log2 L = 1000 x log2 2 = 1000 bpsNote here, bps = bit per second20110916Example 2A signal has four data levels with a pulse duration of 1 m

6、s. We calculate the pulse rate and bit rate as follows:Pulse Rate = = 1000 pulses/sBit Rate = Pulse Rate x log2 L = 1000 x log2 4 = 2000 bps00011011Four levels, each level can represent TWO bits.Figure 4.3 DC component2011091620110916SynchronizationSynchronization is very important for digital signa

7、ls.If there is no information about when to start a new bit and when to stop, the receiver will have difficulty in converting signal into digital data bits.The clock rate at receiver should be EXACTLY the same as that in sender. Otherwise, there will be ERROR at receiver.Line coding should contain i

8、nformation for self-synchronization.Here is an example for lack of sync between sender and receiver.Figure 4.4 Lack of synchronizationThe waveform is the same, while we have different results at receiver.2011091620110916Example 3In a digital transmission, the receiver clock is 0.1 percent faster tha

9、n 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? SolutionAt 1 Kbps:1000 bits sent 1001 bits received1 extra bpsAt 1 Mbps: 1,000,000 bits sent 1,001,000 bits received1000 extra bps 20110916Figure 4.5 Line cod

10、ing schemesLine Coding SchemesThere are three BASIC types of line coding.Unipolar 单极性码Polar 双极性码Bipolar 极性码20110916Unipolar encoding uses only one voltage level. The other one is 0 or ground level. Note:Disadvantages:There is a DC componentThere is no synchronisation information0 VFigure 4.6 Unipola

11、r encodingData 0 is represented by a waveform of pulse with 0 V; Data 1 is represented by a pulse of +A V.+A V2011091620110916Polar encoding uses two voltage levels (positive and negative).Note:The encoding can be arranged to cancel out any DC component.Figure 4.7 Types of polar encodingNRZ = Non-Re

12、turn to Zero 非归零码NRZ-L = Non-Return to Zero level 非归零-电平码NRZ-I = Non-Return to Zero Invert 非归零-反转码RZ = Return to Zero 归零码2011091620110916In NRZ-L the level of the signal is dependent upon the state of the bit.Long streams of 1s or 0s cause problems because there is no sync. informationNote:In NRZ-I

13、the signal is inverted if a 1 is encountered.Figure 4.8 NRZ-L and NRZ-I encoding0 = +A V; 1 = -A V0 = no level change; 1 = has level change2011091620110916A good encoded digital signal must contain a provision for synchronization, i.e. the signal level should change at least once per bit.Note:Figure

14、 4.9 RZ encodingThe bit value is given by the transition at the centre of the bit:Logic 1 = +V 0Logic 0 = -V 0both bits return to zeroDisadvantage: 3 signal levels required 2 transitions per bit, more bandwidth required20110916Figure 4.10 Manchester encoding 曼彻斯特编码Manchester encoding achieves same l

15、evel of sync. as RZ, but with only 2 signal levelsNOTE the definition of Manchester coding here. Zero data bit is represented by transition from high to low level, while One data bit is a transition from low to high.It is possible to define another set of Manchester coding.2011091620110916In Manches

16、ter encoding, the transition at the middle of the bit is used for both synchronization and bit representation.Note:20110916In 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

17、 at the beginning of the bit.Note:Figure 4.11 Differential Manchester encoding 差分曼彻斯特编码0 = no level change; 1 = has level change2011091620110916In bipolar encoding 双极性编码, we use three levels: positive, zero, and negative.Unlike RZ, zero level is used to represent logic zeroNote:Figure 4.12 Bipolar A

18、MI encodingAMI = Alternate mark inversion 交替极性反转码Mark传号 is an old telegraphy term meaning logic 1Space空号 is logic 0Note: no synchronisation in a stream of zeroes20110916Figure 4.13 2B1QOther Schemes: 2 binary, one quaternary, 2B1QQuaternary means four levels.20110916201109164.2 Block CodingSteps in

19、TransformationSome Common Block Codes20110916Why need Block CodingTo improve performance of line codingTwo goals:Redundancy for synchronizationRedundancy for error detectionThere are many forms of Block coding in use, for example, 4B/5B, 8B/10B.20110916Steps in TransformationSTEP 1 DivisionDivide th

20、e bit stream into groups of m bitsSTEP 2 SubstitutionConvert m bits into n (nm) bitsSTEP 3 Line CodingChange bits into signals using line coding schemesFigure 4.15 Block codingStep 1Step 2Step 320110916Figure 4.16 Substitution in block codingBlock codes contain extra redundant bits to aid error dete

21、ction (only certain patterns are allowed) but this costs bandwidth20110916Table 4.1 4B/5B encodingDataCodeDataCode0000111101000 10010000101001100110011001010100101010110001110101101110111010001010110011010010101011110111011011001110111011100011101111111111101Note: No more than 1 leading zero, or 2 t

22、railing zeroesMax number of consecutive zeroes is therefore = 320110916Table 4.1 4B/5B encoding (Continued)DataCodeQ (Quiet)00000I (Idle) 11111H (Halt) 00100J (start delimiter)11000K (start delimiter)10001T (end delimiter)01101S (Set)11001R (Reset)00111Control CharactersNote: rule about leading and

23、trailing zeroes is not followed20110916Figure 4.17 Example of 8B/6T encoding8B/10B schemes exists and provide better performance than 4B/5B8B/6T:Block of 8 Bits coded to 6 ternary (multiple) levelsRequires increased SNR, but reduced bandwidth 20110916201109164.3 SamplingPulse Amplitude Modulation -

24、PAMPulse Code Modulation - PCMSampling Rate: Nyquist TheoremHow Many Bits per Sample?Bit RateFigure 4.18 PAMPAM only digitalizes 数字化 signal in the time (horizontal) axis.2011091620110916Pulse amplitude modulation 脉冲幅度调制 has some applications, but it is not used by itself in data communication. Howev

25、er, it is the first step in another very popular conversion method called pulse code modulation脉冲编码调制.Note:Figure 4.19 Quantized PAM signalQuantized 量化 PAM signal digitalizes both time and amplitude axes.20110916Figure 4.20 Quantizing by using sign and magnitudeNote: Sign + Magnitude, NOT 2s complem

26、ent20110916Figure 4.22 From analog signal to PCM digital code2011091620110916According to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequency.Note:orSampling frequencySampling rate根据奈奎斯特定理，采样率至少必须是信号最高频率的二倍。Figure 4.23 Nyquist theorem1/2xNote the error in textbook20

27、11091620110916Example 4What sampling rate is needed for a signal with a bandwidth of 10,000 Hz (1000 to 11,000 Hz)? SolutionThe sampling rate must be twice the highest frequency in the signal: Sampling rate = 2 x (11,000) = 22,000 samples/s20110916Note that we can always change a band-pass signal to

28、 a low-pass signal before sampling. In this case, the sampling rate is twice the bandwidth.Note:20110916How Many Bits per Sample?We can decide the sampling rate using the Nyquist Theorem.After quantization, samples can be converted into levels of signal, which can be further represented by bits.How

29、many bit needed for each sample? That depend on the precision required.The more precise, the more bits.20110916Example 5A 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?SolutionWe need 4 bits; 1 bit for th

30、e 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 + sign is not enough since 22 = 4. A 4-bit value + sign is too much because 24 = 16. 20110916Example 6We want to digitize the human voice. What is the bit rate, ass

31、uming 8 bits per sample?SolutionThe human voice normally contains frequencies from 0 to 4000 Hz. Sampling rate = 4000 x 2 = 8000 samples/sBit rate = sampling rate x number of bits per sample = 8000 x 8 = 64,000 bps = 64 kbps201109164.4 Transmission ModeParallel TransmissionSerial TransmissionFigure 4.24 Data transmissionTransmission Mode20110916并行同步串行异步串行Figure 4.25 Parallel transmissionIn parallel mode, multiple bits are sent with each clock tick.Multiple bits can be sent simultaneously over several wires.Advantage: Speed; Disadvantage: CostE