Without gaps and start and stop bits, there is no built-in mechanism to help thereceiving device adjust its bit synchronization midstream. Timing becomes very important,therefore, because the accuracy of the received information is completely dependenton the ability ofthe receiving device to keep an accurate count of the bits as they come in.SECTION 4.5 KEY TERMS 135The advantage of synchronous transmission is speed. With no extra bits or gaps tointroduce at the sending end and remove at the receiving end, and, by extension, withfewer bits to move across the link, synchronous transmission is faster than asynchronoustransmission. For this reason, it is more useful for high-speed applications such asthe transmission of data from one computer to another. Byte synchronization is accomplishedin the data link layer.We need to emphasize one point here. Although there is no gap between charactersin synchronous serial transmission, there may be uneven gaps between frames.IsochronousIn real-time audio and video, in which uneven delays between frames are not acceptable,synchronous transmission fails. For example, TV images are broadcast at the rateof 30 images per second; they must be viewed at the same rate. If each image is sentby using one or more frames, there should be no delays between frames. For this typeof application, synchronization between characters is not enough; the entire stream ofbits must be synchronized. The isochronous transmission guarantees that the dataarrive at a fixed rate.4.4 RECOMMENDED READINGFor more details about subjects discussed in this chapter, we recommend the followingbooks. The items in brackets [...] refer to the reference list at the end of the text.BooksDigital to digital conversion is discussed in Chapter 7 of [Pea92], Chapter 3 of[CouOl], and Section 5.1 of [Sta04]. Sampling is discussed in Chapters 15, 16, 17, and18 of [Pea92], Chapter 3 of [CouO!], and Section 5.3 of [Sta04]. [Hsu03] gives a goodmathematical approach to modulation and sampling. More advanced materials can befound in [Ber96].4.5 KEY TERMSadaptive delta modulationalternate mark inversion (AMI)analog-to-digital conversionasynchronous transmissionbaselinebaseline wanderingbaud ratebiphasebipolarbipolar with 8-zero substitution (B8ZS)bit rateblock codingcompanding and expandingdata elementdata rateDC componentdelta modulation (DM)differential Manchesterdigital-to-digital conversiondigitization136 CHAPTER 4 DIGITAL TRANSMISSIONeight binary/ten binary (8B/lOB)eight-binary, six-ternary (8B6T)four binary/five binary (4B/5B)four dimensional, five-level pulseamplitude modulation (4D-PAM5)high-density bipolar 3-zero (HDB3)isochronous transmissionline codingManchestermodulation ratemultilevel binarymultiline transmission, 3 level (MLT-3)nonreturn to zero (NRZ)nonreturn to zero, invert (NRZ-I)nonreturn to zero, level (NRZ-L)Nyquist theoremparallel transmissionpolarpseudoternarypulse amplitude modulation (PAM)pulse code modulation (PCM)pulse ratequantizationquantization errorreturn to zero (RZ)samplingsampling ratescramblingself-synchronizingserial transmissionsignal elementsignal ratestart bitstop bitsynchronous transmissiontransmission modetwo-binary, one quaternary (2B I Q)unipolar4.6 SUMMARYo Digital-to-digital conversion involves three techniques: line coding, block coding,and scrambling.o Line coding is the process of converting digital data to a digital signal.o We can roughly divide line coding schemes into five broad categories: unipolar,polar, bipolar, multilevel, and multitransition.o Block coding provides redundancy to ensure synchronization and inherent errordetection. Block coding is normally referred to as mB/nB coding; it replaces eachm-bit group with an n-bit group.o Scrambling provides synchronization without increasing the number of bits. Twocommon scrambling techniques are B8ZS and HDB3.o The most common technique to change an analog signal to digital data (digitization)is called pulse code modulation (PCM).o The first step in PCM is sampling. The analog signal is sampled every Ts s, where Tsis the sample interval or period. The inverse of the sampling interval is called thesampling rate or sampling frequency and denoted by fs, where fs =lITs. There arethree sampling methods-ideal, natural, and flat-top.o According to the Nyquist theorem, to reproduce the original analog signal, onenecessary condition is that the sampling rate be at least twice the highest frequencyin the original signal.SECTION 4.7 PRACTICE SET 137o Other sampling techniques have been developed to reduce the complexity of PCM.The simplest is delta modulation. PCM finds the value of the signal amplitude foreach sample; DM finds the change from the previous sample.o While there is only one way to send parallel data, there are three subclasses ofserial transmission: asynchronous, synchronous, and isochronous.o In asynchronous transmission, we send 1 start bit (0) at the beginning and 1 ormore stop bits (1 s) at the end of each byte.o In synchronous transmission, we send bits one after another without start or stopbits or gaps. It is the responsibility of the receiver to group the bits.o The isochronous mode provides synchronized for the entire stream of bits must. Inother words, it guarantees that the data arrive at a fixed rate.4.7 PRACTICE SETReview Questions1. List three techniques of digital-to-digital conversion.2. Distinguish between a signal element and a data element.3. Distinguish between data rate and signal rate.4. Define baseline wandering and its effect on digital transmission.5. Define a DC component and its effect on digital transmission.6. Define the characteristics of a self-synchronizing signal.7. List five line coding schemes discussed in this book.8. Define block coding and give its purpose.9. Define scrambling and give its purpose.10. Compare and contrast PCM and DM.11. What are the differences between parallel and serial transmission?12. List three different techniques in serial transmission and explain the differences.Exercises13. Calculate the value of the signal rate for each case in Figure 4.2 if the data rate is1 Mbps and c = 1/2.14. In a digital transmission, the sender clock is 0.2 percent faster than the receiver clock.How many extra bits per second does the sender send if the data rate is 1 Mbps?15. Draw the graph of the NRZ-L scheme using each of the following data streams,assuming that the last signa11evel has been positive. From the graphs, guess thebandwidth for this scheme using the average number of changes in the signal level.Compare your guess with the corresp.onding entry in Table 4.1.a. 00000000b. 11111111c. 01010101d. 00110011138 CHAPTER 4 DIGITAL TRANSMISSION16. Repeat Exercise 15 for the NRZ-I scheme.17. Repeat Exercise 15 for the Manchester scheme.18. Repeat Exercise 15 for the differential Manchester scheme.19. Repeat Exercise 15 for the 2B1Q scheme, but use the following data streams.a. 0000000000000000b. 1111111111111111c. 0101010101010101d. 001100110011001120. Repeat Exercise 15 for the MLT-3 scheme, but use the following data streams.a. 00000000b. 11111111
c. 01010101
d. 00011000
21. Find the 8-bit data stream for each case depicted in Figure 4.36.
đang được dịch, vui lòng đợi..
