# Chapter 4 Digital Transmission - PowerPoint PPT Presentation

Chapter 4 Digital Transmission

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Chapter 4 Digital Transmission

## Chapter 4 Digital Transmission

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1. Chapter 4 Digital Transmission

2. Data Analog: Continuous value data (sound, light, temperature) Digital: Discrete value (text, integers, symbols) Signal Analog: Continuously varying electromagnetic wave Digital: Series of voltage pulses (square wave) Components of Data Communication

3. Analog data to analog signal Inexpensive, easy conversion (eg telephone) Used in traditional analog telephony Analog data to digital signal Requires a codec (encoder/decoder) Allows use of digital telephony, voice mail Analog Data-->Signal Options

4. Digital data to analog signal Requires modem (modulator/demodulator) Necessary when analog transmission is used Digital data to digital signal Less expensive when large amounts of data are involved More reliable because no conversion is involved Digital Data-->Signal Options

5. 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 Line Coding SchemesBlock Coding Scrambling

6. Figure 4.1 Line coding and decoding

7. Figure 4.2 Signal element versus data element r = number of data elements / number of signal elements

8. Data Rate Vs. Signal Rate • Data rate: the number of data elements (bits) sent in 1s (bps). It’s also called the bit rate • Signal rate: the number of signal elements sent in 1s (baud). It’s also called the pulse rate, the modulation rate, or the baud rate. • We wish to: • 1. increase the data rate (increase the speed of transmission) • 2. decrease the signal rate (decrease the bandwidth requirement) • Worst case, best case, and average case of r • S = c * N / r baud

9. Baseline wandering Baseline: running average of the received signal power DC Components Constant digital signal creates low frequencies Self-synchronization Receiver Setting the clock matching the sender’s

10. Figure 4.3 Effect of lack of synchronization

11. Figure 4.4 Line coding schemes

12. Figure 4.5 Unipolar NRZ scheme

13. Most common, easiest method is different voltage levels for the two binary digits Typically, negative=1 and positive=0 Known as NRZ-L, or nonreturn-to-zero level, because signal never returns to zero, and the voltage during a bit transmission is level Digital Encodingof Digital Data

14. Differential version is NRZI (NRZ, invert on ones) Change=1, no change=0 Advantage of differential encoding is that it is more reliable to detect a change in polarity than it is to accurately detect a specific level Differential NRZ

15. Difficult to determine where one bit ends and the next begins In NRZ-L, long strings of ones and zeroes would appear as constant voltage pulses Timing is critical, because any drift results in lack of synchronization and incorrect bit values being transmitted Problems With NRZ

16. Figure 4.6 Polar NRZ-L and NRZ-I schemes

17. Figure 4.7 Polar RZ scheme

18. Transition in the middle of each bit period Transition provides clocking and data Low-to-high=1 , high-to-low=0 Used in Ethernet Manchester Code

19. Midbit transition is only for clocking Transition at beginning of bit period=0 Transition absent at beginning=1 Has added advantage of differential encoding Used in token-ring Differential Manchester

20. Figure 4.8 Polar biphase: Manchester and differential Manchester schemes

21. High=0, Low=1 • No change at begin=0, Change at begin=1 • H-to-L=0, L-to-H=1 • Change at begin=0, No change at begin=1

22. Bipolar schemes: AMI (Alternate Mark Inversion) and pseudoternary

23. Multilevel Schemes • In mBnL schemes, a pattern of m data elements is encoded as a pattern of n signal elements in which 2m ≤ Ln • m: the length of the binary pattern • B: binary data • n: the length of the signal pattern • L: number of levels in the signaling • B for l=2 binary • T for l=3 ternary • Q for l=4 quaternary

24. Figure 4.10 Multilevel: 2B1Q scheme Used in DSL

25. Figure 4.11 Multilevel: 8B6T scheme

26. Figure 4.13 Multitransition: MLT-3 scheme

27. Block Coding • Redundancy is needed to ensure synchronization and to provide error detecting • Block coding is normally referred to as mB/nB coding • it replaces each m-bit group with an n-bit group • m < n

28. Figure 4.14 Block coding concept

29. Table 4.2 4B/5B mapping codes

30. Figure 4.16 Substitution in 4B/5B block coding

31. Figure 4.17 8B/10B block encoding

32. Scrambling • It modifies the bipolar AMI encoding (no DC component, but having the problem of synchronization) • It does not increase the number of bits • It provides synchronization • It uses some specific form of bits to replace a sequence of 0s

33. Figure 4.19 Two cases of B8ZS scrambling technique B8ZS substitutes eight consecutive zeros with 000VB0VB

34. Figure 4.20 Different situations in HDB3 scrambling technique HDB3 substitutes four consecutive zeros with 000V or B00V depending on the number of nonzero pulses after the last substitution.

35. 4-2 ANALOG-TO-DIGITAL CONVERSION The tendency today is to change an analog signal to digital data. In this section we describe two techniques, pulse code modulationanddelta modulation.

36. Figure 4.21 Components of PCM encoder

37. According to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequency contained in the signal. What can we get from this: 1. we can sample a signal only if the signal is band-limited 2. the sampling rate must be at least 2 times the highest frequency, not the bandwidth

38. Figure 4.25 Sampling of a clock with only one hand

39. Example An example related 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 under-sampling creates the impression of a backward rotation.

40. 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.

41. Example A complex bandpass signal has a bandwidth of 200 kHz. 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.

42. Figure 4.26 Quantization and encoding of a sampled signal

43. Contribution of the quantization error to SNRdb SNRdb= 6.02nb + 1.76 dB nb: bits per sample (related to the number of level L) What is the SNRdB in the example of Figure 4.26? Solution We have eight levels and 3 bits per sample, so SNRdB = 6.02 x 3 + 1.76 = 19.82 dB Increasing the number of levels increases the SNR.

44. Example A telephone subscriber line must have an SNRdB 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.

45. PCM decoder: recovers the original signal

46. The minimum bandwidth of the digital signal is nb times greater than the bandwidth of the analog signal. Bmin= nb x Banalog We have a low-pass analog signal of 4 kHz. If we send the analog signal, we need a channel with a minimum bandwidth of 4 kHz. If we digitize the signal and send 8 bits per sample, we need a channel with a minimum bandwidth of 8 × 4 kHz = 32 kHz.