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  1. 1. 4.1Chapter 4Digital Transmission
  2. 2. 4.2Components of DataCommunication Data Analog: Continuous value data (sound, light,temperature) Digital: Discrete value (text, integers,symbols) Signal Analog: Continuously varyingelectromagnetic wave Digital: Series of voltage pulses (squarewave)
  3. 3. 4.3Analog Data-->Signal Options 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
  4. 4. 4.4Digital Data-->Signal Options 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 dataare involved More reliable because no conversion isinvolved
  5. 5. 4.54-1 DIGITAL-TO-DIGITAL CONVERSION4-1 DIGITAL-TO-DIGITAL CONVERSIONIn this section, we see how we can represent digitalIn this section, we see how we can represent digitaldata by using digital signals. The conversion involvesdata by using digital signals. The conversion involvesthree techniques:three techniques: line codingline coding,, block codingblock coding, and, andscramblingscrambling. Line coding is always needed; block. Line coding is always needed; blockcoding and scrambling may or may not be needed.coding and scrambling may or may not be needed.Line CodingLine Coding SchemesBlock CodingScramblingTopics discussed in this section:Topics discussed in this section:
  6. 6. 4.6Figure 4.1 Line coding and decoding
  7. 7. 4.7Figure 4.2 Signal element versus data elementr = number of data elements / number of signal elements
  8. 8. 4.8Data Rate Vs. Signal Rate•Data rate: the number of data elements (bits) sentin 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 modulationrate, or the baud rate.We wish to:1. increase the data rate (increase the speed oftransmission)2. decrease the signal rate (decrease the bandwidthrequirement)3. Worst case, best case, and average case of r4. S = c * N / r baud
  9. 9. 4.9Baseline wanderingBaseline: running average of thereceived signal powerDC ComponentsConstant digital signal creates lowfrequenciesSelf-synchronizationReceiver Setting the clock matching thesender’s
  10. 10. 4.10Figure 4.3 Effect of lack of synchronization
  11. 11. 4.11Figure 4.4 Line coding schemes
  12. 12. 4.12Figure 4.5 Unipolar NRZ scheme
  13. 13. 4.13Digital Encodingof Digital Data Most common, easiest method is differentvoltage levels for the two binary digits Typically, negative=1 and positive=0 Known as NRZ-L, or nonreturn-to-zerolevel, because signal never returns to zero,and the voltage during a bit transmission islevel
  14. 14. 4.14Differential NRZ Differential version is NRZI (NRZ, inverton ones) Change=1, no change=0 Advantage of differential encoding is that itis more reliable to detect a change inpolarity than it is to accurately detect aspecific level
  15. 15. 4.15Problems With NRZ Difficult to determine where one bit endsand the next begins In NRZ-L, long strings of ones and zeroeswould appear as constant voltage pulses Timing is critical, because any drift resultsin lack of synchronization and incorrect bitvalues being transmitted
  16. 16. 4.16Figure 4.6 Polar NRZ-L and NRZ-I schemes
  17. 17. 4.17Figure 4.7 Polar RZ scheme
  18. 18. 4.18Manchester Code Transition in the middle of each bit period Transition provides clocking and data Low-to-high=1 , high-to-low=0 Used in Ethernet
  19. 19. 4.19Differential Manchester Midbit transition is only for clocking Transition at beginning of bit period=0 Transition absent at beginning=1 Has added advantage of differentialencoding Used in token-ring
  20. 20. 4.20Figure 4.8 Polar biphase: Manchester and differential Manchester schemes
  21. 21. 4.21• High=0, Low=1• No change at begin=0, Change atbegin=1• H-to-L=0, L-to-H=1• Change at begin=0, No change atbegin=1
  22. 22. 4.22Bipolar schemes: AMI (Alternate Mark Inversion) and pseudoternary
  23. 23. 4.23Multilevel Schemes• In mBnL schemes, a pattern of mdata elements is encoded as apattern of n signal elements inwhich 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. 24. 4.24Figure 4.10 Multilevel: 2B1Q schemeUsed in DSL
  25. 25. 4.25Figure 4.11 Multilevel: 8B6T scheme
  26. 26. 4.26Figure 4.13 Multitransition: MLT-3 scheme
  27. 27. 4.27Table 4.1 Summary of line coding schemesPolar
  28. 28. 4.28Block Coding• Redundancy is needed to ensuresynchronization and to provideerror detecting• Block coding is normally referred toas mB/nB coding• it replaces each m-bit group with ann-bit group• m < n
  29. 29. 4.29Figure 4.14 Block coding concept
  30. 30. 4.30Figure 4.15 Using block coding 4B/5B with NRZ-I line coding scheme
  31. 31. 4.31Table 4.2 4B/5B mapping codes
  32. 32. 4.32Figure 4.16 Substitution in 4B/5B block coding
  33. 33. 4.33Figure 4.17 8B/10B block encoding
  34. 34. 4.34Scrambling• It modifies the bipolar AMI encoding(no DC component, but having theproblem of synchronization)• It does not increase the number ofbits• It provides synchronization• It uses some specific form of bits toreplace a sequence of 0s
  35. 35. 4.35Figure 4.19 Two cases of B8ZS scrambling techniqueB8ZS substitutes eight consecutive zeros with 000VB0VB
  36. 36. 4.36Figure 4.20 Different situations in HDB3 scrambling techniqueHDB3 substitutes four consecutive zeros with 000V or B00V dependingon the number of nonzero pulses after the last substitution.
  37. 37. 4.374-2 ANALOG-TO-DIGITAL CONVERSION4-2 ANALOG-TO-DIGITAL CONVERSIONThe tendency today is to change an analog signal toThe tendency today is to change an analog signal todigital data.In this section we describe two techniques,In this section we describe two techniques,pulse code modulationpulse code modulation andand delta modulationdelta modulation..
  38. 38. 4.38Figure 4.21 Components of PCM encoder
  39. 39. 4.39According to the Nyquist theorem, thesampling rate must be at least 2 timesthe highest frequency contained in thesignal.What can we get from this:1. we can sample a signal only if the signal isband-limited2. the sampling rate must be at least 2 times thehighest frequency, not the bandwidth
  40. 40. 4.40Figure 4.23 Nyquist sampling rate for low-pass and bandpass signals
  41. 41. 4.41Figure 4.24 Recovery of a sampled sine wave for different sampling rates
  42. 42. 4.42Figure 4.25 Sampling of a clock with only one hand
  43. 43. 4.43An example related is the seemingly backward rotation ofthe wheels of a forward-moving car in a movie.This can be explained by under-sampling. A movie isfilmed at 24 frames per second. If a wheel is rotatingmore than 12 times per second, the under-samplingcreates the impression of a backward rotation.Example
  44. 44. 4.44A complex low-pass signal has a bandwidth of 200 kHz.What is the minimum sampling rate for this signal?SolutionThe 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 thehighest frequency (200 kHz). The sampling rate istherefore 400,000 samples per second.Example
  45. 45. 4.45A complex bandpass signal has a bandwidth of 200 kHz.What is the minimum sampling rate for this signal?SolutionWe cannot find the minimum sampling rate in this casebecause we do not know where the bandwidth starts orends. We do not know the maximum frequency in thesignal.Example
  46. 46. 4.46Figure 4.26 Quantization and encoding of a sampled signal
  47. 47. 4.47What is the SNRdB in the example of Figure 4.26?SolutionWe have eight levels and 3 bits per sample, soSNRdB = 6.02 x 3 + 1.76 = 19.82 dBIncreasing the number of levels increases the SNR.Contribution of the quantization error to SNRdbSNRdb= 6.02nb + 1.76 dBnb: bits per sample (related to the number of level L)
  48. 48. 4.48A telephone subscriber line must have an SNRdB above40. What is the minimum number of bits per sample?SolutionWe can calculate the number of bits asExampleTelephone companies usually assign 7 or 8 bits persample.
  49. 49. 4.49PCM decoder: recovers the original signal
  50. 50. 4.50We have a low-pass analog signal of 4 kHz. If we send theanalog signal, we need a channel with a minimumbandwidth of 4 kHz. If we digitize the signal and send 8bits per sample, we need a channel with a minimumbandwidth of 8 × 4 kHz = 32 kHz.The minimum bandwidth of the digital signal is nbtimes greater than the bandwidth of the analogsignal.Bmin= nb x Banalog
  51. 51. 4.51DM (delta modulation) finds the change from theprevious sampleNext bit is 1, if amplitude of the analog signal is largerNext bit is 0, if amplitude of the analog signal is smaller
  52. 52. 4.52Figure 4.29 Delta modulation components
  53. 53. 4.53Figure 4.30 Delta demodulation components
  54. 54. 4.544-3 TRANSMISSION MODES4-3 TRANSMISSION MODES1. The transmission of binary data across a link can1. The transmission of binary data across a link canbe accomplished in either parallel or serial accomplished in either parallel or serial mode.2. In parallel mode, multiple bits are sent with each2. In parallel mode, multiple bits are sent with eachclock tick.clock tick.3. In serial mode, 1 bit is sent with each clock tick.3. In serial mode, 1 bit is sent with each clock tick.4. there are three subclasses of serial transmission:4. there are three subclasses of serial transmission:asynchronous, synchronous, and isochronous.asynchronous, synchronous, and isochronous.
  55. 55. 4.55Figure 4.31 Data transmission and modes
  56. 56. 4.56Figure 4.32 Parallel transmission
  57. 57. 4.57Figure 4.33 Serial transmission
  58. 58. 4.58Asynchronous transmission1. We send 1 start bit (0) at the beginning and 1 or more stop bits(1s) at the end of each byte.2. There may be a gap between each byte.3. Extra bits and gaps are used to alert the receiver, and allow it tosynchronize with the data stream.4. Asynchronous here means “asynchronous at the byte level,”but the bits are still synchronized, their durations are the same.
  59. 59. 4.59Synchronous transmissionIn synchronous transmission, we send bits one afteranother without start or stop bits or gaps. It is theresponsibility of the receiver to group the bits.