Digital transmission


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Digital transmission

  1. 1. Data Communications By: Faiza Tariq Week 4 – Digital Transmission
  2. 2. References  DCN By Behrouz Forouzan  DCN By: William Stallings  Computer Networks By: Andrew Tannenbaum
  3. 3. 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 Schemes Block Coding Scrambling 4.# 3
  4. 4. Line Encoding  Process of converting binary data – a sequence of numbers into digital signal  Signal Level vs. Data Level Signal Level (Level of voltage e.g. +9,0, -3 are three signal levels  Data Level means ( One level for binary 0 another for binary 1 total two levels) 
  5. 5. Line coding and decoding
  6. 6. Signal element versus data element Data elements are data bits carried while signal element is the signal which is carrier of information like car and passenger. In some cases on driver can drive car + trailer.
  7. 7. Note Although the actual bandwidth of a digital signal is infinite, the effective bandwidth is finite.
  8. 8. Pulse Rate vs. Bit Rate  Pulse Rate/Signal Rate/Baud Rate/Modulation Rate  No. of signal elements or pulses per second  Bit Rate  No. of bits per second  One goal in data communications is to increase the data rate (increase transmission speed) while decreasing the baud rate (decrease the bandwidth requirement). More people in fewer vehicles to prevent traffic jam – limited roads - Bw  If one pulse carries one bit then bit rate and pulse rate is same  Bit Rate = Pulse Rate X log2L  Where L is data level
  9. 9. Example 1 A signal has 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/s Bit Rate = Pulse Rate x log2 L = 1000 x log2 2 = 1000 bps
  10. 10. Example 2 A signal has four data levels with a pulse duration of 1 ms. We calculate the pulse rate and bit rate as follows: Pulse Rate = = 1000 pulses/s Bit Rate = PulseRate x log2 L = 1000 x log2 4 = 2000 bps
  11. 11. Synchronization  To correctly interpret the signals received from the sender, the receiver’s bit interval must correspond exactly to the sender’s bit intervals  If receiver clock is faster or slower, the bit intervals are not matched and receiver may interpret signal differently.  Synchronization is required to solve this problem  Self Synchronizing digital signal includes timing information in the data being transmitted, which can be achieved if there are transitions in the signal that alert the receiver to the beginning, middle or end of the pulse.  If receiver clock is out of sync, these points can reset the clock.
  12. 12. Example 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: 1000 bits sent 1001 bits received1 extra bps At 1 Mbps: 1,000,000 bits sent 1,001,000 bits received1000 extra bps
  13. 13. Data Encoding & Modulating  Digital to digital conversion To transfer digital data (in binary form) from one place to another, must be converted into a digital signal.  Also called encoding digital data into a digital signal.  Each pulse is a signal element  Binary data encoded into signal elements  For example a print job  Position of signal high (1) or low (0) 
  14. 14. 1011001 Digital / Digital Encoding (Digital Transmitter) 1 0 1 1 0 0 1 Time
  15. 15. Figure 4.4 Line coding schemes
  16. 16.  Unipolar Signal:  In unipolar transmission only one kind polarity (either positive or negative) is used. • Unipolar encoding uses only one voltage level.  Polar Signal:  One logic state is represented by positive voltage level and other by negative.  Signal’s amplitude varies between positive and negative voltage level.
  17. 17. Unipolar NRZ scheme
  18. 18. Types of polar encoding • Polar encoding uses two voltage levels (positive and negative). • In NRZ-L the level of the signal is dependent upon the state of the bit.
  19. 19. Non-return to zero (NRZ)  In NRZ encoding, the level of the signal is always either positive or negative  Non – Return to Zero Level (NRZ – L) In NRZ – L encoding, the level of the signal depends on the type of bit it represents.  A positive voltage usually means the bit is 0  A negative voltage usually means the bit is 1 or vice versa 
  20. 20. In NRZ-L the level of the voltage determines the value of the bit. In NRZ-I the inversion or the lack of inversion determines the value of the bit.
  21. 21. Polar NRZ-L and NRZ-I schemes •One level represents 1 another 0. •Signal does not return to zero in the middle of each bit In NRZ-I the signal is inverted if a 1 is encountered.
  22. 22. In NRZ-I the signal is inverted if a 1 is encountered.
  23. 23. Return to Zero (RZ) Main problem with NRZ was that sender and receiver clocks are not synchronized. Receiver does not know when bit is starting and ending. RZ signal changes in the middle during bit, signal goes to the 0 in the middle and stays there until start of the next bit
  24. 24. Manchester Encoding Combination of NRZ-L and RZ, where duration of the bit is divided into two halves. The voltage remains at one level during the first half and moves to the other level during second half In Manchester encoding, the transition at the middle of the bit is used for both synchronization and bit representation.
  25. 25. Differential Manchester  Combines NRZ-I and RZ. There is always transition at the middle of each bit and the bit values are determined at the beginning of the bit.  If there is next bit 0, there is transition, if next bit is 1 there is no transition
  26. 26. Note: A good encoded digital signal must contain a provision for synchronization.
  27. 27. Manchester Encoding In Manchester encoding, the transition at the middle of the bit is used for both synchronization and bit representation.
  28. 28. In 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 at the beginning of the bit.
  29. 29. Note how with a Differential Manchester code, every bit has at least one signal change. Some bits have two signal changes per bit (baud rate is twice the bps).
  30. 30. Bipolar AMI (Alternate Mark Inversion) encoding  In bipolar encoding, we use three levels: positive, zero, and negative
  31. 31. Multilevel: 2B1Q scheme
  32. 32. Multi - Level Encoding 2B1Q (Two binary, one quaternary)  Uses four voltage levels. Each pulse can then represent two bits, making each pulse more efficient
  33. 33. Multi Transitions MLT-3 (Multiline Transmission, Three Level)  Very similar to NRZ-I, but it uses three levels of signals (+ve, zero and –ve)  Transition at beginning of 1 and no transition at beginning of zero
  34. 34. Block Coding Steps in Transformation Some Common Block Codes
  35. 35. 4B/5B Digital Encoding Yet another encoding technique that converts four bits of data into five-bit quantities. The five-bit quantities are unique in that no five-bit code has more than 2 consecutive zeroes. The five-bit code is then transmitted using an NRZ-I encoded signal.
  36. 36. Block coding
  37. 37. Figure 4.16 Substitution in block coding
  38. 38. Table 4.1 4B/5B encoding Data Code Data Code 0000 11110 1000 10010 0001 01001 1001 10011 0010 10100 1010 10110 0011 10101 1011 10111 0100 01010 1100 11010 0101 01011 1101 11011 0110 01110 1110 11100 0111 01111 1111 11101
  39. 39. Table 4.1 4B/5B encoding (Continued) Data Code Q (Quiet) 00000 I (Idle) 11111 H (Halt) 00100 J (start delimiter) 11000 K (start delimiter) 10001 T (end delimiter) 01101 S (Set) 11001 R (Reset) 00111
  40. 40. Figure 4.17 Example of 8B/6T encoding
  41. 41. Analog to digital  If a signal is sampled at regular intervals at a rate higher than twice the highest signal frequency, the samples contain all the information of the original signal  Voice data limited to below 4000Hz  Require 8000 sample per second  Analog samples (Pulse Amplitude Modulation, PAM)  Each sample assigned digital value
  42. 42. Analogue to digital conversion  To send voice over long distance, we need to digitise signal, since digital signals are less prone to noise.  Sampling:  Measuring amplitude of the signal at equal intervals.
  43. 43. Pulse Amplitude Modulation (PAM)  First step in analog to digital conversion  Takes analogue signal, samples it and generates a series of pulses based on the results of sampling.  The PAM signal is still analog since its amplitude can vary over the amplitude range.
  44. 44. Pulse Code Modulation (PCM)  PCM modifies the pulses created by PAM to create a complete digital signal.  PAM is converted into an all digital form by quantizing each pulse into its equivalent binary form.  8 bits are used to quantize each PAM signal, which include one sign bit. + 024 + 038 00011000 00100110 -015 -080 10001111 11010000
  45. 45. Pulse Amplitude Modulation (PAM) Pulse amplitude modulation has some applications, but it is not used by itself in data communication. However, it is the first step in another very popular conversion method called pulse code modulation. PAM
  46. 46. 4.3 Sampling Pulse Amplitude Modulation Pulse Code Modulation Sampling Rate: Nyquist Theorem How Many Bits per Sample? Bit Rate
  47. 47. Figure 4.18 PAM
  48. 48. Note: Pulse amplitude modulation has some applications, but it is not used by itself in data communication. However, it is the first step in another very popular conversion method called pulse code modulation.
  49. 49. Figure 4.19 Quantized PAM signal
  50. 50. Figure 4.20 Quantizing by using sign and magnitude
  51. 51. Figure 4.21 PCM
  52. 52. Figure 4.22 From analog signal to PCM digital code
  53. 53. Converting Analog Data into Digital Signals To convert analog data into a digital signal, there are two basic techniques: • Pulse code modulation (used by telephone systems) • Delta modulation
  54. 54. Pulse Code Modulation The analog waveform is sampled at specific intervals and the “snapshots” are converted to binary values.
  55. 55. Pulse Code Modulation When the binary values are later converted to an analog signal, a waveform similar to the original results.
  56. 56. Pulse Code Modulation The more snapshots taken in the same amount of time, or the more quantization levels, the better the resolution.
  57. 57. Pulse Code Modulation Since telephone systems digitize human voice, and since the human voice has a fairly narrow bandwidth, telephone systems can digitize voice into either 128 levels or 256 levels. These levels are called quantization levels. If 128 levels, then each sample is 7 bits (2 ^ 7 = 128). If 256 levels, then each sample is 8 bits (2 ^ 8 = 256).
  58. 58. Pulse Code Modulation How fast do you have to sample an input source to get a fairly accurate representation? Nyquist says 2 times the bandwidth. Thus, if you want to digitize voice (4000 Hz), you need to sample at 8000 samples per second.
  59. 59. Note: According to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequency.
  60. 60. Figure 4.23 Nyquist theorem
  61. 61. Sampling Rate  Nyquist theorem: Sampling rate must be at least two times the highest frequency  Telephone frequency range = ?  Highest frequency of the range = ?  Sampling rate = 2 X highest frequency samples/second 
  62. 62. Question  Highest frequency of a signal is: 2100 Hz and lowest frequency 200 Hz. Calculate bandwidth and sample rate.  Bw = fH – fL = 2100 – 200 = 1900 Hz  Sample rate = 2 X fH = 2 X 2100 = 4200 Hz
  63. 63. Bits per sample  Number of bits are chosen such that the original signal can be reproduced with the desired precision in amplitude.  Bit Rate:  Bit rate = sampling rate X number of bits per sample
  64. 64. Example 4 What sampling rate is needed for a signal with a bandwidth of 10,000 Hz (1000 to 11,000 Hz)? Solution The sampling rate must be twice the highest frequency in the signal: Sampling rate = 2 x (11,000) = 22,000 samples/s
  65. 65. Example 5 A 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? Solution We need 4 bits; 1 bit for the 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 is not enough since 22 = 4. A 4-bit value is too much because 24 = 16.
  66. 66. Example 6 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. Sampling rate = 4000 x 2 = 8000 samples/s Bit rate = sampling rate x number of bits per sample = 8000 x 8 = 64,000 bps = 64 Kbps
  67. 67. Note: Note that we can always change a band-pass signal to a low-pass signal before sampling. In this case, the sampling rate is twice the bandwidth.
  68. 68. 4.4 Transmission Mode Parallel Transmission Serial Transmission
  69. 69. Figure 4.24 Data transmission
  70. 70. Figure 4.25 Parallel transmission
  71. 71. Figure 4.26 Serial transmission
  72. 72. Note: In asynchronous transmission, we send 1 start bit (0) at the beginning and 1 or more stop bits (1s) at the end of each byte. There may be a gap between each byte.
  73. 73. Note: Asynchronous here means “asynchronous at the byte level,” but the bits are still synchronized; their durations are the same.
  74. 74. Figure 4.27 Asynchronous transmission
  75. 75. Note: In synchronous transmission, we send bits one after another without start/stop bits or gaps. It is the responsibility of the receiver to group the bits.