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Computer Networks Dr. Miled M. Tezeghdanti October 19, 2010 Dr. - PowerPoint PPT Presentation

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Computer Networks Dr. Miled M. Tezeghdanti October 19, 2010 Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 1 / 79 Syllabus Basic Concepts OSI Model Data-Link Layer Local Area Networks Network Layer Transport Layer

  1. Packet Switching Packet Switching from: h1@lan1.net to: h5@lan5.net subject: MS Email uses MS . . 1−>5,5 Email uses MS 1−>5,4 subject: MS 1−>5,3 to: h5@lan5.net 1−>5,2 from: h1@lan1.net 1−>5,1 Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 20 / 79

  2. Packet Switching Packet Switching from: h1@lan1.net to: h5@lan5.net subject: MS Email uses MS . . 1−>5,5 Email uses MS 1−>5,4 subject: MS 1−>5,3 to: h5@lan5.net 1−>5,2 from: h1@lan1.net 1−>5,1 Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 20 / 79

  3. Packet Switching Packet Switching from: h1@lan1.net to: h5@lan5.net subject: MS Email uses MS . from: h1@lan1.net 1−>5,1 . 1−>5,5 Email uses MS 1−>5,4 subject: MS 1−>5,3 to: h5@lan5.net 1−>5,2 Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 20 / 79

  4. Packet Switching Packet Switching from: h1@lan1.net to: h5@lan5.net subject: MS Email uses MS . from: h1@lan1.net 1−>5,1 to: h5@lan5.net 1−>5,2 . 1−>5,5 Email uses MS 1−>5,4 subject: MS 1−>5,3 Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 20 / 79

  5. Packet Switching Packet Switching from: h1@lan1.net to: h5@lan5.net subject: MS Email uses MS . from: h1@lan1.net 1−>5,1 to: h5@lan5.net 1−>5,2 1−>5,3 subject: MS . 1−>5,5 Email uses MS 1−>5,4 Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 20 / 79

  6. Packet Switching Packet Switching from: h1@lan1.net to: h5@lan5.net subject: MS Email uses MS . from: h1@lan1.net 1−>5,1 to: h5@lan5.net 1−>5,2 1−>5,3 subject: MS Email uses MS 1−>5,4 . 1−>5,5 Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 20 / 79

  7. Packet Switching Packet Switching from: h1@lan1.net to: h5@lan5.net subject: MS Email uses MS . from: h1@lan1.net 1−>5,1 to: h5@lan5.net 1−>5,2 1−>5,3 subject: MS Email uses MS 1−>5,4 . 1−>5,5 Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 20 / 79

  8. Packet Switching Packet Switching from: h1@lan1.net from: h1@lan1.net to: h5@lan5.net to: h5@lan5.net subject: MS subject: MS Email uses MS Email uses MS . . from: h1@lan1.net 1−>5,1 to: h5@lan5.net 1−>5,2 1−>5,3 subject: MS Email uses MS 1−>5,4 . 1−>5,5 Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 20 / 79

  9. Packet Switching Packet Switching from: h1@lan1.net from: h1@lan1.net to: h5@lan5.net to: h5@lan5.net subject: MS subject: MS Email uses MS Email uses MS . . Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 20 / 79

  10. Cell Switching Similar to packet switching Messages/Packets are segmented into cells Cells have fixed size Padding bytes in the last cell Good for real time traffic (transmission time is fixed for each cell) Example: ATM Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 21 / 79

  11. Transmission Link Characteristics Simplex Transmission in one direction Radio, TV Half-Duplex Transmission in both directions, but in only one direction at a given time Walkie Talkie Full-Duplex Transmission in both directions simultaneously Telephone Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 22 / 79

  12. Signal A signal is a varying quantity (voltage, air pressure,) that can be expressed as a continuous function of an independent variable usually time Used for data representation Digital Signal Discrete time signal Discrete values (+5V and 5V) Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 23 / 79

  13. Analog Signal Continuous time signal Amplitude varies continuously Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 24 / 79

  14. Digital Signal Discrete time signal Discrete values (+5V and 5V) Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 25 / 79

  15. Fourier Analysis A periodic signal g ( t ) with frequency f can be written as follows: ∞ ∞ g ( t ) = c � � 2 + a n sin (2 π nft ) + b n cos (2 π nft ) n =1 n =1 � T c = 2 g ( t ) dt T 0 � T a n = 2 g ( t ) sin (2 π nft ) dt T 0 � T b n = 2 g ( t ) cos (2 π nft ) dt T 0 if the signal is not periodic, we can apply Fourier on portions of the signal. Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 26 / 79

  16. Fourier Analysis � T a n = 2 g ( t ) sin (2 π nft ) dt T 0 T = 2 � 2 sin (2 π nft ) dt T 0 � T = 2 � 1 2 2 π nf cos (2 π nft ) − T 0 = 2 � 1 cos (2 π nf T �� � 2 ) − cos (0) − T 2 π nf = − 2 1 �� �� cos (2 π nf T 2 ) − cos (0) T 2 π nf = − 1 � � cos ( n π ) − 1 n π = 1 � � 1 − cos ( n π ) n π Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 27 / 79

  17. Fourier Analysis � T b n = 2 g ( t ) cos (2 π nft ) dt T 0 T = 2 � 2 cos (2 π nft ) dt T 0 � T = 2 � 1 2 2 π nf sin (2 π nft ) T 0 = 2 � 1 �� sin (2 π nf T � 2 ) − sin (0) 2 π nf T = 2 1 sin (2 π nf T � � 2 ) − sin (0) T 2 π nf = 1 � � sin ( n π ) − 0 n π = 0 Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 28 / 79

  18. Fourier Analysis � T c = 2 g ( t ) dt T 0 � T = 2 � 2 t T 0 = 2 � T � 2 − 0 T = 2 T T 2 = 1 Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 29 / 79

  19. Fourier Analysis ∞ ∞ g ( t ) = c � � 2 + a n sin (2 π nft ) + b n cos (2 π nft ) n =1 n =1 a n = 1 � � 1 − cos ( n π ) n π b n = 0 c = 1 ∞ g ( t ) = 1 1 � � � 2 + 1 − cos ( n π ) sin (2 π nft ) + 0 n π n =1 g ( t ) = 1 2 + 2 π sin (2 π ft ) + 2 3 π sin (6 π ft ) + 2 5 π sin (10 π ft ) + ... Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 30 / 79

  20. Fourier Analysis Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 31 / 79

  21. Fourier Analysis Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 32 / 79

  22. Fading/Attenuation Fading/Attenuation Diminution of the amplitude of the signal Depends on: Frequency of the signal Transmission media Circuit length Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 33 / 79

  23. Fading/Attenuation Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 34 / 79

  24. Distortion Distortion Signal deformation The signal is constituted by many harmonics with different frequencies Harmonics are transmitted with different speeds Received signal will be distorted Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 35 / 79

  25. Distortion Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 36 / 79

  26. Noise Noise Presence of parasite signal Gaussian Noise Random motion of electrons Emission of electromagnetic waves Constant signal Its power is proportional to temperature Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 37 / 79

  27. Distortion Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 38 / 79

  28. Signal Transmission Analog Transmission Analog signal is used to transmit Information Digital Transmission Digital signal is used to transmit Information Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 39 / 79

  29. Analog Transmission Analog Signal over Analog Channel Signal directly transmitted (base band) Analog Modulation (broadband) Digital Signal over Analog Channel Modem: modulator demodulator Amplitude Modulation Frequency Modulation Phase Modulation Combined Modulation Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 40 / 79

  30. Digital Transmission Digital Signal over Digital Channel Manchester Code Bit 1: top-down transition Bit 0: bottom-up transition Analog Signal over Digital Channel Codec PCM : Pulse Code Modulation Sampling Quantization Coding Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 41 / 79

  31. Sampling Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 42 / 79

  32. Nyquist Theorem Nyquist Theorem A signal with a maximal frequency H must be sampled at a frequency 2 H Maximum rate: 2 Hlog 2 Vbit / s V : number of discrete levels of the signal Example: A modem uses AM-PSK modulation (Phase Shift Key) with 8 levels, PSTN bandwidth is 3100 Hz C = 2 Hlog 2 V C = 2 ∗ 3100 ∗ log 2 8 C = 2 ∗ 3100 ∗ 3 C = 18600 bit / s Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 43 / 79

  33. Analog Transmission Analog Transmission Communication by exchanging analog signals Amplifier to amplify the signal Original signal can not be reconstituted Noise signal is also amplified! Signal looses its quality with distance Fading/Attenuation and noise dont affect so much voice transmission but data transmission may be seriously affected (data corruption) Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 44 / 79

  34. Digital Transmission Digital signal with two states on and off Digital transmission is done by impulsions Repeater to regenerate the signal Initial signal is reconstituted exactly Noise is eliminated Fading/Attenuation does not affect so much digital signal An faded/attenuated signal has always a series of on and off pulses Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 45 / 79

  35. Digital Transmission Radio, TV have analog transmission systems Why use digital transmission? Cheap LSI/VLSI technology Data Integrity Efficiency Best use of bandwidth Easy multiplexing with digital techniques Security Cryptography and authentication Integration Similar processing of analog and digital data Good quality (Noise elimination) Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 46 / 79

  36. Digital Encoding Manchester Encoding 0: Bottom-Up Transition 1: Top-Down Transition Bipolar Encoding 0: -V 1: +V NRZ Encoding (No Return to Zero) 0: -V 1: +V NRZI Encoding (No Return to Zero Inverted) 0: Transition 1: No transition Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 47 / 79

  37. Manchester Encoding Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 48 / 79

  38. Bipolar Encoding Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 49 / 79

  39. NRZ Encoding Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 50 / 79

  40. NRZI Encoding Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 51 / 79

  41. Modulation Modification of the characteristics of the carrier using the amplitude of base band signal Process allowing the transmission Analog signal with a higher frequency Digital signal over analog channel Carrier P(t) = A sin(2wFt + P) A : amplitude F : frequency P : phase Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 52 / 79

  42. Modulation Amplitude Modulation Two different amplitudes are used to represent bit 0 and bit 1 Frequency Modulation Two different frequencies are used to represent bit 0 and bit 1 Phase Modulation Two different phases are used to represent bit 0 and bit 1 Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 53 / 79

  43. Amplitude Modulation Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 54 / 79

  44. Frequency Modulation Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 55 / 79

  45. Phase Modulation Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 56 / 79

  46. Combined Modulation Simultaneous use of two or more of former modulation methods Amplitude Modulation + Frequency Modulation Amplitude Modulation + Phase Modulation Frequency Modulation + Phase Modulation All the three methods Transmission of many bits simultaneously Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 57 / 79

  47. Combined Modulation Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 58 / 79

  48. Capacity Baud Number of signal transitions per second Unity: baud Bits per second Number of transmitted bits per second Unity: bit/s 2 different levels ( 0 and 1) Capacity ( bits / s ) = capacity ( bauds ) 4 different levels (00, 01, 10, 11) Capacity ( bits / s ) = 2 ∗ capacity ( bauds ) Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 59 / 79

  49. Capacity Shannon Theorem C = Blog 2 (1 + S N ) C : Capacity B : Bandwidth SNR : Signal/Noise Ratio SNR = 10 log 10 ( S N ) N = 10 ( SNR S 10 ) Example: Twisted Pair SNR = 20 dB Bandwidth 3000 Hz 10 ) = 10 2 = 100 N = 10 ( 20 S C = 3000 log 2 (1 + 100) C = 19963 bit / s Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 60 / 79

  50. Multiplexing Goal Transmission of many signals over one transmission media Analog Multiplexing Frequency Division Multiplexing (FDM) Wave-length Division Multiplexing (WDM) Code Division Multiplexing (CDM) Digital Multiplexing Time Division Multiplexing (TDM) Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 61 / 79

  51. Frequency Division Multiplexing Frequency Division Multiplexing Frequency Bandwdith Time Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 62 / 79

  52. Frequency Division Multiplexing Frequency Division Multiplexing Frequency Bandwdith Time Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 62 / 79

  53. Frequency Division Multiplexing Frequency Division Multiplexing Frequency Bandwdith Time Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 62 / 79

  54. Frequency Division Multiplexing Frequency Division Multiplexing Frequency Bandwdith Time Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 62 / 79

  55. Frequency Division Multiplexing Frequency Division Multiplexing Frequency Bandwdith Time Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 62 / 79

  56. Frequency Division Multiplexing Frequency Division Multiplexing Frequency Bandwdith Time Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 62 / 79

  57. Time Division Multiplexing Time Division Multiplexing Frequency Bandwdith Time Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 63 / 79

  58. Time Division Multiplexing Time Division Multiplexing Frequency Bandwdith Time Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 63 / 79

  59. Time Division Multiplexing Time Division Multiplexing Frequency Bandwdith Time Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 63 / 79

  60. Time Division Multiplexing Time Division Multiplexing Frequency Bandwdith Time Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 63 / 79

  61. Time Division Multiplexing Time Division Multiplexing Frequency Bandwdith Time Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 63 / 79

  62. Time Division Multiplexing Time Division Multiplexing Frequency Bandwdith Time Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 63 / 79

  63. Time Division Multiplexing Time Division Multiplexing Frequency Bandwdith Time Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 63 / 79

  64. Time Division Multiplexing Time Division Multiplexing Frequency Bandwdith Time Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 63 / 79

  65. Time Division Multiplexing Time Division Multiplexing Frequency Bandwdith Time Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 63 / 79

  66. Code Division Multiplexing - Frequency Hopping Code Division Multiplexing Frequency Bandwdith Time Dr. Miled M. Tezeghdanti () Computer Networks October 19, 2010 64 / 79

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