Computer Networks - ddddddddddpart 2.pdf
- 1. COMPUTER NETWORKS – PART 2 University Of Kurdistan Dept. Of Engineering (Computer Group) By: Khatereh Ahmadi Khahmadi.uok@gmail.com
- 2. SOFTWARE OF COMPUTER NETWORKS Reference model TCP/IP 30
- 3. SOFTWARE OF COMPUTER NETWORKS TCP/IP ➢ Older than OSI (Introduced in 1960) ➢ Simple and more practical design ➢ Reference model used by Internet 31
- 4. SOFTWARE OF COMPUTER NETWORKS TCP/IP Network Access (Network Interface): Physical and datalink tasks transmission channels (physical media) ➢ Telephone lines ➢ Twisted Pair cables ➢ Coaxial Cables ➢ Optical Fiber ➢ Satellite channels 32
- 5. SOFTWARE OF COMPUTER NETWORKS Reference model – Network Interface TCP/IP Twisted Pair cable A type of cable made using pairs of copper wires. These cables are used in communication networks and data transmission. Two main categories: STP, UTP 33
- 6. SOFTWARE OF COMPUTER NETWORKS Reference model – Network Interface TCP/IP Coaxial cable A shielded type of cable that has an inner conductor surrounded by concentric insulating layers that are surrounded by a conductive shielding. 34
- 7. SOFTWARE OF COMPUTER NETWORKS Reference model – Network Interface TCP/IP Optical Fibers The technology that transmits information as light pulses along a glass or plastic fiber. Fiber optic cables are commonly used because of their advantages over copper cables including higher bandwidth and transmit speeds. 35
- 8. SOFTWARE OF COMPUTER NETWORKS Reference model – Network Interface TCP/IP Optical Fibers: Total internal reflection phenomenon. 36
- 9. SOFTWARE OF COMPUTER NETWORKS Reference model – Network Interface TCP/IP Common parameters Bitrate: The number of bits that are conveyed or processed per unit of time. Bandwidth: A measurement indicating the maximum capacity of a wired or wireless communications link to transmit data over a network connection in a given amount of time. 37
- 10. SOFTWARE OF COMPUTER NETWORKS Reference model – Network Interface TCP/IP Common parameters Multiplexing: A technique that allows a number of lower bandwidth communication channels to be combined and transmitted simultaneously over one higher bandwidth channel Two types of multiplexing: ➢ FDM (Frequency Division Multiplexing) ➢ TDM (Time Division Multiplexing) 38
- 11. SOFTWARE OF COMPUTER NETWORKS Reference model – Network Interface TCP/IP Common parameters Error Control Checksum – error detection method based on data integrity CRC – Cyclic Redundancy Check 39
- 12. PHYSICAL LAYER
- 13. PHYSICAL LAYER DATA COMMUNICATION BASICS 41
- 14. PHYSICAL LAYER DATA COMMUNICATION BASICS 𝒙 𝒕 = 𝒙 𝒕 + 𝑻 𝒇 = 𝟏 𝑻 𝝎 = 𝟐𝝅𝒇 42
- 15. PHYSICAL LAYER DATA COMMUNICATION BASICS Fourier analysis 𝒙 𝒕 = 𝟏 𝟐 𝒄 + σ𝒏=𝟏 ∞ 𝒂𝒏𝒔𝒊𝒏(𝟐𝝅𝒏𝒇𝒕) + σ𝒏=𝟏 ∞ 𝒃𝒏𝒄𝒐𝒔(𝟐𝝅𝒏𝒇𝒕) 𝒂𝒏 , 𝒃𝒏 : amplitudes of 𝑛𝑡ℎ harmony of signals 43
- 16. PHYSICAL LAYER DATA COMMUNICATION BASICS 44
- 17. PHYSICAL LAYER DATA COMMUNICATION BASICS 45
- 18. PHYSICAL LAYER DATA COMMUNICATION BASICS Bitrate, Frequency and Bandwidth Consider a telephone line with the following features: Bitrate=300 b/s T (for 1 byte) = 8/300 = 0.02667 (sec) f = 1/T = 1/0.02667 = 37.5 (Hz) Bandwidth= f (high) – f (low) = ? > 300b/s 46
- 19. PHYSICAL LAYER DATA COMMUNICATION BASICS Bitrate Vs Baud rate Bitrate: number of bits transmitted per second Baud rate: number of signal changes per second 47
- 20. PHYSICAL LAYER DATA COMMUNICATION BASICS Channel capacity - Nyquist’s Theorem: (Noiseless channels) C = 𝟐 × 𝑩𝒂𝒏𝒅𝒘𝒊𝒅𝒕𝒉 × log𝟐 𝑳 Example #1: Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. What can be the maximum bit rate? C = 2 * 3000 * log2(2) = 6000bps Example #2: We need to send 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How many signal levels do we need? 265000 = 2 * 20000 * log2(L) → log2(L) = 6.625 L = 26.625 = 98.7 levels 48
- 21. PHYSICAL LAYER DATA COMMUNICATION BASICS Channel capacity - Shannon's Theorem (Noisy channel) 𝑪 = 𝐵𝑎𝑛𝑑𝑤𝑖𝑑𝑡ℎ × log𝟐(1 + 𝑺 𝑵 ) , SNR (db) = 10 × log𝟏𝟎( 𝑺 𝑵 ) Example #1: A telephone line normally has a bandwidth of 3000 Hz (300 to 3300 Hz) assigned for data communication. The Signal to noise ratio is usually 3162. What will be the capacity for this channel? C = 3000 * log2(1 + 𝑺 𝑵 ) = 3000 * 11.62 = 34860 bps Example #2: Assume that SNR(dB) is 36 and the channel bandwidth is 2 MHz Calculate the theoretical channel capacity. SNR(dB) = 10 * log10(S/N) → 36 = 10 * log10(S/N) log10(S/N) = 3.6 → S/N = 103.6= 3981 → 𝐶 = 2 × 106 × 𝑙𝑜𝑔2( 3982) 49
- 22. PHYSICAL LAYER DATA COMMUNICATION BASICS Modulation ➢ Frequency Modulation (FM) ➢ Phase Modulation (PM) ➢ Amplitude Modulation (AM) ➢ Quadrature amplitude modulation (QAM) – AM + PM 50
- 23. PHYSICAL LAYER DATA COMMUNICATION BASICS Modulation 51
- 24. PHYSICAL LAYER DATA COMMUNICATION BASICS Phase Modulation 52
- 25. PHYSICAL LAYER DATA COMMUNICATION BASICS QAM Modulation 53