1. Introduction to Communication Networks 2/2006 1 Introduction to Communication Networks Lecture 2 Physical Layer

2. Introduction to Communication Networks 1/2006 2 The OSI Model (7 layers) Application Physical Link Network Transport Session Presentation The 7-layer OSI Model

3. Introduction to Communication Networks 1/2006 3 Transmission of Information Well-understood basics: – From physics Energy Electromagnetic wave propagation – From mathematics Coding theory Fourier Analysis

4. Introduction to Communication Networks 1/2006 4 Signals To be transmitted, data must be transformed to electromagnetic signals. Signals can be analog or digital. – Analog signals can have an infinite number of values in a range; – digital signals can have only a limited number of values

5. Introduction to Communication Networks 1/2006 5 Comparison of analog and digital signals Vertical axis – value or strength Horizontal axis – passage of time

6. Introduction to Communication Networks 1/2006 6 Analog Signals Simple – cannot be decomposed into simpler signals (sine wave) Composite – composed of multiple sine waves S(t) = A sin (2  ft +  ) A – peak amplitude f – frequency  - phase

7. Introduction to Communication Networks 1/2006 7 Amplitude Highest intensity – for electric signals, usually measured in volts

8. Introduction to Communication Networks 1/2006 8 Phase Phase describes the position of the waveform relative to time zero – measured in degrees and/or radians

9. Introduction to Communication Networks 1/2006 9 Period and frequency f = 1/T T = 1/f Period – amount of time to complete 1 cycle Frequency - # of periods in one second Frequency is the rate of change with respect to time. Change in a short span of time means high frequency. Change over a long span of time means low frequency

10. Introduction to Communication Networks 1/2006 10 Sine wave examples

11. Introduction to Communication Networks 1/2006 11 Sine wave examples

12. Introduction to Communication Networks 1/2006 12 Sine wave examples

13. Introduction to Communication Networks 1/2006 13 Analog Signals – cont ’ An analog signal is best represented in the frequency domain.

14. Introduction to Communication Networks 1/2006 14 Signals – cont ’ A single-frequency sine wave is not useful in data communications; – we need to change one or more of its characteristics to make it useful When we change one or more characteristics of a single-frequency signal, it becomes a composite signal made of many frequencies. According to Fourier analysis, any composite signal can be represented as a combination of simple sine waves with different frequencies, phases, and amplitudes

15. Introduction to Communication Networks 1/2006 15 Square wave

16. Introduction to Communication Networks 1/2006 16 Three harmonics

17. Introduction to Communication Networks 1/2006 17 Adding first three harmonics

18. Introduction to Communication Networks 1/2006 18 Digital Signals

19. Introduction to Communication Networks 1/2006 19 Bit rate and bit interval Bit interval – time required to send 1 bit Bit rate – number of bits sent in 1 sec – expressed in bits per second (bps)

20. Introduction to Communication Networks 1/2006 20 Bandwidth The bandwidth is a property of a medium: It is the difference between the highest and the lowest frequencies that the medium can satisfactorily pass. We use the term bandwidth to refer to the property of a medium or the width of a single spectrum.

21. Introduction to Communication Networks 1/2006 21 Bandwidth Maximum rate that the hardware can change a signal Measured in cycles per seconds or Hertz (Hz) The bit rate and the bandwidth are proportional to each other The bit rate and the bandwidth are proportional to each other

22. Introduction to Communication Networks 1/2006 22 Example Question: – If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is the bandwidth? Draw the spectrum, assuming all components have a maximum amplitude of 10 V. Answer: – B = f h - f l = 900 - 100 = 800 Hz – The spectrum has only five spikes, at 100, 300, 500, 700, and 900

23. Introduction to Communication Networks 1/2006 23 Example – cont ’

24. Introduction to Communication Networks 1/2006 24 Standard Transmission Rates Low Rates – 300 bps and multiplies – 1200, 2400, 9600, 19200 LANs – 2, 4, 10, 16, 100 Mbps Wireless LANS – 11 Mbps, 54 Mbps WirelessMAN – 32 Mbps Usually depends on the used frequency bandwidth

25. Introduction to Communication Networks 1/2006 25 Digital Transmission System Measures - Delay Propagation delay – Time required for signal to travel across media – Example: electromagnetic radiation travels through space at the speed of light (C=3x10 8 meters/sec) Transmission time – Time required to transmit N bits at a given transmission rate

26. Introduction to Communication Networks 1/2006 26 Propagation and Transmission Time Example – Message length: 500 Bytes – Bit transmission Rate: 4 Mb/s – Optical fiber length: 2 Km – Speed of information in the fiber: 66% C ~ 2*10 8 m/s Transmission time: (500*8bits) / (4*10 6 b/s) = 10 -3 s = 1 ms Propagation delay (2000 m) / (2*10 8 m/s) = 0.01 ms

27. Introduction to Communication Networks 1/2006 27 the ‘ Big Issues ’ bandwidth and noise dominate the design and performance of all communication systems: – bandwidth how quickly can you change the signal – noise what is the smallest signal change you can detect

28. Introduction to Communication Networks 1/2006 28 Methods of communication 1 Binary signalling using a single cable transmission rate determined by how fast the voltage (or other symbol type) can be varied on the channel before the frequency content (as predicted by the Fourier series expansion) is so high that the filtering of the channel attenuates and distorts too much of the signal - limited by the bandwidth of the channel Binary signalling using many parallel cables multiple cables increase throughput or allow same throughput at lower bandwidth which is probably cheaper [parallel printer ports / cables]

29. Introduction to Communication Networks 1/2006 29 Methods of communication 2 Multi-level signalling using a single cable data transmission need not be limited to binary (two symbol state) format over a channel - any number of voltage levels or symbol types could be used. For example using 4 voltage levels we can uniquely encode 2 bits into each of the levels (00=level A, 01=level B, 10=level C, 11=level D). Each time we change symbol state, two bits of information are conveyed compared with only one for a binary system. We thus send information twice as fast for a given bandwidth Multi-level signalling using many parallel cables can use multi-level signalling over many cables if so desired with consequent increase in throughput

30. Introduction to Communication Networks 1/2006 30 Multi-level symbol operation – in principle can use any number of symbols (symbol states) for conveying digital information, e.g. 1024 different voltage levels conveying log 2 1024 = 10 bits. – practical limit depends on our ability to resolve the individual states (voltages, frequencies, light intensities etc) accurately at the receiver; this will depend on the level of noise and distortion in the channel.

31. Introduction to Communication Networks 1/2006 31 Information transfer rate Information transfer rate for a data channel is defined as the speed at which binary information can be transferred from source to destination. Units of Information transfer rate – > bits/second Related to underlying hardware bandwidth (maximum times per second the signal can change) Information transfer rate = 6 bits/6ms = 1000 bits/second

32. Introduction to Communication Networks 1/2006 32 Symbol rate (baud rate) – Symbol Rate (sometimes called baud rate) is the rate at which the signal state changes when observed in the communications channel - not necessarily equal to the information transfer rate units of symbol rate – > symbols/second (baud) example if 4 frequencies convey pairs of bits and the frequency (symbol) is changed every 0.5 ms, then: – symbol rate = 1/0.5 = 2000 symbols/second (2000 baud) – (note: the information transfer rate is 4000 bits/second)

33. Introduction to Communication Networks 1/2006 33 Symbol rate (baud rate) don ’ t confuse information transfer rate with the rate at which symbols are varied to convey the binary information over the channel. We already know that we can encode several bits in each symbol. – bandwidth efficiency of a communications link is a measure of how well a particular modulation format makes use of the available bandwidth. – bandwidth efficiency -> bits/second/Hz – example a system requires 4 kHz of bandwidth continuously to send 8000 bps of data - bandwidth efficiency = 8000bps/4000Hz = 2 bits/second/Hz

34. Introduction to Communication Networks 1/2006 34 Multi-level signalling The relationship between bits and symbols – the number of symbol states needed uniquely to represent any pattern of n bits is given by the simple expression: M = 2 n symbol states – Thus a group of 3 bits can be represented by M = 2 3 = 8 symbol states 4 bits by M = 2 4 = 16 symbol states 5 bits by M = 2 5 = 32 symbol states

35. Introduction to Communication Networks 1/2006 35 Symbol Constellations typically these are combinations of amplitude and phase as plotted below

36. Introduction to Communication Networks 1/2006 36 Advantages of M-ary signalling – a higher information rate is possible for a given symbol rate and corresponding channel bandwidth or – a lower symbol rate can be obtained leading to a reduced bandwidth requirements

37. Introduction to Communication Networks 1/2006 37 Maximum Data Rate of a Channel Nyquist Sample Theorem: Relation between digital throughput and bandwidth D = 2*B*log 2 K D: maximum data rate, in b/s B: system bandwidth K: possible values of voltages (binary K = 2) Absolute maximum

38. Introduction to Communication Networks 1/2006 38 Nyquist ’ s Theorem: particular cases For RS-232 – K is 2 because RS-232 only uses two values, +15 or -15 volts, to encode data bits - log 2 2 = 1 – D = 2 B For phase-shift encoding – Suppose K is 8 (possible shifts) - log 2 8=log 2 2 3 =3 – D = 6 B

39. Introduction to Communication Networks 1/2006 39 Noise Undesired signal associated with the transmission May produce error information Noise Level: signal-to-noise ratio S/N (SNR) – S: average signal power – N: Noise signal power – Measured in decibels: 10log 10 S/N

40. Introduction to Communication Networks 1/2006 40 The Effect of Noise Shannon ’ s Theorem: Gives capacity in presence of noise. C = B*log 2 (1 + S/N) C: effective limit on the channel capacity (b/s) B: hardware bandwidth S/N: signal to noise ratio

41. Introduction to Communication Networks 1/2006 41 The Bottom Line Nyquist ’ s + Shannon ’ s Theorems – No amount of clever engineering can overcome the fundamental physical limits of a real transmission system – Relates throughput to bandwidth – Encourages engineers to use complex encoding Increase the value of K – finding a way to encode more bits per cycle improves the data rate Adjusts for noise Specifies limits on real transmission systems

42. Introduction to Communication Networks 1/2006 42 Example We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signal-to-noise ratio is usually 30dB. For this channel the capacity is calculated as C = B log 2 (1 + SNR) = 3000 log 2 (1 + 10 3 ) C = 29.901 kbps

43. Introduction to Communication Networks 1/2006 43 Example We have a channel with a 1 MHz bandwidth. The SNR for this channel is 63; what is the appropriate bit rate and signal level? C = B log 2 (1 + SNR) = 10 6 log 2 (1 + 63) = 10 6 log 2 (64) = 6 Mbps Then we use the Nyquist formula to find the number of signal levels. 6 Mbps = 2  1 MHz  log 2 L  L = 8 First, we use the Shannon formula to find our upper limit.

44. Introduction to Communication Networks 1/2006 44 Transmission media Copper wires – Twisted pairs – Coaxial Cables Glass Fibers Satellites – Geo-synchronous – Low orbit Radio – Microwave – Infrared – Laser beam

45. Introduction to Communication Networks 1/2006 45 Copper Wires Untwisted pair Twisted Pair Coaxial Cable

46. Introduction to Communication Networks 1/2006 46 Micro Waves

47. Introduction to Communication Networks 1/2006 47 Transmission Performance

48. Introduction to Communication Networks 1/2006 48 Encoding Schemes Evaluation factors – Signal spectrum: less bandwidth, no dc component, shape of spectrum (better to center in the middle of bandwidth) – Clocking: self-clocking capability is desired for synchronization – Error detection: better to have error-detection capability – Signal interference and noise immunity – Cost and complexity

49. Introduction to Communication Networks 1/2006 49 RZ & NRZ RZ(Return to Zero) – Signal returns to zero after each encoded bit – 0 / 1: positive / negative pulse NRZ(Nonreturn to Zero) – Voltage level is constant during bit interval (no return to a zero voltage level) NRZ-L(NRZ Level) – 0: positive voltage – 1: negative voltage NRZ-I(NRZ Inverted) – Differential encoding scheme – 0: no transition – 1: transition NRZ is simple, and efficient NRZ limitations – presence of dc component – lack of synchronization capability

50. Introduction to Communication Networks 1/2006 50 Modulation Techniques Problem: How can we encode our signals when we can effectively use only a single frequency (or better: small frequency range)? Answer: Apply modulation techniques: Amplitude Modulation Frequency Modulation Phase Modulation

51. Introduction to Communication Networks 1/2006 51 Modulation A binary signal Amplitude Modulation Frequency Modulation Phase Modulation

52. Introduction to Communication Networks 1/2006 52 Multiplexing: FDM Problem:Considering that the bandwidth of a channel can be huge, wouldn’t it be possible to divide the channel into sub-channels? Frequency Division Multiplexing: Divide the available bandwidth into channels through frequency filtering, and apply modulation techniques per channel:

53. Introduction to Communication Networks 1/2006 53 Multiplexing: TDM Time Division Multiplexing:Simply merge/split streams of digital data into a new stream. Data is handled in frames – a fixed series of consecutive bits:

54. Introduction to Communication Networks 1/2006 54 Example Question What is the Round Trip Time when sending data to a satellite which is 32000 km away and the transmission rate is 1 Mb/s?

55. Introduction to Communication Networks 1/2006 55 Answer Remember: T=S/V Round trip time (RTT) is the time it takes the signal to travel back and forth RTT=2*(32000 km )/(3*10 8 m/s )=213.ms

56. Introduction to Communication Networks 1/2006 56 Next Lecture Data link layer (layer 2) – Framing – Error detection – Multiple access protocols  MAC Layer FDMA/TDMA/CDMA Random access Reliable Data transfer  ARQ