1. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/1 COM342 Networks and Data Communications Ian McCrumRoom 5B18 Tel: 90 366364 voice mail on 6 th ring Email: IJ.McCrum@Ulster.ac.uk Web site: http://www.eej.ulst.ac.uk Lecture 2: Fundamentals and Theory at the Physical layer

2. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/2 Keypoints from last lecture Read chapter 1 of the text A distributed system is not a “Computer Network Layers are used to manage complexity. Only bottom layers can communicate across machines Protocol stacks are used to help manage complexity Different topologies exist: Bus or Ring Broadcast and Peer to Peer WANs, MANs and LANs… linking using routers/switches Connection orientated or connectionless Reliable or unreliable, both are useful

3. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/3 Fundamentals of communications Voltages as Data, a way of passing information timing considerations data transfer Synchronous and asynchronous communications redundancy genesis of a protocol. Serial or parallel Broadcast or peer to peer Direction; Duplex or Simplex

4. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/4 Data Transfer Device 1Device 2 Fig 1.1 simple serial connection 0 1 0 1 1 0 1 0

5. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/5 Voltages 0V 5V1 0 logic 5V 0V -5V -15V 15V 0 1 1 0 Internal Voltages RS232C Voltages

6. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/6 Voltages as data 0 1 0 0 1 0 1 +12V -12V time

7. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/7 Timing of signals Consider that each state was of 10ms in duration this means that 100 such states could be set in 1s or expressed as 100bits per second (100bps) i.e.period / number of bits consider 28,000bps what it the time needed for each bit What are the problems?

8. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/8 Problems How does the receiver know when to start listening? How can the receiver recognise the boundaries between individual bits being sent? Can it cope with errors arising, thereby corrupting bits?

9. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/9 Asynchronous transfer 1 0 Start bitStop bit 8 data bits Send to receiver Idle state

10. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/10 Bits; sent in the reverse way to the way humans read binary numbers!! Start bitStop bit Send to receiver 0 1 0 0 0 0 0 1 0 1 lsbmsb parity bit 1 0 1 0 0 0 0 0 1 0 Note: Start and stop bits are of opposite values…why Note: Least significant bit sent first (after start), parity usually sent after MSB

11. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/11 Timing In order to receive a data transmission correctly the receiver must be clocking at the same rate. This must be established before data is transferred. Some systems over-sample at 16 time the bit rate in order to avoid the transitions. The receiver will also check the framing i.e. check that the stop appears where it should.

12. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/12 Wee sums Consider 5,000 characters to be sent this is 50,000 bits since 5000 * (1start + 8 data + 1stop) bits at a data rate of 9600bps the time taken for transmission is 50000/9600 s or 5.2 seconds this assumes no time delay between characters being transmitted

13. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/13 Some common speeds Data rate bits/sCharacter rate chars/s 11010 30030 1200120 2400240 4800480 9600960 192001920 280002800 556005560

14. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/14 Implications Note the redundant bits in the transmission for every 10 bits sent only 8 are data or 25% overhead whenever one considers that 8th bit is parity this rises to 30% or the useful information is only 70% of that transferred. There may be a variable time delay between characters… e.g if someone is typing on a keyboard; because the delay can be random, we describe this as asynchronous communications Do an example of parity for letter f 1100110 p185(3 rd ed) rem rev order and parity on rhs. do odd and even

15. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/15 Parity What are the limitations of parity? How many errors can it detect? Can it correct an error in one codeword? Could we use parity to detect an error? Consider Chinese whispers. How do we correct errors in English?

16. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/16 Synchronous transfer Data to be transferred Synchronising bits Start of text (STX)End of text (ETX) Send to receiver

17. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/17 A protocol 2 * SYN, STX, (1024 chars), ETX efficiency is 1024*8 /1028*8 = 99.7% compared with 1024*8 / 1024*10 = 80% for asynchronous however as block size decreases efficiency falls. At what block length does synchronous cease to improve over asynchronous transfer? Assuming the protocol above. What should limit the size of the data segment of this protocol

18. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/18 What did we learn? Binary data can be sent serially in many forms receivers and transmitters must be in agreement about timing timing information can be included within each character or across a group of characters. data can be protected against errors protocols can be efficient.

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20. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/20 Theoretical Communication Rate What form the signal takes. Mathematical description Nyquist’s limitation Shannon’s theorem Carrying data using sine waves

21. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/21 Fourier Series g(t) a function which varies with time and has a period T such that f = 1/T which is the fundamental frequency a n and b n the amplitudes of the sine and cosine of the nth harmonic. These generate terms that may go on to infinity, though typically by then their amplitudes are weak… This is known as the Fourier series and from it the original function can be reconstructed. In other words, if you have a repetitive waveform of arbitrary shape, it can be made or reconstructed from adding together a collection of sine waves; pure frequencies, and these are all multiplies of each other varying just in amplitude and phase/timing. The problem is that the higher harmonics aren’t passed by some communication channels.

22. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/22 Fourier and binary communciations If you send a ascii character ‘b’ then you send a binary pattern 01100010 This is an arbitrary waveform and can be represented using a fourier series. Comprised of lots of harmonics. The “Root Mean Square” amplitudes of each term gives us the energy in each harmonic =  (a n 2 + b n 2 ) So this particular code has energies in some higher harmonics; but a real data channel will attenuate high frequencies much more then low frequencies; (e.g <3kHz) Distortion is the result, not clean square waves any more!

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24. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/24 Bit Pattern representation Fig 2.1 Tanenbaum (a) depicts the binary pattern and its waveform of voltage against time. The lower traces shows root mean square amplitudes for each harmonic number. (b) shows how the waveform would look if only frequencies up to and including the first harmonic would get through. (c ) the minimum frequency now is about the second harmonic and more information is getting through but still cannot reconstruct the original. (d) and (e) more and more harmonics added until with 8 added it roughly approximates the transmitted signal. See Fig 2.2

25. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/25 Bits/SecT(msec)1 St Harmonic#Harmonics sent 30026.6737.580 60013.337540 12006.6715020 24003.3330010 48001.676005 96000.8312002 192000.4224001 384000.2148000 Fig 2-2: Relationship- data rate & harmonics

26. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/26 Baud v bits/sec 1 baud is a change of state per second not to be confused with 1 bit/sec which is 1 bit/sec if a change of state represents only one bit then 1 baud is 1 bit/sec many cases a change of state represents 2, 3, or more bits. If the voltages 0,1,2,3,4,5,6 and 7 were used, each signal change conveys 3 bits, the bit rate is three times the baud rate.

27. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/27 one change of state corresponding to 3 bits of data 7654321076543210 Volts 010111 Bits

28. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/28 Maximum Data rate of a channel (Nyquist) Nyquist’s theorem more generally with V discrete levels bits/sec H is the bandwidth, the maximum range of frequencies Available. With V= number of discrete levels. Thus with 3kHz channel and 2 level signals the data rate cannot exceed 6000bps

29. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/29 Channels with Noise Shannon’s theorem gives the following outcome; Thus a channel with a signal to noise ratio of 30dB and bandwidth of 3,000 Hz will have a maximum capacity of ~ 30,000 bps 30dB is 1000:1, 20dB is 100:1 (logs!)

30. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/30 pure sine

31. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/31 pure sine + harmonic

32. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/32 Modems Modulator and Demodulator. –Modulation is the act of changing a carrier wave so that it can represent a signal. –Demodulation is the act of measuring the changes made to a carrier wave and thereby recovering the original modulating signal. Consider AM Radio broadcast. –RTE 567kHz and Radio Ulster 1341kHz bandwidth 9kHz and FM Radio broadcast. –Classic FM 102MHz bandwidth 25kHz

33. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/33 Sine wave X varies in time Amplitude A Frequency f time t theta the phase 1/3sec 1/5sec

34. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/34 For binary 1 For binary 0 Thus the amplitude for 0 is half that for 1, this could be chosen to be zero if so desired. Amplitude Modulation

35. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/35 Frequency Modulation For binary 1 For binary 0 Thus the frequency for 0 is different than that for 1, called frequency shift keying (tone dialing on telephone).

36. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/36 Phase Modulation For binary 1 For binary 0 Thus the phase for 0 is shifted 180 degrees than that for 1, called phase shift keying (tone dialing on telephone).

37. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/37 Data rate and Baud rate for PSK Baud rate is the number of changes of state per second if a change of state can be made to represent more than one bit ie two then the data rate is twice the baud rate time Amplitude

38. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/38 Note how the addition of 2 levels of AM enable another bit to be encoded with each baud

39. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/39 Data Communications and telephone systems Existing telephones developed to carry voice. Data to communicated must taken on voice type properties –Must use frequencies between 300 and 3,400 Hz

40. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/40 Do the following simple exercises What is the ratio represented by 30dB? What is the bandwidth where f upper is 3400Hz? What is the bandwidth where f lower is 300Hz? What the Information capacity of such a channel?

41. 21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/41 What we covered Serial data transfer ; Voltage/Times/Start of characters Sync and Async, start and end bits start and end charactsers Speed, Chars/sec, efficiency, overhead Baud vs Bits/Sec Maximum data rate of a channel Modems: AM/FM/PM… telphones more later! Fridays class will be tutorials on sums…