21/09/04 ian/modules/COM342/COM342_L2.ppt L2/1 COM342 Networks and Data Communications Ian McCrumRoom 5B18 Tel: 90 366364 voice mail

21/09/04 www.eej.ulster.ac.uk/~ian/modules/COM342/COM342_L2.ppt L2/1

COM342Networks and Data Communications

Ian McCrum Room 5B18

Tel: 90 366364 voice mail on 6th ring

Email: [email protected]

Web site: http://www.eej.ulst.ac.uk

Lecture 2: Fundamentals and Theory at the Physical layer


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


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


Data Transfer

Device 1 Device 2

Fig 1.1 simple serial connection

0 1 0 1 1 0 1 0


Voltages

0V

5V 1

0

logic

5V

0V

-5V

-15V

15V

0

1

1

0

Internal Voltages RS232C Voltages


Voltages as data

0 1 0 0 1 0 1

+12V

-12V

time


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?


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?


Asynchronous transfer

1

0

Start bit Stop bit

8 data bits

Send to receiver

Idle state


Bits; sent in the reverse way to the way humans read binary numbers!!

Start bit Stop bit

Send to receiver

0 1 0 0 0 0 0 1 0 1

lsb msb

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


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.


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


Some common speedsData rate bits/s Character rate chars/s

110 10300 301200 1202400 2404800 4809600 96019200 192028000 280055600 5560


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(3rd ed) rem rev order and parity on rhs. do odd and even


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?


Synchronous transfer

Data to be transferred

Synchronising bits

Start of text (STX) End of text (ETX)

Send to receiver


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


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.



Theoretical Communication Rate

• What form the signal takes.

• Mathematical description

• Nyquist’s limitation

• Shannon’s theorem

• Carrying data using sine waves


Fourier Series g t c a nft b nftn

n

n

n

12

1 1

2 2sin cos

g(t) a function which varies with time and has a periodT such that f = 1/T which is the fundamental frequency

an and bn 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.


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

=(an2 + bn

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!



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


Bits/Sec T(msec) 1St Harmonic #Harmonics sent

• 300 26.67 37.5 80• 600 13.33 75 40• 1200 6.67 150 20• 2400 3.33 300 10• 4800 1.67 600 5• 9600 0.83 1200 2• 19200 0.42 2400 1• 38400 0.21 4800 0

Fig 2-2: Relationship- data rate & harmonics


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.


one change of state corresponding to 3 bits of data

76543210

Vol

ts

010 111 Bits


Maximum Data rate of a channel (Nyquist)

• Nyquist’s theorem more generally with V discrete levels

I H Vmax log2 2 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 signalsthe data rate cannot exceed 6000bps


Channels with Noise

Shannon’s theorem gives the following outcome;

I HS

Nmax log

2 1

Thus a channel with a signal to noise ratioof 30dB and bandwidth of 3,000 Hz willhave a maximum capacity of ~ 30,000 bps

30dB is 1000:1, 20dB is 100:1 (logs!)


pure sine


pure sine + harmonic


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


Sine wave

)2cos()( cctfAtx

X varies in time Amplitude AFrequency ftime ttheta the phase

3 Hz Sine Wave

5 Hz Sine Wave

1/3sec

1/5sec


)2cos()( cctfAtx

)2cos(2

)( cctfA

tx

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.

1 1 1 110 0Data

Amplitude Modulation


Frequency Modulation)2cos()( 1 ctfAtx

)2cos()( 0 ctfAtx

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).

0 1 10Data

0 0 0


Phase Modulation

)2

2cos()( tfAtx c

)2

32cos()(

tfAtx cFor 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).


Data rate and Baud rate for PSKBaud rate is the number of changes of state per secondif a change of state can be made to represent more thanone bit ie two then the data rate is twice the baud rate

time

Am

plit

ude

01 11Data

11 11


Note how the addition of 2 levels of AM enable another bit to be

encoded with each baud

000 111 110


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

HzAnalogue

PSTN

Sender

Receiver

MODEM

MODEM


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?


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…

Documents

21/09/04 ian/modules/COM342/COM342_L2.ppt L2/1 COM342 Networks and Data Communications Ian McCrumRoom 5B18 Tel: 90 366364 voice mail