bst handout e07

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DWR/AGS Digital Data Comms

ROYAL SCHOOL OF ARTILLERY

BASIC SCIENCE & TECHNOLOGY SECTION

Digital Data Communication

INTRODUCTION

Computer-based systems are widespread inweapons systems and many weapons systemsnow have several computer systems embedded in

them. The two systems - weapon and computer - are

inextricable linked and, probably, neither would function

without the other. The computer functions are often

shared between a number of distinct IT sub-systems as

this gives resilience. There is also a requirement for

these sub-systems to communicate not only with each

other but also with the main controlling system and with

the human operator(s). Further data communicationsmight be required between one weapons system and

another.

This handout describes the various techniques used

to communicate data between computers and other

computers or computer-based equipment. Many of the

techniques to be described are also used in PCs, fac-

simile machines and teleprinters.

BINARY DATA

Digital systems store data using binary numbers. Theprefix bi-, as in bicycle, means two of somethingand in computers it means two possible states of a sig-nal. These two states are designated Zero and One and

some of the ways commonly used to represented them

are described in the following:

Voltage: Zero Volts represents the Zero state and

Five Volts represents the One state. This might be

used for used when storing data in a memory chip.

Magnetism: a South Pole represents the Zerostate

and a North Pole represents the One state. This

might be used when storing data on a floppy disk.

Frequency: a low frequency represents the Zero

state and a high frequency represents the One

state. This might be used to send data or fax signalsalong a telephone line.

Light: A change from high brightness to low bright-

ness represents the Zero state and a change from

low brightness to high brightness represents the

Onestate. This might be used to send data along a

fibre-optic cable.

Phase: a phase change of +90 represents a One

and a phase change of 90 represents a Zero.

This might be used when sending data along a tele-

phone line.

The binary representation is used because it is rela-tively easy and very fast for an electronic circuit to dis-

tinguish between just two states. Any binary digit is

either one or zero, with no halfway.

Binary numbers usually have more than one digit,

just like any other number system, and each digit is

stored and transmitted separately from the others.

Place Value: the dials on the odometer (mileometer)

of a car might read 07352 and we know that the 7digit,

because of its position, means seven thousand whereas

the 3 digit means three hundred. This illustrates what

we mean by place value and, in decimal numbers the

place values are as follows:

105 104 103 102 101 100

| | | | | |100k 10k 1k 100 10 Units

The place values are based on powers of ten and the

least significant digit (the units) is ten to the power Zero.

Binary Numbers: these follow the same pattern as

decimal numbers except that they are based on powers

of two, rather than ten. The place values are:

27 26 25 24 23 22 21 20

| | | | | | | |

128 64 32 16 8 4 2 Units

Thus, the binary number 01000110 represents 64 +

4 +2 or the decimal number 70. The decimal number 47

can be expressed in binary as 00101111 or 32 +8 +4 +

2 +1, remembering to use a zero digit in the place val-

ues that are not used.

Digit Grouping: a single-digit number is limited in

what it can store and most numbers require several dig-

its, depending on their purpose. The Trip Meterof a car

generally has three digits for miles (and a further one for

tenths of a mile), so it can display up to 999.9 miles

(effectively, this is one-thousand miles). The odometer

of a car often has six digits and can display up to one-

million miles. People generally group denary digits inthrees but binary numbers are grouped as required by

the data processing system in use. The most common

grouping is the Byte.

Byte: a group of eight binary digits is called a byte

and this is the standard unit of computer data storage.

A byte can store numbers between 00000000 (Zero)

and 11111111 (255) so numbers outside that range will

need more than a single byte to store them. A byte has

the capacity to store 26 capital letters, 26 lower-case let-

ters, punctuation, numbers and many special characters

(e.g. , , , ). A byte can be thought of as the amount

of data generated by one key-press.Hexa-decimal : representing decimal digits directly

in bytes (binary-coded decimal - BCD) is not very effi-

cient because a byte can have values from zero to 255

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Digital Data Comms DWR/AGS

whereas a decimal digit uses only zero to nine. If a byte

is divided into two sets of four-digit numbers (called

nybbles) then the situation is slightly better, because

each half can hold numbers up to fifteen - but there is

still some inefficiency if decimal digits are used as these

do not exceed nine. Hexa-decimal numbers are based

on sixteen and have values from zero to fifteen - requir-

ing four bits, that can be fitted, without waste, exactly

two to each byte.

The hexa-decimal notation (or hex) requires digits

from zero to fifteen, and uses the same zero to nine as

decimal digits, but also letters A to F to represent digitsten to fifteen. This is illustrated in Figure One.

The place values for hex are based on sixteen, in the

sam way that the place values for decimal numbers are

based on ten, as follows:

163 162 161 160

| | | |

4096 256 16 Units

Thus, the hex number AB represents a ten in the six-

teensplace and an eleven in the units place. That cor-

responds to decimal 160 + 11 = 171. The decimal

number 124 can be broken down into seven sixteens(112) and twelve ones to give the hex number 7C. (The

letters in hex numbers may be typed in either lower-

case or upper-case.)

Words: groups of binary digits may also be called

Words and they might contain any number of bits. In

Windows, a Word is sixteen bits (2 Bytes) and a DWord

(double-word) is 32 bits (4 Bytes). A word is a group of

bits that are treated as a single unit but the particular

system in use.

Octal: when bits are treated in groups of three then

the numbers that can be stored in each digit range from

zero to seven (eight values), and place values arebased on powers of eight. This grouping is called Octal

and implies that the bits are processed in threes. In this

system, a Word is often 24 bits as this conveniently

contains eight octals as well as three bytes. IFF codes

use Octal encoding.

REPRESENTING TEXT IN BINARY

Messages containing text and numbers are repre-sented in digital systems as a series of byteswhere each byte corresponds to a letter, digit, punctua-

tion mark, etc. There are several different coding meth-

ods but the ASCII (pronounced ass-key) Code is very

common and will be used here. Each datum to be codedhas a number in the range that one byte can store (i.e.

Zero - 255). For example, the letter Ahas code number

65 (six tens and five units) in decimal, 01000001 (one

sixty-four and one unit) in binary or 41 (four sixteens and

one unit) in hexadecimal whilst the letter ahas the code

01100001b or 97d or 61h where the subscripts b, d and

h are used to identify binary, denary and hex numbers,

respectively. Often, the number in a byte will represent

something other than a letter, for example,t he bright-

ness of a pixel in an image.

UNICODE

To enable computer font files (e.g. Times NewRoman) to contain letters and symbols from manylanguages (e.g. Cyrilic, Greek, Chinese) then the alloca-

tion of one byte to store the code for a symbol is insuffi-

cient, as this only permits 255 different symbols.

Unicode, which uses two bytes to store the code for

each of the symbols. Unicode allows up to 65535 differ-

ent codes and each font file can store every character

from every human language - and still have spare

capacity. In Unicode, each keypress generates two

bytes of data - compared to one byte when using ASCII.

E07 Digital Data Comms.QXD E07-2 7 J ul 05

Decimal Hex Binary

0 0 0000

1 1 0001

2 2 0010

3 3 0011

4 4 0100

5 5 0101

6 6 01107 7 0111

8 8 1000

9 9 1001

10 A 1010

11 B 1011

12 C 1100

13 D 1101

14 E 1110

15 F 1111

Figure 1: Conversion between Decimal and Hex

01011001 = 89 denary (64+16+8+1)

0101 1001 = 59 hex (516) +9

0101 1001 = 59 BCD (510) +9

001 011 001 = 131 Octal (164) +(38) +1

Figure 2: Types of Binary Number


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TRANSMITTING DATA IN BINARY - PARALLEL

To send a digital representation of a letter Afrom oneplace to another then parallel transmission can beused. This allocates one wire or track on a printed circuit

board to each binary digit and sets it to 0 V or 5 V

depending on the binary digit. Thus, eight parallel wires

are required to carry the data and, to avoid interference

between the signals in the parallel wires (cross-talk),

eight earth wires are often used, placed one between

each data wire. This is illustrated in Figure Three.

In addition to the sixteen wires that might be used to

carry the data there are additional wires used for sig-

nalling, for example:

a signal that the receiving end is ready for the data

to be sent.

a signal that the transmitting end is ready to send.

a signal that the data have been sent.

a signal that the data have been received.

The cable used to transfer data between a PC and aprinter has 36 conductors, including one to signal when

the printer has run out of paper.

Ribbon cables, used inside computers and other

electronic equipment to convey digital data from one

place to another, might have fifty or sixty parallel wires.

Some parallel connections allow for a number of

devices to be connected. These are often called a bus

from the latin omnibus, which translates as for all. A

bus requires additional wires to identify which device is

to handle the data and whether it is to send data or

receive it. The wires that carry the data form the data

bus; those that identify the location of the data form theaddress busand those that carry the signalling data are

the control bus.

Parallel transfer is a short-range system as the

expense of the cables and connectors increases rapidly

over long distances and for complex systems. There are

additional problems when attempting to use it at high

speeds or over long distances, because minor differences

between the wires can cause some signals to arrive later

than others. However, for short distances, parallel trans-

mission offers medium to high speed as many bits are

sent simultaneously (as many as 64, or 8-bytes, at a

time). Parallel transfer is used in PCs to transfer data

between motherboard and hard disk (ribbon cable) andfrom CPU to memory (parallel tracks on the mother-

board), for example.

Types of parallel systems that you might encounter

are: IDE, E-IDE, ISA, E-ISA, VESA, SCSI, PCI, AGP,

Centronics, IEE488 and Euro-Bus. Unfortunately, most

of them can only be directly connected to another bus of

the same type and conversion from one to another can

be difficult. All parallel systems share the same basic

properties as outlined above.

TRANSMITTING DATA IN BINARY - SERIAL

For long transmission distances, or where there isinsufficient space for a large cable, it becomesimpractical to send each bit along its own, dedicated

wire as in parallel transmission, and an alternative

called serial transmission, is used. In the serial system,

the bits are sent one-by-one along a single pair of wires

and re-assembled at the far end.

For serial transmission, each Byte is placed in a shift

register. This is a special type of memory in which the bits

move one position along in response to a shift signal

applied to one of its control lines. (Often, the bits circulate

so that they are all back in their correct places after eightshifts.) The sequence of events, might be as follows and

illustrated by Figure four.:

The byte to be transmitted is moved into the shift

register.

The bit on the end of the shift register is sent along

the wire.

The Byte is shifted one place, the (new) bit on the

end is sent along the wire.

This process is repeated until all eight bits have

been sent.

Meanwhile, at the far end of the line, the receiveraccepts each bit in turn and shifts it into its register.

Thus, after eight operations of the shift register, a

Byte has been sent. Obviously, some means is required

of synchronising the shift registers at each ends and that

will be covered later.

The waveform of the signal that was sent is illus-

trated at the bottom of Figure Four. This is a rectangular


00

00

00

11

00

0000

11

0V

0V

0V

0V

0V

0V

A

Figure 3: Parallel Transmission of One Byte

00

00

00

00

0 0 0 0

11

1 1 11

11 11

11

Transmit

OUTIN

Figure 4: Serial Transmission of One Byte


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wave where consecutive signals (e.g. 111) are merged

into one long pulse and there is no return to zero

between such pulses. This is called a non-return to zero

(NRZ) waveform.

PROPERTIES OF SERIAL TRANSMISSION

T

he advantage of serial communications is that only

one pair of wires or one communications link is

required to carry the information. It seems that, under

identical conditions, the transmission of a Byte by serial

means would take at least eight times longer than if a

parallel link were used to transmit the same byte.

However, a serial link can be operated at a much higher

speed than a parallel link because all the bits travel

down the same piece of wire and, therefore, arrive at the

other end in the correct sequence. With a parallel link,

there might be sixteen, different wires (running side-by-

side) and the bits can travel at different speeds in adja-

cent wires. This limits the speed at which parallel links

can be operated, because the system has to allow time

for all bits from one set to arrive before sending the nextset. Consequently, few parallel links operate faster than

about 100 MHz whereas serial links of 10 GHz are avail-

able.

When information is to be transmitted in both direc-

tions then this normally requires a minimum of three

conductors:

an earth or common wire.

a wire to carry data out of the data transmitter

a wire to carry data into the data receiver.

Depending on the methods used to transmit the data,it is possible to use the same wire to send and receive

data - and to do this simultaneously - so just two wires

can be used. PC modem communications to the

Internet, that use ordinary telephone lines, transmit and

receive data using just two wires.

Simplex: communications links that are capable of

conveying data in only one direction obviously need two

channels to conduct a two-way conversation - one for

each direction. These are called Simplex channels.

Half-Duplex: where the channel is bi-directional, but

can only be used in one direction at a time then it is

called Half-Duplex- many radio nets use this technique

and the person transmitting says Over when he has fin-ished so that the other person knows that he can begin

to transmit.

Duplex: when a channel can be used in both direc-

tions at once (e.g. telephone cable) then it is called

Duplex or Full-Duplex. (Mobile phones use two sim-

plex channels - on different radio frequencies - to simu-

late a duplex system.)

Control: additional wires can be employed for hand-

shaking - to assist in controlling the flow of data and for

signalling tasks (this is called hardware flow control).

For example, a Modem that answers an incoming call

uses a dedicated line to signal to the computer that ithas detected an incoming call and various other lines to

keep the computer informed about the status of the call.

Alternatively, some of the 255 codes that can be accom-

modated in one byte can be allocated to commands

(e.g. stop transmitting) so the data lines can be used to

carry control signals (this is called software flow control

and the codes are called X-ON and X-OFF).

Examples of serial transmission that you might

encounter in computing are:

IEE-1394 (Fire-wire): used for high-speed data, such

as digital video

USB (Universal Serial Bus): used to connect exter-

nal devices (e.g. scanner, modem) to a computer.

SATA (Serial ATA - used to connect a computer

motherboard to a hard disk, inside the computer.

RS-232, RS332, RS422: the serial port on a com-

puter (obsolete, as USB replaces it).

PCI-Express: the latest replacement for PCI and

AGP connections on computer motherboards.

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PROPERTIES OF COMMUNICATIONS LINES

Communications links, be they radio, telephone ordedicated cable, all have their practical limitations.You are no doubt aware of the differences between the

sound of a piece of music played on a Hi-Fi and the

same music heard on the telephone. This is mainly due

to differences in bandwidth between the two systems.

The Hi-Fi has a bandwidth ranging from about 20 Hz to

20 kHz whereas the telephone only allows 300 Hz to

3,400 Hz. Since audible sounds range from 20 Hz to

20 kHz, then the telephone gives only an impression of

the sound - sufficient for voice communications but inad-

equate for music.

Bandwidth. The bandwidth of many communications

lines does not extend to Zero Hz (dc) - this means that

dc signals (steady values that change relatively slowly,

if at all) cannot be communicated down such links in

their original form. The bandwidth of a signal always has

an upper limit (for telephone signals it is about 3.4 kHz,

for television signals it is about 5.5 MHz) and the

expense of setting up and running the communicationslink increases as the bandwidth increases. If the band-

width of the computers signals exceeds the bandwidth

available in the cable, radio link or telephone system

then the signals will be unlikely to arrive at the far end in

a usable form.

Capacitance. Many communications links contain

capacitors and, when transmitting ordinary voice signals

(with equal positive and negative parts of the signal) the

average charge in these capacitors is zero so they do

not cause problems. However, you should be able to

see from Figure Three that the example of a computer

data signal is never negative - always zero or positive -and this causes the capacitors to gain charge during the

transmission of the data signal. A long series of binary

Ones will tend to charge the capacitors towards 5 V

whilst a long string of Zeroes would tend to discharge

them. On average they might reach half-way.

Distortion. The shape of the wave that arrives at the

far end of a communications link should be the same as

the shape of the original wave. Even if there is sufficient

bandwidth, the shape of the wave can be altered . This

is because capacitance and inductance in the system

reduce the amplitude of the some frequencies by a

greater amount than they reduce the the amplitude of

others. This distortion can cause problems when dataare transmitted.

Noise and Interference. This is an irregular signal

that arises from a variety of sources in the communica-

tions channel. In telephony, it can be heard as a hiss-

ing or rushing sound; a noisy television picture is

indistinct with a granular appearance. Impulsive interfer-

ence, produced by electrical machinery (even light

switches) could exceed 2.5 V in our computer data link

and cause a signal that had been sent as a Zero to be

received as a Oneand vice-versa. This is illustrated in

Figure Five.

Dispersion . When white light passes through a prismthen the individual colours appear as a rainbow. This

splitting arises because the different frequencies of light

(which we see as different colours) travel at different

speeds in the glass. This effect is called dispersion. The

different frequencies of electrical signals travel along a

cable at different speeds and, as in the beam of white light

example, they spread out as a result. In a comms link, a

single pulse (e.g. a binary One), that contains a mixture

of harmonic frequencies, gets separated into its various

harmonics as it proceeds along the cable.

The problem of dispersion worsens as the length of

the cable increases until some of the slower-moving fre-

quencies from one pulse fall behind and get mixed up

with some of the faster-moving frequencies from the fol-

lowing pulse - the pulses tend to merge together and the

computer cannot recognise the data.

At tenuat ion. As the signal travels along the cable or

radio link then its power reduces as some is absorbed

by resistance and other losses in the line. A short cable

(few metres) often has negligible loss (attenuation) but

the losses mount as the length is increased - and they

get worse at higher frequencies (i.e. at higher data

rates). Amplifiers (called line drivers) can be used to

boost the signal before it enters the cable but they dontcure dispersion and bandwidth problems. Repeater

amplifiers can be used at intervals along a cable but

they will also amplify any noise and distortion present so

they have limited value. Digital lines can use re-genera-

tors - these must be located sufficiently close together

along the line so that the data signal is still detectable.

The re-generator reads the data and uses switching cir-

cuits to re-form the pulses into the correct shape before

forwarding them to the next part of the line.

Reflections. Electrical signals travelling along wires

will reflect, either wholly or partly, whenever they

encounter a change in the electrical properties of theline. An open-ended cable or a socket without a cable

plugged into it are typical examples of discontinuities

that can cause reflections. It can become impossible to

receive any data at all when reflections arise as for each

pulse that was transmitted there might be several dupli-

cates produced by reflections. Correct termination of

cables is essential - co-axial cables typically have an

electrical impedance between 50 and 75 so a 50

or 75 resistor is normally connected at the end of

every bus, cable, etc., to prevent reflections. [Some very


0

0

0

0

0

0

0

0

1

1

1

1

1

1

1

1

Figure 5: Noise and Distortion Added to a SIgnal


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fast systems, in modern PCs, have more complex

means of eliminating reflections at the end of cables,

called Active Terminations. Without these, the bus fails

to transfer data reliably.]

The combined effect of the above problems is illus-trated in Figure Six which shows how a byte of serial

data might get distorted whilst passing along a comms

link. The effect of interference and distortion is usually to

flip one or more bits in the signal. In plain text, a letter b

could become a r but the operator would probably

notice this and correct a spelling error. If the signal

were from a cash machine to your bank to inform it that

you had taken out 25 then you would not be impressed

if some electrical interference changed the sum involved

to 65! In a weapons system, data corrupted during

transmission is most undesirable too.

Errors can be detected - so that a re-transmissioncan be made - or detected and corrected - so that no re-

transmission is needed. This will be covered later.

REDUCING INTERFERENCE

Afibre-optic cable is immune to interference becauseinterference is electrical and fibre-optics uses lightsignals - not use electrical signals. At some point, the

light signals will probably be converted back into electri-

cal ones but this should take place in a shielded enclo-

sure to minimise interference.

Where copper cables are used then a method called

differential signalling can be used. When an ordinary

signal passes along a wire as, for example, a positivesignal - this is a single-ended system as the common, or

earth, wire remains at zero volts and only the livewire

has the signal on it. A differential transmission system

uses two wires (a pair) - neither is earthed - and anti-

phase signals are sent along the pair to the other end

which only responds to differences between the wires.

This can also be called a balanced system of transmis-

sion.

For example, a 5 V signal would be sent as +2.5 V

on one wire of the pair and -2.5 V on the other. At the far

end, the difference between the wires is taken as the

signal. If any interference enters the system then it willaffect both wires equally so that, for example, 1 V of

interference would add to each wire and make one

+3.5V and the other -1.5V. Nevertheless, the difference

remains at 5 V so the interference is rejected by the

receiver. The signal is said to operate in differential

mode (opposite signals on each wire) whilst the interfer-

ence is common mode (same signal on each wire). This

is illustrated in Figure Seven, which shows the same

signals as Figure Six but with differential signalling. The

interference that corrupted the bit in Figure Six now has

no effect on the same bit in Figure Eight.

Under normal conditions and using similar signals and

cables, an unbalanced system, such as RS-232, can

operate up to about 10 m whilst a balanced system,

such as RS-422, can operate up to about 1 km. Theextra range is due to the resistance to interference pro-

vided by the balanced system. The unbalanced system

needs three wires whilst the balanced system needs

four wires well worth the expense for the benefits

achieved.

BANDWIDTH

The digital signals described above contain harmon-ics that cover the frequency range from zero (dc) toseveral times the pulse repetition rate. Most transmis-

sion paths (e.g. radio, telephone, satellite) do not permit

the passage of dc and they are not very effective at car-

rying the sharp edges of rectangular waves either. Inother words, you cant send digital data signals down a

transmission link using the waveforms illustrated above

because their bandwidth does not match.

One solution to the mismatch is to use a carrier sig-

nal that is compatible with the transmission link and to

change the carrier (modulate it) using the computer sig-

nals. One simple example of this is frequency modula-

tion. In its simplest form, a binary digit of Zero causes a

low frequency (e.g. 1 200 Hz) to be transmitted whilst a

binary digit of One causes a high frequency (e.g.

1 400 Hz) to be transmitted. This is also called fre-

quency-shift keying (FSK) and was used to transmitteleprinter data.


This bit changes to Zero!

0 1 0 1 1 0 00

0 0 1 0 1 1 1 0

Figure 6: Data Changed by Noise & Interference

This bit no longer changes to Zero!

Signal 1

Signal 1

Signal 2

Signal 2

1

1

0

0

1

1

0

0

1 1 1 01 0

Figure 7: Differential Signalling to Reduce theEffects of Interference


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A sample waveform is shown in Figure Eight. In the

Figure, the byte 00111001 is shown being transmitted

using simple FSK.

The digital data may also be used to modulate the

amplitude or phase of the carrier signal, as shown in

Figure Eight.

ASYNCHRONOUS TRANSMISSION

Most data are not sent continuously - rather they aresent as and when necessary, with breaks inbetween. (A teleprinter is an example of this where acharacter is transmitted each time the operator presses

a key.) The time between one character and the next is

variable and unknown and this is called asynchronous

transmission. Much serial traffic passes in this way,

including that sent via a PCs Modem to the Internet, etc.

It relies on using special indicators to signal the start and

stop of the binary data being transmitted.

Recognising the start of the data. The digital signals

from each byte contain Ones and Zeroes. If a signal of

00010000 were to be sent then how would the receiver

know that the signal had begun? The arrival of the One

digit in the middle is a clear indication that data are pres-ent but the receiver has already missed the start as it

contained several Zeroes - and there is no way of going

back. Clearly, some method is required of signalling that

data are about to be sent in order to allow the receiver

time to prepare to accept it.

Recognising the end of the data. If the cable

became disconnected in the middle of sending some

data then the receiver might interpret the missing data

as all Zeroes (since the voltage would fall to zero). A sig-

nal is needed at the end of the data so that the receivercan be sure that correct number of bits was received.

Some slower systems, e.g. mechanical teleprinters

require time to get ready for the next character and this

time influences the duration of the stop signal.

The Ready Signal. In between transmissions, the

receiver needs to be aware that the transmitter is pow-

ered up and liable to send at any moment. AZero signal

on the line would not suffice as a disconnected line

would give this. A signal that says idle but ready is

required. In many systems this is a steady Onesignal.

If ever you have dialled a fax machine then you will have

heard the high-pitched whine that it produces - this is thecarrier signal - it contains no data yet, but you know

that it is there! Some communications links (those with

dedicated wires) send a steady 5 V or 12 V, instead of

an audible tone, to indicate that the transmitting device

is powered up and ready.

The message (data). The data to be sent fit in

between the start and stop signals. The data can be of

any length but many are quite short for reasons that will

be covered later

Frame: the entire signal is called a frame. The start

signal is a change from the ready state (binary one) to

Zero. This lasts for one bit. The next eight bits are the

data and the final bit, always a binary One, marks theend of the frame of data. The stop bit must be a One

because this re-sets the line for the next start bit. The

line may remain idle (set to One) for an indefinite time

until the next byte is ready for transmission.

Baud Rate: the rate at which the pulses are sent

down the line is called the Baud Rate. A signalling rate

of 2 400 Baud is possible on a telephone line. This cor-

responds to the number of changes of signal - per sec-

ond - that can be sent down the transmission link.


100 0 01 1 1

Direction of travel

Figure 8: Types of Modulation

(Upper) Frequency Modulation (FSK)

(Middle) Amplitude Modulation (ASK)

(Lower) Phase-Modulation (PSK)

1000

START

STOP

0 0 11 1 1

Direction of travel

Figure 9: A Frame, with Start & Stop Bi ts


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DECODING STRATEGY

The receiver requires timing information in order todecode the incoming signal. The standard strategyis as follows:

The receiver monitors the oncoming carrierwaiting

for it to change from One to Zero- the start bit.

The receiver knows the parameters of the transmis-

sion because the operator has set them (e.g. 1 200

Baud, 8 data bits, one stop-bit) or because the

receiver has previously performed test communica-

tions with the transmitting device to determine them.

Upon receipt of the start bit, the receiver waits for 1.5

pulse-periods and then tests the signal on the line.

At 1 200 Baud, this will be 1.5/1200 or 1.25 ms. This

should be right in the middle of the first data bit. This

bit, zero or one, is shifted into its receiving register.

The receiver continues to test the signal on the line

at intervals of one pulse-period, 1/1200 s or 833 s,

until eight bits (one byte) have been received.

The receiver then tests for the stop bit by waiting onefurther pulse-period before testing the line for a One.

Errors. If the stop bit is not received when expected

then a Framing Error is generated. This usually occurs

when the receiver is set to the wrong Baud rate and it

ends up looking for the stop bit at the wrong time. It can

also occur if the cable is shorted or dis-connected dur-

ing transmission.

Timing Accuracy. The receiver examines the incom-

ing signal in the middle of each (expected) bit-period.

The message lasts for ten bit-periods and this is the time

during which the receiver must remain synchronised tothe bit stream. Atiming error of two percent will not affect

the accuracy of reception because, even after ten bit-

periods have elapsed, as this would only amount to

twenty percent of a bit-period - not enough to miss the

stop bit as this is tested at its expected 50% point (mid-

point), whilst the error would test at the 70% point or

30% point - still well within the safe time. In this exam-

ple, a timing error of 5% would be the greatest consis-

tent error that could be tolerated. A random error greater

than this might not interfere with communications as it

might consist of some gains and some losses of time

which could partly cancel each other.

SYNCHRONOUS TRANSMISSION

In weapons systems, the data do not usually travelalong a telephone line although the same considera-tions apply for most cables and radio links. Where the

computer system has its own dedicated cables, not

shared with telephony signals, then a different system

can be used. Computers can easily generate data at a

rate of Mega-bits per second whilst ordinary communi-

cations channels are several hundred times slower.

Bigger Packets: when a dedicated cable is used

then the bandwidth can be much greater so the data canbe transmitted at a much higher rate. This also enables

the use of much bigger frames (packets) of data - con-

taining 128 or 256 bits, for example. Fewer start and

stop bits are needed so more data can be transmitted.

Eliminating DC. To eliminate any stray dc signals, the

connection between the cable and the computer system

can be made using a transformer (which does not pass

dc) or opto-couplers (which pass the data using light sig-

nals) and capacitor filters (to remove dc). This means

that the signalling method chosen must produce signals

with an average value of Zero Volts - i.e. with no dc con-

tent.Reducing Spurious Detection. In Manchester

Encoding, a binary One is represented by a signal that

is positive for half of the bit-period and then changes to


100 0 0 0 11 1 1

Direction of travel

Start

This bit is a Zero becauseit goes from neg to pos halfway along.

This bit is a One because itgoes from pos to neg half

way along.

etc....

Pos

Neg

Figure 10: Illustrating One Type of Manchester Encoding for Synchronous Transmission


9/24


negative half-way through. Abinary Zerois negative for

the first half of the bit-period and then changes to posi-

tive half way through. The average value of each bit is

zero because each is encoded with a 50% positive and

50% negative signal. If the change does not occur close

to the middle of the bit-period then an error is flagged.

Long Strings of Identical Bits. If a data message

contains lots of Ones or lots of Zeroesthen it becomes

difficult to identify where one ends and the next one

starts. (This is a problem with the non-return-to-zero

(NRZ) signal - if the next bit has the same value as the

current bit then the signal transmitted remains constant.

This is shown in Figure Ten where three successive

Ones appear in the middle of the signal.) A string of

thirty successive Ones might be decoded as thirty-one

or twenty-nine because the receiver loses synchronisa-

tion.

Embedded Timing: In Manchester Encoding, there is

a change from positive to negative (or vice-versa) in

every bit - the receiver can easily identify each bit

because the timing information is included in the data.Longs strings of ones or zeroes cannot merge together

in this system.

Start Bit. This signal must identify clearly the start of

a message and, therefore, it is usually formed differently

from the message that follows it. In FSB2, the start-bit

(also called a sync-bit) spans two or three bit-periods

(depending on the type of message that follows it) and it

only changes once during that time. This marks the

beginning of a transmission in a distinct way as the ini-

tial pulse is two or three times longer than any other in

the system. Since the start bit is used to synchronise the

receiver to the transmitter then it is often called thesync signal. Typical Manchester Encoded signals are

shown in Figure Ten.

Stop Bit: A useful check that the entire packet was

received - however, this can also be accomplished by

counting the number of changes that were identified in

the packet.

Padding: when the amount of data to be sent is less

than a complete packet then the packet can be padded

with zeroes (or ones) to ensure that it is the correct

length.


100

STAR

STO

PARIT

0 0 01 1 1

Direction of travel

Figure 11: A Complete Data Frame, with StartBit, Stop Bit and Parity Bit


10/24


THE IDEAL TRANSMISSION

The main ideals of message transmission are to sendthe information as fast as possible down a channelthat has a given bandwidth, using the least amount of

power with no errors in the received message. In any

comms system there will be noise and interference

which degrade the signal and will cause errors. There

are two main ways to reduce the frequency of bit errors.

One method is to increase the signal to noise ratio,which may cost in the form of higher power delivery, or

the receiver electronics could be made more sophisti-

cated and again this will cost money. The second

method is to add additional parity bits in order to perform

an error check on the data bits. This second method

obviously costs too, because if the rate of transmission

is not altered, the data will take longer to send. If the

data rate is to be maintained to the speed before adding

the parity bits, then the bandwidth must increased

accordingly.

ERROR DETECTION - PARITY

You have already seen, in Figure Four, how a noiseor interference signal can change a bit and, there-fore, corrupt the signal. When noise and interference

levels are fairly low then this is a rare event and the vast

majority of bits arrive at the destination. However, errors

can occur during transmission and it is often necessary

to be able to detect them.

When an error is detected then the receiver can sig-

nal for a re-transmission sometimes called Automatic

Repeat on Request (ARQ) and it requires a duplex

channel. When many errors are detected then the

receiver might signal that a lower Baud rate should be

used - this narrows the bandwidth and reduces theamount of noise and interference.

Errors can be detected using a simple system called

parity where the transmitter counts the number of

Ones in a frame of data and ensures that this number

is always even. This works as follows:

The frame is extended by one bit - the parity bit.

If a frame with an odd number of Ones is about to be

transmitted then a One is placed into the parity bit.

There will now be an even number of Ones.

If the frame already has an even number of Ones

then the parity bit is set to Zero. The start and stop bits do not count as they are not

part of the data

Thus, the byte 00111001 would be sent as:

Start Data Bits Parity Stop

0 00111001 0 1

The nine bits, data plus parity, contain four Ones -

an even number and this is illustrated in Figure Ten. If a

bit is changed during transmission then the receiver will

decode nine bits that have either three or five Ones.These are odd numbers so a parity error will be recog-

nised. The transmission must be repeated until a correct

frame is received because simple parity does not iden-

tify which bit is wrong.

The byte 00100110, which currently has an odd

number of Ones needs a parity bit of Oneand would

be sent as:

Start Data Bits Parity Stop

0 00100110 1 1


Data

Source

Data

Sink

Noise / Interference + Errors

Parity or:

Encoding

Parity check or:

Decoder

detect / correct

Comms Channel

or storage device

Figure 11A: Error correction principle

CHECKSUM COMPUTATION

The following example uses decimal numbers for

simplicity; normally it would be done with hexadeci-

mal numbers and many more digits.

The message to be sent is a stream of eight num-

bers and we want to ensure that no changes occur

during transmission. The numbers are:

0 3 2 4 9 8 5 6

Compute a checksum by multiplying the first digit

by 9, the next by 8, the next by 7, etc. and add togetherall the results to get: 166. Next, divide this by 25 and

find the remainder (16). Add one to this to get 17 and

find the 17th letter of the alphabet Q - this is the

checksum and it is placed at the end of the digits to be

sent. Thus, the numbers are sent as:

0 3 2 4 9 8 5 6 Q

Upon receipt of the message, the receiver repeats

the calculation and verifies that it comes to Q - the

checksum that is included in the message. Any

changes to the data will produce the wrong checksum.In practice, more complicated methods may be

used to calculate the checksum, but the principle is

generally the same as that described here.


11/24


We now require a frame of duration eleven bits in order

to transmit eight bits of data - an efficiency of8/11 or 73%.

Some systems use odd parity, where the number of

Ones is made up to an odd number. Parity fails when

two bits have been changed - but that is very rare and it

would represent a situation where practically every

frame would have been corrupted. In that case, commu-

nication would be practically impossible at the selected

Baud rate as re-transmitted frames would be clogging

up the system.

Checksum: parity works well when there are few

errors and short frames. For longer frames, with many

more bits, then there is a significant chance that two or

more bits might be altered during transmission - and

parity might fail to detect this. Any long stream of data

(e.g. a computer file saved on a disk or a long packet

sent using synchronous transmission) has a checksum

figure that is calculated by the originator of the data.

This might be a 32-bit number calculated by multiplying

the bytes of the message by a series of numbers and

adding them all together. The resulting checksum isadded to the message.

When the data are received then the receiver re-cal-

culates the checksum and compares it with the original.

Any changes to the data will alter the checksum and it is

most unlikely that several changes could cancel out and

restore the original checksum. This method is also

called a cyclic redundancy check.

Checksums are also used by some computer pro-

grams to determine whether a file has been tampered

with, for example by a virus, as it is very difficult to

change a file whilst keeping the same checksum - espe-

cially if the location of the checksum in the file is not

known or when the checksum is stored elsewhere.

FORWARD ERROR CORRECTION (FEC)

The key to error correction is redundancy. Indeed, the

simplest error-correcting code is simply to repeat every-

thing several times. If, for example, you anticipate no

more than one error to occur in transmission, then

repeating each bit three times and using "majority vote"

at the receiving end will guarantee that the message is

heard correctly (e.g., 111 000 011 111 will be correctly

heard as 1011). In general, n errors can be compen-

sated for by repeating things 2n +1 times. Simple par-

ity, as described above, checks the validity of a

message but it cannot determine which bit or bits are in

error and it cannot correct the problem. Simple parity is

an error detection method. More advanced methods can

not only detect an error but also correct it immediately.

This is achieved by increasing the number of bits in the

message and, therefore, reducing the efficiency of

transmission. There are three common methods usedfor error correction: block parity, Hamming and Reed-

Solomon.

Block Parity. Consider a set of eight Bytes with sim-

ple parity that are to be transmitted. The set of eight is

considered to be a blockof data and can be thought of

as eight rows of data. As each Byte has its own parity

then each row has nine bits. If a ninth row were added

and its bits used as parity bits for the columns of the

block then we will have both horizontal (row) parity and

vertical (column) parity. When an error occurs then two

parity bits will identify it as the bit at the intersection of

the row and column with the error. This is illustrated inFigure Eleven. The block of data has one error and the

row parity is wrong in row 5 whilst the column parity is

wrong in column 4 (numbering columns from the right,

starting at Zero). To correct the error, bit 4 of Byte 5

needs to be changed.

Efficiency. The block parity method works well when

there a few errors but it is inefficient unless there are

eight bytes ready to be sent. If fewer bytes are to be

sent then the unused ones are filled with zeroes and this

represents an inefficient use of the system. When all

eight bytes are in use then there are 64 data bits and 17

parity bits - an efficiency of 64/81 or 79%. This falls to

just 10% if only one byte contains real data.

BIT ERROR RATE PROBABILITY

When a coin is tossed, there are two possible out-comes (Head or Tail) and these outcomes areequally likely. The probability in laymans terms is 50-50.

Mathematically speaking, the probability of getting a

head is 1/2 =0.5 or 50%. All three methods of quoting

the result are valid. In Data Comms, the mathematical

probability of an error occurring cannot be determined,

but we can make an estimate by counting the number of

errors that occur over a given time interval. The estimate

is determined like this:

Bit Error Rate probability =Number of error bits

Total Number of bits sent


TRANSMITTED DATA BLOCK

Data Byte Parity

Row1 01100110 0

Row2 11011010 1

Row3 10000000 1

Row4 01111111 1

Row5 00000000 0

Row6 11111111 0

Row7 10110111 0

Row8 00100011 1

Parity 00101000 0

RECEIVED DATA BLOCK

Data Byte Parity

Row1 01100110 0

Row2 11011010 1

Row3 10000000 1

Row4 01111111 1

Row5 00010000 0 x

Row6 11111111 0

Row7 10110111 0

Row8 00100011 1

Parity 00101000 0x

Figure 12: Block Parity Check to Correct an Error


12/24


The longer the period of time over which this experiment

is performed, the more accurate the estimate will be.

The resultant (BERP) will lie between 0 and 1. Stated

below are bench mark values in comms. Of course the

qualitative nature of these benchmarks depend on the

criticality of the data.

BERP =0.1 Very poor (one in ten bits are errors)

BERP =0.001 Average (one in 1000 bits are errors)

BERP =0.000 001 Good (one in a million bits are

errors).

In scientific notation the above are written like this

101, 103, 106

The computer industry standard allows for an error rate

of 1013 , that is less than one uncorrectable error in

1012 (one million million) bits. The audio industry is a lit-

tle more forgiving. DAT assumes a data error rate of 105

(at 48kHz) which works out at about two errors every

second. Optical media has a high error rate of 1 bit out

of every 100 000, but magnetic disk has a much lower

error rate of one bit out of every thousand million.

SIGNAL TO NOISE RATIO

Agood Comms system has a high signal to nose ratioand is usually expressed in dB. Here are someexamples of what is means: For each dB quoted, the

factor is the number of times more powerful that the sig-

nal is in relation to the noise.

SN Ratio (dB) Factor

3 2x

6 4x

10 10x

20 100x

40 10 000x

60 1 000 000x

A television engineer will monitor this dB level using a

meter for every channel that your aerial is capable of

receiving. A reading of over 60dB is desirable for a clear

pictures from analogue systems (ie normal terrestrial

television channels from a transmitter like

Mendip that serves the South West. Areading of 30dB

will give a snowy picture. In digital data comms, the

value does not need to be any where near this high. We

will discover below that Signal to Noise Ratios as low as

10dB can give good results.

COMPARISON OF DIFFERENT COMMS SYSTEMS

We will discover that for any Comms system, the BERPwill decrease (improvement) when the Signal to Noise

ratio is increased (improvement). If there is an inherent

low Signal to Noise Ratio, that cannot be improved

upon, the BERP can be improved by the use of a

Hamming Code (see below). If the Signal to Noise Ratio

is high, then it turns out that the BERP is not improved

with the use of a Hamming Code. Hence there will be a

turning point when the two systems give equal BERP.

The choice of system then will be decided on other mat-


Figure 18: Baseline PSK BERP plotted against S-N Ratio


13/24


ters such as cost of manufacture, running costs and

ease of use.

A BASELINE SYSTEM

Lets examine a Binary Phase Shift Keying system for

three levels of Signal to Noise Ratio. This system has no

form of parity bit error checker, so it will act as a base-

line system, to which other systems can be compared.

SN Ratio (dB) BERP3 0.03 (3 errors in a 100)

7.5 0.0006 (6 errors in 10 000)

10 0.000006 (6 in a million)

HAMMING CODING

Now what happens to the BERP when we add paritybits? This will be answered later. Lets examine aspecific arrangement; the (7,4) Hamming code. In this

code, data is taken 4 bits at a time and 3 parity bits

appended to make a "word" of 7 bits, hence the desig-

nation (7,4). Denoting the data bits D0, D1, D2, D3,

each of the parity bits P0, P1, P2 are formed from theoutput of one three-input exclusive-ORs (EOR gate) of

a chosen three of the data bits. The rules for the choice

of combination is shown in figure 14 and the truth table

for an EOR gate is shown in figure 15. For example P0

is formed by putting the inputs of an EOR gate to the bit

values represented by D0, D1, D3. This is written sybol-

lically in Electronics as in the top line of figure 14. The

corresponding values of D0, D1, D3 are respectively 0,

0, 1 for a transmitted binary word (a string of 4 bits in this

example) shown in italics on the upper right of figure 13.

The parity bit P0 is output from the EOR truth table (sec-

ond line of numbers in figure 15). We see that this value

is 1, so P0 =1. The other two parity bits P1 and P2 are

calculated in a similar way using the EOR rule given in

figure 14. All seven bits are then sent down the Comms

channel and when they are received, the parity bits are

calculated from the same set of rules as before. If noerrors have occurred, the same parity bits will be

achieved. Figure 16 shows the case when one error

occurs. The figure shows a word received as 1 1 0 1

when it should be 1 1 0 0. Dont forget here that figure 13

shows the least significant bit is called D0 and the most

significant bit called D3. We can see from figure 16 that

two of the parity bits do not agree with parity bits that

have been sent. Analysis of the make up of the rules for

parity bit calculation, reveals that D0 is common to the

errors in P0 and P1. Hence D0 is the culprit, and its

value will be inverted to D0 =0. Here is a summary:

Transmitted data = 1100 001

Received data = 1101 001

From the first four bits ie data bits received, the parity

bits should be 111, but the received parity bits are in dis-

agreement showing 001. Two errors pinpoint that the

least significant bit D0 of the data block is wrong.


P0 =D0 D1 D3P1 =D0 D2 D3

P2 =D1 D2 D3

D0 =1 (error)D1 =0D2 =1D3 =1

P0 =1 0 1 =0 (wrong)P1 =1 1 1 =1 (wrong)P2 =0 1 1 =0 (correct)

The common factor in the wrong Parity bits is D0.

Figure 16: Pinpointing the error

P0 =D0 D1 D3P1 =D0 D2 D3

P2 =D1 D2 D3

D0 =0D1 =0D2 =1D3 =1

P0 =0 0 1 =1P1 =0 1 1 =0P2 =0 1 1 =0

Figure 14: Hamming Code Formula

Dec Inputs Output

0 0 0 0 0

1 0 0 1 1

2 0 1 0 1

3 0 1 1 0

4 1 0 0 1

5 1 0 1 0

6 1 1 0 0

7 1 1 1 1

DDDD PPP DDDD PPP DDDD PPP DDDD PPP

3210 210 3210 210 3210 210 3210 210

--------------------------------------------------

0000 000 0100 110 1000 111

0001 011 0101 101 1001 100 1101 010

0010 101 0110 011 1010 010 1110 100

0011 110 0111 000 1011 001 1111 111

1100 001

Figure 15: EOR GateTruth Table Figure 13: Sixteen blocks of Data/Parity


14/24


The Bit Error Rate Probabilities for this data comms sys-

tem is show below for three levels of S-N Ratio.

SN Ratio (dB) BERP

3 0.07 (7 errors in a 100)

7.5 0.0006 (6 errors in 10 000)

10 0.000002 (2 in a million)

We observe that Hamming code provides an improve-

ment over the baseline system when the S-N Ratio is

greater than 7.5

CODING GAIN

The fundamental concept of error control coding isthe addition of redundancy to a signal at the trans-mitter, and the exploitation of that redundancy at the

receiver to detect and/or correct errors. The inclusion of

redundancy in the transmitted signal results in a coded

signal consisting of more bits than the original uncoded

signal. The trade-off for this overhead is the ability to

detect, and possibly correct, errors at the receiver. Theperformance improvement that occurs when using error

control coding is often measured in terms of coding

gain. Suppose an uncoded communications system

achieves a given bit error rate (BER) at an SNR of 30

dB. Imagine that an error control coding scheme with a

coding gain of 3 dB was added to the system. This

coded system would be able to achieve the same BER

at the even lower SNR of 27 dB. Alternatively, if the sys-

tem was still operated at an SNR of 30 dB, the BER

achieved by the coded system would be the same BER

that the uncoded system achieved at an SNR of 33 dB.

The power of the coding gain is that it allows a commu-nications system to either maintain a desired BER at a

lower SNR than was possible without coding, or achieve

a higher BER than an uncoded system could attain at a

given SNR. As an example, lets compare the Hamming

(7, 4) system previously discussed, with a Hamming

(32, 6) system used by the Mariner 69 deep space

probe on Mars. (32, 6) means that the data blocks are 6

bits long followed by 26 parity bits. For the (7, 4) system

a S-N Ratio of 10.2 dB is required to maintain a BERP

of 10 -6. The (32, 6) system only requires 6.7dB S-N

Ratio to maintain the same BERP. Hence the Coding

Gain of the Mariner System over the (7, 4) system is

3.5dB. This system was used because transmitterpower to the space probe was limited and there was a

vast distance between the finite antenna size on the

spacecraft and the receiver on Earth.


Random Bit ErrorHello thos is a test transmistion

Hello xgey ts s tejd transmissionBurst Error

Figure 17: Error classification


15/24


TYPES OF DATA ERROR

There are two main types of dataerror, Random Bit Errors andBurst Errors. Our last discussion was

an example of a Random-Bit Error. It is

an error that has no relation to any other

error that may occur in the future. This

type of error occurs singly and are eas-

ily corrected. Burst errors occur when a

sequence of bits is corrupted - e.g. a

long circumferential scratch on a CD.

That is the reason why one should

always use a soft cloth along the radius

of a CD rather than along the perimeter

of the CD. Burst errors can also be

caused by noise spikes, connector

problems, transmission problems in the

connecting cables, defects, dust, etc.

Burst error usually results in data and

redundant data loss (ie the parity bits).

Correction is difficult, and our systemalready discussed would not cope well

with burst error. Burst error is a serious

phenomenon. So much so that systems

are measured by how they react to

burst errors. Look at figure 17, the burst

error has caused the main meaning of

the message to be lost. The maximum

number of contiguous error bits that can

be corrected is a measure of the quality

of a system. Also, a system must be

able to correct random and burst errors

simultaneously. The problem with BERPas a measure of quality, is that it counts

large and small burst errors equally

without taking account of the distribution

of the errors. Reed-Solomon encoding

is a technique discussed later which can

deal with burst errors up to a specified

amount.

CONVOLUTIONAL CODING

Returning to the subject of channel coding; theschemes outlined so far are examples of "BlockCoding", because data

is transmitted andprocessed in discrete

blocks. But there is

another method called

" C o n v o l u t i o n a l

Coding". In this the

parity bits and data

bits are interwoven

(often alternately); the

parity bits are formed

from the exclusive-OR

of selected preceding

data bits on a continu-ous basis.


% freq

Morse

Code

Number

of

Symbols Huffman Code

Number

of

Symbols

A 6.22 .- 2 1011 4

B 1.32 -... 4 10100 5

C 3.11 -.-. 4 10101 5

D 2.97 -.. 3 1011 4

E 28.00 . 1 1 1

F 1.68 ..-. 4 110001 6

G 1.65 --. 3 110000 6

H 3.63 .... 4 11001 5

I 6.14 .. 2 1001 4

J 0.06 .--- 4 1010111011 10

K 0.31 -.- 3 1010110 7

L 3.07 .-.. 4 10100 5

M 2.48 -- 2 11 2N 5.73 -. 2 100 3

O 6.06 --- 3 1000 4

P 1.87 .--. 4 0 1

Q 0.10 --.- 4 101011100 9

R 5.87 .-. 3 111 3

S 5.81 ... 3 110 3

T 7.68 - 1 1101 4

U 2.27 ..- 3 10 2

V 0.70 ...- 4 101010 6

W 1.13 .-- 3 11 2

X 0.25 -..- 4 10101111 8

Y 1.07 -.-- 4 10 2

Z 0.06 --.. 4 101011101011 10

99.24 Average 3.15 Average 4.65

Figure 19: Huffman: assigning symbol length to character

Figure 20: Huffman:

10)0.0006(-logS

0.0006p

0.00060.06%z)(letter1Example

1

)0.28(-logS

0.28p

0.2828%e)(letter1Example

appearingcharacterthatof

yprobabilittheispwhere

)(-logS

S)Character(ofLengthSymbol

2

2

2

=

=

=

=

=

=

=

=

=

S

S

p

Figure 21: Symbol Length of character:


16/24


17/24


The decoder can correct any 16 symbol errors in the

code word: i.e. errors in up to 16 bytes anywhere in the

codeword can be automatically corrected.

Maximum Codeword Length

Given a symbol size s, the maximum codeword length

(n) for a Reed-Solomon code is n =2s 1

For example, the maximum length of a code with 8-bit

symbols (s =8) is 255 bytes.

SHANONS LAW: CAPACITY OF A CHANNEL

This law relates the maximum data capacity of atransmission system, C , to its bandwidth, W, andits signal-to-noise ratio S/N. The law is:

C = 3.32 W Log ( 1 +S/N )

Where Log is the logarithm to the base ten of the

contents of the bracket (use the Log button on the cal-

culator. For example, a channel with a bandwidth of

3 kHz and S/N ratio of 30 db (30 dB =103 =1,000), the

maximum theoretical capacity is:

C = 3.32 3000 Log ( 1001 )

= 29.9 kbits per second

Real systems can never exceed this limit and practi-

cal systems only approach it (due to cost). A high-qual-

ity line is one with a wide bandwidth and low noise.

Figures 23 and 24 show two examples of how to calcu-

late the capacity of a communication channel

DATA TRANSMISSION RATE

The bandwidth of the comms link determines themaximum rate at which data may be sent. The fre-quency at which the signal on the line changes is called

the Baud rate (typical telephone lines can work at rates

up to about 2 400 Baud).

Some advanced techniques can send more than

one bit per signal so that the bit rate can be as much as

56 k bits per second along a telephone line (V90

modem) even though it is still operating at 2 400 Baud.

When a short copper cable or a medium/long fibre-optic

cable is used than the signalling rate can easily exceed

several million Bauds. The time taken to send one bit

might vary between several milli-seconds to less than a

micro-second. Modern Giga-Byte Ethernet can send

10 000 Million Bits per second.

Di-bits: to understand how we can send data faster

than the Baud rate the following simple example is

included. The data stream is one that we have used

before: 00101110. At 1 000 Baud, each bit takes 1 ms

(1/1000 s) to send so the whole byte takes 8 ms (not

including the start, stop and parity bits). If we break the

message into four pairs of bits then it becomes:

00 10 11 10

Instead of defining the line signals as 0 V =Zero and

5 V =One we change this so that there are four levels

and each one corresponds to a pair of bits (a di-bit) as

follows:

0 V =00, 1.7 V =01, 3.4 V =10, 5 V =11

The before and after signals are shown in Figure

25. You will see that the signal sent down the line con-

tinues to change one thousand times each second (1

000 Baud) but each change sends two bits. The bytenow takes 4 ms - half the previous time - to send.

You can probably see that the new signal will be

more susceptible to noise and interference as a result of

the change to di-bits. This is because although the

changes on the line still occur at the Baud rate, they are

now smaller changes than before and, therefore, more

likely to be affected by noise and interference.

In practice, these techniques are used in conjunction

with frequency-shift keying in computer modems. The

amplitude, phase and frequency of the carrier are


( )( )

skbitC

C

NSWC

/27

1001log432.3

1log32.3

4kHzBandwidth

100factor20dBS/N

ExampleTelephone

=

+=

+=

=

==

( )( )

skbitC

C

NSWC

/50

035.01log100032.3

1log32.3

greater)(much1000kHzBandwidth

low)(very0.035factor14.5dB-S/N

ExampleSpectrumSpread

=

+=

+=

=

==

Figure 23:Shannons Law Capacity of a telephone channel

Figure 24:Shannons Law Capacity using spread spectrum

0 0 0 01 1

1 ms

11

00 10 1011

Figure 25: Using Di-Bits to Double the Data-Capacity of a Transmission System


18/24


changed so that there can be, for example, as many as

sixteen combinations. This allows the bit rate to be six-

teen times greater than the Baud rate when line condi-

tions are good. An ideal telephone line can transmit data

at 56 kbits per second, but few real telephone lines

achieve more that 49 kbits per second in practice.

When line conditions are poor then the bit rate must

be reduced until the number of errors is small. Modern

modems adjust their bit rate during transmission to

accommodate changes in levels of noise and interfer-

ence during the data transfer.

ENCRYPTION

Suppose an 8 bit message needs to be sent securelydown a comms channel. The message is 11010110.Consider a bit stream 01011011 as an encrypter. One

method of encryption is to use these two data streams

as inputs to an Exclusive-OR gate. The output of the

EOR gate is the encrypted message. Working from the

least significant bit with the message bit placed first, the

result will be

0 1 =1 (LSB)

1 1 =0

1 0 =1

0 1 =1

1 1 =0

0 0 =0

1 1 =0

1 0 =1(MSB)

The encrypted message is therefore 10001101 and this

is sent down the channel to a decoder which knows theencrypter data stream 01011011. If these two are input

to an EOR gate, the original message is revealed as fol-

lows:

1 1 =0 (LSB)

0 1 =1

1 0 =1

1 1 =0

0 1 =1

0 0 =0

0 1 =1

1 0 =1(MSB)

the original data 11010110 has been decrypted

CODE DIVISION MULTIPLEXING (CDM)

Code division multiplexing (CDM) allows signals froma series of independent sources to be transmittedat the same time over the same frequency band. This is

accomplished by using orthogonal codes (see later) to

spread each signal over a large, common frequency

band. At the receiver, the appropriate orthogonal code is

then used again to recover the particular signal intended

for a particular user.

The key principle of CDM is Spread Spectrum.Spread Spectrum is a means of communication with the

following features:

1. Each information-bearing signal is transmitted with a

bandwidth in excess of the minimum bandwidth nec-

essary to send the information.

2. The bandwidth is increased by using a spreading

code that is independent of the information.

3. The receiver has advance knowledge of the spread-

ing code and uses this knowledge to recover the

information from the received, spread-out signal.

Lets begin by considering a technique for spreading

the spectrum of an information signal. Suppose we have

a series of information bitse.g, 110110that we

want to transmit at a particular speede.g, rb =1000

bits/sec. We know that the bandwidth required to trans-

mit this signal will be proportional to the transmission

speed (If PSK is used and we need 90% in-band power,

the signal will require a bandwidth of 2rb =2 kHz). Now

consider the circuit shown in Figure 26, which exclusive-

ORs the information bits from the source with a second

sequence of bits known as a spreading code. The

spreading code is being clocked at a rate three times asfast as the source is outputting information. If the source

information and the spreading code are synchronized,

the sequence at the output of the exclusive-OR gate

also has a rate of 3rb . Lets use rss to symbolize the rate

of the spreading code, which we call the chipping rate.

The circuit in Figure 26 (which we can call a

spreader) converts the information sequence, which is

being output by the source at a rate of rb , into a longer,

faster sequence of bits being output at a rate of rss. This

longer, faster sequence of bits will require more band-

width to transmit (again, if PSK is used and we need

90% in-band power, the new sequence will require abandwidth of 2rss =6 kHz). The technique of exclusive-

ORing an information sequence with a faster spreading

code is known as direct sequence spread spectrum

The general spreading and despreading processes

in the frequency domain is shown in Figures 27 and 28.

Spreading increases the bandwidth by a factor of rss /rb,

and despreading reduces the bandwidth back to that of

the original signal. This factor, rss /rb, is known as the

processing gain and is symbolized Gp.

Weve now shown that spreading and despreading

are mathematically valid operations, but we have yet to

show why they are useful. In fact, as mentioned earlier,

these operations seem counterintuitive because theyincrease the bandwidth of the transmitted signal

something communication systems engineers instinc-

tively avoid. In order to see practical applications for

spreading and despreading, we must investigate the

properties of the spreading code. any sequence of 1s

and 0s can be used as a spreading code. Practical

spreading codes must look like a sequence of random,

independent, equiprobable bits. These codes are called

pseudo-random (PN)rather than truly random

because the transmitter and receiver must generate the

same sequence (otherwise despreading will not work.)

The bits in a PN code are essentially uncorrelated.Thus, when a PN spreading code is applied to additive

white Gaussian noise, the power spectral density of the

noise remains flat and unchanged. Figure 28 shows the



19/24


spreading, transmitting, and despreading processes for

a general signal corrupted in the channel by additive

white Gaussian noise. A PN spreading code is used.

Note that the spreading process flattens (as well as

spreads) the spectrum of the information signal, and

that the spread transmitted signal is essentially buried in

the noise by the time it arrives at the receiver. The

despreading process

narrows and height-ens the spectrum of

the received signal,

yet leaves the spec-

trum of the received

noise essentially

unchanged. The

receiver can now

extract the despread

signal from the noise

using a bandpass fil-

ter. One advantage of

spread spectrum isthat the transmitted

signal is buried in the

noise and is virtually

undetectable by other

receivers that do not know the spreading code.

This provides security. Also, since a spread

spectrum signal contains low energy spread

over a large bandwidth, the signal can be trans-

mitted on a channel containing other communi-

cations without causing significant interference

to the other communications (i.e., the spread

spectrum signal will simply look like a very

small amount of additional noise)

The Pseudo-Random code must satisfy a

number of properties. One particular property is

that in any block of spreading code, the number

of zeroes and ones must not differ by more than one.

The second propery puts a restriction of the sequence of

identical bits and the third property is to do with correla-

tion which is beyond the scope of this handout. If these

conditions are met, the codes are orthogonal codes


Figure 26: Speading using a chipping code

Figure 27: Spreading and De-spreading

Figure 28 Code Division Multiplexing


20/24


SELF ASSESSMENT QUESTIONS

Q1. When a byte of data is carried by a parallel bus

then:

a. No handshaking is required.

b. Each bit has its own conductor.

c. A shift register is required to queue the bits.

d. Frequency-shift Keying is used.

Q2. A communications channel that is used to carry sig-

nals in both directions simultaneously is called:

a. Full-Simplex.

b. Simplex.

c. Half-Duplex.

d. Duplex.

Q3. Which one of the following is NOT a feature of a par-

allel data bus?

a. Requires start and stop bits.b. Gives high-speed data transfer.

c. Transfers several bits simultaneously.

d. Requires handshaking signals.

Q4. When serial transmission is required:

a. A telephone link cannot be employed.

b. A modem is necessary.

c. A shift-register is required.

d. A cable with at least 50 conductors is normally

used.

Q5. The rawdata from a computer cannot be transmit-

ted directly over a long distance, e.g. 5 km, because:

a. There is no dc component in the waveform.

b. There will be too many reflections from the other

end.

c. The signal will be attenuated and distorted in

transit.

d. There are no parity bits.

Q6. Frequency Shift Keying is used because:

a. It is needed for error detection and/or correction.

b. It produces a non-return to zero (NRZ) signal

waveform.

c. Fewer conductors are required in the comms

cable.

d. It changes the frequencies of the data waveform

to lie within the bandwidth of the datalink.

Q7. The use of differential transmission on a bal-

anced line reduces problems caused by:

a. Poor bandwidth.b. Dispersion.

c. Reflections.

d. Interference.

Q8. Which one of the following binary numbers has the

highest bandwidth?

a. 10101010

b. 11001100

c. 00110011

d. 11111111

Q9. The following binary data, 8 data bits and one par-

ity bit, have been transmitted with even parity. Which

one was corrupted during transit?

a. 100100110

b. 111111111

c. 000000000

d. 010101010

Q10. Why does Manchester encoding allow long mes-

sages to be sent asynchronously?

a. Differential transmission, on balanced wires, isused.

b. Error detection is used.

c. The message carries timing signals.

d. Ones and Zeroes are encoded in different ways.

Q11. A byte is transmitted with a bit-period of 10 ms

(100 Baud) using one start bit, 8 data bits, even parity

and one stop bit. How long after the arrival of the start

bit does the receiver examine the parity bit?

a. 100 ms

b. 95 msc. 15 ms

d. 10 ms

Q12. In frequency shift keying, which of the following

statements would be incorrect?

a. A Oneis sent as a loud tone and a Zerois sent

as a quiet tone.

b. Ones and Zeroes are encoded using different

frequencies.

c. A One is sent as a high-pitched tone and a

Zerois sent as a low-pitched tone.

d. For use on a telephone circuit, the tones must beaudible.

Q13. One method that could be used to correct most

errors that arise during transmission of data is:

a. One parity bit for each byte.

b. Manchester encoding.

c. Differential transmission on balanced wires.

d. Block parity.



21/24


Q14. When communicating over a poor quality datalink,

the system is likely to:

a. increase the Baud rate.

b. reduce the bit-rate.

c. encode more bits into each pulse.

d. stop sending the parity bits.

Q15. The hex number 6B is equivalent to the decimal

number:

a. 71

b. 611

c. 107

d. 17

Q16. The decimal number 25 is equivalent to the hex

number:

a. 15

b. 19c. FA

d. A1

Q17. The binary number 10000011 is equivalent to the

decimal number:

a. 255

b. 128

c. 131

d. 111

Q18. When a byte is sent asynchronously, with parity,then the number of bits per frame is:

a. 8

b. 9

c. 10

d. 11

19. One difference between synchronous and asynchro-

nous transmission of data is that synchronous transmis-

sion has:

a. fewer errors.

b. smaller frames.c. greater throughput.

d. frames containing a single byte.

Q20. One feature of a non-return to zero system is that:

a. consecutive Onesmerge into a long pulse.

b. parity bits are not needed.

c. its average value is zero.

d. timing signals are embedded in the data.

Q21. Achannel with a bandwidth of 10 kHz and a signal-

to-noise ratio of 50 db (105) has a maximum data-carry-

ing capacity of:

a. 50 kbits per sec.

b. 133 kbits per sec.

c. 10 kbits per sec.

d. 500 kbits per sec

Q22. A Bit Error Rate Probability of 0.000001 implies

that one error will occur in the following number of bits:

a. 100

b. 1000

c. 10000

d. 1000000

Q23. A Comms channel uses Hamming Code (11,8)

This will mean that the number of data bits and parity

bits are respectively:

a. 11 and 8

b. 11 and 3

c. 8 and 3

d. 3 and 8

Q24. If the Signal to Noise ratio in a comms channel is

one, then if 500 bits are sent, the likely number of erro-

neous bits will be:

a. 500

b. none

c. oned. 250

Q25. An uncoded communications system achieves a

bit error rate (BER) of 0.000000001 at an SNR of 20 dB.

An error control coding scheme with a coding gain of 4

dB is added to the system which improves the BER. To

what level could the S-N ratio drop for the BER to fall to

the same value as before :

a. 16dB

b. 5dB

c. 10dB

d. 19dB



22/24



1. Each bit has its own wire (b)

2. Duplex channels allow bi-directional signals (d)

3. Parallel data does not require Start/Stop bits (a)

4. The shift reg. converts par-ser & v.v. (c)

5. Attenuation and distortion damage the signal (c)

6. FSK puts all harmonics in the comms bandw (d)

7. Diff trans reduces effects of interference (d)

8. 10101010 has most changes - 8 of them (a)

9. 111111111 has nine Ones - should be even (b)

10. Every pulse has a change - for timing (c)

11. Parity is 10th pulse - 90 - 100 ms after start (b)

12. FSK does not change the amplitude (a)

13. Block parity detects & corrects errors (d)14. Data rate - bit rate - reduces over poor lines (b)

15. Six 16 plus Eleven 1 =96 +11 =107 (c)

16. 25d =16d +9d =One 16 plus Nine 1 =19h (b)

17. 128 +2 +1 =131 (c)

18. Start +8 Data +Parity +Stop =11 bits (d)

19. Bigger frames (packets) in async - more through(c)

20. NRZ allows pulses to merge - lose timing (a)

21. 3.32 10 k Log ( 10 001) =133 k (appx) (b)

22. 0.000001 is 1 in a million (d)

23. (11,8) Hamming code uses 8 data +3 parity (c)

24. S-N=1 BERP =0.5. 0.5 x 500 =250 (d)

25. Subtract the coding gain from original S-NR (a)

Answers


23/24

DWR Digital Data Comms


Teaching Objective Comments

E.07.01 Use Simple Binary and Hexadecimal Notations

E.07.01.01 Define the terms: Byte (8-bits), Word (n-Bytes) and aNybble (4-bits)

E.07.01.02 Convert numbers between zero and 255 from decimal tobinary and vice-versa.

E.07.01.03 Convert numbers between zero and 255 from decimal tohexadecimal and vice-versa.

E.07.02 Describe how a Byte is transmitted using Parallel Transmission

E.07.02.01 State that each bit is encoded as, say, 0 V for Zero and5 V for One.

E.07.02.02 State that each bit has its own wire from end to end. Ribbon or multi -core cables.

E.07.02.03 Describe parallel communication as efficient only forshort distances.

E.07.02.04 State that additional lines are required to control the flowof data.

E.07.02.05 State that parallel communication offers a highthroughput, n-times faster than serial.

E.07.03 Describe how a Byte is transmitted using Serial Transmission

E.07.03.01 State that each bit must be transmitted in turn along asingle wire or pair.

Diagram with shift registers.

E.07.03.02 Describe serial communication as efficient for longdistances by cable, radio, telephone and satellite.

E.07.03.03 State that additional lines or flow-control codes arenecessary to control the flow.

E.07.03.04 Sketch a waveform showing how a Byte might beencoded using 0 V and 5 V signals.

E.07.03.05 Define the terms: Simplex, (Full -) Duplex & Half-Duplex.

E.07.04 Describe the effects of communications links on signals

E.07.04.01 Sketch a signal before and after transmission, showing theeffects of limitations of Bandwidth, Noise &Interference, Dispersion and Attenuation in transmissionlinks .

E.07.04.02 Describe differential transmission as a means of reducingthe effects of interference.


24/24

Digital Data Comms DWR

Teaching Objective Comments

E.07.05 Describe Asynchronous Transmission

E.07.05.01 State that data are sent at irregular intervals, often onebyte at a time.

E.07.05.02 Describe a 'frame' of start, data and stop bits.

E.07.05.03 State that the timebase is re-set at the start of each frame.

E.07.05.04 Describe the non-return-to-zero system of transmittingdata.

E.07.05.05 Describe the frequency-shift encoding (fsk) methods usedby modems (for example).

E.07.06 Describe Synchronous Transmission

E.07.06.01 State that data are sent in long sequences, of 128 or 256bits or more.

E.07.06.02 Describe a frame of Sync and data bits.

E.07.06.03 Describe the main features of the Manchester Encodingsystem used in Rapier comms.

E.07.07 Describe methods of error detection and correction

E.07.07.01 Describe the method of odd or even parity used inasynchronous transmission to detect single-bit errors.

E.07.07.02 Describe the method of block parity used to correctsingle-bit errors.

E.07.07.03 Describe the checksum method used to detect errors insynchronous transmission.

E.07.07.04 Identify Reed-Solomon and Hamming as error correctionmethods.

Bit Error Rate Probability and Coding gain

E.07.07 Describe the factors affecting data transmission rate

E.07.07.01 Define the term baud rate as the maximum number oftransitions per second on the transmission link.

Not usually equal to the bit -rate in modernsystems.

E.07.07.02 Define the term bit rate as the maximum number of bitstransmitted per second.

Might be as much as 20x the baud rate.

E.07.07.03 Describe the use of multiple-bit encoding to increase thebit-rate.

Simple example using di-bits and 4-level line.

E.07.07.04 Describe the use of compression to increase the bit rate Same techniques used for computer data (e.g.like 'zip' files)

bst handout e07

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