MSc in Industrial Computing Systems Computer Communications Lecture 1.

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MSc in Industrial Computing SystemsComputer
CommunicationsLecture 1.

• Lecturer Dr. David Al-Dabass
• Room N315
• Tel. 6015
• email david.al-dabass_at_doc.ntu.ac.uk

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Computer Communications
Week 1. Information channel theory (Comer21,
Halsall3241) character framing (Halsall97),
Asynchronous/Synchronous communication (Comer15,
Halsall101, 107, 121). Week 2. The Telephone
network (Halsall57, 68), Modulation techniques,
Modems (Comer25, Halsall81). Week 3.
Character Networks multiplexing techniques
(Comer31, Halsall72, 158, 477). Week 4.
Message-based systems ( Comer37, Halsall7, 162,
219, 222, 229) Error control (Comer55,
Halsall125, 130, 867) ARQ, generating
polynomial, circuits, Idle RQ (Halsall169-188)
Continuous RQ (Halsall189, 203) GBN
(Halsall195) Selective repeat
(Halsall90). Week 5. Flow control
(Halsall198) Computer networks, circuits
(Halsall458) Messages (Halsall786,) Packet
switching (Halsall429). Week 6. Radio nets
(Halsall30) CSMA (Comer60, Halsall280)
satellite nets (Halsall29, 52-54) LANs
(Comer53, Halsall271) WANs (Comer119,
Halsall423) X25 (Comer134, Halsall429).
Recommended Books
1. Computer Networks and Internets, Douglas E.
Comer, with CD-ROM, Prentice Hall, ISBN
0-13-599010-6. 2. Data Communications, Computer
Networks and Open Systems, 4th edition, by Fred
Halsall, Addison Wesley, ISBN 0-201-42293-X. 3.
Computer Networks and Internets with Internet
Applications, Douglas E. Comer, 3rd edition with
CD-ROM, Prentice Hall, 2001, ISBN
0-13-091449-5. 4. INTRODUCTION TEXT Computer
Communications, 2nd edition, Robert Cole,
Macmillan Computer Science Series.
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Section 1 Introduction

• 1.1 Scope
• Concerned with Transfer of computer-based data
  over long distances.

Not concerned (in this course) with bus
communications inside the computer.
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Cost sophistication of Computers
Communication Networks

• Ever more powerful computers at less cost enable
• Ever more powerful computers

More efficient use of communication devices, the
cost of which is rising to cope with higher
transfer rates demanded by

• In Computer-based Communication Systems
  computers are used within the network to operate
  it more efficiently. (computers are embedded
  within the network)
• In Communication-based Computer Systems a
  distributed (multi) computer system relies on
  communication networks to perform its overall
  objective.
• Typical Applications of Communication-based
  Computer Systems
• Banks and Building Societies.
• Airlines
• Etc.

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1.2 Information channel capacity and bandwidth.
Sine Wave Signal
Information in digital systems
Frequency number of complete waveforms per
second
Information-carrying waves change according to
data content. A regular square wave represents
the highest possible number of changes
Regular square wave
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A square wave can be thought of having a
fundamental frequency plus harmonics
f f fundamental 3f high order 5f harmonics
Higher order harmonics
Bandwidth is the difference between the highest
and lowest frequency that can be transmitted
through the channel.
Human ear 20Hz - 20kHz
Telephone (analogue) 300Hz - 3400Hz
bandwidth 3100 Hz.
To transmit a perfect square wave we need an
infinite bandwidth. For a limited bandwidth
channel, the signal received will be a smoothed
wave.
Smaller bandwidths will result in larger
differences between sent and received waves.
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The minimum bandwidth for a square wave to be
still recognisable as a square wave is 2 x
freq.of sq. wave.
Capacity of Communication Systems.
The Maximum rate of information transfer is
related to bandwidth w and signal-to-noise ratio
it is termed the Capacity of the System and is
given by the Shannon theorem as
C w log2(1 S/N) bits/sec
The Signal to noise ratio is usually given in
decibel units dB which is related to the
arithmetic ratio S/N byand conversely
(S/N)db 10 Log10(S/N) db
S/N 10(S/N)dB/10
e.g. 20 dBs means S/N 10 20/10 102 100
Example calculate the channel capacity when the
bandwidth 3000 Hz and the signal-to-noise ratio
is 100.
C w Log2(1S/N) 3000 x Log2(1 100) 3000 x
Log2101 3000 x 6.658 19,975 bits/sec
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Channel organisation.
Problems with Parallel Transmission.
1. Cost more lines means more expensive
communication system. 2. Skew due to tolerances
in components and wires, signals sent out at the
same time instant do not arrive at the
destination at exactly the same time. Problem
gets worse with distance and/or with increasing
the transmission rate, e.g. at several miles or
at several mega bits/sec, skew causes unreliable
results. Answer Serial Transmission, i.e. one
data bit is sent in one time unit slow, but more
reliable, and cheap for long distances
Coding data in signals
1 0
Two level code.
20 milliseconds
Max. no of bits/sec 1000/20 50 bits/sec Baud
rate number of level changes / sec (also called
signalling rate)
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Four level code. Not used much in practice, but
useful for comparison
00 01 10 11
In one time unit any one of 4 symbols can be sent
( 00, 01, 10, 11) Therefore the data
transmission rate is doubled (compared to 2-level
code). Number of level changes (Baud rate) is
still the same, but bits/sec has doubled.
Modes of operation of a transmission channel.
1. Simplex data transfer in one direction only
one sends, the other receives. 2. Half Duplex 2
directional data transfer, but not at the same
time, i.e. one direction at a time direction is
reversed by switching between sending and
receiving equipment (over). 3. Full Duplex 2
directional transfer at all time.
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Questions
1. The channel capacity given by the Shannon
theorem include the variables C, w, S and N.
State the theorem and define the variables. 2.
The bandwidth of a given channel is 2 kHz and S/N
is known to be 50. Calculate the capacity of the
channel. 3. What would be the capacity if the
signal to noise ratio was doubled? 4. Re-arrange
the above equation to give the signal to noise
ratio explicitly in terms of the capacity and
bandwidth. 5. By how much should the S/N ratio
be improved in order to double the capacity? (S/N
and w have the values as those in 3. above). 6.
State the decibel equivalent of S/N. 7. Assuming
that S/N gtgt 1, derive an expression for C, w, and
S/N dB which does not involve logarithmic
function.
Answers
Note log2(x) log10(x)/log10(2) 3.322
log10(x).
2. 11.345 b/s 3. For S/N 100 C w log2(101)
13317 4. C/w log2(1 S/N) 2C/W 1 S/N
S/N 2C/W - 1 5. For C 2 13317 26634 S/N
226.6/2 - 1 10086 6. S/N dB 10
log10S/N 7. S/N dB / 10 log10S/N S/N 10
S/NdB/10 C/w log2(S/N) log2 10 S/NdB/10
S/N dB / 10 log2 10 0.332S/NdB
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Bit Coding
International standard called V24 uses a coding
method called Voltage Amplitude System For
full duplex mode of communications it uses 3
wires
Signal A to B ( to - V)
A
B
Common 0 V
Signal A to B ( to - V)
0 bit is signalled by 12 V (mark) 1 bit by
-12 v (space) Example 110100
12 v 0 v -12 v
1 1 0 1 0
0
0 v on the line means break condition, I.e.
power is off or break key pressed.
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Character Framing
This refers to methods of indicating the start
and finish of one or more complete characters.
Two methods Asynchronous
Synchronous.
Speed and Applications of
Synchronous - Buffered Devices - Intelligent
Devices - Remote Job Entry 2400 - 19200
bps - Computer to Computer communicactions gt
9600
Asynchronous - Mechanical Terminals Slow
Printers 75 - 300 bps - Unbuffered
devices Terminals 300 - 9600 bps
Transmission line speeds (bits per second) are
chosen from a list of preferred speeds
(internationally agreed) 75, 110, 300, 600,
1200, 2400, 4800, 9600, 19200, 28800, 38400,
56600..
Asynchronous character framing is used when
communication is intermittent (irregular time
periods between chars)
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Time
12 v 0 v -12 v
8 Data bits
2 Start bits Warn receiver that a new character
is coming start the local clock
1 Stop bit
a) mechanical terminals/slow printers use 2
start bits to allow mechanism to settle down
before new char (110 bps) b) Stop bit uses
opposite polarity to start bit. Parameters
included in the interface design 1. Number of
bits in char (could be 7 or 8). 2. Bit unit
time, e.g. 300 bps gives a bit unit time 3.3
milliseconds. 3. Number of stop bits default
110 bps (or less) uses 2 stop bits, all other
(higher) rates use 1 stop bit.
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Synchronous character framing is used for large
volume of data transfer such as whole files
between computers where higher speeds are needed.
A number of chars (block of bits) are sent in one
continuous stream.
Special SYN (synchronisation) character is used,
ASCII code 00010110 to identify start of stream.
Receiver
i) Stream of bits is received and shifted until
the SYN bit pattern is recognised. This is taken
as the character defining the boundary ot the
stream being sent, I.e. the next 8 bits will be a
whole character. ii) A local clock is used for
capturing the bit stream and is synchronised to
the sender clock. iii) Receiver clock is
re-synchronised frequently by the SYN character
between blocks
re-synchronisation gaps
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Send and Receive Equipment
- Parallel-to-serial conversion out of the
computer and then Serial-to-parallel conversion
at the other end. - LSI circuits do one of 3
types of tasks Asynchronous framing only
Synchronous framing only Both - UART
Universal Asynchronous Receiver and
Transmitter - USART Universal
Synchronous/Asynchronous Receiver and
Transmitter - UART performs Serial to
Parallel conversion Frame generation and
checking Programmed for 5, 6, 7, 8 bit
characters Parity even, odd, none. Line
bit rate using an external clock.
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UART Block Diagram
1. While one character is being transmitted from
the transmission register, the next character can
be loaded into the buffer register, i.e.
continuous transmission is possible. 2. Once
data is in transmission register, a start bit is
sent followed by the character bits, shifted out
one bit at a time. A stop bit is sent after the
last character bit. 3. A new character is then
loaded from the transmission buffer.
Stop bit
Input sampled when counter reaches zero. UART
clock rate 16 x bit rate, e.g. for 300 bps a
4800 Hz clock is used. Start bit V edge starts a
counter of 8 decrementing (half bit time), then
the levels checked when counter 0. If V level,
then it must be start bit, then set counter 16
and retake level every time counter 0 ( and
reset counter to 16).
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Asynchronous/Synchronous communication comparison
Efficiency Asynchronous Each character (8
bits) needs 2 overhead bits (start/stop) max
efficiency 8/10 80. Synchronous Higher
efficiency, e.g. for 96 character block needs
only 4 extra characters (SYNC, block length
etc.) efficiency 96.
Synchronisation Method Asynchronous Receiving
clock is re-synchronised on every character, i.e.
clocks are in synch during character
transfer only and only for that character.
Synchronous Receiving clock and character
frame are re-synchronised for every block of
characters, sender and receiver clocks are
always in synch.
Communication Type Asynchronous Few
characters at intermittent intervals with
arbitrary long periods of idle time, e.g.
keyboard to CPU. Synchronous To transfer large
amounts of data continuously, e.g. whole files in
go..
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Questions
1. Explain how the full duplex mode of
communication operates within V24 standard.
Illustrate your answer by representing the
sequence 11010110 and discuss the function of the
BREAK key. 2. There are two types of character
framing techniques. What are they? Discuss their
applications and transmission line speeds
used. 3. Draw a diagram of a typical
asynchronous character transmission sequence and
show the polarity of the start and stop bits, the
number of bits and typical bit-unit time. 4.
What is the main reason for using synchronous
character framing? How is the synchronisation
character used to recognise the start of a
block? 5. What are UART and USART? Use diagram
to explain the operation of a device associated
with these terms and comment on the ratio between
clock rate and baud rate. 6. Compare and
contrast the asynchronous and synchronous modes
of data transfer in terms of efficiency,
synchronisation method and type of communication.