Chapter Contents 04103 Multimedia Information Networking

boilermakerwrapperΗλεκτρονική - Συσκευές

8 Νοε 2013 (πριν από 4 χρόνια και 2 μέρες)

79 εμφανίσεις

Chapter Contents

04103

Multimedia Information Networking


Chapter 1: An overview of multimedia information networking


Chapter 2: Multimedia information representation


Chapter 3: Data communication principles


Chapter 4: Data communication protocols


Chapter 5: Networking fundamentals


Chapter 6: Multimedia applications






























2

User
-
Machine
inter
face

Machine
-
information
interface

Information Network ( Infonet)

Chapter (1)

An Overview of Multimedia Information Networking


1.
Describe the

concept of information networking and information universe.

(5 marks)



Information networking is the process of connecting a large number of
information sources by a computer network.



The sum total of information that exists within a specified boundary is called its
Information universe.


2.
With the current tec
hnology, how would and information network is created. (10
marks)



With t
he current technology, an Information Network (Infonet) can be created by
connecting a computer
-
based workstation to the Information Universe.


The user workstation can be
a personal computer, a terminal connected to a mainframe
computer, a laptop computer, a network computer, or a television connected to a service
provider via a set
-
top
-
box (STB). The connection between the user workstation and the
Information Universe can
be any one of the various options.


It could be
a direct connection between the personal computer and a Local Area
Network (LAN). From the home, the user may use the phone line to establish a connection
with a server at the workplace, or to an Internet Ser
vice Provider (ISP). Mobile computing
involves connecting a hand
-
held computer
-
based sys
tem to a service provider via a mobile
phone.














Fig: A conceptual
view of Information Networking and user interface to the Information
Universe







User

Workstation

Information
Universe


3

3.

(a)
Define an infonet and information server.

(
b) The internet can be classified as an infonet. However, other forms

of networking
are needed to cater to environments and applications. Why?

(20 marks)



(a) Infonet




Infonet is a collect
ion of a large number of information servers and user
workstations connected by a high
-
transmission capacity computer network.



Information server




An information server is a computer system used to store information.
This information may be stored by u
sing one or more information storage techniques. The
most widely used technique for information storage is a text
-
based database. In a Multimedia
Infonet, the information servers must be capable of storing multimedia information.





(b) The

internet can
be classified as an infonet. It connects thousands of servers to
million of users. The biggest advantage of the Internet is that no single organization owns it.
It is an advantage because anyone with the required hardware, software and access
mechanism can

connect to the Internet. This is one of its biggest disadvantages as well.
Because no single organization owns it, there is no guarantee of performance and security.
Thus, other forms of Information Networking to cater to environments and applications tha
t
require better performance and security than that provided by the Internet.


4.

In order to provide better performance and security. Infonets are classified into
three types.


-
Universal Infonets


-
Corporate Infonets


-

Application specific Info
nets



(20 marks)

(a)
Discuss each of these types by giving an example

a
pplication scenario
.

(b)

Make a comparison of the above three types of infonets by considering
important parameter such as accessibility, number of users, number of
servers, network sp
an, ownership, performance guarantee and security.


Infonets are classified into three types.

-

Universal infonets

-

Corporate infonets ad

-

Application specific infonets.


Universal infonets


Universal infonet is an infonet that is available universally.
Almost anyone on
paying a fee if required can gain access to a universal infonet. Users connected to a
universal infonet can make their own information available to other users. The internet
is the most widely used universal infonet.


4


Corporate infonet


A
corporate

infonet is a collection of LANs, MANs and and WANs

linking all
sites and services of a specific corporation. This corporate infonet is likely to have
access to a universal infonet through a gateway.


For example, all campuses of Victoria universi
ty in Melbourne and its suburbs
are connected by a corporate infonet. From their desktop computer or workstations,
university staff can access information about the university library holdings and
student records etc.


Application specific infonet



An app
lication specific infonet is a collection of LANs,
MANs, and WANs
linking all sites and services of a specific application domain. An application specific infonet
is much like a corporate infonet but the common factor is not a cor
poration.



For example, t
hey exist a network of international libraries
. The
r
e

could be a
doctor’s infonet, access to which is available only to practicing doctors.


A comparison of the three types of infonets


Important
Parameters

Universal Infonet

Corporate
Infonet

Application s
pecific
infonet

Accessibility

Universal

Highly restricted

Restricted

No. of users

> 10,000s

100s
-
1,000s

100s
-
1,000s

No. of servers

> 1,000s

10s
-
1,000s

10s
-
1,000s

Network span

Global

Local/Global

Local/Global

Ownership

Not well defined

Well defined

Lo
osely defined

Performance

Not guaranteed

Guaranteed

Ensured

Security

Low to very low

High

Low to high



5.
Define virtual reality.






(10 marks)


The aim of a

virtual reality system is to create a very close interaction between the
user’s senses an
d the computer system. The user will usually have a head
-

mounted display
system and various types of clothing with electronic interfaces to monitor
the user’s
responses. User commands are input through the “electronic clothing “ and the computer
generates

moving pictures and sound through the head
-
mounted display system can be used
for training as well as entertainment.










5

6. To facilitate a systematic study, multimedia systems can be divided into two areas:
Multimedia Hardware and Multimedia So
ftware. Provide a classification structure of
multimedia systems and draw a schematic diagram showing hardware components of
complete multimedia system.

(20 marks)


CPU Display
Keyboard
Mouse

Real
-
purpose H/W

Multimedia system

Multimedia Hardware

Multimedia Software

Special
-
purpose H/W

Software tools

Multimedia content

CD
-
ROM

Video
card camera,
monito
r Audio
card, speaker,
microphone

C
ontent creation tools Content
p
resentation tools

C
ontent structure

Presentation

D
ocuments

Objects

C
ontent type

T
ext


Still images
Moving
images video
animation
Audio
connected by
Interactive
links


6



7. Compare and contrast between c
lient
-
server computing and distributed computing.








(10 marks)


Client
-
server computing


In the new corporate computing environment, users work on their personal computers
or workstations and communicate with departmental minicomputers distributed ove
r the
various sites of the corporation. Appropriate data networks


LANs, MANs or WA
Ns are
used for establishing the connections. The user workstations are called the clients and the
minicomputers are the servers. A number of client workstations connected
to a number of
servers via a data network is called client
-
server computing.


Distributed computing


Distributed computing refers to a computing paradigm in which the various computing
tasks being carried out are distributed to multiple, geographically dis
persed processors. Thus,
distributed computing is a departure from single
-
site mainframe computing. Just like client
-
server computing, distri
buted computing relies on data networks for communication between
the distributed processors.


One of the main moti
vations in moving from single
-
site mainframe computing to
distributed computing is to avoid the possibility of single point failures. This implies that
there should be no single entity in the system, a failure of which would lead to the entire
system.


***
*******************************


CD
-
ROM
CPU Disk




Keyboard

Mou
se

Scanner

Video
monitor

Video
card


Audio card

Bus (internal)

Camera

Speakers

Microphone


7

Chapter (2)

Multimedia Information Representation


8. Text as well as multimedia information is stored on computer disks as files. Thus, the
file formats also play an important role in the storage and transmission of such
in
formation. Name two commonly used tagged file formats. Why are the tagged file
formats more advantageous than the sequential file formats for

storing mul
timedia
information? (10 marks)


Tagged Image File Format (TIFF)


The traditional file structure uses
a sequential storage method. The file beings with a
fixed length header, which contains information about the format of the data, followed by the
data. This sequential storage method works well for small file. When editing a sequential
file, the data as we
ll as the header’s contents are first loaded from the disk to the main
memory, modified, and then stored back on the disk. However, a multimedia file that
contains text, sound, still images, an
d

moving images is substantially larger than a simple
text file
. Also, creating additional space

for modified multimedia objects become difficult
with a single header and fixed
-
length fields. The limitation is overcome in the Tagged Image
File Format.


The most important property of the tagged file structure is that t
he tags are located
with the help of pointers.



9. Audio CDs use a sampling rate of 44.1 kHz. 16 bit resolution and two channels to
store the stereo sound.


(a) Calculate the required digital bandwidth.


(b) What is the highest frequency of the anal
og signal that can be r
educe from the
digitized signal?


(c) Calculate the SNR of quantization and comment upon the quality of sound
reproduction that can be obtaine
d from such a system. (20 marks)


Solution


Sampling rate = 44.1 kHz



Resolution =
16 bit


Number of channels = 2

(a) Calculate required digital bandwidth


Digital bandwidth = bit/sec = bps



= sample/sec x bit/sample x
channel


= 44.1 x 10
3

x 16 x 2

= 1.4

Mbps

(b) Highest frequency of analog signal


Sampling Rate = 2 x frequency


44.1 x 10
3

= 2 x f


f = 22.05 kHz


8

17

=
s
=
䠠e‴⼳
=
14”
=
s
=
䠠e‴⼳
=

(c) SNR of quantization


SNR = 20 log 2
b

= 20 log 2
16

= 96 dB


SNR value of 96 dB, which
is very close to the best possible SNR value 100. The quality
of sound is good.




10.
An super VGA

(SVGA) display has 1024 dots per scan line and 768 horizontal scan
lines in the system. Compare the image quality on a 14” screen with that on 17” s
creen
for SVGA frame formed. Take a dot pitch of 0.28 mm to 0.30 m
m in both screen
.




(20 marks)







14
2

=

(4/3 x)
2

+ x
2







196 = (4/3 x)
2

+ x
2








x =


8.4”

Vertical







Horizontal = 4/3 x =11.2”





Horizontal dot pitch =

11
.2”/1024 * 25.4 =0.28 mm





Vertical dot pitch = 8.4”/768 * 25.4 =0.28 mm





The image
will loose accuracy.













17
2

=

(4/3 x)
2

+ x
2







289 = (4/3 x)
2

+ x
2








x =


10.2”

Vertical







Horizontal = 4/3 x =13
.6”





Horizontal dot pitch =

13.6”/1024 * 25.4 =0.34 mm





Vertical dot pitch = 10.2”/768 * 25.4 =0.34 mm





The image quality is good.







11. Determine the data rate required for super VGA (1024.768) resolution color screen
at a re
fresh rate of 25Hz.




(10 marks)


Data rate = ?

Resolution = 1024 x 768

1024 dots per scan line



Horizontal

768 scan lines


Vertical


9

Refresh rate = 25 Hz = 25 frames/sec

Time required for one frame =
1/25 sec

Time required for each line =
1/25/768= 52.0833 µsec

Time required for each pixel =

52.0833/1024 = 0.05086µsec

Required sa
mpling rate =

1/0.05086µ


=19.66 MHz


12. (a) Two electronic systems are connected in series. The amplification of the first
system is 20dB, and that of the second system is
-
10dB. What is the overall
amplification of t
he two systems? Give the answer as a dB value and as a ratio.








(10 marks)

(b) A computer monitor has a resolution of 640
*480 and can show 256 color.
Calculate the size of each frame buffer required for this monitor.









(10 marks)

12.(a
)






overall amplification = 20 dB+ (
-
10d B ) = 10 dB

amplification a (f) = So( f
)





Si ( f )


amplification a (f ) in dB = 20 log So( f )





Si ( f )




10

= 20 log So( f )





Si ( f
)


antilog y/ 2

=
So( f )







Si ( f )






So( f ) = 3.16





Si ( f )


12. (b)


resolution = 640 * 480

color level = 256 color = 2
8

no. of bit per pixel = 8 bit

size of frame buffer = 640 *

480 * 8





= 2457.6 K bit


= 307.2 K byte.

****************************

20d B


-
10d B

1 st system

2 nd system


10

Chapter (3)

Data Communication Principles



13. Explain with the help of diagram
s and examples
,

the three modes of
communication: simplex, ha
lf duplex and full duplex. (20 marks)


Simplex Transmission


In the connection shown in figure, data transmission can take place only in one
direction; left to right. If transmission can take place in one direction only, it is called
simplex transmissio
n. Examples of simplex transmission are radio transmission, television
transmission and a public address system.


Half Duplex Transmission




A transmitter and a receiver at both end
s make

it possible to have transmission in both
directions
.

T
r
an
smission

can take place in only one direction at a time. Bidirectional
transmission, in which information can flow in only one directional at a time, is called half
duplex transmission. Example of a half duplex communication system is a walkie
-
talkie.



Fu
ll Duplex Transmission


If two signal lines and two return lines are provided, data can be transmitted in both
directions simultaneously. If data can be transmitted in both directions simultaneously, it is
called full duplex transmission in 4
-
wires full du
plex transmission. It is not necessary to have
four wires for full duplex transmission. Full duplex transmission over two wires can be
achieved by using modulation techniques.



11










Receiver

R
x

Signal

Line

Return
Line

Transmitter

(a)

T
x

R
x

T
x

T
x

R
x

Signal Line

Return
Line

(b)


T
x

Signal Line

Retur
n
Line

Signal Line

Return
Line

R
x


R
x

T
x

Fig: (a) Simplex transmission; (b) Half duplex transmission;


(c) Full duplex transmission.

(c)


12

14. Explain, with the help of diagrams, th
e working of the various

Rz

encoding
schemes. (20marks)























Unipolar Rz


In the unipolar Rz code, the signal is logical AND of the NRZ, L signal and clock
signal. The resultant signal will be ze
ro for a 0 data bit and will give a clock pulse for a 1
data bit. This code will not be self
-
clocking.


Manchester Encoding


In the Manchester encoding scheme, a 1 data bit is sent as uninverted clock pulse and
0 data bit is sent as inverted clock pulse. T
his makes
self
-
clocking because there is always a
1 to 0 or 0 to 1 transition in the middle of the bit frame. Therefore clock information can
always be derived from signal itself.


Differential Manchester Encoding


Differential Manchester encoding combines

two concepts: Manchester encoding and
invert on space.


If the next bit to be encoded is a 0, then the signal transition is inserted at the
beginning of the bit cell. Also the signal is made to toggle in the middle of the bit cell. This
makes the code sel
f
-
clocking.



Clock

Unipolar Rz

Differential
Manchester
encoded
signal

Raw Digital

Signal

Manchester
encoded
signal


13

15. Explain how bit synchronization, character synchronization and frame
synchronization are achieved in


(a) character
-
oriented synchronous transmission and

(b) bit
-
or
iented synchronous transmission
. (20 marks)



Character
-
oriented

schemes are used transmitting files that contain printable
characters. Synchronization is achieved by sending synchronous idle (SYN) character on the
line before sending data.


For bit synchronization


The electronic hardware circuit such as DPLL locks in
to the SYN character bit stream.
There are many 0 to 1 and 1 to 0 transitions in this stream. This allows the DPLL to lock
onto the signal and synchronize its internal clock.


Character synchronization


Character synchronization is achieved by the receiver
. By looking for the bit pattern
00010110 in the incoming stream. As soon as the receiver detects that eight consecutive bits
form this pattern, it assumes it to be a SYN character. It looks for another SYN character to
ascertain that the4 first match. Onl
y on receiving two consecutive

SYN character does the
receiver

confirm the character boundary.


SYN SYN SYN STX DATA COMMUNICATIONS ETX


Frame synchronization


To achieve frame synchronization, the receiver must find the start of frame cha
racter
STX. On receiving STX character, it achieves frame synchronization. It reads the sequent
characters as data until it encounters the ETX character. This scheme works fine as long as
binary, this can lead to misinterpretation of information, we did wi
th asynchronous
transmission by using DLE character.


SYN SYN DLE STX ……… DLE DLE
…….. ETX


Bit
-
oriented scheme


In the bit oriented scheme, a special flag byte 0111 1110 is used to indicate the
beginning and end of a frame. The transmitte
r starts transmission by sending idle byte
1111111


The general pattern of a data frame


11111111 11111111 01111110 …. data .. 01111110 11111111


Idle idle flag flag idle


We need to guard against a sequence of six 1s occurring in the data stream. This is
achieved by a process called bit stuffing.






14

Bit stuffing


In the bit stuffing process, the data stream is examined to

detect if five 1s occur in a
sequence. As soon as

five consecutive 1s are detected, 0 is inserted after them. In the
receiving end, as soon as five 1s are detected, the following 0 is removed from the data
stream when the receiver get six consecutive 1s followed by 0, it knows this is the ending
flag.




16. Explain Hardware Handshake and software Handshake with the aid of schematic
diagram and by giving an example application scenario implemented in real computer
communication system.



(20 marks)

Hardware handshake


In a hardware hand
shake, a separate signal line is required for each of the handshake
primitives. This technique is used mainly for parallel data transfer. For example, in the RS
-
232 standard, a hardware handshake

is use
d over the connection between the computer and
the modern. Signals such as request to send, clear to send, data set ready, and data terminal
ready, which are used in the RS
-
232 standard, are examples of hardware handshake signals.


Software handshake


In a

software handshake, a separate control code is used for each of the handshake
primitives. ASCII control codes are used for a software handshake. Software handshake is
used for the section of the link that runs over a long distance.


A point
-
to
-
point seria
l connection between two computers is shown in fig. The two
compu
ters are connected to two modem
s, and th
e two modem
s are connected via a telephone
exchange system. The connections betwee
n a computer and the local modem use a hardware
handshake. Software h
andshake is used over the serial link connection the two modems. The
end
-
to
-
end connection between the two computers is also a serial connection. A software
handshake must be used for actions such as file transfer between the two computers.





15



17. Describe the signals that used between
exchange and the phone equipment

exchange
system.


(20 marks)



The signals that use between the exchange and the phone equipment are


1.

O
ff
-
hook



This signal is sent by the telephone to the local exchange. As the handset is lifted
off the hook, the hook
-
switch closes, causing a small current to flow through the local loop.
This current is detected by the local exchange and
interpreted as a signal to indicate that the
user wishes to place a call.


2.

Dial tone


After receiving an off
-
hook signal from at telephone, if the exchange is able to
provide the service, it sends

back a special tone called the dial tone.


3.
Phone number





On getting a dial tone, the caller must provide the

exchange with the number of the other phone. There are two types of dialing schemes: pulse

dialing and tone dialing.

Telephone
Exchange
System

Telephone Line

Remote modem

Local modem

Local PC

Software
handshake

Software
handshake

RS
-
232
-

D cable

Fig: Connecting two computers
using

modems

Remote PC

RS
-
232
-

D cable

Hardware handshake

Hardware hands
hake


16



Pulse dialing uses a rotary mechan
ism, which generates
current pulses by
opening and closing the hook
-
switch. For dialing a number n, which lies between 1and 9, n
current pulses are sent from the phone to the local exchange.


Tone dialing uses a set of standard tones or f
requencies to pass the phone number
to the exchange. This method is called touch
-
tone dialing or dual tone mutifrequency
(DTMF) dialing. The number keypad on a

DTMF phone machine is organized as a matrix
with three rows and four columns.


4.
Ringing signa
ls







If the exchange is able
to establish a connection between the calling phone and
the called phone, it sends a special voltage, called the ringing voltage, to the called phone.


5.
Busy ton
e

If the called phone cannot be reached or is

taking another

call, the exchange sends
backs busy tone to the caller.


6.

On
-
h
ook


To

terminate a call, the caller places the handset back on the hook or the cradle.


7
. Howler tone


If the phone is left off
-
hook for so
me time without placing a call, the exchange
sends a howler tone back to the phone machine.



8.
Call waiting tone


This tone is sent by the exchange to the user to indicate that another call is
waiting on the line.


9
. Flash signal



The flash signal is a combination of on
-
hook and off
-
hook
signals, often used to
take a waiting call.









******************








17

Chapter (4)

Data Co
mmunications Protocols


18. What is meant by a communication protocol and why d
o we need it?



(5 marks)

Solution


A communications protocol consi
sts of a set of rules tat specify the various aspects of
data interchange/. For example, the stop and wait automatic repeat request protocol states:
Send one packet wait for acknowledgement if the packet is successfully acknowledged send
next packet.


The
aim of any data communications system is to transfer data from the transmitter to
the receiver. In achieving successful data transfer data from the transmitter to the receiver. In
achieving successful data transfer, the two ends need to set matching parame
ters coordinate
the tasks involved in this transfer, and ove
rcome the effect of error on an
coordinate
communications media.


19. What is the main idea on which forward error correction techniques are based?
(
10 marks
)


Solution


Forward error correction uses error
-
correcting codes, which allows data to be
recovered even from some faulty frames. This required that the sequence of bits actually
transmitted have enough redundancy to extract the correct data from the bits

received even
when some of the bits have been corrupted.


A simple way to do this would be to use triple redundancy; i.e., to transmit each data
byte three times. In case of an error in one of the three bytes, if the other two bytes match
their value woul
d be assumed to be correct. Overhead for such a scheme would be very high
Hamming code
s are a more robust

and more

economical means of forward error correction.



20. Explain the main idea behind reverse error correction techniques, as three step
process.

(10 marks)



Solution


The basic principle of reverse error correction works as follows:

Step 1: On the transmitter, a well
-
defined function is applied to the data bits to calculate a
of check bits. These check

bits are then attached to the data bits.

Step 2: Data bits and check bits are transmitted in a single frame.

Step 3: On the receiver, the data bits are used once again to calculate the check bits using the
same function as was used on the transmitter. T
he calculated check bits are compared with
the set of check bits received on the communication line. If the two sets of c
heck bits match,
is
that no error occurred.


18

21. Briefly explain the principle concepts of error control strategy: automatic repeat
requ
est method. (10 marks)


Solution


In the automatic repeat request (ARQ) methods, a control message is returned to
request retransmission of the corrupted data. ACK is sent to acknowledge correct reception
of a data frame. NAK is used

to (automatically) request retransmission of the corrupted data
frame. There are two ARQ techniques: idle RQ (also called send
-
and
-
wait or stop
-
or
-
wait)
and continuous RQ. The idle RQ technique is suitable only for low data transfer rates. When
high data
transfer rates are required, and especially with high round
-
rip delay values, a
continuous RQ protocol is used to ensure that the bandwidth of the communication channel
is fully utilized.


22. Describe the main advantages and disadvantages of Idle RQ prot
ocol. Draw a
series of graph to show how the efficiency of the
Idle RQ protocol varies with
transmission rate and distance between nodes. Why this protocol unsuitable for satellite
based link.

(20 marks)

Solution


In the idle protocol, the primary (P) send
s one frame of information and waits for is
acknowledgment form the secondary (S) before sending the next frame. The main advantage
of the idle RQ protocol is that is requires minimum buffering. Therefore, it is suitable for
dumb terminals connected over s
hort distance. The main disadvantage of this protocol is that
it has rather low efficiency for high data rates and long distances. This is clearly illustrated
by the graph shown in figure.


On a high
-
speed link with a long propagation delay, many packets
could has been
transmitted in the time it takes for one packet to reach the destination and its
acknowledgment to reach back to the source. Therefore, with the idle RQ protocol the link
lies idle for most of the time, leading to low utilization.





100%












1


10


100



1,000

10,000 100,000



Fig: Efficiency of the idle RQ protocol vs distance

33%

66
%

Utilizati
on at 1 Mbps

Utilization at 1 Kbps

0%

Distance in Km

Effic
i
ency


19

23. What are the main ideas on which the selection retransmission technique and Go
-
Back
-
N technique are based? Compare the adva
ntages and disadvantages of each
technique. (20 marks)


Solution


In the selective retransmission technique, the main ide
a is retransmit only corrupted
INF
-
frames. To be able to retransmit a packet, it must be available in the retransmission list.

Theref
ore, a packet is removed from the retransmission list only after it has been
acknowledged. The selective retransmission protocol is used when the sequence of the INF
-
frames is not important, as the retransmitted frames will arrive out of sequence. With
sel
ective retransmission, the buffer requirement can become excessive, especially in the case
of a noisy line. Also, the fixed size buffer is not sufficient in all circumstances.


In the Go
-
Back
-
N transmission technique, when an error is detected in a partic
ular
frame, retransmission starts form that frame. This method is used to overcome the above
-
mentioned two problems of the selective retransmission protocol. The Go
-
Back
-
N protocol is
less efficient than the selective retransmission protocol, but it reduce
s the amount of buffer
required, and frames are always receive in the correct sequence.


24. Briefly explain the main factors that influence for design in a communication
protocol.





(20 marks)


Solution


The main factors that guide the development of a
communication protocol are
described as follows.

Application Requirements: Some of the main factors used in developing a new
communications protocol are the requirement of the application, such as bandwidth, delay,
and the ability to handle out
-
of
-
order p
ackets. For example, file transfer protocol can access
reasonable delays and out
-
of
-
order packets. Transaction processing applications require low
delays but can accept out
-
of
-
order packets. Real
-
time networked multimedia applications
such as video confere
ncing require high bandwidth, very low delay, and packets in the
correct order. Thus protocol developed for file transfer may not work for transaction
processing and protocol that work for transaction processing may not work for video
conferencing.



Commu
nication environment: The environment in which communication take place
includes factor such as the physical media and the error rate encounter on this media. For
example, the properties of copper media and the electrica
l signals carry on them are quite
d
ifferent from the properties of optical signals.


The error rates encounter on optical media are much smaller than the error rates
encounter on a satellite channel. Thus, the protocol for an optical fiber
-
based network need
not have as robust error contro
l as a satellite channel
-
based communication system must.



20

Network Architecture: The network architecture includes factors such as the link being
point
-
to
-
point or multipoint. In a point
-
to
-
point, the entire bandwidth, or a pre
-
allocated
proportion of the

same, is available for a given communication session. In a multipoint link,
the various nodes connected to the media content for the bandwidth of the channel. Thus, a
protocol designed for a bus topology network must allow for the possibility of the signa
l
colliding with one another.





Fig:
C
ontinuous RQ, selective

repeat














P

1(N)

1(N+1)

1(N+2)

1(N+3)

1(N+4)

1(N+
1
)

Lost

ACK (N)

ACK (N+1)

not sent

N

N+1

N

N+2

N+1


N

N+3

N+2

N+1

N+4

N+3

N+2

N+1

N+1

N+4

N+3

N+1

ACK (N+2)

ACK (N+3)

S


21



Fig: Continuous RQ ,Go
-
back
-
N for corrupted INF frame




25. Describe the frame structure and control characters
of the BSC protocol.







(20 marks)

BSC is a half duplex, stop and wait ARQ protocol designed by IBM. It can be
used in point
-
to point communications as well as in multipoint communication. Point
-
t
-
point and multipoint configuration are shown in figure.
There are two types of BSC
frames: data frames and control frame. The structure of there frames are shown in
figure. The components of the BSC frame are given in Table 1. Table 2 gives some of
the control characters used in BSC protocol.



Figure (a) Point
-
to
-
point configuration


(b) Multipoint configuration

N

N+1


N

N+2

N+1


N

N+3

N+2

N+1

N+4

N+3

N+2

N+1

N+4

N+3

N+2

N+1

N+4

N+3

N+2

N+1

1(N)

1(N+1)

1(N+2)

1(N+3)

1(N+4)

1(N+1)

1(N+2)

S

Accept

Corrupted

Discard

Discard

Discard

Accept

Accept

ACK (N)

NAK

(N)

NAK

(N)

Host

Host

Terminal

Terminal

Terminal

Control station

Tributary stations

Terminal

(a)

(b)


22


Control Frame


Data Frame




Figure
:

Control and data frames in BSC



Table 1 Components of the BSC fram
e


PAD


SYN


SOH

STX

ETX

BCC

A special pattern of ones and zeros used to allow time for the
transmission line to stabilize.

Another special bit pattern. Two SYN characters are used by the
receiver to synchronize its internal clock to that of the transmitte
r.

Start of
H
eader

Start of

T
ext

End of
T
ext

Block
C
heck
C
haracter



Table 2 Some of control

ch
aracters used in BSC protocol



ACK 0

NAK

WACK

EOT

ENO

Device ready to receive data

Device not ready to receive data

Device ready to receive data, but temporarily busy

End of transmission

Enquiry used in multipoint communication for polling and selecting.
In a polling operation, the control station asks the tributary stations if
they have data to send. In the selecting operation, the control station
allows the tributary station to send data one at a time






******************






PAD

SYN

SYN

control


information

PAD

PAD

SYN

SYN

SOH

Header


STX

Data

ETX

BCC

BCC

PAD


23

Chapter (5)

Networking
Fundamentals


26. List the three types of communication media used in data networks.







(10 marks)

27. List the three types of
optical fibers commonly used in

data communication system.

Compare and contrast their modes of operation and properties.





(
20 marks)

28. Compare the properties of three basic network topologies: mesh, star and bus.







(20 marks)

29. Explain the various communication casting modes with the help of some real life
examples.





(10 marks)

30. Explain the following statement wi
th the help of examples: Multiplexing is another
term for sharing.


(10 marks)



31. Explain the operation of FDM, TDM, STDM with the help of diagram.








(20 marks)

Frequency Division Multiplexing


Frequency Division M
ultipl
exing (FDM) divides the bandwidth of the trunk channel
into many smaller bandwidth channels and carries these simultaneously on the trunk line. A
graphical representation of this process is shown in figure. The complete bandwidth of the
channel (
BHz) is di
vided into many channels of smaller bandwidth, namely b
1
, b
2
, b
3
, b
4
, etc.
Frequency gaps, called guard bands, are left between adjoining frequency channels to avoid
cross
-
talk. Cross
-
talk occurs when the signals on one channel infiltrate another channel.


Figure (a) gives a graphical representation of the operation FDM. It depicts the total
bandwidth (BHz) as a conduit between the trunk ports of the two multiplexers. This conduit
is divided into the required number of channels, such that one channel can be

allocated per
branch port. Some frequency separation (i.e. a guard band) is kept between these channels to
avoid cross
-
talk. Figure (B
) shows a frequency response plot of the channels between the
two multiplexes.




BHz = Bandwidth of the trunk port









(a)


Bandwidth of a
single channel

Mux

Guard band

Mux

b
2
Hz

b
4
Hz

Guard

band


24













(b)


Figure: Frequency division multiplexing



Time Division Multiplexing


Time Division Multiplexing (TDM) divides time into distinct time slot and shares
these time
-
slots between the branch ports. A graphical r
epresentation of this process is
shown in figure. Each time
-
slot is allocated to one branch port. Consecutive time
-
slots are
cycled from branch port. Consecutive time
-
slots are cycled from branch port b
1

to b
2

to b
3

to
b
4

etc.


Two techniques, called bit i
nterleaving and character interleaving, are used in TDM.
In the bit interleaving technique, the data from each branch port are divided down to a
sequence of bits. Then each time
-
slot is allocated to one bit from one branch port, as shown
in figure (a).


In

the character interleaving technique, the data on each branch port are divided down
to the level of characters only. Then eac
h time
-
slot is allocated to one character, as shown in
figure (b).


If the time
-
slots are allocated in a fixed cyclic order, the t
echnique is called pure TDM
or synchronous TDM. The main disadvantage of pure TDM is that any time
-
slot not used by
a branch port goes to waste.


Statistical Time Division Multiplexing


A Statistical Time Division Multiplexing (STDM) system allocates time
-
slots to the
branch ports in a statistical manner; that is, it allocates more time
-
slots to a busy branch

port
and fewer time
-
slots to a sparingly used branch port. This improves th
e utilization of the
trunk line. IN STDM, the order of time
-
slot allocation

is not fixed; therefore, each data unit
transmitted in a time
-
slot must carry the address of the originating branch port. Only then can
this data be passed on to the matching branch port on the other end of the link.




b
1

b
2

b
3


25











(a)









(b)


Figure: Time division multiplexing


32. Explain the general concept of circuit switching. (10 marks)



In the circuit switching technique, a dedicated end
-
to
-
end circuit is established
between the two communicat
ing nodes by using physical connections in the switching nodes.
The links between the various user nodes and

switching nodes are fixed, i.e.,

static. The
connections within the switching nodes are dynamic and need to be switched to establish the
required e
nd
-
to
-
end circuit.


When information is to be passed between any two ports in a circuit switching node,
first a physical connection is established between these ports. As shown in figure, the Conn
(P
1

< H > P
2
) connection is formed in the switching node
S
to establish a circuit between user
nodes U
1

and U
2
. To establish a circuit between user nodes U
2

and U
3
, the physical
connection in the switching node S1 is switched to Conn (P
2

< H > P
3
). In a large network,
many such circuit switching are performed to e
stablish a connection from one user node to
another.


Communication using circuit switching requires a three
-
phase sequence. These three
phases are:


1. establish a circuit


2. transfer information


3. disconnect the circuit


0

0

1

1

Time
-
slots

A=1000001

B
=100001
0

C
=100001
1

D
=10001
00

b
1

b
2

b
3

b
4

1

2

3

4

b
1

b
2

b
3

b
4

1

2

3

4

D

B

C

A

Time
-
slots

EA

FB

GC

HD

b
1

b
2

b
3

b
4

1

2

3

4

b
1

b
2

b
3

b
4

1

2

3

4


26













Figure: A circuit

switching node



33. Compare the operation

of the diagram and virtual circuit approaches for packet
switching.



(20 marks)


Circuit Switching

Packet Switching
Datagram

Packet Switching: Virtual
Circuit Cell Relay


ATM

End
-
to
-
end physical path.


Continu
ous transmission
of the information is
possible.

No storage of information
on the switching nodes.



End
-
to
-
end connection
lasts for the entire session.


Involves a circuit setup
delay. Negligible
propagation delay.

A busy signal returned if
the called par
ty is busy.


Traffic overload blocks
new calls. No additional
delay on established calls.



No dedicated physical
path.

Information must be
broken into packets.


Information packets
stored on the switching
nodes.


Route determined for
each packet
independe
ntly.

No circuit setup delay.
Appreciable propagation
delay.

Transmission protocol
must inform the sender of
any undelivered packets.

Traffic overload increases
packet delivery delays.




No dedicated physical
path.

Information broken into
small packets ca
lled cells.


Information packets
stored on the switching
nodes.


Same route followed by
all packets in a session.


Involves a route setup
delay and some
propagation delay.

Sender notified if a route
for a virtual circuit cannot
be found.

Traffic overload m
ay
block new calls and
increase packet delivery
delays.


U
1

U
3

Switching
Node

S
1

P
2

P
1




P
3


27


Circuit Switching

Packet Switching
Datagram

Packet Switching: Virtual
Circuit Cell Relay
-

ATM

Speed and code
conversion generally not
available.

Packs, if used, are
delivered in the correct
seque
nce.

Fixed rate service:
Bandwidth fixed for all
connections.



Multirate
service:
Bandwidth fixed for a
connection, but can be
different from one
connection to another.

Minimum protocol
overhead in each packet.

Speed and code
conversion generally
availabl
e.

Packets may be delivered
out of sequence.


No fixed bandwidth, but
there is an upper limit on
the throughput.








Packet overhead highest
of the three techniques.

Speed and code
conversion generally
available.

Packets are delivered in
the correct seq
uence.


No fixed bandwidth, but
there is an upper limit on
the throughput.
Throughput is higher than
in datagram technique.






Packet overhead lower
than in datagram
technique.





34. Explain the general concept of packet switching. (10 marks)


Packet
Switching


Most data communications protocols break up the data files into packets before
transmissions. Each packet comprises a packet header and a fixed or variable length data
section, followed by a
CRC for error detection.


The packet switching techniq
ue treats each packet as a separate entity for the entire
duration of its transmission from one user node to the other. The operation of a packet
switching node is shown in fig. The three ports P1, P2 and P3 are connected by three logical
connections, C (P
1

< L > P
2
), C (P
2

< L > P
3
) and C (P
1

< L > P
3
).


Physical port connections are created by having a direct electrical connection between
two ports. In a logical connection, there is no direct electrical connection between two ports.
Two ports are connecte
d by a logical link can pass packets to each other via buffers. On a
packet switching node, many logical connections can exit from one port to other ports.


There are two main approaches to packet switching: data
gram and virtual circuit. Each
packet is giv
en a sequence number. In the datagram approach, each packet is treated as an
independent entity and transmitted through the network.


28


Virtual circuit: The virtual circuit technique combines the datagram idea and the
circuit switching technique. An end
-
to
-
end path is selected and a virtual circuit is established
before any data packets are transmitted. All packets take the same path over the network, this
path works virtually like a circuit.


Two variations of the original packet switching techniques are ca
lled frame relay and
cell relay.


Frame Relay: The frame relay techniques use variable length data packets called
frames. These frames reduce the protocol overhead by reducing the level of error checking
with respect to the traditional packet switching se
rvices.


Cell Relay
-

ATM: The cell relay technique uses small, fixed length data packets
called cells for information transmission. Cell relay use virtual channels to route cells from
the source to the destination.


The different components of a multim
edia presentation may have different
requirements in terms of bandwidth, error control, synchronization and transmission delays.


35. Calculate the round trip delay between two stations that are 3000 km apart if

(a) they are connected by a terrestrial mic
rowave link.

(b) they are connected by a satellite link.


(10 marks)



All electromagnetic

waves travel close to the speed of light, which is 300,000 km per
second. If we neglect processing delays, the round trip delay will consist only of the
propagation
delay. The round trip delays can then be calculated as follows.

(a) For terrestrial microwave link


Total distance covered = 2 x distance between the two stations


= 2 x 3000 = 6000 km


Round trip
delay = 6000 / 300,000 = 0.02 sec = 20 milliseconds

(b) For satellite link


Total distance covered = 4 x satellite altitude


= 4 x 36,000 = 144,000 km


Round trip delay = 36,000
/ 300,000 = 0.48 sec = 0.5 sec















29

Chapter (6)

Multimedia Applications


36. What are the main characteristics of networked multimedia application
s in terms of
bandwidth delay and synchronization?

(10 marks)


Multimedia applications

generally require high bandwidth, low end
-
to
-
end delay and
synchronization. Many multimedia applications are driving the development of new
technology, and any more applications are becoming viable
because of the technological
advances. Many applications
that did not use multimedia content in their earlier versions
presently include multimedia, because multimedia makes a product more attractive and
marketable.


As computer become more powerful, it becomes easier to run good
-
quality
multimedia presentations
. A new generation of computers comes out about every 18 months.
Multimedia presentations that were unacceptable slow on 80386
-
based computer became
presentable on 80486
-
based machines. On Pentium
-
based computers, the same presentations
run smoothly. On sy
stems with MMX architecture CPUs, it is possible to run even more
demanding multimedia presentations.


Various types of activities are performed by the users of multimedia systems, such as
entering commands, reading text, viewing pictures, browsing, naviga
ting, and responding to
questions. Different user activities are accompanied by different types of interactions and
transmission of different types of multimedia objects. Different types of activities and
interactions demand different services from the net
work.


Many user activities and interactions are common over the entire range of
applications. The most common activities and interactions are performed by the users of
multimedia systems.


37. List the various user activities and interactions involved in
s
tandalone and network
multimedia applications. For each one of these
,

give at least two examples.(20 marks)

Table:
User Activities and Interactions

User Activities

Enter commands:

Enter text:

Read text:

Create graphics:

View graphics:

Enter responses:


Interaction Modes

User
-
to
-
computer


User
-
to
-
user




From command line, using menus, via touch screen, etc.

Compose messages, etc.

Browse through text, navigate using hyperlink, etc.

Still images, animation, and moving images.

Navigate through and view stil
l and moving images.

Click on buttons, enter text, etc.



The user interacts with an application residing either on a
standalone computer or on a network server.

Two or more users interact with each other, generally via a
network.


30

38. Describe the various

types of equipment use for video conferencing.










(10 marks)

Television
-
b
ased System


A television
-
based video conferencing system uses TV monitors, video cameras, and
associated hardware. This type of video conferencing system one end to the other
may be
analog or digital.


For digital transmission, the analog video and audio signals must first be converted to
digital format by using an ADC. The digital signals may then be encoded or compressed. On
the receiving end, the digital signal must be decom
pressed, decoded, and converted back to
the original analog signals with the help of DAC.


Computer
-
based System


Computer
-
based (desktop) video conferencing uses computers connected to digital
cameras. On a computer monitor, it is much easier to combine l
ive video and other types of
multimedia information. It is very easy to create multiple windows to display different
activities simultaneously. A desktop computer used in this manner is hence forth referred to
as a conference workstation.


One major advant
age of using a conference workstation is that end
-
to
-
end digital
transmission can be used.



39. Explain the operation of collaborative video conferencing. (10 marks)


Video
-
Only Conference


In a conventional video conference, only video and audio signals
are transmitted. The
participants exchange information primarily via voice; facial expressions and hand gestures
also pay a part in the conversion. No data are transmitted in a video
-
only conference.


Collaborative Conference


In the current business envir
onment, people want not only to talk to one another but
also to exchange multimedia information to collaborate in the development of joint projects.
This collaboration is done by sharing a digital workspace. The digital shared workspace is a
screen area on

each participant’s desktop computer. When one participant draws on the
shared workspace of his screen, the image appears on the shared workspace of all
participants. The concepts of collaborating via
desktop video conferencing are

also called
concurrent
e
ngineering when used in engineering projects.










31

40. Explain the three types of networking technologies that can be used for video
conferencing.






(20 marks)

( or )

List the various networking options available for networked multimedia applications
,
particular for video conferencing.


Three networking categories are
:

Circuit Switching


Video information can be transmitted as a continuous stream of analog or digital
signals. A switched circuit with the required bandwidth can be used to transmit these

signals.
Many video conferencing systems are based on circuit switched networks such as POSTS,
ISDN, and T
-
1 lines. A POSTS connection is adequate for a video phone full
-
fledged video
conferencing system require higher bandwidth circuit
-
switched networks
such as ISDN or T
-
1 lines.


Packet Switching


Digitized and compressed video can be transmitted over packet switching networks as
well. The

advantages of packet switching

can be applied to video conferencing by sending
video signals over packet switching a
nd cell switching (ATM) networks. Using packet
switching becomes even more attractive when multimedia content other than video is
included for collaborative conferencing.


Internet


The internet is the most widely used Universal Information Network. It can

be used as
a networking mechanism for video and collaborative conferencing. Two systems for
providing video conferencing over the Internet are worth a special mention: MBone and CU
-
SeeMe.


MBone

stands for multicast backbone. It provide
s

mechanisms for mu
lticasting high
-
bandwidth traffic over the Internet.


CU
-
SeeMe

is a software tool for video conferencing over the Internet, developed by
Comell University and its collaborators. It allows a user to become either receiver and
transmitter or receiver only. I
t uses a proprietary protocol for multicasting. Almost any
digital camera can be used in conjunction with it to create a video conference over the
Internet.












32

Content of Programming Methodo
logy


CHAPTER









CONTENT


PART (1)


Chapter (7)
------
------------
-






10


Chapter (8)
-------------------






10


Chapter (9)
-------------------







9


Chapter (10)
------------------







7


Chapter (11)
------------------







2


PART (2)


Chapter (2)
--------------------







3


Chapter (3)
-----
---------------







6


Chapter (4)
--------------------






11


Chapter (5)
--------------------







9


Chapter (6)
--------------------








3


Chapter (7)
--------------------







3


Chapter (8)
--------------------







3











Total


75