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ES154 Communications and Media 2001-2002
Questions and worked examples
UNIVERSITY OF WARWICK
School of Engineering
Lectures 11 to 20
D.D.Udrea
Aims and Objectives
- Give examples of questions theory
- Give worked examples of numeric questions
Q1
What are the uncompressed bit rates for the following multimedia sources :
a) Telephone voice signal with a bandwidth between 200 and 3400 Hz, sampled with 8000 samples/s
and 12 bit/sample.
b) Wideband speech signal with a bandwidth between 50 and 7000 Hz, sampled with 16000
samples/s and 14 bit/sample.
c) Wideband stereo audio signal with a bandwidth between 20 and 20000 Hz, sampled with 44100
samples/s and 16 bit/sample/channel
d) Colour image signal with a resolution of 512 x 512 pixels and a colour depth of 24 bit/pixel
e) CCIR TV image signal with a resolution of 750 x 576 pixels and a colour depth of 24 bit/pixel, at
a frame rate of 30 frames/s
f) HDTV image signal with a resolution of 1280 x 720 pixels and a colour depth of 24 bit/pixel, at a
frame rate of 60 frames/s
g)
Answer
a) 8000 x 12 = 96 kbit/s
b) 16000 x14 = 224 kbit/s
c) 2 x 44100 x 16 = 1412 kbit/s
d) 512 x 512 x 24 = 6300 kbit
e) 750 x 576 x 24 x 30 = 311000 kbit/s
f) 1280 x 720 x 24 x 60 = 1327000 kbit/s
Q2
How long it takes to transmit a file of 100 Mbit over the following lines:
a) PSTN (14.4, 28.8, 33.6 or 56 kbit/s)
b) ISDN (64, 128 kbit/s)
c) ADSL (16-640 kbit/s upstream, 1.544-8.448 Mbit/s downstream)
d) CATV(20-40 Mbit/s)
e) Ethernet (10, 100, 1000 Mbit/s)
f)
Answer
transfer time = file size / bit rate
a) 100 000 / 14.4 = 6945 s
100 000 / 28.8 = 3472 s
100 000 / 33.6 = 2976 s
100 000 / 56 =1786 s
b) 100 000 / 64 = 1562.5
100 000 / 128 = 781 s
c) 100 000 / 640 = 156 s upstream
100 / 8.448 = 11.84 s downstream
d) 100 / 20 = 5 s
100 / 40 = 2.5 s
e) 100 / 10 =10 s
100 / 100 = 1 s
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ES154 Communications and Media 2001-2002
Questions and worked examples
UNIVERSITY OF WARWICK
School of Engineering
Lectures 11 to 20
D.D.Udrea
100 / 1000 = 0.1 s
Q3
Express the following 32-bit integer as a binary and as a decimal IP address: 2215445022
Answer
2215445022 = 231 + 226 + 219 + 218 + 216 + 29 + 24 + 23 + 22 + 21
in binary 10000100 00001101 00000010 00011110
in decimal 132.13.2.30
Q4
How many hosts can a network have if the network IP address is 132.xxx.xxx.xxx.
What if the address is 137.205.xxx.xxx ? What then if the address is 137.205.144.xxx ?
Answer
224-2 = 16777214 hosts
In the second case 216-2 = 65534 hosts
In the third case 28-2 = 254 hosts
Q5
Determine the propagation delay associated with the following communication channels:
a) a connection through a private telephone network of 1 km,
b) a connection through a PSTN of 200 km,
c) a connection over a satellite channel of 50000 km.
Assume that the velocity of propagation of a signal in the case of (a) and (b) is 2 × 108 ms–1 and in the
case of (c) 3 × 108 ms–1.
Answer
Propagation delay Tp = physical separation/velocity of propagation
a) 1000 / (2 x 108) = 5x10–6 s
b) 200 x 1000 / (2 x 108) = 10–3 s
c) 50000 x 1000 / (2 x 108) = 1.67 x 10–1 s
Q6
Determine the rate of the sampler and the bandwidth of the bandlimiting filter in an encoder which is to
be used for the digitization of an analog signal which has a bandwidth from 15 Hz through to 10 kHz,
assuming the digitized signal:
a) is to be stored within the memory of a computer,
b) is to be transmitted over a channel which has a bandwidth from 200Hz through to 3.4 kHz.
Answer
The Nyquist sampling rate must be at least twice the highest frequency component of the signal or
transmission channel. Hence:
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ES154 Communications and Media 2001-2002
Questions and worked examples
UNIVERSITY OF WARWICK
School of Engineering
Lectures 11 to 20
D.D.Udrea
a) The sampling rate must be at least 2 × 10 kHz = 20 kHz or 20 ksamples/s and the bandwidth of
the bandlimiting filter is from 0 Hz through to 10 kHz.
b) The sampling rate must be at least 2 × 3.4 kHz = 6.8 kHz or 6.8 ksamples/s and the bandwidth of
the bandlimiting filter is from 0 Hz through to 3.4 kHz.
In practice, it should be noted that, because of imperfections in filters, some higher frequency components
above the filter cut-off frequency may be passed and hence the sampling rate is normally higher than the
two derived values. In the case of (b), for example, it is common to assume that frequency components of
up to 4 kHz may be passed by the bandlimiting filter and hence a sampling rate of 8ksamples/s is
normally used.
Q7
An analog signal has a dynamic range of 40 dB. Determine the magnitude of the quantization noise
relative to the minimum signal amplitude if the quantizer uses,
a) 6 bits
b) 10 bits
You assume that the maximum noise is ½ of a quantization step.
Comment on whether the level of quantization noise is acceptable or not.
Answer
Quantization noise = ±q/2, where q is the quantization interval. q = Vmax / (2n-1), where n is the number
of quantization levels.
The dynamic range is D = 20 log10 (Vmax/Vmin) = 40 dB. Hence Vmax/Vmin = 102 = 100 .
It follows that Vmax = 100 Vmin
a) for n = 6 bits, q = Vmax / (24-1) = 100 Vmin / 63 = 1.6 Vmin . The error in this case is ±q/2 = ±0.8
Vmin.. Not acceptable because is very close to the minimum signal.
b) for n = 10 bits, q = Vmax / (210-1) = 100 Vmin / 1023 = 0.1 Vmin . The error in this case is ±q/2 =
±0.05 Vmin.. Acceptable level, an order of magnitude smaller than the smallest signal.
Q8
Derive the time to transmit the following digitized images at both 64 kbit/s and 1.5 Mbit/s:
a) a 640 × 480 × 8 VGA-compatible image,
b) a 1024 × 768 × 24 SVGA-compatible image.
Comment on the possibility of an interactive service which uses these transmission times.
Answer
a) The size of the image in bits is: VGA = 640 × 480 × 8 = 2457600 bits
Hence the time to transmit the image at 64 kbps is: 2457600 / 64 000 = 38.4 s
At 1.5 Mbit/s the time is 2457600 / 1 500 000 = 1.64 s
b) The size of the image in bits is: SVGA = 1024 × 768 × 24 = 18874368 bits
Hence the time to transmit the image at 64 kbps is: 18 874 / 64 = 295s
At 1.5 Mbit/s the time is 18.87 / 1.500 000 = 12.6 s
As we can see, the times to transmit a signal image at 64 kbps are such that interactive access would not
be feasible, nor at 1.5Mbps with the higher-resolution SVGA image.
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ES154 Communications and Media 2001-2002
Questions and worked examples
UNIVERSITY OF WARWICK
School of Engineering
Lectures 11 to 20
D.D.Udrea
Q9
Assuming the bandwidth of a speech signal is from 50 Hz through to 10kHz and that of a music signal is
from 15 Hz through to 20 kHz, derive the bit rate that is generated by the digitization procedure in each
case
assuming the Nyquist sampling rate is used with 12 bits per sample for the speech signal and 16 bits per
sample for the music signal. Derive the memory required to store a 10 minute passage of stereophonic
music.
Answer
Bit rates: Nyquist sampling rate = 2 fmax.
Speech:
Nyquist rate = 2 × 10 kHz = 20 kHz or 20 ksamples/s.
Hence with 12 bits per sample, the bit rate generated = 20 k × 12 = 240 kbit/s
Music:
Nyquist rate = 2 × 20 kHz = 40 kHz or 40 ksamples/s
Hence bit rate generated = 40 k × 16 = 640 kbit/s (mono)
or 2 × 640k = 1280 kbit/s (stereo)
Memory required: Memory required = bit rate (bit/s) × time (s)/8 bytes
Hence at 1280 kbps and 600 s
Memory required = 1280 × 103 × 600 / 8 = 96 Mbytes
Q10
Assuming the CD-DA standard is being used, derive:
a) the storage capacity of a CD-ROM to store a 60 minute multimedia title,
b) the time to transmit a 30 second portion of the title using a transmission channel of bit rate:
64 kbps and 1.5Mbps.
Answer
a) The CD-DA digitization procedure yields a bit rate of 1.411Mbps.
Hence storage capacity for 60 minutes = 1.411 × 60 × 60 Mbits
= 5079.6 Mbits or 634.95 Mbytes
b) One 30 second portion of the title = 1.411 × 30 = 42.33 Mbits
Hence time to transmit this data:
At 64 kbps
(42.33 × 106)/( 64 × 103) = 661.4 s (about 11 minutes)
At 1.5Mbps (42.33 × 106)/( 1.5 × 106) = 28.22 s
Q11
Derive the scaling factors used for both the U and V (as used in PAL) and I and Q (as used in NTSC)
colour difference signals in terms of the three R, G, B colour signals, knowing that for
PAL :
Y = 0.299R + 0.587G + 0.114B
U = 0.493 (B – Y) and V = 0.877 (R – Y)
NTSC
Y = 0.299R + 0.587G + 0.114B
I = 0.74 (R – Y) – 0.27 (B – Y)
Q = 0.48 (R – Y) + 0.41 (B – Y)
.
Answer
PAL
U = 0.493B – 0.493 (0.299R + 0.587 G + 0.114B) = –0.147R – 0.289G + 0.437B
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ES154 Communications and Media 2001-2002
Questions and worked examples
UNIVERSITY OF WARWICK
School of Engineering
Lectures 11 to 20
D.D.Udrea
V = 0.877R – 0.877 (0.299R + 0.587G + 0.114B)= 0.615R – 0.515G – 0.100B
NTSC
I = 0.74R – 0.27B – 0.47Y = 0.599R – 0.276G – 0.324B
Q = 0.48R + 0.41B – 0.89Y = 0.212R – 0.528G + 0.311B
Q12
Derive the bit rate and the memory requirements to store each frame that result from the digitization of
both a 525-line and a 625-line system assuming a 4:2:2 format. Also find the total memory required to
store a 1.5 hour movie/video. Remember that the digital format uses 480, respectively 576 visible lines
for the two systems, with 720 pixels/line and the sampling rate for luminance is 13.5 MHz with 8
bits/sample.
Answer
525-line system:
The number of samples per line is 720 and the number of visible lines is 480.
Hence the resolution of the luminance (Y) and two chrominance (Cb and Cr) signals are:
Y = 720 × 480
Cb = Cr = 360 × 480
Bit rate: Line sampling rate is fixed at 13.5 MHz for Y and 6.75 MHz for both Cb and Cr, all with 8 bits
per
sample.
Hence: Bit rate = 13.5 × 106 × 8 + 2 (6.75 × 106 × 8) = 216 Mbit/s
Memory required: Memory required per line
= 720 × 8 + 2 (360 × 8)
= 11 520 bits or 1440 bytes
Hence memory per frame, each of 480 lines = 480 × 11520
= 5.5296 Mbits or 691.2 kbytes
and memory to store 1.5 hours assuming 60 frames per second:
= 691.2 × 60 × 1.5 × 3600 kbytes
= 223.9488 Gbytes
625-line system:
Resolution: Y = 720 × 576
Cb = Cr = 360 × 576
Bit rate
= 13.5 × 106 × 8 + 2 (6.75 × 106 × 8)
= 216Mbps
Memory per frame
= 576 × 11 520
= 6.63555 Mbits or 829.44 kbytes
and memory to store 1.5 hours assuming 50 frames per second:
= 829.44 × 50 × 1.5 × 3600kbytes
= 223.9488 Gbytes
It should be noted that, in practice, the bit rate figures are less than the computed values since they
include samples during the retrace times when the beam is switched off. Nevertheless, as we can deduce
from the computed values, both the bit rate and the memory requirements are very large for both systems
and it is for this reason that the various lower resolution formats have been defined.
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ES154 Communications and Media 2001-2002
Questions and worked examples
UNIVERSITY OF WARWICK
School of Engineering
Lectures 11 to 20
D.D.Udrea
Q13
Explain the concept of a network and classify networks on the following criteria:
a) number of direct connections
b) type of connection
c) type of access
d) type of data transmission
e) area span
Answer
Lecture 11, Networks
Q14
Give examples of typical LAN topologies and explain their characteristics. Use diagrams in your
explanations. Consider for each topology possible advantages and disadvantages.
Answer
Lecture 11, LAN topologies
Q15
Give examples of wireless physical media used for networking and potential applications. Consider for
each medium possible advantages and disadvantages.
Answer
Lecture 11, Network media
Q16
Give three examples of devices used as network interfaces and indicate what type of network they are
suitable for.
Answer
Lecture 11, Network interfaces
Q17
Explain the meaning of communication protocols and give some examples, indicating (wherever
possible) whether they are low level or high level.
Answer
Lecture 11, Network software
Lecture 13 TCP/IP, UDP (Universal Datagram Protocol)
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ES154 Communications and Media 2001-2002
Questions and worked examples
UNIVERSITY OF WARWICK
School of Engineering
Lectures 11 to 20
D.D.Udrea
Q18
Define the term multimedia and give examples of multimedia information content.
Answer
Lecture 12, Multimedia
Q19
Draw a simple functional diagram of a multimedia network and explain the role of its components.
Answer
Lecture 12, Multimedia communication networks
Q20
Give examples of multimedia networks and indicate, in each case, what the Source and destination
terminals, the access and delivery networks and the backbone network are.
Answer
Lecture 12, Multimedia communication networks
Lecture 12, POTS
Lecture 12, Telephony and data N-ISDN
Lecture17, Satellite/Braoadcast networks
Lecture17, Cable networks
Q21
Give examples of services that can be run over PSTN (Public Switched Telephone Network) and discuss
the requirements of the terminals.
Answer
Lecture 12, POTS
Q22
Give examples of services that can be run over ISDN (Integrated Services Digital Network) and discuss
the requirements of the terminals.
Answer
Lecture 12, ISDN
Q23
Explain what the Internet is.
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ES154 Communications and Media 2001-2002
Questions and worked examples
UNIVERSITY OF WARWICK
School of Engineering
Lectures 11 to 20
D.D.Udrea
Answer
Lecture 13, Internet Defined
Q24
Give examples of services that can be run over the Internet and discuss the requirements of the terminals.
Answer
Lecture 13, Internet Services
Q25
Explain the terms ‘packet switching’ and ‘circuit switching’ and give examples of networks that use
them.
Answer
Lecture 13, Packet switching vs. Circuit switching
Q26
Explain what an IP address is and how it is used.
Answer
Lecture 13, IP Addresses
Q27
Explain what a routing table is and how it is used.
Answer
Lecture 13, IP Addresses
Q28
Explain what a domain name is and how it is used.
Answer
Lecture 13, Domain Name Service (DNS) & Host IP Resolution
Q29
Explain what a domain name is and how it is used.
Answer
Lecture 13, Domain Name Service (DNS) & Host IP Resolution
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ES154 Communications and Media 2001-2002
Questions and worked examples
UNIVERSITY OF WARWICK
School of Engineering
Lectures 11 to 20
Q30
Explain what the connection is between an IP address and a Domain name.
Answer
Lecture 13, Domain Name Service (DNS) & Host IP Resolution
Q31
Explain how a client-server application works. Use diagrams for your explanations.
Answer
Lecture 13, Client-Server models
Q32
Explain the terms HTTP and HTML in the context of World Wide Web applications.
Answer
Lecture 14, World wide web
Q33
Explain what a URL is and what its components represent.
Answer
Lecture 14, Uniform Resource Locator (URL)
Q34
Give examples of Internet services and indicate the type of high level protocol they use.
Answer
Lecture 14, Protocol Service description table
Q35
Describe the role and the components of a web browser.
Answer
Lecture 14, Components of a web browser
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D.D.Udrea
ES154 Communications and Media 2001-2002
Questions and worked examples
UNIVERSITY OF WARWICK
School of Engineering
Lectures 11 to 20
D.D.Udrea
Q36
Explain what is meant by static, dynamic and active web documents.
Answer
Lecture 14, Web documents
Q37
Explain how a static / dynamic /active web document is generated.
Answer
Lecture 14, Static/Dynamic/active document generation
Q38
Give examples of how digital images can be generated.
Answer
Lecture 15, CCD, C-MOS, scanning diode, scanner
Q39
Explain how a charged coupled device (CCD camera) works.
Answer
Lecture 15, CCD camera
Q40
Give examples of compression formats for still images and indicate whether they are lossy or lossless.
Answer
Lecture 15, Image formats
Q41
Explain how an image is formed on a CRT(cathode ray tube) display.
Answer
Lecture 15, CRT
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ES154 Communications and Media 2001-2002
Questions and worked examples
UNIVERSITY OF WARWICK
School of Engineering
Lectures 11 to 20
D.D.Udrea
Q42
Explain how an image is formed on an LCD (Liquid Crystal Display).
Answer
Lecture 15, LCD
Q43
Give an example of microphone technology and explain, with the aid of diagrams, how a microphone
works.
Answer
Lecture 16, Microphone
Q44
Explain what is understood by ‘aliasing’ and how it can be avoided.
Answer
Lecture 16, Sampling and quantization
Q45
Explain briefly the principle of PCM (Pulse Code Modulation) and what are the advantages it brings.
Answer
Lecture 16, Coding and Decoding -PCM
Q46
Give an example of speaker technology and explain, with the aid of diagrams, how a speaker works.
Answer
Lecture 16, Speaker
Q47
Give examples of services that can be run over a TV Broadcast Network and discuss the requirements of
the terminals.
Answer
Lecture 17, Satellite/Broadcast networks
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ES154 Communications and Media 2001-2002
Questions and worked examples
UNIVERSITY OF WARWICK
School of Engineering
Lectures 11 to 20
D.D.Udrea
Q48
Give examples of services that can be run over a Cable TV Network and discuss the requirements of the
terminals.
Answer
Lecture 17, Cable networks
Q49
Explain the concept of colour mixing and the difference between additive and subtractive colour schemes.
Answer
Lecture 17, Colour theory
Q50
Explain briefly, in a qualitative manner, the similarities and the differences between the two analog TV
formats PAL and NTSC.
Answer
Lecture 17, Analog TV signal: PAL/NTSC
Q51
Explain briefly, in a qualitative manner, the similarities and the differences between the two digital TV
formats 4:2:2 and 4:2:0.
Answer
Lecture 17, Digital TV signal
Q52
Explain the similarities and the differences between DAT (Digital Audio Tape) and CD (Compact Disk)
when used for the storing of audio.
Answer
Lecture 18, DAT/CD
Q53
Explain, with the aid of diagrams, how the information is read from a CD.
Answer
Lecture 18, CD-optical reading
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ES154 Communications and Media 2001-2002
Questions and worked examples
UNIVERSITY OF WARWICK
School of Engineering
Lectures 11 to 20
D.D.Udrea
Q54
Explain, with the aid of diagrams, how the information is written to and read from a minidisk.
Answer
Lecture 18,Minidisk:magneto-optical recording, optical reading
Q55
Explain the similarities and the differences between CD (Compact Disk) and Minidisk.
Answer
Lecture 18,CD/Minidisk
Q56
Explain, with the aid of diagrams, how the information is read from a double layer DVD.
Answer
Lecture 18, DVD: optical reading
Q57
Explain the similarities and the differences between CD (Compact Disk) and DVD (Digital Versatile
Disk)
Answer
Lecture 18,DVD vs CDROM
Q58
Explain the role of a Modem in multimedia communications.
Answer
Lecture 18, Modems
Q59
Give examples of wireless multimedia interfaces and suggest advantages and drawbacks for each of them
Answer
Lecture 18, Wireless interfaces
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ES154 Communications and Media 2001-2002
Questions and worked examples
UNIVERSITY OF WARWICK
School of Engineering
Lectures 11 to 20
D.D.Udrea
Q60
Suggest a lossless compression algorithm for the following string of characters : BANANARAMA. The
ASCII codes (in decimal representation) for each letter are as follows: A = 65, B = 66, M = 77, N = 78
and R = 82. Calculate the compression rate obtained.
Answer
Lecture 19, Text compression: entropy encoding
BANANARAMA = 66 65 78 65 78 65 82 65 77 65 (decimal) = 42 41 4E 41 4E 41 52 41 4D 41 (hex) =
1000010 1000001 1001110 1000001 1001110 1000001 1010010 1000001 1001101 1000001
Total uncompressed bits= 7 bit/character x 10 characters = 70 bit
Proposed compression algorithm
A=1; N=01; B=001, R=0001, M=00001
BANANARAMA = 001 1 01 1 01 1 0001 1 00001
Total compressed bits= 20 bit
Compression rate = total compressed bits/total uncompressed bits = 20/70 = 0.28
Or %compression = size difference/total uncompressed = (70-20)/70 *100 = 71%
Q61
Explain how dictionary compression works for a text. Give a short example
Answer
Lecture 19, Text compression: dictionary encoding
Q62
Explain how run length encoding works for an image. Give a short example.
Answer
Lecture 19, Image compression: run-length encoding
Q63
Calculate the first four DCT (Discrete Cosine Transform) coefficients of the signal block with the
amplitude sampled as s = [4 2 2 0 0 2 2 4], knowing that
2 M 1
1 M 1
 (2k  1)u 
C ( 0) 
s (k ) u =1,2,… C (u ) 
s(k ) cos 



M k 0
M k 0
 2M

Answer
Lecture 19, Image compression: DCT
1 7
1
1
C (0) 
s (k ) 
( s (0)  s(1)  s (2)  ...  s (7)) 
(4  2  2  0  0  2  2  4)  5.66

8 k 0
8
8
2 7

3
5
15
 (2k  1)  1
C (1) 
s (k ) cos 
 ( s (0) cos( )  s (1) cos( )  s (2) cos( )  ...  s (7) cos(
)) 


16
16
16
16
8 k 0
 16  2
1
(4  0.98  2  0.83  2  0.55  2  0.55  2  0.83  4  0.98)  0
2
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ES154 Communications and Media 2001-2002
Questions and worked examples
C ( 2) 
2
7
UNIVERSITY OF WARWICK
School of Engineering
Lectures 11 to 20
D.D.Udrea

3
15
 (2k  1)2  1
 ( s (0) cos( )  s (1) cos( )  ...  s (7) cos(
)) 

16
8
8
8
 2
 s(k ) cos 
8
k 0
1
(4  0.92  2  0.38  2  0.38  2  0.38  2  0.38  4  0.92)  3.7
2
2 7
3
9
45
 (2k  1)3  1
C (3) 
s (k ) cos 
 ( s (0) cos( )  s (1) cos( )  ...  s (7) cos(
)) 


16
16
16
16
8 k 0

 2
1
(4  0.83  2  0.19  2  0.98  2  0.98  2  0.19  4  0.83)  0
2
Because C(1) and C(3) are zero, compression occurs by sending only C(0) and C(2) instead of all four
coefficients.
Q64
Explain the difference between PCM (Pulse Code Modulation) and D-PCM (Differential-PCM).
Answer
Lecture 19, Audio compression: D-PCM
Lecture 16, Coding and Decoding -PCM
Q65
Explain what is understood by multimedia streaming and what the benefits of this technique are
Answer
Lecture 19, Streaming
Q66
Suggest a security measure for each of the following network problems: Virus attack; Improper or
accidental access to a system; Data intercepted in transit by unauthorized persons.
Answer
Lecture 20, Security measures
Q67
Suggest an encryption technique (and the keys) based on substitution and transposition for the following
plaintext:
This is a lovely day
Answer
Lecture 20, Encryption techniques
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ES154 Communications and Media 2001-2002
Questions and worked examples
UNIVERSITY OF WARWICK
School of Engineering
Lectures 11 to 20
D.D.Udrea
References
DDU – ES154 Communication and Media , Lectures 11 to 20, Lecture Slides 11 to 20.
Halsall, Fred – Multimedia communications: Applications, Networks, Protocols and Standards – Addison Wesley, 2001, ISBN
0-201-39818-4 QA 78.51.H2
Comer, Douglas - Computer networks and Internets with Internet applications / Douglas E. Co. - 3rd ed. - Upper Saddle River,
N.J.; London: Prentice Hall, 2001. – ISBN 0-13-091449-5 QA 74.2.C6
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