Tai lieu on thi mon multimedia

ES154 Communications and Media 2001-2002 Questions and worked examples UNIVERSITY OF WARWICK School of Engineering Lectures 11 to 20 D.D.Udrea Vào link tham khảo thêm bà http://www.eng.warwick.ac.uk/~es786/ES154/ Test: http://highered.mheducation.com/sites/844815617x/ student_view0/chapter29/multiple_choice_quiz.htm l http://www.slideshare.net/kmanishuli/inter-quiz 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) 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) mqz1529462202.doc Page of 15 ES154 Communications and Media 2001-2002 Questions and worked examples UNIVERSITY OF WARWICK School of Engineering Lectures 11 to 20 D.D.Udrea 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 = s 100 / 40 = 2.5 s e) 100 / 10 =10 s 100 / 100 = s 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 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 × 10 ms–1 and in the case of (c) × 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 mqz1529462202.doc Page of 15 ES154 Communications and Media 2001-2002 Questions and worked examples UNIVERSITY OF WARWICK School of Engineering Lectures 11 to 20 D.D.Udrea 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: a) The sampling rate must be at least × 10 kHz = 20 kHz or 20 ksamples/s and the bandwidth of the bandlimiting filter is from Hz through to 10 kHz b) The sampling rate must be at least × 3.4 kHz = 6.8 kHz or 6.8 ksamples/s and the bandwidth of the bandlimiting filter is from 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 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) 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 = V max / (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 = bits, q = V max / (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 = V max / (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 × 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 × = 2457600 bits Hence the time to transmit the image at 64 kbps is: 2457600 / 64 000 = 38.4 s mqz1529462202.doc Page of 15 ES154 Communications and Media 2001-2002 Questions and worked examples UNIVERSITY OF WARWICK School of Engineering Lectures 11 to 20 D.D.Udrea At 1.5 Mbit/s the time is 2457600 / 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 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 = f max Speech: Nyquist rate = × 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 = × 20 kHz = 40 kHz or 40 ksamples/s Hence bit rate generated = 40 k × 16 = 640 kbit/s (mono) or × 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 / = 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) mqz1529462202.doc Page of 15 ES154 Communications and Media 2001-2002 Questions and worked examples NTSC Answer PAL NTSC UNIVERSITY OF WARWICK School of Engineering Lectures 11 to 20 D.D.Udrea 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) U = 0.493B – 0.493 (0.299R + 0.587 G + 0.114B) = –0.147R – 0.289G + 0.437B V = 0.877R – 0.877 (0.299R + 0.587G + 0.114B)= 0.615R – 0.515G – 0.100B 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 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 bits per sample Hence: Bit rate = 13.5 × 106 × + (6.75 × 106 × 8) = 216 Mbit/s Memory required: Memory required per line = 720 × + (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 × + (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 mqz1529462202.doc Page of 15 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) Q18 Define the term multimedia and give examples of multimedia information content mqz1529462202.doc Page of 15 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 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 Answer Lecture 13, Internet Defined mqz1529462202.doc Page of 15 ES154 Communications and Media 2001-2002 Questions and worked examples UNIVERSITY OF WARWICK School of Engineering Lectures 11 to 20 D.D.Udrea 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 mqz1529462202.doc Page of 15 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 Q36 Explain what is meant by static, dynamic and active web documents Answer Lecture 14, Web documents mqz1529462202.doc Page of 15 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 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 Q42 Explain how an image is formed on an LCD (Liquid Crystal Display) Answer Lecture 15, LCD mqz1529462202.doc Page 10 of 15 ES154 Communications and Media 2001-2002 Questions and worked examples UNIVERSITY OF WARWICK School of Engineering Lectures 11 to 20 D.D.Udrea 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 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 mqz1529462202.doc Page 11 of 15 ES154 Communications and Media 2001-2002 Questions and worked examples UNIVERSITY OF WARWICK School of Engineering Lectures 11 to 20 D.D.Udrea 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 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 mqz1529462202.doc Page 12 of 15 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 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 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= bit/character x 10 characters = 70 bit Proposed compression algorithm A=1; N=01; B=001, R=0001, M=00001 BANANARAMA = 001 01 01 0001 00001 Total compressed bits= 20 bit mqz1529462202.doc Page 13 of 15 ES154 Communications and Media 2001-2002 Questions and worked examples UNIVERSITY OF WARWICK School of Engineering Lectures 11 to 20 D.D.Udrea 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 0 2 4], knowing that M1 C (0)   s(k ) u =1,2,… M k 0 C (u )  M M1  (2k  1)u   2M  s(k ) cos k 0 Answer Lecture 19, Image compression: DCT 1 C (0)  s(k )  ( s (0)  s (1)  s (2)   s (7))  (4        4) 5.66  k 0 8  3 5 15  (2k  1)  C (1)  s(k ) cos   ( s(0) cos( )  s (1) cos( )  s(2) cos( )   s (7) cos( ))    16 16 16 16 k 0  16  (4 0.98  0.83  0.55  0.55  0.83  0.98) 0 2  3 15  (2k  1)2  C ( 2)  s (k ) cos   ( s (0) cos( )  s (1) cos( )   s (7) cos( ))    16 8 8 k 0   (4 0.92  0.38  0.38  0.38  0.38  0.92) 3.7 2 3 9 45  (2k  1)3  C (3)  s(k ) cos   ( s (0) cos( )  s(1) cos( )   s(7) cos( ))    16 16 16 16 k 0   (4 0.83  0.19  0.98  0.98  0.19  0.83) 0 Because C(1) and C(3) are zero, compression occurs by sending only C(0) and C(2) instead of all four coefficients mqz1529462202.doc Page 14 of 15 ES154 Communications and Media 2001-2002 Questions and worked examples UNIVERSITY OF WARWICK School of Engineering Lectures 11 to 20 D.D.Udrea 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 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 mqz1529462202.doc Page 15 of 15 ... 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... 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. .. 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