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SCHOOL OF ENGINEERING MODULAR HONOURS DEGREE COURSE LEVEL 3

EO321 1 / 14

SEMESTER 2

2005/2006

AUDIO SYSTEMS
Examiners: Dr. D. S. Gill/Dr. S. C. Busbridge

Attempt FIVE questions only

Time allowed : 3 hours Total number of questions = 8

All questions carry equal marks

The figures in brackets indicate the relative weightings of parts of a question.

You may find the following information useful: Speed of sound at 20 C = 340 m s-1. The density of air at 20 C = 1.2 kg m-3. Threshold of hearing = 2 x 10-5 Pa

EO321 2 / 14 Data supplied: PX4 and 12B4A data sheets 1) a) Starting from the definition of the bulk modulus of a fluid, K, show that plane pressure vibrations, p, propagate through the fluid according to:

2 p  2 p  x 2 K t 2
Where  is the density of the fluid. (8)

b) Show that the following is a solution of the plane wave equation:
x  p  po sin   t    c

What is the physical meaning of c? c) A bass loudspeaker is required to generate a sound pressure of 100 dB at a frequency of 20 Hz at a distance of 3 m away. Estimate the amplitude of the displacement of the loudspeaker cone, ignoring room reflections. Assume that the acoustic centre of the loudspeaker is 10 cm behind the cone.

(4)

(8)

EO321 3 / 14 2) a) Explain what is meant by the terms interference and diffraction. b) Show that the intensity ratio from a slit of width w and infinite length is given by:
I sin 2   Io 2

(2)

where:
  w sin  

 is the wavelength of the sound being diffracted and  is the angle of
diffraction in a plane perpendicular to the length of the slit.

Assume that the incident sound is normal to the slit.

(8)

c) Four loudspeakers are equally spaced 0.68 m apart in a straight line. They are connected in parallel to an oscillator that produces a signal of frequency 500 Hz. Considering a plane containing the loudspeakers and the far-field sound: i) Calculate the positions of zero sound pressure

ii) Sketch the (far-field) sound pressure pattern for 0 to 360 degrees as a polar diagram. Assume that the loudspeakers radiate sound omnidirectionally. (10)

EO321 4 / 14 3) a) Explain what is meant by the term loudness by reference to the FletcherMunson curves.

What is the principal reason why these curves exhibit a resonance? Convert the sound pressure of a 1 kHz tone of rms value 8.3 x 10-2 N m-2 into a loudness level in phons. (6)

b) Sketch a diagram of the human auditory system. Explain the roles of each of the outer, middle and inner ears in the transduction process.

Explain how the auditory system codes frequency, time (phase) and intensity in a manner which the brain can understand. c) A straight road carries 2000 vehicles hr-1 whose average speed is 50 km hr-1. If each vehicle has a noise output of 0.06 W, calculate the intensity level in dB at a distance of 30 m from the road assuming that the traffic stream is continuous and acts as a hemi-cylindrical source and that the vehicles are equally spaced. (8) (6)

EO321 5 / 14 4) a) Explain what is meant by the Miller effect.

How does the Miller effect limit the performance of an amplifier?

Use the data of the 12B4A triode supplied to calculate the bandwidth of the amplifier shown below in Figure Q4.1.

Figure Q4.1 (8) b) Describe a small-signal equivalent circuit for a triode valve.

Hence show that an unbypassed cathode bias resistor of value Rk appears reflected in the anode circuit in the gain expression of a common cathode amplifier as a resistor of value ( + 1)Rk where  is the amplification factor of the triode. (12)

EO321 6 / 14 5) a) Explain what is meant by the following terms when applied to audio power amplifiers: class A, class B, class C and class D. (4)

b) A single-ended power amplifier is formed with a single triode anodetransformer coupled to a load of resistance R. Show that the theoretical greatest rms power transferred into the load is 50% of the quiescent dissipation in the triode and occurs when rB = n2 R, where rB is the ratio of the quiescent voltage to that of the quiescent current and n is the primary:secondary turns ratio of the transformer. (6)

c) Using the supplied data sheet for the PX4 valve, draw a load line for the triode working into an 8  loudspeaker which is coupled to the anode circuit of the valve via a 25:1 ratio transformer.

The anode voltage is 250 V DC and the grid bias voltage (relative to the filament) is – 35 V.

Hence determine the maximum undistorted power obtainable from the amplifier operating in class A1. Determine also the maximum undistorted power obtainable if the amplifier were to be operated in class A2, pointing out what specific precautions would have to be taken. (8)

d) What type of distortion is principally generated by single ended amplifiers?

(2)

EO321 7 / 14 6) a) Identify the main characteristics of the Huffman code and Hamming codes, highlighting their differences. (4)

b) A signal source has four states A, B, C and D with probabilities of 0.4, 0.3, 0.2 and 0.1 respectively. Construct a Huffman code for this source, with a diagram showing the code reductions, and evaluate the efficiency of your code. (8)

c) Two eight-bit words are to be transmitted along a transmission link and require error correction to be used. Each of the eight-bit words will need to be split up into two four-bit parts before being encoded using the Hamming code (7,4).

i) Show with working what each of the codes required are, using the following eight-bit words (displayed in hexadecimal format): (5)

A4 9E ii) Show how a one-bit error can be detected at the receiver. Assume that bit I2 of the codeword for the nibble “E” above is in error. (3)

The following may be useful: C1 = I1  I2  I4 C2 = I1  I3  I4 C3 = I2  I3  I4 The order of bits for the code word is as follows: C1 C2 I1 C3 I2 I3 I4

EO321 8 / 14 7) a) Discuss how digital audio signals are stored onto a magnetic tape using the DAT (Digital Audio Tape) technology. Your answer should focus on how the magnetic medium is used to encode the data, the channel code used, the use of azimuth recording and the rationale for the use of interleaving. (6)

b) The channel code used for CD (Compact Disc), EFM, is different from that used in DAT. Explain why this particular code is suitable for recording digital audio on CD. (4)

c) What is meant by DSV (Digital Sum Value) when applied to CD systems?

Show on a diagram how the DSV changes when the channel code for CD is applied to the following 16-bit audio sample (shown in hexadecimal format):

58 9C

Assume that the initial DSV level is zero. You must justify the choice of the coupling bits used. You will find the details of the EFM code in Appendix Two. You may if you wish use the template given in Appendix One. Please ensure that you write your student number on the sheet and ensure that you attach it to your answer script as soon as you have completed your answer. (7)

d) An audio signal with a maximum peak-to-peak voltage of 100 mV is to be encoded digitally onto a CD. A 16-bit Analogue-to-Digital Converter is used. What would the maximum quantisation error be when the signal is reconstituted using a 16-bit Digital-to-Analogue Converter? Your answer must include all working. (3)

EO321 9 / 14 8) a) Compare and contrast the AES/EBU digital audio standard (AES-3-1992) and the MADI standard. Your answer will focus on the following: frame format, data rate and sample rate. (4)

b) Explain how the Professional-Channel-Status block is obtained from the subframes as specified in the AES/EBU digital audio standard (AES-3-1992). Hence explain the significance of the use of the pre-ambles for each sub-frame. (6)

c) Estimate, with justification, from the sequence shown in Figure Q8.1 where frame 1 is located (assume that the first frame is identified as frame 0). You will note that each time-period is numbered and that you should identify the start of frame 1 by its time-period number. State any assumptions made,

including any assumptions of what the sequence might look like beyond that shown. (2)

d) Digital audio compression has been defined in the MPEG set of standards. Explain, with the aid of diagrams, how the implementation for Layer one is different from Layers two and three. Your answer should focus on explanations where required of the psychoacoustic models used, the data rate used, the different stages involved in the bit-rate-reduction process and appropriate block diagrams. (8)

EO321 10 / 14

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Clock Signal Preamble patterns defined in the AES/EBU standard. X Y Z

Figure Q8.1 Part of a waveform captured on an oscilloscope.

EO321 11 / 14 APPENDIX One Attach to your answer script. Student Registration number: ____ / ________________

Answer sheet for Question 7 – Optional Template for students to use for 7 (c) to sketch the DSV (Digital Sum Value) variations.
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15 14 13 12 11 10 9 8 7 6 5 4 3 2 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10 -11 -12 -13 -14 -15 -16

1 0

EO321 12 / 14
APPENDIX Two Eight-to-Fourteen Modulation Conversion Table

Decimal Hex 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 20 21 22 23 24 25 26 27 28 29 2A

Binary 00000000 00000001 00000010 00000011 00000100 00000101 00000110 00000111 00001000 00001001 00001010 00001011 00001100 00001101 00001110 00001111 00010000 00010001 00010010 00010011 00010100 00010101 00010110 00010111 00011000 00011001 00011010 00011011 00011100 00011101 00011110 00011111 00100000 00100001 00100010 00100011 00100100 00100101 00100110 00100111 00101000 00101001 00101010

EFM 01001000100000 10000100000000 10010000100000 10001000100000 01001000000000 00000100010000 00010000100000 00100100000000 01001001000000 10000001000000 10010001000000 10001001000000 01000001000000 00000001000000 00010001000000 00100001000000 10000000100000 10000010000000 10010010000000 00100000100000 01000010000000 00000010000000 00010010000000 00100010000000 01001000010000 10000000010000 10010000010000 10001000010000 01000000010000 00001000010000 00010000010000 00100000010000 00000000100000 10000100001000 00001000100000 00100100100000 01000100001000 00000100001000 01000000100000 00100100001000 01001001001000 10000001001000 10010001001000

Decimal Hex 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 80 81 82 83 84 85 86 87 88 89 8A 8B 8C 8D 8E 8F 90 91 92 93 94 95 96 97 98 99 9A 9B 9C 9D 9E 9F

Binary

EFM

10000000 01001000100001 10000001 10000100100001 10000010 10010000100001 10000011 10001000100001 10000100 01000100100001 10000101 00000000100001 10000110 00010000100001 10000111 00100100100001 10001000 01001001000001 10001001 10000001000001 10001010 10010001000001 10001011 10001001000001 10001100 01000001000001 10001101 00000001000001 10001110 00010001000001 10001111 00100001000001 10010000 10000000100001 10010001 10000010000001 10010010 10010010000001 10010011 00100000100001 10010100 01000010000001 10010101 00000010000001 10010110 00010010000001 10010111 00100010000001 10011000 01001000000001 10011001 10000010010000 10011010 10010000000001 10011011 10001000000001 10011100 01000010010000 10011101 00001000000001 10011110 00010000000001 10011111 00100010010000 00001000100001 10000100001001 01000100010000 00000100100001 01000100001001 00000100001001 01000000100001 00100100001001 01001001001001 10000001001001 10010001001001

A0 10100000 A1 10100001 A2 10100010 A3 10100011 A4 10100100 A5 10100101 A6 10100110 A7 10100111 A8 10101000 A9 10101001 AA 10101010

EO321 13 / 14 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 2B 2C 2D 2E 2F 30 31 32 33 34 35 36 37 38 39 3A 3B 3C 3D 3E 3F 40 41 42 43 44 45 46 47 48 49 4A 4B 4C 4D 4E 4F 50 51 52 53 54 55 56 57 58 59 00101011 00101100 00101101 00101110 00101111 00110000 00110001 00110010 00110011 00110100 00110101 00110110 00110111 00111000 00111001 00111010 00111011 00111100 00111101 00111110 00111111 01000000 01000001 01000010 01000011 01000100 01000101 01000110 01000111 01001000 01001001 01001010 01001011 01001100 01001101 01001110 01001111 01010000 01010001 01010010 01010011 01010100 01010101 01010110 01010111 01011000 01011001 10001001001000 01000001001000 00000001001000 00010001001000 00100001001000 00000100000000 10000010001000 10010010001000 10000100010000 01000010001000 00000010001000 00010010001000 00100010001000 01001000001000 10000000001000 10010000001000 10001000001000 01000000001000 00001000001000 00010000001000 00100000001000 01001000100100 10000100100100 10010000100100 10001000100100 01000100100100 00000000100100 00010000100100 00100100100100 01001001000100 10000001000100 10010001000100 10001001000100 01000001000100 00000001000100 00010001000100 00100001000100 10000000100100 10000010000100 10010010000100 00100000100100 01000010000100 00000010000100 00010010000100 00100010000100 01001000000100 10000000000100 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 AB AC AD AE AF 10101011 10101100 10101101 10101110 10101111 10001001001001 01000001001001 00000001001001 00010001001001 00100001001001

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

EO321 14 / 14 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 5A 5B 5C 5D 5E 5F 60 61 62 63 64 65 66 67 68 69 6A 6B 6C 6D 6E 6F 70 71 72 73 74 75 76 77 78 79 7A 7B 7C 7D 7E 7F 01011010 01011011 01011100 01011101 01011110 01011111 01100000 01100001 01100010 01100011 01100100 01100101 01100110 01100111 01101000 01101001 01101010 01101011 01101100 01101101 01101110 01101111 01110000 01110001 01110010 01110011 01110100 01110101 01110110 01110111 01111000 01111001 01111010 01111011 01111100 01111101 01111110 01111111 10010000000100 10001000000100 01000000000100 00001000000100 00010000000100 00100000000100 01001000100010 10000100100010 10010000100010 10001000100010 01000100100010 00000000100010 01000000100100 00100100100010 01001001000010 10000001000010 10010001000010 10001001000010 01000001000010 00000001000010 00010001000010 00100001000010 10000000100010 10000010000010 10010010000010 00100000100010 01000010000010 00000010000010 00010010000010 00100010000010 01001000000010 00001001001000 10010000000010 10001000000010 01000000000010 00001000000010 00010000000010 00100000000010 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 DA DB DC DD DE DF 11011010 11011011 11011100 11011101 11011110 11011111 10010000010001 10001000010001 01000000010001 00001000010001 00010000010001 00100000010001

E0 11100000 01000100000010 E1 11100001 00000100000010 E2 11100010 10000100010010 E3 11100011 00100100000010 E4 11100100 01000100010010 E5 11100101 00000100010010 E6 11100110 01000000100010 E7 11100111 00100100010010 E8 11101000 10000100000010 E9 11101001 10000100000100 EA 11101010 00001001001001 EB 11101011 00001001000010 EC 11101100 01000100000100 ED 11101101 00000100000100 EE 11101110 00010000100010 EF 11101111 00100100000100 F0 F1 F2 F3 F4 F5 F6 F7 F8 F9 FA FB FC FD FE FF 11110000 00000100100010 11110001 10000010010010 11110010 10010010010010 11110011 00001000100010 11110100 01000010010010 11110101 00000010010010 11110110 00010010010010 11110111 00100010010010 11111000 01001000010010 11111001 10000000010010 11111010 10010000010010 11111011 10001000010010 11111100 01000000010010 11111101 00001000010010 11111110 00010000010010 11111111 00100000010010

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