THE SCIENCE OF MUSIC (PC1327/GEK1519)
End-of-Term Class Test, Semester 1, 2016/17
This is an open book test. The test is one hour long.
Give your answers to ALL 25 questions on the computer-readable sheet provided, using a soft (2B)
pencil to shade the appropriate choice for each question.
1. Which of the following is the least appropriate example of science or technology in music?
(a) A chemist rehearsing for a concert on a new type of oboe which uses new synthetic materials
for the oboe reeds.
(b) An organist inventing a new type of display screen for the musical score on an electronic
piano.
(c) A trumpeter comparing different trumpet mouthpieces of the same shape but using different metals, to find out how this affects the trumpet’s sound quality.
(d) A civil engineer creating a new type of organ using concrete as the material for the large
organ pipes.
Answer: (a) The organist and civil engineer are performing essentially technological activities, and the trumpeter is performing an essentially scientific activity. Only the chemist is
performing an essentially musical activity.
2. The musical score of a piece for the organ is written on two staves. The upper staff starts with
a treble clef and the lower staff starts with a bass clef. The organ piece has a key signature
with four flats on both staves. A certain chord in this organ piece has its lowest note written
on the second lowest line of the five lines of the lower staff, and the highest note of the chord
written on the highest space of the four spaces of the upper staff. What is the ratio of the
interval between the lowest note and the highest note of this chord? (Assume that the chord
is played on an organ which is tuned correctly to the Equal-tempered scale.)
√ 41
(a) ( 12 2) .
√ 29
(b) ( 12 2) .
√ 17
(c) ( 12 2) .
(d) None of the above.
Answer: (b) The key signature has four flats, and hence all the notes named B, E, A and D
in the organ piece should be played as B flat, E flat, A flat and D flat respectively. The second
lowest line of the lower staff is the note B2 which should therefore be played as Bflat2, and the
highest space of the upper staff is the note E5, which should be played as Eflat5. Therefore
the interval from Bflat2 to Bflat4 is equal to 2 octaves or 24 equal-tempered semitones, and
the interval from Bflat4 to Eflat5 is equal to 5 equal-tempered semitones. The interval from
Bflat2 to Eflat5 thus consists of 29 Equal-tempered semitones, and the ratio of the interval
√ 29
between these two notes is therefore given by ( 12 2) .
3. A clarinet player is playing in an Orchard Road underpass. She plays a note on her clarinet,
and a man walking by then sings a note which is two octaves and a Just third below the
clarinet’s note. A young girl also passing by hears the man’s note and then sings a note which
is three octaves and a Just sixth above the man’s note. If the note sung by the girl has a
frequency of 1,320 Hz, what is the frequency of the clarinet’s note?
1
(a)
(b)
(c)
(d)
990 Hz.
825 Hz.
495 Hz.
None of the above.
Answer: (c) The girl’s note has a frequency of 1,320 Hz, and the note a Just sixth below this
will have a frequency given by 1,320 Hz divided by 53 which is the same as 1,320 Hz times 35
i.e. 792 Hz. The note three octaves below this will have a frequency of 792 Hz divided by 2
three times, i.e. 99 Hz. The note a Just third above this has a frequency given by 99 Hz times
5
i.e. 123.75 Hz. The clarinet’s note is thus two octaves above this and its frequency is equal
4
to 123.75 Hz times 2 twice i.e. 495 Hz.
4. The beginning of a piece of music for solo viola has a time signature of 57/16, and a particular
bar of this piece starts with 18 demisemiquavers and ends with 5 semiquavers and two dotted
quavers. Which of the following combinations of notes will fit exactly into the middle of this bar
in accordance with the time signature? (A demisemiquaver is half the duration of a semiquaver.
A note or rest which has a dot has its duration value increased by 50%.)
(a)
(b)
(c)
(d)
A dotted minim rest, 26 demisemiquavers and 12 semiquavers.
Two minims, 20 demisemiquavers and 10 semiquavers.
Three dotted quavers, 21 semiquavers and two crotchets.
None of the above.
Answer: (a) Since the time signature is 57/16, each bar of the piece must contain the equivalent of 57 semiquavers. The start of the bar already has 18 demisemiquavers which is equivalent
to 9 semiquavers, and the end of the bar has 5 semiquavers and two dotted quavers equivalent
to 6 semiquavers. The bar thus already has the equivalent of 20 semiquavers, and hence the
middle of the bar requires the equivalent of 37 semiquavers. A dotted minim rest is equivalent to 12 semiquaverrs, 26 demisemiquavers are equivalent to 13 semiquavers, and with 12
semiquavers this is a total of 37 semiquavers.
5. An open pipe labelled A is sliced into 9 short open pipes labelled B1, B2, B3, B4, B5, B6,
B7, B8 and B9, all of equal length. B1, B2, B3, B4 and B5 are joined up to make an open
pipe labelled C. A closed pipe labelled D is 50% longer than the open pipe A and is joined
up with B6, B7, B8 and B9 to make another closed pipe labelled E. What is the ratio of the
interval between the frequency of C when it vibrates with 6 nodes between its two ends and
the frequency of E when it vibrates with 4 nodes between its two ends (not counting the node
at one end)?
(a) 73 .
(b)
51
.
5
14
.
3
(c)
(d) None of the above.
Answer: (c) Let the fundamental frequency of A be f Hz and let its length be k cm. C is
five-ninths the length of A and hence its fundamental frequency must be given by f Hz times 95
i.e 9f5 Hz. When C has 6 nodes it will be at its 6th harmonic and its frequency will be equal to
9f
Hz times 6 i.e. 54f
Hz. The length of D is given by k cm times 1.5 i.e. 1.5k cm. Therefore
5
5
the closed pipe E has a length given by 1.5k cm plus 4k
cm i.e. 13.5k+4k
cm i.e. 17.5k
cm. An
9
9
9
9f
9
open pipe of this length will have a fundamental frequency given by f Hz times 17.5 i.e. 17.5
Hz. E is a closed pipe of this length so its fundamental frequency will be half of this i.e. 9f
35
2
Hz. When E has 4 nodes it will be at its 9th harmonic, and its frequency will be given by 9f
35
54f
Hz.
The
ratio
of
the
interval
between
this
frequency
and
Hz
is
given
Hz times 9 i.e. 81f
35
5
81f
70
14
by 54f
Hz
divided
by
Hz
i.e.
which
can
be
reduced
to
.
5
35
15
3
6. A closed pipe with a fundamental frequency of 240 Hz vibrates with 3 nodes between its two
ends (not counting the node at one end). An open pipe vibrates with 5 antinodes between its
two ends (not counting the antinodes at both ends) with a frequency equal to the frequency
of the closed pipe vibrating with 3 nodes. When the open pipe is then made to vibrate with 7
antinodes between its two ends (not counting the antinodes at both ends), beats of 8 Hz are
heard between the open pipe and a string 71 cm long which is vibrating at its fundamental
frequency. When the string is slightly shortened, the beat frequency increases (without passing
through zero Hz). If the string is shortened to 70.25 cm and is still vibrating at its fundamental
frequency, what is the beat frequency when the note from the string combines with the note
from the open pipe vibrating with 7 antinodes?
(a) 16 Hz.
(b) 24 Hz.
(c) 32 Hz.
(d) None of the above.
Answer: (c) Since the closed pipe has 3 nodes, it is vibrating at its 7th harmonic, and its
frequency is given by 240 Hz times 7 i.e. 1,680 Hz. The open pipe has 5 antinodes and 6 nodes,
and is therefore at its 6th harmonic. Since it is also vibrating at 1,680 Hz, its fundamental
frequency is equal to 1,680 Hz divided by 6 i.e. 280 Hz. If the open pipe is vibrating with 7
antinodes, it will have 8 nodes and will thus be at its 8th harmonic. Its frequency will then
be equal to 280 Hz times 8 i.e. 2,240 Hz. The beat frequency is 8 Hz and hence the frequency
of the string is either 2,240 Hz minus 8 Hz i.e. 2,232 Hz, or 2,240 Hz plus 8 Hz i.e. 2,248 Hz.
If the string is shortened its frequency will increase. The beat frequency increases and hence
the string’s frequency was initially greater than 2,240 Hz, i.e. it was equal to 2,248 Hz. If the
string is shortened from 71 cm to 70.25 cm, its fundamental frequency will change to 2,248 Hz
71
times 70.25
i.e. 2,272 Hz, and the beat frequency will therefore change to 2,272 Hz minus 2,240
Hz i.e. 32 Hz.
7. When a closed pipe vibrates with 5 nodes between its two ends (not counting the node at one
end) it has a frequency of 3,960 Hz. This closed pipe then vibrates with 2 nodes between its
two ends (not counting the node at one end), and its frequency is equal to the frequency of an
open pipe which is vibrating with 8 antinodes between its two ends (not counting the antinodes
at both ends). What is the frequency of the 13th line from the left in the spectrum of the open
pipe when it vibrates at its fundamental frequency? (The spectrum shows the fundamental
frequency and harmonics of the open pipe as vertical lines with the frequency increasing from
left to right.)
(a) 2,200 Hz.
(b) 2,288 Hz.
(c) 2,925 Hz.
(d) None of the above.
Answer: (d) Since the closed pipe is vibrating with 5 nodes, it is at its 11th harmonic and its
fundamental frequency is hence given by 3,960 Hz divided by 11 i.e. 360 Hz. When this closed
pipe vibrates with 2 nodes, it will be at its 5th harmonic and its frequency is given by 360 Hz
times 5 i.e. 1,800 Hz. Since the open pipe has 8 antinodes, it will have 9 nodes and is therefore
3
at its 9th harmonic. Its fundamental frequency is therefore equal to 1,800 Hz divided by 9 i.e.
200 Hz. The 13th line from the left in the spectrum of this open pipe when it is vibrating at
its fundamental frequency is its 13th harmonic, whose frequency is given by 200 Hz times 13
i.e. 2,600 Hz.
8. A violinist is tuning the E string of her violin with the aid of an electronic tuner. The tuner
is producing a musical note which has a frequency of 660 Hz, and the violinist then uses her
violin bow to play the E string of her violin without placing her finger on the fingerboard of the
violin. In this way, the E string is played as an open string to enable its full length to vibrate.
The 660 Hz note from the electronic tuner and the note from the violin’s open E string are
heard to produce beats of 3 Hz. The violinist then tightens the E string while still playing
the E string with her bow, and the beats then decrease in frequency to 2 Hz (without passing
through 0 Hz). What was the frequency of the violin’s D string when the beat frequency was
3 Hz? (Assume that all the strings of the violin were tuned in Just fifths in relation to each
other as for a normal violin, when the beat frequency was 3 Hz.)
(a) 292 Hz.
(b) 293.3 Hz.
(c) 294.7 Hz
(d) None of the above.
Answer: (a) When beats of 3 Hz were heard, the frequency of the E string of the violin was
either 660 Hz minus 3 Hz i.e. 657 Hz, or 660 Hz plus 3 Hz i.e. 663 Hz. When the violinist
tightened the E string, its frequency would have increased. As the beat frequency was heard
to decrease to 2 Hz, the frequency of the E string was therefore initially below 660 Hz, and its
frequency was thus equal to 657 Hz. Since the frequency of the violin’s D string is two Just
fifths below the frequency of the E string, its frequency is equal to 657 Hz divided by 32 twice,
which is the same as 657 Hz multiplied by 32 twice i.e. 292 Hz.
9. A piano is very slightly flat, and the piano’s strings are tuned relative to each other in agreement
with the Equal-tempered scale. A pianist then plays the note C3 on this piano, and the 5th
harmonic of this C3 note combines with the note played on the E string of a violin by a violinist
to produce beats of 8 Hz. The A string of the violin is tuned to 440 Hz and all the strings of
the violin are tuned relative to each other in Just fifths as is normal for a violin. Which of the
following is closest to the fundamental frequency of the A2 note on the piano? (Assume that
the Equal-tempered semitone has a ratio equal to 1.05946.)
(a) 109.65 Hz.
(b) 112.34 Hz.
(c) 219.31 Hz.
(d) 224.69 Hz.
Answer: (a) Since the violin’s A string which is the A4 note is tuned to a frequency of 440
Hz, the frequency of its E string which is the E5 note is equal to 440 Hz times 32 i.e. 660 Hz.
The piano is very slightly flat and since the beat frequency is 8 Hz, the 5th harmonic of the
piano’s C3 note has a frequency equal to 660 Hz minus 8 Hz i.e. 652 Hz. The fundamental
frequency of the piano’s C3 note is thus given by 652 Hz divided by 5 i.e. 130.4 Hz. The A2
note on the piano is 3 Equal-tempered semitones below the C3 note, and hence the frequency
of the A2 note is equal to 130.4 Hz divided by 1.05946 three times i.e. approximately 109.65
Hz.
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10. A closed pipe which is vibrating with 2 nodes between its two ends (not counting the node
at one end) forms beats of 5 Hz with an open pipe vibrating with 4 antinodes between its
two ends (not counting the antinodes at both ends). What are the two possible values of the
fundamental frequency of the closed pipe, if the fundamental frequency of the open pipe is 390
Hz?
(a)
(b)
(c)
(d)
311 Hz or 313 Hz.
389 Hz or 391 Hz.
518.3 Hz or 521.7 Hz.
None of the above.
Answer: (b) Since the open pipe has 4 antinodes it will have 5 nodes and it will be vibrating
at its 5th harmonic. If its fundamental frequency is 390 Hz, its frequency of vibration will be
equal to 390 Hz times 5 i.e. 1,950 Hz. Since the beat frequency is 5 Hz, the frequency of the
closed pipe is either 1,950 Hz minus 5 Hz i.e. 1,945 Hz, or 1,950 Hz plus 5 Hz i.e. 1,955 Hz.
The closed pipe has 2 nodes and is thus at its 5th harmonic, and its fundamental frequency is
thus either 1,945 Hz divided by 5 i.e. 389 Hz, or 1,955 Hz divided by 5 i.e. 391 Hz.
11. Which of the following is the most suitable description of a Cristofori piano?
(a)
(b)
(c)
(d)
Wind instrument.
MIDI instrument.
Percussion instrument.
Electronic instrument.
Answer: (c) A Cristofori piano is not a wind instrument, and it is also certainly not a MIDI
or electronic instrument, having been built long before both MIDI or electronics existed. But
it is definitely a percussion instrument since its hammers strike the strings.
12. When one of the keys on its keyboard is struck downwards, the action on a Cristofori piano
enables the corresponding hammer to strike the corresponding string thus producing a musical
note. The first and second levers of the action for a particular key on the keyboard amplify
the distance moved by the effort by factors of 1 and 2.5 times respectively. The first, second
and third levers acting together cause the corresponding hammer to rise upwards with a speed
of 59.64 cm per second when the corresponding key is struck downwards with a speed of 4.2
cm per second. The second lever is then modified so that it amplifies the distance moved by
2.8 times instead of 2.5 times. What would be the speed of the hammer when it rises upwards
if the key is struck downwards with a speed of 4.5 cm per second?
(a)
(b)
(c)
(d)
59.64 cm per second.
66.7968 cm per second.
71.568 cm per second.
None of the above.
Answer: (c) The three levers acting together have a combined multiplication factor given by
the speed of the hammer divided by the speed of the key. This multiplication factor is equal
to 59.64 cm per second divided by 4.2 cm per second i.e. 14.2 times. Therefore the third lever
by itself amplifies the distance moved by a factor equal to 14.2 times divided by 2.5 i.e. 5.68
times. After the second lever is modified so that it amplifies the distance moved by 2.8 times
instead of 2.5 times, the multiplication factor for the three levers acting together is given by 2.8
times 5.68 i.e. 15.904 times. If the key moves downwards with a speed of 4.5 cm per second,
the hammer will move upwards with a velocity given by 4.5 cm per second times 15.904 i.e.
71.568 cm per second.
5
13. A standard grand piano has three pedals which function as usual for a grand piano. These
are a soft pedal on the left, a sostenuto pedal in the middle, and a sustain or loud pedal on
the right. A chord is played by a pianist on the grand piano who depresses the appropriate
keys on the keyboard, and also depresses one of the three pedals. Of the following sequence of
actions, which sequence would not sustain the sound of the chord?
(a) He depresses the sostenuto pedal, then depresses the keys and then releases the keys,
keeping the sostenuto pedal depressed.
(b) He depresses the keys, then depresses the sostenuto pedal and then releases the keys,
keeping the sostenuto pedal depressed.
(c) He depresses the sustain pedal, then depresses the keys, then releases the keys and keeps
the sustain pedal depressed.
(d) He depresses the sustain pedal, then depresses the keys, then releases the sustain pedal
and keeps the keys depressed.
Answer: (a) If he depresses the sostenuto pedal after the keys are depressed, the dampers will
be kept lifted off only the corresponding strings even after the keys have been released, as long
as the sostenuto pedal is kept depressed. If the keys are depressed after the sostenuto pedal
is depressed this will not happen and hence the chord will not be sustained. Since the sustain
pedal lifts all the dampers off the strings as long as it is depressed, the chord will be sustained
even when the keys have been released, but the sustain pedal has to remain depressed to keep
the chord sustained. If the keys remain depressed, the chord will be sustained irrespective of
any pedal action.
14. An electronic keyboard which only has a MIDI in IK and a MIDI out OK sends MIDI messages
to a notebook computer, through a MIDI interface box with a MIDI in IB and a MIDI out
OB, to enable a piece of music to be composed on the computer. The piece of music is then to
be played on the electronic keyboard and a tone generator which has a MIDI in IG, a MIDI
out OG and a MIDI thru TG. Which of the following connections is not a proper connection
for the notebook computer, electronic keyboard and tone generator to function as required?
(a)
(b)
(c)
(d)
OK to IB.
OB to IK.
OB to IG.
TG to IK.
Answer: (b) The notebook computer will receive MIDI messages from the electronic keyboard’s MIDI out OK through the MIDI interface box’s MIDI in IB. To play the piece, MIDI
messages should be sent from the notebook computer through OB to the electronic keyboard
and the tone generator. These MIDI messages should be sent first to the tone generator’s
MIDI in IG and then sent out of the tone generator’s MIDI thru TG into the MIDI in IK of
the electronic keyboard. It is not possible to connect the electronic keyboard directly to the
MIDI interface box as the electronic keyboard does not have a MIDI thru to send the MIDI
messages to the tone generator.
15. A notebook computer sends a MIDI message to an electronic synthesiser. This MIDI message
is supposed to instruct the electronic synthesiser to turn off the musical note D6 in the lowest
numbered MIDI channel as slowly as possible. What is the correct sequence of numbers (in
decimal form) in this MIDI message?
(a) 8, 0, 86, 127.
(b) 8, 0, 74, 0.
6
(c) 9, 0, 86, 0.
(d) None of the above.
Answer: (d) The first number of the MIDI message (in decimal) is 8, telling the electronic
synthesizer to turn a note off; the second number is 0, telling the electronic synthesizer that
the MIDI message is for the lowest numbered MIDI channel; the third number is 86, telling
the synthesizer that the note to be turned off is D6; and the fourth number is 0, telling the
electronic synthesizer to turn off the note as slowly as possible.
16. The frequency of the 8th line from the left in the spectrum of a square wave is 2,520 Hz. A low
pass filter removes all the harmonics of the square wave above the frequency 2,500 Hz from the
square wave spectrum, and a high pass filter removes all the harmonics below the frequency
1,500 Hz. Which of the lines in the square wave spectrum are left remaining after the square
wave has passed through the low pass and high pass filters?
(a) Only the 4th, 5th, 6th and 7th lines from the left.
(b) Only the 6th, 7th and 8th lines from the left.
(c) Only the 5th, 6th and 7th lines from the left.
(d) None of the above.
Answer: (c) The square wave only has odd harmonics in its spectrum and hence the 8th line
from the left in its spectrum is its 15th harmonic. The fundamental frequency of the square
wave is thus equal to 2,520 Hz divided by 15 i.e. 168 Hz. The 1st, 2nd, 3rd, 4th, 5th, 6th,
7th, 8th, 9th, 10th and 11th lines in the square wave’s spectrum are its fundamental frequency
and its 3rd, 5th, 7th, 9th, 11th, 13th, 15th, 17th, 19th and 21st harmonics respectively. These
harmonics have the following frequencies respectively: 168 Hz, 504 Hz, 840 Hz, 1,176 Hz, 1,512
Hz, 1,848 Hz, 2,184 Hz, 2,520 Hz, 2,856 Hz, 3,192 Hz and 3,528 Hz respectively. The low pass
filter removes all the frequencies of the square wave which are above 2,500 Hz, leaving the 13th
and lower harmonics. The high pass filter removes all the harmonics which are below 1,500
Hz, leaving only the 9th, 11th and 13th harmonics i.e. the 5th, 6th and 7th lines from the left
will remain in the spectrum.
17. A well-established technique of waveform synthesis generates a required waveform by producing its required harmonic using only two starting simple sine waveforms, with one waveform
affecting the frequency of the other waveform. What is this technique of waveform synthesis
usually known as?
(a) Additive synthesis.
(b) Physical modelling synthesis.
(c) Amplitude Modulation synthesis.
(d) Frequency Modulation synthesis.
Answer: (d) In this technique, one waveform modifies or modulates the frequency of the other
waveform, and hence this technique is usually known as frequency modulation synthesis.
18. A musical waveform is being synthesised by FM synthesis, with the frequency of the carrier
waveform equal to 16,800 Hz and the frequency of the modulator waveform equal to 345 Hz.
Of the following frequencies, which one is a valid frequency of one of the harmonics in the
spectrum of the musical waveform which is being synthesized?
(a) 19,915 Hz.
(b) 17,135 Hz.
7
(c) 16,465 Hz.
(d) 15,765 Hz.
Answer: (d) Since the carrier frequency is 16,800 Hz and the modulator frequency is 345 Hz,
the harmonics generated are evenly separated and spaced out from the carrier frequency by
multiples of 345 Hz. Therefore the frequencies of the 9 harmonics which are just below 16,800
Hz are 16,455 Hz, 16,110 Hz, 15,765 Hz, 15,420 Hz, 15,075 Hz, 14,730 Hz, 14,385 Hz, 14,040 Hz
and 13,695 Hz. The frequencies of the 9 harmonics which are just above 16,800 Hz are 17,145
Hz, 17,490 Hz, 17,835 Hz, 18,180 Hz, 18,525 Hz, 18,870 Hz, 19,215 Hz, 19,560 Hz and 19,905
Hz. The frequency 15,765 Hz is the only one which is a frequency of one of the harmonics in
the FM spectrum of the waveform.
19. The digital broadcast of a live concert by the NUS Symphony Orchestra has a sampling rate of
39,680 samples per second. According to the Nyquist criterion, what is the highest frequency
which can be preserved in this digital broadcast?
(a) 19,840 Hz.
(b) 39,680 Hz.
(c) 79,360 Hz.
(d) None of the above.
Answer: (a) According to the Nyquist theorem or criterion, the sampling frequency of the
digital broadcast should be double that of the highest frequency to be preserved in the digital
broadcast. Since the sampling frequency of this digital broadcast is 39,680 samples per second,
half of this frequency which is 19,840 Hz is the highest frequency which can be preserved in
the digital broadcast.
20. When the digitization process of a digital recording takes place, the bit length of the digital
samples produced determines the signal-to-noise (S/N) ratio of the digital recording. This is
due to the fact that the number of quantization levels of the digital recording which directly
determines the S/N ratio depends on the bit length. In a certain digital recording, if the S/N
ratio is 78 dB, how many quantization levels are there in this digital recording? (Assume that
for each bit of the bit length, 6 dB is added to the S/N ratio.)
(a) 4,096 quantization levels.
(b) 8,192 quantization levels.
(c) 16,384 quantization levels.
(d) None of the above.
Answer: (b) The bit length of the digital samples is given by 78 dB divided by 6 i.e. 13 bits.
The number of quantization levels in this digital recording is thus given by 2 to the power of
13, which is equal to 8,192. Therefore the digital recording has 8,192 quantization levels.
21. A digital broadcast of a concert by the Singapore Symphony Orchestra is being transmitted
over a microwave link, preserving audio frequencies in the broadcast up to 20,800 Hz. If this
microwave link can transmit at most 924,000 bits per second, what is the best possible signalto-noise (S/N) ratio in this digital broadcast? (Assume that the digital broadcast is in stereo
sound, with two audio channels of equal bit rates in the broadcast. Assume also that each bit
of the sample bit length contributes 6 dB to the S/N ratio.)
(a) 54 dB.
(b) 60 dB.
8
(c) 72 dB.
(d) None of the above.
Answer: (d) By the Nyquist criterion or theorem, the sampling rate of the digital broadcast
is equal to 20,800 Hz times 2 i.e. 41,600 samples per second. The broadcast is in stereo, so
for each of the two audio channels, the bit rate is half of 924,000 bits per second i.e. 462,000
bits per second, and the bit length of the digital samples is equal to 462,000 bits per second
divided by 41,600 samples per second i.e. approximately 11.11 bits. The bit length is therefore
equal to 11 bits as it must be an integer, and 12 bits would give a bit rate above the maximum
allowed of 462,000 bits per second. The S/N noise ratio is thus given by 6 dB times 11 i.e. 66
dB.
22. A digital broadcast of a rock concert at the Indoor Stadium is being transmitted over the
Internet, and the digital broadcast can transmit at most only 970,984 bits per second. It is
desired that the signal-to-noise (S/N) ratio of this digital Internet broadcast be at least 82
dB. What is the highest frequency which can be preserved in the digital Internet transmission
of the rock concert? (Assume that the digital transmission of the concert is in stereo sound,
with two audio channels of equal bit rates. Assume also that each bit of the sample bit length
contributes 6 dB to the S/N ratio.)
(a)
(b)
(c)
(d)
17,339 Hz.
18,673 Hz.
34,678 Hz.
None of the above.
Answer: (a) The maximum bit rate for each audio channel of the digital transmission over
the Internet is 970,984 bits per second divided by 2 i.e. 485,492 bits per second. The bit
length of the digital samples of the digital transmission must be equal to 82 dB divided by
6 i.e. approximately 13.67 bits,and hence the bit length which has to be an integer must be
14 bits, since a bit length of 13 bits would give a S/N ratio which is below 82 dB. Therefore
the sampling frequency for each channel is equal to 485,492 bits per second divided by 14 bits
i.e. 34,678 samples per second. Hence the highest frequency which can be be preserved in the
transmission is half of this, i.e. 17,339 Hz.
23. A heavy metal concert is being digitally transmitted over a satellite link. The signal-to-noise
(S/N) ratio of this digital transmission is 78 dB, and its highest audio frequency being preserved
is 17,563 Hz. It so happens that the bit rate being produced by this digital transmission is
exactly the maximum bit rate which the satellite link can transmit. The following week, a
string orchestra concert is digitally transmitted over the same satellite link, and the highest
frequency to be preserved in this string orchestra concert is 18,560 Hz. What is the best
signal-to-noise (S/N) ratio which the digital transmission of the string orchestra concert can
have? (Assume that both concerts are being transmitted in stereo sound i.e. with two audio
channels of equal bit rates. Assume also that each bit of the sample bit length contributes 6
dB to the S/N ratio.)
(a)
(b)
(c)
(d)
66 dB.
78 dB.
84 dB.
None of the above.
Answer: (d) According to the Nyquist criterion, for the heavy metal concert the sampling
rate is equal to 17,563 Hz times 2 i.e. 35,126 samples per second. The bit length of the digital
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samples must be given by 78 bits divided by 6 i.e. 13 bits. For one audio channel, the bit
rate is given by 35,126 samples per second times 13 bits i.e. 456,638 bits per second. For each
channel of the string orchestra concert transmission, this is also the maximum bit rate. The
sampling rate for the string orchestra concert is 18,560 Hz times 2 i.e. 37,120 samples per
second, and hence for the string orchestra concert transmission the sample bit length is equal
to 456,638 bits per second divided by 37,120 samples per second i.e. approximately 12.30 bits.
Therefore the bit length must be 12 bits, because a bit length of 13 bits would give a bit rate
exceeding the maximum bit rate of 456,638 bits per second for each channel. The S/N ratio
for the string orchestra concert is therefore given by 12 times 6 dB i.e. 72 dB.
24. A gamelan orchestra concert to be broadcast on the radio is to be recorded on the hard disk
of your notebook computer. The highest frequency to be preserved in the digital recording
of the concert is 15,840 Hz. If your notebook computer has exactly 325,600,000 bytes for the
digital recording to be stored, calculate the highest possible signal-to-noise (S/N) ratio which
the digital recording of the gamelan concert can achieve, assuming that you must record at
least 45 minutes of the concert. (Assume the recording is in stereo i.e. there are two audio
channels to be recorded, and that each audio channel has the same bit rate. Assume also that
each bit of the sample bit length contributes 6 dB to the S/N ratio.)
(a)
(b)
(c)
(d)
84 dB.
90 dB.
96 dB.
None of the above.
Answer: (b) Since the highest frequency to be preserved is 15,840 Hz, according to the
Nyquist criterion the sampling frequency must be double this frequency i.e. 31,680 samples per
second. You would like to record at least 45 complete minutes of the concert on your notebook
computer, so you need to store at least 45 times 60 times 31,680 samples i.e. 85,536,000 samples
on the hard disk of the computer. However, there are only 325,600,000 bytes on the hard disk,
which in terms of bits is equal to 325,600,000 times 8 bits i.e. 2,604,800,000 bits. Since each
of the two audio channels can store half of this i.e. 1,302,400,000 bits, the bit length of the
digital samples is given by 1,302,400,000 bits divided by 85,536,000 samples i.e. approximately
15.23 bits. However, bit length should be an integer which means that the bit length should
be 15 bits. This is because a bit length of 16 bits would require more bits to be stored on the
hard disk than the available capacity of 1,302,400,000 bits per channel. If the bit length is 15
bits, the S/N ratio must be equal to 15 times 6 dB i.e. 90 dB.
25. In the digital stereo recording of a band performance, the highest frequency to be preserved
in the recording is 18,000 Hz. The digital stereo recording is then compressed into an MP3
file whose bit rate is 192,000 bits per second. If the compression ratio of the conversion from
the file of the original digital recording to the MP3 file was 5.25 to 1, what was the signal-tonoise (S/N) ratio achieved by the original digital recording? (Assume that the concert is being
recorded in stereo with two channels of equal bit rate, and assume also that each bit of the
sample bit length contributes 6 dB to the S/N ratio.)
(a)
(b)
(c)
(d)
72 dB.
78 dB.
84 dB.
None of the above.
Answer: (c) If the compression ratio was 5.25 to 1, this means that the uncompressed bit
rate of the digital stereo file was equal to 192,000 bits per second times 5.25 i.e. 1,008,000 bits
10
per second. Hence the bit rate for each audio channel of the stereo recording must be equal to
half of this i.e. 504,000 bits per second. If the highest frequency to be preserved in the digital
recording was 18,000 Hz, the sampling rate of the digital recording must have been equal to
double this frequency i.e 36,000 samples per second. The bit length of the digital samples must
thus be given by 504,000 bits per second divided by 36,000 samples per second i.e. 14 bits.
The S/N ratio is thus equal to 14 times 6 dB i.e. 84 dB.
END OF TEST PAPER
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