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MMB > MMB-MusicPerception_1a_Acoustics > Flashcards

MMB-MusicPerception_1a_Acoustics Flashcards

Understand the basics of music perception

1
Q
  1. What’s (auditory) perception for?
A
  • Sample current information from the environment
  • Integrate together with context and long-term knowledge (e.g. to interpret ambiguous information)
  • Complement concurrent information from other senses
  • Construct a model of the world which we can act upon
  • Communication and expression (speech, music)
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2
Q
  1. What is Sound?
A
  • Air consists of molecules colliding constantly and randomly
            ⇒static pressure, uniform in all directions
  • Pressure depends on
         – Density of medium (air: 100,000 N/m squared)

     – Temperature
  • Sound: Variations in pressure in a medium
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3
Q
  1. Where does sound come from?
A
  • Interaction of an event and an object that results in pressure variations in the air (and hence at our eardrum)
      Example:

  – tuning fork (object) being struck (event)
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4
Q
  1. How does sound propogate?
A

The movement produces alternate condensation and rarefaction of air molecules causes
– travelling wave of air pressure

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5
Q
  1. Draw a model of The spring-and-golf-ball model of sound propagation
A
  • Sound waves are longitudinal waves where air molecules travel in the direction of the wave
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6
Q
  1. A sine wave is characterised by which three measurements?
A
  • A: (Peak) Amplitude of the pressure variation
  • Frequency (or period) of the occilation
  • Phase
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7
Q
  1. Give the wave function to describe amplitude variation over time
A
  • x(t) = Asin(2πft +φ)

A: peak amplitude
f: Frequency in Hertz t: Time in seconds

t: Time in seconds

φ: Phase in degrees or multiples of π

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8
Q
  1. Describe Frequency (f):
A
  • number of cycles of condensation and rarefaction per second (Hz)
  • related to the percept of pitch (more cycles per unit time = higher pitch)
  • Related: Period = 1/f => time taken to complete one wave cycle, measured in seconds per cycle
  • Related: wave lengths (λ) => distance covered by one complete cycle, measured in metres
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9
Q
  1. If a sine wave has a frequency of 50Hz how many ms per cycle?
A

Sine wave with frequency of 50Hz = 50 cycles per second


Period: 1/50cycles per second = 1000ms/50cycles = **20ms/cycle **

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10
Q
  1. State the equation for finding Wave length (λ)
A

• Wave length (λ) is speed of sound (c) divided by frequency (f).

λ= c / f

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11
Q
  1. What is the wave length for sine wave with 50Hz frequency at 20oC (i.e. speed of sound 344m per second)
A

Wave length (λ) is speed of sound (c) divided by frequency (f).

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12
Q
  1. Exercise 1: Calculate the period of a wave with frequency 440Hz
A

Frequency of 440Hz = 440 cycles per second

Period: 1/440cycles per second = 1000ms/440cycles = 2.27ms per cycle

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13
Q
  1. Calculate the frequency of a wave with period 100ms
A

frequency = 1/period

1000/100 = 10Hz

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14
Q
  1. Calculate the frequency of a wave with period 15ms
A

frequency = 1/period

1000/15 = 66.67Hz

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15
Q
  1. Calculate the frequency of a wave with period 10ms
A

frequency = 1/period

               1000/10 = 100Hz
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16
Q
  1. Calculate the frequency of a wave with period 5ms
A

frequency = 1/period

1000/5 = 200Hz

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17
Q
  1. – Calculate the wave length of a wave with a frequency of 100Hz
A

λ = wavespeed/frequency

λ = wavespeed/100Hz

λ = (344m/s)/100Hz = 3.44m/cycle

18
Q
  1. Calculate the wavelength of a frequency at 800Hz
A

λ = wavespeed/frequency

λ = wavespeed/800Hz

λ = (344m/s)/800Hz = 0.43m/cycle

19
Q
  1. Calculate the wavelength of a frequency at 18kHz
A

λ = wavespeed/frequency

λ = wavespeed/800Hz

λ = (344m/s)/18kHz = 344m/s)/18000 = 0.0191m/cycle

20
Q
  1. The amplitude measurements for a period of a sine wave are (in an arbitrary unit): 0,7,10,7,0,-7,-10,-7

Claculate the rms amplitude

A

0,7,10,7,0,-7,-10,-7

– Square all amplitude values in time window

0,49,100,49,0,49,100,49

– Take the mean of all squared values

Mean = total / 8 = 49.5

– Take square root of that mean

rms = 7.035624

21
Q
  1. Describe Amplitude:
A

Amplitude = magnitude of pressure variation relative to atmospheric pressure => measured in Newton/m2

• related to the percept of loudness (greater amplitude = louder sound)

Measures:
– Peak or peak to peak pressure
– Root mean square (rms) amplitude

22
Q
  1. What are the steps to compute RMS amplitude?
A

– Pick a period of time window for which you want to calculate the rms amplitude

– Square all amplitude values in time window

– Take the mean of all squared values

– Take square root of that mean

23
Q
  1. State some key points about sound intensity
A
  • Sound intensity is a more common concept in psychoacoustics
  • Intensity = energy passing through unit area (m2) per second => measured in Watts/ m2
  • Intensity is proportionally related to square of rms pressure
  • Intensity is expressed relative to absolute hearing threshold
  • The reference intensity is 10-12 W/m2
24
Q
  1. How much greater is the sound pain threshold than sound at absolute threshold hearing?
A

Our ears cover huge dynamic range of intensities: Sound at pain threshold is over 10 pascals or 1 trillion times louder than sound at absolute threshold of hearing

25
Q
  1. Describe the decibel scale.
A

When the intensity of a sound is expressed in dB, this is referred to as the sound pressure level, written e.g. 10dB SPL

26
Q
  1. Explain how the decibel scale works.
A
  • Sound intensities are expressed in logarithmic units called decibels (dB) and are called sound levels or sound pressure level (SPL)
  • A dB number is the logarithm of the ratio of a given sound intensity and the reference intensity

On the decibel scale, the smallest audible sound (near total silence) is 0 dB. A sound 10 times more powerful is 10 dB. A sound 100 times more powerful than near total silence is 20 dB. A sound 1,000 times more powerful than near total silence is 30 dB.

27
Q
  1. Calculate the sound level of a sound with an intensity of 0.001 W/m2
A

10×log10 (0.00001/0.000000000001)

= 10×log10 (10-5/10-12)

= 10×log10 (107)

= 10x7

= 70dB

28
Q
  1. Calculate the sound level of a sound with an intensity of 0.05 W/m2
A

10×log10 (0.05/0.000000000001)

= 10×log10 (0.05/10-12)

= 10×log10 (510)

= 10x20.69897

= 106.99dB

29
Q
  1. Calculate the sound level of a sound with an intensity of 0.0002 W/m2
A

10×log10 (0.0002/0.000000000001)

= 10×log10 (0.0002/10-12)

= 10×log10 (8.30102996)

= 10x8.30102996

= 83.01dB

30
Q
  1. List SPL frequency trhesholds.
A
31
Q
  1. Phase =
A

The point reached on the pressure cycle at a particular time, measured in degrees or multiples of π

Example: two sinusoids have same frequency and

amplitude but are out of phase by 180 degrees, i.e. 1π

32
Q
  1. Calculate the amplitude of a sine wave of 50Hz with φ = 0 and a peak amplitude of 1 at 20ms
A

x(t) = Asin(2πft +φ) = Asin(wt +φ)

x =1sin(2*3.14*50*0.02+0)

x=1sin(6.283185307

x=1(0)

x = 0

33
Q
  1. Calculate the amplitude of a sine wave of 100Hz with φ = 0 and a peak amplitude of 1 at 1 second
A

x(t) = Asin(2πft +φ) = Asin(wt +φ)

x=1sin(2*3.14*100*0.1ms+0)

x=1sin(62.83185307)

x=1(0)

x = 0

34
Q

34 Calculate the amplitude of a sine wave of 440Hz with φ = π/2 and a peak amplitude of 2 at 0.5 seconds

A

x(t) = Asin(2πft +φ) = Asin(wt +φ)

x =2sin(2*3.14*440*0.5ms+(3.14/2))

x =2sin(1383.871564)

x =2(1)

x = 2

35
Q
  1. Explain Sinusoids vs Complex Tones
A
  • Sinusoids (pure tones) are simple.
  • Many real world sounds (speech, music) are regularly repeating (periodic) but are complex rather than sinusoidal.
  • Complex sounds: made up of a number of different sinusoids of given frequency, amplitude and phase.
36
Q
  1. What is Fourier Analysis?
A
  • Fourier analysis is a mathematical technique to derive the sinusoidal components that make up a complex waveform
  • Each component is described by its relative contribution (magnitude) and its phase
  • Every complex waveform can be decomposed into a set of sinusoids (Fourier theorem)
  • The overall waveform repeats at frequency of the lowest sinusoidal component => gives rise to the perception of a fundamental pitch.
37
Q
  1. Describe and draw different frequency spectra.
A

Sounds with the same fundamental frequency have the same pitch, even though their frequency spectra differ

38
Q
  1. What is the missing fundamental?
A

Removing the fundamental frequency does not change the percept of pitch

A harmonic sound is said to have a missing fundamental, suppressed fundamental, or phantom fundamental when its overtones suggest a fundamental frequency but the sound lacks a component at the fundamental frequency itself.

39
Q

39 A complex sound has harmonics of 800,1000 and 1200 Hz. What is its fundamental frequency?

A

Rule: fundamental is the largest integer that divides into all the harmonic components.

Fundamental frequency of the sound is 200: (800 = 200 x 4; 1000 = 200 x 5; 1200 = 200 x 6)

40
Q

41 What are** aperiodic** sounds?

A
  • Some sounds do not regularly repeat. They are non-periodic or aperiodic.
  • Decomposition gives a continuous frequency spectrum with more or less equal energy at all component frequencies
  • White Noise: Sounds where all frequency components are equally strong

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