MMB-MusicPerception_1a_Acoustics

Question 1

1. What’s (auditory) perception for?

Answer 1

• 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)

Question 2

2. What is Sound?

Answer 2

• 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

Question 3

3. Where does sound come from?

Answer 3

• 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)

Question 4

4. How does sound propogate?

Answer 4

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

Question 5

5. Draw a model of The spring-and-golf-ball model of sound propagation

Answer 5

• Sound waves are longitudinal waves where air molecules travel in the direction of the wave

Question 6

6. A sine wave is characterised by which three measurements?

Answer 6

• A: (Peak) Amplitude of the pressure variation
• Frequency (or period) of the occilation
• Phase

Question 7

7. Give the wave function to describe amplitude variation over time

Answer 7

• 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

Question 8

8. Describe Frequency (f):

Answer 8

• 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

Question 9

9. If a sine wave has a frequency of 50Hz how many ms per cycle?

Answer 9

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


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

Question 10

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

Answer 10

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

λ= c / f

Question 11

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

Answer 11

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

Question 12

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

Answer 12

Frequency of 440Hz = 440 cycles per second

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

Question 13

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

Answer 13

frequency = 1/period

1000/100 = 10Hz

Question 14

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

Answer 14

frequency = 1/period

1000/15 = 66.67Hz

Question 15

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

Answer 15

frequency = 1/period

               1000/10 = 100Hz

Question 16

16. Calculate the frequency of a wave with period 5ms

Answer 16

frequency = 1/period

1000/5 = 200Hz

Question 17

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

Answer 17

λ = wavespeed/frequency

λ = wavespeed/100Hz

λ = (344m/s)/100Hz = 3.44m/cycle

Question 18

18. Calculate the wavelength of a frequency at 800Hz

Answer 18

λ = wavespeed/frequency

λ = wavespeed/800Hz

λ = (344m/s)/800Hz = 0.43m/cycle

Question 19

19. Calculate the wavelength of a frequency at 18kHz

Answer 19

λ = wavespeed/frequency

λ = wavespeed/800Hz

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

Question 20

20. 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

Answer 20

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

Question 21

21. Describe Amplitude:

Answer 21

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

Question 22

22. What are the steps to compute RMS amplitude?

Answer 22

– 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

Question 23

23. State some key points about sound intensity

Answer 23

• 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

Question 24

24. How much greater is the sound pain threshold than sound at absolute threshold hearing?

Answer 24

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

Question 25

25. Describe the decibel scale.

Answer 25

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

Question 26

26. Explain how the decibel scale works.

Answer 26

* 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.

Question 27

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

Answer 27

10×log₁₀ (0.00001/0.000000000001) = 10×log₁₀ (10^-5/10^-12) = 10×log₁₀ (10⁷) = 10x7 **= 70dB**

Question 28

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

Answer 28

10×log₁₀ (0.05/0.000000000001) = 10×log₁₀ (0.05/10^-12) = 10×log₁₀ (5¹⁰) = 10x20.69897 **= 106.99dB**

Question 29

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

Answer 29

10×log₁₀ (0.0002/0.000000000001) = 10×log₁₀ (0.0002/10^-12) = 10×log₁₀ (8.30102996) = 10x8.30102996 **= 83.01dB**

Question 30

30. List SPL frequency trhesholds.

Answer 30

Question 31

31. Phase =

Answer 31

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

Question 32

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

Answer 32

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

Question 33

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

Answer 33

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

Question 34

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

Answer 34

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

Question 35

35. Explain Sinusoids vs Complex Tones

Answer 35

* 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.

Question 36

36. What is Fourier Analysis?

Answer 36

* **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.

Question 37

37. Describe and draw different frequency spectra.

Answer 37

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

Question 38

38. What is the missing fundamental?

Answer 38

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.

Question 39

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

Answer 39

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)

Question 40

41 What are** aperiodic** sounds?

Answer 40

* 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