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Acoustics > Sem1Acoustics > Flashcards

Sem1Acoustics Flashcards

1
Q

What is the difference between a sawtooth and square wave?

A
  • Sawtooth is more ‘buzzy’, nicer sounding
  • Sawtooth has progressive summation of 3 components identical starting phases (of odd and integer multiples of f0)
  • odd and even integers
  • -6dB per octave of amplitude envelope
  • Square waves have progressive summation of 4 components with identical starting phases
  • Square wave: each component is an ODD integer multiple of f0
  • Square waves - add odd integer multiples to f0 if have same starting phase
  • -16dB per octave on amplitude envelope
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2
Q

Give an example of a transient sound and explain whether they are periodic or aperiodic sounds…

A

“clicks” - Aperiodic

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3
Q

For a sawtooth wave describe:

  • waveform
  • amplitude spectrum
A
  • Change in amplitude (positive to negative instantaneously) with a gradual change in time.
  • Relative amplitude decrease at a rate of 6dB per octave.
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4
Q

What is the amplitude spectrum?

A

Shows amplitude as a function of frequency. Shows all component frequencies present as well as their relative amplitudes.

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5
Q

What is an octave?

A

A doubling or halving of frequency. Frequency ratio of 2:1/1:2
i.e. •250 Hz is one octave above 125 Hz
•1000 Hz is one octave below 2000 Hz

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6
Q

How do you determine SNR?

A
  • Calculate the intensities of the level of the signal and the noise
  • Subtract the intensities to determine s/n
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7
Q

What type of waveform and spectral envelope of the VFs represent?

A

Sawtooth waveform, relatively smooth envelope, no prominences, -12dB per octave

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8
Q

Define natural frequency

A

The rate at which a system vibrates with the least amount of energy applied to it

A system will vibrate MAXIMALLY at its preferred frequency

Governed by the MASS and STIFFNESS/elasticity of the system.

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9
Q

Describe the phenomenon of resonance

A

For a given amplitude of forcing motion, the vibration of the mass is largest when the driving frequency equals fnat of the system

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10
Q

What would happen if a tuning fork was applied to a hard surface?

A

Hard surface is an elastic system, therefore it is forced to vibrate at the frequency of applied force/the same freq as the tuning fork, not at its own fnat.
The closer the freq of applied force is to the fnat of the system, the greater the amplitude of vibration- louder.

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11
Q

How do you calculate the first frequency in a resonant series? 1/4 wavelength system

A

Speed of sound / 4x Length of the tube

334m/s divided by 4x L of Tube

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12
Q

How do you add sinusoidal waves together?

A

Related to adding their pressure waves.
Dependent on whether they are correlated (same waveform - slightly different phase or starting phase) or uncorrelated (completely different.
Correlated - can ADD (+) instantaneous pressures.
e.g: time + amplitude, 1+1 = 2
A double in amplitude, a double in pressure = 6dB increase in dB SPL

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13
Q

When add two pure tones of the same frequency the resulting signal is dependent on what?

A

Phase of the component signals

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14
Q

Pure tones of different frequencies can be added together? T or F

A

True -

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15
Q

What is Fourier Analysis?

A

Decomposing of complex waves to determine the amplitude, frequency, phase of the sinusoidal component

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16
Q

Each sinusoid in a series must be an integer (whole number) multiple of the highest in the series? T or F

A

False - Whole number integer of the LOWEST in the series.

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17
Q

All sound waves can be classified with reference to which 2 things?

A
  • Is periodicity present?

- How complex is the wave?

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18
Q

A sawtooth wave is a type of sinusoidal wave T or F?

A

False

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19
Q

What are the frequencies of the first five harmonics? T = 8ms

A

T = 8ms (a period)… 8ms = 0.008SECONDS

f = 1/T …. f0 = 125Hz
125x2= 250
125x3= 375
125x4= 500 Hz
125x5= 625 Hz
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20
Q

What does the amplitude spectrum plot?

A

Amplitude as a function of frequency

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21
Q

Describe the amplitude envelope for a sawtooth wave…

A

Relative amplitude decrease of 6dB per octave of frequency.

i.e. For every doubling of frequency (octave) a 6dB decrease in amplitude

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22
Q

The waveform of VF vibration is a square wave? T or F

Line spectrum of VF vibration is a smooth envelope with no sharp prominences>

A

False - it is sawtooth in shape
Complex, periodic

True, and slope around -12dB per octave

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23
Q

What is volume velocity?

A

Particle velocity of molecules flowing through an area of 1mcubed/s

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24
Q

Broad resonance response curve is more sought after than a narrow resonance curve for speech recognition - HAs? T or F

A

True,

So it can produce frequencies low and high and everything in between.

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25
Q

Explain what is occurring when a tuning fork is pushed against a desk?

A
  • Periodic force (tuning fork) applied to an elastic system (hard surface)
  • Elastic system is forced to vibrate at frequency of applied force, not at its own fnat
  • The closer the frequency of the applied force to the fnat of the system, the greater the amplitude of vibration (loudness increases)
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26
Q

At what region do we have the greatest amplitude?

A

Compression - sound is loudest

In a closed tube system, this occurs at the opening of the tube, this is where the resonance occurs.

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27
Q

Resonant frequency is _____________to the _________ of the tube, thus the _______tube = ________ resonances

A

Resonant frequency is “inversely proportional” to the “length” of the tube – “longer” tube = “lower” resonances

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28
Q

What are types of speech sounds?

A

Complex tones
Continuous noise
Transients noise

29
Q

What is a filter?

A

An acoustical system that changes the spectrum of a sound
“frequency-selective” system - transfer function (change the input to a particular output, depending on the properties of the transfer function)
Amplitude coming out of filter / amplitude going in (at each freq)

30
Q

Gain

A

= Amplitude at output / Amplitude at input

= Pressure at output / Pressure at input

Gain (dB) = 20 log Pout/Pin

> 1 = amplification - positibe
<1 = attenuation - negative

Gain (dB) = SPLout – SPLin
*therefore SPLout = SPLin + gain (dB)

31
Q

What is the difference between the lower cut of frequency and the higher cut off frequency

A

Bandwidth (width of the filter)

32
Q

Where will the lower cut off frequency be?

A

Frequency below fC 3dB decrease compared to the fC (centre frequency)
i.e. the amplitude of the response is 3dB less than the response of the fC
“shoulders”

33
Q

Where will the upper cut off frequency be?

A

frequency above fC
3dB decrease compared to the fC (centre frequency)
i.e. the amplitude of the response is 3dB less than the response of the fC
“shoulders”

34
Q

Passband

A

Determines whether the filter is narrow or broad the bandwidth is
Difference between the lower and upper cut frequencies

35
Q

What is an ideal passband?

A

A steep cutoff (attenuation rate) - roll off

36
Q

Attenuatuon rate

A

E.g. in filter A, 500Hz it is at -20, up in one octave has resulted in a 10dB drop per octave.
Therefore, from 1000Hz to 500Hz (dropping an octave)
Also dropped 10dB in amplitude
Attenuation rate of Filter A = 10dB/per octave

37
Q

Difference between idealised and realisation filters

A

Specification of attenuation rate reveals how much the realised filter departs from the idealised (rectangular) one. There is still skirt.

38
Q

What energy does a band pass filter pass?

What are its parameters

A 
Energy between fL and fU
attenuates energy below fL and above fU
-  fL
- fU
- attenuation rate
- fC
- difference btw lower and upper frequencies
39
Q

High pass filter

A

Pass energy above fL, attenuates energy below fL

40
Q

Low pass filter

A

Passes energy below fU, attenuates energy above fU

41
Q

Describe band-reject filter

A

Rejects energy between fL and fU

42
Q

Constant percentage of bandwidth filter:
Suppose constant % is 70.7%
The fC is?

A

0.707

43
Q

The octave filter fC is…

A

∆f = 0.707 fC

The bandwidth of an octave filter is always 70.7% of the centre frequency

44
Q

Bandwidth is given by…

A

∆f = fU-fL

45
Q

The centre frequency is given by…

A

fC = fL x √2

46
Q

The octave filter

A

= √ fL x 2fL

47
Q

fL equation

A

fL = .707 x fC

48
Q

fU equation

A

fU = 1.414 x fC

49
Q

fU is always ?:? re fL (one octave)

A

2:1

50
Q

Three concepts of sound transmission

A
  • sound fades away over time
  • Sound fades away over distance
  • sound waves encounter obstacles
51
Q

When sound is propagated in a free, unbounded medium (no obstacles), intensity (watt/m2) decreases in a lawful way - is the

A

inverse square law

52
Q

If power remains constant but surface area increases, intensity must

A

decrease

53
Q

Relationship btw area and energy/intensity

A

As A increases, intensity decreases.
i.e. if A increases by 4:1
Intensity decreases by 1:4

54
Q

If there were an obstacle that a sound wave runs into, IF the inverse square law was used - how would it have predicted the sound level

A

The intensity decrease would be LESS than what inverse square law wouldve predicted

55
Q

With reference to reflection, the intensity of the reflected wave is _____ than the intensity of the incident wave at _____ distance

A

Intensity of reflected wave is “less” than the intensity of the incident wave at “equal” distance

56
Q

Convex surfaces diverge or converge sounds reflected?

A

Diverge - energy is scattered

57
Q

Concave surfaces diverge or converge sounds reflected from them?

A

Converge - energy is concentrated to one area (milk frother)

e.g. radar dishes,

58
Q

At the node it is a point of…

A

No vibration - displacement = 0

Remain the same

59
Q

At teh antinode it is a point of…

A

MAximal vibration

  • Magnitude of displacement dependent on the r/ship btw the incident and reflected waves
  • Always 1/2 length away from eachother
60
Q

When does destructive/cancellation interference occur

A

180 degrees out of phase, peaks and troughs cancel each other out

61
Q

Different frequencies have the same absorption coefficients on the same suface T or F

A

False -

E.g. carpet will absorb different frequencies than others, glass will absorb some frequencies than others

62
Q

Reverberation time

A

• Reverberation time is the time required for the sound energy to decay by 60dB (“T60”)
depends on room volume and materials present
T60 = kV/A
T60 = 0.16 x V/A

63
Q

What is the aim for a reverberation time for an imparied listener?

A

<500ms

To ensure the vowels will not impact the following consonants and other reasons

64
Q

Refraction

A

When a wave moves to another medium, or encounters a change in the medium, the speed of propagation changes and rays are bent

65
Q

Diffraction

A

bending or scattering of a sound wave around or through an obstacle

• Long wavelengths (low frequencies) are more likely to be diffracted around obstacles than shorter wavelengths

i.e. Why you hear the base of a party down the street
Can be plane progressive:

Plane progressive (flat wave front) sound waves encountering a barrier (A) or an opening in a wall (B). Some energy is reflected back, the wave fronts scatter or bend around the obstacle, then the wave fronts reform and continue as plane wave fronts

66
Q

Sound behaves in a more predictable many in the Near Field

A

No, Sound behaves in a more predictable many in the DIRECT FIELD

67
Q

Calculating the direct field?

A

dBSPL > dBHL > subtract MAF at frequency (e.g. 1kHZ - using chart)

68
Q

How to calibrate a room?

A
  1. SLM reading (dBA) Needs to be in this
    1. Subtract A-weighting to give dBSPL
    2. Subtract MAF (minimum audible field) to give dBHL.
      dBHL from SLMdBHL on audiometer?

Acoustics (2 decks)

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