Audition

Question 1

Frequency, amplitude and timbre can also be described as:

Answer 1

Pitch, loudness and complexity (colour)

Question 2

What is the function of the middle ear?

Answer 2

The middle era contains the eardrum (timpanic membrane) and the ossicles. When the eardrum vibrates the ossicles also vibrate, causing the stirrup (stapes is at the base of the stirrup) to push against the soft membrane of the oval window. This process amplifies the signal.

Question 3

When the stapes (of the stirrup) pushes on the soft membrane of the round window of the middle ear, what does this cause in the inner ear? How does this assist in the ultimate function of the inner ear?

Answer 3

When the stapes on the stirrup pushes against the soft membrane of the round window, this causes ripples in the liquid of the cochlear which is in the inner ear.

The inner contains the auditory receptors, a frequency analysis system, and ultimately transduces these signals into neural impulses

Question 4

Describe the tonotopic map and the movement of fluid around the basilar membrane inside the cochlear. What kind of waves does this cause?

Answer 4

Low frequencies will causes vibrations at the tip of the basilar, which is in the middle of the spiral. The fluid circulates around the basilar, which does not extend o the absolute limit inside the cochlea. The basilar is thereby vibrated from both direction as standing waves (oscillations which go up and down, not a moving wave) vibrate the fluid.

Question 5

How does the middle ear achieve impedance matching, since less than 1% of sound energy passes from air into water?

Name 2 ways that the malleus and the stapes (2 of the ossicles), amplify the signal from the eardrum. Note that the malleus doubles the signal, which then ends up being 20x louder after being transmitted to the stapes.

Answer 5

Sound is amplified in 2 ways:

1. Sound is collected from the eardrum, and then travels through the malleus, which is approximately twice as long. Because of the length and mechanical leverage, the eardrum signal is doubled.
2. This doubled signal is then transmitted to the stapes, which transmits it to the (soft) round window, which is about 10-20x smaller than the eardrum. This further amplifies the signal (assume by 10x), resulting in a total amplification between middle and inner ear of approximately 20x.

Question 6

What is the lowest frequency in a sound called, the one which would be perceived as the note being played on an instrument?

Answer 6

The lowest frequency in a complex wave form is called F0, the fundamental frequency

Question 7

On top of the fundamental frequency, how do the harmonics multiply?

Answer 7

The first harmonic, F1 is double F0, and each successive harmonic is a double of the last

Question 8

Which cilia bundle together to form the auditory nerve?

Answer 8

The inner hair cells bundle together to form the auditory nerve

Question 9

How are the cilia triggered to move?

Answer 9

Cilia are triggered by movement of the basilar membrane; it moves up and down in standing waves

Question 10

Relative to the basilar membrane, where is the tectorial membrane, and what does it contain?

Answer 10

The tectorial membrane sits on top of the basilar membrane and contains the cilia which protrude from hair cells which carry the first neural signals from the inner ear.

Question 11

What is the pattern of inner and outer hair cell rows on the basilar membrane? What is a good way of thinking about why this might be the case? (Hint - amplification)

Answer 11

There’s one row of inner hair cells (IHC) at the top and 3 rows of outer hair cells (OHC) at the bottom.

OHCs amplify low level signals which vibrate in standing waves on the basilar, whilst IHCs transduce the signals to the brain.

Question 12

Which hair cells bundle together to form the auditory nerve?

Answer 12

The IHCs bundle together to form the auditory nerve

Question 13

What is the effect of the sheering of IHC cilia, which takes place as a result of the basilar membrane’s standing waves?

Answer 13

The sheering of the cilia is transduced into neural signal when a mechanical gate opens and closes ion channels, generating neural impulses.

Question 14

We have about 15000 hair cells in the inner ear.

How many of these are IHC vs OHC?

Answer 14

We have about 3500 inner hair cells, which don’t reproduce, and about 12000 OHC.

Question 15

How do the OHCs break, causing partial hearing loss?

Answer 15

When music is too loud, the standing waves of the basilar membrane get so big that the sheering of the cilia causes them to break each other!

Question 16

Why is rate coding (also phase locking to spike timing) more relevant for lower frequencies (below 1000Hz), and place coding more relevant for higher frequencies (above 800Hz), ?

How does the Volley theory (frequency (temporal) theory) get around the limitations of phase locking?

Answer 16

Rate coding / phase locking/ spike timing = IHC spikes occur at the peak of every sine wave modulation. This is an effective theory for lower frequencies (below 1000Hz, or one spike every millisecond), because it would not be possible for higher frequencies (above 800Hz) - sine wave modulations are faster than a neuron can fire.

Place coding applies to higher frequencies because localised modulation of OHCs corresponds with tonotopic standing waves of the basilar. Cilia which are closer to the apex (centre of cochlea) code for lower frequencies than those closer to the base (outer) of the cochlea.

Frequency (temporal) theory makes sense for lower frequencies, below 1000 Hz. Phase locked signal spiking occurs in conjunction with sine wave peaks. The volley theory explains how this could still apply for higher frequencies.

Question 17

What is the Volley Theory, and how does it account for frequencies which occur at higher rapidity than neurons can fire at?

Answer 17

Not every sine wave peak will be pass threshold to cause firing of every neuron, but across the population of neurons every peak will incur a spike.

This is especially helpful for higher frequencies above 800 Hz because so many peaks are missed, due to the rapid oscillation of the wave.

Question 18

Which theories best explain frequencies of around 125 to 4000 Hz?

Answer 18

Frequencies of around 125 to 4000 Hz can be best explained by both frequency and place theory, this is the overlapping range explained by both theories.

Question 19

How does frequency theory support the missing fundamental phenomenen?

Answer 19

When F0 is removed, the peaks of the wave still exist (remaining harmonics). When we hear frequencies with missing fundamentals, it still sounds like F0 is there.

This supports the frequency theory because it’s the wave peaks which cause APs at the frequency that we hear.

Question 20

At what point do the auditory nerve fibres cross the midline, therefore becoming a ‘stereo’ signal?

Answer 20

The auditory fibres cross the midline after the cochlea nucleus.

Question 21

The auditory nerve carries tonotopic information; how is this possible?

Answer 21

The auditory nerve splays out along the cochlea, corresponding with the basilar membrane’s tonotopic standing waves (which cause IHC signals).

Each of these splayed parts samples a narrow selection along the basilar membrane.

Question 22

What is masking, and how does it select for the width of auditory filters?

What is the standard frequency range that auditory nerve fibres are tuned to?

Answer 22

To test a frequency, for instance 1000 Hz, the frequency is played at softer and softer volumes until threshold is determined.

Then white noise is introduced at a low level such as 10 bandwidths, and this is broadened gradually by adding more frequencies.

As more and more frequencies are added, broadening the bandwidth of the white noise, the frequency being tested will need to be turned up louder and louder.

The limit of the auditory filter is found when adding more frequencies doesn’t cause any loss of perception of the frequency.

For example, frequencies could be added to the mask causing a need for volume increase up until 25 Hz above and below the frequency being tested, then adding more frequencies makes no difference. This is the breadth of the auditory nerve fibre (1000 Hz +/- 25 Hz).

Auditory fibre bandwidths are tuned to about an 8th of the frequency.

Question 23

Auditory nerve fibres have very finely tuned filters. What is the usual bandwidth of a frequency?

Answer 23

The usual bandwidth is about an 8th of the frequency, which works out as 12.5% (f/8 = 12.5%). This will be spread evenly over the peak sensitivity (eg 1000 Hz), at 6.25% each side. In the example of 1000Hz that’s about 60Hz each side so about 940-162Hz

Question 24

What kind of HCs are damaged?

Loss of signal specificity due to a broadened bandwidth and reduced volume. The amplitude threshold will be higher therefore the bandwidth is broader because it’s not as pointy at the bottom.

Increased bandwidth means less selectivity; sounds are blurred and less distinct

Answer 24

OHC damage causes decreased sensitivity and broader bandwidth tuning

Question 25

What is the difference in auditory tuning curves between OHC damage, caused by oversized basilar standing waves, and IHC damage, caused by disease or infection?

Answer 25

IHC damage causes raised thresholds; meaning that we’re more sensitive to sound, but at the same time we can still appreciate different frequencies because the bandwidth remains the same. (Bandwidth encompasses audible frequencies for that tuning curve).

Question 26

What is sensori-neural hearing loss, as opposed to conductive hearing loss?

Answer 26

Sensori-neural hearing loss encompasses both IHC and OHC damage, and is associated with tinnitus and reduced increased threshold. Frequency bandwidth is also increased for OHC but not IHC damage.

Conductive hearing loss originates at the ossicles, for instance they could be too brittle, and there is little signal to transfer to the inner ear.

Question 27

What is the difference in loss of frequency range between sensori-neural VS conductive hearing loss?

Answer 27

Sensori-neural loss involves a loss mostly of higher frequencies, whereas conductive hearing loss affects all frequencies in uniform (signal from ossicles is muffled).

Question 28

What is industrial, or band-specific hearing loss?

Answer 28

Industrial hearing loss occurs as a result of too much exposure to sound at a particular frequency, often happens to tradies. The loss will occur at that specific frequency only.

Question 29

How drastic can industrial hearing loss be? Compare a 25 y/old carpenters hearing to a 50 y/olds.

Answer 29

A 25 y/old carpenters hearing will be equal to a healthy 50 y/olds.

Question 30

How many electrode channels do modern cochlear implants have?

Answer 30

Cochlear implants now have around 28 channels, previously it was 20.

Question 31

Why does the basilar membrane’s tonotopic map receive lower frequency waves at the apex and higher frequency at the tip?

Answer 31

The basilar membrane’s tonotopic map is a result of it’s physical characteristics; it’s thicker at the base where the higher frequencies are encoded and thinner at the top.

Question 32

Although sensori-neural hearing loss usually results from a combination of OHC and IHC damage, why is it mostly caused by damage to OHCs?

Answer 32

Because OHCs mechanically amplify frequencies over the tonotopic map of the basilar membrane. This is why certain frequencies are often affected more than others.

Question 33

At which point on the auditory pathway do the nerves cross over so that we have stereo sound?

What kind of information is each fibre carrying?

Answer 33

The auditory nerves cross at the cochlear nucleus, so after that point we have stereo sound.

Fibres in the auditory nerve carry frequency signals - because of their correspondence to the tonotopic basilar membrane.

Question 34

What’s the difference for the auditory tuning curve between OHC and IHC damage?

Answer 34

OHC damage causes raised threshold (need to turn volume up to detect signal), and increased bandwidth (detecting a broader range of frequencies - sound becomes ‘blurred’).

IHC damage causes raised threshold but bandwidth remains the same.

Question 35

Using an audiogram allows detection of either sensorineural versus conductive hearing loss. What would be the discerning feature of each?

Answer 35

Sensorineural hearing loss (damage to OHCs / IHCs) results in a dip in threshold mostly for the higher frequencies, whereas conductive hearing loss displays reduced threshold in all frequencies.

Question 36

Why does sound decrease in intensity over distance?

Answer 36

Sound waves become dispersed by r squared over distance; the same amount of stimulus is spread over successively louder areas (eg r, 2r, 3r or I, I/4, I/9)

Question 37

Why do low frequencies travel better than high frequencies?

Answer 37

Sound energy is absorbed by air at a higher rate for high frequencies, low waves survive for longer.

Absorption increases with frequency

Question 38

The Mel scale is a scale of perceived pitch-change steps. Equal Mel steps are perceived as equal intervals, however, this is not the case for frequency steps above 1000Hz (below about 1000Hz the Mel and Hz steps coincide and are equal - linear relation).

What happens to the Hz steps above 500Hz when the function compresses, and what does this mean about the ratio of transduction?

Answer 38

Above 500Hz the physical frequency steps increase in order to perceive the same Mel step.

This shows that transduction doesn’t have a 1:1 relationship.