Sound localization

Question 1

what are sound waves?

Answer 1

pressure waves that we are embedded into ⇒ how to figure out how we can use waves to localize objects

Question 2

how is an owl skull shaped?

Answer 2

there is an enormous cavity for the owls eyes (½ head) and they have a smaller brain
- The owls must move their entire head and they have a flexible/fast neck to rotate their head 270 degrees
- This is also how they stabilize their head position to concentrate high acuity in a particular point of interest as they hunt

Question 3

if a sound is loud in one ear but not the other what does that tell you?

Answer 3

if you have two sensors (2 ears), then if a sound is particularly loud in one ear and softer in another then that suggests it is closer to the louder ear and would help localize in that dimension
- that angle is called azimuth

Question 4

what waves can our head block? what frequencies can we hear with only sound intensity?

Answer 4

your head actually blocks the wave so there is a lower intensity as a result of your head blocking it on the far ear ⇒ this only works if the wavelength is less than your head size
- for long wavelengths there is not a lot of intensity difference between the two ears so if all we used was intensity difference there would be a whole range of frequencies we could not localize because intensity difference is absent => any frequency below 1700 Hz we would not be able to localize

Question 5

what are the 2 cues for sound localization?

Answer 5

1. Binaural intensity differences
2. Binaural time differences

Question 6

Binaural intensity differences

Answer 6

mainly elevation and critical wavelength

Question 7

Binaural time differences

Answer 7

mainly azimuth microsecond resolution required ⇒ if pressure wave that gets to one ear happens prior to getting to the other ear which is a very tiny difference that we can compute
- Other cues are frequency spectrum

Question 8

how much time does it take for a sound wave to reach the far ear from one side?

Answer 8

it takes 0.2/340 which is on the order of less than a ms (100 microseconds) ⇒ this is a very short amount of time to look at interaural time differences and figure out where something is coming form
- Requires a very precise time mechanism ⇒ auditory transduction needs to be temporally precise and propagated in a precise way
- We also need a neural structure able to compute time differences on the order of microseconds

Question 9

what is the facial and auditory structuring that allows barn owls to hear?

Answer 9

barn owls have a facial disk which is a pattern of roughage that looks like an antenna or parabolic dish ⇒ designed to funnel sound waves into the aural canals under the spatial disk
- there is a difference from the sound that come from above and below when they hit the auditory canals
- One is aimed up and the other is aimed down so they can use sound differences in the two ears to compute azimuth and elevation

Question 10

how do humans account for both azimuth and elevation sound differences?

Answer 10

there needs to be some asymmetry between up and down with regard to how sound is propagated into the ears => the pinna is asymmetric so there is slight differences coming from above and below and how they are conducted in the outer ear is how we localize things in the elevation dimension

Question 11

what do the signal maps look like for measuring audio information in barn owls?

Answer 11

there are vertical lines that corresponds with a 30 microsecond difference and this is the range of positions in which there was exact synchrony in terms of microseconds to reach the same ears
- There are sets of positions toward the right around 20 degrees azimuth and a whole range of elevation ⇒ range of position for a 90 microsecond difference
- when we go to a higher frequency and due to the facial disk there is some distortion

Question 12

what happens to the line of constant interaural distances at higher frequencies?

Answer 12

The line of constant interaural distance is slightly non vertical at higher frequencies
- things are hard to analyze at higher frequencies
- the maps are a little bit irregular so the barn owl nervous system must understand these in terms of intensity and time differences to determine where an object is ⇒ this is a computational map

Question 13

what are the 4 steps in auditory processing?

Answer 13

1. Cochlear nucleus
2. Nucleus laminaris
3. Inferior colliculus –central core
4. Inferior colliculus –external shell

Question 14

what does the cochlear nucleus do?

Answer 14

–monoaural ⇒ 2 of these that have a systematic frequency map due to elasticity of the basilar membrane
- These must be combined into 2 different structures associated with the interaural time cues

Question 15

what does the ventral nucleus of the lateral lemniscus do?

Answer 15

Ventral nucleus of the lateral lemniscus looks at intensity differences in the barn owl
- Lateral superior olive is associated with intensity in primates

Question 16

what does the nucleus laminaris do?

Answer 16

The nucleus laminaris is for interaural time difference computations
- Medial superior olive in primates

Question 17

what does the central core of the inferior colliculus do?

Answer 17

the two cues get integrated here which is a binaural structure with a map of frequency and position propagated into a pure map of space

Question 18

what does the external shell of the inferior colliculus do?

Answer 18

this is where the auditory map of space is finally realized
- This gets propagated into the optic tectum for birds and the superior colliculus for humans/primates

Question 19

what does the tectum/superior colliculus do?

Answer 19

This is responsible for rapid orienting ⇒ systematic map of space and needs to have the map constructed according to vision and audition that are consistent with one another

Question 20

what is the Jeffrey model?

Answer 20

it is a delay line model in the nucleus laminaris
- soundwaves coming in toward the right and another coming toward the left and if they arrive at the same time at the ears then this means they will reach C at the same time
- The solid lines is they are actually axons that propagate activation associated with the sound signal but the action potentials take time to travel from A to B to C to D to E

Question 21

what does the position of coactive stimulation activity tell us in the Jeffrey model?

Answer 21

If you orient the system so you have sound coming in from opposite directions then the position of coactive simultaneous activity is actually going to tell you the interaural time difference which infers the position
- if a sound arrives at the right ear first that means there is a strong likelihood that it is coming from the right side ⇒ this signal propagates down and it takes a while
- Finally the sound wave reaches the other ear but by the time it reached the other ear the action potentials have traveled to a particular point so simultaneous activation is point (neuron) E ⇒ only simultaneously activates if a stimulus hits the right ear prior to left and A would be the opposite situation from leftward stimuli

Question 22

what does the Jeffrey model depend on in terms of detection?

Answer 22

depends on coincidence detection from the two ears
- and some axonal arborizations so that conduction velocity is on the order of microseconds

Question 23

where does fine processing in the nuclear laminaris happen?

Answer 23

fine processes are entering from opposite directions and interdigitating with each other
- Tracing of two neurons from cochlear nucleus (CN) to the laminaris
- there are physiological recordings of sine waves where you can look at a field potential at particular locations and see this is preferential activated for particular ipsilateral and contralateral delays ⇒ this is predicted by the Jeffreys model

Question 24

what auditory structure is responsible for intensity differences?

Answer 24

lateral superior olive
- the model for intensity differences happens in the ventral nucleus of the lateral lemniscus or in the lateral superior olive has neurons that have slightly different properties of inhibition

Question 25

what differences in dB excitation/inhibition can we see in the superior olive?

Answer 25

the ipsilateral side may have excitation and the contralateral side may be inhibition that is delayed and a different magnitude from the excitation - inhibition doesn't wipe out excitation until you get a higher stimulus => At 0 and minus 10 there is no response as inhibition has matched or exceeded excitation - This creates a cell that responds to a particular range of intensity differences by virtue of changing the timing of inhibition to a system where excitatory influences from ipsilateral side and inhibitory influences from the contralateral side are integrated