How to calculate interaural phase difference and interaural time difference ?

[ttmike42]

How to calculate interaural phase difference and interaural time difference ?

Here is what I am given:
A sound reaches one ear with a mean phase of 338 degrees and vector strength of 0.67
It reaches the other ear with a phase of 294 degrees and vector strength of 0.60

The sound frequency is 4409 hz.

I need to find out what is the best interaural phase difference and interaural time difference of the neuron that is being looked at.



I'd also love it if someone could explain the recurring peaks shown in this plot:

Also, what would the above curve look like if a frequency of 5102hz was used instead? (assume the neuron responds equally well to each frequency)


Lastly, (I don't even understand what this question is asking, so the more explanation you could give, the better...) let's assume the above figure has a best frequency of 4409hz. Draw the ITD curve that you'd expect from using a broadband stimulus in which all frequencies within the audible range of the owl were present (hint: the basilar membrane acts like a Fourier transformer.) Draw the ITD curve that would result if the positive and negative terminals of the inputs to the headphone on the right side were reversed. Label your diagram clearly to indicate the consequences. This polarity reversal will invert the sine wave by 180 degrees in the right headphone.

I have no idea what is going on with any of this and just need to get through this question... so any help you could offer would be AWESOME!

Thanks in advance...

 

[jbsteele]

Interaural Phase Difference: The interaural phase difference is the difference between the mean phases of the two ears. In this case, it would be 338-294=44 degrees.Interaural Time Difference: The interaural time difference is the difference in the time it takes for the sound to reach each ear. This can be calculated by multiplying the frequency by the phase difference. In this case, 4409Hz x (44/360) = 24.7 ms.Recurring Peaks: The recurring peaks shown in the plot represent the neuronal response to the sound frequency. The neurons fire in response to the frequency, creating peaks that correspond to the frequency.If a frequency of 5102Hz was used instead, the plot would look similar but would have peaks corresponding to the new frequency.ITD Curve with Broadband Stimulus: The ITD curve with a broadband stimulus would depict the neuron's response to all frequencies within the audible range of the owl. The curve would include peaks corresponding to each frequency as well as a broad peak representing the combined response of all the frequencies.ITD Curve with Reversed Polarity: The ITD curve with reversed polarity would be similar to the one with the broadband stimulus, but with all the peaks inverted by 180 degrees. The peak corresponding to the 4409Hz frequency would be shifted by 180 degrees, while the other peaks would remain unchanged.

 

[nlightner]

To calculate the interaural phase difference, first we need to understand what phase and vector strength mean in this context. Phase refers to the position of a sound wave within a cycle, measured in degrees. Vector strength is a measure of how well a neuron responds to a specific frequency, with a higher value indicating a stronger response.

In this case, the sound reaches one ear with a mean phase of 338 degrees and a vector strength of 0.67, and the other ear with a phase of 294 degrees and a vector strength of 0.60. This means that the sound is stronger and reaches its peak earlier in the cycle in the first ear, compared to the second ear.

To calculate the interaural phase difference, we need to find the difference between the mean phase values of the two ears. In this case, it would be 338 degrees - 294 degrees = 44 degrees. This means that the sound reaches the first ear 44 degrees earlier in the cycle compared to the second ear.

To calculate the interaural time difference, we need to find the difference in the time it takes for the sound to reach each ear. This can be calculated using the formula: time difference = interaural phase difference / (frequency / 360). In this case, the frequency is 4409 Hz, so the time difference would be 44 degrees / (4409 Hz / 360) = 0.032 seconds.

The recurring peaks shown in the plot are due to the sound reaching the ears at regular intervals, which corresponds to the frequency of the sound. As the sound waves reach each ear at different times, the peaks appear at different locations on the plot.

If a frequency of 5102 Hz was used instead, the plot would look similar but with peaks appearing at a shorter interval, as the sound waves would reach the ears more frequently.

For the last part of the question, we are asked to draw an ITD curve for a broadband stimulus, where all frequencies within the audible range of the owl are present. This means that the sound would reach both ears at the same time, resulting in an ITD of 0 seconds. The curve would be a straight line at 0 seconds.

If the positive and negative terminals of the inputs to the headphone on the right side were reversed, the polarity of the sound waves would be inverted. This would result in an ITD curve that is the mirror image of the original curve,