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How the cochlea computes (2024)

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475 points izhak | 2 comments | 30 Oct 25 17:01 UTC | HN request time: 0.477s | source
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adornKey ◴[30 Oct 25 18:14 UTC] No.45763186[source]▶
This subject has bothered me for a long time. My question to guys into acoustics was always: If the cochlea performs some kind of Fourier transform, what are the chances, that it uses sinus waves as a base for the vector-space? - if it did anything like that it could just as good use any slightly different wave-forms as a base for transformation. Stiffness and non-linearity will for sure take care that any ideal rubber model in physics will in reality be different from the perfect sinus.
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empiricus ◴[30 Oct 25 20:52 UTC] No.45765185[source]▶
well, cochlea is working withing the realm of biological and physical possibilities. basically it is a triangle through which waves are propagating, and sensors along the edge. smth smth this is similar to a filter bank of gabor filters that respond to rising freq along the triangle edge. ergo you can say fourier, but it only means sensors responding to different freq becasue of their location.
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1. adornKey ◴[30 Oct 25 21:10 UTC] No.45765401[source]▶
Yeah, but not only the frequency is important - the wave-form is very relevant. For example if your wave-form is a triangle, listerners will tell you that it is very noisy compared to a simple sinus. If you use sinus as a base of your vector space triangles really look like a noisy mix. My question is, if the basic elements are really sinus, or if the basic Eigen-Waves of the cochlea are other Wave-Forms (e.g. slightly wider or narrower than sinus, ...). If physics in the ear isn't linear, maybe sinus isn't the purest wave-form for a listener.

Most people in Physics only know sinus and maybe sometimes rectangles as a base for transformations, but mathematically you could use a lot of other things - maybe very similar to sinus, but different.

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2. kragen ◴[31 Oct 25 06:03 UTC] No.45768860[source]▶
But if you apply a frequency-dependent phase shift to the triangle wave, nobody will be able to tell the difference unless the frequency is very low.