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I am a biologist studying frogs, and as part of my research I use hundreds of thousands of audio recordings (.wav) from autonomous recording units. I use various software to create recognizers that pick out the frog call of interest based on acoustic pattern using Markov Chain models or based on the occupancy of sound in a certain frequency. All of these methods do require me to input the frequency of my target sound. Typically, the frog of interest has a higher frequency (5500 - 7200 Hz) and can be easily distinguished by the computer from other calls. However, louder (an increase in dB) calls from other species can end up in this frequency range and trigger my recognizer since it's based on SNR dB thresholds. I'm trying to explain this concept with physics - what is causing those louder calls to cover a larger frequency range than quieter (lower dB) sounds?

I keep getting close to understanding and I'm looking at various versions of the Fletcher curve but I am a biologist and I need someone to dumb it down for me, please.

EDIT: I'm not sure now that the Fletcher Munson curve is relevant as it is focused on perceived sound. I am looking at spectrograms of sound.

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  • $\begingroup$ Is it possible that louder calls are no wider in frequency distribution, but still have a sizeable component in the frequency range of interest just by virtue of being loud? Is the louder call lower or higher in frequency than this range? Maybe it happens to have a harmonic in this range. Do you have any measurements of the spectral distribution of the calls? These could be obtained by a Fourier transform on recordings. $\endgroup$
    – Puk
    Commented May 7, 2023 at 3:58
  • $\begingroup$ The call of interest ranges from 5500 - 7200 Hz, while the other calls are typically in the 3500-4000 range. In the case of one of the overlapping calls (Spring Peeper) it definitely appears to be a harmonic based on the parallel lines in the spectrogram at even distances. For the other (Southern Chorus Frog), it literally looks like the call has been stretched along the frequency axis when the dB increases. $\endgroup$
    – Anna
    Commented May 7, 2023 at 4:50
  • $\begingroup$ I also suspect the Fletcher Munson curves may not be relevant here. I'm not aware of some sort of law that says frequency spectra tend to broaden for louder sources, but there may well be one. When you talk about dB increasing, does that mean the call is actually louder? You don't see the spectra broadening when e.g. the microphone is closer to the frog? $\endgroup$
    – Puk
    Commented May 7, 2023 at 4:55
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    $\begingroup$ I noticed there is a page on the Florida Museum of Natural History website called Florida frog calls which provides samples of frog calls, among them the Southern Chorus frog. I'm assuming the provided samples are recordings of calls by individual frogs. The recordings you are working with, are those recordings of the sound production of individuals, or of a population (that is close to the recording unit)? $\endgroup$
    – Cleonis
    Commented May 7, 2023 at 6:14
  • $\begingroup$ I am recording entire soundscapes, my microphone is in the breeding pond and every frog calling in that pond simultaneously is picked up. Here are some more thoughts I am considering 1. Multiple frogs calling simultaneously could have an additive effect on dB, correct? 2. Individual frogs may also be calling closer to the recording unit, I have no way to be sure. Given this, is it possible that the incr in frequency range is because louder calls allow more of the call to be detected? eg, at a distance I can only hear the mid freq of the call [which would be connected to fletcher munson] $\endgroup$
    – Anna
    Commented May 7, 2023 at 15:52

3 Answers 3

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Probably just a nonlinarity in the frogs vocal cord, eg. by a non-stiff fixations of the ends or just by the nonconstant mass density of the cord if elaongated.

The greater the amplitude, the less the amplitudes of the harmonics of the ground tone falls off to infinity. Not to speak about multiple mixed spectra of 2-dimensional cords.

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There's a lot things that may be going on here. First of all, I have to note that I am neither a biologist nor a bioacoustics expert so I will just try to provide my view here in case it can help.

Fletcher-Munson curves

I can't find a connection of the curves of equal loudness (as the Fletcher-Munson curves are also termed) to your problem. These are related to the response of the human hearing system/mechanism to "simple" (co)sinusoidal excitation (this is how originally the curves were created although refinements and newer results may have altered or even confirmed their validity with other signals too).

The curves connect the objective sound parameters (amplitude and frequency) to the subjective feeling of loudness. In contrast, your problem, in my opinion, is of pure objective measures.

Increased bandwidth with increased amplitude

I believe that in order to reach the conclusion that higher amplitudes result in higher frequency bandwidths at the source (this is the frog) you have to perform some measurements at the source, which of course is these other species. In any other case you can't really say whether increased volume results in higher bandwidth.

From my limited knowledge on human voice (I am completely ignorant to bioacoustics and I will use human voice to make my point), the way to increase your bandwidth is not through the volume. At least to an approximation, human voice is modelled as a linear system (an example is a pipe with non-constant diameter excited by some wide-band signal from the vocal tracts) which, due to its nature (linear) cannot introduce frequencies that do not exist in the excitation.

I do know of specific case where the person deliberately can change the frequency spectrum of their voice but most often this is a result of training and not something common. Of course, this may not be the case in the frog species of interest.

Finally, there may be some non-linear effects taking place in the vocal generation mechanism of the frogs, or other species of interest. These may well introduce additional frequencies in the spectra with increasing volume.

Possible explanation in the linear case

Although you may very well argue that the data (the spectrograms and your detector) show evidence of extension in the bandwidth of various species with increasing vocalisation volume, there is still a possibility that the extension is not present.

Assuming that the species that are not of interest do have energy in the band of interest (the band where the frog vocalisation is most prominent, $5.5 ~ kHz - 7.2 ~ kHz$ according to your data), when their volume is increased, the spectrum will just rise in a "kind-of" linear way (this is rarely the case in my limited experience) with all frequencies rising pretty much equally.

If the energy in the band of interest exceeds the threshold of your detector (here I assume you are using some kind of energy detector) then it could be triggered providing a false positive.

In the case of your spectrograms, it very well depends on the parameters set on the plotting algorithm. Blanking may be used for very low values, or even logarithmic scale, or non-linear logarithmic scaling (this could range from using logarithms of different bases for different value ranges, to applying scaling - like blanking - after the conversion to the logarithmic scale), or any other set of parameters that do are not appropriate for the data at hand could very well hide the energy in this band up until it starts to become significant.

This is one of the reasons I stated that, in my opinion, you should measure the interfering sounds at different volumes to see whether there is indeed energy in the band of interest irrespective of the volume or there is indeed some mechanism in the voicing of the species that results in bandwidth extension. Nevertheless, this could very well be the topic of a completely different project out of scope of the one you are working on at the moment.

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This is a long comment.

Researching the concept of sound wave packet has not given me a useful link,

This is the definition of a wave packet for solutions of wave equations. What you are picking up are wave packets of sound, as seen here. What you need is the frequency distribution of a sound wave packet, or a series of wave packets coming from the frogs. I suspect that when the intensity of a sound wave packet is increased more frequencies appear that add up to the sound curve, but I am not up to the mathematics needed to demonstrate this.

On this type of reasoning:

enter image description here, more wavelengths enter for narrower pulses, and as the link states

The actual numbers involved depend upon the definition of the pulse width, but creating short pulses inherently requires a large frequency bandwidth.

It is just a guess.

An alternative (not wave packets) connecting amplitude with frequency in fig 9, showing growth in amplitude .

( https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwjo8ZrUmuP-AhU37LsIHcxHAU8QFnoECCwQAQ&url=https%3A%2F%2Fcran.r-project.org%2Fweb%2Fpackages%2Fseewave%2Fvignettes%2Fseewave_analysis.pdf&usg=AOvVaw3KRqDxDgjGJTLTJUzDhJ8y )

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    $\begingroup$ I am not sure about that but I believe that wave packets are mostly related to the frequency content of the packet in respect to the duration of the packet (and the envelope in many cases) constituting a form of the "uncertainty principle" of waves. In linear acoustics though, this shouldn't have any connection with the amplitude of the packet. Of course, there's the possibility of the inclusion of non-linear effects in OP's problem but I still can't really see a connection to the wave packets. Would you care to elaborate a bit? $\endgroup$
    – ZaellixA
    Commented May 7, 2023 at 10:58
  • $\begingroup$ see my edit. It is just a comment not a rigorous answer. $\endgroup$
    – anna v
    Commented May 7, 2023 at 12:28
  • $\begingroup$ I am not sure I can really see how the spectrum of a part of a recording is connected to the question but I am sure I'm missing something. Since I can't make a connection I won't argue about it and leave it here. $\endgroup$
    – ZaellixA
    Commented May 7, 2023 at 19:44

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