Undergrad Interpretation of the Fourier Transform of a Cauchy Distribution

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SUMMARY

The discussion centers on the Fourier transform of the Cauchy distribution, also known as the Lorentzian function, and its implications for sound waves. The inverse Fourier transform results in an exponentially decaying sinusoid, raising questions about the auditory experience when continuously playing a sound with a Cauchy frequency distribution. Key insights reveal that while the initial sound amplitude decays exponentially, over time, the phases of the frequencies can realign, leading to oscillating amplitude rather than perpetual decay. This understanding is crucial for interpreting sound wave behavior in practical applications like Fourier Transform Infrared (FT-IR) Spectroscopy.

PREREQUISITES
  • Understanding of Fourier transforms and their applications
  • Familiarity with Cauchy distributions and Lorentzian functions
  • Basic knowledge of sound wave propagation and amplitude modulation
  • Concepts of Fourier Transform Infrared (FT-IR) Spectroscopy
NEXT STEPS
  • Explore the mathematical properties of the Cauchy distribution and its Fourier transform
  • Learn about the principles of sound wave interference and phase alignment
  • Investigate the applications of FT-IR Spectroscopy in various scientific fields
  • Study the effects of frequency distribution on sound perception and amplitude behavior
USEFUL FOR

Mathematicians, physicists, audio engineers, and anyone interested in the intersection of sound theory and Fourier analysis.

Jim Fowler
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Hi,

I'm struggling with a conceptual problem involving the Fourier transform of distributions. This could possibly have gone in Physics but I suspect what I'm not understanding is mathematical.

The inverse Fourier transform of a Cauchy distribution, or Lorentian function, is an exponentially decaying sinusoid. What I don't get is this...

Can't I, in principle, play a sound through a speaker that has any frequency distribution I like? If I choose to continuously play such a sound with a Cauchy distribution of frequencies, what will I hear? Does the sound decay exponentially? If I'm continuously sending that combination of frequencies to my speaker, that doesn't make sense to me.

Any insights about what it is I'm missing would be most welcome.

Thanks in advance.
 
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Jim Fowler said:
Hi,

I'm struggling with a conceptual problem involving the Fourier transform of distributions. This could possibly have gone in Physics but I suspect what I'm not understanding is mathematical.

The inverse Fourier transform of a Cauchy distribution, or Lorentian function, is an exponentially decaying sinusoid. What I don't get is this...

Can't I, in principle, play a sound through a speaker that has any frequency distribution I like? If I choose to continuously play such a sound with a Cauchy distribution of frequencies, what will I hear? Does the sound decay exponentially? If I'm continuously sending that combination of frequencies to my speaker, that doesn't make sense to me.

Any insights about what it is I'm missing would be most welcome.

Thanks in advance.
Don't forget that the phase information in your signal. At time zero, all of the frequencies are in phase and the signal amplitude is at a maximum. As time goes on the signal decays as the phases move farther and farther out of alignment.
 
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Thanks, that does make sense, and I'm quite happy with that in a different scenario. Let me explain...What got me thinking about this was trying to get to grips with Fourier Transform Infrared (FT-IR) Spectroscopy. In an FT-IR instrument there's a Michelson interferrometer and a light source with a relatively broad spectrum of frequencies. The beam is split, a path length difference introduced and the two beams are recombined The signal that is detected depends on how the two beams interfere, which changes as you vary the path length difference, called the retardation. (You probably know more about spectrometers than me, I'm just recapping so we're all on the same page for the below...)

In spectroscopy it is normal to describe frequencies/wavelengths by wavenumber in inverse cm. The signal is a function of the retardation, not the time, and it's inverse cosine Fourier transform is the light source spectrum as a function of wavenumber. So the Fourier transform pair is between cm-1 and cm.

So far so good.

Now, it makes sense to me in this case that a spectrum with a frequency distribution (rather than discrete spectral lines) would give an eponential decay. As you say, at zero retardation all of the frequencies are in phase and construct. As you move the mirror in your interferrometer and so sweep the retardation (cm), the further you go the more wavenumbers start to deconstruct until at a large enough distance you have every phase difference under the sun and everything cancels.

That makes sense. There is still a constant light source but it all cancels out. I've seen it enough in the instrument data, it's referred to as the "centre burst" where nearly all the data is very close to zero retardation and drops off exponentially. But this is easy to picture what's going on. I can imagine running the mirror backwards from far to near, sweeping retardation from large to zero, and reversing the signal. I can picture that at a fixed retardation all the different sinusoidal waves have traveled slightly further in one arm than the other so for each wavenumber there is a phase difference in the two beams that depends on the wavenumber and the retardation. I can picture it because for a fixed retardation the signal is time invariant so I have 'time' to let the mental picture wander around the interferrometer. I struggle to have a similar picture in the case of sound

Mathematically, transforming wavenumber(cm-1) - retardation(cm) should be analagous to transforming frequency(s-1) - time(s) so I don't dispute the result but I'm struggling to have any intuition about what's going on in the case of sound waves. If the source is a single frequency then the signal would be a single sine wave that lasts as long as you're driving the speaker. If the source broadens just a little bit, but you're still driving the speaker indefinitely, does the volume of the sound produced really decay exponentially with time? I might have to do the experiment before I'll believe it :-)
 
Jim Fowler said:
Can't I, in principle, play a sound through a speaker that has any frequency distribution I like? If I choose to continuously play such a sound with a Cauchy distribution of frequencies, what will I hear? Does the sound decay exponentially? If I'm continuously sending that combination of frequencies to my speaker, that doesn't make sense to me.
You don't feed frequencies into a speaker, you send a time signal.
 
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Thinking a bit more about it, I might have found what is bugging you.

Imagine that you don't have a speaker, but an instrument that can produce many individual tones, close enough in frequency to be almost continuous. You then excite the tones with a frequency-dependent amplitude that is Lorentzian, leading initially to an exponentially decaying sound amplitude. But for a finite frequency range, after a while the signal will start to rephase, so the result will be an oscillating amplitude, not a forever exponentially decreasing amplitude.
 
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