3 ms·

Nope. Intermodulation distortion for out-of-range frequencies is inaudible. The 1hz beat you are hearing is not a 1hz sound wave, it's a 1000.5hz sound wave bec

Nope. Intermodulation distortion for out-of-range frequencies is inaudible. The 1hz beat you are hearing is not a 1hz sound wave, it's a 1000.5hz sound wave becoming louder and softer once per second. The comparison of a 1hz beat to a 1hz sound should be absurd on its face: you need about 20-30hz to become audible, and it's a low rumble more felt than heard. Very low frequencies sound absolutely nothing like intermodulation beats.

tonecluster 11y ago
"...and cause inter-modulation products in the audible range." AFAIK, this is true in acoustic environments under conditions as described in the original post.
- darkmighty 11y ago
  There are no inter-modulation products in the sense rplst8 expects. The interaction of waves of 1000Hz and 1001Hz doesn't produce 1Hz waves in linear media. Even if it did produce some new frequencies through some non-linearity (which is negligible in most environments afaik), the recording equipment would capture the low frequency waves produced by those interactions. So the only question is whether there are significant non-linearities in our hearing system, and the overwhelming evidence is no again afaik.
  - rplst8 11y ago
    The recording equipment won't necessarily pick it up. If the microphones are place very close the instruments relative to the other instruments creating the interaction.
    - darkmighty 11y ago
      If no recording equipment would pick it up, no human would pick it up either. Quoting xiphmont > Once you're driving air so hard it becomes nonlinear, thus introducing intermodulation distortion in the air, that distortion produces actual audible-range distortion products. And because the distortion you're hearing is in the audible range, a recording will sample and reproduce it accurately. > You're hearing the audible _result_ of IMD, you're not somehow listening to the distortion curve itself.
      - TillE 11y ago
        > If no recording equipment would pick it up, no human would pick it up either. Have you been to a concert? There's no recording/playback technology that can reproduce anything close to the sound of a full orchestra. It's all lossy.
        darkmighty 11y ago
        That's an entirely different discussion having nothing to do with intermodulation products. Have you read the article/quantization discussion on what is meant by lossy? If your recording equipment is good and your reproduction equipment is good, 16/44 is enough to reproduce the concert sound perfectly (as far as human hearing goes). What you do not experience is everything else but the sound -- the vibration of the super loud bass on your skin, the energy of the public, the beauty of the venue.
hamiltonkibbe 11y ago
Intermodulation creates both sum and difference frequencies, the latter can certainly fall into the audible spectrum, assuming the ultrasonic frequencies are within the passband of your recording medium. The sum component can also alias back into the passband as well...
thescriptkiddie 11y ago
Is there a difference? Audio is one-dimensional, frequency is just the derivative of amplitude. An arbitrarily high frequency sound wave becoming louder and softer 440 times per second is just as much an A as a 440 Hz sound wave at constant volume. A lot of cheap audio gear even uses a "1-bit DAC" that is just very high frequency PWM.
- hamiltonkibbe 11y ago
  The signal you describe in your comment is very much NOT an A. look at the fourier transform of a 1kHz signal modulated by a 440Hz signal, and you won't see any frequency component at 440Hz, nor any integer multiple 440Hz! You can look at your "A440 at a constant volume" example as a 0Hz(DC) signal getting louder and softer 440 times a second, but this is the only case in which your example holds. Amplitude modulation creates sum and difference frequencies, so the A that you hear is 440Hz + 0Hz. if you change that 0Hz to 1kHz, you get a signal thats the sum of a sinewave at 560Hz and a sinewave at 1.44kHz, neither of which are an A. The distinction is that the 1Hz signal is modulating the audible signal, not adding to it, if you look at the spectrum of the sum of those frequencies there is no 1Hz component, whereas if you added a 1Hz signal you'd get something completely different. And in this case the amplitude of the signal is always changing faster than 1Hz. Edit: another way of looking at it: You wouldn't say you can "hear DC" because you can hear an A440 played at a perfectly constant volume.
  - thescriptkiddie 11y ago
    All of that makes sense if we consider the case of amplitude modulation, which is multiplication. But if we are talking about the interference patterns caused by two overlapping audio signals, that is addition, is it not?
    - hamiltonkibbe 11y ago
      The two signals which are "interfering" are added together. The amplitude of the resultant signal varies sinusoidally, as the instantaneous phase difference between the two signals goes from 0->2pi. One way of describing the signal would be that it is a separate tone (sitting in the middle of the two frequencies) being amplitude modulated by a signal at the beat frequency... which is what you hear and why you "hear" the beat frequency) I went with this way of describing the signal because you were talking about "an arbitrarily high frequency signal getting louder and softer 440 times per second" which is the definition of amplitude modulation. Counterintuitively, there is no frequency component generated at the beat frequency when you sum a 1kHz and 1.001kHz signal, its easy to test that out with matlab, octave, scipy/numpy/matplotlib, etc. Generate the two signals, add them together, and look at the Fourier transform, you'll see two components, one at 1kHz and one at 1.001kHz (assuming you take a long enough window to have that type of resolution) and no component at the beat frequency. A third sinewave doesn't just jump out of nowhere when you add two separate sinewaves together. If you take the sum of those two signals and run them through an ideal brickwall highpass filter at 999Hz so there are no frequency components below 999Hz, you'll still "hear" the beat frequency because it isn't a separate spectral component, its just the two signals slowly going out of phase, cancelling eachother out, and then going back in phase and boosting the amplitude.

tonecluster 11y ago

"...and cause inter-modulation products in the audible range." AFAIK, this is true in acoustic environments under conditions as described in the original post.

darkmighty 11y ago
There are no inter-modulation products in the sense rplst8 expects. The interaction of waves of 1000Hz and 1001Hz doesn't produce 1Hz waves in linear media. Even if it did produce some new frequencies through some non-linearity (which is negligible in most environments afaik), the recording equipment would capture the low frequency waves produced by those interactions. So the only question is whether there are significant non-linearities in our hearing system, and the overwhelming evidence is no again afaik.
- rplst8 11y ago
  The recording equipment won't necessarily pick it up. If the microphones are place very close the instruments relative to the other instruments creating the interaction.
  - darkmighty 11y ago
    If no recording equipment would pick it up, no human would pick it up either. Quoting xiphmont > Once you're driving air so hard it becomes nonlinear, thus introducing intermodulation distortion in the air, that distortion produces actual audible-range distortion products. And because the distortion you're hearing is in the audible range, a recording will sample and reproduce it accurately. > You're hearing the audible _result_ of IMD, you're not somehow listening to the distortion curve itself.
    - TillE 11y ago
      > If no recording equipment would pick it up, no human would pick it up either. Have you been to a concert? There's no recording/playback technology that can reproduce anything close to the sound of a full orchestra. It's all lossy.
      - darkmighty 11y ago
        That's an entirely different discussion having nothing to do with intermodulation products. Have you read the article/quantization discussion on what is meant by lossy? If your recording equipment is good and your reproduction equipment is good, 16/44 is enough to reproduce the concert sound perfectly (as far as human hearing goes). What you do not experience is everything else but the sound -- the vibration of the super loud bass on your skin, the energy of the public, the beauty of the venue.

hamiltonkibbe 11y ago

Intermodulation creates both sum and difference frequencies, the latter can certainly fall into the audible spectrum, assuming the ultrasonic frequencies are within the passband of your recording medium. The sum component can also alias back into the passband as well...

thescriptkiddie 11y ago

Is there a difference? Audio is one-dimensional, frequency is just the derivative of amplitude. An arbitrarily high frequency sound wave becoming louder and softer 440 times per second is just as much an A as a 440 Hz sound wave at constant volume. A lot of cheap audio gear even uses a "1-bit DAC" that is just very high frequency PWM.

hamiltonkibbe 11y ago
The signal you describe in your comment is very much NOT an A. look at the fourier transform of a 1kHz signal modulated by a 440Hz signal, and you won't see any frequency component at 440Hz, nor any integer multiple 440Hz! You can look at your "A440 at a constant volume" example as a 0Hz(DC) signal getting louder and softer 440 times a second, but this is the only case in which your example holds. Amplitude modulation creates sum and difference frequencies, so the A that you hear is 440Hz + 0Hz. if you change that 0Hz to 1kHz, you get a signal thats the sum of a sinewave at 560Hz and a sinewave at 1.44kHz, neither of which are an A. The distinction is that the 1Hz signal is modulating the audible signal, not adding to it, if you look at the spectrum of the sum of those frequencies there is no 1Hz component, whereas if you added a 1Hz signal you'd get something completely different. And in this case the amplitude of the signal is always changing faster than 1Hz. Edit: another way of looking at it: You wouldn't say you can "hear DC" because you can hear an A440 played at a perfectly constant volume.
- thescriptkiddie 11y ago
  All of that makes sense if we consider the case of amplitude modulation, which is multiplication. But if we are talking about the interference patterns caused by two overlapping audio signals, that is addition, is it not?
  - hamiltonkibbe 11y ago
    The two signals which are "interfering" are added together. The amplitude of the resultant signal varies sinusoidally, as the instantaneous phase difference between the two signals goes from 0->2pi. One way of describing the signal would be that it is a separate tone (sitting in the middle of the two frequencies) being amplitude modulated by a signal at the beat frequency... which is what you hear and why you "hear" the beat frequency) I went with this way of describing the signal because you were talking about "an arbitrarily high frequency signal getting louder and softer 440 times per second" which is the definition of amplitude modulation. Counterintuitively, there is no frequency component generated at the beat frequency when you sum a 1kHz and 1.001kHz signal, its easy to test that out with matlab, octave, scipy/numpy/matplotlib, etc. Generate the two signals, add them together, and look at the Fourier transform, you'll see two components, one at 1kHz and one at 1.001kHz (assuming you take a long enough window to have that type of resolution) and no component at the beat frequency. A third sinewave doesn't just jump out of nowhere when you add two separate sinewaves together. If you take the sum of those two signals and run them through an ideal brickwall highpass filter at 999Hz so there are no frequency components below 999Hz, you'll still "hear" the beat frequency because it isn't a separate spectral component, its just the two signals slowly going out of phase, cancelling eachother out, and then going back in phase and boosting the amplitude.