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I am certain that for all frequencies above about 440 Hz, there is only about a few cm of listening zone within which the polarity of the audio makes any difference.

Am I wrong?

If you connect a sound speaker cable black to red, it will reverse the + and - of the sound pressure waveforms.

440 Hz is about 78 cm wave periods, 34 cm from max + the min -, and 4 kHz sound is about 8.6 cm, 4.3 cm from the wave peak to trough another, then by being to close to one speaker you change all the sound range completely just as much as using reverse polarity on one of the speakers?

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    \$\begingroup\$ You're thinking in term of 'pure' sine-wave tones. Audio is way more complex than that, so your argument only holds for the simplistic case. \$\endgroup\$
    – brhans
    May 31, 2018 at 11:36
  • \$\begingroup\$ Add to that the reflections, even if you are outside, in the field, not to mention other, lesser effects, such as mechanical distortion, noise (you are not in an anechoic room), etc. The plot thickens. \$\endgroup\$
    – a concerned citizen
    May 31, 2018 at 11:50
  • \$\begingroup\$ Look up acoustic diffusers, absorbers, etc. Speaker interactions with each other and the ambient are quite complex, not only due to the science in itself but also due to subjective issues. (i.e. if you just use absorbers in an environment, I think people tend to not enjoy it as much if you use some diffraction panels - maybe it has to do with how we perceive space) \$\endgroup\$
    – Wesley Lee
    May 31, 2018 at 12:34

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Sound propagates at about 3 ms per meter. 440 Hz therefore has a wavelength of about 760 mm, which fits with what you said.

If you had two speakers pointing at each other and you were directly in line between them and they were both emitting the same 440 Hz signal, then you'd perceive nulls and peaks in the sound with apparent 380 mm (15 inch) repetition.

However, speakers in a normal listening environment aren't opposite each other, and they don't usually play a single frequency. The pattern of apparent inversions is different for every frequency. The distances between inversions are also larger when you are not directly in line between two speakers.

Therefore, in the general case, you can't compensate for the polarity of one speaker being flipped by moving the listening position a little bit.

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  • \$\begingroup\$ The signals are also often out of phase with each other so the nulls and peaks will move about. \$\endgroup\$
    – ratchet freak
    May 31, 2018 at 11:51

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