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How to avoid first wavefront compression/rarefaction problem?
Old 17th September 2012
  #1
Here for the gear
 
🎧 5 years
How to avoid first wavefront compression/rarefaction problem?

I was thinking about an "ideal" listening position and it didn't make theoretical sense to me, for this simple reason:

The compression of X wave might be the rarefaction of Y wave.

For example, I worked out that the compression of 63 Hz is 4.48 ft. However, 4.48 ft is also the rarefaction of 189.17 Hz.

My math:
?/4 of 63Hz = 1130 x 1/63 x 1/4 = 4.48'
3/4? of 189.17Hz = 1130 x 1/189.17 x 3/4 = 4.48'

So, if you have a true room (EQ is uniform throughout) and your speakers are 4.48ft apart and you 4.48ft away from each monitor (getting that equilateral orientation) then you will be hearing the greatest amplitude component of 63Hz and will barely hear 189.17 Hz.

Do you see my problem?

Room treatment makes no difference because I'm talking about the direct wavefront. Indeed, you could avoid the effects of unwanted peaks and nulls with the room via treatment, but this implies taking reflections into consideration. I'm not. I'm considering the first direct wavefront: there will be some frequencies that are louder than others depending on how far away you are from the source.

How do you avoid/get around this?

If my reasoning is wrong somewhere, please let me know.
Old 17th September 2012
  #2
Lives for gear
 
🎧 5 years
The amplitudes will still be the same. With no interference, the sound field is true for close proximities. I want to say there is a phase shift, but I am not sure.
Old 18th September 2012
  #3
Lives for gear
 
Dange's Avatar
 
🎧 10 years
If you have two sources (loudspeakers) the you will see an interference pattern that will vary in frequency with different distances from the sources. Reflections from boundaries can also be seen as other sources.

But you're measuring with a mic so you're taking the response at a single point. Your ears aren't at a single point, they're spaced apart.

If you have one ideal source (phase linear and flat in frequency response) in free space (no boundaries) then all frequencies will be at the same level (relative to each other, i.e. flat frequency response) regardless of distance.
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