Re: how much noise will three 8dB(A) fans make?
- From: Randy Yates <yates@xxxxxxxx>
- Date: Sun, 06 Jan 2008 23:42:56 -0500
"Arny Krueger" <arnyk@xxxxxxxxxx> writes:
"William Sommerwerck" <grizzledgeezer@xxxxxxxxxxx> wrote in
message news:EoSdnRAKoayQJeLanZ2dnUVZ_jOdnZ2d@xxxxxxxxxxx
"Arny Krueger" <arnyk@xxxxxxxxxx> wrote in message
news:2J6dnRmQUvXMOeLanZ2dnUVZ_qGknZ2d@xxxxxxxxxxxxxx
"William Sommerwerck" <grizzledgeezer@xxxxxxxxxxx> wrote
in message
news:-NudnUqAxKNWCOLanZ2dnUVZ_jWdnZ2d@xxxxxxxxxxx
Assuming they're identical and near each other, three
times the noise. That's about 5dB more.
True only if they are phase-locked. If they are AC
powered, there's a pretty good chance of that. If they
are DC powered,
then they will be turning at similar but not identical
speeds.
If the fans can be modelled as statistically independent
acoustical sources, then you add their power using the
square root of the sum of the squares.
I think you need to re-think this.
Nope. I learned all about this, both theory and lab experiments, back in the
early 1970s.
I was assuming random noise (primarily from air motion).
In such cases, the powers simply add.
Nope. Please see:
http://www.rose-hulman.edu/~black/Noise/img011.gif
The general rule for adding signals is that they add algebraically only if
they have identically the same frequency.
Most noise signals have anything but the same identically same frequency.
Then they are best added using the square root of the sum of the
squares.
I think you're confusing adding voltages with adding powers. If two
signals are uncorrelated and zero-mean (and WSS), then their powers
add algebraically (P1 + P2) but their voltages add as the sum of
squares ((V1^2 + V2^2)^(1/2)).
--
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%%%% <yates@xxxxxxxx> % 'Shangri-La', *A New World Record*, ELO
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