Could some expert out there please verify whether my basic assumptions are
correct or not?

1) 3db increase in volume is a preceivable increase (i.e. slightly louder)

2) 10db increase in volume is preceived as twice as loud

3) each 3db increase in volume requires double the amplifier power

4) If a speaker was producing 100db at1 meter with 1 watt of input then
adding a second identical speaker, also with 1 watt of input, would only
raise the volume to 103db at 1 meter. 4 identical speakers would yeild 106
db, and 8 identical speakers would yeild 109 db

5) phasing issues between 2 speakers can produce "dead spots" and "super
spots" throughout a room. In theory the dead spots could go to zero and the
super spots could be twice as loud (+10db?)

I'm sure theres a mistake here somewhere, as #4 says a second speaker can
only add 3db and #5 says a second speaker can add 10db (in certain spots).
Could someone help?

Thanks

On Thu, 2 Dec 2004 10:57:57 -0600, "Dan K"
wrote:

Could some expert out there please verify whether my basic assumptions are
correct or not?

1) 3db increase in volume is a preceivable increase (i.e. slightly louder)

2) 10db increase in volume is preceived as twice as loud

3) each 3db increase in volume requires double the amplifier power

4) If a speaker was producing 100db at1 meter with 1 watt of input then
adding a second identical speaker, also with 1 watt of input, would only
raise the volume to 103db at 1 meter. 4 identical speakers would yeild 106
db, and 8 identical speakers would yeild 109 db

5) phasing issues between 2 speakers can produce "dead spots" and "super
spots" throughout a room. In theory the dead spots could go to zero and the
super spots could be twice as loud (+10db?)

I'm sure theres a mistake here somewhere, as #4 says a second speaker can
only add 3db and #5 says a second speaker can add 10db (in certain spots).
Could someone help?

Thanks



Yes, the mistake is to neglect the change in dispersion pattern caused
by adding speakers. If it remained the same then your maths would
apply, but because the source gets wider the beam gets narrower. This
means that directly in front of the speakers you get a rather bigger
loudness increase, while to the sides you get less. The net power of
course is the sum of all the inputs.

d
Pearce Consulting
http://www.pearce.uk.com

"Dan K" wrote in message


Could some expert out there please verify whether my basic
assumptions are correct or not?

Listen for yourself by downloading musical samples recorded at different
levels from

http://www.pcabx.com/technical/levels/index.htm


1) 3db increase in volume is a preceivable increase (i.e. slightly
louder)

Yes.

1 dB is pretty readily audible if you can do close comparisons.

0.4 dB differences might be audible as some kind of an indistinct but
noticable change under ideal conditions.

2) 10db increase in volume is preceived as twice as loud

Usual rule of thumb, YMMV

3) each 3db increase in volume requires double the amplifier power

Yes, presuming the loudspeaker is linear which isn't always true.

4) If a speaker was producing 100db at1 meter with 1 watt of input
then adding a second identical speaker, also with 1 watt of input,
would only raise the volume to 103db at 1 meter. 4 identical
speakers would yeild 106 db, and 8 identical speakers would yeild 109
db

Depends on how the sound from the speakers combines which relates to the
directional patterns of the speakers and the room this all happens in.

5) phasing issues between 2 speakers can produce "dead spots" and
"super spots" throughout a room.

You don't need two speakers for this to happen, a speaker and a flat surface
that reflects its sound can suffice.

In theory the dead spots could go to zero

Yes, and can come startlingly close to zero (-40 dB) in the real world.

and the super spots could be twice as loud (+10db?)

Make that +3 dB and add caveats from the question about speaker output
summing (4).

I'm sure theres a mistake here somewhere, as #4 says a second speaker
can only add 3db and #5 says a second speaker can add 10db (in
certain spots). Could someone help?

I tried! ;-)

2) 10db increase in volume is preceived as twice as loud

This is true around 1kHz, but at lower frequencies, for example, this number
tends to decrease. Do a google search for "Robinson-Dadson".

3) each 3db increase in volume requires double the amplifier power

....and a speaker that can handle this increase in power. And this assumes
that power compression is not playing a role, which for most speakers is
probably not a safe assumption.

"Dan K" wrote in message
...
Could some expert out there please verify whether my basic assumptions are
correct or not?

1) 3db increase in volume is a preceivable increase (i.e. slightly
louder)

2) 10db increase in volume is preceived as twice as loud

3) each 3db increase in volume requires double the amplifier power

4) If a speaker was producing 100db at1 meter with 1 watt of input then
adding a second identical speaker, also with 1 watt of input, would only
raise the volume to 103db at 1 meter. 4 identical speakers would yeild
106
db, and 8 identical speakers would yeild 109 db

5) phasing issues between 2 speakers can produce "dead spots" and "super
spots" throughout a room. In theory the dead spots could go to zero and
the
super spots could be twice as loud (+10db?)

With respect to point "5", this is called comb filtering.
It induces horrible, violent frequency response variations that vary from
centimeter to centimeter (distance depending upon the frequency) as you move
your head.
However, psychoacoustic research has revealed that the ear-brain system, for
some reason, filters out the effect of comb filtering when a stereo signal
is played. When two speakers are used for monophonic playback, it is quite
audible and irritating.

"Dan K" writes:
[...]
4) If a speaker was producing 100db at1 meter with 1 watt of input then
adding a second identical speaker, also with 1 watt of input, would only
raise the volume to 103db at 1 meter.

That could be right or wrong depending on several things:

1. If you are in the far field and the second source is completely
correlated with the first, i.e., producing the same in-phase signal at
the receiving position, then you get a 6 dB increase, not 3 dB. I
would say that this is pretty close to what you'll find in actual
practice.

The reason for 6 dB instead of 3 dB is that you're doubling the sound
pressure, which is analogous to voltage in an electrical circuit, and
the log relationship for pressures (and voltages) is 20*log10(y/x). If
y/x = 2, then the resulting increase is (approximately) 6 dB.

2. If you are in the near field and the second source is completely correlated,
then you get 3 dB. This is due to the way sound waves propagate in the near-field.

3. If you are in the far field, the second source is uncorrelated to
the first, and the second source is the same power as the first, then
you get a 3 dB increase. Here's why.

Let V = X + Y, where V is the sound pressure at the receiver, X is the sound pressure
of one source, and Y is the sound pressure of the other source. Then

E[V^2] = E[(X+Y)^2]
= E[X^2 + 2XY + Y^2]
= E[X^2] + 2E[XY] + E[Y^2].

If X and Y are uncorrelated, then E[XY] = 0. Then

E[V^2] = E[X^2] + E[Y^2],

which means that if the two sources are the same power levels, the result
is twice the power, or 3 dB higher.
--
% Randy Yates % "...the answer lies within your soul
%% Fuquay-Varina, NC % 'cause no one knows which side
%%% 919-577-9882 % the coin will fall."
%%%% % 'Big Wheels', *Out of the Blue*, ELO
http://home.earthlink.net/~yatescr

Randy Yates writes:
[...]
E[V^2] = E[(X+Y)^2]
= E[X^2 + 2XY + Y^2]
= E[X^2] + 2E[XY] + E[Y^2].

If X and Y are uncorrelated, then E[XY] = 0. Then

E[V^2] = E[X^2] + E[Y^2],

which means that if the two sources are the same power levels, the result
is twice the power, or 3 dB higher.

I should have stated the following:

1. "E[Z]" denotes the expected value of the random variable
underlying the random process Z, where Z is ergodic.

2. We assume X and Y are zero-mean processes.

3. Given assumption 2, E[Z^2] is the variance of Z, which is the
same as the power in the signal.

--RY


--
% Randy Yates % "My Shangri-la has gone away, fading like
%% Fuquay-Varina, NC % the Beatles on 'Hey Jude'"
%%% 919-577-9882 %
%%%% % 'Shangri-La', *A New World Record*, ELO
http://home.earthlink.net/~yatescr