[Don Pearce[_3_]]

On Sat, 31 Mar 2012 12:23:00 +0100,
lid (Adrian Tuddenham) wrote:

Don Pearce wrote:

On Sat, 31 Mar 2012 09:35:08 +0100,
lid (Adrian Tuddenham) wrote:

Don Pearce wrote:

On Fri, 30 Mar 2012 20:22:10 GMT, (Don Pearce) wrote:

On Fri, 30 Mar 2012 16:03:29 -0400, Matt Faunce
wrote:

On 3/30/12 12:24 PM, Mike Rivers wrote:
On 3/30/2012 8:37 AM, Matt Faunce wrote:

.. So late last night I tested a 0.5 inch
cardioid mike I have (AKG blueline) by placing it one inch
away from my metronome speaker which clicks at two different
pitches. I used the metronome because it has a single small
speaker, so direction will be more precise than a
multi-speaker speaker. I had to make two takes, one for each
mike position, then for playback I boosted the gain on the
sideways position take to match the other. The bass boost
seemed gone at the sideways position.

How much bass do you get out of a metronome?

This is another question that I'm not sure I know enough to even
articulate. I still hear a fattening effect. Maybe the mike isn't
boosting the bass frequencies of the metronome, but fattening the
attack. Or is it converting the attack into a bass frequency?

I haven't gotten the physics of proximity effect straight in my head
yet. So Don, I'll be looking forward to going back your article.

Matt

The fattening is because you are hearing more of the reverberant sound
from the room, and less of the dry, direct sound.

I have to reconstruct a web site, and that isn't going to happen until
tomorrow. I'll let you know.

d

OK, it all went a little quicker than I expected. Here you go:

http://www.soundthoughts.co.uk/read/mic/


The critical point, which is missing from most descriptions of the
Proximity Effect, is that it is caused by the expansion of a spherical
wavefront.

If we use a ribbon mic as an example, the ribbon is driven by a
difference in pressure between its two sides which is cause by a
difference in the path length to the front and back as the sound
wavefront passes. The electrical ouput is proportional to the velocity
of the ribbon movement, but the ribbon resonance is below audio
frequency; so, as the frequency goes down, the ribbon velocity and
output would increase if the pressure difference remained constant.

With a plane wave (from a distant source) the path difference between
the front and back of the ribbon becomes a smaller and smaller
proportion of a cycle as the frequency goes down, so the pressure
difference falls with falling frequency. This exactly compensates for
the increasing sensitivity of the ribbon system and the result is a flat
frequency response.

Independently of this, a spherical wavefront expands and falls in
pressure as its radius increases. It will give different pressures on
each side of the ribbon because it will have expanded more and dropped
in pressure by the time it reaches the back of the ribbon. This effect
does not decrease with frequency, so the rising sensitivity of the
ribbon system will give increasing output as the frequency falls.

The ratio of spherical to plane wavefront effect is what determines the
frequency below which the 'expansion' response becomes larger than the
'phase difference' response and bass tip-up begins to occur.


Even in an excellent book such as Robertson's "Microphones", this
information is buried in a load of mathematics in a theoretical
discussion and not presented clearly as a practical explanation. I have
never seen a satisfactory explanation in any practical microphone book,
although I have seen several disgracefully incorrect ones.

I've seen this explanation before, and I really don't buy it. The
distance between the front and back of a ribbon is essentially zero,
...[...]

It's easy enough to check: take two miniature pressure capsules and
wire them in antiphase. Place them touching and expose them to sound;
with a bit of balancing, the electrical output can be reduced to zero.

http://www.poppyrecords.co.uk/other/images/MicTest1.gif


Now pull them apart by various small distances and see what effect you
get with sound which reaches one after the other. Alternatively,
interpose obstacles of various sizes and see what the increased path
length does. You will find that increasing the path length by either
method increases the sensitivity of the doublet. You will also find
that the plane-wave response sounds very harsh because it is caused by
the 'phase difference' effect which increases at 6dB/octave. The
response to a spherical wave will be flat.

If you now integrate the signal by putting a 'large' capacitor across
the amplifier terminals, the response to a distant source will become
flat (up to the point where the path difference between the capsules
approaches half a wavelength) and the spherical response will increase
with decreasing frequency. This is equivalent to the condition brought
about by working the ribbon above its resonance frequency.

Perhaps many aspects of this demonstration fit with your theory too, but
the path-difference explanation certainly covers all the observed
phenomena.

Sure, I understand that interposing the obstacles causes a phase shift
between the two, resulting in poor cancellation. And the 6dB per
octave also makes sense, up to the point where a half wave is being
approached and the effect starts to turn over.

But that isn't the same as a velocity-gradient induced pressure
difference between two points a thousandth of an inch apart.

d