In article , Andy wrote:
Loudspeakers are generally described and designed around response
characteristics showing the ratio of output SPL / input power (in dB) as
a function of frequency.

In particular for sealed systems these curves show a predominant role of
the box volume affecting the overall resonance frequency. Also it is
well-known that below the resonance frequency the output falls off with
a certain asymptotic slope.

I was wondering what the response looks like if you plot
output SPL / diaphragm displacement vs. frequency instead.

Below resonance, in the stiffness controlled region of
operation, excursion is independent of frequency. Above
resonance in the mass controlled region, excursion goes as the
reciprocal square of frequency.

Now, compare that to the SPL, where below resonance, the SPL
goes as the square of frequency and above resonance, it is
constant with frequency.

If your asking what does ratio of SPL to excursion look like,
well, below resonance, it's proportional to the the square of
frequency divided by a constant, so it goes as the
square of frequency. Above resonance, it's a constant divided by
the inverse square of the frequency, which is the square of the
frequency.

This is a somehwta indirect way of saying that SPL is a function
of linear displacement times the square of the frequency.

Regardless of how you would achieve displacement as 'input' rather than
electrical power (e.g. by motion feedback), which driver or box
parameters does the response now depend on?

Which "response?" SPL vs frequency?

:My guess is that it now
depends only on cone area and possibly on the shape of the box. Or does
box size or volume still play a part?

Your question is a little to imprecise to answer. Which response
are you talking about?

--
| Dick Pierce |
| Professional Audio Development |
| 1-781/826-4953 Voice and FAX |
| |

In article , Andy wrote:
Richard D Pierce wrote:

Below resonance, in the stiffness controlled region of
operation, excursion is independent of frequency. Above
resonance in the mass controlled region, excursion goes as the
reciprocal square of frequency.

Does this mean stiffness = system reacts like a spring, mass = system reacts
like mass?) The point is that this standard analysis seems to include the
driver in the system.

A louspeaker driver is a mechanical resonant system. The
combination of the driver suspension's mechanical stiffness and
the stiffness of the enclosed air together form the stuffness
portion of the system, and the effective mass of the driver is,
well, the mass.

In any such mechanical resonant system, the constraint to
motion, be it stiffness, mass or losses, depends upon what
frequency you are operating at. Below resonance, it is the
system stiffness that dominates. Above resonance, it is the
moving mass. AT resonance, it is the combined mechanical losses
of the system that dominate.

Below resonance, in the stiffness region, an applied force will
only push the driver so far before the restoring force equals
the applied force and the driver stops. Thus, excursion is
constant. Above resonance, an applied force will only accelrate
the the mass of the driver so much, thus the driver is mass/
acceleration limited and the excursion goes as the inverse
square of frequency (or, equivalently, the square of time,
since x = 1/2 a t^2).


Which "response?" SPL vs frequency?

By regarding displacement as input I'm trying to eliminate the
electro/mechanical part of the system from consideration. Another way of
putting it is: how does the SPL go with frequency for constant displacement?
(Rather than constant input voltage of the driver)
Your analysis is based on taking the resonance frequency as divider between low
and high frequencies. I'm asking how to compute this resonance frequency for
constant displacement input, i.e. what resonance does the cone/air part of the
system exhibit without the electromechanical part? Does it only depend on cone
area and properties of air?

Remember that things like cone mass and driver resonance are now eliminated.

ANY fixed-size diaphragm will produce a sound level which is
proportional to the area, proportional to the displacement, and
proportional to the SQUARE of frequency.

Thus, REGARDLESS of resonance, a constant excursion independent
of frequency will result in a response that rise at 12
dB/octave.

The issue of the resonance is important in conventional speakers
because the excursion "conveniently" decreases at -12 dB/octave
above resonace, thus resulting in an output which is flat.

--
| Dick Pierce |
| Professional Audio Development |
| 1-781/826-4953 Voice and FAX |
| |

OK so you're saying:

always SPL ~ x times omega^2 .

Above resonance:
x ~ a omega^-2,
a ~ force,
therefore SPL = constant for force input

Below resonance:
x ~ force,
therefore SPL ~ omega^2 times force, ie. -12dB/oct for decreasing frequencies for
force input.

So a constant x input would give -12dB/oct towards zero frequency (same as force
input) but would continue with +12dB/oct above resonance.

Parameters such as sealed box volume enter in the stiffness below resonance which
leads to more and more power required to achieve x.

Thanks for the help,
Andy


Richard D Pierce wrote:

ANY fixed-size diaphragm will produce a sound level which is
proportional to the area, proportional to the displacement, and
proportional to the SQUARE of frequency

Below resonance, in the stiffness region, an applied force will
only push the driver so far before the restoring force equals
the applied force and the driver stops. Thus, excursion is
constant.

Above resonance, an applied force will only accelrate
the the mass of the driver so much, thus the driver is mass/
acceleration limited and the excursion goes as the inverse
square of frequency (or, equivalently, the square of time,
since x = 1/2 a t^2).