[PD] overdriven speaker

Martin Peach martin.peach at sympatico.ca
Fri Oct 22 01:17:48 CEST 2010

A perfect speaker will reproduce the sound exactly by transforming the 
instantaneous voltage to a displacement in or out. That's called 
'compliance' in the speaker biz. The ideal speaker has zero mass and is 
totally rigid.
So nonlinearities will show up:
1> when the speaker is massive and can't reach the ideal position 
quickly enough.
2> when the speaker deforms as it is accelerated from the centre but 
lags at the edges.

Also the stationary magnet is driving a coil in the speaker cone. When 
the speaker is overdriven the coil will be pushed away from the field of 
the magnet on the outward stroke and so the coil will be less able to 
move the speaker, so a kind of soft clipping will occur.
But on the inward stroke the coil will bottom out and slam into the 
support structure, giving a hard clipping and possibly some bouncing.

I don't think the nonlinearities of the air are relevant here, or the 
doppler effect of the moving speaker.

Cavitation might occur with an underwater speaker, where bubbles are 
formed by the negative pressure on the trailing side of the cone.


On 2010-10-21 11:48, Martin Schied wrote:
> Hi!
>>> Looking at power and air pressure - we don't have to care about them
>>> as long as we don't want to include thermal effects
>> Isn't the heat proportional to the mean power ? Then you just do [*~]
>> with itself and then some kind of [rpole~] to account for the
>> accumulation thereof. After that I don't really know what to do with that.
> one could feed the output of this [rpole~] into a [*~] to the input
> signal (or at any place later, but then it has to be cared for the delay
> of the doppler vd~ too, so better do it first). The exact parameters can
> be found by experiments, but as written in an earlier mail the signal
> reduction can be up to 7dB (I would use slightly more). I don't have
> time right now, but maybe I'll experiment  later...
>>> or nonlinearities of the air I think.
>> I don't know them at all. I've never heard of anyone taking them into
>> account.
> I think there aren't many too. The only case I can imagine and I heard
> of is for huge negative signals, where you can't go beyond 0 pascal of
> air pressure and the signal is clipped physically - but I doubt this
> ever happens in small speakers. Also I'm not sure if this only happens
> in compression speakers.
>>> The pressure directly in front of the cone is related to the
>>> acceleration I think, but I'm not sure about that.
>> It has to : the speaker makes a sound by pushing and pulling on the
>> air, and that changes the pressure.
> yes, but it could also be seen in that way: the cone's travel compresses
> air in an enclosed volume. when the cone is at its travel peak (the
> acceleration is 0) the air is compressed at maximum. And I'm not sure if
> this is the case for a speaker in "free field" (not sure if that's the
> exact term) too. I know that there's something called radiation
> resistance - but I know nothing useful about it yet.
>>> Can anybody confirm that? I think that's not trivial to answer
>>> anyways, because already 10cm farther from the speaker the pressure
>>> and air velocity are different.
>> It has to... the reason why you hear the sound and why sound has a
>> speed, is because neighbouring pressure differences cause pressure
>> differences to propagate. It's a second-order differential equation,
>> as the position of air particles is proportional to their
>> acceleration. The Laplacian of the wave function along x,y,z is
>> proportional to the 2nd derivative of the wave function along t... I'd
>> write it like :
>>   D[D[f,x],x] + D[D[f,y],y] + D[D[f,z],z] = D[D[f,t],t] / v²
>> Where v² is the square of the speed of sound.
>> With a slight coordinate change using imaginary numbers, you can see
>> it as a Laplacian along x,y,z,t instead, in 4-dimensional spacetime,
>> and the Laplacian is equal to zero. But that's only if the air is
>> considered frictionless :)
> sorry, I don't understand that equation I think (I'm not understanding
> difference equations very well in general) and also I have no idea of
> Laplacians until now (I'll read about it and maybe understand later)
>> Ah, that means that the mechanical amplitude (travel) of the wave is
>> much smaller for treble than bass, is that right ?
> exactly.
> cheers
> Martin
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