[PD] overdriven speaker

Martin Schied crinimal at gmx.net
Thu Oct 21 17:48:02 CEST 2010

>> 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 ?


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