[PD] overdriven speaker
matju at artengine.ca
Fri Oct 22 07:35:30 CEST 2010
On Thu, 21 Oct 2010, Martin Schied wrote:
> 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).
Why would you be doing the doppler first ? The heat is generated first, in
the moving coil, but the doppler is relative to the observer, who comes
at the very end in the data flow from the amp to the ear. Thus it seems to
me that the [vd~] should come at the very end.
How does one take the réactance into account, again ?... then we'd have to
change the first [*~] to account for ampères not following volts, and do
we have to change the other [*~] too ?
> 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.
I doubt that it (getting close to 0 pascal) happens at all. It sounds more
like a weapon of mass destruction, than like something for listening to.
Though... in some ways, it does happens, at a very small scale. What's the
speed of air molecules, and how much time do they take to fill the void
made by the speaker moving ? What happens if the speaker moves faster than
that ?... (and is that actually the Doppler effect said using different
>> It has to : the speaker makes a sound by pushing and pulling on the
>> air, and that changes the pressure.
To be clear, by pulling I just mean the apparent pulling that happens when
air randomly expands to fill the void introduced by the speaker... it's
this «pulling» that I'm talking about now.
> 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.
I don't know what you mean here by «enclosed volume», «free field», nor
> 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)
I just mean that if you make a map of the pressure in x,y,z and t, you
will find that normal wave transmission means that the sum of the 3 second
spatial derivatives is proportional to the second temporal derivative.
This can also apply if you remove the z and also if you remove the y, so
perhaps it's easiest for you to first see that it works with just x and t.
The notation D[f,t] means something like df/dt but I don't have the
partial derivative symbol nor the other funny notations used in math
texts (indices, multiple-level division).
| Mathieu Bouchard ------------------------------ Villeray, Montréal, QC
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