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Which way is up? (Which way does a loudspeaker driver move?)

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René - Acculution.com

René - Acculution.com

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So when we have room gain (lets say < 80Hz) pressure is still in phase? How does a microphone measure this than?
The microphone does not care how the pressure was generated, so if it is designed correcly it will give you the correct phase. (Its amplifier might flip the polarity, but that is another matter). If you relate the microphone voltage phase back to the loudspeaker behavior, it will be in-phase with the displacement of the driver at very low frequencies. It is 'syringe' with the plunger moving slowly, so the impedance is that of a compliance only. That is also what is seen in my animation.
 

Hayabusa

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The microphone does not care how the pressure was generated, so if it is designed correcly it will give you the correct phase. (Its amplifier might flip the polarity, but that is another matter). If you relate the microphone voltage phase back to the loudspeaker behavior, it will be in-phase with the displacement of the driver at very low frequencies. It is 'syringe' with the plunger moving slowly, so the impedance is that of a compliance only. That is also what is seen in my animation.
Ok, so even if we have a 'perfect' speaker with perfect phase response we would measure a phase shift of 180 degrees going to the higher frequencies?
Is this phase then going up or down?
 
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René - Acculution.com

René - Acculution.com

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Ok, so even if we have a 'perfect' speaker with perfect phase response we would measure a phase shift of 180 degrees going to the higher frequencies?
Is this phase then going up or down?
Between the displacement and the resulting pressure yes, but we of course need to take into account the room modes going from the very low/small rooms to the very high frequencies/large rooms.
 

pjug

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I don't know this stuff but I found another graphic that seems to be opposite from what you show in your simulation (or maybe I just don't understand what you are presenting). Below is a screen shot of a piston moving forward passing through the zero position where its velocity would be maximum, and aligning with the max positive pressure. So this is wrong?
(Screen shot below, moving forward (to the right) at zero position. Moving image is here: https://resource.isvr.soton.ac.uk/spcg/tutorial/tutorial/Tutorial_files/Web-basics-sound.htm)
1678279428114.png
 
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René - Acculution.com

René - Acculution.com

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I don't know this stuff but I found another graphic that seems to be opposite from what you show in your simulation (or maybe I just don't understand what you are presenting). Below is a screen shot of a piston moving forward passing through the zero position where its velocity would be maximum, and aligning with the max positive pressure. So this is wrong?
(Screen shot below, moving forward (to the right) at zero position. Moving image is here: https://resource.isvr.soton.ac.uk/spcg/tutorial/tutorial/Tutorial_files/Web-basics-sound.htm)
View attachment 270237
An infinite tube has a real impedance ('resistor'), so it falls inbetween a free field (mass-like, 'inductor') and a small enclosure (compliance, 'capacitor'), and so the pressure will correlate with velocity in that situation. It all fits nicely.
 

pjug

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An infinite tube has a real impedance ('resistor'), so it falls inbetween a free field (mass-like, 'inductor') and a small enclosure (compliance, 'capacitor'), and so the pressure will correlate with velocity in that situation. It all fits nicely.
Thank you. It seems I have to go deeper if I want to get a grasp on this.
 

pjug

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If we look at the pressure generated in free field from a flat piston in a baffle, we can calculate analytically calculate the pressure generated via the so-called First Rayleigh integral [Fourier Acoustics, E.G. Williams].
View attachment 269995
The pivotal point here is the sign on the righthand side. With w(Q) being the outwards displacement in a point Q on the piston, it is clearly seen that for a piston radiating into free space (mass-like impedance), this displacement is in anti-phase with p(P); the pressure in a measurement point P. This is really all we need to see. There, in general, is a 180 degree phase difference between the outwards displacement of a loudspeaker and the resulting pressure. Which of course means that when it moves outwards, defined as a positive displacement, we clearly get a negative pressure!
Looking at this equation, lets assume position w(Q) is a sine function. The integral of sin(x) = -cos(x) + C. So the overall pressure result is a +cos function.
Velocity is the derivitive of the position function. Since the derivitive of a sin(x) is a cos(x), V of the piston also a +cos function. So then pressure is in phase with velocity, not 180 degrees out of phase.

Or I don't know how to do the math. Sorry to be the annoying student!
 

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I think the integral in the equation is a surface integral over the front of the piston, not an integral with respect to time.

...but this is why all terms should always be specified.
 

mhardy6647

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He seems very cosy and nice, but his explanations are typically not very good.
This is anecdotal, but many of his video "explanations" I've seen were patently incorrect. I stopped watching him/them a long time ago.
I used to think reasonably highly of PS's products, but seeing him in action torpedoed that impression for me, probably irretrievably.
 

pjn

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Yes. Positive displacement is outwards, and at it’s maximum value the velocity is zero. When the driver comes to a stop, the adjacent particles stop with it, but further out the particles still have an outward displacement so you get rarefaction. This is opposite of what many would intuitively say.
Fantastic explanation - as for many things, intuition is only a vague approximation.
 
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René - Acculution.com

René - Acculution.com

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I think the integral in the equation is a surface integral over the front of the piston, not an integral with respect to time.

...but this is why all terms should always be specified.
It is clearly seen to be a surface integral. They Rayleigh integral is a pillar in acoustics, and easy to look up for those interested. This forum is not great for greek symbols and so have look at the text books instead. The details are not too important to know here, as the sign remains for different situations.
 
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René - Acculution.com

René - Acculution.com

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This is anecdotal, but many of his video "explanations" I've seen were patently incorrect. I stopped watching him/them a long time ago.
I used to think reasonably highly of PS's products, but seeing him in action torpedoed that impression for me, probably irretrievably.
This goes for GR Research too; very low level but high confidence. REL has some of the worst explanations (but probably great subwoofers nonetheless...).
 

fpitas

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This is anecdotal, but many of his video "explanations" I've seen were patently incorrect. I stopped watching him/them a long time ago.
I used to think reasonably highly of PS's products, but seeing him in action torpedoed that impression for me, probably irretrievably.
And now, we're left to decide if he's a con man, or just hopelessly confused.
 
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René - Acculution.com

René - Acculution.com

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And now, we're left to decide if he's a con man, or just hopelessly confused.
I think he is confused, but without realizing it. Many unknown unknowns, which is typical for the typical HiFi manufacturer. The designs are probably mainly done by Chris, and he seems very competent.
 

ebslo

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I don't know this stuff but I found another graphic that seems to be opposite from what you show in your simulation (or maybe I just don't understand what you are presenting). Below is a screen shot of a piston moving forward passing through the zero position where its velocity would be maximum, and aligning with the max positive pressure. So this is wrong?
(Screen shot below, moving forward (to the right) at zero position. Moving image is here: https://resource.isvr.soton.ac.uk/spcg/tutorial/tutorial/Tutorial_files/Web-basics-sound.htm)
Things change dramatically when the driver velocity approaches the speed of sound, as it appears to in that animation. Real drivers don't move anywhere near that fast.
 
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René - Acculution.com

René - Acculution.com

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Things change dramatically when the driver velocity approaches the speed of sound, as it appears to in that animation. Real drivers don't move anywhere near that fast.
Animations are tricky. You need to exagerate things to see anything. My piston moves way away from the equilibrium so that we can see its displacement, but in the simulation there is no real displacement, as it is completely linear analysis.
 

ebslo

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Animations are tricky. You need to exagerate things to see anything. My piston moves way away from the equilibrium so that we can see its displacement, but in the simulation there is no real displacement, as it is completely linear analysis.
I wasn't referring to your animation, but the one linked in the post I quoted.
 

ebslo

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I know, but the same goes for that. For realistic displacements it can difficult to get the point across.
Yes, for sure. And that animation isn't really trying to show the phase relationship to the driver, it seems to be trying more to show the relation of particle movement to pressure propagation. But if someone were to infer something about the driver phase relation from it (as @pjug seemed to be doing), then it will be quite misleading as the surface of the driver appears to keep up with the positive pressure region as it propagates.
 
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René - Acculution.com

René - Acculution.com

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I wasn't referring to your animation, but the one linked in the post I quoted.
I know, but the goes for that. For realistic displacements it can difficult to get the point across.
Yes, for sure. And that animation isn't really trying to show the phase relationship to the driver, it seems to be trying more to show the relation of particle movement to pressure propagation. But if someone were to infer something about the driver phase relation from it (as @pjug seemed to be doing), then it will be quite misleading as the surface of the driver appears to keep up with the positive pressure region as it propagates.
It shows that the pressure peaks as the velocity peaks. Which is correct for the resistive loading seen by an infinite tube.
 
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