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

[NTK]

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.

The acoustic particle velocity is a lot lower than you'd expect. Another one of the non-intuitive aspects of acoustics. Below is taken from An Introduction to Acoustics. In the example shown, at 1 kHz 140 dB SPL, for a plane wave, the amplitude of the acoustic particle velocity is only 0.5 m/s, with a displacement amplitude of only 80 μm. Also notice the relationship between velocity and pressure does not dependent on frequency.

 

[ebslo]

It shows that the pressure peaks as the velocity peaks. Which is correct for the resistive loading seen by an infinite tube.

Ok, so now I'm curious. Is there a difference between a finite-diameter driver in an infinitely long tube vs. a driver with infinite diameter in a free field?

ps. And of course by "infinite" in both cases, I mean much greater (by an order of magnitude or two) than the wavelength.

 

[René - Acculution.com]

Ok, so now I'm curious. Is there a difference between a finite-diameter driver in an infinitely long tube vs. a driver with infinite diameter in a free field?

ps. And of course by "infinite" in both cases, I mean much greater (by an order of magnitude or two) than the wavelength.

I would think not. You are forcing plane waves in both cases for a constant displacement over the pistons, but the plane wave does not satisfy Sommerfeld's radiation condition, so it is a thought experiment, not something physically realizable.

 

[ebslo]

I would think not. You are forcing plane waves in both cases for a constant displacement over the pistons, but the plane wave does not satisfy Sommerfeld's radiation condition, so it is a thought experiment, not something physically realizable.

So as frequency (or driver diameter) increases and the beamwidth narrows, does pressure remain in phase with driver acceleration, or does it shift towards the phase of driver velocity as it would in the hypothetical infinite driver diameter case?

 

[charleski]

Well, that is not really too relevant, as the phase change coming from the propagation is always taken out anyway to not obscure the phase response. So it holds generally that the pressure phasor with that phase removed will be in anti-phase with displacement.

Well there's a lot of jargon there that would need to be picked through. If you don't want to introduce a bunch of other equations to deal with that I think it's really just easier to say that we're looking at the case where R->0 (i.e. right next to the surface of the driver).

 

[René - Acculution.com]

So as frequency (or driver diameter) increases and the beamwidth narrows, does pressure remain in phase with driver acceleration, or does it shift towards the phase of driver velocity as it would in the hypothetical infinite driver diameter case?

There is a big difference between a large piston and an infinite piston. The acceleration will drive the pressure for the finite piston.

 

[René - Acculution.com]

Well there's a lot of jargon there that would need to be picked through. If you don't want to introduce a bunch of other equations to deal with that I think it's really just easier to say that we're looking at the case where R->0 (i.e. right next to the surface of the driver).

Anyone who gets inspired by what I write, and need more information is free to ask for a quotation. The sign is the important thing to note here, and the rest can be found by looking up the integral. Without the integral, people could say the simulation is wrong, but the equation is seen to back up the simulation, regardless of the many details that I could add from there.

 

[LTig]

Well, once again I learned something new. ASR is a great place! Thank you @René - Acculution.com !

 

[René - Acculution.com]

And in the infinite room, free-field, resistive external air-load only?

Sorry?

 

[pjug]

Anyone who gets inspired by what I write, and need more information is free to ask for a quotation. The sign is the important thing to note here, and the rest can be found by looking up the integral. Without the integral, people could say the simulation is wrong, but the equation is seen to back up the simulation, regardless of the many details that I could add from there.

The reference in your original post is available online on a European acoustics laboratory site. I hesitate to post this because I am not sure whether they are OK in doing that. Mods can take this post down if it is encouraging theft of the book.

 

[sarumbear]

In a recent video by PS Audio/Paul, he mentions the sentiment that many probably share with him, that a loudspeaker driver works by compressing air as it moves outwards. This illustrates perfectly that you can work with loudspeakers for decades without ever understanding how a loudspeaker actually works. As we will see in a minute, there is rarefaction as the driver moves outwards, and compression as it moves inwards.

Let us first discuss why intuition tells us to agree with Paul. It could be that you think about sticking your hand out the car window and feeling the pressure pushing back on your hand, and so there effectively is 'positive' or forwards displacement with your hand to the air molecules resulting in a positive pressure. Or you might think about the driver moving outwards for a positive DC voltage from a battery, and that is what is tripping you up. Or maybe you are thinking about the pressure in an enclosure, not necessarily a loudspeaker but a bike pump, where you push the piston in a 'positive' direction and get a positive pressure. But we are talking about acoustics with wave propagation here, and not fluid flow. And how a driver behaves at DC is not how it behave above its characteristic frequency. And free field is different than an enclosure.

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!

You can also show this with a lumped model, but here you need to be careful with how you map vector velocities to current directions. And it also needs the Raleigh integral or some other integral anyway, as the pressure is not found directly in the lumped circuit, only the velocity of the piston is there, and then you calculate from there.

So, simulation is probably the way to go, since here we can clearly see both the displacement and the pressure vary in 'real-time'. We should remember that we are looking steady-state, as we most often do when looking at loudspeakers, but I am working on an article where I show transient behavior of loudspeakers and bass ports, so hang tight. The simulation below was somewhat constructed, since you will actual see for more realistic enclosures that the sound field inside is not constant, but in general the following holds:

"When the driver moves outwards, it generates a negative(!) sound pressure on the exterior side (playing into an acoustic mass; Za=i*omega*f*acoustic mass), and an negative sound pressure also inside the enclosure (playing into an acoustic compliance; Za=1/(i*omega*f*acoustic compliance))".

And that is exactly what we see:

View attachment 270001

Mic drop.

A major issue with learning from experience and not from the theory is that when you relate the acoustic pressure back to the voltage, both in principle complex values, you will see a certain phase relation between the two (I have discussed that in detail in my video on Loudspeaker Phase), but you don't see the phase related to the displacement. And since you don't consider the solid mechanics, you are skipping both the transductance from EM to Mech, and from Mech to Acou, thinking (wrongly) that the driver moves outwards for a positive voltage phasor, and thinking (wrongly) that this outward movement creates a positive pressure. So two wrongs here make a right in some sense, and you can go about your day. But this can be detrimental in other products such a hearing aids, where you main concern is stability, and since you have microphones affected both by the hearing aid moving and the moving housing creating sound, if you don't understand the transductance, you can make a problem much worse than it was before with your design changes.

Now, we should of course move on to more details, such as the transient behavior; what is more exactly the radiation impedance; what happens just behind the dust cap vs behind the surround; what is going on in the enclosure and on and on. But the basics has to be understood before moving on, and the right way to think about the function of a loudspeaker for an elevator pitch, is exactly the opposite of what Paul and so many others believe. Also remember that we can decompose any relevant signal into sinousoids that are steady-state in their very nature, so the above animation pretty much shows us what we need to know.

Let me know if you have a topic with acoustics or signal processing that you just cannot wrap your head around, and let us deal with it.

René Christensen, PhD
Acculution ApS

The sound we hear is a pressure wave. In which case why would we be interested on which way the cone moves? Shouldn’t it move in and out to fallow the electrical signal? Any difference is distortion.

What am I missing?

 

[audiofooled]

Great thread Rene, as always!
Many things are counterintuitive, so understanding some science is always beneficial. If I may, this cool 3d animation, starting from 2:37 shows what is going on. Yes, clearly it isn't like Mr. Mcg. explains it:

As a suggestion, at least to me it would be interesting to see some science behind low frequency wave propagation. That is, since pistons are too small for long wavelengths, how many times does a driver have to move back and forth to create what we would audibly interpret as a, say, single 40 Hz bass transient attack?

Also, if this would be a complex waveform of a recorded kick drum, what would be the order of things when it comes to drivers duty cycle vs. what we hear? I mean, in room, do we first hear mid bass and then the lower frequency thump? With respect to room gain, Is this more driver displacement or room dependent?

 

[René - Acculution.com]

The sound we hear is a pressure wave. In which case why would we be interested on which way the cone moves? Shouldn’t it move in and out to fallow the electrical signal? Any difference is distortion.

What am I missing?

The sign of any variable is important. If Paul were correct we would need to modify how we think of polarity of a speaker. Which I think is a big deal. More importantly it illustrates why it is better to work from first principles instead of analogy and intuition.

 

[René - Acculution.com]

Great thread Rene, as always!
Many things are counterintuitive, so understanding some science is always beneficial. If I may, this cool 3d animation, starting from 2:37 shows what is going on. Yes, clearly it isn't like Mr. Mcg. explains it:

As a suggestion, at least to me it would be interesting to see some science behind low frequency wave propagation. That is, since pistons are too small for long wavelengths, how many times does a driver have to move back and forth to create what we would audibly interpret as a, say, single 40 Hz bass transient attack?

Also, if this would be a complex waveform of a recorded kick drum, what would be the order of things when it comes to drivers duty cycle vs. what we hear? I mean, in room, do we first hear mid bass and then the lower frequency thump? With respect to room gain, Is this more driver displacement or room dependent?

Yes it would be good to show the transient response also. Too many articles in the works already…

 

[sarumbear]

The sign of any variable is important. If Paul were correct we would need to modify how we think of polarity of a speaker. Which I think is a big deal. More importantly it illustrates why it is better to work from first principles instead of analogy and intuition.

I now understood what you want to show but in music playback why would overall polarity matter? Audio signal goes through many stages of unknown polarity until it reaches us.

Has there been a study of the audible effect of polarity on a system where the sound wave polarity on a microphone replicated on a loudspeaker?

 

[restorer-john]

If Paul ea correct we would need to modify how we think of polarity of a speaker. Which I think is a big deal.

If you send an assymetrically clipped sine through a system and mic/display the output on a scope, that clipped sine is correct polarity when the driver is wired positive to excursion.

 

[René - Acculution.com]

I now understood what you want to show but in music playback why would overall polarity matter? Audio signal goes through many stages of unknown polarity until it reaches us.

Has there been a study of the audible effect of polarity on a system where the sound wave polarity on a microphone replicated on a loudspeaker?

The text books would need to be rewritten. His explanation goes entirely against the actual function of the loudspeaker. A sign can be extremely important when solving the problems that I solve.

 

[René - Acculution.com]

If you send an assymetrically clipped sine through a system and mic/display the output on a scope, that clipped sine is correct polarity when the driver is wired positive to excursion.

If by wired positive to excursion you mean with a battery as source then yes but the phase at dc is not the phase above the characteristic frequency.

 

[sarumbear]

The text books would need to be rewritten. His explanation goes entirely against the actual function of the loudspeaker. A sign can be extremely important when solving the problems that I solve.

I’m not interested in his explanation. He is a charlatan as far as I’m concerned. Your explanation is correct (as expected from you). My question was a technical question: Does overall polarity from microphone to speaker matter?

 

[sarumbear]

If you send an assymetrically clipped sine through a system and mic/display the output on a scope, that clipped sine is correct polarity when the driver is wired positive to excursion.

Unless there’s a circuit like that somewhere in the amplifier chain.