# How does a loudspeaker cone generate acoustical pressure?

#### jackocleebrown

##### Member
Technical Expert
Audio Company
Introduction

This post is inspired the thread "Which way is up", started by @René - Acculution.com, and especially by a comment from @BeerBear. In "Which way is up" René shows that when replaying a sine wave, the maximum pressure coincides with the point when the loudspeaker cone is at maximum negative excursion. This is irrefutably true, but on the face of it seems very confusing and counter-intuitive. Apologies for singling out @BeerBear, as he was not alone is asking for a simple explanation of why this happens. There were also a few people asking about the time domain too. I've prepared some animations below.... I hope this adds some clarity and doesn't create yet more confusion but let's see. This is intended to be an accessible post, so I'll try my best to minimise jargon and absolutely no equations. In a couple of places I adjusted the animations to make the physical concepts clearer.

After some thought, I ended up thinking that the most helpful approach would be to look at the situation when the cone starts at rest position and then steps forwards by a set amount (+20mm in this case). Here's how that looks in two different ways. On the left side is a very simple animation of the cone movement, shown as a thick vertical blue line. You can think of this as a cross-section through a flat disc. On the right, the cone position on the vertical axis with time shown on the horizontal axis. I've used a smooth Gaussian step. Labelled on the graph are five phases that the cone goes through in travelling from rest to the eventual 20mm displacement. One key concept that's a prerequisite for understanding this post is that displacement, acceleration and velocity are all ways of describing the motion of a body and they're not independent from one another.

That's the cone movement sorted, so what pressure does it create...? This is the part where it gets slightly complicated, because the resulting acoustical pressure depends on what the cone is driving. We're going to look at three cases, starting with the simplest and most intuitive and ending with the one that gives the surprising result that René showed.

Case 1: The cone drives a enclosed volume of air
If the cone drives a totally sealed volume of air then the acoustical pressure inside the volume increases uniformly with the distance the cone moves*. This makes total sense because no air can escape from the enclosure, and therefore less volume is available when the cone moves forwards. The same amount of air is inside a smaller volume, the "air particles" are forced closer together, and the pressure increases. This behaviour is illustrated below. The little black dots represent air particles and it's easy to see that they're packed more tightly together after the cone has moved forwards.

Case 2: The cone drives an extremely long tube of air

If the cone is connected to an extremely long tube then the air is no longer fully enclosed. This means that eventually the entire column of air in the tube is pushed away from the cone and the air returns to the ambient pressure. This means that rather than seeing a step in the pressure, we instead see that the pressure peaks and then reduces again. The maximum peak in the pressure corresponds to the highest velocity of the cone. In-fact the pressure is directly related to the cone velocity in this situation. Note that there is still some restriction of the air; the side walls of the tube along with the fact that the tube is exactly the same size as the cone mean that the air particles only move from left to right and the size and shape of the wave doesn't change, it simply travels along the tube.

Case 3: The cone is mounted in an infinite baffle
This situation is the closest to what happens with a loudspeaker playing in a room and it's the most difficult case to understand. As the cone pushes forwards, air moves away from the cone in all directions. Initially, during the cone acceleration phase, positive acoustical pressure is generated. However, the critical thing is that the wavefront expands into a larger space and this causes the pressure to reduce quickly as the wave travels away from the cone. In-fact once the cone stops accelerating the pressure returns to zero even though the cone is still in motion, and when the cone decelerates the pressure goes negative. Overall the result is that positive pressure is associated with acceleration of the cone and negative pressure is associated with deceleration of the cone. Note that in the graphic below I've magnified the particle movement to make it visible. Here we see René's result, at maximum cone displacement, negative pressure was generated.

Summary
• In all cases the initial acoustical pressure has positive polarity corresponding to forward movement of the cone from rest position.
• When a speaker drives a sealed volume, the acoustic pressure in the volume is proportional to the cone displacement
• When a speaker drives a long tube, the acoustic pressure in the tube is proportional to the cone velocity.
• When a speaker drives an open space, the acoustic pressure in the space is proportional to the cone acceleration.
• Only the first case, the enclosure, can result in a static change in the acoustic pressure.
The final question: given the above, how on earth can a speaker replicate the audio signal?
The outcome of this is that the cone acceleration needs to replicate the recorded audio signal. This is where things get very fortunate because, provided the cone mass dominates the mechanical behaviour (as opposed to the stiffness or damping of the surround/suspension) and the driver motor converts the audio signal directly to a force signal, this will naturally result in cone acceleration that matches the audio signal.**

EDIT: 05/10/2023
To drill down a bit more into the reason why case 2 and case 3 are different I've added two small cyan boxes to the case 2 and case 3 animations above. These boxes represent "volume elements" and we can use them to get a better understanding of the physics of what's going on. I'm hopeful that gives a bit more of an intuitive view of why it's the expansion of the sound-wave that causes the different behaviour in case 2 and case 3.

The pressure inside these volume elements is approximately uniform, so that means it only depends on how the volume element edge lengths are changing as the wave passes. Zooming in on the volume element in case 1 we can see that whole volume element is pushed to the right by the cone movement. The only movement is left to right. The vertical edges of the element are perpendicular to the cone and wave movement and always remain the same length. The horizontal edges are parallel to the cone and wave moment and they are compressed as the wave passes. The compression of the horizontal edges means that the volume element is smaller and hence the pressure is higher during this period.

Case 3 behaves differently in a significant way. The motion of the air particles is not parallel. The air particles "spread out" as they are pushed away from the cone. This means that the vertical edges of the volume element are stretched as the wave passes. It is this stretching motion that leads to negative pressure as soon as the cone starts to decelerate. The final shape of the volume element is no longer square. The final volume of the volume element is the same as the initial volume and hence the initial and final pressure are also the same, but the vertical edges have been stretched and the horizontal edges have been compressed.

*To help keep this simple I've ignored second order effects like standing waves.
**Yes, there's a lot more to it than that, but it's beyond the scope of this post to discuss it here.

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#### René - Acculution.com

##### Senior Member
Technical Expert
Good stuff! Thanks for this, Jack. And yes, hope this clarifies and not confuses anyone.

#### Purité Audio

##### Master Contributor
Industry Insider
Barrowmaster
Forum Donor
Educational, informative and entertaining super posting.
Thank you.
Keith

#### Grotti

##### Addicted to Fun and Learning
Forum Donor
Is all of this also true for a flexible plate diaphragm like the Manger transducer?

I am asking because it seems to create positive and negative air pressure nearly at the same time:

Btw: thank you very much for your posting, learned one or two things

OP

#### jackocleebrown

##### Member
Technical Expert
Audio Company
Is all of this also true for a flexible plate diaphragm like the Manger transducer?

I am asking because it seems to create positive and negative air pressure nearly at the same time:

Btw: thank you very much for your posting, learned one or two things
You're welcome.

Yes, the same applies but with the Manger (or a conventional cone in breakup) some parts of the cone are moving forwards while others are moving backwards. Even though it's a much more complicated situation, the acceleration of the different parts of the cone determine the radiated pressure in the same way as case 3.

#### Curvature

##### Major Contributor
The final question: given the above, how on earth can a speaker replicate the audio signal?
The outcome of this is that the cone acceleration needs to replicate the recorded audio signal. This is where things get very fortunate because, [1] provided the cone mass dominates the mechanical behaviour (as opposed to the stiffness or damping of the surround/suspension) and [2] the driver motor converts the audio signal directly to a force signal, this will naturally result in cone acceleration that matches the audio signal.
Why are 1 and 2 necessary to mention? Are there speaker drivers where this is not the case?

Thanks.

#### Music1969

##### Major Contributor
Some manufacturers are scared of ASR and distance themselves from this site.

KEF are not afraid, their products demonstrate solid engineering - kudos to KEF Team

Keep up the great work

OP

#### jackocleebrown

##### Member
Technical Expert
Audio Company
Why are 1 and 2 necessary to mention?
Because they aid understanding of why it can't be cone displacement that determines pressure in case 3.
Are there speaker drivers where this is not the case?
This is a property of the radiation condition. Speakers with horns/waveguides have slightly different behaviour but the same principles apply; the key is how restricted the sound waves are.

#### MCH

##### Major Contributor
Thanks for the explanation, very cool graphs.
I believe for some people like me it might be easier to understand if explained in terms of energy: it all comes from the fact that there is no energy cost in keeping something (the speaker membrane and the gas around it) moving at a constant velocity but there will be energy involved when the acceleration is not zero.

I would love someone to expand the explanation on how all this relates to the electrical signal. For instance, in simple a sine wave signal, at what point is the acceleration=0? At the intersection of 0V? at the peaks and valleys? Never?

#### Curvature

##### Major Contributor
Because they aid understanding of why it can't be cone displacement that determines pressure in case 3.
Sorry, I thought there was chance I would confise after I posted. You mentioned two new points not covered before in youre conclusion that i labelled 1/2 in the snippet I quoted. Should have used A/B and wrote more clearly.

" [1] provided the cone mass dominates the mechanical behaviour (as opposed to the stiffness or damping of the surround/suspension) and [2] the driver motor converts the audio signal directly to a force signal"

#### René - Acculution.com

##### Senior Member
Technical Expert
Thanks for the explanation, very cool graphs.
I believe for some people like me it might be easier to understand if explained in terms of energy: it all comes from the fact that there is no energy cost in keeping something (the speaker membrane and the gas around it) moving at a constant velocity but there will be energy involved when the acceleration is not zero.

I would love someone to expand the explanation on how all this relates to the electrical signal. For instance, in simple a sine wave signal, at what point is the acceleration=0? At the intersection of 0V? at the peaks and valleys? Never?
At low frequencies below the driver's fundamental resonance (we are talking standard electrodynamic speakers), the phase of the displacement will be that of the voltage, whereas at higher frequencies it will be the phase of the acceleration being dictated by the voltage. This will also be seen in the pressure phase; it does not just simply follow the voltage phase, but varies across the frequency range.

#### Scgorg

##### Active Member
Great post. Having these easily digestible texts with animations is really great for furthering people's understanding of loudspeaker behaviour. I am personally guilty of reading a lot of AES papers where some of the math presented goes above my head (an issue which luckily has lessened with time), so having the concepts themselves so concisely explained is wonderful. Thank you!

OP

#### jackocleebrown

##### Member
Technical Expert
Audio Company
Sorry, I thought there was chance I would confise after I posted. You mentioned two new points not covered before in youre conclusion that i labelled 1/2 in the snippet I quoted. Should have used A/B and wrote more clearly.

" [1] provided the cone mass dominates the mechanical behaviour (as opposed to the stiffness or damping of the surround/suspension) and [2] the driver motor converts the audio signal directly to a force signal"
Oh, sorry for the misunderstanding. Case 3 says we should try and design a driver where the acceleration of the cone is dictated by the input signal. This sounds complicated but there's a great way to do it. Newton's laws of motion say that if we apply a force to a mass it will accelerate. F=ma. This equation is valid if force varies with time and in this case the mass acceleration would vary with time according to the same signal as the force. So it gives a nice route to designing a speaker. We need a motor to convert the Audio signal to a force signal, we use that to push a mass and we get the Audio signal replicated as an acceleration signal! More or less every driver uses this principle in the main working band.

#### René - Acculution.com

##### Senior Member
Technical Expert
You still in principle want a constant surface acceleration to drive the acoustic pressure assuming a free-field load, but now it is a different transductance principle plus you have a perforated plate right up against the membrane, so the load is actually different and the membrane is probably relatively lightweight with the air coupling being more important.

#### pjn

##### Active Member
Thanks! More complex than one would intuitively think. I would guess that original sound reproduction devices ran more on intuitive design and then later, smart people went "how does that actually work". Very cool.

#### audiofooled

##### Addicted to Fun and Learning
Thanks! More complex than one would intuitively think. I would guess that original sound reproduction devices ran more on intuitive design and then later, smart people went "how does that actually work". Very cool.

Very complex indeed and at times even counter intuitive. I've learned this the hard way when trying to build a "better" sub. The equations are way over my head but my intuition at least improved with a bit of cost of killing an amp in the process. Science matters

This video may or may not improve someone's intuition, but I think it's worth looping starting from this sequence:

If you think about it, cone needs to move forward to initially push the adjacent particles closer together but it's the cone's acceleration backwards that really puts things in motion. In the end, it's the pressure that counts.

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pjn

#### Curvature

##### Major Contributor
@jackocleebrown @René - Acculution.com

Similar thing happens in acoustics, where air velocity decreases as it approaches a wall but pressure increases, max pressure/zero velocity being on the wall.

Is there some analogy between that and a driver approaching x-max? Could you please comment?

#### Jorge

##### Member
... The outcome of this is that the cone acceleration needs to replicate the recorded audio signal. This is where things get very fortunate because, provided (1) the cone mass dominates the mechanical behaviour (as opposed to the stiffness or damping of the surround/suspension) and (2) the driver motor converts the audio signal directly to a force signal, this will naturally result in cone acceleration that matches the audio signal.**
(My markings in boldface)
Thanks @jackocleebrown for your great educational post. Just some comments to try to complete the exposition (if you find these too basic or obvious, just ignore this post):
1.
Cone and suspension form an elastic system, and all elastic systems have mass, stiffness and dumping, with all three effects contributing in some amount to the "force" in Newton law during the cone movement.
With this in mind, could you explain more precisely what is intended by the highlighted "mass dominates..."?
I know that you wanted to keep equations to a minimum, but given that you already mentioned the Newton law in the "F=Ma" form, perhaps it would be pedagogically useful to include here also the "Ku" and "Cv" terms. Standard texts on vibrations present this as:
F_ext = Ma+Cv+Ku
where a, v, u are respectively acceleration, velocity and displacement and, M, C, K are mass, damping and stiffness of the structure or medium. I like to interpret this form of the Newton law as "Inertial forces are the difference between external forces and internal forces", where internal forces are the forces due to the stiffness and damping of the medium.
Again, in this context, how should one interpret the "mass dominates..." quote?
2.
Regarding the three great looking animations in your original post, we see that in the first one, compression of the enclosed air is a static process, while in the two subsequent ones (long tube and emission to a room) there is wave propagation. So, what determines if wave propagation will or won't occur in a certain medium? The concept of "wavelength" and wavelength-to-enclosure-size "ratio" play a role here. If the box size is less than the wavelength, then simply there is not enough space to be travelled by a wave, and the compression looks as a static process. On the contrary, wave propagation may eventually occur if the box size is larger than the wavelength (and a room is just a larger box), as there is travel distance available for a wave and also, because the very same Newton law imposes, when all terms are considered, that force applied at a certain point in space takes some time and distance to produce an effect at a remote location.
3.
Electrostats and Manger-style drivers have been mentioned in the comments. In these, and also in cones, wave propagation may occur in the same manner as described in point 2., except that in 2. the medium is enclosed air and here, it is the elastic solid material of the membrane or cone. Elastic properties and density of these materials determine their wave propagation velocity, while this velocity together with frequency of an excitation determines wavelength. Finally relationship between wavelength and medium size (cone or membrane size) determine if wave propagation will occur inside the membrane or cone. If this happens, it is informally known as membrane or cone ringing. If not, it is said, roughly, that the behavior of the membrane or cone is pistonic. Most manufacturers try to avoid ringing, but some try to use it in some way (remember reading something to this effect in some white paper or web info from Mark Audio drivers, but not truly sure, maybe it was Jordan?). Is ringing used or avoided in Manger or electrostats drivers?

Regards and thanks again for your post,
Jorge

pjn

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