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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].

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:

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

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].

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:

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

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