Where Does the Energy Go in Noise Cancellation and Sound Wave Interference?

[Jallu]

Noise cancellation through interference; I guess most people here know what I am talking about.

My question is: where does the energy go?

In a (sound)wave there is constant exchange between kinetic and potential energy, or at least that's how I see it.
Now the wave gets canceled through interference and the result is a 'wave' with 0 amplitude. Is all the air now standing still and has all the energy dissipated into heat?

 

[Xezlec]

This happens to be one of my favorite questions.

First, you have to picture the circumstances under which waves cancel. Let's say I have 2 speakers in basically the same spot, emitting opposite waveforms. The waves will (mostly) cancel from the perspective of an observer a reasonable distance away from the speakers.

So where does the energy go? What energy? It simply doesn't take any energy (ideally) to vibrate the speakers in that way. Neither speaker will (ideally) produce any load on the driving circuit. Of course, in reality it takes some energy because the speakers, air right in between the speakers, and electrons in the wires still have to be shoved around. And of course there's still a tiny bit of radiated sound regardless, since the speakers must be some distance apart. But let's not count that.

If you want a more detailed explanation of what's going on, you have to actually think about how the acoustic impedance seen by the speaker changes when you add the other speaker, and therefore how the electrical impedance seen by the circuit changes in response. Both kinds of impedance will drop like crazy.

Alternate explanation: if, instead, you'd rather think in terms of the two systems separately and then add them together, then you have to remember that each speaker is also a microphone, and therefore an amount of power is generated by receiving the other speaker's sound, which happens to be just equal to the amount of power being radiated by this speaker.

Is that enough explanations for you?

 

[da_willem]

If you want a more detailed explanation of what's going on, you have to actually think about how the acoustic impedance seen by the speaker changes when you add the other speaker, and therefore how the electrical impedance seen by the circuit changes in response. Both kinds of impedance will drop like crazy.

But the (at least my) intuitive problem is that it takes time for the two speakers to notice each other, what happens (locally) with the energy emitted when the two speakers didn't yet noticed each others presence?

And in equilibrium, how does this impedance thing you mention work. Doesn't the driver still moves in the same way, converting kinetic energy into pressure waves, in the presence of the other speaker? Where does the sound energy go, or is there in some way less sound energy emitted?

 

[Xezlec]

But the (at least my) intuitive problem is that it takes time for the two speakers to notice each other, what happens (locally) with the energy emitted when the two speakers didn't yet noticed each others presence?

As I said, you still have to expend a tiny bit of energy to move the air that sits in between the speakers (assuming the air has some amount of friction going on), but of course the waves don't "cancel" right in between. In the small space in between the speakers, they actually add constructively, giving you a louder sound!

And in equilibrium, how does this impedance thing you mention work. Doesn't the driver still moves in the same way, converting kinetic energy into pressure waves, in the presence of the other speaker? Where does the sound energy go, or is there in some way less sound energy emitted?

I don't know what you mean by equilibrium in this case. But you sound like you didn't understand my explanation, which I thought directly asked and answered both of those questions. I can try to reword it, but I don't see which part you're not understanding. Did you read the whole thing? Sorry if I'm being annoying.

I'll try to come up with a better answer by the end of the day though, if no one else chimes in.

 

[Jallu]

I think I understand the phenomenon in case of waves which are traveling in the same direction.

But is cancellation possible when waves travel in opposite direction, e.g. after (delayed?) reflection of a sound wave? First I thought this was possible, but now it seems to me that in this situation cancellation only occurs at the anti nodes resulting wave, whereas amplification occurs at the antinodes. Is that true?

 

[da_willem]

As I said, you still have to expend a tiny bit of energy to move the air that sits in between the speakers (assuming the air has some amount of friction going on), but of course the waves don't "cancel" right in between. In the small space in between the speakers, they actually add constructively, giving you a louder sound!



I don't know what you mean by equilibrium in this case. But you sound like you didn't understand my explanation, which I thought directly asked and answered both of those questions. I can try to reword it, but I don't see which part you're not understanding. Did you read the whole thing? Sorry if I'm being annoying.

I'll try to come up with a better answer by the end of the day though, if no one else chimes in.

It sounds like you didn't understand my objections; I do understand the thing you're saying but I'm wondering about the underlying mechanism.

Imagine this: due to some sound barrier the sound can only move in one direction. i.e. instead of some spherical like wave it is more or less a plane wave which is for the two closely lying speakers almost parallel. After some distance these waves start to overlap, which will give desctructive interference given the right phase difference in the signal. So before they started to overlap there was some sound energy, which is gone now when the two waves overlap. Is this energy dissipated in the air?

For the speaker the motion of the driver remains the same you say, well without the other speaker some of the kinetic energy is taken away by the pressure field of the surrounding air. Now when the second speaker is present, this energy transport reduces right, due to destructive interference?

So now either
-The energy of the two sound waves is dissipated in the air, where there is destructive interference, which means the speaker will emit the same sound energy.
-Or, the impedence changes such that there is less sound energy emitted. Now the kinetic energy of the driver (if this remains the same) is dissipated in the circuit.

With the second option I have the problem that, looking back at the first situation I sketched, initially the two speakers didn't know the other was there. Now, how can less sound energy be emitted even though the effect of cancellation is yet to come?

Maybe (probably?) the cancellation in this circumstance can't be perfect because of the two source not truly overlapping, thus causing always regions of constructive and destructive interference, with the total sound energy unchanged?!

 

[Jallu]

I think that noise cancellation does not work with parallel speakers, only with speakers which are mounted 'in series'.

Theoretically, noise cancellation with 2 parallel speakers is only possible if the cones are vibrating at the same spot but in opposite direction = no movement of the cone. Otherwise there will be an interference pattern rather than noise cancellation. Of course, interference means that in certain regions (the nodes) there will be no movement of air, but all-in-all there will still be a wave. The energy is transferred to other particles (at the antinodes) now moving with bigger amplitudes.

Only when 2 speakers are mounted 'in series', a wave can be canceled completely by superposition. Still not sure where the energy goes though in this case, though.

 

[Jallu]

In other words, in case of parallel sources, complete noise cancellation is only possible by vibration cancellation, which is preventing actual noise from coming into being.

Cancelling an existing (sound)wave is only possible by cancelling the vibration of an object which is (supposed to be) transfering the wave (a 2nd speaker, a vibrating wall or whatever) towards the observer. The 'serial' version.

 

[da_willem]

This might very well be the case; there is only local noise cancellation and overall the square of the amplitude integrated over space doesn't change. But then there would be no effect whatsoever in the used power of a speaker.

 

[Jallu]

I still don't think it affects the energy emitted from the speakers.

Now I also realize that the initial sound energy (kinetic and potential) is not dissipated as heat by noise cancellation.
Simply because in that case it would be the same as 'damping'.

 

[Xezlec]

I apologize for the delay. I've been ridiculously busy.

I think I understand the phenomenon in case of waves which are traveling in the same direction.

But is cancellation possible when waves travel in opposite direction, e.g. after (delayed?) reflection of a sound wave?

No, absolutely not. In fact, waves of the same frequency propagating in opposite directions create what is known as a http://en.wikipedia.org/wiki/Standing_wave" .

First I thought this was possible, but now it seems to me that in this situation cancellation only occurs at the anti nodes resulting wave, whereas amplification occurs at the antinodes. Is that true?

Yes.

I think that noise cancellation does not work with parallel speakers, only with speakers which are mounted 'in series'.

Theoretically, noise cancellation with 2 parallel speakers is only possible if the cones are vibrating at the same spot but in opposite direction = no movement of the cone. Otherwise there will be an interference pattern rather than noise cancellation. Of course, interference means that in certain regions (the nodes) there will be no movement of air, but all-in-all there will still be a wave.

Unless the speakers are at exactly the same point in 3D space (which makes no sense), there will always be some pieces of air with sound. If the speakers are nearby, though, and if their phases are aligned the correct way (depending on the directions that the speakers are pointed), there will be very little sound far away from the two speakers. I mean far away relative to how far apart they are spaced.

Think about it this way: let's say we have two speakers pointed in the same direction, perpendicular to the line between them. They have opposite phase. Now, the air that is "swished forward" by one speaker just gets "swished back" by the other (in a circular way). Far away from the speakers, then, it doesn't look like any air is coming toward you or away from you. The speakers are mostly just moving the air around in a little circle, roughly speaking. Hardly any of the energy propagates outward.

There's no pressure wave coming toward you because it's all sabotaged by the "vacuum wave" from the other speaker which is "about the same distance away". Of course, if the frequency being played is so high that by the time the wave propagates across the distance between the speakers, a significant portion of the cycle has gone by, then weirder things happen. I'm assuming the speakers are much closer together than one wavelength of the sound in air. A wavelength of normal sound in air is pretty long. I guess I shouldn't have assumed that was obvious.

The energy is transferred to other particles (at the antinodes) now moving with bigger amplitudes.

Oh, now that's certainly not true. Just because waves cancel somewhere doesn't mean they get "exactly that much bigger somewhere else", if that's what you're thinking. You can put 2 speakers very close together and almost all of their output will cancel. In a few tiny places, the sound will increase, but never more than double. The total energy output definitely goes way, way down.

Only when 2 speakers are mounted 'in series', a wave can be canceled completely by superposition.

If by "in series" you mean "facing along the same line", you're not correct that they completely cancel. You're thinking 1-dimensionally. We have 3 dimensions to play with. But regardless of whether they face along the same line, if they have opposite phases, they will cancel in many places. And when I say cancel, I don't just mean they create standing waves with nodes and antinodes. I mean the speakers will output far less energy.

Still not sure where the energy goes though in this case, though.

I keep wanting to answer this, but my answer would be exactly the same as what I said before. What energy are you talking about? There's no extra energy that needs to go anywhere. To move the speaker cone the same distance, the speakers have to do much less work (work = force x distance) if they are just shuffling a tiny piece of air back and forth, than they do if they want to move back and forth a whole universe of air. It takes more force to accelerate lots of air than it does to accelerate the little bit of air between two speakers. I just can't visualize why you expect the speakers to draw the same amount of power in these two cases.

Imagine a (strong!) record player turntable that always spins at the same speed. If I put a cow on top of the turntable, do you not think the power consumption goes up?

And I still don't see what was wrong with my superposition explanation: if you are really imagining superposition of the two "speaker + air" systems, then you have to also remember that a speaker is equivalent to a microphone, and that the speaker will "pick up" almost as much electrical energy from the sound coming from the other speaker as it "lays down" itself. Superpose the electrical systems too. Power flowing out + slightly less power flowing in = only a little power flowing out. Both acoustically and electrically speaking.

I realize I'm just repeating myself, but I don't know what else to say.

I still don't think it affects the energy emitted from the speakers.

For God's sake... I promise you it does! Get a bleedin' voltmeter, a bleedin' function generator, and a couple speakers. I mean, I swear to ya...

 

[Xezlec]

It sounds like you didn't understand my objections

I agree completely! I will keep trying.

Imagine this: due to some sound barrier the sound can only move in one direction. i.e. instead of some spherical like wave it is more or less a plane wave which is for the two closely lying speakers almost parallel. After some distance these waves start to overlap,

Maybe I'm not reading this right, but I'm not sure what you mean by this. I'm going to guess that you are describing 2 speakers separated by a partition, and when you talk about the waves overlapping you just mean when the "initial trajectories" meet at the ends of the partition. Of course, the waves superpose (add together) everywhere. Sound bends around barriers. It's a wave after all.

which will give desctructive interference given the right phase difference in the signal. So before they started to overlap

"Before"? Are we talking about time or space? I'm assuming space, and that you are just talking about the regions where there is some sound, around the partition.

there was some sound energy, which is gone now when the two waves overlap. Is this energy dissipated in the air?

It sounds like you are confusing time and space. Some energy is being expended to move the air in the loop around the partition (and, in this case, also to radiate sound waves in the direction perpendicular to the partition, I believe). Almost no energy is radiated in the direction parallel to the partition. There is no energy here that exists at one time and disappears at another time. There is energy at one point in space, and none at another point. There's nothing wrong with that.

You might as well picture a spinning wheel in a frictionless world, and ask "there's energy here in the wheel, but no energy a foot away from the wheel. Where did the energy go?" It didn't go anywhere, it is simply in the wheel and not anywhere else, and no energy is leaving the wheel.

Similarly, you have speakers spinning a ring of air back and forth, and none of it is propagating in some directions because it is just getting jerked around in a circle.

For the speaker the motion of the driver remains the same you say, well without the other speaker some of the kinetic energy is taken away by the pressure field of the surrounding air. Now when the second speaker is present, this energy transport reduces right, due to destructive interference?

So now either
-The energy of the two sound waves is dissipated in the air, where there is destructive interference, which means the speaker will emit the same sound energy.
-Or, the impedence changes such that there is less sound energy emitted.

Yes! The second one.

Now the kinetic energy of the driver (if this remains the same) is dissipated in the circuit.

huh?

The kinetic energy of the driver is supplied by the circuit. It is dissipated (spread outward) by the air. But since we have less acoustic impedance in the cancellation case, and since acoustic impedance is what determines how fast (if at all) the air dissipates (spreads outward) the energy, it dissipates at a slower rate than in other cases. Since the energy dissipates more slowly, the circuit doesn't need to replenish it as quickly.

With the second option I have the problem that, looking back at the first situation I sketched, initially the two speakers didn't know the other was there. Now, how can less sound energy be emitted even though the effect of cancellation is yet to come?

Of course some energy has to be supplied to get the air sloshing around in the first place. Once it's moving, though, it only takes a little energy to keep it going.

Maybe (probably?) the cancellation in this circumstance can't be perfect because of the two source not truly overlapping, thus causing always regions of constructive and destructive interference, with the total sound energy unchanged?!

This might very well be the case; there is only local noise cancellation and overall the square of the amplitude integrated over space doesn't change. But then there would be no effect whatsoever in the used power of a speaker.

This just isn't true. I don't know how to convince you.

And I fail to see how two sound waves would add to produce an amplitude greater than twice the amplitude of the original.

 

[Jallu]

What if I put two gasoline motors close enough together? Will there be noise cancellation?

Can you refer to any scientific articles in which 3D-noise cancellation from 2 speakers or other sources is discussed? I didn't know it was possible, and I am still not convinced.

 

[Jallu]

@Xezlec, can you refer to an article supporting your theory or did you develop this all by yourself?