1/4 wavelength confusion

[Vaughan Odendaa]

Hi everyone,

I bought myself the Master Handbook of Acoustics and I have learned so much in three weeks. In fact I can't believe how much I've learned. Nevertheless, I am confused on a few details.

When it comes to absorbtion, 1/4 wavelength placement is supposed to be ideal because particle velocity is greatest at that wavelength, and of course, pressure is zero.

Now in the book I am given hints that peaks form at 1/4 wavelengths. Or at least that they do not say that they form nulls. Perhaps someone could nudge me in the right direction so that I can see where I've gone wrong (perhaps cite a paragraph from the book).

But when referencing how the ear mechanisms work, for example, the ear boosts frequencies at 1/4 wavelength in the ear canal. I am given reason to believe that 1/4 wavelengths (from boundries) will cause peaks in the response.

Please, those knowledgeable in this area, please correct me if I'm wrong.

Then nulls. I am not entirely sure at what distance from the boundry nulls form at. Ethan Winer has said on his website (and videos) that nulls are formed at 1/4 wavelength from the boundry. So I'm a little confused now.

Perhaps someone can set this straight for me. Thanks.

--Sincerely,

 

[Savant]

Vaughn,

When a sound wave interferes with a boundary, the reflecting wave (assuming normal, 0°, incidence) interferes with the incident wave, forming particle pressure maxima at at nλ/4 distances from the boundary, where n is zero or an even integer greater than zero and λ is the wavelength. Particle pressure minima (ideally, 0) are formed at mλ/4 distances, where m is a non-zero, positive, odd integer. This model assumes the boundary is a perfect reflector (no damping) at the frequency of interest.

Everywhere particle pressure is a maximum, particle velocity is a minimum. Therefore, (ideally) particle velocity is zero at the wall, a minimum at nλ/4, and a maximum at mλ/4.

For absorbing problems in a room, resistive absorbers, such as foam or fibrous panels, are a common tool. Since a resistive absorber acts on particle velocity, it would be ideal to place the absorber at the mλ/4 distance from the wall at the frequency of interest. The other kind of absorber is a reactive type - typically one of the various "Helmholtz resonator" variety. These reactive absorbers act on particle pressure and thus should be placed at the nλ/4 points for the greatest effect at the frequency of interest.

Now, this all works very well in the theoretical world. The real world presents a different set of conditions. One is no longer interested a single frequency, but in many (all) frequencies. One is no longer dealing with perfect boundaries that reflect all the energy back into the room. One is faced with many locally reacting boundaries - wall, ceiling, floors, soffits, coffee-tables, etc. - as opposed to a single reflector. When the theory is applied to a practical situation, what was an absolute rule becomes a guideline. Instead of resistive absorber placement at mλ/4 distance for a single frequency, placement at mλ/4 for the lowest frequency one wishes to treat is a good guideline. In practical experiments (again, based on theory), resistive absorbers placed a distance equal to their thickness away from the reflective surface maximizes their performance at the lowest frequencies. I.e., place a 2" glass fiber panel 2" from the wall.

As for your ears, instead of a boundary, the ear canal can be viewed as a tube closed on one end. The basic mechanical resonance of such a device is at frequencies where the length of the tube corresponds to mλ/4 as defined above.

I hope this helps!

 

[Ethan Winer]

Vaughan,

> Ethan Winer has said on his website (and videos) that nulls are formed at 1/4 wavelength from the boundry. <

Yes, though as Jeff explained it's exactly 1/4 wavelength only when waves strike the boundary straight on. But the main point is this:

An absorber does the most good at a given frequency when it's 1/4 wavelength away, but it doesn't have to be that precise distance to still make a meaningful improvement. It's not necessary for a panel to absorb 100 percent to help either. For example, if a material absorbs only 0.25 at a given frequency, that reduces the strength of the reflection by 3 dB. So a 1/4 wavelength null that had been infinitely deep (in theory) is now only 11 dB deep. This is a very big improvement!

--Ethan

 

[Savant]

-∞ + 3 = -11?

...or...

∞ = 14?

 

[Vaughan Odendaa]

Savant, in my book it says that particle velocity is maximum before a sound wave hits the wall and pressure is maximum upon impact. And vice-versa. Everast states this.

Perhaps you can explain. He also states that when particle velocity is greatest that is when you want your absorbing material to be absorbed. The faster the particle is traveling, the more heat it will cause because of vibrating particles.

So I am a little confused. I don't know if you have the book I'm refering to, but if it's possible, could you please post references to what you are talking about.

Thanks.

Hi Ethan,

I am still a little confused. I know that it has been said on several occasions that nulls occur at 1/4 wavelength, and I'm not disagreeing, it's just that I am getting confused here because in the book
(Master handbook of Acoustics) it tells me or at least insinuates that 1/4 wavelength even frequencies are peaks, or boosts. Perhaps 1/4 wavelength odd multiples of frequencies are nulls. I'm not sure on that.

Unless I've missed something here. I don't want to be an annoyance, but I would really appreciate it if you could help me to understand what Everast is saying.

Thanks.

--Sincerely,

 

[Ethan Winer]

Jeff,

> -∞ + 3 = -11? <

In this particular case, Yes.

This is from a post I made at the SOS forum when the guys there were arguing about whether a closed glass window might help to absorb bass by reducing the strength of the reflections:

If a reflection is 100 percent it will create a null of infinite depth 1/4 wavelength away from the boundary.

If you reduce the reflection by 1 dB the null is now only 19 dB deep.

Reduce it another dB and now the null is only 14 dB deep.

Reduce it again to -3 dB and now the null is only 11 dB deep.

If you reduce it to -6 dB the null is now only 6 dB deep.

I determined these values empirically using sine waves in Sound Forge. I created a sine wave, made a copy, reversed the polarity, and combined the copy and original with varying level differences.

The point is that reducing the strength of an out-of-phase reflection even a little can make a big change in the depth of the null. If you're familiar with a Twin-T notch filter circuit, it's the same principle. If this still isn't clear let me know and I'll find another way to explain it. I'm sure my expert friend Bill can show me how to express this in math terms you'll understand.

--Ethan

 

[Ethan Winer]

Vaughan,

> (Master handbook of Acoustics) it tells me or at least insinuates that 1/4 wavelength even frequencies are peaks, or boosts. Perhaps 1/4 wavelength odd multiples of frequencies are nulls. <

Assuming the sound waves arrive perpendicular to the boundary:

1/4 wavelength away from the boundary is a null.
2/4 wavelength away from the boundary is a peak.
3/4 wavelength away from the boundary is a null.
4/4 wavelength away from the boundary is a peak.
5/4 wavelength away from the boundary is a null.

And so forth. There's a program you can download from my company's site that accepts a frequency and displays all the 1/4 wavelength distances, or takes a distance and gives all the related frequencies.

Is this what you're asking? If it helps, everything you said so far is correct. Porous absorbers do the most work 1/4 wavelength away, and pressure absorbers like panel traps work best right at the boundary. So it seems to me you understand it perfectly.

--Ethan

 

[Vaughan Odendaa]

Ethan, thank you for attempting to clear this up for me. Perhaps you could explain to me why nulls occur at 1/4 wavelength. The theory behind that would benefit me most.

You know, I thought I would have understood it, but somehow something is not clicking. I think the book has made me, if anything, even more confused. Perhaps other members who have this book could reference a paragraph or two that would elucidate the matter.

As always, thank you.

--Sincerely,

 

[Sarius]

Ethan, thank you for attempting to clear this up for me. Perhaps you could explain to me why nulls occur at 1/4 wavelength. The theory behind that would benefit me most.

Think of the signal as a continuous chain of peaks and valleys hitting the wall and bouncing off. At 1/4 wavelength the reflected peak hit the incoming valley and fills it in causing the null. At 1/2 wavelength the reflected peak and the incoming peak match up and become twice as big. The peaks and nulls thus occur every 1/4 wavelength as the peaks and valleys alternately fill and reinforce each other. Now you need to understand that nothing is 'destroyed' during the fill-in or reinforcement process, the reflected and incoming waves just keep moving through each other and the process continues every 1/4 wavelength as they meet.

To best understand this, take a piece of paper, draw a vertical line and a sinewave across the sheet just touching it. The just below it, draw another sinewave that is going the other way- that is it's going down where the original is going up. Remember that a full wavelength is the complete up and down cycle, that is, both the zero to peak cycle plus the zero to valley cycle. Thus, going from zero to the maximum (or minimum) amplitude is 1/4 wavelength.

Now realizing that you're looking at your wall with the incoming and reflected waves, it should become pretty obvious as to why you get the peaks and nulls.

Hope this helps.

 

[Vaughan Odendaa]

I know why there are peaks and nulls, it's just that I never understood the reasons behind 1/4 wavelength. I think that Everast is one of the most knowledgeable acousticions who ever lived.

Perhaps I'm wrong. But he makes several references that 1/4 wavelengths will result in peak amplitudes and that 1/2 wavelength will resut in zero amplitude.

Could someone please set this straight for me ? I think it would be best if someone has the book so that they can reference the relevant parts so that I can see where I'm going wrong.

I really appreciate your explanation, Sarias, I guess I'm confused because of seemingly conflicting opinions here. Thank you again.

--Sincerely,

 

[Ethan Winer]

Vaughan,

> I know why there are peaks and nulls, it's just that I never understood the reasons behind 1/4 wavelength ... he makes several references that 1/4 wavelengths will result in peak amplitudes and that 1/2 wavelength will resut in zero amplitude. <

Ah, now I see the confusion! The reason you get a null 1/4 wavelength away is because the total round trip from that distance to the wall and back is one half wavelength. So the total round trip of half a wavelength equals 180 phase shift, and thus a null. At half wavelength away the total round trip is 360 degrees, which causes the original and reflected waves to reinforce each other fully - a peak.

--Ethan

 

[Vaughan Odendaa]

Thank you for explaining this to me, Ethan. I hope someone here can also reference a paragraph or two from the Master Handbook of Acoustics explaining this, because to me at least, he says the opposite.

Perhaps I'm wrong on that, but if someone could bring up statements in the book surrounding this issue I would really appreciate it.

Again, thank you for explaining this to me. I guess I'm scratching my head as to why the book would say other things instead. Again, I could be wrong and may have been mistaken.

--Sincerely,

 

[Ethan Winer]

Vaughan,

> I hope someone here can also reference a paragraph or two from the Master Handbook of Acoustics explaining this, because to me at least, he says the opposite. <

I'm sure he doesn't say the opposite!

I have the fourth edition of that book, so tell me what page number and paragraph you're referring to, and I'll do my best to straighten you out. If you have other than the fourth edition, tell me what chapter and section so I can find it.

--Ethan

 

[Vaughan Odendaa]

Dang. I forgot the exact page because I was recalling off the top of my head what I thought he said. I will have to go look and see if I can find the exact page.

I might be awhile.

--Sincerely,

 

[Ethan Winer]

Vaughan,

> I will have to go look and see if I can find the exact page. <

Okay, in the mean time click HERE to see a graph that accompanied an article I wrote for EQ magazine showing the comb filtering peaks and nulls 20 inches away from a sheet rock wall. All of the peaks and nulls align with that 20 inch distance.

Rather than rely on this book or that expert, do the math yourself using the round trip distance and you'll see why the frequencies are as I've stated.

Also, here's an article by studio designer Wes Lachot that explains this:

www.overdublane.com/wes/database/pagemaker.cgi?1071600431

--Ethan

 

[Savant]

Vaughn,

I think you and I basically said the same things in posts #2 and #5. Granted, I think your way is easier to understand!

Just like the ear canal resonating at mλ/4 values, wave interference at m(or n)λ/4 distances from a (e.g.) wall results from the respective condition-specific solutions to the wave equation. The former results from the condition of a tube closed at one end. The latter results from the condition of a normally incident wave reflecting from a single, perfectly reflective boundary and interfering with the incident wave.

There are many references I can point you to. Everest is not one of them. (His is a great book, but not very theoretical.) The one that comes to mind is Room Acoustics by Heinrich Kuttruff. The price for it is a bit steep - you might want to check the library. Of course, if you're way into this acoustics bidniss and don't mind a little higher mathematics, I highly recommend owning it.

 

[Savant]

Jeff,

> -∞ + 3 = -11? <

In this particular case, Yes.

I understand, for this particular case. However...

This is from a post I made at the SOS forum when the guys there were arguing about whether a closed glass window might help to absorb bass by reducing the strength of the reflections:

If a reflection is 100 percent it will create a null of infinite depth 1/4 wavelength away from the boundary.

If you reduce the reflection by 1 dB the null is now only 19 dB deep.

Reduce it another dB and now the null is only 14 dB deep.

Reduce it again to -3 dB and now the null is only 11 dB deep.

If you reduce it to -6 dB the null is now only 6 dB deep.

I determined these values empirically using sine waves in Sound Forge. I created a sine wave, made a copy, reversed the polarity, and combined the copy and original with varying level differences.

I believe this part of the post can be a little misleading. Those are not empirical values - they are only obtained by beginning the process with a 0 dB signal in Sound Forge. For me, personally, this example can lead to confusion. Acoustics doesn't quite work this way. I.e., an absorber that lowers the level of the reflection by 1 dB does just that - lowers the level of the reflection by 1 dB. It does not bring up an infinitely low null to (an absolute) -19 dB. For starters, the null is not "infinitely" low - it is simply 0. -∞ is a decibel reference. It's unitless since the decibel only expresses a ratio in a more manageable form. -∞ only exists due to the fact that the pressure - or voltage in the case of the Sound Forge process in question - is 0. The log of 0 is "-∞."

But I digress (a little)...

In practice, it is impossible to obtain an infinitely low null on a decibel scale since we can never truly create an acoustical situation where the waves perfectly cancel out - i.e., a perfectly reflective surface. And if an absorber is added to the surface, it is not the null that is "brought up" but the level of the reflection that is reduced, thereby creating less interference. I.e., nothing is "added." What is there is simply what was always there, but with less interference.

I think this explanation conveys the same information. Putting things in such finite terms, however, can be misleading since the amount the null is "increased" is relative to the starting point. I hope this helps.

 

[Ethan Winer]

Jeff,

> Acoustics doesn't quite work this way. <

Why do you believe that sound waves will not combine exactly the same as electrical waves?

> For starters, the null is not "infinitely" low - it is simply 0. <

Sure, no null is absolute. But it doesn't matter in the context of my example. And it was just an example.

> it is not the null that is "brought up" but the level of the reflection that is reduced, thereby creating less interference. <

Well okay, you say Tomatoe and I say Tomahto. Either way, the result is the same.

Again, you had questioned my logic that reducing a reflection by 3 dB will raise a null much more than 3 dB. Which it does.

--Ethan

 

[Savant]

> Acoustics doesn't quite work this way. <

Why do you believe that sound waves will not combine exactly the same as electrical waves?

I don't.

> For starters, the null is not "infinitely" low - it is simply 0. <

Sure, no null is absolute. But it doesn't matter in the context of my example. And it was just an example.

I guess I misunderstood. When you claimed the quoted values were determined "empirically," I probably took it too literally.

> it is not the null that is "brought up" but the level of the reflection that is reduced, thereby creating less interference. <

Well okay, you say Tomatoe and I say Tomahto. Either way, the result is the same.

Again, you had questioned my logic that reducing a reflection by 3 dB will raise a null much more than 3 dB. Which it does.

...sometimes. I don't disagree with your experiment. I just feel it's necessary to point out that the results quoted are not necessarily indicative of results a real person in a real room would obtain by reflecting a real sound wave off a real wall with a real absorber on it. I.e., don't expect a null to suddenly rise by 10, 20, 30 dB simply by placing an absorber on the wall. It's more complicated than that is all I was pointing out. Sound in a room does not behave quite as cleanly and conveniently as sine waves in Sound Forge.

 

[Ethan Winer]

Jeff,

> I guess I misunderstood. When you claimed the quoted values were determined "empirically," I probably took it too literally. <

Yes, too literally.

Also, you had said:

the null is not "infinitely" low - it is simply 0

In fact, it doesn't matter what level I started with in Sound Forge because it's all relative. As long as the polarity-reversed version is the same level as the original, the null will be infinite. Then making one sine wave 1 or 2 dB less than the other will "bring back up" that infinite null.

> the results quoted are not necessarily indicative of results a real person in a real room would obtain by reflecting a real sound wave off a real wall with a real absorber on it. <

Sure, because in a small room there are multiple waves all arriving from different directions. But the basic concept of opposing waves creating or "relieving" a null is as I stated. And it does indeed apply acoustically. If you were to do this experiment outdoors in a parking lot with an elevated loudspeaker pointing down, an absorber on the ground would bring back up a null exactly as I described.

I'm sure this principle applies at least to some extent in small rooms, because that explains why a four inch thick bass trap can make a meaningful improvement as low as 50 Hz even though its absorption is down around 0.25. It also explains why an absorber doesn't need to be 1/4 wavelength thick to be effective. I see that myth repeated fairly often.

--Ethan