Amar Bose: Lecture 23

AMAR BOSE: OK. What do you think would happen if I had a good, rigid jug that connected to the piping, the faucet, and turned the faucet on? Water goes in, yeah, we agree about that. What happens eventually?

AUDIENCE: [INAUDIBLE]

AMAR BOSE: Hmm? No, it can't overflow. It's sealed, except for the pipe coming in.

AUDIENCE: Water will stop coming in.

AMAR BOSE: This is the right course in acoustics, by the way. Just in case. Yeah?

AUDIENCE: [INAUDIBLE]

AMAR BOSE: The water would stop, unless this was some sort of an ideal source, In which case, you have the same problem as a volted source across a short circuit, no solution. Now, what do you think would happen in a room? Suppose we had a room. The room, itself, was ideal. There was no absorption of sound. And we put a loudspeaker in the room and turned it on.

This one is easy to see. I'm want you to think about it. Ideal room, no absorption. Only thing in the room is the loudspeaker. And if you feed a signal to it. And, let's say, you put random noise in it, anything, it doesn't matter. It's a real loudspeaker. Yeah?

AUDIENCE: Maybe it would just get louder and louder and louder until, pretty soon, all of the waves will be going every which way. And it would just, finally, heat up.

AMAR BOSE: What would heat up?

AUDIENCE: Just the air molecules.

AMAR BOSE: Yeah.

AUDIENCE: The source?

AMAR BOSE: Think a little bit closer about this model, and then you'll be able to get the limit. What I'm really interested in is what happens eventually.

AUDIENCE: Smell smoke?

AMAR BOSE: Besides smoke, yeah. Yeah?

AUDIENCE: Eventually, the speaker won't be able to drive the system anymore.

AMAR BOSE: So it'll stop putting out energy?

AUDIENCE: Yeah.

AMAR BOSE: Exactly like this faucet did, coming in here. In other words, the pressure built up until the pressure here matched the pressure here. And that's all over. It stops. That's exactly what would happen in the room if you had no absorption.

Turn on the loudspeaker, which initially, to you, would sound just like a normal level loudspeaker playing. Then we keep building up, building up, building up, until it got so high that the pressure on the cone was such that it couldn't radiate anymore, just like the pressure on the faucet.

That gives us a good introduction to a very important calculation. Namely, we want to know what the sound level would be, in this room, for a source. It might be myself, or a loudspeaker, or anything else. The sound level, if you think about it for a second, is very different than it would be outside. If you took away the walls of this room and the ceiling, you would have quite a hard time, in the back of the row, hearing me, outside. As you know, it's a pretty good, spherical source, and 1 over R goes down. So the reason you're hearing me now is because of all the walls. With all the nasty things that they do, as we learned with the reflections, the millions and millions of reflections, but without them, we wouldn't hear much.

So now we want to find out what level you do hear based on how much power the source radiates. And we want to, particularly, do something which distinguishes the direct field from the reverberant field.

Now let's define these. The direct field is the field that comes from the source to your ears, without intersecting any other object in the room, or any of the walls of the room. Pretty clear definition, direct field means direct from the source to you, without hitting anything. And, by definition, we break the world up into the remainder part, which we call the reverberant field. The reverberant field is the field that occurs after one or more reflections. It's everything but the direct field.

So the total field then, we will think of that you're experiencing right now, is the direct field from me to you plus the reverberant field. And since I already told you, and you could experience, that, if we were outdoors, you'd have a hard time. I would sound much less, having less volume. You know, right away, that the direct field is going to be much lower than the reverberant field, where you're sitting.

Well there are an enormous number of implications of this, which we'll get to in a few minutes. Yes?

AUDIENCE: Does the direct field include issues relating to reflections? Suppose you're sitting in a null, in a room where--

AMAR BOSE: Yeah.

AUDIENCE: Would the direct field include that null?

AMAR BOSE: No, because there are no nulls for the direct field. The direct field, remember, is the field that came from the source to the person. So it doesn't know that there's a rest of the room out there yet. It hasn't reflected. So there are no nulls. In fact, that gets right to the essence of the calculation of the direct field. When we go to calculate that, we will assume that we have a source sitting in the middle of a room, sitting in the middle of space, because you don't care about the room boundaries.

So let's look for the reverberant field first. And we'll call this-- Well, let's do the following thing. Let's imagine that we have some huge room here. And we have, somewhere in the middle of it, we have a spherical source radiating w watts total. And we want to find out what the reverberant field is and the direct field. Right now, we'll go for the reverberant field first, then we'll do the direct.

Reverberant field is very much like our example of the water. The only thing is, imagine that you have a pot, now, with the faucet coming in here and some holes. Not so big holes as I've drawn, but some thin holes in the thing.

Well, you want to find out what level the water comes to. You can do that. It's coming in here. It's coming in here at a given rate. The way you find the level is, you find out, as a function of the height, here, how much leaks out? And, when you get to a certain height, the amount leaking out, when that is equal to the amount coming in, that's the height that will be equilibrium. It will always stay there, because there's so many gallons per minute coming in here, so many gallons per minute going out here.

And, of course, the higher this level, the faster it would go out. The lower the level, the less it would go out. And so more is coming in, it'll come right to equilibrium. Any questions about that? This is the same. You've probably seen it in physics, with a little thing, a whole, or something in the bottom and find the equilibrium level. It's the level for which output is equal to input.

So now, that's exactly what we do in the room. So this source, sitting here, is radiating w watts. How much of that w watts is going to go into the reverberant field? See if you can just get that, right out, from definitions that I gave you, in the last few minutes. The reverberant field is the pot. w watts is coming out from the spherical source.

Well, let's see if we can. Anybody want to guess? Just a guess, I won't ask you why. No guesses, wow. Yeah?

AUDIENCE: The energy that isn't absorbed by the walls?

AMAR BOSE: Yes, the energy that-- Remember, the reverberant field was the field that has experienced one or more reflections from the wall. So, after the first reflection from the wall, all that energy goes into the reverberant field, by definition, because the direct field is that which hasn't collided yet.

So, let's see. That is actually enough to guess at the answer. But since there are no guesses, let's go and derive it. And then, I think, you'll see that the guess is obvious. Well, no, I take it back. What's going into the reverberant field? This is the power, which is energy per unit time. After the first collision with the walls, how much is absorbed of W? alpha bar.

So how much goes into the reverberant field? w times 1 minus alpha bar into reverberant field. w is power that it's radiating, watts, power. So that's the input. That's the faucet that goes into the pot. Now we just have to find out how much goes out. If we have an average.

Now, remember, in this pot, when you're in equilibrium, this level stays here all the time. So when you compute what's going out per second, if you wish, don't worry that this is going down. We're at equilibrium. Because equilibrium is, such that, what comes in here is what goes out through all the holes.

So we can think of an energy density, d average. Energy density of sound that's in the room. d average times the volume of the room. This is the volume of the room. Should I put an r here? d average times the volume of the room is the total energy in the room. This was energy per unit volume. This is total energy.

In our statistical model, every mean free path , that thing loses alpha. Because once it goes through a mean free path, it collides with a wall, and that's where alpha goes out. So it loses alpha bar per collision. It goes out. Now, all we have to do is find out how many collisions per second. We have the power flowing out. And we can equate the power flowing out to the power coming in. Not so complicated.

So collisions per second, we knew that the mean free path, remember, in the room was for volume of the room over the surfaces in the room. And since we know the velocity, and then the time between collisions, or the time to travel one mean free path, is d over the velocity. So t prime, time between collisions, is equal to d over c, which is equal to 4v over cs.

So now, this is how many collisions per second we have. Oops, 1 over that is collisions per second. So finally, we take, this is the energy per collision that we multiplied by 1 over t prime, which is collisions per second, and we have the power. We have energy going out of the room per second.

So d average, v room, alpha bar, is the energy going out per second, divided by t prime, is power out equals d average, vr, alpha bar. And we'll put all this stuff in now. 4 v over cs. Let's see, I think that looks OK. We divided by t prime, yes. v, this is the v room, also. So the power flowing out, then, is alpha bar s times c times d average over 4.

Now, all we have to do is equate that, as we did in the pot with the faucet coming in, to the power going in. And that will relate d average, the average energy density, to the power that the source is radiating. And from the average energy density, we know our relationship between that and pressure. Ones proportional to the square of the other.

And so we can get the pressure in the reverberant field, in terms of the power that this source, sitting here, is radiating. So let's do it. The power coming in was w times 1 minus alpha bar. That must be the power going out, which is alpha bar s c d average over four, or d average-- I'll go over here-- which is the thing I want at the moment, because from that I'm going to get pressure. d average is equal to 1 minus alpha bar over alpha bar s times 4 over c. 4 over c times w.

Now, this fellow here. And this is just extra baggage. But it's so that, in case you look at expressions in books and whatnot, you'll know what it is. This is, in acoustics, given a term, for this whole part. It involves the room, obviously. And the area of the room, in particular, as well as the average absorption coefficient of the room. And it's called the room constant. Let's see, room constant is defined as. Room constant, I think, it is alpha s bar over-- I think it's the upside down of this-- alpha s over 1. So d average, then, we can express as 4 over c, times the room constant, times w.

Now, d average, we can relate to the pressure average. The energy density, when we use the term d average here, the energy density is, d without the average, is clearly jumping up and down, all around the room. Because at any frequency you have all these normal modes, and the energy density is proportional to the square of the pressure. So you have a thing that's a big mountain running around, all over the room. But you average all of that stuff, and you call it d average. Because, in your statistical model, you're only dealing, as we said before, with averages.

And you know that all the energy in the room, on the average, is going to collide with one of the surfaces every mean free path. So the relationship that we had derived earlier between d average and p was p squared. Our relationship, where these were complex amplitudes, p squared over 2 rho 0 c squared. Let's see. I think that was it. The 2 was in there.

This is what we derived earlier in the subject, if it wasn't, we'll find out later. This was d average. Yeah, I think so. Sorry, this was d, and, now, I'm averaging both sides. I'm going to do the same thing with the pressure that I did with the energy.

So, finally, let's now get an expression for p squared, in terms of w. And that's where we really want to be, for the reverberant field. So p squared. I'm going to now say, because I'm going to have a direct field in this room and a reverberant field, let's just put an r sub on that p, meaning that we're dealing with the reverberant part of the field, now. So this isn't generally so, but I'm going to deal with the reverberant part of the field, p squared average.

I want to get this. Let's see. I'm substituting this in for here. I have to take up the rest of this stuff. 2 rho 0 c squared. On the other side, well here, I can do it here. 2 rho 0 c squared over c room constant times w. That's in terms of complex amplitudes, here. Wait a minute. What happened? I wrote a c squared up here. And I can cancel this fellow.

AUDIENCE: [INAUDIBLE]

AMAR BOSE: A 4 here? Let's see, there's a 4 here. Oh, that's an 8. No, wait a minute, let's be careful. I'm going to substitute this into here, so there's a 2 downstairs. It comes up as an 8 up here. Everything else OK? Yep. So rho 0 c over r times w. That is the square of the reverberant field, the sound pressure.

Now, let's get the direct field. That one is easy, relatively speaking. Sitting here, it goes large compared to a sixth of a wavelength. They go a wavelength out. We remember, we had a pulsating sphere in space, and it becomes a planewave out here, basically, the equivalent of a planewave. So all that you have to do is get the intensity out here and multiply it by 4 pi r squared. And that gives the total power that's coming out.

So you don't have to worry about the room at all, because the definition of the direct field is that field which has not met anything in the room. And so it might as well be outside. The direct field, in this room, to you, is exactly what it would be outside.

So let's see, how do we do that? Anybody want to give me an idea? Remember, if you had a source over there and you're here, the expression for the intensity, the power per unit area, p times one half p times u. Over here, on scratch paper, 1/2 p times u. And u could be expressed as p over rho 0 c. So 1/2 p squared over a rho 0 c. This was the intensity that we derived before.

So all I have to do is multiply all of that, this business, by 4 pi r squared. So we have p direct, p direct squared. 1/2 p direct squared divided by rho 0 c. p direct squared, 2 rho 0 c times 4 pi r squared, the distance that you're away from the source. You're out here at r, let's say. 4 pi c. Well, maybe I'll stop a second. The reverberant field, we only dealt with the d average. It didn't make any difference. And this is a good point to make at this point in time, because it'll seem odd to you a little bit later. The reverberant field, right next to me, when I'm speaking, is the same as it is back there, in the back of the room. We're dealing with something that came into the room, went all around in a giant mixing, and its existing all over. We neglect the ups and downs, because we're taking an average. But the reverberant field, in this room, even with the ups and downs, is the same, where they are, is the same as it is back there, the same intensity. Yes?

AUDIENCE: Why is it that, if there were no absorption, why wouldn't you reach some maximum energy in a room? Why couldn't it just keep getting bigger and bigger?

AMAR BOSE: It couldn't get bigger and bigger, because I said the room was ideal, in the sense it had no absorption, but the source was real. And so what would happen is the pressure would build up so much that the cone wouldn't move anymore in the loudspeaker. If you said that the source was ideal, also, then you have the classic question of a volted source across a short circuit. What gives? I mean, in the ideal world, nothing gives, and you have infinite pressure. Yeah?

AUDIENCE: One other question, to get through all of these, you've done a lot of manipulations like finding average power by multiplying average d by average alpha, things like that. And generally, that doesn't seem to work.

AMAR BOSE: Why not?

AUDIENCE: Well, if I had two, say, random variables, and I wanted the average of their product is not equal to the product of their averages.

AMAR BOSE: Yes, OK. Wait till the end, and ask me the question again. If I had dentures, by now, they would be bouncing on the front seat. I mean, that is a problem, and I've hidden it. And so ask it to me at the end, and I'll tell it to you. I mean, not after the lecture, in. Let's see, so we have, this was what now? This was the intensity. We multiply by that. And that must've been then pd squared. That was equal to w.

Now, because w is entering the direct field. w times 1 minus alpha bar enters the reverberant field. But w came into the direct field. From this, we would have pd squared is equal to. Let's see, pd squared is equal to 2 rho 0 c. Looks OK. No, no, no, upside down. pd squared, I have 2 pi r here, and a rho 0c here. rho 0 c times w over 2 pi. I think that's right. 2 pi r. Yeah, squared, thank you.

So that looks something like what we want. Now, I should say, here, I hate to do it, because it looks like you're changing a variable. It tends to mix people up, but everywhere you see these quantities, they will be in RMS quantities and not complex amplitudes. Because the meters that read the sound pressure level are reading things, just like an AC voltmeter, they read RMS. Which is, as you know, 1 over the square root of 2 times the complex amplitude. So these expressions, when written, the way you will see them, would be written with a pd. Ah god, I have to put a different symbol for the thing. Well, yeah, put an RMS. Let's write it the other way. p direct RMS is 1 over square root of 2 p direct. So p direct squared, if I put that as pRMS, I get p direct RMS squared is equal to-- let's see, get the darn thing in the right direction-- 2 p direct squared. So I put in p direct and that's equal to 2 times p. Yes. rho 0 c over 4pi r squared times w.

That's what you'll normally see in the books, everything, because the instruments read that way. And a similar thing happens for the other one here. Where did we have the expression for the reverberant field? p reverberant, this fellow here. p reverberant squared is pRMS times 2. So, if I wrote this expression, I would now have p reverberant RMS. That is equal to 4 rho 0 c over the room constant times w.

Now, still hiding the problem of what do you do when you're measuring squares of things and adding them, I will do the following. p total RMS is equal to-- I'll just say it is, and we'll justify this later-- p direct RMS squared plus p reverberant RMS squared. And these are average. Yes, this is an average. This is going to get really complicated, when I write the average. p reverberant average RMS. This one's not an average, because wherever you are away from the source, that's it.

So that should be the sum of this and this-- where in the heck-- and this one. Let's do it. rho 0 c over 4pi r squared, w plus 4 rho 0 c over r times w. I can take rho 0c times w out. And we have left, 1 over 4 pi r squared, plus 4 over the room constant.

Now, this expression, you'll see entabulated in books, all over. But it ends there, how to calculate the reverberant field and the direct field. And it turns out that this is extremely significant to a lot of things that today are not understood in sound, particularly in the recording and reproduction of sound.

So everything I'm going to tell you from now is not in the books. Let's take a look at this expression. Everything that we'll talk about for a while has to do just with this.

The first part goes down as a 1 over r squared, the second part doesn't matter where you are in the room. The reverberant field, it's a constant. If you look at these, just the direct field and the logarithm of it, the pressure would be the square root, of course. And that takes away that. It's a 1 over r. And so it would fall. You'll see them plotted in books, very often, this way, logarithmic. This is distance away from the source, and this is pressure, SPO.

You'll see this go down at 6 dB per octave in distance. And then you'll see the reverberant field here. And this is the direct. Those are asymptotes, of course. But let's suppose your direct field was looking like this. It's the whole thing for the direct field. And your reverberant field is here. What it tells you is what you expect. When you're very close to the source, you're getting mostly direct field.

Anybody that was listening to me here, at this distance, it wouldn't make any difference if you change the absorption in the room, or changed the dimensions, or the shape of the room. If you were listening before these two things crossed over. If you're listening back here, it doesn't vary with distance at all. Now, that's obvious from the curves. What I'd like to point out now is that we're entering the 21st century, and nobody knows how to record a single voice or instrument. And when you read all of the articles on this, they talk about better microphones and whatnot. And that's not a limit at all.

Those things are performing as well as-- you go to the hi-fi store, and some hi-fi people will pay much more for .001% distortion then they would pay 0.1% distortion. And nobody alive, in music or speech, can hear the difference. So the microphone, it does its job, as it's supposed to do. That isn't the problem in how to record.

And to find out what is the problem, let's think about the following thing. I can tell you, everybody, well, with the possible exception of the first row or two in this room, is in the reverberant field. When I say, you're in the reverberant field, I mean, you're beyond this point. So the reverberant field has more energies, more sound for you than the direct field.

Now when I speak, if I speak, and I turn around like this while I'm talking, you don't seem to have a problem. Afterwards, when you go outside, try that with somebody in the direct field. The outside, it is a direct field, except for the ground. And what you'll find out, right away, is it that all the fricatives of the voice, the sh, ss, ff kind of thing, they get attenuated, rapidly, when the person turns around. If you say one, two, three, four, five, one, two, three, four, five, one, two, three, four, five, back there, it won't make any difference. Well, that's because the energy from a source gets directional. Now here, it didn't matter. So I turned around, hit there, and it came out. So what you're responding to, in the reverberant field, is the total energy that's radiated from the source.

You are not responding to the frequency response on axis. This has, not only, implications for sound recording, as we'll see, but certainly for loudspeaker design. They tell you to design these things to have a certain frequency response on the axis, and yet, when you sit in your room, if you're lucky, you're in the reverberant field. We can calculate these distances. I'll tell you a little bit more about that in a minute. This is a very critical distance, where the reverberant field and the direct field are equal in magnitude.

If you're in the reverberant field, you're hearing the total energy. Now you could get a frequency response on axis that was perfectly flat, and the thing beamed, like this, forward. And, of course, at low frequencies, regardless of what the thing looks like, it's omnidirectional. If a speaker were flat on axis, let's say, it had this pattern at the high frequencies, and this pattern at the low frequencies, at which frequency does it radiate more power? Flat frequency response here.

Yeah, sure, at low frequencies, it's radiating power all over the place. At high frequencies, it's only radiating forward. So the total, if you were to plot the energy, the power, at the different frequencies that was coming out of the loudspeaker, total power over the sphere, it's going to be much larger at low frequencies then at high frequencies.

And if you are in the reverberant field, you are responding to the total energy, not the frequency response on axis. So when I turn around, now, you still hear me. And you hear the fricatives and everything else. So here, you respond to total power. By that, I mean in all directions, the total power coming out of the source. Here, you respond to the on axis.

Well, I shouldn't say, on axis, on any axis. The direct field, if I'm pointing this direction, and you're sitting over here, you get this much of it. This is the polar pattern, let's say. You got that much on axis. And so you respond to, on the axis that you're sitting, it's not the principal axis, on axis power, on axis radiation, or whatever you want to call it, power, intensity.

So this then says that the notion which is in all the text books, for example, that you would design a loudspeaker to have a flat frequency response on axis, we spoke about that a little later, is all part of the emperor's new clothes. It's a natural thing. It came up when loudspeakers were first developed. Seemed good. Engineers love things like flat frequency response. They're taught that from the sophomore year. It's great. And so, it's like motherhood and apple pie, you do it. It turns out it's wrong. And if you do it that way, there will be a harshness and a shrillness.

It's very interesting, over the years, that's the problem that got me actually interested in this whole field, after I graduated. Because I bought this product, a hi-fi system, and, god, it was horrible. I mean, I had studied music for a number of years. And I listened to this, and when violins went up on the E-string, it sounded like they were made of steel, or the whole instrument was steel. And couldn't understand why, because it measured well. And I believed all this stuff too at the time. Flat frequency response was it. I mean, if it measured that way, that was good. And so that's actually what caused me to get involved in another field, also.

If you're sitting out here, you respond to the total power. So the total power of the source, if you were talking about reproduction, is what you want to make have the right spectrum. So you'd want the total power of the loudspeaker, if it was trying to reproduce me for you in the reverberant field, to equal the total power that I radiate. Well, the total power that I radiate you can't measure, and you can't measure it as a signal and reproduce it.

There's a little detail that powers the portion of the square of pressure, no? Try to take the square of something and then reproduce the sound, see what you get. You have no way, we have no way, of measuring the total power out. So I can't have one microphone or 1,000 microphones, even, that measured the total power, and then put it into a hi-fi set. And after I've done all squaring, and then the square root of the sum of the squares, that's the problem I've been hiding. And that would, of course, give you nothing but distortion.

So that's a minor problem for sound recording and reproduction. Now in this thing, you respond, as we saw, to the power on axis. Now, if you want to make, for example, you can do this, if you want to make a loudspeaker that will reproduce the human voice so accurately that, if it's your friend and somebody sneaks up behind you and puts this loudspeaker and a recording on, you will respond to the person. It's that good.

What you need to do is, you need to make a ball, roughly the size of the head, with a speaker, about the size of the mouth, and so, if the dimensions are right, the directional response will be correct. So it has the right directional response, and then you just equalize it to have the same balance of frequencies that you recorded. Now you have to watch out. How in the heck do you record that. If you record on axis, and if the speaker has the same polar patterns, at each frequency, that the person had, then you could design it on any one axis, and it'll come out right for all of it.

So you could have this dummy, and it would sit here, and, for all the people in the room, it would sound just like the person, very accurate, no problem. But that isn't where you're looking.

Now in a concert hall, for example in Symphony Hall, you can compute where these things are equal. That's very easy to do, because you have this expression. All you have to do is set this equal to this. This is the room constant. We had it up here somewhere. It's s alpha over 1 minus alpha bar. And that solves for r, where the direct and reverberant fields are equal. Now, if you go into Symphony Hall, this turns out to be about 19 feet from the instrument.

So most of the people, the vast majority of the people in the hall, in all concert halls, are over here. And, by the way, in your living room, you'll find out this is for a spherical source. Obviously, this would change. This changes a bit-- and you can see that in Beranek-- if the source were directional. Obviously, if you had a piercing source, it'll go much further before the reverberant field will take it over. So the crossover point would be out here. But, for many instruments, this is a good approximation.

So music was composed, basically, for you to be in the reverberant field and listen to it. Now, the reverberant field in your living room, you can compute it. It'll probably turn out to be a couple of feet, 2 feet or so, in a much smaller room. So you're in the reverberant field. It's not too often that you're sitting there listening to your radio or hi-fi with your ear 2 feet away. So, basically, you've got a loudspeaker, which was designed, in most cases, for the direct field, playing, to you, in the reverberant field.

Now, what does all this have to say about recording of sound? It brings back an interesting memory. I think I told you this. The PolyGram people, it's a consortium of Deutsche Grammaphon, and various ones. When I told the recording engineers that, at the beginning of the talk, to get them interested to follow, that they didn't know how to record a single voice or an instrument, obviously, that's their paid occupation. So they were really angry. And they paid great attention, because they were wanting to bury me at any point. Well, at the end, we asked, does anybody think that they know how to record a single instrument or voice, and not one hand went up. People just don't realize the problem.

What do you do with the microphones? Turns out that the people that record, and are paid the most to record, never saw the inside of a university, unless they were recording there, let alone any course in acoustics. They are people who have been thought to have golden ears, by the executives of the company. Because why? They made a recording that sold very well. That's generally how it goes. Now, to a certain extent, these people, they don't have golden ears, but they do have experience. And they converge on things.

And they'll go into a hall, and they'll just look around. And the orchestra is on the stage, and they will decide that they're going to place a microphone here and a microphone there and whatnot. And the good ones, most of the time, will get some good sound. The poor ones, we don't hear about.

So, basically, what happens, and it became of great interest, at one point, for me to find out, what's going on? What do people really do with these microphones? Well, it turns out, and it's very hard to get in to see what they do, because each company keeps this great secret of ignorance from the other company. In fact, that reminds me, I was once working for Philco in radar in Philadelphia, and the boss announced, at the end of the day, now, make sure-- this was a long time ago-- make sure that everything is locked up. God forbid we should let the Russians find out how far behind we are. So this is almost like the recording industry. Basically, I got in, as a professor of MIT who wanted to study some of this, into the Boston Symphony one day, back in the days when Leontyne Price was singing. And what I saw was something that I never expected I would ever see. All the seats in the Symphony Hall, here, were removed. This is a recording session of RCA. The orchestra was placed on the floor, not on the stage, in three groups, like 75 feet apart. Now, 75 feet is 75 milliseconds.

Now, Leontyne Price was there trying to sing, and she, of course, couldn't because the delays from the orchestra, she couldn't get synchronized. So they had headphones on her. When you put headphones on, you sound different. Imagine how you sing when you have headphones on. We're talking about, you want a natural performance, and here you put the person with headphones on. And then the audio mixers mixed the three groups and her altogether. Unbelievable.

However, what was interesting, what I wanted to know, was where did they place the microphones. Because that's what their experience has told them over time. Now, they don't know anything about direct fields, reverberant fields, or anything else. But they, with time, if they survive, they know where to place microphones.

So it turns out, what they were doing, if you look at the microphones, because some the microphones were directional. you have to calculate that into the issue, which we did. It turns out that they place the microphones, without knowing anything about direct or reverberant field, very close to this point, where the reverberant field is equal to the direct field, and there's a reason for it.

Let's just say how they probably converge in this. If you put the microphone too close to the instrument, try this some time. All you got to do is go in an anechoic chamber. Take a violin in there and put the microphone a foot or so from the bridge. We actually, in the experiments with Boston Symphony musicians, we actually did this. And, one day, one of the musicians, George Zazofsky came over early, and we were playing back a recording that we had made. And he heard it. He recognized he had played it. And he got very serious. He said look, I want to make one thing clear. I'm here for research purposes alone. You have no right to release these things. He said, this would kill me if this ever came out. Because it sounded so terrible.

Well what happens is, take many instruments, but a violin, if you measure it, for example, you'll have a polar pattern at, let's say some harmonics of the open E-string. A polar pattern that looks like this, in which you change by 5 degrees and the intensity drops by 30 dB. So, if you record close to the instrument, at some notes, you're going to be down here null, and at the next note, when the polar pattern juggles all around, you're up at a peak. And so you get the wildest spectrum out of this thing that you can ever imagine, when you record here.

So the recording people have learned, with time, that if you go this way, too far, you get very unusual results, very unpleasant results. So you'd say, well fine, let's go the other way. Let's get what you get out there in the audience.

Now, if you place a microphone out in the audience of any hall, and we've done that a number of times, and then you try to play it back in the room, then you get something very bad. It's like the whole thing is sounding like it's coming back through a barrel. And this will be related to the experiment that you'll hear on the last day. Now why is that? That's because, remember, that the reverberant field, while it has the right average energy of the source. Remember, we said it didn't matter which way I stand for the reverberant field, because it mixes everything. What goes out, if I'm turned to the board, it takes that sound and projects it out there. And it's like a giant mixing pot. So it has the right average spectrum, the right balance of frequencies that came out as a total energy from the source.

However, it has all the normal modes in it. These millions and millions of normal modes and all, and when you go through those things, it's like very high q filters. In fact, if we get a chance, we can show you what the bandwidth of those normal modes is. So if you record out there, it's like recording through a giant comb filter that has peaks and dips in the tens of millions.

So one recording out there, too far, and the person knows to stay away from that forever. So what happens is he, through his experience of developing, he's probably made errors out here, errors out here. He's come back here, and he's converged. And this is where he stays, without knowing anything about this. So you can actually go into a room, knowing only the simple things that are in this expression, and you can tell where to put the microphones, without ever having made a recording before, and you'll come out well.

So the average spectrum out here is OK. Because that's what you listen to when you go to the music. That's what the music was composed for, to sound good in this part of it. Here, the spectrum is all screwed up because it's only on axis, so it has no meaning in there. So this part here, is an attempt to get some of the sound that has the right average spectrum, but not too many normal modes. In other words, you have to move this way to get out of this part where the spectrum is totally wrong, the violin that's recorded too close, and the polar pattern that makes everything bad. And you move as far as you can out here to get the right spectrum, until you get into the normal modes, then you quit.

So, basically, that's what you find them doing. If I were to be a little bit more precise, it would probably be a band like this. They seldom go out that direction, I think. Questions so far, yeah?

AUDIENCE: Why is it that we don't hear the normal part? Why doesn't the audience hear the normal modes when they sit in the audience?

AMAR BOSE: They do, but that's what music was composed for. In other words--

AUDIENCE: But why is it bad when you play it back?

AMAR BOSE: Ah, OK. Because you now have gone through the normal modes of your living room. For example, if I take a binaural head, a binaural head is a dummy head with two microphones in the ear, and I place that anywhere in this room, in the back, and make a recording of music or me, and play that to you through headsets. What went into the left ear of the dummy goes into your left ear, right ear of the dummy into your right ear. It's fine. It'll sound as close as you can get. The only problem with binaural sound is localization is all in the head. In other words, you don't sense an environment outside, you sense the whole environment in the head. But other than that, the intelligibility, everything else is excellent, because you haven't gone through the normal modes twice. You only went through them once. But the minute you play the loudspeaker in a room, you've gone through a whole set of normal modes again, and you can't take that. As you'll see in the demonstration on the last day. Yes?

AUDIENCE: So why can't you take a violin, put it in an anechoic chamber with several microphones and average the signals from each so that you can even out the--

AMAR BOSE: Yeah, why can't you average the signals from each microphone? That is not getting the power. That's not the same thing as the power, because you'd have to average the square of the microphones. And there is some thought that by putting a lot of microphones around, you can go to at least a better thing. You cannot get the total energy that I'm sending out, because that involves a square. But at least people can get a feeling that, OK, if I happen to be directional here, and I put a microphone here, and here, and here, and here, at least I'll be picking up some more of the bass. And I'll be working in the right direction. And that certainly can happen. It's not a thing without problems either, because the sum of the squares isn't the square of the sum. But it is something, that in some circumstances, helps, to just put a lot of microphones around. Just so long as we know that, by doing that, we're not getting the total energy out. Because that involved sum of the squares.

Maybe I should take some time. The direct field has another property that's very important. That's how you localize. In other words, when I'm speaking to you, the direct field gets to you first. Because the reverberant, obviously, didn't go straight from me to you, it hit something else, and, therefore, it had to go a longer path. So it is the first arrival at your ear of a complex sound. And I don't mean when I start speaking. I mean, as I'm speaking continuously, every sound that's coming out, there's a part of it that goes directly to you. There's a part that goes reverberant. Most of it's going reverberant, of course. It shouldn't be surprising that the reverberant field, here, is dominant, because any source, even that violin, is radiating in all directions.

And think of the solid angle that is subtended from my mouth to your ear, up there. It's a very small, solid angle. And so all the rest of the energy was going out in all directions. It's not surprising that the reverberant field is well above the direct field, where you're sitting out there. You shouldn't be surprised.

Now, I forgot what I was. I wanted to make a point about this. Darn, I had something in mind. Yeah? Oh, localization, thanks. It's on the direct field that you localize. The first arrival. You couldn't localize on the reverberant field. It's coming equally from all directions. In fact, if you go out into the concert hall, and you measure-- that's the other thing we should. I'm happy we're talking about that.

This field, here, arrives from all directions. All directions equal probability. This field arrives from the source. So the only hope you have to localize on anything is the wave that's coming from the source. And I think we may have mentioned, I don't know, but that for localization you use two mechanisms that you may not be conscious of, but you do. Below 1 and 1/2 kilohertz-- and it is the crossover between the two. I think we mentioned it. You localize on direction by the time difference between years. But that gets to be sort of irrelevant when the sound doesn't bend around the head.

So at 1 and 1/2 kilohertz, which is wavelengths like that, you shift, and you localize on intensity. If a person makes a sound, sss, you know that it's over here, because it's louder here then it is here. That's how you've come to learn where the sound source is.

So this direct field, now, is you're sitting out here somewhere, the direct field is way down under this one, under the reverberant field. But you can still use that for localization. You don't have all the localization, when you go to a concert, that the stereo buffs would have, separating the violas from the violins and all this baloney. If you were blindfolded, you'd have a heck of a time doing this kind of thing.