Amar G. Bose: 6.312 Lecture 20

AMAR G. BOSE: Room acoustics, the usual approach to this is to break up, generally, the rooms into two categories, regular rooms and the rest of the world, irregular rooms. As we will see when we move along, it would be much better-- sorry about that, this is supposed to be the signal to start the cameras-- it would be much better if there were more irregular rooms in the world and less regular ones. But you'll see why as we go along.

So the first part is regular rooms. The normal way of doing this, the way you'll see if you look in textbooks, is take the three-dimensional wave equation for regular rooms. You do this in rectangular coordinates. And you just get a solution for that wave equation subject to the boundary conditions, that the normal velocity at each surface is zero, and that's the result.

Now we're not going to do it that way. We're going to start with a one-dimensional room that you've seen many, many times, a tube. And you can get an incredible amount of information and insight as to what's going on the total room, if you understand what's going on in the tube.

And then, from the tube, we will be able to build up without actually going to the three-dimensional wave equation. We'll be able to build up what actually is going on in this room, and I think it will surprise you what actually is happening. By the way, there's a lecture demo that comes up for-- yeah, is it next time, or the time after?

SPEAKER 1: Thanksgiving Day.

AMAR G. BOSE: Thanksgiving Day, huh?

SPEAKER 1: [INAUDIBLE].

AMAR G. BOSE: Yeah, OK. So let's go back to the one-dimensional tube first, the one we have seen, open at one end, closed at the other, x, 0. And we had for this, the P of x and omega was equal to P plus e to the minus j omega x/c plus P minus e to the plus j omega x/c. One-dimensional, complex amplitude for the wave moving to the right and for the wave moving to the left.

Similarly for the velocity, for u plus times this plus u minus times this. But P plus-- u plus was P plus over z zero e to the minus j omega x/c. u minus was minus P minus over z zero e to the plus j omega x/c.

And we impose the boundary condition, u of zero and x is equal to zero. Right here, no normal velocity. When we did that, we got P plus equals P minus right from here. At zero, these are unity. And that's zero, so P plus must be equal to P minus.

And from that, we got-- well, you can see it begins to look like sines and cosines. And so what we got was P of x. And omega 1, we solved this, was 2 p plus cosine omega s/c. And u of x and omega was equal to minus 2 P plus over z zero-- I think that was right-- sine omega x/c.

If I would draw the cosine 1, this looked like pressure that was maximum here would come down here. Maximum again. Come down, et cetera. This was the envelope. Remember the sine waves? This is just the complex amplitude and it's times the cosine omega t. So this looked like the pressure envelope.

And this, well let me do it this way. This is 2P plus here. And that's the envelope for P of x and omega. And the envelope for the velocity then, of course, was the sine wave, which is orthogonal to that. And it came like so.

OK, now as far as that problem, as far as we dealt with it, as far as that's concerned, you could put any frequency you want in there. It's just a question of where this cross-over, it would go to higher frequency, if it moves that way. It'll go to a lower frequency, if it moves that way. But there's no constraint on frequency. The only constraint introduced by the boundary condition was that the reflected wave, in this case, equaled the incident wave.

Now let's suppose we take the first step toward moving into a room and we put a boundary on here. Namely, we close the tube. But don't worry, for the moment, that there's no source there.

You can kick the end of the tube and there will be a source. There will be energy in there and it'll go. And there's no loss in our model. So if you had this tube sealed at both ends and you gave it a good whack, theoretically, the sound waves would bounce around there forever.

But now that puts another constraint on. And what's it do? It doesn't put any constraint at all on pressure. But it does put the same kind of constraint that you had here at the other end. But since this is the most general thing that can exist in here, we can satisfy this boundary condition just by looking at this.

And we don't even have to look at the pressure. We can look at the velocity here because that's the one that a boundary condition is imposed on. And that boundary condition that this wall imposes is at this end, if this tube is of length l, let's say.

At x equals minus l, this must be zero. So if that's going to be satisfied, the only way to satisfy this thing is if this is an integral number of pi. So in other words, if omega x-- or l, rather-- I don't care whether l is positive or negative here, still it's going to be an integral number of pi-- is equal to n pi because, whenever this argument is an integer, where n is an integer times pi, the sine function is zero.

So this really is the result of second boundary, this condition. So that means that any omega-- so let's just write this with an omega. The omega on the left-hand side is n/l c pi. Or omega is 2 pi f, so f is equal to c/2 n/l. So any integer of n that I plug into here gives me an allowable frequency in this tube. And that's the first time we've used that notion of allowable frequency.

In other words, a normal mode of frequency at which the closed tube could sustain a wave. It's only at those frequencies that it could sustain such a wave because any other frequency that wasn't one of the ones obtained when you put an integer in here wouldn't meet the boundary conditions. That's all there is to it. So it could not be a mode.

Think again in terms of the lumped parameter. A normal mode of that circuit, a coil and a capacitor put in parallel is the frequency which, if you once got some energy in here, you could do it by starting with a capacitor that was charged and putting an inductor across it, for example. And then it'll start going at the normal mode of this circuit, which is 1 over frequency, 1 over 2 pi, root of lc.

Well the same thing with the waves, exactly the same thing. The frequency at which you can have oscillation or have a signal, sustained signal-- well I shouldn't say sustained because this has a normal mode. In other words, some frequency that dies out too has a normal mode.

Even if you had a loss in this tube, as you would, really, in viscous loss, as we'll see when we get to air loss, you would still call it a normal mode. And so it's a frequency at which things can go without any excitation, if you want to say it in a broad sense.

Now let's look a little bit more closely at this. It's true, we can find all the frequencies. Well, in fact, let's look at an interesting problem.

Suppose that you had an open tube here, first, like we did when we started. And you were putting some loudspeaker or something at the end. And you had some amazing waveform you managed to generate in here. And it looked like this.

Whoop, that doesn't work for a time waveform. Does it? It goes backwards on itself. Forget it. It looked some terrible thing. And you closed the door. In other words, you were having a loudspeaker on there. And then somehow you, bloop, you made it rigid, where I've told you that only these frequencies can exist. And the thing that I was putting in there before I closed the end, gees, it had everything in it.

It seems like a contradiction. I told you that only these particular frequencies here could exist. Now I tell you, aha, I put a loudspeaker on there. And it creates this amazing waveform that's-- well, in fact, let's not make it so amazing. Let's just say it was this.

I put a waveform in which was, somehow-- let's see, what shall I draw here? I'll draw a velocity waveform. No matter what I did, that would always have to be zero.

Let's suppose I had a waveform that looked like this. It had this velocity at this end. And it was linear. Had this velocity at that end, time, time waveform.

And at the instant when I had this for the velocity, u of x and t, I closed the door. Now that thing, we know, has all frequencies in it. I close the door. I can only have those frequencies?

SPEAKER 2: [INAUDIBLE].

AMAR G. BOSE: For a series. That's the key. This was the waveform that was in here, or some arbitrary thing. You bang the door shut with that in there. Now immediately, when you did that, you brought this down to zero here.

Now the problem is can you make up that waveform by a set of sinusoids? That's the velocity. Here's all the can exist in the tube, sinusoids with frequency given by that expression.

This was your x equals zero. It doesn't matter which way you want to think about this. But if you think about it as a tube of twice the length-- this was l. This is l. And this is the waveform. You could make a Fourier series of sine functions about this point here.

This is an odd function now, this. So it would contain sines, no cosines. And the period would be 2l. And you could make a whole bunch of sines that satisfied exactly this because the period of the sine wave that satisfies this is 2l. And you could get all of the coefficients of the different sine terms that would result in that waveform by making a Fourier series that came to 2l.

So any waveform that you think of or can draw can be represented by these normal modes just by a Fourier series. And the reason you took the Fourier series about here with a double period was then you have only the sine functions, which are the modes that you know you have in there. Otherwise, you'll have cosines and everything else. But the cosines wouldn't meet the boundary conditions.

So you can look at that then as being able to support any kind of a waveform. But that kind of a waveform can be represented by sine waves. And of course, the whole concept behind Fourier series is that these sine waves have been going on forever. And so that whole set of sine waves going on forever at t equals zero would represent this function.

Now that function that we had drawn in there isn't going to stay there. The frequency components are going to stay there. But the waveform, with time, will bounce all around, of course. The different sine functions that make it up, if you take their arguments in t, you'll see how it moves.

OK, now let's take another look at this tube, which will be very useful for us when we go to the more dimensional case. If you take this relationship here and you use lambda f equals c, the relationship between wavelength and frequency, substitute for f, you get c over lambda equals c over 2n/l. Or lambda equals 2-- sorry, what I want to get here is n/2 times lambda is equal to l, where n is an integer.

Now that means there's an integral number of half wavelengths in that tube. And when we have the envelope, like this, of standing waves of velocity, if I were to just draw a line through here to indicate that that's a null of velocity, these nulls occur at lambda over two, every lambda over 2, of course. And this tells us that the allowable waves that can be in there, each one of them, is such that there are an integral number of lambda over 2s. Maybe if I write this this way, it'd be clearer. n lambda over 2, just put the 2 under the lambda, must be equal to the whole thing.

So let's see, if I start from one end, I had n equals 1. I have lambda over 2 is the first one that could exist. This is for n equals 1. I'd count the integral number of half wavelengths that'll fit in here. This is a half wavelength for n equals 1.

For n equals 2, I'd have two of those. And that would be for n equals 2. That's what we call the second normal mode, if you wish, that can exist in here.

For n equals 3, we would have three half wavelengths in here, something like this. And let's see, let's draw it. I have to divide this space equally into about three. So it comes here.

And that would look like this. I'll draw it with a different amplitude, just to distinguish, so I don't get the curves all over each other. n equals 3.

So it's nice to think of this space here as a space in which you can look at it this way, as lines. The n equals 3 has three equally spaced lines in it, which are troughs of the velocity. They're the lines here such that, if you have a plane wave coming in here, there will be zero velocity all the time for this orange one here, for n equals 3. There will be zero velocity at this point in the tube. Zero velocity all along here as the n equals 3 mode propagates back and forth in here with no loss.

The n equals 2 one would just look like that. And the higher numbers just has a lot of these lines in here. So you can see how they fit.

An integral number of half wavelengths fit into the tube constrained by the second boundary. Without the second boundary, anything fits. It's not a question of fitting. Anything can be shoved into it. There's no constraint on the other end.

Now it turns out that this picture is very useful as we move to the two-dimensions. And that's the way we're going to do it. We'll move from the one into two and on up.

However, before doing that, I'd like to give you a feeling for what all of this means in the time domain. This is in frequency. You can see sine waves put in there. I feel it gives you some insight. But I think the time domain picture will relate more to experiences that you've actually had.

Let's do the following thing. Let's suppose that you're sitting here with a microphone. It could be measuring pressure, let's say. You could pick a velocity microphone to measure voltage. I mean, whoop, to measure velocity.

And you give this thing a sharp kick at the end. And that,

let's say, you've kicked it very, very carefully so that you introduce a pulse like this of height P0 that starts traveling down here, height P0 in some narrow width. The question is-- oh, ha ha. Let me-- well, let's leave the problem like this. What do you think would be seen as a function of time over here? Let's say it's a pressure microphone. So it's reading P of t at this microphone point here. And let's say the tube is l in length, just like the other one. Anybody want to-- Yeah?

SPEAKER 3: A periodic train of impulses?

AMAR G. BOSE: Yep, a periodic train of impulses. Ah, ha, OK, when would the first one occur? This would be height P0. It's this one who's traveling down here. It would occur at--

SPEAKER 4: [UNINTELLIGIBLE]?

AMAR G. BOSE: [UNINTELLIGIBLE] I mean l, where would it occur?

SPEAKER 4: [UNINTELLIGIBLE].

AMAR G. BOSE: Hm?

SPEAKER 4: [UNINTELLIGIBLE].

AMAR G. BOSE: l over 2c. That's the time it takes for the wave to travel to the middle, where we said the microphone was. OK, next one would come back, when?

SPEAKER 5: 3l over 2c.

AMAR G. BOSE: 3l over 2c. ha, ha! Et cetera. And these would be then spaced l/c. l over-- wait a minute. l over 2c. Just a minute. Yeah. No, l/c. It takes them l over 2c to go here. It takes l over 2c to come back. So they're spaced l/c apart. And they would go on and on. You see, this is a funny kind of a situation because I have half the space here that I have here. And that's just how I happened to start it up.

If I call t equals zero at this point, I just measured t equals zero, not when I kicked the thing, but afterward, then I would have another one that I could-- either one is just as good. But if I kicked it and measured, I started the clock at this point, then I would have the first one at zero, of course.

And the second one would come along when this one went down here and came back, which was l/c. And the second one would be 2l over c, et cetera, either way you pick the coordinate system.

Now just to make sure that everybody's happy with the fact that the same pulse keeps coming back, what happens when it gets down here? You get twice the pressure at this point. If I were to measure the pressure here, it would go up to 2P.

But remember, P minus is equal to P plus due to the boundary condition that's here. And so what comes back is exactly the same wave. And it keeps going back and forth.

Now let's look at this. And this will be an introduction to something that will come a little later. Let's look at what happens if you have a little bit of loss in here. In other words, somehow you manage to have a tiny bit of loss in here.

Remember the reflection coefficient, which was P minus over P plus? For us, that reflection coefficient was equal to 1 in the case we had. You could write the reflection coefficient we derived, Zl/Z0 minus 1 Zl/Z0 plus 1.

Now let's suppose that Zl is equal to some R. And that's much, much greater than Z0. In that case, then gamma is equal to, essentially, Zl/Z0 is much, much greater than 1. Zl/Z0, OK.

So I could essentially say that gamma was very, very close to 1 in that case. And this is the resistance. A very large real number minus 1, a very large real number plus 1, so gamma is approximately equal to 1 in that situation, but a little less than 1.

Now we want to see how this picture would change if that were the case. I had pulses going back and forth there that were always the same. I'm going to choose the origin now, in time, when the first pulse hit the microphone. So I'll have them spaced l/c all the time.

So I have P0 here. This is time. This is the pressure, P of t read by the microphone. OK, second one would be at l/c. And how high would it be?

SPEAKER 7: P0.

AMAR G. BOSE: P0 times?

SPEAKER 7: Gamma.

AMAR G. BOSE: P0 times gamma. P0 times gamma because P minus over P plus, it's P minus that comes back when the first one went by. So P minus comes back. It's gamma times P plus. And P plus was P0.

The next one comes in, height? P0 gamma squared, et cetera. P0 gamma cubed.

Now I want to take a look at the envelope of this thing. And the real reason I want to look at this envelope-- you all can see it on the board. It's going down. We know that. But the reason I want to look at it is simply because I want you to be familiar with, if you're not, ways of looking at the envelope of something that only occurs discreetly in time. And that's a useful technique that'll come up again and again.

So we could write for this waveform here P of n delta T. Delta T is this distance over here, which happens to be l/c. But at n delta T, I'm at the nth one of these. At 1 delta T, I'm here. At 2 delta T, I'm here, 3 delta T, I'm here. So it's clear that that is P0 gamma cubed-- oh gees, number 3, P0 gamma to the n.

Now this is a discrete function which describes this. And to look at the envelope of it, if you wish, let n delta T equal a variable T sub n. When n is 3, T sub n is 3 delta T. But then I can look at T sub n as if it was continuous later and then get the envelope.

So if we do this, that'll become P of Tn is equal to P0 gamma. Now gamma to the n would be T sub n over delta T, T sub n over delta T. That's the exponent.

Is that clear? Maybe I should write it a little lower, so it doesn't get confused with that. P of T sub n equals P0 gamma to the n, which is T sub n over delta T. This is an exponent on that.

OK. Now that's OK. I can use this expression. But there's an expression which you would much rather use and which you know by now what's going to happen. But let's get this in the form of an exponential. Something that goes with a power like that can always be expressed as an exponential. So we will note that gamma was very, very close to 1. And when that happens, since gamma is approximately equal to 1, we can represent gamma as e to the minus 1 minus gamma, approximately.

And that you get right from the Taylor series, expansion of an exponential, which is, if I try to make that expansion, it's e to the x is equal to 1 plus x, plus x squared over 2 factorial, plus x cubed over 3 factorial, do the whole thing. So if I take 1 plus x, x is minus 1 plus x. x is minus 1 minus gamma, which is minus 1 plus gamma.

So to a first order, when gamma is very, very close to 1, then I can represent it as an exponential like that. And so I finally get P of tn is equal to, approximately if you want, P0 e to the minus 1 minus gamma tn over delta t. So now, if I think of this, for the moment, as something that's not discrete, but as tn as a continuous variable, this is exactly the curve I get for the decay.

As gamma gets closer and closer to 1, the decay gets less. As gamma gets smaller, 1 minus gamma is bigger, and you go down like this. So looking at the envelope helps us.

Now if you go into a room-- this room almost would classify, except, well, the people that are in it. I can't do it down here because it's absorbing material. But if you go,

[HAND CLAP]

AMAR G. BOSE: I don't know if these walls are exactly parallel. Somebody, just one of you in the middle of the room, clap your hands and see if you hear anything.

[HAND CLAP]

AMAR G. BOSE: Yep. I could even hear it. Did you hear the-- see, if I do it down here, it probably won't be as much because you will be--

[HAND CLAP]

AMAR G. BOSE: I can't even hear it.

[HAND CLAP]

AMAR G. BOSE: I just hear one clap.

[HAND CLAP]

AMAR G. BOSE: Try that again.

[HAND CLAP]

AMAR G. BOSE: Oh, no. Hit it--

[HAND CLAP]

AMAR G. BOSE: --sharp.

[HAND CLAP]

[LAUGHTER]

[HAND CLAP]

[HAND CLAP]

[BOOM]

[HAND CLAP]

AMAR G. BOSE: Oh, you can hear it?

SPEAKER 8: Yeah.

AMAR G. BOSE: Oh, that's interesting. You see, I was standing up here before. I can't hear it at all. I only hear one clap standing back here. Do it again.

[HAND CLAP]

AMAR G. BOSE: Oh, yeah. Sorry, it was not you. It was me.

[LAUGHTER]

AMAR G. BOSE: OK, yeah. So you almost think you can count those things. You hear them, brr drr drr drr drr, like that. It goes down. And you will see these high-paid acousticians, architectural fellows coming in. They'll go--

[HAND CLAP]

[LAUGHTER]

AMAR G. BOSE: Nod head. And now you know how to do that.

[LAUGHTER]

AMAR G. BOSE: So we'll come back to this thing when we go to more dimensions. OK, now I want to make sure that we have a very good understanding of this. And I'm willing to review any part that we need because, if we really understand this, going to two-dimension isn't a big step. Three-dimensions is almost trivial. And then, we can get all the meat out that we need. Yes?

SPEAKER 9: If you were to let gamma go significantly away from 1, wouldn't your boundary condition change? All those pulse spacings would come back and probably start to smear out or something. Would it still look like this decaying exponential envelope?

AMAR G. BOSE: Yeah, it would still look like the decaying exponential. When they smear out, would be when you had a frequency-dependent termination. Maybe I should take a couple of minutes and do that. You can look at this whole thing-- this periodic pulse has a heck of a lot of sine waves. And as long as the termination is-- let me back up.

All these sine waves have been going on forever. And they make this pulse. And then, a little later, they make the pulse which goes down here, et cetera.

Well when they hit a reflection, if all the sine waves are reflected in the same phase, then you get that same kind of pulse going back. If there's any phase difference between them, or amplitude difference for that matter, then the whole thing begins to smear out. So only in the case that it's real does it stay. Yes?

SPEAKER 10: When you derived that continuous equation, you used the assumption that gamma was, approximately, 1. Then you went back and said, well, let's make n very small now. and we get this quick decay. How much air is there?

AMAR G. BOSE: Yeah. You have to actually just evaluate that from here to find out how valid that expression is. That's a good point. You can't get very small and have that approximation that I made still work because the approximation depended, as you saw, that the exponential, this is only good-- I can only take this first term if 1 minus gamma is very small.

And the answer is, I don't know. I don't know, but you can calculate it right out of there. Other questions? Yes?

SPEAKER 11: Going back to the arbitrary waveform when you sealed the other end of the tube, that waveform will change shape over time? How many cycles? Is it only being made up of the normal modes?

AMAR G. BOSE: Yeah, it's always made up of those normal modes playing back and forth, all the sine functions that will fit in there playing back and forth. And the waveform will change with time. And it's not a question of how many cycles, every instant it'll look different.

SPEAKER 11: Because it will never achieve a steady state?

AMAR G. BOSE: No.

SPEAKER 11: [UNINTELLIGIBLE PHRASE].

AMAR G. BOSE: No, no. They keep on going. I guess, in the frequency domain, the most important thing we want is that there are n of these things that you can draw. And these, we're going to call them velocity troughs that you can draw in here, the reason that there are n of them, where n is an integer, is that the velocity here and the air must be zero.

And so the way you make that up is by putting an integral number of the half wavelengths in the velocity envelope into this space. I'm going to give you some time to think. See if you can ask any other questions.

OK, let's see if we can't move to two-dimensions. Now again, in two dimensions, the way you would normally go about this is to just get the two-dimensional solution of the wave equation, put in the boundary conditions here, here, here and here. That normal velocity is zero, and out pops a solution.

But we're going to look at it building it up. Now the way we're going to look at it now is, I would say, very definitely not the way you would do it if you were doing it the first time. It's with a knowledge of what the answer is that you come back and get some insight on another way to do it. And I guess I can say, at least, with my own experience, that, if I did it the way I'm going to do it now the first time, I wouldn't be too confident that it really was all true until I verified it with a general solution of the wave equation.

OK, first we know that, in a single dimension, one-dimensional tube, in order to get a standing wave, remember we had to have a wave going opposite. Couldn't get a standing wave unless we had one going this way and one going this way. Remember they didn't have to be equal in magnitude, though.

If they weren't equal in magnitude, instead of the envelope coming down to zero like this, we found out that we could get things like this for standing waves. So equal magnitude was not necessary. But that does happen when you have hard surfaces, because there's no absorption in here, so nothing dies out.

But we needed a wave going back on itself. So first thing we have to find out now is in this-- this is a two-dimensional room only on the blackboard-- in this two-dimensional situation, if you launch a wave in any direction, arbitrarily-- don't worry how you do this now. Imagine that there's a plane wave it you can launch going like this.

The question is will that create a wave going back exactly in the same but opposite direction? If it doesn't, then that wave can't create a standing wave in the room, anymore than, if you didn't get a wave going backward here, you could create a standing wave in the tube. So the first thing we have to find out is that this might happen.

So if we take a wave, let's say, that's coming like this-- maybe this is launched at an angle to the bottom here of, say, theta, then this is theta. Let's see, have we had a problem on oblique? No, no? OK.

OK, what happens when a wave hits at an angle here? This is, let's say, theta. And let's call this one phi, which 90 degrees, the sum of the two.

Now when this hits here, you can think of the velocity, and you only have to worry about the velocity, just like we did there, because velocity is the thing that's constrained by the hard boundary.

So this has a normal component of velocity. The wave is going in this direction. I can break up its velocity into a normal component, to a component normal to the wall and a component tangential to the wall.

This fellow is killed by the wall, for sure. You can't absorb him, so the way he gets killed is by a wave going backward. We don't know which direction yet, but it has a normal component of velocity exactly equal to that, the reflected wave, so that the normal components are zero here.

Now tangential components, this fellow here-- let's call this u sub-normal and u sub-tangential, if you want. u sub-normal generated 1 in the reflected wave. This is the incident wave. This is the reflected. And it generated a u sub-normal for that, that was equal and opposite, which satisfied the boundary condition, tangential one, no constraint at all.

So the wave going off here, when this hits, what comes off has a normal velocity in this direction whose magnitude was exactly the same as this and a tangential one which has the same magnitude. So this comes off at an angle like that. And therefore, phi, the angle that it comes in at to this, is the angle then it goes out.

So that's the normal. Same thing in light, angle of incidence is equal to angle of reflection, which you've seen since early days looking into a mirror. And hard walls to sound do exactly what mirrors do to light.

OK, so this comes off here. This was theta. Oh, how did I get this to be theta? This is some phi where phi and theta are the complements. This comes off here at phi, at an angle phi.

Since this is a right angle in here, then this must be theta, OK? The sum of these two must be 90 degrees, just as the sum of these two were 90 degrees. And this comes, then, off here at an angle of theta. Oops, it comes off here at an angle of theta. So this is then parallel to this.

So that only happens, mind you of course, when the angle here is 90 degrees. So interestingly enough, any angle that you launch a wave in this room, it didn't matter what theta was, it comes back on theta.

So number one condition is satisfied. Namely, that we launch a wave down here, and we get a wave coming back parallel to it. Same amplitude. There's no loss. Has the same projection this way and this way, same magnitude, therefore. So that condition is met. But that wasn't enough over here. This wall gave us a wave that went back, a condition that's nice. We have to have it, or no standing wave.

But then the thing had to fit in here when the other side was closed. And that's the thing we have to worry about over here. How do we get a wave that will, so to speak, fit in here? Namely, fit means meet the boundary conditions of all the walls.

OK. By the way, just from what we've been talking about this morning and just now, once we know angle of incidences, angle of reflection, we can realize that, if this room had walls that, for example, instead of being parallel were like that, in other words, walls like here's the front, if it was like this and you clapped your hands here, a wave would go across here.

And angle of incidence, angle of reflection, it would come back here. It would go like this. It would go like this. It would have many reflections before it got back to you.

What happened in this room was you clapped your hands and the thing went, boom, boom, boom, boom, boom. Now as elementary as it is, that would make a huge improvement in the acoustics of all rooms if you had non-parallel walls. Because that thing--

[HAND CLAP]

AMAR G. BOSE: --you heard when-- I can't do it. But the thing you heard before when he clapped his hands, you heard, brr drr dtt, like that. If you listen closely and you make enough experiments, you can actually pick that out in sound in music. There's like a buzz. It's almost like a buzz that you can hear.

There have been many studies made why people, for example, cannot tolerate reproduced sound even from the best of sound systems. Say an orchestra reproduced in your living room at the same level that you hear the live performance, people will tire much more rapidly if you give them the same level in their living room as they had in the live concert hall.

They won't want to listen to the equivalent of a whole performance. They will want the volume down. And that's one of the things that causes it, that brr brr brr, that buzz that comes onto the sound. Now I don't want to go any further than that. Otherwise, I will be guilty of dealing in the folklore of Hi-Fi.

But if the walls were-- for example, if your living room, let's say it's 15 feet or so or 20 feet, if you had just six inches difference in the wall at one end from the other, probably nobody would notice it either. Just like just a tiny bit. Right away, the acoustics of the room would improve enormously.

But when you build your own house one day, I challenge you to get an architect who will design that and a construction firm who will build it. I tried.

[LAUGHTER]

AMAR G. BOSE: And they, what's the matter, you crazy? If anybody sees this, I'll lose my license forever, for god's sake. I build crooked walls!

The best I could do is get him to build 45 degrees. Literally. I mean, they weren't going to construct anything like that that was oddball because it was clearly going to be seen by anybody as an error.

So homes would be so much better if your rooms were just slightly non-parallel. And if you want to see how bad they are, just go into a sample home that's not furnished yet and do the same thing. And boy, oh boy, it is unbelievable. The plaster on the wall has a very, very high reflection coefficient. Yes?

SPEAKER 13: A big improvement is just doing one of the surfaces because it's relatively common to get a house with a sloped ceiling.

AMAR G. BOSE: Yeah. Yeah, that's a help. But that's a help primarily for those modes that are hitting the ceiling. That definitely helps, but if you could take just two of the other walls and-- and-- and--

[LAUGHTER]

AMAR G. BOSE: Then you'll have something. OK, let's see. OK, we wanted to go to the two-dimensional case. Let's see how much of this I need. I'd like to start here and use more. Let's see what's used here. Ah, good.

OK, big room here. We'll try as big as we can draw it. We will launch a wave, let's say, in this direction. Theta to the wall. And just as we did for the thing that I erased, just as we did for the single dimension case, we'll draw the troughs of the velocity, the lines from zero velocity.

And we already know, from what we talked about here, that whatever I have drawn up there, there's a reflected wave that comes back on it. So I do have these velocity troughs that correspond to a standing wave. The only thing I don't know is will they, at the particular angle I chose and the particular frequency I chose, i.e. wavelength, will this thing fit in here? And that's what I've got to determine now.

So let's draw some velocity troughs for this thing. That starts out that way. Well lambda over 2 is this. In other words, this fellow here has a frequency f, and therefore, a wavelength given by lambda f/c. Lambda f equals c.

So we have lambda over 2 are the troughs, just as they were in the one-dimension. And we now have the condition that we need in order to get no, we have-- in the other problem, we just needed that there was no normal velocity at one surface. Now we have to have no normal velocity to this surface and to this surface.

And in order to do that, we want to have an integral number of these intersection lines fall in lx-- this is the x dimension of the room, this is the ly direction-- because, if they didn't, there could be a normal velocity in the standing wave pattern. In other words, what I've got to establish here is that there's no normal velocity this direction and there's no normal velocity this direction.

Well if you chase through the angles, as we did up here, what you find out is that, when this wave hits here and goes off again at phi, where phi plus theta was 90 degrees, this fellow generates another set of velocity troughs, if you want, because there's a wave that comes right back on him. In other words, any wave, I go in any direction in this room, will develop one that comes opposite to it.

And so what happens is you now have two sets of troughs, if you wish, you want to call them that. How do I best illustrate this? Let's see, it's very hard to do this without making a total mess of the drawing.

I will have a set that looks like this. And just by tracing the angles through, it turns out-- just like we did up here== that these will fall at the same points here. These are the velocity troughs associated with the reflected wave that goes out here.

And again, only because the room is 90 degrees. Otherwise, you might wonder why in the heck do we bother analyzing this thing, if the rooms have to be just 90 degrees? Well one could argue that, unfortunately, there are a lot of rooms that do that. But the other point is that you can get a lot of insight and useful information here that we can even actually use in irregular rooms.

OK, so what you have to have, you have a wave going this way. You have a wave coming this way. In order to get no normal velocity here or here at this surface, you need an integral number of these troughs in each direction. I need an integral number here.

I need an integral number along here, so nx, let's call it, times this distance. And now here, nx times that distance must be equal to lx. But here I'm going to call that distance a very special name.

And all the ears, all eyes, because this is the place where the sock comes down if you miss something. Namely, I'm going to call this thing-- and it has no relationship whatsoever to a real wavelength or a real frequency-- lambda x, that distance. I'm going to call this thing, this distance between the troughs up here, lambda y. No real wavelength, whatsoever.

In fact, there's only one real wavelength in this two-dimensional room, it is lambda and the corresponding frequency that goes with that. So lambda x is equal to lambda over cosine theta, OK? In general, this is lambda over 2. I can put lambda over 2 both sides there.

But if this is lambda over 2, this is a bigger distance here when you project it this way. And these two are related by the cosign. In other words, this distance is this times cosine theta. So the lambda x is lambda over cosine theta.

The lambda y is lambda over cosine phi, OK? The lambda y is the distance between these troughs. The angle is phi. And so this distance is this distance, which is, again, this real distance here is lambda over 2. That's the only real frequency in there. And we'll call this lambda y. No relation at-- well, I can't say no relation, but not a real wavelength.

Now just like we had over here, n times this lambda over 2 had to fit into l. So the condition we constructed this on was that n, we'll call it nx. In this case, nx times lambda x/2 is equal to lx. And ny times lambda y over 2 is equal to ly.

So those are the conditions now that must exist for a frequency having a wavelength lambda to be launched at an angle theta in the room and satisfy the condition that it fits. It's a normal mode that it meets all the boundary conditions.

So now this thing, of course, you can put in here for lambda x. You can put lambda over cosine theta. If you do that, lambda is the frequency you're thinking about. Theta is the angle you must launch that frequency in order to satisfy this equation. Yes? Yeah?

SPEAKER 14: Why is lambda x equal to just lambda over cosine theta and not lambda over 2 cosine theta?

AMAR G. BOSE: Lambda x, this is-- Oh, you are exactly right. You are exactly right. Lambda y over 2. Sorry about that. Yeah, these are the distances, the projected distances, from the half wavelength. And so I forgot to put the factors of 2 under there. Thank you, very much.

And this is OK. These are n times these distances. If you want me say more about that, that's a major screw up, so if that's caused people problems that need to be further straightened out-- OK.

OK, so all we've done so far now is taken from one dimension into two the condition that a wave-- you get standing waves that they consist of waves always going back on each other, whichever direction they're going. If a wave's going here, you get a wave coming back on it. Wave's going here, you get a wave coming back. That's always true for any rectangular room.

That's just like it's always true for any frequency that goes in here. You get a wave coming back, but now you have to meet the boundary condition. And we have to meet the boundary conditions now in two places on perpendicular walls. And very similar to what happened over here, n times lambda over 2, fit in here. n times the projection over here fits here.

Now what you're going to see is that this room can support many more waves than you would have thought from a first guest. First guess might have been that theta equals zero. You get all the waves which just bang back and forth here.